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6. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}5|x|+12|y|=60, \\ x^{2}+y^{2}-2 y+1-a^{2}=0\end{array} \quad\right.$ a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a|=4$; b) $|a| \in\left\{\frac{48}{13}\right\} \cup\left(4 ; \frac{72}{13}\right) \cup\{\sqrt{145}\}$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced by $-y$. Therefore, the set of points defined by the first equation is symmetric with respec... | )||=4;b)||\in{\frac{48}{13}}\cup(4;\frac{72}{13})\cup{\sqrt{145}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,124 |
7. In triangle $ABC$, the median $BM$ is drawn; $MD$ and $ME$ are the angle bisectors of triangles $AMB$ and $CMB$ respectively. Segments $BM$ and $DE$ intersect at point $P$, and $BP=1$, $MP=3$.
a) Find the segment $DE$.
b) Suppose it is additionally known that a circle can be circumscribed around quadrilateral $ADE... | Answer. a) $D E=6$; b) $R=3 \sqrt{65}$.
Solution. a) By the property of the angle bisector of a triangle, we get $A D: D B=A M: M B, C E: E B=C M: M B$, and since $A M=C M$, it follows that $A D: D B=C E: E B$, so $A C \| D E$. Then $\angle P D M=\angle A M D=\angle B M D$, which means that triangle $P D M$ is isoscel... | 3\sqrt{65} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,125 |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 13, 13, and 35, respectively. Find the smallest possible value of $f(x)$. | Answer: $\frac{41}{4}$.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\mathrm{B}}$, then $x_{\mathrm{B}}=n+0.5$, and thus $f(x)$ can be represente... | \frac{41}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,126 |
2. Solve the inequality $x^{2}-2 x+1-\left|x^{3}-1\right|-2\left(x^{2}+x+1\right)^{2} \geqslant 0$. | Answer. $x \in\left[-1 ;-\frac{1}{2}\right]$.
Solution. The given inequality is equivalent to the inequality
$$
|x-1|^{2}-|x-1|\left(x^{2}+x+1\right)-2\left(x^{2}+x+1\right)^{2} \geqslant 0
$$
Let $|x-1|=u, x^{2}+x+1=v$ (note that $u \geqslant 0, v>0$, since $v$ is a quadratic trinomial with a negative discriminant)... | x\in[-1;-\frac{1}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,127 |
3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=12, P_{1} S_{1}=3$. Find the area of triangle $... | Answer: $\frac{225}{2}$.
Solution: Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=9$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of he... | \frac{225}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,128 |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $x^{2}+x y=30000000$. | Answer: 256.
Solution. By factoring the left and right sides of the equation, we get $x(x+y)=$ $3 \cdot 2^{7} \cdot 5^{7}$. Then, if $x>0$, $x$ is one of the divisors of the right side. The right side has a total of $2 \cdot 8 \cdot 8=128$ divisors (since any divisor can be represented as $3^{a} \cdot 2^{b} \cdot 5^{c... | 256 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,129 |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and at the same time,
the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b \vdots 5$, so one of the numbers $a$ or $b$ mus... | 1260 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,130 |
6. In triangle $A B C$, it is known that $A B=3, A C=4, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$ ( $O$ - the center of the circumcircle of triangle $\left.A B C\ri... | Answer. $S_{\triangle O A_{2} C}=\frac{13 \sqrt{3}}{12} ; S_{\triangle A_{1} A_{2} C}=\frac{13 \sqrt{3}}{21}$.
Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+9-2 \cdot \frac{1}{2} \cdot 3 \cdot 4=13$, $B C=\sqrt{13}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C... | S_{\triangleOA_{2}C}=\frac{13\sqrt{3}}{12};S_{\triangleA_{1}A_{2}C}=\frac{13\sqrt{3}}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,131 |
7. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
3|y|-4|x|=6 \\
x^{2}+y^{2}-14 y+49-a^{2}=0
\end{array}\right.
$$
a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a| \in\{5 ; 9\} ;$ b) $|a| \in\{3\} \cup(5 ; 9)$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ by $-y$. Therefore, the set of points defined by the first equation is symmetric with respect to both coordinate axes. In the first quadrant, we get a part o... | )||\in{5,9};b)||\in{3}\cup(5,9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,132 |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 14, and 14, respectively. Find the greatest possible value of $f(x)$. | Answer: 15.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text{v}}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,133 |
2. Solve the inequality $x^{2}+2 x+1-\left|x^{3}+1\right|-2\left(x^{2}-x+1\right)^{2} \leqslant 0$. | Answer. $x \in\left(-\infty ; \frac{1}{2}\right] \cup[1 ;+\infty)$.
Solution. The given inequality is equivalent to the inequality
$$
|x+1|^{2}-|x+1|\left(x^{2}-x+1\right)-2\left(x^{2}-x+1\right)^{2} \leqslant 0
$$
Let us denote here $|x+1|=u, x^{2}-x+1=v$ (note that $u \geqslant 0, v>0$, since $v$ is a quadratic tr... | x\in(-\infty;\frac{1}{2}]\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,134 |
3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=3, P_{1} S_{1}=9$. Find the area of triangle $A... | Answer: 72.
Solution. Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=6$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of the height to t... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,135 |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $y^{2}-x y=700000000$. | Answer: 324.
Solution. Factoring the left and right sides of the equation, we get $y(y-x)=$ $7 \cdot 2^{8} \cdot 5^{8}$. Then if $y>0$, $y$ is one of the divisors of the right side. The right side has a total of $2 \cdot 9 \cdot 9=162$ divisors (since any divisor can be represented as $7^{a} \cdot 2^{b} \cdot 5^{c}$, ... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,136 |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,137 |
6. In triangle $A B C$, it is known that $A B=4, A C=6, \angle B A C=60^{\circ}$. The extension of the angle bisector $A A_{1}$ intersects the circumcircle of triangle $A B C$ at point $A_{2}$. Find the areas of triangles $O A_{2} C$ and $A_{1} A_{2} C$. ( $O$ - the center of the circumcircle of triangle $\left.A B C\r... | Answer: $S_{O A_{2} C}=\frac{7}{\sqrt{3}} ; S_{A_{1} A_{2} C}=\frac{7 \sqrt{3}}{5}$.
