problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $\sqrt{6}+\sqrt{2}$. The distances from the center of the inscribed circle of triangle $D E F$ to points $A$ and $C$ are 2 and $2 \sqrt{2}$, respectively. Find the radius of the circumscribed circle around triangle $D E F$. (16... | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=2, C O=2 \sqrt{2}$. ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ of a circular track, a car and a motorcycle started simultaneously and in the same direction. The car drove two laps without stopping in one direction. At the moment when the car caught up with the motorcyclist, the motorcyclist turned around and increased his speed by $16 \mathrm{~km} / \mathrm{u}$, ... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $y$ (km/h) be the speed of the car, and $S$ (km) be the distance the motorcyclist travels before turning around. Then the total length of the track is $2 S + 5.25$. We have $\frac{S}{x} = \frac{3 S + 5.25}{y}$, $\frac{3 x}{8} + 6 = S$, $\frac{3 y}{8} = S + 5.25... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given 2019 indistinguishable coins. All coins have the same weight, except for one, which is lighter. What is the minimum number of weighings required to guarantee finding the lighter coin using a balance scale without weights? | Solution. $\quad$ We will prove the following statement by induction on $k$: if there are $N$ visually identical coins, with $3^{k-1}<N \leq 3^{k}$, and one of them is lighter, then it can be found in $k$ weighings. Base case: $k=0, N=1$, no weighing is needed for a single coin. Inductive step: suppose the statement is... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$. | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiply them by 307, and consider linear combinations for integer $t$, we get values in natural numbers
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t \in\{... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) The creative competition at the institute consisted of four tasks. In total, there were 70 applicants. The first test was successfully passed by 35, the second by 48, the third by 64, and the fourth by 63 people, with no one failing all 4 tasks. Those who passed both the third and fourth tests were admit... | Solution. 1st and 2nd tasks were solved by at least $35+48-70=13$ people. 3rd and 4th - at least $64+63-70=57$ people. No one failed all tasks, so 1st and 2nd were solved by 13 people, 3rd and 4th - 57 people.
Answer: 57 people.
Criteria.
| Points | Conditions for awarding |
| :--- | :--- |
| 10 points | Justified s... | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) In a convex quadrilateral $A B C D, A B=10, C D=15$. Diagonals $A C$ and $B D$ intersect at point $O, A C=20$, triangles $A O D$ and $B O C$ have equal areas. Find $A O$.
# | # Solution:
From the equality of the areas of triangles $A O D$ and $B O C$ and the equality of angles $\angle A O D=\angle B O C$, it follows that $\frac{A O \cdot O D}{B O \cdot O C}=1$ (by the theorem on the ratio of areas of triangles with one equal angle). From this, we get $\frac{A O}{O C}=\frac{B O}{O D}$. Addi... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) In one of the regions on the planet, seismic activity was studied. 80 percent of all days were quiet. The instrument predictions promised a calm situation in 64 out of 100 cases; moreover, in 70 percent of all cases when the day was quiet, the instrument predictions came true. What percentage of days wit... | Solution. Let the total number of observed days be x. The number of actually quiet days was $0.8x$, and seismically active days were $0.2x$. The predictions of quiet days matched the actually quiet days: $0.7 \cdot 0.8x = 0.56x$. Then the number of active days that did not match the predictions was $0.64x - 0.56x = 0.0... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A car left point A for point B, and a second car left with some delay. When the first car had traveled half the distance, the second car had traveled $26 \frac{1}{4}$ km, and when the second car had traveled half the distance, the first car had traveled $31 \frac{1}{5}$ km. After overtaking the first car, the second... | Solution. S - the distance between points A and B.
$$
\frac{S-2-S / 2}{S+2-26.25}=\frac{S-2-31.2}{S+2-S / 2}, \quad 5 S^{2}-383 S+5394=0, \quad \sqrt{D}=197, \quad S=58
$$
Answer: 58. | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest integer \( a \) such that the expression
\[
a^{2}-15 a-(\tan x-1)(\tan x+2)(\tan x+5)(\tan x+8)
\]
is less than 35 for any value of \( x \in (-\pi / 2, \pi / 2) \).
(6 points) | Solution. Let's make the substitution $t=\operatorname{tg} x$. We need to determine for which values of $a$ the inequality $a^{2}-15 a-(t-1)(t+2)(t+5)(t+8)a^{2}-15 a-35,\left(t^{2}+7 t-8\right)\left(t^{2}+7 t+10\right)>a^{2}-15 a-35$
$z=t^{2}+7 t+1,(z-9)(z+9)>a^{2}-15 a-35, z^{2}>a^{2}-15 a+46$, $0>a^{2}-15 a+46, \sqr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given six socks, all of different colors and easily stretchable. You cannot turn them inside out. In how many ways can you put on 3 socks on each foot, considering which one to put on earlier and which one later? | Solution. There is a sequence of 6 sock puttings on: $\mathrm{C}_{6}^{3}=20$ ways to choose which puttings are for the right foot. For each such choice, there are $6!=720$ ways to choose which sock to take for each putting.
Answer: 14400. | 14400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x, y, z$ be the roots of the equation $t^{3}-2 t^{2}-9 t-1=0$. Find $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$.
(12 points) | Solution. Let's bring the desired expression to a common denominator: $\frac{y^{2} z^{2}+x^{2} z^{2}+x^{2} y^{2}}{x y z}$. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem $x+y+z=2, x y+x z+y z=-9, x y z=1$.
$$
\begin{aligned}
& x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the plane $x O y$, the lines $y=3 x-3$ and $x=-1$ intersect at point $\mathrm{B}$, and the line passing through point $M(1 ; 2)$ intersects the given lines at points A and C respectively. For what positive value of the abscissa of point A will the area of triangle $\mathrm{ABC}$ be the smallest?
