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141. If $D$ is the diameter of a sphere, then $\frac{D^{3}}{2}+\frac{1}{21} \cdot \frac{D^{3}}{2}$ is the measure of the volume of the sphere. Determine the value of $\pi$ for which this statement is true. | 141. We have $\frac{D^{3}}{2}+\frac{1}{21} \cdot \frac{D^{3}}{2}=\frac{\pi D^{3}}{6}$, or $\frac{\pi}{6}=\frac{1}{2}+\frac{1}{21} \cdot \frac{1}{2}$, from which
$$
\pi=3+\frac{1}{7}=\frac{22}{7}
$$ | \frac{22}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,783 |
144. Solve the equation in rational numbers:
$$
a x+b y+c=x y
$$
## A modern Indian problem. | 144. The equation $a x+b y+c=x y$ can be transformed as: $c=x y-a x-b y$.
Adding $a b$ to both sides:
$$
a b+c=x y-a x-b y+a b
$$
or
$$
a b+c=y(x-b)-a(x-b)=(x-b)(y-a)
$$
Bhaskara concludes that in the case of rational $x$ and $y$, one should take:
$$
\begin{gathered}
x=b+n \\
y=a+\frac{a b+c}{n}
\end{gathered}
$$ | b+n,\quad+\frac{+}{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,786 |
145. Three friends had a monkey. They bought a certain number of mango fruits and hid them. At night, one of the friends wanted to indulge, went to the storeroom, and wanted to take a third of the fruits, but it turned out that one fruit was left over, and he gave it to the monkey, and ate his share. After some time, t... | 145. This problem was proposed by an Indian professor to Delbo.
Solution: Let $x$ be the number of fruits each received in the morning. Then $3x + 1$ is what remained in the morning. The third one took $\frac{1}{2}(3x + 1)$, so he found $\frac{3}{2}(3x + 1) + 1$. The second one took $\frac{1}{2}\left[\frac{3}{2}(3x + ... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,787 |
147. In this treatise, it is stated that the perimeter of a square circumscribed around a circle is a quarter larger than the circumference of that circle. Determine the approximation of $\pi$ used in this case by the Jews.
## Problem from the treatise "Mena hot."
Talmud. | 147. The tractate "Ohalot" (Tents) contains a number of geometric questions. In this problem, if we call the side of the square $S$, and the circumference of the circle $C$, we have $4 S - C = S$, i.e., $C = 3 S$; since $S$ is equal to the diameter of the inscribed circle, then $2 \pi R = 6 R$, hence, $\pi = 3$. | \pi=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,788 |
151. Solve the equation $x^{2}+4 x=77$. | 151. $x_{1}=7 ; x_{2}=-11$ (solved geometrically by Al-Khwarizmi).
152 The general solution: $z^{2}$ $-14 z+48=0 ;$ from which $z_{1}=8 ; z_{2}=6$. Al-Khwarizmi constructs a rectangle $A D B C=$ $=48$ and a line $C F=14$ (Fig. 43). Then he constructs a square $K C F H$ on the latter. It is clear that $G H E B=48$, and... | x_{1}=7;x_{2}=-11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,792 |
153. $\left\{\begin{array}{c}x^{2}+y^{2}=a^{2} \\ x y=b\end{array}\right.$ and $\left\{\begin{array}{c}x^{2}+y^{2}=13 \\ x y=6 .\end{array}\right.$ | 153. Doubling the second equation term by term, we have:
$$
(x+y)^{2}=a^{2}+2 b,(x-y)^{2}=a^{2}-2 b
$$
therefore,
$$
x+y=\sqrt{a^{2}+2 b} \quad \text { and } \quad x-y=\sqrt{a^{2}-2 b}
$$
from here we find $x$ and $y$. For the numerical equation $x=3 ; y=2$. | 3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,793 |
154. $\left\{\begin{aligned} x^{2}+y^{2} & =a^{2} \\ x-y & =b\end{aligned}\right.$ | 154. The second equation is squared:
$$
x^{2}+y^{2}-2 x y=b^{2}
$$
therefore,
$$
a^{2}-b^{2}=2 x y ; 2 a^{2}-b^{2}=(x+y)^{2} ; \quad x+y=\sqrt{2 a^{2}-b^{2}}
$$
from here we find $x$ and $y$. | x+y=\sqrt{2a^2-b^2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,794 |
155. $\left\{\begin{array}{c}x+y=a \\ x^{2}-y^{2}=b^{2}\end{array}\right.$
## Problems of Aben-Dreata.
Joannes Hispalensis. | 155. From the second equation $(x+y) \quad(x-y)=b^{2}$ we get $x-y=\frac{b^{2}}{a} ;$ the rest is obvious.
Aben-Dreant, a Spanish rabbi of the 12th century, who converted to Christianity under the name Joannes Hispalensis. Author of the treatise "Liber Algorismi," which represents an extraction from the "Algebra" of t... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,795 | |
156. What number, when added to nine, gives its own square root?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is the final version:
156. What number, when added to nine, gives its own square root? | 156. $x^{2}+9=6 x ; x=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
156. $x^{2}+9=6 x ; x=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,796 |
158. What number is equal to its tripled root, added to four?
## Problem of Levi ben Gerson. | 158. $x^{2}=3 x+4 ; x_{1}=4 ; x_{2}=-1$.
Levi ben Gerson (1288--1344) from Avignon. | x_1=4,x_2=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,798 |
161. In an isosceles triangle with a side length of 10 and a base of 12, inscribe a square. | 161. Height $B G=\sqrt{B C^{2}-G C^{2}}=$ $=\sqrt{100-36}=8 ; \triangle B D E \subset \triangle A B C$ (fig. 44 ).
If $D E=\boldsymbol{x}$, then $x:(8-x)=12: 8$; $20 x=96 ; x=4 \frac{4}{5}$.
114 | 4\frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,801 |
162. Determine the segments into which the base $A C$ is divided by the perpendicular dropped from the opposite vertex $B$ of triangle $A B C$, if the sides of the triangle are given: $A B=15 ; B C=13$; $A C=14$. | 162. Let $O C=x$ (Fig. 45), $O B^{2}=169-x^{2}$;
$$
A O^{2}=(14-x)^{2}=196-28 x+x^{2}
$$
$O B^{2}=225-A O^{2}=225-196+28 x-x^{2}=29+28 x-x^{2}$, therefore, $169-x^{2}=29+28 x-x^{2} ; \quad 140=28 x ; \quad x=5$.
$A O=9 ; \quad O B=\sqrt{169-25}=12$.
