# 1952 AHSME Problems/Problem 21

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## Problem

The sides of a regular polygon of $n$ sides, $n>4$, are extended to form a star. The number of degrees at each point of the star is:

$\textbf{(A) \ }\frac{360}{n} \qquad \textbf{(B) \ }\frac{(n-4)180}{n} \qquad \textbf{(C) \ }\frac{(n-2)180}{n} \qquad$

$\textbf{(D) \ }180-\frac{90}{n} \qquad \textbf{(E) \ }\frac{180}{n}$

## Solution

$[asy] import olympiad; pair A,B,C,D,E; A=(-1.5,1.5sqrt(3)); B=(-4.5,1.5sqrt(3)); C=origin; D=(1.5,1.5sqrt(3)); E=3*dir(-120); draw(A--D,dashed); draw(D--E,dashed); draw(B--A--C--E); markscalefactor=0.075; draw(anglemark(C,A,D)); draw(anglemark(D,C,A)); [/asy]$

The measure of each angle, in degrees, of an $n$-sided regular polygon is $\frac{(n-2)180}{n}$. Hence, the two base angles of each triangle formed measure $180-\frac{(n-2)180}{n}$ degrees, and the vertex angle measures $180-2\left(180-\frac{(n-2)180}{n}\right)$ degrees. This simplifies to $\frac{180n-720}{n}$, or $\boxed{\textbf{(B)}\ \frac{(n-4)180}{n}}$.