problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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3. In $\triangle A B C$, let $A D$ be the altitude, and $B E$ be the angle bisector. If $B C=6, C A=7, A B=8$, then $D E=$ $\qquad$ | 3. As shown in the figure, from the given information, we have $\frac{B}{A}=\frac{BC}{AB}$. Therefore, we can derive
$$
\begin{array}{l}
\frac{CE}{CE+EA}=\frac{BC}{BC+AB}. \\
\therefore CE=7 \times \frac{6}{14}=3. \\
\text{Also, } \cos C \\
=\frac{AC^2+BC-AB^2}{2 \cdot AC \cdot BC} \\
=\frac{7^2+6^2-8^2}{2 \times 7 \ti... | \frac{\sqrt{151}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,979 |
4. Place two circular paper pieces with a radius of 5 and one circular paper piece with a radius of 8 on the table, so that they are pairwise externally tangent. If a large circular paper piece is used to completely cover these three circular paper pieces, then the minimum radius of this large circular paper piece equa... | 4. As shown in the figure, let the radius of $\odot O_{1}$ be $8$; the radii of $\odot O_{2}$ and $\odot O_{3}$ be $5$, with the tangent point at $A$. By symmetry, the center $O$ of the smallest circular paper piece that can cover these three circles must lie on the line of symmetry $O_{1} A$, and it must be internally... | 13 \frac{1}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,980 |
一、(20 points) As shown in the figure. In $\triangle A B C$, $A B=A C$. Extend $C A$ to $P$ arbitrarily, and extend $A B$ to $Q$. So that $A P=B Q$. Prove: The circumcenter $O$ of $\triangle A B C$ and $A, P, Q$ are concyclic. | Proof 1: Connect $O A$, $O C$, $O P$, $O Q$. In $\triangle O C P$ and $\triangle O A Q$, $O C = O A$. Given that $C A = A B$, $A P = B Q$.
$$
\therefore C P = A Q.
$$
Also, since $O$ is the circumcenter of $\triangle A B C$,
$$
\therefore \angle O C P = \angle O A C.
$$
Since the circumcenter of an isosceles triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,981 |
II. (20 points) Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it does, please prove how many there are? | There exists such a right-angled triangle, and there is only one. Let the hypotenuse of this right-angled triangle be $c$, the two legs be $a, b$, and the area be $S$, then
$$
\left\{\begin{array}{l}
a \leqslant b < c < a + b, \\
a + b + c = 6, \\
a^{2} + b^{2} = c^{2}, \\
S = \frac{1}{2} a b \text{ is an integer. }
\e... | a = \frac{5 - \sqrt{7}}{3}, b = \frac{5 + \sqrt{7}}{3}, c = \frac{8}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 706,982 |
Three. (20 points) A certain mathematics competition had a total of 15 questions. The table below shows the statistics for the number of people who got $n (n=0,1,2, \cdots, 15)$ questions correct.
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people wh... | Three, Solution 1 From the statistical table, we know: The total number of people who got $0-3$ questions correct is $7+8+10+21=40$ people, and the total number of questions they got correct is $7 \times 0+8 \times 1+10 \times 2+21 \times 3=91$ (questions); The total number of people who got $12-15$ questions correct i... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,983 |
1. If $\frac{100}{x}>\sqrt{1994}-25>\frac{125}{x+2}$, then the positive integer $x$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | $-1 . B$
From $44^{2}=193620$, when $x=6,7$, the left side $=\frac{100}{x}<19$, it is easy to verify that the inequality holds when $x=5$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 706,984 |
2. Given the sequence: $4,5,6,7,8,9,13,14,15$, $25,26,27, \cdots$, according to its inherent pattern, the 16th number should be ( ).
(A) 93
(B) 95
(C) 97
(D) 99 | 2. C
In this sequence, every three consecutive natural numbers can be divided into a group. Starting from the second group, the middle number of each group is twice the first number of the previous group. Therefore, the 5th group should be 49, 50, 51, and the 6th group should be 97, 98, 99. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,985 |
4. The system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-(7+m) x+7 m \leqslant 0, \\
x^{2}-(n+1) x+n \geqslant 0
\end{array}\right.
$$
has the solution set $5 \leqslant x \leqslant 7$. Then the values of $m, n$ are ( ).
