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34. a) Express the condition of perpendicularity of two vectors if their coordinates are known: $\overrightarrow{r_{1}}\left(x_{1} ; y_{1}\right), \overrightarrow{r_{2}}\left(x_{2} ; y_{2}\right)$.
b) By what formula is the distance between points with the same coordinates expressed? How to calculate the angle between... | 34. a) Since in the case of perpendicular vectors $\overrightarrow{r_{1}} \vec{r}_{2}=0$, the condition of perpendicularity can be expressed by the formula:
$$
x_{1} x_{2}+y_{1} y_{2}=0
$$
b) To determine the distance between points, we will square the scalar difference $\vec{d}=\vec{r}_{1}-\vec{r}_{2}$:
$$
\begin{g... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,237 |
35. On the coordinate plane, there is a line $l, O \perp l$. Consider a vector $\overrightarrow{O P}=\vec{p}$, where $P \in l$; then $p=|\vec{p}|$ is the distance from the line to the origin. Denote by $\varphi$ the angle that $\vec{e}=\frac{\vec{p}}{p}$, the unit vector of $\vec{p}$, forms with the unit vector $\vec{i... | 35. Since vectors $\vec{p}$ and $\vec{r}-\vec{p}$ are perpendicular, we have: $(\vec{r}-\vec{p}) \vec{p}=0$. From this, given that $\vec{p}=\overrightarrow{p e}$, we obtain:
$$
\overrightarrow{r e}=p
$$
Here $\vec{r}=x \vec{i}+y \vec{j}, \vec{e}=\vec{i} \cos \varphi+\vec{j} \sin \varphi$. Multiplying, we get:
$$
x \... | x\cos\varphi+y\sin\varphi=p | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,238 |
36. Prove that the equation of the form
$$
\vec{r} \vec{a}=m
$$
where $\vec{r}$ is a variable radius vector, $\vec{a}$ is a fixed vector, and $m$ is a real number, is the equation of a line. Convert this equation to the normal form: $r \boldsymbol{c}=p$. Prove that in coordinate form, equation (1) has the form:
$$
A... | 36. An equation of the form $\overrightarrow{r a}=m$ is an equation of a straight line, since, setting $\vec{a}=\vec{a}$, where $a=|\vec{a}|$, we can rewrite it as are $=m$, from which $\vec{r} \vec{r} \vec{e}=\frac{m}{a}$, or $\stackrel{+\vec{r}}{r}=p$. It should be noted that $p \geqslant 0$, so if it turns out that ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,239 |
37. Prove that any linear equation of the form $A x + B y + C = 0$ is the equation of a straight line. How to convert it to the vector form $\vec{r} \vec{a} = m$ and to the normal form:
$$
x \cos \varphi + y \sin \varphi = p ?
$$ | 37. To prove that any linear equation of the form $A x+B y+C=0$ is the equation of a straight line, we will move $C$ to the right side and assume that $-C \geqslant 0$. If $C$ does not satisfy this condition, then we will change the sign of each term in the equation to the opposite. Now let: $\vec{r}=: x \vec{i}+y \vec... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,240 |
38*. Prove: if the coordinates $x_{1}$, $y_{1}$ of an arbitrary point $M$ are substituted into the left side of the normal equation of a line in the form $x \cos \varphi + y \sin \varphi - p = 0$, the resulting number is equal in absolute value to the distance from the point $M$ to the given line. What will be the sign... | 38. Suppose that by substituting the coordinates $x_{1}, y_{1}$ of the given point $M$ into the left side of the considered equation, we obtain the number $k$:
$$
x_{1} \cos \varphi + y_{1} \sin \varphi - p = k
$$
By moving $p$ to the right side of the equation, we get:
$$
x_{1} \cos \varphi + y_{1} \sin \varphi = p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,241 |
39. Given equations of two lines:
$$
A_{1} x+B_{1} y+C_{1}=0 \text { and } A_{2} x+B_{2} y+C_{2}=0
$$
a) Calculate the coordinates of the intersection point of these lines and the angle formed by them. Determine the conditions for the parallelism and perpendicularity of these lines.
b)* By converting the given equat... | 39. a) By solving the system of equations of two lines by eliminating one of the variables, we find the formulas for the coordinates ( $x_{1} ; y_{1}$ ) of the intersection point of these lines:
$$
x_{1}=\frac{B_{1} C_{2}-B_{2} C_{1}}{A_{1} B_{2}-A_{2} B_{1}} ; \quad y_{1}=\frac{A_{2} C_{1}-A_{1} C_{2}}{A_{1} B_{2}-A_... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,242 |
40. Write the general equation of a line passing through a given point $M$. Write the equation of a line passing through two given points $M$ and $N$. The points are given by coordinates:
$$
M\left(x_{1} ; y_{1}\right), N\left(x_{2} ; y_{2}\right)
$$ | 40. If the point \( M\left(x_{1} ; y_{1}\right) \) belongs to the line \( A x + B y + C = 0 \), then its coordinates satisfy the equation of this line, i.e.,
\[
A x_{1} + B y_{1} + C = 0
\]
Subtracting the second equation from the first, we get:
\( A\left(x - x_{1}\right) + B\left(y - y_{1}\right) = 0 \) - the gener... | \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,243 |
41. Under what condition do three lines, given by the equations: $A_{1} x+B_{1} y+C_{1}=0 ; \quad A_{2} x+B_{2} y+C_{2}=0 \quad$ and $\quad A_{3} x+B_{3} y+C_{3}=0$, pass through the same point? | 41. Finding the coordinates of the intersection point of the second and third line (see problem 39), we get the following formulas:
$$
x=\frac{\left|\begin{array}{ll}
B_{2} & C_{2} \\
B_{3} & C_{3}
\end{array}\right|}{\left|\begin{array}{ll}
A_{2} & B_{2} \\
A_{3} & B_{3}
\end{array}\right|} ; \quad y=\frac{\left|\beg... | |\begin{pmatrix}A_{1}&B_{1}&C_{1}\\A_{2}&B_{2}&C_{2}\\A_{3}&B_{3}&C_{3}\end{pmatrix}|=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,244 |
42. A circle is defined by its center $C$, which is determined by the radius vector $\overrightarrow{O C}=\vec{c}$, and its radius $R$. Write the equation of the circle in vector and coordinate form, if the coordinates of the point $C$ are $(a ; b)$. How do these equations look depending on the position of the center: ... | 42. Let $\vec{r}$ be a variable vector. The magnitude of the distance $|\vec{r}-\vec{c}|$ is constant and equal to the radius $R$ of the circle. From this, we obtain:
$$
(\vec{r}-\vec{c})^{2}=R^{2}
$$
This is the equation of a circle with center $C(O \vec{C}==\vec{c})$ and radius $R$. If $\vec{r}=x \vec{i}+y \vec{j} ... | x^{2}-2+y^{2}-2=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,245 |
43. Under what conditions does the general second-order equation:
$$
a_{11} x^{2}+2 a_{12} x y+a_{22} y^{2}+2 a_{13} x+2 a_{23} y+a_{33}=0
$$
represent a circle on the coordinate plane? | 43. If we expand the parentheses in the canonical equation of a circle, we get:
$$
x^{2}-2 a x+a^{2}+y^{2}-2 b y+b^{2}-R^{2}=0
$$
Compared to the general second-order equation, this equation has two peculiarities: 1) the coefficients of $x^{2}$ and $y^{2}$ are equal; 2) there is no term containing $x y$. Consider a g... | ^{2}+b^{2}-\frac{p}{k}=R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,246 |
44. Prove the identity:
$$
\stackrel{+\vec{a}}{\vec{b}}=\left(\frac{\vec{a}+\vec{b}}{2}\right)^{2}-\left(\frac{\vec{a}-\vec{b}}{2}\right)^{2}
$$
Apply this identity to prove the following statement: if a circle with center $C$ and radius $R$ is given, $[A B]$ is its diameter, $M$ is an arbitrary point on the plane, a... | 44. To prove the identity, we need to expand the brackets and combine like terms.