. We complete the trapezoid to form triangle $DSC$. The height $SE=10$ (from the similarity of triangles $ABS$ and $DSC$). The volume of the truncated pyramid can be found as the difference between the volume of the comple... | 93\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,803 |
164. Multiply: $\left(3 x^{2}+2 x+4\right) \cdot\left(2 x^{2}+3 x+5\right)$. | 164. Answer: $6 x^{4}+13 x^{3}+29 x^{2}+22 x+20$. | 6x^{4}+13x^{3}+29x^{2}+22x+20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,804 |
165. Al-Karhi's rule for approximating the square root. If $a^{2}$ is the greatest square contained in a given number $N$, and $r$ is the remainder, then
$$
\sqrt{N}=\sqrt{a^{2}+r}=a+\frac{r}{2 a+1}, \text{ if } r<2 a+1
$$
Explain how Al-Karhi might have derived this rule. Evaluate the error by calculating $\sqrt{415... | 165. If \(a^{2}\) is the greatest square contained in a given number \(N\), then let \(\sqrt{N}=\sqrt{a^{2}+r}=a+x^{2}\), where \(x<1\). Then \(a^{2}+r=a^{2}+2 a x+x^{2}\), or \(r=2 a x+x^{2}=x(2 a+x)\), from which
\[
x=\frac{r}{2 a+x}
\]
$8^{*}$
Al-Karhi replaces \(x\) in the denominator with 1 and obtains:
\[
\be... | 20.366 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,805 |
168. Find the number which, when multiplied by $3+\sqrt{5}$, gives one. | 168. Al-Karaji believes:
\[
\begin{gathered}
x(3+\sqrt{5})=1 ; 3 x+x \sqrt{5}=1 ; 3 x+\sqrt{5 x^{2}}=1 \\
\sqrt{5 x^{2}}=1-3 x ; 5 x^{2}=(1-3 x)^{2}=1-6 x+9 x^{2} \\
4 x^{2}-6 x+1=0 ; x=\frac{3 \pm \sqrt{5}}{4}
\end{gathered}
\]
Al-Karaji takes the root \( x_{2}=\frac{3-\sqrt{5}}{4} \) as the desired one \((x_{1}>1)\... | \frac{3-\sqrt{5}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,807 |
169. $x^{2}+10 x=39$.
169. $x^{2}+10 x=39$.
(Note: The equation is the same in both languages, so the translation is identical to the original text.) | 169. Al-Karaji is looking for a number which, when added to $x^{2}+10 x$, would make a complete square. Such a number is 25. Then
$x^{2}+10 x+25=(x+5)^{2}=39+25=64 ; x+5=8$ and $x=3$.
He calls this method the "method of solving in the manner of Diophantus". | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,808 |
172. Find the last term and the sum of 20 terms of the progression:
$$
3,7,11,15 \ldots
$$ | 172. $a_{20}=19 \cdot 4+3=79 ; S_{20}=\frac{(3+79) \cdot 20}{2}=820$. | 820 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,810 |
173. Find the sum of the first ten numbers in the natural number series. | 173. $S_{10}=\frac{(1+10) \cdot 10}{2}=55$.
The above text is translated into English as follows, retaining the original text's line breaks and format:
173. $S_{10}=\frac{(1+10) \cdot 10}{2}=55$. | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,811 |
176. Find a square number that is equal to the sum of the squares of two numbers. | 176. We need to solve the equation: $x^{2}+y^{2}=z^{2}$. Al-Karaji assumes: $y=x+1 ; z=n x-1$.
$x^{2}+x^{2}+2 x+1=n^{2} x^{2}-2 n x+1 ; x\left(n^{2}-2\right)=2(n+1)$ and $x=\frac{2(n+1)}{n^{2}-2} ; z=\frac{(n+1)^{2}+1}{n^{2}-2} ; y=\frac{n(n+2)}{n^{2}-2}$. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,813 | |
178. If $a=m n$, then show that the numbers $\left(\frac{m-n}{2}\right)^{2}+a$ and $\left(\frac{m+n}{2}\right)^{2}-a$ are always squares. | 178.
$$
\sqrt{\left(\frac{m+n}{2}\right)^{2}-a}=\sqrt{\frac{m^{2}+n^{2}+2 m n}{4}-a}=
$$
$=\sqrt{\frac{m^{2}+n^{2}-2 m n}{4}}=\frac{m-n}{2}$.
Similarly, $\sqrt{\left(\frac{m-n}{2}\right)^{2}+a}=\frac{m+n}{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,815 |
179. On two opposite banks of the river, there stands a palm tree on each. The height of one is 20 cubits, and the other is 30. The width of the river is 50 cubits. At the top of each palm tree, there sits a bird. Both birds see a fish in the river and simultaneously fly in a straight line to it, reaching the water sur... | 179. Al-Karhi's Solution: Let $x$ be the distance from the meeting point to the roots of the taller palm, square it, you get $x^{2}$. Add to this 900, i.e., the square of the height of the taller palm, and set this sum equal to the square of $50-x$, i.e., $2500+x^{2}-100 x$, increased by the square of the height of the... | 20, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,816 |
180. Find the sides of a rectangle whose area is numerically equal to the sum of its perimeter and diagonal, and the base is three times the height.
32 | 180. If the height is $x$, then the base is $3 x$, the area is $3 x^{2}$. The sum of the perimeter and the diagonal is $8 x+x \sqrt{10}$. [By condition
$$
3 x^{2}=8 x+x \sqrt{10} ; x=2 \frac{2}{3}+\frac{1}{3} \sqrt{10}
$$
The base $3 x=8+\sqrt{10}$. | 8+\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,817 |
181. Find the area of a rectangle where the base is twice the height, and the area is numerically equal to the perimeter. | 181. If the width is $x$, then the length is $2 x$. Thus, the area is $2 x^{2}$, and the perimeter is $6 x$. According to the condition, $2 x^{2}=6 x$, therefore, $x=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,818 |
182. Find the diameter of a circle with an area of 100. | 182. Al-Karhi's Solution: Let the diameter be $x$, the square of it $x^{2}$. Subtract $\frac{1}{7}+\frac{1}{7} \cdot \frac{1}{2}$ of the square of the diameter. The remainder
$$
\frac{5}{7} x^{2}+\frac{1}{7} \cdot \frac{1}{2} x^{2}=100 ; x^{2}=127 \frac{3}{11}=\frac{1400}{11}
$$
Common solution: $\frac{\pi d^{2}}{4}=... | \frac{1400}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,819 |
183. To determine the side of a regular dodecagon inscribed in a circle.
Al-Karhi gives a verbal expression, according to which:
$$
a_{12}^{2}=\left[\frac{d}{2}-\sqrt{\left(\frac{d}{2}\right)^{2}-\left(\frac{d}{4}\right)^{2}}\right]^{2}+\left(\frac{d}{4}\right)^{2}
$$
Verify the validity of this expression.
