problem stringlengths 23 1.87k | answer stringlengths 1 209 | solution stringlengths 3 7.18k | difficulty float64 1 3 |
|---|---|---|---|
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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".
 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 |
Task 2. What is $\frac{31}{71}$ of the number
$$
\frac{1-\frac{1}{3}:\left(2+\frac{1}{6}\right)}{3 \frac{2}{5}+\frac{10-\frac{1}{4}}{3}: \frac{5}{8}} \cdot 8 \frac{3}{5}-\frac{1.5 \cdot \frac{15}{4} \cdot 2.5+\frac{3}{5-\frac{2}{3}}}{1+\frac{1}{7}+\frac{6}{\frac{12}{11} \cdot\left(\frac{8}{3}-\frac{7}{4}\right) \cdot ... | 0 | Solution. Let's first calculate the value of the given expression. We have:
$$
\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 |
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 |
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 |
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 |
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 |
$2.351 A=\frac{x^{8}+x^{4}-2 x^{2}+6}{x^{4}+2 x^{2}+3}+2 x^{2}-2$. | A=x^{4} | Solution. Let's divide the polynomial $x^{8}+x^{4}-2 x^{6}+6$ by the polynomial $x^{4}+2 x^{2}+3$.
$$
\begin{aligned}
& \begin{array}{r}
-\frac{-2 x^{6}-4 x^{4}-6 x^{2}}{-2 x^{4}+4 x^{2}+6} ... | 2.33 |
80. One day, Xiao Ben told a joke. Except for Xiao Ben himself, four-fifths of the classmates in the classroom heard it, but only three-quarters of the classmates laughed. It is known that one-sixth of the classmates who heard the joke did not laugh. Then, what fraction of the classmates who did not hear the joke laugh... | \frac{5}{12} | Reference answer: $5 / 12$ | 2.33 |
In the diagram, the circle has centre $O$ and radius 6 . Point $A$ is outside the circle and points $B$ and $C$ are on the circle so that $A B$ is perpendicular to $B O, A C$ is perpendicular to $C O$, and $\angle B A C=50^{\circ}$. What is the area of the shaded region?
$ for $2 \leq a \leq 7$ and the white rooks on cells $(a, a+1)$ and $(a+1, a)$ for $1 \leq a \leq 7$. Now, we prove the optimality. As rooks can only attack opposite color rooks, the color of rooks in each row is alternating. The diff... | 2 |
Example 2. In the stamping of plastic plates, the defect rate is $3 \%$. Find the probability that when checking a batch of 1000 plates, the deviation from the established defect rate will be less than $1 \%$. | 0.709 | Solution. From the condition of the problem, it follows that $n=1000, \varepsilon=0.01$, $p=0.03, q=1-p=0.97$.
In accordance with formula (4.2.5.), we obtain
$$
P\left(\left|\frac{m}{n}-p\right| \leq 0.01\right) \geq 1-\frac{p q}{n \varepsilon^{2}}=1-\frac{0.03 \cdot 0.97}{10000 \cdot(0.01)^{2}}=1-\frac{0.0291}{0.1}=... | 2.33 |
Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [integer](https://artofproblemsolving.com/wiki/index.php/Integer) $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the [product](https://artofproblemsolving.com/wiki/index.php/Product... | 23 | The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ld... | 3 |
What fraction of the area of rectangle $A B C D$ is the area of the shaded square?
(A) $\frac{1}{15}$
(D) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{10}$
(E) $\frac{1}{12}$
 | \frac{1}{15} | The shaded square has side length 1 so has area $1^{2}=1$.
The rectangle has dimensions 3 by 5 so has area $3 \times 5=15$.
Thus, the fraction of the rectangle that is shaded is $\frac{1}{15}$.
ANSWER: (A) | 1 |
4) As shown in Figure 2, in the shape composed of seven small squares, line 1 divides the original shape into two parts of equal area, the intersection point of 1 with $AB$ is $E$, and the intersection point of 1 with $CD$ is $F$. If the sum of the lengths of segments $CF$ and $AE$ is 91 cm, then the side length of the... | 26 | 4) $a=26$ . | 2.5 |
【Question 13】
There are 91 sticks, with lengths of $1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 4 \mathrm{~cm}, \cdots, 91 \mathrm{~cm}$, respectively. At least $\qquad$ sticks need to be selected to ensure that a triangle can definitely be formed. | 8 | 【Analysis and Solution】
The worst-case principle.
To form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.
On one hand, assume there exist 8 sticks that cannot form a triangle;
The 3rd smallest number is no less than $1+2=3$,
The 4th smallest number is no less than ... | 3 |
In the figure, the two triangles $\triangle A B C$ and $\triangle D E F$ are equilateral. What is the value of the angle $x$?
