It is a truism to say that once you know the answer to a puzzle it becomes trivially easy. Yet this statement is especially true when it comes to geometrical problems, which are often solved with a single insight that, once you know it, seems so forehead-slappingly obvious.
Both of today’s two puzzles are unravelled with a joyous ‘aha’. I hope you find them for yourselves.
1. A 12cm x 12cm square piece of paper is marked as below.
If you fold along the dotted lines and position the white triangles upwards so they touch, you get a pyramid with height h.
What is h?
2. The second problem involves the equilateral triangle below. It has side length 6cm, and is cut by two straight lines through the triangle’s centre. One of them cuts the bottom edge 1cm along, and they intersect at a 60 degree angle.
What is the ratio of the total area of the two red regions to the total area of the two blue regions?
Both of today’s puzzles were written by Manfred Pietsch, a retired maths teacher from Düren in Germany. Danke!
NO SPOILERS. I’ll be back at 5pm UK with the solutions.
UPDATE: Read the solutions here.
Before I go, two updates on the locks problem from two columns ago.
Here’s the original problem:
The three directors of a bank are deeply suspicious of one another, and agree a system of locks and keys for the bank’s safe, such that:
No single director can open the safe alone.
Any two directors can open the safe by pooling their keys.
What is the smallest number of locks and keys they need to open the safe, and how do they distribute them?
My answer was three locks and six keys (two per lock). If the locks are A, B and C, then the keys need to be distributed as follows: A and B to one director, B and C to another, and A and C to the third. In this way, no single director can open the safe, but any combination of two directors can.
Last column I featured a better solution using three locks and three keys, which was based on the Borromean rings. Today, two more ingenious solutions sent in by readers: another one with three locks and three keys, and one with two locks and three keys.
Richard Seaby proposes a system in which a bar runs though two sets of hoops, with three padlocks: at either end and in the middle, as shown below. (The locks are the red cable ties.) If only one padlock is removed, the bar cannot be taken out. But if any two padlocks are, the bar can be taken out. (The bar would have to be short enough that it can be taken out when both side locks are removed.) Nice!
Even simpler is the following solution from Joffrey Joly, who pointed out that the puzzle does not specify that the keys or locks needed to be different. If they can be identical, then all you need is two identical locks, and three identical keys, such that when a key is in a lock it cannot be removed without relocking the safe. In this case, a director on their own cannot open the safe, but two together can.
This ruse, however, only works if the directors are not able to duplicate the keys. The set-up made no mention of key duplication, so I am happy to accept Joffrey’s solution.
More lock puzzle news as it breaks…
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
If you like geometry puzzles, I have a chapter of them in my most recent puzzle book, So You Think You’ve Got Problems. Perfect reading matter for staycations!
Thanks to Manfred Pietch for today’s puzzles.
