# 1994 AHSME Problems/Problem 18

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$. If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ $[asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy]$ $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$

## Solution

We solve for $\angle A$ as follows: $$4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.$$ That means that minor arc $\widehat{BC}$ has measure $40^\circ$. We can fit a maximum of $\frac{360}{40}=\boxed{\textbf{(C) }9}$ of these arcs in the circle.

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