problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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53*. How to place $n$ points $(n=2,3,4,5,6)$ on a sphere so that the distance between the nearest two of these $n$ points is the greatest? | 53. Let us consider the cases $n=2,3,4$, 5 and 6 in sequence.
$1^{\circ} n=2$. Obviously, in this case, the most advantageous placement will be to place two points at diametrically opposite points on the sphere (Fig. 140, a). The distance between these points will be equal to the diameter $2 R$ of the sphere.
$2^{\ci... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,315 |
54. What is the smallest number of circles with which a circle of twice the radius can be completely covered? | 54. Since the diameter of the smaller circles is equal to the radius $R$ of the larger circle, each such circle intersects the circumference of the larger circle at points no more than $R$ apart, and thus covers an arc of this circumference no greater than $60^{\circ}$. It follows that at least six smaller circles are ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,316 |
55. How to place three ice cream kiosks on a circular square in the most advantageous way, i.e., so that the greatest distance from points on the square to the nearest kiosk is as small as possible?[^0] | 55. Let the radius of the square be $R$, and the maximum distance from a point on the square to the nearest kiosk be $r$. In this case, every point on the square will be at a distance not greater than $r$ from at least one kiosk, i.e., three circles of radius $r$ with centers at the points of the kiosk locations will c... | \frac{R\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,317 |
58. Given a circle $C$ (the center of which is not specified) and a point $M$. Using only a straightedge, draw a line through point $M$ that is perpendicular to the given diameter $A B$ of circle $C$. Consider separately the cases where point $M$ is located on the circle $C$ or on the diameter $A B$. | 58. Let point \( M \) be located neither on the circle nor on the diameter. Connect point \( M \) with points \( A \) and \( B \). Let \( A_{1}, B_{1} \) be the points of intersection of the lines \( M B \) and \( M A \) with the circle (Fig. 151, a, \(\sigma\), b). Then the angle \( A B_{1} B = 90^{\circ} \), because
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,320 |
59. In a plane, given a circle $C$ with center $O$, a line $l$, and a point $M$. Using only a straightedge, draw through point $M$
a) a line parallel to $l$;
b) a line perpendicular to $l$. | 59. a) If a line $l$ intersects a given circle at points $A$ and $B$ (Fig. $153, \alpha$), then the diameters $A O$ and $B O$ intersect the circle again at points $A^{\prime}$ and $B^{\prime}$ such that $B^{\prime} A^{\prime}$ is parallel to $B A$. This observation reduces our problem to problem 57 b).
If the line $l$... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,321 |
60. In a plane, there is a square $A B C D$, a line $l$, and a point $M$. Using only a ruler,
a) draw a line through point $M$ parallel to $l$;
b) drop a perpendicular from point $M$ to the line $\boldsymbol{l}$. | 60. a) Let $O$ be the center of the square $ABCD$, $K$ and $L$ be the points of intersection of line $l$ with sides $AB$ and $DC$ of the square, respectively, $K'$ be the point of intersection of $KO$ with side $DC$, and $L'$ be the point of intersection of $LO$ with side $AB$ (Fig. 154, a). In this case, $L'K' \parall... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,322 |
62. Using only one compass
a) increase the given segment $A B$ by $n$ times ${ }^{1}$ );
b) divide the given segment $A B$ into $n$ equal parts;
c) determine the center of the given circle $C$. | 62. a) Draw a circle with center at point $B$ and radius $BA$. Then, using the radius equal to $BA$, make marks on the circle $AA_1 = BA$, $A_1A_2 = BA$, $A_2A' = BA$ (Fig. 158). Obviously, in this case, $A'$ is the second end of the diameter $AA'$ of the circle, i.e., $AA' = 2AB$. Repeating this construction,
 without a compass, construct a perpendicular from point $M$ to the line $l$. | 64. Draw an arbitrary line \( m \) through point \( M \) (Fig. 162) and using a double-edged ruler, draw lines \( m_{1} \) and \( m_{2} \), parallel to \( m \) and at an equal distance \( d \) from it, equal to the width of the ruler. Let \( N \) be the point of intersection of \( m_{1} \) and \( l \). Now, place our r... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,326 |
75. Given a circle with center $O$, two points $A$ and $B$ on it, and a line $l$. Find a point $X$ on the circle such that the lines $A X$ and $B X$ intercept on the line $l$ a segment:
a) having a given length $a$;
b) bisected by a given point $C$ on the line $l$. | 75. a) Suppose the problem is solved; \( M N = a \) (Fig. 183). Draw a segment \( B B_{1} \) from point \( B \), equal and parallel to
Fig. 183. the segment \( M N \). Then the quadrilatera... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,333 |
76. Given angle $L A K$ and point $M$. Draw a line through point $M$ such that it cuts off a triangle $A B C$ from the angle
a) of a given perimeter $2 p$;
b) of the smallest possible perimeter. | 76. a) Let $L A K$ be the given angle, $M$ - the given point, and $2 p$ - the given perimeter. On the sides of the angle, lay off segments $A Q$ and $A P$, each equal to $p$ (Fig. 185). Draw a circle $S$ tangent to the sides of the angle at points $Q$ and $P$, and finally, draw a line $M B C$ through point $M$, tangent... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,334 |
82. a) On the surface of a sphere, three points are given. Draw a circle on the sphere passing through these three points.
b) On the surface of a material sphere, two points are given, which are not the ends of the same diameter. Draw a great circle passing through these points. | 82. a) Let $A, B$ and $C$ (Fig. $197, a$) be given points. We construct a triangle $A^{\prime} B^{\prime} C^{\prime}$, equal to triangle $A B C$, and describe a circle around it (Fig. 197, b).
 the difference of whose distances from the lines $a$ and $b$ is equal to a given value $s ;$
b) the sum of whose distances from the lines $a$ and $b$ is equal to a given value $s$. | 83. a) Draw lines $b^{\prime}$ and $b^{\prime \prime}$, parallel to $b$ and at a distance $s$ from it (Fig. 199). They intersect line $a$ at points $P_{1}$ and $P_{4}$, which belong to the desired geometric locus. Proceed similarly with line $a$. Then on line $b$, find two more points $P_{2}$ and $P_{3}$ of the geometr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,339 |
87. A right triangle $ABC$ changes in such a way that the vertex $A$ of the right angle does not change its position, while vertices $B$ and $C$ slide respectively along fixed circles $S_{1}$ and $S_{2}$, which touch each other externally at point $A$ (Fig. 18). Find the geometric locus of the feet $D$ of the altitudes... | 87. Let $ABC$ (Fig. $205, a$) be some position of the triangle. Draw the radii $O_{1} B=R$ and $O_{2} C=r$ and consider the isosceles triangles $O_{1} A B$ and $O_{2} A C$. Clearly, $\begin{array}{ll}\angle B A O_{1}+\angle C A O_{2}=2 d-d=d . & \text { Hence } \quad \angle B A O_{1}+ \\ +\angle A B O_{1}+\angle C A O_... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,342 |
88. A right triangle $ABC$ ($A$ - the right angle) moves on a plane in such a way that vertices $B$ and $C$ slide along the sides of a given right angle $P$ (Fig. 19). Prove that the geometric locus of points $A$ is a segment. Determine the length of this segment. | 88. Let $B$ lie on side $l$, and $C$ - on side $m$ of a right angle with vertex $P$ (see Fig. 206, a). Construct a circle $O$ with $BC$ as its diameter. Since $\angle A = \angle P = d$, points $A$ and $P$ lie on this circle. $\angle APC = \angle ABC$, as they subtend the same arc. Therefore, point $A$ must always lie o... | BC-AB | Geometry | proof | Yes | Yes | olympiads | false | 28,343 |
90*. a) Find the geometric locus of points $M$ such that the triangle, whose vertices are the projections of point $M$ onto the sides of a given triangle $ABC$, has a given area $\sigma$.
