problem
stringlengths 15
4.7k
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stringlengths 2
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| answer
stringclasses 51
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stringclasses 8
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stringclasses 4
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stringclasses 1
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|---|---|---|---|---|---|---|---|---|
105. Is it true that there exists a number $C$ such that for all integers $k$ the inequality
$$
\left|\frac{k^{8}-2 k+1}{k^{4}-3}\right|<C ?
$$
|
105. Let's see how the expression $\left|\frac{k^{3}-2 k+1}{k^{4}-3}\right|$ behaves for large (in absolute value) values of $k$. Clearly, in the numerator, the term $k^{3}$ plays the main role, and in the denominator, $k^{4}$. Therefore, we can expect that for large values of $k$, our expression is approximately equal to $\left|\frac{k^{3}}{k^{4}}\right|=\frac{1}{|k|}$. Now let's investigate how much the exact value of our expression differs from the found approximate value. For this, we will perform the following transformation:
$$
\begin{aligned}
&\left|\frac{k^{3}-2 k+1}{k^{4}-3}\right|=\left|\frac{k^{3}\left(1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right)}{k^{4}\left(1-\frac{3}{k^{4}}\right)}\right|= \\
&=\frac{1}{|k|} \frac{\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right|}{\left|1-\frac{3}{k^{4}}\right|}
\end{aligned}
$$
Let $|k| \geqslant 2$; then
$$
\begin{gathered}
\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right| \leqslant 1+\frac{2}{k^{2}}+\frac{1}{|k|^{3}} \leqslant \\
\leqslant 1+\frac{1}{2}+\frac{1}{8}\frac{1}{2}
\end{gathered}
$$
Therefore, for $|k| \geqslant 2$, the inequality
$$
\frac{1}{|k|} \frac{\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right|}{\left|1-\frac{3}{k^{4}}\right|}<\frac{1}{2} \cdot \frac{2}{\frac{1}{2}}=2
$$
holds. Thus, for $|k| \geqslant 2$, our expression does not exceed 2. It remains to see what values it takes for $k=$ $=-1,0,1$. These values are respectively $1,1 / 3,0$.
Answer: the number $C$ exists, for example, we can take
$$
C=2
$$
Remark. In the same way, we can obtain more accurate estimates for our expression and verify that the maximum value, equal to 1, it takes at $k=-1$. Do this independently.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
148. To compute the square root of a positive number $a$, one can use the following method of successive approximations. Take any number $x_{0}$ and construct a sequence according to the following rule:
$$
x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)
$$
Prove that if $x_{0}>0$, then $\lim _{n \rightarrow \infty} x_{n}=\sqrt{a}$, and if $x_{0}<0$, then $\lim _{n \rightarrow \infty} x_{n}=-\sqrt{a}$. (By $\sqrt{a}$ we denote the arithmetic square root of $a$.)
How many successive approximations (i.e., how many terms of the sequence $\left\{x_{n}\right\}$ need to be computed) are required to find the value of $\sqrt{10}$ with an accuracy of 0.00001, if the initial value is taken as $x_{0}=3$?
|
148. First, let's prove that if the limit of $\{x_{n}\}$ exists, then it equals $\pm \sqrt{a}$. Indeed, let $\lim _{n \rightarrow \infty} x_{n}=b$. Then $\lim _{n \rightarrow \infty} \frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)=\frac{1}{2}\left(b+\frac{a}{b}\right)$. We obtain the equation $b=\frac{1}{2}\left(b+\frac{a}{b}\right)$, from which $b^{2}=a, b= \pm \sqrt{a}$.
It remains to note that if $x_{0}>0$, then all terms of the sequence are positive; if $x_{0}>0$ (Fig. 47). Let us denote by $y_{n}$ the difference between $x_{n}$ and $\sqrt{a}$, divided by $\sqrt{a}$. Substituting the expression $x_{n}=\sqrt{a}\left(1+y_{n}\right)$ into the equality $x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)$, we get
$$
\begin{aligned}
& \sqrt{a}\left(1+y_{n+1}\right)= \\
& \quad=\frac{1}{2}\left(\sqrt{a}\left(1+y_{n}\right)+\frac{a}{\sqrt{a}\left(1+y_{n}\right)}\right)
\end{aligned}
$$
Fig. 47
from which
$$
y_{n+1}=\frac{y_{n}^{2}}{2\left(1+y_{n}\right)}
$$
We need to prove that the sequence $\{y_{n}\}$ tends to zero. Note first that since
$$
1+y_{0}=1+\frac{x_{0}-\sqrt{a}}{\sqrt{a}}=\frac{x_{0}}{\sqrt{a}}>0
$$
all numbers $y_{n}$ for $n \geqslant 1$ are positive. Therefore,
$$
\left|y_{n+1}\right|=y_{n+1}=\frac{y_{n}^{2}}{2\left(1+y_{n}\right)}<\frac{y_{n}^{2}}{2}
$$
If $x_{0}=3$ and $a=10$, then $\sqrt{10}<3.2$.
Therefore, $\left|y_{0}\right|=\left|\frac{3-\sqrt{10}}{\sqrt{10}}\right|<\frac{0.2}{3}=\frac{1}{15}$ and, hence,
$$
\left|y_{1}\right|=\frac{y_{0}^{2}}{2\left(1+y_{0}\right)}<\frac{\left(\frac{1}{15}\right)^{2}}{2\left(1-\frac{1}{15}\right)}<\frac{1}{400}
$$
Further,
$$
\left|y_{2}\right|=\frac{y_{1}^{2}}{2\left(1+y_{1}\right)}<\frac{\left(\frac{1}{400}\right)^{2}}{2}<\frac{1}{320000}
$$
Therefore, $\left|x_{2}-\sqrt{10}\right|=y_{2} \sqrt{10}<\frac{\sqrt{10}}{320000}<$ $<0.00001$. Thus, to find $\sqrt{10}$ with an accuracy of 0.00001, it is sufficient to find the term $x_{2}$. We have
$$
\begin{aligned}
x_{1}=\frac{1}{2}\left(x_{0}+\frac{10}{x_{0}}\right)= & \frac{1}{2}\left(3+\frac{10}{3}\right)= \\
= & 3 \frac{1}{6}=3.166666 \ldots \\
x_{2}=\frac{1}{2}\left(x_{1}+\frac{10}{x_{1}}\right)= & \frac{1}{2}\left(3 \frac{1}{6}+\frac{10}{3 \frac{1}{6}}\right)= \\
& =3 \frac{37}{228}=3.162280 \ldots
\end{aligned}
$$
In fact, $\sqrt{10}=3.16227765 \ldots$
As you can see, the value we found indeed differs from the true value by less than 0.00001.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
228. Given an angle MAN and a point $O$ not lying on the side of the angle. Draw a line through $O$ intersecting the sides of the angle at points $X$ and $Y$, such that the product $O X \cdot O Y$ has a given value $k$.
|
228. Suppose the problem is solved. From OX. $O Y=k$, it follows that $X$ is obtained from point $Y$ by the inversion with center $O$ and power $k$; therefore, $X$ lies on the circle $S$, which is obtained from the line $A N$ by the inversion with center $O$ and power $k$ (i.e., $k$), i.e., $X$ is the intersection point of the line $A M$ and the circle $S$ (which can be constructed). The problem can have up to four solutions.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. (GDR, 74). What is greater: $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}$ or 0?
|
7.1. Using Theorem 4, we get
$$
\begin{aligned}
& \sqrt{4+\sqrt{7}}=\sqrt{4-\sqrt{7}}-\sqrt{\overline{2}}= \\
& =\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}+\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)- \\
& -\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}-\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)-\sqrt{2}= \\
& =\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}-\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}-V^{\overline{2}}=0
\end{aligned}
$$
i.e., the given numbers are equal to each other.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. (England, 75). Solve the equation
$$
[\sqrt[3]{1}]+[\sqrt[3]{2}]+\ldots+\left[\sqrt[3]{x^{3}-1}\right]=400
$$
in natural numbers.
|
8.2. Note that the relation $[\sqrt[3]{m}]=k$, where $m, k \in \mathrm{N}$, is equivalent to the inequality $k^{3} \leqslant m \leqslant(k+1)^{3}-1$. The number of natural numbers $m$ satisfying this condition (for a fixed $k$) is $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$. Therefore, the left side of the equation is $\sum_{k=1}^{x-1} S_{k}$, where
$$
S_{k}=k\left(3 k^{2}+3 k+1\right)
$$
Since $S_{k}>0$ for $k \in \mathrm{N}$, $S_{1}=1 \cdot 7 = 7$, $S_{2}=2 \cdot 19 = 38$, $S_{3}=3 \cdot 37 = 111$, $S_{4}=4 \cdot 61 = 244$, and $S_{1}+S_{2}+S_{3}+S_{4}=400$, the original equation has a unique solution in natural numbers: $x=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.16. (England, 66). Find the number of sides of a regular polygon if for four of its consecutive vertices \( A, B, C, D \) the equality
\[
\frac{1}{A B}=\frac{1}{A C}+\frac{1}{A D}
\]
is satisfied.
|
11.16. Let a circle with center 0 and radius $R$ be circumscribed around a polygon (Fig. 58). Denote $\alpha=\angle A O B$, then $0<\alpha<$ $<120^{\circ}$ and
$A B=2 R \sin (\alpha / 2), A C=2 R \sin \alpha$,
$$
A D=2 R \sin (3 \alpha / 2)
$$
from which we have
$$
\frac{1}{\sin (\alpha / 2)}=\frac{1}{\sin \alpha}+\frac{1}{\sin (3 \alpha / 2)}
$$
Therefore, we obtain
$$
\begin{array}{r}
0=\sin \alpha \sin \frac{3 \alpha}{2}-\left(\sin \alpha+\sin \frac{3 \alpha}{2}\right) \times \\
\times \sin \frac{\alpha}{2}=\frac{1}{2}\left(\cos \frac{\alpha}{2}-\cos \frac{5 \alpha}{2}\right)-\frac{1}{2}\left(\cos \frac{\alpha}{2}-\cos \frac{3 \alpha}{2}\right)- \\
-\frac{1}{2}(\cos \alpha-\cos 2 \alpha)=\frac{1}{2}\left(\left(\cos \frac{3 \alpha}{2}+\cos 2 \alpha\right)-\left(\cos \alpha+\cos \frac{5 \alpha}{2}\right)\right)= \\
=\cos \frac{7 \alpha}{4}\left(\cos \frac{\alpha}{4}-\cos \frac{3 \alpha}{4}\right)=2 \cos \frac{7 \alpha}{4} \sin \frac{\alpha}{4} \sin \frac{\alpha}{2}
\end{array}
$$
from which
$$
7 \alpha / 4=90^{\circ} \text { and } \alpha=360^{\circ} / 7
$$
i.e., the original polygon has seven sides.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12.10. (SFRY, 76). Find all values of $n \in \mathbf{N}$, greater than 2, for which it is possible to select $n$ points on a plane such that any two of them are vertices of an equilateral triangle, the third vertex of which is also one of the selected points.
|
12.10. We will prove that the condition of the problem is satisfied only by the value $n=3$ (in which case the points can be placed at the vertices of an equilateral triangle). Indeed, suppose that it is possible to arrange $n \geqslant 4$ points in the manner specified in the problem. We select two points $A$ and $B$, the distance between which is maximal, and a third point $C$, such that the triangle $A B C$ is equilateral. Then all the other points lie within the figure $M$, which is the intersection of three circles of radius $A B$ centered at points $A, B$, and $C$ (Fig. 68). If point $O$ is the center of the triangle $A B C$, then the segments $A O, B O$, and $C O$ divide the set $M$ into three equal parts, none of which can contain any of the $n$ points, except those already mentioned. Indeed, suppose, for example, that part $M_{B C}$, consisting of the triangle $B O C$ and the segment of the circle centered at $A$ with the arc $B C$, contains another point $D$. Then there exists a point $D^{\prime}$ such that the triangle $A D D^{\prime}$ is equilateral. Consequently, when rotating (in a certain direction) around point $A$ by an angle of $60^{\circ}$, point $D$ transitions to point $D^{\prime}$, which thus lies simultaneously in the image $M_{B C}^{\prime}$ of part $M_{B C}$ under this rotation and in the set $M$. But since $\angle B A C=60^{\circ}$, either $C^{\prime}=B$ or $B^{\prime}=C$. For definiteness, assume that $B^{\prime}=C$. Then the sets $M_{B C}^{\prime}$ and $M$ lie in different half-planes relative to the line $B^{\prime} O^{\prime}$, because
$$
\angle B B^{\prime} O^{\prime}=\angle B C A+\angle A B^{\prime} O^{\prime}=60^{\circ}+30^{\circ}=90^{\circ},
$$
i.e., the line $B^{\prime} O^{\prime}$ is tangent to the arc $A C$. Therefore, $M_{B C}^{\prime}$ and $M$ have only one common point $B^{\prime}$, from which it follows that $D^{\prime}=B^{\prime}$ and $D=B$, which contradicts the choice of point $D$. The statement is proved.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12.11. (CSSR, 80). The set $M$ is obtained from the plane by removing three distinct points $A, B$, and $C$. Find the smallest number of convex sets whose union is the set $M$.
|
12.11. Let points $A, B$, and $C$ initially lie on the same line. Then these points divide the line into 4 intervals, and no points from different intervals can lie in the same convex set. Therefore, the number of required sets cannot be less than 4. The number 4 is achieved if the set $M$ is divided into parts as shown in Fig. 69.
Fig. 69
Now, let points $A, B$, and $C$ not lie on the same line. Then points $B$ and $C$ divide the line $BC$ into 3 intervals, and points from different intervals must lie in different convex sets. Therefore, the number of required sets cannot be less than 3, and 3 sets can be chosen as shown in Fig. 70.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12.12. (Jury, SRP, 79). Prove that for any value of $n \in \mathbf{N}$, greater than some number $n_{0}$, the entire plane can be divided into $n$ parts by drawing several lines, among which there must be intersecting ones. Find the smallest such value of $n_{0}$.
|
12.12. Among the lines drawn according to the condition, there are necessarily intersecting lines that already divide the plane into 4 parts. If another line is drawn, then, as a simple case analysis shows, the number of parts will increase by at least 2. Therefore, it is impossible to get exactly 5 parts, from which it follows that \( n_{0} \geqslant 5 \). On the other hand, any number \( n > 5 \) of parts can be obtained in the required way: if \( n = 2k \) ( \( k \in \mathbf{N} \) ), then the division of the plane can be constructed as shown in Fig. 71; if \( n = 4k + 3 \), see Fig. 72; and if \( n = 4k + 5 \), see Fig. 73. Thus, the smallest value of \( n_{0} \) is 5.
Fig. 70
Fig. 71
Fig. 72
Fig. 73
|
5
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
15.7. (NBR, 68). Inside the tetrahedron $A B C D$ is a point $O$ such that the lines $A O, B O, C O, D O$ intersect the faces $B C D, A C D, A B D, A B C$ of the tetrahedron at points $A_{1}, B_{1}$, $C_{1}, D_{1}$ respectively, and the ratios
$$
\frac{A O}{A_{1} O}, \frac{B O}{B_{1} O}, \frac{C O}{C_{1} O}, \frac{D O}{D_{1} O}
$$
are all equal to the same number. Find all possible values that this number can take.
|
15.7. Let \( V \) be the volume of the tetrahedron \( ABCD \), and \( k \) be the desired number. Then we have
\[
\frac{V}{V_{OBCD}} = \frac{AA_1}{OA_1} = \frac{AO}{A_1O} + \frac{OA_1}{OA_1} = k + 1
\]
\[
\frac{V}{V_{OACD}} = \frac{V}{V_{OABD}} = \frac{V}{V_{OABC}} = k + 1,
\]
from which
\[
k + 1 = \frac{4V}{V_{OBCD} + V_{OACD} + V_{OABD} + V_{OABC}} = \frac{4V}{V} = 4, \text{ i.e., } k = 3
\]
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16.15. (England, 68). Find the maximum number of points that can be placed on a sphere of radius 1 so that the distance between any two of them is: a) not less than $\sqrt{2} ;$ b) greater than $\sqrt{2}$.
|
16.15. a) Let's prove that if points $A_{1}, A_{2}, \ldots, A_{n}$ are located on a sphere with center $O$ and radius 1 such that the distance between any points $A_{i}, A_{j} (i \neq j)$ is at least $\sqrt{2}$, then $n \leqslant 6$. Indeed, let $n>6$. By the cosine theorem, we have
$$
A_{i} A_{j}^{2}=2-2 \cos \angle A_{i} O A_{j} \geq 2
$$
from which $\angle A_{i} O A_{j} \geqslant 90^{\circ}$ and $\overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}} \leq 0$. Let's choose a rectangular coordinate system in space with the origin at point $O$ as follows. First, note that among the vectors $\overrightarrow{O A}_{i}$, there must be three non-coplanar vectors (otherwise, all $n>4$ vectors would lie in the same plane, and the angle between some two of them would be acute). Without loss of generality, we can assume that the vectors $\overrightarrow{O A}_{1}, \overrightarrow{O A}_{2}, \overrightarrow{O A}_{3}$ are non-coplanar. The axis $O X$ is directed along the line $O A_{i}$ so that the abscissa of point $A_{1}$ is 1. Next, the axis $O Y$ is directed so that point $A_{2}$ lies in the plane $X O Y$ and has a positive ordinate. Finally, the axis $O Z$ is directed so that the applicate of point $A_{3}$ is positive. Let $x_{i}, y_{i}, z_{i}$ be the coordinates of point $A_{i}$. Then
$$
\overrightarrow{O A}_{1}=(1 ; 0 ; 0), \overrightarrow{O A}_{2}=\left(x_{2} ; y_{2} ; 0\right), \overrightarrow{O A}_{3}=\left(x_{8} ; y_{8} ; z_{3}\right),
$$
where $y_{2}>0, z_{3}>0$, so for all values of $i>1$ we have
$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A}_{1}=x_{i} \leqslant 0
$$
further, for all values of $i>2$ we have
$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A_{2}}=x_{i} x_{2}+y_{i} y_{2} \leqslant 0
$$
from which $y_{i} \leqslant 0$ (since $x_{2} \leqslant 0, x_{i} \leqslant 0, y_{2}>0$); finally, for all values of $i>3$ we have
$$
\overrightarrow{O A}_{i} \cdot \overrightarrow{O A_{3}}=x_{i} x_{3}+y_{i} y_{3}+z_{i} z_{3} \leqslant 0
$$
from which $z_{i} \leqslant 0$ (since $x_{3} \leqslant 0, x_{i} \leqslant 0, y_{3} \leqslant 0, y_{i} \leqslant 0, z_{3}>0$). Among the four vectors $\overrightarrow{O A}_{4}, \overrightarrow{O A}_{5}, \overrightarrow{O A}_{8}, \overrightarrow{O A}_{7}$, at least two have the same negative coordinate, and therefore their scalar product is positive. The obtained contradiction proves that $n \leqslant 6$. Since 6 points with coordinates
$$
( \pm 1 ; 0 ; 0),(0 ; \pm 1 ; 0),(0 ; 0 ; \pm 1)
$$
satisfy the condition of part a), the maximum number of points is 6. 276
b) Let's prove that if points $A_{1}, A_{2}, \ldots, A_{n}$ are located on the same sphere such that the distance between any points $A_{i}, A_{j} (i \neq j)$ is greater than $\sqrt{2}$, then $n \leq 4$. Then $\overrightarrow{O A}_{i} \cdot \overrightarrow{O A}_{j} < 0$ for all $i \neq j$. For $n \geq 3$ we will have $x_{i} < 1 / 2$ and $n < 6$
(the last inequality also follows from the fact that the dihedral angle $\angle A_{1} H A_{2} = 360^{\circ} / n$ at the lateral edge $O A_{0}$ is greater than the plane angle $\angle A_{1} A_{0} A_{2} = 60^{\circ}$ of the pyramid's base), and for the other values of $n=2,3,4,5$ we have
$$
r_{n}=\sqrt{1-\frac{1}{4 \sin ^{2}\left(180^{\circ} / n\right)}}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17.3. (New York, 74). Let
$$
a_{n}=\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{2 \cdot 4 \cdot 6 \ldots 2 n}, n \in \mathrm{N}
$$
Find $\lim a_{n}$.
