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 $M$ be a point inside $\triangle ABC$, and let $M A^{\prime}$ be perpendicular to $BC$, with the foot of the perpendicular being $A^{\prime}$. Similarly, define $B^{\{\prime\}}$ on $CA$ and $C^{\prime}$ on $AB$. Let
$$
P(M)=\frac{M A^{\prime} \cdot M B^{\prime} \cdot M C^{\prime}}{M A \cdot M B \cdot M C} .
$$
... | Let $\angle M A B=\alpha_{1}, \angle M A C=\alpha_{2}, \angle M B C=\beta_{1}$, $\angle M B A=\beta_{2}, \angle M C A=\gamma_{1}, \angle M C B=\gamma_{2}$. Then
$$
\begin{array}{l}
\frac{M B^{\prime} \cdot M C^{\prime}}{M A^{2}}=\sin \alpha_{1} \cdot \sin \alpha_{2} \\
=\frac{1}{2}\left[\cos \left(\alpha_{1}-\alpha_{2}... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,291 |
5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle ... | Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,292 |
6. Let $P$ be a point outside $\triangle A B C$ on the same plane. Suppose $A P, B P, C P$ intersect the sides $B C, C A, A B$ or their extensions at $D, E, F$ respectively. If the areas of $\triangle P B D, \triangle P C E, \triangle P A F$ are all equal, prove: these areas are equal to the area of $\triangle A B C$.
... | Prove: As shown in Figure 8, let $S_{\triangle P B D}=S_{\triangle Y E}=S_{\triangle P 4 F}=x, S_{\triangle A K E}$ prove $S_{\triangle A B C}=u+w=x$.
Since each ratio $\frac{B D}{D C}, \frac{C E}{E A}, \frac{A F}{F B}$ can be calculated in two different ways, we have
$$
\begin{array}{l}
\frac{x}{2 x+w}=\frac{x-u-v}{x... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,293 |
7. Let $O$ be a point inside acute triangle $ABC$, and let $OA_1$ be perpendicular to $BC$, with the foot of the perpendicular being $A_1$. Similarly, define $B_1$ on $CA$ and $C_1$ on $AB$. Prove that $O$ is the circumcenter of $\triangle ABC$ if and only if the perimeter of $\triangle A_1B_1C_1$ is not less than the ... | Prove: If $O$ is the circumcenter of $\triangle ABC$, then $A_{1}, B_{1}, C_{1}$ are the midpoints of $BC, CA, AB$ respectively. Thus, $p_{A_{1}B_{1}C_{1}} = p_{AB_{1}C_{1}} = p_{BC_{1}A_{1}} = p_{CA_{1}B_{1}}$, where $p_{XYZ}$ denotes the perimeter of $\triangle XYZ$.
$$
\begin{array}{l}
\angle CAB = \alpha, \angle CA... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,294 |
1. Prove: There does not exist a positive integer $n$, such that for all $k=$ $1,2, \cdots, 9,(n+k)$! the leading digit (the leftmost digit in decimal representation) equals $k$. | Proof: For each positive integer $m$, define
$$
N(m)=\frac{m}{10^{d(m)-1}},
$$
where $d(m)$ is the number of digits in $m$, then $1 \leqslant N(m)<10$.
For any positive integers $l$ and $m$, since $lm$ has at least $d(l)+d(m)-1$ digits, we have
$$
d(l)-1+d(m)-1 \leqslant d(lm)-1.
$$
Therefore, $\frac{lm}{10^{d(l)+d(m... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,295 |
3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$
Find $a_{4^{4}}$. | For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,297 |
4. Let $p \geqslant 5$ be a prime number. Prove: There exists an integer $a$ such that $a^{p-1}-1$ and $(a+1)^{p-1}-1$ are not divisible by $p^{2}$, where $1 \leqslant a \leqslant p-2$. | Proof: Let $S=\{1,2, \cdots, p-1\}, A=\left\{a \in S, a^{p-1}\right.$ $\not \equiv 1\left(\bmod p^{2}\right) \mid$. We prove: $|A| \geqslant \frac{p-1}{2}$, where $|A|$ denotes the number of elements in the set $A$.
In fact, if $1 \leqslant a \leqslant p-1$, by the binomial theorem, we have $(p-a)^{p-1}-a^{p-1} \equiv-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,298 |
6. Can we find 100 positive integers not exceeding 25000, such that the sum of any two of them is unique. | Proof: We first prove a lemma.
Lemma For any odd prime $p$, there exist $p-1$ positive integers not exceeding $2 p^{2}$, such that the sums of any two of these positive integers are all different.
Proof of the lemma: Consider $f_{n}=2 p n+\left(n^{2}\right), n=1,2, \cdots, p-1$, where $\left(a^{2}\right)$ denotes the ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,299 |
1. Let $a=(x+1)(x+2)(x+3)(x+4)$, $b=(x-4)(x-3)(x-2)(x-1)$. Then $a-b$ equals ( ).
(A) $20 x^{3}+50 x$
(B) $2 x^{3}+5 x$
(C) $20 x^{4}+100 x^{2}$
(D) $20 x^{3}+100 x$ | \begin{array}{l}-.1 .(\mathrm{D}) . \\ \because a=(x+1)(x+4) \cdot(x+2)(x+3) \\ =\left(x^{2}+5 x+5\right)^{2}-1 \text {. } \\ b=(x-4)(x-1) \cdot(x-3)(x-2) \\ =\left(x^{2}-5 x+5\right)^{2}-1 \text {, } \\ \therefore a-b=\left(x^{2}+5 x+5\right)^{2}-\left(x^{2}-5 x+5\right)^{2} \\ =20 x^{3}+100 x \text {. } \\\end{array} | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,300 |
Example 1 Haha
Zhang classmate laughed "haha" twice and quickly covered her mouth to hold back her laughter. Li classmate teased her, "Ha ya, ha!" Zhang classmate couldn't help it and finally burst into laughter, "hahahaha" continuously. Using this plot, form a multiplication problem:
```
哈哈
x 哈呀哈
```
In the above prob... | Explanation: Each digit of the four-digit product is "Ha"
(We can start from here). Thus, we have
HaHaHaHa $=$ Ha $\times 1111$
$$
\begin{array}{l}
=\text { Ha } \times 11 \times 101 \\
=\text { HaHa } \times 101 .
\end{array}
$$
By comparing with the original equation, we know that “HaYaHa” $=101$, which means “Ha” $... | 11 \times 101=1111 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,301 |
Example 2 Green Mountains and Clear Waters
After the summer vacation group tour, the teachers talked about their impressions of the tour. The Chinese teacher said: "Green mountains, clear waters, strong pines, and a thousand beautiful peaks." The foreign language teacher said: "Beautiful peaks, a thousand strong pines... | Explanation: Observe the first digit of the addition equation (the first digit of an addition equation is usually the easiest to start with). From the equation, we know that the difference between "秀" and "青" is at least 8, which has reached the maximum possible limit. Therefore, "秀" must be 9, and "青" must be 1.
