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Example 6 Let $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \\
\geqslant \frac{3}{1+a b c} .
\end{array}
$$
(2000, National Training Team Practice Problem) | Explanation: When $a b x=1$, the fractional substitution can transform the inequality into a homogeneous one. For example, the 2000 IMO problem:
Given $a, b, c > 0, a b c = 1$. Prove:
$$
\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1.
$$
Perform the fractional substitu... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,119 |
Conclusion 1 Draw the perpendicular from vertex $A$ of $\triangle A B C$ to $B C$, with the foot of the perpendicular being $D$. Take any point $H$ on $A D 1$. The line $B H$ intersects $A C$ at $E$, and $C H$ intersects $A B$ at $F$. Then $A D$ bisects the angle formed by $E D$ and $D F$. | Proof: As shown in Figure 1, first establish a Cartesian coordinate system. Let \( A(0, a) \), \( B(b, 0) \), \( C(c, 0) \), and \( H(0, h) \). Then the equations of \( AC \) and \( BH \) are
\[
\frac{x}{c} + \frac{y}{a} = 1, \quad \frac{x}{b} + \frac{y}{h} = 1.
\]
Therefore, the equation of the line system passing th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,123 |
Example 1 On one side of line $l$, draw a semicircle $\Gamma$. $C$ and $D$ are points on $\Gamma$. The tangents to $\Gamma$ at $C$ and $D$ intersect $l$ at $B$ and $A$, respectively. The center of the semicircle lies on segment $BA$. $E$ is the intersection of segments $AC$ and $BD$, and $F$ is a point on $l$ such that... | Proof: As shown in the figure]
2. Let the intersection of $AD$ and $BC$ be $P$, and draw $PF' \perp AB$ at $F'$. From the fact that $P$, $D$, $O$, and $F'$ are concyclic, we have
$$
\begin{array}{l}
AD \cdot AP \\
=AO \cdot AF' .
\end{array}
$$
Similarly, $BC \cdot BP = BF' \cdot BO$.
Dividing the two equations and us... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,126 |
Example 3: Five teachers, A, B, C, D, and E, predict the rankings of five students, A, B, C, D, and E, in a competition:
Teacher A predicts: B 3rd, C 5th;
Teacher B predicts: E 4th, D 5th;
Teacher C predicts: A 1st, E 4th;
Teacher D predicts: C 1st, B 2nd;
Teacher E predicts: A 3rd, D 4th.
The competition results show ... | Explanation: Fill in the guesses of the 5 teachers in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline Rank & 1 & 2 & 3 & 4 & 5 \\
\hline Jia & & & $B$ & & $C$ \\
\hline Yi & & & & $E$ & $D$ \\
\hline Bing & $A$ & & & $E$ & \\
\hline Ding & $C$ & $B$ & & & \\
\hline Wu & & & $A$ & $D$ & \\
\hline
\end{tabular}
T... | C 1st, B 2nd, A 3rd, E 4th, D 5th | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,127 |
Example 3 Let $BC$ be the diameter of $\odot O$. Take two points $D$ (between $B$ and $C$) and $P$ on the line $BC$, satisfying $\frac{BD}{CD} = \frac{BP}{CP}$. Draw a secant line $PEF$ through point $P$ to $\odot O$. Prove:
(1) $\angle BDF = \angle CDE$;
(2) $\angle OFD = \angle OED = \angle FPO$.
---
Here is the tr... | Proof: (1) As shown in Figure 4, extend $BF$ and $CE$ to intersect at point $A$. By Conclusion 4 and the fact that $BE$ and $CF$ are altitudes, we have $AD \perp BC$, thus,
$$
\angle ADF = \angle ADE.
$$
Therefore, $\angle BDF = \angle CDE$.
(2) From the fact that $B, F, H, D$ and $C, E, H, D$ are concyclic, we get $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,128 |
Example 4 Given that $A$ is an intersection point of two circles $\odot O_{1}$ and $\odot O_{2}$ with unequal radii, and the two external common tangents $P_{1} P_{2}$ and $Q_{1} Q_{2}$ touch the two circles at $P_{1}$, $P_{2}$, $Q_{1}$, and $Q_{2}$ respectively. $M_{1}$ and $M_{2}$ are the midpoints of $P_{1} Q_{1}$ a... | Prove: As shown in Figure 5, let $P_{1} P_{2}$ and $Q_{1} Q_{2}$ intersect $O_{1} O_{2}$ at $H$, and $O_{1} O_{2}$ intersects the larger circle at $D$ and $E$. Then
$$
H P_{1}^{2}=H E \cdot H D=H M_{1} \cdot H O_{1} .
$$
Further, we can get $\frac{D M_{1}}{M_{1} E}=\frac{D H}{H E}$.
Therefore, $D, M_{1}, E, H$ form a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,129 |
Example 5 Given that $\odot O_{1}$ and $\odot O_{2}$ are externally separated, the two common tangents touch $\odot O_{1}$ at $A_{1}, B_{1}$, and touch $\odot O_{2}$ at $A_{2}, B_{2}$. Chords $A_{1} B_{1}$ and $A_{2} B_{2}$ intersect line $O_{1} O_{2}$ at $M_{1}$ and $M_{2}$, respectively. $\odot O$ passes through $A_{... | Proof: When the radii of the two circles are equal, the proposition is obviously true.
As shown in Figure 6, when the radii of the two circles are unequal, let $A_{1} A_{2}$ and $B_{1} B_{2}$ intersect $O_{1} O_{2}$ at $S$.
Similarly to Example 4, we can obtain
$$
\angle O_{1} P_{1} M_{1}=\angle O_{1} S P_{1}.
$$
Conn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,130 |
Example 6 As shown in Figure 7, in quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$. Take a point $E$ on $CD$, and let $BE$ intersect $AC$ at $F$. Extend $DF$ to intersect $BC$ at $G$. Prove:
$$
\angle GAC = \angle EAC \text{.}
$$
(1999, National High School Mathematics Competition) | Prove: When $A B=A D$, quadrilateral $A B C D$ is a kite, the conclusion holds.
When $A B \neq A D$, draw a perpendicular from $A$ to $A C$ intersecting the extensions of $C B$ and $C D$ at points $M$ and $N$, respectively.
By Conclusion 2, when $\angle B A C=\angle D A C$, the intersection point $H$ of $B N$ and $D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,131 |
Example 7 Let $H$ be any point on the altitude $CP$ of acute $\triangle ABC$, and let the lines $AH$ and $BH$ intersect $BC$ and $AC$ at points $M$ and $N$ respectively.
(1) Prove that $\angle NPC = \angle MPC$;
(2) Let $O$ be the intersection of $MN$ and $CP$. An arbitrary line through $O$ intersects the sides of quad... | Proof:
(1) This follows directly from Conclusion 1.
(2) As shown in Figure 8,
Extend \( PE \) to intersect \( AC \) at \( G \), and connect \( PD \) to intersect \( AM \) at \( K \).
In \( \triangle G E N \) and \( \triangle K D M \), the intersection point of \( GE \) and \( KD \) is \( P \), the intersection point o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,132 |
Example 8 Let the extensions of the two pairs of opposite sides of a convex quadrilateral $ABCD$ meet at points $E$ and $F$, and let the diagonals intersect at point $P$. Draw $PO \perp EF$ at $O$. Prove that:
$$
\angle BOC = \angle AOD.
$$
(2002, IMO China National Training Team Selection Exam) | First, prove a lemma.
Lemma: In quadrilateral $A E O D$, the extensions of two pairs of opposite sides intersect at points $M$ and $F$, respectively. $B$ is any point on $A E$, and $B F$ intersects $D E$ at point $C$. $A C$ intersects $B D$ at point $P$. Connecting $O P$ intersects $A B$ and $B F$ at points $Q$ and $T$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,133 |
Question 1
(1) In a $7 \times 7$ square grid, $k$ cells' centers are colored red, such that no 4 red points form the 4 vertices of a rectangle with sides parallel to the grid lines. Find the maximum value of $k$;
(2) For a $13 \times 13$ square grid, solve the same problem;
(3) For a $21 \times 21$ square grid, solve t... | Solution: (1) Let the $i$-th row of the grid table be marked with $x_{i}$ red dots, then we have $x_{1}+x_{2}+\cdots+x_{7}=k$.
According to the given condition, if there are 2 red dots in a certain row, then in the remaining rows, there cannot be 2 red dots in the same columns as the first 2 red dots. This means that ... | 21, 52, 105 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,134 |
Question 4: A spatial figure is formed by $n$ points and $l$ lines connecting these points, where $n=q^{2}+q+1, l \geqslant \frac{1}{2} q(q+1)^{2}+1, q \geqslant 2, q \in \mathbf{N}$. It is known that no four points in this figure are coplanar, each point has at least one line, and there exists at least one point with ... | Prove: Draw an $n \times n$ grid and label the given $n$ vertices as $A_{1}, A_{2}, \cdots, A_{n}$. If there is a line connecting $A_{i}$ and $A_{j}$, then color the centers of the two cells in the $i$-th row and $j$-th column and the $j$-th row and $i$-th column of the grid red. In this way, there are $2l$ red dots in... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,136 |
Question 5 In a $12 \times 9$ rectangular grid, $k$ cells' centers are colored red, such that no 4 red points form the 4 vertices of a rectangle with sides parallel to the grid lines. Find the maximum value of $k$.
