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3. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good".
It is known that the number 1 is "good". Find the nearest "good" number to 2015.
|
Answer: 2047. Solution. Note that good numbers have the form $2^{n}-1$. Indeed, let a number have the form $M=2^{n}+k$, where $k=0,1, \ldots, 2^{n}-2$. Then from such a number, you can subtract $k+1 \leqslant \frac{1}{2}\left(2^{n}+k\right)=\frac{M}{2}$. On the other hand, from a number of the form $N=2^{n}-1$, you need to subtract at least $2^{n-1}$, which exceeds $\frac{N}{2}=2^{n-1}-\frac{1}{2}$. Therefore, the closest to 2015 will be the number $2047=2^{11}-1$
|
2047
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Calculate
$$
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}} \quad \text { for } \quad a=-1, \underbrace{5 \ldots 5}_{2010} 6, \quad b=5, \underbrace{4 \ldots 44}_{2011}
$$
Answer: 343.
|
Solution. Given $a$ and $b$, we get
$$
\begin{gathered}
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}}=2 a b(a-b)-(a-b)\left(a^{2}+b^{2}\right)= \\
\quad=-(a-b)\left(a^{2}-2 a b+b^{2}\right)=-(a-b)^{3}=-(-7)^{3}=343
\end{gathered}
$$
|
343
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University).
|
Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1755 \Leftrightarrow 1000 a+100 b+10 c+d=a+b+c+d+1755 \Leftrightarrow \\
\Leftrightarrow 999 a+99 b+9 c=1755 \Leftrightarrow 111 a+11 b+c=195
\end{gathered}
$$
From the last equation, we conclude that the only possible value for the digit $a$ is 1. Next, we get the equation $11 b+c=84$, which means the digit $b$ can only be 7. Finally, we need to find the value of $c=84-11 \cdot 7=7$. Therefore, the desired number is 1770, as it is the smallest among all numbers of the form $\overline{177 d}$.
|
1770
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 9.
Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$.
|
Answer: 12045
Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest odd divisors of the numbers from $n+1$ to $2n$ are $n$ different odd numbers that do not exceed $2n$. Thus, these numbers are $1, 3, 5, \ldots, 2n-1$. If we add the number 220 to the set of numbers, the desired sum will be $1+3+5+\ldots+219-A(220)=\frac{1+219}{2} \cdot 110-55=110^2-55=12045$.
## B-2
Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(113)+A(114)+\ldots+A(222)+A(223)$.
Answer: 12537
## B-3
Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(115)+A(116)+\ldots+A(226)+A(227)$.
Answer: 12939
## B-4
Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(117)+A(118)+\ldots+A(230)+A(231)$.
Answer: 13427
|
12045
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Two trains, each containing 15 identical cars, were moving towards each other at constant speeds. Exactly 28 seconds after the first cars of the trains met, passenger Sasha, sitting in the third car, passed passenger Valera from the oncoming train, and another 32 seconds later, the last cars of these trains had completely passed each other. In which car, by count, was Valera traveling?
Answer: 12.
|
Solution. Since 60 seconds have passed from the moment the "zero" cars parted ways (which is also the moment the first cars met) to the moment the 15th cars parted ways, the next cars parted ways every $60: 15=4$ seconds. Therefore, after 28 seconds, the 7th cars of the trains had just parted ways, meaning the 7th car of one train had just aligned with the 8th car of the other. At this moment, Sasha's 3rd car aligned with Valery's car, which has the number $8+(7-3)=12$.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find the area of the figure defined on the coordinate plane by the system
$$
\left\{\begin{array}{l}
\sqrt{1-x}+2 x \geqslant 0 \\
-1-x^{2} \leqslant y \leqslant 2+\sqrt{x}
\end{array}\right.
$$
|
Answer: 4.
Solution. The system is defined under the condition $x \in[0,1]$, under which the first inequality of the system is satisfied. Since the functions $y=x^{2}$ and $y=\sqrt{x}$ are inverses of each other for $x \in[0,1]$, the area of the figure defined by the conditions $x \in[0,1],-1-x^{2} \leqslant y \leqslant 2+\sqrt{x}$, consists of the same parts as the area of the rectangle defined by the conditions $x \in[0,1], y \in[-2,2]$, i.e., it is equal to 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Two circles touch each other internally at point K. A chord $AB$ of the larger circle touches the smaller circle at point $L$, and $AL=10$. Find $BL$, if $AK: BK=2: 5$.
|
Answer: $B L=25$.
Solution. Draw the chord $P Q$ and the common tangent $M N$ (see fig.), then
$$
\angle P Q K=\frac{1}{2} \smile P K=\angle P K M=\frac{1}{2} \smile A K=\angle A B K
$$
therefore $P Q \| A B$ and $\smile P L=\smile Q L$. Consequently, $\angle P K L=\angle Q K L$, i.e., $K L$ is the bisector of triangle $A K B$, from which we obtain
$$
\frac{A L}{B L}=\frac{A K}{B K}=\frac{2}{5} \Longrightarrow B L=10 \cdot \frac{5}{2}=25
$$
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. What is the smallest (same) number of pencils that need to be placed in each of 6 boxes so that in any 4 boxes there are pencils of any of 26 predefined colors (there are enough pencils available)?
|
Answer: 13.
Solution. On the one hand, pencils of each color should appear in at least three out of six boxes, since if pencils of a certain color are in no more than two boxes, then in the remaining boxes, which are at least four, there are no pencils of this color. Therefore, the total number of pencils should be no less than $3 \cdot 26$.
On the other hand, if we take 3 pencils of each of the 26 colors and arrange them so that no pencils of the same color are in the same box, then a pencil of any color will be found in any four boxes (since there are only two other boxes). The required arrangement of pencils can be achieved, for example, by dividing all the pencils into 3 groups of 26 pencils of all colors and distributing each group evenly (13 pencils) into two boxes.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3.
In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$, if $A B=\sqrt{13}$.
|
Answer: 12
Solution. Let $A B=c, B E=A D=2 a$. Since triangle $A B D$ is isosceles (the bisector is perpendicular to the base, $A B=B D=c, B C=2 c$), then by the formula for the length of the bisector (where $\angle A B C=\beta$) $2 a=\frac{4 c^{2}}{3 c} \cos \frac{\beta}{2} \Leftrightarrow \frac{3 a}{2}=c \cos \frac{\beta}{2}$. Consider triangle $A B F$ (where $F$ is the midpoint of segment AD and the intersection point of the bisector and the median). We have $a=c \sin \frac{\beta}{2}$, $c=\frac{a}{\sin \frac{\beta}{2}}$. Therefore, $\operatorname{tg} \frac{\beta}{2}=\frac{2}{3}$ and $\cos \frac{\beta}{2}=\frac{3}{\sqrt{13}}, \sin \frac{\beta}{2}=\frac{2}{\sqrt{13}}, \sin \beta=\frac{12}{13}$. The area of triangle $A B C$ is $S_{A B C}=c^{2} \sin \beta=\frac{12}{13} c^{2}$.
## B-2
In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $A B=\sqrt{26}$.
Answer: 24
## B-3
In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=4$.
Answer: 12
## B-4
In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=6$.
Answer: 27
## Lomonosov High School Mathematics Olympiad
Preliminary stage 2020/21 academic year for 9th grade
## B-1
#
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4.
Masha tightly packed 165 identical balls into the shape of a regular triangular pyramid. How many balls are at the base?
|
Answer: 45
Solution. We will solve the problem in general for $\frac{1}{6} n(n+1)(n+2)$ balls. At the base of a triangular pyramid of height $n$ rows lies the $n$-th triangular number of balls $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$. Then the total number of balls in the pyramid is $\sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{1}{6} n(n+1)(n+2)$. The problem can be solved by enumeration.
## B-2
Masha tightly packed 220 identical balls into a regular triangular pyramid. How many balls are at the base?
Answer: 55
## B-3
Masha tightly packed 286 identical balls into a regular triangular pyramid. How many balls are at the base?
Answer: 66
## B-4
Masha tightly packed 364 identical balls into a regular triangular pyramid. How many balls are at the base?
Answer: 78
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary Stage 2020/21 academic year for 9th grade
#
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6.
B-1
Find the minimum value of the expression $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}$ given that $x>4$ and $y>5$.
|
Answer: 71
Solution. $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}=4 x-16+\frac{4}{4 x-16}+9 y-45+\frac{9}{9 y-45}+61 \geq 2 \sqrt{4}+2 \sqrt{9}+61=$ 71 (inequality of means). Equality is achieved when $4 x-16=2,9 y-45=3$, that is, when $x=\frac{9}{2}, y=\frac{16}{3}$.
B-2
Find the minimum value of the expression $9 x+4 y+\frac{1}{x-3}+\frac{1}{y-4}$ given that $x>3$ and $y>4$.
Answer: 53
## B-3
Find the minimum value of the expression $4 x+9 y+\frac{1}{x-1}+\frac{1}{y-2}$ given that $x>1$ and $y>2$.
Answer: 32
## B-4
Find the minimum value of the expression $9 x+4 y+\frac{1}{x-5}+\frac{1}{y-6}$ given that $x>5$ and $y>6$.
