Did you solve it? The Riddle of Randall Monroe | Mathematics

I set up the following two puzzles earlier today. Here they are again with solutions.

The first is webcomic cartoonist Randall Monroe, author of the bestselling book xkcd What does that mean?

1. All together now

Find all the possible ways, if any, to fill in the three blanks to make the three statements true.

solution There are three ways, some more interesting than others!

Can represent voids x, Y And z.

If we consider only the numbers in the blanks – there is only one solution x = y = z = oh The 'joke' is to imply that we can add the numbers of the panels. In this case, there are two solutions.

The panels become three equations: x = 5 + y + z, Y = √3xzAnd z = min (x, y, 1)

Let's say minute (x, y, 1) = Y. Therefore z = Y, Thus x = 2Y + 5, and y = √3xy. It makes it easier Y2 + 3Y = 0. We know that Y is greater than or equal to 0 because it is the value of a square root Y= Oh. gives us a solution: x = 5, y = z = oh

Now let's say minute (x, y, 1) = x Thus z = xAnd y = –5, There is a contradiction from Y Greater than or equal to 0.

min (x, y, 1) = xWe get the second solution: x = 12, y = 6, z = 1.

2. One for a friend

The four squares below form a Venn diagram with ten regions labeled A through J. Each of the letters A to J represents a number from 1 to 10, meaning that no two parts share the same number and all numbers from 1 to 10 are used. . The areas in each square add up to the same number. What is the value of the intersection D of the four squares?

The sums of the areas in each square are equal, so A + B + C + D + E (red square) = C + D + F + G + H (yellow square) = green square = blue square Photo: Alex Bellows

solution: D = 7

Since the sum of the areas for each square is the same, when two squares intersect, the areas that do not intersect must be equal. So, for the red and green squares illustrated below, A = G + H + I.

Both shaded areas should be equal

Likewise, F = B + E + I, and J = B + C + G.

Thus A+F+J = G+H+I+B+E+I+B+C+G

= 2(G+I+B) +C+E+H.

The maximum possible value of A+F+G is 10 + 9 + 8 = 27.

The lowest value of 2(G+I+B) +C+E+H is 2(1 + 2 + 3) + 4 + 5 + 6, which is also 27.

Both these values ​​are 27. The only number is 7, and the only label is D, so D = 7.

I hope you enjoyed today's puzzles. I will be back in two weeks.

Today's two puzzles are taken from The Mathematical Playground: People and Problems from 31 Years of Math Horizons. Math Horizons is the undergraduate journal of the Mathematical Society of America, and the book, with its collection of great puzzles, is full of great material. See here for more information.

What does that mean? (10th Anniversary Edition) by Randall Munroe is out on November 14 in the UK.

My latest book Think Twice: Solve Simple Puzzles That (Almost) Everyone Gets Wrong. (In the US it's called Puzzle Me Twice and is out tomorrow.) To support the Guardian and Observer, order your copy at guardianbookshop.com. Delivery charges may apply.

I've been doing a puzzle here on Mondays since 2015. I'm always looking for great puzzles. Email me if you want to recommend one.