problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. If $100a+64$ and $201a+64$ are both four-digit numbers, and both are perfect squares, then the value of the integer $a$ is | 4.17.
Let $100 a+64=m^{2}, 201 a+64=n^{2}$, then
$$
32 \leqslant m, n<100 \text {. }
$$
Subtracting the two equations gives
$$
101 a=n^{2}-m^{2}=(n+m)(n-m) \text {. }
$$
Since 101 is a prime number, and $-101<n-m<101, 0<n$ $+m<200$, so, $n+m=101$.
Thus, $a=n-m=2 n-101$.
Substituting $a=2 n-101$ into $201 a+64=n^{2}$... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,743 |
One, (20 points) Let $m, n$ be positive integers, and $m \neq 2$. If for all real numbers $t$, the quadratic function
$$
y=x^{2}+(3-m t) x-3 m t
$$
has its graph intersecting the $x$-axis at two points whose distance is no less than $|2 t+n|$, find the values of $m, n$. | Given the quadratic equation
$$
x^{2}+(3-m t) x-3 m t=0
$$
has roots $m t$ and $-3$, the graph of the quadratic function
$$
y=x^{2}+(3-m t) x-3 m t
$$
intersects the $x$-axis at two points, and the distance between these points is $|m t+3|$. According to the problem, $|m t+3| \geqslant|2 t+n|$, which implies
$$
\begi... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,744 |
II. (25 points) As shown in Figure 1, quadrilateral $ABCD$ is a trapezoid, and point $E$ is a point on the upper base $AD$. The extension of $CE$ intersects the extension of $BA$ at point $F$. A line through point $E$ parallel to $BA$ intersects the extension of $CD$ at point $M$, and $BM$ intersects $AD$ at point $N$.... | Let $M N$ and $E F$ intersect at point $P$.
Since $N E \parallel B C$, we have
$\triangle P N E \sim \triangle P B C$.
Thus, $\frac{P N}{P B}=\frac{P E}{P C}$, which means $P B \cdot P E=P N \cdot P C$.
Similarly, $P B \cdot P E=P M \cdot P F$.
Therefore, $P N \cdot P C=P M \cdot P F$.
Hence, $\frac{P M}{P N}=\frac{P C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,745 |
Three. (25 points) Given that $a$ is a positive integer. If the roots of the equation
$$
x^{3}+(a+17) x^{2}+(38-a) x-56=0
$$
are all integers, find the value of $a$ and the integer roots of the equation. | Three, it is easy to observe that the equation has an integer root $x_{1}=1$.
The left side of the equation can be factored as
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Since $a$ is a positive integer, the quadratic equation in $x$
$$
x^{2}+(a+18) x+56=0
$$
has a discriminant $\Delta=(a+18)^{2}-224>0$, s... | a=12, \text{ roots } 1, -2, -28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,746 |
One. (20 points) Let $m, n$ be positive integers, and $m \neq 2$. The quadratic function $y=x^{2}+(3-m t) x-3 m t$ intersects the $x$-axis at two points with a distance of $d_{1}$ between them. The quadratic function $y=-\dot{x}^{2}+(2 t-n) x+2 n t$ intersects the $x$-axis at two points with a distance of $d_{2}$ betwe... | $$
\begin{array}{l}
\text { Because the roots of the quadratic equation } \\
x^{2}+(3-m t) x-3 m t=0 \\
\text { are } m t \text { and } -3, \text { so } d_{1}=|m t+3|; \\
\text { and the roots of the quadratic equation } \\
-x^{2}+(2 t-n) x+2 n t=0 \\
\text { are } 2 t \text { and } -n, \text { so } d_{2}=|2 t+n|. \\
\... | m=3, n=2 \text{ or } m=6, n=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,747 |
Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.
---
The above text has been translat... | Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$
which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersect... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,748 |
Three. (25 points) Let $a$ be a positive integer. If the quadratic function $y=2 x^{2}+(2 a+23) x+10-7 a$ and the reciprocal function $y=\frac{11-3 a}{x}$ have a common integer point (a point where both the x-coordinate and y-coordinate are integers), find the value of $a$ and the corresponding common integer point.
-... | Three, combining the two equations and eliminating $y$ yields
$$
2 x^{2}+(2 a+23) x+10-7 a=\frac{11-3 a}{x} \text {, }
$$
which simplifies to $2 x^{3}+(2 a+23) x^{2}+(10-7 a) x+3 a-11=0$.
Factoring gives
$$
(2 x-1)\left[x^{2}+(a+12) x+11-3 a\right]=0 \text {. }
$$
If the graphs of the two functions have a common inte... | a=12, (1,-25), (-25,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,749 |
2. Given $a=\sqrt{2006}-\sqrt{2005}$,
$$
\begin{array}{l}
b=\sqrt{2007}-\sqrt{2006}, \\
c=\sqrt{2008}-\sqrt{2007} .
\end{array}
$$
Then the correct conclusion is ( ).
(A) $a>b>c$
(B) $c>b>a$
(C) $b>a>c$
(D) $b>c>a$ | 2.A.
Since $a=\frac{1}{\sqrt{2006}+\sqrt{2005}}$,
$$
b=\frac{1}{\sqrt{2007}+\sqrt{2006}}, c=\frac{1}{\sqrt{2008}+\sqrt{2007}},
$$
Therefore, $a>b>c$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,751 |
3. The blood types of parents and the possible blood types of their children have the relationship shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Parents' \\
Blood Type
\end{tabular} & 0,0 & $\mathrm{O}, \mathrm{A}$ & $\mathrm{O}, \mathrm{B}$ & $\mathrm{O}, \mathrm{AB}$ & $\mathrm{A... | 3. D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 718,752 |
Example 7 Several people gather, some of whom know each other. It is known that if two people have the same number of acquaintances among the gatherers, then they do not have any common acquaintances. Prove: If there is someone among the gatherers who has at least 2008 acquaintances, then there must also be someone who... | Explanation: Consider the person with the most acquaintances (if there are several such people, choose any one of them) and denote this person as $A$. Suppose $A$ knows $n$ people, who are sequentially denoted as $B_{1}, B_{2}, \cdots, B_{n}$.
Since any two people $B_{i}$ and $B_{j}$ among $B_{1}, B_{2}, \cdots, B_{n}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,753 |
6. Given that the three sides of a triangle $a, b, c$ are all integers, and $a \leqslant b < c$. If $b=7$, then the number of such triangles is ( ).
(A) 21
(B) 28
(C) 49
(D) 54 | 6. A.
When $a=2$, there is 1; when $a=3$, there are 2; when $a=4$, there are 3; when $a=5$, there are 4; when $a=6$, there are 5; when $a=7$, there are 6. In total, there are 21. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,756 |
7. As shown in Figure 1, the line $l_{1}: y=x+1$ and the line $l_{2}: y=-x-\frac{1}{2}$ divide the Cartesian coordinate plane into four parts. Then the point $\left(-\frac{4}{5}, \frac{4}{7}\right)$ is in ( ).
(A) the first part
(B) the second part
(C) the third part
(D) the fourth part | $$
\begin{array}{l}
\text { 7.B. } \\
\text { Since }\left(-\frac{4}{5}\right)+1=\frac{1}{5}<\frac{4}{7}, \text { and } \\
-\left(-\frac{4}{5}\right)-\frac{1}{2}=\frac{3}{10}<\frac{4}{7},
\end{array}
$$
Therefore, the point $\left(-\frac{4}{5}, \frac{4}{7}\right)$ is above $l_{1}$ and $l_{2}$, i.e., in the second regi... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,757 |
8. Given that the real number $a$ satisfies
$$
|2006-a|+\sqrt{a-2007}=a \text {. }
$$
Then, the value of $a-2006^{2}$ is ( ).
(A) 2005
(B) 2006
(C) 2007
(D) 2008 | 8.C.
Since $a \geqslant 2007$, we have $a-2006+\sqrt{a-2007}=a$. Therefore, $\sqrt{a-2007}=2006$, which means $a-2006^{2}=2007$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,758 |
9. Suppose the fraction $\frac{n-13}{5 n+6}(n \neq 13)$ is not in its simplest form. Then, the smallest positive integer value of $n$ can be ( ).
(A) 84
(B) 68
(C) 45
(D) 115 | 9.A.
Let $d$ be a common divisor of $n-13$ and $5 n+6$, then
$$
\begin{array}{l}
d|(n-13), d|(5 n+6) \\
\Rightarrow d|[(5 n+6)-5(n-13)] \Rightarrow d| 71 .
\end{array}
$$
Since 71 is a prime number, then $d=71$.
As $d \mid(n-13)$, so, $n-13 \geqslant 71$.
Hence $n \geqslant 84$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,759 |
10. As shown in Figure $2, P$ is a point inside $\triangle A B C$, and the extensions of $B P$, $C P$, and $A P$ intersect $A C$, $A B$, and $B C$ at points $E$, $F$, and $D$ respectively. Consider the following three equations:
(1) $\frac{S_{\triangle A B P}}{S_{\triangle A P C}}=\frac{B D}{C D}$;
(2) $\frac{S_{\trian... | 10.D.