The ... | 123456789 + 864197532 = 987654321 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,302 |
Example 1 In the tetrahedron $A B C D$, $A D=D B=A C$ $=C B=1$, then its maximum volume is | Analysis: As shown in Figure 2, let $E$ and $F$ be the midpoints of $CD$ and $AB$ respectively, then it is easy to know that $AB \perp CD$, and $EF$ is the common perpendicular of $AB$ and $CD$. Let $AB = 2x, CD = 2y$. From the fact that $\triangle ACD$, $\triangle BCD$, and $\triangle EAB$ are all isosceles triangles,... | \frac{2 \sqrt{3}}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,305 |
Example 2 In a regular triangular frustum $A B C-A_{1} B_{1} C_{1}$, $A_{1} B_{1}: A B=5: 7$, the section $A B C_{1}$ and $A_{1} B C_{1}$ divide the frustum into three triangular pyramids $C_{1}-A B C$, $C_{1}-A B A_{1}$, and $B-A_{1} B_{1} C_{1}$. Then, the ratio of their volumes $V_{1}$, $V_{2}$, and $V_{3}$ is $\qqu... | Analysis: This problem is the best illustration of Basic Conclusion 4: Let the height of the frustum of a triangular pyramid be $h$, and the areas of the upper and lower bases be $S_{1}$ and $S_{2}$, respectively. Then $V_{1}=\frac{h}{3} S_{2}, V_{2}=\frac{h}{3} \sqrt{S_{1} S_{2}}, V_{3}=\frac{h}{3} S_{1}$, i.e., $V_{1... | 49: 35: 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,306 |
Example 3 Given that the base of the tetrahedron $S-ABC$ is an equilateral triangle, the projection of point $A$ on the side face $SBC$ is $H$, which is the orthocenter of $\triangle SBC$, the plane angle of the dihedral angle $H-AB-C$ is $30^{\circ}$, and $SA = 2\sqrt{3}$. Then, the volume of the tetrahedron $S-ABC$ i... | Analysis: The given segment length seems insufficient. From the basic conclusion 7, we know that the three pairs of opposite edges are mutually perpendicular, so the projection $O$ of $S$ on plane $A B C$ is also the orthocenter of $\triangle A B C$. Furthermore, since $\triangle A B C$ is an equilateral triangle, $O$ ... | \frac{9 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,307 |
Example 4 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. At this moment, the volume of the tetrahedron $A-B C M$ is $\qquad$ . | Analysis: The first difficulty is how to make a more intuitive three-dimensional figure? Before folding, $M A=M B$ $=M C=A C=2$, these lengths will not change after folding into a pyramid. From the basic conclusion 5, we know that the projection of point $M$ on plane $A B C$ after folding is the circumcenter of $\trian... | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,308 |
Example 5 The base of the pyramid is an isosceles triangle, with the base and the equal sides being 12 and 10, respectively. Moreover, the projection of the pyramid's apex onto the base lies within the triangle, and the dihedral angles between each lateral face and the base are all $30^{\circ}$. Then the height of this... | Analysis: From Basic Conclusion 6, we know that the projection of the apex of the cone on the base is the incenter of the base triangle. Then, given that the dihedral angle between the lateral face and the base is $30^{\circ}$, we only need to first find the inradius of the base triangle. Using plane geometry methods, ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,309 |
Example 6 If three lines $a, b, c$ in space are pairwise skew lines, then the number of lines that intersect $a, b, c$ is ( ).
(A)0
(B) 1
(C) more than 1 but finite
(D) infinitely many | Analysis: If $a$, $b$, and $c$ are not parallel to the same plane, by Basic Conclusion 8, we can construct a parallelepiped $ABCD-A_{1}B_{1}C_{1}D_{1}$, such that $a$, $b$, and $c$ are its non-parallel edges $AB$, $B_{1}C_{1}$, and $DD_{1}$. Through a point $E$ on $DD_{1}$, different from $D$ and $D_{1}$, and the line ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,310 |
Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is | Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed s... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,311 |
Example 8 Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and points $M, N$ are the midpoints of edges $C C_{1}, A_{1} D_{1}$, respectively. Then the volume of the tetrahedron $O-M N B_{1}$ is ( ).
(A) $\frac{1}{6}$
(B) $\frac{5}{48}$
(C) $\frac{1}{8}$
(D) $\f... | Analysis: As shown in Figure 6, the four faces of the required pyramid are oblique sections of a cube, making it difficult to find the volume. However, using Basic Conclusion 14, the volume of the tetrahedron $O-M N B_{1}$ can be converted to a multiple of the volume of the triangular pyramid $C_{1}-M N B_{1}$. The key... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,312 |
Example 9 Let points $E$, $F$, $G$ be the midpoints of the edges $AB$, $BC$, $CD$ of the regular tetrahedron $ABCD$. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\arctan \sqrt{2}$
(1) $\pi-\operatorname{arccot} \f... | Analysis: It is easy to know that $AC$ is parallel to plane $EFG$ and $AC \perp FG$ (the intersection line of plane $EFG$ and plane $BCD$). Therefore, by Basic Conclusion 12, the required dihedral angle is equal to the supplementary angle of the angle formed by $AC$ and plane $BCD$. Draw $AO \perp$ plane $BCD$, with th... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,314 |
Example 10 In the pyramid $P-ABC$, $PA \perp PB$, $PB \perp PC$, $PC \perp PA$. If $M$ is a point inside the base $\triangle ABC$, and it is known that $\sin \angle APM=\frac{4}{5}, \cos \angle BPM=\frac{\sqrt{3}}{3}$. Then the value of $\cos \angle CPM$ is | Analysis: This cone has one vertex where two edges are mutually perpendicular. By cutting out a rectangular parallelepiped with $PM$ as the diagonal, this can be achieved by drawing planes through point $M$ parallel to the faces $PBC$, $PCA$, and $PAB$. At this point, $\angle APM$, $\angle BPM$, and $\angle CPM$ become... | \frac{\sqrt{69}}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,315 |
1. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with $AB$ as the hypotenuse, $SA=SB=SC=2, AB=2$. Suppose points $S, A, B, C$ are all on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$ (1997, National High School Mathematics Competition) | (Tip: From the basic conclusion 5, we know that the projection of $S$ on the plane $ABC$ is $M$, which is the circumcenter of $\triangle ABC$, i.e., $M$ is the midpoint of the hypotenuse of $\triangle ABC$ and the radius of the circumcircle is $MC=1$. Also, $MS=\sqrt{3}$. From $OA=OB=OC$, we know that the projection of... | \frac{2\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,316 |
3. In tetrahedron $ABCD$, the lengths of edges $AB$ and $CD$ are $a$ and $b$ respectively, and the distance between the midpoints of these two edges is $d$. Then the maximum volume of tetrahedron $ABCD$ is $\qquad$ (1994, Shanghai High School Mathematics Competition for Grade 10) | Answer: $\frac{1}{6} a b d$ | \frac{1}{6} a b d | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,318 |
The first question As shown in Figure 1, in $\angle ABC$, $\angle A=60^{\circ}, AB > AC$, point $O$ is the circumcenter, the two altitudes $BE$ and $CF$ intersect at point $H$, points $M$ and $N$ are on segments $BH$ and $HF$ respectively, and satisfy $BM = CN$. Find the value of $\frac{MH + NH}{OH}$. | ```
Solution 1: Connect $O B, O C$. Let the circumradius of $\triangle A B C$ be $R$.
By the properties of the circumcenter, we know that $\angle B O C$ is
By the properties of the circumcenter, we know that $\angle B H C=180^{\circ}-\angle A=120^{\circ}$.
Therefore, points $B, C, H, O$ are concyclic. By Ptolemy's ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,320 |
Third Question: Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in ... | Solution: Let the times for $A_{1}, A_{2}, A_{3}, A_{4}$ be $7 k_{1}, 7 k_{2}$, $7 k_{3}, 7 k_{4}$, and their sum be $x: A_{5}, A_{6}, A_{7}$ have times $13 k_{5}$, $13 k_{6}, 13 k_{7}$, and their sum be $y$.
Then the original problem is to find the number of positive integer solutions $\left(k_{1}, k_{2}, \cdots, k_{... | 42244 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,322 |
Example 1 As shown in Figure 1, in the convex quadrilateral $ABCD$, the extensions of $AB$ and $DC$ meet at $E$, and the extensions of $AD$ and $BC$ meet at $F$. $P$, $Q$, and $R$ are the midpoints of $AC$, $BD$, and $EF$ respectively. Prove that $P$, $Q$, and $R$ are collinear. | Proof: Let $AB = a, AD = b, BE = \lambda a, DF = \mu b, EC = mE, FC = nFB$, then we have
$$
\begin{aligned}
AC & = AE + EC = AE + m(AD - AE) \\
& = (1 + \lambda)(1 - m)a + mb.
\end{aligned}
$$
Also, $AC = AF + FC = AF + n(AB - AF)$
$$
= (1 + \mu)(1 - n)b + na,
$$
Thus, we have the system of equations:
$$
\left\{\begi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,323 |
Example 4 Competition Scores
At the New Year's Eve party, the school organized a knowledge competition on Chinese, Mathematics, Foreign Language, Olympic Games, and daily life common knowledge, setting the full score for each subject at 40 points, followed by 30 points, 20 points, 10 points, and 0 points, a total of 5 ... | Explanation: According to condition (1), in each vertical column, the five groups' scores are all different.
