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 |
|---|---|---|---|---|---|---|---|---|---|
2. Let $n$ be a positive integer, and let $A_{1}, A_{2}, \cdots, A_{2n+1}$ be subsets of some set $B$. Suppose
(a) each $A_{i}$ contains $2 n$ elements,
(b) $A_{i} \cap A_{j} (1 \leqslant i<j \leqslant 2 n+1)$ contains exactly one element,
(c) every element of $B$ belongs to at least two subsets $A_{i}$.
Question: For... | 2. We first prove that conditions $(a), (b), (c)$ together can imply a stronger condition than (c), namely
( $C^{*}$ ) Each element in $B$ belongs to exactly two subsets $A_{i}$. First, we have the equation
$$
\begin{array}{l}
A_{1}=\left(A_{2} \cap A_{1}\right) \cup\left(A_{3} \cap A_{1}\right) \cdots \\
\cup\left(A_... | n \text{ is even} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,816 |
In the tetrahedron $ABCD$, let $K$ and $L$ be the midpoints of edges $AB$ and $CD$ respectively. Prove: every plane containing the line $FL$ divides the tetrahedron into two equal volume parts. | 6. (Czechoslovakia 3)
Solution: Let $\omega$ denote the plane containing line $A B$ and parallel to $C D$. In space, define the point transformation $f: X \rightarrow X^{\prime}$: If $X \in K L$ then $X^{\prime}=\lambda^{\prime}$, if $X \in K L$, let $p$ be the intersection line of the plane $X K L$ and the plane thro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,817 |
11. (France 3) Let $u_{1}, u_{2}, \cdots, u_{a}$ be vectors in the plane, each with length $\leqslant 1$, and their sum is zero. Prove: $u_{1}, u_{2}, \cdots, u_{2}$ can be rearranged into a sequence $v_{1}, v_{2}, \cdots, v_{n}$, such that each partial sum $v_{1}, v_{1}+v_{2}, v_{1}+v_{2}+v_{3}, \cdots, v_{1}+v_{2}+\c... | 3. (France 3)
Xin: Considering the situation 1, i.e., these closed stars are all on a straight line $\mathrm{L}$, at this time they are multiples of a unit vector $e$, we can write them as $u_{1}=x_{1} e, u_{2}=x_{2} e$, $\cdots, u_{m}=x_{m} e$, where $x_{1}, \cdots, x_{m}$ are numbers, their sum is 0, and $|x| \leqsl... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,819 |
18. (GDR 1) Let $N=\{1, 2, \cdots \cdots, n\}, n \geqslant 2$.
$F=\left\{A_{1}, \cdots, A_{\mathrm{t}}\right\}, A_{i} \subseteq N, i=1,2$. $\cdots, t$. If for every pair of elements $\{x, y\}$ in $N$, there exists $A_{i} \in F$ such that $A_{i} \cap\{x, y\}$ contains exactly one element, then $F$ is called separating. ... | 10. (Ji Zhu Germany 1)
Solution:
$$
f(m)=\left\{\begin{array}{ll}
{\left[\log _{2} n\right]+1,} & \text { if } n=2^{r} \\
{\left[\log _{2} n\right],} & \text { if } n \neq 2^{r}
\end{array}\right.
$$
There is a set $F=\left\{A_{1}, \cdots, A_{1}\right\}$ that is separating for an $n$-element set $N$ only when $n \leq... | f(n)=\left\{\begin{array}{ll}
{\left[\log _{2} n\right]+1,} & \text { if } n=2^{r} \\
{\left[\log _{2} n\right],} & \text { if } n \neq 2^{r}
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,820 |
31 (Hungary 1) For which $n$, does there exist an $n \times n$ matrix composed of -1, 0, or 1, such that the sums of the elements in each row and the sums of the elements in each column result in $2n$ distinct sums? | 14. (Hungary 1)
Solution: Consider a matrix with the mentioned properties, and let $r_i$ denote the sum of the elements in the $i$-th row. Let $c_j$ denote the sum of the elements in the $j$-th column. Clearly,
$$
\sum_{r}\left|+\Sigma_{c j}\right| \geqslant \sum_{\mathrm{k}=-\mathrm{n}}^{N}|k|-n=n^{2} .
$$
Since in ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,823 |
34 (Ice Body 1) Let $ABC$ be an acute triangle. Three lines $L_{\mathrm{A}}, L_{\mathrm{B}}, L_{\mathrm{C}}$ pass through vertices $A, B, C$ respectively in the following manner: Let $H$ be the foot of the perpendicular from $A$ to $BC$: $S_{\mathrm{A}}$ is the circle with diameter $AH$; $S_{\mathrm{A}}$ intersects sid... | 15. (Iceland 1)
Proof: Let $A, B, C$ denote the three angles, and $a, b, c$ denote the lengths of the corresponding opposite sides. By the construction of $L_{\mathrm{A}}$, we have
$$
\begin{aligned}
\angle A M N & =\angle A H N=90^{\circ}-\angle H A C \\
& =C .
\end{aligned}
$$
Similarly, $\angle A N M=B$. Therefore... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,824 |
47. (Israel 2) In the convex pentagon $ABCDE$, sides $BC, CD, DE$ are equal, and each diagonal is parallel to one side (e.g., $AC$ is parallel to $DE$, $BD$ is parallel to $AE$, etc.). Prove that $ABCDE$ is a regular pentagon. | 17. (Israel 2)
Proof: Let $A C, A D$ intersect $B E$ at $R, S$, and let $\angle C B D=\alpha$, then
$$
\begin{aligned}
a & =\angle B D A=\angle C D B=\angle D B E \\
& =\angle A E B=\angle D A E .
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
\angle A C E & =\angle D E C=\angle D C E=\angle C E B \\
& =\angle B A C=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,825 |
49. (感洏㢦1) Let $f(n)$ be a function defined on the set of positive integers, with values also in this set. If for all positive integers $n, m, f(f(n)+f(m))=n+n$ holds, find all possible values of $f(1938)$.
The text above is translated into English, preserving the original text's line breaks and format. Here is the tr... | 19. (Jiangxi 1)
Solution: We prove that for all $n$, $f(n)=n$, so $f(1983)=1988$ is the only possible value.
Let $r=f(1), s=f(2)$. For simplicity, we denote the equations as follows:
$$
\begin{array}{l}
f(f(n)+f(m))=n+m, \quad (1) \\
f(1)=r, \quad (2) \\
f(2)=s. \quad (3)
\end{array}
$$
From (1) and (3), $f(2s)=4$.
... | 1938 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,826 |
3. $N$ is the set of positive integers. Define the function $f$ on $N$ as follows:
$f(1)=1, f(3)=3$, and for $n \in N$ we have $f(2 n)=f(n)$,
$f(4 n+1)=2 f(2 n+1)-f(n)$,
$f(4 n+3)=3 f(2 n+1)-2 f(n)$.
Question: How many $n \in N$, and $n \leqslant 1988$ such that $f(n) = n$? | 3. According to these formulas, we can find
The pattern shown in this table seems to be
$$
\begin{array}{l}
f\left(2^{k}\right)=1, f\left(2^{k}-1\right)=2^{k}-1, \\
f\left(2^{k}+1\right)=2^{k}+1 .
\end{array}
$$
This suggests that we should consider the "binary" expansion of natural numbers. Our conjecture is:
$f(n)$... | 92 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,827 |
54. (Mongolia 4) Find the smallest natural number $\boldsymbol{n}$ such that if the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two non-intersecting subsets, then one of the subsets contains 3 different numbers, where the sum of two of them equals the third. | 20. (Mongolia 4)
Solution: The required minimum value is 96.
Let $A_{\mathrm{s}}=\{1,2, \cdots, n\}$ be divided into two subsets $B_{\mathrm{n}}$ and $C_{\mathrm{s}}$, and neither $B_{\mathrm{s}}$ nor $C_{\mathrm{s}}$ contains three distinct numbers where the product of two equals the third. If $n \geqslant 96$, then ... | 96 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,828 |
61. (Poland 4) 49 students solved 3 problems. Each problem's score can be one of 0, 1, 2, 3, 4, 5, 6, 7. Prove: For each problem, there are two students $A$ and $B$ such that $A$'s score is at least as much as $B$'s. | 21. (Boss 4)
Solution: Let $X$ be the set of all ordered triples $\left(a_{1}, a_{2}, a_{3}\right)$, where $a_{i} \in \{0,1,2,3,4,5, 6,7\} (i=1,2,3)$. For elements $a=\left(a_{1}, a_{2}, a_{3}\right), b=\left(b_{1}, b_{2}, b_{3}\right)$ in $X$, if $a_{i} \leqslant b_{i}$ $(i=1,2, 3)$, and $a \neq b$, we write $a \prec... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,829 |
For consecutive integers, prove that there do not exist integers $x_{1}, x_{2}, \cdots, x_{p}$, such that:
$$
\sum_{i=1}^{p} x_{i}^{2}-\frac{4}{4 p+1}\left(\sum_{i=1}^{p} x_{i}\right)^{2}=1 \text {. }
$$
Or prove that there are only finitely many values of $p$, for which there exist integers $x_{1}, x_{2}, \cdots, x_{... | 22. (South Korea 2)
Proof: Let $j=n(n+1)(n \geqslant 3), \sum_{\mathrm{i}=1}^{p} x_{\mathrm{i}} = X$, since $4 p+1=(2 n+1)^{2}$, the given equation becomes
$$
(2 n+1)^{2}\left(\sum_{\mathbf{i}=1}^{\mathbf{p}} x_{\mathbf{i}}^{2}-1\right)=4 x^{2}.
$$
Since $x_{i}^{2} \equiv x_{i} (\bmod 2), \sum_{i=1}^{p} x_{i}^{2} \eq... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,830 |
74. (Sweden 2) Let $\left\{a_{k}\right\}$ be a sequence of non-negative real numbers such that for $k=1,2, \cdots$,
$$
a_{k}-2 a_{k+1}+a_{k+2} \geqslant 0, \sum_{i=1}^{k} a_{j} \leqslant 1 .
$$
Prove: $0 \leqslant\left(a_{k}-a_{k+1}\right)<\frac{2}{k^{2}}, k=1,2, \cdots$. | 24. (Hudian 2)
Proof: By the monotonicity condition, $a_{k}-a_{k+1}$ is a decreasing sequence. If for all $k, a_{k}-a_{k+1}=-d$, $\delta>0$, then for all $m>k, a_{m}-a_{m+1}1$, which is a contradiction. Therefore, for all $k, a_{k}-a_{k+1} \geqslant 0$.