Note. P... | 183. Examining the drawing, the reader will easily find that
\[
\begin{gathered}
a_{12}^{2}=\left[\frac{d}{2}-\sqrt{\frac{3}{16} a^{2}}\right]^{2}+\left(\frac{d}{4}\right)^{2}=\frac{d^{2}}{4}+{ }_{16}^{3} d^{2}-d_{4}^{2} \sqrt{3}+ \\
+\frac{d^{2}}{16}=d^{2}\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)=\frac{d^{2}}{4}(2-... | a_{12}=r\sqrt{2-\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,820 |
184. If the numbers: $p=3 \cdot 2^{n}-1 ; q=3 \cdot 2^{n-1}-1$ and $r=$ $=9 \cdot 2^{2 n-1}-1$ are prime, show that the numbers $A$ and $B$ will be amicable if $A=2^{n} \cdot p \cdot q$, and $B=2^{n} \cdot r$. Apply to the special case $n=2$.
Note. Amicable numbers are a pair of numbers with the property that the firs... | 184. The divisors of the number $A$ will be $1,2,2^{2}, \ldots, 2^{n}$, the sum of which is obviously $2^{n+1}-1$, then $p\left(2^{n+1}-1\right)$ will be the sum of the divisors containing $p ; q\left(2^{n+1}-1\right)$ will be the sum of the divisors containing $q$, finally, $p q\left(2^{n}-1\right)$ - the sum of the d... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,821 |
185. Divide 10 into two parts such that the sum, composed of the sum of the squares of these parts and the quotient of dividing the larger part by the smaller part, equals 72 (solved by elementary means).
Translate the above text into English, please retain the original text's line breaks and format, and output the tr... | 185. The question boils down to solving the equation:
$$
(10-x)^{2}+x^{2}+\frac{10-x}{x}=72
$$
or
$$
2 x^{3}+27 x+10=20 x^{2}
$$
This equation was first solved by the Arab mathematician Abu al-Jud (11th century). Al-Kuhi himself could not solve it, although it is solved quite simply:
$$
\begin{gathered}
2 x^{3}-20... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,822 | |
186. Construct a circle that touches two given straight lines, so that the center lies on a given straight line.
## Problems of Abul Wafa.
From the "Treatise on Geometric Constructions". | 186. Obviously, the center of the sought circle lies at the intersection of the given line with the bisector of the angle formed by the given lines. If the lines are parallel, the center is located at the midpoint of the segment cut off by the parallels from the given line.
Abu'l-Wafa, a remarkable Arab mathematician ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,823 |
191. If a number, when divided by 9, gives a remainder of 1 or 8, then the square of this number, when divided by 9, gives a remainder of 1. | 191. Let $N=9 k+1$, then
$$
N^{2}=81 k^{2}+18 k+1=\text { sq. } \cdot 9+1
$$
Let $N_{1}=9 k+8$, then
$$
N_{1}^{2}=81 k^{2}+144 k+64=9\left(9 k^{2}+16 k+7\right)+1
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,828 |
192. If a number, when divided by 9, gives a remainder of 2 or 7, then the square of the number, when divided by 9, gives a remainder of 4. | 192. Let $N=9 k+2, \quad$ then $N^{2}=81 k^{2}+18 k+4=$ $=$ multiple of 9 +4 ; let $N_{1}^{2}=9 k+7$, then $N_{1}^{2}=81 k^{2}+$ $+126 k+49=9\left(9 k^{2}+14 k+5\right)+4$. | 4 | Number Theory | proof | Yes | Yes | olympiads | false | 37,829 |
193. If a number, when divided by 9, gives a remainder of 1, 4, or 7, then the cube of the number, when divided by 9, gives a remainder of 1. | 193. Let $N=9 k+1$, then $N^{3}=(9 k)^{3}+3 \cdot(9 k)^{2}+3 \cdot 9 k+$ $+1=$ multiple of 9 +1 ; let $N_{1}=9 k+4$, then $N_{1}^{3}=(9 k)^{3}+$ $+3 \cdot(9 k)^{2} \cdot 4+3 \cdot 9 k \cdot 16+64=$ multiple of $9 \cdot$ +1 ; let $N_{2}=9 R+7$, then $N_{2}^{3}=(9 k)^{3}+3 \cdot(9 k)^{2} \cdot 7+3 \cdot 9 k \cdot 49+343=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,830 |
196. Given a square $A B C D$ with side $A B=10$, divided in half by line $Z E$, parallel to $A B$. Cut off a triangle $K T Z$ by a secant passing through vertex $A$, such that the area of the triangle is in the given ratio to the area of the given square. (Ratio $2: 25$).
## Problem of Hassan ibn al-Haytham. | 196. Let $K Z=X ; T Z=y ; x: y=B K: A B=x+5: 10$; $10 x=y x+5 y ; \quad x y=K Z \cdot T Z=16 \quad$ (since $\frac{x y}{2}: 100=2: 25$ ); $5 y+16=10 x, y=2 x-3 \frac{1}{5} ; x y=2 x^{2}-3 \frac{1}{5} x=16$. $x^{2}=8+1 \frac{3}{5} x ; x^{2}-\frac{8}{5} x-8=0 ; x=\frac{4}{5} \pm \sqrt{\frac{10}{25}+8}=K Z$.
![](https://c... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,833 | |
197. Given an irregular quadrilateral. Determine whether a circle can be circumscribed around it without measuring the angles.
## Problems by Al-Kalasadi.
From the treatise "Unveiling the Secrets of the Science of Gobar"
| 197. Solution. Given a quadrilateral $A B D C$. Extend $D C$ to point $G$. Lay off $B K = C H$ and $B I = C T$. If it turns out that $T H = K I$, then the quadrilateral can be inscribed in a circle (for then $\Varangle A B D + \Varangle A C D = 186^{\circ}$). Otherwise, it cannot be inscribed. (This problem is placed i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,834 |
199. In the expression $\frac{a}{m+\sqrt{n}}$, rationalize the denominator. | 199. $\frac{a(m-\sqrt{n})}{(m+\sqrt{n})(m-\sqrt{n})}=\frac{a(m-\sqrt{n})}{m^{2}-n}$. | \frac{(-\sqrt{n})}{^{2}-n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,836 |
200. Al-Kalsadi's Rule. If $\boldsymbol{a}^{2}$ is the greatest square contained in a number, and $r$ is the remainder, then
$$
\sqrt{a^{2}+r}=a+\frac{r}{2 a}
$$
34
if $r \gtrless a$. If $r > a$, a more accurate value is given:
$$
\sqrt{a^{2}+r}=a+\frac{r+1}{2 a+2}
$$
Determine the path the Arab mathematician follo... | 200. Let $n=a^{2}+r$. If $r \leqslant a$, the author considers $\sqrt{a^{2}+r}$ to be equal to $a+\frac{r}{2 a}$. If $r>a$, he gives a more accurate value:
$$
\sqrt{a^{2}+r}=a+\frac{r+1}{2 a+2}
$$
i.e., Al-Kalasadi knows that when $r>a$
$$
\sqrt{a^{2}+r}2 a$, i.e., at least $r=2 a+1$, then $n=a^{2}+r=a^{2}+2 a+1=$ $... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,837 |
202. $x^{2}+10 x=39$ (solve geometrically). | 202. Omar Khayyam's solution: "The square and ten roots are equal to thirty-nine. Multiply half the roots by itself. Add this product to the number (i.e., to 39), subtract half the number of roots from the square root. The remainder will be equal to the root of the square:
$$
\left(x=\sqrt{\left(\frac{10}{2}\right)^{2... | \sqrt{25+39}-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,839 |
203. $\frac{1}{x^{2}}+2 \frac{1}{x}=1 \frac{1}{4}$.
