problem stringlengths 2 1.73k | answer stringlengths 1 313 | solution stringlengths 3 9.24k | difficulty float64 1 10 |
|---|---|---|---|
6. In the Kingdom of Mathematics, the denominations of banknotes in circulation are 1 yuan, 5 yuan, 10 yuan, 20 yuan, 50 yuan, and 100 yuan. One day, two customers each bought a chocolate bar worth 15 yuan at the same grocery store. One of them paid with two 10-yuan banknotes, while the other paid with a 20-yuan and a ... | 8 | 【Answer】D
【Analysis】Considering the four cases respectively, when the gum costs 8 yuan, one person pays 10 yuan (2 five-yuan notes), and another person pays 13 yuan (1 ten-yuan note, 3 one-yuan notes), the first person's 1 five-yuan note can be given as change to the second person, and the second person's 2 one-yuan no... | 4.67 |
4. Triples of natural numbers $\left(a_{i}, b_{i}, c_{i}\right)$, where $i=1,2, \ldots, n$ satisfy the following conditions:
1) $a_{i}+b_{i}+c_{i}=2017$ for all $i=1,2, \ldots, n$;
2) if $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. What is the maximum possible value of $n$? (M. Popov) | 1343 | Answer: 1343.
Solution: Note that
$$
\sum_{i=1}^{n} a_{i} \geqslant \sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$
A similar inequality is written for the sums $b_{i}$ and $c_{i}$. Adding the three obtained inequalities, we get
$$
\begin{gathered}
3 \cdot \frac{n(n+1)}{2} \leqslant \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}... | 4.67 |
5. What is the maximum number of rooks that can be placed on the cells of a $300 \times 300$ board so that each rook attacks no more than one other rook? (A rook attacks all cells it can reach according to chess rules, without passing through other pieces.)
# | 400 | # Answer: 400
First solution. We will prove that no more than 400 rooks can be placed on the board. In each row or column, there are no more than two rooks; otherwise, the rook that is not at the edge will attack at least two other rooks. Suppose there are $k$ columns with two rooks each. Consider one such pair. They ... | 5 |
4. A kilo of sausages was placed on a straight line between a dog in a kennel and a cat. The animals simultaneously rushed to the sausages. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and ate an equal amount. It is known that the cat could eat a... | 1.4 | Answer: 1.4 times closer to the dog than to the cat.
## Solution:
Let $v$ be the running speed of the dog, $u$ be the eating speed of the cat, and the volume of sausages eaten by each animal be 1.
Then, $2 v$ is the running speed of the cat, and $2 u$ is the eating speed of the dog.
Let the distance from the cat to... | 2 |
Compute the smallest positive integer that does not appear in any problem statement on any round at HMMT November 2023. | 22 | The number 22 does not appear on any round. On the other hand, the numbers 1 through 21 appear as follows. \begin{tabular}{c|c|c} Number & Round & Problem \\ \hline 1 & Guts & 21 \\ 2 & Guts & 13 \\ 3 & Guts & 17 \\ 4 & Guts & 13 \\ 5 & Guts & 14 \\ 6 & Guts & 2 \\ 7 & Guts & 10 \\ 8 & Guts & 13 \\ 9 & Guts & 28 \\ 10 ... | 1.6 |
I2.3 Determine the smallest positive integer $\gamma$ such that the equation $\sqrt{x}-\sqrt{\beta \gamma}=4 \sqrt{2}$ has an integer solution in $x$. | 3 | $$
\begin{array}{l}
\sqrt{x}-\sqrt{24 \gamma}=4 \sqrt{2} \\
\sqrt{x}=2 \sqrt{6 \gamma}+4 \sqrt{2}
\end{array}
$$
The smallest positive integer $\gamma=3$ | 1.6 |
## Task 1.
If $a_{1}, a_{2}, \ldots, a_{2000}$ is a sequence of 2000 positive real numbers, for how many indices $i \in$ $\{1,2, \ldots, 2000\}$ can the equality
$$
a_{i} a_{i+3}=a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} ?
$$
hold? We consider that $a_{j+2000}=a_{j}$ for $j \in\{1,2,3\}$. | 999 | ## Solution.
For an index $i$, we say it is good if it satisfies the equality in the problem statement.
Assume there exist two consecutive good indices $i, i+1$. Then we have
$$
\begin{aligned}
a_{i} a_{i+3} & =a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} \\
a_{i+1} a_{i+4} & =a_{i+1} a_{i+2}+a_{i+2} a_{i+3}+a_{i+3... | 5.5 |
13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points ... | 20 | Answer: 20.
Solution. Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when ... | 5.75 |
7. (10 points) The last 8 digits of $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$ are | 87654321 | 【Analysis】By applying the commutative and associative laws of multiplication to integrate $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$, then organize it into $11111111 \times 111111111111$, we can derive that the last 8 digits of $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$ a... | 4 |
As shown in the figure, $\triangle A B C$ is an isosceles right triangle, $A B=28 \mathrm{~cm}$. A semicircle is drawn with $B C$ as the diameter, and point $D$ is the midpoint of the semicircle arc. Try to find the area of the shaded part. (Take $\pi=\frac{22}{7}$.) | 252 \text{ cm}^2 | Geometry, cleverly finding area, cutting and supplementing.
(Method 1)
Take the midpoint $E$ of $B C$, connect $D E$; connect $B D$;
$S_{\triangle A B D}=A B \times B E \div 2=28 \times 14 \div 2=196$ square centimeters;
$S_{\text {sector } B E D}=\frac{1}{4} \times \pi \times B E^{2}=\frac{1}{4} \times \pi \times 14^{... | 3 |
14. (12 points) Use 36 solid rectangular prisms of size $3 \times 2 \times 1$ to form a large cube of size $6 \times 6 \times 6$. Among all possible arrangements, the maximum number of small rectangular prisms that can be seen from a point outside the large cube is $\qquad$. | \documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
31
\end{document} | 【Answer】Solution: As shown in the figure,
To see the maximum number from the outside, it is necessary to make the rectangles seen from the outside as "deeply" inside the square as possible, the result is as follows: a total of $6 \times 3+3 \times 4+3 \times 1+1=31$ (pieces). Therefore, the answer is: 31. | 4 |
11. (3 points) There are 20 points below, with each adjacent pair of points being equidistant. By connecting four points with straight lines, you can form a square. Using this method, you can form $\qquad$ squares.
The text above has been translated into English, preserving the original text's line breaks and format... | 20 | 【Answer】Solution: The number of squares with a side length of 1 unit is 12;
The number of squares with a side length of 2 units is 6;
The number of squares with a side length of 3 units is 2;
The maximum side length is 3 units, any larger and it would not form a square;
In total, there are squares: $12+6+2=20$ (squares... | 1 |
## Task B-2.1.
For which values of the real parameter $m$ does the inequality $\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2}<47$ hold if $x_{1}$ and $x_{2}$ are the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$? | m \in (2,4) | ## Solution.
For the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$, we have
$$
x_{1}+x_{2}=2^{m-1}-5 \quad \text{and} \quad x_{1} x_{2}=1
$$
Therefore,
$$
\begin{aligned}
\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2} & =\frac{x_{1}^{4}+x_{2}^{4}}{x_{1}^{2} x_{2}^{2}}=\fr... | 3.33 |
The elements of the sequence $\mathrm{Az}\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence. | x_n = n | The elements of the sequence $\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence.
