Proposed Problem
Click the figure below to see the complete problem 304 about Triangle, Angle bisector of 120 degrees.
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Complete Problem 304
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Sunday, June 14, 2009
Problem 304. Triangle, Angle bisector of 120 degrees.
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Applying sine rule to triangles ADB and ADC we get: x/b+x/c = sin(C)/sin(B+60)+ sin(B)/sin(C+60)
ReplyDelete= [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
Erratum in line 3 above: sin(C) = sin(B+120) since C and B+120 are supllementary and thus
ReplyDeletesin(C)= - sin(B)/2 + V3cos(B)/2.
Vihaan
draw DE // AB, win tr equilateral ADB, => AD=DE=AE=x
ReplyDeleteTr. 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
another solution
ReplyDelete[ABC]=[ACD]+[ACB]
then
bcsin120=bxsin60+cxsin60
sin120=sin60
dividing both sides by bcxsin60 gives the solution
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).
ReplyDeleteSince 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)