(12 points) | # Solution.
$A C: \quad y=k x+d, \quad M \in A C \Rightarrow d=2-k$
$A(a ; 3 a-3) \in A C \Rightarrow 3 a-3=k a+2-k \Rightarrow a=\frac{5-k}{3-k}$,
$C(-1 ; c) \in A C \Rightarrow c=-2 k+2$,
$S_{A B C}=\frac{1}{2}(c+6) \cdot(a+1)=\frac{2(k-4)^{2}}{3-k}$
$S^{\prime}=\frac{2(k-4)(2-k)}{(3-k)^{2}}=0, k_{\min }=2, \qua... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Specify the smallest integer value of \(a\) for which the system has a unique solution
\[
\left\{\begin{array}{l}
\frac{y}{a-\sqrt{x}-1}=4 \\
y=\frac{\sqrt{x}+5}{\sqrt{x}+1}
\end{array}\right.
\] | # Solution.
Solving the system by substitution, we arrive at an equation with constraints on the unknown quantity ${ }^{x}$.
$$
\left\{\begin{array} { l }
{ \frac { y } { a - \sqrt { x } - 1 } = 4 } \\
{ y = \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ x \geq ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the midline of which is equal to $5 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $\mathrm{ABCD}$, into which it is divided by the midline, is $7: 13$. All lateral faces of the pyramid $\mathrm{TABCD}$ are inclined ... | # Solution.
Let $TO$ be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, $O$ is the center of the circle inscribed in the base. Let $MP$ be the midline of the trapezoid, $\quad AD=a, BC=b. \quad$ According to the problem, we have $S_{MB CP}=7x, S_{\text{AMPD}}=13x, \quad \... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A group of schoolchildren heading to a school camp was to be seated in buses so that each bus had the same number of passengers. Initially, 22 people were to be seated in each bus, but it turned out that three schoolchildren could not be seated. When one bus left empty, all the schoolchildren were able to sit evenly... | Solution. Let $\mathrm{n}$ be the number of buses, $\mathrm{m}$ be the number of schoolchildren in each bus, and $\mathrm{S}$ be the total number of schoolchildren.
We have
$$
S=22 n+3, \quad S=(n-1) m, n \leq 10, m \leq 36, \quad 22 n+3=(n-1) m, \quad n=1+\frac{25}{m-22}
$$
Considering the constraints on $\mathrm{n... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x^{2}-x y-6 y^{2}-11=0$. For each pair $(x, y)$ found, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $x^{2}-x y-6 y^{2}-11=0,(x-3 y)(x+2 y)=11$. Since $x$ and $y$ are integers, we have four cases:
$\left\{\begin{array}{c}x-3 y=11 \\ x+2 y=1\end{array} \Leftrightarrow\left\{\begin{array}{c}y=-2 \\ x=5\end{array}\right.\right.$
2)
$\left\{\begin{array}{c}x-3 y=-11 \\ x+2 y=-1,\end{array} \Leftrightarrow\left\{... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can the line $x \sin \sqrt{16-x^{2}-y^{2}}=0$ be drawn without lifting the pencil and without retracing any part of the line? (12 points) | Solution. Since $\pi^{2}<16<(2 \pi)^{2}$, the given line consists of 2 circles with radii 4 and $\sqrt{16-\pi^{2}}$ and a vertical segment.
This line is unicursal, as it has only 2 odd poin... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. For how many two-digit natural numbers n are exactly two of these three statements true: (A) n is odd; (B) n is not divisible by $3 ;$ (C) n is divisible by 5? | Solution. We can consider the first 30 two-digit numbers (from 10 to 39), and then multiply the result by 3, since the remainders when dividing by 2, 3, and 5 do not change when shifted by 30 or 60. There are three mutually exclusive cases.
1) (A) and (B) are satisfied, and (C) is not. From (A) and (B), it follows tha... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Find all integer values of the parameter \(a\) for which the system has at least one solution
\[
\left\{\begin{array}{l}
y-2=x(x+2) \\
x^{2}+a^{2}+2 x=y(2 a-y)
\end{array}\right.
\]
In the answer, specify the sum of the found values of the parameter \(a\). | Solution. Transform the system
$$
\left\{\begin{array} { l }
{ y - 1 = ( x + 1 ) ^ { 2 } , } \\
{ ( x + 1 ) ^ { 2 } + ( y - a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y-1=(x+1)^{2} \\
y-2+(y-a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$$
y^{2}-y(2 a-1)+a^{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the length of the larger base $A D$ of which is $12 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $A B C D$, into which it is divided by the midline, is $5: 7$. All lateral faces of the pyramid TABCD are inclined to... | # Solution.
Let TO be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, O is the center of the circle inscribed in the base. Let MP be the midline of the trapezoid, $A D=a=12 \sqrt{3}, B C=b$.
According to the problem, we have
$S_{\text {MBCP }}=5 x, S_{\text {AMPD }}=7 x... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If a two-digit natural number is decreased by 54, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the median of the sequence of all such numbers.
