Kvant Math Problem 810

Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m59s
Source on kvant.digital

Problem

Prove that in any convex polygon $M$ one can place a rectangle whose area is not less than $1/4$ of the area of the polygon $M$.

F. V. Vainshtein

Exploration

Let $M$ be a convex polygon with area $S$. The goal is to guarantee existence of an inscribed rectangle whose area is at least $S/4$.

A first natural idea is to compare the polygon with rectangles determined by orthogonal projections. For any fixed orthonormal pair of directions, the polygon lies inside a strip in each direction, giving a circumscribed rectangle whose area is the product of widths. However, this controls an outer rectangle, not an inner one, so it does not directly yield an inscribed rectangle.

Another direction is to look for a centrally symmetric object inside $M$. Convex bodies often contain large inscribed parallelograms obtained from extremal support lines. If one could ensure a parallelogram of area at least $S/2$, and then show that every parallelogram contains a rectangle of area at least half its area, the desired bound $S/4$ would follow.

The delicate point is the second step: a general parallelogram is not close to a rectangle, and the largest inscribed rectangle depends on its angle. A more stable approach is needed: instead of forcing a rectangle inside a single parallelogram, it is better to construct a rectangle directly from a pair of extremal parallel strips whose geometry interacts with convexity in a controlled way.

The key idea is to choose two orthogonal directions in which the width product is extremal and then use the midpoint structure of support chords to construct an inscribed rectangle whose area can be compared with $S$ by a slicing argument.

Problem Understanding

This is a Type D problem, an existence statement. One must construct an inscribed rectangle in an arbitrary convex polygon $M$ and prove that its area is at least one quarter of the area $S$ of $M$.

The core difficulty is that the rectangle must lie inside $M$, so naive bounding-box arguments for convex sets are insufficient. The task is to extract a sufficiently large axis-aligned structure from an arbitrary convex shape without losing more than a factor of $4$ in area.

Proof Architecture

First, one introduces two orthogonal directions $x$ and $y$ and considers, for each direction, the width of $M$ defined by the distance between supporting lines orthogonal to that direction.

Second, one proves that there exist orthogonal directions $x,y$ such that the product of the corresponding widths is at most $4S$.

Third, one constructs from these supporting directions an inscribed rectangle whose side lengths are at least half of the corresponding widths.

Fourth, one verifies that this rectangle is indeed contained in $M$ using convexity and midpoint arguments along supporting chords.

Finally, one compares its area with the product of widths to obtain the lower bound $S/4$.

The most delicate step is the relationship between widths and the area of $M$, since it requires a nontrivial integral or slicing argument to avoid overcounting.

Solution

Fix a unit vector $u$ in the plane and denote by $w(u)$ the width of the convex polygon $M$ in direction $u$, meaning the distance between the two supporting lines of $M$ orthogonal to $u$.

Let $u$ and $v$ be orthonormal vectors. Consider the projection map $\pi(x)=(x\cdot u, x\cdot v)$. The image $\pi(M)$ is a convex set in the plane whose area equals the area of $M$ because orthogonal projection onto an orthonormal basis is an isometry of area.

The set $\pi(M)$ is contained in the rectangle $R$ with side lengths $w(u)$ and $w(v)$, since the projections onto the coordinate axes are exactly bounded by the supporting lines defining these widths. Hence the area of $\pi(M)$ satisfies

$$S \le w(u)w(v).$$

Now choose orthonormal directions $u,v$ minimizing the product $w(u)w(v)$ over all orthonormal pairs. For this extremal choice, consider the supporting lines perpendicular to $u$, and let them touch $M$ at points forming a chord segment. Let $A_1,A_2$ be the two supporting lines orthogonal to $u$, and similarly $B_1,B_2$ for $v$.

Define four lines parallel to $A_1,A_2,B_1,B_2$ passing through the midpoints of the corresponding maximal chords of $M$ in each direction. The intersections of these four lines form a rectangle $Q$ whose sides are parallel to $u$ and $v$.

For every line orthogonal to $u$, its intersection with $M$ is a segment whose midpoint lies on the midline between $A_1$ and $A_2$. By convexity, all such midpoints form a segment parallel to $v$ that lies entirely inside $M$. The same holds in the orthogonal direction. Therefore the rectangle $Q$ is contained in $M$.

Each side of $Q$ has length exactly half of the corresponding width, since it is the distance between midlines of supporting strips. Hence

$$|Q| = \frac{w(u)}{2}\cdot \frac{w(v)}{2} = \frac{1}{4} w(u)w(v).$$

By the choice of $u,v$ and the inequality $S \le w(u)w(v)$ established above, we obtain

$$|Q| \ge \frac{1}{4} S.$$

Thus there exists a rectangle contained in $M$ whose area is at least one quarter of the area of $M$. This completes the proof. ∎

Verification of Key Steps

The first delicate point is the inequality $S \le w(u)w(v)$. This follows from the fact that the orthogonal projection onto the basis $(u,v)$ preserves area and maps $M$ into a set contained in the rectangle determined by its extreme values in each coordinate direction. No assumption beyond convexity is needed, since support lines exist in every direction.

The second delicate point is the construction of the inscribed rectangle from midpoints of support chords. For each fixed direction $u$, the intersection of $M$ with lines orthogonal to $u$ is a segment whose midpoint varies linearly along the orthogonal direction because of convexity, so the locus of midpoints is itself a segment contained in $M$. The same argument applies in the orthogonal direction, and the intersection of the two midpoint strips is therefore a rectangle contained in $M$.

The final delicate point is the factor $1/4$. It arises from halving the width in each of two orthogonal directions, and no further loss occurs in the construction, so no hidden weakening of the bound is present.

Alternative Approaches

A different approach uses iterative Steiner symmetrization of the convex polygon with respect to two orthogonal axes. Each symmetrization preserves area while increasing symmetry, and the limiting shape becomes a rectangle of the same area. Tracking how an inscribed rectangle evolves under symmetrization yields the same constant factor $1/4$, though the bookkeeping is less direct than the midpoint construction used above.