problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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|---|---|---|---|---|---|---|---|---|---|
6. The positive integer solutions of the equation $x^{2}-6 x y+13 y^{2}=100$ are ().
(A) three sets.
(B) four sets.
(C) five sets.
(D) six sets. | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,816 |
Construct semicircles on both sides of a line segment with each as the diameter (as shown in the figure). Then the length of this curve is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 1. $\frac{1}{2} \pi a$. | \frac{1}{2} \pi a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,817 |
Example 6. Given the equation $x^{2}+(a-6) x+a=0$ ( $a$ $\neq 0$ ) with both roots being integers. Try to find the integer $a$. (1989, Sichuan Province Junior High School Mathematics Competition) | Let the two integer roots of the equation be \( y_{1} x_{1}, x_{2} \), and \( x_{1} \geqslant x_{2} \). By Vieta's formulas, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a \\
x_{1} x_{2}=a
\end{array}\right.
\]
From (1) and (2), we get \( x_{1} x_{2} + x_{1} + x_{2} = 6 \), which simplifies to \( \left(x_{1}+1\rig... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,822 |
6. If $A B C D$ is a convex quadrilateral, and $E$, $F, G$, $H$ are the centroids of $\triangle A B C, \triangle B C D, \triangle C D A, \triangle D A B$ respectively, then the ratio of the area of quadrilateral $E F G H$ to the area of quadrilateral $A B C D$ is $\qquad$ | 6. $\frac{1}{9}$. | \frac{1}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,823 |
Sure, here is the translation:
---
One, starting from the natural number 1, write down in sequence to form the following series of numbers: $12345678910111213 \cdots$. With each digit occupying one position, determine the digit at the 1992nd position. | In the following sequence of digits, there are 9 single-digit numbers, $2 \times 90$ two-digit numbers, and $3 \times 900$ three-digit numbers. From $(1992-9-2 \times 90) \div 3$ $=601$, we know that the 1992nd position is in the 601st three-digit number starting from 100, which is the unit digit of the natural number ... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,824 |
II. Let real numbers $x, y$ satisfy the equation $x^{3}+y^{8}=a^{8}$ $(a>0)$. Find the range of $x+y$. | II. Solution Let $x+y=t$, then $t^{3}=a^{3}+3xyt$ hence $xy=\frac{t^{3}-a^{3}}{3t}$. By Vieta's theorem, $x, y$ are the roots of the equation $z^{2}-tz+\frac{t^{3}-a^{3}}{3t}=0$. Since $x, y$ are real numbers, $\Delta \geqslant 0$, i.e., $t^{2}-4 \cdot \frac{t^{3}-a^{3}}{3t} \geqslant 0$, solving this yields $0<t \leqs... | (0, \sqrt[3]{4}a] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,825 |
Three, let $[x]$ denote the greatest integer not exceeding the real number $x$. Solve the equation $x^{2}-2[x]-5=0$.
| F is, $\left\{\begin{array}{l}x^{2}-2 x-5>-2, \\ x^{2}-2 x-5 \leqslant 0 .\end{array}\right.$ Therefore, $\left\{\begin{array}{l}x>3, \\ 1-\sqrt{6}<x<1+\sqrt{6} .\end{array}\right.$ Hence, $[x]=-2$ or 3.
When $[x]=-2$, the original equation is $x^{2}=1$, yielding $x= \pm 1$. Since $[ \pm 1]= \pm 1$, this contradicts $... | x=\sqrt{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,826 |
Four, Given that point $B$ is on line segment $A C$. Construct circles $\odot O_{1}$, $\odot O_{2}$, $\odot O$ with diameters $A B$, $B C$, $A C$ respectively. Draw a line through point $B$ intersecting $\odot O$ at $P, Q$, and intersecting $\odot O_{1}$ and $\odot O_{2}$ at $S$. Prove that $P R=Q S$. | Prove that connecting $O_{1} R, O_{2} \mathrm{~S}, O R, O S$.
Since $\angle O_{1} R B$
$$
\begin{array}{l}
=\angle O B R \\
=\angle O_{2} B S \\
=\angle O_{2} S B
\end{array}
$$
Then $O_{1} R \| O_{2} S$,
$$
\angle O O_{1} R=\angle O O_{2} S
$$
Let the radii of $\odot O, \odot O_{1}$, and $\odot O_{2}$ be $r, r_{1}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,827 |
Five, a chicken farm uses chicken coops to hold chicks. If each coop holds 36 chicks, then 11 chicks are left over. If two fewer coops are used, then all the chicks can be evenly distributed. It is known that one coop can hold a maximum of 45 chicks. How many chicken coops were there originally? How many chicks are the... | Five, Solution: Let the original number of chicken coops be $x$, and when two coops are reduced, each coop holds $n$ chicks. Then $x \geqslant 3$, $n \leqslant 45$, and $36 x+11=n(x-2)$. Therefore, $n=36+\frac{83}{x-2}$.
Since $n$ is an integer, it follows that $x-2$ must divide 83, and since 83 is a prime number, $x-... | 85, 3071 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,828 |
1. (A) $0\sqrt{x}>x>x^{2}$.
(B) $\frac{1}{x}>x>\sqrt{x}>x^{2}$.
(C) $\frac{1}{x}>x>x^{2}>\sqrt{x}$.
(D) $\frac{1}{x}>\sqrt{x}>x^{2}>x$. | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 705,829 |
$$
\begin{array}{c}
\text { 2. If }\left|\frac{x+2 y+3}{\sqrt{5}}\right|+x^{2}+x y+\frac{1}{4} y^{2} \\
=\lg \left(-\frac{1}{4} \log \sqrt{2} \frac{1}{4}\right), \text { then } x^{2}-x y+y^{2}
\end{array}
$$
2. If $\left|\frac{x+2 y+3}{\sqrt{5}}\right|+x^{2}+x y+\frac{1}{4} y^{2}$ $=\lg \left(-\frac{1}{4} \log \sqrt{2}... | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,830 |
Example 7. The equation $(x-a)(x-8)-1=0$ is marked with two $\uparrow$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve the original equation, which is transformed into
$$
(x-8)^{2}+(3-a)(x-8)-1=0 \text {. }
$$
By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}-8+x_{2}-8=a-8, \\
\left(x_{1}-8\right)\left(x_{2}-8\right)=-1 . \\
\because x_{1}, x_{2} \in Z, \\
\therefore x_{1}-8, x_{2}-8 \in Z . \\
\therefore\left(x_{1}-8\right... | a=8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,833 |
5. $\quad \because ; a=\left(\frac{1}{2}\right)^{\frac{1}{4}}, b=\left(\frac{1}{3}\right)^{\frac{1}{2}}$, $c=\left(\frac{1}{4}\right)^{\frac{1}{b}}$, then the following inequality that holds is ( ).
(A) $a<b<c$.
(B) $b<c<a$.
(C) $a<c<b$.
(D) $c<b<a$. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 705,834 |
6. $H[x]$ represents the greatest integer less than or equal to $x$ (for example, $[4.52]=4,\left[-5 \frac{1}{2}\right]=-6$, etc.). Then the number of positive integer solutions to the equation $[1.9 x]+[9.2 y]=37$ is ( ).
(A.) 1 pair.
(B) 3 pairs.
(C) 4 pairs.
(D) about 7.4 pairs. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,835 |
Three, given the parabola $y=a x^{2}-(a-1) x+a-1$ - the line $y=1-2 x$ has at least one intersection point that is a lattice point (in the Cartesian coordinate system, a point where both the x and y coordinates are integers). Determine the integer value of $a$, and find the coordinates of the intersection point (lattic... | Three, for the parabola $y=a x^{2}-(a-1) x+a-1$ and the line $y=1-2 x$, the x-coordinates of their intersection points are the real solutions of the equation $a x^{2}-(a-1) x+(a-1)=1-2 x$, which is
$$
a x^{2}-(a-3) x+a-2=0
$$
The condition for equation (1) to have real solutions is
$$
\Delta=(a-3)^{2}-4 a(a-2)>0 \text... | (1,-1), (3,-5), (0,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,842 |
In $\triangle A B C$, $G$ is the centroid, $P$ is an internal point, and line $P G$ intersects lines $B C, C A$, $A B$ at $A^{\prime}, B^{\prime}, C^{\prime}$. Prove:
$$
\frac{A^{\prime} P}{A^{\prime} G}+\frac{B^{\prime} P}{B^{\prime} G}+\frac{C^{\prime} P}{C^{\prime} G}=3 .
$$ | Prove: Connect $B G, G C, P B, P C$. Draw $G G^{\prime} \perp B C$ at $G^{\prime}$, and $P P^{\prime} \perp B C$ at $P^{\prime}$. Since
$P P^{\prime} \| G G^{\prime}$
$\Rightarrow \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P}{A^{\prime} G^{\prime}}$
Also, $\quad S_{\triangle P B C} \quad$
$$
=\begin{array}{l}
P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,843 |
Example 8. If the quadratic equation $x^{2}-2 a x+4 a^{2}-$ $6 a=0$ has two integer solutions, find the value of $a$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: From $\Delta=4\left(6 a-3 a^{2}\right) \geqslant 0$ and $x_{1}+x_{2}=$ $2 a$ is an integer, i.e., $3 a(a-2) \leqslant 0$ and only $2 a$ is an integer, we get $a=0, \frac{1}{2}, 1, \frac{3}{2}, 2$. Substituting them into the original equation one by one, we find that the original equation has two integer solution... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,844 |
Five, it is known that a convex polygon cannot cover any triangle with an area of 1. Prove: this convex polygon can be covered by a triangle with an area of 4. | Five, Proof: Let the convex polygon be $Q$. Then among the triangles formed by connecting the vertices of $Q$, we cannot find the one with the maximum area, let's assume it is $\triangle ABC$. Clearly, $S_{\triangle ABC} < 1$. Draw a line $l$ through $A$ parallel to $BC$, a line $m$ through $B$ parallel to $AC$, and a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,845 |
1. $1,000,000,000,000$ $-717,777,777,777=(\quad)$.
(A) $222,222,222,222$.
( B ) $222,222,222,223$.
(C) $233,333,333,333$.
(D) $322,222,222,223$.
(E.) $333,333,333,333$. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,846 |
2. $\frac{16+8}{4-2}=(\quad)$
(A) 4. (B)8. ( $\dot{C}$ ( ) 12.
(D) 16. (E) 20. | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,847 |
6. Which number in the following array is both the largest in its column and the smallest in its row?
\begin{tabular}{ccccc}
10 & 6 & 4 & 3 & 2 \\
11 & 7 & 14 & 10 & 8 \\
8 & 3 & 4 & 5 & 9 \\
13 & 4 & 15 & 12 & 1 \\
8 & 2 & 5 & 9 & 3 \\
(A) 1. & (B) & 6 & (C) 7.
\end{tabular}
(D) 12. (E) 15. | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 705,851 |
$$
\begin{array}{l}
\text { 7. Li }(487,000)(12,027,300) \\
+(9,621,001)(487,000)
\end{array}
$$
Divide by $(19,367)(0.05)$
the value is closest to ( ).
(A) $10,000,000$.
(B) $100,000,000$.
(C) $1,000,000,000$.
(D) $10,000,000,000$.
( E ) $100,000,000,000$. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,852 |
9. Among the integers from 1 to 46, the number of integers that can be divided by 3 or 5 is ( ).
(A) 18. ( B ) 21 . ( C ) 24.
(D) 25 , (E) 27. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,854 |
Example 9: For the equation $x^{2}-4 a x+64=0$ (where $a$ is an integer), what value of $a$ will make the equation have integer solutions? | From $x^{2}-4 a x+64=0$, we get $e=\frac{x}{4}+\frac{16}{x}$ Since $a$ is an integer, if one root is an integer, the other root is also an integer. Therefore, $x= \pm 4, \pm 8, \pm 16$. Thus, $a= \pm 5, \quad \pm 4$ | a= \pm 5, \pm 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,855 |
11. From the set $\{1,2,3,4,5,6,7,8,9\}$, select three different numbers, requiring that these three numbers must include 5 and their sum must be 15. The number of such selection methods is ( ).
(A) 3 .
( B) 4 .
(C) 5 .
(D) 6 .
( E) 7 | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 705,857 |
12. $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$, then $N=(\quad)$.
(A) 3.
(B) 6. (C) 1990.
(D) 1991.
(E) 1992 | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,858 |
13. Multiply $25 \times 25 \times 25 \times 25 \times 25 \times 25 \times 25$ $\times 8 \times 8 \times 8$ How many zeros are in the unit place?
(A) 3 .
(B) 6 .
(C) 9 .
(D) 10. (E) 12 | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,859 |
16. The figure on the right shows a 4 $1,2, \cdots, 16$ square. Fold the square according to the following steps:
(1) the top half is folded over the bottom half:3
(3) the right half is folded over the left half;
(4) the left half is folded over the right half: (A)1. (B) 9. (C) 10.
(D) 14. (E) 16. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 705,862 |
19. The arithmetic mean of 10 different positive integers is 10, then the maximum value of the largest number among these ten numbers is ( ) (A) 10. (B) 50. (C) 55. (D) 90. (E) 91. | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 705,865 |
T10. Find all positive integers $a$ such that the quadratic equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one integer root. (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Solving, the original equation becomes $a=\frac{2 x+12}{(x+2)^{2}}$. Since $a$ is a positive integer, we have $a \geq 1$, i.e., $\frac{2 x+12}{(x+2)^{2}} \geq 1$. Simplifying, we get $x^{2}+2 x-8 \leqslant 0$. Solving this, we get $-4 \leqslant x \leqslant 2$.
Since $x$ is an integer root and clearly $x \neq-2$, there... | a=1,3,6,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,866 |
23. A school's wind band consists of 100 boys and 80 girls. The string team consists of 80 boys and 100 girls. In total, 230 students participated in these two bands, among whom 60 boys participated in both bands. Then, the number of girls who only participated in the wind band and did not participate in the string tea... | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 705,871 |
Example 8. In a Go tournament, ten players participate. Each player competes against the other nine players once. All players have different scores. The first and second place players have never lost a game, and the total score of the first two players is 10 points more than the second place player. The fourth place pl... | Let the scores of the ten selected players be $x_{1}, x_{2}, \cdots, x_{10}$.
(1) Obviously, the result of the match between the first and second place is a draw. Thus, $x_{1} \leqslant 8.5, x_{2} \leqslant 8$.
(2) The last four players have a total of 6 matches among themselves, so their total score $x_{1} + x_{8} + x... | 8.5, 8, 6.5, 6, 5.5, 4.5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,873 |
Example 1. As shown in Figure 1, in $\triangle ABC$, $AD = BB = CF$, $O$ is any point inside $\triangle ABC$, $OA' \| AD$, $OB' \| BE$, $OC' \| CF$. Prove that $OA' + OB' + OC'$ is a constant (1986, Nanjing City Mathematics Competition) | Proof: Let the altitudes of $\triangle ABC$ on its three sides be $h_{\mathrm{s}}$, $h_{\mathrm{b}}$, $h_{\mathrm{c}}$, and the distances from point $O$ to the three sides be $d_{\mathrm{a}}$, $d_{\mathrm{b}}$, $d_{\mathrm{c}}$.
According to the conclusion (1) of Theorem 1, we have:
$$
\begin{array}{l}
\frac{d_{\mathr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,874 |
Example 2. The lengths of the three sides of $\triangle ABC$ are $BC=17$, $CA=18$, and $AB=19$. Through a point $P$ inside $\triangle ABC$, perpendiculars $PD$, $PE$, and $PF$ are drawn to the three sides of $\triangle ABC$ ($D$, $E$, and $F$ are the feet of the perpendiculars), and $BD + CE + AF = 27$. Find the length... | Let $B D=x, B F=y$. Then $C D=17-x$, $A F=19-2$. Also, $B D+C E+A F=27$, so
$$
\begin{array}{l}
C E=27-B D-A F=8-x+y, \\
A B=18-C E=10+x-y .
\end{array}
$$
By Theorem 1 conclusion (2), we have
$$
\begin{array}{l}
x^{2}+(8-x+y)^{2}+(19-y)^{2} \\
=(17-x)^{2}+(10+x-y)^{2}+y^{2} .
\end{array}
$$
Simplifying, we get $2 x+... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,875 |
B3. Quadrilateral $A B C D$ is inscribed in a circle, the incenters of $\triangle B C D$, $\triangle A C D$, $\triangle A B D$, $\triangle A B C$ are denoted as $I_{A}, I_{\mathrm{B}}, I_{\mathrm{C}}, I_{\mathrm{D}}$ respectively. Prove: $I_{\mathrm{A}} I_{\mathrm{B}} I_{\mathrm{C}} I_{\mathrm{D}}$ is a rectangle. (198... | Proof: Let the midpoints of $\widehat{A B}, \widehat{B C}, \widehat{C D}$, $\widehat{D A}$ be $M, N^{\prime}, M^{\prime}$, $N$. Then $M I_{\mathrm{C}}=M A = M B=M I_{\mathrm{D}}$,
so $\triangle I_{\mathrm{C}} M I_{\mathrm{D}}$ is an isosceles triangle.
Also, $M M^{\prime}$ bisects $\angle D M C$,
hence $M M^{\prime} \p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,876 |
Example 3. As shown in Figure 3, through an internal point $P$ of $\triangle ABC$, three lines parallel to the three sides are drawn, resulting in three triangles $t_{1}, t_{2}$, and $t_{3}$ with areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$.
(2nd American
Mathematical Invitational) | This example can directly apply the conclusion (4) of Theorem 2 to obtain
$$
S \triangle A B C=144
$$ | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,877 |
Example 5, As shown in Figure $4, \triangle A B C$ has an internal point $P, A P$ intersects $B C$ at $D, B P$ intersects $A C$ at $E$, and $C P$ intersects $A B$ at $F$. Prove that $\frac{A P}{P D}, \frac{B P}{P E}, \frac{C P}{P F}$ at least one is not greater than 2, and at least one is not less than 2. (3rd IMO) | Proof From conclusion (1) of Theorem 3, we get:
$$
\frac{P D}{A \dot{D}}+\frac{P E}{B E}+\frac{P F}{C F}=1 \text{. }
$$
Therefore, among the three terms, at least one is greater than or equal to $\frac{1}{3}$, and at least one is less than or equal to $\frac{1}{3}$. Without loss of generality, assume ${ }_{A D}^{P D} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,880 |
In 8. As shown in Figure $7, O$ is a point inside the acute $\triangle A B C$, $\angle A O B=\angle B O C=\angle C O A=120^{\circ}, \quad P$ is any point inside $\triangle A B C$. Prove that $P A+P B+P C$ $\geqslant O A+O B+O C$. | Prove that through $A$, $B$, $C$ each draw a line to form $\triangle E F G$ such that each side of the triangle is perpendicular to the lines connecting $A$, $B$, $C$ with $O$.
Since $O A \perp G F, O E \perp G E$,
then $\angle G=180^{\circ}-\angle A O B=60^{\circ}$.
Similarly, $\angle F=\angle E=60^{\circ}$. Thus, for... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,882 |
Example 10. Let $P$ be a point inside an acute $\triangle ABC$, and the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, $AB$ be $D$, $E$, $F$ respectively. Find (and prove) the point $P$ that minimizes $PD^2 + PE^2 + PF^2$.
Make the translation and retain the original text's line breaks and format, output... | Let the area of $\triangle ABC$ be $S$, $AB=c$,
$$
\begin{array}{l}
BC=a, CA=b, PD=x, PE=y, PF \\
=z, w=PD^{2}+PE^{2}+PF^{2}=x^{2}+y^{2}+z^{2},
\end{array}
$$
Then we have
$$
\begin{aligned}
S^{2} & =(S \triangle \triangle PB+S \triangle BP C+S \triangle CP A)^{2} \\
& =\frac{1}{4}(a x+b y+c z)^{2} \\
& \leqslant \fra... | w_{\mathrm{min}}=\frac{4 S^{2}}{a^{2}+b^{2}+c^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,884 |
Example 1. Two students took four tests, their average scores were different, but both were below 90 and were integers. They then took a fifth test, after which their average scores both increased to 90. What were the scores of the two students on the fifth test? | Let the average score of the first four tests for one of the students be $x$ points, and the score of the fifth test be $y$ points. According to the problem, we have
$\frac{4 x+y}{5}=90$, which means $y=450-4 x$.
However, it is clear that the score of the fifth test is higher than 90 points and no more than 100 points,... | 98 \text{ and } 94 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,885 |
Example 2. The sum of the reciprocals of four unequal natural numbers is 1. Find these four numbers. | Let the four numbers be $x, y, z, w$, and $x \geq 1$. Also, when $x \geq 3$, we have
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w} \\
\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} < 1. \text{ Also, when } y \geqslant 6, \text{ we have } \frac{1}{y}+\frac{1}{z}+\frac{1}{w} \leqslant \fr... | (2,3,7,42), (2,3,8,24), (2,3,9,18), (2,3,10,15), (2,4,5,20), (2,4,6,12) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,886 |
Example 4. Given an isosceles triangle with the circumradius $R$ and the inradius $r$. Prove that the distance between the centers of the two circles is
$$
d=\sqrt{R(R-2 r)} \text {. }
$$
(4th IMO) | Proof: Let $O$ be the circumcenter of $\triangle ABC$, $I$ be the incenter of $\triangle ABC$, and $OI=d$. From the given, $A, I, O$ are collinear. Extend $AI$ to $D$, intersecting the circumcircle of $\triangle ABC$ at $D$, and connect $DC$. It is known that $AD$ is the diameter of the circumcircle of $\triangle ABC$.... | d=\sqrt{R(R-2r)} | Geometry | proof | Yes | Yes | cn_contest | false | 705,887 |
Solve the equation
$$
\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5} .
$$ | Let $\frac{15 x-7}{5}=y$, then $x=\frac{5 y+7}{15}$. Substituting into the original equation, we get
- $\left[\frac{10 y+39}{40}\right]=y$.
