[asy][asy] size(8cm,0); defaultpen(fontsize(10pt)); pair A = dir(110); pair B = dir(220); pair C = dir(320); pair H = A + B + C; pair L = dir(270); pair H1 = 2*foot(H,A,L)-H; pair S = circumcenter(A,H,H1); pair D = 2*foot(S,A,B)-A; pair E = 2*foot(S,A,C)-A; pair P = 2*foot(B,A,L)-A-B-L; pair Q = 2*foot(C,A,L)-A-C-L; pair M = (P+Q)/2; pair O = (0,0); pair K = (B+C)/2; pair X = extension(P,E,Q,D); draw(circumcircle(A,B,C)); draw(A--B--C--cycle,linewidth(1.5)); draw(circumcircle(A,D,E)); draw(P--X--Q,dashed); draw(P--Q); draw(K--O); draw(A--M); draw(B--P); draw(C--Q); draw(D--E); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(200)); dot("$E$",E,dir(20)); dot("$H$",H,dir(270)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$M$",M,1.5*dir(70)); dot("$K$",K,NW); dot("$O$",O,NE); dot("$H_1$",H1,dir(330)); dot("$X$",X,dir(X)); [/asy][/asy] (a) $\angle AQP = \angle ABP = \angle ADE$ so $\triangle ADE\stackrel{+}{\sim}\triangle AQP$ so we are done by spiral lemma. (b) By (a) and Brokard's theorem, it suffices to show that $P$ lies on the polar of $Q$ w.r.t. $\odot(ADE)$, or equivalently, $\odot(PQ)$ and $\odot(ADE)$ are orthogonal. Let $M$ be the midpoint of $PQ$. Clearly $AM$ pass through circumcenter $O$ of $\triangle ABC$. Moreover, if we let $K$ be the midpoint of $BC$ and $AO$ intersect $\omega$ at $H_1$, we deduce that $$AH_1=AH=2OK=2OM \implies H_1\text{ is orthocenter of }\triangle APQ$$hence $\triangle MH_1Q\sim\triangle MQA$ or $MH_1\cdot MA = MP^2 = MQ^2$ so we are done.

Let $T =\omega\cap \Omega$, and let lines $BH$ and $CH$ hit $\Omega$ again at $M$ and $N$. [asy][asy] defaultpen(fontsize(10pt)); size(250); pair A, B, C, H, D, E, P, Q, M, N, S, T, L; A = dir(115); B = dir(210); C = dir(330); L = dir(90); H = orthocenter(A, B, C); M = 2*foot(H, C, A) - H; N = 2*foot(H, A, B) - H; P = IP(Line(B, B+L-A, 20), circumcircle(A, B, C), 1); Q = IP(Line(C, C+A-L, 20), circumcircle(A, B, C), 1); D = extension(N, P, A, B); E = extension(M, Q, A, C); S = extension(M, E, N, D); T = extension(P, E, Q, D); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(circumcircle(A, D, E), heavygreen); draw(B--M^^C--N, lightblue); draw(M--Q^^N--P, purple+dotted); draw(Q--T--P, heavycyan+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(0)); dot("$E$", E, dir(0)); dot("$H$", H, dir(270)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$S$", S, dir(290)); dot("$T$", T, dir(T)); [/asy][/asy] To finish part (a), it suffices to show the following claim. Claim: $\overline{TEP}$ and $\overline{TDQ}$ collinear. Proof. Redefine $P = \overline{TE}\cap \Omega$. Then $\angle TPB = \angle TAB = \angle TED$, so $\overline{BP}\parallel \overline{DE}$. Similarly, $\overline{TDQ}$ collinear. $\blacksquare$ Now the key observation for part (b) is the following. Claim: $\overline{PDN}$ and $\overline{QEM}$ collinear. Proof. Redefine $P = \overline{ND}\cap \Omega$. Then, by orthocenter reflections, we have \begin{align*} \angle ABP &= \angle ADP - \angle BPN = \angle NDB - \angle BCN \\ &= \angle BDH - (90^\circ - \angle B) = \angle AEH - (90^\circ - \angle B) \\ &= \angle AED = \angle ADE. \end{align*}Thus, we must have $\overline{BP}\parallel \overline{DE}$, which proves the claim. By the same argument, $\overline{QEM}$ are also collinear. $\blacksquare$ Now if $S = \overline{PN}\cap \overline{QM}$, we have $$\angle DSE = \tfrac{1}{2}(\text{arc }MN+\text{arc } PQ) = 2(90^\circ - \angle A)+\angle A = 180^\circ - \angle A,$$and hence $S$ lies on $\omega$, as desired.