Now suppose for some $k, a_{k}-a_{k+1} \geqslant \frac{2}{k^{2}}$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,832 |
76.(UK 1) A positive integer written in decimal notation is called a "composite palindrome" if it consists of a segment of digits not starting with 0 followed immediately by the same segment of digits: for example, 360360 is a composite palindrome, but 36036 is not. Prove that there are infinitely many composite palind... | 25. (UK 1)
Proof: Note that 7 divides $1001=10^{3}+1, 10^{3}$ $=-1 \div 2 \cdot 143$. By the binomial theorem, $10^{21}=-1$ $+7 \cdot 7 \cdot 143+(\text{multiple of } 7^{2})$, i.e., $10^{21}=-1$ $(\bmod 49)$. Analyze this: $10^{21 \pi} \equiv-1(\bmod 49)$ (for odd number $n$).
Therefore, $9\left(10^{21 n}+1\right) / ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,833 |
79. (UK 4) $\triangle A B C$ is an acute-angled triangle, $L$ is any line in the plane of the triangle, and $u$, $v$, $\omega$ are the lengths of the perpendiculars from $A$, $B$, $C$ to $L$. Prove that:
$$
u^{2} \tan A + v^{2} \tan B + \omega^{2} \tan C \geqslant 2 \Delta \text{, }
$$
and determine the line $L$ for w... | 27. (UK 4)
Solution: Take the foot of the perpendicular from $A$ to $BC$ as the origin. Since $\triangle ABC$ is an acute triangle, the origin $O$ is between $B$ and $C$. Take the line $OC$ as the $Ox$ axis and the line $OA$ as the $Oy$ axis. Let $A(0, \alpha), B(-\beta, 0)$, and $C(\gamma, 0)$ ($\alpha, \beta, \gamma... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,834 |
83. (USA 3) Several signal lights are placed at intervals along a single-track railway and are marked in order $1$, $2$, $\cdots$, $N(N \geqslant 2)$. According to safety operation rules, if a train is running between a signal light and the next one, no other trains are allowed to pass through this signal light. Howeve... | 29. (USA 3)
Proof: Let the first train depart from signal 1 at time 0, and let $T_{\mathrm{i}}$ represent the time it takes for the $j$-th train to travel from one signal to the next. By induction on $k$, we prove that the time for train $k$ to reach signal $N$ is $S_{\mathrm{k}}+(N-2) M_{\mathrm{k}}$, where $S_{\math... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,836 |
84. (USSR 1) On the side $AC$ of $\triangle ABC$, choose a point $M$ such that the incircles of triangles $ABM$ and $BMC$ have equal radii. Prove:
$$
BM^{2}=\triangle \operatorname{ctg}\left(\frac{B}{2}\right),
$$
where $\triangle$ is the area of triangle $ABC$. | 30. (Su Kuan 1)
Proof: Let $\triangle_{1}$ be the area of $\triangle A B M$, $s_{1}$, the semi-perimeter of $\triangle A B M$, and $r_{1}$ the radius of the incircle of $\triangle A B M$: $\triangle_{2}$ is the area of $\triangle B M C$, $s_{2}$ is the semi-perimeter of $\triangle B M C$, and $r_{2}$ is the radius of ... | BM^{2}=\triangle \operatorname{ctg}\left(\frac{B}{2}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 703,837 |
35 . (Soviet 2) An even number of people are sitting around a round table discussing. After a break, they sit around the table again in a different order. Prove: at least two people will have the same number of people sitting between them as before. | 1. (Soviet 2)
Proof: Consider $2n$ vectors starting from the center of a circle (table), dividing the circumference into $2n$ equal segments. Let the endpoint of each vector determine the seat of a participant, and each participant corresponds to a vector.
After a break, each vector rotates by a positive angle to some... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,839 |
15. Let $S$ be an $n$-element set, and $P_{\mathbf{I}}(k)$ be the number of permutations of set $S$ that have exactly $k$ fixed elements. Prove:
a) $\sum_{k=0}^{n} k P_{0}(k)=n!$;
b) $\sum_{k=0}^{n}(k-1)^{2} P_{\mathbf{A}}(k)=n$!. | 15. $a$) is the first problem of the first test of the 28th 1MD, so the proof is omitted.
$b$) For each permutation of $S$, we define an $n$-dimensional vector $\left(d_{1}, d_{2}, \cdots, d_{n}\right)$. If $\mathbf{i}$ is one of the $k$ fixed points of this permutation, then $d_{i}=k$, otherwise $d_{1}=0(1<i \leqslant... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,843 |
16. Prove: There exists a method to color the set $M=\{1, 2, \cdots, 1987\}$ with four colors such that any 10-term arithmetic sequence composed of elements from $M$ is not monochromatic.
Note: The equivalent statement of this proposition is:
Let $M=\{1,2, \cdots, 1987\}$. Prove that there exists a function $f: M \righ... | 16. The number of ways to color set $M$ with 4 colors is equal to $4^{1087}$. Let $A$ be the number of arithmetic sequences of 10 elements in $M$, and the number of colorings containing a monochromatic arithmetic sequence of 10 elements is less than $4 A \cdot 4^{1087-10}$. If it is proven that $A \cdot 4^{1087-0}<4^{1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,844 |
18. Let $\alpha, \beta, \gamma$ be positive real numbers satisfying $\alpha+\beta+\gamma$. Prove: the segments of length $\sin \alpha, \sin \beta, \sin \gamma$ can form a triangle, and its area is no greater than
$$
{ }_{8}^{1}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) .
$$ | 18. According to the conditions given in the proposition, a tetrahedron can be constructed with dihedral angles at the vertices being $2 \alpha, 2 \beta, 2 \gamma$, and side edges equal to 1. The base of this tetrahedron consists of $-\infty$ triangles, with side lengths of $2 \sin \alpha, 2 \sin \beta, 2 \sin \gamma$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,846 |
19. Let $f(x)=x^{2}+x+p, p \in N$. If $f(\mathrm{j}(\mathrm{B})), f(\mathrm{i}), \cdots, f\left(\sqrt{\frac{p}{3}}\right)$ are all prime numbers, then all the numbers $f(0), f(1), \cdots, f(p-2)$ are also prime. | 19. Let $y$ be the smallest non-negative integer satisfying the inequality $y \leqslant p-2$. And let $f(y)$ be a composite number. Let $q$ be the smallest prime factor of $f(y)$.
We first prove that $q>2 y$. Otherwise, assume $q \leqslant 2 y$, and analyze the difference
$$
f(y)-f(x)=(y-x)(y+x+1).
$$
If $x$ changes ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,847 |
5 . In right triangle $A B C$, $A D$ is the altitude on the hypotenuse $B C$, and the line connecting the incenter of triangle $A B D$ with the incenter of triangle $A C D$ intersects sides $A B$ and $A C$ at points $K$ and $L$, respectively. The areas of triangles $A B C$ and $A K L$ are denoted as $S$ and $T$, respec... | 5. Let the incenter of $\triangle A B D$ be $M$, and the incenter of $\triangle A C D$ be $N$. Since $\triangle A D B \sim \triangle C D A$, note that $D M$ and $D N$ are two external angle bisectors from $D$, hence we have
$$
\frac{n M}{D N}=\frac{B D}{A D},
$$
and it is clear that $\angle M D N=90^{\circ}$. Therefor... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,849 |
22. Does there exist a function $f: N \rightarrow N$, such that for any natural number $n, f(f(n))=n+1987$? | 22. The answer is negative. That is, "there does not exist a function $f: N \rightarrow N, N=\{0,1,2,3$, $\cdots, n, \cdots\}$, such that for any $n \in N, f[f(n)]$ $=n+1987$." | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,851 |
2. Given six numbers that sum up to 63 consecutive unique natural numbers, find these 6 numbers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | We will arrange the required numbers in a non-decreasing sequence:
$x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{0}$. The sum of some of the numbers $x_{1}, x_{2}, \cdots, x_{8}$ can be expressed as:
$$
\varepsilon_{1} x_{1}+\cdots+\varepsilon_{e} x_{0} \text {. }
$$
where $\varepsilon_{\mathrm{i}}=0,1$, and at... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 703,853 |
3. Given a regular heptagon $A_{1} A_{2} \cdots A_{7}$, prove:
$$
\frac{1}{A_{1} A_{2}}=\frac{1}{A_{1} A_{3}}+\frac{1}{A_{1} A_{4}} .
