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 |
|---|---|---|---|---|---|---|---|---|---|
7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$ | ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$ | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,418 |
8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest inte... | 8.3.
Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,419 |
9. Given that $x$ and $y$ are real numbers, satisfying $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $\left|x^{2}+2 x y-y^{2}\right|$ is $\qquad$ . | 9. $\sqrt{2}$.
Let $x=r \cos \theta, y=r \sin \theta(0 \leqslant r \leqslant 1,-\pi<\theta \leqslant \pi)$,
then
$$
\begin{aligned}
& \left|x^{2}+2 x y-y^{2}\right| \\
& =r^{2}\left|\cos ^{2} \theta+2 \sin \theta \cdot \cos \theta-\sin ^{2} \theta\right| \\
= & r^{2}|\sin 2 \theta+\cos 2 \theta| \\
= & \sqrt{2} r^{2}\... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,420 |
10. If each element in set $A$ can be expressed as the product of two different numbers from 1, $2, \cdots, 9$, then the maximum number of elements in set $A$ is $\qquad$. | 10.31.
From $1,2, \cdots, 9$ each time take a pair of numbers to form a product, a total of $C_{G}^{2}=$ 36 values are obtained. However, there are repetitions, the repeated cases are
$$
\begin{array}{l}
1 \times 6=2 \times 3,1 \times 8=2 \times 4,2 \times 9=3 \times 6, \\
2 \times 6=3 \times 4,3 \times 8=4 \times 6,
... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,421 |
11. Draw the midlines of each face (triangle) of a regular tetrahedron, resulting in 12 line segments. Among these line segments, the number of pairs of skew lines is $\qquad$.
Make the above text in English, please keep the original text's line breaks and format, and output the translation result directly. | 11. 24 .
Consider any median line $AB$. The face containing $AB$ has no line segments that are skew to $AB$; the other face containing point $A$ has exactly one median line that is skew to $AB$; the other face containing point $B$ has exactly one median line that is skew to $AB$; the face not containing $A$ or $B$ has... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,422 |
12. Use five different colors to color the five vertices of the "pentagram" in Figure 1 (each vertex is colored with one color, and some colors may not be used), so that the two vertices on each line segment are of different colors. Then the number of different coloring methods is . | 12.1020.
Convert it into a coloring problem of a disk with 5 sectors, each to be colored with one of 5 colors, such that no two adjacent sectors have the same color (as shown in Figure 4).
Let the number of ways to color a disk with $k$ sectors using 5 colors be $x_{k}$. Then, we have $x_{k}+x_{k-1}=5 \times 4^{k-1}$... | 1020 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,423 |
Four. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
a_{0}=b_{0}=1, a_{n+1}=5 a_{n}+7 b_{n}, b_{n+1}=7 a_{n}+10 b_{n} \text {. }
$$
Prove: For any $m, n \in \mathbf{N}$,
$$
a_{m+n}+b_{m+n}=a_{m} a_{n}+b_{m} b_{n} .
$$ | For any $k \in \mathbf{N}, 0 \leqslant k \leqslant m+n$, we have
$$
a_{m+n}+b_{m+n}=a_{m+n-k} a_{k}+b_{m+n-k} b_{k} .
$$
We use mathematical induction on $k$.
When $k=0$, the conclusion is obvious.
Assume that for $k=r$, equation (1) holds, i.e.,
$$
\begin{array}{l}
a_{m+n}+b_{m+n}=a_{m+n-r} a_{r}+b_{m+n-r} b_{r} \tex... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,424 |
13. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ non-negative real numbers. Prove: $\sum_{i=1}^{n} \sqrt{\sum_{j=i}^{n} a_{j}} \geqslant \sqrt{\sum_{k=1}^{n} k^{2} a_{k}}$. | Three, 13. Induction on $n$.
When $n=1,2$, the conclusion holds.
Assume that for $n=k$, the conclusion holds for any $k$ non-negative real numbers.
When $n=k+1$, for any $k+1$ non-negative real numbers $a_{1}$, $a_{2}, \cdots, a_{k}, a_{k+1}$, consider $a_{k}+a_{k+1}$ as one number. Using the induction hypothesis, we ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,425 |
14. As shown in Figure 2, in $\triangle A B C$, $A B=A C$,
$M$ is the midpoint of $B C$, $D$,
$E$, and $F$ are points on sides $B C$,
$C A$, and $A B$ respectively, and
$A E=A F$, the circumcircle of $\triangle A E F$ intersects line segment $A D$ at
point $P$. If point $P$ satisfies
$P D^{2}=P E \cdot P F$, prove: $\a... | 14. In the circumcircle of $\triangle AEF$, since $AE=AF$, we have
$$
\begin{array}{l}
\angle APE=\angle APF=\frac{1}{2}\left(180^{\circ}-\angle A\right) \\
=\angle ABC=\angle ACB.
\end{array}
$$
Therefore, $P, D, B, F$ and $P, D, C, E$ are concyclic, respectively.
Thus, $\angle PDB=\angle PFA=\angle PEC$.
As shown in... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,426 |
15. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{0}=1, a_{n}=\left[\sqrt{S_{n-1}}\right]$ $(n=1,2, \cdots,[x]$ represents the greatest integer not greater than $x$, $\left.S_{k}=\sum_{i=0}^{k} a_{i}(k=0,1, \cdots)\right)$. Find the value of $a_{2006}$. | 15. Observe the initial terms of the sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{n}$ & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 4 & 4 \\
\hline$S_{n}$ & 1 & 2 & 3 & 4 & 6 & 8 & 10 & 13 & 16 & 20 & 24 \\
\hline \hline$n$ & 11 &... | 998 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,427 |
1. If $a>b>c>0$, then $b>a-c$ is ( ) for $a$, $b$, and $c$ to be the lengths of the sides of a triangle.
(A) a sufficient but not necessary condition
(B) a necessary but not sufficient condition
(C) a sufficient and necessary condition
(D) neither a sufficient nor a necessary condition | -1.C.
The necessity of the condition is obvious.
From $b>a-c$, we get $b+c>a$. Since $a$ is the largest, this ensures that the sum of any two lengths is greater than the third length. Thus, we know the condition is sufficient. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,428 |
2. Let the complex number $z$ satisfy $(z-\mathrm{i})(\bar{z}+\mathrm{i}) \leqslant 1$. Then, the set of points corresponding to $z$ in the complex plane forms the figure ( ).
(A) a circle with center at $(1,0)$ and radius 1, and its interior
(B) a circle with center at $(0,1)$ and radius 1, and its interior
(C) a circ... | 2.B.
Let $z=a+b \mathrm{i}$, then we have
$$
\begin{array}{l}
(z-\mathrm{i})(\bar{z}+\mathrm{i})=(a+b \mathrm{i}-\mathrm{i})[a-(b \mathrm{i}-\mathrm{i})] \\
=a^{2}+(b-1)^{2} \leqslant 1 .
\end{array}
$$
Therefore, its graph is a circle with center at $(0,1)$ and radius 1, including the interior of the circle. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,429 |
3. Let the set $A=\left\{\left.x|| x\right|^{2}-3|x|+2=0\right\}$, $B=\{x \mid(a-2) x=2\}$.
Then the number of values of $a$ that satisfy $B \varsubsetneqq A$ is ( ) .
(A) 2
(B) 3
(C) 4
(D) 5 | 3. D.
$$
A=\{x|| x|=1,| x \mid=2\}=\{-2,-1,1,2\} \text {. }
$$
It is easy to see that when $a$ takes the values $0,1,3,4$, $B$ is respectively $\{-1\},\{-2\},\{2\}$, $\{1\}$. Also, when $a=2$, $B$ is an empty set, $B \subseteq A$, so, there are 5 values for $a$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,430 |
4. Given the power function $y=x^{m^{2}-m-6}(m \in \mathbf{Z})$ whose graph has no common points with the $x$-axis. Then the range of $m$ is ( ).
(A) $\{m \mid-2<m<3, m \in \mathbf{Z}\}$
(B) $\{m \mid-2 \leqslant m \leqslant 3, m \in \mathbf{Z}\}$
(C) $\{m \mid-3<m<2, m \in \mathbf{Z}\}$
(D) $\{m \mid-3 \leqslant m \le... | 4.B.
The function $y=x^{m^{2}-m-6}(m \in \mathbf{Z})$ has no common points with the $x$-axis $\Leftrightarrow y$ is never zero $\Leftrightarrow$ the function $y$ is undefined at $x=0$. Therefore, $m^{2}-m-6 \leqslant 0$. Solving this, we get $-2 \leqslant m \leqslant 3$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,431 |
6. The necessary and sufficient condition for the skew diagonals of two lateral faces of a regular triangular prism to be perpendicular to each other is: the ratio of the side length of its base to the side length of its lateral edge is ( ).
(A) $\sqrt{3}: 1$
(B) $1: \sqrt{3}$
(C) $\sqrt{2}: 1$
(D) $1: \sqrt{2}$ | 6.C.
