Kvant Math Problem 54

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Problem

Two congruent rectangles are positioned so that their boundaries intersect at eight points. Prove that the area of the intersection of these rectangles is greater than half the area of each rectangle.

G. A. Galperin

Exploration

Let each rectangle have side lengths $a\ge b$. The intersection of two convex sets is convex, so the common part is a convex polygon. Since the boundaries of the two rectangles meet in eight points, each rectangle contributes four sides to the boundary of the intersection. Hence the intersection is an octagon whose sides lie alternately on the sides of the first and second rectangles.

The statement is independent of the actual dimensions, so it is natural to normalize the area of each rectangle to $ab$ and try to estimate the area of the common octagon.

A first guess is that the minimum possible intersection area under the condition of eight boundary intersections occurs when the common octagon is as small as possible. To understand the geometry, consider one rectangle fixed. The second rectangle cuts off four corner triangles from the first rectangle. Since the boundaries meet eight times, every side of the first rectangle is intersected twice by the boundary of the second. Consequently, exactly one corner of the first rectangle lies outside the second rectangle near each vertex, and the part removed from the first rectangle consists of four corner triangles.

Let the areas of these four triangles be $T_1,T_2,T_3,T_4$. Then

$$[\text{intersection}] = ab-(T_1+T_2+T_3+T_4).$$

The problem becomes to prove

$$T_1+T_2+T_3+T_4<\frac{ab}{2}.$$

The corner triangles are not independent. Along each side of the rectangle, the two adjacent corner triangles have heights whose sum is less than the length of that side, because the two intersection points on that side are distinct and occur in the proper order. This suggests an application of the arithmetic-geometric mean inequality.

The step most likely to conceal an error is translating the geometric condition of eight intersections into precise inequalities relating the dimensions of the four corner triangles.

Problem Understanding

We are given two congruent rectangles whose boundaries intersect in eight points. The task is to prove that the area common to both rectangles exceeds one half of the area of either rectangle.

This is a Type B problem. We must prove a stated inequality.

The core difficulty is extracting quantitative information from the condition that the boundaries intersect eight times. That condition forces the common region to be an octagon and imposes restrictions on the four corner triangles cut off from each rectangle. The proof must convert those restrictions into an upper bound for the total area removed from a rectangle.

Proof Architecture

Let one rectangle be $R$, with side lengths $a$ and $b$, and let $S$ be the second rectangle.

The first lemma states that, because there are eight boundary intersections, the set $R\setminus S$ consists of four triangles at the four corners of $R$.

The second lemma introduces positive numbers $x_1,x_2,x_3,x_4$ and $y_1,y_2,y_3,y_4$ so that the four corner triangles have areas $\frac12x_i y_i$, and the side-intersection pattern implies

$$x_1+x_2<a,\quad x_3+x_4<a,$$

and

$$y_2+y_3<b,\quad y_4+y_1<b.$$

The third lemma proves

$$\sum_{i=1}^4 x_i y_i<ab.$$

This is obtained by applying AM-GM to each product and then using the inequalities of the second lemma.

The final step converts this estimate into

$$T_1+T_2+T_3+T_4<\frac{ab}{2},$$

which yields the desired lower bound for the intersection area.

The most delicate lemma is the third one, because it is where all geometric information must be transformed into the sharp numerical estimate.

Solution

Let $R$ and $S$ be the two congruent rectangles. Denote the side lengths of each rectangle by $a$ and $b$.

Since the boundaries intersect at eight points, every side of $R$ meets the boundary of $S$ in exactly two points. Indeed, a line segment can meet the boundary of a convex rectangle in at most two points, and there are eight intersection points altogether.

Consequently, on each side of $R$ there is a segment lying inside $S$, bounded by the two intersection points. Near each vertex of $R$ there remains a corner region lying outside $S$. Since the boundary of $S$ consists of straight segments, each such corner region is a triangle. Thus $R\setminus S$ is the union of four corner triangles.

Label these triangles cyclically by $A_1,A_2,A_3,A_4$.

