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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
$$
\left\{\begin{array} { l }
{ y - x = 7 , } \\
{ 5 x - y = 1 , \text { or } }
\end{array} \left\{\begin{array} { l }
{ y - x = - 7 , } \\
{ 5 x - y = - 1 , \text { or } }
\end{array} \left\{\begin{array} { l }
{ y - x = 1 , } \\
{ 5 x - y = 7 , \text { or } }
\end{array} \quad \left\{\begin{array}{l}
y-x=-1 \\
5 x-y=-7
\end{array}\right.\right.\right.\right.
$$
the second and fourth systems do not have natural solutions. The solution to the first system is the pair $(2 ; 9)$, and the third is $(2 ; 3)$. Therefore, $x+y=11$ or $x+y=5$. The smallest natural number that has 5 and 11 as divisors is 55.
Answer: 55.
|
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 points)
|
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, $A O=2, C O=2 \sqrt{2}$. Let $O E=x, A E=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=4, \\ (\sqrt{2}+\sqrt{6}-y)^{2}+x^{2}=8 .\end{array}\right.$ Solving the system, we get $y=\sqrt{2}, x=\sqrt{2}$. Then $\quad \angle D A C=\angle B C A=45^{\circ}, \quad B C=2 \sqrt{3}$,
$\angle F C A=\operatorname{arctg} \frac{1}{\sqrt{3}}=30^{\circ}, \quad \angle A B E=30^{\circ}, \quad A B=2 \sqrt{2}$. $\angle D F E=90^{\circ}, D E=A B \cos 45^{\circ}=2 . \quad R_{o n .}=D E / 2=1$.
## Answer: 1.
|
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 $16 \mathrm{~km} / \mathrm{u}$, and after $3 / 8 \mathrm{~h}$ after the turn, arrived at point $A$ simultaneously with the car. Find the entire distance (in km) traveled by the motorcyclist, if this distance is 5.25 km shorter than the entire track. Provide the answer as a number without specifying the unit.
(5 points)
|
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 $\frac{S}{x} = \frac{3 S + 5.25}{y}$, $\frac{3 x}{8} + 6 = S$, $\frac{3 y}{8} = S + 5.25$. This leads to the quadratic equation $4 S^2 - 36 S - 63 = 0$, whose positive solution is $S = 10.5$, and the entire distance traveled by the motorcyclist is $2 S = 21$.
Answer: 21.
|
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 true for $0,1,2, \ldots, k$. Now let $3^{k}<N \leq 3^{k+1}$. Place no less than $N / 3$ coins, but no more than $3^{k}$ coins, on the left pan, and the same number on the right pan. If the left pan is lighter, then the lighter coin is on it; if the right pan is lighter, then the lighter coin is on it; if the scales are balanced, then the lighter coin is among the remaining coins, the number of which is less than or equal to $N / 3 \leq 3^{k}$. As a result, we need to find the lighter coin among no more than $3^{k}$ coins, and it will take no more than $k$ additional weighings. Since $3^{6}<2019 \leq 3^{7}$, the number of weighings $k$ is 7.
Answer: 7.
|
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
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t \in\{123,124,125,126\} \Rightarrow\right.$
$(58,1),(41,6),(24,11),(7,16)$
$59+47+35+23=164$
Answer: 164.
|
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 admitted to the institute. How many were admitted?
|
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 solution |
| 5 points | An arithmetic error or insufficiently justified solution in an otherwise justified solution. |
| 0 points | Any other situation |
|
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 $\frac{A O}{O C}=\frac{B O}{O D}$. Additionally, $\angle A O B=\angle D O C$ as vertical angles. Therefore, triangles $A O B$ and $C O D$ are similar. $\frac{A B}{B C}=\frac{A O}{O C}=\frac{2}{3}$, which means $A O=8$.
Answer: 8.
Criteria:
| Points | Conditions for awarding |
| :---: | :--- |
| 15 points | Complete, justified solution |
| 12 points | The solution approach is correct, but a computational error has been made. |
| 10 points | Similarity of triangles $A O B$ and $C O D$ is proven, but further actions are not performed or are performed incorrectly. |
| 5 points | The theorem on the ratio of areas of triangles with one equal angle is correctly applied, but further actions are not performed or are performed incorrectly. |
| 0 points | The solution does not meet the criteria listed above |
|
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 with increased seismic activity are those in which the predictions did not match reality?
|
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 $0.64x - 0.56x = 0.08x$, and $\frac{0.08x}{0.2x} \cdot 100\% = 40\%$.
Answer: $40\%$
## Criteria
| Points | Conditions for Awarding |
| :---: | :--- |
| 10 points | Fully justified solution |
| 5 points | An arithmetic error was made with correct reasoning or insufficiently justified solution |
| 0 points | Incorrect reasoning or only the answer is written. |
|
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 car arrived at point B, immediately turned around, and after traveling 2 km, met the first car. Find the distance between points A and B. Provide the answer as a number without specifying the unit.
|
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 $a^{2}-15 a-(t-1)(t+2)(t+5)(t+8)a^{2}-15 a-35,\left(t^{2}+7 t-8\right)\left(t^{2}+7 t+10\right)>a^{2}-15 a-35$
$z=t^{2}+7 t+1,(z-9)(z+9)>a^{2}-15 a-35, z^{2}>a^{2}-15 a+46$, $0>a^{2}-15 a+46, \sqrt{D}=\sqrt{41},(15-\sqrt{41}) / 2<a<(15-\sqrt{41}) / 2, \Rightarrow a=10$.
Answer: 10.
|
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 $x+y+z=2, x y+x z+y z=-9, x y z=1$.
$$
\begin{aligned}
& x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}=(x y+x z+y z)^{2}-2\left(x^{2} y z+y^{2} x z+z^{2} x y\right) \\
& =(x y+x z+y z)^{2}-2 x y z(x+y+z)=81-2 * 1 * 2=77
\end{aligned}
$$
Answer: 77.
|
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.
$A C: \quad y=k x+d, \quad M \in A C \Rightarrow d=2-k$
$A(a ; 3 a-3) \in A C \Rightarrow 3 a-3=k a+2-k \Rightarrow a=\frac{5-k}{3-k}$,
$C(-1 ; c) \in A C \Rightarrow c=-2 k+2$,
$S_{A B C}=\frac{1}{2}(c+6) \cdot(a+1)=\frac{2(k-4)^{2}}{3-k}$
$S^{\prime}=\frac{2(k-4)(2-k)}{(3-k)^{2}}=0, k_{\min }=2, \quad a_{\text {min }}=3$.
Answer: 3.
|
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 ${ }^{x}$.
$$
\left\{\begin{array} { l }
{ \frac { y } { a - \sqrt { x } - 1 } = 4 } \\
{ y = \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ x \geq 0 } \\
{ \sqrt { x } \neq a - 1 } \\
{ \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } = 4 ( a - \sqrt { x } - 1 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \geq 0 \\
\sqrt{x} \neq a-1 \\
\sqrt{x}+5=4(a-\sqrt{x}-1)(\sqrt{x}+1)
\end{array}\right.\right.\right.
$$
First, let's check for which values of the parameter the case $\sqrt{x}=a-1$ is possible. $4(a-1)^{2}+(9-4 a)(a-1)+(9-4 a)=0 \Rightarrow 4(a-1)^{2}+9 a-4 a^{2}=0 \Rightarrow a=-4$.
For this value of the parameter, we get a root equal to a negative number, which is impossible.
We solve the quadratic equation $4 x+(9-4 a) \sqrt{x}+(9-4 a)=0$.
$D=(4 a-9)(4 a+7) \Rightarrow \sqrt{x}_{1,2}=\frac{4 a-9 \pm \sqrt{(4 a-9)(4 a+7)}}{8}$.
The only non-negative solution will be under the conditions
$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
(4 a-9)(4 a+7)=0 \\
\frac{4 a-9}{8} \geq 0
\end{array}\right. \\
\left\{\begin{array} { l }
{ a \in ( - \infty , - 7 / 4 ) \cup ( 9 / 4 , + \infty ) } \\
{ [ \begin{array} { l }
{ \frac { 9 - 4 a } { 4 } < 0 }
\end{array} } \\
{ \{ \begin{array} { l }
{ x _ { 1 } x _ { 2 } = \frac { 9 - 4 a } { 4 } = 0 } \\
{ x _ { 1 } + x _ { 2 } = \frac { 4 a - 9 } { 4 } < 0 }
\end{array} }
\end{array} \Rightarrow \left[\begin{array}{l}
a=9 / 4 \\
a \in(9 / 4,+\infty)
\end{array}\right.\right.
\end{array}\right.
$$
Choosing the smallest integer value of the parameter, we get $a=3$.
