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

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Problem

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 1

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

Use coordinates. 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 $\boxed{021}$ .

Solution 2 (Trig)

Angle chasing yields that both triangles $PAF$ and $PQR$ are $75$ - $60$ - $45$ triangles. First look at triangle $PAF$ . Using Law of Sines, we find:

$\frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1} = \frac{\frac{\sqrt{2}}{2}}{PA}$

Simplifying, we find $PA = \sqrt{3} - 1$ . Since $\angle{Q} = 60^\circ$ , WLOG assume triangle $BQC$ is equilateral, so $BQ = 1$ . So $PQ = \sqrt{3} + 1$ .

Apply Law of Sines again,

$\frac{\frac{\sqrt{2}}{2}}{\sqrt{3} + 1} = \frac{\frac{\sqrt{3}}{2}}{PR}$

Simplifying, we find $PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})$ .

$[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot sin 75^\circ$ .

Evaluating and reducing, we get $\frac{9 + 5\sqrt{3}}{4},$ thus the answer is $\boxed{021}$

@@ Line 1: / Line 1: @@
+==Problem==
 == Problem 12 ==
 Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>.

2013 AIME I (Problems • Answer Key • Resources)
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