Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{a^2-b^2}=1$ with foci $(\pm b,0)$. Then, let $P=(m,n)$, where $n^2=\frac{(a^2-m^2)(a^2-b^2)}{a^2}$. The coordinates of $A$ satisfy \begin{align*} \frac{x^2}{a^2}+\frac{y^2}{a^2-b^2}&=1\\ y&=\frac n{m+b}(x+b)\\ \frac{x^2}{a^2}+\frac{n^2(x+b)^2}{(m+b)^2(a^2-b^2)}&=1. \end{align*} Since $x=m$ satisfies the equation, we get that the other root is $-\frac{\frac{2bn^2}{(a^2-b^2)(m+b)^2}}{\frac1{a^2}+\frac{n^2}{(a^2-b^2)(m+b)^2}}-m=-\frac{2a^2b+a^2m+b^2m}{a^2+b^2+2bm}$ by Vieta. Therefore, the coordinates of $A$ are $$\left(-\frac{2a^2b+a^2m+b^2m}{a^2+b^2+2bm},\frac n{m+b}\left(-\frac{2a^2b+a^2m+b^2m}{a^2+b^2+2bm}+b\right)\right)=\left(-\frac{2a^2b+a^2m+b^2m}{a^2+b^2+2bm},\frac{n(b^2-a^2)}{a^2+b^2+2bm}\right).$$Similarly, the coordinates of $B$ are $$\left(-\frac{2a^2b-a^2m-b^2m}{a^2+b^2-2bm},\frac{n(b^2-a^2)}{a^2+b^2-2bm}\right).$$The equation of the tangent at $A$ is $$\frac{-\frac{2a^2b+a^2m+b^2m}{a^2+b^2+2bm}}{a^2}x+\frac{\frac{n(b^2-a^2)}{a^2+b^2+2bm}}{a^2-b^2}y=1$$or $$-\frac{2a^2b+a^2m+b^2m}{a^2}x-ny=a^2+b^2+2bm.$$Similarly, the equation of the tangent at $B$ is $$-\frac{-2a^2b+a^2m+b^2m}{a^2}x-ny=a^2+b^2-2bm.$$Therefore, we get $-4bx=4bm$, so $x=-m$. Then, we get $$y=\frac{m(2a^2b+a^2m+b^2m)-a^4-a^2b^2-2a^2bm}{a^2n}=\frac{a^2m^2+b^2m^2-a^4-a^2b^2}{a^2n}=\frac{(a^2+b^2)(m^2-a^2)}{a^2n}.$$ Therefore, the midpoint of $PQ$ is $$\left(0,\frac{(a^2+b^2)(m^2-a^2)+a^2n^2}{2a^2n}\right)=\left(0,\frac{(a^2+b^2)(m^2-a^2)+(a^2-m^2)(a^2-b^2)}{2a^2n}\right)=\left(0,\frac{b^2(m^2-a^2)}{a^2n}\right).$$ Now, the circumcenter of $PF_1F_2$ is $\left(0,\frac{m^2+n^2-b^2}{2n}\right)$, and the circumradius is $$\sqrt{b^2+\frac{(m^2+n^2-b^2)^2}{4n^2}}.$$Therefore, we need to show $$\sqrt{b^2+\frac{(m^2+n^2-b^2)^2}{4n^2}}=\frac{m^2+n^2-b^2}{2n}-\frac{b^2(m^2-a^2)}{a^2n}$$or $$\sqrt{4n^2b^2+(m^2+n^2-b^2)^2}=m^2+n^2-b^2-\frac{2b^2(m^2-a^2)}{a^2}$$which directly simplifies to $n^2=\frac{(a^2-m^2)(a^2-b^2)}{a^2}$. Therefore, the midpoint of $PQ$ lies on the circumcircle of $PF_1F_2$.

