HMMT February

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a rectangle and $E$ be a point on segment $AD$ . We are given that quadrilateral $BCDE$ has an inscribed circle $\omega_{1}$ that is tangent to $BE$ at $T$ . If the incircle $\omega_{2}$ of $ABE$ is also tangent to $BE$ at $T$ , then find the ratio of the radius of $\omega_{1}$ to the radius of $\omega_{2}$ .

Solution

Solution:
Let $\omega_{1}$ be tangent to $AD$ , $BC$ at $R$ , $S$ and $\omega_{2}$ be tangent to $AD$ , $AB$ at $X$ , $Y$ . Let $AX = AY = r$ , $EX = ET = ER = a$ , $BY = BT = BS = b$ . Then noting that $RS \parallel CD$ , we see that $ABSR$ is a rectangle, so $r + 2a = b$ . Therefore $AE = a + r$ , $AB = b + r = 2(a + r)$ , and so $BE = (a + r) \sqrt{5}$ . On the other hand, $BE = b + a = r + 3a$ . This implies that $a = \frac{1 + \sqrt{5}}{2} r$ . The desired ratio is then $\frac{RS}{2AY} = \frac{AB}{2r} = \frac{a + r}{r} = \frac{3 + \sqrt{5}}{2}$ .

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