b) Find the geometric locus of points $M$ such that the polygon, whose vertices are the projections of point $M$ onto the sides of ... | 90. a) Let's describe a circle around triangle $ABC$. For definiteness, let point $M$ lie inside this circle, and $P, Q, R$ be the projections of this point onto the sides of the triangle (see Fig. 209, a). We connect point $M$ with the vertices $A, B$, and $C$ of the triangle; let $D$ and $E$ be the points of intersec... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,345 |
91. On the sides of a parallelogram, outside of it, squares are constructed (Fig. 21). Prove that their centers form a square.
Fig. 21.
. The triangles $AO_{1}O_{4}, BO_{1}O_{2}, CO_{2}O_{3},$ and $DO_{3}O_{4}$ are equal, since
$$
AO_{1}=BO_{1}=CO_{3}=DO_{3}; \quad AO_{4}=BO_{2}=CO_{2}=D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,346 |
93. Side $AD$ of parallelogram $ABCD$ is divided into $n$ equal parts. The first division point $P$ is connected to vertex $B$. Prove that line $BP$ cuts off a segment $AQ$ on diagonal $AC$ which is equal to $\frac{1}{n+1}$ of the entire diagonal. | 93. Draw a line through point $P$ parallel to $A B$ (Fig. 216); let $R$ be the point of intersection of this line with $A C$. Then $A R=\frac{A C}{n}$ and $P R=\frac{D C}{n}=\frac{A B}{n}$. From the similarity of triangles $A Q B$ and $P Q R$ we have
$$
\frac{Q R}{A Q}=\frac{P R}{A B}=\frac{1}{n}, \quad Q R=\frac{A Q}... | AQ=\frac{AC}{n+1} | Geometry | proof | Yes | Yes | olympiads | false | 28,348 |
94. Prove that in any triangle, the bisector lies between the median and the altitude drawn from the same vertex. | 94. First solution. Let in triangle $A B C$ points $H, D, M$ be the feet of the altitude, bisector, and median, respectively, drawn from vertex $B$ (Fig. 217). If $A B = B C$, then the triangle is isosceles and points $H, D$, and $M$ coincide. Now let $A B \angle C$ and, therefore, $\angle A B H < \angle H B C$. Hence,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,349 |
95. Determine the angles of a triangle in which the median, bisector, and altitude divide an angle into four equal parts. | 95. Let $ABC$ be a given triangle, $BH$, $BD$, and $BM$ the altitude, bisector, and median of this triangle, respectively, which divide angle $B$ into four equal parts (Fig. 219). Describe a circle $\mathcal{S}$ around triangle $ABC$ and extend the bisector of the triangle to intersect the circle at point $D_{1}$; conn... | \angleBCA=2230',\angleBAC=6730' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,350 |
96. Prove that if two angle bisectors in a triangle are equal, then the triangle is isosceles.
 | 96. We will prove that if in two triangles $A B C$ and $A_{1} B_{1} C_{1}$ (Fig. 220) the bases $A C$ and $A_{1} C_{1}$, the angles $B$ and $B_{1}$, and the angle bisectors $B P$ and $B_{1} P_{1}$ of these angles are equal, then the triangles are equal. For this, we will describe a circle around triangle $A B C$ and dr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,351 |
97. The angles of an arbitrary triangle $ABC$ are divided into three equal parts by the lines $AY, AZ; BZ, BX; CX, CY$ (Fig. 23). Prove that triangle $XYZ$ is equilateral. | 97. First solution. Let the angles of triangle $ABC$ be denoted as $3\alpha, 3\beta,$ and $3\gamma$ (Fig. 222); since
Fig. 222.
$3\alpha + 3\beta + 3\gamma = 180^\circ$, it follows that $\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,352 |
98. Prove that in any triangle, the point of intersection of the medians, the orthocenter (the point of intersection of the altitudes), and the center of the circumscribed circle lie on one line (Euler's line). | 98. Consider the triangle $A_{1} B_{1} C_{1}$ formed by the midlines of $\triangle A B C$ (Fig. 223). $\triangle A_{1} B_{1} C_{1}$ is similar to $\triangle A B C$ and
Fig. 223. the coeffici... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,353 |
99. Prove that in any triangle, the three midpoints of the sides, the three feet of the altitudes, and the three points that bisect the segments of the altitudes from the orthocenter to the vertices, lie on one circle (the nine-point circle). | 99. Let $O$ be the center of the circumscribed circle, and $H$ the orthocenter of triangle $ABC$ (Fig. 224). We will prove that the center of the desired circle is the midpoint $O'$ of segment $OH$, and the radius is half the radius $R$ of the circumscribed circle around triangle $ABC$. Indeed, if $B_1$ is the midpoint... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,354 |
100. To each side of the circular triangle \(ABC\) (i.e., the figure formed by three intersecting arcs of circles; see Fig. 24), we assign a sign "plus" or "minus" depending on whether the corresponding arc of the circle is convex to the exterior or the interior of the triangle. Prove that the sum of the angles of the ... | 100. Let $A M_{1}, A M_{2}, B N_{1}, B N_{2}, C P_{1}, C P_{2}$ be the tangents to the sides of the circular triangle $A B C$ at its vertices $A, B$, and $C$ (Fig. 225). Clearly, we have
$$
\begin{gathered}
\angle M_{1} A B=\angle N_{2} B A=\frac{\cup A B}{2}, \angle N_{1} B C=\angle P_{2} C B=\frac{\cup B C}{2} \\
\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,355 |
101. Let point $A_{1}$ be symmetric to some point $A$ in the plane with respect to a given point $O_{1}$; point $A_{2}$ is symmetric to point $A_{1}$ with respect to another point $O_{2}$; point $A_{8}$ is symmetric to point $A_{2}$ with respect to a third point $O_{8}$. Further, let point $A_{4}$ be symmetric to $A_{3... | 101. Quadrilateral $A A_{4} A_{1} A_{3}$ (Fig. 226) is a parallelogram because its diagonals are bisected at the point of intersection $O_{1}$. Therefore, $A_{3} A$ is parallel, equal, and directed in the same direction as $A_{1} A_{4}$. Quadrilateral
. Now suppose that $M$ does not lie on any of the three medians of triangle $ABC$.
Let $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6},... | 7 | Geometry | proof | Yes | Yes | olympiads | false | 28,357 |
103. Point $Q$ moves straight towards vertex $A$ of triangle $A B C$. At the halfway point, it turns and moves towards vertex $B$, at the halfway point it turns and moves towards vertex $C$; at the halfway point of this path, it turns again towards $A$, and so on (Fig. 25).
Prove that there exists a triangle towards w... | 103. Let's first carry out some auxiliary constructions. On the side $CB$ of the given triangle $ABC$, we lay off the segment $CM = \frac{1}{3} CB$, and on the side $BA$ - the segment $BN = \frac{1}{3} BA$ (Fig. 228, a) and connect points $M$ and $N$ with $A$ and $C$, respectively. Let $C_{1}$ be the point of intersect... | \frac{1}{7}S | Geometry | proof | Yes | Yes | olympiads | false | 28,358 |
105. Prove that the point of intersection of the diagonals of a quadrilateral circumscribed around a circle coincides with the point of intersection of the diagonals of the quadrilateral whose vertices are the points of tangency of the sides of the first quadrilateral with the circle (Fig. 27).
 (Fig. 232) be a quadrilateral circumscribed around a circle, and let \(E, F, G, K\) be the points of tangency of the sides of this quadrilateral with the circle, and \(M\) be the point of intersection of \(AC\) and \(EG\).
 Prove that in any triangle, the inequality $R \geqslant 2 r$ holds (where $R$ and $r$ are the radii of the circumscribed and inscribed circles), and equality holds only for an equilateral triangle.
b) Conversely, if segments $R$ and $r$ are related by the inequality $R \geqslant 2 r$, then they are the radii o... | 106. a) Let $S_{1}$ and $S_{2}$ be the inscribed and circumscribed circles of triangle $ABC$ (see diagram 233, a). Draw a line through each vertex of $\triangle ABC$ parallel to the opposite side. This will form a new triangle $A_{1} B_{1} C_{1}$, similar to $\triangle ABC$ and with sides twice as long. Each side of $\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,360 |
107. What dependence should exist between the radii $R$ and $r$ and the distance $d$ between the centers of two circles, so that a triangle can be constructed that is inscribed in the first circle and circumscribed around the second circle | 107. Let $O$ and $o$ be the centers of the circles with radii $R$ and $r$, respectively, and $Oo = d$. Let $ABC$ be a triangle inscribed in the first circle and circumscribed around the second circle (Fig. 235). Extend $Oo$ to intersect the circumcircle of triangle $ABC$ at points $M$ and $N$. Then $OM = R + d$, $ON = ... | ^2=R^2-2Rr | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,361 |
109. Prove that if in a hexagon the opposite sides are parallel and the three diagonals connecting opposite vertices are equal to each other, then a circle can be circumscribed around this hexagon. | 109. Let $A B C D E F$ be a hexagon satisfying the conditions of the problem (Fig. 237). The quadrilateral $A B D E$ is a trapezoid, the diagonals $A D$ and $B E$ of which are equal to each other. Hence, it is isosceles. (Indeed, dropping perpendiculars $D D_{1}$ and $E E_{1}$ from points $D$ and $E$ to $A B$, we obtai... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,363 |
110. Prove that if three diagonals of an inscribed hexagon serve as diameters of the circumscribed circle, then the area of this hexagon is equal to twice the area of the triangle with sides equal to the other three diagonals. | 110. Let $A B C D E F$ (Fig. 238) be the hexagon in question. Note that
$$
S_{\triangle D O B}=S_{\triangle O D E}=S_{\triangle O A B}
$$
since any two adjacent of these triangles have equal bases and heights. Similarly,
$$
\begin{aligned}
& S_{\triangle D O F}=S_{\triangle O C D}=S_{\triangle O F A} \\
& S_{\triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,364 |
111. In an arbitrary convex hexagon, the midpoints of the sides are connected every other one. Prove that the points of intersection of the medians of the two triangles formed coincide. | 111. Let us denote the midpoints of the sides of the convex hexagon $A B C D E F$ sequentially as $K, L, M, N, P, Q$ (Fig. 239). We will draw the median $K R$ of triangle $K M P$ and the median $N S$ of triangle $N Q L$ and prove that at the intersection point $O$ these two segments are divided in the ratio $K O: O R =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,365 |
113. Given $n$ circles $C_{1}, C_{2}, \ldots, C_{n}$, passing through one point $O$ (Fig. 28). The second points of intersection of $C_{1}$ with $C_{2}$, $C_{2}$ with $C_{3}, \ldots, C_{n}$ with $C_{1}$ are denoted respectively by $A_{1}$, $A_{2}, \ldots, A_{n}$. Take an arbitrary point $B_{1}$ on $C_{1}$, different fr... | 113. Let us first prove the following well-known proposition: if two circles $C^{\prime}$ and $C^{\prime \prime}$ with centers $O^{\prime}$ and $O^{\prime \prime}$ intersect at points $A$ and $O$ (Fig. 244) and a secant is drawn through point $A$, intersecting circles $C^{\prime}$ and $C^{\prime \prime}$ at points $B^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,367 |
118. a) On the circle circumscribed around an equilateral triangle \(ABC\), a point \(M\) is taken. Prove that the greatest of the segments \(MA\), \(MB\), and \(MC\) is equal to the sum of the other two.
b)* Let \(A_1 A_2 A_3 \ldots A_n\) be a regular polygon with an odd number of sides, and \(M\) be an arbitrary poi... | 118. a) First solution. Let $M B$ be the largest of the segments $M A, M B$, and $M C$ (Fig. 251). On $M B$, we lay off the segment $M K = M A$. Since $\angle K M A = \angle B C A = 60^{\circ}$ (angles subtending the same arc), triangle $A M K$ is equilateral, so $A K = A M$. Further, triangles $A M C$ and $A K B$ are ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,369 |
119. Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ be a regular heptagon. Prove that
$$
\frac{1}{A_{1} A_{2}}=\frac{1}{A_{1} A_{3}}+\frac{1}{A_{1} A_{4}}
$$ | 119. Apply Ptolemy's theorem to the quadrilateral $A_{1} A_{3} A_{4} A_{5}$ (Fig. 253); we get
$$
A_{1} A_{3} \cdot A_{4} A_{5} + A_{2} A_{4} \cdot A_{1} A_{5} = A_{1} A_{4} \cdot A_{3} A_{5}
$$
Dividing both sides of this equation by $A_{1} A_{2} \cdot A_{1} A_{3} \cdot A_{1} A_{4}$ and using the fact that $A_{4} A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,370 |
120. Prove that the algebraic sum of the distances from the center of the circumscribed circle around a triangle to its sides is equal to the sum of the radii of the circumscribed and inscribed circles (in this case, if the center of the circumscribed circle lies on the same side of a certain side as the triangle itsel... | 120. Let $ABC$ (Fig. $254a, \sigma$) be an arbitrary triangle and $O$ the center of the circumscribed circle around it.
Fig. 254.