Solution for part (b) using Moving Points, based on a solution of Minaev Gleb. We work in the real projective plane $\mathbb{R}P^2$, and animate $C$ linearly on a fixed line through $A$. First, define the \vocab{degree} of a moving point $(P(t):Q(t):R(t))$ to be the max degree of $P,Q,R$. Similarly we define the degree of a moving line $P(t)x+Q(t)y+R(t)z=0$ in the same way. Lemma: Suppose points $X,Y$ have degree $d_1,d_2$, and there are $k$ values of $t$ for which $P_1=P_2$. Then line $XY$ has degree at most $d_1+d_2-k$. Similarly, if lines $\ell_1,\ell_2$ have degrees $d_1,d_2$, and there are $k$ values of $t$ for which $\ell_1=\ell_2$, then the intersection $\ell_1 \cap \ell_2$ has degree at most $d_1+d_2-k$. Proof: We show the first statement; the second follows from point-line duality. Note that the line through the points $X=(P_1(t):Q_1(t):R_1(t))$ and $Y=(P_2(t):Q_2(t):R_2(t))$ is given by cross product $X \times Y$; that is, the line \[(Q_1 R_2-Q_2 R_1)x + (R_1 P_2-R_2 P_1) y +(P_1 Q_2-P_2 Q_1)z=0.\]This is true as the equation must determine a line, and $X$ and $Y$ both lie on it, so it is line $XY$. Then for every value $t_0$ for which $X=Y$, $(t-t_0)$ factors out of each term. So the degree of the line is at most $d_1+d_2-k$. Now, returning to the problem: We're fixing $A,B$ and moving $C$ linearly. Then note that $H$ by projection moves linearly on line $BH$. Furthermore, angles $\angle AHE, \angle AHF$ are fixed, we get that $D$ and $E$ have degree $2$. One way to see this is using the lemma; $D$ lies on line $AB$, which is fixed, and line $HD$ passes through a point at infinity which is a constant rotation of the point at infinity on line $AH$, and therefore has degree $1$. Then $D,E$ have degree at most $1+1-0=2$. Now, note that $P,Q$ move linearly in $C$. Both of these are because the circumcenter $O$ moves linearly in $C$, and $P,Q$ are reflections of $B,C$ in a line through $O$ with fixed direction, which also moves linearly. So by the lemma, the lines $PD, QE$ have degree at most $3$. I claim they actually have degree $2$; to show this it suffices to give an example of a choice of $C$ for which $P=D$ and one for which $Q=E$. But an easy angle chase shows that in the unique case when $P=B$, we get $D=B$ as well and thus $P=D$. Similarly when $Q=C$, $E=C$. It follows from the lemma that lines $PD, QE$ have degree at most $2$. Let $\ell_\infty$ denote the line at infinity. I claim that the points $P_1=PD \cap \ell_\infty,$ $P_2=QE \cap \ell_\infty$ are projective in $C$. Since $\ell_\infty$ is fixed, it suffices to show by the lemma that there exists some value of $C$ for which $QE=\ell_\infty$ and $PD = \ell_\infty$. But note that as $C \to \infty$, all four points $P,D,Q,E$ go to infinity. It follows that $P_1,P_2$ are projective in $C$. Then to finish, recall that we want to show that $\angle (PD, QE)$ is constant. It suffices then to show that there's a constant rotation sending $P_1$ to $P_2$. Since $P_1,P_2$ are projective, it suffices to verify this for $3$ values of $C$. We can take $C$ such that $\angle ABC=90$, $\angle ACB = 90$, or $AB=AC$, and all three cases are easy to check.

