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values | question_type stringclasses 4
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77. On a circular highway 300 km long, cars of the same make are placed. The total amount of gasoline in them is enough for 301 km. The driver chooses one of the cars at his discretion. Upon reaching another car, he refuels, taking all of its gasoline. Prove that the driver can drive around the highway without resortin... | 77. Next to each car on the highway, we will write a number that is equal to the difference between the number of meters this car can travel without refueling and the distance to the next (clockwise) car. The sum of all these numbers is positive. According to problem 76, there will be such a number that it itself, the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,105 |
78. Prove that for any natural $n$
$$
1+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}} \geqslant \sqrt{n}
$$ | 78. Obviously, if $n \geqslant k$, then $\sqrt{n} \geqslant \sqrt{k}$, and therefore $\frac{\sqrt{n}}{\sqrt{k}} \geqslant 1$. Hence, $\quad \frac{\sqrt{n}}{\sqrt{1}}+\frac{\sqrt{n}}{\sqrt{2}}+\ldots+\frac{\sqrt{n}}{\sqrt{n-1}}+\frac{\sqrt{n}}{\sqrt{n}} \geqslant n$ and dividing both sides by $\sqrt{n}$, we obtain the d... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 30,106 |
79. For which $n$ do there exist positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, satisfying the equations
$$
x_{1}+x_{2}+\ldots+x_{n}=3, \quad \frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} \geq 3 ?
$$ | 79. Use the fact that $x+\frac{1}{x} \geqslant 2$ for $x>0$, with equality holding only when $x=1$. Add the equations
$$
\begin{array}{r}
x_{1}+x_{2}+\ldots+x_{n}=3 \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}=3
\end{array}
$$
and deduce that $n \leqslant 3$. It is then easy to show that for $n=3$
$$
x_... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 30,107 | |
80. The glass has the shape of a square with an area of 1. On both sides of this glass, maps are drawn, with five countries on each map. The countries on one side of the glass are colored with five different colors. It is required to color the countries on the opposite side of the glass with the same five colors so tha... | 80. Always. To prove this, let's number the countries on the colored side of the glass with the numbers $1,2,3,4,5$, and on the uncolored side with the numbers $6,7,8,9,10$. Let $s_{m}^{n}$ denote the area of the glass segments where the territory of the $m$-th country is depicted on the colored side ($\leqslant m \leq... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,108 |
82a*. Five complex numbers $c_{1}, c_{2}, c_{3}, c_{4}$, and $c_{5}$ are related by the equation
$$
\frac{1}{c_{1}}+\frac{1}{c_{2}}+\frac{1}{c_{3}}+\frac{1}{c_{4}}+\frac{1}{c_{5}}=0
$$
Prove that if the points in the plane representing the complex numbers $c_{1}, c_{2}, c_{3}, c_{4}$, and $c_{5}$ lie at the vertices ... | 82a. Vectors representing complex numbers $z$ and $1 / z$ form equal and symmetrically located angles with the $O x$ axis. (To verify this, it is sufficient to write $z$ and $1 / z$ in trigonometric form.) Further, if the vectors representing complex numbers lie on one side of a line passing through 0, then their sum i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,110 |
84. A traveler wishes to visit five cities $A, B, C, D$ and $E$ in sequence. All segments $A B, B C, C D$ and $D E$ have different lengths and each is served by four different types of aircraft $K, L, M$ and $N$ with varying speeds. Prove that if the traveler wants to fly on each of these aircraft and spend the least a... | 84. For brevity, let's denote the lengths of segments $A B, B C$, $C D$ and $D E$ as $a, b, c, d$. Without loss of generality, we can assume that $a>b>c>d$. Suppose the traveler used an airplane on segment $A B$ with speed $v_{1}$, on $B C$ with speed $v_{2}$, on $C D$ with speed $v_{3}$, and finally on $D E$ with spee... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,111 |
85. Bacteria have the following development rule: each one lives for 1 hour and every half hour it produces one new one (a total of two during its life). What will be the offspring of one bacterium 6 hours after its birth? | 85. Let's call bacteria "young" if their age does not exceed half an hour, and "old" if they have lived for more than half an hour. Let's take half an hour as the unit of time. Denote the number of bacteria at moment $n$ by $u_{n}$. At moment $n-1$, $u_{n-1}$ bacteria are born, so by moment $n$ there are $u_{n-1}$ youn... | 377 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,112 |
86. At some point on a straight line, there is a particle. In the first second, it splits in half, and the halves move in opposite directions to a distance I from the previous position. In the next second, the resulting particles again split in half, and the halves move in opposite directions to a distance 1 from their... | 86. The number and position of particles at moments $t=0$, $1,2,3$ and 4 sec are shown in Fig. 44. We will prove that after
Fig. 44.
$2^{n}-1$ sec there will be $2^{n}$ particles, arranged i... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,113 |
87. Prove that there do not exist integers $k, l, m$ and $n$ satisfying the following equations:
$$
\begin{aligned}
& \text { klmn }-k=1966, \\
& \text { klmn }-l=966, \\
& k l m n-m=66, \\
& \text { klmn }-n=6 .
\end{aligned}
$$ | 87. Let's rewrite these equalities as follows:
$$
\begin{aligned}
& -k=1966-k l m n \\
& -l=966-k l m n \\
& -m=66-k l m n \\
& -n=6-k l m n
\end{aligned}
$$
Multiplying them and denoting $\mathrm{klmn}=x$, we get
$$
x=(1966-x)(966-x)(66-x)(6-x)
$$
If the original system had a solution in integers, then the equatio... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,114 |
88*. Find all real solutions of the system
$$
x^{4}-17=y^{4}-7=z^{4}+19=u^{4}+5=x y z u
$$
89*. Solve the equation
$$
5^{x}+12^{x}=13^{x}
$$ | 88. Suppose the required $x, y, z, u$ exist. Adding the equations
$$
\begin{aligned}
x^{4}-17=x y z u, & z^{4}+19=x y z u \\
y^{4}-7=x y z u, & u^{4}+5=x y z u
\end{aligned}
$$
we get
$$
x^{4}+y^{4}+z^{4}+u^{4}=4 x y z u
$$
But
$$
\begin{gathered}
x^{4}+y^{4}+z^{4}+u^{4}-4 x y z u=x^{4}+y^{4}-2 x^{2} y^{2}+z^{4}+ ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,115 |
90*. Solve the equation
$$
x=1-1967\left(1-1967 x^{2}\right)^{2}
$$ | 90. Let
$$
z=1-1967 x^{2}
$$
Then the original equation reduces to the system
$$
\left\{\begin{array}{l}
x=1-1967 z^{2} \\
z=1-1967 x^{2}
\end{array}\right.
$$
Subtracting the second equation from the first, we get
$$
x-z=1967\left(x^{2}-z^{2}\right)
$$
From this, either $z=x$, or $z=\frac{1}{1967}-x$. In the fir... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,116 | |
91*. Each real number $x$ is associated with a real number $A(x)$. It is known that for any $x$ and $y$:
a)
\[
\begin{aligned}
A(x+y) & =A(x)+A(y) \\
A(x y) & =A(x) \cdot A(y) .
