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10.039. The lines containing the non-parallel sides of an isosceles trapezoid intersect at a right angle (Fig. 10.39). Find the lengths of the sides of the trapezoid if its area is \(12 \mathrm{~cm}^{2}\) and the height is \(2 \mathrm{~cm}\).
. Find the length of the shorter lateral side of the trapezoid if its midline is 10 cm, and one of the bases is $8 \mathrm{~cm}$. | ## Solution.
Given $\angle B C A=30^{\circ}, \angle A B C=90^{\circ}, K M=10$ cm, $D E=8$ cm, $\frac{8+A C}{2}=10, A C=12$ cm (since $K M$ is the midline). Since $\triangle D B E \sim \triangle A B C$, then $\frac{A B}{D B}=\frac{A C}{D E} ; \frac{x+D B}{D B}=\frac{12}{8}$ (where $A D=x$ ), $x=D B \cdot \frac{1}{2}$. ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.045. The common chord of two intersecting circles is seen from the centers at angles of $90^{\circ}$ and $60^{\circ}$. Find the radii of the circles if the distance between their centers is $\sqrt{3}+1$. | Solution.
Let $r$ be the radius of one circle, and $R$ be the radius of the other circle. According to the problem, $\angle A O_{1} B=90^{\circ}, \angle A O_{2} B=60^{\circ}, O_{1} O_{2}=\sqrt{3}+1 ; A B=r \sqrt{2} ;$ $O_{1} E=A B / 2 ; O_{1} E=r \frac{\sqrt{2}}{2}, A B=2 R \cdot \sin 30^{\circ}=R\left(\angle A O_{2} ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.051. Three circles of different radii touch each other pairwise. The segments connecting their centers form a right triangle. Find the radius of the smallest circle, if the radii of the largest and medium circles are 6 and 4 cm. | Solution.
Fig. 10.51
Fig. 10.52
Let $r$ be the radius of the s... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.064. The lateral side of an isosceles triangle is $10 \mathrm{~cm}$, and the base is $12 \mathrm{~cm}$. Tangents are drawn to the inscribed circle of the triangle, parallel to the height of the triangle, and cutting off two right triangles from the given triangle. Find the lengths of the sides of these triangles. | ## Solution.
The area of the triangle is found using the formula (Fig. 10.63):
$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{16(16-10)(16-10)(16-12)}=48
$$
The radius of the inscribed circle $r=\frac{S}{p}=\frac{48}{16}=3$ (cm); $r=D H . H C=D C-$ $-D H=6-3=3 \text{ cm} ; \triangle B C D$ and $\triangle F C H$ are similar, so ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.096. The area of an isosceles trapezoid circumscribed around a circle is $32 \sqrt{3} \mathrm{~cm}^{2}$ (Fig. 10.92). Determine the lateral side of the trapezoid, given that the acute angle at the base is $\pi / 3$. | Solution.
$\triangle C E D-$ is a right triangle and $\sin 60^{\circ}=\frac{2 R}{C D}$. Therefore, $2 R=\frac{\sqrt{3}}{2} C D$, $C D=x \Rightarrow 2 R=\frac{\sqrt{3}}{2} x$. The area of the trapezoid $S=\frac{a+b}{2} h=2 R x=\frac{\sqrt{3}}{2} x^{2}$. According to the condition $\frac{\sqrt{3}}{2} x^{2}=32 \sqrt{3} \... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.142. A circle is inscribed in an equilateral triangle, and a regular hexagon is inscribed in the circle. Find the ratio of the areas of the triangle and the hexagon. | ## Solution.
Let the side of the equilateral triangle be $a$. Then its area $S_{1}=\frac{a^{2} \sqrt{3}}{4} \cdot$ The radius of the circle inscribed in the triangle, $r=\frac{a \sqrt{3}}{6}$.
=32$. The perimeter of one triangle is $p_{1}=b+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, and the second is $p_{2}=a+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, where $d_{1}, d_{2}$ are the diagonals of the parallelogram. The differen... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.157. Through points $R$ and $E$, belonging to sides $A B$ and $A D$ of parallelogram $A B C D$, and such that $A R=(2 / 3) A B, A E=(1 / 3) A D$, a line is drawn. Find the ratio of the area of the parallelogram to the area of the resulting triangle. | ## Solution.
Let $h$ be the height of parallelogram $ABCD$, and $h_{1}$ be the height of triangle $ARE$ (Fig. 10.134). Then $S_{ABCD} = AD \cdot h$, and $S_{\triangle ARE} = \frac{1}{2} AE \cdot h_{1}$. But $\frac{h}{h_{1}} = \frac{AB}{AR} = \frac{3}{2}$. Therefore, $\frac{S_{ABCD}}{S_{\triangle ARE}} = \frac{AD \cdot... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.189. A square is inscribed in a segment whose arc is $60^{\circ}$. Calculate the area of the square if the radius of the circle is $2 \sqrt{3}+\sqrt{17}$. | Solution.
Let $B C=C D=D A=A B=x$ (Fig. 10.156). Consider $\triangle O N C$,
Fig. 10.156
. Therefore, $H=\sqrt{a^... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.073. The measurements of a rectangular parallelepiped are 2, 3, and 6 cm. Find the length of the edge of a cube such that the volumes of these bodies are in the same ratio as their surface areas. | Solution.
The volume of a rectangular parallelepiped is $V_{\text {par }}=2 \cdot 3 \cdot 6=36 \mathrm{~cm}^{3}$. The volume of a cube is $V_{\mathrm{x}}=a^{3}$. The area of the complete surface of a rectangular parallelepiped is calculated as follows: $S_{\text {par }}=S_{\text {side }}+2 S_{\text {base }}=P H+2 S_{\... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.011. Due to the reconstruction of equipment, the labor productivity of a worker increased twice during the year by the same percentage. By what percentage did the labor productivity increase each time, if during the same time a worker used to produce goods worth 2500 rubles, and now produces goods worth 2809 rubles? | Solution.
Let a worker produce a parts in 8 hours of work. Then the rate is $\frac{2500}{a}$ rubles per part, and the labor productivity is $\frac{a}{8}$ parts per hour. After the first increase in productivity by $x \%$, the worker started producing $\frac{a}{8}+\frac{x \cdot a}{100 \cdot 8}$ parts per hour; after th... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.019. Two snow-clearing machines are working on snow removal. The first can clear the entire street in 1 hour, while the second can do it in $75\%$ of this time. Starting the cleaning simultaneously, both machines worked together for 20 minutes, after which the first machine stopped. How much more time is needed for ... | ## Solution.
Let's consider the entire volume of work as 1. The productivity of the first machine is 1 (per hour), and the second machine's productivity is $1: \frac{3}{4}=\frac{4}{3}$ (per hour). Working together for $\frac{1}{3}$ of an hour, they will complete $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{4}{3} = \... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.020. The sum of the first three terms of the proportion is 58. The third term is $2 / 3$, and the second term is $3 / 4$ of the first term. Find the fourth term of the proportion and write it down. | Solution.
Let the proportion be $\frac{a}{b}=\frac{c}{d}$. Given that $a+b+c=58$; $c=\frac{2}{3} a$; $b=\frac{3}{4} a$. Therefore, $a+\frac{3}{4} a+\frac{2}{3} a=58$, from which $a=24$; $c=16$; $b=18$. Thus, $d=\frac{b c}{a}=12$.