## Answer: 3.
|
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 to the base plane at an angle of $30^{\circ}$. Find the volume of the pyramid TAKND, where points $\mathrm{K}$ and $\mathrm{N}$ are the midpoints of edges $\mathrm{TB}$ and $\mathrm{TC}$, respectively, and AD is the larger base of the trapezoid ABCD.
|
# 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 $S_{MB CP}=7x, S_{\text{AMPD}}=13x, \quad \frac{7}{13}=\frac{b+5\sqrt{3}}{a+5\sqrt{3}}, a+b=10\sqrt{3}$, $a=8\sqrt{3}, b=2\sqrt{3}$. Since a circle can be inscribed in the trapezoid $ABCD$, then $AB+CD=a+b, \quad AB=CD=5\sqrt{3}$. We calculate the height of the trapezoid $h=\sqrt{AB^2-(a-b)^2/4}=4\sqrt{3}$. Draw a line through point $O$ perpendicular to the bases of the trapezoid and intersecting these bases at points $Q$ and $R, OR=h$. Since the lateral faces are inclined to the base plane at an angle of $30^\circ$, the height of the pyramid $TO=\frac{1}{2} QR \tan 30^\circ=2$.
Let $TF$ be the height of the pyramid TAKND, where $TF$ is the perpendicular dropped from point $T$ to the line $QS$, and $S$ is the midpoint of $KN$. We calculate the volume of the pyramid TAKND: $V_{\text{TAKND}}=\frac{1}{3} \cdot \frac{AD+KN}{2} \cdot QS \cdot TF$. The area of triangle $TQS$ can be calculated in two ways: $S_{\mathrm{TQS}}=\frac{QR \cdot TO}{4}, S_{\mathrm{TQS}}=\frac{QS \cdot TF}{2}, QS \cdot TF=\frac{QR \cdot TO}{2}$, $V_{\text{TAKND}}=\frac{1}{6} \cdot \frac{AD+KN}{2} \cdot QR \cdot TO$. From this, we get $V_{\text{TAKND}}=18$.
Answer: 18.
## First (online) stage of the academic competition "Step into the Future" for school students in the subject "Mathematics", autumn 2019, 11th grade
#
|
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 in the remaining buses. How many schoolchildren were in the group, given that no more than 18 buses were allocated for transporting the schoolchildren, and no more than 36 people can fit in each bus? Provide the answer as a number without specifying the unit.
$(2$ points)
|
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 $\mathrm{n}$ and $\mathrm{m}$, we get the only possible case: $m=27, \quad n=6, \quad S=135$.
Answer: 135.
|
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)
$\left\{\begin{array}{c}x-3 y=-11 \\ x+2 y=-1,\end{array} \Leftrightarrow\left\{\begin{array}{c}y=2 \\ x=-5\end{array}\right.\right.$
$\left\{\begin{array}{c}x-3 y=1, \\ x+2 y=11,\end{array} \Leftrightarrow\left\{\begin{array}{l}y=2 \\ x=7\end{array}\right.\right.$
$\left\{\begin{array}{c}x-3 y=-1 \\ x+2 y=-11\end{array} \Leftrightarrow\left\{\begin{array}{l}y=-2 \\ x=-7\end{array}\right.\right.$
Answer: 8.
$$
\begin{aligned}
& \text { 3. Let } \\
& g(x)=\frac{2}{x^{2}-8 x+17} . \text { Find all possible } \\
& a^{2}+6 a+\frac{727}{145} \leq g\left(g^{4}(x)\right) \leq 10 a^{2}+29 a+2
\end{aligned}
$$
|
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 points $\mathrm{A}(0 ; 4)$ and $\mathrm{B}(0 ;-4)$. You should start drawing the line from one of these points and finish at the other (2 options). Next, there are 3! ways to choose the order in which to pass the left arc, the right arc, and the diameter. When passing the diameter, there are also 3! ways to choose the order in which to pass the arcs and the diameter of the inner circle. In total, the number of ways is $2 * 3! * 3! = 72$.
Answer: 72.
|
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 that \( n \) is divisible by 6 with a remainder of 1 or 5, and \( n \) should not be divisible by 5. There are eight such numbers: 11, 13, 17, 19, 23, 29, 31, 37.
2) (A) is satisfied, (B) is not, and (C) is satisfied. Then \( n \) is divisible by 6 with a remainder of 3, and \( n \) is divisible by 5. There is one such number: 15.
3) (A) is not satisfied, (B) and (C) are satisfied. Then \( n \) is divisible by 6 with a remainder of 2 or 4, and \( n \) is divisible by 5. There are two such numbers: 10 and 20.
In total: \( 3 * (8 + 1 + 2) = 33 \) numbers.
Answer: 33.
|
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
$$
y^{2}-y(2 a-1)+a^{2}-2=0, \quad D=(2 a-1)^{2}-4\left(a^{2}-2\right)=9-4 a
$$
For the existence of a solution, the following conditions must be satisfied
$$
\left\{\begin{array} { l }
{ D = 9 - 4 a \geq 0 } \\
{ [ \begin{array} { l }
{ f ( 1 ) = a ^ { 2 } - 2 a > 0 } \\
{ \frac { 2 a - 1 } { 2 } > 1 } \\
{ f ( 1 ) = a ^ { 2 } - 2 a \leq 0 }
\end{array} }
\end{array} \Rightarrow \left\{\begin{array}{l}
a \leq 9 / 4 \\
{\left[\begin{array}{l}
\left\{\begin{array}{l}
a-2)>0 \\
2 a-1>2
\end{array}\right. \\
a(a-2) \leq 0
\end{array}\right.}
\end{array} \Rightarrow a \in[0,9 / 4]\right.\right.
$$
Summing the integer values of the parameter, we get $0+1+2=3$.
Answer: 3.
|
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 the base plane at an angle of $30^{\circ}$. Find the volume of the pyramid SAKND, where points $\mathrm{K}$ and $\mathrm{N}$ are the midpoints of edges ТВ and $\mathrm{TC}$ respectively, and point $\mathrm{S}$ lies on edge $\mathrm{TD}$, such that $\mathrm{TS}: \mathrm{SD}=1: 2$.
|
# 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
$S_{\text {MBCP }}=5 x, S_{\text {AMPD }}=7 x, \quad \frac{5}{7}=\frac{b+(a+b) / 2}{a+(a+b) / 2}=\frac{3 b+12 \sqrt{3}}{b+36 \sqrt{3}}, \quad b=6 \sqrt{3}$. Since a circle can be inscribed in trapezoid ABCD, then $A B+C D=a+b, \quad A B=C D=19 \sqrt{3}$. We calculate the height of the trapezoid $h=\sqrt{A B^{2}-(a-b)^{2} / 4}=6 \sqrt{6}$. Draw a line through point O perpendicular to the bases of the trapezoid, intersecting these bases at points $\mathrm{Q}$ and $\mathrm{R}, \mathrm{OR}=\mathrm{h}$. Since the lateral faces are inclined to the plane
of the base at an angle of $30^{\circ}$, the height of the pyramid $T O=\frac{1}{2} Q R \operatorname{tg} 30^{\circ}=3 \sqrt{2}$.
Let TF be the height of the pyramid TAKND, TF is the perpendicular dropped from point $\mathrm{T}$ to line $\mathrm{QL}$, where $\mathrm{L}$ is the midpoint of $\mathrm{KN}$. We calculate the volume of the pyramid
SAKND:
$$
V_{S A K N D}=\frac{1}{3} \cdot \frac{A D+K N}{2} \cdot Q S \cdot \frac{2}{3} T F
$$
we can
$S_{\mathrm{TQS}}=\frac{Q R \cdot T O}{4}, S_{\mathrm{TQS}}=\frac{Q S \cdot T F}{2}, \quad Q S \cdot T F=\frac{Q R \cdot T O}{2}$,
$V_{\text {and }} V_{S K N D}=\frac{1}{6} \cdot \frac{A D+K N}{2} \cdot \frac{2}{3} Q R \cdot T O$.
From this, we get $V_{\text {SAKND }}=90$.
Answer: 90.
|
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 the tens digit equal to 6.
| $\boldsymbol{X}$ | equation | $\boldsymbol{y}$ | number |
| :---: | :---: | :---: | :---: |
| 6 | $60+y=10 y+6+54$ | $y=0$ | 60 does not fit the condition |
| 7 | $70+y=10 y+7+54$ | $y=1$ | 71 |
| 8 | $80+y=10 y+8+54$ | $y=2$ | 82 |
| 9 | $90+y=10 y+9+54$ | $y=3$ | 93 |
These could be the numbers $71,82,93$. The median of the series is 82.
Answer: 82.
|
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, he scored 26 points, which is only possible with the unique combination of numbers $7,8,9: 8+9+9=26$. Therefore, the shooter hit the seven once, the eight twice, and the nine three times.
