Difference between revisions of "1995 AHSME Problems/Problem 17"
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<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> | ||
| − | ==Solution== | + | ==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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | 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>. | ||
| + | |||
| + | Since <math>\angle ABC=\angle BCD=108^\circ</math> and <math>\angle OAB=\angleODC=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 04:00, 5 July 2014
Contents
Problem
Given regular pentagon 
, a circle can be drawn that is tangent to 
at 
and to 
at 
. The number of degrees in minor arc 
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]](http://latex.artofproblemsolving.com/6/7/9/67955d58098fd50e5d4caa0fd57150a361ae17af.png)
Solution 1
Define major arc DA as 
, and minor arc DA as 
. Extending DC and AB to meet at F, we see that 
. We now have two equations: 
, and 
. Solving, 
and 
.
Solution 2
Let 
be the center of the circle. Since the sum of the interior angles in any 
-gon is 
, the sum of the angles in 
is 
.
Since 
and $\angle OAB=\angleODC=90^\circ$ (Error compiling LaTeX. Unknown error_msg), it follows that the measure of 
, and thus the measure of minor arc , equals 
.
See also
| 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. 
