A geometry problem by Mehul Chaturvedi

Due to symmetry, $\triangle ABC$ is an isosceles triangle with $AB=AC$ . Let the coordinates be $O(0,0)$ , $A(0,a)$ , $B(-1,0)$ and $C(1,0)$ .

The straight line $AC$ is given by:

$\dfrac {y-0}{x-1} = \dfrac {a-0}{0-1} \quad \Rightarrow y = a(1-x)\quad ...(1)$

Straight line $BQ$ is given by:

$\dfrac {y-0}{x+1} = \dfrac {1}{b} \quad \Rightarrow y = \dfrac {1+x}{a}\quad ...(2)$

Therefore the coordinates of $Q$ are:

$y_Q = a(1-x_Q) = \dfrac {1+x_Q}{a} \quad \Rightarrow x_Q = \frac {a^2-1}{a^2+1}\quad \Rightarrow y_Q = \frac {2a}{a^2+1}$

Let the heights of the orthocenter of $\triangle ABC$ and $\triangle APQ$ be $h$ and $h'$ respectively. Then we note that:

$\dfrac {h'} {\frac {a^2-1}{a^2+1}} = \dfrac {h} {1} = \dfrac {1}{a}\quad \Rightarrow \dfrac {1-\frac {2a}{a^2+1}} {\frac {a^2-1}{a^2+1}} = \dfrac {1}{a} \quad \Rightarrow a(a-1)^2 = a^2 -1$

$\Rightarrow a(a-1) = a+1\quad \Rightarrow a^2-2a-1 = 0\quad \Rightarrow a = 1 + \sqrt{2}$

We note that $\angle BAC = 2\tan^{-1} {\frac{1}{a}} = 2\tan^{-1} {(\sqrt{2}-1)} = \boxed{45}^\circ$

Let the orthocentre of the circle be D And consider the centre of the circle to be O

Now Join OP and OQ

Now,let the angle POQ=2x

Therefore,angle PDQ=180 - x

Now, Notice that angle PAQ = x

Also, angle POB + angle QOC = 180 - 2x

Therefore, angle PBO +angle QCO=90 + x And finally.

angle BAC + angle ACB +angle ABC = 90 + x + x

Therefore,

x = 45 = angle BAC