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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
|
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
|
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
|
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
|
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
|
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
|
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
|
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
|
# 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
|
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 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
|
7. What is the minimum number of cells that need to be painted in a square with a side of 35 cells (a $35 \times 35$ square, with a total of 1225 cells), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
#
|
# Solution.
You should color every 3rd cell diagonally (see the figure). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be colored. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be colored. $\left[\frac{35^{2}}{3}\right]=408$.
Answer: 408.
|
408
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Find the sum of all integer values of c for which the equation $15|p-1|+|3 p-| p+c||=4$ has at least one root with respect to p
#
|
# Solution:
Consider the function $\mathfrak{1} \mathbb{R}=15|\mathrm{p}-1|+|3 \mathrm{p}-| \mathrm{p}+\mathrm{c}||-4 \mathrm{p}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{p}$. $15>4+1+3$. Therefore, on all intervals up to $\mathrm{p}=1$, the coefficient of the linear increment is negative, and after $\mathrm{p}=1$ it is positive. $\mathrm{P}=1$ is a point of minimum. For the equation $\mathrm{f}(\mathrm{p})=0$ to have at least one root, it is necessary and sufficient that the inequality:
$\mathrm{f}(1) \leq 0$
$\Rightarrow$ Let $|\mathrm{c}+1|=\mathrm{t} ;|3-\mathrm{t}|-4 \leq 0$
$(3-t)^{2}-4^{2} \geq 0$
$(7-t)(-1-t) \geq 0$
$\mathrm{t} \in[-1 ; 7]$
$|c+1| \leq 7$
Answer: -15.
|
-15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. It is known that the number of occurrences of a certain character in the text is from $10.5\%$ to $11\%$ of the length of the text (the length of the text is understood as the total number of characters in the text). Find the minimum possible length of the text.
#
|
# Solution:
Let the length of the text be L. Let a character appear in the text $x$ times. The problem can be reformulated as: find the smallest natural number $\mathrm{L}$, for which there exists a natural number $x$ such that $\frac{10.5}{100}19$ when $x \geq 3$.
Answer: 19.
## Solution variant № 2
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the arithmetic progression $\left(a_{n}\right)$, $a_{1}=1$, $d=3$.
Calculate $A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}$.
In the answer, write the smallest integer greater than $A$.
|
Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\sqrt{a_{1}}}{a_{2}-a_{1}}+\frac{\sqrt{a_{3}}-\sqrt{a_{2}}}{a_{3}-a_{2}}+\ldots+\frac{\sqrt{a_{1580}}-\sqrt{a_{1579}}}{a_{1580}-a_{1579}}=
\end{aligned}
$$
$=\frac{\sqrt{a_{2}}-\sqrt{a_{1}}}{d}+\frac{\sqrt{a_{3}}-\sqrt{a_{2}}}{d}+\ldots+\frac{\sqrt{a_{1580}}-\sqrt{a_{1579}}}{d}=$
$=\frac{\sqrt{a_{1580}}-\sqrt{a_{1}}}{d}=\frac{\sqrt{a_{1}+1579 d}-\sqrt{a_{1}}}{d}=\frac{\sqrt{1+1579 \cdot 3}-\sqrt{1}}{3}=\frac{\sqrt{4738}-1}{3}$
We estimate the value of $A: \quad 22<\frac{\sqrt{4738}-1}{3}<23$, and write the smallest integer greater than $A$ in the answer.
Answer: 23.
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In rectangle $A B C D$, $A B: A D=1: 2$. Point $M$ is the midpoint of $A B$, and point $K$ lies on $A D$ and divides it in the ratio $3: 1$ starting from point $A$. Find the sum of $\angle C A D$ and $\angle A K M$.
#
|
# Solution:
Complete the rectangle $A B C D$ to a square $A E F D$ with side $A D$.
$$
\begin{aligned}
& \text { Let } L \in E F, E L: L F=1: 3, \\
& \triangle M E L=\triangle A K M \Rightarrow \angle E M L=\angle A K M \\
& N \in F D, F N=N C, M R\|A D \Rightarrow M N\| A C \Rightarrow \\
& \Rightarrow \angle N M R=\angle C A D . \\
& \triangle E M L=\Delta L F N, M L=L N, \angle M L N=90^{\circ} \Rightarrow \\
& \Rightarrow \angle L M N=45^{\circ} \\
& \angle C A D+\angle A K M=\angle N M R+\angle E M L= \\
& =90^{\circ}-\angle L M N=90^{\circ}-45^{\circ}=45^{\circ} .
\end{aligned}
$$
Answer: 45.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. What is the minimum number of cells that need to be painted in a square with a side of 35 cells (a total of $35 \times 35$ cells, which is 1225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess?
|
# Solution.
Cells should be shaded in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be shaded. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table (larger than $4 \times 4$) such that the knight visits each cell exactly once (see the 5x5 table). If these moves are numbered, it is clear that fewer than $\left[\frac{N^{2}}{2}\right]$ cells cannot be shaded because then there would necessarily be two consecutive unshaded cells in this sequence, i.e., a move from one to the other would be possible. The $35 \times 35$ table needs to be divided into 49 tables of $5 \times 5$, and numbered alternately, starting with the first $5 \times 5$ table. $\left[\frac{35^{2}}{2}\right]=612$.
| 21 | 16 | 5 | 10 | 23 | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 11 | 22 | 17 | 4 | |
| 1 | 20 | 15 | 24 | 9 | 26 |
| 12 | 7 | 18 | 3 | 14 | |
| 19 | 2 | 13 | 8 | 25 | |
| | | | | | |
Answer: 612.
|
612
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. A table consisting of 2019 rows and 2019 columns is filled with natural numbers from 1 to 2019 such that each row contains all numbers from 1 to 2019. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the filling of the table is symmetric with respect to this diagonal.
#
|
# Solution:
We will show that all numbers from 1 to 2019 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 2019\}$ is not on the diagonal. Then, due to the symmetry of the table, the number $a$ appears an even number of times. On the other hand, since the number $a$ appears once in each row, the total number of $a$ in the table is odd (2019). This is a contradiction.
There are 2019 cells on the diagonal, so each number from the set $\{1,2, \ldots 2019\}$ will appear exactly once on the diagonal. By calculating the sum of the arithmetic progression, we find the answer.
Answer: 2039190.
|
2039190
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Calculate: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}$.
|
Solution: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}=$
$$
\begin{aligned}
& =\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\ldots+\left(2015^{2}-2016^{2}\right)+2017^{2}= \\
& =(1-2)(1+2)+(3-4)(3+4)+\ldots+(2015-2016)(2015+2016)+2017^{2}= \\
& =-(1+2)-(3+4)-\ldots-(2015+2016)+2017^{2}= \\
& =-(1+2+3+4+\ldots+2015+2016)+2017^{2}= \\
& =-\frac{1+2016}{2} \cdot 2016+2017^{2}=-2017 \cdot 1008+2017^{2}=2017 \cdot(2017-1008)= \\
& =2017 \cdot 1009=2035153
\end{aligned}
$$
Answer: 2035153.
|
2035153
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The train is traveling at a speed of 60 kilometers per hour, making stops every 48 kilometers. The duration of each stop, except the fifth, is 10 minutes, and the fifth stop is half an hour. How far has the train traveled if it departed at noon on September 29 and arrived at its destination on October 1 at 10:00 PM?
|
Solution: The train was on the way for 58 hours.
The train covers a section of 48 kilometers in $\frac{4}{5}$ of an hour.
Let the train make $N$ stops during its journey, then the time of its movement will be $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$, where $t$ is the time to cover the last section of the path, which is more than 0 km but no more than 48 km, so $t \in\left(0 ; \frac{4}{5}\right]$. $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$ $\frac{29}{30} N+\frac{1}{3}+t=58$ $t=57 \frac{2}{3}-\frac{29}{30} N$ $0<57 \frac{2}{3}-\frac{29}{30} N \leq \frac{4}{5}$ $56 \frac{13}{15} \leq \frac{29}{30} N<57 \frac{2}{3}$ $58 \frac{24}{29} \leq N<59 \frac{19}{29}$
We get that $\mathrm{N}=59, t=57 \frac{2}{3}-\frac{29}{30} \cdot 59=\frac{19}{30}$.
The distance covered by the train, $N \cdot 48+t \cdot 60=59 \cdot 48+\frac{19}{30} \cdot 60=2870$.
Answer: 2870 km.
|
2870
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Calculate $f(2)$, if $25 f\left(\frac{x}{1580}\right)+(3-\sqrt{34}) f\left(\frac{1580}{x}\right)=2017 x$. Round the answer to the nearest integer.
|
Solution: Let's make the substitution: $\mathrm{t}=\frac{x}{1580}$, then the equation will take the form:
$25 f(t)+(3-\sqrt{34}) f\left(\frac{1}{t}\right)=2017 \cdot 1580 \cdot t \quad(1)$,
substitute $\frac{1}{t}$ for $t$ in the equation, we get
$$
25 f\left(\frac{1}{t}\right)+(3-\sqrt{34}) f(t)=2017 \cdot 1580 \cdot \frac{1}{t}
$$
Solve the system of equations
$$
\begin{aligned}
& \left\{\begin{array}{l}
25 f(t)+(3-\sqrt{34}) f\left(\frac{1}{t}\right)=2017 \cdot 1580 \cdot t \\
25 f\left(\frac{1}{t}\right)+(3-\sqrt{34}) f(t)=2017 \cdot 1580 \cdot \frac{1}{t}
\end{array}\right. \\
& \left\{\begin{aligned}
25^{2} f(t)+25(3-\sqrt{34}) f\left(\frac{1}{t}\right) & =25 \cdot 2017 \cdot 1580 \cdot t \\
(3-\sqrt{34})^{2} f(t)+25(3-\sqrt{34}) f\left(\frac{1}{t}\right) & =2017 \cdot 1580 \cdot(3-\sqrt{34}) \cdot \frac{1}{t}^{, \text { then }}
\end{aligned}\right. \\
& \left(25^{2}-(3-\sqrt{34})^{2}\right) f(t)=25 \cdot 2017 \cdot 1580 \cdot t-2017 \cdot 1580 \cdot(3-\sqrt{34}) \cdot \frac{1}{t} \\
& f(t)=\frac{2017 \cdot 1580\left(25 t-(3-\sqrt{34}) \cdot \frac{1}{t}\right)}{\left(25^{2}-(3-\sqrt{34})^{2}\right)}, \text { then } \\
& f(2)=\frac{2017 \cdot 1580\left(50-(3-\sqrt{34}) \cdot \frac{1}{2}\right)}{\left(25^{2}-(3-\sqrt{34})^{2}\right)}=\frac{2017 \cdot 1580 \cdot(97+\sqrt{34})}{2 \cdot(582+6 \sqrt{34})}=\frac{2017 \cdot 1580}{2 \cdot 6} \\
& f(2)=\frac{2017 \cdot 395}{3}=265571 \frac{2}{3} \approx 265572
\end{aligned}
$$
Answer: 265572.
|
265572
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. Buratino, Karabas-Barabas, and Duremar are running along a path around a circular pond. They start simultaneously from the same point, with Buratino running in one direction and Karabas-Barabas and Duremar running in the opposite direction. Buratino runs three times faster than Duremar and four times faster than Karabas-Barabas. After Buratino meets Duremar, he meets Karabas-Barabas 150 meters further. What is the length of the path around the pond?
|
# Solution:
Let the length of the path be S.
Since Buratino runs three times faster than Duremar, by the time they meet, Buratino has run three-quarters of the circle ($3 / 4$ S), while Duremar has run one-quarter. Since Buratino runs four times faster than Karabas-Barabas, by the time they meet, Buratino has run four-fifths of the circle, while Karabas-Barabas has run one-fifth ($1 / 5$ S).
The distance between the meeting points is 150 meters. We get the equation:
$$
\begin{aligned}
& \frac{3}{4} S + 150 + \frac{1}{5} S = S \\
& S = 3000
\end{aligned}
$$
Answer: 3000 meters.
|
3000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. At the same time, a cyclist leaves point $A$ for point $B$ at a speed of $15 \mathrm{km} / \mathrm{u}$, and a tourist leaves point $B$ for point $C$ at a speed of 5 km/h. After 1 hour and 24 minutes from the start of their journey, they were at the minimum distance from each other. Find the distance between the points if all three points are equidistant and connected by straight roads. (8 points)
|
Solution: Let $AB = BC = AC = S$. Denote the distance between the cyclist and the tourist as $r = r(t)$, where $t$ is the time from the start of the movement. Then, by the cosine theorem, we have: $r^{2} = (S - 15t)^{2} + (5t)^{2} - 5t(S - 15t)$. To find the time when the distance between the cyclist and the tourist was the smallest, we compute the derivative of the function $r^{2} = r^{2}(t)$:
between the start and the movement.