Solution. By the cosine theorem for triangle $A B C$, we find that $B C^{2}=16+36-2 \cdot \frac{1}{2} \cdot 6 \cdot 4=28$, $B C=\sqrt{28}$. Then, by the sine theorem, the radius of the circle $R$ is $\frac{B C}{2 \sin 60^{\circ}}=\fra... | S_{OA_{2}C}=\frac{7}{\sqrt{3}};S_{A_{1}A_{2}C}=\frac{7\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,138 |
7. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{l}
5|x|-12|y|=5 \\
x^{2}+y^{2}-28 x+196-a^{2}=0
\end{array}\right.
$$
a) has exactly 3 solutions; b) has exactly 2 solutions. | Answer. a) $|a| \in\{13 ; 15\} ;$ b) $|a| \in\{5\} \cup(13 ; 15)$.
Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced by $-y$. Therefore, the set of points defined by the first equation is symmetric with respect to both coordinate axes. In the first quadrant,... | )||\in{13,15};b)||\in{5}\cup(13,15) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,139 |
2. Solve the equation $\left(\frac{3 x}{2}\right)^{\log _{3}(8 x)}=\frac{x^{7}}{8}$. | Answer: $x=\frac{729}{8}, x=2$.
Solution. By taking the logarithm to base 3, we obtain an equation equivalent to the original:
$$
\log _{3}(8 x) \cdot \log _{3}\left(\frac{3 x}{2}\right)=\log _{3}\left(\frac{x^{7}}{8}\right) \Leftrightarrow\left(\log _{3} x+3 \log _{3} 2\right)\left(1+\log _{3} x-\log _{3} 2\right)=7... | \frac{729}{8},2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,140 |
5. On the edge $A A_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $A T: A_{1} T=4: 1$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{\frac{5}{3}}$; b) $V=\frac{1280 \pi}{29 \sqrt{29}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each... | \frac{1280\pi}{29\sqrt{29}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,141 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x=|y-b|+\frac{3}{b} \\
x^{2}+y^{2}+32=a(2 y-a)+12 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in(-\infty, 0) \cup\left[\frac{3}{8} ;+\infty\right)$.
Solution. The second equation of the system can be transformed into the form $(x-6)^{2}+(y-a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center $(6 ; a)$. For all possible $a \in \mathbb{R}$, these circles sweep out the strip $4 \leq x \... | b\in(-\infty,0)\cup[\frac{3}{8};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,142 |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $BAD$ and $BCD$ respectively, with the first touching side $AD$ at point $K$, and the second touching side $BC$ at point $T$.
a) Find the r... | Answer. a) $r=4$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{5}-1}{2}$ or $\angle B D C=\pi-\operatorname{arctg} \frac{\sqrt{5}+1}{2}$.
Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (the center of the circle inscribed in an angle lies on the bisector of this angle... | )r=4,b)\angleBDC=\operatorname{arctg}\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,143 |
2. Solve the equation $\left(\frac{x}{243}\right)^{\log _{2}\left(\frac{9 x}{4}\right)}=\frac{729}{x^{4}}$. | Answer: $x=\frac{243}{4}, x=\frac{1}{9}$.
Solution. By taking the logarithm with base 2, we obtain an equation equivalent to the original:
$$
\log _{2}\left(\frac{x}{243}\right) \cdot \log _{2}\left(\frac{9 x}{4}\right)=\log _{2}\left(\frac{729}{x^{4}}\right) \Leftrightarrow\left(\log _{2} x-5 \log _{2} 3\right)\left... | \frac{243}{4},\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,144 |
5. On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 5$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{\frac{7}{3}} ;$ b) $V=\frac{3500 \pi}{37 \sqrt{37}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to eac... | \frac{3500\pi}{37\sqrt{37}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,145 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x=|y+a|+\frac{4}{a} \\
x^{2}+y^{2}+24+b(2 y+b)=10 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in(-\infty, 0) \cup\left[\frac{2}{3} ;+\infty\right)$.
“Phystech-2015”, 11th grade, solutions for ticket 2
Solution. The second equation of the system can be transformed into the form $(x-5)^{2}+(y+b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center $(5 ;-b)$. For all possible $b \in \math... | \in(-\infty,0)\cup[\frac{2}{3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,146 |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $ABC$ and $ADC$ respectively, with the first touching side $BC$ at point $K$, and the second touching side $AD$ at point $T$.
a) Find the r... | Answer. a) $r=3$, b) $\angle B D C=30^{\circ}$.
Solution. a) Segments $B O_{1}$ and $D O_{2}$ are the angle bisectors of angles $A B C$ and $A D C$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite an... | )r=3,b)\angleBDC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,147 |
2. Solve the equation $\left(\frac{x}{400}\right)^{\log _{5}\left(\frac{x}{8}\right)}=\frac{1024}{x^{3}}$. | Answer: $x=\frac{8}{5}, x=16$.
Solution. By logarithmizing with base 3, we obtain an equation equivalent to the original:
$$
\log _{5}\left(\frac{x}{400}\right) \cdot \log _{5}\left(\frac{x}{8}\right)=\log _{5}\left(\frac{1024}{x^{3}}\right) \Leftrightarrow\left(\log _{5} x-2-4 \log _{5} 2\right)\left(\log _{5} x-3 \... | \frac{8}{5},16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,148 |
5. On the edge $C C_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $C T: C_{1} T=1: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{2}$; b) $V=\frac{576 \pi \sqrt{3}}{11 \sqrt{11}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that these three vertices are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each o... | \frac{576\pi\sqrt{3}}{11\sqrt{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,149 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x=\frac{7}{b}-|y+b| \\
x^{2}+y^{2}+96=-a(2 y+a)-20 x
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in\left(-\infty ;-\frac{7}{12}\right] \cup(0 ;+\infty)$.
Solution. The second equation of the system can be transformed into the form $(x+10)^{2}+(y+a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center $(-10 ;-a)$. For all possible $a \in \mathrm{R}$, these circles sweep out the strip $-12 \... | b\in(-\infty;-\frac{7}{12}]\cup(0;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,150 |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $BAD$ and $BCD$ respectively, with the first touching side $AB$ at point $L$, and the second touching side $BC$ at point $F$.
a) Find the r... | Answer. a) $r=2$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{3}-1}{\sqrt{2}}$.
Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a ci... | )r=2,b)\angleBDC=\operatorname{arctg}\frac{\sqrt{3}-1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,151 |
2. Solve the equation $\left(\frac{x}{4}\right)^{\log _{5}(50 x)}=x^{6}$. | Answer: $x=\frac{1}{2}, x=2500$.
Solution. By logarithmizing with base 5, we obtain an equation equivalent to the original:
$$
\log _{5}(50 x) \cdot \log _{5}\left(\frac{x}{4}\right)=\log _{5}\left(x^{6}\right) \Leftrightarrow\left(\log _{5} x+2+\log _{5} 2\right)\left(\log _{5} x-2 \log _{5} 2\right)=6 \log _{5} x
$... | \frac{1}{2},2500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,152 |
3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. Answer: 4000. | Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.
a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathrm{Z}$. Then we get $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrightarrow(89 p+87)(p+... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,153 |
5. On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.
a) Find the ratio of the height of the prism to the edge... | Answer. a) $\sqrt{5}$; b) $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.
Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that the three vertices of the prism are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equa... | \frac{180\pi\sqrt{3}}{23\sqrt{23}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,154 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x=\frac{6}{a}-|y-a| \\
x^{2}+y^{2}+b^{2}+63=2(b y-8 x)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in\left(-\infty ;-\frac{2}{3}\right] \cup(0+\infty)$.
“Phystech-2015”, 11th grade, solution to ticket 4
Solution. The second equation of the system can be transformed into the form $(x+8)^{2}+(y-b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center at $(-8 ; b)$. For all possible $b \in \mat... | \in(-\infty;-\frac{2}{3}]\cup(0+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,155 |
7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $ABC$ and $ADC$ respectively, with the first touching side $BC$ at point $F$, and the second touching side $AD$ at point $P$.
a) Find the r... | Answer. a) $r=\sqrt{6}$, b) $\angle B D C=30^{\circ}$.
Solution. a) Segments $B O_{1}$ and $D O_{2}$ are the bisectors of angles $A B C$ and $A D C$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite a... | )r=\sqrt{6},b)\angleBDC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,156 |
1. Solve the inequality $\frac{\log _{3}\left(x^{4}\right) \cdot \log _{\frac{1}{3}}\left(x^{2}\right)+\log _{3}\left(x^{2}\right)-\log _{\frac{1}{3}}\left(x^{4}\right)+2}{\left(\log _{\frac{1}{3}}\left(x^{2}\right)\right)^{3}+64} \leq 0$.
“Phystech-2015”, 11th grade, solutions for ticket 5 | Answer. $x \in(-9 ;-3] \cup\left[-\frac{1}{\sqrt[4]{3}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[4]{3}}\right] \cup[3 ; 9)$.
Solution. The given inequality is equivalent to the following:
$$
\frac{2 \log _{3} x^{2} \cdot\left(-\log _{3} x^{2}\right)+\log _{3} x^{2}+2 \log _{3} x^{2}+2}{64-\log _{3}^{3} x^{2}} \leq 0
$... | x\in(-9;-3]\cup[-\frac{1}{\sqrt[4]{3}};0)\cup(0;\frac{1}{\sqrt[4]{3}}]\cup[3;9) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,157 |
2. Solve the equation $\left(\frac{7}{2} \cos 2 x+2\right) \cdot|2 \cos 2 x-1|=\cos x(\cos x+\cos 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\cos x(\cos x+\cos 5 x)=\cos x \cdot 2 \cos 3 x \cos 2 x=\cos 2 x \cdot(2 \cos x \cos 3 x)=\cos 2 x \cdot(\cos 2 x+\cos 4 x)=
$$
$$
=\cos 2 x\left(2 \cos ^{2} 2 x+\cos 2 x-1\right)=\cos 2 x(2 \cos... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,158 |
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=-\frac{2}{15} \\ \frac{1}{y}+\frac{1}{x+z}=-\frac{2}{3}, \\ \frac{1}{z}+\frac{1}{x+y}=-\frac{1}{4}\end{array}\right.$, | Answer. $(5; -1; -2)$.
Solution. Multiplying both sides of the first equation by $-\frac{15}{2} x(y+z)$, both sides of the second by $-\frac{3}{2} y(x+z)$, and the third by $-4 z(x+y)$, we obtain the system
“MIPT-2015”, 11th grade, solutions for ticket 5
$$
\left\{\begin{array}{l}
-7.5(x+y+z)=xy+xz \\
-1.5(x+y+z)=xy... | (5;-1;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,159 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a+y-x)=49 \\
y=15 \cos (x-b)-8 \sin (x-b)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in[-24 ; 24]$.
Solution. The first equation of the system can be transformed into the form $(x-a)^{2}+(y+a)^{2}=7^{2}$, hence it represents a circle of radius 7 with center $(a ;-a)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=17 \cos (x-b-\theta)$. ... | \in[-24;24] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,160 |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $A C=4$ and $\angle D B C=30^{\circ}$. A sphere passes through the vertices $D, A, B, B_{1}, C_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $16 \pi$, b) $90^{\circ}$, c) $48 \sqrt{3}$.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, we get that the acute angle of the rhombus is $60^{\circ}$. In the cross-section of the sphere by the plane $B C D$, we obtain a circle circumscribed around the triangle $A B D$. The c... | )16\pi,b)90,)48\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,161 |
1. Solve the inequality $\frac{125+\left(\log _{\frac{1}{2}}\left(x^{4}\right)\right)^{3}}{\log _{2}\left(x^{4}\right) \cdot \log _{2}\left(x^{2}\right)+6 \log _{\frac{1}{2}}\left(x^{4}\right)+17 \log _{2}\left(x^{2}\right)-3} \geq 0$. | Answer. $x \in[-2 \sqrt[4]{2} ;-\sqrt[4]{2}) \cup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup(\sqrt[4]{2} ; 2 \sqrt[4]{2}]$.
Solution. The given inequality is equivalent to the following:
$$
\frac{125-\left(2 \log _{2}\left(x^{2}\right)\right)^{3}}{2 \log _{2}\left(x^{2}\right... | x\in[-2\sqrt[4]{2};-\sqrt[4]{2})\cup(-\frac{1}{2\sqrt{2}};0)\cup(0;\frac{1}{2\sqrt{2}})\cup(\sqrt[4]{2};2\sqrt[4]{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,162 |
2. Solve the equation $\left(\frac{7}{4}-2 \cos 2 x\right) \cdot|2 \cos 2 x+1|=\cos x(\cos x-\cos 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in Z$.
Solution. Transform the right side of the equation:
$\cos x(\cos x-\cos 5 x)=\cos x \cdot(2 \sin 2 x \sin 3 x)=\sin 2 x \cdot(2 \sin 3 x \cos x)=\sin 2 x \cdot(\sin 4 x+\sin 2 x)=$
$$
=\sin 2 x(2 \sin 2 x \cos 2 x+\sin 2 x)=\sin ^{2} 2 x(1+2 \cos 2 x)=\left(1-... | \\frac{\pi}{6}+\frac{k\pi}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,163 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 b(b-x+y)=4 \\
y=5 \cos (x-a)-12 \sin (x-a)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in[-15 ; 15]$.
