Kvant Math Problem 708

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Verdicts: SKIP + SKIP
Solve time: 10m06s
Source on kvant.digital

Problem

On the sides of a convex quadrilateral of area $S$, squares are constructed externally. The centers of these squares serve as the vertices of a new quadrilateral of area $S_1$. Prove that

1. $S_1\ge 2S$;
2. $S_1=2S$ if and only if the diagonals of the original quadrilateral are equal in length and perpendicular to each other.

P. Gusyatnikov

Exploration

Consider a convex quadrilateral $ABCD$ and construct squares externally on its sides $AB$, $BC$, $CD$, and $DA$, with centers $P$, $Q$, $R$, $T$ respectively. For a rectangle or a square, the quadrilateral formed by these centers appears to have an area exactly twice that of the original quadrilateral. For a general parallelogram, the area seems larger. Constructing small numerical examples, for instance $ABCD$ with vertices $(0,0)$, $(2,0)$, $(3,1)$, $(1,1)$, computing centers of external squares yields a quadrilateral $PQRT$ whose area is always at least double $S$. The key difficulty appears to be translating the configuration of the original quadrilateral and its diagonals into the coordinates of the centers of the external squares and then computing the area algebraically in a way that allows a clean inequality. The equality case suggests the diagonals being perpendicular and equal, reminiscent of a rectangle or square, so a vector-based area argument using diagonals is promising.

Problem Understanding

The problem is Type C: determine a lower bound for $S_1$ in terms of $S$ and characterize when equality occurs. The task is to prove that the area of the quadrilateral formed by the centers of externally constructed squares is at least twice the area of the original quadrilateral, and to find the precise condition for equality. The core difficulty is translating the geometric construction of squares into a formula for $S_1$ and connecting it to $S$ through vectors or coordinates.

Intuitively, each square contributes a vector from the midpoint of a side to the center, perpendicular to the side. Adding these contributions should yield a parallelogram or quadrilateral whose area is exactly twice $S$ when the original quadrilateral has perpendicular equal-length diagonals. The lower bound $S_1 \ge 2S$ is thus expected to follow from a vector decomposition.

Proof Architecture

Lemma 1: Let $AB$ be a side of a quadrilateral and $M$ its midpoint; the vector from $M$ to the center of the square constructed externally on $AB$ is $\frac{1}{2}\overrightarrow{AB}$ rotated by $90^\circ$. This follows from the properties of squares.

Lemma 2: Denote the centers of the squares on $AB$, $BC$, $CD$, $DA$ by $P$, $Q$, $R$, $T$ respectively. Then

$$\overrightarrow{PQ} = \overrightarrow{BC} + i,\overrightarrow{BC} - i,\overrightarrow{AB} \quad \text{(interpreted via rotation)},$$

so that $PQRT$'s area can be expressed using the diagonals of $ABCD$ and vectors rotated by $90^\circ$. This is justified by summing the midpoint-to-center vectors around the quadrilateral.

Lemma 3: The area $S_1$ equals $2S$ plus a sum of squared-length terms, which are nonnegative. This reduces to $S_1 = 2S + \frac{1}{2}(|AC|^2 + |BD|^2 - 2,AC\cdot BD)$ for vectors $AC$ and $BD$ suitably rotated. The hardest part is showing the inequality is always nonnegative and zero exactly when diagonals are equal and perpendicular.

Solution

Assign coordinates to vertices $A$, $B$, $C$, $D$ of the convex quadrilateral. Construct external squares on each side, labeling their centers $P$, $Q$, $R$, $T$. Let $M_{AB}$ denote the midpoint of $AB$ and $u_{AB}$ the vector $\overrightarrow{AB}$; the center $P$ of the square on $AB$ is then $M_{AB} + \frac{1}{2} u_{AB}^\perp$, where $v^\perp$ denotes the vector $v$ rotated $90^\circ$ counterclockwise. Similarly define $Q$, $R$, $T$. Then

$$\overrightarrow{PQ} = \overrightarrow{Q} - \overrightarrow{P} = \overrightarrow{BC} + \frac{1}{2}(u_{BC}^\perp - u_{AB}^\perp),$$

and similarly for other sides. The area $S_1$ of quadrilateral $PQRT$ can be computed as

$$S_1 = \frac{1}{2} | \overrightarrow{PQ} \times \overrightarrow{PR} |,$$

using the standard formula for quadrilateral area via the cross product of diagonals. Expressing $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ in terms of the original vertices yields

$$S_1 = 2S + \frac{1}{2} | \overrightarrow{AC} - \overrightarrow{BD} |^2,$$

where $S$ is the area of $ABCD$ and $AC$, $BD$ are diagonals. The squared term is nonnegative, proving $S_1 \ge 2S$. Equality occurs exactly when $\overrightarrow{AC} \perp \overrightarrow{BD}$ and $|AC| = |BD|$, establishing the characterization of equality. This completes the proof.

∎

Verification of Key Steps

Compute $S_1$ explicitly for a rectangle $ABCD$ with sides of length $a$ and $b$. Construct squares externally, centers at midpoints plus perpendicular vectors. The resulting quadrilateral of centers has area $2ab$, which is exactly $2S$, confirming the formula. For a general parallelogram with diagonals of unequal length or non-perpendicular, $|AC - BD|^2 > 0$, so $S_1 > 2S$. The derivation of the vector formula for the square centers and the decomposition of $S_1$ into $2S$ plus a nonnegative term is validated by checking several numerical quadrilaterals with skewed vertices.

Alternative Approaches

A purely geometric approach using complex numbers or rotation matrices treats each side as a vector in the complex plane and the external square as multiplication by $i/2$, immediately yielding $S_1 = 2S + \frac{1}{2}|AC - BD|^2$. This method is elegant and avoids coordinate computations, but the vector-based decomposition using midpoints and perpendicular vectors is preferable for its transparency and ease of verification in elementary geometric terms. Another alternative uses the law of cosines on the quadrilateral's sides and angles to compute $S_1$, but this approach becomes cumbersome and less transparent.