$\left(r^{2}\right)^{\prime} = -2 \cdot 15(S - 15t) + 2 \cdot 25t - 5(S - 15t) + 15 \cdot 5t = -35S + 2(225 + 25 + 75)t = 0$.
Since the minimum value of the function $r^{2} = r^{2}(t)$ during the movement is achieved at $t = 1.4$, then $-35S + (225 + 25 + 75) \cdot 2.8 = 0$, and $S = 26$.
Answer: 26 km.
|
26
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. At the same time, a car departs from point $A$ to point $B$ at a speed of 80 km/h, and from point $B$ to point $C$ - a train at a speed of $50 \kappa м / ч$. After 7 hours of travel, they were at the shortest distance from each other. Find the distance between the points, if all three points are equidistant from each other and are connected by straight roads.
|
Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 80t)^{2} + (50t)^{2} - 50t(S - 80t)$. To find the time at which the distance between the car and the train was the smallest, we will compute the derivative of the function $r^{2} = r^{2}(t):$
$\left(r^{2}\right)^{\prime} = -2 \cdot 80(S - 80t) + 2 \cdot 2500t - 50(S - 80t) + 80 \cdot 50t = -210S + 2(6400 + 2500 + 4000)t = 0$.
Since the function $r^{2} = r^{2}(t)$ takes its minimum value during the motion at $t = 7$, then $-105S + (6400 + 2500 + 4000) \cdot 7 = 0$, and $S = 860$.
Answer: $860 \kappa m$.
|
860
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. At the same time, a car departs from point $A$ to point $B$ at a speed of $90 \mathrm{~km} / \mathrm{h}$, and from point $B$ to point $C$ - a train at a speed of $60 \mathrm{~km} /$ h. After 2 hours of travel, they found themselves at the minimum distance from each other. Find the distance between the points, if all three points are equidistant from each other and are connected by straight roads. (8 points)
|
Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 90t)^{2} + (60t)^{2} - 60t(S - 90t)$. To find the time at which the distance between the car and the train
was the smallest, we will compute the derivative of the function $r^{2} = r^{2}(t):$
$\left(r^{2}\right)^{\prime} = -2 \cdot 90(S - 90t) + 2 \cdot 3600t - 60(S - 90t) + 90 \cdot 60t = -240S + 2(8100 + 3600 + 5400)t = 0$.
Since the function $r^{2} = r^{2}(t)$ takes its minimum value during the motion at $t = 2$, then $-120S + (8100 + 3600 + 5400) \cdot 2 = 0$, and $S = 285$.
Answer: 285 km.
|
285
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. For chemical experiments, two identical test tubes were taken, each containing 200 ml of a liquid substance. From the first test tube, $1 / 4$ of the content was poured out and the same amount of water was added, then this procedure was repeated 3 more times, each time pouring out a quarter of the content of the test tube and adding the same amount of water. A similar procedure was performed twice for the second test tube, each time pouring out a certain amount of the content of the test tube and adding the same amount of water. In the end, the concentration of the resulting mixtures in the first and second test tubes became related to each other as 9/16. Determine how much of the mixture was poured out each time from the second test tube.
(12 points)
|
Solution. The initial amount of the substance is $-V$. After pouring out $a$ part, the concentration of the substance in the test tube becomes $\frac{V-a}{V}$. After the second time, the concentration is $\left(\frac{V-a}{V}\right)^{2}$, and after the fourth time, it is $\left(\frac{V-a}{V}\right)^{4}$. Substituting the data from the problem, we get the ratio $\left(\frac{200-50}{200}\right)^{4}:\left(\frac{200-x}{200}\right)^{2}=\frac{9}{16}$, or $\left(\frac{3}{4}\right)^{4}:\left(\frac{200-x}{200}\right)^{2}=\frac{9}{16} \Rightarrow\left(\frac{3}{4}\right)^{2}:\left(\frac{200-x}{200}\right)^{2}=1 \Rightarrow \frac{200-x}{200}=\frac{3}{4} \Rightarrow x=50$
Answer: 50 ml.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find the smallest natural number $m$, for which the expression $148^{n}+m \cdot 141^{n}$ is divisible by 2023 for any odd natural $n$. (16 points)
|
Solution. $2023=7 \cdot 289$, GCD $(7 ; 289)=1$. Since $n-$ is an odd number, then $148^{n}+m \cdot 141^{n}=(289-141)^{n}+m \cdot 141^{n}=289 l+(m-1) 141^{n}, l \in \square$. Then $(m-1) 141^{n}$ must be divisible by 289. Since 289 and 141 are coprime, then $m-1=289 k, k \in\{0\} \cup \square$. On the other hand $148^{n}+m \cdot 141^{n}=148^{n}+m \cdot(148-7)^{n}=7 p+(m+1) 148^{n}, p \in \square$. Since 7 and 148 are coprime, then $m+1=7 s, s \in \square$. Then $m=289 k+1$ and $m=7 s-1$, and $289 k+1=7 s-1$, $289 k+2=7 s, \quad 287 k+2 k+2=7 s, \quad 2 k+2=7(s-41 k)$. The number $(s-41 k)=2 q, q \in \square$, and $k=7 q-1, k \geq 6$. For $k=6$ we have $m=289 \cdot 6+1=1735$, and $s=248$.
Answer: 1735.
|
1735
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. In triangle $\mathrm{KLM}$ with angle $\mathrm{L}=120^{\circ}$, the angle bisectors LA and $\mathrm{KB}$ of angles KLM and LKM are drawn respectively. Find the measure of angle KBA.
|
# Solution.
1). Let KS be the extension of KL beyond point L. Then LM is the bisector of angle MLS, since $\angle M L S = \angle M L A = \angle A L K = 60^{\circ}$. Point B is the intersection of the bisectors LM and KB of angles SLA and SKM, respectively $\Rightarrow$ the intersection of the bisectors of the external angles of triangle KLA $\Rightarrow \mathrm{AB}$ is the bisector of angle LAM.
2). Similarly, it can be proven that AC is the bisector of angle LAK, but angles LAM and LAK are adjacent $\Rightarrow \angle C A B = 90^{\circ}$.
3). In triangle KLA: E is the point of intersection of the bisectors $\Rightarrow$ $\angle K E A = \frac{1}{2} \angle K L A + 90^{\circ} = \frac{1}{2} 60^{\circ} + 90^{\circ} = 120^{\circ} \Rightarrow \angle B E A = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
4). In triangle ВЕА: $\angle A = 90^{\circ} \Rightarrow \angle B = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Answer: $\angle K B A = 30^{\circ}$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A pedestrian left point $A$ for point $B$, and after some delay, a second pedestrian followed. When the first pedestrian had walked half the distance, the second had walked 15 km, and when the second pedestrian had walked half the distance, the first had walked 24 km. Both pedestrians arrived at point $B$ simultaneously. 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 $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$ and $\frac{s-24}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24 ;$ $s^{2}-52 s+480=0 ; s_{1,2}=26 \pm 14 . s_{1}=40, s_{2}=12$ does not satisfy the conditions $s>15, s>24$.
Answer: 40 km.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. The base of the pyramid is a rectangle with sides $AB=24$ and $BC=30$, and the lateral edge of the pyramid $TA=16$ is perpendicular to the plane of the base. What is the minimum area that the section of the pyramid by a plane passing through the center of symmetry of the base $O$, the vertex of the pyramid, and a point $M$ lying on the side $BC$ can have? In this case, into what parts does point $M$ divide the edge $BC$?
|
# Solution:
Regardless of the position of point $M$ on side $B C$, the face $T A B$ is the orthogonal projection of the section $T M N$. The area of the section will be the smallest if the angle between the cutting plane and the face $T A B$ is the smallest. Since the cutting plane passes through the center of symmetry of the base $O$ and the vertex of the pyramid $T$, the segment $O T$ is the inclined line to the plane of the face $T A B$, and the smallest possible angle will be $\angle O T F$, where $O F \perp T A B, F \in A B$. The line of intersection of the cutting plane and the plane of the face $T A B \quad T K \perp T F$ and intersects the line $A B$ at point $K$. If the condition $T K \perp T F$ is not met, then $F T_{1}\operatorname{tg} \alpha, \cos \alpha_{1}\frac{S_{\triangle A T B}}{\cos \alpha}$.
The line passing through points $K$ and $O$ intersects the edge $A D$ at point $N$ and the edge $B C$ at point $M, \triangle N T M$ is the desired section.
If we denote $\quad A B=a, B C=b, T A=c, \quad$ then $\quad T F=\sqrt{(a / 2)^{2}+c^{2}}$;
$T O=\sqrt{T F^{2}+O F^{2}}=\sqrt{(a / 2)^{2}+c^{2}+(b / 2)^{2}}=\sqrt{a^{2}+b^{2}+4 c^{2}} / 2, \quad \cos \angle O T F=\frac{T F}{T O}=\frac{\sqrt{a^{2}+4 c^{2}}}{\sqrt{a^{2}+b^{2}+4 c^{2}}}$.
$S_{\triangle N T M}=\frac{S_{\triangle A T B}}{\cos \alpha}=\frac{a c \sqrt{a^{2}+b^{2}+4 c^{2}}}{2 \sqrt{a^{2}+4 c^{2}}}=\frac{a c \sqrt{(a / 2)^{2}+(b / 2)^{2}+c^{2}}}{2 \sqrt{(a / 2)^{2}+c^{2}}}$.
In $\triangle K T F \angle K T F=90^{\circ}, A K=A T^{2} / A F=2 c^{2} / a . B K=A K+A B=\frac{2 c^{2}}{a}+a$.
Point $M$ divides the segment $B C$ in the ratio $\frac{B M}{M C}=\frac{B M}{A N}=\frac{B K}{A K}=\frac{a^{2}+2 c^{2}}{2 c^{2}}=1+\frac{1}{2} \frac{a^{2}}{c^{2}}$.
| $A B$ | $B C$ | $T A$ | $T F$ | $T O$ | $\cos \alpha$ | $S_{N T M}$ | $B M: M C$ | $B M$ | $M C$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 24 | 30 | 16 | 20 | 25 | $4 / 5$ | 240 | $17: 8$ | $102 / 5$ | $48 / 5$ |
First (qualifying) stage of the academic competition
"Step into the Future" School Olympiad in the subject of "Mathematics", autumn 2016
|
240
|
Geometry
|
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 still had 24 km to go, and when the second cyclist had covered half the distance, the first cyclist still had 15 km to go. 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 $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}} \quad$ and $\quad \frac{s-15}{v_{1}}=\frac{s}{2 v_{2}} . \quad$ From this, $\quad \frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; \quad s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24 ;$ $s^{2}-52 s+480=0 ; s_{1,2}=26 \pm 14 . s_{1}=40, \quad s_{2}=12$ does not satisfy the conditions $s>15, s>24$.
Answer: 40 km.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A batch of shoes, purchased for 180 thousand rubles, was sold in the first week at a price higher than the purchase price by $25 \%$, then the markup was reduced to $16 \%$ of the purchase price; and the entire batch of shoes was sold for $20 \%$ more than it was purchased for. For what amount was the shoes sold in the first week?
|
# Solution:
$x$ thousand rubles - the purchase cost of shoes sold in the first week, $y$ - the remainder.
$$
\left\{\begin{array} { c }
{ x + y = 180 } \\
{ 25 x + 16 y = 20 ( x + y ) ; }
\end{array} \left\{\begin{array} { c }
{ 5 x = 4 y , } \\
{ x + 5 / 4 x = 180 ; }
\end{array} \left\{\begin{array}{l}
x=80 \\
y=100
\end{array}\right.\right.\right.
$$
Revenue from sales in the first week is $80 \cdot 1.25=100$ thousand rubles.
Answer: 100 thousand rubles
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. What is the greatest value that the sum of the first $n$ terms of the arithmetic progression $113,109,105, \ldots$ can take?
#
|
# Solution:
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ takes its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. Since $a_{n}=a_{1}+d(n-1)$, from the inequality $113-4(n-1)>0$ we find $n=[117 / 4]=29$.
Then $\max S_{n}=S_{29}=0.5 \cdot(113+113-4 \cdot 28) \cdot 29=1653$.