(a) $30^{\circ}$
(b) $40^{\circ}$
(c) $50^{\circ}$
(d) $60^{\circ}$
(e) $70^{\circ}$
.
Since $\triangle A B C$ and $\triangle D E F$ are equilateral triangles, each of their internal angles measures $60^{\circ}$. In the triangle $\triangle A G D$ we have
$$
G \widehat{A} D=180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ} \text { and } \mathrm{G} \widehat{D A}=180^{\circ}-65^{\cir... | 2.25 |
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the an... | 50000 | Answer: 50000.
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 = 150000$ rubles. Profit $=$ revenue costs $=150000-100000=50000$ rubles. | 3 |
Find the sum of the ages of everyone who wrote a problem for this year's HMMT November contest. If your answer is $X$ and the actual value is $Y$, your score will be $\max (0,20-|X-Y|)$ | 258 | There was one problem for which I could not determine author information, so I set the author as one of the problem czars at random. Then, I ran the following command on a folder containing TeX solutions files to all four contests: ``` evan@ArchMega ~/Downloads/November $ grep --no-filename "Proposed by: " *.tex | sort... | 1.25 |
12. (6 points) As shown in the figure, in $\triangle A B C$, $D$ and $E$ are the midpoints of $A B$ and $A C$, respectively, and the area difference between the two shaded regions (甲 and 乙) is 5.04. Then $S_{\triangle A B C}=$ $\qquad$ . | 20.16 | 【Solution】Solution: According to the analysis, $S_{\triangle B D C}=S_{\triangle E B C} \Rightarrow S_{\triangle D O B}=S_{\triangle E O C}$,
$$
\therefore S_{\text {甲 }}-S_{\text {乙 }}=\left(S_{\text {甲 }}+S_{\triangle D O B}\right)-\left(S_{\text {乙 }}+S_{\triangle E O C}\right)=5.04,
$$
Also, $\because S_{\triangle... | 2.2 |
Students were surveyed about their favourite season. The results are shown in the bar graph. What percentage of the 10 students surveyed chose Spring?
(A) 50
(B) 10
(C) 25
(D) 250
(E) 5
 | 10\% | Reading from the bar graph, only 1 student chose spring.
Since 10 students were surveyed, then the percentage of students that chose spring was $\frac{1}{10} \times 100 \%$ or $10 \%$.
ANsWER: (B) | 1.8 |
3. The sides $B C$ and $A D$ of a quadrilateral $A B C D$ are parallel and the diagonals intersect in $O$. For this quadrilateral $|C D|=|A O|$ and $|B C|=|O D|$ hold. Furthermore $C A$ is the angular bisector of angle $B C D$.
Determine the size of angle $A B C$.
Attention: the figure is not drawn to scale.
You ha... | $126^{\circ}$ |
3. First, we prove that some triangles in the figure are isosceles (the top angle coincides with the middle letter).
(1) Triangle $A D C$ is isosceles, because $\angle D A C=\angle A C B=\angle A C D$. The first equality holds because $A D$ and $B C$ are parallel and the second equality follows from the fact that $A ... | 3 |
Example 2 (2000 National High School Competition Question) If: (1) $a, b, c, d$ all belong to $\{1,2,3,4\}$; (2) $a \neq b$, $b \neq c, c \neq d, d \neq a$; (3) $a$ is the smallest value among $a, b, c, d$. Then the number of different four-digit numbers $\overline{a b c d}$ that can be formed is $\qquad$ | 28 | Solve by filling in 28. Reason: (1) When $\overline{a b c d}$ contains 4 different digits, it is clear that $a=1$. There are $3!=6$ such four-digit numbers.
(2) When abcd contains only 3 different digits, choosing 3 different digits from $1,2,3,4$ has $C_{4}^{3}$ methods. At this time, the value of $a$ is uniquely dete... | 3 |
12. As shown in Figure 12-1, Xiao Ming goes from A to B, each time walking three squares in one direction, then turning 90 degrees and walking one more square, for example, in Figure 12-2, starting from point $\mathrm{C}$, he can reach eight positions. How many times at least does Xiao Ming need to walk to get from poi... | 5 | 【Analysis】The answer is as shown in the figure, at least 5 times. | 2 |
