problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. Find the value of the expression $4 \sin 40^{\circ}-\operatorname{tg} 40^{\circ}$. | Answer: $\sqrt{3}$.
Solution. $4 \sin 40^{\circ}-\operatorname{tg} 40^{\circ}=\frac{4 \sin 40^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\frac{2 \sin 80^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\frac{\left(\sin 80^{\circ}-\sin 40^{\circ}\right)+\sin 80^{\circ}}{\cos 40^{\circ}}=$ $=\frac{2 \sin 20^{\ci... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,735 |
4. In the number $2016^{* * * *} 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,736 |
6. A circle $\Omega$ is circumscribed around an acute isosceles triangle $A D E$ with base $A D$. The distance from the midpoint of the arc $D E$, not containing point $A$, to the side $D E$ is 5, and the distance from the midpoint of the arc $A D$, not containing point $E$, to the side $A D$ is $\frac{1}{3}$. Find the... | Answer: $R=6, S=\frac{35 \sqrt{35}}{9}$.
Solution. Let $M$ and $T$ be the midpoints of arcs $A D$ and $D E$; $B$ and $H$ be the midpoints of segments $D E$ and $A D$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O T \perp D E, O M \perp A D ; O B=O T-T B=R-5, O H=O M-M H=R-\frac{1}{3}$.
From the... | R=6,S=\frac{35\sqrt{35}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,738 |
1. Find all values of $p$, for each of which the numbers $p-2$, $3 \cdot \sqrt{p}$, and $-8-p$ are respectively the first, second, and third terms of some geometric progression. | Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(p-2)(-8-p)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+15 p-16=0
\end{array} \Leftrightarrow p=1\right.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,739 |
2. Plot on the plane $(x ; y)$ the set of points satisfying the equation $\left|9-x^{2}-y^{2}-2 y\right|+|-2 y|=9-x^{2}-y^{2}-4 y$, and find the area of the resulting figure. | Answer: $S=10 \pi+3-10 \operatorname{arctg} 3$.
Solution. Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system of inequaliti... | 10\pi+3-10\operatorname{arctg}3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,740 |
3. Find the value of the expression $\operatorname{ctg} 70^{\circ}+4 \cos 70^{\circ}$. | Answer: $\sqrt{3}$.
Solution. $\operatorname{ctg} 70^{\circ}+4 \cos 70^{\circ}=\frac{\cos 70^{\circ}+4 \sin 70^{\circ} \cos 70^{\circ}}{\sin 70^{\circ}}=\frac{\cos 70^{\circ}+2 \sin 140^{\circ}}{\sin 70^{\circ}}=\frac{(\cos 70^{\circ}+\cos 50^{\circ})+\cos 50^{\circ}}{\sin 70^{\circ}}=$ $=\frac{2 \cos 60^{\circ} \cos ... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,741 |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | # Answer: 864.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits a... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,742 |
5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}2 x-\sqrt{x y}-4 \sqrt{\frac{x}{y}}+2=0, \\ 2 x^{2}+x^{2} y^{4}=18 y^{2} .\end{array}\right.$ | Answer: $(2 ; 2),\left(\frac{\sqrt[4]{286}}{4} ; \sqrt[4]{286}\right)$.
Solution. Let $\sqrt{\frac{x}{y}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$ ). Then $\quad u v=\sqrt{\frac{x}{y}} \cdot \sqrt{x y}=\sqrt{x^{2}}=|x|=x$,
$\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{x}{y}}=\sqrt{y^{2}}=|y|=y$, since $x$ and $y$ are posi... | (2;2),(\frac{\sqrt[4]{286}}{4};\sqrt[4]{286}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,743 |
6. A circle $\Omega$ is circumscribed around an acute isosceles triangle $BCD$ with base $CD$. The distance from the midpoint of the arc $BD$, not containing point $C$, to the side $BD$ is 3, and the distance from the midpoint of the arc $CD$, not containing point $B$, to the side $CD$ is 0.5. Find the radius of the ci... | Answer: $R=4, S=\frac{15 \sqrt{15}}{4}$.
Solution. Let $Q$ and $T$ be the midpoints of arcs $CD$ and $BD$; $A$ and $H$ be the midpoints of segments $BD$ and $CD$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $OT \perp BD, OQ \perp CD; OA=OT-TA=R-3, OH=OQ-QH=R-\frac{1}{2}$.
From the Pythagorean th... | R=4,S=\frac{15\sqrt{15}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,744 |
1. Find all values of $p$, for each of which the numbers $-p-8$, $3 \cdot \sqrt{p}$, and $p-7$ are the first, second, and third terms, respectively, of some geometric progression. | Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(-p-8)(p-7)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+10 p-56=0
\end{array} \Leftrightarrow p=4\right.
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,745 |
2. Plot on the plane $(x ; y)$ the set of points satisfying the equation $\left|4-2 x-x^{2}-y^{2}\right|+|-2 x|=4-4 x-x^{2}-y^{2}$, and find the area of the resulting figure. | Answer: $S=5 \pi-5 \pi \operatorname{arctg} 2+2$.
Solution. Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system of inequali... | 5\pi-5\pi\operatorname{arctg}2+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,746 |
3. Find the value of the expression $\operatorname{ctg} 50^{\circ}-4 \cos 50^{\circ}$. | Answer: $-\sqrt{3}$.
Solution. $\operatorname{ctg} 50^{\circ}-4 \cos 50^{\circ}=\frac{\cos 50^{\circ}-4 \cos 50^{\circ} \sin 50^{\circ}}{\sin 50^{\circ}}=\frac{\cos 50^{\circ}-2 \sin 100^{\circ}}{\sin 50^{\circ}}=\frac{(\cos 50^{\circ}-\cos 10^{\circ})-\cos 10^{\circ}}{\sin 50^{\circ}}=$ $=\frac{-2 \sin 20^{\circ} \si... | -\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,747 |
4. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 8$ (4 ways).
To ensure divisibility by 3, we proceed as follows. Select three digits arbitrari... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,748 |
5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}3 y-\sqrt{\frac{y}{x}}-6 \sqrt{x y}+2=0 \\ x^{2}+81 x^{2} y^{4}=2 y^{2} .\end{array}\right.$, Answer: $\left(\frac{1}{3} ; \frac{1}{3}\right),\left(\frac{\sqrt[4]{31}}{12} ; \frac{\sqrt[4]{31}}{3}\right)$ | Solution. Let $\sqrt{\frac{y}{x}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$,
$\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system takes the form
$$
\left\{\begin{array} ... | (\frac{1}{3};\frac{1}{3}),(\frac{\sqrt[4]{31}}{12};\frac{\sqrt[4]{31}}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,749 |
6. Around an acute isosceles triangle $A M T$ with base $M T$, a circle $\Omega$ is described. The distance from the midpoint of the arc $A T$, not containing point $M$, to the side $A T$ is 3, and the distance from the midpoint of the arc $M T$, not containing point $A$, to the side $M T$ is 1.6. Find the radius of th... | Answer: $R=5, S=\frac{168 \sqrt{21}}{25}$.
