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In an acute triangle ABC, the base BC has
the equation $4x – 3y + 3 = 0$. If the coordinates of
the orthocentre (H) and circumcentre (P) of the
triangle are $(1, 2)$ and $(2, 3)$ respectively, then the
radius of the circle circumscribing the triangle is
$dfrac{sqrt m}{n}$ , where m and n are relatively prime. Find the
value of (m+ n).

(You may use the fact that the distance between
orthocentre and circumcentre of the triangle is
given by $R sqrt{1 – 8cos Acos Bcos C}$)

Attempt:
I found $R$ by taking reflection ($A$) of $H$ about $BC$ and then finding the distance between $P$ and $A$. But, I cannot figure out how to solve the problem by using the hint given.

Solution using the given hint:

Distance of orthocentre (H) from side BC is $2Rcos Bcos C$ and that of circumcentre (P) from BC is $Rcos A$ where R is the circumradius. Finding these distances using coordinate geometry and equating we get:
$$2Rcos Bcos C=frac{1}5$$
$$Rcos A=frac{2}5$$
From this we get:
$$cos Acos Bcos C=frac{1}{25R^2}$$

Distance between P and H is $sqrt{2}$. Thus,
$$R sqrt{1 – 8cos Acos Bcos C}=sqrt{2}$$
$$R=frac{sqrt{58}}{5}$$
$$m=58,n=5$$

As you are looking for a different way to solve your question, here is one.

Take a look at the following figure : it provides positions for $A,B,C$ that fullfill all the constraints :

Fig. 1.

How is it possible to obtain these points (out of which the strange result you are asked is easy to compute) ?

First of all, let us transform the implicit equation of straight line $BC$ into a parametric form ; as it passes through point $binom{0}{1}$ with directing vector $binom{3}{4}$, we can write :

$$binom{x}{y}=binom{0}{1}+tbinom{3}{4} iff begin{cases}x&=&3t\y&=&1+4tend{cases}tag{1}$$

In particular, the coordinates of $B$ and $C$ are resp.

$$begin{cases}x_B&=&3b\y_B&=&1+4bend{cases} text{and} begin{cases}x_C&=&3c\y_C&=&1+4cend{cases}tag{2}$$

for specific values $b, c$ of parameter $t$.

Besides, $A$ belongs to the straight line passing by $H$ and orthogonal to $BC$ ; the parametric form of this straight line is easily found to be :

$$binom{x}{y}=binom{1}{2}+tbinom{4}{-3} iff begin{cases}x_A&=&1+4a\y_A&=&2-3aend{cases}tag{3}$$

Now, let us locate the constraints : as we need three precise values for $a,b,c$, we need three constraints. Here they are :

$$PA^2=PB^2=PC^2 text{and} overrightarrow{AB} perp overrightarrow{CH},tag{4}$$

giving rise, using (2) and (3), to the following system :

$$(1+4a-2)^2+(2-3a-3)^2=(3b-2)^2+(1+4b-3)^2=(3c-2)^2+(1+4c-3)^2$$
$$(3b-1)(1+4a-3c)+(1+4b-2)((2-3a)-(1+4c))=0tag{5}$$

that is easily solved by a CAS giving the following values of parameters $a,b,c$ (please note that $b$ and $c$ can be exchanged) :

$$a=frac{4}{25}, b=frac{14+3sqrt{6}}{25}, c=frac{14-3sqrt{6}}{25}tag{6}$$

Here is the Matlab program that has given these values and Fig. 1 :

% First part : solving constraints in order to obtain parameters values

syms a b c : % symbolic variables

% 3 equations (5) :

eq1=(1+4*a-2)^2+(2-3*a-3)^2==(3*b-2)^2+(1+4*b-3)^2; 

eq2=(1+4*a-2)^2+(2-3*a-3)^2==(3*c-2)^2+(1+4*c-3)^2; 

eq3=(3*b-1)*(1+4*a-3*c)+(1+4*b-2)*((2-3*a)-(1+4*c))==0; 

[A,B,C]=solve([eq1,eq2,eq3],a,b,c)

%

% second part : plotting the only significant result :

clear all;close all;hold on;axis equal;grid on;

xH=1;yH=2;plot(xH,yH,'*r');text(xH+0.1,yH,'H');

xP=2;yP=3;plot(xP,yP,'*r');text(xP-0.2,yP,'P');

a=4/25;b=(14+3*sqrt(6))/25;c=(14-3*sqrt(6))/25;%param. values obtained in the first part;

xA=@(a)(1+4*a);yA=@(a)(2-3*a);plot(xA(a),yA(a),'ob');text(xA(a),yA(a)-0.2,'A')

xB=@(b)(3*b);yB=@(b)(1+4*b);plot(xB(b),yB(b),'ob');text(xB(b),yB(b)+0.1,'B')

xC=@(c)(3*c);yC=@(c)(1+4*c);plot(xC(c),yC(c),'ob');text(xC(c)-0.2,yC(c),'C')

plot([xC(c),xA(a),xB(b),xC(c)],[yC(c),yA(a),yB(b),yC(c)]);

Remarks :

1) Relationships (4) take into account the characteristic properties

• of the circumcenter (the unique point at equal distances from each vertex),




• of the orthocenter (the intersection of 2 altitudes ; don't forget that we have previously considered $A$ to belong to the altitude opposite to $BC$).

2) There are other solutions for $a,b,c$ than (6) ; but none is satisfying ; some are with complex numbers (!), some others generating non acute triangles, and in particular a flat triangle.

3) $b$ and $c$ are roots of the same quadratic equation. It is very understandable because $B$ and $C$ play interchangeable roles.

This is another way (still w/o using the hint).

begin{align}
text{Orthocenter of $triangle ABC$: }quad
H&=(1,2)
,\
text{Circumcenter of $triangle ABC$: }quad
O&=(2,3)
,\
text{Centroid of $triangle ABC$: }quad
M&=tfrac13(H+2O)=(tfrac53,tfrac83)
.
end{align}

The line through $BC$:
begin{align}
y&=tfrac43x+1
,
end{align}

the line $ODperp BC$:
begin{align}
y&=-tfrac34 x+tfrac92
.
end{align}

The point $D$ is the middle
of the side $BC$:

begin{align}
D&=(tfrac{42}{25},tfrac{81}{25})
.
end{align}

Now the vertex $A$ of the $triangle ABC$
can be found as
begin{align}
A&=M+2(M-D)
=(tfrac{41}{25},tfrac{38}{25})
,
end{align}

and the radius of the circumscribed circle is

begin{align}
R&=|O-A|=
sqrt{
left(frac{50-41}{25}right)^2
+left(frac{75-38}{25}right)^2
}
=
frac{sqrt{58}}5
.
end{align}

Given $R=frac{sqrt{m}}n$, $m$ and $n$ can be found as

begin{align}
m&=58
,\
n&=5
.
end{align}