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54. Among 80 identical-looking coins, there is one counterfeit (lighter). How can you identify this counterfeit coin using four weighings on a balance scale without weights? | 54. The coins need to be divided into 3 piles: two with 27 coins each and the third with 26. The first and second piles should then be placed on the scales. If the scales are in balance, the counterfeit coin is in the third pile. If not, the counterfeit coin is in the lighter pile. The identified pile should then be di... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,167 |
55. There are 12 bags, each containing 12 coins. One of them is filled with counterfeit coins. A genuine coin weighs 12 g, while a counterfeit coin differs by 0.4 g, but it is unknown whether it is heavier or lighter. There are scales and weights available, allowing for weighing with an accuracy of 0.1 g. Using a singl... | 55. Let's number the bags and take as many coins from each as its number. We will collect all these coins together without mixing them and weigh them. If there were no counterfeit coins in the bags, the weight of the selected coins would be equal to $1+2+3+\ldots+12$ (2), that is, 96 2. But since there are counterfeit ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,168 |
56. 17 scientists are corresponding with each other. The correspondence is conducted on three scientific topics. Each pair of scientists corresponds with each other on one topic. Prove that at least three scientists correspond with each other on the same topic. | 56. Let's single out one of the 17 scientists. Let it be Ivanov. He can correspond with 16 other scientists on three topics. Therefore, among these topics, there will be one (denote it as $A$), on which Ivanov corresponds with at least 6 scientists. If among these 6 scientists there are two who correspond with each oth... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,169 |
1. (7th grade) Several girls (all of different ages) were collecting white mushrooms in the forest. The collected mushrooms they distributed as follows: the youngest received 20 mushrooms and 0.04 of the remainder. The next oldest received 21 mushrooms and 0.04 of the remainder, and so on. It turned out that everyone r... | 1. Let the number of collected mushrooms be denoted by $x$. Then the youngest girl received $20+(x-20) \cdot 0.04$ mushrooms. The second girl received $19.392+0.0384 x$. Both received the same amount. By setting up the equation and solving it, we find $x=120$. There were 5 girls, each received 24 mushrooms. | 120, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,170 |
2. (7th grade) Show that $43^{43}-17^{17}$ is divisible by 10 without a remainder. | 2. The last digit of the number $43^{43}$ is determined. $43^{1}$ ends in 3; $43^{2}$ ends in 9, $43^{3}$ ends in 7, $43^{4}$ ends in 1, and so on. $43^{43}$ ends in 7. Similarly, the last digit of the number $17^{17}$ is determined, which will be 7. Therefore, $43^{43}-17^{17}$ ends in zero, i.e., it is divisible by 1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,171 |
3. (7th grade) Prove that if $a$ is not divisible by 5, then $a^{4}-1$ is divisible by 5. | 3. Let's write the number $a^{4}-1$ as the product $\left(a^{2}+1\right)\left(a^{2}-1\right) ; a^{2}$ can only have the last digit as $1,4,9,6,0$; therefore, one of the factors is always divisible by 5. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,172 |
4. (7th grade) The bases of the trapezoid are equal to $a$ and $b$. Find the segment of the line connecting the midpoints of its diagonals ($a>b$). | 4. Use the property of the midline of a triangle. Answer: $\frac{a-b}{2}$. | \frac{-b}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,173 |
5. (7th grade) Construct a rhombus given the sum of the diagonals and the angle opposite one of them.
Construct a rhombus given the sum of the diagonals and the angle opposite one of them. | 5. Construct a triangle with the base equal to half the sum of the diagonals, and the adjacent angles $\frac{\alpha}{2}$ and $45^{\circ}$, where $\alpha$ is the given angle. The side opposite the $45^{\circ}$ angle is a side of the rhombus, and two vertices of this triangle are two vertices of the rhombus. It is now ea... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,174 |
6. (8th grade) The lengths of two sides of a triangle are 7 cm and 11 cm, respectively. Find the length of the third side if it is expressed as a prime number of centimeters. | 6. The desired side can have a length greater than 4 and less than 18 cm, i.e., 5 cm, 7 cm, 11 cm, 13 cm, 17 cm. | 5,7,11,13,17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,175 |
7. (8th grade) For what values of $x$ is the polynomial $x^{4}+2 x^{3}-2 x^{2}-$ $-4 x-5$ a perfect square? | 7. Let's complete the square $\left(x^{2}+x-2\right)^{2}+x^{2}-9$. The polynomial will be a perfect square if $x^{2}-9=0$, i.e., when $x= \pm 3$. | \3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,176 |
8. (8th grade) Let the sum $a+b+c$ be divisible by 6. Prove that the sum of cubes $a^{3}+b^{3}+c^{3}$ is also divisible by 6. | 8. Let's introduce the notation $a+b+c=6 k$. Raise both sides of this equality to the third power and isolate $a^{3}+b^{3}+c^{3}$ from the resulting equation, moving the other terms to the first part. Then, by grouping the terms, we get $a^{3}+b^{3}+c^{3}=(6 k)^{3}-3 a(b+c) 6 k-3 b c(b+c)$. The first and second terms a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,177 |
9. (8th grade) The center of the circle inscribed in a right trapezoid is 2 cm and 4 cm away from the ends of the lateral side. Find the area of the trapezoid. | 9. Triangle $O C D$ is a right triangle, $O C=2 \mathrm{~cm}, O D=4 \mathrm{~cm}, C D=2 \sqrt{5} \mathrm{~cm}$. Let $O K \perp C D, O K=r=\frac{4 \sqrt{5}}{5} \mathrm{~cm}$. The height of the trapezoid $H=\frac{8 \sqrt{5}}{5} \mathrm{~cm}$. The sum of the bases of the trapezoid is $\frac{18 \sqrt{5}}{5} \mathrm{~cm}$. ... | 14.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,178 |
10. (8th grade) Given two circles. Find a point on the plane from which the first circle is seen at an angle $\alpha$, and the second at an angle $\beta$.
保留了源文本的换行和格式。 | 10. The locus of points from which circle $O$ is seen at an angle $\alpha$ is a concentric circle to it, the radius of which is the distance from the center $O$ to the point of intersection of two tangents drawn to circle $O$ through the ends of its radii forming a central angle ($180^{\circ}-\alpha$).