$$ | Proof: Let
$$
\begin{array}{l}
A_{1} A_{2}=a, A_{1} A_{3} \\
=b, A_{\mathrm{T}} A_{4}=c,
\end{array}
$$
and draw the diagonals (Figure 1). Let B be the intersection point of the diagonals $A_{1} A_{\text { and }}$ $A_{3} . A_{7}$. It is easy to show that $A_{1} A_{2} A_{3}$
is a trapezoid,
$$
A_{1} A_{4}=A_{4} A_{7} \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,854 |
4. In a "Battleship" game on a grid with $7 \times 7$ cells, what is the minimum number of shots needed to guarantee hitting a four-deck ship? It is known that the ship (1) has shape $A$ (Figure 2); (2) consists of four adjacent cells, but is not a square. | Solution: (1) Since 12 ships of shape $A$ can be placed in a $7 \times 7$ grid such that no two ships share a common cell, the number of shots required is at least 12. From figures $3, a, b$, it is shown that 12 shots are sufficient.
(2) 20 steps.
As shown in Figure 4, 20 shots can complete the task. Suppose we shoot ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 703,855 |
7. The vertex $B$ of $\angle ABC$ is outside the circle, and the rays $BA$ and $BC$ intersect the circle. A line perpendicular to the angle bisector is drawn through the intersection point $K$ of ray $BA$ and the circle, intersecting the circle at point $P$ and ray $BC$ at point $M$. Prove that the segment $PM$ is twic... | Proof: Let $O^{\prime}$ be the point symmetric to the center $O$ with respect to the angle bisector $l$ of $\angle A B C$ (Figure 8). Points $K$ and $M$ are also symmetric with respect to $K O, P O$, and $M O^{\prime}$. Clearly, the intersection point $Q^{\prime}$ of $K O$ and $M O^{\prime}$ lies on $l$. From the isosc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,858 |
6. Positive integers $a$ and $b$ make $a b+1$ divide $a^{2}+b^{2}$. Prove that $\quad \frac{a^{2}+b^{2}}{a b+1}$ is the square of some positive integer. | 6. Since $a, b$ are symmetric in the expression $\frac{a^{2}+b^{2}}{a b+1}$, without loss of generality, assume $a \geqslant b$. When $a=b$, there exists a positive number $q$ such that
$$
\frac{2 a^{2}}{a^{2}+1}=q \text {. }
$$
This implies
$$
(2-q) a^{2}=\hat{q} \text {. }
$$
From this, it is clear that $a=1$. In t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,860 |
Place one more of these shapes inside the square.
Keep the original text's line breaks and format, and output the translation result directly.
Note: The second sentence is a directive for the translation task and should not be included in the translation output. Here is the translation:
Place one more of these shap... | Solution: For each $2 \times 2$ square, at least two cells should be covered. An $8 \times 8$ square can be divided into 16 non-overlapping $2 \times 2$ squares. Therefore, at least 11 figures should be covered. With 11 figures, it is feasible. | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,861 |
4. On the plane, there are two closed broken lines, each with an odd number of segments. Each segment of these broken lines lies on a straight line, none of which coincide, and no three lines intersect at the same point: Prove that from each broken line, one can select a segment such that they form a pair of opposite s... | First, we will prove two auxiliary conclusions.
Lemma 1. The regions into which the plane is divided by a closed broken line satisfying the conditions of the problem can be colored with two colors such that adjacent regions are colored differently (if two regions of the plane have a common boundary containing a line se... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,863 |
5. Uncle Qiel Yuemuer every night selects 9 or 10 warriors from 33 warriors to be on duty according to his own wishes. At least how many days are needed to ensure that each warrior has the same number of duty shifts? | The minimum number of days is 7 days, during which each warrior is on duty 2 times.
Let the required minimum number of days be $N$, and each warrior is on duty $n$ times within these $N$ days. We number the warriors sequentially as $1,2, \cdots, 33$, and introduce the number $a_{1 \times}(i=1,2, \cdots, N ; k=1,2, \cd... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,864 |
2. Prove: In a convex pentagon $A B C D E$, if $\angle A B C=\angle A D E, \angle A E C=\angle A D B$. Then $\angle B A C=\angle D A E$.
保留源文本的换行和格式，直接输出翻译结果。 | Proof: Suppose the lines $B D$ and $C E$ intersect at point $F$. Since $\angle A E F = \angle A D F$, then points $A, F, D, E$ are concyclic. From this, we deduce that $\angle A B C = \angle A D E = \angle A F E$ and $\angle A B C + \angle A F C = 180^{\circ}$. Therefore, points $A, B, C, F$ are also concyclic. Consequ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,869 |
3. Find all values of $\alpha$ such that all terms in the sequence $\cos \alpha$, $\cos 2 \alpha, \cos 4 \alpha, \cos 8 \alpha, \cdots$ are negative. | The solution for the value of $\alpha$ is $\pm \frac{2 \pi}{3}+2 \pi m, m \in Z$. In fact, if $\alpha$ satisfies the conditions of the problem, then $\cos \alpha \leqslant -\frac{1}{4}$ (otherwise $-\frac{1}{4} < 0$). For the same reason, for any $n \in N$, we have $\cos 2^{2} \alpha \leqslant -\frac{1}{4}$, thus $\cos... | \pm \frac{2 \pi}{3}+2 \pi m, m \in Z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,870 |
4 . In a square grid with $1987 \times 1987$ cells, remove any one cell. Try to prove: the remaining part can be divided into several L-shaped pieces (Figure 2) consisting of three cells. | Notice that $1987=6 \cdot 331 + 1$, we use mathematical induction to prove the conclusion for the general case of a $(6n+1) \times (6n+1)$ square, where $n \in \mathbb{N}$.
（1）When $n=1$, i.e., for a $7 \times 7$ square, the conclusion holds. In fact, by symmetry, it is sufficient to consider the following cases, i.e.... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 703,872 |
5 . Two people take turns writing numbers on the blackboard that do not exceed $n$. The game rules state that it is not allowed to write a number that is a factor of any number already written on the blackboard. The player who has no numbers left to write loses.
（1）Determine who has a winning strategy when $n=10$, and ... | Solution: The first player has a winning strategy.
(1) The first player writes 6 on the first turn, after which the other player can only write one of the numbers $4,5,7,8$, 9, 10. We pair these numbers as (4, 5), $(7,9),(8,10)$, and whenever it's the first player's turn, they just write the number from the pair of the... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,873 |
6. The function $y=f(x)$ is defined over the entire real axis, and its graph remains unchanged after a rotation of $\frac{\pi}{2}$ around the origin.
(1) Prove that the equation $f(x)=x$ has exactly one solution.
(2) Provide an example of such a function. | Proof: (1) Let $f(0)=y_{0}$, then the point $\left(0, y_{0}\right)$ on the graph of the function $y=f(x)$, after two rotations (each rotation by $\frac{\pi}{2}$), results in the point $\left(0,-y_{0}\right)$, which is also on the graph. Therefore, $y_{0}=-y_{0}$, so $f(0)=0$. On the other hand, for any solution $x=x_{0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,874 |
7. All faces of a convex polyhedron are triangles. Prove: each edge can be painted either red or blue, such that from any vertex to any other vertex, one can walk only along red edges, and also only along blue edges. | Proof: We take all the faces
that have a common vertex $A$, and color the edges of this part as shown in Figure 18 (using lines of the same thickness to represent edges of the same color). This way, for the vertices of the colored part, the necessary conditions of the problem can be satisfied. Add faces that share a co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,875 |
Let $\triangle A B C$ have three sides $a, b, c$ of integer lengths, and its inradius is 1.
Prove: $\triangle A B C$ is a right triangle. | Proof: Let the incenter of $\triangle ABC$ be $O$, and the points of tangency on sides $BC$, $CA$, and $AB$ be $A_1$, $B_1$, and $C_1$ respectively, with $CA_1 = CB_1 = x$, $AB_1 = AC_1 = y$, and $BC_1 = BA_1 = z$. Then,
$$
x + y = b, \quad y + z = c, \quad z + x = a.
$$
$$
\therefore \quad x = \frac{a + b - c}{2}, \qu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,877 |
Given $\{a$.$\}$ is a sequence of positive integers,
$$
\begin{aligned}
& a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{0}\right), \quad(a \in N) . \\
a_{0}= & 2238 .
\end{aligned}
$$
Find $a_{1}, a_{2}, a_{3}$. | Solution: $\because a_{4}=a_{3}\left(a_{2}+2 a_{1}\right)$,
$a_{5}=a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{3}\right)$
$\therefore a_{6}=a_{3}^{2}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)$
- $\left(a_{2}+2 a_{1}+2\right)$.
Also, $a_{0}=2288=2^{4} \times 11 \times 13$.
