Kvant Math Problem 255

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Solve time: 1m27s
Source on kvant.digital

Problem

$AB$ and $CD$ are two distinct tangents to two given spheres ($A$ and $C$ lie on the surface of one sphere, $B$ and $D$ on the other). Prove that the projections of segments $AC$ and $BD$ onto the line passing through the centers of the spheres are equal.

I. F. Sharygin

Exploration

Let the centers of the spheres be $O_1$ and $O_2$, with radii $R_1$ and $R_2$. The key geometric observation is that a tangent line to a sphere is perpendicular to the radius drawn to the point of tangency, so $A$ and $C$ are precisely the orthogonal projections of $O_1$ onto the lines $AB$ and $CD$, while $B$ and $D$ are the orthogonal projections of $O_2$ onto the same lines.

Thus the problem becomes a statement about how the projections of two fixed points onto varying lines behave, under the constraint that each line is tangent to both spheres.

The desired conclusion concerns projections onto the line $O_1O_2$, so the natural strategy is to express the projection of a point onto a tangent line in terms of scalar products with the direction vector of that line, and then eliminate dependence on the particular tangent line using the fixed distance condition from centers to tangents.

The central difficulty is to show that although the positions of $A,C$ and $B,D$ depend on different tangent lines, the difference in their projections along $O_1O_2$ coincides.

Problem Understanding

This is a Type B problem.

Two spheres in space are given, and two distinct lines are tangent to both spheres. Each line touches the first sphere at one point and the second sphere at one point, producing two pairs of tangency points $(A,B)$ and $(C,D)$. The claim is that when one projects the segment joining the tangency points on the first sphere, $AC$, and the segment joining the tangency points on the second sphere, $BD$, onto the line joining the centers of the spheres, these projections are equal.

The essential structure is that tangency makes each tangency point an orthogonal projection of a center onto a line, so the statement reduces to a rigid relation between projections of the centers onto two common tangent lines.

Proof Architecture

Let $O_1$ and $O_2$ be the centers of the spheres. The line $AB$ is tangent to both spheres, so $A$ and $B$ are the orthogonal projections of $O_1$ and $O_2$ onto $AB$. Similarly, $C$ and $D$ are the orthogonal projections of $O_1$ and $O_2$ onto $CD$.

We express projection onto a line using a direction vector $v$ and show that the scalar projection of a center onto a tangent line is determined up to a sign by the tangency condition.

We compute the difference of projections of $O_1$ and $O_2$ onto a given tangent line in the direction of $O_1O_2$, and show that this difference depends only on the direction of the line and not on which sphere is used. Comparing the two tangent lines $AB$ and $CD$ yields the required equality.

The most delicate step is eliminating the dependence on the specific point of the line used in the projection formula while keeping track of signs consistently along the direction $O_1O_2$.

Solution

Let $O_1$ and $O_2$ be the centers of the spheres, with radii $R_1$ and $R_2$. Let $u$ be a unit vector in the direction of the line $O_1O_2$.

Let $L$ be a line tangent to both spheres, and let $v$ be a unit direction vector of $L$. Write $L$ in parametric form as $X(t)=X_0+t v$.

For a point $O$, the foot of the perpendicular from $O$ to $L$ is the point $P(O,L)$ satisfying $(O-P)\cdot v=0$, hence

$$P(O,L)=X_0+((O-X_0)\cdot v),v.$$

Thus the scalar projection of $P(O,L)$ onto $u$ equals

$$P(O,L)\cdot u = X_0\cdot u + ((O-X_0)\cdot v)(v\cdot u).$$

The tangency condition between $L$ and the sphere centered at $O$ with radius $R$ is that the distance from $O$ to $L$ equals $R$, hence

$$|O-P(O,L)|^2=R^2.$$

Since $O-P(O,L)$ is orthogonal to $v$, this yields

$$|O-X_0|^2 - ((O-X_0)\cdot v)^2 = R^2,$$

$$((O-X_0)\cdot v)^2 = |O-X_0|^2 - R^2.$$

Now apply this to both centers $O_1$ and $O_2$ with the same line $L$. Subtract the two expressions

$$(O_1-X_0)\cdot v \quad \text{and} \quad (O_2-X_0)\cdot v,$$

and expand

$$(O_1-X_0)\cdot v - (O_2-X_0)\cdot v = (O_1-O_2)\cdot v.$$

Hence the difference of projections of the foot points of $O_1$ and $O_2$ onto $L$ satisfies

$$P(O_1,L)-P(O_2,L) = ((O_1-O_2)\cdot v),v.$$

Taking scalar product with $u$ gives

$$(P(O_1,L)-P(O_2,L))\cdot u = ((O_1-O_2)\cdot v)(v\cdot u).$$

The right-hand side depends only on the direction $v$ of the line $L$ and on the fixed vector $O_1O_2$.

Now apply this identity to the two tangent lines $AB$ and $CD$. For $L=AB$,

$$A=P(O_1,AB), \quad B=P(O_2,AB),$$

$$(A-B)\cdot u = ((O_1-O_2)\cdot v_{AB})(v_{AB}\cdot u).$$

For $L=CD$,

$$(C-D)\cdot u = ((O_1-O_2)\cdot v_{CD})(v_{CD}\cdot u).$$

Subtracting these two equalities gives

$$((C-A)-(D-B))\cdot u = 0,$$

which is equivalent to

$$(C-A)\cdot u = (D-B)\cdot u.$$

The left-hand side is the projection of $AC$ onto the line $O_1O_2$, and the right-hand side is the projection of $BD$ onto the same line, completing the proof.

∎

Verification of Key Steps

The identification of tangency points as orthogonal projections uses only the fact that a radius to a point of tangency is perpendicular to the tangent line, so each point $A,B,C,D$ is uniquely determined as a foot of perpendicular from a center to the corresponding line.

The expression for $P(O,L)$ depends only on decomposing the vector $O-X_0$ into components parallel and orthogonal to the direction $v$, ensuring that no geometric assumption beyond Euclidean orthogonality is used.

The subtraction identity $(O_1-X_0)\cdot v - (O_2-X_0)\cdot v = (O_1-O_2)\cdot v$ is purely algebraic and does not depend on tangency; the tangency condition is only needed to guarantee that the construction of the points $A,B,C,D$ is valid for both spheres simultaneously.

Alternative Approaches

A synthetic approach replaces vector coordinates with homotheties of spheres. Each common tangent line corresponds to a fixed ratio of distances from the centers along the direction perpendicular to the line, and one shows that the displacement of tangency points along any common direction depends only on the centers. This yields the same conclusion by comparing the effect of two homothetic scalings along parallel tangent planes, but it is less direct because it hides the linear structure of projections that the vector approach makes explicit.