Solution. We wi... | 5 |
Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $n$ to $5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the numb... | 35 | Consider the number of beans Alice has in base 5. Note that $2008=31013_{5}, 42=132_{5}$, and $100=400_{5}$. Now, suppose Alice has $d_{k} \cdots d_{2} d_{1}$ beans when she wins; the conditions for winning mean that these digits must satisfy $d_{2} d_{1}=32, d_{k} \cdots d_{3} \geq 310$, and $d_{k} \cdots d_{3}=4i+1$ ... | 5 |
18. (2 marks) Let $A_{1} A_{2} \cdots A_{2002}$ be a regular 2002-sided polygon. Each vertex $A_{i}$ is associated with a positive integer $a_{i}$ such that the following condition is satisfied: If $j_{1}, j_{2}, \cdots, j_{k}$ are positive integers such that $k<500$ and $A_{j_{1}} A_{j_{2}} \cdots A_{j_{k}}$ is a regu... | 287287 | 18. Since $2002=2 \times 7 \times 11 \times 13$, we can choose certain vertices among the 2002 vertices to form regular 7-, 11-, 13-, $\cdots$ polygons (where the number of sides runs through divisors of 2002 greater than 2 and less than 1001). Since $2002=7 \times 286$, so there are at least 286 different positive int... | 6.33 |
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov) | 6 | Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, ... | 6 |
2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card? | 8 | Answer: 8.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear th... | 2.2 |
2. (10 points) Five pieces of paper are written with $1$, $2$, $3$, $4$, and $5$, facing up from smallest to largest, stacked in a pile. Now, the 1, 3, and 5 are flipped to their backs, and still placed in their original positions. If the entire stack of paper is split at any one piece of paper into two stacks, and the... | 5 | 2. (10 points) Five pieces of paper are written with $1, 2, 3, 4, 5$ respectively, facing upwards from smallest to largest, stacked into one pile. Now, the $1, 3,$ and $5$ are flipped to their backs and placed back in their original positions. If the entire stack is split at any one piece of paper into two stacks, and ... | 2.25 |
Given two positive integers $n$ and $k$. In the plane, there are $n$ circles ($n \geq 2$) such that each circle intersects every other circle at two points, and all these intersection points are pairwise distinct.
Each intersection point is colored with one of $n$ colors such that each color is used at least once and o... | 2 \leq k \leq n \leq 3 \text{ or } 3 \leq k \leq n | The answer is: $2 \leq k \leq n \leq 3$ or $3 \leq k \leq n$.
Obviously, $k \leq n$ according to the problem statement, and $k \geq 2$, because for $k=1$ all points would have the same color, while the number $n$ of colors should be $\geq 2$. We number the circles and the colors from 1 to n and denote by $F(i, j)$ the ... | 5 |
[ $\underline{\text { Classical Inequalities (Miscellaneous) })]}$
$$
\sqrt{x_{1}^{2}+\left(1-x_{2}\right)^{2}}+\sqrt{x_{2}^{2}+\left(1-x_{3}\right)^{2}}+\ldots+\sqrt{x_{2 n}^{2}+\left(1-x_{1}\right)^{2}}
$$ | \frac{n}{\sqrt{2}} | According to the inequality between the quadratic mean and the arithmetic mean
$$
\begin{aligned}
& \sqrt{2}\left(\sqrt{x_{1}^{2}+\left(1-x_{2}\right)^{2}}+\sqrt{x_{2}^{2}+\left(1-x_{3}\right)^{2}}+\ldots+\sqrt{x_{n}^{2}+\left(1-x_{1}\right)^{2}}\right) \geq \\
& \geq\left|x_{1}\right|+\left|1-x_{2}\right|+\left|x_{2}... | 4.33 |
10,11
There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be broken so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 150 g can be formed (a broken link also weighs 1 g)? | \text{4} | Answer: 4 links. According to the solution of problem 5 for grades $7-8$, for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to unfasten 4 links. | 4.2 |
$12.23 y=\frac{x}{\ln x}$.
The above text is translated into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a mathematical format which is universal and does not require translation, the translation is as f... | (e, e) | 12.23 The function is defined for $x>0$ and $x \neq 1$. We have
$$
y^{\prime}=\frac{\ln x - x \cdot \frac{1}{x}}{\ln ^{2} x}=\frac{\ln x - 1}{\ln ^{2} x}
$$
Thus, $y^{\prime}=0$ when $\ln x - 1 = 0$, i.e., when $x=e$. Let's construct a table:
| Interval | $(0,1)$ | $(1, e)$ | $e$ | $(e, \infty)$ |
| :---: | :---: | ... | 3.5 |
6. A factory produces sets of $n>2$ elephants of different sizes. According to the standard, the difference in mass between adjacent elephants within each set should be the same. The inspector checks the sets one by one using a balance scale without weights. For what smallest $n$ is this possible? | $n=5$ | - Omвem: for $n=5$. Let's number the elephants in ascending order of their size, then it is enough to verify that $C_{1}+C_{4}=C_{2}+C_{3}, C_{1}+C_{5}=C_{2}+C_{4}$ and $C_{2}+$ $C_{5}=C_{3}+C_{4}$. These equalities are equivalent to $C_{4}-C_{3}=C_{2}-C_{1}$, $C_{2}-C_{1}=C_{5}-C_{4}$ and $C_{5}-C_{4}=C_{3}-C_{2}$, th... | 5 |
7.214. $9^{x}+6^{x}=2^{2 x+1}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.214. $9^{x}+6^{x}=2^{2 x+1}$. | 0 | Solution.
Rewrite the equation as $3^{2 x}+2^{x} \cdot 3^{x}-2 \cdot 2^{2 x}=0$ and divide it by $2^{2 x} \neq 0$. Then $\left(\frac{3}{2}\right)^{2 x}+\left(\frac{3}{2}\right)^{x}-2=0 \Rightarrow\left(\left(\frac{3}{2}\right)^{x}\right)=-2$ (no solutions) or $\left(\left(\frac{3}{2}\right)^{x}\right)_{2}=1 \Rightarro... | 3 |
After a fair die with faces numbered 1 to 6 is rolled, the number on the top face is $x$. What is the most likely outcome? | x > 2 | With a fair die that has faces numbered from 1 to 6, the probability of rolling each of 1 to 6 is $\frac{1}{6}$. We calculate the probability for each of the five choices. There are 4 values of $x$ that satisfy $x>2$, so the probability is $\frac{4}{6}=\frac{2}{3}$. There are 2 values of $x$ that satisfy $x=4$ or $x=5$... | 1.2 |
25 Two equilateral triangles overlap to form a six-pointed star (as shown in the figure). The first 12 positive integers 1, $2, \cdots, 12$ are to be placed at the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Find the minimum sum of the numbers placed at the six vert... | 24 | (1) For any filling method that satisfies the conditions, the number filled at point $a_{i}$ is still denoted as $a_{i}, i=1,2, \cdots, 12$. If the sum of the four numbers on each line is $s$, then from $6 s=2(1+2+\cdots+12)$, we get $s=26$;
In $\triangle a_{1} a_{3} a_{5}$, on the three sides, we have
$$
\begin{array}... | 5.2 |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq \pi, 0 \leq y \leq 2 \pi\}$ satisfy the system of equations $\left\{\begin{array}{c}\sin x+\sin y=\sin 2 \\ \cos x+\cos y=\cos 2\end{array}\right.$. How many such points are there in the square? Find the coordinates $(x ; y)$ of the point ... | \left(2+\frac{\pi}{3}, 2-\frac{\pi}{3}\right) | Answer: 1) two points
$$
\text { 2) } x=2+\frac{\pi}{3}, y=2-\frac{\pi}{3}
$$ | 2.25 |
9. Let $a+b=1, b>0, a \neq 0$, then the minimum value of $\frac{1}{|a|}+\frac{2|a|}{b}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2 \sqrt{2}-1 | Solve $2 \sqrt{2}-1$.
