problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
4. Let $a, b$ be any positive real numbers such that $a^{2}+b^{2}=1$. Find the smallest possible value of the expression
$$
\frac{a^{2}\left(a+b^{3}\right)}{b-b^{3}}+\frac{b^{2}\left(b+a^{3}\right)}{a-a^{3}}
$$
and determine for which pairs $(a, b)$ this value is achieved. (Tomáš Bárta) | SOLUTION. Using the condition $a^{2}+b^{2}=1$, we first simplify the first fraction of the given expression:
$$
\frac{a^{2}\left(a+b^{3}\right)}{b-b^{3}}=\frac{a^{2}\left(a+b^{3}\right)}{b\left(1-b^{2}\right)}=\frac{a^{2}\left(a+b^{3}\right)}{b \cdot a^{2}}=\frac{a+b^{3}}{b}=\frac{a}{b}+b^{2} .
$$
Similarly, for the ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A $3 \times 3$ table is filled with distinct natural numbers such that in each row and column, the sum of the outer numbers equals the number written between them. Determine the smallest number that can be written in the center of the table.
(Tomáš Jurík) | Solution. Let's denote the numbers in the corners of the table as $a, b, c, d$ (from left to right, top to bottom). These four numbers uniquely determine all other numbers in the table, as we can sequentially calculate the numbers between them and finally the number $a+b+c+d$ in the center of the table.
| $a$ | | $b$... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find all tetrahedra that have a net in the shape of a deltoid and exactly four edges of a given length a. (A deltoid is understood to be a convex quadrilateral symmetric with respect to only one of its diagonals; thus, neither a square nor a rhombus is considered a deltoid.) | SOLUTION. In the first (more significant) part of the solution, we will find all tetrahedra that have a net in the shape of a kite; then it will be relatively easy to determine which of the found tetrahedra have exactly four equal edges.
Consider, therefore, an arbitrary tetrahedron $ABCD$ and describe the lengths of ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Assume that for positive real numbers $a, b, c, d$ the following holds:
$$
a b+c d=a c+b d=4 \quad \text { and } \quad a d+b c=5 .
$$
Find the smallest possible value of the sum $a+b+c+d$ and determine which quadruples $a, b, c, d$ achieve it.
The regional round of category A takes place
## on Tuesday, January 1... | 4. Since the given equations contain mixed products of variables, it will be advantageous to examine the square of the sum $a+b+c+d$. By rearrangement and substitution of the given equations, we get
$$
\begin{aligned}
(a+b+c+d)^{2} & =a^{2}+b^{2}+c^{2}+d^{2}+2(ab+cd+ac+bd+ad+bc)= \\
& =a^{2}+b^{2}+c^{2}+d^{2}+2(4+4+5)... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On the board, we will write ten different natural numbers. In each step, we first underline each number that is not the sum of any two different numbers written on the board, and then erase all underlined numbers. For example:
 According to the problem statement, all numbers on the board are positive and distinct. In each step, we will certainly underline the smallest number on the board, and - if it is not the only one - also the second smallest number, as it cannot equal the sum of two different numbers on the board. Since we e... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
3. Find all integers $n>2$ such that the number $n^{n-2}$ is the $n$-th power of an integer.
(Patrik Bak) | SOLUTION. We will show that only $n=4$ satisfies the condition. For brevity, we will use "n-th power" instead of "n-th power of an integer" throughout the solution.
A positive integer is an $n$-th power if and only if each prime number in its prime factorization appears with an exponent that is a multiple of $n$. Ther... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. If a, b, c are distinct positive real numbers, what is the smallest possible number of distinct numbers among the numbers $a+b, b+c, c+a, ab, bc, ca, abc$?
(Patrik Bak) | SOLUTION. Since $a, b, c$ are distinct positive numbers, so are the numbers $a b, b c, c a$, because, for example, from $a b=b c$ it follows that $a=c$ (thanks to $b \neq 0$). We see, then, that in the examined septet of numbers $a+b, b+c, c+a, a b, b c, c a, a b c$ there are at least 3 different values. We will first ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For any four-element subset $\mathrm{P}$ of the set $\{1,2,3, \ldots, 12\}$, denote
$$
\mathrm{Q}=\{3 x: x \in \mathrm{P}\} \quad \text { and } \quad \mathrm{R}=\{4 x: x \in \mathrm{P}\} \text {. }
$$
Determine the number of such sets $P$ for which the numbers from $P, Q, R$ give all possible non-zero remainders w... | Solution. We will calculate with all numbers modulo 13, i.e., as remainders upon division by 13. Furthermore, it is clear that not only the set $P$, but also the derived sets $Q$ and $R$ are four-element sets. Therefore, if the three mentioned sets are to contain all non-zero remainders modulo 13, of which there are tw... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Determine the number of all triples of two-digit natural numbers $a, b, c$, whose product $abc$ has a representation in which all digits are the same. Triples differing only in the order of the numbers are considered the same, i.e., they are counted only once. | SOLUTION. For two-digit numbers $a, b, c$, the product $abc$ is a four-digit, five-digit, or six-digit number. If all the digits of the number $abc$ are equal to the same digit $k$, one of the equalities $abc = k \cdot 1111, abc = k \cdot 11111$, or $abc = k \cdot 111111, k \in \{1,2, \ldots, 9\}$ holds.
The numbers $... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given a trapezoid $A B C D$ with bases $A B$ and $C D$. Let $E$ be the midpoint of side $A B$, $F$ the midpoint of segment $D E$, and $G$ the intersection of segments $B D$ and $C E$. Express the area of trapezoid $A B C D$ in terms of its height $v$ and the length of segment $F G$, assuming that points $A, F, C$ li... | SOLUTION. According to the problem, angles $E F D$ and $A F C$ are right angles, so we have (Fig. 3)
$$
\begin{aligned}
& |\Varangle C D F|=|\Varangle A E F| \quad \text { (alternate interior angles), } \\
& |\Varangle C F D|=|\Varangle A F E| \quad \text { (vertical angles). }
\end{aligned}
$$
Furthermore, point $F$... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. For any natural number $n$, let us form all possible "words" of length $n$ from the letters $A$ and $B$, and let $p_{n}$ denote the number of those words that do not contain the quadruple $AAAA$ of consecutive letters $A$, nor the triple $BBB$ of consecutive letters $B$. Determine the value of the expression
$$
\fr... | Solution. Let the number of suitable words of length $n$ ending with the letter $A$, respectively $B$, be denoted by $a_{n}$, respectively $b_{n}$. Then
$$
p_{n}=a_{n}+b_{n} \text {. }
$$
Let $n \geqq 4$. A suitable word ending with the letter $A$ has one of the endings $B A, B A A$, or $B A A A$. The number of words... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a segment $A B$. Construct a point $C$ such that the area of triangle $A B C$ equals $1 / 8$ of the area $S$ of a square with side $A B$, and the sum of the areas of squares with sides $A C$ and $B C$ equals $S$. How many solutions does the problem have for a given placement of segment $A B$ in the plane?
