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 3 of 5: Force equality in every paired difference

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In plain words

A sum of nonnegative terms can vanish only when every term vanishes.

a_{j+1}\ge a_{n-j+1},\quad\sin(j\theta)>0\Longrightarrow a_{j+1}=a_{n-j+1}\quad(1\le j\le\lfloor(n-1)/2\rfloor)

Detailed analysis

For the displayed range of $j$ , $0<j\theta<\pi$ , so $\sin(j\theta)>0$ . The ordering gives $a_{j+1}\ge a_{n-j+1}$ , hence every summand is nonnegative. Since their sum is $0$ , all paired lengths are equal; in particular $a_2=a_n$ .