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# HELP PLS

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In △ABC, AB=2, D is a point on BC, such that AD bisects ∠BAC. If $$\frac{BD}{CD} = \frac{1}{3}$$, ∠BAC=30∘, the value of BC^2 can be expressed as $$p + q\sqrt{r}$$, where p,q, and r are integers, such that r is a positive integer free of square factors larger than 1. Find the value of p+q+r.

Jun 28, 2022

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B

D

2

A                                           C

Note that angles BAD and DAC  = 15°

By the Law of Sines

sin BDA / 2  =  sin BAD / BD

sin BDA / 2  = sin 15°/ BD

sin BDA =    2sin 15° / BD        (1)

AC / DC =  AB/ BD →  AC / 3BD = 2 /BD   →   AC =  3BD * 2 / BD =  6

And

sin CDA = sin  BDA    so

sin CDA / 6 =  sin DAC / 3BD

sin BDA / 6 =  sin 15°  / 3BD   →  sin BDA = 6sin 15° / 3BD     (2)

Equating (1) and (2)

2 sin 15° / BD  =   6sin 15° / 3BD

3BD / BD =  6sin 15° / 2sin  15°

So

BC =   3BD + BD

BC^2  = (3B  + BD)^2

BC^2 =   ( 6sin 15° + 2sin 15°)^2   =

(sin 15° ( 6 + 2) )^2  =

sin^2 (15°) * 8^2  =

(sqrt 3 - 1)^2 / 8 * 64  =

(3 - 2sqrt 3 + 1) * 8  =

(4 - 2sqrt (3) ) * 8  =

32 - 16sqrt (3)

p = 32   q = -16     r =   3

p + q + r  =   19

Jun 28, 2022