In a $12 \times 9$ rectangular grid, $k$ cells' centers are colored red, such that no 4 red points form ... | Solution: Let the $i$-th row of the grid be marked with $x_{i}$ red dots, then we have $x_{1}+x_{2}+\cdots+x_{12}=k$.
Similarly to the first problem, we have
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{12}}^{2} \leqslant \mathrm{C}_{9}^{2}=36 \text {. }
$$
Simplifying and using the Cauchy-Sc... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,137 |
Example $4 A, B, C, D, E, F$ six players are participating in a round-robin table tennis singles tournament (each player plays one match against every other player). Every day, three matches are held simultaneously on three tables. It is known that on the first day, $B$ played against $D$; on the second day, $C$ played... | Explanation: In fact, this problem requires listing the entire match schedule for this competition (Table 2).
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline & Day 1 & Day 2 & Day 3 & Day 4 & Day 5 \\
\hline Court 1 & $B-D$ & $C-E$ & $D-F$ & $B-C$ & $1-?$ \\
\hline Court 2 & (1) & (2) & (3) & (1) & (1) \\
\hline Court 3 ... | A-B, E-F, C-D | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,138 |
Question 7 A certain unit has a total of 15 guards, who are divided into 5 groups of 3 people each day, and the 5 groups are assigned to 5 different locations for duty for one day. On the second day, the groups are reorganized for duty. Can the groups be appropriately divided so that within 7 consecutive days, every tw... | Solution: Figure 11 provides an example of a maximum of 105 red dots in a $35 \times 15$ grid, where no four red dots form the four vertices of a rectangle with sides parallel to the grid lines. Each row contains 3 red dots, forming exactly 3 "red dot pairs" in the same row, totaling $35 \times 3 = 105 = C_{15}^{2}$. T... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,139 |
1. Prove: There do not exist positive integers $a, b$ that satisfy the equation $4 a(a+1)=b(b+3)$. | 1. The given equation is equivalent to
$$
4 a^{2}+4 a-\left(b^{2}+3 b\right)=0 \text {. }
$$
For a given integer $b$, the above equation is a quadratic equation in $a$, with the discriminant
$$
\Delta=4^{2}+4^{2}\left(b^{2}+3 b\right)=4^{2}\left(b^{2}+3 b+1\right) \text {. }
$$
If for some integer $b$, there exists a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,141 |
2. Find the ordered integer pairs $(x, y)$ that satisfy $(|x|-2)^{2}+(|y|-2)^{2}<5$.
untranslated text remains unchanged. | 2. Clearly, $(|x|-2)^{2}$ and $(|y|-2)^{2}$ are both perfect squares.
Since the sum of these two perfect square terms is less than 5, each term must be no greater than $2^{2}=4$.
If $(|x|-2)^{2}=0$, then $|x|=2$. Thus, $x= \pm 2$. Therefore, $x$ has two possible values.
If $(|x|-2)^{2}=1$, then $|x|=1$ or $|x|=3$. T... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,142 |
3. Find all real number triples $(x, y, z)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
{[x]+\{y\}=z,} \\
{[y]+\{z\}=x,} \\
{[z]+\{x\}=y}
\end{array}\right.
$$
(Note: $x=[x]+\{x\},[x]$ is an integer, $0 \leqslant\{x\}<1$ ). | 3. Rewrite the original system of equations as
$$
\left\{\begin{array}{l}
{[x]+\{y\}=[z]+\{z \mid .} \\
{[y]+|z|=[x]+\mid x\} . \\
{[z]+\mid x\}=[y]+|y| .
\end{array}\right.
$$
Since for any real number $x, x=[x]+\{x\}$, from equation (1) we get $[x]=[z]$, from equation (2) we get $[y]=[x]$, and from equation (3) we g... | (t, t, t), t \in \mathbf{R} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,143 |
4. Given that the area of $\triangle A B C$ is 2000, points $P$, $Q$, and $R$ are the midpoints of $B C$, $C A$, and $A B$ respectively, points $U$, $V$, and $W$ are the midpoints of segments $Q R$, $R P$, and $P Q$ respectively, and the lengths of segments $A U$, $B V$, and $C W$ are $x$, $y$, and $z$ respectively. Pr... | 4. As shown in Figure 1, since $Q$ and $R$ are the midpoints of $CA$ and $AB$ respectively, $\triangle AQR \sim \triangle ACB$, with point $A$ as the similarity center and a similarity ratio of $\frac{1}{2}$.
Since $P$ is the midpoint of $BC$, and $U$ is the midpoint of $RQ$, it is easy to see that $U$ is also the mid... | 375 | Geometry | proof | Yes | Yes | cn_contest | false | 717,144 |
1. On a highway, starting from $3 \mathrm{~km}$, a speed limit sign is encountered every $4 \mathrm{~km}$, and starting from $10 \mathrm{~km}$, a speed monitoring device is encountered every $9 \mathrm{~km}$. If a car first encounters both facilities simultaneously at $19 \mathrm{~km}$, then the kilometer number where ... | -.1.C.
Since the least common multiple of 4 and 9 is $36, 19+36=55$, therefore, the second time they pass both facilities simultaneously is at $55 \mathrm{~km}$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,145 |
2. Given $m=1+\sqrt{2}, n=1-\sqrt{2}$, and
$$
\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8 \text {. }
$$
Then the value of $a$ is ( ).
(A) -5
(B) 5
(C) -9
(D) 9 | 2.C.
From the given, we have $m^{2}-2 m=1, n^{2}-2 n=1$.
Also, $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$, so, $(7+a)(3-7)=8$.
Solving for $a$ gives $a=-9$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,146 |
4. A square piece of paper is cut into two parts along a straight line that does not pass through any vertex; take one of the parts and cut it into two parts along a straight line that does not pass through any vertex; then take one of the resulting three parts and cut it into two parts along a straight line that does ... | 4. B.
According to the problem, when cutting along a line that does not pass through the vertices with the front knife, each cut increases the sum of the interior angles of each part by \(360^{\circ}\). Thus, after \(k\) cuts, \(k+1\) polygons can be obtained, and the sum of the interior angles of these polygons is \(... | 2005 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,148 |
Example 5 In an exam with a full score of 100, $A, B, C, D, E$ five people participated in this exam. $A$ said: “I got 94 points.” $B$ said: “I scored the highest among the five of us.” $C$ said: “My score is the average of $A$ and $D$.” $D$ said: “My score is exactly the average of the five of us.” $E$ said: “I scored... | Explanation: Let $b$, $c$, $d$, $e$ represent the scores of $B$, $C$, $D$, $E$ respectively. We will try to reduce the number of unknowns based on their conversation:
According to $E$'s statement, the scores of $B$, $C$, $D$, $E$ are $b$, $c$, $d$, $c+2$ respectively.
According to $C$'s statement, the scores of $B$, ... | 94, 98, 95, 96, 97 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,149 |
5. As shown in Figure 1, square $ABCD$ is inscribed in $\odot O$, point $P$ is on the arc $\overparen{AB}$, and $DP$ intersects $AC$ at point $Q$. If $QP=QO$, then the value of $\frac{QC}{QA}$ is ( ).
(A) $2 \sqrt{3}-1$
(B) $2 \sqrt{3}$
(C) $\sqrt{3}+\sqrt{2}$
(D) $\sqrt{3}+2$ | 5.D.
As shown in Figure 4, let the radius of $\odot O$ be $r$, $QO = m$, then $QP = m$, $QC = r + m$, $QA = r - m$.
In $\odot O$, according to the intersecting chords theorem, we have
$$
QA \cdot QC = QP \cdot QD,
$$
which is $(r-m)(r+m) = m \cdot QD$.
Therefore, $QD = \frac{r^2 - m^2}{m}$.
Connecting $DO$, by the P... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,150 |
6. Given that $a$, $b$, and $c$ are integers, and $a+b=2006$, $c-a=2005$. If $a<b$, then the maximum value of $a+b+c$ is $\qquad$ | II, 6.5 013.
From $a+b=2006, c-a=2005$, we get $a+b+c=a+4011$.
Since $a+b=2006(a<b, a$ is an integer $)$, therefore, the maximum value of $a$ is 1002.
Thus, the maximum value of $a+b+c$ is 5013. | 5013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,151 |
7. As shown in Figure 2, the area of the square $D E F G$ inscribed in the equilateral triangle $\triangle A B C$ with area 1 is $a \sqrt{b}-c$, where $a, b, c$ are integers, and $b$ cannot be divided by the square of any prime number. Then $\frac{a-c}{b}$ | 7. $-\frac{20}{3}$.
Let the side length of square DEFG be $x$, and the side length of equilateral $\triangle ABC$ be $m$. Then $m^{2}=\frac{4}{\sqrt{3}}$.
From $\triangle ADG \backsim \triangle ABC$, we get
$$
\frac{x}{m}=\frac{\frac{\sqrt{3}}{2} m-x}{\frac{\sqrt{3}}{2} m} .
$$
Solving this, we get $x=(2 \sqrt{3}-3) ... | -\frac{20}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,152 |
8. Given that the perimeter of the regular pentagon square $A B C D E$ is $2000 \mathrm{~m}$, two people, A and B, start from points $A$ and $C$ respectively at the same time, walking around the square in the direction of $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow A \rightarrow \cdots$. Pers... | 8. 104 .
Suppose that when person A has walked $x$ edges, A and B start walking on the same edge for the first time. At this point, A has walked $400 x \mathrm{~m}$, and B has walked $46 \times$ $\frac{400 x}{50}=368 x \mathrm{~m}$. Therefore,
$$
368(x-1)+800-400(x-1)>400 \text {, }
$$
and $(368 x+800)-400 x \leqslan... | 104 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,153 |
9. [ $x$ ] represents the greatest integer not exceeding $x$ (for example, $[3.2]=3$). Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
| 9.334.
Let $\left[\frac{n}{6}\right]=m$. Then $\frac{n}{6}=m+\alpha(0 \leqslant \alpha<1)$. When $0 \leqslant \alpha<\frac{1}{2}$, we have
$$
\frac{n}{3}=2 m+2 \alpha,\left[\frac{n}{3}\right]=2 m \text {. }
$$
According to $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$, we get
$$
2 m+m=3 m+3 \alpha \... | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,154 |
10. Xiaoming's home phone number was originally a six-digit number. The first upgrade was to add the digit 8 between the first and second digits, making it a seven-digit phone number; the second upgrade was to add the digit 2 before the first digit, making it an eight-digit phone number. Xiaoming found that the eight-d... | 10.282500.
Let the original six-digit telephone number be $\overline{a b c d e f}$. Then, after two upgrades, the eight-digit telephone number is $\overline{2 a 8 b c c l e f}$.
According to the problem, we have $81 \times \overline{a b c d e f}=\overline{2 a 8 b c d e f}$. Let $x=b \times 10^{1}+c \times 10^{3}+d \ti... | 282500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,155 |
11. Given $x=\frac{b}{a}, a, b$ are coprime positive integers, and $a \leqslant 8, \sqrt{2}-1<x<\sqrt{3}-1$.
(1) Write down one value of $x$ that satisfies the condition;
(2) Find all values of $x$ that satisfy the condition. | 3.11. (1) $x=\frac{1}{2}$ satisfies the condition.
(2) Since $x=\frac{b}{a}(a, b$ are coprime positive integers, and $a \leqslant$ $8)$, therefore,
$$
(\sqrt{2}-1) a<b<(\sqrt{3}-1) a .
$$
When $a=1$, $(\sqrt{2}-1) \times 1<b<(\sqrt{3}-1) \times 1$, there is no such positive integer $b$;
When $a=2$, $(\sqrt{2}-1) \tim... | \frac{1}{2}, \frac{2}{3}, \frac{3}{5}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{5}{8} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,156 |
12. Let $a$, $b$, and $c$ be distinct real numbers, and satisfy the following equations:
$$
\begin{array}{l}
b^{2}+c^{2}=2 a^{2}+16 a+14, \\
b c=a^{2}-4 a-5 .
\end{array}
$$
Find the range of values for $a$. | 12. From (1) $-2 \times$ (2), we get
$$
(b-c)^{2}=24(a+1)>0 \text {, }
$$
Therefore, $a>-1$.
When $a>-1$, we have
$$
b^{2}+c^{2}=2 a^{2}+16 a+14=2(a+1)(a+7)>0 \text {. }
$$
Furthermore, when $a=b$, from equations (1) and (2) we get
$$
\begin{array}{l}
c^{2}=a^{2}+16 a+14, \\
a c=a^{2}-4 a-5 .
\end{array}
$$
Squaring... | a>-1 \text{ and } a \neq -\frac{5}{6}, a \neq \frac{1 \pm \sqrt{21}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,157 |
13. As shown in Figure $3, P$ is a point outside $\odot O$. Two tangents are drawn from point $P$ to $\odot O$, touching the circle at points $A$ and $B$. A line parallel to $PB$ is drawn through point $A$, intersecting $\odot O$ at point $C$. Line $PC$ intersects $\odot O$ at point $E$, and line $AE$ is extended to in... | 13. Since $A C / / P B$, therefore, $\angle K P E=\angle A C E$.
Since $P A$ is a tangent to $\odot O$, thus,
$\angle K A P=\angle A C E$.
Hence $\angle K P E=\angle K A P$.
f: Therefore, $\triangle K P E \backsim \triangle K A P$, so $\frac{K P}{K A}=\frac{K E}{K P}$.
Thus, $\boldsymbol{K} P^{2}=K E \cdot K A$.
By th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,158 |
14.2006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find the minimum value of $a_{1}+a_{2}+\cdots+$ $a_{2006}$. | 14. First prove the proposition:
For any 119 positive integers $b_{1}, b_{2}, \cdots, b_{119}$, there must exist some (at least one, or possibly all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
$$
b_{1}, b_{1}+b_{2}, \cdots, b_{1}+b_{2}+\cdots+b_{119} \text {. }
$$
If one of... | 3910 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,159 |
1. $a, b$ are real numbers, the set $M=\left\{\frac{b}{a}, 1\right\}, P=$ $\{a, 0\}, f: x \rightarrow x$ represents mapping the element $x$ from set $M$ to set $P$ as $x$. Then the value of $a+b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) $\pm 1$ | -1.C.
From the given, we have $M=P$, thus $\frac{b}{a}=0, a=1$, which means $a=1$, $b=0$. Therefore, $a+b=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,161 |
2. Given the function $f(x)$ satisfies $f\left(\frac{2}{x+|x|}\right)=$ $\log _{2} \sqrt{x|x|}$, then the analytical expression of $f(x)$ is ( ).
(A) $\log _{2} x(\mathrm{~B})-\log _{2} x(\mathrm{C}) 2^{-x}(\mathrm{D}) x^{-2}$ | 2. B.
From $x|x|>0$, we get $x>0$. Therefore, $f\left(\frac{1}{x}\right)=\log _{2} x$. Thus, $f(x)=\log _{2} \frac{1}{x}=-\log _{2} x$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,162 |
3. If the equation $\left(\frac{3}{2}\right)^{x}=\frac{2+3 a}{5-a}$ has a negative root for $x$, then the range of real number $a$ is ( ).
(A) $\left(-\infty,-\frac{2}{3}\right) \cup(5,+\infty)$
(B) $\left(-\infty,-\frac{3}{4}\right) \cup(5,+\infty)$
(C) $\left(-\frac{2}{3}, 5\right)$
(D) $\left(-\frac{2}{3}, \frac{3}{... | 3. D.
Since $x<0$, we have $0<\left(\frac{3}{2}\right)^{x}<1$. Therefore, $0<\frac{2+3a}{5-a}<1$. Solving this, we get $-\frac{2}{3}<a<\frac{3}{4}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,163 |
4. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, their partial sums up to the $n$-th term are $A_{n}$ and $B_{n}$ respectively. Let $c_{n}=a_{n} B_{n}+b_{n} A_{n}-a_{n} b_{n}(n \geqslant 1)$. Then the sum of the first 10 terms of the sequence $\left\{c_{n}\right\}$ is ( ).
(A) $A_{10}+B_{10}$
(... | 4.C.
When $n \geqslant 2$,
$$
\begin{array}{l}
c_{n}=\left(A_{n}-A_{n-1}\right) B_{n}+\left(B_{n}-B_{n-1}\right) A_{n}- \\
\quad\left(A_{n}-A_{n-1}\right)\left(B_{n}-B_{n-1}\right) \\
=A_{n} B_{n}-A_{n-1} B_{n-1} . \\
\text { Therefore, } c_{1}+c_{2}+\cdots+c_{10} \\
=A_{1} B_{1}+\left(A_{2} B_{2}-A_{1} B_{1}\right)+\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,164 |
5. As shown in Figure 1, let $P$ be a point inside $\triangle A B C$, and $A P$ $=\frac{2}{5} A B+\frac{1}{5} A C$. Then $\frac{S_{\triangle B P}}{S_{\triangle A B C}}=(\quad)$.
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{1}{4}$
(D) $\frac{1}{3}$ | 5.A.
As shown in Figure 6, let $A M=\frac{2}{5} A B$,
$$
A N=\frac{1}{5} A C \text {. }
$$
Then $A P=A M+A N$.
By the parallelogram rule, $N P / / A B$, so,
$$
\frac{S_{\text {SUHP }}}{S_{\triangle M K C}}=\frac{|A N|}{|A C|}=\frac{1}{5} .
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,165 |
6. If $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta(0 \leqslant \theta$ $<2 \pi)$, then the range of $\theta$ is ().
(A) $\left[0, \frac{\pi}{4}\right]$
(B) $\left[\frac{\pi}{4}, \pi\right]$
(C) $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$
(D) $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ | 6.C.
Solution 1: From $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta$, we get $(\sin \theta-\cos \theta)\left(2+\frac{1}{2} \sin 2 \theta\right) \geqslant 0$.
Since $2+\frac{1}{2} \sin 2 \theta>0$, we have $\sin \theta-\cos \theta \geqslant 0$.
Solving this, we get $\frac{\pi}{4} \leqslant \thet... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,166 |
7. A bag contains $m$ red balls and $n$ white balls $(m > n \geqslant 4)$. Now, two balls are drawn at random. If the probability of drawing two balls of the same color is equal to the probability of drawing two balls of different colors, then the number of arrays $(m, n)$ that satisfy the relation $m+n \leqslant 40$ i... | 7.A.
Let “drawing two red balls” be event $A$, “drawing two white balls” be event $B$, and “drawing one red and one white ball” be event $C$, then
$$
P(A)=\frac{\mathrm{C}_{m}^{2}}{\mathrm{C}_{m+n}^{2}}, P(B)=\frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{m+n}^{2}}, P(C)=\frac{\mathrm{C}_{m}^{1} \cdot C_{n}^{1}}{\mathrm{C}_{m+... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,167 |
8. Given a quadratic equation with real coefficients
$$
x^{2}+(1+a) x+a+b+1=0
$$
with two real roots $x_{1}, x_{2}$, and $0<x_{1}<1<x_{2}$. Then the range of $\frac{b}{a}$ is $(\quad)$.
(A) $\left(-1,-\frac{1}{2}\right]$
(B) $\left(-1,-\frac{1}{2}\right)$
(C) $\left(-2,-\frac{1}{2}\right]$
(D) $\left(-2,-\frac{1}{2}\r... | 8.D.
Let $f(x)=x^{2}+(1+a) x+a+b+1$. From $01$, we get
$$
\left\{\begin{array}{l}
f(0)=a+b+1>0 . \\
f(1)=2 a+b+3<0 .
\end{array}\right.
$$
In the right-angle coordinate plane $a O b$,
the region represented by the above inequalities
is shown as the shaded part in Figure 7 (excluding the boundaries). The intersection ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,168 |
9. As shown in Figure 2, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $P$ is a point on edge $A B$. A line $l$ is drawn through point $P$ in space such that $l$ makes a $30^{\circ}$ angle with both plane $A B C D$ and plane $A B C_{1} D_{1}$. The number of such lines $l$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 9. B.
Since the dihedral angle $C_{1}-A B-D$ has a plane angle of $45^{\circ}$, within this dihedral angle and its "opposite" dihedral angle, there does not exist a line passing through point $P$ that forms a $30^{\circ}$ angle with both plane $A B C D$ and plane $A B C_{1} D_{1}$. Instead, consider its supplementary ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,169 |
10. As shown in Figure 3, from the left focus $F$ of the hyperbola $\frac{x}{a^{2}}-\frac{y}{b^{2}}=1(a>0, b>0)$, a tangent line is drawn to the circle $x^{2}+y^{2}=a^{2}$. The point of tangency is $T$, and the extension of $F T$ intersects the right branch of the hyperbola at point $P$. If $M$ is the midpoint of line ... | 10.B.
As shown in Figure 8, let the right focus of the hyperbola be $F^{\prime}$, and connect $P F^{\prime}$, $O T$. In the right triangle $\triangle O T F$, given $|F O|=c,|O T|=a$ (where $c$ is the semi-focal distance of the hyperbola), we get $|T F|=b$. Therefore, according to the midline theorem of triangles and t... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,170 |
Example 7: Xiao Li and his brother attended a gathering, along with two other pairs of brothers. After meeting, some people greeted each other with handshakes, but no one shook hands with their own brother, and no one shook hands with the same person twice. At this point, Xiao Li noticed that, apart from himself, every... | Explanation: From the problem, it is easy to see that the maximum number of handshakes for each person is 4, and the minimum is 0. Also, except for Xiao Li, each person has a different number of handshakes. Therefore, the 5 people must have handshakes 0 times, 1 time, 2 times, 3 times, and 4 times respectively.
Using ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,171 |
11. Given that $\theta$ is an acute angle, and $\frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $\frac{\sin 3 \theta}{\sin \theta}=$ $\qquad$ | $=.11 \cdot \frac{7}{3}$.
Solution 1: From the given and the triple angle cosine formula, we get $4 \cos ^{2} \theta-3=\frac{1}{3}$, which means $4 \cos ^{2} \theta=\frac{10}{3}$.
Therefore, $\frac{\sin 3 \theta}{\sin \theta}=3-4 \sin ^{2} \theta=4 \cos ^{2} \theta-1=\frac{7}{3}$.
Solution 2: Let $\frac{\sin 3 \theta}{... | \frac{7}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,172 |
12: Use 6 equal-length thin iron rods to weld into a regular tetrahedral frame, ignoring the thickness of the iron rods and welding errors. Let the radius of the largest sphere that this frame can contain be $R_{1}$, and the radius of the smallest sphere that can enclose this frame be $R_{2}$. Then $\frac{R_{1}}{R_{2}}... | 12. $\frac{\sqrt{3}}{3}$.
According to the problem, $R_{1}$ is the radius of the edge-tangent sphere of this regular tetrahedron frame, and $R_{2}$ is the radius of the circumscribed sphere. Suppose the edge length of the regular tetrahedron is 1, it is easy to know that the diameter of the edge-tangent sphere is the ... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,173 |
13. Let $f(x)$ be an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=3$. If $\sin \alpha=\frac{\sqrt{5}}{5}$, then the value of $f(4 \cos 2 \alpha)$ is | 13. -3.
Since $\sin \alpha=\frac{\sqrt{5}}{5}$, therefore,
$$
\cos 2 \alpha=1-2 \sin ^{2} \alpha=\frac{3}{5} \text{. }
$$
Also, $f(x)$ is an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=$
3, so,
$$
f(4 \cos 2 \alpha)=f\left(\frac{12}{5}\right)=f\left(\frac{2}{5}\right)=-f\left(-\frac{2}{5}\right)=... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,174 |
14. If $a, b, c$ form an arithmetic sequence, then the midpoint of the line segment cut by the line $a x + b y + c = 0$ on the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{8} = 1$ has the trajectory equation $\qquad$. | $14.2\left(x-\frac{1}{2}\right)^{2}+\frac{(y+1)^{2}}{2}=1$.
From $a-2 b+c=0$, we know that the line $a x+b y+c=0$ passes through the fixed point $P(1,-2)$. Since point $P$ lies on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{8}=1$, $P$ is one endpoint of the intercepted segment. Let the other endpoint be $Q\left(x_{1}, y_... | 2\left(x-\frac{1}{2}\right)^{2}+\frac{(y+1)^{2}}{2}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,175 |
15. Let $x>1, y>1, S=\min \left\{\log _{x} 2, \log _{2} y\right.$ , $\left.\log _{y}\left(8 x^{2}\right)\right\}$. Then the maximum value of $S$ is $\qquad$ . | 15.2.
From the problem, we have $\log _{x} 2 \geqslant S, \log _{2} y \geqslant S, \log _{r}\left(8 x^{2}\right) \geqslant S$, then $S \leqslant \log ,\left(8 x^{2}\right)=\frac{3+2 \log _{2} x}{\log _{2} y}=\frac{3+\frac{2}{\log _{x} 2}}{\log _{2} y} \leqslant \frac{3+\frac{2}{S}}{S}$. Therefore, $S^{3}-3 S-2 \leqsla... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,176 |
One. (20 points) Let $P\left(x+a, y_{1}\right)$, $Q\left(x, y_{2}\right)$, and $R\left(2+a, y_{3}\right)$ be three distinct points on the graph of the inverse function of $f(x)=2^{x}+a$, and they satisfy $y_{1}+y_{3}=2 y_{2}$. Find the range of real numbers $a$ for which there is exactly one real number $x$ that satisf... | $-f(x)=2^{x}+a$ has the inverse function $f^{-1}(x)=\log _{2}(x-a)$.
Then $y_{1}=\log _{2} x, y_{2}=\log _{2}(x-a), y_{3}=1$.
From $y_{1}+y_{3}=2 y_{2}$, we get $1+\log _{2} x=2 \log _{2}(x-a)$. This equation is equivalent to $\left\{\begin{array}{l}x>a, \\ 2 x=(x-a)^{2}\end{array}\right.$.
(1) $\Delta=0$, i.e., $a=-\f... | \left\{-\frac{1}{2}\right\} \cup(0,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,177 |
II. (20 points) Given that $x, y, z$ are all positive numbers.
(1) Prove: $\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy} \geqslant \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$;
(2) If $x + y + z \geqslant xyz$, find the minimum value of $u = \frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}$. | (1) Since $x, y, z$ are all positive numbers, we have:
$$
\frac{x}{y z}+\frac{y}{z x}=\frac{1}{z}\left(\frac{x}{y}+\frac{y}{x}\right) \geqslant \frac{2}{z} \text {. }
$$
Similarly, $\frac{y}{z x}+\frac{z}{x y} \geqslant \frac{2}{x}, \frac{z}{x y}+\frac{x}{y z} \geqslant \frac{2}{y}$.