Answer: 79
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary Stage 2020/21 Academic Year for 9th Grade
## B-1
#
|
71
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959.
|
Solution. Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^{7} \cdot 10-2^{5}\right) \cdot 10^{105}=$ $1248 \cdot 10^{105}$. Therefore, the number $A$ is $1248 \cdot 10^{105}-1=1247 \underbrace{999 \ldots 999}_{105 \text { digits }}$, and the sum of its digits is $1+2+4+7+9 \cdot 105=959$.
|
959
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Each Kinder Surprise contains exactly 3 different gnomes, and there are 12 different types of gnomes in total. In the box, there are enough Kinder Surprises, and in any two of them, the triplets of gnomes are not the same. What is the minimum number of Kinder Surprises that need to be bought to ensure that after they are opened, there is at least one gnome of each of the 12 types?
|
Answer: 166.
Solution. From 11 different gnomes, a maximum of $C_{11}^{3}=\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}=165$ different Kinder can be formed, so if you take one more Kinder, there cannot be fewer than 12 different gnomes among them.
|
166
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Points $A$ and $B$ lie on a circle with center $O$ and radius 6, and point $C$ is equidistant from points $A, B$, and $O$. Another circle with center $Q$ and radius 8 is circumscribed around triangle $A C O$. Find $B Q$. Answer: 10
|
Solution. Since $A Q=O Q=8$ and $A C=O C$, triangles $\triangle A C Q$ and $\triangle O C Q$ are equal. Therefore, points $A$ and $O$ are symmetric with respect to the line $C Q$, so $A O \perp C Q$.
Next, $\angle B O Q=\angle B O C+\angle C O Q=\angle A O C+\angle O C Q=\pi / 2$. Therefore, $B Q=\sqrt{B O^{2}+O Q^{2}}=\sqrt{6^{2}+8^{2}}=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A motorcycle and a quad bike are driving on a circular road, a quarter of which passes through a forest, and the remaining part - through a field. The motorcycle's speed when driving through the forest is 20 km/h, and through the field - 60 km/h. The quad bike's speed when driving through the forest is 40 km/h, and through the field - 45 km/h. The quad bike and the motorcycle enter the forest at the same time. Which of the vehicles will overtake the other first, and on which lap will this happen?
|
Answer: The quad will overtake the motorcycle on its $10-\mathrm{th}$ lap while the quad is on its 11-th lap.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. It is known that $m, n, k$ are distinct natural numbers greater than 1, the number $\log _{m} n$ is rational, and, moreover,
$$
k^{\sqrt{\log _{m} n}}=m^{\sqrt{\log _{n} k}}
$$
Find the minimum of the possible values of the sum $k+5 m+n$.
|
Answer: 278.
Solution. Transform the original equation, taking into account that the numbers $m, n, k$ are greater than 1 and the corresponding logarithms are positive:
$$
\begin{aligned}
& k \sqrt{\log _{m} n}=m \sqrt{\log _{n} k} \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\log _{n} k} \Leftrightarrow \\
& \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\frac{\log _{m} k}{\log _{m} n}} \Leftrightarrow \\
& \Leftrightarrow \sqrt{\log _{m} k} \cdot \log _{m} n=1 \Leftrightarrow \log _{m} k \cdot \log _{m}^{2} n=1
\end{aligned}
$$
By performing the substitution $k=n^{p}, m=n^{q}$, where $p \neq 0, q \neq 0$, we get
$$
\frac{p}{q} \cdot \frac{1}{q^{2}}=1 \Leftrightarrow p=q^{3}
$$
Thus, all triples $(k, m, n)$ that satisfy the condition of the problem have the form $\left(n^{q^{3}}, n^{q}, n\right)$.
It is clear that $q>0$ : otherwise, from the naturality of the number $n$ it follows that the number $n^{q}$ lies in the interval $(0 ; 1)$ and is not a natural number. The case $q=1$ contradicts the condition of the problem. If $q \in(0 ; 1)$, then $q=1 / r$, where $r>1$, and then the triple of natural numbers $\left(n^{q^{3}}, n^{q}, n\right)$ can be written in the form $\left(l, l^{r^{2}}, l^{r^{3}}\right)$ with a natural $l$. The sum $k+5 m+n$ for this triple will be greater than the same sum for the numbers in the triple $\left(l^{r^{3}}, l^{r}, l\right)$. Therefore, $q>1$.
For integer $q$, the triple of numbers with the minimum sum $k+5 m+n$ is obtained when $n=2, q=2$ : these are the numbers $k=2^{8}, m=2^{2}, n=2$, and for them the sum $k+5 m+n$ is 278. If $q$ is rational but not an integer, then for the number $n^{q^{3}}$ to be an integer, it is necessary that $n$ be at least the eighth power of an integer not less than 2, and, of course, the sum $n^{q^{3}}+5 n^{q}+n$ will be greater than 278 (since $n \geqslant 256$).
|
278
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. To guard the object, the client agreed with the guards on the following: all of them will indicate the time intervals of their proposed shifts with the only condition that their union should form a predetermined time interval set by the client, and he will choose any set of these intervals that also satisfies the same condition, and pay for the work at a rate of 300 rubles per hour for each guard. What is the longest time interval the client can set to ensure that he definitely stays within 90000 rubles?
|
6. $\frac{90000}{2 \cdot 300}=150 \text{h}$.
|
150
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. Find the sum of all integer values of $a$ from the interval $[-2012 ; 2013]$, for which the equation $(a-3) x^{2}+2(3-a) x+\frac{a-7}{a+2}=0$ has at least one solution.
|
Answer: 2011. Solution. When $a=3$ there are no solutions. For other $a$, it should be $\frac{D}{4}=(a-3)^{2}-\frac{(a-3)(a-7)}{a+2} \geq 0$. The last inequality (plus the condition $a \neq 3$) has the solution $a \in(-\infty ;-2) \bigcup\{1\} \cup(3 ;+\infty)$. The desired sum is: $-2012-2011-\ldots-5-4-3+1+4+5+\ldots 2013=-3+1+2013=2011$.
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Among the numbers exceeding 2013, find the smallest even number $N$ for which the fraction $\frac{15 N-7}{22 N-5}$ is reducible.
|
Answer: 2144. Solution: The presence of a common factor in the numbers $15 N-7$ and $22 N-5$ implies that the same factor is present in the number $(22 N-5)-(15 N-7)=7 N+2$, and subsequently in the numbers $(15 N-7)-2 \cdot(7 N+2)=N-11,(7 N+2)-7 \cdot(N-11)=79$. Since 79 is a prime number, the fraction is reducible by 79, so $N-11=79 m, N=11+79 m$. Given that $N$ is even, we have $N=90+158 p$. The required value is achieved at $p=13$.
|
2144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. Find the area of the figure defined on the coordinate plane by the inequality $\sqrt{\arcsin \frac{x}{3}} \leq \sqrt{\arccos \frac{y}{3}} \cdot$ In your answer, specify the integer closest to the found value of the area.
|
Answer: 16. Solution. The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin \frac{x}{3} \geq 0, \\ \arccos \frac{y}{3} \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}0 \leq x \leq 3, \\ -3 \leq y \leq 3 .\end{array}\right.\right.$ If $-3 \leq y \leq 0$, then $\arccos \frac{y}{3} \geq \frac{\pi}{2} \geq \arcsin \frac{x}{3}$. Therefore, the entire square $\{0 \leq x \leq 3,-3 \leq y \leq 0\}$ with an area of 9 is included in the desired set.
On the remaining set $\{0 \leq x \leq 3,0 \leq y \leq 3\}$, the inequalities $0 \leq \arcsin \frac{x}{3} \leq \frac{\pi}{2}, 0 \leq \arccos \frac{y}{3} \leq \frac{\pi}{2}$ hold, and since the function $\sin t$ is increasing for $t \in\left[0 ; \frac{\pi}{2}\right]$, we obtain equivalent inequalities $\sin \left(\arcsin \frac{x}{3}\right) \leq \sin \left(\arccos \frac{y}{3}\right) \Leftrightarrow \frac{x}{3} \leq \sqrt{1-\left(\frac{y}{3}\right)^{2}} \Leftrightarrow x^{2}+y^{2} \leq 9$. Therefore, here the desired set includes a quarter circle, the area of which is $\frac{\pi \cdot 3^{2}}{4}$.
In total, the desired area is $9\left(1+\frac{\pi}{4}\right) \approx 16.069 \ldots$
|
16
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Given two different geometric progressions, the first terms of which are equal to 1, and the sum of the denominators is 3. Find the sum of the fifth terms of these progressions, if the sum of the sixth terms is 573. If the answer to the question is not unique, specify the sum of all possible values of the desired quantity.
|
Answer: 161. Solution. Let the denominators of the progressions be $p$ and $q$. According to the condition, we get: $\left\{\begin{array}{c}p+q=3, \\ p^{5}+q^{5}=573\end{array}\right.$. We need to find $p^{4}+q^{4}$. Denoting $p+q=a, p q=b$, we express:
$$
\begin{gathered}
p^{4}+q^{4}=\left((p+q)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=\left(a^{2}-2 b\right)^{2}-2 b^{2}=a^{4}-4 a^{2} b+2 b^{2} ; \\
p^{5}+q^{5}=(p+q)\left(p^{4}-p^{3} q+p^{2} q^{2}-p q^{3}+q^{4}\right)=(p+q)\left(p^{4}+q^{4}-p q\left(p^{2}+q^{2}\right)+p^{2} q^{2}\right) \\
=a\left(a^{4}-4 a^{2} b+2 b^{2}-b\left(a^{2}-2 b\right)+b^{2}\right)=a\left(a^{4}-5 a^{2} b+5 b^{2}\right) .