Answer: 3.
|
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 period of time during which both the wolf and the roe deer make a certain whole number of jumps. Each time, it turns out that the wolf has made $t\%$ more jumps than the roe deer. Find the greatest integer value of $\mathrm{t}$, for which the wolf will not be able to catch the roe deer.
|
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 able to catch up with the roe deer if the distance covered by the roe deer over the specified time interval $x y$ - is not less than the distance covered by the wolf over the same time interval: $0.78 x y\left(1+\frac{t}{100}\right)$. We form the inequality: $0.78 x y\left(1+\frac{t}{100}\right) \leq x y$; $1+\frac{t}{100} \leq \frac{50}{39} ; t \leq \frac{1100}{39}$. The maximum value of $t$ that satisfies this inequality, $t=28 \%$. Answer: 28.
|
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 $\{70 ; 61; 20\}$? If the question allows multiple answers, write them without spaces in ascending order.
|
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 $O O_{3}$.
|
Solution. Point $O$ is the intersection point of the angle bisectors of triangle $ABC$, therefore, the bisector $BO$ is perpendicular to the line $O_{1} O_{2}$ (as the bisectors of adjacent angles of the triangle).
Point $O_{3}$, being equidistant from the lines $BA$ and $BC$, lies on $BO$. Therefore, the line $OO_{3}$, which coincides with $BO$, is perpendicular to the line $O_{1} O_{2}$.
Answer: 90 .
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 3 and 4, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AE}$, a point $\mathrm{D}$ is marked such that $\mathrm{AD}: \mathrm{DE}=3: 1$; on $\mathrm{AD}$, a point $\mathrm{F}$ is marked such that $\mathrm{AF}: \mathrm{FD}=2: 1$; on $\mathrm{BD}$, a point $\mathrm{G}$ is marked such that $\mathrm{BG}: \mathrm{GD}=1: 2$. Determine the degree measure of angle $\mathrm{CFG}$.
|
Solution. Construct the height $I H$ such that $G \in I H$ and connect points C and G.
1) $\triangle I G C = \triangle G F H$ - by two legs, since $I C = G H = 2, I G = H F = 1$, therefore $F G = G C$, $\angle I C G = \angle F G H = \alpha$, and $\angle H F G = \angle I G C = 90^{\circ} - \alpha$.
2) $\triangle F G C$ is a right isosceles triangle, since $F G = F D$,
$$
\begin{aligned}
& \angle H G F + \angle F G C + \angle I G C = 180^{\circ} \\
& \angle F G C = 180^{\circ} - \angle H G F - \angle I G C \\
& \angle F G C = 180^{\circ} - \alpha - 90^{\circ} + \alpha = 90^{\circ}
\end{aligned}
$$
Thus, $\angle C F G = 45^{\circ}$.
Answer: 45.
|
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 that point $A_{1}$ lies on the bisector of $\angle B A C$, and thus is equidistant from its sides, we get that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, and therefore lies on the bisector of $\angle B B_{1} C$. Similarly, we prove that $B_{1} C_{1}$ is the bisector of $\angle A B_{1} B$. Therefore, $\angle C_{1} B_{1} A_{1}=90^{\circ}$, as the angle between the bisectors of adjacent angles.
In triangle $B B_{1} C$, point $M$ is the intersection of the bisectors $B_{1} A_{1}$ and $C C_{1}$, and thus $B M$ is also the bisector of $\angle B_{1} B C$, so $\angle B_{1} B M=\angle M B C=30^{\circ}$. Further, $\angle A B M=\angle C_{1} B B_{1}+\angle B_{1} B M=60^{\circ}+30^{\circ}=90^{\circ}$, which means a circle can be circumscribed around quadrilateral $B M B_{1} C_{1}$. Therefore, $\angle B_{1} M C_{1}=\angle B_{1} B C_{1}=60^{\circ}$, as they subtend the same arc.
Answer: 60.
|
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. Determine how many liters of liquid the student poured out the first time, if the tank's volume is 1000 liters.
#
|
# 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 $(1000-x) \cdot \frac{100-p}{100}$ water.
3) After adding $x$ liters of water, there became: 1000 liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100}+x$ liters of water.
4) At the end of all the pouring, there became: 1000 liters of the solution with a syrup content of $\frac{3 p}{25} \%$,
$$
3 p
$$
that is, $1000 \cdot \frac{\overline{25}}{100}=\frac{30 p}{25}$ liters of syrup and $1000-\frac{30 p}{25}$ liters of water.
5) Then before the last addition of $2 x$ liters of water, there was $(1000-2 x)$ liters of the solution in the tank, and in it $\frac{30 p}{25}$ liters of syrup and $1000-\frac{30 p}{25}-2 x$ water. This is the same liquid as in point 3) of the solution, so the ratio of syrup to liquid in it is the same. We set up the equation:
$$
\frac{(1000-x) \cdot \frac{p}{100}}{1000}=\frac{\frac{30 p}{25}}{1000-2 x} \Leftrightarrow 2 x^{2}-3000 x+\frac{22}{25} \cdot 1000^{2}=0 \Leftrightarrow
$$
$\Leftrightarrow x \in\{1100 ; 400\}, 1100$ liters does not satisfy the condition of the problem.
Answer: 400.
## Solution variant № 2
|
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), $x = 8 t - 1998$.
Substitute into (1), $y = 2 \cdot 1998 - 15 t$.
Since $x > 0, y > 0$, then $250 \leq t \leq 266$. To have the smallest number of 8 kg coffee bags, then $t = 266$. Then accordingly $x = 130, y = 6$.
130 and 6 bags, a total of 136 bags.
Answer: 136.
|
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 the cans. To Ivan Ivanovich's surprise, the cans filled up simultaneously. How many times more water does the stronger stream provide compared to the weaker one?
|
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 equation: $\frac{4}{x}=\frac{10-x}{4} ; x^{2}-10 x+16=0$, which has two roots $x_{1}=2, x_{2}=8$. The two roots of the equation correspond to two possibilities: placing the smaller can under the stronger or the weaker stream first. However, in both cases, the answer is the same: one stream provides twice as much water as the other.
Answer. 2.
|
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 that point $A_{1}$ lies on the bisector of $\angle B A C$, and thus is equidistant from its sides, we get that $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, and therefore lies on the bisector of $\angle B B_{1} C$. Thus, $B_{1} A_{1}$ is the bisector of $\angle B B_{1} C$.
In triangle $B B_{1} C$, point $M$ is the intersection of the bisectors $B_{1} A_{1}$ and $C C_{1}$, and therefore $B M$ is also the bisector of $\angle B_{1} B C$, so $\angle B_{1} B M = \angle M B C = 30^{\circ}$.
Answer: 30.
|
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 chosen such that $\mathrm{BM}=\mathrm{MC}$. Determine the degree measure of angle КМТ.
|
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) $\Delta T P M = \Delta K O M$, by two sides and the angle between them, since
$$
\begin{gathered}
A O = \frac{1}{2} A C = K O = P M \\
A P = \frac{1}{2} A B = T P = O M \\
\angle T P M = \angle T P B + \angle B P M = \angle C O K + \angle C O M = \angle K O M
\end{gathered}
$$
Thus, TM = MK, $\angle P T M = \angle K M O$, and $\angle P M T = \angle M K O$.
2) Find the sum of angles $\angle P M T$ and $\angle K M O$:
$$
\angle P M T + \angle K M O = \angle M K O + \angle K M O = 180^{\circ} - \angle M O K
$$
In turn, $\angle M O K = \angle K O C + \angle M O C = 60^{\circ} + \angle B A C$.
3) $\angle K M T = \angle P M O + (\angle P M T + \angle K M O) = \angle B A C + (180^{\circ} - \angle M O K) = \angle K M T = \angle B A C + 180^{\circ} - 60^{\circ} - \angle B A C = 120^{\circ}$.
Answer: 120.
|
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 side $C J$ such that $C B: B J=2: 3$. We obtain equal triangles: $\triangle A K D=\triangle A L H=\triangle H J N$.
Next, $\quad A H=H N, \angle A H N=90^{\circ}$, therefore, $\quad \angle H A N=45^{\circ} . \quad$ From the equality of triangles $M D C$ and $N A B$, we get $\angle N A B=\angle M C D$.
Next, $\angle L A B=\angle L A H+\angle H A N+\angle N A B=90^{\circ}$.
From this, finally, $\angle L A H+\angle N A B=\angle A K D+\angle M C D=45^{\circ}$.
Answer: 45.
|
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 $l_{i} \in\left\{0,1, \ldots, k_{i}\right\}, i=1, \ldots, m$. The number of divisors of the number $n$ is $\left(k_{1}+1\right)\left(k_{2}+1\right) \cdots\left(k_{m}+1\right)=70$. We decompose the number 70 into non-unit factors in all possible ways and choose the smallest number $n$. Since $70=2 \cdot 5 \cdot 7$, we have five cases:
1) $70=70$, the smallest number $n=2^{69}>40000$;
2) $70=35 \cdot 2$, the smallest number $n=2^{34} \cdot 3^{1}>40000$;
3) $70=14 \cdot 5$, the smallest number $n=2^{13} \cdot 3^{4}>40000$;
4) $70=10 \cdot 7$, the smallest number $n=2^{9} \cdot 3^{6}=512 \cdot 81>40000$;
5) $70=7 \cdot 5 \cdot 2$, the smallest number $n=2^{6} \cdot 3^{4} \cdot 5^{1}=25920$.