As shown in Figure 4, let's assume the base edge length is $a>0$, and the side edge length is 1. According to the problem, we have
$$
\begin{array}{l}
B_{1} A \cdot B C_{1}=0 . \\
\text { Since } B_{1} A=B_{1} A_{1}+B_{1} B, \\
B C_{1}=B B_{1}+B_{1} C_{1}
\end{array}
$$
Then
$$
\begin{aligned}
& B_{1} \boldsymbo... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,433 |
$\begin{array}{l}\text { Five. (20 points) Let } x, y, z \in \mathbf{R}_{+}, x y+y z+x z \\ =1 \text {. Prove: } \\ \frac{(x y)^{2}}{z}+\frac{(x z)^{2}}{y}+\frac{(y z)^{2}}{x}+6 x y z \geqslant x+y+z .\end{array}$ | $$
\begin{array}{l}
x+y+z=(x+y+z)(x y+y z+x z) \\
=x y(x+y)+y z(y+z)+x z(x+z)+3 x y z,
\end{array}
$$
That is, we need to prove
$$
\frac{(x y)^{2}}{z}+\frac{(x z)^{2}}{y}+\frac{(y z)^{2}}{x}+3 x y z
$$
$$
\geqslant x y(x+y)+y z(y+z)+x z(x+z).
$$
By symmetry, assume $x \geqslant y \geqslant z$. Then
$$
\begin{array}{l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,435 |
8. Given a function $y=f(x)$ defined on $\mathbf{R}$ that satisfies: $f(x)+f(-x)=0$, and when $x \leqslant 0$, we have
$$
y=x^{2 n+1}+x^{2 n}+\cdots+x \text {. }
$$
Then the expression for $f(x)$ on $\mathbf{R}$ is ( ).
(A) $x\left(|x|^{2 n}+|x|^{2 n-1}+\cdots+1\right)$
(B) $|x|\left(x^{2 n}+x^{2 n-1}+\cdots+1\right)$... | 8.C.
Since $f(x)=-f(-x)$, i.e., $y=f(x)$ is an odd function, thus, if when $x \leqslant 0$, we have $y=\varphi(x)$, then
$$
y=f(x)=\left\{\begin{array}{ll}
-\varphi(-x), & x>0 ; \\
\varphi(x), & x \leqslant 0 .
\end{array}\right.
$$
When $x \neq 0$, this piecewise expression can be uniformly represented as
$$
y=f(x)=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,436 |
9. For the parabola $y^{2}=2 p x(p>0)$, a line $l$ passing through its focus $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|F A| \cdot |F B|$ is ( ).
(A) $\frac{p^{2}}{4}$
(B) $p^{2}$
(C) $4 p^{2}$
(D) does not exist | 9. B.
Focus $F\left(\frac{p}{2}, 0\right)$. When $l$ is not perpendicular to the $x$-axis, we can set the equation of the line $l$ as $y=k\left(x-\frac{p}{2}\right)$.
Substituting into the parabola equation and eliminating $y$, we get
$$
k^{2} x^{2}-p\left(k^{2}+2\right) x+\frac{k^{2} p^{2}}{4}=0 \text {. }
$$
The tw... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,437 |
10. 5 people participate in 4 groups, each group has two people, and each person must participate in at least one group. Then, the number of different groupings is ( ).
(A) 135
(B) 130
(C) 125
(D) 120 | 10.A.
5 people can form $C_{5}^{2}=10$ pairs. Every 4 pairs are called a grouping method. A valid grouping method should include all 5 people. Since fewer than 4 people cannot form 4 pairs, all invalid grouping methods, i.e., those containing only 4 people, total $\mathrm{C}_{5}^{4} \mathrm{C}_{6}^{4}=75$ (each 4 peopl... | 135 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,438 |
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then the value of $\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}}$ is . $\qquad$ | $$
\begin{array}{l}
\text { Since } a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}, \text { then, } \\
\frac{3^{n}}{a_{n}}=\frac{3^{n}}{\mathrm{C}_{n}^{2} \cdot 3^{n-2}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} . \\
\text { Therefore, } \frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{18}}{a_{18}} \\
=18\left(\fr... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,439 |
13. As shown in Figure 1, $A^{\prime}(2,1)$, $B^{\prime}(6,3)$, and $C^{\prime}(3,5)$ are three points on the coordinate plane $x O y$. Points $A$, $B$, and $C$ are the midpoints of line segments $C C^{\prime}$, $A A^{\prime}$, and $B B^{\prime}$, respectively. The coordinates of the centroid of $\triangle A B C$ are $... | 13. $\left(\frac{11}{3}, 3\right)$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) 、 C\left(x_{3}, y_{3}\right)$. By the midpoint formula, we have
$$
\begin{array}{l}
2 x_{1}=x_{3}+3,2 x_{2}=x_{1}+2,2 x_{3}=x_{2}+6, \\
2 y_{1}=y_{3}+5,2 y_{2}=y_{1}+1,2 y_{3}=y_{2}+3 .
\end{array}
$$
Solving these equati... | \left(\frac{11}{3}, 3\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,441 |
14. A line passing through the left focus $F_{1}$ of a hyperbola and perpendicular to the real axis of the hyperbola intersects the hyperbola at points $A$ and $B$. If there exists a point $C$ on the line of the imaginary axis of the hyperbola such that $\angle A C B=90^{\circ}$, then the range of the eccentricity $e$ ... | 14. $e \geqslant \frac{\sqrt{5}+1}{2}$.
Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. Its real axis is on the $x$-axis, and its imaginary axis is on the $y$-axis. The left focus is $F_{1}(-c, 0) \left(c = \sqrt{a^{2}+b^{2}}\right)$, and the line through points $A$ and $B$ ... | e \geqslant \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,442 |
15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$
is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$. | Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,443 |
16. (12 points) As
shown in Figure 2, given the dihedral angle
$$
\begin{array}{l}
M-C D-N \text { is } \theta \\
\left(0^{\circ}<\theta<90^{\circ}\right), A \in \\
M, B \in N, A B \perp \\
C D, A B \text { forms a } 30^{\circ} \\
\text { angle with plane } N
\end{array}
$$
If the area of $\triangle A C D$ is $S$, fo... | 16. As shown in Figure 5, draw $A E \perp$ plane $N$, with $E$ as the foot of the perpendicular. Connect $B E$ and extend it to intersect $C D$ at point $F$, then connect $A F$.
Since $A E \perp$ plane $N$, $B E$ is the projection of $A B$ on plane $N$, and $A E \perp C D$.
Also, $A B \perp C D$, so $C D \perp$ plane ... | 2S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,444 |
17. (12 points) The rules of the Chinese Men's Basketball Premier League stipulate: Each game, the winner gets 2 points, the loser gets 1 point (each game, even if it goes to overtime, must have a winner). A certain men's premier basketball team is very strong, with a probability of winning each game being $\frac{3}{4}... | 17. Let the probability of the team scoring $n$ points after several matches be $p_{n}$, then we have
$$
p_{1}=\frac{1}{4}, p_{2}=p_{1}^{2}+\frac{3}{4}=\frac{13}{16} .
$$
When $n>2$, we have
$$
\begin{array}{l}
p_{n}=\frac{1}{4} p_{n-1}+\frac{3}{4} p_{n-2}(2S$, then due to the limitation to $S$ matches, the cases wher... | p_{n}=\frac{1}{4} p_{n-1}+\frac{3}{4} p_{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,445 |
One, (50 points) As shown in Figure 2, in $\triangle PAB$, $E$ and $F$ are points on sides $PA$ and $PB$ respectively. On the extensions of $AP$ and $BP$, take points $C$ and $D$ such that $PC = AE$ and $PD = BF$. Let $M$ and $N$ be the orthocenters of $\triangle PCD$ and $\triangle PEF$ respectively. Prove: $MN \perp ... | As shown in Figure 8, let the midpoints of line segments $DE$, $CF$, and $PF$ be $G$, $H$, and $K$ respectively, then $K$ is also the midpoint of $BD$. According to the Midline Theorem:
In $\triangle BDE$, $KG \parallel BE$, $KG = \frac{1}{2} BE$;
In $\triangle PCF$, $KH \parallel PC$, $KH = \frac{1}{2} PC$, i.e., $K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,446 |
18. (15 points) Given the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, construct an isosceles right triangle $\triangle A B C$ inscribed in the ellipse with the right angle at the endpoint $B(0,1)$ of the minor axis. If there is exactly one such triangle, find the range of values for $a$.
---
The above text has been t... | 18. Let the equation of line $AB$ be $y=kx+1 (k>0)$, then the equation of line $BC$ is $y=-\frac{1}{k}x+1$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y=kx+1,\end{array}\right.$ we get $x_{A}=\frac{-2a^{2}k}{a^{2}k^{2}+1}$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y=-\frac{1}{k}x+1... | 1<a \leqslant \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,447 |
19. (15 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{n+1} a_{n}+3 a_{n+1}+a_{n}+4=0 \text {. }
$$
If $a_{2006}$ is the minimum term of the sequence $\left\{a_{n}\right\}$, find the range of the first term $a_{1}$. | 19. Solution 1: From the problem condition, we get $a_{n+1}=\frac{-a_{n}-4}{a_{n}+3}$.