\end{aligned}
\]
Prove that either $A(x)=x$, or $A(x)=0$. | 91. Let $A(0)=b$. From the first equality, we get
$$
A(0+0)=A(0)+A(0)=2 b
$$
from which $b=A(0)=0$. Let, further, $A(1)=a$. Then
$$
a=A(1)=A(1 \cdot 1)=A(1) \cdot A(1)=a^{2}
$$
from which either $a=0$ or $a=1$. If $a=A(1)=0$, then for any $x$, $A(x)=A(1 \cdot x)=A(1) \cdot A(x)=0$. It remains to consider the second... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,117 |
92. All natural numbers are arbitrarily divided into two groups. Prove that in at least one of them, there will be three numbers, one of which is the arithmetic mean of the other two. | 92. 1-st m e t h o d. Let's try to find all partitions of numbers from 1 to 9 into two groups such that there are no three numbers in one group where one of them is the arithmetic mean of the other two. We will represent the numbers in the group containing 1 with white circles and the numbers in the second group with b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,118 |
94. In any infinite increasing arithmetic progression composed of natural numbers, there will be a number that starts with the digits 10. Prove it. | 94. Let the first term of the progression be $a_{1}=a$, and the common difference be $d$. Choose a natural number $n$ such that $a+10d < 10^{n}$. Let $a_{k}$ and $a_{k+1}$ be two consecutive terms of the arithmetic progression for which the inequalities $a_{k}<10^{n}$, $a_{k+1} \geqslant 10^{n}$ hold. Clearly,
$$
10^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,120 |
95. Prove that there exists a perfect square starting with any combination of digits. | 95. Let $\sqrt{\overline{a b c \ldots f}}$ be the set of digits given in the problem. Consider two numbers:
$$
N_{\mathbf{1}}=\underbrace{\overline{a b c \ldots f} \underbrace{00 \ldots 0}_{3 k}}_{k} \text { and } \quad N_{2}=\underbrace{\overline{a b c \ldots f} f \underbrace{99 \ldots 9}_{3^{k}}}_{k}
$$
Let $n^{2}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,121 |
96**. Prove that among the numbers $0,1,2, \ldots, 3^{k}-1$ one can find $2^{k}$ different numbers such that none of them is the arithmetic mean of two others. | 96. To solve this, we will need binary and ternary number systems. Any natural number \( n \) can be uniquely represented in the form
\[
n=2^{k}+c_{1} 2^{k-1}+c_{2} 2^{k-2}+\ldots+c_{k}
\]
where \( c_{i} \) take values 0 or 1. The numbers \( 1, c_{1}, c_{2}, \ldots \) are called the digits of the binary expansion. We... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,122 |
97**. Prove that among 79 consecutive natural numbers, there will always be a number whose sum of digits is divisible by 13. Provide an example of 78 consecutive numbers whose sums of digits are not divisible by 13.
$98^{* *}$. Mark on the number line in red all points corresponding to numbers of the form $81 x+100 y$... | 97. First, note that the sum of the digits of the number $n+1$ increases by one compared to the sum of the digits of the number $n$ if the last digit of the number $n$ is not 9, and decreases by $9s-1$ if the number $n$ ends with $s$ nines. Consider two cases.
A. Among the numbers $n, n+1, n+2, \ldots, n+78$, there is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,123 |
99. There is a certain number of cards. Each card has a natural number written on it. For each $n \leqslant 1968$, there are exactly $n$ cards on which the numbers are divisors of $n$. Prove that the number $2^{10}=$ $=1024$ is written on at least one card. | 99. We will prove by induction the following statement: the number $2^{n}$ (where $n \leqslant 10)$ is written on exactly $2^{n-1}$ cards. For $n=1$, the statement is trivial. Consider the number $2^{n}$. Its divisors are written on $2^{n}$ cards. But the divisors of the number $2^{n}$ are it itself and the divisors of... | 1024 | Number Theory | proof | Yes | Yes | olympiads | false | 30,124 |
101**. Let $n$ be an arbitrary natural number, and $p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots p_{k}^{n_{k}}$ be its prime factorization (where $p_{1}, p_{2}, \ldots, p_{k}$ are prime numbers, and $p_{1}<p_{2}<\ldots<p_{k})^{*}$. Prove that the number $\varphi(n)$ of irreducible proper fractions with denominator $n$ is given b... | 101. The validity of the formula to be proved follows from the following observations:
a) If $n=p^{k}$, then $\varphi(n)=(p-1) p^{k-1}$. This statement is easily proved by induction (see problem 99, where the case $p=2$ is discussed).
b) If the numbers $n$ and $m$ are coprime, then $\varphi(n m)=$ $=\varphi(n) \varph... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,126 |
102. At each vertex of a triangle, a non-negative number is written, and the sum of all these numbers is 3000. Each number is replaced by the arithmetic mean of the numbers at the adjacent vertices, and this operation is repeated 10 times. Prove that after this, each of the numbers will be less than 1002.
$103^{**}$. ... | 102. Let the numbers at the vertices of the triangle after the $n$-th operation be denoted by $a_{n}, b_{n}$, and $c_{n}$. According to the problem, $a_{0}+b_{0}+c_{0}=3000$ and since the numbers $a_{0}, b_{0}, c_{0}$ are non-negative, each of them does not exceed 3000. The sum
$a_{n}+b_{n}+c_{n}$ does not change from ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,127 |
104. On the blackboard, the numbers $1,2,3, \ldots$ 1966 are written. It is allowed to erase any two numbers and write their difference instead. Obviously, after 1965 repetitions of this operation, one number will remain on the board. Prove that this number is odd. | 104. If two numbers have the same parity, their difference is an even number; if one number is even and the other is odd, their difference is an odd number. Therefore, after each operation, the number of odd numbers either remains the same or decreases by two.
At the beginning, there are 983 odd numbers on the board. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,128 |
105. On the board, several plus and minus signs are written. It is allowed to erase any two signs, writing a plus instead of two identical signs, and a minus instead of different ones. Prove that the last sign left on the board does not depend on the order in which the signs are erased.[^1] | 105. Let's agree to write the number +1 instead of the sign «+», and the number -1 instead of the sign "—". In this case, our operation will obviously not change the product of all the numbers written on the board, so the last remaining number, regardless of the order in which we erase the numbers, will be equal to the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,129 |
106. Several circles, squares, and triangles are drawn on the board. It is allowed to erase any two figures and draw a third one in their place according to the following rules: instead of a pair of circles - one circle, instead of a pair of squares - a triangle, instead of a pair of triangles - a square, instead of a ... | 106. Let's denote the circle, square, and triangle by the letters $O, K, T$. The operation of replacing two figures with a third will be denoted by circles. The condition of the problem can now be written in the following form:
$$
\left.\begin{array}{rl}
O \circ O=O, & O \circ K=K \circ O=K \\
K \circ K=T, & O \circ T... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,130 |
108. Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an irrational number. | 108. Let's conduct a proof by contradiction. Suppose the number $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is rational, i.e.,
$$
\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{p}{q}
$$
( $p$ and $q$ - integers). Then
$$
\sqrt{2}+\sqrt{5}=\frac{p}{q}-\sqrt{3}
$$
Squaring both sides of this equation, we get:
$$
5+2 \sqrt{10}+2=\frac{p^{2}}{q^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,132 |
109**. Prove that among the first ten million digits of the decimal expansion of $\sqrt{2}$, no digit will repeat 5000001 times in a row. | 109. Suppose that in the decimal representation of the number $\sqrt{2}$, starting from the $n$-th place ($n \leqslant 4999999$), the same digit $a$ repeats 5000001 times consecutively. Then
$$
\sqrt{2}=\frac{N_{1}}{10^{n-1}}+\frac{a}{10^{n}}+\cdots+\frac{a}{10^{n+5000000}}+\frac{r}{10^{n+5000000}}
$$
where $N_{1}$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,133 |
Example 1. Solve the linear system
$$
\left\{\begin{aligned}
x_{1}+2 x_{2}-3 x_{3}-x_{4} & =10 \\
-2 x_{1}-3 x_{2}+7 x_{3} & =-23 \\
2 x_{1}+6 x_{2}-5 x_{3}-5 x_{4} & =18 \\
-x_{1}+3 x_{3}-4 x_{4} & =-11
\end{aligned}\right.
$$ | Solution. We have $m=4, n=4$.
