Problem:
Square ABCD has center O,AB=900,E and F are on AB with AE<BF and E between A and F,m∠EOF=45∘, and EF=400. Given that BF=p+qr, where p,q, and r are positive integers and r is not divisible by the square of any prime, find p+q+r.
Solution:
Let G be the midpoint of AB, let α=m∠EOG, and let β=m∠FOG. Then OG=450,EG=450tanα,FG=450tanβ, and α+β=45∘. Therefore 450(tanα+tanβ)=400, so tanα+tanβ=8/9. Notice that tanβ=tan(45∘−α)=1+tanα1−tanα. Hence tanα+1+tanα1−tanα=98. Simplify to obtain 9tan2α−8tanα+1=0, and conclude that {tanα,tanβ}={(4±7)/9}. Because BF>AE, conclude that EG>FG, and so α>β. Then tanα=(4+7)/9 and tanβ=(4−7)/9. Thus
Draw AO and BO. Then m∠OAB=45∘=m∠EOF, and m∠OEF=m∠OAB+m∠AOE=45∘+m∠AOE=m∠AOF. Therefore △AFO∼△BOE, so AFAO=BOBE. Let BF=x. Then AF=900−x and BE=400+x. Thus
900−x4502=4502400+x, which yields 2⋅4502=360000+500x−x2, and then x2−500x+45000=0
Use the Quadratic Formula to obtain x=250±507. Recall that BF>AE, and so x>(900−400)/2=250. Then BF=x=250+507, and p+q+r=307.