problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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7.4. There are three piles of stones, and it is allowed to add stones to any pile, with the number of stones added must equal the sum of the number of stones in the other two piles at that time; it is also possible to remove stones from any pile when possible, with the number of stones removed equal to the sum of the n... | 7.4. Cannot.
Let the method of adding stones be $A$, and the method of reducing stones be $B$. For any given three piles of stones, let the $i$-th operation be $c_{i}(i=1,2, \cdots)$. To obtain the pile with the fewest stones, since $A$ operations are limited, there must be a case where “$c_{i}=A, c_{i+1}=B$”. In this... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,369 |
8.2. In quadrilateral $A B C D$, $E$ and $F$ are the midpoints of sides $B C$ and $D C$ respectively. Segments $A F$ and $A E$ intersect diagonal $B D$ at points $K$ and $M$ respectively. It is known that $D K=K M=M B$. Prove: Quadrilateral $A B C D$ is a parallelogram. | 8.2. Connect $M C$ and $K C$, it is easy to see that $M C // A F, K C // A E$. Therefore, quadrilateral $A K C M$ is a parallelogram. Connect the diagonal $A C$, it is easy to see that it intersects $B D$ at its midpoint. Thus, the two diagonals of quadrilateral $A B C D$ bisect each other. Therefore, quadrilateral $A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,370 |
8.3. Let a $6 \times 6$ square be divided into 9 rectangles with integer side lengths. Prove: there are 2 rectangles that are congruent. | 8.3 Write out 9 rectangles with side lengths not exceeding 6 and the smallest area:
$$
\begin{array}{lllll}
1 \times 1, & 1 \times 2, & 1 \times 3, & 1 \times 4, & 2 \times 2, \\
1 \times 5, & 1 \times 6, & 2 \times 3, & 2 \times 4 . &
\end{array}
$$
Their total area is 39. This shows that the sum of the areas of 9 no... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,371 |
8.4. Find at least one solution to the equation $\left[x^{2}\right]-[x]^{2}=1993$ and verify it. Here $[A]$ denotes the greatest integer not exceeding $A$ (for example, $\left[-\frac{5}{2}\right]=-3,[\sqrt{2}]=1$). | $\begin{array}{l}\text { 8.4. For example } x=1993.5 \text {. } \\ \text { Given }\left[x^{2}\right]-[x]^{2}=1993 \\ \Rightarrow\left[2[x] \cdot\{x\}+\left\{\left.x\right|^{2}\right]=1993 \text {. }\right. \\ \text { where }\{x\}=x-[x] .\end{array}$ | 1993.5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,372 |
6.1. Seven people are sitting around a round table, each of whom is either a truth-telling honest person or a liar who always lies. It is known that each of them says: “One of my two neighbors is an honest person, and the other is a liar.” Prove: They are all liars. | 6.1. Number 7 people in sequence. Assume there are honest people among them, let's say No. 2 is an honest person, so, one of his two neighbors is an honest person, and the other is a scoundrel. Let's assume No. 1 is an honest person. Then, No. 3 is a scoundrel. In this case, No. 4 must be an honest person, and so on, w... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 715,373 |
6.2. Write the positive integers $1,2, \cdots, 100$ in increasing order on a blackboard in a row. Each time, it is allowed to erase several consecutive numbers and write down the number of erased numbers in their place (for example, erasing $13,14,15$, and writing 3). Can it be done after several such operations, so th... | 6.2. Not possible.
The key lies in 50, 51 not being added numbers. Since it is impossible to erase 101 numbers, 51 must have been on the board initially, and all numbers to its right should have been erased, which is impossible. | Not possible | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,374 |
6.3. Several boys and 5 girls are sitting around a round table, and there are 30 pieces of bread on the plate on the table. Each girl takes 1 piece of bread from the plate for each boy she knows, and then each boy takes 1 piece of bread from the plate for each girl he does not know, at which point the bread on the plat... | 6.3. There are $n$ boys, so there are $5 n$ different “boy-girl” pairs. In each such “pair”, 1 piece of bread was taken, because if the 2 people in the pair know each other, the girl took 1 piece of bread for the boy; and if they do not know each other, the boy took 1 piece of bread for the girl. Therefore, $5 n=$ $30,... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,375 |
6.4. Does there exist 5 distinct positive integers, where the product of the 2 largest numbers equals the sum of all 5 numbers? | 6.4 does not exist.
Assume there exist such 5 positive integers $a>b>c>d>e$. Then, $b \geqslant 4$.
If $b \geqslant 5$, then we have
$$
a b \geqslant 5 a > a + b + c + d + e,
$$
which contradicts the problem statement.
Therefore, $b=4$.
Thus, $c=3, d=2, e=1$, and $4a = ab = a + b + c + d + e = a + 10$, which simplifi... | 6.4 does not exist | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,376 |
6.5. A balance has 3 pans. Each time after weighing 3 objects, the pan with the object of median weight is put down. Given 7 objects of different weights, how can you find the object with the median weight among them through 8 weighings? In each weighing, place 1 object in each pan.
In each weighing, place 1 object in... | 6.5. After one weighing, 3 objects can be arranged in increasing or decreasing order of weight (but it cannot be determined whether it is increasing or decreasing). By weighing the 4th object with the first 2 objects and the last 2 objects once each, the 4 objects can be arranged in the same increasing or decreasing or... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,377 |
6.6. There are 1993 triangles distributed on a plane, and in each triangle, there are at least 4 vertices of other triangles. Prove: There must be 3 triangles that have a common point. | 6.6. Suppose no 3 triangles have a common point. Then, any point inside a triangle can be a vertex of at most 1 triangle, and any vertex of a triangle can belong to at most 1 other triangle. Let's calculate the total number of vertices.
On the one hand, the total number is obviously no more than $3 \times 1993$;
On the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,378 |
Example 1: Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, A, B, and C, according to the signed contracts. They are to supply 45 tons to A, 75 tons to B, and 40 tons to C. Base A can arrange 60 tons, and Base B can arrange 100 tons. The distances in kilometers between A and B... | Solution: Let the base B supply $x \mathrm{t}$ to $A$, $y \mathrm{t}$ to $B$, and $[100-(x+y)] \mathrm{t}$ to $C$, then base A supplies $(45-x) \mathrm{t}$ to $A$, $(75-y) \mathrm{t}$ to $B$, and $[40-(100-x-y)] \mathrm{t}=(x+y-60) \mathrm{t}$ to $C$.
Let the total transportation cost be $W$ yuan. According to the prob... | 960 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,379 |
Example 2 As shown in Figure 2,
there are soil piles $A$ and $B$, a depression $E$, and a pond $F$ at a construction site.
The volumes of soil piles $A$ and $B$ are $781 \mathrm{~m}^{3}$ and $1584 \mathrm{~m}^{3}$, respectively. Depression $E$ requires $1025 \mathrm{~m}^{3}$ of soil to be filled, and pond $F$ can a... | Solution: Let " $\mathrm{m}^{3} \cdot \mathrm{m}$ " be the unit for the labor cost of transporting soil. Suppose the volume of soil transported from $A$ to $E$ is $x$, and the volume of soil transported to $F$ is $781-x$; the volume of soil transported from $B$ to $E$ is $1025-x$, and the volume of soil transported to ... | 207170 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,380 |
Example 3 In a cube with edge length 1, let the volume of the right prism with the midpoints of the edges of the top and bottom faces as vertices be $V_{1}$, and the volume of the right prism with the midpoints of the edges of the left and right faces as vertices be $V_{2}$. Try to find the volume $V$ of the common par... | Solution: As shown in Figure 4, let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be the given cube, and $P Q R S-P_{1} Q_{1} R_{1} S_{1}$ and $E P H P_{1}-F R G R_{1}$ be the right square prisms $V_{1}$ and $V_{2}$, respectively.
