**2013 AIME I 12.** Let be a triangle with and . A regular hexagon with side length 1 is drawn inside so that side lies on , side lies on , and one of the remaining vertices lies on . There are positive integers , , , and such that the area of can be expressed in the form , where and are relatively prime and is not divisible by the square of any prime. Find .

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

Since is on and is on , we can extend and to meet at :

must go through one of and . Since , it means that the line is “steeper” than the line ( would make just ). In order for the hexagon to be inside , we must have passing through :

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

We don’t seem to know much about the side , whereas parts of and are known. Hence, in calculating the area of , **it seems like using the formula**

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

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

From the diagram, we know that . We can use the sine rule again to find the value of :

Hence, we have

* It remains to get rid of the term.* , and we can use the double-angle formula to relate to :

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

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

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

Since the area is supposed to be in the form , we have , , , , and

**Done!**