Library / /1266 of 1394

, 2024

Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:

In triangle $ABC$ , a circle $\omega$ with center $O$ passes through $B$ and $C$ and intersects segments $\overline{AB}$ and $\overline{AC}$ again at $B'$ and $C'$ , respectively. Suppose that the circles with diameters $BB'$ and $CC'$ are externally tangent to each other at $T$ . If $AB = 18$ , $AC = 36$ , and $AT = 12$ , compute $AO$ .

Solution

Solution:

By Radical Axis Theorem, we know that $AT$ is tangent to both circles. Moreover, consider power of a point $A$ with respect to these three circles, we have $AB \cdot AB' = AT^2 = AC \cdot AC'$ . Thus $AB' = \frac{12^2}{18} = 8$ , and $AC' = \frac{12^2}{36} = 4$ .

Consider the midpoints $M_B, M_C$ of segments $\overline{BB'}$ , $\overline{CC'}$ , respectively. We have $\angle OM_BA = \angle OM_CA = 90^\circ$ , so $O$ is the antipode of $A$ in $\left(AM_BM_C\right)$ .

Notice that $\triangle AM_BT \sim \triangle AOM_C$ , so $\frac{AO}{AM_C} = \frac{AM_B}{AT}$ .

Now, we can do the computations as follows:
$\begin{aligned} AO & = \frac{AM_B \cdot AM_C}{AT} \\ & = \left(\frac{AB + AB'}{2}\right)\left(\frac{AC + AC'}{2}\right) \frac{1}{AT} \\ & = \left(\frac{8 + 18}{2}\right)\left(\frac{36 + 4}{2}\right) \frac{1}{12} = \frac{65}{3} \end{aligned}$

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