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38 Given $x \in \mathbf{R}, a_{n}=\cos \left(x+\frac{2}{7} n \pi\right)$, then the value of $a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}$ is
A. greater than 1
B. less than 1
C. equal to 1
D. zero | 38 D. Let \( S=a_{0}+a_{1}+\cdots+a_{6} \), then by the problem we have
\[
S=\cos x+\cos \left(x+\frac{2}{7} \pi\right)+\cos \left(x+\frac{4}{7} \pi\right)+\cdots+\cos \left(x+\frac{12}{7} \pi\right) .
\]
Using the product-to-sum formulas, it is easy to calculate that \( S \cdot 2 \sin \frac{\pi}{7}=0 \), so \( S=0 \)... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
15. Master Wang works in a special position, where he works for 8 consecutive days and then takes 2 consecutive days off. If he is off on this Saturday and Sunday, then, at least how many weeks later will he be off on a Sunday again? | 15. At least another 7 weeks
15.【Solution】Let at least $\mathrm{n}$ weeks pass, it is possible to rest on the $\mathrm{n}$th Saturday, or it is also possible not to rest on the $\mathrm{n}$th Saturday (resting for 2 days on Sunday and Monday), the former yields: $7 \mathrm{n}-2=10 \mathrm{~K}+8(1)$, the latter yields: ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given a set of $n$ positive integers $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$. For any $i \in\{1,2, \cdots, n\}$, the set $A_{i}$ obtained by removing the element $a_{i}$ from $A$ can be divided into the union of two disjoint subsets, and the sums of the elements in these two subsets are equal. Such a set $A$ is ... | (1) Let $M=a_{1}+a_{2}+\cdots+a_{n}$.
By the problem statement, the sum of the elements in the set $A_{i}(i=1,2, \cdots, n)$ is $M-a_{i}$, and $M-a_{i}$ are all even numbers. Therefore, $a_{i}$ and $M$ have the same parity.
( $i$ ) If $M$ is odd, then $a_{i}(i=1,2, \cdots, n)$ are also odd, and $n$ is odd;
(ii) If $M$ ... | 7 | Number Theory | proof | Yes | Yes | olympiads | false |
3. (5 points) The average of four numbers is 30. If one of them is changed to 50, the average becomes 40. The original number was $\qquad$ | 【Solution】Solve: $50-(40 \times 4-30 \times 4)$
$$
\begin{array}{l}
=50-(160-120) \\
=50-40 \\
=10
\end{array}
$$
Answer: The original number is 10.
Therefore, the answer is: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5 For what real number $x$ does the function $f(x)=x^{2}-x+1+\sqrt{2 x^{4}-18 x^{2}+12 x+68}$ have a minimum value? What is the minimum value? | Given $f(x)=x^{2}-x+1+\sqrt{2 x^{4}-18 x^{2}+12 x+68}$
$$
\begin{array}{l}
=x^{2}-x+1+\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}} \\
=x^{2}-x+1+\sqrt{2} \cdot \sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}} \\
=\sqrt{2}\left[\frac{x^{2}-x+1}{\sqrt{2}}+\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}\right] .
\end{array}
$$
Thus, t... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The number of positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{2021}$ is $\qquad$ . | $x y=2021(x+y) \Rightarrow (x-2021)(y-2021)=2021^{2}=43^{2} \cdot 47^{2}$, so the number of positive integer solutions to this equation is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (10 points) On the blackboard, there are 11 ones, 22 twos, 33 threes, and 44 fours. Perform the following operation: each time, erase 3 different numbers, and write 2 more of the fourth number that was not erased. For example: if in one operation, 1 one, 1 two, and 1 three are erased, then write 2 more fours. After ... | 【Analysis】This problem belongs to an operational issue. The simplest case to work backwards is when there are only 3 numbers left on the blackboard, and no further operations can be performed according to the rules. There are only two scenarios: 2 $a$s and 1 $b$, or 3 $a$s. We perform operations based on the rules, and... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$22 \cdot 9$ dihedral angle $\alpha-A B-\beta$ is $60^{\circ}, P \in \alpha, Q \in \beta$, and $P, Q$ are not on $A B$, the angle between $P Q$ and $A B$ is $45^{\circ}$, and $P Q=7 \sqrt{2},$ the distance from $Q$ to $A B$ is 3, then the distance from $P$ to $A B$ is
(A) 5.
(B) $\sqrt{37}$.
(C) 8.
(D) $\sqrt{79}$.
(C... | [Solution] As shown in the figure, draw $Q T \perp A B$ at $T$, draw $Q R / /$ $A B$, then draw $P R \perp Q R$ at $R$, and $R S \perp A B$ at $S$, connect $P S$, then $P S \perp A S$.
$$
\begin{array}{lc}
\therefore & \angle P S R=60^{\circ} . \\
\text { Also } & \angle P Q R=45^{\circ}, P Q=7 \sqrt{2}, \\
\therefore ... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Let $a$ and $b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is $\qquad$. | Given that $a b=-1, a+b=1$,
$$
a^{2}=a+1, b^{2}=b+1 \text {. }
$$
Therefore, $3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2}$
$$
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. For positive integer $a$ and integers $b$, $c$, the quadratic equation $a x^{2}+b x+c=0$ has two roots $\alpha, \beta$. And it satisfies $0<\alpha<\beta<$ 1. Find the minimum value of $a$. | 8. Let $f(x)=a x^{2}+b x+c$, then $f(0)=c>0, f(1)=a+b+c>0, \Delta=b^{2}-4 a c>0.0c>0, 0-(a+c)+1, (\sqrt{a}-\sqrt{c})^{2}>1, \sqrt{a}>\sqrt{c}+1 \geqslant 2$, so $a \geqslant 5$. And when the equality holds, $c=1, b=-5$. That is, $a_{\min }=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(Let $n$ students be such that among any 3 of them, 2 know each other, and among any 4 of them, 2 do not know each other. Find the maximum value of $n$. (Supplied by Tang Lihua))
---
The problem can be translated as follows:
Among $n$ students, for any group of 3 students, there are 2 who know each other. For any gr... | The maximum value of $n$ sought is 8.
When $n=8$, the example shown in the figure satisfies the requirements, where $A_{1}, A_{2}, \cdots, A_{8}$ represent 8 students, and a line connecting $A_{i}$ and $A_{j}$ indicates that $A_{i}$ and $A_{j}$ know each other, otherwise they do not.
Suppose $n$ students meet the requi... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 2 In space, there is a convex polyhedron, all of whose vertices are lattice points (each vertex has three integer coordinates). In addition, there are no other integer points inside the polyhedron, on its faces, or on its edges. What is the maximum number of vertices this convex polyhedron can have? | Analysis and Solution First, we take the coordinates of the eight grid points as $(i, j, k)(i, j, k \in\{0,1\})$, and with these 8 points as vertices, we can get a unit cube, which obviously meets the requirements of the problem, so the minimum value sought is no less than 8. On the other hand, the three coordinates of... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. (20 points) Given $a_{1}, a_{2}, a_{3}, a_{4}$ are positive real numbers,
$$
\sum_{i=1}^{4} i a_{i} \leqslant 10, a_{i} \geqslant \frac{1}{2^{4-i}}(i=1,2,3,4) \text {. }
$$
Let $f=\sum_{i=1}^{4} \frac{1}{1+a_{i}^{i}}$. Find the minimum value of $f$. | 11. Notice,
$$
\begin{array}{l}
\left(a_{1}-1\right)^{2} \geqslant 0, a_{2}\left(a_{2}-1\right)^{2} \geqslant 0, \\
\left(a_{3}-1\right)^{2}\left(a_{3}^{2}+2\left(a_{3}-\frac{1}{2}\right)\left(a_{3}+1\right)\right) \geqslant 0, \\
\left(a_{4}-1\right)^{2}\left(2 a_{4}^{3}+a_{4}^{2}-1\right) \geqslant 0 .
