Is there PDF of IMEO 2020 somewhere?

Firstly restate the problem in terms of $\triangle{ABC}$. It somewhat gets restated as this Restated problem wrote: Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $A'$ be the reflection of $A$ in $\overline{BC}$ and let $X,Y$ be the intersections of the perpendicular bisectors of $\overline{AC},\overline{AB}$ with $\overline{A'B}, \overline{A'C}$ respectively. Let $\overline{OA'}$ intersect $\overline{BC}$ at a point $D$ then show that $\overline{AD} \perp \overline{XY}$ Proof: Firstly note that clearly $\overline{OX} \perp \overline{AC}$ and $\overline{OY} \perp \overline{AB}$ and so $(XA'YO)$ is cyclic. Now notice that we just need to show that $$\angle DAA'=90^\circ-\angle (\overline{AA'},\overline{XY})$$where $\angle (\overline{AA'},\overline{XY})$ denotes the acute angle between the lines. In other words we need to show that $$\angle OAA'=90^\circ-\angle (\overline{AA'},\overline{XY})$$But for this just notice that \begin{align*} & \angle(\overline{AA'}, \overline{XY}) \\ &=\angle XYA'+\angle AA'Y \\ &=\angle XYA'+90^\circ-\angle ACB \\ &=\angle XOA'+90^\circ-\angle ACB \\ \end{align*}So in other words we just need to show that $$\angle XOA'+\angle OAA'=\angle ACB$$But for thisjust note that $$\angle XOA'+\angle OAA'=\angle (\overline{XO},\overline{AA'})=\angle ACB$$and hence we are done $\qquad \blacksquare$

[asy][asy] size(8cm); import geometry; import olympiad; pair O = (0,0), A = dir(115), B = dir(195), C = dir(-15); pair A1 = reflect(B, C)*A, O1 = reflect(B, C)*O; pair P = intersectionpoint(bisector(A1, C), line(A, B)), Q = intersectionpoint(bisector(A1, B), line(A, C)); pair E = foot(P, A1, C), F = foot(Q, A1, B); string[] names = {"$O$", "$A$", "$B$", "$C$", "$A'$", "$O'$", "$P$", "$Q$", "$E$", "$F$"}; pair[] points = {O, A, B, C, A1, O1, P, Q, E, F}; pair[] ll = {dir(0), A, B, C, A1, dir(-110), P, dir(45), E, F}; for (int i=0; i<points.length; ++i) {dot(names[i], points[i], dir(ll[i]));} draw(B--C--A--B--A1--C, blue); draw(P--E^^Q--F, grey); markscalefactor = 0.0075; draw(rightanglemark(Q, F, A1)^^rightanglemark(P, E, A1)); draw(circumcircle(A, P, Q), dashed+red); draw(A--O1^^P--Q, magenta); [/asy][/asy] Let $O'$ be the circumcenter of $\triangle A'BC$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ and $Q$ to $A'C$ and $A'B$, respectively. Then $APO'Q$ is evidently cyclic. The rest is just angle chasing. We have \begin{align*} \measuredangle(A'O, PQ) &= \measuredangle(A'O, BC) + \measuredangle(BC, AO') + \measuredangle(AO', PQ) \\ &= 2\measuredangle(BC, AO') + \measuredangle AQP + \measuredangle O'AQ \end{align*}We now compute $\measuredangle AQP$ and $\measuredangle O'AQ$ separately. We have \begin{align*} \measuredangle AQP &= \measuredangle CQP \\ &= \measuredangle FQP + \measuredangle CQF \\ &= \measuredangle O'AB + \measuredangle QCA' + \measuredangle CA'F + \measuredangle A'FQ \\ &= \measuredangle(O'A, BC) + \measuredangle CBA + 2\measuredangle ACB + \measuredangle BAC + 90^{\circ} \\ &= \measuredangle(O'A, BC) + \measuredangle ACB + 90^\circ \end{align*}and \[ \measuredangle O'AQ = \measuredangle(O'A, BC) + \measuredangle BCA \]Therefore, \begin{align*} \measuredangle(A'O, PQ) &= 2\measuredangle(BC, AO') + \measuredangle(O'A, BC) + \measuredangle ACB + 90^{\circ} + \measuredangle(O'A, BC) + \measuredangle BCA \\ &= 90^{\circ} \end{align*}