Notice that $E F$ intersects $Q Q_{1}$ at $K$, and $P, R$ are also common points of the plane $E F R... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,381 |
Four, (15 points) As shown in Figure 4, $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$, respectively. It is known that $\angle OIB = 30^{\circ}$. Prove that $\angle BAC = 60^{\circ}$. | As shown in Figure 7, connect $A I$ and extend it to intersect $\odot 0$ at $D$, then connect $B D$. Then $D$ is the midpoint of $\overparen{B C}$. Thus,
$$
\begin{array}{l}
\angle I B D \\
=\angle I B C+\angle C B D \\
=\frac{\angle A B C}{2}+\angle C A D \\
=\frac{\angle A B C}{2}+\frac{\angle B A C}{2},
\end{array}
... | 60^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 715,382 |
Five. (15 points) Given the sequence $\left\{a_{n}\right\}:$
$$
a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}(n=1,2, \cdots),
$$
where $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Prove:
(1) For any positive integer $n$, we have
$$
a_{n+2}=a_{n+1}+a_{n} \text {; }
$$
(2) All terms in the sequence $\l... | (1) Since $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$, according to Vieta's formulas, we have
$$
\begin{array}{l}
\alpha+\beta=1, \alpha \beta=-1 . \\
\text { Therefore, } \alpha^{n+2}-\beta^{n+2} \\
=(\alpha+\beta)\left(\alpha^{n+1}-\beta^{n+1}\right)-\alpha \beta\left(\alpha^{n}-\beta^{n}\right) \\
=\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,383 |
2. The monotonic decreasing interval of the function $y=\log _{\frac{1}{2}}\left(2 x-x^{2}\right)$ is ( ).
(A) $(0,2)$
(B) $[1,+\infty)$
(C) $[1,2)$
(D) $(0,1]$ | 2.D.
Determine the conclusion based on the monotonicity of composite functions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,385 |
3. Given a set $S \subseteq\{1,2, \cdots, 26\}$, and it satisfies that the sum of any 2 elements in $S$ cannot be divisible by 5. Then the maximum number of elements in the set $S$ is ( ) .
(A) 10
(B) 11
(C) 12
(D) 13 | 3.C.
The number of elements in $S=\{1,2,5,6,7$, $11,12,16,17,21,22,26$ and so on that satisfy the condition is the largest. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,386 |
5. As shown in Figure 1, in the cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, there is a point $M$ on the side face $A A^{\prime} B^{\prime} B$ such that the distances from $M$ to the two lines $A B$ and $B^{\prime} C^{\prime}$ are equal. Then, the trajectory of $M$ is ( ).
(A) a part of a parabola
(B) a ... | 5.A.
Since $B^{\prime} C^{\prime} \perp$ plane $A A^{\prime} B^{\prime} B$, it follows that $M B^{\prime} \perp B^{\prime} C^{\prime}$. Therefore, the distance from $M$ to point $B^{\prime}$ is equal to the distance from $M$ to line $A B$. Thus, the locus of $M$ is part of a parabola. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,388 |
6. Given $A(a, b), B(c, d)$, and
$$
(a-c)^{2}+(b-d)^{2} \neq 0,
$$
point $P_{n}\left(n \in \mathbf{N}_{+}\right)$ satisfies $A P_{1}=\frac{1}{2} A B, B P_{2}=$ $\frac{1}{2} B P_{1}, P_{n} P_{n+2}=\frac{1}{2} P_{n} P_{n+1}$. Then $\lim _{n \rightarrow \infty} A P_{n}$ is ( ).
(A) $\left(\frac{c-a}{3}, \frac{d-b}{3}\rig... | 6.C.
It is easy to see that the limit position of $P_{n}$ is at the point dividing segment $A B$ in the ratio $2: 1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,389 |
8. The terms of a sequence are all 3 or 5, the first term is 3, and there are $2^{k-1}$ fives between the $k$-th 3 and the $(k+1)$-th 3, i.e., $3,5,3,5,5,3,5,5,5,5,3, \cdots$. Then the sum of the first 2004 terms of this sequence $S_{2004}=$ $\qquad$ . | 8. 9998 .
The $k$-th 3 and the $2^{k-1}$ 5s following it, totaling $\left(2^{k-1}+1\right)$ terms, are considered as one group. Suppose the 2004th term is in the $k$-th group, then $k$ is the smallest positive integer satisfying
$$
k+1+2+\cdots+2^{k-1} \geqslant 2004
$$
It is easy to find that $k=11$.
Thus, $S_{2004}... | 9998 | Other | math-word-problem | Yes | Yes | cn_contest | false | 715,391 |
Example 4 Given that the base of the quadrilateral pyramid $P-ABCD$ is a parallelogram, with height $h$, a section is made through the base side $BC$ intersecting the side face $PAD$ at $EF$. If the section divides the pyramid into two parts of equal volume, find the distance from $EF$ to the base.
---
The translatio... | Solution 1: Let the distance between $EF$ and the base be $x$, then $x$ is the height of the tetrahedrons $E-ABD$ and $F-BCD$. Let $S$ be the area of the base, then
$$
V_{E-ABD}=V_{F-BCD}=\frac{1}{3} \times \frac{1}{2} S x=\frac{x S}{6}.
$$
Since the tetrahedrons $B-ADE$ and $B-DEF$ have the same height, and $\triangl... | \frac{3-\sqrt{5}}{2} h | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,392 |
9. Let the line $l: y=x+b\left(0<b<\frac{1}{2}\right)$ intersect the parabola $y^{2}=2 x$ at points $A$ and $B$, and let $O$ be the origin. Then, when the area of $\triangle A O B$ is maximized, the equation of the line $l$ is $\qquad$ | 9. $y=x+\frac{1}{3}$.
$|A B|=2 \sqrt{2} \cdot \sqrt{1-2 b}$, the distance from $0$ to $A B$ is $d=\frac{b}{\sqrt{2}}$, so $S_{\triangle \triangle B B}=b \sqrt{1-2 b}$.
When $b=\frac{1}{3}$, the maximum value is $\frac{\sqrt{3}}{9}$. | y=x+\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,393 |
10. A regular tetrahedron has three side edges of length 1, and they are pairwise perpendicular. If this tetrahedron is rotated $60^{\circ}$ around its height, then the volume of the common part of the rotated tetrahedron and the original tetrahedron is | 10. $\frac{1}{9}$.
The common part is a regular hexagonal pyramid with a base area equal to $\frac{2}{3}$ of the original base area and the height unchanged. | \frac{1}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,394 |
11. A school is building a teaching building and needs 400 pieces of glass of size A and 500 pieces of glass of size B. The store has two types of glass, A and B. It is known that each piece of type A glass costs 48 yuan and can be cut into 4 pieces of size A and 6 pieces of size B; each piece of type B glass costs 58 ... | 11.4720 .
Let the number of glass panels of specifications $A$ and $B$ to be purchased be $x$ and $y$ respectively. The constraints are
$$
\left\{\begin{array}{l}
x \geqslant 0, y \geqslant 0, \\
4 x+5 y \geqslant 400, \\
6 x+5 y \geqslant 500 .
\end{array}\right.
$$
The cost paid by the school $P=48 x+58 y$. | 14720 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,395 |
13. Given that $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies $x, y \in (-1,1)$, we have $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.
