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$$
\begin{array}{l}
\text { Three. (50 points) Given } \\
a_{n}=\sum_{i=0}^{n-1} 10^{i}(n=1,2, \cdots) .
\end{array}
$$
Prove: there exist infinitely many positive integers $n$, such that the remainders of $a_{1}, a_{2}$, $\cdots, a_{n}$ when divided by $n$ are all distinct. | From the given, we have $a_{n} \sum_{i=0}^{n-1} 10^{i}=\underbrace{11 \cdots 1}_{n \uparrow}$.
We will prove that $n=3^{m} (m=1,2, \cdots)$ satisfies the condition.
First, we prove that $3^{m} \| a_{3 m} (m=1,2, \cdots)$.
We use mathematical induction on $m$.
When $m=1$, $a_{3}=111=3 \times 37$, so $3 \| a_{3}$.
Assume... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,722 |
Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, fin... | Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$
Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.
We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from th... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,723 |
As shown in Figure 1, three circles with radius $r$ are externally tangent to each other and internally tangent to $\odot O$ at points $A$, $B$, and $C$. The tangents to $\odot O$ at $A$ and $B$ intersect at point $P$. Find the area of the shaded region formed by the two tangents and the arc between the points of tange... | Solve As shown in Figure 1, connect $O_{1} O_{2}$, $O_{2} O_{3}$, $O_{1} O_{3}$, $O A$, $O B$, $O P$, and let $O P$ intersect $\odot O$ at point $D$.
Since $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ are pairwise externally tangent, $\triangle O_{1} O_{2} O_{3}$ is an equilateral triangle with side length $2r$, an... | \left[\frac{12+7 \sqrt{3}}{3}-\frac{(7+4 \sqrt{3}) \pi}{9}\right] r^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,724 |
As shown in Figure 2, given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $CD$ is the external common tangent of the two circles, with points of tangency at $C$ and $D$. Line $BC$ intersects $AD$ and $BD$ intersects $AC$ at points $E$ and $F$ respectively. Prove that $\frac{EA}{AF}=\frac{FB}{BE}... | Connect $A B$ and extend it to intersect $C D$ at point $M$, connect $E F$ to intersect $A B$ at point $N$.
By the secant-tangent theorem, we have
$$
M C^{2}=M B \cdot M A=M D^{2} \Rightarrow M C=M D \text {. }
$$
By the inscribed angle theorem, we have
$$
\angle M C B=\angle B A C, \angle M D B=\angle B A D \text {. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,725 |
Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$
If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\... | The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contrad... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,726 |
Given $P$ is a convex $n$-sided polygon inscribed in a given circle, divide $P$ arbitrarily into $n-2$ triangles using $n-3$ non-intersecting diagonals within $P$. Prove: The sum of the radii of the incircles of these $n-2$ triangles does not depend on the division method (i.e., for any division method, the sum of the ... | Proof of the lemma first.
Lemma For $\triangle ABC$, let the radii of its circumcircle and incircle be $R$ and $r$ respectively. Then
$$
\frac{r}{R}=\cos A+\cos B+\cos C-1.
$$
Proof In fact,
$$
\begin{array}{l}
\frac{r}{R}=\frac{2 S}{(a+b+c) R}=\frac{2 \cdot \frac{a b c}{4 R}}{(a+b+c) R} \\
=\frac{a b c}{2(a+b+c) R^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,727 |
Example 2: Using $1,2,3,4,$ and $5$, how many $n$-digit numbers can be formed where each pair of adjacent digits differ by exactly 1?
untranslated text:
例 2 用 $1,2,3,4 、 5$ 可以构成多少个各相邻数字恰好相差 1 的 $n$ 位数?
translated text:
Example 2: Using $1,2,3,4,$ and $5$, how many $n$-digit numbers can be formed where each pair of ... | Let the number of numbers that satisfy the condition be $a_{n}$.
Let the numbers that start with $1,2,3,4,5$ and satisfy the condition be $y_{n}, z_{n}, u_{n}, v_{n}, w_{n}$ respectively. Then,
$$
a_{n}=y_{n}+z_{n}+u_{n}+v_{n}+w_{n},
$$
and $y_{n}=w_{n}, z_{n}=v_{n}$.
When starting with $1$ or $2$, we have $y_{n}=z_{n... | a_{2 n}=8 \times 3^{n-1}, \quad a_{2 n+1}=14 \times 3^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,728 |
Example 3 Let $a_{i}=0,1(i=1,2, \cdots, n)$. Try to find the number of sequences $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy
$$
a_{1} \leqslant a_{2}, a_{2} \geqslant a_{3}, a_{3} \leqslant a_{4}, a_{4} \geqslant a_{5}, \cdots \cdots .
$$ | Let $f_{n}$ be the number of sequences that satisfy the condition. Then $f_{1}=2, f_{2}=3, f_{3}=5$.
Let $b_{n}$ be the number of sequences that satisfy the condition when $a_{n}=0$; and $c_{n}$ be the number of sequences that satisfy the condition when $a_{n}=1$.
Then $f_{n}=b_{n}+c_{n}$.
It is easy to see that,
when ... | f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,729 |
Example 4 In a $2 \times n$ grid, color the $2 n$ cells with red and white, each cell being colored with only one color. If it is required that no two adjacent cells are both colored red, how many different coloring methods are there? | Let the number of coloring methods that satisfy the conditions be $a_{n}$.
Let the number of coloring methods where the first column of the $2 \times n$ grid is colored red on top and white on bottom be $b_{n}$.
In $a_{n}$, the first column can be both white or one red and one white, thus, we have $a_{n}=a_{n-1}+2 b_{... | \frac{1}{2}\left[(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,730 |
For example, a $53 \times 2 n$ chessboard can be covered with $3 n$ rectangular pieces of size $1 \times 2$. Question: How many ways are there to completely cover the chessboard? | Let the number of covering methods that satisfy the conditions be $a_{n}$ (each two columns form a group, for a total of $n$ groups).
As shown in Figure 1, in a $3 \times 2n$ grid, when the first two columns are covered with three $1 \times 2$ rectangular pieces placed horizontally, the number of covering methods that... | a_{n}=\frac{\sqrt{3}+1}{2 \sqrt{3}}(2+\sqrt{3})^{n}+\frac{\sqrt{3}-1}{2 \sqrt{3}}(2-\sqrt{3})^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,731 |
Example 6 As shown in Figure 2, let $A$ and $E$ be opposite vertices of a regular octagon. A frog starts jumping from vertex $A$. Except for vertex $E$, the frog can jump to any of the two adjacent vertices from any vertex of the octagon. When it jumps to vertex $E$, it stops there. Let the number of ways the frog can ... | Solution: Clearly,
$$
a_{1}=a_{2}=a_{3}=a_{2 m-1}=0\left(m \in \mathbf{N}_{+}\right), a_{4}=2 .
$$
Let the number of ways to jump from vertex $C$ to $E$ in $n$ steps be $b_{n}$. Then $b_{2}=1$.
For $a_{n}$, starting from vertex $A$ and jumping two steps can be divided into:
(1) Reaching point $C$ or $G$;
(2) Jumping f... | a_{2 m}=\frac{1}{\sqrt{2}}(2+\sqrt{2})^{m-1}-\frac{1}{\sqrt{2}}(2-\sqrt{2})^{m-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,732 |
Example 7 A password lock's password setting involves assigning one of the two numbers, 0 or 1, to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex with one of two colors, red or blue, such that for any two adjacent vertices, at least one of the number or color is the same... | Let the total number of settings that satisfy the conditions be $a_{n}$.
Let the number of settings where the values and colors at points $A_{1}$ and $A_{2}$ are completely the same be $b_{n}$; the number of settings where only one of the value or color at points $A_{1}$ and $A_{2}$ is the same be $c_{n}$; and the numb... | a_{n}=\left\{\begin{array}{ll}3^{n}+1, & n=2 k+1 \text {; } \\ 3^{n}+3, & n=2 k .\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,733 |
Example 8 A rectangular building consists of two rows of square rooms, with $n$ rooms in each row (like a $2 \times n$ chessboard). Each room has three doors leading to 1, 2, or 3 adjacent rooms (doors facing outside the building are not counted). If one can travel from any room to any other room through these doors. H... | Let $a_{n}$ be the number of ways to satisfy the conditions for a $2 \times n$ rectangular building.
Let $b_{n}$ be the number of ways where the building is not fully accessible when the middle wall of the $n$-th column has no door, but becomes fully accessible when the middle wall has a door.