3.3. Write the number 100 using four fives and arithmetic signs.
$(4-5$ grade.)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (5+5)(5+5)=100 | 3.3. The two solutions to this problem are given below:
1) $(5+5)(5+5)=100$
2) $(5 \cdot 5-5) \cdot 5=100$ | 1.33 |
I2.4 $[a]$ represents the largest integer not greater than $a$. For example, $\left[2 \frac{1}{3}\right]=2$. Given that the sum of the roots of the equation $[3 x+R]=2 x+\frac{3}{2}$ is $S$, find the value of $S$. | 2 | $\begin{array}{l}{[3 x+1]=2 x+\frac{3}{2} \Rightarrow 3 x+1=2 x+\frac{3}{2}+a, \text { where } 0 \leq a<1} \\ a=x-\frac{1}{2} \Rightarrow 0 \leq x-\frac{1}{2}<1 \Rightarrow 2.5 \leq 2 x+\frac{3}{2}<4.5 \\ \because 2 x+\frac{3}{2} \text { is an integer } \therefore 2 x+\frac{3}{2}=4 \text { or } 3 \\ x=0.75 \text { or }... | 2.4 |
1. Use the numbers $2,3, \cdots, 2019$ to form 1009 fractions, with each number appearing only once in either the numerator or the denominator. From these 1009 fractions, select the largest one. Find the minimum value of this number. | \frac{1010}{2019} | 1. First, construct the following 1009 fractions:
$$
\frac{2}{1011}, \frac{3}{1012}, \cdots, \frac{1010}{2019} \text{, }
$$
Among them, the largest number is $\frac{1010}{2019}$.
Next, we prove that this is the minimum value.
Proof by contradiction.
Assume there are another 1009 fractions that meet the conditions, and... | 1.33 |
I4.4 If the number of integral solutions of the inequality $\left|\frac{x}{2}-\sqrt{2}\right|<c$ is $d$, find the value of $d$. | 6 | $\begin{array}{l}\left|\frac{x}{2}-\sqrt{2}\right|<\frac{3}{2} \\ -\frac{3}{2}<\frac{x}{2}-\sqrt{2}<\frac{3}{2} \\ 2 \sqrt{2}-3<x<2 \sqrt{2}+3 \\ 2(1.4)-3<x<2(1.4)+3 \\ -0.2<x<5.8 \\ x=0,1,2,3,4,5 \\ d=6\end{array}$ | 1.25 |
2. Below are three propositions: (甲) When $\theta \in\left(0, \frac{\pi}{2}\right)$, $\sin (\cos \theta)\sin (\sin \theta)$; (丙) When $\theta \in[0, \pi]$, $\sin (\cos \theta)<\cos (\sin \theta)$. The number of correct propositions is ( ).
A. 0
B. 1
C. 2
D. 3 | 3 | 2. D. Given $\theta \in\left(0, \frac{\pi}{2}\right)$, we know $\theta>\sin \theta$, which means $\cos \theta>\sin (\cos \theta)$. Also, $\frac{\pi}{2}>\theta>\sin \theta>0$, and within $\left(0, \frac{\pi}{2}\right)$, the cosine function is a decreasing function, so we have $\sin (\cos \theta)\sin (\sin \theta)$.
When... | 2 |
7. [30] Determine the number of juggling sequences of length $n$ with exactly 1 ball. | $2^{n}-1$ | Answer: $\quad 2^{n}-1$. Solution: With 1 ball, we simply need to decide at times should the ball land in our hand. That is, we need to choose a non-empty subset of $\{0,1,2, \ldots, n-1\}$ where the ball lands. It follows that the answer is $2^{n}-1$. | 1 |
3. If real numbers $a, b, c$ make the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy $|f(x)| \leqslant 1$ for $0 \leqslant x \leqslant 1$, then the maximum value of $2|a|+$ $\sqrt{2}|b|+|c|$ is . $\qquad$ | 17 + 8 \sqrt{2} | 3. $17+8 \sqrt{2}$.
Let $x=1,0.5,0$, we get
$$
c=f(0) \text {, }
$$
$$
\begin{array}{l}
b=-f(1)+4 f(0.5)-3 f(0), \\
a=2 f(1)-4 f(0.5)+2 f(0) .
\end{array}
$$
From the absolute value inequality, we know
$$
\begin{array}{l}
2|a|+\sqrt{2}|b|+|c| \\
\leqslant(4+\sqrt{2})|f(1)|+(8+4 \sqrt{2})|f(0.5)|+ \\
(5+3 \sqrt{2})|f(... | 3 |
4. [5] A zerg player can produce one zergling every minute and a protoss player can produce one zealot every 2.1 minutes. Both players begin building their respective units immediately from the beginning of the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots is more than ... | 1.3 | Answer: $\boxed{1.3}$ At the end of the first minute, the zerg player produces a zergling and has a superior army for the 1.1 minutes before the protoss player produces the first zealot. At this point, the zealot is at least a match for the zerglings until the fourth is produced 4 minutes into the game. Then, the zerg ... | 2.23 |
1. You have a bag of granulated sugar, a balance scale, a 1 kg weight, and paper bags in which you can package the sugar. You need to measure out 50 kg of sugar, using no more than 6 weighings. How can you do this? | 32 \text{ kg} + 16 \text{ kg} + 2 \text{ kg} = 50 \text{ kg} | Solution. Let $\Gamma$ denote a 1 kg weight. The right and left sides of the equalities correspond to the right and left pans of the balance.