# | # Solution.
Let $\overline{x y}=10 x+y$ be the original two-digit number, then $\overline{y x}=10 y+x$ is the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$. From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 9 points, if there were 4 tens, and the results of the hits were sevens, eights, and nines. There were no misses at all. | Solution: Since the soldier scored 90 points and 40 of them were scored in 4 attempts, he scored 50 points with the remaining 6 shots. Since the soldier only hit the seven, eight, and nine, let's assume that in three shots (once each in seven, eight, and nine), he scored 24 points. Then, for the remaining three shots, ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let $x$ be the length of the roe deer's jump, then $0.78 x$ is the length of the wolf's jump; $y$ - the number of jumps the roe deer makes over the time interval specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes over the same time interval. The wolf will not be a... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61; 20\}$?... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$B=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given triangle $A B C$. Lines $O_{1} O_{2}, O_{1} O_{3}, O_{3} O_{2}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $O$ is the center of the inscribed circle of triangle $A B C$. Find the angle in degrees between the lines $O_{1} O_{2}$ and $O O_{3}$.
.
1) $\triangle I G C = \triangle G F H$ - by two legs, since $I C = G H = 2, I G = H F = 1$, therefore ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} M C_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering t... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{25}{3}$ times. Deter... | # Solution.
1) Let the syrup content in the initial solution be $p \%$ and let $x$ liters of the solution were poured out the first time.
2) Then after pouring out the liquid, there remained $(1000-x)$ liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A warehouse has coffee packed in bags of 15 kg and 8 kg. How many bags of coffee in total does the warehouseman need to prepare to weigh out 1998 kg of coffee, with the number of 8 kg bags being the smallest possible? | Solution. Let $x$ be the number of bags weighing 15 kg, and $y$ be the number of bags weighing 8 kg. We get the equation $15 x + 8 y = 1998$. $8(x + y) + 7 x = 1998$, let $x + y = k$,
$8 k + 7 x = 1998$,
$7(k + x) + k = 1998$, let $k + x = t$,
$7 t + k = 1998$, $k = 1998 - 7 t$. Substitute into (2), $x = 8 t - 1998$... | 136 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Ivan Ivanovich approached a source with two empty cans, one with a capacity of 10 liters, and the other - 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched t... | Solution. Let $x$ liters of water fill the larger can while 4 liters fill the smaller can. After the switch, $(10-x)$ liters fill the larger can, and 4 liters fill the smaller can again. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equa... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} B M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B C A=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=60^{\circ}$. On the side $\mathrm{BC}$, a point $\mathrm{M}$ is ch... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\Delta T P M = \Delta K O M$... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Right triangles $M D C$ and $A D K$ have a common right angle $D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$, counting from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of the degree measures of angles $A K D$ and $M C D$, if $A D: C D=2: 5$. | Solution. Extend triangle $A D C$ to form a square $L J C D$.
Choose point $H$ on side $L J$ such that $L H: H J=2: 3$, point $N$ on side $C J$ such that $C N: N J=3: 2$, and point $B$ on si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
(16 points) | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}^{m}}$, where $l_{i} \in\left\{0,1, \l... | 25920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N, \quad$ is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N$ (12 points) | Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$.
For the second sequence $\quad b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}, \quad$ the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.
We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, dark blue - 87, purple - 91, black -... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text{ times}}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few value... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. (Option 2).
How to build a highway? (an old problem) From a riverside city A, goods need to be transported to point B, located $a$ kilometers downstream and $d$ kilometers from the riverbank. How should the highway be built from B to the river so that the transportation of goods from A to B is as cost-effec... | Solution: Let the distance $AD$ be denoted by $x$ and the length of the highway $DB$ by $y$: by assumption, the length of $AC$ is $a$ and the length of $BC$ is $d$.
Since transportation along the highway is twice as expensive as along the river, the sum $x + 2y$ should be the smallest according to the problem's requir... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 6. (Option 2).
Translate the text above into English, preserving the original text's line breaks and formatting, and output the translation result directly. | Solution: A total of $68+59+46=173$ "yes" answers were given to the three questions. Note that since each resident of Kashino lives in exactly one district, if every resident were a knight (i.e., told the truth only), the number of "yes" answers would be equal to the number of residents in the city (a knight says "yes"... | 12 | Number Theory | proof | Yes | Yes | olympiads | false |
1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$
considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)... | Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.
Grading criteria.
| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but there ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. According to the inverse theorem of Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) Solve the equation
$$
\sqrt{7-x^{2}+6 x}+\sqrt{6 x-x^{2}}=7+\sqrt{x(3-x)}
$$ | # Solution:
The domain of the variable x in our problem is the interval [0;3]. By completing the square in the expressions under the square roots on the left side of the equation or by plotting the graphs, we notice that the values of the first root do not exceed 4, and the second does not exceed 3, with the minimum v... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of the expression $2 a-\left(\frac{2 a-3}{a+1}-\frac{a+1}{2-2 a}-\frac{a^{2}+3}{2 a^{2-2}}\right) \cdot \frac{a^{3}+1}{a^{2}-a}+\frac{2}{a}$ when $a=1580$. | Solution:
1) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)(a+1)} \cdot \frac{(a+1)\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
2) $2 a-\frac{2(a-1)(2 a-3)+(a+1)(a+1)-\left(a^{2}+3\right)}{2(a-1)} \cdot \frac{\left(a^{2}-a+1\right)}{a^{2}-a}+\frac{2}{a}$
3) $2 a-\frac{\left(-4 a+2+2 a^{2}\right)}{(a... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For a rectangle, the sum of two sides is 11, and the sum of three sides is 19.5. Find the product of all possible different values of the perimeter of such a rectangle. | Solution:
Let the sides of the rectangle be $a$ and $b$.