Answer: 1653.
|
1653
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 2$) in a total of 100 pieces, with at least three flasks of each size. The lab assistant randomly selects three flasks in sequence, and fills the first one with an 80% salt solution, the second one with a 50% salt solution, and the third one with a 20% salt solution. Then he pours the contents of these three flasks into one dish and determines the percentage of salt in it. For what minimum number of large flasks $N$ will the event "the percentage of salt in the dish is between $45\%$ and $55\%$ inclusive" occur less frequently than the event "when two fair coins are tossed, one head and one tail appear (in any order)"? Justify your answer. (16 points)
|
Solution. If $N$ is the number of large flasks in the laboratory, $N=3,4, \ldots, 97$, then $n=100-N$ is the number of small flasks in the laboratory, $n=3,4, \ldots, 97$. For the event $A=\{$ the salt content in the dish is between $45 \%$ and $55 \%$ inclusive $\}$, it is necessary to find the smallest $N$ such that the probability $\mathrm{P}(A)45<N<55$. Therefore, $N_{\min }=46$.
Answer: when $N=46$.
|
46
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Clowns Plukha and Shmyaka have six pairs of valenki (traditional Russian felt boots) between them. Each pair of valenki is painted in a unique color, and the valenki in a pair are identical (they are not distinguished as left or right). In how many ways can both clowns be simultaneously wearing mismatched valenki? (16 points)
|
Solution. We can use valenki (traditional Russian felt boots) from two, three, or four pairs.
1) We choose two pairs of valenki $C_{6}^{2}=15$ ways. Each clown puts on one valenok from different pairs, choosing which one for which foot. This gives us $15 \cdot 2 \cdot 2=60$ ways.
2) We use three pairs of valenki. We choose one complete pair of valenki, and two other pairs from which we will take one valenok each. This can be done $6 \cdot C_{5}^{2}=60$ ways. Each clown puts on one valenok from the pair, choosing which foot to put it on. This can be done in 4 ways. Then each clown chooses one of the two different valenki in 2 ways. This gives us $60 \cdot 4 \cdot 2=480$ ways.
3) We use four pairs of valenki. We choose four pairs from which we will take one valenok each. This can be done $C_{6}^{4}=15$ ways. The four chosen valenki of different colors can be put on the four feet of the clowns in 4! ways. This gives us $15 \cdot 4! =360$ ways.
Finally, we have $60+480+360=900$.
## Answer: 900.
|
900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. Option I.
Two different natural numbers are written on the board, the larger of which is 2015. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board.
|
Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $2015-1024=991$.
Answer: 991.
|
991
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. II variant.
Two different natural numbers are written on the board, the larger of which is 1580. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board.
|
Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $1580-1024=556$.
Answer: 556.
|
556
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1.
A librarian at a physics and mathematics school noticed that if the number of geometry textbooks in the school library is increased by several (integer) times and the number of algebra textbooks is added to the resulting number, the total is 2015. If the number of algebra textbooks is increased by the same factor and the number of geometry textbooks is added to the resulting number, the total is 1580. How many algebra textbooks are in the library?
|
Solution. Let the number of geometry textbooks be $x$, and the number of algebra textbooks be $y$. We can set up the system
$$
\left\{\begin{array}{l}
x n+y=2015 \\
y n+x=1580
\end{array}\right.
$$
We can write an equivalent system, where the equations represent the sum and difference of the equations in the original system:
$\left\{\begin{array}{l}x(n+1)+y(n+1)=3595, \\ x(n-1)-y(n-1)=435\end{array} \Leftrightarrow\left\{\begin{array}{l}(x+y)(n+1)=5 \cdot 719, \\ (x-y)(n-1)=3 \cdot 5 \cdot 29 .\end{array}\right.\right.$
The factor $(n+1)$ of the number $5 \cdot 719$ must be 2 more than the factor $(n-1)$ of the number $3 \cdot 5 \cdot 29$. Clearly, this is only satisfied when $n=4$. Then
$$
\left\{\begin{array} { l }
{ x + y = 7 1 9 } \\
{ x - y = 1 4 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=432 \\
y=287
\end{array}\right.\right.
$$
Answer: 287.
|
287
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6.
A $10 \times 10$ square was cut into rectangles, the areas of which are different and expressed as natural numbers. What is the maximum number of rectangles that can be obtained?
|
Solution. The area of the square is 100. If we represent 100 as the sum of natural numbers, the number of addends will be the largest if the difference between the numbers is one. Let's take rectangles with areas of $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. Their total area is 55. Therefore, the sum of the areas of the remaining rectangles is 45.
Notice that if the area of a rectangle is greater than 10, it cannot be
a prime number, otherwise such a rectangle would have a side greater than 10 and would not fit into a $10 \times 10$ square. Composite numbers greater than ten are $12, 14, 15, 16, 18$, $19, \ldots$ Any four of them sum to a number greater than 45. Sums equal to 45 can be given, for example, by such three numbers: $12, 15, 18$ or $14, 15, 16$.
Thus, the number of rectangles is less than or equal to 13. An example of a possible arrangement for 13 rectangles is shown in the figure.
## Evaluation criteria for the correspondence round tasks for 9th grade
#
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The production of ceramic items consists of 3 sequential stages: forming a ceramic item on a potter's wheel for 15 minutes, drying for 10 minutes, and firing for 30 minutes. It is required to produce 75 items. How should 13 masters be distributed between molders and firers to work on stages 1 and 3 respectively for the entire duration of the stage (drying does not require workers), to complete the work in the shortest time possible? In the answer, write the shortest time (in minutes) required to complete the entire job. Provide the answer as a number without indicating the unit. (5 points)
|
Solution. Molders - 4, decorators - 8. The thirteenth master can work at any stage, or not participate in the work at all. In this case, the working time is 325 minutes $\left(55+\left(\left[\frac{75}{4}\right]+1\right) \cdot 15=325\right)$. We will show that with other arrangements, the working time is longer. Suppose there are fewer than 4 molders (3, 2, or 1), then the time for molding is no less than 375 minutes. If there are fewer than 8 decorators, then the time for the third stage is no less than 330 minutes. Answer: 325
|
325
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find all natural values of $n$ for which
$$
\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9}, \text { and } \log _{2}^{2} n+45<\log _{2} 8 n^{13}
$$
In your answer, write the sum of the obtained values of $n$.
(6 points)
|
# Solution.
$$
\begin{aligned}
& \cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9} \quad \Leftrightarrow 2 \sin \frac{\pi}{9} \cos \frac{2 \pi}{9}+2 \sin \frac{\pi}{9} \cos \frac{4 \pi}{9}+\cdots+ \\
& 2 \sin \frac{\pi}{9} \cos \frac{2 \pi n}{9}=2 \sin \frac{\pi}{9} \cos \frac{\pi}{9} \Leftrightarrow
\end{aligned}
$$
$\sin \frac{3 \pi}{9}-\sin \frac{\pi}{9}+\sin \frac{5 \pi}{9}-\sin \frac{3 \pi}{9}+\cdots+\sin \frac{\pi}{9}(1+2 n)-\sin \frac{\pi}{9}(2 n-1)=\sin \frac{2 \pi}{9} \Leftrightarrow$ $\sin \frac{\pi}{9}(1+2 n)-\sin \frac{\pi}{9}=\sin \frac{2 \pi}{9} \Leftrightarrow \sin \frac{\pi}{9}(1+2 n)=\sin \frac{2 \pi}{9}+\sin \frac{\pi}{9} \Leftrightarrow \sin \frac{\pi}{9}(1+2 n)=$ $2 \sin \frac{\pi}{6} \cos \frac{\pi}{18} \Leftrightarrow \sin \frac{\pi}{9}(1+2 n)=\sin \frac{4 \pi}{9} \Leftrightarrow 2 \sin \left(\frac{\pi n}{9}-\frac{\pi}{6}\right) \cos \left(\frac{\pi n}{9}+\frac{5 \pi}{18}\right)=0 \Leftrightarrow \frac{\pi n}{9}-\frac{\pi}{6}=$ $\pi k, k \in \mathbb{Z}, \quad \frac{\pi n}{9}+\frac{5 \pi}{18}=\frac{\pi}{2}+\pi m, m \in \mathbb{Z}, \Leftrightarrow 2 n=3(6 k+1), k \in \mathbb{Z}, \quad n=2+$
$9 m, m \in \mathbb{Z}, m \geq 0$. Since $n$ is a natural number, the first relation does not hold (an even number cannot be equal to an odd number). Therefore, $n=2+9 m, m \in \mathbb{Z}, m \geq 0$.
Solve the inequality $\log _{2}^{2} n+45<\log _{2} 8 n^{13} \Leftrightarrow \log _{2}^{2} n-13 \log _{2} n+42<0$, $\Leftrightarrow$ $\left(\log _{2} n-6\right)\left(\log _{2} n-7\right)<0, \Leftrightarrow 2^{6}<n<2^{7}, \Leftrightarrow 64<2+9 m<128, m \in \mathbb{Z}, m \geq$ $0 \Leftrightarrow 6 \frac{8}{9}<m<14, m \in \mathbb{Z}, m \geq 0 \Leftrightarrow m=7,8,9,10,11,12,13 . S_{7}=\frac{65+119}{2}$. $7=644$.
## Answer: 644
|
644
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In how many ways can a bamboo trunk (a non-uniform natural material) 4 m long be sawn into three parts, the lengths of which are multiples of 1 dm, and from which a triangle can be formed?
(12 points)
|
Solution.
Let $A_{n}$ be the point on the trunk at a distance of $n$ dm from the base. We will saw at points $A_{p}$ and $A_{q}, p<q$. To satisfy the triangle inequality, it is necessary and sufficient that each part is no longer than 19 dm:
$$
p \leq 19, \quad 21 \leq q \leq p+19
$$
Thus, the number of ways to choose $p$ and $q$ will be
$$
\sum_{p=1}^{19}(p+19-20)=0+1+2+\ldots+18=\frac{18 \cdot 19}{2}=171
$$
## Answer: 171
|
171
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. How many solutions in natural numbers $x, y$ does the inequality $x / 76 + y / 71 < 1$ have?
(12 points)
|
# Solution.
All solutions lie in the rectangle
$$
\Pi=\{0<x<76 ; \quad 0<y<41\}
$$
We are interested in the number of integer points lying inside Π below its diagonal $x / 76 + y / 41=1$. There are no integer points on the diagonal itself, since 76 and 41 are coprime. So we get half the number of integer points inside Π:
$$
\frac{75 \cdot 40}{2}=1500
$$
Answer: 1500
|
1500
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A workshop produces items of types $A$ and $B$. For one item of type $A$, 10 kg of steel and 23 kg of non-ferrous metals are used, and for one item of type $B-70$ kg of steel and 40 kg of non-ferrous metals are used. The profit from selling one item of type $A$ is 80 thousand rubles, and for type $B-100$ thousand rubles. The shift fund for steel is 700 kg, and for non-ferrous metals, it is 642 kg. How many items of types $A$ and $B$ should be produced per shift to maximize the profit from selling the items, given that resource consumption should not exceed the allocated shift funds? Write the maximum profit (in thousands of rubles) that can be obtained under these conditions. Provide the answer as a number without specifying the unit. $\quad$ (5 points)
|
Solution. Let $x$ be the number of items of type $A$, and $y$ be the number of items of type $B$. Then the profit per shift is calculated by the formula $D=80 x+100 y$, with the constraints $10 x+70 y \leq 700$, $23 x+40 y \leq 642$, and $x$ and $y$ are non-negative integers.
$10 x+70 y \leq 700 \quad \Leftrightarrow \quad y \leq 10-\frac{x}{7}, \quad 23 x+40 y \leq 642 \quad \Leftrightarrow \quad y \leq 16.05-\frac{23 x}{40}$.
The intersection point of the lines $x+7 y=70$ and $23 x+40 y=642$ is the point with coordinates $x=14, y=8$. In this case, $D=80 \cdot 14+100 \cdot 8=1920$, but when $x=\frac{642}{23}=27 \frac{21}{23}, y=0$, we have the maximum value of the function $D=80 x+100 y$ for non-negative $x$ and $y$ that satisfy the inequalities $10 x+70 y \leq 700, 23 x+40 y \leq 642$. This value is $2233 \frac{1}{23}$. Since $x$ and $y$ are integers, when $y=0, x=27$, the profit $D=2160$; when $y=1, x \leq \frac{602}{23}, x=26, D=2180$; when $y=2, x \leq \frac{562}{23}, x=24, D=2120$. Therefore, the maximum profit is 2180 thousand rubles, given the production of one item of type $A$ and 26 items of type $B$.
## Answer: 2180
|
2180
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In how many ways can the number 210 be factored into the product of four natural numbers? The order of the factors does not matter.
$(12$ points)
|
# Solution.
Factorize 210 into a product of prime numbers: $2 \cdot 3 \cdot 5 \cdot 7$. Let's see how 4 prime
#### Abstract
divisors can be distributed among the desired factors
1) $4+0+0+0-1$ way.
