1992 AMC 12 Problem 20

Question

Part of an “

n

-pointed regular star” is shown. It is a simple closed polygon in which all

2n

edges are congruent, angles

A_1,

A_2,

\ldots,

A_n

are congruent, and angles

B_1,

B_2,

\ldots,

B_n

are congruent. If the acute angle at

A_1

is

10^\circ

less than the acute angle at

B_1,

then

n=

12

18

24

36

60

Let the acute angles at the

A

- and

B

-vertices be

\alpha

and

\beta

At each inward

B

-vertex, the polygon’s interior angle is

360^\circ-\beta

Let the acute angles at the tips and notches be

\alpha

and

\beta,

respectively, so

\beta-\alpha=10^\circ.

The

2n

-gon has

n

interior angles

\alpha

and

n

reflex interior angles

360^\circ-\beta.

Therefore

\begin{gathered} n\alpha+n(360^\circ-\beta)\\ {}=(2n-2)180^\circ. \end{gathered}

Substituting

\beta-\alpha=10^\circ

gives

350n=360n-360,

so

n=36.

Thus the correct answer is D.

Problem 20 in other years