if such three condition find the $sin{angle OPA}$
$begingroup$
Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$dfrac{AP}{PD}=dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=dfrac{3}{2}CD$$find $sin{angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=dfrac{1}{4}sqrt{dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$cos{(angle APB+angle CPB+angle CPD)}=0$$
or $angle APB=angle 1,angle CPB=angle 2,angle CPD=angle 3$
$$cos{angle 1}cos{angle 2}cos{angle 3}=sin{angle 1}sin{angle 2}cos{angle 3}+sin{angle 1}sin{angle 3}cos{angle 2}+sin{angle 2}sin{angle 3}cos{angle 1}$$
geometry contest-math
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
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$begingroup$
Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$dfrac{AP}{PD}=dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=dfrac{3}{2}CD$$find $sin{angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=dfrac{1}{4}sqrt{dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$cos{(angle APB+angle CPB+angle CPD)}=0$$
or $angle APB=angle 1,angle CPB=angle 2,angle CPD=angle 3$
$$cos{angle 1}cos{angle 2}cos{angle 3}=sin{angle 1}sin{angle 2}cos{angle 3}+sin{angle 1}sin{angle 3}cos{angle 2}+sin{angle 2}sin{angle 3}cos{angle 1}$$
geometry contest-math
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
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2
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Where did this problem come from?
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– rogerl
Dec 5 '18 at 17:34
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What happened to Blue’s answer?
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– D.R.
Dec 10 '18 at 3:49
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what happend to Blue's answer?
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– communnites
Dec 10 '18 at 4:07
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Yeah i was wondering why he deleted his answer too.
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– dezdichado
Dec 10 '18 at 4:47
1
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yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
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– dezdichado
Jan 1 at 9:35
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show 4 more comments
$begingroup$
Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$dfrac{AP}{PD}=dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=dfrac{3}{2}CD$$find $sin{angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=dfrac{1}{4}sqrt{dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$cos{(angle APB+angle CPB+angle CPD)}=0$$
or $angle APB=angle 1,angle CPB=angle 2,angle CPD=angle 3$
$$cos{angle 1}cos{angle 2}cos{angle 3}=sin{angle 1}sin{angle 2}cos{angle 3}+sin{angle 1}sin{angle 3}cos{angle 2}+sin{angle 2}sin{angle 3}cos{angle 1}$$
geometry contest-math
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
$endgroup$
Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$dfrac{AP}{PD}=dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=dfrac{3}{2}CD$$find $sin{angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=dfrac{1}{4}sqrt{dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$cos{(angle APB+angle CPB+angle CPD)}=0$$
or $angle APB=angle 1,angle CPB=angle 2,angle CPD=angle 3$
$$cos{angle 1}cos{angle 2}cos{angle 3}=sin{angle 1}sin{angle 2}cos{angle 3}+sin{angle 1}sin{angle 3}cos{angle 2}+sin{angle 2}sin{angle 3}cos{angle 1}$$
geometry contest-math
geometry contest-math
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
edited Dec 6 '18 at 0:37
communnites
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
asked Dec 5 '18 at 13:34
communnitescommunnites
1,1731433
1,1731433
2
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Where did this problem come from?
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– rogerl
Dec 5 '18 at 17:34
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What happened to Blue’s answer?
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– D.R.
Dec 10 '18 at 3:49
$begingroup$
what happend to Blue's answer?
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– communnites
Dec 10 '18 at 4:07
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Yeah i was wondering why he deleted his answer too.
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– dezdichado
Dec 10 '18 at 4:47
1
$begingroup$
yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
$endgroup$
– dezdichado
Jan 1 at 9:35
|
show 4 more comments
2
$begingroup$
Where did this problem come from?
$endgroup$
– rogerl
Dec 5 '18 at 17:34
$begingroup$
What happened to Blue’s answer?
$endgroup$
– D.R.
Dec 10 '18 at 3:49
$begingroup$
what happend to Blue's answer?
$endgroup$
– communnites
Dec 10 '18 at 4:07
$begingroup$
Yeah i was wondering why he deleted his answer too.
$endgroup$
– dezdichado
Dec 10 '18 at 4:47
1
$begingroup$
yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
$endgroup$
– dezdichado
Jan 1 at 9:35
2
2
$begingroup$
Where did this problem come from?
$endgroup$
– rogerl
Dec 5 '18 at 17:34
$begingroup$
Where did this problem come from?
$endgroup$
– rogerl
Dec 5 '18 at 17:34
$begingroup$
What happened to Blue’s answer?
$endgroup$
– D.R.
Dec 10 '18 at 3:49
$begingroup$
What happened to Blue’s answer?
$endgroup$
– D.R.
Dec 10 '18 at 3:49
$begingroup$
what happend to Blue's answer?
$endgroup$
– communnites
Dec 10 '18 at 4:07
$begingroup$
what happend to Blue's answer?
$endgroup$
– communnites
Dec 10 '18 at 4:07
$begingroup$
Yeah i was wondering why he deleted his answer too.
$endgroup$
– dezdichado
Dec 10 '18 at 4:47
$begingroup$
Yeah i was wondering why he deleted his answer too.
$endgroup$
– dezdichado
Dec 10 '18 at 4:47
1
1
$begingroup$
yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
$endgroup$
– dezdichado
Jan 1 at 9:35
$begingroup$
yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
$endgroup$
– dezdichado
Jan 1 at 9:35
|
show 4 more comments
5 Answers
5
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$$textbf{Edition of 11.01.2019}$$
$color{brown}{textbf{The issue data}}$
Let
$$angle OPA=varphi,quad angle DAB = alpha,quad angle CDA=delta,quad AD=d=13x,quad PB=u,quad PC=v.tag1$$
Denote
begin{vmatrix}
AB & BC & CD & AC & BD & PA & PD & AO\
a & b & c & f & g & - & - & R\
dfrac{8x+u}3 & dfrac{u+v}2 & dfrac23(5x+v) & - & - & 8x & 5x & R\
end{vmatrix}
$$ $$ $color{brown}{textbf{Triangles AOP and POD}}$
$$AO = OD = R,quad AP=8x,quad PD=5x,quad angle OPA=varphi.$$
Let
$$OP=y,quad angle OAP = angle ODP = psi.$$
Using cosine theorem, one can get
$$
begin{cases}
y^2 = (8x)^2 + R^2 - 16Rxcospsi\[4pt]
y^2 = (5x)^2 + R^2 - 10Rxcospsi\[4pt]
R^2 = (5x)^2+y^2+10xycosvarphi
end{cases}rightarrow
begin{cases}
y^2=R^2-40x^2,\[4pt]
cosvarphi = dfrac{R^2-y^2-25x^2}{10xy},
end{cases}
$$
$$cosvarphi = dfrac{3x}{2sqrt{R^2-40x^2}}.tag2$$
$color{brown}{textbf{Triangle ABP}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad PA=8x,quad PB=u.$$
Cosine theorem leads to the equation
$$u^2=a^2+(8x)^2-16axcosalpha,$$
$$cosalpha = dfrac{a^2+(8x)^2-u^2}{16ax}=dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)}
=dfrac{8x+u+9(8x-u)}{48x},$$
$$cosalpha = dfrac{10x-u}{6x},quad sinalpha = dfrac1{6x}sqrt{(6x)^2-(10x-u)^2}tag3.$$
$color{brown}{textbf{Triangles ABD and BCD}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad AD=13x,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac23(5x+v),$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle BCD = pi-alpha,quad cos(angle BCD) = -cosalpha.$$
Cosine theorem for $triangle ABD$ and $triangle BCD$ gives
$$f^2 = a^2+d^2-2adcosalpha = b^2+c^2+2bccosalpha,$$
$$f^2=dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = dfrac1{36x}left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)right),$$
$$f^2=dfrac19(14u^2-10ux+545x^2),tag4$$
$$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2)
= 0.tag5$$
$triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac f{2sinalpha}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.tag6$$
$color{brown}{textbf{Triangle PCD}}$
$$angle PDC = delta,quad CD=c=dfrac{10x+2v}3,quad PD=5x,quad PC=v.$$
Cosine theorem leads to the equation
$$v^2=c^2+(5x)^2-10cxcosdelta,$$
$$cosdelta = dfrac{c^2+(5x)^2-v^2}{10cx}=dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$
$$cosdelta = dfrac{13x-v}{12x},quad sindelta = dfrac1{12x}sqrt{(12x)^2-(13x-v)^2}.tag7$$
$color{brown}{textbf{Triangles ACD and ABC}}$
$$angle ADC = delta,quad AB=a=dfrac{8x+u}3,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac{10x+2v}3,quad AD=d=13x.$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle ABC = pi-delta,quad cos(angle ABD) = -cosdelta.$$
Cosine theorem for $triangle ACD$ and $triangle ABC$ gives
$$g^2 = c^2+d^2-2cdcosdelta = a^2+b^2+2abcosdelta,$$
$$g^2=dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = dfrac1{36x}left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)right),$$
$$g^2=dfrac19(17v^2-64vx+776x^2),tag8$$
$$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3
=0.tag9$$
$triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac g{2sindelta}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.tag{10}$$
$color{brown}{textbf{Constraints to the radius}}$
Equations $(6),(10)$ form the system
$$R^2 = dfrac{14u^2-10u+545}{-u^2+20u-64}=dfrac{68v^2-256v+3104}{-v^2+26v-25}.tag{11}$$
Easy to show that
$$minlimits_{uin(4,16)} R^2 = dfrac{169}4quadtext{at}quad u=dfrac{38}5$$
(see also Wolfram Alpha),
$$minlimits_{vin(1,25)} R^2 = dfrac{169}4quadtext{at}quad v=dfrac{43}7$$
(see also Wolfram Alpha).
I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$
At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.
$color{brown}{textbf{Calculation of u and v}}$
If assume $x=1$, then will be obtained solution for $dfrac ux, dfrac vx.$
Equations $(4),(9)$ under this condition form the system
$$begin{cases}
u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{12}$$
Summation of equations $(12)$ with factors $1$ and $(-4)$ gives
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{13}$$
Taking in account that
$$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\
=(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\
+ 37v^3+191v^2-3720v+54112,$$
this leads to the system with the explicit expresssion for $p,$
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0,
end{cases}tag{14}$$
so
$$begin{cases}
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\[4pt]
222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0end{cases}tag{15}$$
wherein
$$222=2cdot3cdot37,quad 152739664 = 2^4cdot7^2cdot11cdot89cdot199.tag{16}$$
The integer roots are dividers of $152739664.$
Horner's scheme for them is
begin{vmatrix}
v & mathbf{222} & mathbf{8426} & mathbf{-64431} & mathbf{-137056} & mathbf{3387595} & mathbf{-40679520} & mathbf{152739664}\
2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\
-2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \
4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\
-4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\
color{brown}{mathbf{7}} & mathbf{222} & mathbf{9980} & mathbf{5429} & mathbf{-99053} & mathbf{2694224} & mathbf{-21819952} & 0 \
end{vmatrix}
The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.
After the integer roots should be checked rational roots, using dividers of 222 as denominators.
