problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
1. In triangle $\mathrm{ABC}$, angle $\mathrm{C}$ is a right angle, and $\mathrm{CD}$ is the altitude. Find the length of the radius of the circle inscribed in triangle $\mathrm{ABC}$, if the lengths of the radii of the circles inscribed in triangles $\mathrm{ACD}$ and $\mathrm{BCD}$ are 6 and 8, respectively. | Answer: 10
## №4: Progression.
The numbers $5 \mathrm{x}-\mathrm{y} ; 2 \mathrm{x}+3 \mathrm{y} ; \mathrm{x}+2 \mathrm{y}$ are consecutive terms of an arithmetic progression. The numbers $(\mathrm{y}+1)^{2} ; \mathrm{xy}+1 ;(\mathrm{x}-1)^{2}$ are consecutive terms of a geometric progression. Find the numbers x and y... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,945 |
3. Find all values of the parameter $a$ for which the equation
$$
\frac{4 a \sin ^{2} t+4 a(1+2 \sqrt{2}) \cos t-4(a-1) \sin t-5 a+2}{2 \sqrt{2} \cos t-\sin t}=4 a
$$
has exactly two distinct solutions in the interval $(0 ; \pi / 2)$. | Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1, x \in(0 ; 1), y \in(0 ; 1)$. Then the equation will have the form
$$
\begin{gathered}
\frac{4 a y^{2}+4 a(1+2 \sqrt{2}) x-4(a-1) y-5 a+2}{2 \sqrt{2} x-y}=4 a \\
4 a y^{2}+4 a(1+2 \sqrt{2}) x-4(a-1) y-5 a+2=4 a 2 \sqrt{2} x-4 a y \\
4 a y^{2}+4 a x+4 y-5 a+2=0, \quad 2 ... | \in(6;18+24\sqrt{2})\cup(18+24\sqrt{2};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,946 |
3. Find all values of the parameter $a$ for which the equation
$$
\frac{4 a \cos ^{2} t+4 a(2 \sqrt{2}-1) \cos t+4(a-1) \sin t+a+2}{\sin t+2 \sqrt{2} \cos t}=4 a
$$
has exactly two distinct solutions in the interval $(-\pi / 2 ; 0)$.
(12 points) | Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1, x \in(0 ; 1), y \in(-1 ; 0)$. Then the equation will have the form
$$
\begin{gathered}
\frac{4 a x^{2}+4 a(2 \sqrt{2}-1) x+4(a-1) y+a+2}{2 \sqrt{2} x+y}=4 a \\
4 a x^{2}+4 a(2 \sqrt{2}-1) x+4(a-1) y+a+2=4 a 2 \sqrt{2} x+4 a y \\
4 a x^{2}-4 a x-4 y+a+2=0, \quad 2 \sqrt... | \in(-\infty;-18-24\sqrt{2})\cup(-18-24\sqrt{2};-6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,947 |
3. Find all values of the parameter $a$ for which the equation
$$
\frac{|\cos t-0.5|+|\sin t|-a}{\sqrt{3} \sin t-\cos t}=0
$$
has at least one solution on the interval $[0 ; \pi / 2]$. Specify the number of distinct solutions of this equation on the interval $[0 ; \pi / 2]$ for each found value of the parameter $a$. ... | Solution. Let $x=\cos t, y=\sin t, x^{2}+y^{2}=1$, $x \in[0 ; 1], y \in[0 ; 1]$. Then the equation will have the form
$$
\frac{|x-0.5|+|y|-a}{\sqrt{3} y-x}=0
$$
$$
|x-0.5|+|y|-a=0, \sqrt{3} y-x \neq 0
$$
In the end, we have the system:
$\left\{\begin{array}{c}|x-0.5|+|y|=a, \\ \sqrt{3} y-x \neq 0, \\ x^{2}+y^{2}=1,... | \in[0.5;1.5],for\in[0.5;\sqrt{3}/2]\cup(\sqrt{2}-0.5;1.5]theequationhas1solution,for\in(\sqrt{3}/2;\sqrt{2}-0.5)theequationhas3solutions,for | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,948 |
6. (20 points) Each of the two baskets contains white and black balls, and the total number of balls in both baskets is 25. One ball is randomly drawn from each basket. It is known that the probability that both drawn balls will be white is 0.54. Find the probability that both drawn balls will be black. | Answer: 0.04.
Solution. Let in the $i$-th basket there be $n_{i}$ balls, among which $k_{i}$ are white, $i=1,2$. Then $\frac{k_{1}}{n_{1}} \cdot \frac{k_{2}}{n_{2}}=0.54=\frac{27}{50}$. Therefore, for some natural number $m$, the equalities $k_{1} \cdot k_{2}=27 m, n_{1} \cdot n_{2}=50 m$ hold. One of the numbers $n_{... | 0.04 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,952 |
1. Find the value of the expression $\left(\left(\frac{3}{a-b}+\frac{3 a}{a^{3}-b^{3}} \cdot \frac{a^{2}+a b+b^{2}}{a+b}\right)\right.$ ? $\left.\frac{2 a+b}{a^{2}+2 a b+b^{2}}\right) \cdot \frac{3}{a+b}$ when $a=2023, b=2020$ | Solution:
$$
\begin{aligned}
& \left(\left(\frac{3}{a-b}+\frac{3 a}{(a-b)(a+b)}\right) \text { back } \frac{2 a+b}{(a+b)^{2}}\right) \cdot \frac{3}{a+b}=\left(\left(\frac{3(a+b)+3 a}{(a-b)(a+b)}\right) \cdot \frac{(a+b)^{2}}{2 a+b}\right) \cdot \frac{3}{a+b}=\left(\frac{3(2 a+b)}{(a-b)} \cdot\right. \\
& \left.\frac{(... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,953 |
2. Inside the square $A B C D$ with a side length of 5, there is a point $X$. The areas of triangles $A X B, B X C$, and $C X D$ are in the ratio $1: 5: 9$. Find the sum of the squares of the distances from point $X$ to the sides of the square. | Solution:
Let the side of the square be $a=5$, and the distances from point $X$ to sides $AB$, $BC$, $CD$, and $DA$ be $h_{1}$, $h_{2}$, $h_{3}$, and $h_{4}$, respectively. Since the area of a triangle is $S=\frac{1}{2} a h$, we conclude that $h_{1}: h_{2}: h_{3}=1: 5: 9$ or $h_{1}=x$, $h_{2}=5 x, h_{3}=9 x$. However,... | 33 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,954 |
3. The boat traveled 165 km upstream against the current and then returned. On the return trip, it took 4 hours less than the trip there. Find the boat's own speed if the river's current speed is 2 km/h. | # Solution:
$$
\begin{aligned}
& v_{\text {down}}=v_{\mathrm{c}}+v_{\text {current}} \\
& v_{\text {up}}=v_{\mathrm{c}}-v_{\text {current}}
\end{aligned}=>v_{\text {down}}-v_{\text {up}}=2 v_{\text {current}}=2 \cdot 2=4
$$
Then $\quad v_{\text {up}}=x$ km/h, $v_{\text {down}}=x+4$ km/h
Preliminary (online) stage of... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,955 |
4. In the box, there are 3 white cups, 3 red cups, and 2 black cups. Sonya took out 5 cups at random. What is the probability that she took out 2 white, 2 red, and 1 black cup? (Round the answer to the nearest hundredth). | Solution:
Let white cups be denoted as W, red cups as R, and black cups as B.
