Popular Mechanics (South Africa)

# We spent all day arguing about this triangle brain teaser. Can you solve it?

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NOTHING HALTS PRODUCTIVI­TY AMONG co-workers quite like a maddening brain teaser. The latest to ensnare the Popular Mechanics editors and readers: How many triangles are in this drawing? When I posed the problem to our team, responses ranged from four triangles all the way to 22. Most people saw 18. One wise guy counted the triangles in the A-letters in the question itself, while another seemed to have an existentia­l crisis: ‘None of these lines are truly straight, just curves – thus you cannot define any of them as a triangle,’ he said. ‘There are no triangles in this photo. Life has no meaning.’

I could’ve listened to my colleagues’ questionab­le processes all day, but instead, I reached out to geometry experts to see if we could arrive at a consensus. All of the mathematic­ians I contacted found the same solution – but not all of them figured it out in the same way.

‘I would approach this like one approaches any mathematic­al problem: reduce it and find structure,’ says Sylvester Eriksson-Bique, PhD, a postdoctor­al fellow with the University of California Los Angeles’s mathematic­s department.

The only way to form triangles in the drawing, Eriksson-Bisque says, is if the top vertex (corner) is part of the triangle. The base of the triangle will then have to be one of the three levels below. ‘There are three levels, and on each you can choose a base among six different ways. This gives 18, or 3 times 6 triangles.’

Look at the triangle again: ‘It’s convenient to generalise to the case where there are ‘n’ lines passing through the top vertex, and ‘p’ horizontal lines,’ says Francis Bonahon, PhD, a professor of mathematic­s at the University of Southern California.

In our case, n = 4, and p = 3. Any triangle we find in the drawing should have one top vertex and two others on the same horizontal line, so for each horizontal line, the number of triangles is equal to the number of ways we can choose two distinct vertices on that line out of n total points, Bonahon says – or ‘n choose 2.’

That’s n(n – 1)/2. And since there are p horizontal lines, this gives p × n(n – 1)/2 possible triangles. In our case, that’s 3 × 4(4 – 1)/2 = 18.

Here’s a handy breakdown of how to find each possible triangle:

When we published the problem on popularmec­hanics.com, readers started sending in their own solutions. Software engineer Slobodan Jaksic came up with a mathematic­al proof to demystify the triangle teaser. It goes like this:

Two lines meet at the top vertex denoted by ‘A’, and the segment line connecting those two lines is the ‘base’. The new polygon is the fundamenta­l triangle. We can place an arbitrary number of points on the base and connect them to the top vertex A.

Assuming there are ‘n’ vertices on the base, where n is a natural number greater than or equal to 2, we will prove that the new shape – let’s call it ‘pyramid’ – contains a number of triangles equal to S(n) = n(n – 1) × ½. (See Figure 1.) For n = 2, S(2) = (2 × 1) × ½ = 1. It is correct.

Assuming the formula S(n) holds for n = k, with k greater than or equal to 2, S(k) = k(k – 1) × ½. We will prove that S(n), where n = k + 1, holds as well. (See Figure 2.)

In the modified pyramid (see Figure 3), there are ‘k’ additional new triangles. Therefore, S(k + 1) = S(k) + k = (k(k – 1) × ½) + k = (k + 1) × k × ½.

Modify the pyramid by extending both edge lines, and all lines between them connect to the top vertex A, beyond the base, in the direction opposite of the top vertex. Create new bases by connecting one or more segments parallel to the starting base.

Assuming there are a total of ‘m’ bases in the pyramid, we can prove the formula calculatin­g the number of triangles in the structure is: S(n, m) = n(n – 1) × ½ × m.

Let’s prove that the formula holds for m = p + 1. (See Figure 3.) The number of new triangles can be given as: (n – 1) + … + 2 + 1 = n(n – 1) × ½. Consequent­ly, S (n, p + 1) = S(n, p) + (n(n – 1) × ½) = (n(n – 1) × ½) × p + (n(n – 1) × ½) = (n(n – 1) × ½) (p + 1).

According to the Principles of Mathematic­al Induction, we just proved the total number of triangles in the pyramid is given through the formula S(n, m) = n (n – 1) × ½ × m, where n and m are natural numbers, n is greater than or equal to 2, n is an arbitrary number of vertices, and m is an arbitrary number of bases.

Hat tip to Jaksic. Here I was just trying to annoy my co-workers. / BY ANDREW DANIELS /  