problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. Let non-zero complex numbers $z_{1}, z_{2}$ satisfy $z_{1}^{2}-z_{1} z_{2}+x_{2}^{2}=0$, and construct a circle $\odot C$ with the origin as the center and $R=\left|z_{1}-x_{2}\right|$ as the radius. Then the relationship between $z_{1}, z_{2}$ and $\odot C$ is ( ).
(A) Both points are outside the circle
(B) Both po... | 4. C
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 | 707,606 |
6. Arrange all natural numbers that are coprime with 70 in ascending order, forming a sequence $\left\{a_{n}\right\}, n \geqslant 1$. Then $a_{200}$ equals ( ).
(A) 568
(B) 583
(C) 638
(D) 653 | 6. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,608 |
2. Let $f(x)=\frac{1}{1+a^{x}}-\frac{1}{2}, a>0, a \neq 1$, and $[m]$ denotes the greatest integer not exceeding $m$. Then the range of $[f(x)]+[f(-x)]$ is $\qquad$ | 2. $\{-1,0\}$ | \{-1,0\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,610 |
3. In the regular triangular frustum $A^{\prime} B^{\prime} C^{\prime}-A B C$, $A^{\prime} B^{\prime} : A B = 5 : 7$, the sections $A B C^{\prime}$ and $A^{\prime} B C^{\prime}$ divide the frustum into three pyramids $C^{\prime}-A B C$, $B-A^{\prime} B^{\prime} C^{\prime}$, and $C^{\prime}-A A B A^{\prime}$. The volume... | 3. $49: 25: 35$ | 49: 25: 35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,611 |
Example 8. For a regular $n$-sided polygon $(n \geqslant 5)$ centered at point $O$, let two adjacent points be $A, B$. $\triangle X Y Z$ is congruent to $\triangle O A B$. Initially, let $\triangle X Y Z$ coincide with $\triangle O A B$, then move $\triangle X Y Z$ on the plane such that points $Y$ and $Z$ both move al... | Solve: Since $\angle Y X Z+$
$$
\begin{array}{l}
\angle Y B Z=\angle A O B+ \\
\angle A B C=\frac{n-2}{n} \pi+\frac{2}{n} \pi \\
=\pi,
\end{array}
$$
$\therefore X, Y, B, Z$ are concyclic, thus $\angle X B Y=$ $\angle X Z Y=\angle O B Y$, so $X$ lies on the extension of $B O$.
$$
\begin{aligned}
\because S_{\triangle X... | d=\frac{a\left(1-\cos \frac{\pi}{n}\right)}{\sin ^{2} \frac{\pi}{n}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,614 |
6. Let $a>1, m>p>0$. If the solution to the equation $x+\log _{a} x=m$ is $p$. Then the solution to the equation $x+a^{x}=m$ is | 6. $-p+m$ | -p+m | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,615 |
Three, (Full marks 20 points) On the circle $C_{:}(x-1)^{2}+y^{2}=2$ there are two moving points $A$ and $B$, and they satisfy the condition $\angle A) B=90^{\circ}(O$ being the origin). Find the equation of the trajectory of point $P$ of the parallelogram $O A P B$ with $O A, O B$ as adjacent sides. | Three, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), P(x, y)$. According to the problem, we have
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+y_{1}^{2}=2, \\
\left(x_{2}-1\right)^{2}+y_{2}^{2}=2 .
\end{array}
$$
By adding (1) and (2), we get
$$
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2\left(x_{1}+x_{2}\... | (x-1)^{2}+y^{2}=(\sqrt{3})^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,616 |
Connect it with the opposite side of the triangle, obtaining 4 line segments. Prove that these 4 lines intersect at one point.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text{Given: } M \text{ is the midpoint of } CD, \text{ connect } AM, BM, \text{ where } P, Q \text{ are points that divide } BM \text{ and } AM \text{ in the ratio } 2:1. \\
\text{Thus, } P \text{ and } Q \text{ are the centroids of } \triangle BCD \text{ and } \triangle ACD, \text{ respectively. C... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,617 |
Record, (Full marks 20 points) Let the complex number $z=x+i y(x, y \in$ $R)$ satisfy $|z-i| \leqslant 1$. Find the maximum and minimum values of $A=x\left(|z-i|^{2}-1\right)$ and the corresponding $z$ values. | Let $z-i=x+(y-1) i=r(\cos \theta+i \sin \theta)$, then
$$
|z-i|=r, 0 \leqslant r \leqslant 1 \text {. }
$$
So $x=r \cos \theta$.
Then $A=r \cos \theta\left(r^{2}-1\right)=r\left(r^{2}-1\right) \cdot \cos \theta$,
$$
-r\left(1-r^{2}\right) \leqslant A \leqslant r\left(1-r^{2}\right),
$$
Thus $A^{2} \leqslant \frac{1}{... | A_{\text {max }}=\frac{2 \sqrt{3}}{9}, \, A_{\text {min}}=-\frac{2 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,618 |
(1) (This question is worth 35 points) Let the set $A=\{1,2,3, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17|(a $+b$), then $B$ is said to have property $P$.
(1) Find the number of all binary subsets of $A$ that have property $P$;
(2) Find the number of all pairwise disjoint binary subsets of $A$ ... | (1) $a+b$ is divisible by 17 if and only if $a+b \equiv 0 \pmod{17}$, i.e., $\square$
$$
\begin{array}{l}
a \equiv k \pmod{17}, b \equiv 17-k \pmod{17}, \\
k=0,1,2, \cdots, 16 .
\end{array}
$$
Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: [0], [1], ..., [16]. Since $366=17 \time... | 3928 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,619 |
II. (This question is worth 35 points) Given the sequence $\left\{a_{n}\right\}(n=1,2, \cdots)$ where $a_{1}=1, a_{2}=3, a_{3}=6$, and for $n>3$, $a_{n}=3 a_{n-1}-$ $a_{n-2}-2 a_{n-3}$. Prove that for all natural numbers $n>3$, $a_{n}>3 \times$ $2^{n-2}$.
Translate the above text into English, please retain the origin... | From the recursive relation, we have
$$
a_{3}=6, a_{4}=13, a_{5}=27, a_{6}=56 \text {, }
$$
Thus, we conjecture: when $n>3$, $a_{n}>2 a_{n-1}$.
We prove this by induction:
When $n=4$, $a_{4}=13>2 a_{3}$, i.e., $a_{4}>2 a_{3}$ holds.
Assume that for $n \leqslant k$, the conjecture holds, i.e.,
$$
a_{k}>2 a_{k-1}, a_{k-... | null | Algebra | proof | Yes | Yes | cn_contest | false | 707,620 |
Three. (This question is worth $35^{\circ}$ points) In the plane quadrilateral $A B C D$, $A B=1, B C=\frac{\sqrt{10}}{2}, C D=\frac{3 \sqrt{2}}{2}, D A=\sqrt{3}$, $\angle D=75^{\circ}$. Find the minimum value of the sum of the distances from a point inside the quadrilateral $A B C D$ to the midpoints of the four sides... | Three, connect $A C, B D$ intersecting at $O$, let $\angle A O D=\theta, \angle A D O=\alpha$. From $A B^{2}+C D^{2}=A D^{2}+B C^{2}=\frac{11}{2}$, we know $\theta=90^{\circ}$, see the figure.
Thus, $O D=\sqrt{3} \cos \alpha$
$$
=\frac{3 \sqrt{2}}{2} \cos \left(75^{\circ}-\alpha\right) \text {. }
$$
From (1) and $\sin... | \frac{1}{2} \sqrt{28+6 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,621 |
1. Let $a, b$ be real numbers, $\cos a=b, \cos b=a$. Then the relationship between $a$ and $b$ is ( ).
(A) $a>b$
(B) $a<b$
(C) $a=b$
(D) cannot be determined | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,622 |
2. The number of different real numbers $a$ that make the roots of the quadratic equation $x^{2}-a x+9 a=0$ both positive integers is ( ).
(A) 3
(B) 5
(C) 6
(D) 10 | 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 | 707,623 |
3. Let the complex number $z$ satisfy the equation $|z+3-4i|=A$, then the maximum value of $|z|$ is ( ).
(A) 10
(B) 0
(C) 8
(D)? | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,624 |
Example 9. Given a convex polygon that cannot cover any triangle of area 1. Prove: this polygon can be covered by a triangle of area 4. (Eighth All-Union Mathematical Competition) | Proof: Let the convex polygon be $Q$, then among the inscribed triangles of $Q$, there must be one with the largest area. Let's assume it is $\triangle ABC$, clearly $S_{\triangle ABC}S_{\triangle ABC}$, which contradicts the assumption that $\triangle ABC$ is the inscribed triangle with the largest area in $Q$.
Simil... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,625 |
$a_{1} x+a_{2} x^{2}+\cdots+a_{100} x^{100}$, the number of terms with rational coefficients is ( ).
(A) 14
(B) 15
(C) 16
(D) 17 | 4. 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 | 707,626 |
6. $f(x)=\sin ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\sqrt{3} \sin x$
(A) Strictly increasing function
(C) Non-constant periodic function
(D) Constant function that does not change with $x$ | 6. 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 | 707,628 |
7. Let $a, b, c$ be the lengths of the three sides of a triangle, $t=\frac{c}{a+b}+$ $\frac{a}{b+c}+\frac{b}{c+a}$. Then, regardless of the shape of the triangle, we have
(A) $t \leqslant \frac{3}{2}$
(B) $\frac{3}{2}<t \leqslant 2$
(C) $\frac{3}{2} \leqslant t<2$
(D) $t \geqslant 2$ | 7. 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 | 707,629 |
1. Let $z=\frac{1-t i}{1+t i}, t \in R, i=\sqrt{-1}$. Then the trajectory of $z$ in the complex plane is $\qquad$ | 1. The trajectory is a unit circle centered at the origin, but $z \neq-1$ | z \neq -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,631 |
3. The range of the function $f(x)=2 \sqrt{x^{2}-4 x+3}+\sqrt{3+4 x-x^{2}}$ is | 3. $[\sqrt{6}, \sqrt{30}]$ | [\sqrt{6}, \sqrt{30}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,633 |
5. The solution $(x, y)$ of the equation $x^{2}+2 x \sin (x y)+1=0$ in the real number range is $\qquad$ . | 5. $x=1$
or $-1, y=2 k \pi+\frac{3}{2} \pi, k \in Z$ | x=1 \text{ or } -1, y=2 k \pi+\frac{3}{2} \pi, k \in Z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,635 |
6. The volume of a regular $n$-sided pyramid $(n \geqslant 3)$ is $V$, and the area of each side triangular face is $S$. Point $O$ is located inside the base. Then the sum of the distances from point $O$ to each side face is $\qquad$ . | 6. $\frac{3 V}{S}$ | \frac{3 V}{S} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,637 |
Three. (Full marks 20 points) Let $a_{1}, a_{2}, \cdots$ and $b_{1}, b_{2}, \cdots$ be two arithmetic sequences, and their sums of the first $n$ terms are $A_{n}$ and $B_{n}$, respectively. It is known that for all $n \in N$, $\frac{A_{n}}{B_{n}}=\frac{2 n-1}{3 n+1}$. Try to write the expression for $\frac{a_{n}}{b_{n}... | Three、 $\frac{4 n-3}{6 n-2}$ | \frac{4 n-3}{6 n-2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,640 |
Five. (Full marks 20 points) In the convex quadrilateral $ABCD$, $\angle BAD = \angle BCD = 90^{\circ}$, ray $AK$ bisects $\angle BAD$. It is known that $AK \parallel BC$, $AK \perp CD$, and $AK$ intersects $BD$ at point $E$.
Prove: $AE < \frac{1}{2} CD$.
---
Translate the above text into English, please retain the o... | Five, there are many ways to prove this.
Trigonometric Proof: Let the intersection of $A K$ and $C D$ be $F$ (as shown in the figure). Then, by the problem, $\triangle A F D$ is a right isosceles triangle, so $\angle A D F = 45^{\circ}$.
Let $\angle A D B = \theta$, then $0^{\circ}0$, it is sufficient to prove
$$
\cos ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,642 |
9. 1 Proof: If $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)$ $=1$, then $x+y=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.1 If we multiply both sides of the given equation by $x-\sqrt{x^{2}+1}$, we get $-y-\sqrt{y^{2}+1}=x-\sqrt{x^{2}+1}$. If we multiply both sides by $y-\sqrt{y^{2}+1}$, we get $-x-\sqrt{x^{2}+1}=y-\sqrt{y^{2}+1}$. Adding the two resulting equations, we get $-(x+y)=x+y$. From this, we know that $x+y=0$. | null | Algebra | proof | Yes | Yes | cn_contest | false | 707,643 |
9.2 Circles $S_{1}$ and $S_{2}$ are externally tangent at point $F$, line $l$ is tangent to $S_{1}$ and $S_{2}$ at points $A$ and $B$ respectively. A line parallel to $l$ is tangent to $S_{2}$ at point $C$, and intersects $S_{1}$ at two points. Prove: Points $A, F$, and $C$ are collinear. | 9. 2 Solution
Xuan 1. Since
the two tangents to circle $S_{2}$
at points $B$
and $C$ are
parallel, $BC$ is the diameter of $S_{2}$. Therefore, $\angle B F C=90^{\circ}$. We will prove that $\angle A F B=90^{\circ}$. Draw the common tangent of the two circles through point $F$ (as shown in the figure), and let it inter... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,644 |
9.4 Mark $n$ different blue points and $n$ different red points on a straight line. Prove: the sum of the distances between points of the same color does not exceed the sum of the distances between points of different colors. | 9.4 We will prove the proposition in a more general setting, assuming that the marked points can coincide. Suppose there are $N$ distinct points among the $2n$ marked points, and we will use induction on $N$.
When $N=1$, the inequality is obviously true.
When there are $N$ distinct points, let $S_{1}^{N}$ denote the s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,646 |
Example 1. As shown in the figure, in $\triangle A B C$, $A C=B C, \angle C$ $=20^{\circ}$. Also, $M$ is on $A C$, $N$ is on $B C$, and it satisfies $\angle B A N=50^{\circ}, \angle A B M=60^{\circ}$. Then $\angle N M B=$ $\qquad$ . (1991, Qinfen Cup Mathematics Invitational Competition) | Draw a line through $M$ parallel to $AB$ intersecting $BC$ at $K$. Connect $AK$ intersecting $MB$ at $P$, and connect $PN$. It is easy to know that $\triangle A P R$ and $\triangle M K P$ are both equilateral triangles.
$$
\begin{array}{l}
\because \angle B A N=50^{\circ}, \quad A C=B C, \angle C=20^{\circ}, \\
\theref... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,647 |
Example 2. As shown in the figure, in $\triangle A B C$, $A B<A C<$ $B C, D$ is a point on $B C$, $E$ is a point on the extension of $B A$, and $B D=B E=A C, \triangle B D E$'s circumcircle intersects with $\triangle A B C$'s circumcircle at point $F$. Prove: $B F=A F+C F$.