2) $3+1+0+0-C_{4}^{3}=4$ ways.
3) $2+2+0+0-C_{4}^{2} / 2=3$ ways.
4) $2+1+1+0-C_{4}^{2}=6$ ways.
5) $1+1+1+1-1$ way.
Answer: 15.
## Answer: 15
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The sequence is defined recursively:
$x_{0}=0, x_{n+1}=\frac{\left(n^{2}+n+1\right) x_{n}+1}{n^{2}+n+1-x_{n}} . \quad$ Find $x_{8453}$. (12 points)
|
Solution.
$$
\text { We calculate } x_{1}=\frac{1}{1}=1, x_{2}=\frac{4}{2}=2, x_{3}=\frac{15}{5}=3 \text {, a hypothesis emerges: } x_{n}=n \text {. }
$$
Let's verify by induction:
$$
x_{n+1}=\frac{\left(n^{2}+n+1\right) n+1}{n^{2}+n+1-n}=\frac{n^{3}+n^{2}+n+1}{n^{2}+1}=n+1
$$
## Answer: 8453
|
8453
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. What is the greatest area that a rectangle can have, the coordinates of whose vertices satisfy the equation $|y-x|=(y+x+1)(5-x-y)$, and whose sides are parallel to the lines $y=x$ and $y=-x$? Write the square of the found area in your answer. $\quad(12$ points $)$
#
|
# Solution.
Substitution: $x_{1}=x+y, y_{1}=y-x$. This substitution increases all dimensions by a factor of $\sqrt{2}$. We have $\left|y_{1}\right|=\left(x_{1}+1\right)\left(5-x_{1}\right), \quad S\left(x_{1}\right)=4\left(x_{1}-2\right)\left(x_{1}+1\right)\left(5-x_{1}\right), x_{1} \in(2 ; 5)$. $S^{\prime}\left(x_{1}\right)=4\left(\left(x_{1}+1\right)\left(5-x_{1}\right)+\left(x_{1}-2\right)\left(5-x_{1}\right)-\left(x_{1}-2\right)\left(x_{1}+1\right)\right)=$ $=-12\left(x_{1}^{2}-4 x_{1}+1\right)=-12\left(x_{1}-2-\sqrt{3}\right)\left(x_{1}-2+\sqrt{3}\right)=0$.
The maximum area is achieved at the point $x_{1}=2+\sqrt{3}$. In the coordinate system $\mathrm{Ox}_{1} \mathrm{y}_{1}$, the maximum area is
$$
S(2+\sqrt{3})=4 \sqrt{3}(3+\sqrt{3})(3-\sqrt{3})=24 \sqrt{3}
$$
In the original coordinate system, $S_{\max }=12 \sqrt{3}$. $\left(S_{\max }\right)^{2}=432$. Answer: 432
Answer: 432
|
432
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. The lateral face of a regular triangular pyramid $S A B C$ is inclined to the base plane $A B C$ at an angle $\alpha=\operatorname{arctg} \frac{3}{4}$. Points $M, N, K$ are the midpoints of the sides of the base $A B C$. Triangle $M N K$ is the lower base of a right prism. The edges of the upper base of the prism intersect the lateral edges of the pyramid $S A B C$ at points $F, P$, and $R$. The total surface area of the polyhedron with vertices at points $M, N, K, F, P, R$ is $53 \sqrt{3}$. Find the side of triangle $A B C$. (16 points)
|
# Solution.
The height of the pyramid $SO = h$. The side of the base of the pyramid $AC = a$.
The height of the prism is $3h/4$, and the sides of the base of the prism are $a/2$.
The area of triangle $MNK$:
$$
S_{MNK} = \frac{a^2 \sqrt{3}}{16}
$$
The area of triangle $FPR$:
$$
S_{FPR} = \frac{a^2 \sqrt{3}}{64}
$$
The area of triangle $MPN$:
$$
S_{MPN} = \frac{1}{2} \cdot \frac{a}{2} \cdot \frac{3h}{4}, \quad S_{MPN} = S_{NPK} = S_{KRM}
$$
The area of triangle $FPM: S_{FPN} = \frac{1}{2} \cdot \frac{a}{4} \cdot \sqrt{\frac{9h^2}{16} + \frac{3a^2}{64}}, \quad S_{FPN} = S_{PRK} = S_{RFM}$.
Since the radius of the circle inscribed in triangle $ABC$ is $\frac{a \sqrt{3}}{6}$, and all lateral faces of the pyramid are inclined to the base plane at an angle $\alpha = \operatorname{arctg} \frac{3}{4}$, then $h = \frac{a \sqrt{3}}{8}$.
According to the problem, the total surface area of the polyhedron with vertices at points $M, N, K, F, P, R$ is $53 \sqrt{3}$, i.e.,
$$
\begin{gathered}
S_{\mathrm{MH}} = S_{MNK} + S_{FPR} + 3 S_{MPN} + 3 S_{FPN} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9ah}{16} + \frac{3a}{8} \sqrt{\frac{9h^2}{16} + \frac{3a^2}{64}} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9a^2 \sqrt{3}}{128} \\
\quad + \frac{3a}{8} \sqrt{\frac{27a^2}{16 \cdot 64} + \frac{3a^2}{64}} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9a^2 \sqrt{3}}{128} + \frac{15a^2 \sqrt{3}}{256} \\
= \frac{53a^2 \sqrt{3}}{16^2} = 53 \sqrt{3}
\end{gathered}
$$
Therefore, $a = 16$.
Answer: 16
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events
$$
A=\{\text { There is a cat }\} \text { and } B=\{\text { There is a dog }\} .
$$
| | There is a dog | There is no dog |
| :--- | :---: | :---: |
| There is a cat | $a$ | $b$ |
| There is no cat | $c$ | $d$ |
These events are independent if the proportion of "cat owners" among "dog owners" is the same as among "non-dog owners," i.e., the conditional probability of event $A$ given $B$ is equal to the probability of event $A$ given $\bar{B}: \mathrm{P}(A \mid B)=\mathrm{P}(A \mid \bar{B})$. We get ${ }^{1} \frac{a}{a+c}=\frac{b}{b+d}$, from which $a d=b c$.
Due to random variability in the data, this equality may not be exact, but it should hold at least approximately. Suppose that all numbers in the Scientist's report are correct except for $a$. Then $a$ is approximately (rounded to the nearest integer) $\frac{b c}{d}=\frac{1110 \cdot 978}{121} \approx 8972$, and the total number of respondents is approximately $8972+1110+978+121=11181$. This is much greater than the number of residents in the city, so this scenario is implausible.
Suppose now that the number $b$ is incorrect. In this case, $b \approx \frac{a d}{c}=\frac{765 \cdot 121}{978} \approx 95$, and the total number of respondents is approximately $765+95+978+121=1959$, which is much less than 3000. Similarly, if the number $c$ is incorrect, then $c \approx \frac{a d}{b}=\frac{765 \cdot 121}{1110} \approx 83$, and the total number of respondents is close to $765+1110+83+121=2079$, which is also too small.
If the incorrect number is $d$, then $d \approx \frac{b c}{a}=\frac{1110 \cdot 978}{765} \approx 1419$, and the total number of survey participants is approximately $765+1110+978+1419=4272$. This is possible.
|
Answer: approximately 4272 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a guess about which number is incorrect) | 1 point |
| Solution is incorrect or missing | 0 points |
|
4272
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events $A=\{$ Has a card $\}$ and $B=\{$ Makes online purchases $\}$. These events are independent if the proportion of cardholders among online shoppers is the same as among those who do not make online purchases, that is,
| | Has a card | No card |
| :--- | :---: | :---: |
| Buys online | $a$ | $b$ |
| Does not buy online | $c$ | $d$ |
the conditional probability of event $A$ given $B$ is equal to the probability of event $A$ given $\bar{B}: \mathrm{P}(A \mid B)=\mathrm{P}(A \mid \bar{B})$. We get $\frac{a}{a+b}=\frac{c}{c+d}$, from which $a d=b c$.
Due to random variability in the data, this equality may not be exact, but it should at least hold approximately. Suppose that all numbers in the Scientist's report are correct except for $a$. Then the number $a$ should approximately (rounded to the nearest integer) equal $\frac{b c}{d}=\frac{245 \cdot 1142}{535} \approx 523$, and the number of respondents will be approximately $523+1142+535+245=2445$. This is much greater than 2000. The assumption is implausible.
Suppose now that the incorrect number is $b$. Then $b \approx \frac{a d}{c}=\frac{81 \cdot 535}{1142} \approx 38$, and the total number of survey participants is close to $1142+535+81+38=1796$. This is possible.
If the incorrect number is $c$, then $c \approx \frac{a d}{b}=\frac{81 \cdot 535}{245} \approx 177$, and the approximate total number of respondents is $177+535+81+245=1038$. This is too small: according to the condition, the sample size is more than 1500 people.
If the incorrect number is $d$, then $d \approx \frac{c b}{a}=\frac{1142 \cdot 245}{81} \approx 3454$, and the total sample size is approximately $1142+245+81+3454=4922$. This is also implausible.
|
Answer: approximately 1796 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a guess about which number is incorrect) | 1 point |
| Solution is incorrect or missing | 0 points |
|
1796
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. Minimum Sum (9th grade, 4 points). There are a lot of symmetric dice. They are thrown simultaneously. With some probability $p>0$, the sum of the points can be 2022. What is the smallest sum of points that can fall when these dice are thrown with the same probability $p$?
|
Answer: 337.
Solution. The distribution of the sum rolled on the dice is symmetric. To make the sum $S$ the smallest possible, the sum of 2022 should be the largest possible. This means that the sum of 2022 is achieved when sixes are rolled on all dice. Therefore, the total number of dice is $2022: 6=337$. The smallest sum of 337 is obtained if ones are rolled on all dice.
|
337
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. The King's Path (from 7th grade, 2 points). A chess king is on the a1 square of a chessboard and wants to move to the h8 square, moving right, up, or diagonally up-right. In how many ways can he do this?
|
Solution. Instead of letter designations for the columns, we use numerical ones. Then, instead of al, we can write $(1,1)$. Let $S_{m, n}$ be the number of ways to get from the cell $(1,1)$ to the cell $(m, n)$. We are interested in $S_{8,8}$.
Obviously, $S_{1, n}=1$ and $S_{m, 1}=1$ for all $m$ and $n$. In particular, we can consider that $S_{1,1}=1$, since the king can get to the cell $(1,1)$ in only one way: by standing in place and doing nothing.
If $m>1$ and $n>1$, then the cell $(m, n)$ can be reached only from one of the three cells $(m-1, n),(m, n-1)$, and $(m-1, n-1)$. This gives the recurrence relation $S_{m, n}=S_{m-1, n}+S_{m, n-1}+S_{m-1, n-1}$.
The calculations can be easily performed in Excel (Fig. 2).
| INDEX | | | $-x \vee f=C 5+C 6+D 6$ | | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 4 | A | B | C | D | E | F | G | H | I |
| | 8 | 1 | 15 | 113 | 575 | 2241 | 7183 | 19825 | 48639 |
| | 7 | 1 | 13 | 85 | 377 | 1289 | 3653 | 8989 | 19825 |
| | 6 | 1 | 11 | 61 | 231 | 681 | 1683 | 3653 | 7183 |
| | 5 | 1 | 9 | 41 | 129 | 321 | 681 | 1289 | 2241 |
| | 4 | 1 | 7 | $=\mathrm{C} 5+$ | +D6 | 129 | 231 | 377 | 575 |
| | 3 | 1 | 5 | 13 | 25 | 41 | 61 | 85 | 113 |
| | 2 | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| | | a | b | c | d | e | f | g | h |
Fig. 2. Solution to problem 8 in Excel
Answer: 48639.
|
48639
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. Happy Sums (from 8th grade, 4 points). In the "Happy Sum" lottery, there are $N$ balls numbered from 1 to $N$. During the main draw, 10 balls are randomly drawn. During the additional draw, 8 balls are randomly selected from the same set of balls. The sum of the numbers on the drawn balls in each draw is announced as the happy sum, and players who predicted this sum win a prize.
Can it be that events $A$ "the happy sum in the main draw is 63" and $B$ "the happy sum in the additional draw is 44" are equally likely? If yes, under what condition?
|
Solution. In the main draw, there are $C_{N}^{10}$ possible combinations, and in the additional draw, there are $C_{N}^{8}$ combinations. Let $S_{63,10}$ be the number of combinations in the first draw where the sum is 63. The addends are different natural numbers from 1 to $N$. If the smallest number is 2 or more, then the sum of all ten addends is no less than
$$
2+3+\ldots+11=\frac{2+11}{2} \cdot 10=13 \cdot 5=65
$$
Thus, the smallest number is 1. By removing one from each addend, i.e., subtracting 10 from the sum, we find that the remaining sum of 53 is obtained by nine different natural addends, each not exceeding the number $N$. In other words, the number of ways to get a sum of 63 with ten addends is the same as the number of ways to get a sum of 53 with nine addends: $S_{63,10}=S_{53,9}$.