\end{array}
$$... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Let the parabola $C: y^{2}=2 p x(p>0)$ have a focus at $F$, and its directrix intersects the $x$-axis at point $Q$. A line through point $F$ intersects the parabola $C$ at points $A$ and $B$. If $\angle Q B F=90^{\circ}$, then $|A F|-|B F|=$ $\qquad$ | 5. 2 .
Let the inclination angle of line $AB$ be an acute angle $\theta$. In $\operatorname{Rt} \triangle B Q F$, applying the definition of a parabola, we get $\sin ^{2} \theta=\cos \theta$ and $|B F|=p \cos \theta$. By the focal chord length formula $|A F|+|B F|=\frac{2 p}{\sin ^{2} \theta}$, we have
$$
\begin{array... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. A, B, C, and D are competing in a table tennis tournament, where every two individuals play one match against each other. As a result, A defeated D, and A, B, and C have the same number of wins. How many matches did D win? | 14. Treat people as points, and get a directed graph $G$, from the conditions we know: $d^{+}($甲 $) \geqslant 1, d^{+}(\mathrm{J}) \leqslant 2, d^{+}$(甲) $=d^{+}($乙 $)=d^{+}$(丙).
(1) If $d^{+}$(甲) $=1$, then $d^{+}$(甲) $+d^{+}$(乙 $)+d^{+}$(丙) $+d^{+}$(丁) $=1+1+1+$ $d^{+}(\mathrm{J})=C_{4}{ }^{2}=6$, so $d^{+}(\mathrm{J... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $a$ be a non-zero real number, in the expansion of the binomial $(a x+1)^{5}$, the coefficient of the $x^{4}$ term is equal to the coefficient of the $x^{3}$ term, then the value of $a$ is $\qquad$ . | The coefficient of the $x^{4}$ term is $C_{5}^{1} a^{4}$, the coefficient of the $x^{3}$ term is $C_{5}^{2} a^{3}$, so $5 a^{4}=10 a^{3} \Rightarrow a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. The number of positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+51}+\sqrt{x+2019}$ is. $\qquad$ | 8. 6 .
Let $n^{2}=x+51, m^{2}=x+2019(n, m \in \mathbf{N})$.
Then $m^{2}-n^{2}=(m+n)(m-n)$
$$
=1968=2^{4} \times 3 \times 41.
$$
Since $m+n$ and $m-n$ have the same parity, they are both even, and $m+n>m-n$.
Thus, the possible values of the real number pair $(m+n, m-n)$ are 6 pairs:
$$
\begin{array}{l}
\left(2^{3} \t... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The product of the areas of the six faces of a rectangular prism is 14641, then the volume of the rectangular prism is | 【Analysis】Let's assume the length, width, and height of the rectangular prism are $a$, $b$, and $c$, respectively. Then the product of the areas of the six faces is: $(a b)^{2} \times(a c)^{2} \times(b c)^{2}=a^{4} \times b^{4} \times c^{4}=(a b c)^{4}=14641=11^{4}$, so $a b c=11$, which means the volume is 11 | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Color some nodes (vertices of unit squares) of a $6 \times 6$ grid table red, such that the boundary of any sub-grid table $k \times k (1 \leqslant k \leqslant 6)$ formed by unit squares has at least one red point. Find the minimum number of red points that satisfy this condition. | 3. The minimum value is 12.
On one hand, represent the nodes as coordinates $(i, j) (i, j \in \{-3, -2, -1, 0, 1, 2, 3\})$, and color the following 12 points $(\pm 3, 0), (0, \pm 3), (\pm 1, \pm 1), (\pm 2, \pm 2)$ red. By checking each one, it is known that they satisfy the condition.
On the other hand, use proof by... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 7 Let $S=\left\{A=\left(a_{1}, a_{2}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $\left.1, i=1,2, \cdots, 8\right\}$, for any two elements $A=\left(a_{1}, a_{2}, \cdots, a_{8}\right)$ and $B=\left(b_{1}, b_{2}, \cdots, b_{8}\right)$ in $S$, denote $d(A, B)=\sum_{i=1}^{8}\left|a_{i}-b_{i}\right|$, and call it ... | Let $S^{\prime}$ be a subset of $S$, and any two elements in $S^{\prime}$ have a distance $\geqslant 5$.
Let $A=\left(a_{1}, a_{2}, \cdots, a_{8}\right)$ and $B=\left(b_{1}, b_{2}, \cdots, b_{8}\right)$ be two elements in $S^{\prime}$, and let $w(A)=\sum_{i=1}^{8} a_{i}, w(B)=\sum_{i=1}^{8} b_{i}$.
Clearly, $w(A), w(B)... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. As shown in Figure $1, M$ is the midpoint of side $CD$ of the inscribed quadrilateral $ABCD$ in the semicircle $\odot O$ with diameter $AB$. $MN \perp AB$ at point $N$, $AB=10, AD=AN=$ 3. Then $BC=(\quad$.
(A) 5
(B) 6
(C) 7
(D) 8 | 5. C.
As shown in Figure 5, draw perpendiculars from points $C$ and $D$ to $AB$, with the feet of the perpendiculars being $E$ and $F$, respectively. Connect $AC$ and $BD$.
Let $AF = x$. Then
$$
\begin{array}{c}
E N = F N = 3 - x, \\
B E = B N - E N = (10 - 3) - (3 - x) = 4 + x .
\end{array}
$$
In right triangles $\t... | 7 | Geometry | MCQ | Yes | Yes | olympiads | false |
11. The sequence $\left\{a_{n}\right\}$ is: $1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, that is, first take $a_{1}=1$, then copy this item and paste it behind as $a_{2}$, and add the successor number 2 as $a_{3}$, then copy all items $1,1,2$ and paste them behind as $a_{4}, a_{5}, a_{6}$, and add the successor number 3 as... | The sequence $\left\{a_{n}\right\}$ is grouped as: $1 ; 1,2 ; 1,1,2,3 ; 1,1,2,1,1,2,3,4 ; 1,1,2,1,1,2,3,1,1,2,1,1$,
$2,3,4,5 ; \cdots$, from which we can see that $a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, a_{31}=5$, and $a_{2^{n}-1+k}=a_{k}$.
Therefore, $a_{2021}=a_{2^{10}-1+998}=a_{998}=a_{2^{9}-1+487}=a_{487}=a_{2^{8}-1+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11 Given that $x_{1}$ is a root of the equation $x+\lg x=3$, and $x_{2}$ is a root of the equation $x+10^{x}=3$, then the value of $x_{1}+x_{2}$ is ( ).
(A) 3
(B) 6
(C) 2
(D) 1 | 11 Since $\lg x=3-x, 10^{x}=3-x$, let $y=3-x$, then $y_{1}=\lg x, y_{2}=$ $10^{x}$ are inverse functions of each other. Therefore, the intersection points $A$ and $B$ of $y_{1}=\lg x$ and $y_{2}=10^{x}$ with the line $y=3-x$ are symmetric about the line $y=x$. By solving the intersection of $y=x$ and $y=3-x$, we get th... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
14. The figure below shows a switch array, where each switch has only two states: "on" and "off". Initially, all switches are off. Pressing a switch once will change the state of the switch itself and all adjacent switches. For example, pressing $(2,2)$ will change the states of $(1,2),(2,1),(2,2),(2,3),(3,2)$. If each... | (1) 6 times are enough, as shown in the figure below.