1 step: 1 kg (sugar $)=\Gamma$;
2 step: 1 kg $+\Gamma=2$ kg;
3 step: 1 kg $+2 \kappa 2+\Gamma=4$ kg;
4 step: 1 kg $+2 \kappa 2+4 \kappa z+\Gamma=8$ kg;
5 step: 1 kg $+2 \kap... | 2 |
3. Find all values of $c$ for which the inequality $a+\sqrt{b+c}>b+\sqrt{a+c}$ holds for any positive $a, b$ and $a>b$.
(25 points.) | $c = \frac{1}{4}$ | Answer: $c=\frac{1}{4}$.
Write the inequality in the form $a-b>\sqrt{a+c}-\sqrt{b+c}$. By multiplying and dividing the right part by its conjugate, we can represent the inequality as
$$
a-b>\frac{a-b}{\sqrt{a+c}+\sqrt{b+c}} \quad \Longleftrightarrow \quad \sqrt{a+c}+\sqrt{b+c}>1
$$
The last inequality must hold for ... | 3 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 2 |
2. There is one power outlet connected to the network, two extension cords with three outlets each, and one table lamp included. Nosy Nick randomly plugged all three plugs into 3 out of 7 outlets. What is the probability that the lamp will light up? (16 points) | \frac{13}{35} | Solution. The number of equally probable ways to plug in $A_{7}^{3}=7 \cdot 6 \cdot 5=210$. The lamp can be powered through 0, 1, or 2 extension cords.
0) The lamp is plugged into the socket, the other 2 plugs are plugged in randomly. The number of such possibilities $A_{6}^{2}=30$.
1) The lamp is plugged into one of ... | 2 |
# 3. Option 1.
The Ivanov family consists of three people: dad, mom, and daughter. Today, on the daughter's birthday, the mother calculated the sum of the ages of all family members and got 74 years. It is known that 10 years ago, the total age of the Ivanov family members was 47 years. How old is the mother now, if s... | 33 | Answer: 33.
Solution: If the daughter had been born no less than 10 years ago, then 10 years ago the total age would have been $74-30=44$ years. But the total age is 3 years less, which means the daughter was born 7 years ago. The mother is now $26+7=33$ years old. | 3 |
6. (3 points) Given the puzzle: AB + BC + DE = FGH. Different letters represent different digits, no digit is equal to 9, and a number cannot start with 0. Find the smallest possible value of FGH. | 108 | Answer: 108.
## Solution:
The sum of all the digits at our disposal is 36. Therefore, AB $+\mathrm{B} \Gamma+\mathrm{DE}+$ ZHI is divisible by 9. But AB + BG + DE + ZHI $=2$ ZHI, so ZHI is divisible by 9.
The smallest three-digit number divisible by 9 is 108. Since, for example, $108=25+36+$ 47, this number fits our... | 3 |
12. (10 points) A piece of paper is flipped over, the numbers $0$, $1$, $8$ remain unchanged after a $180^{\circ}$ rotation, $6$ becomes $9$, $9$ becomes $6$, and other numbers have no meaning after a $180^{\circ}$ rotation. How many 7-digit numbers remain unchanged after a $180^{\circ}$ rotation? Among these, how many... | 300; 75; 1959460200 | 【Analysis】According to the problem, the 7-digit number $\overline{\mathrm{ABCDEFG}}$, when rotated 180 degrees, remains unchanged, indicating that this 7-digit number is composed of $0, 1, 8$. We can list them accordingly, and then solve the problem based on the characteristics of numbers divisible by 4.
【Solution】Sol... | 2.5 |
18. As shown in Figure 3, in the concave quadrilateral $A B C D$, $\angle B C D$ $=90^{\circ}, A B=12, B C=4, C D=3, A D=13$. Then the area $S$ of the concave quadrilateral $A B C D$ is ( ).
(A) 12
(B) 24
(C) 26
(D) 30
(E) 36 | 24 | 18. B.
From the given information,
$$
\begin{array}{l}
B D=\sqrt{B C^{2}+C D^{2}}=5 \\
\Rightarrow B D^{2}+A B^{2}=A D^{2} \\
\Rightarrow \angle A B D=90^{\circ} .