Now consider a circle with center $C$ and radius $R$ (Fig. 31); let $[A B]$ be its diameter, and $M$ be an arbitrary point on the plane. In the notation of the problem, we have:
$$
\overrightarrow{M C}=\vec{m}=\frac{\vec{a}+\vec{b}}{2},... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,247 |
45*. Prove that if the coordinates $x_{1}, y_{1}$ of an arbitrary point $M$ are substituted into the left side of the equation of a circle of the form
$$
(x-a)^{2}+(y-b)^{2}-R^{2}=0
$$
the resulting number will be the power of the point $M$ with respect to the given circle. If the point $M$ lies outside the circle, t... | 45. If we substitute $x_{1}$ and $y_{1}$ - the coordinates of an arbitrary point $M$ - into the equation of the circle $(x-a)^{2}+(y-b)^{2}-R^{2}=$ $=0$, we get: $\left(x_{1}-a\right)^{2}+\left(y_{1}-\right.$ $-b)^{2}-R^{2}=m^{2}-R^{2}= \pm p^{2}$, since the sum of the first two terms on the left side of the equation g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,248 |
46**. a) Prove that if two non-concentric circles are given, then the set of points having the same power with respect to both circles is a straight line, which is called the radical axis of these circles.
b) If the centers of three circles do not lie on the same straight line, then the three radical axes of these cir... | 46. a) By equating the left sides of the equations of two circles, we get:
$$
\left(x-a_{1}\right)^{2}+\left(y-b_{1}\right)^{2}-R_{1}^{2}=\left(x-a_{2}\right)^{2}+\left(y-b_{2}\right)^{2}-R_{2}^{2}
$$
or
$$
2\left(a_{2}-a_{1}\right) x+2\left(b_{2}-b_{1}\right) y+R_{2}^{2}-R_{1}^{2}=0
$$
This equation represents a l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,249 |
47**. Let's agree to denote by $K$ the expression standing on the left side of the equation of a circle of the form
$$
(x-a)^{2}+(y-b)^{2}-R^{2}=0
$$
Prove that the family of equations of the form $K_{1}+t K_{2}=0$, where $t$ is a variable parameter, defines a so-called pencil of circles - a family of circles having ... | 47. In the equation of the form $K_{1} + t K_{2}=0$, where $K_{1}$ and $K_{2}$ are abbreviated notations for the left-hand sides of the equations of two circles, the coefficients of $x^{2}$ and $y^{2}$ are the same and equal to $1+t$. In this equation, the term containing $x y$ is absent. Therefore, the equation
$$
K_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,250 |
1**. Prove that a necessary and sufficient condition for a given triangle to be a right triangle is the equality:
$$
2 R+r=p
$$
1where $R$ is the radius of the circumscribed circle, $r$ is the radius of the inscribed circle, and $p$ is the semiperimeter of the triangle. | 1. The condition is necessary. In the right-angled triangle \(ABC\), angle \(C\) is a right angle (Fig. 34). Therefore, the lengths of the segments of the tangents from vertex \(C\) to the points of tangency with the inscribed circle are equal to \(r\) - the radius of this circle. It is known that the lengths of the se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,252 |
6**. Given two circles with centers $O_{1}$ and $O_{2}$ and radii $R_{1}$ and $R_{2}, R_{1}>R_{2}$, lying outside each other or touching each other externally. Let $S$ be the center of homothety with a positive coefficient that transforms these circles into each other. Suppose a secant line, drawn from point $S$, inter... | 6. Let circles with centers $O_{1}$ and $O_{2}$ be mapped onto each other by a homothety with a positive coefficient and center $S$ (Fig. 36). Let a circle with center $O$ touch them externally at points $T_{1}$ and $T_{2}$. Our
, A_{2}\left(x_{2} ; y_{2}\right)$, $A_{3}\left(x_{3} ; y_{3}\right)$. (The vertices are numbered in the order of traversal around the triangle's perimeter counterclockwise.) | 8. Let's first consider the case when vertex $A_{3}$ coincides with the origin, and points $A_{1}$ and $A_{2}$ are positioned such that $A_{1} \widehat{O X}\beta$, then our formula gives the area of triangle $O A_{1} A_{2}$ with a negative sign.
It is easy to see that the area $S$ of triangle $A_{1} A_{2} A_{3}$ with ... | \frac{1}{2}(|\begin{pmatrix}x_{2}&y_{2}\\x_{3}&y_{3}\end{pmatrix}|+|\begin{pmatrix}x_{3}&y_{3}\\x_{1}&y_{1}\end{pmatrix}|+|\begin{pmatrix}x_{1}&y_{1}\\x | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,257 |
9. Derive the formula for calculating the area of an $n$-sided polygon $A_{1} A_{2} \ldots A_{n}$, if its vertices are numbered in the same order as in the previous problem, and their coordinates are given: $A_{1}\left(x_{1} ; y_{1}\right), A_{2}\left(x_{2} ; y_{2}\right), \ldots, A_{n}\left(x_{n} ; y_{n}\right)$. | 9. For the area $S$ of a $p$-sided polygon, just like for a triangle, the formula is:
$$
\begin{aligned}
S= & \frac{1}{2}\left(\left|\begin{array}{ll}
x_{1} & y_{1} \\
x_{2} & y_{2}
\end{array}\right|+\left|\begin{array}{ll}
x_{2} & y_{2} \\
x_{3} & y_{3}
\end{array}\right|+\cdots+\right. \\
& \left.+\left|\begin{arra... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,258 | |
10. Prove that the area of any quadrilateral is equal to half the product of the lengths of its diagonals and the sine of the angle between them. | 10. Consider a quadrilateral $ABCD$ with diagonals $|BD|=m$ and $|AC|=n$, intersecting at point $O$, and the angle between them $\varphi$ (Fig. 38). Let the parallel translation $\overrightarrow{OD}$ map points $A$ and $C$ to $A'$ and $C'$. Then we have:
$$
\begin{gathered}
S_{ADB}=S_{A'BD}=\frac{1}{2} m n' \sin \varp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,259 |
11. To conduct an experiment with fertilizers, an agronomist needed to allocate a plot from a field that has the shape of a convex quadrilateral without parallel sides, occupying one-fifth of the field's area. For this, he divided two opposite boundaries of the field into 5 equal-length segments and connected two pairs... | 11. In quadrilateral $ABCD$, sides $AB$ and $CD$ are divided into five congruent parts: $\left|A_{1} B_{1}\right|=\frac{1}{5}|AB| ;\left|C_{1} D_{1}\right|=\frac{1}{5}|CD| ;$ $A_{1} B_{1} C_{1} D_{1}$ - the highlighted segment (Fig. 39). We draw diagonal $AC$ and divide it into five congruent parts. The midpoints of di... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,260 |
12. Divide the given convex pentagon into two equal-area parts with a straight line passing through a given point on one of the pentagon's sides. | 12. Let point $P$ be on side $AB$ of pentagon $ABCDE$ (Fig. 40). Connect point $P$ to vertices $C$ and $E$, and move vertex $B$ parallel to line $PC$ to point $B'$ on line $CD$, and move vertex $A$ parallel to line $PE$ to point $A'$ on line $DE$. This transforms pentagon $ABCDE$ into an equal-area quadrilateral $PB'DA... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,261 | |