Reasoning similarly, we get: $S_{63,10}=S_{53,9}=S_{44,8}$. Now we can find the ratio of probabilities:
$$
\frac{\mathrm{P}(B)}{\mathrm{P}(A)}=\frac{S_{44,8} \cdot C_{N}^{10}}{C_{N}^{8} \cdot S_{63,10}}=\frac{C_{N}^{10}}{C_{N}^{8}}=\frac{8!(N-8)!}{10!(N-10)!}=\frac{(N-8)(N-9)}{90}
$$
Let's find for which $N$ the obtained fraction equals one:
$$
\frac{(N-8)(N-9)}{90}=1 ; N^{2}-17 N-18=0, \text { from which } N=18
$$
Answer: This is possible if there are 18 balls in the lottery.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In the village of Znoynoye, there are exactly 1000 residents, which exceeds the average population of villages in the valley by 90 people.
How many residents are there in the village of Raduzhny, which is also located in Sunny Valley?
|
Solution. Let $x$ be the total number of residents in all other villages except Znoynoye. Then the average population is
$$
\frac{1000+x}{10}=100+0.1 x=910, \text{ hence } x=8100
$$
Thus, the average population in 9 villages, except Znoynoye, is 900 people. If there are more than 900 residents in Raduzhny, there must be a village with fewer than 900 residents, but then the number of residents in Raduzhny would differ from the number of residents in Znoynoye by more than 100. This is a contradiction. Therefore, there are no more than 900 residents in Raduzhny. It is obvious that there are no fewer than 900 residents in Raduzhny. Thus, there are exactly 900 residents.
Answer: 900.
|
900
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Promotion. The store "Crocus and Cactus" announced a promotion: if a customer buys three items, the cheapest one of them is free. The customer took ten items, all of different prices: 100 rubles, 200 rubles, 300 rubles, and so on - the most expensive item cost 1000 rubles.
a) (from 6th grade. 1 point). What is the smallest amount the customer will pay if they arrange their purchases skillfully before the cashier?
b) (from 6th grade. 1 point). What is the largest amount the customer will pay if they arrange the purchases poorly?
c) (from 9th grade. 3 points). Find the expected value of the amount the customer will pay if they arrange the purchases randomly.
|
Solution. a) The cheapest item in each trio should be as expensive as possible. If the buyer distributes the purchases into trios as follows (indicating only the prices):
$(1000,900,800),(700,600,500),(400,300,200)$ and a separate item for $100 \mathrm{p}.$, then he will pay $1000+900+700+600+400+300+100=4000$ rubles.
b) The worst distribution is when the buyer gets the three cheapest items for free under the promotion, which are 100, 200, and 300 rubles. In this case, the buyer will pay 4900 rubles. For example, the distribution could be:
$(1000,900,100),(800,700,200),(600,500,300)$ and a separate item for $400 \mathrm{p}$.
c) Let's take an item priced at $100 k$ rubles ( $k=1,2, \ldots, 10)$. Introduce a random variable - the indicator $I_{k}$, which is 1 if the buyer did not have to pay for this item, and 0 if they did. Obviously, $I_{10}=I_{9}=0$.
In other cases, $I_{k}=1$ only if, along with the item priced at $100 k$, two of the $10-k$ more expensive items are in the same trio. Such a combination can be formed in $C_{10-k}^{2}$ ways. The probability of such a combination is $\frac{C_{10-k}^{2}}{C_{9}^{2}}=\frac{(10-k)(9-k)}{72}$. The number $C_{9}^{2}$ in the denominator of the fraction is the total number of ways to select two items from the remaining nine to form a trio with the item priced at $100 k$.
We get the distribution
$$
I_{k} \sim\left(\begin{array}{cc}
0 \\
-\frac{k^{2}-19 k+18}{72} & \frac{k^{2}-19 k+90}{72}
\end{array}\right)
$$
The expected value: $\mathrm{E} I_{k}=\frac{k^{2}-19 k+90}{72}$. The total amount the buyer will save is
$$
S=100 I_{1}+200 I_{2}+\ldots+1000 I_{10}=100\left(I_{1}+2 I_{2}+\ldots+8 I_{8}\right)=100 \sum_{k=1}^{8} k I_{k}
$$
Transitioning to expectations, we get:
$$
\begin{gathered}
\mathrm{E} S=100 \sum_{k=1}^{8} \frac{k\left(k^{2}-19 k+90\right)}{72}= \\
=100\left(\frac{72}{72}+2 \cdot \frac{56}{72}+3 \cdot \frac{42}{72}+4 \cdot \frac{30}{72}+5 \cdot \frac{20}{72}+6 \cdot \frac{12}{72}+7 \cdot \frac{6}{72}+8 \cdot \frac{2}{72}\right)=916 \frac{2}{3} .
\end{gathered}
$$
The expected savings have been found. Therefore, the expected amount to be paid is
$$
\left.\frac{100+1000}{2} \cdot 10-916 \frac{2}{3}=5500-916 \frac{2}{3}=4583 \frac{1}{3} \text { (rubles }\right),
$$
which is 4583 rubles and 33 kopecks.
Answer: a) 4000 rubles; b) 4900 rubles; c) 4583 rubles 33.
|
4583
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Stubborn Squares (from 7th grade. 2 points). Given 100 numbers. 2 was added to each of them. The sum of the squares of the numbers did not change. 2 was added to each of the resulting numbers again. How did the sum of the squares change now?
|
Solution. Let the given numbers be $x_{1}, x_{2}, \ldots, x_{100}$. By adding 2 to each number, we get the set $y_{j}=x_{j}+2$. Adding 2 again, we get the set $z_{j}=y_{j}+2=x_{j}+4$. However, the variances of all three sets are equal, that is,
$$
\overline{x^{2}}-\bar{x}^{2}=\overline{y^{2}}-\bar{y}^{2}=\overline{z^{2}}-\bar{z}^{2}
$$
By the condition, $\overline{x^{2}}=\overline{y^{2}}$ and, obviously, $\bar{y}=\bar{x}+2$. Therefore, $\bar{x}^{2}=(\bar{x}+2)^{2}$, from which $\bar{x}=-1$. Then $\bar{z}=-1+4=3$. Therefore,
$$
\overline{z^{2}}=\overline{x^{2}}+\bar{z}^{2}-\bar{x}^{2}=\overline{x^{2}}+9-1=\overline{x^{2}}+8
$$
Since there are 100 numbers in total, the sum of the squares has increased by $8 \cdot 100=800$.
Answer: the sum of the squares increased by 800.
|
800
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Retro Collection (recommended from 8th grade, 2 points). Vitya collects toy cars from the "Retro" series. The problem is that the total number of different models in the series is unknown - it is the biggest commercial secret, but it is known that different cars are produced in the same edition, and therefore it can be assumed that all models are uniformly and randomly distributed across different online stores.
On different websites, Vitya found several offers, but upon closer inspection, it turned out that among the offered cars, there were only 12 different ones. Vitya is almost convinced that there are only 12 cars in total and that further search is futile. But who knows?
How many more offers from other stores should Vitya consider to be convinced that there are only 12 models in the series? Vitya considers himself convinced of something if the probability of this event is higher than 0.99.
|
Solution. Let there be a series of $n \geq 13$ cars. The probability that among the next $k$ offers there will only be the 12 models that Vitya has already seen is
$$
\left(\frac{12}{n}\right)^{k} \leq\left(\frac{12}{13}\right)^{k}
$$
We form the inequality: $\left(\frac{12}{13}\right)^{k}\log _{12 / 13} 0.01=\frac{\ln 100}{\ln 13-\ln 12}=57.53 \ldots
$$
The smallest integer value of $k$ is 58.
Answer: 58.
Comment. The smallest integer solution to the inequality can be found without logarithms by trial and error using a calculator or computer.
|
58
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. The Right Stars (from 8th grade, 3 points). Let natural numbers $k$ and $n$ be coprime, with $n \geq 5$ and $k < n / 2$. A proper $(n ; k)$-star is defined as a closed broken line that results from replacing every $k$ consecutive sides in a regular $n$-gon with a diagonal having the same endpoints. For example, the $(5 ; 2)$-star shown in the figure has 5 points of self-intersection, which are the bold points in the figure. How many self-intersections does the star $(2018 ; 25)$ have?
|
Solution. We will solve the problem in general for the $(n ; k)$-star. Let the vertices of the $n$-gon be denoted by $A_{0}, A_{1}$, and so on up to $A_{n-1}$. Take two consecutive segments of the star: $A_{0} A_{k}$ and $A_{1} A_{k+1}$. They intersect at a circle of radius $r_{1}$ with the center at the center of the $n$-gon. By rotating by $360\%$, we obtain another $n-1$ points lying on the same
circle. In total, there are $n$ points on this circle, located at the vertices of a smaller $n$-gon.
Now take the pair of segments $A_{0} A_{k}$ and $A_{2} A_{k+2}$. They have a common point on a circle of radius $r_{2} > r_{1}$. There are $n$ such points.
In the diagram, the case $k=3$ is shown: the intersection point of the segments $A_{0} A_{3}$ and $A_{1} A_{4}$ lies on a circle of radius $r_{1}$. On this same circle, the intersection point of the segments $A_{1} A_{4}$ and $A_{2} A_{5}$ is also located. The intersection point of the segments $A_{0} A_{4}$ and $A_{2} A_{5}$ lies on a larger circle of radius $r_{2}$.
Proceeding in this manner, we obtain a total of $k-1$ circles of different radii, each of which contains exactly $n$ intersection points. Therefore, no two points coincide. This way, all intersection points of the segments are enumerated, and there are a total of $n(k-1)$ of them.
For $n=2018, k=25$, we get that the number of self-intersections is $2018 \cdot 24 = 48432$.
Other solutions are possible, but in any case, it is necessary to explain why no two intersection points coincide.
Answer: 48432.
|
48432
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Target (from 8th grade, 2 points). A target is hanging on the wall, consisting of five zones: a central circle (the bullseye) and four colored rings (see figure). The width of each ring is equal to the radius of the bullseye. It is known that the number of points for hitting each zone is inversely proportional to the probability of hitting that zone, and that the bullseye is worth 315 points. How many points is a hit in the blue (second-to-last) zone worth?
|
Solution. Suppose, for definiteness, that the apple has a radius of 1. Then the blue zone is enclosed between circles with radii 3 and 4. The probability of hitting the blue zone relative to the probability of hitting the apple is the ratio of the areas of these zones:
$$
\frac{p_{g}}{p_{c}}=\frac{1}{4^{2}-3^{2}}=\frac{1}{7}
$$
Then the number of points \( x \) for the blue zone can be found from the proportion
$$
\frac{x}{315}=\frac{p_{g}}{p_{c}}=\frac{1}{7}
$$
from which \( x = 315 : 7 = 45 \).
Answer: 45.
## $\mathrm{XI}$ Internet Olympiad in Probability Theory and Statistics. Main (correspondence) round. $15.12 .77-21.01 .18$
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed that each time he gets a random airplane. Out of curiosity and habit, the Absent-Minded Scientist records the name of the airplane he flies on each time. In his fifteenth flight, the Scientist found himself on board an airplane proudly named "Siméon Denis Poisson." After takeoff, the Scientist took out his notebook to record the name of the airplane and discovered that he had already flown on "Poisson" once before, and there were no other repetitions before that. Estimate the number of airplanes in the airline.
|
Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.
Let's find $\mathrm{E} X$. Denote $n$ as the number of planes in the airline. Consider the event $A_{j}$ "there were no repetitions up to and including the $j$-th flight" $(j=1, \ldots, n)$. Let $I_{j}$ be the indicator of the event $A_{j}$. Then
$$
X=1+I_{1}+I_{2}+\ldots+I_{n}
$$
Let's find the expected values of the indicators.
$$
\mathrm{EI}_{j}=\mathrm{P}\left(A_{j}\right)=\frac{n(n-1)(n-2) \ldots(n-j+1)}{n^{j}}=\frac{n!}{(n-j)!\cdot n^{j}}
$$
Therefore,
$$
\mathrm{E} X=\mathrm{E}\left(1+I_{1}+I_{2}+\ldots+I_{n}\right)=\sum_{j=0}^{n} \frac{n!}{(n-j)!\cdot n^{j}}=\frac{n!}{n^{n}} \sum_{j=0}^{n} \frac{n^{n-j}}{(n-j)!}=\frac{n!}{n^{n}} \sum_{k=0}^{n} \frac{n^{k}}{k!}
$$
In the last transformation, we made the substitution $k=n-j$.
