Estonian Mathematical Olympiad

Let $O$ be the origin of coordinates. Let $A_0$ be the starting point of the beetle's journey, $A_1$ the first turning point, $A_2$ the second turning point, $A_3$ the third turning point and $A_4$ the endpoint (see figure below).



The right triangles $O{A}_0{A}_1$ and $O{A}_1{A}_2$ are similar with ratio $x$ because $\angle O{A}_1{A}_2=90^{\circ }-\angle O{A}_1{A}_0=\angle O{A}_0{A}_1$ and $\frac{O{A}_1}{O{A}_0}=x$. So are the triangles $O{A}_1{A}_2$ and $O{A}_2{A}_3$ similar with ratio $x$ since $\frac{O{A}_2}{O{A}_1}=x$ by similarity of the triangles $O{A}_0{A}_1$ and $O{A}_1{A}_2$. Analogously, the triangles $O{A}_2{A}_3$ and $O{A}_3{A}_4$ are similar with ratio $x$. Consequently, $O{A}_2=x^2$, $O{A}_3=x^3$ and $O{A}_4=x^4$.

By the Pythagorean theorem, the length of the beetle's journey is $\sqrt{1+x^2}+\sqrt{x^2+x^4}+\sqrt{x^4+x^6}+\sqrt{x^6+x^8}$, or equivalently, $(1+x+x^2+x^3)\sqrt{1+x^2}$ which equals $(x^4-1)\cdot \frac{\sqrt{x^2+1}}{x-1}$ if $x\ne 1$. On the other hand, the distance between $A_0$ and $A_4$ is $|x^4-1|$.

a) Taking $x=\frac{4}{3}$, the number $x^4-1$ is rational, since $x$ is rational. As $\sqrt{1+x^2}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$, the number $(x^4-1)\cdot \frac{\sqrt{x^2+1}}{x-1}$ is also rational.

b) Assume that the distance between $A_0$ and $A_4$ is an integer; then $x^4$ is an integer. Suppose that $(x^4-1)\cdot \frac{\sqrt{x^2+1}}{x-1}$, the length of the beetle's journey, is also an integer (one can assume $x\ne 1$ since otherwise the length of the journey is $4\sqrt{2}$ that is not an integer). As $x^4-1$ is an integer, the number $\frac{\sqrt{x^2+1}}{x-1}$ must be rational. Consider three cases.

1) If $x$ is an integer then $x-1$ is integer whence $\sqrt{x^2+1}$ must be rational and $x^2+1$ must be a perfect square. This is impossible as two consecutive positive integers cannot be perfect squares.

2) Suppose that $x$ is irrational and $x^2$ is an integer. As $\frac{x^2+1}{(x-1{)}^2}$ is the square of a rational number, so is also $\frac{(x-1{)}^2}{x^2+1}$. Hence $\frac{2x}{x^2+1}$ must be rational. But this is impossible, since $2x$ is irrational and $x^2+1$ is an integer.

3) Let $x^2$ be irrational. Similarly to the previous case we see that $\frac{2x}{x^2+1}$ is rational. Hence $\frac{4x^2}{(x^2+1{)}^2}$ is the square of a rational number, implying that $\frac{(x^2+1{)}^2}{4x^2}$ is rational. The latter in turn implies that $\frac{x^4+1}{4x^2}$ must be rational. This is impossible as $4x^2$ is irrational and $x^4+1$ is a positive integer.