Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD=BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at $F$, and line $PF$ meets $\Gamma$ again at $Q$. Line $AC$ meets the circumcircle of $\triangle PNQ$ again at $E$. Prove that $\angle DQE = 90^{\circ}$.
Problem
Source: Philippine Mathematical Olympiad 2020/4
Tags: PMO, geometry
19.01.2020 16:28
........hard
30.01.2020 07:24
Any solution?
01.02.2020 10:05
Can someone help
01.02.2020 11:52
Here is my solution: Draw line parallel to $BC$ passing through $A$ and let the intersection of this line and $DN$ be $S$.Then we get $\frac{AN}{NC}=\frac{AD}{DC}=\frac{BC}{DC}$ So $\frac{AN}{BC}=\frac{NC}{DC}=\frac{AN}{AS}$ There we get $BC=AS$ Now let the feet of perpendicular from $S$ to $BC$ be $S'$. Then since $BC=AS=DS'$, $S'DAS$ is a square. Therefore we get $BM \times MC=DM \times MS'$. But we also have $BM\times MC=AM \times MT$ so we get that $(A,S,S',T,D)$ is concyclic. So $DF \times FS=AF \times FT=BF \times FQ$ and $(P,D,Q,S)$ is concyclic. On the other hand, $AP=BC=AS$, $PC=AB=BC$, so $P$ is the reflection of $S$ wrt line $AC$. Now, let $PB \cap DQ=X$, the intersection of $AC$ and the line perpendicular to $DQ$ and passing $Q$ be $E'$ By angle chasing $\angle PQD= \angle PBD= \angle NPB$ implies that the circumcircle of $PQX$ is tangent to $PN$. So now we have $\angle NQE'= \angle QXP= \angle NPQ$ so the problem is proved
03.12.2020 19:10
I and my friend solution
We can prove in 4 ways.
@above. nice solution. shorter and better than mine.
12.06.2021 17:02
Let $D'$ be the reflection of $D$ over $\overline{AC}$, and let $K=\overline{DD'}\cap\overline{AC}$. Let $\omega$ be the circle through $D$ centered at $A$, and let $l$ be the radical axis between it and $(ABC)$. Claim 1: $P,Q,F,K$ all lie on $l$. Proof. It is obvious that $P$ lies on $\omega$ and hence on $l$ because $AP=BC=AD$. Also, $k$ lies on $l$ because \[\text{Pow}(K,(ABC))=KA\cdot KC=KD\cdot KD'=\text{Pow}(K,\omega).\]Therefore, it suffices to show that $F$ lies on $l$. Let $AD=p$, $MD=q$, and let $f$ be the function defined by $f(X)=\text{Pow}(X,\omega)-\text{Pow}(X,(ABC))$. It is known that $f$ is linear, and we wish to show that $f(F)=0$. This is equivalent to showing that $p\cdot f(M)+q\cdot f(A)=0$ which is true since \[p\cdot f(M)+q\cdot f(A)=p(q^2+q(p-q))+q(-p^2)=p^2q-p^2q=0. \quad\square\] Claim 2: $P,N,D'$ are collinear. Proof. Notice that $\triangle PAN\sim\triangle DCN$ since $PA:AN=DA:AN=DC:CN$ and $\angle PAN=\angle DCN$. Therefore, $\overline{AC}$ is the external angle bisector of $\angle PND$ and the result follows. $\square$ Now consider an inversion centered at $K$ with radius $-\sqrt{KA\cdot KC}$. This sends $\{P,N,D'\}$ to $\{Q,E,D\}$ so $K,Q,E,D$ are concyclic. This means that $\angle DQE=\angle DKE=90^\circ$ as desired. $\blacksquare$ [asy][asy] defaultpen(fontsize(10pt)); size(11cm); pen mydash = linetype(new real[] {5,5}); pair B = (-1,0); pair C = (1,0); real s = .3; pair D = s*B+(1-s)*C; pair M = midpoint(D--C); pair A1 = D+C-B; pair A = rotate(90,D)*A1; pair F = extension(D,incenter(D,A,C),A,M); pair N = extension(D,incenter(D,A,C),A,C); pair K = foot(D,A,C); pair D1 = 2*K-D; pair PQ[] = intersectionpoints(circle(A,abs(D-A)), circumcircle(A,B,C)); pair P = PQ[0]; pair Q = PQ[1]; pair E = 2*foot(circumcenter(P,Q,N),A,C)-N; draw(A--B--C--cycle, black+1); draw(A--M); draw(D--N); draw(D--D1); draw(P--Q, royalblue+1); draw(P--D1, mydash); draw(C--E); draw(A--D); draw(circumcircle(P,N,Q)); draw(circle(A,abs(D-A))); draw(circumcircle(A,B,C)); clip((-1.5,2.4)--(2,2.4)--(2,-1.3)--(-1.5,-1.3)--cycle); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$D$", D, dir(270)); dot("$E$", E, dir(0)); dot("$P$", P, dir(180)); dot("$Q$", Q, dir(-20)); dot("$N$", N, dir(30)); dot("$M$", M, dir(270)); dot("$D'$", D1, dir(315)); dot("$F$", F, dir(135)); dot("$K$", K, dir(135)); [/asy][/asy]