$\because\left\{a_{n}\right\}$ is... | a_{1}=5, a_{2}=1, a_{3}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,878 |
Three, given a chessboard consisting of $16 \times 16$ small squares, these small squares are alternately colored black and white (as shown in the figure). All the small squares in the $i$-th row and the $j$-th column are inverted in color, meaning black squares become white and white squares become black, for $1 \leqs... | Solution: Perform the above operation on each small square where $i+j=$ even (i.e., the small squares marked in the figure) to turn all small squares white. Or perform the above operation on each small white square where $i+j=$ odd (i.e., the small squares marked in the figure) to turn all small squares black.
Proof: ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,879 |
7. We call fractions with a numerator of 1 and a denominator of a natural number greater than 1 unit fractions. If the unit fraction $\frac{1}{6}$ is expressed as the sum of two different unit fractions, then all possible representations are | 7. Let $\frac{1}{6}=\frac{1}{x}+\frac{1}{y}, x<y, x, y \in N$. Then we have $\frac{1}{x}<\frac{1}{6}<\frac{2}{x}$, which means $6<x<12$. When $x=7$, from $\frac{1}{y}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}$, we get $y=42$; when $x=8$, from $\frac{1}{y}=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}$, we get $y=24$; when $x=9$, fro... | \frac{1}{6}=\frac{1}{7}+\frac{1}{42}=\frac{1}{8}+\frac{1}{24}=\frac{1}{9}+\frac{1}{18}=\frac{1}{10}+\frac{1}{15} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,887 |
8. A number plus 79 becomes a perfect square, and when this number plus 204 is added, it becomes another perfect square. Then this number is $\qquad$ . | $$
\begin{array}{c}
204=y^{2}, \text { where } x \text { and } y \text { are non-negative integers. Thus, } \\
y^{2}-x^{2}=125=5^{3}, \text { i.e., } (y+x)(y-x)=5^{3}. \\
\therefore \quad\left\{\begin{array} { l }
{ y + x = 125, } \\
{ y - x = 1; }
\end{array} \text { or } \left\{\begin{array} { l }
{ y + x = 25, } \... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,888 |
9. Let $E(n)$ denote the greatest integer $k$ that is a divisor of the product $1^{1}, 2^{2}, 3^{3} \cdots n^{n}$. Then $E(150)=$ | $9.1,2, \cdots, 150$ contains 30 multiples of 5, each having one factor of 5 removed, along with their respective exponents, the exponent of 5 is
$$
\begin{array}{l}
5+5 \cdot 2+5 \cdot 3+\cdots+5 \cdot 30=5(1+2+3 \\
+\cdots+30)=2325 .
\end{array}
$$
1. $2, \cdots, 150$ contains 6 multiples of $5^{2}$, each having one ... | 2975 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,889 |
10. Let $M$ be the set of all $\triangle ABC$ satisfying the conditions $BC=1, \angle A=2 \angle B$. If $\triangle ABC \in M$, and the angle bisector $CD$ of $\angle C$ divides the opposite side $AB$ into two segments in the ratio $AD:DB=t$, then $\triangle ABC$ is associated with the real number $t$. Determine the set... | 10. Let $\triangle A B C \in M$. Draw the angle bisector $A E$ of $\angle A$, then!! According to the given, $\angle 1=\angle 2=\angle B$. Therefore, $\triangle A E C \sim \triangle B A C$. Combining with the properties of the angle bisector of a triangle,
we get $t=\frac{A D}{D B}=\frac{A C}{D C}=\frac{E C}{A C}=\fra... | \left(\frac{1}{2}, 1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,890 |
$$
\begin{array}{l}
\text { 11. Let } A=\{x \mid-2 \leqslant x \leqslant a\}, \\
B=\{y \mid y=2 x+3, x \in A\}, \\
C=\left\{z \mid z=x^{2}, x \in A\right\} \text {, and } C \subseteq B, \text { then }
\end{array}
$$ | $$
\begin{array}{l}
\text { 11. } B=\{y \mid y=2 x+3, x \in A\} \\
=[-1,2 a+3] \\
\text { When } a \geqslant 2 \text {, } C=\left\{2 \mid z=x^{2}, \quad x \in A\right\} \\
=\left[0, a^{2}\right], \| C=E \text { when } a^{2} \leqslant 2 a+3 \text {, } \\
\end{array}
$$
$$
2 \leqslant 0<3 \text {. } 0 \leqslant a<2 \text... | \left[\frac{1}{2}, 3\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,891 |
4. (Canada 1) ～Let triangle $ABC$ be inscribed in a circle, and the angle bisectors of angles $A, B, C$ intersect this circle at points $A^{\prime}, B^{\prime}, C^{\prime}$, respectively.
Prove: The area of triangle $A^{\prime} B^{\prime} C^{\prime}$ is greater than or equal to the area of triangle $ABC$. | 3. (Canada 1)
Proof: Let $R, r, s, \triangle$ denote the circumradius, inradius, semiperimeter, and area of $\triangle ABC$, respectively, and let $\triangle^{\prime}$ denote the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then,
$$
\begin{array}{l}
\triangle=\frac{a b c}{4 R}=2 R^{2} \sin A \sin B \sin C, \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,893 |
5. The area enclosed by the curve $x^{2}+y^{2}-|x|-|y|=0$ is $\qquad$ . | 5. When $x \geqslant 0, y \geqslant 0$, the given equation is $x^{2}+y^{2}-x-y=0$, which can be rewritten as $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} = \frac{1}{2}$. The corresponding curve in the first quadrant (including endpoints) is a semicircle with its center at $\left(\frac{1}{2}, \frac{1}{... | 2+\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,898 |
6. In $\triangle A B C$, if $2 \cos A+\cos B$ $+\cos C=2$, then the equal quantitative relationship satisfied by the three sides $a, b, c$ is | $\begin{array}{l}6. \text { From } 2 \cos A+\cos B+\cos C=2 \text { we get } \\ \cos B+\cos C=2(1-\cos A), \\ 2 \cos \frac{B+C}{2} \cos \frac{B-C}{2}=4 \sin ^{2} \frac{A}{2}. \because \frac{A}{2}+\frac{B}{2} \\ +\frac{C}{2}=\frac{\pi}{2}, \therefore \cos \frac{B+C}{2}=\sin \frac{A}{2}, \sin \frac{B+C}{2} \\ =\cos \frac... | b+c=2a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,899 |
7. All ordered integer pairs $(x, y)$ that satisfy the equation $x^{2}+2 x y+3 y^{2}-2 x$ $+y+1=0$ are | 7. The original equation is $x^{2}+2(y-1) x+\left(3 y^{2}\right.$ $+y+1)=0 . \because x \in R, \quad \therefore \Delta=4\left[(y-1)^{2}\right.$ $\left.-\left(3 y^{2}+y+1\right)\right] \geqslant 0, -\frac{3}{2} \leqslant y \leqslant 0$, but $y$ is an integer, so $y$ can only take the values $0, -1$.
When $y=0$, $x^{2}-... | (1,0),(1,-1),(3,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,900 |
8 . All three-digit numbers $a b c$ that satisfy $a b c=(a+b+c)^{3}$ are $\qquad$ | 8. $\because 100 \leqslant a b c=(a+b+c)^{3} \leqslant 999$,
$\therefore 5<a+b+c$ S 9 . Directly calculating $5^{3}=125$, $6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729$, we can see that only the three-digit number 512 meets the conditions of the problem. | 512 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,901 |
9. The central angle of sector $OAB$ is $45^{\circ}$, with a radius of 1. Rectangle $PQMN$ is inscribed in this sector (as shown in the figure). The minimum length of the diagonal $PM$ of rectangle $PQMN$ is | $\begin{array}{l}\text { 9. In the figure } O M \text {, let } \angle N P M=\alpha, P M \\ =l \text {, then } P Q=N M=l \sin \alpha, Q M=P N \\ =l \cos \alpha . \quad \because \angle C=45^{\circ}, \quad \therefore O P=P N \\ =l \cos \alpha \text {. In } \triangle O M Q \text {, by the Pythagorean theorem we get } \\ 1=... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,902 |
10. In a regular $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively. Fold $\triangle A B C$ along $D E$ to form a dihedral angle $A-D E-C B=60^{\circ}$. If $B C=10 \sqrt{13}$, then the distance between the skew lines $A E$ and $B D$ is $\qquad$ | 10. Let $M$ be the midpoint of $BC$, and the median $AM$ of $\triangle ABC$ intersects $DE$ at $N$. After folding into a dihedral angle, $\angle ANM$ is the plane angle of the dihedral angle $A-DE-CB$, i.e., $\angle ANM = 60^{\circ}$.