shame this didn't get on the test The key leap of faith is the claim that said midpoint is the midpoint of minor arc $F_1F_2$ in $(PF_1F_2)$. We will prove separately that $PQ$ bisects $\angle F_1PF_2$ and that $PQ$ is bisected by the perpendicular bisector of $F_1F_2$. Let $I,J$ be the circular points at infinity. Part 1: $PQ$ bisects $\angle F_1PF_2$. By Desargues's Involution Theorem on $P$ and $\{IF_1, JF_1, JF_2, IF_2\}$, there is an involution on the pencil of lines through $P$ fixing $PP$, swapping $PI \leftrightarrow PJ$, and swapping $PF1 \leftrightarrow PF_2$. The first two imply that this is reflection over $PP$, but $(PP, PQ; PF1, PF2) = -1$ since $Q = AA \cap BB$, so $PQ \perp PP$ and $PQ$ bisects $\angle F_1PF_2$. Part 2: the perpendicular bisector of $F_1F_2$ bisects $PQ$. Let $L$ be the point at infinity along the perpendicular bisector of $F_1F_2$. Let $C = PP \cap AA$, and let $D = PP \cap BB$. Then, by Desargues's Involution Theorem on $L$ and $\{QC, CP, PD, DQ\}$, there is an involution on the pencil of lines through $L$ swapping the two tangents from $L$ to $\mathcal E$, swapping $LC \leftrightarrow LD$, and swapping $LP \leftrightarrow PQ$. But $C$ and $D$ lie on the polars of $F_1$ and $F_2$, respectively, by La Hire, i.e. the two directrices of $\mathcal E$, so the first two pairs imply that this involution is the reflection over the perpendicular bisector of $F_1F_2$, as desired. Combining the two parts, we are done. $\blacksquare$

Step 1 $PQ$ bisects $\angle F_1PF_2$ Proof :-Let the tangent at $P$ to ellipse meet $QA,QB$ at $N,M$ respectively. By property of tangent of ellipse we know that $MN$ is external angle bisector of $\angle F_1PF_2$.By brianchon theorem we know that $AM,BN,PQ$ concur so $(MN,PQ;PA,PB)=-1$. So $PQ$ is internal angle bisector of $\angle F_1PF_2$. Let $C$ be the center of the ellipse. $R$ be the point symmetric to $P$ wrt perpendicular bisector of $F_1F_2$. $P'$ be the reflection of $P$ across $C$. Step 2 $R$,$P'$,$Q$ are collinear. Proof By symmetry $R \in (PF_1F_2)$ and $PR \parallel F_1F_2$.$-1=(F_1,F_2;C,\infty_{F_1F_2})\stackrel{P}{=}(A,B;P',R)$. $ABP'R$ is a harmonic quadrilateral so $R$,$P'$,$Q$ are collinear. Let $D$ be the midpoint of arc $F_1F_2$ not containing $P$. Step 3 $D$ is midpoint of $PQ$. Proof $CD$ is perpendicular bisector of $F_1F_2$ so $CD \perp PR$. $CP=CR=CP'$ so $P'R \perp PR$.We have $CD \parallel P'Q$. So $CD$ is midline in $PP'Q$ implying $D$ is midpoint of $PQ$.

Unironically my favorite geo in the entire shortlist!! Let $M$ be the midpoint of $PQ$. First, observe that $AF_1 + AF_2 = BF_1+BF_2$, so there exists a circle $\Omega$ tangent to lines $PA, PB, AF_2, BF_1$. Moreover, by optical property of ellipse, $AQ$ externally bisects $\angle PAF_2$ and $BQ$ externally bisects $PBF_1$. Thus, $Q$ is actually the center of $\Omega$, implying that $PQ$ bisects $\angle APB$. Now, let $\Omega$ be tangent to $PA$ and $PB$ at $A'$ Moreover, let the projection of $M$ onto $PF_1$ is $M'$. Then, we have $$PM' = \frac{PA'}{2} = \frac{PA + AF_2+PF_2}{4} = \frac{PF_1 + PF_2}{2}.$$Thus, if $F_2'$ is the reflection of $F_1$ across $M'$, then we have $PF_2'=PF_2$. This implies that $M$ is the circumcenter of $\triangle F_2'F_1F_2$, which is well-known (by Fact 5) to pass through the incenter of $\triangle PF_1F_2$. Hence, we have that $M$ is the midpoint of arc $F_1F_2$ in $\odot(PF_1F_2)$.

