problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8. The solution set of the inequality $\left|x^{2}-2\right| \leqslant 2 x+1$ is | 8. $\{x \mid \sqrt{2}-1 \leqslant x \leqslant 3\}$
---
8. $\{x \mid \sqrt{2}-1 \leqslant x \leqslant 3\}$ | \{x \mid \sqrt{2}-1 \leqslant x \leqslant 3\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,918 |
9. The license plates of motor vehicles in a certain city are numbered consecutively from "10000" to "99999". Then, among these 90000 license plates, the number of plates where the digit 9 appears at least once, and the sum of the digits is a multiple of 9, is $\qquad$ . $\qquad$ | 9. 4168 | 4168 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,919 |
10. If $0<a, b, c<1$ satisfy the condition $ab+bc+ca=1$, then the minimum value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is $\qquad$ | 10. $\frac{3(3+\sqrt{3})}{2}$ | \frac{3(3+\sqrt{3})}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,920 |
11. Given a regular quadrilateral pyramid $V-A B C D$ with all edges equal to $a$, the midpoints of the lateral edges $V B$ and $V D$ are $H$ and $K$, respectively. If the plane passing through points $A$, $H$, and $K$ intersects the lateral edge $V C$ at $L$, then the area of quadrilateral $A H L K$ is $\qquad$ | 11. $\frac{\sqrt{5}}{6} a^{2}$ | \frac{\sqrt{5}}{6} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,921 |
13. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}=2 a_{n-1}+n-2(n \geqslant 2) \text {. }
$$
Find the general term $a_{n}$. | Three, 13. From the given information,
$$
a_{n}+n=2\left[a_{n-1}+(n-1)\right](n \geqslant 2) \text {. }
$$
Let $b_{n}=a_{n}+n$, then
$$
b_{1}=a_{1}+1=2, b_{n}=2 b_{n-1}(n \geqslant 2) \text {. }
$$
Thus, $b_{n}=2 \times 2^{n-1}=2^{n}$, which means
$$
a_{n}+n=2^{n}(n \geqslant 2) \text {. }
$$
Therefore, $a_{n}=2^{n}... | a_{n}=2^{n}-n(n \geqslant 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,923 |
14. In Figure $1, O$ and $H$ are the circumcenter and orthocenter of acute $\triangle ABC$, respectively. $D$ is the midpoint of side $BC$. Perpendiculars are drawn from $H$ to the internal and external angle bisectors of $\angle A$, with the feet of the perpendiculars being $E$ and $F$, respectively. Prove that $D$, $... | 14. As shown in Figure 2, connect $O A$ and $O D$, and extend $O D$ to intersect the circumcircle of $\triangle A B C$ at $M$. Then $O D \perp B C$, and $\overparen{B M} = \overparen{M C}$. Therefore, $A$, $E$, and $M$ are collinear.
Also, $A E$ and $A F$ are the angle bisectors of $\angle A$ and its exterior angle in... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,924 |
15. Let $A=\{1,2, \cdots, n\}$, and let $S_{n}$ denote the sum of the elements in all non-empty proper subsets of $A$, and $B_{n}$ denote the number of subsets of $A$. Find the value of $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}$. | 15. It is known that $B_{n}=2^{n}$.
First, consider the number of times the digit 1 appears in all non-empty proper subsets of $A$. It is known that it appears 1 time in the one-element sets, i.e., $C_{n-1}^{0}$ times, in the two-element sets $C_{n-1}^{1}$ times, in the three-element sets $C_{n-1}^{2}$ times, ..., and... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,925 |
16. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the moving circle $\Gamma: x^{2}+y^{2}=R^{2}$, where $b<R<a$. If $A$ is a point on the ellipse $E$, and $B$ is a point on the moving circle $\Gamma$, and the line $A B$ is tangent to both the ellipse $E$ and the moving circle $\Gamma$, fin... | 16. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the equation of line $AB$ be $y=k x+m$.
Since $A$ is on both the ellipse $E$ and the line $AB$, we have
$$
\left\{\begin{array}{l}
y_{1}=k x_{1}+m, \\
\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1 .
\end{array}\right.
$$
Substituting (1) in... | a-b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,926 |
9.1. Each integer point on the coordinate plane is colored with one of three colors, and points of all three colors exist. Prove: A right-angled triangle can be found whose three vertices are points of three different colors. | 9.1. We call the integer points on the plane nodes. If all nodes on every vertical line are of the same color, then we choose one of them (let it be color 1). Draw 2 perpendicular lines through it, making them intersect the vertical direction at $45^{\circ}$ angles, and take 1 point of color 2 and one point of color 3 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,927 |
Example 3 Choose several colors from the given 6 different colors to color the 6 faces of a cube, with each face being colored with exactly 1 color, and any 2 faces sharing a common edge must be colored differently. Then the number of different coloring schemes is $\qquad$ (Note: If we color two identical cubes and can... | Solution: This problem is relatively complex and requires a classification discussion based on the number of colors used.
(1) If only 3 colors are used, there are $\mathrm{C}_{6}^{3}$ ways to choose 3 colors from 6 different colors. Since every 2 faces sharing a common edge are painted in different colors, the opposite... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,928 |
9.2. Quadrilateral $A B C D$ is circumscribed around a circle, the external angle bisectors of $\angle A$ and $\angle B$ intersect at point $K$, the external angle bisectors of $\angle B$ and $\angle C$ intersect at point $L$, the external angle bisectors of $\angle C$ and $\angle D$ intersect at point $M$, and the ext... | 9. 2. As shown in Figure 1, let the incenter of quadrilateral $ABCD$ be denoted as $O$. Since the internal angle bisectors and the external angle bisectors are perpendicular to each other, we have:
$$
\begin{array}{l}
O A \perp N K, \\
O B \perp K L .
\end{array}
$$
Since the altitude $A K_{1}$ of $\triangle A B K$ is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,929 |
9.3. There are 2004 small boxes on the table, each containing 1 ball. It is known that some of the balls are white, and there are an even number of white balls. You are allowed to point to any 2 boxes and ask: "Do they contain at least 1 white ball?" How many times do you need to ask, at a minimum, to determine 2 boxes... | 9.3. 4005.
We will number the boxes (and the balls inside them) from 1 to 2004, and refer to the problems by the numbers of the box pairs. We will call all non-white balls black balls.
First, we prove that 2 white balls can be found with 4005 questions. We ask the questions $(1,2),(1,3), \cdots,(1,2004)$, $(2,3),(2,4... | 4005 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,930 |
9.4. Let $n (n>3)$ be a natural number, and $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ positive numbers, whose product equals 1. Prove:
$$
\frac{1}{1+x_{1}+x_{1} x_{2}}+\frac{1}{1+x_{2}+x_{2} x_{3}}+\cdots+\frac{1}{1+x_{n}+x_{n} x_{1}}>1 .
$$ | $\begin{array}{l}\text { 9.4. } \sum_{k=1}^{n} \frac{1}{1+x_{k}+x_{k} x_{k+1}}>\sum_{k=1}^{n} \frac{1}{1+\sum_{i=0}^{n-2} \prod_{j=k}^{k+i} x_{j}} \\ =\sum_{k=1}^{n} \frac{x_{n} \prod_{j=1}^{k-1} x_{j}}{x_{n} \prod_{j=1}^{k-1} x_{j}\left(1+\sum_{i=0}^{n-2} \prod_{j=k}^{k+i} x_{j}\right)} \\ =\sum_{k=1}^{n} \frac{x_{n} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,931 |
9.5. Do there exist pairwise distinct positive integers $m, n, p, q$ such that
$$
m+n=p+q, \sqrt{m}+\sqrt[3]{n}=\sqrt{p}+\sqrt[3]{q}>2 \text { 004? }
$$ | 9.5. Existence.
We are looking for positive integers of the following form:
$$
m=a^{2}, n=b^{3}, p=c^{2}, q=d^{3},
$$
where $a, b, c, d$ are positive integers.
Note that the conditions in the problem can be transformed into
$$
a+b=c+d, a^{2}+b^{3}=c^{2}+d^{3},
$$
which implies $a-c=d-b$,
$$
(a-c)(a+c)=(d-b)\left(d^{... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,932 |
9.6. In the office, there are 2004 telephones, each of which is connected to any other by a wire of one of 4 colors. It is known that wires of all 4 colors are used. Is it necessarily possible to find some telephones such that the wires connecting them use exactly 3 different colors? | 9.6. It is definitely possible.
Construct a graph where the vertices are telephones and the edges are wires. Consider the smallest set of vertices such that the wires connecting them contain all 4 colors. Remove any one vertex (called $A$) from this set. Since the original set is the smallest set with the described pr... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,933 |
9.7. Positive integers from 1 to 100 are arranged in a circle in such a way that each number is either greater than both of its neighboring numbers or less than both of its neighboring numbers. A pair of adjacent numbers is called "good" if, after removing them, the above property still holds. How many "good" pairs of ... | 9.7. The minimum possible number of "good" neighboring pairs is 51.
