How many kisses do you get if every person in a room full of ten people kisses every other person once? The first person kisses 9 others, the second one kisses 8 others (having already kissed the first), the third kisses 7 others, and so on, with the 10th person not having any new person to kiss at all. So in total you get

9+8+7+ …. +2+1 = 45

kisses. You can also look at this another way. Every person kisses 9 others and there are 10 people, giving 10×9=90 kisses. But this counts every kiss twice, so we have to divide by 2 to get a total of

10×9/2=45 kisses.

This is neat: imagine you have n+1 people, then using the two ways of counting tells us that

n+(n-1)+(n-2)+ … +2+1=(n+1)n/2

In other words, the kissing argument gives us a quick formula for summing the first n whole numbers: no need to tap them all into your calculator.

If you prefer your maths visually, there is another way of keeping tally of the kisses: ask everyone to make a cross on a piece of paper for each, as yet uncounted, kiss and make sure they arrange them as in the picture :

Once all the kissing has been done the shape we see is… a triangle! It doesn’t matter what the value of n is, we will always get a triangle since we’ve been stacking kisses like oranges in a grocery shop. This is why numbers of the form (n+1)n/2, numbers that are the sum of the first n whole numbers, are called triangular numbers. Substituting n=1, n=2, n=3, and so on gives the sequence

1, 3, 6, 10, 15, 21, 28…

which adorns one of the faces of your Cube. The nth triangular number gives you the number of kisses between n+1 people.

If we go up a dimension, the kiss-stacking problem leads us straight to some cutting edge mathematics. Let’s swap kisses for oranges and ask: what is the most efficient way of stacking the oranges? Efficient in the sense that it leaves the smallest gaps possible? The question was posed in the 16th century by Sir Walter Raleigh (who also brought Europe the potato) to his assistant Thomas Harriot, although Raleigh was thinking of canon balls rather than oranges. Harriot eventually forwarded the question to Johannes Kepler, best known for his laws of planetary motion. Kepler suggested that one most efficient way of stacking the balls is the one commonly used by greengrocers. Create the first layer of oranges in a similar way as we created our kiss-triangle, with oranges in one row snuggling into the dip between two oranges in the adjacent row:

Then put the oranges in the next layer into the dip between three oranges in the layer below. Using this method you can fill just over 74% of space with oranges, canon balls, or whatever balls you want to consider.

Kepler’s conjecture is straightforward and appeals to common sense but it has landed mathematicians in a difficult philosophical debate. For nearly 500 years the conjecture remained unproved, until in 1998 the mathematician Thomas Hales announced that he had cracked it. The trouble was that his proof involved an exhaustive search that could only be completed by a computer. It involved over three gigabytes of computer code and data which made it humanly impossible to check the proof. Mathematicians have been quoted as saying they are “99% sure” that the proof is correct, but 99% is not good enough in mathematics. The question whether computer assisted proofs should be allowed remains a hot one in mathematics and, assuming that they are not, Kepler’s conjecture remains open.

But back to our kissing situation, we might consider a slightly different problem: imagine each person in turn kisses all the other people, regardless of whether they have kissed them before, and to top it off, they also kiss themselves. Then clearly for n people you get n2 kisses. Arranging the kisses on paper you simply get a square.

 

You can also get such a square by taking the triangle corresponding to the nth triangular number and then gluing to it, at its base, the triangle corresponding to the n-1st triangular number: