Pythagorean Cuts – Semicircles: A Special Case

Let triangle ABC be a 30-60 right triangle with hypotenuse AC and shortest side AB \(= 1\) (see Figure 8, below). Let semicircles be constructed on each side of triangle ABC.

Figure 8: A special case of semicircles. The Pythagorean cut is semicircle LMK, shown in red.

The semicircle on AB has radius \(\frac{1}{2}\) and so has area \({{\frac{1}{2}}\pi\left({{\frac{1}{2}}}\right)^2,}\) or \(\frac{\pi}{8}.\)  Similarly, the area of the semicircle on side BC is \(\frac{3\pi}{8}\) and the area of the semicircle on hypotenuse AC is \(\frac{\pi}{2}\).  Since \(\frac{\pi}{8}+\frac{3\pi}{8}=\frac{\pi}{2},\) the Pythagorean relationship holds:

Area of semicircle on AB + Area of semicircle on BC = Area of semicircle on AC.

Now let LK be the segment of length 1 perpendicular to AC and let the semicircle be constructed having diameter LK as shown.  Since AB = LK, the semicircles on AB and on LK must have the same area of \(\frac{\pi}{8}.\) Moreover, since quarter circle AKL has area one-half the area of semicircle AKC, the area of quarter circle AKL is \(\frac{\pi}{4}.\)  Thus,

Area of region ALMK \(=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}\) = Area of semicircle on side AB,

and so semicircle LMK is a Pythagorean cut for semicircles built on the 30-60 right triangle. A GeoGebra applet showing the Pythagorean cut for semicircles with their diameters along the sides of this special right triangle is shown in Figure 9, below.

Figure 9. Semicircles with diameters along the sides of a 30-60 right triangle