1995 AHSME Problems/Problem 17
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("$A$", 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]](http://latex.artofproblemsolving.com/f/c/7/fc77566234f916102bc0b6392710e2ff38a5b3ad.png)
Solution
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 
.
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 | ||
