problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|---|
43. Show the incompatibility of the following conditions: polyhedron, all polyhedral angles are trihedral, the number of vertices is odd. | 43. Let $B$ be the number of vertices, $P$ be the number of edges of the polyhedron. Let's calculate the number of edges. Since all polyhedral angles are trihedral, three edges come out of each vertex. But in this way, we count each edge twice (it connects two vertices). Therefore, $3 B=\frac{P}{2}$, from which $2 P=3 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,458 |
81. Find at least two different definitions of a circle. | 81. 82) The geometric locus of points on a plane that are at a given distance from a given point.
2) A closed plane curve of constant curvature. (Other definitions are possible.) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,462 |
86. Indicate logical errors and omit unnecessary words in the following sentences: 1) a ray is a straight line limited on one side; 2) the magnitude of an angle does not depend on the magnitude of its sides; 3) a secant is an infinite straight line passing through any two points of a circle; 4) a segment is a straight ... | 86. 87) A ray is not a line, but a part of a line; 2) the sides of an angle are rays, so the statement makes no sense; 3) the word "infinite" should be omitted; 4) a segment is not a line, but a part of a line; 5) in a rectangle, the diagonals are equal; 6) in a rhombus, the diagonals are perpendicular to each other; 7... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,466 |
96. "Two planes are called mutually perpendicular if, when intersecting, they form right dihedral angles." Which word here is redundant? | 96. The word "intersecting".
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。
注意：最后一句是说明，不是需要翻译的部分，因此不翻译。 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,472 |
97. Find the logical error in the definition: «A dihedral angle is an angle formed by two half-planes originating from the same line». | 97. The concept of "angle" mentioned in the "definition" is not actually a generic concept for "dihedral angle": a "dihedral angle" refers to a spatial figure, while an "angle" refers to a planar figure (before defining the dihedral angle, we know of no other angles except the usual "planar" ones).
(Provide a correct ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,473 |
120. Read the following equalities from left to right and from right to left and formulate the corresponding rules for operations with radicals:
1) $\sqrt[n]{a_{1} a_{2} \ldots a_{k}}=\sqrt[n]{a_{1}} \sqrt[n]{a_{2}} \ldots \sqrt[n]{a_{k}}$
2) $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$. | 120. 121) The root of a product is equal to the product of the roots of the same degree of each factor. The product of roots of the same degree is equal to the root of the same degree of the product of the radicands.
2) The root of a quotient is equal to the root of the same degree of the numerator divided by the root ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,482 |
121. $\sin \left(180^{\circ}-\alpha\right)=\sin \alpha(\alpha-$ an arbitrary angle $)$. What does this statement assert? | 121. The given statement expresses the theorem: “If two angles sum up to $180^{\circ}$, then the sines of these angles are equal.” | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,483 |
122. Which theorems are expressed by the equalities:
$$
\sin (-x)=-\sin x ? \cos (-x)=\cos x ?
$$ | 122. $y=\sin x$ is an odd function; $y=\cos x$ is an even function. | \sinx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,484 |
123. How can the following sentences be briefly recorded using mathematical symbols?
1) None of the numbers $a, b, c$ are equal to zero.
2) At least one of the numbers $a, b, c$ is equal to zero.
3) Either the numbers $a$ and $b$ have opposite signs, or at least one of them is equal to zero. | 123. 124) $a b c \neq 0$; 2) $a b c=0$; 3) $a b \leqslant 0$. | \leqslant0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,485 |
126. In the following sentences, the word "может" (may/can) has different meanings. Explain this difference.
1) The area of a triangle can be found if its base and height are known.
2) The point of intersection of the heights of a triangle may lie outside the triangle. | 126. 127) The word "always" is implied before the word "may". 2) The word "sometimes" is implied before the word "may". | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,487 |
133. Replace the following sentences with pairs of simpler sentences equivalent to them:
1) If $A$ or $B$, then $C$.
2) If $A$, then $B$ and $C$. | 133. 134) If $A$, then $C$; if $B$, then $C$.
2) If $A$, then $B$; if $A$, then $C$. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,489 |
134. Explain the difference between the statements:
1) If $A$ or $B$, then $C$.
2) If $A$ and $B$, then $C$. | 134. In the first case, $C$ is a consequence of both $A$ and $B$. In the second case, for $C$ to be present, both conditions $A$ and $B$ must be satisfied simultaneously. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,490 |
146. Symmetrical figures are equal. Is the converse statement true: "Any two equal figures located in a plane are symmetrical with respect to some axis"? For which figures is the converse statement true? | 146. No. (provide an example!) The converse statement is true, for example, for circles. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,494 |
169. $A$ is a necessary condition for $B$. Formulate a sufficient condition for the negation of $B$. | 169. For $B$ to be false, it is sufficient that $A$ is false. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,505 |
198. What is the necessary and sufficient condition for segments $a, b, c$ to form a triangle? | 198. $a+b>c, \quad|a-b|<c$.