Analysis $\frac{1}{|a|}+\frac{2|a|}{b}=\frac{a+b}{|a|}+\frac{2|a|}{b}=\frac{a}{|a|}+\left(\frac{b}{|a|}+\frac{2|a|}{b}\right) \geq \frac{a}{|a|}+2 \sqrt{\frac{b}{|a|} \cdot \frac{2|a|}{b}}=2 \sqrt{2}+\frac{a}{|a|}$, where the equality holds when $\frac{b}{|a|}=\frac{2|a|}{b}$, i.e., $b^{2}=2 a^{2}... | 9 |
25. A scout is in a house with four windows arranged in a rectangular shape. He needs to signal to the sea at night by lighting a window or several windows. How many different signals can he send? | 10 | 25. Let's schematically represent the windows and measure them:
a) Lighting all four windows gives one signal;
b) Lighting one of the windows is perceived as one signal, as in the dark, th... | 1.6 |
6.014. $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$. | $x=0$ | ## Solution.
Domain: $x \in R$.
$\frac{2 x^{4}+9 x^{2}}{\left(x^{2}+4\right)\left(x^{2}+5\right)}=0 \Leftrightarrow 2 x^{4}+9 x^{2}=0 \Leftrightarrow x^{2}\left(2 x^{2}+9\right)=0$,
$x^{2}=0, x_{1}=0$ or $2 x^{2}+9=0, x_{2,3} \in \varnothing$.
Answer: $x=0$. | 2 |
6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers. | 4060224 | Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\... | 8 |
3. Given $n^{2}$ points arranged in the plane in the form of a square grid $n \times n$. A broken line of length $l$ is the union of closed segments $A_{0} A_{1}$, $A_{1} A_{2}$, $A_{2} A_{3}, \ldots, A_{l-1} A_{l}$ in that plane. Determine the smallest natural number $l$ for which there exists a broken line of length ... | $2n-2$ | 3. Let the desired value of the number $l$ be denoted by $l_{n}$. Trivially, $l_{1}=1$ and $l_{2}=3$. The examples in the figure show that $l_{3} \leqslant 4$ and, in general, $l_{n} \leqslant 2 n-2$ for $n \geqslant 3$.
Assume that the broken line $\mathcal{C}$ of length $l$ covers all $n^{2}$ points. Let there be $a... | 3.4 |
2. Find all positive integers $n$ $(n \geqslant 3)$ such that there exists a set $M$ with $n$ elements, where the elements are distinct non-zero vectors of equal length, and the following conditions are satisfied: $\sum_{u \in M} u=0$, and for any $v, w \in M$, $\boldsymbol{v}+\boldsymbol{w} \neq \mathbf{0}$. | \{n \in \mathbf{N} \mid n \geqslant 3, n \neq 4\} | 2. First, for any odd number $n$ not less than 3, take $M$ as the $n$ distinct complex roots of the equation $z^{n}-1=0$. Clearly, the set $M$ satisfies the requirements.
Next, consider even numbers $n$ not less than 6.
Let $\frac{n}{2}=k$, first consider the decomposition of $\frac{1}{2}$.
From $\frac{1}{n}=\frac{1}{n... | 3 |
Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$?
... | 14.6 | First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\le... | 2.75 |
12. Given that $m, n, t (m<n)$ are all positive integers, points $A(-m, 0), B(n, 0), C(0, t)$, and $O$ is the origin. Suppose $\angle A C B=90^{\circ}$, and
$$
O A^{2}+O B^{2}+O C^{2}=13(O A+O B-O C) \text {. }
$$
(1) Find the value of $m+n+t$;
(2) If the graph of a quadratic function passes through points $A, B, C$, f... | $y = -\frac{1}{3} x^{2} + \frac{8}{3} x + 3$ | 12. (1) According to the problem, we have
$$
O A=m, O B=n, O C=t \text {. }
$$
From $\angle A C B=90^{\circ}, O C \perp A B$, we get
$$
O A \cdot O B=O C^{2} \Rightarrow m n=t^{2} \text {. }
$$
From the given equation, we have
$$
\begin{array}{l}
m^{2}+n^{2}+t^{2}=13(m+n-t) . \\
\text { Also, } m^{2}+n^{2}+t^{2} \\
=... | 3 |
497. Shuffling Cards. An elementary method of shuffling cards consists of taking a deck face down in the left hand and transferring the cards one by one to the right hand; each successive card is placed on top of the previous one: the second on top of the first, the fourth on top of the third, and so on until all the c... | 14 | 497. To shuffle 14 cards in the manner described above and return them to their original order, it takes 14 shuffles, although in the case of 16 cards, only 5 are required. We cannot delve into the nature of this phenomenon here, but the reader may find it interesting to conduct an independent investigation of this que... | 4 |
5. The number of real roots of the equation $\frac{\left(x^{2}-x+1\right)^{3}}{x^{2}(x-1)^{2}}=\frac{\left(\pi^{2}-\pi+1\right)^{3}}{\pi^{2}(\pi-1)^{2}}$ with respect to $x$ is exactly $($ )
A. 1
B. 2
C. 4
D. 6 | 6 | 5. D
From the given conditions, it is clear that $x=\pi$ is a root of the original equation. Let's denote the left side as $f(x)$. It is easy to verify:
$$
f\left(\frac{1}{x}\right)=f(x), f(1-x)=f(x), f\left(\frac{1}{1-x}\right)=f(x), f\left(1-\frac{1}{x}\right)=f
$$
$(x), f\left(\frac{x}{x-1}\right)=f(x)$. Therefore,... | 5 |
7. From the sides and diagonals of a regular 12-sided polygon, three different segments are randomly selected. The probability that their lengths are the side lengths of a triangle is $\qquad$ | \frac{223}{286} | 7. $\frac{223}{286}$.