Answer: $12 ; \frac{24}{18}=\frac{16}{12}$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.021. One brigade can harvest the entire field in 12 days. The second brigade needs $75\%$ of this time to complete the same work. After the first brigade worked alone for 5 days, the second brigade joined, and together they finished the work. How many days did the brigades work together? | ## Solution.
Let the teams work together for $x$ days. The productivity of the first team is $-\frac{1}{12}$; the productivity of the second team is $-\frac{1}{12 \cdot 0.75}=\frac{1}{9}$. According to the condition, $\frac{1}{12} \cdot 5+\left(\frac{1}{12}+\frac{1}{9}\right) \cdot x=1$, from which $x=3$ days.
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.026. A tourist traveled 90 km by boat along the river and walked 10 km on foot. For the walking route, 4 hours less were spent compared to the river journey. If the tourist had walked for as long as he had rowed, and rowed for as long as he had walked, these distances would have been equal. How long did he walk and ... | ## Solution.
Let $x$ be the number of hours the tourist walked. Then he spent $(x+4)$ hours on the boat. The tourist's walking speed is $\frac{10}{x}$ km/h; the speed on the boat is $-\frac{90}{x+4}$ km/h. According to the problem, $\frac{10}{x}(x+4)=\frac{90}{x+4} \cdot x$, from which $x=2$ hours.
Answer: 2 and 6 ho... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.032. In 3.5 hours of operation, one stamping press can produce $42\%$ of all ordered parts. The second press can produce $60\%$ of all parts in 9 hours, and the work speeds of the third and second presses are in the ratio of $6:5$. How long will it take to complete the entire order if all three presses work simultan... | Solution.
Let $x$ parts be the entire order. Then the working speed of the first press is $-\frac{0.42 x}{3.5}$ parts/hour; the second press is $\frac{0.6 x}{9}$ parts/hour; the third press is $\frac{6}{5} \cdot \frac{0.6 x}{9}$ parts/hour. Therefore, all three presses working simultaneously will complete the entire o... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.033. Each of the two typists retyped a manuscript of 72 pages. The first typist retyped 6 pages in the same time it took the second to retype 5 pages. How many pages did each typist retype per hour, if the first finished the work 1.5 hours faster than the second? | Solution.
Let $x$ pages per hour be the typing speed of the first typist, then the second typist's speed is $\frac{5}{6} x$. The first typist worked $\frac{72}{x}$ hours; the second - $\frac{72}{5}$ hours. According to the problem, $\frac{72}{x}=\frac{72}{\frac{5}{6} x}-1.5$, from which $x=9.6($ pages/hour); the secon... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.051. Two cylinders roll down an inclined board 6 m long, one of which has a circumference of 3 dm, and the other 2 dm. Can the circumferences of both cylinders be increased by the same amount so that on the same path one of them makes 3 more revolutions than the other? | Solution.
Assume that the lengths of the circumferences can be increased by $x$ dm. Then the first cylinder will make $\frac{60}{3+x}$ revolutions, and the second will make $-\frac{60}{2+x}$. According to the condition, $\frac{60}{3+x}+3=\frac{60}{2+x}$, from which we get $x^{2}+5 x-14=0$. Solving this equation, we fi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.061. To pay for the delivery of four parcels, 4 different postage stamps were needed for a total of 84 kopecks. Determine the cost of the stamps purchased by the sender, if these costs form an arithmetic progression, and the most expensive stamp is 2.5 times more expensive than the cheapest one. | Solution.
Let $x$ kop. - the cost of the cheapest stamp, $x+d, x+2 d$, $x+3 d$ - the costs of the other stamps. By condition
$$
\left\{\begin{array}{l}
x+x+d+x+2 d+x+3 d=84, \\
x+3 d=2.5 x,
\end{array} \text { from which } x=12 ; d=6\right. \text {. Costs of the }
$$
stamps: 12 kop., 18 kop., 24 kop, 30 kop.
Answer... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.064. The hay reserve is such that 96 kg can be issued daily for all the horses. In fact, the daily portion for each horse could be increased by 4 kg because two horses were sold. How many horses were there originally? | Solution.
Let there initially be $x$ horses. The daily portion for each horse was $\frac{96}{x}$ kg. Since two horses were sold to another collective farm, the number of horses became $x-2$, and the daily portion for each horse became $\frac{96}{x}+4$ kg. According to the condition, $\left(\frac{96}{x}+4\right)(x-2)=9... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.069. In one pool there are 200 m $^{3}$ of water, and in another - 112 m $^{3}$. Taps are opened, through which the pools are filled. After how many hours will the amount of water in the pools be the same, if 22 m $^{3}$ more water is poured into the second pool per hour than into the first? | Solution.
Let the amount of water in the pools become equal after $x$ hours. Let $y$ m$^{3}$ of water be added to the first pool per hour, and $y+22$ m$^{3}$ - to the second. According to the condition, $200+x y=112+x(y+22)$, from which $x=4$.
Answer: in 4 hours. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.082. Two groups of tourists must walk towards each other from tourist bases $A$ and $B$, the distance between which is 30 km. If the first group leaves 2 hours earlier than the second, they will meet 2.5 hours after the second group leaves. If the second group leaves 2 hours earlier than the first, the meeting will ... | Solution.
Let $x$ km/h be the speed of the first group, $y$ km/h be the speed of the second. If the first group starts earlier than the second, then $2.5 y + (2.5 + 2) x = 30$. If the second group starts earlier than the first, then $3 x + (3 + 2) y = 30$. Solving the system $\left\{\begin{array}{l}2.5 y + 4.5 x = 30,... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.085. Two brothers took their bicycles and set off simultaneously with the intention of riding $42 \mathrm{km}$. The older brother maintained the same speed throughout the journey; while the younger brother fell behind by 4 km every hour. However, since the older brother rested for a whole hour during the trip, and t... | ## Solution.
Let $x$ km/h be the speed of the older brother, then $(x-4)$ km/h is the speed of the younger brother. The older brother was on the road for $t_{\text {old }}=\frac{42}{x}+1$, and the younger brother for $t_{\text {young }}=\frac{42}{x-4}+\frac{1}{3}$. According to the problem, $t_{\text {old }}=t_{\text ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.086. A positive integer is thought of. To its notation, the digit 7 is appended on the right, and from the resulting new number, the square of the thought number is subtracted. The remainder is reduced by $75\%$ of this remainder, and the thought number is subtracted again. In the final result, zero is obtained. Wha... | Solution.
Let the number be $x$. Consider the numbers $10 x+7, 10 x+7-x^{2}$, and the remainder $\frac{25}{100}\left(10 x+7-x^{2}\right)$. Then $\frac{1}{4}\left(10 x+7-x^{2}\right)-x=0, x^{2}-6 x-7=0 \Rightarrow$ $\Rightarrow x=7$
Answer: 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.087. A positive integer is thought of. The digit 5 is appended to its right, and from the resulting new number, the square of the thought number is subtracted. The difference is divided by the thought number, and then the thought number is subtracted, and the result is one. What number is thought of? | ## Solution.
Let $x$ be the number thought of. Then the new number can be represented as $10 x + 5$. According to the condition, $\frac{10 x + 5 - x^2}{x} - x = 1$, from which $x = 5$.
