IMO 2016 Problem 3:

Let $P = A_1 A_2 \cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, \ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.

I tried using the formula for area of a polygon in cartesian plane, after assuming coordinates, but to no avail.


1 Answer 1


Incomplete proof

Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.

Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.

If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.

Please figure out the case with $P$ truncated by some non-reducible triangles.


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