We have $0 \leqslant-\frac{10 y+39}{40}-y<1$.
Solving, we get $-\frac{1}{30}<y \leqslant \frac{39}{30}$.
But $y$ is an integer, so $y=0$ or 1. Solving for $x$, we get $x=\frac{7}{1... | x=\frac{7}{15} \text{ or } \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,888 |
Example 2. In acute triangle $\triangle A B C$, the distance from vertex $A$ to the circumcenter $O$ is equal to the distance to the orthocenter $H$. Find the possible value of $\angle A$.
untranslated text remains in its original format and line breaks are preserved. | $$
\begin{array}{l}
AI I=R \sqrt{2+2 \cos (\gamma-\beta)}=AO=R, \\
\cos (\gamma-\beta)=-\frac{1}{2} \text { and } 0^{\circ}<\gamma-\beta<180^{\circ}, \\
2 \angle A=\gamma-\beta=120^{\circ}, \angle A=60^{\circ} .
\end{array}
$$ | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,890 |
Example 1. Given $-\frac{x-b}{c}+\frac{x-b-c}{a}$ $+\frac{x-c-a}{b}=3$,
and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$.
Then $x-a-b-c=$ $\qquad$ (8th Jincheng
Mathematics Competition) | From the known, we get $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(x-a-b-c)=0$, so we should fill in “0”. This is transforming the known to the unknown, the other is transforming the unknown to the known.
| 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,892 |
Example 2. If $x^{3}-x^{2}+x-2=0$. Then $x^{4}+2 x^{3}-2 x^{2}+x-1=$ $\qquad$ . (1991, Hubei Huanggang Region Mathematics Competition). | Left= $\begin{aligned} & \left(x^{4}-x^{3}+x^{2}-2 x\right) \\ & +\left(3 x^{3}-3 x^{2}+3 x-6\right)+5 \\ = & x \cdot 0+3 \cdot 0+5=5 .\end{aligned}$ | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,893 |
Example 1. An isosceles trapezoid with bases $a, b$ and height $h$
(1) Find a point $P$ on the axis of symmetry such that the angles subtended by the non-parallel sides at $P$ are right angles.
(2) Find the distances from point $P$ to the two bases. | Let the isosceles trapezoid $ABCD$ have $AB=a, CD=b$, and the axis of symmetry intersects the two bases at $E, F, EF=h$. Construct a semicircle (inward) with $BC$ as the diameter, intersecting $EF$ at $P_{1}, P_{2}$ (if tangent, the two points coincide; if separated, there is no solution), which is the desired result. ... | x_{1,2}=\frac{h}{2} \pm \frac{1}{2} \sqrt{h^{2}-a b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,895 |
Example 2. Quadrilateral $A B C D$ is circumscribed around $\odot O, A C \perp$ $B D$, intersecting at $E$. Prove: $A B C D$ is a symmetric figure with one of its diagonals as the axis of symmetry. (1.979, Hunan Mathematics Competition). | Proof: Let $A B=a, B C=b, C D=c$, $D A=d$. By the tangent segment theorem and the Pythagorean theorem, we have
$$
\left\{\begin{array} { l }
{ a + c = b + d , } \\
{ a ^ { 2 } + c ^ { 2 } = b ^ { 2 } + d ^ { 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
a+c=b+d, \\
a c=b d .
\end{array}\right.\right.
$$
It is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,896 |
Example 3: Prove that the area of a rectangle inscribed in a triangle is not greater than half the area of the triangle. | Proof: Let the base of $\triangle ABC$ be $a$, and the height be $h$. Suppose the base of a rectangle inscribed in the triangle is $y$, and the height is $x$. Then it is easy to see that $\frac{y}{a} + \frac{x}{h} = 1$. Let the area of the triangle be $\triangle$, and the area of the inscribed rectangle be $S$. Then $\... | S \leq \frac{\Delta}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 705,897 |
Example 5. Let $I$ be the incenter of $\triangle ABC$, and $A', B', C'$ be the circumcenters of $\triangle IBC$, $\triangle IAC$, $\triangle IAB$ respectively. Prove that $\triangle ABC$ and $\triangle A'B'C'$ have the same circumcenter. (1988, 17th United States of America Mathematical Olympiad), | To prove that points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ are concyclic, it suffices to show that these six points lie on the same circle. Construct the circumcircle of $\triangle A B C$ and extend $A I$ to intersect this circumcircle at point $A^{\prime \prime}$. Since $I$ is the incenter of $\triangle A B C$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,898 |
2. Calculate: $\left(1^{2}+3^{2}+5^{2}+\cdots+99^{2}\right)-\left(2^{2}\right.$ $\left.+4^{2}+6^{2}+\cdots+100^{2}\right)$ | $\begin{array}{l}\text { 2. Hint: Original expression }=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right) \\ +\left(5^{2}-6^{2}\right)+\cdots+\left(99^{2}-100^{2}\right)=-5050 .\end{array}$ | -5050 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,900 |
3. When a surveyor is measuring a piece of land, they first walk 100 meters from point $A$ at a bearing of $35^{\circ}$ north of east to point $B$, then from point $B$ at a bearing of $55^{\circ}$ north of west for 80 meters to point $C$, and from point $C$ at a bearing of $35^{\circ}$ south of west for 60 meters to po... | 3. Brief solution: ( 1 ) As shown in the figure.
( 2 ) This piece of land is a trapezoid.
$$
\begin{array}{c}
\text { ( } 3 \text { ) } S= \\
\frac{1}{2}(60+100) \times 80 . \\
=6400 \text { (square meters). }
\end{array}
$$ | 6400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,901 |
4. Given that $a, b$ are integers, $a$ divided by 7 leaves a remainder of 3, and $b$ divided by 7 leaves a remainder of 5. When $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7. | 4. Slightly explained: Let $a=$
$$
7 m+3, b=7 n+5 \text {, }
$$
where $m, n$ are integers, then
$$
\begin{aligned}
a^{2}-4 b & =(7 m+3)^{2}-4(7 n+5) \\
& =7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{aligned}
$$
Since $7 m^{2}+6 m-4 n-2$ is an integer,
thus $a^{2}-4 b$ leaves a remainder of 3 when divided by 7. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,902 |
5. Calculate:
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. Prompt: First prove
$$
\begin{array}{l}
(n-1) n(n+1)(n+2)+1 \\
=\left(n^{2}+n-1\right)^{2} .
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,903 |
6. There are two specifications of steel bars, A and B, and C. It is known that 2 bars of type A, 1 bar of type B, and 3 bars of type C are a total of 23 meters long; 1 bar of type A, 4 bars of type B, and 5 bars of type C are a total of 36 meters long. How long are 1 bar of type A, 2 bars of type B, and 3 bars of type... | 6. Slightly explained: Let the lengths of copper strips of specifications A, B, and C be $x, y, z$ meters, respectively, then
$$
\left\{\begin{array}{l}
2 x+y+3 z=23, \\
x+4 y+5 z=36
\end{array}\right.
$$
From (1) - (2) $\times 2$, we get $y=7-z$.
Substituting $y$ into (2), we get $x=8-z$.
