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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Suppose $ABCD$ is an isosceles trapezoid in which $\overline{AB} \parallel \overline{CD}$ . Two mutually externally tangent circles $\omega_1$ and $\omega_2$ are inscribed in $ABCD$ such that $\omega_1$ is tangent to $\overline{AB}$ , $\overline{BC}$ , and $\overline{CD}$ while $\omega_2$ is tangent to $\overline{AB}$ , $\overline{DA}$ , and $\overline{CD}$ . Given that $AB=1$ , $CD=6$ , compute the radius of either circle.

Solution

Solution:

Let the radius of both circles be $r$ , and let $\omega_1$ be centered at $O_1$ . Let $\omega_1$ be tangent to $\overline{AB}$ , $\overline{BC}$ , and $\overline{CD}$ at $P$ , $Q$ , and $R$ respectively. Then, by symmetry, $PB = \frac{1}{2} - r$ and $RC = 3 - r$ . By equal tangents from $B$ and $C$ , $BQ = \frac{1}{2} - r$ and $QC = 3 - r$ .

Now, $\angle BO_1C$ is right because $m \angle O_1BC + m \angle BCO_1 = \frac{1}{2}(m \angle PBC + m \angle BCR) = 90^\circ$ .

Since $\overline{O_1Q} \perp \overline{BC}$ , $r^2 = O_1Q^2 = BQ \cdot QC = \left(\frac{1}{2} - r\right)(3 - r) = r^2 - \frac{7}{2}r + \frac{3}{2}$ .

Solving, we find $r = \frac{3}{7}$ .

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