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Task 1. Prove that the perimeter of any section of the triangular pyramid $D A B C$ by a plane does not exceed the largest of the perimeters of its faces. | Solution.
The above text has been translated into English, preserving the original text's line breaks and formatting. Here is the direct output of the translation result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,630 |
Task 2. Check whether the statement is true: "The area of any section of the triangular pyramid $D A B C$ by a plane does not exceed the largest of the areas of its faces." | Solution. 1. Let the face $ABC$ have the largest area $S$. If the section $A_{1} B_{1} C_{1}$ is parallel to the plane $ABC$, then triangle $A_{1} B_{1} C_{1}$ is similar to triangle $ABC$. The similarity coefficient of triangles $A_{1} B_{1} C_{1}$ and $ABC$ is less than one. Therefore, the area of triangle $A_{1} B_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,631 |
Task 1. Prove that all solutions of the equation $\left(2 \sqrt{2} \cos 25^{\circ}-1\right) \operatorname{tg} x^{\circ}=\left(2 \sqrt{2} \sin 25^{\circ}-1\right) \operatorname{tg} 3 x^{\circ}$
- are integers. | Solution. The main period of equation (1) is $180^{\circ}$. Therefore, we first investigate this equation on the interval $\left[-90^{\circ} ; 90^{\circ}\right]$.
We transform equation (1) into the form:
$$
\frac{\operatorname{tg} 3 x^{\circ}}{\operatorname{tg} x^{\circ}}=\frac{2 \sqrt{2} \cos 25^{\circ}-1}{2 \sqrt{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,634 |
Problem 3. Solve the equation
$$
\arcsin 2 x+\arcsin x=\frac{\pi}{3}
$$
152 | The first solution. Since the function $\arcsin x$ is defined on the interval $[-1 ; 1]$, the function $\arcsin 2 x$ is defined on the interval $[-0.5 ; 0.5]$. Therefore, the monotonic (increasing) and continuous function $f(x) = \arcsin 2 x + \arcsin x$ is defined on the interval $[-0.5 ; 0.5]$. We find:
$$
\begin{al... | x_{1}=0.5\sqrt{\frac{3}{7}}\approx0.32732683 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,635 |
Problem 4. Solve the equation
$$
\cos \frac{4}{3} x=\cos ^{2} x
$$ | Solution. We find the main period of equation (1). The main period of the function $y=\cos ^{2} x$ is $\pi$. The main period of the function $y=\cos \frac{4}{3} x$ is $1.5 \pi$. 154
The least common multiple of the numbers $\pi$ and $1.5 \pi$ is $3 \pi$. Therefore, the main period of equation (1) is $3 \pi$. Let $T=[0... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,636 | |
P r o b l e m 1. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 1), with an edge length of 1. A sphere is drawn through the vertices $A$ and $C$ and the midpoints $F$ and $E$ of the edges $B_{1} C_{1}$ and $C_{1} D_{1}$, respectively. Find the radius $R$ of this sphere. | Solution. The center $M$ of the sphere is equidistant from points $A$ and $C$, so it lies in the plane of symmetry of these points, i.e., the plane $B D D_{1}$. For the same reason, point $M$ also lies in the plane $A_{1} C_{1} C$ (the plane of symmetry of points $F$ and $E$). Therefore, point $M$ lies on the line $O O... | \frac{\sqrt{41}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,638 |
Problem 2. Through each vertex of a triangular pyramid, a plane is drawn containing the center of the circle circumscribed around the opposite face and perpendicular to the opposite face. Prove that these four planes intersect at one point $X$.
.
What is good about it for forming a working hypothesis?
Well, it is easy to construct the centers of the circles circumscribed around the faces and the bases of it... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,639 |
Problem 3. The base of the pyramid $DABC$ is an equilateral triangle $ABC$, with the length of each side being $4\sqrt{2}$. The lateral edge $DC$ is perpendicular to the plane of the base and has a length of 2. Find the measure of the angle and the distance between the skew lines, one of which passes through point $D$ ... | Solution. The required angle is $P M D$. Therefore, the problem has been reduced to solving the right trihedral angle $M C P D$ (with $M$ as its vertex, $C M$ as the edge of the right dihedral angle).
The length of the segment $C H$ is equal to the distance between the skew lines $D M$ and $C K$, because the line $C K... | \frac{2}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,640 |
Problem 4. Given a triangular pyramid $A B C D$, where angles $B D C, C D A$, and $A D B$ are right angles and $D A=a$, $D B=b, D C=c$. Point $M$, belonging to the face $A B C$, is equidistant from the planes of all other faces of the pyramid. Calculate the length of the segment $D M$. | Solution. Let, for definiteness, $a \leqslant b \leqslant c$. We extend the given pyramid $A B C D$ to a cube with bases $A^{\prime} D B^{\prime} K^{\prime}$ and $A_{1} C B_{1} K_{1}$ (Fig. 4). Each point on the diagonal $D K_{1}$ of this cube is equidistant from the planes $C D B, C D A$, and $A D B$. Therefore, the p... | \frac{\sqrt{3}}{++} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,641 |
Problem 5. $DABC$ is a triangular pyramid (Fig. 5). $\overline{AK}=\overline{KD}, \overline{BP}=\overline{PC}, \overline{DM}=0.4 \overline{DC}$. Find the area $S$ of the section of the pyramid by the plane $KMP$, if the vertex $A$ is at a distance $h=1$ from the plane $KMP$ and the volume of the pyramid $DABC$ is 5. | Solution. We construct the section of the pyramid $DABC$ using the method of parallel projections. We choose the line $KP$ as the direction of projection. In this case, the pyramid $\dot{D}ABC$ is represented by the parallelogram $ABDC$ (Fig. 6), and the cutting plane is represented by the line $KM$, which intersects t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,642 |
Problem 7. The base of the pyramid $HABCD$ is a square $ABCD$, and the edge $HA$ is perpendicular to the base plane. $AB=3, HA=4$. Prove that a sphere can be inscribed in the pyramid, and find its radius.
For comparison, we will provide two solutions to this problem. | The first solution. First, we find the radius of the sphere. Then we prove its existence. For this, we apply the formula $3 v = S r$. Clearly, $V = \frac{1}{3} \cdot 3 \cdot 3 \cdot 4 = 12$; $H B = 5$; $S(ABCD) = 9$; $S(BAH) = S(HAD) = 0.5 \cdot 3 \cdot 4 = 6$; $S(HBC) = S(HDC) = 0.5 \cdot 3 \cdot 5 = 7.5$. Therefore, ... | 1 | Geometry | proof | Yes | Yes | olympiads | false | 37,644 |
Problem 8. The base of the triangular pyramid $D A B C$ is triangle $A B C$, in which angle $A$ is a right angle, angle $A C B$ is $30^{\circ}, B C=2 \sqrt{2}$. The lengths of the edges $A D, B D, C D$ are equal to each other. A sphere with radius 1 touches the edges $A D, B D$, the extension of edge $C D$ beyond point... | Solution. The base of the given pyramid is a right-angled triangle $ABC$, and its lateral edges are equal, so the height of the pyramid is the segment $DH$ (where $H$ is the midpoint of the hypotenuse $BC$ of triangle $ABC$).
The given sphere touches the edges $AD$ and $BD$, so its center $O$ lies in the plane of symm... | \sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,645 |
Problem 9. At the base of the pyramid $DABC$ lies an isosceles triangle $ABC (AB = AC = 1$, $\angle ABC = \alpha)$. The plane $DBC$ forms an angle of $\alpha$ with the plane $ABC$; $\angle DAB = \angle DAC = \alpha$. The height $DH$ of the pyramid is located inside it (Fig. 11). Find the volume of the pyramid $KDP C$, ... | Solution. For the pyramid $DABC$ to exist, it is necessary:
$$
\text { 1) } \begin{aligned}
\angle DAB + \angle DAC & \geqslant \angle BAC, \text { i.e., } 45^{\circ} < \alpha < \\
& < 90^{\circ}.
\end{aligned}
$$
2) $\angle DAM + \angle AMD < 180^{\circ}$ (this inequality is a consequence of condition (1), since $\a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,646 |
5. For what value of $z$ does the function $h(z)=$ $=\sqrt{1.44+0.8(z+0.3)^{2}}$ take its minimum value? | Solution. The function $h(z)$ takes the smallest value if $z+0.3=0$, i.e., when $z=-0.3$. It is clear that at this value of $z$, points $E$ and $H$ coincide, i.e., points $E$ and $H$ coincide with point $Q$.
It is also clear that if $z=-0.3$, then the plane $K N_{1} M$ is perpendicular to the plane of the rectangle $N... | -0.3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,651 |
1. On this cuneiform tablet, divisors and quotients are placed, which, according to the sexagesimal system used by the Babylonians, is written as:
$$
60^{8}+10 \cdot 60^{7}
$$
Express this number in the decimal system.
## 1 -st Senkere table. | 1. 195955200000000 . | 195955200000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,653 |
4. The following table contains a series of numbers written in the sexagesimal system: $1.36 ; 1.52 ; 2.8 ; 2.24 ; 2.40 ; 2.56$; $3 \cdot 12 ; 3 \cdot 28 ; 3 \cdot 44 ; 4$
Show that these numbers represent an arithmetic progression, and write it in the decimal system. | 4. In the decimal system, we will have the progression: $\therefore 96,112,128,144 \ldots$ | 96,112,128,144\ldots | Algebra | proof | Yes | Yes | olympiads | false | 37,659 |
10. Determine the volume of a truncated square pyramid if its height is 6, the side of the lower base is 4, and the upper base is 2. | 10. Using the formula for the volume of a truncated pyramid
$$
V=\frac{H}{3}(B+b+\sqrt{B b})
$$
we get:
$$
V=\frac{6}{3}(16+4+\sqrt{16 \cdot 4})=2(16+4+8)=56
$$
It is interesting to see the solution of the Egyptian mathematician. We will translate the text from the papyrus verbatim:
"Problem to solve the pyramid (... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,664 |
11. Determine the lengths of the sides of a rectangle if their ratio and the area of the figure are known.
## Problems from the Kahun Papyrus. | 11. If the sides of the rectangle are $x$ and $y$, then the problem reduces to solving the system of equations:
$$
\frac{x}{y}=\frac{m}{n} \text { and } x y=S
$$
Multiplying these equations gives:
$$
x^{2}=\frac{m}{n} \cdot S
$$ | x^{2}=\frac{}{n}\cdotS | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,665 |
13. Divide a volume of 120 cubic cubits into 10 parts, each with a height of 1 cubit, and with a rectangular base whose sides are in the ratio of $1: \frac{3}{4}$.
## Problems from the Berlin Papyrus. | 13. One part $=12$ cubic cubits. Area of the base 12 sq. cubits. If the ratio of the sides $1: \frac{3}{4}$, then the measurements of the rectangular base are 4 and 3. | 43 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,667 |
14. If it is said to you: divide 100 square cubits into 2 unknown parts and take $\frac{3}{4}$ of the side of one as the side of the other, give me each of the unknown parts (i.e., divide an area of 100 sq. cubits into 2 squares, the sides of which are in the ratio $\left.1: \frac{3}{4}\right)$.
## Problems from the R... | 14. The problem reduces to the system of equations:
$$
x: y=1: \frac{3}{4} ; x^{2}+y^{2}=100
$$
It is not hard to see that $x=8 ; y=6$.
$$
\text { 15. } \begin{gathered}
\frac{2}{2 n+1}=\frac{2(n+1)}{2 n+1}:(n+1)=\left(1+\frac{1}{2 n+1}\right):(n+1)= \\
=\frac{1}{n+1}+\frac{1}{(n+1)(2 n+1)} \text {. }
\end{gathered}... | 8,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,668 |
17. Two thirds in addition (to the unknown), $\frac{1}{3}$ in subtraction, 10 remains (i.e., two thirds of the unknown are added to it, and one third is subtracted from the resulting sum. The remainder is 10). | 17. The equation will be:
$$
x+\frac{2}{3} x-\frac{1}{3}\left(x+\frac{2}{3} x\right)=10 ; x=9
$$
$\begin{array}{ll}\text { 18. } x=\frac{3}{10} ; x=\frac{15}{53} & \text { 19. } 4 \frac{1}{6} \text {. }\end{array}$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,670 |
20. The Egyptians, replacing the area of a circle with the area of an equal-sized square, took the side of the latter as $\frac{8}{9}$ of the diameter of the circle. Find the approximate value of $\pi$ from this. | 20. Since $\left(\frac{8}{9} d\right)^{2}=\frac{\pi d^{2}}{4}$, then $\pi=3.16$. | 3.16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,671 |
23. In one of the problems, the first term of a decreasing arithmetic progression is given by an expression that, in modern notation, is represented by the following formula:
$$
a=\frac{S}{n}+(n-1) \frac{d}{2}
$$
where \(a\) is the first term, \(S\) is the sum of the terms, \(n\) is the number of terms, and \(d\) is ... | 23. We have:
\[
\begin{aligned}
& 2 a n=2 S+(n-1) d n \\
& 2 a n-(n-1) d n=2 S \\
& n[2 a-(n-1) d]=2 S
\end{aligned}
\] | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,674 |
25. Seven people have seven cats; each cat eats seven mice, each mouse eats seven ears of barley, and from each ear can grow seven measures of grain. What are the numbers in this series and what is their sum? (The problem is reconstructed according to Rodet.)
