Problem:
In △ABC,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 qp, 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 △ABC and △ADE are isosceles, M and N are the midpoints of BC and AE, respectively. Thus CM=6, so if x=AD=DE=EC, then AN=210−x. Let θ=∠CAM. Because AM=102−62=8, it follows that cosθ=54. Note that cos(∠BAC)=cos(2θ)=2cos2θ−1=2⋅(54)2−1=257. Thus
cos(∠DAN)=x(210−x)=257
Solving this equation yields x=39250. The requested sum is 250+39=289.
By the Pythagorean Theorem the altitude of △ABC to BC has length 8 . Let points N and P lie on AC and AB, respectively, so that DN⊥AC and CP⊥AB. Because 2[ABC]=12⋅8=CP⋅AB, it follows that CP=548 and AP=AC2−CP2=514. Because DA=DE, it follows that
AN=NE=2AC−CE=210−AD.
Right triangles ADN and ACP are similar to each other. Because APAC=ANAD,
