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

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==Problem==
 
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 1==
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<asy>
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pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0);
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pair M = (1, 0);
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pair D = (2/3, 0), E = (4/3, 0);
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draw(A--B--C--cycle);
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label("$A$", A, N);
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label("$B$", B, SW);
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label("$C$", C, SE);
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label("$D$", D, S);
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label("$M$", M, S);
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label("$E$", E, S);
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draw(A--D);
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draw(A--M);
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draw(A--E); </asy>
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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 right, 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) = \sqrt{1-\sin(\angle EAM)^2}=\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 = \boxed{020}</math>.
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== Solution 2 ==
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We find that, as before, <math>AE = \sqrt{7}</math>, and also the area of <math>\Delta DAE</math> is 1/3 the area of <math>\Delta ABC</math>. Thus, using the area formula, <math>1/2 \cdot 7 \cdot \sin(\angle EAD) = 3\sqrt{3}/4</math>, and  <math>\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}.</math>
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== Solution 3 ==
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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>
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== Solution 4 ==
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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>
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== Solution 5 (Vectors)==
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Setting up a convinient coordinate system, we let <math>A</math> be at point <math>(0, 0)</math>, <math>B</math> be at point <math>(3, 3\sqrt3)</math>, and <math>C</math> be at point <math>(6, 0)</math>. Then <math>D</math> and <math>E</math> will be at points <math>(4, 2\sqrt3)</math> and <math>(5, \sqrt3)</math>. Then <math>\cos(\angle DAE) = \frac{\vec{AD}\cdot\vec{AE}}{\|\vec{AD}\| \|\vec{AE}\|} = \frac{4\cdot5 + 2\sqrt{3}\cdot\sqrt{3}}{28}=\frac{13}{14}</math>. From here, we see that <math>\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}</math>
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==See Also==
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{{AIME box|year=2013|n=II|num-b=4|num-a=6}}
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{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

Latest revision as of 18:46, 2 January 2024

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("$M$", M, S); label("$E$", E, S); draw(A--D); draw(A--M); draw(A--E); [/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, $a + b + c = \boxed{020}$

Solution 5 (Vectors)

Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$, $B$ be at point $(3, 3\sqrt3)$, and $C$ be at point $(6, 0)$. Then $D$ and $E$ will be at points $(4, 2\sqrt3)$ and $(5, \sqrt3)$. Then $\cos(\angle DAE) = \frac{\vec{AD}\cdot\vec{AE}}{\|\vec{AD}\| \|\vec{AE}\|} = \frac{4\cdot5 + 2\sqrt{3}\cdot\sqrt{3}}{28}=\frac{13}{14}$. From here, we see that $\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$

See Also

2013 AIME II (Problems • Answer Key • Resources)
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Problem 6
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