USA IMO 2003

Let $\mathcal{P}=A_1{A}_2\dots A_n$, 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 $A_i{A}_j$ ($1\le i<j\le n$) be a diagonal of $\mathcal{P}$. Assume that $C_1,C_2,\dots ,C_m$ are the consecutive division points on diagonal $A_i{A}_j$ (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 $A_i{A}_j$ meets diagonal $A_s{A}_t$. Then quadrilateral $A_i{A}_s{A}_j{A}_t$ satisfies the conditions of the Lemma. Consequently, segments $A_i{C}_{\ell }$ and $C_{\ell }{A}_j$ have rational lengths. Therefore, segments $A_i{C}_1,A_i{C}_2,\dots ,A_j{C}_m$ all have rational lengths. Thus, $C_{\ell }{C}_{\ell +1}=A{C}_{\ell +1}-A{C}_{\ell }$ 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 $$\begin{array}{rlrlrl}AB& =\sqrt{a^2+b^2},& AC& =c,& AD& =\sqrt{d^2+e^2},\\ BC& =\sqrt{(a-c{)}^2+b^2},& BD& =\sqrt{(a-d{)}^2+(b-e{)}^2},& & \\ CD& =\sqrt{(d-c{)}^2+e^2},& & & & \end{array}$$ are rational. In particular, $$B{C}^2-A{B}^2-A{C}^2=(a-c{)}^2+b^2-(a^2+b^2)-c^2=-2ac$$ is rational. Because $c\ne 0$, $a$ is rational. Likewise, $d$ is rational.

Now we have that $b^2=A{B}^2-a^2$, $e^2=A{D}^2-d^2$, and $(b-e{)}^2=B{D}^2-(a-d{)}^2$ 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.