Proof Contest Day 4

In the triangle $ABC$ the point $J$ is the center of the excircle opposite to $A$ . This excircle is tangent to the side $BC$ at $M$ , and to the lines $AB$ and $AC$ at $K$ and $L$ respectively. The lines $LM$ and $BJ$ meet at $F$ , and the lines $KM$ and $CJ$ meet at $G$ . Let $S$ be the point of intersection of the lines $AF$ and $BC$ , and let $T$ be the point of intersection of the lines $AG$ and $BC$ . Prove that $M$ is the midpoint of $ST$ .

Not original

#Geometry

Easy Math Editor

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`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

This is 2012 IMO Problem 1. No chance that I can solve it :P

A Former Brilliant Member - 5 years, 5 months ago

Upload the solution

Department 8 - 5 years, 5 months ago

I did not get what you mean. I have not solved the problem. Just copy and paste this problem in google and you will find the solutions when you click the Aops if you were asking me for that.

A Former Brilliant Member - 5 years, 5 months ago

Hint: There are cyclic shapes in the diagram. Prove that $J$ is the circumcenter of $AST$

Xuming Liang - 5 years, 5 months ago

@Xuming Liang, gave a nice hint but here is the official solution

Department 8 - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$