Problem 3

In an

n

-gon all interior angles are equal, and the lengths of consecutive sides satisfy

a_1\ge a_2\ge\cdots\ge a_n

. Prove that

a_1=a_2=\cdots=a_n

Step 2 of 5: Project the closed side-vector sum

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In plain words

The sine projection pairs directions on opposite sides of the polygon; the side-length ordering makes every paired difference nonnegative.

\sum_{i=1}^{n}a_i\sin((i-1)\theta)=0\Longrightarrow\sum_{j=1}^{\lfloor(n-1)/2\rfloor}(a_{j+1}-a_{n-j+1})\sin(j\theta)=0

Detailed analysis

Orient the first side along the $x$ -axis. Closure of the polygon gives $\sum_i a_i(\cos((i-1)\theta),\sin((i-1)\theta))=0$ . Taking the sine component and pairing the terms for directions $j\theta$ and $(n-j)\theta$ gives the displayed sum; the remaining direction (when $n$ is even) has sine $0$ .