Suppose that a rectangle with sides $ a$ and $ b$ is arbitrarily cut into $ n$ squares with sides $ x_{1},\ldots,x_{n}$. Show that $ \frac{x_{i}}{a}\in\mathbb{Q}$ and $ \frac{x_{i}}{b}\in\mathbb{Q}$ for all $ i\in\{1,\cdots, n\}$.

Click for solution 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.

Find all $x$ and $y$ which are rational multiples of $\pi$ with $0<x<y<\frac{\pi}{2}$ and $\tan x+\tan y =2$.

Let $ \alpha$ be a rational number with $ 0 < \alpha < 1$ and $ \cos (3 \pi \alpha) + 2\cos(2 \pi \alpha) = 0$. Prove that $ \alpha = \frac {2}{3}$.

Suppose that $\tan \alpha =\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Prove the number $\tan \beta$ for which $\tan 2\beta =\tan 3\alpha$ is rational only when $p^2 +q^2$ is the square of an integer.

Click for solution If $ \tan\alpha =\frac{p}{q}$ then the equation reads $ \frac{2\tan\beta}{1-\tan^{2}\beta}=\frac{p(p^{2}-3q^{2})}{q(3p^{2}-q^{2})}$, solving to $ \tan\beta$ gives a quadratic with discriminant $ (p^{2}+q^{2})^{3}$, so $ \tan\beta\in\mathbb{Q}$ iff $ p^{2}+q^{2}$ is a rational square. But $ p^{2}+q^{2}$ is an integer, so is the square of an integer, QED.

Prove that there is no positive rational number $x$ such that \[x^{\lfloor x\rfloor }=\frac{9}{2}.\]

Click for solution Put $ x=\frac{p}{q}$ with $ p,q$ relatively prime, and $ \lfloor x\rfloor=n$. Then we want that $ 2p^{n}=9q^{n}$, so $ 2|q$, for $ n>1$ this means the RHS has more than one factor $ 2$ so $ 2|p$, contradiction with relative prime. So $ n=1$ (since $ n=0$ is impossible), so $ 2p=9q$, so $ x=\frac{9}2$, but then $ n>1$, contradiction.

Let $x, y, z$ non-zero real numbers such that $xy$, $yz$, $zx$ are rational. Show that the number $x^{2}+y^{2}+z^{2}$ is rational. If the number $x^{3}+y^{3}+z^{3}$ is also rational, show that $x$, $y$, $z$ are rational.

Click for solution Part a) is trivial: $x^2 = \frac{xy \cdot zx}{yz}$ Then $x^2,y^2,z^2$ are rational numbers. Let $k = x^3+y^3+z^3 \in \mathbb{Q}$. Then $(k-x^3)^2 = (y^3+z^3)^2 = (y^2)^3 + (z^2)^3 + 2(yz)^3 \in \mathbb{Q}$ $k^2 + x^6 -2kx^3 \in \mathbb{Q}$, but $k, x^2 \in \mathbb{Q}$. Therefore $2kx^3 \in \mathbb{Q} \Rightarrow x^3 \in \mathbb{Q} \Rightarrow x = \frac{x^3}{x^2} \in \mathbb{Q}$. In the same way we prove that $y,z \in \mathbb{Q}$.

If $x$ is a positive rational number, show that $x$ can be uniquely expressed in the form \[x=a_{1}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where $a_{1}a_{2},\cdots$ are integers, $0 \le a_{n}\le n-1$ for $n>1$, and the series terminates. Show also that $x$ can be expressed as the sum of reciprocals of different integers, each of which is greater than $10^{6}$.

Find all polynomials $W$ with real coefficients possessing the following property: if $x+y$ is a rational number, then $W(x)+W(y)$ is rational.

Prove that every positive rational number can be represented in the form \[\frac{a^{3}+b^{3}}{c^{3}+d^{3}}\] for some positive integers $a, b, c$, and $d$.

Click for solution For a positive rational $\frac{m}{n}$, find another positive rational $\frac{s}{t}$ such that $\frac{t^3m}{s^3n}=\frac{p}{q}$ lies between $\frac12$ and $2$. Such a rational clearly exists. Then $\frac{p}{q}=\frac{(p+q)^3+(2p-q)^3}{(p+q)^3+(2q-p)^3}$, and all four cubes are positive. Then just multiply back, giving $\frac{m}{n}=\frac{(s(p+q))^3+(s(2p-q))^3}{(t(p+q))^3+(t(2q-p))^3}$.

The set $ S$ is a finite subset of $ [0,1]$ with the following property: for all $ s\in S$, there exist $ a,b\in S\cup\{0,1\}$ with $ a, b\neq s$ such that $ s =\frac{a+b}{2}$. Prove that all the numbers in $ S$ are rational.

Click for solution Assume there was an irrational number in $ S$, only finitely much, and number them $ x_{1}\le\ldots\le x_{n}$. Now $ x_{1}=\frac{a+x_{k}}{2}$ with $ a<x_{1}$ rational. So, we can express $ x_{1}$ as a (nonnegative) linear combination of rationals and irrationals with higher indices (over $ \mathbb{Q}$). Now let $ t<n$ be the smallest index for which this is impossible, then $ x_{t}=\frac{a+b}{2}$ with $ a>x_{t}>b$ and necessarily $ b$ irrational (else $ a$ irrational but must have higher index). But then $ b\in\{x_{1},...,x_{t-1}$, which is a nonnegative linear combination of $ x_{t},...,x_{n}$. Now, if the coefficient in that combination of $ x_{t}$ is not $ 1$, then we can solve it for $ x_{t}$ as a (nonnegative) linear combination of $ x_{t+1},\ldots,x_{n}$, contradiction. So $ x_{t}$ appears in coefficient $ 1$ and, by nonnegativity, the rest appears in coefficient $ 0$, so $ b=x_{t}$, contradiction. So $ t=n$ and thus every $ x_{i}$ is a rational multiple of $ x_{n}$. But then $ x_{n}=\frac{a+b}{2}$ with $ b>x_{n}$ rational and $ a<x_{n}$ irratiional. But then $ a=qx_{n}$, and thus $ b=(2-q)x_{n}$ would be irrational, contradiction. So there can be no irrational number in $ S$.

Let $S=\{x_0, x_1, \cdots, x_n\} \subset [0,1]$ be a finite set of real numbers with $x_{0}=0$ and $x_{1}=1$, such that every distance between pairs of elements occurs at least twice, except for the distance $1$. Prove that all of the $x_i$ are rational.

Does there exist a circle and an infinite set of points on it such that the distance between any two points of the set is rational?

Click for solution We consider the set of points given by $P_s:=\left(1,\frac{2s}{s^2-1}\right)$ for all rational $s \neq \pm 1$. Their distances from the origin are rational by $\sqrt{1^2+\left(\frac{2s}{s^2-1}\right)^2} = \sqrt{\frac{s^4-2s^2+1+4s^2}{(s^2-1)^2}}=\frac{s^2+1}{s^2-1}$, and trivially their pairwise distances are rational. If we invert on the unit circle and $A,B$ are points with images $A^\prime , B^\prime$, we have $\overline{A'B'}=\frac{\overline{AB}}{\overline{A0} \cdot \overline{B0}}$. Especially all $P_s$ have pairwise rational distances again and get, together with the whole line $x=1$, mapped to the circle around $\left(\frac{1}{2},0\right)$ with radius $\frac 12$. Thus their images solve the problem.

Prove that numbers of the form \[\frac{a_{1}}{1!}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where $0 \le a_{i}\le i-1 \;(i=2, 3, 4, \cdots)$ are rational if and only if starting from some $i$ on all the $a_{i}$'s are either equal to $0$ ( in which case the sum is finite) or all are equal to $i-1$.

Let $k$ and $m$ be positive integers. Show that \[S(m, k)=\sum_{n=1}^{\infty}\frac{1}{n(mn+k)}\] is rational if and only if $m$ divides $k$.

Find all rational numbers $k$ such that $0 \le k \le \frac{1}{2}$ and $\cos k \pi$ is rational.

Click for solution Sorry, let me kill the little errors: The primitive $ n$-th roots of unity have algebraic degree $ \varphi(n)$ (showing this requires us to prove the irreducibility if the cyclotomic polynomials, but I will skip it now). Let $ k = \frac {2m}n$ with $ m,n$ coprime integers and let $ \zeta = e^{\frac {2\pi m i}{n}}$ (a primitive $ n$-th root of unity). Since $ c: = 2\cdot \cos \left( \frac {2m}{n} \pi \right) = \zeta + \zeta^{ - 1}$, we get that $ \zeta$ is a root of $ x^2 - 2c x + 1 = 0$. Especially, the dimension $ [\mathbb{Q}[\zeta]: \mathbb{Q}[c]]$ is $ 1$ or $ 2$. If $ n > 2$, then $ \zeta$ is nonreal, but $ c$ is real, giving that the degree is in fact $ 2$. Thus the algebraic degree of $ c$ is $ \frac {\varphi(n)}2$ if $ n \geq 3$, thus $ c$ is rational iff $ n = 1,2,3,4,6$.

Prove that for any distinct rational numbers $a, b, c$, the number \[\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}\] is the square of some rational number.

Click for solution Quite easy $\sum \frac{1}{x^2}=\frac{\sum x^2y^2}{x^2y^2z^2}=(\frac{\sum xy}{xyz})^2$(notice that $x+y+z=0$