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Iranian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Iran

Triangle ABC is an acute-angled triangle with orthocenter H. Point P is inside triangle $BHC$ such that $3\angle HBC = \angle HPC$ and $3\angle HCB = \angle HPB$ . Denote by X and Y, the reflection of P with respect to $BH$ and $CH$ , respectively. If S is the circumcenter of triangle $AXY$ , prove that $\angle CAP = \angle BAS$ .

Solution

Let's assume that the feet of $A$ , $B$ , $C$ correspond to points $D$ , $E$ , $F$ , respectively. We apply an inversion with radius $\sqrt{AH \cdot AD}$ . In this inversion, points $D$ , $H$ , as well as $F$ , $B$ , and $E$ , $C$ , interchange with each other. We aim to prove that point $P$ goes to the nine point center of $ABC$ ( $N_9$ ). We use directed angles modulo $180^\circ$ . Let's suppose that the image of $P$ after inversion is $P'$ and $M$ , $N$ be the midpoints of the sides $AC$ and $AB$ .

$\angle CPH + \angle HAC = \angle DP'E = 4\angle CBH$
$M$ is the center of the cyclic quadrilateral $ABDE$ , so:
$\angle DME = 2\angle HBC$
Also $N_9$ is the circumcenter of $MDE$ . Which implies:
$\angle DN_9E = 2\angle DME = \angle DP'E$
In the same way we obtain:
$\angle FN_9D = \angle FP'D$
Finally we can conclude $P' = N_9$ .

Now we will find the images of $X$ , $Y$ after inversion. Let $X'$ , $Y'$ be these images. As $CH$ is the perpendicular bisector of $PY$ , after the inversion the circumcircle of $ADE$ is the $A$ -Apollonian circle in the triangle $AY'N_9$ . The center of this circle is $M$ so $M$ , $N_9$ , $Y'$ are collinear and:
$MA^2 = MN_9 \cdot MY'$
$AS$ becomes the $A$ -altitude in $AX'Y'$ .

We claim $ABC \sim AY'X'$ . First we calculate the ratio $\frac{AY'}{AX'}$ .
$AMN_9 \sim Y'MA \implies \frac{AY'}{AN_9} = \frac{AM}{MN_9}$
Similarly
$\frac{AN_9}{AX'} = \frac{NN_9}{AN} \implies \frac{AY'}{AX'} = \frac{AM}{AN} \cdot \frac{NN_9}{MN_9} = \frac{AB}{AC}$
Then we calculate the angle $X'AY'$ .
$\angle X'AY' = \angle Y'AB + \angle X'AC - \angle BAC = \angle AN_9M + \angle AN_9N - \angle BAC = \angle BAC.$
Hence the claimed similarity is proven. Now let the line perpendicular to $X'Y'$ be $\ell$ . We use again directed angles modulo $180^\circ$ .
$\begin{aligned} \angle \ell, AC &= \angle AX', AC = \angle ACB - 90^\circ + \angle NN_9A \\ &= \angle ACB + \angle NN_9, AC + \angle AC, AN_9 - 90^\circ = \angle BAC + \angle CAN = \angle BAN_9 \end{aligned}$
The final equality proves the desired result. ■

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