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 |
|---|---|---|---|---|---|---|---|---|---|
Three. (50 points) There are 10 schools, each sending several boys and several girls to participate in a checkers competition. Contestants from the same school do not compete, while contestants from different schools, regardless of gender, must compete in a match between any two of them. The total number of matches bet... | Three, let the $i$-th school send $x_{i}$ boys and $y_{i}$ girls to participate in the competition $(i=1,2, \ldots, 10)$. Then the total number of matches between two boys or two girls is
$$
\sum_{1 \leqslant i<j \leqslant 10}\left(x_{i} x_{j}+y_{i} y_{j}\right) ,
$$
and the total number of matches between a boy and a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,309 |
Example 3 Determine whether the following equations have real solutions. If they do, find the real solutions.
(1) $\sum_{0}^{5}\left[2^{i} x\right]=12345$;
(2) $\sum_{0}^{5}\left[2^{i} x\right]=123456$. | Solution: (1) $\because \sum_{0}^{5} 2^{i}=63$,
$$
\therefore p=\left[\frac{12345}{63}\right]=195 \text{. }
$$
$\because 12345=195 \times 63+60$,
$\therefore$ take $k=31<2^{5}$, then $31+\left[\frac{31}{2}\right]+\left[\frac{31}{2^{2}}\right]+\left[\frac{31}{2^{3}}\right]+\left[\frac{31}{2^{4}}\right]=57$.
$$
\sum_{0}^... | 1959 \frac{21}{32} \leqslant x < 1959 \frac{22}{32} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,310 |
Initially $103 \odot O$ is the circumcircle of $\triangle A B C$, which is an equilateral triangle. $A D$ is the diameter of $\odot O$. Take any point $P$ on $\overparen{B C}$ $(P \neq B, P \neq C)$, and let $E$ and $F$ be the incenters of $\triangle P A B$ and $\triangle P A C$, respectively. Prove: $P D$ $=|P E-P F|$... | Proof: As shown in the figure, without loss of generality, assume $P$ is inside $\overparen{B D}$. Take the midpoints $M$ and $N$ of $\overparen{A B}$ and $\overparen{A C}$, respectively.
$\because A D$ is the common axis of symmetry for $\odot O$ and the equilateral $\triangle A B C$, and $M$ and $N$ are two symmetric... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,311 |
In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$. Try to find the maximum value of $n$.
The difference between any two adjacent interior angles of a convex $n$-sided polygon is $18^{\circ}$. Try to find the maximum value of $n$. | Solution: Complete in three steps.
(1) Prove that $n$ is even. Let the convex $n$-gon be $A_{1} A_{2} A_{3} \cdots$
$A_{n}$, then $A_{2}=A_{1} \pm 18^{\circ}$, (where “ + ” or “-” is chosen)
$A_{3}=A_{2} \pm 18^{\circ}=A_{1} \pm 18^{\circ} \pm 18^{\circ}$, (choose “ + ” or “ - ” as before) $\qquad$
$$
\begin{array}{l}
... | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,312 |
103 Given $a, b, c \in \mathbf{R}^{+}, a^{x}=b^{y}=c^{x}, \frac{1}{x}+$ $\frac{1}{y}+\frac{1}{z}=0$. Prove: $a b c=1$. | Proof: If $a, b, c$ have one that is 1, then the other two are also 1, at this time, $abc=1$;
If $a, b, c$ are all not 1, then from $a^{x}=b^{y}$ we can get
$$
x=\log _{a} b^{y}=y \log _{a} b .
$$
Similarly, we get $z=y \log _{c} b$.
$$
\text { Then } \frac{1}{x}=\frac{1}{y} \log _{b} a, \frac{1}{z}=\frac{1}{y} \log _... | abc=1 | Algebra | proof | Yes | Yes | cn_contest | false | 712,313 |
Given $a>1, b>1, c>1$. Prove:
$$
\frac{a^{5}}{b^{3}-1}+\frac{b^{5}}{c^{3}-1}+\frac{c^{5}}{a^{3}-1} \geqslant \frac{5}{2} \sqrt[3]{50} .
$$ | Proof: Given $b>1$, we have $b^{3}-1>0$,
then $\left(b^{3}-1\right)\left(b^{3}-1\right)\left(b^{3}-1\right) \times \frac{3}{2} \times \frac{3}{2}$
$\leqslant\left[\frac{3\left(b^{3}-1\right)+2 \times \frac{3}{2}}{5}\right]^{5}=\frac{3^{5}}{5^{5}} \cdot b^{15}$,
that is
$\left(b^{3}-1\right)^{3} \leqslant \frac{4 \times... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,314 |
Example 2 In a football invitational tournament, 16 cities participate, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. After several days of the tournament, it was found that except... | Solution: Consider the general case: Suppose there are $n$ cities participating, and let the number of matches played by Team B from City A be $a_{n}$. Clearly, $a_{1}=$ 0.
Since there are $n$ cities participating, according to the competition rules, each team can play at most $2(n-1)$ matches. Given the conditions, t... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,315 |
Example 3 There is a type of sports competition with $M$ events, and athletes $A$, $B$, and $C$ participate. In each event, the first, second, and third places receive $p_{1}$, $p_{2}$, and $p_{3}$ points respectively, where $p_{1}$, $p_{2}$, $p_{3} \in \mathbf{Z}^{+}$, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 2... | Solution: Consider the total score of three people, we have
$$
\begin{array}{l}
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40 . \\
\because p_{1}, p_{2}, p_{3} \in \mathbf{Z}^{+}, \text { and } p_{1}>p_{2}>p_{3}, \\
\therefore p_{1}+p_{2}+p_{3} \geqslant 3+2+1=6 .
\end{array}
$$
Thus, $6 M \leqslant 40, M \leqslant 6$.
Si... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,316 |
Example 4 Two seventh-grade students are allowed to participate in the chess competition for eighth-grade students. Each pair of players competes once, with a win earning 1 point, a draw earning half a point, and a loss earning zero points. The two seventh-grade students scored a total of 8 points, and each eighth-grad... | Solution: Let there be $n$ eighth-grade students participating, each scoring $k$ points, so the total score of the seventh and eighth-grade students is $8+k n$ points. Since the sum of the scores of both sides in each match is definitely 1 point, it follows that the number of matches played among the participants shoul... | 7 \text{ and } 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,317 |
Example 5 In a certain competition, each player plays exactly one game against each of the other players. A win earns the player 1 point, a loss 0 points, and a draw 0.5 points for each player. After the competition, it is found that exactly half of each player's points were earned in games against the 10 lowest-scorin... | Solution: Let there be a total of $n$ players, then they collectively score $\mathrm{C}_{n}^{2}$ points. The 10 players with the lowest scores collectively score $\mathrm{C}_{10}^{2}=45$ points through their matches against each other, which is half of their total score, so these 10 players collectively score 90 points... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,318 |
Example 6 In a tennis club of 20 members in a certain area, 14 singles matches are held, with each member playing at least once. Prove that there must be 6 matches whose 12 participants are all different.
(18th American High School Mathematics Examination) | Prove: Construct a graph $G$ with 20 vertices representing 20 players and 14 edges representing 14 matches.
Let the degree of vertex $A_{i}$ be $d_{i}(1 \leqslant i \leqslant 20)$. By the given conditions, we have
$$
d_{i} \geqslant 1, \sum_{i=1}^{20} d_{i}=14 \times 2=28 .
$$
Now, remove $d_{i}-1$ edges at each vert... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,319 |
In a football round-robin tournament with 18 teams, after 8 rounds, each team has played 8 different teams. Prove that it is always possible to find 3 teams that have not played against each other yet.
（15th All-Soviet Union Mathematical Olympiad） | Proof: Since each team has already played against the other 8 teams, and for each team, there are 17 other teams, for any one team $A$, it is possible to find 9 teams that have not yet played against it.