198. $a+b>c, \quad|a-b|<c$. | +b>,\quad|-b|< | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,514 |
200. Break down each of the following sentences into two sentences so that one expresses the direct theorem, and the other expresses the opposite theorem:
1) the geometric locus of points equidistant from two given points is a perpendicular line drawn to the segment connecting these points through its midpoint;
2) for ... | 200. 201) Direct theorem: "If a point lies on the perpendicular drawn to a line segment through its midpoint, then it is equidistant from the endpoints of the segment." Converse theorem: "If a point does not lie on the perpendicular drawn to a line segment through its midpoint, then it is not equidistant from the endpo... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,516 |
211. How, without measuring angles (measuring only distances), can one determine whether a given quadrilateral is a square? Formulate the corresponding criterion for a square.
3 Order 3323
## III. PROOFS
Just as concepts are used to form judgments, judgments are used to construct inferences, which are chains of judg... | 211. You need to measure all sides and diagonals of the quadrilateral. If it turns out that all sides are equal to each other, and also equal to the diagonals, the quadrilateral will be a square. Thus, a rhombus with equal diagonals is a square. | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,518 |
215. Sometimes it is easier to prove a more general theorem than a less general one. Give at least one such example. | 215. "Consider the following stereometric problem: 'A regular octahedron and a straight line occupy a fixed position in space. Find the plane,
## 72
passing through the given line and dividing the octahedron into two equal parts.' This problem may seem complex; however, a slight familiarity with the shape of a regula... | proof | Other | proof | Yes | Yes | olympiads | false | 45,522 |
216. Proving the existence of some object can be easier than constructing (indicating) it. Provide an example. | 216. A classic example is Euclid's proof of the theorem on the infinity of the sequence of prime numbers: for any $n$ prime numbers $2,3,5$, ..., $p_{n}$, we take the number $2 \cdot 3 \cdot 5 \cdot \ldots \cdot p_{n} + 1$, which is not divisible by any of them and, therefore, has a prime divisor that does not coincide... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 45,523 |
223. How to prove the statement: "Not every triangle is acute"? | 223. By giving an example.
The translation is provided as requested, maintaining the original text's line breaks and format. | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,529 |
231. Show that any proof can be given the form of a proof by contradiction if desired. | 231. This is achieved in the following trivial way: assume the opposite of what has been proven and repeat the "direct" proof. The contradiction is evident. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 45,536 |
232. Prove the "Hauber's Theorem": "Suppose several theorems have been proved, the conditions of which are $A_{1}, A_{2}, \ldots, A_{n}$, and the conclusions - respectively $B_{1}, B_{2}, \ldots, B_{n}$. If the conditions $A_{1}, A_{2}, \ldots, A_{n}$ exhaust all possible cases, and each of the conclusions is incompati... | 232. Suppose that at $B_{1}$, $A_{1}$ does not hold; then one of the conditions $\boldsymbol{A}_{2}, A_{3}$, ... must hold - these are the only possible hypotheses by the condition. But from $A_{2}$ follows $B_{2}$, which is incompatible with $B_{1}$; hence, $A_{2}$ cannot hold. Similarly, $A_{3}, A_{4}, \ldots, A_{n}$... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 45,537 |
236. Consider the problem: prove that among the three numbers $\overline{a b c^{1}}, \overline{b c a}, c a b$ either all are divisible by 37, or none of them are divisible by 37.
$1 \overline{a b c}$ - this is the notation for a three-digit number using the digits $a, b, c$. 26
Show that the proposed problem can be r... | 236. Let the truth of the statement "If $\overline{a b c}$ is divisible by 37, then the numbers $\overline{b c a}$ and $\overline{c a b}$ are also divisible by 37" be proven. Due to the "symmetry" (equality) of the numbers $\overline{a b c}, \overline{b c a}, \overline{c a b}$ in this formulation, the statements "If $\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,540 |
245. Prove the following statements by contradiction:
1) For no integer $n$ are the fractions $\frac{n-6}{15}$ and $\frac{n-5}{24}$ simultaneously integers.
2) If a proper fraction is irreducible, then the fraction that complements it to one is also irreducible.
3) There does not exist a polyhedron with an odd number o... | 245. 246) Assuming that for some $n$ simultaneously
$$
\frac{n-6}{15}=A \text { and } \frac{n-5}{24}=B,
$$
where $A$ and $B$ are integers, we get: $n=15 A+6$ and $n=$ $=24 B+5$, that is, $15 A+6=24 B+5$, or $3(8 B-5 A)=1$, which cannot hold for integers $A$ and $B$.