Similar to above. Nice problem, I hope ellipses become more popular in olympiads in the near future! We have $AF_1+AF_2=BF_1+BF_2$, so by the converse of Pitot's theorem there exists a circle $\gamma$ externally tangent to $PA, PB, AF_2, AF_1$. Since $AQ$ and $BQ$ are the external angle bisectors of $\angle F_1AF_2$ and $\angle F_1BF_2$, the center of $\gamma$ is $Q$. Let $\gamma$ be tangent to $PA, PB$ at $A', B'$, and let $M$ be the midpoint of $PQ$. I claim $M$ is the center of the spiral similarity taking $A'F_1P\to PF_2B'$, which would imply that $PF_1F_2M$ is cyclic and thus solve the problem. To prove this, note that $$F_2B'= PB'- PF_2 = \frac{1}{2} (PF_1+PB+F_1B)-PF_2 =\frac{1}{2} (PF_1+PF_2+F_2B+F_1B)-PF_2 = PF_1+PF_2 - PF_2=F_1P,$$so $F_1$ and $F_2$ correspond in the similarity centered at $M$ taking $A'P\to PB'$, and thus we are done. $\square$

It's not hard but it's a very revolutionary attempt in creating the pioneers of geo problems. Let the midpoint be $M$. By the 1st definition of eclipse the quad $F_1BF_2A$ has an excircle and by light properties $Q$ is the center of it, so $PQ$ bisects $\angle F_1PF_2$. Let $\angle F_1PQ=\theta$. $PQ=\frac{PX+PY}{2cos\theta}=\frac{PF_1+PF_2+F_1X+F_2Y}{2cos\theta}=\frac{PF_1+PF_2+F_1A+F_2A}{2cos\theta}$, so $PM=\frac{PF_1+PF_2}{2cos\theta}$, and by 3-chord theorum $F_1PF_2M$ is concyclic.

this is probably one of the best problems ive ever done [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = linewidth(2) + black; /* point style */ real xmin = -14.207372658051753, xmax = 35.82212281709306, ymin = -22.13942457541276, ymax = 6.7680417327393; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw figures */draw(shift((3,0))*rotate(0)*xscale(5.246942056458684)*yscale(4.304695220783334)*unitcircle); draw((-0.956924680748209,-2.8269562077893053)--(4.613464538036539,-8.074988809159727)); draw((4.613464538036539,-8.074988809159727)--(7.849970775728293,-1.6425174329537393)); draw((1.38653546196346,4.096116560033611)--(4.613464538036539,-8.074988809159727)); draw((-0.956924680748209,-2.8269562077893053)--(6,0)); draw((0,0)--(6,0)); draw((7.849970775728293,-1.6425174329537393)--(0,0)); draw(circle((4.613464538036539,-8.074988809159727), 6.958970713738725)); draw(circle((3,1.2672294033544131), 3.256665527917471)); draw((0,0)--(3,-1.989436124563058)); draw((3,-1.989436124563058)--(6,0)); draw((1.38653546196346,4.096116560033611)--(10.61346453803654,-4.096116560033611)); draw((4.613464538036539,-8.074988809159727)--(10.61346453803654,-4.096116560033611)); draw((-1.38653546196346,-4.096116560033611)--(4.613464538036539,-8.074988809159727)); draw((-1.38653546196346,-4.096116560033611)--(10.61346453803654,-4.096116560033611)); draw((4.613464538036539,-8.074988809159727)--(9.233765053128298,-2.8711348003585497)); draw((-1.9781066587723886,-5.843741948753609)--(1.38653546196346,4.096116560033611)); draw((-1.9781066587723886,-5.843741948753609)--(4.613464538036539,-8.074988809159727)); /* dots