Find all possible values of an angle
$begingroup$
Let $ABC$ be an acute triangle with altitudes $BB_1$ and $CC_1$. Let $B_2$ and $C_2$ be the midpoints of $AC$ and $AB$, respectively. Find all possible values of $angle BAC$ if $B_1C_2$ and $B_2C_1$ are perpendicular.
I have angle chased this problem many times, but I always keep getting $angle BAC = 90^{circ}$, contrary to the hypotesis of the problem that $ABC$ is an acute triangle.
geometry
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be an acute triangle with altitudes $BB_1$ and $CC_1$. Let $B_2$ and $C_2$ be the midpoints of $AC$ and $AB$, respectively. Find all possible values of $angle BAC$ if $B_1C_2$ and $B_2C_1$ are perpendicular.
I have angle chased this problem many times, but I always keep getting $angle BAC = 90^{circ}$, contrary to the hypotesis of the problem that $ABC$ is an acute triangle.
geometry
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be an acute triangle with altitudes $BB_1$ and $CC_1$. Let $B_2$ and $C_2$ be the midpoints of $AC$ and $AB$, respectively. Find all possible values of $angle BAC$ if $B_1C_2$ and $B_2C_1$ are perpendicular.
I have angle chased this problem many times, but I always keep getting $angle BAC = 90^{circ}$, contrary to the hypotesis of the problem that $ABC$ is an acute triangle.
geometry
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
$endgroup$
Let $ABC$ be an acute triangle with altitudes $BB_1$ and $CC_1$. Let $B_2$ and $C_2$ be the midpoints of $AC$ and $AB$, respectively. Find all possible values of $angle BAC$ if $B_1C_2$ and $B_2C_1$ are perpendicular.
I have angle chased this problem many times, but I always keep getting $angle BAC = 90^{circ}$, contrary to the hypotesis of the problem that $ABC$ is an acute triangle.
geometry
geometry
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
edited Nov 5 '14 at 20:05
ajotatxe
53.8k23890
53.8k23890
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
asked Nov 5 '14 at 19:58
AnglesrbadAnglesrbad
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Without loss of generality, we can place the intersection of $B_1C_2$ and $B_2C_1$ at the origin of a Cartesian grid.
Further, since these are perpendicular, we can without loss of generality, choose to place $B_1$ at $(0,-a)$, $B_2$ at $(+b,0)$, $C_1$ at $(-d, 0)$, and $C_2$ at $(0,+c)$, iwth $a,b,c,d ? 0$. Then if you draw the diagram, $A$ liles on the intersection point of $B_1B_2$ and $C_1C_2$ (which will exist unless $(ad-bc = 0$).
The sine of the angle bewtween the lines define by $B_1B_2$ and $C_1C_2$ is given by the dot prodoct of the two difference vectors, over the product of magnitudes. That is,
$$
sin A = frac{(b,a)cdot (d,c)}{sqrt{b^2+a^2}sqrt{c^2+d^2}}
$$
Squaring, dividing the numerator by $abcd$, and letting $frac{b}{a}=u$ and $frac{c}{d}=v$ we get
$$
sin^2 angle BAC = frac{uv + frac{1}{uv}+2} {uv + frac{1}{uv}+frac{u}{v}+frac{v}{u}}
$$
To explore the range of possible values of $sin^2 angle BAC$ we can fixate on the case where $uv=1$ (we will see that this gives us a full range of values).
When $u=1$ we see that $sin^2 angle BAC = 1$, and this is not allowed since the triangle is required to be acute. But when $1=1+epsilon$, for arbitrarily small $epsilon$. $sin^2 angle BAC approx 1 - epsilon^2$ so we can come arbitrarily close to $90^circ$.
When $u$ is large, $$sin^2 angle BAC approx frac{4}{u^2}$$ which can be made arbitrarily small.
So the ranve of possible values of $angle BAC$ is $(0^circ, 90^circ)$.
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
$endgroup$
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
add a comment |
$begingroup$
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2angle A$
By angle sum of triangle, $180^0 = angle BKC + x + y$
$180^0 = angle BKC + [angle B – (180^0 – 2A)] + [angle C – (180^0 – 2A)]$
:
$360^0 = angle BKC + 3angle A$
By corresponding angles between parallels, $angle BKC = angle B_1HC1 = angle B_3KC_3 = 90^0$
∴ $angle A$ has to be $90^0$
This means that requiring $triangle ABC$ to be acute should be re-phrased as $triangle ABC$ is non-obtuse.
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
$endgroup$
add a comment |
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$begingroup$
Without loss of generality, we can place the intersection of $B_1C_2$ and $B_2C_1$ at the origin of a Cartesian grid.