## Problems of Begh-Edin.
From the treatise "The Essence of the Art of Calculation". | 203. Omar's Solution: $z=\frac{1}{x} ; \quad z^{2}+2 z=\frac{5}{4} ;$
$$
\begin{aligned}
& z^{2}+2 z+1=\frac{9}{4} ;(z+1)^{2}=\frac{9}{4} ; z+1=\frac{3}{2} ; z=\frac{1}{2} \\
& z^{2}=\frac{1}{4}, \text { therefore, } x^{2}=4 \text { and } x=2
\end{aligned}
$$
Bega Ed-Din (16th century), a Persian mathematician, autho... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,840 |
204. Divide the number 10 into 2 parts, the difference of which is 5. | 204. Let the smaller part be $x$, the larger $5+x$. Then, by the condition $2 x+5=10 ; 2 x=5$ and $x=2 \frac{1}{2}$. | 2\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,841 |
205. It is required to find a number which, when multiplied by itself, added to two, then doubled, added to three again, divided by 5, and finally multiplied by 10, results in 50. | 205. $\frac{50}{10}=5 ; 5 \cdot 5=25 ; 25-3=22 ; 11-2=9 ; \sqrt{9}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,842 |
206. Find the number which, when increased by two thirds of itself and one unit, gives 10.
Note. The last two problems are solved by the method of inversion or, alternatively, the method of reverse actions. | 206. $x+\frac{2}{3} x+1=10 ; \frac{5}{3} x=9 ; x=5 \frac{2}{5}$.
But Bega Ed-Din solves it by the "rule of the balance" (regula falsi), found in the works of ibn-Ezra, ibn-Albanna, and Al-Kalsadi. The essence of the method is as follows: let the equation be $a x+b=0$. Substitute arbitrary values $z_{1}$ and $z_{2}$ fo... | \frac{27}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,843 |
207. Zaimu is promised a reward in the form of the larger of two parts, which add up to 20, and the product of these parts is 96. How great is the reward? | 207. The ordinary solution: Let the larger part be $x$. Then the smaller one is $20-x$. According to the condition, $x(20-x)=96$, or $x^{2}-20 x+96=0 ; x=10 \pm \sqrt{100-96}=10 \pm 2 ; x_{1}=12 ; x_{2}=8$. But Al-Baghdadi's solution is simpler: put one number as $10+x$, then the other is $10-x$. Their product is $100-... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,844 |
208. Given: height $h$ and radii $r$ and $R$ of the upper and lower bases of a frustum of a cone, find the height of the corresponding complete cone. | 208. By completing the given truncated cone (Fig. 52), we have $x: r = x+h: R$, from which $x = \frac{r h}{R-r}$, therefore, $x+h = \frac{R h}{R-r}$. This expression was known even to Al-Karaji.
![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-126.jpg?height=417&width=386&top_left_y=1616&top_left_x... | \frac{R}{R-r} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,845 |
212. Divide one hundred measures of wheat among one hundred persons so that each man receives 3, each woman 2, and each child $\frac{1}{2}$ measure. How many men, women, and children? | 212. We have the system: $3 x+2 y+\frac{1}{2} z=100$ and $x+y+z=100$. Eliminating $2:$
$$
200-6 x-4 y=100-x-y ; 100=5 x+3 y
$$
or $20=x+\frac{3 y}{5}$. Since $y$ is an integer, it must be a multiple of 5, for example $5 u$, then $y=5 u$ and $20=x+3 u$, from which $x=20-3 u$. It is clear that $u$ can take the values $... | 11,15,74 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,847 |
213. Three heirs received 21 barrels: 7 - full, 7 - filled to half, and 7 - empty. Divide the inheritance so that each heir receives as much wine as barrels.
Problem from the manuscript collection "Problems for the Sharpening of the Mind." | 213. Since each should have $3 \frac{1}{2}$ barrels of wine, it is clear that there are two solutions:
| | $\\| \mathrm{y} 1-$ | $\sqrt{2-1}$ | $3-$ | $\sqrt{1-1}$ | $2-1$ | $3-\mathrm{r} 0$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Full barrels . . . . . ” half-full barrels „empty barrels . . .... | 3fullbarrels,1half-fullbarrel,3emptybarrels\quador\quad1\text | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,848 |
214. A dog is chasing a rabbit that is 150 feet ahead of it. It makes a leap of 9 feet every time the rabbit jumps 7 feet. How many jumps must the dog make to catch the rabbit?
## Herbert's Problem. | 214. With each jump, the dog gains 2 feet, therefore, the entire distance of 150 feet will be covered in 75 jumps.
Herbert (died 1003), bishop of Reims, later Pope Sylvester II, author of two arithmetical treatises ("Calculations with the Abacus" and "On the Division of Numbers"). | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,849 |
215. Given the area of a right-angled triangle and its hypotenuse, find the legs.
Problems by Leonardo Fibonacci from "Liber Abaci".
Problem "De duobus hominibus" (About two men). | 215. If $x$ and $y$ are the legs, $a$ is the hypotenuse, and 1 is the area, we have: $x y=2 \Delta ; x^{2}+y^{2}=a^{2}$.
Herbert's Solution:
$$
\begin{gathered}
2 x y=4 \Delta \\
x^{2}+y^{2}=a^{2}
\end{gathered}
$$
Adding and subtracting give:
$$
\begin{aligned}
& (x+y)^{2}=a^{2}+4 \Delta ; x+y=\sqrt{a^{2}+4 \Delta... | \begin{aligned}&\frac{\sqrt{^{2}+4\Delta}+\sqrt{^{2}-4\Delta}}{2}\\&\frac{\sqrt{^{2}+4\Delta}-\sqrt{^{2}-4\Delta}}{2}\end{aligned} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,850 |
216. One says to the other: "Give me 7 denarii, and I will be 5 times richer than you." And the other says: "Give me 5 denarii, and I will be 7 times richer than you": How much does each have? | 216. The question boils down to solving the system:
$$
x+7=5(y-7) ; y+5=7(x-5)
$$
from which $x=7 \frac{2}{17} ; y=9 \frac{14}{17}$. | 7\frac{2}{17};9\frac{14}{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,851 |
218. Someone bought 30 birds for 30 coins, among these birds, for every 3 sparrows, 1 coin was paid, for every 2 doves, also 1 coin, and finally, for each pigeon - 2 coins. How many birds of each kind were there?