Let $a_{n+1}-\lambda=\frac{\alpha\left(a_{n}-\lambda\right)}{a_{n}+3}$, i.e.,
$$
a_{n+1}=\frac{(\alpha+\lambda) a_{n}-\alpha \lambda+3 \lambda}{a_{n}+3} .
$$
By comparing coefficients, we get
$$
\left\{\begin{array}{l}
\alpha+\lamb... | -\frac{4011}{2005}<a_{1}<-\frac{4013}{2006} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,448 |
For any given positive integer $n(n \geqslant 2)$, there exist $2 n+1$ distinct positive integers $x_{1}, x_{2}, \cdots, x_{n}$, $y_{1}, y_{2}, \cdots, y_{n}, z$, such that
$$
x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=\cdots=x_{n}^{2}+y_{n}^{2}=z^{2} \text {. }
$$ | Prove: Take $n$ sets of Pythagorean triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2, \cdots, n)$ satisfying
$$
a_{i}=2(i+2), b_{i}=(i+2)^{2}-1, c_{i}=(i+2)^{2}+1 .
$$
Then $a_{i}x_{2}>\cdots>x_{n}, y_{1}<y_{2}<\cdots<y_{n}, x_{i}<y_{i}, \\
x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=\cdots=x_{n}^{2}+y_{n}^{2}=z^{2} .
\end... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,449 |
Given an acute triangle $\triangle ABC$ and its circumcircle, $AM$ is the median of side $BC$. Tangents to the circumcircle are drawn through points $B$ and $C$, intersecting at point $N$. $T$ is a point on $AM$ such that $\angle ATC = \angle ABN$. Prove that $\frac{AB}{AC} = \frac{TB}{TC}$. | Proof: As shown in Figure 2, draw \( CD \perp AB \) at point \( D \), and connect \( DM \) and \( AN \). Then,
\[
\begin{array}{l}
DM = \frac{1}{2} BC = CM. \\
\text{Also, } \angle BAC = \angle BCN,
\end{array}
\]
Thus,
\[
\begin{array}{l}
\frac{AD}{AC} = \cos \angle BAC \\
= \cos \angle BCN \\
= \frac{CM}{CN} = \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,450 |
Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$
a_{n}=a_{1}+\left[\frac{n}{2}\right] d_{1}+\left[\frac{n-1}{2}\right] d_{2} \text {, }
$$
the sum of the first $n$ terms is $S_{n}, n \in \mathbf{N}_{+}, d_{1} 、 d_{2}$ are constants, and $[x]$ denotes the greatest integer not exceeding $x$. Pr... | Proof: Let $m \in \mathbf{N}_{+}$, from the general term formula of $\left\{a_{n}\right\}$ we have
$$
\begin{array}{l}
a_{2 m+1}=a_{1}+\left[\frac{2 m+1}{2}\right] d_{1}+\left[\frac{2 m}{2}\right] d_{2} \\
=a_{1}+m d_{1}+m d_{2}, \\
a_{2 m}=a_{1}+\left[\frac{2 m}{2}\right] d_{1}+\left[\frac{2 m-1}{2}\right] d_{2} \\
=a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,451 |
For the cubic function $y=$ $x^{3}+b x^{2}+c x+d$ whose graph is shown in Figure 3, the line $B D / / A C$, and the line $B D$ is tangent to the graph of the cubic function at point $B$ and intersects it at point $D$. The line $A C$ is tangent to the graph of the cubic function at point $C$ and intersects it at point $... | Proof: Let the expansion of the cubic function $y=f(x)$ at point $B\left(x_{B}, f\left(x_{B}\right)\right)$ be
$$
y=\left(x-x_{B}\right)^{3}+b_{1}\left(x-x_{B}\right)^{2}+c_{1}\left(x-x_{B}\right)+f\left(x_{B}\right) .
$$
Then $f^{\prime}(x)=3\left(x-x_{B}\right)^{2}+2 b_{1}\left(x-x_{B}\right)+c_{1}$,
$$
f^{\prime}\l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,452 |
(50 points) Color the space in five colors in any way (each point belongs to one color, and each color has points).
(1) Prove: There exists a plane that contains at least four different colored points;
(2) Does there necessarily exist a plane with points of all five colors? | (1) If there exists a four-color line $l$, then the plane containing $l$ is the one we are looking for. If there exists a three-color line $l$, then we can take another point $M$ of the fourth color outside the line $l$, and the plane passing through point $M$ and line $l$ is the one we are looking for.
Suppose that n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,453 |
Three. (50 points) A rhombus $ABCD$ with side length $n$ has its apex angle $\angle A$ as $60^{\circ}$. Using three sets of equally spaced parallel lines, each parallel to $AB$, $AD$, and $BD$ respectively, the rhombus is divided into $2n^2$ equilateral triangles with side length 1 (as shown in Figure 3). Try to find t... | Three, since any two line segments in Figure 3 are either parallel or intersect at a $60^{\circ}$ acute angle, all trapezoids formed by the line segments in the figure are isosceles trapezoids with base angles of $60^{\circ}$. For such trapezoids, if the intersection point of the extensions of the two non-parallel side... | \frac{n\left(n^{2}-1\right)(2 n+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,454 |
3. If in the three common logarithms $\lg a \sqrt{ } \lg b \sqrt{\lg c}$, the sum of the mantissas of any two is greater than the mantissa of the third, then the three positive numbers $a, b, c$ are said to form a “logarithmic triangle”. Now, choose three distinct integers from the set
$$
M=\{7,8,9,10,11,12,13,14\}
$$
... | 3.A.
Obviously, the mantissa of $\lg 10$ is 0, so we only need to consider the ternary groups that can form a triangle from the logarithms of the numbers in $M_{1}=\{7,8,9, 1.1,1.2,1.3,1.4\}$. We divide $M_{1}$ into two subsets:
$$
A=\{7,8,9\} \text { and } B=\{1.1,1.2,1.3,1.4\} \text {. }
$$
Note that, if $1\lg c$, ... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,457 |
3. As shown in Figure 1, in the spatial quadrilateral $ABCD$, $E$ and $F$ are points on $AB$ and $CD$ respectively, such that $\frac{AE}{EB}=\frac{DF}{FC}=2007$. Then $\boldsymbol{EF}=(\quad)$. | 3.A.
Let $\frac{A E}{E B}=2007=a, \frac{F C}{D F}=\frac{1}{2007}=b$, then $a b=1$. Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{1}{a+1}+\frac{1}{\frac{1}{a}+1}=\frac{1}{a+1}+\frac{a}{a+1}=1$.
Thus, $E F=E B+B C+C F$
$$
\begin{array}{l}
=\frac{A B}{a+1}+B C+\frac{C D}{b+1} \\
=\frac{A B}{a+1}+\left(\frac{B C}{a+1}+\fr... | \frac{A C+2007 B D}{2008} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,458 |
4. Given an ellipse centered at the origin, with foci on the $x$-axis and a focal distance of 4, which is tangent to the line $x+y=9$ at point $P$. Then, the coordinates of point $P$ are ( ).
(A) $\left(\frac{35}{6}, \frac{19}{6}\right)$
(B) $\left(\frac{21}{9}, \frac{60}{9}\right)$
(C) $\left(\frac{85}{18}, \frac{77}{... | 4.C.
As shown in Figure 2, let $A$ and $B$ be the foci of the ellipse, and $P$ be any point on the line $x + y = 9$. Then, $P$ is outside the ellipse or coincides with point $P$. Suppose $P'A$ intersects the ellipse at point $Q$, then
$$
P'A + P'B = P'B + PQ + QA \geqslant QB + QA = PA + PB.
$$
Thus, point $P$ is the... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,459 |
5. In $\triangle A B C$, $\angle A \leqslant \angle B \leqslant \angle C$, $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3}$.
Then the range of $\angle B$ is ( ).
(A) $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
(B) $\left(0, \frac{\pi}{2}\right)$
(C) $\frac{\pi}{3}$
(D) $\left(\frac{\pi}{4}, \frac{\pi}{3}\r... | 5.C.
From the condition, we have
$$
\begin{array}{l}
\sin A + \sin B + \sin C = \sqrt{3} (\cos A + \cos B + \cos C) \\
\Rightarrow 2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2} + \sin B \\
\quad = \sqrt{3} \left(2 \cos \frac{A+C}{2} \cdot \cos \frac{A-C}{2} + \cos B\right) \\
\Rightarrow \left(2 \sqrt{3} \cos \frac{A+... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,460 |
6. Let $S_{1}, S_{2}, S_{3}$ be infinite subsets of the set of integers $\mathbf{Z}$, and satisfy:
(1) For any permutation $i, j, k$ of $1, 2, 3$, if $x \in S_{i}, y \in S_{j}$, then $x-y \in S_{k}$;
(2) There exist $1 \leqslant i<j \leqslant 3$ such that $S_{i} \cap S_{j}=\varnothing$.
Then, the number of such triplet... | 6.D.
For any positive integer $k$, take
$$
\begin{array}{l}
S_{1}=S_{2}=\{x \mid x \equiv k(\bmod 2 k)\}, \\
S_{3}=\{x \mid x \equiv 0(\bmod 2 k)\} .
