This problem was proposed by me Here's the first solution I found. It's quite silly.

If this problem is too easy for you, try the following one: Quote: Let $P$ and $Q$ be points inside triangle $ABC$. Let $\omega$ be the circumcircle of $ABC$. Let $A_1$ and $A_2$ be points on $\omega$ satisfying $\measuredangle BA_1P = \measuredangle QA_1C$ and $\measuredangle BA_2P = \measuredangle QA_2C$. Points $B_1$, $B_2$, $C_1$, $C_2$ are defined similarly. Prove that $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent.

Solved with L567 Suppose that the inellipse of $\triangle ABC$ with foci $P,Q$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Claim 1 (known): $AD, BE, CF$ are concurrent Proof: By pascal on $DDEEFF$, $DD \cap EF$, $EE \cap FD$, and $FF \cap DE$ are collinear. Taking the polar of these three points proves the desired lemma. Claim 2: $\frac{BA_1}{CA_1} = \frac{BD}{CD} \frac{AB}{AC}$ Proof: By DDIT, $A_1D$ and $A_1A$ are isogonal in $\angle BA_1C$. This implies $\triangle BA_1A \sim \triangle DA_1C$. Note that $BA_1 = A_1D \frac{BA}{DC}$. Since we similarly have that $CA_1 = A_1D \frac{CA}{DB}$, the claim follows. Finish: Let the bisector of $\angle BAC$ meet $BC$ and $(ABC)$ at $X$ and $U$ respectively. Let $D_1$ and $A_2$ be the reflections of $D$ and $A_1$ across the perpendicular bisector of $BC$. \[ (BC; UA_2) = (BC; A_1U) \stackrel{\text{Claim 2}}{=} \frac{BA}{CA} \frac{BD}{CD} = (BC; XD_1)\]Hence, we have that $AD_1A_2$ are collinear. Thus the required point is the isogonal conjugate of the isotomic conjugate of the concurrency in Claim 1. $\blacksquare$

i found a sol to the generalization i think : D We claim the following even further generalization: Quote: Let $P$ and $Q$ be points inside circle $\omega$ (conics work too!). Let $\ell$ be a fixed line. There is some fixed point $Z$ such that for all $X$ on $\ell$, if $A_1,A_2$ are the two choices of $A$ such that the line through the second intersections of $AP$ and $AQ$ with $\omega$ passes through $X$, then $A_1A_2$ passes through $Z$. Taking $\ell$ to be the line at $\infty$ and picking the cases where $X$ is the point at infinity on $BC$, $CA$, and $AB$ recovers the original result.

We proceed by geometry, or something like that.

I have a significantly worse solution that is sort of similar to 2above, after getting a major hint from L567 Do a $\sqrt{bc}$ inversion, then let $P,Q$ go to $P',Q'$ respectively. Then $\overline{PQ}||\overline{P'Q'}$ and $A-P-Q',A-Q-P'$. $A_1$ goes to the point $A_1'$ on $\overline{BC}$ which, after a little angle chasing, can be shown to have the property that $\overline{BC}$ is an angle bisector of $\overline{P'A_1},\overline{Q'A_1}$. $P',Q'$ are isogonal conjugates, so $A_1'$ is the point on $\overline{BC}$ that is also on the inellipse of $\triangle ABC$ with $P',Q'$ as foci. Let the inellipse with $P,Q$ as foci touch $\overline{BC}$ at $D$. These 2 inellipses are homothetic by $\overline{PQ}||\overline{P'Q'}$ and $A-P-Q',A-Q-P'$, so when the inellipse with foci $P,Q$ is affine transformed to a circle, we see that $BC$ and $DA_1'$ have the same midpoint. Thus $\overline{AA_1},\overline{BB_1},\overline{CC_1}$ concur at the isogonal conjugate of the isotomic conjugate of the intersection of $\overline{AD}$ and its 2 other corresponding lines.