2) Suppose that the fraction $\frac{b-a}{b}$, whic... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,546 |
246. Find the flaw in the indirect proof given below.
Theorem. If a line $A O$ (Fig. 8), intersecting a plane $M$, is perpendicular to any two lines $O B$ and $O C$ drawn in this plane through the point of intersection $O$ of the given line and the plane, then it is perpendicular to any third line $O D$ drawn in the p... | 246. The existence of a perpendicular to a plane is established only after this theorem has been proved. Meanwhile, in the given proof, it is assumed in advance that such a perpendicular must exist. The question about the line $A O$ remains unresolved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,547 |
252. The method of mathematical induction, which uses the transition "from $n$ to $n+1$", is not only used for proofs but also for defining (specifying) certain objects. For example, the sequence
$$
a, a+d, a+2 d, \ldots, a+(n-1) d, \ldots
$$
can be defined using the equations:
40
$$
\begin{gathered}
u_{1}=a \\
u_{... | 252. 253) $u_{1}=a, u_{n+1}=u_{n} \cdot q$
2) $u_{1}=1, u_{n+1}=u_{n}(n+1)$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,552 |
254. When applying the method of mathematical induction, it is always necessary to carefully ensure that the parts of the proof are connected and that the transition from $n$ to $n+1$ is justified. Otherwise, one can arrive at an absurdity. An example of this is the following "proof":
"All natural numbers are equal to... | 254. The flaw in the reasoning is that the first and second parts of the proof are not connected: the transition from $n$ to $n+1$ only makes sense for $n \geqslant 2$. (If we split a group of two numbers $a_{1}$ and $a_{2}$ into two parts, there will be no common elements in the groups, and we cannot draw any conclusi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,553 |
256. Can any true statement be proven by the method of mathematical induction?
Can a statement be refuted by the method of mathematical induction? | 256. 257) To refute the statement, it is sufficient to show that for any $n$, $1+2+\ldots +n<2^{n}$.
2) Using the method of mathematical induction, it needs to be proven that for all $n \geqslant 5$, $n^{2}<2^{n}$. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,555 |
260. There exists an irregular pentagon, each diagonal of which is parallel to one of the sides. Prove this statement in two ways: 1) using general reasoning, employing the property of parallel projection; 2) by constructing at least one such pentagon. | 260. 261) It is known that in a regular pentagon, the diagonals are parallel to the corresponding sides. During parallel projection, a regular pentagon will transform into an irregular one, but the parallelism of sides and diagonals will be preserved.
2) It is easy to obtain the required irregular pentagon by compressi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,557 |
262. Find the error in the following reasoning.
We will prove that the sum of the angles of a triangle is $180^{\circ}$.
For this, let's take an arbitrary triangle $ABC$ (Fig. 9) and divide it into two triangles with the segment $AD$. Denoting the sum of the angles of a triangle by $x$, we get:
$$
x=\angle 1+\angle ... | 262. Denoting the sum of the angles of different triangles by the same letter $x$ is equivalent to the statement,
6 Order 3323
81
that this sum is the same for all triangles. However, this statement (incidentally, equivalent to the one being proved) itself requires proof! | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,559 |
263. Problem. Someone thought of a number. He subtracted 3 from it, divided the result by 2, then added 3 and multiplied by 2. After this, the resulting number was 6 more than the thought number. What number was thought of?
Solution. Let the desired number be $x$. We get the equation
$$
2\left(\frac{x-3}{2}+3\right)=... | 263. We have obtained a contradictory equality. The reasoning contained no errors. Therefore, we must have used an incorrect assumption somewhere. But the only assumption was the assumption of the existence of the desired number $x$. Hence, it was incorrect, and the problem has no solution at all. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,560 |
265. Are the following answers of the student to the teacher's questions correct?
Teacher. Does a triangle with sides $2,3,7$ exist?
Student. It does not exist, because in a triangle any side must be less than the sum of the other two sides, and in our case $7>2+3$.
Teacher. Does a triangle with sides $5,5,5$ exist?... | 265. The student's answer to the first question is correct. As for the second question, the existence of a triangle with sides $5,5,5$ follows not from the fact that any side of a triangle is less than the sum of the other two sides, and that we have $5<5+5$, but from the possibility of constructing an equilateral tria... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,562 |
268. If in triangle $ABC (AB)^{2} = (BC)^{2} + (AC)^{2}$, 44
then by the Pythagorean theorem, triangle $ABC$ is a right triangle. Is this reasoning correct? | 268. The statement that triangle $A B C$ is a right triangle follows not from the Pythagorean theorem, but from its converse. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,565 |
281. Some proof problems can be solved by computation. Prove in this way that the volume of the body obtained by rotating a circular segment with chord $a$ about a diameter parallel to this chord does not depend on the radius of the circle. | 281. Calculating the volume of the body, we obtain the formula $V=\frac{\pi a^{s}}{6}$. The radius of the circle does not enter into the volume expression - this proves the required. | \frac{\pi^{3}}{6} | Calculus | proof | Yes | Yes | olympiads | false | 45,573 |
282. Prove that the fifth power of any natural number ends in the same digit as the number itself. | 282. $0^{5}=0,1^{5}=1,2^{5}=32,3^{5}=243,4^{5}=1024,5^{5}=3125$, $6^{5}=777 \overline{6}, 7^{5} \equiv 1 \overline{6} 80 \overline{7}, 8^{5}=3 \overline{2} 7 \overline{68}, 9^{5}=59 \overline{0} \overline{=} 9$.