Then for $a_{n}$, it can... | a_{n}=\frac{1}{\sqrt{17}}\left[\left(\frac{5+\sqrt{17}}{2}\right)^{n}-\left(\frac{5-\sqrt{17}}{2}\right)^{n}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,734 |
1. Find the number of sequences that satisfy the following conditions: the number of terms is $n$, each term is 0 or 1 or 2, and 0 cannot be the preceding term or the following term of 2. | Let the number of sequences that satisfy the condition be $f_{n}$. Let the number of sequences ending with 0 be $a_{n}$, the number of sequences ending with 1 be $b_{n}$, and the number of sequences ending with 2 be $c_{n}$. Then
$$
f_{n}=a_{n}+b_{n}+c_{n} .
$$
It is easy to see that $a_{n}=a_{n-1}+b_{n-1}$,
$$
\begin... | \frac{1}{2}\left[(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,735 |
Example 3 Given that $AB$ is the diameter of semicircle $\odot O$, two chords $AC$ and $BD$ intersect at point $E$, and the tangents to the semicircle at points $C$ and $D$ intersect at point $P$. Prove: $PE \perp AB$.
---
The translation maintains the original text's format and line breaks as requested. | 【Analysis】As shown in Figure 5, extend $P E$ to intersect $A B$ at point $F$, and connect $O C$, $O D$, $C D$, and $B C$.
To prove $P E \perp A B$, it suffices to prove
$$
\angle A + \angle A E F = 90^{\circ} \text{.}
$$
Notice that
$$
\begin{array}{l}
\angle A = \angle A C O, \angle A E F = \angle P E C, \\
\angle A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,736 |
2. Color some cells in a $2 \times n$ strip of cells so that no $2 \times 2$ square is completely colored (at most 3 cells are colored). Let the number of different colorings that satisfy this condition be $P_{n}$. Prove that $P_{1989}$ is divisible by 3, and find the highest power of 3 that divides $P_{1989}$. [2] | Let $a_{n}$ be the number of ways that satisfy the condition and the last column has both cells colored; and $b_{n}$ be the number of ways that the last column has at most one cell colored (in three ways). Then
$$
P_{n}=a_{n}+b_{n} \text {. }
$$
It is easy to see that $a_{n}=b_{n-1}, b_{n}=3\left(a_{n-1}+b_{n-1}\right... | 3^{944} | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,737 |
Question: For the set of positive integers $M=\{1,2, \cdots, 99\}$, color the elements with three colors: red, blue, and green, such that each color has 33 numbers, which is called a "three-color partition" of set $M$. For the following three propositions, please provide a proof or refute them:
(1) There exists a three... | ```
Note that if positive integers }a,b,c\mathrm{ form a Pythagorean
c=k(m
where , , b can be expressed as }k(\mp@subsup{m}{}{2}-\mp@subsup{n}{}{2})、k\cdot2mn(
\in\mp@subsup{\mathbf{Z}}{*}{},m,n\mathrm{ are coprime, one odd and one even, }m>n>0).
Considering the structure of }c\mathrm{ from largest to smal... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,738 |
Example 1 Let $n$ be a positive integer. There is a rectangle $A B C D$ with side lengths $A B=90 n+1, B C=90 n+5$. Using horizontal and vertical lines, the rectangle is divided into $(90 n+1) \times(90 n+5)$ unit squares, and $S$ is the set of all vertices of these unit squares. Prove: The number of lines passing thro... | Prove first a lemma.
Lemma Given positive odd numbers $m, m^{\prime}\left(m\frac{m}{2} \text { or } q>\frac{m^{\prime}}{2} .
\end{array}\right.
$
(3) The slope of the line is
$$
-\frac{p}{q}\left((p, q)=1,1 \leqslant p \leqslant m, 1 \leqslant q \leqslant m^{\prime}\right)
$$
The slope of the line symmetric to the lin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,739 |
Example 2 Let $n, k (n \geqslant k \geqslant 1)$ be positive integers. There are $n$ lamps placed on a circle, all of which are initially off. Each time, you can change the on or off state of any $k$ consecutive lamps. For the following three cases:
(1) $k$ is an odd prime,
(2) $k$ is an odd number,
(3) $k$ is an even ... | Consider $n$ lamps arranged in a circle, numbered as 1, 2, ..., $n$.
In a state that can be achieved through several operations, assign a value $a_{i} \in \{0,1\}$ to the $i$-th lamp ($i=1,2, \cdots, n$), where 0 indicates the lamp is off and 1 indicates the lamp is on.
Assign a value $b_{i} (i \in \{1,2, \cdots, n\})... | 2^{n-d}, & k_{0} \text{ is even; } \\ 2^{n-d+1}, & k_{0} \text{ is odd, } | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,740 |
410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two t... | Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,742 |
Example 5 A strip composed of $n$ squares, these $n$ squares are sequentially numbered as $1,2, \cdots, n$. Initially, one square is empty, while each of the other squares contains a piece. When a square contains a piece, and the two adjacent squares have one with a piece and the other without a piece, the second piece... | To facilitate the description, the conditions and operations of the problem are redefined as follows, without changing their essence.
Let the situation where the $t$-th (1 ≤ t ≤ n) square is empty and the rest of the squares in a strip of length $n (n \geq 3)$ are occupied by chess pieces be denoted as $(n, t)$.
Simi... | 2, 5, 2004, 2007 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,743 |
Example 6 Given a convex quadrilateral $A B C D$. Prove: There exists a point $P$ inside the quadrilateral $A B C D$ such that
$$
\begin{array}{l}
\angle P A B+\angle P D C=\angle P B C+\angle P A D \\
=\angle P C D+\angle P B A \\
=\angle P D A+\angle P C B=90^{\circ}
\end{array}
$$
if and only if the diagonals $A C$ ... | Proof As shown in Figure 3, first prove:
$$
\begin{array}{l}
\angle P A B + \angle P D C = 90^{\circ} \\
\Leftrightarrow \angle A P D = \angle A B C + \angle B C D - 90^{\circ} \text{.} \\
\angle A P D = 180^{\circ} - (\angle P A D + \angle P D A) \\
= 360^{\circ} - (\angle P A D + \angle P D A) - 180^{\circ} \\
= \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,744 |
Example 7 Given that $B E, C F$ are altitudes of acute $\triangle A B C$, and two circles passing through points $A, F$ are tangent to line $B C$ at points $P, Q$ respectively, and point $B$ is between $C, Q$. Prove: The intersection of $P E, Q F$ lies on the circumcircle of $\angle A E F$.
---
The translation mainta... | As shown in Figure 4, let $A D \perp B C$,
connect $A Q$.
From the given conditions, we have
$$
\begin{array}{l}
B Q^{2}=B F \cdot B A=B P^{2} \\
\Rightarrow \triangle A Q B \backsim \triangle Q F B \\
\Rightarrow \angle A Q B=\angle Q F B \\
=\angle A F S .
\end{array}
$$
Then $A, S, E, F$ are concyclic
$$
\begin{arr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,745 |
Let $x_{1}, x_{2}, x_{3}$ be positive numbers, and $x_{1} x_{2} x_{3}=1$. Prove:
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)\left(x_{3}+1\right) \geqslant 8.
$$
This inequality can also lead to the following two generalizations.
Generalization 1 If $x_{1}, x_{2}, \cdots, x_{n}>0(n \geqslant 2)$, and $n$ is a positive ... | Let $\prod_{i=1}^{n} x_{i}=S$, and $\sum_{1 \in i \in n} \prod_{i}^{k} x_{i}$ denote the sum of the products of all $k$ distinct elements from $x_{1}, x_{2}, \cdots, x_{n}$. It is easy to see that there are $\mathrm{C}_{n}^{k}$ terms in the sum.
$$
\text { For example, } \sum \prod_{1 \leqslant i \leqslant n}^{2} x_{i}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,746 |
Example 4 As shown in Figure 6, in $\triangle A B C$, it is known that $\angle B A C$ $=\angle B C A=44^{\circ}, M$ is a point inside $\triangle A B C$, such that $\angle M C A=30^{\circ}$, $\angle M A C=16^{\circ}$. Find the degree measure of $\angle B M C$. | Solve as shown in Figure 6, construct a regular $\triangle ABD$ with $AB$ as a side, and connect $CD$.
From the problem, we have $BA = BC$, $\angle ABC = 92^{\circ}$.
Thus, $BA = BC = BD$.
According to Theorem 1, $B$ is the circumcenter of $\angle ACD$.