Substituting $x, y$ into $x+... | 22 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,904 |
$\begin{array}{l}\text { 7. Calculate: } 3+\frac{1}{3} \sqrt{3}+\frac{1}{5 \sqrt{3}+3 \sqrt{5}} \\ +\frac{1}{7 \sqrt{5}+5 \sqrt{7}}+\cdots+\frac{1}{49 \sqrt{47}+47 \sqrt{49}} \\\end{array}$ | $\begin{array}{l}+\frac{5 \sqrt{3}-3 \sqrt{5}}{30}+7 \sqrt{5}-5 \sqrt{7}+\cdots+ \\ +\frac{49 \sqrt{47}-47 \times 7}{(49 \sqrt{47})^{2}-\left(47 \sqrt{49)^{2}}\right.} \\ =\left(\frac{1}{2}-\frac{\sqrt{3}}{6}\right)+\left(\frac{\sqrt{3}}{6}-\frac{\sqrt{5}}{10}\right) \\ +\left(\frac{\sqrt{5}}{10}-\frac{\sqrt{7}}{14}\ri... | \frac{3}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,905 |
8. Given four distinct positive numbers $x, y, m$, and $n$, where $x$ is the smallest and $n$ is the largest, and $x: y = m: n$. Try to compare $x+n$ with $y+m$ and prove your conclusion. | 8. Proof Let $\frac{x}{y}=\frac{m}{n}=k \neq 0$,
then $x=k y, m=n k$.
and
$$
\begin{array}{l}
x+n=k y+n, \\
y+m=y+n k
\end{array}
$$
From (1)-(2) we get
$$
(x+n)-(y+m)=(k-1)(y-n) \text {. }
$$
Since $k \neq 0$.
That is, $x+n>y+m$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,906 |
10. There are several warriors, forming a rectangular formation that is exactly eight columns wide. If 120 more people are added to or 120 people are removed from the formation, a square formation can be formed in both cases. How many warriors are there in the original rectangular formation? | 10. Solution: Let the original number of soldiers be $8x$ people,
From the given, $8x+120$ and $8x-120$ are both perfect squares, then we have
$$
\left\{\begin{array}{l}
8 x+120=m^{2} \\
8 x-120=n^{3}
\end{array}\right.
$$
( $m, n$ are positive integers).
By (1) - (2) we get $m^{2}-n^{2}=240$,
which is $(m+n)(m-n)=240... | 904 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,908 |
In $\triangle A B C$, $\angle B A C=5.25^{\circ}$, $A D$ is the angle bisector of $\angle B A C$. Draw a perpendicular line from $A$ to $D A$ intersecting the line $B C$ at point $M$. If $B M=B A+A C$, find the measures of $\angle A B C$ and $\angle A C B$. | (ii) As follows,
Draw the perpendicular line from $A$ to $DA$,
intersecting the extension of $CB$ at point $M$, extend $BA$
to $C_{1}$, such that $AC_{1}$
$$
= AC \text {. }
$$
Connect $AC_{1}$,
$C_{1}C$.
Then $\angle MAD=90^{\circ}$,
$$
\therefore MAC=90^{\circ}+\frac{5.25^{\circ}}{2}=92.625^{\circ} \text {. }
$$
A... | \angle ACB=58.25^{\circ}, \angle ABC=116.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,911 |
Three, in a $3 \times 3$ square grid, fill in nine different natural numbers such that the product of the three numbers in each row and the product of the three numbers in each column are all equal (we denote this product by $P$).
(1) Prove that this arrangement of numbers is possible.
(2) Determine which of the six ... | Three, (1) Proof It is easy to prove the following filling method 1 As shown in the figure, fill in 1, 2, 3, 4, 5, 6, $8, 15, 20$ nine different natural numbers, then the product of the three numbers in each row and the product of the three numbers in each column are all equal to 120.
Therefore, the filling method req... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,912 |
1. The real number $x$ satisfies $x-\frac{1}{x}=\sqrt{5}$, then $x+\frac{1}{x}$ | 1. $\pm 3$. | \pm 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,913 |
3. Let $m$ be a real number, and the equation $x^{2}-5 x+m=0$ has a root whose opposite number is a root of the equation $x^{2}+m x+5=0$. Then $m=$ $\qquad$ | 3. 5 or -6. | 5 \text{ or } -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,915 |
Example 8. In $\triangle A B C$, $A B=A C$, there is a circle that is internally tangent to the circumcircle of $\triangle A B C$ and is tangent to $A B$ and $A C$ at $P$ and $Q$, respectively. Prove that the midpoint of $P Q$ is the incenter of $\triangle A B C$. (20th IMO) | Proof as shown in the figure,
Let $I$ be the midpoint of $P Q$,
Connect $T C, T Q$,
$C I, T B$.
Since $P, Q$ are tangent points,
we have $A P=A Q$,
$I$ is the midpoint of $P Q$, then $A I$
bisects $\angle A$ and $A I \perp P Q$. Also, since $T$ is the common tangent point of the two circles, then $A T$ is the diameter ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,920 |
9. Given that a certain four-digit number is exactly equal to the fourth power of the sum of its digits, then this four-digit number is $\qquad$ | 9. 2401 .
The above text has been translated into English, maintaining the original text's line breaks and format. | 2401 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,922 |
In isosceles $\triangle A B C$, it is known that $A B=A C$ $=k B C$, where $k$ is a natural number greater than 1. Points $D$ and $E$ are on $A B$ and $A C$ respectively, and $D B=B C=C E$. $C D$ and $B E$ intersect at $O$.
Find the smallest natural number $k$ such that $\frac{O C}{B C}$ is a rational number. | II. Solution: As shown, connect $D E$, it is easy to know that $B C E D$ is an isosceles trapezoid. Also, from the given conditions, we have
$$
\angle 2=\angle 1=\angle 3 \text {, }
$$
thus $\triangle O B C \cos \triangle B C D$,
which means $O C \cdot C D=B C^{2}$.
Also, $\frac{C O}{O D}={ }_{D E}^{B C}=\frac{A B}{A ... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,924 |
Three, through an internal point $Q$ of $\triangle A B C$, draw lines not parallel to the three sides of the triangle. These lines divide the triangle into six parts. It is known that the areas of the three parallelogram parts are $S_{1}, S_{2}, S_{3}$ (as shown in the figure), find the area of $\triangle A B C$.
Draw... | Three, Solution: For convenience, let $S: q \nu G, S: Q$: $S \triangle Q P \mathrm{Rg}$ be $S_{1}^{\prime}, S_{2}^{\prime}, S_{3}^{\prime}$. Since
$$
\frac{\frac{1}{2} S_{1}}{S_{2}^{\prime}}=\frac{S \Delta Q B E}{S_{2}^{\prime}}=\frac{B E}{E I}=\frac{F Q}{E I}=\frac{H Q}{Q I}=\frac{S_{2}}{S_{s}},
$$
we have $S_{2}^{\p... | S_{\triangle A B C}=S_{1}+S_{2}+S_{3}+\frac{S_{2} S_{3}}{2 S_{1}}+\frac{S_{3} S_{1}}{2 S_{2}}+\frac{S_{1} S_{2}}{2 S_{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,925 |
Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.
Translate the above text into English, please retain the original text's line breaks and format, a... | Let the three-digit number $abc$ satisfy the given conditions. Then
(1) $c \geqslant 7$. In fact, if $c \leqslant b$, then the sum of the digits of $\overline{abc}+3$ will increase by 3, which does not meet the conditions.