Equality holds in all three inequa... | \sqrt{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,178 |
Three. (20 points) Given $\sin (2 \alpha+\beta)=3 \sin \beta$. Let $\tan \alpha=x, \tan \beta=y$, and denote $y=f(x)$.
(1) Find the expression for $f(x)$;
(2) Define the sequence of positive numbers $\left\{a_{n}\right\}: a_{1}=\frac{1}{2}, a_{n+1}^{2}=$ $2 a_{n} f\left(a_{n}\right)\left(n \in \mathbf{N}_{+}\right)$. F... | (1) From $\sin (2 \alpha+\beta)=3 \sin \beta$, we get $\sin [(\alpha+\beta)+\alpha]=3 \sin [(\alpha+\beta)-\alpha]$,
which means $\sin (\alpha+\beta) \cdot \cos \alpha=2 \cos (\alpha+\beta) \cdot \sin \alpha$.
Therefore, $\tan (\alpha+\beta)=2 \tan \alpha$.
Thus, $\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan ... | a_{n}=\sqrt{\frac{2^{n-2}}{2^{n-1}+1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,179 |
Four. (30 points) As shown in Figure 4, the incenter of $\triangle ABC$ is $I$. Draw a perpendicular line from point $A$ to line $BI$, and let the foot of the perpendicular be $H$. Let $D$ and $E$ be the points where the incircle $\odot I$ touches sides $BC$ and $CA$, respectively. Prove that points $D$, $H$, and $E$ a... | Proof 1: As shown in Figure 9, let the line $B I$ intersect side $C A$ at point $K$, and connect $A I$, $D I$, $E I$, $D H$, and $E H$.
Since $\angle B D I = \angle A H B$ and $\angle I B D = \angle A B H$, we have
$\triangle I B D \sim \triangle A B H$.
Thus, $\frac{B D}{B H} = \frac{I B}{A B}$, which means
$\frac{B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,180 |
Five. (30 points) As shown in Figure 5, given the parabola $C: y^{2} = 4 p x (p > 0)$, $F$ is the focus of $C$, $l$ is the directrix, and the intersection point of $l$ with the $x$-axis is $E$. Draw any line through point $F$ intersecting the parabola $C$ at points $A$ and $B$.
(1) If $\boldsymbol{A F} = \lambda \bolds... | (1) Solution 1: The coordinates of point $F$ are $(p, 0)$. Let the equation of the line passing through point $F$ be $x=m y+p$. Substituting into $y^{2}=4 p x$, we get
$$
y^{2}-4 p m y-4 p^{2}=0 \text {. }
$$
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then $y_{1}$ and $y_{2}$ are the two roots ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,181 |
Example 1 Let the side lengths of $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ be $a, b, c, a_{1}, b_{1}, c_{1}$, and their areas be $S, S_{1}$. Then
$$
\begin{array}{l}
H \equiv \sum a_{1}^{2}\left(b^{2}+c^{2}-a^{2}\right) \\
\geqslant 8\left(\frac{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}{a^{2}+b^{2}+c^{2}} \cdot S^{2}+\... | $$
\begin{array}{l}
16 S^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right) . \\
\text { Let } \lambda=\frac{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}{a^{2}+b^{2}+c^{2}}, \\
D_{1}=\sqrt{\lambda} a^{2}-\sqrt{\lambda^{-1}} a_{1}^{2}, \\
D_{2}=\sqrt{\lambda} b^{2}-\sqrt{\lambda^{-1}} b_{1}^{2}, \\
D_{3}=\sqrt{\la... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,182 |
1. The size relationship of the following three numbers $\frac{3}{2}, \log _{16} 82, \log _{27} 124$ is ( ).
(A) $\frac{3}{2}<\log _{16} 82<\log _{27} 124$
(B) $\frac{3}{2}<\log _{27} 124<\log _{16} 82$
(C) $\log _{27} 124<\frac{3}{2}<\log _{16} 82$
(D) $\log _{27} 124<\log _{16} 82<\frac{3}{2}$ | - 1.C.
Since $\log _{16} 82>\log _{16} 81=\log _{2} 81^{\frac{1}{4}}=\log _{2} 3$ $\log _{7} 124\frac{3}{2}$. Let $y=\log _{3} 5$, then $3^{y}=5$.
And $3^{\frac{3}{2}}=\sqrt{27}>5=3^{y}$, so, $y<\frac{3}{2}$.
In summary, $\log _{27} 124<\frac{3}{2}<\log _{16} 82$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,183 |
2. Given two points $A(1,2)$ and $B(3,1)$, the distances from these points to line $l$ are $\sqrt{2}$ and $\sqrt{5}-\sqrt{2}$, respectively. The number of lines $l$ that satisfy the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2.C.
Given $|A B|=\sqrt{5}$, construct two circles with $A$ and $B$ as centers and $\sqrt{2}$ and $\sqrt{5}-\sqrt{2}$ as radii, respectively. Then the two circles are externally tangent, and there are three common tangents. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,184 |
3. Let $f(n)$ be the sum of the squares of the digits of the positive integer $n$ (in decimal), for example, $f(123)=1^{2}+2^{2}+3^{2}=14$. Let
$$
\begin{array}{l}
f_{1}(n)=f(n), \\
f_{k+1}(n)=f\left(f_{k}(n)\right)(k=1,2, \cdots) .
\end{array}
$$
Then $f_{2006}(2006)=(\quad)$.
(A) 20
(B) 4
(C) 42
(D) 145 | 3.D.
Let $f(2006)=40$ be denoted as $2006 \rightarrow 40$, then we have
$$
\begin{array}{l}
2006 \rightarrow 40 \rightarrow 16 \rightarrow 37 \rightarrow 58 \rightarrow 89 \rightarrow 145 \rightarrow 42 \rightarrow 20 \rightarrow 4 \\
\rightarrow 16 \rightarrow \cdots
\end{array}
$$
Starting from 16, $f_{n}(2006)$ is... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,185 |
4. Let in the $x O y$ plane, the area enclosed by $0<y \leqslant x^{2}, 0 \leqslant x \leqslant 1$ be $\frac{1}{3}$. Then the intersection of the sets
$$
\begin{array}{l}
M=\{(x, y)|| y|-| x \mid \leqslant 1\}, \\
N=\left\{(x, y)|| y \mid \geqslant x^{2}+1\right\}
\end{array}
$$
representing the area of $M \cap N$ is ... | 4.B.
$M \cap N$ is symmetric about both the $x$-axis and the $y$-axis in the $x O y$ plane. Therefore, the area of $M \cap N$ can be calculated by multiplying the area in the first quadrant by 4. For this purpose, we only need to consider the area in the first quadrant. According to the problem, the area of $M \cap N$ ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,186 |
5. In a regular 2006-gon, the number of diagonals that are not parallel to any side is ( ).
(A) 2006
(B) $1003^{2}$
(C) $1003^{2}-1003$
(D) $1003^{2}-1002$ | 5.C.
For a regular $2n$-gon $A_{1} A_{2} \cdots A_{2 n}$, the number of diagonals is
$$
\frac{1}{2} \times 2 n(2 n-3)=n(2 n-3) \text {. }
$$
To calculate the number of diagonals parallel to side $A_{1} A_{2}$, since $A_{1} A_{2} \parallel A_{n+1} A_{n+2}$, the endpoints of the diagonals parallel to $A_{1} A_{2}$ can ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,187 |
6. The function $f(x)=\frac{\sin x+\cos x}{\sin x+\tan x}+$ $\frac{\tan x+\cot x}{\cos x+\tan x}+\frac{\sin x+\cos x}{\cos x+\cot x}+\frac{\tan x+\cot x}{\sin x+\cot x}$ has a minimum value of $(\quad$ when $x \in\left(0, \frac{\pi}{2}\right)$.
(A) 2
(B) 4
(C) 6
(D) 8 | 6. B.
$$
\begin{array}{l}
f(x)=(\sin x+\cos x)\left(\frac{1}{\sin x+\tan x}+ \\
\left.\quad \frac{1}{\cos x+\cot x}\right)+(\tan x+\cot x) . \\
\left(\frac{1}{\cos x+\tan x}+\frac{1}{\sin x+\cot x}\right) \\
\geqslant(\sin x+\cos x)\left(\frac{4}{\sin x+\tan x+\cos x+\cot x}\right)+ \\
\quad(\tan x+\cot x)\left(\frac{4... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,188 |
7. The dial of a watch lies in a plane. The 12 numbers 1, 2, $\cdots$, 12 are evenly distributed on the circumference of a circle with radius $\frac{\sqrt{2}}{2}$. The vector from number $i$ to number $(i+1)$ is denoted as $\boldsymbol{t}_{i} \boldsymbol{t}_{i+1}$. Then $\boldsymbol{t}_{1} \boldsymbol{t}_{2} \cdot \bol... | ii. $7.6 \sqrt{3}-9$.
The line segments connecting adjacent scales form a regular 12-sided polygon inscribed in a circle with a radius of $\frac{\sqrt{2}}{2}$. The angle between two adjacent sides, which is the exterior angle of the 12-sided polygon, is $30^{\circ}$. The length of each side is
$$
2 \times \frac{\sqrt{2... | 6 \sqrt{3}-9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,189 |
9. In $1,2, \cdots, 2006$, the probability of randomly selecting three numbers that can form an increasing arithmetic sequence is $\qquad$ . | 9. $\frac{3}{4010}$.