\end{gathered}
$$
$$
\text { According to the condition }\left\{\begin{array} { c }
{ a = 3 , } \\
{ a ( a ^ { 4 } - 5 a ^ { 2 } b + 5 b ^ { 2 } ) = 5 7 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ a = 3 , } \\
{ 8 1 - 4 5 b + 5 b ^ { 2 } = 1 9 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a=3, \\
b^{2}-9 b-22=0
\end{array} .\right.\right.\right.
$$
From here, $b=11$ or $b=-2$. Since the system $\left\{\begin{array}{c}p+q=3, \\ p q=11\end{array}\right.$ has no solutions, then $b=-2$.
Therefore, $a=3$ and $b=-2$. Thus, $p^{4}+q^{4}=a^{4}-4 a^{2} b+2 b^{2}=81+72+8=161$.
|
161
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Chords $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ of a sphere intersect at a common point $S$. Find the sum $S A^{\prime}+S B^{\prime}+S C^{\prime}$, if $A S=6, B S=3, C S=2$, and the volumes of pyramids $S A B C$ and $S A^{\prime} B^{\prime} C^{\prime}$ are in the ratio $2: 9$. If the answer is not an integer, round it to the hundredths.
|
Answer: 18. Solution. If $a, b, c$ are the given edges, $x, y, z$ are the corresponding extensions of these edges, and $k$ is the given ratio, then by the theorem of intersecting chords and the lemma on the ratio of volumes of pyramids with equal (vertical) trihedral angles at the vertex, we have: $\left\{\begin{array}{l}a x=b y=c z \\ a b c: x y z=k\end{array} \Rightarrow \sqrt[3]{a b c x y z}=\sqrt[3]{\frac{(a b c)^{2}}{k}} \Rightarrow x=\sqrt[3]{\frac{b^{2} c^{2}}{a k}}\right.$, from which we get $p=x+y+z=x\left(1+\frac{a}{b}+\frac{a}{c}\right)=\sqrt[3]{\frac{b^{2} c^{2}}{a k}}\left(1+\frac{a}{b}+\frac{a}{c}\right)$.
Here $a=6, b=3, c=2, k=\frac{2}{9} \Rightarrow p=\sqrt[3]{\frac{3^{2} \cdot 2^{2} \cdot 9}{6 \cdot 2}}\left(1+\frac{6}{3}+\frac{6}{2}\right)=18$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find the sum of all such integers $a \in[0 ; 400]$, for each of which the equation $x^{4}-6 x^{2}+4=\sin \frac{\pi a}{200}-2\left[x^{2}\right]$ has exactly six roots. Here the standard notation is used: $[t]$ - the integer part of the number $t$ (the greatest integer not exceeding $t$).
|
Answer: 60100. Solution. Let $b=\sin \frac{\pi a}{200}, t=x^{2}$. Using the equality $[t]=t-\{t\}$ (here $\{t\}$ is the fractional part of $t$), we get that the original equation is equivalent to the equation $t^{2}-6 t+4=b-2 t+2\{t\}$. Note that any positive $t$ corresponds to two different $x$, two different positive $t$ correspond to four different $x$, three different positive $t$ correspond to six different $x$, and so on.
Solve the equation:
$$
t^{2}-4 t+4=b+2\{t\}, b \in[-1 ; 1]
$$
Since the right-hand side of the equation belongs to the interval $[-1 ; 3)$, the left-hand side does not exceed this interval when $(t-2)^{2}<3$, i.e., when $t \in(2-\sqrt{3} ; 2+\sqrt{3})$.
Consider four cases:
1) Let $t \in(2-\sqrt{3} ; 1)$. Then $\{t\}=t$, and the original equation takes the form: $t^{2}-6 t+4=b$, from which $t_{1}^{ \pm}=3 \pm \sqrt{5+\sin b}$. Of these roots, only $t_{1}^{-}$ belongs to the interval $(2-\sqrt{3} ; 1)$ for $b \in(-1 ; 1]$.
2) Let $t \in[1 ; 2)$. Then $\{t\}=t-1$, and the original equation takes the form: $t^{2}-6 t+6=b$, from which $t_{2}^{ \pm}=3 \pm \sqrt{3+\sin b}$. Of these roots, only $t_{2}^{-}$ belongs to the interval $[1 ; 2)$ for any $b \in[-1 ; 1]$.
3) Let $t \in[2 ; 3)$. Then $\{t\}=t-2$, and the original equation takes the form: $t^{2}-6 t+8=b$, from which $t_{3}^{ \pm}=3 \pm \sqrt{1+\sin b}$. Of these roots, only $t_{3}^{-}$ belongs to the interval $[2 ; 3)$ for $b \in[-1 ; 0]$.
4) Let $t \in[3 ; 2+\sqrt{3})$. Then $\{t\}=t-3$, and the original equation takes the form: $t^{2}-6 t+10=b$, from which $(t-3)^{2}+(1-b)=0$. This equation has a unique solution $t_{4}=3$ for $b=1$ (recall that $b \in[-1 ; 1]$).
In the end, we get:
for $b=-1$ the equation (1) has one positive solution for $t$, namely $t_{2}^{-}$, from which the original equation has two different solutions for the variable $x$;
for $b \in(0 ; 1)$ the equation (1) has two different positive solutions for the variable $t$, namely $t_{1}^{-}, t_{2}^{-}$, from which the original equation has four different solutions for the variable $x$;
for $b \in(-1 ; 0] \cup\{1\}$ the equation (1) has three different positive solutions for the variable $t$, namely $t_{1}^{-}, t_{2}^{-}, t_{3}^{-}$ for $b \in(-1 ; 0]$ and $t_{1}^{-}, t_{2}^{-}, t_{4}$ for $b=1$, from which the original equation has six different solutions for the variable $x$.
Since $b=\sin \frac{\pi a}{200}$, then for $a \in[0 ; 400]$ we get that for $a \in\{0\} \bigcup\{100\} \bigcup[200 ; 300) \bigcup(300 ; 400]$ the original equation will have six solutions.
We get the answer: $0+100+(200+201+\ldots+400)-300=\frac{200+400}{2} \cdot 201-200=60100$.
|
60100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a box, there are a hundred colorful balls: 28 red, 20 green, 13 yellow, 19 blue, 11 white, and 9 black. What is the smallest number of balls that need to be pulled out without looking into the box to ensure that there are at least 15 balls of the same color among them?
|
Answer: 76. Solution. Worst-case scenario: 14 red, 14 green, 13 yellow, 14 blue, 11 white, and 9 black balls will be drawn - a total of 75 balls. The next ball will definitely be the 15th ball of one of the colors: either red, green, or blue.
Answer of variant 152: 109.
Answer of variant 153: 83.
Answer of variant 154: 98.
|
76
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a convex quadrilateral $A B C D$, diagonals $A C$ and $D B$ are perpendicular to sides $D C$ and $A B$ respectively. A perpendicular is drawn from point $B$ to side $A D$, intersecting $A C$ at point $O$. Find $A O$, if $A B=4, O C=6$.
|
Answer: 2. Solution. Since $\angle A B D=\angle A C D=90^{\circ}$, the quadrilateral is inscribed in a circle with diameter $A D$. Therefore, $\angle A C B=\angle A D B$ - as angles subtending the same arc. Since $B H$ is the altitude in the right $\triangle A B D$, then $\angle A D B=\angle A B H$. Then triangles $A B O$ and $A C B$ are similar by two angles, and the following ratios hold: $\frac{A B}{A O}=\frac{A C}{A B} \Leftrightarrow \frac{4}{A O}=\frac{A O+6}{4}$ $\Leftrightarrow A O^{2}+6 A O-16=0 \Leftrightarrow A O=-8$ or $A O=2$. Answer: $A O=2$.
## Answer for variant 152: 9.
Answer for variant 153: 1.
Answer for variant 154: 12.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Masha, bored during a math lesson, performed the following operation on a certain 2015-digit natural number: she discarded the last digit of its decimal representation and added twice the discarded digit to the number obtained by multiplying the remaining number by 3. She then performed the same operation on the resulting number, and so on. After applying this operation multiple times, the numbers Masha obtained stopped changing, and she stopped.
a) What number did Masha end up with?
b) What is the smallest number Masha could have started with (indicate its last two digits)
|
Answer: a) 17; b) 09 (the number $100 \ldots 0009$).
Solution. a) Let this number be $n$, ending in the digit $y$. Then $n=10 x+y$ after the next operation will become $3 x+2 y$. The equation $10 x+y=3 x+2 y$ is equivalent to $7 x=y$ and, since $y$ is a digit, then $y=7, x=1$. Therefore, $n=17$.
b) Note that if the initial number is not equal to 17, then it must decrease: $10 x+y>3 x+2 y \Leftrightarrow 7 x>y$ (which for $x \neq 0$ and $x \neq 1$ is always true). From the relation $2(10 x+y)=17 x+3 x+2 y$ it follows that the number $10 x+y$ is divisible by 17 if and only if $3 x+2 y$ is divisible by 17. Since the stabilization of the operation occurs at the number 17, the initial number must also be divisible by 17.