Among these 9 teams, there must be two teams that have not played against each other. In fact, these 9 teams have a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,320 |
Example 8 After 20 figure skaters have performed, 9 judges each assign them ranks from 1 to 20. It is known that for each skater, the difference between any two ranks does not exceed 3. If the sum of the ranks each skater receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant ... | If 9 judges all give a certain athlete the first place, then $C_{1}=9$. If two athletes are both judged as the first place, then one of them gets no less than 5 first places, and the other 4 ranks no higher than fourth. Therefore, $C_{1}=5 \times 1+$ $4 \times 4=21$.
If 3 athletes all get the first place, then their o... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,321 |
Example 9 In a certain competition, if there are $a$ contestants and $b$ judges, where $b \geqslant 3$ and is odd. Each judge can rate each contestant as either "pass" or "fail". Let $k$ be an integer that satisfies the following condition: any two judges can give the same rating to at most $k$ contestants. Prove: $\fr... | Proof: First, if two judges give the same evaluation to a contestant, we call it an "agreement." From the given information, any two judges can have at most $k$ "agreements." Thus,
the total number of "agreements" $\leqslant k \mathrm{C}_{b}^{2}$.
On the other hand, for any one contestant, suppose $A$ judges pass them... | \frac{k}{a} \geqslant \frac{b-1}{2 b} | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,322 |
Example 10 In a certain sports competition, each pair of players plays 1 match, and outstanding players are determined through the matches. Player $A$ is determined to be an outstanding player under the condition that: for any other player $B$, either $A$ beats $B$, or $A$ indirectly beats $B$, i.e., there exists a pla... | Proof: First, we prove the existence of an excellent player. Since there are only a finite number of participants, there must be one with the most wins. Let $A$ be one of the players with the most wins.
If $A$ wins against all other players, then $A$ is certainly an excellent player. Otherwise, let $A$ win against $B_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,323 |
Example 3. Given a $4 \times 4$ number table (as shown below), changing the sign of all numbers in any row or any column is called one operation. Can all the numbers in the table be made positive after several operations?
$$
\left[\begin{array}{cccc}
1 & -3 & 4.7 & -12 \\
-7.8 & 41.4 & -9.6 & -7.6 \\
3.4 & -12.1 & 32 &... | Solution: It is impossible to meet the requirements of the problem.
Since each operation changes the signs of 4 numbers in the table, we first study the situation of the sign changes of these 4 numbers.
Let there be $k$ positive numbers in any row (or any column), then there are $4-k$ negative numbers.
Before the ope... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,324 |
1. $n(n \geqslant=2)$ players participate in a competition that lasts for $k$ days, with each player scoring exactly $1,2, \cdots, n$ points each day (no two players have the same score). At the end of $k$ days, each player's total score is 26. Find all possible values of $(n, k)$.
(22nd Canadian Intermediate Mathemati... | (Cue: Calculate the total score of $n$ players, with $k(1+2+\cdots+n)=26n$, then $k(n+1)=52$. And $52=2^{2} \times 13$, which has 6 factors: $1,2,4,13,26,52$. Also, $n+1 \geqslant 3, k \geqslant 2$. Thus, we get three solutions, namely $(n, k)=(25,2),(12,4),(3,13)$.) | (n, k)=(25,2),(12,4),(3,13) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,325 |
2. In the final stage of a professional bowling tournament, the top five players compete as follows: First, the fifth-place player competes with the fourth-place player, the loser gets fifth place, the winner competes with the third-place player; the loser gets third place, the winner competes with the first-place play... | (Due to determining the top five rankings, 4 races were held, and each race has two possible outcomes. By the multiplication principle, there are $2^{4}=16$ possible ranking orders for the top five winners.) | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,326 |
3. In a cross-country race, there are two teams participating, each with 5 members. When the athletes reach the finish line, their ranks are recorded, and the team of the $n$-th arriving athlete scores $n$ points, with the team having the lower total score winning. Now, assuming no two team members arrive at the finish... | (The hint: The sum of the scores of 10 athletes is $1+2+\cdots+10$ $=55$, thus the total score of the winning team does not exceed $\left[\frac{55}{2}\right]=27$. Also, the total score of the winning team is no less than $1+2+\cdots+5=15$, it is easy to see that the score of the winning team can take all values from $1... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,327 |
4. After the completion of the Russian Football League's double round-robin tournament, it was found that all teams had different points. At the same time, the total points of the 6 teams from Moscow equaled the total points of the remaining 12 teams. Prove that among the Moscow teams, there must be a tournament winner... | (Since each match results in a total of 2 points, the 18 teams together score $18 \times 17 \times 2=612$ points after all matches are completed. The Moscow teams score a total of 300 points, with the highest scorer among them not less than 54. Assuming none of the Moscow teams are winners, the total points of the top ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,328 |
5. $n(n>3)$ table tennis players have played several singles matches. After the matches, any two players have played against a set of opponents that are not exactly the same. Try to prove: it is always possible to remove one player such that among the remaining players, any two players have still played against a set o... | (提示: Let $n$ players be $A_{1}, A_{2}, \cdots, A_{n}$, and the set of opponents that $A_{i}$ has played against be $B_{i}$. It is easy to know that $B_{i} \subseteq\left\{A_{k}\right\}_{k=1,2, \cdots, n}, i=1$, $2, \cdots, n, B_{i} \neq B_{j}(i \neq j)$.
Construct a graph $G$, with $\left\{B_{i}\right\}_{i=1, \cdots, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,329 |
Let $I$ be the incenter of $\triangle ABC$, and let the incircle of $\triangle ABC$ touch the sides $BC$, $CA$, $AB$ at points $K$, $L$, $M$ respectively. The line through $B$ parallel to $MK$ intersects lines $LM$ and $LK$ at points $R$ and $S$ respectively. Prove that $\angle RIS$ is acute. | Proof: Establish a rectangular coordinate system as shown in Figure 1. Let the inradius of $\triangle ABC$ be 1, then $I\left(\csc \frac{B}{2}, 0\right)$.
It is easy to see that
$$
\angle B S K=90^{\circ}-\angle \frac{A}{2}.
$$
In $\triangle B S K$, by the Law of Sines,
$$
\frac{B K}{\sin \left(90^{\circ}-\frac{A}{2}\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,331 |
Two circles $\Gamma_{1}$ and $\Gamma_{2}$ are contained within circle $\Gamma$, and are tangent to circle $\Gamma$ at two distinct points $M$ and $N$. $\Gamma_{1}$ passes through the center of $\Gamma_{2}$. The line through the two intersection points of $\Gamma_{1}$ and $\Gamma_{2}$ intersects $\Gamma$ at points $A$ a... | Proof: The following only proves the case where $\Gamma_{1}$ and $\Gamma_{2}$ are both internally tangent to $\Gamma$.
Establish a Cartesian coordinate system as shown in Figure 2, and let the centers of $\Gamma$, $\Gamma_{1}$, and $\Gamma_{2}$ be $F$, $F_{1}$, and $F_{2}$, with radii $r$, $r_{1}$, and $r_{2}$, respec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,332 |
Example $1\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}=$ $\qquad$
(1999, National Junior High School Mathematics League Wuhan Selection Competition) | Let $3^{999}=a, 5^{099}=b, 7^{999}=c$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{c}{a} \sqrt{\frac{a^{2}+(a b)^{2}}{c^{2}+(b c)^{2}}} \\
=\frac{c}{a} \sqrt{\frac{a^{2}\left(1+b^{2}\right)}{c^{2}\left(1+b^{2}\right)}} . \\
=\frac{c}{a} \cdot \frac{a}{c}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,333 |
Example 2 Let the parabola
$$
y=x^{2}+(2 a+1) x+2 a+\frac{5}{4}
$$
intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+323 a^{-6}$.