For example, first arrange the numbers from 1 to 100 in a clockwise direction in increasing order on a circle, then swap the positions of 2 and 3, 4 and 5, ..., 98 and 99. In the resulting arrangement \(1, 3, 2, 5, 4, \cdots, 99, 98, 100\), there are ... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,934 |
9.8. Let $\triangle A B C$ be an acute triangle, with its circumcenter denoted as $O$. Denote the circumcenter of $\triangle A O C$ as $T$, and the midpoint of side $A C$ as $M$. Take points $D, E$ on sides $A B, B C$ respectively, such that $\angle B D M=\angle B E M=\angle A B C$. Prove: $B T \perp D E$. | 9.8. Since points $D$ and $E$ are on the sides $AB$ and $BC$ of $\triangle ABC$, respectively, $\angle ABC$ is the largest angle in the triangle. Therefore, $\angle AOC = 2 \angle ABC > 120^{\circ}$, and points $O$ and $T$ are on opposite sides of $AC$.
Let lines $ME$ and $MD$ intersect lines $AB$ and $BC$ at points $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,935 |
10.2. There are 2004 small boxes on the table, each containing 1 ball. It is known that some of the balls are white, and there are an even number of white balls. You are allowed to point to any 2 boxes and ask: "Do they contain at least 1 white ball?" How many times do you need to ask, at a minimum, to determine a box ... | 10.2. 2003.
Number the boxes (and the balls inside them) from 1 to 2004, and refer to the problems by the numbers of the box pairs, and call all non-white balls black balls.
First, we prove that a white ball can be found with 2003 questions. Ask the questions $(1,2),(1,3), \cdots,(1,2004)$ in sequence. If all the ans... | 2003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,936 |
10.3. Quadrilateral $ABCD$ is both circumscribed around a circle and inscribed in a circle, and the incircle of quadrilateral $ABCD$ is tangent to sides $AB, BC, CD, AD$ at points $K, L, M, N$ respectively. The external angle bisectors of $\angle A$ and $\angle B$ intersect at point $K'$, the external angle bisectors o... | 10.3. If $A B C D$ is a trapezoid, and $A B / / C D$. At this time, the lines $L L^{\prime}$ and $N N^{\prime}$ have a common intersection point with the perpendicular bisector of $A B$, and points $K, K^{\prime}, M, M^{\prime}$ are all on this perpendicular bisector. Therefore, the conclusion in the problem holds.
If... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,937 |
10.6. A country has 1001 cities, and there is a one-way road between every 2 cities. Each city has exactly 500 outgoing roads and 500 incoming roads. The country designates a region that contains 668 cities. Prove: from each city in this region, it is possible to reach any other city in the region without crossing the ... | 10.6. Suppose the conclusion of the problem does not hold, for example, city $X$ in the region cannot reach another city $Y$ through the roads within the region. Let the set of all cities in the region that can be reached from city $X$ via the roads within the region be denoted as $A$ (which includes city $X$ itself); ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,940 |
10.7. Triangle $T$ is contained within a centrally symmetric convex polygon $M$, and triangle $T^{\prime}$ is the reflection of triangle $T$ about a point $P$ inside $T$. Prove: Triangle $T^{\prime}$ has at least 1 vertex located inside polygon $M$ or on its boundary. | 10.7. Proof 1: Let $O$ be the center of symmetry of the convex polygon $M$, and let $A, B, C$ be the three vertices of triangle $T$, while $A', B', C'$ are the corresponding vertices of triangle $T'$. Let $\triangle A_0 B_0 C_0$ be the symmetric image of $\triangle ABC$ with respect to $O$. Clearly, $\triangle A_0 B_0 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,941 |
10.8. Is there a positive integer $n>10^{1000}$, which is not a multiple of 10, and can have two of its non-zero digits in its decimal representation swapped, such that the set of prime factors of the resulting number is the same as the set of prime factors of the original number. | 10.8. Existence.
First, give an example of such a positive integer.
Let $n=13 \times 11 \cdots 1=144 \cdots 43$, where the number of 1s is to be determined. If the positions of 1 and 3 are swapped, then we get $344 \cdots 41=31 \times 11 \cdots 1$. Thus, as long as the $11 \cdots 1$ here can be divided by $13 \times 3... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,942 |
11.2. Let the circumcircle of $\triangle A B C$ be denoted as $\Omega$, and point $P$ lies on the circumference of $\Omega$. Let the excenters of $\triangle A B C$ opposite to sides $B C$ and $C A$ be denoted as $I_{A}$ and $I_{B}$, respectively. Prove that the midpoint of the line segment connecting the circumcenters ... | 11.2. Let the midpoint of segment $I_{A} I_{B}$ be $M$. Since the internal and external angle bisectors are perpendicular to each other, we have $A I_{A} \perp A I_{B}$. In the right triangle $\triangle A I_{A} I_{B}$, the midpoint $M$ of the hypotenuse $I_{A} I_{B}$ is equidistant from all vertices, so,
$$
M A=\frac{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,943 |
11.3. Given polynomials $P(x)$ and $Q(x)$. It is known that for some polynomial $R(x, y)$, the equation
$$
P(x) - P(y) = R(x, y)[Q(x) - Q(y)]
$$
holds. Prove: There exists a polynomial $S(x)$ such that
$$
P(x) = S(Q(x)) \text{. }
$$ | 11.3. Proof by contradiction.
Assume there exist polynomials $P(x)$ and $Q(x)$ such that the statement of the problem does not hold. We will examine the polynomial $P(x)$ with the lowest degree among them. Let the degrees of $P(x)$ and $Q(x)$ be $n$ and $k$, respectively. Clearly, multiplying a polynomial by a non-zer... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,944 |
11.4. There is a $9 \times 2004$ grid, in which the positive integers from 1 to 2004 are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. Find the minimum possible value of the sum of the numbers in the first row. | 11.4. $C_{2003}^{2}+1=2005004$.
Consider a $9 \times n$ grid, where the integers from 1 to $n$ are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. We will prove by mathematical induction that the sum of the numbers in the first row is not less than $C_{n-1}^{2}+1$.
Whe... | 2005004 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,945 |
11.6. Proof: There does not exist a finite set $G$ of more than $2 n (n>3)$ pairwise non-parallel vectors in the plane with the following properties:
(1) For any $n$ vectors in this set, it is possible to find another $n-1$ vectors from the set such that the sum of these $2 n-1$ vectors is 0;
(2) For any $n$ vectors in... | 11.6. Assume the conclusion of the problem is not true.
Choose a line $l$ such that it is not perpendicular to any vector in the set $G$. Thus, there are at least $n$ vectors in $G$ whose projections on the line $l$ point in the same direction, let them be $e_{1}, e_{2}, \cdots, e_{n}$. Choose a direction on the line ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,947 |
11.7. A country has several cities and $k$ different airlines. Any 2 cities are either connected by a single bidirectional direct flight belonging to one of the airlines, or they are not connected by any flight. It is known that any 2 flights of the same airline have a common endpoint. Prove: It is possible to divide a... | 11.7. We prove the conclusion of the problem by induction on $k$.
When $k=0$, the conclusion is obviously true, because there are no airlines at this time.
Construct a graph, where the vertices are the cities of the country, and the edges are the air routes. Let $E_{1}, E_{2}, \cdots, E_{k}$ denote the sets of edges ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,948 |
11.8. A section in the shape of a hexagon is made on a rectangular parallelepiped. It is known that this hexagon can fit into a rectangle II. Prove: A face of the rectangular parallelepiped can fit into the rectangle $\Pi$. | 11.8. Let the rectangular prism be denoted as $A B C D-A_{1} B_{1} C_{1} D_{1}$, and set $A B=a, A D=b, A A_{1}=c$, where $a \leqslant b \leqslant c$. Without loss of generality, we can assume that the section $K L M N P Q$ is positioned such that $K \in A D, L \in A B, M$ $\in B B_{1}, N \in B_{1} C_{1}, P \in C_{1} D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,949 |
1. The number of integer pairs $(k, b)$ that make the parabola $y=x^{2}$ and the line $y=k x+b$ always have one intersection point is ( ).
(A) 1
(B) 2
(C) 3
(D) Infinite | -1.D.
Substitute $y=x^{2}$ into $y=k x+b$, and rearrange to get
$$
x^{2}-k x-b=0 \text {. }
$$
According to the condition given in the problem, equation (1) should have a unique solution, thus
$$
\Delta=k^{2}+4 b=0 \text {, }
$$
which means $k^{2}=-4 b$.
There are infinitely many integer pairs $(k, b)$ that satisfy e... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,950 |
2. A certain product at the same price underwent 2 price adjustments in 3 stores. Store A first decreased the price by a percentage of $a$, then increased it by a percentage of $b$; Store B first decreased the price by a percentage of $\frac{a+b}{2}$, then increased it by a percentage of $\frac{a+b}{2}$; Store C first ... | 2.A.
