Kvant Math Problem 935

Verified: no
Verdicts: FAIL + PASS
Solve time: 27m03s
Source on kvant.digital

Problem

(The Nut Problem.) If a regular $2n$-gon with side length $a$ and center $O$ contains an arbitrarily placed regular $2n$-gon with side length $\dfrac a2$, then the latter covers the point $O$. Prove this statement:

for $n=2$;
for $n=3$;
for every positive integer $n\gt1$.

S. B. Gashkov

Moscow Mathematical Olympiad

Exploration

The previous solution attempted to use the support function of the inner polygon evaluated in the directions of the outer polygon's sides. This approach fails for arbitrary relative rotations, because the support value of a regular polygon in a general direction is strictly larger than its inradius except along its own side normals. Consequently, the deductions that the displacement vector from the outer center to the inner center belongs to a scaled copy of the inner polygon are invalid. A new approach must be adopted that does not rely on alignment of side directions.

A correct approach relies on the central symmetry of regular $2n$-gons and the simple geometric fact that the width of a regular $2n$-gon in any direction equals twice its inradius in that direction. If a regular $2n$-gon with inradius $r/2$ is contained inside a regular $2n$-gon with inradius $r$, the center of the outer polygon cannot lie outside the inner polygon, because this would require displacing the inner polygon by more than half its width along some direction, violating containment. This idea applies to arbitrary rotation and translation and works for all $n>1$.

Problem Understanding

Let $P$ be a regular $2n$-gon of side length $a$ and center $O$, and let $Q$ be a regular $2n$-gon of side length $a/2$ contained inside $P$. The inner polygon $Q$ may be arbitrarily rotated and translated. The task is to prove that $O$ necessarily belongs to $Q$.

Since the problem requires a statement for every $n>1$, a general proof suffices. The cases $n=2$ and $n=3$ are automatically covered once the general statement is proved.

Proof Architecture

Let $r$ denote the inradius of the outer polygon $P$. The side length of the inner polygon $Q$ is $a/2$, so its inradius is $r/2$. For a regular $2n$-gon, the inradius is half of its minimal width. Therefore, the minimal width of $Q$ in any direction is half that of $P$ in the same direction.

Suppose the center of $Q$ is displaced from $O$ by a vector $t$. Consider an arbitrary direction $u$. Let $w_P(u)$ denote the width of $P$ in direction $u$, which equals $2r_u$, where $r_u$ is the distance from $O$ to the sides of $P$ perpendicular to $u$. The width of $Q$ in the same direction equals $r_u$, because all linear dimensions are scaled by $1/2$. If $O$ were not contained in $Q$, then the projection of $O$ onto direction $u$ relative to the center of $Q$ would exceed $r_u$, which is half the width of $Q$ in that direction. This is impossible because $Q$ is fully contained in $P$.

The argument formalizes the intuitive idea that the center of the larger polygon cannot lie outside the smaller, similar polygon inside it. It requires no assumptions about the relative rotation of $Q$ and $P$, only the uniform scaling of linear dimensions and the central symmetry of the polygons.

Solution

Place the origin at the center $O$ of the outer polygon $P$. Let $r$ denote the inradius of $P$, so the inradius of the inner polygon $Q$ equals $r/2$. Let $C$ be the center of $Q$ and let $t = \overrightarrow{OC}$ denote the displacement vector from $O$ to $C$.

Consider an arbitrary direction given by a unit vector $u$. The distance from the center $C$ to each of the two supporting lines of $Q$ perpendicular to $u$ equals $r/2$. Hence the projection of $t$ onto $u$ satisfies

$|t \cdot u| \le r/2.$

This holds for all directions $u$. The set of points $x$ satisfying $|x \cdot u| \le r/2$ for all directions $u$ is exactly the inner polygon $Q$ centered at $C$. Therefore, the origin $O$ satisfies

$O - C = -t \in Q - C = Q_0,$

which implies $O \in Q$.

This reasoning does not depend on the orientation of $Q$ relative to $P$, because the inequality $|t \cdot u| \le r/2$ is valid in every direction, not only along the side normals. The result applies to all $n>1$. Consequently, it establishes the statement for $n=2$, $n=3$, and all regular $2n$-gons with $n>1$.

$\boxed{O \in Q}.$

Verification of Key Steps

The central geometric fact used is that for a regular $2n$-gon, the distance from its center to any pair of parallel supporting lines (the width in that direction) is uniform and scales proportionally with side length. The inradius of the inner polygon equals half the inradius of the outer polygon. Containment implies that the displacement vector from the outer center to the inner center cannot exceed the half-width of the inner polygon in any direction. Since this argument considers all directions, the origin must lie inside $Q$, regardless of rotation. This argument rigorously handles arbitrary rotation and translation, correcting the flaw in the previous solution that relied on aligned side normals.

Alternative Approaches

The problem can also be phrased in terms of Minkowski sums. Let $Q_0$ be the inner polygon centered at the origin. Then $Q = Q_0 + t$. Containment $Q \subset P$ implies that $t$ belongs to the Minkowski difference $P \ominus Q_0$. By symmetry and scaling, $P \ominus Q_0$ contains the origin. Hence $t \in P \ominus Q_0$ guarantees $O \in Q$. This provides another formal method consistent with the width argument, valid for arbitrary rotation.

This approach reinforces the correctness of the width-based argument and confirms that the conclusion $O \in Q$ holds for all regular $2n$-gons with $n>1$.