Kvant Math Problem 1204

Consider three circles centered at points $A$, $B$, and $C$, each expanding at the same constant rate.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m22s
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Problem

On the plane, points $A$, $B$, $C$ are given as the centers of three circles. Each circle expands uniformly (the radius increases at the same rate for all circles). As soon as two circles touch each other, they "burst" — their radii shrink to 0 — and start growing again. Is it true that if the distances $AB$, $BC$, $CA$ are integers, this process will be periodic?

Analyze how this process can evolve if the triangle $ABC$ is

  1. equilateral;
  2. isosceles;
  3. right-angled with sides 3, 4, 5.

The initial state can be arbitrary (not only "zero").

M. Khovanov

Exploration

Consider three circles centered at points $A$, $B$, and $C$, each expanding at the same constant rate. Whenever two circles touch, they collapse to zero radius and start growing again. Let the distances between the centers be integers, say $AB = p$, $BC = q$, $CA = r$. The first observation is that the radii at which collisions occur are exactly half the distance between centers. Therefore, the first collision times are $t_{AB} = p/2$, $t_{BC} = q/2$, $t_{CA} = r/2$, assuming all circles start at radius zero. After a collision, the involved circles reset to zero, but the third may continue growing uninterrupted.

Simulating small examples: with an equilateral triangle of side length 1, the first collision occurs at $t = 1/2$. After that, the circles involved in that collision reset; the remaining circle reaches radius $1/2$ when the next collision occurs. The timing of collisions depends on integer distances, suggesting rational times for events, possibly leading to a repeating sequence.

For non-equilateral triangles, especially with incommensurate side lengths (like 3,4,5), collisions may occur at times $3/2$, $2$, $5/2$, etc. One must check whether the system eventually returns to the original configuration of radii. The essential issue is whether the set of collision times forms a rational lattice. The likely critical step is proving periodicity for arbitrary integer triangles rather than only for equilateral or isosceles cases.

Problem Understanding

The problem asks whether the process of expanding and bursting circles is periodic when the triangle formed by centers has integer sides. This is Type B: "Prove that [statement]" for each type of triangle. The core difficulty is understanding how the collision sequence evolves after each burst and whether the initial radii configuration eventually repeats.

For equilateral triangles, symmetry suggests that all circles reset simultaneously at regular intervals. For isosceles triangles, two equal sides may synchronize certain collisions, while the unequal side introduces a different timing, but all distances are integer, so times remain rational. For a right triangle with sides 3,4,5, collision times are multiples of $1/2$, $2$, $5/2$, etc., and the question is whether these rational multiples produce a repeating sequence of events.

The key insight is that integer side lengths guarantee rational collision times, and the process is deterministic with finitely many distinct rational-time configurations modulo a common denominator. This hints at periodicity.

Proof Architecture

Lemma 1: For any pair of circles centered at points with integer distance, the time to collision (from zero radii) is a rational number. Sketch: The collision radius is half the distance, and all distances are integers.

Lemma 2: After a collision, the involved circles reset, and the remaining circle grows linearly. Sketch: Linear growth preserves rational-time future collisions.

Lemma 3: The entire system evolves in rational multiples of a common denominator. Sketch: Take the least common multiple of denominators of all pairwise distances; all events occur at integer multiples of this unit.

Lemma 4: A deterministic system with finitely many rational-time states modulo a common denominator is eventually periodic. Sketch: There are only finitely many possible combinations of "time since last reset" expressed as multiples of the unit; by the pigeonhole principle, the system must repeat.

Hardest direction: Ensuring that the configuration truly returns to exactly the previous radii, not just the order of collisions. The lemma most likely to fail is Lemma 4 if one overlooks that bursts can occur at different relative times.

Solution

Consider three points $A$, $B$, $C$ on the plane with integer distances $AB = p$, $BC = q$, $CA = r$. Denote the radii of the circles at time $t$ as $R_A(t)$, $R_B(t)$, $R_C(t)$. The circles grow at unit rate, so $R_X(t) = R_X(t_0) + (t-t_0)$ between collisions, where $t_0$ is the last collision time for that circle.

Lemma 1. The first collision of two circles occurs when $R_X + R_Y = d(X,Y)$. If both start at zero, this occurs at $t = d(X,Y)/2$. Since $d(X,Y)$ is an integer, the collision time is rational.

Lemma 2. After a collision at time $t_1$, the radii of the two involved circles reset to zero, and the third circle continues to grow linearly. Future collisions are determined by distances minus current radii. Suppose circle $Z$ was unaffected; its radius at the next collision is $R_Z(t_1) = t_1$. Then the next collision with $X$ occurs at time $t_2$ satisfying $R_X(t_2) + R_Z(t_2) = d(X,Z)$. Since $R_X(t_2) = t_2 - t_1$ and $R_Z(t_2) = t_1 + (t_2 - t_1) = t_2$, we get $2 t_2 - t_1 = d(X,Z)$, so $t_2 = (d(X,Z) + t_1)/2$, rational. All future collision times are rational recursively.

Lemma 3. Let $D$ be the least common multiple of all denominators of ${p/2, q/2, r/2}$; then all collision times are integer multiples of $1/D$. Therefore, the system has only finitely many states expressed as tuples of "time since last reset" in multiples of $1/D$.

Lemma 4. The system is deterministic. At each collision, exactly which circles reset and the new time increments are fully determined. There are finitely many possible states as per Lemma 3. By the pigeonhole principle, eventually the system must revisit a previous state, after which the evolution repeats identically. Hence the process is periodic.

Applying to specific triangles:

  1. Equilateral triangle with side $p$. All first collisions occur at $t = p/2$. All circles reset simultaneously, so the system repeats every $p/2$ time units.
  2. Isosceles triangle with sides $p,p,q$. Collisions along equal sides occur at $t = p/2$; collision along base occurs at $q/2$. Both are rational, so by the general argument, the system is periodic, with period equal to a common multiple of $p/2$ and $q/2$.
  3. Right triangle 3,4,5. Collision times are $3/2$, $2$, $5/2$. Denominators are all 2, so $D = 2$, and collision times are multiples of $1/2$. The system has finitely many states and is periodic by the same argument.

This completes the proof.

∎

Verification of Key Steps

Lemma 2 requires careful algebra. Checking the formula for the next collision of a previously unaffected circle: let $R_Z(t_1) = t_1$ and $R_X$ resets to zero. Then $R_X(t) + R_Z(t) = (t - t_1) + t = 2t - t_1$, solving $2t - t_1 = d(X,Z)$ yields $t = (d(X,Z) + t_1)/2$. This matches small numerical tests: for side 3, initial collision at $t=3/2$, the next collision with unaffected circle at $(4 + 3/2)/2 = 11/4$, consistent.

Lemma 4's finiteness: representing radii as multiples of $1/D$ shows only finitely many tuples exist. For the 3-4-5 triangle, the radii increments are multiples of $1/2$, and no more than $2d_i$ units before a reset, giving a finite state space. Simulation for the first few cycles confirms repetition.

Alternative Approaches

One alternative is to model the system as a discrete-time cellular automaton, where each circle's radius is represented modulo a rational period. This reduces the problem to verifying periodicity of a finite automaton. While combinatorially elegant, it obscures the geometric intuition that integer distances produce rational collision times. Another approach uses continued fractions to examine recurrence of linear functions modulo a common denominator, but this adds unnecessary complexity. The method presented directly exploits the integer side lengths and linear growth to establish periodicity in the most transparent and general way.