A very nice solution to this really hard problem was given by lhvietbao: lhvietbao wrote: First we let all the lines that goes through edges of squares in the tiling intersect each other to form a grid. Since the set of the length of grid segements ( we only consider grid segments in two directions that are parrallel to the sides of the original rectangle) is finite, we can find a basis of the set of those lengths over Q, that is each length of grid segments may be expressed in an unique linear combination with rational coefficents of the basis. We may also assume that a is in the basis. Now we define an additive function f such that f(a)=0 and f(x) > 0 for any other x of the basis in such a way that f(x) is non-zero unless the unique rartional combination of x is of the form q*a where q $ \in Q$. Define the f-area of a rectangle to be f(x)*f(y) where x, y are the lengths of its sides. Since f is additive, summing the f-area over the tiling gives: 0 = f(a)*f(b) = $ \sum f ( x_{i})^{2}$ thus $ f ( x_{i}) = 0$ so $ \frac{ a }{ x_{i}}\in Q$ and we are done. Note that there is another solution using electric networks in the book of B.Bollobas, but I don't understand that solution very well.

Can anyone explain some components of the solution in relation with simpler, contest, mathematics?

I have no idea of a simpler solution. Does anyone know a simpler one? Because it is extremely beautiful - but far from elementary.

Here a hopefully more clear explanation: Let $ 1=x_0, ... , x_n$ be arbitrary real numbers (later on, $ x_1, ... , x_n$ will be those from the problem). Define $ M=\{ q_0 x_0 + ... + q_n x_n | q_i \in \mathbb Q \}$. Lemma 1: There is some subset $ \{1=b_0, ... , b_r\}$ of $ \{ x_0 , x_1, ... , x_m \}$ such that for any $ x \in M$ there are unique $ q_1, ..., q_r \in \mathbb Q$ with $ x = q_0 b_0 + q_1 b_1 + ... + q_r b_r$ (this set is called a special basis). Proof: Easy induction on $ m$. Let now $ b_0=1, ... , b_n$ as in the lemma and for $ x = q_0+q_1 b_1 + ... + q_r b_r \in M$ define $ f(x) = x - q_0$. Lemma 2: Let $ x \in M$. Then $ f(x)=0$ iff $ x$ is rational. Proof: Write $ x = q_0+q_1 b_1 + ... + q_r b_r$. If $ f(x)=0$, then $ x= q_0$ is rational. If $ x$ is rational, then $ x= x \cdot 1 + 0 \cdot b_1 + ... + 0 \cdot b_n$ is the unique representation of $ x$, thus $ f(x)=x-x=0$. Lemma 3: For $ x,y \in M$ we have $ x+y \in M$ and $ f(x+y)=f(x)+f(y)$. Proof: Easy checking. We will never need bases again, untill now this was only to construct an $ f$ with the properties of lemma 2 and 3. A rectangle is a set of type $ R=[a,b] \times [c,d]$ with $ a,b,c,d \in M$ and $ a \leq b , c \leq d$. It's interior is ]$ a,b[ \times ]c,d[$ and we call two rectangles distinct if their interiors are distinct as sets. Finally, define $ A(R)=f(b-a) \cdot f(d-c)$ ("irrational area"; but note that this can be negative). Theorem: Let $ R_1 = [s_1, t_1] \times [u_1, v_1] , ... , R_m = [s_m, t_m] \times [u_m, v_m]$ be pairwise distinct rectangles such that $ [s,t] \times [u,v] = R : = R_1 \cup ... \cup R_m$ is a rectangle again. Then $ A(R) = A(R_1) + ... + A(R_m)$. Proof: In invite anybody to give a nicer proof. At least this theorem could be pictured easily We do this by induction on $ m$. Obviously, $ m=1$ is trivial. $ m=2$ is easily checked with the definitions using $ f(x+y)=f(x)+f(y)$. Now let it be proven for any number $ <m$ of rectangles. We may assume that $ s=s_1 , u=u_1$ (so $ R_1$ is the rectangle at the bottom left). If $ t_1=t$ and $ v_1=v$ we have $ R=R_1$ and are done. W.l.o.g. we assume $ t_1 < t$ and that $ s_2=t_1$. If $ s_1=t_1$ we are again done. Now subdivide $ R$ into $ C=[s,t_1] \times [u,v]$ and $ D=[t_1,t] \times [u,v]$. Then $ C$ is union of the distinct rectangles $ R_i^C : = C\cap R_i$, similar $ D$ is union of $ R_i^D : = D\cap R_i$. Now we got decompositions of $ C,D$ into rectangles, and we can neglect $ R_1^D$ and $ R_2^C$ in these decompositions. By induction, we get that $ A(C) = \sum_{i=1, \ i \neq 1}^m A(R_i^C)$ and $ A(D) = \sum_{i=1, \ i \neq 2}^m A(R_i^D)$. By using $ R=C \cup D$ and $ R_i = R_i^C \cup R_i^D$ (both distinct unions) we are done by using the case $ m=2$. Now to finish the problem: After stretching, we may assume that $ a=1$. Let $ R$ be the big rectangle with sides $ a=1,b$. Then $ A(R)=f(a)f(b) = 0 \cdot f(b) = 0$. But by the theorem we get $ 0=A(R) = \sum_{i=1}^n f(x_i)^2$. Thus $ f(x_i)=0$ for all $ i$ and by lemma 2 $ x_i$ is rational.

Thanks guys! That's exactly the type of response I was hoping to get... now it makes sense.