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Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:

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

Solution

Solution:

First, note that point $X$ can be constructed from intersection of $\odot(DOF)$ and side $\overline{BC}$ . Thus, if there is a unique equilateral triangle, then we must have that $\odot(DOF)$ is tangent to $\overline{BC}$ . Furthermore, $\odot(DOF)$ is tangent to $DE$ , so by equal tangents, we have $CD = CX$ .

We now compute the answer. Let $x = AZ = CX = CD = BF$ . Then, by power of point,
$BF \cdot BD = BX^2 \Longrightarrow BD = \frac{(1-x)^2}{x}$
Thus, by law of cosine on $\triangle BDC$ , 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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.