\end{array}
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,461 |
1. A chord $P Q$ is drawn through the focus $F$ of the parabola $y=a x^{2}(a<0)$. If $|P F|=2007,|Q F|=10$, then, $a=$ $\qquad$ | $$
\text { II.1. }-\frac{10280}{2017} \text {. }
$$
Let the distance from the focus to the directrix be $p$, and
$$
|P F|=2007=x,|Q F|=10=y \text {. }
$$
From the similarity of the two triangles in Figure 3, we have
$$
\begin{array}{l}
\frac{x-p}{p-y}=\frac{x}{y} . \\
\text { Therefore, } p=\frac{2 x y}{x+y} \\
=\fra... | -\frac{10280}{2017} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,462 |
2. Let $u$, $v$, $w$ be complex numbers, where $w=a+b \mathrm{i}$ $\left(a, b>3, a^{2}+b^{2}=25\right), u-w=3 v$. If $|v|=$ 1, then when the principal value of the argument of $u$ is the smallest, the value of $\frac{u}{w}$ is $\qquad$ | 2. $\frac{16}{25}-\frac{12}{25}$ i.
Since $|v|=1$, we have $|u-w|=3|v|=3$.
Thus, the point $P$ corresponding to $u$ lies on the circle $C$ with the point $M$ corresponding to $w$ as the center and a radius of 3.
When the principal value of the argument of $u$ is the smallest, $OP$ is tangent to the circle $C$. Given ... | \frac{16}{25}-\frac{12}{25} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,463 |
3. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any $n \in \mathbf{N}_{+}$, $a_{n}+a_{n+1}=2007$, and $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$. Then the sum of all different values of the sum of the first 2007 terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$. | 3. 1345368366 .
From $31\left(a_{n}^{2}+a_{n} a_{n+1}+a_{n+1}^{2}\right)$, we get
$$
31\left[a_{n}^{2}+a_{n+1}\left(a_{n}+a_{n+1}\right)\right] \text {. }
$$
Also, $31\left(a_{n}+a_{n+1}\right)$, so, $31 a_{n}$.
Since $a_{1}=2007-a_{2}<2006$, we have
$$
\begin{array}{l}
a_{1} \leqslant 2004, a_{1}=3 k(k=1,2, \cdots, ... | 1345368366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,464 |
4. There is a math competition problem, the probabilities of A, B, and C solving it alone are $\frac{1}{a} 、 \frac{1}{b} 、 \frac{1}{c}$, respectively, where $a 、 b 、 c$ are all positive integers in the range of 1 to 9. Now A, B, and C are solving this problem independently at the same time. If the probability that exac... | 4. $\frac{4}{15}$.
According to the problem, we have
$$
\frac{1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{1}{c}=\frac{7}{15},
$$
which simplifies to $15[(a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)]=7 a b c$.
Thus, $5 \mid a b c$.... | \frac{4}{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,465 |
5. The number of prime pairs $(p, q)$ that satisfy $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ is $\qquad$ . | 5.2.
(1) When $p=2$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1$, which is not a prime number, contradiction.
(2) When $p=3$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1+1=2$ is a prime number.
(3) When $p=5$, $q=\left[\frac{p}{2}\right]+\left[\fr... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,466 |
6. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f\left(\frac{\pi}{4}\right)=0$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right) .
$$
Then $f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left... | 6.0 .
Let $\frac{x-y}{2}=\frac{\pi}{4}$, then
$$
\begin{array}{l}
f(x)+f\left(x-\frac{\pi}{2}\right)=2 f\left(x-\frac{\pi}{4}\right) f\left(\frac{\pi}{4}\right)=0 . \\
\text { Therefore } f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)=0 .... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,467 |
4. For $\triangle A B C$, the tangent values of the three interior angles are all integers. If similar triangles are counted as only one case, then the total number of cases that meet the conditions is ( ).
(A) impossible
(B) only one
(C) infinitely many
(D) at least two, with a finite number | 4.B.
Let $\tan A=x, \tan B=y, \tan C=z, x, y, z$ be non-zero integers, and at most one of them is negative. From the identity
$$
\cot A \cdot \cot B + \cot B \cdot \cot C + \cot C \cdot \cot A = 1,
$$
we get $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} = 1$,
which simplifies to $x y z = x + y + z$.
If $x, y, z$ inc... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,468 |
Three, (20 points) Let $x, y, z \geqslant 0, x+y+z=1$. Find the maximum value of $A=\left(x-\frac{1}{6}\right)\left(y-\frac{1}{6}\right)\left(z-\frac{1}{6}\right)^{2}$.
| (1) When $x-\frac{1}{6}$ and $y-\frac{1}{6}$ have opposite signs, $A \leqslant 0$.
(2) When $x-\frac{1}{6}0, y-\frac{1}{6}>0, z-\frac{1}{6}\frac{1}{3}, 0 \leqslant z0, y-\frac{1}{6}>0, z-\frac{1}{6}>0$,
$$
\begin{array}{l}
A=\left(x-\frac{1}{6}\right)\left(y-\frac{1}{6}\right)\left(z-\frac{1}{6}\right)\left(z-\frac{1}{... | \frac{25}{1296} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,469 |
Four. (20 points) Let $\alpha, \beta \in \mathbf{C}, \alpha, \beta \neq 0,$ and the principal values of the arguments of $\alpha$ and $\beta$ are different. Prove:
$$
\left|\frac{|\alpha| \beta-\alpha|\alpha \beta|}{\beta|\alpha|-\alpha|\beta|}\right| \geqslant \frac{1+|\alpha|}{2} .
$$ | To prove:
$$
\frac{\left|\frac{\beta}{|\beta|}-\alpha\right|}{\left|\frac{\beta}{|\beta|}-\frac{\alpha}{|\alpha|}\right|} \geqslant \frac{1+|\alpha|}{2}.
$$
Thus, it suffices to prove: when $|z|=1$,
$$
\frac{|z-\alpha|}{\left|z-\frac{\alpha}{|\alpha|}\right|} \geqslant \frac{1+|\alpha|}{2}.
$$
When $|\alpha|=1$, ineq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,470 |
Five. (20 points) On the circumference, there are $n$ points $A_{1}$, $A_{2}, \cdots, A_{n}$ in sequence. Now, $k$ points are randomly selected as vertices to form a convex $k$-gon $B_{1} B_{2} \cdots B_{k}$, with the possibility of selection being the same. Try to find the probability that for each $i \in\{1,2, \cdots... | Let $A_{i_{1}} A_{i_{2}} \cdots A_{i_{k}}$ be a convex $k$-gon that meets the conditions, and let $P=\left\{A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k}}\right\}$. Suppose there are $x_{t}$ points in $X$ between $A_{i_{t}}$ and $A_{i_{t+1}}$ that do not belong to $P \left(t=1,3, \cdots, k, \text{ with } A_{i_{k+1}}=A_{i_{1}}... | \frac{\mathrm{C}_{n-s(r)-1}^{k-1}}{\mathrm{C}_{n-1}^{k-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,471 |
One, (50 points) Let $D$ be a moving point on side $BC$ of a given $\triangle ABC$, and $P$ divides $AD$ in a fixed ratio $\lambda$. Let $BP$ intersect $CD$ at $E$, and $CP$ intersect $AB$ at $F$. Find the maximum value of the area of $\triangle AEF$.
---
Translate the above text into English, preserving the original... | $$
\text { I. Let } \frac{A F}{F B}=u, \frac{A E}{E C}=v \text {. }
$$
By Ceva's Theorem, we have
$$
\begin{array}{l}
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1 \\
\Rightarrow \frac{B D}{D C}=\frac{E A}{C E} \cdot \frac{F B}{A F}=\frac{v}{u} \\
\Rightarrow \frac{B C}{D C}=\frac{C D+D B}{D C}=1+\frac... | \left(\frac{\lambda}{\lambda+2}\right)^{2} S_{\triangle A B C} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,472 |
II. (50 points) Find the maximum value of a prime number $p$ with the following property: there exist two permutations (which can be the same) of $1, 2, \cdots, p$, $a_{1}, a_{2}, \cdots, a_{p}$ and $b_{1}, b_{2}, \cdots, b_{p}$, such that the remainders of $a_{1} b_{1}$, $a_{2} b_{2}, \cdots, a_{p} b_{p}$ when divided... | Lemma (Wilson's Theorem): For any prime $p$, we have $(p-1)! \equiv -1 \pmod{p}$.
Proof of the Lemma: When $p=2,3$, equation (1) is obviously true.
When $p>3$, we prove: For any $2 \leqslant k \leqslant p-2$, there must exist $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant p-2, k^{\prime} \neq k\right)$, such that $... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,473 |
Three, (50 points) $n$ people exchange greetings by phone during a holiday. It is known that each person makes calls to at most three friends, any two people make at most one call to each other, and among any three people, at least two of them, one person has called the other. Find the maximum value of $n$.
| Three, first prove the lemma.