$$
n \rightarrow \infty
$$
|
17.3. Since
$$
\begin{aligned}
a_{n}^{2}=\frac{1^{2} \cdot 3^{2} \cdot \ldots \cdot(2 n-1)^{2}}{2^{2} \cdot 4^{2} \cdot \ldots \cdot(2 n)^{2}} & = \\
& =\frac{1 \cdot 3}{2^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdots \frac{(2 n-1)(2 n+1)}{(2 n)^{2}} \cdot \frac{1}{2 n+1}<\frac{1}{2 n+1}
\end{aligned}
$$
for any $n \in \mathbf{N}$, then
$$
0<a_{n}<1 / \sqrt{2 n+1} \quad \text { and } \quad \lim _{n \rightarrow \infty} a_{n}=0
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21.5. (Austria, 83). Find all values of $a$ for which the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy the equation
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0
$$
|
21.5. Let's make the substitution $y=x-3$, then the numbers $y_{1}=x_{1}-3, y_{2}=$ $=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$
By Vieta's theorem, we have the equalities
$$
\begin{aligned}
& y_{1}+y_{2}+y_{3}=-3 \\
& y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}=a-9 \\
& y_{1} y_{2} y_{3}=27-4 a
\end{aligned}
$$
and in addition, the relation $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=0$ must hold. By direct verification, we confirm the validity of the identity $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=$
$$
\equiv\left(y_{1}+y_{2}+y_{3}\right)^{3}-3\left(y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}\right)\left(y_{1}+y_{2}+y_{3}\right)+3 y_{1} y_{2} y_{3}
$$
from which we obtain the necessary and sufficient condition for $a$ :
$$
0=(-3)^{3}-3(a-9) \cdot(-3)+3(27-4 a)=-27-3 a
$$
i.e. $a=-9$.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
23.13*. (CSSR, 74). Let $M$ be the set of all polynomials of the form
$$
P(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \in \mathbf{R})
$$
satisfying the inequality $|P(x)| \leqslant 1$ for $x \in[-1 ; 1]$. Prove that some number $k$ provides the estimate $|a| \leqslant k$ for all polynomials $P(x) \in M$. Find the smallest value of $k$.
|
23.13. The polynomial $P_{0}(x)=4 x^{3}-3 x$ belongs to the set $M$, since $P_{0}(-1)=-1, P_{0}(1)=1$, and at its extremum points we have $P_{0}(-1 / 2)=1, P_{0}(1 / 2)=-1$. Let's prove that for any polynomial $P(x) \in M$ the estimate $|a| \leqslant 4$ holds. Suppose, on the contrary, that there exists a polynomial $P(x)=a x^{3}+b x^{2}+c x+d$, satisfying the inequalities $|a|>4$ and $|P(x)| \leq 1$ for $|x| \leq 1$. Then consider the non-zero polynomial
$$
Q(x)=P_{0}(x)-\frac{4}{a} P(x)
$$
the degree of which does not exceed two. Since $\left|\frac{4}{a} P(x)\right|0, Q(1 / 2)0$, and thus the polynomial $Q(x)$ has at least three roots. The obtained contradiction proves that the desired number $k$ is 4.
|
4
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
25.6. (USSR, 81; USA, 81). In a certain country, any two cities are directly connected by one of the following means of transportation: bus, train, or airplane. It is known that there is no city provided with all three types of transportation, and at the same time, there do not exist three cities such that any two of them are connected by the same means of transportation. Find the largest possible number of cities in this country.
|
25.6. Suppose there are five cities connected in the manner indicated in the problem. First, let's prove that no city has three lines of the same type of transport leading out of it. Let city \( A \) be connected to cities \( B, C \), and \( D \), for example, by airplane. Then, according to the condition, no pair of cities \( B, C \), and \( D \) can be connected by airplane. Suppose \( B \) and \( C \) are connected, for example, by train. Cities \( C \) and \( D \) cannot be connected by bus, as otherwise city \( C \) would have all three types of transport. Therefore, \( C \) and \( D \) are connected by train. For the same reasons, cities \( B \) and \( D \) are also connected by train. We have obtained that \( B, C \), and \( D \) are pairwise connected by train. Contradiction. Thus, from each city, two transport lines of one type and two transport lines of another type lead out. Then each city is served by exactly two types of transport. Therefore, at least one type of transport serves no more than three cities (otherwise, there would be at least \( 4 \cdot 3 / 2 = 6 \) cities in total). If it serves exactly two cities, then from each of these cities, only one transport line of this type leads out, which is impossible. If it serves exactly three cities, then they must be pairwise connected by this type of transport, which is also impossible. Thus, we have proved that the condition of the problem is unfulfillable for five cities. It is even more unfulfillable for a larger number of cities. If, however, we consider four cities \( A, B \), \( C \), and \( D \), which are connected as follows: \( A \) with \( B \) by train, \( C \) with \( D \) by bus, and all other pairs by airplane, then all conditions of the problem will be met. Therefore, the maximum number of cities is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27.5. (Jury, Brazil, 82; Australia, 83). An urn contains $n$ white and $m$ black balls, and next to the urn is a box with a sufficiently large number of black balls. The following operation is performed: a pair of balls is randomly drawn from the urn; if they are of the same color, a black ball from the box is moved to the urn; if they are of different colors, the white ball is returned to the urn. The operation is repeated until only one ball remains in the urn. What is the probability that it will be white?
|
27.5. Since the parity of the number of white balls contained in the urn does not change after each operation, the last ball will be white if and only if the number $n$ is odd. Therefore, the desired probability is either 1 (if $n$ is odd), or 0 (if $n$ is even).
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
27.11. (Belgium, 77). Three shooters $A, B, C$ decided to duel simultaneously. They positioned themselves at the vertices of an equilateral triangle and agreed on the following: the first shot is taken by $A$, the second by $B$, the third by $C$, and so on in a circle; if one of the shooters is eliminated, the duel continues between the two remaining. It is known that shooter $A$ hits the target with a probability of 0.3, shooter $C$ with a probability of 0.5, and shooter $B$ never misses. Each shooter aims at one of the two others or at the air in such a way as to maximize their probability of winning the duel. Where should shooter $A$ direct his first shot: 1) at shooter $C$; 2) at shooter $B$; 3) at the air?
|
27.11. Let's consider three events that may occur after the first shot of shooter $A$.
1) $C$ is hit. Then with probability 1, shooter $A$ will be hit by the first shot of $B$.
2) $B$ is hit. Then: either with probability 0.5, shooter $C$ will hit $A$ with his first shot, or with probability $0.5 \cdot 0.3$, shooter $A$ will hit $C$ with his second shot, or with probability $0.5 \cdot 0.7 \cdot 0.5$, shooter $C$ will hit $A$ with his second shot, or with probability $0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3$, shooter $A$ will hit $C$ with his third shot, and so on. Therefore, the probability for $A$ to win the duel in this case is
$$
\begin{aligned}
& 0.5 \cdot 0.3 + 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3 + 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3 + \ldots = \\
& \quad = 0.15 \left(1 + 0.35 + 0.35^2 + \ldots \right) = 0.15 \frac{1}{1 - 0.35} = \frac{15}{100} \cdot \frac{100}{65} = \frac{3}{13}
\end{aligned}
$$
3) No one is hit. After this, $B$ will shoot at $C$ (as the more accurate of his opponents) and hit him. Then $A$ with probability 0.3 will hit $B$, winning the duel.
Thus, the most advantageous situation for shooter $A$ is when no one is hit after his shot. Therefore, he should shoot into the air the first time.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest are duralumin (heavier). How can you determine the number of duralumin cubes using no more than 11 weighings on a balance scale without weights?
Note. The problem assumes that all cubes can be aluminum, but they cannot all be duralumin (since if all the cubes were of the same weight, we would have no way to determine whether they are aluminum or duralumin without this condition).
|
14. Let's place one cube on each pan of the scales (first weighing). In this case, two different scenarios may occur.
$1^{\circ}$. During the first weighing, one pan of the scales tips. In this case, one of the two weighed cubes is definitely aluminum, and the other is duralumin. Next, we place these two cubes on one pan of the scales, and on the other pan, we sequentially place pairs of the remaining cubes (we randomly divide the 18 remaining cubes into 9 pairs). If any pair of cubes tips the scales, it means that both cubes in the second pair are duralumin; if the first pair tips the scales, it means that both cubes in the second pair are aluminum; if both pairs have the same weight, it means that the second pair also contains one aluminum and one duralumin cube. Thus, in case $1^{\circ}$, we can determine the number of duralumin cubes using 10 weighings (one weighing and 9 more).
$2^{\circ}$. During the first weighing, the scales remain in balance. In this case, the cubes of the first pair are either both aluminum or both duralumin. Next, we place these two cubes on one pan of the scales, and on the other pan, we sequentially place pairs of cubes from the remaining 18. Suppose the first $k$ of these pairs have the same weight as the initial pair, while the $(k+1)$-th pair has a different weight. (If $k=9$, all the cubes have the same weight, and thus there are no duralumin cubes at all; the case $k=0$ is no different from the general case). For definiteness, let's assume that the $(k+1)$-th pair is heavier than the initial pair (the reasoning would be similar if the $(k+1)$-th pair were lighter). In this case, the first two cubes, and therefore the cubes of the $k$ pairs that have the same weight as them, are definitely aluminum. Thus, we have performed $1+(k+1)=k+2$ weighings and identified $k+1$ pairs of aluminum cubes. Now, we place one cube from the last weighed pair on each pan of the scales (the $(k+3)$-th weighing). If both cubes have the same weight, they must both be duralumin; otherwise, one is aluminum and the other is duralumin. In both cases, after $k+3$ weighings, we can identify a pair of two cubes, one of which is aluminum and the other is duralumin. Using this pair, we can determine the number of duralumin cubes among the remaining $20-2(k+2)=16-2k$ cubes with $8-k$ weighings, similar to how we proceeded in case $1^{\circ}$. The total number of weighings in case $2^{\circ}$ will be $k+3+(8-k)=11$.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. a) One day, a guest came to the hotel owner, K, without money but with a silver chain consisting of seven links. The owner agreed to keep the guest for a week on the condition that the guest would give him one of the chain links as payment each day. What is the minimum number of links that need to be cut so that the chain owner can pay the hotel owner daily for seven days (possibly taking back previously given links and giving other ones in exchange)?
b) The chain consists of 2000 links. What is the minimum number of links that need to be cut so that any number of links from 1 to 2000 can be collected by taking some number of the resulting parts?
|
16. a) It is sufficient to saw off one third link; in this case, the chain will break into two parts, containing 2 and 4 links respectively, and one separate (sawed) link. On the first day, the guest will give this link; on the second day, he will take it back and give in exchange the part of the chain consisting of two links; on the third day, he will add the sawed link again; on the fourth day, he will take back everything he gave earlier and hand over the part of the chain consisting of four links; on the fifth day, he will add the sawed link again; on the sixth day, he will take back this link and give in exchange the part of the chain consisting of two links; on the seventh day, he will give the last link.
b) It is convenient to first solve the following problem: for what maximum $n$ is it sufficient to saw $k$ links of an $n$-link chain so that any number of links from 1 to $n$ can be obtained by taking some of the resulting parts? To solve this problem, let's consider the most advantageous arrangement of $k$ sawed links. Since after sawing $k$ links, we will have $k$ separate (sawed) links, we can already collect any number of links from 1 to $k$ from them. But we cannot get $k+1$ links if we do not have a part consisting of $k+1$ or fewer links. It is clear that it is most advantageous to have a part exactly of $k+1$ links; then from this part and $k$ separate links, we can collect any number from 1 to $2k+1$. To be able to get $2k+2=2(k+1)$ links, we need to have a part containing $2(k+1)$ or fewer links; the most advantageous for us will be if this part contains exactly $2(k+1)$ links. Now we can already form all numbers from 1 to $2k+1+2(k+1)=4k+3$; the next largest part we need is a part containing $4(k+1)$ links. Continuing to reason in the same way, we will see that the most advantageous will be if the $k+1$ parts, obtained after we saw $k$ links (the $k$ separate links obtained in this process are not counted here as parts), will have the following numbers of links:
$$
k+1, 2(k+1), 4(k+1), 8(k+1), \ldots, 2^k(k+1)
$$
In this case, any number of links from 1 to
$$
\begin{aligned}
n=k+\{k+1+2(k+1)+ & \left.4(k+1)+\ldots+2^k(k+1)\right\}= \\
& =k+\left(2^{k+1}-1\right)(k+1)=2^{k+1}(k+1)-1
\end{aligned}
$$
can be formed from the parts of the chain.
Thus, if $2^k k \leqslant n \leqslant 2^{k+1}(k+1)-1$, then $k$ cuts of the chain are sufficient, but $k-1$ cuts are not sufficient. In particular,
| for $-2 \leqslant n \leqslant 7$ | $k=1$, | for $160 \leqslant n \leqslant 383$ | $k=5$, |
| :--- | :--- | :--- | :--- |
| for $8 \leqslant n \leqslant 23$ | $k=2$, | for $384 \leqslant n \leqslant 895$ | $k=6$, |
| for $24 \leqslant n \leqslant 63$ | $k=3$, | for $896 \leqslant n \leqslant 2047$ | $k=7$. |
| for $64 \leqslant n \leqslant 159$ | $k=4$, | | |
Thus, we see that for $n=2000$, the minimum number of sawed links is 7. The conditions of the problem will be met if these links are chosen so that the 8 resulting parts of the chain (7 separate links are not counted here) have 8, 16, 32, 64, 128, 256, 512, and 977 links respectively.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
28. Two brothers sold a flock of sheep that belonged to both of them, taking as many rubles for each sheep as there were sheep in the flock. The money received was divided as follows: first, the elder brother took ten rubles from the total amount, then the younger brother took ten rubles, after that the elder brother took another ten rubles, and so on. In the end, the younger brother did not have enough ten rubles; therefore, he took all the small change left after the division, and in addition, to make the division fair, the elder brother gave the younger brother his pocket knife. What was the pocket knife valued at?
|
28. Let the number of sheep in the flock be denoted by $n$; in this case, the brothers received $n$ rubles for each sheep, and thus the total amount they received is $N=n \cdot n=n^{2}$ rubles. Let $d$ be the number of whole tens in the number $n$, and $e$ be the number of units; then $n=10 d+e$ and
$$
N=(10 d+e)^{2}=100 d^{2}+20 d e+e^{2}
$$
From the conditions of the division, it follows that the elder brother received one ten more than the younger brother, i.e., that the total sum $N$ contains an odd number of tens (and some remainder). But $100 d^{2}+20 d e=20 d(5 d+e)$ is divisible by 20 (contains an even number of tens); therefore, the number $e^{2}$ must contain an odd number of tens. Since $e$ is less than 10 ( $e$ is the remainder of the division of the number $n$ by 10 ), then $e^{2}$ can only have one of the following values:
$$
1,4,9,16,25,36,49,64,81
$$
But of these numbers, only 16 and 36 contain an odd number of tens; therefore, $e^{2}$ is equal to 16 or 36. Both these numbers end in 6; hence, the remainder that the younger brother received in place of the 10 rubles he was short is 6 rubles, and the elder received 4 rubles more than the younger. Therefore, for the division to be fair, the elder should still pay the younger 2 rubles. Thus, the pocketknife was valued at 2 rubles.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
51. What remainders can the hundredth power of an integer give when divided by 125?
|
51. Any integer either is divisible by 5 or can be represented in one of the following four forms: $5k+1, 5k+2, 5k-2$, or $5k-1$. If a number is divisible by 5, then its hundredth power is clearly divisible by $5^3 = 125$. Further, using the binomial theorem, we get:
$$
(5k \pm 1)^{100} = (5k)^{100} \pm \ldots + \frac{100 \cdot 99}{1 \cdot 2}(5k)^2 \pm 100 \cdot 5k + 1
$$
where all the omitted terms contain a factor of $5k$ raised to a power of at least 3, and thus are divisible by 125. Similarly,
$$
(5k \pm 2)^{100} = (5k)^{100} \pm \ldots + \frac{100 \cdot 99}{1 \cdot 2}(5k)^2 \cdot 2^{98} \pm 100 \cdot 5k \cdot 2^{99} + 2^{100}
$$
But $\frac{100 \cdot 99}{1 \cdot 2}(5k)^2 = 125 \cdot 990k^2$ and $100 \cdot 5k = 125 \cdot 4k$ are divisible by 125. As for the number $2^{100}$, it can be represented as
$$
(5-1)^{50} = 5^{50} - \ldots + \frac{50 \cdot 49}{1 \cdot 2} \cdot 5 - 50 \cdot 5 + 1
$$
from which it is clear that it gives a remainder of 1 when divided by 125.
Thus, the hundredth power of a number divisible by 5 is divisible by 125; the hundredth power of a number not divisible by 5 gives a remainder of 1 when divided by 125.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
54*. Let $N$ be an even number not divisible by 10. What will be the tens digit of the number $N^{20}$? What will be the hundreds digit of the number $N^{200}$?
|
54. Let's find the last two digits of the number \(N^{20}\). The number \(N^{20}\) is divisible by 4 (since \(N\) is even). Further, the number \(N\) is not divisible by 5 (otherwise it would be divisible by 10) and, therefore, can be represented in the form \(5k \pm 1\) or in the form \(5k \pm 2\) (see the solution to problem 51). But the number
\[
(5k \pm 1)^{20} = (5k)^{20} \pm 20(5k)^{19} + \ldots + \frac{20 \cdot 19}{1 \cdot 2}(5k)^{2} \pm 20 \cdot 5k + 1
\]
gives a remainder of 1 when divided by 25, and the number
\[
\begin{aligned}
&(5k \pm 2)^{20} = (5k)^{20} \pm 20(5k)^{19} \cdot 2 + \ldots \\
& \ldots + \frac{20 \cdot 19}{1 \cdot 2}(5k)^{2} \cdot 2^{18} \pm 20 \cdot 5k \cdot 2^{19} + 2^{20}
\end{aligned}
\]
gives the same remainder when divided by 25 as the number \(2^{20} = (2^{10})^2 = (1024)^2 = (1025-1)^2\), i.e., also a remainder of 1. From the fact that the number \(N^{20}\) gives a remainder of 1 when divided by 25, it follows that the last two digits of this number can only be 01, 26, 51, or 76. Considering that \(N^{20}\) must be divisible by 4, we get that the last two digits of this number can only be 76. Therefore, the tens digit of the number \(N^{20}\) is 7.
Now let's find the last three digits of the number \(N^{200}\). The number \(N^{200}\) is divisible by 8. Further, since \(N\) is coprime with 5, \(N^{100}\) gives a remainder of 1 when divided by 125 (see the solution to problem 51): \(N^{100} = 125k + 1\). But then \(N^{200} = (125k + 1)^2 = 125^2 k^2 + 250k + 1\) also gives a remainder of 1 when divided by 125. Therefore, \(N^{200}\) can end in the digits 126, 251, 376, 501, 626, 751, or 876; but since \(N^{200}\) is divisible by 8, it must end in the digits 376. Therefore, the hundreds digit of the number \(N^{200}\) is 3.
Note. It is easy to see that the number \(N^{100}\) must already end in the three digits 376.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
57. The number $123456789(10)(11)(12)(13)(14)$ is written in the base-15 numeral system, i.e., this number is equal to
(14) $+(13) \cdot 15+(12) \cdot 15^{2}+(11) \cdot 15^{3}+\ldots+2 \cdot 15^{12}+15^{13}$. What remainder does it give when divided by 7?
|
57. The number 15 gives a remainder of 1 when divided by 7. Therefore, it follows that
$$
15^{2}=(7 \cdot 2+1) \cdot(7 \cdot 2+1)=7 n_{1}+1
$$
gives a remainder of 1 when divided by 7,
$$
15^{3}=15^{2} \cdot 15=\left(7 n_{1}+1\right) \cdot(7 \cdot 2+1)=7 n_{2}+1
$$
gives a remainder of 1 when divided by 7, and generally, any power of the number 15 gives a remainder of 1 when divided by 7. If we now subtract from the given number the sum \(1+2+3+4+\ldots+14=105\), then, by grouping the terms appropriately, we get:
$$
\begin{aligned}
& 13(15-1)+12\left(15^{2}-1\right)+ \\
& \quad+11\left(15^{3}-1\right)+\ldots+2\left(15^{12}-1\right)+1\left(15^{13}-1\right)
\end{aligned}
$$
i.e., a number that is divisible by 7. But from the fact that the difference between the given number and the number \(105=7 \cdot 15\) is divisible by 7, it follows that the original number is also divisible by 7.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
65. Find the remainder when the number
$$
10^{10}+10^{\left(10^{0}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7.
|
65. First of all, note that \(10^{6}-1=999999\) is divisible by 7 (since \(999999=7 \cdot 142857\)). From this, it easily follows that \(10^{\mathrm{N}}\), where \(N\) is any integer, gives the same remainder when divided by 7 as \(10^{r}\), where \(r\) is the remainder from dividing \(N\) by 6. Indeed, if \(N=6k+r\), then
\[
\begin{aligned}
& 10^{k}-10^{r}=10^{6k+r}-10^{r}=10^{r}\left(10^{6k}-1\right)= \\
& =10^{r} \cdot \left(10^{6}-1\right)\left(10^{6k-6}+10^{6k-12} \ldots+10^{6}+1\right)
\end{aligned}
\]
is divisible by 7.
