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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value | source stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
669. For which numbers $x$ does the equation $x+|x|=0$ hold? | $\triangleright$ This equality means that $|x|=-x$, which occurs when $x \leqslant 0$. (Note that the boundary point 0 is also included.) $\triangleleft$ | x\leqslant0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,160 |
688. Where are the points $M(x, y)$ located, for which $x=y$ (the first coordinate is equal to the second)? | $\triangle$ These points are equidistant from the axes, so they lie on the bisector of the angle between the axes. $\triangleleft$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,163 |
689. Where are the points $M(x, y)$ located, for which $y>x$ (the second coordinate is greater than the first)? | $\triangleright$ Such points are obtained if we take the second coordinate equal to the first (from the previous problem), and then increase it. When the second coordinate is increased (the first one remains unchanged), the point moves vertically upwards. Therefore, the desired points are located above the line $y=x$.
... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 43,164 |
691. Point $M$ has coordinates $(x, y)$. Find the coordinates of point $M^{\prime}$, symmetric to point $M$ with respect to the x-axis. The same question for the point symmetric to $M$ with respect to the y-axis, and also for the point symmetric to $M$ with respect to the origin. | $\triangleright$ Points $M$ and $M^{\prime}$ lie on the same vertical line, so their abscissas are the same. They are at the same distance from the x-axis but on opposite sides of it, so their ordinates are equal in magnitude but opposite in sign. Answer: $(x,-y)$. (Note that this answer holds true for $y=0$ and $y<0$ ... | (x,-y),(-x,y),(-x,-y) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,166 |
694. Find the slope of a line that forms an angle of $45^{\circ}$ with the horizontal. | $\triangleright$ We already know that such a line is given by the equation $y=x$, so the slope is $1 . \triangleleft$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,169 |
695. Find the slope of the line that forms an angle of $60^{\circ}$ with the horizontal. | $\triangle$ In a right triangle with an angle of $60^{\circ}$, the smaller leg is equal to half the hypotenuse (complete to an equilateral triangle, problem 167). By taking the smaller leg to be 1, we can find the larger leg using the Pythagorean theorem: it is equal to $\sqrt{2^{2}-1^{2}}=\sqrt{3}$. Thus, the slope of... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,170 |
696. Find the slope of the line that forms an angle of $30^{\circ}$ with the horizontal. | $\Delta$ The same proportion between the legs of a right triangle with angles of $30^{\circ}$ and $60^{\circ}$, applied in reverse, gives the answer: $k=1 / \sqrt{3}$. $\triangleleft$
There are other "convenient" angle values where the slope can be expressed by a formula (with square roots). This can be done for an an... | \frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,171 |
697. Where are the points $(x, y)$ for which $y=2 x+1$? | $\triangleright$ To get such a point, you can first take a point $(x, y)$ where $y=2x$, and then increase $y$ by 1. In other words, you need to take the line $y=2x$ and then move all its points up by exactly 1, meaning the entire construction on the grid paper is simply shifted up by one cell. $\triangleleft$
In gener... | 2x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,172 |
698. On the coordinate plane, several lines (marked with numbers) are depicted. Each of them is defined by the equation $y=k x+b$ with some $k$ and $b$. List the lines (a) in ascending order of the coefficients $k$; (b) in ascending order of the values of $b$. | $\Delta$ The coefficient $k$ (slope) is the greater, the steeper the line rises when moving from left to right. Line № 2 has a negative slope, and in ascending order come lines 1, 3, 4.
To order the lines by increasing $b$, you need to look at their points of intersection with the y-axis and list the lines from bottom... | 2,1,3,4\text | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,173 |
699. Form the equation for the coordinates of points $(x, y)$ that are equidistant from $(-2,2)$ and $(2,0)$. | $\triangleright$ We know that such points lie on the perpendicular bisector, which can be drawn on graph paper and then its equation can be found.
But we can also act algebraically: the equality of distances is equivalent to the equality of their squares, so the condition of equidistance can be rewritten as $(x+2)^{2}... | 2x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,174 |
700. Find the distance from the point $(x, y)$ to the line $y=x$. | $\triangleright$ Vertically, this distance is equal to $|y-x|$ (see the figure). But the shortest distance will be the perpendicular dropped to the line. In a right triangle with angles of $45^{\circ}$, the leg is $\sqrt{2}$ times shorter than the hypotenuse, so the answer will be $|y-x| / \sqrt{2}$. $\triangleleft$
!... | \frac{|y-x|}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,175 |
718. Let's return to the example where the common measure fits into one segment 15 times, and into another 12 times. Can we find a larger segment that would also be a common measure? | $\triangleright$ Yes: $15=3 \cdot 5$, and $12=3 \cdot 4$, so if we take a common measure three times longer, it will fit (respectively) 5 times and 4 times. $\triangleleft$
In this example, we managed to find a larger common measure; a smaller one can always be found (it is enough to divide the common measure into sev... | 3\cdot53\cdot4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,176 |
719. A rectangle was cut into squares by drawing lines parallel to its sides. Prove that all the squares are equal and that their side is a common measure of the sides of the rectangle. | $\Delta$ Indeed, adjacent squares have the same size, since they share a common side. Therefore, all squares have the same size (one can follow a chain). This means the segment fits into each of the sides of the rectangle an integer number of times. $\triangleleft$
How to find the common measure of two segments, knowi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,177 |
721. Find the common measure of two segments of length $1 / 5$ m and $1 / 3$ m. | $\triangleright$ A segment of length $1 / 15$ m fits three times in the first and five times in the second. $\triangleleft$
In arithmetic, a similar operation is called reducing fractions to a common denominator. The common denominator can be taken as the product of the denominators of the fractions, if nothing smalle... | \frac{1}{15} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,179 |
722. Find the common measure of segments of length $\frac{4}{15}$ and $\frac{8}{21}$. | $\triangleright$ We can take a segment of length $\frac{1}{15 \cdot 21}=\frac{1}{315}$ as a common measure. It is clear that
$$
\frac{4}{15}=\frac{4 \cdot 21}{15 \cdot 21} \quad \text { and } \quad \frac{8}{21}=\frac{15 \cdot 8}{15 \cdot 21}
$$
so the proposed segment fits $4 \cdot 21$ times in the first segment and ... | \frac{1}{105} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,180 |
725. Prove that two segments are commensurable if and only if the ratio of their lengths is rational (expressed as a fraction with integer numerator and denominator). | $\triangleright$ If the common measure fits into the first segment $m$ times, and into the second segment $n$ times, then the ratio of the lengths is $m / n$. Conversely: if the ratio of the lengths $a / b$ is equal to the rational number $m / n$, then $a / m = b / n$, and a segment of this length fits into $a$ exactly... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,181 |
726. Given two segments: the longer one ($a$) and the shorter one ($b$). Then we can consider their difference $c = a - b$ (cutting from $a$ a segment equal to $b$ and taking the remainder). Prove that the common measures of $a$ and $b$ are the same as those of $b$ and $c$. | $\triangle$ We need to prove two things: (1) if some segment $d$ fits into $a$ and $b$ an integer number of times, then it fits into $c$ an integer number of times. And conversely: (2) if it fits into $b$ and $c$ an integer number of times, then it fits into $a$ an integer number of times. But this is almost obvious: t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,182 |
727. Self-taught inventor Euclid Algorithm made a machine that works as follows: you can insert a paper rectangle into it, and it will cut off a square (making a cut parallel to the shorter side). Then the remaining rectangle can be inserted into the machine again, another square will be cut off, and so on. What will h... | $\triangle$ First, the machine will cut out a square $14 \times 14$, leaving a rectangle $14 \times 22$. Then another square $14 \times 14$ will be cut out, leaving a rectangle $14 \times 8$. On the next step, a square $8 \times 8$ will be cut out, leaving a rectangle $8 \times 6$; from it, a square $6 \times 6$ will b... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,183 |
730. Prove that if the Euclidean algorithm finishes its work (leads to a pair of equal segments), then these segments will be a common measure of the two original segments. | $\Delta$ Problem 726 states that in the course of the algorithm, common measures (if they exist) remain the same: for example, the common measures of segments of length 14 and 36 are the same as those of segments of length 14 and 22, the same as those of segments $14,8, \ldots$, the same as those of segments $(2,2)$. E... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,184 |
731. Prove that if two segments have a common measure, then Euclid's algorithm will, after several steps, lead to equal segments (and stop). | $\Delta$ Let's accept the common measure as a unit of length. Then the lengths of the segments are expressed by positive integers. When one segment is subtracted from another, the numbers remain whole and positive, and this cannot continue indefinitely (the number of steps cannot exceed the initial length of the larger... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,185 |
732. A grasshopper jumps along a straight line to the left and right, and it can make large jumps of 36 cm and small jumps of 14 cm (to the left or to the right). Can it move 3 cm from its initial position in several jumps? 2 cm? 1234 cm? | $\Delta$ He cannot move 3 cm, as all his jumps shift him by an even number of centimeters (and by adding or subtracting even shifts, we cannot get an odd number: any shift will be a multiple of 2 cm).