Answer: 25920.
|
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 - 128. If the initial color of the pixel has a number \( n \leq 19 \), then the student's program recolors it to the color with the number \( n+4 \), and if the initial color of the pixel has a number \( n \geq 20 \), then the pixel is recolored to the color with the number \( |129-2n| \). Initially, the pixel was red. The student applied his program 2019 times sequentially. What color did the pixel end up as a result?
|
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 values: $f(5)=9, f^{[2]}(5)=13, f^{[3]}(5)=17$, $f^{[4]}(5)=21, f^{[5]}(5)=87, f^{[6]}(5)=45, f^{[7]}(5)=39, f^{[8]}(5)=51, f^{[9]}(5)=27, f^{[10]}(5)=75$, $f^{[11]}(5)=21=f^{[4]}(5)$. We have obtained a cycle of length 7 operations. Therefore, for any natural value of $k$ and any $r=0,1, \ldots, 6$, we have $f^{[4+7 k+r]}(5)=f^{[r]}(21)$. Since $2019=4+287 \cdot 7+6$, then $r=6$, and $f^{[2019]}(5)=f^{[6]}(21)=75$. The pixel will be blue. Answer: blue.
|
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-effective as possible, if the freight rate per ton-kilometer by river is half that of the highway?
|
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 requirement. Let this smallest value be denoted by $m$.
We have the equation $x + 2y = m$. But $x = a - DC$, and $DC = \sqrt{y^2 - d^2}$; our equation becomes $a - \sqrt{y^2 - d^2} + 2y = m \quad \Rightarrow \quad 3y^2 - 4(m-a)y + (m-a)^2 + d^2 = 0$. $y = \frac{2}{3}(m-a) \pm \frac{\sqrt{(m-a)^2 - 3d^2}}{3}$
For $y$ to be real, $(m-a)^2$ must be at least $3d^2$.
The smallest value of $(m-a)^2$ is $3d^2$, and then $m - a = d\sqrt{3}$, $y = \frac{2(m-a) + 0}{3} = \frac{2d\sqrt{3}}{3} ; \sin \angle BDC = \frac{d}{y} = d : \frac{2d\sqrt{3}}{3} = \frac{\sqrt{3}}{2}$.
The angle whose sine is $\sqrt{3} / 2$ is $60^\circ$. Therefore, the highway should be built at an angle of $60^\circ$ to the river, regardless of the distance $AC$.
The solution only makes sense under certain conditions. If the point is located such that the highway, built at an angle of $60^\circ$ to the river, passes on the other side of city $A$, then the solution is invalid. In such a case, the point $B$ should be directly connected to city $A$ by a highway, without using the river for transportation at all.
|
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" only once when the question is about the district where they live). However, if every resident were a liar, the number of "yes" answers would be exactly twice the number of residents in the city (since each liar resident says "yes" exactly twice, when the question is about the districts where they do not live). Therefore, the number of liars in Kashino is the difference between the number of "yes" answers and the number of residents in the state, i.e., 173-120=53. After Doctor Aibolit visited the Mankino district, all residents of the district started telling the truth, so if they previously answered "yes" to the question "Do you live in Ovsyanikino?", they would now answer "no". Therefore, the number of liars in the Mankino district is the difference between the number of "yes" answers in the first and second surveys (i.e., 59-34=25). Similarly, the number of liars in Grechkin is $59-43=16$. Therefore, the total number of liars in the Mankino and Grechkin districts is $25+16=41$. Since there are 53 liars in total in the city, 53-41=12 liars live in the Ovsyanikino district. Answer: 12 liars
## Criteria for checking tasks
|
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 $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)=m^{2}+2 m+1-4(m-1)=m^{2}-2 m+5=(m-1)^{2}+4$
From which $y=(m-1)^{2}+4$ and $y_{\text {min }}=4$ when $m=1$.
|
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 is an error in the transformation of the expression. |
| 5 points | Vieta's theorem (considering D) is correctly written. |
| 0 points | Other solutions that do not meet the above criteria. |
|
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 $b$: $a-b=32$.
|
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 points | The solution does not meet any of the criteria listed above |
|
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 values being reached at $\mathrm{x}=3$. The value of the left side of the equation does not exceed 7, while the right side is no less than 7. Therefore, equality is possible only when both sides of the equation are simultaneously equal to 7, which is achieved at $x=3$.
Answer: 3
|
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) $2 a-\frac{\left(-4 a+2+2 a^{2}\right)}{(a-1)} \cdot \frac{\left(a^{2}-a+1\right)}{a \cdot(a-1)}+\frac{2}{a}$
4) $2 a-\frac{2(a-1)^{2}}{(a-1)} \cdot \frac{\left(a^{2}-a+1\right)}{a \cdot(a-1)}+\frac{2}{a}$
5) $2 a-\frac{2 a^{2}-2 a+2}{a}+\frac{2}{a}$
6) $\frac{2 a^{2}-2 a^{2}+2 a-2+2}{a}=\frac{2 a}{a}$
We get the answer to the problem: $\frac{2 a}{a}=2$
Answer: 2
|
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 number 11 is equal to the sum of opposite sides, then there are two possible cases $\left\{\begin{array}{l}2 a=11 \\ 2 a+b=19.5\end{array}\right.$ or $\left\{\begin{array}{l}2 a=11 \\ a+2 b=19.5\end{array}\right.$.
The solution to the first system is $a=5.5, b=8.5$, the perimeter of the rectangle is $P_{2}=28$.
The solution to the second system is $a=5.5, b=7$, the perimeter of the rectangle is
$P_{3}=25$. The product of the three possible perimeter values is $P_{1} \cdot P_{2} \cdot P_{3}=22 \cdot 28 \cdot 25=15400$.
Online (correspondence) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
Answer: 15400.
|
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 equal by two sides and the angle between them, hence the corresponding sides and angles are equal: $B K=A C=3, \quad \angle M K B=\angle M A C$. Then we get that triangle $K B N$ is isosceles $\angle B N M=\angle B K N$ and $B N=B K=3$.
Answer: 3.
|
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:
$\left[\begin{array}{l}b=3 \\ b=4\end{array} \Rightarrow\left[\begin{array}{c}a^{2}=4 \\ a^{2}=22\end{array} \Rightarrow\left[\begin{array}{c}a=2 \\ a=-2, \notin N \\ a= \pm \sqrt{22, \notin N}\end{array}\right.\right.\right.$, i.e. $b=3, a=2$.
We get the answer to the problem: $a+b=2+3=5$
Answer: 5
|
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 $\triangle ADC$). $\Rightarrow MP \| HK$, $MP = HK \Rightarrow MPKH$ is a parallelogram.
Since $MK = PH$, the quadrilateral $MPKH$ is a rectangle, with sides parallel to the diagonals $AC$ and $BD$ of the given quadrilateral $ABCD$, therefore $AC \perp BD$. This means that $S_{ABCD} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \cdot 18 \cdot 7 = 63$ (sq.units).
Answer: 63.
| Criterion Description | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 15 |
| The solution contains a computational error, possibly leading to an incorrect answer, but the solution has a correct sequence of all steps. | 10 |
| In the solution, one or two formulas are correctly written, which could be the beginning of a possible solution. | 5 |
| The solution does not meet any of the criteria described above. | 0 |
| Maximum score | 15 |
|
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 for equation (2):
1) the linear case, if the only root is greater than $(-1)$;
2) $D=0 ; t_{b}>-1$;
3) equation (2) has two distinct roots, one of which is greater than (-1), and the other is less than (-1).
The case where one root is greater than (-1) and the other is equal to (-1) does not suit us, as in this case there will be three solutions. Let's investigate the above cases. 1) $a=-1 ; 8 t-2=0 ; t=\frac{1}{4}>-1$ therefore, this value of the parameter is included in the answer.
2) $\frac{D}{4}=(a-3)^{2}-(a+1)(a-1)=10-6 a=0 ; \quad a=\frac{5}{3}$. The only root $-t_{b}=\frac{a-3}{a+1}$; $t_{6}\left(\frac{5}{3}\right)=\frac{\frac{5}{3}-3}{\frac{5}{3}+1}=-\frac{1}{2}>-1, a=\frac{5}{3}$ is included in the answer.
