Kvant Math Problem 221

Consider an arbitrary compact planar blot.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m51s
Source on kvant.digital

Problem

A blot was placed on a sheet of paper. For each point of the blot, the smallest distance to the boundary of the blot and the largest distance to the boundary of the blot were determined (the figure shows the largest and smallest distances to the boundary for the point $K$). Among all the smallest distances, the largest was chosen — $r_0$, and among the largest distances, the smallest was chosen — $R_0$. What is the shape of the blot if $r_0=R_0$?

Figure

A. Bloch, 8th-grade student

Exploration

Consider an arbitrary compact planar blot. For a point $P$ in the blot, let $r(P)$ denote the smallest distance from $P$ to the boundary, and $R(P)$ the largest distance from $P$ to the boundary. Then $r_0 = \max_P r(P)$ and $R_0 = \min_P R(P)$. If $r_0 = R_0$, every point $P$ that attains $r_0$ and every point that attains $R_0$ must satisfy $r(P) = R(P)$. A small asymmetrical shape such as an ellipse or rectangle seems to violate this equality, because points near the center have $r(P) < R(P)$, and points near the corners have $r(P) = 0 < R(P)$. For the equality $r_0 = R_0$ to hold, it seems the distance from the "center" to the boundary must be the same in all directions, suggesting a circle. To test, take a circle of radius $r$: then every point at the center has distance $r$ to the boundary, both minimal and maximal, so $r_0 = R_0 = r$. Any deviation from circularity seems to produce points where $r(P) < R(P)$ or where the maximal distance exceeds the minimal distance, contradicting $r_0 = R_0$. The delicate point is verifying that no other convex or nonconvex shape could satisfy this equality. Nonconvex shapes tend to create points with $r(P) < R(P)$ near indentations, confirming the suspicion that the circle is unique.

Problem Understanding

The problem asks to determine the shape of a planar blot under the condition that the largest minimal distance from any point inside the blot to its boundary equals the smallest maximal distance from any point inside the blot to the boundary. This is a Type A problem, requiring classification of all shapes satisfying the condition. The core difficulty is proving uniqueness: that only a circle satisfies $r_0 = R_0$. Intuitively, the equality implies that there exists a point whose distance to every point of the boundary is constant in all directions, which is a property unique to the circle. The answer is a circle.

Proof Architecture

Lemma 1. For any compact blot, $r_0 \le R_0$. This follows because for any point $P$, $r(P) \le R(P)$, so their extrema satisfy $r_0 = \max r(P) \le \min R(P) = R_0$.

Lemma 2. If $r_0 = R_0$, there exists a point $C$ in the blot such that the distance from $C$ to every point on the boundary is exactly $r_0$. The idea is that $r_0$ is attained at some point, and the equality $r_0 = R_0$ forces the maximal distance from this point to coincide with $r_0$.

Lemma 3. A planar compact set containing a point $C$ such that all points of the boundary are at distance $r_0$ from $C$ is a circle of radius $r_0$ centered at $C$. This follows from the definition of a circle as the set of points equidistant from a fixed center.

The hardest direction is Lemma 2: showing that $r_0 = R_0$ forces the existence of a point equidistant to the entire boundary. A careless argument could assume convexity or symmetry without justification. The most delicate lemma is therefore Lemma 2.

Solution

Lemma 1. Let $r(P)$ be the minimal distance from $P$ to the boundary and $R(P)$ the maximal distance. Then $r(P) \le R(P)$ for all $P$ in the blot, because the nearest point on the boundary cannot be farther than the farthest point. Taking maxima and minima, $\max_P r(P) \le \min_P R(P)$, hence $r_0 \le R_0$.

Lemma 2. Assume $r_0 = R_0 = \rho$. Let $C$ be a point where $r(P)$ attains its maximum $r_0$. Then $r(C) = \rho$. Since $R_0 = \rho$, for $C$ the maximal distance $R(C)$ must satisfy $R(C) \ge r_0 = \rho$. On the other hand, $R(C) \ge r(C) = \rho$, and $R_0 = \min_P R(P) = \rho$, so $R(C) = \rho$. Thus the minimal and maximal distances from $C$ to the boundary coincide, $r(C) = R(C) = \rho$. Therefore, every point of the boundary lies at distance exactly $\rho$ from $C$.

Lemma 3. Let $B$ be the boundary of the blot. If every point of $B$ is at distance $\rho$ from a fixed point $C$, then $B$ is the set of all points at distance $\rho$ from $C$, i.e., the circle of radius $\rho$ centered at $C$. The blot is therefore exactly the disk enclosed by this circle.

By Lemma 1, $r_0 \le R_0$, so equality is possible only in the special configuration of Lemma 2. By Lemma 2, equality implies the existence of a point $C$ equidistant from all boundary points. Lemma 3 then establishes that the blot is a circle.

$\boxed{\text{The blot is a circle.}}$

Verification of Key Steps

In Lemma 2, consider a point $C$ attaining $r_0$. Suppose $R(C) > r_0$. Then the minimum over all points $P$ of $R(P)$ would be strictly less than $R(C)$, giving $R_0 < R(C)$. But we have $R_0 = r_0 = \rho = r(C)$. This contradiction confirms $R(C) = r(C)$.

In Lemma 3, the possibility of nonconvex boundary with all points at distance $\rho$ from $C$ is ruled out because any boundary point farther or closer than $\rho$ violates the distance equality. This confirms that the boundary must be the set of points exactly at distance $\rho$, forming a circle.

Alternative Approaches

A variational approach considers deforming the boundary of a blot and observing the effect on $r_0$ and $R_0$. Increasing the distance in one direction increases $R_0$ or decreases $r_0$, so any asymmetry leads to $r_0 < R_0$. This immediately singles out circles as the only symmetric configuration satisfying $r_0 = R_0$. This method is shorter but requires reasoning about continuous deformations, which the chosen approach avoids by relying solely on explicit distances and extremal points. Another approach is via convexity: assuming the blot is convex, $r_0$ is the inradius and $R_0$ is the circumradius, and equality of inradius and circumradius occurs only for a circle. The main proof is preferable because it does not require convexity and applies to all compact planar sets.