## 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.