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Difference between revisions of "1994 AHSME Problems/Problem 26"

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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 <math>216/2=108</math> degrees. Using the previous formula, <math>n=5</math> <math>\boxed{\textbf{(A) }5}</math>
 
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 <math>216/2=108</math> degrees. Using the previous formula, <math>n=5</math> <math>\boxed{\textbf{(A) }5}</math>
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- proto-solution by Drakodin -
 

Revision as of 19:04, 14 February 2017

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

Note: might be a horrible solution but:

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, $n=5$ $\boxed{\textbf{(A) }5}$

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