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

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label("$C$", C, SE);
 
label("$C$", C, SE);
 
label("$D$", D, S);
 
label("$D$", D, S);
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label("$M$", M, S);
 
label("$E$", E, S);
 
label("$E$", E, S);
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label("$M$", M, S);
 
 
draw(A--D);
 
draw(A--D);
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draw(A--E);
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draw(A--M);
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draw(A--M);</asy>
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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.  
 
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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== Solution 2 ==
 
== 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 * 7 * \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>
+
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>
  
 
== 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>.
+
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>
+
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>
  
−
== Solution 5 ==
+
== Solution 5 (Vectors)==
−
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(/sqrt(7))(/sqrt(7))=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}{169}</math> So <math>\sin(\angle DAE)=\frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
+
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>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2013|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[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 4 		Followed by
Problem 6
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