IMO 1963 · Problem 2

Point \(A\) and segment \(BC\) are given. Determine the locus of points in space which are vertices of right angles with one side passing through \(A\), and the other side intersecting the segment \(BC\).

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

Answer

Let \(K_B\) and \(K_C\) be the closed balls with diameters \(AB\) and \(AC\), respectively. The locus is \[ \boxed{\bigl(K_B\setminus\operatorname{int}K_C\bigr) \;\cup\; \bigl(K_C\setminus\operatorname{int}K_B\bigr).} \] Thus a point must lie in one closed ball and outside the interior of the other. All points on either sphere are included.

Here sides of angles are rays containing their vertices, so an incidence at the vertex is allowed. The modification when both incidences must be away from the vertex is given below.

Solution

For any points \(P,D\), define \[ F_D(P)=(P-A)\cdot(P-D). \] Writing \(O_D=(A+D)/2\), we obtain \[ F_D(P)=|P-O_D|^2-\frac{|A-D|^2}{4}. \] Consequently, \(F_D(P)\leq0\) means that \(P\) belongs to the closed ball with diameter \(AD\), while \(F_D(P)\geq0\) means that \(P\) is outside its interior. These statements remain valid when \(A=D\), in which case the ball is the singleton \(\{A\}\).

We shall prove that the required condition is precisely \[ \boxed{F_B(P)F_C(P)\leq0.} \tag{1} \]

Suppose first that \(P\) is a vertex of a suitable right angle, and let \(X\in BC\) belong to its other side. The vectors \(A-P\) and \(X-P\) lie along perpendicular rays, or one of them is zero. Hence \[ F_X(P)=(P-A)\cdot(P-X)=0. \] Write \[ X=(1-t)B+tC,\qquad 0\leq t\leq1. \] Then \[ 0=F_X(P)=(1-t)F_B(P)+tF_C(P). \] A convex combination of two strictly positive numbers is positive, and a convex combination of two strictly negative numbers is negative. Therefore \(F_B(P)\) and \(F_C(P)\) cannot have the same strict sign, proving (1).

Conversely, suppose (1) holds. There is some \(t\in[0,1]\) for which \[ (1-t)F_B(P)+tF_C(P)=0: \] choose an endpoint if one of the two values is zero; otherwise they have opposite signs, and take \[ t=\frac{F_B(P)}{F_B(P)-F_C(P)}\in(0,1). \] For \(X=(1-t)B+tC\), we consequently have \[ (P-A)\cdot(P-X)=0. \]

If \(P\ne A\) and \(P\ne X\), the rays \(PA\) and \(PX\) form the required right angle. The remaining cases also give genuine right angles:

  • If \(P=A\ne X\), use the ray \(PX\) and any perpendicular ray from \(P\).
  • If \(P=X\ne A\), use the ray \(PA\) and any perpendicular ray from \(P\).
  • If \(P=A=X\), use any two perpendicular rays from \(P\).

Thus (1) is both necessary and sufficient. Finally, (1) is equivalent to \[ \bigl(F_B(P)\leq0,\ F_C(P)\geq0\bigr) \quad\text{or}\quad \bigl(F_C(P)\leq0,\ F_B(P)\geq0\bigr), \] which gives exactly the stated locus.

No assumption about the relative positions of \(A,B,C\) was used. In particular, collinear configurations are covered. If \(B=C\), condition (1) becomes \(F_B(P)=0\), giving the sphere with diameter \(AB\); if \(A=B=C\), it gives just \(\{A\}\).

For completeness, under the convention that both incidences must occur away from the vertex, delete \(A\) and all of \(BC\) from the displayed locus, with one possible exception. When \(B\ne C\), let \(H\) be the perpendicular projection of \(A\) onto the line \(BC\); retain \(H\) if \(H\in BC\) and \(H\ne A\). Indeed, for distinct \(P,X\in BC\), the condition \(PA\perp PX\) holds exactly when \(P=H\). Conversely, such an \(H\) works by choosing any \(X\in BC\setminus\{H\}\). Outside \(\{A\}\cup BC\), the preceding construction already has both incidences away from the vertex.

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