[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:

AIME_I_2013_12_01

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

AIME_I_2013_12_02

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:

AIME_I_2013_12_03

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

AIME_I_2013_12_04

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:

AIME_I_2013_12_05

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!

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