2009 AMC 12A Problems/Problem 19

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Problem

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were $A$ and $B$ , respectively. Each polygon had a side length of $2$ . Which of the following is true?

$\textbf{(A)}\ A = \frac {25}{49}B\qquad \textbf{(B)}\ A = \frac {5}{7}B\qquad \textbf{(C)}\ A = B\qquad \textbf{(D)}\ A$ $= \frac {7}{5}B\qquad \textbf{(E)}\ A = \frac {49}{25}B$

Solution

In any regular polygon with side length $2$ , consider the isosceles triangle formed by the center of the polygon $S$ and two consecutive vertices $X$ and $Y$ . We are given that $XY=2$ . Obviously $SX=SY=r$ , where $r$ is the radius of the circumcircle. Let $T$ be the midpoint of $XY$ . Then $XT=TY=1$ , and $TS=\rho$ , where $\rho$ is the radius of the incircle.

Applying the Pythagorean theorem on the triangle $STX$ , we get that $\rho^2 + 1 = r^2$ .

Then the area between the circumcircle and the incircle can be computed as $\pi r^2 - \pi \rho^2 = \pi r^2 - \pi (r^2 - 1) = \pi$ .

Hence $A=\pi$ , $B=\pi$ , and therefore $\boxed{A=B}$ .

2009 AMC 12A (Problems • Resources)
Preceded by Problem 18	Followed by Problem 20
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

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2009 AMC 12A Problems/Problem 19

From AoPSWiki

Problem

Solution

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