1994 AHSME Problems/Problem 26

Problem

A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$ .) If $m=10$ , what is the value of $n$ ? $[asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy]$ $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$

Solution

To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.

The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{\textbf{(A) }5}$

1994 AHSME (Problems • Answer Key • Resources)
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1994 AHSME Problems/Problem 26

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