Let's find the smallest 2015-digit number that is divisible by 17. For convenience, let's list the remainders of the powers of 10 when divided by 17 (see table):
| Power | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Remainder | 10 | 15 | 14 | 4 | 6 | 9 | 5 | 16 | 7 | 2 | 3 | 13 | 11 | 8 | 12 | 1 | 10 |
Since the remainders repeat every 17 powers, and $10^{16} \equiv 1(\bmod 17)$ (this notation means that $10^{16}$ and 1 give the same remainder when divided by 17), and $2014=16 \cdot 125+14$, then $10^{2014} \equiv 10^{14} \equiv 8(\bmod 17)$, so $10^{2014}+9$ is divisible by 17. This will be the smallest 2015-digit number that is divisible by 17. It ends in 09.
Answer for variant 152: a) 17; b) 99 (the number 999...9999).
Answer for variant 153: a) 17; b) 05 (the number 100...0005).
Answer for variant 154: a) 17; b) 88 (the number 999...9988).
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. Find the largest four-digit number in which all digits are different, and moreover, no two of them can be swapped to form a smaller number.
|
Answer: 7089.
Solution. If the digit 0 is not used, then the digits in each such number must be in ascending order, and the largest of such numbers is 6789. Since a number cannot start with zero, using 0, one can obtain additional suitable numbers where 0 is in the hundreds place, and the other digits are in ascending order. The largest of such numbers is 7089.
|
7089
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be?
|
Answer: 6218.
Solution. Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers decreases by 1, and their sum decreases by 2. Therefore, one number will remain on the board after 55 minutes and will be equal to $6328-55 \cdot 2=6218$.
|
6218
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Masha has seven different dolls, which she arranges in six different doll houses so that each house has at least one doll. In how many ways can Masha do this? It is important which doll ends up in which house. It does not matter how the dolls are seated in the house where there are two of them.
|
Answer: 15120.
Solution. In each such seating arrangement, in one of the cabins there will be two dolls, and in the other cabins - one each. First, we choose which two dolls will sit in one cabin. This can be done in $7 \cdot 6: 2=21$ ways. Then we choose which cabin to seat these two dolls in. This can be done in 6 ways. Finally, we choose how to seat the remaining 5 dolls in the remaining 5 cabins. This can be done in $5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways. Therefore, the number of seating arrangements is $21 \cdot 6 \cdot 120=15120$.
|
15120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{32}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$
ANSWER: 60.
|
Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$.
|
63
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. There are 5 identical buckets, each with a maximum capacity of some integer number of liters, and a 30-liter barrel containing an integer number of liters of water. The water from the barrel was distributed among the buckets, with the first bucket being half full, the second one-third full, the third one-quarter full, the fourth one-fifth full, and the fifth one-sixth full. How many liters of water were in the barrel?
ANSWER: 29.
|
The solution $-\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=\frac{87 x}{60}=\frac{29 x}{20}$ will be an integer only when $x$ is a multiple of 20. But $x>20$ cannot be taken, as the sum will exceed 30 liters.
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Two oligarchs, Alejandro and Maximilian, looted their country in 2012. It is known that Alejandro's wealth at the end of 2012 equals twice Maximilian's wealth at the end of 2011. And Maximilian's wealth at the end of 2012 is less than Alejandro's wealth at the end of 2011. Which is greater: Maximilian's wealth or the national wealth of the country?
|
Answer: Maximilian's state is greater.
Solution: Consider the table where $z, x-$ are the states of Alejandro and Maximilian in 2011:
| | 2011 | 2012 |
| :--- | :--- | :--- |
| $\mathrm{A}$ | $z$ | $2 x$ |
| $\mathrm{M}$ | $x$ | $y$ |
Then $N=(2 x+y)-(x+z)-$ are the national riches of the country. Subtracting $x$ from them, we get $N-x=y-z<0$, hence they are less than Maximilian's state. 2. In the Formula-2013 race, there are 2 drivers. The first driver completed 4 laps in the time it took the second driver to complete 3. Over the next three laps, due to fast driving, the first driver had to drive an additional 20 meters to the pit stop (without stopping). It is known that when the second driver completed 6 laps, the first driver completed 7.75 laps. Find the length of the lap. (The speeds of the drivers are constant.)
Answer: 80 meters.
Solution: If it were not for the pit stop, the first driver would have completed exactly 8 laps in the time it took the second driver to complete 6 laps. Therefore, the pit stop is exactly $\frac{1}{4}$ of a lap. Then the entire lap is 80 meters.
|
80
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table?
|
Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the table. Therefore, we get 19 girls + 16 boys $=35$ people.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
|
Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$.
|
18108
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. A trip in Moscow using the "Troyka" card in 2016 costs 32 rubles for one subway ride and 31 rubles for one trip on surface transport. What is the minimum total number of trips that can be made under these rates, spending exactly 5000 rubles?
|
Answer: 157.
Solution: If $x$ is the number of trips by surface transport and $y$ is the number of trips by subway, then we have $31x + 32y = 5000$, from which $32(x + y) = 5000 + x \geqslant 5000$. Therefore, $x + y \geqslant 5000 / 32 = 156 \frac{1}{4}$. The smallest integer value of $x + y$ is 157, which is achieved when $x = 24, y = 133$.
|
157
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432.
|
Solution. The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0 ; 2]$, $\delta \in[0 ; 1], \varphi \in[0 ; 1]$. The total number of such divisors is $(8+1) \cdot 4 \cdot(2+1)(1+1)(1+1)=432$.
|
432
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Solve the inequality
$$
8 \cdot \frac{|x+1|-|x-7|}{|2 x-3|-|2 x-9|}+3 \cdot \frac{|x+1|+|x-7|}{|2 x-3|+|2 x-9|} \leqslant 8
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<120$.
|
Answer: 6 (solution of the inequality: $\left.\left[\frac{3}{2} ; 3\right) \cup\left(3 ; \frac{9}{2}\right]\right)$.
Solution. Let $a$ and $b$ be the first and second terms in the left part of the inequality, respectively. Then $b>0$ and
$$
a b=24 \cdot \frac{(x+1)^{2}-(x-7)^{2}}{(2 x-3)^{2}-(2 x-9)^{2}}=24 \cdot \frac{8 \cdot(2 x-6)}{6 \cdot(4 x-12)}=16
$$
for $x \neq 3$, from which $a=\frac{16}{b}$. The inequality $\frac{16}{b}+b \leqslant 8\left(\Leftrightarrow\left(\frac{4}{\sqrt{b}}-\sqrt{b}\right)^{2} \leqslant 0\right)$ is satisfied only when $b=4$. It remains to solve the equation
$$
3 \cdot \frac{|x+1|+|x-7|}{|2 x-3|+|2 x-9|}=4
$$
If $x \geqslant 7$ or $x \leqslant-1$, the expression on the left side equals $\frac{3}{2}$. If $-1<x<\frac{3}{2}$ or $\frac{9}{2}<x<7$, it equals $\frac{6}{|3-x|}<4$. For all $\frac{3}{2} \leqslant x \leqslant \frac{9}{2}$, the equation turns into a true equality. Thus, taking into account the condition $x \neq 3$, we obtain the solution of the original inequality: $\left[\frac{3}{2} ; 3\right) \cup\left(3 ; \frac{9}{2}\right]$. The integer solutions are 2 and 4, both of which satisfy the condition $|x|<120$, their sum is 6.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.3. Solve the inequality
$$
12 \cdot \frac{|x+10|-|x-20|}{|4 x-25|-|4 x-15|}-\frac{|x+10|+|x-20|}{|4 x-25|+|4 x-15|} \geqslant-6
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<100$.
|
Answer: 10 (solution of the inequality: $\left.\left[\frac{15}{4} ; 5\right) \cup\left(5 ; \frac{25}{4}\right]\right)$.)
|
10
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.4. Solve the inequality
$$
9 \cdot \frac{|x+4|-|x-2|}{|3 x+14|-|3 x-8|}+11 \cdot \frac{|x+4|+|x-2|}{|3 x+14|+|3 x-8|} \leqslant 6
$$
In the answer, write the sum of its integer solutions that satisfy the condition $|x|<110$.
|
Answer: -6 (solution of the inequality: $[-4 ;-1) \cup(-1 ; 2]$ ).
|
-6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,4,6,9,10,11,15,16,20,22$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: -4914 (numbers on the board: $-3,2,7,9,13$).
