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 "2012 AIME I Problems/Problem 13"

(→‎Solution)
m (→‎Problem 13)
Line 1: Line 1:
 
==Problem 13==
 
==Problem 13==
−
Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \frac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math>
+
Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \tfrac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 21:59, 17 March 2012

Problem 13

Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution

Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^o$ clockwise rotation about vertex $A$ maps $X$ to $X'$ and $C$ to $C'.$ Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XB = 3$ and $X'B = 4$ which tells us that angle $XBX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120,$ $\angle XCA + \angle XBA = 90,$ $\angle XCB+\angle XBC = 30,$ and $\angle BXC = 150.$ Applying the law of cosines on triangle $BXC$ yields

\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 3^2+4^2-24 \cdot \cos(150) = 25+12\sqrt{3}\]

and thus the area of $ABC$ equals \[BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.\]

so our final answer is $3+4+25+9 = \boxed{041.}$

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_13&oldid=45622"