## Morley's Theorem

### Areas

Given an initial parallelogram with sides $$a$$ and $$b$$ forming an angle that measures $$\alpha \in ]0,\pi[$$, its area is $$ab \sin \alpha$$. Regarding the area of Morley’s parallelogram, its value is not determined by the area of the initial parallelogram, similar to what happens with Morley’s triangle. However, its value is also a function of $$a$$, $$b$$ and $$\alpha$$. In fact, it can be proved that its value is given by : $\begin{array}{lll} A & = & \left|\frac{2}{3}ab\sin\alpha-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\right|\\ & = & \frac{2}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right| \end{array}$

Note that we may have $$A=0$$, which in this case means that the intersection points of the angle trisectors are collinear, not being the vertices of a parallelogram. Let us see when this happens: $\begin{array}{lll} A=0 & \Leftrightarrow & \sin\frac{\alpha}{3}=0\lor\sin\frac{\pi-\alpha}{3}=0\lor\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right|=0\\ & \Leftrightarrow & a^{2}+b^{2}=\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3} \end{array}$

Taking $$t=\frac{2}{\sqrt{3}} \sin \frac{\alpha + \pi}{3}$$, we have: $\begin{array}{lll} A=0 & \Leftrightarrow & \left(\frac{b}{a}\right)^{2}-2t\left(\frac{b}{a}\right)+1=0\\ & \Leftrightarrow & \frac{b}{a}=t\pm\sqrt{t^{2}-1}\\ & \Leftrightarrow & b=b_{1}=a\left(t-\sqrt{t^{2}-1}\right)\lor b=b_{2}=a\left(t+\sqrt{t^{2}-1}\right) \end{array}$

In relative terms, the ratio between the areas is: $\begin{array}{lll} r & = & \frac{\left|\frac{2}{3}ab\sin\alpha-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\right|}{ab\sin\alpha}\\ & = & \left|\frac{2}{3}-\frac{2\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}}{\sqrt{3}ab\sin\alpha}\right|\\ & = & \left|\frac{2}{3}-\frac{2\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}}{4\sqrt{3}ab\sin\frac{\alpha}{3}\sin\frac{\alpha+\pi}{3}\sin\frac{\alpha+2\pi}{3}}\right|\\ & = & \left|\frac{2}{3}-\frac{a^{2}+b^{2}}{2\sqrt{3}ab\sin\frac{\alpha+\pi}{3}}\right| \end{array}$

Let us now fix the length $$a$$ and the measure of the angle $$\alpha$$. Unlike what happened with the triangles, the ratio of the areas, $$r$$, has no absolute maximum (the analysis of the given formula shows that when $$b$$ tends to infinity, so does $$r$$). However, there is a local maximum when $$b=a$$, that is, when the initial parallelogram is a rhombus. Note that this local maximum is the geometric average between the values of $$b_{1}$$ and $$b_{2}$$, since $\begin{array}{lll} \sqrt{b_{1}b_{2}} & = & \sqrt{a^{2}\left(t-\sqrt{t^{2}-1}\right)\left(t+\sqrt{t^{2}-1}\right)}\\ & = & a\sqrt{t^{2}-\left(t^{2}-1\right)}\\ & = & a \end{array}$

In fact, there is also a local maximum for the absolute value of the area of the parallelogram obtained by the intersection of the adjacent angle trisectors, $$A$$, which is reached when $$b=at$$, and this value is the arithmetic mean between $$b_{1}$$ and $$b_{2}$$, since $\begin{array}{lll} \frac{b_{1}+b_{2}}{2} & = & \frac{a\left(t-\sqrt{t^{2}-1}\right)+a\left(t+\sqrt{t^{2}-1}\right)}{2}\\ & = & \frac{2at}{2}\\ & = & at \end{array}$

Since the geometric mean of two positive numbers is always less then its arithmetic mean, and both lie between the two initial values, then we have the following inequality: $b_{1}<a<at<b_{2}$

Therefore, the local maximum of $$r$$ is reached before the local maximum of $$A$$, and both occur between the two values for which the area is zero. Confirm that this happens with the following interactive applet: :

Click on the red points to change the shape of the initial parallelogram and on the white rectangles to see the corresponding situation