The perpendicular bisector of $A'C$ meet $AB$ at $P$ and the perpendicular bisector of $A'B$ meet $AC$ at $Q$.$X,Y$ be the midpoint of $A'B$ and $A'C$.$O'=YP\cap XQ$ is the centre of $A'BC$.Since $\angle XO'Y=180^{\circ}=\angle A$ so $APO'Q$ is cyclic. Now $P',Q',O$ be the reflection of $P,Q,O'$ in $BC$.$O$ will be centre of $(ABC)$ and $P',Q'$ will lie on $A'B$ and $A'C$. By symmetry we have $A'P'OQ'$ will be cyclic. So $\angle P'A'O=\angle P'Q'O=\angle PQO'$. Last equality by symmetry. Let $R=A'O\cap PQ$. Hence, $\angle XAO=\angle P'A'O=\angle PQO'=\angle RQX$. So, $XA'QR$ is concyclic.Now as $\angle A'XQ=90^{\circ}$ so $\angle A'RQ=90^{\circ}$.$\blacksquare$

Quote: Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $P$ and $Q$ be points on $AB$ and $AC$, respectively, such that $PA'=PC$ and $QA'=QB$. Prove that the perpendicular from $A'$ to $PQ$ passes through the circumcenter of $\triangle ABC$. Let $A"$ be the orthogonal projection of $A' $ on $PQ$, and define $B''$ and $C''$ cyclically. The lines $AA''', BB'', CC''$ concur in X(31392). (See Antreas Hatzipolakis and Angel Montesdeoca, Hyacinthos 28850, and also El centro del triángulo X(31392) -- In Spanish--)

Solved with eisirrational and goodbear. First check $\measuredangle BPO' = \measuredangle(PO', BC) + \measuredangle CBA = \measuredangle CBA + \measuredangle A'CB + 90^\circ = \measuredangle CBA + \measuredangle BCA + 90^\circ = \measuredangle CQO'$, so $APO'Q$ is cyclic. Finally, \begin{align*} \measuredangle(PQ,BC) &= \measuredangle PQA + \measuredangle ACB + \measuredangle PO'A + \measuredangle ACB \\ &= \measuredangle (PO',BC) + \measuredangle (BC, O'A) + \measuredangle (ACB) \\ &= \measuredangle (BC, O'A) + 90^\circ \\ &= \measuredangle (OA', BC) + 90^\circ, \end{align*}so $\measuredangle (PQ, OA') = 90^\circ$ as desired. [asy][asy] import olympiad; size(6cm); defaultpen(fontsize(11pt)); pair A = dir(110); pair B = dir(195); pair C = dir (345); pair Ap = 2foot(A,B,C)-A; pair Q = extension((B+Ap)/2,(B+Ap)/2+rotate(90)*(B-(B+Ap)/2),A,C); pair P = extension((C+Ap)/2,(C+Ap)/2+rotate(90)*(C-(C+Ap)/2),A,B); pair O = (0,0); pair Op = B+C-O; draw(A--B--C--cycle); draw(B--Ap--C); draw(Q--(Ap+B)/2,dashed+gray(0.5)); draw(P--(Ap+C)/2,dashed+gray(0.5)); draw(circumcircle(A,P,Q)); dot("$A$", A, dir(90)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$A'$", Ap, dir(0)); dot("$O$", O, dir(0)); dot("$O'$", Op, dir(270)); dot("$Q$", Q, dir(0)); dot("$P$", P, dir(180)); [/asy][/asy]

We use complex numbers with $(A'BC)$ as the unit circle. For convenience's sake, swap the labels of $A$ and $A'$. As usual, let lowercase letters denote the complex coordinates of the point represented by their respective uppercase letter unless otherwise stated. Because $A'$ is the reflection of $A$ over $BC$, we get $a' = b + c - \frac{bc}{a}$. $P$ lies on a diameter $UV$ of $(ABC)$ with $uv = -ac$. Thus \begin{align*} p &= \frac{ (\overline{u}v - \overline{v} u)(a' - b)- (u - v)(\overline{a'} b - a \overline{b})}{(\overline{u} - \overline{v})(a'-b) - (u-v)(\overline{a'} - \overline{b})} \\ &= \frac{ (v^2 - u^2)(a' - b) - uv(u-v)(\overline{a'}b - a \overline{b})}{(v - u)(a'-b) - uv(u-v)(\overline{a'} - \overline{b})} \\ &= \frac{uv( \overline{a'} b - a \overline{b})}{ a' - b + uv(\overline{a'} - \overline{b})} \\ &= \frac{ -ac((\frac1b + \frac1c - \frac{a}{bc})b - (b + c - \frac{bc}{a} )\frac{1}{b})}{b + c - \frac{bc}{a} - b -ac( \frac1b + \frac1c - \frac{a}{bc} - \frac{1}{b})} \\ &= \frac{a(abc + ac^2 - bc^2 - abc - ab^2 - ab^2)}{abc - b^2c - a^2b + a^3} \\ &= \frac{a(a-b)(c^2 + ab)}{(a-b)(a^2 + bc)} = \frac{a(c^2+ab)}{a^2+bc}. \end{align*}By symmetry, we may obtain $q$ from swapping $b$ and $c$, so $q = \frac{a(c^2+ab)}{a^2+bc}$. Now we just want to check that \[ \frac{p-q}{a-o} = - \overline{\left( \frac{p-q}{a-o} \right)} . \]Call the expression on the left $z$. Because $O$ is the reflection of the circumcenter of $(ABC)$ over $BC$, we have that $o = b+c$. First we simplify $z$: \begin{align*} z &= \frac{ \frac{a(b^2+ac)}{a^2+bc} - \frac{a(c^2+ab)}{a^2+bc}}{a - b - c} \\ &= \frac{a(c-b)(a - b - c)}{(a-b-c)(a^2 + bc)} = \frac{a(c-b)}{a^2 + bc}. \end{align*}Then \[ \overline{z} = \frac{ \frac1a (\frac1c - \frac1b)}{\frac1{a^2} + \frac1{bc}} = \frac{a(b-c)}{a^2+bc} = -z \]as desired.

