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Problem 852

AIME late

Geometry Difficulty 5.7 Find the answer

9-5. A circle is divided into 100 equal arcs by 100 points. Next to the points, numbers from 1 to 100 are written, each exactly once. It turns out that for any number $k$ , if a diameter is drawn through the point with the number $k$ , then the number of numbers less than $k$ on either side of this diameter will be equal. What number can be written at the point diametrically opposite the point with the number $83?$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Answer: Only 84.

Solution: Consider the odd number $2 m+1$ . Let's mentally discard it and the number diametrically opposite to it. According to the condition, among the remaining numbers, all numbers less than $2 m+1$ are divided into two groups of equal size. Therefore, among the remaining numbers, there is an even number of numbers less than $2 m+1$ . In total, the number of numbers less than $2 m+1$ is also even - there are $2 m$ of them. From this, it follows that the number diametrically opposite to $2 m+1$ must be greater than it!

Then, opposite 99 can only be 100, opposite 97 - only 98 (since the numbers 99 and 100 are already opposite each other), opposite 95 - only 96 (since all larger numbers are already opposite each other), and so on. Therefore, opposite 83 stands 84.

Comment: Note that any variant where for all $m=1,2, \ldots, 50$ the numbers $2 m-1$ and $2 m$ are in diametrically opposite positions, is suitable.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.