Problem:
Let ABC be a triangle where M is the midpoint of AC, and CN is the angle bisector of ∠ACB with N on AB. Let X be the intersection of the median BM and the bisector CN. In addition △BXN is equilateral and AC=2. What is BN2 ?
Answer Choices:
A. 710−62
B. 92
C. 852−33
D. 62
E. 533−4 Solution:
Let α=∠ACN=∠NCB and x=BN. Because △BXN is equilateral it follows that ∠BXC=∠CNA=120∘,∠CBX=∠BAC=60∘−α, and ∠CBA=∠BMC=120∘−α. Thus △ABC∼△BMC and △ANC∼△BXC. Then
2BC=ACBC=BCMC=BC1,
so BC=2; and
2CX+x=ACCN=BCCX=2CX
so CX=(2+1)x.
Let P be the midpoint of XN. Because △BXN is equilateral, the triangle BPC is a right triangle with ∠BPC=90∘. Then by the Pythagorean Theorem,
