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Difference between revisions of "2013 AIME II Problems/Problem 5"

(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 17:49, 4 April 2013

In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$.

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 $M$ be the midpoint of $\overline{DE}$. Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$, $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$. Since the triangle $\Delta AME$ is isosceles, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$. Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \dfrac{3\sqrt{3}}{2\sqrt{7}}$. So we can use the double angle formula for sine, $sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$. Therefore, $a + b + c = 20$.

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