Lemma In a simple graph $G$ of order $n$ without $K_{3}$, then
$$
f(G) \leqslant\left[\frac{n^{2}}{4}\right],
$$
where $f(G)$ represents the number of edges in $G$.
Proof of the lemma: Let $A$ be the vertex with the maximum degree, and let the set of vertices adjacent to $A$ be $M=\left\{... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,474 |
1. The function $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{|x+a|-a}$ is an odd function if and only if ( ).
(A) $a<0$
(B) $a>0$
(C) $-1<a<0$ or $0<a<1$
(D) $-1 \leqslant a<0$ or $0<a \leqslant 1$ | ,$- \mathrm{C}$.
Sufficiency. If $a>0$, then the domain of $f(x)$ is $[-a, 0) \cup (0, a]$. In this case, $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{x}$, which is clearly an odd function.
Necessity. If $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{|x+a|-a}$ is an odd function, then $a \neq 0$ (otherwise, the domain of $f(x)$ is an empty set).... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,475 |
2. Given the sides of $\triangle A B C$ are $a, b, c$ and satisfy $a^{2}-a-2 b-2 c=0$, and $a+2 b-2 c+3=0$. Then the largest interior angle of $\triangle A B C$ is ( ).
(A) $150^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $60^{\circ}$ | 2.B.
From the given conditions, we have
$$
\begin{array}{l}
a+2 b+2 c=a^{2}, a+2 b-2 c=-3 \\
\Rightarrow(a+2 b+2 c)(a+2 b-2 c)=-3 a^{2} \\
\Rightarrow(a+2 b)^{2}-4 c^{2}=-3 a^{2} \\
\Rightarrow c^{2}=a^{2}+b^{2}+a b \\
\Rightarrow \cos C=-\frac{1}{2} \Rightarrow \angle C=120^{\circ} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,476 |
3. A line $l$ is drawn through the fixed point $P(2,1)$, intersecting the positive directions of the $x$-axis and $y$-axis at points $A$ and $B$, respectively. If the area of $\triangle A O B$ (where $O$ is the origin) is to be minimized, then the equation of $l$ is ( ).
(A) $x+y-3=0$
(B) $x+3 y-5=0$
(C) $2 x+y-5=0$
(D... | 3.D.
From the given, we know that $S_{\triangle A O B}=\frac{1}{2} a b$, and $\frac{2}{a}+\frac{1}{b}=1$.
Then $a b=a+2 b \geqslant 2 \sqrt{a \cdot 2 b}=2 \sqrt{2} \cdot \sqrt{a b}$,
which means $\sqrt{a b} \geqslant 2 \sqrt{2}, a b \geqslant 8$.
Equality holds if and only if $\left\{\begin{array}{l}a=2 b, \\ 2 b+a=a ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,477 |
4. For all real numbers $x$, the values of all quadratic functions $f(x) = ax^2 + bx + c (a \leqslant b)$ are non-negative real numbers. Then the maximum value of $\frac{b-a}{a+b+c}$ is $(\quad)$.
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) 3 | 4.C.
From the problem, we know $a>0$, and $b^{2}-4 a c \leqslant 0$, so $c \geqslant \frac{b^{2}}{4 a}$. Let $b-a=t(t \geqslant 0)$, i.e., $b=a+t$. Therefore,
$$
\begin{array}{l}
M=\frac{b-a}{a+b+c} \leqslant \frac{t}{a+(a+t)+\frac{(a+t)^{2}}{4 a}} \\
=\frac{4 a t}{(3 a+t)^{2}} \leqslant \frac{4 a t}{(2 \sqrt{3 a t})^... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,478 |
5. The number of real numbers $x$ that satisfy the equation $36 x^{2}-140[x]-21=0$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 5.C.
Let $x=k+\alpha(0 \leqslant \alpha < 1, k \geqslant 0)$, i.e., the solution of the equation is a positive number.
According to equation (1), we have
$$
\begin{array}{l}
21=36(k+\alpha)^{2}-140 k>36 k^{2}-140 k \\
=36 k\left(k-\frac{35}{9}\right)=36\left(k-\frac{35}{18}\right)^{2}-36\left(\frac{35}{18}\right)^{2} ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,479 |
5. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left focus is $F_{1}$, the vertices are $A_{1}$ and $A_{2}$, and $P$ is any point on the hyperbola. Then the two circles with diameters $P F_{1}$ and $A_{1} A_{2}$ respectively must ( ).
(A) intersect
(B) be tangent
(C) be separate
(D) any of ... | 5.B.
As shown in Figure 1, let point $P$ be on the right branch of the hyperbola. Taking the midpoint $M$ of $P F_{1}$, then $|O M|$ is the distance between the centers of the two circles, $\frac{|P F_{1}|}{2}$ and $a$ are the radii of the two circles, respectively. In $\triangle P F_{1} F_{2}$, $|O M|=\frac{1}{2}|P F... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,480 |
6. In the complex plane, the number of intersection points between the curve $z^{4}+z=1$ and the circle $|z|$ $=1$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6.A.
Since $|z|=1$, we can let $z=\cos \theta+\mathrm{i} \sin \theta(0 \leqslant \theta<2 \pi)$.
From $z^{3}+1=\bar{z}$
$$
\begin{array}{l}
\Rightarrow 1+\cos 3 \theta+\mathrm{i} \sin 3 \theta=\cos \theta-\mathrm{i} \sin \theta \\
\Rightarrow \cos 3 \theta=\cos \theta-1, \sin 3 \theta=-\sin \theta \\
\Rightarrow(\cos ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,481 |
1. If $f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-$ $2 \sqrt{2007} x^{2}+2 x-\sqrt{2006}$, then $f(\sqrt{2006}+\sqrt{2007})=$ $\qquad$ | $$
\begin{array}{l}
\text { 2. } \sqrt{2007}. \\
\text { Consider the quadratic equations } \\
x^{2}-2 \sqrt{2006} x-1=0, \\
x^{2}-2 \sqrt{2007} x+1=0 .
\end{array}
$$
From equation (1), we get $x^{2}-2 \sqrt{2006} x+2006-2007=0$, which is
$$
(x-\sqrt{2006}+\sqrt{2007})(x-\sqrt{2006}-\sqrt{2007})=0.
$$
From equation ... | \sqrt{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,482 |
2. Given that the first $n$ terms sum of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $S_{n}$ and $T_{n}$ respectively, and $\frac{S_{n}}{T_{n}}=\frac{3 n+2}{2 n+1}$. Then $\frac{a_{7}}{b_{5}}=$ $\qquad$ . | 2. $\frac{41}{19}$.
From $\frac{S_{n}}{T_{n}}=\frac{3 n+2}{2 n+1}$, we get $\frac{S_{n}}{T_{n}}=\frac{\left(3 n^{2}+2 n\right) k}{\left(2 n^{2}+n\right) k}$.
Let $S_{n}=\left(3 n^{2}+2 n\right) k, T_{n}=\left(2 n^{2}+n\right) k$.
Then $a_{7}=S_{7}-S_{6}=41 k, b_{5}=T_{5}-T_{4}=19 k$.
Therefore, $\frac{a_{7}}{b_{5}}=\f... | \frac{41}{19} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,483 |
3. Let $x+y+z=1$. Then the minimum value of the function $u=2 x^{2}+3 y^{2}$ $+z^{2}$ is . $\qquad$ | 3. $\frac{6}{11}$.
From the given conditions and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
1=x+y+z=\left(\frac{1}{\sqrt{2}} \cdot \sqrt{2} x+\frac{1}{\sqrt{3}} \cdot \sqrt{3} y+1 \cdot z\right) \\
\leqslant\left(\frac{1}{2}+\frac{1}{3}+1\right)^{\frac{1}{2}}\left(2 x^{2}+3 y^{2}+z^{2}\right)^{\frac{1}{2}}... | \frac{6}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,484 |
4. Given a line $l$ passing through the point $F(a, 0)$ (not perpendicular to the $x$-axis) intersects the curve $C: y^{2}=4 a x$ at points $A$ and $B$. Let point $K(-a, 0)$, and the angle between $\boldsymbol{K A}$ and $\boldsymbol{K B}$ be $\theta$. Then the range of $\theta$ is $\qquad$. | 4. $\left(0, \frac{\pi}{2}\right)$.
Let the equation of $l$ be $y=k(x-a)$.
From $\left\{\begin{array}{l}y^{2}=4 a x, \\ y=k(x-a),\end{array}\right.$ we get $y^{2}-\frac{4 a}{k} y-4 a^{2}=0$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
y_{1} y_{2}=-4 a^{2}, \boldsymbol{K A... | \left(0, \frac{\pi}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,485 |
5. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$ | 5.1.
As shown in Figure 2, draw $M P \perp A B$ at $P$, and connect $P N$. It can be proven that $P N \perp A B$. Thus, $\angle M P N=120^{\circ}$.
Let $A M=F N=x$.
Therefore, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$,
which means $M P=\frac{\sqrt{2}}{2} x$.
Hence, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}\left(\frac{P N}{1}=\frac... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,486 |
6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$ | 6.1
From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\co... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,487 |
Three. (20 points) Given the function $f(x)$ for any real numbers $x, y$ has
$$
f(x+y)=f(x)+f(y)+2(x+y)+1,
$$
and $f(1)=1$.
If $x \in \mathbf{N}_{+}$ and $x \geqslant 2$, the inequality
$$
f(x) \geqslant(a+7) x-(a+10)
$$
always holds, try to find the range of the real number $a$. | Let $y=1$, then
$$
f(x+1)=f(x)+f(1)+2(x+1)+1,
$$
which means $f(x+1)-f(x)=2 x+4$.
When $x \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
f(2)-f(1)=2 \times 1+4, \\
f(3)-f(2)=2 \times 2+4, \\
\cdots \cdots \\
f(x)-f(x-1)=2(x-1)+4 .
\end{array}
$$
Adding the above equations, we get
$$
f(x)-f(1)=2 \times \frac{1+(x-1)... | (-\infty, 2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,488 |
Four. (20 points) Given the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>n$ $>0)$ passes through the fixed point $P(a, b)(a, b$ are real numbers, and $a b \neq$ $0,|a| \neq|b|)$. Find the minimum value of $m+n$.
Translate the above text into English, please retain the original text's line breaks and format, an... | From the given, we have $\frac{a^{2}}{m^{2}}+\frac{b^{2}}{n^{2}}=1$, which means
$$
\left(\frac{a}{m}\right)^{2}+\left(\frac{b}{n}\right)^{2}=1 \text {. }
$$
Given $a b \neq 0$, by symmetry, we can assume $a, b \in \mathbf{R}_{+}$.
From equation (1), let $\frac{a}{m}=\cos \alpha, \frac{b}{n}=\sin \alpha$ (where $\alph... | (m+n)_{\min }=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,489 |
6. As shown in Figure 1, fold up the five "corners" (isosceles small triangles) of the regular pentagram along their bases, so that they form right dihedral angles with their original planes. In the resulting spatial figure, there are ( ) pairs of skew line segments.
(A) 45
(B) 50
(C) 55
(D) 60 | 6. B.
The outer boundary of the pentagram is a closed zigzag line composed of 10 line segments, which are alternately colored red and blue (the inner small pentagon is not colored), as shown in Figure 4. Among these 10 line segments, any pair of the same color are skew, while any pair of different colors are coplanar,... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,490 |
Five. (20 points) Given
$$
\frac{\cos x+\cos y+\cos z}{\cos (x+y+z)}=\frac{\sin x+\sin y+\sin z}{\sin (x+y+z)}=a \text {. }
$$
Find the value of $\cos (y+z)+\cos (z+x)+\cos (x+y)$. | Let $S=x+y+z$.
Also, $\mathrm{e}^{\mathrm{i} x}=\cos x+\mathrm{i} \sin x, \mathrm{e}^{\mathrm{i} y}=\cos y+\mathrm{i} \sin y$, $\mathrm{e}^{\mathrm{i} z}=\cos z+\mathrm{i} \sin z$,
then
$$
\begin{array}{l}
\mathrm{e}^{\mathrm{i} x}+\mathrm{e}^{\mathrm{i} y}+\mathrm{e}^{\mathrm{i} z} \\
=(\cos x+\cos y+\cos z)+\mathrm{i... | a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,491 |
One. (50 points) Given that $\triangle ABC$ is inscribed in $\odot O$ at points $D$, $E$, and $F$. Extend $CO$ to intersect $EF$ at $M$, and extend $BO$ to intersect $EF$ at $G$. Prove: $S_{\triangle BOC} = S_{\text{quadrilateral } AMOG}$. | As shown in Figure 3, connect $A O$, $O D$, $O F$, and $B M$. From the given conditions, we have:
$$
\begin{array}{l}
\angle B O C \\
=180^{\circ}-(\angle 1+\angle 2) \\
\angle 1+\angle 2 \\
=90^{\circ}-\frac{1}{2} \angle B A C .
\end{array}
$$
Also, $\angle A F M$
$$
=90^{\circ}-\frac{1}{2} \angle B A C \text {, }
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,492 |
Sure, here is the translated text:
```
II. (50 points) Let the real-coefficient polynomial
$$
f(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}
$$
have roots $b_{1}, b_{2}, \cdots, b_{n}$, where $n \geqslant 2$. Prove:
For $x>\max \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, we have
$$
f(x+1) \geqslant \frac{2 n^{2}}{\frac{1}{x-... | When $n \geqslant 2$, $n(2-n) \leqslant 0 \Rightarrow -n^{2}+2 n \leqslant 0$
$$
\Rightarrow n^{2}-2 n^{2}+2 n \leqslant 0 \Rightarrow n^{2}-2 n(n-1) \leqslant 0 \text {. }
$$
Therefore, for any real number $t(t>0)$, we have
$$
\frac{n(n-1)}{2} t^{2}-n t+1 \geqslant 0 \text {. }
$$
Thus, $(1+t)^{n} \geqslant 1+n t+\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,493 |
Three. (50 points) $f(x)$ is a function defined on the set of natural numbers, satisfying
$$
f(1)=\frac{4}{3},
$$
and $2 f(k)=f(k-1)-k \cdot 5^{k}+2^{k}(k \geqslant 2, k \in \mathbf{N})$. Try to find the analytical expression of the function $f(x)$. | Given the known condition $2 f(k)=f(k-1)-k \cdot 5^{k}+2^{k}$, construct the following equation:
$$
2\left[f(k)-\frac{2^{k+1}}{3}\right]=\left[f(k-1)-\frac{2^{k}}{3}\right]-k \cdot 5^{k} .
$$
Divide both sides by $5^{k}$:
$$
\frac{2\left[f(k)-\frac{2^{k+1}}{3}\right]}{5^{k}}=\frac{1}{5} \cdot \frac{\left[f(k-1)-\frac{... | f(x)=\frac{5^{x+1}}{81}\left(8 \cdot 10^{1-x}-9 x+1\right)+\frac{2^{x+1}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,494 |
1. For $a \in \mathbf{R}$, the sets are:
$$
\begin{array}{l}
M=\{x|| x \mid<a, x \in \mathbf{R}\}, \\
N=\{x \mid-a<x<a, x \in \mathbf{R}\} .
\end{array}
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M \supset N$
(B) $M \subset N$
(C) $M=N$
(D) Cannot be determined, depends on the sign of $a$ | -、1.C.
When $a>0$, it is well known that $|x|<a \Leftrightarrow -a<x<a$, so $M=$ $N \neq \varnothing$. When $a \leqslant 0$, by the non-negativity of the absolute value, we know that $M=\varnothing$. And from $0 \leqslant -a<x<a \leqslant 0$ which is a contradiction, we know that $N=\varnothing$. Therefore, $M=N$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,495 |
2. In the tetrahedron $A-B C D$,
$$
\begin{array}{l}
\sin \angle B A D=\cos \angle B C D, \\
\sin \angle A B D=\cos \angle C B D, \\
\sin \angle A D B=\cos \angle C D B .
\end{array}
$$
Then $\triangle A B D$ is ( ).
(A) an obtuse triangle
(B) a right triangle
(C) an acute triangle
(D) cannot be determined, depends on... | 2. A.
As shown in Figure 1, from the fact that $\angle B A D$, $\angle A B D$, and $\angle A D B$ are the interior angles of $\triangle A B D$, we have
$$
\begin{array}{l}
\cos \angle B C D \\
=\sin \angle B A D>0, \\
\cos \angle C B D \\
=\sin \angle A B D>0, \\
\cos \angle C D B=\sin \angle A D B>0 .
\end{array}
$$
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,496 |
3. Given the sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ where each term is not equal to 0 or 1, and the sum $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$ satisfies $2 S_{n}=a_{n}-a_{n}^{2}(1 \leqslant n \leqslant 5)$. Then the number of sequences that satisfy the condition is ( ) .
(A) 2
(B) 6
(C) 8
(D) 16 | 3. B.
Substitute $n=1$ into $2 S_{n}=a_{n}-a_{n}^{2}$, we get $2 a_{1}=a_{1}-a_{1}^{2}$, which is $a_{1}^{2}+a_{1}=0$.
But $a_{1} \neq 0$, so $a_{1}=-1$.
For $n \geqslant 2$, from $2 S_{n}=a_{n}-a_{n}^{2}$, we have
$$
2 S_{n+1}=a_{n+1}-a_{n+1}^{2} \text {. }
$$
Thus, $2 a_{n+1}=2\left(S_{n+1}-S_{n}\right)$
$$
=a_{n+1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,497 |
4. Given that $a$ and $b$ are non-zero and non-collinear vectors, let condition $M: b \perp (a-b)$;
condition $N$: for all $x \in \mathbf{R}$, the inequality $|a-x b|$ $\geqslant |a-b|$ always holds.
Then $M$ is ( ) of $N$.
(A) necessary but not sufficient condition
(B) sufficient but not necessary condition
(C) suffic... | 4.C.
As shown in Figure 2, let $O A=a$, $O B=b$, then $x b$ represents any vector collinear with $O B$, $|a-x b|$ represents the distance $A C$ from point $A$ to any point $C$ on the line $O B$, and $|a-b|$ represents the distance from point $A$ to $B$.
