In a regular pentagon, all diagonals are drawn, as shown. Label each vertex of the pentagon and each intersection of the diagonals with the number 1. Now: In one step you can change the signs of all the numbers on a side or on a diagonal. Is it possible, by a sequence of such steps, to convert all the labels in the diagram to -1?
No. At the start, the number of -1s on the pentagon’s perimeter is 0, an even number. Each step will change two of these numbers, so their quantity will always be even. Our goal would make it odd. So the task is impossible.
From Arthur Engel’s Problem-Solving Strategies, 2003.
The names of 13 Jane Austen characters are hidden in the following lines as anagrams of complete consecutive words. For example, “was ill” yields WALLIS. (The names to be found are women’s first names and men’s surnames, as in Austen.) In most cases the anagrams are hidden in two words, but twice they’re in three, once in four, and once in a single word. What are they?
The other day when I was ill
And not a soul I knew came nigh,
Jane Austen was my daily fare —
I rather liked to be laid by.
Each line or page enthralls me quite,
I there can let no man deride;
I may be ill as a wight can be,
But, Jane with me, am satisfied.
In bed my ease is nil, yet I’ll
Be lying therein at any rate
Content. With Jane to chortle at
How can I rail at Fate?
A problem from the qualification round of the 2004/2005 Swedish Mathematical Contest, via the April 2009 issue of Crux Mathematicorum:
The cities A, B, C, D, and E are connected by straight roads (more than two cities may lie on the same road). The distance from A to B, and from C to D, is 3 km. The distance from B to D is 1 km, from A to C it is 5 km, from D to E it is 4 km, and finally, from A to E it is 8 km. Determine the distance from C to E.
This solution is by Titu Zvonaru of Comănești, Romania. The sum of the distances AB, BD, and DE (3 + 1 + 4) equals the distance AE (8), so cities A, B, D, and E lie on the same straight road. The square of distance AC (25) equals the sum of the squares of CD and AD (9 + 16), so by the converse of the Pythagorean theorem CD is perpendicular to AD. This means that CD and DE are the legs of a right triangle, and Pythagoras shows that hypotenuse CE is 5 km.
Note that the king can’t choose any other square. The eighth rank must remain free for the queen; e7 and f6 are vulnerable to knight checks; 1. Ke6 would block the queen’s check in the second variation; and 1. Kg6 would block the bishop’s diagonal in the first and second.
This divides the plane into three regions, here colored red, yellow, and blue. Placing the third house into any of these regions denies it access to the correspondingly colored utility. So the task is impossible.
Pleasingly, the task can be accomplished on a Möbius strip:
08/31/2025 UPDATE: Reader Guy Bolton King points out that Mathsgear sells a mug embossed with the puzzle. The joke here is that this makes the puzzle solvable — like the torus, the mug is of topological genus 1, “a blob with 1 hole in it,” so it admits the same solution.
And reader Shane Speck writes, “My sneaky solution … has always hinged on the fact that in reality, houses don’t all have separate pipes, and popping on a shared water pipe instantly reduces the problem to the status of incredibly trivial”:
“The problem doesn’t, after all, say you can’t do that… :)”