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5. Find all values of the parameter $b$, for which for any value of the parameter $a \in[-1 ; 2]$ the inequality $\operatorname{tg}^{2} x+4(a+b) \operatorname{tg} x+a^{2}+b^{2}-18<0$ is satisfied for each $x \in[-\pi / 4 ; \pi / 4]$. | Solution. Let's make the substitution $y=\operatorname{tg} x, \quad y \in[-1 ; 1]$. We get $y^{2}+4(a+b) y+a^{2}+b^{2}-18<0$. Consider the function $f(y)=y^{2}+4(a+b) y+a^{2}+b^{2}-18$. Its graph is a parabola with branches directed upwards, with the vertex at the point with abscissa $y_{0}=-2(a+b)$. Let's determine fo... | (-2;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,284 |
6. The base of the pyramid $T A B C D$ is a rhombus $A B C D$. The height of the pyramid $T K$ is 5, point $K$ lies on the line containing the diagonal of the base $A C$, and $K C = K A + A C$. The lateral edge $T C$ is $6 \sqrt{5}$, and the lateral faces are inclined to the plane of the base at angles of $30^{\circ}$ ... | # Solution.
1) $\begin{aligned} & K N \perp A D, N=K N \cap A D, \\ & K M \perp C B, M=K M \cap C B .\end{aligned} \quad K N=\frac{T K}{\tan 60^{\circ}}=\frac{T K \sqrt{3}}{3}$,
$K M=\frac{T K}{\tan 30^{\circ}}=T K \sqrt{3}, \quad N M=\frac{2 \sqrt{3}}{3} T K$.
$K C=\sqrt{T C^{2}-T K^{2}}$,
$M C=\sqrt{K C^{2}-K M^{... | \frac{31\sqrt{5}}{12},\quad\arcsin\frac{4\sqrt{15}}{31} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,285 |
6. Given two natural numbers $K$ and $L$. The number $K$ has $L$ divisors, and the number $L$ has $\frac{K}{2}$ divisors. Determine the number of divisors of the number $K+2L$.
(20 points) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,286 | ||
9. In a regular quadrilateral pyramid $T A B C D$, a section is made by a plane through the center of the base $A B C D$ parallel to the median $A M$ of the lateral face $T A B$ and the apothem $T K$ of the lateral face $T C D$. Find the volume of the pyramid with vertex at point $T$ and the base being the above-mentio... | # Solution Variant № 1 (11th Grade, Qualifying Stage) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,288 |
1. From point A to point B, which are 12 km apart, a pedestrian and a bus set out simultaneously. Arriving at point B in less than one hour, the bus, without stopping, turned around and started moving back towards point A at a speed twice its initial speed. After 12 minutes from its departure from point B, the bus met ... | Solution. Let $x$ be the pedestrian's speed (in km/h), $y$ be the car's speed (in km/h) on the way from $A$ to $B$, $2y$ be the car's speed on the way from $B$ to $A$, and $t$ be the time (in hours) the car spends traveling from $A$ to $B$.
$\left\{\begin{aligned} y t & =12, \\ x(t+0.2) & =12-0.4 y, \\ t & <1,\end{ali... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,289 |
2. How many roots does the equation
$\sqrt[3]{|x|}+10[x]=10 x$ ? ( $[x]$ - the integer part of the number $x$, i.e., $[x] \in Z,[x] \leq x<[x]+1$).
(5 points) | # Solution:
$\frac{\sqrt[3]{|x|}}{10}=\{x\},\{x\}=x-[x]$ Since $\{x\} \in[0 ; 1)$, then $x \in(-1000 ; 1000)$.
On the interval $[0 ; 1)$, the equation has 2 roots
$\sqrt[3]{x}=10 x, x=1000... | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,290 |
3. Solve the inequality $2 \sqrt{x^{2}} \arcsin \left(\frac{x}{2}\right)-\frac{\pi}{6 x^{2}}+\frac{\pi}{2} \geq \frac{\pi}{3}|x|-\left(\frac{1}{x^{2}}-3\right) \arcsin \left(\frac{x}{2}\right)$.
In your answer, write the sum of all integer values of $x$ that satisfy this inequality. (6 points) | Solution. $\left(\arcsin \frac{x}{2}-\frac{\pi}{6}\right)\left(2|x|+\frac{1}{x^{2}}-3\right) \geq 0$, | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,291 |
4. In how many ways can seven different items (3 weighing 2 tons each, 4 weighing 1 ton each) be loaded into two trucks with capacities of 6 tons and 5 tons, if the arrangement of the items inside the trucks does not matter? (12 points)
# | # Solution.
Solution. The load can be distributed as $6+4$ or $5+5$.
| 6 tons | 4 tons | number of ways | 5 tons | 5 tons | number of ways |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $2+1+1+1+1$ | $2+2$ | $C_{3}^{1}=3$ | $2+1+1+1$ | $2+2+1$ | $C_{3}^{1} \cdot C_{4}^{3}=12$ |
| $2+2+1+1$ | $2+1+1$ | $C_{3}^{... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,292 |
5. The numbers from 100 to 999 are written without spaces. What is the remainder when the resulting 2700-digit number is divided by 7? | Solution.
Since 1001 is divisible by 7, we get $1000 \equiv-1(\bmod 7) \Longrightarrow 1000^{n} \equiv(-1)^{n}(\bmod 7)$. Therefore, the given number is congruent modulo 7 to the number
$$
999-998+997-996+\ldots+101-100=450 \equiv 2(\bmod 7)
$$
Answer: 2 | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,293 |
6. What is the minimum distance that can be between two points, one of which lies on the graph of the function $y=x^{2}$, and the other - on the curve defined by the equation $4 x^{2}+4 y^{2}-48 x-24 y+163=0 . \quad$ In your answer, write the square of the found distance. (12 points)
# | # Solution.
Transform the equation of the second curve by completing the squares: $(x-3)^{2}+(y-6)^{2}=17 / 4$. The second curve is a circle with center at point $(6 ; 3)$ and radius $\sqrt{17} / 2$. We need to find the minimum distance from the center of this circle to a point lying on the parabola $\quad y=x^{2} . \... | 4.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,294 |
7. In triangle $A B C$, with an area of $180 \sqrt{3}$, the angle bisector $A D$ and the altitude $A H$ are drawn. A circle with radius $\frac{105 \sqrt{3}}{4}$ and center lying on line $B C$ passes through points $A$ and $D$. Find the radius of the circumcircle of triangle $A B C$, if $B H^{2}-H C^{2}=768$.
# | # Solution.
Let $B C=a, A C=b, A B=c, \angle B A D=\angle C A D=\alpha, \angle A D C=\beta$, and the radius of the given circle is $r$. Note that $\beta>\alpha$ (as the external angle of tri... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,295 |
1. Svetlana takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet on the 1580th step of applying this rule, if the initial triplet of numbers was $\{80 ; 71... | # Solution:
Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b . B=80-20=60$.
Answer: 60 . | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,297 |
2. Two balls, the sizes of which can be neglected in this problem, move along a circle. When moving in the same direction, they meet every 20 seconds, and when moving in opposite directions - every 4 seconds. It is known that when moving towards each other along the circle, the distance between the approaching balls de... | # Solution:
Let the speed of the faster ball be $v$, and the slower one be $u$. When moving in the same direction, the faster ball catches up with the slower one when the difference in the distances they have traveled equals the length of the circle. According to the problem, we set up a system of two linear equations... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,298 |
3. In triangle $A B C$, side $B C$ is 19 cm. The perpendicular $D F$, drawn from the midpoint of side $A B$ - point $D$, intersects side $B C$ at point $F$. Find the perimeter of triangle $A F C$, if side $A C$ is $10 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 29 \text{ cm}$.
Answer: 29 cm. | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,299 |
4. Sasha bought pencils in the store for 13 rubles each and pens for 20 rubles each, in total he paid 350 rubles. How many items of pencils and pens did Sasha buy in total?
