1952 AHSME Problems/Problem 21

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 $\boxed{\textbf{(B)}\ \frac{(n-4)180}{n}}$ .

1952 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1952 AHSME Problems/Problem 21

Problem

Solution

See also