## APMO 2016 #3

Discussion on Asian Pacific Mathematical Olympiad (APMO)
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### APMO 2016 #3

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 $\omega$.

nahin munkar
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### Re: APMO 2016 #3

$OK.$
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 $\omega$ $as follows$ $.........[proved]$
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss