Q:

Suppose that \(A,B,C,D\) are distinct points, no three of which lie on a line, in the Euclidean plane. Show that if the squares of the lengths of the line segments \(AB,AC,AD,BC,BD,CD\) are rational numbers, then the quotient

\[ \frac{\text{area}(\triangle ABC)}{\text{area}(\triangle ABD)} \]

is a rational number.

A:

The hardest part about any problem is convincing yourself that the problem is solvable. Once you’ve got that part, the rest follows.

Step 1

Believe. And before you know it, we’re getting somewhere.

Some Definitions

We’re going to translate this problem into a vector geometry problem. We define the following vectors:

\[ \boldsymbol a := \begin{pmatrix}a_1 \\ a_2 \end{pmatrix} := AB \] \[ \boldsymbol b := \begin{pmatrix}b_1 \\ b_2 \end{pmatrix} := BC \] \[ \boldsymbol c := \begin{pmatrix}c_1 \\ c_2 \end{pmatrix} := CD \] \[ \boldsymbol d := \begin{pmatrix}d_1 \\ d_2 \end{pmatrix} := DA \]

For the diagonals, we define:

\[ \boldsymbol e := \begin{pmatrix}e_1 \\ e_2 \end{pmatrix} := CA \] \[ \boldsymbol f := \begin{pmatrix}f_1 \\ f_2 \end{pmatrix} := BD \]

We immediately note from the problem statement:

\[ a_1^2 + a_2^2 = \lVert \boldsymbol a \rVert^2 \in \mathbb{Q} \] \[ \lVert \boldsymbol b \rVert^2 \in \mathbb{Q} \] \[ \lVert \boldsymbol c \rVert^2 \in \mathbb{Q} \] \[ \lVert \boldsymbol d \rVert^2 \in \mathbb{Q} \]

Using the fact that the squares of the diagonals are also rational, we have

\[ \mathbb{Q} \ni \lVert \boldsymbol e \rVert^2 = \lVert \boldsymbol c + \boldsymbol d \rVert^2 = \langle \boldsymbol c + \boldsymbol d, \boldsymbol c + \boldsymbol d \rangle = \langle \boldsymbol c, \boldsymbol c \rangle + 2 \langle \boldsymbol c, \boldsymbol d \rangle + \langle \boldsymbol d, \boldsymbol d \rangle = \lVert \boldsymbol c \rVert^2 + \lVert \boldsymbol d \rVert^2 + 2 \langle \boldsymbol c, \boldsymbol d \rangle \] \[ \implies \langle \boldsymbol c, \boldsymbol d \rangle \in \mathbb{Q} \]

where \(\langle \cdot, \cdot \rangle\) is the inner product on \(\mathbb{R}^2\), i.e. \(\langle \boldsymbol v, \boldsymbol w \rangle = \boldsymbol v \cdot \boldsymbol w = v_1 w_1 + v_2 w_2\).

By the symmetry of the above argument, we can make the following observation:

Observation 1

All inner products of vectors corresponding to consecutive sides of \(ABCD\) are rational, i.e.

\[ \langle \boldsymbol a, \boldsymbol b \rangle, \langle \boldsymbol b, \boldsymbol c \rangle, \langle \boldsymbol c, \boldsymbol d \rangle, \langle \boldsymbol d, \boldsymbol a \rangle \in \mathbb{Q} \]

Observation 2

All dot products of vectors corresponding to opposite sides of \(ABCD\) are rational, i.e.

\[ \langle \boldsymbol a, \boldsymbol c \rangle, \langle \boldsymbol b, \boldsymbol d \rangle \in \mathbb{Q} \]

Proof Observation 2

We note first that

\[ \boldsymbol a + \boldsymbol b + \boldsymbol c + \boldsymbol d = 0 \]

