When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

Sorry if this is considered reviving an old, dead thread, but this was a Contest Problem without a solution...

We can also prove the lemma as follows : By Law of Cosines, $\cos BAC = \frac{AB^{2}+AC^{2}-BC^{2}}{2 AB \cdot AC}$, which is rational. Similarly, $\cos CAD$ and $\cos BAD = \cos BAC \cos CAD-\sin BAC \sin CAD$ are rational. It follows that $\sin BAC \sin CAD$ is rational, as well as $\sin^{2}CAD = 1-\cos^{2}CAD$, so $\frac{\sin BAC \sin CAD}{\sin^{2}CAD}= \frac{\sin BAC}{\sin CAD}$ is rational. Since $\angle BAC = \angle BAO$ and $\angle CAD = \angle OAD$, $\frac{AB \sin BAO}{AD \sin OAD}= \frac{\tfrac{1}{2}AB \cdot AO \cdot \sin BAO}{\tfrac{1}{2}AO \cdot AD \cdot \sin OAD}= \frac{[BAO]}{[AOD]}= \frac{BO}{OD}$ is rational. Since $BO+OD= BD$ is also rational, $BO$ and $OD$ must also be rational. The rest follows by symmetry. $\blacksquare$

Suppose $AB$ is a side of a polygon in the dissection, lying on diagonal $XY$, with $X$, $A$, $B$, $Y$ in that order. Then $AB = XY - XA - YB$. In this way, we see that it actually just suffices to prove the result for a quadrilateral. To do this, we apply barycentric coordinates. Consider quadrilateral $ABDC$, with $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Let $D = (x,y,z)$, with $x+y+z=1$. By hypothesis, each of the numbers \begin{align*} -a^2yz + b^2(1-x)z + c^2(1-x)y &= AD^2 \\ a^2(1-y)z + b^2zx + c^2(1-y)x &= BD^2 \\ -a^2(1-z)y - b^2(1-z)x + c^2xy &= CD^2 \end{align*}is rational. Let $W = a^2yz + b^2zx + c^2xy$. Then, \begin{align*} b^2z + c^2y &= AD^2 + W \\ a^2z + c^2x &= BD^2 + W \\ a^2y + b^2x &= CD^2 + W. \end{align*}This implies that $AD^2 + BD^2 + 2W - c^2 = 2S_C z$ and cyclically (as usual $2S_C = a^2+b^2-c^2)$. If any of $S_A$, $S_B$, $S_C$ are zero, then we deduce $W$ is rational. Otherwise, we have that \[ 1 = x+y+z = \sum_{\text{cyc}} \frac{AD^2 + BD^2 + 2W - c^2}{2S_C} \]which implies that $W$ is rational, because it appears with coefficient $\frac{1}{S_A} + \frac{1}{S_B} + \frac{1}{S_C} \neq 0$ (since $S_{BC} + S_{CA} + S_{AB}$ is actually the area of $ABC$). Hence from the rationality of $W$, we deduce that $x$ is rational as long as $S_A \neq 0$, and similarly for the others. So at most one of $x$, $y$, $z$ is irrational, but since $x+y+z=1$ this implies they are all rational. Finally, if $P = \overline{AD} \cap \overline{BC}$ then $AP = \frac{1}{y+z} AD$, so $AP$ is rational too, completing the proof.

Clearly, it suffices to show that for points $A,B,C,D$ with all side lengths and diagonals between them rational and $AC,BD$ intersecting at $P$, $AP$ is rational. But since $\frac{AP}{BP}=\frac{AD}{BC}$, $BP$ is a rational multiple of $AP$. Similarly $CP$ is a rational multiple of $BP$, so $CP+AP=AC$ is a rational multiple of $AP$. But $AC$ is rational, so $AP$ is as well. EDIT: Oops, totally misread, thought it was cyclic. The same lemma should work, I'll see if I can bash out the details later (should be much worse now :/)

pi37 wrote: $\frac{AP}{BP}=\frac{AD}{BC}$ Unless I'm misunderstanding, this isn't true in general? For example, take $ABCD$ to be a parallelogram which is not a rectangle.

One can also use cartesian coordinate to prove the same lemma.