3) The general case - the roots are on different sides of (-1) - is described by the inequality $(a+1) f(-1) the correct answer is insufficiently justified. | 15 |
| The substitution of the variable is made and the correct restrictions for the new variable are defined. Attempts are made to write down some restrictions for the coefficients in connection with the conditions for the new variable, but they are only partially correct. | 10 |
| The substitution of the variable is made, the problem is reduced to the study of a quadratic trinomial with a parameter, but the reasoning is limited to considering the discriminant. | 5 |
| The solution does not correspond to any of the criteria listed above. | 0 |
| Maximum score | 20 |
|
-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 $L K$ is $5$, and $\angle E B C=15^{0}$. Find the radius of the circumcircle of triangle $E F K$.
#
|
# 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.
$\triangle B L F$ is isosceles, as $\angle F L B = \angle F B L = \angle E B K = \alpha$.
Then $\angle B F L = 180 - 2 \alpha \Rightarrow \sin \angle B F L = \sin 2 \alpha = \sin 30^{\circ} = \frac{1}{2} . \quad 2 R = \frac{L K}{\sin \angle L F K} = 10 \Rightarrow R = 5$.
Answer: 5.
| Criterion | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 20 |
| It is proven that $\triangle B L F$ is isosceles. | 15 |
| It is proven that $\angle E B K = \angle F D E = \alpha$. | 10 |
| It is proven that $\angle F L E = \angle F D E = \angle F K E = \alpha$. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |
|
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 triangle $ABC$. The line passing through point $C$ and perpendicular to $BC$ intersects the line $MN$ at point $L$. Find the radius of the circumcircle of triangle $MKL$, if $\cos \angle ABK = \frac{2}{3}$.
|
# 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.
$$
\begin{gathered}
\angle A H M = \angle H M N = \angle M N H = \angle N H C = \angle C H K = 180^{\circ} - \angle N C K \\
\angle M A C = \angle A C N = \angle N C A
\end{gathered}
$$
$$
\angle K H B = \angle K H C + 90^{\circ} = 180^{\circ} - \angle N C K + 90^{\circ} = \angle K C L,
$$
$\angle A M K = \angle A C K = \angle H N C$ - as angles of equal triangles and as angles subtending the same arc. Therefore, $\triangle H N C$ is similar to $\triangle H C K$ and $\triangle B H C$ is similar to $\triangle N L C$ by two angles. $\frac{K C}{K H} = \frac{N C}{C H} = \frac{L C}{H B}$ and considering that $\angle K H B = \angle K C L$, we get the similarity of $\triangle K C L$ and $\triangle B H K$ $\Rightarrow \angle C L K = \angle H B K \Rightarrow \angle M B K = \angle M L K$, which means points $K, M, B, L$ lie on the same circle and $R = \frac{M K}{2 \sin \angle A B K} = \frac{2 \sqrt{5}}{2 \sqrt{5} / 3} = 3$.
Answer: 3.
| Criterion | Points |
| :--- | :---: |
| The problem is solved correctly. | 20 |
| It is proven that points $K, M, B, L$ lie on the same circle. | 17 |
| It is proven that $\triangle K C L$ is similar to $\triangle B H K$. | 15 |
| It is proven that $\triangle B H C$ is similar to $\triangle N L C$. | 12 |
| It is proven that $\triangle A M H = \Delta H N C$ or $\triangle H N C$ is similar to $\triangle H C K$. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |
|
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 (moving in the same direction). 42nd second - the red elevator passed the green one (moving in opposite directions), 48th second - the red elevator passed the yellow one, 51st second - the yellow elevator passed the blue one, 54th second - the yellow elevator caught up with the green elevator. At what second from the start of the count will the green elevator pass the blue one, if during the observation period the elevators did not stop and did not change their direction of movement?
(12 points)
|
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 move in the same direction, and the red elevator is catching up with the blue one, so $k_{1} \cdot k_{2}>0, k_{1}>k_{2}$. For definiteness, let $k_{1}>0$, then $k_{2}>0$. The green and yellow elevators move in the opposite direction to the first two, and the yellow elevator is catching up with the green one, so $k_{3}<0, \quad k_{4}<0, \quad k_{3}<k_{4} \cdot$ Let's plot the graphs of the functions $S_{i}(t)$, according to the problem. We need to determine the abscissa of the point $M$ of intersection of the second graph with the third. Point $M$ is the point of intersection of the medians of triangle $A B C$. From this, we have $\frac{x-36}{51-x}=\frac{2}{1}, x=46$.
Answer: at 46 seconds.
|
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 all real $x$, then $a>0, \quad D=b^{2}-4 a c \leq 0 \Rightarrow c \geq b^{2} / 4 a$.
Then $\quad \frac{a+2 b+4 c}{b-2 a} \geq \frac{a+2 b+b^{2} / a}{b-2 a}=\frac{a\left(1+2 b / a+b^{2} / a^{2}\right)}{a(b / a-2)}=\frac{t^{2}+2 t+1}{t-2}=\frac{(t+1)^{2}}{t-2}, \quad$ where $\quad t=b / a, \quad t>2$. Consider the function $g(t)=\frac{(t+1)^{2}}{t-2}$ and find its minimum value for $t>2$. $g^{\prime}(t)=\frac{2(t+1)(t-2)-(t+1)^{2}}{(t-2)^{2}}=\frac{(t+1)(2(t-2)-(t+1))}{(t-2)^{2}}=\frac{(t+1)(t-5)}{(t-2)^{2}}$, at $t=5$ the derivative $g^{\prime}(t)$ is 0 and, passing through this point, changes sign from "minus" to "plus", therefore, $t_{\min }=5, \quad g_{\min }=g(5)=12$.
Answer: 12.
|
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 $\angle BAO=\angle ABO, \beta=\alpha / 2$. Triangles $AOC$ and $BOC$ are isosceles, and $\angle ACO=3 \alpha / 4, \angle BCO=\alpha / 4$. Since $\angle A+\angle B+\angle C=180^{\circ}$, then $\alpha+\alpha / 2+3 \alpha / 4+\alpha / 4=180^{\circ}$, and $\alpha=72^{\circ}$. Triangle $ABC$ is isosceles, $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$.
$\frac{x}{y}=\frac{x+y}{x}, \quad x^{2}-x y-y^{2}=0$.
By the condition $x=\sqrt{5}+1, y=x(\sqrt{5}-1) / 2, \quad y=2$.
Answer: 2.
|
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 thousand rubles. The shift fund of iron is 481 kg, and wire is 301 kg. How many transformers of types $A$ and $B$ should be produced per shift to maximize the profit from selling the products, if the resource consumption should not exceed the allocated shift funds? What will be the maximum profit in this case?
$(12$ points)
|
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
$5 x+3 y \leq 481, \quad 3 x+2 y \leq 301, x$ and $y$ - non-negative integers.
$5 x+3 y \leq 481 \Leftrightarrow y \leq 160 \frac{1}{3}-\frac{5 x}{3}, \quad 3 x+2 y \leq 301 \Leftrightarrow y \leq 150.5-\frac{3 x}{2}$. The intersection point of the lines $5 x+3 y=481$ and $3 x+2 y=301$ is the point with coordinates $x=59, y=62$. In this case, $D=12 \cdot 59+10 \cdot 62=1328$, but for $x=0, y=150.5$ we have the maximum value of the function $D=12 x+10 y$ for non-negative $x$ and $y$ satisfying the inequalities $5 x+3 y \leq 481, 3 x+2 y \leq 301$. This value is 1505. Since $x$ and $y$ are integers, for $x=0, y=150$ the revenue $D=1500$, for $x=1, y \leq 150.5-1.5, y=149, D=1502$, for $x=2, y \leq 150.5-3, y=147, D<1502$.
Thus, the maximum revenue is 1502 thousand rubles, provided that one transformer of type $A$ and 149 transformers of type $B$ are produced.
Answer: 1 transformer of type $A$ and 149 transformers of type $B$, 1502 thousand rubles.
Solution. $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$,
$\left(1+(\sin (x+y)+1)^{2}\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$. Since $1+(\sin (x+y)+1)^{2} \geq 1$, and $3^{x}+3^{-x} \geq 2$, $\log _{2}\left(3^{x}+3^{-x}\right) \geq 1$ for any values of $x$ and $y$, then $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \geq 1$, and the inequality $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$ is valid only for those $x$ and $y$ for which $\left\{\begin{array}{c}\sin (x+y)=-1, \\ 3^{x}+3^{-x}=2,\end{array}\left\{\begin{array}{c}y=-\pi / 2+2 \pi n, n \in N, \\ x=0 .\end{array}\right.\right.$
Answer: $x=0, y=-\pi / 2+2 \pi n, n \in N$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. From point A to point B, which are 12 km apart, a pedestrian and a bus set out simultaneously. Arriving at point B in less than one hour, the bus, without stopping, turned around and started moving back towards point A at a speed twice its initial speed. After 12 minutes from its departure from point B, the bus met the pedestrian. Determine the greatest possible integer value of the pedestrian's speed (in km/h), and for this value of the pedestrian's speed, determine the initial speed of the bus (in km/h). In the answer, write the sum of the found values of the pedestrian's and the bus's speeds. (5 points)
|
Solution. Let $x$ be the pedestrian's speed (in km/h), $y$ 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 $B$.