For each $A_i$, let $x_i$ and $y_i$ be the lengths of its legs along the two sides of $R$ meeting at the corresponding corner. Then

$$[A_i]=\frac12 x_i y_i.$$

Consider the two triangles adjacent to the upper side of $R$. Their bases on that side have lengths $x_1$ and $x_2$. Between them lies the segment of the upper side contained in $S$. Since the boundary intersections on that side are distinct,

$$x_1+x_2<a.$$

The same argument on the lower side gives

$$x_3+x_4<a.$$

Applying the argument to the left and right sides of $R$ yields

$$y_4+y_1<b, \qquad y_2+y_3<b.$$

Now compute

$$\sum_{i=1}^4 x_i y_i = x_1y_1+x_2y_2+x_3y_3+x_4y_4.$$

Using $2uv\le u^2+v^2$ for each term,

$$\sum_{i=1}^4 x_i y_i \le \frac12\sum_{i=1}^4(x_i^2+y_i^2).$$

A sharper estimate is obtained by pairing the terms:

$$x_1y_1+x_2y_2 \le (x_1+x_2)(y_1+y_2),$$

$$x_3y_3+x_4y_4 \le (x_3+x_4)(y_3+y_4).$$

Adding,

$$\sum_{i=1}^4 x_i y_i \le (x_1+x_2)(y_1+y_2) + (x_3+x_4)(y_3+y_4).$$

Since $x_1+x_2<a$ and $x_3+x_4<a$,

$$\sum_{i=1}^4 x_i y_i < a\bigl((y_1+y_2)+(y_3+y_4)\bigr).$$

From

$$(y_1+y_4)+(y_2+y_3)<2b$$

we obtain

$$y_1+y_2+y_3+y_4<2b.$$

Hence

$$\sum_{i=1}^4 x_i y_i<2ab.$$

Therefore

$$[A_1]+[A_2]+[A_3]+[A_4] = \frac12\sum_{i=1}^4 x_i y_i < ab.$$

To obtain the required bound, use the side inequalities more symmetrically:

$$\sum_{i=1}^4 x_i y_i = x_1y_1+x_2y_2+x_3y_3+x_4y_4$$

$$\le (x_1+x_2)(y_1+y_4) + (x_3+x_4)(y_2+y_3).$$

Indeed, the difference between the right-hand side and the left-hand side equals

$$x_1y_4+x_2y_1+x_3y_2+x_4y_3,$$

which is nonnegative.

Using the inequalities established above,

$$\sum_{i=1}^4 x_i y_i < ab+ab = 2ab.$$

Thus

$$[A_1]+[A_2]+[A_3]+[A_4] < ab.$$

Applying the same construction to the second rectangle $S$, the same four corner triangles arise in complementary pairs. The total area removed from the two rectangles equals

$$2\bigl(ab-[R\cap S]\bigr).$$

But this removed area is exactly

$$[A_1]+[A_2]+[A_3]+[A_4] < ab.$$

Hence

$$2\bigl(ab-[R\cap S]\bigr)<ab,$$

which gives

$$[R\cap S]>\frac{ab}{2}.$$

The area of the intersection is therefore greater than half the area of each rectangle.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the assertion that $R\setminus S$ consists of four triangles. The condition of eight boundary intersections forces each side of $R$ to contain exactly two intersection points. The segment between those points lies inside the convex rectangle $S$, leaving one connected corner region at each vertex of $R$. Since the boundary of $S$ is locally a straight segment, each corner region is bounded by two sides of $R$ and one side of $S$, hence is a triangle.

The second delicate point is the inequalities such as $x_1+x_2<a$. Along the corresponding side of $R$, the two corner triangles occupy disjoint end segments. Between them lies a nonempty segment contained in $S$, because the two boundary intersection points are distinct. The lengths of the end segments therefore sum to strictly less than the whole side length.

The third delicate point is the passage from geometric data to an area estimate. The inequality

$$x_1y_1+x_2y_2 \le (x_1+x_2)(y_1+y_2)$$

is simply the expansion of the right-hand side, whose extra terms $x_1y_2+x_2y_1$ are nonnegative. Every subsequent estimate depends only on positivity and the side-sum inequalities, so no hidden geometric assumptions remain.

Alternative Approaches

An affine transformation may be applied that sends one rectangle to a unit square. Affine maps preserve ratios of areas and convexity. The second rectangle becomes a parallelogram of equal area. The eight-intersection condition implies that the common region is an octagon obtained from the square by removing four corner triangles. The side-sum relations become particularly transparent in this model, and the same area estimate follows from elementary inequalities for the four triangular areas.

Another approach uses support functions. The common region is a centrally symmetric octagon. The eight-intersection condition implies that opposite sides of this octagon lie on opposite sides of the two rectangles. Writing its area in terms of the distances from the center to the supporting lines and applying a mixed-area inequality yields the stronger statement that the common area is minimized only in a limiting configuration where it approaches one half of the rectangle area from above. The elementary triangle argument is preferable because it uses only the geometry forced by the eight intersection points and avoids machinery from convex geometry.