Connect $EM, AM, DM. \because EM \parallel DB$, $\therefore$ the distance $d$ betwee... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,903 |
Chessboard $(n \geqslant 2)$, each square is placed with one of the numbers $1,2, \cdots$, $n^{2}$ (each number appears). Prove: in two adjacent squares (with a common side), their numbers differ by at least $n$. | 4. (Czechoslovakia 1)
Solution: Assume that the difference between any two adjacent cells is at most $n-1$. For $k=1,2, \cdots, n^{2}-n$, let $A_{k}$ denote the set of cells containing $1,2, \cdots, k$, $B_{k}$ denote the set of cells containing $k+n, \cdots, n^{2}$, and $C_{k}$ denote the set of all other cells. By t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,904 |
11. In $\triangle A B C$, $A B=2, B C=3$, and there exists a point $D$ inside $\triangle A B C$ such that $C D=2, \angle A D C = 180^{\circ}-\angle B$. For such a triangle, the range of $\cos B$ is | 11. Fold $\triangle A C D$ to the exterior of $\triangle A C B$ to get $\triangle A C D^{\prime}$ (as shown in the figure). That is, $C D^{\prime}=2, \angle D^{\prime}+\angle B$ $=\angle A D C+\angle B=180^{\circ}, \therefore A, B, C, D^{\prime}$ are concyclic $\because C D^{\prime}=A B, \quad \therefore A D^{\prime} \... | \frac{1}{4}<\cos B<\frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,905 |
12. Divisible by 3, and the digits of each number are limited to $1, 2, 3$ (1, 2, 3 do not have to be all used) all natural numbers less than 200000 are $\qquad$
$\qquad$ | 12. The only one-digit number that can be divided by 3 and is composed of only the digits 1, 2, 3 is 3.
The two-digit numbers that can be divided by 3 and are composed of only the digits $1, 2, 3$ can be obtained as follows: arbitrarily write one of $1, 2, 3$ in the tens place, then configure the units digit based on ... | 202 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,906 |
Example 1. The vertex of the parabola $y=f(x)$ is $(-2,3)$, and the difference between the two roots of $\mathrm{f}(\mathrm{x})=0$ is 2. Find $\mathrm{f}(\mathrm{x})$. | $$
\begin{array}{l}
\text { Solve for } f(x)=a x^{2}+bx+c, \text { then from the given, we have } \\
-\frac{b}{2 a}=-2, \\
\left\{\begin{array} { l }
{ \frac { 4 a c - b ^ { 2 } } { 4 a } = 3 , } \\
{ \sqrt { \Delta } = 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-3, \\
b=-12, \\
c=-9
\end{array}\right.\righ... | f(x)=-3 x^{2}-12 x-9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,907 |
Example 2. Try to sketch the graph of $\mathrm{y}=3 \mathrm{x}^{2}+2 \mathrm{x}-1$.
In many cases, it is sufficient to know the quadrant in which the vertex of the graph lies, the approximate positions of the coordinate intercepts, and the direction in which the graph opens, without needing to draw a precise image. Th... | $$
\because a=3>0, b=2>0, c=-10 . \\
\text { Also } \because a, b \text { have the same sign, }
\end{array}
$$
$\therefore$ the vertex is to the left of the $\mathrm{y}$-axis.
$\because \Delta>0$,
$\therefore$ the graph intersects the $x$-axis at two points,
$\because a>0$,
$\because c=-1$,
$\therefore$ the graph inter... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,908 |
Example 11. Let $\mathrm{a}$ be a real number, find the minimum value of the quadratic function
$$
y=x^{2}-4 a x+5 a^{2}-3 a
$$
denoted as $\mathrm{m}$.
When $a$ varies in $0 \leqslant a^{2}-4 a-2 \leqslant 10$, find the maximum value of $m$. | $$
\text{Solve} \begin{aligned}
y & = x^{2} - 4ax + 5a^{2} - 3a \\
& = (x - 2a)^{2} + a^{2} - 3a
\end{aligned}
$$
The above expression achieves its minimum value when $x = 2a$, so
$$
m = a^{2} - 3a = \left(a - \frac{3}{2}\right)^{2} - \frac{9}{4}.
$$
Furthermore, $0 \leqslant a^{2} - 4a - 2 \leqslant 10$ implies
$$
0... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,909 |
Example 12. In $\triangle \mathrm{ABC}$, $\angle \mathrm{B}=30^{\circ}, \angle \mathrm{C}=60^{\circ}$, $AC=a$, points $\mathrm{P}$ and $\mathrm{Q}$ start moving from $\mathrm{A}$ along the perimeter of the triangle. The speed of $\mathrm{Q}$ is three times the speed of $\mathrm{P}$.
(1) Let the length of $AP$ be $x$, a... | Given $\mathrm{AC}=\mathrm{a}, \angle \mathrm{B}=30^{\circ}, \angle \mathrm{C}=60^{\circ}$, so!
$$
A B=\sqrt{3} a, B C=2 a \text {. }
$$
(1) When $0 \leqslant x \leqslant \frac{a}{3}$, $\mathrm{P}$ is on $\mathrm{AB}$, $\mathrm{Q}$ is on $\mathrm{AC}$.
So A - A, ACi-3 $\alpha$,
$$
\begin{aligned}
\therefore y^{\prime}... | y_{\max }=\frac{3}{16} \mathrm{a}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,910 |
Example 2. Each vertex of a regular pentagon corresponds to an integer, such that the sum of these five integers is positive. If three consecutive vertices correspond to integers $x, y, z$, and the middle one $y<0$, then the following operation is performed: the integers $x, y, z$ are replaced by $x+y, -y, z+y$, respec... | Let the integers assigned to the five vertices be $x$, $y$, $z$, $u$, $v$.
$$
\begin{array}{c}
f(x, y, z, u, v)=|x|+|y|+|z| \\
+|u|+|v|+|x+y|+|y+z|+|z+u| \\
+|u+v|+|v+x|+|x+y+z|+|y+z \\
+u|+| z+u+v|+| u+v+x|+| v+x+y | \\
+|x+v+z+u|+|y+z+u+v|+| z+u \\
+v+x|+| u+v+x+y|+| v+x+y+z | .
\end{array}
$$
When $y < 0$ and $y > ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,912 |
Example 3. Define the function $f(\boldsymbol{n})$ and $F(k)$ for the set of natural numbers $N \rightarrow N$ as follows:
$$
\begin{array}{l}
f(n)=\left\{\frac{n(3-\sqrt{2})}{2},\right. \\
F(3)=\min \left\{n \in N \mid f^{k}(n)>0\right\}
\end{array}
$$
where $j^{\circ}=f \circ f \circ f \circ \cdots \circ f$, is the ... | $$
\begin{aligned}
& \text{Proof: Since } 00, \\
& f^{k+1}(G(F(k))-1) \\
= & f^{k} \cdot f(G(F(k))-1) \\
= & f^{k}(F(k)-1) \leqslant 0 .
\end{aligned}
$$
Thus, by the definition of $F$,
$$
F(k+1)=G(F(k)) .
$$
Therefore, from (7), (6), and (4), we get
$$
\begin{aligned}
& F(k+2) \\
= & G(F(k+1)) \\
= & 3 F(k+1)-f(F(k+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,913 |
Let $G$ be a non-empty set of non-constant functions of the form $f(x) = ax + b$, where $a, b$ are real numbers and $a \neq 0$, and $x$ is a variable. If $G$ has the following properties:
(1) If $f, g \in G$, then $g \circ f \in G$, where $(g \circ f)(x) = g(f(x))$;
(2) If $f \in G$ and $f(x) = ax + b$, then the invers... | Prove that obviously $I(x)=x \in G$. Consider the set $G \backslash\{I\}$, and prove by contradiction.
Suppose there are $f, g \in G \backslash\{I\}, x_{f} \neq x_{5}$, let
$$
f(x)=a x+b, g(x)=a^{\prime} x+b^{\prime},
$$
Then by $f\left(x_{1}\right)=x_{1}$ and $g\left(x_{5}\right)=x_{5}$, it is easy to get
$$
x_{1}=\f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,914 |
Let $f$ and $g$ be real functions defined on $(-\infty,+\infty)$, and for all $x$ and $y$ the following functional equation holds:
$$
f(x+y)+f(x-y)=2 f(x) g(y) .
$$
Prove: If $f(x) \equiv 0$, and $|f(x)| \leqslant 1$ for all $x$, then $|g(y)| \leqslant 1$ for all $y$. | Let $M=\sup |f(x)|$, then by the problem statement $M>0$, for any $\delta, 0 < \delta < M$. At this point, for any $y$, we have
$$
\begin{aligned}
2 M & \geqslant|f(x+y)|+|f(x-y)| \\
& \geqslant|f(x+y)+f(x-y)| \\
& =2|f(x)||g(y)| \\
& >2(M-\delta)|g(y)| .
\end{aligned}
$$
Thus,
$$
|g(y)|<\frac{M}{M-\delta}=1+\frac{\de... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,915 |
Example 3. (IMO-19-6)
Let $f(n)$ be a function defined on the set of natural numbers and taking values in this set. Prove that if for each value of $n$, the inequality $f(n+1)>f[f(n)]$ holds, then for every value of $n$, $f(n)=n$. | Prove $A_{\text {. }}$ first by induction:
When $m \geqslant n$, $f(m) \geqslant n$.
It is clear that $A_{1}$ is true. Suppose $A_{n-1}$ is true, then for any $m \geqslant n>1, f(m-1) \geqslant n-1$,
thus, $f(m)>f(f(m-1)) \geqslant n-1$.
Therefore, $f(m) \geqslant n$, which means $A_{\mathrm{n}}$ is true.
Hence, $f(n) ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,916 |
4. $(1 \mathrm{MO}-14-3)$
$m, n$ are arbitrary non-negative integers. Prove that, when $0!=1$,
$$
\frac{(2 m)!(2 n)!}{m!n!(m+n)!}
$$
is an integer. | Xunming Let $f(m, n)=\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$, by the product of $m$ with $n$, we only need to verify that for any $n$, $f(m, n)$ is always an integer.
