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8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=57
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 24
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=48$, $B C^{2}=9, A C^{2}=57$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,712 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4000$, if $f(0)=1$ and for any $x$ the equality $f(x+4)=f(x)+3 x+4$ holds. | Answer: 5998001
Solution: In the equation $f(x+4)-f(x)=3 x+4$, we will substitute for $x$ the numbers $0,4,8, \ldots, 3996$. We get:
$$
\begin{aligned}
& f(4)-f(0)=3 \cdot 0+4 \\
& f(8)-f(4)=3 \cdot 4+4
\end{aligned}
$$
$$
f(4000)-f(3996)=3 \cdot 3996+4
$$
Adding the equations, we get: $f(4000)-f(0)=3 \cdot(0+4+8+\... | 5998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,714 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 36 positive and 36 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=36+36$, which means $x=9$.
2) Let there be $y$ people with "positive te... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,715 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 12 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-13.5,13.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 756
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-28 \leqslant y-1 \leqslant 26$ and $-25 \leqslant 2-x \leqslant 29$. Therefore, $(y-1)(2-x)+2 \leqslant 26 \cdot 29+2=756$. ... | 756 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,717 |
6. (8 points) On the board, 25 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 25 minutes? | Answer: 300.
Solution: Let's represent 25 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,718 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=4 \\
z^{2}+x z+x^{2}=79
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 20
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=4, A C^{2}=79$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,720 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
5|x-4 a|+|x-a^{2}|+4 x-3 a=0
$$
has no solution. | Answer: $(-\infty ;-9) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 2. | (-\infty,-9)\cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,725 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
7|x-4 a|+|x-a^{2}|+6 x-2 a=0
$$
has no solution. | Answer: $(-\infty ;-18) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 3. | (-\infty;-18)\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,729 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
6|x-4 a|+\left|x-a^{2}\right|+5 x-4 a=0
$$
has no solution. | Answer: $(-\infty ;-12) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 4. | (-\infty,-12)\cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,733 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
8|x-4 a|+\left|x-a^{2}\right|+7 x-3 a=0
$$
has no solution. | Answer: $(-\infty ;-21) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 5. | (-\infty;-21)\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,737 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
9|x-4 a|+|x-a^{2}|+8 x-2 a=0
$$
has no solution. | Answer: $(-\infty ;-26) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 6. | (-\infty,-26)\cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,741 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
8|x-4 a|+|x-a^{2}|+7 x-2 a=0
$$
has no solution. | Answer: $(-\infty ;-22) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 7. | (-\infty;-22)\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,744 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
7|x-4 a|+|x-a^{2}|+6 x-3 a=0
$$
has no solution. | Answer: $(-\infty ;-17) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 8 | (-\infty;-17)\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,747 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
6|x-4 a|+\left|x-a^{2}\right|+5 x-3 a=0
$$
has no solution. | Answer: $(-\infty ;-13) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 9. | (-\infty,-13)\cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,750 |
10. (20 points) Find all values of the parameter $a$ for which the equation
$$
5|x-4 a|+\left|x-a^{2}\right|+4 x-4 a=0
$$
has no solution. | Answer: $(-\infty ;-8) \cup(0 ;+\infty)$
## Innopolis Open
Mathematics Olympiad
Innopolis University
## Qualifying (correspondence) stage in mathematics, December 2, 2018
## 11th grade, variant 10. | (-\infty,-8)\cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,753 |
1. (5 points) How many roots does the equation
$$
\arcsin (\sin x)=\frac{\pi x}{m} ?
$$ | Solution: Due to the oddness of the functions on both sides of the equation, the number of positive roots is equal to the number of negative roots, and the zero solution is present. The function $y=\arcsin (\sin x)$ takes the maximum value $\frac{\pi}{2}$ at points $\frac{\pi}{2}+2 \pi n$, the line $y=\frac{\pi x}{m}$ ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,758 |
3. (7 points) A sequence of real numbers $a_{k}=\frac{1}{k^{2}+k}$ is given for any natural $k$. It turns out that
$$
a_{m}+a_{m+1}+\ldots+a_{n-1}=\frac{1}{p}, \quad \text { where } p \in \mathbb{P}
$$
for some natural numbers $m$ and $n(m<n)$. Find $m+n$. | Solution: Notice that
$$
a_{k}=\frac{1}{k^{2}+k}=\frac{1}{k}-\frac{1}{k+1}
$$
Therefore,
$$
\begin{aligned}
\frac{1}{p} & =a_{m}+a_{m+1}+\ldots+a_{n-1}=\left(\frac{1}{m}-\frac{1}{m+1}\right)+\ldots+\left(\frac{1}{n-1}-\frac{1}{n}\right)= \\
& =\frac{1}{m}-\left(\frac{1}{m+1}-\frac{1}{m+1}\right)-\ldots-\left(\frac{1... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,760 | |
4. (7 points) Solve the equation $x^{4}-(a+b) x^{3}+(a b+2 c) x^{2}-c(a+b) x+c^{2}=0$. In your answer, write the sum of the squares of the real roots of the equation. | Solution: $x=0$ is not a solution to the equation, so we can divide by $x^{2}$. Grouping terms, we get
$$
\begin{gathered}
\left(x^{2}+2 c+\frac{c^{2}}{x^{2}}\right)-(a+b) \cdot\left(x+\frac{c}{x}\right)+a b=0 \\
\left(x+\frac{c}{x}\right)^{2}-(a+b) \cdot\left(x+\frac{c}{x}\right)+a b=0 \\
\left(x+\frac{c}{x}-a\right)... | ^{2}-2c | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,761 |
5. (8 points) Chord $CD$ is parallel to the diameter $AB$ of a circle with center at point $O$ and radius $r$. The tangent to this circle at point $A$ intersects the lines $BC$ and $BD$ at points $E$ and $F$ respectively. Find $AE \cdot AF$. | # Solution:
Since in the inscribed quadrilateral $ABDC$, the opposite sides $AB$ and $CD$ are parallel, $ABDC$ is an isosceles trapezoid ($BD = AC$). Therefore, $\angle DBA = \angle BAC$. Si... | 4r^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,762 |
6. (8 points) In a basket, there are $2 n$ identical oranges, $2 n$ identical apples, and $2 n$ identical pears. In how many ways can we divide all the fruits between two children so that each gets $3 n$ fruits? | Solution: Let's fix a child and denote by $x$ the number of oranges given to him, $y$ - apples, and $z$ - pears. We get two essentially different cases:
1) $0 \leqslant x \leqslant n$, then $n-x \leqslant y \leqslant 2 n$ and $z$ is uniquely determined. The number of such ways
$$
(n+1)+(n+2)+\ldots+(2 n+1)=\frac{(2 n... | 3n^{2}+3n+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,763 |
7. (10 points) The correspondence stage of the Innopolis University Olympiad was held among 9th, 10th, and 11th grade students, and among them, $a \%$, $b \%$, and $c \%$ respectively advanced to the in-person stage. Moreover, among the 9th and 10th grade students together, $k \%$ advanced to the in-person stage, and a... | Solution: Let $x$ be the number of 9th grade students, $y$ be the number of 10th grade students, and $z$ be the number of 11th grade students. Then
$$
k=\frac{a x+b y}{x+y} \Longrightarrow \frac{y}{x}=\frac{k-a}{b-k} \quad \text { and } \quad l=\frac{b y+c z}{y+z} \Longrightarrow \frac{z}{y}=\frac{l-b}{c-l}
$$
Theref... | \frac{(b-k)(-)+(k-)(-b)}{(b-k)(-)+(k-)(-b)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,764 |
8. (10 points) In a regular hexagonal pyramid $S A B C D E F$, a point $M$ is chosen on the diagonal of the base $A D$, dividing it in the ratio $A M: M D=n: m$ $(n<m)$. A section of the pyramid is made through point $M$ by a plane parallel to the face $S A B$. Find the ratio of the area of the section to the area of t... | # Solution:
We project point $S$ and our section onto the plane $A B C D E F$. Let's describe how to find the projection points of the section. Since the section is parallel to the plane $S ... | 1+\frac{n(4-3n)}{(+n)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,765 |
9. (20 points) For which $n>1$ do there exist such distinct natural numbers $a_{1}, a_{2}, \ldots, a_{n}$ that
$$
\operatorname{LCM}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}+a_{2}+\ldots+a_{n}
$$ | Solution: Let there exist such different natural numbers $a$ and $b$ that $\operatorname{HOK}(a, b)=a+b$. Since
$$
\operatorname{HOK}(a, b) \vdots a \quad \text { and } \quad \operatorname{HOK}(a, b) \vdots b
$$
then
$$
a+b \vdots a \quad \text { and } \quad a+b \vdots b
$$
Therefore, $a \vdots b$ and $b \vdots a$,... | Forallnaturaln>2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,766 |
10. (20 points) $A A_{1}$ and $B B_{1}$ are the bisectors of an isosceles triangle $A B C$. The perpendicular bisector of segment $B B_{1}$ intersects line $A A_{1}$ at point $P$. On segment $B B_{1}$, a point $R$ is taken such that $P B_{1} \| R A_{1}$. Prove that $A R = R A_{1}$.
# | # Solution:
Consider the circumcircle of triangle $B A B_{1}$. Let $P^{\prime}$ be the midpoint of the arc $B B_{1}$, not containing point $A$. On one hand, $P^{\prime}$ lies on the angle b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,767 |
9. (20 points) A circle with its center at the midpoint of the base $B C$ of an isosceles triangle $A B C$ touches the lateral sides of the triangle. A tangent to this circle intersects the sides $A B$ and $A C$ at points $P$ and $Q$ respectively. Prove that
$$
B P \cdot C Q=\frac{B C^{2}}{4}
$$ | Solution. Let $M$ be the midpoint of side $BC$. $PM$ and $QM$ are the bisectors of angles $BPQ$ and $CQP$ respectively. Therefore, $\angle BPM = \angle QPM = \alpha$ and $\angle CQM = \angle PQM = \beta$. Also, by the condition, $\angle ABC = \angle ACB = \gamma$. Then the sum of the angles of quadrilateral $BPQC$ is $... | BP\cdotCQ=\frac{BC^2}{4} | Geometry | proof | Yes | Yes | olympiads | false | 19,768 |
10. (20 points) Given various natural numbers $a, b, c, d$, for which the following conditions are satisfied: $a>d, a b=c d$ and $a+b+c+d=a c$. Find the sum of all four numbers. | Answer: 12
Solution. From the relations $a>d$ and $ab=cd$, we have the inequality $ba+d$ and $\frac{ac}{2} \geqslant 2c > b+c$. Adding them, we get a contradiction with the condition.
Therefore, without loss of generality, assume that $a \geqslant 3$. Since $a$ is a natural number, then $a \in\{1,2,3\}$. Moreover, $d... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,769 |
2. (5 points) When dividing the numbers 312837 and 310650 by a certain three-digit natural number, the remainders were the same. Find this remainder. | Answer: 96
Solution. $312837-310650=2187=3^{7}$. The three-digit divisors of this number are also powers of three, i.e., $3^{6}=729,3^{5}=243$. By dividing these numbers by 729 and 243 with a remainder, we find that the remainder is always 96. | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,770 |
4. (10 points) In a lottery, 6 numbers out of 36 are drawn each week. What is the probability that at least one of the numbers drawn this week will be the same as one of the numbers drawn last week? Round the result to the nearest thousandth. | Answer: 0.695
Solution. Total outcomes: $C_{36}^{6}$; outcomes in which no number matches: $C_{30}^{6}$. Then the probability is $1-\frac{C_{30}^{6}}{C_{36}^{6}}$. | 0.695 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,771 |
5. (10 points) How many ways are there to tile a $2 \times 12$ rectangle with non-overlapping $1 \times 2$ dominoes?
## Answer: 233 | Solution. Let $d_{n}$ be the number of ways to tile a $2 \times n$ rectangle with non-overlapping $1 \times 2$ dominoes. If the first domino is placed vertically, the number of ways is $d_{n-1}$; if the first domino is placed horizontally, the number of ways is $d_{n-2}$. Then $d_{n}=d_{n-1}+d_{n-2}, d_{1}=1, d_{2}=2$. | 233 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,772 |
6. (10 points) The area of a sector of a circle is 100. For what value of the radius of the circle will the perimeter of this sector be minimal? If the answer is not an integer, round to the tenths.