Solution. Let $Q$ and $L$ be the midpoints of arcs $M T$ and $A T$; $B$ and $H$ be the midpoints of segments $A T$ and $M T$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O L \perp A T, O Q \perp M T ; O B=O L-L B=R-3, O H=O Q-Q H=R-\frac{8}{5}$.
From t... | R=5,S=\frac{168\sqrt{21}}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,750 |
1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+\operatorname{tg} \alpha+\frac{5}{2} \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=-\frac{1}{2}$. Find $\operatorname{ctg} \alpha$. | Answer: $\frac{1}{3}$ or $-\frac{6}{7}$.
Solution. $\operatorname{tg} 2 \gamma=\frac{2 \operatorname{tg} \gamma}{1-\operatorname{tg}^{2} \gamma}=\frac{-1}{3 / 4}=-\frac{4}{3}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alp... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,751 |
6. An isosceles triangle $ABC$ with base $BC$ is inscribed in a circle $\Omega$. Chords $LM$ and $PQ$, parallel to the line $BC$, intersect side $AB$ at points $D$ and $T$ respectively, and $AD = DT = TB$. Find the radius of the circle $\Omega$ and the area of triangle $ABC$, if $LM = \frac{10}{\sqrt{3}}, PQ = \frac{2 ... | Answer: $R=\frac{37}{12}, S=6$.
Solution. The line $A O$ is perpendicular to the chords $L M, P Q, B C$ and bisects each of them. Let points $N$, $H$, and $E$ be the midpoints of $L M, P Q$, and $B C$. Denote the radius of the circle $\Omega$ as $R ; A N=N H=H E=x$.
Then $O H=A H-O A=2 x-R, O N=O A-A N=R-x$ and by th... | R=\frac{37}{12},S=6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,752 |
1. It is known that $\operatorname{tg}(2 \alpha-\beta)+6 \operatorname{tg} 2 \alpha+\operatorname{tg} \beta=0, \operatorname{tg} \alpha=2$. Find $\operatorname{ctg} \beta$. | Answer: 1 or $\frac{1}{7}$.
Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{4}{1-4}=-\frac{4}{3}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatornam... | \operatorname{ctg}\beta=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,753 |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|3 x|+|4 y|+|48-3 x-4 y|=48$, and find the area of the resulting figure.
# | # Answer: 96.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,754 |
6. An isosceles triangle $P Q T$ with base $P Q$ is inscribed in a circle $\Omega$. Chords $A B$ and $C D$, parallel to the line $P Q$, intersect the side $Q T$ at points $L$ and $M$ respectively, and $Q L=L M=M T$. Find the radius of the circle $\Omega$ and the area of triangle $P Q T$, if $A B=2 \sqrt{14}, C D=2 \sqr... | Answer: $R=\frac{15}{4}, S=18$.
Solution. The line $T O$ is perpendicular to the chords $C D, A B, P Q$ and bisects each of them. Let points $N$, $H$, and $E$ be the midpoints of $C D, A B$, and $P Q$. Denote the radius of the circle $\Omega$ as $R ; T N=N H=H E=x$.
Then $O H=T H-O T=2 x-R, O N=O T-T N=R-x$ and by th... | R=\frac{15}{4},S=18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,755 |
1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+2 \operatorname{tg} \alpha-4 \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=\frac{1}{3}$. Find $\operatorname{ctg} \alpha$. | Answer: 2 or $\frac{1}{3}$.
Solution. $\operatorname{tg} 2 \gamma=\frac{2 \operatorname{tg} \gamma}{1-\operatorname{tg}^{2} \gamma}=\frac{2 / 3}{1-1 / 9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \opera... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,756 |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|5 x|+|12 y|+|60-5 x-12 y|=60$, and find the area of the resulting figure.
# | # Answer: 30.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,757 |
6. An isosceles obtuse triangle $P Q T$ with base $P T$ is inscribed in a circle $\Omega$. Chords $A B$ and $C D$, parallel to the line $P T$, intersect the side $Q T$ at points $K$ and $L$ respectively, and $Q K=K L=L T$. Find the radius of the circle $\Omega$ and the area of triangle $P Q T$, if $A B=2 \sqrt{66}, C D... | Answer: $R=12.5, S=108$.
Solution. The line $Q O$ is perpendicular to the chords $A B, C D, P T$ and bisects each of them. Let points $M$, $E$, and $H$ be the midpoints of $A B, C D$, and $P T$. Denote the radius of the circle $\Omega$ as $R; Q M=M E=E H=x$.
Since the triangle is obtuse, the center of its circumscrib... | R=12.5,S=108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,758 |
1. It is known that $\operatorname{tg}(2 \alpha-\beta)-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0, \operatorname{tg} \alpha=-3$. Find $\operatorname{ctg} \beta$. | Answer: -1 or $\frac{4}{3}$.
Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{-6}{1-9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorna... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,759 |
6. An isosceles triangle \(ABC\) with base \(AC\) is inscribed in a circle \(\Omega\). Chords \(DN\) and \(LT\), parallel to the line \(AC\), intersect the side \(BC\) at points \(F\) and \(H\) respectively, and \(BF = FH = HC\). Find the radius of the circle \(\Omega\) and the area of triangle \(ABC\), if \(DN = 2\sqr... | Answer: $R=6.5, S=54$.
Solution. The line $B O$ is perpendicular to the chords $A C, D N, L T$ and bisects each of them. Let points $P$, $Q$, and $E$ be the midpoints of $D N, L T$, and $A C$. Denote the radius of the circle $\Omega$ as $R; B P=P Q=Q E=x$.
Then $O Q=O B-B Q=R-2 x, O P=O B-B P=R-x$ and by the Pythagor... | R=6.5,S=54 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,760 |
1. [4 points] On the table lies a piece of sugar, around which an ant and a beetle are crawling along two circles at the same speed. A rectangular coordinate system is introduced on the table plane, in which the sugar (the common center of the circles) is located at point $O(0 ; 0)$. The ant moves clockwise, while the ... | Answer: $(2 ; 2 \sqrt{3}),(-4 ; 0),(2 ;-2 \sqrt{3})$.
Solution. Let's denote the points where the ant and the beetle are located as $M(\alpha)$ and $N(\beta)$ respectively, where $\alpha$ and $\beta$ are the angles that the radius vectors of points $M$ and $N$ form with the positive direction of the x-axis. Note that ... | (2;2\sqrt{3}),(-4;0),(2;-2\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,761 |
2. [4 points] Find all pairs of real parameters $a$ and $b$, for each of which the system of equations
$$
\left\{\begin{array}{l}
3(a+b) x+12 y=a \\
4 b x+(a+b) b y=1
\end{array}\right.
$$
has infinitely many solutions. | Answer: $(1 ; 3),(3 ; 1),(-2-\sqrt{7} ; \sqrt{7}-2),(\sqrt{7}-2 ;-2-\sqrt{7})$.