He can move 2 cm. Indeed, the grasshopper can move 22 cm if he jumps 36 cm in one direction and then returns 14 cm in ... | 2\cdot36-5\cdot14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,186 |
734. Prove that Euclid's algorithm, when applied to a pair of segments that have a common measure, yields their greatest common measure, and this greatest common measure is a multiple of any other common measure. | $\triangleright$ Indeed, for the pairs of segments that appear during the algorithm, the common measures are the same (Problem 726), and for a pair of equal segments, their length will obviously be the greatest common measure, a multiple of any other.
If the lengths of the segments are integers, then Euclid's algorith... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,187 |
735. Find the greatest divisor of the pair of numbers 123456789 and 987654321 (in other words, the greatest common measure of segments of such lengths). | $\triangleright$ In principle, this problem can be solved by brute force (and even in a reasonable time if you have a computer at hand), trying all numbers from 1 to 123456789 (larger numbers clearly do not fit).
However, it can also be done without a computer, using the Euclidean algorithm. The number 987654321 can b... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,188 |
758. Prove that $\sin \left(90^{\circ}-\alpha\right)=\cos \alpha$. | $\triangleright$ The cathetus opposite one acute angle will be adjacent to the other, which is precisely $90^{\circ}-\alpha$ (where $\alpha$ is the first angle). $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,193 |
759. What is $\operatorname{tg}\left(90^{\circ}-\alpha\right)$, if $\operatorname{tg} \alpha=a$? | $\Delta$ Sine and cosine switch places, so that $\operatorname{tg}\left(90^{\circ}-\alpha\right)=1 / a . \triangleleft$ | \frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,194 |
760. What are the sines, cosines, and tangents of the angles $30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$? | $\triangle$ In a right-angled triangle with an angle of $30^{\circ}$, the smaller leg is half the hypotenuse (completing to an equilateral triangle, problem 167), so $\sin 30^{\circ}=1 / 2$. The larger leg can be found using the Pythagorean theorem: if the hypotenuse is 1, then it is
$$
\cos 30^{\circ}=\sqrt{1^{2}-\le... | \begin{aligned}&\sin30=\frac{1}{2},\quad\cos30=\frac{\sqrt{3}}{2},\quad\tan30=\frac{1}{\sqrt{3}}\\&\sin45=\frac{1}{\sqrt{2}},\quad\cos45^ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,195 |
761. It is known that $\operatorname{tg} \alpha=2$ (for the acute angle $\alpha$). Find $\sin \alpha$ and $\cos \alpha$. | $\triangleright$ Let's take the adjacent side as a unit, then the opposite side will be 2. By the Pythagorean theorem, the hypotenuse is $\sqrt{1^{2}+2^{2}}=\sqrt{5}$, so $\sin \alpha=2 / \sqrt{5}$ and $\cos \alpha=1 / \sqrt{5} . \triangleleft$
In general: if $\operatorname{tg} \alpha=a$, then the hypotenuse will be $... | \sin\alpha=\frac{2}{\sqrt{5}},\cos\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,196 |
764. Prove that if the sides of a parallelogram are equal to $a$ and $b$, and the angle between them is $\gamma$, then the area of this parallelogram is $a b \sin \gamma$. | $\Delta$ The area of a parallelogram is equal to the product of the base and the height. Taking $a$ as the base, we see that the height dropped to it is equal to $b \sin \gamma . \triangleleft$
This formula works not only for an acute angle $\gamma$, but also for an obtuse angle. For a right angle, it turns into the f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,198 |
765. Two sides of a triangle are equal to $a$ and $b$, and the angle between them is $\gamma$. What is the area of this triangle | $\Delta$ The area of a triangle is half the area of the parallelogram from the previous problem (half the product of the base and height), that is, $\frac{1}{2} a b \sin \gamma \cdot \triangleleft$
Notice that in the formula for the area of a triangle, different angles (and corresponding sides) can be used. For exampl... | \frac{1}{2}\sin\gamma | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,199 |
766. An inscribed angle $\alpha$ in a circle of radius $R$ subtends a chord of length $a$. Prove that $a=2 R \sin \alpha$. | $\triangleright$ First, note that all inscribed angles subtending the same chord are equal. Therefore, when calculating the length of the chord, we can take the angle that is most convenient. Let's take the angle where one of the sides is the diameter of the circle. This forms a right triangle with the hypotenuse $2 R$... | 2R\sin\alpha | Geometry | proof | Yes | Yes | olympiads | false | 43,200 |
767. Prove the Law of Sines using the previous problem and establish that the ratio mentioned in it is equal to the diameter of the circumscribed circle around the triangle. | $\triangleright$ Indeed, the previous problem shows that $a=2 R \sin \angle A$, which means that
$$
\frac{a}{\sin \angle A}=2 R
$$
Similarly for the other two sides of the triangle. $\triangleleft$
The Law of Sines can be useful for finding the elements of a triangle. For example, if we know all three sides of the t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,201 |
768. Prove the Law of Cosines. | $\Delta$ In essence, we have already encountered this theorem in problems 595 and 596, but now we can prove it very simply using the formula for distance in coordinates. Let's place our triangle so that point $C$ falls at the origin, side $a$ lies along the horizontal axis (and side $b$ as it turns out, at an angle $\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,202 |
1. Find the minimum value of the expression $2 x+y$, defined on the set of all pairs $(x, y)$ satisfying the condition
$$
3|x-y|+|2 x-5|=x+1
$$ | 1.I.1. The following figure shows a set defined by the equation $3|x-y|+|2 x-5|=x+1$.