The first block of the Gauss table for this system has the following form («free terms» means the free terms of the system of equations, the vertical line corresponds to the equality signs):
| $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ | free terms | | |
| :---: | :---: | :---: | :---: | :-... | (-2,2,-3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,136 |
Example 2. Solve the linear system
$$
\left\{\begin{aligned}
-x_{1}-2 x_{2}-6 x_{3}+3 x_{4} & =-1 \\
2 x_{1}+5 x_{2}+14 x_{3}-7 x_{4} & =3 \\
3 x_{1}+7 x_{2}+20 x_{3}-10 x_{4} & =4 \\
-x_{2}-2 x_{3}+x_{4} & =-1
\end{aligned}\right.
$$ | Solution. Each time, we will take the simplest coefficient as the leading one, i.e., either 1 or -1. Let's emphasize that the goal of the transformations is to obtain zeros in the leading column. How to get zero coefficients in the unit column is clear from the solution of example 1. For this, the leading row needs to ... | X_{0}=(-1-2\alpha+\beta,1-2\alpha+\beta,\alpha,\beta),X_{ch1}=(-1,1,1,2),X_{ch2}=(2,4,-1,1),X_{6}=(-1,1,0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,137 |
Example 3. Solve the system of equations
$$
\left\{\begin{aligned}
& x_{1}+2 x_{2}+3 x_{3}+x_{4}=1 \\
& 3 x_{1}+13 x_{2}+13 x_{3}+5 x_{4}=3 \\
& 3 x_{1}+7 x_{2}+7 x_{3}+2 x_{4}=12 \\
& x_{1}+5 x_{2}+3 x_{3}+x_{4}=7 \\
& 4 x_{1}+5 x_{2}+6 x_{3}+x_{4}=19
\end{aligned}\right.
$$ | Solution. Instead of Gauss's table, we will use another, more compact interpretation of its blocks. The vertical line in the blocks corresponds to the equality signs in the equations of the system. The sign (read as "tilde") between two adjacent blocks means that the systems corresponding to these blocks are equivalent... | X_{0}=(4-\alpha,2,\alpha,-7-2\alpha),\quadX_{6}=(4,2,0,-7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,138 |
Example 4. Solve the system of equations
$$
\left\{\begin{aligned}
x_{1}+x_{3}+2 x_{4}+3 x_{5}-4 x_{6}= & 20 \\
2 x_{1}+x_{2}-3 x_{3}+x_{5}+2 x_{6}= & -13 \\
5 x_{1}-x_{2}+x_{3}+2 x_{4}+6 x_{5} & =20 \\
2 x_{1}-2 x_{2}+3 x_{3}+2 x_{5}+2 x_{6} & =13
\end{aligned}\right.
$$ | Solution. In the previous examples of Jordan-Gauss transformations, the operations reduced to actions on the equations of the system, or the rows of the table, because all the leading coefficients were equal to 1. If the leading coefficients are different from 1, then operations on the rows can cause difficulties, and ... | \begin{aligned}&x_{1}=t_{1},\\&x_{2}=-\frac{26}{3}+2t_{2}+\frac{1}{3}t_{3},\\&x_{3}=t_{2},\\&x_{4}=\frac{17}{3}-\frac{5}{2}t_{1}+\frac{1}{2}t_{2}-\frac{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,139 |
Example 1. Solve the system of equations
$$
\left\{\begin{array}{l}
2 x_{1}-x_{2}=3 \\
3 x_{1}+4 x_{2}=7
\end{array}\right.
$$ | Solution.
$$
\begin{gathered}
\Delta=\left|\begin{array}{rr}
2 & -1 \\
3 & 4
\end{array}\right|=2 \cdot 4-3 \cdot(-1)=8+3=11 \\
\Delta_{1}=\left|\begin{array}{rr}
3 & -1 \\
7 & 4
\end{array}\right|=12-(-7)=19, \quad \Delta_{2}=\left|\begin{array}{rr}
2 & -3 \\
3 & 7
\end{array}\right|=14-9=5
\end{gathered}
$$
By Cram... | (\frac{19}{11},\frac{5}{11}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,141 |
Example 2. Solve the system
$$
\left\{\begin{aligned}
3 x_{1}-2 x_{2}+x_{3} & =-10 \\
2 x_{1}+3 x_{2}-4 x_{3} & =16 \\
x_{1}-4 x_{2}+3 x_{3} & =-18
\end{aligned}\right.
$$ | Solution. We have:
$$
\begin{gathered}
\Delta=\left|\begin{array}{rrr}
3 & -2 & 1 \\
2 & 3 & -4 \\
1 & -4 & 3
\end{array}\right|=3 \cdot\left|\begin{array}{rr}
3 & -4 \\
-4 & 3
\end{array}\right|+2 \cdot\left|\begin{array}{rr}
2 & -4 \\
1 & 3
\end{array}\right|+\left|\begin{array}{rr}
2 & 3 \\
1 & -4
\end{array}\right... | (-1,2,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,142 |
Example 3. Solve the system
$$
\left\{\begin{aligned}
x_{1}-2 x_{2}+3 x_{3} & =5 \\
2 x_{1}+3 x_{2}-x_{3} & =7 \\
3 x_{1}+x_{2}+2 x_{3} & =10
\end{aligned}\right.
$$ | Solution. The calculation of the following determinants is based on properties 2) and 5) from point $6^{\circ}$. We have
$$
\Delta=\left|\begin{array}{rrr}
1 & -2 & 3 \\
2 & 3 & -1 \\
3 & 1 & 2
\end{array}\right| \quad \downarrow=\left|\begin{array}{rrr}
1 & -2 & 3 \\
3 & 1 & 2 \\
3 & 1 & 2
\end{array}\right|=0
$$
Th... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,143 |
Example 4. Solve the system
$$
\left\{\begin{aligned}
x_{1}-2 x_{2}+3 x_{3} & =5 \\
2 x_{1}+3 x_{2}-x_{3} & =7 \\
3 x_{1}+x_{2}+2 x_{3} & =12
\end{aligned}\right.
$$ | Solution. We have (we suggest verifying this independently):
$$
\begin{aligned}
& \Delta=\left|\begin{array}{rrr}
1 & -2 & 3 \\
2 & 3 & -1 \\
3 & 1 & 2
\end{array}\right|=0, \quad \Delta_{1}=\left|\begin{array}{rrr}
5 & -2 & 3 \\
7 & 3 & -1 \\
12 & 1 & 2
\end{array}\right|=0 \\
& \Delta_{2}=\left|\begin{array}{rrr}
1 ... | X_{0}=(\frac{29}{7}-\alpha,-\frac{3}{7}+\alpha,\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,144 |
Example 5. Solve the system
$$
\left\{\begin{aligned}
2 x_{1}+2 x_{2}-x_{3}+x_{4}+4 x_{6} & =0 \\
x_{1}+2 x_{2}+2 x_{3}+3 x_{5}+x_{6} & =-2 \\
x_{1}-2 x_{2}+x_{4}+2 x_{5} & =0
\end{aligned}\right.
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,145 | |
Example 1 (to $6^{\circ}$ ). Solve the system $\left\{\begin{array}{l}3 x_{1}-5 x_{2}=2, \\ 2 x_{1}+4 x_{2}=5 .\end{array}\right.$ | Solution. We have
$$
A=\left(\begin{array}{rr}
3 & -5 \\
2 & 4
\end{array}\right), \quad \Delta=\left|\begin{array}{rr}
3 & -5 \\
2 & 4
\end{array}\right|=22,
$$
$A_{11}=4, A_{12}=-2, A_{21}=+5, A_{22}=3$
$A^{-1}=\frac{1}{22}\left(\begin{array}{rr}4 & 5 \\ -2 & 3\end{array}\right)=\left(\begin{array}{rr}2 / 11 & 5 /... | (3/2,1/2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,146 |
Example 2 (to item $6^{\circ}$). Solve the system
$$
\left\{\begin{aligned}
3 x_{1}-x_{2}+3 x_{3} & =5 \\
2 x_{1}-x_{2}+4 x_{3} & =5 \\
x_{1}+2 x_{2}-3 x_{3} & =0
\end{aligned}\right.
$$ | Solution.
$$
A=\left(\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right), \quad \operatorname{det} A=\left|\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right|=-10
$$
Therefore, $A$ is a non-singular matrix, so it has an inverse matrix $A^{-1}$.
