problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
316. $\left\{\begin{array}{l}(x+y)\left(x^{2}+y^{2}\right)=539200 \\ (x-y)\left(x^{2}-y^{2}\right)=78400 .\end{array}\right.$ | 316. Instruction. Dividing the 1st equation by the 2nd:
$$
\frac{x^{2}+y^{2}}{(1-y)^{2}}=\frac{337}{49}
$$
from which
$$
674 x y=288 x^{2}+288 y^{2}
$$
or
$$
98 x y=288(x-y)^{2}
$$
$$
\frac{(x-y)^{2}}{x y}=\frac{49}{144}=\frac{x}{y}+\frac{y}{x}-2
$$
$$
\frac{x}{y}+\frac{y}{x}=\frac{337}{144}
$$
let
$$
\frac{x}... | 64;36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,930 |
318. $\left\{\begin{array}{l}(x+y)\left(x^{2}-y^{2}\right)=675 \\ (x-y)\left(x^{2}+y^{2}\right)=351 .\end{array}\right.$
## Problems by M. Stifel. | 318. We provide the genuine solution.
We multiply
$$
\begin{aligned}
(x+y)\left(x^{2}-y^{2}\right)= & x^{3}+x^{2} y-x y^{2}-y^{3}=675 \\
(x-y)\left(x^{2}+y^{2}\right)= & x^{3}-x^{2} y+x y^{2}-y^{3}=351 \\
& \frac{675}{351}=\frac{25}{13}
\end{aligned}
$$
Therefore,
$$
\begin{gathered}
\left(x^{3}+x^{2} y-x y^{2}-y^{... | x=9,y=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,932 |
332. Divide: $12 x^{3}+18 x^{2}+27 x+17$ by $4 x+3$.
Problems of Peter Ramus. | 332. $3 x^{2}+2 \frac{1}{4} x+5 \frac{1}{16}$. Remainder $1 \frac{13}{16}$.
Peter Ramus (Pierre de la Ramée) (1515-1572), professor at the University of Paris, mathematician, and one of the earliest historians of mathematics. Author of the treatise "Scholarum mathematicarum" ("Mathematical Readings").
333: $63 x^{3}-... | 3x^{2}+2\frac{1}{4}x+5\frac{1}{16}. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,945 |
334. Multiply: $(\sqrt{8}-\sqrt{2}) \cdot(\sqrt{7}-\sqrt{3})$.
## Problems by N. Tartaglia from "Generale trattato". | 334. $\sqrt{5} \overline{6}-\sqrt{24}-\sqrt{14}+\sqrt{6}$.
Simplification gives: $\sqrt{14}-\sqrt{6}$.
Niccolò Tartaglia (1499-1557), a famous Italian mathematician, provided a formula for expressing the roots of cubic
. | 339. Adding $3 x^{2}$ to both sides:
$$
\begin{gathered}
16 x^{2}=x^{4}+2 x^{3}+3 x^{2}+2 x+1 \\
16 x^{2}=x^{4}+2 x^{3}+2 x^{2}+x^{2}+2 x+1=\left(x^{2}+x+1\right)^{2} \\
4 x=x^{2}+x+1 \\
x^{2}-3 x+1=0 ; x=\frac{3}{2} \pm \frac{\sqrt{5}}{2}
\end{gathered}
$$
Cardano only has these two roots. The others are found from ... | \frac{3}{2}\\frac{\sqrt{5}}{2},\frac{-5\\sqrt{21}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,951 |
340. Find the positive root of the equation by construction:
$$
x^{2}+6 x=91
$$ | 340. Cardano's Solution: Let the square $F D$ (Fig. 74) be $x^{2}$, hence its side $F H=x$. $D G=D B=3$ (half the coefficient of $x$). Construct the square $A F E C$. The rectangle $A D$ equals the rectangle $D E$, i.e., equals $3 x$. The sum of the square $F D$ and the two rectangles equals $x^{2}+6 x$, which, by the ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,952 |
341. Construct a common tangent to two given circles.
## Benediktis Problems. | 341. Cardano first gave a complete solution to this problem. First case. The tangent is on one side of the line of centers. From the center $O$ of the larger radius circle (Fig. 75), we describe a circle with radius $R-r$. From point $O_{1}$, we draw a tangent $O, A$ to it. We connect point $A$ with $O$ and extend $O A... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,953 |
342. Solve the system:
$$
\left\{\begin{array}{l}
x+y=a \\
y+z=b \\
z+x=c
\end{array}\right.
$$
Interpret the roots of this system geometrically, assuming $a, b, c$ are the sides of a triangle. | 342. Adding the first two equations gives:
$$
x+2 y+z=a+b
$$
but
$$
x+z=c
$$
therefore,
$$
y=\frac{a+b-c}{2}
$$
Similarly:
$$
x=\frac{a+c-b}{2} ; z=\frac{b+c-a}{2}
$$
Benedetti also provides a geometric interpretation: consider a triangle with sides $a, b, c$ (Fig. 77). If a circle is inscribed in it, then at t... | \frac{+-b}{2},\frac{+b-}{2},\frac{b+-}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,954 |
343. Given two lines, find a third such that its product with the sum of itself and the first given line equals the square of the second given line.
## Problems of Christopher Clavius. | 343. If the sought value is $x$, and the given values are $a$ and $b$, then according to the condition $x(x+a)=b^{2}$, or $x^{2}+a x=b^{2}$, i.e., a quadratic equation.
Fig. 78.
From this,... | -\frac{}{2}+\sqrt{\frac{^{2}}{4}+b^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,955 |
344. Solve the equation:
$$
\frac{4 x+18}{x}=\frac{12 x-58}{2}
$$ | 344. $x_{1}=6 ; x_{2}=-\frac{1}{2}$. | x_{1}=6,x_{2}=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,956 |
346. Solve the equation $x^{2}+p x+q=0$ by the substitution $\boldsymbol{x}=\boldsymbol{y}+\boldsymbol{z}$ and defining $z$ (taking advantage of its arbitrariness) so as to obtain a pure quadratic equation (i.e., one not containing the first power of the unknown). | 346. Given the equation:
\[
x^{2}+p x+q=0
\]
Let
\[
x=y+z
\]
Then
\[
x^{2}=y^{2}+2 y z+z^{2}
\]
Therefore, the equation becomes:
\[
\begin{aligned}
& y^{2}+2 y z+z^{2}+p y+p z+q=0 \\
& y^{2}+y(2 z+p)+z^{2}+p z+q=0
\end{aligned}
\]
154
Choose \( z \) such that the coefficient of \( y \) in the first degree disa... | -\frac{p}{2}\\sqrt{\frac{p^{2}}{4}-q} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,958 |
347. If given a cubic equation:
$$
x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x=a b c
$$
then \(a, b\) and \(c\) are the roots of this equation.
