Author Archives: Jim Stevenson

Wisdom of Old

Here is another Brainteaser from the Quantum magazine.

“King Arthur ordered a pattern for his quarter-circle shield. He wanted it to be painted in three colors: yellow, the color of kindness; red, the color of courage: and blue the color of wisdom. When the artist brought in his work, the king’s armor-bearer said there was more courage than wisdom on the shield. But the artist managed to prove that the proportions of both virtues were equal. Can you tell how? (A. Savin)”

This is another relatively simple problem, though it may look a bit daunting at first.

See Wisdom of Old

Alcuin’s Corn Problem

Alcuin of York (735-804) had a series of similar problems involving the distribution of corn among servants.  Since the three propositions were the same format with only the numbers changing, I thought I would present them in a more concise form:


A certain head of household had a number of servants, consisting of men, women, and children, among whom he wished to distribute quantities, modia, of corn.  The men should receive three modia; the women, two; and the children, half a modium.

(a)  If the head of household has 20 servants and wished to distribute 20 modia of corn among them, let him say, he who can, How many men, women and children must there have been.

(b)  If the head of household has 30 servants and wished to distribute 30 modia of corn among them, let him say, he who can, How many men, women and children must there have been.

(c)  If the head of household has 100 servants and wished to distribute 100 modia of corn among them, let him say, he who can, How many men, women and children must there have been.”

I will give Alcuin’s solutions first, followed by my more expansive solutions that rely on our familiar symbolic algebra that was not available in Alcuin’s time.

See Alcuin’s Corn Problem

The Staircase Race

This is a classic type of puzzle from Henry Dudeney.

“This is a rough sketch of the finish of a race up a staircase in which three men took part. Ackworth, who is leading, went up three steps at a time, as arranged; Barnden, the second man, went four steps at a time, and Croft, who is last, went five at a time. Undoubtedly Ackworth wins. But the point is, how many steps are there in the stairs, counting the top landing as a step?

I have only shown the top of the stairs. There may be scores, or hundreds, of steps below the line. It was not necessary to draw them, as I only wanted to show the finish. But it is possible to tell from the evidence the fewest possible steps in that staircase. Can you do it?”

See the Staircase Race

Triangle Stripes Problem

This is a fairly straight-forward problem from Presh Talwalkar.

“A triangle is divided by 8 parallel lines that are equally spaced, as shown below. Starting from the top small triangle, color each alternate stripe in blue and color the remaining stripes in red. If the blue stripes have a total area of 145, what is the total area of the red stripes?”

See the Triangle Stripes Problem

Triangle Projection Problem

This is a Maths Item of the Month (MIOM) problem that seems opaque at first.  (“The Maths Item of the Month is a monthly problem aimed at teachers and students of GCSE and A level Mathematics.”)

“Two fixed circles, C1 and C2, intersect at A and BP is on C1PA and PB produced meet C2 at A’ and B’ respectively.  How does the length of the chord A’B’ change as P moves?”

Just start noticing relationships and the answer falls out nicely.

(MIOM problems often appear on MathsMonday and are also produced by Mathematics Education Innovation (MEI).)

See the Triangle Projection Problem

Broken Diagonal Problem

This is a nice problem from the UKMT Senior Mathematics Challenge for 2022:

“Five line segments of length 2, 2, 2, 1 and 3 connect two corners of a square as shown in the diagram. What is the shaded area?

A 8____B 9____C 10____D 11____E 12”

The pleasure of solving this problem may be lessened if one is under a time crunch, as is the case with all these timed tests.

See the Broken Diagonal Problem

Road Construction Problem

This is an interesting problem from the Scottish Mathematics Council (SMC) 2014 Senior  Math Challenge .

“Two straight sections of a road, each running from east to west, and located as shown, are to be joined smoothly by a new roadway consisting of arcs of two circles of equal radius. The existing roads are to be tangents at the joins and the arcs themselves are to have a common tangent where they meet.  Find the length of the radius of these arcs.”

See the Road Construction Problem

Falling Sound Problem

This math problem from Colin Hughes’s Maths Challenge website ( hearkens back to basic physics.

“A boy drops a stone down a well and hears the splash from the bottom after three seconds.  Given that sound travels at a constant speed of 300 m/s and the acceleration of the stone due to gravity is 10 m/s2, how deep is the well?”

See the Falling Sound Problem

Neuberg’s Theorem

This turned out to be a challenging puzzle from the 1980 Canadian Math Society’s magazine, Crux Mathematicorum.

Proposed by Leon Bankoff, Los Angeles, California.

Professor Euclide Paracelso Bombasto Umbugio has once again retired to his tour d’ivoire where he is now delving into the supersophisticated intricacies of the works of Grassmann, as elucidated by Forder’s Calculus of Extension. His goal is to prove Neuberg’s Theorem:

If D, E, F are the centers of squares described externally on the sides of a triangle ABC, then the midpoints of these sides are the centers of squares described internally on the sides of triangle DEF.  [The accompanying diagram shows only one internally described square.]

Help the dedicated professor emerge from his self-imposed confinement and enjoy the thrill of hyperventilation by showing how to solve his problem using only highschool, synthetic, Euclidean, ‘plain’ geometry.”

Alas, my plane geometry capability was inadequate to solve the puzzle that way, so I had to resort to the sledge hammer of analytic geometry, trigonometry, and complex variables.

See Neuberg’s Theorem