From $\frac{S_{\triangle A B P}}{S_{\triangle B P D}}=\frac{S_{\triangle A P C}}{S_{\triangle P C D}}=\frac{A P}{P D} \Rightarrow \frac{S_{\triangle A P B}}{S_{\triangle A P C}}=\frac{S_{\triangle B P D}}{S_{\triangle P C D}}=\frac{B D}{C D}$, therefore
(1) is correct;
From $\frac{A B}{B F}=\frac{S_{\triangle A ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,760 |
11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$
always holds. Then the maximum value that $m$ can take is $\qquad$ . | When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$,
$$
|x+1|+\sqrt{x-1}+|x-2|
$$
the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,761 |
13. The military training base purchases apples to comfort the trainees. It is known that the total number of apples, when represented in octal (base 8), is $\overline{a b c}$, and when represented in septenary (base 7), it is $\overline{c b a}$. Therefore, the total number of apples, when represented in decimal (base ... | 13.220.
$$
\begin{array}{l}
\text { Given } 1 \leqslant a, b, c \leqslant 6, \\
a \times 8^{2}+b \times 8+c \\
=c \times 7^{2}+b \times 7+a, \\
63 a+b-48 c=0, \\
b=3(16 c-21 a),
\end{array}
$$
Therefore, $b=0,3,6$.
Upon verification, $b=3$ meets the condition. Hence, $b=3, c=4, a=3$. Thus, $3 \times 8^{2}+3 \times 8+4... | 220 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,763 |
Example 8 Given $n$ points in the plane, not all on a straight line. Then there must be a straight line that passes through exactly two of these $n$ points. | Explanation: Since the $n$ points on the plane do not all lie on the same line, for any two points among these $n$ points, there are other points among these $n$ points that are not on the line determined by these two points. For each line, calculate the distance from the points not on this line to the line. There may ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,764 |
14. A heptagonal board is shown in Figure 4, with 7 vertices numbered from 0 to 6, referred to as seven cells. A piece is placed in cell 0. Now, move this piece counterclockwise, moving 1 cell the first time, 2 cells the second time, $\cdots \cdots$ $n$ cells the $n$-th time. Then the cells where the piece does not sto... | 14.2,4,5.
Try to find:
(1) The squares where the chess piece never stops are $2,4,5$;
(2) The number of the square where the chess piece stops repeats every 7 moves (i.e., the $k$-th move and the $(k+7)$-th move stop at the same square).
Proof: On the $k$-th move, the number of squares moved is
$$
1+2+\cdots+k \text {... | 2,4,5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,765 |
15. There are 40 sets of CASIO cards, each set consisting of C, A, S, I, O five cards stacked in the order of C, A, S, I, O from top to bottom. Now, these 40 sets of cards are stacked together from top to bottom, and then the first card is discarded, the second card is placed at the bottom, the third card is discarded,... | Three, 15. (1) 40 sets of CASIO cards total 200 cards. Number the 200 cards from top to bottom as $1,2, \cdots, 200$.
According to the operation rules, when 100 cards are discarded, the remaining card numbers are $2,4, \cdots, 200$. If another 12 cards are discarded, the involved cards are 24, with numbers $2,4, \cdot... | 22, \text{I of the 29th set} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,766 |
16. As shown in Figure 5, in $\triangle ABC$, $D$ is a point inside $\triangle ABC$. Extend $BA$ to point $E$, and extend $DC$ to point $F$, such that $AE = CF$. $G$, $H$, and $M$ are the midpoints of $BD$, $AC$, and $EF$, respectively. If $G$, $H$, and $M$ are collinear, prove that $AB = CD$. | 16. As shown in Figure 7, take the midpoint $T$ of $BC$, the midpoint $S$ of $AF$, and connect $GT$, $TH$, $HS$, $SM$.
Since $G$, $H$, and $M$ are the midpoints of $BD$, $AC$, and $EF$ respectively, we have:
$$
\begin{array}{l}
MS \perp \frac{1}{2} AE, \\
HS \perp \frac{1}{2} CF, \\
HS = SM.
\end{array}
$$
Thus, $\an... | AB = CD | Geometry | proof | Yes | Yes | cn_contest | false | 718,767 |
Given that $AB$ is a chord of $\odot O$, $M$ is the midpoint of arc $\overparen{AB}$, $C$ is any point outside $\odot O$, and from point $C$ two tangents $CS$ and $CT$ are drawn to $\odot O$. Connect $MS$ and $MT$ to intersect $AB$ at points $E$ and $F$ respectively. Draw perpendiculars from $E$ and $F$ to $AB$, inters... | As shown in Figure 1, first connect $O M$. By the perpendicular diameter theorem, it is easy to see that $\triangle X E S$ and $\triangle O M S$ are similar, so $\triangle X E S$ is an isosceles triangle. Therefore, a circle can be drawn with $X$ as the center and $X E$ and $X S$ as the radii, which is tangent to the c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,768 |
II. A rational number $x$ is called "good" if it satisfies the following conditions: $x=\frac{p}{q}>1$, where $p, q$ are coprime positive integers, and there exist constants $\alpha, N$ such that for any positive integer $n \geqslant N$, we have
$$
\left|\left\{x^{n}\right\}-\alpha\right| \leqslant \frac{1}{2(p+q)},
$$... | Obviously, every integer greater than 1 is good.
Next, we prove that every good rational number must also be an integer greater than 1.
Let \( m_{n} = \left[x^{n+1}\right] - \left[x^{n}\right] \). When \( n \geqslant N \),
\[
\begin{array}{l}
\left|(x-1) x^{n} - m_{n}\right| = \left|\left\{x^{n+1}\right\} - \left\{x^{n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,769 |
Find all functions $f: \mathbf{Q}_{+} \rightarrow \mathbf{Q}_{+}$, such that
$$
f(x)+f(y)+2 x y f(x y)=\frac{f(x y)}{f(x+y)},
$$
where $\mathbf{Q}_{+}$ denotes the set of positive rational numbers.
(Li Shenghong) | (1) Proof: $f(1)=1$.
On the one hand, in equation (1), let $y=1$, and denote $f(1)=a$, then $f(x)+a+2 x f(x)=\frac{f(x)}{f(x+1)}$,
which means $f(x+1)=\frac{f(x)}{(1+2 x) f(x)+a}$.
(2)
Then $f(2)=\frac{a}{4 a}=\frac{1}{4}, f(3)=\frac{1}{5+4 a}$,
$f(4)=\frac{1}{7+5 a+4 a^{2}}$.
On the other hand, in equation (1), take $... | f(x)=\frac{1}{x^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,771 |
Five, let $x_{1}, x_{2}, \cdots, x_{n}$ be $n(n \geqslant 2)$ real numbers, satisfying $A=\left|\sum_{i=1}^{n} x_{i}\right| \neq 0$,
$$
B=\max _{1 \leqslant i<j \leqslant n}\left|x_{j}-x_{i}\right| \neq 0 .
$$
Prove that for any $n$ vectors $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \alpha_{n}$ in the ... | Let $\left|\boldsymbol{\alpha}_{k}\right|=\max _{1 \leqslant i \leqslant n}\left|\boldsymbol{\alpha}_{i}\right|(k \in\{1,2, \cdots, n\})$.
It suffices to prove:
$$
\max _{\left(k_{1}, k_{2}, \cdots, k_{n}\right) \in S_{n}}\left|\sum_{i=1}^{n} x_{k_{i}} \boldsymbol{\alpha}_{i}\right| \geqslant \frac{A B}{2 A+B}\left|\bo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,772 |
Six, let $n$ be a positive integer, $A \subseteq\{1,2, \cdots, n\}$, and the least common multiple of any two numbers in $A$ does not exceed $n$. Prove:
$|A| \leqslant 1.9 \sqrt{n}+5$.
(Supplied by Chen Yonggao) | For $a \in(\sqrt{n}, \sqrt{2 n})$, we have
$$
[a, a+1]=a(a+1)>n.
$$
Then $|A \cap(\sqrt{n}, \sqrt{2 n}]| \leqslant \frac{1}{2}(\sqrt{2}-1) \sqrt{n}+1$.
For $a \in(\sqrt{2 n}, \sqrt{3 n}]$, we have
$[a, a+1]=a(a+1)>n$,
$[a+1, a+2]=(a+1)(a+2)>n$,
$[a, a+2] \geqslant \frac{1}{2} a(a+2)>n$.
Then $|A \cap(\sqrt{2 n}, \sqrt... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,773 |
1. As shown in Figure 1, in the Cartesian coordinate system, the graph of the quadratic function $y=a x^{2}+m c(a \neq$ $0)$ passes through three vertices $A$, $B$, and $C$ of the square $A B O C$, and $a c=-2$. Then the value of $m$ is ( ).
(A) 1
(B) -1
(C) 2
(D) -2 | -1.A.