By Property 2(1), we know
$$
\begin{array}{l}
\angle DCA = \frac{1... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,747 |
Example 1 Let $a, b$ be positive constants, $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $n \geqslant 2$ be a positive integer. Find:
$$
y=\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}
$$
the maximum value. | Notice
$$
\begin{array}{c}
y=\frac{\frac{1}{b}}{\left(\frac{a}{x_{1}}+1\right)\left(\frac{x_{1}}{x_{2}}+1\right) \cdots\left(\frac{x_{n-1}}{x_{n}}+1\right)\left(\frac{x_{n}}{b}+1\right)} . \\
\text { Let } a_{1}=\frac{a}{x_{1}}, a_{2}=\frac{x_{1}}{x_{2}}, \cdots \cdots a_{n}=\frac{x_{n-1}}{x_{n}},
\end{array}
$$
$a_{n+... | \frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,748 |
Example 2 Given $2 n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}$,
$$
\begin{array}{l}
b_{2}, \cdots b_{n} \text { and } \\
\quad\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right)=10 .
\end{array}
$$
Find the maximum value of $y=\sqrt[n]{a_{1} a_{2} \cdots a_{n}}+\sqrt[n]{b_{1} ... | From Corollary 2, we know
$$
\begin{array}{l}
\sqrt[n]{a_{1} a_{2} \cdots a_{n}}+\sqrt[n]{b_{1} b_{2} \cdots b_{n}} \\
\leqslant \sqrt[n]{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right)} \\
=\sqrt[n]{10} .
\end{array}
$$
Equality holds, and $y$ attains its maximum value of $\sqrt[n]{10}... | \sqrt[n]{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,749 |
1. A. Let $a=\sqrt{7}-1$. Then the value of the algebraic expression $3 a^{3}+12 a^{2}-$ $6 a-12$ is ( ).
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$ | ,- 1. A. A.
Notice
$$
a=\sqrt{7}-1 \Rightarrow a^{2}+2 a-6=0 \text {. }
$$
Then $3 a^{3}+12 a^{2}-6 a-12$
$$
=\left(a^{2}+2 a-6\right)(3 a+6)+24=24 .
$$ | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,750 |
1. B. Let $x=\frac{\sqrt{5}-3}{2}$. Then the value of the algebraic expression
$$
x(x+1)(x+2)(x+3)
$$
is ( ).
(A) 0
(B) 1
(C) -1
(D) 2 | 1. B. C.
From the given, we have $x^{2}+3 x+1=0$. Therefore,
$$
\begin{array}{l}
x(x+1)(x+2)(x+3) \\
=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right) \\
=\left(x^{2}+3 x+1\right)^{2}-1=-1 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,751 |
2. A. For any real numbers $a, b, c, d$, define the operation “ $\otimes$ ” between the ordered pairs of real numbers $(a, b)$ and $(c, d)$ as:
$$
(a, b) \otimes(c, d)=(a c+b d, a d+b c) \text {. }
$$
If for any real numbers $u, v$, we have
$$
(u, v) \otimes(x, y)=(u, v) \text {, }
$$
then $(x, y)$ is ( ).
(A) $(0,1)... | 2. A. B.
According to the defined operation rules, we have
$$
\left\{\begin{array} { l }
{ u x + v y = u , } \\
{ v x + u y = v }
\end{array} \Rightarrow \left\{\begin{array}{l}
u(x-1)+v y=0, \\
v(x-1)+u y=0
\end{array}\right.\right.
$$
This holds for any real numbers $u, v$.
By the arbitrariness of real numbers $u,... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,752 |
2. B. Given that $x, y, z$ are real numbers, and satisfy
$$
x+2 y-5 z=3, x-2 y-z=-5 \text {. }
$$
Then the minimum value of $x^{2}+y^{2}+z^{2}$ is ().
(A) $\frac{1}{11}$
(B) 0
(C) 5
(D) $\frac{54}{11}$ | 2. B. D.
From the given, we have $\left\{\begin{array}{l}x=3 z-1 \\ y=z+2\end{array}\right.$, thus, $x^{2}+y^{2}+z^{2}=11 z^{2}-2 z+5$. Therefore, when $z=\frac{1}{11}$, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\frac{54}{11}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,753 |
3. A. If $x>1, y>0$, and satisfy
$$
x y=x^{y}, \frac{x}{y}=x^{3 y} \text {, }
$$
then the value of $x+y$ is ( ).
(A) 1
(B) 2
(C) $\frac{9}{2}$
(D) $\frac{11}{2}$ | 3. A. C.
From the given equations, multiplying them yields
$$
x^{2}=x^{4 y} \Rightarrow y=\frac{1}{2} \text {. }
$$
Furthermore, $x=4$.
Therefore, $x+y=\frac{9}{2}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,754 |
3. B. Given that $\angle A, \angle B$ are two acute angles, and satisfy
$$
\begin{array}{l}
\sin ^{2} A+\cos ^{2} B=\frac{5}{4} t, \\
\cos ^{2} A+\sin ^{2} B=\frac{3}{4} t^{2} .
\end{array}
$$
Then the sum of all possible real values of $t$ is ().
(A) $-\frac{8}{3}$
(B) $-\frac{5}{3}$
(C) 1
(D) $\frac{11}{3}$ | 3. B. C.
Adding the two equations yields $3 t^{2}+5 t=8$.
Solving gives $t=1, t=-\frac{8}{3}$ (discard).
When $t=1$, $\angle A=45^{\circ}, \angle B=30^{\circ}$ satisfies the given equations.
Therefore, the sum of all possible values of the real number $t$ is 1. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,755 |
4. A. Given points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$, respectively, and $BE$ intersects $CD$ at point $F$. Let
$S_{\text{quadrilateral EADF}}=S_{1}, S_{\triangle BDF}=S_{2}$,
$S_{\triangle BCF}=S_{3}, S_{\triangle CEF}=S_{4}$.
Then the relationship between $S_{1} S_{3}$ and $S_{2} S_{4}$ is ... | 4. A. C.
As shown in Figure 8, connect $D E$.
Let $S_{\triangle D E F}=S_{1}^{\prime}$. Then,
$$
\frac{S_{1}^{\prime}}{S_{2}}=\frac{E F}{B F}=\frac{S_{4}}{S_{3}} \text{. }
$$
Thus, $S_{1}^{\prime} S_{3}=S_{2} S_{4}$.
Since $S_{1}>S_{1}^{\prime}$, we have,
$$
S_{1} S_{3}>S_{2} S_{4} .
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,756 |
4. B. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfying
$$
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{10} \text {, }
$$
and no three of these numbers can be the lengths of the sides of a triangle. Then the minimum value of $\frac{a_{10}}{a_{1}}$ is ( ).
(A) 34
(B) 55
(C) 89
(D) 144 | 4. B. B.
Notice
$$
\begin{array}{l}
a_{10} \geqslant a_{9}+a_{8} \geqslant 2 a_{8}+a_{7} \geqslant 3 a_{7}+2 a_{6} \\
\geqslant 5 a_{6}+3 a_{5} \geqslant 8 a_{5}+5 a_{4} \geqslant \cdots \\
\geqslant 34 a_{2}+21 a_{1} \geqslant 55 a_{1} . \\
\text { Therefore, } \frac{a_{10}}{a_{1}} \geqslant 55 .
\end{array}
$$
When... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,757 |
Example 5 Let $P$ be a point inside $\triangle A B C$ such that $\angle P B C = 30^{\circ}, \angle P B A = 8^{\circ}$, and $\angle P A B = \angle P A C = 22^{\circ}$. Question: What is the measure of $\angle A P C$ in degrees? | Solve as shown in Figure 7, take
the circumcenter $D$ of $\triangle P B C$, and
connect $D P, D C, D B$.
Then $D P=D C=D B$,
$\angle P D C$
$=2 \angle P B C=60^{\circ}$.
Therefore, $\triangle D P C$ is an equilateral triangle.
Thus, $P C=D P=D B, \angle D P C=60^{\circ}$.
It is easy to see that $A B>A C$.
Therefore, we... | 142^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,758 |
5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 5. A. A.
When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$
Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$
Therefore, $... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,759 |
6. A. If the equation with respect to $x$
$$
(x-2)\left(x^{2}-4 x+m\right)=0
$$
has three roots, and these three roots can exactly serve as the three side lengths of a triangle, then the range of values for $m$ is | 2,6. A. $30)$. Then $x_{1}+x_{2}=4, x_{1} x_{2}=m$.
From the given condition we know
$$
\begin{array}{l}
0 \leqslant x_{1}-x_{2}<2 \Rightarrow 0 \leqslant 16-4 m<4 \\
\Rightarrow 3<m \leqslant 4 .
\end{array}
$$ | 3<m \leqslant 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,761 |
6. B. Given that the lengths of the two legs are integers $a$ and $b$ $(b<2011)$. Then the number of right triangles with the hypotenuse length $b+1$ is | 6. B. 31.