## Problem of John of Palermo. | 218. We have the system: $x+y+z=30 ; \frac{1}{3} x+\frac{1}{2} y+2 z=30$, where $x$ is the number of sparrows, $y$ is the number of magpies, and $z$ is the number of pigeons. Eliminating $z$: $10 x+9 y=180$ and $y=20-\frac{10}{9} x$. For $x$ the only value is 9. Consequently, $y=10$ and $z=11$. | 9,10,11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,853 |
219. Three men have a certain sum of money, the part of the first being half, the second a third, and the third a sixth of the entire sum.
Wishing to save part of the money, each takes from the total sum as much as he can carry, after which the first deposits half, the second a third, and the third a sixth of what eac... | 219. John of Palermo, master, court mathematician, who proposed this problem to Leonardo Fibonacci at a tournament.
Then we have the system:
$$
\begin{aligned}
& \frac{t}{2}=\frac{x}{2}+\f... | 23\frac{1}{2},15\frac{2}{3},7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,854 |
226. $\left\{\begin{array}{l}x+y=10 \\ \frac{x}{y}+\frac{y}{x}=\sqrt{5}\end{array}\right.$ | 226. Leonardo's solution. Excluding y, we get:
$$
x^{2}+\sqrt{50000}-200=10 x
$$
from which $x=5-\sqrt{225-\sqrt{50000}}$.
Further simplification is possible:
$$
x=15-\sqrt{125}, \quad y=\sqrt{125}-5
$$
Leonardo provides another method. Let $\frac{y}{x}=z$. Then $\frac{x}{y}=$ $=\sqrt{5}-z$. Since $\frac{y}{x} \cd... | 15-\sqrt{125},\quad\sqrt{125}-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,857 |
227. $\left\{\begin{array}{l}x+y+z=10 \\ x z=y^{2} \\ z^{2}+y^{2}=x^{2}\end{array}\right.$
38
Problems from "Practica geometriae" by Leonardo Fibonacci. | 227. Leonardo's solution (using false position). He assumes $z=1$. Then
$$
\begin{aligned}
& x_{1}+y_{1}=9 ; x_{1}=y_{1}^{2} ; 1+y_{1}^{2}=y_{1}^{4} \text { and } x_{1}=\sqrt{\frac{5}{4}}+\frac{1}{2} ; \\
& y_{1}=\sqrt{\sqrt{\frac{5}{4}}+\frac{1}{2}}
\end{aligned}
$$
Instead of 10, the sum of the unknowns is now
$$
... | 5-\sqrt{\sqrt{3125}-50} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,858 |
228. $3 x+4 \sqrt{x^{2}+3 x}=20$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
228. $3 x+4 \sqrt{x^{2}+3 x}=20$. | 228. $4 \sqrt{x^{2}-3 x}=20-3 x$; squaring both sides; after simplification we get:
$$
7 x^{2}+72 x-400=0 ; \quad x_{1}=4 ; \quad x_{2}=-\frac{100}{7}
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,859 | |
231. Prove that the medians of a triangle intersect at one point. | 231. Draw two medians $A a$ and $B b$, and let $O$ be their point of intersection (Fig. 53). Connect points $a$ and $b$. $\triangle a O b c \sim \triangle A O B$; $O a: a b = A O: A B$, but $a b = \frac{1}{2} A B$, therefore:
$$
O a = \frac{1}{2} A O
$$
If we draw medians $C c$ and $A a$, then for their point of inte... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,862 |
232. Inscribe a square in an equilateral triangle so that one of its sides lies on the base of the triangle. | 232. 233) Let the square $D E F H$ be the one sought (Fig. 54). Let its side be $D E = x$. We have $D E: B K = A C: B O$; if $A C = a$, then $B O = \frac{a}{2} \sqrt{3}$, hence
$$
x: \left(\frac{a \sqrt{3}}{2} - x\right) = a: \frac{a \sqrt{3}}{2} = 1: \frac{\sqrt{3}}{2},
$$
from which
$$
x = a(2 \sqrt{\overline{3}} ... | (2\sqrt{3}-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,863 |
233. Prove that the square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions.
Problems of John of Palermo, solved by Leonardo. | 233. Let $A B=p ; \quad B C=q$; $B b=r$ (fig. 55). From the right triangle $A B D$ :
$$
D B^{2}=p^{2}+q^{2}
$$
From the right triangle $D B b$ :
$$
D B^{2}+B b^{2}=D b^{2}=p^{2}+q^{2}+r^{2}
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,864 |
234. Find a square number which, when increased or decreased by 5, remains a square number. | 234. Leonardo's Solution.
Fig. 55. Let the required number be \( x^{2} \), then by the condition:
\[
x^{2}+5=u^{2} \text { and } x^{2}-5=v^{2}
\]
from which
\[
\begin{aligned}
u^{2}-v^{2... | (\frac{41}{12})^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,865 |
237. $\left\{\begin{array}{l}x+y+\sqrt{x^{2}+y^{2}}=\frac{x y}{2} \\ x y=48 .\end{array}\right.$ | 237. $x+y+\sqrt{x^{2}+y^{2}}=24 ; \sqrt{x^{2}+y^{2}}=24-(x+y)$.
By squaring, we find: $x+y=14$; then by the sum and product, we find $x=8$ and $y=6$. | 8,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,868 |
250. Of all triangles with a common base inscribed in a circle, the isosceles has the greatest area.
Problems of Regiomontanus (Johann Müller). | 250. The area of a triangle is equal to $\frac{b h}{2}$, therefore, the maximum area will be at the maximum height. It is clear that the point - the vertex of the isosceles triangle $A B C$ - is the farthest from the base.
Johann Müller, nicknamed Regiomontanus after his birthplace (in Monte Regio, Königsberg) (1436-1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,876 |
251. $16 x^{2}+2000=680 x$. | 251. $x=\frac{85 \pm 5 \sqrt{209}}{4}$. | \frac{85\5\sqrt{209}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,877 |
252. $\frac{x}{10-x}+\frac{10-x}{x}=25$. | 252. $10 x=x^{2}+\frac{100}{27} ; x=5-\sqrt{21 \frac{8}{27}}$. But Regiomontanus also gives another solution, assuming $\frac{x}{10-x}=y$. Then $y+\frac{1}{y}=25 ; y=\frac{25}{2}-\sqrt{\frac{621}{4}}$. In both cases, Regiomontanus takes the roots with a minus sign. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,878 |
253. Find a number which, when divided by 17, 13, and 10, gives remainders of 15, 11, and 3, respectively.
From the treatise "De triangulis omnimodis libri quinque". | 253. The problem is reduced to solving the system:
$$
17 x+15=13 y+11=10 z+3
$$
The smallest values: $x=64 ; y=84 ; z=110$.
The sought number is 1103.