But any integer power of 10 gives a remainder of 4 when divided by 6; indeed, \(10^{n}-4=\underset{(n-1) \text{ times }}{999 \ldots 9}\) is always divisible by \(2 \cdot 3=6\) (due to the divisibility rules for 2 and 3). Thus, all the complex exponents of the terms in our sum give a remainder of 4 when divided by 6. Consequently, each of these 10 terms gives the same remainder when divided by 7 as the number \(10^{4}\), and the entire sum gives the same remainder as the number
\[
\begin{aligned}
10^{4}+10^{4}+10^{4}+10^{4}+10^{4}+10^{4}+10^{4} & +10^{4}+10^{4}+10^{4}= \\
& =10^{5}=100000=7 \cdot 14285+5
\end{aligned}
\]
Thus, the desired remainder is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
68. a) What is the last digit of the number
$$
\left(\ldots\left(\left(\left(7^{7}\right)^{7}\right)^{7}\right)^{\ldots 7}\right)
$$
(raising to the power of 7 is repeated 1000 times)? What are the last two digits of this number?
b) What is the last digit of the number
$$
7\left(7^{\left(.7^{\left(7^{7}\right)}\right)} \ldots\right)
$$
written using 1001 sevens, similar to the number in problem a), but with a different order of exponentiation? What are the last two digits of this number?
|
68. a) If you multiply two numbers, one of which ends in the digit $a$, and the second in the digit $b$, then their product will end in the same digit as the product $a b$. This observation allows us to easily solve the given problem. We will sequentially raise to powers, keeping track only of the last digit of the number: $7^{2}$ ends in the digit $9, 7^{3} = 7^{2} \cdot 7$ ends in the digit $3, 7^{4} = 7^{3} \cdot 7$ ends in the digit 1, and $7^{7} = 7^{4} \cdot 7^{3}$ ends in the digit 3.
Next, in the same way, we find that $\left(7^{7}\right)^{7}$ ends again in the digit 7 (indeed, $\left(7^{7}\right)^{2}$ ends in the digit $9, \left(7^{7}\right)^{3}$ ends in the digit $7, \left(7^{7}\right)^{4}$ ends in the digit 1, and $\left(\left(7^{7}\right)^{7}\right)$ ends in the digit 7). From this, it follows that the number $\left(\left(7^{\prime}\right)^{1}\right)^{\top}$ ends in the same digit as the number $7^{7}$, i.e., the digit 3, the number $\left.\left(\left(7^{7}\right)^{7}\right)^{7}\right)^{7}$ ends again in the digit 7, and so on. Continuing in the same way, after an odd number of exponentiations by 7, we will always arrive at a number ending in the digit 3, and after an even number of exponentiations by 7, a number ending in the digit 7. Since 1000 is an even number, the number of interest to us ends in the digit 7.
If one number ends in a two-digit number $A$, and the second in a two-digit number $B$, then their product ends in the same two digits as the product $A \cdot B$. This allows us to find the last two digits of the number of interest to us. As before, we check that $7^{7}$ ends in the two digits 43, and $\left(7^{7}\right)^{7}$ ends in the same digits as $43^{7}$, namely the digits 07. From this, it follows that, raising the numbers 7, 77, $\left(7^{7}\right)^{7}, \ldots$ sequentially to the power of 7, we will after an odd number of exponentiations end up with a number ending in the digits 43, and after an even number of exponentiations, a number ending in the digits 07. Therefore, the number of interest to us ends in the digits 07.
b) In the solution to part a), we saw that $7^{4}$ ends in the digit 1. From this, it follows that $7^{4 k} = \left(7^{4}\right)^{k}$ also ends in the digit 1 and $7^{4 k + l}$, where $l$ is one of the numbers $0, 1, 2$ or 3, ends in the same digit as $7^{l} \left(7^{4 k + l} = 7^{4 k} \cdot 7^{l}\right)$. Thus, the problem reduces to determining the remainder when the exponent, to which 7 is raised to obtain the number in the problem, is divided by 4.
The exponent to which 7 is raised in the problem itself is 7 raised to a very large power; we need to determine the remainder when this power of 7 is divided by 4. But $7 = 8 - 1$; from this, it follows that $7^{2} = (8 - 1) \cdot (8 - 1)$ gives a remainder of 1 when divided by 4, $7^{3} = 7^{2} \cdot (9 - 1)$ gives a remainder of -1 (or, equivalently, a remainder of 3) when divided by 4, and generally, every even power of 7 gives a remainder of 1 when divided by 4, and every odd power gives a remainder of -1 (i.e., +3). But the power of 7 of interest to us in this case is necessarily an odd number (since it itself is a power of 7), and therefore, the number in the problem has the form $7^{4 k + 3}$ and, consequently, ends in the same digit as $7^{3}$, i.e., the digit 3.
Since $7^{4}$ ends in the digits 01, $7^{4 k + l}$ ends in the same two digits as $7^{l}$. Therefore, the number of interest to us ends in the same two digits as the number $7^{3}$, i.e., the digits 43.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
70. For which natural numbers $n$ is the sum $5^{n}+n^{5}$ divisible by 13? What is the smallest $n$ that satisfies this condition?
|
70. First, let's find the remainders of the division by 13 of the numbers \(5^n\) and \(n^5\) for the first few values of \(n = 0, 1, 2, \ldots\). It is more convenient to start with the numbers \(5^n\):
\(n\)
| 0 | 1 | 2 | 3 | 4 | . | . | . |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 1 | 5 | 25 | 125 | 625 | . | . | . |
| 1 | 5 | -1 | -5 | 1 | . | . | . |
[Here we write the remainder -1 instead of "remainder 12" and "remainder -3" instead of "remainder 8", which allows us to easily find all subsequent remainders: if \(5^n\) gives a remainder of -1 when divided by 13, i.e., \(5^n = 13k - 1\), where \(k\) is an integer, then \(5^{n+1} = 5^n \cdot 5 = (13k - 1) \cdot 5 = 13(5k) - 5\) gives a remainder of -5 when divided by 13; similarly, if \(5^m\) gives a remainder of -5 when divided by 13, i.e., \(5^m = 13l - 5\), then \(5^{m+1} = 5^m \cdot 5 = (13l - 5) \cdot 5 = 13(5l) - 25 = 13(5l - 2) + 1\) gives a remainder of 1 when divided by 13.] There is no need to continue the table further: since \(5^4 = 13q + 1\), then \(5^5 = 5^4 \cdot 5 = (13q + 1) \cdot 5\) gives the same remainder as the number 5 (remainder 5) when divided by 13; the number \(5^6 = 5^4 \cdot 5^2 = (13q + 1) \cdot 5^2\) gives the same remainder as the number \(5^2\) (remainder -1) when divided by 13, and so on; thus, in our "sequence of remainders," the numbers \(1, 5, -1, -5\) alternate sequentially.
Similarly, we can create a table of remainders from the division by 13 of the numbers \(n^5\), where \(n = 0, 1, 2, \ldots\), and so on. But the number \((13p + r)^5 = \frac{(13p + r)(13p + r) \ldots (13p + r)}{5 \text{ factors}}\) gives the same remainder as the number \(r^5\) when divided by 13; therefore, it is sufficient to limit ourselves to the values \(n = 0, 1, 2, 3, \ldots, 12\). Further, if the number \(n\) is equal to (or gives a remainder of) \(s\) when divided by 13, and the number \(n^2\) gives a remainder of \(t\) when divided by 13, then the number \(n^5 = n^2 \cdot n^2 \cdot n\) gives the same remainder as the product \(t \cdot t \cdot s\) when divided by 13—this fact can simplify the creation of the required table for the values \(n = 4, 5, 6\). Finally, note that if the number \(n^5\) gives a remainder of \(u\) when divided by 13, then the number \((13 - n)^5 = \underbrace{(13 - n)(13 - n) \ldots (13 - n)}\) gives the same remainder as the number \((-n)^5\), i.e., the remainder -u, or \(13 - u\).
Now we can create our table:
(The remainders corresponding to the values \(n = 7, 8, \ldots, 12\) are written based on the fact that the numbers \(n^5\) and \((13 - n)^5\) give "complementary" remainders \(u\) and \(-u\).) After the value \(n = 12\) in our table, the same remainders \(0, 1, 6, -4, -3, 5, 2, -2, -5, 3, 4, -6, -1\) will "periodically" repeat.
Since the first table has a "period" of 4, and the second has a "period" of 13, the value \(4 \cdot 13 = 52\) is the "period" of the combination of both tables: increasing \(n\) by 52 (or any multiple of 52) does not change the remainders of the division by 13 of both numbers \(5^n\) and \(n^5\). Further, we are only interested in those columns of the second table that correspond to remainders \(\pm 1\) and \(\pm 5\) (since in the first table, only the remainders \(1, 5, -1, -5\) alternate). But within the range from \(n = 0\) to \(n = 51\), the remainder 1 in the second table corresponds to the values \(n = 1, 1 + 13 = 14\), \(1 + 2 \cdot 13 = 27\), and \(1 + 3 \cdot 13 = 40\); of these four numbers, only 14 has the form \(4x + 2\), which in the first table corresponds to the remainder -1; thus, the number \(n = 14\) satisfies the required condition \((5^{14} + 14^5\) is divisible by 13\). Similarly, the remainder -1 in the second table within the same range corresponds to the values \(n = 12, 12 + 13 = 25\), \(12 + 2 \cdot 13 = 38\), and \(12 + 3 \cdot 13 = 51\); of these numbers, only 12 has the form \(4y\), which in the first table leads to the remainder 1. Similarly, the remainder 5 in the second table corresponds to the values \(n = 5, 5 + 13 = 18\), \(5 + 2 \cdot 13 = 31\), and \(5 + 3 \cdot 13 = 44\), and the remainder -5 to the values \(n = 8, 8 + 13 = 21\), \(8 + 2 \cdot 13 = 34\), and \(8 + 3 \cdot 13 = 47\); but of the first four numbers, only 31 has the form \(4z + 3\), ensuring a remainder of -5 from the division of \(5^{31}\) by 13, and of the last four numbers, only 21 has the form \(4\tau + 1\), ensuring a remainder of 5 from the division of \(5^{21}\) by 13. Thus, within the range \(0 \leq n \leq 52\), the required condition is satisfied by the following four natural numbers: \(n = 12, 14\), 21, and 31; the entire set of natural numbers satisfying the condition of the problem consists of the following four "series" of numbers:
\[
\begin{gathered}
n = 52m + 12, \quad n = 52m + 14 \text{ (i.e., } n = 26(2m) + 12 \quad n \\
= 26(2m + 1) - 12)
\end{gathered}
\]
\[
n = 52m + 21 \quad \text{and} \quad n = 52m + 31 \text{ (i.e., } n = 52m \pm 21 \text{);}
\]
here \(m = 0, 1, 2, \ldots\) (and in the formula \(n = 52m - 21\) we must have \(m > 0\)).
Clearly, the smallest \(n\) that satisfies the conditions of the problem is \(n = 12\).
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
81. The square of an integer ends with four identical digits. Which ones?
|
81. A perfect square can only end in the digits 0, 1, 4, 9, 6, and 5. Furthermore, the square of any even number is clearly divisible by 4, while the square of any odd number gives a remainder of 1 when divided by 4 \(\left((2 k)^{2}=4 k^{2},(2 k+1)^{2}=4\left(k^{2}+k\right)+1\right)\); therefore, the square of no number can end in the digits 11, 99, 66, or 55 (numbers ending in two digits 11, 99, 66, or 55 give remainders of 3, 3, 2, and 3, respectively, when divided by 4). Finally, let's consider what remainders the square of an integer can give when divided by 16. Any integer can be represented in one of the following forms:
$$
8 k, 8 k \pm 1,8 k \pm 2,8 k \pm 3 \text { or } 8 k+4 \text {; }
$$
the squares of these numbers are of the form
$$
\begin{gathered}
16 \cdot\left(4 k^{2}\right), \quad 16\left(4 k^{2} \pm k\right)+1, \quad 16\left(4 k^{2} \pm 2 k\right)+4 \\
16\left(4 k^{2} \pm 3 k\right)+9 \quad \text { or } \quad 16\left(4 k^{2}+4 k+1\right) .
\end{gathered}
$$
Thus, we see that the square of an integer is either divisible by 16 or gives a remainder of 1, 4, or 9 when divided by 16. A number ending in the digits 4444 gives a remainder of 12 when divided by 16 and, therefore, cannot be a perfect square.
Therefore, if a perfect square ends in four identical digits, these digits must be four zeros (for example, \(100^{2}=10000\)).
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
105. All integers are written in a row, starting from one. Determine which digit stands at the $206788-\mathrm{th}$ place.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
105. There are nine single-digit numbers, $99-9=90$ two-digit numbers, $999-99=900$ three-digit numbers, and generally $9 \cdot 10^{n-1}$ $n$-digit numbers.
Single-digit numbers will occupy nine places in the sequence we have written, two-digit numbers $90 \cdot 2=180$ places, three-digit numbers $900 \cdot 3=2700$ places, four-digit numbers $9000 \cdot 4=36000$ places, and five-digit numbers $90000 \cdot 5=$ $\Leftarrow 450000$ places. From this, it is clear that the digit we are interested in will belong to a five-digit number.
Digits belonging to numbers with no more than four digits will have numbers from 1 to $9+180+2700+36000=38889$. To find out how many five-digit numbers will fit in the interval from the 38889th place to the 206788th, we need to divide the difference $206788-38889=167899$ by 5 (division with remainder):
$$
206788-38889=5 \cdot 33579+4
$$
Thus, the digit we are looking for will belong to the 33579th five-digit number, i.e., the number 43579 (since the first five-digit number is 10000). In this number, the digit we are interested in is in the 4th place. Therefore, the digit we are looking for is 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
107. Each of the integers from one to a billion inclusive is replaced by the sum of the digits of the number; single-digit numbers, of course, do not change, while the others decrease. Then each of the newly obtained numbers is again replaced by the sum of its digits, and so on until a billion single-digit numbers are obtained. Which numbers will be more among the resulting numbers - ones or twos?
|
107. It is a good fact to know that every positive integer $N$ gives the same remainder when divided by 9 as the sum of its digits. (This follows from the fact that the digit $a_{k}$ in the $(k+1)$-th position from the end in the decimal representation of $N$ represents the term $a_{k} \cdot 10^{k}$ in the expansion of $N$ in powers of ten; but $a_{k} \cdot 10^{k} = a_{k} \cdot (99 \ldots 9 + 1) = 99 \ldots 9 a_{h} + a_{h}$ gives the same remainder when divided by 9 as $a_{h}$.) From this, it follows that if the number $N$ gives a remainder $r \neq 0$ when divided by 9, then by repeatedly replacing the number with the sum of its digits, we will eventually arrive at a single-digit number $r$, and if $N$ is divisible by 9, we will arrive at the number 9. Thus, among our billion single-digit numbers, the ones will correspond to the numbers $1, 10, 19, 28, \ldots, 1000000000$, which give a remainder of 1 when divided by 9, and the twos will correspond to the numbers $2, 11, 20, 29, \ldots, 999999992$, which give a remainder of 2 when divided by 9. But the first set of numbers is obviously one more than the second set; therefore, we will get one more one than two.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
153. a) Find all three-digit numbers that are equal to the sum of the factorials of their digits.
b) Find all integers that are equal to the sum of the squares of their digits.
|
153. a) Let the hundreds, tens, and units digits of the desired number \( N \) be denoted by \( x, y \), and \( z \), respectively, so that \( N = 100x + 10y + z \). In this case, the condition of the problem gives
\[
100x + 10y + z = x! + y! + z!
\]
Note that \( 7! = 5040 \) is a four-digit number; therefore, no digit of the number \( N \) can exceed 6. Consequently, the number \( N \) itself does not exceed 700, from which it follows that no digit of \( N \) can exceed 5 (since \( 6! = 720 > 700 \)). At least one digit of the number \( N \) is 5, because even \( 3 \cdot 4! = 72 < 100 \), and \( N \) is a three-digit number. The digit \( x \) cannot be 5, since even \( 3 \cdot 5! = 360 < 500 \). It follows from this that \( x \) does not exceed 3. Furthermore, we can assert that \( x \) does not exceed 2, since even \( 3! + 2 \cdot 5! = 246 < 300 \). But the number 255 does not satisfy the condition of the problem, and if only one digit of the desired number is 5, then \( x \) cannot be greater than 1, because even \( 2! + 5! + 4! = 146 < 200 \). Moreover, since \( 1! + 5! + 4! = 145 < 150 \), we conclude that \( y \) cannot exceed 4; therefore, \( z = 5 \), since at least one digit of the number \( N \) must be 5. Thus, we have \( x = 1 \), \( 4 \geqslant y \geqslant 0 \), and \( z = 5 \); this allows us to easily find the only solution to the problem: \( N = 145 \).
b) The desired number \( N \) cannot have more than three digits, since even \( 4 \cdot 9^2 = 324 \) is a three-digit number. Therefore, we can write \( N = 100x + 10y + z \), where \( x, y, z \) are the digits of the number; here, \( x \) or even \( x \) and \( y \) simultaneously can be zero.
From the condition of the problem, we have \( 100x + 10y + z = x^2 + y^2 + z^2 \), from which
\[
(100 - x)x + (10 - y)y = z(z - 1)
\]
From the last equation, it follows first of all that \( x = 0 \), otherwise the number on the left side of the equation is not less than \( 90 \) (since \( x \geqslant 1 \), \( 100 - x \geqslant 90 \), \( (10 - y)y \geqslant 0 \)), and the number on the right side is not more than \( 9 \cdot 8 = 72 \) (since \( z \leqslant 9 \)). Therefore, the equation (") has the form \( (10 - y)y = z(z - 1) \). It is easy to check that this last equation is not satisfied for any positive integers \( x \) and \( y \) not exceeding 9, unless \( y \neq 0 \). If \( y = 0 \), then we have the only obvious solution \( z = 1 \).
Thus, the condition of the problem is satisfied by the only number \( N = 1 \).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
169. In a chess tournament, two 7th-grade students and a certain number of 8th-grade students participated. Each participant played one game with every other participant. The two 7th-graders together scored 8 points, and all the 8th-graders scored the same number of points (in the tournament, each participant earns 1 point for a win and $1 / 2$ point for a draw). How many 8th-graders participated in the tournament?
|
169. Let $n$ be the number of eighth-graders, and $m$ be the number of points earned by each of them. In this case, the number of points scored by all participants in the tournament is $m n + 8$. This number is equal to the number of games played. Since the number of participants in the tournament is $n + 2$ and each played once with each of the $n + 1$ others, they played a total of $\frac{(n+2)(n+1)}{2}$ games (in the product $(n+2)(n+1)$, each game is counted twice). Thus, we have:
$$
m n + 8 = \frac{(n+2)(n+1)}{2}
$$
or, after some transformations,
$$
n(n + 3 - 2m) = 14
$$
Here, $n$ is an integer; the number in parentheses is also an integer (since $m$ is either an integer or a fraction with a denominator of 2).
Since $n$ is a divisor of 14, $n$ can be one of the numbers $1, 2, 7, 14$. The values $n = 1$ and $n = 2$ should be discarded, as in these cases, the total number of participants does not exceed 4, and two seventh-graders could not have scored a total of 8 points.
This leaves $n = 7$ and $n = 14$.
If $n = 7$, then $7(7 + 3 - 2m) = 14; m = 4$.