When $b \perp(a-b)$, $A B \perp O B$.
By the shortest distance fr... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,498 |
5. The triangle with vertices $A(-3,0)$, $B(0,-3)$, and $C\left(\frac{15}{7}, \frac{24}{7}\right)$ has no common points with the circle $x^{2}+y^{2}=R^{2}(R>0)$. Then the range of the radius $R$ is ( ).
(A) $\left(\frac{3 \sqrt{2}}{2}, 3\right)$
(B) $\left(0, \frac{3 \sqrt{2}}{2}\right) \cup(3,+\infty)$
(C) $\left(\fra... | 5.D.
As shown in Figure 3, from the given information we have
$$
\begin{array}{l}
|O A|=3, \\
|O B|=3, \\
|O C|=\frac{3 \sqrt{89}}{7} .
\end{array}
$$
When $R > |O C|$, the triangle and the circle have no intersection points.
The equations of the lines $A B$, $A C$, and $B C$ are respectively
$$
y=-x-3, y=\frac{2}{3... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,499 |
1. Given the function $f(x)=x^{3}+a x^{2}+b x+a^{2}$ has an extremum of 10 at $x=1$. Then the real number pair $(a, b)$ is $\qquad$ . | II. 1. $(4,-11)$.
Given that the function has an extremum of 10 at $x=1$, we have
$$
\left\{\begin{array}{l}
f^{\prime}(1)=0, \\
f(1)=10,
\end{array} \text { i.e., } \left\{\begin{array}{l}
3+2a+b=0, \\
1+a+b+a^{2}=10 .
\end{array}\right.\right.
$$
Solving, we get $\left\{\begin{array}{l}a_{1}=4, \\ b_{1}=-11 ;\end{ar... | (4,-11) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,502 |
$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$ | 2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\ci... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,503 |
3. From the 6 face diagonals of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, two tetrahedra $A-B_{1} C D_{1}$ and $A_{1}-B C_{1} D$ can be formed. If the combined volume of the two tetrahedra is 1 (overlapping parts are counted only once), then the volume of the rectangular prism is $\qquad$ . | 3.2.
Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively, then the volume of the rectangular prism $V_{0}=a b c$. The volumes of the two tetrahedra (as shown in Figure 4) are both
$$
\begin{array}{l}
V_{1}=V_{0}-V_{A_{1}-A B D}-V_{A_{1}-B B C_{1}}-V_{A_{1}-D_{1} D C_{1}}-V_{C_{... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,504 |
4. There are 7 athletes who respectively won first, second, and third prizes in a competition. It is known that the number of people who won the first prize is no less than 1, the number of people who won the second prize is no less than 2, and the number of people who won the third prize is no less than 3. Then the pr... | 4. $\frac{6}{13}$.
In the order of first, second, and third prizes, there are three scenarios for the number of winners: $(1,2,4),(1,3,3),(2,2,3)$.
When $(1,2,4)$, the number of ways to award prizes is
$\mathrm{C}_{7}^{1} \mathrm{C}_{6}^{2} \mathrm{C}_{4}^{4}=7 \times \frac{6 \times 5}{2} \times 1=105$ (ways);
When $(1... | \frac{6}{13} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,505 |
5. Arrange all positive divisors of 8128 in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$, then $\sum_{k=1}^{n} k a_{k}=$ $\qquad$ . | 5.211335 .
Given $8128=2^{6}\left(2^{7}-1\right)$, we have
$$
\begin{array}{l}
a_{n}=\left\{\begin{array}{ll}
2^{n-1}, & 1 \leqslant n \leqslant 7 ; \\
2^{n-8}\left(2^{7}-1\right), & 8 \leqslant n \leqslant 14 .
\end{array}\right. \\
\text { Then } \sum_{k=1}^{n} k a_{k}=\sum_{k=1}^{14} k \cdot 2^{k-1}-\sum_{k=1}^{7}(... | 211335 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,506 |
6. Each face of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is covered with two congruent isosceles right triangles that share a common hypotenuse, one of which is red and the other is yellow. If the sum of the red interior angles at each vertex of the cube is equal to the sum of the yellow interior angles, and it is kn... | 6. $\triangle A D C 、 \triangle A A_{1} D_{1} 、 \triangle C_{1} C B 、 \triangle C_{1} D_{1} D 、$ $\triangle C_{1} B_{1} A_{1}$ (the answer is not unique).
As shown in Figure 6, take two opposite vertices $A$ and $C_{1}$ and draw 3 face diagonals from each: $A C, A B_{1}, A D_{1}, C_{1} B, C_{1} D, C_{1} A_{1}$. Then c... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,507 |
Three. (20 points) In the parallelepiped $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, the base $A B C D$ is a rhombus (but not a square). Please provide a sufficient condition for $A C_{1} \perp$ plane $A_{1} B D$, and prove it.
---
Given the parallelepiped $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, where the base $A B C D$ is a... | Three, as shown in Figure 7, let \( AB \)
\[
= a, AD = b, AA_1 =
\]
\(\boldsymbol{c}\), then \( AC_1 = a + b + c \).
To make \( AC_1 \perp \) plane \( A_1BD \),
the necessary and sufficient condition is that \( AC_1 \) is perpendicular to two intersecting lines in the plane, denoted as
\[
\begin{array}{l}
AC_1 \perp BD... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,508 |
Four. (20 points) Given that $a, b, c$ are distinct complex numbers, satisfying $abc \neq 0, \frac{a+b}{a-b}=\frac{b+c}{b-c}=\frac{c+a}{c-a}$. Prove:
$$
\left|a^{2007}\right|=\left|b^{2007}\right|=\left|c^{2007}\right| \text {. }
$$ | Given the conditions, the complex numbers $a, b, c$ are pairwise distinct and none of them are zero. Using the properties of ratios, we have
$$
\frac{a}{b}=\frac{b}{c}=\frac{c}{a} .
$$
Let $k=\frac{a}{b}=\frac{b}{c}=\frac{c}{a} \neq 1$, then
$$
c=a k, b=a k^{2}, a=a k^{3} \text {. }
$$
Since $a \neq 0$, it follows th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,509 |
Five. (20 points) Through a point $P$ on the line $l: 5 x-7 y-70=0$, draw tangents $P M$ and $P N$ to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, with points of tangency $M$ and $N$, respectively, and connect $M N$.
(1) When point $P$ moves on the line $l$, prove: the line $M N$ passes through a fixed point $Q$;
... | Let $P\left(x_{0}, y_{0}\right) 、 M\left(x_{1}, y_{1}\right) 、 N\left(x_{2}, y_{2}\right)$. Then the tangent line equations of the ellipse passing through points $M$ and $N$ are respectively
$$
\frac{x_{1} x}{25}+\frac{y_{1} y}{9}=1, \frac{x_{2} x}{25}+\frac{y_{2} y}{9}=1 \text {. }
$$
Since both tangent lines pass th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,510 |
(50 points) In an equilateral $\triangle ABC$, take any point $P$ such that $\angle APB > 150^{\circ}$. Connect $PA$, $PB$, and $PC$, and let $PA = a$, $PB = b$, $PC = c$.
(1) Prove that a triangle can be formed with sides $a$, $b$, and $c$, and that this triangle, denoted as $\triangle A_{1} B_{1} C_{1}$, is an obtuse... | (1) As shown in Figure 8, rotate $\triangle P A B$ clockwise around point $B$ by $60^{\circ}$ to get $\triangle P_{1} C B \cong \triangle P A B$.
Connect $P P_{1}$, then $P_{1} C = a$, and $\triangle B P_{1} P$ is an equilateral triangle, so
$$
P_{1} P = B P_{1} = P B = b.
$$
Thus, a triangle $\triangle P_{1} P C$ can... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,511 |
2. Let $x, y$ be real numbers. If for any real numbers $\alpha, \beta$ satisfying $\cos \alpha - \cos \beta \neq 0$, we have
$$
\frac{\sin \left(\alpha+\frac{\pi}{6}\right)+\sin \left(\beta-\frac{\pi}{6}\right)}{\cos \alpha-\cos \beta}=x \cot \frac{\alpha-\beta}{2}+y
$$
then $(x, y)$ | 2. $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
Given $\cos \alpha - \cos \beta \neq 0$, we know $\sin \frac{\alpha - \beta}{2} \neq 0$.
Therefore, all involved expressions are meaningful.