(1) The sequence $\{x_n\}$ satisfies
$$
x_1 = \frac{1}{2}, \quad x_{n+1} = \frac{2x_n}{1 + x_n^2}.
$$
Let $a_n = f(x_n)$. Find the general term formula ... | $$
\begin{array}{l}
\text { Therefore, } 1+f\left(\frac{1}{b_{1}}\right)+f\left(\frac{1}{b_{2}}\right)+\cdots+f\left(\frac{1}{b_{2000}}\right)+f\left(\frac{1}{2004}\right) \\
=1+\left[f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)\right]+\left[f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)\right]+ \\
\quad \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,397 |
14. As shown in Figure 2, points $M$ and $N$ are the midpoints of sides $AB$ and $CD$ of quadrilateral $ABCD$, respectively. $BN$ intersects $MC$ at point $P$, and $AN$ intersects $MD$ at point $Q$. Prove that: $S_{\text{quadrilateral } MNPQ} = S_{\triangle BCP} + S_{\triangle ADQ}$. | 14. From $S_{\triangle D Q}=\frac{1}{2} S_{\triangle A D B}-S_{\triangle Q N}$,
$S_{\triangle C P}=\frac{1}{2} S_{\triangle A C B}-S_{\triangle L S P}$,
we know that $S_{\triangle D Q}+S_{\triangle B C P}$
$$
=\frac{1}{2} S_{\triangle A D B}+\frac{1}{2} S_{\triangle A C B}-S_{\triangle A Q W}-S_{\triangle M R P} \tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,398 |
15. A and B take turns tossing a fair coin, and the one who tosses heads first wins. They play several rounds, and it is stipulated that the loser of the previous round tosses first in the next round. If A tosses first in the 1st round, what is the probability that A wins in the 6th round? | 15. The probability of the first player to tie winning is
$$
\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}=\frac{2}{3} \text {. }
$$
The probability of the second player to tear winning is $1-\frac{2}{3}=\frac{1}{3}$.
$\{$ Player A wins ... | \frac{364}{729} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,399 |
1. Given $a+b+c=0$, let
$$
\begin{array}{l}
M=\frac{a^{2}}{2 a^{2}+b c}+\frac{b^{2}}{2 b^{2}+a c}+\frac{c^{2}}{2 c^{2}+a b}, \\
N=\frac{b c}{2 a^{2}+b c}+\frac{a c}{2 b^{2}+a c}+\frac{a b}{2 c^{2}+a b} .
\end{array}
$$
Then the relationship between $M$ and $N$ is ().
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determi... | -1. B.
Given $a+b+c=0$, we have
$$
\begin{array}{l}
2 a^{2}+b c=2 a^{2}+b(-a-b) \\
=a^{2}-a b+a^{2}-b^{2}=(a-b)(a-c) .
\end{array}
$$
Similarly, $2 b^{2}+a c=(b-c)(b-a)$,
$$
2 c^{2}+a b=(c-a)(c-b) \text {. }
$$
Thus, $M=-\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}{(a-b)(b-c)(c-a)}$.
And $N=\frac{b c}{(a-b)(a-c)}+\frac{a ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,400 |
2. In the range of real numbers, let
$$
\begin{array}{c}
M=[\sqrt{(a+1)(|a|-2)}+\sqrt{(a+1)(2-|a|)}+ \\
\sqrt{(a-1)(|a|+2)}]^{2005} .
\end{array}
$$
Then the unit digit of $M$ is ( ).
(A) 2
(B) 3
(C) 5
(D) 1 | 2. A.
From the problem, we have
$$
\left\{\begin{array}{l}
(a+1)(|a|-2) \geqslant 0, \\
(a+1)(2-|a|) \geqslant 0 .
\end{array}\right.
$$
Therefore, $(a+1)(|a|-2)=0$.
Solving this, we get $a=-1$ or $a= \pm 2$.
Since $a=-1$ or $a=-2$ do not satisfy the conditions, we have $a=2$.
At this point, $M=2^{2005}=2 \times 2^{4... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,401 |
3. In Rt $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the altitude on the hypotenuse $A B$, and the inradii of Rt $\triangle A B C$, Rt $\triangle A C D$, and Rt $\triangle B C D$ are $r_{1}$, $r_{2}$, and $r_{3}$, respectively. Which of the following conclusions is correct? ( ).
(A) $r_{1}>r_{2}+r_{3}$
(B) $r_{1... | 3.C.
As shown in Figure 4, because
$$
\begin{array}{l}
r_{1}=\frac{A C+B C-A B}{2}, \\
r_{2}=\frac{A D+C D-A C}{2}, \\
r_{3}=\frac{B D+C D-B C}{2},
\end{array}
$$
$$
\begin{array}{l}
\text { hence } r_{3}+r_{2}-r_{1} \\
=(A B+C D)- \\
(A C+B C) .
\end{array}
$$
By the Pythagorean theorem and the area formula, we kno... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,402 |
Example 5 As shown in Figure 6, the base of the pyramid $S-ABCD$ is a parallelogram. A plane passing through vertex $A$ and the midpoint $K$ of edge $SC$ intersects edges $SB$ and $SD$ at points $M$ and $N$, respectively. Prove:
$$
\frac{1}{3} \leqslant \frac{V_{S-AMKN}}{V_{S-ABCD}} \leqslant \frac{3}{8} \text {. }
$$ | Solution: From $\frac{V_{S-A M K}+V_{S-A N K}}{V_{S-A B C}}=\frac{V_{S-A M N}+V_{S-M A K}}{V_{S-A B D}}$, we can obtain $\frac{x}{2}+\frac{y}{2}=x y+\frac{1}{2} x y$, which means
$$
y=\frac{x}{3 x-1} \text {. }
$$
Given $0<x, y \leqslant 1$, thus, $0<\frac{x}{3 x-1} \leqslant 1$.
Therefore, $x \in\left[\frac{1}{2}, 1\... | \frac{1}{3} \leqslant \frac{3 x^{2}}{4(3 x-1)} \leqslant \frac{3}{8} | Geometry | proof | Yes | Yes | cn_contest | false | 715,403 |
4. If $a$, $b$, and $c$ are all non-zero real numbers, and $a+b+c = abc = a^3$, then the minimum value of $ab + bc + ca$ is ( ).
(A) 6
(B) 8
(C) 9
(D) 13 | 4.C.
From the given conditions, we have
$$
\begin{array}{l}
b+c=a b c-a=a(b c-1)=a\left(a^{2}-1\right), \\
b c=a^{2} .
\end{array}
$$
Therefore, $b$ and $c$ are the two roots of the equation $x^{2}-\left(a^{3}-a\right) x+a^{2}=0$.
Since $a$, $b$, and $c$ are real numbers, we have
$$
\Delta=\left(a^{3}-a\right)^{2}-4 ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,404 |
5. A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits minus 1. Then the number of four-digit numbers with the abo... | 5.D.
Let the first two digits of this four-digit number be $c_{1}$, and the last two digits be $c_{2}$, then $10 \leqslant c_{1} \leqslant 99,1 \leqslant c_{2} \leqslant 99$. Thus, the original four-digit number can be represented as $100 c_{1}+c_{2}$. According to the problem, we have
$$
100 c_{1}+c_{2}=\left(c_{1}-1... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,405 |
6. As shown in Figure 1, in the right trapezoid $A B C D$, $\angle B=\angle C$ $=90^{\circ}, \angle A M C=120^{\circ}$, $A M=B C=6, C D=\sqrt{3}$. Let the circumcenters of $\triangle M C D$, $\triangle D M A$, and $\triangle A B M$ be $O_{1}$, $O_{2}$, and $O_{3}$, respectively. Then the area of $\triangle O_{1} O_{2} ... | 6.A.
As shown in Figure 5,
$$
\begin{array}{l}
\angle A M C=120^{\circ} \\
\Rightarrow \angle A M B=60^{\circ}
\end{array}
$$
$\Rightarrow$ In the right triangle $\triangle A B M$,
$$
\begin{aligned}
& \angle M A B=30^{\circ} \\
\Rightarrow & M B=3=M C \\
\Rightarrow & D M=2 \sqrt{3}
\end{aligned}
$$
$\Rightarrow$ In ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,406 |
1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is | From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$
Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,407 |
2. The sum of the sum and the product of two positive integers is 2005, and one of them is a perfect square. Then the difference between the larger number and the smaller number is | $2.1001,101$.
Let the larger of these two positive integers be $x$, and the smaller one be $y$, then
$$
x y + x + y = 2005 \text{, }
$$
which means $(x+1)(y+1)=2006$.