The... | 3. The condition that the area of triangle $ABC$ is to equal $\frac{1}{8}$ of the area $S$ of the square with side $AB$ means that the height of triangle $ABC$ to side $AB$ has a length of $\frac{1}{4}|AB|$, so point $C$ must lie on one of the two lines parallel to line $AB$ and at a distance of $\frac{1}{4}|AB|$ from ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let $\tau(k)$ denote the number of all positive divisors of a natural number $k$, and suppose the number $n$ is a solution to the equation $\tau(1.6 n)=1.6 \tau(n)$. Determine the value of the ratio $\tau(0.16 n): \tau(n)$. | SOLUTION. If the prime factorization of the number $n$ is $n=\prod_{i=1}^{k} p_{i}^{s_{i}},$ where $p_{1}, \ldots, p_{k}$ are distinct primes and $s_{1}, \ldots, s_{k}$ are non-negative integers, the number of its positive divisors is $\tau(n)=\prod_{i=1}^{k}\left(s_{i}+1\right)$. If $n=2^{\alpha} 5^{\beta} n^{\prime}$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let a regular n-gon be denoted as $A_{1} A_{2} \ldots A_{n}$. The point $A_{3}$ is reflected over the axis $A_{2} A_{4}$, resulting in the point $A_{3}^{\prime}$. Then, the point $A_{3}^{\prime}$ is reflected over the axis $A_{1} A_{3}$, resulting in the point $A_{3}^{\prime \prime}$. For which $n \geqq 4$ is the po... | SOLUTION. Let $S$ be the center of the circle circumscribed around the $n$-gon $A_{1} A_{2} \ldots A_{n}$. Since the central angle $A_{i} S A_{i+1}$ is equal to $\frac{360^{\circ}}{n}$ for each $i=1, \ldots, n-1$, all the acute inscribed angles over the chords $A_{i} A_{i+1}$ have a size of $\frac{180^{\circ}}{n}$. Fou... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For the sequence $\left(a_{n}\right)_{n=1}^{\infty}$, we know that for all natural numbers $n$,
$$
a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6} .
$$
a) Find all values of $a_{1}$ for which this sequence is constant.
b) Let $a_{1}=5$. Determine the greatest integer not exceeding $a_{2018}$. (Vojtech Bálint) | SOLUTION. a) Suppose the sequence $\left(a_{n}\right)_{n=1}^{\infty}$ is constant. Then it must hold that $a_{2}=a_{1}$, which we can write using the relation from the problem as
$$
a_{1}=\frac{a_{1}^{2}}{a_{1}^{2}-4 a_{1}+6}
$$
We can easily rearrange this equation to $a_{1}\left(a_{1}-2\right)\left(a_{1}-3\right)=0... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For any three-digit number, we determine its remainders when divided by the numbers 2, 3, 4, ..., 10 and then sum the nine resulting numbers. Find the smallest possible value of such a sum. | SOLUTION. Let $S(n)$ denote the sum of the mentioned remainders of the three-digit number $n$. We will explain why $S(n) \geqq 3$.
- For odd $n$, $S(n) \geqq 5$ (consider the remainders when dividing by even numbers $2,4,6,8$, 10). Hence, let $n$ be even from now on.
- If $4 \nmid n$, then $S(n) \geqq 4$ ( $n$ gives a... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the real number $p$ such that the equation
$$
x^{2}+4 p x+5 p^{2}+6 p-16=0
$$
has two distinct roots $x_{1}, x_{2}$ and the sum $x_{1}^{2}+x_{2}^{2}$ is minimized. | 1. For the roots $x_{1}, x_{2}$ of the given quadratic equation (if they exist), according to Vièta's formulas, the following equalities hold:
$$
x_{1}+x_{2}=-4 p \quad \text { and } \quad x_{1} x_{2}=5 p^{2}+6 p-16
$$
from which we can compute the desired sum
$$
\begin{aligned}
x_{1}^{2}+x_{2}^{2} & =\left(x_{1}+x_... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the smallest values that the expression $V=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ can take, given that the real numbers $a$, $b$, $c$ satisfy the pair of conditions
$$
\begin{aligned}
a+3 b+c & =6 \\
-a+b-c & =2
\end{aligned}
$$ | SOLUTION. By adding both equations, we find that $b=2$. Substituting $b$ into one of them yields $c=-a$. Therefore, $V=(a-2)^{2}+(2+a)^{2}+(-2 a)^{2}$. After squaring and adding, we find that $V=6 a^{2}+8 \geqq 8$. Equality occurs precisely when $a=0, b=2$ and $c=0$.
The sought minimum value of the expression $V$ is t... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On the board, there is a positive integer $n$. In one step, we can erase the number on the board and write either its double, or its double increased by 1. For how many initial numbers $n$ different from 2019 can we achieve that the number 2019 appears on the board after a finite number of steps?
(Josef Tkadlec) | 1. In one step, the current natural number $k$ is increased either to the even number $m=2k$, or to the odd number $m=2k+1$. Depending on the parity of the new number $m$, we can reconstruct the previous number $k$: either $k=m/2$, or $k=(m-1)/2$ — depending on whether $m$ is even or odd.
The odd number 2019 will thus... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A $10 \times 10$ table is filled with numbers 1 and -1 such that the sum of the numbers in each row, except one, is equal to zero, and the sum of the numbers in each column, except one, is also equal to zero. Determine the maximum possible sum of all the numbers in the table.
(Patrik Bak) | SOLUTION. If we add up all the numbers in the considered table row by row, we will conclude that the total sum is equal to the sum of the numbers in the exceptional row where it is not zero. This sum is at most 10, and it equals 10 if all the numbers in the given row are ones. (We reach the same conclusion if we add up... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In a square table $11 \times 11$, we have written the natural numbers $1,2, \ldots, 121$ sequentially from left to right and from top to bottom. Using a square tile $4 \times 4$, we covered exactly 16 cells in all possible ways. How many times was the sum of the 16 covered numbers a perfect square? | SOLUTION. Let the number covered by the top left corner of the tile be denoted as $z$. The entire tile must lie within the given table, so the values of $z$ can only be the numbers written in the first eight rows and the first eight columns of the table (for example, if $z=10$, the tile would protrude, and thus could n... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a $4 \times 4$ table, positive real numbers are written such that the product in each pentad of the shape 母 is equal to 1. Determine the maximum number of different numbers written in the table.