13**. In the plane of triangle $A B C$, a point $M$ is given. This point starts moving in a straight line towards vertex $A$, but stops halfway and begins moving towards vertex $B$, again stopping halfway and moving towards vertex $C$, stopping halfway again, and continues this movement towards vertex $A$ and so on. Pr... | 13. Let's try to construct a triangle \(A_0 B_0 C_0\) inside triangle \(ABC\) such that point \(A_0\) lies on the midpoint of \([A C_0]\), point \(B_0\) on the midpoint of \([B A_0]\), and point \(C_0\) on the midpoint of \([C B_0]\). Suppose the required triangle is already constructed (see Fig. 41), and let \(X = (A ... | \frac{1}{7} | Geometry | proof | Yes | Yes | olympiads | false | 44,262 |
14. A convex $n$-gon is inscribed in a circle, containing the center of the circle. Tangents to the circle are drawn through each of its vertices; they form an $n$-gon circumscribed around the given circle. A point is taken on the considered circle. Prove that the product of the distances from this point to all the tan... | 14. The proof of the statement of the problem is based on the following theorem: the distance from any point on a circle to a line,
Fig. 42
intersecting the circle, is the mean proportiona... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,263 |
15. A regular $n$-sided polygon is inscribed in a circle. By connecting the midpoint of each arc to the two nearest vertices of the $n$-sided polygon with segments, we obtain a regular $2n$-sided polygon. a) Prove that, by sequentially connecting the midpoints of the sides of this $2n$-sided polygon, we obtain a new $2... | 15. After performing the construction specified in the problem (Fig. 43), we obtain an $n$-sided polygon and two $2n$-sided polygons. If $[AB]$ is a side of the $n$-sided polygon (in Fig. 43, $n=4$), and $M$ is a vertex of the $2n$-sided polygon, then in triangle $AMB$, the segment $A_1B_1$ is the midline, therefore $\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,264 |
16*. Repeating the construction of the previous problem, we obtain from a regular $2n$-gon a regular $4n$-gon with the same perimeter, then an $8n$-gon, and so on. Prove that the apothems and radii obtained in this way can be successively calculated by the formulas:
$$
h_{p+1}=\frac{h_{p}+r_{p}}{2}, r_{p+1}=\sqrt{h_{p... | 16. The reasoning we used in the previous problem can obviously be applied to derive similar transition formulas from any regular inscribed polygon to a polygon with the same perimeter and double the number of sides, and formulas (1) and (2) from the previous problem give:
$$
h_{p+1}=\frac{h_{p}+r_{p}}{2}
$$
and
$$
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,265 |
17*. From the formula for the circumference: $C=2 \pi r$, we get: $\pi=$ $=\frac{C}{2 r}$ or $\frac{1}{\pi}=\frac{2 r}{C}$. Consider a square with a perimeter of 2 and sequentially find the values of $h_{p}$ and $r_{p}$ using the formulas from the previous problem. By repeatedly performing the calculations, achieve tha... | 17. We will start from a square with a perimeter equal to 2. Under these conditions, we get $h_{1}=\frac{1}{4}, r_{1}=h_{1} \sqrt{2}=\frac{\sqrt{2}}{4}$. Further calculations are conveniently recorded in the following table:
| $p$ | $h_{p}\left(h_{p+1}=\frac{h_{p}+r_{p}}{2}\right)$ | $r_{p}\left(r_{p+1}=\sqrt{\left.\o... | 0.3183098 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,266 |
18. Given a circle with center $O$, diameter $[C D]$, and tangent $t$ at point $D$. The arc $D A$ corresponds to a central angle of measure $\varphi$.
Let the circle with center $C$ and radius $[C D]$ intersect the ray $C A$ at point $A_{1}$. Prove that the arcs $D A$ and $D A_{1}$ have the same length and that the ra... | 18. The length of the arc $D A$ of a circle with radius $r$ and a central angle of magnitude $\varphi$ (in radians) is given by the formula: $\breve{D A}=r \varphi$ (Fig. 44). The arc $D A_{1}$ corresponds to a radius that is twice as large, but with an angle that is half the size. Therefore, the lengths of the arcs $D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,267 |
19. Using the construction from the previous problem, on the same drawing construct the arc $D A_{2}$, which has the same length as the arcs $D A$ and $D A_{1}$, using a circle with a radius twice the length of $|C D|$. Continuing this sequence of constructions becomes difficult due to the excessively large radii of th... | 19. The construction of points $A_{1}, A_{2}$, $A_{3}, \ldots$ can be carried out without drawing circles. It is sufficient to draw the bisector of the angle between the chord of the arc and the tangent, and then draw a perpendicular to the chord through the end of the arc. The intersection of this perpendicular with t... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,268 |
20*. By sequentially calculating the lengths of the chords $\left[D A||, D A_{1} \mid\right.$, $\left[D A_{2}\right] \ldots$ (see problems $\left.18-19\right)$, derive Euler's formula, which gives the length of the arc $D A$:
$$
\breve{D A}_{\infty}=r \varphi=\lim _{n \rightarrow \infty} \frac{r \sin \varphi}{\cos \fr... | 20. Let's draw $[A B] \perp[C D]$ (Fig. 44). We will get:
$|A B|=r \sin \varphi ;|A D|=\frac{|A B|}{\cos \frac{\varphi}{2}} ;\left|A_{1} D\right|=\frac{|A D|}{\cos \frac{\varphi}{4}} ;\left|A_{2} D\right|=\frac{\left|A_{1} D\right|}{\cos \frac{\varphi}{8}}$
and so on. Substituting these formulas into each other, we w... | \frac{\pi}{2}=\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\cdots | Geometry | proof | Yes | Yes | olympiads | false | 44,269 |
21. Two plane figures are called equicomposite if each can be divided into parts that are respectively congruent to the parts of the other figure. Obviously, equicomposite figures are equal in area, i.e., have equal areas. For polygons, the converse is also true: equal-area polygons are always equicomposite. Prove this... | 21. a) Figure 45 shows how to cut two equal-area parallelograms with congruent bases into respectively congruent parts (the slanted lines of the first parallelogram are parallel to the sides of the second parallelogram).
Note for further, that the relation of equal composition is transitive, i.e., two figures, equally... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,270 |
22*. Above (see § 2, problem 4) we introduced the symbol $(A B C)$ to denote a simple relation of three points $A, B, C, C \in (A B)$: $(A B C) = \frac{\overrightarrow{C A}}{\overrightarrow{C B}}$. It is convenient to introduce a similar symbol $(a b c)$ for three lines $a, b, c$ passing through a single point $S$ (lin... | 22. Considering the homothety with center $S$, mapping point $P$ to some other point $P^{\prime} \in c$, we get: $\left(\frac{m^{\prime}}{m}=\frac{n^{\prime}}{n}\right) \Rightarrow$ $\Rightarrow\left(\frac{m^{\prime}}{n^{\prime}}=\frac{m}{n}\right)$. Further, since $m=|S P| \sin (\widehat{c, a}), n=|S P| \times$ $\time... | ()=\frac{\sin(\widehat{,})}{\sin(,\widehat{b})} | Geometry | proof | Yes | Yes | olympiads | false | 44,271 |
23. Prove: for the lines $x, y, z$, passing through the vertices $A, B, C$ of triangle $ABC$, to intersect at one point or be parallel to each other, it is necessary and sufficient that the following equality holds:
$$
(b c x)(c a y)(a b z)=-1, \text{ where } a=(B C), b=(C A), c=(A B).