# | # Solution.
Let $x$ be the number of pencils, $y$ be the number of pens. We get the equation $13 x+20 y=355$
$13(x+y)+7 y=355$, let $x+y=t(1)$
$13 t+7 y=355$
$7(t+y)+6 t=355$, let $t+y=k(2)$
$7 k+6 t=355$
$6(k+t)+k=355$, let $k+t=n(3)$
$6 n+k=355$
$k=355-6 n$. Substitute into (3), $t=7 n-355$
Substitute into (... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,300 |
5. A boy wrote the first twenty natural numbers on a piece of paper. He did not like how one of them was written and crossed it out. It turned out that among the 19 remaining numbers, there is a number equal to the arithmetic mean of these 19 numbers. Which number did he cross out? If the problem has more than one solu... | # Solution:
The sum of the numbers on the sheet, initially equal to $1+2+3+\ldots+20=210$ and reduced by the crossed-out number, is within the range from 210-20=190 to 210-1=209. Moreover, it is a multiple of 19, as it is 19 times one of the addends. Since among the numbers $190,191,192, \ldots 209$ only 190 and 209 a... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,301 |
6. A family of beekeepers brought containers of honey to the fair with volumes of $13,15,16,17,19,21$ liters. In August, three containers were sold in full, and in September, two more, and it turned out that in August they sold twice as much honey as in September. Determine which containers were emptied in August. In y... | # Solution:
A total of $13+15+16+17+19+21=101$ liters of honey were brought. The amount of honey sold is divisible by three. Therefore, the volume of the unsold container must give a remainder of 2 when divided by 3 (the same as 101), i.e., 17 liters. Thus, 101-17=84 liters were sold, with one-third of 84 liters, or 2... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,302 |
7. In rectangle $A B C D$, point $E$ is the midpoint of side $C D$. On side $B C$, point $F$ is taken such that angle $A E F$ is a right angle. Find the length of segment $F C$, if $A F=7, B F=4$. | # Solution:
Let the lines $A E$ and $B C$ intersect at point $K$, then triangles $A E D$ and $K C E$ are equal ( $\angle A E D = \angle C E D$ as vertical angles, $C E = E D, \angle A D E = \... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,303 |
8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find $\angle C_{1} B_{1} A_{1}$. | # Solution:
Extend side $A B$ beyond point $\mathrm{B}$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, which means it is equidistant from its sides.
We get that $A... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,304 |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-4|=p(x)$, where $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, $p(x)=a$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}+3 x}{x+3}-\frac{x^{2}-4 x+4}{2-x}\right|$, we get $f(x)=|2 x-2|$, where $x \neq-3, x \neq 2$.
Solve the equation || $2 x-2|-4|=a$, where $x \neq-3, x \neq 2$ graphically in the system $x O a$.
The equation has three solutions when $a=2$.
The product is 2.
 they will coincide for the 19th time. If necessary, round the answer to the hundredths place according to rounding rules. | # Solution:
The minute hand passes 1 circle in an hour, while the hour hand passes $1 / 12$ of a circle, so their closing speed is 11/12 of a circle per hour, and one closing takes $1 /(11 / 12)=12 / 11$ hours or $720 / 11$ minutes. 19 closings is $19 * 720 / 11=13680 / 11$ minutes.
Answer: $1243.64$. | 1243.64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,306 |
2. Masha, Dasha, and Sasha are tasked with harvesting all the currants from the bushes on the garden plot. Masha and Dasha together can collect all the berries in 7 hours 30 minutes, Masha and Sasha - in 6 hours, and Dasha and Sasha - in 5 hours. How many hours will it take the children to collect all the berries if th... | Solution: Let the entire work be 1. Let $x$ (parts of the entire work per hour) be Masha's labor productivity, $y$ be Dasha's, and $z$ be Sasha's. Since the labor productivity adds up when working together, we can set up a system of equations based on the problem's conditions: $\left\{\begin{array}{l}(x+y) \cdot 7.5=1 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,307 |
3. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are marked respectively such that $\angle CMA = \angle ANC$. Segments $MC$ and $AN$ intersect at point $O$, and $ON = OM$. Find $BC$, if $AM = 5 \, \text{cm}, BM = 3 \, \text{cm}$. | # Solution:
Triangles $A M O$ and $C N O$ are congruent ($\angle A M O=\angle C N O, O N=O M, \angle M O A=\angle N O C$, as vertical angles). From the congruence of the triangles, it follow... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,308 |
4. If a two-digit number is decreased by 36, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting number sequence.
# | # Solution.
$\overline{x y}=10 x+y-$ the original two-digit number, then $\overline{y x}=10 y+x-$ the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+36$
From the equation, it is clear that the two-digit number is greater than 36. Let's start the investigation with the ten... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,309 |
5. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 7 points, if he hit the bullseye 4 times, and the results of the other hits were sevens, eights, and nines? There were no misses at all. | # Solution:
Since the soldier scored 90 points and 40 of them were scored in 4 shots, he scored 50 points with the remaining 6 shots. As the soldier only hit the 7, 8, and 9, let's assume that in three shots (once each in 7, 8, and 9), he scored 24 points. Then, for the remaining three shots, he scored 26 points, whic... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,310 |
6. Boys are dividing the catch. The first one took $r$ fish and a seventh part of the remainder; the second - $2 r$ fish and a seventh part of the new remainder; the third - $3 r$ fish and a seventh part of the new remainder, and so on. It turned out that in this way all the caught fish were divided equally. How many b... | # Solution:
Let $x$ be the number of boys; $y$ be the number of fish each received. Then the last boy took $x r$ fish (there could be no remainder, otherwise there would not have been an even distribution), so $y=x r$. The second-to-last boy took $(x-1) r+\frac{x r}{6}=y$; i.e., $x r$ is $\frac{6}{7}$ of the second-to... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,311 |
7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=5, A E=2$. | # Solution:
Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A$, $\angle C E B=\angle C A F$. Considering that $B C=C E$, we get that triangle $FCA$ ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,312 |
8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $\mathrm{O}$ is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $O_{1} O_{2}$ an... | # Solution:
We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \)... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,313 |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $f(x)=p(x)$ have one solution, if $f(x)=\left|\frac{2 x^{3}-x^{2}-18 x+9}{(1.5 x+1)^{2}-(0.5 x-2)^{2}}\right|, p(x)=|-2 x+2|+a$. If there are more than one value of the parameter, indicate their sum in the answer. | # Solution.
Simplify $f(x)=\left|\frac{2 x^{3}-x^{2}-18 x+9}{(1.5 x+1)^{2}-(0.5 x-2)^{2}}\right|$, we get $f(x)=|x-3|$, where $x \neq 0.5, x \neq -3$.
Solve the equation $|x-3|=|-2 x+2|+a$, where $x \neq 0.5, x \neq -3$ graphically in the system $x O a$.
1) $\left\{\begin{array}{c}x<1 \\ -x+3=-2 x+2+a\end{array} \qu... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,314 |
1. Task: Compare the numbers $\left(\frac{2}{3}\right)^{2016}$ and $\left(\frac{4}{3}\right)^{-1580}$. | Solution: $\left(\frac{2}{3}\right)^{2016}<\left(\frac{2}{3}\right)^{1580}<\left(\frac{3}{4}\right)^{1580}=\left(\frac{4}{3}\right)^{-1580}$
Answer: $\left(\frac{2}{3}\right)^{2016}<\left(\frac{4}{3}\right)^{-1580}$ | (\frac{2}{3})^{2016}<(\frac{4}{3})^{-1580} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,315 |
2. Task: Calculate $\sqrt[3]{\frac{x}{2015+2016}}$, where $x$ is the harmonic mean of the numbers
$a=\frac{2016+2015}{2016^{2}+2016 \cdot 2015+2015^{2}}$ and $b=\frac{2016-2015}{2016^{2}-2016 \cdot 2015+2015^{2}}$.