We prove this by induction on $m$. When $m=0$, $f(m, n)=C_{2 \pi}^{2}$ is an integer. It is easy to prove the following iterative relationship:
$$
f(m+1, n)=4 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,917 |
Example 5. (1MO-23-1)
The function $f(n)$ is defined for all positive integers $n$, taking non-negative integer values. For all positive integers $m, n$, $f(m+n)-f(m)-f(n)=0$ or 1;
and $f(2)=0, f(3)>0, f(9999)=3333$. Find $f(1982)$. | Solve $0=f(2) \geqslant 2 f(1)$, we get $f(1)=0$.
From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$.
Given $f(3)>0$, we get $f(3)=1$.
First, we prove that for $k3333,
\end{aligned}
$$
Contradicting the given condition, so it must be that $f(3k)=k$.
It is clear that $660 \leqslant \hat{j}(1332) \leqs... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,918 |
Example?. The following are rough sketches of the graphs of two quadratic functions:
$$
\begin{array}{l}
y=a x^{2}+b x+c \\
y=p x^{2}+q x+r
\end{array}
$$
(i) Based on the graphs, determine $p, q, r, a+c, pa^{2}+qa$
(ii) Let the x-coordinates of the intersection points of the two graphs be $\boldsymbol{\alpha}, \boldsy... | (i)(1) is an opening question about the "line",
$$
\therefore \mathrm{a}0 \text {. }
$$
Since (1) intersects the $y$-axis in the negative half-axis, $c0$. Therefore, $\mathrm{a}+\mathrm{c}0$. The vertex of the parabola (2) has a positive $x$-coordinate, so $-\frac{\mathrm{q}}{2 \mathrm{p}}>0$, but $\mathrm{p}>0$, henc... | \alpha + \beta = -\frac{b-q}{a-p}, \alpha \beta = \frac{c-r}{a-p} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,919 |
Example 6. (IMO- $22-6$ )
The function $f(x, y)$ satisfies for all non-negative integers $x, y$
$$
\begin{array}{l}
\text { (1) } f(0, y)=y+1 \text {, } \\
\text { (2) } f(x+1,0)=f(x, 1) \text {. } \\
\text { (3) } f(x+1, y+1)=f(x, f(x+1, y)) \text {. }
\end{array}
$$
Determine $f(4,1981)$. | $$
\begin{aligned}
f(1, n) & =f(0, f(1, n-1)) \\
& =f(1, n-1)+1
\end{aligned}
$$
Given $f(1,0)=f(0,1)=2$,
we get $\quad f(1, n)=n+f(1,0)=n+2$.
Also,
$$
\begin{aligned}
f(2, n) & =f(1, f(2, n-1)) \\
& =f(2, n-1)+2=2 n+f(2
\end{aligned}
$$
Since $f(2,0)=f(1,1)=3$,
we have $f(2, n)=2 n+3$.
Furthermore,
$$
\begin{aligned... | 2^{1984}-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,920 |
Example 7. (IMO-20-3)
Let $f, g: Z^{+} \rightarrow Z^{+}$ be strictly increasing functions, and $f\left(Z^{+}\right) \cup g\left(Z^{+}\right)=Z^{+}, f\left(Z^{+}\right) \cap g\left(Z^{+}\right)=\phi$, $g(n)=f[f(n)]+1$. Find $f(240)$. Here $Z^{+}$ is the set of positive integers. | Let
$$
F=\{f(n)\}, G=\{g(n)\} .
$$
If $F \cup G=Z^{+}, F \cap G=\phi$, then $F$ and $G$ are complementary.
We prove $f(n)=(\sqrt{5}+1) \cdots$ by induction. For: $g(n)=f(f(n))+1>1$. So we can set $f(1)=1=\left(\frac{\sqrt{5}+1}{2}\right)$, and $g(1)$ $=2$.
If $v_{n}<m$, then $f(n)=\left\{\frac{\sqrt{5}+1}{2}\right\}... | 388 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,921 |
Example 8. (I $150-17-3$ )
Find all polynomials in two variables that satisfy the following conditions:
1) $P$ is a homogeneous polynomial of degree $n$, i.e., for all real numbers $t$, $x$, $y$, we have
$$
P(t x, t y)=t^{\mathbb{n}} P(x, y) \text {; }
$$
2) For all real numbers $a, b, c$ we have
$$
\begin{array}{l}
P(... | Solving H constant multiple must not satisfy 2) and 3) simultaneously, hence $n \geqslant 1$.
In 2), let $a=b, c=-2a$, then by the homogeneity of $P$ we have
$$
\begin{aligned}
0 & =P(2a, -2a) + P(-a, a) + P(-a, a) \\
& =\left[(-2)^{n} + 2\right] P(-a, a).
\end{aligned}
$$
If $n>1$, then $P(-a, a)=0$, implying that $... | P(x, y) = (x+y)^{n-1}(x-2y) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,922 |
Example 10. (IMO-27-5)
$f$ is a function defined on the non-negative real numbers and taking non-negative real values. Find all functions $f$ that satisfy the following conditions:
( i ) $f(x f(y)) f(y)=f(x+y)$;
( ii ) $f(2)=0$;
(iii) $f(x) \neq 0$, when $0 \leqslant x<2$.
| If $w>2$, then in (i) let $y=2, x=w-2$, we get
$$
f(w)=f((w-2) f(2)) f(2)=0 .
$$
Thus,
$$
f(x)=0 \Leftrightarrow x \geqslant 2 .
$$
In (i), let $x>0, 0 \leqslant y < 2$, then when $x>0$, we have
$$
x f(y) \geqslant 2 \Leftrightarrow x+y \geqslant 2 .
$$
This is equivalent to
$$
\begin{array}{l}
f(y) \geqslant \frac{... | f(y)=\left\{\begin{array}{cl}
\frac{2}{2-y}, & \text { when } 0 \leqslant y<2 ； \\
0 . & \text { when } y \geqslant 2 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,924 |
$$
\begin{array}{l}
f(x+y)=\frac{f(x) f(y)-1}{f(x)+f(y)}, \\
x, y, x+y \in(0, b) .
\end{array}
$$
Then $f(x)=\cot a x, x \in(0, b),(17)$ where $a$ is a constant and $|a|<\frac{\pi}{6}$. | Prove that when $x \in\left(0, \frac{b}{2}\right)$, by (16) we have
$$
f(2 x)=\frac{f^{2}(x)-1}{2 f(x)} \text {. }
$$
$$
g(x)=\frac{1}{f}(x), x \in\left(0, \frac{b}{2}\right) .