We will compute 9 alge... | {\begin{pmatrix}x_{1}=1\\x_{2}=1\\x_{3}=1\end{pmatrix}.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,147 |
Example 3 (to $7^{\circ}$ ). Find $A^{-1}$, if
$$
A=\left(\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right)
$$ | Solution. In the left part of the first block of the Gaussian table, we enter the elements of matrix $A$. In the right part of the block, we write the identity matrix of the third order. The transition from one block to the next is carried out using the Jordan-Gauss formulas. The leading coefficients are outlined. The ... | A^{-1}=(\begin{pmatrix}1/2&-3/10&1/10\\-1&6/5&3/5\\-1/2&7/10&1/10\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,148 |
Example 1 (to item $1^{\circ}$). Given the matrix
$$
A=\left(\begin{array}{rrrrr}
1 & 2 & 3 & 5 & 8 \\
0 & 1 & 4 & 6 & 9 \\
0 & 0 & 1 & 7 & 10
\end{array}\right)
$$
Determine its rank. | Solution. We have
$$
M_{1}=|1| \neq 0, \quad M_{2}=\left|\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right| \neq 0, \quad M_{3}=\left|\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 4 \\
0 & 0 & 1
\end{array}\right| \neq 0
$$
Minors of higher orders cannot be formed.
Answer: $\operatorname{rank} A=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,149 |
Example 2 (to item $2^{\circ}$). Find the rank of the matrix
$$
A=\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
7 & -1 & 2 & 1 & -12 \\
11 & -1 & 3 & 0 & -16 \\
10 & -2 & 3 & 3 & -20
\end{array}\right)
$$ | Solution. After subtracting the first row from all the others (from the last one with a factor of 2), we obtain the equivalent matrix
$$
A \sim\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
4 & 0 & 1 & -1 & -4 \\
8 & 0 & 2 & -2 & -8 \\
4 & 0 & 1 & -1 & -4
\end{array}\right) \sim\left(\begin{array}{rrrrr}
3 & -1 & 1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,150 |
Example 3 (to item $4^{\circ}$). Determine whether the system is solvable:
$$
\left\{\begin{aligned}
x_{1}+2 x_{2}-x_{3}+3 x_{4}-x_{5} & =0 \\
2 x_{1}-x_{2}+3 x_{3}+x_{4}-x_{5} & =-1 \\
x_{1}-x_{2}+x_{3}+2 x_{4} & =2 \\
4 x_{1}+3 x_{3}+6 x_{4}-2 x_{5} & =5
\end{aligned}\right.
$$ | Solution. We will write the augmented matrix and obtain as many unit columns as possible in it. Each time, we will circle the leading coefficient:
$$
\begin{aligned}
& \widehat{A}=\left(\begin{array}{rrrrr|r}
1 & 2 & -1 & 3 & -1 & 0 \\
2 & -1 & 3 & 1 & -1 & -1 \\
1 & -1 & 1 & 2 & 0 & 2 \\
4 & 0 & 3 & 6 & -2 & 5
\end{a... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,151 |
Example 4 (to $5^{\circ}$). Solve the system
$$
\left\{\begin{aligned}
x_{1}+2 x_{2}-x_{3}+x_{4}-2 x_{5} & =-3 \\
x_{1}+2 x_{2}+3 x_{3}-x_{4}+2 x_{5} & =17 \\
2 x_{1}+4 x_{2}+2 x_{3} & =14
\end{aligned}\right.
$$ | Solution. First, we solve the homogeneous system
$$
\left\{\begin{aligned}
x_{1}+2 x_{2}-x_{3}+x_{4}-2 x_{5} & =0 \\
x_{1}+2 x_{2}+3 x_{3}-x_{4}+2 x_{5} & =0 \\
2 x_{1}+4 x_{2}+2 x_{3} & =0
\end{aligned}\right.
$$
Subtract the sum of the first two equations from the third equation. We get a trivial equation, which we... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,152 | |
Example 1. Segment $AB$ is divided by point $C(-3,0)$ in the ratio $\lambda=\frac{AC}{CB}=\frac{2}{3}$. Find the length of $AB$, if point $A(-5,-4)$ is given. | Solution. 1) To find the required length using the formula from point $2^{\circ}$, it is necessary to know the coordinates of point $B\left(x_{2}, y_{2}\right)$, which we will determine using the formulas from point $3^{\circ}$.
2) We have:
$$
-3=\frac{-5+\frac{2}{3} x_{2}}{1+\frac{2}{3}}, \quad 0=\frac{-4+\frac{2}{3... | 5\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,153 |
Example 1. Construct the curve $r=\varphi$ (linear function). | Solution. It is clear that $\varphi$ is measured in radians, or $\varphi$ is a number, otherwise $r=\varphi$ makes no sense. The function $r=\varphi$ is defined only for $\varphi \geqslant 0$, and $\varphi$ can vary from 0 to $+\infty$. Points with polar coordinates $(\varphi, \varphi)$ and $(\varphi+2 \pi n, \varphi+2... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,154 |
Example 2. Construct the curve $r=2 \sin 3 \varphi$. | Solution. Let's conduct an analysis of this function.
1) This function is odd, so we can limit ourselves to values $\varphi \geqslant 0$, and then $0 \leqslant r \leqslant 2$.
2) Since $r\left(\varphi+\frac{2 \pi}{3}\right)=2 \sin 3\left(\varphi+\frac{2 \pi}{3}\right)=2 \sin (3 \varphi+2 \pi)=2 \sin 3 \varphi$, then $... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,155 |
Example 1. Given the equation of a straight line $\frac{3 x-2}{4}-\frac{2 y-1}{2}=1$, find its
1) general equation;
2) slope-intercept equation;
3) intercept form equation
4) normal equation | Solution. 1) By bringing to a common denominator, we obtain the general equation of the line (point $1^{\circ}$) $3 x-4 y-4=0$.
2) From this, it is easy to obtain the equation of the line with the slope $y=\frac{3}{4} x-1$.
3) The equation in segments can be obtained from the general equation $3 x-4 y=4$ by term-by-te... | 3x-4y-4=0,\quad\frac{3}{4}x-1,\quad\frac{x}{4/3}+\frac{y}{-1}=1,\quad\frac{3}{5}x-\frac{4}{5}y-\frac{4}{5}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,157 |
Example 2. Form the equation of the line passing through the intersection point of the lines $x+y-2=0$ and $3x+2y-5=0$ and perpendicular to the line $3x+4y-12=0$. | Solution. 1) The coordinates of the point $M_{0}$ of intersection of the lines are found by solving the system
$$
\left\{\begin{array}{l}
x+y=2 \\
3 x+2 y=5
\end{array} \quad \Rightarrow \quad M(1,1)\right.
$$
2) The slopes of perpendicular lines are related (point $3^{\circ}$) as follows: $k_{2}=-1 / k_{1}$. The slo... | 4x-3y-1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,158 |
Example 4. Given a triangle with vertices $A(7,0), B(3,4)$, $C(2,-3)$. Find the equations of side $A B$, altitude $C D$, bisector $B E$, their lengths, and angle $A$. Determine the type of the triangle by its angles. Describe the triangle using a system of inequalities. Make a drawing. | Solution. The drawing is constructed (Fig. 2.16).