Results of the searching allow to write the system in the form of
$$begin{cases}
(v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\[4pt]
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944},
end{cases}$$
with the positive solutions
$$binom{u/x}{v/x} = left{binom{157/19}7, binom7{199/37}right}.tag{17}$$
Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$
$$left(dfrac Rxright)^2 = dfrac{14left(dfrac uxright)^2-10left(dfrac uxright)+545}{36-left(left(dfrac uxright)-10right)^2},tag{18}$$
$$cosvarphi = dfrac{3}{2sqrt{left(dfrac Rxright)^2-40}}.tag{19}$$
From $(18)$ should
$$dfrac ux in(4,16).$$
$color{brown}{textbf{Results}}$
If $dfrac ux = 7,$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
If $dfrac ux=frac{157}{19}approx 8.26316$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
Therefore, both of the possible solutions leads to the similar angle $varphi.$
$color{brown}{textbf{Coordinates}}$
If $O=(0,0),quad x=1,quad u=7,quad v=dfrac{199}{37},quad R=sqrt{43},quad b=dfrac{229}{37},quad c=dfrac{256}{37},quad$ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-cos^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
5 & dfrac12 & dfrac{sqrt3}2 & -dfrac{13}2 & -4 & 3sqrt3 &dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
dfrac{256}{37} & dfrac{47}{74} & dfrac{33}{74}sqrt3 & left(dfrac32,dfrac{sqrt3}2right) & left(dfrac{13}2,dfrac{sqrt3}2right) & left(dfrac{5765}{2738},dfrac{9817}{2738}sqrt3right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tandfrac{P_Y}{P_X}=dfrac{sqrt3}3.$$
If $O=(0,0),quad x=1,quad u=dfrac{157}{19},quad v=7,quad R=sqrt{43},quad b=dfrac{145}{19},quad c = 8,quad $ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-sin^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
dfrac{103}{19} & dfrac{11}{38} & dfrac{21sqrt{3}}{38} & -dfrac{13}2 & -dfrac{1780}{361} & dfrac{901}{361}sqrt3 &-dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
8 & dfrac12 & dfrac{sqrt3}2 & left(dfrac32,-dfrac{sqrt3}2right) & left(dfrac{13}2,-dfrac{sqrt3}2right) & left(dfrac52,dfrac{7sqrt3}2right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tanleft|dfrac{P_Y}{P_X}right|=dfrac{sqrt3}3.$$
It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
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very nice. did you try to calculate $u$ and $v$ by hands?
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– dezdichado
Dec 24 '18 at 17:26
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@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
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– Yuri Negometyanov
Dec 24 '18 at 20:42
add a comment |
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(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)
Ignoring reflections in $overline{AD}$, there are two viable quadrilaterals. They're shown here as $square AB^prime CD$, and $square ABC^prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $square ABCD$ that traces two equilateral triangles
$square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:
$$angle OPA = 30^circ quadtoquad sin angle OPA = frac12$$
Currently, I don't have a straightforward way to construct $B^prime$ and $C^prime$, but their specifications appear below.
$$begin{array}{c:cccc:cccc:cc}
square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & cos APB & cos DPC \ hline \
AB^prime C D & 8 & dfrac{157}{19} & 7 & 5 & dfrac{103}{19} & dfrac{145}{19} & 8 & 13 & dfrac{4643}{5966} & dfrac17 \ \
ABC^prime D & 8 & 7 & dfrac{199}{37} & 5 & 5 & dfrac{229}{37} & dfrac{256}{37} & 13 & dfrac{11}{14} & dfrac{829}{7363} \ \ hline
end{array}$$
(In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^prime$ or $C^prime$.)
My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using
$$|AB|=p(|PA|+|PB|) qquad |BC|=q(|PB|+|PC|) qquad |CD|=r(|PC|+|PD|)$$
but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $sin angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)
For the derivation, define
$$a := |PA| quad b := |PB| quad c := |PC| quad d := |PD|$$
$$alpha := angle APB qquad beta := angle BPC qquad gamma := angle CPD$$
where we may as well take $d=5$. The problem conditions tell us
$$a = frac85 d = 8 qquad |AB| = frac1{3}( b + 8 ) qquad |BC| = frac12(b+c) qquad |CD| = frac23(c+5) tag{1}$$
By the Law of Cosines and (1), we have
$$cosalpha = frac1{2ab}left(-|AB|^2+a^2+b^2right) = frac{1}{18b}left(b^2 - 2 b + 64right) tag{2a}$$
also
$$cosbeta = frac{1}{8bc} left(3 b^2 - 2 b c + 3 c^2right) qquad
cosgamma = frac{1}{18c} left(c^2 - 8 c + 25right) tag{2b}$$
Now, since $alpha+beta+gamma=180^circ$, one readily shows
$$1 - 2 cosalpha cosbeta cosgamma - cos^2alpha - cos^2beta - cos^2gamma = 0 tag{3}$$
Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...
$$begin{align}
0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \
&- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \
&+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \
&- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \
&+ 29200 b^2 - 12800 b c + 84736 c^2
end{align} tag{4}$$
To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:
Lemma. Given $triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by
$$p^2-r^2 = - xyz frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ y z sinangle YPZ + z x sinangle ZPX + x y sinangle XPY }
tag{$star$}$$
where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.
Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have
$$p^2-r^2 = - frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ dfrac{1}{x} sinangle YPZ + dfrac{1}{y} sinangle ZPX + dfrac{1}{z} sinangle XPY }$$
which is somewhat intriguing. But I digress ...
In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $triangle ABC$, getting
$$-40 = - a b c frac{ a sinbeta + b sin(180^circ+gamma) + c sinalpha }{ a b sinalpha + b c sinbeta + c a sin(180^circ+gamma)}
= - 8 b c frac{ 8 sinbeta - b singamma + c sinalpha }{ 8 b sinalpha + b c sinbeta - 8 c singamma}$$
Since $gamma = 180^circ-alpha-beta$, we can manipulate above into the form
$$sinalphaleft(-b c^2 + 40 b + ccosbetaleft( b^2 - 40 right)right) = c sinbeta left(-3 b + cosalphaleft( b^2 - 40right)right) tag{5}$$
Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find
$$0 = (b + c) (b + 8) left(begin{align}
&phantom{+} 4 b^4 c^2 -
80 b^3 c^2 \
&- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \
&- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \
&+ 9175 b^2 + 11230 b c + 2775 c^2 \
&- 128000 b + 409600 end{align}right) tag{6}$$
From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $square$
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
$endgroup$
$begingroup$
This problem is something else.
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– dezdichado
Dec 8 '18 at 1:01
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It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
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– Blue
Dec 8 '18 at 12:06
$begingroup$
Also, it seems that $Q$ lies on the segment $B'C'$.
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– timon92
Dec 8 '18 at 12:07
1
$begingroup$
You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
|
show 3 more comments
$begingroup$
Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that
begin{eqnarray}
AP & = & frac{8}{13}x\
DP & = & frac{5}{13}x\
AB & = & frac{8x+13y}{39}\
BC & = & frac{y+z}{2}\
CD & = & frac{2(5x+13z)}{39}
end{eqnarray}
For the angle at $A$, we have that
$$
cos(angle BAP) = frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)}
$$
where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$:
$$f(x,y,z)=
-frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x left(8 x^2+18 x y+5 x z+13 y z+13 z^2right)}=0.
$$
In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$:
$$g(x,y,z)=
-frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y
z^2}{156 x left(20 x^2+8 x y+60 x z+13 y^2+13 y zright)} = 0.
$$
Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution:
$$
z = frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 left(5280 x^2+4277 x
y-7267 y^2right)}.
$$
If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain
$$
frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2} = 0
$$
or
$$
-frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2}=0.
$$
Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results:
begin{eqnarray}
& y=frac{157}{247}x qquad & z=frac{7}{13}x\
& y=frac{7}{13}x qquad & z=frac{199}{481}x\
& y=frac{70}{13}x qquad & z=-frac{44}{13}x
end{eqnarray}
and hence we have to discard the third solution.
Let $O=(0,0)$ be the centre of the circle, and choose $A=(-frac{x}{2},h)$ and $B=(frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by
begin{eqnarray}
B & = & P + y (-cos angle APB,sin angle APB)\
C & = & P + z ( cos angle CPD,sin angle CPD)
end{eqnarray}
With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{157}{19 sqrt{43}},frac{7}{sqrt{43}},-frac{sqrt{3}}{2sqrt{43}})
$$
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{7}{sqrt{43}},frac{199}{37 sqrt{43}},frac{sqrt{3}}{2sqrt{43}})
$$
In either case one finds that $sin angle OPA = frac{1}{2}$.
It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
$endgroup$
add a comment |
$begingroup$
Here is a start: Let $R = 1$ and $AP = 8x, PD = 5x.$ Then, by the power of point $P$ with respect to our circle, you have:
$$R^2-OP^2=APcdot PDimplies OP = sqrt{1-40x^2}.$$ This means that by the theorem of cosines:
$$cosangle OPA=dfrac{24x^2+1-1}{2cdot 8xsqrt{1-40x^2}} = dfrac{3x}{2sqrt{1-40x^2}}.$$
So the challenge now is to find $x$ using the other information provided. This should be a chain of straightforward calculations; however, if you cannot finish from here, let me know in comments.
Update: For the time being, let $BP = y$ and $CP=z.$ Also, let $angle BAD = t.$ Then $AB = frac{y+8x}{3}$ and from the theorem of cosines in $triangle BAP:$
$$cos t = dfrac{frac{(y+8x)^2}{9} - (y+8x)(y-8x)}{2cdot 8xcdotfrac{y+8x}{3}} = dfrac{10x-y}{6x}quad (1)$$
Find $BD$ using Stuart's theorem in $triangle BAD:$
$$BD^2 = dfrac{1}{AP}(APcdot PDcdot AD +BP^2cdot AD - PDcdot AB^2)=$$
$$= dfrac{14y^2-10xy+545x^2}{9}quad (2)$$
By the Theorem of Sines in $triangle BDA$ and using $(1), (2):$
$$R^2=1=dfrac{BD^2}{4sin^2 t}=dfrac{BD^2}{4(1-cos^2 t)}iff $$
$$y^2(14x^2+1) - 10y(x^3+2x)+(545x^4+x^2) = 0.$$
Now this looks ugly at first, but it turns out to have somewhat nice discriminant:
$$frac yx = dfrac{5x^2+10+3sqrt{(5x^2+1)(2-13x)(2+13x)}}{14x^2+1}.$$
If you mimic the same idea for $triangle CDA,$ you will get the following:
$$frac zx = dfrac{128x^2+13+6sqrt{(2-13x)(2+13x)(32x^2+1)}}{68x^2+1}.$$
So at this point, we have not used the property $PB+PC = 2BC$, which is now obviously used to find $x.$ I will be busy until next week, so if you can finish it from here, let me know.
@Blue seems to have answered the question, but I still hope to finish computation here. So I will keep this attempt up for a while.
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
$endgroup$
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
add a comment |
$begingroup$
I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-sqrt{1 - a^2}, -a)$ and $D(sqrt{1 - a^2}, -a)$
Using the ratio formula $P(frac{3sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = frac{3}{13}sqrt{1 - a^2} + sqrt{1 - a^2} = frac{16}{13}sqrt{1 - a^2}$
$PD = sqrt{1 - a^2} - frac{3}{13}sqrt{1 - a^2} = frac{10}{13}sqrt{1 - a^2}$
Also let B be $(x_1, sqrt{1 - x_1^2})$ and C be $(x_2, sqrt{1 - x_2^2})$
Hence $AB = sqrt{(x_1 + sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$PB = sqrt{(x_1 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$BC = sqrt{(x_1 - x_2)^2 - (sqrt{1 - x_1^2} - sqrt{1 - x_2^2})^2}$
$PC = sqrt{(x_2 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
$CD = sqrt{(x_2 - sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$sinangle OPA = frac{13}{sqrt{frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $sin angle OPA = 0.399$.
