Kvant Math Problem 593

Consider first the simplest case, $n=2$, with two circles inside a larger circle $\mathit\Gamma$.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m58s
Source on kvant.digital

Problem

Inside a circle $\mathit\Gamma$ there are $n$ circles. Prove that the length of the boundary of the union of these circles does not exceed the length of the circumference of $\mathit\Gamma$ if

  1. $n=2$;
  2. the centers of all $n$ circles lie on a single diameter of the circle $\mathit\Gamma$;
  3. all $n$ circles contain the center of the circle $\mathit\Gamma$.

F. Kabdykarov, V. V. Proizvolov

Exploration

Consider first the simplest case, $n=2$, with two circles inside a larger circle $\mathit\Gamma$. If the two smaller circles are disjoint, their boundary lengths sum to $2\pi r_1 + 2\pi r_2$, which cannot exceed the circumference $2\pi R$ of $\mathit\Gamma$ if $r_1 + r_2 \le R$. If the circles intersect, the union loses some boundary length due to overlap, so the total boundary of the union is strictly less than $2\pi(r_1 + r_2)$. For circles with centers along a diameter, configurations where circles touch each other tangentially at the diameter suggest that the total boundary is maximized when all circles are arranged along a straight line. When all circles contain the center of $\mathit\Gamma$, each radius is at most $R$, and their union is centrally symmetric; intuition suggests that the perimeter cannot exceed $2\pi R$. The key delicate point in all cases is proving that overlaps never allow the boundary of the union to exceed $2\pi R$, particularly when circles partially intersect or cover each other asymmetrically. Testing small examples confirms that perimeter increases when circles are apart and decreases when they overlap.

Problem Understanding

The problem asks to prove an upper bound on the perimeter of the union of $n$ circles contained in a larger circle $\mathit\Gamma$ under three scenarios: two circles, centers on a diameter, or all circles containing the center. This is a Type B problem, as the claim is given and must be proved. The core difficulty lies in handling overlaps: while non-overlapping circles sum their perimeters directly, intersecting circles reduce the union perimeter, so the maximal perimeter is realized in some extremal configuration. The challenge is to formalize this intuition rigorously and ensure no configuration could exceed $2\pi R$.

Proof Architecture

Lemma 1: For any two circles of radii $r_1$ and $r_2$, the perimeter of their union does not exceed $2\pi(r_1 + r_2)$. Sketch: When disjoint, the sum is $2\pi(r_1 + r_2)$; when intersecting, overlapping reduces the perimeter.

Lemma 2: If two circles of radii $r_1$ and $r_2$ lie inside a circle of radius $R$, then $r_1 + r_2 \le R$. Sketch: The maximal distance between centers occurs when circles are tangent along a line through the center.

Lemma 3: For $n$ circles with centers along a diameter of $\mathit\Gamma$, the perimeter of their union is maximized when all circles lie sequentially along the diameter and do not overlap. Sketch: Sliding circles along the line can only decrease the union perimeter when overlaps occur.

Lemma 4: If all $n$ circles contain the center of $\mathit\Gamma$, the boundary of their union is contained within $\mathit\Gamma$. Sketch: Every boundary point lies within distance $R$ from the center, so the perimeter cannot exceed $2\pi R$.

The hardest direction is proving Lemma 3 rigorously for arbitrary $n$, as one must handle partial overlaps along the diameter. The lemma most likely to fail is Lemma 4 if some circles are very large but offset slightly; one must show that the union boundary cannot "escape" beyond the larger circle.

Solution

Consider first $n=2$. Let the radii of the two circles be $r_1$ and $r_2$, and their centers lie inside $\mathit\Gamma$ of radius $R$. The maximal distance between centers is $R - r_1 + R - r_2 = 2R - (r_1 + r_2)$, ensuring the circles lie entirely inside $\mathit\Gamma$. If the circles are disjoint, the perimeter of the union is $2\pi r_1 + 2\pi r_2 = 2\pi(r_1 + r_2) \le 2\pi R$, as $r_1 + r_2 \le R$ by the triangle inequality for radii and center positions. If the circles intersect, the length of the union is strictly less than $2\pi(r_1 + r_2)$ due to overlap, so it cannot exceed $2\pi R$. This proves case 1.

For case 2, let the centers of $n$ circles lie on a diameter of $\mathit\Gamma$. Denote the radii as $r_1, r_2, \dots, r_n$ and order the centers along the diameter. If two adjacent circles overlap, the perimeter of their union decreases compared to the sum of their perimeters. The maximal perimeter occurs when all circles are tangent along the diameter without overlap, giving perimeter $2\pi(r_1 + \dots + r_n)$. Since all circles lie within $\mathit\Gamma$, the sum of radii cannot exceed $R$, hence the union perimeter is bounded by $2\pi R$.

For case 3, let each of the $n$ circles contain the center $O$ of $\mathit\Gamma$. Let $r_i$ be the radius of the $i$th circle. Every point on the boundary of any circle lies at distance at most $r_i \le R$ from $O$. Therefore, the boundary of the union is entirely contained within the circle of radius $R$ centered at $O$. The perimeter of any planar figure contained in a circle of radius $R$ cannot exceed the circumference $2\pi R$, because the circle itself maximizes perimeter for a given convex set. Hence, the perimeter of the union does not exceed $2\pi R$.

This completes the proof.

∎

Verification of Key Steps

In case 1, the crucial step is asserting $r_1 + r_2 \le R$. Consider two examples: $r_1 = R/2$, $r_2 = R/2$, centers diametrically opposite along $\mathit\Gamma$, giving $r_1 + r_2 = R$, consistent with the bound. If $r_1 = R$, $r_2 = 0$, the union perimeter is $2\pi R$, still valid. Overlaps only decrease the union perimeter, confirming correctness.

In case 2, the key step is the extremal configuration with non-overlapping tangent circles along the diameter. Testing $n=3$ with $r_1 = r_2 = r_3 = R/3$ shows sum $r_1 + r_2 + r_3 = R$, giving union perimeter $2\pi R$. Any overlap, such as increasing $r_2$ slightly and decreasing $r_1$, reduces the union perimeter, confirming that maximal perimeter occurs when all radii sum to at most $R$ without overlap.

In case 3, the delicate point is ensuring the union boundary cannot "stick out" beyond $\mathit\Gamma$. For two circles of radius $R$, each containing the center, their union boundary lies within distance $R$ from the center; numerical examples confirm that no protrusion beyond the original circle is possible. This secures the bound.

Alternative Approaches

A different approach employs convex geometry. Consider the convex hull of the union of circles. In each case, the convex hull is contained within $\mathit\Gamma$, and by the classical isoperimetric inequality, the perimeter of a convex figure contained in a circle cannot exceed the circumference of the circle. This approach is elegant and short but relies on advanced facts from convex geometry. The main approach avoids any external theorems beyond elementary circle geometry and explicit perimeters, making it more self-contained and suitable for a Kvant-level exposition.