By the Pythagorean theorem, we have
$$
a^{2}=(b+1)^{2}-b^{2}=2 b+1 \text{. }
$$
Given $b<2011$, we know that $a$ is an odd number in the interval $(1, \sqrt{4023})$, so $a$ must be $3, 5, \cdots, 63$.
Therefore, there are 31 right-angled triangles that satisfy the conditions. | 31 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,762 |
7. A. A fair cube die has the numbers $1,2,2,3,3,4$ on its six faces; another fair cube die has the numbers $1,3, 4, 5, 6, 8$ on its six faces. When these two dice are rolled simultaneously, the probability that the sum of the numbers on the top faces is 5 is | 7. A. $\frac{1}{9}$.
Among the 36 possible outcomes, there are 4 pairs
$$
(1,4) 、(2,3) 、(2,3) 、(4,1)
$$
that sum to 5. Therefore, the probability that the sum of the numbers on the top faces is 5 is $\frac{4}{36}=\frac{1}{9}$. | \frac{1}{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,763 |
7. B. A fair cube die has the numbers $1,2,2,3,3,4$ on its six faces; another fair cube die has the numbers $1, 3, 4, 5, 6, 8$ on its six faces. When these two dice are rolled simultaneously, the probability that the sum of the numbers on the top faces is 7 is | 7. B. $\frac{1}{6}$.
As above, the probability that the sum of the numbers on the top faces is 7 is $\frac{6}{36}=\frac{1}{6}$. | \frac{1}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,764 |
8. A. As shown in Figure 1, points $A$ and $B$ lie on the line $y=x$. Parallel lines to the $y$-axis through $A$ and $B$ intersect the hyperbola $y=\frac{1}{x} (x>0)$ at points $C$ and $D$. If $BD=2AC$, then the value of $4OC^2-OD^2$ is $\qquad$ | 8. A. 6 .
Let points $C(a, b), D(c, d)$.
Then points $A(a, a), B(c, c)$.
Since points $C, D$ are on the hyperbola $y=\frac{1}{x}$, we have $a b=1, c d=1$.
Given $B D=2 A C$
$$
\begin{array}{l}
\Rightarrow|c-d|=2|a-b| \\
\Rightarrow c^{2}-2 c d+d^{2}=4\left(a^{2}-2 a b+b^{2}\right) \\
\Rightarrow 4\left(a^{2}+b^{2}\rig... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,765 |
8. B. As shown in Figure 2, given the hyperbola $y=\frac{2}{x}(x>0)$ intersects the sides $C B$ and $B A$ of rectangle $O A B C$ at points $E$ and $F$ respectively, and $A F=B F$, connect $E F$. Then the area of $\triangle O E F$ is $\qquad$ | 8. B. $\frac{3}{2}$.
Let point $B(a, b)$. Then point $F\left(a, \frac{b}{2}\right) 、 E\left(\frac{2}{b}, b\right)$.
Since point $F$ is on the hyperbola $y=\frac{2}{x}$, we have $a b=4$.
Therefore, $S_{\triangle O E F}=S_{\text {rectangle } OFBC }-S_{\triangle O E C}-S_{\triangle F B E}$ $=\frac{1}{2}\left(\frac{b}{2}+... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,766 |
9. A. If $y=\sqrt{1-x}+\sqrt{x-\frac{1}{2}}$ has a maximum value of $a$ and a minimum value of $b$, then the value of $a^{2}+b^{2}$ is $\qquad$. | 9. A. $\frac{3}{2}$.
From $1-x \geqslant 0$ and $x-\frac{1}{2} \geqslant 0$, we get
$$
\frac{1}{2} \leqslant x \leqslant 1 \text {. }
$$
Then $y^{2}=\frac{1}{2}+2 \sqrt{-x^{2}+\frac{3}{2} x-\frac{1}{2}}$ $=\frac{1}{2}+2 \sqrt{-\left(x-\frac{3}{4}\right)^{2}+\frac{1}{16}}$.
Since $\frac{1}{2}<\frac{3}{4}<1$, therefore... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,767 |
Example 6 In $\triangle A B C$, it is known that $A C=B C, \angle C=$ $20^{\circ}, D 、 E$ are points on sides $B C 、 A C$ respectively. If $\angle C A D$ $=20^{\circ}, \angle C B E=30^{\circ}$, find the degree measure of $\angle A D E$. | Solve as shown in Figure 8, take point $F$ on $BC$ such that $AF=AB$.
It is easy to know that $\angle ABC=80^{\circ}$.
Notice that
$$
\begin{array}{l}
\angle AEB \\
=\angle ACB + \angle EBC \\
=50^{\circ} \\
=\angle ABC - \angle EBC \\
=\angle ABE. \\
\text{Therefore, } AB=AE.
\end{array}
$$
Thus, $A$ is the circumcen... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,769 |
10. A. As shown in Figure 3, in the right triangle $\triangle ABC$, the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$ | 10. A. 84.
Let $BC = a, AC = b$. Then,
$$
a^{2} + b^{2} = 35^{2} = 1225.
$$
Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$, we have,
$$
\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.
$$
Thus, $12(a + b) = ab$.
From equations (1) and (2), we get
$$
\begin{array}{l}
(a + b)^{2} = a^{2}... | 84 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,770 |
10. B. Let the four-digit number
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$
Then the number of such four-digit numbers is $\qquad$ | 10. B. 5 .
From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,771 |
11. A. Given the quadratic equation in $x$
$$
x^{2}+c x+a=0
$$
has two integer roots that are each 1 more than the roots of the equation
$$
x^{2}+a x+b=0
$$
Find the value of $a+b+c$. | Three, 11. A. Let the equation $x^{2}+a x+b=0$ have two roots $\alpha, \beta (\alpha, \beta$ are integers, and $\alpha \leqslant \beta)$. Then the roots of the equation $x^{2}+\alpha x+a=0$ are $\alpha+1, \beta+1$. From the problem, we have $\alpha+\beta=-a, (\alpha+1)(\beta+1)=a$.
Adding the two equations gives $\alph... | -3 \text{ or } 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,772 |
11. B. Given that for any real number $x$, the quadratic function $y=a x^{2}+b x+c$ satisfies
$$
x^{2}+2 x+2 \leqslant y \leqslant 2 x^{2}+4 x+3 \text {, }
$$
and when $x=9$, $y=121$. Find the value of $a+b+c$. | 11. B. From the given conditions, we have
$$
(x+1)^{2}+1 \leqslant y \leqslant 2(x+1)^{2}+1 \text {. }
$$
Then \( y=a(x+1)^{2}+1 \) where \( 1 \leqslant a \leqslant 2 \).
When \( x=9 \), \( y=121 \), we get
$$
100 a+1=121 \Rightarrow a=\frac{6}{5} \text {. }
$$
Thus, \( y=\frac{6}{5}(x+1)^{2}+1=\frac{6}{5} x^{2}+\fra... | \frac{29}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,773 |
12. A. As shown in Figure 4, given that point $H$ is the orthocenter of $\triangle A B C$, the circle $\odot O_{1}$ with $A B$ as its diameter intersects the circumcircle $\odot O_{2}$ of $\triangle B C H$ at point $D$. Extending $A D$ intersects $C H$ at point $P$. Prove: $P$ is the midpoint of $C H$. | 12. A. As shown in Figure 9, extend $A P$ to intersect $\odot O_{2}$ at point $Q$, and connect $A H, B D, Q B, Q C, Q H$.
From $A B$ being the diameter of $\odot O_{1}$, we know
$\angle A D B = \angle B D Q = 90^{\circ}$.
Therefore, $B Q$ is the diameter of $\odot O_{2}$.
Thus, $C Q \perp B C, B H \perp H Q$.
Since $H$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,774 |
12. B. As shown in Figure 5, given that the side length of square $A B C D$ is $1$, and $P, Q$ are two points inside it, and
$$
\angle P A Q=\angle P C Q=45^{\circ} \text {. }
$$
Find the value of $S_{\triangle P A B}+S_{\triangle P C Q}+S_{\triangle Q A D}$. | 12. B. As shown in Figure 10, rotate $\triangle A Q D$ 90° clockwise around point $A$ to $\triangle A Q^{\prime} B$, and rotate $\triangle C Q D$ 90° counterclockwise around point $C$ to $\triangle C Q^{\prime \prime} B$. Connect $P Q^{\prime}$ and $P Q^{\prime \prime}$. Then
$$
\begin{array}{l}
\triangle A P Q^{\prime... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,775 |
13. A. As shown in Figure 6, point $A$ is on the positive $y$-axis, and points $A$ and $B$ are symmetric with respect to the $x$-axis. A line passing through $A$ intersects the parabola $y=\frac{2}{3} x^{2}$ at points $P$ and $Q$.
