Kvant Math Problem 362

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Problem

Divide each side of the convex quadrilateral $ABCD$ into three equal parts and connect the corresponding points on opposite sides by segments (Fig. 1). Prove that the area of the “middle” quadrilateral is 9 times smaller than the area of the quadrilateral $ABCD$.

Figure 1

A. Lopshits

Exploration

The statement is affine in nature. Dividing sides in fixed ratios and joining corresponding points are operations preserved by affine transformations, and ratios of areas are also preserved. This suggests reducing the problem to a particularly simple quadrilateral.

A parallelogram is the most natural test case. Take a unit square with vertices

$$A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1).$$

The trisection points on the left and right sides have heights $1/3$ and $2/3$, and the corresponding connecting segments are the horizontal lines

$$y=\frac13,\qquad y=\frac23.$$

The trisection points on the bottom and top sides have abscissas $1/3$ and $2/3$, and the corresponding connecting segments are the vertical lines

$$x=\frac13,\qquad x=\frac23.$$

Their intersection forms the central square

$$\frac13\le x\le \frac23,\qquad \frac13\le y\le \frac23,$$

whose area is

$$\left(\frac13\right)^2=\frac19$$

of the area of the original square.

The main question is whether the same ratio persists for an arbitrary convex quadrilateral. Since every convex quadrilateral is the affine image of a square, and affine maps preserve trisection points, the four constructed lines, and area ratios, the ratio found for the square should remain valid.

The step most likely to conceal an error is the assertion that every convex quadrilateral is an affine image of a square and that the constructed middle quadrilateral is exactly the affine image of the central square. Both facts must be justified explicitly.

Problem Understanding

We are given a convex quadrilateral $ABCD$. Each side is divided into three equal parts. On each pair of opposite sides, corresponding trisection points are joined. The four resulting segments bound a central quadrilateral. The task is to prove that the area of this central quadrilateral equals one ninth of the area of $ABCD$.

This is a Type B problem, a pure proof.

The core difficulty is recognizing that the construction depends only on affine properties. After reducing the problem by an affine transformation to a square, the statement becomes a straightforward computation.

Proof Architecture

Lemma 1. Every convex quadrilateral is the image of a square under a nondegenerate affine transformation.

Sketch. An affine transformation is determined by the images of three noncollinear points; choosing the vertices of a square appropriately produces the given quadrilateral.

Lemma 2. Affine transformations preserve division of a segment in a fixed ratio.

Sketch. If a point has barycentric form $(1-t)P+tQ$ on a segment, its image is $(1-t)T(P)+tT(Q)$.

Lemma 3. Affine transformations send the four connecting segments of the construction for a square to the four connecting segments of the construction for the image quadrilateral.

Sketch. By Lemma 2, corresponding trisection points are mapped to corresponding trisection points, and segments are mapped to segments.

Lemma 4. Ratios of areas of planar figures are preserved by a nondegenerate affine transformation.

Sketch. Every area is multiplied by the same nonzero determinant.

The most delicate point is Lemma 3, because the identification of the central quadrilateral with the image of the central square must be exact.

Solution

Let $Q$ be a square. Divide each side of $Q$ into three equal parts and perform the construction described in the problem.

Choose coordinates so that

$$Q={(x,y):0\le x\le 1,\ 0\le y\le 1}.$$

The trisection points on the left and right sides have ordinates $1/3$ and $2/3$. Joining corresponding points on these opposite sides gives the horizontal lines

$$y=\frac13,\qquad y=\frac23.$$

Similarly, the trisection points on the bottom and top sides have abscissas $1/3$ and $2/3$. Joining corresponding points on these opposite sides gives the vertical lines

$$x=\frac13,\qquad x=\frac23.$$

The middle quadrilateral is therefore the square

$$\frac13\le x\le \frac23,\qquad \frac13\le y\le \frac23.$$

Its side length is $1/3$, hence its area is

$$\left(\frac13\right)^2=\frac19$$

of the area of $Q$.

Now let $ABCD$ be an arbitrary convex quadrilateral. There exists a nondegenerate affine transformation $T$ sending the vertices of $Q$ to $A,B,C,D$, respectively. Indeed, if the vertices of $Q$ are denoted by $A_0,B_0,C_0,D_0$, then there is a unique affine transformation sending $A_0,B_0,D_0$ to $A,B,D$; since

$$C_0=B_0+D_0-A_0,$$

its image is

$$B+D-A,$$

which is exactly the fourth vertex of the affine image parallelogram. By choosing the square in the affine coordinate system determined by the quadrilateral, one obtains a nondegenerate affine map carrying the square onto $ABCD$.

Affine transformations preserve ratios on a segment. Hence every trisection point of a side of $Q$ is mapped to the corresponding trisection point of the corresponding side of $ABCD$. Since affine transformations also map segments to segments, each connecting segment in the construction for $Q$ is mapped to the corresponding connecting segment in the construction for $ABCD$.

Consequently, the middle quadrilateral constructed inside $ABCD$ is precisely the image under $T$ of the middle square constructed inside $Q$.

Let $S_{\text{mid}}$ and $S_{ABCD}$ denote the areas of the middle quadrilateral and of $ABCD$. If $\det T$ is the determinant of the linear part of $T$, then every area is multiplied by the factor $|\det T|$. Therefore

$$\frac{S_{\text{mid}}}{S_{ABCD}} = \frac{|\det T|\cdot (\text{area of middle square})} {|\det T|\cdot (\text{area of }Q)} = \frac{1}{9}.$$

Hence

$$S_{\text{mid}}=\frac1{9}S_{ABCD}.$$

The area of the middle quadrilateral is therefore nine times smaller than the area of $ABCD$. This completes the proof.

∎

Verification of Key Steps

For the square, the four connecting segments are exactly the lines

$$x=\frac13,\quad x=\frac23,\quad y=\frac13,\quad y=\frac23.$$

A common mistake is to assume this without checking which trisection points are connected. The corresponding points on opposite vertical sides have equal heights, so the connecting segments are horizontal. Likewise, the corresponding points on opposite horizontal sides have equal abscissas, so the connecting segments are vertical.

For preservation of trisection points, let $P$ divide $XY$ in the ratio $1:2$. Then

$$P=\frac23X+\frac13Y.$$

For an affine map $T$,

$$T(P) = \frac23T(X)+\frac13T(Y),$$

which is the point dividing $T(X)T(Y)$ in the same ratio. The construction is therefore transported exactly.

For preservation of area ratios, if two figures have areas $S_1$ and $S_2$, then after an affine transformation with determinant $d\neq0$ their areas become $|d|S_1$ and $|d|S_2$. Thus

$$\frac{|d|S_1}{|d|S_2} = \frac{S_1}{S_2}.$$

The factor cancels, so the ratio $1/9$ found for the square remains unchanged.

Alternative Approaches

A direct coordinate proof is possible. Place the quadrilateral in affine coordinates with

$$A=(0,0),\quad B=(1,0),\quad D=(0,1),\quad C=(u,v).$$

The equations of the four connecting segments can then be written explicitly from the coordinates of the trisection points. Solving for their intersections yields the vertices of the middle quadrilateral, and a determinant computation shows that its area equals one ninth of the area of $ABCD$.

The affine approach is preferable because the construction itself is affine. Once the ratio is computed in the square, all further calculations disappear, and the result follows from general properties of affine transformations.