(1998, National Junior High School Mathematics Competition) | Solution: (1) From the given condition, the equation
$$
x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0
$$
has two equal real roots, hence
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$
which simplifies to $a^{2}-a-1=0$.
Solving for $a$, we get $a=\frac{1 \pm \sqrt{5}}{2}$.
(2) From (1), we know $a^{2}-a-1=0$, thus
$$
\... | 5796 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,334 |
Example 4 Starting from the three numbers $3,4,12$, at each step, arbitrarily select two numbers $a$ and $b$, and replace them with $0.6 a-0.8 b$ and $0.8 a \div 0.6 b$. Can the three numbers be transformed into 4, 6, 12 after several such operations? | Solution: No. Notice that $0.8^{2}+0.6^{2}=1$, so we can study the characteristic of the sum of squares of the numbers before and after the operation.
Since $(0.6 a-0.8 b)^{2}+(0.8 a+0.6 b)^{2}$ $=a^{2}+b^{2}$,
the sum of the squares of the three numbers remains unchanged before and after the operation.
Initially, the... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,335 |
Example 3 Given $a^{2}+a+1=0$. Then, $a^{1992}+$ $a^{322}+a^{2}=$ $\qquad$
(Harbin 15th Junior High School Mathematics Competition) | Given the condition $a \neq 1$, from $(a-1)(a^2 + a + 1) = 0$, we get $a^3 - 1 = 0$, hence $a^3 = 1$.
Substituting 1 for $a^3$ and 0 for $a^2 + a + 1$, we get
$$
\begin{aligned}
\text { Original expression } & =\left(a^{3}\right)^{664}+\left(a^{3}\right)^{107} a+a^{2} \\
& =1^{664}+1^{107} a+a^{2} \\
& =1+a+a^{2}=0 .
\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,336 |
Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
(Jiangsu Province 14th Junior High School Mathematics Competition) | Solution: From the conditions, we have
$$
\begin{array}{l}
x y+(x+y)=71, \\
x y(x+y)=880,
\end{array}
$$
Therefore, $x y$ and $x+y$ are the two roots of the equation
$$
z^{2}-71 z+880=0
$$
Solving this equation, we get
$$
x+y=16, x y=55,
$$
or $x+y=55, x y=16$.
When $x+y=55, x y=16$, $x$ and $y$ are the two roots of... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,337 |
Example 5 Given that $x, y$ are real numbers, and $x^{2}+x y+$ $y^{2}-2=0$. Then the range of values for $x^{2}-x y+y^{2}$ is $\qquad$
(1996, Huanggang Region, Hubei Province Junior High School Mathematics Competition) | Solution: Let $x^{2}-x y+y^{2}=t$, then from the given conditions we have
$$
t=2-2 x y \text {, }
$$
i.e., $x y=\frac{2-t}{2}$;
$$
(x+y)^{2}=2+x y=2+\frac{2-t}{2}=\frac{6-t}{2} .
$$
Since $(x+y)^{2} \geqslant 0$,
$$
(x+y)^{2}-4 x y=(x-y)^{2} \geqslant 0 \text {. }
$$
In the above two equations, substitute $\frac{6-t... | \frac{2}{3} \leqslant x^{2}-x y+y^{2} \leqslant 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,338 |
Example 6 Given that $x, y, z$ are real numbers, and $x+y+z=5, xy+yz+zx=3$. Try to find the maximum and minimum values of $z$.
(Canadian 10th High School Mathematics Competition) | Solution: From the conditions, we have
$$
\begin{array}{l}
x+y=5-z, \\
x y=3-(x+y) z=3-(5-z) z \\
=z^{2}-5 z+3.
\end{array}
$$
And $(x+y)^{2}-4 x y=(x-y)^{2} \geqslant 0$,
Substituting $5-z$ and $z^{2}-5 z+3$ for $x+y$ and $x y$ respectively in the above equation, we get
$$
(5-z)^{2}-4\left(z^{2}-5 z+3\right) \geqslan... | -1 \leqslant z \leqslant \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,339 |
Example 1 Let $f(x)=\sum_{0}^{5}\left[2^{i} x\right]$. Find $f(3.14)$. | Given: $\because x=3.14, \frac{4}{2^{5}} \leqslant 0.14<\frac{5}{2^{5}}$, so take $p=3, k=4$.
$$
\begin{aligned}
\therefore f(3.14) & =\sum_{0}^{5} 2^{i} \times 3+\sum_{0}^{5}\left[\frac{4}{2^{i}}\right] \\
& =3 \times 63+7=196 .
\end{aligned}
$$ | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,340 |
Example 2 Find $\sum_{0}^{1000}\left[\frac{1}{3} \times 2^{i}\right]$. (2000, Russian Mathematical Olympiad) | Analysis: According to Proposition 1, here $p=0$, the key is to find a natural number $k$, such that $\frac{1}{3} \in\left[\frac{k}{2^{1000}}, \frac{k+1}{2^{1000}}\right)$,
$$
\because \frac{1}{3}=\frac{\frac{1}{3} \times 2^{1000}}{2^{1000}}, k \leqslant \frac{1}{3} \times 2^{1000}<k+1 \text {, }
$$
$\therefore$ take $... | \frac{1}{3}\left(2^{1001}-2\right)-500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,341 |
1. Let the circumcenter of acute triangle $\triangle ABC$ be $O$. Draw the altitude from $A$ to $BC$, and let the foot of the perpendicular be $P$, and $\angle BCA \geqslant \angle ABC + 30^{\circ}$. Prove:
$$
\angle CAB + \angle COP < 90^{\circ} \text{.}
$$ | $$
\begin{aligned}
& \text { Prove: Let } \alpha=\angle C A B, \beta=\angle A B C, \gamma=\angle B C A, \delta \\
= & \angle C O P .
\end{aligned}
$$
Let $K$ and $Q$ be the reflections of points $A$ and $P$ across the line $B C$, and let $R$ be the circumradius of $\triangle A B C$. Then
$$
O A=O B=O C=O K=R \text {. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,342 |
2. For all positive real numbers $a, b, c$, prove:
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .
$$ | Prove: Let $x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, z=\frac{c}{\sqrt{c^{2}+8 a b}}$, then $x, y, z \in \mathbf{R}^{+}, x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, i.e., $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y^{2}}-1=\frac{8 a c}{b^{2}}, \frac{1}{z^{2}}-1=\frac{8 a b}{c^{2}}$.
Thus,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,343 |
3.21 girls and 21 boys participate in a math competition.
(1) Each participant solved at most 6 problems;
(2) For every girl and every boy, there is at least one problem that both of them solved.
Prove: There is a problem that was solved by at least 3 girls and at least 3 boys. | Proof: Assume each problem is solved by at most 2 boys or at most 2 girls.
Let \( A = \{A_1, A_2, \cdots, A_k\} \) be the set of all problems solved by at most 2 girls,
and \( B = \{A_{k+1}, A_{k+2}, \cdots, A_{k+m}\} \) be the set of problems not in \( A \) and solved by at most 2 boys.
21 boys are denoted as \( p_1,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,344 |
Example 5 There are 1993 glass pieces, each coated with one of the three colors: red, yellow, or blue. The following operation is performed: two pieces of different colors are wiped clean, and then the third color is applied (for example, a blue glass piece and a red glass piece have their blue and red wiped off, and t... | Proof: Let the number of red, yellow, and blue glass pieces be \( x, y, z \) respectively. Then,
\[
x + y + z = 1993.