## Problems from the Akhmim Papyrus. | 25. $7,49,343,2401,16807$, sum 19607. | 19607 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,676 |
26. Someone sowed 7 artabs, another 8, the third 9 and from them a tax of $3 \frac{1}{2} \frac{1}{4}$ (i.e., $\left.3 \frac{3}{4}\right)$ was taken. How much will the seventh, eighth, and ninth (i.e., the share of seven, the share of eight, the share of nine, in other words: how great is the tax taken from each) be? | 26. From the first, $1 \frac{3}{32}$ is taken, from the second $1 \frac{1}{4}$, from the third $1 \frac{13}{32}$. | 1\frac{3}{32},1\frac{1}{4},1\frac{13}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,677 |
27. Someone took $\frac{1}{13}$ from the treasury. From what was left, another took $\frac{1}{17}$, and he left 150 in the treasury. We want to find out how much was originally in the treasury? | 27. In the treasury, there was $172 \frac{21}{32}$. In the papyrus, the fractional part is represented as the sum of unit fractions: $\frac{1}{2}+\frac{1}{8}+\frac{1}{48}+\frac{1}{96}$. | 172\frac{21}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,678 |
28. In the Akhmim papyrus, the area of a circle, the circumference of which is the arithmetic mean of two given circumferences, is taken as the arithmetic mean of their areas. Show that this is incorrect, and find how large the error is in percentage, where the radii of the given circles \( r=5 \); \( R=10 \).
Greek mathematics reached its peak by the 3rd century BC (the Alexandrian period). From this time, a number of mathematical treatises (... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,679 |
30. Any odd number is the difference of two squares. | 30. It is clear that $2 n+1=(n+1)^{2}-n^{2}$ (this can be easily proven geometrically). | 2n+1=(n+1)^{2}-n^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 37,681 |
31. The Pythagorean rule for calculating the sides of a right-angled triangle is based on the identity:
$$
(2 n+1)^{2}+\left(2 n^{2}+2 n\right)^{2}=\left(2 n^{2}+2 n+1\right)^{2}
$$
Using this identity, calculate the sides of the right-angled triangles for \( n=1,2,3,4,5 \). | 31. According to the Pythagorean rule, for the smaller leg, we take an odd number $2 n+1$. If we square it, subtract one, and divide the remainder by two, we get the larger leg:
$$
(2 n+1)^{2}=4 n^{2}+4 n+1, \frac{4 n^{2}+4 n}{2}=2 n^{2}+2 n
$$
By adding one to the obtained result, we find the hypotenuse $2 n^{2}+2 n... | \begin{aligned}&forn=1:3,4,5\\&forn=2:5,12,13\\&forn=3:7,24,25\\&forn=4:9,40,41\\&forn=5: | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,682 |
32. The so-called "Plato's rule". If one of the legs is taken as an even number $2 p$, then the other leg will be $p^{2}-1$, and the hypotenuse $p^{2}+1$. Verify and calculate the sides of the triangles for $p=2,3,4,5$.
## Euclid's Problems.
From the treatise "Elements". | 32. Platon's rule is easily found by doubling the numbers that Pythagoras took.
$$
\begin{array}{rccrrr}
\text { For } & p=2 & \text { we will } & \text { have } & 4, \quad 3,5 \\
n & p=3 & & & & \\
n & & 6, & 8,10 \\
n & p=4 & n & n & 8,15,17 \\
n & p=5 & n & n & 10,24,26 .
\end{array}
$$
Euclid of Alexandria (3rd c... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,683 | |
33. On the given ray $A B$, construct an equilateral triangle. | 33. The first proposition of the first book. From point $A$ as center, with radius $A B$ equal to the given segment, describe the circle $B C D$. From point $B$ as center, with the same
F... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,684 |
34. Divide a straight angle into two equal parts. | 34. Given angle $B A C$. Take an arbitrary point $D$ on side $A B$ and lay off segment $A E = A D$ on side $A C$. Connecting points $D$ and $E$, construct an equilateral triangle $D E F$ on $D E$. Connect point $F$ with the vertex of the angle $A$. Line $A F$ bisects the angle. The proof is evident from the diagram. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,685 |
35. Construct a parallelogram whose sides are inclined at a given angle, so that it is equal in area to a given triangle. | 35. Given a triangle $ABC$ and an angle $D$. We divide $BC$ at point $E$ into two equal parts and construct an angle at point $E$ equal to $D$. Through point $C$, we draw a line $CG \parallel FE$, and through point $A$, we draw a line $AG \parallel BC$. The parallelogram $ECFG$ is the desired one.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,691 |
43. Inscribe in the larger of two given concentric circles a regular polygon with an even number of sides, which would not touch the smaller circle. | 43. Proposition XVI of the twelfth book. Let us draw the diameter $B D$ and at the point $G$ of intersection with the smaller circle, draw the tangent $A C$. Let us bisect the semicircle $B A D$ and continue this division until we reach a part $H D$ which is less than $A D$. Drop a perpendicular $H K$ from point $H$ to... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,694 |
44. The square constructed on the side of the inscribed pentagon is equal to the sum of the squares of the sides of the inscribed regular hexagon and decagon in the same circle.
From an Arabic manuscript of Euclid's treatise "On Divisions of Figures." | 44. Proposition $\mathrm{X}$ of the thirteenth book. Euclid geometrically proves that $a_{5}^{2}=a_{10}^{2}+a_{6}^{2}$; but $a_{5}=\frac{R}{2} \sqrt{10-2 \sqrt{5}}$, $a_{10}=\frac{R}{2}(\sqrt{5}-1)$ and $a_{6}=R$, where $R$ is the radius of the circumscribed circle. Therefore, we have:
$$
\frac{R^{2}}{4}\left(10-2 \sq... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,695 |