(1) Prove:
$$
\angle A B P=\angle A B Q \text {; }
$$
(2) If point $A(0,1)$, and $\angle ... | 13. A. (1) Draw perpendiculars from points $P$ and $Q$ to the $y$-axis, with feet at $C$ and $D$ respectively.
Let point $A(0, t)$. Then point $B(0, -t)$.
Let the equation of line $PQ$ be $y = kx + t$, and let $P(x_P, y_P)$ and $Q(x_Q, y_Q)$ with $x_P < 0 < x_Q$.
From $\left\{\begin{array}{l}y = kx + t, \\ y = \frac{2}... | y = -\frac{\sqrt{3}}{3} x + 1 \text{ or } y = \frac{\sqrt{3}}{3} x + 1 | Geometry | proof | Yes | Yes | cn_contest | false | 723,776 |
13. B. If five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are randomly selected from $1,2, \cdots, n$, and one of these integers is always a prime number, find the maximum value of $n$. | 13. B. When $n \geqslant 49$, take the integers $1, 2^{2}, 3^{2}, 5^{2}, 7^{2}$. These five integers are five pairwise coprime distinct integers, but none of them are prime.
When $n=48$, in the integers $1, 2, \cdots, 48$, take any five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. If $a_{1},... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,777 |
14. A. As shown in Figure 7, in $\triangle ABC$, it is given that $\angle BAC=60^{\circ}, AB=$ $2AC$, point $P$ is inside $\triangle ABC$, and $PA=\sqrt{3}, PB=$ $5, PC=2$. Find the area of $\triangle ABC$. | 14. A. As shown in Figure 11, construct
$\triangle A B Q$ such that
$$
\begin{array}{l}
\angle Q A B=\angle P A C, \\
\angle A B P=\angle A C P .
\end{array}
$$
Then $\triangle A B Q$
$$
\backsim \triangle A C P \text {. }
$$
Since $A B=2 A C$, the similarity ratio is 2.
Thus, $A Q=2 A P=2 \sqrt{3}, B Q=2 C P=4$.
The... | \frac{6+7\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,778 |
14. B. Given $a_{i}>0(i=1,2, \cdots, 2011)$, and $a_{1}<a_{2}<\cdots<a_{2011}$.
Prove: $a_{1}, a_{2}, \cdots, a_{2011}$ must contain two numbers $a_{i} 、 a_{j}(i<j)$, such that
$$
a_{j}-a_{i}<\frac{\left(1+a_{i}\right)\left(1+a_{j}\right)}{2010} .
$$ | 14. B. Let $x_{i}=\frac{2010}{1+a_{i}}(i=1,2, \cdots, 2011)$.
Then $0<x_{2011}<x_{2010}<\cdots<x_{1}<2010$.
Therefore, there must exist $1 \leqslant k \leqslant 2010$, such that.
$$
x_{k}-x_{k+1}<1 \text {, }
$$
which means $\frac{2010}{1+a_{k}}-\frac{2010}{1+a_{k+1}}<1$.
Thus, $a_{k+1}-a_{k}<\frac{\left(1+a_{k}\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,779 |
1. Given $I$ is the incenter of acute $\triangle A B C$, and $A_{1} 、 B_{1}$ 、 $C_{1}$ are the reflections of point $I$ over sides $B C 、 C A 、 A B$ respectively. If point $B$ lies on the circumcircle of $\triangle A_{1} B_{1} C_{1}$, then $\angle A B C$ equals $(\quad)$.
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\cir... | Prompt: From Conclusion 1, we know that $I$ is the circumcenter of $\triangle A_{1} B_{1} C_{1}$.
Answer: $\mathrm{C}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,780 |
1. The domain of the function $y=\arccos \left(\frac{\sqrt{12+4 x-x^{2}}-2}{4}\right)$ is $\qquad$ , and the range is $\qquad$ . | $$
-1 .[-2,6],\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] \text {. }
$$
From $12+4 x-x^{2} \geqslant 0$, we get $-2 \leqslant x \leqslant 6$. At this point,
$$
\begin{array}{l}
0 \leqslant \sqrt{12+4 x-x^{2}}=\sqrt{-(x-2)^{2}+16} \leqslant 4 \\
\Rightarrow-\frac{1}{2} \leqslant \frac{\sqrt{12+4 x-x^{2}}-2}{4} \leqslan... | [-2,6],\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,781 |
2. If $n$ positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the equation
$$
\sum_{k=1}^{n}\left|\lg x_{k}\right|+\sum_{k=1}^{n}\left|\lg \frac{1}{x_{k}}\right|=\left|\sum_{k=1}^{n} \lg x_{k}\right|
$$
then the values of $x_{1}, x_{2}, \cdots, x_{n}$ are $\qquad$ | 2. $x_{1}=x_{2}=\cdots=x_{n}=1$.
The original equation is
$$
\begin{array}{l}
2 \sum_{k=1}^{n}\left|\lg x_{k}\right|=1 \sum_{k=1}^{n} \lg x_{k} \mid \\
\Rightarrow 2 \sum_{k=1}^{n}\left|\lg x_{k}\right| \leqslant \sum_{k=1}^{n}\left|\lg x_{k}\right| \\
\Rightarrow \sum_{k=1}^{n}\left|\lg x_{k}\right| \leqslant 0 .
\en... | x_{1}=x_{2}=\cdots=x_{n}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,782 |
3. In a rectangular coordinate system, there is $\triangle A B C$, with vertices $A$ $(3,0)$, $B(0,2)$, and $C(0,-1)$. The area of the common part between $\triangle A B C$ and the two parallel lines $x=t$, $x=\frac{t}{2}(0<t \leqslant 3)$ is denoted as $S(t)$. Then, as $t$ varies, the maximum value of $S(t)$ is . $\qq... | 3. $\frac{3}{2}$.
Notice that
$$
\begin{array}{l}
l_{A B}: \frac{x}{3}+\frac{y}{2}=1(0 \leqslant x \leqslant 3), \\
l_{A C}: \frac{x}{3}-y=1(0 \leqslant x \leqslant 3) .
\end{array}
$$
As shown in Figure 3, let the intersection points of the line $x=t$ with segments $A B$ and $A C$ be $F$ and $E$ respectively; the in... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,783 |
4. Suppose Bag A contains 4 white balls, 5 red balls, and 6 black balls; Bag B contains 7 white balls, 6 red balls, and 2 black balls. If one ball is drawn from each bag, then the probability that the two balls are of different colors is
(use the simplest fraction as the answer). | 4. $\frac{31}{45}$.
The probability that the two balls have the same color is
$$
P=\frac{4 \times 7+5 \times 6+6 \times 2}{15 \times 15}=\frac{14}{45},
$$
Therefore, the probability that the two balls have different colors is $1-\frac{14}{45}=\frac{31}{45}$. | \frac{31}{45} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,784 |
5. Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $x^{2}-y^{2}=1$, $M$ is a point on the right branch of the hyperbola, and $O$ is the origin. If $\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{|M O|}=\sqrt{6}$, then the coordinates of point $M$ are $\qquad$ | 5. Let $M(x, y)(x \geqslant 1)$. It is easy to know that the foci of the hyperbola are $F_{1}(-\sqrt{2}, 0)$ and $F_{2}(\sqrt{2}, 0)$. Then
$$
\begin{array}{l}
\left|M F_{1}\right|=\sqrt{(x+\sqrt{2})^{2}+y^{2}}, \\
\left|M F_{2}\right|=\sqrt{(x-\sqrt{2})^{2}+y^{2}}, \\
|M O|=\sqrt{x^{2}+y^{2}} . \\
\text { Therefore, }... | M\left(\frac{\sqrt{6}}{2}, \pm \frac{\sqrt{2}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,785 |
7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and there exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010 .
$$
Then the maximum value of $n$ is | 7.3.
Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,
where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,787 |
8. As shown in Figure 1, in $\triangle A B C$, it is known that $A B=5, B C$ $=8, A C=7$. Points $P$ and $Q$ are moving points on sides $A B$ and $A C$, respectively, such that the circumcircle of $\triangle A P Q$ is tangent to $B C$. Then the minimum length of line segment $P Q$ is $\qquad$ | 8. $\frac{30}{7}$.
Since the three side lengths of $\triangle ABC$ are known, the three interior angles of $\triangle ABC$ are fixed.
By $\frac{PQ}{\sin A} = 2R$ (where $R$ is the circumradius of $\triangle APQ$),
we get $PQ = 2R \sin A$.
To minimize $PQ$, $R$ must be minimized.