\]
Since 1993 is not divisible by 3, the remainders of \( x, y, z \) when divided by 3 cannot all be the same, nor can they all be different. Therefore, there must be exactly two of \( x, y, z \) that ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,345 |
4. Let $n$ be an odd number greater than $1$, and $k_{1}, k_{2}, \cdots, k_{n}$ be given integers. For each permutation $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $1,2, \cdots, n$, denote $S(a)=\sum_{i=1}^{n} k_{i} a_{i}$. Prove that there exist two permutations $b$ and $c, b \neq c$, such that $S(b)-S(c)$ is divi... | Proof: Assume for any two distinct $b$ and $c$, we have
$$
S(b)-S(c) \neq 0(\bmod n!) \text {. }
$$
Then when $a$ takes all $n!$ permutations of $1,2, \cdots, n$, $S(a)$ traverses a complete residue class modulo $n!$, and each residue class is visited exactly once. Therefore,
$$
\begin{array}{l}
\sum_{a} S(a) \equiv 1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,346 |
5. In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. It is known that $\angle B A C=60^{\circ}$ and $A B+B P=A Q+Q B$. What are the possible values of the angles of $\triangle A B C$? | Solution: Let $A=\angle BAC, B=\angle ABC, C=\angle ACB$.
On the extension of $AQ$, construct $E$ such that $QB=QE$, and connect $BE$ to intersect line $AP$ at $F$. On the extension of $AB$, construct $D$ such that $BD=BP$. Then,
$$
\angle BDP=\frac{1}{2} \angle ABC=\angle ABQ.
$$
Therefore, $BQ \parallel DDP$.
Thus, ... | B=80^{\circ}, C=40^{\circ}, A=60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,347 |
6. Let $a, b, c, d$ be integers, $a>b>c>d>0$, and $a c+b d=(b+d+a-c)(b+d-a+c)$. Prove: $a b+c d$ is not a prime. | Proof: By contradiction, if $p=ab+cd$ is a prime, it will lead to a contradiction.
Since $ac+bd=(b+d+a-c)(b+d-a+c)$
$$
=(b+d)^{2}-(a-c)^{2},
$$
then $a^{2}-ac+c^{2}=b^{2}+d^{2}+bd$.
Substituting $a=\frac{p-cd}{b}$ into the above equation, we get
$$
\frac{(p-cd)^{2}}{b^{2}}-\frac{p-cd}{b} \cdot c+c^{2}=b^{2}+d^{2}+bd,
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,348 |
2. A stair-shaped brick is composed of 12 unit cubes, with a width of 2 and 3 steps (as shown in Fig. 1). Find all positive integers $n$, such that the brick can be used to form a cube with edge length $n$.
保留源文本的换行和格式，翻译结果如下：
2. A stair-shaped brick is composed of 12 unit cubes, with a width of 2 and 3 steps (as sho... | Because the volume of a single brick is 12. Let a cube with edge length $n$ be composed of $m$ bricks, from $12 m=n^{3}$ we know that $n$ is a multiple of 6. Let $n=6 l$, where $l$ is a positive integer. On the other hand, two bricks can form a $2 \times 3 \times 4$ rectangular prism, and several such prisms can form a... | l \text{ is even} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,349 |
3. Let $n \geqslant 4$ be a positive integer, and let $S$ be the set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane such that: no three points are collinear, and no four points are concyclic. Let $a_{t}(1 \leqslant t \leqslant n)$ denote the number of circles $P_{i} P_{j} P_{k}$ that contain $P_{t}$, and let ... | Prove: When $n=4$, if $P_{1}, P_{2}, P_{3}, P_{4}$ are not the vertices of a convex quadrilateral, assume without loss of generality that $P_{4}$ is inside $\triangle P_{1} P_{2} P_{3}$, then $a_{1} =a_{2}=a_{3}=0, a_{4}=1$, so $m(S)=1$.
Assume $P_{1}, P_{2}, P_{3}, P_{4}$ are the vertices of a convex quadrilateral, i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,350 |
4. Let $n$ and $k$ be positive integers, satisfying $\frac{1}{2} n < k \leqslant \frac{2}{3} n$. Find the smallest number $m$, such that placing each of $m$ "pawns" in one square of an $n \times n$ chessboard ensures that no row or column contains $k$ consecutive squares without a "pawn". | Solution: If there does not exist a $k \times 1$ or $1 \times k$ block of squares without a "pawn" on the chessboard, then the placement of "pawns" on the chessboard is called "good". Let the rows and columns be from the 0th row to the $(n-1)$th row and from the 0th column to the $(n-1)$th column, respectively.
An ide... | 4(n-k) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,351 |
5. On the plane, there are $n$ rectangles with sides parallel to each other, and the different sides lie on different lines. The perimeters of these rectangles divide the plane into several connected regions. If a region has at least one vertex on its boundary that is a vertex of one of the $n$ rectangles, then this re... | Proof: For each vertex of a good region, if the interior angle formed by the vertex is $90^{\circ}$, then the vertex is called convex; if it is $270^{\circ}$, then the vertex is called concave (as shown in Figure 3), where the shaded part is the interior of the region.
For a simple closed curve $C$ that forms the peri... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,352 |
$f(x+g(y))=x f(y)-y f(x)+g(x) . \quad(*)$ | Proof: First, prove the existence of $\alpha$ such that $g(\alpha)=0$.
If $f(0)=0$, substitute $y=0$ into equation (*), we have
$$
f(x+g(0))=x f(0)+g(x)=g(x) \text {. }
$$
Let $x=-g(0)$, then $g(-g(0))=f(0)=0$, i.e., $\alpha=-g(0)$.
If $f(0)=b \neq 0$, set $g(0)=a$, and substitute $x=0$ into equation (*), we have
$$
f(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,355 |
1. In a $3 \times 3$ square grid, fill in the numbers as shown in the table below. The operation on the table is as follows: each operation involves adding a number to two adjacent numbers in the grid (adjacent means two small squares that share a common edge).
\begin{tabular}{|l|l|l|}
\hline 0 & 3 & 2 \\
\hline 6 & 7 ... | (Tip: (1) After 5 operations, all cells can be 0.
(2) As shown in the table below, the invariant $S=a-$ $b+c-d+e-f+g-h+k$.
\begin{tabular}{|l|l|l|}
\hline$a$ & $b$ & $c$ \\
\hline$d$ & $e$ & $f$ \\
\hline$g$ & $h$ & $k$ \\
\hline
\end{tabular}
The initial state is
$$
S=0-3+2-6+7-0+4-9+5=0,
$$
The target state is
$$
S... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,356 |
1. A bus starts a 100-kilometer journey at 12:20 PM. There is a computer on the bus that at 1:00 PM, 2:00 PM, 3:00 PM, 4:00 PM, 5:00 PM, and 6:00 PM, says: “If the average speed in the future is the same as the average speed in the past, then it will take one more hour to reach the destination.” Is this possible? If so... | 1. The computer is correct.
At $k: 00$ PM, where $1 \leqslant k \leqslant 6$, the distance the car has traveled is $\frac{60 k-20}{60 k+40}$ of the total distance, and the average speed of the car up to this point is $\frac{1}{60 k+40}$. If it continues to travel the remaining $\frac{60}{60 k+40}$ at the same speed, i... | 85 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,357 |
2. The cube of an $n$-digit number is an $m$-digit number, can $n+m=$ 2001? | 2. Notice that $10^{500}$ has 501 digits, and $\left(10^{500}\right)^{3}$ has 1501 digits, totaling 2002 digits.
For any positive number $a>10^{500}, a$ and $a^{3}$ have at least 2002 digits combined.
If $a<10^{500}$, it has at most 500 digits, and $a^{3}<10^{1500}$ has at most 1500 digits. Therefore, the sum of the ... | 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,358 |
3. In $\triangle A B C$, $X$ is a point on $A B$, $Y$ is a point on $B C$, and line segments $A Y$ and $C X$ intersect at $Z$. If $A Y=Y C$ and $A B=Z C$, prove that $B$, $X$, $Z$, and $Y$ are concyclic. | 3. First, note that $C Y=A YA Y$. In the latter case, $C Z>C Y$.