45. Divide in half the area bounded by an arc of a circle and two intersecting straight lines.
## Archimedes' Problems.
From the treatise "On the Quadrature of the Parabola". | 45. This problem is found in an Arabic manuscript discovered by the orientalist mathematician Bönk: in the Paris Library. In it, Euclid is named as the author of the treatise "On Divisions of Figures." Euclid's commentator Proclus mentions this work, but until the second half of the 16th century, this treatise was unkn... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,696 |
46. Find the sum of the infinite decreasing geometric progression:
$$
1+\frac{1}{4}+\left(\frac{1}{4}\right)^{2}+\left(\frac{1}{4}\right)^{3}+\ldots
$$
From the treatise "On the Sphere and Cylinder". | 46. Since $\lim S=\frac{a}{1-q}$, then for $a=1$ and $q=\frac{1}{4}$ we have 6 Ppopyov. sororiiik task. 81
$\lim S=\frac{4}{3}$. Archimedes' proof is interesting: we need to find the sum of the terms of an infinitely decreasing progression $a+b+$ $+c+d+\ldots$, if the common ratio $=\frac{1}{4}$. We have:
$$
\begin{ga... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,697 |
47. The surface of a spherical segment is equal to the area of a circle having as its radius the straight line drawn from the vertex of the segment to the circumference which forms its base. | 47. It is known that the surface area of a segment is $2 \pi R \cdot H$, where $R$ is the radius of the sphere, $H$ is the height. If $b$ is the line connecting the vertex of the segment with the circumference of the base, then $b^{2}=2 R H$, therefore, $S=\pi b^{2}$. | \pib^{2} | Geometry | proof | Yes | Yes | olympiads | false | 37,698 |
48. A cylinder whose base is the great circle of a sphere and whose height is the diameter of this sphere has a volume equal to $\frac{3}{2}$ the volume, and a surface area equal to $\frac{3}{2}$ the surface area of the sphere.
## From the treatise "Lemmas". | 48. Volume of the cylinder:
$$
V_{u}=\pi R^{2} \cdot 2 R=2 \pi R^{3}=\frac{3}{2}\left(\frac{4}{3} \pi R^{3}\right)=\frac{3}{2} V_{u}
$$
Surface area of the cylinder (total) $S_{u}=2 \pi R \cdot 2 R+2 \pi R^{2}=$
 $A F D H C B$ is equal to the area of a circle with diameter $D... | 50. Let's call the area of the arbelos $S$. It is easy to see that
$$
S=\frac{\pi}{8}\left[A C^{2}-\left(A D^{2}+C D^{2}\right)\right]
$$
but $A C=A D+C D$, therefore, $S=\frac{\pi \cdot A D \cdot C D}{4}$; since
$$
A D \cdot D C=B D^{2}, \text { then } S=\pi \cdot \frac{B D^{2}}{4}
$$ | \pi\cdot\frac{BD^{2}}{4} | Geometry | proof | Yes | Yes | olympiads | false | 37,701 |
51. If a circle is circumscribed around a square, and another is inscribed in it, then the circumscribed circle is twice as large as the inscribed one. | 51. If the side of the square is denoted by $2 a$, then the radius of the inscribed circle is $a$, and the radius of the circumscribed circle is $a \sqrt{2}$; the area of the inscribed circle is $\pi a^{2}$, and that of the circumscribed circle is $2 \pi a^{2}$. | 2\pi^{2} | Geometry | proof | Yes | Yes | olympiads | false | 37,702 |
53. If in a circle the chords $A B$ and $C D$ intersect at point $E$ at a right angle, then the sum of the squares of the segments $A E, B E, C E, D E$ is equal to the square of the diameter. | 53. Archimedes' Solution. Draw the diameter $A F$ and the chords $A C, A D, C F$
Fig. 2. and $B D$. Since $\angle A E D$ is a right angle, it equals $\angle A C F$. But $\angle A D C = \an... | DE^{2}+BE^{2}+AE^{2}+CE^{2}=AF^{2} | Geometry | proof | Yes | Yes | olympiads | false | 37,704 |
54. If from an external point to a circle one secant is drawn through the center, and another so that the external segment equals the radius of the circle, then the angle between the secants will be measured by one third of the larger of the arcs intercepted between its sides.
From the treatise "On the Sphere and Cyli... | 54. $\angle B O A=\angle B A O=\alpha$, if $A B=r$. But $\angle C B O=2 \alpha$ as an external angle relative to triangle $A O B$. Triangle $B O C$ is isosceles, therefore, $\angle B C O=2 \alpha$. But $\angle C O D$ is an external angle relative to triangle $A O C$, therefore, $\angle C O D=\angle C A O+\angle A C O=\... | 3\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,705 |
55. Find a sphere that has a volume equal to that of a given cone or cylinder. | 55. The volume of a cone is $\frac{\pi}{3} \cdot r^{2} h$; if the radius of the sphere is $R$, then $r^{2} h=4 R^{3}$, and therefore, $R=\sqrt[3]{\frac{r^{2} h}{4}}$.
Fig. 24.
$6^{*}$
T... | \sqrt[3]{\frac{r^{2}}{4}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,706 |
56. To cut off from a sphere a segment by a plane such that the ratio of its volume to the volume of a cone having the base and height of the segment is a given ratio.
$$
\text { From the treatise "On Spirals". }
$$ | 56. $V_{\text {segm }}=V_{\text {sect }}-V_{\text {cone }}$,
$$
V_{ce2M}=\frac{2}{3} \pi r^{2} h-\frac{\pi r^{2}}{3}(R-h)=\pi r^{2}\left(h-\frac{R}{3}\right)
$$
Fig. 25.
Let the ratio be ... | n<3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,707 |
57. Find the sum of the squares of the first $n$ numbers of the natural series.
## Heron's Problem. From the treatise
Find the sum of the squares of the first $n$ numbers of the natural series. | 57. Archimedes solved this problem in the following form.
It goes as follows: if the excess is equal to the smallest of all, and if other lines, each equal to the largest of the lines of th... | 1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,708 |
58. Determine the area of a triangle if its three sides are given:
1) $a=13 ; b=14 ; c=15 ; 2) a=5 ; b=12 ; c=13$.
## Task by Hypsicles. | 58. Answer. 84; 30 (the second triangle is right-angled).
Hypsicles of Alexandria (150 BC), a geometer to whom the XIV book of Euclid's "Elements" (on the properties of regular polyhedra) is attributed. Problem 59 is found in the work "Anaphorikos." | 84;30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,709 |
59. Prove that in an arithmetic progression with an even number of terms, the sum of the terms of the second half exceeds the sum of the terms of the first half by a number that is a multiple of the square of half the number of terms.
## Problems of Damascius.
From the additions to Euclid's "Elements," Book XV. | 59. Solution. We have an arithmetic progression consisting of an even number ( $2 n$ ) of terms: $-\div a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}, \ldots, a_{2 n}$.