1) The probability of drawing 2 white, then 2 red, and then 1 black cup, in the order WWR RB: $\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{2}{4}=\frac{3}{280}$
2) The number of ways to rearrange the letter... | 0.32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,956 |
5. . Find all values of the parameter $a$ for which the equation
$$
\left(\left|\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}\right|-a\right) \cdot|4 a-3 x-19|=0
$$
has one solution. In your answer, write the largest value of the parameter $a$ | Solution:
Simplify $\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}=\frac{a x^{2}-x^{2}+(-a x+x)+(-12 a+12)}{a(x+3)-(x+3)}=$
$=\frac{x^{2}(a-1)-x(a-1)-12(a-1)}{(x+3)(a-1)}=\frac{(a-1)\left(x^{2}-x-12\right)}{(x+3)(a-1)}=$
Preliminary (correspondence) online stage of the "Step into the Future" School Students' Olympi... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,957 |
6. In the right-angled triangle $ABC$ with the right angle at vertex $B$, on the longer leg $BC$, a segment $BD$ equal to the shorter leg is laid off from point $B$. Points $K$ and $N$ are the feet of the perpendiculars dropped from points $B$ and $D$ to the hypotenuse, respectively. Find the length of segment $BK$, if... | Solution:
Drop a perpendicular from point $D$ to segment $B K$, let the foot of this perpendicular be point $M$. Then the right triangles $A K B$ and $B M D$ are equal by hypotenuse and acute angle, therefore, $B M=A K=2$. Quadrilateral $M K N D$ is a rectangle and $M K=N D=2$, then $B K=B M+M K=2+2=4$.
Answer: 4. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,958 |
7. The production of x thousand units of a product costs $q=0.5 x^{2}-2 x-10$ million rubles per year. At a price of p thousand rubles per unit, the annual profit from selling this
Preliminary (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
product (in million rubl... | # Solution:
Let the annual profit $f(x)=p x-q=p x-0.5 x^{2}+2 x+10=$
$-0.5 x^{2}+x(p+2)+10=-0.5\left(x^{2}-2 x(p+2)-20\right)=$
$=-0.5\left(x^{2}-2 x(p+2)+(p+2)^{2}-(p+2)^{2}-20\right)=$
$=-0.5(x-(p+2))^{2}+(p+2)^{2} / 2+10$
The quadratic trinomial $\mathrm{f}(\mathrm{x})$ reaches its maximum value at $\mathrm{x}=... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,959 |
8. In triangle $A B C$, sides $A B, A C, B C$ are equal to 5, 6, and 7 respectively. On the median $A M$, segment $A K$ equal to 2 is laid off. Find the ratio of the areas of triangles $A B K$ and $A C K$. Write the answer as $\frac{S_{A B K}}{S_{A C K}}$. | # Solution:
The median of a triangle divides the triangle into two equal-area (equal in area) triangles. In triangle $ABC$, the areas of triangles $ABM$ and $ACM$ are equal because $AM$ is its median. In triangle $KBC$, the segment $KM$ is the median, so the areas of triangles $KBM$ and $KCM$ are equal. $S_{ABK} = S_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,960 |
9.1. Solve the equation $\left|x^{2}-100\right|=2 x+1$. | Answer. $x_{1}=1+\sqrt{102}, x_{2}=9$. Solution. If $x^{2}-100 \geq 0$, i.e., under the condition $|x| \geq 10$, we have the equation $x^{2}-2 x-101=0$, its roots are $x=1 \pm \sqrt{102}$. The root $1+\sqrt{102}$ satisfies the condition $|x| \geq 10$, while the root $1-\sqrt{102}$ does not. If $|x|<10$, then we have th... | x_{1}=1+\sqrt{102},x_{2}=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,962 |
9.3. Given two positive numbers. It is known that their sum, as well as the sum of their cubes, are rational numbers. Can we assert that a) the numbers themselves are rational? b) the sum of their squares is a rational number? | Answer. a) No. b) Yes, it is possible. Solution. a) As an example, we can take the numbers $a=2+\sqrt{2}$, $b=2-\sqrt{2}$, then $a+b=4$ and it is easy to calculate (using the formula for the cube of a sum): $a^{3}+b^{3}=40$ b) Let the numbers $x=a+b$ and $y=a^{3}+b^{3}$ be rational. Then $x^{3}=a^{3}+b^{3}+3 a b(a+b)=y... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,964 |
9.4. Inside triangle $ABC$, an arbitrary point $M$ is taken. Through the vertices of the triangle and this point, three segments are drawn until they intersect the opposite sides. Prove that among these segments, two can be chosen such that point $M$ divides one of them (measured from the vertex) in the ratio $\geq 2$,... | Solution. Let the areas of triangles $AMB$, $BMC$, and $AMC$ be ordered as follows:
$$
S_{AMB} \leq S_{BMC} \leq S_{AMC}
$$
Then $S_{AMB} \leq S / 3$, and $S_{AMC} \geq S / 3$, where $S$ is the area of triangle $ABC$. Suppose, for the sake of contradiction, that at point $M$ all the specified segments are divided in ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,965 |
9.3. How many solutions in integers $x, y$ does the inequality $|y-x|+|3 x-2 y| \leq a$ have a) for $a=2$; b) for $a=20$? | Answer: a) 13; b) 841. Solution. Let $m=y-x, n=3 x-2 y$, then (expressing $x$, y from these equations) we get $x=2 m+n$ and $y=3 m+n$ and thus, for any integers m, n there correspond integers $x, y$ (and vice versa), i.e., there is a one-to-one correspondence between ordered pairs (m.n) and ( $x, y$ ). Therefore, we ne... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,968 |
9.1. In a three-digit number, the first digit was crossed out, resulting in a two-digit number. If the original number is divided by this two-digit number, the quotient is 9 and the remainder is 8. Find the original number. (Provide all possible solutions.) | Answer: 4 possible numbers: $224 ; 449 ; 674 ; 899$. Solution. Let $x$ be the first digit of the original number, $y$ be the two-digit number after erasing $x$. Then we have the equation $100 x+y=9 y+8 \Leftrightarrow 2 y+2=25 x$. Therefore, $x$ is an even number: $x=2 z$ for some $z(z \leq 4)$. Hence, $y+1=25 z$, and ... | 224,449,674,899 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,970 |
9.2. In trapezoid $A B C D$, point $N$ is the midpoint of the lateral side $C D$. It turned out that $\angle A N B=90^{0}$. Prove that $A N$ and $B N$ are the bisectors of angles $A$ and $B$ respectively. | Let $M$ be the midpoint of $AB$. Consider $\triangle BMN$. We have $MB=MN$ (by the property of the median of the right triangle $ANB$), so $\angle NBM = \angle BNM$. Furthermore, $\angle BNM = \angle NBC$, since the midline of the trapezoid is parallel to the bases. Therefore, $\angle NBM = \angle NBC$. Similarly, we o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,971 |