If $n = 14$, then $14(14 + 3 - 2m) = 14; m = 8$.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
256. Let all numbers $a_{1}, a_{2}, \ldots, a_{n}$ (where $n \geqslant 2$) be positive; how many real solutions does the system of equations have:
$$
x_{1} x_{2}=a_{1}, x_{2} x_{2}=a_{2}, \ldots, x_{n-1} x_{n}=a_{n-1}, x_{n} x_{1}=a_{n} ?
$$
|
256. Let's consider two cases separately.
$1^{\circ} . n$ is even. By multiplying the "odd" (i.e., $1$-st, $3$-rd, ..., $(n-1)$-th) and "even" equations of our system, we get:
$$
x_{1} x_{2} x_{3} \ldots x_{n}=a_{1} a_{3} a_{5} \ldots a_{n-1} \text { and } x_{1} x_{2} x_{3} \ldots x_{n}=a_{2} a_{4} a_{6} \ldots a_{n}
$$
from which it is clear that if $a_{1} a_{3} a_{5} \ldots a_{n-1} \neq a_{2} a_{4} a_{6} \ldots a_{n}$, the system has no solutions. If, however, $a_{1} a_{3} a_{5} \ldots a_{n-1}=a_{2} a_{4} a_{6} \ldots a_{n}$, then by taking any value for $x_{1}$ (certainly different from 0), we can subsequently find from the 1-st, 2-nd, ..., $(n-1)$-th equations of the system:
$$
x_{2}=\frac{a_{1}}{x_{1}}, \quad x_{3}=\frac{a_{2}}{x_{2}}, \ldots, \quad x_{n}=\frac{a_{n-1}}{x_{n-1}}
$$
Substituting all these values into the last equation shows that it is also satisfied.
$2^{\circ} . n$ is odd. Dividing the product of the "odd" (1-st, 3-rd, ..., $n$-th) equations by the product of the "even" equations, we get:
$$
x_{1}^{2}=\frac{a_{1} a_{3} a_{5} \ldots a_{n}}{a_{2} a_{4} \ldots a_{n-1}}, \quad \text { hence } x_{1}= \pm \sqrt{\frac{a_{1} a_{3} a_{5} \ldots a_{n}}{a_{2} a_{4} a_{6} \ldots a_{n-1}}}
$$
(we recall that all $a_{i}$ are positive). Then from the 1-st, 2-nd, ..., $(n-1)$-th equations, we sequentially find:
$$
x_{2}=\frac{a_{1}}{x_{1}}, \quad x_{3}=\frac{a_{2}}{x_{2}}, \ldots, \quad x_{n}=\frac{a_{n-1}}{x_{n-1}}
$$
Verification shows that, due to (*), the last equation of the system is also satisfied.
Answer: 0, if $n$ is even and $a_{1} a_{3} \ldots a_{n-1} \neq a_{2} a_{4} \ldots a_{n}$; infinitely many, if $n$ is even and $a_{1} a_{3} \ldots a_{n-1}=a_{2} a_{4} \ldots a_{n}$; 2, if $n$ is odd.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
299. When dividing the polynomial $x^{1051}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$, a quotient and a remainder are obtained. Find the coefficient of $x^{14}$ in the quotient.
|
299. The polynomial $x^{4}+x^{8}+2 x^{2}+x+1$ can be factored; it equals $\left(x^{2}+1\right)\left(x^{2}+x+1\right)$. From this, it is easy to see that this polynomial is a divisor of the polynomial
$$
x^{12}-1=\left(x^{6}-1\right)\left(x^{6}+1\right)=\left(x^{3}-1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)
$$
specifically:
$$
\begin{aligned}
x^{4}+x^{3}+2 x^{2}+x+1= & \frac{x^{12}-1}{(x-1)\left(x^{3}+1\right)\left(x^{4}-x^{2}+1\right)}= \\
& =\frac{x^{12}-1}{x^{3}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1}
\end{aligned}
$$
Dividing $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$ is the same as dividing $x^{1951}-1$ by $x^{12}-1$, and then multiplying the result by $x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1$. But it is easy to see that
$$
\frac{x^{1951}-1}{x^{12}-1}=x^{1939}+x^{1927}+x^{1915}+x^{1903}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}
$$
(it is easy to verify this by performing polynomial division or by noting that $x^{1951}-1=x^{7}\left[\left(x^{12}\right)^{162}-1\right]+x^{7}-1$ and using the known formula for dividing the difference of even powers of two monomials by the difference of their bases). From this, it follows that the desired coefficient coincides with the coefficient of $x^{14}$ in the product
$$
\begin{aligned}
\left(x^{1939}+x^{1927}+\ldots+x^{31}+\right. & \left.x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}\right) \times \\
& \times\left(x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1\right)
\end{aligned}
$$
which is equal to -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
53. Prove that for any convex polyhedron with a sufficiently large number of faces, at least one face has no fewer than six neighbors. Find the minimum number $N$ such that for any convex polyhedron with a number of faces not less than $N$, this statement is true.
|
53. Let a certain convex polyhedron $W$ have the number of neighbors of each face less than six, i.e., not exceeding five.
It is easy to see that in each vertex of such a polyhedron, no more than five edges (and thus faces) can meet. To show this, consider any face of the polyhedral angle. This face has at least three sides. Two of these sides, adjacent to the vertex of the considered angle, are adjacent to the faces of this angle, and the others are adjacent to other faces. If six or more edges met at the vertex of the polyhedral angle, then with our face, there would be five or more faces of the angle and at least one more face, i.e.,
the number of its neighbors would be at least six, which contradicts the made assumption. We will show that if in some vertices of the polyhedron $W$ more than three edges meet, then $W$ can be replaced by a polyhedron $W_{1}$, in which exactly three edges meet at each vertex and the number of neighbors of each face is also no more than five. For this, we will cut off all vertices where four or more edges meet with planes, so that the cutting planes intersect only the faces of the corresponding angle (Fig. 108). In this process, new faces will appear, which, according to the above remark, will have four or five sides. Since all newly formed polyhedral angles will be trihedral, the new faces will have no more than five neighbors. The number of neighbors of each face can only decrease. Indeed, if in one of the vertices of a face of the polyhedron $W$ four or five faces met, then only two of them could share a common edge with this face, consequently, at least one face from the number of faces meeting at this vertex was adjacent to the first face only at the vertex. After cutting, this face (or several such faces) ceased to be a neighbor, while one (and only one) new face became adjacent. Thus, the number of neighbors of the old faces did not increase.
Thus, we have transitioned from the polyhedron $W$ to the polyhedron $W_{1}$, where the number of neighbors of each face did not increase, the number of faces only increased, and in each vertex of $W_{1}$, three edges meet.
Let $\mathscr{B}$, $\mathscr{P}$, and $\mathscr{F}$ be the number of vertices, edges, and faces of the polyhedron $W_{1}$. Since three edges meet at each vertex and each edge has ends in two vertices, we have
$$
\mathscr{P} = \frac{3}{2} \mathscr{B}
$$
By assumption, each face of the polyhedron $W$ had no more than five neighbors, so each face of $W_{1}$ also has no more than five neighbors and, consequently, no more than five vertices. Therefore, the sum of the number of vertices of all faces does not exceed $5 \mathscr{F}$. Since each vertex of the polyhedron is counted three times, we have the inequality
$$
\mathscr{B} \leqslant \frac{5}{3} \mathscr{F}
$$
Now, let's use Euler's theorem, which states that $\mathscr{B} - \mathscr{P} + \mathscr{F} = 2$. Considering the relations (*) and (**), we get sequentially $\mathscr{B} - \frac{3}{2} \mathscr{B} + \mathscr{F} = 2$ or $\mathscr{F} = \frac{\mathscr{B}}{2} + 2$ and, further, $\mathscr{F} \leqslant \frac{5}{6} \mathscr{B} + 2$, from which
$$
\mathscr{F} \leqslant 12 \text{. }
$$
Thus, the number of faces of the polyhedron $W_{1}$, and therefore of $W$, does not exceed 12, i.e., any polyhedron, each face of which has no more than five neighbors, contains no more than 12 faces. At the same time, a dodecahedron has exactly 12 faces, and each of them has five neighbors. Therefore, the desired number $N$ is 12.
|
12
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
85. The diagonals of a convex 17-gon, drawn from one vertex, divide it into 15 triangles. Can a convex 17-gon be cut into 14 triangles?
What about a non-convex 17-gon? What is the smallest number of triangles into which a 17-gon can be cut?
|
85. Each of the angles of a convex polygon is less than $2 d$, so the vertices cannot lie on the sides of the triangles of the partition. When a convex polygon is divided into triangles (not necessarily by diagonals), all the angles of the polygon will be divided into parts, which will
serve as angles of the triangles of the partition (Fig. 154). It follows that the sum of the angles of all the triangles of the partition must be no less than the sum of the angles of the polygon. But the sum of the angles of 14 triangles is $28 d$, while the sum of the angles of a 17-sided polygon is $30 d$. Therefore, a convex 17-sided polygon cannot be divided into 14 triangles.
It is easy to construct a non-convex 17-sided polygon that can be divided into six triangles (see, for example, Fig. 155, a, σ).
We will now show that no 17-sided polygon (convex or non-convex) can be cut into five triangles (if the 17-sided polygon does not have angles of $180^{\circ}$ ).
Fig. 155.
Indeed, each vertex of the 17-sided polygon either coincides with the vertices of several (possibly one) triangles of the partition, or lies on the side of a triangle of the partition; in the latter case, the corresponding angle of the 17-sided polygon is obviously greater than $180^{\circ}$. Therefore, the remaining part of this angle must be filled by at least one more triangle of the partition; hence, in this case, the vertex also coincides with the vertex of at least one triangle. It follows that the total number of vertices of all the triangles of the partition is no less than the number of vertices of the 17-sided polygon, i.e., 17.
Therefore, if $T$ is the number of triangles in the partition, then
$$
3 T \geqslant 17
$$
from which
$$
T \geqslant 5 \frac{2}{3}
$$
Since $T$ is an integer, $T$ cannot be less than six.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
92. Cut a rectangular parallelepiped with side ratios $8: 8: 27$ into four parts, from which a cube can be assembled.
|
92. The given parallelepiped is cut into two equal stepped parts, as shown in Fig. 165, $a$ (the height of the step is 9, and the width is 4). From these parts, we form a new
Fig. 165.
parallelepiped with the ratio of sides 12:8:18 (Fig. 165, b). This parallelepiped is then cut into stepped bodies, as shown in Fig. 165, c (the height of the step is 6, and the width is 4). From such parts, we form a cube with a side of 12 (Fig. 165, d). In Fig. 165, $\partial$ the four parts into which the original parallelepiped splits during such cutting are shown separately.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
71. A segment $AB$ of unit length, which is a chord of a sphere with radius 1, is positioned at an angle of $\pi / 3$ to the diameter $CD$ of this sphere. The distance from the end $C$ of the diameter to the nearest end $A$ of the chord $AB$ is $\sqrt{2}$. Determine the length of the segment $BD$.
|
71. Draw a line through $C$ parallel to $A B$, and take a point $E$ on it such that $|C E|=|A B|$, making $A B E C$ a parallelogram. If $O$ is the center of the sphere, then since $\widehat{O C E}=\pi / 3$ and $|C E|=1$ (as follows from the condition), $\triangle O C E$ is equilateral. Therefore, point $O$ is equidistant from all vertices of the parallelogram $A B E C$. This implies that $A B E C$ is a rectangle, the projection of $O$ onto the plane $A B E C$ is the point $K$ - the center of $A B E C$, and $|B D|=2|O K|=$ $=2 \sqrt{|O C|^{2}-\frac{1}{4}|B C|^{2}}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
104. The centers of three spheres, with radii of 3, 4, and 6, are located at the vertices of an equilateral triangle with a side length of 11. How many planes exist that are tangent to all three spheres simultaneously?
|
104. Any tangent plane divides space into two parts, and either all three spheres are located on one side, or two are on one side and one on the other. It is obvious that if a certain plane is tangent to the spheres, then the plane symmetric to it relative to the plane passing through the centers of the spheres is also tangent. We will show that there is no plane tangent to the given spheres such that the spheres of radii 3 and 4 are on one side of it, and the sphere of radius 6 is on the other.
Let the centers of the spheres of radii 3, 4, and 6 be at points $A, B$, and $C$. A plane tangent to the given spheres in the manner described above will divide the sides $A C$ and $B C$ in the ratios $1: 2$ and $2: 3$, i.e., it will pass through points $K$ and $L$ on $A C$ and $B C$ such that $|C K|=22 / 3$, $|C L|=33 / 5$. It is not difficult to find the distance from $C$ to $K L$. It is equal to $33 \sqrt{3 / 91}<6$. Therefore, it follows that a plane tangent to the sphere of radius 6 with center at $C$ cannot be drawn through $K L$. It can be shown that all other tangent planes exist, and there will be 6 of them in total.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
108. The volume of the tetrahedron \(ABCD\) is 5. A plane is drawn through the midpoints of the edges \(AD\) and \(BC\), intersecting the edge \(CD\) at point \(M\). The ratio of the length of segment \(DM\) to the length of segment \(CM\) is \(2/3\). Calculate the area of the section of the tetrahedron by the specified plane, given that the distance from this plane to vertex \(A\) is 1.
|
108. Let $K$ and $L$ be the midpoints of edges $AD$ and $BC$, and let $N$ and $P$ be the points of intersection of the plane with lines $AB$ and $AC$ (Fig. 16). We need to find the ratios $\frac{|PA|}{|PC|}$ and $\frac{|PK|}{|PM|}$. Draw $KQ$ and $AR$ parallel to $DC$, with $Q$ being the midpoint of $AC$.
$$
|AR|=|DM|, \quad \frac{|PA|}{|PC|}=\frac{|AR|}{|MC|}=\frac{|DM|}{|MC|}=\frac{2}{3}
$$
$$
\frac{|PK|}{|PM|}=\frac{|KQ|}{|MC|}=\frac{|DC|}{2|MC|}=\frac{5}{6}
$$
Next, we find
$$
\begin{gathered}
\frac{|AN|}{|NB|}=\frac{2}{3}, \quad \frac{|PN|}{|PL|}=\frac{4}{5} \\
\frac{V_{PAKN}}{V_{ABCD}}=\frac{|PA| \cdot|AK| \cdot|AN|}{|AC| \cdot|AD| \cdot|AB|}=\frac{2}{5}
\end{gathered}
$$
i.e., $V_{PAKN}=2$. Since the height dropped from $A$ to $PNK$ is $1$, and $S_{PNK}=6$,
$$
\frac{S_{PML}}{S_{PNK}}=\frac{|PK| \cdot|PN|}{|PM| \cdot|PL|}=\frac{3}{2}, \quad S_{PML}=9
$$
Thus, the area of the section will be $S_{PML}-S_{PNK}=3$.
Fig. 16.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
156. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Points $M$ and $N$ are taken on segments $A A_{1}$ and $B C_{1}$ such that the line $M N$ intersects the line $B_{1} D$. Find
$$
\frac{\left|B C_{1}\right|}{|B N|}-\frac{|A M|}{\left|A A_{1}\right|}
$$
|
156. Projecting a cube onto a plane perpendicular to $B_{i} D_{\text {, }}$ we obtain a regular hexagon $A B C C_{1} D_{1} A_{1}$ (Fig. 35) with side
Fig. $35_{s}$ equal to $\sqrt{\frac{2}{3}} a=b$, where $a$ is the edge of the cube (the equilateral triangle $B C_{\overline{\mathbf{1}}} A_{\mathbf{1}}$ projects into an equal triangle, as the plane $B C_{1} A_{\mathbf{i}}$ is perpendicular to $B_{1} D$). Consider $\triangle K A C_{i}$, in it $|K A|=\left|A C_{\mathbf{i}}\right|=2 b$, the line $N M$ passes through the midpoint of $A C_{\mathbf{i}}$. Let
$$
\frac{|A M|}{\left|A A_{1}\right|}=x
$$
Draw $C_{\overline{1}} L$ parallel to $M N_{9}$, we will have:
$$
\frac{|K N|}{\left|K C_{1}\right|}=\frac{|M L|=|A M|}{|K L|}=\frac{\mid K+x}{2+2 x}
$$
then
$$
\begin{aligned}
\frac{|B N|}{\left|B C_{1}\right|}=\frac{2\left(|K N|-\left|B C_{1}\right|\right)}{\left|K C_{1}\right|} & = \\
& =2 \frac{|K N|}{\left|K C_{1}\right|}-1=\frac{2+x}{1+x}-1=\frac{1}{1+x}
\end{aligned}
$$
Thus,
$$
\frac{\left|B C_{1}\right|}{|B N|}-\frac{|A M|}{\left|A A_{1}\right|}=1+x-x=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
195. Given a tetrahedron $A B C D$. In the planes defining its faces, points $A_{i}, B_{i}, C_{i}, D_{i}$ are taken such that the lines $A A_{i}, B B_{i}, C C_{i}, D D_{i}$ are parallel to each other. Find the ratio of the volumes of the tetrahedrons $A B C D$ and $A_{1} B_{1} C_{1} D_{\mathrm{i}}$.
|
195. Let $M$ be the point of intersection of the lines $C B_{1}$ and $C_{1} B$. The vertex $A$ lies on $D M$. We draw a plane through the points $D, D_{1}$, and $A$. Denote by $K$ and $L$ its points of intersection with $C_{1} B_{i}$ and $C B$, and by $A_{2}$ the point of intersection of the line $A A_{\hat{1}}$ with $D_{\mathrm{i}} K$ (Fig. 43). From the fact that ${ }^{A_{2}} C_{1} B_{1} B$ is a trapezoid and $K L$ passes through the intersection of its diagonals, it follows that $|K M|=|M L|$. Further, considering the trapezoid $D_{\mathrm{i}} K L D$, we will prove that $\left|A A_{1}\right|=\frac{1}{2}\left|A A_{2}\right|$. Therefore,
$$
v_{A B C D}=\frac{1}{3} v_{A_{2} B C D}
$$
But from the previous problem, it follows that $V_{A_{2} B C D}=V_{A_{1} B_{1} C_{1} D_{1}}$. Thus, the ratio of the volumes of the pyramids $A_{1} B_{1} C_{1} D_{i}$ and $A B C D$ is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
212. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{i}$, the diagonal $A C_{1}$ of which is equal to $d$, and the volume $V$. Prove that from the segments equal to the distances from the vertices $A_{i}, B$ and $D$ to the diagonal $A C_{1}$, a triangle can be constructed and that if $s$ is the area of this triangle, then $V=2 d s$.
|
212. Let $M$ be the point of intersection of the diagonal $A C_{i}$ with the plane $A_{1} B D$. Then $M$ is the point of intersection of the medians of triangle $A_{1} B D$, and, moreover, $M$ divides the diagonal $A C_{1}$ in the ratio $1: 2$, i.e., $|A M|=\frac{1}{3}d$.
Consider the pyramid $A B A_{1} D$ (Fig. 45). Take a point $K$ on the line $B M$ such that $|M K|=|B M|$, and construct the prism $M K D A N P$. It is not difficult to note that the distances between the lateral edges of this prism are equal to the corresponding distances from the points $A_{\mathrm{i}}, B$ and $D$ to $A M$. Therefore, the sides of the section perpendicular to the lateral edges of the prism $M K D A N P$ are equal to these distances. Furthermore, the volume of the pyramid $A B A_{1} D$ is equal to the volume of the constructed prism and is $\frac{1}{6}$ of the volume of the parallelepiped, i.e., $\frac{1}{6} V=\frac{1}{3} d S, V=2 d S$.
Fig. 45.
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
306. The sum of the planar angles of a trihedral angle is $180^{\circ}$. Find the sum of the cosines of the dihedral angles of this trihedral angle.
|
306. Consider a tetrahedron, all faces of which are equal triangles, the angles of which are respectively equal to the plane angles of a given trihedral angle. (Prove that such a tetrahedron exists.) All trihedral angles of this tetrahedron are equal to the given trihedral angle. The sum of the cosines of the dihedral angles of such a tetrahedron (see problem 304) is 2. Therefore, the sum of the cosines of the dihedral angles of the given trihedral angle is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
335. What is the greatest number of balls of radius 7 that can simultaneously touch, without intersecting, a ball of radius 3?
Transition to space.
|
335. We will prove that there cannot be more than six such spheres. Suppose there are seven. We connect the centers of all seven spheres with the center of the given sphere and denote by $O_{1}, O_{2}, \ldots, O_{7}$ the points of intersection of these segments with the surface of the given sphere. For each point $O_{i}$, we consider the set of points on the sphere for which the distance (along the surface of the sphere) to the point $O_{i}$ is no greater than the distance to any other point $O_{k}, k \neq i$. The sphere will be divided into seven spherical polygons. Each polygon is the intersection of six hemispheres containing the point $O_{i}$, the boundaries of which are great circles, along which the plane passing through the midpoint of the segment $O_{i} O_{k}$ and perpendicular to it intersects the sphere.
Each of the resulting polygons contains a circle inside it, the spherical radius of which is seen from the center of the original sphere at an angle $\alpha, \sin \alpha=0.7$.
