## [Soln] 2013 AIME I Problem 12

2013 AIME I 12. Let $\Delta PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\Delta PQR$ 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 $\Delta PQR$ 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 $a + b + c + d$.

As I was drawing the diagram, I realised that it was easier to start by drawing the hexagon instead of $\Delta PQR$. Start with the hexagon:

Since $AB$ is on $PQ$ and $CD$ is on $QR$, we can extend $AB$ and $CD$ to meet at $Q$:

$PR$ must go through one of $E$ and $F$. Since $\angle P = 75^\circ$, it means that the line $PR$ is “steeper” than the line $EF$ ($EF$ would make $\angle P$ just $60^\circ$). In order for the hexagon to be inside $\Delta PQR$, we must have $PR$ passing through $F$:

Now we label some sides and angles based on what the problem gives us:

We don’t seem to know much about the side $PR$, whereas parts of $PQ$ and $QR$ are known. Hence, in calculating the area of $\Delta PQR$, it seems like using the formula

$\text{Area of } \Delta PQR = \displaystyle\frac{1}{2} |PQ||QR|\sin PQR = \frac{1}{2} |PQ||QR| \sin 60^\circ$

could work. Let’s label the diagram further to give more info on $PQ$ and $QR$:

We have 2 unknown quantities we would like to find: $|PQ|$ and $|QR|$. We can use the sine rule to reduce that to just finding $|PQ|$:

\begin{aligned} \displaystyle\frac{|PQ|}{\sin 45^\circ} &= \frac{|QR|}{\sin 75^\circ}, \\ |QR| &= \frac{|PQ|}{\sin 45^\circ} \sin 75^\circ = \sqrt{2}|PQ| \sin 75^\circ. \end{aligned}

From the diagram, we know that $|PQ| = 2 + |AP|$. We can use the sine rule again to find the value of $|AP|$:

\begin{aligned} \displaystyle\frac{|AP|}{\sin 45^\circ} &= \frac{|AF|}{\sin 75^\circ}, \\ |AP| &= \frac{1}{\sin 75^\circ} \sin 45^\circ = \frac{1}{\sqrt{2}\sin 75^\circ}. \end{aligned}

Hence, we have

\begin{aligned} \text{Area} &= \displaystyle\frac{1}{2}|PQ| \cdot \sqrt{2} |PQ| \sin 75^\circ \cdot \sin 60^\circ \\ &= \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} \sin 75^\circ |PQ|^2 \\ &= \frac{\sqrt{6}}{4} \sin 75^\circ \left( 2 + \frac{1}{\sqrt{2}\sin 75^\circ} \right)^2 \\ &= \frac{\sqrt{6}}{4}\sin 75^\circ \left( 4 + \frac{4}{\sqrt{2}\sin 75^\circ} + \frac{1}{2\sin^2 75^\circ} \right) \\ &= \sqrt{6}\sin 75^\circ + \sqrt{3} + \frac{\sqrt{6}}{8 \sin 75^\circ}. \end{aligned}

It remains to get rid of the $\sin 75^\circ$ term. $\sin 75^\circ = \cos (90^\circ - 75^\circ) = \cos 15^\circ$, and we can use the double-angle formula to relate $\cos 15^\circ$ to $\cos 30^\circ$:

\begin{aligned} \cos 30^\circ &= 2 \cos^2 15^\circ - 1, \\ \displaystyle\frac{\sqrt{3}}{2} &= 2\cos^2 15^\circ -1, \\ \cos^2 15^\circ &= \frac{2 + \sqrt{3}}{4}, \\ \cos 15^\circ &= \frac{1}{2}\sqrt{2 + \sqrt{3}}. \end{aligned}

Great, but not good enough: we can’t have the “square root in a square root” show up in the final value for the triangle’s area. Could we transform the stuff inside the first square root to a perfect square? Our first try could be

$(1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2 \sqrt{3},$

which is really close: it’s twice the value in our square root! Hence,

\begin{aligned} \cos 15^\circ &= \frac{1}{2} \sqrt{\frac{1}{2}(1 + \sqrt{3})^2} \\ &= \frac{1 + \sqrt{3}}{2\sqrt{2}}. \end{aligned}

Substituting into the expression we have for the triangle’s area:

\begin{aligned} \text{Area} &= \sqrt{6}\sin 75^\circ + \sqrt{3} + \frac{\sqrt{6}}{8 \sin 75^\circ} \\ &= \frac{\sqrt{6}(1+\sqrt{3})}{2\sqrt{2}} + \sqrt{3} + \frac{\sqrt{6}(2\sqrt{2})}{8(1+\sqrt{3})} \\ &= \frac{\sqrt{3}(1 + \sqrt{3})}{2} + \sqrt{3} + \frac{4\sqrt{3}(\sqrt{3} -1)}{8 (\sqrt{3}+1)(\sqrt{3}-1)} \\ &= \frac{\sqrt{3} + 3}{2} + \sqrt{3} + \frac{3 - \sqrt{3}}{2(3 - 1)} \\ &= \frac{\sqrt{3} + 3}{2} + \sqrt{3} + \frac{3 - \sqrt{3}}{4} \\ &= \frac{2(\sqrt{3}+3) + 4\sqrt{3} + (3 - \sqrt{3})}{4} \\ &= \frac{9 + 5\sqrt{3}}{4}. \end{aligned}

Since the area is supposed to be in the form $\frac{a + b\sqrt{c}}{d}$, we have $a = 9$, $b = 5$, $c = 3$, $d = 4$, and

$a + b + c + d = 9 + 5 + 3 + 4 = \fbox{21}.$

Done!