For a single subject, all possible different scores are $0, 10, 20, 30,$ and $40$, only 5 kinds. The total scores of the five groups in each subject are
$$
5 \times (0 + 10 + 20 + 30 + 40) = 500.
$$
Subtractin... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,324 |
For example, let the circumcenter of $\triangle ABC$ be $O$, the centroid be $G$, and the orthocenter be $H$. Prove that $O$, $G$, and $H$ are collinear, and $OG = \frac{1}{3} OH$.
| $$
\begin{array}{l}
\text { Proof: } \because O G=O A+A G=O B+B G= \\
O C+C G \text {, } \\
\therefore 3 O G=(O A+O B+O C)+(A G+B G \\
+\boldsymbol{C} \boldsymbol{G} \text {. } \\
\therefore A G+B G+C G=0 \text {, } \\
\end{array}
$$
$\because G$ is the centroid of $\triangle A B C$,
i.e., $O G=\frac{1}{3}(\boldsymbol{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,325 |
Example 3 As shown in Figure 3, in $\triangle ABC$, $O$ is the circumcenter, the three altitudes intersect at $H$, and $D$, $E$, and $F$ are the feet of the perpendiculars. Line $ED$ and $AB$ intersect at $M$, and line $FD$ and $AC$ intersect at $N$. Prove:
(1) $OB \perp DF, OC \perp DE;$
(2) $OH \perp MN$ | Proof: (1) Let the circumradius of $\triangle ABC$ be $R$, then we have
$$
\begin{aligned}
& \boldsymbol{O B} \cdot \boldsymbol{D F}=\boldsymbol{O B} \cdot(\boldsymbol{D B}+\boldsymbol{B F}) \\
= & \boldsymbol{O B} \cdot \boldsymbol{D} \boldsymbol{B}+\boldsymbol{O B} \cdot \boldsymbol{B F} \\
= & \boldsymbol{B O} \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,326 |
Example 4 As shown in Figure 4, on the sides $BC$ and $CD$ of $\square ABCD$, construct squares $BCNM$ and $CDPQ$ outward. Prove that $AC \perp QN$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 4 As shown in Figure 4, on the sides $BC$ and $CD$ of $\square ABCD$, construct squares $BCNM$ and $CDPQ$ outward. Prove that $AC \perp Q... | Proof: $A C \cdot Q N$
$$
\begin{aligned}
= & (A B+B C) \cdot \\
& (Q C+C N) \\
= & A B \cdot Q C+A B \cdot C N+B C \cdot Q C+B C \cdot C N \\
= & A B \cdot C N+B C \cdot Q C \quad(\because A B \perp Q C, B C \perp C N) \\
= & C B \cdot C Q-C D \cdot C N \\
= & |C B| \cdot|C Q| \cdot \cos \angle B C Q-|C D| . \\
& |C N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,327 |
Find all positive integers $n$ such that $20n+2$ divides $2003n+2002$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Obviously, $21 n$. Let $n=2 m\left(m \in \mathbf{N}^{*}\right)$, then
$(20 m+1) 1(2003 m+1001)$.
Since $2003 m+1001=100(20 m+1)+3 m+901$,
thus $(20 m+1) \mid(3 m+901)$.
It is easy to see that $\frac{3 m+901}{20 m+1}=1,2,3,4$ have no positive integer solutions. Therefore, $\frac{3 m+901}{20 m+1} \geqslant 5$, which impl... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,328 |
(1) Ask: When $n=3$, does there exist an arrangement that meets the requirements? Prove your conclusion;
(2) Prove: $n$ is an odd number. | (1) When $n=3$, there exists an arrangement that meets the requirements. The specific arrangement is as follows (label the 9 female students as $1,2, \cdots, 9$):
$$
\begin{array}{l}
(1,2,3),(1,4,5),(1,6,7),(1,8,9),(2,4,6), \\
(2,7,8),(2,5,9),(3,4,8),(3,5,7),(3,6,9), \\
(4,7,9),(5,6,8) .
\end{array}
$$
(2) Arbitrarily ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,329 |
Three, try to find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$. | Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,330 |
Four, $\odot O_{1}$ and $\odot O_{2}$ intersect at points $B$ and $C$, and $BC$ is the diameter of $\odot O_{1}$. A tangent to $\odot O_{1}$ is drawn through point $C$, intersecting $\odot O_{2}$ at another point $A$. Connect $AB$, intersecting $\odot O_{1}$ at another point $E$. Connect $CE$ and extend it to intersect... | As shown in Figure 2, since $BC$ is the diameter of $\odot O_{1}$, and $AC$ is tangent to $\odot O_{1}$ at $C$, we have $\angle BEC = \angle FEA = \angle BCA = \angle BCD = 90^{\circ}$.
Let $\angle ABC = \alpha$, $\angle CBD = \beta$, then $\angle AFC = \alpha$, $\angle CEG = \beta$.
According to the Law of Sines, we ... | \frac{AH}{HF}=\frac{AC}{CD} | Geometry | proof | Yes | Yes | cn_contest | false | 713,331 |
Let $P_{1}, P_{2}, \cdots, P_{n}(n \geqslant 2)$ be any permutation of $1,2, \cdots, n$. Prove that:
$$
\begin{array}{l}
\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}} \\
+\frac{1}{P_{n-1}+P_{n}}>\frac{n-1}{n+2} .
\end{array}
$$ | $$
\begin{array}{l}
\left(\frac{1}{P_{1}+P_{2}}+\cdots+\frac{1}{P_{n-1}+P_{n}}\right)\left[\left(P_{1}+P_{2}\right)+\cdots\right. \\
\left.+\left(P_{n-1}+P_{n}\right)\right] \geqslant(n-1)^{2} . \\
\text { Then } \frac{1}{P_{1}+P_{2}}+\cdots+\frac{1}{P_{n-1}+P_{n}} \\
\geqslant \frac{(n-1)^{2}}{2\left(P_{1}+\cdots+P_{n... | \frac{n-1}{n+2} | Inequalities | proof | Yes | Yes | cn_contest | false | 713,332 |
Six, find all positive integer pairs $(x, y)$ that satisfy $x^{y}=y^{x^{-y}}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Six, if $x=1$, then $y=1$; if $y=1$, then $x=1$; if $x=y$, then $x=y=1$. Therefore, we only need to discuss the case where $x>y \geqslant 2$. From the equation, we get
$$
12 y, y \mid x$.
Let $x=k y$, then $k \geqslant 3, k \in \mathbf{N}$. Thus, $k^{y}=y^{(k-2)}$.
This gives $k=y^{k-2}$.
Since $y \geqslant 2$, we have... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,333 |
Seven, In an acute triangle $\triangle A B C$, the three altitudes are $A D, B E, C F$. Prove: The perimeter of $\triangle D E F$ does not exceed half the perimeter of $\triangle A B C$.
保留源文本的换行和格式，直接输出翻译结果如下：
Seven, In an acute triangle $\triangle A B C$, the three altitudes are $A D, B E, C F$. Prove: The perimete... | Proof 1: Since points $D, E, A, B$ are concyclic, and $AB$ is the diameter of this circle, according to the Law of Sines, we have
$$
\frac{DE}{\sin \angle DAE} = AB = c,
$$
which means $DE = c \sin \angle DAE$.
Also, $\angle DCA = 90^{\circ} - \angle DAC$, so $DE = c \cos C$.
Similarly, $DF = b \cos B$.
Therefore, $DE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,334 |
Example 5 Lift Your Veil
A 101-digit natural number $A=\underbrace{88 \cdots 8}_{\text {S0 digits }} \square \underbrace{99 \cdots 9}_{\text {S0 digits }}$ is divisible by 7. What is the digit covered by $\square$? | Explanation: “ $\square$ ” is a veil, covering the number to be found. To lift the veil, whether approaching from the front or the back, there are 50 digits in between, which is too many. The key is to find a way to reduce the number of digits.
The difference between two multiples of 7 is still a multiple of 7.
We alr... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,335 |
Eight, let $A_{1}, A_{2}, \cdots, A_{8}$ be any 8 points taken on a plane. For any directed line $l$ taken on the plane, let the projections of $A_{1}, A_{2}, \cdots, A_{8}$ on this line be $P_{1}, P_{2}, \cdots, P_{8}$, respectively. If these 8 projections are pairwise distinct and are arranged in the direction of lin... | Proof 3: First, prove that $\triangle D E F$ is the triangle with the shortest perimeter among all inscribed triangles in the acute $\triangle A B C$.
Let point $D^{\prime}$ be any fixed point on side $B C$, as shown in Figure 4. Construct the reflection of point $D^{\prime}$ about $A B$ and $A C$.
Assume the number o... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,336 |
Let convex quadrilateral $ABCD$ have the lines of its two pairs of opposite sides intersect at points $E$ and $F$, respectively, and the intersection of the diagonals be point $P$. Draw $PO \perp EF$ at $O$. Prove that $\angle BOC = \angle AOD$.
---
The translation maintains the original text's formatting and line br... | Solution: As shown in Figure 1, it suffices to prove that $O P$ is both the angle bisector of $\angle A O C$ and the angle bisector of $\angle D O B$.