(1991, National Junior High School Mathematic... | Prove that extending $CF$ to $G$ such that $FG = FA$, then $\angle GFA = \angle EBD$, and $\triangle AFG$ is an isosceles triangle.
$$
\because \triangle BED \text{ is an isosceles triangle,}
$$
$$
\therefore \angle BDE = \angle G.
$$
Connecting $EF$, then
$$
\begin{array}{c}
\angle BFE = \angle BDE. \\
\because \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,648 |
Example 3. An $n \times n$ chessboard $(n \geqslant 2)$, each square is filled with one of the numbers $1,2, \cdots, n^{2}$, and each of these $n^{2}$ numbers appears exactly once on the chessboard. Prove that there must exist two adjacent squares (i.e., squares sharing a common edge) such that the numbers in these squ... | Prove that in any two adjacent squares, the numbers differ by at least $n-1$.
For $k=1,2, \cdots, n^{2}-n$, consider:
The set $A_{k}$ of squares filled with numbers $1,2, \cdots, k$, the set $B_{k}$ of squares filled with numbers $k+n, k+n+1, \cdots, n^{2}$, and the set $C_{k}$ of squares filled with numbers $k+1, k+2,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,649 |
Example 4. Suppose there are $n$ gas stations on a circular highway, each station storing a certain number of barrels of gasoline (some stations may store none), and the total amount of gasoline stored at the $n$ stations is sufficient for a car to travel around the highway once. Now, let a car that originally has no g... | Let the total amount of gasoline required to travel around the circular highway be $a$ barrels, and let the $n$ stations be denoted in a counterclockwise direction as $A_{1}$, $A_{2}$, ..., $A_{n}$. Suppose station $A_{i} (i=1,2, \cdots, n)$ has $k_{i}$ barrels of gasoline.
We will use mathematical induction to prove ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 707,650 |
Example 5. Given $n$ points in the plane, they are connected by $m n\left(m \leqslant \frac{n-1}{2}\right)$ line segments. Prove that there exists a sequence of distinct points $V_{0}, V_{1}, \cdots, V_{m}$ such that for $1 \leqslant i \leqslant m$, the points $V_{i-1}$ and $V_{i}$ are connected by a given line segment... | Prove by fixing $m$ and using induction on $n$.
Given $m \leqslant \frac{n-1}{2}$, we have $n \geqslant 2 m+1$.
(1) When $n=2 m+1$, $m n=\frac{n(n-1)}{2}=$ $C_{n}^{2}$. Therefore, if every pair of $n$ points is connected by a line segment, there are exactly $m n$ line segments, and the conclusion is clearly true.
(2) A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,651 |
Example 6. If in the prime factorization of a positive integer, the exponent of each prime factor is greater than 1, then this positive integer is called a good number, for example \(2^{2} 3^{5}, 5^{3} 7^{4}\) are good numbers. Prove: There exist infinitely many consecutive good numbers. | First, note that 8 and 9 are a pair of consecutive good numbers, because \(8=2^{3}, 9=3^{2}\).
Assume \(a_{n}\) and \(a_{n}+1\) are the \(n\)-th pair of good numbers, then set
\[
a_{n+1}=4 a_{n}\left(a_{n}+1\right) \text{, }
\]
Therefore, \(a_{n+1}+1=4 a_{n}\left(a_{n}+1\right)+1\)
\[
=\left(2 a_{n}+1\right)^{2} \text... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,652 |
Example 7. On a certain day, several readers visited the library individually, and among any three readers, at least two of them met in the library on that day. Prove: The librarian can definitely find two moments to issue a verbal announcement, so that the readers who visited the library that day can hear it at least ... | Prove that the librarian can choose such extreme moments: the moment when the first reader $A$ leaves the library and the moment when the last reader $B$ enters the library.
If the last reader $B$ enters the library before the first reader $A$ leaves, then at this moment all readers are inside the library and will def... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,653 |
Example 8. In a tennis club of 20 members in a certain area, 14 singles matches are held, with each person playing at least once. Prove: there must be six matches, the 12 participants of which are all different.
(1989, American Mathematics Competition) | Let the participants in the $j$-th match be denoted as $\left(a_{j}, b_{j}\right)$, and let
$$
S=\left\{\left(a_{j}, b_{j}\right) \mid j=1,2,3,4, \cdots, 14\right\}.
$$
Let $M$ be a subset of $S$. If the players appearing in the pairs in $M$ are all distinct, then $M$ is called a "good" subset of $S$.
Clearly, there ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,654 |
9. Let there be $n$ people $A_{1}, A_{2}, \cdots, A_{n}$, some of whom know each other. Prove: it is possible to divide them into two groups in such a way that each person has at least half of their acquaintances in the other group. | Proof: Let $A_{i}(i=1,2, \cdots, n)$ have $c_{i}$ people, among which $d_{i}$ do not belong to the same group as $A_{i}$. Here, $d_{i}$ varies with the different groupings.
This problem is equivalent to proving: there exists an appropriate grouping method such that for all $i=1,2, \cdots, n$, we have $d_{i} \geqslant ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,655 |
Example 1. From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions? (1989, National High School Competition) | Destruct the following model:
Take 10 identical white balls and arrange them in a row, take 5 identical black balls and divide them into 3 groups in the order of $2,2,1$, then insert them into the 10 gaps between the white balls, from left to right, excluding the left end but including the right end. There are $C_{10}^... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,657 |
Example 3. As shown in the figure, in $\triangle ABC$, $AB > AC$, the external angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at point $E$. A perpendicular line $EF$ is drawn from $E$ to $AB$, with the foot of the perpendicular being $F$. Prove that: $2 AF = AB - AC$.
(1989, National High S... | $$
\begin{array}{l}
\because \angle H A E=\angle G A E, \\
\therefore \angle G A E=\angle E G A . \\
\therefore \angle H A E=\angle E G A, \angle E A C=\angle E G B . \\
\because \angle E B C=\angle E A H, \angle E C B=\angle E A B, \\
\therefore \angle E B C=\angle E C B, E C=E B . \\
\text { Also } \angle 1=\angle 2,... | 2 AF = AB - AC | Geometry | proof | Yes | Yes | cn_contest | false | 707,659 |
Example 4. Team A and Team B each send out 7 players to participate in a Go competition according to a pre-arranged order. Both sides start with Player 1 competing, the loser is eliminated, and the winner then competes with the next (Player 2) of the losing side, $\cdots \cdots \cdots$, until one side is completely eli... | Let's first assume that Team A wins. Then, the 7 members of Team A need to eliminate the 7 members of Team B, which means they need to win a total of 7 games. This involves distributing these 7 games among the 7 members of Team A, with each distribution corresponding to a possible match process.
Thus, the problem is e... | 2 C_{13}^{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,661 |
Example. Find the range of the function $y=4 \sqrt{8-16 x}$ $+3 \sqrt{9 x-\frac{9}{2}}$. | The function's domain is $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$. The function can be transformed as follows:
$$
\begin{array}{l}
y=16 \sqrt{\frac{1}{2}-x}+9 \sqrt{x+\frac{1}{2}} . \\
\text { Let } x_{0}=\sqrt{\frac{1}{2}-x} \quad(0 \leqslant x_{0} \leqslant 1), \\
y_{0}=\sqrt{x+\frac{1}{2}} \quad\left(0 \leqs... | y \in [9, \sqrt{337}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,662 |
1. Given $a=3^{55}, b=4^{44}, c=5^{33}$, then ( ).
(A) $a<b<c$
(B) $c<b<a$
(C) $c<a<b$
(D) $a<c<b$ | \begin{array}{l}-1 . \text { (C). } \\ \because a=3^{55}=\left(3^{5}\right)^{11}=243^{11}, \\ b=4^{44}=\left(4^{4}\right)^{11}=256^{11}, \\ c=5^{33}=\left(5^{3}\right)^{11}=125^{11}, \\ \therefore c<a<b, \text { choose (C). }\end{array} | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,663 |
3. If the three roots of the equation $(x-1)\left(x^{2}-2 x+m\right)=0$ can be the lengths of the three sides of a triangle, then the range of the real number $m$ is ( ).
