Found problems: 31
The square $ABCD$ has to be decomposed into $n$ triangles (which are not overlapping) and which have all angles acute. Find the smallest integer $n$ for which there exist a solution of that problem and for such $n$ construct at least one decomposition. Answer whether it is possible to ask moreover that (at least) one of these triangles has the perimeter less than an arbitrarily given positive number.
For an integer $m\geq 1$, we consider partitions of a $2^m\times 2^m$ chessboard into rectangles consisting of cells of chessboard, in which each of the $2^m$ cells along one diagonal forms a separate rectangle of side length $1$. Determine the smallest possible sum of rectangle perimeters in such a partition.
[i]Proposed by Gerhard Woeginger, Netherlands[/i]
Consider decompositions of an $8\times 8$ chessboard into $p$ non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If $a_i$ is the number of white squares in the $i$-th rectangle, then $a_1<a_2<\ldots <a_p$.
Find the maximum value of $p$ for which such a decomposition is possible. For this value of $p$, determine all possible sequences $a_1,a_2,\ldots ,a_p$.
Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.
Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.
Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).
Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.
Prove that a regular polygon with an odd number of edges cannot be partitioned into four pieces with equal areas by two lines that pass through the center of polygon.
For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?
For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?
A figure with area $1$ is cut out of paper. We divide this figure into $10$ parts and color them in $10$ different colors. Now, we turn around the piece of paper, divide the same figure on the other side of the paper in $10$ parts again (in some different way). Show that we can color these new parts in the same $10$ colors again (hereby, different parts should have different colors) such that the sum of the areas of all parts of the figure colored with the same color on both sides is $\geq \frac{1}{10}.$
Could the three-dimensional space be expressed as the union of disjoint circumferences?
Prove that a regular polygon with an odd number of edges cannot be partitioned into four pieces with equal areas by two lines that pass through the center of polygon.
For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?
Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).
For an integer $m\geq 1$, we consider partitions of a $2^m\times 2^m$ chessboard into rectangles consisting of cells of chessboard, in which each of the $2^m$ cells along one diagonal forms a separate rectangle of side length $1$. Determine the smallest possible sum of rectangle perimeters in such a partition.
[i]Proposed by Gerhard Woeginger, Netherlands[/i]
A figure with area $1$ is cut out of paper. We divide this figure into $10$ parts and color them in $10$ different colors. Now, we turn around the piece of paper, divide the same figure on the other side of the paper in $10$ parts again (in some different way). Show that we can color these new parts in the same $10$ colors again (hereby, different parts should have different colors) such that the sum of the areas of all parts of the figure colored with the same color on both sides is $\geq \frac{1}{10}.$
$(GDR 4)$ Find the maximal number of regions into which a sphere can be partitioned by $n$ circles.
The square $ABCD$ has to be decomposed into $n$ triangles (which are not overlapping) and which have all angles acute. Find the smallest integer $n$ for which there exist a solution of that problem and for such $n$ construct at least one decomposition. Answer whether it is possible to ask moreover that (at least) one of these triangles has the perimeter less than an arbitrarily given positive number.
Let $K$ and $K'$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_1, T_2, \ldots, T_p$ with mutually disjoint interiors and find translations $t_1, t_2, \ldots, t_p$ such that
\[K'=\bigcup_{i=1}^{p} t_i(T_i) \ ? \]
A rectangle $ D$ is partitioned in several ($ \ge2$) rectangles with sides parallel to those of $ D$. Given that any line parallel to one of the sides of $ D$, and having common points with the interior of $ D$, also has common interior points with the interior of at least one rectangle of the partition; prove that there is at least one rectangle of the partition having no common points with $ D$'s boundary.
[i]Author: Kei Irie, Japan[/i]
We consider the division of a chess board $8 \times 8$ in p disjoint rectangles which satisfy the conditions:
[b]a)[/b] every rectangle is formed from a number of full squares (not partial) from the 64 and the number of white squares is equal to the number of black squares.
[b]b)[/b] the numbers $\ a_{1}, \ldots, a_{p}$ of white squares from $p$ rectangles satisfy $a_1, , \ldots, a_p.$ Find the greatest value of $p$ for which there exists such a division and then for that value of $p,$ all the sequences $a_{1}, \ldots, a_{p}$ for which we can have such a division.
[color=#008000]Moderator says: see [url]https://artofproblemsolving.com/community/c6h58591[/url][/color]
$(GDR 4)$ Find the maximal number of regions into which a sphere can be partitioned by $n$ circles.
For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos? When it is possible, what is the minimum number of trominos needed?
Let $K$ and $K'$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_1, T_2, \ldots, T_p$ with mutually disjoint interiors and find translations $t_1, t_2, \ldots, t_p$ such that
\[K'=\bigcup_{i=1}^{p} t_i(T_i) \ ? \]