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 "1994 AIME Problems/Problem 8"

(Added a solution.)
m (→‎Solution Two: Fixed grammar and verbosity.)
Line 18: Line 18:
 
Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> can lead you down the path of using simple vectors.
 
Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number <math>\sqrt{3}</math> can lead you down the path of using simple vectors.
  
−
Let's begin: we drop a perpendicular from <math>O</math> to <math>AB</math>. Call the point <math>M</math>, as it is the midpoint of <math>AB</math> as well. Thus, <math>M=(\frac{a+b}{2}, 24)</math>. Now we can use perpendicularity and slope to find that of <math>OM</math> first: we get <math>\frac{48}{a+b}</math>. Its direction is <math>[\frac{a+b}{2}, 24]</math>. Meanwhile from point <math>M</math> we can use a vector with <math>\frac{\sqrt{3}}{3}</math> the distance; we have to switch the <math>x</math> and <math>y</math> directions to get a displacement of <math>[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]</math>. (Do you see why we had to switch <math>x</math> and <math>y</math> due to the rotation?)
+
First, drop a perpendicular from <math>O</math> to <math>AB</math>. Call this midpoint of <math>AB M</math>. Thus, <math>M=(\frac{a+b}{2}, 24)</math>. Now we can use perpendicularity and slope. That of <math>OM</math> is <math>\frac{48}{a+b}</math>. The vector is <math>[\frac{a+b}{2}, 24]</math>. Meanwhile from point <math>M</math> we can use a vector with <math>\frac{\sqrt{3}}{3}</math> the distance; we have to switch the <math>x</math> and <math>y</math> and our displacement is <math>[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]</math>. (Do you see why we switched <math>x</math> and <math>y</math> due to the rotation of 90 degrees?)
  
 
We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>.
 
We see this displacement from <math>M</math> to <math>A</math> is <math>[\frac{a-b}{2}, 13]</math> as well. Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}</math>. And the answer is <math>\boxed{315}</math>.
  
 
'''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
 
'''Note''': This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".
 +
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=7|num-a=9}}
 
{{AIME box|year=1994|num-b=7|num-a=9}}

Revision as of 16:11, 3 September 2017

Problem

The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$.

Solution

Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:

\[(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]

Equating the real and imaginary parts, we have:

\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}

Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$. Thus, the answer is $\boxed{315}$.

Note: There is another solution where the point $b+37i$ is a rotation of $-60$ degrees of $a+11i$; however, this triangle is just a reflection of the first triangle by the $y$-axis, and the signs of $a$ and $b$ are flipped. However, the product $ab$ is unchanged.

Solution Two

Using the Pythagorean theorem doesn't seem promising (you can look at the beastly numbers). It's better to use some properties of equilateral triangles. Thinking about the number $\sqrt{3}$ can lead you down the path of using simple vectors.

First, drop a perpendicular from $O$ to $AB$. Call this midpoint of $AB M$. Thus, $M=(\frac{a+b}{2}, 24)$. Now we can use perpendicularity and slope. That of $OM$ is $\frac{48}{a+b}$. The vector is $[\frac{a+b}{2}, 24]$. Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]$. (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)

We see this displacement from $M$ to $A$ is $[\frac{a-b}{2}, 13]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$. Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$. And the answer is $\boxed{315}$.

Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_8&oldid=87352"