Problem:
Let ABC be a triangle where M is the midpoint of AC, and CN is the angle bisector of β ACB with N on AB. Let X be the intersection of the median BM and the bisector CN. In addition β³BXN is equilateral and AC=2. What is BN2 ?
Answer Choices:
A. 710β62ββ
B. 92β
C. 852ββ33ββ
D. 62ββ
E. 533ββ4β Solution:
Let Ξ±=β ACN=β NCB and x=BN. Because β³BXN is equilateral it follows that β BXC=β CNA=120β,β CBX=β BAC=60ββΞ±, and β CBA=β BMC=120ββΞ±. Thus β³ABCβΌβ³BMC and β³ANCβΌβ³BXC. Then
2BCβ=ACBCβ=BCMCβ=BC1β,
so BC=2β; and
2CX+xβ=ACCNβ=BCCXβ=2βCXβ
so CX=(2β+1)x.
Let P be the midpoint of XN. Because β³BXN is equilateral, the triangle BPC is a right triangle with β BPC=90β. Then by the Pythagorean Theorem,