Check on the equations:
$$
\begin{aligned}
& \text { 1) } x^{3}-6 x^{2}+11 x-6=0 \\
& \text { 2) } x^{3}-4 x^{2}-4 x+16=0
\end{aligned}
$$ | 347. Indeed, let $x=a$. Then
$$
\begin{aligned}
& a^{3}-(a+b+c) a^{2}+(a b+a c+b c) a=a b c \\
& a^{3}-a^{3}-a^{2} b-a^{2} c+a^{2} b+a^{2} c+a b c=a b c
\end{aligned}
$$
We obtain the identity $a b c=a b c$.
The first equation has roots $1,2,3$
The second: $-2,+2,+4$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,959 | |
348. If $a$ and $b$ are the legs of a right triangle, then $\sqrt{a^{2}+b^{2}}$ represents the hypotenuse, and if $a$ is a leg and $c$ is the hypotenuse, then $\sqrt{c^{2}-a^{2}}$ represents the other leg. Using this, construct: $a \sqrt{5}, a \sqrt{11}, a \sqrt{13}, a \sqrt{17}, a \sqrt{26}, a \sqrt{34}$. | 348. By the Pythagorean theorem (Fig. 81)
$$
c=\sqrt{a^{2}+b^{2}}
$$
and
$$
b=\sqrt{c^{2}-a^{2}}
$$
thus, knowing $a$ and $b$, for example, one can
Fig. 81. not only calculate $c$, bu... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,960 |
349. Solve the equation:
$$
\frac{a x}{b}+\frac{a x-a c}{d}=a
$$ | 349. $x=\frac{b(c+d)}{b+d}$. | \frac{b(+)}{b+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,961 |
350. Solve the equation:
$$
y^{6}+2 b^{3} y^{3}=a^{6}
$$
## Bombelli's Problems. | 350. We assume:
$$
y^{3}=z ; z^{3}+2 b^{3} z-a^{6}=0 ; \quad z=-b^{3} \pm \sqrt{b^{6}+a^{6}}
$$
Knowing $z_{1}$ and $z_{2}$, we obtain two cubic binomial equations:
$$
y^{3}=z_{1} ; y^{3}=z_{2}
$$
Rafael Bombelli (1530-?), an Italian mathematician, author of the treatise on algebra (in Italian) "L'algebra parte mag... | y^{3}=-b^{3}\\sqrt{b^{6}+^{6}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,962 |
351. Show that
$$
\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}=4
$$ | 351. $\sqrt[3]{2+v \overline{-121}}=\sqrt[3]{2+11 i}=\sqrt[3]{8+12 i-6-i}=$
$$
\begin{aligned}
& =\sqrt[3]{2^{3}+3 \cdot 2^{2} i+3 \cdot 2 \cdot i^{2}+i^{3}}=\sqrt[3]{(2+i)^{3}}=2+i \\
& \sqrt[3]{2-\sqrt{-121}}=\sqrt[3]{2-11 i}=\sqrt[3]{8-12 i-6+i}= \\
& =\sqrt[3]{2^{3}-3 \cdot 2^{2} \cdot i+3 \cdot 2 i^{2}-i^{3}}=\sq... | 4 | Algebra | proof | Yes | Yes | olympiads | false | 37,963 |
353. Multiply: $(x+b)(x+c)(x-d)$. | 353. $x^{3}+x^{2}(b+c+d)+x(b c-b d-c d)-b c d$.
354.
$$
\begin{gathered}
x^{3}-4 x^{2}+4 x^{2}-16 x+13 x-52=0 \\
x^{2}(x-4)+4 x(x-4)+13(x-4)=0 \\
(x-4)\left(x^{2}+4 x+13\right)=0 \\
x_{1}=4 ; x_{2,3}=-2 \pm i \sqrt{3}
\end{gathered}
$$
Albert Girard (1590-1632), a Flemish mathematician, author of the book "Invention... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,964 | |
358. Solve the equation:
$$
\begin{aligned}
& 5 x-5 x^{3}+x^{5}=0 . \\
& \text { Problems by S. Stevin. }
\end{aligned}
$$ | 358. $x^{4}-5 x^{2}+5=0$
$$
x= \pm \sqrt{\frac{5 \pm \sqrt{5}}{2}}
$$
Stevin Simon (1548-1620), a Dutch engineer who can justly be considered the inventor of decimal fractions (treatise "La Disme", 1585). | \\sqrt{\frac{5\\sqrt{5}}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,967 |
359. Multiply: $2 x^{3}-4 x^{2}+3 x$ and $2 x^{4}+3 x^{3}$.
Examples of multiplication and division of polynomials. | 359. $4 x^{7}-2 x^{6}-6 x^{5}+9 x^{4}$. 360. 1) $\frac{6 x^{3}}{y^{2} z^{3}}$; 2) $3 x^{2} y z^{2}$. | 4x^{7}-2x^{6}-6x^{5}+9x^{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,968 |
362. $(\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}})^{2}=$ ? | 362. $10+\sqrt{88}$.
Simplification:
$$
10+2 \sqrt{22}
$$ | 10+2\sqrt{22} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,969 |
363. Solve the equation:
$$
x^{9}=3 x^{6}+5 x^{3}
$$ | $$
\begin{gathered}
x^{6}-3 x^{3}-5=0 \\
x^{3}=z ; z^{2}-3 z-5=0 \\
z=\frac{3 \pm \sqrt{29}}{2}
\end{gathered}
$$ | \frac{3\\sqrt{29}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,970 |
364. Solve the equation:
$$
x^{3}=6 x+40
$$
(elementary method).
## Wallis's Problem. | 364.
$$
x^{3}-6 x-40=0
$$
$$
\begin{gathered}
x^{3}-4 x^{2}+4 x^{2}-16 x+10 x-40=0 \\
x^{2}(x-4)+4 x(x-4)+10(x-4)=0 ; x_{1}=4 \\
x^{2}+4 x+10=0 ; x_{2}=-2 \pm i \sqrt{6}
\end{gathered}
$$
John Wallis (1616-1703), a prominent English mathematician; his main work "Arithmetica infinitorum" (1655). In another treatise ... | x_1=4,\quadx_2=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,971 |
365. Show algebraically and geometrically that among rectangles of the same perimeter, the square has the largest area (solved elementarily).
## Problems from "Thaumaturgus mathematicus“. | 365. This problem can be solved in several ways quite elementarily.
1st method. If one side of the rectangle is $x$, then the other is $p-x$, where $p$ is the semi-perimeter.
Area
$$
x(p-x)=S
$$
or
$$
x^{2}-p x+S=0
$$
158
from which
$$
x=\begin{aligned}
& p \\
& 2
\end{aligned} \pm \sqrt{\frac{p^{2}}{4}-S}
$$
O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,972 |
376. Prove that if an integer $N$ is not divisible by a prime number $p$, then $N^{p-1}-1$ is divisible by $p$. | 376. This is the so-called "Fermat's theorem".