Let $d_{1}, d_{2}$, and $d_{3}$ be the distances $OM$, $ON$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,371 |
121*. In a cyclic polygon, non-intersecting diagonals are drawn in such a way that the polygon is divided into triangles (one of such partitions is shown in Fig. 32). Prove that the sum of the radii of all the circles inscribed in these triangles does not depend on the partition. | 121. It is easy to see that a polygon can be divided into $n-2$ triangles, where $n$ is the number of sides of the polygon. This fact can be easily proven by induction. Indeed, the statement is true for $n=3$. Suppose it is true for all polygons with the number of sides not exceeding $n$, and we will show that it will ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,372 |
122. Let $A B C A^{\prime} B^{\prime} C^{\prime}$ be a hexagon inscribed in a circle. The sides and diagonals $A A^{\prime}, B B^{\prime}, C C^{\prime}$ of this hexagon are denoted as shown in Fig. 33. Prove that
$$
e f g = a a^{\prime} e + b b^{\prime} f + c c^{\prime} g + a b c + a^{\prime} b^{\prime} c^{\prime}
$$
... | 122. Let the diagonals $A C, C B^{\prime}, B^{\prime} A$ and $A^{\prime} B$ of the hexagon be denoted by $x, y, z$ and $u$ respectively (Fig. 256). Applying Ptolemy's theorem (Problem 117) to the quadrilaterals $A B C A^{\prime}$ and $B C A^{\prime} B^{\prime}$, we get
$$
a c + b^{\prime} f = u x \text{ and } a^{\prim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,373 |
124. a) Prove that the bases of the perpendiculars dropped from any point on the circle to the sides of the inscribed triangle lie on one straight line (Fig. $35, a$).
Fig. 35.
b) On three ... | 124. a) Let $ABC$ be an arbitrary triangle, and $X, Y$, and $Z$ be the feet of the perpendiculars dropped from an arbitrary point $P$ on the circumcircle of this triangle to its sides (Fig. 261). The quadrilaterals $PXBZ$ and $PX Y C$, each having two opposite right angles,
Fig. 268. lie on the circumcircle of triangle $A B C$, and let $Z, Y, X$ be the feet of the perpendiculars dropped from point $P$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,377 |
128. On a plane, there are three lines intersecting at one point, and three points. Construct a triangle whose vertices lie on the given lines, and whose sides pass through the given points. | 128. Let $M, N, P$ be given points; $O$ be the point of intersection of the given lines $l, m, n$; $A B C$ be the desired triangle; $X Y Z$ be an arbitrary triangle whose vertices lie on the lines $l, m$, and $n$ and whose two sides $X Y$ and $X Z$ pass through the points $M$ and $N$ (Fig. 272). Since the lines connect... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,379 |
131. Theorem of Menelaus. Let $A_{1}, B_{1}$, and $C_{1}$ be three points lying on the sides $B C, C A$, and $A B$ of triangle $A B C$, respectively. For these points to lie on one straight line, it is necessary and sufficient ${ }^{1}$ ) that the following relation holds:
$$
\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1... | 131. Let the sides of triangle $ABC$ be intersected by the secant $A_{1} C_{1} B_{1}$ (Fig. $276, a$). Draw an arbitrary line in the plane of the triangle. $P Q \parallel B_{1} A_{1}$. Let it intersect $B_{1} A_{1}$ at point $O$. Project the vertices of triangle $ABC$ parallel to the line $B_{1} A_{1}$ onto the line $P... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,381 |
132. Derive from Menelaus' theorem
a) the theorem about a complete quadrilateral (see problem 114);
b) Desargues' theorem (see problem 127);
c) Pascal's theorem (see problem 129). | 132. a) Let $A B C D E F$ be a complete quadrilateral, the diagonals of which $A B, C D, E F$ have midpoints at points $L$, $M$, and $N$ (Fig. 277). Consider the triangle $A C E$, formed by three sides of the quadrilateral, and let $a, c, e$ be the midpoints of sides $C E, E A, A C$. The line $ce \parallel C E$ passes ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,382 |
133. Theorem of Ceva. Let $A_{1}, B_{1}$ and $C_{1}$ be three points lying on the sides $B C, C A$ and $A B$ of triangle $A B C$, respectively. For the lines $A A_{1}, B B_{1}$ and $C C_{1}$ to intersect at one point or be all parallel, it is necessary and sufficient that the following relation holds:
$$
\frac{A C_{1}... | 133. Let $ABC$ be an arbitrary triangle, and let the lines $AA_1$, $BB_1$, and $CC_1$ intersect at a single point $K$ (Fig. 280, a). Draw lines through points $A$ and $C$ parallel to $BB_1$, intersecting $CC_1$ and $AA_1$ at points $P$ and $F$, respectively. Then
$$
\frac{AB_1}{B_1C} = \frac{PK}{KC}
$$
Since triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,383 |
138. Let $A_{1}, B_{1}$, and $C_{1}$ be points lying on the sides $B C, C A$, and $A B$ of triangle $A B C$; $A_{1}^{\prime}, B_{1}^{\prime}$, and $C_{1}^{\prime}$ be points symmetric to the first three points with respect to the midpoints of the corresponding sides of the triangle. Prove that
a) if points $A_{1}, B_{... | 138. We have
$$
A B_{1}=B_{1}^{\prime} C, \quad B_{1} C=A B_{1}^{\prime}
$$
hence,
$$
\frac{A B_{1}}{B_{1} C}=\frac{C B_{1}^{\prime}}{B_{1}^{\prime} A}
$$
analogously,
$$
\frac{C A_{1}}{A_{1} B}=\frac{B A_{1}^{\prime}}{A_{1}^{\prime} C}, \frac{B C_{1}}{C_{1} A}=\frac{A C_{1}^{\prime}}{C_{1}^{\prime} B}
$$
The pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,388 |
139. Let $A A_{1}, B B_{1}$, and $C C_{1}$ be three lines passing through the vertices of triangle $A B C$, and $A A_{1}^{\prime}, B B_{1}^{\prime}$, and $C C_{1}^{\prime}$ be the lines symmetric to the first three lines with respect to the angle bisectors of the corresponding angles of the triangle (points $A_{1}$ and... | 139. Let the sides $BC, AC$, and $AB$ of triangle $ABC$ (Fig. 291, a, b) be equal to $a, b$, and $c$ respectively. Let the line $AA_1$, passing through vertex $A$, form angles $\alpha_1$ and $\alpha_2$ with sides $AB$ and $AC$ of the triangle, respectively. In this case, it is obvious that
$$
\frac{BA_1}{A_1C} = \frac... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,389 |
141*. Given two triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$. Prove that if the lines drawn through the vertices of triangle $A^{\prime} B^{\prime} C^{\prime}$ perpendicular to the corresponding sides
 ${ }^{1}$ ). We will prove that triangle $L M N$ has the smallest area. Draw any other line through point $A$; let it, for example, intersect $M L$ at point $L^{\prime}$ and the extension of $M N$ at point $N^{\prime}$. Then $L^{\prime} A$
$>S_{M N A L^{\prime}}+S_{\triangle A L^{\prime... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,392 |
143. a) Given a segment $M N$ and a line $p$. Find a point $X$ on the line $p$ from which the segment $M N$ is seen at the largest angle.
b) Given a segment $M N$ and a circle $\mathcal{S}$. Find points $X$ and $Y$ on the circle $\mathcal{S}$ from which the segment $M N$ is seen at the largest and the smallest angle. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,393 | |
144. Find the triangle of the smallest possible perimeter, given:
a) two vertices $A$ and $B$ of the triangle and a line $l$ on which the third vertex lies;
b) vertex $A$ and two lines $l$ and $m$ on which vertices $B$ and $C$ lie;
c) three lines $l, m, n$ on which vertices $A$, $B$ and $C$ lie. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,394 | |
145. Find a point in a triangle where the sum of the distances from this point to the vertices of the triangle is the smallest. | 145. First solution. Construct equilateral triangles $A B C_{1}$, $A C B_{1}$, and $B C A_{1}$ outside the given triangle $A B C$ (Fig. 330). Then, as shown in the solution to problem 68b), the lines $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect at a single point $O$, from which each side of triangle $A B C$ is seen at... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,395 |
146. Find a point in a triangle where the sum of the squares of the distances from this point to the vertices of the triangle is the smallest. | 146. First solution. Let $A D, B E$ and $C F$ be the medians of triangle $A B C$, $I$ be the point of intersection of the medians, and $P$ be the foot of the perpendicular dropped from an arbitrary point $M$ to the median $A D$ (Fig. 332). Connect point $M$ with points $A, B, C, D$, and $I$. For definiteness, let $\ang... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,396 |
148. Given two points $A$ and $B$ and a line $P Q$. Prove that the sum $b \cdot A M + a \cdot B M$, where $a$ and $b$ are given positive numbers, has the smallest value for a point $M$ on the line $P Q$ such that
$$
\frac{\cos \angle A M P}{\cos \angle B M Q} = \frac{a}{b}
$$
Note. The present problem can have the fo... | 148. Suppose first that points $A$ and $B$ lie on opposite sides of the line $P Q$, and let $M$ be a point on this line such that
$$
\frac{\cos \angle A M P}{\cos \angle B M Q}=\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}
$$
For simplicity, let $\angle A M P=\alpha, \angle B M Q=\beta$ (Fig. 337).
 holds:
$$
1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}
$$ | 1. We will solve the problem using the method of mathematical induction.
a) For $n=1$ the equality is true.
b) Suppose the equality is true for $n=k$, where $k$ is an arbitrary number. We will prove that in this case it is also true for $n=k+1$, i.e., that
$$
\begin{gathered}
1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,401 |
2. Prove that for all natural $n$ the equality holds
$$
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}
$$ | 2. First solution. We apply the principle of mathematical induction.
a) By direct verification, we confirm the validity of the equality for $n=1$.
b) Suppose the equality holds for some natural number $k$, i.e.,
$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{k(k+1)}=\frac{k}{k+1}$.
We ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,403 |
3. Compute the sum
$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 n-2)(3 n+1)}
$$ | 3. First solution. We will prove by mathematical induction that the equality
$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}
$$
a) For $n=1$, the equality holds.
b) Suppose this equality holds for a natural number $k$, i.e., suppose
$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot ... | \frac{n}{3n+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,404 |
4. Prove that the equality
$$
\begin{aligned}
& \frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\ldots+ \\
& \quad+\frac{1}{(a+n-1)(a+n)}=\frac{n}{a(a+n)}
\end{aligned}
$$
is valid for all natural $n$ and all $a$ that are not zero or negative integers. | 4. Let's present one of the possible solutions based on the principle of mathematical induction.
a) It is easy to verify by direct inspection that the equality holds for $n=1$.
b) Suppose the equality holds for $n=k:$
$\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\ldots+$
$$
+\frac{1}{(a+k-1)(a+k)}=\frac{k}{a(a+k)}
$$
We ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,405 |
5. Prove the equality $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!=(n+1)!-1$ ( $n$ - any natural number), | 5. Let's solve the problem using mathematical induction.
a) For $n=1$ the equality is valid (check it!).
b) Suppose the equality holds for $n=k$:
$$
1 \cdot 1!+2 \cdot 2!+\ldots+k \cdot k!=(k+1)!-1
$$
We will show that in this case it will also hold for $n=k+1$. Indeed,
$$
\begin{array}{r}
1 \cdot 1!+2 \cdot 2!+\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,406 |
6. Prove that the equality
$$
\begin{array}{r}
\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{n^{2}}\right)= \\
=\frac{n+1}{2 n}
\end{array}
$$
is valid for all integers $n \geqslant 2$. | 6. First solution. We will prove our equality by the method of mathematical induction.
a) For $n=1$, a direct check shows the validity of the equality.
b) Suppose the equality is valid for some natural $k \geqslant 2$:
$$
\begin{array}{r}
\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,407 |
7. Prove the equality
\[
\begin{aligned}
& 1-2^{2}+3^{2}-4^{2}+\ldots+(-1)^{n-1} n^{2}= \\
&=(-1)^{n-1} \frac{n(n+1)}{2}
\end{aligned}
\] | 7. We will apply the method of mathematical induction.
a) For $n=1$ the formula is valid (check it!).
b) Suppose the formula is valid for some natural number $k$, i.e., suppose
$$
\begin{aligned}
& 1-2^{2}+3^{2}-\ldots+(-1)^{k-1} k^{2}= \\
&=(-1)^{k-1} \frac{k(k+1)}{2}
\end{aligned}
$$
We will show that it is also ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,408 |
9. Prove that any amount of money greater than 7 kopecks can be paid without change using only three- and five-kopeck coins (Fig. 3). | 9. a) For $n=8$, it is obvious that 8 kopecks can be represented by three- and five-kopeck coins: $8=5+3$.
b) Let us assume that we can represent $k$ kopecks as the sum of three- and five-kopeck coins. We will show that the same can be done for $k+1$ kopecks. Consider two cases: there is at least one five-kopeck coin ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,410 |
10. Prove that a plane divided into parts by $n$ lines can be painted black and white so that any two parts sharing a common side will be painted in different colors (such a coloring is called proper, see Fig. 4). | 10. a) If a plane is divided by only one line, it can be painted as required by the problem. For this, it is enough to paint one half-plane in one color and the second in another.
b) Suppose we have painted the plane, on which $n$ lines are drawn, in the required manner. Let's draw the $(n+1)$-th line and show that th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,411 |
11. Prove that the sum of the cubes of three consecutive natural numbers is divisible by 9. | 11. We need to prove that for any natural number $n$, the number
$$
n^{3}+(n+1)^{3}+(n+2)^{3}
$$
is divisible by 9.
a) For $n=1$, the statement is true, since $1^{3}+2^{3}+3^{3}=36$ is divisible by 9.
b) Suppose the statement is true for some natural number $k$, i.e., suppose $k^{3}+(k+1)^{3}+(k+2)^{3}$ is divisibl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,412 |
12. Prove that for any integer $n \geqslant 0$.
$$
11^{n+2}+12^{2 n+1}
$$
is divisible by 133. | 12. a) Let's check our statement for $n=0$. In this case
$$
11^{n+2}+12^{2 n+1}=11^{2}+12=121+12=133
$$
is divisible by 133.
b) Suppose our statement holds for some $k \geqslant 0$, i.e., suppose $11^{k+2}+12^{2 k+1}$ is divisible by 133. We will show that in this case it also holds for $n=k+1$. We have
$$
\begin{a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,413 |
13. Prove that for any $n \geqslant 2$ the inequality
$$
(1+a)^{n}>1+n a
$$
holds if we assume that $a>-1$ and $a \neq 0$. | 13. a) For $n=2$ the inequality is valid, since $(1+a)^{2}=1+2 a+a^{2}>$ $>1+2 a$, because $a^{2}>0$ when $a \neq 0$.