Let the original price of this commodity be 1. After 2 price adjustments, the prices of this commodity in stores A, B, and C are respectively
$$
\begin{array}{l}
M_{\text {A }}=(1-a)(1+b)=1-a+b-a b ; \\
M_{\text {B }}=\left(1-\frac{a+b}{2}\right)\left(1+\frac{a+b}{2}\right)=1-\left(\frac{a+b}{2}\right)^{2} ; \\
M... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,951 |
3. Given a convex quadrilateral $A B C D$ where $A D / / B C$, and $A B+B C=C D+A D$. Then the size relationship between $A D$ and $B C$ is ( ).
(A) $A D>B C$
(B) $A D<B C$
(C) $A D=B C$
(D) Any of the above three situations is possible | 3.C.
As shown in Figure 3, draw a line through $C$ parallel to $AB$, intersecting the extension of $AD$ at $E$. Then, quadrilateral $ABCE$ is a parallelogram. Therefore,
$$
A B+B C=C E+A E .
$$
Since $A B+B C=C D+A D$, we have
$$
C E+A E=C D+A D \text {. }
$$
Thus, $D$ must coincide with $E$, meaning quadrilateral $... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,952 |
4. Given the parabola $y=x^{2}+a x+b$ intersects the $x$-axis at two different points $A$ and $B$, and the distances from the origin to points $A$ and $B$ are both greater than 1 and less than 2, and the lengths of the two legs of a right triangle are $a$ and $b$. Then the range of the hypotenuse $c$ is ( ).
(A) $4<c<2... | 4.D.
Let the two roots of the equation $x^{2}+a x+b=0$ be $x_{1}$ and $x_{2}$, and $a>0, b>0$. Then
$$
\begin{array}{l}
10 .
\end{array}
$$
Thus, $x_{1}<0, x_{2}<0, -2<x_{1}<-1, -2<x_{2}<-1$.
And $a^{2}+b^{2}=\left(x_{1}+x_{2}\right)^{2}+\left(x_{1} x_{2}\right)^{2}$
$$
=\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\righ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,953 |
5. In the plane of equilateral $\triangle A B C$, the condition that line $m$ satisfies is: the distances from the 3 vertices of equilateral $\triangle A B C$ to line $m$ take only 2 values, and one of these values is twice the other. The number of such lines $m$ is ( ).
(A) 16
(B) 18
(C) 24
(D) 27 | 5.C.
As shown in Figure 4, there are a total of $8 \times 3=24$ lines. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,954 |
6. A container is filled with an alcohol solution. The first time, $10 \mathrm{~L}$ is poured out and then water is added to fill it up; the second time, $6 \mathrm{~L}$ is poured out and then water is added to fill it up again. At this point, the volume ratio of alcohol to water in the container is 7: 13. Then the cap... | 6. B.
Let the volume of the container be $x \mathrm{~L}$. Then
$$
\frac{x-10-\frac{x-10}{x} \times 6}{x}=\frac{7}{20} \text {. }
$$
Solving, we get $x_{1}=20, x_{2}=\frac{60}{13}$ (rejected). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,955 |
1. Given $m=\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+$ $\frac{1}{\sqrt{2003}+\sqrt{2005}}$. Then $50[2 m]-142=$ $\qquad$ .
(Where $[x]$ denotes the greatest integer not exceeding $x$) | 2. 2008 .
From the known equation, we get
$$
m=\frac{\sqrt{2005}-1}{2} \text {. }
$$
Thus, $[2 m]=43$.
Therefore, $50[2 m]-142=2008$. | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,956 |
2. As shown in Figure 1, PB is a secant of circle $\odot O$ with radius 5, and the lengths of $PA$ and $PB$ are the two roots of the equation $x^{2}-10 x+16=0$ $(PA<PB)$, $PC$ is a tangent to $\odot O$, and $C$ is the point of tangency. Then the area of quadrilateral $PAOC$ is $\qquad$ | 2.14.
As shown in Figure 5, connect $O P$, and draw $O D \perp A B$, with $D$ being the foot of the perpendicular. Solving the equation $x^{2}-10 x+16=0$, we get $P A=2, P B=8$. Thus, $A D=3$, $O D=4$, and $P C=\sqrt{P A \cdot P B}=$
4. Therefore,
$S_{\text {quadrilateral PAOC }}$
$$
\begin{array}{l}
=S_{\triangle P O... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,957 |
3. Given $|y-2| \leqslant 1$, and $x+y=3$. Then the minimum value of $2 x^{2}-$ $12 x+y^{2}+1$ is $\qquad$ .
| 3. -14 .
It is easy to get $1 \leqslant y \leqslant 3, y=3-x$.
Also, $M=2(x-3)^{2}+y^{2}-17=3 y^{2}-17$, it is clear that when $y=1$, $M$ is the smallest, and the minimum value is -14. | -14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,958 |
4. One morning at 9:00, three people, A, B, and C, set off simultaneously and drove south along the Jingzhu Expressway. At the start, C was $180 \mathrm{~km}$ ahead of A, and B was $100 \mathrm{~km}$ behind C. The speeds of A, B, and C were $120 \mathrm{~km} / \mathrm{h}$, $100 \mathrm{~km} / \mathrm{h}$, and $80 \math... | 4.2 Points or 1:20 PM.
Let $t$ be the hours after which the distance between A and B, C is exactly the same. Then:
(1) When B catches up with C, the distance between A and B, C is of course the same, at this time, $t=\frac{100}{100-80}=5$.
(2) When A surpasses B and is between B and C, the condition is also met. For t... | 4.2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,959 |
Example 6 In rectangle $A B C D$, $A B=20 \mathrm{~cm}, B C=$ $10 \mathrm{~cm}$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value. | Solution: As shown in Figure 9, construct the symmetric point $B^{\prime}$ of point $B$ with respect to $A C$, and connect $A B^{\prime}$. Then the symmetric point of point $N$ with respect to $A C$ is point $N^{\prime}$ on $A B^{\prime}$. At this time, the minimum value of $B M + M N$ is equal to the minimum value of ... | 16 \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,960 |
Example 7 In $\triangle A B C$, $\angle A$ is the smallest angle, and points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point of the arc with endpoints $B$ and $C$ (not containing $\angle A$). The perpendicular bisectors of segments $A B$ and $A C$ intersect $A U$ at point... | Proof: It is easy to prove that when $\angle A$ is the smallest angle in $\triangle ABC$, point $T$ is inside $\triangle ABC$.
As shown in Figure 10, let lines $BV$ and $CW$ intersect the circumcircle of $\triangle ABC$ at $B_{1}$ and $C_{1}$, respectively. According to the symmetry of the circle about the perpendicul... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,961 |
Example 6 Let $n$ be a given natural number. Find all positive pairs $(a, b)$ such that $x^{2}+a x+b$ is a factor of $a x^{2 n}+$ $(a x+b)^{2 n}$.
(1992, Shanghai High School Mathematics Competition) | Let $x_{0}$ be a root of the equation $x^{2} + a x + b = 0$. From $b > 0$, we know $x_{0} \neq 0$, and $x_{0}$ is also a root of the equation
$$
a x^{2 n} + (a x + b)^{2 n} = 0
$$
Thus, $a x_{0}^{2 n} + \left(-x_{0}^{2}\right)^{2 n} = 0$, which gives $x_{0}^{2 n} = -a$. Solving this, we get
$$
x_{0} = \sqrt[2 n]{a} \l... | a = \left[2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2 n}{2 n - 1}}, \quad b = \left[2 \cos \frac{(2 k + 1) \pi}{2 n}\right]^{\frac{2}{2 n - 1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,962 |
8.5. Given positive integers $k$ and $n$, their difference is greater than 1. It is known that $4 k n + 1$ is divisible by $k + n$. Prove: $2 n - 1$ and $2 k + 1$ have a common divisor greater than 1. | 8.5. Since $4 k(k+n)$ and $4 n(k+n)$ are divisible by $k+n$, combining this with the conditions given in the problem, we know that $4 n^{2}-1$ and $4 k^{2}-1$ are also divisible by $k+n$, i.e., $(2 n-1)(2 n+1)$ and $(2 k-1)(2 k+1)$ are divisible by $k+n$. Let $(a, b)$ denote the greatest common divisor of the positive ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,963 |
$8.6 .7 \times 7$ grid squares are colored with two different colors. Prove: at least 21 rectangles can be found, whose vertices are the centers of squares of the same color, and whose sides are parallel to the grid lines. | 8.6. Consider any one column and estimate the number of same-color "square pairs" in it. Suppose there are $k$ squares of one color and $7-k$ squares of the other color in this column. Then, the total number of same-color "square pairs" in this column is
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}+\mathrm{C}_{7-k}^{2}=\frac... | 21 | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,964 |
1. Real numbers $a, b$ satisfy $ab=1$, let
$$
M=\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}, N=\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}} \text {. }
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determined | 1. B.
$$
\begin{array}{l}
M=\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}} \\
=\frac{a^{2} b^{2}}{a^{2} b^{2}+a^{2}}+\frac{a^{2} b^{2}}{a^{2} b^{2}+b^{2}}=\frac{b^{2}}{1+b^{2}}+\frac{a^{2}}{1+a^{2}}=N .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,965 |
2. If $a, b, c$ are 3 distinct numbers, and $\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=1$, then
$$
\left(\frac{1}{a-b}\right)^{2}+\left(\frac{1}{b-c}\right)^{2}+\left(\frac{1}{c-a}\right)^{2}
$$
equals ( ).
