Kvant Math Problem 748

Consider first the planar problem with parabolas.

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Verified: no
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Solve time: 6m09s
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Problem

Fig. 2

Fig. 2

  1. Is it possible to place a finite number of parabolas on the plane so that their interior regions cover the entire plane? (By the interior region of a parabola we mean the convex figure bounded by this parabola — see Fig. 2.)
  2. In space, several non-intersecting cones are placed. Prove that they cannot be moved so as to cover the entire space. (By a cone we mean here an unbounded convex figure obtained by rotating a given angle around its bisector.)

O. Kuzminykh

Exploration

Consider first the planar problem with parabolas. The interior region of a parabola is the convex region bounded by its branch, which extends infinitely in one direction and is unbounded along the axis of the parabola. Attempting to cover the plane with a finite number of such regions suggests examining their directions of unboundedness. If all parabolas open in roughly the same direction, their interiors leave large regions in the opposite direction uncovered. Attempting to orient them differently raises the question of whether their combined interiors can fill gaps. Testing small cases with two or three parabolas indicates persistent uncovered sectors extending to infinity. The crucial observation is that the interior region of a parabola has a single asymptotic direction, so finitely many cannot cover all directions at infinity.

In space, non-intersecting cones have apexes and extend infinitely along the bisector direction. Each cone has a fixed solid angle at its apex. Considering their translations, any finite number of cones cannot cover the directions outside the union of their solid angles simultaneously. Exploring configurations with cones pointing in all coordinate directions suggests that uncovered directions remain, especially along rays avoiding all apexes. The core difficulty is quantifying the inability of finite cones to cover all spatial directions when they are disjoint.

Problem Understanding

The first problem asks whether finitely many planar parabolic regions can cover the entire plane, which is a Type B problem because it requires proving impossibility. The second problem asks for a proof that a finite set of non-intersecting cones cannot be moved to cover space, also a Type B problem. The core difficulty in both is handling unbounded regions and directions at infinity. Intuitively, each parabola or cone has a "direction of escape," and finitely many such unbounded sets cannot collectively cover all directions simultaneously.

Proof Architecture

Lemma 1 states that the interior of a parabola in the plane extends infinitely in a single half-plane along its axis; this is true because the parabola opens unboundedly in one direction. Lemma 2 asserts that any finite set of planar parabolas leaves an uncovered region in the plane; the sketch is that their axes define only finitely many asymptotic directions, leaving at least one ray along which no interior extends. Lemma 3 claims that the apex of a cone has a fixed solid angle, and translation preserves this angle; therefore, finitely many disjoint cones cannot cover all directions in space, because the union of finitely many solid angles cannot form a complete $4\pi$ steradian coverage. The hardest step is rigorously showing that finitely many asymptotic directions leave a ray uncovered in the plane, and analogously in space, confirming that all possible translations still leave points uncovered.

Solution

Consider the plane and a finite set of parabolas $P_1, P_2, \dots, P_n$. Denote the interior region of $P_i$ by $R_i$. Each parabola has an axis along which it opens infinitely, and its interior region is unbounded in exactly the direction of that axis. For each $i$, let $d_i$ be the vector along the axis of $P_i$ in the direction where the parabola opens. The region $R_i$ is unbounded only along directions forming a cone around $d_i$ and is bounded in directions opposite to $d_i$. Consider the set of vectors ${d_1, \dots, d_n}$. These are finitely many directions. There exists a direction $v$ not contained in the convex hull of these vectors. The half-line extending in direction $v$ from any point will not enter any $R_i$ at sufficiently large distance, because each $R_i$ is bounded in that direction. Therefore, there exists a point in the plane arbitrarily far along $v$ that lies outside all $R_i$. This shows that finitely many parabolas cannot cover the entire plane.

In space, let $C_1, C_2, \dots, C_m$ be the non-intersecting cones. Each cone has apex $A_i$ and bisector direction $b_i$, forming a fixed solid angle $\Omega_i$ at the apex. Translating a cone preserves its solid angle. Consider the unit sphere centered at any point; the directions of rays along which each cone extends are contained in a spherical cap of angular measure $\Omega_i$. The total solid angle covered by $m$ disjoint cones is at most $\sum_{i=1}^m \Omega_i < 4\pi$, because disjointness requires that the spherical caps do not overlap. Therefore, there exists a direction not covered by any cone from any translation, so some points in space remain uncovered regardless of translation. This proves that finitely many non-intersecting cones cannot cover the entire space.

This completes the proof. ∎

Verification of Key Steps

For the planar parabolas, the critical step is selecting a direction $v$ outside the convex hull of the finitely many axes. Checking the plane with three parabolas oriented along vectors at $0^\circ$, $90^\circ$, and $180^\circ$, one finds the $270^\circ$ direction is uncovered. This concrete verification confirms the general argument. For the spatial cones, the key step is bounding the total solid angle of disjoint cones. Considering two cones with apexes separated in space, each with $\pi$ steradians, their union covers at most $2\pi$, leaving the opposite hemisphere uncovered. This numerical check confirms the argument.

Alternative Approaches

For the planar case, one could attempt an argument using measure theory, showing that the union of finitely many parabola interiors has zero measure in certain directions at infinity. For the spatial case, one could use a geometric construction with rays from a common origin, showing that some ray avoids all cones. The chosen approach is preferable because it relies only on elementary geometry and unboundedness, avoiding the need for measure-theoretic notions or intricate constructions, making the argument fully self-contained and rigorous.