Let $ABC$ be an acutangle triangle inscribed in a circle $\Gamma$ of center $O$. Let $D$ be the height of the vertex $A$. Let E and F be points over $\Gamma$ such that $AE = AD = AF$. Let $P$ and $Q$ be the intersection points of the $EF $ with sides $AB$ and $AC$ respectively. Let $X$ be the second intersection point of $\Gamma$ with the circle circumscribed to the triangle $AP Q$. Show that the lines $XD$ and $AO $ meet at a point above sobre
Problem
Source: Brazil National Olympiad 2019 - level 2 - #3
Tags: euclidean geometry, geometry, Brazilian Math Olympiad
Euler365
07.12.2019 09:09
Quite Easy Problem. Lets perform inversion with $(A,AD)$. Under this $\Gamma$ goes to $DE$. So $B,C$ go to $P,Q$ repectively. Angles get flipped under inversion. So $\angle APD=\angle ADB=90^{\circ}$. Similarly $\angle AQD=90^{\circ}$. Now consider quad. $APDQ$. $\angle APD=\angle AQD=90^{\circ}$. So it is cyclic. $X$ lies on $\odot(APDQ)$. So $\angle AXD=90^{\circ} \implies DX$ passes through antipode of $A$ w.r.t. $\Gamma$ as desired.
Steve12345
10.12.2019 22:44
Can be done without inversion. $\angle AEF= \angle AFE = \angle ABE$ so $AE$ is tangent to the circumcircle of $\triangle EBP$ or $AE^2=AP\cdot AB= AD^2$ hence $\triangle APD \sim ADB$ so $PD$ is perpendicular to $AB$. Similarly $\angle AQD=90^{\circ}$ so $APDQ$ is cyclic. Now, $\angle AXD = 90^{\circ}$ since $X$ belongs to the circle with diameter $AD$. So if $S$ is the meeting point of ray $XD$ and $\Gamma$ we have $AS$ is a diameter of $\Gamma$. So we are done.
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hydo2332
20.07.2020 16:55
Suffices to prove $\angle{ DPA} =90$. But since $P$ lies on the radical axis of $\Gamma$ and the circle that has center $A$ and radius $AF$, we have $AP.BP = AD^2 - AP^2$ $AD^2 = AP(AP + PB)$ $AD^2 = AP.AB $ and hence we are done since $D$ is the foot of the perpendicular.