# Difference between revisions of "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 $\frac{9 + 5\sqrt{3}}{4}, thus the answer is \boxed{021}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 