If the sum of adjacent sides is 11, then the system describing the condition of the problem is $\left\{\begin{array}{l}a+b=11 \\ 2 a+b=19.5\end{array}\right.$, its solution is $a=8.5, b=2.5$, the perimeter of the rectangle is $P_{1}=22$.
If, however, the numbe... | 15400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In a convex quadrilateral $A B C D$, the angles at vertices $B, C$, and $D$ are $30^{\circ}, 90^{\circ}$, and $120^{\circ}$ respectively. Find the length of segment $A B$, if $A D=C D=2$. | Solution:
Extend lines $A B$ and $C D$ to intersect at point $E$, the resulting triangle $A D E$ is equilateral, so $E D=E A=2$, the leg $E C=E D+D C=2+2=4$, since it lies in the right triangle $B C E$ opposite the angle $30^{\circ}$, then the hypotenuse $B E=8$, and the segment $A B=B E-E A=8-2=6$.
Answer: 6. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In an acute-angled triangle $ABC$ with sides $AB=4, AC=3$, a point $N$ is marked on the median $AM$ such that $\angle BNM = \angle MAC$. Find the length of the segment $BN$. | Solution:
Let's make an additional construction, doubling the median $A M$ beyond point $M$, thereby obtaining point $K$ such that $K \in A M, K M=A M$. Triangles $K M B$
Preliminary (correspondence) online stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
and $A M C$ are eq... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. Solve the equation $a^{2}+2=b$ !, given that a, b belong to N. In the answer, indicate the sum of the product of all possible a and the product of all possible b (if the equation has no solutions, indicate 0; if there are infinitely many solutions, indicate 1000). | Solution:
$b!-2=a^{2} ; x, y \in N$
$a \geq 1$, i.e. $a^{2} \geq 1 \Rightarrow b!\geq 3$, i.e. $b \geq 3$
If $x \geq 5$, then $x!$ ends in 0, then $y^{2}$ ends in 8, but there is no number whose square ends in 8, i.e. $x<5$.
This gives us:
$\left[\begin{array}{l}b=3 \\ b=4\end{array} \Rightarrow\left[\begin{array}... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) Find the area of a convex quadrilateral with equal diagonals, if the lengths of the segments connecting the midpoints of its opposite sides are 13 and 7. | Solution. Let $MK$ and $PH$ be segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$, with $MK = PH$, $AC = 18$, and $BD = 7$.
We have: $MP \| AC$, $MP = \frac{1}{2} AC$ (as the midline of $\triangle ABC$); $HK \| AC$, $HK = \frac{1}{2} AC$ (as the midline of $\triangle ADC$). $\Rightarr... | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) For what values of the parameter $a$ does the equation
$$
(a+1)(|x-2.3|-1)^{2}-2(a-3)(|x-2.3|-1)+a-1=0
$$
have exactly two distinct solutions? | Solution. Let $|x-2.3|-1=t$ (1), then the original equation will take the form: $(a+1) t^{2}-2(a-3) t+a-1=0$ (2). Let's analyze equation (1): when $t>-1$, it corresponds to two different values of x. Thus, the original equation can have from zero to four solutions. It has two distinct roots in the following three cases... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $E$, and the rays $D A$ and $C B$ intersect at point $F$. The ray $B A$ intersects the circumcircle of triangle $D E F$ at point $L$, and the ray $B C$ intersects the same circle at point $K$. The length of segment $L K$ ... | # Solution.
$\angle F L E = \angle F D E = \angle F K E = \alpha$, since these angles subtend the arc $F E$.
$\angle E B K = \angle F D E = \alpha$, since quadrilateral $A B C D$ is cyclic... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A circle passes through the vertices $A$ and $C$ of an isosceles triangle $ABC (AB = BC)$ and intersects the sides $AB$ and $BC$ at points $M$ and $N$, respectively. $MK$, a chord of this circle, equal in length to $2 \sqrt{5}$, contains point $H$, which lies on $AC$ and is the foot of the altitude of tr... | # Solution.
Quadrilateral $A M N C$ is an isosceles trapezoid. $\triangle A M H = \Delta H N C$ - by two sides and the angle between them.
$$
\begin{gathered}
\angle A H M = \angle H M N =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Four elevators of a skyscraper, differing in color (red, blue, green, and yellow), are moving in different directions and at different but constant speeds. Observing the elevators, someone started a stopwatch and, looking at its readings, began to record: 36th second - the red elevator caught up with the blue one (m... | Solution. Let's number the elevators: red - first, blue - second, green - third, yellow - fourth. The elevators move at constant speeds, so the distance traveled $S_{i}, i=1,2,3,4$, in some coordinate system depends on time according to the law $S_{i}=k_{i} t+b_{i}$. According to the problem, the red and blue elevators... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the minimum value of the expression $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}$, if $f(x)=a x^{2}+b x+c$ is an arbitrary quadratic function satisfying the condition $b>2 a$ and taking non-negative values for all real $x$.
(12 points) | Solution. We have $f(1)=a+b+c, \quad f(0)=c, \quad f(-1)=a-b+c, f(-2)=4 a-2 b+c$, $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}=\frac{3(a+b+c)+6 c-a+b-c}{c-4 a+2 b-c}=\frac{2 a+4 b+8 c}{2 b-4 a}=\frac{a+2 b+4 c}{b-2 a}$.
Since $f(x)=a x^{2}+b x+c \quad-$ is an arbitrary quadratic function that takes non-negative values for ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, the bisector $A D$ is drawn. It is known that the centers of the inscribed circle of triangle $A B D$ and the circumscribed circle of triangle $A B C$ coincide. Find $C D$, if $A C=\sqrt{5}+1$. The answer should not include trigonometric function notations or their inverses.
