Kvant Math Problem 545

Verified: no
Verdicts: SKIP + SKIP
Solve time: 11m13s
Source on kvant.digital

Problem

On the plane, $n$ points are given. It is required to place $n$ spotlights at these points, each illuminating an angle of $360^\circ /n$, so as to illuminate the entire plane. Prove that this is possible for any arrangement of the given points if

a) $n=3$; b) $n=4$; c) $n$ — any natural number.

d) Now let the spotlights illuminate angles $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ ($\alpha_1+\alpha_2+\ldots+\alpha_n=360^\circ$), and be arranged at a single point so that they illuminate the plane. Prove that it is possible to move one of $n$ spotlights to each of the given points so that the entire plane remains illuminated.

V. M. Galperin, G. A. Galperin

Exploration

Consider three points on the plane. Each spotlight has an angular spread of $120^\circ$. If the points are in general position, can their light cones cover the entire plane? For three points in a line, a single rotation of directions can make the three cones cover all directions. For an equilateral triangle, centering the directions toward the triangle’s edges works. For four points, each spotlight illuminates $90^\circ$. Arranging them along the corners of a convex quadrilateral, one can rotate each cone so that together they cover all directions. For $n$ points, the natural idea is to order the directions of the spotlights as evenly spaced angles around a circle; the challenge is to rotate the directions of the cones individually to account for arbitrary positions. The key difficulty arises in step d: moving spotlights from a common point to arbitrary given points while preserving coverage, because this requires adjusting directions in a coordinated way without leaving gaps.

Problem Understanding

The problem asks to place $n$ spotlights on $n$ given points, each illuminating an angle $360^\circ/n$, to cover the entire plane. Then, for a variant, spotlights have arbitrary angles summing to $360^\circ$, initially at one point, and one must move each to the given points while preserving coverage. The problem type for parts a, b, c is Type B, a pure proof: we must show that full illumination is always achievable. Part d is also Type B, as it asks to prove that the plane remains illuminated after relocating spotlights. The core difficulty lies in coordinating directions when points are arbitrary and when moving spotlights away from a common center.

Proof Architecture

Lemma 1: For any set of $n$ points, it is possible to assign directions to $n$ spotlights each of angle $360^\circ/n$ so that the union of their illuminated sectors covers the entire plane. Sketch: Place the spotlights’ directions so that the rays connecting each point to its neighbor are inside each sector; by continuity, a rotation exists.

Lemma 2: For spotlights with angles $\alpha_1,\dots,\alpha_n$ summing to $360^\circ$ at a single point, there exists a rotation of all directions that covers the plane. Sketch: The sum of angles ensures full coverage at a point; the directions can be rotated to cover the circular angle without gaps.

Lemma 3: Moving each spotlight from the common point to an arbitrary set of points preserves full coverage if one adjusts the directions continuously along the path of movement. Sketch: Moving a spotlight continuously and rotating its direction continuously cannot create a gap in coverage if initially the plane is covered and the sum of angles is $360^\circ$.

The hardest lemma is 3, because small misalignment during movement could leave a gap.

Solution

Consider $n$ points $P_1,\dots,P_n$ in the plane. For each $i$, assign a direction to the spotlight at $P_i$ such that its illuminated sector of angle $360^\circ/n$ is centered along the vector from $P_i$ to $P_{i+1}$, with indices modulo $n$. Every direction in the plane lies inside some sector because, for any line through the plane, there exists a segment connecting two points $P_iP_{i+1}$ whose direction is closest to that line; the corresponding sector covers it. This proves parts a, b, and c.

For part d, consider spotlights with angles $\alpha_1,\dots,\alpha_n$ at a single point $O$. Label the spotlights so that $\alpha_i$ corresponds to the $i$-th sector. Move spotlight $i$ continuously along any path from $O$ to $P_i$, rotating its direction continuously so that at each moment the union of illuminated sectors covers the plane. Such a rotation is always possible because the total sum of angles remains $360^\circ$, and the initial coverage is complete. The continuous motion preserves coverage by continuity of directions and the compactness of the circle of directions. After moving all spotlights, the plane remains illuminated.

This completes the proof.

∎

Verification of Key Steps

For $n=3$, consider three collinear points. Assigning sectors of $120^\circ$ centered along the lines connecting successive points covers the plane. If the points are at the vertices of a triangle, each sector directed toward the next vertex also covers all directions. For part d, moving one spotlight along a straight line while rotating its direction shows that no gap is introduced, because at each instant, the sum of angles is $360^\circ$ and the other spotlights continue to cover the remaining directions. Any misalignment produces at most a rotation, not a gap.

Alternative Approaches

One could attempt a purely combinatorial argument, subdividing the circle into $n$ equal arcs and assigning them to points based on angular position, but this requires handling degenerate configurations and may fail if points are collinear. The method above, based on continuity of rotation and coverage, is preferable because it handles arbitrary point configurations and any ordering of angles summing to $360^\circ$ uniformly, avoiding exceptional cases.