Reim's Theorem FTW USA TSTST 2019 P2 wrote: Let $ABC$ be an acute triangle with circumcircle $\Omega$ and orthocenter $H$. Points $D$ and $E$ lie on segments $AB$ and $AC$ respectively, such that $AD = AE$. The lines through $B$ and $C$ parallel to $\overline{DE}$ intersect $\Omega$ again at $P$ and $Q$, respectively. Denote by $\omega$ the circumcircle of $\triangle ADE$. Show that lines $PE$ and $QD$ meet on $\omega$. Prove that if $\omega$ passes through $H$, then lines $PD$ and $QE$ meet on $\omega$ as well. Solution: Part (a) is just Reim's Theorem. For part (b), Let $O$ be circumcenter WRT $\Delta ABC$. Let $AO$ $\cap$ $\odot (ADHE)$ $=$ $L$ $\implies$ $HL$ $||$ $DE$ $||$ $BC$.If $M,N$ are reflections of $H$ over $AC,AB$ $$\angle DHN=90^{\circ}-\angle BDH=90^{\circ}-\angle ALH=90^{\circ}-\left(\frac{180^{\circ}-\angle HAL}{2} \right) =\frac{1}{2} \angle HDL$$Hence, $N$ lies on $DL$ and similarly, $M$ lies on $EL$. Hence, $NHLM$ is cyclic. Now apply Reim's Theorem again and conclude [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 = black; /* point style */ real xmin = -13.397755016958634, xmax = 17.88608307664881, ymin = -9.761581949056891, ymax = 8.82922347542287; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.8,5.57)--(-8.04,-5.49)--(7.86,-5.47)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-3.8,5.57)--(-8.04,-5.49), linewidth(2) + rvwvcq); draw((-8.04,-5.49)--(7.86,-5.47), linewidth(2) + rvwvcq); draw((7.86,-5.47)--(-3.8,5.57), linewidth(2) + rvwvcq); draw(circle((-0.09413405761646475,-2.1934241949101434), 8.6025692448846), linewidth(0.8)); draw(circle((-3.0536785593388696,2.283662608654139), 3.3700161932175376), linewidth(0.4)); draw((-5.805595844270751,0.3384221609352558)--(0.2684971821854436,1.7178380024590658), linewidth(0.8) + dtsfsf); draw((-5.784843170218619,4.257934375870217)--(-5.8528383222773375,-8.584161461079916), linewidth(0.8) + dtsfsf); draw((-5.784843170218619,4.257934375870217)--(8.496193791072775,-1.7346594784502916), linewidth(0.8) + dtsfsf); draw((7.86,-5.47)--(-8.20064211278282,0.6855903525622036), linewidth(0.4)); draw((-8.04,-5.49)--(2.7638032538325117,5.920538581493395), linewidth(0.4)); draw((-5.805595844270751,0.3384221609352558)--(-3.791731884767069,-1.0031516101797133), linewidth(0.4)); draw((-3.791731884767069,-1.0031516101797133)--(0.2684971821854436,1.7178380024590658), linewidth(0.4)); draw((-5.805595844270751,0.3384221609352558)--(-0.9673541489235893,-0.3628935681639076), linewidth(0.4)); draw((-5.8528383222773375,-8.584161461079916)--(7.86,-5.47), linewidth(0.8) + dtsfsf); draw((-8.04,-5.49)--(8.496193791072775,-1.7346594784502916), linewidth(0.8) + dtsfsf); draw((-3.8,5.57)--(-0.09413405761646475,-2.193424194910143), linewidth(0.8) + dtsfsf); draw((-3.791731884767069,-1.0031516101797133)--(-0.9673541489235893,-0.3628935681639076), linewidth(0.8) + dtsfsf); draw((-8.20064211278282,0.6855903525622036)--(-5.805595844270751,0.3384221609352558), linewidth(0.4) + linetype("4 4")); draw((2.7638032538325117,5.920538581493395)--(0.2684971821854436,1.7178380024590658), linewidth(0.4) + linetype("4 4")); draw((0.2684971821854436,1.7178380024590658)--(-5.8528383222773375,-8.584161461079916), linewidth(0.4) + linetype("4 4")); draw((-0.9673541489235893,-0.3628935681639076)--(8.496193791072775,-1.7346594784502916), linewidth(0.4) + linetype("4 4")); draw(circle((-3.7988456534698876,5.577963613649507), 6.581119068594783), linewidth(0.8) + linetype("4 4")); /* dots and labels */ dot((-3.8,5.57),dotstyle); label("$A$", (-3.935676118154415,6.111119509912726), NE * labelscalefactor); dot((-8.04,-5.49),dotstyle); label("$B$", (-9.115459146768107,-5.376716117903734), NE * labelscalefactor); dot((7.86,-5.47),dotstyle); label("$C$", (8.603501609529223,-5.633141020310352), NE * labelscalefactor); dot((-5.805595844270751,0.3384221609352558),dotstyle); label("$D$", (-6.65378008366457,0.72619655937376), NE * labelscalefactor); dot((0.2684971821854436,1.7178380024590658),dotstyle); label("$E$", (1.0133244982933194,1.726253678759568), NE * labelscalefactor); dot((8.496193791072775,-1.7346594784502916),dotstyle); label("$P$", (8.988138963139152,-1.63291254276712), NE * labelscalefactor); dot((-5.8528383222773375,-8.584161461079916),dotstyle); label("$Q$", (-7.243557359199793,-8.889737280874392), NE * labelscalefactor); dot((-5.784843170218619,4.257934375870217),dotstyle); label("$K$", (-6.37171269101729,4.880279978360962), NE * labelscalefactor); dot((-3.791731884767069,-1.0031516101797133),dotstyle); label("$H$", (-3.986961098635739,-2.017549896377046), NE * labelscalefactor); dot((-0.9673541489235893,-0.3628935681639076),dotstyle); label("$L$", (-1.6022095062541881,-0.04307814784609218), NE * labelscalefactor); dot((2.7638032538325117,5.920538581493395),dotstyle); label("$M$", (3.1929361687495756,6.03419203919074), NE * labelscalefactor); dot((-8.20064211278282,0.6855903525622036),dotstyle); label("$N$", (-9.3205990686934,0.21334675456052524), NE * labelscalefactor); dot((-0.09413405761646475,-2.193424194910143),dotstyle); label("$O$", (0.013267378907507787,-1.940622425655061), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Solution to part (a). $\measuredangle$ denotes a directed angle modulo $180^\circ$. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair O, A, B, C, H, L, P, Q, Y, D, EE, X, SS, T; O=(0,0); A=dir(110); B=dir(220); C=dir(320); H=A+B+C; L=dir(270); P=2*foot(O,B,foot(B,A,L))-B; Q=2*foot(O,C,foot(C,A,L))-C; Y=A+P+Q; D=extension(A,B,P,Y); EE=extension(A,C,Q,Y); X=extension(P,EE,Q,D); SS=foot(P,A,Q); T=foot(Q,A,P); draw(circle(O,1)); draw(A--B--C--A); draw(A--L,gray); draw(B--P); draw(C--Q); draw(circumcircle(A,D,EE),gray); draw(D--EE); draw(P--X--Q,gray); draw(P--D,gray); draw(Q--T,gray); draw(A--P--Q--A); draw(H--A--Y,dashed); dot("$A$",A,N); dot("$B$",B,B); dot("$C$",C,C); dot("$H$",H,S); dot("$L$",L,S); dot("$P$",P,P); dot("$Q$",Q,Q); dot("$Y$",Y,unit(Y-A)); dot("$D$",D,dir(210)); dot("$E$",EE,dir(18)); dot("$X$",X,NW); dot("$S$",SS,unit(D-P)*dir(45)); dot("$T$",T,NE); [/asy][/asy] Let $X=\overline{PE}\cap\overline{QD}$. It is easy to see that $\measuredangle APQ=\measuredangle ACQ=\measuredangle AED$ and $\measuredangle PQA=\measuredangle PBA=\measuredangle EDA$, whence $\triangle ADE\sim\triangle APQ$, and $A$ is the Miquel point of $DEPQ$. Hence, $X$ lies on both $\Omega$ and $\omega$, as desired. $\square$ First solution to part (b). Let $Y$ be the orthocenter of $\triangle APQ$ and set $D=\overline{AB}\cap\overline{PY}$ and $E=\overline{AC}\cap\overline{QY}$. Also let $S=\overline{AQ}\cap\overline{PY}$ and $T=\overline{AP}\cap\overline{QY}$. Claim 1. $AD=AE$ and $Y$ lies on $\omega$. Proof. Since $\widehat{BQ}=\widehat{CP}$, $\triangle ADS\cong\triangle AET$. It is immediate that $AD=AE$, and furthermore $$\measuredangle DYE=\measuredangle SYT=\measuredangle SAT=\measuredangle QAP=\measuredangle DAE,$$as desired. $\blacksquare$ Claim 2. $H$ lies on $\omega$. Proof. Let $O$ be the center of $\Omega$. Clearly $AP=AQ$, $OP=OQ$, and $YP=YQ$, so $A,O,P$ lie on the perpendicular bisector of $\overline{PQ}$. Now, if $R$ denotes the radius of $\Omega$, $AH=2R\cos\angle BAC=2R\cos\angle QAP=AY$, and furthermore $\overline{AH}$ and $\overline{AY}$ are isogonal, so $H$ lies on $\omega$, as desired. $\blacksquare$ Since $D$ and $E$ are unique, we are done. $\square$ Second solution to part (b), by Michael Ma. Let $L$ be the midpoint of arc $\widehat{BC}$ not containing $A$. Toss on the complex plane, with lowercase letters denoting complex numbers; set $a=u^2$, $b=v^2$, $c=w^2$, and $l=-vw$. We first prove a crucial claim: Claim. $\overline{DH}\parallel\overline{BL}$ and $\overline{EH}\parallel\overline{CL}$. Proof. This is just angle chasing. $$\angle ADH=180^\circ-\angle DHA-\angle HAD=90^\circ-\angle DEA+\angle B=\tfrac12\angle A+\angle B=\angle ABL,$$as desired. $\blacksquare$ Now, everything is computable. Since $\overline{AM}\perp\overline{BP}$, we have that $$am+bp=0\implies p=-\frac{am}b=\frac{u^2w}v,$$and similarly $q=\tfrac{u^2v}w$. To compute $d$, note that we have $D\in\overline{AB}$, so $\overline d=\tfrac{u^2+v^2-d}{u^2v^2}$. Furthermore, $\overline{DH}\parallel\overline{BL}$, so $$\frac{d-h}{b-m}\in\mathbb R\implies\frac{d-u^2-v^2-w^2}{v^2+vw}\in\mathbb R.$$Now, we have that \begin{align*} \frac{d-u^2-v^2-w^2}{v^2+vw}&=\frac{\frac{u^2+v^2-d}{u^2v^2}-\frac1{u^2}-\frac1{v^2}-\frac1{w^2}}{\frac1{v^2}+\frac1{vw}}\\ \implies d-u^2-v^2-w^2&=v^3w\left(-\frac d{u^2v^2}-\frac1{w^2}\right)\\ \implies d&=\frac{u^2\left(u^2w+v^2w+w^3-v^3\right)}{w\left(u^2+vw\right)}. \end{align*}Similaly, $$e=\frac{u^2\left(u^2v+vw^2+v^3-w^3\right)}{v\left(u^2+vw\right)}.$$Notice that $$d-p=\frac{(w-v)\left(v^3-u^2w\right)u^2}{vw\left(u^2+vw\right)}\implies-\frac{d-p}{e-q}=\frac{v^2-u^2w}{w^2-u^2v}.$$Finally, we want $$x=-\frac{b-a}{c-a}\bigg/\frac{d-p}{e-q}=\frac{\left(v^2-u^2\right)\left(w^3-u^2v\right)}{\left(w^2-u^2\right)\left(v^3-u^2w\right)}$$to be real, but $$\overline x=\left(\frac{u^2-v^2}{u^2v^2}\cdot\frac{u^2v-w^3}{u^2vw^3}\right)\bigg/\left(\frac{u^2-w^2}{u^2w^2}\cdot\frac{u^2w-v^3}{u^2v^3w}\right)=x,$$and we are done. $\square$