The harmonic mean of two positive numbers $a$ and $b$ is the number $c$ such that $\frac{1}{c}=\frac{1}... | Solution:
Let's calculate
$\frac{1}{a}+\frac{1}{b}=\frac{2016^{2}+2016 \cdot 2015+2015^{2}}{2016+2015}+\frac{2016^{2}-2016 \cdot 2015+2015^{2}}{2016-2015}=$
$=\frac{(2016-2015)\left(2016^{2}+2016 \cdot 2015+2015^{2}\right)+(2016+2015)\left(2016^{2}-2016 \cdot 2015+2015^{2}\right)}{(2016+2015)(2016-2015)}$
$=\frac{2... | \frac{1}{2016} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,316 |
3. Task: Plot the figure on the coordinate plane defined by the inequality, and calculate its area $x^{2}+y^{2}+4(x-|y|) \leq 0$. | Solution: $x^{2}+y^{2}+4(x-|y|) \leq 0$
$$
\begin{aligned}
& x^{2}+4 x+y^{2}-4|y| \leq 0 \\
& x^{2}+4 x+4+y^{2}-4|y|+4 \leq 8 \\
& (x+2)^{2}+(|y|-2)^{2} \leq(2 \sqrt{2})^{2}
\end{aligned}
$$
The inequality defines a set of points symmetric with respect to the $O x$ axis. For $y \geq 0$, this is the interior of a circ... | 12\pi+8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,317 |
4. Task: On an $8 \times 8$ chessboard, 64 checkers numbered from 1 to 64 are placed. 64 students take turns approaching the board and flipping only those checkers whose numbers are divisible by the ordinal number of the current student. A "Queen" is a checker that has been flipped an odd number of times. How many "Que... | Solution: It is clear that each checker is flipped as many times as the number of divisors of its number. Therefore, the number of "queens" will be the number of numbers from 1 to 64 that have an odd number of divisors, and this property is only possessed by perfect squares. That is, the numbers of the "queens" remaini... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 10,318 |
# 5. Problem: Solve the equation
$$
\left(x^{3}+x^{2}+x+1\right)\left(x^{11}+x^{10}+\ldots+x+1\right)=\left(x^{7}+x^{6}+\ldots+x+1\right)^{2}
$$ | Solution: Multiply both sides of the equation by $(x-1)^{2}$, then the equation will take the form:
$\left(x^{4}-1\right)\left(x^{12}-1\right)=\left(x^{8}-1\right)^{2}$, expand the brackets:
$x^{16}-x^{12}-x^{4}+1=x^{16}-2 x^{8}+1$
$x^{12}-2 x^{8}+x^{4}=0$
$x^{4}\left(x^{8}-2 x^{4}+1\right)=0$
$x^{4}\left(x^{4}-1\... | -1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,319 |
6. Problem: For what values of the parameter $a$ does the equation have a unique solution $\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}+x=a ?$ | Solution: Notice that the expression under the square root can be represented as a complete square.
$x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=x+\frac{1}{4}+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+\frac{1}{4}=\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}$, then the equation will take the form $\sqrt{\left(\sqrt{x+\... | \in[\frac{1}{4};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,320 |
7. Problem: Prove the inequality $\sqrt{2}+\sqrt{4}+\sqrt{6}+\ldots+\sqrt{4030} \leq 2015 \sqrt{2016}$.
# | # Solution:
$\frac{1}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2} \leq x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}, \quad$ based on the Cauchy inequality:
$2 x_{i} x_{j} \leq x_{i}^{2}+x_{j}^{2}$, for any positive numbers $x_{i}, x_{j}(1 \leq i, j \leq n)$. Therefore:
$\sqrt{2}+\sqrt{4}+\sqrt{6}+\ldots+\sqrt{4030} \leq \sq... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 10,321 |
8. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}=75^{\circ}$. A point K is chosen on side $\mathrm{AC}$. Circles are circumscribed around triangles $\mathrm{ABK}$ and $\mathrm{CBK}$ with centers $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$, respectively. Find the radius of the circumcircle of triangle $\mathrm... | # Solution:
Since VK is the common chord of the two given circles, $O_{1} O_{2}$ is the perpendicular bisector of VK. Therefore, $\angle O_{2} O_{1} K = \frac{1}{2} \angle B O_{1} K = \angl... | 2\sqrt{6}-2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,322 |
3. If $a-$ is the first term and $d-$ is the common difference of an arithmetic progression, $\left\{\begin{array}{l}a+16 d=52, \\ a+29 d=13\end{array} \Leftrightarrow d=-3, a=100\right.$.
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ reaches its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. ... | Answer: 1717
Consider the equation of the system $\sqrt{2} \cos \frac{\pi y}{8}=\sqrt{1+2 \cos ^{2} \frac{\pi y}{8} \cos \frac{\pi x}{4}-\cos \frac{\pi x}{4}}$. Given the condition $\sqrt{... | 1717 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,323 |
5. Let's solve the inequality $\quad \frac{\sqrt{x^{2}+x}-\sqrt{4-2 x}}{2 x+5-2 \sqrt{x^{2}+5 x+6}} \leq 0$.
Domain of definition for the numerator and denominator: $x(x+1) \geq 0, \quad 4-2 x \geq 0, x^{2}+5 x+6 \geq 0, \Rightarrow$ $x \in(-\infty ;-3] \cup[-2 ;-1] \cup[0 ; 2]$.
On the domain of definition, the orig... | Answer: $x \in(-\infty ;-4] \cup[-2 ;-1] \cup[0 ; 1]$ | x\in(-\infty;-4]\cup[-2;-1]\cup[0;1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,324 |
6. Let's find the range of the function $f(x)=g\left(g^{2}(x)\right)$, where $g(x)=\frac{3}{x^{2}-4 x+5}$.
The function $g(x)=\frac{3}{x^{2}-4 x+5}=\frac{3}{(x-2)^{2}+1}$ is defined for all real numbers and takes all values from the interval $(0 ; 3]$. The function $g(x)$ reaches its maximum value at the point $x=2$, ... | Answer: $E_{f}=\left[\frac{3}{50} ; 3\right]$ | [\frac{3}{50};3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,325 |
7. $\mathrm{ABCD}$ is an isosceles trapezoid, $AB = CD$, $\angle A$ is acute, $AB$ is the diameter of a circle, $M$ is the center of the circle, $P$ is the point of tangency of the circle with the side $CD$. Let the radius of the circle be $x$. Then $AM = MB = MP = x$. Let $N$ be the midpoint of side $CD$, then triangl... | Answer: $1,49,30,30$ or $\frac{3}{4}, \frac{147}{4}, 30,30$ | 1,49,30,30or\frac{3}{4},\frac{147}{4},30,30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,326 |
8. Let's form the equation of the common tangent to the graphs of the functions $y=1+x-x^{2}$ and $y=0.5\left(x^{2}+3\right)$. Let $y=a x+b-$ be the equation of the common tangent. Then,
$1+x-x^{2}=a x+b, x^{2}+(a-1) x+b-1=0$.
$D=(a-1)^{2}-4 b+4=0, a^{2}-2 a-4 b+5=0$.
$0.5\left(x^{2}+3\right)=a x+b, x^{2}-2 a x-2 b+... | Answer: $y=x+1, y=-\frac{1}{3} x+\frac{13}{9}$ | x+1,-\frac{1}{3}x+\frac{13}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,327 |
1. One worker in two hours makes 5 more parts than the other, and accordingly spends 2 hours less to manufacture 100 parts. How much time does each worker spend on manufacturing 100 parts?
# | # Solution:
Let the second worker make 100 parts in $x$ hours, and the first worker in $x-2$ hours.
$\frac{100}{x-2}-\frac{100}{x}=\frac{5}{2} ; \frac{40}{x-2}-\frac{40}{x}=1 ; x^{2}-2 x-80=0, x=1+9=10$. Answer: 8 and 10 hours. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,330 |
2. How many consecutive terms of the arithmetic progression $32,28,24, \ldots$, starting from the first, must be summed to get a sum equal to 132?
