The ordinary tetrahedron, or triangular pyramid, has no diagonals — every pair of vertices is joined by an edge. How many other polyhedra have this feature? In 1949, Hungarian topologist Ákos Császár found the specimen above, which has 7 vertices, 14 faces, and 21 edges.

But so far these two are the only residents in this particular zoo. The next possible such creature would have 44 faces and 66 edges, but this isn’t realizable as a polyhedron. Whether there’s anything beyond that is not known.

Suppose you hire two proofreaders to go through the same manuscript independently. The first reports A mistakes, the second reports B mistakes, and C mistakes are reported by both. How can you estimate how many errors remain undiscovered?

Let M be the total number of mistakes in the manuscript. Then the number undiscovered by the two proofreaders is M – (A + B – C). Let p and q be the probabilities that the first and second proofreaders, respectively, notice any given mistake. Then A ≈ pM and B ≈ qM. And because they work independently, the chance that they both find a given mistake is C ≈ pqM.

But now

and the number of misprints that remain unnoticed is just

This means that as long as the proofreaders work independently, you can estimate the number of errors they’ve overlooked without even knowing how skillful they are. If they find a large number of mistakes in common but relatively few independently, then the manuscript is probably relatively clean. But if they generate large independent lists of errors with few in common, there are probably many mistakes remaining to be found (which matches our intuition).

(George Pólya, “Probabilities in Proofreading,” American Mathematical Monthly 83:1 [January 1976], 42.)

If an equilateral triangle is inscribed in, and has a common vertex with, a rectangle, as shown above, then areas A + B = C.

If a triangle with angles α, β, γ is inscribed in, and has a common vertex with, a rectangle, as shown below, and if the right triangles opposite α, β, γ have areas A, B, C, respectively, then A cot α + B cot β = C cot γ.

Somewhat related: A Curious Equality.

(Tom M. Apostol and Mamikon Mnatsakanian, “Triangles in Rectangles,” Math Horizons 5:3 [February 1998], 29-31.)

Humans are bad at evaluating complex data, but we’re good at reading faces. So in 1973 Stanford statistician Herman Chernoff proposed using cartoon faces to encode information. He found that up to 18 different data dimensions can be represented in a computer-drawn face, mapping one variable to the length of the nose, another to the space between the eyes or the position of the mouth, and so on. This produces an array of faces that we can assess quickly using the brain’s natural talent for reading features. (The example above shows lawyers’ ratings of state judges in U.S. Superior Court.)

“This approach is an amusing reversal of a common one in artificial intelligence,” Chernoff noted. “Instead of using machines to discriminate between human faces by reducing them to numbers, we discriminate between numbers by using the machine to do the brute labor of drawing faces and leaving the intelligence to the humans, who are still more flexible and clever.”

(Herman Chernoff, “The Use of Faces to Represent Points in K-Dimensional Space Graphically,” Journal of the American Statistical Association 68:342 [June 1973], 361-368.)

Nepal’s constitution contains complete instructions for drawing its flag:

(A) Method of Making the Shape Inside the Border

(1) On the lower portion of a crimson cloth draw a line AB of the required length from left to right.
(2) From A draw a line AC perpendicular to AB making AC equal to AB plus one third AB. From AC mark off D making line AD equal to line AB. Join BD.
(3) From BD mark off E making BE equal to AB.
(4) Touching E draw a line FG, starting from the point F on line AC, parallel to AB to the right hand-side. Mark off FG equal to AB.
(5) Join CG.

(B) Method of Making the Moon

(6) From AB mark off AH making AH equal to one-fourth of line AB and starting from H draw a line HI parallel to line AC touching line CG at point I.
(7) Bisect CF at J and draw a line JK parallel to AB touching CG at point K.
(8) Let L be the point where lines JK and HI cut one another.
(9) Join JG.
(10) Let M be the point where line JG and HI cut one another.
(11) With centre M and with a distance shortest from M to BD mark off N on the lower portion of line HI.
(12) Touching M and starting from O, a point on AC, draw a line from left to right parallel to AB.
(13) With centre L and radius LN draw a semi-circle on the lower portion and let P and Q be the points where it touches the line OM respectively.
(14) With centre M and radius MQ draw a semi-circle on the lower portion touching P and Q.
(15) With centre N and radius NM draw an arc touching PNQ [sic] at R and S. Join RS. Let T be the point where RS and HI cut one another.
(16) With Centre T and radius TS draw a semi-circle on the upper portion of PNQ touching it at two points.
(17) With centre T and radius TM draw an arc on the upper portion of PNQ touching at two points.
(18) Eight equal and similar triangles of the moon are to be made in the space lying inside the semi-circle of No. (16) and outside the arc of No. (17) of this Schedule.

(C) Method of Making the Sun

(19) Bisect line AF at U and draw a line UV parallel to line AB touching line BE at V.
(20) With centre W, the point where HI and UV cut one another and radius MN draw a circle.
(21) With centre W and radius LN draw a circle
(22) Twelve equal and similar triangles of the sun are to be made in the space enclosed by the circles of No. (20) and of No. (21) with the two apexes of two triangles touching line HI.

(D) Method of Making the Border

(23) The width of the border will be equal to the width TN. This will be of deep blue colour and will be provided on all the sides of the flag. However, on the five angles of the flag the external angles will be equal to the internal angles.
(24) The above mentioned border will be provided if the flag is to be used with a rope. On the other hand, if it is to be hoisted on a pole, the hole on the border on the side AC can be extended according to requirements.

Explanation: The lines HI, RS, FE, ED, JG, OQ, JK and UV are imaginary. Similarly, the external and internal circles of the sun and the other arcs except the crescent moon are also imaginary. These are not shown on the flag.

That’s a good thing — it’s the only national flag that’s not a quadrilateral. The two pennants represent different branches of a ruling dynasty in the 19th century. The nation signaled its pride in the new design last February by setting a world record for the largest human flag — 35,000 Nepalese gathered in Kathmandu to break Pakistan’s record and to demonstrate their own national unity. I wonder how they worked out the geometry:

In the 17th century, Italian mathematician Evangelista Torricelli experimented with a figure known as Gabriel’s Horn. Rotate the function y = 1/x about the x-axis for x ≥ 1. The resulting figure has finite volume but infinite surface area — it’s sometimes said that, while the horn could be filled up with π cubic units of paint, an infinite number of square units of paint would be needed to cover its surface.

English cosmologist John D. Barrow describes an infinite wedding cake in which each tier is a solid cylinder 1 unit high; the bottom tier has radius 1, the second radius 1/2, the third radius 1/3, and so on. Now the total volume of the cake is π³/6, but the area of its surface is infinite. Barrow writes, “Our infinite cake recipe requires a finite volume of cake to make but it can never be iced because it has an infinite surface area!”

Mike Steuben, a correspondent of Martin Gardner, imagined a set of boxes, each with area 1 × 1. If the height of the first box is 1, the second 1/2, the third 1/4, and so on, then the total volume of the group is 2 cubic units, but the length and the total area of the tops are infinite.

(Barrow’s example is from 100 Essential Things You Didn’t Know You Didn’t Know About Math and the Arts, 2014.)