From the problem, we know $A(0, m c), O A=m c=B C$.
Also, $B$ and $C$ are symmetric with respect to the $y$-axis, so $C\left(\frac{1}{2} m c, \frac{1}{2} m c\right)$.
Thus, $\frac{1}{2} m c=a\left(\frac{1}{2} m c\right)^{2}+m c$, which gives $a m c=-2$.
Also, $a c=-2$, so $m=1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,774 |
1. Let $n$ be a natural number greater than 2, and $a_{1}<a_{2}<\cdots<a_{k}$ be all positive integers less than $n$ and coprime with $n$. Prove: Among these $k$ numbers, there must be a prime number. | (Obviously, $a_{1}=1$. Also, $n>2$, so $k>2$ (why?). Prove that $a_{2}$ is a prime number. Since $a_{2}$ is the smallest positive integer other than 1, less than $n$, and coprime with $n$, if $a_{2}$ is not a prime, because $a_{2} > 2$, then $a_{2}=x y$, where $x$ and $y$ are both positive integers greater than 1. Sinc... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,775 |
2. Roll two fair six-sided dice numbered $1 \sim 6$ (integers) with one red and one green. The numbers facing up on the red and green dice are used as the coefficients of the linear term and the constant term, respectively, in the quadratic function $y=x^{2}+m x+n$. The probability that the quadratic function $y=x^{2}+... | 2.C.
By operation, 36 quadratic functions can be obtained. According to the problem, $\Delta=m^{2}-4 n>0$, which means $m^{2}>4 n$. Through enumeration, it is known that there are 17 pairs of $m, n$ that satisfy the condition. Therefore, $P=\frac{17}{36}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,776 |
3. If a positive integer can be expressed as the difference of squares of two consecutive even numbers, it is called a "mysterious number" (for example, 4 $=2^{2}-0^{2}, 12=4^{2}-2^{2}, 20=6^{2}-4^{2}$). Which of the following statements about mysterious numbers is correct? ( ).
(1) 2008 is a mysterious number;
(2) The... | 3. B.
Let two consecutive even numbers be $2k+2$ and $2k$, then
$$
(2k+2)^2 - (2k)^2 = 4(2k+1).
$$
Since $2k+1$ is an odd number, the mysterious number is a multiple of 4 but not a multiple of 8.
Let any two positive odd numbers be $2m+1$ and $2n+1$, then $(2m+1)^2 - (2n+1)^2 = 4(m+n+1)(m-n)$.
Since $m+n+1$ and $m-n$... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,777 |
4. In $\triangle A B C$, $A B=A C=b, B C=a$. If $\frac{a}{b}=\frac{\sqrt{5}-1}{2}$, then $\triangle A B C$ is called a "golden triangle". In a golden $\triangle A B C$, the measure of $\angle A$ is ( ).
(A) $36^{\circ}$
(B) $54^{\circ}$
(C) $72^{\circ}$
(D) $108^{\circ}$ | 4. A.
As shown in Figure 6, on $AC$, intercept $AD = BC = a$, then $CD = b - a$.
Given $\frac{a}{b} = \frac{\sqrt{5} - 1}{2}$, we have
$$
2a + b = \sqrt{5}b.
$$
Squaring both sides, we get
$$
\begin{array}{l}
4a^2 + 4ab + b^2 = 5b^2 \\
\Rightarrow a^2 + ab = b^2 \Rightarrow \frac{b - a}{a} = \frac{a}{b} \Rightarrow \... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,778 |
5. As shown in Figure 2, points $E$ and $F$ are on the sides $AB$ and $BC$ of rhombus $ABCD$, respectively, and $\frac{AE}{BE} = \frac{BF}{CF}$. If $\triangle DEF$ is an equilateral triangle, then the degree measure of $\angle BAD$ is ( ).
(A) $40^{\circ}$
(B) $60^{\circ}$
(C) $80^{\circ}$
(D) $100^{\circ}$ | 5.B.
As shown in Figure 7, from the problem, we know that $B E = C F$. Take a point $K$ on side $A B$ such that $A K = B E$. It is easy to see that
$\triangle A K D \cong \triangle C F D$.
Thus, $D K = D F = D E$.
Therefore, $\triangle D K E$ is an isosceles triangle, and we have
$$
\angle D K E = \angle D E K \text{.... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,779 |
6. As shown in Figure 3, the graph of the function $y=m x-4 m$ intersects the $x$-axis and $y$-axis at points $N$ and $M$, respectively. Points $A$ and $B$ on $MN$ have projections on the $x$-axis at $A_{1}$ and $B_{1}$, respectively. If $O A_{1}+O B_{1}>4$, then the relationship between the area $S_{1}$ of $\triangle ... | 6.A.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
y_{1}=m x_{1}-4 m, y_{2}=m x_{2}-4 m . \\
\text { Also, } S_{1}=\frac{1}{2} O A_{1} \cdot A A_{1}=\frac{1}{2} x_{1}\left(m x_{1}-4 m\right), \\
S_{2}=\frac{1}{2} O B_{1} \cdot B B_{1}=\frac{1}{2} x_{2}\left(m x_{2}-4 m\right),
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,780 |
1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$ | $=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \time... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,781 |
3. Given real numbers $a, b, c$ and $\lambda>0$ satisfy
$$
\begin{array}{l}
|b+\lambda c| \geqslant(\lambda+1)|a|, \\
|c+\lambda a| \geqslant(\lambda+1)|b|, \\
|a+\lambda b| \geqslant(\lambda+1)|c| .
\end{array}
$$
Then the relationship between $a, b, c$ is $\qquad$ | 3. $a=b=c$.
Squaring both sides of the three inequalities, we have
$$
\begin{array}{l}
b^{2}+2 \lambda b c+\lambda^{2} c^{2} \geqslant\left(\lambda^{2}+2 \lambda+1\right) a^{2}, \\
c^{2}+2 \lambda c a+\lambda^{2} a^{2} \geqslant\left(\lambda^{2}+2 \lambda+1\right) b^{2}, \\
a^{2}+2 \lambda a b+\lambda^{2} b^{2} \geqsl... | a=b=c | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,783 |
One, (20 points) Given $m>0$, and $m$ is a rational number, $\sqrt{m}$ is an irrational number. Find all positive integers $a, b, c$, such that $\frac{\sqrt{m} a+b}{\sqrt{m} b+c}$ is a rational number, and $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=3$. | Given that $\sqrt{m} b - c$ is an irrational number, therefore, $\sqrt{m} b - c \neq 0$.
$$
\begin{array}{l}
\text { Hence } \frac{\sqrt{m} a + b}{\sqrt{m} b + c} = \frac{(\sqrt{m} a + b)(\sqrt{m} b - c)}{m b^2 - c^2} \\
= \frac{(m a b - b c) + \sqrt{m}(b^2 - a c)}{m b^2 - c^2}.
\end{array}
$$
Since the above expressi... | (1, 2, 4), (4, 2, 1), (3, 3, 3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,785 |
2. In a round-robin table tennis tournament, among $n(n \geqslant 3)$ players, there is no one who wins all the matches. Prove: it is always possible to find three players $A$, $B$, and $C$, such that $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$. | (Tip: Take the player with the most wins among $n$ players and denote this player as $A$ (consider the extreme case). Since $A$ has not won all the matches, there exists a player $C$ who has defeated $A$. Considering all the players defeated by $A$, there must be a player $B$ who has defeated $C$ (otherwise, all the pl... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,786 |
II. (25 points) As shown in Figure 5, in $\triangle ABC$, $P$ is a point on the angle bisector of $\angle BAC$, $M$ (different from $A, B$) is a point on side $AB$, and lines $AP, CP, MP$ intersect side $BC$, $AB$, $AC$ at points $D, E, N$ respectively. If $\angle MPB = \angle PCN$, $\angle NPC = \angle MBP$, prove tha... | Let $\angle M P B=\alpha, \angle N P C=\beta, A B=c, B C=a$, $C A=b, r$ be the inradius of $\triangle A B C$, $h_{a} 、 h_{c}$ be the altitudes to sides $a 、 c$ respectively, and $S$ be the area of $\triangle A B C$. Then we have
$$
\begin{array}{l}
180^{\circ}=\angle B A C+\angle A M N+\angle A N M \\
=\angle B A C+2(\... | S_{\text{quadrilateral } ACDE} = 2 S_{\triangle APC} | Geometry | proof | Yes | Yes | cn_contest | false | 718,787 |
Three. (25 points) The parabola $y=a x^{2}+b x+c(a>$ $0)$ passes through points $A(-3 \sqrt{3}, 0)$ and $B(\sqrt{3}, 0)$, and intersects the $y$-axis at point $C$, with $\angle A C B \geqslant 90^{\circ}$. Let the vertex of the parabola be $D$, and in $\triangle B C D$, the altitude from $C$ to $D$ is $h$. Find the ran... | Three, because $A(-3 \sqrt{3}, 0)$ and $B(\sqrt{3}, 0)$ are on $y=a x^{2}+b x+c$, then
$$
y=a(x+3 \sqrt{3})(x-\sqrt{3})=a\left(x^{2}+2 \sqrt{3} x-9\right) \text {. }
$$
Thus, $c=-9 a$.