But if $\alpha$ is the last digit of a number, then the numbers $n^{5}$ and $\alpha^{5}$ end in the same di... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,574 |
283. Prove the equivalence of the statements:
1) each side of a triangle is less than the sum of the other two;
2) any one side of a triangle is less than the sum of the other two sides and greater than their difference;
3) the largest side of a triangle is less than the sum of the other two sides. | 283. The equivalence of the first and third statements is evident. If each, then, consequently, the larger side of the triangle is less than the sum of the other two sides. Conversely, if the larger side is less than the sum of the other two sides, then it is even more so for the smaller sides. Let's prove the equivale... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,575 |
286. Which of the following statements are consequences of the judgment: "Only some $A$ are $B$ "?
1) Not every $B$ is $A$. 2) Not every $A$ is $B$. 3) Some $B$ are not $A$. 4) Some $A$ are not $B$. 5) Some $B$ are $A$. 6) Some $A$ are $B$. 7) Some non-$B$ are $A$. 8) Some non-$A$ are $B$. 9) All $B$ are $A$. | 286. 2) 4) 5) 6) and 7). | 2)4)5)6)7) | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 45,578 |
306. How to prove the falsity of the identity
$$
\cos (\alpha+\beta)=\cos \alpha+\cos \beta ?
$$ | 306. It is sufficient to provide one counterexample, say,
$$
\begin{aligned}
& \cos \left(45^{\circ}+45^{\circ}\right) \neq \\
& \neq \cos 45^{\circ}+\cos 45^{\circ}
\end{aligned}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,583 |
307. Construct the bisector of an angle whose vertex is inaccessible. Find at least three different solutions. | 307. Let's consider three different solutions to the problem.
Solution 1 (Fig. 25). Inside the given angle $A$, at the same distance $d$ from the sides, we draw lines $A_{1} B_{1}$ and $A_{1} C_{1}$, parallel to $A B$ and $A C$; $d$ can always be chosen such that the point $A_{1}$, the intersection of lines $A_{1} B_{... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,584 |
320. Prove the theorem: "If a line forms right angles with two intersecting lines of a certain plane, then it intersects the given plane."
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 320. Let the line $A B$ form right angles with two intersecting lines $a$ and $b$ of the plane $S$. Suppose that $A B$ is parallel to $S$. Draw the plane $P$ through $A B$ and the point of intersection of the lines $a$ and $b$. The line $C D$ of intersection of the planes $S$ and $P$ is parallel to $A B$, and therefore... | Geometry | MCQ | Yes | Yes | olympiads | false | 45,590 | |
321. Can three circles have one common point? two common points? three common points? four common points? | 321. The answers to the first two questions are positive, to the third one - negative. (Provide an explanation.) | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,591 |
329. What is the greatest number of equal segments that the set $a_{1}, a_{2}, \ldots, a_{n}$ can have, if $a_{1} \neq a_{2}$, $a_{2} \neq a_{3}, \ldots, a_{n-1} \neq a_{n}$? | 329. When $n=2 k$ the maximum number of equal segments that a given set can have is $k$, and when $n=2 k+1$ it is $k+1$. (Prove.) | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,596 |
330. What type of triangle has one angle that is greater than the sum of the other two angles? equal to the sum of the other two angles? | 330. Obtuse triangle; right triangle. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,597 |
332. What type of triangle is it if each of its angles is less than the sum of the other two angles? | 332. Acute-angled triangle. | Acute-angledtriangle | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,599 |
333. Does there exist a triangle in which the sum of any two angles is less than $120^{\circ}$? more than $120^{\circ}$? | 333. The answers to both questions are negative. (Prove it.) | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,600 |
334. Does there exist a point inside a rectangle that is equidistant from all its sides?
*335. Does there exist a quadrilateral, different from a square and an isosceles trapezoid, around which a circle can be circumscribed and into which a circle can also be inscribed?