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
$endgroup$
add a comment |
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$begingroup$
$$textbf{Edition of 11.01.2019}$$
$color{brown}{textbf{The issue data}}$
Let
$$angle OPA=varphi,quad angle DAB = alpha,quad angle CDA=delta,quad AD=d=13x,quad PB=u,quad PC=v.tag1$$
Denote
begin{vmatrix}
AB & BC & CD & AC & BD & PA & PD & AO\
a & b & c & f & g & - & - & R\
dfrac{8x+u}3 & dfrac{u+v}2 & dfrac23(5x+v) & - & - & 8x & 5x & R\
end{vmatrix}
$$ $$ $color{brown}{textbf{Triangles AOP and POD}}$
$$AO = OD = R,quad AP=8x,quad PD=5x,quad angle OPA=varphi.$$
Let
$$OP=y,quad angle OAP = angle ODP = psi.$$
Using cosine theorem, one can get
$$
begin{cases}
y^2 = (8x)^2 + R^2 - 16Rxcospsi\[4pt]
y^2 = (5x)^2 + R^2 - 10Rxcospsi\[4pt]
R^2 = (5x)^2+y^2+10xycosvarphi
end{cases}rightarrow
begin{cases}
y^2=R^2-40x^2,\[4pt]
cosvarphi = dfrac{R^2-y^2-25x^2}{10xy},
end{cases}
$$
$$cosvarphi = dfrac{3x}{2sqrt{R^2-40x^2}}.tag2$$
$color{brown}{textbf{Triangle ABP}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad PA=8x,quad PB=u.$$
Cosine theorem leads to the equation
$$u^2=a^2+(8x)^2-16axcosalpha,$$
$$cosalpha = dfrac{a^2+(8x)^2-u^2}{16ax}=dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)}
=dfrac{8x+u+9(8x-u)}{48x},$$
$$cosalpha = dfrac{10x-u}{6x},quad sinalpha = dfrac1{6x}sqrt{(6x)^2-(10x-u)^2}tag3.$$
$color{brown}{textbf{Triangles ABD and BCD}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad AD=13x,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac23(5x+v),$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle BCD = pi-alpha,quad cos(angle BCD) = -cosalpha.$$
Cosine theorem for $triangle ABD$ and $triangle BCD$ gives
$$f^2 = a^2+d^2-2adcosalpha = b^2+c^2+2bccosalpha,$$
$$f^2=dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = dfrac1{36x}left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)right),$$
$$f^2=dfrac19(14u^2-10ux+545x^2),tag4$$
$$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2)
= 0.tag5$$
$triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac f{2sinalpha}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.tag6$$
$color{brown}{textbf{Triangle PCD}}$
$$angle PDC = delta,quad CD=c=dfrac{10x+2v}3,quad PD=5x,quad PC=v.$$
Cosine theorem leads to the equation
$$v^2=c^2+(5x)^2-10cxcosdelta,$$
$$cosdelta = dfrac{c^2+(5x)^2-v^2}{10cx}=dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$
$$cosdelta = dfrac{13x-v}{12x},quad sindelta = dfrac1{12x}sqrt{(12x)^2-(13x-v)^2}.tag7$$
$color{brown}{textbf{Triangles ACD and ABC}}$
$$angle ADC = delta,quad AB=a=dfrac{8x+u}3,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac{10x+2v}3,quad AD=d=13x.$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle ABC = pi-delta,quad cos(angle ABD) = -cosdelta.$$
Cosine theorem for $triangle ACD$ and $triangle ABC$ gives
$$g^2 = c^2+d^2-2cdcosdelta = a^2+b^2+2abcosdelta,$$
$$g^2=dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = dfrac1{36x}left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)right),$$
$$g^2=dfrac19(17v^2-64vx+776x^2),tag8$$
$$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3
=0.tag9$$
$triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac g{2sindelta}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.tag{10}$$
$color{brown}{textbf{Constraints to the radius}}$
Equations $(6),(10)$ form the system
$$R^2 = dfrac{14u^2-10u+545}{-u^2+20u-64}=dfrac{68v^2-256v+3104}{-v^2+26v-25}.tag{11}$$
Easy to show that
$$minlimits_{uin(4,16)} R^2 = dfrac{169}4quadtext{at}quad u=dfrac{38}5$$
(see also Wolfram Alpha),
$$minlimits_{vin(1,25)} R^2 = dfrac{169}4quadtext{at}quad v=dfrac{43}7$$
(see also Wolfram Alpha).
I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$
At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.
$color{brown}{textbf{Calculation of u and v}}$
If assume $x=1$, then will be obtained solution for $dfrac ux, dfrac vx.$
Equations $(4),(9)$ under this condition form the system
$$begin{cases}
u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{12}$$
Summation of equations $(12)$ with factors $1$ and $(-4)$ gives
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{13}$$
Taking in account that
$$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\
=(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\
+ 37v^3+191v^2-3720v+54112,$$
this leads to the system with the explicit expresssion for $p,$
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0,
end{cases}tag{14}$$
so
$$begin{cases}
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\[4pt]
222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0end{cases}tag{15}$$
wherein
$$222=2cdot3cdot37,quad 152739664 = 2^4cdot7^2cdot11cdot89cdot199.tag{16}$$
The integer roots are dividers of $152739664.$
Horner's scheme for them is
begin{vmatrix}
v & mathbf{222} & mathbf{8426} & mathbf{-64431} & mathbf{-137056} & mathbf{3387595} & mathbf{-40679520} & mathbf{152739664}\
2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\
-2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \
4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\
-4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\
color{brown}{mathbf{7}} & mathbf{222} & mathbf{9980} & mathbf{5429} & mathbf{-99053} & mathbf{2694224} & mathbf{-21819952} & 0 \
end{vmatrix}
The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.
After the integer roots should be checked rational roots, using dividers of 222 as denominators.
Results of the searching allow to write the system in the form of
$$begin{cases}
(v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\[4pt]
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944},
end{cases}$$
with the positive solutions
$$binom{u/x}{v/x} = left{binom{157/19}7, binom7{199/37}right}.tag{17}$$
Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$
$$left(dfrac Rxright)^2 = dfrac{14left(dfrac uxright)^2-10left(dfrac uxright)+545}{36-left(left(dfrac uxright)-10right)^2},tag{18}$$
$$cosvarphi = dfrac{3}{2sqrt{left(dfrac Rxright)^2-40}}.tag{19}$$
From $(18)$ should
$$dfrac ux in(4,16).$$
$color{brown}{textbf{Results}}$
If $dfrac ux = 7,$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
If $dfrac ux=frac{157}{19}approx 8.26316$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
Therefore, both of the possible solutions leads to the similar angle $varphi.$
$color{brown}{textbf{Coordinates}}$
If $O=(0,0),quad x=1,quad u=7,quad v=dfrac{199}{37},quad R=sqrt{43},quad b=dfrac{229}{37},quad c=dfrac{256}{37},quad$ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-cos^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
5 & dfrac12 & dfrac{sqrt3}2 & -dfrac{13}2 & -4 & 3sqrt3 &dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
dfrac{256}{37} & dfrac{47}{74} & dfrac{33}{74}sqrt3 & left(dfrac32,dfrac{sqrt3}2right) & left(dfrac{13}2,dfrac{sqrt3}2right) & left(dfrac{5765}{2738},dfrac{9817}{2738}sqrt3right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tandfrac{P_Y}{P_X}=dfrac{sqrt3}3.$$
If $O=(0,0),quad x=1,quad u=dfrac{157}{19},quad v=7,quad R=sqrt{43},quad b=dfrac{145}{19},quad c = 8,quad $ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-sin^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
dfrac{103}{19} & dfrac{11}{38} & dfrac{21sqrt{3}}{38} & -dfrac{13}2 & -dfrac{1780}{361} & dfrac{901}{361}sqrt3 &-dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
8 & dfrac12 & dfrac{sqrt3}2 & left(dfrac32,-dfrac{sqrt3}2right) & left(dfrac{13}2,-dfrac{sqrt3}2right) & left(dfrac52,dfrac{7sqrt3}2right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tanleft|dfrac{P_Y}{P_X}right|=dfrac{sqrt3}3.$$
It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
1
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
add a comment |
$begingroup$
$$textbf{Edition of 11.01.2019}$$
$color{brown}{textbf{The issue data}}$
Let
$$angle OPA=varphi,quad angle DAB = alpha,quad angle CDA=delta,quad AD=d=13x,quad PB=u,quad PC=v.tag1$$
Denote
begin{vmatrix}
AB & BC & CD & AC & BD & PA & PD & AO\
a & b & c & f & g & - & - & R\
dfrac{8x+u}3 & dfrac{u+v}2 & dfrac23(5x+v) & - & - & 8x & 5x & R\
end{vmatrix}
$$ $$ $color{brown}{textbf{Triangles AOP and POD}}$
$$AO = OD = R,quad AP=8x,quad PD=5x,quad angle OPA=varphi.$$
Let
$$OP=y,quad angle OAP = angle ODP = psi.$$
Using cosine theorem, one can get
$$
begin{cases}
y^2 = (8x)^2 + R^2 - 16Rxcospsi\[4pt]
y^2 = (5x)^2 + R^2 - 10Rxcospsi\[4pt]
R^2 = (5x)^2+y^2+10xycosvarphi
end{cases}rightarrow
begin{cases}
y^2=R^2-40x^2,\[4pt]
cosvarphi = dfrac{R^2-y^2-25x^2}{10xy},
end{cases}
$$
$$cosvarphi = dfrac{3x}{2sqrt{R^2-40x^2}}.tag2$$
$color{brown}{textbf{Triangle ABP}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad PA=8x,quad PB=u.$$
Cosine theorem leads to the equation
$$u^2=a^2+(8x)^2-16axcosalpha,$$
$$cosalpha = dfrac{a^2+(8x)^2-u^2}{16ax}=dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)}
=dfrac{8x+u+9(8x-u)}{48x},$$
$$cosalpha = dfrac{10x-u}{6x},quad sinalpha = dfrac1{6x}sqrt{(6x)^2-(10x-u)^2}tag3.$$
$color{brown}{textbf{Triangles ABD and BCD}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad AD=13x,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac23(5x+v),$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle BCD = pi-alpha,quad cos(angle BCD) = -cosalpha.$$
Cosine theorem for $triangle ABD$ and $triangle BCD$ gives
$$f^2 = a^2+d^2-2adcosalpha = b^2+c^2+2bccosalpha,$$
$$f^2=dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = dfrac1{36x}left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)right),$$
$$f^2=dfrac19(14u^2-10ux+545x^2),tag4$$
$$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2)
= 0.tag5$$
$triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac f{2sinalpha}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.tag6$$
$color{brown}{textbf{Triangle PCD}}$
$$angle PDC = delta,quad CD=c=dfrac{10x+2v}3,quad PD=5x,quad PC=v.$$
Cosine theorem leads to the equation
$$v^2=c^2+(5x)^2-10cxcosdelta,$$
$$cosdelta = dfrac{c^2+(5x)^2-v^2}{10cx}=dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$
$$cosdelta = dfrac{13x-v}{12x},quad sindelta = dfrac1{12x}sqrt{(12x)^2-(13x-v)^2}.tag7$$
$color{brown}{textbf{Triangles ACD and ABC}}$
$$angle ADC = delta,quad AB=a=dfrac{8x+u}3,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac{10x+2v}3,quad AD=d=13x.$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle ABC = pi-delta,quad cos(angle ABD) = -cosdelta.$$
Cosine theorem for $triangle ACD$ and $triangle ABC$ gives
$$g^2 = c^2+d^2-2cdcosdelta = a^2+b^2+2abcosdelta,$$
$$g^2=dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = dfrac1{36x}left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)right),$$
$$g^2=dfrac19(17v^2-64vx+776x^2),tag8$$
$$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3
=0.tag9$$
$triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac g{2sindelta}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.tag{10}$$
$color{brown}{textbf{Constraints to the radius}}$
Equations $(6),(10)$ form the system
$$R^2 = dfrac{14u^2-10u+545}{-u^2+20u-64}=dfrac{68v^2-256v+3104}{-v^2+26v-25}.tag{11}$$
Easy to show that
$$minlimits_{uin(4,16)} R^2 = dfrac{169}4quadtext{at}quad u=dfrac{38}5$$
(see also Wolfram Alpha),
$$minlimits_{vin(1,25)} R^2 = dfrac{169}4quadtext{at}quad v=dfrac{43}7$$
(see also Wolfram Alpha).
I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$
At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.
$color{brown}{textbf{Calculation of u and v}}$
If assume $x=1$, then will be obtained solution for $dfrac ux, dfrac vx.$
Equations $(4),(9)$ under this condition form the system
$$begin{cases}
u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{12}$$
Summation of equations $(12)$ with factors $1$ and $(-4)$ gives
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{13}$$
Taking in account that
$$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\
=(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\
+ 37v^3+191v^2-3720v+54112,$$
this leads to the system with the explicit expresssion for $p,$
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0,
end{cases}tag{14}$$
so
$$begin{cases}
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\[4pt]
222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0end{cases}tag{15}$$
wherein
$$222=2cdot3cdot37,quad 152739664 = 2^4cdot7^2cdot11cdot89cdot199.tag{16}$$
The integer roots are dividers of $152739664.$
Horner's scheme for them is
begin{vmatrix}
v & mathbf{222} & mathbf{8426} & mathbf{-64431} & mathbf{-137056} & mathbf{3387595} & mathbf{-40679520} & mathbf{152739664}\
2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\
-2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \
4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\
-4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\
color{brown}{mathbf{7}} & mathbf{222} & mathbf{9980} & mathbf{5429} & mathbf{-99053} & mathbf{2694224} & mathbf{-21819952} & 0 \
end{vmatrix}
The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.