Answer: 5.
, (2), ..., (12):
| Tourist | Before Meeting | | | After Meeting | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | speed, km $/ \mathbf{4}$ | time, h | distance, | speed, km $/ \mat... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.103. From the railway station to the tourist base, one can walk along the highway or a path, with the path being 5 km shorter. Two friends agreed that one would walk along the highway at a constant speed of $\mathrm{v}$ km/h, while the other would take the path at a speed of 3 km/h. The second one arrived at the tou... | Solution.
Let $x$ km be the distance from the station to the tourist base by road, and $x-5$ km by trail. According to the problem, $\frac{x}{v}-\frac{x-5}{3}=1$, from which $x=\frac{2 v}{v-3}$. The expression makes sense for $v>3$. By trial, we find: at $v=4$ km/h, $x=8$ km. Other solutions to this equation do not sa... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.109. Two cyclists set off simultaneously towards each other from two places, the distance between which is 270 km. The second cyclist travels 1.5 km less per hour than the first, and meets him after as many hours as the first cyclist travels in kilometers per hour. Determine the speed of each cyclist. | ## Solution.
Let $x$ km/h be the speed of the first cyclist, and $x-1.5$ km/h be the speed of the second cyclist. Before they meet, the first cyclist traveled $x \cdot x$ km, and the second cyclist traveled $x(x-1.5)$ km. According to the problem, $x^{2}+x(x-1.5)=270$, from which $x=12$ km/h; $12-1.5=10.5$ km/h.
Answ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.112. Two bodies move towards each other from two places, the distance between which is 390 km. The first body traveled 6 m in the first second, and in each subsequent second, it traveled 6 m more than in the previous one. The second body moved uniformly at a speed of $12 \mathrm{~m} / \mathrm{c}$ and started moving ... | ## Solution.
Let $t$ s be the time of movement of the first body until the meeting; $t-5$ s of the second. The first traveled until the meeting $\frac{a t^{2}}{2}=\frac{6 t^{2}}{2}=3 t^{2}$ km; the second $12(t-5)$ km. According to the condition $3 t^{2}+12(t-5)=360$, from which $t=10$ s.
Answer: in $10 \mathrm{s}$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.148. A photograph measuring $12 \times 18$ cm is inserted into a frame of constant width. Determine the width of the frame if its area is equal to the area of the photograph itself. | ## Solution.
Let $x$ cm be the width of the frame (Fig. 13.9). Then its area is $2 \cdot 12 x + 2(18 + 2 x) x$ cm. According to the condition, $2 \cdot 12 x + 2(18 + 2 x) x = 12 \cdot 18$, from
. What are the actual dimensions of the sports field?
(y-4)=x y-1012$, from which $x y-4(x+y)=$ $=x y-1028 ; x+y=257 ; y=257-x$. Solving the quadratic equation $x^{2}+(257-x)^{2}-185^{2}=0$, we find $x_{1}=104, x_{2}=153$.
Answer: $100... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.156. A motorboat with a speed of 20 km/h traveled the distance between two points along the river and back without stopping in 6 hours and 15 minutes. The distance between the points is 60 km. Determine the speed of the river current. | Solution.
Let $v_{\text {r }}$ km/h be the speed of the river current; the boat traveled 60 km downstream in $\frac{60}{20+v_{\mathrm{r}}}$ hours, and upstream in $\frac{60}{20-v_{\mathrm{r}}}$ hours. According to the problem, $\frac{60}{20+v_{\mathrm{r}}}+\frac{60}{20-v_{\mathrm{r}}}=6.25$, from which $v_{\mathrm{r}}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.162. A certain product was purchased in the fall for 825 rubles. A kilogram of this product in the fall was 1 ruble cheaper than in the spring, and therefore, for the same amount in the spring, 220 kg less was purchased. How much does 1 kg of the product cost in the spring and how much of it was purchased in the fal... | ## Solution.
Let $x$ kg of the product be purchased in the fall. In the spring, $x-220$ kg of the product was purchased for the same amount of money. The cost of 1 kg in the fall is $\frac{825}{x}$ rubles, and in the spring it is $\frac{825}{x-220}$ rubles. According to the condition, $\frac{825}{x-220}-\frac{825}{x}=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.168. A one-digit number was increased by 10 units. If the resulting number is increased by the same percentage as the first time, the result is 72. Find the original number. | Solution.
Let $x$ be the number we are looking for, which was increased by $y \cdot 100\%$. Then $x+10=y x$, from which $y=\frac{x+10}{x}$. According to the condition $(x+10) \cdot \frac{x+10}{x}=72$, from which $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.181. A team of workers was supposed to manufacture 7200 parts per shift, with each worker making the same number of parts. However, three workers fell ill, and therefore, to meet the entire quota, each of the remaining workers had to make 400 more parts. How many workers were in the team? | Solution.
Let there be $x$ workers in the team. Each worker made $\frac{7200}{x}$ parts. After the team size was reduced to $x-3$ workers, each worker started making $\frac{7200}{x}+400$ parts. According to the condition, $\left(\frac{7200}{x}+400\right)(x-3)=7200$, from which $x=9$.
## Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.191. Two forces are applied to a material point, the angle between which is $30^{\circ}$. The magnitude of one of the applied forces is $7 \sqrt{3}$ times the magnitude of the other, and the magnitude of the resultant force is $24 \mathrm{N}$ greater than the magnitude of the smaller force. Determine the magnitude o... | ## Solution.
Let $x \mathrm{H}$ be the modulus of the smaller force (Fig. 13.11). By the cosine rule: $\quad(24+x)^{2}=x^{2}+(7 \sqrt{3} x)^{2}-2 x \cdot 7 \sqrt{3} x \cdot \cos \left(180^{\circ}-30^{\circ}\right)$, from which $x=2 \mathrm{H}$ is the modulus of the smaller force; $24+2=26 \mathrm{H}$ is the modulus of... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.193. During exercises, a reconnaissance boat approached the lead ship of the squadron and received an order to conduct reconnaissance ahead of the squadron in the direction of its movement at a distance of 70 km. Determine how long it will take for the boat to return to the lead ship of the squadron, which continues... | Solution.
Let the boat return to the lead ship after $x$ hours. In this time, the squadron will travel $14 x$ km. Therefore, the boat must travel $70+70-14 x$ km. According to the condition, $140-14 x=28 x$, from which $x=3 \frac{1}{3}$ hours.
Answer: in 3 hours 20 minutes. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.194. The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $1 / 4$ of its length, and the circumference of the rear wheel is increased by $1 / 5$ of its length, then over the same distance, the front wheel will... | Solution.
The circumference of the wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $\mathrm{Cn}=s$. We will fill in the table of values for these quantities in the order indicated by the numbers (1), (2), ..., (12):
| Wheel | Before change | | | After change | | |
| :---:... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.197. A cinema hall has two doors, a wide one and a narrow one. After a screening, the audience exits the hall through both doors in 3 minutes and 45 seconds. If the audience is let out through only the wide door, it takes 4 minutes less than if they are let out through only the narrow door. How much time is required... | ## Solution.
