## Sunday, June 14, 2009

### Problem 304. Triangle, Angle bisector of 120 degrees.

Proposed Problem
Click the figure below to see the complete problem 304 about Triangle, Angle bisector of 120 degrees.

Complete Problem 304
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. Applying sine rule to triangles ADB and ADC we get: x/b+x/c = sin(C)/sin(B+60)+ sin(B)/sin(C+60)
= [sin(B)+sin(C)]/sin(B+60)
Now, sin(C) = sin(B+60)= -sin(B)/2+V3cos(B)/2
while sin(B+60) = sin(B)/2 + V3cos(B)/2.
Thus,x/b+x/c=[sin(B)-sin(B)/2+V3cos(B)/2]/sin(B+60)= [sin(B)/2+V3cos(B)/2]/)/2/[sin(B)/2+V3cos(B)/2]= 1 which gives us: 1/b + 1/c = 1/x
Vihaan Uplenchwar: vihaanup@gmail.com

2. Erratum in line 3 above: sin(C) = sin(B+120) since C and B+120 are supllementary and thus
sin(C)= - sin(B)/2 + V3cos(B)/2.
Vihaan

Tr. DEC and tr ABC are similar (DE//AB)
=> DE/AB = EC/AC, or x/c = b-x/b, => bx = cb - cx,
=> bx+cx = cb, => (b+c)x = cb, => x = cb/b+c,
=> 1/x = b+c/cb, => 1/x = 1/c + 1/b

P.s

4. another solution
[ABC]=[ACD]+[ACB]
then
bcsin120=bxsin60+cxsin60
sin120=sin60
dividing both sides by bcxsin60 gives the solution

5. Set up inversion on point A, with a suitable radius such that A'B'C'D' forms a cyclic quadrilateral with A'D' as a diameter. (The line BCD would becomes the circle).
Since inversion preserve angles, ∠B'A'D' = 60, ∠A'C'D' = 60 and
B',C' lies on the circle, ∠A'B'D' = 90 ; ∠A'C'D' = 90.
So as a result,
A'D' = A'C' + A'B'
=> (1/AD) = (1/AC) + (1/AB)