(A) 1
(B) 2
(C) 3
(D) $\frac{3}{2}$ | 2.A.
$$
\begin{array}{l}
\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^{2} \\
=\left(\frac{1}{a-b}\right)^{2}+\left(\frac{1}{b-c}\right)^{2}+\left(\frac{1}{c-a}\right)^{2}+ \\
2\left(\frac{1}{a-b} \cdot \frac{1}{b-c}+\frac{1}{b-c} \cdot \frac{1}{c-a}+\frac{1}{c-a} \cdot \frac{1}{a-b}\right) . \\
\text { and } ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,966 |
3. As shown in Figure 1, in equilateral
triangle $\triangle A B C$, $M$ and $N$ are the midpoints of sides $A B$ and $A C$, respectively. $D$ is any point on $M N$, and the extensions of $B D$ and $C D$ intersect $A C$ and $A B$ at points $E$ and $F$, respectively. If $\frac{1}{C E}+\frac{1}{B F}=3$, then $S_{\triangl... | 3.D.
As shown in Figure 4, draw $DS \parallel BM$ and $DT \parallel CN$ intersecting $BC$ at $S$ and $T$. Since $M$ and $N$ are the midpoints of sides $AB$ and $AC$ respectively, and $MN = \frac{1}{2} BC$. Therefore, quadrilaterals $BSDM$ and $TCND$ are both parallelograms. Hence,
$$
MD = BS, DN = TC.
$$
Thus, $ST = ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,967 |
4. A fruit tree planting base has adopted two measures, expanding the planting scale and scientific management, to continuously meet market demand. In recent years, the amount of fruit supplied to the market has been: 46 t in 1998, 58 t in 2000, and 75 t in 2002. According to predictions, within a certain period, the f... | 4.C.
For convenience, let 1998 be the starting year for calculation, with the year number as 0, and assume the predicted fruit supply in 2004 is $M \mathrm{t}$, and the quadratic function we are looking for is $y=a x^{2}+b x+c$.
From the problem, we know that the points $(0,46) 、(2,58) 、(4,75) 、(6, M)$ all lie on the... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,968 |
5. Let the line $k x+(k+1) y-1=0(k$ be a positive integer) intersect the coordinate axes to form the figure $S_{k}, k=1$, $2, \cdots, 2$ 004. Then, $2\left(S_{1}+S_{2}+S_{2004}\right)$ equals ( ).
(A) $\frac{1001}{2004}$
(B) $\frac{1001}{1002}$
(C) $\frac{2004}{2005}$
(D) $\frac{1002}{2005}$ | 5.C.
It is known that the intersection points of the line $k x+(k+1) y-1=0$ with the two coordinate axes are $\left(0, \frac{1}{k+1}\right)$ and $\left(\frac{1}{k}, 0\right)$. Therefore,
$$
\begin{array}{l}
S_{k}=\frac{1}{2} \cdot \frac{1}{k(k+1)}, k=1,2, \cdots, 2004 . \\
2\left(S_{1}+S_{2}+\cdots+S_{2004}\right) \\
... | \frac{2004}{2005} | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,969 |
6. Given the equation
$$
2(a-b) x^{2}+(2 b-a b) x+(a b-2 a)=0
$$
has two equal real roots. Then $\frac{2}{b}-\frac{1}{a}=(\quad)$.
(A) 0
(B) $\frac{1}{2}$
(C) 1
(D) 2 | 6. B.
Since $\Delta=4 b^{2}-4 a b^{2}+a^{2} b^{2}-8(a-b)(a b-2 a)=0$, after rearrangement we get
$$
4 b^{2}+4 a b^{2}+a^{2} b^{2}-8 a^{2} b+16 a^{2}-16 a b=0 .
$$
Given that $a \neq 0, b \neq 0$. Therefore, we have
$$
\begin{array}{l}
\frac{4}{a^{2}}+\frac{4}{a}+1-\frac{8}{b}+\frac{16}{b^{2}}-\frac{16}{a b}=0 \\
\Rig... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,970 |
1. Let $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}, y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. Then $x^{2}+x y$ $+y^{2}+1=$ $\qquad$ | 1.100 .
$$
\begin{array}{l}
\text { Given } x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=(\sqrt{3}+\sqrt{2})^{2}=5+2 \sqrt{6}, \\
y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=(\sqrt{3}-\sqrt{2})^{2}=5-2 \sqrt{6},
\end{array}
$$
Therefore, $x^{2}+x y+y^{2}+1=(x+y)^{2}-x y+1=100$. | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,971 |
2. If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ . | 2.0.
Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,972 |
Example 7 Given that $a, b, c, d$ take certain real values, the equation $x^{4}+a x^{3}+b x^{2}+c x+d=0$ has 4 non-real roots, where the product of 2 of the roots is $13+i$, and the sum of the other 2 roots is $3+4i$, where $i$ is the imaginary unit. Find $b$.
(13th American Invitational Mathematics Examination) | Explanation: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the 4 roots of the equation. By the problem, we can assume $x_{3}=\overline{x_{1}}, x_{4}=\overline{x_{2}}$, then we have $x_{1} x_{2}=$
$$
\begin{aligned}
13+\mathrm{i}, & x_{3}+x_{4}=3+4 \mathrm{i} \text {. Therefore, } \\
b & =x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2}... | 51 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,973 |
3. As shown in Figure 2, in the equilateral $\triangle A B C$, side $A B$ is tangent to $\odot O$ at point $H$, and sides $B C$ and $C A$ intersect $\odot O$ at points $D, E, F, G$. Given that $A G=2, G F=6, F C$ $=1$. Then $D E=$ $\qquad$ | 3. $\sqrt{21}$.
By the secant-tangent theorem, we know $A H^{2}=A G \cdot A F=16$, which means $A H=4$.
Also, $A C=A G+G F+F C=9$, so, $A B=9$.
Therefore, $B H=5$.
Let $B D=x, C E=y$.
By the secant-tangent theorem, we know $B H^{2}=B D \cdot B E$, that is,
$$
25=x(9-y) \text {. }
$$
Also, $C E \cdot C D=C F \cdot C G... | \sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,974 |
4. A bookstore is holding a summer book fair, where book buyers can enjoy the following discounts:
(1) For a single purchase not exceeding 50 yuan, no discount is given;
(2) For a single purchase exceeding 50 yuan but not exceeding 200 yuan, a 10% discount is given on the marked price;
(3) For a single purchase exceedi... | 4.3.
Obviously, 81 and 126 are both the prices after the discount. If not discounted, the combined price should be
$$
(81+126) \div 0.9=230 \text { (yuan). }
$$
Since 230 yuan exceeds 200 yuan, according to (3), it should be
$$
200 \times 0.9+30 \times 0.8=204 \text { (yuan) } \text {. }
$$
Thus, we have $81+126-204... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,975 |
One. (20 points) Given the equations about $x$: $2 x^{2}-4 n x$ $-2 n=1$ and $x^{2}-(3 n-1) x+2 n^{2}-3 n=2$. Does there exist a value of $n$ such that the sum of the squares of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain th... | Given $\Delta_{1}=16 n^{2}-8(-2 n-1)$
$$
=16 n^{2}+16 n+8=(4 n+2)^{2}+4>0
$$
we know that for any real number $n$, the first equation has real roots.
Let the two roots of the first equation be $x_{1}$ and $x_{2}$, then we have
$$
x_{1}+x_{2}=2 n, x_{1} x_{2}=-\frac{2 n+1}{2} \text {. }
$$
Thus, $x_{1}^{2}+x_{2}^{2}=\... | n=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,976 |
II. (25 points) As shown in Figure 3, in the right triangle $\triangle ABC$, $\angle B=90^{\circ}$, $AB=BM=12$, $DM \parallel AB$. Also, $N$ is the midpoint of $BM$, and $\angle ADN=\angle BAD$.
(1) Prove:
$$
CN \cdot AB=CB \cdot DN \text {; }
$$
(2) Find $S_{\triangle DNM}$. | II. As shown in Figure 5, construct $\angle T B A = \angle C$.
Since $\angle A D N = \angle B A D$, then $\angle C D N = \angle B A T$.
Therefore, $\triangle C D N \sim \triangle B A T$.