# | # Solution:
$\frac{32+32-4(n-1)}{2} \cdot n=132, n^{2}-17 n+66=0, n=\frac{17 \pm 5}{2}=\left[\begin{array}{c}6, \\ 11 .\end{array}\right.$ Answer: 6 or 11. | 6or11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,331 |
3. Solve the equation $9^{1+\sqrt{x}}+3^{1-2 \sqrt{x}}=28$.
# | # Solution:
$9^{1+\sqrt{x}}+3^{1-2 \sqrt{x}}=28 \Leftrightarrow 9^{1+2 \sqrt{x}}-28 \cdot 9^{\sqrt{x}}+3=0 \Leftrightarrow 9 \cdot\left(9^{\sqrt{x}}\right)^{2}-28 \cdot 9^{\sqrt{x}}+3=0 ; 9^{\sqrt{x}}=\frac{14 \pm 13}{9}$.
1) $9^{\sqrt{x}}=\frac{1}{9}$, no solutions; 2) $9^{\sqrt{x}}=3, \sqrt{x}=\frac{1}{2}, x=\frac{... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,332 |
4. Solve the system of equations $\left\{\begin{array}{l}2 \cos ^{2} x+2 \sqrt{2} \cos x \cos ^{2} 4 x+\cos ^{2} 4 x=0, \\ \sin x=\cos y .\end{array}\right.$.
# | # Solution:
Solve the 1st equation of the system:
$$
\begin{aligned}
& 2 \cos ^{2} x+2 \sqrt{2} \cos x \cos ^{2} 4 x+\cos ^{2} 4 x=0 \quad \Leftrightarrow\left(\sqrt{2} \cos x+\cos ^{2} 4 x\right)^{2}+\cos ^{2} 4 x \sin ^{2} 4 x=0 \\
& \left\{\begin{array} { l }
{ \sqrt { 2 } \cos x + \cos ^ { 2 } 4 x = 0 , } \\
{ \... | (\frac{3\pi}{4}+2\pik,\\frac{\pi}{4}+2\pin),(-\frac{3\pi}{4}+2\pik,\\frac{3\pi}{4}+2\pin),\quadk,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,333 |
5. Solve the inequality $\sqrt{(x+2)|x+1|+|x|} \geq x+2$.
# | # Solution:
$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
x+2<0, \\
(x+2)(-x-1)-x \geq 0 ;
\end{array}\right. \\
\left\{\begin{array} { l }
{ x + 2 \geq 0 , } \\
{ ( x + 2 ) | x + 1 | + | x | \geq ( x + 2 ) ^ { 2 } . }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
x<-2, \\
x^{2}+4 x+2... | x\in[-2-\sqrt{2};-1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,334 |
6. Find the set of values of the function $f(x)=2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right)$.
# | # Solution:
$f(x)=2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right)$ Since $\sqrt{x-3}+2 x+2 \in[8 ; \infty)$, then $\sin (\sqrt{x-3}+2 x+2) \in[-1 ; 1], \quad \frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2) \in[-\pi / 4 ; \pi / 4]$. Therefore, $\cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right) \in[\sqrt{2} / 2 ; ... | [\sqrt{2};2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,335 |
7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is a median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the area of triangle $A B C$ if the length of segment $D E$ is $4(\sqrt{2}-1)$. | # Solution:
By the property of tangents to a circle, we have:
$$
A G=A K=x, C G=C D=y
$$
$C E=C F=z, B F=B H=u$,
$D M=\frac{A B}{2}-A K=\frac{A B}{2}-x$,
$M E=\frac{A B}{2}-B H=\frac{A B}{2}-u$,
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,336 |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 24$, passing through the point $M(4 ;-2 \sqrt{3})$.
# | # Solution:
$$
y=\frac{x^{2}}{8 \sqrt{3}}, M(4 ;-2 \sqrt{3}) \cdot y=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(x-x_{0}\right) ;-2 \sqrt{3}=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(4-x_{0}\right) ;
$$
$x_{0}^{2}-8 x_{0}-48=0 ; x_{0}=4 \pm 8 ;\left(x_{0}\right)_{1}=12,\left(x_{0}\righ... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,337 |
10. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, the diagonal $C A_{1}$, equal to $d$, is inclined to the base plane at an angle of $60^{\circ}$ and forms an angle of $45^{\circ}$ with the plane passing through the diagonal $A C_{1}$ and the midpoint of the lateral edge $B B_{1}$. Find the area of... | # Solution:
If $P$ and $Q$ are the midpoints of the lateral edges $D D_{1}$ and $B B_{1}$, then $P Q \| B D$ and $\left(A Q C_{1}\right) \|(B D)$.
Draw $(M N) \| B D, A \in(M N) ; C K \pe... | \frac{\sqrt{3}^{2}}{8\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,340 |
2. In an arithmetic progression, there are 12 terms, and their sum is 354. The sum of the terms with even indices is to the sum of the terms with odd indices as 32:27. Determine the first term and the common difference of the progression.
# | # Solution:
$$
\begin{aligned}
& S_{\text {even }}=\frac{32}{59} \cdot 354=192 ; S_{\text {odd }}=\frac{27}{59} \cdot 354=162 . \\
& \frac{a+d+a+11 d}{2} \cdot 6=192 ; \frac{a+a+10 d}{2} \cdot 6=162 \cdot\left\{\begin{array} { l }
{ a + 6 d = 3 2 } \\
{ a + 5 d = 2 7 }
\end{array} \Leftrightarrow \left\{\begin{array}... | =2,=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,341 |
3. Solve the equation $9^{1+2 \sqrt{x}}-28 \cdot 9^{\sqrt{x}}+3=0$. | # Solution:
1) $9^{\sqrt{x}}=\frac{1}{9}$, no solutions;
2) $9^{\sqrt{x}}=3, \sqrt{x}=\frac{1}{2}, x=\frac{1}{4}$.
Answer: $x=\frac{1}{4}$. | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,342 |
4. Solve the system of equations $\left\{\begin{array}{l}2 \sin ^{2} x+2 \sqrt{2} \sin x \sin ^{2} 2 x+\sin ^{2} 2 x=0, \\ \cos x=\cos y .\end{array}\right.$.
# | # Solution:
Solve the 1st equation of the system:
$$
\begin{aligned}
& 2 \sin ^{2} x+2 \sqrt{2} \sin x \sin ^{2} 2 x+\sin ^{2} 2 x=0 \Leftrightarrow\left(\sqrt{2} \sin x+\sin ^{2} 2 x\right)^{2}+\sin ^{2} 2 x \cos ^{2} 2 x=0 \Leftrightarrow \\
& \left\{\begin{array} { l }
{ \sqrt { 2 } \sin x + \sin ^ { 2 } 2 x = 0 ... | (2\pik,2\pin),\quad(\pi+2\pik,\pi+2\pin),\quad(-\frac{\pi}{4}+2\pik,\\frac{\pi}{4}+2\pin),\quad(-\frac{3\pi}{4}+2\pik,\\frac{3\pi}{4}+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,343 |
5. Solve the inequality $x+6-\sqrt{(x+6)|x+5|+|x+4|} \geq 0$.
# | # Solution:
$$
\sqrt{(x+6)|x+5|+|x+4|} \leq x+6 \Leftrightarrow\left\{\begin{array}{l}
x+6 \geq 0 \\
(x+6)|x+5|+|x+4| \leq(x+6)^{2} \\
(x+6)|x+5|+|x+4| \geq 0
\end{array}\right.
$$
Given $x+6 \geq 0$, the inequality $(x+6)|x+5|+|x+4| \geq 0$ holds, and the system is equivalent to $\quad$ the following $\quad\left\{\b... | x\in[-5,+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,344 |
6. Find the set of values of the function $f(x)=2 \sin ((\pi / 4) \sin (\sqrt{x-2}+x+2)-(5 \pi / 2))$. | Solution:
$f(x)=2 \sin \left(\frac{\pi}{4} \sin (\sqrt{x-2}+x+2)-\frac{5 \pi}{2}\right)=-2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-2}+x+2)\right)$ Since $\sqrt{x-2}+x+2 \in[4 ; \infty)$, then $\sin (\sqrt{x-2}+x+2) \in[-1 ; 1], \quad \frac{\pi}{4} \sin (\sqrt{x-2}+x+2) \in[-\pi / 4 ; \pi / 4]$.
