# Thread: Aime 2006 II #6

1. ## Aime 2006 II #6

"Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD}$, respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}$. The length of a side of this smaller square is $\displaystyle \frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$is not divisible by the square of any prime. Find $a+b+c$."

I've tried to draw my version of this diagram, but I'm not so good at paint, so I'll try to do my best in explaining how I interpret this problem.

There is this square with an inscribed equilateral triangle, and a smaller square. Because the smaller square's sides are parallel to that of the larger one, and $B$ is a vertex for the smaller square as well as a point on $\overline{AE}$, then I get the following equalities.

Let $x$ be one side of the smaller square and let $y$ be one side of the inscribed triangle. Then it follows that:

(1) $x^2+1=y^2$
(2) $(1-x)^2+(1-x)^2=y^2$

Equating these two expressions and solving for $x$, I get that $x=\frac{4-\sqrt{10}}{2}$, which is incorrect. Thoughts?

2. Originally Posted by Jameson
...
(1) $x^2+1=y^2$
(2) $(1-x)^2+(1-x)^2=y^2$

Equating these two expressions and solving for $x$, I get that $x=\frac{4-\sqrt{10}}{2}$, which is incorrect. Thoughts?
Hello,

using your system of equations I get x= 2+sqrt(3) or x = 2- sqrt(3). Both solutions are positive. Which one fits best to your problem I don't know.

Greetings

EB

3. I made a small error. I should have solved the system and gotten that $x=\frac{4-\sqrt{12}}{2}$, which reduces down to $\frac{2-\sqrt{3}}{1}$, thus leading to the correct answer of 006. Thanks for pointing out that error.