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# 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... | 12045 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,319 |
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 comp... | 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 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,320 |
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 \leqsla... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,321 |
3. What is the smallest radius of a sphere from which a regular quadrilateral pyramid with a base edge of 14 and an apothem of 12 can be cut? | Answer: $7 \sqrt{2}$.
Solution. On the one hand, the diameter of the sphere cannot be less than the diagonal $14 \sqrt{2}$ of the base (a square with side 14) of the pyramid contained in it, so the radius of the desired sphere is not less than $7 \sqrt{2}$. On the other hand, a sphere with radius $7 \sqrt{2}$ and cent... | 7\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,322 |
4. Solve the inequality
$$
\log _{5}\left(5 x^{2}+2 x\right) \cdot \log _{5}\left(5+\frac{2}{x}\right)>\log _{5} 5 x^{2}
$$ | Answer: $\left(\frac{-1-\sqrt{2}}{5} ;-\frac{2}{5}\right) \cup\left(\frac{-1+\sqrt{2}}{5} ; \infty\right)$.
Solution.
$$
\begin{aligned}
\log _{5}\left(5+\frac{2}{x}\right) \cdot \log _{5}\left(5 x^{2}+2 x\right)>\log _{5} 5 x^{2} & \Leftrightarrow \log _{5}\left(5+\frac{2}{x}\right)\left(\log _{5}\left(5+\frac{2}{x}... | (\frac{-1-\sqrt{2}}{5};-\frac{2}{5})\cup(\frac{-1+\sqrt{2}}{5};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,323 |
5. Two circles touch each other internally at point K. A chord $AB$ of the larger circle touches the smaller circle at point $L$, and $AL=10$. Find $BL$, if $AK: BK=2: 5$. | Answer: $B L=25$.
Solution. Draw the chord $P Q$ and the common tangent $M N$ (see fig.), then
$$
\angle P Q K=\frac{1}{2} \smile P K=\angle P K M=\frac{1}{2} \smile A K=\angle A B K
$$
th... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,324 |
6. For what values of $a, b$ and $c$ does the set of real roots of the equation
$$
x^{5}+2 x^{4}+a x^{2}+b x=c
$$
consist exactly of the numbers -1 and $1?$ | Answer: $a=-6, b=-1, c=-4$.
Solution. Since $x=1, x=-1$ are roots of the equation $x^{5}+2 x^{4}+a x^{2}+b x-c=0$, the numbers $a, b, c$ satisfy the system
$$
\left\{\begin{array} { l }
{ 3 + a + b - c = 0 , } \\
{ 1 + a - b - c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
b=-1, \\
c=a+2
\end{array} \Righta... | =-6,b=-1,=-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,325 |
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... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,326 |
8. The function $y=f(t)$ is such that the sum of the roots of the equation $f(\sin x)=0$ on the interval $[3 \pi / 2, 2 \pi]$ is $33 \pi$, and the sum of the roots of the equation $f(\cos x)=0$ on the interval $[\pi, 3 \pi / 2]$ is $23 \pi$. What is the sum of the roots of the second equation on the interval $[\pi / 2,... | Answer: $17 \pi$.
Solution. Let the equation $f(\sin x)=0$ on the interval $\left[\frac{3 \pi}{2}, 2 \pi\right]$ have $k$ roots $x_{1}, x_{2}, \ldots, x_{k}$. This means that the equation $f(t)=0$ on the interval $[-1 ; 0]$ has $k$ roots $t_{1}, \ldots, t_{k}$. Since $\arcsin t_{i} \in$ $\left[-\frac{\pi}{2}, 0\right]... | 17\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,327 |
# Problem 1.
B-1
How many times is the second of the numbers $\frac{x}{2}, 2 x-3, \frac{18}{x}+1$ greater than the first, if it is known that it is as many times less than the third? Round the answer to two decimal places. | Answer: $\frac{52}{25}=2.08$
Solution. From the condition $\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}$ we get $a_{1} a_{3}=a_{2}^{2}$ or $\frac{x}{2} \cdot\left(\frac{18}{x}+1\right)=(2 x-3)^{2}$. From this, $8 x^{2}-25 x=0$. For $x=0$, the domain is violated, and for $x=\frac{25}{8}$, we get $a_{1}=\frac{25}{16}, a_{2}=... | 2.08 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,329 |
