Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5)
(→‎Solution 5 (Vectors))
 
(12 intermediate revisions by 6 users not shown)
Line 13: Line 13:
 
label("$C$", C, SE);
 
label("$C$", C, SE);
 
label("$D$", D, S);
 
label("$D$", D, S);
 +
label("$M$", M, S);
 
label("$E$", E, S);
 
label("$E$", E, S);
−
label("$M$", M, S);
 
 
draw(A--D);
 
draw(A--D);
−
draw(A--E);
+
draw(A--M);
−
draw(A--M);</asy>
+
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.  
Line 24: Line 24:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
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 <math>\dfrac{1}{3}</math> the area of <math>\Delta ABC</math>. Thus, using the area formula, <math>\dfrac{1}{2} \cdot 7 \cdot \sin(\angle EAD) = \dfrac{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 ==
−
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 (Vectors)==
 +
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>
 +
 
 +
== Solution 6 ==
 +
We first drop the altitude from <math>A</math> to <math>BC</math>, and that evaluates to a length of <math>5</math>. Now, we can find <math>AD</math> by Pythag. We have <math>AD = \sqrt{7}</math>. Now, we can set <math>\angle{DAE}</math> to be <math>\theta</math>, and now we we can solve for <math>\sin{\theta}</math>. We will take advantage of the different ways to find the area of a triangle. We have the area of the triangle is <math>\dfrac{\dfrac{1}{2} \cdot \dfrac{3\sqrt3}{2}}{2} = \dfrac{3\sqrt3}{4}</math>. We can also express it as <math>\dfrac{1}{2} \cdot 7 \cdot \sin{\theta}</math>, and solving for <math>\sin{\theta}</math>, we get <cmath>\dfrac{3\sqrt3}{14} \Longrightarrow \boxed{020}</cmath>
  
−
== Solution 5 ==
+
~jb2015007
−
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>
 
  
 
==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 21:13, 16 April 2025

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 $\dfrac{1}{3}$ the area of $\Delta ABC$. Thus, using the area formula, $\dfrac{1}{2} \cdot 7 \cdot \sin(\angle EAD) = \dfrac{3\sqrt{3}}{4}$, 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 $\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$

Solution 6

We first drop the altitude from $A$ to $BC$, and that evaluates to a length of $5$. Now, we can find $AD$ by Pythag. We have $AD = \sqrt{7}$. Now, we can set $\angle{DAE}$ to be $\theta$, and now we we can solve for $\sin{\theta}$. We will take advantage of the different ways to find the area of a triangle. We have the area of the triangle is $\dfrac{\dfrac{1}{2} \cdot \dfrac{3\sqrt3}{2}}{2} = \dfrac{3\sqrt3}{4}$. We can also express it as $\dfrac{1}{2} \cdot 7 \cdot \sin{\theta}$, and solving for $\sin{\theta}$, we get \[\dfrac{3\sqrt3}{14} \Longrightarrow \boxed{020}\]

~jb2015007

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_5&oldid=246830"