$\left\{\begin{aligned} y t & =12, \\ x(t+0.2) & =12-0.4 y, \\ t & <1,\end{aligned}\right.$
$D=x^{2}-600 x+3600 \geq 0 ; \quad x^{2}-600 x+3600=0, D / 4=86400, x_{1 / 2}=300 \pm 120 \sqrt{6}$.
Since
$$
\frac{x}{5}<x\left(t+\frac{1}{5}\right)<12 \Rightarrow x<60, \quad \text { then } \quad x^{2}-600 x+3600 \geq 0 \quad \text { when }
$$
$x \leq 300-120 \sqrt{6} \approx 6.06$. Therefore, the greatest possible integer value of the pedestrian's speed $x=6$. Let's find $y$ when $x=6: 2 y^{2}-54 y+3600=0, y_{1}=12, y_{2}=15$. Since $t<1$, then $y<12 \Rightarrow y=15$.
## Answer: 21
|
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
$\sqrt[3]{x}=10 x, x=1000 x^{3}, x=0, x=\sqrt{0.001}$. On all other intervals $[n, n+1)$, where $n=-1000, \ldots,-1,1, \ldots, 999$, the equation has one root. In total, we have 2000 roots.
Answer: 2000
|
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 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $2+1+1+1+1$ | $2+2$ | $C_{3}^{1}=3$ | $2+1+1+1$ | $2+2+1$ | $C_{3}^{1} \cdot C_{4}^{3}=12$ |
| $2+2+1+1$ | $2+1+1$ | $C_{3}^{2} \cdot C_{4}^{2}=18$ | $2+1+2+1$ | $2+1+1+1$ | $C_{3}^{2} \cdot C_{4}^{1}=12$ |
| $2+2+2$ | $1+1+1+1$ | 1 | $2+2+1$ | | |
## Answer: 46
|
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 triangle $A D C$). From the condition, it follows that $c>b$, so $\angle B\alpha$, then point $C$ lies between $D$ and $O$. Therefore, $\angle C A O=\beta-\alpha$. From this, it follows that triangles $A B O$ and $C A O$ are similar by two angles. Thus, $B O: A O=A O: C O$, or
$$
\frac{r+B D}{r}=\frac{r}{r-C D}
$$
Since $A D$ is the angle bisector of triangle $A B C$, we have
$$
B D=\frac{a c}{b+c}, \quad C D=\frac{a b}{b+c}
$$
Substituting the found expressions into equation (1), we get
$\left(r+\frac{a c}{b+c}\right)\left(r-\frac{a b}{b+c}\right)=r^{2}$, or $r\left(c^{2}-b^{2}\right)=a b c$.
By the cosine theorem, we have
$$
\begin{gathered}
b^{2}=c^{2}+a^{2}-2 a c \cos (A B C), c^{2}=b^{2}+a^{2}-2 a b \cos (A C B) \\
c^{2}-b^{2}=a(c \cos (A B C)-b \cos (A C B))=a(B H-H C)=B C(B H-H C)=768 \\
R=\frac{a b c}{4 S}=\frac{105 \sqrt{3} \cdot 768}{16 \cdot 180 \sqrt{3}}=28
\end{gathered}
$$
## Answer: 28
|
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 $\{80 ; 71 ; 20\}$? If the question of the problem allows for multiple answers, then specify them all in the form of a set.
|
# 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 decreases by 75 cm every 3 seconds (until they meet). Find the speed of the slower ball (in cm/sec).
|
# 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.
$$
\left\{\begin{array}{l}
(u+v) \cdot 4=(v-u) \cdot 20 \\
(u+v) \cdot 3=75
\end{array} . \text { From which } u=\frac{2}{3} v, v=15 \text{ cm/s }; u=10 \text{ cm/s}\right.
$$
Answer: 10.
|
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 (2), $y=710-13 n$
Substitute into (1), $x=20 n-1065$. Since $x>0, y>0$, then $n=54$. Therefore, $x=8, y=15$.
8 pencils and 15 pens, a total of 23 items.
Answer: 23.
|
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 solution, write the sum of these numbers in the answer.
|
# 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 are multiples of 19, either the number $20=210-190$ or $1=210-209$ was erased. In both cases, the arithmetic mean of the numbers remaining on the sheet does not match the erased number.
Answer: 21.
|
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 your answer, indicate the largest volume. #
|
# 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 28 liters, sold in September. This corresponds to the 13 and 15-liter containers. In August, the containers of 16, 19, and 21 liters were sold. The largest of these is 21 liters.
Answer: 21.
|
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 $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, which means it lies on the bisector of $\angle B B_{1} C$. Similarly, we prove that $B_{1} C_{1}$ is the bisector of $\angle A B_{1} B$.
Therefore, $\angle C_{1} B_{1} A_{1}=90^{\circ}$, as the angle between the bisectors of adjacent angles.
Answer: $\angle C_{1} B_{1} A_{1}=90^{\circ}$.
|
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.
## Answer: 2.
## Solution for Variant #2
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Masha, Dasha, and Sasha are tasked with harvesting all the currants from the bushes on the garden plot. Masha and Dasha together can collect all the berries in 7 hours 30 minutes, Masha and Sasha - in 6 hours, and Dasha and Sasha - in 5 hours. How many hours will it take the children to collect all the berries if they work together?
|
Solution: Let the entire work be 1. Let $x$ (parts of the entire work per hour) 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: $\left\{\begin{array}{l}(x+y) \cdot 7.5=1 \\ (x+z) \cdot 6=1 \\ (y+z) \cdot 5=1\end{array} ;\left\{\begin{array}{l}(x+y)=\frac{2}{15} \\ (x+z)=\frac{1}{6} . \\ (y+z)=\frac{1}{5}\end{array}\right.\right.$. By adding all the equations in the system, we get: $2(x+y+z)=\frac{2}{15}+\frac{1}{6}+\frac{1}{5}=\frac{15}{30}=\frac{1}{2} ; x+y+z=\frac{1}{4}$. Therefore, the three children will gather all the currants together in 4 hours.
Answer: 4.
|
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 follows that $A M=N C, \angle M A O=\angle N C O$ and $O A=O C$. Therefore, triangle $A O C$ is isosceles, which means angles $C A O$ and $A C O$ are equal. Hence, triangle $A B C$ is isosceles and $A B=B C=8 \text{ cm}$.
Answer: 8 cm.
|
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 tens digit equal to 4.
| $\boldsymbol{X}$ | equation | $\boldsymbol{y}$ | number |
| :---: | :---: | :---: | :---: |
| 4 | $40+y=10 y+4+36$ | $y=0$ | 40 does not fit the condition |
| 5 | $50+y=10 y+5+36$ | $y=1$ | 51 |
| 6 | $60+y=10 y+6+36$ | $y=2$ | 62 |
| 7 | $70+y=10 y+7+36$ | $y=3$ | 73 |
| 8 | $80+y=10 y+8+36$ | $y=4$ | 84 |
| 9 | $90+y=10 y+9+36$ | $y=5$ | 95 |
These could be the numbers $51,62,73,84,95$. The arithmetic mean is 73.
Answer: 73.
|
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, which is only possible with the unique combination of numbers $7, 8, 9$: $8+9+9=26$. Therefore, the shooter hit the 7 once.
## Answer: 1.
|
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 boys were there?
|
# 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-last remainder, so the last remainder is $\frac{7}{6} x r$, and its seventh part is $\frac{x r}{6}$.
Then $(x-1) r+\frac{x r}{6}=x r ; 6 r=x r ; x=6$.
Answer: 6.
|
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 $FCA$ is isosceles, hence $F C=A C$ and $F B=A E$. Triangles $F B A$ and $A E F$ are congruent, as $F B=A E, \angle A F B=$ $\angle F A E, A F$ is common. We obtain that $\angle F B A=\angle A E F=90^{\circ}$, from which $\angle F E C=90^{\circ}$. Triangle $F C E$ is right-angled and $M C=M E$, so $\mathrm{FM}=\mathrm{MC}$ and $F M=F B+B M=A E+B M=M C=7$.
Answer: 7.
|
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 $O_{1} O_{2}$ and $O O_{3}$.
#
|
# 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 \( B C, A C \) and \( A B \) respectively. Then triangles \( A H O_{1} \) and \( B F O_{1} \), \( B F O_{1} \) and \( B G O_{1} \) are right triangles and are equal by the hypotenuse and acute angle. From the equality of the triangles, it follows that \( H O_{1} = G O_{1} \). Therefore, \( O_{1} \) is equidistant from the sides of angle \( C \), which means \( C O_{1} \) is the bisector of angle \( C \). Similarly, it can be proven that \( A O_{2} \) and \( B O_{3} \) are the bisectors of angles \( A \) and \( B \). The point \( O \) is the point of intersection of the bisectors of triangle \( \mathrm{ABC} \). The angle \( O_{1} A O_{2} \) is a right angle because it is formed by the bisectors of adjacent angles. Similarly, the angles \( O_{1} B O_{3} \) and \( O_{1} C O_{2} \) are right angles, and therefore, \( O \) is the point of intersection of the altitudes of triangle \( O_{1} O_{2} O_{3} \). Thus, \( O_{3} O \) is perpendicular to \( \mathrm{O}_{1} \mathrm{O}_{2} \).