9.3. $\quad$ Find all quadratic trinomials $P(x)=x^{2}+b x+c$ such that $P(x)$ has integer roots, and the sum of its coefficients (i.e., $1+b+c$) is 10. | Answer: $(x-2)(x-11),(x-3)(x-6), x(x+9),(x+4)(x+1) . S o l u t i o n . ~ L e t ~ x_{1}<x_{2}$ be the roots. Then $P(x)=\left(x_{1}\right.$ $-x)\left(x_{2}-x\right)$ and $P(1)=10=\left(x_{1}-1\right)\left(x_{2}-1\right)$. From the factorizations of the number 10 into two factors, we obtain the possible values of the fac... | (x-2)(x-11),(x-3)(x-6),x(x+9),(x+4)(x+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,972 |
7.1. A column of cars is moving along the highway at a speed of 80 km/h with a distance of 10 m between the cars. Passing a speed limit sign, all cars reduce their speed to 60 km/h. What will be the distance between the cars in the column after the sign? | Answer: 7.5 m. Solution. Let the speeds be $v_{1}=80$ km/h, $v_{2}=60$ km/h, and the distance $a=10$ m $=0.01$ km. The second car passes the sign later than the first by $\frac{a}{v_{1}}$ (hours). During this time, the first car will travel a distance of $v_{2} \cdot \frac{a}{v_{1}}=7.5 \text{ m}$, with this distance t... | 7.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 21,974 |
7.2. In 7a grade, there are 33 students. At the beginning of the school year, two clubs were organized in the class. According to school rules, a club can be organized if at least $70 \%$ of all students in the class sign up for it. What is the smallest number of students who could have signed up for both clubs simulta... | Answer: 15 students. Solution. In each club, there should be no less than $33 \cdot 0.7=23.1$ people, which means no less than 24 people. Let $n_{1}, n_{2}$ be the number of students who signed up for the first and second club, respectively, and $n$ be the number of students who signed up for at least one club. Obvious... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,975 |
7.3. In a three-digit number, the first digit was crossed out to obtain a two-digit number. If the original number is divided by the obtained number, the quotient is 8 and the remainder is 6. Find the original number. | Answer: 342. Solution. Let $a$ be the first digit of the original number, and $b$ be the two-digit number formed by the last two digits. According to the condition, $100 a+b=8 b+6 \Leftrightarrow 7 b=2(50 a-3)=2 \cdot 49 a+2(a-3)$. Thus, the number $a-3$ must be divisible by 7. Considering that $0<a \leq 9$, we get tha... | 342 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,976 |
7.4. Find the smallest natural number with the sum of its digits equal to 2014. | Answer: 799... 9 (223 nines). Solution. This answer is intuitively clear: you need to use as many nines as possible, and write the missing digit to 2014 first. Let's justify this strictly. If the desired number had fewer than 224 digits, the sum of the digits would not exceed $223 \cdot 9=2007$; if the first digit of t... | 799..9(223nines) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,977 |
7.5. There are 200 matches. How many ways are there to form, using all the matches, a square and (separately) an equilateral triangle? (Different ways differ in the sizes of the square and the triangle). | Answer: 16. Solution. Let $x$ (matches) be the side length of the square, and $y$ (matches) be the side length of the triangle. Then $4 x+3 y=200 \Leftrightarrow 3 y=4(50-x)$. Thus, for divisibility by 3, we need to consider all natural numbers up to 50 as $x$ that give a remainder of 2 when divided by 3 (i.e., the sam... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,978 |
8.4. In triangle $A B C$, the bisector $A M$ is perpendicular to the median $B K$. Find the ratios $B P: P K$ and $A P: P M$, where $P$ is the point of intersection of the bisector and the median. | Answer: $\quad B P: P K=1, A P: P M=3: 1$. Solution. Triangles $A B P$ and $A K P$ are equal (side $A P$ is common, and the adjacent angles are equal by condition). Therefore, $A B=A K=K C, B P=P K$. Hence, triangles $A B M$ and $A K M$ are equal (by two sides and the angle $\frac{\angle A}{2}$ between them). Then $S_{... | BP:PK=1,AP:PM=3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,981 |
9.2. The numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x+y=a \\
x^{2}+y^{2}=-a^{2}+2
\end{array}\right.
$$
What is the greatest and the least value that the product $x y$ can take? | Answer: The maximum value is $1 / 3$, the minimum is -1.
Solution. We have $2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=a^{2}-\left(-a^{2}+2\right)=2\left(a^{2}-1\right)$, i.e., $x y=a^{2}-1$. The system $\left\{\begin{array}{l}x+y=a \\ x y=a^{2}-1\end{array}\right.$ is equivalent to the original system (since from it, u... | [-1;\frac{1}{3}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,983 |
9.3. How many points on the hyperbola $y=\frac{2013}{x}$ have integer coordinates $(x ; y)$? | Answer: 16. Solution. Integer points in the first quadrant correspond to the natural divisors of the number $2013=3 \cdot 11 \cdot 61$. The number of such divisors is 8 (they can be listed directly or the formula $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{k}+1\right)$ for the number of natu... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,984 |
9.4. In triangle $A B C$, the bisector $A M$ is perpendicular to the median $B K$. Find the ratios $B P: P K$ and $A P: P M$, where $P$ is the point of intersection of the bisector and the median. | Answer: $\quad B P: P K=1, A P: P M=3: 1$. See problem 8.4.
## 10th grade | BP:PK=1,AP:PM=3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,985 |
10.3. Does there exist a number $x$ for which both numbers $(\sin x+\sqrt{2})$ and ( $\cos x-\sqrt{2}$ ) are rational? | Answer: Does not exist. Solution. Suppose, for the sake of contradiction, that $\sin x+\sqrt{2}=p$, $\cos x-\sqrt{2}=q$, where $p$ and $q$ are rational numbers. Then $1=\sin ^{2} x+\cos ^{2} x=(p-\sqrt{2})^{2}+(q+\sqrt{2})^{2}=$ $\left(p^{2}+q^{2}+4\right)-2(q-p) \sqrt{2}$. If $q-p \neq 0$, then this immediately leads ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,988 |
10.4. Given a rectangle for which the numerical value of the area is greater than the perimeter. Prove that the perimeter of the rectangle is greater than 16. | Solution. Let $a, b$ be the sides of the rectangle. From the condition of the problem, $ab > 2a + 2b$ $\Leftrightarrow (a-2)(b-2) > 4(*)$
First, let's check that both factors $(a-2)$ and $(b-2)$ are positive. Indeed, otherwise from (*) it follows that $a-2 < 0$ or $b-2 < 0$, which contradicts the positivity of the sid... | P>16 | Geometry | proof | Yes | Yes | olympiads | false | 21,989 |
11.1. Solve the equation $2 \cos ^{2} x+\sqrt{\cos x}=3$. | Answer: $\quad x=2 \pi k, k \in Z$.
Solution. Since $2 \cos ^{2} x+\sqrt{\cos x} \leq 2 \cdot 1+1=3$, the equality can only hold if $\cos x=1$, from which the answer follows. | 2\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,990 |
11.2. A rhombus is inscribed in the given rectangle (each side of the rectangle contains one vertex of the rhombus). Prove that the ratio of the diagonals of the rhombus is equal to the ratio of the sides of the rectangle. | Solution. First, we prove that the center of the rhombus and the center of the rectangle coincide. Let $\mathrm{O}$ be the center of the rhombus. Since the diagonals of a parallelogram bisect each other, the distance from O to the opposite sides of the rectangle is equal. Therefore, point O lies at the intersection of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,992 |
11.3. Find all values of the parameter $a$ for which the equation has a unique solution: $\boldsymbol{a}$ ) $\left.a x^{2}+\sin ^{2} x=a^{2}-a ; \boldsymbol{6}\right) a x^{2}+\sin ^{2} x=a^{3}-a$. | Answer: a) $a=1$; b) $a=1, a=-1$. Solution. a) See problem 10.3. b) As in the solution of problem 10.3, the question reduces to checking the values of $a$ for which $a^{3}-a=0$. For $a=0$ and $a=1$, the result has already been obtained in problem 10.3. For $a=-1$, we have the equation $\sin ^{2} x=x^{2} \Leftrightarrow... | )=1;b)=1,=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,993 |
11.4. Prove that the set of rational numbers $x$ for which the number $\sqrt{x-1}+\sqrt{4 x+1}$ is rational, is infinite. | Solution. We will prove that there exists an infinite set of numbers $x$ for which $x-1=a^{2}$ and $4 x+1=b^{2}$ for some rational numbers $a, b$. From these equations, we have $b^{2}-4 a^{2}=5 \Leftrightarrow (b-2 a)(b+2 a)=5$. Let $b-2 a=t$ and $b+2 a=\frac{5}{t}$, where $t$ is a rational number. Then $a=\frac{5-t^{2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,994 |
9.1. Prove that for any natural $n$ the number $n^{3}+6 n^{2}+12 n+16$ is composite. | Solution. The result follows from the formulas for the cube of a sum and the sum of cubes: $n^{3}+6 n^{2}+12 n+16=$ $(n+2)^{3}+8=(n+4)\left((n+2)^{2}-2(n-2)+4\right)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,995 |
9.2. a) Given the quadratic equation $x^{2}-9 x-10=0$. Let $a$ be its smallest root. Find $a^{4}-909 a$. b) For the quadratic equation $x^{2}-9 x+10=0$, where $b$ is the smallest root, find $b^{4}-549 b$. | Answer: a) 910; b) -710. Solution. a) Let's solve the problem in general. Let $x^{2}-c x+d=0-$ be a quadratic equation, and $a-$ be its root. Then $a^{4}=(c a-d)^{2}=c^{2} a^{2}-2 a c d+d^{2}=$ $c^{2}(c a-d)-2 a c d+d^{2}=a\left(c^{3}-2 c d\right)+d^{2}-c^{2} d$. Therefore, $a^{4}-\left(c^{3}-2 c d\right) a=d^{2}-c^{2}... | 910 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,996 |
9.5. In a square with side 1, 53 points are marked, four of which are the vertices of the square, and the remaining 49 points (arbitrary) lie inside. Prove that there exists a triangle with marked vertices having an area of no more than 0.01. | Solution. Let $n$ points be marked in a square: 4 vertices of the square and $n-4$ points inside ($n>4$). We will prove by induction that the square can be divided into triangles with marked vertices, and the number of triangles is $2n-6$ (it is not difficult to arrive at such an expression by considering the values $n... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,999 |
11.1. Find all values of the parameter $a$ for which the equation $\left|x^{3}+1\right|=a(x+1)$ has three roots. | Answer: $3 / 40$ for all $x$, the equation can be written in the form $|x+1|\left(x^{2}-x+1\right)=a(x+1)$. The root $x=-1$ exists for any $a$. For $x \neq-1$, reduce the equation by $(x+1)$. We get the equation $x^{2}-x+1=a \cdot \operatorname{sign}(x+1)$, where $\operatorname{sign}($ ) denotes the sign of the number ... | \frac{3}{4}<3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,000 |
11.3. On the lateral edges $A D, B D$ and $C D$ of the tetrahedron $A B C D$, points $A_{1}, B_{1}, C_{1}$ are taken, respectively, such that the plane $A_{1} B_{1} C_{1}$ is parallel to the base $A B C$. Point $D_{1}$ lies in the base. Prove that the volume of the tetrahedron $A_{1} B_{1} C_{1} D_{1}$ does not exceed ... | Solution. From the condition of parallelism of planes $A_{1} B_{1} C_{1}$ and $A B C$, it follows that the tetrahedra $A_{1} B_{1} C_{1} D$ and $A B C D$ are similar. Let $x$ be the similarity coefficient of these tetrahedra ($x<1$) and $h-$ the height of the tetrahedron $A B C D$ from point $D$. Then $h-x h-$ is the h... | \frac{4}{27} | Geometry | proof | Yes | Yes | olympiads | false | 22,002 |
11.5. Does there exist a real $\alpha$ such that both numbers $2 \sin \alpha + \sqrt{3}$ and $2 \cos \alpha - \sqrt{3}$ are rational? | Answer: There is no Solution. Suppose, to the contrary, that for some $\alpha$ the following holds: $2 \sin \alpha+\sqrt{3}=a$ and $2 \cos \alpha-\sqrt{3}=b$, where $a, b-$ are rational numbers. Then $2 \sin \alpha=a-\sqrt{3}, 2 \cos \alpha=b+\sqrt{3}$. Squaring and adding these relations, we get $4=(a-\sqrt{3})^{2}+(b... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,004 |
7.1 Vasya and Tolya exchanged badges. Before the exchange, Vasya had 5 more badges than Tolya. After Vasya exchanged $24\%$ of his badges for $20\%$ of Tolya's badges, Vasya had one fewer badge than Tolya. How many badges did the boys have before the exchange? | Answer. Toly had 45 badges, and Vasya had 50 badges. Solution. Let Toly have $x$ badges before the exchange, then Vasya had $(x+5)$ badges. After the exchange, Toly had $x-\frac{x}{5}+(x+5) \cdot \frac{6}{25}$, and Vasya had $x+5-(x+5) \cdot \frac{6}{25}+\frac{x}{5}$. Solving the equation
$$
x-\frac{x}{5}+(x+5) \cdot ... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,005 |
7.2. Are there fractional (non-integer) numbers \( x, y \) such that both numbers \( 13x + 4y \) and \( 10x + 3y \) are integers? | Answer. They do not exist. Solution. Let $13 x+4 y=m, 10 x+3 y=n$, where $m$ and $n-$ are integers. Solve this system of equations by multiplying the first equation by 3, and the second by 4. Subtracting the equations, we get $x=-3 m+4 n$, i.e., $x$ is an integer. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,006 |
7.3. Among the first 500 natural numbers $1,2, \ldots, 500$, is there a sequence consisting of consecutive a) nine composite numbers; b) eleven composite numbers? | Answer: a) yes; b) yes. Solution. We can provide the desired series of 11 composite numbers: 200, 201, $\ldots$, 210. First, let's explain how to find a similar series of 9 composite numbers. There are 4 prime numbers less than 10: these are $2,3,5,7$. Their product is 210. Therefore, for any integer $k$, each of the t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,007 |
7.4. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are taken, respectively. It turns out that the perimeter of $\triangle AMC$ is equal to the perimeter of $\triangle CNA$, and the perimeter of $\triangle ANB$ is equal to the perimeter of $\triangle CMB$. Prove that $\triangle ABC$ is isosceles. | Solution. Let's denote the perimeter by the letter $P$. From the condition of the problem, we have $P(\triangle A M C)+P(\triangle C M B)=P(\triangle C N A)$ $+P(\triangle A N B)$. Therefore, $P(\triangle A B C)+2 \cdot C M=P(\triangle A B C)+2 \cdot A N$. This means $C M=A N$. From this relation, considering the equal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,008 |
8.2. Do there exist natural numbers $m, n$ such that $m^{2}=n^{2}+2014$? | Answer: No. Solution. One way to solve this is by factoring the difference of squares into factors that are clearly of the same parity, but this contradicts the fact that 2014 is divisible by 2 but not by 4. Another solution involves considering the remainders when dividing by 4: squares of integers give a remainder of... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,010 |
8.3. Can the numbers $1,2,3, \ldots, n$ be rearranged so that adjacent numbers in the sequence differ by either 3 or 5, if a) $n=25 ;$ b) $n=1000$. | Answer: a) possible; b) possible. Solution. The numbers can be rearranged as follows:
a) $\mathbf{1}, 4,7,2,5,8,3,6,9,12,15,10,13,16,11,14,17,20,23,18,21,24,19,22, \mathbf{2 5}$. b) The solution to part a) shows how to sequentially rearrange octets of numbers of the form $8 k+1,8 k+2, \ldots, 8 k+8$, specifically: $8 ... | )possible;b)possible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,011 |
9.3. There are 12 matches, each 2 cm long. Can a polygon with an area of 16 cm $^{2}$ be formed from them? (The matches cannot be broken, and all matches must be used.) | Answer: It is possible. Solution. The result follows from the Pythagorean theorem (since $10^{2}=6^{2}+8^{2}$) and construction (see figure). The area of the polygon is $\frac{6 \cdot 8}{2}-8=16$.
