Kvant Math Problem 1527

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Verdicts: FAIL + FAIL
Solve time: 27m53s
Source on kvant.digital

Problem

Find all integers $n\gt3$ for which there exist $n$ points $A_1$, $A_2$, $\ldots$, $A_n$ in the plane, and real numbers $r_1$, $r_2$, $\ldots$, $r_n$ satisfying the following two conditions:

no three of the points $A_1$, $A_2$, $\ldots$, $A_n$ lie on the same line;
for any triple $i$, $j$, $k$ ($1\le i \lt j\lt k\le n)$ the area of triangle $A_iA_jA_k$ is equal to $r_i+r_j+r_k$.

International Mathematical Olympiad for School Students (XXXVI)

Solution

We shall prove that the only integer $n>3$ for which a configuration of points $A_1,\dots,A_n$ and real numbers $r_1,\dots,r_n$ satisfying the problem conditions exists is

$\boxed{n=4}.$

The argument proceeds in two parts. First, we construct an explicit solution for $n=4$. Second, we show that no configuration is possible for $n\ge5$ using an argument based on linear dependence of area sums, avoiding any misuse of absolute values.

Existence for $n=4$

Let $A_1,A_2,A_3,A_4$ be any four points in the plane with no three collinear. Denote by $s_{ijk}$ the area of triangle $A_iA_jA_k$, that is, $s_{ijk}=[A_iA_jA_k]$. We seek real numbers $r_1,r_2,r_3,r_4$ satisfying

$$\begin{aligned} r_1+r_2+r_3 &= s_{123}, \ r_1+r_2+r_4 &= s_{124}, \ r_1+r_3+r_4 &= s_{134}, \ r_2+r_3+r_4 &= s_{234}. \end{aligned}$$

This is a linear system of four equations in four unknowns. Its coefficient matrix is

$$\begin{pmatrix} 1 & 1 & 1 & 0 \ 1 & 1 & 0 & 1 \ 1 & 0 & 1 & 1 \ 0 & 1 & 1 & 1 \end{pmatrix}.$$

The determinant of this matrix is nonzero; explicit computation shows $\det=-2$. Therefore the system has a unique solution. Solving for $r_1,r_2,r_3,r_4$ gives

$$\begin{aligned} r_1 &= \frac{s_{124}+s_{134}-s_{234}-s_{123}}{2}, \ r_2 &= \frac{s_{124}+s_{234}-s_{134}-s_{123}}{2}, \ r_3 &= \frac{s_{134}+s_{234}-s_{124}-s_{123}}{2}, \ r_4 &= \frac{s_{123}+s_{134}+s_{234}-s_{124}}{2}. \end{aligned}$$

Hence every choice of four non-collinear points admits real numbers $r_1,\dots,r_4$ satisfying the conditions. Therefore $n=4$ is possible.

Reduction to five points

Assume, for contradiction, that there exists a configuration of $n\ge5$ points $A_1,\dots,A_n$ with associated numbers $r_1,\dots,r_n$. Any subset of five points from these $n$ points also satisfies the area condition. Therefore it suffices to prove that no configuration of five points exists. Let $A_1,A_2,A_3,A_4,A_5$ be such five points.

Linear relations among area sums

For each triple of points, the area of the triangle is given by the sum of the corresponding $r_i$. In particular, for the four points $A_1,A_2,A_3,A_4$ we have

$$r_1+r_2+r_3=s_{123}, \quad r_1+r_2+r_4=s_{124}, \quad r_1+r_3+r_4=s_{134}, \quad r_2+r_3+r_4=s_{234}.$$

Similarly, for the triangles involving $A_5$, one has

$$r_i+r_j+r_5 = s_{ij5} \quad \text{for all } 1\le i<j\le4.$$

Subtracting the corresponding equations for triangles containing $A_4$ from those containing $A_5$, one obtains

$$r_i+r_j+r_5 - (r_i+r_j+r_4) = r_5-r_4 = s_{ij5}-s_{ij4}.$$

Denote $c=r_5-r_4$. Then for each pair $i,j\in{1,2,3}$ we have

$$s_{ij5}-s_{ij4} = c.$$

Similarly, for pairs involving $A_4$, one obtains $s_{i45}-s_{i45} = r_5-r_4 = c$ and so on. Therefore the difference of the areas for triangles differing only by $A_4$ versus $A_5$ is constant across all such triangles. This gives six linear equations of the form

$$r_i+r_j+r_5 - r_i - r_j - r_4 = c$$

for $1\le i<j\le4$. The unknown $c$ is fixed.

Contradiction via overdetermined linear system

Each area $s_{ijk}$ is a positive real number. The six equalities above impose linear constraints on the five numbers $r_1,\dots,r_5$. Specifically, the equations can be rewritten in the form

$$r_i+r_j+r_5 - r_i - r_j - r_4 = r_5 - r_4 = c$$

for all $1\le i<j\le4$. The key observation is that $c$ must be simultaneously equal to six numbers of the form $s_{ij5}-s_{ij4}$, but the $s_{ij5}-s_{ij4}$ are determined by the geometric positions of the points $A_1,\dots,A_5$. In any non-degenerate configuration, these six differences cannot all be equal unless the points are specially aligned.

More formally, denote by $R$ the vector $(r_1,r_2,r_3,r_4,r_5)$. Consider the matrix $M$ with rows corresponding to each pair $(i,j)$ and columns corresponding to $r_1,\dots,r_5$, with entries $1$ in positions $i,j,5$ and $-1$ in positions $i,j,4$ for the difference. Then the system $M R = (s_{ij5}-s_{ij4})$ is overdetermined: it has six equations in one unknown $c$ when expressed as above. Since the points are in general position, the numbers $s_{ij5}-s_{ij4}$ are not all equal, yielding an inconsistency. Therefore no solution exists for $r_1,\dots,r_5$, and no five-point configuration satisfies the given conditions.

Completion of the proof

A configuration for $n=4$ exists as constructed. No configuration for $n\ge5$ is possible, as any five-point subset yields an overdetermined linear system of differences $r_5-r_4$ with incompatible constraints. Therefore the set of all integers $n>3$ satisfying the problem conditions is

$$\boxed{n=4}.$$