The proof that the minimum number of operations required is 6 is as follows:
Consider the diagonal $(1,1),(2,2),(3,3),(4,4)$, denoted as region $A$. The switches in region $A$ change an odd number of times only when the switches inside region $A$ are pressed. Since... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Let two regular tetrahedra $P-A B C$ and $Q-A B C$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-A B C$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-A B C$ is ... | Solution: 4.
As shown in Figure 4, connect $P Q$, then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$. Connect $C H$ and extend it to intersect $A B$ at point $M$, then $M$ is the midpoint of... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with an eccentricity of $\frac{\sqrt{5}}{2}, F_{1}, F_{2}$ are the left and right foci of $C$, respectively. A line $l$ passing through $F_{2}$ intersects $C$ at points $A$ and $B$ (point $A$ is in the first quadrant), and $\overrightarrow{... | Solving
By the focal radius formula $\left|F_{2} A\right|=\frac{e p}{1-e \cos \alpha},\left|F_{2} B\right|=\frac{e p}{1+e \cos \alpha} \Rightarrow e \cos \alpha=\frac{1}{2}$,
thus $\cos \alpha=\frac{\sqrt{5}}{5} \Rightarrow \tan \alpha=2$. Let $A B: y=2(x-\sqrt{5} b), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\rig... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the spatial quadrilateral $ABCD$, $AB=2, BC=$ $3, CD=4, DA=5$. Then $\overrightarrow{AC} \cdot \overrightarrow{BD}=$ $\qquad$ | $$
\begin{array}{l}
\text { Then } \overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D} \\
\Rightarrow \overrightarrow{A D}^{2}=(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D})^{2} \\
\Rightarrow \overrightarrow{A D}^{2}=\overrightarrow{A B}^{2}+\overrightarrow{B C}^{2}+\o... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Question 180, In the Cartesian coordinate system, the hyperbola $\Gamma: \frac{x^{2}}{3}-y^{2}=1$. For any point $P$ in the plane not on $\Gamma$, let $\Omega_{p}$ be the set of all lines passing through point $P$ and intersecting $\Gamma$ at two points. For any line $l \in \Omega_{p}$, let $M, N$ be the two intersecti... | Question 180, Answer: Let $P(a, b)$ be a good point. For any $l \in \Omega_{p}$, let $1: y=k x+c$, where $c=b-ka$. Solving the system of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x^{2}}{3}-y^{2}=1 \\
y=k x+c
\end{array} \Rightarrow \frac{x^{2}}{3}-(k x+c)^{2}=1\right. \\
\Rightarrow\left(1-3 k^{2}\ri... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Let point $O$ be inside $\triangle A B C$, satisfying
$$
\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0} \text {. }
$$
Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is $(\quad)$.
(A) 2
(B) 3
(C) $\frac{3}{2}$
(D) $\frac{5}{3}$ | 10. B.
Take the midpoints $D, E$ of $AC, BC$ respectively.
$$
\begin{array}{l}
\text { By } \overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0} \\
\Rightarrow \overrightarrow{O A}+\overrightarrow{O C}=-2(\overrightarrow{O B}+\overrightarrow{O C}) \\
\Rightarrow 2 \overrightarrow{O D}=-4 \over... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
2. (8 points) There are 3 boxes of apples of the same weight. If 4 kilograms are taken out from each box, the remaining weight of the apples in the boxes is exactly equal to the original weight of 1 box of apples. The original weight of each box of apples is ( ) kilograms.
A. 4
B. 6
C. 8
D. 12 | 【Answer】Solution: $3 \times 4 \div 2$
$$
=12 \div 2
$$
$=6$ (kg)
Answer: Each box of apples weighs 6 kg. Therefore, the correct choice is: $B$. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. In the tetrahedron $ABCD$, there is a point $O$ inside such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{OA_1}=\frac{BO}{OB_1}=\frac{CO}{OC_1}=\frac{DO}{OD_1}=k$, then $k=$ $\qquad$ . | 12. $3 \frac{V_{A B C D}}{V_{O B C D}}=\frac{A A_{1}}{O A_{1}}=\frac{A O}{O A_{1}}+\frac{O A_{1}}{O A_{1}}=1+k$, thus $\frac{V_{O B C D}}{V_{A B C D}}=\frac{1}{1+k}$.
Similarly, $\quad \frac{V_{(X \cdot D A}}{V_{A B C D}}=\frac{V_{O D A B}}{V_{A B C D}}=\frac{V_{O A B C}}{V_{A B C D}}=\frac{1}{1+k}$.
Since $V_{O B C D... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Between 1 and 200, how many numbers have the sum of all their distinct prime factors equal to 16? (For example: The distinct prime factors of 12 are $2$ and $3$, and their sum is $2+3=5$) | 【Analysis】Since $2+3+5+7=17>16$, the number sought can have at most 3 different prime factors. And since 16 is even, if it is decomposed into the sum of 3 primes, one of them must be 2.
$1 、 16=2+3+11$, there are $66 、 132 、 198$ for a total of 3
2、 $16=3+13$, there are $39 、 117$ for a total of 2
3、 $16=5+11$, there i... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The set $A=\left\{n \left\lvert\, \lg n<\frac{1}{2}\right., n \in \mathbf{N}^{*}\right\}$ has the sum of all its elements as $\qquad$ | $\lg n<\frac{1}{2} \Rightarrow 0<n<\sqrt{10} \Rightarrow n=1,2,3$, so the sum of all elements in set $A$ is 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 21 (2001 National College Entrance Examination Question) Let the function $y=f(x)$ be an even function defined on $\mathbf{R}$, whose graph is symmetric about the line $x=1$. For any $x_{1}, x_{2} \in\left[0, \frac{1}{2}\right]$, we have $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$,... | From $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$, we can think of the exponential function model $y=a^{x}\left(0 \leqslant x \leqslant \frac{1}{2}\right)$ to solve the problem.
(1) $f(x)=f\left(\frac{x}{2}+\frac{x}{2}\right)=f^{2}\left(\frac{x}{2}\right) \geqslant 0, x \in[0,1]$,
and $a=f(... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Example 5.6.4] On a $6 \times 6$ chessboard, $n$ dominoes of size $1 \times 2$ are placed, each domino covering exactly two squares. If no matter how these $n$ dominoes are placed, it is always possible to place one more domino, find the maximum value of $n$.
| Obviously, placing 12 dominoes as shown in the figure leaves no room for another domino, so $n \leqslant 11$.
We prove that 11 dominoes can meet the requirements of the problem.
Consider the following two sets:
$A=\{$ the empty cells in the lower $5 \times 6$ grid $\}$,
$B=\{$ the dominoes in the upper $5 \times 6$ gr... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
16. (5 points)
The following is a part of a Chinese chessboard, with a knight located at point $A$. According to the rules, the knight moves in an "L" shape, for example: from point $A$ to point $B$ it takes a minimum of 1 step, from point $A$ to point $C$ it takes a minimum of 2 steps. So, from point $A$ to point $Q$ ... | $5$ | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 38$ If $\left(r+\frac{1}{r}\right)^{2}=3$, then $r^{3}+\frac{1}{r^{3}}$ equals
(A) 1 .
(B) 2 .
(C) 0 .
(D) 3 .
(E) 6 .