b) Let the inequality be valid for some natural number $k \geqslant 2$, i.e., let
$$
(1+a)^{k}>1+k a
$$
We will prove that it is also valid for $n=k+1$. Indeed, by the condition $1+a>0$. Therefore, mu... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,414 |
14. Prove that for any $n \geq 2$
$$
1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}
$$
Now we move on to problems about sequences. It is useful to look again at the definition of a sequence and examples on pages 7-10. | 14. a) For $n=2$ the inequality holds, since $1+\frac{1}{\sqrt{2}}>\sqrt{2}$.
b) Let the inequality hold for $n=k$, i.e., let
$$
1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k}
$$
We will show that the inequality also holds for $n=k+1$. For this, it is sufficient to prove the validity of t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,415 |
15. Let the sequence be defined as follows:
$$
\begin{gathered}
u_{1}=2 ; \quad u_{2}=3 ; \quad u_{n}=3 u_{n-1}-2 u_{n-2} \\
\text { for } n>2
\end{gathered}
$$
Prove that the formula $u_{n}=2^{n-1}+1$ holds. | 15. a) For $n=1$ and 2, the formula is obvious.
b) Suppose the formula is valid for all $n \leqslant k$, where $k \geqslant 2$. We will show that it is also valid for $n=k+1$. Indeed,
\[
\begin{aligned}
& u_{k+1}=3 u_{k}-2 u_{k-1}=3\left(2^{k-1}+1\right)- \\
& -2\left(2^{k-2}+1\right)=3 \cdot 2^{k-1}+3-2^{k-1}-2= \\
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,416 |
16. Let $u_{1}=1 ; \quad u_{n+1}=u_{n}+8 n$, $n$ - a natural number. Prove that $u_{n}=(2 n-1)^{2}$.
Let a sequence $u_{1}$, $u_{2}, \ldots$ be given. We will form a new sequence $v_{1}, v_{2}, \ldots$ as follows: $v_{n}=u_{n+1}-u_{n}$ for all n. This sequence is called the sequence of differences of the sequence $u_{... | 16. a) For $n=1$ the formula is valid.
b) Let the formula be valid for $n=k$, i.e., let
$$
u_{k}=(2 k-1)^{2}
$$
We will show that it is also valid for $n=k+1$. Indeed,
$$
\begin{aligned}
& u_{k+1}=u_{k}+8 k=(2 k-1)^{2}+8 k= \\
& \quad=4 k^{2}-4 k+1+8 k=4 k^{2}+4 k+1=(2 k+1)^{2}
\end{aligned}
$$
By applying the pri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,417 |
17. Let $u_{n}=n^{2}$. Find the formula for the general term of the sequence of differences. | 17. $\quad \Delta u_{n}=u_{n+1}-u_{n}=(n+1)^{2}-n^{2}=$ $=2 n+1$. | 2n+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,418 |
18. Given two sequences $u_{1}$, $u_{2}, \ldots$ and $v_{1}, v_{2}, \ldots$ with the same sequence of differences, i.e., such that $\Delta u_{n}=\Delta v_{n}$ for all $n$.
a) Can we assert that $u_{n}=v_{n}$?
b) Can we assert that $u_{n}=v_{n}$ if it is additionally known that $u_{1}=v_{1}^{n}$? | 18. Let's show that for all $n$ the formula
$$
u_{1}+\Delta u_{1}+\Delta u_{2}+\ldots+\Delta u_{n-1}=u_{n}
$$
is valid. Indeed, recalling the definition of the sequence, we have
$$
\begin{aligned}
u_{1}+\Delta u_{1}+\Delta u_{2} & +\ldots+\Delta u_{n-1}=u_{1}+\left(u_{2}-u_{1}\right)+ \\
& +\left(u_{3}-u_{2}\right)+... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,419 |
19. Let each term of the sequence $w_{1}, w_{2}, \ldots$ be equal to the sum of the corresponding terms of the sequences $u_{1}, u_{2}, \ldots$ and $v_{1}, v_{2}, \ldots$, i.e., $w_{n}=$ $=u_{n}+v_{n}$ for all $n$. Prove that in this case $\Delta w_{n}=\Delta u_{n}+\Delta v_{n}$ for all $n$. | 19. The task is solved by direct verification
$$
\begin{aligned}
\Delta w_{n}= & \left(w_{n+1}-w_{n}\right)=\left(u_{n+1}+v_{n+1}\right)-\left(u_{n}+v_{n}\right)= \\
& =\left(u_{n+1}-u_{n}\right)+\left(v_{n+1}-v_{n}\right)=\Delta u_{n}+\Delta v_{n} .
\end{aligned}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,420 |
20. Given the sequence $u_{n}=n^{k}$. Prove that the general term of its sequence of differences is expressed by a polynomial of degree ( $k-1$ ) in $n$, and find the leading coefficient of this polynomial. | 20. We have $\Delta u_{n}=(n+1)^{k}-n^{k}$. Let's show that
$$
(n+1)^{k}=n^{k}+k n^{k-1}+\ldots
$$
where the dots denote terms containing $n$ in powers less than $k-1$.
We will prove this using induction on the number $k$.
a) For $k=1$, the statement is true.
b) Suppose the statement is true for some $k=k_{0}$, i.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,421 |
22. Given the sequence $u_{n}=n^{k}$, where $k$ is a fixed positive number. The sequence of differences is formed from it, and from the resulting sequence, another sequence of differences is formed, and this process is repeated $k$ times (Fig. 5). Prove that the resulting sequence will have all terms equal to the same ... | 22. Let's show that the $k$-th differences of the sequence $u_{n}=n^{k}$ for all $k$ are equal to $k$!. We will use induction on $k$.
a) It is clear that for $k=1$ the statement is true, since if $u_{n}=n$, then $\Delta u_{n}=1=1!$.
b) Suppose the statement is proven for all $k \leqslant k_{0}$. We will prove it for ... | k! | Algebra | proof | Yes | Yes | olympiads | false | 28,423 |
23. Given the sequence $v_{n}=n^{k}$. Prove that there exists a sequence $u_{1}, u_{2}, \ldots$, whose general term is expressed by a polynomial of degree $(k+1)$ in $n$ and for which $\Delta u_{n}=v_{n}$. Also find the leading coefficient of this polynomial. | 23. Let's solve the problem by induction on $k$.
a) Let $k=0$, i.e., $v_{n}=1$ for all $n$. Clearly, in this case, the desired sequence exists, and it is $u_{n}=n$. We will denote this sequence by $\left\{u_{n}^{(0)}\right\}$.
