Kvant Math Problem 1082

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Problem

In a convex quadrilateral $ABCD$, the diagonals intersect at point $O$. Prove that the equality $$AB^2+BC^2+CD^2+DA^2=2(AO^2+BO^2+CO^2+DO^2)$$ holds if and only if either the diagonals $AC$ and $BD$ are perpendicular, or one of them is bisected by point $O$.

A. P. Savin

Exploration

The given equality resembles the identity for the sum of squares of the sides of a quadrilateral. Let

$$\vec a=\overrightarrow{OA},\quad \vec b=\overrightarrow{OB},\quad \vec c=\overrightarrow{OC},\quad \vec d=\overrightarrow{OD}.$$

Then

$$AB^2=|\vec a-\vec b|^2,\quad BC^2=|\vec b-\vec c|^2,$$

and similarly for the other sides. Expanding the sum of the four side squares gives

$$AB^2+BC^2+CD^2+DA^2 = 2\sum |\vec a|^2 -2(\vec a!\cdot!\vec b+\vec b!\cdot!\vec c+\vec c!\cdot!\vec d+\vec d!\cdot!\vec a).$$

Hence the required equality is equivalent to

$$\vec a!\cdot!\vec b+\vec b!\cdot!\vec c+\vec c!\cdot!\vec d+\vec d!\cdot!\vec a=0.$$

The vertices are not arbitrary. Since $O$ is the intersection of the diagonals, points $A,O,C$ are collinear and points $B,O,D$ are collinear. Thus there exist positive numbers $x,y,u,v$ such that

$$\vec c=-x\vec a,\qquad \vec d=-y\vec b.$$

Substituting into the dot-product condition gives

$$(1-x)(1-y),\vec a!\cdot!\vec b=0.$$

This factorization is the central point. Geometrically,

$$1-x=0 \iff OA=OC,$$

so $O$ bisects $AC$; similarly $1-y=0$ means that $O$ bisects $BD$. Finally,

$$\vec a!\cdot!\vec b=0$$

means the lines $AC$ and $BD$ are perpendicular, because $\vec a$ and $\vec b$ lie on those diagonals.

Everything reduces to proving the factorization carefully.

Problem Understanding

We are given a convex quadrilateral $ABCD$ with diagonals intersecting at $O$. We must prove that

$$AB^2+BC^2+CD^2+DA^2 = 2(AO^2+BO^2+CO^2+DO^2)$$

holds exactly in the following situations: either the diagonals $AC$ and $BD$ are perpendicular, or $O$ is the midpoint of one of the diagonals.

This is a Type B problem. We must prove an equivalence.

The core difficulty is to convert the metric equality involving four side lengths into a condition expressed in terms of the diagonals. The decisive step is to use vectors centered at $O$ and exploit the collinearity relations imposed by the diagonals.

Proof Architecture

First, express the sum $AB^2+BC^2+CD^2+DA^2$ in vector form and show that the given equality is equivalent to

$$\vec a!\cdot!\vec b+\vec b!\cdot!\vec c+\vec c!\cdot!\vec d+\vec d!\cdot!\vec a=0.$$

This follows from expanding the four squared distances.

Next, use the fact that $A,O,C$ are collinear and $B,O,D$ are collinear to write

$$\vec c=-x\vec a,\qquad \vec d=-y\vec b,$$

with $x,y>0$.

Then prove that the dot-product condition becomes

$$(1-x)(1-y),\vec a!\cdot!\vec b=0.$$

This is obtained by direct substitution.

Finally, interpret each factor geometrically. The equation $\vec a!\cdot!\vec b=0$ means the diagonals are perpendicular. The equation $x=1$ means $OA=OC$, hence $O$ is the midpoint of $AC$. The equation $y=1$ means $O$ is the midpoint of $BD$.

The hardest direction is showing that the original metric equality forces one of these geometric conditions. The factorization above is the crucial lemma.

Solution

Let

$$\vec a=\overrightarrow{OA},\qquad \vec b=\overrightarrow{OB},\qquad \vec c=\overrightarrow{OC},\qquad \vec d=\overrightarrow{OD}.$$

Then

Also,

Expanding these four squares yields

$$\begin{aligned} AB^2+BC^2+CD^2+DA^2 &= (|\vec a|^2+|\vec b|^2-2\vec a!\cdot!\vec b) \ &\quad+ (|\vec b|^2+|\vec c|^2-2\vec b!\cdot!\vec c) \ &\quad+ (|\vec c|^2+|\vec d|^2-2\vec c!\cdot!\vec d) \ &\quad+ (|\vec d|^2+|\vec a|^2-2\vec d!\cdot!\vec a). \end{aligned}$$