(A) $0 \leqslant m \leqslant 1$
(B) $m \geqslant \frac{3}{4}$
(C) $\frac{3}{4}<m \leqslant 1$
(D) $\frac{3}{4} \leqslant m \leqslant 1$ | 3. (C).
Let $\alpha, \beta$ be the roots of the equation $x^{2}-2 x+m=0$. From $\Delta=4-4 m \geqslant 0$, we get $m \leqslant 1$. Clearly, $\alpha+\beta=2>1$. Also, by the triangle inequality, which states that the difference of any two sides is less than the third side, we know $|\alpha-\beta|<\frac{3}{4}$. Therefor... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,665 |
4. If the sides of a cyclic quadrilateral are 25, 39, 52, and 60, then the circumference of the circle is ( ).
(A) $62 \pi$
(B) $63 \pi$
(C) $64 \pi$
(D) $65 \pi$ | 4. (D).
Let $A B C D$ be a cyclic quadrilateral, and $A B=25$,
$$
B C=39, C D=52, D A
$$
$=60$. By the property of a cyclic quadrilateral, $\angle A=180^{\circ}-\angle C$. Connecting $B D$. By the cosine rule,
$$
\begin{aligned}
B D^{2} & =A B^{2}+A D^{2}-2 A B \cdot A D \cos \angle A \\
& =C B^{2}+C D^{2}-2 C B \cdot... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,666 |
5. Let $AB$ be a chord of $\odot O$, and $CD$ be a diameter of $\odot O$ that intersects chord $AB$. Let $M=$
$|S_{\triangle CAB} - S_{\triangle DAB}|$, $N=$ $2 S_{\triangle OAB}$, then ( ).
(A) $M>N$
D-
(B) $M=N$
(C) $M<N$
(D) The relationship between $M$ and $N$ is uncertain | 5. (B).
As shown in the figure, construct $DE \perp AB, CF \perp AE$ (the feet of the perpendiculars are $E, F$), and let the extension of $CF$ intersect the circle at $G$. Connect $DG$. Since $\angle DGC=90^{\circ}$ (as $CD$ is the diameter), it follows that
$EFCD$ is a rectangle, hence $DE = GF$. Without loss of gen... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,667 |
1. Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place total $\qquad$.
untranslated part: $\qquad$ | In $1^{2}, 2^{2}, \cdots, 10^{2}$, it is found through calculation that the tens digit is odd only for $4^{2}=16, 6^{2}=36$. A two-digit square number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2},
$$
Therefore, $b$ can only be 4 or 6. That is, in every 10 consecutive numbers, there are two numbers whose ... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,669 |
2. Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$, then the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$ is. $\qquad$ . | 2. 20 .
Factorization yields
$$
\begin{array}{l}
\alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right), \\
\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}=\alpha^{2}(\alpha-1)(\alpha+1)^{2} .
\end{array}
$$
$\alpha$ satisfies $\alpha^{2}+\alpha-\frac{1}{4}=0$, hence $\alpha \neq 1$.
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\al... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,671 |
3. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ . | 3. 1.
The formula yields
$$
\begin{aligned}
y & =(x-1)^{2}+x+\frac{1}{x}-1 \\
& =(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1 .
\end{aligned}
$$
When $x=1$, both $(x-1)^{2}$ and $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ simultaneously take the minimum value of 0, so the minimum value of $y=x^{2}-x+\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,672 |
4. Given a line segment $A B$ as the diameter of a semicircle, with the center at $O$, and $C$ is a point on the semicircle. If $O C^{2}=A C \cdot B C$, then $\angle C A B=$ $\qquad$ . | 4. $75^{\circ}$ pear $155^{\circ}$
Draw $\% A B$, with the foot of the perpendicular at D, we have
$A C \cdot B C=O C^{2}=$ $\frac{1}{4} A B^{2}$,
On the other hand, by the area formula
$$
A C \cdot B C=C D \cdot A B \text {, }
$$
Therefore, $C D \cdot A B=\frac{1}{4} A B^{2}$, which means $C D=\frac{1}{4} A B$, thus... | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,673 |
One, (20 points) As shown in the figure, it is known that $\angle A C E=\angle C D E=90^{\circ}$, point $B$ is on $C E$, $C A=C B=C D$, the circle passing through points $A, C, D$ intersects $A B$ at point $\tilde{r}$. Prove that $F$ is the incenter of $\triangle C D E$.
---
Note: The symbol $\tilde{r}$ in the origin... | $$
\begin{array}{c}
\text { -、Proof } \because A C= \\
\therefore \angle C A B= \\
\angle C B A=45^{\circ}, \\
\text { Also } \because A, C, F, D \text { are four }
\end{array}
$$
$$
\text { 1. Proof } \because A C=B C, \angle A C B \text { is a right angle, }
$$
points on a circle,
$\therefore \angle C D F=$ $\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,674 |
II. (25 points) On the coordinate plane: Points whose both coordinates are integers are called integer points. Try to find all integer points $(x, y)$ on the graph of the quadratic function $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$ that satisfy $y \leqslant|x|$, and explain your reasoning. | II. Solve $\because y \leqslant|x|$, i.e., $\frac{x^{2}-x+18}{10} \leqslant|x|$,
Thus, $x^{2}-x+18 \leqslant 10|x|$.
(*)
When $x \geqslant 0$, equation (*) becomes
$$
x^{2}-x+18 \leqslant 10 x \text {, }
$$
which simplifies to $x^{2}-11 x+18 \leqslant 0$.
Therefore, $2 \leqslant x \leqslant 9$.
In this case, the point... | (2,2),(4,3),(7,6),(9,9),(-6,6),(-3,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,675 |
Three, (25 points) Prove that every natural number $n$ greater than 6 can be expressed as the sum of two natural numbers greater than 1 and coprime to each other.
| (i) If $n$ is odd, let $n=2k+1$, where $k$ is an integer greater than 2. Then write $n=k+(k+1)$, and it is clear that $(k, k+1)=1$, so this representation meets the requirement.
(ii) If $n$ is even, then we can set $n=4k$ or $4k+2$, where $k$ is a natural number greater than 1.
When $n=4k$, we can write $n=(2k-1)+(2k+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,676 |
1. The "Monkey Cycling" performance in the circus is carried out by 5 monkeys using 5 bicycles, with each monkey riding at least once, but no monkey can ride the same bicycle more than once. After the performance, the 5 monkeys rode the bicycles $2, 2, 3, 5, x$ times respectively, and the 5 bicycles were ridden $1, 1, ... | ,- 1 (B).
Every time the monkey rides a bike, the monkey and the bike each get one count, so the number of times the monkey rides a bike = the number of times the bike is ridden, i.e., $2+2+3+5+x=1+1+2+4+y$. Simplifying, we get $x+4=y$. From $x \geqslant 1$, we get $y \geqslant 5$. Also, from the problem, $y \leqslant ... | 6 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,677 |
2. Given the quadratic trinomial in $x$ with integer coefficients $a x^{2}+b x+c$, when $x$ takes the values $1, 3, 6, 8$, a student recorded the values of this quadratic trinomial as $1, 5, 25, 50$. Upon checking, only one of these results is incorrect. The incorrect result is ( ).
(A) When $x=1$, $a x^{2}+b x+c=1$
(B... | 2. (C).