b) Suppose the statement is proved for all $k \leqslant k_{0}$. This means that there exist... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,424 |
24. Given a sequence $u_{1}$, $u_{2}, \ldots$ The sequence of differences was formed for it. For the resulting sequence, the sequence of differences was formed again. When this was done four times, a sequence of all zeros was obtained. Prove that the original sequence is defined by a polynomial of the 3rd degree. | 24. Let us introduce the notations $v_{n}=\Delta u_{n}$, $w_{n}=\Delta v_{n}$, $x_{n}=\Delta w_{n}$, and $y_{n}=\Delta x_{n}$. From the problem statement, it is known that $y_{n}=0$ for all $n$. From this, it follows that $x_{n}$ are the same for all $n$. Hence, it is clear that $w_{n}$ for all $n$ is expressed as a po... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,425 |
25. For a certain sequence $u_{1}, u_{2}, \ldots$, it is known that its sequence of differences $v_{1}, v_{2}, \ldots$ is expressed by a polynomial of degree $k$. Prove that the sequence $u_{1}$, $u_{2}, \ldots$ itself can be defined by a polynomial of degree $(k+1)$. | 25. Let the polynomial $\sigma_{n}$ have the form
$$
v_{n}=a_{0} n^{k}+a_{1} n^{k-1}+\ldots+a_{k-1} n+a_{k}
$$
Then consider the sequence
$$
w_{n}=a_{0} u_{n}^{(k)}+a_{1} u_{n}^{(k-1)}+\ldots+a_{k-1} u_{n}^{(1)}+a_{k} u_{n}^{(0)}
$$
where $u_{n}^{(k)}, u_{n}^{(k-1)}, \ldots, u_{n}^{(1)}, u_{n}^{(0)}$ are sequences ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,426 |
26. Calculate the sum $1^{2}+2^{2}+\ldots+n^{2}$, | 26. Let $u_{n}=1+2^{2}+\ldots+(n-1)^{2}+n^{2}$. Then it is clear that $\Delta u_{n}=n^{2}$. From problem 24, it follows that $u_{n}$ can be expressed as a polynomial of the 3rd degree, i.e.,
$$
u_{n}=a_{0} n^{3}+a_{1} n^{2}+a_{2} n+a_{3}
$$
Now let's find $a_{0}, a_{1}, a_{2}, a_{3}$. For this, substitute the numbers... | \frac{n(n+1)(2n+1)}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,427 |
27. Calculate the sum
$$
1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)
$$
An arithmetic progression is a sequence in which $u_{n+1}=u_{n}+d$ for all $n$. The number $d$ is called the common difference of the progression.
A geometric progression is a sequence in which $u_{n+1}=u_{n} \cdot q$ for all $n$. The number $q$... | 27. Let $u_{n}=1 \cdot 2+2 \cdot 3+\ldots+n(n+1)$.
Then $\Delta u_{n}=n(n+1)$ is a polynomial of the second degree. Therefore, from the result of problem 24, it follows that $u_{n}$ is a polynomial of the 3rd degree. Let
$$
u_{n}=a_{0} n^{3}+a_{1} n^{2}+a_{2} n+a_{3}
$$
To find $a_{0}, a_{1}, a_{2}, a_{3}$, substitu... | \frac{n(n+1)(n+2)}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,428 |
28. a) Express the $n$-th term of an arithmetic progression in terms of the first term $u_{1}$ and the common difference $d$.
b) Express the $n$-th term of a geometric progression in terms of the first term $u_{1}$ and the common ratio $q$. | 28. a) From the definition of an arithmetic progression, we have
$$
\begin{aligned}
u_{n}=u_{n-1}+d=u_{n-2}+2 d & =\ldots= \\
& =u_{1}+(n-1) d
\end{aligned}
$$
b) From the definition of a geometric progression, we have
$$
u_{1}=u_{n-1} q=u_{n-2} q^{2}=\ldots=u_{1} q^{n-1}
$$ | u_{n}=u_{1}+(n-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,429 |
29. a) Find the formula for the sum $S_{n}$ of the first $n$ terms of an arithmetic progression.
b) Find the formula for the product $P_{n}$ of the first $n$ terms of a geometric progression. | 29. a) We need to find the following sum:
$$
S_{n}=u_{1}+u_{2}+\ldots+u_{n-1}+u_{n}
$$
From problem 28 a), we know that this sum can be written as follows:
$$
\begin{aligned}
S_{n}=u_{1} & +\left(u_{1}+d\right)+\left(u_{1}+2 d\right)+\ldots+ \\
& +\left[u_{1}+(n-2) d\right]+\left[u_{1}+(n-1) d\right]
\end{aligned}
$... | S_{n}=u_{1}n+\frac{n(n-1)}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,430 |
30. a) Find the sum of the first 15 terms of an arithmetic progression where the first term is 0, and the common difference is $\frac{1}{3}$.
b) Find the product of the first 15 terms of a geometric progression where the first term is 1, and the common ratio is $\sqrt[3]{10}$. | 30. a) According to the formula obtained in problem $29 a)$
$$
S_{n}=u_{1} n+\frac{n(n-1)}{2} d=\frac{15 \cdot 14}{2} \cdot \frac{1}{3}=35
$$
b) According to the formula obtained in problem 29 b),
$$
\begin{aligned}
& P_{n}=u_{1}^{n} q^{\frac{n(n-1)}{2}}=(\sqrt[3]{10})^{\frac{15 \cdot 14}{2}}= \\
& =10^{\frac{15 \cd... | 10^{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,431 |
31. a) The third term of an arithmetic progression is equal to 0. Find the sum of the first 5 terms.
b) The third term of a geometric progression is equal to 4. Find the product of the first 5 terms. | 31. a) We know that for an arithmetic progression $u_{1}=u_{3}-2 d ; u_{2}=u_{3}-d$; $u_{4}=u_{3}+d ; \quad u_{5}=u_{3}+2 d$. Therefore, the sum of the first five terms is
$$
\begin{aligned}
& \delta_{5}=u_{1}+u_{2}+u_{3}+u_{4}+u_{5}=\left(u_{3}-2 d\right)+ \\
& +\left(u_{3}-d\right)+u_{3}+\left(u_{3}+d\right)+ \\
& +... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,432 |
32. Prove that the sum $S_{n}$ of the first $n$ terms of a geometric progression is expressed by the formula $S_{n}=a_{1} \frac{q^{n}-1}{q-1}$. | 32. We need to find the sum of the first n terms of a geometric progression, i.e., the quantity
$$
\begin{aligned}
& S_{n}=u_{1}+u_{2}+u_{3}+\ldots+u_{n}= \\
& \quad=u_{1}+u_{1} q+u_{1} q^{2}+\ldots+u_{1} q^{n-1}
\end{aligned}
$$
Consider the quantity $q S_{n}-S_{n}$ :
$$
\begin{gathered}
q S_{n}-S_{n}=q\left(u_{1}+... | S_{n}=\frac{u_{1}q^{n}-u_{1}}{q-1} | Algebra | proof | Yes | Yes | olympiads | false | 28,433 |
33. Someone arrives in the city with very interesting news and after 10 minutes reports it to two people. Each of the newly informed people, after 10 minutes, reports it to two more (who have not yet heard it) and so on. *). How long will it take for the entire city to know this news, if the city has three million resi... | 33. It is easy to see that the number of people who were told the news at the end of the 1st ten-minute period is 2, the number of people who were told the news after the 2nd ten-minute period is 4, and generally, the number of people who were told the news at the end of the $k$-th ten-minute period is $2^{k}$.
The nu... | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,434 |
35. Find the sum of all positive odd numbers not exceeding a thousand. | 35. We need to find the sum $1+3+$ $+5+\ldots+997+999$. It is clear that this is the sum of the first 500 terms of an arithmetic progression with the first term 1 and common difference 2. Therefore, our sum is
$1 \cdot 500+\frac{500 \cdot 499}{2} \cdot 2=500+500 \cdot 499=$
$$
=500 \cdot 500=250000
$$ | 250000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,436 |
36. Find the sum of all positive three-digit numbers that are not divisible by 2 or 3. | 36. Subtract from the sum of all positive three-digit numbers $S^{(1)}$ the sum of positive three-digit numbers divisible by 2 ($S^{(2)}$), and the sum of positive three-digit numbers divisible by 3 ($S^{(3)}$). In this process, numbers divisible by both 2 and 3, i.e., divisible by 6, are subtracted twice.