This problem was proposed by Regiomontanus to the astronomer Bianchini. The latter, by trial and error, found two solutions: 1103 and 3313.
 parallels to the sides opposite these vertices. In the resulting triangle $M N P$, points $A, B$, and $C$ are the midpoints of sides $M P, M N$, and $P N$. Perpendiculars from these points to the sides of triangle $M N P$ will intersect at on... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,880 |
255. In a triangle, the difference between two lateral sides is 3, the height is 10, and the difference between the segments of the base formed by the foot of the height is 12. Find the sides. | 255. Let $A C - A B = F C = 3 ; \quad A D = 10 ; \quad D C - B D = E C = 6 ; \quad B D = \frac{B C - 6}{2} = \frac{B C}{2} - 3$ (Fig. 58).
If $B C = x$, then $B D = \frac{x}{2} - 3$. Further,
$$
A C - A B = 3 = \frac{6}{2} = \frac{1}{2}(D C - B D)
$$
$A D^{2} = A C^{2} - D C^{2} = A B^{2} - B D^{2}$, from which $A C... | \sqrt{\frac{427}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,881 |
256. Find the radius of the circle circumscribed around a triangle, given its sides.
## Task of Campano of Navarre.
From a comment on the Latin translation from Arabic of Euclid's "Elements".
## | 256. Let the sides of the triangle be \( AC = b, BC = a, AB = c \) (Fig. 59). The height \( h = BE \) is unknown. By drawing the diameter \( BD \), we have from the similarity of triangles \( ABD \) and \( EEC \):
}
$$
and of an icosahedron:
$$
2 R^{2}(5 \sqrt{3}-\sqrt{15})
$$
Similarly, the volume of a dodecahedron is:
$$
\frac{2}{9} R^{3} \sqrt{30(3+V \overline{\overline{5}})}
$$
and of an icosahedron:
$$
\frac{2}{3} R^{3} V^{10+2 \s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,883 |
258. $\left(a^{m}\right)^{\frac{1}{n}} ;\left(a^{\frac{1}{n}}\right)^{\frac{n}{m}} ;\left(a^{n} b\right)^{\frac{1}{n}} ;\left(a^{n} b^{m}\right)^{\frac{1}{m n}} ;\left(\frac{a^{n}}{b^{m}}\right)^{\frac{1}{m n}}$.
Problems from an anonymous Italian manuscript of the 14th century. | 258. $a^{\frac{m}{n}} ; a^{\frac{1}{m}} ; a b^{\frac{1}{n}} ; a^{\frac{1}{m}} \cdot b^{\frac{1}{n}} ; \frac{a^{\frac{1}{m}}}{b^{\frac{1}{n}}}$.
258. $a^{\frac{m}{n}} ; a^{\frac{1}{m}} ; a b^{\frac{1}{n}} ; a^{\frac{1}{m}} \cdot b^{\frac{1}{n}} ; \frac{a^{\frac{1}{m}}}{b^{\frac{1}{n}}}$.
The text is already in a form ... | ^{\frac{}{n}};^{\frac{1}{}};^{\frac{1}{n}};^{\frac{1}{}}\cdotb^{\frac{1}{n}};\frac{^{\frac{1}{}}}{b^{\frac{1}{n}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,884 |
261. Given the three sides of a right triangle, determine the radius of the circle inscribed in this triangle. | 261. Let the sides of triangle $ABC$ be given (Fig. 60): the hypotenuse $AC=b$; the legs $AB=c$ and $BC=a$. Clearly, if we inscribe a circle of radius $r$ in the triangle, we will have:
$$
AB = AE + r \quad BC = CF + r
$$
or
$$
c + a = AE + CF + 2r
$$
But
$$
AE = AD \text{ and } CF = CD \text{. }
$$
Adding gives:... | \frac{b}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,886 |
262. Inscribe the largest square in the given semicircle. | 262. If the side of the sought square is $x$, and the radius of the given semicircle is $r$, then
$$
r^{2}=x^{2}+\frac{x^{2}}{4}=\frac{5}{4} x^{2}
$$
or
$$
\frac{4 r^{2}}{5}=x^{2}, \text { i.e. }{ }_{5}^{4} r: x=x: r
$$
thus, the sought side can be constructed as the mean proportional between the radius and its fou... | \sqrt{\frac{4r^2}{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,887 |
263. The sides of the triangle are $13, 14, 15$. Taking the last side as the base, find the height and the segments of the base | 263. If the height is $h$, and the segments of the base are $m$ and $15-m$, then
$$
\begin{gathered}
13^{2}=h^{2}+m^{2} \\
14^{2}=h^{2}+(15-m)^{2}
\end{gathered}
$$
Subtracting term by term, we find: $196-169=2!5-3$ ) m, therefore $m=6 \frac{3}{5}$;
$$
\begin{gathered}
\text { then } 15-m=8 \frac{2}{5} \\
h^{2}=169-... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,888 | |
264. Given an equilateral triangle. Determine the radius of the inscribed and circumscribed circles. | 264. Let the side of the triangle be $a$ (Fig. 63). Then it is clear that $r=\frac{R}{2}$, and also, $R^{2}=r+\frac{a^{2}}{4}=\frac{R^{2}}{4}+\frac{a^{2}}{4}$, from which
\[
\begin{gathered}
3 R^{2}=a^{2} ; R^{2}=\frac{a^{2}}{3} ; R \frac{a}{\sqrt{3}}=\frac{a \sqrt{3}}{3} . \\
r=\frac{R}{2}=\frac{a \sqrt{3}}{6} .
\end... | R=\frac{\sqrt{3}}{3},\quadr=\frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,889 |
265. If $h$ is the height of a triangle, $b$ is the base, and $x$ is the smaller of its segments, then the radius of the circumscribed circle according to Wiedmann is expressed by the formula:
$$
r=\sqrt{\binom{b}{2}^{2}+\left(\frac{h^{2}+\left(\frac{b}{2}-x\right)^{2}-\frac{b^{2}}{4}}{2 h}\right)^{2}}
$$
Verify the ... | 265. From the drawing (Fig. 64), it is clear that \( r^{2}=y^{2}+\frac{b^{2}}{4} \). Further,
\[
r^{2}=(h-y)^{2}+\left(\frac{b}{2}-x\right)^{2}=h^{2}-2 h y+y^{2}+\left(\frac{b}{2}-x\right)^{2}
\]
thus,
\[
y^{2}+\frac{b^{2}}{4}=h^{2}-2 h y+y^{2}+\left(\frac{b}{2}-x\right)^{2}
\]
138
from which
\[
2 h y=h^{2}+\left(... | \sqrt{(\frac{b}{2})^{2}+[\frac{^{2}+(\frac{b}{2}-x)^{2}-\frac{b^{2}}{4}}{2}]^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 37,890 |
266. $43 x+41=39 y+33=35 z+25=31 u+17$.