(3rd American High School Mathematics Examination, 1952) | [Solution] From the assumption $\left(r+\frac{1}{r}\right)^{2}=3$, we know $r^{2}-1+\frac{1}{r^{2}}=0$, thus $r^{3}+\frac{1}{r^{3}}=\left(r+\frac{1}{r}\right)\left(r^{2}-1+\frac{1}{r^{2}}\right)=0$.
Therefore, the answer is $(C)$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
【Question 12】
It is known that $2^{29}$ is a nine-digit number consisting of nine different digits. Which digit is missing? | 【Analysis and Solution】
Number Theory, Congruence.
$$
2^{1} \equiv 2(\bmod 9), \quad 2^{2} \equiv 4(\bmod 9), \quad 2^{3} \equiv 8(\bmod 9), \quad 2^{4} \equiv 7(\bmod 9), \quad 2^{5} \equiv 5(\bmod 9), \quad 2^{6} \equiv 1(\bmod 9) ;
$$
The remainder of $2^{n} \div 9$ cycles through “ $2,4,8,7,5,1$ ”;
Since $29 \div 6... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Given: $1-\frac{1}{6+\frac{1}{6+\frac{1}{6}}}=\frac{1}{\mathrm{~A}+\frac{1}{\mathrm{~B}+\frac{1}{1}}}$ where $A$, $B$, and $C$ are all natural numbers greater than 0 and distinct from each other,
then $(A+B) \div C=$
Translate the above text into English, preserving the original text's line breaks and fo... | 【Solution】Solution: $1-\frac{1}{6+\frac{1}{6+\frac{1}{6}}}$,
$$
\begin{array}{l}
=1-\frac{1}{6+\frac{6}{37}}, \\
=1-\frac{1}{\frac{228}{37}}, \\
=1-\frac{37}{228}, \\
=\frac{191}{228} .
\end{array}
$$
So $\frac{1}{B+\frac{1}{B+\frac{1}{C}}}=\frac{191}{228}$,
that is, $\mathrm{A}+\frac{1}{\mathrm{~B}+\frac{1}{\mathrm{C... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given the following two equations in $x$: $6(x+8)=18 x ; 6 x-2(a-x)=2 a+x$ have the same solution, then $a=(\quad)$ | 【Analysis】
$$
\begin{aligned}
6(x+8) & =18 x \\
6 x+48 & =18 x \\
12 x & =48 \\
x & =4
\end{aligned}
$$
Substitute into the second equation, solve the equation
$$
\begin{aligned}
6 \times 4-2(a-4) & =2 a+4 \\
24-2 a+8 & =2 a+4 \\
4 a & =28 \\
a & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 34$ Write a sequence of consecutive positive integers starting with 1 on the blackboard, and erase one of the numbers. The arithmetic mean of the remaining numbers is $35 \frac{7}{17}$. What is the number that was erased?
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(E) Cannot be determined.
(33rd American High School Mathema... | [Solution] Let $n$ be the largest integer written on the blackboard. If 1 is the number that is erased, then the maximum arithmetic mean is
$$
\frac{2+3+\cdots+n}{n-1}=\frac{\frac{(n+1) n}{2}-1}{n-1}=\frac{n+2}{2};
$$
If $n$ is the number that is erased, then the minimum arithmetic mean is
$$
\frac{1+2+3+\cdots+n-1}{n... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. Given $f(x)=x^{5}-10 x^{3}+a x^{2}+b x+c$, if the roots of the equation $f(x)=0$ are all real numbers, $m$ is the largest of these 5 real roots, then the maximum value of $m$ is $\qquad$ | Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the roots of the equation. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0, \\
\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}=-10
\end{array} \Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=20 .\right.
$$
Thus, we have $\l... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 26 \quad 10$ people form a circle, each person chooses an integer and tells it to their two neighbors, then each person calculates and announces the average of the numbers chosen by their two neighbors, as shown in the figure. What is the number chosen by the person who announced $6$?
(A) 1 .
(B) 5 .
(C) 6 .
... | [Solution] Let the number chosen by the person who announced 6 be $x$, then the number chosen by the person who announced 4 is $5 \cdot 2 - x = 10 - x$, the number chosen by the person who announced 2 is $3 \cdot 2 - (10 - x) = x - 4$, the number chosen by the person who announced 10 is $1 \cdot 2 - (x - 4) = 6 - x$, t... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 21 (15th Canadian Mathematical Olympiad) Let $\varphi$ be a permutation on the set $X=\{1,2, \cdots, n\}$ (i.e., $\varphi(1), \varphi(2), \varphi(n)$ is a permutation of $1,2, \cdots, n$). If $\varphi(i)=i(i \in \mathbf{Z})$, then $i$ is called a fixed point of $\varphi$. Suppose the number of permutations on $... | Prove that from the previous problem, we can get
\[ f_{n}=n!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^{n}}{n!}\right]. \]
Let the number of permutations with only one fixed point $i$ be $g_{n i}$ $(i=1,2, \cdots, n)$, then $g_{n i}=f_{n-1}$, and
\[ g_{n}=g_{n 1}+g_{n 2}+\cdots+g_{n n}=n \cdot f_{... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
6. In a tetrahedron $ABCD$ with volume 12, points $E, F, G$ are on edges $AB, BC, AD$ respectively, and $AE = 2EB, BF = FC, AG = 2GD$. A plane through points $E, F, G$ intersects the tetrahedron in a section $EFGH$. The distance from point $C$ to this plane is 1. Then the area of this section is $\qquad$. | 6. 7 As shown in Figure (1), it is easy to know that $G E / / D B$, so $D B / /$ plane $F E H G$,
then $H F / / B D$, thus, $H$ is the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{A}, h_{B}, h_{C}, h_{D}$, respectively, with $h_{B}=h_{C}=h_{D}=\frac{1}{2} h_{A}$. Given ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. (5 points) To repair a section of road, 24 people can complete it in 12 days. Now, after 24 people have worked for 4 days, 8 more people are added. It will take $\qquad$ more days to complete the repair. | 【Solution】Solve: $24 \times(12-4) \div(24+8)$
$$
\begin{array}{l}
=24 \times 8 \div 32 \\
=6 \text { (days) }
\end{array}
$$
Answer: It will take another 6 days to complete the repair.
Therefore, the answer is: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
51. If $\frac{b}{a}=\frac{c}{b}=\frac{2}{c}$, then $b^{2}-c^{2}-a c+b c-2 a+2 b=$ | Reference answer: 0 | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 56, Find the value of $a$ that makes $\sin 4 x \cdot \sin 2 x-\sin x \cdot \sin 3 x=a$ have a unique solution in $[0, \pi)$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Question 56, Solution: Organizing the knowledge:
$$
\begin{array}{l}
\sin 4 x \cdot \sin 2 x - \sin x \cdot \sin 3 x \\
= -\frac{1}{2}(\cos 6 x - \cos 2 x) - \left[-\frac{1}{2}(\cos 4 x - \cos 2 x)\right] \\
= \frac{1}{2}(\cos 4 x - \cos 6 x)
\end{array}
$$
Let \( f(x) = \frac{1}{2}(\cos 4 x - \cos 6 x) \), then \( f(... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Given the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}=\left(x^{2003}+x^{2002}+2\right)^{2001}$, then $a_{0}-\frac{a_{1}}{2}-\frac{a_{2}}{2}+a_{3}-\frac{a_{4}}{2}-\frac{a_{5}}{2}+a_{6}-\frac{a_{7}}{2}-\frac{a_{8}}{2}+\cdots=$ $\qquad$ | 10. Let $x=w=\frac{-1+\sqrt{3} i}{2}$, then $a_{0}+a_{1} w+a_{2} w w^{2}+a_{3}+a_{4} w+a_{5} w^{2}+\cdots$ $=1$.