When $\angle A C B=90^{\circ}$, by the projection theorem we get
$$
O C^{2}=O A \cdot O B=3 \sqrt{3} \times \sqrt{3}=9 \Rightarrow O C... | 0<a \leqslant \frac{1}{3}, 0<h \leqslant \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,788 |
2. Given the inequality $2^{2 x}+a 2^{x}+b \geqslant 0$ has a minimum solution of 2. Then the range of values for $a$ is ( ).
(A) $(-\infty, 0)$
(B) $[0,+\infty)$
(C) $[2,+\infty)$
(D) $[-4,+\infty)$ | 2.D.
It is known that 2 should be a root of the equation $2^{2 x}+a 2^{x}+b=0$.
We can obtain $b=-4 a-16$.
Thus, $2^{2 x}+a 2^{x}+b=\left(2^{x}-4\right)\left(2^{x}+4+a\right) \geqslant 0$.
Then $a \geqslant-\left(4+2^{x}\right)$ for $x \in \mathbf{R}-\{2\}$ always holds.
Also, $-\left(4+2^{x}\right)<-4$, therefore, $a ... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,790 |
3. Given the regions $x^{2}+y^{2} \leqslant 1$ and $[x]^{2}+[y]^{2}=1$ with areas $S_{1}$ and $S_{2}$ respectively (where $[x]$ denotes the greatest integer less than or equal to $x$). Then ( ).
(A) $S_{1}>S_{2}$
(B) $S_{1}=S_{2}$
(C) $S_{1}<S_{2}$
(D) Cannot compare the sizes of $S_{1}$ and $S_{2}$ | 3.C.
From the problem, we know $S_{1}=\pi \times 1^{2}=\pi$.
Since $[x], [y] \in \mathbf{Z}, [x]^{2}+[y]^{2}=1$, therefore, $\left\{\begin{array}{l}[x]=0, \\ [y]= \pm 1\end{array}\right.$ or $\left\{\begin{array}{l}[x]= \pm 1, \\ [y]=0 . .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}0 \leqslant x<1, \\ ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,791 |
4. Given $\alpha, \beta, \alpha+\beta \in\left(0, \frac{\pi}{2}\right)$,
$$
A=\frac{1}{\sin 2 \alpha}+\frac{1}{\sin 2 \beta}, B=\frac{1}{\sin \alpha}+\frac{1}{\sin \beta} \text {. }
$$
Then ( ).
(A) $AB$
(D) $A \geqslant B$ | 4.A.
From $\alpha, \beta, \alpha+\beta \in \left(0, \frac{\pi}{2}\right)$, we get
$$
\begin{array}{l}
\sin (\alpha+\beta)=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\
\frac{\sin (\alpha+\beta)}{\sin \alpha \cdot \sin \beta}-\frac{2 \sin (\alpha+\beta) \cdot \cos (\alpha-\beta)}{\sin 2 \alpha \cdot \sin... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,792 |
5. Given the real-coefficient quadratic equation $x^{2}+p x+q=0$ with real solutions $\alpha, \beta$ satisfying $|\alpha|+|\beta| \leqslant 1$. Then the range of $m=$ $p^{2}+4 q$ is ( ).
(A) $[-1,1]$
(B) $[-1,2]$
(C) $[0,1]$
(D) $[0,2]$ | 5.B.
By Vieta's formulas, we have $\alpha+\beta=-p, \alpha \beta=q$.
Then $m=(\alpha+\beta)^{2}+4 \alpha \beta=2(\alpha+\beta)^{2}-(\alpha-\beta)^{2}$.
$$
\begin{array}{l}
\text { Also }|\alpha+\beta| \leqslant|\alpha|+|\beta| \leqslant 1, \\
|\alpha-\beta| \leqslant|\alpha|+|\beta| \leqslant 1,
\end{array}
$$
Theref... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,793 |
6. Figure 1 is a chess endgame where the Red side (KaiTi font) is certain to win. If the Red side does not move the horse, but only moves the pawn, then, the number of shortest ways to checkmate the Black king is ( ).
(A) 12
(B) 15
(C) 17
(D) 21 | 6.C.
According to the rules, the red pawn can only checkmate the black general by moving to the position of the black advisor on the right. The numbers in Figure 4 represent the number of shortest moves for the red pawn to reach that intersection. Therefore, the number of shortest moves is 17. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,794 |
2. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with left and right foci $F_{1}$ and $F_{2}$, respectively, and $P$ as any point on the ellipse $C$, $M=|| P F_{1}|-| P F_{2}||+\left|P F_{1}\right| \cdot\left|P F_{2}\right|$. Then the maximum value of $M$ is $\qquad$ | 2. $M=\left\{\begin{array}{ll}a^{2}+1, & c \geqslant 1 ; \\ 2 c+b^{2}, & 0<c<1 .\end{array}\right.$
Let $P(x, y)$, then
$$
\begin{array}{l}
\left|P F_{1}\right|=a+e x,\left|P F_{2}\right|=a-e x . \\
\text { Hence } M=2 e|x|+a^{2}-e^{2} x^{2} \\
=a^{2}+1-(e|x|-1)^{2} \\
=a^{2}+1-e^{2}\left(|x|-\frac{1}{e}\right)^{2} .
... | M=\left\{\begin{array}{ll}a^{2}+1, & c \geqslant 1 ; \\ 2 c+b^{2}, & 0<c<1 .\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,796 |
Example 1 Let $x, y, z$ be three real numbers, not all zero. Prove:
$$
\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sqrt{5}}{2} \text {. }
$$
Analysis: For a fraction, if the denominator is a quadratic term and the numerator is the product of the elements of the denominator, one can try using the basic inequal... | Prove: Introduce a parameter $\lambda$, and transform $x^{2}+y^{2}+z^{2}$ into $x^{2}+\lambda y^{2}+(1-\lambda) y^{2}+z^{2}$. Notice that
$$
\begin{array}{l}
x^{2}+\lambda y^{2} \geqslant 2 \sqrt{\lambda} x y, \\
(1-\lambda) y^{2}+z^{2} \geqslant 2 \sqrt{1-\lambda} y z,
\end{array}
$$
then
$$
x^{2}+y^{2}+z^{2} \geqsla... | \frac{x y+2 y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sqrt{5}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,797 |
Example 2 Let $x, y, z, w$ be real numbers, not all zero. Prove that:
$$
\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}} \leqslant \frac{\sqrt{2}+1}{2} \text {. }
$$
Analysis: This problem is slightly different from Example 1. Due to the absence of terms like $x z, x w$, etc., we consider splitting $y^{2}, z^{2}$ into t... | Prove: Introducing positive parameters $\alpha, \beta, \gamma$, then we have
$$
\begin{array}{l}
\alpha^{2} x^{2}+y^{2} \geqslant 2 \alpha x y, \beta^{2} y^{2}+z^{2} \geqslant 2 \beta y z, \\
\gamma^{2} z^{2}+w^{2} \geqslant 2 \gamma z w .
\end{array}
$$
Adding the above three inequalities yields
$$
\begin{array}{l}
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,798 |
Example 2 As shown in Figure 3, on a $2 \times 3$ rectangular grid paper, the vertices of each small square are called grid points. Then the number of isosceles right triangles with grid points as vertices is ( ).
(A) 24
(B) 38
(C) 46
(D) 50
This problem can be solved using the labeling method, making the solution clea... | Solution: As shown in Figure 4, at each grid point, mark the number of isosceles right triangles with it as the right-angle vertex. For example, $A_{2+0}$ indicates that there are 2 upright and 0 slanted isosceles right triangles with $A$ as the right-angle vertex; $B_{3+2}$ indicates that there are 3 upright and 2 sla... | 50 | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,799 |
Example 1 Find the range of the function $y=\sqrt{x-4}+\sqrt{15-3 x}$.
Analysis: The general approach is: squaring, rearranging, isolating the radical, squaring again, and converting to a rational expression for solving. Since $4 \leqslant x \leqslant 5$, a trigonometric substitution can be used. | Solution: From $\left\{\begin{array}{l}x-4 \geqslant 0, \\ 15-3 x \geqslant 0,\end{array}\right.$ we get $4 \leqslant x \leqslant 5$.
Let $x=4+\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then
$y=\sin \theta+\sqrt{3} \cos \theta=2 \sin \left(\theta+\frac{\pi}{3}\right)$.
Since $\theta \in\l... | [1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,802 |
Example 2 Given $00$. Find the minimum value of $y=\frac{a^{2}}{x}+\frac{b^{2}}{1-x}$.
