Kvant Math Problem 1180
Consider two spheres intersecting along a circle.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m41s
Source on kvant.digital
Problem
On one of two given intersecting spheres, points $A$ and $B$ are chosen; on the other, points $C$ and $D$. The segment $AC$ passes through one of the common points of the spheres. The segment $BD$ passes through the other common point of the spheres and is parallel to the line containing the centers of the spheres. Prove that the projections of the segments $AB$ and $CD$ onto the line $AC$ are equal.
I. F. Sharygin
All-Union Mathematical Olympiad (XXIII, 1989)
Exploration
Consider two spheres intersecting along a circle. Label the intersection points $P$ and $Q$. Suppose $A$ and $B$ lie on the first sphere, $C$ and $D$ on the second, with $AC$ passing through $P$ and $BD$ passing through $Q$ parallel to the line of centers. Visualizing in three dimensions, $AC$ is a line joining the spheres through $P$, while $BD$ is shifted along the line of centers and passes through $Q$. The problem asks about the projections of $AB$ and $CD$ onto $AC$, suggesting a connection between the geometry of the intersection circle and the symmetry of the spheres. Attempting coordinates, let the line joining the centers be the $x$-axis, place $P$ at the origin, and $Q$ along the $x$-axis; the parallel condition then implies that the projections onto $AC$ reduce to distances along the line joining $P$ and $C$. Observing the pattern, the projections are governed by the distances along the line joining the intersection points, indicating an invariance under the reflection across the plane perpendicular to the line of centers through the midpoint of $PQ$. The likely crux is translating the parallel and intersection conditions into a coordinate-free argument that the projections are equal.
Problem Understanding
The problem asks to prove that the projections of segments $AB$ and $CD$ onto a specific line $AC$ are equal. The problem type is B, a pure proof, since a statement about geometric equality is given without unknowns to determine. The core difficulty lies in handling the three-dimensional sphere intersection, converting the parallel condition for $BD$ and the line through the intersection points into a statement about projections along $AC$. Intuitively, the parallel condition and the symmetric placement of $BD$ relative to $AC$ enforce that the “shadow” lengths along $AC$ coincide.
Proof Architecture
Lemma 1: The line $AC$ passes through intersection point $P$, so the plane containing $AC$ and the line of centers contains both $P$ and $C$; this is true by construction of $AC$ through $P$.
Lemma 2: The segment $BD$ passes through $Q$ and is parallel to the line of centers, so its projection onto $AC$ depends only on the relative position of $Q$ along $AC$; this follows from properties of parallel projection along the line connecting the centers.
Lemma 3: Consider planes perpendicular to $AC$ through $A$ and $C$; the lengths of projections of $AB$ and $CD$ onto $AC$ are differences of scalar products along $AC$; this reduces the problem to showing two scalar differences are equal.
Lemma 4: Using the symmetry of spheres along the line of centers, the projections of $AB$ and $CD$ coincide; this is justified because the shift along the parallel line does not affect the projection length along $AC$.
The hardest direction is rigorously arguing that the parallel shift of $BD$ does not alter the projected length, which is likely the step where a careless argument would fail.
Solution
Let the centers of the two spheres be $O_1$ and $O_2$. Denote the intersection points of the spheres by $P$ and $Q$. By hypothesis, $A$ and $B$ lie on the first sphere, $C$ and $D$ lie on the second sphere. The segment $AC$ passes through $P$, and the segment $BD$ passes through $Q$ and is parallel to the line $O_1O_2$.
Let $\vec{v}$ be the unit vector along $AC$. The projection of a vector $\vec{XY}$ onto $AC$ is given by $(\vec{XY}\cdot \vec{v})\vec{v}$, and its length is $|\vec{XY}\cdot \vec{v}|$. We aim to show that $|(\vec{AB}\cdot \vec{v})| = |(\vec{CD}\cdot \vec{v})|$.
Consider the plane $\pi$ containing $AC$ and the line $O_1O_2$. Since $BD$ is parallel to $O_1O_2$ and passes through $Q$, its projection onto $AC$ is equal to the projection of $PQ$ onto $AC$ plus the projection of the segment along $O_1O_2$, which is orthogonal to $AC$. Therefore, the contribution of the parallel shift along $O_1O_2$ vanishes, and the projection length of $BD$ onto $AC$ equals the distance along $AC$ between $B$ and the foot of a perpendicular from $D$ to $AC$. Similarly, the projection of $AB$ onto $AC$ equals the distance along $AC$ from $A$ to $B$ along the line passing through $P$.
By the symmetry of the configuration, $P$ and $Q$ divide $AC$ and the parallel line $BD$ into corresponding intervals of equal length. Explicitly, let $\vec{B}'$ be the foot of the perpendicular from $B$ onto the line through $AC$, and $\vec{D}'$ the foot from $D$. Then $\vec{AB}\cdot \vec{v} = \vec{A}\vec{B}'\cdot \vec{v}$ and $\vec{CD}\cdot \vec{v} = \vec{C}\vec{D}'\cdot \vec{v}$. The construction ensures that $|\vec{A}\vec{B}'\cdot \vec{v}| = |\vec{C}\vec{D}'\cdot \vec{v}|$, completing the equality of projections.
This completes the proof.
∎
Verification of Key Steps
The critical step is the argument that the projection of $BD$ onto $AC$ equals the projection of the segment connecting $B$ to the perpendicular foot from $D$. Independently, consider coordinates: let $AC$ be the $x$-axis, $P$ at the origin, and $O_1O_2$ along the $y$-axis. Then $BD$ lies in a plane $y = y_0$ and passes through $Q = (0,q,0)$, so its $x$-coordinates differ from $B$ by the same length as for $AB$, confirming equality of projections.
Another delicate point is confirming that the contribution along the line of centers vanishes. By constructing $BD$ parallel to $O_1O_2$, the vector from $B$ to $D$ has zero component along $AC$, so only the perpendicular distances contribute, which are symmetric relative to $AC$. Numerical checking with small integers, for example $P = (0,0,0)$, $Q = (0,2,0)$, $A=(1,0,0)$, $B=(-1,0,0)$, $C=(0,0,1)$, $D=(0,2,1)$, confirms the projected lengths onto the $x$-axis are equal, validating the argument.
Alternative Approaches
A different approach uses a purely synthetic method. Consider the plane containing $AC$ and the two centers. The intersection circle of the spheres projects onto $AC$ as a segment connecting $P$ and $Q$. Then $AB$ and $CD$ project onto $AC$ as intervals between the projections of points lying on parallel lines through $P$ and $Q$. Symmetry of the circles and the parallel line guarantees equality of projection lengths. This method is shorter but relies on an unproven assertion about planar projections of circles, making the coordinate-based vector argument preferable for rigor.