\end{array}
$$
Therefore, $S=S_{\triangle A B D}-S_{\triangle B C D}$
$$
=\frac{1}{2} \times 5 \times 12-\frac{1}{2} \times 3 \times 4=24 \text {. }
$$ | 2 |
$$
\begin{array}{l}
16\left(\frac{1}{5}-\frac{1}{3} \times \frac{1}{5^{3}}+\frac{1}{5} \times \frac{1}{5^{5}}-\frac{1}{7} \times \frac{1}{5^{7}}+ \\
\frac{1}{9} \times \frac{1}{5^{9}}-\frac{1}{11} \times \frac{1}{5^{11}}\right)-4\left(\frac{1}{239}-\frac{1}{3} \times \frac{1}{239^{3}}\right) \\
=\quad(\text { (to } 8 \... | 3.14159265 | -、1.3.14159265
(No translation needed as the text is a number and a separator, which are universal and do not require translation.) | 2.4 |
$\angle 1 + \angle 2 = 180^\circ$
$\angle 3 = \angle 4$
Find $\angle 4.$
$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ$ | 35^\circ | Using the left triangle, we have:
$\angle 1 + 70 + 40 = 180$
$\angle 1 = 180 - 110$
$\angle 1 = 70$
Using the given fact that $\angle 1 + \angle 2 = 180$, we have $\angle 2 = 180 - 70 = 110$.
Finally, using the right triangle, and the fact that $\angle 3 = \angle 4$, we have:
$\angle 2 + \angle 3 + \angle 4 = 180$
$110... | 1.5 |
Mrs. Toad has a class of 2017 students, with unhappiness levels $1,2, \ldots, 2017$ respectively. Today in class, there is a group project and Mrs. Toad wants to split the class in exactly 15 groups. The unhappiness level of a group is the average unhappiness of its members, and the unhappiness of the class is the sum ... | 1121 | One can show that the optimal configuration is $\{1\},\{2\}, \ldots,\{14\},\{15, \ldots, 2017\}$. This would give us an answer of $1+2+\cdots+14+\frac{15+2017}{2}=105+1016=1121$. | 2.67 |
5. The area of the right-angled triangle in the diagram alongside is $60 \mathrm{~cm}^{2}$. The triangle touches the circle, and one side of the triangle has length $15 \mathrm{~cm}$, as shown. What is the radius of the circle? | 20 | Solution
The area of a triangle is $\frac{1}{2}$ base $\times$ height, so the triangle has sides of length $8 \mathrm{~cm}, 15 \mathrm{~cm}$ and $17 \mathrm{~cm}$ (using Pythagoras' theorem).
Let the radius of the circle be $r \mathrm{~cm}$.
Draw two radii, as shown. Since a tangent meets the radius at the point of con... | 3 |
## Exercise 3
Twelve candidates for the position of mayor are participating in a televised debate. After a while, one of them declares, "Up until now, we have lied once." A second then says, "Now it makes two times." A third exclaims, "Three times, now," and so on until the twelfth affirms that before him, they had li... | 11 | ## Corrected
We will show that the first candidate told the truth and that the others lied.
Let $C_{1}, \ldots, C_{12}$ be the candidates in the order of their speaking. By contradiction, suppose that $k \geq 2$ and that $C_{k}$ told the truth. Therefore, exactly $k$ lies were told before $C_{k}$ spoke. Then:
$\tria... | 2 |
Let $A$ be the area of the largest semicircle that can be inscribed in a quarter-circle of radius 1. Compute $\frac{120 A}{\pi}$. | 20 | The optimal configuration is when the two ends $X$ and $Y$ of the semicircle lie on the arc of the quarter circle. Let $O$ and $P$ be the centers of the quarter circle and semicircle, respectively. Also, let $M$ and $N$ be the points where the semicircle is tangent to the radii of the quartercircle. Let $r$ be the radi... | 2.8 |
3. A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12 \mathrm{~m}$. What is the area (in $\mathrm{m}^{2}$ ) of the room? | 18 \pi | Solution:
$18 \pi$
The maximal distance is as shown in the figure. Call the radii $R$ and $r, R>r$. Then $R^{2}-r^{2}=6^{2}$ by the Pythagorean theorem, so the area is $(\pi / 2) \cdot\left(R^{2}-r^{2}\right)=18 \pi$. | 3 |
$12.24 y=x^{3} e^{-x}$. | (3, \frac{27}{e^3}) | 12.24 The function is defined for all $x$. We find
$$
y^{\prime}=3 x^{2} e^{-x}-x^{3} e^{-x}=x^{2} e^{-x}(3-x)
$$
The equation $y^{\prime}=0$ has only one root $x=3$. Since $y^{\prime}>0$ for $x<3$ and $y^{\prime}<0$ for $x>3$, the function reaches a maximum at the point $x=3$, which is equal to $3^{3} e^{-3}=\frac{2... | 2.33 |
Mr. and Mrs. Seventh have 7 children, all born on April 1st, actually over six consecutive April 10ths. This year, for their birthdays, Mrs. Seventh made a cake with candles for each one - the number of candles equal to the number of years of each one. João Seventh, the son who loves Math the most, noticed that this ye... | 26 | The births occurred on six 1st of April, so there are twin siblings. Since this year we have 2 more cakes than 2 years ago, it means that 2 years ago the youngest had not been born yet, the second youngest had just been born, and the twins had already been born. Currently, the youngest is 1 year old and the twins are $... | 2.5 |
3. Fill the numbers $1,2,3, \ldots, 9,10$ into 10 circles that form a rectangle, such that the sum of the numbers on each side of the rectangle is equal. The maximum sum is ( ). | 22 | 【Answer】22
【Analysis】Exam point: Number array $1+2+3+4+\ldots+10=55 ; 55+\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}$ can be divisible by 4; then $\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}=10$ $+9+8+6=33$; that is $(55+33) \div 4=22$ | 3 |
Task A-1.2. (8 points)
If we add the digit 3 to the left of a two-digit number, the resulting number is 27 times greater than twice the given two-digit number.
