Spiral similarity geo

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Source: Bulgarian Winter Tournament 2024 12.2

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#1 h Jan 28, 2024, 8:42 AM • 3 Y

Y by Spiritpalm, Assassino9931, Rounak_iitr

Let $ABC$ be scalene and acute triangle with $CA>CB$ and let $P$ be an internal point, satisfying $\angle APB=180^{\circ}-\angle ACB$ ; the lines $AP, BP$ meet $BC, CA$ at $A_1, B_1$ . If $M$ is the midpoint of $A_1B_1$ and $(A_1B_1C)$ meets $(ABC)$ at $Q$ , show that $\angle PQM=\angle BQA_1$ .

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#2 h Jan 28, 2024, 9:26 AM • 1 Y

Y by GeoKing

Note $P$ lie on $(A_1B_1C)$ . As $BA_1 \cap AB_1 =C$ and $Q=(A_1B_1C) \cap (ABC)$
We get spiral similarity at $Q$ send $A_1B_1 \rightarrow AB$ hence if $N$ is midpoint of $AB$
$\angle NQM = \angle BQA_1$ We prove $\overline{Q-P-N}$

Let $X =QP \cap (ABC)$
$\angle QCB = \angle QXB = \angle QCA_1 = \angle QPA_1 = \angle XPA$
hence $PA \parallel BX$ and similarly $PB \parallel AX$ and hence we get $N= PX \cap AB$

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#3 h Jan 28, 2024, 6:30 PM

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A great problem for JBMO training! (Writing $N$ for the midpoint of $A_1B_1$ instead of $M$ , as I like it more.)

Angle chasing gives $\triangle ABQ \sim \triangle B_1A_1Q$ , while $\angle PQN = \angle BQA_1$ is equivalent to $\angle A_1QN = \angle BQP$ . Hence if $QP \cap AB = M$ , by the similarity it suffices to show that $M$ is the midpoint of $AB$ . (In particular, $N$ is no longer needed in the diagram.)

Note that $\angle APM = \angle QPA_1 = \angle QCA_1 = \angle QCB = \angle QAB$ , so $MA^2 = MP \cdot MQ$ . Analogously $MB^2 = MP \cdot MQ$ and we are done.

This post has been edited 1 time. Last edited by Assassino9931, Jan 28, 2024, 6:51 PM

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#4 h Jan 29, 2024, 4:09 AM

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Note that $P$ lies on the reflection of $(ABC)$ over $\overline{AB}$ . For spiral similarity reasons it suffices to show that $\overline{PQ}$ passes through the midpoint of $\overline{AB}$ . If $P'$ is the second intersection of $\overline{PQ}$ with $(ABC)$ then $\measuredangle QPB_1=\measuredangle QA_1B_1=\measuredangle QBA=\measuredangle QP'A$ , so $\overline{P'A} \parallel \overline{PB_1}$ . Likewise, $\overline{P'B} \parallel \overline{PA_1}$ . Thus $APBP'$ is a parallelogram so $\overline{PP'}$ bisects $\overline{AB}$ as desired.

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#5 h Jun 14, 2024, 1:44 PM

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VicKmath7 wrote:

Let $ABC$ be scalene and acute triangle with $CA>CB$ and let $P$ be an internal point, satisfying $\angle APB=180^{\circ}-\angle ACB$ ; the lines $AP, BP$ meet $BC, CA$ at $A_1, B_1$ . If $M$ is the midpoint of $A_1B_1$ and $(A_1B_1C)$ meets $(ABC)$ at $Q$ , show that $\angle PQM=\angle BQA_1$ .

Let $N$ be the intersection of lines $QP$ , $AB$
Cuz $\angle APB=180^{\circ}-\angle ACB$ so $Q$ , $A_1$ , $B_1$ , $C$ , $P$ are concyclic
Lead to $\angle NPB = \angle B_1PQ = \angle B_1CQ = \angle ABQ$
Therefore, $\triangle NBP \sim \triangle NBQ$ , $NB^2 = NP. NQ$ . Similar, $NA^2 = NP. NQ$
So we can see that $N$ is the midpoint of segment $AB$
In the order hand, since $(A_1B_1C)$ meets $(ABC)$ at $Q$ , we have $Q$ is the center of the spiral similarity carries $AB$ to $A_1B_1$ .
Which means $\triangle QMN \sim^+ \triangle QA_1B \sim^+ \triangle QB_1A$
Thus $\angle PQM = \angle NQM = \angle BQA_1$ , done

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