Answer: 10 | Solution. The area of the sector corresponding to $\alpha$ radians $S_{\alpha}=\frac{1}{2} R^{2} \alpha=100$. The length of the perimeter is $L(R)=2 R+R \alpha=2 R+\frac{200}{R}$, which reaches a minimum at the point $R=10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,773 |
7. (12 points) Solve the system of equations. Write the greatest value of $y$ in your answer.
$$
\left\{\begin{array}{l}
3 x^{2}-x y=1 \\
9 x y+y^{2}=22
\end{array}\right.
$$ | Answer: 5.5
Solution. Multiply the first equation by 3 and add to the second
$$
\left\{\begin{array} { l }
{ 3 x ^ { 2 } - x y = 1 , | \cdot 3 } \\
{ 9 x y + y ^ { 2 } = 2 2 }
\end{array} \Longleftrightarrow \left\{\begin{array} { l }
{ 3 x ^ { 2 } - x y = 1 , } \\
{ 9 x ^ { 2 } + 6 x y + y ^ { 2 } = 2 5 }
\end{arr... | 5.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,774 |
8. (13 points) On the coordinate plane, all points whose coordinates satisfy the conditions
$$
\left\{\begin{array}{l}
|2 x+3 y|+|3 x-2 y| \leq 13 \\
2 x^{2}-3 x y-2 y^{2} \leq 0
\end{array}\right.
$$
are shaded. Find the area of the resulting figure.
# | # Answer: 13
Solution. The solution to the first inequality defines the square $ABCD$, the solution to the second inequality is the shaded part of the plane between the perpendicular lines $2x + y = 0$ and $x - 2y = 0$.
 A tetrahedron has all faces as triangles with sides 8, 9, and 10. Can such a tetrahedron be packed into a box with internal dimensions $5 \times 8 \times 8 ?$ | Solution. It can. From the condition, it follows that the opposite edges of the tetrahedron are equal. Any tetrahedron can be inscribed in a parallelepiped as shown in the figure.
For our pa... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,776 |
1. Alice and Bob play a game. The gaming field is a checked line of size $1 \times 2022$. Players take turns (Alice starts) writing any of the letters O and G in an empty cell. The player who, after their move, has the letters OGO in three adjacent cells wins. If all cells are filled and there is no OGO, the game ends ... | 1. Alice and Bob play a game. The gaming field is a checked line of size $1 \times 2022$. Players take turns (Alice starts) writing any of the letters O and G in an empty cell. The player who, after their move, has the letters OGO in three adjacent cells wins. If all cells are filled and there is no OGO, the game ends ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,777 |
4. Alice and Bob play a game. The gaming field is a checked line of size $1 \times 777$. Players take turns (Alice starts) writing any of the letters O and G in an empty cell. The player who, after their turn, has the letters OGO in three adjacent cells wins. If all cells are filled and the word OGO does not appear, th... | 4. Alice and Bob play a game. The gaming field is a checked line of size $1 \times 777$. Players take turns (Alice starts) writing any of the letters O and G. The player who, after their move, has the letters OGO in three adjacent cells wins. If all cells are filled without the word OGO appearing, the game ends in a dr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,780 |
9. (20 points) In triangle $ABC$, the altitude $AD$ is drawn. A circle is tangent to $BC$ at point $D$, intersects side $AB$ at points $M$ and $N$, and side $AC$ at points $P$ and $Q$. Prove that
$$
\frac{A M+A N}{A C}=\frac{A P+A Q}{A B}
$$ | Solution. Let $O$ be the center of the circle in the problem. Then $O$ lies on $AD$. Drop perpendiculars $OE$ and $OF$ from $O$ to sides $AB$ and $AC$ respectively. Then points $E$ and $F$ are the midpoints of segments $MN$ and $PQ$ respectively, so $AE=\frac{1}{2}(AM+AN)$ and $AF=\frac{1}{2}(AP+AQ)$. $\angle OEB=\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,782 |
10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2(x+7)^{2}+(y-4)^{2}=c \\
(x+4)^{2}+2(y-7)^{2}=c
\end{array}\right.
$$ | Answer: $c=6.0$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$
\begin{aligned}
& \left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right) \geqslant(|x+\alpha|+|y-\beta|)^{2} \\
& \left(1+\frac{1}{2}\right)\left((x+\beta)^{2}+2(y-\alpha)^{2}\right) \geqslant(|x+\beta|+|y-\alpha|)^{2}
\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,783 |
2. (6 points) It is known that no digit of a three-digit number is zero and the sum of all possible two-digit numbers formed from the digits of this number is equal to this number. Find the largest such three-digit number. | # Answer: 396
Solution. If there are identical digits, then the sum of three two-digit numbers is less than 300. If the digits are different, then $\overline{a b c}=\overline{a b}+\overline{a c}+\overline{b a}+\overline{b c}+\overline{c a}+\overline{c b}=22(a+b+c)$. The sum of the digits gives the same remainder when... | 396 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,784 |
3. (8 points) In a circle, chords $A B$ and $A C$ are drawn, with $A B=2, A C=1$, and $\angle C A B=120^{\circ}$. Find the length of the chord of the circle that bisects angle $C A B$.
Answer: 3 | Solution. $A E=A C=1$, triangles $A C E$ and $B C D$ are equilateral. When rotated $60^{\circ}$ around point $C$, triangle $A B C$ transforms into triangle $E D C$. Therefore, $A B=E D=2$.
 | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,785 |
4. (10 points) A square is divided into 2016 triangles, with the vertices of no triangle lying on the sides or inside another triangle. The sides of the square are sides of some triangles in the partition. How many points, which are vertices of the triangles, are located inside the square? | Answer: 1007
Solution. Let us have $k$ points inside the square. Then the sum of the angles of all triangles is $360 k+4 \cdot 90=180 \cdot 2016$ degrees. | 1007 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,786 |
5. (10 points) Anton, Boris, Vadim, Gena, Dima, and Egor gathered at the cinema. They bought 6 seats in a row. Anton and Boris want to sit next to each other, while Vadim and Gena do not want to. In how many ways can the boys sit in their seats considering these preferences?
# | # Answer: 144
Solution. The total number of seating arrangements where Anton and Boris sit next to each other is $2 \cdot 5!=240$. The number of seating arrangements where the pairs Anton-Boris and Vadim-Gena end up next to each other is $2 \cdot 2 \cdot 4!=96$. Subtracting the second set from the first gives the answ... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,787 |
7. (10 points) Solve the system of equations. In your answer, write the greatest value of $y$.
$$
\left\{\begin{array}{l}
x^{2}+3 x y-y^{2}=27 \\
3 x^{2}-x y+y^{2}=27
\end{array}\right.
$$
Answer: 3 | Solution. Subtracting the first equation from the second, we get $2 x^{2}-4 x y+2 y^{2}=0$, from which $x=y$. Substituting into the first equation, we get the answer. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,788 |
8. (12 points) On the coordinate plane, all points whose coordinates satisfy the condition
$$
|| x|-2|+|y-3| \leq 3
$$
are shaded. Find the area of the resulting figure. | Answer: 34
Solution. The polygon that results from solving the equation ||$x|-2|+$ $|y-3|=3$ consists of segments of straight lines, glued together at special points: $x=0, x=2, x=$ $-2, y=3$.