Hence

where

$$S= \vec a!\cdot!\vec b +\vec b!\cdot!\vec c +\vec c!\cdot!\vec d +\vec d!\cdot!\vec a.$$

Therefore the equality

$$AB^2+BC^2+CD^2+DA^2 = 2(AO^2+BO^2+CO^2+DO^2)$$

is equivalent to

$$S=0.$$

Since $O$ is the intersection of the diagonals, points $A,O,C$ are collinear and points $B,O,D$ are collinear. Because $O$ lies between the endpoints of each diagonal in a convex quadrilateral, there exist positive numbers $x$ and $y$ such that

$$\vec c=-x\vec a, \qquad \vec d=-y\vec b.$$

Substituting these expressions into $S$ gives

$$\begin{aligned} S &= \vec a!\cdot!\vec b +\vec b!\cdot(-x\vec a) +(-x\vec a)!\cdot(-y\vec b) +(-y\vec b)!\cdot\vec a \ &= \bigl(1-x+xy-y\bigr),\vec a!\cdot!\vec b \ &= (1-x)(1-y),\vec a!\cdot!\vec b. \end{aligned}$$

Thus the given equality holds if and only if

$$(1-x)(1-y),\vec a!\cdot!\vec b=0.$$

There are three possibilities.

$$\vec a!\cdot!\vec b=0,$$

then the vectors $\vec a$ and $\vec b$ are perpendicular. Since $\vec a$ lies on the diagonal $AC$ and $\vec b$ lies on the diagonal $BD$, the diagonals are perpendicular.

$$1-x=0,$$

then $x=1$, hence

$$\vec c=-\vec a.$$

Consequently,

$$OC=|\vec c|=|\vec a|=OA.$$

Since $A,O,C$ are collinear and $O$ lies between $A$ and $C$, point $O$ is the midpoint of $AC$.

$$1-y=0,$$

then $y=1$, and similarly $O$ is the midpoint of $BD$.

Conversely, if the diagonals are perpendicular, then $\vec a!\cdot!\vec b=0$, so $S=0$, and the required equality follows. If $O$ is the midpoint of $AC$, then $x=1$, so again $S=0$. If $O$ is the midpoint of $BD$, then $y=1$, so $S=0$.

Thus the equality

$$AB^2+BC^2+CD^2+DA^2 = 2(AO^2+BO^2+CO^2+DO^2)$$

holds if and only if either the diagonals are perpendicular, or $O$ bisects one of the diagonals.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the expansion of the side squares. A sign error would invalidate the whole argument. Recomputing,

$$|\vec a-\vec b|^2 = |\vec a|^2+|\vec b|^2-2\vec a!\cdot!\vec b,$$

and analogous formulas hold for the other three sides. Each vector norm appears exactly twice, producing the coefficient $2$ in front of

$$|\vec a|^2+|\vec b|^2+|\vec c|^2+|\vec d|^2.$$

The second delicate step is the representation

$$\vec c=-x\vec a,\qquad \vec d=-y\vec b.$$

Because $O$ lies on both diagonals and is between the endpoints of each diagonal in a convex quadrilateral, the vectors from $O$ to opposite vertices point in opposite directions. Hence the coefficients are positive and the minus signs are mandatory.

The third delicate step is the factorization. Substituting gives

$$1-x+xy-y.$$

Factoring incorrectly as $(1-x-y)(1+y)$ would lead to false conclusions. Direct multiplication shows

$$(1-x)(1-y)=1-x-y+xy,$$

which matches the obtained coefficient exactly.

Alternative Approaches

A coordinate proof can be arranged by placing the origin at $O$ and choosing the coordinate axes along the diagonals. Then

$$A=(a,0),\quad C=(-c,0),\quad B=(0,b),\quad D=(0,-d)$$

when the diagonals are perpendicular. Computing the two sides of the required equality immediately shows equality. For the general case one may use oblique coordinates adapted to the diagonals and derive the same factorization in coordinate form.

Another approach starts from the identity

$$AB^2+BC^2+CD^2+DA^2 = AC^2+BD^2+4,\overrightarrow{OA}!\cdot!\overrightarrow{OB} -4,\overrightarrow{OC}!\cdot!\overrightarrow{OD},$$

and then uses the collinearity relations on the diagonals. This eventually reduces to the same product

$$(1-x)(1-y),\vec a!\cdot!\vec b.$$

The vector method used in the main proof is preferable because the relevant factorization appears immediately after a single expansion, and every geometric condition corresponds directly to one factor.