$$
(This is an analog of Menela... | 23. We will consider the case when the lines $x, y, z$ intersect at point $S$, and $S \notin a \bigcup b \cup c$. Let $m, n$, and $p$ be the distances from point $S$ to the lines $a, b$, and $c$. If $S$ lies inside the triangle $A B C$, then
$$
(b c x)=-\frac{n}{p},(c a y)=-\frac{p}{m},(a b z)=-\frac{m}{n}
$$
thus,
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,272 |
24*. Let the lines $x, y, z$, as in the previous problem, pass through the vertices of triangle $A B C$ and intersect at a point $S$, which does not lie on the lines $A B, B C, C A$. Consider the lines $x^{\prime}, y^{\prime}, z^{\prime}$ passing through $A, B, C$ and symmetric to the lines $x, y, z$ respectively, with... | 24. Let
$$
(b c x)(c a y)(a b z)=\frac{\sin (\widehat{x, b})}{\sin (\widehat{x, c})} \cdot \frac{\sin (\widehat{y, c})}{\sin (\widehat{y, a})} \cdot \frac{\sin (\widehat{z, a})}{\sin (\widehat{z, b})}=-1
$$
Since, by symmetry, $\left(\widehat{x^{\prime}, b}\right)=(\widehat{x, c}),\left(\widehat{x^{\prime}, c}\right)... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,273 |
25*. Given a triangle $A B C$ and through its vertices $A, B$ and $C$ are drawn respectively the lines $x, y, z$, intersecting at point $S$. Denote $(B C)=a,(C A)=b,(A B)=c$. Draw the lines $a^{\prime}\left\|x, \quad b^{\prime}\right\| y, \quad c^{\prime} \| z ;$ these lines bound the triangle $A^{\prime} B^{\prime} C^... | 25. Due to the congruence of angles with correspondingly parallel sides, we have: $\left(\widehat{a^{\prime}, y^{\prime}}\right)=(\widehat{x, b}),\left(a^{\prime}, z^{\prime}\right)=(\widehat{x, c}),\left(\widehat{b^{\prime}, z^{\prime}}\right)=$ $=(\widehat{y, c}),\left(\widehat{b^{\prime}}, x^{\prime}\right)=(\wideha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,274 |
26*. Prove Ceva's theorem¹: if points $X, Y, Z$ are given on the sides $(B C),(C A)$, $(A B)$ of triangle $A B C$ respectively, then for the lines $(A X),(B Y)$ and $(C Z)$ to intersect at one point or be parallel, it is necessary and sufficient that the condition $(B C X)(C A Y)(A B Z)=-1$ is satisfied. | 26. Let the distances from point $X \in (BC) = a$ to the lines $AC = b$ and $AB = c$ be $n$ and $p$. Denoting the angles of triangle $ABC$ at vertices $A, B, C$ by $\alpha$, $\beta$, and $\gamma$ respectively, and the lines $XA, YB$, and $ZC$ by $x, y, z$, we get:
$$
|XB| = \frac{p}{\sin \beta}, \quad |XC| = \frac{n}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,275 |
27*. In triangle $ABC$, a circle is inscribed, touching the sides $[\bar{B} C], [C A]$, and $[A B]$ at points $M, N$, and $P$ respectively. Prove that the lines $A M, B N$, and $C P$ pass through one point.
28*. On the lines $B C, C A$, and $A B$, which contain the sides of triangle $A B C$, points $X, Y, Z$ are taken... | 27. It is easy to see that $|A N|=|A P|=p-a,|B P|==|B M|=$ $=p-b,|C M|=|C N|=p-c$, where $p$ is the semiperimeter of triangle $A B C$, and $a, b, c$ are the lengths of its sides. From this, we obtain:
$$
(B C M)(C A N)(A B P):=\left(-\frac{p-b}{p-c}\right) \cdot\left(-\frac{p-c}{p-a}\right)\left(-\frac{p-a}{p-b}\right... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,276 |
29*. On the plane, two rectangular coordinate systems $O x y$ and $O^{\prime} x^{\prime} y^{\prime}$ are chosen, such that points $O$ and $O^{\prime}$ do not coincide, the axis $O x$ is not parallel to the axis $O^{\prime} x^{\prime}$, and the units of length in these systems are different.
Does there exist a point on... | 29. Let us be given two frames: $O X Y$, where $|O X|=|O Y|$, and $O^{\prime} X^{\prime} Y^{\prime}$, where $\left|O^{\prime} X^{\prime}\right|=\left|O^{\prime} Y^{\prime}\right|$. These two frames represent two similar and identically oriented figures. According to the statement of problem 19 from § 2, there exists a ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,277 |
31**. In triangle $ABC$, it is given: $\widehat{A}=60^{\circ}, \widehat{B}=45^{\circ}$. Circles are constructed on the medians $[A M]$ and $[B N]$ as diameters. In what ratio does the radical axis of these circles divide the midline of the triangle, parallel to side $[A B]$? | 31. Let the circle constructed on the diameter $[A M]$ intersect the side $[B C]$ at point $P$, and the circle constructed on
Fig. 53 diameter $[B N]$, intersect the side $[A C]$ at point $... | |HB|:|HA|=\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,279 |
32. Can a closed broken line be constructed so that each of its segments intersects with some other segment (at an internal point) and the intersecting segments are mutually perpendicular? Can such a line have 315 segments? Can it be symmetric relative to some axis?
$33^{*}$. Given a scalene triangle $A B C$, divide i... | 32. The answer is given in figure 54. The broken line is symmetrical, the number of its segments is even, but not a multiple of 4 (why?). Such a broken line consisting of 315 segments does not exist. (Try to construct a similar broken line with 6 segments, with 14 segments.) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,280 |
35*. From the given angle, cut off a triangle of given area and given perimeter. The perimeter is given by a segment of length $2 p$, and the area is given as the area of a square with side length $q$.
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | 35. If the perimeter is $2 p$ and the area is $q^{2}$, it is not difficult to construct $r$ - the radius of the inscribed circle of the desired triangle - from the formula $q^{2}=r p$ or from the proportion $r: q==q: p$. As is known, the semiperimeter $p$ is the distance from the vertex of the triangle $A$ to the point... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 44,282 | |
36. Can the plane be "tiled" with congruent quadrilaterals (convex or non-convex)? | 36. Let $A B C D$ be the given quadrilateral. Applying to it the compositions of parallel translations $\overrightarrow{A C}$ and $\overrightarrow{C A}$, we obtain a "strip" of quadrilaterals, the vertices of which are located on three parallel lines. After this, apply to the resulting strip the compositions of transla... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,283 |
37. Can we find a point in this convex polygon such that all projections of it onto the lines containing the sides of the polygon fall not on the sides themselves, but only on their extensions? | 37. Such a point does not exist. To prove this, consider an arbitrary point \( P \) belonging to the convex polygon \( A_{1} A_{2} \ldots A_{n} \). Among the lines containing the sides of the polygon, choose the one closest to point \( P \) - let this be the line \( A_{2} A_{3} \). If we assume that the projection of p... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,284 |
38. On one side of the angle with vertex $O$ there is a point $A$ and outside the angle there is a point $B$ (Fig. 2). Find a point $M$ on the other side of the angle such that the intersection of the given angle with the angle $A M B$ is an isosceles triangle with vertex $M$. | 38. Suppose the problem is solved and point $M$ is the desired one (Fig. 58). Let $\beta$ be the angles at the base of the isosceles triangle $A M C$. Then $\widehat{M}=\pi-2 \beta$. Let point $A^{\prime}$ be symmetric to point $A$ with respect to the line $O M$. We will compute $A^{\prime} \widehat{M} B$. Since $\angl... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,285 |
39. A trapezoid $A B C D$ is inscribed in a circle, with $[B C] \|[A D]$ and $|B C|<|A D|$. A diameter $[C M]$ is drawn through the center of the circle $O$. The line $B M$ intersects the side $[A D]$ at point $K$. Point $L$ is the projection of point $K$ onto the diameter $[C M]$. Calculate $|E F|$, if it is known tha... | 39. Let's connect points $M$ and $D$ (Fig. 59) and denote the angles of triangle $MBD$ as $\alpha$, $\beta$, and $\gamma$. Let $|OL| = m; |OL| = |LC| - |OC|$. In triangle $EOF$, we have: $|OE| = R \cos \alpha, |OF| = \frac{R}{2}, \widehat{EOF} = \alpha + 2\gamma$. From this, we get:
$$
\begin{gathered}
|EF|^2 = R^2 \c... | \frac{3\sqrt{2}}{4}R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,286 |
40. Given a chord $[A B]$ of a circle, point $M$ is its midpoint. On the smaller arc $A B$, arbitrary points $C$ and $E$ are taken. Line $C M$ intersects the circle at point $D$, line $E M$ intersects the circle at point $F$, $(E D) \cap (A B) = P, (C F) \cap (A B) = Q$. Prove that $|A P| = |B Q|$. | 40. Construct circles on segments $O P$ and $O Q$ as diameters (Fig. 60). The first will intersect the chord [ $E D$ ] at point $K$, and the second will intersect the chord $[C F]$ at point $L$. Angles $O K E$ and $O L C$ are right angles (they subtend the diameters), so (OK) $\perp(D E)$ and
 (Fig. 63) such that the sum of its distances from the three vertices of the triangle is minimal. Take an arbitrary point \(M\) inside the triangle... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,290 |
44**. Through two given points, draw a circle that intercepts a chord of given length on a given line. | 44. Let $l$ be a given line, $A$ and $B$ be given points, and $p$ be the length of the chord that needs to be constructed (Fig. 64). $(AB) \cap l = S$, and the desired circle intersects the line $l$ at points $M$ and $N$. Then, by the property of the power of a point with respect to a circle, we have: $|SA| \cdot |SB| ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,291 |
45*. For a triangle to be isosceles, it is necessary and sufficient that two of its angle bisectors are congruent. Prove it. | 45. The condition is necessary. Let in triangle $ABC$ (Fig. 65) $|AB|=|AC|$, $[BM]$ and $[CN]$ be the angle bisectors. Then the symmetry with respect to the angle bisector of $\angle A$ maps point $B$ to point $C$, and point $C$ to point $B$. The lines $(AB)$ and $(AC)$ are also symmetric, as are the bisectors $[BM)$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,292 |
46**. In a plane, there are $n$ arbitrary points, not lying on the same straight line. A line is drawn through each pair of them. Prove that among the drawn lines, there is at least one that passes through exactly two points.
