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

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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 = \boxed{021}</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{021}</math>
  
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==Cartesian Variation Solution==
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Also use a coordinate system, but upon drawing the diagram call <math>Q</math> the origin and <math>QP</math> be on the x-axis. It is easy to see that <math>F</math> is the vertex on <math>RP</math>. After labeling coordinates (noting additionally that <math>QBC</math> is an equilateral triangle), we see that the area is <math>QP</math> times <math>0.5</math> times the ordinate of <math>R</math>. Draw a perpendicular of <math>F</math>, call it <math>H</math>, and note that <math>QP = 1 + \sqrt{3}</math> after using the trig functions for <math>75</math> degrees.
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 +
Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our answer is <math>\boxed{021}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:50, 23 January 2018

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 $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$

Cartesian Variation Solution

Also use a coordinate system, but upon drawing the diagram call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$. After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the ordinate of $R$. Draw a perpendicular of $F$, call it $H$, and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.

Now, get the lines for $QR$ and $RP$: $y=\sqrt{3}x$ and $y=-(2+\sqrt{3})x + (5+\sqrt{3})$, whereupon we get the ordinate of $R$ to be $\frac{3+2\sqrt{3}}{2}$, and the area is $\frac{5\sqrt{3} + 9}{4}$, so our answer is $\boxed{021}$.

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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