$\underline{\text { Frame }}$ D:

On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that the polygon has no parallel sides?
Let $A_{1}, A_{2}, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=A_{2}$). We will divide them into quadruples

$\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right),\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \ldots,\left(A_{1005}, A_{1006}, A_{2011}, A_{2012}\right)$. If among the selected $k$ points, all points of some quadruple ($A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}$) are encountered, then in the resulting polygon, there will be two sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which are symmetric with respect to the center of the circle and therefore parallel. Thus, in each of the 503 quadruples, no more than three vertices will be marked, that is, $k \leq 503 \cdot 3 = 1509$.

An example of a 1509-gon without parallel sides with vertices at the marked points: $A_{1} A_{2} \ldots A_{1006} A_{1008} A_{1010} \ldots A_{2012}$ (the vertices are all points with numbers from 1 to 1006 and all points with even numbers from 2008 to 2012). Indeed, the sides $A_{2012} A_{1}, A_{1} A_{2}, \ldots, A_{1005} A_{1006}$ lie on one side of the diameter $A_{2012} A_{1006}$ and are therefore not parallel; similarly, the sides $A_{1006} A_{1008}, \ldots, A_{2010} A_{2012}$ are pairwise not parallel. Finally, the small diagonal $A_{j} A_{j+2}$ of a regular 2012-gon is not parallel to its sides; hence, no two sides of the form $A_{i} A_{i+1}$ and $A_{j} A_{j+2}$ can be parallel.

## [ Inscribed sphere in a tetrahedron $\quad$ ] Problem 116816 Topics: [ Inequality problems. Case analysis ] Difficulty: $4-$ [Examples and counterexamples. Constructions ]

Author: Shapovalov A.v.

Prove that it is possible to write a non-negative number on each edge of an arbitrary tetrahedron such that the sum of the numbers on the sides of each face is numerically equal to its area.

## Solution

Let the area of the smallest face be $s$, the largest face be $S$, and the two remaining faces be $a$ and $b$. Write the number $s$ on the side common to the smallest and largest faces, and write zero on the other two sides of the smallest face. Write the numbers $1 / 2(S-b+a-s)$ and $1 / 2(S-a+b-s)$ (each of which is non-negative as the sum of two non-negative numbers) on the other two sides of the largest face. Finally, write the number $1 / 2(a+b+s-S)$ (which is non-negative because the projections of three faces cover the fourth, and the area of a face is not less than the area of its projection onto another face) on the only remaining edge. It is easy to verify that the condition is satisfied.

Submit a comment