Part II.
STEREOMETRY
| 46. Currently, several solutions to this rather difficult problem are known. We will present the most elementary one.
Let us have $n$ arbitrary points: $P_{1}, P_{2}, \ldots, P_{n}$. The maximum possible number of lines is the number of combinations of these $n$ points taken two at a time, i.e., $\frac{n(n-1)}{2}$.
N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,293 |
1*. Into how many regions do $n$ planes divide space if every three of them have exactly one common point, and no four of them have a common point? | 1. The problem is solved similarly to problem 3 from § 1. To the question: into how many separate parts does $n$ lines divide the plane if every two of them have a common point, and every three do not have a common point? - The answer obtained there was: $F_{2}(n)=\frac{n(n+1)}{2}+1$.
Let now $n$ planes divide the spa... | F_{3}(n)=C_{n}^{3}+C_{n}^{2}+C_{n}^{1}+C_{n}^{0} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,294 |
2. Given two intersecting lines $a$ and $b$ and a point $P$ not lying on them, how can one determine (construct) a line passing through this point and intersecting lines $a$ and $b$? | 2. A line $a$ and a point $P$ define a plane $\alpha$ containing them; a line $b$ and a point $P$ define a plane $\beta$ containing them. Planes $\alpha$ and $\beta$, having a common point $P$, must intersect along a line $l$, which may intersect both lines $a$ and $b$, in which case $l$ is the desired line. If it turn... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,295 |
3. Prove that for two lines to be parallel, it is necessary and sufficient that any plane intersecting one of them also intersects the other (intersection at only one point is meant). Derive from this the theorem on the transitivity of the parallelism relation of lines in space. | 3. The condition is necessary. Let $a \| b$, and $a \subset \gamma, b \subset \gamma$. A plane $\alpha$, intersecting $a$, will also intersect the plane $\gamma$. Let $\alpha \cap \gamma = l$. The line $l$, intersecting one of the parallel lines - $a$, must also intersect the other - $b$. The intersection point of line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,296 |
4. Two triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ lie in different planes, such that $(B C) \cap\left(B^{\prime} C^{\prime}\right)=X, \quad(C A) \cap\left(C^{\prime} A^{\prime}\right)=Y, \quad(A B) \cap$ $\cap\left(A^{\prime} B^{\prime}\right)=Z$. Prove that a) $X \in(Y Z)$; b) either $\left(A A^{\prime}\... | 4. a) Since the planes of the triangles are different, and points $X$, $Y, Z$ belong to both planes simultaneously, all of them lie on the line of intersection of these planes, and $X \in (YZ)$.
b) The lines $BC$ and $B' C'$ with a common point $X$ define a plane $\alpha$, the lines $CA$ and $C' A'$ define a plane $\b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,297 |
5*. Statements similar to those in problem 4 are also valid for triangles lying in the same plane.
a) Prove that if two triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ lie in the same plane and satisfy the condition: $(B C) \cap\left(B^{\prime} C^{\prime}\right)=X$, $(C A) \cap\left(C^{\prime} A^{\prime}\righ... | 5. a) Outside the given plane, construct triangle \(A_{1} B_{1} C_{1}\) such that the lines \(B_{1} C_{1}\), \(C_{1} A_{1}\), and \(A_{1} B_{1}\) pass through points \(X\), \(Y\), and \(Z\) respectively, and then apply the conclusions from the previous problem to it.
To construct triangle \(A_{1} B_{1} C_{1}\), take a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,298 |
6**. The set of lines and points featured in Desargues' theorem is an example of what is known as a configuration: on each line there is the same number of points, and through each point there passes the same number of lines. In the Desargues configuration, we have 10 points: $A, B, C, A^{\prime}, B^{\prime}, C^{\prime... | 6. Let's denote the points of the pentahedron with the numbers: $1,2,3,4,5$. These points define 10 lines: (12), (13), (14), (15), (23), (24), (25), (34), (35), (45) and 10 planes: (123), (124), (125), (134), (135), (145), (234), (235), (245), (345). At the intersection of these lines and planes with a secant plane, we... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,299 |
7. a) In the plane $\alpha$, there are two lines $a$ and $b$, the point of intersection of which is inaccessible. Outside this plane, there is a point $P$. How can you determine the line that contains point $P$ and the inaccessible intersection point of lines $a$ and $b$?
b*) Solve the same problem for the case when $... | 7. a) If one plane is drawn through $P$ and $a$, and another through $P$ and $b$, then the line of intersection of these planes $l$ will also pass through the point of intersection of all three planes, i.e., through the point $S=a \cap b$.
b) Let $a, b$ and $P$ lie in the same plane (Fig. 68). Intersect $a$ and $b$ wi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,300 |
8. What does the set of points belonging to lines that intersect one of two given skew lines and are parallel to the second one represent? | 8. Let $a$ and $b$ be given skew lines. Through the point $A_{1} \in a$, draw a line $a_{1} \| b$. The intersecting lines $a$ and $a_{1}$ define a plane $\alpha$ to which they belong. On the line $a$, take another point $A_{2}$ and draw $a_{2} \| b$ through it. Since $b \| \alpha$, then $a_{2} \subset \alpha$ (if the l... | \alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,301 |
9*. Given a plane $\alpha$ and a triangle $A B C$, the plane of which is not parallel to the plane $\alpha$. Find a point $S$ in space such that the lines $S A, S B, S C$ intersect the plane $\alpha$ at points $A^{\prime}, B^{\prime}, C^{\prime}$, respectively, and the triangle $A^{\prime} B^{\prime} C^{\prime}$ is con... | 9. Let the lines $B C, C A$ and $A B$ intersect the given plane $\alpha$ at points
Fig. 68
 also intersects the other. Derive from this the theorem on the transitivity of the relation of parallelism of planes. | 10. The condition is necessary. Let plane $\alpha$ be parallel to plane $\beta$ and line $l$ have only one common point $A$ with plane $\alpha$. Take an arbitrary point $B$ in plane $\beta$ and draw plane $\gamma$ through line $l$ and point $B$, which will intersect planes $\alpha$ and $\beta$ along parallel lines $a$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,303 |
11. There exists one and only one pair of planes parallel to each other, each containing one of two given skew lines. Prove it. | 11. Let lines $a$ and $b$ be skew. Take a point $A$ on line $a$ and draw a line $b^{\prime} \| b$ through it. Take a point $B$ on line $b$ and draw a line $a^{\prime} \| a$ through it. Planes $\alpha$ and $\beta$ can be drawn through lines $a$ and $b^{\prime}$ and through lines $b$ and $a^{\prime}$, respectively - thes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,304 |
12. Three planes parallel to each other determine two pairs of proportional segments on two lines intersecting these planes. Prove. | 12. Consider planes $\alpha\|\beta\| \gamma$, intersected by two lines $l$ and $l^{\prime}, l \cap \alpha=A, l \cap \beta=B, l \cap \gamma=C, l^{\prime} \cap \alpha=A^{\prime}, l^{\prime} \cap \beta=$ $=B^{\prime}, l^{\prime} \cap \gamma=C^{\prime} \quad$ (Fig. 70).