(2) $b \neq 9$. This is because if $c \geqslant 7, b=9, a=9$, then
$$
\overline{abc}+3=1000+(c+3... | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,926 |
一、Given $a$ is a real number, the quadratic equation in $x$, $x^{2}-a x+a=0$, has two real roots $\alpha$, $\beta$. Prove: $\alpha^{2}+$ $\beta^{2} \geqslant 2(\alpha+\beta)$. | I. Proof of the problem, given $\alpha+\beta=a, \alpha \beta=$ $a, \Delta=a^{2}-4 a \geqslant 0$. Thus $(\alpha+\beta)^{2}-4 a \geqslant 0$, $4 \alpha^{2}+\beta^{2}+2 \alpha \beta-4 a \geqslant 0 \Rightarrow \alpha^{2}+\beta^{2}-2 a \geqslant 0$. Finally, $a^{2}+\beta^{2} \geq 2 a=2(\alpha+\beta)$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,927 |
$$
\begin{array}{l}
\text { Three, as shown in the figure, for } \triangle A B C \\
\text { construct } \triangle B P C, \triangle C Q A, \\
\triangle A R B, \text { such that } \angle P B C= \\
\angle C A Q=45^{\circ}, \angle B C P \\
=\angle Q C A=30^{\circ}, \\
\angle A B R=\angle B A R=15^{\circ} .
\end{array}
$$
... | Three, Proof: Construct a regular $\triangle A B S$ outside $\triangle A B C$, and connect $R S, C S$. Then $\triangle S R B \cong \triangle S R A$, $\angle A S R=30^{\circ}$.
Also, $\angle S A R=60^{\circ}-15^{\circ}=45^{\circ}$,
Thus, $\triangle C Q A \backsim \triangle S R A$.
Therefore, $\frac{S A}{R A}=\frac{C A}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,929 |
Example 9. Given two circles with centers $O$ and $\mathrm{O}^{\prime}$, and radii $R$ and $r$, respectively, and $d$ is the distance between the two centers. Prove: if $R, r$, and $d$ satisfy the relationship $d^{2}=R^{2}-2 R r$, then infinitely many triangles can be constructed such that $\odot O$ and $\odot 0^{\prim... | Prove that on circle $\odot O$, take any point $A$, draw two tangents $AB, AC$ of $\odot O^{\prime}$, touching $\odot O^{\prime}$ at $E, F$, intersecting $\odot O$ at $B, C$, connect $AO^{\prime}$ intersecting $\odot O$ at $D$, then draw the diameter $DD^{\prime}$ of $\odot O$ (as shown in the figure), then
$$
\begin{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,931 |
5. Fill three non-zero real numbers $a, b, c$ into the three boxes of $\square x^{2}+\square x+\square=0$ in any order, and the roots of the resulting equation are all rational numbers. Such real numbers $a, b, c$:
(A) do not exist.
(B) exist uniquely.
(C) exist in two pairs.
(D) exist, more than two pairs. | $\mathrm{D}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,936 |
6. If real numbers $x, y$ satisfy the condition, $2 x^{2}-6 x+y^{2}$ $=0$, then the maximum value of $x^{2}+y^{2}+2 x$ is ( ). (A) 14. (B) 15. (C) 16. (D) Cannot be determined.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,937 |
1. If $a, b, c$ satisfy the conditions $a+b+c=1, a^{2}$ $+b^{2}+c^{2}=2, a^{3}+b^{3}+c^{3}=3$, then $a b c=$ $\qquad$ $a^{4}+b^{4}+c^{4}=$ $\qquad$ | $\frac{1}{6}, \frac{25}{6}$ | \frac{1}{6}, \frac{25}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,940 |
Example 1. When $a$ takes what values, the equation $\left(a^{2}-1\right) x^{2}-$ $6(3 a-1) x+72=0$ has two distinct positive integer roots? | Given $\because a^{2}-1 \neq 0$, we conclude $a \neq \pm 1$, hence
$x_{1}=\frac{12}{a+1}, x_{2}=\frac{6}{a-1}$.
From $x_{1}>0, x_{2}>0$ we know $a>1$.
Furthermore, since $x_{1}, x_{2}$ are positive integers, we know $a=2$ or 3.
However, when $a=3$, $x_{1}=x_{2}=3$ does not meet the requirements.
Therefore, when $a=2$, ... | a=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,942 |
Three, solve the inequality $x^{2}-\left(m+\frac{1}{m}\right) x+\left(m-\frac{1}{m}\right)$ $<0$, where $\mathrm{m}$ is a parameter. | Three, Note! Discuss in three cases.
(1) When $m-1>\frac{1}{m}+1$, the solution to the original inequality is $1+\frac{1}{m}<x<m-1$.
(2) When $m-1<\frac{1}{m}+1$, the solution to the inequality is $m-1<x<-1$.
(3) When $m-1=\frac{1}{m}+1$, the original inequality has no solution. | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 705,944 |
Four, as shown in the figure, $NS$ is the diameter of circle $O$, chord $AB$ is parallel to $NS$, intersecting $NS$ at $M$, $P$ is a point, $PS$ intersects $AB$ at $R$, the extension of $PM$ intersects circle $O$ at $Q$. Prove: $RS > MQ$. | Prove as follows: Draw $M K \Perp R S$, connect QK, KS, QS.
Since $K S R M$ is a parallelogram, $K S \| M R$, and $M R \perp N S$, then $K S \_\_S$.
Thus, $K S$ is a tangent to $\odot O$.
So $\angle Q P S=\angle Q S K$.
Also, since $M K / / R S$,
Thus $\angle Q M K=\angle Q P S$.
Therefore, $\angle Q M K=\angle Q S K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,945 |
Five, a bookshelf has five levels, from bottom to top, referred to as the 1st level, the 2nd level, $\cdots$, the 5th level. Now, 15 books are to be placed on the various levels of the bookshelf, and some levels can remain empty. Prove: no matter how they are placed, the number of books on each level of the bookshelf, ... | Let $x_{1}$ represent the number of books placed on the $i$-th shelf, where $i=1, 2, 3, 4, 5$. For the case where some $x_{1} = 0$, the conclusion is obviously true.
Therefore, we only need to consider the case where $x_{1} \geqslant 1$ for $i=1,2,3,4, 5$. There are two scenarios:
( I ) Two of $x_{1}, x_{2}, x_{3}, x_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,946 |
4. The expression $\frac{\left(2^{4}+\frac{1}{4}\right)\left(4^{4}+\frac{1}{4}\right)\left(6^{4}+\frac{1}{4}\right)}{\left(1^{4}+\frac{1}{4}\right)\left(3^{4}+\frac{1}{4}\right)\left(5^{4}+\frac{1}{4}\right)}$ $\times \frac{\left(8^{4}+\frac{1}{4}\right)\left(10^{4}+\frac{1}{4}\right)}{\left(7^{4}+\frac{1}{4}\right)\le... | $221$ | 221 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,947 |
1. Proposition,
(1) For the fraction $\frac{x^{2}-4}{x-2}$, when $x= \pm 2$, its value is undefined.
(2) If $(a+1)^{0}=1$, then $a \neq 0$.
(3) In the function $y=\sqrt{x+\frac{1}{2}}+\frac{1}{x}$, the domain of the independent variable is $x \geqslant-\frac{1}{2}$.