(20 points) | Solution: Let $\angle A=\alpha, \quad B=\beta$. Point $O$ is the center of the inscribed circle of triangle $ABD$. $\angle BAO=\alpha / 4, \angle ABO=\beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO=\angle ABO, \beta=\alpha / 2$. Triangl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A workshop produces transformers of types $A$ and $B$. For one transformer of type $A$, 5 kg of transformer iron and 3 kg of wire are used, and for a transformer of type $B$, 3 kg of iron and 2 kg of wire are used. The profit from selling a transformer of type $A$ is 12 thousand rubles, and for type $B$ it is 10 tho... | Solution. Let $x$ be the number of transformers of type $A$, and $y$ be the number of transformers of type $B$. Then the profit per shift is calculated by the formula $D=12 x+10 y$, with the conditions
, $y$ be the car's speed (in km/h) on the way from $A$ to $B$, $2y$ be the car's speed on the way from $B$ to $A$, and $t$ be the time (in hours) the car spends traveling from $A$ to $B$.
$\left\{\begin{aligned} y t & =12, \\ x(t+0.2) & =12-0.4 y, \\ t & <1,\end{ali... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. How many roots does the equation
$\sqrt[3]{|x|}+10[x]=10 x$ ? ( $[x]$ - the integer part of the number $x$, i.e., $[x] \in Z,[x] \leq x<[x]+1$).
(5 points) | # Solution:
$\frac{\sqrt[3]{|x|}}{10}=\{x\},\{x\}=x-[x]$ Since $\{x\} \in[0 ; 1)$, then $x \in(-1000 ; 1000)$.
On the interval $[0 ; 1)$, the equation has 2 roots
$\sqrt[3]{x}=10 x, x=1000... | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can seven different items (3 weighing 2 tons each, 4 weighing 1 ton each) be loaded into two trucks with capacities of 6 tons and 5 tons, if the arrangement of the items inside the trucks does not matter? (12 points)
# | # Solution.
Solution. The load can be distributed as $6+4$ or $5+5$.
| 6 tons | 4 tons | number of ways | 5 tons | 5 tons | number of ways |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $2+1+1+1+1$ | $2+2$ | $C_{3}^{1}=3$ | $2+1+1+1$ | $2+2+1$ | $C_{3}^{1} \cdot C_{4}^{3}=12$ |
| $2+2+1+1$ | $2+1+1$ | $C_{3}^{... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The numbers from 100 to 999 are written without spaces. What is the remainder when the resulting 2700-digit number is divided by 7? | Solution.
Since 1001 is divisible by 7, we get $1000 \equiv-1(\bmod 7) \Longrightarrow 1000^{n} \equiv(-1)^{n}(\bmod 7)$. Therefore, the given number is congruent modulo 7 to the number
$$
999-998+997-996+\ldots+101-100=450 \equiv 2(\bmod 7)
$$
Answer: 2 | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, with an area of $180 \sqrt{3}$, the angle bisector $A D$ and the altitude $A H$ are drawn. A circle with radius $\frac{105 \sqrt{3}}{4}$ and center lying on line $B C$ passes through points $A$ and $D$. Find the radius of the circumcircle of triangle $A B C$, if $B H^{2}-H C^{2}=768$.
# | # Solution.
Let $B C=a, A C=b, A B=c, \angle B A D=\angle C A D=\alpha, \angle A D C=\beta$, and the radius of the given circle is $r$. Note that $\beta>\alpha$ (as the external angle of tri... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Svetlana takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet on the 1580th step of applying this rule, if the initial triplet of numbers was $\{80 ; 71... | # Solution:
Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b . B=80-20=60$.
Answer: 60 . | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two balls, the sizes of which can be neglected in this problem, move along a circle. When moving in the same direction, they meet every 20 seconds, and when moving in opposite directions - every 4 seconds. It is known that when moving towards each other along the circle, the distance between the approaching balls de... | # Solution:
Let the speed of the faster ball be $v$, and the slower one be $u$. When moving in the same direction, the faster ball catches up with the slower one when the difference in the distances they have traveled equals the length of the circle. According to the problem, we set up a system of two linear equations... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C$, side $B C$ is 19 cm. The perpendicular $D F$, drawn from the midpoint of side $A B$ - point $D$, intersects side $B C$ at point $F$. Find the perimeter of triangle $A F C$, if side $A C$ is $10 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 29 \text{ cm}$.
Answer: 29 cm. | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Sasha bought pencils in the store for 13 rubles each and pens for 20 rubles each, in total he paid 350 rubles. How many items of pencils and pens did Sasha buy in total?
# | # Solution.
Let $x$ be the number of pencils, $y$ be the number of pens. We get the equation $13 x+20 y=355$
$13(x+y)+7 y=355$, let $x+y=t(1)$
$13 t+7 y=355$
$7(t+y)+6 t=355$, let $t+y=k(2)$
$7 k+6 t=355$
$6(k+t)+k=355$, let $k+t=n(3)$
$6 n+k=355$
$k=355-6 n$. Substitute into (3), $t=7 n-355$
Substitute into (... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A boy wrote the first twenty natural numbers on a piece of paper. He did not like how one of them was written and crossed it out. It turned out that among the 19 remaining numbers, there is a number equal to the arithmetic mean of these 19 numbers. Which number did he cross out? If the problem has more than one solu... | # Solution:
The sum of the numbers on the sheet, initially equal to $1+2+3+\ldots+20=210$ and reduced by the crossed-out number, is within the range from 210-20=190 to 210-1=209. Moreover, it is a multiple of 19, as it is 19 times one of the addends. Since among the numbers $190,191,192, \ldots 209$ only 190 and 209 a... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A family of beekeepers brought containers of honey to the fair with volumes of $13,15,16,17,19,21$ liters. In August, three containers were sold in full, and in September, two more, and it turned out that in August they sold twice as much honey as in September. Determine which containers were emptied in August. In y... | # Solution:
A total of $13+15+16+17+19+21=101$ liters of honey were brought. The amount of honey sold is divisible by three. Therefore, the volume of the unsold container must give a remainder of 2 when divided by 3 (the same as 101), i.e., 17 liters. Thus, 101-17=84 liters were sold, with one-third of 84 liters, or 2... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find $\angle C_{1} B_{1} A_{1}$. | # Solution:
Extend side $A B$ beyond point $\mathrm{B}$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, which means it is equidistant from its sides.
