Kvant Math Problem 21

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Problem

Inside a square of side length 1, several circles are placed whose total circumference is 10. Prove that there exists a line intersecting at least four of these circles.

Evening Mathematical School at Moscow State University (MSU)

Exploration

Let the circles have radii $r_1,r_2,\dots,r_n$. Since their total circumference is $10$,

$$2\pi\sum_{i=1}^n r_i=10, \qquad \sum_{i=1}^n r_i=\frac5\pi.$$

The statement concerns lines intersecting circles. A natural idea is to average over a family of lines. Since all circles lie inside the unit square, every vertical line has equation $x=t$ with $0\le t\le1$.

For a circle of radius $r$, the set of vertical lines meeting it is exactly the interval of $x$-coordinates between the leftmost and rightmost points of the circle. Its length is $2r$. Hence, if we count for each vertical line how many circles it meets, then the average over all vertical lines should be the sum of these lengths, namely

$$\sum 2r_i=\frac{10}{\pi}.$$

Numerically,

$$\frac{10}{\pi}\approx3.183.$$

If every vertical line met at most three circles, the average number of intersections would be at most $3$, contradicting $3.183>3$. This already suggests the result.

The only step that might hide an error is the averaging argument. One must justify carefully that integrating the number of circles met by a vertical line indeed equals the sum of the widths $2r_i$ of the circles.

Problem Understanding

Several circles are contained in a unit square. The sum of their circumferences equals $10$. We must prove that some line intersects at least four of the circles.

This is a Type B problem. The task is to prove the existence of a line with a certain property.

The core difficulty is to convert the global information about the total circumference into information about how many circles a typical line intersects. An averaging argument over all vertical lines in the square is the natural mechanism for doing this.

Proof Architecture

The first lemma states that for a circle of radius $r$, the set of vertical lines intersecting the circle has total parameter length $2r$; this follows because such lines correspond exactly to the horizontal projection of the circle.

The second lemma states that if $N(x)$ denotes the number of circles intersected by the vertical line $x=\text{constant}$, then

$$\int_0^1 N(x),dx=\sum_{i=1}^n 2r_i.$$

This follows by summing the contributions of the individual circles.

The third lemma states that

$$\int_0^1 N(x),dx=\frac{10}{\pi}>3.$$

This is obtained from the given total circumference.

The final step argues by contradiction. If every vertical line met at most three circles, then $N(x)\le3$ for all $x$, giving

$$\int_0^1 N(x),dx\le3,$$

which contradicts the previous lemma.

The most delicate point is the second lemma, where the integral of the counting function must be identified with the sum of the widths $2r_i$.

Solution

For each $x\in[0,1]$, let $N(x)$ denote the number of given circles intersected by the vertical line

$$\ell_x:\ x=\text{constant}.$$

Consider one of the circles, with center $(a,b)$ and radius $r$.

The line $\ell_x$ intersects this circle exactly when the horizontal distance from the center to the line does not exceed the radius:

$$|x-a|\le r.$$

Thus the set of values of $x$ for which $\ell_x$ meets this circle is the interval

$$[a-r,a+r].$$

Its length is $2r$.

Let $I_i(x)$ be the indicator function of the event that $\ell_x$ intersects the $i$-th circle. Then

$$N(x)=\sum_{i=1}^n I_i(x).$$

Integrating over $[0,1]$ and using linearity of the integral,

$$\int_0^1 N(x),dx = \sum_{i=1}^n \int_0^1 I_i(x),dx.$$

Since every circle lies inside the unit square, the entire interval $[a_i-r_i,a_i+r_i]$ is contained in $[0,1]$. Therefore

$$\int_0^1 I_i(x),dx=2r_i.$$

Hence

$$\int_0^1 N(x),dx = \sum_{i=1}^n 2r_i.$$

The total circumference of the circles is $10$, so

$$\sum_{i=1}^n 2\pi r_i=10.$$

Therefore

$$\sum_{i=1}^n 2r_i = \frac{10}{\pi}.$$

Consequently,

$$\int_0^1 N(x),dx=\frac{10}{\pi}.$$

Since

$$\frac{10}{\pi}>3,$$

the average value of $N(x)$ on the interval $[0,1]$ is greater than $3$.

Assume, for contradiction, that every line $\ell_x$ intersects at most three circles. Then

$$N(x)\le3 \qquad\text{for all }x\in[0,1].$$

Integrating,

$$\int_0^1 N(x),dx\le3,$$

which contradicts

$$\int_0^1 N(x),dx=\frac{10}{\pi}>3.$$

Hence there exists some $x_0\in[0,1]$ such that

$$N(x_0)>3.$$

Since $N(x_0)$ is an integer, we have $N(x_0)\ge4$. The vertical line $\ell_{x_0}$ therefore intersects at least four of the circles.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the claim that a circle of radius $r$ contributes exactly $2r$ to the integral. Let the center be $(a,b)$. The line $x=t$ meets the circle if and only if

$$(t-a)^2\le r^2.$$

This is equivalent to

$$a-r\le t\le a+r.$$

The set of admissible parameters has length $2r$, so the contribution is exactly $2r$.

The second delicate step is the passage from individual circles to the counting function. For each $x$, the quantity $N(x)$ equals the sum of the indicators $I_i(x)$. Integrating first and summing afterwards gives

$$\int_0^1 N(x),dx = \sum_i \int_0^1 I_i(x),dx.$$

A careless argument might count overlapping circles incorrectly, but the indicator formulation shows that every circle contributes independently and exactly once.

The third delicate step is the contradiction from the average. Since the interval has length $1$, the average value of $N(x)$ equals its integral. If every value were at most $3$, the average could not exceed $3$. The computed average is $10/\pi\approx3.183$, forcing at least one value strictly larger than $3$.

Alternative Approaches

One may phrase the argument probabilistically. Choose a vertical line uniformly at random from all vertical lines crossing the square. For a circle of radius $r$, the probability that the chosen line intersects it is $2r$. The expected number of circles hit is therefore

$$\sum_i 2r_i=\frac{10}{\pi}>3.$$

If every vertical line intersected at most three circles, the expectation could not exceed $3$. Hence some vertical line must intersect at least four circles.

This is essentially the same averaging argument expressed in the language of expectation. The integral formulation used in the main proof is preferable because it avoids introducing probabilistic terminology and keeps the geometric counting completely explicit.