Let $C=2 x+y$. The problem requires finding the minimum value of $C$ for all points in the depicted set... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,203 |
2. Find all values of the parameter $a$ for which the set of solutions to the inequality $x^{3}-\left(a^{2}+a+1\right) x^{2}+\left(a^{3}+a^{2}+a\right) x-a^{3} \geqslant 0$ is some ray. | Solution 1. Factorize the left side of the given inequality,
$$
x^{3}-\left(a^{2}+a+1\right) x^{2}+\left(a^{3}+a^{2}+a\right) x-a^{3}=(x-1)(x-a)\left(x-a^{2}\right)
$$
If among the numbers $1, a$, and $a^{2}$ there are no identical ones, then the solution set of the given inequality is the union of a ray and a segmen... | \1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 43,204 |
3. Find the minimum value of the expression
$$
\sqrt{a^{2}+1}+\sqrt{b^{2}+9}+\sqrt{c^{2}+25}
$$
if it is known that $a+b+c=12$. | Solution 1. Introduce vectors $\boldsymbol{m}(a ; 1), \boldsymbol{n}(b ; 3)$ and $\boldsymbol{p}(c ; 5)$ and set $\boldsymbol{q}=\boldsymbol{n}+\boldsymbol{m}+\boldsymbol{p}$. Since $a+b+c=12$, then $\boldsymbol{q}(12 ; 9)$. Since $|\boldsymbol{q}|=|\boldsymbol{n}+\boldsymbol{m}+\boldsymbol{p}| \leqslant|\boldsymbol{n}... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,205 |
4. In the tetrahedron $ABCD$, edge $CD$ is perpendicular to the plane $ABC$. Point $N$ is the midpoint of edge $AB$, point $M$ is the midpoint of edge $BD$, and point $K$ divides edge $DC$ in the ratio $DK: KC = 2: 1$. Prove that the line $CN$ is equidistant from the lines $BK$ and $AM$. | 1.I.4. Let's project this tetrahedron onto the plane $\alpha$, perpendicular to the line $C N$ (figure).
The distance from the line $C N$ to the line $B K$ and the distance from the line $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,206 |
5. Natural numbers $a, b$ and $c$ are pairwise coprime. The sequence of numbers $a, 2a, 3a, \ldots, b, 2b, 3b, \ldots, c, 2c, 3c, \ldots$ is ordered in ascending order. A new finite sequence is then constructed from the ordered sequence, consisting of ones and twos according to the following rule. Instead of a number t... | 1.I.5. Let $a<b<c$ be the desired numbers. We should consider all numbers less than $abc$. Among them, there are $bc-1$ numbers divisible by $a$, $ac-1$ numbers divisible by $b$, and $ab-1$ numbers divisible by $c$. Thus,
$$
bc-1+ac-1+ab-1=356+2 \cdot 36, \text{ or } ab+bc+ac=431.
$$
Similarly, there are $c-1$ number... | =7,b=9,=23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,207 |
6. Prove that if the equation $x^{4}-a x^{3}+2 x^{2}-b x+1=0$ has a real root, then $a^{2}+b^{2} \geqslant 8$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution 1. Since $x=0$ is not a root of the given equation, we are entitled to move to the equation
$$
x^{2}-a x+2-\frac{b}{x}+\frac{1}{x^{2}}=0
$$
transforming the left side as follows.
$$
\begin{aligned}
& x^{2}-a x+2-\frac{b}{x}+\frac{1}{x^{2}}= \\
&=x^{2}-a x+\frac{a^{2}}{4}+\frac{1}{x^{2}}-\frac{b}{x}+\frac{b^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,208 |
7. Find all natural $n$ and $k$ such that for all $a, b \geqslant 0$ the inequality
$$
a^{k} b^{k}\left(a^{2}+b^{2}\right)^{n} \leqslant \frac{(a+b)^{2 k+2 n}}{2^{2 k+n}}
$$
## Methodological Block
A. Below are the solutions to problems $8-10$. Evaluate each solution and the answers obtained. Indicate all errors and... | 1.I.7. We will prove that the given inequality is valid for all $a, b \geqslant 0$ if and only if $k \geqslant n$. Let $u=(a+b)^{2}$ and $v=2 a b$. Then $a^{2}+b^{2}=u-v$, and $u \geqslant 2 v$. The given inequality can be rewritten as
$$
v^{k}(u-v)^{n} \leqslant\left(\frac{u}{2}\right)^{k+n}, \text { or }\left(\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,209 |
8. Find the minimum value of the function
$$
f(x)=4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5
$$
Solution: Since
$$
\begin{aligned}
4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5=2^{2 x}+ & 2^{-2 x}-2\left(2^{x}+2^{-x}\right)+5= \\
=2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x} & -2-2\left(2^{x}+2^{-x}\right)+5= \\
& =\left(2^{x}+2^{-x}\right)^{2... | 1.I.8. Comment. The following error was made in the given solution. Since $2^{x}>0$, the range of the function $y=2^{x}+2^{-x}$ is the interval $[2 ;+\infty)$. Therefore, it is necessary to find the minimum value of the function $g(t)=t^{2}-2 t+3$ on the interval $[2 ;+\infty)$. On this interval, the function $g(t)$ is... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,210 |
9. Solve the equation $\sin x-\cos x-3 \sin 2 x+1=0$.
Solution: Let $t=\sin x-\cos x$. Since $t^{2}=(\sin x-\cos x)^{2}=$ $1-\sin 2 x$, we get the equation $t-3\left(1-t^{2}\right)+1=0$, or $3 t^{2}+t-2=0$. The roots of this equation are $t=-1 ; \frac{2}{3}$. Substituting the found values into the formula $\sin 2 x=1-... | 1.I.9. Comment. The following error is made in the given solution. The equation $\sin 2 x=t^{2}-1$ is not equivalent to the equation $\sin x-\cos x=t$, but is a consequence of it. Therefore, the answer contains numbers that are not solutions to the original equation. | \frac{\pik}{2};\frac{(-1)^{k}}{2}\arcsin\frac{5}{9}+\frac{\pik}{2},k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,211 |
10. Does there exist a function $f$ such that $\sin 4 x=f(\sin x)$?
Solution: Since $\sin 4 x=2 \sin 2 x \cos 2 x=4 \sin x\left(1-2 \sin ^{2} x\right) \cos x$, then
$\sin 4 x=\left\{\begin{array}{l}4 \sin x\left(1-2 \sin ^{2} x\right) \sqrt{1-\sin ^{2} x}, \text { if } x \in\left[-\frac{\pi}{2}+2 \pi k ; \frac{\pi}{2... | 1.I.10. Comment. The following error is made in the given justification. The formula expressing $\sin 4 x$ in terms of $\sin x$ does not define a function $f(x)$ such that $\sin 4 x=f(\sin x)$, since the conditions in this formula are set not on the values of $\sin x$, but on the argument of the sine. Moreover, such a ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,212 |
11. Compare the numbers $\sqrt{2005}+\sqrt{2007}$ and $2 \sqrt{2006}$. | Solution 1. From the inequality between the arithmetic mean and the quadratic mean, it follows that
$$
\frac{\sqrt{2005}+\sqrt{2007}}{2}<\sqrt{\frac{2005+2007}{2}}=\sqrt{2006}
$$
from which the desired inequality follows.