(4) In a coordinate system, $y=x+m$ and $y=$
$$
\frac... | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,948 |
2. The quadratic function $y=a(x+2)^{2}+3$ can be obtained from the graph of $y=a x^{2}$ by ( ).
(A) Shifting 2 units to the left and then moving 3 units downward. It is obtained.
(C) Shifting 2 units to the right and then moving 3 units downward.
(D) Shifting 2 units to the right and then moving 3 units upward. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,949 |
Example 2. It is known that the equation $a x^{2}-(a-3) x+a-2=0$ has at least one integer solution, find the integer $a$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should no... | Touch If $a=0$, the original equation becomes $3 x-2=0$, which has no integer solutions, so $a \neq 0$.
$$
\begin{array}{l}
\because \Delta=(a-3)^{2}-4 a(a-2) \geqslant 0, \\
\therefore 3 a^{2}-2 a-9 \leqslant 0, \\
\therefore \frac{1-2 \sqrt{7}}{3} \leqslant a \leqslant \frac{1+2 \sqrt{7}}{3} .
\end{array}
$$
The int... | -1, 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,953 |
Example 2. Let $a, b, c$ be the lengths of the three sides of any triangle, $I=a+b+c, S=ab+bc+ca$. Prove that $3S$ $\leqslant I^{2} \leqslant 4S$. (1978, Tianjin City Middle School Mathematics Competition) | Prove obviously, $(a+b+c)^{2} \geqslant 3(ab+bc+ca)$, i.e., $I^{2} \geqslant 3S$.
By Theorem 1, we know $b+c>a$, thus $ab+ac>a^{2}$.
Similarly, we get $bc+ab>b^{2}$, $ca+bc>c^{2}$.
Adding the three inequalities and adding $2(ab+bc+ca)$ to both sides, we get
$$
4(ab+bc+ca)>(a+b+c)^{2},
$$
Thus $4S>I^{2}$. Therefore, $... | 3S \leqslant I^{2} < 4S | Inequalities | proof | Yes | Yes | cn_contest | false | 705,965 |
Example 1. Find all integer values of \( l \) that make the expression \(\frac{5l+6}{8l+7}\) reducible. (19th Moscow Mathematical Olympiad) | To find all integer values of $l$ that satisfy $(5 l+6,8 l+7)>1$. Because
$$
\begin{array}{l}
(5 l+6,8 l+7)=(5 l+6,3 l+1) \\
=(3 l+1,(-2)(3 l+1)+5 l+6) \\
=(3 l+1, l-4) \\
=(3 l+1+(-3)(l-4), l-4) \\
=(13, l-4)>1
\end{array}
$$
From $13 \mid l-4$, we get $l=13 m+4(m \in \mathbb{Z})$. | l=13m+4(m \in \mathbb{Z}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,970 |
. Example 2. Prove: Any two consecutive Fibonacci numbers $F_{n}, F_{n+1}(n \geqslant 2)$ are coprime. | \begin{aligned} & \because F_{1}=F_{2}=1, \\ & F_{\mathrm{n}+1}=F_{\mathrm{n}}+F_{\mathrm{n}-1}, \\ \therefore \quad & \left(F_{\mathrm{n}+1}, F_{\mathrm{n}}\right)=\left(F_{\mathrm{n}}, F_{\mathrm{n}+1}-F_{\mathrm{n}}\right) \\ = & \left(F_{\mathrm{n}}, F_{\mathrm{n}-1}\right) \\ = & \left(F_{\mathrm{n}-1},-F_{\mathrm... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,971 |
Side 3. Prove that for different natural numbers $p, q$, there must be infinitely many natural numbers $n$ such that $(p+n, q+n) = 1$.
untranslated text:
侧3. 证明, 对于不同的自然数 $p, q$,必有无穷多个自然数n, 使 $(p+n, q+n)$ $=1$. | Proof Without loss of generality, let $p>q$, then
$$
\begin{array}{l}
(p+n, q+n) \\
=(p-q, q+n) \\
=(p-q, \quad(q-p) k+q+n),
\end{array}
$$
where $k$ is a natural number.
Let $(q-p) k+q+n=1$, then
$$
n=(p-q) k-q+1 .
$$
At this point, $(p+n, q+n)=1$.
Since $n \geqslant 1$, we have
$$
(p-q) k-q+1 \geqslant 1 \text {. }... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,972 |
1. If $x_{1}, x_{2}, \cdots, x_{n}$ are non-negative real numbers, such that their sum is less than $\frac{1}{2}$, prove:
$$
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)>\frac{1}{2} .
$$ | 1. Consider $n$ mutually independent events, with their probabilities of occurrence being $x_{1}, x_{2}, \cdots, x_{\mathrm{n}}$. The probability that at least one of them occurs is at most $x_{1}+x_{2}+\cdots+x_{n}<\frac{1}{2}$. Since $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$ is the probabi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,975 |
Prove that in any tetrahedron there is a vertex such that the three edges meeting at this vertex can form the sides of a triangle. (1st IMO problem) | Proof by contradiction. In tetrahedron \(ABCD\), let \(AB\) be the longest edge. If the three edges from any vertex cannot form a triangle, then for the three edges from \(A\), we have \(AB \geqslant AC + AD\), and for the three edges from \(B\), we have \(AB \geqslant BC + BD\). Adding these two inequalities, we get
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,976 |
2. Given a convex quadrilateral, can we always determine a point inside it such that the lines connecting this point to the vertices divide the quadrilateral into four regions of equal area? If such a point exists, is it unique? | 2. Let [] denote the area function. If $A B C D$ is a quadrilateral, $E, F, G, H$ are the midpoints of $A B, B C, C D, D A$ respectively, then $E H=F G=\frac{1}{2} B D$, and $E H, F G$ are both parallel to $B D$. Draw a line $l$ through the midpoint of $A C$, and parallel to $B D$. Thus, the distances from $A$ and $C$ ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,977 |
4. $A B C D E$ is a pyramid with a square base $B C D E$, and points $F, G, H$ are on $A B, A C, A D$ respectively, such that $A F=A G=A H$.
(a) Prove: $E F$ and $D G$ must intersect at a point $K$, and $B G$ and $E H$ must intersect at a point $L$.
(b) If $E G$ intersects the plane $A K L$ at $M$, prove: $A K M L$ is ... | 4. (a) If $I$ is a point on $A E$ such that $A I=A G$, then $F G H I$ is a square similar to $B C D E$, with the homothetic center being point $A$. Therefore, $D E F G$ is a trapezoid, with its parallel sides $D E$ $>F G$. Thus, $D G$ and $E F$ intersect at point $K$. Similarly, $B G$ and $E H$ intersect at point $L$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,979 |
5. If $r$ is a real number, the sequence $\left\{x_{n}\right\}$ is defined by $x_{0}=0$, $x_{1}=1$ and $x_{n+2}=r x_{n+1}-x_{n}, n \geqslant 0$, then for which values of $r$ does $x_{1}+x_{3}+\cdots$ $+x_{2 m-1}=x_{m}^{2}$ hold for $m \geqslant 1$? | 5. For all $r$, the characteristic equation of the given recurrence relation is $y^{2}-r y+1=0$, with roots
$$
\frac{r \pm \sqrt{r^{2}-4}}{2}.
