IMO 1964 - Problem 5

Problem: Five points are given in the plane. Among the lines that connect these five points, no two coincide and no two are parallel or perpendicular. Through each point we construct an altitude to each of the other lines. What is the maximal number of intersection points of these altitudes (excluding the initial five points)>
Solution: Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to 3.6=18 intersection points. Each perpendicular from B to the 3 lines not containing C can intersect at most 5 of the perpendicular passing through C, which counts 3.5=15 intersection points. Thus there are 18+15=33 intersection points corresponding to B,C.
It follows that the required total number is at most 10.33=330. But some of these points, namely the orthcenters of the triangles with vertices at the given points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330-2.10=310.
Remark. The jury considered only the combinaorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points A(1,1),B(e,$\pi$),C($e^2,\pi^2$,D($e^3,\pi^3$),E($e^4,\pi^4$)) works.