and labels */ dot((6,0),dotstyle); label("$F_{2}$", (6.417125614712827,-0.06840129535594472), NE * labelscalefactor); dot((0,0),dotstyle); label("$F_{1}$", (-0.7786452814359955,-0.22042462414782016), NE * labelscalefactor); dot((1.38653546196346,4.096116560033611),dotstyle); label("$P$", (1.4003557645809013,4.568310232796256), NE * labelscalefactor); dot((7.849970775728293,-1.6425174329537393),dotstyle); label("$B$", (7.9626961240969045,-1.4366112544828236), NE * labelscalefactor); dot((-0.956924680748209,-2.8269562077893053),dotstyle); label("$A$", (-1.234715267811625,-2.3740884486993887), NE * labelscalefactor); dot((4.613464538036539,-8.074988809159727), dotstyle); label("$Q$", (4.476069115162364,-8.759708941608672), NE * labelscalefactor); dot((3,-1.989436124563058),dotstyle); label("$M$", (3.300647374479358,-2.424762891630014), NE * labelscalefactor); dot((-1.38653546196346,-4.096116560033611),dotstyle); label("$F_1'$", (-2.172192462028197,-3.7676356292915805), NE * labelscalefactor); dot((10.61346453803654,-4.096116560033611),dotstyle); label("$F_2'$", (10.724453263815995,-3.843647293687518), NE * labelscalefactor); dot((9.233765053128298,-2.8711348003585497),dotstyle); label("$S$", (9.330906083223793,-2.6781351062831398), NE * labelscalefactor); dot((1.9937065867787627,-1.6279630087231483),dotstyle); label("$X$", (1.4003557645809013,-1.4619484759481363), NE * labelscalefactor); dot((6.038685742834014,-1.2635265643793243),dotstyle); label("$Y$", (6.138416178594386,-1.056552932503135), NE * labelscalefactor); dot((-1.9781066587723886,-5.843741948753609),dotstyle); label("$R$", (-2.5269135625425756,-5.4905666889328355), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] In fact, I claim that $M$, the midpoint of $PQ$, is the midpoint of arc $F_1F_2$ in $(PF_1F_2)$. WLOG $PF_1 \le PF_2$ so I don't have to deal with annoying config issues later. Claim: $Q$ lies on the angle bisector of $\angle APB$ (similarly, $M$ lies on this bisector). Proof. Note that $BF_1 + BF_2 = AF_1 + AF_2$, so by the converse of Pitot's, there exists a circle tangent to the self-intersecting quadrilateral $F_1AF_2B$ - let this circle be $\omega$. It is well known that $BQ$, $AQ$ are external angle bisectors with respect to $\angle PAF_2$ and $\angle PBF_1$ respectively. Thus, $Q$ is the $P$-excenter in $PF_2A$, which trivially lies on the angle bisector of $\angle APB$ (and thus $M$ does as well). $\square$ Note that $Q$ being the $P$-excenter in $PAF_2$ implies that $Q$ is the center of $\omega$. Now, let $\omega$ be tangent to $PA$, $PB$, $F_2A$, $F_1B$ at $R$, $S$, $X$, $Y$ respectively. Claim: $PF_1 = F_2S$. Similarly, $PF_2 = F_1R$. Proof. We have that \begin{align*} F_2S = PS - PF_2 = \frac{PR + PS}{2} - PF_2 &= \frac{PF_1 + F_1R + PF_2 + F_2B + BS}{2} - PF_2 \\ &= \frac{PF_1 + F_1Y + PF_2 + F_2B + BY}{2} - PF_2 \\ &= \frac{PF_1 + PF_2 + F_1B + F_2B}{2} - PF_2 \\ &= PF_1 + PF_2 - PF_2 = PF_1 \end{align*}as desired; the other equality follows analogously. $\square$ Now to finish, let $F_1'$ and $F_2'$ be the reflections of $P$ around $F_1$ and $F_2$, respectively. Then $RF_1' = SF_2' = PF_2 - PF_1$, which when coupled with $RQ = QS$ gives $F_1'Q = QF_2'$. Hence $PF_1'QF_2'$ is cyclic with the arc midpoint of $F_1'F_2'$ being $Q$; taking a homothety at $P$ with ratio $\frac{1}{2}$ finishes. $\blacksquare$

Since $F_1A+F_2A=F_1B+F_2B$, there exists a circle tangent to segments $AF_2$ and $BF_1$ and rays $F_1A$ and $F_2B$. It is well known that $\overline{AQ}$ bisects $\angle F_1AF_2$ and $\overline{BQ}$ bisects $\angle F_1BF_2$, so this circle is centered at $Q$. Suppose that this circle is tangent to lines $PA$, $PB$, $AF_2$, and $BF_1$ at $K$, $L$, $M$, and $N$, respectively. Let $R$ be the midpoint of arc $F_1F_2$ in the circumcircle of $PF_1F_2$ opposite $P$, which lies on $\overline{PQ}$. Let the homothety centered at $P$ mapping $R$ to $Q$ map $F_1$ to $F_1'$ and $F_2$ to $F_2'$. Notice that $\angle PKQ=\angle PLQ=90^\circ$, $QK=QL$, and $\angle QF_1'K=\angle QF_2'L$, so $\triangle QKF_1 \cong \triangle QLF_2$, which means $PF_1'+PF_2'=PK+PL$. We also have \begin{align*} &PK+PL=PA+AK+PF_2+F_2L=PA+AM+PF_2+F_2M \\ &=F_1P+F_2P+F_1A+F_2A=2(PF_1+PF_2), \end{align*}so the homothety has scale factor $2$. Thus, $R$ is the midpoint of $\overline{PQ}$, as desired. $\square$

solvedy with Kanav patrick bateman approves

Solution from Twitch Solves ISL: Let $\rho = PF_1 + PF_2 = AF_1 + AF_2 = BF_1 + BF_2$. We start with the following claim: Claim: The $P$-excircles of triangle $PAF_2$ and $PF_1B$ coincide. Proof. This follows from the fact that $\triangle PAF_2$ and $\triangle PB_1$ have the same perimeter, namely $2\rho$. $\blacksquare$ Claim: That common excircle has center $Q$. Proof. The tangency of $\overline{AQ}$ to the ellipse at the vertex of $\angle F_1AF_2$ implies that $\overline{AQ}$ is the external bisector of $\angle F_1AF_2$ (by the famous ``river problem''). Similarly, $\overline{BQ}$ is the external bisector of $\angle F_2BF_1$. $\blacksquare$ Let $A_1$ and $A_2$ denote the reflections of $P$ over $F_1$ and $F_2$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair F_1 = (-1.,0.); pair F_2 = (1.,0.); pair P = (-0.5,1.5); pair A = (-1.35410,-1.06230); pair B = (1.67082,-0.67082); pair Q = (0.5,-2.73606); pair A_1 = (-1.5,-1.5); pair A_2 = (2.5,-1.5); pair T_1 = (-1.67082,-2.01246); pair T_2 = (2.11803,-1.11803); import graph; size(10cm); pen ubqqys = rgb(0.29411,0.,0.50980); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen qqffff = rgb(0.,1.