Notice that, the only possibilities are
$$
(x+1)(y+1)=2 \times 1003 = 17 \times 118 \text{, }
$$
Therefore, $x-y=1001$ or $x-y=101$. | 1001, 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,408 |
3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ . | 3.4 .
From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$
Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadri... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,409 |
4. As shown in Figure 3, given that the area of the equilateral $\triangle A B C$ is $S$, $M$ and $N$ are the midpoints of sides $A B$ and $A C$ respectively, $M E \perp B C$ at $E$, and $E F \perp A C$ at $F$. Let the areas of $\triangle A M N$, $\triangle B E M$, $\triangle E F C$, and quadrilateral $M E F N$ be $S_{... | 4. $\frac{11}{21}$.
As shown in Figure 6, connect $B N$, and draw the altitude $A H$. Since $\triangle A B C$ is an equilateral triangle, we know that $B N \perp A C$.
From $\triangle M B E \backsim \triangle A B H$, we get
$$
\frac{S_{2}}{S}=\frac{1}{8},
$$
which means $S_{2}=\frac{1}{8} S$.
Since $\frac{B E}{B H}=... | \frac{11}{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,410 |
One, (20 points) Renovate a pedestrian path, changing the ground originally paved with $n^{2}-96$ identical square tiles to be paved with larger identical square tiles, requiring a total of $5 n+51$ tiles to complete the paving. If $n^{2}-96$ is divisible by $5 n+51$, find the value of the positive integer $n$. | Given the problem, let $\frac{n^{2}-96}{5 n+51}=k, k$ be a positive integer, then
$$
\begin{array}{l}
n^{2}-5 k n-(51 k+96)=0 . \\
\text { Therefore, } \Delta=25 k^{2}+4(51 k+96) \\
=25 k^{2}+204 k+384 .
\end{array}
$$
Since $n$ is an integer, $\Delta$ must be a perfect square.
$$
\text { And }(5 k+19)^{2}<25 k^{2}+20... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,411 |
II. (25 points) In the right triangle $\triangle ABC$, $AD$ is the altitude on the hypotenuse $BC$, and the angle bisector of $\angle C$ intersects $AB$ and $AD$ at points $E$ and $F$, respectively. The circle passing through points $A$, $E$, and $D$ intersects $AC$ at point $H$. Prove that $HF \parallel BC$.
---
Ple... | II. As shown in Figure 7, connect $DE$ and $DH$. Since $CE$ bisects $\angle ACB$, we have $\angle 1 = \angle 2$. Therefore,
$$
\frac{AF}{FD} = \frac{AC}{DC}.
$$
Also, $\angle 1 + \angle 3 = \angle 2 + \angle 5$, and $\angle 4 = \angle 5$, so,
$$
\angle 3 = \angle 4, \quad AE = AF.
$$
Since points $A$, $E$, $D$, and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,412 |
Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.
The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ an... | Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,413 |
Example 6 Given that the distances between the three pairs of opposite edges $AB$ and $CD$, $AD$ and $BC$, $AC$ and $BD$ of tetrahedron $ABCD$ are $d_{1}$, $d_{2}$, $d_{3}$ respectively. Prove: the volume $V$ of the tetrahedron is not less than $\frac{1}{3} d_{1} d_{2} d_{3}$. | Solution: In the plane $A^{\prime} D B^{\prime} C$, construct $E F \perp C A^{\prime}, E \in A^{\prime} C, F \in D B^{\prime}$, then the length of $E F$ is no less than the distance between the plane $A^{\prime} A C_{1} C$ and the plane $D D_{1} B B^{\prime}$, i.e., $E F \geqslant d_{3}$.
Since $C A^{\prime} \geqslant ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,414 |
1. Given the function $f(x)=-2 x^{3}-x$. If $x_{1} 、 x_{2}$ 、 $x_{3} \in \mathbf{R}$ and $x_{1}+x_{2}>0, x_{2}+x_{3}>0, x_{3}+x_{1}>0$. Then $f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)(\quad)$.
(A) greater than zero
(B) less than zero
(C) equal to zero
(D) greater than zero or less than zero | $-1 . B$.
It is known that $f(x)$ is an odd function and a decreasing function on $\mathbf{R}$, so $f\left(x_{1}\right)<f\left(-x_{2}\right)$,
which means $f\left(x_{1}\right)+f\left(x_{2}\right)<0$.
Similarly, $f\left(x_{2}\right)+f\left(x_{3}\right)<0, f\left(x_{3}\right)+f\left(x_{1}\right)<0$.
Therefore, $f\left(x... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,415 |
2. In $\triangle A B C$, $a^{2}+b^{2}=7 c^{2}$. Then $(\cot A+\cot B) \tan C=(\quad)$.
(A) $\frac{1}{3}$
(B) 3
(C) -3
(D) $-\frac{1}{3}$ | 2.A.
$$
\begin{array}{l}
(\cot A+\cot B) \tan C=\left(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}\right) \frac{\sin C}{\cos C} \\
=\frac{\sin (A+B) \cdot \sin C}{\sin A \cdot \sin B \cdot \cos C}=\frac{c}{a} \cdot \frac{c}{b} \cdot \frac{2 a b}{a^{2}+b^{2}-c^{2}} \\
=\frac{2 c^{2}}{a^{2}+b^{2}-c^{2}}=\frac{1}{3} .
\end... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,416 |
3. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its major axis endpoints at $A$ and $B$. If there is a point $P$ on the ellipse $C$ such that $\angle A P B=120^{\circ}$, then the minimum value of the eccentricity of the ellipse $C$ is $($.
(A) $\frac{1}{3}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\f... | 3. D.
Let $P\left(x_{0}, y_{0}\right)$, then $\tan 120^{\circ}=\frac{2 a\left|y_{0}\right|}{x_{0}^{2}+y_{0}^{2}-a^{2}}=-\sqrt{3}$.
Substitute $x_{0}^{2}-a^{2}=-\frac{a^{2}}{b^{2}} y_{0}^{2}$ into the above equation, we get
$\left|y_{0}\right|=\frac{2 a b^{2}}{\sqrt{3}\left(a^{2}-b^{2}\right)}$.
Also, $-b \leqslant y_{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,417 |
4. Given a regular quadrilateral pyramid $S-A B C D$ with a height of 2 and a base edge length of $\sqrt{2}$, points $P$ and $Q$ lie on segments $B D$ and $S C$ respectively. The shortest distance between $P$ and $Q$ is $(\quad)$.
(A) 1
(B) $\frac{2 \sqrt{3}}{3}$
(C) $\frac{2 \sqrt{5}}{5}$
(D) $\frac{\sqrt{10}}{5}$ | 4.C.
The shortest distance between $P$ and $Q$ is the distance between the skew lines $B D$ and $S C$. Let the center of the square base $A B C D$ be $O$, then $B D \perp$ plane $S O C$. Draw $O M \perp S C$ at $M$, then $O M$ is the required distance.
Since $S O=2, O C=1$, hence $S C=\sqrt{5}$.
From $S O \cdot O C=S C... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,418 |
5. Among the natural numbers from 1 to 300, choose three different numbers such that their sum is divisible by 3. There are ( ) ways to do this.
(A) $\mathrm{C}_{1 \infty}^{3}$
(B) $\mathrm{C}_{1 \infty}^{1} \cdot \mathrm{C}_{100}^{1} \cdot \mathrm{C}_{1 \infty}^{1}$
(C) $3 C_{1 \infty}^{3}$
(D) $3 C_{1 \infty}^{3}+\ma... | 5.D.
Classify the natural numbers from 1 to 300 by modulo 3. Taking one number from each class or taking three numbers from the same class can satisfy the condition. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,419 |
6. Given that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$. Then the minimum value of the positive integer $n$ is ( ).
(A) 14
(B) 15
(C) 16
(D) 1599 | 6.B.
When $x$ is even, $x^{4} \equiv 0(\bmod 16)$;
When $x$ is odd, $x^{2} \equiv 1(\bmod 8), x^{4} \equiv 1(\bmod 16)$.