Topic: "Inequalities between means" | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 43,213 |
12. Given an isosceles triangle with base $a$, side $b$, and the angle at the vertex equal to $12^{\circ}$. Prove that $b < 5a$.
$$
\text { II (onsite) round }
$$ | Solution 1. We will complete the given triangle $A B C$ to a regular triangle $A B C_{4}$ by sequentially attaching triangles equal to the given one to the sides (figure).
The length of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,214 |
1. Given the function $f(x)=\frac{x^{3}}{1+x^{3}}$. Find the value of the sum
$$
\begin{aligned}
f\left(\frac{1}{1}\right)+f\left(\frac{2}{1}\right)+\ldots+f\left(\frac{2007}{1}\right) & +f\left(\frac{1}{2}\right)+f\left(\frac{2}{2}\right)+\ldots+f\left(\frac{2007}{2}\right)+ \\
& +\ldots+f\left(\frac{1}{2007}\right)+... | 1.II.1. Note that $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x^{3}}}{1+\frac{1}{x^{3}}}=\frac{1}{1+x^{3}}$, from which it follows that
$$
f(x)+f\left(\frac{1}{x}\right)=\frac{x^{3}}{1+x^{3}}+\frac{1}{1+x^{3}}=1 .
$$
Therefore, for each pair of distinct numbers $k$ and $n$ from the set $\{1,2, \ldots, 2007\}$, the equa... | \frac{2007^2}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,215 |
2. Peter and Pavel are playing the following game: they take turns placing coefficients in the equation
$$
x^{3}+\ldots x^{2}+\ldots x+\ldots=0
$$
First, Peter places a coefficient in any of the three free spots, then Pavel in any of the two remaining spots, and then Peter in the last remaining spot. Prove that for a... | 1.II.2. Let Peter's first move be to place -1 as the coefficient of $x$. If Pavel places any number $a \neq \pm 1$ in one of the two remaining positions, then Peter can place $-a$ in the last position. As a result, we get a polynomial of the form
$$
x^{3}+b x^{2}-x-b=(x-1)(x+1)(x+b)
$$
Since by assumption $b \neq \pm... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,216 |
3. Find the number of solutions to the equation $x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]$ (here, as usual, $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$). | 1.II.3. Let $x=30 k+d$, where $k \in \mathbb{Z}$, and $d \in\{0,1, \ldots, 29\}$. Substituting this expression into the given equation, we get
$$
30 k+d=15 k+\left[\frac{d}{2}\right]+10 k+\left[\frac{d}{3}\right]+6 k+\left[\frac{d}{5}\right]
$$
or $k=d-\left[\frac{d}{2}\right]-\left[\frac{d}{3}\right]-\left[\frac{d}{... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,217 |
4. Let $p(x)$ be a polynomial, all of whose roots are real and distinct. Prove that $\left(p^{\prime}(x)\right)^{2} \geqslant p(x) p^{\prime \prime}(x)$. | Solution 1. We will use induction on the degree of the polynomial. The base case is obvious. Suppose the statement is true for a polynomial of degree $n-1$. Let $p(x)$ be a polynomial of degree $n$, and write it as $p(x)=\left(x-x_{1}\right) q(x)$. By the inductive hypothesis, $\left(q^{\prime}(x)\right)^{2} \geqslant ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,218 |
5. Point $A_{1}$ lies on side $B C$ of triangle $A B C$, point $B_{1}$ lies on side $A C$. Let $K$ be the intersection point of segments $A A_{1}$ and $B B_{1}$. Find the area of quadrilateral $A_{1} C B_{1} K$, given that the area of triangle $A_{1} B K$ is 5, the area of triangle $A B_{1} K$ is 8, and the area of tri... | 1.II.5. We will use the following consideration. If a line passes through the vertex of a triangle, then the ratio of the areas of the parts into which it divides the triangle is equal to the ratio of the lengths of the segments into which it divides the opposite side, $\frac{S_{1}}{S_{2}}=\frac{m}{n}$. Let's connect v... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,219 |
6. Solve the following problem using as many methods as possible. Methods are considered different if they use different mathematical ideas, as well as various technical approaches to implementing the same idea. Indicate the place of each method used in the school mathematics curriculum.
Determine (depending on the va... | Solution 1. The given function is defined and continuous on the interval $[0 ; 3]$. Let's compute its derivative
$$
f^{\prime}(x)=\frac{1}{2 \sqrt{x}}-\frac{1}{\sqrt{6-2 x}}=\frac{\sqrt{6-2 x}-2 \sqrt{x}}{2 \sqrt{x(6-2 x)}}.
$$
By solving the inequality $\sqrt{6-2 x} \geqslant 2 \sqrt{x}$, we get that $x \leqslant 1$... | \in[\sqrt{3},\sqrt{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,220 |
7. Evaluate the solution and the answer provided below. Indicate all errors and shortcomings.
Solve the equation $a\left(\frac{1}{\cos x}-\operatorname{tg} x\right)=1$.
Solution: Domain of definition: $x \neq \frac{\pi}{2}+\pi k, k \in \mathbb{Z}$. It is clear that when $a=0$ the equation
has no solutions. When $a \n... | 1.II.7. Comment. The given solution contains two errors. Firstly, the transition from the equation $a \sin x+\cos x=a$ to the equation $\cos \left(x-\arccos \frac{1}{\sqrt{a^{2}+1}}\right)=\frac{a}{\sqrt{a^{2}+1}}$ is incorrect. The issue is that $\sin \left(\arccos \frac{1}{\sqrt{a^{2}+1}}\right)$ equals $\frac{a}{\sq... | \arccos\frac{2}{^{2}+1}+2\pik,k\in\mathbb{Z},for||\geqslant1;-\arccos\frac{2}{^{2}+1}+2\pik,k\in\mathbb{Z},for\in(-1;0)\cup( | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,221 |
1. The coefficients of the quadratic equation $a x^{2}+b x+c=0$ satisfy the condition $2 a+3 b+6 c=0$. Prove that this equation has a root in the interval $(0 ; 1)$. | Solution 1. Let $f(x)=a x^{2}+b x+c$ and assume that the equation $a x^{2}+b x+c=0$ has no roots in the interval $(0 ; 1)$. Then the numbers $f(0)=c, 4 f\left(\frac{1}{2}\right)=a+2 b+4 c$ and $f(1)=a+b+c$ have the same sign (or the first or third of them are zero). Therefore, their sum, which is $2 a+3 b+6 c$, cannot ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,222 |
2. Prove that for all $0<x<\frac{\pi}{2}$ the inequality $2^{\sin x}+2^{\tan x} \geqslant 2^{x+1}$ holds. | 2.I.2. The inequality holds
$$
2^{\sin x}+2^{\tan x} \geqslant 2 \sqrt{2^{\sin x} \cdot 2^{\tan x}}=2 \cdot 2^{\frac{\sin x+\tan x}{2}}.