Let $x$ min be necessary to release the audience only through the wide door, $(x+4)$ min - only through the narrow door. In one minute, $\frac{1}{x}$ people exit through the wide door, and $\frac{1}{x+4}$ people exit through the narrow door. According to the condition: $\frac{1}{x}+\frac{1}{x+4}=\frac{1}{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.322 A=\sqrt{y^{2}-6 y+9}-|y-9|+2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.322 A=\sqrt{y^{2}-6 y+9}-|y-9|+2$. | Solution. $A=\sqrt{(y-3)^{2}}-|y-9|+2=|y-3|-|y-9|+2$.
Consider 3 cases.
1) $\left\{\begin{array}{l}y9, \\ A=y-3-(y-9)+2\end{array} \Leftrightarrow\left\{\begin{array}{l}y>9, \\ A=8 .\end{array}\right.\right.$
Answer: if $y>9$, then $A=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.275 $\frac{2+x}{2-x}+\sqrt{x}=1+x$ | Solution. $\frac{2+x}{2-x}-1=x-\sqrt{x} \Leftrightarrow \frac{2 x}{2-x}=\sqrt{x}(\sqrt{x}-1)$. Let $y=\sqrt{x}$, then $\frac{2 y^{2}}{2-y^{2}}=y(y-1) \Leftrightarrow\left[\begin{array}{l}y=0, \\ y \neq \sqrt{2}, \\ y^{3}-y^{2}+2=0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left[\begin{array}{l}y=0 . \\ (y+1)\l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.277 \sqrt{x-1}+\sqrt{x+3}+2 \sqrt{(x-1)(x+3)}=4-2 x
$$
6.277 \sqrt{x-1}+\sqrt{x+3}+2 \sqrt{(x-1)(x+3)}=4-2 x
$$ | Solution. Let $u=\sqrt{x-1}, v=\sqrt{x+3}$, then $u+v+2uv=-(u^2+v^2)+6 \Leftrightarrow$ $(u+v)^2+(u+v)-6=0 \Leftrightarrow\left[\begin{array}{l}u+v=-3, \\ u+v=2\end{array} \Leftrightarrow\left[\begin{array}{l}\varnothing, \\ \sqrt{x-1}+\sqrt{x+3}=2 .\end{array}\right.\right.$
Since $x \geq 1$, then $\sqrt{x+3} \geq 2 ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.278 \sqrt{2 x+3}+\sqrt{x+1}=3 x+2 \sqrt{2 x^{2}+5 x+3}-16$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.278 \sqrt{2 x+3}+\sqrt{x+1}=3 x+2 \sqrt{2 x^{2}+5 x+3}-16$. | Solution. $\sqrt{2 x+3}+\sqrt{x+1}=3 x-16+2 \sqrt{(2 x+3)(x+1)}$.
Let $u=\sqrt{2 x+3}, v=\sqrt{x+1}$, then $u+\dot{v}=u^{2}+v^{2}-20+2 u v \Leftrightarrow$
$$
\Leftrightarrow(u+v)^{2}-(u+v)-20=0 . \Leftrightarrow\left[\begin{array}{l}
u+v=5, \\
u+v=-4
\end{array} \Leftrightarrow \sqrt{2 x+3}+\sqrt{x+1}=5 \Leftrightar... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.279 \sqrt[4]{x+8}-\sqrt[4]{x-8}=2$ | Solution. Let $u=\sqrt[4]{x+8}, v=\sqrt[4]{x-8}$, then $\left\{\begin{array}{l}u-v=2, \\ u^{4}-v^{4}=16\end{array} \Rightarrow\right.$ $\Rightarrow(v+2)^{4}-v^{4}=16 \Leftrightarrow\left((v+2)^{2}-v^{2}\right)\left((v+2)^{2}+v^{2}\right)=16 \Leftrightarrow(v+1)(v+2 v+2)=2$.
Since $v \geq 0$, then $v+1 \geq 1, v^{2}+2 ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$. | Solution. $\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4} \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ x+x+9+2 \sqrt{x(x+9)}=x+1+x+4+2 \sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ 2+\sqrt{x(x+9)}=\sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.282 \sqrt{x^{2}-x-1}+\sqrt{x^{2}+x+3}=\sqrt{2 x^{2}+8}, x>0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.282 \sqrt{x^{2}-x-1}+\sqrt{x^{2}+x+3}=\sqrt{2 x^{2}+8}, x>0$. | Solution. Raise both sides to the square:
$\left\{\begin{array}{l}x^{2}-x-1 \geq 0, ; \\ 2 x^{2}+2+2 \sqrt{\left(x^{2}-x-1\right)\left(x^{2}+x+3\right)}=2 x^{2}+8, \\ x>0\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x^{2}-x-1 \geq 0, \\ \sqrt{\left(x^{2}-x-1\right)\left(x^{2}+x+3\right)} \\ x>0\end{array... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.288 \frac{x^{2}}{\sqrt{2 x+15}}+\sqrt{2 x+15}=2 x$. | Solution.
$$
\begin{aligned}
& \frac{x^{2}}{\sqrt{2 x+15}}+\sqrt{2 x+15}-2 x=0 \Leftrightarrow\left(\frac{x}{\sqrt[4]{2 x+15}}-\sqrt[4]{2 x+15}\right)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{x}{\sqrt[4]{2 x+15}}=\sqrt[4]{2 x+15} \Leftrightarrow x=\sqrt{2 x+15} \Leftrightarrow\left\{\begin{array}{l}
x \geq 0, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6.290 \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=x-1$. | Solution. $\sqrt{(x-1)+2 \sqrt{x-1}+1}+\sqrt{(x-1)-2 \sqrt{x-1}+1}=x-1 \Leftrightarrow$ $\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}}=x-1 \Leftrightarrow \sqrt{x-1}+1+|\sqrt{x-1}-1|=x-1 \Leftrightarrow$ $\Leftrightarrow \sqrt{x-1}+|\sqrt{x-1}-1|=x-2$.
Since $x \geq 2$, then $\sqrt{x-1} \geq 1$. ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.293
$$
\frac{\sqrt[7]{x-\sqrt{2}}}{2}-\frac{\sqrt[7]{x-\sqrt{2}}}{x^{2}}=\frac{x}{2} \cdot \sqrt[7]{\frac{x^{2}}{x+\sqrt{2}}}
$$ | Solution. $\frac{\sqrt[7]{x^{2}-2}}{2}-\frac{\sqrt[7]{x^{2}-2}}{x^{2}}=\frac{x}{2} \cdot \sqrt[7]{x^{2}} \Leftrightarrow \sqrt[7]{x^{2}-2} \cdot\left(x^{2}-2\right)=x^{3} \cdot \sqrt[7]{x^{2}} \Leftrightarrow$ $\Leftrightarrow\left(x^{2}-2\right)^{8}=x^{23} \Leftrightarrow\left|x^{2}-2\right|=x^{\frac{23}{8}}$.
Consid... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.300 \sqrt{x-2}+\sqrt{4-x}=x^{2}-6 x+11 | Solution. $\sqrt{(x-3)+1}+\sqrt{1-(x-3)}=\left(x^{2}-6 x+9\right)+2$.
Let $y=x-3$, then $\sqrt{y+1}+\sqrt{1-y}=y^{2}+2 \Leftrightarrow$ $y^{4}+4 y^{2}+2\left(1-\sqrt{1-y^{2}}\right)=0$.
