# Child's play

How many different types of candy do these children have in total?

*Source: I once met an axiomatic projective plane.*

**Seven.**

This situation is an axiomatic projective plane in disguise. Let each child be a point, let each type of candy be a line and let a point be incident with a line if and only if that child has candy of the type corresponding with that line. The first property states that every two distinct points lie on a unique line; the second property states that every two distinct lines intersect in a unique point. Finally, the fact that every point is contained in at least two lines indirectly implies that there are at least four points, no three of which are collinear (convince yourself of this). Then one can refer to the known fact that each such axiomatic projective plane has equally many points as there are lines.

To have a more elementary proof, one can prove that each two lines contain equally many points. Take two arbitrary lines \(l_1\) and \(l_2\); by the axioms, these intersect in a point. Suppose all points of the plane are contained in \(l_1\cup l_2\). Then, if one of these lines contain at most two points, we cannot find four points, no three of which are collinear. Hence, each of the lines must contain at least three points. By cleverly connecting points lying on different lines, we find (intersection) points outside \(l_1\cup l_2\), which is a contradiction. Thus, we can find a point \(P \notin l_1\cup l_2\). Now we can define a bijection between the points of \(l_1\) and the points of \(l_2\): connect each point of \(l_1\) with \(P\), the line arising from this will intersect \(l_2\) in a unique point. In conclusion, these arbitrary chosen lines contain equally many points.

By taking the dual of the previous reasoning, one can prove that each two points are incident with equally many lines.

Moreover, the (constant) number of points on a line equals the (constant) number of lines through a point. Indeed, take a line and a point not contained in that line. Then we can define a bijection between the points on the line and the lines through the point by considering their incidence with each other.

Finally, by double counting the pairs \((Q,l)\), with \(Q\in l\), one can convince him-/herself that the number of lines has to equal the number of points in the aximatic projective plane.

Note that the choice of the number seven is completely random; you can generalise this riddle by considering \(n\geqslant2\) children instead of seven.

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