Therefore, $\cos \left(... | [-2;-\sqrt{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,345 |
7. In triangle $A B C$, angles $A$ and $B$ are $45^{\circ}$ and $30^{\circ}$ respectively, and $C M$ is the median. The incircles of triangles $A C M$ and $B C M$ touch segment $C M$ at points $D$ and $E$. Find the radius of the circumcircle of triangle $A B C$ if the length of segment $D E$ is $4(\sqrt{2}-1)$. | # Solution:
By the property of tangents to a circle, we have:
$$
A G=A K=x, C G=C D=y, C E=C F=z, B F=B H=u, D M=\frac{A B}{2}-A K=\frac{A B}{2}-x
$$
$M E=\frac{A B}{2}-B H=\frac{A B}{2}-u$,
Then $D E=z-y, D E=D M-M E=u-x$. Therefore, $2 D E=z-y+u-x=C B-$ AC. Let $C B=a, A C=b$. Then $a-b=8(\sqrt{2}-1)$. By the Law... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,346 |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 6$, passing through the point $M(1 ;-\sqrt{3} / 2)$. | Solution:
$$
y=\frac{x^{2}}{2 \sqrt{3}}, M\left(1 ;-\frac{\sqrt{3}}{2}\right) \cdot y=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(x-x_{0}\right) ;-\frac{\sqrt{3}}{2}=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(1-x_{0}\right) ; x_{0}^{2}-2 x_{0}-3=0
$$
$; x_{0}=1 \pm 2 ;\left(x_{0}\right)_{1}... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 10,347 |
1. One car covers a distance of 120 km 18 minutes faster than the other. If the first car reduced its speed by 12 km/h, and the second car increased its speed by $10 \%$, they would spend the same amount of time on the same distance. Find the speeds of the cars. | Solution: Let $x$ be the speed of the first car, $y$ be the speed of the second car $\frac{120}{x}+\frac{18}{60}=\frac{120}{y}+5, \quad x-12=1.1 y, \quad \frac{40}{x}+\frac{1}{10}=\frac{44}{x-12}, \quad x^{2}-52 x-4800=0, \quad x=100, \quad y=80$ Answer: $100 \mathrm{km} / \mathrm{h}, 80 \mathrm{km} / \mathrm{h}$ | 100\mathrm{}/\mathrm{},80\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,350 |
2. Indicate all values of $n$ for which the sum of $n$ consecutive terms of the arithmetic progression $25,22,19, \ldots$, starting from the first, is not less than 66. | Solution: For the given arithmetic progression, we have: $a_{1}=25, d=-3$. The inequality $S_{n} \geq 66$ must be satisfied, i.e., $\frac{2 a_{1}+d(n-1)}{2} \cdot n \geq 66$. Substituting $a_{1}$ and $d$ into the last inequality, we get
$$
\frac{50-3(n-1)}{2} \cdot n \geq 66 \Leftrightarrow 53 n-3 n^{2} \geq 132 \Left... | 3,4,5,\ldots,14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,351 |
3. Solve the equation $2^{1-2|x|}+2 \cdot 4^{1+|x|}=17$. | Solution:
$$
2^{2|x|}=t \geq 1, \quad 8 t^{2}-17 t+2=0, \quad t=2, \quad 2^{2|x|}=2, \quad 2|x|=1, \quad x= \pm \frac{1}{2} \quad \text{Answer}: x= \pm \frac{1}{2}
$$ | \\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,352 |
4. Solve the system of equations $\left\{\begin{array}{l}2 \cos ^{2} x-2 \sqrt{2} \cos x \cos ^{2} 8 x+\cos ^{2} 8 x=0, \\ \sin x=\cos y .\end{array}\right.$ | Solution:
Solve the 1st equation of the system:
$2 \cos ^{2} x-2 \sqrt{2} \cos x \cos ^{2} 8 x+\cos ^{2} 8 x=0 \Leftrightarrow\left(\sqrt{2} \cos x-\cos ^{2} 8 x\right)^{2}+\cos ^{2} 8 x \sin ^{2} 8 x=0 \Leftrightarrow$ $\left\{\begin{array}{l}\sqrt{2} \cos x-\cos ^{2} 8 x=0, \\ \cos ^{2} 8 x \sin ^{2} 8 x=0\end{arra... | (\frac{\pi}{4}+2\pik,\\frac{\pi}{4}+\pin),(-\frac{\pi}{4}+2\pik,\\frac{3\pi}{4}+\pin),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,353 |
5. Solve the inequality $\frac{x+3-3 \sqrt{x+1}}{x^{2}-4 x}>0$ (10 points) | Solution:
Domain of the function: $x \in[-1 ;+\infty), \quad x \neq 0, \quad x \neq 4$.
Factorize the numerator:
$x+3-3 \sqrt{x+1}=x+1-3 \sqrt{x+1}+2=(\sqrt{x+1})^{2}-3 \sqrt{x+1}+2=(\sqrt{x+1}-1)(\sqrt{x+1}-2)$
We have:
$$
\frac{(\sqrt{x+1}-1)(\sqrt{x+1}-2)}{x(x-4)}>0 \Leftrightarrow \frac{(x+1-1)(x+1-4)}{x(x-4)}... | x\in[-1;0)\cup(0;3)\cup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,354 |
6. Find the set of values of the function $f(x)=2 \cos \left(\frac{\pi}{4} \sin \left(x^{2}+2 x+2+\cos x\right)\right)$.
# | # Solution:
$f(x)=2 \cos \left(\frac{\pi}{4} \sin \left(x^{2}+2 x+2+\cos x\right)\right)=2 \cos \left(\frac{\pi}{4} \sin \left((x+1)^{2}+1+\cos x\right)\right)$. Since $g(x)=(x+1)^{2}+1+\cos x \subset[0 ; \infty)$, and the range of this function contains a numerical ray, then
$\varphi(x)=\sin \left((x+1)^{2}+1+\cos x... | [\sqrt{2};2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,355 |
7. On the side $B C$ of triangle $A B C$, a point $K$ is marked.
It is known that $\angle B+\angle C=\angle A K B, \quad A K=5, \quad B K=6, \quad K C=2$.
Find the area of the circle inscribed in triangle $A B C$. | Solution:
$\angle B+\angle C=\angle A K B, \quad \angle A K C=\angle A, \quad \angle K A C=\angle B$,
$\triangle A B C \approx \triangle A K C \Rightarrow$
$\frac{A B}{A K}=\frac{A C}{K C}=\frac{B C}{A C} \Rightarrow \frac{A B}{5}=\frac{A C}{2}=\frac{8}{A C} \Rightarrow$
$A C=4, \quad A B=10$.
$ to the parabola $8 y=(x-3)^{2}$. Determine the angle between the tangents. Find the area of the triangle $A B M$, where $A$ and $B$ are the points of tangency. | # Solution:
$8 y=(x-3)^{2} ; M(0 ;-2)$.
$y=\frac{\left(x_{0}-3\right)^{2}}{8}+\frac{\left(x_{0}-3\right)}{4}\left(x-x_{0}\right)$
$y=\frac{9}{2}+\frac{\left(x_{0}-3\right)}{4} x-\frac{x_{0}^{2}}{8}$
$-2=\frac{9}{8}+\frac{\left(x_{0}-3\right)}{4} \cdot 0-\frac{x_{0}^{2}}{8}, x_{0}^{2}-25=0$,
$\left(x_{0}\right)_{1}... | -2x-2,\frac{1}{2}x-2;90;\frac{125}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,357 |
9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=8(2 y-x+a-2), \frac{1-\sqrt{y}}{1-\sqrt{x / 2}}=1$ has at least one solution, and solve it for each $a$.