Since $1 \geq \sqrt{1-y^{2}}$, the last equation is equivalent to the system:
$$
\left\{\begin{array}{l}
y^{4}+4 y^{2}=0 \\
1=\sqrt... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.365 Solve the equation $\left(x^{3}+x^{-3}\right)+\left(x^{2}+x^{-2}\right)+\left(x+x^{-1}\right)=6$. | Solution. $\left(x^{3}-2+\frac{1}{x^{3}}\right)+\left(x^{2}-2+\frac{1}{x^{2}}\right)+\left(x-2+\frac{1}{x}\right)=0 \Leftrightarrow \frac{\left(x^{3}-1\right)^{2}}{x^{3}}+\frac{\left(x^{2}-1\right)^{2}}{x^{2}}+\frac{(x-1)^{2}}{x}=0$ $(x-1)^{2}\left(\left(x^{2}+x+1\right)^{2}+x(x+1)^{2}+x^{2}\right)=0 \Leftrightarrow$ $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.370 Solve the equation $\sqrt[4]{x^{4}+x-2}+\sqrt{x^{4}+x-2}=6$ given that $\boldsymbol{x}>\mathbf{0}$. | Solution. Let $y=\sqrt[4]{x^{4}+x-2}$, then $y^{2}+y-6=0 \Leftrightarrow \sqrt[4]{x^{4}+x-2}=2 \Leftrightarrow$ $\Leftrightarrow x^{4}+x-18=0 \Leftrightarrow(x-2)\left(x^{3}+2 x^{2}+4 x+9\right)=0$.
Since $x>0$, then $x^{3}+2 x^{2}+4 x+9>0 \Rightarrow x=2$.
Answer: $x=2$.
## pRoGReSSiOnS | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.084 In a volleyball competition, $n$ teams participated. Each team played with all the others once. For each game, the winning team was awarded one point, and no points were awarded for a loss, as there are no draws in volleyball. At the end of the competition, it turned out that the points scored by the teams formed... | Solution. Let the points scored by the teams form a non-decreasing arithmetic progression $a_{1}, a_{2}, \ldots, a_{n}$ with a common difference $d$. A total of $\frac{n(n-1)}{2}$ games were played. Therefore, the total number of points scored by all teams is $\frac{n(n-1)}{2}$, from which we have $a_{1}+a_{2}+\ldots+a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.296
$2 \log _{a}^{\frac{1}{2}} b \cdot\left(\left(\log _{a} \sqrt[4]{a b}+\log _{b} \sqrt[4]{a b}\right)^{\frac{1}{2}}-\left(\log _{a} \sqrt[4]{\frac{b}{a}}+\log _{b} \sqrt[4]{\frac{a}{b}}\right)^{\frac{1}{2}}\right), a, b>1$. | Solution.
1) $\left(\log _{a} \sqrt[4]{a b}+\log _{b} \sqrt[4]{a b}\right)^{\frac{1}{2}}=\left(\frac{1}{4}\left(\log _{a} b+1+\log _{b} a+1\right)\right)^{\frac{1}{2}}=$
$=\left(\frac{1}{4}\left(2+\log _{a} b+\frac{1}{\log _{a} b}\right)\right)^{\frac{1}{2}}=\frac{1}{2}\left(\frac{\left(\log _{a} b+1\right)^{2}}{\log... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.308 \quad(\sqrt[3]{0.5}+\sqrt[3]{4})^{x}=13.5$. | ## Solution.
$$
\left(\frac{1}{\sqrt[3]{2}}+\sqrt[3]{4}\right)^{x}=\frac{27}{2} \Leftrightarrow\left(\frac{3}{\sqrt[3]{2}}\right)^{x}=\frac{27}{2} \Leftrightarrow\left(\frac{27}{2}\right)^{\frac{x}{3}}=\left(\frac{27}{2}\right)^{1} \Leftrightarrow x=3
$$
Answer: $x=3$.
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3092 \log _{9}^{2} x=\log _{3} x \cdot \log _{3}(\sqrt{2 x+1}-1) | Solution. $\frac{\log _{3}^{2} x}{2}=\log _{3} x \cdot \log _{3}(\sqrt{2 x+1}-1) \Leftrightarrow \log _{3} x \cdot\left(\log _{3} x-2 \log _{3}(\sqrt{2 x+1}-1)\right)=0$
Answer: $x_{1}=1, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.313 \log _{x+1}\left(x^{3}-9 x+8\right) \cdot \log _{x-1}(x+1)=3$. | ## Solution.
$\log _{x+1}\left((x-1)\left(x^{2}+x-8\right)\right) \cdot \log _{x-1}(x+1)=3 \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\log _{x+1}\left((x-1)\left(x^{2}+x-8\right)\right)=3 \log _{x+1}(x-1), \\ x \neq 2\end{array} \Leftrightarrow\left\{\begin{array}{l}x-1>0, \\ x \neq 2, \\ (x-1)\left(x^{2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.317 \log _{4} x+\log _{x} 2-\log _{4} \sqrt{x}=1$ | Solution. $\frac{1}{2} \log _{2} x+\frac{1}{\log _{2} x}-\frac{1}{4} \log _{2} x=1 \Leftrightarrow \log _{2}^{2} x-4 \log _{2} x+4=0 \Leftrightarrow$ $\left(\log _{2} x-2\right)^{2}=0 \Leftrightarrow x=4$.
Answer: $x=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3.457 A=\operatorname{ctg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$
$3.457 A=\cot\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$ | Solution. Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}\left(-\frac{1}{2}\right) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$, $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=2, \operatorname{ctg} \beta=-\frac{1}{2}, \operatorname{ctg} 2 \beta=\frac{\operatorname{c... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.477 Find the minimum value of the expression
$$
A=\frac{\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha}{1+\sin \left(\frac{5 \pi}{2}-8 \alpha\right)} \text { for } 0<\alpha<\frac{\pi}{8}
$$ | ## Solution.
$\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha=\frac{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}{\sin 2 \alpha \cdot \cos 2 \alpha}=\frac{2 \cos 4 \alpha}{\sin 4 \alpha} \Rightarrow A=\frac{2 \cos 4 \alpha}{\sin 4 \alpha \cdot(1+\cos 8 \alpha)}=$ $=\frac{2 \cos 4 \alpha}{2 \sin 4 \alpha \cdot \cos ^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.481 Find the minimum value of the expression
$A=\frac{\operatorname{ctg} \alpha-\operatorname{tg} \alpha}{\cos 4 \alpha+1}$ for $0<\alpha<\frac{\pi}{4}$. | ## Solution.
$\operatorname{ctg} \alpha-\operatorname{tg} \alpha=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\sin \alpha \cdot \cos \alpha}=\frac{2 \cos 2 \alpha}{\sin 2 \alpha} \Rightarrow A=\frac{2 \cos 2 \alpha}{\sin 2 \alpha \cdot 2 \cos ^{2} 2 \alpha}=\frac{2}{\sin 4 \alpha}$.
Since $0<\alpha<\frac{\pi}{4}$, the mi... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 3.485 Find the maximum value of the expression
$$
A=\frac{1}{\sin ^{6} \alpha+\cos ^{6} \alpha} \text { for } 0 \leq \alpha \leq \frac{\pi}{2}
$$ | ## Solution.