Let the diameter of the circumscribed circle of a regular 12-sided polygon be 1. Then, the lengths of all $\mathrm{C}_{12}^{2}=66$ line segments fall into 6 categories, which are
$$
\begin{array}{l}
d_{1}=\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}, \\
d_{2}=\sin 30^{\circ}=\frac{1}{2}, \\
d_{3}=... | 7 |
Riley has 64 cubes with dimensions $1 \times 1 \times 1$. Each cube has its six faces labelled with a 2 on two opposite faces and a 1 on each of its other four faces. The 64 cubes are arranged to build a $4 \times 4 \times 4$ cube. Riley determines the total of the numbers on the outside of the $4 \times 4 \times 4$ cu... | 49 | We can categorize the 64 small cubes into four groups: 8 cubes that are completely in the interior of the larger cube (and are completely invisible), $4 \times 6=24$ "face cubes" that have exactly one of their faces showing, $2 \times 12=24$ "edge cubes" that have exactly two faces showing, and 8 "corner cubes" that ha... | 4 |
16. Given the six-digit number $\overline{9786 \square}$ is a multiple of $\mathbf{99}$, the quotient when this six-digit number is divided by $\mathbf{99}$ is ( ). | 6039 | $\begin{array}{l}\text { [Analysis] Let } 99 \mid \overline{A 9786 B} \text {, sum of pairs from right to left } \\ 99 \mid \overline{A 9}+78+\overline{6 B} \text {, i.e., } 99|78+69+\overline{A B} \Rightarrow 99| 48+\overline{A B} \\ \overline{A B}=51 \text {, i.e., } \mathrm{A}=5, \mathrm{~B}=1 \\ 597861 \div 99=6039... | 1.5 |
The base of the right prism $A B C A_{1} B_{1} C_{1}$ is an isosceles triangle $A B C$, where $A B=B C=5$, $\angle A B C=2 \arcsin 3 / 5$. A plane perpendicular to the line $A_{1} C$ intersects the edges $A C$ and $A_{1} C_{1}$ at points $D$ and $E$ respectively, such that $A D=1 / 3 A C, E C_{1}=1 / 3 A_{1} C_{1}$. Fi... | \frac{40}{3} | Let $M$ be the midpoint of the base $AC$ of the isosceles triangle $ABC$ (Fig. 1). From the right triangle $AMB$ we find that
Fig. 1
+17=42
$$
and when $x=4, y=2$, the equality can be achieved, so the maximum value of $x y+8 x+y$ is 42.
Alternatively,... | 5 |
a) We have 3 initially empty urns. One of them is chosen at random with equal probability (1/3 for each). Then, a ball is placed inside the chosen urn. The process is repeated until one of the urns has two balls. What is the probability that when the process ends, the total number of balls in all urns is equal to 2?
b... | \frac{2013}{2014} \cdot \frac{2012}{2014} \cdot \frac{2011}{2014} \cdot \frac{2010}{2014} \cdot \frac{2009}{2014} \cdot \frac{2008}{2014} \cdot \frac{2007}{2014} \cdot \frac{2006}{2014} \cdot \frac{9}{2014} | Solution
a) First, a ball is placed in some urn. For the process to end with two balls, it is necessary that, in the next step, the next ball is placed in the same urn where the first ball was placed. This has a probability of 1/3. Therefore, the probability that the process ends with two balls is equal to $1 / 3$.
b... | 5.67 |
4. In acute triangle $A B C, M$ and $N$ are the midpoints of sides $A B$ and $B C$, respectively. The tangents to the circumcircle of triangle $B M N$ at $M$ and $N$ meet at $P$. Suppose that $A P$ is parallel to $B C$, $A P=9$ and $P N=15$. Find $A C$. | 20 \sqrt{2} | Answer: $20 \sqrt{2}$
Solution: Extend rays $P M$ and $C B$ to meet at $Q$. Since $A P \| Q C$ and $M$ is the midpoint of $A B$, triangles $A M P$ and $B M Q$ are congruent. This gives $Q M=M P=P N=15$ and $Q B=A P=9$. Observe that the circumcircle of triangle $B M N$ is tangent to $Q M$, so by power of a point, we com... | 6 |
6. In trapezoid $\mathrm{PQRS}$, it is known that $\angle \mathrm{PQR}=90^{\circ}, \angle Q R S>90^{\circ}$, diagonal $S Q$ is 24 and is the bisector of angle $S$, and the distance from vertex $R$ to line $Q S$ is 5. Find the area of trapezoid PQRS. | \frac{27420}{169} | Answer: $\frac{27420}{169}$.
Solution $\angle \mathrm{RQS}=\angle \mathrm{PSQ}$ as alternate interior angles, so triangle $\mathrm{RQS}$ is isosceles. Its height $\mathrm{RH}$ is also the median and equals 5. Then $S R=\sqrt{R H^{2}+S H^{2}}=13$. Triangles SRH and SQP are similar by two angles, the similarity coeffici... | 6 |
97. There are 5 parts that are indistinguishable in appearance, 4 of which are standard and of the same mass, and one is defective, differing in mass from the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the defective part? | 3 | $\triangle$ Let's number the parts. Now try to figure out the weighing scheme presented below (Fig. 32).
As can be seen from this, it took three weighings to find the defective part.
Answer: in three. | 3 |
12. (6 points) There are 1997 odd numbers, the sum of which equals their product. Among them, three numbers are not 1, but three different prime numbers. Then, these three prime numbers are $\qquad$、$\qquad$、$\qquad$. | 5, 7, 59 | 【Solution】Solution: According to the problem, we have: $1994+a+b+c=a b c$; when $a=3, b=5$, $15 c=c+2002, c=143$, which is not a prime number; when $a=3, b=7$, $21 c=c+2004, c=50 \frac{1}{5}$, which is not an integer; when $a=5, b=7$, $35 c=c+2006, c=59$, which satisfies the condition; therefore, the answer is: $5, 7, ... | 5.5 |
10. Color the 6 regions $A, B, C, D, E, F$ in Figure 1, with each region being colored with 1 color, and no two adjacent regions having the same color. If there are 4 colors available, then there are $\qquad$ different coloring schemes. | 96 | 10. 96
$A, B, C$ have different colors from each other. They have $4 \times$ $3 \times 2=24$ ways of coloring.
For any one of these (let's assume $A-a \quad B-b \quad C-c$. The other color is $d$), then $D$ can be colored with $b, d$, $E$ can be colored with $c, d$, and $F$ can be colored with $a, d$.
Since $D, E, F$ a... | 2.67 |
Given positive integers $a$, $b$, and $c$ satisfy $a^{2}+b^{2}-c^{2}=2018$.
Find the minimum values of $a+b-c$ and $a+b+c$. | 2 \text{ and } 52 | From the problem, we know
$$
a^{2}+b^{2}=c^{2}+2018 \equiv 2 \text { or } 3(\bmod 4)
$$
$\Rightarrow a, b$ are odd, $c$ is even
$$
\begin{array}{l}
\Rightarrow c^{2}+2018=a^{2}+b^{2} \equiv 2(\bmod 8) \\
\Rightarrow c^{2} \equiv 0(\bmod 8) \Rightarrow 4 \mid c .
