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>. | ||
Latest revision as of 20:10, 19 June 2024
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 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
.
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 |
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