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

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

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1995 AHSME Problems/Problem 17

Problem

Solution

See also