Notice that $(n-m) \mid\left[\left(a n^{2}+b n+c\right)-\left(a m^{2}+b m+c\right)\right]$. From $(6-1) \mid (25-1), (8-6) \mid (50-25)$, but $(81) \mid (50-1)$, this leads to the result being $\mathbf{C}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,678 |
3. Given the radius of $\odot O$ is 10, the distance from point $P$ to the center $O$ is 8, the number of chords passing through point $P$ and having integer lengths is ( ) .
(A) 16
(B) 14
(C) 12
(D) 9 | 3. (A).
The diameter passing through point $P$ is 20 units long, and the chord passing through point $P$ and perpendicular to the diameter is between 12.12 and 20 units long. There are 7 more integer values between 12.12 and 20, considering the symmetry, the number of chords sought is $2+7 \times 2=16$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,679 |
4. The side length of square $A B C D$ is $2 \sqrt{15}, E, F$ are the midpoints of $A B, B C$ respectively, $A F$ intersects $D E, D B$ at $M, N$. Then the area of $\triangle D M N$ is
(A) 8
(B) 9
(C) 10
(D) 11 | 4. (A).
It is easy to know that $\triangle A M E \backsim \triangle A B F$, thus,
$$
S_{\triangle A M E}=\left(\frac{A E}{A F}\right)^{2} S_{\triangle A B F}=\frac{1}{5} S_{\triangle A B F}
$$
Since $N$ is the centroid of $\triangle A B C$,
$\therefore S_{\triangle B F N}=\frac{1}{3} S_{\triangle A B F}$.
Since $S_{\t... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,680 |
Example 5. As shown in the figure, in $\triangle A B C$, $A B=A C, D$ is a point on the base $B C$, and $E$ is a point on the line segment $A D$, and $\angle B E D=2 \angle C E D=\angle A$. Prove: $B D=2 C D$. (1992, National Junior High School Competition) | Prove that the bisector of $\angle A E B$ intersects $A B$ at $G$.
Then
$$
\begin{array}{l}
\angle 1=\angle 2=\frac{1}{2}\left(180^{\circ}-\angle B E D\right) \\
\quad=\frac{1}{2}\left(180^{\circ}-\angle A\right). \\
\because \angle B=\angle C=\frac{1}{2}\left(180^{\circ}-\angle A\right), \\
\therefore \angle 1=\angle ... | B D=2 C D | Geometry | proof | Yes | Yes | cn_contest | false | 707,681 |
5. For the range of values of the independent variable $x$, $n \leqslant x \leqslant n+1$ (where $n$ is a natural number), the number of integer values of the quadratic function $y=x^{2}+x+\frac{1}{2}$ is ( ).
(A) $n$
(B) $n+1$
(C) $2 n$
(D) $2(n+1)$ | 5. (D).
The function $y=\left(x+\frac{1}{2}\right)^{2}+\frac{1}{4}$, when $x>-\frac{1}{2}$, the function value increases as the independent variable increases. Therefore, when $n \leqslant x \leqslant n+1$, we have
$$
n^{2}+n+\frac{1}{2} \leqslant y \leqslant(n+1)^{2}+(n+1)+\frac{1}{2} \text {. }
$$
Since $y$ is an i... | 2(n+1) | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,682 |
6. Let $x_{1}, x_{2}, \cdots, x_{62}$ all be natural numbers, $x_{1}<x_{2}<\cdots<$ $x_{62}$, and $x_{1}+x_{2}+\cdots+x_{62}=1994$. Then the maximum value that $x_{62}$ can take is ( ).
(A) 101
(B) 102
(C) 103
(D) 104 | 6. (C).
$$
\begin{array}{l}
x_{1}+\left(x_{2}-1\right)+\left(x_{3}-2\right)+\cdots+\left(x_{62}-61\right) \\
=103,
\end{array}
$$
To maximize $x_{62}$, it is necessary to minimize $x_{1}, x_{2}-1, x_{3}-2, \cdots, x_{61}-60$. For this, we can take $x_{1}=x_{3}-1=x_{3}-2=\cdots=x_{61}-60=1$, then $x_{62}-61=42$. Theref... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,683 |
For each natural number $n$, the parabola $y=\left(n^{2}+n\right) x^{2}$
$-(2 n+1) x+1$ intersects the $x$-axis at points $A_{n}, B_{n}$. Let $\left|A_{n} B_{n}\right|$ denote the distance between these two points. Then $\left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots$ $+\left|A_{1995} B_{1995}\right|$ is $\qq... | $$
\begin{aligned}
& \text { II.1. } 1995 \\
& \because y=(n x-1)[(n+1) x-1], \\
x_{1}= & \frac{1}{n}, x_{2}=\frac{1}{n+1}, \\
& \therefore\left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} .
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
& \left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1005} B_{100... | 1995 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,684 |
4. The ice numbers $a_{1}, u_{2}, \cdots, a_{9}$ can only take two different values, +1 or -1. Then, the maximum value of the expression $a_{1} a_{5} a_{9}-a_{1} a_{6} a_{8}+a_{2} a_{8} a_{7}-$ $a_{2} a_{4} a_{9}+a_{3} a_{4} a_{8}-a_{3} a_{5} a_{7}$ is $\qquad$. | 4. 4 .
Since each term in the sum can only be 11 or -1, the sum is even (the parity of $a-b$ and $a+b$ is the same), so the maximum value of the sum is at most 6. It is easy to see that the sum cannot be 6, because in this case $a_{1} a_{5} a_{9}, a_{2} a_{6} a_{7}, a_{3} a_{6} a_{8}$ should all be +1, making their pr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,687 |
One, (20 points) Let $x$ and $y$ be natural numbers such that the sum of the two fractions $\frac{x^{2}-1}{y+1}+\frac{y^{2}-1}{x+1}$ is an integer. Prove: These two fractions are both integers. | Solution One: Let $\frac{x^{2}-1}{y+1}=u, \frac{y^{2}-1}{x+1}=v$, then $u+v$, $uv$ are both integers. Therefore, $u, v$ are the two roots of a quadratic equation with integer coefficients $z^{2}+mz+n=0$, where $u+v=-m, uv=n$. Since $u, v$ are rational numbers, $\Delta=m^{2}-4n$ is a perfect square, and its parity is th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,688 |
II. (25 points) Let the diagonals of convex quadrilateral $ABCD$ intersect at $O$, and the areas of $\triangle AOB$ and $\triangle COD$ be $S_{1}$ and $S_{2}$, respectively. The area of quadrilateral $ABCD$ is $S$. Prove: $\sqrt{S_{1}} + \sqrt{S_{2}} \leqslant \sqrt{S}$, where equality holds if and only if $AB \paralle... | $$
\begin{array}{l}
S=S_{1}+S_{2}+S_{3} \div S_{4} \text {. } \\
\because \frac{S_{1}}{S_{3}}=\frac{B O}{O D} \text {, } \\
\frac{S_{4}}{S_{2}}=\frac{B O}{O D}, \\
\therefore \frac{S_{1}}{S_{3}}=\frac{S_{4}}{S_{2}} \text {, i.e., } S_{1} S_{2}=S_{3} S_{4} \text {. } \\
\sqrt{S_{1}}+\sqrt{S_{2}} \leqslant \sqrt{S} \\
\L... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,689 |
Three. (25 points) Suppose there are enough unit squares, each of whose four sides is dyed in various ways with red, yellow, white, and blue. Question: For what positive integers $m, n$, can these small squares be used to form an $m \times n$ chessboard, such that
(1) the same side of the chessboard is the same color;
... | If it can be assembled, since each cell has one side of red, yellow, white, and blue, the sum of the lengths of each color's edges is $mn$.