Note: The shape 母 refers to a specific arrangement of five cells in the table, resembling the Chinese character 母. For... | SOLUTION. Let $a, b, c, d, e, f, g, h, i$ be the numbers inscribed in the left upper $3 \times 3$ square of the table (Fig. 1). By comparing the products for pentads of the form $\mathbb{\text { a }}$ and placed in this part of the table, we must have $a b c d e=b d e f g$, or $a c=f g$. Similarly, for pentads $\mathrm... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Determine how many numbers we can select $z$ from the set $\{1,2,3, \ldots, 75599,75600\}$ such that the set includes the number 75600 and for any two selected numbers $a$, $b$, it holds that $a$ is a divisor of $b$ or $b$ is a divisor of $a$. (List all possibilities.) | SOLUTION. Let's consider a set M that satisfies the conditions given in the problem. Since M contains the number 75600, it must be at least a one-element set. Further, notice that if we remove some number $a \neq 75600$ from the set M, we get a set $\mathrm{M}^{\prime} \subset \mathrm{M}$, which also satisfies the give... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On the board, the numbers 1, 2, .., 33 are written. In one step, we choose two numbers on the board, one of which is a divisor of the other, erase both, and write their (integer) quotient on the board. We continue this process until only numbers remain on the board, none of which is a divisor of another. (In one ste... | SOLUTION. The numbers on the board will likely remain only from the set $M=\{1,2, \ldots, 33\}$. The prime numbers 17, 19, 23, 29, and 31 will always be written, each exactly once, because they have no divisors other than 1, and the set $M$ does not contain any of their multiples (so they can never disappear from the b... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The quadratic equation $x^{2}-35 x+334=0$, whose coefficients are written in a number system with base $z(z \geqq 6)$, has two distinct real roots. Determine $z$ and both roots.
School - the written part of the 1st round of category B takes place
## on Tuesday, January 26, 1999
so that it starts in the morning an... | ## 3. Given equation
$$
x^{2}-(3 z+5) x+\left(3 z^{2}+3 z+4\right)=0
$$
has two distinct real roots if and only if its discriminant $D$ is positive,
$$
\begin{aligned}
D & =(3 z+5)^{2}-4\left(3 z^{2}+3 z+4\right)=-3 z^{2}+18 z+9= \\
& =-3\left(z^{2}-6 z-3\right)=-3\left((z-3)^{2}-12\right)>0
\end{aligned}
$$
from w... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number $k$ such that every $k$-element set of pairwise coprime three-digit numbers contains at least one prime number. | SOLUTION. For constructing a set of two-by-two coprime three-digit composite numbers with a large number of elements, we can use the fact that powers of two different prime numbers are coprime. The set
$$
\left\{2^{7}, 3^{5}, 5^{3}, 7^{3}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}\right\}
$$
contains 11 ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A right-angled triangle has integer lengths of sides. Its perimeter is the square of a natural number. We also know that one of its legs has a length equal to the square of a prime number. Determine all possible values of this length.
(Patrik Bak) | SOLUTION. We are looking for all prime numbers $p$ for which there exists a described triangle with one leg of length $p^{2}$. Let the length of its other leg be $b$ and the length of the hypotenuse $c$. Then, according to the Pythagorean theorem, the equation $c^{2}=p^{4}+b^{2}$ holds, which we rearrange into a produc... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Natural numbers $a$, $b$, $c$ are placed in a circle as shown in the figure, with each number being a divisor of the sum of the two numbers adjacent to it. How many of the numbers $a$, $b$, $c$ can be different? (Josef Tkadlec)
^{2}+(b+c)^{2}+(c+a)^{2} .
$$ | Solution. Notice that the expression $a+b+c$ contains only the first powers of the variables, and the expression we need to work with further contains their second powers. By expanding and rearranging the expression $V$ from the problem statement, we get
$$
\begin{aligned}
V & =(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=2\left(a^{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the values that the expression $V=a b+b c+c d+d a$ can take, given that the real numbers $a, b, c, d$ satisfy the following conditions:
$$
\begin{aligned}
& 2 a-5 b+2 c-5 d=4, \\
& 3 a+4 b+3 c+4 d=6 .
\end{aligned}
$$ | 1. For the given expression $V$, we have
$$
V=a(b+d)+c(b+d)=(a+c)(b+d).
$$
Similarly, we can modify both given conditions:
$$
2(a+c)-5(b+d)=4 \quad \text { and } \quad 3(a+c)+4(b+d)=6
$$
If we choose the substitution $m=a+c$ and $n=b+d$, we get the solution of the system (1) as $m=2$ and $n=0$. For the given expres... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On a $7 \times 7$ board, we are playing a game of battleships. There is one ship of size $2 \times 3$ on it. We can ask about any square on the board, and if we hit the ship, the game ends. If not, we ask again. Determine the smallest number of questions we need to ask to be sure to hit the ship. | SOLUTION. According to Fig. 2, we can place 8 disjoint rectangles $2 \times 3$ on the board (the middle cell of the board will remain empty). To definitely hit the ship, we need to ask about at least one cell in each of the eight marked rectangles, so the necessary number of questions is at least 8.
In Fig. 3, an exam... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Let A, B be sets of positive integers such that the sum of any two different numbers from $A$ belongs to $B$ and the quotient of any two different numbers from $B$ (the larger divided by the smaller) lies in $A$. Determine the maximum possible number of elements in the set $A \cup B$.
| 2. First, we will prove that the set $\mathrm{A}$ can contain at most two numbers. Assume that $\mathrm{A}$ contains three numbers $a < b < c$. Then the set $\mathrm{B}$ contains the numbers $a+b < a+c < b+c$, and thus the set $\mathrm{A}$ must contain the number
$$
\frac{b+c}{a+c}=1+\frac{b-a}{a+c}
$$
which is not a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. We need to fill a $3 \times 3$ table with nine given numbers such that in each row and column, the largest number is the sum of the other two. Determine whether it is possible to complete this task with the numbers
a) 1, 2, 3, 4, 5, 6, 7, 8, 9;
b) 2, 3, 4, 5, 6, 7, 8, 9, 10.
If yes, find out how many ways the tas... | SOLUTION. The sum of the numbers in each row (and in each column) is even, equaling twice the largest of the three written numbers. Therefore, the sum of all the numbers in the table must also be even. Since in case a) the sum of the numbers is 45, the task cannot be completed.
Since the sum of the numbers in case b) ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find the largest possible number of elements in the set $\mathrm{M}$ of integers, which has the following property: for every triple of distinct numbers $z \mathrm{M}$, it is possible to select some two of them whose sum is a power of 2 with an integer exponent. | SOLUTION. Let us note at the outset that the mentioned power as an integer sum must have a non-negative exponent. We will only consider such powers of the number 2 further on.
We will prove that the set M can have at most 6 elements, as does, for example, the suitable (as we will verify immediately) set
$$
\{-1,3,5,-... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. At the party, 10 boys and 10 girls gathered. Each boy likes a different positive number of girls. Each girl likes a different positive number of boys. Determine the largest integer n with the following property: It is always possible to form n disjoint pairs where both individuals like each other.
(Josef Tkadlec) | SOLUTION. We will show that the largest $n$ sought is equal to 1.