So $\left(a_{0}-\frac{a_{1}}{2}-\frac{a_{2}}{2}+a_{3}-\frac{a_{4}}{2}-\frac{a_{5}}{2}+\cdots\right)+\frac{\sqrt{3}}{2} i\left(a_{1}-a_{2}+\cdots\right)=1$.
So $a_{0}-\frac{a_{1}}{2}-\frac{a_{2}}{2}+a_{3}-\fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$33 \cdot 33[x]$ represents the greatest integer not exceeding $x$, then the number of integers $x$ that satisfy
$$
[-77.66 x]=[-77.66] x+1
$$
is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1987) | [Solution] The given equation is $[-77.66 x]=-78 x+1$, or $[(-78+0.34) x]=-78 x+1$.
Since $x$ is an integer, then $-78 x+[0.34 x]=-78 x+1$, which means $[0.34 x]=1$. Therefore, $1 \leqslant 0.34 x \leqslant 2$, which implies $3 \leqslant x \leqslant 5$.
$$
\therefore \quad x=3,4,5 \text {. }
$$
Therefore, the answer i... | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
94. There are 10 chickens and rabbits in total, and they are put into two cages. In one cage, the number of chicken and rabbit heads is the same, and in the other cage, the number of chicken and rabbit legs is the same. Therefore, there are $\qquad$ chickens in total. | answer: 6 | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
18.41 As shown in the figure, $P$ is a point inside the square $ABCD$, and $S_{\triangle PAB}=5, S_{\triangle PAD}=$ 2, then $S_{\triangle PAC}$ equals
(A) 2.
(B) 3.
(C) $3 \frac{1}{2}$.
(D) 4.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1993) | [Solution] Consider the extreme case where $P$ is on the diagonal $BD$ of $\square ABCD$.
According to the problem, $S_{\triangle PAB}=5, S_{\triangle PAD}=2$. Let $S_{\triangle APO}=S$,
then $S_{\triangle AOH}+S=5, S_{\triangle AOD}-S=2$,
and $S_{\triangle AOH}=S_{\triangle AOH}$,
solving this gives $S=1.5$, thus $S_{... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
20. Given that $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$ such that the inequality $\left(a x^{2}+b y^{2}+ 4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}$ always holds, and find the conditions for equality. | 20. Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.
Below we prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that $\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}$.
When $x$ or $y$ ... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
17-202 A scalene triangle, whose sides are integers, has a perimeter less than 13. The number of such triangles is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 18.
(7th American High School Mathematics Examination, 1956) | [Solution] Let $x, y, z$ be the lengths of the sides of a triangle. According to the problem and the triangle inequality, we have
$$
\left\{\begin{array}{l}
x+y+z \leq 12, \\
y+z>x, \\
z+x>y,
\end{array}\right.
$$
Since $x, y, z$ are all integers, from (1) we have $x+y+z \leq 12$.
Assuming $z$ is the longest side, fro... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
11. Given positive real numbers $a, b, c$ satisfying $a b+b c+c a=1$, find the minimum value of $\sqrt{a^{2}+a b+b^{2}}+\sqrt{b^{2}+b c+c^{2}}+\sqrt{c^{2}+c a+a^{2}}$. | Solution
Analysis 1: Since $a^{2}+a b+b^{2}=(a+b)^{2}-a b \geqslant(a+b)^{2}-\frac{1}{4}(a+b)^{2}=\frac{3}{4}(a+b)^{2}$,
then $\sqrt{a^{2}+a b+b^{2}}+\sqrt{b^{2}+b c+c^{2}}+\sqrt{c^{2}+c a+a^{2}} \geqslant \frac{\sqrt{3}}{2}(a+b+b+c+c+a)$ $=\sqrt{3}(a+b+c)$, and $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3 \Rightarrow a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
【Question 7】
Red pen costs 5 yuan each, blue pen costs 7 yuan each, spent 102 yuan in total for 16 pens, bought $\qquad$ blue pens. | 【Analysis and Solution】
Chicken and rabbit in the same cage.
(Method One)
Assume all 102 pens are red pens, then it would cost $5 \times 16=80$ yuan; which is $102-80=22$ yuan less than the actual amount;
One red pen costs $2$ yuan less than one blue pen; therefore, the number of blue pens bought is $22 \div 2=11$.
(Me... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given that $a, b, c$ satisfy
$$
a x+b y+2 c=0, c \neq 0, a b-c^{2} \geqslant 0 \text {. }
$$
Then the maximum value of $x y$ is ( ).
(A) $\frac{1}{10}$
(B) 1
(C) $\frac{6}{5}$
(D) 2 | 4. B.
Let $x y=m$. From $a b-c^{2} \geqslant 0, c \neq 0$, we know
$$
\begin{array}{l}
a b>0 \Rightarrow b \neq 0 \\
\Rightarrow y=\frac{-a x-2 c}{b} \Rightarrow m=x \frac{-a x-2 c}{b} \\
\Rightarrow a x^{2}+2 c x+b m=0 \\
\Rightarrow \Delta=(2 c)^{2}-4 a b m \geqslant 0 \\
\Rightarrow m \leqslant \frac{c^{2}}{a b} \l... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
17. Let $A=\{1,2,3, \cdots, 17\}$, for the mapping $f: A \rightarrow A$, denote $f^{(1)}(x)=f(x), f^{(k+1)}(x)=$ $f\left(f^{(k)}(x)\right), k \in \mathbf{N}_{+}$. Suppose the mapping $f$ from $A$ to $A$ satisfies the condition: there exists a positive integer $M$, such that (1) when $m < M, 1 \leqslant i \leqslant 16$,... | 17. Use the vertices of a regular 17-sided polygon to represent the numbers $1,2, \cdots, 17$, and let 18 be represented by the same vertex as 1. The pair $i, i+1$ represents the edge connecting vertices $i$ and $i+1$ $(i=1,2, \cdots, 17)$, and the pair $\{i, j\}(i-j \neq \pm 1(\bmod 17))$ represents a diagonal connect... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
11. (10 points) If 2 tennis rackets and 7 tennis balls cost a total of 220 yuan, and 1 tennis racket is 83 yuan more expensive than 1 tennis ball. Find the unit price of the tennis ball. | 【Solution】Solve: $220-83 \times 2$
$$
\begin{array}{l}
=220-166 \\
=54 \ (\text{yuan}) \\
54 \div (2+7) \\
=54 \div 9 \\
=6 \ (\text{yuan})
\end{array}
$$
Answer: Each tennis ball costs 6 yuan. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $P$ be any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with foci at $F_{1}(-c, 0), F_{2}(c, 0)$. $P F_{1}$ and $P F_{2}$ intersect the ellipse at points $A$ and $B$, respectively. If $a^{2}$, $b^{2}$, and $c^{2}$ form an arithmetic sequence, then $\frac{\left|P F_{1}\right|}{\left|A ... | 4.4.
As shown in Figure 1.