Without loss of generality, let $A C$ intersect $E F$ at $Q$. Considering $\triangle A E C$ and point $F$, by Ceva's Theorem, we have
$$
\frac{E B}{B A} \cdot \frac{A Q... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,337 |
Let $a_{1}=\frac{1}{4}, a_{n}=\frac{1}{4}\left(1+a_{n-1}\right)^{2}, n \geqslant 2$. Find the smallest
real number $\lambda$, such that for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2} 002$, we have
$$
\sum_{k=1}^{2002} A_{k} \leqslant \lambda a_{2002} \text {. }
$$
where $A_{k}=\frac{x_{k}-k}{\left(x_{k... | Let $\delta(k)=\frac{1}{2} k(k-1)$. First, we prove a few lemmas.
Lemma 1 For any real numbers $a \geqslant 0, c>0, b>0$, the function $f(x)=\frac{a}{x+b}+\frac{x-c}{(x+b)^{2}}$.
When $x=\frac{(1-a) b+2 c}{1+a}$, it attains the maximum value $\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c}$.
Proof: Let $y=\frac{1}{x+b}$, then... | \frac{1}{2003 \times 1001+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,338 |
Three, 17 football fans plan to go to Cao Guo to watch the World Cup football matches, and they have selected a total of 17 matches. The booking of tickets meets the following conditions:
(i) Each person can book at most one ticket per match;
(ii) The tickets booked by any two people have at most one match in common;
(... | Solution: Draw a $17 \times 17$ grid, with 17 columns representing 17 matches and 17 rows representing 17 people. If the $i$-th person has booked a ticket for the $j$-th match, then the center of the cell at the intersection of the $i$-th row and the $j$-th column is marked with a red dot. Thus, the problem is transfor... | 71 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,339 |
Let $k$ be a given integer, and $f(n)$ be a function defined on the set of negative integers with integer values, satisfying
$$
\begin{array}{l}
f(n) f(n+1)=(f(n)+n-k)^{2}, \\
n=-2,-3,-4, \cdots .
\end{array}
$$
Find the expression for the function $f(n)$. | Lemma: There exist infinitely many prime numbers that are not of the form $5k \pm 1$.
Proof: For any given positive integer $n$, consider
$N=5[1 \cdot 3 \cdot 5 \cdots \cdots(2n+1)]^{2}+2$.
Since the product of integers of the form $5k \pm 1$ is still of the form $5k \pm 1$, $N \equiv 2(\bmod 5)$, hence $N$ has a prim... | f(n)=(n-k)^{2}, n \leqslant k \text{ and } n<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,341 |
$$
\begin{array}{c}
\text { Six, let } f\left(x_{1}, x_{2}, x_{3}\right)=-2\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}\right)+3\left[x_{1}^{2}\right. \\
\left.\cdot\left(x_{2}+x_{3}\right)+x_{2}^{2}\left(x_{3}+x_{1}\right)+x_{3}^{2}\left(x_{1}+x_{2}\right)\right]-12 x_{1} x_{2} x_{3} \text {. }
\end{array}
$$
For any real num... | Let $x=x_{3}-(r+1)$, then
$f\left(r, r+2, x_{3}\right)$
$=-2\left[r^{3}+(r+2)^{3}+(x+r+1)^{3}\right]$
$+3\left[r^{2}(x+2 r+3)+(r+2)^{2}(x+2 r+1)\right.$
$\left.+(x+r+1)^{2}(2 r+2)\right]-12 r(r+2)(x+r+1)$
$=-2 x^{3}+18 x$.
Let $a=t-(r+1)$. Then
$g(r, s, t)=\max _{u \leqslant x \leqslant a+2} 1-2 x^{3}+18 x+s \mid$.
Sin... | 6 \sqrt{3}-\frac{20 \sqrt{6}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,342 |
1. Given $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}=3$. Then the value of $(a-b)^{2}+(b-c)^{2}+(a-b) \cdot(b-c)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | ,$- 1 .(\mathrm{C})$.
$$
\begin{array}{l}
\because a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
$\therefore$ The original condition equation is $a^{2}+b^{2}+c^{2}-a b-b c-c a=3$. The required algebraic expression, when expanded and combined, is
$$
a^{2}+b^{2}+c^{2}-a ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,343 |
2. Define " $\triangle$ " as an operation on ordered pairs of real numbers, as shown below, $(a, b) \triangle(c, d)=(a c+b d, a d+b c)$. If for any real numbers $a, b$ we have $(a, b) \triangle(x, y)=(a, b)$, then $(x, y)$ is ( ).
(A) $(0,1)$
(B) $(1,0)$
(C) $(-1,0)$
(D) $(0,-1)$ | 2. (B).
By definition, we have
$$
(a, b)=(a, b) \doteq(x, y)=(a x+b y, a y+b x) \text {. }
$$
Then, \(a x+b y=a, a y+b x=b\).
Adding the two equations, we get
$$
(a+b) x+(a+b) y=a+b \text {. }
$$
By the arbitrariness of \(a, b\), we know \(x+y=1\).
Subtracting the two equations, we get \((a-b) x-(a-b) y=a-b\).
Simil... | (B) | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,344 |
3. In $\triangle A B C$, $\frac{2}{a}=\frac{1}{b}+\frac{1}{c}$. Then $\angle A()$.
(A) must be an acute angle
(B) must be a right angle
(C) must be an obtuse angle
(D) none of the above | 3. (A).
Proof by contradiction. If $\angle .4$ is not an acute angle, then $\angle .4 \geqslant 90^{\circ}$. Given $a>b$ and $a>c$, we have $\frac{1}{a}<\frac{1}{b}, \frac{1}{a}<\frac{1}{c}$. Therefore, $-\frac{1}{a}+\frac{1}{a}<\frac{1}{b}+\frac{1}{c}$, which simplifies to $\frac{2}{a}<\frac{1}{b}+\frac{1}{c}$, contr... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,345 |
6. All integer solutions to the equation $x^{2}+y^{2}=2(x+y)+x y$ are ( ).
(A) one set
(B) two sets
(C) two sets
(D) four sets | 6. (C).
The original equation is rearranged to $x^{2}-(y+2) x+\left(y^{2}-2 y\right)=0$.
$$
\begin{array}{l}
\Delta=(y+2)^{2}-4\left(y^{2}-2 y\right) \geqslant 0, \\
\frac{6-4 \sqrt{3}}{3} \leqslant y \leqslant \frac{6+4 \sqrt{3}}{3} .
\end{array}
$$
Since $y$ is a positive integer, we have $1 \leqslant y \leqslant 4... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,349 |
1. A bus, a truck, and a car are traveling in the same direction on the same straight line. At a certain moment, the truck is in the middle, the bus is in front, and the car is behind, and the distances between them are equal. After 10 minutes, the car catches up with the truck; after another 5 minutes, the car catches... | Ni.1.15.
As shown in Figure 4, let the distances between the truck and the bus, and the truck and the car be $S$. The speeds of the car, truck, and bus are approximately $a, b, c$. It is given that the truck catches up with the bus after $x$ minutes. Therefore, we have
$$
\begin{array}{l}
10(a-b)=S, \\
15(a-c)=2S, \\
x... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,350 |
2. If the polynomial $P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b$ +2070, then the minimum value of $P$ is | $$
\begin{array}{l}
2.2002 \\
P=2 a^{2}-8(b+2) a+17 b^{2}-4 b+2070 \\
=2\left[a^{2}-4(b+2) a+4(b+2)^{2}\right]-8(b+2)^{2} \\
\quad+17 b^{2}-4 b+2070 \\
=2(a-2 b-4)^{2}+9 b^{2}-36 b+36+2002 \\
=2(a-2 b-4)^{2}+9(b-2)^{2}+2002
\end{array}
$$
When $a-2 b-4=0$ and $b-2=0$, $P$ reaches its minimum value of 2002. | 2002 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,351 |
3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is | 3.10 .
As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$
which implies
$$
\begin{aligned}
l_{\triangle P... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,352 |
4. Given the quadratic function $y=a x^{2}(a \geqslant 1)$, the x-coordinates of points $A$ and $B$ on its graph are $-1, 2$, respectively, and $O$ is the origin. If $\triangle A O B$ is a right triangle, then the perimeter of $\triangle O A B$ is $\qquad$ . | $$
4.4 \sqrt{2}+2 \sqrt{5}
$$
As shown in Figure 6, $A(-1$,
a), $B(2,4 a)$, then
$$
\begin{array}{l}
O A=\sqrt{a^{2}+1}, \\
O B=\sqrt{4+16 a^{2}}, \\
A B=\sqrt{9+9 a^{2}} .
\end{array}
$$
Since $a \geqslant 1$, $O A$ is the smallest side, so $O A$ cannot be the hypotenuse.