Answer: \( 90^{\circ} \).
|
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 "Queens" will be on the board after the last student has stepped away from it?
|
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" remaining on the board will be \(1, 4, 9, 16, 25, 36, 49\), and 64, which totals 8.
## Answer: 8.
#
|
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 $a_{n+1} \leq 0$. Since $a_{n}=a+d(n-1)$, from the inequality $100-3(n-1)>0$ we find $n=[103 / 3]=34$. Then $\max S_{n}=S_{34}=0.5 \cdot(100+100-3 \cdot 33) \cdot 34=1717$.
|
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 $\sqrt{2} \cos \frac{\pi y}{8} \geq 0$, i.e., $-4+16 k \leq y \leq 4+16 k, k \in \mathbb{Z}$, we have
$2 \cos ^{2} \frac{\pi y}{8}-1-2 \cos ^{2} \frac{\pi y}{8} \cos \frac{\pi x}{4}+\cos \frac{\pi x}{4}=0 \quad \Leftrightarrow \quad\left(2 \cos ^{2} \frac{\pi y}{8}-1\right)\left(1-\cos \frac{\pi x}{4}\right)=0 \quad \Leftrightarrow \sqrt{2} \cos \frac{\pi y}{8}=1$ or $\cos \frac{\pi x}{4}=1 \quad \Leftrightarrow y=2+16 k, y=-2+16 k, k \in \mathbb{Z}$, or $\quad x=8 n, n \in \mathbb{Z}$. The integer solutions of the system will be points $(l ; 2+16 k),(l ;-2+16 k),(8 n ; m), l, k, n, m \in \mathbb{Z}, \quad-4+16 s \leq m \leq 4+16 s, s \in \mathbb{Z}$, lying in the square centered at the point $(0 ; 4)$, with side $4 \sqrt{2}$, diagonals parallel to the coordinate axes, and in the half-plane $y<x+2$. Such points will be $(0 ; 0),(0 ; 1),(1 ; 2),(2 ; 2)$.
Answer: $(0 ; 0),(0 ; 1),(1 ; 2),(2 ; 2)$
|
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:
$$
y=\frac{x^{2}}{8 \sqrt{3}}, M(4 ;-2 \sqrt{3}) \cdot y=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(x-x_{0}\right) ;-2 \sqrt{3}=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(4-x_{0}\right) ;
$$
$x_{0}^{2}-8 x_{0}-48=0 ; x_{0}=4 \pm 8 ;\left(x_{0}\right)_{1}=12,\left(x_{0}\right)_{2}=-4$. Equations of the tangents:
1) $y=6 \sqrt{3}+\sqrt{3}(x-12) ; y=\sqrt{3} x-6 \sqrt{3}, \operatorname{tg} \alpha_{1}=\sqrt{3}, \alpha_{1}=60^{\circ}$;
2) $y=\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}(x+4) ; y=-\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}, \operatorname{tg} \alpha_{2}=-\frac{1}{\sqrt{3}}, \alpha_{2}=-30^{\circ}$.
Angle between the tangents: $\varphi=60^{\circ}-\left(-30^{\circ}\right)=90^{\circ}$,
Answer: $90^{\circ}$.
|
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 of Sines, we have
$\frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 30^{\circ}}$, or $\sqrt{2} a=2 b$,
i.e., $a=\sqrt{2} b$. Thus,
$b=8, a=8 \sqrt{2}$.
The radius of the circumcircle of triangle
$A B C$ is found using the formula
$R=\frac{b}{2 \sin 30^{\circ}}=b=8$.
## Answer: 8.
|
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:
$$
y=\frac{x^{2}}{2 \sqrt{3}}, M\left(1 ;-\frac{\sqrt{3}}{2}\right) \cdot y=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(x-x_{0}\right) ;-\frac{\sqrt{3}}{2}=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(1-x_{0}\right) ; x_{0}^{2}-2 x_{0}-3=0
$$
$; x_{0}=1 \pm 2 ;\left(x_{0}\right)_{1}=3,\left(x_{0}\right)_{2}=-1$. Equations of the tangents:
1) $y=\frac{3 \sqrt{3}}{2}+\sqrt{3}(x-3) ; y=\sqrt{3} x-\frac{3 \sqrt{3}}{2}, \operatorname{tg} \alpha_{1}=\sqrt{3}, \alpha_{1}=60^{\circ}$;
2) $y=\frac{1}{2 \sqrt{3}}-\frac{1}{\sqrt{3}}(x+1) ; y=-\frac{1}{\sqrt{3}} x-\frac{1}{2 \sqrt{3}}, \operatorname{tg} \alpha_{2}=-\frac{1}{\sqrt{3}}, \alpha_{2}=-30^{\circ}$.
Angle between the tangents: $\varphi=60^{\circ}-\left(-30^{\circ}\right)=90^{\circ}$.
Answer: $90^{\circ}$.
|
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 $\mathrm{y} \leq \mathrm{P} ; \kappa \leq \mathrm{P} ; \mathrm{e} \leq \mathrm{P} ; =>$
$$
\mathrm{x}+\mathrm{y}+\mathrm{K}+\mathrm{e} \leq 3.5 \mathrm{P} ; => \mathrm{P} \geq \frac{2}{7}(\mathrm{x}+\mathrm{y}+\mathrm{K}+\mathrm{e})=\frac{2}{7} \cdot 2023=578
$$
$\mathrm{x}=289 ; 2 \mathrm{x}=\mathrm{y}=\mathrm{K}=\mathrm{e}=578 . \mathrm{x}+\mathrm{y}+\mathrm{\kappa}+\mathrm{e}=2023$. $\mathrm{P}=578$. Answer: 578.
## №5: Percentages.
From a bottle filled with a $12 \%$ salt solution, 1 liter was poured out and the bottle was topped up with water, then another 1 liter was poured out and topped up with water again. The bottle ended up with a $3 \%$ salt solution. What is the capacity of the bottle?
|
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:
$$
\begin{aligned}
& 1 \rightarrow 7 \quad 9 \rightarrow 2 \rightarrow 3 \\
& \uparrow \quad \uparrow \\
& 5 \leftarrow 8 \longleftarrow 6 \leftarrow 4
\end{aligned}
$$
Traversing the digits in the row from right to left corresponds to moving against the direction of the arrows. If in this diagram we take steps against the direction of the arrows, starting from the digit 5, we end up at 6 in two steps, and then we will be moving in a cycle, landing back at 6 every 5 steps.
Since \(2017 = 2 + 5 \cdot 403\), the first digit will be 6.
Answer. 6.
|
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:
$$
\left\{\begin{array}{c}
\frac{s}{x}=5 \frac{5}{6} \\
\frac{S}{x+2 y} \geq 2.5 \\
\frac{S}{x+y}-\frac{S}{x+2 y} \geq 1
\end{array}\right.
$$
From the first equation, $S=5 \frac{5}{6} \cdot x=\frac{35}{6} \cdot x$. Substituting this result into the second inequality, we get: $\frac{35}{6} \cdot x \geq 2.5 \cdot x + 5 \cdot y$, from which $20 x \geq 30 y$ and $y \leq \frac{2}{3} x$.
Substituting $S$ into the second inequality, after transforming it: $S \cdot \frac{x+2 y-x-y}{(x+y) \cdot(x+2 y)} \geq 1 ; \quad \frac{35}{6} x y \geq x^{2} + 3 x y + 2 y^{2} ; \quad 6 x^{2} - 17 x y + 12 y^{2} \leq 0 ; \quad \frac{2}{3} \leq \frac{y}{x} \leq \frac{3}{4} ; \quad$ i.e., $y \geq \frac{2}{3} x$. From the two obtained estimates, it follows that $y \geq \frac{2}{3} x$. Let's find the time the tractor spends on the journey: $t=\frac{S}{y}=\frac{\frac{35}{6} x}{\frac{2}{3} x}=\frac{35}{4}=8$ hours 45 minutes. The required time is $9 + (8$ hours 45 minutes $)=17$ hours 45 minutes.
|
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 $\angle \mathrm{MCA}=2 \alpha+\beta=\succ(\square \mathrm{AMC})$
$2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=\succ \alpha+\beta=60^{\circ} \Rightarrow \succ \mathrm{MAF}=60^{\circ} \Rightarrow \succ \mathrm{AMF}=30^{\circ}$.