Solution The numbers divisible by $N$ are, obviously:
$$
N, 2 N, 3 N, \ldots,(p-1) N
$$
dividing them by $p$ :
$$
\begin{gathered}
N=q_{1} p+r_{1} \\
2 N=q_{2} p+r_{2} \\
3 N=q_{3} p+r_{3} \\
\dot{(} \dot{N}=\dot{q_{p-1}}+\dot{p}+r_{p-1}
\end{gathered}
$$
162
Multipl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,980 |
377. Divide the given number into two parts so that their product is the greatest possible (solve in an elementary way). | 377. Let the number be $N$. If one part is $x$, the other is $N-x$. Their product $x(N-x)$ should be the largest. The question reduces to problem 365. Therefore, $x=\frac{N}{2}$. Verify this with the number 10. | \frac{N}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,981 |
378. On the diameter $AB$ of the semicircle $AMB$, we construct a rectangle, the height of which $AC$ is equal to the side of a square inscribed in the circle. If we connect the vertices $C$ and $D$ with any point $M$ on the semicircle by the lines $CM$ and $DM$, intersecting the diameter at points $E$ and $F$, we will... | 378. Extend chords \( M A \) and \( M B \) (Fig. 83) to meet the extensions of \( C D \) at points \( G \) and \( H \). Since angle \( M \) is a right angle, triangles \( A G C \) and \( B D H \) are similar:
\[
\frac{G C}{B D} = \frac{A C}{D H}
\]
but
\[
B D = A C
\]
 mentions this structure without, however, specifying to whom it belongs.
On an arbitrary straight line, from an arbitrary point \( A \), we lay off the given lines \( A B = a \) and \( A C = b \) (Fig. 84). On the same straight line,
, French mathematician, author of "Cours de Mathématique" in 4 volumes and the collection of problems "Recreations mathémati... | 13;7;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,987 |
385. Three people want to buy a house for 26000 pounds. They agreed that the first would give half, the second one third, and the third one quarter. How much will each give?
## Leibniz's Problems. | 385. Since $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ and this amounts to 26000, then $\frac{1}{12}$ equals 2000, therefore, the first gives 12000, the second 8000, and the third 6000 livres.
Leibniz, Gottfried Wilhelm (1646-1716), a famous lawyer, philosopher, and mathematician. Leibniz, along with Newton, i... | 12000,8000,6000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,988 |
391. Solve the system:
$$
\begin{aligned}
& x+y+z=26 \\
& x-y=4 \\
& x-z=6
\end{aligned}
$$ | 391. MacLaurin's solution. He adds all three equations term by term and gets \(3x = 36\), from which \(x = 12\), then from the 2nd \(y = 8\) and from the 3rd \(z = 6\).
\[
\begin{aligned}
& \text{392.} \quad \frac{\sqrt[3]{20}}{\sqrt[3]{4}-\sqrt[3]{2}}=\frac{\sqrt[3]{20}}{\sqrt[3]{4}-\sqrt[3]{2}} \cdot \frac{\sqrt[3]{... | 2\sqrt[3]{5}+\sqrt[3]{20}+\sqrt[3]{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,991 |
393. Several people had dinner together and had to pay 175 shillings for the bill. It turned out that two of them had no money, so each of the rest had to pay 10 shillings more than their share. How many people had dinner? | 393. Let $x$ be the number of people who had lunch. It is easy to form the equation
$$
\frac{175}{x-2}-\frac{175}{x}=10
$$
from which
$$
x^{2}-2 x=35
$$
Maclaurin would solve it as follows:
$$
\begin{gathered}
x^{2}-2 x+1=36 \\
(x-1)^{2}=36 ; x-1= \pm 6 ; x_{1}=7 \text { or } x_{2}=-5
\end{gathered}
$$
The second... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,992 |
394. Solve the equation: $x^{6}-19 x^{3}=216$.
## Lambert's Problems.
From "Beiträge zur Mathematik ${ }^{4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 394. Let
$$
x^{3}=z, z-19 z-216=0
$$
Maclaurin's solution:
$$
\begin{gathered}
z^{2}-19 z+\frac{361}{4}=216+\frac{361}{4}=\frac{1225}{4} \\
z-\frac{19}{2}= \pm \frac{35}{2} ; z=\frac{19 \pm 35}{2}=27 \text { or }-8
\end{gathered}
$$
therefore, $x(=\sqrt[3]{z})$ equals 3 or -2.
Johann Heinrich Lambert (1728-1777), ... | 3or-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,993 |
396. If an odd square, not divisible by 9, is decreased by one, the result will be divisible by 24.
## Varignon's Problem. | 396. Instruction. Every odd square is either divisible by 9 or gives a remainder of 1, 4, or 7 when divided by 9. If you subtract one, the result is either divisible by 9 or gives a remainder of 3 or 6 when divided by 9. In all these cases, it will be divisible by 3. On the other hand, every odd number has the form $4n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,995 |
397. If two adjacent sides of a parallelogram and the diagonal emanating from the same vertex are projected onto some line, then the projection of the diagonal will be equal to the sum or difference of the projections of the sides.
## Clairaut's Problem. | 397. (From the treatise "Projet d'une nouvelle mécanique"). Project the sides $A B, B C$ and the diagonal $A C$ of the parallelogram $A B C D$ (Fig. 87) onto the line $x y$.
pr. $A B=a b$
pr. $B C=b c$
pr. $A B+$ pr. $B C=a b+b c=a c$;
but
pr. $A C=a c$,
therefore,
pr. $A B+$ pr. $B C=$ pr. $A C$.
Similarly:
$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,996 |
398. On an infinite line connecting two light sources, find the point equally illuminated by both sources.
## Simpson's Problem. | 398. This problem, first published in Clairaut's "Algebra," has been passed on to textbooks under the name of the problem of light sources. Let the distance between the sources $A$ and $B$ be $d$. The distance of the sought point $C$ (equally illuminated) from the source $A$ is $x$. Then the distance to the other sourc... | x_1=\frac{\sqrt{}}{\sqrt{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,997 |
399. Show that if perpendiculars are dropped from an arbitrary point on the circumcircle of a triangle to its sides, then their feet lie on one straight line.
52
## Moivre's Problem | 399. Given a triangle \(ABC\) (Fig. 88). From point \(M\), we drop perpendiculars \(MC_1\), \(MB_1\), and \(MA_1\) to its sides. We connect \(B_1\) with \(A_1\), \(B_1\) with \(C_1\), and point \(M\) with vertices \(A\) and \(C\). Since points \(M, A, B, C\) lie on the same circle, \(\angle MAB = \angle MCB\), from whi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,998 |
402. If through the vertices $A, B, C$ of triangle $A B C$ and an arbitrary point $O$ in its plane, lines are drawn intersecting sides $A B, A C, B C$ at points $C_{1}, B_{1}$, and $A_{1}$ respectively, defining two segments on each of them, then the products of each set of three segments that do not share a common ver... | 402. Since triangles $A O B$ and $A O C$ have the same base $A O$, their areas are proportional to their heights or to the segments $B A_{1}, C A_{1}$; the same applies to triangles $B O C$ and $B O A$ and triangles $C O A$ and $C O B$. Therefore,
$$
\begin{aligned}
& A O B: A O C=B A_{1}: C A_{1} \\
& B O C: B O A=C ... | AC_{1}\cdotBA_{1}\cdotCB_{1}=AB_{1}\cdotCA_{1}\cdotBC_{1} | Geometry | proof | Yes | Yes | olympiads | false | 38,001 |
403. Show that if $a, b, c$ are the sides of a triangle, and $R$ and $r$ are the radii of the circumscribed and inscribed circles, respectively, then
$$
R \cdot r=\frac{a \cdot b \cdot c}{2(a+b+c)}
$$
Problem from "Eléments de géométrie" by Legendre. | 403. The author of this book is the English mathematician Chapple. It was published in 1776.