$\sin ^{6} \alpha+\cos ^{6} \alpha=\sin ^{4} \alpha-\sin ^{2} \alpha \cdot \cos ^{2} \alpha+\cos ^{4} \alpha=1-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=$ $=1-\frac{3}{4} \sin ^{2} 2 \alpha=\frac{1+3 \cos ^{2} 2 \alpha}{4} \Rightarrow A=\frac{4}{1+3 \cos ^{2} 2 \alpha}$.
From this, it is clear that $A$ t... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 3.486 Find the maximum value of the expression
$A=\frac{1}{\sin ^{4} \alpha+\cos ^{4} \alpha}$ for $0 \leq \alpha \leq \frac{\pi}{2}$. | Solution. $\sin ^{4} \alpha+\cos ^{4} \alpha=1-2 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=1-\frac{\sin ^{2} 2 \alpha}{2}=\frac{1+\cos ^{2} 2 \alpha}{2} \Rightarrow$
$\Rightarrow A=\frac{2}{1+\cos ^{2} 2 \alpha}$. From this, it is clear that $A$ takes its maximum value when $1+\cos ^{2} 2 \alpha$ takes its minimum value,... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 9.303 Find integer values of $x$ that satisfy the inequality
$$
\log _{0.3}(\sqrt{x+5}-x+1)>0
$$ | Solution. The inequality is equivalent to the system:
$\left\{\begin{array}{l}\sqrt{x+5}-x+10, \\ x-\text { integer }\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{x+5}x-1, \\ x-\text { integer. }\end{array}\right.\right.$
From this, $x>0$ and, consequently, $x \geq 1$ ( $x$ - integer). Therefore,
$\left\{\... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
10.361 In triangle $A B C$, the measure of angle $A$ is twice the measure of angle $B$, and the lengths of the sides opposite these angles are 12 cm and 8 cm, respectively. Find the length of the third side of the triangle. | Solution.
A
Let $A K$ be the bisector of $\angle A$. Then
$\angle K A C = \angle A B C \Rightarrow \triangle A K C \sim \triangle A B C \Rightarrow$
$\Rightarrow \frac{A C}{B C} = \frac{K C}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.368 Calculate the length of the angle bisector of angle $A$ in triangle $A B C$ with side lengths $a=18$ cm, $b=15$ cm, $c=12$ cm. | Solution. Let $A K$ be the bisector of $\angle A$ of the given $\triangle A B C$. By Theorem 4, $\frac{C K}{B K}=\frac{15}{12} \Rightarrow \frac{18-B K}{B K}=\frac{15}{12} \Rightarrow B K=8, C K=10$.
Applying the Law of Cosines to $\triangle A B K$ and $\triangle A C K$, we have:
$\left\{\begin{array}{l}8^{2}=12^{2}+... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.369 In a triangle with a perimeter of 20 cm, a circle is inscribed. The segment of the tangent, drawn parallel to the base and enclosed between the sides of the triangle, measures 2.4 cm. Find the base of the triangle. | ## Solution.
Answer: 4 cm or 6 cm. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.377 The bisector of a triangle's angle divides the opposite side into segments of length 4 and 2 cm, and the height drawn to the same side is $\sqrt{15}$ cm. What are the lengths of the sides of the triangle, given that they are expressed as integers? | Solution. Let the bisector of angle $A$ divide side $B C$ of the given $\triangle A B C$ into segments 2 and 4. Suppose $A Bh=\sqrt{15}(h$ - the height of $\triangle A B C) \Rightarrow \sqrt{15}<n<6 \Rightarrow 4 \leq n \leq 5$.
If $n=5$, then by Heron's formula $S_{A B C}=\sqrt{\frac{21}{2} \cdot \frac{11}{2} \cdot \... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.391 On the segment $A C$ there is a point $B$, and $A B=14 \text{~cm}, B C=28 \text{~cm}$. On the segments $A B, B C$ and $A C$ as diameters, semicircles are constructed in the same half-plane relative to the boundary $\boldsymbol{A} \boldsymbol{B}$. Find the radius of the circle that is tangent to all three semicir... | ## Solution.
Let $O_{1}, O_{2}, O_{3}$ be the centers of semicircles with radii $r_{1}, r_{2}, r_{3}$ and diameters $A B, B C$, and $A C$ respectively; $P$ be the center of the circle with... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.403 In a right triangle $A B C$ ( $\angle C=90^{\circ}$ ), the altitude $CD$ is drawn. The radii of the circles inscribed in triangles $ACD$ and $BCD$ are 0.6 and 0.8 cm, respectively. Find the radius of the circle inscribed in triangle $\boldsymbol{A} \boldsymbol{\text { B }}$. | Solution. Let $x, r_{1}, r_{2}$ be the radii of the circles inscribed in $\triangle A B C$, $\triangle A C D, \triangle B C D$ respectively. $\triangle A C D \sim \triangle B C D \Rightarrow \frac{A C}{B C}=\frac{r_{1}}{r_{2}}=\frac{0.6}{0.8}$ (the radii of the circles inscribed in similar triangles are proportional to... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
### 10.417
In a right triangle $ABC\left(\angle C=90^{\circ}\right)$, a circle is inscribed, touching its sides at points $A_{1}, B_{1}, C_{1}$. Find the ratio of the area of triangle $ABC$ to the area of triangle $A_{1} B_{1} C_{1}$, if $AC=4 \text{ cm}, BC=3 \text{ cm}$. | ## Solution.
Let $O$ be the center of the circle with radius $r$ inscribed in the given $\triangle ABC$.
Then $AB=5, r=\frac{3+4-5}{2}=1$ (Theorem 1).
$S_{A_{1} B_{1} C_{1}}=S_{O A_{1} B_{1}}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.373 A computing machine was given the task to solve several problems sequentially. Registering the time spent on the assignment, it was noticed that the machine spent the same multiple of time less on solving each subsequent problem compared to the previous one. How many problems were proposed and how much time did ... | Solution. Let $n$ be the number of problems, $b$ be the time to solve the first problem, and $q$ be the common ratio of the geometric progression that represents the times to complete the problems. Using the formula for the sum of $k$ terms of a geometric progression, we have:
$$
\begin{aligned}
& \left\{\begin{array}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.384 Five people perform a certain job. The first, second, and third, working together, can complete the entire job in 327.5 hours; the first, third, and fifth together - in 5 hours; the first, third, and fourth together - in 6 hours; and the second, fourth, and fifth together - in 4 hours. In what time interval will... | Solution. Let $x_{i}$ be the productivity of the $i$-th worker, $y$ be the volume of work. Then, according to the problem,
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=\frac{y}{7.5} \\
x_{1}+x_{3}+x_{5}=\frac{y}{5} \\
x_{1}+x_{3}+x_{4}=\frac{y}{6} \\
x_{2}+x_{4}+x_{5}=\frac{y}{4}
\end{array}\right.
$$
Multiplying the... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.389 From two pieces of alloy of the same mass but with different percentage content of copper, pieces of equal mass were cut off. Each of the cut pieces was melted with the remainder of the other piece, after which the percentage content of copper in both pieces became the same. How many times smaller is the cut pie... | Solution. Let $y$ be the mass of each alloy piece, $x$ be the mass of the cut-off piece, $p$ and $q$ be the concentration of copper in the first and second pieces of the alloy. Then $p x+q(y-x)$ and $q x+p(y-x)$ are the amounts of copper in the new alloys, respectively. According to the problem,
$$
\frac{p x+q(y-x)}{y... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.391 Two cars start simultaneously from points A and B and meet at 12 o'clock noon. If the speed of the first car is doubled while the speed of the second car remains the same, the meeting will occur 56 minutes earlier. If, however, the speed of the second car is doubled while the speed of the first car remains the s... | Solution. Let $x$ and $y$ be the initial speeds (km/min) of the cars, and $2z$ be the time (in minutes) after which they met after leaving from A and B. Then $2z(x+y)$ is the distance between A and B, hence
\[
\left\{
\begin{array}{l}
2z(x+y) = (2z-56)(2x+y) \\
2z(x+y) = (2z-65)(x+2y)
\end{array}
\right.
