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Going through the official solution of IMO 2016 Problem G2. Full pdf can be found here: https://www.imo-official.org/problems/IMO2016SL.pdf

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I have a couple of questions regarding the solution.

  1. Can someone prove or point me to a proof of the statement given at the end of the first paragraph starting with "it is well-known that angle BAT ..."? For something that is well-known the proof of it is surely hard to find.

  2. In the second paragraph quadrilateral $SFTE$ is obviously harmonic. Why? I know the definition, but can't prove it.

  3. At the end of the 2nd paragraph they project $T$ to infinity and say that $X$ is thus projected to $M$. Why? I played with the cross-ratio, but can't get anything close to the result.

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    $\begingroup$ For $1.$, it is enough to notice that the reflection $A’$ of $A$ through the perpendicular bisector of $BC$ is on the circumcircle of $ABC$. Indeed, we then have the angle equality $BAT=CA’T_1$ (reflexions preserve angles), then $CA’T_1=CAT_1=CAD_1$. $\endgroup$ – Mindlack Jul 1 at 18:34
  • $\begingroup$ This may just be all the rust on my geometry skills, but how is this solution even addressing the instruction to "prove that lines XD and AM meet on gamma"? I don't see it. $\endgroup$ – Zach Favakeh Jul 1 at 18:38
  • $\begingroup$ It isn’t the full proof, I think, look at the last line. $\endgroup$ – Mindlack Jul 1 at 18:39
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    $\begingroup$ For 2., I would suggest (although I am not sure) it has something to do with the fact that $AE$ and $AF$ are tangent to the circumcenter of the quadrilateral, while $A,S,T$ are collinear. $\endgroup$ – Mindlack Jul 1 at 18:58
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    $\begingroup$ For 3., they consider the projection of $XBTC$ from $T_1$ onto $BC$. Note that actually it is the quadruple $X,T,B,C$ which is harmonic. Since $TT_1$ and $BC$ are parallel, $T$ is mapped to infinity and $B$ and $C$ stay in place. Let $X’$ be the image of $X$: then $X’,\infty,B,C$ must be harmonic (on $BC$), thus $\overline{BX’}=-\overline{CX’}$, therefore $X’=M$. $\endgroup$ – Mindlack Jul 1 at 19:00
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1 and 3 are answered by Mindlack in the comments. His/her intuition how to approach 2 has led me to a proof, which I'm leaving here.

Since $AF$ touches $\omega_A$ at $F$, we have $\angle FTS=\angle SFA$, hence $\triangle AFT \sim \triangle AFS$ and $\frac{FS}{FT}=\frac{AF}{AT}$. Similarly, $\frac{SE}{TE}=\frac{AE}{AT}$. Since $AE=AF$, we get $FS\cdot TE=FT\cdot SE$. Which establishes that $SFTE$ is harmonic.

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