Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.1 AIME, harder Prove it South Africa

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$ . The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac{2}{3}$ , determine the length of $YZ$ .

Solution

Let the length of $YZ$ be $x$ , and let $y$ be the associated height of triangle $XYZ$ . Note that triangles $XYZ$ and $XAB$ are similar. Since corresponding sides $YZ$ and $AB$ have lengths $x$ and $1$ respectively, the heights $y$ and $1+y$ have to satisfy
$\frac{y}{1+y} = \frac{x}{1} = x.$
Hence the area of $XYZ$ is
$\frac{2}{3} = \frac{xy}{2} = \frac{y^2}{2(1+y)}.$

This yields the quadratic equation
$y^2 - \frac{4}{3}y - \frac{4}{3} = \left(y + \frac{2}{3}\right) (y - 2) = 0,$
from which it follows that $y = 2$ and thus
$x = \frac{y}{1+y} = \frac{2}{3}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.