*336. In a quadrilateral around which a circle ... | 334. In a square (and in any rhombus) - yes; in an unequal-sided parallelo-
Fig. 30
 | \frac{\alpha}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,614 |
354. $a+b+c=d$. Can each of the segments $a, b, c$ be incommensurable with $d$? | 354. For example, $a=4-\sqrt{2}, b=3-\sqrt{2}, c=2 \sqrt{2}$. | =4-\sqrt{2},b=3-\sqrt{2},=2\sqrt{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,616 |
355. Can the ratio of the internal angles of a triangle be chosen arbitrarily? Solve this question also for convex quadrilaterals and pentagons. | 355. Each angle in a convex quadrilateral is less than the sum of the other three angles, and in a convex pentagon, it is less than half the sum of the other four angles. Therefore, for example, in a quadrilateral and a pentagon, the angles cannot be in the ratio of $7: 3: 2: 1$ and $5: 4: 3: 2: 1$. In a triangle, the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,617 |
356. What combinations of interior angles are possible in a convex quadrilateral? in a convex pentagon? (The number of acute, right, and obtuse angles is meant.) | 356. In a convex quadrilateral, the following combinations of angles are possible:
4 right angles;
2 right, 1 acute, and 1 obtuse angle;
1 right, 1 acute, and 2 obtuse angles;
92
Fig. 39... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,618 |
357. Determine the smallest (or largest) number of acute (right, obtuse) exterior angles that a convex $n$-gon can have. | 357. The smallest number of acute exterior angles for a triangle is 0, the largest is 1. Acute and right triangles do not have acute exterior angles. An obtuse triangle has only one acute exterior angle.
, $A C=a, B D=b, B A_{1}=h$. If we shift the diagonal $A C$ parallel to the direction $B C$ by the length $A A_{1}$, we will obtain a rectangular trapezoid $A_{1} B C_{1} D$. Obviously,
$$
A D+B C=A_{1} D+B C_{1}
$$
since when the diagonal $A C$ is shifted parallel, ... | \sqrt{^{2}-^{2}}+\sqrt{b^{2}-^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,622 |
364. Prove by contradiction the theorem: "Through a point, only one plane can be drawn parallel to a given plane." | 364. Suppose through point $A$ two planes $M$ and $N$ are drawn, parallel to a given plane $P$. Draw any plane $T$ through $A$ intersecting the given plane. Plane $T$ will intersect $P$, $M$, and $N$ along lines $a$, $b$, and $c$. All three lines lie in the same plane $T$, and both $b$ and $c$ do not intersect $a$. We ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,623 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{3}-(n-1)^{3}}{(n+1)^{4}-n^{4}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n^{3}-(n-1)^{3}}{(n+1)^{4}-n^{4}}=\lim _{n \rightarrow \infty} \frac{n^{3}-n^{3}+3 n^{2}-3 n+1}{\left((n+1)^{2}-n^{2}\right) \cdot\left((n+1)^{2}+n^{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{3 n^{2}-3 n+1}{\left(n^{2}+2 n+1-n^{2}\right)\lef... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,625 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}=\lim _{n \rightarrow \infty} \frac{n^{-\frac{7}{5}}\left(\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}\right)}{n^{-\frac{7}{5}}\left(\sqrt[5]{n^{7}}-\sqrt{n+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,626 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{5}+1\right)\left(n^{2}-1\right)}-n \sqrt{n\left(n^{4}+1\right)}}{n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{\left(n^{5}+1\right)\left(n^{2}-1\right)}-n \sqrt{n\left(n^{4}+1\right)}}{n}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{5}+1\right)\left(n^{2}-1\right)}-n \sqrt{n\left(n^{4}+1\right)}\right)\left(\sqrt{\left(n^{5}+1\right)\le... | -\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,627 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{2}+\sqrt{n}-1}{2+7+12+\ldots+(5 n-3)}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{n^{2}+\sqrt{n}-1}{2+7+12+\ldots+(5 n-3)}=\lim _{n \rightarrow \infty} \frac{n^{2}+\sqrt{n}-1}{\left(\frac{(2+(5 n-3)) n}{2}\right)}=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{2\left(n^{2}+\sqrt{n}-1\right)}{(5 n-1) n}=\lim _{n \rightarrow \infty} \frac{\... | \frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,628 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+3}{n+1}\right)^{-n^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+3}{n+1}\right)^{-n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{n+1+2}{n+1}\right)^{-n^{2}}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2}{n+1}\right)^{-n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+1}{2}\right)}\right)... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,629 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ ):
$$
\lim _{x \rightarrow-4} \frac{2 x^{2}+6 x-8}{x+4}=-10
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,630 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$$
f(x)=-3 x^{2}-5, x_{0}=2