Consider the resulting sum separately:
$$
\sum_{k=0}^{n} \frac{n^{k}}{k!}=e^{n} \sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=e^{n} F(n)
$$
where $F(x)$ is the cumulative distribution function of the Poisson distribution with parameter $n$. The median of the Poisson distribution with an integer parameter $n$ is $n$. This means that
$$
F(n-1) \leq \frac{1}{2} \leq F(n)
$$
therefore
$$
F_{P}(n)=\sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=\frac{1}{2}+\theta \frac{n^{n} e^{-n}}{n!}
$$
where $0 \leq \theta \leq 1$. Setting $\theta=1 / 2$ for definiteness, we get:
$$
\mathrm{E} X=\frac{n!e^{n}}{n^{n}} F_{P}(n) \approx \frac{n!e^{n}}{n^{n}}\left(\frac{1}{2}+\frac{n^{n} e^{-n}}{2 n!}\right)=\frac{n!e^{n}}{2 n^{n}}+\frac{1}{2}
$$
Using Stirling's formula
$$
\mathrm{E} X \approx \frac{\sqrt{2 \pi n} \cdot n^{n} e^{n}}{2 n^{n} e^{n}}+\frac{1}{2}=\sqrt{\frac{\pi n}{2}}+\frac{1}{2}
$$
To find the estimate $x=\hat{n}$, solve the equation $\sqrt{\frac{\pi x}{2}}+\frac{1}{2}=15$. With rounding to the nearest integer, we get: $x \approx 134$.
Answer: approximately 134 (point estimate using the method of moments).
[^0]: ${ }^{1}$ English name - Stem-and-Leaf plot.
[^1]: ${ }^{2}$ The mode is the number that appears in the set more times than any other.
[^2]: ${ }^{3}$ It is believed that this problem was invented by Academician Andrei Dmitrievich Sakharov. I. F. Ginzburg. Academician A.D. Sakharov. Scientific works. Collection. - M.: AOZT "Izdatel'stvo TsentrKom", 1995.
[^3]: ${ }^{4}$ The number $e$ is also called Euler's number.
[^4]: ${ }^{5}$ The median of a random variable $X$ is a number $m$ such that $\mathrm{P}(X \leq m) \geq 0.5$ and $\mathrm{P}(X \geq m) \geq 0.5$. If the random variable has a finite density function $y=f(x)$, then the line $x=m$ divides the figure bounded by the x-axis and the density graph into two figures, each with an area of 0.5.
|
134
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Ministers in Anchuria. In the government cabinet of ministers in Anchuria, there are 100 ministers. Among them, there are crooks and honest ministers. It is known that among any ten ministers, at least one minister is a crook. What is the smallest number of crook ministers that can be in the cabinet?
|
Solution. There are no more than nine honest ministers, otherwise, a group of ten honest ministers would be found, which contradicts the condition. Therefore, the number of minister-cheats is no less than $100-9=91$.
Answer: 91.
|
91
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. For a procrastinator, the waiting time will increase by $a$ minutes.
Therefore, the total wasted time of all standing in the queue will decrease by $b-a$ minutes. We will thus swap people in pairs of "Procrastinator-Hurry" until we get a queue where all the hurries stand before all the procrastinators. In this queue, the wasted time will be minimal.
Let's find the total wasted time. The second person waits for the first, the third waits for the previous two, and so on.
The total time spent waiting for all the hurries in front to finish their operations is
$$
(a+2a+\ldots+(n-1)a)+m \cdot n a=a \cdot \frac{1+n-1}{2} \cdot(n-1)+m n a=a C_{n}^{2}+m n a
$$
Here, the term $m n a$ is the total time spent by all procrastinators waiting for all the hurries.
The total time spent waiting for the procrastinators in front to finish their operations is $b+2b+\ldots+(m-1)b=b C_{m}^{2}$.
The overall minimum waiting time for all customers is
$$
T_{\min }=a C_{n}^{2}+a m n+b C_{m}^{2}
$$
In our specific case, we get:
$$
1 \cdot C_{5}^{2}+1 \cdot 5 \cdot 3+5 \cdot C_{3}^{2}=10+15+15=40
$$[^3]
Similarly, it can be proven that the maximum wasted time will be if all procrastinators stand before all hurries. This time is
$$
T_{\max }=a C_{n}^{2}+b m n+b C_{m}^{2}
$$
With the numerical data from the problem, we get $1 \cdot C_{5}^{2}+5 \cdot 5 \cdot 3+5 \cdot C_{3}^{2}=10+75+15=100$.
b) Consider the $k$-th customer in the queue. Let $X_{k}$ be the number of procrastinators standing before him. Then $X_{k}=I_{1}+I_{2}+\ldots+I_{k-1}$, where the indicator $I_{j}$ takes the value 1 if the $j$-th customer is a procrastinator and 0 if the $j$-th customer is a hurry.
The $j$-th customer can be a procrastinator with probability $\frac{m}{m+n}$ and a hurry with probability $\frac{n}{m+n}$. Therefore, $\mathrm{E} I_{j}=0 \cdot \frac{n}{m+n}+1 \cdot \frac{m}{m+n}=\frac{m}{m+n}$.
Thus, $\mathrm{E} X_{k}=\frac{(k-1) m}{m+n}$.
Let $T_{k}$ be the waiting time of the $k$-th customer. We get:
$$
T_{k}=X_{k} b+\left(k-1-X_{k}\right) a=(b-a) X_{k}+a(k-1)
$$
Therefore,
$$
\mathrm{E} T_{k}=(b-a) \mathrm{E} X_{k}+a(k-1)=(b-a) \cdot \frac{(k-1) m}{m+n}+a(k-1)=\frac{b m+a n}{m+n} \cdot(k-1)
$$
Summing the obtained expression over all customers from 1 to $m+n$-th, we get the expected total wasted time:
$$
\begin{gathered}
\mathrm{E} T=\mathrm{E}\left(T_{1}+T_{2}+\ldots+T_{n}\right)=\frac{b m+a n}{m+n} \cdot(0+1+2+\ldots+(m+n-1))= \\
=\frac{(b m+a n)(n+m-1)}{2}=C_{n+m}^{2} \cdot \frac{b m+a n}{m+n}
\end{gathered}
$$
Substituting the known numbers from the condition: $C_{8}^{2} \cdot \frac{5 \cdot 3+1 \cdot 5}{8}=\frac{7 \cdot 20}{2}=70$.
|
Answer: a) 40 minutes and 100 minutes; b) 70 minutes.
Comment. The expected time turned out to be the arithmetic mean between the maximum and minimum possible. Try to prove the fact $\mathrm{E} T=\frac{T_{\min }+T_{\max }}{2}$ in general, that is, prove the combinatorial identity
$$
C_{n+m}^{2} \cdot \frac{b m+a n}{m+n}=a C_{n}^{2}+\frac{a+b}{2} m n+b C_{m}^{2}
$$
By the way, part b) could have been solved almost without calculations, by noting that each arrangement of slowpokes and hurries can be matched with a symmetrically equal and equally probable arrangement, and the sum of the waiting times for both arrangements will be equal to $a C_{n}^{2}+\frac{a+b}{2} \cdot m n+b C_{m}^{2}$. For this, it is convenient to use the graphical method (see the footnote above).
|
40
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Median. In a set of 100 numbers. If one number is removed, the median of the remaining numbers is 78. If another number is removed, the median of the remaining numbers is 66. Find the median of the entire set.
|
Solution. Arrange the numbers in ascending order. If we remove a number from the first half of the sequence (with a number up to 50), the median of the remaining numbers will be the $51-\mathrm{st}$ number in the sequence. If we remove a number from the second half, the median of the remaining numbers will be the number with the 50th position, but it is not greater than the 51st number. Therefore, the 50th number is 66, and the 51st number is 78. Thus, the median of the entire set is ${ }^{2}$ $\frac{66+78}{2}=72$.
Answer: 72.[^1]
|
72
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
|
Solution. With probability $\frac{1}{3}$, the pedestrian falls into the interval when the green light is on (event $G$), and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\bar{G}$). In this case, the waiting time is uniformly distributed over the interval from 0 to 2 minutes, so the conditional expected value of $T$ given $\bar{G}$ is one minute: $\mathrm{E}(T \mid G)=\frac{0+2}{2}=1$ (min).
We get:
$$
\mathrm{E}(T)=\mathrm{E}(T \mid G) \cdot \mathrm{P}(G)+\mathrm{E}(T \mid \bar{G}) \cdot \mathrm{P}(\bar{G})=\frac{1}{3} \cdot 0+\frac{2}{3} \cdot 1=\frac{2}{3}(\text { min })
$$
Answer: 40 seconds.
|
40
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. There are 25 children in the class. Two are chosen at random for duty. The probability that both duty students will be boys is $\frac{3}{25}$. How many girls are in the class?
|
# Solution.
Let there be $n$ boys in the class, then the number of ways to choose two duty students from them is $\frac{n(n-1)}{2}$, the number of ways to choose two duty students from the entire class is $25 \cdot 24$. The ratio of the two obtained fractions, i.e., $\frac{n(n-1)}{25 \cdot 24}$. It is also equal to $\frac{3}{25}$ according to the condition. We get the quadratic equation $n^{2}-n-72=0$. Its roots are -8 and 9. Only the positive root fits us, i.e., there are 9 boys in the class, and, consequently, 16 girls.
Answer: 16.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young driver Yura spends one minute on the path from $A$ to $B$ or back. Yura also spends one minute on the loop. The loop can only be driven counterclockwise (arrows indicate possible directions of movement). Yura does not turn back halfway and does not stop. The race duration is 10 minutes. Find the number of possible different routes (sequences of passing sections). #
|
# Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ can be reached in one minute either from point $A$ or from point $B$, so $M_{n, B}=M_{n-1}$.
A point $A$ can be reached in one minute only from point $B$, so $M_{n, A}=M_{n-1, B}=M_{n-2}$. Therefore,
$$
M_{n, A}=M_{n-2}=M_{n-2, A}+M_{n-2, B}=M_{n-4}+M_{n-3}=M_{n-2, A}+M_{n-1, A}
$$
Additionally, note that $M_{1, A}=0, M_{2, A}=1$. Thus, the numbers $M_{n, A}$ form the sequence $0,1,1,2,3,5,8,13,21,34, \ldots$
The number $M_{10, A}$ is 34 - the ninth Fibonacci number ${ }^{1}$.
Answer: 34.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed that each time he gets a random airplane. Out of curiosity and habit, the Absent-Minded Scientist records the name of the airplane he flies on each time. In his fifteenth flight, the Scientist found himself on board an airplane proudly named "Siméon Denis Poisson." After takeoff, the Scientist took out his notebook to record the name of the airplane and discovered that he had already flown on the "Poisson" once before, and there were no other repetitions before that. Estimate the number of airplanes in the airline.
|
Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.
Let's find $\mathrm{E} X$. Denote $n$ as the number of planes in the airline. Consider the event $A_{j}$ "there were no repetitions up to and including the $j$-th flight" $(j=1, \ldots, n)$. Let $I_{j}$ be the indicator of the event $A_{j}$. Then
$$
X=1+I_{1}+I_{2}+\ldots+I_{n}
$$
Let's find the expected values of the indicators.
$$
\mathrm{EI}_{j}=\mathrm{P}\left(A_{j}\right)=\frac{n(n-1)(n-2) \ldots(n-j+1)}{n^{j}}=\frac{n!}{(n-j)!\cdot n^{j}}
$$
Therefore,
$$
\mathrm{E} X=\mathrm{E}\left(1+I_{1}+I_{2}+\ldots+I_{n}\right)=\sum_{j=0}^{n} \frac{n!}{(n-j)!\cdot n^{j}}=\frac{n!}{n^{n}} \sum_{j=0}^{n} \frac{n^{n-j}}{(n-j)!}=\frac{n!}{n^{n}} \sum_{k=0}^{n} \frac{n^{k}}{k!}
$$
In the last transformation, we made the substitution $k=n-j$.