In the first part of the solution, we will show that one suitable pair can always be formed. According to the problem statement, the numbers of girls liked by individual boys are all the numbers from 1 to 10 (in some order). The same applies to the numb... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Find the greatest real number $k$ such that the inequality
$$
\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqq \sqrt{a b}
$$
holds for all pairs of positive real numbers $a, b$. | SOLUTION. For $k=2$, the inequality can be simplified to $\frac{1}{2}(a+b) \geqq \sqrt{a b}$, which is the well-known inequality between the arithmetic and geometric means, valid for any positive numbers $a, b$. The largest $k$ we are looking for is therefore at least 2. Let's further examine the given inequality under... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
For real numbers $a, b, c$ it holds that
$$
\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}
$$
Determine all possible values of the expression
$$
\frac{a^{3}+b^{3}+c^{3}}{(b+c)^{3}+(c+a)^{3}+(a+b)^{3}}.
$$
(Michal Rolínek) | SOLUTION. From the problem statement, it follows that the numbers $a, b, c$ satisfy the conditions
$$
b+c \neq 0, \quad c+a \neq 0, \quad a+b \neq 0 .
$$
Under the assumptions $(\mathrm{P})$, we equivalently transform the first of the given equalities:
$$
\begin{gathered}
\frac{a}{b+c}=\frac{b}{c+a}, \\
a(c+a)=b(b+c... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Consider two quadratic equations
$$
x^{2}-a x-b=0, \quad x^{2}-b x-a=0
$$
with real parameters $a, b$. Determine the smallest and the largest values that the sum $a+b$ can take, given that there exists exactly one real number $x$ that satisfies both equations simultaneously. Also, determine all pairs $(a, b)$ of r... | 1. By subtracting the given equations, we obtain the equality $(b-a) x+a-b=0$ or $(b-a)(x-1)=0$, from which it follows that $b=a$ or $x=1$.
If $b=a$, both equations take the form $x^{2}-a x-a=0$. Exactly one solution exists if and only if the discriminant $a^{2}+4 a$ is zero. This holds for $a=0$ and for $a=-4$. Since... | -8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the plane containing the segment $B D$, find the set of all vertices $A$ of convex quadrilaterals $A B C D$ for which the following conditions are simultaneously satisfied:
a) the center $O_{C}$ of the incircle of triangle $B C D$ lies on the circumcircle of triangle $A B D$,
b) the center $O_{A}$ of the incirc... | Solution: Let $\alpha=|\angle B A D|$ and $\gamma=|\angle B C D|$. We have
$$
\left|\angle B O_{C} D\right|=180^{\circ}-\left(\left|\angle O_{C} B D\right|+\left|\angle O_{C} D B\right|\right)=
$$
$$
=180^{\circ}-\frac{1}{2}(|\angle C B D|+|\angle C D B|)=90^{\circ}+\frac{1}{2}
$$
Similarly, $\left|\angle B O_{A} D\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mr. Kutil wanted to paint star ornaments on 20 tiles in the bathroom. On the can of paint, it was written that the paint would cover $750 \mathrm{~cm}^{2}$. How many cans of paint did Mr. Kutil have to buy at least, if one square of the grid has an area of $1 \mathrm{~cm}^{2}$? The ornament on one tile is shown in the ... | The tile is a square with a side of $12 \mathrm{~cm}$, its area is $(12 \mathrm{~cm})^{2}=144 \mathrm{~cm}^{2}$.
Let's calculate the area of the parts that will not be colored:
4 pentagons in the corners have a total area of $4 \cdot 15 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$,
4 isosceles triangles in the middle of th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Anička has 50 Kč, Anežka has 46 Kč, and they want to buy snacks for a family celebration with all their money. They are deciding between cupcakes and windmills: a windmill is 4 Kč more expensive than a cupcake, and with all their money, they could buy a third more cupcakes than windmills.
How much does each snack cost... | Anička and Anežka together have 96 Kč. This amount can only be spent in a few ways, which are derived from expressing the number 96 as a product of two natural numbers:
$$
96=1 \cdot 96=2 \cdot 48=3 \cdot 32=4 \cdot 24=6 \cdot 16=8 \cdot 12 \text {. }
$$
The price of a kite and the price of a cake should correspond t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine for how many natural numbers greater than 900 and less than 1001 the digital sum of the digital sum of their digital sum is equal to 1.
(E. Semerádová)
Hint. What is the largest digital sum of numbers from 900 to 1001? | Among the numbers 900 and 1001, the number with the greatest digit sum is 999, which is 27; we don't need to consider larger sums.
Among the numbers 1 and 27, the number with the greatest digit sum is 19, which is 10; we don't need to consider larger sums.
Among the numbers 1 and 10, the digit sum of 1 is only found ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
King Pan was distributing ducats to his sons. He gave a certain number of ducats to his eldest son, one ducat less to the next younger son, and continued this pattern until he reached the youngest. Then he returned to the eldest son, giving him one ducat less than he had just given to the youngest, and continued distri... | For a specific number of sons, one can vividly test the king's method of distributing ducats. It's enough to proceed from the back: the youngest received one ducat in the second round, the second youngest received two ducats, etc. For example, for two, three, and four sons, the number of ducats in individual rounds wou... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Mom brought 10 snacks of three types: there were fewer coconut snacks than lollipops, and the most were caramel chews. Josef chose two different types of snacks, Jakub did the same, and for Jan, only snacks of the same type were left.
How many coconut snacks, lollipops, and caramel chews did Mom bring?
Hint. Which ty... | When Jan received $\mathrm{k}$ snacks, there were 6 of the same kind, and they were caramel chews - if they were coconuts or laskonky, there would have to be more than 6 chews and more than 10 snacks in total. Therefore, there were at least 6 caramel chews originally, and mom brought
- either 1 coconut, 3 laskonky, an... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Adam and Eve were playing chess.
Adam won and comforted Eve: "You know, I've been playing chess for a long time, twice as long as you!" Eve got upset: "But last time you said you've been playing three times longer!"
Adam was surprised: "Did I say that? And when was that?"
"Two years ago!"
"Well, then I was telling ... | Let's assume that Eva has been playing chess for $x$ years. Then the time data appearing in the problem can be briefly expressed in the following table:
| | last year | today |
| :---: | :---: | :---: |
| Eva | $x-2$ | $x$ |
| Adam | $2 x-2$ | $2 x$ |
Last year, Adam played chess three times longer than Eva, which w... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Grandma had a square garden. She bought a few neighboring plots, thus obtaining a square plot whose side was three meters longer than the side of the original garden. The area of the new plot was nine square meters larger than twice the original area.
How long was the side of the original garden?
(K. Buzáková) | Let $a$ be the length of the side of the original square garden. After the purchase, a new square plot was formed, with the side length of $a+3$. According to the areas of the plots given in the problem, we have
$$
(a+3)^{2}=2 a^{2}+9 \text {. }
$$
By equivalent transformations, we get
$$
\begin{aligned}
a^{2}+6 a+9... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the picture, there is a structure composed of twelve identical cubes. To how many different places can we move the dark cube so that the surface area of the assembled body does not change?