Therefore, $x_{i}^{4} \equiv 0$ or $1(\bmod 16)$.
Since $1599=15(\bmod 16)$, it follows that $n \geqslant 15$.
Also, $1599=5^{4}+12 \times 3^{4}+2 \times 1^{4}$, hence, 1599 can be expressed as the s... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,420 |
1. Let $a$ and $b$ be skew lines, and on line $a$ there are three points $A$, $B$, and $C$, such that $AB = BC$. Through $A$, $B$, and $C$, perpendiculars to line $b$ are drawn, intersecting $b$ at points $D$, $E$, and $F$ respectively. It is given that $AD = \sqrt{15}$, $BE = \frac{7}{2}$, and $CF = \sqrt{10}$. Then t... | $=1 . \sqrt{6}$.
Let the distance between the skew lines $a$ and $b$ be $x$. Construct the common perpendicular segment $MN$ of lines $a$ and $b$, with $N \in a$. Through point $M$, draw line $a' \parallel a$, and let lines $b$ and $a'$ determine plane $a$. By the given conditions, we know $MN = x$, and $AB = BC$, then... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,421 |
2. The equation $x^{2}-2 a \sin (\cos x)+a^{2}=0$ has only one solution. Then the real number $a=$ | 2.0 or $2 \sin 1$.
Let $f(x)=x^{2}-2 a \sin (\cos x)+a^{2}$, then $f(x)$ is an even function.
Thus, $f(0)=0$, that is
$$
a^{2}-2 a \sin 1=0 \text {. }
$$
Solving this, we get $a=0$ or $a=2 \sin 1$. | 2 \sin 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,422 |
3. Let $x$ and $y$ be positive numbers, and $x+y=4$. If the inequality $\frac{1}{x}+\frac{4}{y} \geqslant m$ always holds, then the maximum value of the real number $m$ is $\qquad$ | $\begin{array}{l}\text { 3. } \frac{9}{4} \\ \frac{1}{x}+\frac{4}{y}=\frac{1}{4}(x+y)\left(\frac{1}{x}+\frac{4}{y}\right) \\ =\frac{1}{4}\left(5+\frac{y}{x}+\frac{4 x}{y}\right) \\ \geqslant \frac{1}{4}\left(5+2 \sqrt{\frac{y}{x} \cdot \frac{4 x}{y}}\right)=\frac{9}{4} .\end{array}$ | \frac{9}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,423 |
4. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$ | 4.3.
Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}_{+}, a>b\right)$. Then
$$
a^{3}-98=b^{3} \text {, }
$$
i.e., $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.
Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$
... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,424 |
Example 7 Let $V_{1}$ and $V_{2}$ be the volumes of the circumscribed sphere and the inscribed sphere of a regular quadrilateral pyramid, respectively. Try to find the minimum value of $\frac{V_{1}}{V_{2}}$. | Solution: As shown in Figure 8, let the side length of the base of the regular quadrilateral pyramid be $a$, and the radii of the circumscribed sphere and the inscribed sphere be $R$ and $r$ respectively. Let $2 \theta$ be the angle between the side face and the base, then $r = \frac{a}{2} \tan \theta$. Let $PO_1 = h$,... | 7+5 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,425 |
5. There are 8 small squares on a $4 \times 4$ chessboard to be colored black, such that each row and each column has exactly 2 black squares. Then there are kinds of different coloring methods.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Note: The note above is not part of the translation but is provided to clarify th... | 5.90.
The first row can be colored with 2 black cells in $C_{4}^{2}$ ways.
After the first row is colored, there are three cases:
(1) The black cells in the second row are in the same columns as those in the first row, in which case, the remaining rows have only one way to be colored;
(2) The black cells in the second... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,426 |
6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$
and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$ | 6.0 .
Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$
Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$
That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,427 |
Three. (20 points) Let $1<x_{1}<2$, for $n=1,2$,
$\cdots$, define $x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2}$. When $m \geqslant 3$, prove: $\sum_{i=3}^{m}\left|x_{i}-\sqrt{2}\right|<\frac{1}{4}$. | When $n=2$,
$$
\begin{array}{l}
x_{3}=1+x_{2}-\frac{1}{2} x_{2}^{2}=\frac{3}{2}-\frac{1}{2}\left(x_{2}-1\right)^{2} \\
=\frac{3}{2}-\frac{1}{2}\left(x_{1}-\frac{1}{2} x_{1}^{2}\right)^{2} .
\end{array}
$$
Since the quadratic function $f(x)=-\frac{1}{2} x^{2}+x$ is monotonically decreasing on $(1,2)$, therefore,
$$
\fr... | \frac{1}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 715,428 |
Four. (20 points) Given points $A$ and $B$ are the upper and lower vertices of the ellipse $\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$, and $P$ is a point on the hyperbola $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ in the first quadrant. The lines $PA$ and $PB$ intersect the ellipse at points $C$ and $D$, respe... | (1) Let $P\left(x_{0}, y_{0}\right), A(0, a), B(0,-a)$. Then $\frac{y_{0}^{2}}{a^{2}}-\frac{x_{0}^{2}}{b^{2}}=1$.
Also, the midpoint $D\left(\frac{x_{0}}{2}, \frac{y_{0}-a}{2}\right)$ of $P B$ lies on the ellipse, so we have $\frac{\left(y_{0}-a\right)^{2}}{4 a^{2}}+\frac{x_{0}^{2}}{4 b^{2}}=1$.
From equations (1) an... | \frac{\sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,429 |
Five. (20 points) Given that $a$, $b$, and $c$ are real numbers, and
$$
\sqrt{2 x^{2} + a x + b} > x - c
$$
holds if and only if $x \leqslant 0$ or $x > 1$.
Find the range of values for $c$. | From the problem, we get the inequality
$$
\sqrt{2 x^{2}+a x+b}>x-c
$$
The solution set is $x \leqslant 0$ or $x>1$, and equation (1) is equivalent to
$$
\left\{\begin{array}{l}
2 x^{2}+a x+b \geqslant 0, \\
2 x^{2}+a x+b>(x-c)^{2}
\end{array}\right.
$$
or $\left\{\begin{array}{l}2 x^{2}+a x+b \geqslant 0, \\ x-c(x-c... | 0<c \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,430 |
One, (50 points) As shown in Figure 1, in the right trapezoid $A B C D$, $E$ is the midpoint of $A B$, and $\angle C E D=90^{\circ}$. Let the incenter of $\triangle D A E$, $\triangle C B E$, and $\triangle C E D$ be $O_{1}$, $O_{2}$, and $O_{3}$, respectively. Prove: Quadrilateral $\mathrm{EO}_{1} \mathrm{O}_{3} \math... | As shown in Figure 2, let $CE$ intersect the extension of $DA$ at $F$. Then
$\triangle BEC \cong$
$\triangle AEF$.
Thus, $DE$ is the perpendicular bisector of $CF$.
Therefore, $\angle EDA = \angle EDC$.
Hence, Rt $\triangle ADE \sim$ Rt $\triangle EDC$.
Thus, $\angle O_1DO_3 = \angle ADE$, $\frac{DO_1}{AD} = \frac{DO_3... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,431 |
II. (50 points) Given positive integers $x_{1}<x_{2}<\cdots<x_{n}$, and
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=2003, n \geqslant 2, \\
f(n)=n\left(x_{1}+x_{n}\right) .
\end{array}
$$
Find the minimum value of $f(n)$. | If $x_{1}>1$, let $x_{1}^{\prime}=1, x_{n}^{\prime}=x_{n}+x_{1}-1$, and the other $x_{i}$ values remain unchanged, then
$$
\begin{array}{l}
x_{1}^{\prime}+x_{2}+x_{3}+\cdots+x_{n-1}+x_{n}^{\prime}=2003, \\
f(n)=n\left(x_{1}+x_{n}\right)=n\left(1+x_{n}^{\prime}\right) .