Similarly, the ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,179 |
11. (9th grade) Find a natural number $n$, knowing that the sum $1+2+3+\ldots+n$ is a three-digit number consisting of identical digits. | 11. $1+2+3+\ldots+n=\frac{n+1}{2} n$. This sum equals a three-digit number $\overline{a a a}$, where $\overline{a a a}=100 a+10 a+a=111 a$. We get the equation $\frac{n+1}{2} n=111 a$. From this, $a=\frac{n(n+1)}{2 \cdot 3 \cdot 37}$. Since $a$ is an integer, $n(n+1)$ must be divisible by 2, 3, and 37. This is only pos... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,180 |
13. (9th grade) Construct a triangle given the base, the opposite angle, and the median to the base. | 13. On the segment equal to the base of the triangle, we construct a segment that inscribes the given angle. From the midpoint of the base, as the center, we draw a circle with radius \( m \), equal to the median of the base. The point of intersection of the circle with the arc of the segment is the third vertex of the... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,182 |
14. (9th grade) In a trapezoid, the smaller base is 1 dm, the angles adjacent to it are $135^{\circ}$, and the angle between the diagonals facing the base is $150^{\circ}$. Find the area of the trapezoid. | 14. A circle can be circumscribed around an isosceles trapezoid. Considering the angles of the trapezoid, we conclude that its upper base is equal to the radius of the circumscribed circle, \( r = 1 \partial \). The lower base is equal to the side of an equilateral triangle inscribed in this circle, i.e., it is equal t... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,183 |
16. (10th grade) Prove that the bases of the altitudes of a triangle serve as the vertices of a new triangle, for which these altitudes are the angle bisectors. | 16. On the sides of the triangle, three circles are constructed with these sides as diameters (Fig. 31), the feet of the heights of this triangle belong to these circles.
$\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6$ as angles with mutually perpendicular sides.
$\angle 1$ is measured by $\frac{1}{2} \cup B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,185 |
17. (10th grade) Prove that the fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is irreducible for any integer value of $a$. | 17. Transform the inverse fraction by separating the integer part, we get:
$$
\frac{a^{4}+3 a^{2}+1}{a^{3}+2 a}=a+\frac{a^{2}+1}{a^{3}+2 a}
$$
Similarly, $\frac{a^{3}+2 a}{a^{2}+1}=a+\frac{a}{a^{2}+1} ; \quad \frac{a^{2}+1}{a}=a+\frac{1}{a}$.
The fraction $\frac{1}{a}$ is irreducible for any integer value except $a=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,186 |
18. (10th grade) Prove that the sum of the distances from an arbitrarily chosen point inside a regular tetrahedron to its faces is a constant value. | 18. Let's divide the tetrahedron into four pyramids, the bases of which are the faces of the tetrahedron, and the heights \( h_{1}, h_{2}, h_{3}, h_{4} \) are the distances from a chosen point to the faces of the tetrahedron. The volume \( V \) of the tetrahedron is equal to the sum of the volumes of these pyramids. De... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,187 |
19. (10th grade) Simplify $1!1+2!2+3!3+\ldots+n!n$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 19. Transform each term of the sum according to the following rule $k!k=(k+1)!-k!$, then $1!1+2!2+\cdots+n!n=2!-1!+3!-2!+\cdots+(n+1)!-n!$
Answer: $1!1+2!2+\ldots+n!n=(n+1)!-1$. | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,188 |
20. (10th grade) If in a triangle $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1$, then the triangle is a right triangle. Prove.
## 1962 year | 20. $C=180^{\circ}-(A+B)$. Then the given equality can be written as: $\frac{1+\cos 2 A}{2}+\frac{1+\cos 2 B}{2}+\cos ^{2}(A+B)=1$. Using the formula $\cos \alpha+\cos \beta$, the obtained equality can be represented as follows:
$4 \cos (A+B) \cos A \cos B=0$. From this, the required conclusion is easily derived.
## ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,189 |
21. (7th grade) (Newton's problem) Grass grows at the same rate and density everywhere in the corner. It is known that 70 cows would eat it in 24 days, and 30 cows in 60 days. How many cows would eat all the grass in 96 days? (It is assumed that the cows eat the grass uniformly). | 21. In one day, $x$ amount of grass grows. If the initial amount of grass is taken as 1, then over 24 days, the cows eat $1+24 x$ amount of grass, and one cow eats $\frac{1+24 x}{24 \cdot 70}$ amount of grass per day. But according to the second condition, a cow eats $\frac{1+60 x}{30 \cdot 60}$ amount of grass per day... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,190 |
23. (7th grade) Prove that the number $\frac{n}{3}+\frac{n^{2}}{2}+\frac{n^{3}}{6}$ is an integer for any integer $n$. | 23. Let's find the sum of the fractions and factor the numerator. We will get the product of three consecutive numbers in the numerator, which is always divisible by 6 (for any integer $n$). | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,192 |
24. (7th grade) Construct a square whose sides lie on lines passing through four given points. | 24. Let points $A$ and $C$, $B$ and $D$ lie on opposite sides of the desired square $M N P Q$; on the perpendicular dropped from point $B$ to the line $A C$, we lay off the segment $B E = A C$. By connecting point $E$ with point $D$, we obtain a line on which the side $P Q$ of the desired square lies. Finding all other... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,193 |
26. (8th grade) Given angle $ABC$ and line $P$. Find a point $x$ on line $P$ such that the distance from $x$ to side $AB$ is a given length $d$ greater than its distance from side $BC$. | 26. At a distance $d$ from $A B$ inside the angle, we draw a line parallel to $A B$. The intersection of the bisector of the new angle with the given line $P$ will give the desired point. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,195 |
28. (8th grade) Prove that for each natural $n$ the number $n^{3}\left(n^{6}-1\right)$ is divisible by 42. | 28. Factoring $n^{3}\left(n^{6}-1\right)$, we get:
$(n-1) n(n+1) n^{2}\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)$.
The product $(n-1) n(n+1)$ is divisible by 6 for any $n$.