$$
By (16) and (19), when $x, y, x+y \in (0, \frac{b}{2})$, we have
$$
\begin{array}{l}
g(x+y)=\frac{1}{f(x+y)}=\frac{f(x)-f(y)}{f(x) f(y)-1} ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,927 |
Example 1. $\mathrm{AC}$ and $\mathrm{CE}$ are two diagonals of a regular hexagon $\mathrm{ABCDEF}$, and points $\mathrm{M}$ and $\mathrm{N}$ internally divide $\mathrm{AC}$ and $\mathrm{CE}$, respectively, such that $\mathrm{M}: \mathrm{AC} = \mathrm{CN}: \mathrm{CE} = \mathrm{r}$. If points $\mathrm{B}$, $\mathrm{M}$... | Connect $B D$, $N D$. Since $\triangle M B C$ rotates counterclockwise by $120^{\circ}$ around point $O$ and coincides with $\triangle \mathrm{NDE}$, we know that $\angle B N D=120^{\circ}$. Construct an equilateral triangle $B D G$ with $B D$ as one side towards the direction of point $N$, then $N$,
$$
\mathrm{CN}=\ma... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,929 |
Example 4. Let $f(x)=x^{2}+p x+q$, if $1+p+q<0$, then the quadratic equation $\mathrm{f}(\mathrm{x})=0$ has one real root greater than 1 and one real root less than 1. | Prove $\because f(1)=1+p+q<0$, and when $x \rightarrow \infty$, $x \rightarrow-\infty$ there are $f(x) \rightarrow \infty$,
$\therefore$ the parabola opens upwards and intersects the $\mathrm{x}$-axis, hence $\mathrm{f}(\mathrm{x})=0$ has one real root greater than 1 and one real root less than 1. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,930 |
Example 2. Given that $\mathrm{A}$ is a point on the plane where two circles with unequal radii, circle $\mathrm{O}_1$ and circle $\mathrm{O}_2$, intersect at one point. The two external common tangents of the circles are $\mathrm{P}_1 \mathrm{P}_2$ and $\mathrm{Q}_1 \mathrm{Q}_2$, with points of tangency at $P_1, P_2,... | Let the intersection of $P_{1} D$ and $S Q \mid \bar{x}_{2}$ be $C$, then $C$ is the midpoint of $\mathrm{DM}$. Using $\mathrm{BM}_{2}$, connect $\mathrm{BA}$ and extend it to intersect $\mathrm{P}_{1} \overline{\mathrm{P}}_{2}$ at $\mathrm{N}$, then
$$
P: \mathrm{N}^{2}=\mathrm{PA} \cdot \mathrm{PB}=\mathrm{P}_{2} \ma... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,931 |
Example 3. Given in a convex quadrilateral $\mathrm{ABCD}$, line $\mathrm{CD}$ is tangent to the circle with $\mathrm{AB}$ as its diameter. Prove: the sufficient and necessary condition for line $\mathrm{AB}$ to be tangent to the circle with $\mathrm{CD}$ as its diameter is $\mathrm{BC} \parallel \mathrm{AD}$. | Prove Necessity. Let the midpoints of $\mathrm{AB}$ and $\mathrm{CD}$ be denoted as $\mathrm{O}$ and $\mathrm{O}^{\prime}$, respectively. Extend $\mathrm{O}$ to be tangent to $\mathrm{CD}$ at $\mathrm{E}$, and let the circle $\mathrm{O}^{\prime}$ be tangent to $\mathrm{AB}$ at $\mathrm{F}$. Connect $\mathrm{O}^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,932 |
Example 4. Quadrilateral ABCD is inscribed in a circle, and another circle has its center on side $\mathrm{AB}$ and is tangent to the other three sides. Prove: $A A + \mathrm{BC} = \mathrm{AB}$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 4. Quadrilateral ABCD is inscribed in a circle, and another circle has its center on side $... | $$
\begin{aligned}
\angle \mathrm{CMB} & =\frac{1}{2}\left(180^{\circ}-\angle \mathrm{B}\right)=\frac{1}{2} \angle \mathrm{CDA} \\
& =\angle \mathrm{CDO},
\end{aligned}
$$
Therefore, points C, D, M, and O are concyclic. Thus,
$$
\left.\angle \mathrm{AMD}=\angle \mathrm{DCD}=\frac{1}{2} \angle \mathrm{DCB}=\frac{1}{2}(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,933 |
Example 5. Circle $O$ passes through vertices $\mathrm{A}$ and $\mathrm{C}$ of $\triangle \mathrm{ABC}$ and intersects sides $\mathrm{AB}$ and $\mathrm{BC}$ at points $K$ and $N$ respectively, where $K$ and $N$ are distinct. The circumcircles of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BKN}$ intersect at $B$ and... | Proof B.1 It is easy to prove that the extensions of $\mathrm{AC}$, $\mathrm{KN}$, and $\mathrm{BM}$ intersect at a point $\mathrm{P}$. Construct segments $\mathrm{MK}$, $\mathrm{KO}$, $\mathrm{OC}$, $\mathrm{MC}$, $\mathrm{KC}$, and $\mathrm{MN}$. Since $\angle \mathrm{BMN} = \angle \mathrm{KN} = \angle \mathrm{NCP}$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,934 |
1. Prove: If the hyperbola $x^{2}-y^{2}=a^{2}$ and the ellipse $x^{2}+3 y^{2}=b^{2}+1$ have four intersection points, then the four intersection points are concyclic. | Prove $\because x^{2}-y^{2}=u^{2} \quad x^{2}+3 y^{2}:=b^{2}+1$, $\therefore 4 y^{2}=b^{2}+1-a^{2}$. If $b^{2}+1>a^{2}$, there must be intersection points. And $\left(x^{2}-y^{2}-a^{2}\right)+\lambda\left(x^{2}+3 y^{2}-b^{2}-1\right)=0$ is the equation of the quadratic curve system passing through the intersection poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,937 |
Example 2. Find the equation of the circle passing through the intersection points of the line $2 x+y+4=0$ and the circle $x^{2}+y^{2}+2 x-4 y+1=0$, and satisfying one of the following conditions: (1) passing through the origin; (2) having the minimum area. | The equation of a circle passing through the intersection points of a line and a circle can be written as
$$
\lambda(2 x+y+4)+\left(x^{2}+y^{2}+2 x-4 y+1\right)=0 \text {, }
$$
Rearranging, we get
$$
\begin{array}{l}
x^{2}+y^{2}+2(1+\lambda) x+(\lambda-4) y+(1+4 \lambda) \\
=0 .
\end{array}
$$
(1) Since this circle pa... | x^{2}+y^{2}+\frac{3}{2} x-\frac{17}{4} y=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,938 |
For example. Prove: A circle with the focal chord of the parabola $y^{2}=4 p x$ as its diameter must intersect this parabola
保留了源文本的换行和格式。 | To prove that the chord $AB$ of the focus has the equation
$$
y=k(x-p)
$$
The coordinates of $A$ and $B$ are
$$
\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \text {. Substituting } y=k(x-p)
$$
into $v^{2}=4 b x$ and rearranging for $x$ gives $k^{2} x^{2}-2 p\left(k^{2}+2\right) x+k^{2} p^{2}=0$. Then
$$
x_{1}+x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,939 |
Example 4. Given the hyperbola $x y=1$, find the tangent line that is intercepted by the hyperbola $4 x y=3$ to form the shortest chord. Determine the equation of the tangent line, and under what conditions the chord is the shortest, as well as the length of the shortest chord.
Translate the above text into English, p... | Let the equation of the tangent line be: $y=kx+b$. Substituting into $xy=1$ yields $kx^{2}+bx-1=0$.
$\because$ it is tangent,
$\therefore \triangle=b^{2}+4k=0, k=-\frac{b^{2}}{4}$. Therefore, the equation of the tangent line is $y=-\frac{b^{2}}{4}x+b$.
Substituting $y=kx+b$ into $4xy=3$, we get $b^{2}x^{2}-4bx+3=0$. L... | y=-x \pm 2, |AB|=\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,940 |
Example 5. In the quadratic function $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$, if $\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)\left(\mathrm{x}_{1} \neq \mathrm{x}_{2}\right)$, then $\mathrm{x}=\frac{\bar{x}_{1}+x_{2}}{2}$ is the axis of symmetry of the graph of th... | $$
\begin{array}{l}
\text { Car } \because f\left(x_{1}\right)=f\left(x_{2}\right), \\
\therefore a x_{1}^{2}+b x_{1}+c=a x_{2}+b x_{2}+c, \\
a\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)+b\left(x_{1}-x_{2}\right)=0, \\
\left(x_{1}-x_{2}\right)\left(a x_{1}+a x_{2}+b\right)=0 . \\
\because x_{1} \neq x_{2}, x_{1}-x... | \frac{x_{1}+x_{2}}{2}=-\frac{b}{2a} | Algebra | proof | Yes | Yes | cn_contest | false | 703,941 |
Through point $M(2,1)$, a chord $AB$ is drawn of the ellipse $x^{2}+4 y^{2}=16$ such that $|A M|:|M B|=1: 2$. Find the length of the chord $AB$.
The text is translated while preserving the original text's line breaks and format. | Let the equation of the chord $AB$ be
$$
\left\{\begin{array}{l}
x=2+t \cos \alpha \\
y=1+t \sin \alpha .
\end{array}\right.
$$
Substituting into the ellipse equation and simplifying, we get
$$
\left(3 \sin ^{2} \alpha+1\right) t^{2}+4(\cos \alpha+2 \sin \alpha) t-8=0 \text {. }
$$
By Vieta's formulas, we have $t_{1}... | \frac{6}{165} \sqrt{165(59 \pm 8 \sqrt{7})} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,942 |
Example 6. If $AB$ is any chord of the hyperbola $b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}$, and $AB$ is bisected by the known line $y=k x$, prove that the line $AB$ has a fixed direction. | Let the endpoints of the chord $AB$ have coordinates $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ (these are the "auxiliary points"). Then we have
$$
\begin{array}{l}
b^{2} x_{1}^{2} - a^{2} y_{1}^{2} = a^{2} b^{2}, \\
b^{2} x_{2}^{2} - a^{2} y_{2}^{2} = a^{2} b^{2}.
\end{array}
$$
Subtracting the two e... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,943 |
Example 2. Prove: $C_{\mathrm{n}}^{1}-2 C_{\mathrm{n}}^{2}+3 C_{\mathrm{n}}^{3}-4 C_{\mathrm{n}}^{4}+\cdots$ $+(-1)^{n-1} \cdot n \cdot C_{n}^{n}=0$. | Prove that $k \cdot C_{\mathrm{n}}^{k}=n \cdot C_{\mathrm{n}-1}^{\mathrm{k}-1}$,
$$
\begin{array}{l}
\therefore C_{n}^{1}-2 C_{n}^{2}+3 C_{n}^{3}-4 C_{n}^{4}+\cdots \\
+(-1)^{n-1} \cdot n \cdot C_{n}^{n} \\
=n\left[C_{z}^{0}-1-C_{z}^{1}{ }_{1}+C_{n-1}^{2}-C_{n-1}^{3}+\cdots\right. \\
\left.+(-1)^{n-1} \cdot C_{n-1}^{n-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,945 |
Example 6. Prove: $C_{\mathrm{k}}^{n} \cdot C_{n}^{m}+C_{k}^{1} \cdot C_{n}^{m-1}$
$$
\begin{array}{l}
+C_{k}^{2} \cdot C_{n}^{m-2}+\cdots+C_{k}^{k} \cdot C_{n}^{m-k}=C_{n}^{m}+k \\
(m \leqslant n) .
