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 \(ST\).

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TopNewestThis is 2012 IMO Problem 1. No chance that I can solve it :P – Brilliant Member · 1 year, 8 months ago

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– Lakshya Sinha · 1 year, 8 months ago

Upload the solutionLog in to reply

– Brilliant Member · 1 year, 8 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.Log in to reply

@Xuming Liang, gave a nice hint but here is the official solution – Lakshya Sinha · 1 year, 8 months ago

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here – Brilliant Member · 1 year, 8 months ago

More solutions can be foundLog in to reply

Hint: There are cyclic shapes in the diagram. Prove that \(J\) is the circumcenter of \(AST\) – Xuming Liang · 1 year, 8 months ago

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