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.)
Line 15: Line 15:
 
'''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are flipped. However, the product <math>ab</math> is unchanged.
 
'''Note''': There is another solution where the point <math>b+37i</math> is a rotation of <math>-60</math> degrees of <math>a+11i</math>; however, this triangle is just a reflection of the first triangle by the <math>y</math>-axis, and the signs of <math>a</math> and <math>b</math> are flipped. However, the product <math>ab</math> 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 <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?)
 +
 +
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".
 
== 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 16:08, 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.

Let's begin: we drop a perpendicular from $O$ to $AB$. Call the point $M$, as it is the midpoint of $AB$ as well. Thus, $M=(\frac{a+b}{2}, 24)$. Now we can use perpendicularity and slope to find that of $OM$ first: we get $\frac{48}{a+b}$. Its direction 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 $y$ directions to get a displacement of $[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]$. (Do you see why we had to switch $x$ and $y$ due to the rotation?)

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=87351"