On the other hand, the sum of the lengths of each color's edges inside the chessboard, according to condition (3), is even, and the sum of the lengths of the edges on the chessboa... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,690 |
1. Let $p>3, p$ and $p+4$ are both prime numbers, denote $S_{1}=p(p+$ 4), for $i=1,2, \cdots, k-1, S_{i+1}$ represents the sum of the digits of $S_{i}$. If $S_{\star}$ is a single digit, then $S_{k}=(\quad)$.
(A) 3
(B) 5
(C) 7
(D) 9 | -1. (13)
If $p \equiv (\bmod 3)$, then $p+4 \equiv 0(\bmod 3)$, which contradicts that $p+4$ is a prime number. Therefore, $p \equiv 1(\bmod 3)$, so $3 \mid (p+2)$.
$$
\text { Also, } \begin{aligned}
S_{1} & =p(p+4)=p^{2}+4 p+4-4 \\
& =(p+2)^{2}-9+5,
\end{aligned}
$$
then $p(p+4) \equiv 5(\bmod 9)$. Thus, $S_{k}=5$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,691 |
Example 6. As shown in the figure, on the two legs $AB$, $AC$ of isosceles $\triangle ABC$, points $E$ and $F$ are taken respectively such that $AE=CF$. Given $BC=2$. Prove: $EF \geqslant 1$. (1990, Xi'an Junior High School Mathematics Competition) | Prove that by constructing $A C^{\prime} / / B C, C C^{\prime} / / A B$ through $A, C$ respectively, and $E G / / B C$ through $E$ intersecting $C C^{\prime}$ at $G$, and connecting $F G$, the quadrilaterals $A B C C^{\prime}$ and $E B C G$ are both parallelograms.
$$
\begin{array}{l}
\because A B=A C, A E=C F, \\
\qua... | E F \geqslant 1 | Geometry | proof | Yes | Yes | cn_contest | false | 707,692 |
2. Given that the vertices of $\triangle A B C$ are all lattice points (points with integer coordinates are called lattice points), and $A(0,0), B(36,15)$. Then the minimum value of the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{3}{2}$
(C) $\frac{5}{2}$
(D) $\frac{7}{2}$ | 2. (B).
From the given information, the line $l_{AB}: y=\frac{15}{36} x=\frac{5}{12} x$, which can be rewritten as $5 x-12 y=0$.
Let $C(x, y), x \in \mathbb{Z}, y \in \mathbb{Z}$. The height $h$ from $C$ to the line $l_{AB}$ in $\triangle ABC$ is the distance from $C$ to $l_{AB}$, so $h=\frac{|5 x-12 y|}{13}$.
Also, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,693 |
6. A regular dodecahedron has 20 vertices, 30 edges, and each vertex is the intersection of three edges, the other ends of which are three other vertices of the regular dodecahedron, and these three vertices are called adjacent to the first vertex. Place a real number at each vertex, such that the real number placed at... | 6. (A).
Let the total at $P_{1}$ be $m$, if $P_{2}, P_{3}, P_{4}$ and $P_{2}$ H $_{1}$ jaw, $P_{3}, P_{4}$ have a number greater than $m$, then $P_{1}>m$. Otherwise, $P_{2}, P_{3}$, $P_{4}$ can only have $m$. Considering the points adjacent to $P_{2}, P_{3}, P_{4}$, they also cannot have $m$. Thus, all points can only... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,697 |
1. Let the function $f(x)$ have a domain and range both equal to $R$, and for any $a, b \in R$ there is $f[a f(b)]=a b$. Then the value of $|f(1995)|$ is $\qquad$ | Ni, i. 1995.
For the known functional equation, substituting $a=1, a=b, a=f(b)$, we have $f(f(b))=b, f(b f(b))=b^{2}, f\left[f^{2}(b)\right]=b f(b)$. Applying the function $f$ to the third equation, $f\left\{f\left[f^{2}(b)\right]\right\}=f(b f(b))=b^{2}$. Thus, from $f(J(b))=b$, we have $f^{2}(b)=b^{2}$, i.e., $|f(b)|... | 1995 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,698 |
2. Given the hyperbola $\frac{x^{2}}{t}-\frac{y^{2}}{t}=1$ with the right focus $F$, any line passing through $F$ intersects the right branch of the hyperbola at $M, N$, and the perpendicular bisector of $M N$ intersects the $x$-axis at $P$. When $t$ takes any positive real number except 0, $\frac{|F P|}{|M N|}=$ $\qqu... | 2. $\frac{\sqrt{2}}{2}$.
Obviously, the eccentricity $e=\sqrt{2}$. Let $M\left(x_{1}, y_{1}\right), N\left(x_{3}\right.$, $\left.y_{2}\right)$, the slope angle of $M N$ be $a$, the midpoint of $M N$ be $Q$, and the distance from $M$ to the right directrix be $d$. Then, by the definition of the hyperbola, we have $\fra... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,699 |
4. Let $f(x)=a x^{2}+b x+c(a, b, c \in R, a \neq 0)$. If for $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then for $|x| \leqslant 1$, the maximum value of $|2 a x+b|$ is $\qquad$ | 4. 4 .
Given $f(0)=c, f(1)=a+b+c, f(-1)=a-b+c$,
we solve to get $a=\frac{f(1)+f(-1)-2 f(0)}{2}, b=\frac{f(1)-f(-1)}{2}$.
For $|x| \leqslant 1$, $|f(x)| \leqslant 1$, we have
$$
\begin{array}{l}
|2 a x+b| \\
=\left|[f(1)+f(-1)-2 f(0)] x\right. \\
\left.+\frac{f(1)-f(-1)}{2} \right| \\
=\left|\left(x+\frac{1}{2}\right)... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,701 |
Example 7. As shown in the figure, in the right triangle $\triangle ABC$, $AD$ is the altitude on the hypotenuse $BC$. The line connecting the incenter $M$ of $\triangle ABD$ and the incenter $N$ of $\triangle ACD$ intersects sides $AB$ and $AC$ at points $K$ and $L$, respectively. The areas of $\triangle ABC$ and $\tr... | Prove that connecting AN, AM, $A M, C N$, and extending $B M$ to intersect: $A N$ at ~ extending $C N$ to intersect $A M$ at. $Q$, let $C Q$ intersect $E B^{D}$ at $Y$, then $H$ is the incenter of $\triangle A B C$. Connecting $A M$ and extending it to intersect $L K$ at $E$, then $\angle N A M=$ $45^{\circ}, \angle E ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,703 |
6. Among all tetrahedra with edge lengths $2,3,3,4,5,5$, the maximum volume is $\qquad$ | 6. $\frac{8 \sqrt{2}}{3}$.
Consider a triangle with one side of length 2, the other two sides can only have the following four scenarios: (a) 3,3, (b) 5,5, (c) 4,5; (d) 3,4.
Thus, a tetrahedron with a base of a triangle with side length 2 has only three possible configurations.
(1) Base $2,3,3$, side $2,5,5$. In this... | \frac{8 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,704 |
One, (25 points) $M$ is a point inside $\triangle ABC$ satisfying $\angle AMC = 90^{\circ}, \angle AMB = 150^{\circ}, \angle BMC = 120^{\circ}$. Let $P, Q, R$ be the circumcenters of $\triangle AMC, \triangle AMB$, and $\triangle BMC$ respectively. Prove:
$S_{\triangle PQR} > S_{\triangle ABC}$. | Let $A M \cap P Q=U, C M \cap P R=W, B M \cap R Q=V$.
Since $P, Q, R$ are circumcenters, then $U, W$, and $V$ are the midpoints of $A M, C M$, and $B M$ respectively, and $P Q \perp A M, Q R \perp B M, R P \perp C M$, thus $P Q \perp P R$.