\end{array}
$$
Notice,
$$
(a+b)^{2}-c^{2}>a^{2}+b^{2}-c^... | 3.25 |
The Dingoberry Farm is a 10 mile by 10 mile square, broken up into 1 mile by 1 mile patches. Each patch is farmed either by Farmer Keith or by Farmer Ann. Whenever Ann farms a patch, she also farms all the patches due west of it and all the patches due south of it. Ann puts up a scarecrow on each of her patches that is... | 7 | Whenever Ann farms a patch $P$, she also farms all the patches due west of $P$ and due south of $P$. So, the only way she can put a scarecrow on $P$ is if Keith farms the patch immediately north of $P$ and the patch immediately east of $P$, in which case Ann cannot farm any of the patches due north of $P$ or due east o... | 7 |
Let $x_{1}=y_{1}=x_{2}=y_{2}=1$, then for $n \geq 3$ let $x_{n}=x_{n-1} y_{n-2}+x_{n-2} y_{n-1}$ and $y_{n}=y_{n-1} y_{n-2}- x_{n-1} x_{n-2}$. What are the last two digits of $\left|x_{2012}\right|$ ? | 84 | Let $z_{n}=y_{n}+x_{n} i$. Then the recursion implies that: $$\begin{aligned} & z_{1}=z_{2}=1+i \\ & z_{n}=z_{n-1} z_{n-2} \end{aligned}$$ This implies that $$z_{n}=\left(z_{1}\right)^{F_{n}}$$ where $F_{n}$ is the $n^{\text {th }}$ Fibonacci number $\left(F_{1}=F_{2}=1\right)$. So, $z_{2012}=(1+i)^{F_{2012}}$. Notice ... | 7.4 |
In a round-robin tournament, 23 teams participated. Each team played exactly once against all the others. We say that 3 teams have cycled victories if, considering only their games against each other, each of them won exactly once. What is the maximum number of cycled victories that could have occurred during the tourn... | 506 | Let's solve the problem more generally for $n$ teams, where $n$ is an odd number. We will represent each team with a point. After a match between two teams, an arrow should point from the losing team to the winning team. (In this context, we can assume that there has never been a tie.)
If the teams $A$, $B$, and $C$ h... | 5 |
15. (3 points) There are three different sizes of cubic wooden blocks, A, B, and C, where the edge length of A is $\frac{1}{2}$ of the edge length of B, and the edge length of B is $\frac{2}{3}$ of the edge length of C. If A, B, and C blocks are used to form a large cube with the smallest possible volume (using at leas... | 50 | 15. (3 points) There are three different sizes of cubic wooden blocks, A, B, and C, where the edge length of A is $\frac{1}{2}$ of the edge length of B, and the edge length of B is $\frac{2}{3}$ of the edge length of C. If A, B, and C blocks are used to form a large cube with the smallest possible volume (using at leas... | 3 |
6. Given a string of 2021 letters A and B. Consider the longest palindromic substring. What is its minimum possible length? A palindrome is a string that reads the same from right to left and from left to right. | 4 | Solution. The minimum possible length of the maximum palindrome is 4.
We will prove that it cannot be less than 4. Consider the 5 letters in the center of the string. If these are alternating letters, then it is a palindrome of length 5. Suppose among these five letters there are two identical letters standing next to... | 4 |
10. (5 points) According to the pattern shown in the figure, deduce that $M=$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 1692 | 【Solution】Solution: According to the problem, we have:
First, observe the pattern: $12+3=15$;
$15+5=20 ; \cdots$ Each square plus the circle equals the next number.
At the same time, we notice $20=12+3+5$
$$
27=12+3+5+7
$$
Summarizing the pattern, the numbers in the circles form an arithmetic sequence with the first t... | 6.25 |
Example 78. A triangular pyramid is cut by a plane into two polyhedra. We will find the ratio of the volumes of these polyhedra, given that the cutting plane divides the edges converging at one vertex of the pyramid in the ratio $1: 2, 1: 2, 2: 1$, counting from this vertex.
Construction of the image. Let the quadrila... | 25:2 | Solution. Let $V$ be the volume of the pyramid $SABC$, and $V_{1}$ the volume of the pyramid $PAQR$. Then $V_{1}=V-V_{2}$.
Construct $[SO]$ and assume that $[SO]$ is the image of the height of the pyramid $SABC$ (thus, two parameters are used). Construct $(AO)$ and $[PM] \|[SO]$. Then $M \in (AO)$. To simplify the cal... | 2.67 |
A3. A square with a side length of 2 is divided into 4 triangles (see the figure). All 3 shaded triangles have the same area. What is the area of the white triangle?
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{8}{5}$
(C) 2
(D) $3 \sqrt{5}-5$
(E) $6-2 \sqrt{5}$
.
Let $x = |A E|$. Since the right triangles $A E D$ and $... | 4 |
Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 504 | The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \(1\), and assume the side length of the octagon is \(2\).
Let \(r\) denote the radius of the circle, \(O\) be the center of the circle. Then:
\[r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.\]
No... | 6.8 |
A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of different lines that contain exactly $8$ of these points. | 168 | $Case \textrm{ } 1:$ The lines are not parallel to the faces
A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
We look at the one from $(1,1,1)$ to $(10,10,10)$. ... | 6.6 |
5. (8 points) Fill in the blanks with $1-6$, so that the numbers in each row and each column are not repeated. The two cells occupied by the same symbol in the figure have the same number combination, but the order is uncertain. Therefore, the five-digit number formed by the first five numbers from left to right in the... | 46123 | 46123
【Solution】Solution: According to the problem, we know:
First, the number in the second row and second column can only be 5, and the number in the third row and fourth column can only be 6.
Continuing the reasoning, the solution is as shown in the figure:
Therefore, the answer is: 46123. | 5 |
22. Let $x>1, y>1$ and $z>1$ be positive integers for which the following equation
$$
1!+2!+3!+\ldots+x!=y^{2}
$$
is satisfied. Find the largest possible value of $x+y+z$. | 8 | 22. Answer. 8
Solution. We first prove that if $x \geq 8$, then $z=2$. To this end, we observe that the left hand side of the equation $1!+2!+3!+\ldots+x!$ is divisible by 3 , and hence $3 \mid y^{2}$. Since 3 is a prime, $3 \mid y$. So, $3^{z} \mid y^{z}$ by elementary properties of divisibility.
On the other hand, wh... | 8 |
7.3 Given that a closed broken line is drawn along the grid lines of a square grid paper, consisting of 14 segments and with at most one edge of the broken line on each grid line, how many self-intersection points can this broken line have at most?
untranslated text remains unchanged:
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... | 17 | [Solution] Since the polyline is drawn along the grid lines, its horizontal and vertical edges are alternately arranged, so among the 14 edges, there must be 7 horizontal edges and 7 vertical edges. Number the horizontal edges from top to bottom as 1 to 7. It is easy to see that there are no self-intersection points on... | 6 |
11. Divide the set $M=$ $\{1,2, \cdots, 12\}$ of the first 12 positive integers into four triplets, such that in each triplet, one number is equal to the sum of the other two. Find the number of different ways to do this. | 8 | 11. Let the four subsets be $M_{i}=\left(a_{i}, b_{i}, c_{i}\right)$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}, i=1,2,3,4$.
Given $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from
$$
a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17,
$$
we get $a_{2}=9, a_{1}=8$
$$
\Rightarrow\left(a_{1}, a_{2}, a_{3}, a_{4... | 3 |
11、Given the ellipse $C: \frac{x^{2}}{3}+y^{2}=1$ with its upper vertex $A$, a line $l$ that does not pass through $A$ intersects the ellipse $C$ at points $P$ and $Q$, and $A P \perp A Q$. Find the maximum value of the area of $\triangle A P Q$.
---
Note: The format and line breaks have been preserved as requested. | \frac{9}{4} | 11. Analysis: When the slope of line $P Q$ does not exist, we know that $\angle P A Q$ is an acute angle, which does not meet the problem's requirements. Therefore, we only need to consider the case where the slope exists.
Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and the equation of line $P Q$ be ... | 5 |
G1.1 In the given diagram, $\angle A+\angle B+\angle C+\angle D+\angle E=a^{\circ}$, find $a$. | 180 | $\begin{array}{l}\text { In } \triangle A P Q, \angle B+\angle D=\angle A Q P \ldots \ldots \text { (1) (ext. } \angle \text { of } \triangle) \\ \angle C+\angle E=\angle A P Q \ldots \ldots \text { (2) (ext. } \angle \text { of } \triangle) \\ \begin{array}{l} \angle A+\angle B+\angle C+\angle D+\angle E=\angle A+\ang... | 1 |
6. To reduce heating costs in a residential building, the tenants decided to change the facade, and they were granted non-repayable funds from the Environmental Protection and Energy Efficiency Fund, which cover 60% of all total costs. The total cost for the new facade is 1,200,000 kn. The new facade guarantees that th... | 9 | 6. The cost for the facade is 1200000 kn. The non-refundable portion of funds is $60\%$, which means that the residents have to cover $40\%$ of the costs, amounting to $480000 \mathrm{kn}$.
1 POINT
The average annual heating cost was 168000 kn, and with a savings of 35%, it will amount to 109200 kn.
1 POINT
Let $n$... | 1 |
Three squares are attached to each other by their vertices and to two vertical rods, as shown in the figure. Determine the measure of angle $x$.
# | 39^{\circ} | Solution
In the drawing below, where $AB$ is parallel to $CD$, we will show that the sum of the white angles is equal to the sum of the measures of the gray angles. This result holds for any number of "peaks" in the drawing and is popularly known as the "Theorem of Peaks".
$.
According to the conditions, we have $2 a=|A B|=4$,
$a \pm \sqrt{a^{2}-b^{2}}=|A F|=2+\sqrt{3}$.
Thus, $a=2, b=1$, and $|E F|=2 \sqrt{a^{2}-b^{2}}=2 \sqrt{3}$.
By t... | 4 |
8. There are two teams competing, and the probability of each team winning a match is $\frac{1}{2}$. It is stipulated that a team must win four consecutive matches to end the series. The number of matches played is a random variable $\xi$, then the expected value E $\xi=$ $\qquad$ . | \frac{27}{2} | 8. $\frac{27}{2}$.
Let the probability that the match ends exactly after the $k(k \geqslant 1)$-th game be $p_{k}$. Then
$$
p_{1}=0, p_{2}=0, p_{3}=0, p_{4}=\frac{1}{2^{3}} \text {. }
$$
When $n \geqslant 6$,
$$
\begin{array}{l}
p_{n-1}=\frac{1}{2} p_{n-2}+\frac{1}{2^{2}} p_{n-3}+\frac{1}{2^{3}} p_{n-4} \\
p_{n}=\fra... | 5.5 |
9.
\begin{tabular}{ll}
Across & Down \\
1. A square & 1. Twice a fifth power \\
3. A fourth power & 2. A cube
\end{tabular}
When completed correctly, the cross number is filled with four three-digit numbers. What digit is *?
A 0
B 1
C 2
D 4
E 6 | 4 | Solution
D
When you are faced with a crossnumber, the best strategy is to look for clues where it is easy to find a unique solution.
Among the three-digit integers there are more squares and cubes than fourth and fifth powers. So the best strategy is to begin with 1 Down and 3 Across.
The first few numbers that are tw... | 2 |
9. (12 points) Three people, A, B, and C, depart from location $A$ to location $B$. A departs at 8:00, B at 8:20, and C at 8:30. They all travel at the same speed. 10 minutes after C departs, the distance from A to $B$ is exactly half the distance from B to $B$. At this moment, C is 2015 meters away from $B$. Therefore... | 2418 | 9. (12 points) Three people, A, B, and C, depart from location $A$ to location $B$. A departs at 8:00, B at 8:20, and C at 8:30. They all travel at the same speed. 10 minutes after C departs, the distance from A to $B$ is exactly half the distance from B to $B$. At this time, C is 2015 meters away from $B$, and the dis... | 2.67 |
We create a "die" from a spherical object by cutting off six equal spherical caps in such a way that the circles formed where the spherical caps were removed each touch four of their neighbors. What percentage of the total surface area of the die is the combined area of the six circles?
How many percent of the total s... | 86.08\% | Solution. Let the radius of the sphere be denoted by $R$, its center by $O$, and the radius of the circular faces by $r$. Consider one of the circular faces, and let the two opposite points of tangency on this face be $A$ and $B$.
}{3 \frac{2}{5}+\frac{10-\frac{1}{4}}{3}: \frac{5}{8}} \cdot 8 \frac{3}{5}-\frac{1.5 \cdot \frac{15}{4} \cdot 2.5+\frac{3}{5-\frac{2}{3}}}{1+\frac{1}{7}+\frac{6}{\frac{12}{11} \cdot\left(\frac{8}{3}-\frac{7}{4}\right) \cdot ... | 0 | Solution. Let's first calculate the value of the given expression. We have:
$$
\begin{aligned}
& \frac{1-\frac{1}{3}:\left(2+\frac{1}{6}\right)}{3 \frac{2}{5}+\frac{10-\frac{1}{4}}{3}: \frac{5}{8}} \cdot 8 \frac{3}{5}-\frac{1.5: \frac{15}{4} \cdot 2.5+\frac{3}{5-\frac{2}{3}}}{1+\frac{1}{7}+\frac{6}{\frac{12}{11} \cdot... | 2 |
Define a sequence $a_{i, j}$ of integers such that $a_{1, n}=n^{n}$ for $n \geq 1$ and $a_{i, j}=a_{i-1, j}+a_{i-1, j+1}$ for all $i, j \geq 1$. Find the last (decimal) digit of $a_{128,1}$. | 4 | By applying the recursion multiple times, we find that $a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}$, and $a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2}$. At this point, we can conjecture and prove by induction that $a_{m, n}=\sum_{k=0}^{m-1}\binom{m-1}{k}(n+k)^{n+k}=\sum_{k \geq 0}\binom{m-1}{k}(n+k)^{n+k}$. (The second expressio... | 7.25 |
Example 14 Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be a family of distinct three-element subsets of $I=\{1,2,3, \cdots, 36\}$, satisfying:
(1) For any $1 \leqslant i<j \leqslant n, A_{i} \cap A_{j} \neq \varnothing$,
(2) $A_{1} \cap A_{2} \cap \cdots \cap A_{n}=\varnothing$.