\end{array}
$$
Assume $x_{1}=1$.
When $n=2$, $f(2... | 2254 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,432 |
Three, (50 points) There are 12 football teams participating in a match, with each pair of teams playing one game. The winner gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. How many points does one team need to score at a minimum to ensure that no more than 6 teams score at l... | Three, assuming there are 7 teams with the same score as a certain team, then the maximum score for each match between these 8 teams is 3 points. Therefore, their maximum total score is $3 C_{8}^{2}=84$ points, and the score of each team cannot exceed $\left[\frac{84}{8}\right]=10$ points, so each team can win at most ... | 23 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,433 |
Find the smallest positive integer $a$ that satisfies the following condition: there exists a positive odd number $n$, such that $203^{n} + a \cdot 198^{n}$ is a multiple of 2005. | Solution: Since $2005=401 \times 5$, and
$$
203+198=401,203-198=5 \text {, }
$$
Therefore,
$$
203^{n}+a \cdot 198^{n}=(401-198)^{n}+a \cdot 198^{n}
$$
is a multiple of $401 \times 5$.
Since $n$ is odd, then
$$
-198^{n}+a \cdot 198^{n}=(a-1) \cdot 198^{n}
$$
is a multiple of 401.
Since the greatest common divisor of ... | 1204 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,434 |
In $\triangle A B C$, $\angle A C B=90^{\circ}, A C=B C$, $D$ is a point on side $A B$, and the perpendicular bisector of line segment $C D$ intersects sides $A C$ and $B C$ at points $M$ and $N$, respectively. If $A D=a, B D=b$ (where $a, b$ are given positive integers), try to find the lengths of $C M$ and $C N$ (exp... | Solution: As shown in Figure 2, it is clear that
$A C=B C$
$=\frac{a+b}{\sqrt{2}}$,
$\angle C A B=45^{\circ}$.
By the cosine rule,
$C D^{2}=\left(\frac{a+b}{\sqrt{2}}\right)^{2}+a^{2}-2 \cdot \frac{a+b}{\sqrt{2}} \cdot a \cos 45^{\circ}$
$=\frac{a^{2}+b^{2}}{2}$.
Therefore, $C D=\sqrt{\frac{a^{2}+b^{2}}{2}}$.
Let $C D$... | C M=\frac{a^{2}+b^{2}}{2 \sqrt{2} b}, C N=\frac{a^{2}+b^{2}}{2 \sqrt{2} a}, \sqrt{2}-1 \leqslant \frac{b}{a} \leqslant \sqrt{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,435 |
In a convex quadrilateral $ABCD$, $\angle ABD=$
$$
\begin{array}{l}
\angle CBD=\angle ADC=45^{\circ}, AB=a, BC=b, CD=c, DA \\
=d(a \neq b). \text{ Determine the range of } \frac{a^{2}-b^{2}}{c^{2}-d^{2}}.
\end{array}
$$ | Solution: As shown in Figure 3, connect $AC$, take the circumcenter $O$ of $\triangle ACD$, and connect $OA$, $OB$, $OC$.
Since $\angle AOC = 2 \angle ADC = 90^{\circ}$,
and $\angle ABC = \angle ABD + \angle CBD = 90^{\circ}$, then
$$
\begin{array}{l}
\angle AOC + \angle ABC \\
= 180^{\circ}.
\end{array}
$$
Thus, $O... | 1 < \frac{a^2 - b^2}{c^2 - d^2} < \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,437 |
152 (1) Find the range of positive real numbers $\lambda$ such that the inequality
$$
\left(\frac{a}{a+\lambda 8}\right)^{2}+\left(\frac{b}{b+\lambda a}\right)^{2} \geqslant \frac{2}{(1+\lambda)^{2}}
$$
holds for any positive real numbers $a$ and $b$;
(2) Find the range of positive real numbers $\lambda$ such that the... | Let $x_{1}=\frac{b}{a}, x_{2}=\frac{a}{b}$, then $x_{1}, x_{2}>0$, and $x_{1} x_{2}=1$. Therefore, equation (1) is equivalent to
$$
\begin{array}{l}
\frac{1}{\left(1+\lambda x_{1}\right)^{2}}+\frac{1}{\left(1+\lambda x_{2}\right)^{2}} \geqslant \frac{2}{(1+\lambda)^{2}} \\
\Leftrightarrow(1+\lambda)^{2}\left[\left(1+\l... | \lambda \geqslant \frac{1}{2} \text{ for (1), } 0 < \lambda \leqslant \sqrt{2}-1 \text{ for (2)} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,438 |
Example 9: Prove that for any tetrahedron $A_{1} A_{2} A_{3} A_{4}$, the heights $h_{i}$ and the radii $r_{i}$ of the exspheres $(i=1,2,3,4)$, the following holds:
$$
2 \sum_{i=1}^{4} \frac{1}{h_{i}}=\sum_{i=1}^{4} \frac{1}{r_{i}} .
$$ | Let the volume of the tetrahedron be $V$, and $O_{i}$ and $S_{i}$ be the center of the excircle opposite to point $A_{i}$ and the area of the opposite face, respectively. As shown in Figure 11. Then,
$$
\begin{array}{l}
V=V_{O_{1}-A_{1} A_{3} A_{4}}+ \\
V_{O_{1}-A_{1} A_{2} A_{4}}+ \\
V_{O_{1}-A_{1} A_{2} A_{3}}-V_{O_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,439 |
1. The volume of the solid of revolution obtained by rotating the closed figure formed by the function $y=\sqrt{b^{2}-\frac{b^{2} x^{2}}{a^{2}}}$ and the $x$-axis around the $y$-axis once is $\qquad$ . | (Tip: Similar to the derivation of the volume of a sphere, according to Zu's principle, the volume of this solid of revolution can be regarded as the volume of a cylinder (base radius $a$, height $b$) minus a cone (base radius $a$, height $b$). Answer: $\frac{2}{3} \pi a^{2} b$.) | \frac{2}{3} \pi a^{2} b | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 715,440 |
2. One edge of a tetrahedron is of length $x$, and the other edges are of length 1. If the volume $V$ of the tetrahedron is expressed as a function $f(x)$ of $x$, find the monotonic intervals of the function $f(x)$. | (In the tetrahedron $ABCD$, let the edge $CD=x$, and the other edges are all 1. Let the distance from $D$ to the plane $ABC$ be $H$, then $V=\frac{1}{3} S_{\triangle B C} \cdot H=\frac{\sqrt{3}}{12} H$. Therefore, the monotonic intervals of the functions $V(x)$ and $H(x)$ are the same. Imagine the face $ABD$ rotating a... | \left(0, \frac{\sqrt{6}}{2}\right] \text{ is the increasing interval; } \left[\frac{\sqrt{6}}{2}, \sqrt{3}\right) \text{ is the decreasing interval.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,441 |
3. Prove: For any tetrahedron, the inequality $r < \frac{ab}{2(a+b)}$ holds, where $a$ and $b$ are the lengths of a pair of opposite edges, and $r$ is the radius of its inscribed sphere. | (It is known that $r=\frac{3 V}{S}$. And $V=\frac{1}{6} a b d \sin \theta$, thus, $r \leqslant \frac{1}{2} \cdot \frac{a b d}{S}$, where $S$ is the surface area of the tetrahedron, $d$ and $\theta$ are the distance and the angle between opposite edges, respectively. Compared with the required inequality, it is only nec... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,442 |
Example 3 If a store sells a certain product, which costs 100 yuan, at 120 yuan, it can sell 300 units. If the price of the product is increased by 1 yuan based on 120 yuan, it will sell 10 fewer units, and if the price is reduced by 1 yuan, it will sell 30 more units. Question: To maximize profit, what price should th... | Solution: (1) If sold at 120 yuan each, 300 can be sold, and the profit is $300 \times 20 = 6000$ (yuan).