It's suffice to prove the result for a quadrilateral. Let $P=AC\cap BD$, since $AP+CP$ is rational, we only need to prove $\dfrac {AP}{CP}=\dfrac {[BAD]}{[BCD]}$ is rational. Let $\sqrt {a}=[BAD]$, where $a$ is a rational number, define $\sqrt {b}, \sqrt {c}, \sqrt {d}$ similarly. Since $\sqrt {a}+\sqrt {c}=\sqrt {b}+\sqrt {d}$, so if $\dfrac {\sqrt {a}}{\sqrt {c}}$ is not rational, then we must have $\{ a, c\} =\{ b, d\} $, which means $ABCD$ must be a trapezoid. The result is obviously true for the case $ABCD$ is a trapezoid.$\Box $

Leooooo wrote: $\sqrt {a}+\sqrt {c}=\sqrt {b}+\sqrt {d}$, so if $\dfrac {\sqrt {a}}{\sqrt {c}}$ is not rational, then we must have $\{ a, c\} =\{ b, d\} $

Clearly it suffices to show the result for a quadrilateral $A_1A_2A_3A_4$. Let $X=A_1A_3\cap A_2A_4$, $a_i=A_iA_{i+1}$, and $x_i=XA_i$. Let $\alpha_i=\angle XA_iA_{i+1}$, and $\beta_i=\angle XA_{i+1}A_i$. By the law of cosines, we have $\cos\alpha_i,\cos\beta_i\in\mathbb{Q}$. We have the equations \[x_i\cos\alpha_i+x_{i+1}\cos\beta_i=a_i,\]where the indices are taken mod $4$. We see that $x_1+x_3$ and $x_2+x_4$ are rational, so letting $x_1=p$ and $x_2=q$, it suffices to show that $p$ and $q$ are rational. Indeed, the four equations transform to \begin{align*} p\cos\alpha_1 + q\cos\beta_1 &= r_1 \\ -p\cos\beta_2+q\cos\alpha_2 &= r_2. \end{align*}If the determinant of this system is nonzero, then we're done, so WLOG we may assume that we have \[\cos\alpha_1\cos\alpha_2 + \cos\beta_1\cos\beta_2=0.\]By symmetry then, if this was nonzero if we replaced $1$ with $i$ and $2$ with $i+1$, then we'd also be done, so we may assume that \[\cos\alpha_i\cos\alpha_{i+1} + \cos\beta_i\cos\beta_{i+1}=0\]for all $i$. Note that $\alpha_i+\beta_i+\alpha_{i+1}+\beta_{i+1}=\pi$, so using product to sum, we get \[\cos(\alpha_i-\alpha_{i+1}) + \cos(\beta_i-\beta_{i+1}) = 0.\]For a given $i$, we then either have \[\alpha_i-\alpha_{i+1} + \beta_i-\beta_{i+1} = (2k+1)\pi\]or \[[\alpha_i-\alpha_{i+1} -( \beta_i-\beta_{i+1} )= (2k+1)\pi. \]Letting $\theta=\angle A_iXA_{i+1}$, the first case implies \[\pi-\theta - (\theta) = (2k+1)\pi,\]so $\theta= -k\pi$, which is absurd as $0<\theta<\pi$. Thus, we always have the second case, so \[(\alpha_i-\beta_i)-(\alpha_{i+1}-\beta_{i+1})=(2k+1)\pi\]for some integer $k$. Thus, \[(\alpha_i-\beta_i)-(\alpha_{i+2}-\beta_{i+2}) = 2k\pi\]for some integer $k$. However, $-\pi<\alpha_i-\beta_i<\pi$, so we in fact have \[\alpha_i-\beta_i = \alpha_{i+2}-\beta_{i+2}.\]But we clearly also have $\alpha_i+\beta_i=\alpha_{i+2}+\beta_{i+2}$, so in fact $\alpha_i=\alpha_{i+2}$ and $\beta_i=\beta_{i+2}$. This means that $ABCD$ is a parallelogram. Thus, as long as $ABCD$ is not a parallelogram, we're done. Let's now solve the problem in the case that it is a parallelogram. In this case, $X$ is just the midpoint of $AC$ and $BD$, so $XA=AC/2$ and so on, which are clearly rational.