# 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{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,330 |
# 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... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,331 |
# Problem 5.
B-1
Numbers $x, y, z$ are such that $\frac{x+\frac{53}{18} y-\frac{143}{9} z}{z}=\frac{\frac{3}{8} x-\frac{17}{4} y+z}{y}=1$. Find $\frac{y}{z}$. | Answer: $\frac{352}{305} \approx 1.15$
Solution. We will solve the problem in a general form: find $\frac{y}{z}$, if the numbers $x, y, z$ are such that $\frac{x+A y-B z}{z}=\frac{C x-D y+z}{y}=1$. We will use the fact that if $\frac{a}{b}=\frac{c}{d}=k$, then $\frac{a+n c}{b+n d}=k$ for any $n$ for which the denomina... | \frac{76}{61} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,332 |
# 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+\f... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,333 |
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, t... | 959 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,335 |
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 the... | 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 | 6,336 |
1. Solve the inequality $\operatorname{tg} \arccos x \leqslant \sin \operatorname{arctg} x$. Answer: $\left[-\frac{1}{\sqrt[4]{2}} ; 0\right) \cup\left[\frac{1}{\sqrt[4]{2}} ; 1\right]$. | Solution. Since
$$
\operatorname{tg} \arccos x=\frac{\sin \arccos x}{\cos \arccos x}=\frac{\sqrt{1-x^{2}}}{x}
$$
$\sin \operatorname{arctg} x=\operatorname{tg} \operatorname{arctg} x \cdot \cos \operatorname{arctg} x=\frac{x}{\sqrt{1+x^{2}}}$,
the inequality takes the form
$$
\begin{aligned}
& \frac{\sqrt{1-x^{2}}}... | [-\frac{1}{\sqrt[4]{2}};0)\cup[\frac{1}{\sqrt[4]{2}};1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,338 |
2. Solve the inequality $\operatorname{tg} \arccos x \leqslant \cos \operatorname{arctg} x$. | Answer: $[-1 ; 0) \cup\left[\sqrt{\frac{\sqrt{5}-1}{2}} ; 1\right]$. | [-1;0)\cup[\sqrt{\frac{\sqrt{5}-1}{2}};1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,339 |
1. Points $A$ and $B$ lie on a circle with center $O$ and radius 6, and point $C$ is equidistant from points $A, B$, and $O$. Another circle with center $Q$ and radius 8 is circumscribed around triangle $A C O$. Find $B Q$. Answer: 10 | Solution. Since $A Q=O Q=8$ and $A C=O C$, triangles $\triangle A C Q$ and $\triangle O C Q$ are equal. Therefore, points $A$ and $O$ are symmetric with respect to the line $C Q$, so $A O \perp C Q$.
Next, $\angle B O Q=\angle B O C+\angle C O Q=\angle A O C+\angle O C Q=\pi / 2$. Therefore, $B Q=\sqrt{B O^{2}+O Q^{2}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,340 |
1. Find all values of $a$, for each of which the maximum of the two numbers $x^{3}+3 x+a-9$ and $a+2^{5-x}-3^{x-1}$ is positive for any $x$. Answer: $a>-5$. | Solution. The first function (of $x$) is strictly increasing, while the second is decreasing. Therefore, the requirement for the parameter $a$ specified in the problem means that the intersection point of the graphs of these functions lies above the x-axis, i.e., their value at the root (guessed) of the equation
$$
x^... | >-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,341 |
4. Find all values of $a$, for each of which the minimum of the two numbers
$$
2^{x-1}-3^{4-x}+a \quad \text{ and } \quad a+5-x^{3}-2 x
$$
is negative for any $x$ | Answer: $a<7$.
