Triangles with multiple angles 
Triangles with multiple angles (3)
The geometric argument used before may be repeated for other values of \(n\), although the relation between sides becomes more and more intricate. From this inductive process it follows that, if \(F_{n}(a,b,c)=0\) denotes the relation between sides, of lengths \(a\), \(b\) and \(c\), of a triangle \(\triangle ABC\) such that \(\angle B = n \angle A\), then: \[\begin{array}{ccl} F_{1}(a,b,c) & := & b-a\\ F_{2}(a,b,c) & := & b^{2}-a^{2}-ac\\ F_{3}(a,b,c) & := & \left(b^{2}-a^{2}+c^{2}\right)\left(b-a\right)-bc^{2}\\ & = & \left(b^{2}-a^{2}+c^{2}\right)F_{1}(a,b,c)-bc^{2}\\ F_{4}(a,b,c) & := & \left(b^{2}-a^{2}+c^{2}\right)\left(b^{2}-a^{2}-ac\right)-b^{2}c^{2}\\ & = & \left(b^{2}-a^{2}+c^{2}\right)F_{2}(a,b,c)-b^{2}c^{2}\\ F_{5}(a,b,c) & := & \left(b^{2}-a^{2}+c^{2}\right)F_{3}(a,b,c)-b^{2}c^{2}F_{1}(a,b,c) \end{array}\] and, more generally, for all natural number \(n\geq4,\) \[F_{n+1}(a,b,c):=\left(b^{2}-a^{2}+c^{2}\right)F_{n-1}(a,b,c)-b^{2}c^{2}F_{n-3}(a,b,c).\] Reciprocally, if \(b>a\) and \(F_{n}(a,b,c)=0\), then \( \angle B = n \angle A\).
Then in the family of all triangles with perimeter \(\mathcal{P}>0\), there is one that stands out: the one with that perimeter which is equilateral, because it is the one with the largest area. In fact, by Heron’s formula [3], the area of a triangle with side lengths \(a\), \(b\) and \(c\) such that \(a + b + c = \mathcal{P}\), is given by \[\mathcal{A}\left(a,b,c\right)=\sqrt{\mathcal{S}(\mathcal{S}-a)(\mathcal{S}-b)(\mathcal{S}-c)}\] where \(\mathcal{S}=\frac{\mathcal{P}}{2}\). Therefore, by the inequality between arithmetic and geometric means [3] applied to \(\mathcal{S}-a,\mathcal{S}-b\) and \(\mathcal{S}-c\) whose sum is \(\mathcal{S}\), we deduce \[(\mathcal{S}-a)(\mathcal{S}-b)(\mathcal{S}-c)\leq\left[\frac{(\mathcal{S}-a)+(\mathcal{S}-b)+(\mathcal{S}-c)}{3}\right]^{3}\] and the equality holds if and only if the three numbers are equal. That is, \[\mathcal{A}\left(a,b,c\right)\leq\frac{\mathcal{S}^{2}}{3\sqrt{3}},\] and the maximum value is reached if and only if \(a=b=c=\frac{\mathcal{P}}{3}.\)
For \(n \geq 1\), let \(\mathcal{T}_{n}\) denote the set of triangles \(\triangle ABC\) such that \(\angle B =n \angle A\). When \(n \geq 2\), the equilateral triangle doesn’t belong to \(\mathcal{T}_{n}\), and there are only two isosceles triangles in this set:
- triangle \(T_{1,n}\), with angles \(\frac{\pi}{n+2},\frac{n\pi}{n+2},\frac{\pi}{n+2},\)
- triangle \(T_{2,n}\), with angles \(\frac{\pi}{2n+1},\frac{n\pi}{2n+1},\frac{n\pi}{2n+1}.\)
What is the solution, if any, to the isoperimetric problem in \(\mathcal{T}_{n}\)? By the law of sines, the area of a triangle in \(\mathcal{T}_{n}\) with perimeter \(\mathcal{P}\) is given by \[\mathcal{A}\left(\angle A\right)=\frac{\mathcal{P}^{2}}{2}\frac{\sin\left(\angle A\right)\sin\left(n\angle A\right)\sin\left(\left(n+1\right)\angle A\right)}{[\sin\left(\angle A\right)\sin\left(n\angle A\right)\sin\left(\left(n+1\right)\angle A\right)]^{2}}\] where \(\angle A\in\left]0,\frac{\pi}{n+1}\right[.\)
