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
braverookie
19.01.2020 16:28
........hard
hyx
30.01.2020 07:24
Any solution?
hyx
01.02.2020 10:05
Can someone help
Contradiction
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
k12byda5h
03.12.2020 19:10
I and my friend solution
Since, $AD = BC = AP$. I claim that circle $\omega$ center at $A$ radius $AD$ passes through $P,Q,D$. (The proof is quite long. I will show later.) Let $C'$ lie on $BD$ and $C'A \perp AD$. We get $\square AC'BC$ is parallelogram and $C' \in \omega$. $\angle ACD = \angle BC'A = 45 ^{\circ} +\angle BC'D=45^{\circ} + 90^{\circ}-\angle APD$ and $\angle DNA = 45^{\circ} + \angle ACD$. So, $A,P,D,N$ are concyclic. $\angle PQD = 0.5 \angle PAD = 90^{\circ} - \angle ADP = \angle PNC - 90^{\circ}$. Hence, $\angle PQD + \angle DQE = \angle PQE = \angle PNC = 90^{\circ} + \angle PQD$. done
The claim: Let $\triangle ABC$ (acute triangle) and altitude $AH$ such that $AH = BC$. $D,E$ are points such that $\square ACBD , \square ABCE$ are parallelograms. $\omega$ center at $A$ radius $AH$ interserts $\odot(ABC)$ at $P,Q$ ($P \in BD , Q \in CE , \ D,E \in \omega$). $CD \cap EH = N , BE \cap DH = M$. (By triangle similarly, $AM,AN \cap BC$ at midpoints of $BH,CH$ respectively.) Then, $P,Q,M,N$ are collinear.
We can prove in 4 ways.
Coordinate bashing. Set $\omega$ as the unit circle.
Pascal $DDHHQP$ on $\omega$ we get $X = DD \cap HQ , B,M' = DH \cap PQ$ are collinear. Suffice to show that $X,B,E$ are collinear. (Then $M'=M$.) Let $X',Y'$ be points such that $\square ADX'H , \square AEY'C$ are squares. Equivalent to show that $XX':X'B = XD : DE$. $HQ \cap EY' = Y$.
\begin{align*}
\frac{XX'}{X'B} = \frac{XD}{DE} & \iff \frac{Y'Y}{HC} = \frac{XD}{2AE} \iff \frac{Y'Y}{XD} = \frac{HC}{2AE} \\ &\iff \frac{Y'Y}{EY} = \frac{HC}{2CY'}.
\end{align*}Which is true since $\frac{YY'}{EY} = \frac{HY' \sin \angle Y'HY}{HE \sin \angle EHY}, \frac{HC}{CY'} = \frac{HE \sin \angle HEC}{Y'E \sin \angle Y'EC}$ and $\angle Y'HY = \angle HEC$.
Let $I,J$ be foots of altitude from $H$ to $AB,AC$ respectively. Then, $P,Q,I,J,M,N$ are collinear on a line $\ell$. Let $M',N'$ are midpoints of $BH,CH$ respectively.
$I'$ is the reflection of $H$ across $AB$. Hence $I' \in \omega$. By radical axis of $\omega,\odot(ABC),\odot(AHBI')$ and $\omega,\odot(ABC),\odot(AHCJ')$. Then we get that $P,Q,I,J$ are collinear.
Apply Menelous on $\triangle ABC, \triangle ABN'$.
Let $R,S = HD \cap AB , HE \cap AC$. By Pappus theorem, $DC \cap EB = Z \in IJ$. And again, $Z' = RC \cap SB \in MN$. Let $MN \cap BC = H'$. $$(M,N;Z',H) \overset{B}{=} (E,N;S,H) \overset{C}{=} (E,D;A,\infty_{BC}) = -1$$Hence, $Z' \in AH$. Let $H''$ is a point such that $(B,C;H,H'')=-1$, then, $-1=(H',Z';M,N) \overset{A}{=} (H',H;M',N') \overset{homo \times 2\ \text{at} \ H}{=} (2H'-H,H;B,C)$. So, $H'$ is the midpoint of $HH''$. Since $\square AHIJ , \square BCIJ$ is cyclic quadrilateral ($AB \cdot AI = AC \cdot AJ = AH^2$). That is, $H''' = IJ \cap BC$, $H'''H^2 = H'''I \cdot H'''J = H'''B \cdot H'''C$. Which mean $H'=H''''$.
We'll show that $IJ$ passes through $Z'$. Suffice to show that $(A,B;R,I) = (A,S;C,J)$. $$(A,B;R,I)
= (HA,HB;HR,HI)\overset{\text{Rotate} 90^{\circ} \text{at} A}{=} (A\infty_{DE},AH;AX',AB) = (\infty_{DE},H,X',B) = \frac{BH}{HX'}=\frac{CY'}{HY'}\overset{\text{similarly}}{=}(A,S;C,J)$$and done .
@above. nice solution. shorter and better than mine.
KST2003
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]