Kvant Math Problem 1079
For $n=3$ the problem asks for a single triangle whose three side lengths are irrational and whose area is a nonzero rational number.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m25s
Source on kvant.digital
Problem
Let $n$ be a natural number, $n\ge3$. Is it possible to place $n$ points on the plane so that the distance between any two of them is an irrational number, while the area of the triangle formed by any three of them is a rational number (different from zero)?
International Mathematical Olympiad for School Students (1987)
Exploration
For $n=3$ the problem asks for a single triangle whose three side lengths are irrational and whose area is a nonzero rational number. Such triangles certainly exist. For example, take a right triangle with legs $1$ and $2$. Its area is $1$, and the side lengths are $1,2,\sqrt5$. This does not satisfy the condition because two sides are rational. A better idea is to start from coordinates and force all pairwise distances to be irrational.
Take the points
$$A=(0,0),\qquad B=(1,0),\qquad C=(0,2).$$
The area is $1$, while the distances are $1,2,\sqrt5$, again not all irrational.
To make every distance irrational, it is natural to place points on a line $x=\alpha$ where $\alpha$ is irrational, or to translate rational-coordinate configurations by irrational amounts. Translation does not change distances, so it cannot turn rational distances into irrational ones.
A more promising direction is to look for a large configuration. If all coordinates are rational, then every triangle area is rational because the determinant formula gives a rational number. Thus the area condition becomes automatic. The difficulty is to make every pairwise distance irrational.
Suppose all points have rational coordinates and no two have the same $x$-coordinate. Then the square of the distance between two points is a positive rational number. If we can arrange that this rational number is never a rational square, then the distance itself is irrational.
Consider points on the parabola $y=x^2$ with rational $x$-coordinates. Let
$$P_a=(a,a^2),\qquad P_b=(b,b^2),$$
where $a,b\in\mathbb Q$ and $a\ne b$. Then
$$|P_aP_b|^2=(a-b)^2+(a^2-b^2)^2 =(a-b)^2\bigl(1+(a+b)^2\bigr).$$
If we choose rational numbers $a$ such that every number $1+(a+b)^2$ is not a rational square, then all distances are irrational.
The identity
$$1+t^2=s^2$$
has many rational solutions. Hence arbitrary rational $a$ will not work. We need a more systematic construction.
Take instead points
$$P_k=(k,k^2),\qquad k=0,1,\dots,n-1.$$
Then
$$|P_iP_j|^2=(i-j)^2\bigl(1+(i+j)^2\bigr).$$
Here $i+j$ is an integer. If $m=i+j\ge1$, then
$$m^2< m^2+1 < (m+1)^2,$$
so $m^2+1$ is not a perfect square. Hence $|P_iP_j|$ is irrational whenever $m\ge1$.
The exception is $m=0$, which occurs only for $i=j=0$, impossible for distinct points. Thus every pairwise distance is irrational.
Now check triangle areas. Three points on a parabola are never collinear unless two coincide. For
$$P_a=(a,a^2),\quad P_b=(b,b^2),\quad P_c=(c,c^2),$$
the determinant equals the Vandermonde product
$$(a-b)(b-c)(c-a),$$
so the area is
$$\frac12 |(a-b)(b-c)(c-a)|.$$
For integer parameters this is a nonzero rational number. Thus the construction seems to work for every $n$.
The most delicate point is proving that no three chosen parabola points are collinear and computing the area formula correctly.
Problem Understanding
We must decide whether for each natural number $n\ge3$ there exists a set of $n$ points in the plane such that every distance between two distinct points is irrational, while every triangle determined by three of the points has nonzero rational area.
This is a Type D problem. We must construct such a configuration and verify all required properties.
The core difficulty is to satisfy simultaneously two conditions that pull in opposite directions. Rational coordinates make triangle areas rational, but distances between rational-coordinate points often become rational as well. The construction must force every pairwise distance to be irrational while preserving rational areas.
The answer is yes for every $n\ge3$. Points with integer coordinates on the parabola $y=x^2$ provide the required configuration.
Proof Architecture
Lemma 1. For distinct integers $i$ and $j$,
$$|P_iP_j|^2=(i-j)^2\bigl(1+(i+j)^2\bigr),$$
where $P_k=(k,k^2)$; this follows from direct computation.
Lemma 2. For every integer $m\ge1$, the number $m^2+1$ is not a perfect square; it lies strictly between $m^2$ and $(m+1)^2$.
Lemma 3. Every distance between two distinct points $P_i$ and $P_j$ is irrational; combine Lemmas 1 and 2.
Lemma 4. For distinct numbers $a,b,c$,
$$2[\triangle P_aP_bP_c] = \left| \begin{vmatrix} 1&a&a^2\ 1&b&b^2\ 1&c&c^2 \end{vmatrix} \right| = |(a-b)(b-c)(c-a)|.$$
This is the Vandermonde determinant identity.
