Putnam 2018, Problem A-6

Suppose that $A, B, C$, and $D$ 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 $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$ are rational numbers, then the quotient
\[
\frac{\operatorname{area}(\triangle ABC)}{\operatorname{area}(\triangle ABD)}
\]
is a rational number.

Solution:
**First solution.** Choose a Cartesian coordinate system with origin at the midpoint of $AB$ and positive $x$-axis containing $A$. By the condition on $AB$, we have $A = (\sqrt{a}, 0)$, $B = (-\sqrt{a}, 0)$ for some positive rational number $a$. Let $(x_1,y_1)$ and $(x_2,y_2)$ be the respective coordinates of $C$ and $D$; by computing the lengths of the segments $AC, BC, AD, BD, CD$, we see that the quantities
$$
\begin{aligned}
&(x_1 - \sqrt{a})^2 + y_1^2, \quad (x_1 + \sqrt{a})^2 + y_1^2, \\
&(x_2 - \sqrt{a})^2 + y_2^2, \quad (x_2 + \sqrt{a})^2 + y_2^2, \\
&(x_1 - x_2)^2 + (y_1 - y_2)^2
\end{aligned}
$$
are all rational numbers. By adding and subtracting the first two quantities, and similarly for the next two, we see that the quantities
$$
x_1^2 + y_1^2, \quad x_1 \sqrt{a}, \quad x_2^2 + y_2^2, \quad x_2 \sqrt{a}
$$
are rational numbers. Since $a$ is a rational number, so then are
$$
\begin{aligned}
x_1^2 &= \frac{(x_1 \sqrt{a})^2}{a} \\
x_2^2 &= \frac{(x_2 \sqrt{a})^2}{a} \\
x_1 x_2 &= \frac{(x_1 \sqrt{a})(x_2 \sqrt{a})}{a} \\
y_1^2 &= (x_1^2 + y_1^2) - x_1^2 \\
y_2^2 &= (x_2^2 + y_2^2) - x_2^2.
\end{aligned}
$$
Now note that the quantity
$$
(x_1 - x_2)^2 + (y_1 - y_2)^2 = x_1^2 - 2x_1 x_2 + x_2^2 + y_1^2 - 2y_1 y_2 + y_2^2
$$
is known to be rational, as is every summand on the right except $-2y_1 y_2$; thus $y_1 y_2$ is also rational. Since $y_1^2$ is also rational, so then is $y_1/y_2 = (y_1 y_2)/(y_1^2)$; since
$$
\text{area}(\triangle ABC) = \sqrt{a} y_1, \quad \text{area}(\triangle ABD) = \sqrt{a} y_2,
$$
this yields the desired result.

**Second solution.** (by Manjul Bhargava) Let $\mathbf{b}, \mathbf{c}, \mathbf{d}$ be the vectors $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ viewed as column vectors. The desired ratio is given by
$$
\begin{aligned}
\frac{\det(\mathbf{b}, \mathbf{c})}{\det(\mathbf{b}, \mathbf{d})} &= \frac{\det(\mathbf{b}, \mathbf{c})^T \det(\mathbf{b}, \mathbf{c})}{\det(\mathbf{b}, \mathbf{c})^T \det(\mathbf{b}, \mathbf{d})} \\
&= \det \begin{pmatrix} \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{pmatrix} \det \begin{pmatrix} \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{d} \\ \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{d} \end{pmatrix}^{-1}.
\end{aligned}
$$
The square of the length of $AB$ is $\mathbf{b} \cdot \mathbf{b}$, so this quantity is rational. The square of the lengths of $AC$ and $BC$ are $\mathbf{c} \cdot \mathbf{c}$ and $(\mathbf{c}-\mathbf{b}) \cdot (\mathbf{c}-\mathbf{b}) = \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} - 2\mathbf{b} \cdot \mathbf{c}$, so $\mathbf{b} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{b}$ is rational. Similarly, using $AD$ and $BD$, we deduce that $\mathbf{d} \cdot \mathbf{d}$ and $\mathbf{b} \cdot \mathbf{d}$ is rational; then using $CD$, we deduce that $\mathbf{c} \cdot \mathbf{d}$ is rational.

