A geometry problem by U Z

Geometry Level 1

triangle ABC is inscribed in a unit circle. The 3 bisectors of angles A,B and C are extended to intersect the circle at A 1 , B 1 a n d C 1 A_{1} , B_{1} and C_{1} respectively

then the value of A A 1 c o s ( A / 2 ) + B B 1 c o s ( B / 2 ) + C C 1 c o s ( C / 2 ) s i n A + s i n B + s i n C \frac{AA_{1}cos(A/2) + BB_{1}cos(B/2) +CC_{1}cos(C/2)}{sinA + sinB + sinC}


The answer is 2.

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U Z by U Z , 24, Guyana

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1 solution

U Z
Sep 30, 2014

A A 1 AA_{1} =sin( B + (A/2) ) = 2sin( (A+ B + C)/2 + (B/2) +(C/2)) = 2cos((B/2) - (C/2))

THUS

A A 1 AA_{1} cos(A/2) = 2cos(A/2)cos(B/2 -C/2) = cos(90 -C) +cos(90 - B)

= sinC +sinB

similarly for others

we get the expression as

2 ( s i n A + s i n B + s i n C ) s i n A + s i n B + s i n C \frac{2( sinA + sinB +sinC)}{ sinA +sinB +sinC} = 2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

how did you get the value of AA_1?

prashant sharma - 4 years, 3 months ago

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