## Problem 6 | <7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,342 |
1. The Niva and Toyota cars are driving on a circular test track, a quarter of which is a dirt road, and the remaining part is paved. The Niva's speed on the dirt road is 80 km/h, and on the paved road - 90 km/h. The Toyota's speed on the dirt road is 40 km/h, and on the paved road - 120 km/h. The cars start simultaneo... | Answer: The Niva will overtake the Toyota on its $11-$th lap while the Toyota is on its $10-$th lap. Solution. Let $S$ km be the length of the dirt road section, then $3 S$ km is the length of the asphalt section. The Niva completes one lap in $\frac{S}{80}+\frac{3 S}{90}=\frac{11 S}{240}$ hours, while the Toyota compl... | TheNivawillovertaketheToyotaonits11-lapwhiletheToyotaisonits10-lap | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,343 |
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 ... | 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 | 6,344 |
3. Sasha is riding a scooter, and Galia is riding a gyroscooter. On an uphill, Sasha rides the scooter at a speed of 8 km/h, and downhill at a speed of 24 km/h. Galia rides the gyroscooter uphill at a speed of 16 km/h, and downhill at a speed of 18 km/h. Sasha and Galia are competing on a circular track, a quarter of w... | Answer: On her $11-\mathrm{th}$ lap, Galya will overtake Sasha on his $10-\mathrm{th}$ lap. | Onher11-\mathrm{}lap,GalyawillovertakeSashaonhis10-\mathrm{}lap | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,345 |
4. A snowmobile and an ATV are competing on a winter circular track, a quarter of which is covered with loose snow, and the rest with packed snow. The snowmobile travels at 32 km/h over loose snow and 36 km/h over packed snow. The ATV travels at 16 km/h over loose snow and 48 km/h over packed snow. The snowmobile and t... | Answer: The snowmobile on its $11^{\text{th}}$ lap will overtake the quad on its $10^{\text{th}}$ lap.
Lomonosov Moscow State University
Lomonosov Olympiad in Mathematics for School Students
Final Stage 2019/2020 academic year for 11th grade
## Problem 7 | The\snowmobile\on\its\11^{}\lap\will\overtake\the\quad\on\its\10^{}\lap | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,346 |
1. A cube with edge $a=\sqrt{2+\sqrt{3}}$ is illuminated by a cylindrical beam of light with radius $\rho=\sqrt{2}$, directed along the main diagonal of the cube (the axis of the beam contains the main diagonal). Find the area of the illuminated part of the cube's surface. | Answer: $\sqrt{3}(\pi+3)$.
Solution. Using the scalar product, it is easy to show that the cosine of the angle between the main diagonal (the axis of the beam) and any edge of the cube is $1 / \sqrt{3}$. The same value is the cosine of the angle between the axis of the beam and the normal to any face of the cube.
![]... | \sqrt{3}(\pi+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,347 |
2. A cube with edge $a=1 / \sqrt{2}$ is illuminated by a cylindrical beam of light with radius $\rho=\sqrt{2-\sqrt{3}}$, directed along the main diagonal of the cube (the axis of the beam contains the main diagonal). Find the area of the illuminated part of the cube's surface. | Answer: $\left(\sqrt{3}-\frac{3}{2}\right)(\pi+3)$. | (\sqrt{3}-\frac{3}{2})(\pi+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,348 |
3. A cube with edge $a=\sqrt{2+\sqrt{2}}$ is illuminated by a cylindrical beam of light with radius $\rho=\sqrt{2}$, directed along the main diagonal of the cube (the axis of the beam contains the main diagonal). Find the area of the illuminated part of the cube's surface. | Answer: $\frac{\pi \sqrt{3}}{2}+3 \sqrt{6}$ | \frac{\pi\sqrt{3}}{2}+3\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,349 |
4. A cube with edge $a=1$ is illuminated by a cylindrical beam of light with radius $\rho=\sqrt{2-\sqrt{2}}$, directed along the main diagonal of the cube (the axis of the beam contains the main diagonal). Find the area of the illuminated part of the cube's surface. | Answer: $\frac{(2 \sqrt{3}-\sqrt{6})(\pi+6 \sqrt{2})}{4}$.
Lomonosov Moscow State University
## "Lomonosov" School Students Olympiad in Mathematics
Final Stage 2019/2020 academic year for 11th grade
## Problem 8 | \frac{(2\sqrt{3}-\sqrt{6})(\pi+6\sqrt{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,350 |
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} \Leftrightarro... | 278 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,351 |