Let $M U=a, M W=c, M V=b$, then
$S_{\triangle A B C}=4 S_{\triangle U W V}$
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,705 |
Sure, here is the translation:
```
(25 points) Let $\left\{a_{n}\right\}$ be an arithmetic sequence with common difference $d$, and $a_{1}$ and $d$ are real numbers, $d \neq 0$, and the sum of the first $n$ terms is denoted as $S_{n}$. Let the sets be
$$
\begin{array}{l}
A=\left\{\left.\left(a_{n}, \frac{S_{n}}{n}\rig... | (1) Correct. Proof as follows:
$$
\because S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}, \therefore \frac{S_{n}}{n}=\frac{a_{1}+a_{n}}{2} \text {. }
$$
This shows that $\left(a_{n}, \frac{S_{n}}{n}\right)$ satisfies the equation $y=\frac{a_{1}+x}{2}$. Therefore, for $n \in N$, the point with coordinates $\left(a_{n}, \f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,706 |
Four, (Full marks 35 points) A bridge club stipulates that only four people can play together if no two of them have ever been partners before. At a gathering with 14 participants, each person has previously been partners with 5 others. After playing 3 rounds, they have to stop according to the rules. Just as they are ... | Four, represent members as points. Connect a line between two points if and only if the two have never been partners.
The maximum situation for 14 people is that they have never been partners with each other, thus the maximum number of lines is $C_{14}^{x}=91$.
Since each person has been partners with 5 others, the nu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,708 |
27. Given that line $m$ passes through the center of circle $\odot O$, line $l \perp m$, and $M$ is the foot of the perpendicular. Through two points $A, B$ on line $l$, tangents $A C, B D$ are drawn to $\odot O$, with $C, D$ being the points of tangency.
(1) If $A, B$ are on the same side of point $M$, and $A M > B M$... | Prove (1) On ray $A B$, intercept $A M^{\prime}=A C$, then $B M^{\prime}=B D$. Extend $B D$ to intersect $A C$ at $E$, connect $O C, O D, C M^{\prime}$, $D M^{\prime}$.
Then $O D \perp B E, \triangle C \perp, A C$.
$\therefore O, C, E, \bar{D}$ are concyclic.
Thus $\angle A E B=\angle C O D$.
$$
\begin{aligned}
\becaus... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,709 |
27. If $n$ is a positive integer not less than 2, prove:
$$
\frac{4}{7}<1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}<\frac{\sqrt{2}}{2} .
$$ | $$
\begin{array}{l}
\text { Prove that } \\
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n} \\
=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2 n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) \\
=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2 n}\right)-\left(1+\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,710 |
Example 8. As shown in the figure, given an acute triangle $\triangle ABC$ in the plane, the circle with $AB$ as its diameter intersects the altitude $CC'$ and its extension at $M, N$, and the circle with $AC$ as its diameter intersects the altitude $BB'$ and its extension at $P, Q$. Prove that $M$, $P, N, Q$ are concy... | Prove that connecting $A N, A M, A P, A Q, B M$, $P M$.
$$
\begin{array}{l}
\because A B \text { is the diameter, } C N \perp A B, \\
\therefore A N=A M .
\end{array}
$$
Similarly, $A P=A Q$.
In the right triangle $\triangle A C C^{\prime}$, $\cos A=\frac{A C^{\prime}}{A C}$,
$$
\therefore A C^{\prime}=A C \cos A \tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,712 |
Example 1. In the coordinate plane, does there exist a family of infinitely many lines $l_{1}, l_{2}, \cdots, l_{n}, \cdots$ that satisfy the following conditions:
(1) Point $(1,1) \in I_{n}, n=1,2,3, \cdots$;
(2) $k_{n+1}=a_{n}-b_{n}$, where $k_{n+1}$ is the slope of $l_{n+1}$, $a_{n}$ and $b_{n}$ are the $x$-intercep... | Assume such a family of lines exists, then the equation of $l_{n}$ is $y-1=k_{n}(x-1)$.
When $y=0$, $-1=k_{n}(a_{n}-1), a_{n}=1-\frac{1}{k_{n}}$.
When $x=0$, $b_{n}-1=-k_{n}, b_{n}=1-k_{n}$.
Since $l_{n}$ exists, then $a_{n}, b_{n}$ exist, thus $k_{n} \neq 0, n=1,2,3, \cdots$.
Therefore, $k_{n+1}=a_{n}-b_{n}=k_{n}-\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,713 |
Example 2. Given any 16 points in a plane, the distance between any two points does not exceed 1. Prove: there must exist two points, the distance between which does not exceed $\frac{\sqrt{2}}{4}$.
| Prove that since the distance between any two points does not exceed 1, a square with a side length of 1 can completely cover these 16 points. Divide this square into 16 smaller squares with side lengths of \(\frac{1}{4}\). The diagonal of the smaller squares is \(\frac{\sqrt{2}}{4}\).
(1) If there is a small square c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,714 |
Example 1. As shown in the figure, $B D$: $D C$: $C E$: $E A=2$ : $1, A D, B E$ intersect at $F$, then $A F: F D=$ $\qquad$ | $$
\begin{array}{l}
\because \lambda_{1}=\lambda_{2}=\frac{1}{2}, \\
\therefore \lambda_{3}=\left(1+\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{3}{4} .
\end{array}
$$
In this problem, $\triangle ABC$ only has the characteristics of a basic figure, and $D, F$ are known fixed points dividing $BC, AC$ respectively. Theref... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,715 |
Example 2. Add a figure, in $\triangle A B C$, $D, E$ are the trisection points on $B C$, point $M$ is the midpoint of $A C$, $B M$ intersects $A D$ at $G$,
intersects $A E$ at $H$. Then
$$
\begin{array}{l}
B G: G H: H M \\
=(\quad) .
\end{array}
$$
(A) $3: 2: 1$
(B) $4: 2: 1$
(C) $5:$
(D) $5: 3: 2$ | Analysis: Consider $\triangle A B C$ as the basic triangle. Given that $M$ and $D$ are section points, from (*) we can get
$$
\frac{B G}{G M}=(1+1) \cdot \frac{1}{2}=1,
$$
thus $B G=G M$.
Similarly, in $\triangle A B C$, $\frac{B H}{H M}=(1+1) \cdot 2$ $=4$. Therefore,
$$
\begin{array}{l}
B G+G H=B G+(B G-H M) \\
=2 B... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,716 |
3. Which of the following statements is correct:
(1) $\left[(1-\sqrt{2})^{2}\right]^{\frac{1}{2}}=1-\sqrt{2}$;
(2) In the field of real numbers, $x^{2}+4$ can be factored;
(3) The three sides of one triangle are $12, 18, 27$, and the three sides of another triangle are $6, 4, 9$. Therefore, these two triangles are simi... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,719 |
6. Given
$$
\begin{array}{c}
\frac{a_{2}+a_{3}+a_{1}}{a_{1}}=\frac{a_{1}+a_{3}+a_{4}}{a_{2}}=\frac{a_{1}+a_{2}+a_{1}}{a_{3}} \\
=\frac{a_{1}+a_{2}+a_{3}}{a_{4}}=k .
\end{array}
$$
Then the value of $k$ is ( ).
(A) 3
(B) $\frac{1}{3}$
(C) -1
(D) 3 or -1 | 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 | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,722 |
7. As shown in the figure, in the equilateral $\triangle A B C$, each side is divided into six equal parts, and the number of regular hexagons in the figure is.
(A) 8
(B) 10
(C) 11
(D) 12 | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,723 |
8. In an acute triangle $\triangle A B C$ with all sides of different lengths, $D$ is a point on side $B C$, and $\angle B A D+\angle C=90^{\circ}$. Then $A D$ must pass through the ( ) of $\triangle A B C$.
