Problem:
Let ABC be an equilateral triangle, and let D and F be points on sides BC and AB, respectively, with FA=5 and CD=2. Point E lies on side CA such that ∠DEF=60∘. The area of triangle DEF is 143. The two possible values of the length of side AB are p±qr, where p and q are rational, and r is an integer not divisible by the square of a prime. Find r.
Solution:
In the following, let [XYZ]= the area of triangle XYZ.
In general, let AB=s,FA=a,DC=c,EF=x,FD=y, and AE=t. Then EC=s−t. In triangle AEF, angle A is 60∘, and so [AEF]=21⋅sin60∘⋅AE. AF=43at. Similarly, [BDF]=43(s−a)(s−c),[CDE]=43c(s−t), and [ABC]=43s2. It follows that
The given conditions then imply that 56=5(s−t)+2t−10, or 5(s−t)+2t=66. Because ∠A=∠DEF=60∘, it follows that ∠AEF+∠AFE=120∘=∠AEF+∠CED, implying that ∠AFE=∠CED. Note also that ∠C=∠A. Thus △AEF∼△CDE. Consequently, AFAE=CECD, or 5t=s−t2. Thus t(s−t)=10 or s−t=t10. Substituting this into the equation 5(s−t)+2t=66 gives 5⋅10+2t2=66t. Solving this quadratic equation gives t=233±989, and hence s=t+t10=10231±103989. Repeated applications of the Law of Cosines show that both values of s produce valid triangles. Thus r=989.