After the integer roots should be checked rational roots, using dividers of 222 as denominators.
Results of the searching allow to write the system in the form of
$$begin{cases}
(v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\[4pt]
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944},
end{cases}$$
with the positive solutions
$$binom{u/x}{v/x} = left{binom{157/19}7, binom7{199/37}right}.tag{17}$$
Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$
$$left(dfrac Rxright)^2 = dfrac{14left(dfrac uxright)^2-10left(dfrac uxright)+545}{36-left(left(dfrac uxright)-10right)^2},tag{18}$$
$$cosvarphi = dfrac{3}{2sqrt{left(dfrac Rxright)^2-40}}.tag{19}$$
From $(18)$ should
$$dfrac ux in(4,16).$$
$color{brown}{textbf{Results}}$
If $dfrac ux = 7,$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
If $dfrac ux=frac{157}{19}approx 8.26316$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
Therefore, both of the possible solutions leads to the similar angle $varphi.$
$color{brown}{textbf{Coordinates}}$
If $O=(0,0),quad x=1,quad u=7,quad v=dfrac{199}{37},quad R=sqrt{43},quad b=dfrac{229}{37},quad c=dfrac{256}{37},quad$ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-cos^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
5 & dfrac12 & dfrac{sqrt3}2 & -dfrac{13}2 & -4 & 3sqrt3 &dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
dfrac{256}{37} & dfrac{47}{74} & dfrac{33}{74}sqrt3 & left(dfrac32,dfrac{sqrt3}2right) & left(dfrac{13}2,dfrac{sqrt3}2right) & left(dfrac{5765}{2738},dfrac{9817}{2738}sqrt3right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tandfrac{P_Y}{P_X}=dfrac{sqrt3}3.$$
If $O=(0,0),quad x=1,quad u=dfrac{157}{19},quad v=7,quad R=sqrt{43},quad b=dfrac{145}{19},quad c = 8,quad $ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-sin^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
dfrac{103}{19} & dfrac{11}{38} & dfrac{21sqrt{3}}{38} & -dfrac{13}2 & -dfrac{1780}{361} & dfrac{901}{361}sqrt3 &-dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
8 & dfrac12 & dfrac{sqrt3}2 & left(dfrac32,-dfrac{sqrt3}2right) & left(dfrac{13}2,-dfrac{sqrt3}2right) & left(dfrac52,dfrac{7sqrt3}2right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tanleft|dfrac{P_Y}{P_X}right|=dfrac{sqrt3}3.$$
It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
1
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
add a comment |
$begingroup$
$$textbf{Edition of 11.01.2019}$$
$color{brown}{textbf{The issue data}}$
Let
$$angle OPA=varphi,quad angle DAB = alpha,quad angle CDA=delta,quad AD=d=13x,quad PB=u,quad PC=v.tag1$$
Denote
begin{vmatrix}
AB & BC & CD & AC & BD & PA & PD & AO\
a & b & c & f & g & - & - & R\
dfrac{8x+u}3 & dfrac{u+v}2 & dfrac23(5x+v) & - & - & 8x & 5x & R\
end{vmatrix}
$$ $$ $color{brown}{textbf{Triangles AOP and POD}}$
$$AO = OD = R,quad AP=8x,quad PD=5x,quad angle OPA=varphi.$$
Let
$$OP=y,quad angle OAP = angle ODP = psi.$$
Using cosine theorem, one can get
$$
begin{cases}
y^2 = (8x)^2 + R^2 - 16Rxcospsi\[4pt]
y^2 = (5x)^2 + R^2 - 10Rxcospsi\[4pt]
R^2 = (5x)^2+y^2+10xycosvarphi
end{cases}rightarrow
begin{cases}
y^2=R^2-40x^2,\[4pt]
cosvarphi = dfrac{R^2-y^2-25x^2}{10xy},
end{cases}
$$
$$cosvarphi = dfrac{3x}{2sqrt{R^2-40x^2}}.tag2$$
$color{brown}{textbf{Triangle ABP}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad PA=8x,quad PB=u.$$
Cosine theorem leads to the equation
$$u^2=a^2+(8x)^2-16axcosalpha,$$
$$cosalpha = dfrac{a^2+(8x)^2-u^2}{16ax}=dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)}
=dfrac{8x+u+9(8x-u)}{48x},$$
$$cosalpha = dfrac{10x-u}{6x},quad sinalpha = dfrac1{6x}sqrt{(6x)^2-(10x-u)^2}tag3.$$
$color{brown}{textbf{Triangles ABD and BCD}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad AD=13x,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac23(5x+v),$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle BCD = pi-alpha,quad cos(angle BCD) = -cosalpha.$$
Cosine theorem for $triangle ABD$ and $triangle BCD$ gives
$$f^2 = a^2+d^2-2adcosalpha = b^2+c^2+2bccosalpha,$$
$$f^2=dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = dfrac1{36x}left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)right),$$
$$f^2=dfrac19(14u^2-10ux+545x^2),tag4$$
$$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2)
= 0.tag5$$
$triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac f{2sinalpha}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.tag6$$
$color{brown}{textbf{Triangle PCD}}$
$$angle PDC = delta,quad CD=c=dfrac{10x+2v}3,quad PD=5x,quad PC=v.$$
Cosine theorem leads to the equation
$$v^2=c^2+(5x)^2-10cxcosdelta,$$
$$cosdelta = dfrac{c^2+(5x)^2-v^2}{10cx}=dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$
$$cosdelta = dfrac{13x-v}{12x},quad sindelta = dfrac1{12x}sqrt{(12x)^2-(13x-v)^2}.tag7$$
$color{brown}{textbf{Triangles ACD and ABC}}$
$$angle ADC = delta,quad AB=a=dfrac{8x+u}3,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac{10x+2v}3,quad AD=d=13x.$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle ABC = pi-delta,quad cos(angle ABD) = -cosdelta.$$
Cosine theorem for $triangle ACD$ and $triangle ABC$ gives
$$g^2 = c^2+d^2-2cdcosdelta = a^2+b^2+2abcosdelta,$$
$$g^2=dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = dfrac1{36x}left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)right),$$
$$g^2=dfrac19(17v^2-64vx+776x^2),tag8$$
$$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3
=0.tag9$$
$triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac g{2sindelta}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.tag{10}$$
$color{brown}{textbf{Constraints to the radius}}$
Equations $(6),(10)$ form the system
$$R^2 = dfrac{14u^2-10u+545}{-u^2+20u-64}=dfrac{68v^2-256v+3104}{-v^2+26v-25}.tag{11}$$
Easy to show that
$$minlimits_{uin(4,16)} R^2 = dfrac{169}4quadtext{at}quad u=dfrac{38}5$$
(see also Wolfram Alpha),
$$minlimits_{vin(1,25)} R^2 = dfrac{169}4quadtext{at}quad v=dfrac{43}7$$
(see also Wolfram Alpha).
I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$
At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.
$color{brown}{textbf{Calculation of u and v}}$
If assume $x=1$, then will be obtained solution for $dfrac ux, dfrac vx.$
Equations $(4),(9)$ under this condition form the system
$$begin{cases}
u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{12}$$
Summation of equations $(12)$ with factors $1$ and $(-4)$ gives
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{13}$$
Taking in account that
$$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\
=(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\
+ 37v^3+191v^2-3720v+54112,$$
this leads to the system with the explicit expresssion for $p,$
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0,
end{cases}tag{14}$$
so
$$begin{cases}
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\[4pt]
222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0end{cases}tag{15}$$
wherein
$$222=2cdot3cdot37,quad 152739664 = 2^4cdot7^2cdot11cdot89cdot199.tag{16}$$
The integer roots are dividers of $152739664.$
Horner's scheme for them is
begin{vmatrix}
v & mathbf{222} & mathbf{8426} & mathbf{-64431} & mathbf{-137056} & mathbf{3387595} & mathbf{-40679520} & mathbf{152739664}\
2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\
-2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \
4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\
-4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\
color{brown}{mathbf{7}} & mathbf{222} & mathbf{9980} & mathbf{5429} & mathbf{-99053} & mathbf{2694224} & mathbf{-21819952} & 0 \
end{vmatrix}
The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.
After the integer roots should be checked rational roots, using dividers of 222 as denominators.
Results of the searching allow to write the system in the form of
$$begin{cases}
(v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\[4pt]
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944},
end{cases}$$
with the positive solutions
$$binom{u/x}{v/x} = left{binom{157/19}7, binom7{199/37}right}.tag{17}$$
Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$
$$left(dfrac Rxright)^2 = dfrac{14left(dfrac uxright)^2-10left(dfrac uxright)+545}{36-left(left(dfrac uxright)-10right)^2},tag{18}$$
$$cosvarphi = dfrac{3}{2sqrt{left(dfrac Rxright)^2-40}}.tag{19}$$
From $(18)$ should
$$dfrac ux in(4,16).$$
$color{brown}{textbf{Results}}$
If $dfrac ux = 7,$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
If $dfrac ux=frac{157}{19}approx 8.26316$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
Therefore, both of the possible solutions leads to the similar angle $varphi.$
$color{brown}{textbf{Coordinates}}$
If $O=(0,0),quad x=1,quad u=7,quad v=dfrac{199}{37},quad R=sqrt{43},quad b=dfrac{229}{37},quad c=dfrac{256}{37},quad$ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-cos^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
5 & dfrac12 & dfrac{sqrt3}2 & -dfrac{13}2 & -4 & 3sqrt3 &dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
dfrac{256}{37} & dfrac{47}{74} & dfrac{33}{74}sqrt3 & left(dfrac32,dfrac{sqrt3}2right) & left(dfrac{13}2,dfrac{sqrt3}2right) & left(dfrac{5765}{2738},dfrac{9817}{2738}sqrt3right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tandfrac{P_Y}{P_X}=dfrac{sqrt3}3.$$
If $O=(0,0),quad x=1,quad u=dfrac{157}{19},quad v=7,quad R=sqrt{43},quad b=dfrac{145}{19},quad c = 8,quad $ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-sin^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
dfrac{103}{19} & dfrac{11}{38} & dfrac{21sqrt{3}}{38} & -dfrac{13}2 & -dfrac{1780}{361} & dfrac{901}{361}sqrt3 &-dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
8 & dfrac12 & dfrac{sqrt3}2 & left(dfrac32,-dfrac{sqrt3}2right) & left(dfrac{13}2,-dfrac{sqrt3}2right) & left(dfrac52,dfrac{7sqrt3}2right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tanleft|dfrac{P_Y}{P_X}right|=dfrac{sqrt3}3.$$
It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
$endgroup$
$$textbf{Edition of 11.01.2019}$$
$color{brown}{textbf{The issue data}}$
Let
$$angle OPA=varphi,quad angle DAB = alpha,quad angle CDA=delta,quad AD=d=13x,quad PB=u,quad PC=v.tag1$$
Denote
begin{vmatrix}
AB & BC & CD & AC & BD & PA & PD & AO\
a & b & c & f & g & - & - & R\
dfrac{8x+u}3 & dfrac{u+v}2 & dfrac23(5x+v) & - & - & 8x & 5x & R\
end{vmatrix}
$$ $$ $color{brown}{textbf{Triangles AOP and POD}}$
$$AO = OD = R,quad AP=8x,quad PD=5x,quad angle OPA=varphi.$$
Let
$$OP=y,quad angle OAP = angle ODP = psi.$$
Using cosine theorem, one can get
$$
begin{cases}
y^2 = (8x)^2 + R^2 - 16Rxcospsi\[4pt]
y^2 = (5x)^2 + R^2 - 10Rxcospsi\[4pt]
R^2 = (5x)^2+y^2+10xycosvarphi
end{cases}rightarrow
begin{cases}
y^2=R^2-40x^2,\[4pt]
cosvarphi = dfrac{R^2-y^2-25x^2}{10xy},
end{cases}
$$
$$cosvarphi = dfrac{3x}{2sqrt{R^2-40x^2}}.tag2$$
$color{brown}{textbf{Triangle ABP}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad PA=8x,quad PB=u.$$
Cosine theorem leads to the equation
$$u^2=a^2+(8x)^2-16axcosalpha,$$
$$cosalpha = dfrac{a^2+(8x)^2-u^2}{16ax}=dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)}
=dfrac{8x+u+9(8x-u)}{48x},$$
$$cosalpha = dfrac{10x-u}{6x},quad sinalpha = dfrac1{6x}sqrt{(6x)^2-(10x-u)^2}tag3.$$
$color{brown}{textbf{Triangles ABD and BCD}}$
$$angle DAB = alpha,quad AB=a=dfrac{8x+u}3,quad AD=13x,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac23(5x+v),$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle BCD = pi-alpha,quad cos(angle BCD) = -cosalpha.$$
Cosine theorem for $triangle ABD$ and $triangle BCD$ gives
$$f^2 = a^2+d^2-2adcosalpha = b^2+c^2+2bccosalpha,$$
$$f^2=dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = dfrac1{36x}left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)right),$$
$$f^2=dfrac19(14u^2-10ux+545x^2),tag4$$
$$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2)
= 0.tag5$$
$triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac f{2sinalpha}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.tag6$$
$color{brown}{textbf{Triangle PCD}}$
$$angle PDC = delta,quad CD=c=dfrac{10x+2v}3,quad PD=5x,quad PC=v.$$
Cosine theorem leads to the equation
$$v^2=c^2+(5x)^2-10cxcosdelta,$$
$$cosdelta = dfrac{c^2+(5x)^2-v^2}{10cx}=dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$
$$cosdelta = dfrac{13x-v}{12x},quad sindelta = dfrac1{12x}sqrt{(12x)^2-(13x-v)^2}.tag7$$
$color{brown}{textbf{Triangles ACD and ABC}}$
$$angle ADC = delta,quad AB=a=dfrac{8x+u}3,quad BC=b=dfrac{u+v}2,quad CD=c=dfrac{10x+2v}3,quad AD=d=13x.$$
In the inscribed quadrilateral, the sums of opposite angles are equal to $pi,$ so
$$angle ABC = pi-delta,quad cos(angle ABD) = -cosdelta.$$
Cosine theorem for $triangle ACD$ and $triangle ABC$ gives
$$g^2 = c^2+d^2-2cdcosdelta = a^2+b^2+2abcosdelta,$$
$$g^2=dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = dfrac1{36x}left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)right),$$
$$g^2=dfrac19(17v^2-64vx+776x^2),tag8$$
$$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3
=0.tag9$$
$triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives
$$R = dfrac g{2sindelta}.$$
Using $(3)-(4),$
$$R = xsqrt{dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.tag{10}$$
$color{brown}{textbf{Constraints to the radius}}$
Equations $(6),(10)$ form the system
$$R^2 = dfrac{14u^2-10u+545}{-u^2+20u-64}=dfrac{68v^2-256v+3104}{-v^2+26v-25}.tag{11}$$
Easy to show that
$$minlimits_{uin(4,16)} R^2 = dfrac{169}4quadtext{at}quad u=dfrac{38}5$$
(see also Wolfram Alpha),
$$minlimits_{vin(1,25)} R^2 = dfrac{169}4quadtext{at}quad v=dfrac{43}7$$
(see also Wolfram Alpha).