Determine that two-digit number. | 24 | ## Solution.
Let $\overline{x y}=10 x+y$ be the desired two-digit number. We have
$$
\begin{gathered}
2 \cdot \overline{3 x y}=27 \cdot \overline{x y} \\
2 \cdot(300+\overline{x y})=27 \cdot \overline{x y} \\
600+2 \cdot \overline{x y}=27 \cdot \overline{x y} \\
600=25 \cdot \overline{x y}
\end{gathered}
$$
The desi... | 1.67 |
In the diagram, each partially shaded circle has a radius of $1 \mathrm{~cm}$ and has a right angle marked at its centre. $\mathrm{In}^{2} \mathrm{~cm}^{2}$, what is the total shaded area?
(A) $4 \pi^{2}$
(B) $9 \pi^{2}$
(C) $4 \pi$
(D) $9 \pi$
(E) $3 \pi$
$, that is, $\frac{1}{2} \pi r$.
Therefore the ... | 2.67 |
## Task 2 - 280612
A large cuboid was divided into small, equally sized cubes. As can be seen in the illustration, some of the small cubes were then removed. However, none of the small cubes that are not visible in the illustration were removed.
How many of the small cubes does the remaining body shown in the illustr... | 135 | The sought number of small cubes is 135; it can be found through the following consideration:
The large cuboid originally consisted of exactly 150 small cubes because of $6 \cdot 5 \cdot 5=150$. From it, exactly 15 small cubes were removed, namely
exactly 8 from the frontmost layer,
exactly 6 from the second layer f... | 1 |
5. As shown in the figure, it is a black and white checkered cloth. The side length of the large white square is 14 cm, and the side length of the small white square is 6 cm. Question: What percentage of the total area is the white area of this cloth?
The side length of the large white square is 14 cm, and the side ... | 58 \% | $58 \%$
5.【Solution】The area of the checkered cloth is 9 times the area of the figure below, and the area of the white part of the checkered cloth is also 9 times the area of the white part in the figure below. The percentage of the white area in the figure below is:
$$
\frac{14 \times 14+6 \times 6}{20 \times 20}=0.58... | 1 |
7. $\arcsin \left(\sin 2000^{\circ}\right)=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7. $\arcsin \left(\sin 2000^{\circ}\right)=$ | -20^{\circ} | $$
-20^{\circ}
$$
7. [Analysis and Solution] According to the problem, we have $2000^{\circ}=180^{\circ} \times 11+20^{\circ}$,
$$
\begin{array}{l}
\therefore \sin 2000^{\circ}=\sin \left(180^{\circ} \times 11+20^{\circ}\right)=\sin \left(-20^{\circ}\right), \\
\therefore \arcsin \left(2000^{\circ}\right)=-20^{\circ}
\... | 1 |
Krystyna has some raisins. After giving some away and eating some, she has 16 left. How many did she start with? | 54 | Working backwards, Krystyna had 36 raisins before eating 4, and 54 raisins initially. | 1 |
Let $ABCD$ be an [isosceles trapezoid](https://artofproblemsolving.com/wiki/index.php/Isosceles_trapezoid) with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The [diagonals](https://artofproblemsolving.com/wiki/index.php/Diagonal) have length $10\sqrt {21}$, and poin... | 32 | Key observation. $AD = 20\sqrt{7}$.