Three numbers form an increasing arithmetic sequence, denoted as $a, a+d, a+2d$. According to the problem, it must satisfy $a+2d \leqslant 2006, d \leqslant 1002$. For a given $d$, $a$ can take values $1,2, \cdots, 2006-2d$.
Therefore, the number of increasing arithmetic sequences of three numbers... | \frac{3}{4010} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,191 |
10. Let $a, b$ be non-zero real numbers, $x \in \mathbf{R}$. If $\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}}$,
then $\frac{\sin ^{2008} x}{a^{2006}}+\frac{\cos ^{2008} x}{b^{2006}}=$ | $10 \cdot \frac{1}{\left(a^{2}+b^{2}\right)^{10 \times 13}}$
Given $\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}}$.
Rewrite equation (1) as
$$
\mathrm{I}=\sin ^{4} x+\cos ^{4} x+\frac{b^{2}}{a^{2}} \sin ^{4} x+\frac{a^{2}}{b^{2}} \cos ^{4} x .
$$
And $1=\left(\sin ^{2} x+\cos ^{2} x\right)^... | \frac{1}{\left(a^{2}+b^{2}\right)^{602}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,192 |
Example 2 Let $0<\theta_{i} \leqslant \frac{\pi}{4}, i=1,2,3,4$. Prove: $\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3} \cdot \tan \theta_{4} \leqslant \sqrt{\frac{\sum \sin ^{8} \theta_{i}}{\sum \cos ^{8} \theta_{i}}}$ (2001, China National Training Team Problem) | Explanation: First, let $x_{i}=\sin ^{2} \theta_{i}$, then
$$
0<x_{i} \leqslant \frac{1}{2} \text {. }
$$
Thus, the original inequality is equivalent to
$$
\begin{array}{l}
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \\
\geqslant f\left(1-x_{1}, 1-x_{2}, 1-x_{3}, 1-x_{4}\right),
\end{array}
$$
where, $f\left(x_{1}, x_{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,193 |
$$
\begin{array}{l}
\text { 11. Given } A=\left\{(x, y) \mid x^{2}+y^{2}-2 x \cos \alpha+\right. \\
2(1+\sin \alpha)(1-y)=0, \alpha \in \mathbf{R}\}, \\
B=\{(x, y) \mid y=k x+3, k \in \mathbf{R}\} .
\end{array}
$$
If $A \cap B$ is a singleton set, then $k=$ $\qquad$ | $$
\begin{array}{l}
\text { 11. } \pm \sqrt{3} \text {. } \\
x^{2}+y^{2}-2 x \cos \alpha+2(1+\sin \alpha)(1-y)=0 \\
\Rightarrow(x-\cos \alpha)^{2}+(y-1-\sin \alpha)^{2}=0 \\
\Rightarrow x=\cos \alpha, y=1+\sin \alpha \\
\Rightarrow x^{2}+(y-1)^{2}=1 .
\end{array}
$$
Notice that $\boldsymbol{A} \cap B$ is a singleton s... | \pm \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,194 |
$\begin{array}{l}\text { 12. } \max _{a, b, c \in \mathbf{R}_{+}} \min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\} \\ = \\ .\end{array}$ | 12. $\sqrt{3}$.
Let $t=\min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\}$, then $0<t \leqslant \frac{1}{a}, 0<t \leqslant \frac{1}{b^{2}}, 0<t \leqslant \frac{1}{c^{3}}$.
Thus, $a \leqslant \frac{1}{t}, b^{2} \leqslant \frac{1}{t}, c^{3} \leqslant \frac{1}{t}$.
Therefore, $t \leqslant a+... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,195 |
13. (20 points) Two circles on the same side of the $x$-axis: a moving circle $C_{1}$ and the circle $4 a^{2} x^{2}+4 a^{2} y^{2}-4 a b x-2 a y+b^{2}=0$ are externally tangent $(a, b \in \mathbf{N}, a \neq 0)$, and the moving circle $C_{1}$ is tangent to the $x$-axis. Find
(1) the equation of the locus $\Gamma$ of the ... | 3,13. (1) From $4 a^{2} x^{2}+4 a^{2} y^{2}-4 a b x-2 a y+b^{2}=0$, we can get
$$
\left(x-\frac{b}{2 a}\right)^{2}+\left(y-\frac{1}{4 a}\right)^{2}=\left(\frac{1}{4 a}\right)^{2} .
$$
Given $a, b \in \mathbf{N}$ and the two circles are on the same side of the $x$-axis, it is known that the center of the moving circle ... | a=686, b=784 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,196 |
14. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, $a_{n+1}=a_{n}+2 n(n=1,2, \cdots)$, and the sequence $\left\{b_{n}\right\}$ satisfies $b_{1}=$ $1, b_{n+1}=b_{n}+\frac{b_{n}^{2}}{n}(n=1,2, \cdots)$. Prove:
$$
\frac{1}{2} \leqslant \sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}... | 14. Let $I_{n}=\sum_{k=1}^{n} \frac{1}{\sqrt{a_{k+1} b_{k}+k a_{k+1}-b_{k}-k}}$, then $I_{1}=\frac{1}{2}<I_{2}<\cdots<I_{n}$.
And $I_{n}=\sum_{k=1}^{n} \frac{1}{\sqrt{\left(a_{k+1}-1\right)\left(b_{k}+k\right)}}$
$$
\leqslant \sqrt{\sum_{k=1}^{n} \frac{1}{a_{k+1}-1} \cdot \sum_{k=1}^{n} \frac{1}{b_{k}+k}} \text {. }
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,197 |
15. (20 points) A cube with six faces labeled $1,2,3,4,5$, and 6 is called a die. Answer the following:
（1）How many different dice are there;
(2) The absolute difference between the numbers on two adjacent faces of a die is called the variation between these two faces, and the sum of all variations is called the total ... | 15. (1) Suppose there is a square on the table that is congruent to the side of a die. Placing a die on this square can be done in $6 \times 4$ ways. Therefore, the number of different dice is $\frac{6!}{6 \times 4}=30$.
(2) The variations produced by the six numbers from $1 \sim 6$ are 15 in total, and their sum is
$$... | 30, 32, 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,198 |
1. Let $a>0, f(x)=\frac{x}{\sqrt{a}}-\sqrt{a(x+a)}$ $(x \geqslant-a)$, denote the minimum value of $f(x)$ as $m(a)$. If $m(a)=-2006 \sqrt{a}$, then $a$ equals ( ).
(A) 8020
(B) 8016
(C) 2006
(D) 2005 | -.1.A.
Let $\sqrt{x+a}=t(t \geqslant 0)$, then $x=t^{2}-a$. Hence $f(t)=\frac{t^{2}-a}{\sqrt{a}}-\sqrt{a} t=\frac{1}{\sqrt{a}}\left(t^{2}-a t-a\right)$ $=\frac{1}{\sqrt{a}}\left[\left(t-\frac{a}{2}\right)^{2}-a-\frac{a^{2}}{4}\right]$.
Obviously, when $t=\frac{a}{2}>0$, $f(x)$ reaches its minimum value $m(a)=-\frac{1}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,199 |
2. If $(m+6)(m+5)-(n+6)(n+5)=$ $2006 m(m-1)-2006 n(n-1)$, and $m \neq n$, then among the following four conclusions, the correct one is ( ).
(A) $m+n$ is not necessarily a constant
(B) $m+n=2006$
(C) $m+n=\frac{2017}{4010}$
(D) $m+n=\frac{2017}{2005}$ | 2. D.
From the problem, we have
$$
\begin{array}{l}
(m+6)(m+5)-2006 m(m-1) \\
=(n+6)(n+5)-2006 n(n-1) . \\
\text { Let } f(x)=(x+6)(x+5)-2006 x(x-1) .
\end{array}
$$
Then $f(x)$ is a quadratic function, and
$$
f(m)=f(n)(m \neq n) \text {. }
$$
Therefore, the points $(m, f(m))$ and $(n, f(n))$ must be symmetric about... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,200 |
3. As shown in Figure $1, \odot O$'s chord $A B$ intersects with $C D$ at point $P, A O$ intersects with $C D$ at point $Q$. If $A O \perp C D$, $\angle A O D=2 \angle P A C$, then among the following four conclusions, the correct one is ( ).
(A) $D, O, B$ are collinear
(B) $A, D, O, P$ are concyclic
(C) $P$ is the inc... | 3.B.
Let $\angle A O D=\alpha$. By the perpendicular diameter theorem, we have $\angle A O C=\alpha$.
Thus, $\angle A B C=\frac{1}{2} \angle A O C=\frac{\alpha}{2}=\angle P A C$.
Therefore, $C$ is the midpoint of $\overparen{A B}$, meaning points $A, C$ trisect $\overparen{D B}$. Hence,
$$
\begin{array}{l}
\angle A P ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,201 |
4. The number of integer solutions $(m, n)$ for the equation $m^{2}-n^{2}+3 m-7 n-2=0$ is $($.
(A) 1
(B) 2
(C) 3
(D) 4 | 4.D.