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(P--A_1--A_2--cycle, linewidth(0.6) + zzttqq); draw(Q--A, linewidth(0.6) + ubqqys); draw(Q--B, linewidth(0.6) + ubqqys); draw(P--A, linewidth(0.6)); draw(P--B, linewidth(0.6)); draw(P--Q, linewidth(0.6)); draw(F_1--F_2, linewidth(0.6)); draw(A--F_2, linewidth(0.6)); draw(B--F_1, linewidth(0.6)); draw(P--A_1, linewidth(0.6) + zzttqq); draw(A_1--A_2, linewidth(0.6) + zzttqq); draw(A_2--P, linewidth(0.6) + zzttqq); draw(circle((0.5,-0.5), 2.23606), linewidth(0.6) + linetype("2 2") + qqwuqq); draw(circle(Q, 2.28824), linewidth(0.6) + qqffff); dot("$F_1$", F_1, dir((-19.952, 12.485))); dot("$F_2$", F_2, dir((2.622, 6.059))); dot("$P$", P, dir((2.487, 5.610))); dot("$A$", A, dir((-10.830, 9.473))); dot("$B$", B, dir((2.137, 4.792))); dot("$Q$", Q, dir(-90)); dot("$A_1$", A_1, dir((-28.370, 4.172))); dot("$A_2$", A_2, dir((6.263, -9.265))); dot("$T_1$", T_1, dir((-20.635, 12.188))); dot("$T_2$", T_2, dir((2.398, 4.531))); [/asy][/asy] Claim: The points $P$, $A_1$, $Q$, $A_2$ are concyclic. Proof. Let $T_1$ and $T_2$ denote the tangency points of $(Q)$ to $\overline{PA}$ and $\overline{PB}$. Note that \[ PT_1 + PT_2 = 2PT_1 = 2\rho = 2(PF_1+PF_2) = PA_1 + PA_2 \implies T_1A_1 = T_2A_2 \]so it follows that $\triangle Q T_1 A_1 \cong \triangle Q T_2 A_2$ as right triangles, with the same orientation. In particular, \[ \measuredangle PA_1Q = \measuredangle T_1 A_1 Q = \measuredangle T_2 A_2 Q = \measuredangle T_1 A_1 Q = \measuredangle P_2 A_2 Q. \]$\blacksquare$ Finally, apply a homothety at $P$ with ratio $\frac{1}{2}$ to deduce the problem statement.

Rephrase the problem as follows: Let $ABC$ be a triangle and let $M$, $N$ be the midpoints of $AB$, $AC$ respectively. Let the ellipse with foci $M$, $N$ passing through $A$ intersect $AB$ and $AC$ again at points $X$, $Y$, and let the tangents at $X$ and $Y$ to the ellipse intersect at $R$. Show that $R$ lies on $(ABC)$. Let $Q$ be the arc midpoint of $BC$ on the angle bisector of $ABC$. Let $S$ be the intersection of $(CNQ)$ and $BC$ which isn't $C$, and let $T$ be the intersection of $(BSQ)$ with $AB$ not equal to $B$. $\textbf{Claim.}$ The perpendicular bisector of $NT$ is tangent to the ellipse on line $AB$. Firstly note that from some length equalities we need to prove that $MT = AM + AN$, or $BT = AN$. However, this is not hard: remark that $\providecommand{\dang}{\measuredangle} \dang BSQ = \dang CSQ,$ and $BQ = CQ$, so the circles $(CNQ)$ and $(BSQ)$ are congruent. Moreover, \[ \providecommand{\dang}{\measuredangle} \dang QST = \dang QBT = \dang QBA = \dang QCA = \dang QCN = \dang QSN, \]whence $T$, $S$, $N$ are collinear, and as such, \[ \providecommand{\dang}{\measuredangle} \dang BST = \dang CSN \implies BT = CN = AN. \]Moreover we get the tangency for free since $N$ and $T$ are reflections across the perpendicular bisector of $NT$, and tangents to an ellipse are the angle bisector of the angle formed by the point of tangency and the foci. The other case follows symmetrically. Thus $Q=R$, and we are done.

[asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(10)); real a = 1; real b = 0.8; draw(ellipse((0,0),a,b), fuchsia); pair F1 = (-sqrt(a^2-b^2),0); pair F2 = (sqrt(a^2-b^2),0); pair M = (0,0.4); pair P = intersectionpoints(circumcircle(F1, F2, M), ellipse((0,0),a,b))[0]; pair C = 2*F1-P; pair D = 2*F2-P; pair A = intersectionpoints(P--C,ellipse((0,0),a,b))[0]; pair B = intersectionpoints(P--D,ellipse((0,0),a,b))[0]; pair Q = 2*M-P; pair F1p = 2*foot(F1,Q,B)-F1; pair F2p = 2*foot(F2,Q,A)-F2; draw(F1--P--F2^^A--Q--B^^P--Q, red); draw(F2p--Q--F1--F2p^^Q--C^^F1p--Q--F2--F1p^^Q--D, orange); draw(circumcircle(P,F1,F2), mediummagenta+dashed); dot("$F_1$", F1, dir(M--F1)); dot("$F_2$", F2, dir(M--F2)); dot("$P$", P, dir(Q--P)); dot("$M$", M, dir(-60)); dot("$A$", A, dir(M--A)); dot("$B$", B, dir(M--B)); dot("$Q$", Q, dir(P--Q)); dot("$F_1'$", F1p, dir(F2p--F1p)); dot("$F_2'$", F2p, dir(F1p--F2p)); dot("$C$", C, dir(M--C)); dot("$D$", D, dir(Q--D)); [/asy][/asy] Let ${F_1'}$ and ${F_2'}$ be reflections of ${F_1}$ and ${F_2}$ over $\overline{AQ}$ and $\overline{BQ}$ respectively. Remark that $F_1' \in \overline{BP}$, $F_2' \in \overline{AP}$, and $F_1F_2' = PF_1+PF_2 = F_1'F_2$. Construct ${C}$ and ${D}$ on segments $\overline{F_1F_2'}$ and $\overline{F_1'F_2}$ respectively such that \[ F_1C = PF_1 = F_1'D \quad\text{and}\quad F_2'C = PF_2 = F_2D. \]Then from $QF_1 = QF_1'$ and $QF_2' = QF_2$ there is a rotation centered at ${Q}$ that takes $F_1CF_2'$ to $F_1'DF_2$. This implies that $\angle{F_2'CQ} = \angle{F_2DQ}$, whence $PCQD$ is cyclic. A homothety at ${P}$ with ratio $\tfrac{1}{2}$ gives the desired conclusion. $\blacksquare$

Let the tangent through $P$ cut the tangent through $A$ at $C$ (which are the ex-bisectors) hence $D$ is the $F_2$-excenter of $F_2PA$; let also the (bisector) of $\angle F_2PA$ cuts $AC$ at $Q_1$( which is $F_2$-excenter of $F_2PA$) and cuts again $(PF_1F_2)$ at $D$ applying the same lemma as here we get $ PD'=\frac{F_2P+F_1P}{2}$ where $D'$ is the projection of $D$ on $PA$ , but $PQ'_1=\frac{F_2P+F_1P+F_2A+AF_1}{2}$ where $Q'_1$ is the projection of $Q_1$ on $PA$ hence $D$ is the midpoint of $PQ_1$ idem if we define $Q_2 $ as the intersection of the tangent at $B$ with the bisector of $ \angle BPF_1$ then we get $D$ is the midpoint of $PQ_2\ $ so $\ Q_1=Q_2=Q$ and the result follows . Best regards. RH HAS