 Find all values of $a$ for which the equation
$$
\sin \left(\sqrt{a^{2}-x^{2}-2 x-1}\right)=0.5
$$
has exactly seven distinct solutions. | Answer: $: \frac{17 \pi}{6}$
Solution. The number of solutions to the original equation coincides with the number of solutions to the system
$$
\left\{\begin{array}{l}
y=\sqrt{a^{2}-x^{2}-2 x-1} \\
\sin y=0.5
\end{array}\right.
$$
Which is equivalent to
$$
\left\{\begin{array}{l}
y \geq 0 \\
y^{2}+(x+1)^{2}=a^{2} \... | \frac{17\pi}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,790 |
10. (15 points) Let $S(n)$ denote the sum of the digits of a natural number $n$. Prove that there exists an infinite set of natural numbers $n$, not ending in 0, such that $S\left(n^{2}\right)=S(n)$. | Solution. All numbers of the form $n=10^{k}-1.10^{k}-1=\underbrace{9 \ldots 9}_{k}, S\left(10^{k}-1\right)=9 k$. $n^{2}=\left(10^{k}-1\right)^{2}=10^{2 k}-2 \cdot 10^{k}+1=\underbrace{9 \ldots 9}_{k-1} 8 \underbrace{0 \ldots 0}_{k-1} 1 . S\left(n^{2}\right)=9(k-1)+8+1=9 k$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,791 |
1. In a round-robin table tennis tournament, $2 n$ boys and $n$ girls participated. By the end of the tournament, it turned out that the number of matches won by girls was $\frac{5}{4}$ times the number of matches won by boys (there were no draws). Prove that the number of participants in the tournament is divisible by... | Solution: A total of $3 n$ people participated, so it is enough to prove that $n: 3$. The boys played among themselves $\frac{2 n(2 n-1)}{2}$
matches, the girls played among themselves $\frac{n(n-1)}{2}$, and the number of boy-girl matches was exactly $2 n^{2}$. Let $m$ be the number of boy-girl matches won by the boy... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,792 |
2. Can the letters in the word INNOPOLIS be replaced with digits (the same letters with the same digits, different letters with different digits) so that the resulting number is divisible by 1010?
(M. Popov)
# | # Answer: No.
Solution: It is clear that the letter $\mathrm{C}=0$, since the number 1010 is divisible by 10. Replace the last letter with zero and divide by 10. The remaining number INNOPOLI is divisible by 101, so the alternating sum of two-digit segments is divisible by 101. Therefore,
$$
\text { LI - PO + NO - IN... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,793 |
3. Point $M$ is the midpoint of side $A B$ of a convex quadrilateral $A B C D$. It turns out that $\angle B C A = \angle A C D$ and $\angle C D A = \angle B A C$. Prove that $\angle B C M = \angle A B D$.
# | # Solution:
Let $N$ be the midpoint of side $AD$. Since $\angle BCA = \angle ACD$ and $\angle BAC = \angle ADC$, triangles $BAC$ and $ADC$ are similar by two angles. Therefore, the correspo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,794 |
4. Triples of natural numbers $\left(a_{i}, b_{i}, c_{i}\right)$, where $i=1,2, \ldots, n$ satisfy the following conditions:
1) $a_{i}+b_{i}+c_{i}=2017$ for all $i=1,2, \ldots, n$;
2) if $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. What is the maximum possible value of $n$? (M. Popov) | Answer: 1343.
Solution: Note that
$$
\sum_{i=1}^{n} a_{i} \geqslant \sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$
A similar inequality is written for the sums $b_{i}$ and $c_{i}$. Adding the three obtained inequalities, we get
$$
\begin{gathered}
3 \cdot \frac{n(n+1)}{2} \leqslant \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}... | 1343 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,795 |
5. Find all values of the angle $\alpha$ from the interval $\left[0^{\circ} ; 360^{\circ}\right]$ such that the system
$$
\left\{\begin{array}{l}
y^{3} \geqslant 3 x^{2} y \\
(x-\cos \alpha)^{2}+(y-\sin \alpha)^{2}=\frac{2-\sqrt{3}}{4}
\end{array}\right.
$$
has a unique solution.
(R. Alishev) | # Answer: $\alpha \in\left\{15^{\circ}, 45^{\circ}, 135^{\circ}, 165^{\circ}, 255^{\circ}, 285^{\circ}\right\}$ Solution:
Note that $\sin ^{2} 15^{\circ}=\frac{1-\cos 30^{\circ}}{2}=\frac{... | \alpha\in{15,45,135,165,255,285} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,796 |
1. Vasya wrote natural numbers on the pages of an 18-sheet notebook. On each page, he wrote no fewer than 10 different numbers, and on any three consecutive pages, no more than 20 different numbers. What is the maximum number of different numbers Vasya could have written on the pages of the notebook?
(A. Khryabrov) | # Answer: 190.
Solution. Consider two consecutive pages, neither of which is the first page. Let their numbers be $n$ and $n+1$. Then on page $n-1$, there are at least 10 different numbers, and on pages $n-1, n$, and $n+1$, there are no more than 20 different numbers. Therefore, pages $n$ and $n+1$ add no more than 10... | 190 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,797 |
2. For which natural numbers $n$ will there exist a natural number $k$ such that the number $2 k^{2}+k+2018$ is divisible by $n$ ! (as usual, $n$ ! denotes the product of all natural numbers not exceeding $n$, for example, $4!=1 \cdot 2 \cdot 3 \cdot 4) ?$ | Answer: $n=1,2,3,4$.
