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

$\mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }$

Solution

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 $da=144\Rightarrow \mathrm{(E)}$ .

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1995 AHSME Problems/Problem 17

Problem

Solution

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