Solution. The first equation of the system can be transformed into the form $(x-b)^{2}+(y+b)^{2}=2^{2}$, hence it represents a circle of radius 9 with center $(-b ;-b)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=13 \cos (x-a-\theta)$.... | b\in[-15;15] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,164 |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $C D=3$ and $\angle A B D=30^{\circ}$. A sphere passes through the vertices $D, C, B, B_{1}, A_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $9 \pi$, b) $90^{\circ}$, c) 81.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, we get that the acute angle of the rhombus is $60^{\circ}$. In the cross-section of the sphere by the plane $A C D$, we obtain a circle circumscribed around the triangle $B C D$. The center of thi... | )9\pi,b)90,)81 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,165 |
1. Solve the inequality $\frac{\log _{2}\left(x^{6}\right) \cdot \log _{\frac{1}{2}}\left(x^{2}\right)-\log _{\frac{1}{2}}\left(x^{6}\right)-8 \log _{2}\left(x^{2}\right)+2}{8+\left(\log _{\frac{1}{2}}\left(x^{2}\right)\right)^{3}} \leq 0$. | Answer. $x \in(-2 ;-\sqrt[6]{2}] \cup\left[-\frac{1}{2} ; 0\right) \cup\left(0 ; \frac{1}{2}\right] \cup[\sqrt[6]{2} ; 2)$.
Solution. The given inequality is equivalent to the following:
$$
\frac{3 \log _{2} x^{2} \cdot\left(-\log _{2} x^{2}\right)+3 \log _{2} x^{2}-8 \log _{2} x^{2}+2}{8-\log _{2}^{3} x^{2}} \leq 0
... | x\in(-2;-\sqrt[6]{2}]\cup[-\frac{1}{2};0)\cup(0;\frac{1}{2}]\cup[\sqrt[6]{2};2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,166 |
2. Solve the equation $\left(3 \cos 2 x+\frac{9}{4}\right) \cdot|1-2 \cos 2 x|=\sin x(\sin x-\sin 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$\sin x(\sin x-\sin 5 x)=\sin x \cdot(-2 \sin 2 x \cos 3 x)=-\sin 2 x \cdot(2 \sin x \cos 3 x)=-\sin 2 x \cdot(\sin 4 x-\sin 2 x)=$
$$
=-\sin 2 x(2 \sin 2 x \cos 2 x-\sin 2 x)=\sin ^{2} 2 x(1-2 \cos ... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,167 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 a(a-x-y)=64 \\
y=8 \sin (x-2 b)-6 \cos (x-2 b)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \in[-18 ; 18]$.
Solution. The first equation of the system can be transformed into the form $(x-a)^{2}+(y-a)^{2}=8^{2}$, hence it represents a circle of radius 8 with center $(a ; a)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=10 \cos (x-2 b-\theta)$... | \in[-18;18] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,168 |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $B D=12$ and $\angle B A C=60^{\circ}$. A sphere passes through the vertices $D, A, B, B_{1}, C_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing th... | Answer. a) $48 \pi$, b) $90^{\circ}$, c) $192 \sqrt{3}$.
Solution. a) Since the diagonals of a rhombus are the bisectors of its angles, the acute angle of the rhombus is $60^{\circ}$, and $A C=4 \sqrt{3}$. In the cross-section of the sphere by the plane $A_{1} C_{1} B_{1}$, we obtain a circle circumscribed around the ... | )48\pi,b)90,)192\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,169 |
1. Solve the inequality $\frac{64+\left(\log _{\frac{1}{5}}\left(x^{2}\right)\right)^{3}}{\log _{\frac{1}{5}}\left(x^{6}\right) \cdot \log _{5}\left(x^{2}\right)+5 \log _{5}\left(x^{6}\right)+14 \log _{\frac{1}{5}}\left(x^{2}\right)+2} \leq 0$. | Answer. $x \in[-25 ;-\sqrt{5}) \cup\left(-\frac{1}{\sqrt[3]{5}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[3]{5}}\right) \cup(\sqrt{5} ; 25]$.
Solution. The given inequality is equivalent to the following:
$$
\frac{64-\left(\log _{5}\left(x^{2}\right)\right)^{3}}{-3 \log _{5}\left(x^{2}\right) \cdot \log _{5}\left(x^{2}... | x\in[-25;-\sqrt{5})\cup(-\frac{1}{\sqrt[3]{5}};0)\cup(0;\frac{1}{\sqrt[3]{5}})\cup(\sqrt{5};25] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,170 |
2. Solve the equation $\left(\frac{7}{4}-3 \cos 2 x\right) \cdot|1+2 \cos 2 x|=\sin x(\sin x+\sin 5 x)$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in Z$.
Solution. Transform the right side of the equation:
$\sin x(\sin x+\sin 5 x)=\sin x \cdot 2 \sin 3 x \cos 2 x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)=$
$$
=\cos 2 x\left(-2 \cos ^{2} 2 x+\cos 2 x+1\right)=\cos 2 x(1-\cos 2 x)(2 \cos... | \\frac{\pi}{6}+\frac{k\pi}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,171 |
4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=3: 7$. The bisector $B L$ of the given triangle and the segment $A M$ intersect at point $P$ at an angle of $90^{\circ}$.
a) Find the ratio of the area of triangle $A B P$ to the area of quadrilateral $L P M C$.
b) On the segment $M C$... | Answer. a) $39: 161$, b) $\arccos \sqrt{\frac{13}{15}}$.
Solution. a) In triangle $A B M$, segment $B P$ is both a bisector and an altitude, so triangle $A B M$ is isosceles, and $B P$ is also its median. Let $B M=3 x$, then $A B=3 x, M C=7 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=3 x: 10 ... | )\frac{39}{161},b)\arccos\sqrt{\frac{13}{15}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,172 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 b(b+x+y)=81 \\
y=4 \cos (x+3 a)-3 \sin (x+3 a)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \in[-14 ; 14]$.
Solution. The first equation of the system can be transformed into the form $(x+b)^{2}+(y+b)^{2}=9^{2}$, hence it represents a circle of radius 9 with center $(-b ;-b)$.