1) $(A B): \frac{x-7}{3-7}=\frac{y-0}{4-0} ; x+y-7=0$
2) $|A B|=\sqrt{(3-7)^{2}+(4-0)^{2}}=\sqrt{4^{2}+4^{2}}=4 \sqrt{2}$.
3) The slope $k_{A B}=-1$.
4) $C D \perp A B \Rightarrow k_{C D}=1, \quad(C D): y+3=1(x-2) ; x-y-5=0$.
5) To derive the equation of the bisector $... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,160 | |
Example 5. The polar equation $r=\frac{7}{2 \cos \varphi-5 \sin \varphi}$ can be written in rectangular coordinates. | Solution. Let's rewrite the given equation as $2 r \cos \varphi - 5 r \sin \varphi = 7$ and use the formulas: $r \cos \varphi = x, r \sin \varphi = y$. We obtain the equation of the line: $2 x - 5 y = 7$.
## Exercises
1. Form the equation of the line passing through the point of intersection of the medians of a trian... | 3x-3y-8=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,161 |
Example 1. Form the equation of a circle whose diameter is the segment cut off by the coordinate axes from the line $3 x-$ $-2 y+12=0$. | Solution. In Fig. 2.17, the line $3x - 2y + 12 = 0$ is depicted. It intersects the coordinate axes at points $A(-4,0)$ and $B(0,6)$.
Fig. 2.17
^{2}+(y-3)^{2}=13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,162 |
Example 2. Derive the equation of an ellipse that is symmetric with respect to the coordinate axes and passes through the points $A\left(3,-\frac{16}{5}\right)$ and $B\left(-4,-\frac{12}{5}\right)$. Find the distances from point $A$ to the foci. Find the eccentricity of the ellipse. Derive the equations of its directri... | Solution. 1) The parameters $a$ and $b$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ will be found by substituting the coordinates of points $A$ and $B$ into this equation. This leads to the system
$$
\left\{\begin{array} { l }
{ \frac { 3 ^ { 2 } } { a ^ { 2 } } + \frac { ( - \frac { 1 6 } { 5 } ) ^ { ... | \frac{x^{2}}{25}+\frac{y^{2}}{16}=1,\varepsilon=0.6,r_{1}=6.8,r_{2}=3.2,\\frac{25}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,163 |
Example 3. Derive the equation of an ellipse symmetric with respect to the coordinate axes, passing through the point $A(-3,1.75)$ and having an eccentricity $\varepsilon=0.75$. | Solution. We have a system of equations with respect to the parameters $a, b, c=\sqrt{a^{2}-b^{2}}:$
$$
\left\{\begin{array}{l}
\left.\frac{(-3)^{2}}{a^{2}}+\frac{1.75^{2}}{b^{2}}=1 \quad \text { (the ellipse passes through point } A\right) \\
\frac{\sqrt{a^{2}-b^{2}}}{a}=0.75, \quad \text { or } \frac{a^{2}-b^{2}}{a^... | \frac{x^{2}}{16}+\frac{y^{2}}{7}=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,164 |
Example 5. Write the polar equation $r=\frac{2}{4-\sin \varphi}$ in rectangular coordinates | Solution. First, rewrite the given equation in the form $4 r = r \sin \varphi + 2$ and use the formulas (substitutions) $r = \sqrt{x^{2} + y^{2}}$, $r \sin \varphi = y$. We get: $4 \sqrt{x^{2} + y^{2}} = y + 2 \ (y + 2 \geqslant 0)$. Next, by squaring this equality and after transformations and completing the square, w... | \frac{x^{2}}{(\frac{2}{\sqrt{15}})^{2}}+\frac{(y-\frac{2}{15})^{2}}{(\frac{8}{15})^{2}}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,166 |
Example 6. On the hyperbola with the equation $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, find the point $M$ such that $M F_{1}=2 M F_{2}$. Write the equations of the asymptotes and the directrices of the hyperbola. Find its eccentricity. Make a sketch. | Solution. 1) We have $a=4, b=3, c^{2}=a^{2}+b^{2}, c=5$. We construct the hyperbola as follows (Fig. 2.22): in a rectangle with sides $x= \pm a$ and $y= \pm b$
Fig. 2.22 (i.e., $x= \pm 4, y=... | M_{1}(\frac{48}{5},\frac{3}{5}\sqrt{119}),\quadM_{2}=(\frac{48}{5},-\frac{3}{5}\sqrt{119}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,167 |
Example 7. On the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, find the point $M(x, y)$ such that its distance to one asymptote is three times greater than the distance to the other asymptote. | Solution. 1) Let's make a symbolic drawing of the hyperbola (Fig. 2.22) and its asymptotes. It depicts two different possible situations that satisfy the conditions of the problem: the distance from point $M$ to asymptote $l_{2}$ is three times greater than the distance to asymptote $l_{1}$; for point $M_{\mathrm{I}}-$... | M(\frac{8}{\sqrt{3}},\sqrt{3}),M^{\}(-\frac{8}{\sqrt{3}},-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,168 |
Example 8. Determine the coordinates of the point of intersection of two mutually perpendicular lines passing through the foci of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, given that the point $A(6,-2)$ lies on the line passing through its right focus. | Solution. 1) Let's make a drawing (Fig. 2.24) and list the parameters of the hyperbola. We have \(a=4, b=3, c=5, F_{1}(-5,0), F_{2}(5,0)\). We proceed to the calculations.
Fig. 2.24
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,169 |
Example 9. Derive the equation of a parabola symmetric with respect to the $O y$ axis, given that it passes through the points of intersection of the line $x-y=0$ and the circle $x^{2}+y^{2}-6 y=0$. | Solution. The equation of the desired parabola must have the form $x^{2} = 2 p y$; it is depicted in Fig. 2.26. Let's find the points of intersection of the given line and circle:
\[
\begin{aligned}
& \left\{\begin{array}{l}
y=x \\
x^{2}+y^{2}-6 y=0
\end{array}\right. \\
& 2 x^{2}-6 x=0 ; x_{1}=0, x_{2}=3
\end{aligned... | x^{2}=3y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,170 |
Example 10. Derive the equation of a parabola symmetric with respect to the $O x$ axis, with its vertex at the origin, given that the parabola passes through the point $A(2,2)$.
Find the length of the chord passing through the point $M(8,0)$ and inclined to the $O x$ axis at an angle of $60^{\circ}$. | Solution. 1) Let's make a drawing (Fig. 2.27).
2) The canonical equation of such a parabola is $y^{2}=2 p x$. The unknown parameter $p$ will be determined from the condition that the parabola passes through the point $A(2,2)$:
$$
2^{2}=2 \cdot p \cdot 2 \Rightarrow p=1
$$
Thus, the equation of the parabola is $y^{2}... | \frac{28}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,171 |
Example 11. Write the equation of the parabola $y^{2}=5 x$ in polar coordinates | Solution. Substitute the given equation with $x=r \cos \varphi, y=$ $=r \sin \varphi$. For $x \geqslant 0$, i.e., $\varphi \in\left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$, we get $r^{2} \sin ^{2} \varphi=5 r \cos \varphi$, or $r=\frac{5 \cos \varphi}{\sin ^{2} \varphi}(\varphi \neq 0)$.
## Exercises
1. Form the equat... | \frac{5\cos\varphi}{\sin^2\varphi} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,172 |
Example 1. Reduce the following second-order equations to canonical form: 1) $8 x^{2}+4 x y+5 y^{2}-56 x-32 y+80=0$; 2) $3 x^{2}+$ $+10 x y+3 y^{2}-2 x-14 y-13=0$; 3) $x^{2}-4 x y+4 y^{2}-2 x-6 y+2=0$; 4) $x^{2}+y^{2}-3 x+2 y=0$; 5) $3 x^{2}-4 x y+3 y^{2}+20=0$.