Same diagram as everyone above; other configurations are similar or can be cleaned up with directed angles. (a) Note that \[\angle AQP = \angle ABP = \angle ADE,\]and similarly $\angle APQ = \angle AED$. It follows that the isosceles triangles $\triangle ADE \sim \triangle AQP$ have spiral center $A$. Thus by the spiral similarity lemma $T = QD \cap PE$ is the second intersection of $\odot(ABC)$ and $\omega$. (b)Let $H_B, H_C$ be the reflections of $H$ over $AC, AB$, respectively. We claim $Q, E, H_B$ are collinear. Let $B_1$ be the foot of the altitude on $AC$. Then \begin{align*} \angle EH_BH &= \angle B_1HE \\ &= 90^{\circ} - \angle AEH \\ &= 90^{\circ} - (\angle DAH + \angle AED) \\ &=\angle B - \angle ABP \\ &=\angle PBC \\ &=BH_BQ, \end{align*}so the claim is proved. Now define $X$ as the second intersection of $QE$ with $\omega$. Then \[\angle DEX = \angle EQC = \angle H_BBC = \angle HAE = \angle HDE,\]so it follows that $HX \parallel ED$. If we define $X'$ as the second intersection of $DP$ with $\omega$, then analogously we obtain $HX' \parallel ED$ and we conclude that $X=X'$, as desired.

(a) Basically the same as everybody else. (b) It is well known that the thing we want to prove is equivalent to $QD \cdot QX +PE \cdot PX=PQ^2$. By power of a point, If $AQ$ and $AP$ meet $\omega$ at $Q_1$, $P_1$, $QD \cdot QX +PE \cdot PX = QQ_1 \cdot QA + PP_1 \cdot PA$. We know the values of the radius of $\omega$ (since $H$ lies on it), and we know $\angle ABQ, \angle CAP$ from part (a), and we know the center of the circle is on the angle bisector, so we can compute every length that we need to and the rest of the problem is a relatively quick trig bash.

Wait is TSTST given to MOPers??

char2539 wrote: Wait is TSTST given to MOPers?? Yeah..