Analysis: Since $0<x<1$ and $x+(1-x)=1$, we can use trigonometric substitution. | Solution: Let $x=\sin ^{2} \theta, \theta \in\left(0, \frac{\pi}{2}\right)$, then
$$
\begin{array}{l}
y=\frac{a^{2}}{x}+\frac{b^{2}}{1-x}=\frac{a^{2}}{\sin ^{2} \theta}+\frac{b^{2}}{\cos ^{2} \theta} \\
=a^{2}\left(1+\cot ^{2} \theta\right)+b^{2}\left(1+\tan ^{2} \theta\right) \\
=a^{2}+b^{2}+(a \cot \theta)^{2}+(b \ta... | (a+b)^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,803 |
Example 3 Given $a, b \in \mathbf{R}$, and
$$
a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}=1 \text{. }
$$
Prove: $a^{2}+b^{2}=1$.
(3rd "Hope Cup" National Mathematics Invitational Competition)
Analysis: From the given, we can deduce that $|a| \leqslant 1,|b| \leqslant 1$, so we use trigonometric substitution. | Solution: Let $a=\sin \alpha, b=\sin \beta$, where $\alpha, \beta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then we have
$\sin \alpha \cdot \cos \beta + \sin \beta \cdot \cos \alpha = 1$.
Therefore, $\sin (\alpha + \beta) = 1$ and $\alpha + \beta \in [-\pi, \pi]$.
Thus, $\alpha + \beta = \frac{\pi}{2}$.
Hence, $... | a^2 + b^2 = 1 | Algebra | proof | Yes | Yes | cn_contest | false | 718,804 |
Example 4 Real numbers $x, y$ satisfy
$$
\frac{(x+2)^{2}}{9}+\frac{(y-2)^{2}}{16}=1 \text {. }
$$
If $x+y-k<0$ always holds, find the range of $k$. Analysis: From $\frac{(x+2)^{2}}{9}+\frac{(y-2)^{2}}{16}=1$, we know $-1 \leqslant \frac{x+2}{3} \leqslant 1,-1 \leqslant \frac{y-2}{4} \leqslant 1$.
Therefore, trigonomet... | Solution: Let $\frac{x+2}{3}=\cos \theta, \frac{y-2}{4}=\sin \theta$, where $\theta \in \mathbf{R}$, then
$$
\begin{array}{l}
x=3 \cos \theta-2, y=4 \sin \theta+2 . \\
\text { Substituting } x+y-k3 \cos \theta+4 \sin \theta=5 \sin (\theta+\varphi)$, where, $\tan \varphi=\frac{3}{4}$.
Since $\theta \in \mathbf{R}$, then... | k \in (5, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,805 |
Example 5 Find the maximum and minimum values of the function $y=\frac{x}{x^{2}+1}$. Analysis: Notice that its structure is basically consistent with the universal formula
$$
\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}
$$
Therefore, a trigonometric substitution can be used. | Solution: Note that $x \in \mathbf{R}$, we can set
$$
x=\tan \theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text {. }
$$
Thus, $y=\frac{1}{2} \sin 2 \theta$.
Since $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then
$$
-\pi<2 \theta<\pi \text {. }
$$
Therefore, $-1 \leqslant \sin 2 \theta \leq... | y_{\min }=-\frac{1}{2}, y_{\max }=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,806 |
Example 6 The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=\frac{1}{3}, a_{n}=\sqrt{\frac{1+a_{n-1}}{2}}(n=1,2, \cdots) \text {. }
$$
Prove: $\left\{a_{n}\right\}$ is a monotonic sequence.
(15th All-Russian High School Mathematics Olympiad)
Analysis: From $a_{n}=\sqrt{\frac{1+a_{n-1}}{2}}$, it is easy to see tha... | Proof: Let $a_{0}=\cos \theta=\frac{1}{3}, \theta \in\left(0, \frac{\pi}{2}\right)$.
Since $a_{n}=\sqrt{\frac{1+a_{n-1}}{2}}\left(n \in \mathbf{N}_{+}\right)$, then
$$
\begin{array}{l}
a_{1}=\sqrt{\frac{1+\cos \theta}{2}}=\cos \frac{\theta}{2}, \\
a_{2}=\cos \frac{\theta}{4}, \\
\cdots \cdots \\
a_{n}=\cos \frac{\theta... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,807 |
Example 7 Given any 7 real numbers, prove: there are at least 2 real numbers $x, y$ that satisfy $0 \leqslant \frac{x-y}{1+x y} \leqslant \frac{\sqrt{3}}{3}$.
(16th Canadian Mathematical Olympiad)
Analysis: The structure of the expression $\frac{x-y}{1+x y}$ to be proved is exactly the same as the structure of the tang... | Proof: Let $x=\tan \alpha, y=\tan \beta$, where $\alpha, \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) (\alpha \geqslant \beta)$. Then
$$
\frac{x-y}{1+x y}=\tan (\alpha-\beta) \text{. }
$$
Since $\alpha, \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ can ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,808 |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} \geqslant 1.
$$
Analysis: This inequality is a cyclic symmetric inequality, and we can consider constructing an expression that sums to 1. However, this expression cannot be specified, so we introduce a para... | Prove: Introduce the parameter $\lambda$, set
$$
\frac{1}{1+2 a} \geqslant \frac{a^{\lambda}}{a^{\lambda}+b^{\lambda}+c^{\lambda}},
$$
i.e., $b^{\lambda}+c^{\lambda} \geqslant 2 a^{1+\lambda}$.
By the mean inequality, we have
$$
b^{\lambda}+c^{\lambda} \geqslant 2 \sqrt{(b c)^{\lambda}}=2 \sqrt{\left(\frac{1}{a}\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,809 |
Example 8 Find the real number pairs $(x, y, z)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
y=4 x^{3}-3 x, \\
z=4 y^{3}-3 y, \\
x=4 z^{3}-3 z
\end{array}\right.
$$
Analysis: Each equation in the system has the same structure as the triple angle formula $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \the... | Solution: If $|x|>1$, then
$$
y=4 x^{3}-3 x=x\left(4 x^{2}-3\right),
$$
we get $|y|>|x|$.
Similarly, $|z|>|y|,|x|>|z|$, which is contradictory.
Therefore, $|x| \leqslant 1$.
Similarly, $|y| \leqslant 1,|z| \leqslant 1$.
Let $x=\cos \theta, \theta \in[0, \pi]$, then the original system of equations becomes
$$
\left\{\b... | (x, y, z)=(\cos \theta, \cos 3 \theta, \cos 9 \theta) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,810 |
Question 1 Given the sequence $\left\{a_{n}\right\}, a_{1}=8, a_{2}=20$,
$$
a_{n+2}=a_{n+1}^{2}+12 a_{n+1} a_{n}+a_{n+1}+11 a_{n} \text {. }
$$
Prove: No term in $\left\{a_{n}\right\}$ can be expressed as the sum of the seventh powers of three integers. | Proof: Analyze with modulus 29
For any $n \in \mathbf{Z}, n^{7} \equiv 0, \pm 1, \pm 12(\bmod 29)$.
List the first few terms of $\left\{a_{n}\right\}$
$a_{2}=20 \equiv 20(\bmod 29)$,
$a_{3} \equiv 20^{2}+12 \times 20 \times 8+20+11 \times 8$
$\equiv 21(\bmod 29)$
$\equiv 9(\bmod 29)$
$a_{5} \equiv 9^{2}+12 \times 9 \ti... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,811 |
Question 2 Given the sequence $\left\{a_{n}\right\}, a_{1}=49, a_{2}=48$, and satisfies
$$
a_{n+2}=a_{n+1} a_{n}^{2}+2 a_{n+1}^{2}+49 a_{n}^{2}+102 a_{n+1}+147 .
$$
Prove: No term in $\left\{a_{n}\right\}$ can be expressed as the sum of 47 integers raised to the 48th power.
Analysis: This problem requires the use of ... | Proof: Analyze modulo 97.
For any \( n \in \mathbf{Z}, n^{48} \equiv 0, \pm 1(\bmod 97) \). Transform the right side of the recurrence formula to get
\[
a_{n+2}=a_{n+1}+\left(a_{n+1}+49\right)\left(a_{n}^{2}+2 a_{n+1}+3\right) \text {. }
\]
It can be seen that when \( a_{n} \equiv 48(\bmod 97) \),
\[
a_{n+1} \equiv 48... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,812 |
1. If $a, b$ are real numbers, satisfying $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then the value of $(1+a+b)(1-a-b)$ is $(\quad)$.
(A) -1
(B) 0
(C) 1
(D) 2 | - 1. C.
From the given condition, we know that $1+a+b+ab=1-a-b+ab$ then $a+b=0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,814 |
2. Let $p$ be a positive odd number. Then the remainder when $p^{2}$ is divided by 8 is ( ).
(A) 1
(B) 3
(C) 5
(D) 7 | 2.A.
Since $p$ is a positive odd number, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, then $p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1$.
Since $k-1$ and $k$ must be one odd and one even, $(k-1)k$ must be even. Therefore, the remainder when $p^{2}$ is divided by 8 is 1. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,815 |
3. Given in $\triangle A B C$, $A B=A C=4 \sqrt{3}$, and the altitude $A D=4$. Then the circumradius of $\triangle A B C$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. D.
