Kvant Math Problem 227

Let the parallelogram be mapped by an affine transformation to the unit square, since affine maps preserve parallelism, ratios of areas, and the condition of a point lying on a segment.

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Problem

On each side of a parallelogram, a point is chosen. The area of the quadrilateral with vertices at these points is equal to half the area of the parallelogram. Prove that at least one of the diagonals of the quadrilateral is parallel to one of the sides of the parallelogram.

E. V. Sallinen

Exploration

Let the parallelogram be mapped by an affine transformation to the unit square, since affine maps preserve parallelism, ratios of areas, and the condition of a point lying on a segment. Under this normalization the parallelogram becomes a square with vertices $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$.

Let the chosen points on the sides be

$E(a,0)\in AB$, $F(1,b)\in BC$, $G(c,1)\in CD$, $H(0,d)\in DA$ with $a,b,c,d\in[0,1]$.

The quadrilateral is $EFGH$. The condition on a diagonal being parallel to a side reduces to a simple coordinate constraint: a diagonal is parallel to a side of the square only if it is horizontal or vertical. The segment $EG$ has vertical displacement $1$, so it can never be horizontal; hence it is parallel to a side exactly when it is vertical, which forces $a=c$. Similarly, $FH$ is parallel to a side exactly when $b=d$.

Thus the problem reduces to proving $a=c$ or $b=d$ from the area condition.

The key step is to express the area of $EFGH$ in terms of $a,b,c,d$ and analyze the resulting algebraic constraint.

Problem Understanding

This is a Type B problem. One must prove that if four points are chosen on the sides of a parallelogram and the resulting quadrilateral has area equal to half the parallelogram, then one of its diagonals is parallel to a side of the parallelogram.

After affine normalization, the task becomes a coordinate geometry statement in the unit square. The essential difficulty is translating the area condition into an algebraic identity that forces equality of opposite parameters.

Proof Architecture

First, the parallelogram is mapped to the unit square using an affine transformation, preserving all relevant properties.

Second, coordinates of the four boundary points are introduced as $E(a,0)$, $F(1,b)$, $G(c,1)$, $H(0,d)$.

Third, the area of quadrilateral $EFGH$ is computed using the shoelace formula, producing an expression in $a,b,c,d$.

Fourth, the condition that this area equals $1/2$ is converted into an equation involving $(a-c)(b-d)$.

Fifth, this equation is analyzed to show that it forces $a=c$ or $b=d$.

The most delicate step is the algebraic reduction of the area condition and the exclusion of the second algebraic case using boundedness of parameters.

Solution

An affine transformation reduces the parallelogram to a unit square without changing parallelism or the ratio of areas. Hence it suffices to consider the square with vertices $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$.

Let $E(a,0)\in AB$, $F(1,b)\in BC$, $G(c,1)\in CD$, $H(0,d)\in DA$, where $a,b,c,d\in[0,1]$.

A diagonal of $EFGH$ is parallel to a side of the square if and only if it is horizontal or vertical. The vector $EG=(c-a,1)$ is never horizontal, so it is parallel to a side exactly when it is vertical, which is equivalent to $c-a=0$, hence $a=c$. The vector $FH=(-1,d-b)$ is never vertical, so it is parallel to a side exactly when it is horizontal, which is equivalent to $d-b=0$, hence $b=d$. It remains to prove $a=c$ or $b=d$.

The area of $EFGH$ is computed by the shoelace formula in the cyclic order $E,F,G,H$:

$$S_1 = a b + 1 + c d,$$

$$S_2 = 0\cdot 1 + b c + 1\cdot 0 + d a = bc + ad.$$

Hence the area is

$$[EFGH] = \frac12 |S_1 - S_2| = \frac12 |1 + ab + cd - bc - ad|.$$

The condition $[EFGH]=\frac12$ implies

$$|1 + ab + cd - bc - ad| = 1.$$

Let

$$T = 1 + ab + cd - bc - ad.$$

Then $|T|=1$, so either $T=1$ or $T=-1$.

If $T=1$, then

$$ab + cd - bc - ad = 0.$$

Grouping terms gives

$$a(b-d) + c(d-b) = (a-c)(b-d) = 0,$$

so $a=c$ or $b=d$.

If $T=-1$, then

$$1 + ab + cd - bc - ad = -1,$$

so

$$ab + cd - bc - ad = -2.$$

The left-hand side equals $(a-c)(b-d)$, hence

$$(a-c)(b-d) = -2.$$

Since $a,c,b,d\in[0,1]$, each factor lies in $[-1,1]$, so their product lies in $[-1,1]$, contradicting $(a-c)(b-d)=-2$. Hence this case is impossible.

Therefore $a=c$ or $b=d$, which implies that $EG$ or $FH$ is parallel to a side of the square, and thus to a side of the original parallelogram.

This completes the proof. ∎

Verification of Key Steps

The affine reduction preserves the condition because affine maps preserve collinearity, parallelism, and ratios of areas, so the equality of areas is preserved up to a constant factor that cancels when comparing to half the total area.

In the shoelace computation, each vertex contributes exactly one product of the form $x_i y_{i+1}$ and one of the form $y_i x_{i+1}$ in cyclic order, and all zero terms arise from boundary coordinates lying on axes, leaving only the four nonzero contributions recorded.

The identity $ab + cd - bc - ad = (a-c)(b-d)$ follows by direct expansion, and the contradiction in the second case relies only on the fact that $a-c$ and $b-d$ each lie in $[-1,1]$, so their product cannot equal $-2$.

Alternative Approaches

A synthetic approach avoids coordinates by interpreting the area condition as a linear dependence relation between oriented subareas obtained by projecting the quadrilateral onto directions parallel to the sides of the parallelogram. This leads to the same algebraic constraint implicitly through signed area decomposition. The coordinate method is preferable because it reduces the condition to a single explicit equation whose factorization immediately forces the required degeneracy.