It is known that $R=\frac{a b c}{4 \Delta}$ and $r=\frac{\Delta}{p}=\frac{2 \Delta}{a+b+c}$, therefore, $R \cdot r=\frac{a b c}{2(a+b+c)}$.
Legendre, Adrien-Marie (1752 - 1833), a famous French mathematician, wrote several se... | R\cdotr=\frac{}{2(+b+)} | Geometry | proof | Yes | Yes | olympiads | false | 38,002 |
404. Determine the area of a quadrilateral inscribed in a circle given the sides $a, b, c, d$.
## Euler's Problems. | 404. Let the diagonals of quadrilateral $ABCD$ (Fig. 89) be $AC = x$; $BD = y$. According to the theorem of Hipparchus, also known as Ptolemy's theorem, we have:
$$
x y = ac + bd
$$
Moreover,
$$
\frac{x}{y} = \frac{ad + bc}{ab + cd}
$$
From this,
$$
\begin{aligned}
& x = \sqrt{\frac{(ac + bl)(ad + bc)}{ab + cd}} \... | \sqrt{(p-)(p-b)(p-)(p-)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,003 |
406. Rationalize $y=\sqrt{(a+b x)(c+d x)}$ by the substitution $y=(a+b x) z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 406. Squaring: $c+d x=(a+b x) z^{2}$, from which
$$
\begin{gathered}
x=\frac{c-a z^{2}}{b z^{2}-d} \\
a+b x=\frac{b c-a d}{b z^{2}-d}
\end{gathered}
$$
therefore, $y=(a+b x) z=\frac{(b c-a d) z}{b z^{2}-d}$. | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,004 | |
407. Rationalize $y=\sqrt{a^{2}+b x+c x^{2}}$ by the substitution $y=a+x z$. | 407. We square:
$$
a^{2}+b x+c x^{2}=a^{2}+2 a x z+x^{2} z^{2}
$$
from which
$$
x=\frac{b-2 a z}{z^{2}-c}
$$
therefore, $a+x z=\frac{b z-a z^{2}-a c}{z^{2}-c}$. | +x\frac{-^{2}-}{z^{2}-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,005 |
409. Several merchants contributed to a common fund 100 times as many rubles as there were merchants. They sent a trusted person to Venice, who received a number of rubles from each hundred rubles, twice the number of merchants. The question is: how many merchants were there, if the trusted person received 2662 rubles? | 409. Let the number of merchants be $x$. Each contributed $100 x$, and the total capital is $100 x^{2}$. The profit on the capital is $2 x^{3}=2662$, from which $x^{3}=1331$, hence $x=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,006 |
411. Determine the rational values for $x$ and $y$ in the equation: $\lambda^{y}=y^{x}$. | 411. Let $y=a x$, i.e., $(a x)^{x}=x^{a x}$ or $a x=x^{a} ; a=x^{a-1}$, from which $x=a^{\frac{1}{a-1}}$, then $y=a^{\frac{a}{a-1}} ;$ if we set $\frac{1}{a-1}=n$, we get $a=\frac{n+1}{n}$, consequently.
$$
x=\left(\frac{n+1}{n}\right)^{n} \text { and } y=\left(\frac{n+1}{n}\right)^{n+1}
$$
12 According to the proble... | (\frac{n+1}{n})^{n}\text{}(\frac{n+1}{n})^{n+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,008 |
412. Show that in a triangle, the distance between the centers of the circumscribed and inscribed circles is the mean proportional between the radius of the former and the excess of it over the diameter of the latter. | 412. Let $ABC$ be a triangle inscribed in a circle with center $O$ and radius $R$ (Fig. 90). Point $I$ is the center of the inscribed circle, with radius $IE = r$. Line $DD_1$ is a diameter perpendicular to $BC$. Draw the diameter $FG$ of the circumscribed circle through the centers $O$ and $I$, connect point $C$ with ... | OI^2=R^2-2Rr | Geometry | proof | Yes | Yes | olympiads | false | 38,009 |
413. Show that if $N_{1}$ and $N_{2}$ are coprime numbers, $k$ is the number of integers less than $N_{2}$ and coprime with it, then $N_{1}^{k}-1$ is divisible by $N_{2}$. | 413. Let the numbers be prime with $N_{2}$ and less than $N_{2}$, namely
$$
\alpha, \beta, \gamma, \ldots, N_{2}-1
$$
Then
$$
\begin{aligned}
& N_{1} \alpha=\text { multiple of } N_{2}+r_{1} \\
& N_{1} \beta=\text { multiple of } N_{2}+r_{2} \\
& N_{1}\left(N_{2}-1\right)=\text { multiple of } N_{2}+r_{k} .
\end{ali... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,010 |
415. If $A$ is the number of edges, $F$ is the number of faces, and $S$ is the number of vertices of a closed polyhedron, then show that it always holds:
$$
A+2=F+S
$$ | 415. Let us remove several adjacent faces of a polyhedron and call such a figure an "open" polyhedron. For it, the following equality holds:
$$
A+1=F+S
$$
Indeed, it is valid for a figure with one face, representing a polygon with $n$ sides. In this case, $F=1$, and the number of vertices $S$ and the number of edges ... | A+2=F+S | Geometry | proof | Yes | Yes | olympiads | false | 38,012 |
416. The product of two numbers, each of which is the sum of four squares, is also equal to the sum of four squares.
Euler's identity. | 416. Let's take the identity:
$$
\frac{1}{a-b}-\frac{1}{a-d}=\left(\frac{1}{a-b}-\frac{1}{a-c}\right)+\left(\frac{1}{a-c}-\frac{1}{a-d}\right)
$$
By performing the subtractions and eliminating the denominators, we get:
$$
(b-d)(a-c)=(b-c)(a-d)+(c-d)(a-b)
$$
Now let:
$$
a=\frac{p+q i}{r+s i} ; b=\frac{p_{1}+q_{1} i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,013 |
418. If the human race after the flood multiplied from 6 people and if we assume that two hundred years later the number of people increased to a million, the question is: by what fraction should the population have increased annually?
b4
## Malfatti's Problem. | 418. This problem is found in "Introductio in Analysin Infinitorum". Euler's solution: "Suppose the population increased annually by $\frac{1}{x}$ of its own part, then after 200 years we get:
$$
\left(\frac{1+x}{x}\right)^{200} \cdot 6=1000000
$$
from which $\frac{1+x}{x}=\left(\frac{1000000}{6}\right)^{\frac{1}{200... | x\approx16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,015 |
419. Given a triangle. It is required to construct three circles such that each of them touches the other two circles and two sides of the triangle.