Construct the geometric representation of each equation. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,173 | |
Example 1. Prove that the triangle with vertices at points $A(1,2), B(2,5), C(3,4)$ is a right triangle. | Solution. We will construct vectors coinciding with the sides of the triangle (see point $1^{\circ}$): $\overrightarrow{A B}=\{1,3\}, \overrightarrow{A C}=\{2,2\}, \overrightarrow{B C}=\{1,-1\}$ (Fig. 3.4).
$. | Solution. Let point $B$ have coordinates $B(x, y, z)$ (Fig. 3.6). Then the coordinates of the vector (point $1^{\circ}$)
$$
\overrightarrow{A B}=\vec{a}=\{x-2, y-(-1), z-1\}=\{3,-4,2\}
$$
... | B(5,-5,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,176 |
Example 5. The vector $\vec{d}=\{-9,2,25\}$ is to be decomposed into the vectors $\vec{a}=$ $=\{1,1,3\}, \vec{b}=\{2,-1,-6\}$ and $\vec{c}=\{5,3,-1\}$. | Solution. It is necessary to find such numbers $x, y, z$ that $x \vec{a}+y \vec{b}+z \vec{c}=\vec{d}$, i.e., $\{x, x, 3 x\}+\{2 y,-y,-6 y\}+\{5 z, 3 z,-z\}=\{-9,2,25\}$. Considering that when adding vectors, their coordinates are added and equal vectors have equal coordinates, we arrive at the system of equations
$$
\... | \vec{}=2\vec{}-3\vec{b}-\vec{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,177 |
Example 6. Show that the system of vectors $\overrightarrow{a_{1}}=\{2,1,3\}, \overrightarrow{a_{2}}=$ $=\{1,0,-1\}$ and $\overrightarrow{a_{3}}=\{0,0,1\}$ is linearly independent. | Solution. In this case, equality (1) is $\lambda_{1}\{2,1,3\}+$ $+\lambda_{2}\{1,0,-1\}+\lambda_{3}\{0,0,1\}=0$, or $\left\{2 \lambda_{1}+\lambda_{2}, \lambda_{1}, 3 \lambda_{1}+\lambda_{3}\right\}=0$. From this, we obtain the system of equations
$$
\left\{\begin{array}{l}
2 \lambda_{1}+\lambda_{2}=0 \\
\lambda_{1}=0 ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,178 |
Example 7. Show that the system of vectors $\overrightarrow{a_{1}}=\{2,1,3\}$, $\overrightarrow{a_{2}}=\{1,1,-1\}$ and $\overrightarrow{a_{3}}=\{1,-1,9\}$ is linearly dependent. | Solution. Equation (1) is equivalent to the system of equations
$$
\left\{\begin{array}{l}
2 \lambda_{1}+\lambda_{2}+\lambda_{3}=0 \\
\lambda_{1}+\lambda_{2}-\lambda_{3}=0 \\
3 \lambda_{1}-\lambda_{2}+9 \lambda_{3}=0
\end{array}\right.
$$
It has a non-zero solution, for example, $\lambda_{1}=2, \lambda_{2}=-3, \lambd... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,179 |
Example 1. Are the vectors $\vec{p}$ and $\vec{q}$ perpendicular, if $\vec{p}=$ $=\{2,0,-3\}, \vec{q}=\{3,4,2\}$? | Solution. The condition of perpendicularity of vectors (point $2^{\circ}$ ) $\vec{p} \vec{q}=0$; in our case
$$
\vec{p} \vec{q}=2 \cdot 3+0 \cdot 4-3 \cdot 2=0 \Rightarrow \vec{p} \perp \vec{q}
$$
Answer. Yes. | Yes | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,180 |
Example 4. Find the coordinates of the vector $\vec{x}$, if $\vec{x} \perp \vec{c}_{1}, \vec{x} \perp \vec{c}_{2}$, where $\overrightarrow{c_{1}}=\{2,-3,2\}, \overrightarrow{c_{2}}=\{-1,3,0\}$ and $|\vec{x}|=7$. | Solution. In Fig. 3.9, we have $\vec{x} \perp \vec{c}_{1}, \vec{x} \perp \vec{c}_{2}$.
From the conditions of vector perpendicularity (point $2^{\circ}$), we have $\vec{x} \cdot \vec{c}_{1}=0 ; \vec{x} \cdot \vec{c}_{2}=0$. Let $\vec{x}=\{x, y, z\}$. The condition of the problem can be rewritten as a system:
$$
\left... | \vec{x}={\6,\2,\3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,182 |
Example 1. Find the area of the triangle whose vertices are located at points $A(1,2,3), B(3,2,1), C(1,0,1)$. | Solution. Let's find the coordinates of the vectors $\overrightarrow{A B}=\{2,0,-2\}, \overrightarrow{A C}=$ $=\{0,-2,-2\}$. We will determine the coordinates of the vector product $\overrightarrow{A B} \times$ $\times \overrightarrow{A C}($ Fig. 3.12 )
$$
\begin{aligned}
\overrightarrow{A B} \times \overrightarrow{A ... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,183 |
Example 1. Find the volume of the parallelepiped constructed on the vectors $\vec{a}\{1,2,3\}, \vec{b}\{0,1,1\}, \vec{c}\{2,1,-1\}$. | Solution. The desired volume $V=|\vec{a} \cdot \vec{b} \cdot \vec{c}|$. Since
$$
\left|\begin{array}{rrr}
1 & 2 & 3 \\
0 & 1 & 1 \\
2 & 1 & -1
\end{array}\right|=-4
$$
then $V=4$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,184 |
Example 2. The vertices of a pyramid are located at points $O(0,0,0), A(5,2,0), B(2,5,0)$, and $C(1,2,4)$. Calculate its volume, the area of face $A B C$, and the height of the pyramid dropped to this face. | Solution. 1) Let's make a schematic drawing (Fig. 3.15).