Consider the resulting sum separately:
$$
\sum_{k=0}^{n} \frac{n^{k}}{k!}=e^{n} \sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=e^{n} F(n)
$$
where $F(x)$ is the cumulative distribution function of the Poisson distribution with parameter $n$. The median of the Poisson distribution with an integer parameter $n$ is $n$. This means that
$$
F(n-1) \leq \frac{1}{2} \leq F(n)
$$
therefore
$$
F_{P}(n)=\sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=\frac{1}{2}+\theta \frac{n^{n} e^{-n}}{n!}
$$
where $0 \leq \theta \leq 1$. Setting $\theta=1 / 2$ for definiteness, we get:
$$
\mathrm{E} X=\frac{n!e^{n}}{n^{n}} F_{P}(n) \approx \frac{n!e^{n}}{n^{n}}\left(\frac{1}{2}+\frac{n^{n} e^{-n}}{2 n!}\right)=\frac{n!e^{n}}{2 n^{n}}+\frac{1}{2}
$$
Using Stirling's formula
$$
\mathrm{E} X \approx \frac{\sqrt{2 \pi n} \cdot n^{n} e^{n}}{2 n^{n} e^{n}}+\frac{1}{2}=\sqrt{\frac{\pi n}{2}}+\frac{1}{2}
$$
To find the estimate $x=\hat{n}$, solve the equation $\sqrt{\frac{\pi x}{2}}+\frac{1}{2}=15$. With rounding to the nearest integer, we get: $x \approx 134$.
Answer: approximately 134 (point estimate using the method of moments).
[^0]: ${ }^{1}$ English name - Stem-and-Leaf plot.
[^1]: ${ }^{2}$ The mode is the number that appears in the set more times than any other.
[^2]: ${ }^{3}$ It is believed that this problem was invented by Academician Andrei Dmitrievich Sakharov. I. F. Ginzburg. Academician A.D. Sakharov. Scientific works. Collection. - M.: AOZT "Izdatel'stvo TsentrKom", 1995.
[^3]: ${ }^{4}$ The number $e$ is also called Euler's number.
[^4]: ${ }^{5}$ The median of a random variable $X$ is a number $m$ such that $\mathrm{P}(X \leq m) \geq 0.5$ and $\mathrm{P}(X \geq m) \geq 0.5$. If the random variable has a finite density function $y=f(x)$, then the line $x=m$ divides the figure bounded by the x-axis and the density graph into two figures, each with an area of 0.5.
|
134
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. For a procrastinator, the waiting time will increase by $a$ minutes.
Therefore, the total wasted time of all standing in the queue will decrease by $b-a$ minutes. We will thus swap people in pairs of "Procrastinator-Hurry" until we get a queue where all the hurries stand before all the procrastinators. In this queue, the wasted time will be minimal.
Let's find the total wasted time. The second person waits for the first, the third waits for the previous two, and so on.
The total time spent waiting for all the hurries in front to complete their operations is
$$
(a+2a+\ldots+(n-1)a)+m \cdot n a=a \cdot \frac{1+n-1}{2} \cdot(n-1)+m n a=a C_{n}^{2}+m n a
$$
Here, the term $m n a$ is the total time spent by all procrastinators waiting for all the hurries.
The total time spent waiting for the procrastinators in front to complete their operations is $b+2b+\ldots+(m-1)b=b C_{m}^{2}$.
The overall minimum waiting time for all customers is
$$
T_{\min }=a C_{n}^{2}+a m n+b C_{m}^{2}
$$
In our specific case, we get:
$$
1 \cdot C_{5}^{2}+1 \cdot 5 \cdot 3+5 \cdot C_{3}^{2}=10+15+15=40
$$[^3]
Similarly, it can be proven that the maximum wasted time will be if all procrastinators stand before all hurries. This time is
$$
T_{\max }=a C_{n}^{2}+b m n+b C_{m}^{2}
$$
With the numerical data from the problem, we get $1 \cdot C_{5}^{2}+5 \cdot 5 \cdot 3+5 \cdot C_{3}^{2}=10+75+15=100$.
b) Consider the $k$-th customer in the queue. Let $X_{k}$ be the number of procrastinators standing before him. Then $X_{k}=I_{1}+I_{2}+\ldots+I_{k-1}$, where the indicator $I_{j}$ takes the value 1 if the $j$-th customer is a procrastinator and 0 if the $j$-th customer is a hurry.
The $j$-th customer can be a procrastinator with probability $\frac{m}{m+n}$ and a hurry with probability $\frac{n}{m+n}$. Therefore, $\mathrm{E} I_{j}=0 \cdot \frac{n}{m+n}+1 \cdot \frac{m}{m+n}=\frac{m}{m+n}$.
Thus, $\mathrm{E} X_{k}=\frac{(k-1) m}{m+n}$.
Let $T_{k}$ be the waiting time of the $k$-th customer. We get:
$$
T_{k}=X_{k} b+\left(k-1-X_{k}\right) a=(b-a) X_{k}+a(k-1)
$$
Therefore,
$$
\mathrm{E} T_{k}=(b-a) \mathrm{E} X_{k}+a(k-1)=(b-a) \cdot \frac{(k-1) m}{m+n}+a(k-1)=\frac{b m+a n}{m+n} \cdot(k-1)
$$
Summing the obtained expression over all customers from 1 to $m+n$-th, we get the expected total wasted time:
$$
\begin{gathered}
\mathrm{E} T=\mathrm{E}\left(T_{1}+T_{2}+\ldots+T_{n}\right)=\frac{b m+a n}{m+n} \cdot(0+1+2+\ldots+(m+n-1))= \\
=\frac{(b m+a n)(n+m-1)}{2}=C_{n+m}^{2} \cdot \frac{b m+a n}{m+n}
\end{gathered}
$$
Substituting the known numbers from the condition: $C_{8}^{2} \cdot \frac{5 \cdot 3+1 \cdot 5}{8}=\frac{7 \cdot 20}{2}=70$.
|
Answer: a) 40 minutes and 100 minutes; b) 70 minutes.
Comment. The expected time turned out to be the arithmetic mean between the maximum and minimum possible. Try to prove the fact $\mathrm{E} T=\frac{T_{\min }+T_{\max }}{2}$ in general, that is, prove the combinatorial identity
$$
C_{n+m}^{2} \cdot \frac{b m+a n}{m+n}=a C_{n}^{2}+\frac{a+b}{2} m n+b C_{m}^{2}
$$
By the way, part b) could have been solved almost without calculations, by noting that each arrangement of slowpokes and hurries can be matched with a symmetrically equal and equally probable arrangement, and the sum of the waiting times for both arrangements will be equal to $a C_{n}^{2}+\frac{a+b}{2} \cdot m n+b C_{m}^{2}$. For this, it is convenient to use the graphical method (see the footnote above).
|
40
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
|
Solution. With probability $\frac{1}{3}$, the pedestrian falls into the interval when the green light is on (event $G$), and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\bar{G}$). In this case, the waiting time is uniformly distributed over the interval from 0 to 2 minutes, so the conditional expected value of $T$ given $\bar{G}$ is one minute: $\mathrm{E}(T \mid G)=\frac{0+2}{2}=1$ (min).
We get:
$$
\mathrm{E}(T)=\mathrm{E}(T \mid G) \cdot \mathrm{P}(G)+\mathrm{E}(T \mid \bar{G}) \cdot \mathrm{P}(\bar{G})=\frac{1}{3} \cdot 0+\frac{2}{3} \cdot 1=\frac{2}{3}(\text { min })
$$
Answer: 40 seconds.
|
40
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
|
Solution. With probability $\frac{1}{3}$, the pedestrian falls into the interval when the green light is on (event $G$), and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\bar{G}$). In this case, the waiting time is uniformly distributed over the interval from 0 to 2 minutes, so the conditional expected value of $T$ given $\bar{G}$ is one minute: $\mathrm{E}(T \mid G)=\frac{0+2}{2}=1$ (min).
We get:
$$
\mathrm{E}(T)=\mathrm{E}(T \mid G) \cdot \mathrm{P}(G)+\mathrm{E}(T \mid \bar{G}) \cdot \mathrm{P}(\bar{G})=\frac{1}{3} \cdot 0+\frac{2}{3} \cdot 1=\frac{2}{3}(\text { min })
$$
Answer: 40 seconds.
|
40
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young driver Yura spends one minute on the path from $A$ to $B$ or back. Yura also spends one minute on the loop. The loop can only be driven counterclockwise (arrows indicate possible directions of movement). Yura does not turn back halfway and does not stop. The race duration is 10 minutes. Find the number of possible different routes (sequences of passing sections). #
|
# Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ can be reached in one minute either from point $A$ or from point $B$, so $M_{n, B}=M_{n-1}$.
A point $A$ can be reached in one minute only from point $B$, so $M_{n, A}=M_{n-1, B}=M_{n-2}$. Therefore,
$$
M_{n, A}=M_{n-2}=M_{n-2, A}+M_{n-2, B}=M_{n-4}+M_{n-3}=M_{n-2, A}+M_{n-1, A}
$$
Additionally, note that $M_{1, A}=0, M_{2, A}=1$. Thus, the numbers $M_{n, A}$ form the sequence $0,1,1,2,3,5,8,13,21,34, \ldots$
The number $M_{10, A}$ is 34 - the ninth Fibonacci number ${ }^{1}$.
Answer: 34.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entrance to the toll booth, an automatic system is installed that measures the height of the vehicle. If the height of the vehicle is less than a certain threshold value \( h \), the system automatically classifies it as class 1; if it is higher than \( h \), it is classified as class 2. Errors are possible. For example, a low minivan may be classified as class 1 by the system, and its driver will be pleased. A low SUV may be incorrectly classified as class 2, and its driver will not be happy. The driver can file a claim, and the operating company will have to refund 100 rubles.
The company's management tasked the engineers with the goal of configuring the system so that the number of errors is minimized.
For several weeks, the engineers collected data on the height of vehicles passing through the toll booth. Separately for class 1 vehicles and separately for class 2 vehicles (Fig. 5). On the x-axis, the height of the vehicle (in cm) is plotted, and on the y-axis, the average number of vehicles of such height passing through the toll booth per day. For clarity, the points are connected by a smooth line.
Fig. 5. Graphs - the number of class 1 and class 2 vehicles per day
Having gathered all this information and drawn the graphs, the engineers began to think about what to do next: how to determine the threshold value \( h \) so that the probability of error is the smallest possible? Solve this problem for them.
|
Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
We will prove this geometrically. Let's call the point of intersection $C$. Draw any vertical line $x=h$ (for definiteness, to the right of the point of intersection of the graphs). Introduce the notation for a few more points, as shown
Fig. 3. is shown in Fig. 3.
There can be two types of errors. The first type is when the system incorrectly classifies a class 1 car as class 2. The number of such errors is represented by the area of the figure under the graph 1 to the right of the line $x=h$, that is, the area of the curvilinear triangle $G F D$. The second type of error: the system incorrectly classifies class 2 as class 1. The number of such errors is equal to the area of the curvilinear triangle $A E G$ under the graph 2 to the left of the line $x=h$. The figure composed of triangles $F G D$ and $A E G$ has the smallest area when the area of triangle $C E F$ is the smallest. This area is zero only if the line $x=h$ coincides with the line $B C$. Therefore, the value we are looking for is the abscissa of the common point of the graphs ${ }^{2}$: approximately 190 cm.
Answer: approx. 190 cm.
|
190
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Ivan the Tsarevich's Arrows (from 8th grade. 3 points). Ivan the Tsarevich is learning to shoot a bow. He put 14 arrows in his quiver and went to shoot at pine cones in the forest. He knocks down a pine cone with a probability of 0.1, and for each pine cone he knocks down, the Frog-Princess gives him 3 more arrows. Ivan the Tsarevich shoots until he runs out of arrows. Find the expected number of shots Ivan the Tsarevich will make.
|
Solution. First method. Let Ivan have $n$ arrows at the present moment. Let $X_{0}$ be the random variable "the number of shots needed to reduce the number of arrows by one." Ivan makes a shot. Consider the random variable - the indicator $I$ of a successful shot. $I=0$, if the shot is unsuccessful (probability of this is 0.9), or $I=1$, if the pine cone is knocked down (probability 0.1). Then
$$
X_{0}=(1-I) \cdot 1+I \cdot\left(1+X_{1}+X_{2}+X_{3}\right)
$$
where $X_{1}, X_{2}$, and $X_{3}$ are the same random variables as $X_{0}$. Each of them equals the number of shots needed to reduce the number of arrows in reserve by one after receiving three new arrows from the Princess.