(2) If the price is increased by $x (x > 0)$ yuan, then (300 $- 10x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 + x)(300 - 10x) \\
= -10x^2 + 100x + 6000 \\
= -10(x - 5)^2 + 6250 .
\... | 115 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,443 |
4. In a regular tetrahedron with each edge length of 1, there are 13 points, where no three points are collinear and no four points are coplanar. Prove: there must be a tetrahedron with four of these points as vertices, whose volume \( V < \frac{\sqrt{2}}{48} \). | (Tip: Let the regular tetrahedron with edge length 1 be $A-BCD$, and $AO$ is its height. Take a point $O_{1}$ on $AO$ such that $O_{1}A=O_{1}B=O_{1}C=O_{1}D=a$. The four tetrahedra with $O_{1}$ as the vertex and the four faces of $A-BCD$ as the bases are obviously equal in volume, and $V=\frac{1}{4} \cdot \frac{\sqrt{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,444 |
5. The area of a triangle is uniquely determined by the lengths of its three sides. Is the volume of a tetrahedron uniquely determined by the areas of its four faces? | (Construct a tetrahedron $ABCD$ such that $AB=CD$, $AC=AD=BD=CB$. Take the midpoint $E$ of $CD$, and the midpoint $G$ of $AB$, let $BE=a$, $\angle EBC=\theta$, then $EG=a \sqrt{1-\tan ^{2} \theta}$, $CD=AB=2 \tan \theta$. Therefore, $V=\frac{1}{3} S_{\triangle ABE} \cdot CD=\frac{2}{3} \cdot \sqrt{S^{3} \tan \theta\lef... | V \text{ depends on } \theta | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,445 |
4. In a rectangular flower bed, install two identical sprinklers so that the entire flower bed can be watered. It is known that the watering area of each sprinkler is a circle with a radius of $10 \mathrm{~m}$. How should it be designed (find the distance between the two sprinklers and the length and width of the recta... | (The answer: The sides of the rectangular flower bed are designed to be $10 \sqrt{2} \mathrm{~m}$ and $20 \sqrt{2} \mathrm{~m}$, at which point the area of the flower bed is maximized.) | 10 \sqrt{2} \mathrm{~m} \text{ and } 20 \sqrt{2} \mathrm{~m} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,446 |
Example 1 As shown in Figure $2, \odot O_{1} 、 \odot O_{2}$ are externally separated. $A B 、 C D$ are the external and internal common tangents of the two circles, respectively, with $A 、 B 、 C 、 D$ being the points of tangency. Prove: The three lines $A C 、 B D 、 O_{1} O_{2}$ are concurrent. | Let $A C$ intersect $O_{1} O_{2}$ at $P$. Line $C D$ intersects $A B$ at $E$, and connect $O_{1} A$, $O_{1} E$, $O_{2} E$. Then $O_{1} A \perp A B$. It is easy to prove that $O_{1} E \perp A C$, with the foot of the perpendicular being $F$, and $E O_{1}$ bisects $\angle A E C$, $E O_{2}$ bisects $\angle B E D$. Thus, w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,448 |
Example 2 As shown in Figure 3, the convex hexagon $A B C D E F$ is inscribed in a circle. It is known that $\frac{A B}{B D}=\frac{D E}{E A}$, $\frac{A C}{C D}=\frac{D F}{F A}$. Prove: $A D$, $B E$, and $C F$ are concurrent. | Let $A D$ intersect $B E$ at $P$, it is easy to know
$$
\begin{array}{l}
\triangle A B P \backsim \triangle E D P, \\
\triangle B D P \backsim \triangle A E P . \\
\text { By the given } \frac{A B}{B D}=\frac{D E}{E A}, \text { we get } \\
\frac{A B}{D E}=\frac{B D}{E A} . \\
\text { And } \frac{A B}{D E}=\frac{A P}{E ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,449 |
Example 3 As shown in Figure 4, circle $O$ is separated from line $l$, $OM \perp l$ at $M$. Points $A_{1}$ and $A_{2}$ are on $l$, and through $A_{1}$ and $A_{2}$, the tangents $A_{1} B_{1}$, $A_{1} C_{1}$, $A_{2} B_{2}$, and $A_{2} C_{2}$ are drawn to circle $O$, with $B_{1}$, $C_{1}$, $B_{2}$, and $C_{2}$ being the p... | Analysis: Let the radius of $\odot O$ be $R$, and connect $O B_{1}$, $O C_{1}$, $O A_{1}$, $O M$, $O A_{1}$, which intersect $B_{1} C_{1}$ at $N$ and $E$ respectively.
It is easy to prove that $O A_{1} \perp B_{1} C_{1}$, and the intersection point $E_{1}$ is the foot of the perpendicular.
Since $\angle A_{1} M N + \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,450 |
In a convex quadrilateral $ABCD$, diagonal $BD$ is neither the angle bisector of $\angle ABC$ nor $\angle CDA$. Point $P$ is inside quadrilateral $ABCD$ and satisfies $\angle PBC = \angle DBA$ and $\angle PDC = \angle BDA$. Prove that the sufficient and necessary condition for quadrilateral $ABCD$ to be a cyclic quadri... | Prove: As shown in Figure 1, let the circumcircle of $\triangle BCD$ be $\odot O$. Extend $DP$ and $BP$ to intersect $\odot O$ at $N$ and $M$ respectively. Connect $DM$, $MC$, $MN$, $CN$, $BN$, and denote $\angle NMC = \angle NDC = \beta$, $\angle CNM = \angle CBM = \alpha$, $\angle BDN = \theta$. Then $\triangle ABD \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,451 |
Question 1 As shown in Figure 1, given three fixed points on a line in sequence as $A, B, C$, and $\Gamma$ is a circle passing through $A, C$ with its center not on $AC$. The tangents to circle $\Gamma$ at points $A, C$ intersect at point $P$, and $PB$ intersects circle $\Gamma$ at point $Q$. Prove: The intersection po... | Proof: Extend $PB$ to intersect the circle $\Gamma$ at $M$, connect $AM$ and $CM$. Let the angle bisector of $\angle AQC$ intersect $AC$ at $R$ and the circle $\Gamma$ at $S$.
Since $AP$ and $CP$ are tangents to the circle $\Gamma$, we have
$\triangle PAQ \backsim \triangle PMA, \triangle PCQ \backsim \triangle PMC$.
T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,452 |
For any integer $n(n \geqslant 4)$, find the smallest integer $f(n)$, such that for any integer $m$, any $f(n)$-element subset of the set $\{m$, $m+1, \cdots, m+n-1\}$ contains at least 3 pairwise coprime elements. | Proof: First, we present three lemmas.
Lemma 1 In any sequence of 6 consecutive integers, there are exactly 4 numbers that can be divided by 2 or 3.
Lemma 2 Any three numbers of the form $2m-1, 2m, 2m+1$ are pairwise coprime.
Lemma 3 In any sequence of 6 consecutive positive integers, any selection of 5 positive inte... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,453 |
Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how ma... | Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,454 |
2. The number of positive roots of the equation $2 x-x^{2}=\frac{2}{x}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2.A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,456 |
3. In the Cartesian coordinate system, it is known that point $A(-8, 3)$, $B(-4, 5)$, and the moving points $C(0, n)$, $D(m, 0)$. Then, when the perimeter of quadrilateral $ABCD$ is minimized, the ratio $\frac{m}{n}$ is ( ).
(A) $-\frac{2}{3}$
(B) -2
(C) $-\frac{3}{2}$
(D) -3 | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,457 |
4. Let the three sides of a triangle be positive integers $a, n, b$, where $b \leqslant n \leqslant a$. Then for a given side length $n$, the number of such triangles is ( ).
(A) $n$
(B) $n+1$
(C) $n^{2}+n$
(D) $\frac{1}{2} n(n+1)$ | 4.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,458 |
5. Four people, A, B, C, and D, are shooting at a target. Each person shoots 4 times, and the product of the points each person hits is 72 (the highest points for hitting the target is 10), and the total points hit by the four people are 4 consecutive integers. Then, the number of people who hit 4 points is ( ).