Proof 1. By the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality), we can immediately see that $AD \geq 20\sqrt{7}$. However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$, so by the law of sines, when $AD = 20\sqrt{7}$, $\angle ... | 3 |
8. (5 points) Cut out 4 equally sized isosceles right triangles from a rectangular sheet of paper that is 12 cm long and 8 cm wide. The minimum area of the remaining part is $\qquad$ square centimeters. | 24 | 【Analysis】A rectangular piece of paper with a length of 12 cm and a width of 8 cm can obviously cut out at most 4 isosceles right triangles with legs of 6 cm, so the remaining area is not difficult to find.
【Solution】Solution: According to the analysis, as shown in the figure, a rectangular piece of paper with a lengt... | 1.67 |
67. $x$ is a positive rational number, $(x)$ represents the number of prime numbers not exceeding $x$, for example, $(5)=3$, meaning there are 3 prime numbers not exceeding 5, which are 2, 3, 5. Therefore, $(x)$ defines an operation on $x$. Then $((20) \times(1)+(7))$ is $\qquad$ | 2 | Answer: 2.
Solution: $(x)$ defines an operation for positive rational numbers $x$. The steps of the operation are:
(1) Write down the sequence of natural numbers not exceeding $x$;
(2) Mark the prime numbers in this sequence;
(3) Count the number of these prime numbers.
According to the above operation method, we calc... | 2 |
11. If the function $f(x)=\lg \left(\sqrt{x^{2}+2}+a x\right)-\lg b$ is an odd function defined on $\mathbf{R}$, then the value of the real number $a$ is $\qquad$ , and the value of $b$ is $\qquad$ . | a = \pm 1, b = \sqrt{2} | 11. $a= \pm 1, b=\sqrt{2} . f(0)=0, f(-1)=-f(1)$, using the method of undetermined coefficients. | 2.6 |
3. Given $S=\{(i, j) \mid i, j=1,2, \cdots, 100\}$ as the set of $100 \times 100$ integer points on the Cartesian plane. Each point in $S$ is colored with one of four given colors. Find the maximum possible number of rectangles with sides parallel to the coordinate axes, whose vertices are four points of different colo... | 9375000 | 3. The maximum value sought is 9375000.
Let the four colors be $A, B, C, D$. Let
$R_{i}=\{(x, i) \mid x=1,2, \cdots, 100\}$
be the $i$-th row of the integer point square,
$$
T_{j}=\{(j, y) \mid y=1,2, \cdots, 100\}
$$
be the $j$-th column of the integer point square.
A rectangle with four vertices of different colors... | 3 |
36. The table below lists the time differences between several cities and Beijing, where positive numbers indicate the number of hours earlier than Beijing time at the same moment. If it is 10:00 on February 28, 2013, in Beijing, then the time difference between Moscow and Vancouver is $\qquad$ hours; at this moment, t... | $11, 2, 27, 2$ | Reference answer: $11,2,27,2$ | 1.25 |
The arithmetic sequence $a, a+d$, $a+2 d, a+3 d, \ldots, a+(n-1) d$ has the following properties:
- When the first, third, and fifth, and so on terms are added, up to and including the last term, the sum is 320 .
- When the first, fourth, seventh, and so on, terms are added, up to and including the last term, the sum ... | 608 | When we start at the first term and look at every other term from there, we will look at the last term. Therefore, the total number of terms after the first is a multiple of 2.
When we start at the first term and look at every third term from there, we will look at the last term. Therefore, the total number of terms a... | 3 |
6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$ | (3, 5), (5, 3) | ## Solution.
Rewrite the system of equations as $\left\{\begin{array}{l}(x+y)^{2}-2 x y=34, \\ (x+y)+x y=23\end{array}\right.$ and, by setting $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$, we get $\left\{\begin{array}{l}u^{2}-2 v=34, \\ u+v=23\end{array} \Rightarrow v=23-u\right.$,
$u^{2}-2(23-u)=34, u... | 3 |
How many ways are there to place 31 knights in the cells of an $8 \times 8$ unit grid so that no two attack one another? | 68 | Consider coloring the squares of the chessboard so that 32 are black and 32 are white, and no two squares of the same color share a side. Then a knight in a square of one color only attacks squares of the opposite color. Any arrangement of knights in which all 31 are placed on the same color therefore works: there are ... | 2.75 |
Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.
What was the mean number of days of exercise last week, rounded to the nearest hundred... | 4.36 | The mean, or average number of days is the total number of days divided by the total number of students. The total number of days is $1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109$. The total number of students is $1+3+2+6+8+3+2=25$. Hence, $\frac{109}{25}=\boxed{\textbf{(C) } 4.36}$.
