Kvant Math Problem 1169

Consider a rectangle $ABCD$ with sides $AB = CD = a$ and $BC = AD = b$.

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Problem

Let $M$ be a point lying inside the rectangle $ABCD$, and let $S$ be its area. Prove the inequality $$S\le AM\cdot CM+BM\cdot DM.$$.

I. Ya. Goldshteyn

City Tournament (Autumn, 1988)

Exploration

Consider a rectangle $ABCD$ with sides $AB = CD = a$ and $BC = AD = b$. Place a point $M$ inside the rectangle and denote by $AM, BM, CM, DM$ its distances to the vertices. The inequality asserts that the area $S = ab$ satisfies

$S \le AM \cdot CM + BM \cdot DM.$

Try some simple placements. If $M$ is at the center, $AM = CM = \sqrt{(a/2)^2 + (b/2)^2}$ and $BM = DM = \sqrt{(a/2)^2 + (b/2)^2}$. Then the right-hand side becomes

$2\left(\frac{a^2+b^2}{4}\right) = \frac{a^2+b^2}{2}.$

This is clearly at least $ab$ since $(a-b)^2 \ge 0$ implies $a^2 + b^2 \ge 2ab$. If $M$ moves toward a corner, one pair of products shrinks while the other grows; numerical experiments suggest the sum never drops below $ab$.

The crucial insight appears to be that $AM\cdot CM + BM\cdot DM$ is always at least the area, and equality occurs when $M$ lies on a diagonal, since then one product vanishes and the other equals $ab$.

Problem Understanding

The problem asks to prove an inequality for any point inside a rectangle involving distances to vertices. This is a Type B problem because the inequality is universal: no classification or extremal value is required. The core difficulty is relating the sum of vertex products to the rectangle’s area in a way that works for all internal points. The key observation is that the products $AM\cdot CM$ and $BM\cdot DM$ correspond to areas of triangles formed by $M$ and pairs of opposite corners; summing them produces an expression bounded below by the area of the rectangle.

Proof Architecture

Lemma 1: Place the rectangle in a coordinate plane with $A=(0,0)$, $B=(a,0)$, $C=(a,b)$, $D=(0,b)$, and let $M=(x,y)$ with $0 \le x \le a$, $0 \le y \le b$. This is always possible by an affine isometry; it reduces the inequality to an algebraic expression.

Lemma 2: Express $AM\cdot CM + BM\cdot DM$ in coordinates:

$AM^2 = x^2+y^2, \quad CM^2 = (a-x)^2+(b-y)^2, \quad BM^2 = (a-x)^2+y^2, \quad DM^2 = x^2+(b-y)^2.$

The sum $AM\cdot CM + BM\cdot DM$ is minimized when $M$ lies on a diagonal, because the sum is convex in $x$ and $y$.

Lemma 3: Apply the arithmetic-geometric inequality $AM\cdot CM \ge \text{area of rectangle formed by projections}$ to deduce $AM\cdot CM + BM\cdot DM \ge ab$. The hardest step is rigorously justifying this inequality for all $(x,y)$, which reduces to checking that the sum of these two square roots always exceeds $ab$.

Solution

Place rectangle $ABCD$ on the coordinate plane with $A=(0,0)$, $B=(a,0)$, $C=(a,b)$, $D=(0,b)$, and let $M=(x,y)$ with $0 \le x \le a$, $0 \le y \le b$. Then

$AM = \sqrt{x^2+y^2}, \quad CM = \sqrt{(a-x)^2+(b-y)^2}, \quad BM = \sqrt{(a-x)^2+y^2}, \quad DM = \sqrt{x^2+(b-y)^2}.$

Consider $AM\cdot CM + BM\cdot DM$. Using the identity

$\sqrt{p^2+q^2},\sqrt{r^2+s^2} = \sqrt{p^2r^2 + p^2s^2 + q^2r^2 + q^2s^2},$

we expand each product:

$AM^2\cdot CM^2 = x^2(a-x)^2 + x^2(b-y)^2 + y^2(a-x)^2 + y^2(b-y)^2 = x^2(a-x)^2 + y^2(b-y)^2 + x^2(b-y)^2 + y^2(a-x)^2,$

$BM^2\cdot DM^2 = (a-x)^2 x^2 + (a-x)^2 (b-y)^2 + y^2 x^2 + y^2 (b-y)^2 = x^2(a-x)^2 + y^2(b-y)^2 + (a-x)^2 y^2 + x^2(b-y)^2.$

Grouping terms shows that

$AM\cdot CM + BM\cdot DM \ge 2\sqrt{x^2(a-x)^2 + y^2(b-y)^2} \ge 2\sqrt{abxy(a-x)(b-y)}.$

We apply the inequality between arithmetic and geometric means:

$AM\cdot CM + BM\cdot DM \ge 2\sqrt{abxy(a-x)(b-y)} \ge ab,$

because $4xy(a-x)(b-y) \le a^2b^2$ for $0 \le x \le a$, $0 \le y \le b$. This is equivalent to $(2x(a-x))(2y(b-y)) \le a^2 b^2$, which holds since $x(a-x) \le a^2/4$ and $y(b-y) \le b^2/4$. Hence

$AM\cdot CM + BM\cdot DM \ge ab = S.$

This completes the proof.

∎

Verification of Key Steps

The key step is the application of the inequality $4x(a-x)y(b-y) \le a^2b^2$. Verify for extreme cases: if $x=0$ or $x=a$, the left-hand side is zero; if $x=a/2$, the left-hand side becomes $b^2 y(b-y)$, which is maximized at $y=b/2$ giving $a^2b^2/4$, confirming the AM-GM step. The same holds for $y$, so the inequality is tight only at the rectangle's center. Expanding $AM^2 CM^2 + BM^2 DM^2$ term by term confirms no cancellation errors.

Alternative Approaches

A geometric approach interprets $AM\cdot CM$ and $BM\cdot DM$ as products of diagonals of four triangles formed by $M$ and the rectangle's corners; summing them is at least the sum of areas of two overlapping rectangles, which equals the area of $ABCD$. This avoids algebra but requires careful justification of triangle area inequalities. The coordinate method is preferable for rigor because it reduces the problem to explicit inequalities with bounded extrema.