Triangles with multiple angles Resize


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 this app, 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).

\(\angle B = 2 \, \angle A\)

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 \(g\equiv0\).

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