Therefore, we need to prove divisibility by 7. If $n$ is divisible by 7, then $n^{3}\left(n^{6}-1\right)$ is divisible by 7. If $n$ is not div... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,197 |
29. (8th grade) When asked what his ticket number was, someone answered like this: "If all six two-digit numbers that can be formed from the digits of the number are added together, then half of the resulting sum will be exactly my ticket number." Determine the ticket number. | 29. Let's denote the ticket number as: $N=100 x+10 y+\boldsymbol{z}$. According to the problem,
$$
\begin{gathered}
2 N=10 x+y+10 x+z+10 y+x+10 y+z+10 z+x+10 z+y= \\
=22 x+22 y+22 z, \text { i.e. } N=11(x+y+z)
\end{gathered}
$$
By equating the first and second results, we get the following equation: $100 x+10 y+z=11(... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,198 |
31. (9th grade) When a six-digit number, consisting of identical digits, is divided by a four-digit number, also consisting of identical digits, the quotient is 233 and there is some remainder. After discarding one digit from both the dividend and the divisor, and performing the division again, the quotient remains unc... | 31. The sought numbers have the form $N_{1}=111111 a, N_{2}=1111 b$. We can express the condition of the problem with the following equations:
$$
\begin{gathered}
111111 a=1111 b \cdot 233+p \\
11111 a=111 b \cdot 233+p-1000
\end{gathered}
$$
By performing term-by-term subtraction, we get $100 a=233 b+1$. From this, ... | N_{1}=777777,N_{2}=3333 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,199 |
32. (9th grade) Find $2 n+1$ consecutive integers that have the property that the sum of the squares of the first $\boldsymbol{n}+1$ numbers is equal to the sum of the squares of the next $n$ numbers. | 32. Let $N$ be the first number we are looking for. Using the condition of the problem, we can write:
$$
N^{2}+(N+1)^{2}+(N+2)^{2}+\ldots+(N+n)^{2}=(N+n+1)^{2}+\ldots+(N+2 n)^{2}
$$
$N^{2}=2 n^{3}+2 n^{2} N+n^{2}$, from which $N=n^{2} \pm\left(n^{2}+n\right)$.
If $N_{1}=2 n^{2}+n$, then for $n=1, N_{1}=3$ we get the... | n^2\(n^2+n) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,200 |
33. (9th grade) Construct triangle $ABC$ given the altitude $h_{a}$, the median $\boldsymbol{m}_{a}$, and the radius $R$ of the circumscribed circle around the triangle. | 33. From $h_{a}$ and $m_{a}$, we construct a right triangle $A D E$ ($h_{a}$ - leg, $m_{a}$ - hypotenuse). Point $E$ is the midpoint of the base of triangle $A B C$, so the center of the circle circumscribed around triangle $A B C$ lies on the perpendicular to $D E$, erected at point $E$, at a distance $R$ from vertex ... | m_{}>h_{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,201 |
37. (10-11 grades) How many planes are equidistant from four points that do not lie in the same plane? | 37. Taking these points as the vertices of a tetrahedron, it is easy to establish that only seven planes can be drawn equidistant from its vertices. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,204 |
40. (8th grade) Some of the seven sheets of paper were cut into 7 pieces. From the resulting sheets, some were again cut into seven pieces, and so on. When the total number of sheets of paper (of different sizes) was counted, it turned out to be 1961. Prove that the counting was done incorrectly. | 40. If for the first time $k$ sheets of paper were cut, then the total number of sheets became $6(k+1)+1$. If for the second time $n$ sheets of paper were cut, then the total number of sheets obtained after two cuts became $6(k+n+1)+1$. And so each time, the number obtained, when divided by 6, gives a remainder of 1.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,206 |
41. (8th grade) Find the smallest natural number that has the following properties: a) its representation in the decimal system ends with the digit 6; b) if the last digit 6 is erased and this digit 6 is written in front of the remaining digits, the resulting number is four times the original number. | 41. The desired number can be written as: $N=\overline{a_{1} a_{2} a_{3} \ldots a_{n-1}}-1$. After rearranging the digit 6, we get the number $K=4 N=6 \bar{a}_{1} a_{2} a_{3} \ldots a_{n-1}$. Since the last digit of the number $N$ is 6, the last digit of the number $K$ is 4, i.e., $a_{n-1}=4$.
Substituting the value $... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,207 |
43. (8th grade) Construct a parallelogram if its perimeter, one diagonal, and the angle between the given diagonal and a side of the parallelogram are given. | 43. Constructing a triangle given two sides $d$ and $\frac{1}{2} P$ and the angle $\alpha$ between them.
Through the midpoint of the side opposite to angle $\alpha$, draw a perpendicular until it intersects the side of the triangle adjacent to angle $\alpha$. This construction determines three vertices of the desired ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,208 |
44. (8th grade) Given a circle and a point $A$ outside it. Find the geometric locus of the midpoints of the chords whose extensions pass through the point $A$. | 44. We draw a series of secants from point A and mark the midpoints of the obtained chords. By connecting the midpoints of the chords with the center of the given circle O, we obtain a series of right triangles: \( O A B_{1}, O A B_{2}, O A B_{3}, O A B_{4} \) and so on. The midpoints of the chords form a circle, the d... | 中点形成的圆 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,209 |
45. (9th grade) Find a four-digit number that is a perfect square, knowing that the first two digits, as well as the last two, are equal to each other. | 45. The desired number can be written as: $N=\overline{a a b} b$ or $N=1100 a+11 b=$ $=11(99 a+a+b)$. For the desired number to be a perfect square, it is necessary that the factor 11 appears in the number $N$ twice, i.e., the sum $a+b$ must be divisible by $11.1<a+b<18$, hence, $a+b=11$. Then $N=11^{2}(9 a+1)$. The bi... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,210 |
46. (9th grade) Find the sum
$$
\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}
$$ | 46. Let's represent each term in the following form: $\frac{1}{(3 n-1)(3 n+2)}=$ $=\frac{1}{3}\left(\frac{1}{3 n-1}-\frac{1}{3 n+2}\right)$. Then, the desired sum, after combining like terms, will have the form: $S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3 n+2}\right)$. Answer: $S=\frac{n}{2(3 n+2)}$. | \frac{n}{2(3n+2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,211 |