$$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,631 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x^{3}-x^{2}-x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x^{3}-x^{2}-x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{(x-1)\left(x^{2}-1\right)}= \\
& =\lim _{x \rightarrow 1} \frac{x-1}{x^{2}-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(x+1)}= \\
& ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,632 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[7]{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[7]{x}}=\lim _{x \rightarrow 0} \frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{\sqrt[7]{x}(\sqrt{1+x}+\sqrt{1-x})}= \\
& =\lim _{x \rightarrow 0} \frac{1+x-(1-x)}{\sqrt[7]{x}(\sqrt{1+x}+\sqrt{1-x})}=\lim _{x \rightarrow 0... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,633 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{2^{-3 x}-1} \cdot \ln 2$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\arcsin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{-3 x \ln 2}-1 \sim -3 x \ln 2$, as $x \rightarrow 0(-3 x \ln 2 \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\arcsin 2 x}{2^{-3 x}-1} \cdot \ln 2=\... | -\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,634 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin 3(x+\pi)}$ | Solution
$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin 3(x+\pi)}=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin (3 x+3 \pi)}=$
$=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin (3 x+\pi)}=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{-\sin 3 x}=$
Using the substitution of equivalent infinitesimals:
$\arcsin 2 x \s... | -\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,635 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}$ | ## Solution
Substitution:
$x=y^{12} \Rightarrow y=\sqrt[12]{x}$
$x \rightarrow 1 \Rightarrow y \rightarrow 1$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}=\lim _{y \rightarrow 0} \frac{\sqrt[3]{y^{12}}-1}{\sqrt[4]{y^{12}}-1}= \\
& =\lim _{y \rightarrow 0} \frac{y^{4}-1}{... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,636 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow \pi} \frac{\ln (2+\cos x)}{\left(3^{\sin x}-1\right)^{2}}
$$ | ## Solution
## Substitution:
$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{\ln (2+\cos x)}{\left(3^{\sin x}-1\right)^{2}}=\lim _{y \rightarrow 0} \frac{\ln (2+\cos (y+\pi))}{\left(3... | \frac{1}{2\ln^{2}3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,637 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{\sin 3 x-\tan 2 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{\sin 3 x-\tan 2 x}=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)-\left(e^{2 x}-1\right)}{\sin 3 x-\tan 2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{2 x}-1\right)\right)}{\frac{1}{x}(\sin... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,638 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}=\lim _{x \rightarrow 0} \frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{(1-\cos \sqrt{x})(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{1-\cos x}{(1-\cos \sqrt{x})(1+\sqrt{\cos x})}=
\end{aligned}
$$
Using the substitu... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,639 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(3-\frac{2}{\cos x}\right)^{\operatorname{cosec}^{2} x}
$$ | ## Solution
$\lim _{x \rightarrow 0}\left(3-\frac{2}{\cos x}\right)^{\operatorname{cosec}^{2} x}=\lim _{x \rightarrow 0}\left(e^{\ln \left(3-\frac{2}{\cos x}\right)}\right)^{\frac{1}{\sin ^{2} x}}=$
$$
=\lim _{x \rightarrow 0} e^{\frac{\ln \left(3-\frac{2}{\cos ^{2} x}\right)}{\sin ^{2} x}}=\exp \left\{\lim _{x \righ... | e^{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,640 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\cos \left(\frac{x}{\pi}\right)\right)^{1+x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\cos \left(\frac{x}{\pi}\right)\right)^{1+x}=\left(\cos \left(\frac{0}{\pi}\right)\right)^{1+0}=1^{1}=1$
## Problem Kuznetsov Limits 18-22 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,641 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x-1}\right)^{\frac{\sin (x-1)}{x-1-\sin (x-1)}}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x-1}\right)^{\frac{\sin (x-1)}{x-1-\sin (x-1)}}= \\
& =\lim _{y \rightarrow 0}\left(\frac{\sin ((y+1)-1)}{(y+1)-1}\right)^{\frac{\sin ((y+1)-... | e^{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,642 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}(\sqrt{x}+1)^{\frac{\pi}{\operatorname{arctg} x}}$ | ## Solution
$\lim _{x \rightarrow 1}(\sqrt{x}+1)^{\frac{\pi}{\operatorname{arctg} x}}=(\sqrt{1}+1)^{\frac{\pi}{\operatorname{arctg} 1}}=2^{\frac{\pi}{\left(\frac{\pi}{4}\right)}}=2^{4}=16$
## Problem Kuznetsov Limits 20-22 | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,643 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\cos x+\ln (1+x) \sqrt{2+\cos \left(\frac{1}{x}\right)}}{2+e^{x}}$ | ## Solution
Since $\sqrt{2+\cos \left(\frac{1}{x}\right)}$ is bounded, and $\ln (1+x) \rightarrow 0$, as $x \rightarrow 0$, then $\ln (1+x) \sqrt{2+\cos \left(\frac{1}{x}\right)} \rightarrow 0 \quad$, as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{\cos x+\ln (1+x) \sqrt{2+\cos \left(\frac{1}{x}\right)}}{2... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,644 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-9 ; 5 ; 5\}$