$$
Thus, it is sufficient to prove that $\sin x+\tan x \geqslant 2 x$ for $x \in\left[0 ; \frac{\pi}{2}\right)$. Consider the function $f(x)=\sin x+\tan x-2 x$. For all $x \in\left[... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,223 |
3. Calculate $\left[\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\ldots+\sqrt[2009]{\frac{2009}{2008}}\right]$ (here $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$). | 2.I.3. Since each of the numbers is greater than 1, their sum is greater than 2008. Let's prove that this sum is less than 2009. By the inequality between the geometric mean and the arithmetic mean, we have
$$
\sqrt[n+1]{\frac{n+1}{n}}=\sqrt[n+1]{\left(1+\frac{1}{n}\right) \cdot 1^{n}}<\frac{n+1+\frac{1}{n}}{n+1}=1+\f... | 2008 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,224 |
4. Solve the equation $2^{x}=3^{y}+1$ in integers. | Solution 1. Since $3^{y}+1>1$, then $x>0$, which means $x \geqslant 1$. Therefore, $2^{x} \geqslant 2$, from which $3^{y} \geqslant 1$, hence $y \geqslant 0$. When $y=0$, we get $x=1$.
Let $y \geqslant 1$. The number $3^{y}+1$ when divided by 3 has a remainder of 1, so the number $x=2n$ is even. Then $2^{2n}-1=\left(2... | (1;0),(2;1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,225 |
5. Prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{n+1}+\sqrt{n+2}]$ for each natural number $n$ (here $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$). | 2.I.5. It is sufficient to prove that if $k$ is such a natural number that $\sqrt{n}+\sqrt{n+1}<k$, then $\sqrt{n}+\sqrt{n+2}<k$ as well. Let's transform the first inequality.
$$
\begin{gathered}
\sqrt{n+1}<k-\sqrt{n}, \text { from which } n+1<k^{2}+n-2 k \sqrt{n}, \text { or } 2 k \sqrt{n}<k^{2}-1, \\
\text { from wh... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,226 |
6. A segment divides a triangle into two new figures with equal perimeters and areas. Prove that the center of the circle inscribed in the triangle lies on this segment. | Solution 1. Let segment $E F$ divide triangle $A B C$ into two figures of equal perimeters and areas, and let $O$ be the point of intersection of the bisector $C D$ of the triangle with segment $E F$ (figure).
\left(a-x^{2}-2 x\right)=0
$$
has exactly two roots.
Solution: The given equation is equivalent to the system of equations $a-5+|x+1|=0$ and $a-x^{2}-2 x=0$. First, let's investigate the first equation.
Write it as $|x+1|=5-a$. It is clear... | 2.I.8. Comment. The case of the roots of both equations in the system coinciding has not been considered in the given solution.
Correct answer: $(-\infty ;-1) \cup\{3\} \cup(5 ;+\infty)$. | (-\infty;-1)\cup{3}\cup(5;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,229 |
9. Solve the equation
$$
\sqrt[3]{x^{2}+2 x}+\sqrt[3]{3 x^{2}+6 x-4}=\sqrt[3]{x^{2}+2 x-4}
$$
Solution: Let's make the substitution $t=x^{2}+2 x$, as a result of which the equation takes the form $\sqrt[3]{t}+\sqrt[3]{3 t-4}=\sqrt[3]{t-4}$. Further, we have
$$
\begin{aligned}
& \sqrt[3]{t}+\sqrt[3]{3 t-4}=\sqrt[3]{t... | 2.I.9. Comment. The following error was made in the given solution. The equation $\sqrt[3]{t(3 t-4)} \cdot \sqrt[3]{t-4}=-t$ is not equivalent to the original equation, but is only a consequence of it, so its roots may not be roots of the original equation. Indeed, the root $t=2$ is not a root of the original equation.... | {-2;0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,230 |
10. Prove the inequality $a^{2}+b^{2}+c^{2} \leqslant 9 R^{2}$, where $a, b$ and $c$ are the lengths of the sides of a triangle, and $R$ is the radius of the circumscribed circle around it. | Solution 3. Unlike the previous solutions, the third solution is purely algebraic.
Since $R=\frac{a b c}{4 S}$, the given inequality is equivalent to the inequality $16 S^{2}\left(a^{2}+b^{2}+c^{2}\right) \leqslant 9 a^{2} b^{2} c^{2}$. Now let's use the formula expressing the area of a triangle in terms of the length... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,231 |
11. Find all values of $b$ for which the equation
$$
a^{2-2 x^{2}}+(b+4) a^{1-x^{2}}+3 b+4=0
$$
has no solutions for any $a>1$. | Solution 1. Rewrite the equation as $-\frac{(t+2)^{2}}{t+3}=b$ and consider the function $f(t)=-\frac{(t+2)^{2}}{t+3}$. To find the set of its values, we will investigate this function for monotonicity. Since
$$
f^{\prime}(t)=-\frac{2(t+2)(t+3)-(t+2)^{2}}{(t+3)^{2}}=-\frac{(t+2)(t+4)}{(t+3)^{2}}<0 \text { for } t \geq... | [-\frac{4}{3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,232 |
12. The midpoint of each side of the base of a quadrilateral pyramid is connected by a segment to the point of intersection of the medians of the opposite lateral face. Prove that: a) these segments intersect and are divided by the point of their intersection in the ratio $3: 2$, counting from the side of the base; b) ... | Solution 1. Let points $K_{1}, K_{2}, K_{3}, K_{4}$ be the midpoints of sides $A B, B C, C D, D A$ of the pyramid $P A B C D$; points $M_{1}, M_{2}, M_{3}, M_{4}$ be the points of intersection of the medians of the faces $P C D, P A D, P A B, P B C$, and points $O_{1}, O_{2}$, $O_{3}, O_{4}$ divide the segments $K_{1} ... | S_{EFGH}=\frac{1}{72}S_{ABCD} | Geometry | proof | Yes | Yes | olympiads | false | 43,233 |
1. Compute $\operatorname{arctg}\left(\operatorname{tg} 65^{\circ}-2 \operatorname{tg} 40^{\circ}\right)$. | 2.II.1. The answer follows from a chain of trigonometric transformations:
$$
\begin{aligned}
\operatorname{tg} 65^{\circ}-2 \operatorname{tg} 40^{\circ}=\operatorname{ctg} 25^{\circ}-2 \operatorname{ctg} 50^{\circ} & =\frac{1}{\operatorname{tg} 25^{\circ}}-\frac{2}{\operatorname{tg} 50^{\circ}}= \\
& =\frac{1}{\operat... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,234 |
2. Find all values of the parameter $a$ for which the equation $2 x^{2}-a x+2 a=0$ has two distinct integer roots. | 2.II.2. Setting $a=2b$, we get the equation $x^{2}-b x+2 b=0$. If $m$ and $n$ are its integer roots, then $m n=2 b$ and $m+n=b$, so $m n=2 m+2 n$, or $(m-2)(n-2)=4$, from which
$$
(m ; n)=(0 ; 0),(4 ; 4),(6 ; 3),(3 ; 6),(1 ;-2),(-2 ; 1)
$$
Thus, $b$ can take one of the values $0 ;-1 ; 8 ; 9$. Checking shows that the ... | -2;18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,235 |
3. Let $a, b$ and $c$ be natural numbers, each less than 1000000. Prove that the equation $\sqrt[21]{a x^{2}}+\sqrt[21]{b x}+\sqrt[21]{c}=0$ has no real roots. | 2.II.3. As a result of the substitution $t=\sqrt[21]{x}$, we obtain the quadratic equation $\sqrt[21]{a} t^{2}+\sqrt[21]{b} t+\sqrt[21]{c}=0$. Note that $2^{21}>2 \cdot 10^{6}$, so $\sqrt[21]{a} \leqslant \sqrt[21]{10^{6}}1$ and $\sqrt[21]{c}>1$. Therefore, the discriminant $D=\sqrt[21]{a^{2}}-4 \sqrt[21]{b c}$ of the ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,236 |
4. Prove that the volume of a tetrahedron is $\frac{1}{6} a b c \sin \alpha$, where $a$ and $b$ are the lengths of opposite edges of the tetrahedron, $\alpha$ is the angle between them, and $c$ is the distance between the lines passing through these edges. | 2.II.4. Let's draw a plane through each edge of the tetrahedron, parallel to the opposite edge. As a result, we will obtain a parallelepiped, the intersecting diagonals of whose parallel faces are the edges of the given tetrahedron (figure).