... | Originally, on the surface of the assembled body, there are three dark cube walls, and the other three walls touch light cubes. After removing the dark cube, the total surface area of the body does not change (three dark walls are replaced by three light ones).
In order for the surface area of the body to remain uncha... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Eva wrote down consecutive natural numbers: 1234567891011. . What digit did she write on the 2009th place?
(M. Volfová) | There are 9 single-digit numbers (1 to 9) and it takes 9 digits to write them. There are 90 two-digit numbers (10 to 99) and it takes 180 digits to write them. There are 900 three-digit numbers (100 to 999) and it takes a total of 2700 digits to write them. To write all single-digit and two-digit numbers, 189 digits ar... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When Bořek was painting the garage door, he accidentally painted over the scale of the outdoor wall thermometer. However, the mercury tube remained undamaged, so Bořek covered the original scale with a homemade strip. On it, he carefully marked divisions, all of the same size and labeled with numbers. His division, how... | At $11^{\circ} \mathrm{C}$, the thermometer shows 2 boroks. When the temperature drops to $-4{ }^{\circ} \mathrm{C}$, which is a decrease of $15^{\circ} \mathrm{C}$, the thermometer shows -8 boroks, which is 10 boroks less than in the first case. A change of 10 boroks corresponds to a change of $15^{\circ} \mathrm{C}$,... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Jirka drew a square grid with 25 squares, see the image. Then he wanted to color each square so that squares of the same color do not share any vertex.
How many colors did Jirka need at least?
(M. Dillingerová)
 | We will gradually examine the possibilities based on the number of children:
- If she had 1 child, then the dough should yield $1 \cdot 3 + 2 = 5$ and simultaneously $1 \cdot 4 - 1 = 3$ rolls. These values are different, so she did not have 1 child.
- If she had 2 children, then the dough should yield $2 \cdot 3 + 2 =... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Inside the rectangle $A B G H$ are two identical squares $C D E F$ and $I J K L$. The side $C F$ of the first square lies on the side $B G$ of the rectangle, and the side $I L$ of the second square lies on the side $H A$ of the rectangle. The perimeter of the octagon $A B C D E F G H$ is $48 \, \text{cm}$, and the peri... | Significant parts of the perimeters of the octagon $A B C D E F G H$ and the dodecagon $A B C D E F G H I J K L$ are common to both figures. On the other hand, the segments $I J, J K$, and $K L$ are part of only the perimeter of the dodecagon, and the segment $I L$ is part of only the perimeter of the octagon. These fo... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
I want to construct a triangle $ABC$ with sides $|AB|=3 \text{~cm}$ and $|BC|=4 \text{~cm}$. Furthermore, I want to construct all circles, each of which will have its center at one of the vertices of the triangle and will pass through another vertex.
How long must the side $AC$ be so that there are exactly five such c... | If sides $AB$ and $AC$ were of the same length, then the circle centered at point $A$ passing through point $B$ would also pass through point $C$. In such a case, Ema would construct only one circle centered at point $A$.
If sides $AB$ and $AC$ were of different lengths, then the circle centered at point $A$ passing t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mrs. Teacher needed to come up with examples of equations for a written test. Therefore, she wrote down all equations of the form
$$
a \cdot x + b = 13
$$
where $a$ and $b$ are one-digit natural numbers. From all of them, she selected the equations whose root $x$ was 3. She put one equation into each group. What is t... | We know that $x=3$ is a solution to the given equation, so the equality
$$
a \cdot 3 + b = 13 \text{.}
$$
holds. For $a$ and $b$ to be natural numbers, $a$ must be either $1, 2, 3$, or $4$ (for $a=5$, we get $5 \cdot 3 = 15 > 13$ and $b$ would have to be negative, which is not possible). Now we substitute the individ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
We received a cube, the length of whose edge was expressed in centimeters as an integer greater than 2. We painted all its faces yellow and then cut it into smaller cubes, each with an edge length of $1 \mathrm{~cm}$. We sorted these small cubes into four piles. The first pile contained cubes with one yellow face, the ... | Let the length of the edge of the original cube in centimeters be denoted by $a+2$, where $a$ is a natural number. Each face of the original cube corresponds to $a^{2}$ small cubes with exactly one colored face, so there are a total of $6 a^{2}$ such small cubes. On each edge of the original cube, we have $a$ small cub... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two pairs of parallel lines $A B \| C D$ and $A C \| B D$ are given. Point $E$ lies on line $B D$, point $F$ is the midpoint of segment $B D$, point $G$ is the midpoint of segment $C D$, and the area of triangle $A C E$ is $20 \mathrm{~cm}^{2}$.
Determine the area of triangle $D F G$.
(V. Semeráková)
Hint. Compare t... | The area of a triangle depends on the length of its side and the height to this side. Since lines $A C$ and $B D$ are parallel and point $E$ lies on line $B D$, the area of triangle $A C E$ is always the same for any chosen point $E$. In particular, the area of triangle $A C E$ is the same as the area of triangle $A C ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Pat wrote on the board the example:
$$
589+544+80=2013 .
$$
Mat wanted to correct the example so that both sides would actually be equal, and he searched for an unknown number which he then added to the first addend on the left side, subtracted from the second addend, and multiplied the third addend by. After perform... | By adding an unknown number to the first addend and subtracting the same number from the second addend on the left side, the sum of these two numbers will not change and is equal to $589+544=$ $=1133$. This intermediate sum is $2013-1133=880$ less than the number on the right side of the equation. Therefore, the produc... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Jiřka constructed two congruent equilateral triangles as shown in the figure. She further wants to construct all circles that will have their center at one of the vertices and will pass through another vertex of one of the triangles.
Construct and calculate all circles that meet Jiřka's requirements.
(K. Pazourek)
!... | Let's name the vertices as shown in the following image and note that points $A$ and $C$ are always equidistant from the remaining three points (the corresponding segments form the sides of equilateral triangles). Therefore, the circle centered at point $A$ passing through point $B$ also passes through points $C$ and $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Determine which digit is in the 1000th place after the decimal point in the decimal expansion of the number $\frac{9}{28}$.
(M. Krejčová)
Hint. What does the decimal expansion of the given number look like? | The decimal expansion of the rational number $\frac{9}{28}$ is
$$
0.32 \overline{142857}
$$
where the repeating part, consisting of six digits, is marked with a bar.
Six goes into a thousand 166 times with a remainder of four $(1000 = 166 \cdot 6 + 4)$. There are two digits between the decimal point and the repeatin... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A subtraction pyramid is a pyramid formed by non-negative integers, each of which is the difference between the two nearest numbers from the previous level (read from bottom to top). Here is an example of a subtraction pyramid:
## 1
2
2
5 7
4
3
5
8
A significant number is the largest number in the subtraction py... | If it is 0 at the top, there must be two identical numbers in the previous row. Each excellent pyramid must have at least three floors.