Problem from a German algebraic manuscript (No. 14908 of the Munich collection). | 266. Given: $43 x+8=39 y ; 39 y+8=35 z ; 35 z+8=31 u$ or: $43 x+16=35 z=31 u-8$, and therefore, $43 x+24=31 u$.
$$
\begin{gathered}
u=\frac{43 x-24}{31}=x+1+\frac{12 x-7}{31}=x+1+t \\
12 x-7=31 t \\
x=\frac{31 t+7}{12}=2 t+\frac{7 t+7}{12}=2 t+7 t_{1}, \text { where } t_{1}=\frac{t+1}{12} \\
t=12 t_{1}-1 ; \quad x=31 ... | \begin{aligned}&31\cdot39\cdot35-2\\&31\cdot35\cdot43-2\\&31\cdot39\cdot43-2\\&u=35\cdot39\cdot43-2\end{aligned} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,891 |
267. $x+\sqrt{x^{2}-x}=2$.
Задача из собрания примеров на латинском языке из той же рукописи. | 267. $\sqrt{x^{2}-x}=2-x ; \quad x^{2}-x=4+x^{2}-4 x ; \quad 3 x=4$; $x=\frac{4}{3}$ | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,892 |
268. Someone has workers and money. If he gives each worker 5 (coins), he will have 30 left, but if 7, then he will be short of 30. The question is, how many workers does he have?
Problems by N. Chuquet from the treatise “Triparty...”. | 268. If each person is given two more coins, the difference turns out to be 60 coins, therefore, there were 30 workers.
Shyukе Nicolas (died in 1500), French mathematician, author of the treatise "Le Triparty en la science des nombres" (1484), in which zero and negative exponents are used for the first time. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,893 |
269. $3 x^{2}+12=12 x$ | 269. Two equal roots $(x=2)$.
270. $x=5 \pm \sqrt{21}$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,894 |
278. $\sqrt{12 x-x^{2}}+1=\sqrt{36-x^{2}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
278. $\sqrt{12 x-x^{2}}+1=\sqrt{36-x^{2}}$. | 278. Squaring:
$$
\begin{gathered}
12 x-x^{2}+1+2 \sqrt{12 x-x^{2}}=35-x^{2} \\
2 \sqrt{12 x-x^{2}}=35-12 x
\end{gathered}
$$
Squaring again:
$$
4\left(12 x-x^{2}\right)=1225-840 x+144 x^{2}
$$
or
$$
148 x^{2}-888 x+1225=0
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,896 | |
279. $\sqrt{4 x^{2}+4 x}+2 x+1=100$. | 279. $\sqrt{4 x^{2}+4 x}=99-2 x$
$$
\begin{gathered}
4 x^{2}+4 x=99^{2}+4 x^{2}-4 x \cdot 99 \\
4 x(1+99)=99^{2} \\
x=\left(\frac{99}{20}\right)^{2}
\end{gathered}
$$ | (\frac{99}{20})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,897 |
280. $\left\{\begin{array}{l}x+y+z=12 \\ 2 x+y+\frac{1}{2} z=12 .\end{array}\right.$ | 280. $z=2 x ; y=12-3 x . \quad y$ will be an integer and positive when $x=1,2,3,4$ | 1,2,3,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,898 |
283. Multiply: $4 x-5$ and $2 x-3$. | 283. $(4 x-5)(2 x-3)=8 x^{2}-22 x+15$. Interestingly, in the manuscript, the calculation is carried out as follows:
$$
\begin{aligned}
& 4 x \cdot 2 x=8 x^{2} \\
& 3 \cdot 4 x=-12 x \\
& 5 \cdot 2 x=-10 x
\end{aligned}
$$
140 | 8x^{2}-22x+15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,901 |
284. How great is the number that equals the product of $\frac{4}{5}$ of it by $\frac{5}{6}$ of the same number?
Problem from an algebraic Latin manuscript p. 80 (Dresden collection). | 284. In the manuscript, it is said: let the number be $n$. Four fifths of it will be $\frac{4}{5} n$. Five sixths of it will be $\frac{5}{6} n$. The product of $\frac{4}{5} n$ and $\frac{5}{6} n$ equals the number, i.e., $\frac{2}{3} n^{2}=n$. According to the form of the equation $a x^{2}=b x$, we have
$$
1 \cdot n=\... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,902 |
288. Show that every perfect number (except 6) has the form $9n+1$.
## Reticus's Problem. | 288. The general form of a perfect number $4^{k}\left(4^{k} 2-1\right)$. The number $4^{k}$ when divided by 9 gives remainders of 4, 7, or 1; $4^{k} \cdot 2-1$ when divided by 9 gives remainders of 7, 4, or 1, respectively. The products of these numbers give the same remainders when divided by 9 as 28 and 1, i.e., unit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,904 |
289. If the sides of an acute-angled triangle are $a, b, c$, and the perpendicular dropped from the vertex of the angle opposite side $a$ divides this side into segments, of which the smaller is $p$, then
$$
\frac{a}{c+b}=\frac{c-b}{a-2 p}
$$
## Problems by Luca de Burgo (Pacioli). | 289. Since (Fig. 65)
$$
h^{2}=c^{2}-(a-p)^{2}=b^{2}-p^{2}
$$
then
$$
\begin{gathered}
c^{2}-b^{2}=(a-p)^{2}-p^{2} \\
(c+b)(c-b)=(a-p+p) \\
(a-p-p)
\end{gathered}
$$
or
$$
(c+b)(c-b)=a(a-2 p)
$$
^{2}$. | 291. $\sqrt{160}+4$; simplification will give $4 \sqrt{10}+4$. | 4\sqrt{10}+4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,907 |
292. Find the number which, when multiplied by 5, gives as much as its square, added to four. | 292. $x^{2}+4=5 x$
$$
\text { according to Paciolo } x=\sqrt{\left(\frac{5}{2}\right)^{2}-4}+\frac{5}{2} ; x_{1}=4 ; x_{2}=1
$$ | x_1=4,x_2=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,908 |
293. Solve the equation:
$$
x^{4}+2 x^{3}+3 x^{2}+2 x-81600=0
$$
(solved elementarily). | 293. $x^{4}+2 x^{3}+2 x^{2}+x^{2}+2 x+1=81601$;
$$
\begin{gathered}
\left(x^{2}+x+1\right)^{2}=81601 \\
x^{2}+x+1-\sqrt{81601}=0 \\
x=-\frac{1}{2} \pm \sqrt{\frac{1}{4}-1+\sqrt{81601}}= \\
=-\frac{1}{2} \pm \sqrt{\sqrt{81601}-\frac{3}{4}}
\end{gathered}
$$ | -\frac{1}{2}\\sqrt{\sqrt{81601}-\frac{3}{4}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,909 |
294. The radius (4) of the circle inscribed in a triangle and the segments 6 and 8, into which the point of tangency divides one side of the triangle, are given. Find the other two sides. | 294. It is clear (Fig. 66) that $A D=6$ and $E C=8$; let $B D=$ $=B E=x$. On one hand,
Fig. 66. The area of the triangle:
$$
\begin{aligned}
\Delta & =A O C+A O B+B O C= \\
& =[14 \cdot 4... | AB=13;BC=15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,910 |
296. Given the area and the difference of the sides of a rectangle, find its sides. | 296. Paciolo's Solution: Let $a^{2}$ be the area of the rectangle and $d$ be the difference between the sides. The larger side is $x+\frac{d}{2}$, then the smaller side is $x-\frac{d}{2}$, hence $x^{2}-\frac{d^{2}}{4}=a^{2}, \quad$ from which $x=$ $=\sqrt{\frac{d^{2}}{4}+a^{2}}$; this is simpler than taking both sides ... | \sqrt{\frac{^{2}}{4}+^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,912 |
297. Inscribe two equal circles in a triangle so that each touches two sides and they touch each other. | 297. Instruction. We inscribe a circle of arbitrary radius, touching the sides of angle $A$ (Fig. 68), and construct a second circle of the same radius that touches this circle and side $A C$. We draw a tangent $M N$ to the second circle, parallel to $B C$. We connect $O_{1}$ with point $M$ and draw a line $B K$ from $... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,913 |
298. Determine the diameter of a circle that touches two sides of a given triangle and has its center on the third side. | 298. The geometric problem is solved very simply, since the center of the desired circle lies at the intersection of the bisector of angle $A$ with the opposite side (Fig. 69). Dropping a perpendicular from point $O$ to $AB$, we find the radius $OE$.
