Let $ABC$ be a triangle with $\angle A = 120^\circ$, $I$ be the incenter, and $M$ be the midpoint of $BC$. The line passing through $M$ and parallel to $AI$ meets the circle with diameter $BC$ at points $E$ and $F$ ($A$ and $E$ lie on the same semiplane with respect to $BC$). The line passing through $E$ and perpendicular to $FI$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. Find the value of $\angle PIQ$.
Problem
Source: Sharygin Finals 2023 9.8
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023, Angle Chasing
NO_SQUARES
02.08.2023 12:56
Very nice problem! Here is my solution, but some detailes I didn't wrote The key of solving this problem is understand that points $B$ and $Q$ are symmetric onto line $CI$ and analogy with point $C$. To prove it we can say that $C_1=CI \cap (BEC), B_1 = BI \cap (BEC)$, after that let $Q'$ be a point for which $B$ and $Q'$ are symmetric onto line $CC_1$. By angle chasing $CA \parallel MC_1$, so $Q'$ lies on $CA$ ($MC_1$ is a midline in $Q'BC$) Also, easy to see (by $ME \parallel AI$ and $MC_1 \parallel AC$) that $\angle C_1ME = \angle CAI = 60^{\circ}$ (and so $\angle C_1MB_1=\angle C_1MF=\angle B_1MF =120^{\circ}$) and $C_1P'XI$ is inscribed, there $X = FI \cap (BEC)$, because $\angle C_1Q'I=\angle C_1BI = 60^{\circ}=\angle C_1XI$. So, $\angle Q'XI = 90^{\circ}$ and $P', X, E$ are collinear, so $Q'=Q$. After this easy to see that $\angle PIQ = 90^{\circ}$, because $\angle C_1IQ = 30^{\circ}$ and $\angle PIC = 60^{\circ}$.
A comment, perhaps for myself in the future: Maybe it will seem strange, but when solving this problem, I remembered a few lines from the official decision 9.4 of this year's All-Russian Olympiad. In addition, a very important part of the solution was that I "believed" in my picture, which is where the points $P',Q'$ came from. After that, the task was solved quite quickly.
HoRI_DA_GRe8
03.08.2023 17:21
Yayyy this was what gave me the H.M.,insane day 2 I will never forget this day in my life. Here's an overview of the solution I found at the contest which uses five phantom points
Consider $P'$ and $Q'$ as reflection of $C$ and $B$ over $BI$,$CI$ respectively.Note that midpoints of $CP',BQ'$ lie on the given circle and $P'AIC$ and $Q'AIB$ cyclic
Let $O\in P'Q'$ be such a point s.t $IO \perp P'Q'$ and $E'$ be the midpoint of $P'Q'$ ,then $O,E'$ lie on the circle by angle chasing and midpoint theorem.Note that $\angle P'IQ'=90^{\circ}$
Take the antipode $F'$ of $E$ on the given circle,by using angle chasing and the fact that $\overline{O-I-F'}$ it can be proved that $E'F' \parallel AI$ so $E' \equiv E,F' \equiv F$. Hence $P' \equiv P$ and $Q' \equiv Q$ ,so $\angle PIQ=90^{\circ}$ $\blacksquare$
kamatadu
04.08.2023 09:34
I couldn't solve this on the exam cuz I ran out of time but soved it without paper on the way back home. T__T I can never forgive myself for trying the 9.7 for so long instead of trying this one which I could have solved. T__T Didn't even get the HM. T__T Here is a sketch that uses both coordinate and complex bash along with a few redefinitions. Redefine $P$ to be on $AB$ such that $AB=BC$, and similarly redefine $Q$ too. Now toss the numbers onto the complex plane and fix $B=-1$ and $C=1$. This forces $M=0$. Now furthermore, set $\angle ABC=60^{\circ}-\alpha$ and $\angle ACB=\alpha$. Now it is easy to obtain the explicit value of the affixes of $P$ and $Q$ by shifting the origin to $B$ then multiplying $\mathrm{cis}(60^{\circ}-\alpha)$ to the shifted affix of $C$ to get the affix of $P$. Similarly, we can also get the explicit value of the affix of $Q$. Finally, multiply $\mathrm{cis}(-(60^{\circ}+\alpha))$ to the affix of $B$ to get the affix of $E$. Now finally to get the affix of $F$ simply multiply $-1$ to the affix of $E$. Finally, we take all the explicit values of the affixes and move onto the coordinate plane. Now it is easier to calculate the coordinates of $I$ using the equation of line formula. Put everything into the conditions $\overline{P-E-Q}$ are collinear and $FI\perp PQ$ to see that these turn out to be actually true that finishes.
Infinityfun
11.11.2023 15:14
Redefine $P$ and $Q$ on $AC$ and $AB$ such that $IB=BP$ and $CI=IQ$. Redefine $E$ such that $\angle IBE =\angle ICE =30$ It is easy to check that $\angle PIQ =90$ and $BE$ is the perpendicular bisector $IP$ and $CE$ is the perpendicular bisector of $IQ$. $E$ is the center of $(IPQ)$ so $EPQ$ are collinear and $EP=EQ$. Define $X$ such that $BCX$ is equaliteral triangle and $X$ and $A$ lies on the same side of $BC$. Call center of $(BCX)$ point $T$. $T$ is the isogonal conjugate of $I$ in triangle $BCE$. Let $S$ be the reflection of $T$ across $BC$. It is easy to see that $ISF$ are collinear and $SF\parallel ET$ To finish, realize that $PQ$ is parallel to one side of pedal triangle of $I$ in $EBC$. So $PQ \perp ET \parallel IF$ and we are done.