# | # Solution:
The second equation is equivalent to the system: $x \geq 0, x \neq 2, y=x / 2$. Substituting $y=x / 2$ into the first equation, we get: $(x-a)^{2}=8(a-2)$, or $x^{2}-2 a x+a^{2}-8 a+16=0(*)$, which has $D / 4=a^{2}-a^{2}+8 a-16=8 a-16$. The number of solutions to the given system of equations depends on th... | \in(2;10)\cup(10;+\infty),\\sqrt{8-16},(\\sqrt{8-16})/2;\quad=10,18,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,358 |
10. The base of the pyramid $T A B C D$ is a rectangle $A B C D$. The height of the pyramid coincides with the lateral edge $T A$, and the lateral edge $T C$ is inclined to the base plane at an angle of $30^{\circ}$. The plane passing through the edge $T C$ and parallel to the diagonal of the base $B D$ forms an angle ... | # Solution:
Draw $K C \| B D, A K \perp K C, L=A K \cap B D, L S \perp T K$, then $L S=a$. Let $\beta=\angle T K A$. Since $A K=2 L K$, then $A K=2 a / \sin \beta$ and $A T=A K \operatorna... | \frac{4\sqrt{3}}{\sqrt{5}}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,359 |
1. One tourist covers a distance of 20 km 2.5 hours faster than the other. If the first tourist reduced their speed by 2 km/h, and the second increased their speed by 50%, they would spend the same amount of time on the same distance. Find the speeds of the tourists. | Solution: Let $\mathrm{x}$ be the speed of the first tourist, and $y$ be the speed of the second tourist.
\[
\frac{20}{x}+\frac{5}{2}=\frac{20}{y}, \quad x-2=1.5 y, \quad \frac{4}{x}+\frac{1}{2}=\frac{6}{x-2}, \quad x^{2}-6 x-16=0, \quad x=8, \quad y=4
\]
Answer: $8 \mathrm{km} / \mathrm{u}, 4 \mathrm{km} / \mathrm{q... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,360 |
2. Indicate all values of $n$ for which the sum of $n$ consecutive terms of the arithmetic progression $22,19,16, \ldots$, starting from the first, is not less than 52. | Solution: For this arithmetic progression, we have: $a_{1}=22, d=-3$. The inequality $S_{n} \geq 52$ must be satisfied, i.e., $\frac{2 a_{1}+d(n-1)}{2} \cdot n \geq 52$. Substituting $a_{1}$ and $d$ into the inequality, we get
$\frac{44-3(n-1)}{2} \cdot n \geq 52 \Leftrightarrow 47 n-3 n^{2} \geq 104 \Leftrightarrow 3... | 3,4,5,\ldots,13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,361 |
3. Solve the equation $3^{1-2|x|}+3 \cdot 9^{1+|x|}=82$. (4 points) | Solution:
$$
3^{2 x \mid}=t \geq 1, \quad 27 t^{2}-82 t+3=0, \quad t=3, \quad 3^{2|x|}=3, \quad 2|x|=1, \quad x= \pm \frac{1}{2} \quad \text{Answer: } x= \pm \frac{1}{2} .
$$ | \\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,362 |
4. Solve the system of equations $\left\{\begin{array}{l}4 \cos ^{2} x-4 \cos x \cos ^{2} 6 x+\cos ^{2} 6 x=0, \\ \sin x=\cos y .\end{array}\right.$ | # Solution:
Solve the 1st equation of the system:
$4 \cos ^{2} x-4 \cos x \cos ^{2} 6 x+\cos ^{2} 6 x=0 \Leftrightarrow\left(2 \cos x-\cos ^{2} 6 x\right)^{2}+\cos ^{2} 6 x \sin ^{2} 6 x=0 \Leftrightarrow$ $\left\{\begin{array}{l}2 \cos x-\cos ^{2} 6 x=0, \\ \cos ^{2} 6 x \sin ^{2} 6 x=0\end{array} \Leftrightarrow\le... | (\frac{\pi}{3}+2\pik,\\frac{\pi}{6}+2\pin),(-\frac{\pi}{3}+2\pik,\\frac{5\pi}{6}+2\pin),k,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,363 |
5. Solve the inequality $\frac{x+21-7 \sqrt{x+9}}{x^{2}-8 x}>0$ (10 points)
# | # Solution:
Domain: $x \in[-9+\infty), \quad x \neq 0, \quad x \neq 8$.
Factorize the numerator:
$$
x+21-7 \sqrt{x+9}=x+9-7 \sqrt{x+9}+12=(\sqrt{x+9})^{2}-7 \sqrt{x+9}+12=(\sqrt{x+9}-3)(\sqrt{x+9}-4)
$$
We have:
$\frac{(\sqrt{x+9}-3)(\sqrt{x+9}-4)}{x(x-8)}>0 \Leftrightarrow \frac{(x+9-9)(x+9-16)}{x(x-8)}>0 \Leftri... | x\in[-9;0)\cup(0;7)\cup(8;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,364 |
6. Find the set of values of the function $f(x)=4 \cos \left(\frac{\pi}{3} \sin \left(x^{2}+6 x+10-\sin x\right)\right)$.
# | # Solution:
$f(x)=4 \cos \left(\frac{\pi}{3} \sin \left(x^{2}+6 x+10-\sin x\right)\right)=4 \cos \left(\frac{\pi}{3} \sin \left((x+3)^{2}+1-\sin x\right)\right)$. Since $g(x)=(x+3)^{2}+1-\sin x \in[0 ; \infty)$, and the range of this function contains a numerical ray, then $\varphi(x)=\sin \left((x+3)^{2}+1-\sin x\rig... | [2;4] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,365 |
7. On the side $B C$ of triangle $A B C$, a point $K$ is marked.
It is known that $\angle B+\angle C=\angle A K B, \quad A K=4, \quad B K=9, \quad K C=3$.
Find the area of the circle inscribed in triangle $A B C$. | # Solution:
$\angle B+\angle C=\angle A K B, \quad \angle A K C=\angle A, \quad \angle K A C=\angle B$,
$\triangle A B C \approx \triangle A K C \Rightarrow$
$\frac{A B}{A K}=\frac{A C}{K C}=\frac{B C}{A C} \Rightarrow \frac{A B}{4}=\frac{A C}{3}=\frac{12}{A C} \Rightarrow$
$A C=6, \quad A B=8$.
$ to the parabola $8 y=x^{2}+16$. Determine the angle between the tangents. Find the area of triangle $A B M$, where $A$ and $B$ are the points of tangency. | # Solution:
$8 y=x^{2}+16 ; M(3 ; 0)$.
$y=\frac{x_{0}^{2}}{8}+2+\frac{x_{0}}{4}\left(x-x_{0}\right)$
$y=2+\frac{x_{0}}{4} x-\frac{x_{0}^{2}}{8}$.
$0=2+\frac{x_{0}}{4} \cdot 3-\frac{x_{0}^{2}}{8}, x_{0}^{2}-6 x_{0}-16=0$,
$\left(x_{0}\right)_{1}=-2,\left(y_{0}\right)_{1}=2.5, A(-2 ; 2.5)$.
$\left(x_{0}\right)_{2}=... | \frac{125}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,367 |
9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=4(y-x+a-1), \frac{\sqrt{y}-1}{\sqrt{x}-1}=1$ has at least one solution, and solve it for each $a$.
