# 1966 AHSME Problems/Problem 21

## Problem $[asy] draw((0,-5)--(-6,10),black+dashed+linewidth(1)); draw((-6,10)--(10,0),black+dashed+linewidth(1)); draw((10,0)--(-10,0),black+dashed+linewidth(1)); draw((-10,-0)--(10,10),black+dashed+linewidth(1)); draw((10,10)--(0,-5),black+dashed+linewidth(1)); draw((-2,0)--(-10/3,10/3),black+linewidth(2)); draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2)); draw((10/9,50/9)--(90/17,50/17),black+linewidth(2)); draw((90/17,50/17)--(10/3,0),black+linewidth(2)); draw((10/3,0)--(-2,0),black+linewidth(2)); MP("1", (1,0), N); MP("2", (-2.5,2), E); MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N); [/asy]$

An " $n$-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively $1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5$; for all $n$ values of $k$, sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2$; prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5$).

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals: $\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)$

## Solution $\fbox{E}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 