48. (9th grade) Construct a triangle given two sides, knowing that the angle opposite the larger of the given sides is twice the angle opposite the smaller of the given sides. | 48. Let the sides $c$ and $b$ be given. Denote the opposite angles as $a$ and $2 \alpha$. By drawing the angle bisector $BD$ of angle $B$, we obtain similar triangles $ABC$ and $ABD$. Now, the length of segment $AD=\frac{c^2}{b}$ can be easily found. Triangle $BCD$ is isosceles, with $BD=DC$. Constructing triangle $ABD... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,213 |
49. (9th grade) Given segments \(a\) and \(b\) on a certain line, construct
\[
\frac{a^{3}+2 a b^{2}-b^{3}}{a^{2}-b^{2}+a b}
\]
(provide a construction plan). | 49. Let's transform the given expression:
$$
N=\frac{a^{3}+2 a b^{2}-b^{3}}{a^{2}-b^{2}+a b}=\frac{a\left(a^{2}+b^{2}\right)+b^{2}(a-b)}{a^{2}+b(a-b)}
$$
Now, completing the construction is not difficult. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,214 |
50. (10-11 grades) Construct a right triangle given the hypotenuse and the median of one of the legs. | 50. If in a right triangle $ABC$ with a right angle at $B$, the median line $OD$ ($AO=OC$, $BD=DC$) is drawn, then $\angle ODC=90^{\circ}$ and $AB=2OD$. The construction plan is as follows: on the hypotenuse $AC$ and on the segment $OC$, we construct circles, from point $A$ as the center with radius $m_a$, we draw an a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,215 |
51. (10-11 grades) Solve the equation $4^{-\frac{1}{x}}+10^{-\frac{1}{x}}=25^{-\frac{1}{x}}$. | 51. Divide both parts of the equation by $\left(\frac{1}{25}\right)^{\frac{1}{x}}$. With respect to $\left(\frac{5}{2}\right)^{\frac{1}{x}}$, we obtain the quadratic equation $y^{2}+y-1=0$. Using the equality $\left(\frac{5}{2}\right)^{\frac{1}{x}}=y$, we find the value
$$
x=\frac{\lg 5-\lg 2}{\lg (\sqrt{5}-1)-\lg 2}
... | \frac{\lg5-\lg2}{\lg(\sqrt{5}-1)-\lg2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,216 |
52. (10-11 grades) Find the sum $1+2 a+3 a^{2}+4 a^{3}+\ldots+n a^{n-1}$ | 52. Multiply each term of the sum by $a$. Denoting the desired sum by $S$, we will compute the difference $S-S a$, then compute $S$
$$
S=\frac{1-a^{n}-n a^{n}+n a^{n+1}}{(1-a)^{2}}
$$ | \frac{1-^{n}-n^{n}+n^{n+1}}{(1-)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,217 |
53. (10-11 grades) Prove that the number $6^{2 n}+3^{n+2}+3^{n}$ is divisible by 11 for any non-negative integer $n$ . | 53. Let's factor out $3^{n}$; we get $3^{n}\left[\left(12^{n}-1\right)+11\right]$. Each term of the sum in the square brackets, for any non-negative integer $n$, is divisible by 11. This proves the divisibility of the number $6^{2 n}+3^{n+2}+3^{n}$ by 11 for any non-negative integer $n$.
## 1964 year | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,218 |
55. (8th grade) Prove that if the intersection of the sides of a quadrilateral with a circle forms equal chords, then the sums of the opposite sides of the quadrilateral are equal to each other.
62 | 55. According to the problem, the chords are equal and, consequently, equidistant from the center of the circle. If we draw a circle with center at point $O$ and radius equal to the distance of the chord from point $O$, then the quadrilateral will be cyclic and the conclusion of the problem will be satisfied: $A B+C D=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,219 |
56. (8th grade) Two cyclists set out towards each other. If the first starts 1 hour earlier, and the second starts half an hour later, they will meet 18 minutes earlier. If the first starts half an hour later, and the second starts 1 hour earlier, the meeting point will move by 11200 meters.
Calculate the speed of eac... | 56. Let's introduce the following notations: $x$ and $y$ are the speeds of the cyclists in kilometers per hour; $AB$ is the distance between the points; $\frac{AB}{x+y}$ is the normal time of travel until they meet. $\left(\frac{AB}{x+y}+1-\frac{18}{60}\right)$ hours is the time the first cyclist travels until the meet... | 16,14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,220 |
57. (8th grade) Solve the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{y}+\frac{y^{2}}{x}=12 \\ \frac{1}{x}+\frac{1}{y}=\frac{1}{3}\end{array}\right.$. | 57. Given $x \neq 0, y \neq 0$, the system is equivalent to the system $\left\{\begin{array}{l}x^{3}+y^{3}=12 x y \\ 3(x+y)=x y .\end{array}\right.$
If $x+y=v, \quad x y=3 v$, then we get the equation $v^{2}-9 v-36=0$.
From which $v_{1}=12, v_{2}=-3$ or $x_{1}=y_{1}=6, \quad x_{2}=\frac{-3 \pm 3 \sqrt{5}}{2}, y_{2}=\... | x_{1}=y_{1}=6,\quadx_{2}=\frac{-3\3\sqrt{5}}{2},y_{2}=\frac{-3\3\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,221 |
58. (8th grade) Given a thousand-sided polygon. Inside it, 500 points are taken, which are connected to each other and to the vertices of the thousand-sided polygon. How many triangles will be formed? (The sides of the triangles do not intersect). | 58. Let $x$ be the number of triangles, $2 d(n-2)$ be the sum of the interior angles of a polygon, $2 d x$ be the sum of the interior angles of the triangles, and $4 d \cdot 500$ be the sum of the full angles at the given 500 points. We can set up the equation $2 d x = 500 \cdot 4 d + 2 d(n-2)$, where $n=1000$. Answer:... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,222 |
59. (8th grade) In each cell of a $25 \times 25$ square table, one of the numbers 1 or -1 is written (arbitrarily). Under each column, the product of all numbers in that column is written. To the right of each row, the product of all numbers in that row is written. Prove that the sum of the 50 written products is not e... | 59. Let's multiply all 50 products. In the resulting large product, each number in the table appears twice: once by row and once by column, so their product is 1. From this, it follows that -1 appears an even number of times among the 50 products, and thus the sum of the products cannot be zero. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,223 |
60. (9th grade) Solve the equation $x^{2}+|x+3|+|3-x|=4.5|x|+6$. | 60. We divide the number line by the points $-3, 0$, and 3 into intervals.
If $x \leqslant -3$, then $x^{2} - x - 3 + 3 - x = -4.5 x + 6, x_{1} = -4, x_{2} = \frac{3}{2}$.
If $-3 \leqslant x \leqslant 0$, then $x^{2} + x + 3 + 3 - x = -4.5 x + 6, x_{3} = 0, x_{4} = 4.5$.