(A) 0
... | 5.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. The text "5.C" does not require translation as it is already in a form that is the same in both Chinese and ... | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,459 |
10. The use of passwords is extremely important in modern society. There is a plaintext (real text) in which the letters correspond to the 26 natural numbers $1,2, \cdots, 26$ according to the computer keyboard order (from left to right, from top to bottom) (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|... | 10. $\mathrm{CHQ}$ | CHQ | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,464 |
Example 5 A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment, after the equipment is put into use, if the maintenance and repair costs are averaged to each day, the conclusion is: the maintenance and repair cost on the $x$-th day is $\left[\... | Solution: (1) If the equipment is put into use for $x$ days, and the average daily loss is $y$ yuan, then
$$
\begin{aligned}
y= & \left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)+\right. \\
& \left.\cdots+\left(\frac{x-1}{4}+500\right)\right] \div x \\
= & \frac{1}{x}\left[500000... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,465 |
12. As shown in Figure 2, in $\triangle A B C$, $B C=a, A C=$ $b, A B=c$. If the medians $B E$ and $A D$ from $A C$ and $B C$ are perpendicular to each other at point $O$, then $c$ can be expressed as an algebraic expression in terms of $a$ and $b$ | 12. $c=\frac{1}{5} \sqrt{5\left(a^{2}+b^{2}\right)}$ | c=\frac{1}{5} \sqrt{5\left(a^{2}+b^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,467 |
15. For a real number $a$, there is only one real value of $x$ that satisfies the equation
$$
\frac{x+1}{x-1}+\frac{x-1}{x+1}+\frac{2 x+a+2}{x^{2}-1}=0 \text {. }
$$
Find the sum of all such real numbers $a$. | Three, 15. The equation can be transformed into
$$
2 x^{2}+2 x+a+4=0 .
$$
When equation (1) has two equal real roots, we have
$$
\Delta=4-4 \times 2 \times(a+4)=0 \text {. }
$$
From this, we get $a_{1}=-\frac{7}{2}$.
At this point, (1) has one root $x=-\frac{1}{2}$.
Verification shows that $x=-\frac{1}{2}$ satisfies ... | -\frac{31}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,470 |
16. Several workers load and unload a batch of goods, with each worker having the same loading and unloading speed. If these workers work simultaneously, it will take $10 \mathrm{~h}$ to complete the loading and unloading. Now, the loading and unloading method is changed, starting with one person working, and then addi... | 16. (1) Let the loading and unloading work take $x \mathrm{~h}$ to complete, then the first person worked for $x \mathrm{~h}$, and the last person worked for $\frac{x}{4} \mathrm{~h}$. The total work time of the two people is $\left(x+\frac{x}{4}\right) \mathrm{h}$, and the average work time per person is $\frac{1}{2}\... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,471 |
17. Among the following 4 statements:
(1) Two triangles are congruent if they have two sides and the altitude on one of these sides equal;
(2) Two triangles are congruent if they have two sides and the altitude on the third side equal;
(3) Two triangles are congruent if 5 out of the 6 elements (sides and angles) are eq... | 17. Determine that (1), (2), (3), and (4) are all incorrect.
Counterexample for (1):
As shown in Figure 4, in $\triangle ABC$ and $\triangle AB'C$, $AC = AC$, $BC = B'C$, and the altitude $AH = AH$, but the two triangles are not congruent.
Counterexample for (2):
As shown in Figure 5, in $\triangle ABC$ and $\triangle... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,472 |
18. 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, $\cdots \cdots$, and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest ... | 18.9 judges cannot give 1 point to 5 or more athletes, because among 5 or more athletes, at least one athlete must be rated no less than 5 by a judge, while according to the problem, each of these 5 athletes is rated no more than 4 by each judge, which is a contradiction. Therefore, 9 judges can give 1 point to at most... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,473 |
$1 . A$ City, B City, and $C$ City have 10, 10, and 8 units of a certain machine, respectively. Now it is decided to allocate these machines to $D$ City (18 units) and $E$ City (10 units). It is known that the cost to transport one unit from $A$ City to $D$ City and $E$ City is 200 yuan and 800 yuan, respectively; from... | (1) When $x=9$, $W$ reaches its minimum value of 10000 yuan; when $x=5$, $W$ reaches its maximum value of 13200 yuan. (2) When $x=10$, $y=8$, $W$ reaches its minimum value of 9800; when $x=0$, $y=10$, $W$ reaches its maximum value of 14200 yuan. | 10000 \text{ yuan}, 13200 \text{ yuan}, 9800 \text{ yuan}, 14200 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,476 |
3. Let $a, b, c, d$ all be non-zero real numbers. Then $-ab$, $ac$, $bd$, $cd$ ( ).
(A) are all positive
(B) are all negative
(C) are two positive and two negative
(D) are one positive and three negative or one negative and three positive | 3. 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 | 715,477 |
6. As shown in Figure 2, in $\triangle A B C$, $\angle A C B=90^{\circ}$, $\angle A=20^{\circ}$. $\triangle A B C$ is rotated counterclockwise around point $C$ by an angle $\alpha$ to the position of $\triangle A^{\prime} B^{\prime} C$, where $A^{\prime}$ and $B^{\prime}$ are the corresponding points of $A$ and $B$ res... | 6.D
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 | 715,480 |
7. If $x_{0}$ is a root of the quadratic equation $a x^{2}+b x+c=$ $0(a \neq 0)$, then the relationship between the discriminant $\Delta=b^{2}-4 a c$ and the quadratic form $M=\left(2 a x_{0}+b\right)^{2}$ is ( ).
(A) $\Delta>M$
(B) $\Delta=M$
(C) $\Delta<M$
(D) Cannot be determined | 7. 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 | 715,481 |
8. In $\triangle A B C$, $a$, $b$, $c$ are the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively. If $\angle B=60^{\circ}$, then the value of $\frac{c}{a+b}+$ $\frac{a}{c+b}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$ | 8. C
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 | 715,482 |
9. If $x_{1} , x_{2}$ both satisfy the condition
$$
|2 x-1|+|2 x+3|=4 \text{, and } x_{1}<x_{2} \text{, }
$$
then the range of $x_{1}-x_{2}$ is $\qquad$ | 9. $-2 \leqslant x_{1}-x_{2}<0$ | -2 \leqslant x_{1}-x_{2}<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,483 |
13. A and B are two car retailers (hereinafter referred to as A and B) who ordered a batch of cars from a car manufacturer. Initially, the number of cars A ordered was 3 times the number of cars B ordered. Later, due to some reason, A transferred 6 cars from its order to B. When picking up the cars, the manufacturer pr... | Three, 13. Let the total number of cars finally purchased by A and B be $x$ units. In the 6 fewer cars supplied by the factory, A requested $y$ fewer cars $(0 \leqslant y \leqslant 6)$, and B requested $(6-y)$ fewer cars. Then we have
$$
\begin{array}{l}
\frac{3}{4}(x+6)-6-y \\
=2\left[\frac{1}{4}(x+6)+6-(6-y)\right] .... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,488 |
15. From $1,2, \cdots, 2004$, choose $k$ numbers such that among the chosen $k$ numbers, there are definitely 3 numbers that can form the side lengths of a triangle (here it is required that the three side lengths of the triangle are all different). What is the minimum value of $k$ that satisfies the condition? | 15. This problem is equivalent to:
Selecting $k-1$ numbers from $1,2, \cdots, 2004$, such that no three of these numbers can form the lengths of the sides of a triangle (all sides being unequal). What is the maximum value of $k$ that satisfies this condition?
For any array that meets the above conditions,
when $k=4$, ... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,490 |
1. If $x+y=1, x^{2}+y^{2}=3$, then, $x^{3}+$ $y^{3}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | $$
-1 . C
$$
From $2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=-2$, we have $x y=-1$.
Then $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=4$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,491 |
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