IMO 1978 · Problem 2

\(P\) is a given point inside a given sphere. Three mutually perpendicular rays from \(P\) intersect the sphere at points \(U\), \(V\), and \(W\); \(Q\) denotes the vertex diagonally opposite to \(P\) in the parallelepiped determined by \(PU\), \(PV\), and \(PW\). Find the locus of \(Q\) for all such triads of rays from \(P\)

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

Answer

The locus is the entire sphere with center \(O\) and radius \[ \boxed{\sqrt{3R^2-2OP^2}}, \] where \(O\) and \(R\) are the center and radius of the given sphere.

Solution

Take \(O\) as the origin, and let \(\mathbf p,\mathbf q\) be the position vectors of \(P,Q\). Put \[ d=|\mathbf p|<R,\qquad c=R^2-d^2>0. \]

Let \(\mathbf e_1,\mathbf e_2,\mathbf e_3\) be unit vectors along the three rays. They form an orthonormal basis. Write \[ a_i=\mathbf p\cdot\mathbf e_i, \] and let \(t_i>0\) be the distance from \(P\) to the sphere along the \(i\)-th ray. The equation \[ |\mathbf p+t_i\mathbf e_i|^2=R^2 \] gives \[ t_i^2+2a_it_i=c, \qquad t_i=-a_i+\sqrt{a_i^2+c}. \] Since \(c>0\), this is indeed the unique positive root.

The opposite vertex of the parallelepiped therefore satisfies \[ \mathbf q =\mathbf p+\sum_{i=1}^3t_i\mathbf e_i =\sum_{i=1}^3\sqrt{a_i^2+c}\,\mathbf e_i. \] Consequently, \[ |\mathbf q|^2 =\sum_{i=1}^3(a_i^2+c) =d^2+3c =3R^2-2d^2. \] Thus every possible \(Q\) lies on the claimed sphere. Denote its radius by \[ r=\sqrt{3R^2-2d^2}. \]

We must prove that every point of this sphere is attained.

If \(P=O\), rotating any admissible triad about \(O\) rotates its corresponding \(Q\) in the same way. Since \(|OQ|=r\), these rotations give every point of the claimed sphere.

Now suppose \(d>0\), and put \(\mathbf k=\mathbf p/d\). Choose an orthonormal frame satisfying \[ \mathbf e_1+\mathbf e_2+\mathbf e_3=\sqrt3\,\mathbf k. \] Such a frame exists: the sum of any orthonormal frame has length \(\sqrt3\), so a rotation can align that sum with \(\mathbf k\). Taking scalar products with each \(\mathbf e_i\) shows that \[ a_i=\frac d{\sqrt3}. \] For this triad our formula gives \[ \mathbf q =\sqrt{\frac{d^2}{3}+c}\, (\mathbf e_1+\mathbf e_2+\mathbf e_3) =r\mathbf k. \]

Rotate these three rays continuously through an angle \(\theta\), from \(0\) to \(\pi\), about a fixed line through \(P\) perpendicular to \(\mathbf k\). At each angle, intersect the rays with the given sphere anew and form the corresponding vertex \(Q_\theta\).

Let \(H_\theta\) denote the rotation of direction vectors. At \(\theta=\pi\), we have \(H_\pi\mathbf k=-\mathbf k\), so \[ \mathbf p\cdot H_\pi\mathbf e_i=-\frac d{\sqrt3}, \qquad \sum_{i=1}^3H_\pi\mathbf e_i=-\sqrt3\,\mathbf k. \] Our formula therefore gives \[ \mathbf q_0=r\mathbf k,\qquad \mathbf q_\pi=-r\mathbf k. \] That same formula shows that \(\mathbf q_\theta\) depends continuously on \(\theta\). Hence, by the intermediate value theorem, \[ \mathbf q_\theta\cdot\mathbf k \] takes every value in \([-r,r]\).

Finally, let \(X\) be any point on the claimed sphere, with position vector \(\mathbf x\). Choose \(\theta\) such that \[ \mathbf q_\theta\cdot\mathbf k=\mathbf x\cdot\mathbf k. \] Since both vectors have length \(r\), their endpoints lie on the same circle perpendicular to the axis \(OP\). A rotation about \(OP\) carries \(Q_\theta\) to \(X\); at the two poles the points already coincide. Such a rotation preserves \(P\) and the original sphere, so it carries the corresponding triad to an admissible triad whose opposite vertex is \(X\).

Thus every point of the stated sphere is attainable, completing the proof.

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