Library / /658 of 740

, 2022

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Consider parallelogram $ABCD$ with $AB > BC$ . Point $E$ on $\overline{AB}$ and point $F$ on $\overline{CD}$ are marked such that there exists a circle $\omega_1$ passing through $A, D, E, F$ and a circle $\omega_2$ passing through $B, C, E, F$ . If $\omega_1, \omega_2$ partition $\overline{BD}$ into segments $\overline{BX}, \overline{XY}, \overline{YD}$ in that order, with lengths $200, 9, 80$ , respectively, compute $BC$ .

Solution

Solution:

We want to find $AD = BC = EF$ . So, let $EF$ intersect $BD$ at $O$ . It is clear that $\triangle BOE \sim \triangle DOF$ . However, we can show by angle chase that $\triangle BXE \sim \triangle DYF$ :
$\angle BEG = \angle ADG = \angle CBH = \angle DFH$
This means that $\overline{EF}$ partitions $\overline{BD}$ and $\overline{XY}$ into the same proportions, i.e. $200$ to $80$ . Now, let $a = 200$ , $b = 80$ , $c = 9$ to make computation simpler. $O$ is on the radical axis of $\omega_1, \omega_2$ and its power with respect to the two circles can be found to be
$\left(a + \frac{ac}{a+b}\right) \frac{bc}{a+b} = \frac{abc(a+b+c)}{(a+b)^2}$
However, there is now $x$ for which $OE = a x$ , $OF = b x$ by similarity. This means $x^2 = \frac{c(a+b+c)}{(a+b)^2}$ . Notably, we want to find $(a+b)x$ , which is just
$\sqrt{c(a+b+c)} = \sqrt{9 \cdot 289} = 51$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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