Problem:
In β³ABC,AB=AC=10 and BC=12. Point D lies strictly between A and B on AB and point E lies strictly between A and C on AC so that AD=DE=EC. Then AD can be expressed in the form qpβ, where p and q are relatively prime positive integers. Find p+q.
Solution:
Let M be the foot of the perpendicular from A to BC, and let N be the foot of the perpendicular from D to AC. Because β³ABC and β³ADE are isosceles, M and N are the midpoints of BC and AE, respectively. Thus CM=6, so if x=AD=DE=EC, then AN=210βxβ. Let ΞΈ=β CAM. Because AM=102β62β=8, it follows that cosΞΈ=54β. Note that cos(β BAC)=cos(2ΞΈ)=2cos2ΞΈβ1=2β (54β)2β1=257β. Thus
cos(β DAN)=x(210βxβ)β=257β
Solving this equation yields x=39250β. The requested sum is 250+39=289β.
By the Pythagorean Theorem the altitude of β³ABC to BC has length 8 . Let points N and P lie on AC and AB, respectively, so that DNβ₯AC and CPβ₯AB. Because 2[ABC]=12β 8=CPβ AB, it follows that CP=548β and AP=AC2βCP2β=514β. Because DA=DE, it follows that
AN=NE=2ACβCEβ=210βADβ.
Right triangles ADN and ACP are similar to each other. Because APACβ=ANADβ,
