IMO 1964 · Problem 5

Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have.

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{315} \]

Solution

Write the five points as \(P_1,\ldots,P_5\). For distinct indices \(i,j,k\), let \(\ell_{i;jk}\) be the line through \(P_i\) perpendicular to \(P_jP_k\).

Each point gives \(\binom42=6\) perpendiculars, so there are \(30\) constructed lines. No constructed line contains an original point other than its defining point: otherwise a line joining two original points would be perpendicular to another such joining line, contrary to the hypothesis.

For each of the ten joining lines \(P_jP_k\), the perpendiculars from the other three points are parallel and distinct. Perpendiculars associated with different joining lines are not parallel. Thus all \(30\) constructed lines are distinct, and the number of their intersecting pairs is \[ \binom{30}{2}-10\binom32=435-30=405. \]

At each original point, six constructed lines meet. Counting that intersection once instead of \(\binom62=15\) times reduces the count by \(14\).

Furthermore, every triangle formed by three original points has its three altitudes among the constructed lines. Hence its orthocenter is an intersection of at least three constructed lines, requiring a further reduction of at least \(2\).

These ten orthocenters are distinct from the original points and from one another. The first assertion follows from the observation that no constructed line contains another original point. For the second, any two of the ten triangles share a vertex. Their altitudes from that vertex are distinct, since their opposite sides are not parallel. Consequently they cannot also meet at a common orthocenter different from that vertex.

An intersection of \(r\) lines contributes \(\binom r2\) intersecting pairs but only one intersection point. Therefore the number of distinct intersections is at most \[ 405-5(15-1)-10(3-1)=315. \]

We must prove that this bound can be attained, with no additional concurrences.

Choose the original points on a parabola: \[ P_i=(t_i,t_i^2),\qquad t_i>0. \] We will choose the parameters so that they are distinct and the ten sums \(t_j+t_k\), with \(j<k\), are pairwise distinct. The joining lines then have positive, distinct slopes, so none are parallel, perpendicular, or coincident.

A perpendicular through the point with parameter \(a\), whose base joins points with parameter sum \(u\), has equation \[ x+uy=a+ua^2. \] Consider three constructed lines with distinct defining-point parameters \(a,b,c\) and distinct bases, whose parameter sums are \(u,v,w\), respectively. Their concurrence is equivalent to \[ F=(v-w)(a+ua^2)+(w-u)(b+vb^2)+(u-v)(c+wc^2)=0. \] We claim that, unless these are the three altitudes of the triangle with parameters \(a,b,c\), \(F\) is a nonzero polynomial in the five parameters.

Its homogeneous part of degree two is \[ Q=a(v-w)+b(w-u)+c(u-v) =u(c-b)+v(a-c)+w(b-a). \] Suppose that \(Q\) were identically zero. For either parameter \(d\) outside \(\{a,b,c\}\), let \(\varepsilon_u,\varepsilon_v,\varepsilon_w\) indicate whether \(d\) occurs in the corresponding base sum. The coefficient of \(d\) in \(Q\) is \[ \varepsilon_u(c-b)+\varepsilon_v(a-c)+\varepsilon_w(b-a). \] This vanishes identically only when \[ \varepsilon_u=\varepsilon_v=\varepsilon_w. \] Thus each outside parameter must occur in all three base sums or in none.

There are now three possibilities:

  • If both outside parameters occur, all three bases are identical, contrary to our assumption.
  • If exactly one outside parameter \(d\) occurs, each base consists of \(d\) and one of \(a,b,c\). Since the bases are distinct and no base contains its own defining point, the sums must be \[ (u,v,w)=(d+b,d+c,d+a) \quad\text{or}\quad (d+c,d+a,d+b). \] Substitution gives \[ Q=\pm\bigl(a^2+b^2+c^2-ab-bc-ca\bigr), \] which is not the zero polynomial.
  • If neither outside parameter occurs, necessarily \[ u=b+c,\qquad v=a+c,\qquad w=a+b, \] so the three lines are precisely the triangle’s altitudes.

This proves the claim: for every other triple under consideration, \(Q\), and hence \(F\), is a nonzero polynomial.

We can now choose positive parameters avoiding all unwanted concurrences simultaneously. Indeed, the required inequalities consist of finitely many nonzero polynomial conditions: distinct parameters, distinct base sums, and \(F\ne0\) for each unwanted triple. Their product is a nonzero polynomial. A nonzero real polynomial cannot vanish throughout an open box, as follows by applying the corresponding one-variable fact successively to its variables. Thus there are positive parameters satisfying every required inequality.

For this choice, a concurrence involving two lines from the same original point occurs at that point and cannot involve a line from another original point. Every other concurrence of three lines has distinct defining points and distinct bases, so our construction permits it only for the three altitudes of a triangle. In particular, no orthocenter lies on a fourth constructed line.

Consequently the only intersections involving more than two lines are the five original points, with six lines each, and the ten orthocenters, with three lines each. The remaining intersecting pairs give \[ 405-5\binom62-10\binom32=300 \] distinct intersections of exactly two lines. The total is therefore \[ 5+10+300=\boxed{315}. \]

The solution was checked against John Scholes' IMO solution archive (Kalva).