## Mascheroni Problems. | 419. Let the problem be solved. Let point $O$ be the center of the inscribed circle of radius $r$ (Fig. 91). The radii of circles $K$ and $L$ are denoted by $r_{1}$ and $r_{2}$, respectively. Drop perpendiculars from points $K, O, L$ to side $A B$ and thus find points $D, E, F$. Let $A E=s ; B E=t ; A D=x ; B F=y$. If ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,016 |
420. Using only a compass, it is required to construct a fourth proportional segment to three given segments $a, b, c$. | 420. We describe two concentric circles from the center $O$ and make $A A_{1}=B B_{1}$ (Fig. 92). Then $\triangle A O A_{1}=\triangle B O B_{1}$, therefore $\triangle A O B \backsim \triangle A_{1} O B_{1}$, from which: $\triangle O: O A_{1}=A B: A_{1} B_{1}$
Thus, if we describe concentric circles with radii $a$ and ... | x | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,017 |
421. From an arbitrary point $A$ taken on a given circle $O$, using $A$ as the center, we describe an arc with radius $OA$, which intersects the circle at point $B$. From $B$ with the same radius, we describe an arc that intersects the circle at point $C$. From $C$ with the same radius, we describe an arc that intersec... | 421. Let the radius of the circle be one unit. From the right triangle \(A\) ว \(E\) (Fig. 93), where \(A E = A C = \sqrt{3}\) and \(A O = 1\), we find \(O E = \sqrt{2}\). But \(O M = \frac{\sqrt{3}}{2}\), therefore,
\[
E O - O M = E M = \sqrt{2} - \frac{\sqrt{3}}{2}
\]
Since \(B M = \frac{1}{2}\), from the right tri... | 1.5712\ldots | Geometry | proof | Yes | Yes | olympiads | false | 38,018 |
422. Show that a number of the form $n^{4}+4$ is composite ( $n>1$ ).
## Lagrange's Problems. | 422. The expression $n^{4}+4$ factors into:
$$
\left(n^{2}+2 n+2\right)\left(n^{2}-2 n+2\right)
$$
Lagrange, Joseph Louis (1736-1813), one of the greatest mathematicians of all time, whose name is associated with achievements in the analysis of infinitesimals, higher algebra, the theory of series, number theory, mech... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,019 |
423. If three prime numbers form an arithmetic progression, then the difference of this progression is divisible by 6 (the exception is the case when one of these numbers is 3). | 423. Let the difference of the numbers be $d$, and the numbers themselves $p-d, p$, $p+d$, where $p$ has the form $6 m \pm 1$, and $p-d$ and $p+d$ are odd, while $d$ can only be even and will have the value $6 n \pm 2$ or $6 i$.
1) If $d=6 n+2$ and $p=6 m+1$, the prime numbers will be:
$$
6(m-n)-1,6 m+1,6(m+n)+3
$$
T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,020 |
425. Given the equation: $a x^{4}+b x^{3}+c x^{2}+b x+a=0$. Show that by the substitution $y=x+\frac{1}{x}$, the solution of this equation can be reduced to solving two quadratic equations, the roots of which will be the roots of the given equation.
## Problems of Bezout. | 425. Dividing by $x^{2}$ :
$$
\begin{gathered}
a x^{2}+b x+c+\frac{b}{x}+\frac{a}{x^{2}}=0 \\
a\left(x^{2}+\frac{1}{x^{2}}\right)+b\left(x+\frac{1}{x}\right)+c=0
\end{gathered}
$$
since $y^{2}=x^{2}+\frac{1}{x^{2}}+2$, then
$$
a\left(y^{2}-2\right)+b y+c=0
$$
This equation gives two roots $y_{1}$ and $y_{2}$. Then ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,022 |
426. A lazy worker was told that he would receive 24 su for each day he worked, with the condition of deducting 6 su for each day he skipped. After 30 days, it turned out that he did not receive anything. The question is: how many days did he work? | 426. We form the equation $24 x=6(30-x)$, from which $x=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,023 |
428. Someone bought a horse and after some time sold it for 24 pistoles. In this sale, he lost as many percent as the horse cost him. The question is: for what amount did he buy it? | 428. If the horse is bought for $x$ pistoles, then he loses $\frac{x^{2}}{100}$ on the sale, hence, $x-\frac{x^{2}}{100}=24$, from which $x_{1}=40$, $x_{2}=60$. | x_{1}=40,x_{2}=60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,025 |
429. From a point outside a given circle, draw a secant such that its inner part has a given length.
## Rayno's Problem. | 429. In this circle (Fig. 94), draw a chord $M N$ of a given length. Describe a circle from the center of the given circle, tangent to $M N$, and from point $A$, draw a tangent $A C$ to this circle. This will be the required secant.
This method of solution is constructive.
^{2}+(+2r)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,026 |
430. A man, on his death, left a will according to which the eldest son receives 100 francs and one-tenth of the remainder, the second son - 200 francs and one-tenth of the new remainder, the third - 300 francs and one-tenth of the new remainder, and so on until the last. In this way, the shares of all the sons should ... | 430. Let the entire inheritance be $x$ francs. The first one, according to the condition, receives
$100+\frac{1}{10}(x-100)=\frac{1}{10} x+9$.
The remainder will be $\frac{9}{10} x-90$. The second one receives
$$
200+\frac{1}{10}\left(\frac{9}{10} x-290\right)
$$
But the shares of the first and second are equal, th... | 8100, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,027 |
431. From $a$ cards, $b$ piles are formed, with each pile having $c$ points. The first card of each pile counts as 11 points if it is an ace. 10 points if it is a face card or a ten, and so on. Other cards count as one point. If, after forming the piles, we have $d$ cards left over, we need to determine the sum $\bolds... | 431. \( b c \) is the total sum of points of all piles. If you subtract the cards that count as one point, the remainder will be \( x \), but the number of these cards is \( a-d \) without the number of cards counted as more than one point, i.e., \( a-d-b \). Therefore:
\[
x = b c - (a - d - b) = b(c + 1) + d - a
\] | b(1)+ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,028 |
432. Split the fraction $\frac{n}{d}$ into two fractions, the denominator $d$ of which is the product of two numbers $a$ and $b$ that have no common divisor $\left(\frac{n}{d}=\frac{58}{77}\right)$. | 432. We assume $\frac{n}{d}=\frac{x}{b}+\frac{y}{a}$, from which $a x+b y=n$; for the given case $d=77=11 \cdot 7$, therefore, $a=11$ and $b=7$. We obtain the indeterminate equation $11 x+7 y=58 ; x=7 t-3$; 186
$y=13-11$; under the condition of solving in integers $t=1$, $\boldsymbol{x}=4$ and $y=2$. Therefore, $\frac{... | \frac{58}{77}=\frac{4}{7}+\frac{2}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,029 |
433. Several people have to pay 800 francs for the production of a lawsuit. But three of them have no money, so the rest have to add 60 francs to their share. How many participants are there in the payment of legal costs? | 433. We form the equation $\frac{80}{x+3}=\frac{800}{x}-60$, where $x-$ is the number of participants in the payment. $x^{2}+3 x=40$, from which $x=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,030 |
435. Through the point of intersection of two circles, draw a line so that the part of it enclosed within the circles has a given length.