# | # Solution:
The second equation is equivalent to the system: $x \geq 0, x \neq 1, y=x$. Substituting $y=x$ into the first equation, we get: $(x-a)^{2}=4(a-1)$, or $x^{2}-2 a x+a^{2}-4 a+4=0(*)$, for which $D / 4=a^{2}-a^{2}+4 a-4=4(a-1)$. The number of solutions to the given system of equations depends on the number o... | \in(1;5)\cup(5;+\infty),\2\sqrt{-1};\quad=5,9,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,368 |
10. The base of the pyramid $T A B C D$ is a rectangle $A B C D$. The height of the pyramid coincides with the lateral edge $T A$, and the lateral edge $T C$ is inclined to the base plane at an angle of $45^{\circ}$. The plane passing through the edge $T C$ and parallel to the diagonal of the base $B D$ forms an angle ... | Solution:
Draw $K C \| B D, A K \perp K C, L=A K \cap B D, L S \perp T K$, then $L S=a$. Let $\beta=\angle T K A$. Since $A K=2 L K$, then $A K=2 a / \sin \beta$ and $A T=A K \operatorname... | \frac{8a^2}{\sqrt{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,369 |
2. Find the greatest value of the parameter a for which the equation: $(|x-2|+2 a)^{2}-3(|x-2|+2 a)+4 a(3-4 a)=0$ has three solutions. In the answer, specify the greatest of them. | Solution:
Let $|\mathrm{x}-2|+2 \mathrm{a}=\mathrm{t}$, then
$$
t^{2}-3 t+4 a(3-4 a)=0
$$
$t^{2}-2 \cdot \frac{3}{2} t+\frac{4}{9}-\frac{4}{9}+12 a-16 a^{2}=0$
$\left(\mathrm{t}-\frac{3}{2}\right)^{2}-\left(\frac{9}{4}-12 \mathrm{a}+16 a^{2}\right)=0$;
$\left(t-\frac{3}{2}\right)^{2}-\left(4 a-\frac{3}{2}\right)^{... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,370 |
1. Since $\mathrm{x} \leq \mathrm{P}$ and Р:х, let $2 \mathrm{x} \leq \mathrm{P} => \mathrm{x} \leq \mathrm{P} / 2$, and also $\mathrm{y} \leq \mathrm{P} ; \kappa \leq \mathrm{P} ; \mathrm{e} \leq \mathrm{P} ; =>$
$$
\mathrm{x}+\mathrm{y}+\mathrm{K}+\mathrm{e} \leq 3.5 \mathrm{P} ; => \mathrm{P} \geq \frac{2}{7}(\math... | Answer: 2 l
## №6: Planimetry. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,371 |
# 2. Solution:
According to the condition, the center of the second blot is at a distance of no less than 2 cm from the edge of the sheet, i.e., inside a rectangle of 11 cm by 16 cm. Consider the event A "The blots intersect." For this to happen, the center of the second blot must fall within a circle of radius 4 cm w... | Answer. $\frac{\pi}{11}$. | \frac{\pi}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,373 |
Problem 4. (10 points) In a row, 2018 digits are written consecutively. It is known that in this row, every two-digit number formed by two adjacent digits (in the order they are written) is divisible by 17 or 23. The last digit in this row is 5. What is the first digit in the row? Provide a justified answer.
# | # Solution:
All two-digit numbers divisible by 17 or 23:
$$
\begin{aligned}
& 17,34,51,68,85 \\
& 23,46,69,92
\end{aligned}
$$
The following diagram shows with arrows which digit can follow which in the row:
$$
\begin{aligned}
& 1 \rightarrow 7 \quad 9 \rightarrow 2 \rightarrow 3 \\
& \uparrow \quad \uparrow \\
& 5... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,375 |
Task 6. (15 points) On the plane xOy, indicate all points through which no curve passes, given by the equation
$$
\mathrm{a} x^{2}+(1-6 a) x+2-a-2 y+a y^{2}=0
$$ | # Solution:
Let's represent the equation as linear with respect to the parameter a.
$\mathrm{a}\left(x^{2}-6 x-1+y^{2}\right)=-x-2+2 y$. If this equation does not have solutions for any a, we will find the points (x; y) through which none of the curves pass. $\left\{\begin{array}{c}x^{2}-6 x-1+y^{2}=0 \\ -x-2+2 y \ne... | y^{2}+(x-3)^{2}withoutthepoints(0;1)(4;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,377 |
Problem 7. (15 points) Let $\mathbf{S}_{\mathbf{n}}$ be the sum of the first p terms of the arithmetic progression $\left\{\mathbf{a}_{\mathbf{n}}\right\}$. It is known that $\mathbf{S}_{\mathbf{n}+1}=2 \mathbf{n}^{2}-21 \mathbf{n}-23$.
a) State the formula for the p-th term of this progression.
b) Find the smallest ... | # Solution:
a) $a_{n}=4 n-27 . S_{n}=S_{(n-1)+1}=2(n-1)^{2}-21(n-1)-23=2 n^{2}-25 n$;
$S_{n-1}=2(n-1)^{2}-25(n-1)=2 n^{2}-29 n+27 \Rightarrow a_{n}=S_{n}-S_{n-1}=4 n-27$.
b) -12. Since $a_{n}=4 n-270$ for $n \geq 7, S_{n}$ decreases from the value $a_{1}=-23$ to $S_{6}$, then increases.
$S_{n}=2 n^{2}-25 n0$ for $n... | )a_{n}=4n-27;b)-12;)25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,378 |
# 3. Solution:
Domain of Definition (ODZ): $\left\{\begin{array}{c}x^{2}-4 x+3 \geq 0 \\ 3-x \geq 0 \\ x^{2}-6 x+9 \geq 0 \\ x^{2}-8 x+16 \geq 0 \\ x^{2}+2 \cdot \sqrt{x^{2}-8 x+16} \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}(x-1) \cdot(x-3) \geq 0 \\ x \leq 3 \\ (x-3)^{2} \geq 0 \\ (x-4)^{2} \geq 0 \\ x^... | Answer: $(-\infty ;-6] \cup\{3\}$. | (-\infty;-6]\cup{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,382 |
# 5. Solution:
Let $x$ units of distance/hour be the speed of the bus, $y$ units of distance/hour be the speed of the tractor, and $S$ be the length of the path AB. Then the speed of the truck is $-2y$ units of distance/hour. We can set up a system of equations and inequalities:
$$
\left\{\begin{array}{c}
\frac{s}{x}... | Answer: 17 hours 45 minutes. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,386 |
Problem 7. (15 points) Find all values of the parameter $a$ for which the equation $(a+2) x^{2}+(|a+3|-|a+11|) x+a=4$ has two distinct positive roots. | Solution: $(a+2) x^{2}+(|a+3|-|a+11|) x+a-4=0$.
When $a=-2$, the equation is linear and takes the form: $-8 x-6=0$, which does not satisfy the condition of the problem.
When $a \neq-2$, for the condition of the problem to be satisfied, it is necessary and sufficient that the following system is satisfied:
$$
\left\{... | \in(4;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,387 |
Problem 2. (Option 1). Given an acute triangle $\mathrm{ABC}(\mathrm{AB}=\mathrm{BC})$ and $\mathrm{BC}=12$. $A N \perp B C$. On the side $\mathrm{BC}$, a point $M$ (M lies between B and $\mathrm{N}$) is marked such that $\mathrm{AN}=\mathrm{MN}$ and $\angle \mathrm{BAM}=\angle \mathrm{NAC}$. Find $\mathrm{BN}$. | Solution: 1). Let $\angle \mathrm{BAM}=\angle \mathrm{NAC}=\alpha, \angle \mathrm{MAN}=\angle \mathrm{AMN}=\beta=45^{\circ}$, (since triangle MAN is isosceles) $\prec=\succ \angle \mathrm{MAC}=\alpha+\beta$ and $\angle \mathrm{MCA}=2 \alpha+\beta=\succ$ $(\square \mathrm{AMC}) 2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=... | 6\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,389 |
Problem 2. (Option 2). Given an isosceles triangle $ABC (AB=BC)$ on the lateral side $BC$, points $M$ and $N$ are marked (M lies between B and $N$) such that $AN=MN$ and $\angle BAM = \angle NAC$. $MF$ is the distance from point M to the line $AC$. Find $\angle AMF$. | Solution: Let $\angle$ BAM
$=\angle \mathrm{NAC}=\alpha, \angle \mathrm{MAN}=\angle \mathrm{AMN}=\beta \prec=\angle \mathrm{MAC}=\alpha+\beta$ and $\angle \mathrm{MCA}=2 \alpha+\beta=\succ(\square \mathrm{AMC})$
$2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=\succ \alpha+\beta=60^{\circ} \Rightarrow \succ \mathrm{MAF}=60... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,390 |
Task 3. (Option 1)
A cleaning robot moves at a constant speed and is programmed to turn 90 degrees every 15 seconds, and moves in a straight line between turns. Can we expect the robot to return to the starting point after 6 minutes? | Solution: The number of horizontal segments along which the robot moves for 15 seconds, moving away from the initial position, is equal to the number of horizontal segments along which it moves, approaching the initial position (this is easy to see if you draw a movement diagram that satisfies the conditions). Therefor... | therobot'returntotheinitialpositioncanbeexpectedatanywholeof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,391 |
Task 3. (Option 2)
Thirty clever students from 6a, 7a, 8a, 9a, and 10a grades were tasked with creating forty problems for the olympiad. Any two classmates came up with the same number of problems, while any two students from different grades came up with a different number of problems. How many people came up with ju... | Solution: 26 classmates solved 1 problem, the 27th person solved 2, the 28th solved 3, the 29th solved 4, and the 30th solved 5. This solution is immediately apparent. Let's prove that it cannot be otherwise.