If $l$ and $l^{\prime}$ are skew, then through any p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,305 |
13*. A line moves in such a way that it always remains parallel to a given plane $\alpha$ and at the same time intersects two given skew lines $a$ and $b$. What line does a point on such a line describe, which divides in a constant ratio the segment of this line between the points of its intersection with the lines $a$... | 13. Let a line move parallel to the plane $\alpha$ and, intersecting lines $a$ and $b$, successively occupy positions $(A_{1} B_{1})$, $\left(A_{2} B_{2}\right),\left(A_{3} B_{3}\right), \ldots$ (Fig. 71). A point dividing the segments in the same ratio successively occupies positions $M_{1}, M_{2}$, $M_{3}, \ldots$. T... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,306 |
14. Given four points \( A, B, C, D \) that do not lie in the same plane. Let \( M \) be the midpoint of \( [A B] \), \( N \) be the midpoint of \( [C D] \), \( P \) be the midpoint of \( [B C] \), \( Q \) be the midpoint of \( [A D] \), \( R \) be the midpoint of \( [A C] \), and \( S \) be the midpoint of \( [B D] \)... | 14. By the property of the midline of a triangle, segments $M P$ and $Q N$ are parallel to [ $A C$ ] and in length equal to $\frac{|A C|}{2}$, therefore $M P N Q$ is a parallelogram and its diagonals $[M N]$ and $[P Q]$ intersect at point $O$ - their common midpoint. Similarly, $P R Q . S$ is also a parallelogram, and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,307 |
15. Given three pairwise intersecting lines $a, b$ and $c$. How to construct a line that intersects $a$ and $b$ and is parallel to $c$? Does such a line always exist? | 15. Let's construct plane $\alpha$ such that $a \subset \alpha$ and $\alpha \| c$ (see problem 11), and plane $\beta$ such that $b \subset \beta$ and $\beta \perp c$. If planes $\alpha$ and $\beta$ intersect, $\alpha \cap \beta=l$, then $l \| a$ (why?). This line $l$ intersects both $a$ and $b$, because if it turned ou... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,308 |
16. Intersecting lines $a$ and $b$ intersect the plane $\gamma$ at points $A$ and $B$. How to construct points $M \in a$ and $N \in b$ such that $(M N) \| \gamma$ and $|M N|=d$ ( $d$ is given)? | 16. Let the segment $M N$ (Fig. 72) satisfy the conditions of the problem. Draw a line through point $N$ parallel to line $a$. This line will intersect plane $\gamma$ at point $N^{\prime}$. The line $N^{\prime} B$ is the intersection of plane $\beta$ with the plane passing through $b$ and parallel to line $a$. Since $(... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,309 |
17. To represent points $A, B, C, \ldots$ in space for better visualization, a conditional representation of a certain fixed ("main") plane $\alpha$ and the images $A^{\prime}, B^{\prime}, C^{\prime}, \ldots$ of the parallel projections of these points onto the plane $\alpha$ are used. Perform the following constructio... | 17. a) The point $P$ of intersection of the line $A B$ with the plane $\alpha$ is found as the point of intersection of the line $A B$ with its parallel projection $\left(A^{\prime} B^{\prime}\right)$ onto the plane $\alpha$.
 with one of the vertices of the triangle so that this line intersects another side. This is always possible (see problem 4 from 1$). Let, for example, $(P A) \cap(B C)=$ $=M$. Since the ratio of segments on the same line is preserved during projection, we will find point $M^{\prime... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,317 |
25. The parallelogram $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the image of the square $A B C D$. How to construct the image of the circle inscribed in the square? (The image of a circle in a parallel projection is an ellipse. The mutually perpendicular diameters of the circle correspond to the conjugate diamet... | 25. Let the parallelogram $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be the projection of the square $A B C D$ (Fig. 76). The center of the square $O$ is mapped to the center of the parallelogram - $O^{\prime}$. The inscribed circle touches the sides of the square at their midpoints, which correspond to the midpoint... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,318 |
26. Three points not lying on the same line are the images of the three vertices of an equilateral triangle. Construct the image of the circle circumscribed about this triangle. | 26. In an equilateral triangle, the medians coincide with the diameters of the circumscribed circle, and their point of intersection is the center. Therefore, the point $O^{\prime}$ of intersection of the medians of the triangle's projection is the projection of the center of the circle (Fig. 77). Further construction ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,319 |
27. Given three points not lying on the same line - the projections of three non-adjacent vertices of a regular hexagon. Construct the projection of this hexagon, as well as the image of the projection of the circle inscribed in the hexagon. | 27. The projections of non-adjacent vertices of a regular hexagon are at the same time projections of the vertices of an equilateral triangle. How to construct the missing vertices of the hexagon from them can be seen from the drawing of the previous problem. The projection of the inscribed circle is constructed from t... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,320 |
1. Prove that for vectors in space, the properties of vector operations established for vectors in the plane remain valid:
a) the commutative and associative laws of vector addition, the associative property of multiplying a vector by a number, and the distributive laws for multiplying a vector by a sum of numbers and... | 1. a) Commutativity of vector addition, associativity in multiplying a vector by a number, and the distributive laws for multiplying a vector by the sum of numbers and a number by the sum of vectors - all of these are proven in the same way as in the plane, since these facts involve no more than two vectors, which can ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,324 |
3. Given three pairwise intersecting lines $m, n, p$. How to construct a parallelepiped such that three of its edges lie on these three lines? Prove that there is exactly one such parallelepiped. | 3. The construction can be carried out in the following order:
1) Through the line $m$, we draw planes $\beta \| n$ and $\gamma \| p$, assuming that the line $p$ is not parallel to the plane $\beta$.
2) Through the line $n$, we draw $\alpha \| m$ and $\gamma^{\prime} \| p$.