Since the circumcircle of $\triangle A... | \frac{30}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,788 |
9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor? | II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,789 |
10. (14 points) Given the set $A$ consisting of the positive integers $1,2, \cdots, 1000$, first randomly select a number $a$ from set $A$, and after selecting, return $a$ to set $A$; then randomly select another number $b$ from set $A$. Find the probability that $\frac{a}{b}>\frac{1}{3}$. | 10. Solution 1 Note that
$$
P\left(\frac{a}{b}>\frac{1}{3}\right)=1-P\left(\frac{a}{b} \leqslant \frac{1}{3}\right) \text {. }
$$
By $\frac{a}{b} \leqslant \frac{1}{3} \Leftrightarrow a \leqslant \frac{1}{3} b \Leftrightarrow a \leqslant\left[\frac{1}{3} b\right]$.
Then $P\left(\frac{a}{b} \leqslant \frac{1}{3}\right)... | \frac{1667}{2000} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,790 |
2. In $\triangle A B C$, it is known that $\angle A B C=70^{\circ}$, $\angle B A C=80^{\circ}$, and $P$ is a point inside $\triangle A B C$ such that $\angle C B P$ $=\angle B C P=10^{\circ}$. Find $\angle B A P$. | Hint: It is easy to get $B P=C P$,
$$
\angle B A C=\frac{1}{2} \angle B P C \text {. }
$$
From criterion 2, we know that $P$ is the circumcenter of $\triangle A B C$.
Answer: $60^{\circ}$. | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,791 |
11. (16 points) (1) Let $x_{1}, x_{2}, x_{3} \in \mathbf{R}$. Prove:
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \geqslant \sqrt{2}\left(x_{1} x_{2}+x_{2} x_{3}\right) \text {, }
$$
and specify the conditions under which equality holds;
(2) If a real number $a$ ensures that for any real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, the ... | 11. (1) Notice
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\left(x_{1}^{2}+\frac{x_{2}^{2}}{2}\right)+\left(\frac{x_{2}^{2}}{2}+x_{3}^{2}\right) \\
\geqslant \frac{2}{\sqrt{2}} x_{1} x_{2}+\frac{2}{\sqrt{2}} x_{2} x_{3}=\sqrt{2}\left(x_{1} x_{2}+x_{2} x_{3}\right) .
\end{array}
$$
Equality holds if and only if $... | \sqrt{5}-1 | Inequalities | proof | Yes | Yes | cn_contest | false | 723,792 |
12. (16 points) Given a positive integer $n$ that satisfies the following condition: for each positive integer $m$ in the open interval $(0,2009)$, there always exists a positive integer $k$, such that
$$
\frac{m}{2009}<\frac{k}{n}<\frac{m+1}{2010} \text {. }
$$
Find the minimum value of such $n$. | 12. Notice
$$
\begin{array}{l}
\frac{m}{2009}2010 k
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m n+1 \leqslant 2009 k, \\
m n+n-1 \geqslant 2010 k
\end{array}\right. \\
\Rightarrow 2009(m n+n-1) \geqslant 2009 \times 2010 k \\
\geqslant 2010(m n+1) \\
\Rightarrow 2009 m n+2009 n-2009 \\
\geqslant 2010 m ... | 4019 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,793 |
1. Let $k_{1}<k_{2}<\cdots<k_{n}$ be non-negative integers, satisfying $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=227$.
Then $k_{1}+k_{2}+\cdots+k_{n}=$ $\qquad$ | - 1. 19.
Notice that
$$
\begin{array}{l}
227=1+2+32+64+128 \\
=2^{0}+2^{1}+2^{5}+2^{6}+2^{7} .
\end{array}
$$
Therefore, $k_{1}+k_{2}+\cdots+k_{n}=0+1+5+6+7=19$. | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,794 |
2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ . | 2. 2 .
From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,795 |
3. Given that $S_{n}$ is the sum of the first $n$ terms of an arithmetic sequence with a positive common difference $q$. If $\frac{S_{n}+210}{n}$ reaches its minimum value when $n=6$, then the range of values for $q$ is | 3. $[10,14]$.
Let $a_{n}=a_{1}+(n-1) q$. Then $S_{n}=n a_{1}+\frac{n(n-1)}{2} q$.
Thus, $\frac{S_{n}+210}{n}=\frac{q}{2} n+\frac{210}{n}+a_{1}-\frac{q}{2}$. Therefore, $\frac{6 q}{2}+\frac{210}{6} \leqslant \min \left\{\frac{5 q}{2}+\frac{210}{5}, \frac{7 q}{2}+\frac{210}{7}\right\}$. This gives $10 \leqslant q \leqsl... | [10,14] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,796 |
4. Given the function $y=x^{3}$, the tangent line at $x=a_{k}$ intersects the $x$-axis at point $a_{k+1}$. If $a_{1}=1, S_{n}=\sum_{i=1}^{n} a_{i}$, then $\lim _{n \rightarrow \infty} S_{n}$ $=$ . $\qquad$ | 4.3.
It is known that $y^{\prime}=3 x^{2}$.
Therefore, the equation of the tangent line to $y=x^{3}$ at $x=a_{k}$ is $y-a_{k}^{3}=3 a_{k}^{2}\left(x-a_{k}\right)$.
Thus, the above equation intersects the $x$-axis at the point $\left(a_{k+1}, 0\right)$. Hence, $-a_{k}^{3}=3 a_{k}^{2}\left(a_{k+1}-a_{k}\right)$.
From th... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,797 |
5. The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all $x, y, z \in \mathbf{R}$
$$
f(x+y)+f(y+z)+f(z+x) \geqslant 3 f(x+2 y+z) .
$$
Then $f(1)-f(0)=$ $\qquad$ | 5.0.
Let $x=-y=z$, we get
$$
f(2 x) \geqslant f(0) \Rightarrow f(1) \geqslant f(0) \text {. }
$$
Let $x=y=-z$, we get
$$
f(0) \geqslant f(2 x) \Rightarrow f(0) \geqslant f(1) \text {. }
$$
Thus, $f(1)-f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,798 |
6. In the acute triangle $\triangle A B C$, let the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively. If $\frac{b}{a}+\frac{a}{b}=4 \cos C$, then the minimum value of $\frac{1}{\tan A}+\frac{1}{\tan B}$ is $\qquad$ | 6. $\frac{2}{\sqrt{3}}$.
It is known that, $\cos C=\frac{1}{4}\left(\frac{b}{a}+\frac{a}{b}\right) \geqslant \frac{1}{2}$.
Then $\sin C \leqslant \frac{\sqrt{3}}{2}$.
Also, $a^{2}+b^{2}=4 a b \cos C \Rightarrow a^{2}+b^{2}=2 c^{2}$.
Therefore, $\frac{1}{\tan A}+\frac{1}{\tan B}=\frac{\cos B \cdot \sin A+\sin B \cdot \... | \frac{2}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,799 |
7. Given that $P$ is a moving point on the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, and $F_{1} 、 F_{2}$ are the two foci of the ellipse. Then the range of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is $\qquad$ | 7. $[-4,4]$.
Let $P\left(x_{0}, y_{0}\right) 、 F_{1}(-c, 0) 、 F_{2}(c, 0)$. Then $\overrightarrow{P F_{1}}=\left(-x_{0}-c,-y_{0}\right), \overrightarrow{P F_{2}}=\left(c-x_{0},-y_{0}\right)$. Therefore, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x_{0}^{2}-c^{2}+y_{0}^{2}=x_{0}^{2}+y_{0}^{2}-c^{2}$.
Notic... | -4 \leqslant \overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}} \leqslant 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,800 |
8. Color the eight vertices of a cube with three colors, where at least one color is used to color exactly four vertices. Then the probability that the two endpoints of any edge are of different colors is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and outp... | 8. $\frac{1}{35}$.
When one color is used to paint four vertices, the remaining two colors can be used to paint the remaining four vertices in any way. Therefore, the number of coloring methods is
$$
\mathrm{C}_{3}^{1} \mathrm{C}_{8}^{4}\left(\mathrm{C}_{4}^{0}+\mathrm{C}_{4}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{3}\... | \frac{1}{35} | Other | math-word-problem | Yes | Yes | cn_contest | false | 723,801 |
3. As shown in Figure 9, in $\triangle A B C$, it is given that $\angle A C B=90^{\circ}$, $\angle C A D=30^{\circ}$, $A C=B C=A D$. Prove: $B D=C D$.
保留源文本的换行和格式，直接输出翻译结果。 | Prompt: Following Example 5, take the circumcenter $O$ of $\triangle A D C$, then $\triangle O C D$ is an equilateral triangle. It can be proven that
$\triangle A C O \cong \triangle B C D$.