Find the maximum value of $n$, a... | (100, 34 C_{36}^{3}) | Let $X$ satisfy the conditions of the problem, and without loss of generality, let $A_{1}=\{1,2,3\}$. By condition (2), there exists an element of $X$ that does not contain 1, and we can assume $A_{2}=\left\{a_{1}, a_{2}, a_{3}\right\}$, and $1 \notin A_{2}$. Similarly, we can set $A_{3}=\left\{b_{1}, b_{2}, b_{3}\righ... | 9 |
1.3. A game of Jai Alai has eight players and starts with players $P_{1}$ and $P_{2}$ on court and the other players $P_{3}, P_{4}, P_{5}, P_{6}, P_{7}, P_{8}$ waiting in a queue. After each point is played, the loser goes to the end of the queue; the winner adds 1 point to his score and stays on the court; and the pla... | P_{4} | 1.3 Each time a player loses a match, he has to wait six games before his turn comes again. If $x$ is the number of games before his first turn, then the player will win if $x+7 r+7=37$, where $r \geq 0$ is an integer and $0 \leq x \leq 6$. Here $r$ counts the number of times he lost. From this, we obtain $x=2$ and $r=... | 2.67 |
4. Find the product of all roots of the equation $z^{3}+|z|^{2}=10 i$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | -5+10i | 4. From $z^{3}=-|z|^{2}+10 i$ we get $z^{3}=-|z|^{2}-10 i$, then $|z|^{6}=|z|^{4}+100$, which means $|z|^{2}=$ 5 , so the original equation becomes $z^{3}+5-10 i=0$, thus the product of all roots is $-5+10 i$. | 4 |
8. In a regular tetrahedron \(ABCD\) with edge length 1, two tangent spheres are placed inside, one of which is tangent to the three faces passing through point \(A\), and the other is tangent to the three faces passing through point \(B\). Then the minimum value of the sum of their radii is \(\qquad\) | \frac{\sqrt{6}-1}{5} | 8. $\frac{\sqrt{6}-1}{5}$.
Let the centers of the two spheres be $P, Q$, and their radii be $x, y$, respectively. Without loss of generality, assume that sphere $P$ is tangent to the three faces passing through point $A$, and sphere $Q$ is tangent to the three faces passing through point $B$, then $A P=3 x, B Q=3 y$.
... | 8 |
3.2. A quadrilateral $ABCD$ is inscribed in a circle $\omega$ such that $AB = AD$ and $BC = CD$ (a deltoid). It turns out that the radius of the inscribed circle of triangle $ABC$ is equal to the radius of the circle that touches the smaller arc $BC$ of circle $\omega$ and the side $BC$ at its midpoint (see Fig. 3). Fi... | 6 | 3.2. Let $I$ be the center of the inscribed circle of triangle $ABC$, $P$ the point of tangency of this inscribed circle with side $AB$, $M$ the midpoint of side $AB$, and $Q$ the midpoint of arc $AB$ (see Fig. 5).
Note that triangle $IPB$ is isosceles and right-angled. Moreover, by the trident theorem, triangle $IQB$... | 4.24 |
60th Putnam 1999 Problem A4 Let a ij = i 2 j/(3 i (j 3 i + i 3 j )). Find ∑ a ij where the sum is taken over all pairs of integers (i, j) with i, j > 0. | \frac{9}{32} | Let b i = i/3 i , then a ij = b i 2 b j (b i + b j ). Let the sum be k. Then k = ∑ a ij = 1/2 ∑ (a ij + a ji ) = 1/2 ∑ b i b j . We have the familiar 1 + 2x + 3x 2 + ... = 1/(1 - x) 2 , so x + 2x 2 + 3x 3 + ... = x/(1 - x) 2 , and hence ∑ b i = 1/3 (2/3) -2 = 3/4. So k = 1/2 (3/4) 2 = 9/32. 60th Putnam 1999 © John Scho... | 6 |
Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$, $a_2$, and $a_3$ be the medians of the numbers in rows $1$, $2$, and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$. Let $Q$ be the number of arrangements for which $m = 5$. Fi... | 360 | Assume that $5 \in \{a_1, a_2, a_3\}$, $m \neq 5$, and WLOG, $\max{(a_1, a_2, a_3)} = 5$. Then we know that the other two medians in $\{a_1, a_2, a_3\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \dots, 9$, a Contradiction. Thus, if $5 \in \{a_... | 3.4 |
10.4. A $100 \times 100$ grid is given, with cells colored black and white. In each column, there are an equal number of black cells, while in each row, the number of black cells is different. What is the maximum possible number of pairs of adjacent cells of different colors? (I. Bogdanov) | 14751 | Answer. $6 \cdot 50^{2}-5 \cdot 50+1=14751$ pairs.
Solution. Let the side length of the table be $2 n = 100$ (so $n=50$) and number the rows from top to bottom and the columns from left to right with numbers from 1 to $2 n$.
In each row, there can be from 0 to $2 n$ black cells. Since the number of black cells in all... | 8.67 |
14. As shown in the figure, there are four shapes made up of six different building blocks. The six blocks represent six different single-digit numbers. Three blocks forming a shape represent a three-digit number. 523, 426, 376 correspond to the first three figures below (note: they may not correspond in order, i.e., 5... | 325 | Answer: 325
Explanation: The three given numbers are $523, 426, 376$. The tens place has two identical 2s, and the units place has two identical 6s. From the first three figures, the vertical "one" shaped block represents the number 6; the horizontal "L" shaped block represents the number 2. Therefore, the first figure... | 2 |
4. A number, its fractional part, integer part, and itself form a geometric sequence, then the number is
保留源文本的换行和格式，所以翻译结果如下：
4. A number, its fractional part, integer part, and itself form a geometric sequence, then the number is | \frac{1+\sqrt{5}}{2} | $$
x=\frac{1+\sqrt{5}}{2}
$$
4. 【Analysis and Solution】Let the number be $x$, then its integer part is $[x]$, and the fractional part is $x-[x]$. From the given, we have $x=(x-[x])=[x]^{2}$, where $[x]>0, 0<x-[x]<1$. Solving this, we get $x=\frac{1+\sqrt{5}}{2}[x]$. From $0<x-[x]<1$, we know that $0<\frac{\sqrt{5}-1}{2... | 4.4 |
[ Law of Sines ] [ Law of Cosines ] In triangle $A B C$, medians $A N$ and $C M$ are drawn, $\angle A B C=120^{\circ}$. The circle passing through points $A, M$ and $N$ also passes through point $C$. The radius of this circle is 7. Find the area of triangle $A B C$. | 7 \sqrt{3} | Prove that the given triangle is isosceles, find $A N$, and to find the sides of the given triangle, apply the Law of Cosines to triangle $A B N$.
## Solution
By the theorem of the midline of a triangle, $M N \| A C$, so $A M N C$ is a trapezoid. Since the trapezoid is inscribed in a circle, it is isosceles. Since $A... | 4 |
15. Each edge of a triangular pyramid (tetrahedron) is equal to $a$. Find the maximum area that the orthogonal projection of this pyramid onto a horizontal plane can have. | \frac{a^{2}}{2} | 15. The orthogonal projection of a tetrahedron can be either a triangle or a quadrilateral. In the first case, a regular tetrahedron is projected into a triangle that coincides with the projection of one of its faces. The largest projection in this case will be when the plane of the face is parallel to the projection p... | 6.33 |
6. Let $k$ be a real number, in the Cartesian coordinate system $x O y$ there are two point sets $A=\left\{(x, y) \mid x^{2}+y^{2}=\right.$ $2(x+y)\}$ and $B=\{(x, y) \mid k x-y+k+3 \geqslant 0\}$. If $A \cap B$ is a singleton set, then the value of $k$ is $\qquad$ | -2 - \sqrt{3} | The point set $A$ represents the circle $\odot C:(x-1)^{2}+(y-1)^{2}=2$, and the point set $B$ represents the region below the line $y=k(x+1)+3$. Since the line is tangent to $\odot C$, the distance from the center $C(1,1)$ to the line is
$$
d=\frac{|2 k+2|}{\sqrt{k^{2}+1}}=\sqrt{2} \Rightarrow k^{2}+4 k+1=0 \Rightarro... | 6 |
2. (24th Canadian Mathematical Olympiad) Solve the equation $x^{2}+\left(\frac{x^{2}}{x+1}\right)^{2}=3$ in the set of complex numbers. | \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2}, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2} | 2. Since $x^{2}-2 \cdot \frac{x^{2}}{x+1}+\left(\frac{x}{x+1}\right)^{2}=\left(x-\frac{x}{x+1}\right)^{2}=\left(\frac{x^{2}}{x+1}\right)^{2}$, the original equation can be transformed into $x^{2}+\left(\frac{x}{x+1}\right)^{2}-2 \cdot$ $\frac{x^{2}}{x+1}+2 \cdot \frac{x^{2}}{x+1}=3$
which simplifies to $\left(\frac{x^... | 2.6 |
55. In a $10 \times 10$ grid (the sides of the cells have a unit length), $n$ cells were chosen, and in each of them, one of the diagonals was drawn and an arrow was placed on this diagonal in one of two directions. It turned out that for any two arrows, either the end of one coincides with the beginning of the other, ... | 48 | 55. Answer: when $n=48$.
