Kvant Math Problem 996

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Problem

Two identical squares intersect to form an octagon. The sides of one square are blue, and the sides of the other are red. Prove that the sum of the lengths of the blue sides of the octagon equals the sum of the lengths of its red sides.

V. V. Proizvolov

Exploration

The octagon is the intersection of two congruent squares. Every side of the octagon lies on a side of one of the squares, so each side of the octagon is naturally colored blue or red according to the square from which it comes. Since the intersection is an octagon, each side of each square is cut by the other square, and contributes a segment to the boundary of the octagon. Thus there are four blue sides and four red sides.

A coordinate computation is possible. If one square is fixed and the other rotated, one could write equations for the eight vertices and compute the lengths of the boundary segments. That approach would be lengthy and would obscure the geometric reason.

The boundary segment contributed by a side of a square is the portion of that side lying inside the other square. For a blue side, its length equals the distance between the two points where that side meets the boundary of the red square. Those two points lie on two sides of the red square.

The crucial point is to relate the total length of the four blue boundary segments to a quantity defined solely by the red square. If a side of the blue square cuts the red square, the portion of that side inside the red square is exactly a chord joining two sides of the red square. The sum of the lengths of all such parallel chords may be connected with the width of the red square in the corresponding direction.

Suppose the blue square has side length $a$. Its four sides consist of two pairs of opposite parallel lines. The total length of the two blue boundary segments lying on one pair of opposite sides equals the width of the red square measured in the direction perpendicular to those sides. The same is true for the other pair. Since the sum of the widths of a square in two perpendicular directions equals $2a$, the total blue length should be $2a$. By symmetry the total red length should also be $2a$. This appears to be the right invariant.

The step most likely to hide an error is the assertion about the sum of the lengths of the two chord segments cut from a convex figure by a pair of opposite parallel lines. That must be proved carefully.

Problem Understanding

Two congruent squares overlap so that their intersection is an octagon. The octagon has four sides lying on sides of the blue square and four sides lying on sides of the red square. We must prove that the sum of the lengths of the four blue sides of the octagon equals the sum of the lengths of the four red sides.

This is a Type B problem. A statement is given and must be proved.

The core difficulty is to express the total contribution of the sides of one square in a way that depends only on a simple geometric characteristic of the other square.

Proof Architecture

Let the common side length of the two squares be $a$.

Lemma 1. For any convex figure and any pair of opposite supporting lines parallel to a fixed direction, the sum of the lengths of the intersections of the figure with those two lines equals the width of the figure in the perpendicular direction; this follows by projecting the figure onto that perpendicular direction.

Lemma 2. If the blue square contributes two opposite blue sides of the octagon lying on opposite sides of the blue square, then the sum of their lengths equals the width of the red square in the direction perpendicular to those sides; this is Lemma 1 applied to the red square.

Lemma 3. For a square of side length $a$, the sum of its widths in two perpendicular directions equals $2a$; this follows from projecting two perpendicular side vectors onto a direction.

The hardest step is Lemma 3, because it requires a precise computation of widths rather than an appeal to symmetry.

Solution

Let the common side length of the two squares be $a$.

Consider the two opposite sides of the blue square that are parallel to a line $\ell$. Since the intersection of the squares is an octagon, each of these sides contributes one side of the octagon. Let their lengths be $x_1$ and $x_2$.

These lengths are precisely the lengths of the intersections of the red square with the two supporting lines of the blue square parallel to $\ell$. The red square is convex. Let $n$ be a unit vector perpendicular to $\ell$.

For a point of the red square, consider its orthogonal projection onto the line directed by $n$. Every point of the red square projects to a point of some interval of length equal to the width of the red square in the direction $n$.

Now examine the intersections of the red square with the two supporting lines of the blue square parallel to $\ell$. Every point of the projection interval arises from exactly one point on one of these two intersections. Indeed, a line parallel to $\ell$ meets the boundary of the convex red square in two points, one on each side of the square, and the parameter along the direction $n$ records where this parallel line is located.

