Mein gott what have you done to my diagram

by agbdmrbirdyface, Nov 6, 2016, 12:49 AM

I inverted it.
IMO 2015/3 wrote:
Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

I don't know if it's great to write solutions with motivation involved, but that's what I do:
Solution:

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this is basically my solution verbatim, although I think you explained the motivation much better. probably one of the hardest geo i've ever done lol

by aftermaths, Nov 6, 2016, 1:09 AM

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I see you have learned the ways of asy :)

by pinetree1, Nov 7, 2016, 4:59 AM

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darn that inversion that takes the two circles to themselves reminds me of Feuerbach's from AMSP

also birdy since when did you become such a god!?

by shiningsunnyday, Nov 7, 2016, 10:13 AM

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I agree I have learned the ways of asy w/o hard coding points lmao

and ssd I've just been doing problems about inversion and geo for a few weeks now so motivation comes a lot more naturally now

by agbdmrbirdyface, Nov 7, 2016, 4:05 PM

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

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