Library / /1204 of 1394

, 2024

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:
Let $\Omega$ and $\omega$ be circles with radii $123$ and $61$ , respectively, such that the center of $\Omega$ lies on $\omega$ . A chord of $\Omega$ is cut by $\omega$ into three segments, whose lengths are in the ratio $1:2:3$ in that order. Given that this chord is not a diameter of $\Omega$ , compute the length of this chord.

Solution

Solution:
Denote the center of $\Omega$ as $O$ . Let the chord intersect the circles at $W, X, Y, Z$ so that $WX = t$ , $XY = 2t$ , and $YZ = 3t$ . Notice that $Y$ is the midpoint of $WZ$ ; hence $\overline{OY} \perp \overline{WXYZ}$ .

The fact that $\angle OYX = 90^\circ$ means $X$ is the antipode of $O$ on $\omega$ , so $OX = 122$ . Now applying power of point to $X$ with respect to $\Omega$ gives
$245 = 123^2 - OX^2 = WX \cdot XZ = 5t^2 \Longrightarrow t = 7$
Hence the answer is $6t = 42$ .

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