Solution. The sum of the numbers in the obtained set is 112. Each of the original five numbers appears 4 times in this sum. Therefore, the sum of the required numbers is $112: 4=28$. The sum of the two smallest numbers is -1, and the sum of the two largest numbers is 22. Therefore, the middle number (the third largest of the five) is $28-22-(-1)=7$. In the set given in the problem, the second number is equal to the sum of the first and third required numbers, from which the first number is $4-7=-3$, and the second is 2. Similarly, we obtain that the fourth number is 9, and the fifth is 13. Thus, the numbers on the board are $-3,2,7,9,13$, and their product is -4914.
|
-4914
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $3,8,9,16,17,17,18,22,23,31$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: 3360 (numbers on the board: $1,2,7,15,16$).
|
3360
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.3. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $2,6,10,10,12,14,16,18,20,24$. Determine which numbers are written on the board. In your answer, write their product
|
Answer: -3003 (numbers on the board: $-1,3,7,11,13$ ).
|
-3003
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.4. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $6,9,10,13,13,14,17,17,20,21$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: 4320 (numbers on the board: $1,5,8,9,12$ ).
|
4320
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.5. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,5,8,9,11,12,14,18,20,24$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: -2002 (numbers on the board: $-2,1,7,11,13$ ).
|
-2002
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.6. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,8,9,13,14,14,15,17,18,23$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: 4752 (numbers on the board: $2,3,6,11,12$).
|
4752
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.7. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,2,6,7,8,11,13,14,16,20$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: -2970 (numbers on the board: $-3,2,5,9,11$).
|
-2970
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.8. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,9,10,11,12,16,16,17,21,23$. Determine which numbers are written on the board. In your answer, write their product.
|
Answer: 5292 (numbers on the board: $2,3,7,9,14$).
|
5292
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. In the expansion of the function $f(x)=\left(1+x-x^{2}\right)^{20}$ in powers of $x$, find the coefficient of $x^{3 n}$, where $n$ is equal to the sum of all coefficients of the expansion.
|
Answer: 760.
Solution. The expansion in powers of $x$ has the form $f(x)=\sum_{k=0}^{40} a_{k} x^{k}$. The sum of all coefficients is $\sum_{k=0}^{40} a_{k}=f(1)=1$. Therefore, we need to find the coefficient of $x^{3}$. We have:
$$
\begin{aligned}
f(x)=\left(\left(1-x^{2}\right)+x\right)^{20}=(1- & \left.x^{2}\right)^{20}+20\left(1-x^{2}\right)^{19} x+\frac{20 \cdot 19}{2}\left(1-x^{2}\right)^{18} x^{2}+ \\
& +\frac{20 \cdot 19 \cdot 18}{6}\left(1-x^{2}\right)^{17} x^{3}+\text { terms in which } x^{3} \text { does not appear. }
\end{aligned}
$$
In the first and third terms, $x^{3}$ does not appear. In the second term, $x^{3}$ appears with the coefficient $(-20) \cdot 19=-380$; in the fourth term, $x^{3}$ appears with the coefficient $\frac{20 \cdot 19 \cdot 18}{6}=1140$. Therefore, the coefficient of $x^{3}$ is $-380+1140=760$.
|
760
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Find the smallest four-digit number that is not a multiple of 10 and has the following property: if the digits are reversed, the resulting number is a divisor of the original number, and the quotient is different from one.
Answer. 8712.
|
Solution. Let's find all four-digit numbers with the property specified in the problem. Let $\overline{a b c d}$ be the desired number, then $\overline{c b a d}$ is the number obtained from it by reversing the digits, $a, d \neq 0$. According to the condition, $\overline{a b c d}=k \cdot \overline{c b a d}$, where $k$ is a natural number, $k>1$. On one hand, we have $1000 k d k \cdot \overline{d c b a}=\overline{a b c d}>1000 a$, from which $k(d+1)>a, d>\frac{a}{k}-1$. Therefore, $d=\left[\frac{a}{k}\right]$, where $[x]$ denotes the integer part of the number $x$ (i.e., the smallest integer not exceeding $x$). Then $a \geqslant k$, otherwise $d=\left[\frac{a}{k}\right]=0$.
Notice also that the last digit of the number $a k$ must be equal to $d=\left[\frac{a}{k}\right]$. Let's create a table, in the cells of which for each pair $a$ and $k$ we will sequentially write the value $\left[\frac{a}{k}\right]$ (if it is different from zero) and the last digit of the number $a k$.
| $k \backslash a$ | 2 | 3 | | | 4 | | 5 | | | 6 | | 7 | | | 8 | | | 9 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 1 | 4 | 1 | 6 | 2 | 8 | 2 | 0 | 3 | 2 | 3 | 4 | 4 | 6 | 4 | 8 | | | |
| 3 | | | 1 | 9 | 1 | 2 | 1 | 5 | 2 | 8 | 2 | 1 | 2 | 4 | 3 | 7 | | | |
| 4 | | | | | 1 | 6 | 1 | 0 | 1 | 4 | 1 | 8 | 2 | 2 | 2 | 6 | | | |
| 5 | | | | | | 1 | 5 | 1 | 0 | 1 | 5 | 1 | 0 | 1 | 5 | | | | |
| 6 | | | | | | | | | 1 | 6 | 1 | 2 | 1 | 8 | 1 | 4 | | | |
| 7 | | | | | | | | | | 1 | 9 | 1 | 6 | 1 | 3 | | | | |
| 8 | | | | | | | | | | | | 1 | 4 | 1 | 2 | | | | |
| 9 | | | | | | | | | | | | | | | | 1 | 1 | | |
Here, the cells are highlighted where both digits are the same. Thus, the required condition can only be met in two cases: 1) $a=8, k=4$ (then $d=2$) and 2) $a=9, k=9$ (then $d=1$).
In the first case, for the digits $b$ and $c$, we get the equation
$$
8000+100 b+10 c+2=4 \cdot(2000+100 c+10 b+8) \Leftrightarrow 2 b=13 c+1
$$
If $c=0$, then $2 b=1$, which is impossible. For $c \geqslant 2$, we get $b=\frac{13 c+1}{2}>13$, which is also impossible. The only remaining option is $c=1, b=7$.
In the second case, for the digits $b$ and $c$, we get the equation
$$
9000+100 b+10 c+1=9 \cdot(1000+100 c+10 b+9) \quad \Leftrightarrow \quad b=89 c+8
$$
For $c \geqslant 1$, we have $b=89 c+8 \geqslant 97$, which is impossible. Therefore, in this case, $c=0, b=8$.
Thus, the numbers 8712 and 9801 possess the property specified in the problem.
|
8712
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Propose a word problem that reduces to solving the inequality
$$
\frac{11}{x+1.5}+\frac{8}{x} \geqslant \frac{12}{x+2}+2
$$
Write the problem statement, its solution, and the answer.
Example of the required problem. Points A and B are connected by two roads: one is 19 km long, and the other is 12 km long. At 12:00, a pedestrian left point A on the longer road and walked the first 11 km at a constant speed, and then, getting tired, walked the remaining distance to point B at an average speed 1.5 km/h slower. At 14:00, a second pedestrian left point A on the shorter road and walked the entire distance at an average speed 0.5 km/h faster than the first pedestrian initially walked. What was the average speed at which the first pedestrian walked the last 8 km of the journey, if it is known that he arrived at point B no earlier than the second pedestrian?
Brief solution. If \( x \) is the required average speed of the first pedestrian (in km/h) on the second part of the journey, then \( x+1.5 \) is his speed on the first part, and \( (x+1.5)+0.5=x+2 \) is the average speed of the second pedestrian. Then, by calculating the time each pedestrian spent on the journey, we obtain the required inequality. Since by the problem's condition \( x > 0 \), both sides of the inequality can be multiplied by the quantity \( x(x+1.5)(x+2) > 0 \), then we get \( 11x(x+2) + 8(x+1.5)(x+2) \geqslant 12x(x+1.5) + 2x(x+1.5)(x+2) \), or, after transformations, \( (x-4)(x+1)(x+3) \leqslant 0 \). Therefore, \( 0 < x \leqslant 4 \), and all found values of \( x \) satisfy the problem's condition.
|
Answer: no more than 4 km/h.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60.
|
Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$.
|
63
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Construct on the plane the set of points whose coordinates satisfy the inequality $|3 x+4|+|4 y-3| \leq 12$. In your answer, specify the area of the resulting figure.
ANSWER: 24.
|
Solution: We will shift the coordinate axes by $-\frac{4}{3}$ along the Ox axis, and by $\frac{3}{4}$ along the Oy axis. In the new coordinates, the obtained figure is described by the inequality $|3 x|+|4 y| \leq 12$. It is not difficult to show that this will be a rhombus with diagonals of 6 and 8, therefore, its area is 24.
|
24
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the largest ten-digit number of the form $\overline{a_{9} a_{8} a_{7} a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, possessing the following property: the digit equal to $\mathrm{a}_{\mathrm{i}}$ appears in its representation exactly $\mathrm{a}_{9-\mathrm{i}}$ times (for example, the digit equal to $\mathrm{a}_{2}$ appears exactly $\mathrm{a}_{7}$ times).
ANSWER 8888228888.
|
Solution: We will divide all digits into pairs $a_{0}-a_{9}, \ldots, a_{4}-a_{5}$. We will prove that the digit 9 cannot be present in the given number. Suppose $a_{i}=9$, then the paired digit $a_{9-i}$ appears 9 times. If $a_{9-i} \neq 9$, then 9 appears only once, so we get a number consisting of one nine and nine ones, but then two ones would be in the same pair, which contradicts the condition. If $a_{9-i}=9$, then we get a number consisting of nine nines and one digit not equal to 9, and this digit would form a pair with a nine, which is also impossible.