I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$
At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.
$color{brown}{textbf{Calculation of u and v}}$
If assume $x=1$, then will be obtained solution for $dfrac ux, dfrac vx.$
Equations $(4),(9)$ under this condition form the system
$$begin{cases}
u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{12}$$
Summation of equations $(12)$ with factors $1$ and $(-4)$ gives
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0.
end{cases}tag{13}$$
Taking in account that
$$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\
=(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\
+ 37v^3+191v^2-3720v+54112,$$
this leads to the system with the explicit expresssion for $p,$
$$begin{cases}
19u^2+6uv-37v^2+48u+120v-1068 = 0,\
u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0,
end{cases}tag{14}$$
so
$$begin{cases}
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\[4pt]
222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0end{cases}tag{15}$$
wherein
$$222=2cdot3cdot37,quad 152739664 = 2^4cdot7^2cdot11cdot89cdot199.tag{16}$$
The integer roots are dividers of $152739664.$
Horner's scheme for them is
begin{vmatrix}
v & mathbf{222} & mathbf{8426} & mathbf{-64431} & mathbf{-137056} & mathbf{3387595} & mathbf{-40679520} & mathbf{152739664}\
2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\
-2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \
4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\
-4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\
color{brown}{mathbf{7}} & mathbf{222} & mathbf{9980} & mathbf{5429} & mathbf{-99053} & mathbf{2694224} & mathbf{-21819952} & 0 \
end{vmatrix}
The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.
After the integer roots should be checked rational roots, using dividers of 222 as denominators.
Results of the searching allow to write the system in the form of
$$begin{cases}
(v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\[4pt]
u=dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944},
end{cases}$$
with the positive solutions
$$binom{u/x}{v/x} = left{binom{157/19}7, binom7{199/37}right}.tag{17}$$
Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$
$$left(dfrac Rxright)^2 = dfrac{14left(dfrac uxright)^2-10left(dfrac uxright)+545}{36-left(left(dfrac uxright)-10right)^2},tag{18}$$
$$cosvarphi = dfrac{3}{2sqrt{left(dfrac Rxright)^2-40}}.tag{19}$$
From $(18)$ should
$$dfrac ux in(4,16).$$
$color{brown}{textbf{Results}}$
If $dfrac ux = 7,$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
If $dfrac ux=frac{157}{19}approx 8.26316$ then
$$left(dfrac Rxright)^2=43,quad cosvarphi = dfrac{sqrt3}2,$$
$$color{brown}{boxed{mathbf{varphi = dfracpi6}}}.$$
Therefore, both of the possible solutions leads to the similar angle $varphi.$
$color{brown}{textbf{Coordinates}}$
If $O=(0,0),quad x=1,quad u=7,quad v=dfrac{199}{37},quad R=sqrt{43},quad b=dfrac{229}{37},quad c=dfrac{256}{37},quad$ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-cos^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
5 & dfrac12 & dfrac{sqrt3}2 & -dfrac{13}2 & -4 & 3sqrt3 &dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
dfrac{256}{37} & dfrac{47}{74} & dfrac{33}{74}sqrt3 & left(dfrac32,dfrac{sqrt3}2right) & left(dfrac{13}2,dfrac{sqrt3}2right) & left(dfrac{5765}{2738},dfrac{9817}{2738}sqrt3right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tandfrac{P_Y}{P_X}=dfrac{sqrt3}3.$$
If $O=(0,0),quad x=1,quad u=dfrac{157}{19},quad v=7,quad R=sqrt{43},quad b=dfrac{145}{19},quad c = 8,quad $ then
begin{vmatrix}
mathbf{a} & mathbf{cosalpha} & mathbf{sinalpha} & mathbf{A_X} & mathbf{B_X} & mathbf{B_Y} & mathbf{A_Y} \
dfrac{8x+u}3 & dfrac{10x-u}{6x} & sqrt{1-sin^2alpha} & -dfrac{13}2x & A_X+acosalpha & sqrt{R^2-B_X^2} & B_Y-asinalpha \
dfrac{103}{19} & dfrac{11}{38} & dfrac{21sqrt{3}}{38} & -dfrac{13}2 & -dfrac{1780}{361} & dfrac{901}{361}sqrt3 &-dfrac{sqrt3}2\
end{vmatrix}
The other parameters are
begin{vmatrix}
mathbf{c} & mathbf{cosdelta} & mathbf{sindelta} & mathbf{P} & mathbf{D} & mathbf{C} \
dfrac{2(5x+v)}3 & dfrac{13x-v}{12x} & sqrt{1-cos^2delta} & left(dfrac32 x,A_Yright) & left(dfrac{13}2 x, A_Yright) & D + c(-cosdelta, sindelta) \
8 & dfrac12 & dfrac{sqrt3}2 & left(dfrac32,-dfrac{sqrt3}2right) & left(dfrac{13}2,-dfrac{sqrt3}2right) & left(dfrac52,dfrac{7sqrt3}2right) \
end{vmatrix}
One can check that
$$|A| = |B| = |C| = |D| = sqrt{43},quad tanleft|dfrac{P_Y}{P_X}right|=dfrac{sqrt3}3.$$
It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
edited Jan 11 at 22:46
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
answered Dec 22 '18 at 2:33
Yuri NegometyanovYuri Negometyanov
11.1k1728
11.1k1728
1
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
add a comment |
1
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
1
1
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
very nice. did you try to calculate $u$ and $v$ by hands?
$endgroup$
– dezdichado
Dec 24 '18 at 17:26
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
$begingroup$
@dezdichado Thanks! Horner's scheme see in the updated answer. There are a lot of roots. But $(7)$ means $vin(1,25),$ and this simplifies the problem.
$endgroup$
– Yuri Negometyanov
Dec 24 '18 at 20:42
add a comment |
$begingroup$
(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)
Ignoring reflections in $overline{AD}$, there are two viable quadrilaterals. They're shown here as $square AB^prime CD$, and $square ABC^prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $square ABCD$ that traces two equilateral triangles
$square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:
$$angle OPA = 30^circ quadtoquad sin angle OPA = frac12$$
Currently, I don't have a straightforward way to construct $B^prime$ and $C^prime$, but their specifications appear below.
$$begin{array}{c:cccc:cccc:cc}
square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & cos APB & cos DPC \ hline \
AB^prime C D & 8 & dfrac{157}{19} & 7 & 5 & dfrac{103}{19} & dfrac{145}{19} & 8 & 13 & dfrac{4643}{5966} & dfrac17 \ \
ABC^prime D & 8 & 7 & dfrac{199}{37} & 5 & 5 & dfrac{229}{37} & dfrac{256}{37} & 13 & dfrac{11}{14} & dfrac{829}{7363} \ \ hline
end{array}$$
(In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^prime$ or $C^prime$.)
My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using
$$|AB|=p(|PA|+|PB|) qquad |BC|=q(|PB|+|PC|) qquad |CD|=r(|PC|+|PD|)$$
but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $sin angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)
For the derivation, define
$$a := |PA| quad b := |PB| quad c := |PC| quad d := |PD|$$
$$alpha := angle APB qquad beta := angle BPC qquad gamma := angle CPD$$
where we may as well take $d=5$. The problem conditions tell us
$$a = frac85 d = 8 qquad |AB| = frac1{3}( b + 8 ) qquad |BC| = frac12(b+c) qquad |CD| = frac23(c+5) tag{1}$$
By the Law of Cosines and (1), we have
$$cosalpha = frac1{2ab}left(-|AB|^2+a^2+b^2right) = frac{1}{18b}left(b^2 - 2 b + 64right) tag{2a}$$
also
$$cosbeta = frac{1}{8bc} left(3 b^2 - 2 b c + 3 c^2right) qquad
cosgamma = frac{1}{18c} left(c^2 - 8 c + 25right) tag{2b}$$
Now, since $alpha+beta+gamma=180^circ$, one readily shows
$$1 - 2 cosalpha cosbeta cosgamma - cos^2alpha - cos^2beta - cos^2gamma = 0 tag{3}$$
Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...
$$begin{align}
0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \
&- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \
&+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \
&- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \
&+ 29200 b^2 - 12800 b c + 84736 c^2
end{align} tag{4}$$
To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:
Lemma. Given $triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by
$$p^2-r^2 = - xyz frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ y z sinangle YPZ + z x sinangle ZPX + x y sinangle XPY }
tag{$star$}$$
where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.
Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have
$$p^2-r^2 = - frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ dfrac{1}{x} sinangle YPZ + dfrac{1}{y} sinangle ZPX + dfrac{1}{z} sinangle XPY }$$
which is somewhat intriguing. But I digress ...
In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $triangle ABC$, getting
$$-40 = - a b c frac{ a sinbeta + b sin(180^circ+gamma) + c sinalpha }{ a b sinalpha + b c sinbeta + c a sin(180^circ+gamma)}
= - 8 b c frac{ 8 sinbeta - b singamma + c sinalpha }{ 8 b sinalpha + b c sinbeta - 8 c singamma}$$
Since $gamma = 180^circ-alpha-beta$, we can manipulate above into the form
$$sinalphaleft(-b c^2 + 40 b + ccosbetaleft( b^2 - 40 right)right) = c sinbeta left(-3 b + cosalphaleft( b^2 - 40right)right) tag{5}$$
Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find
$$0 = (b + c) (b + 8) left(begin{align}
&phantom{+} 4 b^4 c^2 -
80 b^3 c^2 \
&- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \
&- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \
&+ 9175 b^2 + 11230 b c + 2775 c^2 \
&- 128000 b + 409600 end{align}right) tag{6}$$
From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $square$
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
$endgroup$
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
$endgroup$
– Blue
Dec 8 '18 at 12:06
$begingroup$
Also, it seems that $Q$ lies on the segment $B'C'$.
