APMO 1990 Problem 1

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sourav das
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APMO 1990 Problem 1

Unread post by sourav das » Sat Dec 24, 2011 1:27 pm

Given triangle $ABC$, let $D, E, F$ be the midpoints of $BC, AC, AB$ respectively and let $G$ be the centroid of the triangle. For each value of $\angle BAC$, how many non-similar triangles are there in which $AEGF$ is a cyclic quadrilateral?
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nafistiham
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Re: APMO 1990 Problem 1

Unread post by nafistiham » Wed Dec 28, 2011 6:58 pm

what are all the triangles ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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sourav das
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Re: APMO 1990 Problem 1

Unread post by sourav das » Wed Dec 28, 2011 7:35 pm

You have to find out (with proof) for any value of $\angle A$ how many non similar triangle have the property $AEGF$ is cyclic.
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*Mahi*
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Re: APMO 1990 Problem 1

Unread post by *Mahi* » Wed Dec 28, 2011 11:11 pm

Okay, as no reply is being posted, I'd post mine. This is quite ugly, and I'd love to see a beautiful solution.
Let $H$ be the orthocenter of $ABC$ . By simple angle chasing, it can be shown that $G$ will be on both of the circumcirles of $AEOF$ and $BHC$ if it abides by the given condition. As we know again, $G$ is the point which divides $OH$ in $1:2$ ratio. So the point where $OH$ intersects circle $BHC$ is $G $. Now , If $\angle A >60$, then $2\angle A > \angle B+\angle C$ and thus point $O$ is inside circle $BHC$ as $\angle BOC= 2\angle A$. But then the second intersection of $OH$ with $BHC$ would be outside $ABC$ , but $G$ must be inside $ABC$. Thus $\angle A \leq 60^0$ is a must.
We know that circumcircle of $BHC$ is the reflection of circumcircle of $ABC$ on $BC$. $\forall \angle A \leq 60^0$ , reflection of $O$ on $BC$ would be outside circle $ABC$ . Let it be $O'$. Then we can find two points $G',H'$ on the circle $ABC$ such that $O'G':G'H'=1:2$ in the segment $H'G'O'$. Those will be the only two points which follows the ration as their product $O'H' \cdot O'G'$ is fixed. Thus the reflection of $G',H'$ on $BC$ would be the centroid and orthocenter of $ABC$. As $A,G,H$ can be found for a circle with center $O$ with $B,C$ on it for $\text{arc }BC\leq 60^0$ so for all $\angle A \leq 60^0$, there will be one and only one triangle $ABC$ in which $AEGF$ is cyclic.
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