Problem:
Triangle ABC is isosceles, with AB=AC and altitude AM=11. Suppose that there is a point D on AM with AD=10 and ∠BDC=3∠BAC. Then the perimeter of △ABC may be written in the form a+b, where a and b are integers. Find a+b.
Solution:
Let θ=∠BAD, so that ∠BDM=3θ and ∠ABD=2θ. Combine the Law of Sines and the double-angle formula for the sine function to find that
sinθBD=sin2θAD=2sinθcosθ10
from which cosθ=BD5 follows. Hence
AB=cosθAM=511BD
Apply the Pythagorean Theorem to obtain
(511BD)2−112=BM2=BD2−12
It follows that BD=255, hence that BM=211 and AB=2115. Thus the perimeter of △ABC is 2(AB+BM)=115+11=605+11, and a+b=616.
OR
Let the bisector of ∠ABD intersect AD at E, and let x=BE=AE. By the Pythagorean Theorem,
BM=BE2−EM2=x2−(11−x)2=22x−121
By applying the Pythagorean Theorem two more times, we find that
ABBD=BM2+AM2=22x and =BM2+DM2=22x−120
By the angle-bisector theorem, we have that
BDAB=DEAE
from which
22x−12022x=10−xx
By squaring both sides of this equation and solving for x, we find that x=55/8. Hence BM=11/2 and AB=(11/2)5. The perimeter of the triangle is 2(AB+BM)=115+11=605+11, so a+b=616.
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