Problem:
In △ABC,AB=13,BC=15, and CA=14. Point D is on BC with CD=6. Point E is on BC such that ∠BAE≅∠CAD. Given that BE=p/q, where p and q are relatively prime positive integers, find q.
Solution:
Let m∠BAE=α=m∠CAD, and let β=m∠EAD. Then
DCBD=[ADC][ABD]=(1/2)AD⋅ACsinCAD(1/2)AB⋅ADsinBAD=ACAB⋅sinαsin(α+β)
Similarly,
ECBE=ACAB⋅sinCAEsinBAE=ACAB⋅sin(α+β)sinα
and so
ECBE=AC2⋅BDAB2⋅DC
Substituting the given values yields BE/EC=(132⋅6)/(142⋅9)=169/294. Therefore BE=(15⋅169)/(169+294)=(3⋅5⋅132)/463. Because none of 3,5, and 13 divides 463,q=463.
The problems on this page are the property of the MAA's American Mathematics Competitions