Hint. To calculate the radius from the three sides of triangle $ABC$... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,914 | |
299. Find the sides of a triangle with an area of 84, if it is known that they are expressed by three consecutive integers. | 299. Let the sides be $x-1, x, x+1$. The perimeter is $3x$, therefore,
$$
\begin{aligned}
& 84= \sqrt{\frac{3}{2} x \cdot \frac{1}{2} x \cdot\left(\frac{x}{2}+1\right)\left(\frac{x}{2}-1\right)}= \\
&=\sqrt{\frac{3 x^{2}(x+2)(x-2)}{16}} \\
& 84={ }_{4}^{x} \sqrt{3\left(x^{2}-4\right)} ; 336^{2}=x^{2} 3\left(x^{2}-4\ri... | 13,14,15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,915 |
301. The sum of the squares of two numbers is 20, and their product is 8. Find these numbers.
## Problems by G. Schreiber (Grammateus). | 301. Paciolo's Solution. Let the sum of the numbers $u$ and $y$ be $x$, and their difference $y$, where $x$ and $y$ are the unknowns. Then:
$$
x^{2}+y^{2}=2\left(u^{2}+v^{2}\right)
$$
and
$$
\begin{gathered}
x y=u^{2}-v^{2} \\
u^{2}+v^{2}=10
\end{gathered}
$$
And
$$
u^{2}-v^{2}=8
$$
from which
$$
u^{2}=9 \text {... | 4,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,917 |
304. $2 x^{3}=128$. $\quad 305.5 x^{4}=20480$
Задачи из трактата "Algorithmus de integris et minutis". | 304. $x_{1}=4 ; x_{2,3}=-2(1 \pm i \sqrt{3})$. | x_{1}=4;x_{2,3}=-2(1\i\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,920 |
306. Given the leg and the sum of the other two sides of a right triangle, determine these sides. | 306. This arithmetic book (Teaching on integers and fractions) was published in Leipzig in 1507.
Given the leg $a$ and the sum of the other leg and the hypotenuse
$$
S=b+c
$$
Since
$$
c^{2}=a^{2}+b^{2}
$$
then
$$
c^{2}=\left(S-b\right)^{2}
$$
gives
$$
a^{2}+b^{2}=S^{2}+b^{2}-2 S b
$$
from which
$$
b=\frac{S^{... | \frac{S^{2}-^{2}}{2S} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,921 |
307. Given the hypotenuse and the sum of the legs. Find the legs.
## Adam Riese's Problems. | 307. Similarly: given the hypotenuse $c$ and the sum of the legs
$$
S=a+b
$$
10 Popov. Problem Collection.
Since
$$
c^{2}=a^{2}+b^{2}
$$
and
$$
S^{2}=a^{2}+b^{2}+2 a b
$$
Then
$$
S^{2}-c^{2}=2 a b
$$
from which
$$
a b=\frac{S^{2}-c^{2}}{2}
$$
The legs $a$ and $b$ can be found from the sum and the product.
A... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,922 | |
309. 26 persons together spent 88 coins, with each man spending 6, each woman 4, and each girl 2 coins. How many were men, women, and girls? | 309. 10 solutions; from $x+y+z=26$
and
$$
6 x+4 y+2 z=88
$$
we get:
$$
2 x+y=18
$$
from which
$$
y=18-2 x
$$
therefore,
$$
2 x \leqslant 18
$$
H
$$
x \leqslant 9
$$
T. e
$$
\begin{aligned}
& x=0, \quad 1, \quad 2, \quad 3, \quad 4, \quad 5, \quad 6, \quad 7, \quad 8, \quad 9 \\
& y=18,16,14,12,10, \quad 8, ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,924 |
310. Three men trade a horse for 12 florins, but none of them individually has that amount. The first says to the other two: "Give me half of your money each, and I will buy the horse." The second says to the first and the third: "Give me one third of your money each, and I will acquire the horse." Finally, the third s... | 310. Let the 1st have $x$, the 2nd have $y$, the 3rd have $z$.
Then
$$
\begin{aligned}
& x+\frac{1}{2}(y+z)=12 \\
& y+\frac{1}{3}(x+z)=12 \\
& z+\frac{1}{4}(x+y)=12
\end{aligned}
$$[^0]
Solving the system is not difficult:
$$
x=3 \frac{9}{17} ; y=7 \frac{13}{17} ; z=9 \frac{3}{17}
$$ | 3\frac{9}{17};7\frac{13}{17};9\frac{3}{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,925 |
313. Someone agreed to work on the condition of receiving clothing and 10 florins at the end of the year. But after 7 months, he stopped working and upon settlement received the clothing and 2 florins. What was the value of the clothing? | 313. Answer: $9 \frac{1}{5}$ florin. | 9\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,927 |
314. Divide $\sqrt[3]{216}$ by $\sqrt[4]{16}$. | 314. Rudolf's Solution:
$$
\begin{aligned}
& =\sqrt[12]{531441}=\sqrt[6]{729}=\sqrt[3]{27}=3
\end{aligned}
$$
Indeed: $\quad \sqrt[3]{216}: \sqrt[4]{16}=6: 2=3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,928 |
315. Solve the system of equations: $x-y=a ; x y=b$.
## Problems from "Coss" by Rudolf in the adaptation by Stifel. | 315. Rudolf's Solution: from the first equation $x=a+y$; substituting into the second, we get: $y(a+y)=b ; y^{2}+a y-b=0$.
Rudolf's "Coss" was processed and supplemented by Stifel (see below) in 1552. | y^2+ay-0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,929 |
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