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

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==Solution 1==
 
==Solution 1==
 
Define major arc DA as <math>DA</math>, and minor arc DA as <math>da</math>. Extending DC and AB to meet at F, we see that <math>\angle CFB=36=\frac{DA-da}{2}</math>. We now have two equations: <math>DA-da=72</math>, and <math>DA+da=360</math>. Solving, <math>DA=216</math> and <math>da=144\Rightarrow \mathrm{(E)}</math>.
 
Define major arc DA as <math>DA</math>, and minor arc DA as <math>da</math>. Extending DC and AB to meet at F, we see that <math>\angle CFB=36=\frac{DA-da}{2}</math>. We now have two equations: <math>DA-da=72</math>, and <math>DA+da=360</math>. Solving, <math>DA=216</math> and <math>da=144\Rightarrow \mathrm{(E)}</math>.
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==Solution 2==
 
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Let <math>O</math> be the center of the circle. Since the sum of the interior angles in any <math>n</math>-gon is <math>(n-2)180^\circ</math>, the sum of the angles in <math>ABCDO</math> is <math>540^\circ</math>.
 
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Since <math>\angle ABC=\angle BCD=108^\circ</math> and <math>\angle OAB=\angle ODC= 90^{\circ}</math>, it follows that the measure of <math>\angle AOD</math>, and thus the measure of minor arc <math>AD</math>, equals <math>540^\circ - 108^\circ-108^\circ-90^\circ-90^\circ=\boxed{\mathrm{(E)}144^\circ}</math>.
 
  
 
==See also==
 
==See also==

Revision as of 19:41, 8 March 2024

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

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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