We get that $A... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-4|=p(x)$, where $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, $p(x)=a$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, we get $f(x)=|2 x-2|$, where $x \neq-3, x \neq 2$.
Solve the equation || $2 x-2|-4|=a$, where $x \neq-3, x \neq 2$ graphically in the system $x O a$.
The equation has three solutions when $a=2$.
The product is 2.
 be Masha's labor productivity, $y$ be Dasha's, and $z$ be Sasha's. Since the labor productivity adds up when working together, we can set up a system of equations based on the problem's conditions: $\left\{\begin{array}{l}(x+y) \cdot 7.5=1 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are marked respectively such that $\angle CMA = \angle ANC$. Segments $MC$ and $AN$ intersect at point $O$, and $ON = OM$. Find $BC$, if $AM = 5 \, \text{cm}, BM = 3 \, \text{cm}$. | # Solution:
Triangles $A M O$ and $C N O$ are congruent ($\angle A M O=\angle C N O, O N=O M, \angle M O A=\angle N O C$, as vertical angles). From the congruence of the triangles, it follow... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. If a two-digit number is decreased by 36, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting number sequence.
# | # Solution.
$\overline{x y}=10 x+y-$ the original two-digit number, then $\overline{y x}=10 y+x-$ the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+36$
From the equation, it is clear that the two-digit number is greater than 36. Let's start the investigation with the ten... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 7 points, if he hit the bullseye 4 times, and the results of the other hits were sevens, eights, and nines? There were no misses at all. | # Solution:
Since the soldier scored 90 points and 40 of them were scored in 4 shots, he scored 50 points with the remaining 6 shots. As the soldier only hit the 7, 8, and 9, let's assume that in three shots (once each in 7, 8, and 9), he scored 24 points. Then, for the remaining three shots, he scored 26 points, whic... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Boys are dividing the catch. The first one took $r$ fish and a seventh part of the remainder; the second - $2 r$ fish and a seventh part of the new remainder; the third - $3 r$ fish and a seventh part of the new remainder, and so on. It turned out that in this way all the caught fish were divided equally. How many b... | # Solution:
Let $x$ be the number of boys; $y$ be the number of fish each received. Then the last boy took $x r$ fish (there could be no remainder, otherwise there would not have been an even distribution), so $y=x r$. The second-to-last boy took $(x-1) r+\frac{x r}{6}=y$; i.e., $x r$ is $\frac{6}{7}$ of the second-to... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=5, A E=2$. | # Solution:
Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A$, $\angle C E B=\angle C A F$. Considering that $B C=C E$, we get that triangle $FCA$ ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $\mathrm{O}$ is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $O_{1} O_{2}$ an... | # Solution:
We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \)... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Task: On an $8 \times 8$ chessboard, 64 checkers numbered from 1 to 64 are placed. 64 students take turns approaching the board and flipping only those checkers whose numbers are divisible by the ordinal number of the current student. A "Queen" is a checker that has been flipped an odd number of times. How many "Que... | Solution: It is clear that each checker is flipped as many times as the number of divisors of its number. Therefore, the number of "queens" will be the number of numbers from 1 to 64 that have an odd number of divisors, and this property is only possessed by perfect squares. That is, the numbers of the "queens" remaini... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. If $a-$ is the first term and $d-$ is the common difference of an arithmetic progression, $\left\{\begin{array}{l}a+16 d=52, \\ a+29 d=13\end{array} \Leftrightarrow d=-3, a=100\right.$.
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ reaches its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. ... | Answer: 1717
Consider the equation of the system $\sqrt{2} \cos \frac{\pi y}{8}=\sqrt{1+2 \cos ^{2} \frac{\pi y}{8} \cos \frac{\pi x}{4}-\cos \frac{\pi x}{4}}$. Given the condition $\sqrt{... | 1717 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. One worker in two hours makes 5 more parts than the other, and accordingly spends 2 hours less to manufacture 100 parts. How much time does each worker spend on manufacturing 100 parts?
# | # Solution:
Let the second worker make 100 parts in $x$ hours, and the first worker in $x-2$ hours.
$\frac{100}{x-2}-\frac{100}{x}=\frac{5}{2} ; \frac{40}{x-2}-\frac{40}{x}=1 ; x^{2}-2 x-80=0, x=1+9=10$. Answer: 8 and 10 hours. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 24$, passing through the point $M(4 ;-2 \sqrt{3})$.