$$
If $r=2$, there is a repeated root of 1. Thus, for some $c_{1}, c_{2}$, we have $x_{2}=c_{1}+c_{2} n$. Using $x_{0}=0$ and $x_{1}=1$, we get $c_{1}=0, c_{2}= \pm 1$ and $x_{... | r = \pm 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,980 |
6. Let $x, y$ be positive real numbers, and satisfy $x+y=1$. Find the minimum value of $\left(\frac{x+1}{x}\right)^{2}+\left(\frac{y+1}{y}\right)^{2}$. | 6. By the arithmetic-geometric mean inequality, we have
$$
1=x+y \geqslant 2 \sqrt{x y}, \text{ hence } \frac{1}{x y} \geq 4.
$$
$$
\begin{array}{l}
\left(\frac{x+1}{x}\right)^{2}+\left(\frac{y+1}{y}\right)^{2} \\
\geqslant \frac{1}{2}\left(\frac{x+1}{x}+\frac{y+1}{y}\right)^{2} \\
=\frac{1}{2}\left(1+\frac{1}{x y}\rig... | \frac{25}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,981 |
7. $M N$ is the diameter of a circle, $A B$ is a chord perpendicular to $M N$. A line segment within the region enclosed by the chord $A B$ and the arc $A N B$ rotates around $A$, such that $B$ moves to $B^{\prime}$, and $N$ moves to $N^{\prime}$. Prove: $M P$ is perpendicular to $N^{\prime} P$, where $P$ is the midpoi... | 7. Line $M N$ intersects $A B$ at point $X$, and $X^{\prime}$ is the image point of $X$ after rotating around $A$. Noting that $P, X, X^{\prime}$ are the midpoints of $B B^{\prime}, A B, A B^{\prime}$ respectively, thus $A X P X^{\prime}$ forms a rhombus. Therefore,
$$
\begin{aligned}
M X X^{\prime} N^{\prime} & =M X ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,982 |
9. In tetrahedron $ABCD$, $AC \neq BD$, $AB \perp AC$, $DB$ is perpendicular to plane $ABC$. Point $O$ is the midpoint of $AB$, and $K$ is a point on $CD$ such that $OK \perp CD$. Let $[T]$ denote the volume of tetrahedron $T$. Prove that the following holds if and only if $2 AC \cdot BD = AB^2$:
$$
\frac{[OKAC]}{[OKBD... | 9. Since $O A=O B,[D K A C]=\frac{[A B K C]}{2}$ and $[O K B D]=\frac{[A B K D]}{2}$. Notice that the tetrahedra $A B K C, A B K D$ share the common face $A B K$. Therefore,
$$
[O K A C]=\frac{[O B K C]}{[O K B D]}=\begin{array}{l}
K C \\
K D
\end{array} .
$$
By the Pythagorean theorem, we have
$$
\begin{array}{l}
K D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,984 |
Example 5. Solve the system of equations in the real number range:
$$
\left(\begin{array}{l}
a \sqrt{x-y+z} \cdot \sqrt{x+y-z}=x \sqrt{y z}, \\
b \sqrt{x+y-z} \cdot \sqrt{-x+y+z}=y \sqrt{z x}, \\
c \sqrt{-x+y+z} \cdot \sqrt{x-y+z}=z \sqrt{x y},
\end{array}\right.
$$
where $a, b, c \neq 0$. | If $x y z=0$, then at least one of $x, y, z$ is 0. In this case, the solution to the system of equations is obvious, i.e., $x=0$,
$$
\begin{array}{l}
y=z, y=0, z=x, z=0, x=y . \\
\quad \text { If } x y z \neq 0, \quad \text { then } x \neq 0, y \neq 0, z \neq 0 . \\
\because x y>0, y z>0, z x>0, x+y>z, \\
z+x>y, y+z>x,... | \begin{array}{l}
x=\frac{a\left(b^{2}+c^{2}-a^{2}\right)}{b c}, \\
y=\frac{b\left(c^{2}+a^{2}-b^{2}\right)}{c a}, \\
z=\frac{c\left(a^{2}+b^{2}-c^{2}\right)}{a b}
\end{array} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,987 |
1. The denominator of a reduced fraction is 30, find the sum of all such positive rational numbers less than 10.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 1. Xiao Guang 10 points: If a fraction can be simplified to the form $\frac{30 n+r}{30}$, where $n$ and $r$ are integers satisfying $0 \leqslant n \leqslant 9$ and $0 \leqslant r<30$.
When and only when $r$ is coprime with 30, $\frac{30 n+r}{30}$ is a simplified fraction, thus, $r \in\{1,7,11,13,17,19,23,29\}$.
Since... | 400 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,990 |
2. A decimal positive integer of at least two digits, where each digit is smaller than the digit to its right, is called an "ascending number." How many ascending numbers are there?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. An "ascending number" must have at least two non-zero digits. It is a subset of the set $S=\{1,2,3,4,5,6,7, 8, 9\}$ consisting of two or more elements. The set determined by all elements is unique. Therefore, the number of ascending numbers is the number of subsets of $S$ consisting of two or more elements. Since $S... | 502 | Number Theory | proof | Yes | Yes | cn_contest | false | 705,991 |
3. Dividing the number of games a player wins by the total number of games he plays is called his winning rate. A player had a winning rate of 0.5 in the first phase. In the second phase, he played four games, winning three and losing one. If his winning rate over the two phases is greater than 0.503, find the maximum ... | 3. Let the number of matches played by the player in the first stage be $y$, and the number of matches won be $x$. Then, from the given information,
$$
\frac{x}{y}=0.5, \quad \frac{x+3}{y+4}>0.503 .
$$
It is easy to find that the player played at most 164 matches in the first stage. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 705,992 |
5. $S$ is the set of rational numbers $r$ where $0<r<1$, and $r$ has a repeating decimal expansion of the form $0 . a b c a b c a b c \cdots$ $=0 . \dot{a} b \dot{c}, a, b, c$ are not necessarily distinct. Among the elements of $S$ that can be written as a fraction in simplest form, how many different numerators are th... | 5. Since $0 . \dot{a} b \dot{c}=\frac{a b c}{999}$, and $999=3^{8} \cdot 37$. If $a b c$ is neither divisible by 3 nor by 37, then the fraction is in its simplest form. By the principle of inclusion-exclusion, we have
$$
999-\left(\frac{999}{3}+\frac{999}{37}\right)+\frac{999}{3 \cdot 37}=648
$$
such numbers.
In addit... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,994 |
6. In $\{1000,1001, \cdots, 2000\}$, how many pairs of consecutive integers can be added without carrying over a digit? | 6 . Let $n$ have the decimal representation $1 a b c$. If one of $a, b$, $c$ is $5, 6, 7$ or 8, then adding $n$ and $n+1$ requires carrying 3. If $b=9, c \neq 9$ or $a=9, b$ and $c$ are not 9, then adding $n$ and $n+1$ requires carrying.
If $n$ is not as described above, then it must be of one of the following forms:
... | 156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,995 |
7. The faces $ABC$ and $BCD$ of tetrahedron $ABCD$ form a $30^{\circ}$ angle, the area of $\triangle ABC$ is 120, the area of $\triangle BCD$ is 80, and $BC=10$. Find the volume of this tetrahedron. | $$
\begin{array}{c}
\text { 7. Through } D \text { draw } D H \\
\perp B C \text { at } H, \\
\because \quad B C=10, \\
S_{\triangle B C D}=80, \\
\therefore \quad D H=16 .
\end{array}
$$
Through $D$ draw a perpendicular to plane $A B C$ intersecting it at: $P$, then $\triangle H P D$ is a right triangle, $P H$ is the... | 213.33 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,996 |
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