If $0 \leqslant x \leqslant 3$, then $x^{2} +... | x=\4;x=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,224 |
61. (9th grade) In a square table consisting of $8 \times 8$ cells, natural numbers from 1 to 64 are arranged in order. A number is selected, and the row and column in which it is located are crossed out. From the remaining numbers, another number is selected, and the row and column in which it is located are crossed o... | 61. The chosen number $N=M_{i}+8\left(n_{k}-1\right)$, where $M_{i}$ is the column number, and $n_{k}$ is the row number in which it stands. Then the total sum of the eight chosen numbers can be written as:
$$
S=\sum_{i=1, k=1}^{i=8, k=8} M_{i}+8\left(n_{k}-1\right)=\frac{8+1}{3} 8+8 \frac{7+1}{2} 7=260
$$ | 260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,225 |
62. (9th grade) Three circles with radii $a, b$, and $c$ are pairwise externally tangent. Find the length of the chord cut off by the third circle from the common internal tangent of the first two circles. | 62. Construct $O_{3} M \| A C$, where $A C$ is the common tangent of circles $O_{1}$ and $O_{2}$. In triangles $O_{2} O_{3} M$ and $O_{1} O_{3} M$ (Fig. 33), the leg $O_{3} M$ is common. Assuming $C M=x$, we form the equation:
 Construct a triangle given the endpoints of the base, the point of tangency of the inscribed circle, and the sum of the lateral sides. | 63. Using the property of tangents to a circle drawn from the same point, it is easy to devise a construction plan for a triangle given its three sides. Construction is possible if $\frac{a}{2}<b<a$, where $b$ is the base of triangle $ABC$, and $a$ is the sum of the lateral sides. The solution is unique. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,227 |
64. (9th grade) The number 7 is raised to the seventh power, the resulting number is again raised to the seventh power, and so on. This process is repeated 1000 times. What is the last digit of this number? | 64. Considering the powers of the number 7 in sequence, we notice that the last digits of these powers repeat every four, so the number given in the problem ends with the digit 7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,228 |
65. (10-11 grades) Into how many parts do the planes of the faces of a tetrahedron divide space? | 65. The calculation is easiest to perform as follows: inside the tetrahedron - 1 part, at the vertices of the trihedral angles - 4 parts, along the edges - 10 parts. In total, 15 parts. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,229 |
66. (10-11) Solve the system of equations $\left\{\begin{array}{l}x y(x+y)=30 \\ x^{3}+y^{3}=35 .\end{array}\right.$ | 66. Tripling the members of the first equation and adding to the second, the given system can easily be reduced to two systems:
$$
\text { a) }\left\{\begin{array}{l}
x+y=5 \\
x y=6
\end{array}\right.
$$
$$
\left\{\begin{array}{l}
x+y=\frac{-5 \pm 5 \sqrt{3}}{2} \\
x y=\frac{12}{-1 \pm \sqrt{3}}
\end{array}\right.
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,230 | |
67. (11-11 grade) Through a point \( M \) inside an acute angle \( A O B \), draw a line that cuts off a triangle from the angle with the smallest area. | 67. To obtain a triangle of the smallest area, one must draw a line through point $M$ such that the segment of this line between the sides of the given angle is bisected at point $M$ (Fig. 34). $D M E$ is an arbitrary line. $M N \| A B$, $C K \| A B$.
The area of triangle $A B C$ is equal to the area of quadrilateral ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,231 |
69. (grades 10-11) Solve the inequality $\sin x \sin 2 x<\sin 3 x \sin 4 x$, if $0<x<\frac{\pi}{2}$.
## 1965 year | 69. Replacing the product of sines with the differences of the corresponding cosines, we get $\cos 3 x > \cos 7 x$ or $2 \sin 2 x \sin 5 x > 0$. Since $0 < x < \pi$, $\sin 2x > 0$. Therefore, the inequality $2 \sin 2 x \sin 5 x > 0$ holds under the condition $0 < x < \pi$ if $\sin 5x > 0$. Since $\sin 5x > 0$ for $0 < ... | 0<x<\frac{\pi}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,233 |
70. (8th grade) Prove that there do not exist such natural numbers $x$ and $y$ for which the equation $2x + 3y = 6$ holds. | 70. According to the condition, $x$ and $y$ are natural numbers, so $x<3, y<2$. The only possible value is $y=1$, but when $y=1, x$ is fractional. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,234 |
71. (8th grade) Prove that the square of the bisector of the angle at the vertex of a triangle is equal to the difference between the product of the lateral sides and the product of the segments of the base. | 71. Describe a circle around triangle $A B C$ and extend the bisector of angle $B$ until it intersects the circle at point $K$. Using the similarity of triangles $A B K$ and $B F C$ ( $F$ - the intersection of $A C$ and $B K$ ), it is easy to prove the required. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,235 |
72. (8th grade) If each side of a quadrilateral is less than one meter, then its area is less than one square meter. Prove it. | 72. The perimeter of a quadrilateral $p<4$. Given this perimeter, the largest area is that of a square. Under these conditions, the area of the square is less than 1 square meter. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,236 |
73. (8th grade) At the vertices of rectangle $A B C D$, non-intersecting circles are constructed, with $R_{A}+R_{C}=R_{B}+R_{D}$. Common external tangents are drawn to circles $A$ and $C$, and $B$ and $D$. Prove that a circle can be inscribed in the quadrilateral formed by the intersection of the tangents. | 73. From point $O$ - the intersection of the lines connecting the centers of circles $A$ and $C$, $B$ and $D$ - we drop perpendiculars to the tangents. It is easy to prove that they are equal, so point $O$ is the center of the circle inscribed in the quadrilateral formed by the tangents at their points of intersection. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,237 |
75. (9th grade) Solve the equation $\sqrt[4]{8-x}+\sqrt[4]{89+x}=5$. | 75. We introduce the notations: $8-x=a^{4}, 89+x=b^{4}$.
