In an acute non-isosceles triangle $ABC$, the inscribed circle touches side $BC$ at point $T, Q$ is the midpoint of altitude $AK$, $P$ is the orthocenter of the triangle formed by the bisectors of angles $B$ and $C$ and line $AK$. Prove that the points $P, Q$ and $T$ lie on the same line. (D. Prokopenko)
Problem
Source: 2022 Oral Moscow Geometry Olympiad grades 8-9 p6
Tags: geometry, collinear, collinearity
sbealing
17.04.2022 22:38
Let $I$ be the incentre of $\triangle ABC$ and let the $B$, $C$ angle-bisectors meet $AK$ at $E,F$ respectively. Let $IP$ meet $AK$ at $L$. Angle chasing shows $\triangle IEF$ is similar to the intouch triangle of $\triangle ABC$. Then some trig gives: $$\frac{IP}{IL}=\frac{\sin{\left(\frac{A}{2}\right)}}{\cos{\left(\frac{B}{2}\right)} \cos{\left(\frac{C}{2}\right)}}$$We then have $P,Q,T$ collinear is equivalent to: $$\frac{IP}{IL}=\frac{IT}{QK}=\frac{r}{R \sin{(B)} \sin{(C)}}$$Using our expression for $\frac{IP}{IL}$: $$\cdots \Leftrightarrow \frac{\sin{\left(\frac{A}{2}\right)}}{\cos{\left(\frac{B}{2}\right)} \cos{\left(\frac{C}{2}\right)}}=\frac{r}{R \sin{(B)} \sin{(C)}} \Leftrightarrow r=4R \sin{\left(\frac{A}{2}\right)} \sin{\left(\frac{B}{2}\right)} \sin{\left(\frac{C}{2}\right)}$$Which is well-known to be true.
VicKmath7
18.05.2022 22:17
Here's the nearly impossible to find official solution, which is of course synthetic. Let $M$ be the feet of the perpendicular from $A$ to $CI$, similarly for $N$; let $MA$ and $NA$ meet $BC$ at $X, Y$. Then $I$ is circumcenter of $AXY$ and $Q$ is on $MN$. Also, $I$ is orthocenter for $MNT$. It's easy to see that $MNT$ and the mentioned triangle in the statement are similar. Take the corresponding altitudes $TS$ in $MNT$ and $IL$ in the other triangle. Then $IS/IT=IP/PL=TP/PQ$ so Thales finishes.