Let $x$ be the number of people who solved one problem, $y$ be the number who solved two, $z$ be the number wh... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,392 |
Problem 4. (Option 1)
Let $\min (a ; c)$ denote the smallest of the numbers a and c. Construct the graph of the function $y=\min \left(x+2 ; x^{2}-6 x+8\right), \quad$ and use it to solve the inequality $\min \left(x+2 ; x^{2}-6 x+8\right) \geq 0$. | Solution: 1) Find the points of intersection of the graphs $y=x+2, y=x^{2}-6 x+8$, we get $\mathrm{x}=1$ and $\mathrm{x}=6$: for $x \in(-\infty ; 1) \cup(6 ;+\infty)$ the smallest $y=x+2$; for $x \in[1 ; 6]_{\text {smallest of }}$ the numbers a and c $\left.y=x^{2}-6 x+8 ; 2\right)$ Find the points of intersection of t... | x\in[-2;2]\cup[4;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,393 |
Problem 4. (Option 2)
Let $\max (a ; c)$ denote the smaller of the numbers a and c. Construct the graph of the function $y=\max \left(x-11 ; x^{2}-8 x+7\right)$, and use it to solve the inequality $\max \left(x-11 ; x^{2}-8 x+7\right) \prec 0$.
# | # Solution:
1) Find the points of intersection of the graphs \( y = x - 11 \) and \( y = x^2 - 8x + 7 \), we get \( x = 3 \) and \( x = 6 \): for \( x \in (-\infty; 3) \cup (6; +\infty) \), the largest \( y = x^2 - 8x + 7 \); for \( x \in [3; 6] \), the largest \( y = x - 11 \);
2) Find the points of intersection of t... | x\in(1;7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,394 |
# Problem 5. (Option 1).
Find the form of all quadratic trinomials $f(x)=a x^{2}+b x+c$, where $a, b, c$ are given constants, $\quad a \neq 0$, such that for all values of $x$ the condition $f(3.8 x-1)=f(-3.8 x)$ is satisfied. | # Solution:
$y(0.01 x+1)=y(-0.01 x) \prec=\succ y(1)=y(0) \prec=\succ a+b+k=k \prec=\succ a+b=0 \prec=\succ$ $b=-a \prec=\succ y(x)=a x^{2}-a x+k$.
Verification: $y(0.01 x+1)=y(-0.01 x) \prec=\succ$
$a(0.01 x+1)^{2}-a(0.01 x+1)+k=a(-0.01 x)^{2}-a(-0.01 x)+k$
$\prec \Rightarrow 0.0001 a x^{2}+0.02 a x+a-0.01 a x-a=0... | y(x)=^{2}-+k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,395 |
Problem 5. (Option 2).
Find the form of all quadratic trinomials $f(x)=a x^{2}+b x+c$, where $a, b, c$ are given constants, $\quad a \neq 0$, such that for all values of $x$ the condition $f(3.8 x-1)=f(-3.8 x)$ is satisfied. | # Solution:
$f(3.8 x-1)=f(-3.8 x) \prec=\succ f(-1)=f(0) \prec=\succ a-b+c=c \prec=\succ a-b=0 \prec=\succ b=a$ $\prec=\succ(x)=a x^{2}+a x+c$.
Verification: $\quad f(3.8 x-1)=f(-3.8 x) \prec=\succ$
$a(3.8 x-1)^{2}+a(3.8 x-1)+c=a(-3.8 x)^{2}+a(-3.8 x)+c$
$\prec=14.44 a x^{2}-7.6 a x+a+3.8 a x-a=14.44 a x^{2}-3.8 a ... | f(x)=^{2}++ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,396 |
Problem 6. (Option 2).
Prove that the expression $x^{5}-4 x^{4} y-5 y^{2} x^{3}+20 y^{3} x^{2}+4 y^{4} x-16 y^{5}$ is not equal to 77 for any integer values of $x$ and $y$ | # Solution:
Factorize
$x^{4}(x-4 y)-5 x^{2} y^{2}(x-4 y)+4 y^{4}(x-4 y) \prec=\succ(x-4 y)\left(x^{4}-5 x^{2} y^{2}+4 y^{4}\right)$ $\prec=\succ(x-4 y)(x+2 y)(x-2 y)(x-y)(x+y)$. It is necessary to check that the factors are pairwise distinct. They can coincide when $y=0$. However, $x^{5}$ is not equal to 77 for any i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 10,398 |
1. Two cyclists set off simultaneously from point $A$ to point $B$. When the first cyclist had covered half the distance, the second cyclist had 24 km left to travel, and when the second cyclist had covered half the distance, the first cyclist had 15 km left to travel. Find the distance between points $A$ and $B$. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}}$ and $\frac{s-15}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24$; $s^{2}-52 s+480=0 ;... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,399 |
2. Solve the inequality $\sqrt{\frac{x-24}{x}}-\sqrt{\frac{x}{x-24}}<\frac{24}{5}$. | Solution:
$\sqrt{\frac{x-24}{x}}-\sqrt{\frac{x}{x-24}}<0 ; \frac{1}{y}-y<0 ;$
$y_{1,2}=\frac{-12 \pm 13}{5}, y_{1}=\frac{1}{5}, y_{2}=-5$. Thus $y>\frac{1}{5}, \frac{x}{x-24}>\frac{1}{25} \Leftrightarrow \frac{x+1}{x-24}>0 \Leftrightarrow\left[\begin{array}{l}x<-1 \\ x>24\end{array}\right.$
Answer: $x \in(-\infty ;-... | x\in(-\infty;-1)\cup(24;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 10,400 |
3. What is the greatest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=327$ and the sum $S_{57}=57$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$\left\{\begin{array}{l}\frac{a+a+2 d}{2} \cdot 3=327, \\ \frac{a+a+56 d}{2} \cdot 57=57\end{array} \Leftrightarrow\left\{\begin{array}{l}a+d=109, \\ a+28 d=1\end{array} \Rightarrow 27 d=-108 ; d=-4, a=113\right.\ri... | 1653 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,401 |
4. Solve the equation $\sqrt{1+\operatorname{tg} x}=\sin x+\cos x$. Find all its roots that satisfy the condition $|2 x-5|<2$.
# | # Solution:
Given the condition $\sin x+\cos x \geq 0$, both sides of the equation $\sqrt{1+\operatorname{tg} x}=\sin x+\cos x$ can be squared. Since $\sin x+\cos x=\sqrt{2} \sin (x+\pi / 4)$, we have $-\pi / 4+2 \pi n \leq x \leq 3 \pi / 4+2 \pi n, n \in Z$. Under these constraints, the equation is equivalent to:
$1+... | \frac{3\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,402 |
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