complex

Define instead $A_1' = (BDQ) \cap (ABC)$. I will show that $A_1'=A_1$ ends up being the Miquel point of $BPCQ$. To do this, set $P_1, P_2, Q_1, Q_2$ to be the intersections of $\overline{BP}, \overline{CP}, \overline{BQ}, \overline{CQ}$ with $\omega$ respectively. Then, notice that \begin{align*} \angle BDA_1 &= \angle BDC - \angle Q_1BA_1 \\ &= \frac 12(\widehat{BC}+\widehat{P_1Q_2} - \widehat{A_1Q_1}) \\ &= \frac 12(\widehat{BA_1} + \widehat{AQ_2}) \\ &= \angle PCA_1. \end{align*}Thus $A_1$ lies on $(PCD)$, hence the result. The finish now is easy. Note that $$\frac{\sin \angle BAA_1}{\sin \angle CAA_1} = \frac{BA_1}{CA_1} = \frac{BA_1}{A_1P} \cdot \frac{A_1P}{A_1C} = \frac{BQ}{CP} \cdot \frac{BP}{CQ}.$$This cyclically multiplies to $1$, so we are done by trig Ceva.

Define $X_A = BQ \cap CP$, and utilize the phantom point $A_2 = (BPX_A) \cap (CQX_A)$. Then $A_2 \in (ABC)$, as \begin{align*} &360 - \angle CA_2B = \angle CQX_A + \angle BPX_A \\ =~& \angle A + \angle ABP + \angle ACP + \angle A + \angle ABQ + \angle ACQ \\ =~& \angle A + 180 \implies \angle CA_2B = 180 - \angle A, \end{align*} so we indeed have $A_2 = A_1$. We can then finish with Trig Ceva using similarity ratios, as \[\prod \frac{\sin \angle BAA_2}{\sin \angle CAA_2} = \prod \frac{BA_2}{CA_2} = \prod \sqrt{\frac{\frac{BP}{CQ} \cdot QA_2 \cdot \frac{BQ}{CP} \cdot PA_2}{\frac{CQ}{BP} \cdot PA_2 \cdot \frac{CP}{BQ} \cdot QA_2}} = \prod \frac{BP \cdot BQ}{CP \cdot CQ} = 1.\]

Solved with kotmhn

Perfect six-point set, ellipses, and projective, omfg Let $E,F$ be $BP \cap CQ$, $BQ \cap CP$. Let $A'$ be the intersection of the line between P and the reflection of Q across BC, with $BC$ (By reflection property of an ellipse, A' is the touchpoint of the ellipse with P,Q with foci to side BC). Then we have that $P,Q$ and $E,F$ are isogonal conjugates in degenerate quadrilateral $(AA')(BC)$, so $(AA')(BC)(PQ)(EF)$ is a perfect eight-point set, (i.e have a common Miquel point). Thus, $A'$ is the isogonal conjugate of $A$ in quadrilateral $(BC)(PQ)$. Let $\mathcal{K}$ be the isoptic cubic of this perfect eight-point set. We have that the isoptic cubic will pass through $A',A,B,C,P,Q,D,E$. Let $M$ be the focus of this isoptic cubic. We have that M is the Miquel point of $(BC)(PQ)$ (i.e. the circles $BPE$, $CQE$) concur at $M$, the Miquel point of $(BC)(AA')$ (i.e. the circle through $BA'$ tangent to $AB$, along with the circle through $CA'$ tangent to $AC$, and the circumcircle $(ABC)$). Since $M$ lies on $\mathcal{K} \implies \angle BMP=\angle CMQ$ and $(ABC)$ (which only intersect in four real points by Bezout's theorem, the other three intersections are $A,B,C$), we have that $A_1$ must just be $M$. We can now completely eliminate isogonal conjugates! By Brianchion, we can choose A', B', C' to be any points on the sides of ABC such that A'B'C' is perspective to ABC. Let point $F$ be the perspector, and let $X$ be our desired concurrence. Finally note that as we vary $F$ on a line through G, we have that $A_1$ moves projectively with degree 2 on $(ABC)$ since $A'$ moves with degree 1 on $BC$, so line $AA_1$ moves projectively with degree 1. Therefore for the concurrence to hold we need to check four cases of $F$. When $F=G$ this is obvious as $X$ is the symmedian point (by bc inversion), and when $P$ is on lines $AB, AC, BC$, we get concurrence by degeneracy, so we win.