(i) If $O B$ is the hypotenuse, then $O B^{... | 4 \sqrt{2}+2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,353 |
One, (20 points) Given real numbers $a, b, c$ satisfy the inequalities
$$
|a| \geqslant|b+c|,|b| \geqslant|c+a|,|r| \geqslant|a+b| .
$$
Find the value of $a+1+r$.
| Given $|a| \geqslant|b+c|$, we get $a^{2} \geqslant b^{2}+2 b c+c^{2}$.
Similarly, $b^{2} \geqslant c^{2}+2 c a+a^{2}$, $c^{2} \geqslant a^{2}+2 a b+b^{2}$.
Adding the three equations, we get
$a^{2}+b^{2}+c^{2} \geqslant 2 a^{2}+2 b^{2}+2 c^{2}+2 a b+2 b c+2 c a$,
which simplifies to $a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c ... | a+b+c=0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,354 |
II. (25 points) As shown in Figure 2, point $D$ is on the side $BC$ of $\triangle ABC$, and does not coincide with $B$ or $C$. A line through point $D$ parallel to $AC$ intersects $AB$ at $E$, and a line through $D$ parallel to $AB$ intersects $AC$ at $F$. It is also known that $BC=5$.
(1) Let the area of $\triangle AB... | (ii) $\because D E / / A C, D F / / A B$,
$\therefore \triangle B D E \backsim \triangle B C A \backsim \triangle D C F$.
Let $S_{- \text {B }}=S_{1}$,
$$
S_{\text {IUCF }}=S_{2} .
$$
$$
\begin{aligned}
\therefore S_{1}+S_{2} & =S-\frac{2}{5} S \\
& =\frac{3}{5} S .
\end{aligned}
$$
$$
\frac{\sqrt{S_{1}}}{\sqrt{S}}=\fr... | B D=\frac{5 \pm \sqrt{5}}{2}, E F=\frac{5 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,355 |
Three, (25 points) Given theorem: “For any prime number $n$ greater than 3, $b$ and $c$ satisfy the equation $2a + 5b = c$. Then $a + b + c$ is how many times the integer $n$? Prove your conclusion.”
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | Three, the maximum possible value of $n$ is 9.
First, we prove that $a+b+c$ is divisible by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$.
Thus, $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, with $r_{a} \neq 0$ and $r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, ... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false | 713,356 |
Example 7 Multiplicative Magic Square
Figure 1 shows a partially filled magic square. Fill in the following nine numbers: $\frac{1}{4}, \frac{1}{2}, 1,2,4,8,16,32,64$ in the grid so that the product of the numbers in each row, column, and diagonal is the same. The number that should be filled in the “ $x$ ” cell is $\q... | Explanation: Label the unfilled cells with letters, as shown in Figure 2.
The product of the nine known numbers is
$$
\frac{1}{4} \times \frac{1}{2} \times 8 \times 1 \times 2 \times 32 \times 4 \times 16 \times 64=64^{3} \text {. }
$$
Therefore, the fixed product of the three numbers in each row, each column, and eac... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,357 |
2. The interval of increase for the function $y=\frac{x^{2}}{x-1}(x \in \mathbf{R}, x \neq 1)$ is
(A) $x \geqslant 2$
(B) $x \leqslant 0$ or $x \geqslant 2$
(C) $x \leqslant 0$
(D) $x \leqslant 1-\sqrt{2}$ or $x \geqslant \sqrt{2}$ | 2. (B).
Let $t=x-1(t \neq 0)$, then $y=2+\left(t+\frac{1}{t}\right)(t \neq 0)$. If $t>0$, then $y=\left(\sqrt{t}-\frac{1}{\sqrt{t}}\right)^{2}+4$, at this time, if $t \geqslant 1$, then $\sqrt{t}-\frac{1}{\sqrt{t}} \geqslant 0, y$ is an increasing function of $t$; when $0<t \leqslant 1$, then $y=\left(\frac{1}{\sqrt{t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,359 |
3. A line $l$ is drawn through the fixed point $P(2,1)$, intersecting the positive direction of the $x$-axis and the positive direction of the $y$-axis at points $A$ and $B$, respectively, such that the area of $\triangle A O B$ (where $O$ is the origin) is minimized. The equation of $l$ is ( ).
(A) $x+y-3=0$
(B) $x+3 ... | 3. (D).
It is easy to get $S_{\triangle A n B}=\frac{1}{2} a b$, and we have $\frac{2}{a}+\frac{1}{b}=1$, which means $a b=a+$ $2 b \geqslant 2 \sqrt{a \cdot 2 b}=2 \sqrt{2} \sqrt{a b}$. Therefore, $\sqrt{a b} \geqslant 2 \sqrt{2}, a b \geqslant 8$. The equality holds if and only if $\left\{\begin{array}{l}a=2 b, \\ 2... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,360 |
4. If the equation $\cos 2 x+\sqrt{3} \sin 2 x=a+1$ has two distinct real solutions $x$ in $\left[0, \frac{\pi}{2}\right]$, then the range of the parameter $a$ is ( ).
(A) $0 \leqslant a<1$
(B) $-3 \leqslant a<1$
(C) $a<1$
(D) $0<a<1$ | 4. (A).
The equation can be transformed into
$$
\sin \left(2 x+\frac{\pi}{6}\right)=\frac{a+1}{2} \text {. }
$$
Since $0 \leqslant x \leqslant \frac{\pi}{2}$, we have
$\frac{\pi}{6} \leqslant 2 x+\frac{\pi}{6} \leqslant \pi$
$+\frac{\pi}{6}$. Let $t=2 x+$
$\frac{\pi}{6}$, then $\sin t=\frac{a+1}{2}\left(\frac{\pi}{6}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,361 |
5. The value of the 1000th term in the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6, \cdots$ is ( ).
(A) 42
(B) 45
(C) 48
(D) 51 | 5. (B).
The key is to determine which number segment $a_{1000}$ falls into. Suppose $a_{1000}$ is in the $k$-th number segment, then $a_{100}=k$. Since the $k$-th number segment contains $k$ numbers, $k$ should satisfy the inequality $\frac{(k-1) k}{2}=1+2+3+\cdots+(k-1)<1000 \leqslant 1+2+3+\cdots+k=\frac{k(k+1)}{2}$... | 45 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,362 |
6. In the permutation $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of $1,2,3,4,5$, the number of permutations that satisfy the conditions $a_{1}a_{3}, a_{3}a_{5}$ is ( ).
(A) 8
(B) 10
(C) 14
(D) 16 | 6. (D).
From $a_{1}a_{3}, a_{3}a_{3}$ we know, either $a_{2}=5$, or $a_{4}=5\left(a_{1} 、 a_{3} 、 a_{5}\right.$ cannot be the maximum), and $a_{2} \geqslant 3, a_{4} \geqslant 3, a_{3} \leqslant 3$.
(1) If $a_{2}=5$, then $a_{4}=4$ or 3.
When $a_{4}=4$, there are $3!=6$ ways;
When $a_{4}=3$, there are $2!=2$ ways.
At ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,363 |
1. [ $x$ ] represents the greatest integer not greater than $x$, then the real solution $x$ of the equation $\frac{1}{2} \times\left[x^{2}+x\right]=19 x+99$ is $\qquad$. | $$
=1 .-\frac{181}{38} \text { or } \frac{1587}{38} \text {. }
$$
From the condition, we get $38 x+198=\left[x^{2}+x\right] \in \mathbb{Z}$. Therefore, $38 x \in \mathbb{Z}$. Let $38 x=k \in \mathbb{Z}$, then $x=\frac{k}{38}(k \in \mathbb{Z})$. Substituting into the equation, we get $\left[\frac{k^{2}+38 k}{38^{2}}\ri... | x=-\frac{181}{38} \text{ or } \frac{1587}{38} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,364 |
2. Let $a_{1}=1, a_{n+1}=2 a_{n}+n^{2}$, then the general term formula $a_{n}$ $=$ $\qquad$ | $$
2.7 \times 2^{n-1}-n^{2}-2 n-3 .
$$
Let $a_{n}=b_{n}+s n^{2}+t n+k$ (where $s$, $t$, and $k$ are constants to be determined). Then
$$
\begin{array}{l}
b_{n+1}+s(n+1)^{2}+t(n+1)+k \\
=2 b_{n}+2 s n^{2}+2 t n+2 k+n^{2},
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
b_{n+1}=2 b_{n}+(s+1) n^{2}+(t-2 s) n+k-t-... | 7 \times 2^{n-1}-n^{2}-2 n-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,365 |
3. The remainder when the number $7^{99}$ is divided by 2550 is | 3.343 .
Notice that $2550=50 \times 51$. First, we find the remainders when $7^{*})$ is divided by 5 and 51, respectively.