Thus

\[ \mathbb{Q} \ni \lVert \boldsymbol d \rVert^2 = \langle \boldsymbol d, \boldsymbol d \rangle = \langle - \boldsymbol a - \boldsymbol b - \boldsymbol c, - \boldsymbol a - \boldsymbol b - \boldsymbol c \rangle = \langle \boldsymbol a + \boldsymbol b + \boldsymbol c, \boldsymbol a + \boldsymbol b + \boldsymbol c \rangle \] \[ = \lVert \boldsymbol a \rVert^2 + \lVert \boldsymbol b \rVert^2 + \lVert \boldsymbol c \rVert^2 + 2\langle \boldsymbol a, \boldsymbol b \rangle + 2\langle \boldsymbol b, \boldsymbol c \rangle + 2\langle \boldsymbol a, \boldsymbol c \rangle\] \[ \implies \langle \boldsymbol a, \boldsymbol c \rangle \in \mathbb{Q} \]

Symmetrically, \(\langle \boldsymbol b, \boldsymbol d \rangle \in \mathbb{Q}\).

Observations 1 and 2 together say that all pairwise inner products of \(\boldsymbol a, \boldsymbol b, \boldsymbol c, \boldsymbol d\) are rational.

Solving the Problem

First, we will make some transformations to make the problem easier.

WLOG, we can assume that the side \(AB\) lies on the \(x\)-axis. This is because all rotations and translations of \(\mathbb{R}^2\) preserve side lengths, relative angles, and areas.

In fact, we can also assume WLOG that the length of \(AB\) is exactly \(1\), i.e.

\[ \boldsymbol a = \begin{pmatrix}a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \]

This is because if we scale all the side lengths down by \(\frac{1}{\lVert AB \rVert} = \frac{1}{\lVert \boldsymbol a \rVert}\), the squares of the resulting side lengths will still be rational, since \(\lVert \boldsymbol a \rVert^2\) is rational. In other words, suppose \(\boldsymbol c' = \frac{\boldsymbol c}{\lVert \boldsymbol a \rVert}\). Then \(\lVert \boldsymbol c' \rVert^2 = \frac{\lVert \boldsymbol c \rVert^2}{\lVert \boldsymbol a \rVert^2} \in \mathbb{Q}\).

This kind of scaling transformation is invertible and also preserves ratios of areas, thus if we can prove the case where \(\rVert \boldsymbol a \lVert = 1\), then we will have proved the question statement for all valid quadrilaterals.

Since we are assuming \(AB\) lies on the \(x\)-axis, the triangles \(\triangle ABC\) and \(\triangle ABD\) have very easily calculable areas. \(\triangle ABC\) has base \(AB\) and height \(\lvert b_2 \rvert\), while \(\triangle ABD\) has base \(AB\) and height \(\lvert d_2 \rvert\). Thus we have

\[ \frac{\text{area}(\triangle ABC)}{\text{area}(\triangle ABD)} = \frac{\frac{1}{2} \lvert AB \rvert \lvert b_2 \rvert}{\frac{1}{2} \lvert AB \rvert \lvert d_2 \rvert} = \frac{\lvert b_2 \rvert}{\lvert d_2 \rvert} \]

By the Observation 1, we have

\[ \mathbb{Q} \ni \langle \boldsymbol a, \boldsymbol b \rangle = a_1 b_1 + a_2 b_2 = 1 \cdot b_1 + 0 \cdot b_2 = b_1 \] \[ \implies b_1 \in \mathbb{Q} \]

Similarly,

\[ \mathbb{Q} \ni \langle \boldsymbol a, \boldsymbol d \rangle = a_1 d_1 + a_2 d_2 = d_1 \] \[ \implies d_1 \in \mathbb{Q} \]

By Observation 2, we have

\[ \mathbb{Q} \ni \langle \boldsymbol b, \boldsymbol d \rangle = b_1 d_1 + b_2 d_2 \] \[ \implies b_2 d_2 \in \mathbb{Q} \]

By the rationality of \(\lVert \boldsymbol d \rVert^2\), we have

\[ \mathbb{Q} \ni \lVert \boldsymbol d \rVert^2 = d_1^2 + d_2^2 \] \[ \implies d_2^2 \in \mathbb{Q} \]

Putting these last two observations together, we have

\[ \mathbb{Q} \ni \frac{b_2 d_2}{d_2^2} = \frac{b_2}{d_2} \] \[ \implies \frac{\text{area}(\triangle ABC)}{\text{area}(\triangle ABD)} = \frac{\lvert b_2 \rvert}{\lvert d_2 \rvert} \in \mathbb{Q} \]