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

$\fbox{E}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1966 AHSME Problems/Problem 21

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