 both numbers $19 x+8 y$ and $8 x+3 y$ are integers?; b) both numbers $19 x^{2}+8 y^{2}$ and $8 x^{2}+3 y^{2}$ are integers? | Answer: a) exist; b) do not exist. Solution. a) If we solve the system $\left\{\begin{array}{l}19 x+8 y=1 \\ 8 x+3 y=1\end{array}\right.$, we get $x=\frac{5}{7}, y=\frac{-11}{7}$ (of course, for such an example, the right-hand sides of the system can be chosen in many ways). b) Let $\left\{\begin{array}{l}19 x^{2}+8 y^... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,017 |
10.3. Does there exist a convex hexagon and a point $M$ inside it such that all sides of the hexagon are greater than 1, and the distance from $M$ to any vertex is less than 1? | Answer. Does not exist. Indication. Suppose that such a hexagon $A B C D E F$ exists. Consider 6 triangles, the bases of which are the corresponding sides of the hexagon, and the vertex is point $M$. Then, in at least one of these triangles, the angle at vertex $M$ is not greater than $60^{\circ}$ (since the sum of all... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,019 |
11.2. Graph on the coordinate plane the set of solutions to the system of equations $\left\{\begin{array}{l}x-2 y=1 \\ x^{3}-6 x y-8 y^{3}=1\end{array}\right.$. | Answer. The set of solutions is the line $y=\frac{x-1}{2}$. Solution. The second equation is a consequence of the first, since
$x^{3}-8 y^{3}-6 x y=(x-2 y)\left(x^{2}+2 x y+4 y^{2}\right)-6 x y=$ $=x^{2}+2 x y+4 y^{2}-6 x y=(x-2 y)^{2}=1$. | \frac{x-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,022 |
11.4. Given the sequence $a_{1}=\cos 10^{\circ}, \quad a_{2}=\cos 100^{\circ}, \ldots, \quad a_{n}=\cos \left(10^{n}\right)^{\circ} \quad$ (degrees).
a) Determine the sign of the number $a_{100} ;$ b) Prove that $\left|a_{100}\right|<0.18$. | Answer: a) $a_{100}>0$. Solution. We will show that starting from the third term, the sequence becomes constant: $a_{3}=a_{4}=a_{5}=\ldots=a_{100}$. Indeed, for $n \geq 3: 10^{n+1}-10^{n}=9 \cdot 10^{3} \cdot 10^{n-3}=$ $=360 \cdot 25 \cdot 10^{n-3}$, i.e., the angles differ by a value that is a multiple of $360^{\circ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,024 |
7.1. A truck and a car are moving in the same direction on adjacent lanes at speeds of 65 km/h and 85 km/h, respectively. At what distance from each other will they be 3 minutes after they are side by side? | Answer: 1 km. Solution. The difference in the speeds of the cars is 20 km/h, so in 3 minutes they will be separated by a distance equal to 20 (km/h) $\cdot \frac{1}{20}$ (hour) $=1$ km. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,025 |
7.2. On a line, several points were marked, after which two points were placed between each pair of adjacent points, and then the same procedure (with the entire set of points) was repeated again. Could there be 82 points on the line as a result? | Answer: Yes. Solution. Let there be x points marked on the line initially. Then after the first procedure, $2(x-1)$ points are added to them, and there are a total of $3x-2$ points. After the second procedure, $2(3x-3)$ points are added to these points. Thus, there are a total of $3x-2+2(3x-3)=9x-8$ points on the line.... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,026 |
7.3. Kolya wants to multiply all the natural divisors of the number 1024 (including the number itself) on his calculator. Will he be able to get the result on a screen with 16 decimal places? | Answer: It will not be able to... The natural divisors of the number $1024=2^{10}$ are the numbers $1,2^{1}, 2^{2}, \ldots, 2^{10}$. Their product is $2^{1+2+\ldots+10}=2^{55}$. Since $2^{10}=1024>10^{3}$ and $2^{5}=128>10$, then $2^{55}=\left(2^{10}\right)^{5} \cdot 2^{5}>\left(10^{3}\right)^{5} \cdot 10=10^{16}$, i.e... | 2^{55}>10^{16} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,027 |
7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$ respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$. | Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,028 |
8.4. We have $n$ sticks of lengths $1,2, \ldots, n$. Can a square be formed from these sticks, and if not, what is the minimum number of sticks that can be broken in half to form a square a) for $n=12;$ b) for $n=15?$ | Answer: a) 2 sticks; b) It is possible. Solution. a) Since the sum $1+2+\ldots+12=78$ is not divisible by 4, it is impossible to form a square. The side of the square would be $\frac{78}{4}=19.5$. If only one stick is broken, its parts might end up on two different sides of the square, but the other two sides cannot ha... | )2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,032 |
9.1. Replace the two asterisks with two numbers so that the identity equality is obtained: $(2 x+*)^{3}=5 x^{3}+(3 x+*)\left(x^{2}-x-1\right)-10 x^{2}+10 x$. | Answer: Replace the first asterisk with -1, the second with 1. Solution. Let $a$ be the value of the first asterisk, $b$ be the value of the second. By equating the coefficients of $x^{2}$, $x$, and the constant terms, after combining like terms (and after checking that the coefficients of $x^{3}$ on the right and left... | =-1,b=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,033 |
9.2. In triangle $ABC$, side $BC$ is equal to segment $AM$, where $M$ is the point of intersection of the medians. Find the angle $\angle BMC$. | Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a kno... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,034 |
9.3. What is the smallest number of digits that need to be appended to the right of the number 2014 so that the resulting number is divisible by all natural numbers less than $10?$ | Answer: 4 digits. Solution. If three digits are appended to 2014, the resulting number will be $\leq 2014$ 999. Dividing 2014999 by $2520=$ LCM $(1,2, \ldots, 9)$, we get a quotient of 799 and a remainder of 1519. Since $1519>1000$, there is no integer multiple of 2520 between the numbers 2014000 and 2014999. Therefore... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,035 |
10.1. Replace the two asterisks with two numbers so that the identity equality is obtained: $(2 x+*)^{3}=5 x^{3}+(3 x+*)\left(x^{2}-x-1\right)-10 x^{2}+10 x$ | Answer: replace the first asterisk with -1, the second with 1. Solution. See problem 9.1. | -1,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,037 |
10.3. Plot on the coordinate plane the set of points satisfying the equation $2 x^{2}+y^{2}+3 x y+3 x+y=2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | The sought set of points is the union of two lines $y=-x-2$ and $y=-2x+1$. Solution. Consider the given equation as a quadratic equation in terms of $y: y^{2}+y(3x+1)+2x^{2}+3x-2=0$. Its discriminant is $D=x^{2}-6x+9=(x-3)^{2}$, and the solutions are $y_{1}=-x-2, y_{2}=-2x+1$. These solutions represent two lines on the... | -x-2-2x+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,039 |