2) Introduce vectors $\overrightarrow{A B}=\{-3,3,0\}, \quad \overrightarrow{A C}=\{-4,0,4\}, \quad \overrightarrow{A O}=$ $=\{-5,-2,0\}$. The volume of the pyramid $O A B C$ (tetrahedron) is
$$
V_{T}=\left|\frac{1}{6}\right| \begin{array}{rrr}
-3 & 3 & 0 \\
-4 ... | 14;6\sqrt{3};\frac{7\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,185 |
Example 1. Construct the planes defined by the equations:
a) $3 x+2 y+4 z-8=0$; b) $x-y=0$; c) $2 x+3 y-6=0$; d) $z-2=0$. | Solution. a) The given equation is reduced to the equation of segments:
$$
\frac{x}{8 / 3}+\frac{y}{4}+\frac{z}{2}=1 \quad\left(a=\frac{8}{3}, \quad b=4, \quad c=2\right)
$$
On the $O x$ axis, we lay off the segment $a=\frac{8}{3}$ (from the origin), on the $O y$ axis - the segment $b=4$, on the $O z$ axis - the segm... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,186 |
Example 2. Form the equation of the plane that intercepts segment $O A=3$ on the $O x$ axis and is perpendicular to the vector $\vec{N}=\{2,-3,1\}$. | Solution. According to the condition, the point $A(3,0,0)$ belongs to the desired plane. According to point $3^{\circ}$, the equation of this plane has the form
$$
2(x-3)-3(y-0)+(z-0)=0, \quad \text { or } \quad 2 x-3 y+z-6=0
$$ | 2x-3y+z-6=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,187 |
Example 3. Derive the equation of the plane parallel to the $O z$ axis and passing through the points $(1,0,1)$ and $(-2,1,3)$. | Solution. The equation of a plane parallel to the $O z$ axis has the form $A x+B y+D=0$. Substituting the coordinates of the given points of the plane into this, we obtain a system for determining the coefficients of the equation:
$$
\left\{\begin{array}{l}
A \cdot 1+B \cdot 0+D=0, \\
A \cdot(-2)+B \cdot 1+D=0
\end{ar... | x+3y-1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,188 |
Example 4. Determine that the planes with equations $2 x+3 y-$ $-4 z+1=0$ and $5 x-2 y+z+6=0$ are perpendicular. | Solution. Let's write down the normal vectors of the given planes: $\vec{N}_{1}=\{2,3,-4\}$ and $\vec{N}_{2}=\{5,-2,1\}$. The planes are perpendicular if and only if the scalar product $\vec{N}_{1} \vec{N}_{2}=0$. We have $2 \cdot 5+3 \cdot(-2)+(-4) \cdot 1=0$ (see point $\left.8^{\circ}\right)$ | 0 | Geometry | proof | Yes | Yes | olympiads | false | 30,189 |
Example 5. Find the distance from the point $A(2,3,-4)$ to the plane $2 x+6 y-3 z+16=0$. | Solution. By the formula of item $7^{\circ}$, we have
$$
d=\frac{|2 \cdot 2+6 \cdot 3-3(-4)+16|}{\sqrt{4+36+9}}=7 \frac{1}{7}
$$
Answer. $d=7 \frac{1}{7}$. | 7\frac{1}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,190 |
Example 6. Find the equation of the plane passing through the points $M_{1}(1,2,0), M_{2}(1,-1,2)$ and $M_{3}(0,1,-1)$. | Solution. According to point $4^{\circ}$, the equation of the desired plane is determined by the equality
$$
\left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
0 & -3 & 2 \\
-1 & -1 & -1
\end{array}\right|=0
$$
We expand the determinant (Chapter I) by the elements of the first row:
$$
(x-1) 5-(y-2) 2+z(-3)=0, \quad \text {... | 5x-2y-3z-1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,191 |
Example 1. Reduce the general equations of a line to canonical form
\[
\left\{\begin{array}{l}
2 x-3 y-3 z-9=0 \\
x-2 y+z+3=0
\end{array}\right.
\] | Solution. We will form the canonical equations of the line using two points (as in point $4^{\circ}$). The coordinates of the two points on the line will be found according to the scheme in point $5^{\circ}$.
1) Let, for example, $z_{1}=0$ and solve the system
$$
\left\{\begin{array}{l}
2 x-3 y-9=0, \\
x-2 y+3=0
\end... | \frac{x}{9}=\frac{y}{5}=\frac{z+3}{1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,192 |
Example 2. For the direction vector of the line
$$
\left\{\begin{array}{l}
2 x-3 y-3 z+4=0 \\
x+2 y+z-5=0
\end{array}\right.
$$
find the direction cosines. | Solution. According to point $6^{\circ}$, we find the direction vector $\vec{l}$ of the given line
$$
\vec{l}=\vec{N}_{1} \times \vec{N}_{2}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & -3 & -3 \\
1 & 2 & 1
\end{array}\right|=3 \vec{i}-5 \vec{j}+7 \vec{k}
$$
Find $|\vec{l}|=\sqrt{9+25+49}=\sqrt{83}$. N... | \cos\alpha=\frac{3}{\sqrt{83}},\quad\cos\beta=-\frac{5}{\sqrt{83}},\quad\cos\gamma=\frac{7}{\sqrt{83}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,193 |
Example 3. Form the equations of the line passing through the point $A(-2,3,1)$ parallel to the line
$$
\left\{\begin{array}{l}
x-2 y-z-2=0 \\
2 x+3 y-z+1=0
\end{array}\right.
$$ | Solution. To write the canonical equations of a line (point $2^{\circ}$), we need a direction vector, which we will determine according to point $6^{\circ}$ (see Example 2):
$$
\vec{l}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -2 & -1 \\
2 & 3 & -1
\end{array}\right|=5 \vec{i}-\vec{j}+7 \vec{k}
$$
T... | \frac{x+2}{5}=\frac{y-3}{-1}=\frac{z-1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,194 |
Example 1. Given the vertices of the tetrahedron \( A(2,3,1), B(4,1,-2) \), \( C(6,3,7), D(-5,-4,8) \). Find:
Fig. 4.11
1) the length of the edge \( A B \)
2) the angle between the edges \(... | Solution. The condition of the problem is satisfied by the constructed drawing (Fig. 4.11).
1) $AB$ is calculated by the formula
$$
\begin{aligned}
d & =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\
AB & =\sqrt{(4-2)^{2}+(1-3)^{2}+(-2-1)^{2}}=\sqrt{17}
\end{aligned}
... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,195 |
Example 2. Find the coordinates of point $Q$, symmetric to point $P(-6,7,-9)$ with respect to the plane passing through points $A(1,3,-1), B(6,5,-2)$, and $C(0,-3,-5)$. | Solution. We will use the sketch of the problem (Fig. 4.13).
1) We will write the equation of the plane $\alpha=(A B C)$ passing through three points:
Fig. 4.13 $(\alpha):\left|\begin{arra... | Q(2,-5,7) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,196 |
Example 3. Find the coordinates of point $Q$, symmetric to point $P(1,3,2)$ with respect to the line $A B$, where $A(1,2,-6), B(7,-7,6)$. | Solution. 1) We have $\overrightarrow{A B}=\{6,-9,12\}$. We take $\vec{l}=\{2,-3,4\}$;
$$
(AB): \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+6}{4}
$$
or
$$
x=1+2 t, \quad y=2-3 t, \quad z=-6+4 t
$$
2) The equation of the plane $\alpha$, passing through $P$ and perpendicular to $AB$, is (Fig. 4.14)
$(\alpha): 2(x-1)-3(y-3)... | Q(5,-5,-6) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,197 |
Example 4. Determine the distance from the point $P(-7,-13,10)$ to the line $l: \frac{x-1}{-2}=\frac{y+2}{1}=\frac{z}{0}$. | Solution. 1) Through $P$ we draw a plane $\alpha$ perpendicular to $l$, taking $\vec{N}=\vec{l}=\{-2,1,0\}$. We obtain $(\alpha):-2(x+7)+(y+13)=0$, i.e., $2 x-y+1=0$.
2) We find the coordinates of the point $O$ of intersection of $l$ and $\alpha$. Substituting the expressions $x=1-2 t, y=-2+t, z=0$ into the equation o... | 2\sqrt{83} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,198 |
Example 5. For what values of $B$ and $C$ is the line $\frac{x-3}{1}=\frac{y-2}{B}=$ $=\frac{z+2}{C}$ perpendicular to the plane $3 x-2 y+5 z=0$? | Solution. The condition of perpendicularity of a line and a plane is equivalent to the condition of parallelism of their vectors $\vec{l}=\{1, B, C\}$ and $\vec{N}=$ $=\{3,-2,5\}$. The corresponding coordinates of these vectors must be proportional ( $\$ 3$ ): $\frac{1}{3}=\frac{B}{-2}=\frac{C}{5}$. From this, $B=-\fra... | B=-\frac{2}{3},C=\frac{5}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,199 |
Example 6. Through the line with the general equations
$$
\left\{\begin{array}{l}
3 x-2 y+5 z+3=0 \\
x+2 y-3 z-11=0
\end{array}\right.
$$
and the origin, draw a plane and derive its equation. | Solution. We reduce the problem to constructing a plane through three points ( $\$ 1$, p. $4^{\circ}$ ). Substitute $z=-2$ into the original system and solve it for $x, y$. We get one point $M_{1}(3,1,-2)$ on the given line. To find another point on this line, we set $z=6: M_{2}(-1,15,6)$. It remains to
compose the eq... | 18x-8y+230 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,200 |
Example 7. Find the equation of the plane containing the point $M_{1}(3,1,0)$ and the line $\frac{x-4}{1}=\frac{y}{2}=\frac{z-1}{3}$. | Solution. From the equation of the line, the coordinates of the point $M_{0}(4,0,1)$ on it and the directing vector $\vec{l}=\{1,2,3\}$ are known.