^{2}}{2} \leqslant u^{2}+v^{2}$. Transforming the obtained inequality, we get that $u^{2}-2 u v+v^{2} \geqslant 0$, or $(u-v)^{2} \geqslant 0$.
Topic: "Proof of Inequalities" | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,240 |
1. Solve the equation $(x+2)(x-4)+4(x-4) \sqrt{\frac{x+2}{x-4}}=12$. | 3.I.1. The domain of the inequality is the union of the intervals $(-\infty; -2]$ and $(4; +\infty)$.
Suppose that $x>4$. In this case, the equation transforms into $(x+2)(x-4)+4 \sqrt{(x+2)(x-4)}=12$. By making the substitution $t=$ $\sqrt{(x+2)(x-4)}$, we obtain the equation $t^{2}+4 t-12=0$, from which $t=-6$ or $t... | 1+\sqrt{13},1-3\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,241 |
2. Prove the inequality
$$
\operatorname{tg}^{6} x+\operatorname{ctg}^{6} x \geqslant 2\left(\sin ^{5} x+\cos ^{5} x+2 \sin ^{3} x \cos ^{3} x\right)
$$ | 3.I.2. Since $\operatorname{tg}^{6} x>0$ and $\operatorname{ctg}^{6} x=\frac{1}{\operatorname{tg}^{6} x}$, then $\operatorname{tg}^{6} x+\operatorname{ctg}^{6} x \geqslant 2$. To prove this inequality, it is sufficient to show that its right-hand side does not exceed 2. The inequalities $\sin ^{5} x \leqslant \sin ^{4}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,242 |
4. Lines parallel to the sides of a triangle and tangent to the inscribed circle cut off three smaller triangles from it. Prove that the sum of the areas of the cut-off triangles is not less than one third of the area of the original triangle. | 3.I.4. Let $S$ be the area of triangle $ABC$ and $P$ be its perimeter. The triangles cut off from triangle $ABC$ by lines parallel to its sides are similar to the given triangle. Let the similarity coefficients be denoted by $k_{1}, k_{2}$, and $k_{3}$. Then the perimeters and areas of the cut-off triangles are, respec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,244 |
5. The sphere touches all the edges of a tetrahedron, in which there is a vertex from which the edges emanate pairwise perpendicularly. Find the radius of this sphere if the radius of the sphere circumscribed around this tetrahedron is $3 \sqrt{3}$. | 3.I.5. Let $PABC$ be a tetrahedron in which the edges $PA$, $PB$, and $PC$ are perpendicular to each other. Let $a = PC$, $b = PA$, and $c = PB$ (left figure).
If a sphere can be inscribed... | 6(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,245 |
6. Let $\mathcal{S}$ be the smallest subset of the set of integers satisfying the following properties:
1) $0 \in \mathcal{S}, 2)$ if $x \in \mathcal{S}$, then $3 x \in \mathcal{S}$ and $3 x+1 \in \mathcal{S}$.
Find the number of non-negative integers in the set $\mathcal{S}$ that do not exceed 2009. | Solution 1. We will write the numbers from the set $\mathcal{S}$ in the ternary numeral system. Then, the transition from number $x$ to number $3x$ means appending 0 to the right of the "ternary representation" of the number $x$, and the transition from $x$ to $3x+1$ means appending 1 to the right. Therefore, a number ... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,246 |
7. It is known that the maximum number of regions into which $n$ lines divide a plane is $\frac{1}{2} n(n+1)+1$. Formulate a condition for such a set of lines and prove this formula. Find a formula in the general case for the number of regions into which $n$ lines divide a plane.
## Methodological Block
A. The soluti... | 3.I.7. a) If no two of the $n$ lines are parallel and no three of them pass through the same point, then the total number of regions into which they divide the plane is $\frac{1}{2} n(n+1)+1$. We will prove this statement by induction. One line, of course, divides the plane into two regions.
Assume that this statement... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 43,247 |
10. Solve the equation $x^{2}+\sqrt{x+5}=5$. | Solution 1. Let $y=\sqrt{x+5}$, from which $y^{2}=x+5$. We can rewrite this equation as $x^{2}=5-y$. Subtracting the obtained equations, we get that $(y-x)(y+x)=y+x$, from which it follows that $y=-x$ or $y=x+1$. Thus, $\sqrt{x+5}=-x$ or $\sqrt{x+5}=x+1$.
Solving the first equation. It is clear that $x \leqslant 0$. S... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,250 | |
11. Prove that in any triangle, the length of the bisector does not exceed the length of the median drawn from the same vertex. | Solution 1. It is well known that the bisector divides side $AB$ in the ratio $AD:DB = b:a$, and the median divides it in half. Since $b < a$, then $AD < AM$. Let $E$ be the point of intersection of line $CD$ with the circumcircle of the given triangle (left figure).
^{2}+2\left(x y-\frac{1}{2}\right)^{2}-\frac{1}{2} \geqslant-\frac{1}{2} .
\end{aligned}
$$
Equality holds if $x^{2}=y^{2}$ and $x y=\frac{1}{2}$, for examp... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,252 |
1. Solve the equation $\frac{1}{x+y+z}=\overline{0, x y z}$ (here $x, y$ and $z-$ are some digits). | 3.II.1. The equation $\overline{x y z} \cdot(x+y+z)=1000$. If $x=0$, then $y+z50$, but there are no two-digit divisors of 1000. The number 1000 has five natural three-digit divisors: $100, 125, 200, 250$, and 500. Direct verification shows that the equation is satisfied only by the triplet $(1; 2; 5)$.
