# 1995 AHSME Problems/Problem 17

## Problem

Given regular pentagon $ABCDE$, a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A$. The number of degrees in minor arc $AD$ is $[asy]size(100); defaultpen(linewidth(0.7)); draw(rotate(18)*polygon(5)); real x=0.6180339887; draw(Circle((-x,0), 1)); int i; for(i=0; i<5; i=i+1) { dot(origin+1*dir(36+72*i)); } label("B", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0))); label("A", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72))); label("E", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144))); label("D", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3))); label("C", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4))); [/asy]$ $\mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }$

## Solution 1

Define major arc DA as $DA$, and minor arc DA as $da$. Extending DC and AB to meet at F, we see that $\angle CFB=36=\frac{DA-da}{2}$. We now have two equations: $DA-da=72$, and $DA+da=360$. Solving, $DA=216$ and $da=144\Rightarrow \mathrm{(E)}$.

## Solution 2

Let $O$ be the center of the circle. Since the sum of the interior angles in any $n$-gon is $(n-2)180^\circ$, the sum of the angles in $ABCDO$ is $540^\circ$.

Since $\angle ABC=\angle BCD=108^\circ$ and $\angle OAB=\angle ODC= 90^{\circ}$, it follows that the measure of $\angle AOD$, and thus the measure of minor arc $AD$, equals $540^\circ - 108^\circ-108^\circ-90^\circ-90^\circ=\boxed{\mathrm{(E)}144^\circ}$.

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