# IMO 2016 Problem G2 — projective geometry

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

• 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 $CA’T_1=CAT_1=CAD_1$. – Mindlack Jul 1 at 18:34
• 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. – Zach Favakeh Jul 1 at 18:38
• It isn’t the full proof, I think, look at the last line. – Mindlack Jul 1 at 18:39
• 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. – Mindlack Jul 1 at 18:58
• 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 $X’=M$. – Mindlack Jul 1 at 19:00

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.