Let $K$ and $N$ be the number of sides and vertices of the resulting partition, respectively. (Each side is a common side of two adjacent polygons and is counted once. The same applies to vertices.) It is easy to see that for such a partition, Euler's formula holds (see problem 324). In our case, this gives $K=N+5$. On the other hand, $K \geqslant \frac{3}{2} N$, since at least three sides emanate from each vertex, and each side is counted twice.
It is not difficult to obtain that $K \leqslant 15, N \leqslant 10$. In problem 325, we proved that among all spherical $n$-gons containing a given circle, the smallest area is that of a regular $n$-gon. Moreover, it can be shown that the sum of the areas of regular $n$- and $(n+2)$-gons is greater than twice the area of a regular $n$-gon. (We consider polygons circumscribed around the same circle.) It is also obvious that the area of a regular circumscribed $n$-gon decreases with increasing $n$. From this, it follows that the sum of the areas of the resulting seven polygons cannot be less than the sum of the areas of five regular quadrilaterals and two regular pentagons circumscribed around circles with a spherical radius corresponding to the central angle $\alpha=\arcsin 0.7$. The area of the corresponding regular pentagon will be
$$
s_{5}=9\left[10 \arccos \left(\cos \alpha \sin \frac{\pi}{5}\right)-3 \pi\right]
$$
and the area of the regular quadrilateral
$$
s_{1}=9\left[8 \arccos \left(\frac{\sqrt{2}}{2} \cos \alpha\right)-2 \pi\right]
$$
It is not difficult to prove that $2 s_{5}+5 s_{4}>36 \pi$. Thus, seven spheres of radius 7 cannot simultaneously, without intersecting, touch a sphere of radius 3. At the same time, it is easy to show that six spheres can be arranged in such a way.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. The bus network in the city of Lisse is arranged in such a way that: a) each route has three stops; b) any two routes either have no common stops at all or have only one common stop. What is the maximum number of routes that can exist in this city, given that there are only nine different stops?
|
10. Let's consider some stop $A$. Define how many routes can pass through it. Besides $A$, there are eight other stops in the city. On each route passing through $A_{\text {r }}$, there are two more stops. Since no two of these routes can have common stops other than $A$, a total of no more than $8: 2=4$ routes can pass through $A$.
Let's number all the stops and denote by $a_{i}$ the number of routes passing through the $i$-th stop. Since each route has three stops, we have
$$
a_{1}+a_{2}+\ldots+a_{9}=3 n
$$
where $\eta$ is the total number of routes. By the proven fact, all terms do not exceed four. Therefore,
$$
3 n \leqslant 4 \cdot 9 = 36 \text { and } n \leqslant 12
$$
In Fig. 14, a scheme is shown that satisfies the conditions of the problem and contains 12 routes (eight straight and four curved).
Fig. 14.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14**. Let the game be played until three signs in a row are achieved. What is the minimum number of cells the board should contain so that the first player can win, regardless of how his opponent plays? [Draw a board (of arbitrary shape) with the minimum number of cells and prove that on any board with fewer cells, the second player can avoid losing.]
|
14. On a board of seven cells, shown in Fig. $16, a$, the first player can win regardless of how the opponent plays: first, he places a cross at the intersection of rows, and then on one of the middle cells in the row that does not have a zero.
No matter what move the opponent makes, the first player can win on the third move. If the board consists of no more than six cells, the second player can always prevent the first player from winning. Let's prove this.
We will call a triplet three adjacent cells. In Fig. 16,6 there are four triplets: $(1,2,3)$; $(2,3,4)$; $(3,4,5)$ and $(4,5,6)$, while in Fig. $16, \partial$ there are three triplets: $(1,2,3)$; $(3,5,6)$ and $(1,4,6)$
a)
b)
c)
d)
e)
f)
g)
h)
i)
Fig. 15.
To prevent the first player from winning, the second player only needs to place one zero on each triplet. If there are no more than two triplets, the second player can cover them with two moves.
Suppose there are more than two triplets. If five or six cells are adjacent, all triplets are in this row (Fig. 16, b and c). By placing a zero in the third or
a)
b)
c)
d)
e)
Fig. 16.
fourth cell from the edge, the second player leaves only one uncovered triplet. With the second move, he can cover this triplet as well. If there are four adjacent cells, there is at most one triplet not in this row (Fig. 16, g). By occupying one of the two middle cells of the row of length 4, the second player covers both triplets in this row.
It remains to consider the case where no four cells are in a row (it is easy to see that in this case, six cells can contain no more than three triplets). If there are three triplets, then any two of them must have a common cell (Fig. $16, \partial$) and there are three such cells where two triplets intersect. By placing a zero on one of these cells, the second player can cover two of the three triplets with the first move; the third he will cover with the second move.
|
7
|
Logic and Puzzles
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
30. There are 111 lamps, and each lamp has its own switch. It is allowed to simultaneously switch 13 of them. At the initial moment, some lamps are on, and some are off.
a) Is it possible to turn off all the lamps?
b) How many switches will be required for this if all the lamps were initially on?
|
30. Let's show that we can always turn off all the lamps. We can assume that initially more than 13 lamps were on (otherwise, we would have turned on 13 lamps from the extinguished ones). Moreover, for the same reason, we can assume that the number of extinguished lamps is more than 6. We will select 13 lamps such that 7 are on and 6 are off. We will perform the switch. As a result, the number of burning lamps will decrease by one. By repeating this procedure several times, we will achieve that exactly 13 lamps are burning and we will turn them off with one switch.
Of course, this method is not the most economical. Let's ensure that in task b) 9 switches are sufficient. (8 switches are not enough. Why?) We can act as follows: first, turn off 13 lamps, then turn off another 7 (for which we will switch 10 burning and 3 extinguished lamps). After two stages, 91 burning lamps will remain, which can be turned off with seven switches.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
58. The numbers $1,2,3, \ldots, 1000000000$ are written down. Then each number is replaced by the sum of its digits, and so on, until only single-digit numbers remain in the sequence. Which digit appears more frequently in this sequence: ones or fives?
|
58. From the divisibility rule for 9, which we have used several times, it follows that the numbers in the last row give the same remainders when divided by 9 as the numbers in the first row above them. From this, it easily follows that in the first 999999999 positions of the last row, the digits $1,2,3,4,5,6$, $7,8,9$ repeat sequentially, and in the last position stands 1, i.e., there is one more 1 than 5 in the last row.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
71. Kolya, Lena, and Misha pooled their money and bought a football. The amount of money each of them contributed does not exceed half of the total amount contributed by the other two. How much money did Misha contribute if the ball cost 6 rubles?
|
71. According to the condition, the doubled amount of money invested by each boy does not exceed the sum invested by the other two. If one of the boys had given more than two rubles, then the other two would have given less than four, i.e., less than the doubled amount of money of the first. Therefore, each gave no more than two rubles. Since the ball cost 6 rubles, each gave 2 rubles.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 previous positions. Upon colliding, any two particles are destroyed, so, for example, after two seconds, only two particles remain. How many particles will remain after 129 seconds?
|
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 in a row at a distance of 2 from each other, with the outermost particles being at a distance of $2^{n}-1$ from the initial position, and after $2^{n}$ sec, only two particles will remain at a distance of $2^{n}$ from the initial position. The validity of this statement for $n=1$ and $n=2$ is clear from Fig. 44. Let this statement be already proven for $n=k$. We will prove it for $n=k+1$.
According to the induction hypothesis, at the moment $t=2^{k}$, only two particles will remain at a distance of $2^{k}$ from the initial position. Then, during the time $2^{k}-1$ sec, the "offspring" of the first particle will not interact with the "offspring" of the second, and therefore, as follows from the induction hypothesis, at the moment $t=2^{k}+2^{k}-1=2^{k+1}-1$ sec, $2 \cdot 2^{k}=2^{k+1}$ particles will form, arranged in a row at a distance of 2 from each other. It is clear that at the moment $t+1=2^{k+1}$, only two particles will remain at a distance of $2^{k+1}$ from the initial position.
For $k=7$, we get from this that after 128 sec, there will be two particles at a distance of 128 from the initial position, and the number of particles after 129 sec is 4.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
$$
A \sim\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
4 & 0 & 1 & -1 & -4 \\
8 & 0 & 2 & -2 & -8 \\
4 & 0 & 1 & -1 & -4
\end{array}\right) \sim\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
4 & 0 & 1 & -1 & -4
\end{array}\right)
$$
Since three rows of the intermediate matrix were proportional, we can obtain two zero rows from them, which we have discarded.
It is clear that $\operatorname{rank} A=2$, because $M_{2}=\left|\begin{array}{rr}3 & -1 \\ 4 & 0\end{array}\right|=4$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
|
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 \( A B \) and \( A D \);
3) the angle between the edge \( A D \) and the plane \( A B C \)
4) the volume of the tetrahedron \( A B C D \)
5) the equation of the edge \( A B \)
6) the equation of the plane \( A B C \);
7) the equation of the height dropped from \( D \) to \( A B C \)
8) the projection \( O \) of point \( D \) onto the base \( A B C \)
9) the height \( D O \).
|
Solution. The condition of the problem is satisfied by the constructed drawing (Fig. 4.11).
1) $AB$ is calculated by the formula
$$
\begin{aligned}
d & =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\
AB & =\sqrt{(4-2)^{2}+(1-3)^{2}+(-2-1)^{2}}=\sqrt{17}
\end{aligned}
$$
2) The angle $\varphi=(\widehat{\overrightarrow{AB}, \overrightarrow{AD}})$ is calculated by the formula (Ch. III, § 2)
$$
\cos \varphi=\frac{\overrightarrow{AB} \cdot \overrightarrow{AD}}{|\overrightarrow{AB}| \cdot|\overrightarrow{AD}|}
$$
a) $\overrightarrow{AB}=\{2,-2,-3\}, AB=\sqrt{17}, \overrightarrow{AD}=\{-7,-7,7\}, AD=7 \sqrt{3}$;
b) $\overrightarrow{AB} \cdot \overrightarrow{AD}=-14+14-21=-21$
c) $\cos \varphi=-\frac{21}{\sqrt{17} \cdot 7 \sqrt{3}} \approx-0.42, \quad \varphi \approx 180^{\circ}-65^{\circ}=115^{\circ}$.
3) The sine of the angle $\theta$ between the edge $AD$
Fig. 4.12 and the plane $ABC$ is equal to the cosine of the angle between the edge $AD$ and the normal vector $N$ of the plane $ABC$ (Fig. 4.12). The vector $\vec{N}$ is collinear with the vector product $\overrightarrow{AB} \times \overrightarrow{AC}$ (Ch. III, § 3)
a) $\overrightarrow{AB}=\{2,-2,-3\}, \overrightarrow{AC}=2\{2,0,3\}$
$$
\overrightarrow{AB} \times \overrightarrow{AC}=2 \cdot\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & -2 & -3 \\
2 & 0 & 3
\end{array}\right|=4 \cdot\{-3,-6,2\}
$$
We take $\vec{N}=\{-3,-6,2\}$.
b) $\overrightarrow{AD}=7\{-1,-1,1\}, \quad \sin \theta=\frac{3+6+2}{\sqrt{3} \cdot 7} \approx 0.92 ; \quad \theta \approx 67^{\circ}$.
4) The volume of the tetrahedron $ABCD$ is equal to $1 / 6$ of the modulus of the mixed product of the vectors $(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD}$ (Ch. III, § 4).
We have: $\overrightarrow{AB}=\{2,-2,-3\}, \quad \overrightarrow{AC}=2\{2,0,3\}, \quad \overrightarrow{AD}=-7\{1,1,-1\};$
$$
\left|\begin{array}{ccc}
2 & -2 & -3 \\
2 & 0 & 3 \\
1 & 1 & -1
\end{array}\right|=2 \cdot\left|\begin{array}{cc}
0 & 3 \\
1 & -1
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right|-3\left|\begin{array}{cc}
2 & 0 \\
1 & 1
\end{array}\right|=-22
$$
The desired volume is: $V=\frac{2 \cdot 7 \cdot 22}{6}=\frac{154}{3}$.
5) The equations of the line passing through two points have the form (§ 2) $\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p} ;\{m, n, p\}$ - the direction vector $\vec{l}$ of the line. We take $\vec{l}=\overrightarrow{AB}=\{2,-2,-3\}$. Then
$$
(AB): \quad \frac{x-2}{2}=\frac{y-3}{-2}=\frac{z-1}{-3}
$$
6) The equation of the plane passing through three given points (§ 1):
$$
\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|=0
$$
We have
$$
(ABC):\left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
2 & 0 & 3
\end{array}\right|=0
$$
or, after expanding the determinant: $3 x+6 y-2 z-22=0$.
7) As the direction vector $\vec{l}$ of the line $DO$, we can take the vector $\vec{l}=\vec{N}=\vec{N}(ABC)=\{3,6,-2\}(\S 2)$,
$$
(DO): \frac{x+5}{3}=\frac{y+4}{6}=\frac{z-8}{-2}
$$
or
$$
x=-5+3 t, y=-4+6 t, z=8-2 t
$$
8) The projection of $D$ on $ABC$ is the point $O$ (the intersection point of $DO$ with $ABC$). We substitute the values of $x, y$ and $z$, expressed in terms of the parameter $t$, into the equation of $ABC$. We find the value of $t$ and substitute it back into the expressions for $x, y$ and $z(\oint 3)$.
We get $3(3 t-5)+6(6 t-4)-2(8-2 t)-22=0,49 t=77, t=\frac{11}{7}$.
Then $x=-5+\frac{33}{7}=-\frac{2}{7}, y=\frac{38}{7}, z=\frac{34}{7}, O\left(-\frac{2}{7}, \frac{38}{7}, \frac{34}{7}\right)$
9) The height $DO$ can be calculated as the distance between $D$ and $O$, or as the distance from $D$ to the plane, or using the formula for the volume of the tetrahedron (§ 1).
In any case, we get $H=OD=\sqrt{\frac{33^{2}}{49}+\frac{66^{2}}{49}+\frac{22^{2}}{49}}=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Find the differential of the function $y=e^{x}\left(x^{2}+3\right)$. Calculate the value of the differential at the point $x=0$.
|
Solution. We have:
$$
\begin{gathered}
d y=y^{\prime} d x=\left(e^{x}\left(x^{2}+3\right)+2 x e^{x}\right) d x=\left(x^{2}+2 x+3\right) e^{x} d x \\
d y(0)=\left.\left(x^{2}+2 x+3\right) e^{x}\right|_{x=0} d x=3 d x
\end{gathered}
$$
## Exercises
Find the differentials of the functions.
1. $y=\ln \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \operatorname{arctg} x$.
2. $y=\frac{3 x^{2}-1}{3 x^{3}}+\ln \sqrt{1+x^{2}}+\operatorname{arctg} x$.
3. $y=\operatorname{arctg} \frac{a}{x}+\ln \sqrt{\frac{x-a}{x+a}}$. 4. $y=\frac{1}{3} \ln \frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+x}$.
Calculate the increments and differentials of the functions
5. $y=2 x^{2}-x$ at $x=1, \Delta x=0.01$.
6. $y=x^{3}+2 x, x=-1, \Delta x=0.02$.
Calculate the approximate values.
7. $\sin 60^{\circ} 18^{\prime}$.
8. $\sqrt[6]{735}$
9. $\operatorname{tg} 46^{\circ}$.
10. $\ln 1.07$
## Answers
1. $d y=\frac{x^{2} d x}{1-x^{4}}$. 2. $d y=\frac{x^{5}+1}{x^{6}+x^{4}} d x$. 3. $d y=\frac{2 a^{3}}{x^{4}-a^{4}} d x$. 4. $d y=-\frac{2 d x}{3 \sqrt{x^{2}+1}}$. 5. $\Delta y=0.0302, d y=0.03$. 6. $\Delta y=0.0988, d y=0.1$. 7. 0.86863 . 8. 3.0041 . 9. 1.03553 . 10. 0.0676586 .
## § 6. Derivatives and Differentials of Higher Orders
$1^{\circ}$. Suppose the derivative $y^{\prime}=f^{\prime}(x)$ of the function $y \doteq f(x)$ is a differentiable function. Then its derivative $\left(y^{\prime}\right)^{\prime}$ is called the second derivative of the function $f(x)$ and is denoted by $y^{\prime \prime}=f^{\prime \prime}(x)$. Higher-order derivatives are defined successively: $y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}, y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}$, and so on. Notations can also be: $y^{(1)}, y^{(2)}, y^{(3)}, \ldots, y^{(n)}$
For example, if $y=x^{2}+5 x-6$, then $y^{\prime}=2 x+5, y^{\prime \prime}=2, y^{\prime \prime \prime}=0$.
If $y=\cos x$, then the derivative of any order of this function is found as follows:
$$
\begin{aligned}
& y^{\prime}=(\cos x)^{\prime}=-\sin x=\cos \left(x+\frac{\pi}{2}\right) \\
& y^{\prime \prime}=-\cos x=\cos \left(x+2 \frac{\pi}{2}\right) \\
& y^{\prime \prime \prime}=\sin x=\cos \left(x+3 \frac{\pi}{2}\right) \\
& y^{I V}=\cos x=\cos \left(x+4 \frac{\pi}{2}\right) \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& y^{(n)}=\cos \left(x+n \frac{\pi}{2}\right) ; \quad n=1,2,3, \ldots
\end{aligned}
$$
$2^{\circ}$. Higher-order differentials are defined as follows: $d^{n} y=f^{(n)}(x) d x^{n}$. Hence $f^{(n)}(x)=\frac{d^{n} y}{d x^{n}}=\frac{d}{d x}\left(\frac{d^{n-1} y}{d x^{n-1}}\right)$.
For example, $d^{n}(\cos x)=\cos \left(x+\frac{n \pi}{2}\right) d x^{n}$.
## Exercises
Find the derivatives and differentials of order $n \in \mathbb{N}$.
1. $y=\frac{1}{a x-b}$
2. $e^{3 x}$.
3. $\frac{x^{2}+2 x+3}{x}$
4. $\sqrt{x}$.
5. $x^{\alpha}$.
6. $\ln \sqrt[33]{x+1}$
7. $\sin \alpha x$. 8. $\cos \beta x$.
## Answers
1. $y^{(n)}=(-1)^{n} \frac{a^{n} n!}{(a x-b)^{n+1}}$.
2. $y^{(n)}=3^{n} e^{3 x}$. 3. $y^{\prime}=1-\frac{3}{x^{2}}$, $y^{\prime \prime}=\frac{6}{x^{3}}, \quad y^{(n)}=(-1)^{n} 3 \frac{n!}{x^{n+1}}$
3. $y^{(n)}=(-1)^{n-1} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-3)}{2^{n} \cdot \sqrt{x^{2 n-1}}}$
4. $y^{(n)}=\alpha(\alpha-1) \cdots(\alpha-n+1) x^{\alpha-n}$. 6. $y^{(n)}=\frac{1}{33}(-1)^{n+1} \frac{(n-1)!}{x^{n}}$.
5. $y^{(n)}=\alpha^{n} \sin \left(\alpha x+\frac{\pi}{2} n\right) \cdot 8 \cdot y^{(n)}=\beta^{n} \cos \left(\alpha x+n \frac{\pi}{2}\right)$.
## § 7. Differentiation of Inverse Functions. Differentiation of Functions Given Implicitly and Parametrically
$1^{\circ}$. Let $y=f(x)$ be a function defined and continuous on the interval $[a ; b]$ and $f(a)=A, f
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Check if Rolle's theorem is valid for the function $f(x)=x^{2}+2 x+7$ on the interval $[-6 ; 4]$, and if so, find the corresponding value of c.
|
Solution. $f(x)=x^{2}+2 x+7$ is continuous and differentiable on any interval, in particular, on $[-6 ; 4]$, and $f(6)=f(4)=31$. Therefore, $f^{\prime}(x)=2 x+2=0$ for some $x \in(-6 ; 4)$. We have $x=c=-1$.
