## 2012 AIME 1 Problem #12

I was checking out Mess or Math?’s cool blog and her latest post wonders whether problem 12 on the 2012 AIME 1 has a simple solution. Simple is a subjective term. The problem’s setup certainly suggests invoking things like the angle bisector theorem and if you know that theorem, then a solution is not far around the corner.

But here’s a solution that is simple in the sense that it uses very little machinery, just a little linear algebra and a dash of trig (the dash of trig being that you have to know or be able to figure out the tangents of 30 and 60 degrees). For the problem statement, check out her blog.

We set up the problem as in the figure. Notice that $\tan B = 1/m$. It often comes in handy on contests to know that the slope of a line is the tangent of the angle that line makes with the x-axis. Since A and B are complementary, their tangents are reciprocals of each other. Points D and E are intersections of lines whose equations are known, so we can find their coordinates with a little linear algebra…and, in fact, we only need to determine their x-coordinates because the condition of the problem is equivalent to DE‘ : EC = 8 : 15, which is the same as DC : EC = 23 : 15.

The x-coordinate of the intersection of the lines $y = m(x+a)$ and $y = px$ is easy to find: we set $px = m(x+a)$ and solve for $x$ and find that $x = ma/(p-m)$.

So the x-coordinate of D‘ is $-\frac{ma}{\tan 30 + m}$ and the x-coordinate of E‘ is $-\frac{ma}{\tan 60 + m}$.

Putting these into the condition DC : EC = 23 : 15 yields:

$\frac{ma}{\tan 30 + m}$ : $\frac{ma}{\tan 60 + m}$ = 23: 15.

This simplifies to the linear equation

$23(\tan 30 + m) = 15(\tan 60 + m)$.

Since $\tan 60 = \frac{1}{\tan 30} = \sqrt{3}$, we find that $m = \frac{11\sqrt{3}}{12}$. We just have to remember that we really wanted the reciprocal of m, so that the answer to the problem is $\frac{4\sqrt{3}}{11}$.

Sometimes, coordinate geometry is unpleasant and not very illuminating, but in this case, it yields a rather painless solution.