Difference between revisions of "1995 AHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | Define major arc DA as DA, and minor arc DA as da. 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>. |
==See also== | ==See also== | ||
Revision as of 11:05, 29 September 2008
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
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 | |
| 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 | ||