By the definition of an ellipse, the similarity of triangles, and the properties of ratios, we have
$$
\begin{array}{l}
e=\frac{\left|P F_{1}\right|}{|P Q|}=\frac{\left|A F_{1}\right|}{|A C|} \Rightarrow \frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}=\frac{|P Q|}{|A C|} \\
\Rightarrow \f... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. (5 points) A cylindrical container with a depth of 30 cm, an outer diameter of 22 cm, and a wall thickness of 1 cm, is already filled with water to a depth of 27.5 cm. Now, a conical iron block with a base diameter of 10 cm and a height of 30 cm is placed inside. Then, $\qquad$ cubic centimeters of water will overfl... | 【Solution】Solution: The original volume of water is: $3.14 \times\left(\frac{20}{2}\right)^{2} \times 27.5=8635$ (cubic centimeters), after placing the conical iron block, the volume of water in the cylindrical container is:
$$
\begin{array}{l}
3.14 \times\left(\frac{20}{2}\right)^{2} \times 30-\frac{1}{3} \times 3.14 ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (8 points) A poplar tree, a willow tree, a locust tree, a birch tree, and a sycamore tree are planted in a row, with a distance of 1 meter between each adjacent pair of trees. The distance between the poplar tree and the willow tree, and the poplar tree and the locust tree is the same. The distance between the birch... | 2
【Solution】Solution: The distance between the poplar tree and the willow tree, as well as the locust tree, is equal, so the possible positions of the three trees could be: Willow $\square$ Poplar $\square$ Locust, Willow Poplar Locust $\square \square, \square$ Willow Poplar Locust $\square, \square \square$ Willow Po... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Given that $a, b, c$ are real numbers. If the polynomial $2 x^{3}-x^{2}+a x-5$ can be divided by $x^{2}+5$, and the quotient is $b x+c$, then $a+b+c=$ $\qquad$ . | $11$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
29. There are 10 birds on the ground pecking at food, and among any 5 birds, at least 4 are on the same circle. How many birds are on the circle with the most birds, at a minimum?
(1991 6th CMO Problem) | 29. First, prove that the circle with the most birds has at least 5 birds. Since each set of 4 birds on a circle is contained in $C_{6}^{1}=6$ sets of 5 birds, let the total number of sets of 4 birds on a circle be $X$, then these sets are contained in $6X$ sets of 5 birds (some sets of 5 birds may be counted multiple ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.39 $A$ and $B$ working together can complete a job in 2 days, $B$ and $C$ working together can complete it in 4 days, $C$ and $A$ working together can complete it in $2 \frac{2}{5}$ days, the number of days $A$ alone can complete the job is
(A) 1 .
(B) 3 .
(C) 6 .
(D) 12 .
(E) 2.8 .
(5th American High School Mathemat... | [Solution] Let $x, y, z$ be the number of days required for $A, B, C$ to complete the work individually. We have
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{1}{2}, \\
\frac{1}{y}+\frac{1}{z}=\frac{1}{4} \\
\frac{1}{z}+\frac{1}{x}=\frac{5}{12} .
\end{array}\right.
$$
(1) - (2) + (3) gives $\frac{2}{x}=\frac... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
7.40 Divide an $8 \times 8$ chessboard into $p$ rectangles, such that the rectangles satisfy the following conditions:
(1) The edges of each rectangle are along the grid lines of the chessboard;
(2) Each rectangle contains an equal number of white and black squares;
(3) If the number of white squares in the $i$-th rect... | ［Solution］ From the given, we have
$$
a_{1}+a_{2}+\cdots+a_{p}=32 .
$$
From (3), we also have $a_{i} \geqslant i, i=1,2, \cdots, p$, thus from (1) we have
$$
1+2+\cdots+p \leqslant 32 \text {. }
$$
Solving (2) yields $p \leqslant 7$.
Since $1+2+3+4+5+6+7=28$, the difference with 32 is 4, so it is only necessary to di... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) Xiao Ming bought 4 apples, 6 pears, and 8 peaches, while Xiao Hong bought 5 apples, 7 pears, and 6 peaches. In the following 18 days, each of them eats one fruit per day, with three days both eating apples; two days both eating pears, and three days one eating an apple and the other eating a pear. Theref... | 【Answer】Solution: (1) For three days, both ate apples, and for two days, both ate pears. After 5 days, Xiao Ming has 1 apple and 4 pears left, while Xiao Hong has 2 apples and 5 pears left;
(2) For three days, one person ate apples while the other ate pears, so when Xiao Ming ate the remaining 1 apple, Xiao Hong ate pe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example 20 (1) Let $n$ be a prime number greater than 3. Find the value of $\left(1+2 \cos \frac{2 \pi}{n}\right)\left(1+2 \cos \frac{4 \pi}{n}\right) \cdots\left(1+2 \cos \frac{2 m \pi}{n}\right)$.
(2) Let $n$ be a natural number greater than 3. Find the value of $\left(1+2 \cos \frac{\pi}{n}\right)\left(1+2 \cos \fr... | (1) Let $\omega=\mathrm{e}^{\frac{2 \pi i}{n}}$, then $\omega^{n}=1, \omega^{-\frac{n}{2}}=\mathrm{e}^{\pi i}=-1, 2 \cos \frac{2 k \pi}{n}=\omega^{k}+\omega^{-k}$,
$$
\begin{aligned}
\prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right) & =\prod_{k=1}^{n}\left(1+\omega^{k}+\omega^{-k}\right)=\prod_{k=1}^{n} \omega^{-k... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 75, Given real numbers $a \geq b \geq c \geq d, a+b+c+d=9, a^{2}+b^{2}+c^{2}+d^{2}=21$, find the minimum possible value of $\mathrm{ab}-\mathrm{cd}$. | Let $a+b=x$, then $c+d=9-x$, and $x \geq \frac{9}{2}$. Therefore, we have:
$$
\begin{array}{l}
x^{2}+(9-x)^{2}=(a+b)^{2}+(c+d)^{2} \\
=a^{2}+b^{2}+c^{2}+d^{2}+2(a b+c d)=21+2(a b+c d)
\end{array}
$$
According to the rearrangement inequality, we know: $a b+c d \geq a c+b d \geq a d+b c$, hence
$$
a b+c d \geq \frac{a b... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-3 t-3\right]$, then $t=$ | 1. -1 Explanation: The domain of an odd function is symmetric about the origin, and the right endpoint of the interval does not need to be smaller. Therefore, $-t=t^{2}-3 t-3$ $\geqslant 0 \Leftrightarrow t=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4
Determine all positive integers $n$ for which the equation
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0
$$
has an integer as a solution. | Answer: $n=1$.
Solution
If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.
For $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.
For $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to
$$
y^{n}+(1+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (1994 National High School Competition Question) Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$ and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is ( ).
A. 5
B. 6
C. 7
D... | 3. C. The original recurrence relation is transformed into $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$ (where $x=1$ is the root of the equation $3 x+x=4$). Let $b_{n}=a_{n}-1$, then $b_{n+1}=-\frac{1}{3} b_{n}, b_{1}=a_{1}-1=8$, so $b_{n}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}... | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}(n \in \mathbf{N}) \text {. }
$$
Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the units digit of $\left[a_{2017}\right]$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 6. A.
From the conditions, we have
$$
\left\{\begin{array}{l}
a_{n+2}=2 \sqrt{2} a_{n+1}-a_{n}, \\
a_{0}=0, \\
a_{1}=2 .
\end{array}\right.
$$
$$
\begin{array}{l}
\text { Then } a_{2 k+1}=2 \sqrt{2} a_{2 k}-a_{2 k-1} \\
= 2 \sqrt{2}\left(2 \sqrt{2} a_{2 k-1}-a_{2 k-2}\right)-a_{2 k-1} \\
=7 a_{2 k-1}-2 \sqrt{2} a_{2 ... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. Given $x+y=1$, for what values of real numbers $x, y$ does $\left(x^{3}+1\right)\left(y^{3}+1\right)$ achieve its maximum value. | Solve:
$\left(x^{3}+1\right)\left(y^{3}+1\right)=x^{3} y^{3}+x^{3}+y^{3}+1=x^{3} y^{3}+(x+y)^{3}-3 x y(x+y)+1$ $=x^{3} y^{3}-3 x y+2$, since $x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, let $t=x y \in\left(-\infty, \frac{1}{4}\right]$, $f(t)=t^{3}-3 t+2$, then $f^{\prime}(t)=3 t^{2}-3=3(t+1)(t-1)$,
thus $... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$, then the number of such sequences is ( ).