\end{array}
$$ | Proof: This problem can be constructed as follows: Suppose there are $n+k$ products, among which $k$ are defective. If $m$ items are randomly selected from these $n+k$ products, there are $C_{\mathrm{n}}^{\boldsymbol{n}}+\mathrm{k}$ ways to choose. Among these $\boldsymbol{m}$ items, there may be no defective items, or... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,949 |
Example 8. Prove: $C_{\mathrm{a}}^{1}-3 C_{\mathrm{n}}^{3}+5 C_{\mathrm{a}}^{5}-7 C_{\mathrm{a}}^{7}+\cdots$
$$
\begin{aligned}
= & n \cdot 2^{\frac{n-1}{2}} \cdot \cos \frac{(n-1) \pi}{4}, \\
& 2 C_{n}^{2}-4 C_{n}^{4}+6 C_{n}^{6}-8 C_{n}^{8}+\cdots \\
= & n \cdot 2^{\frac{n-1}{2}} \cdot \sin \frac{(n-1) \pi}{4} .
\end... | Proof: First, by De Moivre's Theorem, we get
$$
\begin{aligned}
& n \cdot(1+i)^{a-1}=n \cdot 2^{\frac{-1}{2}} \cdot\left[\cos \frac{(n-1)}{4}\right) \pi \\
+ & \left.i \sin \frac{(n-1) \pi}{4}\right] .
\end{aligned}
$$
On the other hand, by the Binomial Theorem, we get
$$
\begin{array}{l}
n \cdot(1+i)^{2-1}=n\left(1-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,951 |
Example 1. Solve the absolute value equation
$$
|x+1|+|x+2|+|x-5|=10 .
$$ | $$
\begin{array}{c}
(-2-x)+(-1-x)+(5-x)=10, \\
\therefore=-\frac{8}{3} .
\end{array}
$$
$$
\begin{array}{c}
(x+2)+(2+1)+(5-x)=10, \\
x=2 .
\end{array}
$$
$\therefore$ The solutions are: $x_{1}=-\frac{8}{3}, x_{2}=2$.
Analysis: Transform the problem into visual thinking, which is to find the point $x$ on the number line... | x_{1}=-\frac{8}{3}, x_{2}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,954 |
Solve the equation $|x-| 2 x+1||=3$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $|2 x+1|$ intercepts a line segment of length 3, find the intersection point $x$ with the $x$-axis at this time (as shown in Figure 2).
From Figure 2, it is not difficult to see that when the vertical line moves away from the line $x=-\frac{1}{2}$ to the left and right, the length of the intercepted line segment gradu... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,955 |
For example, solving the equation $\frac{x^{2}-7}{6}=\sqrt{6 x+7}$. This is the result of transitioning from concrete to abstract thinking.
From a geometric perspective, this is equivalent to finding the x-coordinates of the intersection points of the line $y=\frac{x^{2}-7}{6}$ and the curve $y=\sqrt{6 x+7}$. These tw... | Solve $\sqrt{6 x+7}=x$ to get
$$
\begin{array}{l}
x^{2}-6 x-7=0, \\
\therefore x_{1}=-1 \text { (discard), } x_{2}=7 .
\end{array}
$$
Upon verification, the root of the original equation is $x=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,956 |
Example 4. For what value of $\mathrm{k}$ will the roots of the equation $(k+1) x^{2}$ $-2 k x+1=0$ both lie in the interval $(0,1)$? | Solving this problem essentially means that the parabola
$$
\begin{array}{l}
f(x)=x^{2}-\frac{2 k}{k+1} x+\frac{1}{k+1} \\
=\left(x-\frac{k}{k+1}\right)^{2}-\frac{k^{2}-k-1}{(k+1)^{2}}
\end{array}
$$
intersects the $x$-axis at two points that are both between $0$ and $1$ (as shown in Figure 3).
From Figure 3, it is no... | k \in\left[\frac{\sqrt{5}+1}{2}, 2\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,957 |
Example 5. Given $z \in C,|z|=1$. Try to find the maximum and minimum values of $r=\left|z^{3}-3 z-2\right|$.
Analysis:
$$
\begin{array}{l}
r=\left|z^{3}-3 z-2 z \bar{z}\right| \\
=|z|\left|z^{2}-2 \bar{z}-3\right| \\
\quad=\left|z^{2}-2 \bar{z}-3\right|
\end{array}
$$ | $$
\begin{array}{l}
\text { Let } z=\cos \theta+i \sin \theta, \\
r=1(\cos 2 \theta-2 \cos \theta-3)+(\sin 2 \theta \\
+2 \sin \theta) i \mid \text {. } \\
\text { Let }\left\{\begin{array}{l}
x=\cos 2 \theta-2 \cos \theta-3, \\
y=\sin 2 \theta+2 \sin \theta
\end{array}\right. \\
\Rightarrow(x+3)^{2}+y^{2}=5-4 \cos 3 \... | r_{\mathrm{min}}=0, \quad r_{\mathrm{max}}=3 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,958 |
Example 2. From the graphs of $y=\sqrt{x}, y=\frac{1}{x}, y=\log _{2} x$, make the graphs of $y=\sqrt{x-1}+1, y=\frac{x+2}{x+1}$, $y=\log _{2} \frac{1}{(x-1)^{2}}$. | Solve $y=\sqrt{x}-1+1 E y=\sqrt{ }$ (Fig. 1); $y=\frac{x+2}{x+1}=1+\frac{1}{x+1}$ boundary $y=\frac{1}{x}$ single (1?:
$$
\begin{aligned}
y & =\log _{2} \frac{1}{(x-1)^{2}} \\
& =\left\{\begin{array}{ll}
-2 \log _{2}(x-1) & (\text { when } x>1), \\
-2 \log _{2}(1-x) & (\text { when } x<1) .
\end{array}\right.
\end{alig... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,960 |
Example 3. Given the graph of $y=|x|$, draw the graphs of the following functions (or equations):
(1) $y=|x-1|$, (2) $y=|x+1|$,
(3) $|x|+|y|=1$,
(4) $|x-1|+|y-1|=1$,
(5) $y=\frac{\left|1-x^{2}\right|}{1+|x|}$. | $$
\begin{array}{l}
\text { (1) and (2) can be obtained by shifting the graph of } y=|x| \text { one unit to the right and left, respectively, to get the sub-figure (Figure 4). } \\
\text { (3) can be transformed into } |y|=1-|x|, \quad y \geqslant 0 \text { when } y=1-|x|; \text { when } y<0, y=|x|-1 \text { (Figure 5... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,961 |
Example 4. Six cones have a common vertex. At the same time, each cone has a common tangent line with four other cones. Find the ratio of the total volume of the six cones to the volume of the sphere that is tangent to all the bases of the cones.
untranslated text remains unchanged:
例 4. 六个合同的圆锥具有共同的顶点. 同时每个圆锥与另外四个圆雉... | To solve this problem, we choose a cube as the structure. The common vertex of the cones is located at the center of the cube, and the base of the cone is inscribed in the face of the cube (Figure 5, only one cone is drawn). At this time, each of the six cones shares a common tangent line with four other cones, and the... | 3:2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,966 |
1. For what value of $b$ do the equations $1988 x^{2}+b x+8891=0$,
and $\quad 8891 x^{2}+b x+1988=0$ have a common root? | 1. If the two equations have a common root $x$, then we have
$$
1988 x^{2}+8891=8891 x^{2}+1988 .
$$
This is a quadratic equation, and it is evident that $x=1$ or -1. When $x=1$, we can determine
$$
b=-10879,
$$
When $x=-1$, $b=10879$. | b=\pm10879 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,967 |
4. Let $x_{0}=0, x_{1}=1$, and
$$
\begin{array}{l}
x_{n+1}=4 x_{n}-x_{n-1} ; \\
y_{0}=1, y_{1}=2,1 \\
y_{n+1}=4 y_{n}-y_{n-1},
\end{array}
$$
$n=1,2,3, \cdots$. Prove that for all integers $n \geqslant 0$,
$$
y_{\mathrm{n}}^{2}=3 x_{\mathrm{n}}^{2}+1 .
$$ | 4. We use synchronous induction to prove the following two formulas:
(a) $y_{\mathrm{a}}^{2}=3 x_{\mathrm{n}}{ }^{2}+1$,
(b) $y_{\mathrm{B}} y_{\mathrm{s}-1}=3 x_{\mathrm{n}} x_{\mathrm{n}-1}+2$.
It can be directly verified that when $n=1$, both formulas hold.
$$
\begin{array}{l}
\text { We discuss in two steps: } \\
(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,970 |
5. Let $S=\left\{a_{1}, a_{2}, \cdots, a_{\mathrm{r}}\right\}$ be a set of integers, where $r>1$. For a non-empty subset $A$ of $S$, define $p(A)$ as the product of all integers in $A$. Let $m(S)$ denote the arithmetic mean of $p(A)$, where $A$ ranges over all non-empty subsets of $S$.
If $m(S)=13$, and there exists a... | $$
\begin{array}{l}
5. \text{ For any natural number } n \text{ and } A=\left\{a_{1}, a_{2},\right. \\
\left.\cdots, a_{n}\right\}, \text{ since } \\
\quad a_{1}+a_{2}+\cdots+a_{n}+\left(a_{1} a_{2}+\cdots+a_{1} a_{n}+\right. \\
\left.\cdots+a_{n-1} a_{n}\right)+\cdots+a_{1} a_{2} \cdots a_{n} \\
=\left(1+a_{1}\right)\... | a_{1}=1, a_{2}=1, a_{3}=22, a_{r+1}=7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,971 |
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