Problem 3

Let

P=A_1A_2\cdots A_k

be a convex polygon in the plane. The vertices

A_1,A_2,\dots,A_k

have integer coordinates and lie on a circle. Let

S

be the area of

P

. An odd positive integer

n

is given such that the square of the length of each side of

P

is an integer divisible by

n

. Prove that

2S

is an integer divisible by

n

Step 3 of 5: Rewriting the relation with square roots of rationals

Previous step Next step

In plain words

Each term of the Ptolemy relation is, after squaring the numerator (a squared side length divisible by $p^e$ ) and denominator (a product of two radii), a square root of a positive rational number.

\sqrt{q_1}+\sqrt{q_2}+\dots+\sqrt{q_{k-1}}=\sqrt{q},\qquad q_i,q\in\mathbb{Q}_{>0}

Detailed analysis

Each term $\frac{A_iA_{i+1}}{OA_i\cdot OA_{i+1}}$ can be rewritten using $A_iA_{i+1}=\sqrt{A_iA_{i+1}^2}$ as $\sqrt{q_i}$ where $q_i=\dfrac{A_iA_{i+1}^2}{OA_i^2\cdot OA_{i+1}^2}$ is a positive rational number (since all coordinates, hence all squared distances, are rational — indeed integers). Likewise the right-hand side is $\sqrt q$ for a positive rational $q=\dfrac{A_1A_k^2}{OA_1^2\cdot OA_k^2}$ . So the identity becomes $\sqrt{q_1}+\sqrt{q_2}+\dots+\sqrt{q_{k-1}}=\sqrt q$ .