Solution. Consider the remainders of the number $2 k^{2}+k$ when divided by 5: $2 \cdot 0^{2}+0=0$ gives a remainder of $0, 2 \cdot 1^{2}+1=3$ gives a remainder of 3, $2 \cdot 2^{2}+2=10$ gives a remainder of $0, 2 \cdot 3^{2}+3=21$ gives a remainder of 1, and $2 \cdot 4^{2}+4=36$ gives a remainde... | 1,2,3,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,798 |
3. Solve the system of equations:
$$
\left\{\begin{array}{l}
2 x^{2}+3 y+5=2 \sqrt{2 z+5} \\
2 y^{2}+3 z+5=2 \sqrt{2 x+5} \\
2 z^{2}+3 x+5=2 \sqrt{2 y+5}
\end{array}\right.
$$ | Answer: $x=y=z=-0.5$.
Solution. Consider the function $f(t)=2 t^{2}+3 t+5-2 \sqrt{2 t+5}$. It can be rewritten as $f(t)=\frac{1}{2}(2 t+1)^{2}+\frac{1}{2}(\sqrt{2 t+5}-2)^{2}$. Clearly, $f(t) \geqslant 0$ and equality is achieved only at $t=-0.5$. If we add the left and right sides of our system, we get the equation $... | -0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,799 |
4. Let $h$ be the length of the greatest height in a triangle, $R$ be the radius of the circumscribed circle, and $m_{a}, m_{b}$, and $m_{c}$ be the lengths of the medians of the triangle. Prove the inequality $m_{a}+m_{b}+m_{c} \leqslant 3 R+h$.
(A. Khryabrov) | Solution. Consider triangle $ABC$, in which medians $AA_1, BB_1$, and $CC_1$ are drawn. Let the shortest side of the triangle be $AB$. Then $h$ is the height dropped to $AB$. Denote by $O$ the center of the circumscribed circle of the triangle. Then the segment $OA_1$ is perpendicular to side $BC$, the segment $OB_1$ i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,800 |
5. Several people played a round-robin table tennis tournament. At the end of the tournament, it turned out that for any four participants, there would be two who scored the same number of points in the games between these four participants. What is the maximum number of tennis players that could have participated in t... | Answer: 7 tennis players.
Solution. Let $n \geqslant 8$ be the number of tennis players in the tournament. Then the total number of matches played is $\frac{1}{2} n(n-1)$, and thus the total number of victories is also $\frac{1}{2} n(n-1)$. Therefore, there must be a participant who has won at least $\frac{n-1}{2}$ ma... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,801 |
1. (5 points) How many roots does the equation
$$
\arccos (\cos x)=\frac{\pi x}{n} ?
$$ | Solution: The function $y=\arccos (\cos x)$ takes values in the interval $[0 ; \pi]$, and the maximum is achieved at points $\pi+2 \pi k$. The function $y=\frac{\pi x}{n}$ increases uniformly, taking values $[0 ; \pi]$ on the interval $[0 ; n]$.
 Find the number of isosceles triangles formed by the sides and diagonals of a regular $6k$-gon? | Solution: Fix one vertex and look at how many isosceles triangles exist where the legs are adjacent to this vertex. There will be a total of $3 k-1$ such triangles. Summing over all possible fixed vertices, we get $6 k \cdot(3 k-1)$. In this sum, all equilateral triangles were counted 3 times. There are a total of $2 k... | 18k^{2}-10k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,804 |
5. (8 points) On the sides $A B, B C$ and $C A$ of triangle $A B C$, points $C_{1}, A_{1}$ and $B_{1}$ are marked respectively. Segments $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at point $P$. Find the ratio $A P: P A_{1}$, if $A B: A C_{1}=\alpha>1$ and $A C: A B_{1}=\beta>1$. | # Solution (Ceva's and Menelaus' Theorems):
Since \( A B: A C_{1}=\alpha \) and \( A C: A B_{1}=\beta \), then \(\frac{A C_{1}}{C_{1} B}=\frac{1}{\alpha-1}\) and \(\frac{A B_{1}}{B_{1} C}=\... | \frac{1}{\alpha-1}+\frac{1}{\beta-1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,805 |
6. (8 points) Solve the equation
$$
\sqrt{a^{2}+c-1-2 x-x^{2}}+\sqrt{b^{2}+c-1-2 x-x^{2}}=\sqrt{x^{2}+2 x+1-c+(a+b)^{2}}
$$
In your answer, write the sum of the squares of all possible pairwise differences of the real roots of the equation. Numbers $a, b>0$. | Solution: Let's transform our equation and make the substitution $t=x+1$:
$$
\sqrt{a^{2}+c-t^{2}}+\sqrt{b^{2}+c-t^{2}}=\sqrt{t^{2}-c+(a+b)^{2}}
$$
Notice that if $t$ is a root of the equation, then $(-t)$ is also a root. Therefore, it is sufficient to find all non-negative roots. From now on, we will assume that $t \... | 4c | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,806 |
7. (10 points) The correspondence stage of the Innopolis University Olympiad was held among 9th, 10th, and 11th grade students, and among them, $a \%$, $b \%$, and $c \%$ respectively advanced to the in-person stage. Moreover, among the 9th and 10th grade students together, $k \%$ advanced to the in-person stage, and a... | Solution: Let $x$ be the number of 9th grade students, $y$ be the number of 10th grade students, and $z$ be the number of 11th grade students. Then
$$
k=\frac{a x+b y}{x+y} \Longrightarrow \frac{y}{x}=\frac{k-a}{b-k} \quad \text { and } \quad l=\frac{b y+c z}{y+z} \Longrightarrow \frac{z}{y}=\frac{l-b}{c-l}
$$
Expres... | \frac{(b-k)(-)+b(k-)(-)+(} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,807 |
8. (10 points) The volume of tetrahedron $ABCD$ is $4(m+n)$. Point $M$ divides edge $AB$ in the ratio $m: n$. A section is made through point $M$ and the midpoints of edges $BC$ and $AD$. Find the area of this section if the distance from point $D$ to it is $h$. | # Solution:
Let $N$ and $L$ be the midpoints of sides $AD$ and $BC$ respectively. $K$ is the point of intersection of edge $DC$ with the plane of the section. The plane of the section inter... | \frac{6m}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,808 |
9. (20 points) In an acute isosceles triangle $ABC (BC = CA)$, two medians $BB_1$ and $CC_1$ are drawn. The circumcircle of triangle $BB_1C$ intersects the median $CC_1$ at points $C$ and $Q$. Find the radius of the circumcircle of triangle $ABC$ if $CQ = m$.