$endgroup$
– timon92
Dec 8 '18 at 12:07
1
$begingroup$
You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
|
show 3 more comments
$begingroup$
(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)
Ignoring reflections in $overline{AD}$, there are two viable quadrilaterals. They're shown here as $square AB^prime CD$, and $square ABC^prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $square ABCD$ that traces two equilateral triangles
$square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:
$$angle OPA = 30^circ quadtoquad sin angle OPA = frac12$$
Currently, I don't have a straightforward way to construct $B^prime$ and $C^prime$, but their specifications appear below.
$$begin{array}{c:cccc:cccc:cc}
square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & cos APB & cos DPC \ hline \
AB^prime C D & 8 & dfrac{157}{19} & 7 & 5 & dfrac{103}{19} & dfrac{145}{19} & 8 & 13 & dfrac{4643}{5966} & dfrac17 \ \
ABC^prime D & 8 & 7 & dfrac{199}{37} & 5 & 5 & dfrac{229}{37} & dfrac{256}{37} & 13 & dfrac{11}{14} & dfrac{829}{7363} \ \ hline
end{array}$$
(In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^prime$ or $C^prime$.)
My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using
$$|AB|=p(|PA|+|PB|) qquad |BC|=q(|PB|+|PC|) qquad |CD|=r(|PC|+|PD|)$$
but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $sin angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)
For the derivation, define
$$a := |PA| quad b := |PB| quad c := |PC| quad d := |PD|$$
$$alpha := angle APB qquad beta := angle BPC qquad gamma := angle CPD$$
where we may as well take $d=5$. The problem conditions tell us
$$a = frac85 d = 8 qquad |AB| = frac1{3}( b + 8 ) qquad |BC| = frac12(b+c) qquad |CD| = frac23(c+5) tag{1}$$
By the Law of Cosines and (1), we have
$$cosalpha = frac1{2ab}left(-|AB|^2+a^2+b^2right) = frac{1}{18b}left(b^2 - 2 b + 64right) tag{2a}$$
also
$$cosbeta = frac{1}{8bc} left(3 b^2 - 2 b c + 3 c^2right) qquad
cosgamma = frac{1}{18c} left(c^2 - 8 c + 25right) tag{2b}$$
Now, since $alpha+beta+gamma=180^circ$, one readily shows
$$1 - 2 cosalpha cosbeta cosgamma - cos^2alpha - cos^2beta - cos^2gamma = 0 tag{3}$$
Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...
$$begin{align}
0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \
&- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \
&+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \
&- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \
&+ 29200 b^2 - 12800 b c + 84736 c^2
end{align} tag{4}$$
To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:
Lemma. Given $triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by
$$p^2-r^2 = - xyz frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ y z sinangle YPZ + z x sinangle ZPX + x y sinangle XPY }
tag{$star$}$$
where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.
Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have
$$p^2-r^2 = - frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ dfrac{1}{x} sinangle YPZ + dfrac{1}{y} sinangle ZPX + dfrac{1}{z} sinangle XPY }$$
which is somewhat intriguing. But I digress ...
In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $triangle ABC$, getting
$$-40 = - a b c frac{ a sinbeta + b sin(180^circ+gamma) + c sinalpha }{ a b sinalpha + b c sinbeta + c a sin(180^circ+gamma)}
= - 8 b c frac{ 8 sinbeta - b singamma + c sinalpha }{ 8 b sinalpha + b c sinbeta - 8 c singamma}$$
Since $gamma = 180^circ-alpha-beta$, we can manipulate above into the form
$$sinalphaleft(-b c^2 + 40 b + ccosbetaleft( b^2 - 40 right)right) = c sinbeta left(-3 b + cosalphaleft( b^2 - 40right)right) tag{5}$$
Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find
$$0 = (b + c) (b + 8) left(begin{align}
&phantom{+} 4 b^4 c^2 -
80 b^3 c^2 \
&- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \
&- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \
&+ 9175 b^2 + 11230 b c + 2775 c^2 \
&- 128000 b + 409600 end{align}right) tag{6}$$
From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $square$
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
$endgroup$
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
$endgroup$
– Blue
Dec 8 '18 at 12:06
$begingroup$
Also, it seems that $Q$ lies on the segment $B'C'$.
$endgroup$
– timon92
Dec 8 '18 at 12:07
1
$begingroup$
You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
|
show 3 more comments
$begingroup$
(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)
Ignoring reflections in $overline{AD}$, there are two viable quadrilaterals. They're shown here as $square AB^prime CD$, and $square ABC^prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $square ABCD$ that traces two equilateral triangles
$square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:
$$angle OPA = 30^circ quadtoquad sin angle OPA = frac12$$
Currently, I don't have a straightforward way to construct $B^prime$ and $C^prime$, but their specifications appear below.
$$begin{array}{c:cccc:cccc:cc}
square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & cos APB & cos DPC \ hline \
AB^prime C D & 8 & dfrac{157}{19} & 7 & 5 & dfrac{103}{19} & dfrac{145}{19} & 8 & 13 & dfrac{4643}{5966} & dfrac17 \ \
ABC^prime D & 8 & 7 & dfrac{199}{37} & 5 & 5 & dfrac{229}{37} & dfrac{256}{37} & 13 & dfrac{11}{14} & dfrac{829}{7363} \ \ hline
end{array}$$
(In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^prime$ or $C^prime$.)
My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using
$$|AB|=p(|PA|+|PB|) qquad |BC|=q(|PB|+|PC|) qquad |CD|=r(|PC|+|PD|)$$
but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $sin angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)
For the derivation, define
$$a := |PA| quad b := |PB| quad c := |PC| quad d := |PD|$$
$$alpha := angle APB qquad beta := angle BPC qquad gamma := angle CPD$$
where we may as well take $d=5$. The problem conditions tell us
$$a = frac85 d = 8 qquad |AB| = frac1{3}( b + 8 ) qquad |BC| = frac12(b+c) qquad |CD| = frac23(c+5) tag{1}$$
By the Law of Cosines and (1), we have
$$cosalpha = frac1{2ab}left(-|AB|^2+a^2+b^2right) = frac{1}{18b}left(b^2 - 2 b + 64right) tag{2a}$$
also
$$cosbeta = frac{1}{8bc} left(3 b^2 - 2 b c + 3 c^2right) qquad
cosgamma = frac{1}{18c} left(c^2 - 8 c + 25right) tag{2b}$$
Now, since $alpha+beta+gamma=180^circ$, one readily shows
$$1 - 2 cosalpha cosbeta cosgamma - cos^2alpha - cos^2beta - cos^2gamma = 0 tag{3}$$
Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...
$$begin{align}
0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \
&- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \
&+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \
&- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \
&+ 29200 b^2 - 12800 b c + 84736 c^2
end{align} tag{4}$$
To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:
Lemma. Given $triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by
$$p^2-r^2 = - xyz frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ y z sinangle YPZ + z x sinangle ZPX + x y sinangle XPY }
tag{$star$}$$
where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.
Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have
$$p^2-r^2 = - frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ dfrac{1}{x} sinangle YPZ + dfrac{1}{y} sinangle ZPX + dfrac{1}{z} sinangle XPY }$$
which is somewhat intriguing. But I digress ...
In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $triangle ABC$, getting
$$-40 = - a b c frac{ a sinbeta + b sin(180^circ+gamma) + c sinalpha }{ a b sinalpha + b c sinbeta + c a sin(180^circ+gamma)}
= - 8 b c frac{ 8 sinbeta - b singamma + c sinalpha }{ 8 b sinalpha + b c sinbeta - 8 c singamma}$$
Since $gamma = 180^circ-alpha-beta$, we can manipulate above into the form
$$sinalphaleft(-b c^2 + 40 b + ccosbetaleft( b^2 - 40 right)right) = c sinbeta left(-3 b + cosalphaleft( b^2 - 40right)right) tag{5}$$
Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find
$$0 = (b + c) (b + 8) left(begin{align}
&phantom{+} 4 b^4 c^2 -
80 b^3 c^2 \
&- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \
&- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \
&+ 9175 b^2 + 11230 b c + 2775 c^2 \
&- 128000 b + 409600 end{align}right) tag{6}$$
From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $square$
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
$endgroup$
(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)
Ignoring reflections in $overline{AD}$, there are two viable quadrilaterals. They're shown here as $square AB^prime CD$, and $square ABC^prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $square ABCD$ that traces two equilateral triangles
$square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:
$$angle OPA = 30^circ quadtoquad sin angle OPA = frac12$$
Currently, I don't have a straightforward way to construct $B^prime$ and $C^prime$, but their specifications appear below.
$$begin{array}{c:cccc:cccc:cc}
square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & cos APB & cos DPC \ hline \
AB^prime C D & 8 & dfrac{157}{19} & 7 & 5 & dfrac{103}{19} & dfrac{145}{19} & 8 & 13 & dfrac{4643}{5966} & dfrac17 \ \
ABC^prime D & 8 & 7 & dfrac{199}{37} & 5 & 5 & dfrac{229}{37} & dfrac{256}{37} & 13 & dfrac{11}{14} & dfrac{829}{7363} \ \ hline
end{array}$$
(In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^prime$ or $C^prime$.)
My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using
$$|AB|=p(|PA|+|PB|) qquad |BC|=q(|PB|+|PC|) qquad |CD|=r(|PC|+|PD|)$$
but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $sin angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)
For the derivation, define
$$a := |PA| quad b := |PB| quad c := |PC| quad d := |PD|$$
$$alpha := angle APB qquad beta := angle BPC qquad gamma := angle CPD$$
where we may as well take $d=5$. The problem conditions tell us
$$a = frac85 d = 8 qquad |AB| = frac1{3}( b + 8 ) qquad |BC| = frac12(b+c) qquad |CD| = frac23(c+5) tag{1}$$
By the Law of Cosines and (1), we have
$$cosalpha = frac1{2ab}left(-|AB|^2+a^2+b^2right) = frac{1}{18b}left(b^2 - 2 b + 64right) tag{2a}$$
also
$$cosbeta = frac{1}{8bc} left(3 b^2 - 2 b c + 3 c^2right) qquad
cosgamma = frac{1}{18c} left(c^2 - 8 c + 25right) tag{2b}$$
Now, since $alpha+beta+gamma=180^circ$, one readily shows
$$1 - 2 cosalpha cosbeta cosgamma - cos^2alpha - cos^2beta - cos^2gamma = 0 tag{3}$$
Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...
$$begin{align}
0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \
&- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \
&+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \
&- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \
&+ 29200 b^2 - 12800 b c + 84736 c^2
end{align} tag{4}$$
To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:
Lemma. Given $triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by
$$p^2-r^2 = - xyz frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ y z sinangle YPZ + z x sinangle ZPX + x y sinangle XPY }
tag{$star$}$$
where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.
Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have
$$p^2-r^2 = - frac{ x sinangle YPZ + y sinangle ZPX + z sinangle XPY }{ dfrac{1}{x} sinangle YPZ + dfrac{1}{y} sinangle ZPX + dfrac{1}{z} sinangle XPY }$$
which is somewhat intriguing. But I digress ...