(A) orthocenter
(B) incenter
(C) circumcenter
(D) centroid | 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 | 707,724 |
1. Given $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$ $\qquad$ | Two, 1.3
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1.3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,725 |
3. As shown in the figure, the side length of $\triangle A B C$ is $2, F$ is the midpoint of $A B$, extend $B C$ to $D$, such that $C D=B C$, connect $E D$ intersecting $A C$ at point $E$. Then the area of quadrilateral $B C E F$ is | 3. $\frac{2}{3} \sqrt{3}$ | \frac{2}{3} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,727 |
Three, (20 points) The equation $x^{2}+(2 m-1) x+(m-6)=0$ has one root no greater than -1, and the other root no less than 1.
(1) (8 points) Find the range of values for $m$;
(2) (12 points) Find the maximum and minimum values of the sum of the squares of the roots of the equation. | $$
\text { Three, } \begin{aligned}
\because \Delta & =(2 m-1)^{2}-4(m-6) \\
& =4(m-1)^{2}+21>0
\end{aligned}
$$
$\therefore$ The equation always has two distinct real roots.
(1) According to the problem.
Let the quadratic function $y=$ $x^{2}+(2 m-1) x$ $+(m-6)$ have the following properties:
[As shown in the figure, ... | 10 \frac{3}{4} \text{ and } 101 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,730 |
(4) (Total 20 points) In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $M$ is a point on $AB$, and $AM^2 + BM^2 + CM^2 = 2AM + 2BM + 2CM - 3$. If $P$ is a moving point on the segment $AC$, and $\odot O$ is the circle passing through points $P, M, C$, and a line through $P$ parallel to $AB$ intersects $\o... | (1) From the given, we have
$$
(A M-1)^{2}+(B M-1)^{2}+(C M-1)^{2}=0,
$$
which implies $A M=B M=C M=1$, meaning $M$ is the midpoint of $A B$.
(2) From $\left\{\begin{array}{l}M A=M C \Rightarrow \angle A=\angle M C A, \\ P D / / A B: \angle A=\angle C P D\end{array}\right.$
$\Rightarrow \angle M C A=\angle C P D$, thu... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,731 |
Five, (20 points) A mall installs an escalator from the first floor to the second floor, moving at a uniform speed that is twice the speed of a child. It is known that a boy passed 27 steps to reach the top of the escalator, while a girl walked 18 steps to reach the top. (Assume the boy and girl each step on one step a... | (1) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, the boy's speed be $2x$ levels/min, and the stairs have $s$ levels, then
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y}
\end{array}\right. \\
\Rightarrow \frac{13.5}{18}=... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,732 |
3. Given the equation $2 x^{2}-2 a x+3 a-4=0$ has no real roots, then the value of the algebraic expression $\sqrt{a^{2}-8 \rho \div 16}+12-a i$ is ( ).
(A) 2
(B) 5
(C) $2 a-6$
(D) $6-2 a$ | 3. $\mathrm{A}$ (Hint: From $\Delta<0$ we get $2<a<4$) | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,735 |
4. Given $y=|x-1|-2|x|+|x+2|$, and $-2 \leqslant x$ $\leqslant 1$. Then the sum of the maximum and minimum values of $y$ is ( ).
(A) -1
(B) 2
(C) 4
(D) 5 | 4. $\mathrm{B}$ (Hint: It follows that $y=-(x-1)-2|x|+x$ $+2=3-2|x|$.
| B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,736 |
1. Calculate $\frac{22223^{3}+11112^{3}}{22223^{3}+11111^{3}}$ The result is | 二、1.1 $\frac{1}{33334}$ Let $a=22223, b=11112$. Then the original expression $=\frac{a^{3}+b^{3}}{a^{3}+(a-b)^{3}}=\frac{a+b}{2 a-b}$. | \frac{1}{33334} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,740 |
2. As shown in the figure, in $\triangle ABC$, $\angle C=90^{\circ}$, points $P, Q$ are on the hypotenuse $AB$, satisfying the conditions $BP = BC = a$, $AQ = AC = b$, $AB = c$, and $b > a$. Draw $PM \perp BC$ at $M$, $QN \perp AC$ at $N$, and $PM$ and $QN$ intersect at $L$. Given that $\frac{S_{\triangle PQL}}{S_{\tri... | 2. $3: 4: 5$ From similar triangles, we get $5(a+b)=7c$, squaring and rearranging gives $12a^2 - 25ab + 12b^2 = 0$, solving for $\frac{a}{b} = \frac{3}{4}$ or $\frac{4}{3}$ (discard), hence $a: b: c = 3: 4: 5$. | 3: 4: 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,741 |
3. Given in $\triangle A B C$, $\angle A \leqslant \angle C \leqslant \angle B$, and $2 \angle B$ $=5 \angle A$. Then the range of $\angle B$ is $\qquad$ . | 3. $75^{\circ} \leqslant \angle B \leqslant 100^{\circ}$ | 75^{\circ} \leqslant \angle B \leqslant 100^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,742 |
Three. (20 points) Suppose the equations $x^{2}-m x+5+m=0$ and $x^{2}-$ $(17 m+1) x+13 m+7=0$ have at least one common root. Find all possible values of the product of the four roots of these two equations.
Given the equations $x^{2}-m x+5+m=0$ and $x^{2}-$ $(17 m+1) x+13 m+7=0$ have at least one common root, we need ... | Let $\alpha$ be the common root of the two equations, then
$\alpha^{2}-m \alpha+5+m=0$,
$a^{2}-(7 m+1) \alpha+13 m+7=0$.
(1)-(2) gives $(6 m+1) \alpha=12 m+2$.
If $m=-\frac{1}{6}$, the two equations are identical,
thus the required product is $\frac{841}{36}$.
If $m \neq-\frac{1}{6}$, from (3) we get $\alpha=2$, subst... | \frac{841}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,744 |
Four. (25 points) As shown in the figure, let $P$ be any point on the side $BC$ of the equilateral $\triangle ABC$. Connect $AP$, and construct the perpendicular bisector of $AP$ to intersect $AB$ and $AC$ at $M$ and $N$ respectively. Prove that $BP \cdot PC = BM \cdot CN$.
保留源文本的换行和格式，直接输出翻译结果如下：
Four. (25 points) A... | Connect $P M, P N$. Since $M N$ is the perpendicular bisector of $A P$, then $\triangle M P N \cong \triangle M A N$, so $\angle M P N=\angle M A N=60^{\circ}$. It can be deduced that $\angle B M P=\angle N P C$, thus $\triangle B P M \backsim \triangle C N P$. Therefore, $B P \cdot P C=B M \cdot C N$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,745 |
Five. (25 points) For any real numbers $x, y$, define the operation as follows: $x * y=a x+b y+c x y$, where $a, b, c$ are given numbers, and the right-hand side involves the usual addition and multiplication of numbers (for example, when $a=1, b=2$, $c=3$, then $1 * 3=1 \times 1+2 \times 3+3 \times 1 \times 3=16$). It... | Given $x * d=a x+b d+c d x$, for all real numbers $x$, we have $a x+b d+c d x=x$, which means $(a+c d-1) x+b d=0$. Therefore, we have
$$
\left\{\begin{array}{l}
a+c d-1=0, \\
b d=0 .
\end{array}\right.
$$
Since $d \neq 0$, it follows that $b=0$. Thus, $x * y=a x+c x y$. From $1 * 2=3$ and $2 * 3=4$, we get
$\left\{\be... | a=5, b=0, c=-1, d=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,746 |
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