Further, since these are perpendicular, we can without loss of generality, choose to place $B_1$ at $(0,-a)$, $B_2$ at $(+b,0)$, $C_1$ at $(-d, 0)$, and $C_2$ at $(0,+c)$, iwth $a,b,c,d ? 0$. Then if you draw the diagram, $A$ liles on the intersection point of $B_1B_2$ and $C_1C_2$ (which will exist unless $(ad-bc = 0$).
The sine of the angle bewtween the lines define by $B_1B_2$ and $C_1C_2$ is given by the dot prodoct of the two difference vectors, over the product of magnitudes. That is,
$$
sin A = frac{(b,a)cdot (d,c)}{sqrt{b^2+a^2}sqrt{c^2+d^2}}
$$
Squaring, dividing the numerator by $abcd$, and letting $frac{b}{a}=u$ and $frac{c}{d}=v$ we get
$$
sin^2 angle BAC = frac{uv + frac{1}{uv}+2} {uv + frac{1}{uv}+frac{u}{v}+frac{v}{u}}
$$
To explore the range of possible values of $sin^2 angle BAC$ we can fixate on the case where $uv=1$ (we will see that this gives us a full range of values).
When $u=1$ we see that $sin^2 angle BAC = 1$, and this is not allowed since the triangle is required to be acute. But when $1=1+epsilon$, for arbitrarily small $epsilon$. $sin^2 angle BAC approx 1 - epsilon^2$ so we can come arbitrarily close to $90^circ$.
When $u$ is large, $$sin^2 angle BAC approx frac{4}{u^2}$$ which can be made arbitrarily small.
So the ranve of possible values of $angle BAC$ is $(0^circ, 90^circ)$.
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
$endgroup$
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
add a comment |
$begingroup$
Without loss of generality, we can place the intersection of $B_1C_2$ and $B_2C_1$ at the origin of a Cartesian grid.
Further, since these are perpendicular, we can without loss of generality, choose to place $B_1$ at $(0,-a)$, $B_2$ at $(+b,0)$, $C_1$ at $(-d, 0)$, and $C_2$ at $(0,+c)$, iwth $a,b,c,d ? 0$. Then if you draw the diagram, $A$ liles on the intersection point of $B_1B_2$ and $C_1C_2$ (which will exist unless $(ad-bc = 0$).
The sine of the angle bewtween the lines define by $B_1B_2$ and $C_1C_2$ is given by the dot prodoct of the two difference vectors, over the product of magnitudes. That is,
$$
sin A = frac{(b,a)cdot (d,c)}{sqrt{b^2+a^2}sqrt{c^2+d^2}}
$$
Squaring, dividing the numerator by $abcd$, and letting $frac{b}{a}=u$ and $frac{c}{d}=v$ we get
$$
sin^2 angle BAC = frac{uv + frac{1}{uv}+2} {uv + frac{1}{uv}+frac{u}{v}+frac{v}{u}}
$$
To explore the range of possible values of $sin^2 angle BAC$ we can fixate on the case where $uv=1$ (we will see that this gives us a full range of values).
When $u=1$ we see that $sin^2 angle BAC = 1$, and this is not allowed since the triangle is required to be acute. But when $1=1+epsilon$, for arbitrarily small $epsilon$. $sin^2 angle BAC approx 1 - epsilon^2$ so we can come arbitrarily close to $90^circ$.
When $u$ is large, $$sin^2 angle BAC approx frac{4}{u^2}$$ which can be made arbitrarily small.
So the ranve of possible values of $angle BAC$ is $(0^circ, 90^circ)$.
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
$endgroup$
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
add a comment |
$begingroup$
Without loss of generality, we can place the intersection of $B_1C_2$ and $B_2C_1$ at the origin of a Cartesian grid.
Further, since these are perpendicular, we can without loss of generality, choose to place $B_1$ at $(0,-a)$, $B_2$ at $(+b,0)$, $C_1$ at $(-d, 0)$, and $C_2$ at $(0,+c)$, iwth $a,b,c,d ? 0$. Then if you draw the diagram, $A$ liles on the intersection point of $B_1B_2$ and $C_1C_2$ (which will exist unless $(ad-bc = 0$).
The sine of the angle bewtween the lines define by $B_1B_2$ and $C_1C_2$ is given by the dot prodoct of the two difference vectors, over the product of magnitudes. That is,
$$
sin A = frac{(b,a)cdot (d,c)}{sqrt{b^2+a^2}sqrt{c^2+d^2}}
$$
Squaring, dividing the numerator by $abcd$, and letting $frac{b}{a}=u$ and $frac{c}{d}=v$ we get
$$
sin^2 angle BAC = frac{uv + frac{1}{uv}+2} {uv + frac{1}{uv}+frac{u}{v}+frac{v}{u}}
$$
To explore the range of possible values of $sin^2 angle BAC$ we can fixate on the case where $uv=1$ (we will see that this gives us a full range of values).