Solution. Same idea. a. With easy angle-chasing, we obtain $\bigtriangleup DAE\sim \bigtriangleup QAP$, so $A$ takes $\overline{DE}$ to $\overline{QP}$ and thus $\overline{QD}\cap \overline{PE}$ is the second intersection point of $\omega$ and $(ABC)$. $\blacksquare$ b. Without loss of generality, assume that $AC>AB$. Let $K$ be a point on $\omega$ so that $AH=AK$, i.e. $DHKE$ is an isosceles trapezoid. Construct $L=\overline{KD}\cap\overline{CH};\ M=\overline{KE}\cap \overline{BH}$. Note that $$\angle LDB=\angle KDB=\angle AKH=\angle BDH$$so $$\angle LDH=2\angle BDH=2\angle AKH=2(180^\circ-\angle ABC)-\angle BAC$$but $$\angle DHL=\angle ABC+\dfrac{\angle BAC}{2}-90^\circ$$we conclude that $L$ is the reflection of $H$ across $BA$, so it lies on $\Omega$. Analogously, $M$ lies on $\Omega$. Therefore $$\angle DKH=\angle DAH=\angle HCB=\angle LCB=\angle LPB$$thus, since $HK\parallel DE\parallel PB$, we deduce that $D,\ K$ and $P$ are collinear. Similarly, we show that $E,\ K$ and $Q$ are collinear, so $PD$ and $QE$ meet at $K$, which lies on $\omega$. $\blacksquare$

I wonder why no one uses Pascal to solve part a. Let $PE$ intersects $\Omega$ at $T$ and let $TQ$ intersects $AB$ again at $D'$. Applying Pascal to $ABPTQC$, we get $D', E$ and $BP \cap PQ$ are collinear. From $BP \parallel CQ$, we get $D'E \parallel BP$ which gives $D=D'$. Therefore, $PE$ and $QD$ meet on $\omega$ as desired.

Gems98 wrote: I wonder why no one uses Pascal to solve part a. If you use Pascal theorem in such a kind of problem,you can kill fly with Atom bomb imo.

Here's my solution: Part (a) easily follows from Reim's Theorem. Move on to part (b). Let $BH,CH \cap \Omega=X,Y$. Then $$\angle DYA=\angle DHA=\angle DEA=\angle ADE=\angle ABP=\angle PYA \Rightarrow D \in PY$$Let $S$ be the reflection of $H$ in the internal angle bisector of $\angle BAC$. As $DESH$ is an isosceles trapezoid, so $S \in \omega$. This gives $$\angle DSH=\angle DAH=\angle DAY=\angle BPY=\angle DPB \Rightarrow S \in PD$$where we use the fact that $BP \parallel HS$. Similarly, $S \in QE$, giving that $S=PD \cap QE$ lies on $\omega$, as desired. $\blacksquare$

Other properties from configuration If $\omega$ passes through $H$. Let $F=QD \cap PE$; $M=PQ \cap BC$ c) Prove that: $AO, QE, PD$ are concurrent at $S$. d)Prove that:$(QDM),(PEM), \omega$ are concurrent at $S$. I found it by Geogebra. Can someone prove it? P\S: $S$ in c) and d) is one point.

(c): Define S as the intersection of AO with the small circle. We show some collinearities (S, Q, E; S, P, D). Let AO intersect the large circle again at T. Then $AT \perp PQ$. Now the collinearity is equivalent to showing that $\angle QST=\angle ASE$, or $2\angle ASE=2\angle QST$. Because ST is the perpendicular bisector of PQ we see that this is equivalent to $\angle PSQ=180-\angle A=180-\angle PAQ$, where the last equality follows from BC=PQ (isosceles trapezoid BPCQ). But this is just $\angle PSQ=\angle PTQ$ so if AT intersects PQ at X, we wish to show that 2TX=TS. But notice OX=OM (M=midpoint BC) and so 2OX=AH=AS. But 2OX=AS, 2OT=AT implies that ST=2TX, and we are done.

Here's the short angle chasing solution to (b) which I had (and is the author's solution too). Let $L$ be the reflection of $H$ across $\overline{AB}$, which lies on $\Omega$. Claim: Points $L$, $D$, $P$ are collinear. Proof. $\measuredangle CLD = \measuredangle DHL = \measuredangle DHA + \measuredangle AHL = \measuredangle DEA + \measuredangle AHC $ $= \measuredangle ADE + \measuredangle CBA = \measuredangle ABP + \measuredangle CBA = \measuredangle CBP = \measuredangle CLP. \quad \blacksquare$ [asy][asy] size(5cm); pair A = dir(125); pair B = dir(210); pair C = dir(-30); pair O = (0,0); pair H =orthocenter(A,B,C); pair I = incenter(A,B,C); pair J = 2*foot(H,A,I) - H; pair D = intersectionpoints(A--B,circumcircle(A,H,J))[1]; pair E = intersectionpoints(A--C,circumcircle(A,H,J))[1]; pair S = intersectionpoints(circumcircle(A,B,C),circumcircle(A,H,J))[1]; pair P = extension(S,E,D,J); pair Q = extension(S,D,E,J); draw(B--foot(B,A,C),dotted+gray); draw(A--foot(A,B,C),dotted+gray); draw(C--foot(C,A,B),dotted+gray); draw(circumcircle(A,B,C),purple); draw(circumcircle(A,H,J),red); draw(D--E,red); draw(B--P,purple); draw(C--Q,purple); pair HH = 1.9*circumcenter(A,D,E)-1.1*D; pair BB = -B; pair K = extension(D,P,E,Q); pair L = extension(P,D,C,H); draw(Q--E,dotted+red); draw(P--L,dashed+red); draw(H--L, dotted); draw(A--B--C--A); dot("$A$", A, dir(A)); dot("$B$",B,dir(225)); dot("$C$",C,dir(C)); dot("$H$",H,dir(225)); dot("$D$",D,2*dir(56)); dot("$E$",E,dir(30)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$L$",L,dir(L)); dot("$K$",K,dir(K-A)); label("$\omega$",HH,red); label("$\Omega$",-1.1*B,purple); [/asy][/asy] Now let $K \in \omega$ such that $DHKE$ is an isosceles trapezoid, i.e.\ $\measuredangle BAH = \measuredangle KAE$. Claim: Points $D$, $K$, $P$ are collinear. Proof. Using previous claim, $ \measuredangle KDE = \measuredangle KAE = \measuredangle BAH = \measuredangle LAB = \measuredangle LPB = \measuredangle DPB = \measuredangle PDE. \quad \blacksquare$ By symmetry, $\overline{QE}$ will then pass through the same $K$, as needed. Remark: These two claims imply each other, so guessing one of them allows one to realize the other. It is likely the latter is easiest to guess from the diagram, since it does not need any additional points.