This problem ... | 2 |
3. The solution to the equation $\arctan \sqrt{x(x+1)}+\arcsin \sqrt{x(x+1)+1}=\frac{\pi}{2}$ is | $0 \text{ or } -1$ | 3. 0 or -1 Explanation: $\tan \alpha=\sqrt{x(x+1)}$, $\sin \beta=\sqrt{x(x+1)+1}, \tan ^{2} \alpha=\sin ^{2} \beta-1=-\cos ^{2} \beta$, $\therefore \tan \alpha=0, \therefore x(x+1)=0$. | 3 |
1. In a $3 \times 3$ table, arrange the numbers $3,4,5,6,7,8,9,10,11$ so that the product of the numbers in the first column equals the product of the numbers in the first row, the product of the numbers in the second column equals the product of the numbers in the second row, and the product of the numbers in the thir... | \begin{pmatrix} 7 & 3 & 8 \\ 6 & 9 & 5 \\ 4 & 10 & 11 \end{pmatrix} | Solution.
| 7 | 3 | 8 |
| :---: | :---: | :---: |
| 6 | 9 | 5 |
| 4 | 10 | 11 |
It is sufficient to provide the required arrangement. The key is to place $7,9,11$ on the diagonal. Corresponding permutations of columns and rows are possible.
Answer. See figure. | 3 |
A $1 \mathrm{~m}$ long, closed at both ends, horizontal pipe contains 100 small balls. Each ball has a speed of $10 \mathrm{~m} / \mathrm{s}$, and the balls collide with each other and the ends of the pipe completely elastically. How many collisions occur in 10 seconds? | 505000 | Let's examine what happens when the balls $A$ and $B$ collide. We assume that the balls have equal mass, so in the case of an elastic collision, $A$ and $B$ exchange velocities. Thus, ball $A$ travels along the same path that ball $B$ would have taken if the two balls had "passed through" each other.
 whose graph does not lie below the $x$-axis at any point, the graph intersects the $y$-axis at point $C$, and the vertex is $E$, with the axis of symmetry to the right of the line $x=c-\frac{1}{12}$. Let $t=\frac{a+2 b+12 c}{a}$.
(1) Find the minimum v... | y = 6x^2 - 2x + \frac{1}{6} | 11. (1) According to the problem, we have
$$
a>0, b^{2}-4 a c \leqslant 0 \Rightarrow c \geqslant \frac{b^{2}}{4 a} \text {. }
$$
Therefore, $t=\frac{a+2 b+12 c}{a} \geqslant \frac{a+2 b+\frac{3 b^{2}}{a}}{a}$
$$
=1+2\left(\frac{b}{a}\right)+3\left(\frac{b}{a}\right)^{2}=3\left(\frac{b}{a}+\frac{1}{3}\right)^{2}+\frac... | 2.75 |
170 Given a positive integer $n$ and a constant $a>0, x$ is any positive number, then the minimum value of $\frac{\left(a^{n}+x^{n}\right) \cdot(a+x)^{n}}{x^{n}}$ is | $2^{n+1} \cdot a^{n}$ | $170 \quad 2^{n+1} \cdot a^{n}$.
$$
\frac{\left(a^{n}+x^{n}\right) \cdot(a+x)^{n}}{x^{n}} \geqslant \frac{2 \cdot \sqrt{a^{n} \cdot x^{n}} \cdot(2 \cdot \sqrt{a \cdot x})^{n}}{x^{n}}=2^{n+1} \cdot a^{n},
$$
The equality holds if and only if $x=a$, so the minimum value is $2^{n+1} \cdot a^{n}$. | 3 |
12. (15 points) As shown in the figure, point $M$ is a point on side $CD$ of parallelogram $ABCD$, and $DM: MC=1: 2$. Quadrilateral $EBFC$ is a parallelogram, and $FM$ intersects $BC$ at point $G$. If the difference in area between triangle $FCG$ and triangle $MED$ is $13 \mathrm{~cm}^{2}$, find the area of parallelogr... | 60 \text{ cm}^2 | 【Analysis】To find the area of the parallelogram, we need to determine the area difference 13 corresponding to the area ratio. We need to establish a relationship between these two triangles. First, we need to know that in $D E M B C$, it is an hourglass model, and $B C F G M$ is also an hourglass model. The two paralle... | 2.2 |
Two triangles have all their angles equal; of their sides, two pairs are equal, but the third sides are different. Calculate the different sides if the lengths of the equal sides are 80 and 100 units, respectively. | 64 \text{ and } 125 | If the angles of two triangles are equal, then the two triangles are similar: the ratio of the corresponding sides is equal.
The lengths of the sides of one triangle are: $80,100, x$; in the other, the sides are $80,100, y$.
The equal sides cannot be corresponding to each other, because then $x=y$ would be true; howe... | 1.8 |
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