USAMO 2008: Triangulation of a regular polygon

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Thanic Nur Samin
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USAMO 2008: Triangulation of a regular polygon

Unread post by Thanic Nur Samin » Fri Jan 13, 2017 12:21 pm

Let $ \mathcal{P}$ be a convex polygon with $ n$ sides, $ n\ge3$. Any set of $ n - 3$ diagonals of $ \mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $ \mathcal{P}$ into $ n - 2$ triangles. If $ \mathcal{P}$ is regular and there is a triangulation of $ \mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $ n$.
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ahmedittihad
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Re: USAMO 2008: Triangulation of a regular polygon

Unread post by ahmedittihad » Fri Jan 13, 2017 4:43 pm

We'll show that $n = 2^x * (2^y +1)$. Where $(x,y)$ is an integer more than $-1$.

Let us denote the triangle in the triariangulation which has the center of the circumcircle of the polygon * The Big Triangle*. By the condition, atleast 2 sides of the big triangle are the same. Let the vertices of the triangle be $A,B,C$ where $AB=AC$. It's trivial that the number of points on the minor arc $AB$ is equal to the number of points on the minor arc $AC$. We now triangulate one of the minor arc points. Let the other vertex of the other triangle that $AB$ is on in the triangulation be $X$. $X$ must be the midpoint of the minor arc $AB$ to satisfy the condition. By induction, we can say that each minor arc has $2^k - 1$ points on it except $A,B,C$. So, $n = 2^k-1+2^k-1+2^m-1+3$.

The result follows.
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