Solution. If $b=0$, then the second equation of the system takes the form $1=0$, and therefore the system has no solutions. For all other values of the parameters, in each of the equations at least one of the coefficients of the variables ... | (1;3),(3;1),(-2-\sqrt{7};\sqrt{7}-2),(\sqrt{7}-2;-2-\sqrt{7}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,762 |
3. [4 points] Solve the equation $(x+3) \sqrt{x^{3}-x+10}=x^{2}+5 x+6$.
| Answer: $2 ; \frac{\sqrt{13}-1}{2}$.
Solution. Factoring the right side, we get $(x+3) \sqrt{x^{3}-x+10}=(x+$ $3)(x+2)$. From this, there are two possibilities: either $x+3=0$ (then $x=-3$, which does not fit the domain of definition, as the expression under the root is negative), or $\sqrt{x^{3}-x+10}=x+2$. We solve ... | 2;\frac{\sqrt{13}-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,763 |
4. [6 points] Solve the inequality $2 x^{4}+x^{2}-2 x-3 x^{2}|x-1|+1 \geqslant 0$. | Answer: $\left(-\infty ;-\frac{1+\sqrt{5}}{2}\right] \cup\left[-1 ; \frac{1}{2}\right] \cup\left[\frac{\sqrt{5}-1}{2} ;+\infty\right)$.
Solution. The inequality can be rewritten as $2 x^{4}+(x-1)^{2}-3 x^{2}|x-1| \geqslant 0$ or $\mid x-$ $\left.1\right|^{2}-3 x^{2}|x-1|+2 x^{4} \geqslant 0$. To factorize the left sid... | (-\infty;-\frac{1+\sqrt{5}}{2}]\cup[-1;\frac{1}{2}]\cup[\frac{\sqrt{5}-1}{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,764 |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 1400. The answer should be presented as an integer. | Answer: 5880.
Solution. Since $1400=7 \cdot 2^{3} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) three twos, two fives, one seven, and two ones, (b) one four, one two, two fives, one seven, and three ones, or (c) one eight, two fives, one seven, and four ones. We will calculate the number of... | 5880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,765 |
6. [5 points] Two circles of the same radius 9 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 18.
Solution. Let $R=9$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,766 |
7. [6 points] Solve the system
$$
\left\{\begin{array}{l}
|x-3-y|+|x-3+y| \leqslant 6 \\
(|x|-3)^{2}+(|y|-4)^{2}=25
\end{array}\right.
$$ | Answer: $(0 ; 0),(6 ; 0)$.
Solution. Consider the inequality of the system and plot the set of its solutions on the coordinate plane. To remove the absolute values, find the sets of points where the expressions under the absolute values are zero. These are the lines $x-3-y=0$ and $x-3+y=0$. They divide the plane into ... | (0;0),(6;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,767 |
1. [4 points] Around a bird feeder, in the same plane as it, a tit and a bullfinch are flying along two circles at the same speed. In the plane, a rectangular coordinate system is introduced, in which the feeder (the common center of the circles) is located at point $O(0 ; 0)$. The tit is moving clockwise, while the bu... | Answer: $(4 \sqrt{3} ; 0),(-2 \sqrt{3} ; 6),(-2 \sqrt{3} ;-6)$.
Solution. Let the points where the blue tit and the snow bunting are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles that the radius vectors of points $M$ and $N$ form with the positive direction of the x-axi... | (4\sqrt{3};0),(-2\sqrt{3};6),(-2\sqrt{3};-6) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,768 |
2. [4 points] Find all pairs of real parameters $a$ and $b$, for each of which the system of equations
$$
\left\{\begin{array}{l}
2(a-b) x+6 y=a \\
3 b x+(a-b) b y=1
\end{array}\right.
$$
has infinitely many solutions. | Answer: $(-1 ; 2),(-2 ; 1),\left(\frac{3-\sqrt{17}}{2} ;-\frac{3+\sqrt{17}}{2}\right),\left(\frac{3+\sqrt{17}}{2} ; \frac{\sqrt{17}-3}{2}\right)$.
Solution. If $b=0$, then the second equation of the system takes the form $1=0$, and therefore the system has no solutions. For all other values of the parameters, in each ... | (-1;2),(-2;1),(\frac{3-\sqrt{17}}{2};-\frac{3+\sqrt{17}}{2}),(\frac{3+\sqrt{17}}{2};\frac{\sqrt{17}-3}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,769 |
3. [4 points] Solve the equation $\frac{1}{2}(x+5) \sqrt{x^{3}-16 x+25}=x^{2}+3 x-10$. | Answer: $3 ; \frac{\sqrt{13}+1}{2}$.
Solution. Factoring the right side, we get $(x+5) \sqrt{x^{3}-16 x+25}=(x+$ 5) $(2 x-4)$. From here, there are two possibilities: either $x+5=0$ (then $x=-5$, which does not fit the domain of definition, as the expression under the root is negative), or $\sqrt{x^{3}-16 x+25}=2 x-4$... | 3;\frac{\sqrt{13}+1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,770 |
4. [6 points] Solve the inequality $6 x^{4}+x^{2}+2 x-5 x^{2}|x+1|+1 \geqslant 0$. | Answer: $\left(-\infty ;-\frac{1}{2}\right] \cup\left[\frac{1-\sqrt{13}}{6} ; \frac{1+\sqrt{13}}{6}\right] \cup[1 ;+\infty)$.
Solution. The inequality can be rewritten as $6 x^{4}+(x+1)^{2}-5 x^{2}|x+1| \geqslant 0$ or $\mid x+$ $\left.1\right|^{2}-5 x^{2}|x-1|+6 x^{4} \geqslant 0$. To factorize the left-hand side, no... | (-\infty;-\frac{1}{2}]\cup[\frac{1-\sqrt{13}}{6};\frac{1+\sqrt{13}}{6}]\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,771 |
5. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 7000. The answer should be presented as an integer. | Answer: 5600.
Solution. Since $7000=7 \cdot 2^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three twos, three fives, one seven, and one one, (b) four, two, three fives, one seven, and two ones, or (c) eight, three fives, one seven, and three ones. We will calculate the number of variant... | 5600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,772 |
6. [5 points] Two circles of the same radius 7 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a po... | Answer: 14.
Solution. Let $R=7$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,773 |
7. [6 points] Solve the system
$$
\left\{\begin{array}{l}
|x+y+5|+|x-y+5| \leqslant 10 \\
(|x|-5)^{2}+(|y|-12)^{2}=169
\end{array}\right.
$$ | Answer: $(0 ; 0),(-10 ; 0)$.
