Difference between revisions of "1996 AHSME Problems/Problem 30"
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<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math> | <math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math> | ||
− | ==Solution== | + | ==Solution 1== |
All angle measures are in degrees. | All angle measures are in degrees. | ||
Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>. Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>. We look for <math>AD</math>. | Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>. Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>. We look for <math>AD</math>. | ||
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Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath> | Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath> | ||
Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
== See also == | == See also == |
Revision as of 17:25, 1 August 2016
Contents
[hide]Problem
A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to , where
and
are relatively prime positive integers. Find
.
Solution 1
All angle measures are in degrees.
Let the first trapezoid be , where
. Then the second trapezoid is
, where
. We look for
.
Since is an isosceles trapezoid, we know that
and, since
, if we drew
, we would see
. Anyway,
(
means arc AB). Using similar reasoning,
.
Let and
. Since
(add up the angles),
and thus
. Therefore,
.
as well.
Now I focus on triangle . By the Law of Cosines,
, so
. Seeing
and
, we can now use the Law of Sines to get:
Now I focus on triangle .
and
, and we are given that
, so
We know
, but we need to find
. Using various identities, we see
Returning to finding
, we remember
Plugging in and solving, we see
. Thus, the answer is
, which is answer choice
.
Solution 2
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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All AHSME Problems and Solutions |
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