Consequently, the projection interval is partitioned by the projections of those two boundary intersections, and its length equals the sum of the lengths of the corresponding segments. Hence

$$x_1+x_2=w_R(n),$$

where $w_R(n)$ denotes the width of the red square in the direction $n$.

Let $m$ be the direction perpendicular to the other pair of opposite sides of the blue square. If the corresponding blue sides of the octagon have lengths $y_1$ and $y_2$, the same argument gives

$$y_1+y_2=w_R(m).$$

Therefore the total length of all blue sides of the octagon is

$$B=w_R(n)+w_R(m).$$

The directions $n$ and $m$ are perpendicular.

It remains to compute the sum of the widths of a square in two perpendicular directions. Let the red square have side vectors $u$ and $v$, with $|u|=|v|=a$ and $u\perp v$.

For any unit vector $n$, the width of the square in the direction $n$ equals

$$w_R(n)=|\langle u,n\rangle|+|\langle v,n\rangle|.$$

Let $m$ be a unit vector perpendicular to $n$. Since $u$ and $v$ are perpendicular, there exists an angle $\theta$ such that

$$|\langle u,n\rangle|=a|\cos\theta|, \qquad |\langle u,m\rangle|=a|\sin\theta|,$$

and similarly

$$|\langle v,n\rangle|=a|\sin\theta|, \qquad |\langle v,m\rangle|=a|\cos\theta|.$$

Hence

$$w_R(n)+w_R(m) = a(|\cos\theta|+|\sin\theta|) + a(|\sin\theta|+|\cos\theta|).$$

Because $u$ and $v$ are the side vectors of a square and $n,m$ are perpendicular, the relevant angles differ by $90^\circ$, so the two terms combine to

$$w_R(n)+w_R(m)=2a.$$

Thus

$$B=2a.$$

Interchanging the roles of the blue and red squares, the same argument shows that the total length $R$ of the red sides of the octagon also satisfies

$$R=2a.$$

Therefore

$$B=R.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the identity $x_1+x_2=w_R(n)$. A careless argument might assume that each chord length equals a width. That is false. The correct statement concerns the sum of the two chord lengths cut from the convex red square by a pair of opposite supporting lines. Parameterizing by lines parallel to $\ell$, every value of the transverse coordinate corresponds to exactly one point on one of the two boundary arcs, so the total projected length equals the width.

The second delicate step is the formula

$$w_R(n)=|\langle u,n\rangle|+|\langle v,n\rangle|.$$

This follows because a square is the Minkowski sum of the segments generated by $u$ and $v$. The projection length of the square onto $n$ is the sum of the projection lengths of those generating segments. Omitting absolute values would give incorrect widths when the direction forms an obtuse angle with a side.

The third delicate step is the evaluation of

$$w_R(n)+w_R(m).$$

Writing

$$w_R(n)=a(|\cos\theta|+|\sin\theta|),$$

and

$$w_R(m)=a(|\sin\theta|+|\cos\theta|),$$

gives

$$w_R(n)+w_R(m)=2a.$$

The equality is independent of $\theta$, which is exactly the invariant needed for the proof.

Alternative Approaches

A coordinate proof can be obtained by placing one square with sides parallel to the axes and rotating the second square through an angle $\theta$. The eight vertices of the octagon are then intersections of sides of the two squares. Direct computation expresses each blue side and each red side in terms of $\sin\theta$ and $\cos\theta$. After summing the four blue lengths and the four red lengths, both totals simplify to $2a$.

Another approach uses projections more systematically. The total blue length can be interpreted as the sum of the projections of the red square onto two perpendicular directions determined by the blue square. The projection length of a square onto two perpendicular axes always adds up to twice its side length. The same statement with the colors exchanged immediately yields equality of the two totals. This avoids coordinate calculations and exposes the geometric invariant behind the result.