# Task 2.
At the time when a lion cub, located 6 minutes away, set off for a waterhole, the second, having quenched his thirst, started back along the same path 1.5 times faster than the first. At the same time, a turtle, located 32 minutes away, set off for the waterhole along the same path. After some time, the firs... | Answer: 9.57
## B-9
A crocodile swam across a river 42 m wide in a straight line in 6 seconds. During this time, it was carried downstream by 28 m. After resting and regaining strength, it returned to the same point along the same straight line, but this time fighting the current, in 14 seconds. How many seconds woul... | 9.57 | Other | math-word-problem | Yes | Yes | olympiads | false | 6,352 |
3. A rectangle, the ratio of whose sides is 5, has the largest area among all rectangles whose vertices lie on the sides of a given rhombus, and whose sides are parallel to the diagonals of the rhombus. Find the acute angle of the rhombus. | 3. $2 \operatorname{arctg} \frac{1}{5} \equiv \operatorname{arctg} \frac{5}{12}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3. $2 \operatorname{arctan} \frac{1}{5} \equiv \operatorname{arctan} \frac{5}{12}$. | 2\operatorname{arctan}\frac{1}{5}\equiv\operatorname{arctan}\frac{5}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,354 |
4. Find all pairs $(a, b)$ for which the set of solutions to the inequality $\log _{2014}(2 x+a)>x^{2}-2 x-b \quad$ coincides with the interval $(1 ; 2)$. | 4. $a=-\frac{4024}{2013}, b=\log _{2014}\left(\frac{2013}{4028}\right)$. | =-\frac{4024}{2013},b=\log_{2014}(\frac{2013}{4028}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,355 |
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 sa... | 6. $\frac{90000}{2 \cdot 300}=150 \text{h}$. | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,357 |
3. Determine which of the numbers is greater: $11^{\lg 121}$ or $10 \cdot 10^{\lg ^{2}} 11+11$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: the first.
Solution. Since $11^{\lg 121}=\left(11^{\lg 11}\right)^{2}$, and $10 \cdot 10^{\lg ^{2} 11}+11=10 \cdot\left(10^{\lg 11}\right)^{\lg 11}+11=10 \cdot 11^{\lg 11}+11$, we need to compare the numbers $x^{2}$ and $10 x+11$, where $x=11^{\lg 11}$. The expression $x^{2}-10 x-11=(x-11)(x+1)$ at the point $... | thefirst | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,358 |
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... | 2011 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,360 |
3.1. From a vessel filled to the brim with delicious 100% juice, fifth-grader Masha drank 1 liter of juice during the day, and in the evening, she added 1 liter of water to the vessel. The next day, after thorough mixing, she drank 1 liter of the mixture and in the evening added 1 liter of water. On the third day, afte... | Answer: 1.75. Solution. Let the volume of the container in liters be x. After the first day, there will be $(x-1)$ liters of juice left, after the second day - $\frac{(x-1)^{2}}{x}$ liters of juice, and after the third day - $\frac{(x-1)^{3}}{x^{2}}$ liters of juice. According to the condition, we get the equation $x-\... | 1.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,361 |
4.1. Point $O$ is the intersection of the altitudes of an acute-angled triangle $ABC$. Find $OC$, if $AB=4$ and $\sin \angle C=\frac{5}{13}$. | Answer: 9.6. Solution. From the similarity of triangles $A B B_{1}$ and $O C B_{1}$ (here $B B_{1}$ is the height), it follows that $\frac{A B}{O C}=\frac{B B_{1}}{C B_{1}}=\operatorname{tg} C$. Given $\sin C=\frac{5}{13}$, therefore $\cos C=\frac{12}{13}, \operatorname{tg} C=\frac{5}{12} \cdot$ Thus, $O C=\frac{A B}{\... | 9.6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,362 |
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 ... | 2144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,363 |
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$, t... | 16 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,364 |
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 ... | 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}-... | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,365 |
8.1. Find the maximum value of the expression $\frac{\sin \left(x+\frac{\pi}{4}\right)}{2 \sqrt{2}(\sin x+\cos x) \cos 4 x-\cos 8 x-5}$. | Answer: 0.5. Solution. Let's estimate the modulus of the given expression in the condition:
$$
\begin{aligned}
& \left|\frac{\sin \left(x+\frac{\pi}{4}\right)}{2 \sqrt{2}(\sin x+\cos x) \cos 4 x-\cos 8 x-5}\right| \leq \frac{1}{5-2 \sqrt{2}(\sin x+\cos x) \cos 4 x+\cos 8 x} \\
& =\frac{1}{4-4 \sin \left(x+\frac{\pi}{4... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,366 |
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 integ... | 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}... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,367 |
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... | 60100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,368 |
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 1... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,369 |
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... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,371 |
4. Find the maximum value of $x+y$, if the numbers $x$ and $y$ satisfy the inequality
$$
\log _{\frac{x^{2}+y^{2}}{2}} y \geq 1
$$ | Answer: $1+\sqrt{2}$ Solution. The inequality is equivalent to the system
$$
\left\{\begin{array}{c}
\frac{y-\frac{x^{2}+y^{2}}{2}}{\frac{x^{2}+y^{2}}{2}-1} \geq 0 \\
(x ; y) \neq(0 ; 0) \\
x^{2}+y^{2} \neq 2 \\
y>0
\end{array}\right.
