Library / /843 of 1394

, 2024

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a convex trapezoid such that $\angle DAB = \angle ABC = 90^{\circ}$ , $DA = 2$ , $AB = 3$ , and $BC = 8$ . Let $\omega$ be a circle passing through $A$ and tangent to segment $\overline{CD}$ at point $T$ . Suppose that the center of $\omega$ lies on line $BC$ . Compute $CT$ .

Solution

Solution:

Let $A'$ be the reflection of $A$ across $BC$ , and let $P = AB \cap CD$ . Then since the center of $\omega$ lies on $BC$ , we have that $\omega$ passes through $A'$ . Thus, by power of a point, $PT^2 = PA \cdot PA'$ . By similar triangles, we have
$\frac{PA}{AD} = \frac{PB}{BC} \Longrightarrow \frac{PA}{2} = \frac{PA + 3}{8} \Longrightarrow PA = 1$
and $A'P = 1 + 2 \cdot 3 = 7$ , so $PT = \sqrt{7}$ . But by the Pythagorean Theorem, $PC = \sqrt{PB^2 + BC^2} = 4\sqrt{5}$ , and since $T$ lies on segment $CD$ , it lies between $C$ and $P$ , so $CT = 4\sqrt{5} - \sqrt{7}$ .

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