By introducing an auxiliary angle, the second equation of the system can be reduced to the form $y=5 \cos (x+3 a-\theta)$... | b\in[-14;14] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,173 |
7. At the base of the quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ lies a rhombus $A B C D$, in which $B D=3$ and $\angle A D C=60^{\circ}$. A sphere passes through the vertices $D, C, B, B_{1}, A_{1}, D_{1}$.
a) Find the area of the circle obtained by the intersection of the sphere with the plane passing thr... | Answer. a) $3 \pi$, b) $90^{\circ}$, c) $3 \sqrt{3}$.
Solution. a) Since $B D=3$ and the acute angle of the rhombus is $60^{\circ}$, then $A C=\sqrt{3}$. In the section of the sphere by the plane $A_{1} C_{1} D_{1}$, we obtain a circle circumscribed around the triangle $A_{1} B_{1} D_{1}$. The center of this circle is... | 3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,174 |
1. Solve the equation $x^{\log _{2}(8 x)}=\frac{x^{7}}{8}$. | Answer: $x=2, x=8$.
Solution. By logarithmizing with base 2, we get $\log _{2} x \cdot \log _{2}(8 x)=\log _{2} x^{7}-\log _{2} 8$, which is equivalent to the following: $\log _{2}^{2} x+3 \log _{2} x=7 \log _{2} x-3 \Leftrightarrow \log _{2}^{2} x-4 \log _{2} x+3=0$, from which $\log _{2} x=1$ or $\log _{2} x=3 ; x=2... | 2,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,175 |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x+\frac{1}{2}\right|=\sin ^{2} 3 x-\sin x \sin 3 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\begin{gathered}
\sin ^{2} 3 x-\sin x \sin 3 x=\sin 3 x(\sin 3 x-\sin x)=\sin 3 x \cdot 2 \cos 2 x \sin x=\cos 2 x \cdot(2 \sin x \sin 3 x)=\cos 2 x \cdot(\cos 2 x-\cos 4 x)= \\
=\cos 2 x\left(-2 \... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,176 |
3. Find the number of natural numbers $k$, not exceeding 242400, such that $k^{2}+2 k$ is divisible by 303. | Answer: 3200.
Solution. By factoring the dividend and divisor, we get the condition $k(k+2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k+2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+2):(3 \cdot 101) \Leftrightarrow p(101 p... | 3200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,177 |
4. Solve the system $\left\{\begin{array}{l}3 x \geq 2 y+16 \\ x^{4}+2 x^{2} y^{2}+y^{4}+25-26 x^{2}-26 y^{2}=72 x y .\end{array}\right.$ | Answer. $(6 ; 1)$.
Solution. Transform the equation of the system (add $36 x^{2}+36 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+25+10 x^{2}+10 y^{2}=36 x^{2}+36 y^{2}+72 x y \Leftrightarrow\left(x^{2}+y^{2}+5\right)^{2}=(6 x+6 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { ... | (6;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,178 |
5. On the edge $S A$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $K$ is marked such that $A K: K S=2: 3$. The point $K$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $S A B C D$.
a) Find the ratio $C S: C D$.
b) Suppose ... | Answer. a) $\sqrt{3}$, b) $V=\frac{9 \pi}{\sqrt{5}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H K, B H K, C H K, D H K$. They have the side $H K$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H K$ and $D H K$ are right angles, ang... | )\sqrt{3},b)V=\frac{9\pi}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,179 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
y=-b-x^{2}, \\
x^{2}+y^{2}+8 a^{2}=4+4 a(x+y)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \leq 2 \sqrt{2}+\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x-2 a)^{2}+(y-2 a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $(2 a ; 2 a)$. For all possible $a \in \mathrm{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt... | b\leq2\sqrt{2}+\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,180 |
7. Circles with centers \(O_{1}\) and \(O_{2}\) of equal radius are inscribed in the corners \(A\) and \(B\) of triangle \(ABC\), respectively, and point \(O\) is the center of the circle inscribed in triangle \(ABC\). These circles touch side \(AB\) at points \(K_{1}\), \(K_{2}\), and \(K\) respectively, with \(AK_{1}... | Answer. a) $A K=\frac{32}{5}$, b) $\angle C A B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.
Solution. a) The lines $A O_{1}$ and $B O_{2}$ are the angle bisectors of angles $A$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1... | )AK=\frac{32}{5},b)\angleCAB=2\arcsin\frac{3}{5}=\arccos\frac{7}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,181 |
1. Solve the equation $x^{\log _{3}\left(27 x^{2}\right)}=\frac{x^{9}}{81}$. | Answer: $x=3, x=9$.
Solution. By taking the logarithm with base 3, we get $\log _{3} x \cdot \log _{3}\left(27 x^{2}\right)=\log _{3} x^{9}-\log _{3} 81$, which is equivalent to the following: $2 \log _{3}^{2} x+3 \log _{3} x=9 \log _{3} x-4 \Leftrightarrow \log _{3}^{2} x-3 \log _{3} x+2=0$, from which $\log _{3} x=1... | 3,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,182 |
2. Solve the equation $\frac{1}{2}\left|\cos 2 x-\frac{1}{2}\right|=\cos ^{2} 3 x+\cos x \cos 3 x$. | Answer. $x= \pm \frac{\pi}{6}+\frac{k \pi}{2}, k \in \mathrm{Z}$.
Solution. Transform the right side of the equation:
$$
\begin{gathered}
\cos ^{2} 3 x+\cos x \cos 3 x=\cos 3 x(\cos x+\cos 3 x)=\cos 3 x \cdot 2 \cos 2 x \cos x=\cos 2 x \cdot(2 \cos x \cos 3 x)=\cos 2 x \cdot(\cos 2 x+\cos 4 x)= \\
=\cos 2 x\left(2 \c... | \\frac{\pi}{6}+\frac{k\pi}{2},k\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,183 |
3. Find the number of natural numbers $k$, not exceeding 353500, such that $k^{2}+k$ is divisible by 505. | Answer: 2800.
Solution: By factoring the dividend and divisor, we get the condition $k(k+1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k+1)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+1):(5 \cdot 101) \Leftrightarrow p(101 p... | 2800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,184 |
5. On the edge $S B$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $B L: L S=2: 5$. The point $L$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $S A B C D$.
a) Find the ratio $A S: C D$.
b) Suppose ... | Answer. a) $\sqrt{\frac{5}{3}}$, b) $V=\frac{125 \pi}{\sqrt{21}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H L, B H L, C H L, D H L$. They have the side $H L$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $A H L$ and $C H L$ are righ... | \frac{125\pi}{\sqrt{21}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,185 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
y=x^{2}-a \\
x^{2}+y^{2}+8 b^{2}=4 b(y-x)+1
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $a \geq-\sqrt{2}-\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x+2 b)^{2}+(y-2 b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center $(-2 b ; 2 b)$. For all possible $b \in \mathbf{R}$, the graphs of these functions sweep out the strip $-x-\sqrt{2} \... | \geq-\sqrt{2}-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,186 |
7. In the corners $B$ and $C$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $B K_{1}=4, C K_{2}=8$, and $B C=18$.
a) Fin... | Answer. a) $C K=12$, b) $\angle A B C=60^{\circ}$.