## 19th century.
## Problems by Lacroix. | 435. On the line of centers $O O_{1}$ (fig. 96) as a diameter, we describe a semicircle and with a radius equal to half the given length, from the center $O_{1}$, we make a mark intersecting the semicircle at point $A$. By connecting points $O_{1}$ and $A$, we draw a line $N P$ through the point $M$ of intersection of ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,032 |
436. What is the cost of $15 \frac{13}{16}$ ells of material, if one ell costs 42 livres 17 sous 11 deniers?
Note. Livre - an ancient French coin; one livre $=20$ sous; 1 sou $=12$ deniers. | 436. Answer: 678 livres 5 sous $9 \frac{11}{16}$ deniers. | 678 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,033 |
438. How many myriameters will a traveler cover in 17 days, spending 10 hours a day on this, if in 29 days he has already traveled 112 myriameters, spending 7 hours each day on the road? | 438. The desired number of miriameters is equal to $\frac{112 \cdot 17 \cdot 10}{29 \cdot 7}$.
Damn. 97. | 97 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,035 |
439. To erect a perpendicular at the end of a straight line, without extending it beyond this end. | 439. From an arbitrary point $O$ outside the line $A B$, we draw a circle (fig. 97) with radius $O A$, and connect the point $D$ where it intersects the line $A B$ with the center $O$, extending it until it meets the circle again at point $C$. We then connect this point with point $A$. The line $A C$ will be perpendicu... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,036 |
440. Describe the circle passing through two given points and touching a given straight line. | 440. Let a straight line $P Q$ and points $A$ and $B$ outside it be given (Fig. 98).
Connecting $A$ and $B$, extend $A B$ until it meets the straight line $P Q$ at point $C$. On $B C$ describe a semicircle with $B C$ as the diameter. At point $A$, erect a perpendicular $A D$ and with radius $D C$ from point $C$, descr... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,037 |
441. To describe a circle passing through a given point and touching a given circle at a given point on it. | 441. Given a circle with point $B$ on it and point $A$ outside the circle (Fig. 99). The center of the desired circle, on one hand, should lie on the perpendicular bisector of $A B$, and on the other hand, — on the extension of the line $O B$, therefore, it lies at their
}{2}
$$
from which
$$
\frac{1}{r_{1}}=\frac{b+c-a}{2 Q}
$$
similarly
finally
$$
\begin{aligned}
& \frac{1}{r_{2}}=\frac{a+c-b}{2 Q} \\
& \frac{1}{r_{3}}=\frac{a+b-c}{2 Q}
\end{aligned}
$$
$$
\frac{1}{r}=\frac{a+b+c}{2 Q}
$$
the sum of (... | proof | Geometry | proof | Yes | Yes | olympiads | false | 38,041 |
445. If (see the notations of the previous problem) $Q$ is the area of the triangle, then show that
$$
Q^{2}=r r_{1} r_{2} r_{3}
$$
Problems from "Cours de Mathématiques" Allaize, Billy, Puissant et Boudrot. | 445. Multiplying equations (1), (2), (3), and (4) from the previous problem:
$$
\frac{1}{r r_{1} r_{2} r_{3}}=\frac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{16 Q^{4}}
$$
$\mathrm{HO}$
$$
Q=\frac{1}{4} \sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}
$$
190
therefore,
$$
\frac{1}{r r_{1} r_{2} r_{3}}=\frac{16 Q^{2}}{16 Q^{4}}=\frac{1}{Q^{... | Q^{2}=rr_{1}r_{2}r_{3} | Geometry | proof | Yes | Yes | olympiads | false | 38,042 |
448. A regiment of cuirassiers buys a certain number of horses for 11250 francs. A regiment of dragoons buys fifteen more horses for 16000 francs. Each dragoon horse is 50 francs cheaper than a cuirassier horse. Determine the number of horses of each type and their cost. | 448. If there are $x$ cuirassier horses, then there are $x+15$ dragoon horses. The cost of the former is $\frac{11250}{x}$, and the cost of the latter is $\frac{16000}{15+x}$. According to the condition,
$$
\frac{11250}{x}=\frac{16000}{x+15}+50
$$
The authors solve it as follows: after simplification,
$$
x^{2}+110 x... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,044 |
449. Extract the square root of the polynomial:
$$
16 a c + 4 a^2 - 12 a b + 9 b^2 - 24 b c + 16 c^2
$$ | 449. Answer: $2 a-3 b+4 c . \quad$ 450. a) $1 ; b) \frac{1}{2}$; c) $\frac{1}{3}$. | 2a-3b+4c | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,045 |
452. The sum of the three sides of a right-angled triangle is 156 meters. The area is 1014 square meters. Find the sides.
## Garnier Problems. | 452. We have:
$$
\begin{gathered}
x+y+z=156 \\
xy=2028 \\
z^{2}=x^{2}+y^{2}
\end{gathered}
$$
from which
$$
(x+y)^{2}=x^{2}+y^{2}+2xy=z^{2}+4056=(156-z)^{2}
$$
or
$$
4056=156^{2}-312z
$$
from which
$$
z=65
$$
therefore,
$$
x+y=91
$$
By the sum and product, we find:
$$
x=39 ; y=52
$$
Garnier, Jean Guillaume ... | 39,52,65 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,047 |
454. Show that the problem of inscribing a regular heptagon in a circle leads to the formulation of a cubic equation.
## Gauss's Problem. | 454. Let in an isosceles triangle $ABC$ (Fig. 103) each of the angles at the base $AC$ be three times the angle $B$ at the vertex. Since $A+B+C=2d$ and $A=C=3B$, then $7B=2d$ and $B=\frac{2}{7}d$. The arc $AC=\frac{4}{7} \cdot \frac{\pi}{2}=\frac{2\pi}{7}$, i.e., $AC$ will be a side of a regular heptagon inscribed in a... | x^3-ax^2-2a^2x+^3=0 | Algebra | proof | Yes | Yes | olympiads | false | 38,049 |
455. Show that the product of two positive integers, each less than a prime number $p$, is not divisible by $p$.
## Cauchy's Problems. | 455. Let $p$ be a prime number, $a$ be any number less than $p$; it is required to show that for $b$, less than $p$, the product $a b$ will not be divisible by $p$. Consider a series of numbers $b, c, d, \ldots$, each less than $p$, and assume that $a b, a c, a d, \ldots$ are divisible by $p$. Let $b$ be the smallest o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,050 |
456. Decompose the fractions $\frac{2}{x^{2}-1}$ and $\frac{2 r}{x^{2}-1}$ into the sum of two fractions of the form $\frac{A}{x-1}+\frac{B}{x+1}$. | 456. 457) We assume
$$
\frac{2}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1}
$$
Bringing to a common denominator, we have:
$$
\begin{gathered}
A x+A+B x-B=2 \\
x(A+B)+A-B=2
\end{gathered}
$$
Since the left and right sides of the equation must be identically equal for any $x$, then
$$
\begin{aligned}
& A+B=0 \\
& A-B=2
\en... | \frac{2}{x^{2}-1}=\frac{1}{x-1}-\frac{1}{x+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,051 |
458. If you multiply two integers, each of which is the sum of two squares, the resulting product will also be a sum of two squares.