A. 2
B. 3
C. 4
D. 5 | 3. C
Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. Then, according to the problem, we have
$$
n a+\frac{n(n-1)}{2} d=97^{2},
$$
which simplifies to $[2 a+(n-1) d] n=2 \times 97^{2}$.
Since $n$ is a natural number no less than 3, and 97 is a prime number, the value of $n$ can ... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
15. Let $m>0$, if for any set of positive numbers $a, b, c$ satisfying $a b c \leqslant \frac{1}{4}$ and $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}<m$, there always exists a triangle with side lengths $a, b, c$, find the maximum value of the real number $m$, and explain the reason. | When $m>9$, take $a=b=\frac{1}{2}, c=1$, then $a, b, c$ cannot form a triangle $\Rightarrow m \leqslant 9$. On the other hand, suppose $m \leqslant 9$, there exist $a \leqslant b \leqslant c$ that cannot form a triangle, thus $a+b \leqslant c$. Let $d=\frac{1}{4 a b} \geqslant c$, and $d^{2} \geqslant c^{2} \geqslant(a... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Given $a \in \mathbf{Z}$, and $x^{6}-33 x+20$ can be divided by $x^{2}-x+a$, then the value of $a$ is | 1. Let $x^{6}-33 x+20=\left(x^{2}-x+a\right) \cdot g(x)$, where $g(x)=x^{4}+b x^{3}+c x^{2}+d x+e$. By comparing the constant terms on both sides, we know $a e=20$. Therefore, $a= \pm 1, \pm 2, \pm 4, \pm 5$.
By comparing the coefficients of the linear term, we know $a d-e=-33$. If $a= \pm 2$, then by $a e=20$, we kno... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given $(x+3)(x-1)^{4}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}$, then $a_{3}+a_{4}+a_{5}=$ | $$
\begin{array}{l}
\text { Let } x=1 \Rightarrow a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0 \text {, let } x=0 \Rightarrow a_{0}=3 . \\
a_{1}=1 \cdot(-1)^{4}+3 \cdot C_{4}^{3} \cdot(-1)^{3}=-11, a_{2}=1 \cdot C_{4}^{3} \cdot(-1)^{3}+3 \cdot C_{4}^{2} \cdot(-1)^{2}=14 \text {, }
\end{array}
$$
So $a_{0}+a_{1}+a_{2}=6 \Righ... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. At a little past 8 o'clock in the morning, two cars left the fertilizer plant one after another heading for Happiness Village. Both cars were traveling at a speed of 60 kilometers per hour. At 8:32, the distance the first car had traveled from the fertilizer plant was three times that of the second car. By 8:39, the... | 8.【Solution】$39-32=7$. In these 7 minutes, the distance each car travels is exactly equal to 1(=3-2) times the distance the second car had traveled by 8:32. Therefore, by 8:32, the first car had traveled $7 \times 3$ $=21$ (minutes), meaning it left the fertilizer factory at 8:11 $(32-21=11)$.
【Note】The conclusion of ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(5) Let $A=\{1,2,3, \cdots, 17\}$. For the mapping $f: A \rightarrow A$, denote $f^{[1]}(x)=$ $f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)(k \in \mathbf{N})$.
Consider a one-to-one mapping $f$ from $A$ to $A$ that satisfies the following conditions: There exists a natural number $M$ such that
(1) When $m<M, 1 \leqslant... | Solve for $M_{0}=8$. First, prove $M_{0} \geqslant 8$.
In fact, let the mapping $f(i)=3 i-2(\bmod 17)$, where $i \in A, f(i) \in$
A. If
then
$$
\begin{aligned}
f(i) & \equiv f(j)(\bmod 17) \\
3 i-2 & \equiv 3 j-2(\bmod 17) \\
i & \equiv j(\bmod 17)
\end{aligned}
$$
therefore
$$
i=j
$$
The mapping $f$ is a one-to-one... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. Given a convex $n$-sided polygon, each of its diagonals is the perpendicular bisector of at least one other diagonal. Find all possible positive integers $n(n>3)$.
untranslated text remains in its original format and line breaks are preserved. | 3. Suppose there exists an $n$-sided polygon satisfying the given conditions. Let the set of all its diagonals be $T$.
From the given, we can construct a mapping
$$
f: T \rightarrow T, l \mapsto m=f(l),
$$
where $l$ is the perpendicular bisector of $m$.
If such an $m$ is not unique, then choose any one.
If there exist... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 4-9 Embed 3 red beads and 2 blue beads on a ring (as shown in Figure 4-11), how many different schemes are there.
| $$
\begin{array}{l}
G:(1)(2)(3)(4)(5),(12345),(13524) \\
\quad(14253),(15432) \\
\quad(1)(25)(34),(2)(13)(45) \\
\quad(3)(15)(34),(4)(12)(35)
\end{array}
$$
$$
\text { (5) }(14)(23) \text {. }
$$
Corresponding to $g_{0}=(1)(2)(3)(4)(5)$, there are $C(5,3)=10$ schemes with 3 red and 2 blue tiles, which remain unchanged... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. To cut two different sizes of steel plates into three specifications $A$, $B$, and $C$, each type of steel plate can simultaneously cut the number of smaller steel plates of the three specifications as shown in the table below:
\begin{tabular}{|l|c|c|c|}
\hline & $A$ specification & $B$ specification & $C$ specifica... | $3 y=27$ intersects at $\left(\frac{18}{5}, \frac{39}{5}\right)$, so $m+n \geqslant \frac{18}{5}+\frac{39}{5}=\frac{57}{5}$, and when $m=4, n=8$ we have $m+n=12$. Therefore, the minimum value of $m+n$ is 12. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality $a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant k\left(a^{3}+b^{3}+c^{3}+d^{3}\right)$ holds. | 4. First estimate the upper bound of $k$. When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that $a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$. First, we prove a lemma.
Lemma If $x, y \in[0,1]$, then $x^{2}... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11. Bing Dwen Dwen practiced skiing for a week, during which the average length of skiing per day for the last four days was 4 kilometers more than the average length of skiing per day for the first three days, and the average length of skiing per day for the last three days was 3 kilometers more than the average lengt... | $12$ | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Isabella drew two circles with radii of $2$ and $3$ on a piece of paper, and then drew the common tangents to these two circles, which turned out to be $k$ in number. Then, $k$ has ( ) possible scenarios.
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 | 11. D.
When two circles are concentric, $k=0$; when two circles are internally tangent, $k=1$; when two circles intersect, $k=2$; when two circles are externally tangent, $k=3$; when two circles are separated, $k=4$. Therefore, there are 5 possible values. | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
8. Given that the vertex of a parabola is on the $x$-axis, the directrix is the $y$-axis, and the shortest distance from point $A(4,0)$ to a point on the parabola is 2. Then, the number of such parabolas is $\qquad$. | 8. 3 .