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Calculate the limits:
a) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\tan\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$, b) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, c) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x y}{\sqrt{4-x y}-2}$.
|
Solution. Note that the functions $\frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$ and $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ are undefined only at the point $(0,0)$, while $\frac{x y}{\sqrt{4-x y}-2}$ is undefined on the coordinate axes $x=0$ and $y=0$.
a) Transition to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$, $\varphi \in[0,2 \pi)$. Then $x^{2}+y^{2}=r^{2}$ and $(x, y) \rightarrow(0,0)$ implies $r \rightarrow 0$. Using the corollary of the first remarkable limit, we get
$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}=\left(\frac{0}{0}\right)=\lim _{r \rightarrow 0} \frac{\operatorname{tg} r^{2}}{r^{2}}=1
$$
b) If $(x, y) \rightarrow(0,0)$ along different rays $y=k x, k \in \mathbb{R}$, then
$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{x^{2}-k^{2} x^{2}}{x^{2}+k^{2} x^{2}}=\frac{1-k^{2}}{1+k^{2}}
$$
The right-hand side of this equation depends on $k$, and therefore the limit of the given function as $(x, y) \rightarrow(0,0)$ does not exist (the limit must be unique).
c) Multiply the numerator and denominator by the conjugate of the denominator. We get
$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x y(\sqrt{4-x y}+2)}{-x y}=-\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(\sqrt{4-x y}+2)=-4
$$
Note. From the definition of continuity, it follows that the functions
$$
f(x, y)=\left\{\begin{array}{l}
\frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} \quad \text { for }(x, y) \neq(0, y) \text { and }(x, y) \neq(x, 0) \\
1 \quad \text { for }(x, y)=(0, y) \text { or }(x, y)=(x, 0)
\end{array}\right.
$$
and
$$
g(x, y)=\left\{\begin{array}{l}
\frac{x y}{\sqrt{4-x y}-2} \quad \text { for }(x, y) \neq(0,0) \\
-4 \text { for }(x, y)=(0,0)
\end{array}\right.
$$
are continuous on the entire plane $O x y$, while the function $\varphi(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ cannot be extended at the point $(0,0)$ in such a way that it is continuous at this point.
## Exercises
Calculate the limits.
1. $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\operatorname{tg} x y}{y}$. 2. $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 7}} \lg (x+y)$.
2. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\sin x y}{x}$.
3. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}$
4. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(x+y) \sin \frac{1}{x} \cdot \cos \frac{1}{y}$.
5. $\lim _{\substack{x \rightarrow 1 \\ y \rightarrow 2}} \frac{2(x-1)(y-2)}{(x-1)^{2}+(y-2)^{2}}$
6. $\lim _{\substack{x \rightarrow \infty \\ y \rightarrow \infty}} \frac{x^{2}+y^{2}}{x^{4}+y^{4}}$
7. $\lim _{\substack{x \rightarrow \infty \\ y \rightarrow \infty}} \frac{x+y}{x^{2}-x y+y^{2}}$.
8. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2} y}{x^{2}+y^{2}}$
9. $\lim _{\substack{x \rightarrow 1 \\ y \rightarrow 0}} \frac{\ln \left(1+e^{y}\right)}{\sqrt{x^{2}+y^{2}}}$
10. $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow-2}} \frac{2 x+y}{(x-2)^{2}+(y+2)^{2}}$
11. $\lim _{\substack{x \rightarrow-1 \\ y \rightarrow 2}} \frac{x^{2}+2 x+y^{2}-4 y+5}{x+2 y}$.
## Answers
1. 2. 2. 3. 3. 0. 4. Does not exist. 5. 0. 6. Does not exist. 7. 0. 8.0. 9. 0. 10. $\ln 2.11 .+\infty$. 12.0 .
## § 3. Partial Derivatives and Differential of a Function of Two Variables
$1^{\circ}$. The partial increments of the function $z=f(x, y)$ with respect to the independent variables $x$ and $y$ are the differences
$$
\Delta_{x} z=f(x+\Delta x, y)-f(x, y), \quad \Delta_{y} z=f(x, y+\Delta y)-f(x, y)
$$
where $\Delta x$ and $\Delta y$ are the increments of the independent variables $x$ and $y$.
The total increment of the function $z=f(x, y)$ is the difference
$$
\Delta z=f(x+\Delta x, y+\Delta y)-f(x, y)
$$
In general, the total increment does not equal the sum of the partial increments:
$$
\Delta z \neq \Delta_{x} z+\Delta_{y} z
$$
$2^{\circ}$. The partial derivative of the function $z=f(x, y)$ with respect to the variable $x$ or $y$ is the limit of the ratio of the corresponding partial increment $\Delta_{x} z$ or $\Delta_{y} z$ to the increment of the given variable, provided that the increment of the variable tends to zero.
The notations for partial derivatives are
$$
\begin{aligned}
& z_{x}^{\prime}=\frac{\partial z}{\partial x}=\frac{\partial f(x, y)}{\partial x}=f_{x}^{\prime}(x, y)=\lim _{\Delta x \rightarrow 0} \frac{\Delta_{x} z}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x} \\
& z_{y}^{\prime}=\frac{\partial z}{\partial y}=\frac{\partial f(x, y)}{\partial y}=f_{y}^{\prime}(x, y)=\lim _{\Delta y \rightarrow 0} \frac{\Delta_{y} z}{\Delta y}=\lim _{\Delta y \rightarrow 0} \frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}
\end{aligned}
$$
$3^{\circ}$. When finding a partial derivative (differentiation) with respect to a certain variable, one uses the formulas and rules of differentiation of a function of one variable, treating the other variable as fixed, constant.
$4^{\circ}$. The partial derivative $z_{x}^{\prime}$ $\left(z_{y}^{\prime}\right)$ at a given point $\left(x_{0}, y_{0}\right)$ is equal to the tangent of the angle of inclination of the tangent to the curve obtained by the intersection of the surface
Figure 8.2 $z=f(x, y)$ with the plane $y=y_{0}$ $\left(x=x_{0}\right), z_{x}^{\prime}\left(x_{0}, y_{0}\right)=\operatorname{tg} \alpha\left(z_{y}^{\prime}\left(x_{0}, y_{0}\right)=\operatorname{tg} \beta\right)$ (Figure 8.2).
$5^{\circ}$. The partial differentials of the function $z=f(x, y)$ are the quantities
$$
d_{x} z=z_{x}^{\prime} d x, \quad d_{y} z=z_{y}^{\prime} d y, \quad d x=\Delta x, \quad d y=\Delta y
$$
$6^{\circ}$. The total differential of the function $z=f(x, y)$ is the expression
$$
d z=d_{x} z+d_{y} z=z_{x}^{\prime} d x+z_{y}^{\prime} d y=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y
$$
The total differential $d z$ represents the principal linear (with respect to $\Delta x$ and $\Delta y$) part of the total increment of the function $z=f(x, y)$.
## Examples with Solutions
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Find the derivative of the function $z=x^{2}+y^{2}$ at the point $M(3,1)$ in the direction of the point $M_{1}(0,5)$.
|
Solution. We have: $\overrightarrow{M M}_{1}=\{0-3,5-1\}=\{-3,4\},\left|\overrightarrow{M M}_{\mathrm{I}}\right|=$ $=\sqrt{(-3)^{2}+4^{2}}=5, \cos \alpha=-\frac{3}{5}, \sin \alpha=\frac{4}{5}$.
Let $\vec{l}=\left\{-\frac{3}{5}, \frac{4}{5}\right\}$ and find $\frac{\partial z}{\partial l}$.
We have: $z_{x}^{\prime}=2 x, z_{y}^{\prime}=2 y, z_{x}^{\prime}(3,1)=6, z_{y}^{\prime}(3,1)=2$.
Finally, $\frac{\partial z}{\partial l}=6\left(-\frac{3}{5}\right)+2 \cdot \frac{4}{5}=-2$. The negative sign of the derivative value $\frac{\partial z(3,1)}{\partial l}$ indicates that the function is decreasing in the direction of $\vec{l}$.
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Find the direction of the maximum increase of the function $z=3 x^{2}-2 y^{2}$ at the point $M(1,2)$. Also, find the greatest value of the derivatives in different directions at the point $M$.
|
Solution. Let's find the gradient of the function $z$ at the given point $(1,2)$. We have $z_{x}^{\prime}=6 x, z_{x}^{\prime}(1,2)=6, z_{y}^{\prime}=-4 y, z_{y}^{\prime}(1,2)=-8$. The gradient of the field at point $M(1,2)$ is $\overrightarrow{\operatorname{grad}} z=\{6,-8\}$. This vector indicates the direction of the maximum increase of $z$. The maximum value of the derivative at $(1,2)$ is $\sqrt{6^{2}+(-8)^{2}}=10$.
|
10
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1.1. Calculate
$$
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|
$$
|
Solution:
$$
\begin{gathered}
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|=2 \cdot\left|\begin{array}{rr}
2 & 2 \\
-2 & 1
\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right|+1 \cdot\left|\begin{array}{rr}
3 & 2 \\
1 & -2
\end{array}\right|= \\
=2 \cdot(2+4)+1 \cdot(3-2)+1 \cdot(-6-2)=12+1-8=5
\end{gathered}
$$
Let's consider the concept of the inverse matrix.
Definition. Let a matrix $A$ of order $n$ be given. If its product with some matrix $B$ of order $n$ equals the identity matrix $E$, i.e., $A \cdot B=E$ or $B \cdot A=E$, then the matrix $B$ is called the inverse of matrix $A$ and is denoted by $A^{-1}$.
It can be proven that if $A \cdot A^{-1}=E$, then $A^{-1} \cdot A=E$, i.e., mutually inverse matrices are commutative.
Theorem 1. Every square matrix $A$, whose determinant $\Delta \neq 0$, has a unique inverse matrix $A^{-1}=B$, the elements of which $b_{i j}$ are found by the formula
$$
b_{i j}=\frac{A_{j i}}{\Delta}
$$
From the theorem, the following rule follows: to find the inverse matrix of matrix (1.2), where $m=n$, the following transformations must be performed:
1. Calculate the determinant $\Delta$ of matrix $A$ ($\Delta \neq 0$).
2. Replace each element of matrix $A$ with its algebraic complement, i.e., form the matrix ($A_{i j}$).
3. Transpose the matrix ($A_{i j}$), i.e., write the matrix ($A_{j i}$).
4. Multiply the matrix ($A_{j i}$) by $\frac{1}{\Delta}$.
As a result, we obtain the matrix $A^{-1}$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1.17. Form the equation of the line $l$ passing through the point $A(2; -4)$ and being at a distance of 2 units from the origin.
|
Solution. Let the equation of the desired line have the form:
$$
y-y_{A}=k\left(x-x_{A}\right),
$$
or
$$
y+4=k(x-2)
$$
or
$$
k x-y-(4+2 k)=0 .
$$
To determine the slope $k$ of this line, we will use the fact that it is 2 units away from the origin. We will find this distance directly. The equation of the perpendicular dropped from the origin to the line $k x-y-(2+4 k)=0$ is $y=-\frac{1}{k} x$, or $x+k y=0$. Solving the equations of these two lines simultaneously
$$
\begin{aligned}
& k x-y-(2+4 k)=0 \\
& x+k y=0
\end{aligned}
$$
we obtain the coordinates of the point $C$ of their intersection:
$$
x_{C}=\frac{2 k(k+2)}{1+k^{2}} ; y_{C}=-\frac{2(k+2)}{1+k^{2}}
$$
From this, we find the distance from the origin to the line $l$ :
$$
\begin{gathered}
O C=\sqrt{x_{C}^{2}+y_{C}^{2}}=\frac{2}{1+k^{2}} \sqrt{k^{2}(k+2)^{2}+(k+2)^{2}}= \\
=\frac{2(k+2)}{1+k^{2}} \sqrt{k^{2}+1}=\frac{2(k+2)}{\sqrt{k^{2}+1}}
\end{gathered}
$$
On the other hand, by the condition $O C=2$. Thus, we obtain an equation for finding the slope $k$ of the desired line $l$ :
$$
\frac{2(k+2)}{\sqrt{k^{2}+1}}=2 \text { or } k+2=\sqrt{k^{2}+1},
$$
from which $k=-\frac{3}{4}$. Thus, substituting the found value $k=-\frac{3}{4}$ into the equation (*), we get the equation of the line:
$$
-\frac{3}{4} x-y-4-2 \cdot \frac{3}{4}=0
$$
or finally,
$$
3 x+4 y+10=0
$$
In conclusion, by seeking the equation of the line $l$ in the form $y-y_{A}=k\left(x-x_{A}\right)$, we assumed that this line is not parallel to the y-axis. However, it is clear that the line $x=2$ (parallel to the y-axis) also satisfies the condition of the problem, as it passes through the point $A(2 ;-4)$ and is 2 units away from the origin (Fig. 17).
Fig. 17
|
2
|
Geometry
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math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5.2. Using the definition directly, show that the series converges, and find its sum.
\[
\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}+\ldots
\]
|
Solution. By the definition of the partial sum of a series, we have:
$$
\begin{aligned}
& S_{1}=a_{1}=\frac{1}{2} \\
& S_{2}=a_{1}+a_{2}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3} \\
& S_{3}=a_{1}+a_{2}+a_{3}=\frac{2}{3}+\frac{1}{12}=\frac{3}{4} \\
& S_{4}=a_{1}+a_{2}+a_{3}+a_{4}=\frac{3}{4}+\frac{1}{20}=\frac{4}{5}
\end{aligned}
$$
Thus, we obtain the following sequence of partial sums:
$$
\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots
$$
the general term of which is: $\frac{n}{n+1}$. It is clear that this sequence converges and its limit is one:
$$
\lim _{n \rightarrow \infty} S_{n}=\lim _{n \rightarrow \infty} \frac{n}{n+1}=1
$$
This means that the given series converges and its sum is one.
## 5.2. Necessary Condition for Convergence of a Series. Sufficient Conditions for Convergence of Series with Positive Terms
A series can converge only if its general term $a_{n}$ tends to zero as the index $n$ increases without bound: $\lim _{n \rightarrow \infty} a_{n}=0$ - this is the necessary condition for the convergence of a series.
If $\lim _{n \rightarrow \infty} a_{n} \neq 0$, then the series diverges - this is the sufficient condition for the divergence of a series.
For series with positive terms, the following sufficient conditions can be used to determine their convergence or divergence.
1. Comparison Test. If the terms of a positive series
$$
a_{1}+a_{2}+\ldots+a_{n}+\ldots
$$
starting from some index, do not exceed the corresponding terms of the series
$$
b_{1}+b_{2}+\ldots+b_{n}+\ldots
$$
then the convergence of series (2) implies the convergence of series (1), and the divergence of series (1) implies the divergence of series (2).
When investigating series for convergence and divergence using this test, a geometric progression is often used:
$$
a+a q+a q^{2}+\ldots+a q^{n}+\ldots(a>0)
$$
which converges for $|q|<1$ and diverges for $|q|>1$ (the question of convergence remains unresolved for $l=1$).
|
1
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Example 5.9. Based on the D'Alembert's criterion, investigate the convergence of the series
$$
3+\frac{3^{2}}{2^{2}}+\frac{3^{3}}{3^{3}}+\frac{3^{4}}{4^{4}}+\ldots+\frac{3^{n}}{n^{n}}+\ldots
$$
|
Solution. Knowing the $n$-th term of the series $a_{n}=\frac{3^{n}}{n^{n}}$, we write the $(n+1)$-th term: $a_{n+1}=\frac{3^{n+1}}{(n+1)^{n+1}} \cdot$ From this,
\[
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{3^{n+1}}{(n+1)^{n+1}}: \frac{3^{n}}{n^{n}}=\lim _{n \rightarrow \infty} \frac{3^{n+1} \cdot n^{n}}{3^{n}(n+1)^{n+1}}=\lim _{n \rightarrow \infty} \frac{3}{n+1} \cdot \frac{n}{n+1}{ }^{n}= \\
=3 \lim _{n \rightarrow \infty} 1-\frac{n}{n+1}{ }^{n} \cdot \frac{1}{n+1}=3 \cdot e^{-1} \cdot 0=0
\end{gathered}
\]
Since $l=0<1$, the series converges.
## 5.3. Leibniz's Convergence Test
An alternating series is a series of the form
\[
a_{1}-a_{2}+a_{3}-\ldots+(-1)^{n-1} a_{n}+\ldots
\]
where $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are positive numbers.
For alternating series, the following convergence test applies.
Leibniz's Test. The series (1) converges if its terms decrease monotonically in absolute value and the general term tends to zero as $n \rightarrow \infty$.
The application of convergent series to approximate calculations is based on replacing the sum of the series with the sum of its first few terms. The error introduced by this is easily estimated for an alternating series that satisfies Leibniz's test - the error is less than the absolute value of the first of the discarded terms of the series.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5.18. Find the region of convergence of the power series
$$
x+2!x^{2}+3!x^{3}+\ldots+n!x^{n}+\ldots
$$
|
Solution. Here $u_{n}=n!|x|^{n}, u_{n+1}=(n+1)!|x|^{n+1}$.
From this,
$$
l=\lim _{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}}=\lim _{n \rightarrow \infty} \frac{(n+1)!|x|^{n+1}}{n!|x|^{n}}=|x| \lim _{n \rightarrow \infty}(n+1)
$$
Thus,
$$
l=\begin{array}{ccc}
\infty & \text { when } & x \neq 0, \\
0 & \text { when } & x=0 .
\end{array}
$$
Therefore, according to the D'Alembert's ratio test, the series converges only at the point $x=0$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2.1. Find the integral $\iiint_{V} x d x d y d z$, if the body $V$ is bounded by the planes: $x=0, y=0, x+4 y-2 z=0$ and $x+y+z-6=0$.
|
Solution. The body V is bounded by the coordinate planes $x \mathrm{Oz}$ and $y \mathrm{Oz}$, and from below and above by the planes: $z=\frac{1}{2}(x+4 y)$ and $z=6-x-y$. Let's find the line of intersection of these planes:
$$
\begin{aligned}
& z=\frac{1}{2}(x+4 y) \\
& z=6-x-y
\end{aligned}
$$
or
$$
\frac{1}{2}(x+4 y)=6-x-y
$$
or
$$
x+2 y-4=0 .
$$
We have $y=\frac{1}{2}(4-x)$. Let's construct the region $D$ on the plane $x \mathrm{Oy}$ (Fig. 62). We compute:
$$
\begin{gathered}
\iiint_{V} x d x d y d z=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y \int_{\frac{1}{2}(x+4 y)}^{6-x-y} d z=\left.\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y z\right|_{\frac{1}{2}(x+4 y)} ^{6-x-y}= \\
=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y(6-x-y)-\frac{1}{2}(x+4 y)= \\
=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} 6-\frac{3}{2} x-3 y d y=3 \int_{0}^{4} x d x 2 y-\frac{1}{2} x y-\left.\frac{1}{2} y^{2}\right|_{0} ^{\frac{1}{2}(4-x)}= \\
=3 \int_{0}^{4} x d x 2 \cdot \frac{1}{2}(4-x)-\frac{1}{2} x \cdot \frac{1}{2}(4-x)-\frac{1}{2} \frac{1}{2}(4-x)^{2}= \\
=3 \int_{0}^{4} x 2-x+\frac{1}{8} x^{2} d x=3 \int_{0}^{4} 2 x-x^{2}+\frac{1}{8} x^{3} d x= \\
=32 \cdot \frac{x^{2}}{2}-\frac{1}{3} x^{3}+\left.\frac{1}{8} \frac{1}{4} x^{4}\right|_{0} ^{4}=316-\frac{64}{3}+8=8
\end{gathered}
$$
Fig. 62
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the distance $d$ between points $A$ and $B$ in each of the following cases:
1) $A(2), B(3) ; 2) A(-4), B(-8)$.
|
Solution. Applying formula (2), we get:
1) $d=|3-2|=1 ; 2) d=|-8-(-4)|=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
75. Write the equation of a line parallel to the $O x$ axis and cutting off a segment equal to -2 from the $O y$ axis. Check if the points $A(2; -2)$ and $B(3; 2)$ lie on this line.
|
Solution. According to the problem, the slope $k=0$, the initial ordinate $b=-2$, therefore, the equation of the line we are looking for is $y=-2$. It is easy to see that the coordinates of any point $M(x; -2)$ satisfy this equation.
Point $A$ lies on the line we are looking for, since its coordinates satisfy the equation of the line $y=-2$. Point $B$ does not lie on the line, since its coordinates do not satisfy the equation of the line $y=-2$ (Fig. 14).
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
312. Find the scalar square of the vector $\bar{a}=2 \bar{i}-\bar{j}-2 \bar{k}$ and its length.
|
Solution. We will use formulas (10), (13), and (14):
$$
(\bar{a})^{2}=a^{2}=2^{2}+(-1)^{2}+(-2)^{2}=9
$$
from which $a=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
313. Find the scalar product of the vectors
$$
\bar{p}=\bar{i}-3 \bar{j}+\bar{k}, \bar{q}=\bar{i}+\bar{j}-4 \bar{k}
$$
|
Solution. The scalar product of the vectors will be found using formula (12):
$$
\bar{p} \bar{q}=1 \cdot 1+(-3) \cdot 1+1 \cdot (-4)=-6
$$
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
626. Find $f^{\prime}(0)$, if $f(x)=e^{x} \arcsin x+\operatorname{arctg} x$.
|
Solution. $\quad f^{\prime}(x)=\left(e^{x} \arcsin x\right)^{\prime}+(\operatorname{arctg} x)^{\prime}=\left(e^{x}\right)^{\prime} \arcsin x+$ $+e^{x}(\arcsin x)^{\prime}+(\operatorname{arctg} x)^{\prime}=e^{x} \arcsin x+\frac{e^{x}}{\sqrt{1-\lambda^{2}}}+$ $+\frac{1}{x^{2}+1}=e^{x} \frac{1+\sqrt{1-x^{2}} \arcsin x}{\sqrt{1-x^{2}}}+\frac{1}{1+x^{2}}$.