$p=\{4 ; 1 ; 1\}$
$q=\{2 ; 0 ;-3\}$
$r=\{-1 ; 2 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -p-q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,645 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ; 2 ;-3\}$
$b=\{2 ;-1 ;-1\}$
$c_{1}=4 a+3 b$
$c_{2}=8 a-b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
c_{1}=4 a+3 b=\{4 \cdot 1+3 \cdot 2 ; 4 \cdot 2+3 \cdot(-1) ; 4 \cdot(-3)+3 \cdot(-1)\}=\{10 ; 5 ;-15\}
$$
$$
c_{2}=8 a-b=\{8 \c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,646 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-1 ; 2 ;-3), B(3 ; 4 ;-6), C(1 ; 1 ;-1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(3-(-1) ; 4-2 ;-6-(-3))=(4 ; 2 ;-3)$
$\overrightarrow{A C}=(1-(-1) ; 1-2 ;-1-(-3))=(2 ;-1 ; 2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begi... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,647 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p-2 q$
$b=p+5 q$
$|p|=4$
$|q|=\frac{1}{2}$
$(\widehat{p, q})=\frac{5 \pi}{6}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p-2 q) \times(p+5 q)=3 \cdot p \times p+3 \cdot 5 \cdot p \times q-2 \cdot q \... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,648 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ;-3\}$
$b=\{3 ; 2 ; 1\}$
$c=\{2 ; 3 ; 4\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
1 & -1 & -3 \\
3 & 2 & 1 \\
2 & 3 & 4
\end{array}\right|= \\
& =1 \cdot\left|\b... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,649 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ;-1 ; 1)$
$M_{2}(-2 ; 0 ; 3)$
$M_{3}(2 ; 1 ;-1)$
$M_{0}(-2 ; 4 ; 2)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-1 & y-(-1) & z-1 \\
-2-1 & 0-(-1) & 3-1 \\
2-1 & 1-(-1) & -1-1
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y+1 & z-1 \\
-3 & 1 & 2 \\
... | \frac{9}{\sqrt{101}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,651 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-8 ; 0 ; 7)$
$B(-3 ; 2 ; 4)$
$C(-1 ; 4 ; 5)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$\overrightarrow{B C}=\{-1-(-3) ; 4-2 ; 5-4\}=\{2 ; 2 ; 1\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& 2 \cdot(x-(-8... | 2x+2y+z+9=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,652 |
## Problem Statement
Find the angle between the planes:
$$
\begin{aligned}
& 3 x-y+2 z+15=0 \\
& 5 x+9 y-3 z-1=0
\end{aligned}
$$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{3 ;-1 ; 2\}$
$\overrightarrow{n_{2}}=\{5 ; 9 ;-3\}$
The angle $\phi$ between the planes is determined by the formula:
$\cos \phi=\frac{\left(\overr... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,653 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(-1 ;-1 ;-6)$
$C(2 ; 3 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(-1-0)^{2}+(-1-0)^{2}+(-6-z)^{2}}=\sqrt{1+1+36+12 z+z^{2}}=\sqrt{z^{2}+12 z+38}$
$A C=\sqrt{(2-0)^{2}+(3-0)^{2}+(5-z)^{2}}=\sqrt{4+9+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+38}$
Since by the condition of the problem $A B=A C$, then
$$
\begin{aligned}
& \sqrt{... | A(0;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,654 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2 ; 4 ; 1)$
$a: 3 x+y+2 z+2=0$
$k=3$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x+y+2 z+6=0$
Substitute the coordinates of point $A$ into the equat... | 6\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,655 |
## Problem Statement
Write the canonical equations of the line.
\[
\begin{aligned}
& x+y+z-2=0 \\
& x-y-2 z+2=0
\end{aligned}
\] | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}_{\text {- its direction }}$ vector.
Since the line belongs to both planes simultaneously, its dir... | \frac{x}{-1}=\frac{y-2}{3}=\frac{z}{-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,656 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y}{0}=\frac{z+3}{2} \\
& 2 x-y+4 z=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y}{0}=\frac{z+3}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+t \\
y=0 \\
z=-3+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$2(1+t)-0+4(-3+2 t)=0$
$2+2 t+0-12+8 t=0$
$1... | (2;0;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,657 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(1 ; 2 ; 3)$
$\frac{x-0.5}{0}=\frac{y+1.5}{-1}=\frac{z-1.5}{1}$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{0 ;-1 ; 1\}$
Then the equation of the desired plane is:
$0 ... | M^{\}(0;-3;-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,658 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{6 ; 5 ;-14\}$
$p=\{1 ; 1 ; 4\}$
$q=\{0 ;-3 ; 2\}$
$r=\{2 ; 1 ;-1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -2p-q+4r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,659 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a_{\text {and }} b$, collinear?