# | # Solution:
$$
y=\frac{x^{2}}{8 \sqrt{3}}, M(4 ;-2 \sqrt{3}) \cdot y=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(x-x_{0}\right) ;-2 \sqrt{3}=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(4-x_{0}\right) ;
$$
$x_{0}^{2}-8 x_{0}-48=0 ; x_{0}=4 \pm 8 ;\left(x_{0}\right)_{1}=12,\left(x_{0}\righ... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is the median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the radius of the circumcircle of triangle $A B C$ if the length of segment $D E$ is $4(\sqrt{2}-1)$. | # Solution:
By the property of tangents to a circle, we have:
$$
A G=A K=x, C G=C D=y, C E=C F=z, B F=B H=u, D M=\frac{A B}{2}-A K=\frac{A B}{2}-x
$$
$M E=\frac{A B}{2}-B H=\frac{A B}{2}-u$,
Then $D E=z-y, D E=D M-M E=u-x$. Therefore, $2 D E=z-y+u-x=C B-$ AC. Let $C B=a, A C=b$. Then $a-b=8(\sqrt{2}-1)$. By the Law... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 6$, passing through the point $M(1 ;-\sqrt{3} / 2)$. | Solution:
$$
y=\frac{x^{2}}{2 \sqrt{3}}, M\left(1 ;-\frac{\sqrt{3}}{2}\right) \cdot y=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(x-x_{0}\right) ;-\frac{\sqrt{3}}{2}=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(1-x_{0}\right) ; x_{0}^{2}-2 x_{0}-3=0
$$
$; x_{0}=1 \pm 2 ;\left(x_{0}\right)_{1}... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
1. One tourist covers a distance of 20 km 2.5 hours faster than the other. If the first tourist reduced their speed by 2 km/h, and the second increased their speed by 50%, they would spend the same amount of time on the same distance. Find the speeds of the tourists. | Solution: Let $\mathrm{x}$ be the speed of the first tourist, and $y$ be the speed of the second tourist.
\[
\frac{20}{x}+\frac{5}{2}=\frac{20}{y}, \quad x-2=1.5 y, \quad \frac{4}{x}+\frac{1}{2}=\frac{6}{x-2}, \quad x^{2}-6 x-16=0, \quad x=8, \quad y=4
\]
Answer: $8 \mathrm{km} / \mathrm{u}, 4 \mathrm{km} / \mathrm{q... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Since $\mathrm{x} \leq \mathrm{P}$ and Р:х, let $2 \mathrm{x} \leq \mathrm{P} => \mathrm{x} \leq \mathrm{P} / 2$, and also $\mathrm{y} \leq \mathrm{P} ; \kappa \leq \mathrm{P} ; \mathrm{e} \leq \mathrm{P} ; =>$
$$
\mathrm{x}+\mathrm{y}+\mathrm{K}+\mathrm{e} \leq 3.5 \mathrm{P} ; => \mathrm{P} \geq \frac{2}{7}(\math... | Answer: 2 l
## №6: Planimetry. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. (10 points) In a row, 2018 digits are written consecutively. It is known that in this row, every two-digit number formed by two adjacent digits (in the order they are written) is divisible by 17 or 23. The last digit in this row is 5. What is the first digit in the row? Provide a justified answer.
# | # Solution:
All two-digit numbers divisible by 17 or 23:
$$
\begin{aligned}
& 17,34,51,68,85 \\
& 23,46,69,92
\end{aligned}
$$
The following diagram shows with arrows which digit can follow which in the row:
$$
\begin{aligned}
& 1 \rightarrow 7 \quad 9 \rightarrow 2 \rightarrow 3 \\
& \uparrow \quad \uparrow \\
& 5... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 5. Solution:
Let $x$ units of distance/hour be the speed of the bus, $y$ units of distance/hour be the speed of the tractor, and $S$ be the length of the path AB. Then the speed of the truck is $-2y$ units of distance/hour. We can set up a system of equations and inequalities:
$$
\left\{\begin{array}{c}
\frac{s}{x}... | Answer: 17 hours 45 minutes. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. (Option 2). Given an isosceles triangle $ABC (AB=BC)$ on the lateral side $BC$, points $M$ and $N$ are marked (M lies between B and $N$) such that $AN=MN$ and $\angle BAM = \angle NAC$. $MF$ is the distance from point M to the line $AC$. Find $\angle AMF$. | Solution: Let $\angle$ BAM
$=\angle \mathrm{NAC}=\alpha, \angle \mathrm{MAN}=\angle \mathrm{AMN}=\beta \prec=\angle \mathrm{MAC}=\alpha+\beta$ and $\angle \mathrm{MCA}=2 \alpha+\beta=\succ(\square \mathrm{AMC})$
$2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=\succ \alpha+\beta=60^{\circ} \Rightarrow \succ \mathrm{MAF}=60... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (Option 2)
Thirty clever students from 6a, 7a, 8a, 9a, and 10a grades were tasked with creating forty problems for the olympiad. Any two classmates came up with the same number of problems, while any two students from different grades came up with a different number of problems. How many people came up with ju... | Solution: 26 classmates solved 1 problem, the 27th person solved 2, the 28th solved 3, the 29th solved 4, and the 30th solved 5. This solution is immediately apparent. Let's prove that it cannot be otherwise.