Thus, $\frac{C N}{T B} = \frac{C D}{A B}$, which means
$C N \cdot A B = T B \cdot C D$.
Also, $\angle C = \angle C$,
$\angle C N D = ... | 24 | Geometry | proof | Yes | Yes | cn_contest | false | 714,977 |
Three. (25 points) It is known that there is a four-digit number, and the sum of this four-digit number and the sum of its digits is 2004. Find this four-digit number and explain your reasoning.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | Three, let this four-digit number be $\overline{a b c d}$. From the problem, we get
$$
1000 a+100 b+10 c+d+a+b+c+d=2004 \text {, }
$$
which simplifies to $1001 a+101 b+11 c+2 d=2004$.
Clearly, $a=1$ or 2, otherwise $1001 a>3000$.
(1) When $a=1$, subtract 1001 from both sides of equation (1), we get
$$
101 b+11 c+2 d=1... | 1983 \text{ and } 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,978 |
1. Given that $\alpha, \beta$ are both acute angles, and satisfy $\sin ^{2} \alpha=$ $\cos (\alpha-\beta)$. Then the relationship between $\alpha$ and $\beta$ is ( ).
(A) $\alpha\beta$
(D) $\alpha+\beta=\frac{\pi}{2}$ | 1.C.
Since $\cos \alpha>0, \cos \beta>0$, therefore, $\cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta>\sin \alpha \cdot \sin \beta$.
Also, $\sin ^{2} \alpha=\cos (\alpha-\beta)$, hence $\sin ^{2} \alpha>\sin \alpha \cdot \sin \beta$.
Given that $\alpha, \beta$ are acute angles, then $\sin... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,979 |
2. Given $f(x)=\frac{a}{a^{2}-2} \cdot\left(a^{x}-a^{-x}\right)(a>0$ and $a \neq 1)$ is an increasing function on $\mathbf{R}$. Then the range of the real number $a$ is ( ).
(A) $(0,1)$
(B) $(0,1) \cup(\sqrt{2},+\infty)$
(C) $(\sqrt{2},+\infty)$
(D) $(0,1) \cup[\sqrt{2},+\infty)$ | 2.B.
Let $x_{1}a^{x_{2}}, a^{-x_{1}}1$ when $a^{x_{1}}0$, then $a>\sqrt{2}$, which meets the conditions.
In summary, the range of values for $a$ is $(0,1) \cup(\sqrt{2},+\infty)$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,980 |
3. Given the four vertices of a rectangle $A(0,0)$, $B(2,0)$, $C(2,1)$, $D(0,1)$. A particle starts from the midpoint $P_{0}$ of $A B$ and moves in a direction making an angle $\theta$ with $A B$ to hit point $P_{1}$ on $B C$, and then reflects sequentially to points $P_{2}$ on $C D$, $P_{3}$ on $D A$, and $P_{4}$ on $... | 3. C.
As shown in Figure 1, let $P_{1}(2, a)$ $(a>0), \angle P_{1} P_{0} B=\theta$. Then
$$
\begin{array}{l}
\tan \theta=\tan \angle P_{1} P_{0} B \\
=\frac{P_{1} B}{P_{0} B}=a .
\end{array}
$$
From $\angle P_{1} P_{2} C$
$$
\begin{array}{l}
=\angle P_{3} P_{2} D \\
=\angle P_{3} P_{4} A=\theta,
\end{array}
$$
we ge... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,981 |
4. Given in $\triangle A B D$, $A B=m, A D=n$, and an equilateral $\triangle B D C$ is constructed with $D B$ as a side (not overlapping with $\triangle A B D$). When the area of quadrilateral $A B C D$ is maximized, $\angle B A D$ equals ( ).
(A) $60^{\circ}$
(B) $90^{\circ}$
(C) $120^{\circ}$
(D) $150^{\circ}$ | 4.D.
$S_{\triangle A B D}=\frac{1}{2} m n \sin \theta$, where $\theta=\angle B A D$.
In $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos \theta \\
=m^{2}+n^{2}-2 m n \cos \theta .
\end{array}
$$
Also, $S_{\triangle B D C}=\frac{\sqrt{3}}{4} B D^{2}$, from (... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,982 |
5. Given $P$ is a point on the parabola $y^{2}=4 x$, let the distance from $P$ to the directrix of this parabola be $d_{1}$, and the distance from $P$ to the line $x+2 y-12=0$ be $d_{2}$. Then the minimum value of $d_{1}+d_{2}$ is ( ).
(A) $\frac{11 \sqrt{5}}{5}$
(B) $\frac{12 \sqrt{5}}{5}+1$
(C) $2 \sqrt{5}$
(D) does ... | 5.A.
Since point $P$ is on $y^{2}=4 x$, let's set $P\left(t^{2}, 2 t\right)$. Therefore, we have $d_{1}+d_{2}=t^{2}+1+\frac{\left|t^{2}+4 t-12\right|}{\sqrt{5}}$.
(1) When $t \leqslant-6$ or $t \geqslant 2$,
$$
d_{1}+d_{2}=\left(1+\frac{1}{\sqrt{5}}\right) t^{2}+\frac{4}{\sqrt{5}} t+1-\frac{12}{\sqrt{5}} \text {. }
$$... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,983 |
Example 8 The equation about $x$
$$
x^{3}+t x+s=0\left(t \in \mathbf{R}, s \in \mathbf{C}, \arg s=\frac{\pi}{6}\right)
$$
has 3 complex roots, which correspond to the 3 vertices of an equilateral triangle with side length $\sqrt{3}$ in the complex plane. Find the values of $s$ and $t$. | Explanation: Let the complex number corresponding to the center of an equilateral triangle be $z_{0}$, and the complex numbers corresponding to the three roots be $z_{1}$, $z_{2}$, and $z_{3}$, respectively, and
$$
\omega=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3},
$$
then $z_{2}-z_{0}=\left(z_{1}-z_{0}\right) \omega... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,984 |
6: Arrange 4 identical red balls and 4 identical blue balls in a row, with each ball corresponding to a number from 1, $2, \cdots, 8$ from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the numbers corresponding to the 4 red balls is less than the sum of the... | 6. A.
The sum of 8 numbers is 36. Divide the numbers $1,2, \cdots, 8$ into two groups, there are $\frac{C_{8}^{8}}{2}=35$ ways to do so. When the sums of the two groups are not equal, assume the 4 numbers with the smaller sum are the sequence numbers of red balls, and the 4 numbers with the larger sum are the sequence ... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,985 |
1. In tetrahedron $ABCD$, $AB \perp CD$, $AB=a$, $AB$ forms an angle $\alpha$ with the plane $BCD$, and the dihedral angle $A-CD-B$ is $\beta$. Then the minimum value of the perimeter of the triangular section of the tetrahedron passing through edge $AB$ is $\qquad$ | 1. $\frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)]$.
Draw a section $A B E$ through $A B$ such that $C D \perp$ plane $A B E$, then $C D \perp A E, C D \perp B E$.
Therefore, the perimeter of $\triangle A B E$ is minimized.
At this point, $\angle A B E$ is the angle between $A B$ and plane $B C D$, s... | \frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,986 |
2. The infinite sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=3 a_{n}-4$ $\left(n \in \mathbf{N}_{+}\right)$, and $\left\{a_{n}\right\}$ is a bounded sequence. Then the general term of the sequence $a_{n}=$ $\qquad$ . | 2. $a_{n}=2$.
Transform $a_{n+1}=3 a_{n}-4$ into
$$
a_{n+1}-2=3\left(a_{n}-2\right) \text {. }
$$
Assume $a_{1} \neq 2$.
From equation (1), $\left\{a_{n}-2\right\}$ is a geometric sequence with a common ratio of 3.
Therefore, $a_{n}=2+\left(a_{1}-2\right) 3^{n-1}$.
When $n \rightarrow \infty$, $3^{n-1} \rightarrow \i... | a_{n}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,987 |
3. The function
$$
y=\sqrt{12+x-x^{2}}-\sqrt{15-2 x-x^{2}}
$$
has a range of $\qquad$ . | 3. $[-2 \sqrt{3}, \sqrt{6}]$.
Transform the function into
$$
y=\sqrt{(4-x)(x+3)}-\sqrt{(x+5)(3-x)} \text {. }
$$
Let $y_{1}=\sqrt{(4-x)(x+3)}, y_{2}=\sqrt{(x+5)(3-x)}$.
As shown in Figure 1, the graphs of the above two functions are semicircles
$T_{1}$ and $T_{2}$. Semicircle $T_{1}$ has a center at $\left(\frac{1}{2... | [-2 \sqrt{3}, \sqrt{6}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,988 |
4. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the first term $a$ and common ratio $t(t \neq 1)$. Let $b_{n}=1+a_{1}+a_{2}+\cdots+$ $a_{n}, c_{n}=2+b_{1}+b_{2}+\cdots+b_{n}, n \in \mathbf{N}_{+}$. The pairs of real numbers $(a, t)$ that make $\left\{c_{n}\right\}$ a geometric sequence are $\qqu... | 4. $(1,2)$.