$$
The set of solutions to this system consists of pairs of points $(x ; y)$ lying ... | 1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,372 |
5. Segment $A B=8$ intersects plane $\alpha$ at an angle of $30^{\circ}$ and is divided by this plane in the ratio $1: 3$. Find the radius of the sphere passing through points $A$ and $B$ and intersecting plane $\alpha$ in a circle of the smallest radius. | Answer: $2 \sqrt{7}$. Solution. Denoting the intersection point of $A B$ with the plane $\alpha$ as $C$, we get $A C=2, B C=6$. At the intersection of the sphere with the plane, we obtain a certain circle. Draw the diameter $M N$ of this circle through $C$. Then $A B$ and $M N$ are chords of the sphere, and by the prop... | 2\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,373 |
6. For any natural $n$ and for any set of numbers $x_{1}, x_{2}, \ldots, x_{n}$ from the interval $[0 ; 3]$, the equation $\sum_{i=1}^{n}\left|x-x_{i}\right|=a n$ has a solution $x$ belonging to the interval $[0 ; 3]$. Indicate which of the following values of $a$ satisfy this condition:
a) $a=0$, b) $a=\frac{3}{2}$, c... | Answer: only $a=\frac{3}{2}$. Solution. To prove the impossibility of any $a$ except $a=\frac{3}{2}$, we can provide an example: $n=2, x_{1}=0, x_{2}=3$. Then $\sum_{i=1}^{n}\left|x-x_{i}\right|=|x-0|+|x-3|=3$ for any $x \in[0 ; 3]$, and the resulting equation $3=2 a$ has no solutions for all $a$ except for $a=\frac{3}... | \frac{3}{2} | Algebra | MCQ | Yes | Yes | olympiads | false | 6,374 |
7. What is the minimum volume of a pyramid whose base is an equilateral triangle with a side length of 6, and all planar angles at the vertex are equal to each other and do not exceed $2 \arcsin \frac{1}{3}$? | Answer: $5 \sqrt{23}$. Solution. Let's introduce the following notations: $a$ - the length of the base edge, $S A=x$, $S B=y$ and $S C=z$ - the lengths of the lateral edges, $\alpha$ - the magnitude of the dihedral angle at vertex $S$, $S O$ - the height of the pyramid, $h=A A_{1}$ - the height of the base.
From the c... | 5\sqrt{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,375 |
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 res... | 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 i... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,376 |
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... | 7089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,377 |
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 \cd... | 6218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,379 |
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 w... | 15120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,380 |
Problem 5. Koschei the Deathless gave Baba Yaga an electric broom. On the instrument panel of the electric broom, there is a display that shows the time in the format HH:MM (for example, 13:56) and the charge of the broom on a hundred-point scale (in whole numbers from 1 to 100). The broom consumes the charge evenly an... | Answer: $04: 52,05: 43,06: 35,07: 26,09: 09$.
Solution. In 10 hours (= 600 minutes), the charge is completely depleted, so if the charge value decreases, it will decrease by 1 again exactly 6 minutes later. For the first 4 hours, the charge value is $>60$, so during this time, the charge value and minutes definitely d... | 04:52,05:43,06:35,07:26,09:09 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,381 |
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 | 6,385 |
5. Find all natural numbers $N$ such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13. | Solution - As $N$, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot 5^{b}$, where $a=0,1,2,3,4$ and $b=0,1,2,3$. There will be 20 of them in total, but we need to exclude $1,2,4,5,8,10$ and 16. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,386 |
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 ful... | 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 | 6,387 |
5.1. $[5-6.4$ (а) - 10 points, б) - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this if a) there are no more than four parts; b) there are no more than five par... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
.
