Morley's Theorem 
Proof
The area of a triangle is given by \(A=\frac{\mbox{Base}\times\mbox{Height}}{2}\). If the triangle is equilateral, that is, if all sides have the same length \(l\), then the base is \(l\) and the height is \(\frac{\sqrt{3}}{2}l\), whereby \(A=\frac{\sqrt{3}}{4}l^{2}\). In the case of Morley’s triangle, we saw that \(l=8R\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{c}{3}\), then \[A=\frac{\sqrt{3}}{4}\left(8R\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{c}{3}\right)^{2}=\\ =16\sqrt{3}R^{2}\sin^{2}\frac{a}{3}\sin^{2}\frac{b}{3}\sin^{2}\frac{c}{3}.\]
To calculate the area of the original triangle, we will use the law of sines. Calling \(R\) the circumradius of the triangle \([ABC]\), their sides are \(2R\sin a\), \(2R \sin b\) and \(2R \sin c\). Taking the base as \(2R \sin a\), then the height is \(2R \sin b \sin c\), whereby \[A=\frac{2R \sin a \cdot 2R \sin b \sin c}{2}=\\ =2R^2 \sin a \sin b \sin c.\]
The ratio of the areas is therefore equal to: \[r=\frac{16\sqrt{3}R^{2}\sin^{2}\frac{a}{3}\sin^{2}\frac{b}{3}\sin^{2}\frac{c}{3}}{2R^2 \sin a \sin b \sin c}=\\= 8\sqrt{3} \frac{\sin^{2}\frac{a}{3}\sin^{2}\frac{b}{3}\sin^{2}\frac{c}{3}}{\sin a \sin b \sin c}\]
Considering the function \[f(x,y,z)= 8\sqrt{3} \frac{\sin^{2}\frac{x}{3}\sin^{2}\frac{y}{3}\sin^{2}\frac{z}{3}}{\sin x \sin y \sin z}\] restricted to the values \(x\), \(y\) and \(z\) such that \(x,y,z>0\) and \(x+y+z=\pi\), we find, using the method of Lagrange multipliers, that this function has a critical point \((a,b,c)\) when \(g(a)=g(b)=g(c)\), with \(g(x)=2\cot\frac{x}{3}-3\cot x\). The function \(g(x)\) defined in the interval \(]0,\pi[\) is not injective but, however, is continuous and differentiable, and only has a critical point. Then the points \(a\), \(b\) and \(c\) cannot all be different (otherwise, the function \(g(x)\) would have at least two different critical points, which is not true). Assuming, for example, that \(a=b\), then \(c=\pi-a-b=\pi-2a>0\) whereby \(0<a<\frac{\pi }{2}\) and we have \(g(a)=g(\pi -2a)\), being \(\frac{\pi }{3}\) the single zero of the function \(g(x)-g(\pi -2x)\) in the interval \(]0,\frac{\pi }{2}[\).
It is therefore proved that \(a=b=c=\frac{\pi}{3}\), that is, the initial triangle is equilateral, and the maximum value of the ration between the areas is: \[\begin{array}{lll} f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right) & = & 8\sqrt{3}\frac{\left(\sin^{2}\frac{\pi}{9}\right)^{3}}{\left(\sin\frac{\pi}{3}\right)^{3}}\\ & = & 8\sqrt{3}\frac{\sin^{6}\frac{\pi}{9}}{\left(\frac{\sqrt{3}}{2}\right)^{3}}\\ & = & \frac{64}{3}\sin^{6}\frac{\pi}{9} \end{array}\]
To calculate the maximum of the ratio of the perimeters, just note that the perimeter of Morley’s triangle is given by \(3l=24R\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{c}{3}\), while the perimeter of the initial triangle is \(8R\cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}\).
The ratio of the perimeters is therefore given by: \[r=\frac{24R\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{c}{3}}{8R\cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}}=\frac{3\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{c}{3}}{\cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}}\]
Considering the function \[f(x,y,z)=\frac{3\sin\frac{x}{3}\sin\frac{y}{3}\sin\frac{z}{3}}{\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}}\] restricted to the values of \(x\), \(y\) and \(z\) such that \(x,y,z>0\) and \(x+y+z=\pi\), we find, using the method of Lagrange multipliers, that this function has a critical point \((a,b,c)\) when \(g(a)=g(b)=g(c)\), with \(g(x)=2 \cot \frac{x}{3}+3\tan \frac{x}{2}\). The function \(g(x)\) definided in the interval \(]0,\pi [\) is not injective but, however, is continuous and differentiable, and only has a critical point. Then the points \(a\), \(b\) and \(c\) cannot all be different (otherwise, the function \(g(x)\) would have at least two different critical points, which is not true). Assuming, for example, that \(a=b\), then \(c=\pi -a-b= \pi - 2a >0\) whereby \(0<a<\frac{\pi }{2}\) and we have \(g(a)=g(\pi -2a)\), being \(\frac{\pi}{3}\) the single zero of the function \(g(x)-g(\pi -2x)\) in the interval \(]0,\frac{\pi }{2}[\).
It is therefore proved that \(a=b=c=\frac{\pi}{3}\), that is, the initial triangle is equilateral, and the maximum value of the ration between the perimeters is: \[\begin{array}{lll} f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right) & = & \frac{3\left(\sin \frac{\pi}{9}\right)^{3}}{\left(\cos\frac{\pi}{6}\right)^{3}}\\ & = & \frac{3 \sin^{3}\frac{\pi}{9}}{\left(\frac{\sqrt{3}}{2}\right)^{3}}\\ & = & \frac{8}{\sqrt{3}}\sin^{3}\frac{\pi}{9} \end{array}\]
Note that \(\frac{8}{\sqrt{3}} \sin^{3} \frac{\pi}{9}\) is the similarity ratio of Morley's triangle and the inital triangle, whereby the ratio of their areas is \(\left(\frac{8}{\sqrt{3}} \sin^{3} \frac{\pi}{9}\right)^{2}=\frac{64}{3}\sin^{6}\frac{\pi}{9}\), as we proved above.