Solution. Consider the inequality of the system and represent the set of its solutions on the coordinate plane. To expand the absolute values, we need to find the sets of points where the expressions under the absolute values are zero. These are the lines $x+y+5=0$ and $x-y+5=0$. They divi... | (0;0),(-10;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,774 |
1. [4 points] Around a flower, in the same plane as it, a bumblebee and a bee are flying along two circles. The speed of the bee is one and a half times the speed of the bumblebee. In the specified plane, a rectangular coordinate system is introduced, in which the flower (the common center of the circles) is at the poi... | Answer: $(\sqrt{3} ; 1),(-1 ; \sqrt{3}),(-\sqrt{3} ;-1),(1 ;-\sqrt{3})$.
Solution. Let the points where the bee and the bumblebee are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles that the radius vectors of points $M$ and $N$ form with the positive direction of the x-ax... | (\sqrt{3};1),(-1;\sqrt{3}),(-\sqrt{3};-1),(1;-\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,775 |
2. [4 points] Find all integer parameter triples $a, b$, and $c$, for each of which the system of equations
$$
\left\{\begin{array}{l}
a x+2 y+c z=c \\
3 x+b y+4 z=4 b
\end{array}\right.
$$
has no solutions. | Answer: $(-6, -1, -8), (-3, -2, -4), (3, 2, 4)$.
Solution. A system of two linear equations has no solutions if and only if the coefficients of the variables in the equations are proportional to each other, but not proportional to the free terms. It is also impossible for the coefficients of one of the variables to be... | (-6,-1,-8),(-3,-2,-4),(3,2,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,776 |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}-10 x+7}+1\right) \cdot\left|x^{3}-18 x+28\right| \leqslant 0$. | Answer: $x=-1+\sqrt{15}$.
Solution. The first factor on the left side is positive for any admissible value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}-10 x+7 \geqslant 0$) is equivalent to the equation $x^{3}-18 x+28=0$. One of th... | -1+\sqrt{15} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,777 |
4. [5 points] Solve the equation $2 x^{4}+x^{2}-6 x-3 x^{2}|x-3|+9=0$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: $-\frac{3}{2} ; 1 ; \frac{-1 \pm \sqrt{13}}{2}$.
Solution. The equation can be rewritten as $2 x^{4}+(x-3)^{2}-3 x^{2}|x-3|=0$ or $|x-3|^{2}-$ $3 x^{2}|x-3|+2 x^{4}=0$. To factor the left side, note that it represents a quadratic trinomial in terms of $y=|x-3|$ with a discriminant equal to $\left(3 x^{2}\right... | -\frac{3}{2};1;\frac{-1\\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,778 |
5. [5 points] 70 dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) were rolled and the sum of the numbers that came up was counted. Which probability is greater: that the sum is greater than 350, or that the sum is not more than 140? | Answer: It is more likely that the sum is not more than 140.
Solution. The result of rolling the dice can be described by a set of 70 numbers ranging from 1 to 6. Consider any such set. If each number in the set is replaced from $x$ to 7 - $x$, we get a new set consisting of numbers from 1 to 6. In this case, if the s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,779 |
6. [4 points] Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omega_{... | Answer: $\frac{R_{2}}{R_{1}}=\frac{6}{5}$.
Fig. 1: Option 9, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B O_{2}... | \frac{6}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,780 |
7. [7 points] Find all values of the parameter $a$ for which the system of equations
$$
\left\{\begin{array}{l}
a^{2}-2 a x-6 y+x^{2}+y^{2}=0 \\
(|x|-4)^{2}+(|y|-3)^{2}=25
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in(-12 ;-6) \cup\{0\} \cup(6 ; 12)$.
Solution. Consider the second equation of the system. For $x \geqslant 0$ and $y \geqslant 0$, it takes the form $(x-4)^{2}+(y-3)^{2}=25$, and we obtain a circle of radius 5 centered at the point $(4 ; 3)$ (more precisely, its part lying in the first quadrant). Since th... | \in(-12;-6)\cup{0}\cup(6;12) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,781 |
1. [4 points] On the floor, there is a saucer of milk, around which a kitten and a puppy walk along two circles. The kitten's speed is half the puppy's speed. A rectangular coordinate system is introduced on the floor plane, with the saucer (the common center of the circles) located at point $O(0 ; 0)$. The kitten move... | Answer: $(-3 \sqrt{3} ; 3),(3 ; 3 \sqrt{3}),(3 \sqrt{3} ;-3),(-3 ;-3 \sqrt{3})$.
Solution. Let the points where the kitten and the puppy are located be $M(\alpha)$ and $N(\beta)$ respectively, where $\alpha$ and $\beta$ are the angles that the radius vectors of points $M$ and $N$ form with the positive direction of th... | (-3\sqrt{3};3),(3;3\sqrt{3}),(3\sqrt{3};-3),(-3;-3\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,782 |
2. [4 points] Find all integer parameter triples $a, b$, and $c$, for each of which the system of equations
$$
\left\{\begin{array}{l}
2 x-b y+z=2 b \\
a x+5 y-c z=a
\end{array}\right.
$$
has no solutions. | Answer: $(-2 ; 5 ; 1),(2 ;-5 ;-1),(10 ;-1 ;-5)$.
Solution. A system of two linear equations has no solutions if and only if the coefficients of the variables in the equations are proportional to each other, but not proportional to the free terms. Also, note that it is impossible for the coefficients of one of the vari... | (-2;5;1),(2;-5;-1),(10;-1;-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,783 |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}-18 x-5}+2\right) \cdot\left|x^{3}-4 x^{2}-5 x+18\right| \leqslant 0$. | Answer: $x=1-\sqrt{10}$.
Solution. The first factor on the left side is positive for any admissible value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}-18 x-5 \geqslant 0$) is equivalent to the equation $x^{3}-4 x^{2}-5 x+18=0$. One... | 1-\sqrt{10} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,784 |
4. [5 points] Solve the equation $4 x^{4}+x^{2}+6 x-5 x^{2}|x+3|+9=0$.
翻译结果：
4. [5 points] Solve the equation $4 x^{4}+x^{2}+6 x-5 x^{2}|x+3|+9=0$. | Answer: $-\frac{3}{4} ; 1 ; \frac{1 \pm \sqrt{13}}{2}$.[^3]
Solution. The equation can be rewritten as $4 x^{4}+(x+3)^{2}-5 x^{2}|x+3|=0$ or $|x+3|^{2}-5 x^{2} |x+3|+4 x^{4}=0$. To factor the left side, note that it represents a quadratic trinomial in terms of $y=|x+3|$ with a discriminant equal to $\left(5 x^{2}\righ... | -\frac{3}{4};1;\frac{1\\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,785 |
5. [5 points] 60 dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) were rolled and the sum of the numbers that came up was counted. Which probability is greater: that the sum is not less than 300, or that the sum is less than 120? | Answer: It is more likely that the sum is not less than 300.