Substituting the value $x=0$, we get $f^{\prime}(0)=1+1=2$.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
976. Find the following limits:
1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\tan x y}{y}$
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}$.
|
Solution. 1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\operatorname{tg} x y}{y}=\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} x \frac{\operatorname{tg} x y}{x y}=2 \cdot 1=2$,
since
$$
\lim _{\alpha \rightarrow 0} \frac{\operatorname{tg} \alpha}{\alpha}=1
$$
2) Let's use the definition of the limit of a function at a point. Take the sequences $x=\frac{1}{n} \rightarrow 0, y=\frac{1}{n} \rightarrow 0(n \rightarrow \infty)$,
then
$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\frac{1}{n}}{\frac{1}{n}}=2
$$
Now take $x=\frac{1}{n} \rightarrow 0, y=\frac{2}{n} \rightarrow 0(n \rightarrow \infty), \quad$ then
$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\frac{2}{n}}{\frac{1}{n}}=3
$$
Thus, for different sequences of points converging to the point $(0 ; 0)$, we get different limits. This means that the limit of the function at this point does not exist (see point 1).
Note that the given function is not defined at the point $(0 ; 0)$. Find the following limits:
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1020. Investigate the function for extremum
$$
f(x, y)=x^{2}+y^{2}-4 y+4
$$
|
Solution. Let's find the partial derivatives:
$$
f_{x}^{\prime}(x, y)=2 x ; f_{y}^{\prime}(x, y)=2 y-4
$$
Set them equal to zero, we get a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
2 x=0 \\
2 y-4=0
\end{array}\right.
$$
from which $x=0, y=2$.
The function at the critical point $M_{0}(0 ; 2)$ has a minimum $f(0 ; 2)=4-8+4=0$, since near this point for any points $M(x ; y)$, different from $M_{0}(0 ; 2)$, the values of the function are greater than zero. For example, let $x=0, y=1$, then $f(0 ; 1)=1-4+4=1>0$; let $x=-1, y=3$, then $f(-1 ; 3)=1+9-12+4=2>0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.7. Convert to polar coordinates and evaluate the double integrals.
a) $\iint_{\Omega} x y^{2} d x d y$, where the region $\Omega$ is bounded by the circles $x^{2}+(y-1)^{2}=1$ and $x^{2}+y^{2}=4 y$
b) $\iint_{\Omega} e^{-x^{2}-y^{2}} d x d y$, where $\Omega-$ is the circle $x^{2}+y^{2} \leqslant R^{2}$.
|
Solution.
a) First, let's depict the region $\Omega$ in the Cartesian coordinate system (Fig. 1.32).
The equations of the boundary circles in polar coordinates are obtained after the substitution $x=\rho \cos \varphi, y=\rho \sin \varphi$:
$$
\begin{aligned}
x^{2}+(y-1)^{2}=1 \Rightarrow x^{2}+y^{2} & =2 y \Rightarrow \rho=2 \sin \varphi, 0 \leqslant \varphi \leqslant \pi \\
x^{2}+y^{2} & =4 y \Rightarrow \rho=4 \sin \varphi, 0 \leqslant \varphi \leqslant \pi
\end{aligned}
$$
The region $\Omega$ in coordinates $x, y$ is transformed into the region $D$ in coordinates $\varphi, \rho: D=\{(\varphi, \rho): 0 \leqslant \varphi \leqslant \pi ; 2 \sin \varphi \leqslant \rho \leqslant 4 \sin \varphi\}$. We will perform the transformation to coordinates $\varphi, \rho$ and compute the original double integral:
$$
\begin{gathered}
\iint_{\Omega} x y^{2} d x d y=\iint_{D} \rho^{3} \cos \varphi \sin ^{2} \varphi \rho d \rho d \varphi= \\
=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi d \varphi \int_{2 \sin \varphi}^{4 \sin \varphi} \rho^{4} d \rho=\int_{0}^{\pi} \sin ^{2} \varphi \cos \varphi\left(\left.\frac{\rho^{5}}{5}\right|_{2 \sin \varphi} ^{4 \sin \varphi}\right) d \varphi= \\
=\frac{992}{5} \int_{0}^{\pi} \sin ^{7} \varphi \cos \varphi d \varphi=\frac{992}{5} \int_{0}^{\pi} \sin ^{7} \varphi d(\sin \varphi)= \\
=\left.\frac{992}{5} \frac{\sin ^{8} \varphi}{8}\right|_{0} ^{\pi}=0
\end{gathered}
$$
b) Transform the integral to polar coordinates $x=\rho \cos \varphi, y=\rho \sin \varphi:$
$$
\begin{gathered}
\iint_{\Omega} e^{-x^{2}-y^{2}} d x d y=\iint_{D} e^{-\rho^{2}} \rho d \rho d \varphi= \\
=\int_{0}^{2 \pi} d \varphi \int_{0}^{R} e^{-\rho^{2}} \rho d \rho=\left.\int_{0}^{2 \pi}\left(-\frac{e^{-\rho^{2}}}{2}\right)\right|_{0} ^{R} d \varphi=\pi\left(1-e^{-R^{2}}\right)
\end{gathered}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.5. Compute the specified line integrals of the second kind:
a) $\int_{L} 2 y \sin 2 x d x - \cos 2 x d y$, where $L$ is any piecewise-smooth curve connecting the points $A\left(\frac{\pi}{2}, 2\right)$ and $B\left(\frac{\pi}{6}, 1\right)$.
b) $\int_{L} y x e^{x} d x - (x-1) e^{x} d y$, where $L$ is any piecewise-smooth curve connecting the points $A(0,2)$ and $B(1,2)$.
|
Solution.
a) It is not difficult to understand that the integrand is a complete differential of the function $u=-y \cos 2 x$, i.e.
$$
\begin{gathered}
\int_{L} 2 y \sin 2 x d x-\cos 2 x d y=\int_{L} d u=u(B)-u(A)= \\
=-\cos \frac{\pi}{3}+2 \cos \pi=-2-\frac{1}{2}=-\frac{5}{2}
\end{gathered}
$$
b) In this problem, the function $u$ can be taken as the function $u=y(x-1) e^{x}: d u=y x e^{x} d x-(x-1) e^{x} d y$. Therefore,
$$
\int_{L} y x e^{x} d x-(x-1) e^{x} d y=\int_{L} d u=u(B)-u(A)=0 .
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.21. Calculate the circulation of the vector field:
a) $\vec{A}=x^{2} y^{2} \vec{i}+\vec{j}+z \vec{k}$ along the circle $x^{2}+y^{2}=a^{2}, z=0$;
b) $\dot{A}=(x-2 z) \dot{i}+(x+3 y+z) \dot{j}+(5 x+y) \vec{k}$ along the perimeter of the triangle $A B C$ with vertices $A(1,0,0), B(0,1,0), C(0,0,1)$.
|
Solution.
a) For $z=0$ we obtain the planar field $\vec{A}=x^{2} y^{2} \dot{i}+\ddot{j}$, where $P=x^{2} y^{2}, Q=1$. To compute the circulation of this field along the circle $x^{2}+y^{2}=a^{2}$, we can use Green's formula (2.62):
$$
\begin{gathered}
\oint_{x^{2}+y^{2}=a^{2}} \vec{A} \cdot d \vec{l}=\iint_{x^{2}+y^{2} \leqslant a^{2}}\left(Q_{x}-P_{y}\right) d x d y=-2 \iint_{x^{2}+y^{2} \leqslant a^{2}} x^{2} y d x d y= \\
\quad=-2 \int_{-a}^{+a} x^{2} d x \int_{-\sqrt{a^{2}-x^{2}}}^{+\sqrt{a^{2}-x^{2}}} y d y=-\left.2 \int_{-a}^{+a} x^{2} d x y^{2}\right|_{-\sqrt{a^{2}-x^{2}}} ^{\sqrt{a^{2}-x^{2}}}=0 .
\end{gathered}
$$
b) By Stokes' formula, the circulation of the vector field along the triangle $A B C$ is equal to the flux of $\operatorname{rot} \ddot{A}$ through the plane of the triangle $A B C$ with the outward normal (the normal forms an acute angle with the $O Z$ axis), i.e.,
$$
\oint_{A B C} \dot{A} \cdot d \dot{l}=\iint_{S_{A B C}}(\operatorname{rot} \dot{A} \cdot \dot{n}) d S
$$
We will calculate the necessary quantities entering the right-hand side of Stokes' formula:
$$
\begin{aligned}
& \operatorname{rot} \dot{A}=\left|\begin{array}{ccc}
i & j & \dot{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x-2 z & x+3 y+z & 5 x+y
\end{array}\right|=-\dot{j} \cdot 7+\dot{k} ; \\
& S_{A B C}: x+y+z=1 ; \\
& \vec{n}=\frac{1}{\sqrt{3}}(1,1,1) ; \\
& d S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\sqrt{3} d x d y .
\end{aligned}
$$
Thus, the sought circulation is:
$$
\oint_{A B C} \dot{A} \cdot \vec{l}=-6 \int_{\substack{x+y \leqslant 1 \\ x \geqslant 0 \\ y \geqslant 0}} d x d y=-6 \int_{0}^{1} d x \int_{0}^{1-x} d y=-3
$$
General problems related to scalar and vector fields.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.179. $\frac{d x}{d t}=y, \frac{d y}{d t}=-4 x-2 y$.
|
Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(0-\lambda) & 1 \\
-4 & (-2-\lambda)
\end{array}\right|=2 \lambda^{2}+2 \lambda+4=0
$$
The roots of the characteristic equation are complex conjugates, the real part is negative, and the imaginary part is non-zero:
$$
\left.\lambda_{1,2}=-\frac{1}{2} \pm \frac{\sqrt{3}}{2}, \alpha=-\frac{1}{2}<0, \beta=\frac{\sqrt{3}}{2} \neq 0 \text { (case } 2 a\right) .
$$
The equilibrium point is a stable focus.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.180. $\frac{d x}{d t}=x-y, \frac{d y}{d t}=x+y$.
|
Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
(1-\lambda) & -1 \\
1 & (1-\lambda)
\end{array}\right|=\lambda^{2}-2 \lambda+2=0
$$
The roots of the characteristic equation are complex conjugates, with a positive real part and a non-zero imaginary part:
$$
\lambda_{1,2}=1 \pm i (\text{case } 2 \mathrm{~b}: \alpha=1>0, \beta=1 \neq 0).
$$
The equilibrium point is an unstable focus.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.1. $I=\int_{0}^{2} x^{3} d x$
|
Solution. $I=\int_{0}^{2} x^{3} d x=\left.\frac{x^{4}}{4}\right|_{0} ^{2}=\frac{2^{4}}{4}-\frac{0^{4}}{4}=4$.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.2. $I=\int_{0}^{2}(x-1)^{3} d x$.
|
Solution.
$$
\begin{aligned}
I= & \int_{0}^{2}(x-1)^{3} d x=\int_{0}^{2}\left(x^{3}-3 x^{2}+3 x-1\right) d x= \\
& =\int_{0}^{2} x^{3} d x-3 \int_{0}^{2} x^{2} d x+3 \int_{0}^{2} x d x-\int_{0}^{2} d x= \\
& =\left.\frac{x^{4}}{4}\right|_{0} ^{2}-\left.3 \frac{x^{3}}{3}\right|_{0} ^{2}+\left.3 \frac{x^{2}}{2}\right|_{0} ^{2}-\left.x\right|_{0} ^{2}=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.5. $I=\int_{-1}^{1} x|x| d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text provided is already in a form that is commonly used in English for mathematical expressions, the translation is essentially the same:
Example 2.5. $I=\int_{-1}^{1} x|x| d x$.
|
Solution. Since the integrand is an odd function and the limits of integration are symmetric about zero, then
$$
\int_{-1}^{1} x|x| d x=0
$$
In the following two examples, it will be demonstrated that the formal application of the Newton-Leibniz formula (without considering the integrand functions) can lead to an incorrect result.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.13. $\int_{-1}^{1}|x| e^{|x|} d x$.
|
Solution. Here the integrand is an even function, which needs to be utilized, and then apply the integration by parts formula (2.3):
$$
\begin{gathered}
\int_{-1}^{1}|x| e^{|x|} d x=2 \int_{0}^{1} x e^{x} d x=\left|\begin{array}{l}
u=x, d u=d x \\
d v=e^{x} d x, v=\int e^{x} d x=e^{x}
\end{array}\right|= \\
=2\left(\left.x e^{x}\right|_{0} ^{1}-\int_{0}^{1} e^{x} d x\right)=2\left(\left.x e^{x}\right|_{0} ^{1}-\left.e^{x}\right|_{0} ^{1}\right)= \\
=\left.2 e^{x}(x-1)\right|_{0} ^{1}=-2 e^{0}(0-1)=2
\end{gathered}
$$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.14. $\int_{0}^{(\pi / 2)^{2}} \sin \sqrt{x} d x$.
|
Solution. First, let's show an incorrect solution leading to a non-integrable function. Transform the integrand as follows:
$$
\begin{gathered}
\int_{0}^{(\pi / 2)^{2}} \frac{\sin \sqrt{x} \cdot \sqrt{x} d x}{\sqrt{x}}=\left|\begin{array}{l}
u=\sqrt{x}, d u=\frac{d x}{2 \sqrt{x}} \\
d v=\frac{\sin \sqrt{x} d x}{\sqrt{x}}, v=-2 \cos \sqrt{x}
\end{array}\right|= \\
\quad=\left.\sqrt{x}(-2 \cos \sqrt{x})\right|_{0}^{(\pi / 2)^{2}}-\int_{0}^{(\pi / 2)^{2}} \frac{\cos \sqrt{x} d x}{\sqrt{x}}
\end{gathered}
$$
Since $\cos x=1$, and $\frac{1}{\sqrt{x}}$ does not exist at $x=0$, the integrand $\frac{\cos \sqrt{x}}{\sqrt{x}}$ on the interval $\left[0,\left(\frac{\pi}{2}\right)^{2}\right]$ is not integrable in the sense of the definitions provided at the beginning of this section.
To correctly compute the given definite integral, we will use the method of substitution and set $x=z^{2}, d x=2 z d z$, the value $x_{0}=0$ will correspond to $z_{0}=0$, and the value $x_{1}=\left(\frac{\pi}{2}\right)^{2}-z_{1}=\frac{\pi}{2}$. After the specified substitution, the original integral will take the form:
$$
\begin{gathered}
I=\int_{0}^{\pi / 2} \sin z(2 z d z)=\left\lvert\, \begin{array}{l}
u=z, d u=d z \\
d v=\sin z d z, v=-\cos z
\end{array}\right. \\
\quad=2\left(-\left.z \cos z\right|_{0}^{\pi / 2}-\int_{0}^{\pi / 2}(-\cos z) d z\right) \\
=-\left.2(-z \cos z+\sin z)\right|_{0}^{\pi / 2}=2
\end{gathered}
$$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.21. Find the limit $\lim _{x \rightarrow 0}\left(\left(\int_{0}^{x^{2}} \cos x d x\right) / x\right)$.
|
Solution. We will use L'Hôpital's rule, as there is an indeterminate form of type «0/0».
$$
\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos x d x}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x^{2}} \cos x d x\right)^{\prime}}{x^{\prime}}=\lim _{x \rightarrow 0} \frac{\cos \left(x^{2}\right) \cdot 2 x}{1}=0
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.22. Find the limit
$$
\lim _{x \rightarrow 0}\left(\int_{0}^{\operatorname{arctg} x} e^{\sin x} d x / \int_{0}^{x} \cos \left(x^{2}\right) d x\right)
$$
|
Solution. According to L'Hôpital's rule and the rules for differentiating definite integrals, we transform the limit as follows:
$$
\lim _{x \rightarrow 0}\left(\int_{0}^{\operatorname{arctg} x} e^{\sin x} d x\right)^{\prime} /\left(\int_{0}^{x} \cos \left(x^{2}\right) d x\right)^{\prime}=\lim _{x \rightarrow 0} \frac{e^{\sin (\operatorname{arctg} x)}}{\left(1+x^{2}\right) \cos x^{2}}=1
$$
## 2.3 . IMPROPER INTEGRALS
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.25. $\int_{-\infty}^{0} e^{x} d x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 2.25. $\int_{-\infty}^{0} e^{x} d x$.
|
Solution. According to formula (2.6), we get
$$
\int_{-\infty}^{0} e^{x} d x=\lim _{a \rightarrow-\infty} \int_{a}^{0} e^{x} d x=\left.\lim _{a \rightarrow -\infty} e^{x}\right|_{a} ^{0}=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \operatorname{arctg} x d x$.
Translating the above text into English, while preserving the original text's line breaks and format, yields:
Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \arctan x d x$.
|
Solution.
$$
\begin{gathered}
I=\lim _{a \rightarrow \infty} \int_{-a}^{a} \operatorname{arctg} x d x=\left|\begin{array}{l}
\left.u=\operatorname{arctg} x . d u=\frac{d x}{1+x^{2}} \right\rvert\,= \\
d v=x, v=x
\end{array}\right|= \\
=\lim _{a \rightarrow \infty}\left(\left.x \operatorname{arctg} x\right|_{-a} ^{a}-\int_{-a}^{a} \frac{x d x}{1+x^{2}}\right)= \\
=\lim _{a \rightarrow \infty}(a \operatorname{arctg} a-(-a) \operatorname{arctg}(-a)- \\
\left.-0.5 \ln \left(1+a^{2}\right)+0.5 \ln \left(1+(-a)^{2}\right)\right)=0
\end{gathered}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.41. $I=\int_{0}^{1} \ln x d x$.
|
Solution.
$$
\begin{gathered}
I=\left|\begin{array}{l}
u=\ln x, d u=\frac{d x}{x} \\
d v=d x, v=x
\end{array}\right|=\lim _{a \rightarrow+0}\left(\left.x \ln x\right|_{a} ^{1}-\int_{a}^{1} x \frac{d x}{x}\right)= \\
=-\lim _{a \rightarrow+0} \frac{\ln a}{1 / a}-1=[\text{ apply L'Hôpital's rule] }= \\
=\lim _{a \rightarrow+0} \frac{1 / a}{1 / \dot{a}^{2}}-1=-1 .
\end{gathered}
$$
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.45. $I=\mathrm{V} . \mathrm{p} . \int_{1 / e}^{e} \frac{d x}{x \ln x}$. Calculate the improper integral of the second kind in the sense of the principal value.
|
Solution. According to formula (2.18)
$$
\begin{gathered}
I=\lim _{a \rightarrow+0}\left(\int_{1 / e}^{1-a} \frac{d(\ln x)}{\ln x}+\int_{1+a}^{e} \frac{d(\ln x)}{\ln x}\right)= \\
=\lim _{a \rightarrow+0}\left(\ln \left|\ln x\left\|_{1 / e}^{1-a}+\ln \mid \ln x\right\|_{1+a}^{e}\right)=\right. \\
=\lim _{a \rightarrow+0}(\ln |\ln (1-a)|-\ln |\ln 1-\ln e|+\ln |\ln e|-\ln |\ln (1+a)|)= \\
=\lim _{a \rightarrow+0} \ln \left|\frac{\ln (1-a)}{\ln (1+a)}\right|=\ln \left|\lim _{a \rightarrow+0} \frac{(\ln (1-a))^{\prime}}{(\ln (1+a))^{\prime}}\right|= \\
=\ln \left|\lim _{a \rightarrow+0}\left(-\frac{1+a}{1-a}\right)\right|=\ln 1=0 .