In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $triangle ABC$, getting
$$-40 = - a b c frac{ a sinbeta + b sin(180^circ+gamma) + c sinalpha }{ a b sinalpha + b c sinbeta + c a sin(180^circ+gamma)}
= - 8 b c frac{ 8 sinbeta - b singamma + c sinalpha }{ 8 b sinalpha + b c sinbeta - 8 c singamma}$$
Since $gamma = 180^circ-alpha-beta$, we can manipulate above into the form
$$sinalphaleft(-b c^2 + 40 b + ccosbetaleft( b^2 - 40 right)right) = c sinbeta left(-3 b + cosalphaleft( b^2 - 40right)right) tag{5}$$
Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find
$$0 = (b + c) (b + 8) left(begin{align}
&phantom{+} 4 b^4 c^2 -
80 b^3 c^2 \
&- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \
&- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \
&+ 9175 b^2 + 11230 b c + 2775 c^2 \
&- 128000 b + 409600 end{align}right) tag{6}$$
From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $square$
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
edited Jan 18 at 12:35
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
answered Dec 8 '18 at 0:22
BlueBlue
48k870153
48k870153
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
$endgroup$
– Blue
Dec 8 '18 at 12:06
$begingroup$
Also, it seems that $Q$ lies on the segment $B'C'$.
$endgroup$
– timon92
Dec 8 '18 at 12:07
1
$begingroup$
You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
|
show 3 more comments
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
$endgroup$
– Blue
Dec 8 '18 at 12:06
$begingroup$
Also, it seems that $Q$ lies on the segment $B'C'$.
$endgroup$
– timon92
Dec 8 '18 at 12:07
1
$begingroup$
You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
This problem is something else.
$endgroup$
– dezdichado
Dec 8 '18 at 1:01
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
It is interesting to note that in the non-convex case one has $d=AB$ and $a=CD$, while in the convex cases either $d=AB$ or $a=CD$.
$endgroup$
– timon92
Dec 8 '18 at 9:08
$begingroup$
@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
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– Blue
Dec 8 '18 at 12:06
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@timon92: I'd noticed that myself ... It's actually what made me realize the double-equilateral-triangle connection (featured in my edited solution).
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– Blue
Dec 8 '18 at 12:06
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Also, it seems that $Q$ lies on the segment $B'C'$.
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– timon92
Dec 8 '18 at 12:07
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Also, it seems that $Q$ lies on the segment $B'C'$.
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– timon92
Dec 8 '18 at 12:07
1
1
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You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
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– timon92
Dec 8 '18 at 12:15
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You're right, they're not collinear. If they were collinear then the triangles $QBC'$ and $QC'B$ would be similar and we would have had $145/229=B'C/C'B=BQ/QC=5/8$.
$endgroup$
– timon92
Dec 8 '18 at 12:15
|
show 3 more comments
$begingroup$
Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that
begin{eqnarray}
AP & = & frac{8}{13}x\
DP & = & frac{5}{13}x\
AB & = & frac{8x+13y}{39}\
BC & = & frac{y+z}{2}\
CD & = & frac{2(5x+13z)}{39}
end{eqnarray}
For the angle at $A$, we have that
$$
cos(angle BAP) = frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)}
$$
where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$:
$$f(x,y,z)=
-frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x left(8 x^2+18 x y+5 x z+13 y z+13 z^2right)}=0.
$$
In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$:
$$g(x,y,z)=
-frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y
z^2}{156 x left(20 x^2+8 x y+60 x z+13 y^2+13 y zright)} = 0.
$$
Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution:
$$
z = frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 left(5280 x^2+4277 x
y-7267 y^2right)}.
$$
If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain
$$
frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2} = 0
$$
or
$$
-frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2}=0.
$$
Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results:
begin{eqnarray}
& y=frac{157}{247}x qquad & z=frac{7}{13}x\
& y=frac{7}{13}x qquad & z=frac{199}{481}x\
& y=frac{70}{13}x qquad & z=-frac{44}{13}x
end{eqnarray}
and hence we have to discard the third solution.
Let $O=(0,0)$ be the centre of the circle, and choose $A=(-frac{x}{2},h)$ and $B=(frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by
begin{eqnarray}
B & = & P + y (-cos angle APB,sin angle APB)\
C & = & P + z ( cos angle CPD,sin angle CPD)
end{eqnarray}
With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{157}{19 sqrt{43}},frac{7}{sqrt{43}},-frac{sqrt{3}}{2sqrt{43}})
$$
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{7}{sqrt{43}},frac{199}{37 sqrt{43}},frac{sqrt{3}}{2sqrt{43}})
$$
In either case one finds that $sin angle OPA = frac{1}{2}$.
It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
$endgroup$
add a comment |
$begingroup$
Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that
begin{eqnarray}
AP & = & frac{8}{13}x\
DP & = & frac{5}{13}x\
AB & = & frac{8x+13y}{39}\
BC & = & frac{y+z}{2}\
CD & = & frac{2(5x+13z)}{39}
end{eqnarray}
For the angle at $A$, we have that
$$
cos(angle BAP) = frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)}
$$
where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$:
$$f(x,y,z)=
-frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x left(8 x^2+18 x y+5 x z+13 y z+13 z^2right)}=0.
$$
In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$:
$$g(x,y,z)=
-frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y
z^2}{156 x left(20 x^2+8 x y+60 x z+13 y^2+13 y zright)} = 0.
$$
Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution:
$$
z = frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 left(5280 x^2+4277 x
y-7267 y^2right)}.
$$
If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain
$$
frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2} = 0
$$
or
$$
-frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2}=0.
$$
Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results:
begin{eqnarray}
& y=frac{157}{247}x qquad & z=frac{7}{13}x\
& y=frac{7}{13}x qquad & z=frac{199}{481}x\
& y=frac{70}{13}x qquad & z=-frac{44}{13}x
end{eqnarray}
and hence we have to discard the third solution.
Let $O=(0,0)$ be the centre of the circle, and choose $A=(-frac{x}{2},h)$ and $B=(frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by
begin{eqnarray}
B & = & P + y (-cos angle APB,sin angle APB)\
C & = & P + z ( cos angle CPD,sin angle CPD)
end{eqnarray}
With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{157}{19 sqrt{43}},frac{7}{sqrt{43}},-frac{sqrt{3}}{2sqrt{43}})
$$
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{7}{sqrt{43}},frac{199}{37 sqrt{43}},frac{sqrt{3}}{2sqrt{43}})
$$
In either case one finds that $sin angle OPA = frac{1}{2}$.
It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
$endgroup$
add a comment |
$begingroup$
Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that
begin{eqnarray}
AP & = & frac{8}{13}x\
DP & = & frac{5}{13}x\
AB & = & frac{8x+13y}{39}\
BC & = & frac{y+z}{2}\
CD & = & frac{2(5x+13z)}{39}
end{eqnarray}
For the angle at $A$, we have that
$$
cos(angle BAP) = frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)}
$$
where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$:
$$f(x,y,z)=
-frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x left(8 x^2+18 x y+5 x z+13 y z+13 z^2right)}=0.
$$
In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$:
$$g(x,y,z)=
-frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y
z^2}{156 x left(20 x^2+8 x y+60 x z+13 y^2+13 y zright)} = 0.
$$
Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution:
$$
z = frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 left(5280 x^2+4277 x
y-7267 y^2right)}.
$$
If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain
$$
frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2} = 0
$$
or
$$
-frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2}=0.
$$
Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results:
begin{eqnarray}
& y=frac{157}{247}x qquad & z=frac{7}{13}x\
& y=frac{7}{13}x qquad & z=frac{199}{481}x\
& y=frac{70}{13}x qquad & z=-frac{44}{13}x
end{eqnarray}
and hence we have to discard the third solution.
Let $O=(0,0)$ be the centre of the circle, and choose $A=(-frac{x}{2},h)$ and $B=(frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by
begin{eqnarray}
B & = & P + y (-cos angle APB,sin angle APB)\
C & = & P + z ( cos angle CPD,sin angle CPD)
end{eqnarray}
With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{157}{19 sqrt{43}},frac{7}{sqrt{43}},-frac{sqrt{3}}{2sqrt{43}})
$$
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{7}{sqrt{43}},frac{199}{37 sqrt{43}},frac{sqrt{3}}{2sqrt{43}})
$$
In either case one finds that $sin angle OPA = frac{1}{2}$.
It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
$endgroup$
Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that
begin{eqnarray}
AP & = & frac{8}{13}x\
DP & = & frac{5}{13}x\
AB & = & frac{8x+13y}{39}\
BC & = & frac{y+z}{2}\
CD & = & frac{2(5x+13z)}{39}
end{eqnarray}
For the angle at $A$, we have that
$$
cos(angle BAP) = frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)}
$$
where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$:
$$f(x,y,z)=
-frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x left(8 x^2+18 x y+5 x z+13 y z+13 z^2right)}=0.
$$
In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$:
$$g(x,y,z)=
-frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y
z^2}{156 x left(20 x^2+8 x y+60 x z+13 y^2+13 y zright)} = 0.
$$
Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution:
$$
z = frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 left(5280 x^2+4277 x
y-7267 y^2right)}.
$$
If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain
$$
frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2} = 0
$$
or
$$
-frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) left(89 x^2+78 x
y+338 y^2right)}{left(5280 x^2+4277 x y-7267 y^2right)^2}=0.
$$
Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results:
begin{eqnarray}
& y=frac{157}{247}x qquad & z=frac{7}{13}x\
& y=frac{7}{13}x qquad & z=frac{199}{481}x\
& y=frac{70}{13}x qquad & z=-frac{44}{13}x
end{eqnarray}
and hence we have to discard the third solution.
Let $O=(0,0)$ be the centre of the circle, and choose $A=(-frac{x}{2},h)$ and $B=(frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by
begin{eqnarray}
B & = & P + y (-cos angle APB,sin angle APB)\
C & = & P + z ( cos angle CPD,sin angle CPD)
end{eqnarray}
With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{157}{19 sqrt{43}},frac{7}{sqrt{43}},-frac{sqrt{3}}{2sqrt{43}})
$$
$$
(x,y,z,h) = (frac{13}{sqrt{43}},frac{7}{sqrt{43}},frac{199}{37 sqrt{43}},frac{sqrt{3}}{2sqrt{43}})
$$
In either case one finds that $sin angle OPA = frac{1}{2}$.
It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
answered Dec 19 '18 at 23:35
Ronald BlaakRonald Blaak
2,394310
2,394310
add a comment |
add a comment |
$begingroup$
Here is a start: Let $R = 1$ and $AP = 8x, PD = 5x.$ Then, by the power of point $P$ with respect to our circle, you have:
$$R^2-OP^2=APcdot PDimplies OP = sqrt{1-40x^2}.$$ This means that by the theorem of cosines:
$$cosangle OPA=dfrac{24x^2+1-1}{2cdot 8xsqrt{1-40x^2}} = dfrac{3x}{2sqrt{1-40x^2}}.$$
So the challenge now is to find $x$ using the other information provided. This should be a chain of straightforward calculations; however, if you cannot finish from here, let me know in comments.
Update: For the time being, let $BP = y$ and $CP=z.$ Also, let $angle BAD = t.$ Then $AB = frac{y+8x}{3}$ and from the theorem of cosines in $triangle BAP:$
$$cos t = dfrac{frac{(y+8x)^2}{9} - (y+8x)(y-8x)}{2cdot 8xcdotfrac{y+8x}{3}} = dfrac{10x-y}{6x}quad (1)$$
Find $BD$ using Stuart's theorem in $triangle BAD:$
$$BD^2 = dfrac{1}{AP}(APcdot PDcdot AD +BP^2cdot AD - PDcdot AB^2)=$$
$$= dfrac{14y^2-10xy+545x^2}{9}quad (2)$$
By the Theorem of Sines in $triangle BDA$ and using $(1), (2):$
$$R^2=1=dfrac{BD^2}{4sin^2 t}=dfrac{BD^2}{4(1-cos^2 t)}iff $$
$$y^2(14x^2+1) - 10y(x^3+2x)+(545x^4+x^2) = 0.$$
Now this looks ugly at first, but it turns out to have somewhat nice discriminant:
$$frac yx = dfrac{5x^2+10+3sqrt{(5x^2+1)(2-13x)(2+13x)}}{14x^2+1}.$$
If you mimic the same idea for $triangle CDA,$ you will get the following:
$$frac zx = dfrac{128x^2+13+6sqrt{(2-13x)(2+13x)(32x^2+1)}}{68x^2+1}.$$
So at this point, we have not used the property $PB+PC = 2BC$, which is now obviously used to find $x.$ I will be busy until next week, so if you can finish it from here, let me know.
@Blue seems to have answered the question, but I still hope to finish computation here. So I will keep this attempt up for a while.
