My problem! [asy][asy] size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.7; pen bluefill,pinkfill,turquoisedraw,bluedraw,purpledraw,pinkdraw,graydraw; bluefill = RGB(204,233,255); pinkfill = RGB(255,204,255); turquoisedraw = RGB(0,170,153); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); graydraw = RGB(119,119,119); pair A,B,C,M,N,O,T,P,Q,D; A = dir(85); B = dir(195); C = dir(345); O = circumcenter(A,B,C); M = (A+B)/2; N = (A+C)/2; path c,d,e,f; c = circumcircle(A,B,C); T = intersectionpoints(c,11M-10N--N)[0]; d = circle((O+T)/2,distance(O,T)/2); e = circumcircle(A,B,O); f = circumcircle(A,C,O); P = intersectionpoints(d,e)[0]; Q = intersectionpoints(d,f)[0]; D = foot(A,B,C); filldraw(c,bluefill,bluedraw); fill(d,pinkfill); filldraw(circumcircle(A,M,N),pinkfill,pinkdraw); draw(d,pinkdraw); draw(e,purpledraw); draw(f,purpledraw); draw(A--B--C--cycle,bluedraw); draw(A--P,turquoisedraw); draw(N--T,turquoisedraw); draw(A--D,graydraw); dot("$A$",A,1.9*dir(85)); dot("$B$",B,dir(225)); dot("$C$",C,dir(315)); dot("$M$",M,dir(140)); dot("$N$",N,dir(30)); dot("$O$",O,1.9*dir(270)); dot("$T$",T,dir(150)); dot("$P$",P,dir(265)); dot("$Q$",Q,dir(10)); dot("$D$",D,dir(270)); [/asy][/asy] Let $\measuredangle$ denote directed angles modulo $180^\circ$. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$. The crux of the problem is the following claim. Claim: $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$. Proof: Notice that the circumcircle of $AMN$ has diameter $\overline{AO}$, so it is tangent to $\omega$. Thus, the circumcircles of $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$. Also, we have $\measuredangle OTP=\measuredangle OQP=\measuredangle OCA=\measuredangle NAO$, so $N$ and $P$ are symmetric about the perpendicular bisector of $\overline{AT}$. Similarly, $M$ and $Q$ are symmetric. $\square$ Since $A$, $P$, and $Q$ are collinear, $M$, $N$, and $T$ are collinear. Since $\overline{MN}$ is the perpendicular bisector of $\overline{AD}$, we have $TA=TD$. Let $\overline{AP}$ and $\overline{MN}$ intersect at $X$. By symmetry, $XAT$ is isosceles. Thus, we have \[\measuredangle BTD=\measuredangle BTA+\measuredangle ATD=\measuredangle BCA+2\measuredangle ATX=\measuredangle MNA+\measuredangle AXN=\measuredangle PAC\]and we are done. $\square$

Nice problem, here is a solution found while watching Portugal vs Turkey Match lol. Let the tangency point be $T$, let $M,N$ midpoints of $AC, AB$ respectively and let $(OPQ) \cap (ANMO)=G$, note that they are circles with diameters $TO,AO$ respectively and since $TO=AO$ and $\angle TGO=90$ we get $G$ midpoint of $TA$, so these circles are symetric w.t.t. $GO$. Now $\angle APO=\angle ABO=\angle NAO=\angle NMO$ which shows that arcs $QO,NO$ have the same lenght, in the same way arcs $PO,MO$ have the same lenght and thus $(N,Q), (M,P)$ are symetric w.r.t. $GO$, now this means $T,M,N$ colinear. To finish note that $\angle GTP=\angle AQG=\angle TNG=\angle NTB$ (last one is just midbase), therefore $\angle NTD=\angle ATN=\angle BTP$ which means $\angle BTD=\angle PTN=\angle CAP$ as desired thus we are done .