Solution. a) The lines $\mathrm{CO}_{2}$ and $\mathrm{BO}_{1}$ are the bisectors of angles $C$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1}$ and $O_{2}$ be $r$, and the r... | )CK=12,b)\angleABC=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,187 |
1. Solve the equation $x^{\log _{2}\left(0.25 x^{3}\right)}=512 x^{4}$. | Answer: $x=\frac{1}{2}, x=8$.
Solution. By logarithmizing with base 2, we get $\log _{2} x \cdot \log _{2}\left(\frac{1}{4} x^{3}\right)=\log _{2} x^{4}+\log _{2} 512$, which is equivalent to the following: $3 \log _{2}^{2} x-2 \log _{2} x=4 \log _{2} x+9 \Leftrightarrow \log _{2}^{2} x-2 \log _{2} x-3=0$, from which ... | \frac{1}{2},8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,188 |
3. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400. | Solution. Factoring the dividend and divisor, we get the condition $k(k-2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k-2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-2):(3 \cdot 101) \Leftrightarrow p(101 p-2) \vdots 3$. The... | 4400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,190 |
4. Solve the system $\left\{\begin{array}{l}2 x+y+8 \leq 0, \\ x^{4}+2 x^{2} y^{2}+y^{4}+9-10 x^{2}-10 y^{2}=8 x y .\end{array}\right.$ | Answer. $(-3 ;-2)$.
Solution. Transform the equation of the system (add $4 x^{2}+4 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+9-6 x^{2}-6 y^{2}=4 x^{2}+4 y^{2}+8 x y \Leftrightarrow\left(x^{2}+y^{2}-3\right)^{2}=(2 x+2 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l }
{ ... | (-3,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,191 |
5. On the edge $S A$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $K$ is marked such that $A K: K S=1: 4$. The point $K$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $2 A B C D$.
a) Find the ratio $D S: B C$.
b) Suppose ... | Answer. a) $\frac{2}{\sqrt{3}}$, b) $V=\frac{64 \pi}{\sqrt{15}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H K, B H K, C H K, D H K$. They have the side $H K$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H K$ and $D H K$ are right... | \frac{64\pi}{\sqrt{15}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,192 |
6. Find all values of the parameter $b$, for each of which there exists a number $a$ such that the system
$$
\left\{\begin{array}{l}
y=b-x^{2} \\
x^{2}+y^{2}+2 a^{2}=4-2 a(x+y)
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer. $b \geq-2 \sqrt{2}-\frac{1}{4}$.
Solution. The second equation of the system can be transformed into the form $(x+a)^{2}+(y+a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $( -a ;-a )$. For all possible $a \in \mathbb{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt{2} \... | b\geq-2\sqrt{2}-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,193 |
7. In the corners $C$ and $B$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $C K_{1}=3, B K_{2}=7$, and $B C=16$.
a) Fin... | Answer. a) $C K=\frac{24}{5}$, b) $\angle A C B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.
Solution. a) The lines $\mathrm{CO}_{1}$ and $\mathrm{BO}_{2}$ are the angle bisectors of angles $C$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles wi... | )CK=\frac{24}{5},b)\angleACB=2\arcsin\frac{3}{5}=\arccos\frac{7}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,194 |
1. Solve the equation $x^{\log _{5}(0.008 x)}=\frac{125}{x^{5}}$. | Answer: $x=5, x=\frac{1}{125}$.
Solution. By logarithmizing with base 5, we get $\log _{5} x \cdot \log _{5}\left(\frac{x}{125}\right)=\log _{5} 125-\log _{5} x^{3}$, which is equivalent to the following: $\log _{5}^{2} x-3 \log _{5} x=3-5 \log _{5} x \Leftrightarrow \log _{5}^{2} x+2 \log _{5} x-3=0$, from which $\lo... | 5,\frac{1}{125} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,195 |
3. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505. | Answer: 3600.
Solution. By factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.
a) $k \vdots: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow ... | 3600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,197 |
4. Solve the system $\left\{\begin{array}{l}x+3 y+14 \leq 0, \\ x^{4}+2 x^{2} y^{2}+y^{4}+64-20 x^{2}-20 y^{2}=8 x y .\end{array}\right.$ | Answer: $(-2, -4)$.
Solution. Transform the equation of the system (add $4 x^{2}+4 y^{2}$ to both sides):
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+64-16 x^{2}-16 y^{2}=4 x^{2}+4 y^{2}+8 x y \Leftrightarrow\left(x^{2}+y^{2}-8\right)^{2}=(2 x+2 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } ... | (-2,-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,198 |
5. On the edge $S C$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $C L: L S=1: 5$. The point $L$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $2 A B C D$.
a) Find the ratio $A S: A B$.
b) Suppose ... | Answer. a) $\frac{\sqrt{5}}{2}$, b) $V=\frac{125 \pi \sqrt{2}}{3 \sqrt{3}}$.
Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H L, B H L, C H L, D H L$. They have the side $H L$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H L$ and $D H L... | \frac{\sqrt{5}}{2},\frac{125\pi\sqrt{2}}{3\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,199 |
6. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system
$$
\left\{\begin{array}{l}
y=x^{2}+a \\
x^{2}+y^{2}+2 b^{2}=2 b(x-y)+1
\end{array}\right.
$$
has at least one solution $(x ; y)$. | Answer: $a \leq \sqrt{2}+\frac{1}{4}$.
Solution: The second equation of the system can be transformed into $(x-b)^{2}+(y+b)^{2}=1^{2}$, which means it represents a circle with radius 1 and center $(b, -b)$. For all possible $b \in \mathbf{R}$, the graphs of these functions sweep out the strip $-x-\sqrt{2} \leq y \leq ... | \leq\sqrt{2}+\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,200 |
7. In the corners $C$ and $A$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $AC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $C K_{1}=6, A K_{2}=8$, and $A C=21$.
a) Fin... | Answer. a) $C K=9$, b) $\angle A C B=60^{\circ}$.
