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Difference between revisions of "2012 AIME I Problems/Problem 13"

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== Solution ==
 
== Solution ==
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Reinterpret the problem in the following manner. Equilateral triangle ABC has a point X on the interior such that AX = 5, BX = 4, CX = 3. A 60 degree clockwise (or counter 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, 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, angle BXC = 150. Applying the law of cosines on triangle BXC yields
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Reinterpret the problem in the following manner. Equilateral triangle <math>ABC</math> has a point <math>X</math> on the interior such that <math>AX = 5,</math> <math>BX = 4,</math> and <math>CX = 3.</math> A <math>60^o</math> clockwise rotation about vertex <math>A</math> maps <math>X</math> to <math>X'</math> and <math>C</math> to <math>C'.</math> Note that angle <math>XAX'</math> is <math>60</math> and <math>XA = X'A = 5</math> which tells us that triangle <math>XAX'</math> is equilateral and that <math>XX' = 5.</math> We now notice that <math>XB = 3</math> and <math>X'B = 4</math> which tells us that angle <math>XBX'</math> is <math>90</math> because there is a <math>3</math>-<math>4</math>-<math>5</math> Pythagorean triple. Now note that <math>\angle ABC + \angle ACB = 120,</math> <math>\angle XCA + \angle XBA = 90,</math> <math>\angle XCB+\angle XBC = 30,</math> and <math>\angle BXC = 150.</math> Applying the law of cosines on triangle <math>BXC</math> yields
  
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<math>BX^2+CX^2-2*BX*CX*cos(150) = BC^2 = 3^2+4^2-24*cos(150) = 25+12\sqrt{3}</math> and since we are looking for the area we have <math>BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9</math>
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<cmath>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}</cmath>
  
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so our final answer is 3+4+25+9 = <math>\fbox{041}</math>.
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and thus the area of <math>ABC</math> equals <cmath>BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.</cmath>
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so our final answer is <math>3+4+25+9 = \boxed{041.}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2012|n=I|num-b=12|num-a=14}}

Revision as of 18:35, 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 $a + \frac{b}{c} \sqrt{d},$ 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
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