HMMT February

Geometry Difficulty 5.4 AIME, harder Find the answer

Let $A B C$ be an equilateral triangle with side length 1. Points $D, E, F$ lie inside triangle $A B C$ such that $A, E, F$ are collinear, $B, F, D$ are collinear, $C, D, E$ are collinear, and triangle $D E F$ is equilateral. Suppose that there exists a unique equilateral triangle $X Y Z$ with $X$ on side $\overline{B C}, Y$ on side $\overline{A B}$ , and $Z$ on side $\overline{A C}$ such that $D$ lies on side $\overline{X Z}, E$ lies on side $\overline{Y Z}$ , and $F$ lies on side $\overline{X Y}$ . Compute $A Z$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

First, note that point $X$ can be constructed from intersection of $\odot(D O F)$ and side $\overline{B C}$ . Thus, if there is a unique equilateral triangle, then we must have that $\odot(D O F)$ is tangent to $\overline{B C}$ . Furthermore, $\odot(D O F)$ is tangent to $D E$ , so by equal tangents, we have $C D=C X$ . We now compute the answer. Let $x=A Z=C X=C D=B F$ . Then, by power of point, $B F \cdot B D=B X^{2} \Longrightarrow B D=\frac{(1-x)^{2}}{x}$ Thus, by law of cosine on $\triangle B D C$ , we have that $\begin{aligned} x^{2}+\left(\frac{(1-x)^{2}}{x}\right)^{2}+x \cdot \frac{(1-x)^{2}}{x} & =1 \\ x^{2}+\frac{(1-x)^{4}}{x^{2}}+(1-x)^{2} & =1 \\ \frac{(1-x)^{4}}{x^{2}} & =2x(1-x) \\ \frac{1-x}{x} & =\sqrt[3]{2} \\ x & =\frac{1}{1+\sqrt[3]{2}} \end{aligned}$

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