Thus, $B D=A O=O C=C D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,802 |
9. (14 points) Given $\sin \alpha+\sin \beta=\frac{1}{5}, \cos \alpha+\cos \beta=\frac{1}{3}$. Find the value of $\frac{1-\cos 2(\alpha+\beta)+\sin 2(\alpha+\beta)}{1+\cos 2(\alpha+\beta)+\sin 2(\alpha+\beta)}$.
| 9. From $\tan \frac{\alpha+\beta}{2}=\frac{\sin \alpha+\sin \beta}{\cos \alpha+\cos \beta}=\frac{3}{5}$, we get
$$
\tan (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1-\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{15}{8} .
$$
Notice that
$$
\begin{array}{l}
\tan \gamma=\frac{1-\cos 2 \gamma}{\sin 2 \gamma}=\frac{\sin ... | \frac{15}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,803 |
10. (14 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$ $(n \geqslant 3)$. Prove that:
$$
\sum_{k=1}^{n-2} \frac{1}{a_{k}^{2}+a_{k+1}^{2}+a_{k+2}^{2}}>\frac{2(n-2)^{2}}{n(n+1)(2 n+1)} .
$$ | $$
\begin{array}{l}
\text { By Cauchy's inequality, we have } \\
{\left[\sum_{k=1}^{n-2}\left(a_{k}^{2}+a_{k+1}^{2}+a_{k+2}^{2}\right)\right] \sum_{k=1}^{n-2}\left(a_{k}^{2}+a_{k+1}^{2}+a_{k+2}^{2}\right)^{-1}} \\
\geqslant(n-2)^{2} . \\
\text { Then } \sum_{k=1}^{n-2}\left(a_{k}^{2}+a_{k+1}^{2}+a_{k+2}^{2}\right)^{-1}... | \frac{2(n-2)^{2}}{n(n+1)(2 n+1)} | Inequalities | proof | Yes | Yes | cn_contest | false | 723,804 |
11. (18 points) For any positive integer $n$, prove:
$$
\sum_{k=1}^{n} \frac{k}{k^{4}+k^{2}+1}=\frac{1}{n^{2}+n+1} \sum_{k=1}^{n} k .
$$ | $\begin{array}{l}\text { 11. } \sum_{k=1}^{n} \frac{k}{k^{4}+k^{2}+1}=\sum_{k=1}^{n} \frac{k}{\left(k^{2}+1\right)^{2}-k^{2}} \\ =\sum_{k=1}^{n} \frac{k}{\left(k^{2}+1-k\right)\left(k^{2}+1+k\right)} \\ =\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{k^{2}+1-k}-\frac{1}{k^{2}+1+k}\right) \\ =\frac{1}{2}\left(1-\frac{1}{n^{2}... | \frac{1}{n^{2}+n+1} \sum_{k=1}^{n} k | Algebra | proof | Yes | Yes | cn_contest | false | 723,805 |
12. (18 points) Let $S$ be a set of distinct quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, where $a_{i}=0$ or 1 $(i=1,2,3,4)$. It is known that the number of elements in $S$ does not exceed 15, and satisfies:
if $\left(a_{1}, a_{2}, a_{3}, a_{4}\right) 、\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$, then $\lef... | 12. Clearly, there are 16 possible quadruples. Since at least one quadruple is not in $S$, it follows that $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ must have at least one that is not in $S$. Otherwise, by the given conditions, all quadruples would be in $S$.
Assume $(1,0,0,0) \notin S$.
In this case, by the g... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,806 |
1. After adding a monomial to the algebraic expression $x^{4}+x^{2}$, the result is a perfect square of an algebraic expression. Then, the number of such monomials is ( ).
(A) 6
(B) 5
(C) 4
(D) no more than 3 | -1. A.
The monomials that meet the conditions are only
$$
-x^{4},-x^{2}, \frac{1}{4}, \pm 2 x^{3}, \frac{1}{4} x^{6}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,807 |
2. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
2 x-1>3, \\
4 x-2 \geqslant a
\end{array}\right.
$$
the solution set is $x>b$. Then the value of the expression
$$
|a-3 b|-2|a-4 b|
$$
is ( ).
(A) $3 a-11 b$
(B) $5 b-a$
(C) $22-3 a$
(D) $a-10$ | 2. D.
From the given inequality, we get $x>2, x \geqslant \frac{a+2}{4}$.
From the problem, we get $\frac{a+2}{4} \leqslant 2, b=2$.
Then $a \leqslant 6 \leqslant 3 b<4 b$.
Thus $|a-3 b|-2|a-4 b|$
$$
\begin{array}{l}
=(3 b-a)-2(4 b-a) \\
=a-5 b=a-10 .
\end{array}
$$ | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 723,808 |
3. In a certain chess tournament that only junior high and high school students can participate in, the scoring method is: the winner gets 1 point, the loser gets 0 points, and in a draw, both sides get 0.5 points. Each player plays against every other player once. It is known that the number of high school students is... | 3. C.
Let the number of junior high school students and senior high school students be $x$ and $2x$ respectively. According to the problem, the total score of all students is 11 times that of the junior high school students, i.e.,
$$
\mathrm{C}_{3 x}^{2} \geqslant 11 \mathrm{C}_{x}^{2} \Rightarrow x \leqslant 4 \text ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,809 |
4. As shown in Figure 1, in quadrilateral $A B C D$, it is known that $\triangle A B C$ is an equilateral triangle, $\angle A D C=$ $30^{\circ}, A D=3, B D=5$. Then the length of side $C D$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5 | 4. B.
As shown in Figure 4, with
$C D$ as a side, construct
an equilateral $\triangle C D E$ outside
quadrilateral $A B C D$,
and connect $A E$.
$$
\begin{array}{l}
\text { Given } A C \doteq \\
B C, C D=C E, \\
\angle B C D=\angle B C A+\angle A C D \\
=\angle D C E+\angle A C D=\angle A C E \\
\Rightarrow \triangle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,810 |
5. In the Cartesian coordinate system, points with both coordinates being integers are called "integer points". Let $k$ be a non-zero real number. If the intersection point of the two distinct lines $y=k x-\frac{1}{k}$ and $y=\frac{1}{k} x-1$ is an integer point, then the number of possible values of $k$ is ( ).
(A) 1
... | 5. A.
From the problem, we have
$$
\left(k^{2}-1\right) x=1-k \text{. }
$$
If $k=1$, then the equations of the two lines are the same, which does not meet the problem's requirements.
If $k=-1$, then equation (1) has no solution.
If $k \neq \pm 1$, clearly, $x \neq 0, k=-\frac{1}{x}-1$.
Then $y=\frac{1}{k} x-1=-x-\fra... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,811 |
6. As shown in Figure 2, given that $N$ is a point on the semicircle with diameter $AB$, $NM \perp AB$ at point $M$, semicircles $\odot O_{1}$ and $\odot O_{2}$ have diameters $AM$ and $BM$ respectively, $\odot O_{3}$ and $\odot O_{4}$ are both internally tangent to the semicircle $\odot O$, also tangent to $MN$, and e... | 6. D.
As shown in Figure 5, let the radii of $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O$ be
$R_{1}, R_{2}, R_{3}, R_{4},$ and
$R$, respectively. $\odot O_{4}$ is tangent to $MN$ at point $D$. Draw $O_{4}C \perp AB$ at point $C$, and connect
$$
OO_{4}, O_{2}O_{4}, DO_{4}.
$$
It is easy to see that... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,812 |
4. In $\triangle A B C$, it is known that $\angle B A C=100^{\circ}, A B$ $=A C, P$ is a point inside $\triangle A B C$, and $\angle P A C=$ $\angle A C P=20^{\circ}$. Find $\angle P B A$. | Prompt: In $\triangle A B C$, construct $\triangle A Q B \cong \triangle A P C$.
First, prove that $\triangle Q A P$ is an equilateral triangle, then according to criterion 1, $Q$ is the circumcenter of $\triangle A B P$.
Answer $30^{\circ}$. | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,813 |
1. Four points $A, B, C, D$ in the same plane satisfy
$$
A D=B D=C D, \angle A D C=160^{\circ} \text {. }
$$
Then the degree of $\angle A B C$ is | 2. $80^{\circ}$ or $100^{\circ}$.
From the given information, point $D$ is the center of the circumcircle of $\triangle A B C$.
If point $B$ is on the major arc $\overparen{A C}$, then
$$
\angle A B C=\frac{1}{2} \angle A D C=80^{\circ} \text {; }
$$
If point $B$ is on the minor arc $\overparen{A C}$, then
$$
\angle A... | 80^{\circ} \text{ or } 100^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,814 |
2. If the function
$$
y=\frac{x+k}{(k+1) x^{2}+(2 k+2) x+2 k-3}
$$
has the domain of the independent variable $x$ as all real numbers, then the range of values for $k$ is $\qquad$ | 2. $k \leqslant-1$ or $k>4$.