Let $M(x, y, z) \rightarrow$ the current point of the plane (Fig. 4.15). Then the vectors $\overrightarrow{M_{0} M}, \overrightarrow{M_{0} M_{1}}, \vec{l}$ lie in the same ... | 5x+2y-3z-17=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,201 |
Example 8. Find the distance from the point $M_{1}\left(x_{1}, y_{1}, z_{1}\right)$ to the line
$$
\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p} \text {. }
$$ | Solution. The desired distance can be found as the height $h$ of the parallelogram constructed on the vectors (Fig. 4.16) $\overrightarrow{M_{0} M_{1}}=\left\{x_{1}-x_{0}\right.$, $\left.y_{1}-y_{0}, z_{1}-z_{0}\right\}$ and $\vec{l}=\{m, n, p\}$. The area of the parallelogram, as is known (Chapter III, § 3), is equal ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,202 | |
Example 1 (to item $2^{\circ}$). Determine the type of surface and make a drawing: a) $x^{2}-y^{2}=z^{2}$; b) $-2 x^{2}+2 y^{2}+z^{2}=4$; c) $2 x^{2}-y^{2}+z^{2}+2=0$; d) $3 y^{2}+2 z^{2}=6 x$ | Solution. a) Let's write the given equation in the form $y^{2}+z^{2}-x^{2}=0$. By comparing it with 7), we determine that this is a circular ($a=b=c$) cone with its vertex at the origin and the axis of rotation along $O x$ (compare with Fig. 4.23, where the axis of rotation is $O z$).
b) Rewriting the equation of the ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,203 |
Example 2 (to $3^{\circ}$). Determine the type of surface and make a drawing: a) $y^{2}=4 x ;$ b) $z=x^{2}$; c) $y^{2}+z^{2}=4$. | Solution. a) Since the variable $z$ is missing in the equation of the surface, this is a cylinder with generators parallel to the $O z$ axis and the directrix - a parabola (Fig. 4.30) with equations
$$
\left\{\begin{array}{l}
y^{2}=4 x \\
z=0
\end{array}\right.
$$
b) Since the variable $y$ is missing in the equation ... | \ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,204 |
Example 3 (to points $2^{\circ}, 3^{\circ}$). Sketch the body bounded by the given surfaces: a) $2 y=x^{2}+z^{2}, x^{2}+z^{2}=1$; b) $z=0, y+z=2, y=x^{2}$; c) $z=6-x^{2}-y^{2}, x^{2}+y^{2}-z^{2}=0$. | Solution. a) The first surface is an elliptic paraboloid $\frac{x^{2}}{2}+\frac{z^{2}}{2}=y$, the second is a cylinder with generators parallel to the $O y$ axis (Fig. 4.33).
b) $z=0$ is the coordinate plane $O x y ; y+z=2$ is a plane parallel to the $O x$ axis; $y=x^{2}$ is a parabolic cylinder with generators parall... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,205 |
Example 1. Plot the graph of the function $y=\frac{1}{2} x^{2}+2 x-1$. | Solution. Rewrite the given function as $y=\frac{1}{2}(x+2)^{2}-3$.
Fig. 5.12
1) $\Gamma_{1}: y=x^{2}$ - parabola (Fig. 5.12, a) with vertex at point $O(0,0)$.
2) $\Gamma_{2}: \quad y=(x+2)^... | \frac{1}{2}(x+2)^{2}-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,206 |
Example 2. Plot the graph of the function $y=2-\frac{3 x+1}{2 x-2}$. | Solution. Let's represent the given function as $y=-\frac{2}{x-1}+\frac{1}{2}$. We will construct the graph step by step.
1) Let $\Gamma$ be the graph of the function $y=\frac{1}{x}$ - a hyperbola with equal branches (Fig. 5.13, a).
2) $\Gamma_{1}: y=\frac{1}{x-1}$ - a shift of the hyperbola 1 unit to the right along ... | -\frac{2}{x-1}+\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,207 |
Example 3. Plot the graph of the function
$$
y=2 \sin \left(2 x+\frac{1}{2}\right)-\frac{1}{2}
$$ | Solution. Rewrite: $y=2 \sin 2\left(x+\frac{1}{4}\right)-\frac{1}{2}$.
Fig. 5.13
^{k}$. The graph of this function is shown in ... | \ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,209 |
Example 5. Plot the graph of the function $y=\log _{2} \cos x^{2}$. | Solution. The domain of this function is found from the graph of the previous function: $\cos x^{2}>0$ for $x \in\left(-\frac{\sqrt{\pi}}{2}, \frac{\sqrt{\pi}}{2}\right), x \in$ $\in\left(\frac{\sqrt{3 \pi}}{2}, \frac{\sqrt{5 \pi}}{2}\right), x \in\left(-\frac{\sqrt{5 \pi}}{2},-\frac{\sqrt{3 \pi}}{2}\right)$, etc. We c... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,210 |
Example 6. Plot the graph of the function $y=\left|\log _{2}\right| x-2||$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Solution. The graph of the function $y=\log _{2}|x|$ consists of the graph of $y=\log _{2} x(x>0)$ and the graph of $y=\log _{2}(-x), x<0$ symmetric to it with respect to the $O y$ axis (Fig. $5.17, \mathrm{a}$ ).
The graph of the function $y=\left|\log _{2}\right| x||$ is obtained from the branches of the previous gr... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,211 |
Example 7. Construct a line, the coordinates of the points of which satisfy the equation $|x|-|y|=1$. | Solution. For $x \geqslant 0$, the given equation takes the form $|y|=x-1$. This equation makes sense for $x \geqslant 1$ and is represented by two rays: $y=x-1$ and $y=-(x-1)$. These rays form a right angle. The presence of the modulus $|x|$ in the original equation means that its graph is symmetric with respect to th... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,212 |
Example 1 (to $3^{\circ}$). Prove that $\lim _{n \rightarrow \infty} \frac{2 n-3}{n+2}=2$. | Solution. If $\varepsilon>0$ is an arbitrarily small number, then
$$
\begin{aligned}
& \left|\frac{2 n-3}{n+2}-2\right|<\varepsilon \\
& \left|\frac{2 n-3-2 n-4}{n+2}\right|<\varepsilon \\
& \left|\frac{-7}{n+2}\right|<\varepsilon \\
& \frac{7}{n+2}<\varepsilon \\
& n+2>\frac{7}{\varepsilon}, n>\frac{7}{\varepsilon}-2... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,213 |
Example 2 (to $4^{\circ}$). Prove that $\lim _{n \rightarrow \infty} \frac{2 n^{2}-3 n+2}{n+2}=+\infty$. | Solution. If $\varepsilon>0-$ is an arbitrarily large number, then
$$
\begin{aligned}
\frac{2 n^{2}-3 n+2}{n+2}>\varepsilon \Leftrightarrow & \left\{\begin{array}{cc}
2 n^{2}-3 n+2 & \frac{n+2}{2 n-7} \\
\frac{2 n^{2}+4 n}{-7 n+2} & \\
\frac{-7 n-14}{16}
\end{array}\right\} \Leftrightarrow \\
& \Leftrightarrow 2 n-7+\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,214 |
Example 4. Find $\lim _{n \rightarrow \infty}\left(\frac{3 n-5}{3 n+2}\right)^{n+2}$. | \frac{1}{e^{\frac{7}{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,216 |
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