Answer: $\{(1; ... | (1;2;5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,253 |
3. It is known that there exist such numbers $\alpha \neq 0$ and $\beta$, that $f(x+\alpha)=$ $f(x)+\beta$. Prove that the function $f(x)$ can be represented as the sum of a linear function and a periodic function. | 3.II.3. Let's introduce the function $g(x)=f(x)-\frac{\beta}{\alpha} x$. Since $g(x+\alpha)=f(x+\alpha)-\frac{\beta}{\alpha}(x+\alpha)=f(x)+\beta-\frac{\beta}{\alpha} x-\beta=f(x)-\frac{\beta}{\alpha} x=g(x)$, $g(x)$ is an $\alpha$-periodic function. Therefore, $f(x)$ is the sum of the $\alpha$-periodic function $g(x)$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,255 |
4. The radii of the excircles of a certain triangle are 2, 3, and 6 cm. Find the radius of the circle inscribed in this triangle. | 3.II.4. First, let's derive the formula for the radius of the excircle. Let $P$ be the center of the circle with radius $r_{3}$, which touches the side $AB$ and the extensions of the sides $CA$ and $CB$ of triangle $ABC$. From the equality $S_{PAB} + S_{ABC} = S_{PAC} + S_{PBC}$, it follows that
$$
S_{ABC} = S = \frac... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,256 |
5. A plane divides the medians of the faces $A B C$, $A C D$ and $A B D$ of the tetrahedron $A B C D$ emanating from vertex $A$ in the ratios $1: 2, 1: 1$ and $1: 2$, counting from point $A$. Find the ratio of the volumes of the parts into which this plane divides the pyramid. | 3.II.5. Let $ABCD$ be the given tetrahedron, and $AK, AM, AN$ be the medians of its faces $ABC, ACD, ABD$. Denote by $E, F$, and $G$ the points of intersection of the plane $PQR$ with the medians $AK, AM$, and $AN$, respectively (figure).
=x... | Solution 1. The domain of the given function is the interval $[0 ; 2]$. To find the set of its values, we need to find the points where the derivative of this function is zero. Since $f^{\prime}(x)=1+\frac{3-3 x}{\sqrt{6 x-x^{2}}}$, then
$$
f^{\prime}(x)=0 \Longleftrightarrow \sqrt{6 x-3 x^{2}}=3 x-3 \Longleftrightarr... | [-1;2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,259 |
1. Determine the number of solutions of the equation
$$
2^{3 x}+4 a \cdot 2^{2 x}+a^{2} \cdot 2^{x}-6 a^{3}=0
$$
depending on the value of the parameter $a$. | 4.I.1. Setting $t=2^{x}$, we obtain the equation
$$
t^{3}+4 a t^{2}+a^{2} t-6 a^{3}=0
$$
Since $t=a$ is a root of this equation, its left side is divisible by $t-a$. As a result, we get the equation
$$
(t-a)\left(t^{2}+5 a t+6 a^{2}\right)=(t-a)(t+2 a)(t+3 a)=0
$$
Thus, $t=a$, $t=-2 a$, or $t=-3 a$. Since $t>0$, if... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,260 |
2. Prove that the number $5^{n}\left(5^{n}+1\right)-6^{n}\left(3^{n}+2^{n}\right)$ is divisible by 91 for any natural $n$. | 4.I.2. We will prove that the given number is divisible by 7 and 13. We will use the fact that for any natural number $n$ and any integers $a$ and $b$, the number $a^{n}-b^{n}$ is divisible by $a-b$, which follows from the identity
$$
a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+\ldots+b^{n-1}\right)
$$
## Since
$$
5^{n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,261 |
3. Solve the equation $x \sqrt{y-1}+y \sqrt{x-1}=x y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution 3. After the natural substitution $u=\sqrt{x-1}$ and $v=\sqrt{y-1}$, we obtain the equation $u\left(v^{2}+1\right)+v\left(u^{2}+1\right)=\left(u^{2}+1\right)\left(v^{2}+1\right)$. Since $2 u\left(v^{2}+1\right) \leqslant\left(u^{2}+1\right)\left(v^{2}+1\right)$ and $2 v\left(v^{2}+1\right) \leqslant\left(u^{2}... | Algebra | MCQ | Yes | Yes | olympiads | false | 43,262 | |
4. Point $K$ lies on side $B C$ of parallelogram $A B C D$, and point $M$ lies on its side $A D$. Segments $C M$ and $D K$ intersect at point $L$, and segments $A K$ and $B M$ intersect at point $N$. Find the maximum value of the ratio of the areas of quadrilaterals $K L M N$ and $A B C D$. | 4.I.4. Without loss of generality, we can assume that the area of parallelogram $ABCD$ is 1. Let's introduce the following notations: $S_{1}$ - the area of triangle $BKN$, $S_{2}$ - the area of triangle $AMN$, $S_{3}$ - the area of triangle $CKL$, $S_{4}$ - the area of triangle $DLM$, $x$ - the area of triangle $KMN$, ... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,263 |
5. Given three quadratic trinomials with pairwise distinct leading coefficients. The graphs of any two of them have exactly one common point. Prove that all of them have exactly one common point. | Solution 1. Let's introduce the notations: $p_{1}(x)=a_{1} x^{2}+b_{1} x+c_{1}$, $p_{2}(x)=a_{2} x^{2}+b_{2} x+c_{2}$, and $p_{3}(x)=a_{3} x^{2}+b_{3} x+c_{3}$. Suppose, for definiteness, that $a_{1}>a_{2}>a_{3}$. Since the graphs of $y=p_{1}(x)$ and $y=p_{2}(x)$ have a unique common point, the quadratic equation $p_{1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,264 |
6. The section of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ by a plane passing through its vertex $D$ intersects the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ at points $K, M$, and $N$, respectively. Find the ratio of the volume of the pyramid with vertex at point $P$ and base $D K M N$ to ... | 4.I.6. Let $a, b$, and $c$ denote the lengths of the edges $A D, D C$, and $D D_{1}$ of the given parallelepiped, and let $V$ be its volume. According to the property of parallel planes, $K M \| D N$ and $M N \| K D$, so $D K M N$ is a parallelogram (figure).
}V | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,265 |
7. Prove that for no integer values of $x$ and $y$ is the number
$$
x^{8}-x^{7} y+x^{6} y^{2}-x^{5} y^{3}+x^{4} y^{4}-x^{3} y^{5}+x^{2} y^{6}-x y^{7}+y^{8}
$$
a prime number.