As shown in the figure, construct the quadrilateral $A B C D$, and take a point $W$ on the side $A D$ such that $\angle W C Y=\angle A Y C$, then
$$
C Y=C W, C Z=A B=C D,
$$
and $\angle A Y C=\angle W C Y=\angle D W C$.
Since $A B>A Y, \triangle Z Y C \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,359 |
A person's chess piece is a $1 \times 2$ rectangle, and the piece used by the second player is a $2 \times 1$ rectangle: the one who cannot make a move loses. Who has a surefire winning strategy, and what is the method to win? | 4. First, play on a $3 \times 4$ chessboard. The first player can place the first piece anywhere in the middle row to win. This leaves two columns for the second player to place their pieces, while the first player can place two more pieces above and below the first piece. Now, divide the $3 \times 1000$ chessboard int... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,360 |
5. There are 9 points on the surface of a tetrahedron with a side length of one centimeter. Prove: There must be two points whose spatial distance is no more than half a centimeter. | 5. Connecting the midpoints of two edges of a tetrahedron that share a common vertex divides the solid tetrahedron into 4 smaller tetrahedra and 1 central octahedron. If 2 of the 9 points are located on the surface of one of the smaller tetrahedra, the distance between them is no more than $0.5 \mathrm{~cm}$. For this,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,361 |
2. The integer solutions of the equation $\sqrt{x}+\sqrt{y}=\sqrt{2001}$ are $(\quad)$.
(A) do not exist
(B) there is only 1 set
(C) there are exactly 2 sets
(D) there are at least 4 sets | $2 . C$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,363 |
3. Among the following inequalities, the correct one is ( ).
(A) $12340 \times 12352>12346^{2}$
(B) $1.99 \times 1001 \times 1957>1001^{2}+1007^{2}$
(C) $2 \sqrt{5}>\sqrt{3}+\sqrt{7}$
(1) $\frac{1217}{1999}>\frac{1219}{2001}$ | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,364 |
5. A small boat sailing from Port A to Port B against the current takes 2 hours. When the water flow speed doubles, the journey from Port A to Port B takes 3 hours. After the water flow speed increases, the return trip from Port B to Port A will take ( ) hours.
(A) 0.5
(B) 1
(C) 1.2
(D) 1.5 | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,366 |
2. On an $8 \times 8$ grid paper, fill in 64 non-negative integers (one number per cell). You are allowed to select any $3 \times 3$ or $4 \times 4$ subgrid (the selected rows and columns must be contiguous) and add 1 to the 9 or 16 numbers in this subgrid, which is called performing one operation. Can all 64 numbers i... | (Prompt: It may not be possible. Consider the 20 shaded small squares as shown in the right figure. No matter which $3 \times 3$ or $4 \times 4$ sub-square is selected, each operation will always increase the sum of the 20 numbers in the unshaded squares by an even number. Therefore, when the sum of these 20 numbers is... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,367 |
6. Consider the following 4 propositions:
(1) Two isosceles triangles with one angle of $100^{\circ}$ are similar;
(2) Two right triangles with corresponding equal hypotenuses and perimeters are congruent;
(3) A quadrilateral with perpendicular and equal diagonals is a square;
(4) A trapezoid with equal diagonals is an... | 6. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,368 |
9. If $x^{3}+a x^{2}+b x+8$ has two factors $x+1$ and $x+2$, then $a+b=(\quad$.
(A) 7
(B) 8
(C) 13
(L) 21 | 9.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,371 |
11. According to the "Personal Income Tax Law of the People's Republic of China," citizens do not need to pay tax on the portion of their monthly salary and wages that does not exceed 800 yuan. The portion exceeding 800 yuan is the monthly taxable income, which is taxed according to the following table in segments:
\be... | $=、 11.1350$ | 1350 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,373 |
15. Let $a$, $b$, $c$ be real numbers, consider the following propositions:
(1) If $a^{2} + ab + c > 0$, and $c > 1$, then $0 < a < 1$;
(2) If $a > 1$, and $0 < b < 1$, then $a^{2} + ab + c > 0$;
(3) If $0 < a < 1$, and $a^{2} + ab + c > 0$, then $c > 1$.
Determine which propositions are correct and which are incorrec... | Three, 15. Proposition (1) is incorrect.
Counterexample: Let $b=4, c=5$, at this time, $a^{2}+a b+c=a^{2}+$ $4 a+5=(a+2)^{2}+1>0$, and $c>1$, the conditions are satisfied, but the conclusion $01$, and $0\frac{b}{2}>\left(\frac{b}{2}\right)^{2} \Rightarrow c>\frac{b^{2}}{4}$, i.e., $c-\frac{b^{2}}{4}>0$.
Thus, $a^{2}+a ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,377 |
3. Given three numbers $89, 12, 3$, perform the following operations: take any two of these numbers, find their sum and divide by $\sqrt{2}$, and simultaneously, find their difference and divide by $\sqrt{2}$. Question: After several such operations, can we obtain the three numbers 90, 10, 14? Prove your conclusion: | (Tip: The sum of squares before and after the operation remains unchanged, while $89^{2}+12^{2}+$
$$
\left.3^{2} \neq 90^{2}+10^{2}+14^{2} .\right)
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,378 |
16. Given $\triangle A B C$, where $B C>A C$, $C H$ is the altitude on side $A B$, and it satisfies $\frac{A C^{2}}{B C^{2}}=\frac{A H}{B H}$. Prove:
$$
\angle A+\angle B=90^{\circ} \text { or } \angle A-\angle B=90^{\circ} \text {. }
$$ | 16. (1) If the foot of the perpendicular $H$ is on the line segment $AB$, as shown in Figure 5. From
$$
\begin{array}{l}
A H^{2}+C H^{2}=A C^{2}, \\
B H^{2}+C H^{2}=B C^{2},
\end{array}
$$
we get $B H^{2}-A H^{2}$
$$
=B C^{2}-A C^{2} \text {, }
$$
which leads to $(B H+A H)(B H-A H)=B C^{2}-A C^{2}$,
i.e., $A B=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,379 |
1. Simplify $\sqrt[3]{\frac{\sqrt{5}-1}{2}-\left(\frac{\sqrt{5}-1}{2}\right)^{2}}$ to get $\quad$.
(A) $\frac{\sqrt{5}-1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\sqrt{5}$
(D) $\sqrt[3]{5}$ | $-1 .(\mathrm{A})$
Let $\frac{\sqrt{5}-1}{2}=a$, then $a^{2}+a-1=0$, so $a^{3}+a^{2}-a=0$, which means $a^{3}=a-a^{2}$. Therefore, the original expression $=\sqrt[3]{a-a^{2}}=\sqrt[3]{a^{3}}=a$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,380 |
2. Let $a, b$ be the radii of two circles, and $c$ be the distance between their centers. If the equation $x^{2}-2 a x+b^{2}=c(b-a)$ has equal real roots, then the two circles ( ).
(A) intersect
(B) are internally tangent
(C) are equal
(D) are equal or externally tangent | 2.(D).
According to the problem, we have $\Delta=0$, that is, $a^{2}-\left[b^{2}-c(b-a)\right]=0$. Factoring, we get $(a-b)(a+b-c)=0$. When $a-b=0$, the two circles are equal, and when $a+b-c=0$, the two circles are externally tangent. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,381 |
3. When $n$ takes all natural numbers greater than 1, among the following four expressions, the one that never takes a perfect square value is ( ).
(A) $5 n^{2}-5 n-5$
(B) $3 n^{2}-3 n+3$
(C) $9 n^{2}-9 n+9$
(D) $11 n^{2}-11 n-11$ | 3.(C).