Let's see if the number can contain 8. If it is paired with 8, then the number contains eight eights, forming pairs, the largest such number is 8888228888. If the paired digit is not 8, then the number contains 8 digits that are less than 8, i.e., it cannot start with four digits 8, so it is less than 8888228888.
|
8888228888
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. (2 points) On two bookshelves, there are mathematics books, with an equal number on each. If 5 books are moved from the first shelf to the second, then the second shelf will have twice as many books as the first. How many books are there in total on both shelves?
|
Answer: 30.
Solution. If $x$ is the number of books on both shelves, then $\frac{x}{3}+5=\frac{2 x}{3}-5$, from which $x=30$.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in this team?
|
Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. (14 points) Calculate $x^{3}+\frac{1}{x^{3}}$, given that $x+\frac{1}{x}=3$.
|
Answer: 18.
Solution. If $x+\frac{1}{x}=a$, then $a^{3}=\left(x+\frac{1}{x}\right)^{3}=x^{3}+3\left(x+\frac{1}{x}\right)+\frac{1}{x^{3}}$, from which $x^{3}+\frac{1}{x^{3}}=a^{3}-3 a=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers?
|
Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles.
|
520
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. (14 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1} K B_{2}$.
|
Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of intersection of lines $B_{1} C_{2}$ and $C_{1} B_{2}$ lies inside the triangle. Since $C_{1} B_{2}$ is the median of the right triangle $C C_{1} A$, triangle $C_{1} B_{2} A$ is isosceles. Therefore, $\angle A B_{1} C_{2}=35^{\circ}$. Similarly, we get $\angle A C_{1} B_{2}=35^{\circ}$, from which $\angle A B_{2} C_{1}=180^{\circ}-2 \cdot 35^{\circ}=110^{\circ}$. Then $\angle B_{1} K B_{2}=\angle A B_{2} K-\angle A B_{1} K=110^{\circ}-35^{\circ}=75^{\circ}$.
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three residents of the Ukh tribe among us." How many residents from the Ah tribe are in the hut?
|
Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut. The third one is from tribe Ukh, as he said there are five of them in total. But since he said there are no fewer than three residents from tribe Ukh, there are no more than two. Two have already been found (the third and the first), so there are exactly two. Then the number of residents from tribe Ah is 15.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. (14 points) Find the greatest integer value of $a$ for which the equation
$$
(x-a)(x-7)+3=0
$$
has at least one integer root.
|
Answer: 11.
Solution. For the desired values of $a$, the numbers $x-a$ and $x-7$ are integers, and their product is -3, so there are 4 possible cases:
$$
\left\{\begin{array} { l }
{ x - a = 1 , } \\
{ x - 7 = - 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x - a = 3 , } \\
{ x - 7 = - 1 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x - a = - 1 , } \\
{ x - 7 = 3 ; }
\end{array} \quad \left\{\begin{array}{l}
x-a=-3 \\
x-7=1,
\end{array}\right.\right.\right.\right.
$$
from which we find respectively
$$
\left\{\begin{array} { l }
{ a = 3 , } \\
{ x = 4 ; }
\end{array} \quad \left\{\begin{array} { l }
{ a = 3 , } \\
{ x = 6 ; }
\end{array} \quad \left\{\begin{array} { l }
{ a = 1 1 , } \\
{ x = 1 0 ; }
\end{array} \quad \left\{\begin{array}{l}
a=11 \\
x=8
\end{array}\right.\right.\right.\right.
$$
Thus, the desired values are $a=3$ and $a=11$, of which the largest is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The second term of a geometric progression is 5, and the third term is 1. Find the first term of this progression.
|
Answer: 25. Solution. Since $\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=q$, then $a_{1}=\frac{a_{2}^{2}}{a_{3}}=25$.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find the area of a right triangle, one of whose legs is 6, and the hypotenuse is 10.
|
Answer: 24. Solution. According to the Pythagorean theorem, the second leg is equal to $\sqrt{10^{2}-6^{2}}=8$. Therefore, the area of the triangle is $\frac{6 \cdot 8}{2}=24$.
## Main Task
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. Businessmen Ivanov, Petrov, and Sidorov decided to create an automobile enterprise. Ivanov bought 70 identical cars for the enterprise, Petrov - 40 such cars, and Sidorov contributed 44 million rubles to the enterprise. It is known that Ivanov and Petrov can divide this money between themselves so that the contribution of each of the three businessmen to the common cause will be the same. How much money is due to Ivanov? Answer in millions of rubles.
|
Answer: 40. Solution. Method 1. Each of the businessmen should contribute as much as Sidorov, that is, 44 million rubles. If the car costs $x$, then $\frac{70 x+40 x}{3}=44, x=1.2$. It turns out that Ivanov contributed $70 \cdot 1.2=84$ million rubles, so he should get back $84-44=40$ million rubles. Petrov contributed $40 \cdot 1.2=48$ million rubles, so he should get back $48-44=4$ million rubles.
Method 2. Each of the businessmen should contribute as much as Sidorov, that is, 44 million rubles. If Ivanov takes $t$ million rubles, then Petrov will be left with ( $44-t$ ) million rubles. Therefore, $70 x-t=44$ and $40 x-(44-t)=44$ (here $x$ is the price of the car). Solving this system, we get $x=1.2$ and $t=40$.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 377353752?
|
Answer: 1120. Solution. Since the number is divisible by 5, the digit in the 9th place can only be five. After this, we need to distribute the remaining 8 digits across the 8 remaining places: 3 sevens, 3 threes, a five, and a two. The total number of permutations will be 8!, but since there are repeating digits, the answer will be: $\frac{8!}{3!3!}=\frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{2 \cdot 3 \cdot 2 \cdot 3}=4 \cdot 5 \cdot 7 \cdot 8=1120$.
|
1120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.2. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 131152152?
|
Answer: 840. Instructions. The number of numbers will be $\frac{8!}{4!2!}=840$.
|
840
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.3. How many 9-digit numbers divisible by 5 can be formed by rearranging the digits of the number 137153751?
|
Answer: 1680. Instructions. The number of numbers will be $\frac{8!}{3!2!2!}=1680$.
|
1680
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.4. How many 9-digit numbers divisible by 2 can be formed by rearranging the digits of the number 231157152?
|
Answer: 3360. Instructions. The number of numbers will be $\frac{8!}{3!2!}=3360$.
|
3360
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. Determine the total surface area of a cube if the distance between non-intersecting diagonals of two adjacent faces of this cube is 8. If the answer is not an integer, round it to the nearest integer.
|
Answer: 1152. Solution. Let's take as the diagonals specified in the condition the diagonals $A_{1} C_{1}$ and $A D_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Construct two parallel planes $A_{1} C_{1} B$ and $A D_{1} C$, each containing one of these diagonals. The distance between these planes is equal to the given distance $d$ between the diagonals. This distance can be expressed in terms of the edge length $a$ of the cube, for example, by considering the section $B_{1} D_{1} D B$ perpendicular to these two planes. The diagonal of the cube $B_{1} D$ is perpendicular to both planes and is divided by them into three equal parts. We obtain: $d=\frac{a \sqrt{3}}{3}$, that is, $a=d \sqrt{3}$. Then $S=6 a^{2}=6 \cdot 3 d^{2}=18 d^{2}$.
|
1152
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. Find the sum of all roots of the equation $x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)$.
|
Answer: 7. Solution. The original equation is equivalent to the equation $\left(x+2^{x}\right)^{2}-31\left(x+2^{x}\right)+220=0 \Leftrightarrow\left[\begin{array}{l}x+2^{x}=11, \\ x+2^{x}=20 .\end{array}\right.$
Each of the equations in this system has no more than one root, as the function $f(x)=x+2^{x}$ is monotonically increasing. The first equation has a root $x=3$, and the second has a root $x=4$. The sum of the roots is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.2. Find the sum of all roots of the equation $x^{2}-41 x+330=3^{x}\left(41-2 x-3^{x}\right)$.
|
Answer: 5. Instructions. Roots of the equation: 2 and 3.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Among all the simple fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{3}{4}$. In your answer, specify its numerator.
|
Answer: 73. Solution. It is required to find a fraction $\frac{a}{b}$ such that $\frac{a}{b}-\frac{3}{4}=\frac{4 a-3 b}{4 b}$ reaches a minimum. Therefore, the maximum two-digit $b$ is sought, for which $4 a-3 b=1$. If in this case $b \geq 50$, then the fraction $\frac{a}{b}-\frac{3}{4}=\frac{1}{4 b}$ will always be less than any other fraction with a larger integer numerator and a different two-digit $b$.
