Problem 1117

Solution. Let the odd numbers $a$ and $b$ be the lengths of the sides of a given rectangle. Suppose there is a point $T$ inside the rectangle such that the distance from $T$ to each vertex of the rectangle is an integer. Let $x_{1}$ and $x_{2}$ be the distances from point $T$ to the sides of length $b$, and $y_{1}$ and $y_{2}$ be the distances from point $T$ to the sides of length $a$. Then $a=x_{1}+x_{2}, b=y_{1}+y_{2}$, and the numbers

$$d_{i j}=\sqrt{x_{i}^{2}+y_{j}^{2}}, i, j \in\{1,2\}$$

are integers. We introduce the notations: $a_{i}=a b x_{i}, b_{j}=a b y_{j}$, where $i, j \in\{1,2\}$ and

$$A_{1}=a_{1}-a_{2}, A_{2}=a_{1}+a_{2}, B_{1}=b_{1}-b_{2}, B_{2}=b_{1}+b_{2}$$

Then

$$\begin{aligned} & A_{1}=a b\left(x_{1}-x_{2}\right)=b\left(x_{1}^{2}-x_{2}^{2}\right)=b\left(\left(x_{1}^{2}+y_{1}^{2}\right)-\left(x_{2}^{2}+y_{1}^{2}\right)\right) \\ & A_{2}=a b\left(x_{1}+x_{2}\right)=a^{2} b \\ & B_{1}=a b\left(y_{1}-y_{2}\right)=a\left(y_{1}^{2}-y_{2}^{2}\right)=a\left(\left(x_{1}^{2}+y_{1}^{2}\right)-\left(x_{2}^{2}+y_{2}^{2}\right)\right) \\ & B_{2}=a b\left(y_{1}+y_{2}\right)=a b^{2} \end{aligned}$$

so $A_{1}$ and $B_{1}$ are integers, and $A_{2}$ and $B_{2}$ are odd numbers.

Suppose each of the numbers $a_{1}, a_{2}, b_{1}, b_{2}$ is an integer. Since $A_{2}$ and $B_{2}$ are odd numbers, we get that exactly one of the numbers $a_{1}, a_{2}$ and exactly one of the numbers $b_{1}, b_{2}$ is odd. Let, for example, $a_{1}$ and $b_{1}$ be odd numbers. Then

$$\begin{aligned} & a_{1}^{2} \equiv 1(\bmod 4), b_{1}^{2} \equiv 1(\bmod 4) \\ & a_{1}^{2}+b_{1}^{2} \equiv a^{2} b^{2}\left(x_{1}^{2}+y_{1}^{2}\right)=\left(d_{i j} a b\right)^{2} \equiv 2(\bmod 4) \end{aligned}$$

which is a contradiction. Therefore, at least one of the numbers $a_{1}, a_{2}, b_{1}, b_{2}$ is not an integer. Since

$$2 a_{1}=A_{1}+A_{2}, 2 a_{2}=A_{2}-A_{1}, 2 b_{1}=B_{1}+B_{2}, 2 b_{2}=B_{2}-B_{1},$$

we conclude that at least one of the numbers $2 a_{1}, 2 a_{2}, 2 b_{1}, 2 b_{2}$ is odd. Let, for example, $2 a_{1}$ be an odd number. Then

$$\left(2 a_{1}\right)^{2}+\left(2 b_{1}\right)^{2}=4\left(a_{1}^{2}+b_{1}^{2}\right)=4 a^{2} b^{2}\left(x_{1}^{2}+y_{1}^{2}\right)=4 a^{2} b^{2} d_{11}^{2}$$

i.e.

$$\left(2 b_{1}\right)^{2}=4 a^{2} b^{2} d_{11}^{2}-\left(2 a_{1}\right)^{2} \equiv-1(\bmod 4)$$

which is a contradiction. Finally, from the obtained contradiction, the statement of the problem follows.