First, $7^{199}=7 \times 7^{48}=7 \times 49^{19} \equiv 7 \times(-2)^{459}$
$\equiv-14 \times 16^{12} \equiv-14 \times 256^{6} \equiv-14 \times 1^{6}$
$\equiv-14(\bmod 51)$;
Additionally, $7^{(9)}... | 343 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,366 |
4. In $\triangle A B C$, $\angle A=\frac{\pi}{3}, \sin B=\frac{5}{13}$, then the value of $\cos C$ is | 4. $\frac{5 \sqrt{3}-12}{26}$.
Given $\sin B=\frac{5}{13}\pi-\frac{\pi}{6}$.
If $\angle B>\pi-\frac{\pi}{6}$, then
$$
\begin{array}{l}
\angle C=\pi-\frac{\pi}{3}-\angle B<\pi-\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right) \\
=\frac{\pi}{6}-\frac{\pi}{3}<0 .
\end{array}
$$
This is a contradiction. Therefore, it must be ... | \frac{5 \sqrt{3}-12}{26} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,367 |
Example 8 Octagonal Ingenuity
Fill the numbers $1, 2, 3, 4, 5, 6, 7, 8$ into the 8 vertices of the octagon $A B C D E F G H$ (Figure 4), and let $S_{1}, S_{2}, \cdots, S_{8}$ represent the sums of the numbers at the 3 adjacent vertices $(A, B, C), (B, C, D), \cdots, (H, A, B)$ respectively.
(1) Try to give a filling me... | (1) Let the number at point $A$ be denoted as $a$, the number at point $B$ as $b$, and so on.
Pairing large and small numbers, let's take $a=1, b=8$.
From $a+b+c \geqslant 12$, we get $c \geqslant 3$. To be precise, take the smallest possible value, let $c=3$.
Continuing this precise calculation, we sequentially get $... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,368 |
5. Let $k$ and $\theta$ be real numbers such that the roots of the equation $x^{2}-(2 k+$ 1) $x+k^{2}-1=0$ are $\sin \theta$ and $\cos \theta$. Then the range of $\theta$ is $\qquad$. | $5 . \theta=2 k_{1} \pi+\pi$ or $2 k_{1} \pi-\frac{\pi}{2}$ ( $k_{1}$ is any integer $)$.
By Shuda's theorem, we get
$$
\sin \theta+\cos \theta=2 k+1 \cdot \sin \theta \cdot \cos \theta=k^{2}-1 \text {. }
$$
From $\sin ^{2} \theta+\cos ^{2} \theta=1$ we get
$(\sin \theta+\cos \theta)^{2}-2 \sin \theta \cdot \cos \thet... | \theta=2 k_{1} \pi+\pi \text{ or } 2 k_{1} \pi-\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,369 |
6. The unit digit of the number $(5+\sqrt{24})^{2 n}(n \in \mathbf{N})$ is
保留源文本的换行和格式，翻译结果如下：
6. The unit digit of the number $(5+\sqrt{24})^{2 n}(n \in \mathbf{N})$ is | 6. When $21 n$, it is 1; when $2 \times n$, it is 7.
Let $x=(5+\sqrt{24})^{2 n}=A+\sqrt{24} B(A, B \in \mathbf{N})$.
By the binomial theorem, we know
Therefore, $x+y=2 A \in \mathbf{N}$. f is
$$
x=(2 A-1)+1-y(\text { where } 0<1-y<1),
$$
the unit digit of $x$ is the unit digit of $(2 A-1)$.
$$
\begin{array}{l}
\text ... | 1, 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,370 |
Three. (20 points) Given that $x, y, z$ are non-negative real numbers, and $x+y+z=1$. Prove that $x(1-2 x)(1-3 x)+y(1-2 y)(1-3 y)+z(1-2 z)(1-3 z) \geqslant 0$, and determine the conditions under which equality holds. | $$
\begin{array}{l}
\text { Three, Left side }=x\left(1-5 x+6 x^{2}\right)+y\left(1-5 y+6 y^{2}\right) \\
+z\left(1-5 z+6 z^{2}\right) \\
=1-5\left(x^{2}+y^{2}+z^{2}\right)+6\left(x^{3}+y^{3}+z^{3}\right) \\
=1-5\left(x^{2}+y^{2}+z^{2}\right)+6\left[3 x y z+(x+y+z)\left(x^{2}\right.\right. \\
\left.\left.+y^{2}+z^{2}-x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,371 |
4. (20 points) (1) Find all real numbers $a$ such that the equation $x^{2}+(a+2002) x+a=0$ has two integer roots.
(2) Find all real numbers $a$ such that the equation $x^{3}+\left(-a^{2}+2 a+2\right) x-2 a^{2}-2 a=0$ has two integer roots. | (1) Let the two integer roots be $x_{1}$ and $x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-a-2002 . \\
x_{1} x_{2}=a .
\end{array}\right.
$$
From (D) 2 we get
$$
\begin{array}{l}
x_{1} x_{2}+x_{1}+x_{2}=-2002 . \\
\text { Hence }\left(x_{1}+1\right)\left(x_{2}+1\right)=-2001=-3 \times 23 \times 29 .
\end{array... | a=-3,-1,9,11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,372 |
One, (50 points) Let $a, b, c \in \mathbf{R}, b \neq a c, a \neq -c, z$ be a complex number, and $z^{2}-(a-c) z-b=0$. Prove that:
$$
\left|\frac{a^{2}+b-(a+c) z}{a c-b}\right|=1
$$
if and only if $(a-c)^{2}+4 b \leqslant 0$. | If $(a-c)^{2}+4 b \leqslant 0$, then by the quadratic formula we get
$$
z=\frac{a-c \pm \sqrt{-(a-c)^{2}-4 b} i}{2}\left(i^{2}=-1\right) \text {. }
$$
Therefore, $\left|\frac{a^{2}+b-(a+c) z}{a c-b}\right|$
$=\left|\frac{a^{2}+c^{2}+2 b \mp(a+c) \sqrt{-(a-c)^{2}-4 b} \mathrm{i}}{2(a c-b)}\right|$
$=\sqrt{\frac{\left(a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,374 |
II. (50 points) As shown in Figure 1, in $\triangle ABC$, $\angle ABC$ and $\angle ACB$ are both acute angles. $D$ is an internal point on side $BC$, and $AD$ bisects $\angle BAC$. Perpendiculars $DP$ and $DQ$ are drawn from point $D$ to lines $AB$ and $AC$, respectively, with feet of the perpendiculars being $P$ and $... | Certainly, here is the translation of the provided text into English, preserving the original formatting:
```
(1) Clearly, $P D=D Q, A P=A Q$. Without loss of generality, assume $A B \geqslant A C$, then $\angle A D C \leqslant \angle A D B, \angle A D C$ is a right angle or an acute angle.
As shown in Figure 4, draw... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,375 |
Three. (50 points) Given a positive integer $n$, let $n$ real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following $n$ equations:
$$
\sum_{i=1}^{n} \frac{a_{i}}{i+j}=\frac{4}{2 j+1}(j=1,2,3, \cdots, n) .
$$
Determine the value of the sum $S=\sum_{i=1}^{n} \frac{a_{i}}{2 i+1}$ (expressed in the simplest form inv... | $$
\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}}{x+i}-\frac{4}{2 x+1} \\
=\frac{P(x)}{(2 x+1) \prod_{i=1}^{n}(x+i)}=0 .
\end{array}
$$
From the given, the degree of $P(x)$ does not exceed $n$. For $x=1,2,3, \cdots, n$, it holds, so $P(x)=k \cdot \prod_{i=1}^{n}(x-i)$ (where $k$ is a constant to be determined). Therefor... | S=1-\frac{1}{(2 n+1)^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,376 |
In the plane rectangular coordinate system $x O y$, the equation of $\odot A$ is $(x-2)^{2}+(y-2)^{2}=1$. Two circles $\odot O_{1}$ and $\odot O_{2}$, both with radius $r$ and externally tangent to each other, are also externally tangent to $\odot A$. Furthermore, $\odot O_{1}$ and $\odot O_{2}$ are tangent to the $x$-... | Solution: As shown in Figure 2 and Figure 3, since the center of circle $A$, $A(2,2)$, lies on the line $y=x$, and the radii of $\odot O_{1}$ and $\odot O_{2}$ are equal, the entire figure is symmetric about the line $y=x$. Therefore, the tangent point $T$ of $\odot O_{1}$ and $\odot O_{2}$ lies on the line $y=x$. Let ... | 7-4 \sqrt{2} \pm \sqrt{60-42 \sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,378 |
120. In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $1^{\circ}$. Find the maximum value of the difference between the largest and smallest interior angles. | Solution: Let the largest interior angle be $\alpha$, and the smallest interior angle be $\alpha-k$ (omitting the degree symbol for simplicity, the same applies below). Therefore, what we are looking for is the maximum value of $k$. Hence, the interior angles should also include $\alpha-1, \alpha-2, \alpha-3, \cdots$, ... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,381 |
Example 1 Given a linear function $y=kx+b, kb<0$. Then the graph of such a linear function must pass through $\qquad$ quadrants, i.e., the $\qquad$ quadrant.