Solution. The result of rolling the dice can be described by a set of 60 numbers from 1 to 6. Consider any such set. If each number in the set is replaced from $x$ to 7 - $x$, we get a new set consisting of numbers from 1 to 6. In this case, if the sum of th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,786 |
6. [4 points] Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omega_{... | Answer: $\frac{R_{2}}{R_{1}}=\frac{7}{6}$.
Fig. 3: Option 10, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B... | \frac{7}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,787 |
7. [7 points] Find all values of the parameter $a$ for which the system of equations
$$
\left\{\begin{array}{l}
a^{2}-2 a x+10 y+x^{2}+y^{2}=0 \\
(|x|-12)^{2}+(|y|-5)^{2}=169
\end{array}\right.
$$
has exactly two solutions.
 \cup\{0\} \cup(20 ; 30)$.
Solution. Consider the second equation of the system. For $x \geqslant 0$ and $y \geqslant 0$ it takes the form $(x-12)^{2}+(y-5)^{2}=169$, and we obtain a circle of radius 13 centered at the point $(12 ; 5)$ (more precisely, its part lying in the first quadrant). Sin... | \in(-30;-20)\cup{0}\cup(20;30) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,788 |
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additi... | Answer: (a) $90^{\circ} ;$ (b) $\frac{9 \sqrt{7}}{4}$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, so $\angle B P D = \angle B T D = 90^{\circ}$. Therefore, triangles $A D P$ and $D C T$ are right triangles; $P M$ and $T N$ are their medians. Since the median of a right triangle to the hypo... | 90;\frac{9\sqrt{7}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,789 |
2. [6 points] Solve the equation $\sqrt{x+4}-\sqrt{6-x}+4=2 \sqrt{24+2 x-x^{2}}$. | Answer: $5, \frac{2-3 \sqrt{11}}{2}$.
Solution. Let $\sqrt{x+4}-\sqrt{6-x}=t$. Squaring both sides of this equation, we get $(x+4)-2 \sqrt{(x+4)(6-x)}+(6-x)=t^{2}$, from which $2 \sqrt{24+2 x-x^{2}}=10-t^{2}$. The equation becomes $t+4=10-t^{2}$; hence $t^{2}+t-6=0$, i.e., $t=2$ or $t=-3$. We consider each case separa... | 5,\frac{2-3\sqrt{11}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,790 |
3. [6 points] In the plane $O x y$, the equation $26 a^{2}-22 a x-20 a y+5 x^{2}+8 x y+4 y^{2}=0$ defines the coordinates of point $A$, and the equation $a x^{2}+2 a^{2} x-a y+a^{3}+1=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on opposite sides ... | Answer: $(-\infty ;-1) \cup\left(-\frac{1}{3} ; 0\right) \cup\left(\frac{8}{3} ;+\infty\right)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(4 y^{2}+4(2 x-5 a) y\right)+5 x^{2}-22 a x+26 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left((2 y)^{2}+2 \cdot 2 y \cdot(2 x-5 a)+(... | (-\infty;-1)\cup(-\frac{1}{3};0)\cup(\frac{8}{3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,791 |
4. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{2}{x^{2}+y^{2}}+x^{2} y^{2}=2 \\
x^{4}+y^{4}+3 x^{2} y^{2}=5
\end{array}\right.
$$ | Answer: $(1 ; \pm 1),(-1 ; \pm 1)$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form
$$
\left\{\begin{array}{l}
\frac{2}{u}+v=2 \\
u^{2}+v=5
\end{array}\right.
$$
From the first equation $v=2-\frac{... | (1;\1),(-1;\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,792 |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 59),(59 ; 59)$, and $(59 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 370330
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. Each of them contains 58 points inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 116 ways to choose the firs... | 370330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,793 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=3, AD=7$. Find... | Answer: b) $\frac{79}{100}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid or a rectangle, so a circle (let's call it $\Omega$) can be circumscribed around $ABCD$. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle C... | \frac{79}{100} | Geometry | proof | Yes | Yes | olympiads | false | 3,794 |
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additi... | Answer: (a) $90^{\circ} ;$ ( $\mathbf{\text { ( ) }} 5$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, therefore $\angle B P D=\angle B T D=90^{\circ}$. Consequently, triangles $A D P$ and $D C T$ are right-angled; $P M$ and $T N$ are their medians. Since the median of a right-angled triangle... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,795 |
2. [6 points] Solve the equation $\sqrt{x+3}-\sqrt{7-x}+4=2 \sqrt{21+4 x-x^{2}}$. | Answer: $6, \frac{4-3 \sqrt{11}}{2}$.
Solution. Let $\sqrt{x+3}-\sqrt{7-x}=t$. Squaring both sides of this equation, we get $(x+3)-2 \sqrt{(x+3)(7-x)}+(7-x)=t^{2}$, from which $2 \sqrt{21+4 x-x^{2}}=10-t^{2}$. The equation becomes $t+4=10-t^{2}$; hence $t^{2}+t-6=0$, i.e., $t=2$ or $t=-3$. We consider each case separa... | 6,\frac{4-3\sqrt{11}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,796 |
3. [6 points] In the plane $O x y$, the equation $5 a^{2}-4 a y+8 x^{2}-4 x y+y^{2}+12 a x=0$ defines the coordinates of point $A$, and the equation $a x^{2}-2 a^{2} x-a y+a^{3}+3=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on the same side of th... | Answer: $\left(-\frac{5}{2} ;-\frac{1}{2}\right) \cup(0 ; 3)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(y^{2}-4(x+a) y\right)+8 x^{2}+12 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\left(y^{2}-2 \cdot y \cdot(2 x+2 a)+(2 x+2 a)^{2}\right)-(2 x+2 a)^{2}+8 x^{2}... | (-\frac{5}{2};-\frac{1}{2})\cup(0;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,797 |
4. [ $[$ points] Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{6}{x^{2}+y^{2}}+x^{2} y^{2}=10 \\
x^{4}+y^{4}+7 x^{2} y^{2}=81
\end{array}\right.
$$ | Answer: $(\sqrt{3} ; \pm \sqrt{3}),(-\sqrt{3} ; \pm \sqrt{3})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form
$$
\left\{\begin{array}{l}
\frac{6}{u}+v=10 \\
u^{2}+5 v=81
\end{array}\right.
$$
Fro... | (\sqrt{3};\\sqrt{3}),(-\sqrt{3};\\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,798 |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 69),(69 ; 69)$, and ( $69 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of th... | Answer: 601460.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 68 points on each of them inside the square, and there are no duplicates among them (the intersection point of the lines has non-integer coordinates). There are 136 ways to choose the first point, and ... | 601460 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,799 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=2, AD=7$. Find... | Answer: b) $\frac{67}{81}$.