Multiply both sides of the original equation by 4 and complete the square to get
$$
(2 m+3)^{2}-(2 n+7)^{2}=-32 \text {. }
$$
Thus, $(n+m+5)(n-m+2)=2^{3}$.
Since $n+m+5$ and $n-m+2$ have different parities, solving them simultaneously yields
$$
(m, n)=(-5,1),(2,1),(2,-8),(-5,-8) .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,202 |
5. Does there exist a pair of real numbers $(a, b)$ such that "if $\alpha, \beta$ are the two real roots of the quadratic equation $a x^{2}+b x+b-2=0$, then $\alpha^{2}, \beta^{2}$ are also the two real roots of $a x^{2}+b x+b-2=0$"? If it exists, how many such pairs are there? The correct conclusion is ( ).
(A) Such $... | 5.C. $\left\{\begin{array}{l}\alpha=0, \\ \beta=0\end{array}\right.$ or $\left\{\begin{array}{l}\alpha=0, \\ \beta=1\end{array}\right.$ or $\left\{\begin{array}{l}\alpha=1, \\ \beta=1\end{array}\right.$ (assuming $\left.\alpha \leqslant \beta\right)$.
(1) When $\alpha=0, \beta=0$, substituting $\alpha=0$ into the equat... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,203 |
Example 3 Let $a, b, c > 0$. Prove:
$$
f=\sum \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$
(32nd United States of America Mathematical Olympiad) | Explanation: By homogeneity, we may assume $a+b+c=3$. Then
$$
\begin{aligned}
f & =\sum \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}=\sum \frac{a^{2}+6 a+9}{3\left(a^{2}-2 a+3\right)} \\
& =\frac{1}{3} \sum\left(1+\frac{8 a+6}{(a-1)^{2}+2}\right) \\
& \leqslant \frac{1}{3} \sum(4 a+4)=8 .
\end{aligned}
$$ | 8 | Inequalities | proof | Yes | Yes | cn_contest | false | 717,204 |
6. [ $x$ ] represents the greatest integer not greater than $x$. The number of all real solutions to the equation $[2 x]+[3 x]=8 x-\frac{7}{2}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 5 | 6. B.
From the equation, we get $8 x-\frac{1}{2}=3+[2 x]+[3 x] \in \mathbf{Z}$.
Let $8 x-\frac{1}{2}=n(n \in \mathbf{Z})$, then
$$
x=\frac{2 n+1}{16} \text {. }
$$
From $x-1<[x] \leqslant x$, we get
$$
5 x-2<8 x-\frac{7}{2}=[2 x]+[3 x] \leqslant 5 x \text {. }
$$
Therefore, $\frac{1}{2}<x \leqslant \frac{7}{6}$.
Fro... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,205 |
1. The solution to the inequality $\frac{x+1}{2 x-|x-1|} \geqslant 1$ is | $=$. $1 . x>\frac{1}{3}$.
When $x \geqslant 1$, the requirement is satisfied.
When $x<1$, the original inequality is equivalent to $\frac{x+1}{3 x-1} \geqslant 1$.
(1) When $\frac{1}{3}<x<1$, $x+1 \geqslant 3 x-1$, which means $x \leqslant 1$.
So, $\frac{1}{3}<x<1$ satisfies the requirement.
(2) When $x<\frac{1}{2}$, ... | x>\frac{1}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,206 |
2. If the positive integer $n \geqslant 2006$, and 122 divides $91 n-37$, then the minimum value of $n$ is $\qquad$ . | 2.2061 .
Since 122 is even and 91 and 37 are odd, $n$ must be odd.
Let $n=2k-1(k \in \mathbf{Z})$, then
122 | $(91n-37) \Leftrightarrow 122 | [91(2k-1)-37]$
$\Leftrightarrow \frac{182k-128}{122} \in \mathbf{Z} \Leftrightarrow \frac{91k-64}{61} \in \mathbf{Z}$
$\Leftrightarrow \frac{30k-3}{61} \in \mathbf{Z} \Leftright... | 2061 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,207 |
3. As shown in Figure 2, in the sector $A O B$, $\angle A O B=90^{\circ}$, $O A=2$, point $P$ is on the chord $A B$, and $A P=1$, point $Q$ is on $\overparen{A B}$, and $P Q / / O B$. Then $P Q=$ $\qquad$ . | 3. $\frac{\sqrt{8 \sqrt{2}-2}-\sqrt{2}}{2}$.
As shown in Figure 4, extend $Q P$ to intersect $A O$ at point $D$, and connect $O Q$. Then
$$
\begin{array}{l}
A D=D P=\frac{1}{\sqrt{2}}, \\
D O=O A-A D \\
=2-\frac{\sqrt{2}}{2} .
\end{array}
$$
By the Pythagorean theorem, we get
$$
\begin{array}{l}
D Q=\sqrt{Q O^{2}-D O... | \frac{\sqrt{8 \sqrt{2}-2}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,208 |
4. As shown in Figure 3, in $\triangle A B C$, $M$ is the midpoint of side $B C$, and $M D \perp A B, M E \perp A C$, with $D$ and $E$ being the feet of the perpendiculars. If $B D=2, C E=1$, and $D E \parallel B C$, then $D M^{2}$ equals $\qquad$ | 4.I.
Let $D M=x, M E=y, A D=z, A E=w$.
Since $D E / / B C$, we have $\frac{z}{2}=\frac{w}{1}$, i.e., $z=2 w$.
Because $B M=M C$, we have $S_{\triangle A K M}=S_{\triangle A M}$, i.e., $\frac{1}{2} x(z+2)=\frac{1}{2} y(w+1)$.
Thus, $2 x+x z=y+y w$.
By the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
2^{2}+x^{... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,209 |
One, (20 points) Let $a, b$ be real numbers, and
$$
4 a^{2}+16 b^{2}-16 a b+13 a-26 b+3=0 \text {. }
$$
Find the range of $a^{2}+b^{2}$. | i. Completing the square, factoring, and letting $t=a-2 b$, we get
$$
(4 t+1)(t+3)=0 \text {. }
$$
Solving for $t$ yields $t=-3$ or $t=-\frac{1}{4}$.
When $a-2 b=-3$, we have
$$
a^{2}+b^{2}=(2 b-3)^{2}+b^{2}=5 b^{2}-12 b+9,
$$
the minimum value of which is $\frac{9}{5}$;
When $a-2 b=-\frac{1}{4}$, we have
$$
a^{2}+b^... | \left[\frac{1}{80},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,210 |
II. (25 points) In a convex quadrilateral $ABCD$, $\angle BCD = \angle ADC = 90^\circ$. Point $E$ is on side $CD$, and $\angle AEB = 90^\circ$, $AE = BE$. Point $M$ is the midpoint of side $AB$, $MC$ intersects $BE$ at point $P$, $MD$ intersects $AE$ at point $Q$, and $ME$ intersects $PQ$ at point $R$. Prove that $MR \... | As shown in Figure 5, since $\triangle ABE$ is an isosceles right triangle with point $E$ as the right-angle vertex, and $M$ is the midpoint of the hypotenuse, it follows that $ME \perp AB$. Clearly, points $M, B, C, E$ are concyclic, and points $M, A, D, E$ are also concyclic.
$$
\begin{array}{l}
\text{Therefore, } \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,211 |
Three. (25 points) Find all real numbers $a$ such that the quadratic equation $a x^{2}+a x+a+2006=0$ has two distinct integer roots.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Three. (25 points) Find all real numbe... | If $a x^{2}+a x+a+2006=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
a=\frac{-2006}{x_{1}^{2}+x_{1}+1}=-\frac{2006}{t^{2}+t+1}(t \in Z) .
$$
If $a=-\frac{2006}{t^{2}+t+1}$. Then $a x^{2}+a x+a+2006=0$ becomes $-\frac{2006}{t^{2}+t+1}\left(x^{2}+x+1\right)+2006=0$, which is
$$
x^{2}+x+1-\left(t^{2}+t+1\righ... | a=-\frac{2006}{t^{2}+t+1}(t=0, \pm 1, \pm 2, \pm 3, \cdots) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,212 |
1. Given sets $M=\left\{a_{1}, a_{2}, \cdots, a_{2 n+1}\right\}, N$ $=\left\{-2^{2 n},-2^{2 n-1}, \cdots,-2,0,2, \cdots, 2^{2 n}\right\}$. If the injection $f: M \rightarrow N$ satisfies
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{2 n+1}\right)=0,
$$
then the number of such injections $f$ is $(\quad)$.... | -.1.A.