An excellent pyramid with the smallest number of floors and the smallest possible numbers looks like this (except for the order of the numbers on the third line):
## 0
1
0 ..... 2
... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In Mr. Goat's garden, several cherry trees were in bloom. On each cherry tree, three sparrows were sitting, and one more was sitting on the fence. Mr. Goat's dog scared them away and the sparrows flew off. After a while, they all returned and settled on the cherry trees. The cherry tree under which the dog was sleeping... | Let $x$ be the number of cherries in the garden and $y$ be the number of sparrows. If we know that all the sparrows perched on the cherries in groups of three and one remained on the fence, we can set up the following equation:
$$
y=3 x+1
$$
After the sparrows returned to the cherries, they occupied all the cherries ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In Zverimex, they were selling fish from one aquarium. Ondra wanted half of all the fish, but to avoid cutting any fish, he received half a fish more than he requested. Matěj wished for half of the remaining fish, but like Ondra, he received half a fish more than he requested. Finally, Petr wanted half of the remaining... | Let's consider it from the end:
Petr got half a fish more than half of all the fish that remained after Matej. Since the aquarium was then empty, that half a fish more was exactly half of what remained after Matej. Therefore, after Matej's purchase, one fish remained in the aquarium.
Matej got half a fish more than h... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The encryption games involved 168 players in 50 teams, which had two to five members. The most were four-member teams, there were 20 three-member teams, and the games were attended by at least one five-member team. How many two-member, four-member, and five-member teams were there? (M. Mach) | There were 20 three-member teams, which represents 60 players. The remaining 108 players from the total number need to be divided into 30 teams of two, four, and five players. Four-member teams were the most, i.e., at least 21, which represents at least 84 players. The remaining 24 players need to be divided into 9 tea... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The teacher wrote the following numbers on the board:
$$
1,4,7,10,13,16,19,22,25,28,31,34,37,40,43
$$
Any two adjacent numbers differ by the same value, in this case, 3. Then she erased all the numbers from the board except for 1, 19, and 43. She further added several integers between these three numbers such that an... | One way to fill in the numbers is, of course, the one the teacher erased (in this case, the difference between consecutive numbers is 3). Another possible, and probably the simplest, way is to fill in all natural numbers from 1 to 43 (in this case, the difference is 1).
Each filling according to the task is completely... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
We have a square grid with 77 nodal points. Two of them are marked as $A$ and $B$ as shown in the figure. Let $C$ be one of the remaining nodal points. Find all possible positions of point $C$ such that triangle $ABC$ has an area of 6 square units.
(E. Novotná)
The sum of all the given numbers is
$$
4 a ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Vojta bought 24 identical square tiles. The side of each tile measured $40 \mathrm{~cm}$. Vojta wanted to create a rectangular platform in front of the cottage with the smallest possible perimeter. How many meters did the perimeter of the paved rectangle measure, given that no tiles were left over? Vojta did not cut or... | Let's go through all the possibilities; for each, we will provide the dimensions of such a rectangle in "tiles" and further the perimeter in "tiles," resp. in "sides of tiles":
- dimensions: $1 \times 24$, perimeter: $2 \cdot(1+24)=50$,
- dimensions: $2 \times 12$, perimeter: $2 \cdot(2+12)=28$,
- dimensions: $3 \time... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Káta wants to give presents to her friends and is thinking: if I bought each of them a hair clip for 28 Kč, I would still have 29 Kč left, but if it were a teddy bear for 42 Kč, I would be short of 13 Kč. How many friends does Káta have and how much money does she have for gifts?
(M. Volfová) | The difference in the price of a gift for one friend is $42-28=14$ (CZK). The difference in the total price that Káta would pay for gifts for all her friends is $29+13=42$ (CZK). If we multiply the difference in the price of a gift for one friend by the number of friends, we get the difference in the total price. There... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Šárka declared:
"I am the youngest of three sisters. Líba is three years older, and Eliška is eight years older. Our mom loves to hear that the average age of all of us (including her) is 21. When I was born, mom was already 29."
How many years ago was Šárka born?
(M. Volfová) | If we denote Śárka's age in years by $x$, then Líbě is $x+3$, Eliška is $x+8$, and the mother is $x+29$ years old. The average age of all of them is 21 years, i.e.,
$$
(x+(x+3)+(x+8)+(x+29)): 4=21,
$$
after rearrangement
$$
\begin{aligned}
4 x+40 & =84, \\
x & =11 .
\end{aligned}
$$
Śárka was born 11 years ago. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the figure, squares $ABCD$, $EFCA$, $GHEC$, and $IJHE$ are depicted. Points $S$, $B$, $F$, and $G$ are the centers of these squares, respectively. Segment $AC$ is $1 \, \text{cm}$ long.
Determine the area of triangle $IJS$.
(E. Semerádová)
 | We solve the example from the end, using the $\mathrm{k}$ table:
state at the end
left pocket (L) right pocket $(\mathrm{P}) \quad$ difference
move 16 two-crowns from $\mathrm{P}$ to $\mathrm{L}$
32
13
13
move 4 five-crowns from $\mathrm{L}$ to $\mathrm{P}$
12
owe 19
51
1
Conclusion: I had 11 crowns more in... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Karolína wrote all three-digit numbers formed by the digits 1, 2, and 3, in which no digit was repeated and 2 was in the tens place. Nikola wrote all three-digit numbers formed by the digits 4, 5, and 6, in which no digit was repeated. Kuba chose one number from Karolína and one number from Nikola so that the sum of th... | Karolína wrote the numbers
$$
123, \quad 321 .
$$
Nikola wrote the numbers
$$
456, \quad 465, \quad 546, \quad 564, \quad 645, \quad 654 .
$$
Both of Karolína's numbers are odd. For an even sum, Kuba had to choose an odd number from Nikola. The even sums are given by the following cases:
$$
123+465, \quad 123+645,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In rectangle $A B C D$ with side $A D$ of length $5 \mathrm{~cm}$, point $P$ is located such that triangle $A P D$ is equilateral. Ray $A P$ intersects side $C D$ at point $E$, and segment $C E$ measures $5 \mathrm{~cm}$.
How long is segment $A E$ and what is the measure of angle $A E B$?
(L. Hozová) | Triangle $A P D$ is equilateral, so the lengths of all its sides are $5 \mathrm{~cm}$ and the measures of all its internal angles are $60^{\circ}$.
