At the recent IZhO in mathematics there were unusual problems, many of which trolled the majority of participants. P2 was solved by a trick, and many failed to find it. I also did not solve it: during the contest I acted too technically, which did not help. My approach was an attempt to squeeze the desired number between given ones, which is a stronger statement (but, as it turned out, equivalent) that I managed to prove after the contest.
Problem.
Let $$$n$$$ be a positive integer for which there exist positive integers $$$a$$$ and $$$b$$$ such that
$$$ \lfloor a\sqrt{10} \rfloor = n = \lfloor b\sqrt{11} \rfloor. $$$Prove that there exists a positive integer $$$c$$$ such that
$$$ n = \left\lfloor c(11\sqrt{10} - 10\sqrt{11}) \right\rfloor. $$$
Official solution
Denote
$$$ \alpha = \sqrt{10}, \quad \beta = \sqrt{11}, \quad \gamma = 11\sqrt{10} - 10\sqrt{11}. $$$ $$$ \gamma = \frac{\sqrt{110}}{\sqrt{10} + \sqrt{11}} = \frac{1}{\alpha} + \frac{1}{\beta}. $$$ $$$ n \le a\alpha \lt n+1, $$$ $$$ n \le b\beta \lt n+1 $$$ $$$ \frac{a}{n+1} \lt \frac{1}{\alpha} \le \frac{a}{n}, $$$ $$$ \frac{b}{n+1} \lt \frac{1}{\beta} \le \frac{b}{n}. $$$Adding, we obtain
$$$ \frac{a+b}{n+1} \lt \gamma \le \frac{a+b}{n}. $$$ $$$ n \le (a+b)\gamma \lt n+1, $$$from which
$$$ n = \left\lfloor (a+b)(11\sqrt{10} - 10\sqrt{11}) \right\rfloor. $$$
My solution
Idea.
- We look for values $$$x_c$$$ that are squeezed between $$$a\sqrt{10}$$$ and $$$b\sqrt{11}$$$;
- we prove that such a $$$c$$$ exists and is unique;
- we show that for the remaining $$$c$$$ equality of integer parts is impossible.
Denote
$$$ x_c = c(11\sqrt{10} - 10\sqrt{11}) = \frac{c\sqrt{110}}{\sqrt{10} + \sqrt{11}}. $$$
Step 1. Squeezing condition
If
$$$ \min(a\sqrt{10}, b\sqrt{11}) \le x_c \le \max(a\sqrt{10}, b\sqrt{11}), $$$then it immediately follows that
$$$ \lfloor x_c \rfloor = n. $$$This is equivalent to the inequality
$$$ (x_c - a\sqrt{10})(x_c - b\sqrt{11}) \le 0. $$$Compute the differences:
$$$ x_c - a\sqrt{10} = \frac{(c-a)\sqrt{110} - 10a}{\sqrt{10} + \sqrt{11}}, $$$ $$$ x_c - b\sqrt{11} = \frac{(c-b)\sqrt{110} - 11b}{\sqrt{10} + \sqrt{11}}. $$$Since the denominator is positive, we obtain the equivalent condition
$$$ (c - a - a\sqrt{10/11})(c - b - b\sqrt{11/10}) \le 0. $$$Hence,
$$$ c \in [L, R], $$$where
$$$ L = \min(a + a\sqrt{10/11}; b + b\sqrt{11/10}), $$$ $$$ R = \max(a + a\sqrt{10/11}; b + b\sqrt{11/10}). $$$
Step 2. Uniqueness of the integer solution
The length of the segment equals
$$$ R - L = \left(\frac{1}{\sqrt{10}} + \frac{1}{\sqrt{11}}\right) \left|a\sqrt{10} - b\sqrt{11}\right|. $$$From the conditions we have
$$$ |a\sqrt{10} - b\sqrt{11}| \lt 1, $$$therefore
$$$ R - L \lt 1. $$$Thus, at most one integer can lie in $$$[L, R]$$$.
At the same time,
$$$ {L, R} := { a + b - \frac{b\sqrt{11} - a\sqrt{10}}{\sqrt{11}},a + b - \frac{a\sqrt{10} - b\sqrt{11}}{\sqrt{10}}}, $$$from which
$$$ a + b \in [L, R]. $$$Therefore,
$$$ c = a + b $$$is the unique integer solution; the necessary and sufficient conditions are verified.
Step 3. Absence of other solutions
From the inequality $$$b\sqrt{11} - a\sqrt{10} \lt 1$$$ it follows that
$$$ b\sqrt{110} - 10a \lt \sqrt{10}. $$$Then
$$$ (a+b)\frac{\sqrt{110}}{\sqrt{10} + \sqrt{11}} - a\sqrt{10} \lt \frac{\sqrt{10}}{\sqrt{10} + \sqrt{11}}. $$$Consequently,
$$$ (a+b-1)\frac{\sqrt{110}}{\sqrt{10} + \sqrt{11}} \lt a\sqrt{10} - 1, $$$that is,
$$$ \left\lfloor x_{a+b-1} \right\rfloor \le n-1. $$$Similarly,
$$$ \left\lfloor x_{a+b+1} \right\rfloor \ge n+1. $$$Hence, no other values of $$$c$$$ exist.