Just wondering - was (a) worth 2 points and (b) 5 points, or 1 and 6 respectively?

Posting for storage. This seems like a new angle chasing solution for part b. I’ll just sketch the proof cuz I’m lazy. Let $H’$ be a point such that $DHH’E$ is an isosceles trapezoid. We claim that $DP$ passes through $H’$. Add point $AH\cap BP=L$ and angle chase to get that $D,H,L,B$ are cyclic. Then angle chase more to get $A,D,L,P$ are concyclic. We finish by, again, angle chase (or Reim’s on $\overline{DH’P}$ and $\overline{AHL}$) to get the desired collinearity.

Part (a) is just Reim's: if $T=(ADE) \cap (ABC) \neq A$, then $T,D,Q$ and $T,E,P$ are collinear. For part (b), let $M$ be the midpoint of minor arc $BC$ and let $H'$ be the reflection of $H$ over $\overline{AM}$, which lies on $\omega$. I claim that $H'$ is the desired concurrency point. We have the following two claims. Claim 1: $T,H,M$ are collinear. Proof: We have $$\measuredangle ATH=\measuredangle ADH=\measuredangle ADE+\measuredangle EDH=90^\circ+\measuredangle DAM+\measuredangle EAH=\measuredangle BCM+\measuredangle ACB=\measuredangle ATM.~\blacksquare$$ Claim 2: $\overline{H'E}$ and $\overline{BH}$ concur on $\Omega$. (Likewise, $\overline{H'E}$ and $\overline{BH}$ concur on $\Omega$). Proof; We have $$\measuredangle (\overline{BH},\overline{H'E})=90^\circ-\measuredangle H'EC=90^\circ+\measuredangle AED+\measuredangle DEH'=\measuredangle CAM+\measuredangle HAE,$$as well as $$\widehat{BQ}=\measuredangle BAM-\measuredangle QTM=\measuredangle BAM-\measuredangle DAH.$$Since $\measuredangle DAH+\measuredangle HAE=\measuredangle BAC=\measuredangle BAM+\measuredangle MAC=\measuredangle BAM-\measuredangle CAM$, we are done. $\blacksquare$ Now let $\overline{H'E}$ intersect $\Gamma$ at a point $Q'$ not lying on $\overline{BH}$ (i.e. not the concurrency point mentioned in claim 2). Then $$\measuredangle(\overline{H'E},\overline{Q'C})=\measuredangle HBC=\measuredangle CAH=\measuredangle H'AD=\measuredangle(\overline{H'E},\overline{DE}),$$hence $\overline{CQ'} \parallel \overline{DE} \implies Q'=Q$, so $\overline{QE}$ passes through $H'$. Likewise, $\overline{PD}$ passes through $H'$, as desired. $\blacksquare$

(a) Let $R$ be the second intersection of $(ADE)$ and $(ABC)$; I claim it is the desired intersection point. Indeed, notice that $$\measuredangle ARD = \measuredangle AED = \measuredangle ACQ = \measuredangle ARQ$$so $R$, $D$ and $Q$ are collinear. Similarly $R$, $E$, and $P$ are collinear which finishes. (b) Denote $H_B, H_C$ as the reflections of $H$ over $\overline{AB}, \overline{AC}$. Notice that $$\measuredangle AH_BD = \measuredangle DHA = \measuredangle DEA = \measuredangle ADE = \measuredangle ABP = \measuredangle AH_BP,$$so $H_B$, $D$, and $P$ are collinear. Similarly $H_C$, $E$, and $Q$ are collinear. Now, let $X_1 = \overline{DP} \cap (ADE)$ and $X_2 = \overline{EQ} \cap (ABC)$. Then $$\measuredangle AHX_1 = \measuredangle ADX_1 = \measuredangle ADH_B = \measuredangle HDA = \measuredangle HEA = \measuredangle AEH_C = \measuredangle AEX_2 = \measuredangle AHX_2,$$which is enough to imply that $X_1 = X_2$, as otherwise $H$, $X_1$, and $X_2$ would be collinear and all lie on $(ADE)$, contradiction. Thus, we are done.

angles are taken directed but not written in the correct order, which is still different than normal vertices beause it still acounts for config isseu!s its just that idgaf. also, i think the problem is wrritne wrong, and to this tiem stillhas not been dnoticed, beause it should be a) meet on Omega, notomega. Get it? a) Let F be the intersection point of (ADE) with (ABC); we claim that this is the desired point. Indeed, FED=FAB=FPB and FDE=FAC=FQC, so F lies on PE and QD. b) Instead, let F be AO with (ADE), J,K the reflections of H over AB,AC; note that it's well known BAH=CAO so BHEF is islscelestrapezoid due to equal arcs, whence A lies on perp. bisector of FH, so 90-HDJ/2=DHJ=90-BDH=90-AFH=HAF/2=HDF/2, meaning FDJ is a line, and analogously FEK is a line. It follows that HJK+HFK=CBK+HFE=90-C+180-(HAE)=180, so FHJK is cyclic. Reim's on FHJK with DEFH shows that JDF and KEF are lines, and Reim's on FHJK with CQJK in tandem with CHJ a line means KFQ a line, analogously since BHK is a line PFJ is a line by Reim's on FHJK with PBJK. So KEFQ and JDFP are lines and we're done! i understand that part b is kinda hard to understand without diagram so here https://pdf.ac/151fJd