The internal angles of rectangle $A B C D$ are right angles, so the measures of angles $P D E$ and $P A B$ are $30^{\circ}$. Angle $P E D$ is the third angle of triangle $... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mice have built an underground house consisting of chambers and tunnels:
- each tunnel leads from one chamber to another (i.e., none are dead ends),
- from each chamber, exactly three tunnels lead to three different chambers,
- from each chamber, one can reach any other chamber via tunnels,
- the house has exactly one... | We will denote cells with circles and tunnels with lines. We start with a critical tunnel, whose collapse divides the house into two separate parts. If we denote the cells at the ends of this tunnel as $A$ and $B$, then each cell belongs to exactly one of the following two groups:
a) cell $A$ and all cells that can be... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
At the tournament, teams Akouska, Bovenska, Colska, and Demecka met. Each team played against each other exactly once. The winning team received three points, the losing team received no points, and in the event of a draw, each of the drawing teams received one point. After all six matches were played, Akouska had 7 po... | The team names are abbreviated to A, B, C, and D. Each team played three matches against the remaining teams, and could earn 0, 1, or 3 points per match. The distribution of points for teams A, B, and D is therefore uniquely determined, while for team C there are two possibilities:
$$
7=3+3+1, \quad 4=3+1+0, \quad 3=3... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The worm consists of a white head and several segments, see the image.
When the worm is born, it has a head and one white segment. Each day, the worm gains a new segment in one of the follow... | On the first day of its life, a worm has only one white segment (and a head):
On the second day, it can grow a new segment only in the first of the mentioned ways; all worms that are two days... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Iveta gradually listed natural numbers composed of the digits 1, 3, 5, 7. She used no other digits, proceeded in ascending order from the smallest number, and did not omit any number. She wrote the numbers immediately after each other, thus forming one extraordinarily long number:
$$
1357111315173133 \ldots
$$
Which ... | From the given digits, Iveta created 4 one-digit numbers. All these numbers occupy 4 places in Iveta's long number.
$Z$ from the given digits, Iveta created $4^{2}=16$ two-digit numbers. All these numbers occupy $2 \cdot 16=32$ places in Iveta's long number. The last digit of the last two-digit number is at the 36th p... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In five bags, there are 52 balls in total. In no two bags is the number of balls the same, and some bags may be empty. All the balls from any (non-empty) bag can be moved to the other four bags so that they will have the same number of balls.
a) Find some distribution of balls into the bags that has all the mentioned ... | After moving the balls, one bag is empty and the other four contain the same number. In total, there are 52 balls, so after moving, there are 13 balls in each non-empty bag $(52: 4=13)$. Therefore, originally, there could not have been more than 13 balls in any bag.
a) Possible original numbers of balls in the bags co... | 12 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
Half of the children in 5.A attend a dance club. All girls attend and one third of the 18 boys attend.
a) How many children are in 5.A?
b) How many girls are in 5.A? | All girls and one third of 18 boys, i.e., 6 boys, attend the dance club, which is half of the students in 5.A. 12 boys, who do not attend the dance club, make up the other half of the class.
a) There are thus 24 children in class $V$.
b) $24-18=6$. There are 6 girls.
[ 6 boys ... 1 point, 12 boys are half ... 1 point,... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Matěj and his friends went caroling. In addition to apples, walnuts, and gingerbread, each of the boys also received an orange. Jarda got one orange, Milan too. Radek, Patrik, Michal, and Dušan each got two oranges. Matěj got as many as four oranges, which was the most of all the boys. The other boys each got three ora... | Boys whose names we know received a total of $1+1+2+2+2+2+4=14$ oranges. For the boys whose names we do not know, there are $23-14=9$ oranges left. Since each of these boys received three oranges, there must have been $9: 3=3$ of them. We know the names of seven boys, and three more boys whose names we do not know, so ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Friends Jarda, Premek, and Robin were playing marbles. Jarda wasn't doing well, so after the game, he had the fewest marbles of all. The boys felt sorry for him, so Robin gave Jarda half of all his marbles, and Premek gave him a third of his. Then Jarda had the most marbles, so he returned seven marbles to each of his ... | We can solve the task advantageously from the end:
- Before the second (last) exchange round, Premek and Robin had seven marbles less, while Jarda had 14 more. So Premek and Robin had $25-7=18$ marbles, while Jarda had $25+14=39$.
- Before the first exchange round, Robin had double the number of marbles (he gave half ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Péta constructed several plane figures from identical triangles, see the image. The perimeters of the first three are $8 \mathrm{~cm}, 11.4 \mathrm{~cm}$, and $14.7 \mathrm{~cm}$, respectively.
Determine the perimeter of the fourth figure.
(E. Semerádová)
 | If a number is not divisible by 2, then it is also not divisible by 4, 6, and 8. If a number is not divisible by 3, then it is also not divisible by 6 and 9. If a number is not divisible by 4, then it is also not divisible by 8. If a number is not divisible by 6, then it is not divisible by 2 or 3. Therefore, none of t... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Dwarves painted cubic dice with green and white colors so that each face was entirely painted in one of these two colors. After a while, they noticed that some of the painted dice looked exactly the same after a suitable rotation and began to sort them into groups (in the same group are identically painted dice).
What... | A cube has six faces, with each face adjacent to four other faces (they share an edge) and parallel to one face (they share no points).
We can categorize the colorings based on the number of green (or white) faces. This gives us seven possibilities, for each of which we will analyze different types of coloring.
- No ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the parking lot, there were cars and bicycles. If one more car arrived, there would be as many of them as there are bicycles. If five more bicycles arrived, all the bicycles would have as many wheels as all the cars.
How many cars and how many bicycles were in the parking lot?
(M. Dillingerová)
Hint. Imagine a si... | A car has four wheels, a bicycle has two; one car has the same number of wheels as two bicycles.
There was one more bicycle than cars in the parking lot. Consider the situation where there would be five more bicycles in the parking lot than originally, that is, the situation where the number of wheels would match. In ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ivan and Mirka were sharing pears from a plate. Ivan always took two pears, and Mirka took half of what was left on the plate. They proceeded in this manner: Ivan, Mirka, Ivan, Mirka, and finally Ivan, who took the last two pears.
Determine who ended up with more pears and by how many.
(M. Dillingerová)
Hint. How ma... | Ivan took three times two pears, in the end, he had 6 pears. To determine how many pears Mirka ended up with, we will trace back the changes in the number of pears step by step. $\mathrm{K}$ to do this, it is enough to realize that before each of Ivan's takings, there were two more pears on the plate, and before each o... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Arnošt cut out a $4 \times 4$ square from graph paper. Kristián cut out two holes in it, see the two black squares in the picture. Arnošt tried to cut this shape along the marked lines into two identical parts.
Find at least four different ways Arnošt could have done this. (Two cuttings are considered different if the... | One of the simplest possible divisions is indicated in the following figure:
The newly created parts are — like the original shape — symmetric with respect to the center of the bounding squa... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In triangle $ABC$, we denote the midpoints of sides $CB$ and $CA$ by the letters $K$ and $L$. We know that the quadrilateral $ABKL$ has a perimeter of $10 \text{ cm}$ and triangle $KLC$ has a perimeter of $6 \text{ cm}$. Calculate the length of segment $KL$.