The sum of the terms of the first half
$$
S_{1}=\frac{\left(a_{1}+a_{n}\right)}{2} n
$$
The sum of the terms of the second half
$$
S_{2}=\frac{a_{n+1}+a_{2 ... | \cdotn^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 37,710 |
63. Show that if a sequence of consecutive odd numbers is divided into groups such that the number of members in each group increases as a sequence of natural numbers, then the sum of the members of each group will be equal to the cube of the number of members.
## Problems of Diophantus of Alexandria.
From the treati... | 63. Solution.
$$
1+(3+5)+(7+9+11)+(13+15+17+19)+\ldots
$$
It is immediately clear when breaking down into groups that
$$
\begin{aligned}
1 & =1^{3} \\
8 & =2^{3} \\
27 & =3^{3} \\
64 & =4^{3} \text { and so on. }
\end{aligned}
$$
General proof: up to the first number of the $n$-th group, there are
$$
1+2+3+\ldots+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,714 |
64. The number 100 needs to be divided twice so that the larger part from the first division is twice the smaller part from the second division, and the larger part from the second division is three times the smaller part from the first division. | 64. Diophantus' Solution. Let the smaller part of the second division be $x$. Then the larger part of the first division will be $2 x$. The smaller part of the first division is $100-2 x$, and the larger part of the second division is $300-6 x$. Both parts of the second division should add up to 100, hence $300-5 x=$ $... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,715 |
65. Find two numbers whose sum is 20 and product is 96. | 65. Diophantus' Solution. Let $2 n$ be the difference between the two numbers. Then the larger one is $10+n$, and the smaller one is 10 - $n$. The product is $100-n^{2}=96 ; n^{2}=4$ and $n=2$. The required numbers are 12 and 8. | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,716 |
66. $\{x+y=10$,
$\left\{x^{2}+y^{2}=68\right.$ | 66. Diophantus' Solution.
$$
\frac{x+y}{2}=5
$$
Let
$$
\frac{x-y}{2}=d
$$
Adding gives $x=5+d$, subtracting gives $y=5-d .(5+d)^{2}+$ $+(5-d)^{2}=68 ; 50+2 d^{2}=68 ; 2 d^{2}=18 ; d=3 ; x=8 ; y=2$.
Regular solution:
$$
\begin{aligned}
&-x^{2}+y^{2}+2 x y=100 \\
&-x^{2}+y^{2}=68 \\
& 2 x y=32 \\
& x^{2}+y^{2}-2 x ... | 8,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,717 |
67. Find three numbers such that the largest exceeds the middle by a given part $\left(\frac{1}{3}\right)$ of the smallest, the middle exceeds the smallest by a given part $\left(\frac{1}{3}\right)$ of the largest, and the smallest exceeds the number 10 by a given part $\left(\frac{1}{3}\right)$ of the middle number. | 67. The task is reduced to solving the system:
$$
\begin{gathered}
x-y=\frac{1}{3} z ; \quad y-z=\frac{1}{3} x ; \quad z-10=\frac{1}{3} y \\
\quad x=45 ; y=37 \frac{1}{2} ; z=22 \frac{1}{2}
\end{gathered}
$$
$$
\text { from which }
$$ | 45;37\frac{1}{2};22\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,718 |
68. Find two numbers, the ratio of which is 3, and the ratio of the sum of their squares to their sum is 5. | 68. $\frac{x}{y}=3 ; \frac{x^{2}+y^{2}}{x+y}=5$.
By squaring both sides of the first equation term by term, we get:
$$
\frac{x^{2}}{y^{2}}=9 \text { and } \frac{x^{2}+y^{2}}{y^{2}}=10
$$
Comparing with the second equation gives:
$$
\begin{aligned}
10 y^{2} & =5(x+y) \\
2 y^{2} & =x+y
\end{aligned}
$$
but $x=3 y$, ... | 6,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,719 |
70. Find three numbers such that the product of any pair of them, increased by their sum, would be equal to 8, 15, 24, respectively. | 70. We obtain the system:
$$
\begin{gathered}
x y + x + y = 8 \\
y z + y + z = 15 \\
x z + x + z = 24 \\
x = \frac{11}{4} ; y = \frac{7}{5} ; z = \frac{17}{3}
\end{gathered}
$$ | x=\frac{11}{4};y=\frac{7}{5};z=\frac{17}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,721 |
72. Find two numbers such that their product, added to each of the given numbers, forms the cube of some number. | 72. Diophantus's Solution. Let the first number be $8x$. Then, taking the second to be $x^{2}-1$, we will satisfy one of the requirements, since the product of these numbers $8x^{3}-8x$, added to $8x$, will make $8x^{3}$, i.e., the cube of some number. It is necessary that this same product, added to the second number,... | \frac{112}{13},\frac{27}{169} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,723 |
73. Find three numbers such that the sums of all three and each pair are squares. | 73. Diophantus' Solution. Let the sum of all three numbers be $x^{2}+2 x+1=(x+1)^{2}$
Let $\mathrm{I}+\mathrm{II}=x^{2}$, then $\mathrm{III}=2 x+1$.
Let $\mathrm{II}+\mathrm{III}=(x-1)^{2}=x^{2}-2 x+1$, then I will be found by subtracting the sum of II + III, which is $x^{2}-2 x+1$, from the total sum $x^{2}+2 x+1$. ... | I=80,II=320,III=41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,724 |
74. The cathetus of a right-angled triangle is a perfect cube, the other cathetus represents the difference between this cube and its side (i.e., the first power), and the hypotenuse is the sum of the cube and its side. Find the sides.
## Problems of Iamblichus. | 74. Diophantus' Solution. Let the hypotenuse be $x^{3}+x$, and the leg $x^{3}-x$.
Then the second leg will be:
$$
\sqrt{\left(x^{3}+x\right)^{2}-\left(x^{3}-x\right)^2}=2 x^{2}
$$
but $2 x^{2}=x^{3}$, therefore, $x=2$, the hypotenuse is 10, the first leg is 6, the second is 8.
Iamblichus (4th century) - one of the ... | 10,6,8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,725 |
75. Solve the system:
$$
\begin{aligned}
& x+y=2(z+u) \\
& x+z=3(y+u) \\
& x+u=4(y+z)
\end{aligned}
$$ | 75. Solution. Subtracting term by term from the 1st equation $2-\mathrm{e}$, we have:
$$
x-z=2 z+2 u-3 y-3 u, \text{ or } 4 y-3 z=-u
$$
Subtract the 3rd equation from the 1st: $y-u=2 z+2 u-4 y-4 z$, or $5 y+2 z=3 u$. Now we have the system:
$$
\begin{aligned}
4 y+u & =3 z \\
5 y-3 u & =-2 z
\end{aligned}
$$
9)
Equ... | 73,7,17,u=23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,726 |
78. The crown weighs 60 min and consists of an alloy of gold, copper, tin, and iron. Gold and copper make up $\frac{2}{3}$ of the mixture, gold and tin $\frac{3}{4}$, and gold and iron $\frac{3}{5}$ of the total weight. Determine the weight of each metal separately.