The variable $I$ is only related to the nearest shot, while the variables $X_{k}$ for $k=1,2,3$ are only related to subsequent shots. Therefore, the random variables $I$ and $X_{k}$ are independent. We will transition to expectations in the obtained equality:
$$
\mathrm{E} X_{0}=(1-\mathrm{E} I)+\mathrm{E} I \cdot\left(1+\mathrm{E} X_{1}+\mathrm{E} X_{2}+\mathrm{E} X_{3}\right)
$$
Since all variables from $X_{0}$ to $X_{3}$ are identically distributed, their expectations are equal. Let's denote them by $a$. In addition, $\mathrm{E} I=0.1$. Therefore,
$$
a=(1-0.1)+0.1 \cdot(1+3 a), \text { from which } a=\frac{10}{7}
$$
Thus, to get rid of one arrow, Ivan needs on average 10/7 shots. To get rid of all 14 arrows that were initially present, he needs on average $14 \cdot 10 / 7=20$ shots.
Second method. Determine how many fewer arrows Ivan has as a result of the $k$-th shot (taking into account the new ones given by the Princess). Let this random variable be $X_{k}$ ( $k=1,2,3, \ldots$ ). From the condition, it is clear that $X_{k}=1$, if Ivan did not knock down the pine cone, or $X_{k}=-2$ (the number of arrows increased by 2), if Ivan knocked down the pine cone. That is,
$$
X_{k} \sim\left(\begin{array}{cc}
1 & -2 \\
0.9 & 0.1
\end{array}\right)
$$
independently of $k$. Let's find the expectation: $\mathrm{E} X=0.9-0.2=0.7$ : with each shot, Ivan has on average 0.7 fewer arrows.
Introduce indicators of individual shots $I_{k}$, where $k=1,2,3, \ldots$ :
$$
I_{k}=\left\{\begin{array}{l}
0, \text { if there were not enough arrows for the } k \text {-th shot, } \\
1, \text { if the } k \text {-th shot was made. }
\end{array}\right.
$$
The total number of shots is
$$
S=I_{1}+I_{2}+I_{3}+\ldots+I_{k}+\ldots
$$
and the total change (decrease) in the number of arrows is
$$
I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots
$$
By the time all the arrows are gone, this sum equals 14. We form the equation
$$
14=I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots
$$
and transition to the expectation in the right part (the left part is a constant):
$$
14=\mathrm{E}\left(I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots\right)
$$
The random variable $I_{k}$ depends only on how many arrows are left after $k-1$ shots, so the variables $I_{k}$ and $X_{k}$ are independent. Therefore,
$$
\begin{gathered}
14=\mathrm{E} I_{1} \cdot \mathrm{E} X_{1}+\mathrm{E} I_{2} \cdot \mathrm{E} X_{2}+\ldots+\mathrm{E} I_{k} \cdot \mathrm{E} X_{k}+\ldots= \\
=0.7\left(\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{k}+\ldots\right)=0.7 \mathrm{E}\left(I_{1}+I_{2}+I_{3}+\ldots+I_{k}+\ldots\right)=0.7 \mathrm{E} S
\end{gathered}
$$
Therefore, $\mathrm{E} S=14: 0.7=20$.
Answer: 20.
|
20
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?
|
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is
$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$
We need to find the value of $x$ for which the function
$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$
has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$
$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$
therefore, for all $x$
$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$
and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]
Second method. The point of maximum of the function $p(x)$ can be found using the derivative.
Answer: 60 seconds. The probability is $4 / 9$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. The diagram shows a track layout for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they like, returning to point $A$.
The young driver, Yura, spends one minute on the path from $A$ to $B$ or back. Yura also spends one minute on the loop. The loop can only be driven counterclockwise (arrows indicate possible directions of movement). Yura does not turn back halfway and does not stop. The race duration is 10 minutes. Find the number of possible different routes (sequences of passing sections). #
|
# Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ can be reached in one minute either from point $A$ or from point $B$, so $M_{n, B}=M_{n-1}$.
A point $A$ can be reached in one minute only from point $B$, so $M_{n, A}=M_{n-1, B}=M_{n-2}$. Therefore,
$$
M_{n, A}=M_{n-2}=M_{n-2, A}+M_{n-2, B}=M_{n-4}+M_{n-3}=M_{n-2, A}+M_{n-1, A}
$$
Additionally, note that $M_{1, A}=0, M_{2, A}=1$. Thus, the numbers $M_{n, A}$ form the sequence $0,1,1,2,3,5,8,13,21,34, \ldots$
The number $M_{10, A}$ is 34 - the ninth Fibonacci number ${ }^{1}$.
Answer: 34.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entrance to the toll booth, an automatic system is installed that measures the height of the vehicle. If the height of the vehicle is less than a certain threshold value \( h \), the system automatically classifies it as class 1; if it is higher than \( h \), it is classified as class 2. Errors are possible. For example, a low minivan may be classified as class 1, and its driver will be pleased. A low SUV may be incorrectly classified as class 2, and its driver will not be happy. The driver can file a claim, and the operating company will have to refund 100 rubles.
The company's management tasked the engineers with the goal of configuring the system so that the number of errors is minimized.
For several weeks, the engineers collected data on the height of vehicles passing through the toll booth. Separately for class 1 vehicles and separately for class 2 vehicles (Fig. 5). On the x-axis, the height of the vehicle (in cm) is plotted, and on the y-axis, the average number of vehicles of such height passing through the toll booth per day. For clarity, the points are connected by a smooth line.
Fig. 5. Graphs - the number of class 1 and class 2 vehicles per day
Having gathered all this information and drawn the graphs, the engineers began to think about what to do next: how to determine the threshold value \( h \) so that the probability of error is the smallest possible? Solve this problem for them.
|
Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
We will prove this geometrically. Let's call the point of intersection $C$. Draw any vertical line $x=h$ (for definiteness, to the right of the point of intersection of the graphs). Introduce the notation for a few more points, as shown
Fig. 3. as shown in Fig. 3.
There can be two types of errors. The first type is when the system incorrectly classifies a class 1 car as class 2. The number of such errors is represented by the area of the figure under the graph 1 to the right of the line $x=h$, that is, the area of the curvilinear triangle $G F D$. The second type of error: the system incorrectly classifies class 2 as class 1. The number of such errors is equal to the area of the curvilinear triangle $A E G$ under the graph 2 to the left of the line $x=h$. The figure composed of triangles $F G D$ and $A E G$ has the smallest area when the area of triangle $C E F$ is the smallest. This area is zero only if the line $x=h$ coincides with the line $B C$. Therefore, the value we are looking for is the abscissa of the common point of the graphs ${ }^{2}$: approximately 190 cm.
Answer: approximately 190 cm.
|
190
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?
|
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is
$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$
We need to find the value of $x$ for which the function
$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$
has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$
$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$
therefore, for all $x$
$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$
and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]
Second method. The point of maximum of the function $p(x)$ can be found using the derivative.
Answer: 60 seconds. The probability is $4 / 9$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Solution. For clarity, let's assume that when a bite occurs, the Absent-Minded Scholar immediately reels in and casts the fishing rod again, and does so instantly. After this, he waits again. Consider a 5-minute time interval. During this time, on average, there are 5 bites on the first fishing rod and 1 bite on the second. Therefore, in total, there are on average 6 bites on both fishing rods in these 5 minutes. Consequently, the average waiting time for the first bite is $\frac{5}{6}$ minutes.
|
Answer: 50 seconds.
## Grading Criteria
| Solution is correct and well-reasoned | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points |
|
50
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young driver Yura spends one minute on the path from $A$ to $B$ or back. Yura also spends one minute on the loop. The loop can only be driven counterclockwise (arrows indicate possible directions of movement). Yura does not turn back halfway and does not stop. The race duration is 10 minutes. Find the number of possible different routes (sequences of passing sections). #
|
# Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ can be reached in one minute either from point $A$ or from point $B$, so $M_{n, B}=M_{n-1}$.
A point $A$ can be reached in one minute only from point $B$, so $M_{n, A}=M_{n-1, B}=M_{n-2}$. Therefore,
$$
M_{n, A}=M_{n-2}=M_{n-2, A}+M_{n-2, B}=M_{n-4}+M_{n-3}=M_{n-2, A}+M_{n-1, A}
$$
Additionally, note that $M_{1, A}=0, M_{2, A}=1$. Thus, the numbers $M_{n, A}$ form the sequence $0,1,1,2,3,5,8,13,21,34, \ldots$
The number $M_{10, A}$ is 34 - the ninth Fibonacci number ${ }^{1}$.
Answer: 34.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entrance to the toll booth, an automatic system is installed that measures the height of the vehicle. If the height of the vehicle is less than a certain threshold value \( h \), the system automatically classifies it as class 1; if it is higher than \( h \), it is classified as class 2. In this case, errors are possible. For example, a low minivan may be classified as class 1, and its driver will be pleased. A low-profile UAZ passenger car may be incorrectly classified as class 2, and its driver will not be happy. He can file a claim, and the operating company will have to refund him 100 rubles.
The company's management tasked the engineers with the goal of configuring the system so that the number of errors is minimized.
For several weeks, the engineers collected data on the height of vehicles passing through the toll booth. Separately for class 1 vehicles and separately for class 2 vehicles (Fig. 5). On the x-axis, the height of the vehicle (in cm) is plotted, and on the y-axis, the average number of vehicles of such height passing through the toll booth per day. For clarity, the points are connected by a smooth line.
Fig. 5. Graphs - the number of class 1 and class 2 vehicles per day
Having gathered all this information and drawn the graphs, the engineers began to think about what to do next: how to determine the threshold value \( h \) so that the probability of error is the smallest possible? Solve this problem for them.
|
Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
We will prove this geometrically. Let's call the point of intersection $C$. Draw any vertical line $x=h$ (for definiteness, to the right of the point of intersection of the graphs). Introduce the notation for a few more points, as shown
Fig. 3. as shown in Fig. 3.
There can be two types of errors. The first type is when the system incorrectly classifies a class 1 car as class 2. The number of such errors is represented by the area of the figure under the graph 1 to the right of the line $x=h$, that is, the area of the curvilinear triangle $G F D$. The second type of error: the system incorrectly classifies a class 2 car as class 1. The number of such errors is equal to the area of the curvilinear triangle $A E G$ under the graph 2 to the left of the line $x=h$. The figure composed of triangles $F G D$ and $A E G$ has the smallest area when the area of triangle $C E F$ is the smallest. This area is zero only if the line $x=h$ coincides with the line $B C$. Therefore, the value we are looking for is the abscissa of the common point of the graphs ${ }^{2}$: approximately 190 cm.
Answer: approximately 190 cm.
|
190
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?
|
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is
$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$
We need to find the value of $x$ for which the function
$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$
has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$
$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$
therefore, for all $x$
$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$
and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]
Second method. The point of maximum of the function $p(x)$ can be found using the derivative.
Answer: 60 seconds. The probability is $4 / 9$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entrance to the toll booth, an automatic system is installed that measures the height of the vehicle. If the height of the vehicle is less than a certain threshold value \( h \), the system automatically classifies it as class 1; if it is higher than \( h \), it is classified as class 2. Errors are possible. For example, a low minivan may be classified as class 1 by the system, and its driver will be pleased. A low SUV may be incorrectly classified as class 2, and its driver will not be happy. He can file a claim, and the operating company will have to refund him 100 rubles.
The company's management tasked the engineers with the goal of configuring the system so that the number of errors is minimized.
For several weeks, the engineers collected data on the height of vehicles passing through the toll booth. Separately for class 1 vehicles and separately for class 2 vehicles (Fig. 5). On the x-axis, the height of the vehicle (in cm) is plotted, and on the y-axis, the average number of vehicles of such height passing through the toll booth per day. For clarity, the points are connected by a smooth line.
Fig. 5. Graphs - the number of class 1 and class 2 vehicles per day
Having gathered all this information and drawn the graphs, the engineers began to think about what to do next: how to determine the threshold value \( h \) so that the probability of error is the smallest possible? Solve this problem for them.
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Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
We will prove this geometrically. Let's call the point of intersection $C$. Draw any vertical line $x=h$ (for definiteness, to the right of the point of intersection of the graphs). Introduce the notation for a few more points, as shown
Fig. 3. as shown in Fig. 3.
There can be two types of errors. The first type is when the system incorrectly classifies a class 1 car as class 2. The number of such errors is represented by the area of the figure under the graph 1 to the right of the line $x=h$, that is, the area of the curvilinear triangle $G F D$. The second type of error: the system incorrectly classifies class 2 as class 1. The number of such errors is equal to the area of the curvilinear triangle $A E G$ under the graph 2 to the left of the line $x=h$. The figure composed of triangles $F G D$ and $A E G$ has the smallest area when the area of triangle $C E F$ is the smallest. This area is zero only if the line $x=h$ coincides with the line $B C$. Therefore, the value we are looking for is the abscissa of the common point of the graphs ${ }^{2}$: approximately 190 cm.
Answer: approx. 190 cm.
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190
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Other
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math-word-problem
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Yes
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Yes
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olympiads
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