Solve the equation $4 a-3 b=1$. Since $b=\frac{4 a-1}{3}=a+\frac{a-1}{3}-$ is an integer, then $a=1+3 k$, where $k-$ is any integer. Therefore, $b=1+4 k$. The maximum $k$, for which $a$ and $b$ are two-digit, will be $k=24$. Therefore, $b=97$ and $a=73$, that is, the required fraction: $\frac{a}{b}=\frac{73}{97}$.
|
73
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.2. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{5}{6}$. In your answer, specify its numerator.
|
Answer: 81. Instructions. The required fraction: $\frac{81}{97}$.
|
81
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.3. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{5}$. In your answer, specify its numerator.
|
Answer: 77. Instructions. The required fraction: $\frac{77}{96}$.
|
77
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.6. Among all the irreducible fractions, where the numerator and denominator are two-digit numbers, find the smallest fraction greater than $\frac{4}{9}$. In your answer, specify its numerator.
|
Answer: 41. Instructions. The required fraction: $\frac{41}{92}$.
|
41
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\), if the area of triangle \(ABC\) is 8. If the answer is not an integer, round it to the nearest integer.
|
Answer: 11. Solution. Let in triangle $ABC$ the angle $\alpha$ be equal to the difference of angles $\beta-\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, then $\beta-\gamma+\beta+\gamma=180^{\circ}$ and therefore $\beta=90^{\circ}$. If $\beta=90^{\circ}$ is twice one of the angles $\alpha$ or $\gamma$, then the triangle will be isosceles, which contradicts the condition. Let for definiteness $\alpha>\gamma$. Then $\beta=90^{\circ}, \alpha=60^{\circ}, \gamma=30^{\circ}$.
By the inscribed angle theorem:
$\angle MLC = \angle MBC = 45^{\circ}, \angle CLO = \angle CAO = 30^{\circ}$. Therefore, $\angle MLO = 75^{\circ}$. Similarly,
$\angle LOM = \angle LOA + \angle AOM = \angle LCA + \angle ABM = 15^{\circ} + 45^{\circ} = 60^{\circ}$,
$\angle OML = \angle OMB + \angle BML = \angle OAB + \angle BCL = 30^{\circ} + 15^{\circ} = 45^{\circ}$.
Thus, the angles of triangle $LOM$ are $\alpha_{1}=75^{\circ}, \beta_{1}=60^{\circ}, \gamma_{1}=45^{\circ}$.
Using the sine rule, we get the formula for the area of a triangle inscribed in a circle of radius $R: S=\frac{1}{2} a b \sin \gamma=\frac{1}{2} \cdot 2 R \sin \alpha \cdot 2 R \sin \beta \cdot \sin \gamma=2 R^{2} \sin \alpha \sin \beta \sin \gamma$.
Therefore, $\frac{S_{LOM}}{S_{ABC}}=\frac{2 R \cdot \sin \alpha_{1} \sin \beta_{1} \sin \gamma_{1}}{2 R \cdot \sin \alpha \sin \beta \sin \gamma}=\frac{\sin 75^{\circ} \sin 60^{\circ} \sin 45^{\circ}}{\sin 90^{\circ} \sin 60^{\circ} \sin 30^{\circ}}=\sqrt{2} \cdot \sin 75^{\circ}$. Since $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$, we get: $\frac{S_{LOM}}{S_{ABC}}=\frac{\sqrt{3}+1}{2}$.
Therefore, $S_{LOM}=4 \sqrt{3}+4 \approx 11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.4. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\), if the area of triangle \(ABC\) is 20. If the answer is not an integer, round it to the nearest integer.
|
Answer: 27. Instructions. Exact answer: $10 \sqrt{3}+10$.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. Solve the inequality $\sqrt{x^{2}-x-56}-\sqrt{x^{2}-25 x+136}<8 \sqrt{\frac{x+7}{x-8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$.
|
Answer: -285. Instructions. Solution of the inequality: $x \in(-\infty ;-7] \cup(18 ; 20)$.
|
-285
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. Solve the inequality $\sqrt{x^{2}+3 x-54}-\sqrt{x^{2}+27 x+162}<8 \sqrt{\frac{x-6}{x+9}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$.
|
Answer: 290 . Instructions. Solution of the inequality: $x \in(-21 ;-19) \bigcup[6 ;+\infty)$.
|
290
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. Solve the inequality $\sqrt{x^{2}+x-56}-\sqrt{x^{2}+25 x+136}<8 \sqrt{\frac{x-7}{x+8}}$, and find the sum of its integer solutions that belong to the interval $[-25 ; 25]$.
|
Answer: 285. Instructions. Solution of the inequality: $x \in(-20 ;-18) \cup[7 ;+\infty)$.
|
285
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Find the maximum value of the expression $(\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer.
|
Answer: 33. Solution. Let $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2$, $g(y)=3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y$ and estimate these two functions.
1) $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2 \leq \sqrt{8-4 \sqrt{3}} \cdot 1-6 \cdot 0-2=\sqrt{8-4 \sqrt{3}}-2=\sqrt{(\sqrt{6}-\sqrt{2})^{2}}-2$ $=\sqrt{6}-\sqrt{2}-2$. The specified value is achieved, for example, at $x=\frac{\pi}{2}$.
$f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2=2(\sqrt{2-\sqrt{3}} \sin x-3|\cos x|-1) \geq 2\left(-\sqrt{(\sqrt{2-\sqrt{3}})^{2}+3^{2}}-1\right)$ $=-2(\sqrt{11-\sqrt{3}}+1)$. The specified value is achieved, for example, at $x=\arcsin \frac{3}{\sqrt{11-\sqrt{3}}}-\frac{\pi}{2}$.
Thus, $-2(\sqrt{11-\sqrt{3}}+1) \leq f(x) \leq \sqrt{6}-\sqrt{2}-2$. Note also that $-2(\sqrt{11-\sqrt{3}}+1)1$, so on the interval $t \in[-1 ; 1]$ the function $h(t)$ is increasing and
$h(-1) \leq h(t) \leq h(1)$. Further: $h(-1)=-2-2 \sqrt{11-\sqrt{3}}+4=-2(\sqrt{11-\sqrt{3}}-1)$,
$h(1)=-2+2 \sqrt{11-\sqrt{3}}+4=2(\sqrt{11-\sqrt{3}}+1)$. Here, $h(-1)<0<h(1)$.
Thus, $-2(\sqrt{11-\sqrt{3}}-1) \leq g(y) \leq 2(\sqrt{11-\sqrt{3}}+1)$.
3) Therefore, the maximum of the product is $(-2(\sqrt{11-\sqrt{3}}+1)) \cdot(-2(\sqrt{11-\sqrt{3}}-1))$ $=4(11-\sqrt{3}-1)=40-4 \sqrt{3} \approx 33$.
|
33
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{3-\sqrt{2}} \sin x+1) \cdot(3+2 \sqrt{7-\sqrt{2}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer.
|
Answer: -9 . Instructions. Exact answer: $2 \sqrt{2}-12$.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Find the maximum value of the expression $(\sqrt{36-4 \sqrt{5}} \sin x-\sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer.
|
Answer: 27. Instructions. Exact answer: $36-4 \sqrt{5}$.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer.
|
Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$.
|
-27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find the largest integer $k$ such that for at least one natural number $n>1000$, the number $n!=1 \cdot 2 \cdot \ldots \cdot n$ is divisible by $2^{n+k+2}$.
|
Answer: -3. Solution. The highest power of two that divides $n!$ is the finite sum $\left[\frac{n}{2^{1}}\right]+\left[\frac{n}{2^{2}}\right]+\ldots \leq \frac{n}{2^{1}}+\frac{n}{2^{2}}+\ldots<n$, so it does not exceed $n-1$. At the same time, the equality of this power to the value $n-1$ is achieved at powers of two (for example, when $n=2^{10}=1024$). Therefore, $(k+2)_{\max }=-1, k_{\max }=-3$.
|
-3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table?
|
Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the table. Therefore, we get 19 girls + 16 boys $=35$ people.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
|
Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$.
|
18108
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. A set of natural numbers is called bad if it is possible to select several numbers from it such that their sum equals 2012. Find the smallest such $n$ that the numbers $503, 504, \ldots, 2011$ can be partitioned into $n$ sets such that none of these sets are bad.
|
Answer: $n=2$.
Solution: We will show that the given set can be divided into two subsets, neither of which are bad. Consider the sets $M_{1}=\{503,504, \ldots, 671\}$, $M_{2}=\{672,673, \ldots, 1006\}$, $M_{3}=\{1007,1008, \ldots, 1340\}$, $M_{4}=\{1341,1342, \ldots, 2011\}$, and show that $M_{1} \cup M_{3}$ and $M_{2} \cup M_{4}$ are not bad.
We will prove that $M_{1} \cup M_{3}$ is not bad. Consider the cases:
- Four numbers from the set $M_{1}$ give a sum $S \geqslant 503+504+505+506=$ $2018>2012$
- Three numbers from the set $M_{1}$ give a sum $S \leqslant 669+670+671=2010<2012$
- One number from the set $M_{1}$ and one from $M_{3}$ give a sum $S \leqslant 671+1340=$ $2011<2012$;
Similarly, it can be proven for $M_{2} \cup M_{4}$.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1.1. (2 points) Find the sum of the squares of two numbers if it is known that their arithmetic mean is 8, and the geometric mean is $2 \sqrt{5}$.
|
Answer: 216.
Solution. If $a$ and $b$ are the numbers in question, then $a+b=16, a b=(2 \sqrt{5})^{2}=20$, therefore
$$
a^{2}+b^{2}=(a+b)^{2}-2 a b=256-40=216 .
$$
|
216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1.1. (2 points) Find the greatest value of $x$ that satisfies the inequality
$$
\left(6+5 x+x^{2}\right) \sqrt{2 x^{2}-x^{3}-x} \leqslant 0
$$
|
Answer: 1.