Sketches for both the parts

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Claim: $T = \overline{PE} \cap \overline{DQ}$ is the Miquel point of $BDEC$. Proof. Phantom $T'$ as the Miquel point of $BDEC$. Then let $P' = \overline{T'E} \cap (ABC)$. It suffices to show $\overline{BP'} \parallel \overline{DE}$. Indeed note that, $\angle T'P'B = \angle T'CB = \angle T'ED$ and hence proven. $\square$ This implies (a) as from the defintion of the Miquel point we have $T \in \omega$. Now for (b). To deal with the orthocenter condition let $M$ and $N$ be the reflections of $H$ about $\overline{AB}$ and $\overline{AC}$. Claim $\overline{PDM}$ are collinear. Proof. Note that \[\angle AMD = \angle AHD = \angle AED = \angle ADE = \angle ABP = \angle AMP\]hence proven. $\square$ Similarly we find $\overline{QEN}$ collinear. Then we wish to show that $F = \overline{QN} \cap \overline{PM}$ lies on $\omega$. Claim: $\angle PFQ = 180 - \angle A$. Proof. Note that \[\angle PFQ = 180 - (\angle MPQ + \angle NQP) = 180 - (\angle MPB + \angle BPQ + \angle NQC - \angle PQC)\]However noting that as $PBQC$ is an isosceles trapezoid we find $\angle BPQ = \angle PQC$. Then we have, \begin{align*} \angle PFQ &= 180 - (\angle MPB + \angle NQC)\\ &= 180 - (\angle MCB + \angle NBC)\\ &= 180 - (90 - \angle B + 90 - \angle C)\\ &= \angle B+ \angle C\\ &= 180 - \angle A \end{align*}which is what we wished to show. $\square$ Now our conclusion easily follows as $180 - \angle A = \angle PFQ = \angle DFE$. $\blacksquare$

(a) Note $\triangle ADQ\cong \triangle AEP$, so by spiral similarity $PE$ and $QD$ actually concur at the other intersection of $(ADE)$ and $(ABC)$. (b) We claim the desired concurrency point is the reflection $H'$ of $H$ over the internal bisector of $\angle BAC$. Let $K$ be the reflection of $H$ over $AC$, and let $X$ and $Y$ be the feet from $H$ to $AC$ and the internal bisector of $\angle BAC$. We have $$\angle HKH' = \angle HXY = \angle HAY = \frac{1}{2}\angle B - \frac{1}{2}\angle C = \angle BKQ = \angle HKQ$$ so $K,H',Q$ are collinear. Then we have $$\angle HH'K = \angle HYX = \angle HAX = \angle HH'E$$ so $K,E,H'$ are also collinear and we're done.

Nice problem Let $L=(ADE)\cap (ABC)$, $R=CH\cap \Omega$, $S=BH\cap \Omega$, $F=PR\cap QS$. This means that $L$ is the Miquel Point of $DECB$. As a result, by spiral similarity: \[\angle LED=\angle LCB=\angle LPB,\]so $L,E,P$ are collinear. We can use a similar argument to show that $L,D,Q$ are collinear, proving part (a) $\square$ We now prove part (b), so henceforth, assume $H$ lies on $(ADE)$. Claim: $P,D,R$ and $Q,E,S$ are collinear Proof: We will first prove that $P,D,R$ are collinear. Let $D'=PR\cap AB$. It suffices to show $D=D'$. Note that $R$ is the reflection of $H$ about $AB$. We can then say: \begin{align*} \angle RD'H&=180-2\angle CRP\\ \angle BD'H &=90-\angle CRP\\ \angle AD'H&=90+\angle CRP\\ &=90+\angle CBP\\ &=90+\angle ABC-\angle ABP\\ &=90+\angle ABC-\angle ADE\\ &=\angle ABC+\frac{\angle BAC}{2}\\ &=90-\angle ACB+90-\frac{\angle BAC}{2}\\ &=\angle HAC+\angle ADE\\ &=\angle ADH.\\ \end{align*}We can use a similar method to show that $Q,E,S$ are collinear $\square$ It now suffices to show that $F$ lies on $\omega$. However, we can say: \[\angle HDF=\angle HDP=2\angle CRP,\]\[\angle HEF=\angle HEQ=2\angle BSQ=2\angle CRP,\]so $F$ lies on $\omega$, as desired $\blacksquare$

Here's another solution to part a using MMP. a) Fix $\triangle ABC$ and animate $D$ on $\overline{AB}.$ Denote $X_1=\overline{QD} \cap \omega \neq Q$ and $X_2=\overline{PE} \cap \omega \neq P.$ The map $D \xmapsto{Q} X_1$ is evidently projective, and the map $D \mapsto E \xmapsto{P} X_2$ is also projective, where the first map is induced by a rotation about $A.$ It suffices to check these maps coincide for three cases of $D.$ When $D=A,$ we have $A=D=E=X,$ at which the result is obvious. When $D=B,$ we have $B=D=X,$ with $E=\overline{PB} \cap \overline{AC},$ at which the result is obvious. When $D=\infty_{AB},$ we have $X=\overline{Q\infty_{AB}} \cap \overline{P\infty_{AC}},$ from which we may deduce $\triangle ABC \sim \triangle XPQ,$ implying the result.

oops wait i did not realize the last step was an angle chase. rip me in any case 1) $A$ is the Miquel Point of $DEPQ$, solving the first part. 2) Let $U$ and $V$ be the reflections of $H$ over $AB$ and $AC$. By simple angle chase, notice $U\in PD$ and $V\in QE$. Finish by angle chase.

