Problem:
Triangle ABC has AB=13,BC=14, and AC=15. The points D,E, and F are the midpoints of AB,BC, and AC respectively. Let Xξ =E be the intersection of the circumcircles of β³BDE and β³CEF. What is XA+XB+XC?
Answer Choices:
A. 24
B. 143β
C. 8195β
D. 141297ββ
E. 4692ββ Solution:
Because DE is parallel to AC and EF is parallel to AB it follows that β BDE=β BAC=β EFC. By the Inscribed Angle Theorem, β BDE=β BXE and β EFC=β EXC. Therefore β BXE=β EXC. Furthermore BE=EC, so by the Angle Bisector Theorem XB=XC. Note that β BXC=2β BXE=2β BDE=2β BAC, and by the Inscribed Angle Theorem, it follows that X is the circumcenter of β³ABC, so XA=XB=XC=R the circumradius of β³ABC.
Let a=BC,b=AC, and c=AB. The area of β³ABC equals 4R1β(abc), and by Heron's Formula it also equals s(sβa)(sβb)(sβc)β, where s=21β(a+b+c). Thus