2 Problems from a collection of problems.
## Epigra... | 78. Gold and copper make 40 minae, gold and tin 45, gold and iron 36. If added together, the triple amount of gold, tin, copper, and iron makes 121 minae. Therefore, double the amount of gold equals 61 minae and the ordinary amount 30.5 minae. Then it is easy to find that copper was 9.5 minae, tin 14.5 minae, and iron ... | 30.5minae(Gold),9.5\text | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,729 |
79. - Tell me, famous Pythagoras, how many students attend your school and listen to your discourses?
- In all, 一 replied the philosopher: - half study mathematics, a quarter study music, a seventh part remains in silence, and, in addition, there are three women. | 79. We form the equation:
$$
\frac{1}{2} x+\frac{1}{4} x+\frac{1}{7} x+3=x
$$
from which $x=28$. We can solve the following problems arithmetically as well. | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,730 |
80. Find a set of numbers, where the first, added to a third of the third, equals the second, and the second, added to a third of the first, equals the third. The third is also greater than the first by 10. | 80. System of equations:
$$
\begin{array}{r}
x+\frac{1}{3} z=y \\
y+\frac{1}{3} x=z \\
z-x=10
\end{array}
$$
From (1) and (2): $2 x=z$, therefore, from (3) $x=10 ; z=20$; $y=16 \frac{2}{3}$. | 10,20,16\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,731 |
87. Show that in any triangle, a parallelogram constructed on one side of the triangle inward and having the other two vertices outside the triangle is equal in area to the sum of two parallelograms constructed on the other two sides of the triangle such that the sides parallel to the sides of the triangle pass through... | 87. It is clear that $\triangle A^{\prime} C^{\prime} B^{\prime} = \triangle A B C$ (given); parallelograms $A A^{\prime} C^{\prime} C$ and $B B^{\prime} C^{\prime} C$ are equal to the given parallelograms $A D E C$ and $B F G C$, since their heights and bases are respectively equal. If from the figure $A A^{\prime} C^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,735 |
88. Show that in circles the areas of similar segments are as the squares of the chords which form their bases. | 88. Let the areas of sectors $k$ and $k_{1}$, segments $S$ and $S_{1}$, triangles $O B C$ and $O_{1} B_{1} C_{1}$ be $\Delta$ and $\Delta_{1}$, the radii of the circles $r$ and $r_{1}$, and the bases of the segments $a$ and $a_{1}$.
The triangles are similar, therefore
Fig. 33.
The Erfu... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,740 |
94. In a right triangle, the hypotenuse is 25 feet, and the area is 150 square feet. Determine the legs. | 94. Nipsus's Solution. We will indicate the steps of his calculations using modern notation. Let $b$ and $c$ be the legs, $\Delta$ the area, and $a$ the given hypotenuse. We have:
$$
\begin{aligned}
& a^{2}=b^{2}+c^{2} \\
& 4 \Delta=2 b c
\end{aligned}
$$
Adding and subtracting give:
$$
\begin{aligned}
& a^{2}+4 \De... | b=20,=15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,741 |
95. The sides of the triangle are $6, 8, 10$. Find its area (using Heron's formula).
20
## Junius Moderatus Columella's Problem.
From the work "De re rustica". | 95. Solution of Niphus
\[
\Delta=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{12 \cdot 6 \cdot 4 \cdot 2}=24
\]
It would have been simpler, of course, to take half the product of the legs.
Junius Moderatus Columella (1st century AD) in his treatise "On Agriculture," when discussing land surveying, indicates 9
land measures and sol... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,742 |
96. If the side of a regular triangle is $a$, then Khotimella gives the formula for its area as $\frac{1}{3} a^{2}+\frac{1}{10} a^{2}$. Is this formula correct, and how large is the error (in percent) in determining the area of the triangle using this formula?
## Problem from the Chartres Manuscript. | 96. If $a$ is the side of an equilateral triangle, then its area $\Delta=\frac{a^{2}}{4} \sqrt{3}$. If we equate this expression to the one given by Columella, we obtain that $\sqrt{3}=\frac{26}{15}$, or $3=3 \frac{1}{225}=3,00(4)$. The error is $\frac{4}{27} \%$.
The Chartres Manuscript (on parchment, stored in the l... | \frac{4}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,743 |
98. In a cage, there is an unknown number of pheasants and rabbits. It is only known that the entire cage contains 35 heads and 94 feet. The task is to determine the number of pheasants and the number of rabbits. | 98. If there were only pheasants in the cage, the number of legs would be 70, not 94. Therefore, the 24 extra legs belong to the rabbits, 2 per each. It is clear that there were 12 rabbits and, consequently, 23 pheasants. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,745 |
100. 5 oxen and 2 sheep cost 10 taels, while 2 oxen and 8 sheep cost 8 taels. How much does each ox and each sheep cost separately?
Note. All three problems are from the treatise "Nine Sections on the Mathematical Art."
From the treatise "Nine Sections on the Art of Calculation." | 100. We form the system of equations:
$$
5 x+2 y=10 ; 2 x+8 y=8
$$
A sheep costs $1 \frac{7}{9}$ tael, a goat $\frac{5}{9}$ tael
98
The treatise "Nine Chapters on the Art of Calculation" presents a commentary written in the mid-13th century on an older treatise (early 8th century) "Tian Li Shu". | x=\frac{5}{9},y=1\frac{7}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,747 |
101. From three barrels of rice of the same capacity, a certain amount of rice was stolen by three thieves. The total amount was unknown, but it was found that 1 go of rice remained in the first barrel, 1 shing 4 go in the second, and 1 go in the third. The captured thieves confessed: the first that he had scooped rice... | 101. From the conditions of the problem, it follows that
$$
19 x+1=17 y+14=12 z+1
$$
where $x$ is the number of times rice was scooped with a shovel,
$y$ is the number of times with a shoe,
$\boldsymbol{z}$ is the number of times with a bowl.
Since $19 x=12 z$, then $x=\frac{12 z}{19}; x, y$ and $z$ must be intege... | 168,187,266 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,748 |
105. Determine the sides of a right triangle if the area and perimeter are known. | 105. Let $a, b, c$ be the sides, $p$ be the perimeter, and $\Delta$ be the area.
Then $p=a+b+c ; c^{2}=a^{2}+b^{2}$ and $a b=2 \Delta$.
