Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\omega$ with center $O$. The lines $AD$ and $BC$ meet at $E$, while the lines $AB$ and $CD$ meet at $F$. Let $P$ be a point on the segment $EF$ such that $OP \perp EF$. The circle $\Gamma_{1}$ passes through $A$ and $E$ and is tangent to $\omega$ at $A$, while $\Gamma_{2}$ passes through $C$ and $F$ and is tangent to $\omega$ at $C$. If $\Gamma_{1}$ and $\Gamma_{2}$ meet at $X$ and $Y$, prove that $PO$ is the bisector of $\angle XPY$. Proposed by Nikola Velov
Problem
Source: 4th Memorial Mathematical Competition "Aleksandar Blazhevski - Cane"- Senior D1 P3
Tags: geometry, cyclic quadrilateral, complete quadrilateral, tangent circles, angle bisector
v4913
22.12.2022 06:46
an actual gem :0 First of all, by Brocard OP \perp EF so it suffices to show <EPX = <FPY. Now let Q = AC \cap EF. We claim 2 things: QX = QY, and PQYX cyclic. It clearly suffices to show these 2 things. QX = QY: first let E’, F’ = (CFXY), (AEXY) \cap EF respectively, then <AFE = <DCA and <CEF = <CAD, so CAEE’, CAFF’ are cyclic, furthermore if T, U = AF’, AE’ \cap (ABCD) then by angle chasing P, B, T and P, U, D are collinear (since P is the Miquel point of ABCDEF and so PABE, PADF, PDCE, PBCF are cyclic), furthermore <BPE = <DPF so BU || DT || EF. Thus C, T, E’ and C, U, F’ are also collinear and P is also the Miquel point of CATUE’F’. Now we will actually get to proving the claim: Let the homothety at C taking TD to E’F take A to A’. ADT ~ AF’E, so there is a homothety at Q taking A’E’F -> AF’E, and also (CXY) -> (AXY). Let this homothety take X, Y to X’, Y’; then X’, Y’ lie on (AXY) and XYX’Y’ is an isosceles trapezoid with diagonals intersecting at Q, so QX = QY. PQYX cyclic: By Radical Axes on (ABCD), (AXY), (CXY), the tangents to (ABCD) at A and C intersect on XY; this intersection point also lies on EF by Pascal on AADCCB. Let this point be Z; then it suffices to show ZA^2 = ZP * ZQ, or ZPA ~ ZAQ, which is true by angle chasing and the fact that P is the Miquel point of ABCDEF. So XPQY is cyclic and PX = PY so <EPX = <FQY.
VicKmath7
22.12.2022 18:14
I think the above solution is unnecessarily complicated. Here is something much simpler. Obviously $P$ is the Miquel point of $ABCD$ and $AC \cap BD$ lies on $OP$. Invert at $P$ with radius $\sqrt{PA.PC}=\sqrt{PB.PD}$ and then reflect in the angle bisector of $\angle APC$, which also a bisector of $\angle BPD$. It is well-known that this inversion swaps $(A, C)$, $(B, D)$ and $(E, F)$ (see Brazil 2016/6 where the same inversion is applied). Also, $(ABCD)$ is fixed, so this inversion swaps $\Gamma_1, \Gamma_2$, which means that it swaps $(X, Y)$ as well, so $PO$ is bisector of $\angle XPY$, done.