$\because$ When $n$ is a natural number greater than 1, we have $(n-1)^{2}<n^{2}-n+1<n^{2}$,
i.e., $(3 n-3)^{2}<9 n^{2}-9 n+9<\left(3 n^{2}\right)$, $\therefore 9 n^{2}-9 n+9=(3 n-2)^{2}$ or $(3 n-1)^{2}$.
However, there are no positive integer solutions in these cases. But when $n$ is equal to $3, 2, \mathrm{~A... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,382 |
4. Given that $a$ and $b$ are non-zero constants, and $\frac{a^{2}}{\sin ^{2} \theta}+\frac{b^{2}}{\cos ^{2} \theta}=$ $(a+b)^{2}$. Then $\frac{a^{3}}{\sin ^{4} \theta}+\frac{b^{3}}{\cos ^{4} \theta}$ equals ( ).
( A ) $a^{3}+b^{3}$
(B) $(a+b)^{3}$
(C) $a^{2} b+a b^{2}$
(D) $a+b$ | 4.( B ).
Removing the denominator, we get
$$
\begin{array}{l}
a^{2} \cos ^{2} \theta+b^{2}\left(1-\cos ^{2} \theta\right) \\
=(a+b)^{2} \cos ^{2} \theta\left(1-\cos ^{2} \theta\right),
\end{array}
$$
which simplifies to $\left[(a+b) \cos ^{2} \theta-b\right]^{2}=0$.
However, $a+b \neq 0$ (otherwise, by the given equa... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,383 |
6. In $\square A B C$, points $P$ and $Q$ are on sides $A B$ and $A C$ respectively, and $\frac{B P}{A P}+\frac{C Q}{Q A}=1$. Then $P Q$ must pass through the ( ) of the triangle.
(A) orthocenter (B) centroid (C) incenter (D) circumcenter | 6.(B).
As shown in Figure 4, let the median $AD$ intersect $PQ$ at point $G$. Draw perpendiculars from $A, B, C, D$ to the line $PQ$, with the feet of the perpendiculars being $A', B', C', D'$, respectively. Then,
$$
\begin{array}{c}
1=\frac{BP}{AP_{\top}}+\frac{CQ}{QA_{4+0}} \\
=\frac{BB'+CC'}{AA'}=\frac{2DD'}{AA'}=\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,385 |
3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$ | 3.3.
Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.
From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,388 |
4. A circular clock face (with the numbers $1,2,3, \cdots$, 12 written on its circumference) is fixed at the center on a blackboard, and the clock face can rotate by any integer multiple of $30^{\circ}$. Initially, zero is written on the blackboard at each position corresponding to a number on the clock face. The clock... | (提示: 每次转动钟面后,加到黑板上所有数之和为 $1+2+\cdots+12=78$, 因此, 黑板上所有数之和为 78 的倍数, 而 $12 \times 1993$ 不是 78 的倍数.)
(提示: After each rotation of the clock face, the sum of all numbers added to the blackboard is $1+2+\cdots+12=78$, therefore, the sum of all numbers on the blackboard is a multiple of 78, and $12 \times 1993$ is not a mult... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,389 |
II. (25 points) As shown in Figure 3, $\odot A$ and $\odot B$ intersect at $C$ and $D$, and both are internally tangent to $\odot O$, with the points of tangency being $M$ and $N$. Ray $CD$ intersects $\odot O$ at $P$, $PM$ intersects $\odot A$ at $E$, and $PN$ intersects $\odot B$ at $F$. Prove that $EF$ is the common... | II. As shown in Figure 8, connect $A E$, $O P$, $B F$, $M N$, and $O M$. Clearly, $O M$ passes through point $A$.
$$
\begin{array}{l}
\because P E \cdot P M = P C \cdot P D \\
\quad = P F \cdot P N,
\end{array}
$$
$\therefore M$, $E$, $F$, and $N$ are concyclic.
$$
\begin{aligned}
\text { Hence } & \angle F E P = \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,392 |
Three, (25 points) A chemical plant, starting from January this year, if it does not improve its production environment and continues to produce as it is, will earn 700,000 yuan per month. At the same time, it will receive penalties from the environmental protection department, with the first month's penalty being 30,0... | Three, let the cumulative income for $n$ months without modifying the equipment, under the original conditions, be $a(n)$, and let the cumulative income for $n$ months after modifying the equipment be $b(n)$. From the given conditions, we have $a(n)=70 n$, and we can assume
$$
b(n)=a n^{2}+b n+c, \quad n \leqslant 5.
$... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,393 |
1. $k \in \mathbf{R}$ the real roots $x$ of the equation $x^{4}-2 k x^{2}+k^{2}+2 k-3=$ 0 should satisfy ( ).
( A ) $-1 \leqslant x \leqslant 1$
( B ) $0 \leqslant x \leqslant \sqrt{2}$
( C) $-\sqrt{2} \leqslant x \leqslant \sqrt{2}$
(D) $-\sqrt{2} \leqslant x \leqslant 0$ | -、1.( C).
Transform the original equation into an equation about $k$
$$
\begin{array}{l}
k^{2}+2\left(1-x^{2}\right) k+x^{4}-3=0, \\
\because k \in \mathbf{R}, \therefore \Delta \geqslant 0, \text { i.e., } \\
4\left(1-x^{2}\right)^{2}-4\left(x^{4}-3\right) \geqslant 0 .
\end{array}
$$
Solving this yields $-\sqrt{2} \... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,394 |
3. In $\square A B C$, condition Jia: $\angle A \cos 2 B > \cos 2 C^{\prime \prime}$. Then Jia is ( ) of Yi.
(A) necessary but not sufficient condition
(B) sufficient but not necessary condition
(C) necessary and sufficient condition
(D) neither necessary condition, nor sufficient condition | 3.( C ).
Let $a, b, c, R$ be the three sides and the circumradius of $\triangle ABC$, respectively. From $\angle A1-2 \sin ^{2} B>1-2 \sin ^{2} C$, it follows that $\cos 2 A>\cos 2 B>\cos 2 C$.
The above reasoning is reversible. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,396 |
4. Among all lines passing through the point $(\sqrt{2001}, 0)$, the number of lines passing through two different rational points (points with both coordinates being rational numbers) is ( ).
(A)0 lines
(B)infinitely many lines
(C)at least 1 line
(D)exactly 1 line | 4.(D).
Obviously, there are no rational points on the line $x=\sqrt{2001}$, so the slope of the line exists. Let the equation of the line be
$$
y=k(x-\sqrt{2001}) \text {. }
$$
If there exist two different rational points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then $k=\frac{y_{2}-y_{1}}{x_{2}-x_... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,397 |
5. The ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ has $n$ different points $P_{1}, P_{2}, \ldots, P_{n}, F$ is the right focus, $\left\{\left|P_{i} F\right|\right.$ \} forms an arithmetic sequence with a common difference $d>\frac{1}{100}$. Then the maximum value of $n$ is ( ).
(A ) Fang data $) 200$
( C) 99
( D ) 100 | 5.( B ).
$$
\begin{array}{l}
\because\left|P_{n} F\right|=\left|P_{1} F\right|+(n-1) d, \\
\therefore n=1+\frac{\left|P_{n} F\right|-\left|P_{1} F\right|}{d} .
\end{array}
$$
In an ellipse, the maximum value of the focal radius $\left|P_{i} F\right|$ is $a+c$, and the minimum value is $a-c$, so
$$
\begin{array}{l}
\le... | 200 | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,398 |
6. From the infinite sequence $\left\{\frac{1}{2^{n-1}}\right\}$, sequentially extract an infinite geometric sequence such that the sum of all its terms $S$ satisfies $\frac{4}{61}<S<\frac{1}{13}$. The number of such geometric sequences is ( ) .