10.4. Given $\triangle A B C$ and a point $M$ on the segment $A C$ (distinct from its endpoints). Construct points $P$ and $Q$ on the lateral sides $A B$ and $B C$ such that $P Q \| A C$ and $\angle P M Q=90^{\circ}$. | Solution. Let $P, Q$ be the desired points. Under a homothety with center at point $B$ and coefficient $\frac{A B}{B P}=\frac{C B}{B Q}$, segment $P Q$ will transform into $A C$, and point $M$ will transform into some point $N$. Then $\angle A N C=\angle P M Q=90^{\circ}$. Thus, point $N$ lies on the semicircle (on the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,040 |
11.1. Solve the equation $\sin ^{2} x+1=\cos (\sqrt{2} x)$. | The answer is $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k$ - integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, t... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,041 |
11.2. Prove that the radius of the inscribed circle of a Pythagorean triangle is an integer. (A Pythagorean triangle is a right triangle with integer sides.) | Solution. Let $a, b$ be the legs, $c$ be the hypotenuse, and $r$ be the radius of the inscribed circle. Then $r=\frac{a+b-c}{2}$ (this formula follows from considering the segments into which the sides are divided by the points of tangency). Since $c^{2}=a^{2}+b^{2}=(a+b)^{2}-2 a b$, the numbers $c$ and $(a+b)$ have th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,042 |
11.3. Plot on the coordinate plane the set of points satisfying the equation $2 x^{2}+y^{2}+3 x y+3 x+y=2$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Answer: The desired set of points is the union of two lines $y=-x-2$ and $y=-2x+1$. Solution. See problem 10.3. | -x-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,043 |
9.2. Is the number
$$
\sqrt[3]{2016^{2}+2016 \cdot 2017+2017^{2}+2016^{3}}
$$
- rational or irrational? | Answer: the whole rational number 2017. Solution. Let $a=2017, b=2016$. Then $b^{2}+a b+a^{2}+b^{3}=\frac{a^{3}-b^{3}}{a-b}+b^{3}=a^{3}-b^{3}+b^{3}=a^{3}$ (since $a-b=1$ ). Thus, we get $\sqrt[3]{a^{3}}=a=2017$ | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,046 |
9.3. Are the following statements true: a) If for any point $M$ inside triangle $A B C$ the segments $M A, M B$ and $M C$ can form a triangle, then $\triangle A B C$ is equilateral? b) For any point $M$ inside an equilateral triangle $A B C$, the segments $M A, M B$ and $M C$ can form a triangle | Answer: a) fair; b) fair. Solution. a) See problem 8.3. b) Let $A B C$ be an equilateral triangle. Consider $\triangle A B M$ and rotate it by $60^{\circ}$ around point $B$ so that side $A B$ coincides with $A C$. Point $M$ will then occupy the position $M^{\prime}$. Then $\triangle M B M^{\prime}$ is equilateral (sinc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,047 |
7.1. The sides of a rectangle are in the ratio $3: 4$, and its area is numerically equal to its perimeter. Find the sides of the rectangle. | Answer: 7/2, 14/3. Solution. Let $x$ and $y$ be the sides of the rectangle, then the equalities $4 x=3 y$ and $x y=2(x+y)$ hold. Expressing $x$ from the first equality and substituting it into the second, we get (after dividing by $y \neq 0$) $x=7 / 2, \quad y=14 / 3$. | \frac{7}{2},\frac{14}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,049 |
7.3. How many five-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 5625. Solution. If the first digit is even, then it can be chosen in four ways $(2,4,6,8)$. And all subsequent ones can be chosen in five ways $(1,3,5,7,9$ - for the second and fourth digits and $0,2,4,6,8$ - for the third and fifth). In the end (by the rule of product), we have a total of $4 \cdot 5^{4}=2500$ ... | 5625 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,051 |
7.4. Let's call a rook pair a pair of cells on a chessboard that are on the same vertical or horizontal line and have exactly two cells between them. Is it possible to divide the entire board into rook pairs? | Answer: No. Solution. Let's color the board as shown in the figure. Then any rook pair contains cells of the same color. At the same time, there is an odd number of white cells. Thus, it is impossible to divide the entire board into rook pairs. Note. It is also possible to show the impossibility of partitioning without... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,052 |
7.1. The students of class 7a were informed that a drama club would be organized for them if at least 14 people signed up. It turned out that among those who signed up, more than 85% were girls and the list included friends Petya and Dima. Prove that the club will be organized. | Solution. Assume the opposite. Then there are no more than 13 people in the list. Since there are at least two boys in the list, the percentage of boys is not less than $\frac{2}{13} \cdot 100 \%>15 \%$ (since $\frac{2}{13}>\frac{15}{100}$). Therefore, the percentage of girls in the list is less than $85 \%$. The obtai... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 22,053 |
7.2. At the vertices of a cube, integers were placed, and then for each face, the product of the four numbers at the vertices of that face was calculated. Could it happen that all six calculated products are negative? | Answer: It is possible. Solution. Place two negative numbers in two opposite vertices of the cube, and positive numbers in the remaining six vertices. If $A B C D$ is the lower face of the cube, and above it is the face $A_{1} B_{1} C_{1} D_{1}$, then place the two negative numbers in the vertices $A$ and $C_{1}$. Then... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,054 |
7.3. On the board, 100 natural numbers (not necessarily distinct) are written. a) Prove that if the sum of any three numbers on the board is less than the sum of any four of the remaining, then the sum of any two numbers on the board is less than the sum of any three of the remaining. b) Is it true that if the sum of a... | Answer: b) is incorrect. Solution. a) Reasoning by contradiction: let the sum of some two numbers on the board be not less than the sum of some three of the remaining numbers; then take any two numbers $x, y$ from the remaining 95 numbers, and let for definiteness $x \leq y$. Adding $x$ to the initial triplet of number... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22,055 |
7.4. There are 20 sticks of lengths $1, 2, \ldots, 20$. Can they be used to form a) a square; b) an equilateral triangle? (All sticks must be used without breaking them.) | Answer: a) impossible; b) possible. Solution. a) The sum of the lengths of all 20 sticks is 210. This number is not divisible by 4, and therefore it is impossible to form a square. b) It is possible to distribute 210 into three equal parts of 70 each in different ways, for example, as follows: $(20+19+18+13)+(17+16+15+... | ) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,056 |