For any positive integer $m$, we have
$$
2^{m}>2^{m-1}+\cdots+2=2^{m}-2 \text {. }
$$
Thus, the highest power of 2 in $\left|f\left(a_{j}\right)\right| j=1,2, \cdots, 2 n+1$ is the same as the highest power of -2. Otherwise, the sum would not be 0. By canceling these two terms, similarly, $2^{i}$ and $-2^{i}$ e... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,213 |
2. Given $\theta_{1}, \theta_{2}, \cdots, \theta_{n} \in\left(0, \frac{\pi}{2}\right)$, let
$$
\begin{array}{l}
M=\left(\sum_{i=1}^{\prime \prime} \tan \theta_{i}\right)\left(\sum_{i=1}^{\prime \prime} \cot \theta_{i}\right), \\
N=\left(\sum_{i=1}^{\prime \prime} \sin \theta_{i}\right)\left(\sum_{i=1}^{n} \csc \theta_{... | 2. A.
$$
\begin{array}{l}
M-N \\
=\left(\sum_{i=1}^{n} \tan \theta_{i}\right)\left(\sum_{i=1}^{n} \cot \theta_{i}\right)-\left(\sum_{i=1}^{n} \sin \theta_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{\sin \theta_{i}}\right) \\
=\frac{1}{2} \sum_{j=1}^{n} \sum_{i=1}^{n}\left(\tan \theta_{i} \cdot \cot \theta_{j}+\tan \theta_{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,214 |
Example 1 Let $m$ be a real number not less than -1, such that the equation in $x$
$$
x^{2}+2(m-2) x+m^{2}-3 m+3=0
$$
has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$, find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
(2000, N... | Explanation: Since the equation has two distinct real roots, we have
$$
\begin{array}{l}
\Delta=4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
=-4 m+4>0 .
\end{array}
$$
Thus, $m<1$.
Also, $m \geqslant-1$, so, $-1 \leqslant m<1$.
By the relationship between roots and coefficients, we get
$$
x_{1}+x_{2}=-2(m-2), x_{1} x_{2}=m... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,215 |
Example 2 Given that $x$ and $y$ are real numbers, and satisfy $x y + x + y = 17, x^{2} y + x y^{2} = 66$.
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$. (2000, Shandong Province Junior High School Mathematics Competition) | Explanation: According to the structural characteristics of the known equations, they can be transformed into
$$
x y+(x+y)=17, x y(x+y)=66 \text{. }
$$
Thus, a quadratic equation can be constructed with $x y$ and $x+y$ as its roots:
$$
t^{2}-17 t+66=0 \text{. }
$$
Solving this, we get $t_{1}=6, t_{2}=11$.
When $x y=6... | 12499 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,216 |
2. Given that $x, y$ are positive integers. And they satisfy the conditions $x y+x+y=71, x^{2} y+x y^{2}=880$.
Find the value of $x^{2}+y^{2}$.
(1999, Jiangsu Province Junior High School Mathematics Competition) | Answer: 146.) | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,217 |
12. If $\sin \alpha \cdot \cos \beta=1$, then $\cos \alpha \cdot \sin \beta=$ | 12.0.
From $\sin \alpha \cdot \cos \beta=1$ and the boundedness of sine and cosine functions, we know that $\sin \alpha=\cos \beta=1$ or $\sin \alpha=\cos \beta=-1$. Therefore, $\cos \alpha=\sin \beta=0$, which means $\cos \alpha \cdot \sin \beta=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,218 |
13. Simplify $\log _{\sqrt{2}} \sin \frac{7 \pi}{8}+\log _{\sqrt{2}} \sin \frac{3 \pi}{8}$, the result is $\qquad$ . | 13. -3 .
$$
\begin{array}{l}
\text { Original expression }=\log _{\sqrt{2}}\left(\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}\right)=\log _{\sqrt{2}}\left(\frac{1}{2} \sin \frac{\pi}{4}\right) \\
=\log _{\sqrt{2}} \frac{\sqrt{2}}{4}=\log _{2} \frac{1}{2} 2^{-\frac{3}{2}}=-3 .
\end{array}
$$ | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,219 |
14. There are 2006 balls lined up in a row. They are coloured to be red, white, blue and black in the following order: 5 red, 4 white, 1 blue, 3 black, 5 red, 4 white, 1 blue, 3 black, $\cdots$. Then the last ball is coloured to be $\qquad$ | 14.red.
The problem states: There are 2006 balls arranged in a row, each dyed one of four colors: red, white, blue, and black, and they are arranged in the following order: 5 red balls, 4 white balls, 1 blue ball, 3 black balls, 5 red balls, 4 white balls, 1 blue ball, 3 black balls, $\cdots \cdots$. The color of the ... | red | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,220 |
15. Given the sets
$$
\begin{array}{l}
A=\left\{x \mid x=3^{n}, n \in \mathbf{N}_{+}\right\}, \\
B=\left\{x \mid x=4 n+1, n \in \mathbf{N}_{+}\right\} .
\end{array}
$$
Arrange the elements of $A \cap B$ in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{3}=$ $\qquad$ , and the general term formula ... | $$
15.729,9^{n}
$$
By observation and induction, it is easy to obtain. | 729,9^{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,221 |
16. Suppose that the graph of quadratic function $f(x)$ passes through the origin, and there holds $f(x+2)=f(x)+x+2$ for all $x \in \mathbf{R}$. Then the expression of $f(x)$ is . $\qquad$ | 16. $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x$.
The problem states: If the graph of the quadratic function $f(x)$ passes through the origin, and for any $x \in \mathbf{R}$, $f(x+2)=f(x)+x+2$ holds, then the expression for $f(x)$ is $\qquad$ .
From the fact that the graph of $f(x)$ passes through the origin, we know $f(0)=... | f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,222 |
$$
\text { 17. } \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \text { is }
$$
the value of $\qquad$ . | $\begin{array}{l}\text { 17. } \sqrt{3} \text {. } \\ \tan 24^{\circ}+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ}+\tan 36^{\circ} \\ =\tan \left(24^{\circ}+36^{\circ}\right)\left(1-\tan 24^{\circ} \cdot \tan 36^{\circ}\right)+\sqrt{3} \tan 24^{\circ} \cdot \tan 36^{\circ} \\ =\sqrt{3}-\sqrt{3} \tan 24^{\circ} \cdot ... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,223 |
18. Calculate $1^{2}-2^{2}+3^{2}-4^{2}+\cdots+2005^{2}-$ $2006^{2}=$ $\qquad$ . | 18. -2013021 .
$$
\begin{array}{l}
1^{2}-2^{2}+3^{2}-4^{2}+\cdots+2005^{2}-2006^{2} \\
=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\cdots+\left(2005^{2}-2006^{2}\right) \\
=(1+2)(1-2)+(3+4)(3-4)+\cdots+ \\
(2005+2006)(2005-2006) \\
=-(1+2+3+\cdots+2006)=-2013021 .
\end{array}
$$ | -2013021 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,224 |
Mr. Wang took a taxi to the Capital International Airport. The initial fare is 10 yuan (for the first $3 \mathrm{~km}$), and for every additional $0.6 \mathrm{~km}$ beyond $3 \mathrm{~km}$, the fare increases by 1 yuan. When Mr. Wang arrived at the airport, the meter showed a fare of 34 yuan. Let the distance Mr. Wang ... | 19.17.4 $\leqslant s<18$
According to the problem, we can get the equation $10+\left[\frac{s-3}{0.6}\right] \times 1=34$ (where $[x]$ represents the greatest integer not exceeding $x$). Solving this, we get
$$
24 \leqslant \frac{s-3}{0.6}<25 \text {. }
$$
Therefore, $17.4 \leqslant s<18$. | 17.4 \leqslant s < 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,225 |
20. Given
$$
f(x)=\left\{\begin{array}{ll}
\sqrt{2 x+1}, & -\frac{1}{2} \leqslant x<0 ; \\
\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}, & x \geqslant 0 .
\end{array}\right.
$$
Then $f^{-1}(x)=$ $\qquad$ . | 20. $\left\{\begin{array}{ll}\frac{x^{2}-1}{2}, & 0 \leqslant x<1 ; \\ \ln \left(x+\sqrt{x^{2}-1}\right), & x \geqslant 1 .\end{array}\right.$
When $-\frac{1}{2} \leqslant x<0$, we have $y=\sqrt{2 x+1}$, then $0 \leqslant y<1$, solving for $x$ gives $x=\frac{y^{2}-1}{2}$;
When $x \geqslant 0$, we have $y=\frac{\mathr... | \left\{\begin{array}{ll}\frac{x^{2}-1}{2}, & 0 \leqslant x<1 ; \\ \ln \left(x+\sqrt{x^{2}-1}\right), & x \geqslant 1 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,226 |
3. Let $a$, $b$, and $c$ be real numbers, $k$ be a positive constant, and
$$
\left\{\begin{array}{l}
a+b+c=0, \\
a b c=k .
\end{array}\right.
$$
(1) Find the minimum value of $\max \{a, b, c\}$;
(2) Find the minimum value of $|a|+|b|+|c|$. | (提示: (1) According to the problem, $a, b, c$ must be one positive and two negative. Without loss of generality, assume $a>0, b<0, c<0$. Thus, considering $a$ as a parameter, construct a quadratic equation with $b, c$ as its two real roots. According to $\Delta \geqslant 0$, the minimum value of $\max \{a, b, c\}=a$ can... | 2 \sqrt[3]{4 k} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,228 |
22. Try to write two sets of integer solutions for the indeterminate equation $x^{2}-2 y^{2}=1$ as $\qquad$ . | 22 . $(3,2),(17,12)$ etc. | (3,2),(17,12) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,229 |
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