(A)1
(B) 2
(C) 3
(D) infinitely many | 6.( B ).
Let the first term of the required sequence be $\frac{1}{2^{m}}$, and the common ratio be $\frac{1}{2^{n}}$, then it must be that $m \geqslant 0, n \geqslant 1$ and $m, n \in \mathbf{Z}$.
Then $S=\frac{\frac{1}{2^{m}}}{1-\frac{1}{2^{n}}}=\frac{1}{2^{m}-2^{m-n}}$.
$$
\begin{array}{c}
\because \frac{4}{61}13 \t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,399 |
5. Write the numbers 1 and 2 on the blackboard. New numbers can be added in the following way: if the numbers $a$ and $b$ are on the blackboard, then the number $ab + a + b$ can be added. Can the number 12,131 be obtained using these methods? | (Tip: Let $c+1=a b+a+b+1=(a+1)(b+1)$. If we consider adding 1 to the numbers on the blackboard, then each new number is the product of the original numbers, i.e., originally 1 and 2, we consider 2 and 3, so after several operations, it becomes $2^{n} \cdot 3^{m}$. The original number is $2^{n} \cdot 3^{m}-1$, and 12131... | 12131 \text{ does not have the form } 2^n \cdot 3^m - 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,400 |
1. The solution set of the inequality $\arccos x<\arctan x$ is
$\qquad$ | $$
\text { II. 1. }\left\{x \left\lvert\, \sqrt{\frac{\sqrt{5}-1}{2}}0$, both sides of the inequality are acute angles. The original inequality is equivalent to
$\operatorname{tan}(\arccos x)\frac{\sqrt{5}-1}{2}$ or $x^{2}\sqrt{\frac{\sqrt{5}-1}{2}}$.
Also, since $x \in(0,1]$,
$$
\therefore \sqrt{\frac{\sqrt{5}-1}{2}}<... | \sqrt{\frac{\sqrt{5}-1}{2}} < x \leqslant 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,401 |
2. $\alpha, \beta$ are the two imaginary roots of the quadratic equation $x^{2}+2 p x$ $+1=0$ with real coefficients, and the points corresponding to $\alpha, \beta, 1$ in the complex plane form an equilateral triangle. Then the real number $p=$ $\qquad$ . | 2. $\frac{1}{2}$.
Since $\alpha, \beta$ are conjugate complex numbers, we have $|\alpha|=|\beta|$. And $\alpha \beta=1$, then $|\alpha \beta|=1$, which means $|\alpha|=|\beta|=1$. Therefore, the points corresponding to $\alpha, \beta, 1$ form an equilateral triangle inscribed in the unit circle, so
$$
\begin{array}{l}... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,402 |
3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$
we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,403 |
4. In a regular $\square ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively. The eccentricity of the hyperbola with foci at $B$ and $C$ and passing through points $D$ and $E$ is $\qquad$ | 4. $\sqrt{3}+1$.
As shown in Figure 2, with $BC$ as the $x$-axis and the perpendicular bisector of $BC$ as the $y$-axis, establish a coordinate system. Suppose the equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then $C(c, 0), A(0, \sqrt{3} c)$, so $E\left(\frac{c}{2}, \frac{\sqrt{3}}{2} c\ri... | \sqrt{3}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,404 |
5. The distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite face is 4. The maximum volume of this regular tetrahedron is $\qquad$ | 5.18.
As shown in Figure 3, in the regular tetrahedron $P-ABC$, let the centroid of the lateral face $PBC$ be $G$, and extend $PG$ to intersect $BC$ at $D$. Clearly, $D$ is the midpoint of $BC$. Let the projections of $P$ and $G$ on the base be $O$ and $H$, respectively, with the base edge length being $a$ and the hei... | 5.18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,405 |
6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ . | 6.1 .
For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \f... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,406 |
$$
\text { Three. ( } 20 \text { points) Let } f(x)=x^{2}+p x+q, p, q
$$
$\in \mathbf{R}$. If the maximum value of $|f(x)|$ on $[-1,1]$ is $M$, find the minimum value of $M$. | Three, let $M=\max _{-1 \leqslant x \leqslant 1}|f(x)|$, then
$M \geqslant|f(1)|, M \geqslant|f(0)|, M \geqslant|f(-1)|$,
i.e., $M \geqslant|1+p+q|, M \geqslant|q|, M \geqslant|1-p+q|$,
thus $4 M \geqslant|1+p+q|+2|-q|+|1-p+q|$
$$
\begin{array}{l}
\geqslant|(1+p+q)+(-2 q)+(1-p+q)| \\
=2 .
\end{array}
$$
Therefore, $M ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,407 |
Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$. | $$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$
Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$
Then f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,408 |
Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.
---
Translate the above text into English, please retain the original text's line bre... | As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length form... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,409 |
一、(50 分)梯形 $A B C D$ 的两条对角线相交于 $P$,分别以两腰为直径作圆, 点 $P$ 在两圆外. 求证 : $P$ 向两圆所作的切线长相等.
---
I. (50 points) In trapezoid $ABCD$, the two diagonals intersect at $P$. Circles are constructed with the two non-parallel sides as diameters, and point $P$ is outside both circles. Prove that the lengths of the tangents drawn from $P... | As shown in Figure 5, let the two non-parallel sides of trapezoid $ABCD$ be $AB$ and $CD$, and the second intersection points of lines $AC$ and $BD$ with the respective circles be $M$ and $N$. By the power of a point theorem, to prove that the lengths of the tangents from $P$ to the two circles are equal, it suffices t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,410 |
Example 1 A city charges monthly gas fees according to the following rules: if the gas usage does not exceed 60 cubic meters, the fee is 0.80 yuan per cubic meter; if it exceeds 60 cubic meters, the excess part is charged at 1.20 yuan per cubic meter. It is known that a user's average gas fee in April was 0.88 yuan per... | Solution: Since $0.88>0.80$, the user's gas consumption in April exceeds 60 cubic meters. If the consumption is $x$ cubic meters, then
$$
60 \times 0.80 + 1.20 \times (x - 60) = 0.88 x.
$$
Solving for $x$ gives $x = 75$.
Therefore, the gas fee for April for this household is $75 \times 0.88 = 66$ yuan. Hence, the corr... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,411 |
Example 2 A class participated in a math competition, with a total of $a$, $b$, and $c$ three questions. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are worth 25 points each. After the competition, every student answered at least one question corr... | Solution: Let $x, y, z$ represent the number of correct answers for questions $a$, $b$, and $c$, respectively. Then we have
$$
\left\{\begin{array} { l }
{ x + y = 2 9 , } \\
{ x + z = 2 5 , } \\
{ y + z = 2 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=17, \\
y=12, \\
z=8 .
\end{array}\right.\... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,412 |
4. The prices of small goods $A$, $B$, $C$, and $D$ are 0.13 yuan, 0.17 yuan, 0.22 yuan, and 0.35 yuan respectively. Now, 2 yuan is used to buy exactly 10 pieces of small goods. How many of the small goods $A$, $B$, $C$, and $D$ were bought?
(1991 1992, Shanghai "Jinqiao Cup" Middle School Mathematics Knowledge Applica... | 4. Let the number of items bought of products $A, B, C, D$ be $x, y, z, w$ respectively. Then
$$
\left\{\begin{array}{l}
x+y+z+w=10, \\
13 x+17 y+22 z+35 w=200 .
\end{array}\right.
$$
Eliminating $x$, we get $y=\frac{70-9 z-22 w}{4}$.
