What is the smallest side length BC of triangle ABC with fixed angle and area?

[maxkor]

Calculate what is the smallest side length BC of the triangle ABC if the angle BAC is equal alpha and area of the triangle ABC equals S.

 

[HOI]

Are we to assume that ABC is a right triangle?

 

[mrtwhs]

Calculate what is the smallest side length BC of the triangle ABC if the angle BAC is equal alpha and area of the triangle ABC equals S.

Suppose that angle $$\alpha$$ is fixed and that the area of $$\triangle ABC = S$$ is fixed. We wish to find the smallest value of side $$BC=a$$.

If the other two angles are $$\beta$$ and $$\gamma$$ then $$\alpha + \beta + \gamma = 180$$ and $$\gamma = 180 - (\alpha+\beta)$$.

The area of $$\triangle ABC$$ is $$S = \dfrac{1}{2}a^2 \cdot \dfrac{\sin \beta \sin \gamma}{\sin \alpha}$$.

So we have that $$a^2 = \dfrac{2S \sin \alpha}{\sin \beta \sin \gamma} = \dfrac{4S \sin \alpha}{2\sin \beta \sin (\alpha+\beta)}$$.

To minimize $$a$$ we must maximize $$2\sin \beta \sin (\alpha+\beta)$$.

$$y = 2\sin \beta \sin (\alpha+\beta) = \cos \alpha - \cos(\alpha+2\beta)$$.

Standard calculus yields a maximum when $$\alpha+2\beta=180$$. That is, when $$\beta=\gamma$$ and the triangle is isosceles.

Substituting back, the minimum value of $$BC = a$$ = $$\dfrac{\sqrt{2S \sin \alpha}}{\cos \left(\dfrac{\alpha}{2}\right)}$$.

... I think!