(14th Junior High School Mathematics Competition in Jiangsu Province) | Solution: When $k b0, b0$, its graph passes through the first, second, and fourth quadrants.
Therefore, the graph of this linear function must pass through two common quadrants, namely the first and fourth quadrants. | first and fourth quadrants | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,383 |
Example 2 Given $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=$ $\frac{c+a}{b}=p$. Then, the line $y=p x+p$ must pass through ( ).
(A) the first and second quadrants
(B) the second and third quadrants
(C) the third and fourth quadrants
(D) the first and fourth quadrants
(1998, National Junior High School Mathematics... | Solution: From $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=p$, we get $a+b=c p, b+c=a p, c+a=b p$.
Adding the three equations, we get
$$
2(a+b+c)=p(a+b+c) .
$$
(1) When $a+b+c=0$, $p=-1, y=-x-1$, then the line passes through the second, third, and fourth quadrants;
(2) When $a+b+c \neq 0$, $p=2, y=2 x+2$, then the line ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,384 |
3. In the Cartesian coordinate system $x O y$, the moving point $M(x, 0)$ on the $x$-axis has distances $M P$ and $M Q$ to the fixed points $P(5,5)$ and $Q(2,1)$, respectively. When $M P + M Q$ takes the minimum value, the $x$-coordinate of point $M$ is $x=$ $\qquad$ . (2001, TI Cup National Junior High School Mathemat... | (Answer: $\frac{5}{2}$ ) | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,385 |
14. Given $f(x)=x^{2002}-x^{2001}+1$. Prove: For any positive integer $m$, $m, f(m), f(f(m)), f(f(f(m))), \cdots$ are pairwise coprime.
(2002, Croatian National Mathematical Competition) | Proof: Let $P_{n}(x)=f(f(f(\cdots(x) \cdots))$ ) (where $f$ appears $n$ times).
$\because f(0)=1, f(1)=1$, hence for $\forall n \in \mathbf{N}, P_{n}(0)=1$,
$\therefore P_{k}(x)$ is a polynomial containing the constant term $P_{k}(0)=1$.
Therefore, the remainder of $P_{n}(m)$ when divided by $m>1$ is 1.
Suppose an inte... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,386 |
15. Let $p_{n}$ be the $n$-th prime number starting from the smallest prime 2. For example: $p_{1}=2, p_{2}=3, p_{3}=5, \cdots$.
(1) Given $n \geqslant 10, r$ is the smallest integer satisfying $2 \leqslant r \leqslant n-2, n-r+1 < p_{r}$. Define $N_{1}=s p_{1} p_{2} \cdots p_{r-1}-1$, where $s = 1,2, \cdots, p_{r}$. P... | (1) $\because$ For all $k, l\left(1 \leqslant k < l \leqslant n\right)$, $p_{l} > p_{k} \geqslant kn-r+1$,
$\therefore$ There exists $N_{j}\left(1 \leqslant j \leqslant p_{r}\right)$, such that $p_{r}, p_{r+1}, \cdots, p_{n}$ do not divide $N_{j}$.
By the definition of $N_{j}$, $p_{1}, p_{2}, \cdots, p_{r-1}$ also do ... | m \geqslant 4 | Number Theory | proof | Yes | Yes | cn_contest | false | 713,387 |
1. There are three types of school supplies: pencils, exercise books, and ballpoint pens. If you buy 3 pencils, 7 exercise books, and 1 ballpoint pen, it costs 3.15 yuan; if you buy 4 pencils, 10 exercise books, and 1 ballpoint pen, it costs 4.2 yuan. If you buy 1 pencil, 1 exercise book, and 1 ballpoint pen, it will c... | -1. (B).
Let the cost of buying one pencil, one exercise book, and one ballpoint pen be $x$ yuan, $y$ yuan, and $z$ yuan, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=3.15 \\
4 x+10 y+z=4.20 .
\end{array}\right.
$$
Solving the system of equations for $(x+3 y)$ and $(x+y+z)$, we ... | 1.05 | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,388 |
2. The three sides of a triangle $a, b, c$ are all integers, and satisfy $a b c + b c + c a + a b + a + b + c = 7$. Then the area of the triangle is equal to ().
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{2}}{4}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{2}}{2}$ | 2. (C).
From $8=a b c+b c+c a+a b+a+b+c+1$ $=(a+1)(b+1)(c+1)$,
and $8=8 \times 1 \times 1=4 \times 2 \times 1=2 \times 2 \times 2$,
we can deduce that $a+1=b+1=c+1=2 \Rightarrow a=b=c=1$.
Therefore, the area of an equilateral triangle with side length 1 is $\frac{\sqrt{3}}{4}$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,389 |
3. As shown in Figure $1, \triangle A B C$ is an equilateral triangle, $P M \perp A B, P N \perp A C$. Let the perimeters of quadrilateral $A M P N$ and $\triangle A B C$ be $m$ and $n$, respectively. Then ( ).
(A) $\frac{1}{2}<\frac{m}{n}<\frac{3}{5}$
(B) $\frac{2}{3}<\frac{m}{n}<\frac{3}{4}$
(C) $80 \%<\frac{m}{n}<83... | 3. (D).
Let $B M=x, C N=y$. Then
$$
\begin{array}{l}
B P=2 x, P C=2 y, P M=\sqrt{3} x, P N=\sqrt{3} y, \\
A M+A N=2 \cdot B C-(B M+C N)=3(x+y) . \\
\text { Hence } \frac{m}{n}=\frac{(3+\sqrt{3})(x+y)}{3 \times 2(x+y)}=\frac{3+\sqrt{3}}{6} \approx \frac{4.7321}{6} .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,390 |
4. For all real number pairs $(x, y)$ satisfying $(x-3)^{2}+(y-3)^{2}=6$, the maximum value of $\frac{y}{x}$ is ( ).
(A) $3+2 \sqrt{2}$
(B) $4+\sqrt{2}$
(C) $5+3 \sqrt{3}$
(D) $5+\sqrt{3}$ | 4. (A).
Let $\frac{y}{x}=t$, then $\left(t^{2}+1\right) x^{2}-6(t+1) x+12=0$.
We have $\Delta=36(t+1)^{2}-48\left(t^{2}+1\right) \geqslant 0, t^{2}-6 t+1 \leqslant 0$.
Since $t^{2}-6 t+1=[t-(3-2 \sqrt{2})][t-(3+2 \sqrt{2})]$,
we know that the solution set of $t^{2}-6 t+1 \leqslant 0$ is $3-2 \sqrt{2} \leqslant t \leqs... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,391 |
5. Positive real numbers $a, b, c, d$ satisfy $a+b+c+d=1$. Let $p=\sqrt[3]{7 a+1}+\sqrt[3]{7 b+1}+\sqrt[3]{7 c+1}+\sqrt[3]{7 d+1}$. Then $p$ satisfies ( ).
(A) $p>5$
(B) $p<5$
(C) $p<2$
(D) $p<3$ | 5.(A).
From $0a^{2}>a^{3}$.
Then $7 a+1=a+3 a+3 a+1$
$$
>a^{3}+3 a^{2}+3 a+1=(a+1)^{3} .
$$
Thus, $\sqrt[3]{7 a+1}>a+1$.
Similarly, $\sqrt[3]{7 b+1}>b+1, \sqrt[3]{7 c+1}>c+1, \sqrt[3]{7 d+1}$ $>d+1$.
Therefore, $p>(a+b+c+d)+4=5$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,392 |
6. As shown in Figure 2, point $O$ is the center of the regular hexagon $A B C D E F$, $O M$ $\perp C D$, and $N$ is the midpoint of $O M$. Then $S_{\triangle A B N}: S_{\triangle B C N}$ equals ( ).
(A) $9: 5$
(B) $7: 4$
(C) $5: 3$
(D) $3: 2$ | 6. (C).