Answer: $30^{\circ}$
|
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 just one problem?
|
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 who solved three, $q$ be the number who solved four, and $r$ be the number who solved five problems. According to the problem, $x, y, z, q, r$ are all at least 1.
\[
\left\{\begin{array}{c}
x+2 y+3 z+4 q+5 r=40, \\
x+y+z+q+r=30 ;
\end{array}\right.
\]
Subtract the second equation from the first: $y+2 z+3 q+4 r=10$. This equality, given the conditions of our problem, can only be achieved when $y=z=q=r=1$, which means $x=26$.
Answer: 26 people.
|
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, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24$; $s^{2}-52 s+480=0 ; s_{1,2}=26 \pm 14 . s_{1}=40, \quad s_{2}=12$ does not satisfy the conditions of the problem $s>15, s>24$. Answer: 40 km.
|
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,
$\left\{\begin{array}{l}\frac{a+a+2 d}{2} \cdot 3=327, \\ \frac{a+a+56 d}{2} \cdot 57=57\end{array} \Leftrightarrow\left\{\begin{array}{l}a+d=109, \\ a+28 d=1\end{array} \Rightarrow 27 d=-108 ; d=-4, a=113\right.\right.$.
The sum of the first $n$ terms of the arithmetic progression $S_{n}$ reaches its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. Since $a_{n}=a+d(n-1)$, from the inequality $113-4(n-1)>0$ we find $n=[117 / 4]=29$. Then $\max S_{n}=S_{29}=0.5 \cdot(113+113-4 \cdot 28) \cdot 29=1653 . \quad$ Answer: 1653.
|
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
$\frac{T A}{C D}=\frac{A M}{D M}, \frac{c}{a}=\frac{x}{b-x}$,
$x=\frac{b c}{a+c}=A M, \frac{a b}{a+c}=D M$.
Draw $A K \perp T C$ and connect $K$ and $S$.
$A K=\frac{T A \cdot A M}{\sqrt{T A^{2}+A M^{2}}}=\frac{c x}{\sqrt{c^{2}+x^{2}}}=$
$=\frac{b c}{\sqrt{(a+c)^{2}+b^{2}}} \cdot S K=\sqrt{A S^{2}+A K^{2}}=\sqrt{c^{2}+\frac{b^{2} c^{2}}{(a+c)^{2}+b^{2}}}=c \frac{\sqrt{(a+c)^{2}+2 b^{2}}}{\sqrt{(a+c)^{2}+b^{2}}}$.
$M C=\sqrt{D C^{2}+M D^{2}}=\sqrt{a^{2}+\frac{a^{2} b^{2}}{(a+c)^{2}}}=\frac{a \sqrt{(a+c)^{2}+b^{2}}}{a+c}$.
$S_{\triangle M C S}=\frac{1}{2} \cdot M C \cdot S K=\frac{1}{2} \cdot \frac{a c}{a+c} \cdot \sqrt{(a+c)^{2}+2 b^{2}}=\frac{1}{2} \cdot a c \sqrt{1+2\left(\frac{b}{a+c}\right)^{2}}$.
Answer: when $a=1, b=10, c=4 \quad x=8, S_{\triangle M C S}=6$.
|
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, $\frac{s}{2(s-8)}=\frac{s-15}{s} ; s^{2}=2 s^{2}-46 s+240 ; s^{2}-46 s+240=0$;
$s_{1,2}=23 \pm 17 \cdot s_{1}=40, s_{2}=6$ does not satisfy the conditions of the problem $s>8, s>15$. Answer:
40 km.
|
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,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array} \Rightarrow 16 d=48 ; d=3, a=-50\right.\right.
$$
The sum $S_{n}$ of the first $n$ terms of the arithmetic progression takes the minimum value if $a_{n}<0$, and $a_{n+1} \geq 0$. Since $a_{n}=a+d(n-1)$, from the inequality $-50+3(n-1)<0$ we find $n=[53 / 3]=[172 / 3]=17$. Then $\min S_{n}=S_{17}=0.5 \cdot(-50-50+3 \cdot 16) \cdot 17=-442$. -442.
|
-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 $x=a+\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}-3 x=\left(a+\frac{1}{a}\right)^{3}-3\left(a+\frac{1}{a}\right)=a^{3}+\frac{1}{a^{3}}=7+4 \sqrt{3}+\frac{1}{7+4 \sqrt{3}}=\frac{(7+4 \sqrt{3})^{2}+1}{7+4 \sqrt{3}}= \\
& =\frac{98+56 \sqrt{3}}{7+4 \sqrt{3}}=14
\end{aligned}
$$
Answer: 14.
|
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: $t_{1}+t_{2}=2024, t_{1} \cdot t_{2}=10097$, which means both roots of the quadratic equation are positive. Let's make the reverse substitution:
$$
\begin{array}{cc}
(x+2)^{2}=t_{1} & (x+2)^{2}=t_{2} \\
x+2= \pm \sqrt{t_{1}} & x+2= \pm \sqrt{t_{2}} \\
x_{1,2}=-2 \pm \sqrt{t_{1}} & x_{3,4}=-2 \pm \sqrt{t_{2}} \\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=\left(-2+\sqrt{t_{1}}\right)^{2}+\left(-2-\sqrt{t_{1}}\right)^{2}+\left(-2+\sqrt{t_{2}}\right)^{2}+\left(-2+\sqrt{t_{2}}\right)^{2}= \\
=2\left(4+t_{1}\right)+2\left(4+t_{2}\right)=16+2\left(t_{1}+t_{2}\right)=16+2 \cdot 2024=4064
\end{array}
$$
Answer: 4064.
|
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, $t_{1}+t_{2}=1598, t_{1} \cdot t_{2}=15882$, which means both roots of the quadratic equation are positive. Let's make the reverse substitution:
$$
\begin{array}{cc}
(x+3)^{2}=t_{1} & (x+3)^{2}=t_{2} \\
x+3= \pm \sqrt{t_{1}} & x+3= \pm \sqrt{t_{2}} \\
x_{1,2}=-3 \pm \sqrt{t_{1}} & x_{3,4}=-3 \pm \sqrt{t_{2}} \\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=\left(-3+\sqrt{t_{1}}\right)^{2}+\left(-3-\sqrt{t_{1}}\right)^{2}+\left(-3+\sqrt{t_{2}}\right)^{2}+\left(-3+\sqrt{t_{2}}\right)^{2}= \\
=2\left(9+t_{1}\right)+2\left(9+t_{2}\right)=36+2\left(t_{1}+t_{2}\right)=36+2 \cdot 1598=3232
\end{array}
$$
Answer: 3232.
|
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, where $\mathrm{t}$ is an integer. Assuming that all jumps, both of the dog and the fox, are the same, find the minimum value of $t$ for which the fox can escape from the dog.
|
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 $1.23 x\left(1-\frac{t}{100}\right) y < xy$; simplifying, we get $1.23\left(1-\frac{t}{100}\right) < 1$; further simplifying, $\frac{23}{1.23} < \frac{t}{100}$; thus, $t > 18 \frac{86}{123}$, which means $t=19\%$.
Answer: 19.
|
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: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)=$
$$
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{1}{a_{1580}}-\frac{1}{a_{1581}}+\frac{1}{a_{1581}}-\frac{1}{a_{1582}}+\ldots+\frac{1}{a_{2019}}-\frac{1}{a_{2020}}\right)=
$$
$$
\begin{gathered}
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{1}{a_{1580}}-\frac{1}{a_{2020}}\right)=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{a_{2020}-a_{1580}}{a_{2020} \cdot a_{1580}}\right)=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{(2019-1579) d}{a_{2020} \cdot a_{1580}}\right)= \\
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{(2019-1579) d}{a_{2020} \cdot a_{1580}}\right)=99 \cdot 100 \cdot \frac{1}{d} \cdot \frac{440 d}{660 \cdot 440}=15
\end{gathered}
$$
since $a_{2020}=a_{1000}+1020 d=150+1020 \cdot 0.5=660$, and $a_{1580}=a_{1000}+580 d=150+580 \cdot 0.5=440$.
## Answer: 15.
|
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 $A_{1}$ is equidistant from sides $B_{1} B$ and $B_{1} C$, and thus lies on the bisector of $\angle B B_{1} C$. Similarly, we prove that $B_{1} C_{1}$ is the bisector of $\angle A B_{1} B$.
Therefore, $\angle C_{1} B_{1} A_{1}=90^{\circ}$, as the angle between the bisectors of adjacent angles.
Answer: $\angle C_{1} B_{1} A_{1}=90^{\circ}$
|
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, the condition for the minimum of the function is equivalent to the condition for the minimum of M, at which the circles will intersect, and this is the point of tangency of the two circles. That is, the sum of the radii of the two circles should equal the distance between their centers. $\left(10+(M+71)^{0.5}\right)^{2}=(5+7)^{2}+(5+4)^{2}=>(M+71)^{0.5}=10=>\quad M=29$.
Answer: 29.
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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