Since $t \neq 1, b_{n}=1+\frac{a}{1-t}-\frac{a t^{n}}{1-t}$, we have
$$
\begin{array}{l}
c_{n}=2+\left(1+\frac{a}{1-t}\right) n-\frac{a}{1-t}\left(t+t^{2}+\cdots+t^{n}\right) \\
=2+\left(1+\frac{a}{1-t}\right) n-\frac{a\left(t-t^{n+1}\right)}{(1-t)^{2}} . \\
=2-\frac{a t}{(1-t)^{2}}+\left(\frac{1-t+a}{1-t}... | (1,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,989 |
5. Let $\{x\}$ denote the fractional part of the real number $x$. If $a=$ $(5 \sqrt{13}+18)^{2005}$, then $a \cdot\{a\}=$ $\qquad$ | 5.1.
Let $b=(5 \sqrt{13}-18)^{2005}$.
Since $0<5 \sqrt{13}-18<1$, we have $0<b<1$.
By the binomial theorem, we know
$$
\begin{aligned}
a= & (5 \sqrt{13})^{2005}+\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots+18^{2005}, \\
b= & (5 \s... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,990 |
6. Given that the set $M$ consists of all functions $f(x)$ satisfying the following property:
There exists a non-zero constant $T$, for any $x \in \mathbf{R}$, such that $f(x+T)=T f(x)$ holds.
If the function $f(x)=\sin k x \in M$, then the range of the real number $k$ is . $\qquad$ | 6. $\{k \mid k=m \pi, m \in \mathbf{Z}\}$.
When $k=0$, $f(x)=0 \in M$.
When $k \neq 0$, since $f(x)=\sin k x \in M$, there exists a non-zero constant $T$, for any $x \in \mathbf{R}$, such that
$$
\sin (k x+k T)=T \sin k x,
$$
i.e., $\sin k x \cdot \cos k T+\cos k x \cdot \sin k T=T \sin k x$.
Substituting $x=0$ into ... | \{k \mid k=m \pi, m \in \mathbf{Z}\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,991 |
Three. (20 points) For the function $f(x)$, if there exists $x_{0} \in \mathbf{R}$ such that $f\left(x_{0}\right)=x_{0}$, then the point $\left(x_{0}, x_{0}\right)$ is called a fixed point of the function $f(x)$.
(1) If the function $f(x)=a x^{2}+b x-b(a \neq 0)$ has fixed points $(1,1)$ and $(-3,-3)$, find the values ... | (1) By the definition of a fixed point, we have $f(x)-x=0$, that is,
$$
a x^{2}+(b-1) x-b=0 \text {. }
$$
Substituting $x=1$ into equation (1) yields $a=1$.
Substituting $x=-3$ into equation (1) yields
$$
9 a-3(b-1)-b=0 \text {. }
$$
Solving this gives $b=3$.
(2) According to the condition, for any real number $b$, e... | 0 < a < 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,992 |
Four. (20 points) Given that $a$ and $b$ are non-zero constants, and the variable $\theta$ satisfies the inequality system
$$
\left\{\begin{array}{l}
a \sin \theta + b \cos \theta \geqslant 0, \\
a \cos \theta - b \sin \theta \geqslant 0 .
\end{array}\right.
$$
Try to find the maximum value of $\sin \theta$. | Let $x=\cos \theta, y=\sin \theta$. The constraint conditions are transformed into
$$
\left\{\begin{array}{l}
b x+a y \geqslant 0, \\
a x-b y \geqslant 0, \\
x^{2}+y^{2}=1 .
\end{array}\right.
$$
Thus, the problem is transformed into:
Finding the maximum value of the y-coordinate of points in the unit circle (3) withi... | \frac{a}{\sqrt{a^{2}+b^{2}}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,993 |
Five. (20 points) Let $P_{1}, P_{2}, \cdots, P_{n}, \cdots$ be a sequence of points on the curve $y=\sqrt{x}$, and $Q_{1}, Q_{2}, \cdots, Q_{n}, \cdots$ be a sequence of points on the positive half of the $x$-axis. The triangles $\triangle O Q_{1} P_{1}, \triangle Q_{1} Q_{2} P_{2}$, $\triangle Q_{2} Q_{3} P_{3}, \cdot... | As shown in Figure 3, the point \( Q_{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}, 0\right) \) is the point \( Q_{n-1}\left(S_{n-1}, 0\right) \). Therefore, the equation of the line \( Q_{n-1} P_{n} \) is
\[
y=\sqrt{3}\left(x-S_{n-1}\right).
\]
Thus, the coordinates of point \( P_{n} \) satisfy
\[
\left\{\begin{array}{l}
y=\... | \frac{1}{3} n(n+1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,994 |
One. (50 points) Let quadrilateral $ABCD$ be inscribed in $\odot O$. A tangent to $\odot O$ through $A$ intersects the extension of side $BC$ at point $K$, with point $B$ located between points $K$ and $C$; a tangent to $\odot O$ through $B$ intersects the extension of side $DA$ at point $M$, with point $A$ located bet... | As shown in Figure 4, by the secant-tangent theorem, we have
$$
\begin{array}{l}
M B^{2}=M A \cdot M D \\
=\frac{1}{2} M D^{2}, \\
K A^{2}=K B \cdot K C \\
=\frac{1}{2} K C^{2} .
\end{array}
$$
By the Law of Sines, we know
$$
\begin{array}{l}
\frac{\sin \angle A C K}{\sin \angle C A K}=\frac{A K}{K \bar{C}}=\frac{1}{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,996 |
II. (50 points) The real numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfy:
(1) $a_{0}=a_{n}=0$;
(2) For $0 \leqslant k \leqslant n-1$, we have
$$
a_{k}=c+\sum_{i=k}^{n-1} a_{i-k}\left(a_{i}+a_{i+1}\right) \text {. }
$$
Prove: $c \leqslant \frac{1}{4 n}$. | $$
\begin{array}{l}
\text { II. Let } S_{k}=\sum_{i=0}^{k} a_{i}, k=0,1, \cdots, n \text {. Then } \\
S_{n}=\sum_{k=0}^{n} a_{k}=\sum_{k=0}^{n-1} a_{k} \\
=n c+\sum_{k=0}^{n-1} \sum_{i=k}^{n-1} a_{i-k}\left(a_{i}+a_{i+1}\right) \\
=n c+\sum_{i=0}^{n-1} \sum_{k=0}^{i} a_{i-k}\left(a_{i}+a_{i+1}\right) \\
\left.=n c+\sum... | c \leqslant \frac{1}{4 n} | Algebra | proof | Yes | Yes | cn_contest | false | 714,997 |
Does there exist a positive integer $n$ such that the least common multiple of the four numbers $n$, $n+1$, $n+2$, $n+3$ is 3900000? Please explain your reasoning. | Solution: There does not exist a positive integer $n$ that satisfies the condition. The reason is as follows:
Assume there exists such a positive integer $n$. Since among any four consecutive positive integers, at most one number is a multiple of 5, and their least common multiple is $3900000=13 \times 3 \times 2^{5} \... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,999 |
Initially, $146 \leftrightharpoons A B C$ is an acute triangle, $A D$ is the altitude, point $P$ is on $A D$, and rays $B P, C P$ intersect $A C, A B$ at $E, F$ respectively. If points $B, C, E, F$ are concyclic, then $P$ is called a "good point". How many "good points" are there on $A D$? Prove your statement. | Solution: We discuss in two cases:
(1) When $A B=A C$,
there are infinitely many “good points”.
As shown in Figure 1, it is easy to prove
$\therefore B C E \cong \_C B F$.
Thus, quadrilateral $B C E F$ is
an isosceles trapezoid, and at this time, $B$,
$C$, $E$, and $F$ must be concyclic.
The arbitrariness of the posi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,000 |
Let $a, b, c, d \in \mathbf{R}$. And $a+b+c+d=0$. Then we have
$$
1296\left(a^{7}+b^{7}+c^{7}+d^{7}\right)^{2} \leqslant 637\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{7} .
$$ | Proof: By the symmetry of the original expression, without loss of generality, assume $a$ is the largest and $d$ is the smallest.
From the given conditions, we know $a \geqslant 0, d \leqslant 0$. And $d=-(a+b+c) \leqslant 0$, which means $a+b+c \geqslant 0$.
