Problem:
Let △ABC be an equilateral triangle with side length 55. Points D,E, and F lie on BC,CA, and AB, respectively, with BD=7,CE=30, and AF=40. Point P inside △ABC has the property that
∠AEP=∠BFP=∠CDP
Find tan2(∠AEP).
Solution:
Answer (075):
Let X,Y, and Z be the feet of the perpendiculars from P to BC,CA, and AB, respectively. It follows that △PDX,△PEY, and △PFZ are similar right triangles. Note that
(PX+PY+PZ)⋅AB=2⋅ Area (△ABC)=23AB2.
Thus PX+PY+PZ=23AB=2553.
Note that
PB2−PC2=BX2−CX2=(BX+CX)(BX−CX)=55(BX−CX).
Similarly, PC2−PA2=55(CY−AY) and PA2−PB2=55(AZ−BZ). Adding these three equations yields
BX+CY+AZ=CX+AY+BZ=21⋅ Perimeter (△ABC)=2165.
Note that BD+CE+AF=77. Because △PDX,△PEY, and △PFZ are similar, and 77=BD+CE+AF<BX+CY+AZ=82.5, it follows that D,E, and F lie on segments BX,CY, and AZ, respectively. In particular, DX+EY+FZ=2165−77=211.
