Let $H$ be the orthocenter and $M=(A_1QP)\cap BC$, first we prove that $A_1, Q, C_1$ are collinear, indeed it is easy to see that $B_1HQC$ is cyclic, but obviously $B_1HA_1C$ is cyclic, so $B_1, H, Q, A_1, C$ are all concyclic, hence $\angle QA_1C=180^{\circ}-\angle QB_1C=180^{\circ}-\angle C_1AC=\angle C_1A_1C$ because $B_1Q\|AB$ and $AC_1A_1C$ is cyclic. Analogously we get $A_1, P, B_1$ collinear. Now since $A_1QPM$ is cyclic we get $\angle A_1QM=\angle A_1PM$ $\implies$ $\angle C_1QM=\angle B_1PM$, also easy to see that since $Q$ is the reflection of $B_1$ across $CC_1$, $C_1Q=C_1B_1$, similarly $B_1P=B_1C_1$ so $C_1Q=B_1P$. Now angle chase to get $\angle PQM=\angle PA_1C=\angle B_1A_1C=\angle B_1HC=\angle C_1HB=\angle C_1A_1B=\angle QA_1B=\angle QPM$ and we have $MQ=MP$, with this, $\triangle MQC_1 \cong MPB_1$ which now implies $\angle QC_1M=\angle A_1C_1M=\angle PB_1M=\angle A_1B_1M$ $\implies$ $A_1C_1B_1M$ cyclic and the conclusion follows.

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Ricochet wrote: Let $H$ be the orthocenter and $M=(A_1QP)\cap BC$, first we prove that $A_1, Q, C_1$ are collinear, indeed it is easy to see that $B_1HQC$ is cyclic, but obviously $B_1HA_1C$ is cyclic, so $B_1, H, Q, A_1, C$ are all concyclic, hence $\angle QA_1C=180^{\circ}-\angle QB_1C=180^{\circ}-\angle C_1AC=\angle C_1A_1C$ because $B_1Q\|AB$ and $AC_1A_1C$ is cyclic. Analogously we get $A_1, P, B_1$ collinear. Now since $A_1QPM$ is cyclic we get $\angle A_1QM=\angle A_1PM$ $\implies$ $\angle C_1QM=\angle B_1PM$, also easy to see that since $Q$ is the reflection of $B_1$ across $CC_1$, $C_1Q=C_1B_1$, similarly $B_1P=B_1C_1$ so $C_1Q=B_1P$. Now angle chase to get $\angle PQM=\angle PA_1C=\angle B_1A_1C=\angle B_1HC=\angle C_1HB=\angle C_1A_1B=\angle QA_1B=\angle QPM$ and we have $MQ=MP$, with this, $\triangle MQC_1 \cong MPB_1$ which now implies $\angle QC_1M=\angle A_1C_1M=\angle PB_1M=\angle A_1B_1M$ $\implies$ $A_1C_1B_1M$ cyclic and the conclusion follows. BEAUTIFUL SOLUTION!

It is easy to notice that A1-Q-C1 are collinear and A1-P-B1 are collinear (this is just an angle chase). C1Q=C1B1=B1P1 (follows by reflections). MC1=MB1 (since (BC1B1C) has center M). Angle QC1M =angle PB1M (since they are the angles subtended by arc A1M on the nine point circle of ABC). Hence QC1M and MPB1 are congruent. Hence M is the center of spiral similarity sending QP to C1B1. Hence A1QPM is cyclic.

Solved with Isaac Zhu, Jeffrey Chen, Luke Robitaille, and Raymond Feng. Since \(\widehat{HC_1}=\widehat{HQ}\) on \((BH)\), we have \(\overline{A_1H}\) bisects \(\angle C_1A_1Q\), implying \(Q\in\overline{A_1B_1}\). Analogously \(P\in\overline{A_1C_1}\). Now since \(B_1Q=B_1C_1=C_1P\), we have \(\triangle MBQ_1\cong\triangle MC_1P\), implying the conclusion.

Nothing new, I guess. $P\in A_1B_1, Q\in A_1C_1$ (since $B_1B, C_1C$ are bisectors in $\triangle A_1B_1C_1$). Center of rotation, mapping $B_1P$ to $C_1Q$ is on midpoint of arc $B_1A_1C_1$, which is $M$. Therefore, by Miquel we are done.

My solution with complex numbers Let $(ABC)$ be unit , so that $|a|=|b|=|c|=1$ Then we have $$a_1=\frac{a^2+ab+ca-bc}{2a} \wedge \bar{a_1}=\frac{ab+bc+ca-a^2}{2abc}$$$$b_1=\frac{b^2+ab+bc-ca}{2b} \wedge \bar{b_1}=\frac{ab+bc+ca-b^2}{2abc}$$$$c_1=\frac{c^2+bc+ca-ab}{2c} \wedge \bar{c_1}=\frac{ab+bc+ca-c^2}{2abc}$$Now we ready to compute $p$ and $q$ $$q=\frac{(c-c_1)\bar{b_1}+\bar{c}c_1-\bar{c_1}c}{\bar{c}-\bar{c_1}}=\frac{\frac{(c-b)(c-a)(ab+bc+ca-b^2)}{4c^2ab}+\frac{(c-b)(c-a)(c^2-ab)}{2ac^2b}}{\frac{(c-b)(c-a)}{2abc}}=\frac{2c^2-b^2+bc+ca-ab}{2c}$$Analogously $$p=\frac{2b^2-c^2+bc+ab-ca}{2b}$$Let $M$ be midpoint of $BC$ , We need to show $A_1PMQ$ is cyclic $$\frac{(p-a_1)(m-q)}{(m-a_1)(p-q)} \in \mathbb{R} \iff \frac{(b^2-ca)(c+a)}{(bc-a^2)(b+c)} \in \mathbb{R}$$ so we are done

A generalisation: The same problem, instead we define $P$ and $Q$ as points on the Thales-c. of $BH$ and $CH$, such that $\angle HBP=-\angle HCQ$ Can anyone solve this?

combo is hard Using Ricochet's diagram: $P\in A_1B_1$ and $Q\in A_1C_1$ by angle bisectors. $MB_1=MC_1$ and $M\in (A_1B_1C_1)$ $B_1P=B_1C_1=C_1Q$ Hence we may finish as $M$ is the Miquel point of $PQC_1B_1$.