Solution. It is not difficult to show that $ABCD$ is an isosceles trapezoid or a rectangle, so a circle (let's call it $\Omega$) can be circumscribed around $ABCD$. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD =... | \frac{67}{81} | Geometry | proof | Yes | Yes | olympiads | false | 3,800 |
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additi... | Answer: (a) $90^{\circ};$ (b) $\frac{5 \sqrt{13}}{3 \sqrt{2}}$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $BD$, so $\angle BPD = \angle BTD = 90^{\circ}$. Therefore, triangles $ADP$ and $DCT$ are right triangles; $PM$ and $TN$ are their medians. Since the median of a right triangle to the hypote... | \frac{5\sqrt{13}}{3\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,801 |
2. [6 points] Solve the equation $\sqrt{x+2}-\sqrt{3-x}+3=2 \sqrt{6+x-x^{2}}$. | Answer: $2, \frac{1-2 \sqrt{6}}{2}$.
Solution. Let $\sqrt{x+2}-\sqrt{3-x}=t$. Squaring both sides of this equation, we get $(x+2)-2 \sqrt{(x+2)(3-x)}+(3-x)=t^{2}$, from which $2 \sqrt{6+x-x^{2}}=5-t^{2}$. The equation becomes $t+3=5-t^{2}$; hence $t^{2}+t-2=0$, i.e., $t=1$ or $t=-2$. We consider each case separately.
... | 2,\frac{1-2\sqrt{6}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,802 |
3. [6 points] On the plane $O x y$, the equation $5 a^{2}+12 a x+4 a y+8 x^{2}+8 x y+4 y^{2}=0$ defines the coordinates of point $A$, and the equation $a x^{2}-2 a^{2} x-a y+a^{3}+4=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on opposite sides of... | Answer: $(-\infty ;-2) \cup\left(0 ; \frac{2}{3}\right) \cup\left(\frac{8}{7} ;+\infty\right)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(4 y^{2}+4(2 x+a) y\right)+8 x^{2}+12 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left((2 y)^{2}+2 \cdot 2 y \cdot(2 x+a)+(2 x+a)... | (-\infty;-2)\cup(0;\frac{2}{3})\cup(\frac{8}{7};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,803 |
4. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{4}{x^{2}+y^{2}}+x^{2} y^{2}=5 \\
x^{4}+y^{4}+3 x^{2} y^{2}=20
\end{array}\right.
$$ | Answer: $(\sqrt{2} ; \pm \sqrt{2}),(-\sqrt{2} ; \pm \sqrt{2})$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form
$$
\left\{\begin{array}{l}
\frac{4}{u}+v=5 \\
u^{2}+v=20
\end{array}\right.
$$
From t... | (\sqrt{2};\\sqrt{2}),(-\sqrt{2};\\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,804 |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 65),(65 ; 65)$ and ( $65 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | # Answer: 500032.
Solution. There are two possible cases.
1) Both selected nodes lie on the lines specified in the condition. There are 64 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 128 ways to choose t... | 500032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,805 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=2, AD=5$. Find... | Answer: b) $\frac{39}{49}$.
Solution. It is not difficult to show that $ABCD$ is an isosceles trapezoid or a rectangle, so a circle (let's call it $\Omega$) can be circumscribed around $ABCD$. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD =... | \frac{39}{49} | Geometry | proof | Yes | Yes | olympiads | false | 3,806 |
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additi... | Answer: (a) $90^{\circ} ;$ (b) $\frac{\sqrt{35}}{3}$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, so $\angle B P D=\angle B T D=90^{\circ}$. Therefore, triangles $A D P$ and $D C T$ are right triangles; $P M$ and $T N$ are their medians. Since the median of a right triangle to the hypotenus... | 90;\frac{\sqrt{35}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,807 |
2. [6 points] Solve the equation $\sqrt{x+1}-\sqrt{4-x}+3=2 \sqrt{4+3 x-x^{2}}$. | Answer: $3, \frac{3-2 \sqrt{6}}{2}$.
Solution. Let $\sqrt{x+1}-\sqrt{4-x}=t$. Squaring both sides of this equation, we get $(x+1)-2 \sqrt{(x+1)(4-x)}+(4-x)=t^{2}$, from which $2 \sqrt{4+3 x-x^{2}}=5-t^{2}$. The equation becomes $t+3=5-t^{2}$; hence $t^{2}+t-2=0$, i.e., $t=1$ or $t=-2$. We consider each case separately... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,808 |
3. [6 points] On the plane $O x y$, the equation $2 a^{2}-2 a x-6 a y+x^{2}+2 x y+5 y^{2}=0$ defines the coordinates of point $A$, and the equation $a x^{2}+4 a^{2} x-a y+4 a^{3}+2=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on the same side of t... | Answer: $(-2 ; 0) \cup\left(\frac{1}{2} ; 3\right)$.
Solution. The first equation can be transformed as follows:
$$
\begin{aligned}
& \left(x^{2}+2(y-a) x\right)+5 y^{2}-6 a y+2 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\left(x^{2}+2 \cdot x \cdot(y-a)+(y-a)^{2}\right)-(y-a)^{2}+5 y^{2}-6 a y+2 a^{2}=0 \Leftr... | (-2;0)\cup(\frac{1}{2};3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,809 |
4. [5 points] Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x^{2}+y^{2}}+x^{2} y^{2}=\frac{5}{4} \\
2 x^{4}+2 y^{4}+5 x^{2} y^{2}=\frac{9}{4}
\end{array}\right.
$$ | Answer: $\left(\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right)$.
Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form
$$
\left\{\begin{array}{... | (\frac{1}{\sqrt{2}};\\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}};\\frac{1}{\sqrt{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,810 |
5. [5 points] On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 63),(63 ; 63)$, and $(63 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the... | Answer: 453902.
Solution. There are two possible cases.
1) Both selected nodes lie on the specified lines. There are 62 points on each of them inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 124 ways to choose the first point, and... | 453902 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,811 |
6. [7 points] The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$, and triangles $BOC$ and $AOD$ are equilateral. Point $T$ is symmetric to point $O$ with respect to the midpoint of side $CD$.
a) Prove that $ABT$ is an equilateral triangle.
b) Suppose it is additionally known that $BC=2, AD=4$. Find... | Answer: b) $\frac{7}{9}$.