From the 2nd and 3rd equalities, $(a+b)^{2}=4 \Delta+c^{2},(p-c)^{2}=4 \Delta+c^{2}$, hence
$$
c=\frac{\left.p^{2}-4\right\rfloor}{2 p}
$$
Knowing $c$, we have $a+b=p-c$ and $a b=2... | \frac{p^{2}-4\Delta}{2p} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,751 |
106. Given the perimeter and area of a rectangle. Find the sides. | 106. If $a$ and $b$ are the sides of a rectangle, the perimeter $2 p$, and the area $S$, then $a b=S$ and $a+b=p$, from which we can find $a$ and $b$ as the roots of a quadratic equation. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,752 |
107. Inscribing a circle in a right-angled triangle.
Note. All three problems are from the treatise "The Beginnings of the Art of Calculation"
From the treatise "Cheu-pei" and the commentary by Confucius on the classic book "I-Ching". | 107. The center of the circle is at the intersection of the bisectors of two angles of a triangle.
The treatise "Cheu-pei" - one of the most ancient (around 1100 years before our era), is written in a conversational form and deals with the properties of a right-angled triangle with sides $3,4,5$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,753 |
113. Find the number which, when increased by 5 or decreased by 11, becomes a perfect square.
## Problems of Apastamba.
From the collection "Sulva-sutra“.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 113. According to the condition, we have: $N+5=u^{2}$ and $N-11=v^{2}$, from which $16=u^{2}-v^{2}=(u+v)(u-v)$, thus, we get two combinations:
I) $\left.\begin{array}{l}u+v=8 \\ u-v=2\end{array}\right\} u=5 ; v=3$ and $N=20$
II) $\left.\begin{array}{r}u+v=16 \\ u-v=1\end{array}\right\} u=\frac{17}{2} ; \quad v=\frac{1... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,758 |
114. Construct a square equal in area to the sum of two given squares.
Construct a square equal in area to the sum of two given squares. | 114. Apastamba's Rule: ${ }^{2}$ To cut off from the larger one $(A B C D)$ a part equal to the side $E C$ of the smaller one - the rectangle ECFG. The diagonal of it $B E$ gives as much $c$ as its dimensions, i.e., the sides of the given squares" (fig. 35). In these words, we see the formulation of the Pythagorean the... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,759 |
115. Construct a square equal in area to the difference of two given squares.
## Examples from Baudhayana.
From the collection, "Sulva-sutra".
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
115. Construct a square equal ... | 115. The second rule of Apastamba: "From the side $E C$ of the subtracted square, subtract the part (rectangle $H B C E$) from the larger and extend the greater side $H E$ of the cut-off part until it meets the other side $B C$. The segment $K C$ is the desired one" (fig. 36). It is clear that $K C^{2}=K E^{2}-E C^{2}=... | KC^{2}=KE^{2}-EC^{2}=AB^{2}-EC^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,760 |
118. (Rule XXX). Two persons have equal capitals, each consisting of a known number of articles of equal value and a known number of coins. But both the number of articles and the sums of money are different for each. What is the value of the article? | 118. Solution. Let the 1st have $a$ items and $m$ coins, and the 2nd have $b$ items and $p$ coins.
If $x$ is the value of an item, then $a x+m=b x+p$, from which
$$
x=\frac{p-m}{a-b}
$$
102 | \frac{p-}{-b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,762 |
119. (Rule XXXI). Two luminaries, being at a given distance ( $d$ ) from each other, move towards each other with given speeds ( $v$ and $v_{1}$ ). Determine the point of their meeting. | 119. This problem is, in essence, the progenitor of our notorious "courier problems." Clearly, before the meeting, one celestial body will travel the distance $x$ in $\frac{x}{v}$ time, while the other will travel the distance $d-x$ in $\frac{d-x}{v_{1}}$, hence $\frac{x}{v}=\frac{d-x}{v_{1}}$, from which $x=\frac{v d}... | \frac{v}{v+v_{1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,763 |
120. (Rule XX). In this rule, Aryabhata verbally gives an expression for the number of terms of an arithmetic progression, which is determined by the formula:
$$
n=\frac{1}{2}\left[1+\frac{-2 a+\sqrt{(r-2 a)^{2}+8 S r}}{r}\right]
$$
where $n$ is the number of terms, $a$ is the first term, $r$ is the common difference... | 120. Since $S=\frac{n(a+l)}{2}=\frac{n[2 a+r(n-1)]}{2}$,
then $2 S=2 a n+r n^{2}-r n$, or $n^{2} r-n(r-2)-2 S=0$.
The solution of this quadratic equation gives:
$$
n=\frac{r-22+\sqrt{(r-2 a)^{2}+8 s r}}{2 r}
$$
If we discard the negative root and simplify, we get:
$$
n=\frac{1}{2}\left[1+\frac{-2 a+\sqrt{(r-2 a)^{... | \frac{1}{2}[1+\frac{-2+\sqrt{(r-2)^{2}+8Sr}}{r}] | Algebra | proof | Yes | Yes | olympiads | false | 37,764 |
121. (Rule XXI). Determine the number of layers in a triangular heap.
## Problem of Paramadisvara.
From the commentary on the treatise of Aryabhata. | 121. This task represents the summation of so-called triangular numbers of the form: $\frac{n(n+1)}{2}$.
Let's write the natural series: $1,2,3,4,5,6,7,8,9 \ldots$
The series of triangular numbers will be:
$$
1,3,6,10,15,21,28,36,45,55 \ldots
$$
It is clear that the sum of any two consecutive numbers always represe... | 220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,765 |
122. Find the number which, when multiplied by 3, then divided by 5, increased by 6, after which the square root is extracted, one is subtracted, and the result is squared, will give 4.
## Problem of Sridhara.
From the treatise "The Essence of Calculation." | 122. The method of inverse actions involves approaching the unknown by performing actions on numbers in the reverse order of those specified in the problem, and, obviously, all actions are replaced by their inverses. Sometimes this technique is called the "method of inversion."
$$
\sqrt{4}=2 ; 2+1=3 ; 3^{2}=9 ; 9-6=3 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,766 |
123. A fifth part of the swarm of bees is on the kadamba flower, one third on the silindhara flowers. The tripled difference of the last two numbers has gone to the kutaja flowers. And there is still one bee flying to and fro, attracted by the wonderful fragrance of jasmine and pandanus. Tell me, charming one, how many... | 123. We form the equation:
$$
\frac{1}{5} x+\frac{1}{3} x+3\left(\frac{1}{3} x-\frac{1}{5} x\right)+1=x
$$
from which \( x=15 \).
