Library / /768 of 1394

, 2023

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:
Let $A X B Y$ be a cyclic quadrilateral, and let line $A B$ and line $X Y$ intersect at $C$ . Suppose $A X \cdot A Y = 6$ , $B X \cdot B Y = 5$ , and $C X \cdot C Y = 4$ . Compute $A B^{2}$ .

Solution

Solution:
Observe that
$\begin{aligned} & \triangle A C X \sim \triangle Y C B \Longrightarrow \frac{A C}{A X} = \frac{C Y}{B Y} \\ & \triangle A C Y \sim \triangle X C B \Longrightarrow \frac{A C}{A Y} = \frac{C X}{B X} \end{aligned}$
Multiplying these two equations together, we get that
$A C^{2} = \frac{(C X \cdot C Y)(A X \cdot A Y)}{B X \cdot B Y} = \frac{24}{5}$
Analogously, we obtain that
$B C^{2} = \frac{(C X \cdot C Y)(B X \cdot B Y)}{A X \cdot A Y} = \frac{10}{3}$
Hence, we have
$A B = A C + B C = \sqrt{\frac{24}{5}} + \sqrt{\frac{10}{3}} = \frac{11 \sqrt{30}}{15}$
implying the answer.

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