A C = ? AC=?

Geometry Level 4

In △ A B C \triangle ABC if length of B C BC is 1 and sin ⁡ A 2 = x 1 \sin \dfrac A2 = x_1 , sin ⁡ B 2 = x 2 \sin \dfrac B2 = x_2 , cos ⁡ A 2 = x 3 \cos \dfrac A2=x_3 , cos ⁡ B 2 = x 4 \cos \dfrac B2 = x_4 , with ( x 1 x 2 ) 2007 = ( x 3 x 4 ) 2006 \left( \dfrac{x_1}{x_2} \right)^{2007}=\left(\dfrac{x_3}{x_4} \right)^{2006} , find the length of A C AC .


The answer is 1.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Patrick Corn
Dec 7, 2017

Since A / 2 A/2 and B / 2 B/2 are angles in the first quadrant, the x i x_i are all positive, and:

  • if sin ⁡ ( A / 2 ) > sin ⁡ ( B / 2 ) \sin(A/2) > \sin(B/2) then cos ⁡ ( A / 2 ) < cos ⁡ ( B / 2 ) \cos(A/2) < \cos(B/2)

  • if sin ⁡ ( A / 2 ) < sin ⁡ ( B / 2 ) \sin(A/2) < \sin(B/2) then cos ⁡ ( A / 2 ) > cos ⁡ ( B / 2 ) . \cos(A/2) > \cos(B/2).

So if the left side of the equation involving the x i x_i is > 1 , >1, then the right side is < 1. <1. And if the left side is < 1 , < 1, then the right side is > 1. >1. So the only possible solution is when both sides equal 1. 1. This means that sin ⁡ ( A / 2 ) = sin ⁡ ( B / 2 ) \sin(A/2) = \sin(B/2) and cos ⁡ ( A / 2 ) = cos ⁡ ( B / 2 ) . \cos(A/2)=\cos(B/2). Again, since they're in the first quadrant, this implies A / 2 = B / 2. A/2=B/2. So Δ A B C \Delta ABC is isoceles, so A C = B C = 1 . AC=BC=\fbox{1}.

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