JHMT 2011 Geometry Problem 6 (Bad Things Come in Threes)

2011 JHMT Geometry Problem 6

Let $\triangle ABC$ be equilateral. Two points $D$ and $E$ lie on side $BC$ (with order $B$, $D$, $E$, $C$), and satisfy $\angle DAE = 30^\circ$. If $BD = 2$ and $CE = 3$, what is $BC$?

Initial Observations

Now I remember why contest geometry was my worst subject. I always feel like I’m playing roulette. Do I

1. Find helpful auxiliary lines,
2. Use Law of Sines/Cosines, or
3. Coordinate bash.

Quick, you only have 6 minutes per problem. Decide now.

Trig Solution

My idea was to use Law of Cosines three times: once on $\triangle ABD$, once on $\triangle AEC$, and once on $\triangle ADE$. Doing so yields the following three equations:

$$\begin{aligned} AB^2+2^2-2\cdot AB\cdot2\cdot \cos(60^\circ)&=AD^2 \\ AC^2+3^2-2\cdot AC\cdot3\cdot \cos(60^\circ)&=AE^2 \\ AD^2+AE^2-2\cdot AD\cdot AE\cdot \cos(30^\circ)&=DE^2 \end{aligned}$$

Of course $\triangle ABC$ is equilateral so $AB=AC=DE+5$, which gives us three equations with three variables. I don’t know a better way to solve this except for substituting $AD$ and $AE$ into the third equation, so let’s do that. Call the side length of the triangle $x$ for simplicity. Now I can sort of see what’s going to happen: we’re going to have some square roots and the only way to get rid of them is to square both side of an equation. But the other side will already have quadratic terms… In other words we’re going to end up with a quartic. And this is the part that trips me up but maybe not more experienced solvers. In many problems like this a few good things can happen:

1. The quartic terms could end up canceling out, leaving us with a cubic (or even a quadratic) that we might be able to solve by inspecting/guess and check, or
2. The cubic and linear terms might cancel, leaving us with something of the form $ax^4+bx^2+c=0$.

Here’s the resulting quartic.

$$2 x^2-5 x+13-\sqrt{3} \sqrt{x^2-3 x+9} \sqrt{x^2-2 x+4}=(x-5)^2$$

Grouping rooted terms on one side of the equation and non-rooted on the other, squaring and then expanding everything gives us the following equation:

$$2 x^4-25 x^3+56 x^2+30 x-36=0$$

Darn. The quartic terms almost cancelled out. But Wait! We have one more trick up our sleeve, which is we could’ve instead assigned $DE=x$ and $AB=AC=x+5$ instead of the way we did it. Making the substitution $x\rightarrow x+5$:

$$2 x^4+15 x^3-19 x^2-285 x-361=0$$

Arggh… At this point it’s game over for me and I have confirmed my solution is not contest-worthy. Just to sanity check that my solution is correct, I asked Mathematica to factor this quartic and rub the solution in my face:

$$(x^2-19) (2 x^2+15 x+19)=0$$

So $DE=\sqrt{19}$ and our final answer is $\boxed{BC=5+\sqrt{19}}$.

Today’s Moral

I’ve solved plenty of geometry problems that involved setting up a system of equations using the Law of Cosines. However, now that I think about it, it’s always only two Law of Cosines applications. So, I guess my new heuristic generalization is:

Contest Solution

The key is that we don’t immediately know what $\angle BAD$ or $\angle EAC$ is, but we do know their sum! Rotate the whole diagram $60^\circ$ (CCW, for example) around point $A$ so those two angles get combined side-by-side:

Note that $\triangle ADE \cong \triangle AD'E$ by SAS, so $DE=D'E$. Now using Law of Cosines on $\triangle CD'E$ gives:

$$\begin{aligned} D'E^2 &= EC^2+CD'^2-2\cdot EC \cdot CD' \cdot \cos(\angle ECD') \\ &= 3^2+2^2-2\cdot3\cdot2\cdot\cos(120^\circ) \\ &= 19 \end{aligned}$$