Whoever guesses the pun gets a prize

by shiningsunnyday, Sep 25, 2016, 6:08 PM

Vietnam TST 2001 wrote:
In the plane, two circles intersect at $A$ and $B,$ and a common tangent intersects the circles at $P$ and $Q.$ Let the tangents at $P$ and $Q$ to the circumcircle of triangle $APQ$ intersect at $S,$ and let $H$ be the reflection of $B$ across the line $PQ.$ Prove that the points $A, S, H$ are collinear.
Solution
Tidbit
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What the heck
SSD wrote:
Let the tangents at $P$ and $Q$ to the circumcircle of triangle $APQ$ intersect at $P$ and $Q.$ Let the tangents at $P$ and $Q$ to the circumcircle of triangle $APQ$ intersect at $P$ and $Q.$ Let the tangents at $P$ and $Q$ to the circumcircle of triangle $APQ$ intersect at $S,$
I cannot compute

Fixed
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 26, 2016, 5:42 AM

by zephyrcrush78, Sep 25, 2016, 6:15 PM

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The irony... You want to do more projective problems when this problem is in fact projective

note


also what a tough pun -_-

Darn using harmonic quad to prove the symmedian thing - clever! Oops prob haven't done proj for quite a while
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 26, 2016, 7:08 AM

by tastymath75025, Sep 25, 2016, 7:26 PM

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After much thought, I believe the pun is Click to reveal hidden text

Sorry that was terrible

Here's your prize:
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by bluephoenix, Sep 25, 2016, 10:57 PM

To share with readers my favorite problem I came across today :) (Shout for contrib.)

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