Notice that
$$
\begin{array}{l}
f(x)=(k+1) x^{2}+(2 k+2) x+2 k-3 \\
=(k+1)(x+1)^{2}+k-4 .
\end{array}
$$
When $k=-1$, $f(x)=-5 \neq 0$, which satisfies the condition.
When $k \neq-1$,
$$
f(x) \neq 0 \Rightarrow k+1 \text{ and } k-4 \text{ have the same sign. }
$$
Thus, $k4$.
Therefore, t... | k \leqslant-1 \text{ or } k>4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,815 |
3. Given that $a, b$ are real numbers, satisfying
$$
t=\frac{a^{2}}{a^{2}+2 b^{2}}+\frac{b^{2}}{2 a^{2}+b^{2}} \text {. }
$$
Then the minimum value of $t$ is | 3. $\frac{2}{3}$.
Let $x^{2}=a^{2}+2 b^{2}, y^{2}=2 a^{2}+b^{2}$. Then $a^{2}=\frac{-x^{2}+2 y^{2}}{3}, b^{2}=\frac{-y^{2}+2 x^{2}}{3}$.
Thus, $t=\frac{2 y^{2}-x^{2}}{3 x^{2}}+\frac{2 x^{2}-y^{2}}{3 y^{2}}$ $=\frac{2}{3}\left(\frac{y^{2}}{x^{2}}+\frac{x^{2}}{y^{2}}\right)-\frac{2}{3} \geqslant \frac{2}{3}$.
When and o... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,816 |
4. Given that $A D$ is the altitude of $\triangle A B C$,
$$
\angle B A C=135^{\circ}, A D=3, B C=25 \text {. }
$$
Then the perimeter of $\triangle A B C$ is $\qquad$ . | 4. $30+15 \sqrt{2}$.
As shown in Figure 6, with point $A$ as the center and $AD$ as the radius, draw circle $\odot A$. Draw tangents to $\odot A$ from points $B$ and $C$, with the points of tangency being $E$ and $F$ respectively. The tangents intersect at point $G$, and connect $AE$ and $AF$. Let $BD=x$ and $CD=y$. T... | 30 + 15 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,817 |
One. (20 points) Given the equation in terms of $x$
$$
k x^{2}-\left(k^{2}+6 k+6\right) x+6 k+36=0
$$
the roots of which are the side lengths of a certain isosceles right triangle. Find the value of $k$. | When $k=0$, the root of the equation is $x=6$, which meets the requirement.
When $k \neq 0$, the equation can be factored as
$(k x-6)(x-6-k)=0$.
Solving gives $x_{1}=\frac{6}{k}, x_{2}=6+k$.
Clearly, $k>0$.
Since $x_{1}$ and $x_{2}$ are the lengths of two sides of an isosceles right triangle, we have
$\frac{x_{2}}{x_{1... | 0, \sqrt{15}-3, \sqrt{9+3 \sqrt{2}}-3, \sqrt{6}+\sqrt{3}-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,818 |
II. (25 points) As shown in Figure 3, given that quadrilateral $ABCD$ is a cyclic quadrilateral inscribed in $\odot O$, and the midpoint $I$ of diagonal $AC$ is the incenter of $\triangle ABD$. Prove:
(1) $OI$ is the tangent to the circumcircle of $\triangle IBD$;
$$
\begin{array}{l}
\text { (2) } AB + AD \\
= 2BD.
\en... | (1) As shown in Figure 7.
By the properties of the incenter, we have
$$
C B=C I=C D \text{. }
$$
Therefore, $C$ is the circumcenter of $\triangle I B D$.
Since $I$ is the midpoint of $A C$, we have
$$
O I \perp A C \text{, and } A I=I C \text{. }
$$
Thus, $O I \perp C I$.
Hence, $O I$ is the tangent line to the circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,819 |
Three, (25 points) 15 children each have 15 chess pieces. They play a game of "Rock, Paper, Scissors," where each pair of children only has one match, and the loser gives one chess piece to the winner. After the game, the 15 children are divided into two groups, Group A and Group B. The total number of chess pieces in ... | Three, let Group A have $x$ people. Then Group B has $15-x$ people.
Let Group A win $y$ times and Group B win $z$ times when the two groups compete. Then $z=x(15-x)-y$.
According to the problem,
$$
\begin{array}{l}
15 x+y-[x(15-x)-y] \\
=15(15-x)-y+[x(15-x)-y]+63 .
\end{array}
$$
Simplifying, we get $y=-\frac{1}{2} x^... | 29, 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,820 |
1. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with common difference $d \neq 0$, and the sum of the first $n$ terms be $S_{n}$. Then the sequence $\left\{S_{n}\right\}$ is increasing if and only if $\qquad$. | $$
\text { I. 1. } d>0 \text { and } a_{2}>0 \text {. }
$$
Obviously, $\left\{S_{n}\right\}$ is an increasing sequence
$$
\begin{array}{l}
\Leftrightarrow S_{n}0 \\
\Leftrightarrow d n+a_{2}>0 \\
\Leftrightarrow d>0 \text { and } a_{2}>0 .
\end{array}
$$ | d>0 \text { and } a_{2}>0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,821 |
2. Given points
$$
A\left(x-\frac{3-\sqrt{5}}{2} \cdot y, 1\right) 、 B\left(x-\frac{3+\sqrt{5}}{2} \cdot y,-1\right)
$$
satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$. Then the range of $S=x^{2}-x y+y^{2}-1$ is $\qquad$ . | 2. $\left[-\frac{2}{5},+\infty\right)$.
From the given, we have
$$
\begin{array}{l}
\left(x-\frac{3-\sqrt{5}}{2} \cdot y\right)\left(x-\frac{3+\sqrt{5}}{2} \cdot y\right)-1=0 \\
\Rightarrow x^{2}-3 x y+y^{2}-1=0 \\
\Rightarrow 1=(x+y)^{2}-5 x y \geqslant-5 x y \\
\Rightarrow x y \geqslant-\frac{1}{5} .
\end{array}
$$
... | \left[-\frac{2}{5},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,822 |
3. Given a function $f(x)$ defined on the domain $[-1,1]$ that satisfies: for $\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,
$$
f(\sin \theta)+3 f(-\sin \theta)=\cos \theta \text {. }
$$
Then the range of the function $f(x)$ is | 3. $\left[0, \frac{1}{4}\right]$.
In the given equation, replace $\theta$ with $-\theta$ to get
$$
f(-\sin \theta)+3 f(\sin \theta)=\cos \theta \text {. }
$$
Solving the two equations simultaneously, we get
$$
f(\sin \theta)=\frac{1}{4} \cos \theta=\frac{1}{4} \sqrt{1-\sin ^{2} \theta} \text {. }
$$
Therefore, $f(x)... | \left[0, \frac{1}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,823 |
Example 1 As shown in Figure 1, two utility poles $A B$ and $C D$ are erected on a construction site. They are $15 \mathrm{~m}$ apart, and points $A$ and $C$ are $4 \mathrm{~m}$ and $6 \mathrm{~m}$ above the ground, respectively. Steel wires are stretched from these points to points $E$ and $D$, and $B$ and $F$ on the ... | As shown in Figure 1, draw $P H \perp B D$ at point $H$. Then
$A B / / P H / / C D$.
From $\triangle D P H \backsim \triangle D A B \Rightarrow \frac{P H}{A B}=\frac{D H}{B D}$;
From $\triangle B P H \backsim \triangle B C D \Rightarrow \frac{P H}{C D}=\frac{B H}{B D}$.
Adding the two equations yields $\frac{1}{A B}+\... | \frac{12}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,824 |
Example 2 As shown in Figure 2, in quadrilateral $A B C D$, $A C$ intersects $B D$ at point $O$,
line $l / / B D$, and
intersects $A B, D C, B C$,
$A D$ and the extension of $A C$ at
points $M, N, R, S$ and
$P$. Prove:
$P M \cdot P N=P R \cdot P S$.
(1998, Shandong Province Junior High School Mathematics Competition) | 【Analysis】This problem essentially involves the superposition of two basic figures, Figure 3 and Figure 4.
Proof From Figure 3, we have
$$
\begin{array}{l}
\triangle A B O \sim \triangle A M P \Rightarrow \frac{O B}{P M}=\frac{A O}{A P} ; \\
\triangle A D O \backsim \triangle A S P \Rightarrow \frac{O D}{P S}=\frac{A O... | P M \cdot P N=P R \cdot P S | Geometry | proof | Yes | Yes | cn_contest | false | 723,825 |
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