Loony Lunes

Here are the specifications for the lunes, from left to right, in the form $(R, r, B)$ , where $R$ is the radius of the top arc, $r$ is the radius of the bottom arc, and $B$ is the length of the base.

$(108,117, 216)$
$(80,100,160)$
$(75, 100,120)$
$(75, 117, 90)$

Here's how it's done. We consider 3 pairs of "Lunes of Alhazen", in blue, green, and red. The red overlaps the smaller of the blue and green lunes with matching semi-circle top arcs. The total of the each of colored pairs of Lunes of Alhazen is an integer, so by subtracting the red from combined blue and green areas, we're left with an integer area for the four blue lunes shown above. The specifications for the blue, green, red Lunes of Alhazen pairs are

$(108, 117, 216)$
$(45, 117, 90)$

$(80, 100, 160)$
$(60, 100, 120)$

$(60, 75, 120)$
$(45, 75, 90)$

Notice that $(45, 108, 117), \; (60, 80, 100), \; (45, 60, 75)$ are all Pythagorean triples, which can be used to find the radii of the red pair, as well as the minor blue and green lunes. That is, if one has a good hunch, given that radii $(108, 117)$ and $(80, 100)$ are known or surmised for the first two lunes on the left.

A pair of Lunes of Alhazen always have semi-circle top arcs, and the total area for such a pair is half of the product of the bases. So, for blue, green, and red pairs, the areas are $(9720, 9600, 5400)$ , so the sum of the blue and green pairs minus the red comes out to $13920$ .

There exists no lune with integer radii and base that has a rational area. The Lune of Hippocrates has one irrational radius, a square root.

The Lunes of Alhazen explained

Given a right triangle of sides $a, b, c$ , draw a blue circle with diameter $c$ , and two semicircles with diameters $a, b$ . Let $A, B, T$ be the areas of the two lunes formed, and the area of the right triangle. Then it can be seen that the sum of the areas of the two semicircles plus the right triangle is equal to the areas of the two lunes plus the blue semicircle. Hence, the combined area of the two lunes $A, B$ is equal to

$A+B=\pi (\dfrac{a}{2})^2 +\pi (\dfrac{b}{2})^2 + \dfrac{1}{2}ab -\pi (\dfrac{c}{2})^2 = \dfrac{1}{2}ab = T$