Therefore, ... | 164700 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,437 |
37. The sum $S_{n}$ of the first $n$ terms of a sequence is given by the formula $S_{n}=3 n^{2}$. Prove that this sequence is an arithmetic progression. Calculate its first term and common difference.[^0] | 37. For any sequence, the n-th term can be obtained by subtracting the sum of the first $(n-1)$ terms from the sum of the first $n$ terms, i.e.,
$$
u_{n}=S_{n}-S_{n-1}
$$
In our case,
$u_{n}=S_{n}-S_{n-1}=3 n^{2}-3(n-1)^{2}=6 n-3$ Now let's find $u_{n}-u_{n-1}$:
$$
u_{n}-u_{n-1}=(6 n-3)-[6(n-1)-3]=6
$$
We see that... | u_{1}=3,commondifference=6 | Algebra | proof | Yes | Yes | olympiads | false | 28,438 |
38. Does there exist a geometric progression in which the numbers $27 ; 8 ; 12$ are among the terms (these numbers may not necessarily be consecutive and may not necessarily be in the given order)? On which positions can these numbers appear in the progression? | 38. Suppose we have already found the geometric progression of interest. Let the number 27 be in the $m$-th place, 8 in the $n$-th, and 12 in the $p$-th. Denote the first term of the progression by $u_{1}$ and its common ratio by $q$. Then we have
$$
\begin{aligned}
27 & =u_{1} q^{m-1} \\
8 & =u_{1} q^{n-1} \\
12 & =u... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,439 |
39. The same question for the numbers $1 ; 2 ; 5$.
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。
Note: The note at the end is not part of the translation but is provided to clarify that the translation has been done as requested. | 39. Suppose we found a geometric progression in which the numbers 1, 2, and 5 are in the $m-\mathrm{M}, n-\mathrm{m}$, and $p-\mathrm{m}$ positions, respectively. Then
$$
\begin{aligned}
& 1=u_{1} q^{m-1} \\
& 2=u_{1} q^{n-1} \\
& 5=u_{1} q^{p-1}
\end{aligned}
$$
Dividing the second and third equations by the first, ... | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 28,440 |
40. The squares of the 12th, 13th, and 15th terms of an arithmetic progression form a geometric progression. Find all possible common ratios of this progression. | 40. Let's denote by $u_{12}, u_{13}, u_{15}$ the 12th, 13th, and 15th terms of an arithmetic progression, respectively. We know that the numbers $u_{12}^{2}, u_{13}^{2}, u_{15}^{2}$ form a geometric progression, i.e.,
$$
\frac{u_{13}^{2}}{u_{12}^{2}}=\frac{u_{15}^{2}}{u_{13}^{2}}
$$
Taking the square root of both sid... | 4,4-2\sqrt{3},4+2\sqrt{3},1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,441 |
41. Find all geometric progressions in which each term, starting from the third, is equal to the sum of the two preceding ones. | 41. Let us consider such a geometric progression. Then for it, the equality holds
$$
u_{1} q^{n+1}=u_{1} q^{n}+u_{1} q^{n-1} \text { for all } n \text {. }
$$
From this, it follows that
$$
\begin{gathered}
q^{2}=q+1 \\
q=\frac{1 \pm \sqrt{5}}{2}
\end{gathered}
$$
Thus, the common ratio of a geometric progression, i... | \frac{1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,442 |
44. In the kitchen, there are five light bulbs. How many lighting options exist?
Do you understand what lighting options mean? Each lamp can be on or off. Two options are considered different if they differ in the state of at least one lamp (see Fig. 7 and 8). | 44. First solution. The number of ways to illuminate with one lamp out of five is equal to the number of ways to illuminate with four lamps out of five (Fig. 32).
$$
\begin{aligned}
& 000 \sim 0000 \\
& 10000000 \\
& 0000 \text { } \\
& 1000 \\
& 0000 \sim 0000
\end{aligned}
$$
## Fig. 32
Similarly, the number of wa... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,445 |
45. Let there be $n$ lamps. The number of ways to illuminate with $k$ lamps lit is denoted by $C_{n}^{k}$. Prove that
$C_{n}^{1)}+C_{n}^{1}+\ldots+C_{n}^{k}+\ldots+C_{n}^{n}=2^{n}$. Is the notation $C_{n}^{k}$ clear? What does $C_{5}^{1}$ mean? What is the value of $C_{5}^{4}$? | 45. How many ways of lighting are there if there are $n$ lamps? Let's calculate this number using two methods (cf. the solution to problem 44).
First method:
0) It is possible that no lamp is lit. There are $C_{n}^{0}$ such ways of lighting (0 lamps out of $n$ are lit). Clearly, $C_{n}^{0}=1$ for any $n$. Nevertheles... | C_{n}^{0}+C_{n}^{1}+\ldots+C_{n}^{k}+\ldots+C_{n}^{n}=2^{n} | Combinatorics | proof | Yes | Yes | olympiads | false | 28,446 |
46. In the city, there are $n$ traffic lights. Each can be in one of three states (red, yellow, or green). In how many ways can all the traffic lights be lit? | 46. If there is only one traffic light, there are 3 ways to turn it on. Now let's add a second traffic light. From any way of turning on one traffic light, by changing the state of the second, we can get three ways to turn on two. Thus, the number of ways will triple. In total: $3 \cdot 3=3^{2}$ ways. Now let's add a t... | 3^n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,447 |
47. In how many ways can $n$ traffic lights be lit, of which $k$ can be in one of three states, and the remaining $n-k$ in one of two? | 47. Let the first $k$ traffic lights burn in some way (there are $3^{\text {k }}$ such ways, see problem 46), while the rest can be lit in $2^{n-k}$ ways (see problem 45). This means that from each way of turning on the first $k$ traffic lights, one can obtain $2^{n-k}$ ways of turning on all the traffic lights. Theref... | 3^{k}\cdot2^{n-k} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,448 |
49. What is the maximum possible number of different car license plates consisting of four digits followed by three letters (Fig. 9, with a total of 32 letters in the alphabet)?
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。 | 49. Let's correspond to each inhabitant of the kingdom a sequence of 32 zeros and ones according to the following rule: if the inhabitant has the first tooth, we place 1 in the first position of the sequence, and if not, we place 0; if the inhabitant has the second tooth, we place 1 in the second position of the sequen... | Logic and Puzzles | other | Yes | Yes | olympiads | false | 28,450 | |
54. Prove that in Pascal's triangle, the sum of the numbers in the $(n+1)$-th row is equal to $2^{n}$. | 54. We will use the method of mathematical induction. For $n=1$ (the second row), the sum of the numbers is 2. We will prove that the sum of the numbers in the $(n+1)$-th row is twice that of the $n$-th row. For this, we will write the $n$-th and $(n+1)$-th rows of Pascal's triangle in the following form:
![](https://... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,454 |
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