# | # Solution:
Since triangle $ABC$ is isosceles, $CC_1$ is the height, median, and bisector. Therefore, point $Q$ is the midpoint of arc $BB_1$ that does not contain point $C$. Consequently, $... | \frac{2m}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,809 |
10. (20 points) Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
|2 x+y-b|+\sqrt{5(x-2)^{2}+5(y-a)^{2}}=a+4-b \\
(x-2)^{2}+(y-c)^{2}=r^{2}
\end{array}\right.
$$
has a unique solution. | # Solution:
Transform the first equation:
$$
\frac{|2 x+y-b|}{\sqrt{5}}+\sqrt{(x-2)^{2}+(y-a)^{2}}=\frac{2 \cdot 2+a-b}{\sqrt{5}}
$$
$\frac{|2 x+y-b|}{\sqrt{5}}$ defines the distance from point $M(x ; y)$ to the line $n: 2 x+y-b=0$, $\sqrt{(x-2)^{2}+(y-a)^{2}}$ the distance from $M$ to point $C(2 ; a)$, and $\frac{2... | [3;5)\cup{4+\frac{\sqrt{5}}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,810 |
1. Does there exist a number $x$ such that all three numbers $x-\frac{1}{x}$, $\frac{1}{x}-\frac{1}{x^{2}+1}$, and $\frac{1}{x^{2}+1}-2 x$ are integers? | Solution. Suppose all numbers are integers. Then their sum $-x$ is also an integer. But the fractions $\frac{1}{x}$ and $\frac{1}{x^{2}+1}$ cannot both be integers for integer $x$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,811 |
2. The sides of a convex quadrilateral are equal to $6, 7, 8, 9$ in some order. It is known that a circle can be inscribed in this quadrilateral and a circle can be circumscribed around it. Find the area of the quadrilateral. | Solution. A circle can be inscribed in a convex quadrilateral $ABCD$ if and only if $AB + CD = BC + AD$. Therefore, without loss of generality, we can assume that $AB = 6, BC = 8, CD = 9, AD = 7$. By the cosine rule,
$$
BD^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cdot \cos A, \quad BD^2 = 8^2 + 9^2 - 2 \cdot 8 \cdot 9 \cdot... | 12\sqrt{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,812 |
3. Solve the equation $\frac{1}{2} x^{2}+y^{2}+\frac{3}{2} z^{2}+t^{2}+\frac{1}{3}=x y+y z+z t+t$. | Solution. Multiply everything by 2 and rewrite the equality as
$$
\left(x^{2}-2 x y+y^{2}\right)+\left(y^{2}-2 y z+z^{2}\right)+\left(t^{2}-2 z t+z^{2}\right)+\left(t^{2}-2 t+1\right)+z^{2}=\frac{1}{3}
$$
or
$$
(x-y)^{2}+(y-z)^{2}+(t-z)^{2}+(1-t)^{2}+z^{2}=\frac{1}{3}
$$
We will prove the auxiliary inequality: $a^{... | \frac{1}{3},=\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,813 |
4. It is necessary to connect four lamps located at the vertices of a square with a side length of 4 meters into one electrical network. Will 11 meters of wire be enough for this? (In other words, does there exist a connected graph containing the vertices of a square with a side length of 4, the sum of the lengths of w... | Solution. Yes, enough. If connected as shown in the figure, $A E+B E+C F+$ $D F+E F=4 \cdot \frac{4}{\sqrt{3}}+4-\frac{4}{\sqrt{3}}=4+4 \sqrt{3}<11$.
 | 4+4\sqrt{3}<11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,814 |
5. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
\frac{a^{2} x+2 a}{a x-2+a^{2}} \geqslant 0 \\
a x+a>\frac{5}{4}
\end{array}\right.
$$
has no solutions. | Answer: $a \in(-\infty ;-0.5] \cup\{0\}$
Solution. The value $a=0$ satisfies the condition of the problem. For $a \neq 0$, we make the substitution $t=a x$. For each $t$, there is a unique $x=\frac{t}{a}$.
$$
\left\{\begin{array} { l }
{ \frac { a t + 2 a } { t - 2 + a ^ { 2 } } \geqslant 0 , } \\
{ t + a > \frac { ... | \in(-\infty;-0.5]\cup{0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,815 |
10. (20 points) Point $A$ lies on the line $y=\frac{5}{12} x-11$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $6311 / 624$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 2. | \frac{6311}{624} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,817 |
10. (20 points) Point $A$ lies on the line $y=\frac{8}{15} x-10$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $2234 / 255$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 3. | \frac{2234}{255} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,818 |
10. (20 points) Point $A$ lies on the line $y=\frac{12}{5} x-9$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $189 / 65$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 4. | \frac{189}{65} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,819 |
10. (20 points) Point $A$ lies on the line $y=\frac{15}{8} x-8$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $1823 / 544$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 5. | \frac{1823}{544} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,820 |
10. (20 points) Point $A$ lies on the line $y=\frac{5}{12} x-7$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $4007 / 624$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 6. | \frac{4007}{624} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,821 |
10. (20 points) Point $A$ lies on the line $y=\frac{12}{5} x-5$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $89 / 65$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 8. | \frac{89}{65} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,823 |
10. (20 points) Point $A$ lies on the line $y=\frac{15}{8} x-4$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $47 / 32$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 9. | \frac{47}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,824 |
10. (20 points) Point $A$ lies on the line $y=\frac{12}{5} x-3$, and point $B$ lies on the parabola $y=x^{2}$. What is the minimum length of segment $A B$? | Answer: $3 / 5$
## Innopolis Open
Mathematics Olympiad
Innopolis University
Qualifying (correspondence) stage in mathematics, December 2, 2018
9th grade, variant 10. | \frac{3}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,825 |
1. Someone writes the word "mathematics" in a row without spaces. Like this:
MATHEMATICS MATHEMATICS MATHEMATICS...
What will be the $2023^{rd}$ letter in this chain? | Solution. Obviously, the word «mathematics» contains 10 letters. Thus, the number divisible by 10 without a remainder and closest to the number 2023 is the number 2020. After that, we need to count 3 more digits from the beginning of the word - 2023-2020=3. From this, it is clear that the required letter will be the le... | T | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,826 |
2. Cinderella, together with her fairy godmother, released a collection of crystal slippers featuring 7 new models. The fairy tale heroines organized a presentation of the collection for some guests: the audience was supposed to say which slippers they liked. The guests wrote in a questionnaire which models they though... | Solution. Suppose that the guests were given a questionnaire where they could put a "1" next to each pair of shoes they liked and a "0" next to the pair they did not select as the best. Thus, the opinion of each guest can be recorded as a string of "1"s and "0"s. Then, the number of different sets of favorite shoes is ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,827 |
3. The tennis player and the badminton player got tired of sports tournaments and decided to have a mathematical duel. They wrote down the numbers from 25 to 57 inclusive in a row and agreed to place either “+” or “-” (in any order, but only these signs) between them. If the result of the expression turns out to be eve... | Solution. Among all the given numbers, 17 are odd, and the sum of odd numbers is also odd. The overall result of the expression will then always be odd, regardless of what signs are between our numbers. Therefore, the badminton player will win.
