In an acute triangle $ABC$, let a line $\ell$ pass through the orthocenter and not through point $A$. The line $\ell$ intersects line $BC$ at $P(\neq B, C)$. A line passing through $A$ and perpendicular to $\ell$ meets the circumcircle of triangle $ABC$ at $R(\neq A)$. Let the feet of the perpendiculars from $A, B$ to $\ell$ be $A', B'$, respectively. Define line $\ell_1$ as the line passing through $A'$ and perpendicular to $BC$, and line $\ell_2$ as the line passing through $B'$ and perpendicular to $CA$. Prove that if $Q$ is the reflection of the intersection of $\ell_1$ and $\ell_2$ across $\ell$, then $\angle PQR = 90^{\circ}$.
Problem
Source: KMO 2024 P7
Tags: geometry
Seungjun_Lee
09.11.2024 13:25
Let $\omega$ be the circle with diameter of $PR$, while $M$ is the midpoint of $PR$. Also, let $L$ be the intersection of $AR$ and $BC$, and $T$ be the intersection of the lines passing through $A'$ perpendicular to $BC$ and the line passing through $B'$ perpendicular to $CA$. Rename $Q$ as the intersection of $TX$ and $\omega$.
Actually, we can let $P$ be a variable point on line $BC$, and work on $\mathbb{R}\mathbb{P}^2$, using the method of moving points.
kmh1
13.12.2024 21:31
It takes around 30 minutes~1 hour to bash this on the complex plane, which I think is good enough for a test that gives 3 hours for 4 problems.
We use complex coordinates, with the circumcircle of $ABC$ as the unit circle. The orthocenter $H$ has coordinate $h=a+b+c$.
The equation for line $\ell$ is of the form $\frac{\bar{x}-\bar{h}}{x-h}=k$, for some constant $k$ such that $|k|=1$. Therefore the equation for line $\ell$ would be $$\ell : kx-k(a+b+c)=\bar{x}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$We can calculate $p$ easily. $P$ is on line $BC$, so $p+bc\bar{p}=b+c$. Also $P$ is on line $\ell$, so $kp-k(a+b+c)=\bar{p}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$. Solving for $p$ gives $$p=\frac{kabc(a+b+c)-bc}{a(kbc+1)}$$$AR$ is perpendicular to $\ell$, so $\frac{\bar{a}-\bar{r}}{a-r}=-k$. Since $A,R$ are on the unit circle, this gives $$r=\frac{1}{ak}$$For any point $z$ on the plane, if we denote the foot of perpendicular from $z$ to $\ell$ as $u$, we have $\frac{\bar{z}-\bar{u}}{z-u}=-k$ and $ku-k(a+b+c)=\bar{u}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$. Solving for $u$, we get $u=\frac{a+b+c+z}{2}+\frac{\bar{z}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2k}$.
Using this formula, we get $a'=\frac{2a+b+c}{2}-\frac{b+c}{2bck}$ and $b'=\frac{a+2b+c}{2}-\frac{a+c}{2ack}$.
Denote $S$ as the intersecting point of $\ell_1$ and $\ell_2$. We have $\frac{\bar{s}-\bar{a'}}{s-a'}=-\frac{\bar{b}-\bar{c}}{b-c}=\frac{1}{bc}$ and $\frac{\bar{s}-\bar{b'}}{s-b'}=\frac{1}{ac}$. Solving for $s$ gives $$s=\frac{abc(\bar{a'}-\bar{b'})+bb'-aa'}{b-a}$$.
Using the previously calculated formulas for $a'$ and $b'$, we can calculate $a'-b'=(b-a)(\frac{1}{2abk}-\frac{1}{2})$, so $\bar{a'}-\bar{b'}=\frac{b-a}{ab}(\frac{1}{2}-\frac{abk}{2})$, and $bb'-aa'=(b-a)(\frac{2a+2b+c}{2}-\frac{ab+bc+ca}{2abck})$. Therefore $$s=a+b+c-\frac{abck}{2}-\frac{ab+bc+ca}{2abck}$$Denote $T$ as the foot of perpendicular from $S$ to $\ell$, then $t=\frac{a+b+c+s}{2}+\frac{\bar{s}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2k}$. Therefore $q=2t-s=a+b+c+\frac{\bar{s}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{k}$. Using $s=a+b+c-\frac{abck}{2}-\frac{ab+bc+ca}{2abck}$, we get $$q=\frac{a+b+c}{2}-\frac{1}{2abck^2}$$Now we are ready to solve the problem by proving that $\frac{\bar{p}-\bar{q}}{p-q}=-\frac{\bar{q}-\bar{r}}{q-r}$.
$p=a+b+c-\frac{a^2+ab+bc+ca}{a(kbc+1)}$, so $$p-q=\frac{a+b+c}{2}+\frac{1}{2abck^2}-\frac{a^2+ab+bc+ca}{a(kbc+1)}$$$$=\frac{abck^2(kbc+1)(a+b+c)+kbc+1-2bck^2(a^2+ab+bc+ca)}{2abck^2(kbc+1)}$$$$=\frac{ab^2c^2(a+b+c)k^3-bc(a^2+ab+ac+2bc)k^2+kbc+1}{2abck^2(kbc+1)}$$$$=\frac{(kbc-1)(abc(a+b+c)k^2-2kbc-1)}{2abck^2(kbc+1)}$$Let $m=abc(a+b+c)k^2-2kbc-1$, then $\bar{p}-\bar{q}=\frac{(\frac{1}{kbc}-1)\bar{m}}{2\frac{1}{abck^2}(\frac{1}{kbc}+1)}$, therefore $$\frac{\bar{p}-\bar{q}}{p-q}=-a^2b^2c^2k^4\frac{\bar{m}}{m}$$$q-r=\frac{a+b+c}{2}-\frac{1}{2abck^2}-\frac{1}{ak}=\frac{abc(a+b+c)k^2-2kbc-1}{2abck^2}=\frac{m}{2abck^2}$, and $\bar{q}-\bar{r}=\frac{\bar{m}abck^2}{2}$, so $$\frac{\bar{q}-\bar{r}}{q-r}=a^2b^2c^2k^4\frac{\bar{m}}{m}$$Therefore we have established $\frac{\bar{p}-\bar{q}}{p-q}=-\frac{\bar{q}-\bar{r}}{q-r}$($=-a^2b^2c^2k^4\frac{\bar{m}}{m}$), so $\angle PQR = 90^{\circ}$.