Li4 wrote: Let $O$ and $H$ be the circumcenter and the orthocenter, respectively, of an acute triangle $ABC$. Points $D$ and $E$ are chosen from sides $AB$ and $AC$, respectively, such that $A$, $D$, $O$, $E$ are concyclic. Let $P$ be a point on the circumcircle of triangle $ABC$. The line passing $P$ and parallel to $OD$ intersects $AB$ at point $X$, while the line passing $P$ and parallel to $OE$ intersects $AC$ at $Y$. Suppose that the perpendicular bisector of $\overline{HP}$ does not coincide with $XY$, but intersect $XY$ at $Q$, and that points $A$, $Q$ lies on the different sides of $DE$. Prove that $\angle EQF = \angle BAC$. Proposed by Shuang-Yen Lee What is $F$

That should be a typo---it should have been $EQD$ (or $DQE$ I am not too sure about the orientation XD)

GorgonMathDota wrote: What is $F$ Sorry, typo

Suppose the antipodes of $B, C$ are $B', C'$. We show that $DQ \parallel PC'$, for this would lead to the conclusion. Note that $APXY$ is cyclic. Since the Steiner line of $P$ wrt $ABC$ passes through $H$, it is also the Steiner line of $P$ wrt the triangle formed by the lines $AB, AC$ and the perpendicular bisector of $HP$, so $P$ is on the circumcircle of the triangle formed by these lines. Using this, $\angle PXQ = \angle PAC$ and thus $(PXQ)$ passes through the intersection of the perpendicular bisector of $HP$ and $AB$, say $X'$. Define $Y'$ similarly. Now as $D$ moves uniformly on $AB$, $X$ moves uniformly on $AB$ too, and since $\angle PQY' = \angle PXA$, $Q$ moves uniformly on $X'Y'$. When $Q = X'$, $Y = A$, so it suffices to show that $D = X'$, for then $DQ$ would always be parallel to a fixed line. Since the reflections of $H$ over the sides of $AX'Y'$ are on the circumcircle of $ABC$, $O$ is the isogonal conjugate of $H$ wrt $AX'Y'$, and thus $\angle OX'A = \angle Y'X'H = \angle PX'Y' = \angle PAY' = \angle OEC$, so $D = X'$ for this choice of $Q$. So $DQ$ is always parallel to a fixed line. When $D$ is the midpoint of $AB$, $PX \perp AB$, so $\angle X'QP = 90^\circ$, and thus $Q$ is the midpoint of $HP$. By a homothety at $H$ with ratio $\frac12$, $DQ$ the image of $C'P$, and thus we are done.

Here is a slightly different proof using rotational homothety: Let the midpoints of $AB,AC$ be $M_c, M_b$, respectively, and the projections of $P$ onto $AB, AC$ be $P_c, P_b$. Since $ADOE$ are concyclic, we have $\angle DOM_c = \angle EOM_b$. Therefore the rotational homothety $S_O$ w.r.t. $O$ sending $M_c$ to $D$ also sends $M_b$ to $E$. Now by Steiner's theorem, $P_bP_c$ passes through the midpoint $K$ of $PH$. Since $PX$ is parallel to $OD$ and $PY$ is parallel to $OE$, $APXY$ are also concyclic, and thus the rotational homothety $S_P$ w.r.t. $P$ sending $P_c$ to $X$ also sends $P_b$ to $Y$. Now the image of $K$ under $S_P$ lies on the perpendicular bisector of $PH$ since $\angle PKS_P(K) = \angle PP_cX = 90^{\circ}$. The image also lies on $XY$ since $K$ lies on $P_cP_b$, which shows that $S_P(K)=Q$. Now note that the rotational angles and the dilation ratios of $S_O$ and $S_P$ are the same. This is because that $\Delta OM_cD\sim \Delta PP_cX$. Let $S_H$ be the rotational homothety with the opposite rotational angle and the same dilation ratio w.r.t. $H$. Then since $K$ lies on the perpendicular bisector of $PH$, we have $S_H(K)=Q$. Moreover, since the nine-points circle $\Gamma$ is symmetric w.r.t. the perpendicular bisector of $OH$, we have that $S_O(\Gamma) = S_H(\Gamma)$, which shows that the image of $\Gamma$ under $S_H$ passes through $D,E$. Now since $K$ is also in $\Gamma$, we know that the image of $\Gamma$ is the circumcircle of $DEQ$. Therefore, \[\angle EQD = \frac{1}{2}\angle EN'D = \frac{1}{2}\angle M_bN'M_c = \angle M_bM_aM_c = \angle BAC\]where $N$ is the center of the nine-point circle and $N'$ is the center of $\Gamma$, and the second equality comes from $S_O$.

MMP again fix $D,E$ and move $P$ on the $(ABC)$ we claim that $deg(Q) \le 2$ $deg(P)=2$ since $P \rightarrow X$ is projective so $deg(X)=deg(Y)=2$ so $deg (XY) \le deg(X)+deg(Y)-k$ since if we put $P=A$ then $X=Y$ so $K \ge 1$ so $deg(XY) \le 3$ let $d$ be the perpindacular bisector of $PH$ easily get $deg(d)=2$ so $deg(Q) \le deg(XY)+deg(d)-k$ not that there's a $p$ between $B,C$ such that $d=XY$ and also another on between $A,C$ and also between $A,B$ so $k \ge 3$ so $deg(Q) \le 2$ $\blacksquare$ let $Q'$ be the reflection of $Q$ wrt $DE$ so $deg(Q') \le 2$ we claim that $Q'$ moves on $(ADEO)$ so it suffice to consider five cases of $P$ and they are trivial let $P_1,P_2$ on $ABC$ such that $DP_1=DP_2=DH$ and another two cases for $E$ and finally $P_5=AH \cap (ABC)$ (I'm too lazy to write them down )

dynamic diagram: https://youtu.be/kE-AtYVEFls

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