## Schreiner's Problem. | 458. Let's take four complexes, pairwise conjugates:
$$
a+b i, a-b i, a_{1}+b_{1} i, a_{1}-b_{1} i
$$
Multiplying each pair of conjugate complexes, we find in the product of all four
$$
\left(a^{2}+b^{2}\right)\left(a_{1}^{2}+b_{1}^{2}\right)
$$
If we multiply the first by the third and the second by the fourth, th... | (^{2}+b^{2})(a_{1}^{2}+b_{1}^{2})=(a_{1}-bb_{1})^{2}+(b_{1}+a_{1}b)^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 38,053 |
459. If the point $E$ of intersection of the diagonals of a trapezoid is connected to the point $F$ of intersection of its non-parallel sides, then the larger base will be bisected by the line $E F$. | 459. From Fig. 104, it is seen that:
$$
A G: G B=D H: H C
$$
and
$$
A G: G B=H C: D H
$$
Multiplying these gives:
$$
A G^{2}: G B^{2}=1
$$
from which
$$
A G=G B
$$
Fig. 104. | AG=GB | Geometry | proof | Yes | Yes | olympiads | false | 38,054 |
461. Among all triangles of the same perimeter and base, the isosceles triangle has the greatest area. | 461. Let $A D B$ be an isosceles triangle (Fig. 106), such that $A D + D B = A C + C B$. Triangles $A C B$ and $A D B$ have a common part $A E B$, therefore, to prove the proposition, it is sufficient to show that triangle $B E D$ is smaller than triangle $A E C$. But if we lay off on $E A$ and $E C$ respectively $E F ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 38,056 |
462. On a straight line, there are three points $G, D, F$, with $D$ lying midway between the other two; using only a ruler, draw a line through an arbitrary point $H$ parallel to the given line.
## Binet's Problem. | 462. Draw the lines $G H$ and $F H$ (Fig. 107). Take an arbitrary point $A$ on $G H$ and draw the lines $A D, A F$. Through the point $C$ of intersection of the lines $F H$ and $A D$, and through $G$, draw the line $G C l$, intersecting $A F$ at point I. The line $H I$ is the required parallel.
Binet, Jean Philippe (1... | HI | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,057 |
463. Solve the system of equations:
$$
\begin{aligned}
& z + a v + a^{2} x + a^{3} = 0 \\
& z + b y + b^{2} x + b^{3} = 0 \\
& z + c y + c^{2} x + c^{3} = 0
\end{aligned}
$$
## Bourdon Problems. | 463. Binet's Solution. These equations express by their composition that $\quad U^{3}+x U^{2}+y U+z$ becomes zero three times when substituting the quantities $a, b$, and $c$ for $U$, hence it is divisible by the product $(U-a)(U-b)(U-c)$, with the quotient being one, because the first term of the divisor is $U^{3}$. W... | \begin{gathered}-(+b+)\\++\\-\end{gathered} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,058 |
464. A number consists of three digits; the sum of these digits is 11; the digit in the units place is twice the digit in the hundreds place. If you add 297 to the desired number, you get a number written with the same digits as the desired number, but in reverse order. What number has these properties? | 464. The question boils down to solving the system:
$$
\left\{\begin{array}{l}
x+y+z=11 \\
z=2 x \\
100 x+10 y+z+297=100 z+10 y+x
\end{array}\right.
$$
$x=3 ; y=2 ; z=6$. The desired number is 326. | 326 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,059 |
465. Find four terms of a geometric proportion given the sum of the extremes $2 S$, the sum of the means $2 S_{1}$, and the sum of their squares $4 C^{2}$.
The problem was proposed by the Paris Academy commission (Cauchy, Liouville, Sturm, and Arago) to the phenomenal calculator Henri Monde. | 465. Let the proportion be $u: x=y: z$, then by the condition
$$
u+z=2 S ; x+y=2 S_{1} ; u^{2}+x^{2}+y^{2}+z^{2}=4 c^{2}
$$
Bourdon's Solution. Let $u z=p$, then combining with the equation $u+z=2 S$, we have $u$ and $z$ as the roots of a quadratic equation expressed in terms of $p$. Similarly, $x y=p$ and $x+y=2 S_{... | S^2+S_1^2-^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,060 |
466. What are two square numbers whose difference is 133?
## Montferrier's Problems. | 466. Henri Monde, a peasant from a village near the city of Tours. In 1840, he was presented to the president of the Paris Academy of Sciences, the famous French mathematician Poncelet, by the owner of a boarding school in this city. With Poncelet's permission, the assembled members of the Academy asked the boy two que... | 66,676,13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,061 |
467. Divide the given number $p$ into two such parts that the sum of the quotients obtained by dividing the first by $m$, and the second by $n$ equals the given number 9. | 467. We form the system:
$$
\begin{gathered}
x+y=p \\
\frac{x}{m}+\frac{y}{n}=q
\end{gathered}
$$
from which
$$
x=\frac{m(n q-p)}{n-m} ; y=\frac{n(p-m q)}{n-m}
$$
For example, if
$$
p=17 ; m=3 ; n=4 ; q=5
$$
then
$$
x=9 ; y=8
$$ | 9,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,062 |
468. Solve the equation:
$$
x^{6}+3 x^{5}+2 x^{4}-2 x^{2}-3 x-1=0
$$ | 468. It is not difficult to see that the equation is satisfied by the value $x_{1}=1$.