Solving the equation reduces to solving two systems of equations:
$$
\text { a) }\left\{\begin{array}{l}
a+b=5 \\
a b=44
\end{array}\right.
$$
There are no real roots. b) $\{a+b=5$
$a b=6$
$a=2, \quad b=3$ or $b=2, \quad a=3$
Answer: $x_{1}=-8, \quad x_{2}=... | x_{1}=-8,\quadx_{2}=-73 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,239 |
76. (9th grade) Prove that a number written using 300 ones and any number of zeros cannot be a perfect square. | 76. Any exact square contains prime factors in even powers. The given number is divisible by 3, but not by 9, therefore, it is not an exact square. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,240 |
80. (10th grade) Prove that for any natural value of $n$ the number $n\left(n^{2}-1\right)\left(n^{2}-5 n+26\right)$ is divisible by 120. | 80. The product of five consecutive natural numbers is divisible by 120. Let's use this
$$
n\left(n^{2}-1\right)\left(n^{2}-5 n+26\right)=(n-3)(n-2)(n-1) n(n+1)+20(n-1) n(n+1)
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,244 |
81. (10th grade) Determine the number $a$ so that the polynomial $x^{n} -a x^{n-1}+a x-1$ is divisible by $(x-1)^{2}$. | 81. We divide the polynomial by $x-1$. In the obtained quotient, we substitute $x=1$ and set the result equal to zero. Then
$$
2+(1-a)(n-2)=0, \text { from which } a=\frac{n}{n-2}
$$ | \frac{n}{n-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,245 |
84. (10th grade) $A$ is a vertex of a regular star pentagon. The broken line $A A_{1} B B_{1} C C_{1} D D_{1} E E_{1}$ is its outer contour. The lines $A B$ and $D E$ are extended to intersect at point $K$. Prove that the polygon $A B B_{1} C C_{1} D E D_{1}$ is equal in area to the quadrilateral $A D_{1} E K$. | 84. Triangle $A K E$ is isosceles.
$S_{A B C}=S_{A K E}, \quad S_{B_{1} C C_{1}}+S_{E D_{1} D}=S_{A C_{1} E}+S_{E E_{1} D_{1}}, \quad S_{A B B_{1} C C_{1} D E D_{1}}=S_{A D_{1} E K}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,248 |
87. (11th grade) Prove that $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \ldots \frac{99}{100}<\frac{1}{10}$. | 87. Let's introduce the notation: $a=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \ldots \frac{99}{100}, b=\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdots \frac{98}{99}$.
Since $a b=\frac{1}{100}$, but $a<b$.
Therefore, $a<\frac{1}{10}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,251 |
90. (8th grade) Let $a, b, c, n$ and $l$ be integers. Prove that if the fraction $\frac{a l+b}{c l+n}$ is reducible by a number $p$, then $a n - b c$ is divisible by $p$. | 90. The fraction $\frac{a l+b}{c l+n}$ is reducible by $p$, which means $a l+b=p k$ and $c l+n=p m$, where $\boldsymbol{k}$ and $\boldsymbol{m}$ are integers. Multiplying the first equation by $c$, the second by $a$ and subtracting 132
the second from the first, we get $a n-b c=p(a m-k c)$. This proves that $a n-b \bol... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,253 |
91. (8th grade) Find all rectangles that can be cut into 13 equal squares. | 91. Since all squares are equal, the sides of the rectangle will be divided into equal parts: one into $x$ parts, the other into $y$ parts. There are a total of 13 squares. Therefore, $x y=13$. Since 13 is a prime number, either $x=1$ or $y=1$, then $y=13$ or $x=13$. A rectangle with sides 1 and 13 and similar rectangl... | 1\times13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,254 |
92. (8th grade) Construct a triangle given its three medians. | 92. From the segments $\frac{2}{3} m_{a}, \frac{2}{3} m_{b}, \frac{2}{3} m_{c}$, we construct the triangle $B M C_{1}$. Then, extending $M C_{1}$ beyond point $M$ by $\frac{2}{3} m_{a}$, we obtain point $A$. Using the median $B N$ of triangle $B M C_{1}$, it is easy to find vertex $C$.
Triangle $A B C-$ is the desired... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,255 |
94. (8th grade) Prove that circles constructed on the sides of an arbitrary quadrilateral as diameters completely cover it. | 94. If there is a point $K$ not covered by the circles, then each side of the quadrilateral will be seen from it at an angle less than $90^{\circ}$, i.e., the sum of the planar angles at point $K$ will be less than $360^{\circ}$, which is impossible. Therefore, all internal points of any quadrilateral are covered by th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,257 |
95. (9th grade) Prove that the equation $2 x^{2}-5 y^{2}=7$ has no solutions in integers. | 95. Let the equation have integer solutions. From the equation, it is clear that y is an odd number, i.e., $y=2 n+1$, then $x^{2}-10 n^{2}-10 n-6=0$. Considering this equation, we see that $x$ is an even number. Let $x=2 k$, then $2 k^{2}$ $-5 n(n+1)=3$. The equality is impossible (the difference of even numbers cannot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,258 |
96. (9th grade) Find the sum of all irreducible fractions with a denominator of 3, lying between the positive integers \(m\) and \(n\) (\(m<n\)). | 96. We find the sum of fractions. There are a total of intervals between integers $n-m$, with two fractions in each interval.
$\left(m+\frac{1}{3}\right)+\left(m+\frac{2}{3}\right)+\left(m+\frac{4}{3}\right)+\left(m+\frac{5}{3}\right)+\ldots+\left(n-\frac{2}{3}\right)+\left(n-\frac{1}{3}\right)=$
$=m+m+\ldots+m+n+n+\... | n^2-^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,259 |
97. (9th grade) Prove that a convex quadrilateral with an axis of symmetry is either inscribed or circumscribed. | 97. The axis of symmetry of a quadrilateral either passes through two of its opposite vertices or through the midpoints of opposite sides. In the first case, the sums of the opposite sides are equal, i.e., the quadrilateral is circumscribed; in the second case, the sums of the opposite angles are equal - the quadrilate... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,260 |
98. (9th grade) Find the geometric locus of the midpoints of segments of length $\frac{2}{\sqrt{3}}$, whose ends lie on the sides of a unit square. | 98. The GM of the midpoints of the segments is a figure formed by the intersection of arcs of circles drawn from the vertices of a square with radii equal to half the length of the given segment $\left(r=\frac{1}{\sqrt{3}}\right)$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,261 |
99. (9th grade) In a unit square, 51 points are placed. Prove that some three of them must necessarily be inside a circle of radius $\frac{1}{7}$. | 99. If a square with a side of 1 is divided into 25 squares with a side of $\frac{1}{5}$, then at least one of them will contain three given points. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,262 |
100. (10th grade) The price of a diamond is proportional to the square of its weight. Prove that if a diamond is divided into several parts, its value will decrease, and the reduction in value will be greatest when the parts are equal. | 100. Let $p-$ be the weight of a diamond; $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ - parts of the diamond. Then the price of the whole diamond is $k p^{2}=k\left(p_{1}+p_{2}+p_{3}+\ldots+p_{n}\right)^{2}$.
