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