Triangle?

Geometry Level 2

If A , B A,B and C C denotes the angles of a triangle such that sin ( A 2 ) sin ( B 2 ) sin ( C 2 ) = 1 8 \sin \left( \dfrac A2\right)\sin \left( \dfrac B2\right)\sin \left( \dfrac C2\right) = \dfrac18 , then triangle A B C ABC is a/an __________ \text{\_\_\_\_\_\_\_\_\_\_} triangle.

Equilateral Scalene Right Angled Isosceles

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1 solution

Using the identity sin ( x ) sin ( y ) = 1 2 ( cos ( x y ) cos ( x + y ) ) \sin(x)\sin(y) = \dfrac{1}{2}(\cos(x - y) - \cos(x + y)) we see that

S = sin ( A 2 ) sin ( B 2 ) sin ( C 2 ) = 1 8 S = \sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right) = \dfrac{1}{8}

1 2 sin ( A 2 ) ( cos ( B C 2 ) cos ( B + C 2 ) ) 1 8 = 0 \Longrightarrow \dfrac{1}{2}\sin\left(\dfrac{A}{2}\right)\left(\cos\left(\dfrac{B - C}{2}\right) - \cos\left(\dfrac{B + C}{2}\right)\right) - \dfrac{1}{8} = 0 , (equation (i)).

Now since A = π ( B + C ) A = \pi - (B + C) we see that cos ( B + C 2 ) = cos ( π 2 A 2 ) = sin ( A 2 ) \cos\left(\dfrac{B + C}{2}\right) = \cos\left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin\left(\dfrac{A}{2}\right) .

With θ = B C 2 \theta = \dfrac{B - C}{2} , equation (i) then becomes

4 sin 2 ( A 2 ) 4 cos ( θ ) sin ( A 2 ) + 1 = 0 4\sin^{2}\left(\dfrac{A}{2}\right) - 4\cos(\theta)\sin\left(\dfrac{A}{2}\right) + 1 = 0 ,

which is a quadratic in sin ( A 2 ) \sin\left(\dfrac{A}{2}\right) , and thus

sin ( A 2 ) = 4 cos ( θ ) ± 16 cos 2 ( θ ) 16 8 = cos ( θ ) ± cos 2 ( θ ) 1 2 \sin\left(\dfrac{A}{2}\right) = \dfrac{4\cos(\theta) \pm \sqrt{16\cos^{2}(\theta) - 16}}{8} = \dfrac{\cos(\theta) \pm \sqrt{\cos^{2}(\theta) - 1}}{2} .

Now sin ( A 2 ) \sin\left(\dfrac{A}{2}\right) is real, so we must have that cos 2 ( θ ) 1 0 \cos^{2}(\theta) - 1 \ge 0 . But as the range of the cosine function is [ 1 , 1 ] [-1,1] we must have either cos ( θ ) = 1 θ = π \cos(\theta) = -1 \Longrightarrow \theta = \pi or cos ( θ ) = 1 θ = 0 \cos(\theta) = 1 \Longrightarrow \theta = 0 .

But as θ = B C 2 \theta = \dfrac{B - C}{2} the only possible option is that θ = 0 \theta = 0 , i.e., that B = C B = C . This in turn implies that sin ( A 2 ) = cos ( 0 ) 2 = 1 2 A = π 3 \sin\left(\dfrac{A}{2}\right) = \dfrac{\cos(0)}{2} = \dfrac{1}{2} \Longrightarrow A = \dfrac{\pi}{3} radians.

But then since A + B + C = A + 2 B = π A + B + C = A + 2B = \pi we see that B = C = π 3 B = C = \dfrac{\pi}{3} radians, and thus Δ A B C \Delta ABC is an equilateral triangle.

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