Competition problem — Geometry — Maths Olympiad Prep

Library / /162 of 520

Geometry Difficulty 5.0 AIME, harder Find the answer

Four. (25 points) As shown in Figure 2, in the right trapezoid $ABCD$ , $\angle ABC = \angle BAD = 90^{\circ}$ , $AB = 16$ . The diagonals $AC$ and $BD$ intersect at point $E$ . A line $EF \perp AB$ is drawn through $E$ at point $F$ , and $O$ is the midpoint of side $AB$ , with $FE + EO = 8$ . Find the value of $AD + BC$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $O F=x$ .
Then $F B=8-x, F A=8+x$ .
Given that $D A / / E F / / C B$ , we have $\frac{F E}{A D}=\frac{F B}{A B}$ , which means
$A D=\frac{16}{8-x} E F \text {. }$

Similarly, $B C=\frac{16}{8+x} E F$ .
Thus, $A D+B C=\left(\frac{16}{8-x}+\frac{16}{8+x}\right) E F$
$=\frac{16 \times 16}{8^{2}-x^{2}} E F \text {. }$

Also, in $\mathrm{RI} \triangle E F O$ †, $E F^{2}+O F^{2}=O E^{2}$ , which means $E F^{2}+x^{2}=(8-E F)^{2}$ .
Solving this, we get $E F=\frac{8^{2}-x^{2}}{16}$ .
From equation (D).(2), we have $A D+B C=16$ .

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: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.