3) Through the line $p$, we draw $\alpha^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,326 |
4. Prove: The necessary and sufficient condition for points $A, B, C, D$ to belong to the same plane is that in any relation (equality) connecting the vectors $\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O C}, \overrightarrow{O D}$, the sums of the numerical coefficients on the right and left sides of t... | 4. The condition is necessary. Let the vectors $\overrightarrow{O A}=\vec{a}$,
Fig. 79
$\overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}$, and $\overrightarrow{O D}=\vec{d}$. If... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,327 |
5. Prove that if a plane is drawn through the ends of three edges of a parallelepiped emanating from one vertex, then this plane cuts off $\frac{1}{3}$ of the length of the diagonal of the parallelepiped drawn from the same vertex. | 5. In the parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, let $\overrightarrow{D A}=\vec{m}$, $\overrightarrow{D C}=\vec{n}$, $\overrightarrow{D D^{\prime}}=\vec{p}$. Let $Q$ be the point of intersection of the diagonal [DB'] of the parallelepiped with the plane passing through points $A, C, D^{\p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,328 |
6*. Let's assume that in space, as well as in a plane, the center of mass of two points $A_{1}$ and $A_{2}$, with masses $m_{1}$ and $m_{2}$, is defined by the formula $\overrightarrow{O P}=\frac{m_{1} \overrightarrow{O A_{1}}+m_{2} \overrightarrow{A_{2}}}{m_{1}+m_{2}}$, or, in a shortened form, $\vec{P}=\frac{m_{1} \v... | 6. Assuming that point $P$ has a mass of $\left(m_{1}+m_{2}\right)$, and its position is determined by the formula given in the problem, we will find the center of mass of the system of points $P$ (with mass $\left(m_{1}+m_{2}\right)$) and $A_{3}$ (with mass $m_{3}$), using the same formula. We get:
$$
\overrightarrow... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,329 |
7*. In a model of a regular quadrilateral pyramid, all faces are made of uniform plates of the same material (and have the same thickness). The lateral edges of the pyramid are twice as long as the sides of the base. Determine the position of the center of mass of the pyramid, given the radius vectors of the vertex $S$... | 7. As it was done in planimetry, we will assume that the mass of the triangle is concentrated at its centroid (the point of intersection of the medians), and the mass of each face is proportional to its area. The area of the base (square) is taken as a unit, then the area of each lateral face is $\frac{\sqrt{15}}{4}$. ... | \vec{P}=\frac{4\vec{S}\sqrt{15}+(2\sqrt{15}+3)(\overrightarrow{A_{1}}+\overrightarrow{A_{2}}+\overrightarrow{A_{3}}+\overrightarrow{A_{4}})}{12(\sqrt{15}+1)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,330 |
8. Prove that the lateral edge of a regular triangular pyramid is perpendicular to the opposite side of the base. | 8. The lateral edges of a regular pyramid have the same length $a$. Let the corresponding vectors be denoted by $\vec{a}_{1}, \vec{a}_{2}, \vec{a}_{3}$; then the vector corresponding to the base edge opposite to edge $a_{1}$ is $\overrightarrow{a_{2}}-\overrightarrow{a_{3}}$. Noting also that the angles between the lat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,331 |
9. From the given origin $O$, draw 4 rays so that all 6 angles between them are congruent to each other. | 9. Let point $O$ be the origin of four radius vectors $\overrightarrow{O A}_{1}, \overrightarrow{O A}_{2}, \overrightarrow{O A}_{3}, \overrightarrow{O A}_{4}$, which are equal in length and form equal angles with each other. As in the previous problem, it is not difficult to prove that each of these vectors is perpendi... | \varphi\approx10928^{\}16^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,332 |
10. Three pairwise non-collinear and non-zero radius vectors $\vec{A}, \vec{B}, \vec{C}$ are given. A line $p$ forms equal angles with all these vectors. Prove that the line $p$ is perpendicular to the plane $ABC$. | 10. Without loss of generality, we can assume that $|\vec{A}|=|\vec{B}|=|\vec{C}|=q$, and also that $O \in p$. Let $M \in p$ and consider the vector $\overrightarrow{O M}=\vec{r}$. If $\varphi$ is the angle between the vector $\vec{r}$ and the vectors $\vec{A}, \vec{B}, \vec{C}$, then $\vec{r} \cdot(\vec{A}-\vec{B})=r ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,333 |
11. Non-coplanar vectors $\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O C}$ satisfy the condition: $|\overrightarrow{O A}|=|\overrightarrow{O B}|=|\overrightarrow{O C}|$. Point $P$ is the center of the circle passing through points $A, B$, and $C$. Prove that the line $O P$ is perpendicular to the plane... | 11. Let $\overrightarrow{P A}=\vec{a}, \overrightarrow{P B}=\vec{b}, \overrightarrow{P C}=\vec{c}$, and $|\vec{a}|=|\vec{b}|=|\vec{c}|=r$, where $r$ is the radius of the circumcircle of triangle $A B C$ (with center $P$). We also denote: $|\overrightarrow{O A}|=|\overrightarrow{O B}|=|\overrightarrow{O C}|=q, \overrigh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,334 |
12. Prove that if one of two mutually perpendicular (intersecting or skew) lines is parallel to a given plane $\gamma$, and the other is not perpendicular to $\gamma$, then the orthogonal projections of these lines onto the plane $\gamma$ are mutually perpendicular. Is the converse theorem true? | 12. Let $a \| \gamma, b \perp a$. Denote by $a^{\prime}$ the projection of $a$ onto $\gamma$, and by $b^{\prime}$ the projection of $b$ onto $\gamma$. If $a \subset \alpha$ and $a^{\prime} \subset \alpha$, then $\alpha \perp \gamma$. Similarly, if $b \subset \beta$ and $b^{\prime} \subset \beta$, then $\beta \perp \gam... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,335 |
13. Prove that if plane $\beta$ intersects plane $\alpha$ along line $l$ and line $p$ is perpendicular to plane $\beta$, then the projection $p^{\prime}$ of line $p$ onto plane $\alpha$ is perpendicular to line $l$. | 13. If $p \perp \alpha$, then $p \perp l$. But $l \subset \alpha$, therefore $l=l^{\prime}$. Based on the proposition from the previous problem, we conclude: $p^{\prime} \perp l^{\prime}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,336 |
14. Given a plane $\alpha$ and a point $P$. How to draw a line $p$ through point $P$ that is perpendicular to the plane $\alpha$? Prove the uniqueness of such a line. | 14. In the plane $\alpha$, we draw a line $l$. This line and point $P$ define a plane in which we draw a perpendicular $(PA)$ to the line $l$ through point $P$. Through point $A$, we draw in the plane $\alpha$ a perpendicular $a$ to the line $l$. The line $a$ and point $P$ define a plane $\beta$, in which we draw a per... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,337 |
15. Prove the existence of points symmetric to each other relative to a given plane: if a plane $\alpha$ and a point $A$ outside it are given, then there exists a unique point $A^{\prime}$ on the other side of the plane $\alpha$ that satisfies the conditions: a) $\left(A A^{\prime}\right) \perp \alpha$; b) $\left(A A^{... | 15. Let $(A S) \perp \alpha, S \subset \alpha$. By laying off $\left[S A^{\prime}\right] \cong[S A]$ on the line $A S$ on the other side of $\alpha$, we obtain the point $A^{\prime}$. It is unique due to the uniqueness of the perpendicular.
Now let $M \in \alpha$. The plane $A A^{\prime} S M$ intersects $\alpha$ along... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,338 |
16. Prove that the symmetry with respect to a plane (see the previous problem) is a mapping of space onto itself, possessing the following properties:
a) if point $A$ is mapped by the symmetry to $A^{\prime}$, then, conversely, $A^{\prime}$ is mapped to $A$; points of the plane of symmetry are fixed (mapped to themselv... | 16. a) This follows from the equality $|S A|=\left|S A^{\prime}\right|$ and the uniqueness of the perpendicular $(S A)=\left(S A^{\prime}\right)$.
