Western Illinois University mathematician Iraj Kalantari published an unusual puzzle in Math Horizons in February 2019. A sphere B of radius 150 is centered at (150, 150, 0). A sphere M of radius 144, centered on the z-axis, lies entirely below the (x, y)-plane so that the volume of its intersection with B is 1/2. “Can we find a sphere S of radius 73 that has its center on the circle (x – 73)² + (y – 73)² = 150² in the plane z = 73 so that the volume of B minus its intersections with M and S equals the volume of M minus its intersection with B plus the volume of S minus its intersection with B?”

The answer is no, because Vol(B – (M ∪ S)) = Vol((M – B) ∪ (S – B)) if and only if Vol(B) = Vol(M) + Vol(S), and that’s the case if and only if , where r_B, r_M, and r_S are the radii of the three spheres. “[A]nd because the radii are integers, this equality is impossible by Fermat’s last theorem!”

The placement of the spheres and the fact that the values differ by 1 are red herrings.

(Iraj Kalantari, “The Three Spheres,” Math Horizons 26:3 [February 2019]: 13, 25.)

02/28/2026 UPDATE: In my original statement of the problem I left out a vital phrase in Kalantari’s presentation: The sphere S should have its center on the circle (x – 73)² + (y – 73)² = 150² in the plane z = 73.

I’d omitted the last phrase, a condition that guarantees that S lies above the xy plane and so does not intersect sphere M, which is required to deduce the equation involving the volumes. Many thanks to reader Francesco Veneziano for pointing this out.

From Lee Sallows:

(Thanks, Lee!)

This figure contains four “cliques” of four points each, with each of the four points in each clique connected to each of the others, and each pair of cliques intersecting at a single point. Four colors suffice to color all the points so that no two linked points share a color.

Is this always possible? If k cliques, each containing k points, are arranged in similar fashion, can the result always be colored properly with k colors? In 2021, half a century after Paul Erdős first posed the question, Dong Yeap Kang and his colleagues proved that, for sufficiently large k, the conjecture is true.

The Wallace–Bolyai–Gerwien theorem, first proven in 1807, states that any two polygons of equal area must have a common dissection. That is, there’s always a way to cut up the first one and assemble the pieces to form the second.

But what if the pieces must be connected by hinges? In his “haberdasher” puzzle of 1907, Henry Dudeney showed that it’s possible to convert a triangle into a square by cutting it in pieces and turning it “inside out”:

Is it always possible to arrange such a “hinged dissection” between two polygons of equal area? The question remained open until 2007, when Erik Demaine showed that the answer is yes — and provided an algorithm to find it.

02/25/2026 UPDATE: Reader Simon Schneider directed me to this interactive visualization of the Wallace–Bolyai–Gerwien theorem, which lets you draw two polygons and then converts one to the other before your eyes. It’s hypnotizing. Here’s a paper on the dissection algorithm used. (Thanks, Simon.)

The numbers 1-7 are disposed among the regions in this figure such that each of the circular sets yields the same sum. This makes it a “magic Venn diagram,” a concept that occurred to mathematician David Robinson while teaching a course in mathematical logic at the University of West Georgia. His article appears in the December 2025 issue of Recreational Mathematics Magazine.

(David Robinson and Anja Remshagen, “Magic Venn Diagrams,” Recreational Mathematics Magazine 12:21 [December 2025], 25-44.)