Solution. a) It is not difficult to show that $ABCD$ is an isosceles trapezoid or a rectangle, so a circle (let's call it $\Omega$) can be circumscribed around $ABCD$. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD ... | \frac{7}{9} | Geometry | proof | Yes | Yes | olympiads | false | 3,812 |
2. Solve the inequality $\sqrt{\sqrt{x+1}-2}+\sqrt{x+82-18 \sqrt{x+1}}>5$. | Answer. $x \in[3 ; 35) \cup(120 ;+\infty)$.
Solution. Note that the second radicand can be written as $(\sqrt{x+1}-9)^{2}$. The inequality takes the form $\sqrt{\sqrt{x+1}-2}+|\sqrt{x+1}-9|>5$. Let $\sqrt{\sqrt{x+1}-2}=t$. Then we get
$$
t+\left|t^{2}-7\right|>5 \Leftrightarrow\left[\begin{array} { l }
{ t ^ { 2 } -... | x\in[3;35)\cup(120;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,813 |
3. On the sides of triangle $A B C$, points were marked: 10 - on side $A B$, 11 - on side $B C$, 12 - on side $A C$. At the same time, none of the vertices of the triangle were marked. How many triangles exist with vertices at the marked points? | Answer: 4951.
Solution. Three points out of the 33 given can be chosen in $C_{33}^{3}=5456$ ways. In this case, a triangle is formed in all cases except when all three points lie on one side of the triangle. Thus, $C_{12}^{3}+C_{11}^{3}+C_{10}^{3}=220+165+120=505$ ways do not work. Therefore, there are $5456-505=4951$... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,814 |
4. The continuation of the height $B H$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$ (points $B$ and $D$ lie on opposite sides of line $A C$). The degree measures of arcs $A D$ and $C D$, not containing point $B$, are $60^{\circ}$ and $90^{\circ}$, respectively. Determine in what ratio... | Answer: $\sqrt{3}: 1$.
Solution. By the inscribed angle theorem, angle $D C A$ is half of arc $A D$, and angle $D B C$ is half of arc $C D$. Therefore, $\angle D C H=30^{\circ}, \angle H B C=45^{\circ}$. Then triangle $B H C$ is a right and isosceles triangle, $B H=H C$. But $H D=C H \operatorname{tg} 30^{\circ}=\frac... | \sqrt{3}:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,815 |
5. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point $(50 ; 30)$. Find the number of such squares. | Answer: 930.
Solution. Draw through the given point $(50 ; 30)$ vertical and horizontal lines $(x=50$ and $y=30)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 30 ways: $(... | 930 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,816 |
6. a) On the coordinate plane, plot the figure $\Phi$, the coordinates of the points of which satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-y^{2} \leqslant 2(x-y) \\
x^{2}+y^{2} \leqslant 4(x+y-1)
\end{array}\right.
$$
b) Find the area of the figure $\Phi$ and the distance from the point $T(0 ; ... | Answer. $S=2 \pi ; \rho=2 \sqrt{2}-2$.
Solution. Transform the first inequality: $(x-y)(x+y)-2(x-y) \leqslant 0 \Leftrightarrow(x-y)(x+y-2) \leqslant 0$. It defines two vertical angles, the boundaries of which are the lines $y=x$ and $y=2-x$ (the points $(0 ; \pm 10)$ lie inside these angles). The second inequality ca... | S=2\pi;\rho=2\sqrt{2}-2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,817 |
2. Solve the inequality $\sqrt{\sqrt{x-1}-3}+\sqrt{x+99-20 \sqrt{x-1}}>5$. | Answer. $x \in[10 ; 50) \cup(145 ;+\infty)$.
Solution. Note that the second radicand can be written as $(\sqrt{x-1}-10)^{2}$. The inequality takes the form $\sqrt{\sqrt{x-1}-3}+|\sqrt{x-1}-10|>5$. Let $\sqrt{\sqrt{x-1}-3}=t$. Then we get
$$
t+\left|t^{2}-7\right|>5 \Leftrightarrow\left[\begin{array} { l }
{ t ^ { 2 ... | x\in[10;50)\cup(145;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,818 |
4. The continuation of the height $B H$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$ (points $B$ and $D$ lie on opposite sides of line $A C$). The degree measures of arcs $A D$ and $C D$, not containing point $B$, are $120^{\circ}$ and $90^{\circ}$, respectively. Determine in what rati... | Answer: $1: \sqrt{3}$.
Solution. By the inscribed angle theorem, angle $DCA$ is half of arc $AD$, and angle $DBC$ is half of arc $CD$. Therefore, $\angle DCH=60^{\circ}, \angle HBC=45^{\circ}$. Then triangle $BHC$ is a right and isosceles triangle, $BH=HC$. But $HD=CH \operatorname{tg} 60^{\circ}=CH \sqrt{3}$. Therefo... | 1:\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,820 |
5. On the coordinate plane, squares are considered, all vertices of which have natural coordinates, and the center is located at the point $(55 ; 25)$. Find the number of such squares. | Answer: 600.
Solution. Draw through the given point $(55 ; 25)$ vertical and horizontal lines $(x=55$ and $y=25)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 24 ways: $(... | 600 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,821 |
6. a) On the coordinate plane, plot the figure $\Phi$, the coordinates of the points of which satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y^{2}-x^{2} \leqslant 3(x+y) \\
x^{2}+y^{2} \leqslant 6 y-6 x-9
\end{array}\right.
$$
b) Find the area of the figure $\Phi$ and the distance from the point $T(-6 ... | Answer. $S=\frac{9 \pi}{2} ; \rho=3 \sqrt{2}-3$.
Solution. Transform the first inequality: $(y-x)(y+x)-3(y+x) \leqslant 0 \Leftrightarrow(y+x)(y-x-3) \leqslant 0$. It defines two vertical angles, the boundaries of which are the lines $y=x$ and $y=2-x$ (the points $( \pm 10 ; 0)$ lie inside these angles). The second in... | S=\frac{9\pi}{2};\rho=3\sqrt{2}-3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,822 |
2. Solve the inequality
$$
\sqrt{\frac{|x|-12}{2-x}}>x
$$ | Answer. $x \in(-\infty ;-12] \cup(2 ; 3)$.
Solution. The domain of the inequality is determined by the condition $\frac{|x|-12}{2-x} \geqslant 0 \Leftrightarrow \frac{(x-12)(x+12)}{2-x} \geqslant 0$. Using the method of intervals, we get $x \in(-\infty ;-12] \cup(2 ; 12]$. On the interval $x \in(-\infty ;-12]$, the le... | x\in(-\infty;-12]\cup(2;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,823 |
3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be d... | 1260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,824 |
4. In triangle $A B C$, it is known that $A B=3, A C=4, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C \cdot(O$ - the center of the circumcircle of triangle $A B C)$. | Answer: $S_{\triangle O A_{2} C}=\frac{13}{4 \sqrt{3}} ; S_{\triangle A_{1} A_{2} C}=\frac{13}{7 \sqrt{3}}$.
Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+9-2 \cdot \frac{1}{2} \cdot 3 \cdot 4=13$, $B C=\sqrt{13}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C}{... | S_{\triangleOA_{2}C}=\frac{13}{4\sqrt{3}};S_{\triangleA_{1}A_{2}C}=\frac{13}{7\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,825 |
6. Rays $A B$ and $D C$ intersect at point $P$, and rays $B C$ and $A D$ intersect at point $Q$. It is known that triangles $A D P$ and $Q A B$ are similar (vertices are not necessarily in the corresponding order), and quadrilateral $A B C D$ can be inscribed in a circle of radius 7.
a) Find $A C$.
b) Suppose it is a... | Answer. a) $A C=14$; b) $\angle D A C=45^{\circ}, S_{A B C D}=97$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\a... | AC=14;\angleDAC=45,S_{ABCD}=97 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,826 |
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-3 y^{2}+4 x+4} \leqslant 2 x+1 \\
x^{2}+y^{2} \leqslant 4
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of... | Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 3 y ^ { 2 } + 4 x + 4 \leqslant 4 x ^ { 2 } + 4 x + 1 , } \\
{ ( x + 2 ) ^ { 2 } - 3 y ^ { 2 } \geqslant 0 , } \\
{ 2 x + 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,827 |
2. Solve the inequality
$$
\sqrt{\frac{20-|x|}{x-3}}>x
$$ | Answer. $x \in(-\infty ;-20] \cup(3 ; 4)$.
Solution. The domain of the inequality is determined by the condition $\frac{20-|x|}{x-3} \geqslant 0 \Leftrightarrow \frac{(20-x)(20+x)}{x-3} \geqslant 0$. Using the method of intervals, we get $x \in(-\infty ;-20] \cup(3 ; 20]$. On the interval $x \in(-\infty ;-20]$, the le... | x\in(-\infty;-20]\cup(3;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,828 |
3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,829 |
4. In triangle $A B C$, it is known that $A B=4, A C=6, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$. ( $O$ - the center of the circumcircle of triangle $\left.A B C\r... | Answer: $S_{\triangle O A_{2} C}=\frac{7}{\sqrt{3}} ; S_{\triangle A_{1} A_{2} C}=\frac{7 \sqrt{3}}{5}$.
Solution. By the Law of Cosines for triangle $A B C$, we find that $B C^{2}=16+36-2 \cdot \frac{1}{2} \cdot 6 \cdot 4=28$, $B C=\sqrt{28}$. Then, by the Law of Sines, the radius of the circle $R$ is $\frac{B C}{2 \... | S_{\triangleOA_{2}C}=\frac{7}{\sqrt{3}};S_{\triangleA_{1}A_{2}C}=\frac{7\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,830 |
6. Rays $A B$ and $D C$ intersect at point $P$, and rays $B C$ and $A D$ intersect at point $Q$. It is known that triangles $A D P$ and $Q A B$ are similar (vertices are not necessarily in the corresponding order), and quadrilateral $A B C D$ can be inscribed in a circle of radius 4.
a) Find $A C$.
b) Suppose it is a... | Answer. a) $A C=8$; b) $\angle D A C=45^{\circ}, S_{A B C D}=31$.
Solution. a) Similarity of triangles is equivalent to the equality of all their angles. Since the angle at vertex $A$ is common to both triangles, there are two options: either $\angle A B Q=\angle A D P, \angle A Q B=\angle A P D$, or $\angle A B Q=\an... | AC=8;\angleDAC=45,S_{ABCD}=31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,831 |
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{y^{2}-8 x^{2}-6 y+9} \leqslant 3 y-1 \\
x^{2}+y^{2} \leqslant 9
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of... | Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ y ^ { 2 } - 8 x ^ { 2 } - 6 y + 9 \leqslant 9 y ^ { 2 } - 6 y + 1 , } \\
{ ( y - 3 ) ^ { 2 } - 8 x ^ { 2 } \geqslant 0 , } \\
{ 3 y - 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,832 |
11. It is known that the roots of the quadratic polynomial $f(x)=x^{2}+a x+b$ are each 2 more than the roots of the quadratic polynomial param 1. Let $M$ be the sum of the coefficients of $f(x)$, and $N$ be the sum of the coefficients of $g(x)$. What is the greatest value that $|M-N|$ can take?
| param1 | |
| :---: |... | 11. It is known that the roots of the quadratic polynomial $f(x)=x^{2}+a x+b$ are each 2 more than the roots of the quadratic polynomial param 1. Let $M$ be the sum of the coefficients of $f(x)$, and $N$ be the sum of the coefficients of $g(x)$. What is the maximum value that $|M-N|$ can take?
| param1 | Answer |
| :-... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,833 | |
12. Given a regular ragat1. In how many ways can three of its vertices be chosen so that they form a triangle with all sides of different lengths? (Two sets of vertices that differ only in the order of the vertices are considered the same.)
| ragat1 | |
| :---: | :---: |
| $21-$ gon | |
| 27 -gon | |
| 33-gon | |
... | 12. Given a regular ragat1. In how many ways can three of its vertices be chosen so that they form a triangle with all sides of different lengths? (Two sets of vertices that differ only in the order of the vertices are considered the same.)
| ragat1 | Answer |
| :---: | :---: |
| 21-gon | 1134 |
| 27-gon | 2592 |
| 33... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,834 |
15. Inside triangle $A B C$, a point $S$ is chosen. Lines $A S, B S$, and $C S$ intersect sides $B C, C A$, and $A B$ at points $K, L$, and $M$ respectively. Let $R(X Y Z)$ denote the radius of the circumcircle of triangle $X Y Z$. It turns out that $R(A K B): R(A K C)=$ param1, $R(A L B): R(C L B)=$ param 2. Find the ... | 15. Inside triangle $A B C$, a point $S$ is chosen. Lines $A S, B S$, and $C S$ intersect sides $B C, C A$, and $A B$ at points $K, L$, and $M$ respectively. Let $R(X Y Z)$ denote the radius of the circumcircle of triangle $X Y Z$. It turns out that $R(A K B): R(A K C)=$ param1, $R(A L B): R(C L B)=$ param 2. Find the ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,837 | |
16. Let $f(x)$ be a quadratic trinomial. The graph of the parabola $y=f(x)$ touches the lines param1 and param2. Find the smallest possible value of the discriminant of this trinomial.
| param1 | param2 | |
| :---: | :---: | :---: |
| $y=x-7$ | $y=21-3 x$ | |
| $y=x+2$ | $y=-5 x-10$ | |
| $y=-2 x-22$ | $y=x+11$ | ... | 16. Let $f(x)$ be a quadratic trinomial. The graph of the parabola $y=f(x)$ touches the lines param1 and param2. Find the smallest possible value of the discriminant of this trinomial.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $y=x-7$ | $y=21-3 x$ | -3 |
| $y=x+2$ | $y=-5 x-10$ | -5 |
| $y=-2 x-22$ | $y... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,838 |
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