, 2003

Let $\mathcal{P} = A_1A_2\dots A_n$, where $n$ is an integer with $n \ge 3$. The problem is trivial for $n=3$ because there are no diagonals and thus no dissections. We assume that $n \ge 4$. Our proof is based on the following Lemma.

Lemma Let $ABCD$ be a convex quadrilateral such that all its sides and diagonals have rational lengths. If segments $AC$ and $BD$ meet at $P$, then segments $AP$, $BP$, $CP$, $DP$ all have rational lengths.

It is clear by the Lemma that the desired result holds when $\mathcal{P}$ is a convex quadrilateral. Let $A_iA_j$ ($1 \le i < j \le n$) be a diagonal of $\mathcal{P}$. Assume that $C_1, C_2, \dots, C_m$ are the consecutive division points on diagonal $A_iA_j$ (where point $C_1$ is the closest to vertex $A_i$ and $C_m$ is the closest to $A_j$). Then the segments $C_\ell C_{\ell+1}$, $1 \le \ell \le m-1$, are the sides of all polygons in the dissection. Let $C_\ell$ be the point where diagonal $A_iA_j$ meets diagonal $A_sA_t$. Then quadrilateral $A_iA_sA_jA_t$ satisfies the conditions of the Lemma. Consequently, segments $A_iC_\ell$ and $C_\ell A_j$ have rational lengths. Therefore, segments $A_iC_1, A_iC_2, \dots, A_jC_m$ all have rational lengths. Thus, $C_\ell C_{\ell+1} = AC_{\ell+1} - AC_\ell$ is rational. Because $i, j, \ell$ are arbitrarily chosen, we proved that all sides of all polygons in the dissection are also rational numbers.

Now we present two proofs of the Lemma to finish our proof.

* First approach We show only that segment $AP$ is rational, the proof for the others being similar. Introduce Cartesian coordinates with $A = (0, 0)$ and $C = (c, 0)$. Put $B = (a, b)$ and $D = (d, e)$. Then by hypothesis, the numbers
$$\begin{aligned} AB &= \sqrt{a^2 + b^2}, & AC &= c, & AD &= \sqrt{d^2 + e^2}, \\ BC &= \sqrt{(a-c)^2 + b^2}, & BD &= \sqrt{(a-d)^2 + (b-e)^2}, \\ CD &= \sqrt{(d-c)^2 + e^2}, \end{aligned}$$
are rational. In particular,
$$BC^2 - AB^2 - AC^2 = (a-c)^2 + b^2 - (a^2 + b^2) - c^2 = -2ac$$
is rational. Because $c \neq 0$, $a$ is rational. Likewise, $d$ is rational.

Now we have that $b^2 = AB^2 - a^2$, $e^2 = AD^2 - d^2$, and $(b-e)^2 = BD^2 - (a-d)^2$ are rational, and so that $2be = b^2 + e^2 - (b-e)^2$ is rational. Because quadrilateral $ABCD$ is convex, $b$ and $e$ are nonzero and have opposite sign. Hence $b/e = 2be/2b^2$ is rational.

We now calculate
$$P = \left( \frac{bd - ae}{b - e}, 0 \right),$$
so
$$AP = \frac{\frac{b}{e} \cdot d - a}{\frac{b}{e} - 1}$$
is rational.

* Second approach To prove the Lemma, we set $\angle DAP = A_1$ and $\angle BAP = A_2$. Applying the Law of Cosines to triangles $ADC$, $ABC$, $ABD$ shows that angles $A_1, A_2, A_1+A_2$ all have rational cosine values. By the Addition formula, we have
$$\sin A_1 \sin A_2 = \cos A_1 \cos A_2 - \cos(A_1 + A_2),$$
implying that $\sin A_1 \sin A_2$ is rational.
Thus
$$\frac{\sin A_2}{\sin A_1} = \frac{\sin A_2 \sin A_1}{\sin^2 A_1} = \frac{\sin A_2 \sin A_1}{1 - \cos^2 A_1}$$
is rational.

Note that the ratio between the areas of triangles *ADP* and *ABP* is equal to $\frac{PD}{BP}$. Therefore
$$\frac{BP}{PD} = \frac{[ABP]}{[ADP]} = \frac{\frac{1}{2}AB \cdot AP \cdot \sin A_2}{\frac{1}{2}AD \cdot AP \cdot \sin A_1} = \frac{AB}{AD} \cdot \frac{\sin A_2}{\sin A_1},$$
implying that $\frac{PD}{BP}$ is rational. Because $BP + PD = BD$ is rational, both $BP$ and $PD$ are rational. Similarly, $AP$ and $PC$ are rational, proving the Lemma.