**Third solution.** (by David Rusin) Recall that Heron's formula (for the area of a triangle in terms of its side length) admits the following three-dimensional analogue due to Piero della Francesca: if $V$ denotes the volume of a tetrahedron with vertices $A,B,C,D \in \mathbb{R}^3$, then
$$
288 V^2 = \det \begin{pmatrix}
0 & AB^2 & AC^2 & AD^2 & 1 \\
AB^2 & 0 & BC^2 & BD^2 & 1 \\
AC^2 & BC^2 & 0 & CD^2 & 1 \\
AD^2 & BD^2 & CD^2 & 0 & 1 \\
1 & 1 & 1 & 1 & 0
\end{pmatrix}
$$
In particular, the determinant vanishes if and only if $A,B,C,D$ are coplanar. From the identity
$$
\begin{aligned}
&64(4\text{Area}(\triangle ABC)^2 \text{Area}(\triangle ABD)^2 - 9AB^2 V^2) \\
&= (AB^4 - AB^2(AC^2 + AD^2 + BC^2 + BD^2 - 2CD^2) \\
&\quad + (AC^2 - BC^2)(AD^2 - BD^2))^2
\end{aligned}
$$
we see that $\text{Area}(\triangle ABC)\text{Area}(\triangle ABD)$ is rational; since each of the areas has rational square, we deduce the claim.

**Fourth solution.** (by Greg Martin) Define the signed angles $\alpha = \angle BAC, \beta = \angle BAD, \gamma = \angle CAD$, so that $\alpha + \gamma = \beta$. By the Law of Cosines,
$$
\begin{aligned}
2AB \cdot AC \cos \alpha &= AB^2 + AC^2 - BC^2 \in \mathbb{Q} \\
2AB \cdot AD \cos \beta &= AB^2 + AD^2 - BD^2 \in \mathbb{Q} \\
2AC \cdot AD \cos \gamma &= AC^2 + AD^2 - CD^2 \in \mathbb{Q}.
\end{aligned}
$$
In particular, $(2AB \cdot AC \cos \alpha)^2 \in \mathbb{Q}$, and so $\cos^2 \alpha \in \mathbb{Q}$ and $\sin^2 \alpha = 1 - \cos^2 \alpha \in \mathbb{Q}$, and similarly for the other two angles.

Applying the addition formula to $\cos \beta$, we deduce that
$$
2AB \cdot AD \cos \alpha \cos \gamma - 2AB \cdot AD \sin \alpha \sin \gamma \in \mathbb{Q}.
$$
The first of these terms equals
$$
\frac{(2AB \cdot AC \cos \alpha)(2AB \cdot AC \cos \alpha)}{AC^2} \in \mathbb{Q},
$$
so the second term must also be rational. But now
$$
\begin{aligned}
\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ACD)} &= \frac{AB \cdot AC \sin \alpha}{AC \cdot AD \sin \gamma} \\
&= \frac{2AB \cdot AD \sin \alpha \sin \gamma}{2AD^2 \sin^2 \gamma} \in \mathbb{Q}
\end{aligned}
$$
as desired.

**Remark.** Derek Smith observes that this result is Proposition 1 of: M. Knopf, J. Milzman, D. Smith, D. Zhu and D. Zirlin, Lattice embeddings of planar point sets, *Discrete and Computational Geometry* **56** (2016), 693–710.

**Remark.** It is worth pointing out that it is indeed possible to choose points $A,B,C,D$ satisfying the conditions of the problem; one can even ensure that the lengths of all four segments are themselves rational. For example, it was originally observed by Euler that one can find an infinite set of points on the unit circle whose pairwise distances are all rational numbers. One way to see this is to apply the linear fractional transformation $f(z) = \frac{z+i}{z-i}$ to the Riemann sphere to carry the real axis (plus $\infty$) to the unit circle, then compute that
$$
|f(z_1) - f(z_2)| = \frac{2|z_1 - z_2|}{|(z_1 - i)(z_2 - i)|}.
$$
Let $S$ be the set of rational numbers $z$ for which $2(z^2 + 1)$ is a perfect square; the set $f(S)$ has the desired property provided that it is infinite. That can be checked in various ways; for instance, the equation $2(x^2 + 1) = (2y)^2$ equates to $x^2 - 2y^2 = -1$ (a modified Brahmagupta-Pell equation), which has infinitely many solutions even over the integers:
$$
x + y\sqrt{2} = (1 + \sqrt{2})^{2n+1}.
$$