Obviously, $C$, $O$, and $F$ are collinear. As shown in Figure 6, connect $N D$ and $N E$, and draw a line $P Q \perp A B$ through $N$. Since $A B \parallel D E \parallel C F$, then $P Q \perp D E$, $P Q \perp C F$, and $P$, $K$, $Q$ are all feet of the perpendiculars.
By the axial symmetry of the regular hex... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,393 |
1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ . | 二、1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,394 |
4. Prove: Regardless of the value of $k$, the graph of the linear function $(2 k-1) x-$ $(k+3) y-(k-11)=0$ always passes through a fixed point.
(12th Junior High School Mathematics Competition in Jiangsu Province) | (Tip: This problem involves a "pencil of lines." You can start by finding the intersection coordinates of two special lines, then prove that all other lines must pass through this intersection point.)
The above text is translated into English, preserving the original text's line breaks and format. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,396 |
3. If the function $y=\frac{k x+5}{k x^{2}+4 k x+3}$ has the domain of all real numbers, then the range of real number $k$ is $\qquad$ | $3.0 \leqslant k0$ when, from $k x^{2}+4 k x+3=k(x+2)^{2}+3-$ $4 k$, we know that to make it never zero, then $x$ can take any real number.
$\because k(x+2)^{2} \geqslant 0, \therefore$ it is only necessary that $3-4 k>0$, i.e., $k<\frac{3}{4}$, which contradicts $k<0$, indicating that this situation cannot hold. Ther... | 0 \leqslant k < \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,397 |
One. (Total 20 points) $a$ is a three-digit number, $b$ is a one-digit number, and $\frac{a}{b} 、 \frac{a^{2}+b^{2}}{a b+1}$ are both integers. Find the maximum and minimum values of $a+b$. | Given $b \mid a$, and $(a b+1) \mid\left(a^{2}+b^{2}\right)$, we have
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{a b+1}=\frac{a}{b}+\frac{b^{2}-\frac{a}{b}}{a b+1} . \\
\because a b+1>a b>b^{2}>b^{2}-\frac{a}{b}>-1-a b,
\end{array}
$$
$\therefore b^{2}-\frac{a}{b}=0$, which means $a=b^{3}$.
When $b=5$, $a=125$; when $b=9$,... | 738, 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,399 |
II. (Total 25 points) As shown in Figure 5, in $\triangle ABC$, $\angle A=60^{\circ}$, $O$, $I$, and $H$ are the circumcenter, incenter, and orthocenter of the triangle, respectively. Compare the circumcircle of $\triangle IOH$ with the circumcircle of $\triangle ABC$, and prove your conclusion. | II. Construct the symmetric point $\sigma^{\prime}$ of $O$ with respect to $BC$, as shown in Figure 7.
(1) By the properties of the circumcenter, incenter, orthocenter, and the inscribed angle formula, we have
$$
\begin{array}{l}
\angle B O C=2 \angle A \\
=120^{\circ}, \\
\angle B I C=90^{\circ}+\frac{1}{2} \angle A \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,400 |
Three. (Total 25 points) Find all integer solutions to the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{3}+y^{3}+z^{3}=3\end{array}\right.$. | $$
\begin{array}{l}
\text { 3. } x+y=3-z, \\
x^{3}+y^{3}=3-z^{3} .
\end{array}
$$
(1) - (2) gives $xy=\frac{8-9z+3z^{2}}{3-z}$.
Knowing that $x, y$ are the roots of $t^{2}-(3-z)t+\frac{8-9z+3z^{2}}{3-z}=0$.
Solving gives $t_{1,2}=\frac{(3-z) \pm \sqrt{(3-z)^{2}-4 \times \frac{8-9z+3z^{2}}{3-z}}}{2}$.
$$
\begin{aligne... | (1,1,1),(4,-5,4),(-5,4,4),(4,4,-5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,401 |
$1 . a 、 b$ are skew lines, line $c$ forms an angle with $a$ equal to the angle it forms with $b$. Then the number of such lines $c$ is ( ).
(A)1
(B)2
(C)3
(D)infinite | 1.(D).
Let the common perpendicular of $a$ and $b$ be $l$, then any line $c$ parallel to $l$ is perpendicular to both $a$ and $b$, hence there are infinitely many such lines $c$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,402 |
2. If the three sides of $\triangle A B C$ are $a, b, c$ and satisfy
$$
a^{2}-a-2 b-2 c=0 \text { and } a+2 b-2 c+3=0 \text {, }
$$
then the degree of its largest interior angle is ( ).
(A) $150^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $60^{\circ}$ | 2. (B).
From $(a+2 b+2 c)(a+2 b-2 c)=-3 a^{2}$,
we have $a^{2}+b^{2}+a b=c^{2}$, thus $\cos C=-\frac{1}{2}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,403 |
3. For any given natural number $n, n^{6}+3 a$ is a positive integer cube, where $a$ is a positive integer. Then such $a(\quad)$.
(A) has infinitely many
(B) has only finitely many
(C) has only 1
(D) does not exist
Translate the above text into English, please keep the original text's line breaks and format, and outpu... | 3. (A).
$$
\text { Let } \begin{aligned}
& n^{6}+3 a=\left(n^{2}+3 m\right)^{3} \\
& =n^{6}+3\left(3 m n^{4}+9 m^{2} n^{2}+9 m^{3}\right),
\end{aligned}
$$
so $a=3 m n^{4}+9 m^{2} n^{2}+9 m^{3}, m, n$ are natural numbers. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,404 |
4. In the complex plane, the number of intersection points between the curve $z^{4}+z=1$ and the circle $|z|=1$ is ( ).
(A)0
(B) 1
(C) 2
(D) 3 | 4. (A).
Let $z=\cos \theta+\mathrm{i} \sin \theta(0 \leqslant \theta<2 \pi)$, then $z^{3}+1=\bar{z}$. Therefore, $1+\cos 3 \theta+i \sin 3 \theta=\cos \theta-i \sin \theta$, which means $\left\{\begin{array}{l}\cos 3 \theta=\cos \theta-1, \\ \sin 3 \theta=-\sin \theta\end{array}\right.$
$$
\therefore(\cos \theta-1)^{2... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,405 |
5. Satisfy $a>b>2$ and make $a b$, $\frac{b}{a}$, $a-b$, $a+b$ can be arranged in a certain order to form a geometric sequence of real number pairs $(a, b)(\quad)$.
(A) does not exist
(B) exactly 1
(C) exactly 2
(D) infinitely many | 5. (B).
From the problem, we know that $a b$, $\frac{b}{a}$, $a+b$, and $a-b$ are all positive numbers and $a b > a$.
$$
\begin{array}{l}
+b> \frac{b}{a}, a b > a + b > a - b . \\
\because (a+b)^{2} \neq a b \cdot \frac{b}{a},
\end{array}
$$
$\therefore$ The geometric sequence can only be $a b$, $a+b$, $a-b$, $\frac{b... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,406 |
6. A cube with an edge length of 5 is sawed into 125 smaller cubes with an edge length of 1. Then, at least ( ) cuts are needed.
(A) 7 times
(B) 8 times
(C) 9 times
(D) 12 times | 6. (C).
Each time you saw, you can rearrange the pieces that have been sawed. To minimize the number of sawing times, you only need to consider how many more times the largest piece needs to be sawed. First, saw the $5 \times 5 \times 5$ cube once, and the largest piece will be at least $5 \times 5 \times 3$; after sa... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,408 |
1. A triangle with all sides as integers, and the longest side being 11, has
$\qquad$ possibilities. | Ni, 1.36.
Let the three sides be $x, y, z$, and $x \leqslant y \leqslant z, z=11, x+$ $y>z$, then
$$
6 \leqslant y \leqslant 11 .
$$
When $y=6, x=6$, it forms 1 triangle;
When $y=7, x$ is $5, 6, 7$, it forms 3 triangles; .....
When $y=11, x$ is $1, 2, \cdots, 11$, it forms 11 triangles. Therefore, there are a total of... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,409 |
2. If $\left|\log \frac{a}{\pi}\right|<2$, then the number of $a$ for which the function $y=\sin (x+a)+\cos (x-a)(x \in \mathbf{R})$ is an even function is $\qquad$. | 2.10.
$$
\begin{aligned}
\because-2 & <\log \frac{\alpha}{\pi}<2, \therefore \frac{1}{\pi}<\alpha<\pi^{3} . \\
f(x)= & \sin (x+\alpha)+\cos (x+\alpha) \\
& =\sqrt{2} \sin \left(x+\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$
Since $f(x)$ is an even function,
$$
\therefore \sin \left(\frac{\pi}{4}+\alpha+x\right)=\sin ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,410 |
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