HMMT, 2013

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let $ABCD$ be a quadrilateral such that $\angle ABC = \angle CDA = 90^{\circ}$ , and $BC = 7$ . Let $E$ and $F$ be on $BD$ such that $AE$ and $CF$ are perpendicular to $BD$ . Suppose that $BE = 3$ . Determine the product of the smallest and largest possible lengths of $DF$ .

Solution

Solution:

By inscribed angles, $\angle CDB = \angle CAB$ , and $\angle ABD = \angle ACD$ . By definition, $\angle AEB = \angle CDA = \angle ABC = \angle CFA$ . Thus, $\triangle ABE \sim \triangle ADC$ and $\triangle CDF \sim \triangle CAB$ . This shows that

$\frac{BE}{AB} = \frac{CD}{CA} \text{ and } \frac{DF}{CD} = \frac{AB}{BD}$

Based on the previous two equations, it is sufficient to conclude that $3 = EB = FD$ . Thus, $FD$ must equal $3$ , and the product of its largest and smallest length is $9$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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