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

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== Solution 3 ==
 
== Solution 3 ==
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We notice that <math>\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)=\sin(\angle{DAE})</math>. We can find <math>2\sin(\alpha)\cos(\alpha)</math>, to be <math>2(\frac{1}{\sqrt{28}})(\frac{3\sqrt{3}}{\sqrt{28}})=\frac{3\sqrt{3}}{14}</math>, so our answer is <math>3+3+14=\boxed{020}</math>.
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Let A be the origin of the complex plane, B be <math>1+i\sqrt{3}</math>, and C be <math>2</math>. Also, WLOG, let D have a greater imaginary part than E. Then, D is <math>\frac{4}{3}+\frac{2i\sqrt{3}}{3}</math> and E is <math>\frac{5}{3}+\frac{i\sqrt{3}}{3}</math>. Then, <math>\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) = \frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
  
 
== Solution 4 ==
 
== Solution 4 ==
−
Let A be the origin of the complex plane, B be <math>1+i\sqrt{3}</math>, and C be <math>2</math>. Also, WLOG, let D have a greater imaginary part than E. Then, D is <math>\frac{4}{3}+\frac{2i\sqrt{3}}{3}</math> and E is <math>\frac{5}{3}+\frac{i\sqrt{3}}{3}</math>. Then, <math>\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) =  \frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
 
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== Solution 5 ==
 
 
Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, <math>AE^2=1+3^2-2(1)(3)\cos(\angle DAE)</math> or <math>AE=\sqrt7</math> The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, <math>\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14</math> Since <math>\sin^2(\angle DAE)=1-cos^2(\angle DAE)</math> Then <math>\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}</math> So <math>\sin(\angle DAE)=\frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
 
Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, <math>AE^2=1+3^2-2(1)(3)\cos(\angle DAE)</math> or <math>AE=\sqrt7</math> The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, <math>\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14</math> Since <math>\sin^2(\angle DAE)=1-cos^2(\angle DAE)</math> Then <math>\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}</math> So <math>\sin(\angle DAE)=\frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
  

Revision as of 18:56, 23 February 2018

Problem

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 1

[asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, S); label("$M$", M, S); draw(A--D); draw(A--E); draw(A--M);[/asy]

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 right, 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 $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$. 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, $a + b + c = \boxed{020}$.

Solution 2

We find that, as before, $AE = \sqrt{7}$, and also the area of $\Delta DAE$ is 1/3 the area of $\Delta ABC$. Thus, using the area formula, $1/2 \cdot 7 \cdot \sin(\angle EAD) = 3\sqrt{3}/4$, and $\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}$. Therefore, $a + b + c = \boxed{020}.$

Solution 3

Let A be the origin of the complex plane, B be $1+i\sqrt{3}$, and C be $2$. Also, WLOG, let D have a greater imaginary part than E. Then, D is $\frac{4}{3}+\frac{2i\sqrt{3}}{3}$ and E is $\frac{5}{3}+\frac{i\sqrt{3}}{3}$. Then, $\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) = \frac{3\sqrt{3}}{14}$. Therefore, $a + b + c = \boxed{020}$

Solution 4

Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\cos(\angle DAE)$ or $AE=\sqrt7$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14$ Since $\sin^2(\angle DAE)=1-cos^2(\angle DAE)$ Then $\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}$ So $\sin(\angle DAE)=\frac{3\sqrt{3}}{14}$. Therefore, $a + b + c = \boxed{020}$

See Also

2013 AIME II (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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