We will only prove the result for a quadrilateral; it's easy to see that this suffices. Let the quadrilateral be $ABCD$ and $AC\cap BD=P$. Let $\theta$ be the acute angle between $AC$ and $BD$. We will prove the following claim. Claim: $\cos\theta\in\mathbb Q$. Proof: If $AC\perp BD$, the result is obvious. Otherwise, let $H_B, H_D$ be the projections of $B,D$ onto $AC$. Note by Law of Cosines that $\cos\angle BAC\in\mathbb Q$; thus, $$AH_B = AB\cos\angle BAC \in\mathbb{Q}.$$Similarly, $AH_D\in\mathbb Q$, so $H_BH_D\in\mathbb{Q}$. Observe that $BD = H_BH_D\cos\theta$, done. $\blacksquare$ For any positive reals $a,b$, we say that $a\sim b$ iff $a/b\in\mathbb{Q}$. If $\alpha=\angle PAB$, and $\beta=\angle PBA$, we get that $$\mathbb Q\ni \cos\theta = |\cos(\alpha + \beta)| = |\underbrace{\cos\alpha\cos\beta}_{\in\mathbb Q} - \sin\alpha\sin\beta|$$or $\sin\alpha\sin\beta\in\mathbb{Q}$. However, $\sin^2\alpha = 1-\cos^2\alpha\in\mathbb{Q}$. Thus, $$\sin\alpha\sin\beta \sim \sin^2\alpha \implies \sin\alpha\sim\sin\beta.$$Moreover, the law of sine of $\triangle ABD$ gives $$\sin\angle PAB \sim \sin\angle PBA \sim \sin \angle DAB$$Hence, $[ABP]\sim [ABD]$ or $BP\sim BD$. Hence, $BP\in\mathbb{Q}$ as desired.

Without loss of generality, consider any diagonal $u$. If we prove that any other diagonal $v$ that intersects with $u$ at a point which is a rational distance away from the endpoints if $u$, then all segments on $u$ will be rational, which we can extend to all diagonals. Let $A,C$ be endpoints of $u$ and $B,D$ be endpoints of $v$ and we see that we have reduced the problem to the same problem on not a polygon but a quadrilateral $ABCD$. $~$ Let $AB=w,BC=x,CD=y,DA=z,AC=p,BD=q.$ These numbers are all rational numbers. Now, let $AC$ and $BD$ intersect at $E$, and let $AE=p_1,CE=p_2,BE=q_1,DE=q_2.$ By Stewart's theorem, we have the relations \begin{align*} p_1pp_2+q_1^2p &= w^2 p_2 + x^2p_1 \\ p_1pp_2+q_2^2p &= z^2 p_2 + y^2p_1 \\ q_1qq_2+p_1^2q &= w^2 q_2 + z^2q_1 \\ q_1qq_2+q_1^2q &= x^2 q_2 + y^2p_1 \end{align*}If we subtract the first two and the last two, we get \begin{align*} pq(q_1-q_2) &= p_2(w^2-z^2)+p_1(x^2-y^2) \\ pq(p_1-p_2) &= q_2(w^2-x^2)+q_1(z^2-y^2) \end{align*}Now, if we let $c_1=pq,c_2=w^2-z^2,c_3=x^2-y^2,c_4=w^2-x^2,c_5=z^2-y^2$ then we can use the identity $c_2p_2+c_3p_1=\left(\tfrac{c_2+c_3}{2}\right)(p_1+p_2)+\left(\tfrac{c_2-c_3}{2}\right)(p_1-p_2).$ Let $r_1=p_1-p_2,r_2=q_1-q_2$ and let $s=\tfrac{c_2-c_3}{2}=\tfrac{c_4-c_5}{2}$ then \begin{align*} c_1r_1 &= \text{rational} + sr_2 \\ c_1r_2 &= \text{rational} + sr_1 \end{align*}which has a unique solution if and only if $c_1\neq s.$ If it doesn't have a unique solution, it either has no solution, in which case we are fine since the quadrilateral doesn't exist, or it has infinite solutions, in which case we are still fine because all solutions are rational. Therefore, we are done.

We will first solve the problem for $n=4$. WLOG $AC=1$. Assign Cartesian coordinates $A=(0,0)$, $C=(1,0)$, $B=(x,y)$, and $D=(u,v)$. We have that $x^2+y^2$, $(x-1)^2+y^2$, $u^2+v^2$, and $(u-1)^2+v^2$, and $(x-u)^2+(y-v)^2$ are all rational (in fact they are squares of rationals but we don't need that). Subtracting the first two yields that $2x-1$ is rational, so $x$ is rational, Similarly, $u$ is rational. Thus, $y=\sqrt{a}$ and $v=\sqrt{b}$ for some rationals $a,b$. Expanding the last condition and subtracting out everything rational yields that $$xu+yv$$is rational, and thus $yv$ is rational, so $ab$ is a square of a rational. Note that if $\frac{y}{v}$ is rational, then this means that the $AC$ divides $BD$ in a rational ratio, which would suffice. However, $$\frac{y}{v}=\sqrt{\frac{a}{b}}=\frac{\sqrt{ab}}{b},$$and $ab$ is a square of a rational so we are done. Then, consider general $n$. Consider a segment divided into chunks. By our $n=4$ solution, each intersection point in the middle is the small sector of some quadrilateral, so its distance to the endpoints are rational. Since each prefix sum is rational, each small segment is rational, thus done.