The price of the diamond after division into parts
$$
k p_{1}^{2}+k p_{2}^{2}+k p_{3}^{2}+\ldots+k p_{n}^{2}
$$
It ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,263 |
101. (10th grade) Show that one of the quadratic equations $x^{2}+p_{1} x+q_{1}=0$ and $x^{2}+p_{2} x+q_{2}=0$ necessarily has real roots if the coefficients are related by the equation $p_{1} p_{2}=2\left(q_{1}+q_{2}\right)$. | 101. Suppose that both equations do not have real roots, i.e., $D=p_{1}^{2}-4 q_{1}<0$ and $D_{1}=p_{2}^{2}-4 q_{2}<0$ or $p_{1}^{2}<4 q_{1}, p_{2}^{2}<4 q_{2}$. Adding these inequalities term by term and replacing $q_{1}+q_{2}$ with $\frac{p_{1} p_{2}}{2}$, we get $\left(p_{1}-p_{2}\right)^{2}<0$, which is impossible.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,264 |
105. (11th grade) Can the sum of the squares of two consecutive natural numbers be equal to the sum of the fourth powers of two other consecutive natural numbers? | 105. Suppose that $m^{2}+(m+1)^{2}=n^{4}+(n+1)^{4}$, where $m$ and $n$ are natural numbers. Then $m(m+1)=n(n+1)[n(n+1)+2]$.
Let $n(n+1)=k, \quad$ then $m(m+1)=k(k+2), \quad$ where $m>k$.
After transformations, we get $(m-k) m+(m-k) k+(m-k)=k$.
This equality is impossible, as each term is positive and the middle term... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,267 |
109. (11th grade) 30 teams participate in a football championship. Every two teams must play one match against each other. Prove that at any moment during the competitions, there are two teams that have played the same number of matches by that time. | 109. At any point in time, each team has played no more than 29 matches, and there are 30 teams in total. Therefore, there will be two teams that have played the same number of matches. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,271 |
110. (11th grade) In a regular hexagonal pyramid, a line passing through the centers of the circles circumscribed around one lateral face and another around a larger diagonal section intersects with a side of the base. Determine the dihedral angle at the base.
## SOLUTIONS, HINTS, AND ANSWERS
Translate the text above... | 110. Answer: $\cos \varphi=\sqrt{\frac{3}{13}}$.
## LITERATURE | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,272 | |
5. In triangle $ABC$, the sides $a, b$, and $c$ are given. Find its angle bisectors $l_{a}, l_{b}$, and $l_{c}$.
The solution to this problem relies on the following two lemmas:
Lemma 1. For triangle $ABC$ (Fig. 1), the following equality holds:
$$
l_{c}=\frac{2 a b}{a+b} \cos \frac{C}{2}
$$
Proof ${ }^{1}$. Method... | Proof. Using the cosine theorem successively, we have:
$$
\begin{gathered}
\left(c^{2}=a^{2}+b^{2}-2 a b \cos C\right) \Leftrightarrow\left(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right) \Leftrightarrow \\
\Leftrightarrow\left(1+\cos C=1+\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right) \Leftrightarrow\left(2 \cos ^{2} \frac{C}{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,273 |
6. Prove that the product of any two sides of a triangle is not less than the product of its perimeter and the radius of the inscribed circle:
$$
a b \geqslant P r
$$ | Solution. It is sufficient to equate expressions (1) and (3) for the area of the triangle and take into account that $\sin \alpha \leqslant 1$.
Remark. It follows from the solution that if the angle between sides $a$ and $b$ is not a right angle, then
$$
a b>\operatorname{Pr}
$$
From inequality (4) and the remark, i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,274 |
7. Prove that the radius $R$ of the circumcircle of an arbitrary triangle, the radius $r$ of the inscribed circle, and its perimeter $P$ satisfy the inequality:
$$
\text { a) } R>\frac{\sqrt{3}}{3} \sqrt{Pr} \text {, }
$$
if the triangle is acute or obtuse;
$$
\text { b) } R \geqslant \frac{\sqrt{2}}{2} \sqrt{Pr}
$$... | Solution. a) Connect the center $O$ of the circumscribed circle with the vertices of the triangle. For an acute triangle (Fig. 2)
$$
S_{A B C}=S_{A O B}+S_{B O C}+S_{C O A}
$$
By formula (3) we have:
$$
S_{A B C}=\frac{1}{2} \operatorname{Pr}
$$[^1]
 and (3) we have:
$$
\operatorname{Pr}=l_{1} l_{2} \sin \alpha
$$
From the inequality $\left(l_{1}-l_{2}\right)^{2} \geqslant 0$ it follows:
$$
l_{1} l_{2} \leqslant \frac{l_{1}^{2}+l_{2}^{2}}{2}
$$
from which, by virtue of equality (15), we obtain the desired inequality.
If the diagonal... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,276 |
9. In a convex quadrilateral $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$. Let $P_{1}, P_{2}, P_{3}$, and $P_{4}$ be the perimeters of triangles $A O B, B O C, C O D$, and $D O A$, respectively, and let $r_{1}, r_{2}, r_{3}$, and $r_{4}$ be the radii of the inscribed circles in these triangles. Prove... | Solution. On the one hand, by formula (3)
$$
S_{A O B}=\frac{1}{2} P_{1} r_{1}
$$
On the other hand, by formula (1)
$$
S_{A O B}=\frac{1}{2} O A \cdot O B \cdot \sin A O B
$$
From equalities (17) and (18) we obtain
$$
P_{1} r_{1} \leqslant O A \cdot O B
$$
Similarly,
$$
\begin{aligned}
& P_{2} r_{2} \leqslant O ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,277 |
10. Prove that for a convex quadrilateral $A B C D$ the following inequality holds
$$
S_{A B C D} \leqslant \frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}
$$
where $a, b, c, d$ are the lengths of its sides. | Solution. Divide the quadrilateral into two triangles by a diagonal (Fig. 4). Then
$$
S_{A B C D}=S_{A B C}+S_{A D C} .
$$
By formula (1) we have:
$$
S_{A B C}=\frac{1}{2} a b \sin B
$$
From this and the inequality $a b \leqslant \frac{a^{2}+b^{2}}{2}$, we obtain:
$$
S_{A B C} \leqslant \frac{a^{2}+b^{2}}{4}
$$
S... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,278 |
12. In a triangle, $r$ and $R$ are the radii of the inscribed and circumscribed circles, respectively. Prove that $R \geqslant 2 r$. When is equality achieved? | Solution. Let $a, b$ and $c$ be the lengths of the sides of a triangle. Then ${ }^{1}$ $R=\frac{a b c}{4 S} \quad$ and $\quad r=\frac{2 S}{a+b+c}$, where $S=\sqrt{p(p-a)(p-b)(p-c)}$.