As shown in Figure 3, extend $A D$ to intersect the circumcircle $\odot O$ of $\triangle A B C$ at point $E$. Since $A D$ is also a median, $A E$ is the diameter of $\odot O$. Connecting $B E$, we find that $\triangle A B E$ is a right triangle. Therefore, $A B^{2}=A D \cdot A E$,
which means $(4 \sqrt{3})^{2}=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,816 |
4. Let $a$ and $b$ be integers, and one root of the equation $x^{2} + a x + b = 0$ is $\sqrt{4 - 2 \sqrt{3}}$. Then the value of $a + b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | 4.B.
Notice that $\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1$.
According to the problem, we have $(\sqrt{3}-1)^{2}+a(\sqrt{3}-1)+b=0$, which simplifies to $(a-2) \sqrt{3}+4-a+b=0$.
Thus, $a-2=0$ and $4-a+b=0$.
Solving these, we get $a=2, b=-2$. Therefore, $a+b=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,817 |
5. The shape of iron plate A is an isosceles triangle, with its vertex angle being $45^{\circ}$ and the length of the legs being $12 \mathrm{~cm}$; the shape of iron plate B is a right trapezoid, with the lengths of the two bases being $4 \mathrm{~cm}$ and $10 \mathrm{~cm}$, and one interior angle being $60^{\circ}$. I... | 5.A.
As shown in Figure 4, draw $B D \perp A C$ at point $D$, then
$$
B D=12 \sin 45^{\circ}=6 \sqrt{2}=\sqrt{72} \approx 8.5$, so the second board cannot pass through. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,818 |
6. Let the parabola $y=x^{2}+k x+4$ intersect the $x$-axis at two distinct points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$. Then, which of the following conclusions must be true? ( ).
(A) $x_{1}^{2}+x_{2}^{2}=17$
(B) $x_{1}^{2}+x_{2}^{2}=8$
(C) $x_{1}^{2}+x_{2}^{2}8$
Translate the text above into English, p... | 6.D.
From $\Delta=k^{2}-16>0 \Rightarrow k^{2}>16$.
Also, since $x_{1}+x_{2}=-k, x_{1} x_{2}=4$, therefore, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=k^{2}-8>8$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,819 |
Example 4 Let $x, y, z$ be positive real numbers. Prove:
$$
\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}} \geqslant 2 \text {. }
$$
Analysis: This inequality is also a cyclic symmetric inequality, and we can consider constructing an expression that sums to 2. However, this expr... | Prove: Introduce a parameter $\lambda$, set
$$
\frac{x}{\sqrt{y^{2}+z^{2}}} \geqslant \frac{2 x^{\lambda}}{x^{\lambda}+y^{\lambda}+z^{\lambda}} \text {, }
$$
i.e., $\left(x^{\lambda}+y^{\lambda}+z^{\lambda}\right)^{2} \geqslant 4 x^{2 \lambda-2}\left(y^{2}+z^{2}\right)$.
And $\left(x^{\lambda}+y^{\lambda}+z^{\lambda}\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,820 |
1. Given the inequality $a x+3 \geqslant 0$ has positive integer solutions of $1,2,3$. Then the range of values for $a$ is $\qquad$ . | Ni.1. $-1 \leqslant a<-\frac{3}{4}$.
From $a x+3 \geqslant 0$, we get $a x \geqslant-3$.
Obviously, $a<0$ (otherwise, the positive integer solutions of the original inequality would not be just $1,2,3$). Therefore, $x \leqslant-\frac{3}{a}$.
According to the problem, $3 \leqslant-\frac{3}{a}<4$. Solving this, we get $-... | -1 \leqslant a<-\frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,821 |
2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ . | 2.2.
As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,822 |
3. The quality requirements of a product are divided into four different levels from low to high, labeled as $1,2,3,4$. If the working hours remain unchanged, the workshop can produce 40 units of the lowest level (i.e., level 1) product per day, with a profit of 16 yuan per unit; if the level is increased by one, the p... | 3.3.
Let the profit obtained from producing products of the $x$-th grade in the workshop be $y$. According to the problem, we have
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Therefore, when $x=3$, the profit $y$ is maximized, at 648 yuan. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,823 |
4. The number of all different integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=2007$ is $\qquad$ groups.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above this note. | 4.4 .
Let's first assume $x \geqslant y$, the original equation can be transformed into $(2 x+y)(x+2 y)=2007$.
Since $2007=2007 \times 1=669 \times 3=223 \times 9$
$$
\begin{array}{l}
=(-1) \times 2007=(-3) \times(-669) \\
=(-9) \times(-223),
\end{array}
$$
Therefore, $\left\{\begin{array}{l}2 x+y=2007, \\ x+2 y=1 .\... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 718,824 |
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2)$, $B(-1, \sqrt{3})$, and $C(c$, $2-c)$. Find the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$.
| $$
\left\{\begin{array} { l }
{ \sqrt { 3 } + 2 = \sqrt { 3 } a + b , } \\
{ \sqrt { 3 } = - a + b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\sqrt{3}-1, \\
b=2 \sqrt{3}-1 .
\end{array}\right.\right.
$$
Solving, we get $c=\sqrt{3}-2$.
Therefore, $a-b=-\sqrt{3}, b-c=\sqrt{3}+1, c-a=-1$.
Thus, $a^{2}+b^{2}+c^{... | 4+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,825 |
Four. (25 points) As shown in Figure 2, in $\odot O$, chord $CD$ is perpendicular to diameter $AB$, $M$ is the midpoint of $OC$, the extension of $AM$ intersects $\odot O$ at point $E$, and $DE$ intersects $BC$ at point $N$. Prove: $BN = CN$.
untranslated part:
如图 2, 在 $\odot O$ 中, 弦 $C D$垂直于直径 $A B, M$是 $O C$ 的中点, $... | As shown in Figure 7, connect $AC$ and $BD$. Since chord $CD$ is perpendicular to diameter $AB$, then $BC = BD$. Therefore,
$$
\angle BCD = \angle BDC.
$$
Since $OA = OC$, we have
$$
\begin{array}{l}
\angle OCA = \angle OAC. \\
\text{Also, } \angle BDC = \angle OAC,
\end{array}
$$
Thus, $\angle BCD = \angle OCA$.
Hen... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,826 |
Five. (25 points) A math game is played between two very smart students, A and B. The referee first writes the positive integers $2,3, \cdots, 2006$ on the blackboard, then randomly erases one number. Next, B and A take turns erasing one of the numbers (i.e., B erases one number first, then A erases one number, and so ... | Five, the key to winning is to see which number the referee erases.
Notice that $2,3, \cdots, 2006$ contains 1002 odd numbers and 1003 even numbers.
(1) If the referee erases an odd number, at this point, player B will definitely win.
No matter what number player A takes, as long as there are still odd numbers, player ... | \frac{1003}{2005} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,827 |
2. In astronomy, a sidereal year is denoted as $E=365.256$ days, and the time it takes for the Moon to orbit the Earth once is called a sidereal month, represented as $T=27.322$ days. The cycle of the Moon's phases, from full moon to half moon, crescent moon, gibbous moon, new moon, and back to full moon, is called a "... | 2.29.531 | 29.531 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,829 |
Example 5 Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 \text {. }
$$
(42nd IMO) | Prove: Introduce the parameter $\lambda$, set
$$
\frac{a}{\sqrt{a^{2}+8 b c}} \geqslant \frac{a^{\lambda}}{a^{\lambda}+b^{\lambda}+c^{\lambda}},
$$
i.e., $\left(a^{\lambda}+b^{\lambda}+c^{\lambda}\right)^{2} \cdot a^{2} \geqslant\left(a^{\lambda}\right)^{2}\left(a^{2}+8 b c\right)$.
And $\left(a^{\lambda}+b^{\lambda}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,831 |
4. Real numbers $x, y, z (x \neq y)$ satisfy
$$
\begin{array}{l}
5(x-y)+\sqrt{5}(z-y)+(z-x)=0 . \\
\text { Then } \frac{(y-z)(z-x)}{(x-y)^{2}}=
\end{array}
$$ | 4. $\sqrt{5}-5$ | \sqrt{5}-5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,832 |
5. Given
$$
\begin{array}{l}
\left\{(x, y) \mid(x+\sqrt{2})^{2}+y^{2}=a^{2}-3 a+2\right\} \cap \\
\left\{(x, y) \mid x^{2}+(y+\sqrt{2})^{2}<a\right\} \neq \varnothing .
\end{array}
$$
Then the range of real number $a$ is | 5. (2.215 3,5.9717) | (2.215, 5.9717) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,833 |
(20 points)(1) Draw the graph of the function $f(x)=|3| x|-1|$;
(2) If the equation $|3| x|-1|=2^{x}+$ $a$ has 4 real roots with respect to $x$, find the set of values for the real number $a$. | (1) As shown in Figure 2.
(2) Let
$$
y=|3| x|-1| \text{, }
$$
then $y=2^{x}+a$.
