Difference between revisions of "2013 AIME II Problems/Problem 5"
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In equilateral <math>\triangle ABC</math> let points <math>D</math> and <math>E</math> trisect <math>\overline{BC}</math>. Then <math>\sin(\angle DAE)</math> can be expressed in the form <math>\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is an integer that is not divisible by the square of any prime. Find <math>a+b+c</math>. | In equilateral <math>\triangle ABC</math> let points <math>D</math> and <math>E</math> trisect <math>\overline{BC}</math>. Then <math>\sin(\angle DAE)</math> can be expressed in the form <math>\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is an integer that is not divisible by the square of any prime. Find <math>a+b+c</math>. | ||
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+ | == Solution == | ||
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+ | For Problem 5. | ||
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+ | Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier. | ||
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+ | Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Then <math>\Delta MCA</math> is a 30-60-90 triangle with <math>MC = \dfrac{3}{2}</math>, <math>AC = 3</math> and <math>AM = \dfrac{3\sqrt{3}}{2}</math>. Since the triangle <math>\Delta AME</math> is isosceles, then we can find the length of <math>\overline{AE}</math> by pythagorean theorem, <math>AE = \sqrt{7}</math>. Therefore, since <math>\Delta AME</math> is a right triangle, we can easily find <math>\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}</math> and <math>\cos(\angle EAM) = \dfrac{3\sqrt{3}}{2\sqrt{7}}</math>. So we can use the double angle formula for sine, <math>sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = 20</math>. |
Revision as of 18:49, 4 April 2013
In equilateral let points
and
trisect
. Then
can be expressed in the form
, where
and
are relatively prime positive integers, and
is an integer that is not divisible by the square of any prime. Find
.
Solution
For Problem 5.
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let be the midpoint of
. Then
is a 30-60-90 triangle with
,
and
. Since the triangle
is isosceles, then we can find the length of
by pythagorean theorem,
. Therefore, since
is a right triangle, we can easily find
and
. So we can use the double angle formula for sine,
. Therefore,
.