$$
\begin{aligned}
& a=\{5 ; 0 ;-1\} \\
& b=\{7 ; 2 ; 3\} \\
& c_{1}=2 a-b \\
& c_{2}=3 b-6 a
\end{aligned}
$$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{2}=-3(2 a-b)=-3 c_{1}$ for any $a$ and $b$. That is, $c_{1}=-\frac{1}{3} \cdot c_{2}$, which means the vecto... | c_{1}=-\frac{1}{3}\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,660 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2 ; 1 ;-1), B(6 ;-1 ;-4), C(4 ; 2 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(6-2 ;-1-1 ;-4-(-1))=(4 ;-2 ;-3)$
$\overrightarrow{A C}=(4-2 ; 2-1 ; 1-(-1))=(2 ; 1 ; 2)$
We find the cosine of the angle $\phi_{\text{between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrig... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,661 |
## problem statement
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$a=3 p+2 q$
$$
\begin{aligned}
& b=p-q \\
& |p|=10 \\
& |q|=1 \\
& (\widehat{p, q})=\frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p+2 q) \times(p-q)=3 \cdot p \times p-3 \cdot p \times q+2 \cdot q \times p-2 ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,662 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-2 ; 6\}$
$b=\{1 ; 0 ; 1\}$
$c=\{2 ;-6 ; 17\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
1 & -2 & 6 \\
1 & 0 & 1 \\
2 & -6 & 17
\end{array}\right|= \\
& =1 \cdot\left|\... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,663 |
## Task Condition
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(1 ; 2 ; 0)$
$A_{2}(3 ; 0 ;-3)$
$A_{3}(5 ; 2 ; 6)$
$A_{4}(8 ; 4 ;-9)$ | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{3-1 ; 0-2 ;-3-0\}=\{2 ;-2 ;-3\} \\
& \overrightarrow{A_{1} A_{3}}=\{5-1 ; 2-2 ; 6-0\}=\{4 ; 0 ; 6\} \\
& \overrightarrow{A_{1} A_{4}}=\{8-1 ; 4-2 ;-9-0\}=\{7 ; 2 ;-9\}
\end{aligned}
$$
According to the geometric mea... | 7\frac{2}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,664 |
## problem statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 3 ; 0)$
$M_{2}(4 ;-1 ; 2)$
$M_{3}(3 ; 0 ; 1)$
$M_{0}(4 ; 3 ; 0)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$ :
$$
\left|\begin{array}{ccc}
x-1 & y-3 & z-0 \\
4-1 & -1-3 & 2-0 \\
3-1 & 0-3 & 1-0
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-3 & z \\
3 & -4 & 2 \\
2 & -3 & 1
... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,665 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(0; -3; 5)$
$B(-7; 0; 6)$
$C(-3; 2; 4)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{-3-(-7) ; 2-2 ; 4-6\}=\{4 ; 0 ;-2\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$4 \cdot(x-0)+0 \cdot(y-(-3))-2 \cdot(... | 2x-z+5=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,666 |
## Problem Statement
Find the angle between the planes:
$2 x+2 y+z-1=0$
$x+z-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{2 ; 2 ; 1\} \\
& \overrightarrow{n_{2}}=\{1 ; 0 ; 1\}
\end{aligned}
$$
The angle $\phi_{\text{between the planes is determined by... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,667 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(7 ; 0 ;-15)$
$C(2 ; 10 ;-12)$ | ## Solution
Let's find the distances $A B$ and $A C$:
\[
\begin{aligned}
& A B=\sqrt{(7-0)^{2}+(0-0)^{2}+(-15-z)^{2}}=\sqrt{49+0+225+30 z+z^{2}}=\sqrt{z^{2}+30 z+274} \\
& A C=\sqrt{(2-0)^{2}+(10-0)^{2}+(-12-z)^{2}}=\sqrt{4+100+144+24 z+z^{2}}=\sqrt{z^{2}+24 z+248}
\end{aligned}
\]
Since by the condition of the prob... | A(0;0;-4\frac{1}{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,668 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2 ; 3 ;-3)$
$a: 3 x+2 y-z-2=0$
$k=\frac{3}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x+2 y-z-3=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,669 |
## Problem Statement
Write the canonical equations of the line.
$$
\begin{aligned}
& 4 x+y-3 z+2=0 \\
& 2 x-y+z-8=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vect... | \frac{x-1}{-2}=\frac{y+6}{-10}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,670 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+2}{3}$
$4 x+2 y-z-11=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+2}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+2 t \\
y=1-t \\
z=-2+3 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$4(1+2 t)+2(1-t)-(-2+3 t)-11=0$
$4+8 t... | (3;0;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,671 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(3 ; 3 ; 3)$
$\frac{x-1}{-1}=\frac{y-1.5}{0}=\frac{z-3}{1}$ | ## Solution
Find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as the normal vector of the plane:
$\vec{n}=\vec{s}=\{-1 ; 0 ; 1\}$
Then the equation of the desired plane i... | M^{\}(1;0;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,672 |
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