Inside the acute-angled triangle $ABC$, the points $P{}$ and $Q{}$ are chosen so that $\angle ACP = \angle BCQ$ and $\angle CBP =\angle ABQ$. The point $Z{}$ is the projection of $P{}$ onto the line $BC$. The point $Q'$ is symmetric to $Q{}$ with respect to $Z{}$. The points $K{}$ and $L{}$ are chosen on the rays $AB$ and $AC$ respectively, so that $Q'K \parallel QC$ and $Q'L \parallel QB$. Prove that $\angle KPL=\angle BPC$.
Problem
Source: Russian TST 2018, Day 5 P2
Tags: geometry, angles
idkk
10.10.2024 06:59
Very beautifull the addition of points i really liked . let $Q'K$ intersect $BC$ at $H$ and $Q'L$ intersect $BC$ at $I$ . Easy to see $H$ is reflection of $C$ in $Z$ and $I$ is reflection of $B$ in $Z$ . Now $CP$ intersect circumcircle of $ABC$ at $F$ and $BP$ does at $J$ , let $FJ$ intersect $BC$ at $M$ . Now trivial angle chase and u have $\Delta KHB \sim \Delta ACF$ From there try to see $\angle AMF= \angle KPB$ Now do similarly for $\Delta CIL$ and u get equal angle as desired. bbCode
MathLuis
10.10.2024 17:16
We will share a complete solution for this awesome problem. Let $KL \cap BC=J$, perpendiculars from $Q$ to $AB,AC$ be $E,D$ respectively, project $P$ to $AB$ at $F$. Let $AQ' \cap BC=J'$ and let $B',C'$ reflections of $B,C$ over $Z$ respectively. Finally let $ZD \cap (AQ)=G$ and $Q_1$ be the reflection of $Q$ over $AC$, now we start with the claims. Claim 1: $AQ' \parallel QJ$ Proof: From DIT on $AKQ'L$ and line $BC$ we have that $(C', C), (B', B), (J', J)$ are pairs of involution, but this involution is just reflection over $Z$ meaning that $JZ=J'Z$ so $Q'J'QJ$ is a parallelogram giving the desired parallels. Claim 2: $B,Q,G$ are colinear. Proof: From isogonals and diameter we see that $BQ \perp FZ$, and from Reim's theorem we have that $FZ \parallel AG$ so $BQ \perp AG$ but also $\angle AGQ=90$ meaning the desired colinearity is true. Claim 3: $AQ'LQ_1$ is cyclic. Proof: From midbase and Claim 2 chasing angles gives: \[ \angle LAQ_1=\angle QAC=\angle BGZ=\angle(BQ, Q_1Q')=\angle Q_1Q'L \]Which implies the cyclic, thus we also have $\angle AQ_1L+\angle AQ'L=180$. The finish: From the previous claims and parallels: \[ \angle BQJ=\angle(BQ, AQ')=\angle AQ'L=180-\angle AQ_1L=180-\angle AQL \]Which implies that $Q$ has an isogonal conjugate in $ABJL$, and of course it has to be $P$ by definition so it also means that $\angle BPJ+\angle APL=180$ and that $P,Q$ are also isogonal conjugates in $AKJC$, meaning that $\angle KPJ+\angle APC=180$, so substracting both means $\angle BPK=\angle CPL$ therefore $\angle BPC=\angle KPL$ as desired thus we are done .