For each arrow, consider the three-cell corner obtained by removing from a $2 \times 2$ square, centered at the end of the arrow, a $1 \times 1$ square, the diagonal of which is this arrow. Note that such corners do not intersect and are contained within a $12 \times 12$ square. Therefore, the... | 5 |
15. Master Wang works in a special position, where he works for 8 consecutive days and then takes 2 consecutive days off. If he is off on this Saturday and Sunday, then, at least how many weeks later will he be off on a Sunday again? | 7 | 15. At least another 7 weeks
15.【Solution】Let at least $\mathrm{n}$ weeks pass, it is possible to rest on the $\mathrm{n}$th Saturday, or it is also possible not to rest on the $\mathrm{n}$th Saturday (resting for 2 days on Sunday and Monday), the former yields: $7 \mathrm{n}-2=10 \mathrm{~K}+8(1)$, the latter yields: ... | 2.33 |
The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$?
$\text... | 8 | The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$:
If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Th... | 3.75 |
5 Find all real numbers $a$ such that any positive integer solution of the inequality $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)$ are the lengths of the sides of some triangle. | [1, \frac{6}{5}) | 5. Taking $x=2, y=z=1$, we have $a \geqslant \frac{6}{5}$. Therefore, when $a \geqslant \frac{6}{5}$, the original inequality has integer roots $(2,1,1)$, but $(x, y, z)$ cannot form a triangle, so $a<\frac{6}{5}$.
When $a<1$, $x^{2}+y^{2}+z^{2} \leqslant a(x y+y z+z x)<x y+y z+z x$, which is a contradiction! Hence, $... | 5 |
The base of the right parallelepiped $A B C D A 1 B 1 C 1 D 1$ is a square $A B C D$ with a side length of 4, and the length of each lateral edge $A A 1, B B 1, C C 1, D D 1$ is 6. A right circular cylinder is positioned such that its axis lies in the plane $B B 1 D 1 D$, and the points $A 1, C 1, B 1$ and the center $... | $\frac{5}{\sqrt{3}}, 4 \sqrt{\frac{2}{3}}, \frac{20}{3} \sqrt{\frac{2}{11}}$ | 1. Suppose the axis $l$ of the cylinder intersects the line $OB_1$ at some point $K$ (Fig.1), and the line $BB_1$ at point $F$. Let $A_2, B_2, C_2$, and $O_2$ be the projections of points $A, B, C$, and $O$ onto the line $l$, respectively. Then the point $A_2$ coincides with the point $C_2$, and since points $O$ and $B... | 6.5 |
$2.351 A=\frac{x^{8}+x^{4}-2 x^{2}+6}{x^{4}+2 x^{2}+3}+2 x^{2}-2$. | A=x^{4} | Solution. Let's divide the polynomial $x^{8}+x^{4}-2 x^{6}+6$ by the polynomial $x^{4}+2 x^{2}+3$.
$$
\begin{aligned}
& \begin{array}{r}
-\frac{-2 x^{6}-4 x^{4}-6 x^{2}}{-2 x^{4}+4 x^{2}+6} ... | 2.33 |
80. One day, Xiao Ben told a joke. Except for Xiao Ben himself, four-fifths of the classmates in the classroom heard it, but only three-quarters of the classmates laughed. It is known that one-sixth of the classmates who heard the joke did not laugh. Then, what fraction of the classmates who did not hear the joke laugh... | \frac{5}{12} | Reference answer: $5 / 12$ | 2.33 |
Example 3 Given a positive integer $n \geqslant 3$, let $d_{1}, d_{2}, \cdots, d_{n}$ be positive integers, whose greatest common divisor is 1, and $d_{i}$ divides $d_{1}+d_{2}+\cdots+d_{n}(i=1,2, \cdots, n)$. Find the smallest positive integer $k$, such that $d_{1} d_{2} \cdots d_{n}$ divides $\left(d_{1}+\right.$ $\l... | n-2 | Let $d_{1}=1, d_{2}=n-1, d_{3}=d_{4}=\cdots=d_{n}=n$, then $d_{1}+d_{2}+\cdots+d_{n}=1+$ $(n-1)+(n-2) \cdot n=(n-1) n$, which is divisible by every $d_{i}(i=1,2, \cdots, n)$, and the greatest common divisor of $d_{1}, d_{2}, \cdots, d_{n}$ is 1, at this time $d_{1} d_{2} \cdots d_{n}=(n-1) n^{n-2}$. If $d_{1} d_{2} \cd... | 5 |
5. Teresa the bunny has a fair 8 -sided die. Seven of its sides have fixed labels $1,2, \ldots, 7$, and the label on the eighth side can be changed and begins as 1 . She rolls it several times, until each of $1,2, \ldots, 7$ appears at least once. After each roll, if $k$ is the smallest positive integer that she has no... | 104 | Answer:
104
Solution 1: Let $n=7$ and $p=\frac{1}{4}$.
Let $q_{k}$ be the probability that $n$ is the last number rolled, if $k$ numbers less than $n$ have already bee rolled. We want $q_{0}$ and we know $q_{n-1}=1$.
We have the relation
$$
q_{k}=(1-p) \frac{k}{n-1} q_{k}+\left[1-(1-p) \frac{k+1}{n-1}\right] q_{k+1} .
... | 5 |
64. 1 9 Choose 4 different digits from these 9 numbers, form a four-digit number, so that this four-digit number can be divided by the 5 numbers not selected, but cannot be divided by the 4 selected numbers. Then, this four-digit number is . $\qquad$ | 5936 | Answer: 5936 | 4.75 |
Example 3 (1) Given the geometric sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=a>0, b_{1}-a_{1}=1, \\
b_{2}-a_{2}=2, b_{3}-a_{3}=3 .
\end{array}
$$
If the sequence $\left\{a_{n}\right\}$ exists and is unique, find the value of $a$.
(2) Do there exist two geometric sequ... | \frac{1}{3} | 【Analysis】"The common ratio of a geometric sequence cannot be 0" is a common knowledge. If this is asked directly, it would become a "junk question". However, if this concept is appropriately hidden in the problem-solving process, it could become a good question.
This problem is from the 2011 Jiangxi Province College ... | 5 |
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