Solution. a) The lines $\mathrm{CO}_{1}$ and $A O_{2}$ are the bisectors of angles $C$ and $A$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1}$ and $O_{2}$ be $r$, and the radius of ... | )CK=9,b)\angleACB=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,201 |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 700. The answer should be presented as an integer. | Answer: 2520.
Solution. Since $700=7 \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, one seven, and three ones, or (b) one four, two fives, one seven, and four ones. We will calculate the number of variants in each case.
(a) First, we choose two places out of... | 2520 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,202 |
2. [4 points] Given a geometric progression $b_{1}, b_{2}, \ldots, b_{3000}$, all of whose terms are positive, and their sum is $S$. It is known that if all its terms with indices divisible by 3 (i.e., $b_{3}, b_{6}, \ldots, b_{3000}$) are increased by 50 times, the sum $S$ will increase by 10 times. How will $S$ chang... | Answer: It will increase by $\frac{11}{8}$ times.
Solution. Let the common ratio of the geometric progression be $q$. Since all its terms are positive, $q>0$. If $q=1$, then $S=3000 b_{1}$, and when the terms with numbers divisible by 3 are increased by 50 times, we get the sum $S+49\left(b_{3}+b_{6}+\ldots+b_{3000}\r... | \frac{11}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,203 |
3. [4 points] Solve the equation $\left(\frac{x}{\sqrt{2}}+3 \sqrt{2}\right) \sqrt{x^{3}-4 x+80}=x^{2}+10 x+24$. | Answer: $4, \sqrt{13}-1$.
Solution. Factoring the right-hand side and multiplying both sides by $\sqrt{2}$, we get $(x+6) \sqrt{x^{3}-4 x+80}=\sqrt{2}(x+4)(x+6)$. From here, there are two possibilities: either $x+6=0$ (then $x=-6$, which does not fit the domain of definition, as the expression under the square root is... | 4,\sqrt{13}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,204 |
4. [6 points] Solve the inequality $2 x^{4}+x^{2}-4 x-3 x^{2}|x-2|+4 \geqslant 0$. | Answer: $(-\infty ;-2] \cup\left[\frac{-1-\sqrt{17}}{4} ; \frac{-1+\sqrt{17}}{4}\right] \cup[1 ;+\infty)$.
Solution. The inequality can be rewritten as $2 x^{4}+(x-2)^{2}-3 x^{2}|x-2| \geqslant 0$ or $\mid x-$ $\left.2\right|^{2}-3 x^{2}|x+2|+2 x^{4} \geqslant 0$. To factorize the left side, note that it represents a ... | (-\infty;-2]\cup[\frac{-1-\sqrt{17}}{4};\frac{-1+\sqrt{17}}{4}]\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,205 |
5. [5 points] A water strider and a water beetle slide counterclockwise around a float on two circles. A rectangular coordinate system is introduced on the water surface, in which the float (the common center of the circles) is at the point ( $0 ; 0$ ). The speed of the water strider is twice the speed of the beetle. A... | Answer: $\left(\frac{5}{\sqrt{2}}(1-\sqrt{7}) ; \frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right)$, $\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)... | (\frac{5}{\sqrt{2}}(1-\sqrt{7});\frac{5}{\sqrt{2}}(1+\sqrt{7})),(-\frac{5}{\sqrt{2}}(1+\sqrt{7});\frac{5}{\sqrt{2}}(1-\sqrt{7})),(\frac{5}{\sqrt{2}}(\sqrt{7}- | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,206 |
6. [6 points] a) Two circles of the same radius 5 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... | Answer: $C F=10, S_{\triangle A C F}=7$.
Fig. 1: Variant 3, Problem 6
Solution. a) Let $R=5$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. ... | CF=10,S_{\triangleACF}=7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,207 |
7. [6 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y-6-x|+|y-6+x|=12 \\
(|x|-8)^{2}+(|y|-6)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{4 ; 100\}$.
Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To remove the absolute values, we find the sets of points where the expressions under the absolute values are zero. These are the lines $y-6-x=0$ and $y-6+x=0$. They divide th... | \in{4;100} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,208 |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 4900. The answer should be presented as an integer. | Answer: 4200.
Solution. Since $4900=7^{2} \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, two sevens, and two ones, or (b) a four, two fives, two sevens, and three ones. We will calculate the number of variants in each case.
(a) First, we choose two places ou... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,209 |
2. [4 points] Given a geometric progression $b_{1}, b_{2}, \ldots, b_{3000}$, all of whose terms are positive, and their sum is $S$. It is known that if all its terms with indices divisible by 3 (i.e., $b_{3}, b_{6}, \ldots, b_{3000}$) are increased by 40 times, the sum $S$ will increase by 5 times. How will $S$ change... | Answer: It will increase by $\frac{11}{7}$ times.
Solution. Let the common ratio of the geometric progression be $q$. Since all its terms are positive, $q>0$. If $q=1$, then $S=3000 b_{1}$, and when the terms with numbers divisible by 3 are increased by 40 times, we get the sum $S+39\left(b_{3}+b_{6}+\ldots+b_{3000}\r... | \frac{11}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,210 |
3. [4 points] Solve the equation $\left(\frac{x}{2 \sqrt{2}}+\frac{5 \sqrt{2}}{2}\right) \sqrt{x^{3}-64 x+200}=x^{2}+6 x-40$. | Answer: $6, \sqrt{13}+1$.
Solution. Factoring the right-hand side and multiplying both sides by $2 \sqrt{2}$, we get $(x+10) \sqrt{x^{3}-64 x+200}=2 \sqrt{2}(x-4)(x+10)$. From here, there are two possibilities: either $x+10=0$ (then $x=-10$, which does not fit the domain of definition, as the expression under the squa... | 6,\sqrt{13}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,211 |
4. [6 points] Solve the inequality $4 x^{4}+x^{2}+4 x-5 x^{2}|x+2|+4 \geqslant 0$. | Answer: $(-\infty ;-1] \cup\left[\frac{1-\sqrt{33}}{8} ; \frac{1+\sqrt{33}}{8}\right] \cup[2 ;+\infty)$.
Solution. The inequality can be rewritten as $4 x^{4}+(x+2)^{2}-5 x^{2}|x+2| \geqslant 0$ or $\mid x+$ $\left.2\right|^{2}-5 x^{2}|x+2|+4 x^{4} \geqslant 0$. To factorize the left side, note that it represents a qu... | (-\infty;-1]\cup[\frac{1-\sqrt{33}}{8};\frac{1+\sqrt{33}}{8}]\cup[2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,212 |
5. [5 points] Around a hook with a worm, in the same plane as it, a carp and a minnow are swimming along two circles. In the specified plane, a rectangular coordinate system is introduced, in which the hook (the common center of the circles) is located at the point $(0 ; 0)$. At the initial moment of time, the carp and... | Answer: $(\sqrt{2}-4 ;-4-\sqrt{2}),(-4-\sqrt{2} ; 4-\sqrt{2}),(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4)$.