Let $x$ be the number of people who solved one problem, $y$ be the number who solved two, $z$ be the number wh... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Two cyclists set off simultaneously from point $A$ to point $B$. When the first cyclist had covered half the distance, the second cyclist had 24 km left to travel, and when the second cyclist had covered half the distance, the first cyclist had 15 km left to travel. Find the distance between points $A$ and $B$. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}}$ and $\frac{s-15}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24$; $s^{2}-52 s+480=0 ;... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the greatest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=327$ and the sum $S_{57}=57$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$\left\{\begin{array}{l}\frac{a+a+2 d}{2} \cdot 3=327, \\ \frac{a+a+56 d}{2} \cdot 57=57\end{array} \Leftrightarrow\left\{\begin{array}{l}a+d=109, \\ a+28 d=1\end{array} \Rightarrow 27 d=-108 ; d=-4, a=113\right.\ri... | 1653 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Rectangle $ABCD$ with sides $AB=1$ and $AD=10$ serves as the base of pyramid $SABCD$, and the edge $SA=4$ is perpendicular to the base. Find a point $M$ on edge $AD$ such that triangle $SMC$ has the smallest perimeter. Find the area of this triangle.
# | # Solution:
Let's lay off $A T=A S$ on the extension of edge $A B$. For any position of point $M$ on side $A D$, $T M=S M$, so the minimum value of the sum $S M+M C$ will be when $M=T C \cap A D$.
Let's introduce the notation
$A B=a, A D=b, A S=c, A M=x$. From
$\triangle T A M \sim \triangle C D M$ it follows that
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A pedestrian left point $A$ for point $B$. When he had walked 8 km, a second pedestrian set off from point $A$ after him. When the second pedestrian had walked 15 km, the first was halfway through his journey, and both pedestrians arrived at point $B$ at the same time. What is the distance between points $A$ and $B$... | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the pedestrians. Then,
$\frac{s-8}{v_{1}}=\frac{s}{v_{2}}$ and $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$. From this, $\frac{s}{2(s-8)}=\frac{s-15}{s} ; s^{2}=2 s^{2}-46 s+240 ; s^{2}-46 s+240=0$;
$s_{1,2}=23 \pm 17 \cdot... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array}... | -442 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. (Option 1)
Calculate $x^{3}+3 x$, where $x=\sqrt[3]{2+\sqrt{5}}-\frac{1}{\sqrt[3]{2+\sqrt{5}}}$. | Solution: Let $\sqrt[3]{2+\sqrt{5}}=a$, then $x=a-\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}+3 x=\left(a-\frac{1}{a}\right)^{3}+3\left(a-\frac{1}{a}\right)=a^{3}-\frac{1}{a^{3}}=2+\sqrt{5}-\frac{1}{2+\sqrt{5}}=\frac{(2+\sqrt{5})^{2}-1}{2+\sqrt{5}}= \\
& =\frac{8+4 \sqrt{5}}{2+\sqrt{5}}=4
\end{aligned}
$$
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. (Option 2)
Calculate $x^{3}-3 x$, where $x=\sqrt[3]{7+4 \sqrt{3}}+\frac{1}{\sqrt[3]{7+4 \sqrt{3}}}$. | Solution: Let $\sqrt[3]{7+4 \sqrt{3}}=a$, then $x=a+\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}-3 x=\left(a+\frac{1}{a}\right)^{3}-3\left(a+\frac{1}{a}\right)=a^{3}+\frac{1}{a^{3}}=7+4 \sqrt{3}+\frac{1}{7+4 \sqrt{3}}=\frac{(7+4 \sqrt{3})^{2}+1}{7+4 \sqrt{3}}= \\
& =\frac{98+56 \sqrt{3}}{7+4 \sqrt{3}}=14
\end{aligned}
$$
... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (Option 1)
Find the sum of the squares of the roots of the equation $\left(x^{2}+4 x\right)^{2}-2016\left(x^{2}+4 x\right)+2017=0$. | Solution: Let's make the substitution: $x^{2}+4 x+4=t$, then $x^{2}+4 x=t-4$ and the equation will take the form:
$(t-4)^{2}-2016(t-4)+2017=0$
$t^{2}-2024 t+10097=0$
The discriminant of the equation is greater than zero, therefore, the equation has two roots. By Vieta's theorem: $t_{1}+t_{2}=2024, t_{1} \cdot t_{2}=... | 4064 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (Option 2)
Find the sum of the squares of the roots of the equation $\left(x^{2}+6 x\right)^{2}-1580\left(x^{2}+6 x\right)+1581=0$. | Solution: Let's make the substitution: $x^{2}+6 x+9=t$, then $x^{2}+6 x=t-9$ and the equation will take the form:
$$
\begin{aligned}
& (t-9)^{2}-1580(t-9)+1581=0 \\
& t^{2}-1598 t+15882=0
\end{aligned}
$$
The discriminant of the equation is greater than zero, so the equation has two roots. By Vieta's theorem, $t_{1}+... | 3232 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Seeing a fox several meters away, the dog chased after it along a straight dirt road. The dog's jump is $23 \%$ longer than the fox's jump. There is a time interval during which both the fox and the dog make a whole number of jumps. Each time it turns out that the dog manages to make $t \%$ fewer jumps than the fox,... | Solution: Let $\mathrm{x}$ be the length of the fox's jump, and $y$ be the number of jumps it makes in some unit of time. Then $xy$ is the distance the fox covers in this time. The distance covered by the dog in the same time is $1.23 x\left(1-\frac{t}{100}\right) y$. The fox will escape from the dog if $1.23 x\left(1-... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=150, d=0.5$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
## Find $\angle C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $\mathrm{B}$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and is therefore equidistant from its sides, we get that $A_{... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. What is the minimum value that the function $F(x ; y)=x^{2}+8 y+y^{2}+14 x-6$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.
# | # Solution.
$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=$ M, then $(x+7)^{2}+(y+4)^{2}=(M+71)$ - this is a circle with center $(-7 ;-4)$ and radius $(M+71)^{0.5}$.
Since the center of the second circle lies outside the first, ... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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