Points $P$ and $Q$ are chosen on the side $BC$ of triangle $ABC$ so that $P$ lies between $B$ and $Q$. The rays $AP$ and $AQ$ divide the angle $BAC$ into three equal parts. It is known that the triangle $APQ$ is acute-angled. Denote by $B_1,P_1,Q_1,C_1$ the projections of points $B,P,Q,C$ onto the lines $AP,AQ,AP,AQ$, respectively. Prove that lines $B_1P_1$ and $C_1Q_1$ meet on line $BC$.
Problem
Source: IOM 2021 #2
Tags: geometry, concurrency
L567
08.12.2021 11:42
Let $D$ be the foot from $A$ onto $BC$. Note that $ADB_1B$ is cyclic, so $\angle P_1DQ = \angle PAQ = \angle BAB_1 = \angle BDB_1$, so $B_1P_1$ goes through $D$. Similarly $C_1Q_1$ goes through $D$ and we're done.
Bakhtier
08.12.2021 12:07
Let $AD$-height of $\triangle ABC$. Then $ADC_1C$ is cyclic. Since $\angle CAQ=\angle QAP=\angle PAB$, then $$\angle Q_1DP=\angle PAQ=\angle QAC=\angle C_1AC=\angle C_1DC$$. Similarly, $\angle P_1DQ=\angle BDB_1$.
Attachments:
IvoBucata
08.12.2021 17:02
Easy harmonic bundles exercise: Claim: $(A,P;Q_1,B_1)=-1=(A,Q;P_1,Q_1)$ Proof:Because $AP$ is an angle bisector in $BAQ$ , then if $QQ_1$ intersects $AB$ at $X$ we have $XQ_1=Q_1Q$ and $$-1=(X,Q;Q_1,XQ\cap BB_1)=^B (A,P;Q_1,B_1)$$ Now it is well known that because of the claim $B_1P_1; C_1Q_1$ and $PQ$ are concurrent, which implies the problem.
Mutse
15.02.2022 15:33
True by Balkan MO 2019 P3. See https://artofproblemsolving.com/community/c6h1832236p12271234
Steve12345
15.02.2022 15:38
No, it is not. Here the perpendiculars are from P (X) and Q (Y).