7.5. Andrey and Seva are going to visit Borya. Andrey is at point $A$, and Borya is at point $B$, 30 km away from point $A$ along a straight highway. Seva is at point $C$, exactly halfway between $A$ and $B$. The friends decided to leave simultaneously: Andrey on a bicycle, and Seva on foot, but Andrey will leave the b... | Answer: 5 km before point $B$. Solution. Let $a=15$ (km), $u=5$ (km/h), $v=20$ (km/h). Let $x$ (km) be the distance from point $B$ to the place where the bicycle is left. Then Andrei's travel time is $t_{A}=\frac{2 a-x}{v}+\frac{x}{u}$, and Seva's travel time is $t_{C}=\frac{a-x}{u}+\frac{x}{v}$. We need to find such a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,057 |
8.1. Find the value of the expression $\frac{\left(a^{2}+b^{2}\right)^{2}-c^{2}-4 a^{2} b^{2}}{a^{2}+c-b^{2}}$ for $a=2017, b=2016, c=2015$. Justify your result. | Answer: 2018. Solution. The numerator equals $a^{4}+2 a^{2} b^{2}+b^{4}-4 a^{2} b^{2}-c^{2}=\left(a^{2}-b^{2}\right)^{2}-c^{2}=$ $\left(a^{2}-b^{2}+c\right)\left(a^{2}-b^{2}-c\right)$, and after dividing by the denominator we get $a^{2}-b^{2}-c=$ $(a-b)(a+b)-c=2017+2016-2015=2018$. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,058 |
8.2. Can a square be cut into four convex polygons with a different number of sides | Answer: Yes. Solution. See fig.
 | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,059 |
8.3. Given an isosceles triangle \(ABC (AB = BC)\). On side \(BC\), points \(K\) and \(N\) are marked (point \(K\) lies between \(B\) and \(N\)) such that \(KN = AN\). Prove that \(AC < AB\). | Solution. Triangle ANK is isosceles, so the angles at the base are equal. Let $\alpha=\angle K A N=\angle N K A$. Obviously, $\alpha > \angle B$ (since for triangle $A B K$, angle $\angle N K A$ is external). Therefore, $\angle A > \angle B$, from which the result follows (the side opposite the larger angle is longer). | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,060 |
8.4. Several (more than two) consecutive integers are written on the board. a) Prove that one number can be erased so that the arithmetic mean of the remaining numbers is an integer. b) Which number $k$ (satisfying the property in part a)) can be erased if one hundred numbers: 1, 2, ..., 100 are written? Specify all po... | Answer: b) $k=1$ or $k=100$. Solution. a) Let $n$ numbers be written: $a+1, a+2, \ldots, a+n$. If $n$ is odd, i.e., $n=2 m-1$, then the central number is $a+m$, and it can be erased, as it equals the arithmetic mean of the remaining $2 m-2$ numbers (indeed, $(a+1)+(a+2 m-1)=(a+2)+(a+2 m-2)=\ldots=(a+m-1)+(a+m+1)=2(a+m)... | k=1ork=100 | Number Theory | proof | Yes | Yes | olympiads | false | 22,061 |
8.2. Is the given number $N$ prime or composite, if:
a) $N=2011 \cdot 2012 \cdot 2013 \cdot 2014+1$; b) $N=2012 \cdot 2013 \cdot 2014 \cdot 2015+1$? | Answer. a) Composite; b) composite. Solution. a) See the solution to problem 7.1. A more general solution follows from part b).
b) For any natural number $n$ we have:
$$
N=((n+1)(n+2))(n(n+3))+1=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1=M(M+2)+1=(M+1)^{2},
$$
where $M=n^{2}+3 n$. Thus, the number $N$ is compo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,064 |
8.3. A natural number is called curious if, after subtracting the sum of its digits from it, the result is a number consisting of identical digits. How many curious numbers exist in total: a) three-digit curious numbers?; b) four-digit curious numbers? | Answer. a) 30 numbers; b) 10 numbers. Solution. a) see the solution to problem 7.3.
b) Let $\overline{t x y z}$ be a curious four-digit number. Reasoning similarly to the solution of problem 7.3, we get that the number $A$ can equal 999, while the other two cases are impossible (since the numbers 3333 and 6666 are not... | )30;b)10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,065 |
8.4. Find the natural number $x$ that satisfies the equation
$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3}
$$ | Answer: 2012. Solution: Let $a=2011, \quad b=2012$. Then $a^{2}+a b+b^{2}=\frac{b^{3}-a^{3}}{b-a}=b^{3}-a^{3}$ (since $b-a=1$). Therefore, $x^{3}=b^{3}-a^{3}+a^{3}=b^{3}$. Hence, $x=b=2012$. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,066 |
10.2. Solve the equation $4 x=2+\frac{x}{\sqrt{1+x}+1}$. | Answer: $x=\frac{9}{16}$. Solution. By multiplying the fraction on the right side by the expression $(\sqrt{1+x}-1)$ (conjugate to the denominator), after canceling out by $x \neq$ Oh, we get the equation $4 x=1+\sqrt{1+x}$. Then, by canceling out the expression on the right side and transforming it in the same way as ... | \frac{9}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,069 |
10.3. Given a line on a plane and several (more than two) points marked on it. Prove that one can mark another point on the plane (outside the given line) so that among all triangles with the marked vertices, more than half are acute. | Solution. Let $A_{1}, A_{2}, \ldots, A_{n}$ be the marked points in the order on the line. Let $k=\left[\frac{n}{2}\right]$, where $[m]$ is the integer part of the number $m$. Mark a point $B$ such that its projection on the line belongs to the interval $\left(A_{k}, A_{k+1}\right)$ and $\angle A_{1} B A_{n}$ is acute ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,070 |
9.1. In a five-digit number, one of the digits was crossed out, and this four-digit number was subtracted from the original number. As a result, the number 54321 was obtained. Find the original number. | Answer: 60356. Solution. Let $x$ be the four-digit number obtained after crossing out a digit. Note that the crossed-out digit was the last digit of the five-digit number, because otherwise the last digit after subtraction would be zero. Let this digit be $y$. We have the equation $10 x+y-x=54321 \Leftrightarrow 9 x+y=... | 60356 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,073 |
9.2. From a rectangular table of $m \times n$ cells, several squares of different sizes need to be cut out along the grid lines. What is the maximum number of squares that can be cut out if: a) $m=8$, $n=11$; b) $m=8, n=12 ?$ | Answer: a) 5; b) 5. Solution. a) Note that the area of six different squares is no less than $1+4+9+16+25+36=91>88$. Therefore, it is impossible to cut out more than five different squares. A possible example for five squares (even in a rectangle $8 \times 9$) is shown in the figure. b) Suppose, for the sake of contrad... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,074 |
9.3. Find all prime numbers $p$ for which $p^{2}+200$ is a perfect square of an integer. | Answer: $p=5$ or $p=23$. Solution. Let $p^{2}+200=a^{2}$, where $a$ is a natural number. Since $p=2$ does not satisfy the equation, the sought $p$, and thus $a$, are odd numbers. From the equality $(a-p)(a+p)=200$, considering that $a+p>a-p$ and both factors are even numbers, we get three cases: 1) $a+p=100, a-p=2$; 2)... | p=5orp=23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,075 |