Let $S_{k}=a^{k}+b^{k}+c^{k}+d^{k}, k \in \mathbf{N}_{+}$. T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,001 |
Let
$$
D=\left\{(x_{1}, x_{2}) \mid x_{1}>0, x_{2}>0, x_{1}+x_{2}=k\right\} \text {. }
$$
(1) Find the range of real numbers $k$ such that the inequality
$$
\left(\frac{1}{x_{1}}-x_{1}\right)\left(\frac{1}{x_{2}}-x_{2}\right) \geqslant\left(\frac{2}{k}-\frac{k}{2}\right)^{2}
$$
holds for all $\left(x_{1}, x_{2}\right)... | Solution: Note that $\left(x_{1}-x_{2}\right)^{2}=k^{2}-4 x_{1} x_{2}$, then
$$
\begin{aligned}
P & =\left(\frac{1}{x_{1}}-x_{1}\right)\left(\frac{1}{x_{2}}-x_{2}\right)-\left(\frac{2}{k}-\frac{k}{2}\right)^{2} \\
& =\frac{1}{x_{1} x_{2}}+x_{1} x_{2}-\frac{x_{2}}{x_{1}}-\frac{x_{1}}{x_{2}}-\frac{4}{k^{2}}-\frac{k^{2}}{... | 0 < k \leq 1 \text{ for (1), and } k \geq 1 \text{ for (2)} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,002 |
Example: Let $6 \square A B C D$ have an internal point $P$. If the circumcircles of $\triangle P A B$, $\triangle P B C$, $\triangle P C D$, and $\triangle P D A$ are all equal, then $P$ is called a "good point". Try to determine: Does a "good point" exist inside $\square A B C D$? Is it unique? Prove your assertion. | First, analyze Example 5.
See Figure 5. Suppose all 6 circles are pierced at once, and let the pierced point be $O$. Clearly, point $O$ is inside all 6 circles. Let the centers of the 6 circles be $O_{i} (i=1,2, \cdots, 6)$, and construct $\triangle O O_{1} O_{2}$.
Since $O$ is inside $\odot O_{1}$, $O O_{1}$ is less ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,006 |
Example 7 As shown in Figure 7, in the convex quadrilateral $ABCD$, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively, and $AC$ intersects $BD$ at point $P$. If $M$, $P$, and $N$ are collinear, determine the shape characteristics of the convex quadrilateral $ABCD$.
---
The translation preserves the original... | Example 7 was compiled by the author and published in "Mathematics Teaching".
This is a typical open-ended geometry problem, and we handle it using the area method.
Take any point $C^{\prime}$ on $P A$, and draw $C^{\prime} D^{\prime} \parallel C D$ intersecting $P M$ at $N^{\prime}$ and $P B$ at $D^{\prime}$. Clearly... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,007 |
Example 8 As shown in Figure 8, the convex quadrilateral $ABCD$ is inscribed in $\odot O$. Extend $AB$ and $DC$ to meet at point $E$, and extend $BC$ and $AD$ to meet at point $F$. $M$ and $N$ are the midpoints of $AC$ and $BD$ respectively, and $AC > BD$. Prove:
$$
\frac{MN}{EF}=\frac{1}{2}\left(\frac{AC}{BD}-\frac{BD... | The author handled this problem by using three lemmas:
Lemma $1 S_{\triangle E M N}=S_{\triangle S M N}=\frac{1}{4} S_{\text {quadrilateral } A B C D}$.
Lemma 2 If $A C$ intersects $B D$ at $P$, then $O P \perp E F$.
Lemma 3 As shown in Figure 9,
$O P$ is the diameter of $\odot O^{\prime}$,
points $M$ and $N$ are both ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,008 |
Example 8 As shown in Figure 11, on a rectangular billiard table $ABCD$, there are two balls $P$ and $Q$, with $\angle PAB$ and $\angle QAD$ being equal. If ball $P$ is struck so that it hits point $M$ on $AB$ and bounces to hit ball $Q$, the path is denoted as $P \rightarrow M \rightarrow Q$; if ball $Q$ is struck so ... | Proof: As shown in Figure 12, when ball $P$ collides with $AB$ at point $M$ and bounces to hit ball $Q$, it satisfies $\angle P M B = \angle Q M A$. That is, for the path of $P$, make the symmetric point $P_{1}$ of $P$ about $BA$. Connect $P_{1} Q$ to intersect $BA$ at point $M$, then $P \rightarrow M \rightarrow Q$ is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,009 |
Example 1 Determine all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, where $\mathbf{R}$ is the set of real numbers, such that for all $x, y \in \mathbf{R}$, we have
$$
f(x-f(y))=f(f(y))+x f(y)+f(x)-1
$$
(40th IMO) | Let $A=\operatorname{Im} f$ be the set of the image of the function $f$, and $c=f(0)$. For $x=y=0$, we get $f(-c)=f(c)+c-1$, hence $c \neq 0$.
Taking $x=f(y)$, we have
$$
f(x)=\frac{c+1}{2}-\frac{x^{2}}{2},
$$
for any $x \in A$.
Taking $y=0$, we get
$$
\begin{array}{l}
\{f(x-c)-f(x) \mid x \in \mathbf{R}\} \\
=\{c x+f... | f(x)=1-\frac{x^{2}}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 715,010 |
Example 2 Let $a \in \mathbf{N}_{+}$. Find the function $f: \mathbf{N}_{+} \rightarrow \mathbf{R}$, such that for any $x, y \in \mathbf{N}_{+}, x y>a$, we have
$$
f(x+y)=f(x y-a) .
$$ | Solution: For any $t \in \mathbf{N}_{+}$, since $t=(t+a)-a$ and $1+(t+a)=a+1+t$, then
$$
\begin{array}{l}
f(t)=f((t+a) \times 1-a) \\
=f(1+t+a)=f((a+1)+t) \\
=f((a+1) t-a) \\
=f((a+1)(t-1)+1) \\
=f((a+1)(t-1)-a)=\cdots \\
=f((a+1) \times 1-a)=f(1) .
\end{array}
$$
Therefore, the function sought is a constant function,... | f(x)=f(1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,011 |
Example $\mathbf{3}$ Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that it satisfies $f(f(x))=6 x-f(x)$.
(30th IMO Canadian Training Problem) | Solution 1: Clearly, $f(x) \leqslant 6 x$, then
$$
6 x-f(x)=f(f(x)) \leqslant 6 f(x) \text {. }
$$
Therefore, $f(x) \geqslant \frac{6}{7} x$.
Thus, $6 x-f(x)=f(f(x)) \geqslant \frac{6}{7} f(x)$.
Hence, $f(x) \leqslant \frac{6 x}{1+\frac{6}{7}}$.
Therefore, $6 x-f(x)=f(f(x)) \leqslant \frac{6 f(x)}{1+\frac{6}{7}}$.
Sol... | f(x)=2x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,012 |
Example 4 Let $f: \mathbf{R}^{*} \rightarrow \mathbf{R}^{*}$, and satisfy
(1) $f(x f(y)) f(y)=f(x+y)$;
(2) $f(2)=0$;
(3) $f(x) \neq 0$ for all $0 \leqslant x<2$.
Try to find the function $f(x)$ that satisfies the conditions.
(27th IMO)
Analysis: From conditions (1), (2), and $x=x-2+2$, repeatedly applying condition (1... | When $x>2$,
$$
\begin{array}{l}
f(x)=f((x-2)+2) \\
=f((x-2) f(2)) f(2)=0 .
\end{array}
$$
When $0 \leqslant x < 2$, if $f(x) > 0$, then
$$
\begin{array}{l}
f(x) f((2-x) f(x)) \\
=f((2-x)+x)=f(2)=0 .
\end{array}
$$
By (3), we have
$$
f((2-x) f(x))=0.
$$
Therefore, $(2-x) f(x) \geqslant 2$, which means $f(x) \geqslant... | f(x)=\left\{\begin{array}{ll}\frac{2}{2-x}, & 0 \leqslant x<2, \\ 0 & x \geqslant 2 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,013 |
Example 5 Let $f(n)$ be a function defined on $\mathbf{N}_{+}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
f(m+n)-f(m)-f(n)=0 \text{ or } 1, \\
f(2)=0, f(3)>0, f(6000)=2000 . \\
\text{Find } f(5961) .
\end{array}
$$ | Solution: From the analysis, we know that $f(1)=0$.
From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$.
But $f(3)>0$, so $f(3)=1$.
By the problem statement, we have
$$
f(3 n+3)=f(3 n)+f(3)+0 \text { or } 1 \text {, }
$$
which means $f(3 n+3)=f(3 n)+1$ or 2.
Therefore, $f(3(n+1)) \geqslant f(3 n)+1$.
L... | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,014 |
Example 6 Let $f(n)$ be a function defined on the set of positive integers $\mathbf{N}_{+}$ and taking values in $\mathbf{N}_{+}$. Prove: if for every positive integer $n$, the inequality $f(n+1)>f(f(n))$ holds, then for every $n$, the equation $f(n)=n$ holds.
(19th IMO) | Proof: Let the proposition $A_{n}$ be: for $n \in \mathbf{N}_{+}$, if $m \geqslant n$, then $f(m) \geqslant n$.
Clearly, $A_{1}$ is true.