\end{gathered}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.46. Find the average value of the function $u(x)=$ $=1 / \sqrt{x}$ on the half-interval $x \in(0,1]$.
|
Solution.
$$
M(x)=\lim _{a \rightarrow+0}\left(\int_{a}^{1} \frac{d x / \sqrt{x}}{1-0}\right)=\lim _{a \rightarrow+0} \frac{2 \sqrt{1}-2 \sqrt{a}}{1}=2
$$
## 2.4. GEOMETRIC APPLICATIONS OF DEFINITE INTEGRALS
## Area of a Plane Curve
The area of a plane figure bounded by curves given by their equations in Cartesian coordinates: above - $y=y_{1}(x)$, below - $y=y_{2}(x)$, and on the left and right by segments of the lines $x=a$ and $x=b$, is calculated by the formula
$$
S=\left|\int_{a}^{b}\left(y_{1}(x)-y_{2}(x)\right) d x\right|
$$
The area of a plane figure bounded by curves given by their equations in Cartesian coordinates: to the right - $x=x_{1}(y)$, to the left - $x=x_{2}(y)$, and below and above by segments of the lines $y=c$ and $y=d$, is calculated by the formula
$$
S=\left|\int_{c}^{d}\left(x_{1}(y)-x_{2}(y)\right) d y\right|
$$
The area of a closed plane figure bounded by a curve given by parametric equations: $x=x(t)$, $y=y(t), 0 \leqslant t \leqslant T$ can be calculated using one of three formulas (assuming that as $t$ increases from 0 to $T$, the curve is traversed counterclockwise).
$$
S=\int_{0}^{T} y(t) x^{\prime}(t) d t=\int_{0}^{T} x(t) y^{\prime}(t) d t=0.5 \int_{0}^{T}\left(x(t) y^{\prime}(t)+y(t) x^{\prime}(t)\right) d t
$$
The area of a figure bounded by a curve given by an equation in polar coordinates $\rho=\rho(\varphi)$ and segments of two rays $\varphi=\alpha$ and $\varphi=\beta$, is calculated by the formula
$$
S=0.5\left|\int_{\alpha}^{\beta}(\rho(\varphi))^{2} d \varphi\right|
$$
## Arc Length
The length of an arc of a smooth curve given by the equation in Cartesian coordinates $y=y(x)(a \leqslant x \leqslant b)$, is calculated by the formula
$$
L=\left|\int_{a}^{b} \sqrt{1+\left(y^{\prime}(x)\right)^{2}} d x\right|
$$
The length of an arc of a smooth curve given by the equation in Cartesian coordinates $x=x(y)(c \leqslant y \leqslant d)$, is calculated by the formula
$$
L=\left|\int_{c}^{d} \sqrt{1+\left(x^{\prime}(y)\right)^{2}} d y\right|
$$
The length of an arc of a smooth curve given by parametric equations $x=x(t)$ and $y=y(t)\left(t_{1} \leqslant t \leqslant t_{2}\right)$ is calculated by the formula
$$
L=\left|\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t\right|
$$
The length of an arc of a smooth curve given by parametric equations in three-dimensional space $x=x(t), y=y(t)$, $z=z(t)\left(t_{1} \leqslant t \leqslant t_{2}\right)$ is calculated by the formula
$$
L=\left|\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}+\left(z^{\prime}(t)\right)^{2}} d t\right|
$$
The length of an arc of a smooth curve given by a polar equation, $\rho=\rho(\varphi), \alpha \leqslant \varphi \leqslant \beta$ is calculated by the formula
$$
L=\left|\int_{a}^{\beta} \sqrt{\left.\left.(\rho(\varphi))^{2}\right)+\left(\rho^{\prime}(\varphi)\right)^{2}\right)} d \varphi\right|
$$
## Surface Area of Revolution
If a segment of a smooth curve is rotated around an arbitrary axis, the area of the surface formed is calculated by the formula
$$
S=2 \pi\left|\int_{r_{1}}^{r_{2}} G(l(r)) d l(r)\right|
$$
where $d l$ is the differential of the arc $l, G(l(r))$ is the distance from a point on the curve to the axis of rotation, $r_{1}$ and $r_{2}$ are the values of the arc variable $r$ corresponding to the ends of the arc.
If a segment of a smooth curve given by the equation in Cartesian coordinates $y=f(x)(a \leqslant x \leqslant b)$ is rotated around the $O X$ axis, the area of the surface formed is calculated by the formula
$$
S_{x}=2 \pi\left|\int_{a}^{b} f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x\right|
$$
If a segment of a smooth curve given by the equation in Cartesian coordinates $x=g(y)(c \leqslant x \leqslant d)$ is rotated around the $O Y$ axis, the area of the surface formed is calculated by the formula
$$
S_{y}=2 \pi\left|\int_{a}^{b} g(y) \sqrt{1+\left(g^{\prime}(y)\right)^{2}} d y\right|
$$
If a segment of a smooth curve given by parametric equations $x=x(t)$ and $y=y(t)\left(t_{1} \leqslant t \leqslant t_{2}\right)$ is rotated around the $O X$ axis, the area of the surface formed is calculated by the formula
$$
S_{x}=2 \pi\left|\int_{t_{1}}^{t_{2}} y(t) \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t\right|
$$
If a segment of a smooth curve given by parametric equations $x=x(t)$ and $y=y(t)\left(t_{1} \leqslant t \leqslant t_{2}\right)$ is rotated around the $O Y$ axis, the area of the surface formed is calculated by the formula
$$
S_{y}=2 \pi\left|\int_{t_{1}}^{t_{2}} x(t) \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t\right|
$$
If a segment of an arc given by the equation in polar coordinates $\rho=\rho(\varphi)(\alpha \leqslant \varphi \leqslant \beta)$ is rotated around the polar axis, the area of the surface of revolution is calculated by the formula
$$
S_{x}=2 \pi\left|\int_{\alpha}^{\beta} \rho(\varphi) \sin \varphi \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi\right|
$$
Volume of a Solid
Let $S(x)$ be a continuous function on the interval $[a, b]$ which is equal to the area of a cross-section perpendicular to the $O X$ axis of some solid, then the volume of this solid is calculated by the formula
$$
V=\left|\int_{a}^{b} S(x) d x\right|
$$
Let $Q(y)$ be a continuous function on the interval $[c, d]$ which is equal to the area of a cross-section perpendicular to the $O Y$ axis of some solid, then the volume of this solid is calculated by the formula
$$
V=\left|\int_{a}^{b} Q(y) d y\right|
$$
If the curvilinear trapezoid $y=y(x) \geqslant 0, a \leqslant x \leqslant b$ is rotated around the $O X$ axis, the volume of the resulting solid of revolution is found by the formula
$$
V=\pi \int_{a}^{b}(f(x))^{2} d x
$$
If the curve is given parametrically $x=x(t) \geqslant 0, y=$ $=y(t) \geqslant 0, t_{1} \leqslant t \leqslant t_{2}$, and on this segment $x^{\prime}(t) \geqslant 0$, then the volume of the solid of revolution around the $O X$ axis is
$$
V=\left|\pi \
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.48. Calculate the area of the figure bounded by the parabola $y=-x^{2}+3 x-2$ and the coordinate axes.
|
Solution. Since the figure $y=-x^{2}+3 x-2$ is located in the regions with $y \geqslant 0$ and $y<0$ (Fig. 2.3), the area $S$ should be calculated separately for the part $y \geqslant 0$ and the part $y<0$, and then the absolute values of the obtained integrals should be added:
$$
\begin{gathered}
S=\left|\int_{0}^{1}\left(-x^{2}+3 x-2\right) d x\right|+\left|\int_{1}^{2}\left(-x^{2}+3 x-2\right) d x\right|= \\
\left.=\left.\left|\left(-\frac{x^{3}}{3}+\frac{3 x^{2}}{2}-2 x\right)\right|_{0}^{1}|+|\left(-\frac{x^{3}}{3}+\frac{3 x^{2}}{2}-2 x\right)\right|_{1} ^{2} \right\rvert\,= \\
=\left|-\frac{5}{6}\right|+\left|\frac{1}{6}\right|=1
\end{gathered}
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2.58. Compute the length of the arc of the astroid $x=\cos ^{3} t$, $y=\sin ^{3} t, 0 \leqslant t \leqslant 2 \pi$.
|
Solution. From Fig. 2.9, it is clear that due to symmetry, it is sufficient to compute only a quarter of the astroid for
Fig. 2.9.
$0 \leqslant t \leqslant \pi / 2$. We will use formula (2.26):
$$
\begin{gathered}
L=\int_{0}^{\pi / 2} \sqrt{9 \sin ^{2} t \cos ^{4} t+9 \sin ^{4} t \cos ^{2} t} d t= \\
=\int_{0}^{\pi / 2} \sqrt{9 \sin ^{2} t \cos ^{2} t\left(\sin ^{2} t+\cos ^{2} t\right)} d t= \\
=12 \int_{0}^{\pi / 2}|\sin t \cos t| d t=6 \int_{0}^{\pi / 2}|\sin 2 t| d t=\left.6(-\cos (2 t))\right|_{0} ^{\pi / 2}=12
\end{gathered}
$$
Under the integral, it is necessary to write the arithmetic value of the root. If this is not done, an incorrect result can be obtained. In the example considered here, if instead of $3 \int_{0}^{\pi / 2}|\sin t \cos t| d t$ the integral $3 \int_{0}^{\pi / 2} \sin t \cos t d t$ were written, the result would be $\left.\frac{3}{2} \sin ^{2} t\right|_{0} ^{2 \pi}=0-$ a zero arc length.
|
12
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3.3. Find the level surfaces of the function $y=$ $=\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}$ and the value of the function at the point $P(1,1,3)$.
|
Solution. According to the definition of level surfaces, we have: $\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}=C$, where $C \geqslant 0$. From this, it follows that $36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}=C^{2}$, that is, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=$ $=36-C^{2}$ (obviously, $0 \leqslant C \leqslant 6$). The obtained equation represents the equation of a sphere in the rectangular Cartesian coordinate system $O x_{1} x_{2} x_{3}$ with its center at the origin and radius $\sqrt{36-C^{2}}$. When $\overparen{C}=6$, the level surface degenerates into the point 0.
Now let's find the value of the function at the point $P(1,1,3)$. For this, we substitute $x_{1}=1, x_{2}=1, x_{3}=3$ into the analytical expression defining the function. We get $y=\sqrt{36-1^{2}-1^{2}-3^{2}}=$ $=\sqrt{25}=5$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3.9. Find the limit of the function $f(x, y)=\left(x^{2}+\right.$ $\left.+y^{2}\right)^{2} x^{2} y^{2}$ as $x \rightarrow 0$ and $y \rightarrow 0$.
|
Solution. For the calculation of the specified limit, it is more convenient to switch to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$. We obtain
$$
\begin{gathered}
\lim _{\substack{x \rightarrow 0 \\
y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{2 x^{2} y^{2}}=\lim _{r \rightarrow 0}\left(r^{2}\right)^{2 r^{4} \cos ^{2} \varphi \sin ^{2} \varphi}= \\
=\lim _{r \rightarrow 0}\left(r^{2}\right)^{r^{4} \sin ^{2} 2 \varphi / 2}=e^{\lim _{r \rightarrow 0} r^{4} \sin ^{2} 2 \varphi \ln r}= \\
=e^{\sin ^{2} 2 \varphi \lim _{r \rightarrow 0} r^{4} \ln r}=e^{0}=1
\end{gathered}
$$
Note that the equality to zero of the exponent of the exponential in the considered example follows from the following considerations. The exponent of the exponential represents the product of a bounded function $\sin 2 \varphi$ by an infinitesimal quantity, since calculating the limit $\lim _{r \rightarrow 0} r^{4} \ln r$ by L'Hôpital's rule,
$$
\lim _{r \rightarrow 0} r^{4} \ln r=\lim _{r \rightarrow 0} \frac{(\ln r)^{\prime}}{\left(r^{-4}\right)^{\prime}}=\lim _{r \rightarrow 0}\left(-\frac{r^{4}}{4}\right)=0
$$
Therefore, the value of the exponent of the exponential as $r \rightarrow 0$ is zero.
A function $f(P)$ is called continuous at a point $P_{0} \in$ $D(f) \subset E^{n}$ if: 1) it is defined both at the point $P_{0}$ and in some neighborhood of it; 2) if for any $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for all points $P$ in the domain of definition of the function $D(f)$, for which $\rho\left(P, P_{0}\right)$ $>0\left(f\left(P_{0}\right)0(f(P)0$ there exists such $\delta(\varepsilon)>0$ that for any points $P_{1}, P_{2} \in X$, for which $\rho\left(P_{1}, P_{2}\right)<\delta$ the inequality $\left|f\left(P_{1}\right)-f\left(P_{2}\right)\right|<\varepsilon$ holds.
If a function $f(P)$ is continuous on a bounded closed set $X$, then it is uniformly continuous on $X$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 4.37. How many times can the trigonometric series $\sum_{k=1}^{\infty} \frac{\cos k x}{k^{4}}$ be differentiated term by term?
|
Solution. In this example, $a_{k}=\frac{1}{k^{4}}$ and the series with the general term $k^{s} a_{k}$ will converge for $s=1$ and $s=2$ (for $s=3$ we get the harmonic series, and for $s>3$ the general term of the series will be an infinitely large quantity). Therefore, by Theorem 4.28, this series can be term-by-term differentiated twice.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5. Expand the function
$$
f(z)=\frac{z}{z^{2}-2 z-3}
$$
into a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series.
|
Solution. Let's decompose the given function into partial fractions:
$$
\frac{z}{z^{2}-2 z-3}=\frac{1}{4} \frac{1}{z+1}-\frac{3}{4} \frac{1}{z-3}
$$
Transform the right-hand side as follows:
$$
f(z)=\frac{1}{4} \frac{1}{1+z}-\frac{1}{4} \frac{1}{1-\frac{2}{3}}
$$
Using the expansion (12) of the function $\frac{1}{1+z}$, we get
$$
\begin{aligned}
f(z) & =\frac{1}{4}\left(1-z+z^{2}-z^{3}+\ldots\right)-\frac{1}{4}\left(1+\frac{z}{3}+\frac{z^{2}}{9}+\ldots\right)= \\
& =\frac{1}{4}\left(-\frac{4}{3} z+\frac{8}{9} z^{2}-\frac{28}{27} z^{3}+\ldots\right)=-\frac{z}{3}+\frac{2}{3^{2}} z^{2}-\frac{7}{3^{3}} z^{3}+\ldots
\end{aligned}
$$
The nearest singular point to the point $z_{0}=0$ of the given function is the point $z=-1$. Therefore, the radius of convergence of the obtained series is $R=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5. Investigate the convergence of the infinite product
$$
\prod_{k=1}^{\infty}\left(1-\frac{1}{k+1}\right)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{k+1}\right) \ldots
$$
|
Solution. Here all $u_{k}=-\frac{1}{k+1}$ are negative and the series (14)
$$
\sum_{k=1}^{\infty} u_{k}=-\sum_{k=1}^{\infty} \frac{1}{k+1}=-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)
$$
obviously diverges.
Then, by Theorem 4, the infinite product (15) diverges.
Remark. Calculating the $n$-th partial product in (15), we get
$$
p_{n}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{n-1}{n} \cdot \frac{n}{n+1}=\frac{1}{n+1}
$$
Since $p=\lim _{n \rightarrow \infty} p_{n}=\lim _{n \rightarrow \infty} \frac{1}{n+1}=0$, the infinite product (15) diverges.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Find the order of the zero $z_{0}=0$ for the function
$$
f(z)=\frac{z^{8}}{z-\sin z}
$$
|
Solution. Using the Taylor series expansion of the function $\sin z$ in the neighborhood of the point $z_{0}=0$, we obtain
$$
\begin{aligned}
f(z) & =\frac{z^{8}}{z-\sin z}=\frac{z^{8}}{z-\left(z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots\right)}= \\
& =\frac{z^{8}}{\frac{z^{3}}{3!}-\frac{z^{5}}{5!}+\ldots}=\frac{z^{5}}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}=z^{5} \frac{1}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}
\end{aligned}
$$
Let
$$
\varphi(z)=\frac{1}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}
$$
Then $f(z)=z^{5} \varphi(z)$, where $\varphi(z)$ is a function analytic at the point $z_{0}=0$, and $\varphi(0)=6 \neq 0$. Therefore, the point $z_{0}=0$ is a zero of the fifth order for the given function.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 10. Determine the nature of the singular point $z=1$ of the function
$$
f(z)=\frac{\sin \pi z}{2 e^{z-1}-z^{2}-1}
$$
|
Solution. Consider the function
$$
\varphi(z)=\frac{1}{f(z)}=\frac{2 e^{z-1}-z^{2}-1}{\sin \pi z}
$$
The point $z=1$ is a zero of the third order for the numerator
$$
\psi(z)=2 e^{z-1}-z^{2}-1
$$
since
$$
\begin{gathered}
\psi(1)=0 ; \quad \psi^{\prime}(1)=\left.\left(2 e^{z-1}-2 z\right)\right|_{z=1}=0 ; \\
\psi^{\prime \prime}(1)=\left.\left(2 e^{z-1}-2\right)\right|_{z=1}=0 ; \quad \psi^{\prime \prime \prime}(1)=\left.2 e^{z-1}\right|_{z=1}=2 \neq 0
\end{gathered}
$$
The point $z=1$ is a zero of the first order for the denominator $\sin \pi z$ of the function $\varphi(z)$.
Therefore, the point $z=1$ will be a zero of order $3-1=2$ for the function $\varphi(z)$, and thus a pole of the second order for the given function.
## Problems for Independent Solution
Determine the nature of the singular point $z_{0}=0$ for the following functions:
300. a) $\frac{1}{z-\sin z}$
b) $\frac{1}{\cos z-1+\frac{1}{2} z^{2}}$
c) $\frac{1}{e^{-z}+z-1}$
301. a) $\frac{\sin z}{e^{-z}+z-1}$; b) $\frac{\operatorname{sh} z}{z-\operatorname{sh} z}$.
Find the singular points and determine their nature for the following functions:
302. a) $\frac{1}{1-\sin z}$;
b) $\frac{1-\cos z}{z^{2}}$
303. a) $e^{1 /(x+2)}$;
b) $\cos \frac{1}{z}$
304. a) $\frac{z}{z^{5}+2 z^{4}+z^{3}}$
b) $\frac{1}{e^{-z}-1}+\frac{1}{z^{2}}$.
305. a) $e^{-1 / z^{2}}$
b) $\sin \frac{\pi}{z+1}$
c) $\operatorname{ch} \frac{1}{z}$
306. a) $\frac{z^{2}}{\cos z-1}$
b) $\frac{1-\sin z}{\cos z}$
c) $\frac{z-\pi}{\sin ^{2} z}$
The following statements are true.
1. For a point $z_{0}$ to be a removable singular point of the function $f(z)$, it is necessary and sufficient that the Laurent series expansion of $f(z)$ in a neighborhood of the point $z_{0}$ does not contain the principal part.
2. For a point $z_{0}$ to be a pole of the function $f(z)$, it is necessary and sufficient that the principal part of the Laurent series expansion of $f(z)$ in a neighborhood of $z_{0}$ contains only a finite number of terms
$$
f(z)=\frac{c_{-k}}{\left(z-z_{0}\right)^{k}}+\ldots+\frac{c_{-1}}{z-z_{0}}+\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n} \quad\left(c_{-k} \neq 0\right)
$$
The largest of the exponents of the powers of the differences $z-z_{0}$ contained in the denominators of the terms of the principal part of the Laurent series coincides with the order of the pole.
3. A point $z_{0}$ is a singular essential point for the function $f(z)$ if and only if the principal part of its Laurent series expansion in a neighborhood of the point $z_{0}$ contains infinitely many terms.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 4. Find the residue of the function
$$
f(z)=z^{3} \cdot \sin \frac{1}{z^{2}}
$$
at its singular point.
|
Solution. A singular point of the function $f(z)$ is the point $z=0$. It is an essential singular point of the function $f(z)$. Indeed, the Laurent series expansion of the function in the neighborhood of the point $z=0$ is
$$
f(z)=z^{3}\left(\frac{1}{z^{2}}-\frac{1}{3!z^{6}}+\frac{1}{5!z^{10}}-\cdots\right)=z-\frac{1}{3!z^{3}}+\frac{1}{5!z^{7}}-\cdots
$$
i.e., it contains an infinite number of terms in the principal part. The residue of the function at the point $z=0$ is zero, since the coefficient $c_{-1}$ in the Laurent series expansion of $f(z)$ is zero.
If the function $f(z)$ is of the form $f(z)=\frac{\varphi(z)}{\psi(z)}$, where the analytic functions $\varphi(z)$ and $\psi(z)$ have zeros of higher than the first order at the point $z_{0}$, then in this case it is convenient to replace the functions $\varphi(z)$ and $\psi(z)$ with their Taylor series expansions in the neighborhood of the point $z_{0}^{\prime}$.
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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