Lemma 5. Any three distinct points among the chosen parabola points form a triangle of nonzero rational area; apply Lemma 4 to integers.
The hardest part is Lemma 4, because the nonzero area condition depends on the determinant being computed correctly.
Solution
For a given integer $n\ge3$, consider the $n$ points
$$P_k=(k,k^2),\qquad k=0,1,\dots,n-1.$$
We show that they satisfy the required conditions.
First consider the distance between two distinct points $P_i$ and $P_j$. We have
$$|P_iP_j|^2 =(i-j)^2+(i^2-j^2)^2.$$
Since
$$i^2-j^2=(i-j)(i+j),$$
it follows that
$$|P_iP_j|^2 =(i-j)^2\bigl(1+(i+j)^2\bigr).$$
Let
$$m=i+j.$$
Because $i$ and $j$ are distinct nonnegative integers, $m\ge1$. Then
$$m^2<m^2+1<(m+1)^2.$$
Hence $m^2+1$ is not a perfect square.
Since $(i-j)^2$ is a perfect square and $m^2+1$ is not, the rational number
$$(i-j)^2\bigl(m^2+1\bigr)$$
cannot be a perfect square. Therefore $|P_iP_j|^2$ is not the square of a rational number. Consequently $|P_iP_j|$ is irrational.
Thus the distance between every two distinct points of the configuration is irrational.
Now take any three distinct points
$$P_a=(a,a^2),\qquad P_b=(b,b^2),\qquad P_c=(c,c^2),$$
where $a,b,c\in{0,1,\dots,n-1}$ and $a,b,c$ are pairwise distinct.
Twice the area of the triangle equals
$$\left| \begin{vmatrix} 1&a&a^2\ 1&b&b^2\ 1&c&c^2 \end{vmatrix} \right|.$$
Subtracting the first row from the second and third rows gives
$$\begin{vmatrix} 1&a&a^2\ 0&b-a&b^2-a^2\ 0&c-a&c^2-a^2 \end{vmatrix}.$$
Expanding along the first column,
$$\begin{vmatrix} b-a&b^2-a^2\ c-a&c^2-a^2 \end{vmatrix}.$$
Using
$$b^2-a^2=(b-a)(b+a),\qquad c^2-a^2=(c-a)(c+a),$$
we obtain
$$(b-a)(c-a)\bigl((c+a)-(b+a)\bigr) =(b-a)(c-a)(c-b).$$
Hence
$$2[\triangle P_aP_bP_c] = |(a-b)(b-c)(c-a)|.$$
The numbers $a,b,c$ are distinct integers, so each factor is a nonzero integer. Therefore
$$[\triangle P_aP_bP_c] = \frac12 |(a-b)(b-c)(c-a)|$$
is a positive rational number.
Thus every triangle determined by three of the chosen points has nonzero rational area.
The points $P_0,P_1,\dots,P_{n-1}$ therefore satisfy all requirements. Hence such a configuration exists for every natural number $n\ge3$.
$$\boxed{{(k,k^2)\mid k=0,1,\dots,n-1}}$$
Verification of Key Steps
The first delicate step is the irrationality of the distances. From
$$|P_iP_j|^2=(i-j)^2\bigl(1+(i+j)^2\bigr),$$
it is not enough to say that $1+(i+j)^2$ is not a square. One must also check that multiplying by the square $(i-j)^2$ cannot turn a nonsquare into a square. If
$$(i-j)^2\bigl(1+(i+j)^2\bigr)=r^2$$
for some rational $r$, then
$$1+(i+j)^2=\left(\frac{r}{i-j}\right)^2,$$
which would itself be a rational square. Since $1+(i+j)^2$ is an integer lying strictly between consecutive squares, this is impossible.
The second delicate step is the determinant computation. After row subtraction,
$$\begin{vmatrix} b-a&(b-a)(b+a)\ c-a&(c-a)(c+a) \end{vmatrix}$$
must be factored carefully:
$$(b-a)(c-a)\bigl((c+a)-(b+a)\bigr) =(b-a)(c-a)(c-b).$$
A sign error here would affect the area formula, although the absolute value would still remove it. The crucial fact is that the determinant vanishes exactly when two parameters coincide.
The third delicate step is the noncollinearity of every triple. The area formula shows
$$2[\triangle P_aP_bP_c] =|(a-b)(b-c)(c-a)|.$$
Since distinct integers have nonzero pairwise differences, the product is nonzero, so no three selected points are collinear.
Alternative Approaches
A different construction uses rational points on a circle. One can choose rational points whose pairwise chord lengths have squares equal to rational numbers that are not rational squares. Since all coordinates are rational, every triangle area is rational by the determinant formula. The difficulty is arranging the chord lengths simultaneously for all pairs.
The parabola construction is preferable because both required properties emerge from simple algebraic identities. The distance formula factors into a square times $1+(i+j)^2$, and the area formula becomes a Vandermonde determinant. These computations immediately verify all conditions for every $n$.