Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$ and with $AC$ perpendicular to $BD$. Points $X$ and $Y$ lie on the circumcircle of the triangle $BOD$ such that $\angle AXO=\angle CYO=90^{\circ}$. Let $M$ be the midpoint of $AC$. Prove that $BD$ is tangent to the circumcircle of the triangle $MXY$.
Problem
Source: Baltic Way 2024, Problem 11
Tags: geometry, circumcircle, cyclic quadrilateral, geometry proposed
Lil_flip38
17.11.2024 15:03
We claim that the desired tangency point is the intersection of $AC,BD$. As $\measuredangle AMO = 90^\circ = \measuredangle AXO$ so $A,X,M,O$ is cyclic. Also, this circle is tangent to $(ABCD)$ as the centers are collinear with $A$. Thus, $OX,BD$ and the tangent at $A$ concurr at a point $L$ by radical center theorem. Claim: $P,X,Y,M$ cyclic, where $P$ is the intersection of the diagonals. First note $A,L,P,X$ cycli as $\measuredangle LXA = 90^\circ = \measuredangle LPA$. So, $\measuredangle LPX = \measuredangle LAX = \measuredangle AMX$. This implies $\measuredangle PXM=90^\circ$. Similarly, $\measuredangle PYM = 90^\circ$. Now, to finish note that as $PM\perp BD$, the conclusion follows.
Alex-Five
24.11.2024 23:01
For storage
Let $P=AC\cap BD$, $I = OX\cap BD$
First, as $\angle AMO = 90^\circ = \angle AXO$ we have $AXMO$ is cyclic.
We use the radical axis theorem on $(AXO), (ABCD), (BOD)$ to get that $XO$, the tangent at $A$ to $(ABCD)$ and $BD$ concur - ie $AI$ is tangent to $(ABCD)$
Next, as $\angle API= 90^\circ= \angle AXI$ then $IAPX$ is cyclic.
Finally, as $\angle IPX= \angle IAX = \angle PMX $ we see that $IP$ is tangent to $(PXM)$.
Similarly we get $IP$ is tangent to $(PYM)$.
Therefore $BD$ is tangent to $(XPYM)$.