For the given equation, take $\alpha = \frac{\pi}{2}, \beta = \frac{\pi}{6}$, we get $-1 = \sqrt{3} x + y$; take $\alpha = -\f... | \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,512 |
(50 points) Given
$$
x=\left(\frac{n}{n+1}\right)^{n}, y=\left(\frac{n}{n+1}\right)^{n+1}(n=1,2, \cdots) \text {. }
$$
(1) Prove: $x^{x}=y^{y}$;
(2) Prove: If the point $P\left(\left(\frac{n}{n+1}\right)^{n},\left(\frac{n}{n+1}\right)^{n+1}\right)$ lies on the graph of the exponential function $y=a^{x}(a>0, a \neq 1)$,... | (1) From $x^{x}=\left[\left(\frac{n}{n+1}\right)^{n}\right]^{\left(\frac{n}{n+1}\right)^{n}}$
$$
\begin{array}{l}
=\left(\frac{n}{n+1}\right)^{n \cdot \frac{n^{n}}{(n+1)^{n}}}=\left(\frac{n}{n+1}\right)^{\frac{n^{n+1}}{(n+1)^{n}}}, \\
y^{y}=\left[\left(\frac{n}{n+1}\right)^{n+1}\right]^{\left(\frac{n}{n+1}\right)^{n+1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,513 |
Three. (50 points) All non-negative integers $0,1,2, \cdots$, arranged in their natural order, form a decimal number
$$
p=0.12345678910111213141516171819 \cdots \text {. }
$$
Is $p$ an irrational number? Prove your conclusion. | Three, $p$ is an irrational number.
First, from the fact that there are infinitely many natural numbers, we know that $p$ is an infinite decimal, and it only needs to be proven that it is not a repeating decimal (using proof by contradiction).
If not, let $p$ be a rational number, i.e., $p$ is an infinite repeating de... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,514 |
1. Given $\sin 18^{\circ}=\frac{-1+\sqrt{5}}{4}$. Then
$$
\begin{array}{l}
a=\sin 2007^{\circ}, b=\cos 2007^{\circ}, \\
c=-1+\tan 2007^{\circ}, \\
d=\frac{\sin 2007^{\circ}+\cos 2007^{\circ}}{\sqrt{2}}
\end{array}
$$
the size relationship is ( ).
(A) $b<a<d<c$
(B) $b<a<c<d$
(C) $c<b<a<d$
(D) $d<b<c<a$ | -1.D.
Since $2007^{\circ}=5 \times 360^{\circ}+180^{\circ}+27^{\circ}$, we have:
$$
\begin{array}{l}
a=\sin 2000^{\circ}=-\sin 27^{\circ}, b=\cos 2007^{\circ}=-\cos 27^{\circ}, \\
c=-1+\tan 2007^{\circ}=-1+\tan 27^{\circ}, \\
d=\frac{\sin 2007^{\circ}+\cos 2007^{\circ}}{\sqrt{2}}=-\frac{\sin 27^{\circ}+\cos 27^{\circ}}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,515 |
2. If $x, y \in \mathbf{R}$, and $3 x^{2}+2 y^{2}=6 x$, then the range of the bivariate function $f(x, y)=2 x^{2}+3 y^{2}-4 x-6 y$ is ( ).
(A) $\left[\frac{5}{2}-3 \sqrt{6}, \frac{5}{2}+3 \sqrt{6}\right]$
(B) $[5,10]$
(C) $[2 \sqrt{6}-1,2 \sqrt{6}+1]$
(D) $[7-2 \sqrt{6}, 7+2 \sqrt{6}]$ | 2.A.
The equation is obtained as $(x-1)^{2}+\left(\frac{y}{\sqrt{\frac{3}{2}}}\right)^{2}=1$.
Let $x-1=\cos \theta, \frac{y}{\sqrt{\frac{3}{2}}}=\sin \theta(-\pi<\theta \leqslant \pi)$, then $x=1+\cos \theta, y=\sqrt{\frac{3}{2}} \sin \theta$.
Thus, $f(x, y)=\frac{5}{2} \sin \theta\left(\sin \theta-\frac{6}{5} \sqrt{6... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,516 |
3. Let $z$ be a complex number, satisfying $\left|z-\frac{1}{z}\right|=1$. Then the sum of the maximum and minimum values of $|z|$ is ( ).
(A) $-2+2 \sqrt{5}$
(B) $\frac{3+\sqrt{5}}{2}$
(C) $\sqrt{5}$
(D) A different value from the above three options. | 3.C.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, then
$$
\begin{aligned}
1 & =\left|z-\frac{1}{z}\right|=\left|x+y \mathrm{i}-\frac{x-y \mathrm{i}}{x^{2}+y^{2}}\right| \\
& =\sqrt{\left(x-\frac{x}{x^{2}+y^{2}}\right)^{2}+\left(y+\frac{y}{x^{2}+y^{2}}\right)^{2}} \\
& =\frac{\sqrt{x^{2}\left(x^{2}+y^{2}-1\right)^{2}+y... | \sqrt{5} | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,517 |
4. Among all the triangles formed by any three of the 8 vertices of a parallelepiped, the maximum number of acute triangles is ( ).
(A) 28
(B) 32
(C) 34
(D) 36 | 4.B.
Let the parallelepiped be $A B C D-A_{1} B_{1} C_{1} D_{1}$. The number of triangles formed by any three vertices is $\mathrm{C}_{8}^{3}=56$.
In quadrilateral $ABCD$, $\angle B A D$ and $\angle A B C$ are supplementary, so one of them must be a right angle or an obtuse angle. Therefore, among the four triangles ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,518 |
5. As shown in Figure 1, in the equilateral $\triangle ABC$, $O$ is its center. A line through $O$ intersects sides $AB$ and $AC$ at points $D$ and $E$, respectively. Let $BC=a$, $AE=x$. If $AD=\frac{ax}{\sqrt{3}(a-x)}$, then $\frac{DO}{OE}$ equals ( ).
(A) $\sqrt{3}$
(B) $\frac{1+\sqrt{3}}{2}$
(C) $\sqrt{3}-1$
(D) $2 ... | 5.B.
As shown in Figure 5, connect $B O$ and extend it to intersect $A E$ at point $H$, then $A H=\frac{a}{2}, H E=x-\frac{a}{2}$.
Connect $A O$, then $A O$ bisects $\angle D A E$. By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{D O}{O E}=\frac{A D}{A E} \\
=\frac{a}{\sqrt{3}(a-x)} .
\end{array}
$$
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,519 |
6. Among the 8 vertices, 6 centers of the faces (bottom faces), and the center of the body, a total of 15 points, if a plane formed by any 3 different and non-collinear points is perpendicular to a line formed by another 2 different points, then these 5 points are called an "orthogonal 5-point group". Therefore, the to... | 6.D.
As shown in Figure 6, let $O$ be the center of the body, and $O_{i} (i = 1,2, \cdots, 6)$ be the centers of the respective faces.
Classify and discuss according to the plane formed by the 3 points (referred to as the "3-point plane").
Observation shows that there are only 4 possible 3-point planes: side faces (... | 1824 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,520 |
1. In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $M$ is the midpoint of edge $B B_{1}$, point $P$ is on edge $C D$, and $D P$ $=m(0<m<1)$, let the distance from $M$ to line $A P$ be $d$. Then $d=$ $\qquad$ . | ニ、1. $\sqrt{\frac{1}{4}+\frac{1}{m^{2}+1}}$.
Draw $B Q \perp A P$, intersecting $A P$ at point $Q$.
By the theorem of three perpendiculars, we know $M Q \perp A P$.
Clearly, $A P=\sqrt{m^{2}+1}$.
From $A P \cdot B Q=2 S_{\triangle A B P}=1$, we get $B Q=\frac{1}{\sqrt{m^{2}+1}}$.
Thus, $d=M Q=\sqrt{B M^{2}+B Q^{2}}=\sq... | \sqrt{\frac{1}{4}+\frac{1}{m^{2}+1}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,521 |
2. If the real number $a$ ensures that for every real number $z$, the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x+a y=2 z, \\
x y=2 z^{2}+3 z+1
\end{array}\right.
$$
always has real solutions, then the range of values for $a$ is . $\qquad$ | 2. $-4 \leqslant a0$, and the discriminant of equation (2)
$$
\Delta_{2}=9 a^{2}+4 a(1-2 a)=a(a+4) \leqslant 0,
$$
i.e., $-4 \leqslant a \leqslant 0$.
Also, $a \neq 0$, so the range of values for $a$ is $-4 \leqslant a<0$. | -4 \leqslant a < 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,522 |
3. Among all the eight-digit numbers formed by the digits $1,2, \cdots, 8$ without repetition, the number of those divisible by 11 is. $\qquad$ | 3.4608 .
Since there are 4 odd numbers in $1,2, \cdots, 8$, the algebraic sum after arbitrarily adding “+” and “-” signs is always even. Because the difference between the sum of the largest four numbers and the sum of the smallest four numbers in $1,2, \cdots, 8$ is no more than 16, for each eight-digit number that m... | 4608 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,523 |
3. Find the value: $\cos 10^{\circ} \cdot \sin 20^{\circ} \cdot \sin 40^{\circ}=$ | $\begin{array}{l}\text { 3. } \frac{\sqrt{3}}{8} . \\ \cos 10^{\circ} \cdot \sin 20^{\circ} \cdot \sin 40^{\circ} \\ =\cos 10^{\circ} \times \frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \\ =\frac{1}{2} \cos 10^{\circ} \cdot \cos 20^{\circ}-\frac{1}{4} \cos 10^{\circ} \\ =\frac{1}{4}\left(\cos 30^{\circ}+\cos... | \frac{\sqrt{3}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,524 |
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