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BdMO Online Forum • View topic - APMO 2016 #3

APMO 2016 #3

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APMO 2016 #3

Post Number:#1  Unread postby Zawadx » Fri Aug 05, 2016 10:17 am

Let $AB$ and $AC$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $AC$ at $E$ and ray $AB$ at $F$. Let $R$ be a point on segment $EF$. The line through $O$ parallel to $EF$ intersects line $AB$ at $P$. Let $N$ be the intersection of lines $PR$ and $AC$, and let $M$ be the intersection of line $AB$ and the line through $R$ parallel to $AC$. Prove that line $MN$ is tangent to $\omega$.
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Re: APMO 2016 #3

Post Number:#2  Unread postby nahin munkar » Mon Aug 08, 2016 5:49 pm

$ OK. $
First we draw a second tangent from P to $ \omega $ named $ PG $ . Here, $ PG||MR||AE $ (it can be showed easily).
We let, There is $ P_{\infty} $ on the line $ PG $ . Now, $ PMNP_{\infty} $ is a quadrangle & $ E , F $ are the tangent or contact point of $ MP $ & $ NP_{\infty} $ resp to $ \omega $ . So, by the converse of brianchon's quadrangle theorem , we can tell $ PMNP_{\infty} $ is inscribed about $ \odot (O) $ as $ PN,MP_{\infty},EF $ are concurrent . For this, $ MN $ is tangent to $ \omega $ $ as follows $ $ .........[proved] $ :D
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
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