When the graph of the function $y=2^{x}+a$ passes through the point $A(0,1)$, we have $1=2^{0}+a$, which gives $a=0$. At this point, the graph of $y=2^{x}$ intersects the ray $BC$ at one point $A'$, and intersects the ray $DE$ at two point... | \left\{0,-2^{-\frac{1}{3}}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,837 |
Three. (20 points) Let the general term of the sequence $\left\{a_{n}\right\}$ be
$$
a_{n}=n+\frac{2006}{n^{2}}(n=1,2, \cdots) \text {. }
$$
Find the minimum term of the sequence $\left\{a_{n}\right\}$. | $$
\begin{array}{l}
\text { Let } f(x)=x+\frac{2006}{x^{2}}. \\
\text { When } 0<x<15, \text { we have } \\
2006\left(\frac{1}{15 \times 15^{2}}+\frac{1}{15^{2} \times 15}\right)=\frac{2006 \times 2}{15^{3}}>1,
\end{array}
$$
Therefore, when $0<x<15$,
$$
\begin{array}{l}
f\left(x_{2}\right)-f\left(x_{1}\right) \\
=\le... | \frac{6102}{256} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,838 |
Four. (20 points) Let $n$ be a positive integer greater than 1. If there exists a positive integer $i (i < n)$ such that
$$
\sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{n} b^{2}
$$
is a perfect square, then $n$ is called a "TI number." For example, by $\left(1^{2}+2^{2}+\cdots+9^{2}\right)-\left(10^{2}+11^{2}\right)=8^{2}$, we ... | (1) From $\sum_{a=1}^{39} a^{2}-\sum_{b=40}^{46} b^{2}=7569=87^{2}$, we know that 46 is a TI number.
(2) From $\sum_{a=1}^{30} a^{2}-\sum_{b=31}^{36} b^{2}=52^{2}$, we know that 36 is a TI number.
(3) By calculation, when $1 \leqslant i \leqslant 20$,
$$
\sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{26} b^{2}<0 \text {, }
$$
and... | 26 \text{ is not a TI number.} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,839 |
(1) Find $f(2006)$;
(2) If $f(n) \geqslant 30$, find the minimum value of $n$.
For a positive integer $n$, let $f(n)$ be the sum of the digits of $n^2$. | (1) Since $2006^{2}=4024036$, we have
$$
f(2006)=19.
$$
(2) Since $f(n) \geqslant 30$, then $n^{2} \geqslant 3999$.
Thus, $n>63$.
$$
\begin{array}{l}
\text{And } f(64)=19, f(65)=13, f(66)=18, \\
f(67)=25, f(68)=16, f(69)=18, \\
f(70)=13, f(71)=10, f(72)=18, \\
f(73)=19, f(74)=22, f(75)=18, \\
f(76)=25, f(77)=25, f(78)=... | 83 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,840 |
II. (20 points) Four real numbers $a, b, c, d$ satisfy:
(1) $a \geqslant b \geqslant c \geqslant d$;
(2) $a+b+c+d=9$;
(3) $a^{2}+b^{2}+c^{2}+d^{2}=21$.
Prove: (1) $a+b \geqslant 5$;
(2) $ab - cd \geqslant 2$. | (1) If $a+b<5$, then $4<c+d \leqslant a+b<5$.
Therefore, $(a b+c d)+(a c+b d)+(a d+b c)$
$$
=\frac{(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{2}=30 \text {. }
$$
From $(a-d)(b-c) \geqslant 0,(a-b)(c-d) \geqslant 0$, we get
$$
a b+c d \geqslant a c+b d \geqslant a d+b c .
$$
Thus, $a b+c d \geqslant 10$.
From... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,841 |
Example 6 Given non-negative real numbers $x, y, z$ satisfying $x+y+z=1$. Prove:
$$
\sqrt{2-x}+\sqrt{2-y}+\sqrt{2-z} \geqslant 2 \sqrt{2}+1 \text {. }
$$
Analysis: Note that the graph of $y=\sqrt{2-x}$ is easy to draw, and the graph of this function has monotonicity $(0 \leqslant x \leqslant 1)$. Therefore, we can dra... | Prove: Introduce a parameter $\lambda$, set $\sqrt{2-x} \geqslant \sqrt{2}-\lambda x$.
As shown in Figure 1, draw the
curve $C: y=\sqrt{2-x}(0 \leqslant x \leqslant 1)$ and the line $l: y=$
$\sqrt{2}-\lambda x(0 \leqslant x \leqslant 1)$.
Let the line $l$ coincide with the line $A B$, then $\lambda=\sqrt{2}-1$. At thi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,842 |
Three. (20 points) How many real roots does the equation $\sqrt{x} \sin x^{2}-2=0$ have in the interval $[0,20]$? | Three, Solution 1: Let $y=\sqrt{x} \sin x^{2}$.
The solutions to the equation $y=0$ are $x=\sqrt{k \pi} \in[0,20], k \in \mathbf{Z}$. Therefore, $x=\sqrt{k \pi}(k=0,1, \cdots, 127)$.
When $\sqrt{x} \geqslant 2$, the graph of $y=\sqrt{x} \sin x^{2}$ may intersect with the graph of $y=2$, so $x \geqslant 4$.
Also, $\sqr... | 122 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 718,843 |
1. Let the sets be
$$
\begin{array}{l}
M=\{u \mid u=12 m+8 n+4 l, m, n, l \in \mathbf{Z}\}, \\
N=\{u \mid u=20 p+16 q+12 r, p, q, r \in \mathbf{Z}\} .
\end{array}
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M=N$
(B) $M \subset N$
(C) $N \subset M$
(D) $M \not \subset N, N \not \subset M$ | $-1 . A$.
Let $m=n=0$, then $12 m+8 n+4 l=4 l$. Therefore, $M$ is the set of all multiples of 4.
Let $p=0, q=3 k+l, r=-4 k-l$, then $20 p+16 q+12 r=4 l$. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,844 |
2. Let the volume of a tetrahedron be $V_{1}$, and the volume of a convex polyhedron formed by the midpoints of its edges be $V_{2}$. Then $\frac{V_{2}}{V_{1}}$ is $(\quad$.
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) uncertain | 2.A.
The volume of the convex polyhedron $V_{2}$ is the hexahedron obtained by cutting off 4 small tetrahedra from the original tetrahedron, i.e.,
$$
V_{2}=V_{1}-4 V_{1} \times\left(\frac{1}{2}\right)^{3}=\frac{V_{1}}{2} .
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,845 |
3. In the permutations of $1,2,3,4,5$ denoted as $a_{1}, a_{2}, a_{3}, a_{4}$, $a_{5}$, the number of permutations that satisfy $a_{1}a_{3}, a_{3}a_{5}$ is ( ).
(A) 24
(B) 16
(C) 10
(D) 8 | 3. B.
Obviously, $a_{2} 、 a_{4} \notin\{1,2\}$, so $1 、 2 \in\left\{a_{1}, a_{3}, a_{5}\right\}$.
If $a_{1}=1, a_{3}=2$, then $a_{2}$ has 3 choices, at this time $a_{4} 、 a_{5}$ are determined. Hence there are 3 choices.
Similarly, $a_{3}=2, a_{5}=1 ; a_{1}=2, a_{3}=1 ; a_{3}=1, a_{5}=2$ each have 3 choices.
If $a_{... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,846 |
4. Given positive numbers $p, q, a, b, c (p \neq q)$. If $p, a, q$ form a geometric sequence, and $p, b, c, q$ form an arithmetic sequence, then the quadratic equation $b x^{2}-2 a x+c=0$ ( ).
(A) has two equal real roots
(B) has two real roots of the same sign but different values
(C) has two real roots of opposite si... | 4. D.
Given, $a^{2}=p q, b+c=p+q$, and $c$ is between $p$ and $q$.
From the discriminant of the equation $b x^{2}-2 a x+c=0$,
$$
\begin{array}{l}
\Delta=4 a^{2}-4 b c=4\left(a^{2}-b c\right) \\
=4[p q-(p+q-c) c]=4(p-c)(q-c)<0,
\end{array}
$$
we know that the equation has no real roots. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,847 |
5. If $f(x)=\sin (2 x+\theta)+\sqrt{3} \cos (2 x+\theta)$ is an odd function, and it is decreasing in $\left[0, \frac{\pi}{4}\right]$, then a value of $\theta$ is ( ).
(A) $\frac{5 \pi}{6}$
(B) $\frac{2 \pi}{3}$
(C) $-\frac{\pi}{6}$
(D) $-\frac{\pi}{3}$ | 5. B.
Obviously, $f(x)=2 \sin \left(2 x+\theta+\frac{\pi}{3}\right)$.
From $f(x)+f(-x)=0$, we get
$$
\sin \left(2 x+\theta+\frac{\pi}{3}\right)+\sin \left(-2 x+\theta+\frac{\pi}{3}\right)=0 \text {, }
$$
which means $2 \sin \left(\theta+\frac{\pi}{3}\right) \cdot \cos 2 x=0$.