Solved with Rg230403 Let $O'$ be the circumcenter of $\triangle A'BC$ and let $O'D, O'E$ meet $BC$ at $X,Y$. Let $M$ be the midpoint of $A'B$. I'll use $a,b,c$ to denote $\angle BAC, \angle ABC$ and $\angle ACB$ Observe that since $\angle BXA' = \angle XA'C + \angle XCA' = 2 c = \angle BO'A'$ which means $BXO'A'$ is cyclic. Similarly, $CYO'A'$ is cyclic. Now, see that since $\angle DO'E = 180 - \angle XO'M = 180 - (\angle XO'B + \angle MO'B) = a+b- \angle XO'B = a+b - \angle XA'B = a+b-(\angle BA'C - \angle XA'C) = b+c = 180 - \angle DAE$ and so $AEO'D$ is also cyclic. Now, reflect line $DE$ across $BC$ and let it meet $AC$ at $Z$ and $BC$ at $K$. If $O$ is circumcenter of $\triangle ABC$, showing $A'O \perp DE$ is equivalent to showing that $AO' \perp ZK$ But $180 - \angle EZK + \angle O'AC = \angle ZEK + \angle ZKE + \angle O'AC = \angle DEA + 2 \angle CKE + \angle O'AC = \angle DEA + \angle O'AC + 2c - 2 \angle DEA = 2c - \angle DEA + \angle O'AC$ But $\angle O'AC - \angle DEA = \angle DAE - \angle AEO' = a - \angle AEO'$ and $\angle AEO' = c + \angle EYC = c + \angle O'A'C = c + 90 - b$. So, $2c - \angle DEA + \angle O'AC = 2c - (c-a-b+90) = 180 - 90 = 90^\circ$. So, $AO'$ is indeed perpendicular to $ZK$, as desired. $\blacksquare$

Attachments:

Let $O,M$ be the circumcenters of $\triangle ABC,\triangle A'BC$. Note that $APQM$ is cyclic, as $\measuredangle PAQ=\measuredangle BAC=\measuredangle CA'B=\measuredangle (PM,QM)=\measuredangle PMQ$. Now, \begin{align*} \measuredangle A'OM=\measuredangle OA'A=\measuredangle A'AM=\measuredangle BAM-\measuredangle BAA'=\measuredangle PQM-\measuredangle(BC,QM)=\measuredangle (PQ,BC). \end{align*}Because $OM\perp BC$, we get that $A'O\perp PQ$, we are done. [asy][asy] import geometry;size(10cm);defaultpen(fontsize(10pt)); pair A,B,C,O,P,Q,A1,M; A=dir(110);B=dir(195);C=dir(345);O=(0,0);M=2midpoint(B--C)-O;A1=2foot(A,B,C)-A; P=intersectionpoint(perpendicular(midpoint(A1--C),line(A1,C)),line(A,B)); Q=intersectionpoint(perpendicular(midpoint(A1--B),line(A1,B)),line(A,C)); draw(A--B--C--cycle,red+1);draw(B--A1--C,red+1);draw(A--A1,red+0.5);draw(midpoint(A1--B)--Q,dashed+grey+0.5);draw(midpoint(A1--C)--P,dashed+grey+0.5);draw(P--Q,red+1);draw(B--Q--A1,green+0.5);draw(A1--P--C,blue+0.5); draw(A--M,red+0.5);draw(A1--O,red+0.5);draw(O--M,red+0.5); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$O$", O, dir(90)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$A'$", A1, dir(A1)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); [/asy][/asy]