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Difference between revisions of "2013 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
 
First, find that <math>\angle R = 45^\circ</math>.
 
First, find that <math>\angle R = 45^\circ</math>.
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Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = 21</math>
+
Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = \boxed{21}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2013|n=I|num-b=11|num-a=13}}

Revision as of 12:19, 30 March 2013

Problem 12

Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

First, find that $\angle R = 45^\circ$. Draw $ABCDEF$. Now draw $\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$. The height of $ABCDEF$ is $\sqrt{3}$, so the length of base $QR$ is $2+\sqrt{3}$. Let the equation of $\overline{RP}$ be $y = x$. Then, the equation of $\overline{PQ}$ is $y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3$. Solving the two equations gives $y = x = \frac{\sqrt{3} + 3}{2}$. The area of $\bigtriangleup PQR$ is $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{21}$

See also

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