A quadrilateral $ABCD$ is circumscribed around a circle $\omega$ centered at $I$. Lines $AC$ and $BD$ meet at point $P$, lines $AB$ and $CD$ meet at point $£$, lines $AD$ and $BC$ meet at point $F$. Point $K$ on the circumcircle of triangle $E1F$ is such that $\angle IKP = 90^o$. The ray $PK$ meets $\omega$ at point $Q$. Prove that the circumcircle of triangle $EQF$ touches $\omega$.
Problem
Source: 2021 Sharygin Geometry Olympiad Finals grade IX p8
Tags: geometry, tangent circles, tangential quadrilateral
SerdarBozdag
02.08.2021 20:32
Nice one!
Mr.Trick
03.08.2021 17:35
An easy problem! First, note that the only use of $ABCD$ is to show that the polar of $P$ is line $EF$ wrt $\omega$. Let $O$be the circumcenter of triangle $EIF$. The ray $IO$ meets $\odot O$ at point $J$. Obviously $J,P,K$ are colinear. Let $M,N$ be the intersections of $\omega$ and $\odot O$. The line $MN$ and $EF$ meets at point $G$. Wrt $\omega$, the polar of $J$ is line $MN$ and the polar of $P$ is line $EF$. So the polar of $G$ is line $JP$, that is $PK$, which shows that $GQ$ is tangent to $\omega$. Cause $GE\cdot GF=GM\cdot GN=GQ^2$, $GQ$ is also tangent to $\odot (EQF)$. So $\omega$ and $\odot (EQF)$ are tangent a $Q$.
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Aryan-23
01.12.2021 09:52
Cute. The key idea is to invert about $\omega$. Let $W,X,Y,Z$ denote the incircle touch points of $AB,BC,CD,DA$ respectively. Let $E',F'$ denote the midpoints of $XZ,YW$ respectively. Invert about $\omega$. Note that $E\leftrightarrow E'$, $F\leftrightarrow F'$ and $\odot (EIF) \leftrightarrow E'F'$ under this inversion. Note that since $\angle IKP = \angle IP^*K^*= \frac {\pi}2$, implying $K^* \in EF$. ($EF$ is polar of $P$ by brokards). Note that $PK \leftrightarrow \odot (IP^*K^*)$. Hence we must have $\angle IQK^* =\frac {\pi}2$. Finally note that $EF$ is the radical axis of $\odot (IPK)$ and $\omega$. Finish by radical axis on $\omega$, $\odot (E'F'Q)$ and $\odot (IPK)$.
Om245
08.07.2024 15:24
Very cool problem !! Let $K,M,N,L$ be touchpoints with $X = KM \cap LN$ , $Y= KL \cap MN$. Now invert about $\omega$. Let $E \leftrightarrow E'$ and $F \leftrightarrow F'$. Note that $E',F'$ are midpoint of $ML$ and $KN$. Note if $Z \leftrightarrow P$ then $Z \in XY$ as $K,M,N,L$ cyclic. From Brocard’s theorem $E,F$ lie on $XY$. As $(EIF) \leftrightarrow E'F'$. If $K' \leftrightarrow K$ then $\measuredangle IZK' = 90$. As $PK \leftrightarrow (IZK')$ hence $Q$ is point of $\omega$ such that $\measuredangle K'QI = 90$ Which means $K'Q$ is tangent to $\omega$. Obesrve from $$\measuredangle IF'P = \measuredangle IE'P = \measuredangle IKP= 90$$we get $P,E',F',K,I$ cyclic. Now $$K'Q^2 = K'K \cdot K'I = K'E' \cdot K'F'$$Give us $K'Q$ is tangent to $(E'F'Q)$ and $\omega$, hence both circles are tangent to each others. But as $(E'F'Q) \leftrightarrow (EFQ)$, we get $\omega$ tangent to $(EFQ)$ at $Q$.