## Answer: badminton player. | badmintonplayer | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,828 |
4. Fourteen foxes met ninety-two Koloboks, felt sorry for them, and did not eat them. Prove that at least two of the foxes met the same number of Koloboks. (We will also assume that a fox might not have met any Koloboks at all.) | Solution. Suppose all the foxes met a different number of Koloboks. Arrange them in ascending order of the number of Koloboks seen. 1st fox met 0 or more Koloboks, 2nd fox met 1 or more Koloboks, ...
14th fox met 13 or more Koloboks.
Thus, together they met $0+1+2+\ldots+13$ Koloboks, and the total number of Koloboks... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,829 |
5. In the Umnovka River, there live Premudrye (Wise) minnows, and they have a law: as soon as one of them changes the color of their eyes from gray to green, they must swim to the Mozgovka River on the same night they learn about this transformation. There are no mirrors under the water, and the minnows understand ever... | Solution. If there is only one green-eyed minnow, it sees only gray-eyed minnows around and understands that it is green-eyed; then it immediately swims away. If there are two green-eyed minnows, the first green-eyed minnow sees the second among the minnows, and if the second minnow does not swim away on the same night... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,830 |
# Problem 1.
A numerical characteristic $x$ of a certain thermal power process is a root of the equation
$$
x^{3}-3 x=t
$$
where $t$ is the ambient temperature measured in degrees Celsius. For some technological reasons, the root must be unique. For what values of $t$ does the equation have a unique root $x_{0}$? Es... | # Solution.
It is not difficult to plot the graph of the function $y=x^{3}-3 x$, noting that this function is odd, equals zero at exactly three points $x=0, \pm \sqrt{3}$, $(1; -2)$ is a point of minimum, and $(-1; 2)$ is a point of maximum. The function increases without bound for $x>1$ and decreases without bound fo... | ||>2|x|>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,831 |
# Problem 2.
For each integer value of the parameter $K$, solve the system of equations
$$
\left\{\begin{aligned}
2[x]+y & =3 / 2, \\
([x]-x)^{2}-2[y] & =K
\end{aligned}\right.
$$
Here $[x]$ denotes the integer part of the number $x$. | # Solution.
Let
$$
x=M+a, \quad y=N+b, \quad M, N \in \mathbb{Z}, \quad a, b \in[0 ; 1)
$$
From the first equation, we get $2 M+N+b=3 / 2$ and then
$$
b=0.5, \quad N=1-2 M
$$
Substitute these values into the second equation:
$$
(-a)^{2}-2(1-2 M)=K
$$
Then
$$
a=0, \quad K=4 M-2, \quad x=M, \quad y=3 / 2-2 M
$$
... | M,3/2-2MifK=4M-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,832 |
# Problem 3.
Two equal circles intersect at points $P$ and $Q$. An arbitrary line passing through $Q$ intersects the circles again at points $A$ and $B$, and the tangents to the circles at these points intersect at point $C$. Prove that the segments $A Q$ and $C B$ are seen from point $P$ at the same angle. | Solution. By the theorem on the angle between a tangent and a chord, $\angle B A C = \angle Q P A, \angle C B A = \angle B P Q$. Therefore, $\angle B P A + \angle A C B = \angle B A C + \angle A C B + \angle C B A = \pi$, i.e., quadrilateral $A P B C$ is cyclic (Fig.1). Hence, $\angle B P C = \angle B A C = \angle Q P ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,833 |
# Problem 4.
When processing numerical data, it is often necessary to calculate the arithmetic mean
$$
S(x, y)=(x+y) / 2
$$
and solve equations containing the arithmetic mean. Find all finite (consisting of a finite number of elements) numerical sets $X$ such that for any $a$ and $b$ from $X$ the set $X$ contains th... | Solution. We have
$$
S(a, x)=b \Leftrightarrow x=2 b-a
$$
The required property in the problem is possessed by any one-element set
$$
X=\{a\}, \quad a \in(-\infty ; \infty)
$$
since $S(a, a)=a$.
Suppose further that the set $X$ contains at least two distinct elements $c, d$, with $c<d$ (without loss of generality)... | {},\in(-\infty;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,834 |
# Problem 5.
A young hacker wants to change the grades in the electronic journal. But when changing some grades, others change as well, specifically: a) if he increases the number of fives by 2, then the number of twos decreases by 1; b) if he increases the number of fives by 1, then the number of twos increases by 2;... | # Solution.
