Last week’s sphere problem appears to be missing data, but the solution is unique. Due to the conditions of the problem, both 4r² and 4r³ / 3 must be four-digit integers, that is, between 1,000 and 9,999, so 20> r> 15, and since it must be an integer, r can only take the values 16, 17, 18 or 19. On the other hand, for 4r³ / 3 to be an integer, r must be divisible by 3, so the only possibility is r = 18.
What if instead of four digits both whole numbers had five? Can you generalize to numbers?
Regarding the obtuse triangle to be divided into acu-angles, it is easy to reach the false conclusion that it is impossible (Martin Gardner said that in his day he received several “demonstrations” of such impossibility), because no matter how much we divide it, it always ends up being at least a stubborn little obtuse angle, like the one in the lower right corner of the figure.

However, division is possible, and as a clue I will say that it is one of those instructive problems in which, unconsciously, we put ourselves more conditions than they ask of us. The best known of these puzzles is that of the nine grid points that must be joined with only four rectilinear lines without lifting the pencil from the paper, and which we have talked about more than once in this section.

In this case, the self-imposed condition is that the vertices of the broken line of the solution coincide with two points on the grid.
There are numerous verbal problems of this type, which sometimes reveal deep-rooted prejudices. A classic example:
A young woman and her father are in a car accident. The father dies on the spot and the girl has to be operated on urgently. They transfer her to the nearest hospital, but whoever should perform the delicate operation asks someone else to carry it out, claiming that the victim is their daughter. How is it possible, if the father has just died in the accident? Very simple: the surgeon is the girl’s mother. (In English it is much better, when playing with the ambiguity of the term doctor).
As for the palindrome 121121, derived from the publication date of the previous installment (12 11 21), it is a supercapicúa: in addition to being made up of two 121s (which in turn is the square of 11, the smallest of the capicuas) , 121121 = 33³ + 44³, the sum of the cubes of two consecutive spindles. And as if this were not enough, 121121 = 66³ – 55³, the difference between the cubes of two other consecutive capicuas and consecutive of the previous two.
Who leaves and distributes …
An obtuse triangle divided into acute angles, a sphere divided by a plane (see last week’s installment), a capicúa decomposed into cubes … Let’s continue, then, with the interesting issue of partitions.
A reader recently asked how many parts a pizza can be divided into 1, 2, 3, 4… straight cuts. It is evident that with one cut it can only be divided into two parts, with two cuts into four, with three cuts into seven … (And a cake? But does it make sense to differentiate between a pizza and a cake?). Can the problem be generalized in court?
In order to avoid that the party and distributors get the best part, there is a simple procedure, when the parties are two, to ensure that both are satisfied: one part and the other chooses, with which the first takes good care in which the parts are equal. But what if there are three people who want to share something equally? What if there are four, five, six …?
Carlo Frabetti is a writer and mathematician, member of the New York Academy of Sciences. He has published more than 50 popular science works for adults, children and young people, including ‘Damn physics’, ‘Damn maths’ or ‘The great game’. He was a screenwriter for ‘La bola de cristal’.
You can follow MATTER on Facebook, Twitter e Instagram, or sign up here to receive our weekly newsletter.
elpais.com