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
$endgroup$
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
add a comment |
$begingroup$
Here is a start: Let $R = 1$ and $AP = 8x, PD = 5x.$ Then, by the power of point $P$ with respect to our circle, you have:
$$R^2-OP^2=APcdot PDimplies OP = sqrt{1-40x^2}.$$ This means that by the theorem of cosines:
$$cosangle OPA=dfrac{24x^2+1-1}{2cdot 8xsqrt{1-40x^2}} = dfrac{3x}{2sqrt{1-40x^2}}.$$
So the challenge now is to find $x$ using the other information provided. This should be a chain of straightforward calculations; however, if you cannot finish from here, let me know in comments.
Update: For the time being, let $BP = y$ and $CP=z.$ Also, let $angle BAD = t.$ Then $AB = frac{y+8x}{3}$ and from the theorem of cosines in $triangle BAP:$
$$cos t = dfrac{frac{(y+8x)^2}{9} - (y+8x)(y-8x)}{2cdot 8xcdotfrac{y+8x}{3}} = dfrac{10x-y}{6x}quad (1)$$
Find $BD$ using Stuart's theorem in $triangle BAD:$
$$BD^2 = dfrac{1}{AP}(APcdot PDcdot AD +BP^2cdot AD - PDcdot AB^2)=$$
$$= dfrac{14y^2-10xy+545x^2}{9}quad (2)$$
By the Theorem of Sines in $triangle BDA$ and using $(1), (2):$
$$R^2=1=dfrac{BD^2}{4sin^2 t}=dfrac{BD^2}{4(1-cos^2 t)}iff $$
$$y^2(14x^2+1) - 10y(x^3+2x)+(545x^4+x^2) = 0.$$
Now this looks ugly at first, but it turns out to have somewhat nice discriminant:
$$frac yx = dfrac{5x^2+10+3sqrt{(5x^2+1)(2-13x)(2+13x)}}{14x^2+1}.$$
If you mimic the same idea for $triangle CDA,$ you will get the following:
$$frac zx = dfrac{128x^2+13+6sqrt{(2-13x)(2+13x)(32x^2+1)}}{68x^2+1}.$$
So at this point, we have not used the property $PB+PC = 2BC$, which is now obviously used to find $x.$ I will be busy until next week, so if you can finish it from here, let me know.
@Blue seems to have answered the question, but I still hope to finish computation here. So I will keep this attempt up for a while.
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
$endgroup$
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
add a comment |
$begingroup$
Here is a start: Let $R = 1$ and $AP = 8x, PD = 5x.$ Then, by the power of point $P$ with respect to our circle, you have:
$$R^2-OP^2=APcdot PDimplies OP = sqrt{1-40x^2}.$$ This means that by the theorem of cosines:
$$cosangle OPA=dfrac{24x^2+1-1}{2cdot 8xsqrt{1-40x^2}} = dfrac{3x}{2sqrt{1-40x^2}}.$$
So the challenge now is to find $x$ using the other information provided. This should be a chain of straightforward calculations; however, if you cannot finish from here, let me know in comments.
Update: For the time being, let $BP = y$ and $CP=z.$ Also, let $angle BAD = t.$ Then $AB = frac{y+8x}{3}$ and from the theorem of cosines in $triangle BAP:$
$$cos t = dfrac{frac{(y+8x)^2}{9} - (y+8x)(y-8x)}{2cdot 8xcdotfrac{y+8x}{3}} = dfrac{10x-y}{6x}quad (1)$$
Find $BD$ using Stuart's theorem in $triangle BAD:$
$$BD^2 = dfrac{1}{AP}(APcdot PDcdot AD +BP^2cdot AD - PDcdot AB^2)=$$
$$= dfrac{14y^2-10xy+545x^2}{9}quad (2)$$
By the Theorem of Sines in $triangle BDA$ and using $(1), (2):$
$$R^2=1=dfrac{BD^2}{4sin^2 t}=dfrac{BD^2}{4(1-cos^2 t)}iff $$
$$y^2(14x^2+1) - 10y(x^3+2x)+(545x^4+x^2) = 0.$$
Now this looks ugly at first, but it turns out to have somewhat nice discriminant:
$$frac yx = dfrac{5x^2+10+3sqrt{(5x^2+1)(2-13x)(2+13x)}}{14x^2+1}.$$
If you mimic the same idea for $triangle CDA,$ you will get the following:
$$frac zx = dfrac{128x^2+13+6sqrt{(2-13x)(2+13x)(32x^2+1)}}{68x^2+1}.$$
So at this point, we have not used the property $PB+PC = 2BC$, which is now obviously used to find $x.$ I will be busy until next week, so if you can finish it from here, let me know.
@Blue seems to have answered the question, but I still hope to finish computation here. So I will keep this attempt up for a while.
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
$endgroup$
Here is a start: Let $R = 1$ and $AP = 8x, PD = 5x.$ Then, by the power of point $P$ with respect to our circle, you have:
$$R^2-OP^2=APcdot PDimplies OP = sqrt{1-40x^2}.$$ This means that by the theorem of cosines:
$$cosangle OPA=dfrac{24x^2+1-1}{2cdot 8xsqrt{1-40x^2}} = dfrac{3x}{2sqrt{1-40x^2}}.$$
So the challenge now is to find $x$ using the other information provided. This should be a chain of straightforward calculations; however, if you cannot finish from here, let me know in comments.
Update: For the time being, let $BP = y$ and $CP=z.$ Also, let $angle BAD = t.$ Then $AB = frac{y+8x}{3}$ and from the theorem of cosines in $triangle BAP:$
$$cos t = dfrac{frac{(y+8x)^2}{9} - (y+8x)(y-8x)}{2cdot 8xcdotfrac{y+8x}{3}} = dfrac{10x-y}{6x}quad (1)$$
Find $BD$ using Stuart's theorem in $triangle BAD:$
$$BD^2 = dfrac{1}{AP}(APcdot PDcdot AD +BP^2cdot AD - PDcdot AB^2)=$$
$$= dfrac{14y^2-10xy+545x^2}{9}quad (2)$$
By the Theorem of Sines in $triangle BDA$ and using $(1), (2):$
$$R^2=1=dfrac{BD^2}{4sin^2 t}=dfrac{BD^2}{4(1-cos^2 t)}iff $$
$$y^2(14x^2+1) - 10y(x^3+2x)+(545x^4+x^2) = 0.$$
Now this looks ugly at first, but it turns out to have somewhat nice discriminant:
$$frac yx = dfrac{5x^2+10+3sqrt{(5x^2+1)(2-13x)(2+13x)}}{14x^2+1}.$$
If you mimic the same idea for $triangle CDA,$ you will get the following:
$$frac zx = dfrac{128x^2+13+6sqrt{(2-13x)(2+13x)(32x^2+1)}}{68x^2+1}.$$
So at this point, we have not used the property $PB+PC = 2BC$, which is now obviously used to find $x.$ I will be busy until next week, so if you can finish it from here, let me know.
@Blue seems to have answered the question, but I still hope to finish computation here. So I will keep this attempt up for a while.
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
edited Dec 8 '18 at 17:41
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
answered Dec 6 '18 at 0:52
dezdichadodezdichado
6,3711929
6,3711929
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
add a comment |
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
$begingroup$
Now the key How to solve $x?$
$endgroup$
– communnites
Dec 6 '18 at 0:58
add a comment |
$begingroup$
I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-sqrt{1 - a^2}, -a)$ and $D(sqrt{1 - a^2}, -a)$
Using the ratio formula $P(frac{3sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = frac{3}{13}sqrt{1 - a^2} + sqrt{1 - a^2} = frac{16}{13}sqrt{1 - a^2}$
$PD = sqrt{1 - a^2} - frac{3}{13}sqrt{1 - a^2} = frac{10}{13}sqrt{1 - a^2}$
Also let B be $(x_1, sqrt{1 - x_1^2})$ and C be $(x_2, sqrt{1 - x_2^2})$
Hence $AB = sqrt{(x_1 + sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$PB = sqrt{(x_1 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$BC = sqrt{(x_1 - x_2)^2 - (sqrt{1 - x_1^2} - sqrt{1 - x_2^2})^2}$
$PC = sqrt{(x_2 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
$CD = sqrt{(x_2 - sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$sinangle OPA = frac{13}{sqrt{frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $sin angle OPA = 0.399$.
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
$endgroup$
add a comment |
$begingroup$
I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-sqrt{1 - a^2}, -a)$ and $D(sqrt{1 - a^2}, -a)$
Using the ratio formula $P(frac{3sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = frac{3}{13}sqrt{1 - a^2} + sqrt{1 - a^2} = frac{16}{13}sqrt{1 - a^2}$
$PD = sqrt{1 - a^2} - frac{3}{13}sqrt{1 - a^2} = frac{10}{13}sqrt{1 - a^2}$
Also let B be $(x_1, sqrt{1 - x_1^2})$ and C be $(x_2, sqrt{1 - x_2^2})$
Hence $AB = sqrt{(x_1 + sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$PB = sqrt{(x_1 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$BC = sqrt{(x_1 - x_2)^2 - (sqrt{1 - x_1^2} - sqrt{1 - x_2^2})^2}$
$PC = sqrt{(x_2 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
$CD = sqrt{(x_2 - sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$sinangle OPA = frac{13}{sqrt{frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $sin angle OPA = 0.399$.
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
$endgroup$
add a comment |
$begingroup$
I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-sqrt{1 - a^2}, -a)$ and $D(sqrt{1 - a^2}, -a)$
Using the ratio formula $P(frac{3sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = frac{3}{13}sqrt{1 - a^2} + sqrt{1 - a^2} = frac{16}{13}sqrt{1 - a^2}$
$PD = sqrt{1 - a^2} - frac{3}{13}sqrt{1 - a^2} = frac{10}{13}sqrt{1 - a^2}$
Also let B be $(x_1, sqrt{1 - x_1^2})$ and C be $(x_2, sqrt{1 - x_2^2})$
Hence $AB = sqrt{(x_1 + sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$PB = sqrt{(x_1 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$BC = sqrt{(x_1 - x_2)^2 - (sqrt{1 - x_1^2} - sqrt{1 - x_2^2})^2}$
$PC = sqrt{(x_2 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
$CD = sqrt{(x_2 - sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$sinangle OPA = frac{13}{sqrt{frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $sin angle OPA = 0.399$.
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
$endgroup$
I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-sqrt{1 - a^2}, -a)$ and $D(sqrt{1 - a^2}, -a)$
Using the ratio formula $P(frac{3sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = frac{3}{13}sqrt{1 - a^2} + sqrt{1 - a^2} = frac{16}{13}sqrt{1 - a^2}$
$PD = sqrt{1 - a^2} - frac{3}{13}sqrt{1 - a^2} = frac{10}{13}sqrt{1 - a^2}$
Also let B be $(x_1, sqrt{1 - x_1^2})$ and C be $(x_2, sqrt{1 - x_2^2})$
Hence $AB = sqrt{(x_1 + sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$PB = sqrt{(x_1 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_1^2} + a)^2}$
$BC = sqrt{(x_1 - x_2)^2 - (sqrt{1 - x_1^2} - sqrt{1 - x_2^2})^2}$
$PC = sqrt{(x_2 - frac{3}{13}sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
$CD = sqrt{(x_2 - sqrt{1 - a^2})^2 + (sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$sinangle OPA = frac{13}{sqrt{frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $sin angle OPA = 0.399$.
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
edited Dec 15 '18 at 4:03
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
answered Dec 10 '18 at 3:45
KY TangKY Tang
1245
1245
add a comment |
add a comment |
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2
$begingroup$
Where did this problem come from?
$endgroup$
– rogerl
Dec 5 '18 at 17:34
$begingroup$
What happened to Blue’s answer?
$endgroup$
– D.R.
Dec 10 '18 at 3:49
$begingroup$
what happend to Blue's answer?
$endgroup$
– communnites
Dec 10 '18 at 4:07
$begingroup$
Yeah i was wondering why he deleted his answer too.
$endgroup$
– dezdichado
Dec 10 '18 at 4:47
1
$begingroup$
yeah I really really want to see the synthetic answer or at least the motivation as to why/how one can come up with this problem.
$endgroup$
– dezdichado
Jan 1 at 9:35