Let $n_{i}(i=1,2,3,4)$ denote the number of actions of each of the four possible types. We need to solve the system (the first equation corresponds to the change in the number of fives, the second to the number of twos)
$$
\left\{\begin{array}{l}
2 n_{1}+n_{2}-2 n_{3}-n_{4}=30-3=27 \\
-n_{1}+2 n_{2}+n_{3}... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,835 |
Task 1. There are 3 types of installations, no more than 200 in total. The number of installations of type 2 is 4 times the number of type 1, and the number of installations of type 3 is a multiple of the number of installations of type 1. If the number of installations of type 3 were 5 times more, then they would be 9... | Solution. If $x_{1}, x_{2}, x_{3}$ are the quantities of installations of types $1,2,3$, then the conditions are represented by the relations
$$
\begin{gathered}
x_{1}+x_{2}+x_{3} \leq 200 \\
x_{2}=4 x_{1} \\
x_{3}=k x_{1} \\
5 x_{3}=x_{2}+99 \\
k, x_{1}, x_{2}, x_{3} \in \mathbb{N}
\end{gathered}
$$
From (2)-(4) we ... | 9,36,27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,836 |
Problem 2. On the side $AB$ of triangle $ABC$, a point $M$ is taken. It starts moving parallel to $BC$ until it intersects with $AC$, then it moves parallel to $AB$ until it intersects with $BC$, and so on. Is it true that after a certain number of such steps, point $M$ will return to its original position? If this is ... | Solution. Let the length of side $AB$ be 1, and let point $M$ be at a distance $a$ from point $B$. From the properties of the parallelogram, the small triangles are equal, so after 3 steps, point $M$ will be at a distance $a$ from point $A$, which is at a distance $1-a$ from point $B$. After another 3 steps, the point ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,837 |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, \q... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,838 |
Problem 4. The numbers $x, y, z$ are such that the ratios
$$
\frac{x+y}{z}, \quad \frac{x+z}{y}, \quad \frac{y+z}{x}
$$
are equal. Find this value. | Solution. Let $k$ be the value of such a ratio. Then
$$
\frac{x+y}{z}=\frac{x+z}{y}=\frac{y+z}{x}=k
$$
from which $x y z \neq 0$ and
$$
x+y=k z, \quad x+z=k y, \quad y+z=k x
$$
Adding these equations, we get
$$
2(x+y+z)=k(x+y+z) \text {. }
$$
There are two cases:
$$
x+y+z=0
$$
or
$$
x+y+z \neq 0, \quad k=2
$$
... | -1or2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,839 |
Problem 5. Masha, preparing to receive guests, placed 13 oranges and 3 apples in 4 vases, with 4 fruits in each. Then her sister Sasha decided to change the composition of the fruits in the vases. She took one fruit from each vase at the same time and replaced each fruit with the opposite: an apple with an orange, and ... | Solution. If we interpret the vases and fruits in the vases as a $4 \times 4$ table, and assign +1 to each orange and -1 to each apple, then replacing the fruits by Sasha with opposite ones is equivalent to changing the sign of all numbers in one row or one column of the table. In this case, the sign of the product of ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,840 |
1. A bridge over a river connects two different regions of the country. Once, one of the regions repainted the part of the bridge that belonged to it. If the freshly painted part of the bridge were 1.2 times larger, it would make up exactly half of the entire bridge. What part of the bridge needs to be repainted to mak... | # Solution
Let $x$ be the painted part of the bridge. (We take the entire bridge as one unit.) According to the condition,
$$
1.2 x = 1 / 2 \Longrightarrow x = \frac{5}{12}
$$
To complete the painting, we need
$$
\frac{1}{2} - \frac{5}{12} = \frac{1}{12}
$$
Answer: $\frac{1}{12}$ part. | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,841 |
2. A table of numbers consisting of 20 rows and 15 columns is filled with the numbers 2 and 5.
a) Can the products of the numbers in each row be even, and the products of the numbers in each column be odd?
b) Can the products of the numbers in each row be odd, and the products of the numbers in each column be even? | Solution. Consider the product $P$ of all numbers in the table.
If the product of the elements of row $i$ (column $j$) is even, then this row (this column) contains the element $t_{ij}=2$. Then the product of all elements of column $j$ (row $i$) is also even.
Answer: a) - no, b) - no. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,842 |
3. Is it possible to divide the numbers from 1 to 77 into groups such that each group contains at least three numbers, and one of the numbers in each group is equal to the sum of the other numbers in the same group? | Solution. Let $x_{3}=x_{1}+x_{2}$ in each group. Then the sum of all numbers in this group $x_{1}+x_{2}+x_{3}=2\left(x_{1}+x_{2}\right)$ is even. Thus, the sum of all numbers in the set, as the sum of even addends, is even.
On the other hand, it is equal to
$$
1+\cdots+77=\frac{1}{2}((1+77)+(2+77)+\ldots+(77+1))=\fra... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,843 |
4. Two identical right-angled triangles were placed together with their equal legs touching. As a result, a larger triangle was formed, in which one angle is the arithmetic mean of the other two angles.
a) Can the length of one side of the larger triangle be the arithmetic mean of the lengths of the other two sides?
... | Solution.
Let the angles of the original right triangle be denoted by $\alpha$ and $\beta$. Clearly, $\alpha + \beta = 90$ (degrees). Suppose that the angle $\beta$ is doubled in the describe... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,844 |
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x+3}{2}\right]^{2}-x=1$. | # Solution.
From the equation, it follows that $x=\left[\frac{x+3}{2}\right]^{2}-1$ is an integer. Therefore, $x=n \in \mathbb{Z}$, but we need to consider the cases of even and odd $n$ separately.
First, let's move all terms to the left side of the equation.
1) If $x=2 k$.
$$
\left[\frac{2 k+3}{2}\right]^{2}-2 k-1... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,845 |
6. From the equations
$$
x_{1}+x_{2}+x_{3}+x_{4}=S_{0}, \quad x_{2}+x_{3}+x_{4}=S_{m}+3 m
$$
we find $x_{1}$. | Answer 10 and 32 min.
## Problem 11 (8th grade)
When calculating the fraction
$$
\frac{A x+B}{C-B x}
$$
a student made a mistake by replacing $x$ with $-x$. Find all numbers $A, B, C$, for which such a fraction does not change when any admissible number $x$ is replaced by an admissible number $-x$.
## Solution
Th... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,848 |
# Problem 1.
When designing a power plant, it became necessary to solve the equation
$$
4 x^{4}+4 p x^{3}=(p-4) x^{2}-4 p x+p,
$$
where $p$ is an integer parameter specified by the developer. For quick and reliable solutions, it is required that the roots of the equation be rational numbers. For which $p$ is this tr... | # Solution.
The problem boils down to finding the roots of the polynomial
$$
f(x, p)=\left(x^{2}+1\right)\left(4 x^{2}+4 p x-p\right) .
$$
For the discriminant of the quadratic trinomial, it is necessary that $D / 4=$ $4 p(p+1) \geq 0$.
If $p=0$ or $p=-1$, then there exist rational roots 0 and -1, respectively.
If... | 0,-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,849 |
# Problem 2.
In the Kingdom of Magic Energy, an enchanted transformer booth stands on a flat plain: to an observer looking parallel to the ground, it is visible only at an angle of $90^{\circ}$. In cross-section, the booth is square with a side length of $L$ cubits. Describe the geometric locus of points on the plain ... | # Solution.
It is enough to recall that any inscribed angle in a circle that subtends the diameter is a right angle. Therefore, the desired set will be four semicircles (with diameter $L$) subtending the sides of the square (see the figure below). For clarity, one of the right angles from which the tower is visible fr... | \rho_{\}=0;\quad\rho_{\max}=L/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,850 |
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