\]
\[
\left\{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.394 Two brothers had tickets to a stadium located 10 km from their home. At first, they planned to walk to the stadium, but they changed their mind and decided to use a bicycle, agreeing that one would go by bicycle while the other would walk at the same time. After covering part of the distance, the first brother w... | Solution. Since the brothers arrived at the stadium simultaneously (i.e., spent the same amount of time on the journey), they walked (and cycled) the same distance. But in total, they walked 10 km, so each of them walked 5 km. Therefore, the time gain will be $5 \cdot 12=60$ minutes.
Answer: 1 hour. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.395 A sportsman, walking along a highway, noticed that every 6 minutes a trolleybus catches up with him and every 3 minutes a trolleybus passes him in the opposite direction. Find the intervals at which trolleybuses depart from the terminal points and how many times slower the sportsman was walking compared to the t... | Solution. Let $x$ and $y$ be the speeds of the athlete and the trolleybuses, and $z$ be the interval of the trolleybuses' movement. Then $y z$ is the distance between the trolleybuses. According to the problem, $\left\{\begin{array}{l}y z=3(x+y), \\ y z=6(y-x),\end{array} \Rightarrow x=\frac{y}{3} ; z=4\right.$.
Answe... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.399 The train was delayed at the departure station for 1 hour 42 minutes. Upon receiving the departure signal, the driver followed this schedule: on the section constituting 0.9 of the entire route from the departure station to the destination station, he maintained a speed 20% higher than the usual speed, and on 0.... | Solution. Let $z$ be the distance between stations, $x$ be the usual speed of the train.
$$
\begin{aligned}
& \text { Then } \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}-\text { is the travel time of the train, hence } \\
& \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}=\frac{z}{x}-1 \frac{42}{60} \Rightarrow \frac{z}{x}=10 .
\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.405 Two pedestrians set out simultaneously from A to B and from B to A. When the first had walked half the distance, the second had 24 km left to walk to the end, and when the second had walked half the distance, the first had 15 km left to walk to the end. How many kilometers will the second pedestrian have left to... | Solution. Let $x$ be the distance between A and B, $y$ be the ratio of the speeds of the first and second pedestrians. Then
$\left(\frac{x}{2}\right):(x-24)=y,(x-15):\left(\frac{x}{2}\right)=y$, from which $\frac{x}{2(x-24)}=\frac{2(x-15)}{x} \Rightarrow x=40$ (the value $x=12$ is not suitable, as $x>24) \Rightarrow y=... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.408 A material particle $m_{1}$ oscillates between points $A$ and $B$, which are 3.01 m apart. The particle's speed is constant, and it does not stop at the endpoints. After 11 s from the departure of particle $m_{1}$ from point $A$, another particle $m_{2}$ starts moving from point $B$ with a constant but lower spe... | Solution. Let $x$ and $y$ be the speeds (cm/s) of particles $m_{1}$ and $m_{2}$. By the time particle $m_{2}$ exits, the distance between $m_{1}$ and $m_{2}$ is $301-11 x$ cm, hence $301-11 x=10(x+y)$. Let $C$ be the point of the first meeting of the particles. Then $B C=10 y$, and therefore, by the time of the second ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.410 Along the sides of a right angle, towards the vertex, two spheres with radii of 2 and 3 cm are moving, with the centers of these spheres moving along the sides of the angle at unequal but constant speeds. At a certain moment, the center of the smaller sphere is 6 cm from the vertex, and the center of the larger ... | Solution. Let $x$ and $y$ be the speeds of the balls. After 1 second, the distances from the centers of the balls to the corner are $6-x$ and $16-y$, and after 3 seconds, they are $6-3x$ and $16-3y$. By the Pythagorean theorem, we have: $\left\{\begin{array}{l}(6-x)^{2}+(16-y)^{2}=13^{2}, \\ (6-3 x)^{2}+(16-3 y)^{2}=5^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.415 Two cyclists set out simultaneously from points $A$ and $B$ towards each other. Four hours after their meeting, the cyclist who had set out from $A$ arrived at $B$, and nine hours after their meeting, the cyclist who had set out from $B$ arrived at $A$. How many hours was each cyclist on the road? | Solution. Let $x$ and $y$ be the travel times of the first and second cyclists, respectively. Then $x-4=y-9$ is the time each of them traveled until they met. Since each of them traveled a distance after the meeting equal to the distance the other had traveled before the meeting, $\frac{4}{y-9}=\frac{x-4}{9}$.
$$
\tex... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.426 Two pumps were installed to fill a swimming pool with water. The first pump can fill the pool 8 hours faster than the second one. Initially, only the second pump was turned on for a time equal to twice the amount of time it would take to fill the pool if both pumps were working simultaneously. Then, the first pu... | Solution. Let $x$ and $y$ be the pump efficiencies, and $z$ be the volume of work. Then $\frac{z}{x+y}$ is the time it takes to fill the pool when both pumps are working simultaneously. According to the conditions of the problem,
$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=8, \\ \frac{2 z}{x+y} \cdot y+1.5(x+y)=z\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.428 A batch of identical parts was processed on three machines of different designs in the following sequence: first, only the first machine worked for as many hours as it would take for the second and third machines to complete the entire job together; then, only the second machine worked for as many hours as it wo... | Solution. Let $x, y, z$ be the efficiencies of the machines, and $V$ be the volume of work. According to the problem, $\frac{V}{y+z}$ is the time taken by the first machine, $\frac{V}{x+z}$ is the time taken by the second machine, and $\frac{V}{x+y}$ is the time taken by the third machine.
From this, $\frac{V x}{y+z}+... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.430 A passenger can travel from Moscow to city $N$ by train. In this case, he will be on the way for 20 hours. If, however, he waits for the departure of the plane (and he will have to wait more than 5 hours after the train departs), the passenger will reach city $N$ in 10 hours, including the waiting time. How many... | Solution. Let $x$ and $y$ be the speeds (km/h) of the train and the airplane, respectively, and $z$ be the waiting time for the airplane to depart. Then $\frac{8}{9} y$ is the distance the airplane will fly until it meets the train. According to the problem, $\frac{8}{9} y = x \left(z + \frac{8}{9}\right)$. Additionall... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.437 What whole positive number should 180 be divided by, so that the remainder is $25\%$ of the quotient? | Solution. Let $n$ be the required number, $m$ be the remainder of the division of 180 by $n$, and $k$ be the quotient.
Then $\left\{\begin{array}{l}180=n \cdot k+m, \\ 4 m=k, \\ m4 m n>4 m^{2}$, i.e., $m^{2}<45 \Rightarrow m=4 \Rightarrow 4 n+1=45 \Rightarrow n=11$.