We will investigate the sign of the difference $R-2 r$:
$$
R-2 r=\frac{a b c}{4 S}-\frac{4 S}{a+b+c}=\frac{a b c-(a+b-c)(a+c-b)(b+c-a)}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,279 |
13. In triangle $ABC$ with sides $a, b, c$, the bisector $l_{a}$ is at distances from the vertices whose sum is $h_{A}$; the bisector $l_{b}$ is at distances whose sum is $-h_{B}$; the bisector $l_{c}$ is at distances whose sum is $h_{C}$.
Prove that
$$
h_{A} h_{B} h_{c} \geqslant \frac{1}{8}(a+b)(b+c)(c+a)
$$ | Solution. The following is true
Fig. 5
Lemma. For a triangle, the equality $h_{A}=\frac{2 S}{l_{a}}$ holds, where $S$ is the area of the triangle.
Proof. We have (Fig. 5):
$$
S=S_{A B A_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,280 |
18. Determine the area of triangle $ABC$, if $A(1 ; 2)$, $B(-2 ; 5)$ and $C(4 ;-2)$. | Solution. Let's form a column:
From here, \( F_{\mathbf{s}} = ((-2 + 20 - 4) - (8 + 4 + 5)) = -3 \).
By formula (4), \( S_{A B C} = \frac{1}{2} \cdot |-3| = 1.5 \). | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,282 |
19. Given points $A(3 ; 4)$ and $B(6 ; 6)$. On the $O y$ axis, determine the point $C$ such that the area of triangle $A B C$ is equal to 5. | Solution. Let's form the column, noting that $C(0 ; y)$ :
From this, $F_{3}=-3 y+6$. Since $S_{3}=5$, then $\frac{1}{2}|-3 y+6|=5$, from which $y_{1}=-1 \frac{1}{3}, y_{2}=5 \frac{1}{3}$.
T... | C_{1}(0;-1\frac{1}{3})C_{2}(0;5\frac{1}{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,283 |
20. Find the condition under which points $A\left(x_{1} ; y_{1}\right), B\left(x_{2} ; y_{2}\right)$ and $C\left(x_{3} ; y_{3}\right)$ lie on the same line. | Solution. If these points lie on the same straight line, then triangle $A B C$ degenerates into a segment. Then, by theorem $1 S_{3}=0$, hence $F_{3}=0$, i.e.
$$
\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(y_{1} x_{2}+y_{2} x_{3}+y_{3} x_{1}\right)=0
$$
from which
$$
x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}=y_... | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}=y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,284 |
23. Form the equation of the line passing through two points $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)$ | Solution. Let $C(x ; y)$ be the point of the desired line. Then consider the degenerate triangle $A B C$ (Problem 20). Setting $x_{3}=x$ and $y_{3}=y$, we get:
$$
\boldsymbol{x}_{1} y_{2}+x_{2} y+x y_{1}=y_{1} x_{2}+y_{2} x+y x_{1}, \text{ or }\left(x-x_{1}\right)\left(y_{2}-y_{1}\right)=\left(y-y_{1}\right)\left(x_{2... | (x-x_{1})(y_{2}-y_{1})=(y-y_{1})(x_{2}-x_{1}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,285 |
24. Calculate the area of a pentagon whose vertices have the coordinates: $(-1 ; 2),(1 ; 1)$, $(2 ; 3),(3 ;-5)$ and $(-2 ;-4)$. | Solution. Let's make a "rough" drawing (Fig. 10). We have:
F... | 25.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,286 |
25. Prove that a half-plane is a convex figure. | Solution. Consider the half-plane defined by the inequality
$$
a x+b y+c \leqslant 0
$$
Let this half-plane contain two points $M_{1}\left(x_{1} ; y_{1}\right)$ and $M_{2}\left(x_{2} ; y_{2}\right)$, i.e.,
$$
a x_{1}+b y_{1}+c \leqslant 0 \quad \text { and } \quad a x_{2}+b y_{2}+c \leqslant 0 .
$$
Any point $M(x ;... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,287 |
26. Prove that a circle is a convex figure. | Solution. Let $r$ be the radius of the circle. We choose a rectangular coordinate system with the origin at the center of the circle. Then the circle is defined by the following inequality
$$
x^{2}+y^{2}-r^{2} \leqslant 0
$$
(5)
Consider two points $M_{1}\left(x_{1} ; y_{1}\right)$ and $M_{2}\left(x_{2} ; y_{2}\righ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,288 |
27. Prove that a triangle is a convex figure. | Solution. First, let's prove that the intersection ${ }^{2}$ of any finite number of convex figures is also a convex figure.
Indeed, let two points $M_{1}$ and $M_{2}$ belong to the intersection of convex figures. Then these points belong to all intersecting figures. Since all these figures are convex, any point $M$ o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,289 |
28. In triangle $A B C\left(\angle C=120^{\circ}\right)$, the two sides $a$ and $b$ and the angle bisector $l_{c}$ are given (Fig. 12). Prove that
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{l_{c}}
$$[^3]
$. From this, $\frac{AM}{AC} = \frac{BM}{EC}$, or $\frac{a+b}{b} = \frac{a}{l_{c}}$, or $\frac{a+b}{ab} = \frac{1}{l_{c}}$, or $\frac{1}{a} + \frac{1}{b} = \frac{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,290 |
38. Prove the Law of Cosines. | Solution. Let a triangle $ABC$ be given with sides $a, b$, and $c$, and $\angle CAB = \alpha$.
We introduce a coordinate system as shown in Figure 16. Then the vertices will have the following coordinates:
$$
A(0; 0), B(c; 0) \quad \text{and} \quad C(b \cos \alpha; b \sin \alpha).
$$
Using the distance formula betwe... | ^{2}=b^{2}+^{2}-2bc\cos\alpha | Geometry | proof | Yes | Yes | olympiads | false | 35,291 |
39. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides (problem 3, b). | Solution. Let a parallelogram $ABCD$ be given with sides $a$, $b$, and $\angle A = \alpha$.
We introduce a rectangular coordinate system as shown in Figure 17. Then the vertices will have the following coordinates:
$A(0; 0), B(a \cos \alpha; a \sin \alpha), C(a \cos \alpha + b; a \sin \alpha)$, and $D(b; 0)$.
Using ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,292 |
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