## Methodological Block
A. The solutions to problems $8-9$ are provided below. Evaluate each solution and the answers obtained. Indicate all ... | 4.I.7. The factorization is valid:
$$
\begin{aligned}
& x^{8}-x^{7} y+x^{6} y^{2}-x^{5} y^{3}+x^{4} y^{4}-x^{3} y^{5}+x^{2} y^{6}-x y^{7}+y^{8}= \\
& =x^{6}\left(x^{2}-x y+y^{2}\right)-x^{3} y^{3}\left(x^{2}-x y+y^{2}\right)+y^{6}\left(x^{2}-x y+y^{2}\right)= \\
& =\left(x^{2}-x y+y^{2}\right)\left(x^{6}-x^{3} y^{3}+y... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,266 |
10. Determine the number of solutions of the equation $\sqrt{x+3}=a x+2$ depending on the value of the parameter $a$. | Solution 1. The substitution $t=\sqrt{x+3}$ reduces the given equation to the equation $a t^{2}-t+2-3 a=0$, each non-negative solution of which corresponds to a unique solution of the given equation. When $a=0$, the equation is not quadratic and has a single root $t=2$. Let $a \neq 0$. The discriminant of the quadratic... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,269 |
11. Prove that in any quadrilateral, the segments connecting the midpoints of its opposite sides, as well as the segment with endpoints at the midpoints of its diagonals, pass through one point and are bisected by it. | Solution 1. It is known that the midpoints of the sides of any quadrilateral are the vertices of a parallelogram. Thus, the segments $K M$ and $L N$, connecting the midpoints of the opposite sides
80
$A B$ and $C D, B C$ and $A D$ of quadrilateral $A B C D$, are the diagonals of this parallelogram, and therefore are ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,270 |
12. Find all pairs of positive numbers $a$ and $b$, for which the numbers $\sqrt{a b}, \frac{a+b}{2}$, and $\sqrt{\frac{a^{2}+b^{2}}{2}}$ can form an arithmetic progression. | Solution 1. We will use the characteristic property of an arithmetic progression. We have,
$$
\begin{gathered}
\frac{a+b}{2}=\frac{\sqrt{a b}+\sqrt{\frac{a^{2}+b^{2}}{2}}}{2} \Longleftrightarrow(a+b)^{2}=a b+\sqrt{2 a b\left(a^{2}+b^{2}\right)}+\frac{a^{2}+b^{2}}{2} \\
\Longleftrightarrow a^{2}+b^{2}-2 \sqrt{2 a b} \s... | b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,271 |
1. Compute $\operatorname{tg} \alpha$, if $3 \operatorname{tg} \alpha-\sin \alpha+4 \cos \alpha=12$. | 4.II.1. We have,
$$
\begin{aligned}
& 3 \operatorname{tg} \alpha - \sin \alpha + 4 \cos \alpha = 12 \\
& \Longleftrightarrow \operatorname{tg} \alpha (3 - \cos \alpha) - 4 (3 - \cos \alpha) = 0 \Longleftrightarrow \\
& (\operatorname{tg} \alpha - 4)(3 - \cos \alpha) = 0 \Longleftrightarrow \operatorname{tg} \alpha = 4... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,272 |
2. The sequence $\left\{a_{n}\right\}$ is defined by the recurrence relation $a_{n}=q a_{n-1}+d$, where $q \neq 1$. Prove that there exists a number $c$ such that the sequence $b_{n}=a_{n}+c$ will be a geometric progression. | 4.II.2. Since
$$
b_{n}=a_{n}+c=q a_{n-1}+d+c=q\left(b_{n-1}-c\right)+d+c=q b_{n-1}-q c+c+d
$$
then if $c(q-1)=d$, the sequence $b_{n}$ is a geometric progression. Thus, we can take $c=\frac{d}{q-1}$. | \frac{}{q-1} | Algebra | proof | Yes | Yes | olympiads | false | 43,273 |
4. Let $x$ and $y$ be positive numbers whose sum is 2. Find the maximum value of the expression $x^{2} y^{2}\left(x^{2}+y^{2}\right)$. | Solution 1. The idea is suggested by the formula for $(x+y)^{4}$. Since
$(x+y)^{4}-8 x y\left(x^{2}+y^{2}\right)=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}=(x-y)^{4} \geqslant 0$, then $8 x y\left(x^{2}+y^{2}\right) \leqslant(x+y)^{4}$. It is clear that $4 x y \leqslant(x+y)^{2}$. Therefore,
$$
32 x^{2} y^{2}\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,275 |
5. Two triangular pyramids $M A B C$ and $N A B C$ have a common base and no other common points. All vertices of both pyramids lie on the same sphere. Find the lengths of the edges $M A$ and $M B$, given that they are equal to each other, and the lengths of all other edges of these pyramids are $\sqrt{3}$. | 4.II.5. First, let's find the diameter \( d \) of the sphere circumscribed around the regular tetrahedron \( NABC \). Since \( CO = \frac{AB}{\sqrt{3}} = 1 \), then \( NO = \sqrt{3-1} = \sqrt{2} \). In a regular tetrahedron, the center of the circumscribed sphere coincides with its centroid, which lies on the height \(... | AM=BM=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,276 |
7. Solve the problem using as many methods as possible. Methods are considered different if they use different mathematical ideas, as well as various technical implementations of the same idea. Indicate the place of each used method in the school mathematics curriculum.
All vertices of the right triangle \(ABC\) lie o... | Solution 1. Since $A B^{2}=A C^{2}+B C^{2}$, then
$$
4 x_{0}^{2}=\left(x_{0}+x_{1}\right)^{2}+\left(y_{0}-y_{1}\right)^{2}+\left(x_{0}-x_{1}\right)^{2}+\left(y_{0}-y_{1}\right)^{2}
$$
from which $4 x_{0}^{2}=2 x_{1}^{2}+2 x_{0}^{2}+2\left(y_{0}-y_{1}\right)^{2}$, or $2 y_{0}=y_{1}+y_{0}+\left(y_{0}-y_{1}\right)^{2}$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,278 |
1. Find the number of natural solutions to the equation
$$
\left[\frac{x}{2010}\right]=\left[\frac{x}{2011}\right]+1
$$ | 5.I.1. Let's represent the number $x$ in the form $x=2011 k+d$, where $d=0,1, \ldots 2010$. Since $\left[\frac{x}{2010}\right]=\left[\frac{x}{2011}\right]+1=k+1$, then $k+1 \leqslant \frac{x}{2010}<2011$, from which
$$
2010 k+2010 \leqslant 2011 k+d<2010 k+4020
$$
thus, $2010 \leqslant k+d<4020$, or $2010 \leqslant k... | 2010\cdot2011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,279 |
2. Solve the equation $2 \log _{3} \operatorname{ctg} x=\log _{2} \cos x$. | Solution 1. Rewrite the equation as
$$
\log _{3}(\operatorname{ctg} x)=\log _{4}(\cos x)=t
$$
Then $\cos x=4^{t}, \operatorname{ctg} x=3^{t}$ and $\sin x=\left(\frac{4}{3}\right)^{t}$, so $4^{2 t}+\left(\frac{4}{3}\right)^{2 t}=1$. The function on the right side of the obtained equation is increasing, so it has no mo... | \frac{\pi}{3}+2\pik,k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,280 |
3. Let $a>0, b>0, c>0$ and $a+b+c=1$. Prove that
$$
\frac{a-b c}{a+b c}+\frac{b-c a}{b+c a}+\frac{c-a b}{c+a b} \leqslant \frac{3}{2}
$$ | 5.I.3. Let's write the right-hand side of the given inequality as
$$
\begin{aligned}
\frac{a+b c-2 b c}{a+b c}+\frac{b+a c-2 a c}{b+a c}+ & \frac{c+a b-2 a b}{c+a b}= \\
& =3-2\left(\frac{b c}{a+b c}+\frac{a c}{b+a c}+\frac{a b}{c+a b}\right)
\end{aligned}
$$
Thus, we need to prove the inequality
$$
4\left(\frac{b c... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,281 |
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