Solution.
$$
\left[\begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x = 0 ; } \\
{ \{ \begin{array} { c }
{ 2 x ^ { 2 } - x ^ { 3 } - x > 0 ; } \\
{ 6 + 5 x + x ^ { 2 } \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array} { l }
{ - x ( x - 1 ) ^ { 2 } = 0 ; } \\
{ \{ \begin{array} { c }
{ - x ( x - 1 ) ^ { 2 } > 0 ; } \\
{ ( x + 2 ) ( x + 3 ) \leqslant 0 ; }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array}{l}
x=0 \\
x=1 ; \\
\left\{\begin{array}{l}
x<0 ; \\
-3 \leqslant x \leqslant-2
\end{array}\right.
\end{array}\right.\right.\right.
$$
All solutions of the last system are negative. Therefore, the greatest solution of the original inequality will be 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.2.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$
|
Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{10\left(x_{1}^{2}+x_{2}^{2}\right)+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}= \\
=\frac{10\left(x_{1}+x_{2}\right)^{2}-20 x_{1} x_{2}+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}=\frac{10 \cdot 25-20-10}{1}=220
\end{gathered}
$$
|
220
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.3.1. (12 points) Simplify the expression
$$
\left(\frac{2}{\sqrt[3]{3}}+3\right)-\left(\frac{\sqrt[3]{3}+1}{2}-\frac{1}{\sqrt[3]{9}+\sqrt[3]{3}+1}-\frac{2}{1-\sqrt[3]{3}}\right): \frac{3+\sqrt[3]{9}+2 \sqrt[3]{3}}{2}
$$
Answer: 3.
|
Solution. Let's introduce the notation $a=\sqrt[3]{3}$. Then the expression transforms to
$$
\left(\frac{2}{a}+3\right)-\left(\frac{a+1}{a^{3}-1}-\frac{1}{a^{2}+a+1}-\frac{2}{1-a}\right): \frac{a^{3}+a^{2}+2 a}{a^{3}-1}
$$
and after simplifications and performing the division, it transforms to $\left(\frac{2}{a}+3\right)-\frac{2}{a}$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant.
|
Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \frac{1}{6} v_{1}+s_{0}}{v_{2}}$ hours, where $s_{0}=\frac{1}{3}$ km. From this, we find $\frac{1}{4} v_{1}^{2}+v_{1}-24=0, v_{1}=8$ km/h (the second root is negative), $v_{2}=40$ km/h.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.2.1. (12 points) Every morning, each member of the Ivanov family drinks an 180-gram cup of coffee with milk. The amount of milk and coffee in their cups varies. Masha Ivanova found out that she drank $\frac{2}{9}$ of all the milk consumed that morning and $\frac{1}{6}$ of all the coffee consumed that morning. How many people are in this family?
|
Answer: 5.
Solution. Let there be $x$ (for example, grams) of milk and $y$ grams of coffee, and $n$ people in the family. Since each family member drank the same amount of coffee with milk, then $\left(\frac{2 x}{9}+\frac{y}{6}\right) n=$ $=x+y \Leftrightarrow(4 x+3 y) n=18 x+18 y \Leftrightarrow 2 x(2 n-9)=3 y(6-n)$. From this, it follows that $2 n-9$ and $6-n$ have the same sign, that is, $4.5<n<6$. Since $n$ is an integer, then $n=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.3.1. (12 points) On the table, there are 13 weights arranged in a row by mass (the lightest on the left, the heaviest on the right). It is known that the mass of each weight is an integer number of grams, the masses of any two adjacent weights differ by no more than 5 grams, and the total mass of the weights does not exceed 2019 grams. Find the maximum possible mass of the heaviest weight under these conditions.
|
Answer: 185.
Solution. If the mass of the heaviest weight is $m$, then the masses of the other weights will be no less than $m-5, m-10, \ldots, m-60$ grams, and their total mass will be no less than $13 m-390$ grams. Then $13 m-390 \leq 2019$, from which $m \leq 185$. It remains to verify that the set of weights with masses $185,180,175,170,165,160,155,150,145,140,135,130$ and 129 grams satisfies the condition of the problem.
|
185
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.4.1. (12 points) A goat eats 1 hay wagon in 6 weeks, a sheep in 8 weeks, and a cow in 3 weeks. How many weeks will it take for 5 goats, 3 sheep, and 2 cows to eat 30 such hay wagons together?
|
Answer: 16.
Solution. The goat eats hay at a rate of $1 / 6$ cart per week, the sheep - at a rate of $1 / 8$ cart per week, the cow - at a rate of $1 / 3$ cart per week. Then 5 goats, 3 sheep, and 2 cows together will eat hay at a rate of $\frac{5}{6}+\frac{3}{8}+\frac{2}{3}=\frac{20+9+16}{24}=\frac{45}{24}=\frac{15}{8}$ carts per week. Therefore, 30 carts of hay they will eat in $30: \frac{15}{8}=16$ weeks.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1} K B_{2}$.
|
Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of intersection of lines $B_{1} C_{2}$ and $C_{1} B_{2}$ lies inside the triangle. Since $C_{1} B_{2}$ is the median of the right triangle $C C_{1} A$, triangle $C_{1} B_{2} A$ is isosceles. Therefore,
$\angle A B_{1} C_{2}=35^{\circ}$. Similarly, we get $\angle A C_{1} B_{2}=35^{\circ}$, from which $\angle A B_{2} C_{1}=180^{\circ}-2 \cdot 35^{\circ}=110^{\circ}$. Then $\angle B_{1} K B_{2}=\angle A B_{2} K-\angle A B_{1} K=110^{\circ}-35^{\circ}=75^{\circ}$.
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.3.1. (12 points) Among all possible triangles $ABC$ such that $BC=2 \sqrt[4]{3}, \angle BAC=\frac{\pi}{3}$, find the one with the maximum area. What is this area?
|
Answer: 3.
Solution. The locus of points from which the segment $B C$ is "seen" at an angle $\alpha$ consists of arcs of two circles, from the centers of which the segment $B C$ is "seen" at an angle $2 \pi-2 \alpha$. The points on these arcs that are farthest from the segment are their midpoints. Therefore, the desired triangle is isosceles, and its area is $\frac{a^{2}}{4 \operatorname{tg} \frac{\alpha}{2}}$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2.1. (12 points) Solve the system of equations $\left\{\begin{array}{l}9 y^{2}-4 x^{2}=144-48 x, \\ 9 y^{2}+4 x^{2}=144+18 x y .\end{array}\right.$
Given the solutions $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right), \ldots,\left(x_{n} ; y_{n}\right)$, write the sum of the squares
$$
x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}+y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}
$$
|
Answer: 68.
Solution. From the first equation, we get $(3 y)^{2}=(2 x-12)^{2}$ and sequentially substitute into the second equation $y=\frac{2 x-12}{3}$ and $y=\frac{12-2 x}{3}$. The solutions obtained are: $(x ; y)=(0 ; 4),(0 ;-4),(6 ; 0)$. Answer: $6^{2}+4^{2}+(-4)^{2}=68$.
|
68
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.4.1. (12 points) Solve the inequality $\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}-\sqrt{33-x}>4$. In your answer, write the sum of all its integer solutions.
|
Answer: 525.
Solution. Rewrite the inequality as
$$
\sqrt{x-4}+\sqrt{x+1}+\sqrt{2 x}>\sqrt{33-x}+4
$$
Find the domain of the variable $x$:
$$
\left\{\begin{array}{l}
x-4 \geqslant 0 \\
x+1 \geqslant 0 \\
2 x \geqslant 0 \\
33-x \geqslant 0
\end{array}\right.
$$
from which $x \in[4,33]$. We are interested in the integer values from this interval that satisfy the inequality. Note that the left side of the obtained inequality is increasing, while the right side is strictly decreasing. They coincide at $x=8$. For $x \in\{4,5,6,7\}$, the left side is less than the right side, and for $x \in\{9,10, \ldots, 33\}$, the left side is greater than the right side. We calculate the sum. $9+10+\ldots+33=\frac{9+33}{2} \cdot 25=525$.
|
525
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2.1. (12 points) Calculate the value of the expression $\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}}-\arccos \sqrt{\frac{2}{3}}$. Write the obtained expression in the form $\frac{a \pi}{b}$, where $a$ and $b$ are integers, and are coprime, and specify the value of $|a-b|$.
#
|
# Answer: 7.
Solution. Since $\alpha=\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}} \in\left(0 ; \frac{\pi}{2}\right)$ and $\beta=\arccos \sqrt{\frac{2}{3}} \in\left(0 ; \frac{\pi}{2}\right)$, then $A=\alpha-\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$. Therefore, $\sin A=\sin \alpha \cos \beta-\cos \alpha \sin \beta$. After calculations, we get $\sin A=-\frac{1}{2}$, from which it follows that $A=-\frac{\pi}{6}$. Thus, $a=-1, b=6$, and $|a-b|=7$. Note that the answer can also be written as $a=1, b=-6$, but the absolute value of the difference remains the same.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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