Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Greece

Let $AB\Gamma$ be a triangle with circumcircle $\omega(O, R)$ . A circle $\gamma$ passes through $O$ and $B$ and is tangent to the line $AB$ at $B$ . Let the circle $\kappa$ meets the circle $\omega(O, R)$ for a second time at $P \neq B$ . A circle passes through $P$ and $\Gamma$ and is tangent to the line $A\Gamma$ at $\Gamma$ and meets the circle $\gamma$ at $M \neq P$ . Prove that: $MP = M\Gamma$ .

Solution

First we will prove that $M$ belongs to the side $B\Gamma$ . It is enough to prove that
$B\hat{M}P + P\hat{M}\Gamma = \varphi + \theta = A\hat{\Gamma}P + P\hat{\Gamma}x = 180^\circ.$
Then we have:
$M\hat{P}\Gamma = O\hat{P}\Gamma - O\hat{P}M = O\hat{P}P - O\hat{P}M = M\hat{\Gamma}P \\ \Rightarrow \Gamma \text{ isosceles with } MP = M\Gamma.$

Figure 8

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