Difference between revisions of "1995 AHSME Problems/Problem 17"

m (Problem: <asy> by dragon96)
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Given regular pentagon <math>ABCDE</math>, a circle can be drawn that is tangent to <math>\overline{DC}</math> at <math>D</math> and to <math>\overline{AB}</math> at <math>A</math>. The number of degrees in minor arc <math>AD</math> is
 
Given regular pentagon <math>ABCDE</math>, a circle can be drawn that is tangent to <math>\overline{DC}</math> at <math>D</math> and to <math>\overline{AB}</math> at <math>A</math>. The number of degrees in minor arc <math>AD</math> is
  
[[Image:1995 AHSME num.17.png]]
+
<!-- original image at [[File:1995 AHSME num.17.png]] -->
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<asy>size(100); defaultpen(linewidth(0.7));
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draw(rotate(18)*polygon(5));
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real x=0.6180339887;
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draw(Circle((-x,0), 1));
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int i;
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for(i=0; i<5; i=i+1) {
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dot(origin+1*dir(36+72*i));
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}
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label("$A$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0)));
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label("$B$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72)));
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label("$C$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144)));
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label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3)));
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label("$E$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4)));
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</asy>
  
 
<math> \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }  </math>
 
<math> \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }  </math>

Revision as of 20:26, 18 August 2011

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("$A$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0))); label("$B$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72))); label("$C$", 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("$E$", 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 $da=144\Rightarrow \mathrm{(E)}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AHSME Problems and Solutions