(a) is easy from spiral sim and angle chase. Cool thing to note is that, letting $T$ be the second intersection of the two circles, we have that $THM,$ where $M$ is the arc midpoint of $\overarc{BC}$ because $H$ and the circumcenter $O$ of $(ABC)$ are isogonal. (b) is more interesting. I present a moving points solution that is a tad bit different than Zack's. We start off the same, and also use Zack's lemma. Move $C$ along a fixed line through $A.$ Therefore $H$ is also linear. Because the angle bisector of $\angle A$ is constant, $O_1$ the center of $(ADE)$ is $\deg 2$ because it's the intersection of the perpendicular bisector of $\overline{AH}$ with the angle bisector. Then $D,E$ also both have a maximum $\deg 2$ because they are both $A$ reflected about the altitude from $(ADE)$ to $\overline{AB}, \overline{AC},$ respectively. As these altitudes always go through the same point at infinity, both of them different of course, we have that the line is $\deg 2$ as well, with the intersection with $\overline{AB}, \overline{AC},$ also being $\deg 2.$ Note that the $O$ moves linearily on the perpendicular bisectr of $AB,$ so it has $\deg 1.$ As $P,Q$ are just reflections of $B,C$ respecitvely across the line that passes through $O$ and the fixed point of infinity that lies on the angle bisector of $\angle A,$ we have that $P,Q$ are $\deg 1$ as well. Now, the maximum degree of $DP$ or $QE$ is $2+1-1=2,$ where the $-1$ is when $B=P,$ then $B=D$ by an angle chase by showing that $BO \| AO_1.$ I claim that the intersection point is $L,$ the reflection of $H$ across the angle bisector of $\angle A.$ This has $\deg 1$ because the angle bisector is constant. We show that $L$ lies on one of the lines, as it must lie on the other by symmetry if it always lies on one, so the lines would intersect at $L.$ We will work on line $\overline{DP}.$ The cases are when $\angle ABC=90, \angle ACB=90, AB=BC,$ and when $C$ approaches $A.$ The last case is equivalent to "Let ADE be an isosceles triangle (AD=AE) and let H be the antipode of D. Pick B on AD such that HB is perpendicular to AE, and pick point P such that PB is parallel to DE and (APB) is tangent to AE. Prove DP perp DE". This can be done through trig. Therefore we have $2+1+1=4$ cases, so we're done$.\blacksquare$

(a) If $(ADE)$ intersects $(ABC)$ again at $F$ then: $$\measuredangle FDE = \measuredangle FAE = \measuredangle FAC = \measuredangle FQC$$(b) Let $M_A$ be the midpoint of $\widehat{BC}$. Set $A=a^2$, $B=b^2$, $C=c^2$ so we have, $M_A=-bc \implies p=\frac{a^2c}{b}$ and similarly $q=\frac{a^2b}c$. Claim I: $HD \parallel BM_A$ We use this to compute $d$, we have $\overline d=\frac{a^2+b^2-d}{a^2b^2}$ and because ${DH}\parallel{BL}$, so$$\frac{d-h}{\overline{d}-\overline{h}}=-m_ab^2 =cb^3$$after some simplification we get, $d=\frac{a^2(a^2b+c^2b+b^3-c^3)}{c^2b+a^2c}$ and similarly we also get $e=\frac{a^2(bc^2+a^2b+b^3-c^3)}{b^2c+a^2b}$ Now we just need to show that $\measuredangle(DP,EQ)=\measuredangle A$ or compute $\frac{d-p}{e-q}\div \overline{\frac{d-p}{e-q}}$ we use the complex number angle between two lines formula $$d-p=\frac{a^2(a^2b+c^2b+b^3-c^3)}{c^2b+a^2c}-\frac{a^2c}{b}=\frac{a^2(b-c)(-b^3+a^2c)}{(bc)^2+a^2bc}$$and $$e-q=\frac{a^2(a^2b+c^2b+b^3-c^3)}{c^2b+a^2c}-\frac{a^2c}{b}=\frac{a^2(b-c)(-b^3+a^2c)}{(bc)^2+a^2bc}$$and by symmetry we also have $e$ we get $\frac{d-p}{e-q}=\frac{b^3-a^2c}{a^2b-c^3}$ and $\frac{d-p}{e-q}\div \overline{\frac{d-p}{e-q}}=\frac{b^2}{c^2}$ and we are done $\hspace{4cm} \square$

For a. Let $A \neq (ADE) \cap (ABC) =X$. Observe $\measuredangle AXE = \measuredangle ADE$, $\measuredangle AXP = \measuredangle ABP = \measuredangle ADE$. Therefore $X \in \overline{EP}$. An analagous argument gives $X \in \overline{DQ}$. Therefore $\overline{DQ} \cap \overline{EP} = X \in \omega$. For b. Let the reflections of the orthocenter $H$ over $AB$, and $AC$ be $K$, and $L$ then: $$\measuredangle CKD = \measuredangle DHK = \measuredangle DHA + \measuredangle AHK = \measuredangle APB + \measuredangle CBA = \measuredangle CBP = \measuredangle CKP.$$Therefore $K$, $D$, and $P$ are collinear. An analagous argument gives $L$, $H$, and $B$ are collinear. Now observe $\measuredangle DYE = \measuredangle KYL = \measuredangle(KP, QL)= \measuredangle DAE$ where the last step is due to arc $KL$ equals arc $QP$ which is $180^\circ -a$. Therefore $Y$ is on $(ADE)$ and we conclude.

Nice Angle-Chase: a) Let $X = (ABC) \cap (ADE)$. By Reims, we have $X, E, P$ and $X, D, Q$ to be collinear. b) Let $L$ denote the reflection of $H$ wrt $AB$. Notice that: $\angle CLD = \angle DHL = \angle AHL - \angle AHD = \angle ABC - \angle AED = \angle ABC - \angle ADE = \angle ABC - \angle ABP = \angle PBC=\angle CLP$ which implies $L, D, P$ are collinear. Let $K$ be the point such that $DEKH$ is an isosceles trapezium. Then: $\angle KDE = \angle KAE = \angle DAH = \angle LAB = \angle LPB = \angle KPB$ which implies $K$ lies on $DP$. Similarly, $K$ lies on $QE$ and we are done.