Thus, $\sin \left(\theta+\frac{\pi}{3}\ri... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,848 |
6. Given that $x, y$ satisfy $\left\{\begin{array}{l}x-y+6 \geqslant 0, \\ x+y \geqslant 0, \\ x \leqslant 3 .\end{array}\right.$ If $z=$ $a x+y$ has a maximum value of $3 a+9$ and a minimum value of $3 a-3$, then the range of values for $a$ is ( ).
(A) $0 \leqslant a \leqslant 1$
(B) $-1 \leqslant a \leqslant 0$
(C) $... | 6. C.
Obviously, $(x, y)$ lies on the sides and within $\triangle A B C$, where $A(3,9) 、 B(3,-3) 、 C(-3,3)$.
$$
\begin{array}{l}
\text { When } a \geqslant 0 \text {, } z_{\max }=a \times 3+9=3 a+9, \\
z_{\min }=\min \{3 a-3,-3 a+3\}=3 a-3 \\
\Rightarrow-3 a+3 \geqslant 3 a-3 \Rightarrow a \leqslant 1 ;
\end{array}
$... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,849 |
7. Let the circle $C_{k}=\left\{(x, y) \mid\left(x-m_{k}\right)^{2}+(y-\right.$ $\left.\left.m_{k}\right)^{2} \leqslant 2 k^{2}\right\}\left(k \in \mathbf{N}_{+}\right)$, where $m_{k}$ is defined as follows: $m_{1}=0, m_{k+1}=m_{k}+2 k+1(k \geqslant 1)$.
Then the area of $\cup C_{k}$ is $\qquad$ | $$
\text { II. 7. } \frac{\pi}{3} n(n+1)(2 n+1) \text {. }
$$
Using mathematical induction, it can be proven that $m_{k}=k^{2}-1$. It is easy to see that circle $C_{k}$ is externally tangent to circle $C_{k+1}$ $\left(k \in \mathbf{N}_{+}\right)$. Therefore, the area of $\bigcup_{1 \leqslant k \leqslant n} C_{k}$ is
$... | \frac{\pi}{3} n(n+1)(2 n+1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,850 |
8. Given the set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both integers is... | 8.49 .
Given $\left|z_{1}\right|=1$, let $z_{1}=\cos \theta+\mathrm{i} \sin \theta$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right|=\left|\left(z_{1}^{2}-1\right)^{2}\right|=\left|z_{1}^{2}-1\right|^{2} \\
=|(\cos 2 \theta-1)+\mathrm{i} \sin 2 \theta|^{2} \\
=(\cos 2 \theta-1)^{2}+\sin ^{2} 2 \... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,851 |
9. For any 4 vertices $A, B, C, D$ of a cube that do not lie on the same plane, the number of cosine values of the dihedral angle $A-BC-D$ that are less than $\frac{1}{2}$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.4 .
Among the cosines of the dihedral angles with the body diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A_{1} O_{1} C_{1} \\
=\cos 120^{\circ}=-\frac{1}{2} .
\end{array}
$$
Among the cosines of the dihedral angles with the edge as the edge, only 0 is less than $\frac{1}{2}$.
Among the cosines of the dihedr... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,852 |
Example 7 Given non-negative real numbers $x, y, z$ satisfying $x+y+z=1$. Prove:
$$
\sqrt{1-3 x^{2}}+\sqrt{1-3 y^{2}}+\sqrt{1-3 z^{2}} \geqslant 3-\sqrt{3} \text {. }
$$ | Prove: Introduce a parameter $\lambda$, and set
$\sqrt{1-3 x^{2}} \geqslant 1-\lambda x\left(0 \leqslant x \leqslant \frac{\sqrt{3}}{3}\right)$.
As shown in Figure 2, draw the graphs of
the curve $C: y=\sqrt{1-3 x^{2}}$
$\left(0 \leqslant x \leqslant \frac{\sqrt{3}}{3}\right)$ and the line $l$ :
$$
y=1-\lambda x\left(0... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,853 |
10. Let $A=\{1,2, \cdots, n\}(n>1, n \in \mathbf{N})$, mapping $f: A \rightarrow A$. Then the number of $f$ that satisfy $f(1) \leqslant f(2) \leqslant \cdots \leqslant f(n)$, and the image takes exactly $k(1<k \leqslant n)$ different values is $\qquad$ | 10. $\mathrm{C}_{n}^{k} \mathrm{C}_{n-1}^{k-1}$.
First, select $k$ images from $1 \sim n$, there are $\mathrm{C}_{n}^{k}$ ways to do this.
Then divide $1 \sim n$ into $k$ segments, with at least one number in each segment, i.e., sequentially select $k-1$ numbers $a_{1}, a_{2}, \cdots, a_{k-1}$ from the first $n-1$ num... | \mathrm{C}_{n}^{k} \mathrm{C}_{n-1}^{k-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,854 |
11. Let $F$ be the focus of the parabola $y^{2}=2 x-1$, and $Q(a, 2)$ be a point on the line $y=2$. If there is exactly one point $P$ on the parabola such that $|P F|=|P Q|$, then the value of $a$ is $\qquad$ | 11.0 or 1.
It is known that the perpendicular bisector of $F(1,0), Q F$ is
$$
y=\frac{1-a}{2} x+\frac{a^{2}+3}{4} \text {. }
$$
It intersects the parabola at only 1 point.
$$
\left\{\begin{array}{l}
y=\frac{1-a}{2} x+\frac{a^{2}+3}{4}, \\
y^{2}=2 x-1 .
\end{array}\right.
$$
When $a=1$, it is obvious that there is on... | 1 \text{ or } 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,855 |
12. Let $x, y > 0$,
$$
S(x, y)=\min \left\{x, y, \frac{x^{2}}{x^{3}+y^{3}}, \frac{y^{2}}{x^{3}+y^{3}}\right\} \text {. }
$$
Then the maximum value of $S(x, y)$ is $\qquad$ | 12. $\frac{\sqrt{2}}{2}$.
Let $S(x, y)=\min \left\{x, y, \frac{x^{2}}{x^{3}+y^{3}}, \frac{y^{2}}{x^{3}+y^{3}}\right\}=t$, then $t^{4} \leqslant \frac{x^{3} y^{3}}{\left(x^{3}+y^{3}\right)^{2}} \leqslant \frac{1}{4} \Rightarrow t \leqslant \frac{\sqrt{2}}{2}$.
When $x=y=\frac{\sqrt{2}}{2}$, $s\left(\frac{\sqrt{2}}{2},... | \frac{\sqrt{2}}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,856 |
13. Let \(0 < \alpha, \beta, \gamma < \frac{\pi}{2}\) satisfy
\[
\cos ^{2} \alpha + \cos ^{2} \beta + \cos ^{2} \gamma + 2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma = 1.
\]
Prove:
\[
\begin{array}{l}
\left|\frac{\sin \alpha - \sin \beta}{\sin \alpha + \sin \beta} + \frac{\sin \beta - \sin \gamma}{\sin \beta + \si... | Three, 13. From the known conditions, we have
$$
\begin{array}{l}
(\cos \gamma + \cos \alpha \cdot \cos \beta)^{2} \\
= \left(1 - \cos^{2} \alpha\right)\left(1 - \cos^{2} \beta\right) \\
= \sin^{2} \alpha \cdot \sin^{2} \beta .
\end{array}
$$
Given $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$, we get
$\co... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,857 |
14. Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and a line passing through the point $\left(-\frac{p}{2}, 0\right)$ intersects the parabola in the first quadrant at points $C$ and $D$. If there exists a point $E$ on the $x$-axis such that $C E \perp D E$, find the range of the slope $k$ of the line $C D$. | 14. The equation of the line passing through the point $\left(-\frac{p}{2}, 0\right)$ is $y=k\left(x+\frac{p}{2}\right)$.
Substituting into the parabola equation, we get
$$
k^{2} x^{2}+p\left(k^{2}-2\right) x+\frac{k^{2} p^{2}}{4}=0 .
$$
Let the intersection points of the line and the parabola be $C\left(x_{1}, y_{1}\... | 0<k \leqslant \sqrt{\frac{\sqrt{5}-1}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,858 |
15. $A N$ is the angle bisector of $\triangle A B C$, and the extension of $A N$ intersects the circumcircle of $\triangle A B C$ at point $D$. $M$ is a point on $A N$, and the lines $B M$ and $C M$ intersect the circumcircle of $\triangle A B C$ at points $E$ and $F$, respectively. $D F$ intersects $A B$ at point $P$,... | 15. As shown in Figure 2, connect $P M$, $Q M$, and $B D$.
Since $\angle P A D = \angle M A C$,
$$
\angle A D P = \angle A C M,
$$
then $\triangle A D P \backsim \triangle A C M$.
Thus, $A P: A M = P D: M C$.
$$
\begin{array}{l}
\text { Also, } \angle B P D \\
= \angle P A D + \angle A D P \\
= \angle M A C + \angle A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,859 |
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