Wow, this is my first ever geometry problem involving ellipses. They are pretty weird stuff. Needed a small hint to get started. We start off by proving the following important claim. Claim : Point $Q$ is the $P-$excenter of triangles $\triangle PAF_2$ and $\triangle PBF_1$. Proof : To see why, let $R_1 , R_2$ and $S_1,S_2$ be the tangency points of the $P-$excircles of triangles $\triangle PAF_2$ and $\triangle PBF_1$, with $\overline{PA}$ and $\overline{PB}$ respectively. We let $S = PF_1+PF_2=AF_1+AF_2=BF_1+BF_2$ (from the definition of an ellipse). We start off by noting, \[\frac{PF_1+F_1B+BF_2+F_2P}{2} = S = \frac{PF_1+F_1A+AF_2+F_2P}{2}\]so triangles $\triangle PAF_2$ and $\triangle PF_1B$ have the same semiperimeter $S$. Now, \[AR_1 = S- PF_1-AF_1 = S-P_A=AR_2\]and similarly, $AS_1=AS_2$, so points $R_1 \equiv R_2$ and $S_1 \equiv S_2$, implying that triangles $\triangle PAF_2$ and $\triangle PBF_1$ have the same $P-$excircle. Now, we know that $\overline{AQ}$ is the $\angle PAF_2-$external angle bisector and $\overline{BQ}$ is the $\angle PBF_1-$external angle bisector. So, the excenter which lies on both these lines, must simply be $Q$ as claimed. Now, let $R$ and $S$ be the tangency points of the common $P-$excircle of triangles $\triangle PAF_2$ and $\triangle PBF_1$ and let $G_1$ and $G_2$ be the reflections of $A$ across $F_1$ and $F_2$ respectively. We now do another length calculation. Note that, \[RG_1= S - PA - AG_1 = S-2PF_1\]and \[SG_2 = BG_2 - S + PB = 2PF_2-S\]but now, since $S=PF_1+PF_2$, it follows that $RG_1=SG_2$. Further, $RQ=QS$, and $\measuredangle G_1RQ = 90^\circ = \measuredangle G_2SQ$, from which it follows that $\triangle G_1RQ \cong \triangle G_2SQ$. Thus, $\measuredangle QG_1R = \measuredangle QG_2S = \measuredangle QG_2P$ which implies that $PG_1QG_2$ is cyclic. Considering a homothety centered at $P$ with scale factor $\frac{1}{2}$, implies that the midpoint of segment $PQ$ indeed lies on the circumcircle of $\triangle PF_1F_2$ as desired.

Let $PF_1+PF_2=k$, which is constant for all points on the ellipse. Let $X_1$, $X_2$ be the tangency point with line $PF_1$ of the $P$-excircles of $\Delta PF_1B$ and $\Delta PF_2A$ respectively. It is well-known that $PX_1=\frac{PF_1+PB+BF_1}{2}=k=\frac{PF_2+PA+AF_2}{2}=PX_2$, so it follows that $\Delta PF_1B$ and $\Delta PF_2A$ have the same $P$-excentre. It is well-known that $QA$, $QB$ are the external angle bisectors of $\angle PAF_2$ and $\angle PBF_1$ respectively, so $Q$ is the $P$-excentre of $\Delta PF_1B$ and $\Delta PF_2A$. Let $M$ be the midpoint of arc $F_1F_2$ of $(PF_1F_2)$ not containing $P$. We show that $M$ is the midpoint of $PQ$. Firstly, $PQ$ is the internal angle bisector of $\angle F_1PF_2$, so $M$ lies on $PQ$. Let $Y$ and $Z$ be the feet of $M$ on $PF_1$ and $PF_2$ respectively. Since $MF_1=MF_2$, $MY=MZ$, and $\angle MYF_1 = 90^\circ = \angle MZF_2$, we know that $\Delta MF_1Y \cong \Delta MF_2Z$ by RHS. We have $F_1Y=F_2Z$, and also $PY=PZ$, so $PY=\frac{PF_1+PF_2}{2}=\frac{k}{2}$. By taking a homothety with scale factor 2 centred at $P$, $Y$ goes to $X_1$, so $M$ goes to $Q$. It follows that $M$ is the midpoint of $PQ$ so we are done.