(J. Mazák) | Segment $K L$ is the midline of triangle $A B C$ parallel to $A B$, since $K$ and $L$ are the midpoints of sides $B C$ and $A C$. Therefore, $2|K L|=|A B|$ and also $|A L|=|L C|$ and $|C K|=|K B|$.
The perimeter of triangle $K L C$ is $|C K|+|K L|+|L C|=6$. The perimeter of quadrilateral $A B K L$ is
$$
|A B|+|B K|+|... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Adam, Borrek, and Cenda were comparing how many kilograms of chestnuts they had collected. They found that the arithmetic mean of what Adam and Borrek collected is 10 kg more than Cenda's contribution. And the arithmetic mean of what Adam and Cenda collected is 3 kg less than Borrek's contribution.
Determine the diffe... | The amount of chestnuts ( $\mathrm{vg}$ ) collected by Adam, Borek, and Cenda will be denoted by $a, b$, and $c$ respectively. According to the problem statement, the following equations hold:
$$
\frac{a+b}{2}=c+10, \quad \frac{a+c}{2}=b-3 .
$$
We want to determine the difference between $\frac{b+c}{2}$ and $a$. To d... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A group of children organized a tic-tac-toe tournament. Everyone played with everyone else, and a total of 136 games were played. Out of these, exactly 66 games were of the type girl-girl or boy-boy. How many boys and how many girls were in the group?
(Bednářová) | Let's consider a group of $n$ players. Each one plays a match with every other, and we are interested in the total number of matches. Each player will play $\mathrm{s} n-1$ opponents, and there are $n$ players, so in total we have $n \cdot(n-1)$. Now we must realize that we have counted each match twice. Therefore, we ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The average age of the Kebulov family, consisting of a father, a mother, and several children, is 18 years. When the father, who is 38 years old, is not included, the average age of the family is 14 years. How many children do the Kebulovs have?
(L. Hozová) | Let the number of members in this family be $n$. The sum of the ages of all members is equal to the product of the average age of the family and the number of members, that is, $18 \cdot n$. The family without the father has $n-1$ members, and the sum of the ages of these members is $14 \cdot(n-1)$. We know that this s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Honzík received his pocket money and wants to buy something nice with it. If he bought four rolls, he would have 5 Kč left. If he wanted to buy five rolls, he would be short of 6 Kč. If he bought two rolls and three pretzels, he would spend his entire pocket money without any remainder.
How much does one pretzel cost?... | Honzík's pocket money can be expressed in three ways, namely as
- the sum of the price of 4 cakes plus 5 Kč,
- the sum of the price of 5 cakes minus 6 Kč,
- the sum of the prices of 2 cakes and 3 pretzels.
From the first two expressions, it follows that one cake costs $5+6=11$ Kč. From this, we also find that Honzík'... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the homework for calculating the value of the expression
$$
2-3+4-5+6-7+8-9+10=
$$
Radek forgot to write two parentheses, so he got a result that was 18 greater than what he would have obtained if he had written the parentheses. Add the parentheses in two ways and write down what number Radek got and what number h... | Radek calculated
$$
2-3+4-5+6-7+8-9+10=6 \text {. }
$$
He got a result 18 more than it should have been. The result should have been $6-18=-12$. To get a smaller number, it is necessary to subtract a larger value, i.e., place the parentheses after the minus sign. Two required solutions:
$$
\begin{aligned}
& 2-3+(4-5... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mom cooked homemade currant syrup and poured it into bottles. The bottles were of two types: small with a volume of $500 \mathrm{ml}$ and large with a volume of $750 \mathrm{ml}$. In the end, she had 12 empty small bottles left, and all other bottles were completely filled. Then Mom realized that she could have poured ... | Mom has 12 unfilled bottles, each with a volume of $500 \mathrm{ml}$. They could hold $6000 \mathrm{ml}$ of syrup $(12 \cdot 500=6000)$.
The same amount would fit into 8 large bottles $(6000: 750=8)$. If Mom were to pour the syrup the second way, she would be left with 8 empty large bottles.
Note. One large bottle ha... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Míša has five candies. Vojta has fewer candies than Míša. Vendelín has as many candies as Míša and Vojta together. All three of them together have seven times more candies than Vojta.
How many candies does Vendelín have?
(L. Hozová)
Hint. How few and how many candies can they have together at minimum and maximum? | Vojta has fewer candies than Míša, so he can have
$$
0,1,2,3 \text {, or } 4
$$
candies. Vendelín has five more candies than Vojta, so he can have
$$
5,6,7,8, \text { or } 9
$$
candies. All three together have twice as many as Vendelín, so they can have
$$
10,12,14,16 \text {, or } 18
$$
candies.
Among these num... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Two guards are overseeing the order in a room whose shape and dimensions are shown in the diagram. Each two adjacent walls are perpendicular to each other, and the dimensions are given in meters. The guards stand close to the wall at the locations marked with squares. Together, they can oversee the entire room, but onl... | The entire room can be divided into squares with a side of 10 meters as shown in the image. The parts of the room visible to each guard are marked with two types of hatching. The part that both guards can see is thus doubly hatched:
 | For 5 sticks, there are 5 hats. For 1 stick, there is therefore 1 hat, and thus 5 sticks and 1 hat have the same value as 6 sticks.
Furthermore, we know that for 4 sticks, there are 6 coats. For 2 sticks, there are therefore 3 coats, and thus 6 sticks have the same value as 9 coats.
In total, we can see that 5 sticks... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Grandpa keeps geese, pigs, goats, and chickens - a total of 40 animals. For every goat, there are 3 geese. If there were 8 fewer chickens, there would be as many as geese and pigs combined. If Grandpa traded a quarter of the geese for chickens at a ratio of 3 chickens for 1 goose, he would have a total of 46 animals.
... | Let's denote the quantities of each type of animal by their initial letters. The information from the problem can be written step by step as follows:
\[
\begin{aligned}
h + p + k + s & = 40, \\
h & = 3k, \\
s - 8 & = h + p, \\
40 - \frac{1}{4}h + \frac{3}{4}h & = 46.
\end{aligned}
\]
From the last equation, we get \(... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Adam wrote the numbers from 1 to 9 into a $3 \times 3$ table as shown in the image:
| 7 | 6 | 4 |
| :--- | :--- | :--- |
| 1 | 2 | 8 |
| 9 | 3 | 5 |
For this filling, the sum of the numbers in the three cells along each side is always the same. Adam found that the numbers can be filled into the table in other ways wi... | Given that each corner cell appears in two sums, we try to place the smallest possible numbers in these cells and somehow fill in the rest. After some trial and error, the following filling can be revealed, where the sum of the numbers along each side is 12:
| 1 | 9 | 2 |
| :--- | :--- | :--- |
| 8 | 7 | 6 |
| 3 | 5 |... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.