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 β BDC=3β BAC. Then the perimeter of β³ABC may be written in the form a+bβ, where a and b are integers. Find a+b.
Solution:
Let ΞΈ=β BAD, so that β BDM=3ΞΈ and β ABD=2ΞΈ. Combine the Law of Sines and the double-angle formula for the sine function to find that
sinΞΈBDβ=sin2ΞΈADβ=2sinΞΈcosΞΈ10β
from which cosΞΈ=BD5β follows. Hence
AB=cosΞΈAMβ=511βBD
Apply the Pythagorean Theorem to obtain
(511βBD)2β112=BM2=BD2β12
It follows that BD=255ββ, hence that BM=211β and AB=211β5β. Thus the perimeter of β³ABC is 2(AB+BM)=115β+11=605β+11, and a+b=616β.
OR
Let the bisector of β ABD intersect AD at E, and let x=BE=AE. By the Pythagorean Theorem,
BM=BE2βEM2β=x2β(11βx)2β=22xβ121β
By applying the Pythagorean Theorem two more times, we find that
ABBDβ=BM2+AM2β=22xβ and =BM2+DM2β=22xβ120ββ
By the angle-bisector theorem, we have that
BDABβ=DEAEβ
from which
22xβ120β22xββ=10βxxβ
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)5β. The perimeter of the triangle is 2(AB+BM)=115β+11=605β+11, so a+b=616β.
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