Problem:
Square ABCD has center O,AB=900,E and F are on AB with AE<BF and E between A and F,mβ EOF=45β, and EF=400. Given that BF=p+qrβ, 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 Ξ±=mβ EOG, and let Ξ²=mβ FOG. Then OG=450,EG=450tanΞ±,FG=450tanΞ², and Ξ±+Ξ²=45β. Therefore 450(tanΞ±+tanΞ²)=400, so tanΞ±+tanΞ²=8/9. Notice that tanΞ²=tan(45ββΞ±)=1+tanΞ±1βtanΞ±β. Hence tanΞ±+1+tanΞ±1βtanΞ±β=98β. Simplify to obtain 9tan2Ξ±β8tanΞ±+1=0, and conclude that {tanΞ±,tanΞ²}={(4Β±7β)/9}. Because BF>AE, conclude that EG>FG, and so Ξ±>Ξ². Then tanΞ±=(4+7β)/9 and tanΞ²=(4β7β)/9. Thus
Draw AO and BO. Then mβ OAB=45β=mβ EOF, and mβ OEF=mβ OAB+mβ AOE=45β+mβ AOE=mβ AOF. Therefore β³AFOβΌβ³BOE, so AFAOβ=BOBEβ. Let BF=x. Then AF=900βx and BE=400+x. Thus
β900βx4502ββ=4502β400+xβ, which yields 2β 4502=360000+500xβx2, and then x2β500x+45000=0β
Use the Quadratic Formula to obtain x=250Β±507β. Recall that BF>AE, and so x>(900β400)/2=250. Then BF=x=250+507β, and p+q+r=307β.