If $A_{n-1}$ is true, then for any $m \geqslant n, f(m-1) \geqslant n-1$ holds. Therefore,
$$
f(m)>f(f(m-1)) \geqslant n-1.
$$
Thus, $f(m) \geqslant n$. This shows that $A_{n}$ is a... | f(n)=n | Number Theory | proof | Yes | Yes | cn_contest | false | 715,015 |
Example 7 Let the function $f: \mathbf{Z}_{+}^{3} \rightarrow \mathbf{Z}_{+}$ be defined as follows:
(1) $f(a, 1, n)=n+1(a>1)$;
(2) $f(a, 2,1)=a+1, f(a, m+2,1) \geqslant a$ $(m \geqslant 0)$;
$$
\text { (3) } \begin{array}{l}
f(a, m+1, n+1) \\
=f(a, m, f(a, m+1, n)) .
\end{array}
$$
(3) $f(a, m+1, n+1)$
Prove: $f(a, ... | Prove: When $m=1$ and $n$ is arbitrary, we have $f(a, 1, n)=n+1$.
Assume that when $m=k$ and $n$ is arbitrary, equation (1) holds.
To prove that when $m=k+1$ and $n$ is arbitrary, equation (1) also holds, we use induction on $n$.
(i) When $n=1$, we have
$$
f(a, k+1,1) \geqslant a>1 \text {. }
$$
(ii) Assume that when $... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,016 |
Example 6 Real numbers $a, b$ satisfy $ab=1$, and let
$$
M=\frac{1}{1+a}+\frac{1}{1+b}, N=\frac{a}{1+a}+\frac{b}{1+b}.
$$
Then the relationship between $M$ and $N$ is ( )(to avoid ambiguity, add $a \neq-1, b \neq-1$).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain
(1996, National Junior High School Mathematics Competitio... | The answer to this question is (B). Someone (see reference [1]) derived $M=N$ by taking special values of $a$ and $b$, which can only negate (A) and (C), but cannot negate (D). Choosing the correct answer is also "partially correct", with a logical flaw.
The author provides multiple solutions attached to the problem, ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,017 |
Example 7 Given $a \neq b$, and
$$
a^{2}-4 a+1=0, b^{2}-4 b+1=0 \text {. }
$$
Find the value of $\frac{1}{a+1}+\frac{1}{b+1}$. | Solution 1: Since
$$
\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a b+(a+b)+1},
$$
we only need to find the values of $a b$ and $a+b$.
From the given conditions, $a$ and $b$ are the roots of the quadratic equation $x^{2}-4 x+1=0$, so
$$
a+b=4, a b=1 \text {. }
$$
Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{(a+b)+2}{a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,018 |
Given $M, N, P$ are the midpoints of the sides $BC, CA, AB$ of $\triangle ABC$, and $M_{1}, N_{1}, P_{1}$ are on the sides of $\triangle ABC$, such that $M M_{1}, N N_{1}, P P_{1}$ bisect the perimeter of $\triangle ABC$. Prove:
(1) $M M_{1}, N N_{1}, P P_{1}$ intersect at the same point $K$;
(2) Among $\frac{K A}{B C}... | Prove: Let the centroid of $\triangle ABC$ be $G, BC=a, CA=b, AB=c, AM$ be the median of $\triangle ABC$ on side $BC$, as shown in Figure 1. Then we have
$$
\begin{array}{l}
G A+G B+G C \\
=0 .
\end{array}
$$
Also, $\boldsymbol{K A}=\boldsymbol{K} \boldsymbol{G}+\boldsymbol{G A}$,
$$
\begin{array}{l}
K B=K G+G B, \\
K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,019 |
Example $9 . \angle P O Q=30^{\circ}, A$ is a point on $O Q$, $B$ is a point on $O P$, and $O A=5, O B=12$. Take point $A_{1}$ on $O B$, and take point $A_{2}$ on $A Q$, let $l=A A_{1}+A_{1} A_{2}+A_{2} B$. Find the minimum value of $l$. | Solution: As shown in Figure 13, with the line $O P$ as the axis of symmetry, construct the axis-symmetric figure $\angle P O Q_{0}$ of $\angle P O Q$; with the line $O Q$ as the axis of symmetry, construct the axis-symmetric figure $\angle Q O P_{0}$ of $\angle Q O P$. At this point, the symmetric point of point $A$ a... | 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,020 |
Proposition Let $f$ be a bounded function from the set of real numbers $\mathbf{R}$ to itself, $T_{1}$ and $T_{2}$ be two positive numbers, and there exist positive integers $m$ and $n$, such that $m T_{1}=n T_{2}=T$. If for all $x \in \mathbf{R}$, we have
$$
f\left(x+T_{1}+T_{2}\right)+f(x)=f\left(x+T_{1}\right)+f\lef... | Prove: Equation (1) is
$$
f\left(x+T_{2}\right)-f(x)=f\left(x+T_{2}+T_{1}\right)-f\left(x+T_{1}\right) .
$$
Therefore, the function $\varphi(x)=f\left(x+T_{2}\right)-f(x)$ has a period of $T_{1}$.
Since $T=m T_{1}$, $\varphi(x)$ also has a period of $T$. Thus, we have $\varphi(x)=\varphi(x+T)$, i.e.,
$$
\begin{array}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,021 |
Question: As shown in Figure 1, in trapezoid $A B C D$, $A D / / B C$, squares $A B G E$ and $D C H F$ are constructed on the legs $A B$ and $C D$ respectively. Let the perpendicular bisector $l$ of line segment $A D$ intersect line segment $E F$ at point $M$. Prove that point $M$ is the midpoint of $E F$.
保留源文本的换行和格式... | This article provides three methods of proof.
Proof 1: As shown in Figure 2, draw $PA$ and $QD$ perpendicular to $AD$, intersecting $EF$ at $P$ and $Q$ respectively.
On $BC$, mark $BJ$
$$
= AP, KC = DQ \text{, }
$$
Connect $AJ$ and $DK$, then
$DQ \parallel MN \parallel PA$.
Since $AN = ND$, it follows that $PM = MQ$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,022 |
Proposition As shown in Figure $1, I$ is the incenter of $\triangle A B C$. Draw $A A_{1} \perp A I$ to intersect the extension of $B C$ at $A_{1}$, draw $B B_{1} \perp B I$ to intersect the extension of $C A$ at $B_{1}$, and draw $C C_{1} \perp C I$ to intersect the extension of $B A$ at $C_{1}$. Then $A_{1} 、 B_{1} 、... | Proof: As shown in Figure 1, construct the incircle of $\triangle ABC$ touching $BC$ at $A_{2}$, $AC$ at $B_{2}$, and $AB$ at $C_{2}$. Extend $A_{2}B_{2}$ to intersect $BA$ at $C_{3}$, extend $C_{2}B_{2}$ to intersect $BC$ at $A_{3}$, and extend $A_{2}C_{2}$ to intersect $CA$ at $B_{3}$.
It is easy to see that $A A_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,023 |
2. Given the lengths of the four sides of a quadrilateral are $m$, $n$, $p$, and $q$, and they satisfy $m^{2}+n^{2}+p^{2}+q^{2}=2 m n+2 p q$. Then this quadrilateral is ( ).
(A) parallelogram
(B) quadrilateral with perpendicular diagonals
(C) parallelogram or quadrilateral with perpendicular diagonals
(D) quadrilateral... | 2.C.
From the given, we have
$$
(m-n)^{2}+(p-q)^{2}=0 .
$$
Therefore, $m=n, p=q$.
If $m, n$ are one pair of opposite sides of a quadrilateral, then $p, q$ are its other pair of opposite sides, making this quadrilateral a parallelogram; if $m, n$ are one pair of adjacent sides of the quadrilateral, then $p, q$ are als... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,025 |
4. If $x=123456789 \times 123456786, y=$ $123456788 \times 123456787$, then the size relationship between $x$ and $y$ is ( ).
(A) $x=y$
(B) $x<y$
(C) $x>y$
(D) Uncertain | 4.B.
Let $a=123456$ 789, then
$$
\begin{array}{l}
x=a(a-3)=a^{2}-3 a, \\
y=(a-1)(a-2)=a^{2}-3 a+2=x+2 .
\end{array}
$$
Therefore, $x<y$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,027 |
5. As shown in Figure 3, quadrilateral $ABCD$ is a rhombus, $\triangle AEF$ is an equilateral triangle, points $E$ and $F$ are on sides $BC$ and $CD$ respectively, and $AB = AE$. Then $\angle B$ equals ( ).
(A) $60^{\circ}$
(B) $80^{\circ}$
(C) $100^{\circ}$
(D) $120^{\circ}$ | 5.B.
Since quadrilateral $A B C D$ is a rhombus, $A B=A D, \angle B=\angle D$.
Also, since $\triangle A E F$ is an equilateral triangle, then $A E=A F$. Therefore, $\triangle A B E \cong \triangle A D F$. Hence $B E=D F$.
Since $B C=C D$, then $C E=C F$. Therefore, $\angle C E F=\frac{1}{2}\left(180^{\circ}-\angle C\r... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,028 |
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