IMO 1963 Problem 2

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m09s

Problem

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

Exploration

For a point $P$ in space, the condition that the angle with vertex at $P$ has one side passing through $A$ and the other intersecting segment $BC$ means that there exists a point $X \in BC$ such that $A, P, X$ are distinct and the angle $\angle APX$ is a right angle. This is equivalent to the orthogonality relation $PA \perp PX$.

For a fixed point $X$, the set of points $P$ such that $\angle APX = 90^\circ$ is classically the sphere having segment $AX$ as diameter, since in Euclidean space the locus of points subtending a right angle over a fixed segment is a sphere.

As $X$ varies along the segment $BC$, these spheres vary continuously, with centers at the midpoints of $AX$ and radii equal to $\frac{AX}{2}$. The locus sought is therefore the union of a one-parameter family of spheres whose centers move linearly along the segment joining the midpoint of $AB$ to the midpoint of $AC$, while their radii vary accordingly.

A direct description in terms of inequalities or a single algebraic surface is not necessary for the geometric characterization; the essential difficulty lies in correctly translating the existential condition into a union of classical loci and ensuring no additional constraints are inadvertently introduced.

Problem Understanding

A fixed point $A$ and a fixed segment $BC$ in space are given. We are to determine all points $P$ in space for which there exists a point $X$ on the segment $BC$ such that the angle formed at $P$ between the segment $PA$ and the segment $PX$ is a right angle.

This is a Type A problem, since it asks for a complete description of a locus.

The key geometric structure is that a right angle at $P$ involving fixed endpoints $A$ and a variable point $X$ imposes a spherical constraint for each fixed $X$. The challenge is to correctly incorporate the variability of $X$ along a segment rather than a fixed point, and to express the resulting locus without omitting or overcounting configurations.

The locus is the union of all spheres with diameter $AX$ as $X$ ranges over the segment $BC$.

Proof Architecture

Lemma 1 asserts that for distinct points $A, X$ in space, the set of points $P$ such that $\angle APX = 90^\circ$ is the sphere with diameter $AX$.

Lemma 2 asserts that for each point $X \in BC$, the condition in the problem is equivalent to membership of $P$ in the sphere with diameter $AX$.

Lemma 3 asserts that the locus of all such points $P$ is the union of these spheres as $X$ varies over the segment $BC$.

The hardest direction is the correct handling of the existential quantifier over $X$ and ensuring that no additional geometric constraints arise from restricting $X$ to a segment.

Solution

Lemma 1

For distinct points $A, X$ in space, a point $P$ satisfies $\angle APX = 90^\circ$ if and only if $P$ lies on the sphere having segment $AX$ as diameter.

Let $M$ be the midpoint of $AX$. For any point $P$, the condition $\angle APX = 90^\circ$ is equivalent to the vector condition $(A-P)\cdot (X-P)=0$. Expanding,

$$(A-P)\cdot (X-P)=A\cdot X - A\cdot P - P\cdot X + P\cdot P.$$

Rearranging,

$$P\cdot P - P\cdot (A+X) + A\cdot X = 0.$$

Completing the square,

$$|P - M|^2 = \frac{|A-X|^2}{4}.$$

Thus $P$ lies on the sphere centered at $M$ with radius $\frac{AX}{2}$, which is exactly the sphere with diameter $AX$.

This establishes that the right-angle condition is equivalent to a fixed spherical locus determined by $A$ and $X$, and it rules out any dependence on direction beyond orthogonality.

Lemma 2

For a point $P$, the existence of a point $X \in BC$ such that $\angle APX = 90^\circ$ is equivalent to the existence of $X \in BC$ such that $P$ lies on the sphere with diameter $AX$.

If such an $X$ exists, then by Lemma 1 the point $P$ belongs to the sphere determined by $A$ and $X$. Conversely, if $P$ lies on such a sphere for some $X \in BC$, then the defining property of that sphere guarantees $\angle APX = 90^\circ$, so the required right angle exists with side $PX$ meeting $BC$ at $X$.

This establishes a precise equivalence between the existential geometric condition and membership in a family of spheres indexed by $X \in BC$.

Lemma 3

The locus of all points $P$ satisfying the problem condition is the union

$$\bigcup_{X \in BC} \mathcal{S}_X,$$

where $\mathcal{S}_X$ is the sphere with diameter $AX$.

If $P$ lies in the locus, then by definition there exists $X \in BC$ such that $P \in \mathcal{S}_X$, so $P$ lies in the union. Conversely, any point in the union lies in some $\mathcal{S}_X$, and hence satisfies the required right-angle condition by Lemma 1.

This completes the identification of the locus as a precise union of classical geometric surfaces parameterized by the segment $BC$.

Conclusion

The locus of points in space which are vertices of right angles with one side passing through $A$ and the other intersecting the segment $BC$ is

$$\boxed{\bigcup_{X \in BC} {P \in \mathbb{R}^3 : \angle APX = 90^\circ}},$$

equivalently,

$$\boxed{\bigcup_{X \in BC} \text{sphere with diameter } AX}.$$

Verification of Key Steps

The most delicate point is the equivalence between the right-angle condition and the sphere with diameter $AX$. Re-deriving from the condition $PA \perp PX$ yields $(A-P)\cdot(X-P)=0$, which expands to a quadratic equation in the coordinates of $P$ whose completion of squares produces a sphere centered at the midpoint of $AX$. Any attempt to shortcut this step risks missing the exact center and radius relation.

A second delicate point is the treatment of the variable point $X$ restricted to a segment. The condition is existential rather than universal, so the locus is a union rather than an intersection. Confusing these would incorrectly shrink the locus to points satisfying the condition for all $X \in BC$, which is not intended.

A third delicate point is ensuring that every point on each sphere corresponds to a valid configuration with $X$ still lying on the segment, which is guaranteed because each sphere is defined directly from a point $X \in BC$.

Alternative Approaches

A different approach proceeds by fixing $P$ and analyzing the condition $PA \perp PX$ as a quadratic constraint on $X$ constrained to the line through $BC$. This leads to an intersection condition between the line $BC$ and a quadratic surface associated with $P$, characterizing the locus via solvability of a quadratic equation in a parameter on the segment.

Another approach uses inversion centered at $A$, transforming right angles into orthogonality conditions between lines and circles, which converts the problem into describing images of the segment $BC$ under a one-parameter family of transformations. The direct spherical characterization remains more transparent because it avoids parameterizing the segment algebraically.