Solution. Let the points where the carp and the minnow are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles formed by the radius vectors of points $M$ and $N... | (\sqrt{2}-4;-4-\sqrt{2}),(-4-\sqrt{2};4-\sqrt{2}),(4-\sqrt{2};4+\sqrt{2}),(4+\sqrt{2};\sqrt{2}-4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,213 |
6. [6 points] a) Two circles of the same radius 13 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$, ... | Answer: $C F=26, S_{\triangle A C F}=119$.
Fig. 3: variant 4, problem 6
Solution. a) Let $R=13$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$... | CF=26,S_{\triangleACF}=119 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,214 |
7. [6 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
|y+x+8|+|y-x+8|=16 \\
(|x|-15)^{2}+(|y|-8)^{2}=a
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{49 ; 289\}$.
Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To open the absolute values, we need to find the sets of points where the expressions under the absolute values are zero. These are the lines $y+x+8=0$ and $y-x+8=0$. They di... | \in{49;289} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,215 |
1. [5 points] We rolled 80 fair dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) and counted the sum of the numbers that came up. Which of the following probabilities is smaller: the probability that this sum is greater than 400, or the probability that this sum is no... | Answer: It is more likely that the sum is no more than 160.
Solution. The result of rolling the dice can be described by a set of 80 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 the... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,216 |
2. [4 points] Given a finite arithmetic progression $a_{1}, a_{2} \ldots, a_{n}$ with a positive common difference, such that the sum of all its terms is $S$, and $a_{1}>0$. It is known that if the common difference of the progression is tripled, while the first term remains unchanged, then the sum $S$ doubles. By what... | Answer: It will increase by $\frac{5}{2}$ times.
Solution. The sum of the terms of the original progression is determined by the formula $S_{0}=\frac{2 a_{1}+(n-1) d}{2} \cdot n$, where $d-$ is the common difference of the progression. If $d$ is tripled, the sum of the terms will become $S_{1}=\frac{2 a_{1}+(n-1) 3 d}... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,217 |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+2 x-58}+5\right)\left|x^{3}-7 x^{2}+13 x-3\right| \leqslant 0$. | Answer: $2+\sqrt{3}$.
Solution. The first factor on the left side is positive for any valid 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}+2 x-58 \geqslant 0$) is equivalent to the equation $x^{3}-7 x^{2}+13 x-3=0$. One of the ... | 2+\sqrt{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,218 |
4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$.
翻译结果：
4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$. | Answer: $-\frac{4}{3} ; 1 ; \frac{-1 \pm \sqrt{17}}{2}$.
Solution. The equation can be rewritten as $3 x^{4}+(x-4)^{2}-4 x^{2}|x-4|=0$ or $|x-4|^{2}-4 x^{2} \mid x-$ $4 \mid+3 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x-4|$ with a discriminant equal to $\left(4... | -\frac{4}{3};1;\frac{-1\\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,219 |
5. [5 points] A water strider and a water beetle slide counterclockwise around a float along two circles. A rectangular coordinate system is introduced on the water surface, in which the float (the common center of the circles) is at the point ( $0 ; 0$ ). The speed of the water strider is twice the speed of the beetle... | Answer: $\left(\frac{5}{\sqrt{2}}(1-\sqrt{7}) ; \frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right)$, $\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)... | (\frac{5}{\sqrt{2}}(1-\sqrt{7});\frac{5}{\sqrt{2}}(1+\sqrt{7})),(-\frac{5}{\sqrt{2}}(1+\sqrt{7});\frac{5}{\sqrt{2}}(1-\sqrt{7})),(\frac{5}{\sqrt{2}}(\sqrt{7}- | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,220 |
6. [5 points] a) 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 $\omeg... | Answer: $\frac{R_{2}}{R_{1}}=\frac{4}{3}, R_{1}=\frac{\sqrt{3}}{4}, R_{2}=\frac{1}{\sqrt{3}}$.
Fig. 1: Variant 11, Problem 6
Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\om... | \frac{R_{2}}{R_{1}}=\frac{4}{3},R_{1}=\frac{\sqrt{3}}{4},R_{2}=\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,221 |
7. [7 points] Find all values of the parameter $a$, for each of which the system
$$
\left\{\begin{array}{l}
x^{2}+(y-a)^{2}=64 \\
(|x|-6)^{2}+(|y|-8)^{2}=100
\end{array}\right.
$$
has exactly two solutions. | Answer: $a \in\{-8-12 \sqrt{2}\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{12 \sqrt{2}+8\}$.
Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordi... | \in{-8-12\sqrt{2}}\cup(-24;-8]\cup[8;24)\cup{12\sqrt{2}+8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,222 |
1. [5 points] We rolled 90 fair dice (cubes with numbers from 1 to 6 on the faces; the probability of each face landing is the same) and counted the sum of the numbers that came up. Which probability is greater: that this sum is not less than 500, or that this sum is less than 130? | Answer: It is more likely that the sum is not less than 500.
Solution. The result of rolling the dice can be described by a set of 90 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 | 3,223 |
2. [4 points] Given a finite arithmetic progression $a_{1}, a_{2} \ldots, a_{n}$ with a positive common difference, such that the sum of all its terms is $S$, and $a_{1}>0$. It is known that if the common difference of the progression is increased by 4 times, while keeping the first term unchanged, then the sum $S$ inc... | Answer: It will increase by $\frac{5}{3}$ times.
Solution. The sum of the terms of the original progression is determined by the formula $S_{0}=\frac{2 a_{1}+(n-1) d}{2} \cdot n$, where $d$ is the common difference of the progression. If $d$ is increased by four times, the sum of the terms will become $S_{1}=$ $\frac{... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,224 |
3. [4 points] Solve the inequality $\left(\sqrt{x^{3}+x-90}+7\right)\left|x^{3}-10 x^{2}+31 x-28\right| \leqslant 0$. | Answer: $3+\sqrt{2}$.
Solution. The first factor on the left side is positive for any valid value of $x$. Since the second factor is always non-negative, the inequality on the domain of definition (i.e., under the condition $x^{3}+x-90 \geqslant 0$) is equivalent to the equation $x^{3}-10 x^{2}+31 x-28=0$. One of the ... | 3+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,225 |
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