When $u=1$ we see that $sin^2 angle BAC = 1$, and this is not allowed since the triangle is required to be acute. But when $1=1+epsilon$, for arbitrarily small $epsilon$. $sin^2 angle BAC approx 1 - epsilon^2$ so we can come arbitrarily close to $90^circ$.
When $u$ is large, $$sin^2 angle BAC approx frac{4}{u^2}$$ which can be made arbitrarily small.
So the ranve of possible values of $angle BAC$ is $(0^circ, 90^circ)$.
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
$endgroup$
Without loss of generality, we can place the intersection of $B_1C_2$ and $B_2C_1$ at the origin of a Cartesian grid.
Further, since these are perpendicular, we can without loss of generality, choose to place $B_1$ at $(0,-a)$, $B_2$ at $(+b,0)$, $C_1$ at $(-d, 0)$, and $C_2$ at $(0,+c)$, iwth $a,b,c,d ? 0$. Then if you draw the diagram, $A$ liles on the intersection point of $B_1B_2$ and $C_1C_2$ (which will exist unless $(ad-bc = 0$).
The sine of the angle bewtween the lines define by $B_1B_2$ and $C_1C_2$ is given by the dot prodoct of the two difference vectors, over the product of magnitudes. That is,
$$
sin A = frac{(b,a)cdot (d,c)}{sqrt{b^2+a^2}sqrt{c^2+d^2}}
$$
Squaring, dividing the numerator by $abcd$, and letting $frac{b}{a}=u$ and $frac{c}{d}=v$ we get
$$
sin^2 angle BAC = frac{uv + frac{1}{uv}+2} {uv + frac{1}{uv}+frac{u}{v}+frac{v}{u}}
$$
To explore the range of possible values of $sin^2 angle BAC$ we can fixate on the case where $uv=1$ (we will see that this gives us a full range of values).
When $u=1$ we see that $sin^2 angle BAC = 1$, and this is not allowed since the triangle is required to be acute. But when $1=1+epsilon$, for arbitrarily small $epsilon$. $sin^2 angle BAC approx 1 - epsilon^2$ so we can come arbitrarily close to $90^circ$.
When $u$ is large, $$sin^2 angle BAC approx frac{4}{u^2}$$ which can be made arbitrarily small.
So the ranve of possible values of $angle BAC$ is $(0^circ, 90^circ)$.
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
answered Nov 5 '14 at 21:38
Mark FischlerMark Fischler
32.4k12251
32.4k12251
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
add a comment |
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
$begingroup$
This is wrong. hehe
$endgroup$
– Anglesrbad
Nov 5 '14 at 21:50
add a comment |
$begingroup$
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2angle A$
By angle sum of triangle, $180^0 = angle BKC + x + y$
$180^0 = angle BKC + [angle B – (180^0 – 2A)] + [angle C – (180^0 – 2A)]$
:
$360^0 = angle BKC + 3angle A$
By corresponding angles between parallels, $angle BKC = angle B_1HC1 = angle B_3KC_3 = 90^0$
∴ $angle A$ has to be $90^0$
This means that requiring $triangle ABC$ to be acute should be re-phrased as $triangle ABC$ is non-obtuse.
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2angle A$
By angle sum of triangle, $180^0 = angle BKC + x + y$
$180^0 = angle BKC + [angle B – (180^0 – 2A)] + [angle C – (180^0 – 2A)]$
:
$360^0 = angle BKC + 3angle A$
By corresponding angles between parallels, $angle BKC = angle B_1HC1 = angle B_3KC_3 = 90^0$
∴ $angle A$ has to be $90^0$
This means that requiring $triangle ABC$ to be acute should be re-phrased as $triangle ABC$ is non-obtuse.
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2angle A$
By angle sum of triangle, $180^0 = angle BKC + x + y$
$180^0 = angle BKC + [angle B – (180^0 – 2A)] + [angle C – (180^0 – 2A)]$
:
$360^0 = angle BKC + 3angle A$
By corresponding angles between parallels, $angle BKC = angle B_1HC1 = angle B_3KC_3 = 90^0$
∴ $angle A$ has to be $90^0$
This means that requiring $triangle ABC$ to be acute should be re-phrased as $triangle ABC$ is non-obtuse.
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
$endgroup$
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2angle A$
By angle sum of triangle, $180^0 = angle BKC + x + y$
$180^0 = angle BKC + [angle B – (180^0 – 2A)] + [angle C – (180^0 – 2A)]$
:
$360^0 = angle BKC + 3angle A$
By corresponding angles between parallels, $angle BKC = angle B_1HC1 = angle B_3KC_3 = 90^0$
∴ $angle A$ has to be $90^0$
This means that requiring $triangle ABC$ to be acute should be re-phrased as $triangle ABC$ is non-obtuse.
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
edited Nov 6 '14 at 17:24
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
answered Nov 6 '14 at 17:15
MickMick
11.8k21641
11.8k21641
add a comment |
add a comment |
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