Solution. Let $a, b$ and $c$ be the length, width, and height of the second piece of cheese, respectively. Then its volume is $V_{2}=a b c$. According to the problem, the volume of the first piece of cheese is $V_{1}=\frac{3}{2} a \cd... | 19\frac{1}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,389 |
5.1. $[5-6.4$ (а) - 10 points, б) - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this under the condition that a) there are no more than four parts; b) there are... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,390 |
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 th... | 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$ f... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,391 |
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 tabl... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,392 |
4. On a distant island, there live vegetarians who always tell the truth, and cannibals who always lie. One day, a vegetarian and several other islanders lined up, and each one said: "All vegetarians stand from me at a prime number of people apart." How many islanders could have lined up? | Answer: Any quantity.
Solution: Consider the following arrangement of the island's inhabitants (vegetarians are denoted by the letter "V", cannibals by the letter "C"): $V C C V C C V C \underbrace{C \ldots C}_{\text{any quantity}}$. Each vegetarian stands either 2 or 5 positions away from another vegetarian, so the s... | Anyquantity | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,393 |
5. Fashionista Elizabeth has 100 different bracelets, and every day she wears three of them to school. Could it be that after some time, every pair of bracelets has been worn on Liz's wrist exactly once? | Answer: No.
Solution: Consider the first bracelet. It must be paired with each of the other 99 exactly once. Suppose Lisa wears it $n$ days. Then it will be paired with exactly $2 n$ bracelets, which cannot equal 99. | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,394 |
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+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,395 |
7. Sasha and Maxim (out of boredom) wrote non-zero digits in the cells of a $100 \times 100$ table. After this, Sasha said that among the 100 numbers formed by the digits in each row, all are divisible by 9. To this, Maxim replied that among the 100 numbers formed by the digits in each column, exactly one is not divisi... | Solution: If we believe Sasha, then the sum of the digits in each row is divisible by 9, so the total sum of the digits in the table will also be divisible by 9. But if we believe Maksim, then the sum of the digits in all columns except one is divisible by 9, which means the sum of the digits in the table is not divisi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,396 |
8. A mad painter runs through the cells of a $2012 \times 2013$ board, initially painted in black and white. At the very beginning, he runs into a corner cell. After the painter leaves a cell, the color of that cell changes. Can the painter always run across the board and jump off from one of the border cells so that a... | Answer: Yes, always.
Solution: If the painter runs from a corner cell to an arbitrary white cell, then returns to the corner cell along the same route, the specified cell will change color, while all other cells will remain the same color. Thus, the color of all cells except the corner cell can be changed. If the corn... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,397 |
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 achieve... | 157 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,401 |
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... | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,402 |
3.1. Find all roots of the equation $\sin (\pi \cos 2 x)=\cos \left(\pi \sin ^{2} x\right)$, lying in the interval $\left[-\frac{5 \pi}{3} ;-\frac{5 \pi}{6}\right]$. In the answer, record the sum of these roots divided by $\pi$ (in radians), rounding it to two decimal places if necessary. | Answer: $-6.25$.
Solution. Let $t=\pi \sin ^{2} x$. Then $\pi \cos 2 x=\pi-2 t$, so the equation takes the form $\sin (\pi-2 t)=\cos t \Leftrightarrow \sin 2 t=\cos t \Leftrightarrow(2 \sin t-1) \cos t=0 \Leftrightarrow\left[\begin{array}{l}t=\frac{\pi}{2}+\pi k, \\ t=(-1)^{l} \frac{\pi}{6}+\pi l, k, l \in \mathbb{Z} ... | -6.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,403 |
4.1. In a circle, two perpendicular chords $A B$ and $C D$ are drawn. Determine the distance between the midpoint of segment $A D$ and the line $B C$, if $A C=6, B C=5, B D=3$. Round the answer to two decimal places if necessary. | Answer: 4.24 (exact value: $\sqrt{5}+2$).