b) The plane passing through the parallel lines $\left(A A^{\prime}\right) \|\left(B B^{\prime}\right)$ intersects the plane of symmetry along a line $s$ - the axis of symm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,339 |
17. Prove that two non-coinciding rays with a common origin determine a unique plane relative to which they are mutually symmetric. Prove that this plane is the set of points belonging to all rays having the same origin and forming congruent angles with the given rays. | 17. Let rays $O A$ and $O A^{\prime}$ be drawn in plane $\alpha$. We draw the bisector $O N$ of angle $A O A^{\prime}$ and through point $O$ we draw the perpendicular $O P$ to plane $\alpha$. The plane $\sigma$, passing through (OP) and $[O N)$, is the plane of symmetry of rays $O A$ and $O A^{\prime}$. This follows fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,340 |
18. On one side of the plane $\alpha$ are points $A$ and $B$. Find a point $M$ in this plane such that the sum of the distances $|A M|+|M B|$ is the smallest. | 18. Let point $B^{\prime}$ be symmetric to point $B$ with respect to the plane $\alpha$ and $\left(A B^{\prime}\right) \cap \alpha=M$. We have: $|A M|+|M B|=\left|A B^{\prime}\right|$, and, therefore, the length of the path $A M B$ is equal to the length $\left|A B^{\prime}\right|$. For any other point $N$ in the plane... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,341 |
19. Two points $A$ and $B$ are located on opposite sides of the plane $\alpha$. Find a point $M$ in the plane $\alpha$ such that the difference in distances $|AM| - |MB|$ is maximized (in absolute value). | 19. Let's find point $B^{\prime}$, symmetric to point $B$ with respect to $\alpha$, $\left(A B^{\prime}\right) \cap \alpha=M$. We have: ||$A M|-| B M||=|| A M|-| B^{\prime} M \|=\left|A B^{\prime}\right|$. For any other point $N$ in the plane $\alpha$, we will have: $\|A N|-| B N\|=\left\|A N|-| B^{\prime} N\right\|<\l... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,342 |
20. Through the given point $P$ draw a line perpendicular to two skew lines $a$ and $b$. | 20. Through point $P$ and lines $a$ and $b$ pass two planes $\alpha$ and $\beta$. In the first of these, through point $P$ we draw a line $a^{\prime} \| a$, in the second - $b^{\prime} \| b$. Through the lines $a^{\prime}$ and $b^{\prime}$, intersecting at point $P$, we draw a plane $\gamma$ and through point $P$ by th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,343 |
21. A segment of constant length "slides" with its ends along two mutually perpendicular intersecting lines. What line does the midpoint of the segment describe? | 21. Let us draw two parallel planes $\alpha$ and $\beta$ through two intersecting lines $a$ and $b$. The plane $\gamma$, parallel to $\alpha$ and $\beta$ and passing between them at equal distances from them, bisects any segment passing between them (see problem $12, \S 4$). Let $A$ and $B$ be the endpoints of a given ... | |OM|=\frac{1}{2}|A^{\}B^{\}| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,344 |
22. Prove that if two congruent oblique lines are drawn from one point to a given plane, then the magnitude of the angle between the oblique lines is less than the magnitude of the angle between the orthogonal projections of the oblique lines on the given plane. | 22. Let $[P A]=[P B]$ - two oblique lines, $\left[P^{\prime} A\right]$ and $\left[P^{\prime} B\right]$ - their projections on the plane. Triangles $A B P$ and $A B P^{\prime}$ - are isosceles with a common base $[A B]$. In triangle $A B P^{\prime}$, draw the height $\left[P^{\prime} M\right]$ and on the ray $\left[M P^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,345 |
23. On the image of a cube, construct a segment that defines the distance between two intersecting diagonals of two adjacent faces of the cube. Calculate this distance if the length of the edge of the cube is $a$. | 23. Consider $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (Fig. 80) and draw the diagonals $\left[A D^{\prime}\right]$ and $\left[D C^{\prime}\right]$. To find the distance between the lines ( $A D^{\prime}$ ) and ( $D C^{\prime}$ ), draw two planes through these lines that are parallel to each other. Such pla... | \frac{1}{3}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,346 |
24*. Prove that the composition of symmetries with respect to two parallel planes is a vector (parallel translation) directed perpendicular to these planes from the first to the second, and its length is equal to twice the distance between the planes.
Prove the converse statement: any parallel translation can be repre... | 24. Let $\alpha \| \beta$ and point $A$ is mapped to point $A_{1}$ by symmetry with respect to plane $\alpha$, and point $A_{1}$ is mapped to point $A^{\prime}$ by symmetry with respect to $\beta$. Since $\left(A A_{1}\right) \perp \alpha$, $\left(A_{1} A^{\prime}\right) \perp \beta, \alpha \| \beta$, the points $A, A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,347 |
25*. Prove that the composition of three symmetries with respect to three mutually perpendicular planes is a central symmetry. How to find its center? Conversely, how to represent a central symmetry as a composition of three symmetries with respect to planes? | 25. Three mutually perpendicular planes intersect along three mutually perpendicular lines, having a common point $O$. Let us direct three unit vectors along these lines: $\vec{i}$, $\vec{j}$, and $\vec{k}$. Any point $M$ corresponds to the vector $\overrightarrow{O M}$, which can be decomposed into these three vectors... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,348 |
26*. What kind of mapping is the composition of two homotheties in space? If this mapping is a homothety, what will be its coefficient? How to determine the position of its center? | 26. A homothety is fully determined by the assignment of two parallel segments $[A B]$ and $\left[A^{\prime} B^{\prime}\right]$, if it is specified that under the homothety $A$ maps to $A^{\prime}$ and $B$ to $B^{\prime}$. The center $S$ of the homothety is the intersection $S=\left(A A^{\prime}\right) \cap\left(B B^{\... | k_{3}=k_{2}\cdotk_{1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,349 |
27. How to construct a cube given its diagonal? | 27. The diagonal section of each cube is a rectangle in which the larger side is to the smaller as the diagonal of a square is to its side. Let us construct such a rectangle and take one of its vertices as the center of homothety. As the coefficient of homothety, we will take the ratio of the lengths of the diagonals o... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,350 |
31. Express the scalar product $\vec{r}_{1} \cdot \vec{r}_{2}$ in coordinates, if $\vec{r}_{1}=x_{1} \vec{i}+y_{1} \vec{j}+z_{1} \vec{k}$ and $\vec{r}_{2}=x_{2} \vec{i}+y_{2} \vec{j}+z_{2} \vec{k}$. Calculate $\cos \varphi$, where $\varphi$ is the angle between the vectors $\vec{r}_{1}$ and $\vec{r}_{2}$. Write the con... | 31. Using the distributive property of scalar multiplication of vectors, as well as the fact that $\overrightarrow{i^{2}}=\overrightarrow{j^{2}}=\vec{k}^{2}=1, \vec{i} \vec{j}=\vec{j} \vec{k}=$ $=\vec{k} \vec{i}=0, \quad$ we obtain: $\quad \vec{r}_{1} \vec{r}_{2}=r_{1} r_{2} \cos \varphi=x_{1} x_{2}+y_{1} y_{2}+z_{1} z... | x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}=r_{1}r_{2}\cos\varphi,\quad\cos\varphi=\frac{x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}+z_{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,354 |
32. The vector $\vec{r}$ forms angles $\alpha, \beta, \gamma$ with the axes $O x, O y, O z$ respectively. The cosines of these angles are called the direction cosines of the vector $\vec{r}$. Prove the formula: $\cos ^{2} \alpha+\cos ^{2} \beta+$ $+\cos ^{2} \gamma=1$. | 32. Obviously, we have: $\frac{x_{1}}{r_{1}}=\cos \alpha, \frac{y_{1}}{r_{i}}=\cos \beta, \frac{z_{1}}{r_{i}}=\cos \gamma$. Taking $\left|r_{1}\right|=1$, we get: $\vec{r}_{1}=\vec{i} \cos \alpha+\vec{j} \cos \beta+\vec{k} \cos \gamma$. Squaring both sides of the equation scalarly, we find: $\cos ^{2} \alpha+\cos ^{2} ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,355 |
33. The vector $\vec{r}$ forms angles of $60^{\circ}$ with the $O x$ and $O y$ axes. What angle does this vector form with the $O z$ axis? | 33. According to the formula obtained in the previous problem, we have:
$$
\cos ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} \gamma=1
$$
or
$$
\begin{gathered}
\frac{1}{4}+\frac{1}{4}+\cos ^{2} \gamma=1, \cos ^{2} \gamma=\frac{1}{2} \\
\cos \gamma= \pm \frac{\sqrt{2}}{2}, \gamma=45^{\circ} \text { or } \gamma=135^... | \gamma=45\text{or}\gamma=135 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,356 |
34.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 34. According to the condition of the problem for the desired point $M(x ; y ; z)$ we have:
$$
\begin{gathered}
(\overrightarrow{O A}-\overrightarrow{O M})+(\overrightarrow{O B}-\overrightarrow{O M})+(\overrightarrow{O C}-\overrightarrow{O M})+(\overrightarrow{O D}-\overrightarrow{O M})=\overrightarrow{0} \\
\overrigh... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,357 |
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