9.4. Is it true that for any odd $n>3$ it is possible to mark (and denote) points $A_{1}$, $A_{2}, \ldots, A_{n}$ on the plane so that the $n$ triangles $A_{1} A_{2} A_{3}, A_{2} A_{3} A_{4}, \ldots, A_{n} A_{1} A_{2}$ are acute-angled? | Answer: Correct Solution. Let $n=2 k+1, k>1$. Mark the vertices of a regular $(2 k+1)$-gon and number them as follows: take an arbitrary vertex $A_{1}$ and denote by $A_{2}$ the $k$-th vertex clockwise from vertex $A_{1}$ of this polygon. Similarly, $A_{3}$ is the $k$-th vertex clockwise from vertex $A_{2}$, and so on ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,076 |
7.1. Before a running competition, Petya planned to run the entire distance at a constant speed $V$. However, upon learning the results of his competitors, Petya decided to increase his planned speed by $25 \%$. With this increased speed, he ran half the distance, but then got tired, so he ran the second half of the di... | Answer: more than planned. Solution. Let $a$ be the length of the distance. Then the planned time is $\frac{a}{V}$, and the actual time is $\frac{a}{2 \cdot 1.25 V}+\frac{a}{2 \cdot 0.8 V}=\frac{a}{V} \cdot \frac{41}{40}>\frac{a}{V}$. | thanplanned | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,077 |
7.2. Is it possible to place the 12 numbers $1,2, \ldots, 12$ on the edges of a cube such that the product of the four numbers on the top face equals the product of the four numbers on the bottom face? | Answer: It is possible. Solution. An example of such an arrangement can be as follows: on the top face, place the numbers $2,4,9,10$; on the bottom face, place the numbers $3,5,6,8$; the remaining numbers $1,7,11,12$ are placed on the side edges. The products on the top and bottom faces are the same and equal to 720. (... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,078 |
7.3. If the first digit of a two-digit natural number $N$ is multiplied by 4, and the second digit is multiplied by 2, then the sum of the numbers obtained after the multiplication will give $\frac{N}{2}$. Find $N$ (list all solutions). | Answer: three solutions: $N=32 ; 64 ; 96$. Solution. Let $x$ be the first digit, $y$ be the second digit of the number $N$. Then the condition of the problem can be written as $4 x+2 y=\frac{10 x+y}{2} \Leftrightarrow 2 x=3 y$. Therefore, the digit $x$ is divisible by 3 and $x \neq 0$ (since $x$ is the first digit). Fr... | 32;64;96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,079 |
7.4. There are ten coins of different weights and a balance scale without weights. It is required to select the heaviest and the lightest coin. Can this be achieved in 13 weighings? | Answer: It is possible. Solution. First, divide all the coins into 5 pairs and in 5 weighings compare the weight of each pair. By selecting the heavier coin in each pair, we form a "heavy" group of 5 coins. The remaining 5 coins will form the "light" group. Now, in the heavy group, over 4 sequential weighings, select t... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,080 |
7.5. From a right triangle $A B C$ with legs $A C=3$ and $C B=7$, a square with vertex $C$ of the largest possible area needs to be cut out. What is the side length of such a square? | Answer: 2.1. Solution. Introduce a coordinate system: the origin - at vertex $C$, the $x$-axis - along $C A$, the $y$-axis - along $C B$. Then the hypotenuse lies on the line $y=7-\frac{7}{3} x$ (this follows from the meaning of the slope and the y-intercept for the graph of a line). The diagonal of the square can be w... | 2.1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,081 |
8.2. In a five-digit number, one digit was crossed out, and the resulting four-digit number was added to the original. The sum turned out to be 54321. Find the original number. | Answer: 49383. Solution. Note that the crossed-out digit must be the last digit of the number $N$, because otherwise the sum of the two numbers would have an even digit in the last place. Let's denote this crossed-out digit by $x$ and let $y$ be the four-digit number obtained after the crossing out. Then the condition ... | 49383 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,083 |
8.3. In a convex quadrilateral $A B C D$, points $P$ and $Q$ are the midpoints of sides $A B$ and $C D$. It turns out that line $P Q$ bisects diagonal $A C$. Prove that $P Q$ also bisects diagonal $B D$. | Solution. Let $M$ be the intersection point of lines $P Q$ and $A C$, and $N$ be the intersection point of lines $P Q$ and $B D$. By the given condition, $P M$ is the midline of triangle $A B C$, and therefore $P M \| B C$. Then, in triangle $B C D$, segment $N Q$ is parallel to the base $B C$ and passes through the mi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,084 |
8.5. a) A rectangular table of size $4 \times 10$ (cells) is given. What is the maximum number of crosses that can be placed in the cells of this table so that the following condition is met: in each row and each column of the table, there must be an odd number of crosses? b) Is it possible to place several crosses in ... | Answer: a) 30; b) cannot. Solution. a) Let the table have 4 rows and 10 columns. In each column, there is at least one empty (without a cross) cell, otherwise there would be a column with 4 crosses. Therefore, the total number of crosses in the table is no more than $3 \cdot 10=30$. An example with 30 crosses is shown ... | )30;b)cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,086 |
9.2. Given four real numbers $a, b, c, d$, which satisfy two relations: $a+b=c+d$ and $a^{3}+b^{3}=c^{3}+d^{3}$. a) Prove that $a^{5}+b^{5}=c^{5}+d^{5}$; b) Can we conclude that $a^{4}+b^{4}=c^{4}+d^{4}$? | Answer: b) cannot. Solution. a) Let's write the second relation as $(a+b)\left(a^{2}-a b+b^{2}\right)=$ $=(c+d)\left(c^{2}-c d+d^{2}\right)$. Due to the first equation, the linear factors in this equality coincide. If they are equal to 0, then $a=-b, c=-d$ and thus $a^{5}=-b^{5}, c^{5}=-d^{5}$, i.e., $a^{5}+b^{5}=c^{5}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,088 |
9.4. Given a square table, in some cells of which there are crosses. Let's call a row of the table odd if it contains an odd number of crosses. Similarly, in an odd column, there is an odd number of crosses. a) Can it happen that in the table there are exactly 20 odd rows and 15 odd columns? b) Can 126 crosses be place... | Answer: a) cannot; b) can. Solution. a) Suppose, to the contrary, that such an arrangement is possible. First, let's count all the crosses by rows. The sum of twenty odd numbers in the odd rows will give an even number. The other rows of the table are even and will not change the parity of the total number of crosses. ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,090 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.