Binkov A.d:

Does there exist a parallelogram such that all points of pairwise intersection of the angle bisectors lie outside the parallelogram?
Let in parallelogram $A B C D$ side $A D$ is not less than $A B$. We lay off segment $A E=A B$ on $A D$ and draw a line through $E$ parallel to $A B$. We get a rhombus, and in a rhombus, the diagonals are the bisectors of the angles, and their point of intersection is the center of the rhombus. But this point is the intersection of two angle bisectors of the original parallelogram, and the rhombus is contained within it. Therefore, this point belongs to the parallelogram.

## Answer

Does not exist.

## [Inscribed and Circumscribed Polygons] Problem 115770 Topics: [ Regular Polygons ] Difficulty: 3+ Grades: $8,9,10,11$

Author: Frenkin B.R.

A convex polygon is circumscribed around a circle. The points of tangency of its sides with the circle form a polygon with the same set of angles (the order of the angles may be different). Is it true that the polygon is regular?

## Solution

Author: Kairanbay $M$.

Let $A_{1} A_{2} \ldots A_{n}$ be the given polygon, $B_{1}, B_{2}, \ldots, B_{n}$ be the points of tangency of the inscribed circle with the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$. Denote the angles of the first polygon by $a_{1}, \ldots, a_{n}$, and the second by $b_{1}, \ldots, b_{n}$. Then $b_{i}=1 / 2\left(a_{i}+a_{i+1}\right)$. Multiplying $n$ such equalities, we get

$2^{n} a_{1} a_{2} \ldots a_{n}=\left(a_{1}+a_{2}\right) \ldots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)$. But by the Cauchy inequality $a_{i}+a_{i+1} \geq 2 \sqrt{a_{i} a_{i+1}}$. Therefore, from the obtained equality, it follows that all angles of the polygons are equal, and since the polygon $B_{1} \ldots B_{n}$ is inscribed, the polygons are regular.

## Answer

True.