2013 AIME I Problems/Problem 12

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 $\boxed{021}$

Solution 3(Trig with Diagram)

With some simple angle chasing we can show that $\triangle OJL$ and $\triangle MPL$ are congruent. This means we have a large equilateral triangle with side length $3$ and quadrilateral $OJQN$ . We know that $[OJQN] = [\triangle NQL] - [\triangle OJL]$ . Using Law of Sines and the fact that $\angle N = 45^{\circ}$ we know that $\overline{NL} = \sqrt{6}$ and the height to that side is $\frac{\sqrt{3} -1}{\sqrt{2}}$ so $[\triangle NQL] = \frac{3-\sqrt{3}}{2}$ . Using an extremely similar process we can show that $\overline{OJ} = 2-\sqrt{3}$ which means the height to $\overline{LJ}$ is $\frac{2\sqrt{3}-3}{2}$ . So the area of $\triangle OJL = \frac{2\sqrt{3}-3}{4}$ . This means the area of quadrilateral $OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}$ . So the area of our larger triangle is $\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}$ . Therefore $9+5+3+4 = \boxed{021}$

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
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