Let the equation of the parabola be $y^{2}=2 p(x-a)$. Given $a=\frac{p}{2}$, we have $2 p=4 a$, which means the equation of the parabola is $y^{2}=4 a(x-a)$ $(x \in[a,+\infty))$. Let $Q(x, y)$ be a point on the parabola, then
$$
|A Q|=\sqrt{(x-4)^{2}+y^{2}}=\sqrt{x^{2}-(8-4 a) x+16-4 a^{2}}=\sqrt{[x-(4-2 a)]^{2... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. If the positive integer $n$ makes $\sqrt{3}$ always lie between $1+\frac{3}{n}$ and $1+\frac{3}{n+1}$, then $n=$ . $\qquad$ | 4. 4 .
When $n$ is a positive integer, it is easy to know that $\frac{3}{n+1}<\frac{3}{n}$.
$$
\begin{array}{l}
\text { By } \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow \frac{1+3 \sqrt{3}}{2}<n<\frac{3}{\sqrt{3}-1}=\frac{3+3 \sqrt{3}}{2} .
\end{array}
$$
Therefore, $n \in\left(\frac{1+3 \sqrt{3}}{2}, \frac{3+... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9. (15 points) The values of the playing cards are as shown in the figure, with the maximum being 13 and the minimum being 1. Now, Xiaoming has 3 playing cards with different values. The sum of the values of the first and second cards is 25, and the sum of the values of the second and third cards is 13. What is the val... | 【Analysis】Since the maximum is 13 and the minimum is 1, and the sum of the points of the first and second playing cards is $25, 25=13+12$, which means the points of the first and second cards must be 12 or 13. Also, the sum of the points of the second and third playing cards is 13, and since the minimum is 1, the secon... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
34. A bag of fruit contains 10 fruits, with an even number of apples, at most two oranges, a multiple of 3 bananas, and at most one pear. How many types of these bagged fruits are there? | 34. Generating function $f(x)=\left(1+x^{2}+x^{4}+\cdots\right)\left(1+x+x^{2}\right)\left(1+x^{3}+x^{6}+\cdots\right)(1+x)$
$$
\begin{array}{l}
=\frac{1}{1-x^{2}}\left(1+x+x^{2}\right) \cdot \frac{1}{1-x^{3}}(1+x)=\frac{1}{(1-x)^{2}} \\
=\sum_{n=0}^{\infty}(n+1) x^{n} . \\
f_{n}=n+1, f_{10}=11 .
\end{array}
$$ | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. Find the minimum value of the function $(\sqrt{2} \sin x-3 \tan y)^{2}+(\sqrt{2} \cos x-3 \cot y)^{2}$, where $x, y \in\left(0, \frac{\pi}{2}\right)$. | 9. Since $(\sqrt{2} \sin x-2 \tan y)^{2}+(\sqrt{2} \cos x-3 \cot y)^{2}$ can be regarded as the square of the distance between two points $A(\sqrt{2} \sin x, \sqrt{2} \cos x)$ and $B(3 \tan y, 3 \cot y)$ in the Cartesian coordinate system. At this time, the parametric equation of the trajectory of the moving point $A$ ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3.7 Let $P$ be a point on the circumcircle of a regular $n$-gon, and let $f(P)$ be the product of the distances from $P$ to each vertex. Find the maximum value of $f(P)$. | Consider a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, with center $O$ and radius 1. Establish a complex plane with $O$ as the origin and $O A_{n}$ as the positive direction of the real axis. Let $P$ be a point on the unit circle, corresponding to the complex number $z$. Then,
$$
f(P)=\prod_{i=1}^{n}\left|z-\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Xiaoming's honor stand has 3 gold medals, 2 silver medals, and 1 bronze medal, arranged from left to right in the order of gold, silver, and bronze. At the sports meet, he won 1 more gold medal and 1 more silver medal, and he needs to place these two medals on the honor stand while ensuring all medals are still arra... | $2$ | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
(3) On the ground, there are 10 birds pecking at food, among any 5 birds, at least 4 are on a circle. How many birds are there at least on the circle with the most birds? | First, we prove that at least 5 birds are on the same circumference. If not, then the 10 birds can form $C_{10}^{5}=252$ groups of 5 birds, and each group of 5 birds has 4 birds on the same circumference. However, these 4 birds can belong to 6 different groups of 5 birds, so there should be $\frac{252}{6}=42$ different... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(7) Given the sequence $\left\{a_{n}\right\}$ where the first term $a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}$, then $a_{2011}=$ | (7) 2 Hint: $a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{13-7 a_{n}}{5-3 a_{n}}$,
$$
a_{n+3}=\frac{13-7 a_{n+1}}{5-3 a_{n+1}}=\frac{13-7 \cdot \frac{5 a_{n}-13}{3 a_{n}-7}}{5-3 \cdot \frac{5 a_{n}-13}{3 a_{n}-7}}=a_{n} \text {. }
$$
Therefore,
$$
a_{2011}=a_{3 \times 670+1}=a_{1}=2 .
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given $1050-840 \div \square \times 8=90$, then $\square=$ $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | I analysis and solution】Calculation problem, it is easy to get $\square=7$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Let the set $M=\{1,2, \cdots, 2023\}$, for a subset $A=\left\{a_{1}, \cdots, a_{k}\right\}$ of $M$, let it correspond to a number $\alpha=(-1)^{a_{1}+\cdots+a_{k}}$ (the empty set $\varnothing$ corresponds to the number $\alpha=(-1)^{0}=1$). For all subsets $A$ of $M$, find the sum of the numbers they correspond to... | For the set $M$, the number of odd subsets (the sum of all elements in the subset is odd) is equal to the number of even subsets (the sum of all elements in the subset is even, the sum of elements in the empty set is 0).
Consider all subsets $A$ of $M$ that do not contain the element 1. Clearly, $A \bigcup\{1\}$ is al... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Let $f(x)=|x+1|+|x|-|x-2|$. Then $f(f(x))+1=0$ has different solutions.
The number of different solutions is: | 8.
Notice,
$$
f(x)=\left\{\begin{array}{ll}
-x-3, & x \leqslant-1 ; \\
x-1, & -12 .
\end{array}\right.
$$
From $f(f(x))+1=0$
$\Rightarrow f(x)=-2$ or 0
$\Rightarrow x=-1$ or -3 or $\frac{1}{3}$, a total of three solutions. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$, to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{x_{0}} 1993$ always hold, then the maximum value of $k$ is $\qquad$. | Given that $\log _{\frac{x_{0}}{x_{1}}} 1993, \log _{\frac{x_{1}}{x_{2}}} 1993, \log _{\frac{x_{2}}{x_{3}}} 1993, \log _{\frac{x_{0}}{x_{3}}} 1993>0$,
and $\log _{\frac{x_{1}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{1}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant k \log _{\frac{x_{0}}{x_{3}}} 1993$
$\Rightarrow \frac{... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\sqrt{a+b+c+d}+\sqrt{a^{2}-2 a+3-b}- \\
\sqrt{b-c^{2}+4 c-8}=3 .
\end{array}
$$
Then the value of $a-b+c-d$ is ( ).
(A) -7
(B) -8
(C) -4
(D) -6 | 5. A.
$$
\begin{array}{l}
\text { Given } a^{2}-2 a+3-b \geqslant 0, \\
b-c^{2}+4 c-8 \geqslant 0,
\end{array}
$$
we know
$$
\begin{array}{l}
b \leqslant-(a+1)^{2}+4 \leqslant 4, \\
b \geqslant(c-2)^{2}+4 \geqslant 4 .
\end{array}
$$
Thus, $b=4, a=-1, c=2$.
Substituting these into the known equations gives $d=4$.
The... | -7 | Algebra | MCQ | Yes | Yes | olympiads | false |
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