Therefore, dividing by $x-1$, we get:
$$
\begin{gathered}
x^{5}+4 x^{4}+6 x^{3}+6 x^{2}+4 x+1=0 \\
\left(x^{5}+1\right)+4 x^{3}(x+1)+6 x^{2}(x+1)=0
\end{gathered}
$$
from which
$$
x_{2}=-1
$$
and
$$
x^{4}+3 x^{3}+3 x^{2}+3 x+1=... | x_{1}=1,x_{2}=-1,x_{3,4}=\frac{-6+2\sqrt{5}\\sqrt{-2-6\sqrt{5}}}{4},x_{5,6}=\frac{-6-2\sqrt{5}\\sqrt{-2+6\sqrt{5}}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,063 |
469. Solve the equation:
$$
x^{4}+5 x^{3}-3 x^{2}-35 x-28=0
$$
(solved by elementary means). | 469. $x^{4}+x^{3}+4 x^{3}+4 x^{2}-7 x^{2}-7 x-28 x-28=0$;
$$
\begin{gathered}
x^{3}(x+1)+4 x^{2}(x+1)-7 x(x+1)-28(x+1)=0 ; x_{1}=-1 \\
x^{3}+4 x^{2}-7 x-28=0 \\
x^{2}(x+4)-7(x+4)=0 \\
x_{2}=-4 ; \quad x^{2}-7=0 ; \quad x_{3,4}= \pm \sqrt{7}
\end{gathered}
$$ | x_1=-1,x_2=-4,x_{3,4}=\\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,064 |
471. Someone has twelve pints of wine and wants to give away half of it, but he does not have a six-pint container. He has two containers, one holds 8 pints and the other 5 pints; the question is: how can he pour six pints into the eight-pint container?
## Lebesgue's Problem. | 471. Regarding this problem, Arago tells that it decided the fate of Poisson, as, by becoming interested in it, he thereby discovered his vocation and devoted his whole life to mathematics. We have already seen a similar problem (336). This one also has 2 solutions:
| 12 | 8 | 5 |
| ---: | ---: | ---: |
| 12 | 0 | 0 |... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,066 |
472. Solve the system in integers:
$$
\begin{aligned}
& 2 x+3 y+7 z=131 \\
& 2 x+3 y+8 z=140
\end{aligned}
$$
60
## Schläfli Problems. | 472. Immediately find $z=9$ and $2 x+3 y=68$, from which in the usual way: $y=0,2,4, \ldots, 22$. In total, there are 12 solutions.
Schlömilch, Oscar (1823--1901), a well-known German mathematician, whose name is associated with the expression for the remainder term of a Taylor series. Author of the very useful course... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,067 |
473. Solve the cubic equation:
$$
x^{3}+a x^{2}+b x+c=0
$$
if its roots form:
1) an arithmetic progression;
2) a geometric progression;
3) a harmonic series. | 473. It is known that between the roots $x_{1}, x_{2}, x_{3}$ and the coefficients $a, b, c$ of the cubic equation $x^{3}+a x^{2}+b x+c=0$, there exists a dependency:
$$
\begin{gathered}
a=-\left(x_{1}+x_{2}+x_{3}\right) \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3} \\
c=-x_{1} x_{2} x_{3}
\end{gathered}
$$
1) If the root... | \begin{aligned}& | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,068 |
474. Show that for $n>2$ the following inequality holds:
$$
\begin{aligned}
& 12 \cdot 2^{2} \cdot 3^{2} \cdot 4^{2} \cdots n^{2}>n^{n}, \\
& \text { Bertrand's Problems. }
\end{aligned}
$$ | 474. Assuming $n > p + 1$ and multiplying both sides of the inequality by $p$, we have:
$$
n p > p^2 + p
$$
or
$$
n p + n > p^2 + p + n
$$
from which we get
$$
(n - p)(p + 1) > n
$$
Setting $p$ sequentially to $0, 1, 2, \ldots, (n-1)$, we obtain:
$$
\begin{gathered}
n \cdot 1 = n, \\
(n-1) \cdot 2 > n, \\
(n-2) ... | 1^2\cdot2^2\cdot3^2\ldotsn^2>n^n | Inequalities | proof | Yes | Yes | olympiads | false | 38,069 |
475. Prove that
$$
\sqrt[3]{20+1+\sqrt{2}}+\sqrt[3]{20-14 \sqrt{ } 2}=4
$$ | 475. Cubing:
$$
\begin{gathered}
20+14 \sqrt{2}+20-14 \sqrt{2}+ \\
+3 \sqrt[3]{20+14 \sqrt{2}} \sqrt[3]{20-14 \sqrt{2}} \cdot 4=64 \\
40+12 \sqrt[3]{40-392}=40+24
\end{gathered}
$$
Another method:
$$
\begin{gathered}
\sqrt[3]{20+14 \sqrt{2}}=\sqrt[3]{8+12+12 \sqrt{2}+2 \sqrt{2}}= \\
=\sqrt[3]{2^{3}+3 \cdot 2} \cdot(... | 4 | Algebra | proof | Yes | Yes | olympiads | false | 38,070 |
476. Prove that
$$
\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}+\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}
$$ | 476. Transform:
$$
\begin{gathered}
\sqrt{2 \pm \sqrt{3}}=\sqrt{\frac{2+\sqrt{2^{2}-3}}{2}} \pm \sqrt{\frac{2-\sqrt{2^{2}-3}}{2}}= \\
=\frac{\sqrt{3} \pm 1}{\sqrt{2}}
\end{gathered}
$$
Based on this, the given expression will take the form:
$$
\frac{(2+\sqrt{3}) \sqrt{2}}{2+(\sqrt{3}+1)}+\frac{(2-\sqrt{3}) \sqrt{2}}... | \sqrt{2} | Algebra | proof | Yes | Yes | olympiads | false | 38,071 |
478. Solve the equation:
$$
\left(a^{\frac{1}{2}}+x^{\frac{1}{2}}\right)^{\frac{1}{3}}=\left(a^{\frac{1}{3}}+x^{\frac{1}{3}}\right)^{\frac{1}{2}}
$$ | 478. Instruction. First raise to the cube, then to the square; then it will be easy to determine $x$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,073 | |
480. What is the base of the logarithm system in which 6 is the logarithm of 729?
## Bakhvits' Problem. | 480. The base is 3.
Bahwitz - a German mathematician. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,075 | |
481. Show that any integer that can be represented as the sum of three squares can also be represented as the sum of the squares of four fractions.
Problem from "Miscellany of Mathematical Problems". | 481. Let $N=A^{2}+B^{2}+C^{2}$. Multiply $N$ by the sum of three other squares:
$$
\begin{aligned}
& \quad\left(A^{2}+B^{2}+C^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)= \\
& =(A a+B b+C c)^{2}+(A b-B a)^{2}+(B c-C b)^{2}+(C a-A c)^{2} \\
& \text { from which }
\end{aligned}
$$
$$
\begin{gathered}
A^{2}+B^{2}+C^{2}=\fr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,076 |
482. Three peasants, John, Peter, and Alexis, came to the market with their wives, Mary, Kitty, and Jenny. To find out who is married to whom, if it is known that each of these six individuals paid as many pence for each item bought as the number of items they bought. Each man spent 63 pence more than his wife. In addi... | 482. Let each man buy $x$ items, so he paid $x^{2}$ pence for them. His wife bought $y$ items and paid $y^{2}$ pence. According to the condition
$$
x^{2}-y^{2}=(x+y)(x-y)=63
$$
Since $x$ and $y$ are integers, the following combinations are possible:
$$
63 \cdot 1=21 \cdot 3=9 \cdot 7
$$
We have:
1) $x+y=63$
2) $x+... | John'wifewasJenny,Peter'wifewasKitty,Alexis'wifewasMary | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,077 |
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