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 \(\displaystyle \alpha\) is fixed and that the area of \(\displaystyle \triangle ABC = S\) is fixed. We wish to find the smallest value of side \(\displaystyle BC=a\).

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

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

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

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

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

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

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

... I think!