$w$ can only be $0, 1, 2, 3$, and accordingly find the non-negative integer values o... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,413 |
2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ . | 2.5.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$.... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,414 |
3. Given $F(a, \theta)=\frac{a^{2}+2 a \sin \theta+2}{a^{2}+2 a \cos \theta+2}, a, \theta \in \mathbf{R}$. $\alpha \neq 0$. Then, for any $a, 2(a, \theta)$, the maximum and minimum values are respectively
Translate the above text into English, please retain the original text's line breaks and format, and output the tr... | $3.2+\sqrt{3}, 2-\sqrt{3}$.
Let $t=F(a, \theta)$, then
$2 ta \cos \theta - 2 a \sin \theta + (t-1)(a^2 + 2) = 0$.
Therefore, the line $2 a t x - 2 a y + (t-1)(a^2 + 2) = 0$ intersects
the circle $x^2 + y^2 = 1$ at a common point $(\cos \theta, \sin \theta)$. Thus, we have
$\frac{|t-1|(a^2 + 2)}{2|a| \sqrt{t^2 + 1}} \le... | 2 + \sqrt{3}, 2 - \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,415 |
5. Given the set $\{1,2,3, \cdots, 3 n-1,3 n\}$, it can be divided into $n$ mutually disjoint triples $|x, y, z|$, where $x+y=3 z$. The two smallest positive integers $n$ that satisfy the above requirement are $\qquad$. | $5.5,8$.
From the problem, we know the triplet is $\left\{x_{i}, y_{i}, z_{i}\right\}$, and $x_{i}+y_{i}=3 z_{i}$, $i=1,2, \cdots, n$. Therefore,
$$
\sum_{i=1}^{n}\left(x_{i}+y_{i}+z_{i}\right)=4 \sum_{i=1}^{n} z_{i},
$$
which means $4 \sum_{i=1}^{n} z_{i}=\frac{3 n(3 n+1)}{2}$.
Thus, when $2 \mid n$, $8 \mid n$, the ... | 5,8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,417 |
Three. (20 points) The cross-sectional area of a cube through a certain diagonal is $S$. Try to find the value of $\frac{S_{\text {max }}}{S_{\text {min }}}$.
untranslated text:
(20 分)过正方体的某条对角线的截面面积为 $S$. 试求 $\frac{S_{\text {顼大 }}}{S_{\text {散小 }}}$ 之值.
translated text:
(20 points) The cross-sectional area of a cu... | Three, as shown in Figure 1, let $\square E B E_{1} D_{1}$ be the section through the diagonal $B D_{1}$ of the cube. Then $S_{\square E E E_{1} D}=S=2 S_{\triangle A D_{1} E_{1}}=h B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. When $S$ is minimized, $h$ also takes its minimum value $h^{\prime}$. It is... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,419 |
Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$. | Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$
Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$
Then $a_{k+1}=3^{a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,420 |
Five. (20 points) Given $a, b, c \in \mathbf{R}^{+}$, and $a+b+c=1$. Prove:
$$
\frac{13}{27} \leqslant a^{2}+b^{2}+c^{2}+4 a b c<1 .
$$ | $$
\begin{aligned}
& \text{It is known that } a^{2}+b^{2}+c^{2}+4 a b c a^{2}+b^{2}+c^{2}+4 a b c .
\end{aligned}
$$
Now consider the lower bound of $a^{2}+b^{2}+c^{2}+4 a b c$. Without loss of generality, assume $a \geqslant b \geqslant c > 0$, then from $a+b+c=1$ we get $a+b, \frac{1}{2}, c$ form an arithmetic seque... | \frac{13}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,421 |
One, (50 points) Given $\triangle A B C$, with incenter $I$, circumcircle $\odot O$, and the antipode of point $B$ on $\odot O$ is $K$. On the extension of $A B$, take point $N$, and on the extension of $C B$, take point $M$, such that $M C = N A = s$, where $s$ is the semiperimeter of $\triangle A B C$. Prove: $I K \p... | As shown in Figure 2, extend $K I$ to intersect $\odot O$ at $G$, connect $C K, I C, B G$, and draw $I D \perp B C$ intersecting $B C$ at point $D$.
Let $\angle I K C=\alpha, \angle N I V I=\beta$, and the radius of $\odot O$ be $P$. Then
$$
\angle B C K=\angle B G K=90^{\circ}.
$$
In $\triangle I C K$, by the Law of... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,422 |
Ni, (50)M is the set of all points $(x, y)$ on the plane, where $x, y$ are integers, and $1 \leqslant x \leqslant 12, 1 \leqslant y \leqslant 13$. Prove: Every subset of $M$ with no fewer than 49 points must contain the 4 vertices of a rectangle, whose sides are parallel to the coordinate axes. | $$
\begin{array}{l}
\text { Two, let } S \text { be any } 49 \text { -element subset of } M. \text { Let } \\
X_{i}=\{x \mid(x, i) \in S\}, i=1,2, \cdots, 13, \\
\left|X_{i}\right|=x_{i}, \sum_{i=1}^{13} x_{i}=49,0 \leqslant x_{i} \leqslant 12 \text {. Then } \\
P_{i}=\{|r, s| \mid r \neq s,(r, i),(s, i) \in S\}, i=1,2... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,423 |
5. Ports $A$ and $B$ are on the south shore of a large lake, and port $C$ is on the north shore. Ports $A$, $B$, and $C$ form the vertices of an equilateral triangle. Ships from port $A$ and port $B$ start at the same time and travel in a straight line towards port $C$ at a constant speed. When ship B has traveled 40 k... | 5. As shown in the figure, let $A$, $B$, and $C$ be the three vertices of an equilateral triangle with side length $x$ kilometers. Ships from port $A$ and port $B$ start at the same time. When ship B travels 40 kilometers to point $E$, ship A reaches point $D$. When ship A reaches port $C$, ship B reaches point $F$, wh... | 30+10 \sqrt{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,424 |
Three, (50 points) The real-coefficient polynomial $f(x)=x^{3}+a x^{2}+b x$ $+c$ satisfies $b<0, a b=9 c$. Determine whether this polynomial has three distinct real roots, and explain your reasoning. | If $a=0$, then from $a b=9 c$ we get $c=0$, and we have $f(x)=$ $x^{3}+b x$. Therefore, there are three roots:
$$
x=0, x=\sqrt{-b}, x=-\sqrt{-b} .
$$
If $a>0$, then $c=\frac{a b}{9}$. Thus, there is a real number $r_{1}$ in $(0,+\infty)$ such that $f\left(r_{1}\right)=0$.
Also, $f(-a)=-a^{3}+a^{3}-a b+c=-\frac{8 a b}... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,425 |
In $\triangle ABC$, the symmetrical point of the centroid $G$ about $BC$ is $G'$. What necessary and sufficient condition should the three sides of $\triangle ABC$ satisfy so that points $A, B, G', C$ are concyclic? Prove your conclusion. | Given: $A D$, $B E$, and $C F$ are the medians of $\triangle A B C$. Since the centroid $G$ is symmetric to $P C$, we have $\angle B C C = \angle B G C$.
(1) If $A$, $B$, $G^{\prime}$, and $C$ are concyclic, then
$$
\begin{array}{c}
\angle B G^{\prime} C + \angle B A C \\
= 180^{\circ}. \\
\text{Also, } \because \angl... | A B^{2} + A C^{2} = 2 B C^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 712,426 |
Prove that the equation $3 a x^{2}+2 b x-(a+b)=0$ has at least one real root in $(0,1)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove: When $a=0$, the equation becomes $2 b x=b$. Then, regardless of the value of $b$, there is always a root $x=\frac{1}{2} \in(0,1)$.
When $a \neq 0$, let $f(x)=3 a x^{2}+2 b x-(a+b)$.
Below, find $x_{1}<x_{2}$, such that
$$
f\left(x_{1}\right) \cdot f\left(x_{2}\right) \leqslant 0 \text{. . }
$$
$$
\begin{array}{l... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,427 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.