Brahmagupta, a famous Indian mathematician (born in 598 AD). His astronomical treatise "Brahma-sphutasiddhanta" consists of 20 books. Book XII contains arithmetic, and Book XVIII contains... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,767 |
124. Two ascetics live at the top of a sheer cliff of height $h$, at a distance from the neighboring village that is $m$ times greater. One ascetic, to reach the village, descends the cliff and then walks a straight path. The other, meanwhile, rises into the air to some height $x$ and from there heads straight for the ... | 124. If the desired height is $x$, then
$$
x+\sqrt{(x+h)^{2}+m^{2} n^{2}}=h+m h
$$
By moving $x$ to the right side and squaring, after simplification we will have:
$$
x(m+2)=h m
$$
from which
$$
x=\frac{h m}{m+2}
$$ | \frac{}{+2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,768 |
125. Knowing the height of the candle, the height of the gnomon (vertical pole) and the distance between their bases, find the length of the shadow cast by the gnomon. | 125. From the similarity of triangles (Fig. 37) we have:
$$
\frac{x}{x+d}=\frac{h}{H}, \text { hence } x=\frac{h d}{H-h}
$$
Fig. 37.
 (the latter triangle is obtained by drawing the diameter $B E$ from vertex $B$ and connecting points $A$ and $E$). From the similarity of these triangles, $\frac{a}{h} = \frac{B E}{c}$, from which $\frac{a c}{h} = B E$. | \frac{}{}=BE | Geometry | proof | Yes | Yes | olympiads | false | 37,771 |
128. In a quadrilateral whose diagonals are mutually perpendicular, the square root of the sum of the squares of two opposite sides is equal to the diameter of the circle circumscribed around this quadrilateral.
Note. Use problem 53. | 128. This problem is easily solved based on Archimedes' lemma (problem 53).
Indeed (fig. 40):
$$
m^{2}+n^{2}+p^{2}+q^{2}=D^{2}
$$
Fig. 39.
^{2}}=\sqrt{2}+\sqrt{3}+\sqrt{5}
\end{aligned}
$$ | \sqrt{2}+\sqrt{3}+\sqrt{5} | Algebra | proof | Yes | Yes | olympiads | false | 37,774 |
133. From a bunch of pure lotus flowers, one third, one fifth, and one sixth parts were respectively offered to the gods: Shiva, Vishnu, and the Sun. One quarter went to Bhavani. The remaining 6 lotuses were given to the deeply revered teacher. Count the total number of flowers quickly. | 133. We form the equation: $\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}+6=x$, from which $x=120$ | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,775 |
134. Someone said to his friend: "Give me 100 rupees, and I will be twice as rich as you," to which the latter replied: "If you give me just 10 rupees, I will be six times richer than you." The question is: how much did each have? | 134. If the first had $x$, and the second, , then
$$
\begin{aligned}
x+100 & =2(y-100) \\
y+10 & =6(x-10)
\end{aligned}
$$
from which
$$
x=40 ; y=170
$$
But Bhaskara's solution is original: let the first have $2 x-100$, and the second $x+100$, which satisfies the first condition. Then:
$$
6(2 x-110)=x+110
$$
from... | 40;170 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,776 |
135. A troop of monkeys was having fun; the square of one eighth of them was frolicking in the forest, the remaining twelve were shouting at the top of a hillock. Tell me: how many monkeys were there in total? | 135. The equation will be $\left(\frac{1}{8} x\right)^{2}+12=x$, from which $x_{1}=48$; $x_{2}=16$
The equation, ready for solving, has the form: $x^{2}-64 x+768=0$, but Bhaskara writes $x^{2}-64 x=-768$ and adds $32^{2}$ to both sides:
$$
\begin{aligned}
& x^{2}-64 x+32^{2}=-768+1024 \\
& (x-32)^{2}=256 \\
& x-32=\s... | 4816 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,777 |
136. The square root of half the bee swarm flew to a jasmine bush. Eight ninths of the swarm stayed at home. One bee flew after the male, worried by his buzzing in a lotus flower, where he had ended up at night, attracted by the pleasant fragrance, and cannot get out as the flower has closed. Tell me the number of bees... | 136. Bhaskara's Solution. Let the number of bees in the swarm be $2 x^{2}$. Then the square root of half this number is $x$, and $\frac{8}{9}$ of the entire swarm equals $\frac{16}{9} x^{2}$, therefore,
$$
\begin{aligned}
2 x^{2} & =x+\frac{16}{9} x^{2}+2 \\
2 x^{2}-9 x & =18 ; x=6 \text { and } 2 x^{2}=72
\end{aligne... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,778 |
137. Find a number that has the property that when it is multiplied by 12, and added to its cube, it equals six times its square, increased by thirty-five (solved elementarily). | 137. Equation:
$$
x^{3}+12 x=6 x^{2}+35
$$
Bhaskara's solutions:
$$
x^{3}+12 x-6 x^{2}=35
$$
Subtract 8 from both sides:
$$
\begin{aligned}
& x^{3}-6 x^{2}+12 x-8=27 \\
& (x-2)^{3}=27 \\
& x-2=3 \text { and } x=5
\end{aligned}
$$
To find the other two roots (which Bhaskara does not provide), we will solve this eq... | x_{1}=5,\,x_{2,3}=\frac{1\\sqrt{-27}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,779 |
138. Solve the equation: $x^{4}-2 x^{2}-400 x=9999$ (solved elementarily). | 138. $\quad x^{4}-2 x^{2}-400 x=9999$
$x^{4}-11 x^{3}+11 x^{3}-121 x^{2}+119 x^{2}-1309 x+909 x-9999=0$ $x^{3}(x-11)+11 x^{2}(x-11)+119 x(x-11)+909(x-11)=0$
$$
(x-11)\left(x^{3}+11 x^{2}+119 x+909\right)=0
$$
from here $x-11=0$ and, therefore, $x_{1}=11$ is a root, which Bhaskara gives. But we also have:
$$
\begin{... | x_1=11,\,x_2=-9,\,x^2+2x+101= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,780 |
139. Find a right triangle in which the hypotenuse is expressed by the same number as the area. | 139. Bhaskara uses the identity:
$$
\left[\left(m^{2}+n^{2}\right) x\right]^{2}=\left[m^{2}-n^{2}, x\right]^{2}+\left(2 m n x\right)^{2}
$$
and takes $\left(m^{2}+n^{2}\right) x$ as the hypotenuse, and $\left(m^{2}-n^{2}\right) x$ and $2 m n x$ as the legs. Then, by the condition $\left(m^{2}+n^{2}\right) x=m n\left(... | \frac{^{2}+n^{2}}{n(^{2}-n^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,781 |
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