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 $$\angle A\in\left]0,\frac{\pi}{n+1}\right[.$$

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