Answer: 11. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.438 By mixing $2 \mathrm{~cm}^{3}$ of three substances, 16 g of the mixture was obtained. It is known that $4 \mathrm{r}$ of the second substance occupies a volume that is $0.5 \mathrm{~cm}^{3}$ larger than $4 \mathrm{r}$ of the third substance. Find the density of the third substance, given that the mass of the sec... | Solution. Let $x, y, z$ be the densities of three substances. Then, according to the condition $\left\{\begin{array}{l}2 x+2 y+2 z=16, \\ \frac{4}{y}-\frac{4}{z}=0.5, \quad \Rightarrow z=4 . \\ y=2 x\end{array}\right.$
Answer: 4 g $/ \mathrm{cm}^{3}$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.450 If at the beginning of the time measurement there were $m_{0}$ g of substance $A$ and $2 m_{0}$ g of substance $B$, then after any number $t$ years, as a result of the radioactive decay of these substances, there will remain respectively $\boldsymbol{m}=\boldsymbol{m}_{0} \cdot 2^{-\lambda_{1} t}$ and $M=2 m_{0}... | Solution. Let $x$ and $2x$ be the half-lives of the second and first substances, respectively. According to the conditions,
$$
\left\{\begin{array}{l}
m_{0} \cdot 2^{-\lambda_{1} \cdot 2x} = \frac{m_{0}}{2}, \\
2m_{0} \cdot 2^{-\lambda_{2} \cdot x} = m_{0}, \\
m_{0} \cdot 2^{-\lambda_{1} \cdot 20} + 2m_{0} \cdot 2^{-\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 5.080 Find the largest term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$. | Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$.
Then $\frac{T_{k+1}}{T_{k}}=\frac{C_{20}^{k} \cdot(\sqrt{5})^{20-k} \cdot(\sqrt{2})^{k}}{C_{20}^{k-1} \cdot(\sqrt{5})^{21-k} \cdot(\sqrt{2})^{k-1}}=\frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{21-k}{k}$.
The ratio $\frac{T_{k+1}}{T_{k}}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Senderovv B.A.
Find the smallest natural number that cannot be represented in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are natural numbers. | $1=\frac{4-2}{4-2}, \quad 2=\frac{8-4}{4-2}, 3=\frac{8-2}{4-2}, 4=\frac{16-8}{4-2}, 5=\frac{32-2}{8-2}$
$6=\frac{16-4}{4-2}, 7=\frac{16-2}{4-2}, 8=\frac{32-16}{4-2}, 9=\frac{128-2}{16-2}, 10=\frac{64-4}{8-2}$.
Assume that $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$. Without loss of generality, let $a>b, c>d$. Denote $m=a-b,... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the system in positive numbers:
$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$ | First, note that if one of the unknowns is equal to one, then the others are also equal to one.
Indeed, let $x=1$. Then $z=1^{\mathrm{y}}=1, y=z^{\mathrm{x}}=1^{1}=1$. Suppose there exists another solution besides $(1,1,1)$. Let's first consider $x>1$. Then $z=x^{\mathrm{y}}>1, y=z^{\mathrm{x}}>1$. Therefore, $z=x^{\m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Methods for solving problems with parameters]
[Investigation of a quadratic trinomial]
For what positive value of $p$ do the equations $3 x^{2}-4 p x+9=0$ and $x^{2}-2 p x+5=0$ have a common root? | The common root of the given equations must also be a root of the equation $\left(3 x^{2}-4 p x+9\right)-3\left(x^{2}-2 p x+5\right)=0 \Leftrightarrow$ $2 p x-6=0$. Therefore, it equals $3 / p$. Substituting, for example, into the second equation, we get $9 / p^{2}=1$, from which $p=3$.
## Answer
For $p=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Initially on the computer screen - some prime number. Every second, the number on the screen is replaced by the number obtained from the previous one by adding its last digit, increased by 1. What is the maximum time it will take for a composite number to appear on the screen?
# | Let the initial number on the screen be 2, then we get the following sequence: $2-5-11-13-17-25$. The sixth number is composite, so in this case, it will take 5 seconds. We will prove that in other cases, it will take no more than 5 seconds.
Indeed, if the number on the screen was an odd prime number not ending in 9, ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On the board after the class, there was a note left:
"Calculate $t(0)-t(\pi / 5)+t\left((\pi / 5)-t(3 \pi / 5)+\ldots+t\left({ }^{8 \pi} / 5\right)-t(9 \pi / 5)\right.$, where $t(x)=\cos 5 x+* \cos 4 x+* \cos 3 x+* \cos 2 x+$ $*^{\cos x}+* "$.
Seeing it, a math student told a friend that he could calculate this sum e... | Let $t_{k}(x)=\cos k x$ for $k=0,1,2,3,4$. If we consider a similar sum for $t_{5}$ instead of $t$, we will get 10 as the result. We will check that for all other $t_{k}$ the sums are equal to 0. For $k=0$ this is obvious. For $k=1,2,3,4$ we arrive at the same equality
$\cos 0+\cos (2 \pi / 5)+\cos ((\pi / 5) \cos (6 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bogosnov I.I.
Numbers, either red or blue, are arranged in a circle. Each red number is equal to the sum of its neighboring numbers, and each blue number is equal to the half-sum of its neighboring numbers. Prove that the sum of the red numbers is zero.
# | Let $a, b, c$ - be three consecutive numbers. If $b$ is red, then $b=a+c$, and if $b$ is blue, then $2 b=a+c$. We will write such equations for all triples of consecutive numbers and add them. On the right side, we will get twice the sum of all numbers, and on the left side - the sum of red numbers plus twice the sum o... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Rushkin C.
On the surface of a cube, a closed eight-segment broken line is drawn, the vertices of which coincide with the vertices of the cube.
What is the minimum number of segments of this broken line that can coincide with the edges of the cube?
# | The segments of the broken line are either edges of the cube or diagonals of its faces. To do this, let's color the vertices of the cube in two colors in a checkerboard pattern (see fig.). Note that an edge of the cube connects vertices of different colors, while a diagonal connects vertices of the same color. For the ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Rubanov I.S.
In five pots standing in a row, Rabbit poured three kilograms of honey (not necessarily into each and not necessarily equally). Winnie-the-Pooh can take any two adjacent pots. What is the maximum amount of honey that Winnie-the-Pooh can guarantee to eat?
# | Evaluation. Let's assume Winnie-the-Pooh cannot take at least a kilogram of honey. This means that in any pair of adjacent pots, there is less than a kilogram of honey. This is true for both the two rightmost pots and the two leftmost pots. However, then in the middle pot, there must be more than a kilogram of honey (o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Auto: : Bogosnov I.I.
All integers from -33 to 100 inclusive were arranged in some order, and the sums of each pair of adjacent numbers were considered. It turned out that there were no zeros among them. Then, for each such sum, the number reciprocal to it was found. The obtained numbers were added. Could the result b... | Let's consider an example of such an arrangement. Consider the sequence $100, -33, 99, -32, \ldots, 34, 33$. Then, if the first number of the pair is in an odd position, the sum is 67. If it is in an even position, the sum is 66. The reciprocals will be $\frac{1}{67}$ and $\frac{1}{66}$, respectively, with the first oc... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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