Invert at $O$. We have $\overline{p'}=\frac{2t-a-b}{t^2-ab}$, so $p=\frac{t^2-ab}{2t-a-b}$, which means $p-a=\frac{(t-a)^2}{2t-a-b}$. Similarly, $q-a=\frac{(t-a)^2}{2t-a-c}$, so $\frac{2t-a-b}{2t-a-c}$ is real, so $\frac{2t-a-c}{b-c}$ is real, which implies $2t-a-c=-bc\left(\frac2t-\frac1a-\frac1c\right)$, or $$2at^2-(a+c)(a+b)t+2abc=0.$$ We have $d=\frac12\left(a+b+c-\frac{bc}a\right)$. The condition $\angle BTD=\angle CAP$ is equivalent to $\frac{b-t}{d-t}\frac{c-a}{p-a}$ real, or \begin{align*} \frac{btca}{(d-t)\frac{(t-a)^2}{2t-a-b}}&=\frac1{\left(\overline d-\frac1t\right)\frac{\overline{(t-a)^2}}{\overline{2t-a-b}}}\\ \frac{abct(2t-a-b)}{d-t}&=\frac{a^2t^2\left(\frac2t-\frac1a-\frac1b\right)}{\overline d-\frac1t}\\ b^2c(2t-a-b)(t\overline d-1)&=t(d-t)(2ab-at-bt)\\ t^3(a+b)-t^2(ad+bd+2ab+2b^2c\overline d)+t(2abd+2b^2c+ab^2c\overline d+b^3c\overline d)-b^2c(a+b)&=0\\ t^3a(a+b)-t^2\frac12((a+b)(a^2+ab+ac-bc)+4a^2b+2b(bc+ca+ab-a^2))+t\frac12(2ab(a^2+ab+ac-bc)+4ab^2c+b(a+b)(ab+bc+ca-a^2))-ab^2c(a+b)&=0\\ t^32a-t^2(a^2+3ab+ac+bc)+tb(a^2+3ac+ab+bc)-2ab^2c&=0\\ (2at^2-(a+c)(a+b)t+2abc)(t-b)&=0. \end{align*}

Solved with Benjaminxiao, Zhaom. Let $P'$ be the intersection of lines $OP$ and $AB$, and let $Q'$ be the intersection of lines $OQ$ and $\overline{AC}$; in other words, $P'$ and $Q'$ are the images of $P$ and $Q$ under inversion about $\omega$. Therefore, line $P'Q'$ is tangent to $\omega$ at $T$. Also, since $A$, $P$ and $Q$ are collinear, it follows that $(AOP'Q')$ is cyclic. By Simson's theorem, the feet from $O$ onto sides $\overline{AB}$, $\overline{AC}$ and $\overline{P'Q'}$ are collinear; in other words, $T$ lies on the $A$-midline of $\triangle ABC$. Claim: We have $\triangle TPQ \sim \triangle ACB$ (with similarity factor of $\tfrac{1}{2}$.) Proof: We have \[\angle TPQ = \angle TOQ' = 90^{\circ} - \angle OQ'P' = \angle P'AO - 90^{\circ} = \angle C,\]and likewise for $\angle TQP$ and $\angle B$. If we define $P_1$ and $Q_1$ as the reflections of $T$ across $P$ and $Q$, respectively, we have that $\triangle T Q_1 P_1$ and $\triangle ABC$ are reflections (and in particular, $\omega$ is the circumcircle of $\triangle TQ_1 P_1$). From this, it follows by symmetry $\angle PAT = \angle (P_1 Q_1, AT) = \angle (AT, BC)$. Now, we have \[\angle BTD = \angle ATD - \angle C = \angle PAT + \angle (AT, BC) - \angle C = \angle PAT + \angle (AT, AC) = \angle PAT - \angle CAT = \angle CAP.\]

I regret not discovering the following version of the problem (that I found just now), which has a significantly more appealing statement. ELMO SL 2024 G5 but better wrote: Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. A line through $A$ intersects $\overline{BC}$, the circumcircle of $AOB$, and the circumcircle of $AOC$ at $D$, $E \ne A$, and $F \ne A$, respectively. Suppose the circumcircle of $OEF$ is tangent to $\omega$ at $T$. Prove that $\angle ATD=90^\circ$.

Solution