$\underline{\text { Kanel-Belov A.Y. }}$

On a sphere, five points are marked, no three of which lie on a great circle (a great circle is a circle formed by the intersection of the sphere and a plane passing through its center). Two great circles, not passing through the marked points, are called equivalent if one can be continuously moved to the other on the sphere without passing through the marked points during the process.

a) How many great circles can be drawn that do not pass through the marked points and are not equivalent to each other?

b) The same problem for $n$ marked points.
a) Let's move on to dual objects: each circle corresponds to a pair of opposite points on the sphere such that the diameter connecting them is perpendicular to this circle; conversely, each point corresponds to a great circle. Then the problem reduces to a dual one: points are considered equivalent if one can be translated into another without touching five given great circles (no three of which intersect at one point). Obviously, a point can be moved within the regions into which the great circles divide the sphere. Thus, the number of equivalence classes is half the number of parts into which the great circles divide the sphere (opposite parts correspond to one class, as the original great circle in the dual problem corresponds to two diametrically opposite points).

According to problem $60324 n$ of our great circles divide the sphere into $n^{2}-n+2$ parts. In particular, five circles will divide the sphere into 22 parts. And the answer, as shown above, is twice as small.

b) See a).

| Problem 105167 Topics: | Tournaments and tournament tables | Difficulty: $4++$  Classes: 9,10,11 |
| :---: | :---: | :---: |
|  | Order relation |  |
|  | ![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-24.jpg?height=86&width=838&top_left_y=348&top_left_x=475) |  |
|  | [Estimate + example |  |

![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-24.jpg?height=49&width=280&top_left_y=507&top_left_x=0)

On the shore of a round island called Somewhere, there are 20 villages, each inhabited by 20 wrestlers. A tournament was held in which each wrestler met all the wrestlers from all other villages. Village $A$ is considered stronger than village B if at least $k$ matches between wrestlers from these villages end with a victory for a wrestler from village A. It turned out that each village is stronger than the next one in the clockwise direction. What is the maximum value that $k$ can have? (All wrestlers have different strengths, and the stronger one always wins in a match.)

## Solution

Estimate. We will show that such a situation is impossible for $k>290$. We will order the wrestlers in each village by decreasing strength and select the tenth strongest wrestler in each village. We will show that the village where the weakest of the selected wrestlers lives cannot be stronger than the next one. Let's denote the selected wrestlers in our and the next village as $A$ and $B$ respectively. Then in our village, 11 wrestlers are not stronger than $A$, and in the next village, 10 wrestlers are not weaker than $B$. All 110 matches between these wrestlers will end in favor of the second village, meaning that the number of matches won by wrestlers from our village is no more than $20 \cdot 20-110=290$.

Example. Suppose among the wrestlers there are 210 novices and 190 masters (each novice is weaker than any master). We will number the villages counterclockwise. Place the weakest novice and 19 weakest masters in the first village; in the second village, place the two weakest remaining novices and 18 weakest remaining masters; in the third village, place the three weakest remaining novices and 17 weakest remaining masters; ...; in the last village, place the 20 strongest novices. We will show that the $i$-th village is stronger than the $(i-1)$-th for $i>1$. Indeed, it has $i$ novices and $20-i$ masters. The masters of the $i$-th village will defeat all in the $(i-1)$-th, and the novices will defeat the novices, and the total number of victories will be $i(i-1)+\left\{20(20-i)=i^{2}-21 i+400\right\}$. The x-coordinate of the vertex of this parabola is 10.5, so the minimum value at an integer point is achieved at $i=10,11$ and is 290, meaning that the $i$-th village is stronger than the $(i-1)$-th for $k \leq 290$. In addition, the masters of the first village will defeat the novices of the 20th village, and the total number of victories will be $20 \cdot 19=380>290$, meaning all conditions are met.

## Answer

290.