## Triangles with multiple angles

### Triangles with multiple angles (4)

Notice that the critical points of $$\mathcal{A}$$ don’t depend on $$\mathcal{P}$$ and the denominator of $$\mathcal{A}$$ is positive at $$\left]0,\frac{\pi}{n+1}\right[.$$ Besides, $$\sin((n+1)\frac{\pi}{n+1})=0$$ and $\begin{array}{ccl} lim_{\alpha\mapsto0} & & \sin(\alpha)=0\\ lim_{\alpha\mapsto0} & & \frac{\sin(n\alpha)}{\sin(\alpha)+\sin(n\alpha)+\sin((n+1)\alpha)}\\ & = & lim_{\alpha\mapsto0}\frac{n\cos(n\alpha)}{\cos(\alpha)+n\cos(n\alpha)+(n+1)\cos((n+1)\alpha)}\\ & = & \frac{n}{2n+2}\\ \\ lim_{\alpha\mapsto0} & & \frac{\sin((n+1)\alpha)}{\sin(\alpha)+\sin(n\alpha)+\sin((n+1)\alpha)}\\ & = & lim_{\alpha\mapsto0}\frac{(n+1)\cos((n+1)\alpha)}{\cos(\alpha)+n\cos(n\alpha)+(n+1)\cos((n+1)\alpha)}\\ & = & \frac{n+1}{2n+2}\\ & = & \frac{1}{2}. \end{array}$ Then $$lim_{\alpha\mapsto0^{+}}\mathcal{A}\left(\alpha\right)=0$$ and, therefore, the function $$\mathcal{A}$$ may be continuously extended to $$\left[0,\frac{\pi}{n+1}\right]$$ if we define $$\mathcal{A}\left(\frac{\pi}{n+1}\right)=\mathcal{A}\left(0\right)=0$$ although these extreme cases correspond only to degenerate triangles.

Since $$\mathcal{A}$$ is continuous and is $$0$$ at the endpoints of the domain, it attains a maximum value at a point in the interior of $$\left[0,\frac{\pi}{n+1}\right]$$. Since $$\mathcal{A}$$ is differentiable in this set, we can find the maximum value by searching the critical values of $$\mathcal{A}$$. However, the derivative of $$\mathcal{A}$$ is not simple, and that is why it is useful to stop and conjecture about which triangles of $$T_n$$ with perimeter $$\mathcal{P}$$ encompass the largest area. In the webpage webMathematica, we may draw the graph of the area function and its derivative for some particular choices of $$n$$. For instance, for $$n=2$$, it allows us to predict that there is only one non-isosceles triangle with maximum area: it has angle $$\angle A\thicksim41$$ degrees, which is strictly between $$\frac{\pi}{5}$$ and $$\frac{\pi}{4}.$$

To confirm this, it is simpler to use Héron’s formula, which expresses the area in terms of the lengths of the sides and the perimeter, and use Lagrange multipliers [2], taking advantage of the characterization $$F_{2} = 0$$ described before (see example below).

B = 2 A

Semiperimeter $$\mathcal{S} = 8$$
Maximum area for $$A\thicksim 41^{^{\circ}}$$ $$(a\thicksim4.22)$$

With fixed $$\mathcal{P}$$, we want to maximize the function $\mathcal{A}\left(a,b\right)=\sqrt{\mathcal{S}(\mathcal{S}-a)(\mathcal{S}-b)(a+b-\mathcal{S})}$ where $$\mathcal{S}=\frac{\mathcal{P}}{2}$$ in the set $\{(a,b)\in\left(\mathbb{R}^{+}\right)^{2}:g(a,b)=b^{2}-2a\mathcal{S}+ab=0\}.$

Since the two components of the gradient of $$g$$ $\begin{array}{ccccc} \frac{\delta g}{\delta a} & = & \frac{\delta F_{2}}{\delta a} & = & -2S+b\\ \frac{\delta g}{\delta b} & = & \frac{\delta F_{2}}{\delta b} & = & 2b+a \end{array}$ don’t vanish in$$\mathcal{T}_{2}$$ it is enough to find the elements $$(a,b)$$ where the gradients of $$f=\frac{\mathcal{A}^{2}}{S}$$ and $$g$$ Are collinear. This linear dependence means that the maximum area triangle corresponds to a tangent point between a level curve of $$f$$ with the level curve $$g\equiv0$$.

Next page