Solution. Let $AC = a$, $BC = b$, $BD = c$, $N$ be the point of intersection of the chords, $M$ be the midpoint of $AD$, and $H$ be the point of intersection of the lines $MN$ and $BC$. Denote $\angle BAD = \alpha$, $\angle CDA = \beta$, $\alpha + \beta = 90^\circ$. Then $\ang... | 4.24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,404 |
4.2. Quadrilateral $K L M N$ with sides $K L=4, L M=10, M N=12$ is inscribed in a circle. Determine the distance between the midpoint of side $K N$ and the line $L M$, if the lines $K M$ and $L N$ are perpendicular. Round your answer to two decimal places if necessary. | Answer: 6.87 (exact value: $\sqrt{15}+3$ ). | \sqrt{15}+3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,405 |
4.3. Two perpendicular chords $A B$ and $C D$ are drawn in a circle. Determine the distance between the midpoint of segment $A D$ and the line $B C$, if $B D=6, A C=12, B C=10$. Round the answer to two decimal places if necessary. | Answer: 8.47 (exact value: $2 \sqrt{2}+4$ ). | 8.47 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,406 |
4.4. Quadrilateral $K L M N$ with sides $M N=6, K L=2, L M=5$ is inscribed in a circle. Determine the distance between the midpoint of side $K N$ and the line $L M$, if the lines $K M$ and $L N$ are perpendicular. Round the answer to two decimal places if necessary. | Answer: 3.44 (exact value: $\frac{\sqrt{15}+3}{2}$ ). | \frac{\sqrt{15}+3}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,407 |
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 \fr... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,408 |
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 | 6,410 |
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 | 6,411 |
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 i... | -4914 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,412 |
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 | 6,413 |
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 | 6,414 |
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 | 6,415 |
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 | 6,416 |
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 | 6,417 |
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 | 6,418 |
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 | 6,419 |
7.1. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{12}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 374.12 (exact value: $216 \sqrt{3}$).
Solution. Consider the section of the pyramid $S A B C$ passing through the lateral edge $S A$ and the apothem of the opposite face $S D$. Then $S H$ is the height of the pyramid, the distance from $H$ to the line $S D$ is $H N=2 x$, where $x=2$, and the distance from $H$ ... | 216\sqrt{3}\approx374.12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,420 |
7.2. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{11}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 335.64 (exact value: $1584 \sqrt{2.2} / 7$). | 335.64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,421 |
7.3. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{10}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 309.84 (exact value: $80 \sqrt{15}$ ). | 309.84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,422 |
7.4. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{7}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 296.32 (exact value: $112 \sqrt{7}$ ). | 112\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,423 |
7.5. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{6}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 334.63 (exact value: $432 \sqrt{0.6}$ ). | 334.63 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,424 |
7.6. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{5}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary. | Answer: 485.42 (exact value: $720 \sqrt{5 / 11}$ ). | 485.42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,425 |
7.7. The distances from the midpoint of the height of a regular triangular pyramid to the lateral face and to the lateral edge are 2 and $\sqrt{13}$, respectively. Find the volume of the pyramid. If necessary, round your answer to two decimal places. | Answer: 432.99 (exact value: $208 \sqrt{13 / 3}$ ). | 432.99 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,426 |
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:
$$
\begin{aligned}
f(x)=\left(\left(1-x^{2}\right)+x\right)^{20}=(1- & \left.x^{2}\right... | 760 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,428 |
9.1. Find the smallest four-digit number that is not a multiple of 10 and has the following property: if the digits are reversed, the resulting number is a divisor of the original number, and the quotient is different from one.
Answer. 8712. | Solution. Let's find all four-digit numbers with the property specified in the problem. Let $\overline{a b c d}$ be the desired number, then $\overline{c b a d}$ is the number obtained from it by reversing the digits, $a, d \neq 0$. According to the condition, $\overline{a b c d}=k \cdot \overline{c b a d}$, where $k$ ... | 8712 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,429 |
10.1. Propose a word problem that reduces to solving the inequality
$$
\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 1... | Answer: no more than 4 km/h. | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,430 |
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 | 6,433 |
3. An isosceles triangle can be cut into two isosceles triangles (not necessarily equal). Find the values that the smallest angle of such a triangle can take. In the answer, specify the smallest of these values in degrees, multiplied by 6006.
ANSWER: 154440. | Solution: Consider triangle $ABC$ with angles $\angle A=\angle C=\alpha, \angle B=180^{\circ}-2 \alpha$. For the line to divide the triangle into two, it must pass through one of the vertices. Let's consider the case where it passes through vertex A and divides the triangle into two: $ADB$ and $ADC$ (see fig.).
^{2}+$ $(y-1)^{2}=25$. This is possible when one of the terms equals 25 and the other equals 0 (4 cases), or when one equals 16 and the other equals 9 (8 cases). | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,436 |
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