(A) $\left\{\begin{array}{l}m=5, \\ n \leqslant 5\end{array}\right.$ or $\left\{\begin{array}... | 4. A
The system of inequalities can be transformed into
$$
\left\{\begin{array}{l}
(x-7)(x-m) \leqslant 0, \\
(x-1)(x-n) \geqslant 0 .
\end{array}\right.
$$
The solution set of (1) is $A_{1}=[m, 7]$ (when $m \leqslant 7$). Or $A_{2}=[7, m]$ (when $m>7$), and $A_{2}$ should be discarded. When $n \geqslant 1$, the solut... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 706,987 |
5. Let $\frac{x}{x^{2}-p x+1}=1$. Then the value of $\frac{x^{3}}{x^{6}-p^{3} x^{3}+1}$ is ( )
(A) 1
(B) $\frac{1}{3 p^{2}-2}$
(C) $\frac{1}{p^{3}+3^{2}}$
(D) $\frac{1}{3 p^{2}+1}$ | 5. B
From the given, we have $\frac{x^{2}-p x+1}{x}=1$,
which means
$$
\begin{array}{l}
x+\frac{1}{x}=1+p \text {. } \\
\therefore x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right) \\
=(p+1)\left[(p+1)^{2}-3\right] \\
=p^{3}+3 p^{2}-2 \text {. } \\
\text { And } \frac{x^{6}-p^{3} x^{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,988 |
Example 5. Given a parallelogram $A B C D$, $\triangle A B D$ is an acute triangle, and its circumradius is $R$. Circles with radius 1 are drawn with $A, B, C, D$ as centers, denoted as $\odot A, \odot B, \odot C$, ( $\odot$. Prove:
(1) If the parallelogram $A B C D$ is covered by $\odot A, \odot B, \odot C, \odot D$, ... | Proof (1) First, prove that when $\triangle A B D$ is an acute triangle, point $C$ must be outside the circumcircle $\odot O$ of $\triangle A B D$.
Otherwise, if point $C$ is on the circumference of $\odot O$, then quadrilateral $A B C D$ is a cyclic quadrilateral. Thus, $\angle D A B + \angle B C D = 180^\circ$. Sinc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,989 |
6. In $\triangle A B C$, $B D$
$$
\begin{array}{l}
: D C=m: n, A E: E B= \\
p: q, A P: P C=s: t, \\
P M / / E C, P N / / A D, M N
\end{array}
$$
$$
\begin{array}{l}
\text { intersects } A D \text { at } Q \text {. Then } M Q: Q N \\
=(\quad) .
\end{array}
$$
(A) $\frac{m p}{(p+q) n}$
(B) $\frac{s p}{(p+q) t}$
(C) $-\fr... | 6. A
Draw $M H / / A D$ intersecting $B C$ at $H$, then $M Q: Q N$ $=H D: D N$. Let the lengths of sides $B C, C A, A B$ be $a, b, c$ respectively. Then
$$
\begin{array}{l}
D N=\frac{A P}{A C} \cdot D C \\
=\frac{s}{s+t} \cdot \frac{n a}{m+n}, \\
A M=\frac{A P}{A C} \cdot A E \\
=\frac{s}{s+t} \cdot \frac{p c}{p+q} . ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,990 |
7. Given $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+c(a b c \neq 0)$ that are symmetric with respect to the axis of symmetry. Then when $x=x_{1}+x_{2}$, the value of $y$ is ( ).
(A) 0
(B) $\frac{4 a c-b^{2}}{4 a}$
(C) $c$
(D) Can... | 7. C
By symmetry, we know $y_{1}=y_{2}$, that is,
$$
a x_{1}^{2}+b x_{1}+c=a x_{2}^{2}+b x_{2}+c .
$$
Rearranging gives $a\left(x_{1}+x_{2}\right)+b=0, x_{1}+x_{2}=-\frac{b}{a}$. Therefore,
$$
y=a\left(-\frac{b}{a}\right)^{2}+b\left(-\frac{b}{a}\right)+c=c .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,991 |
8. In the cyclic quadrilateral $A B C D$, $\angle A=$
$$
\begin{array}{l}
\angle 52.5^{\circ}, \angle B=97.5^{\circ}, \\
\angle A O B=120^{\circ}, A B=a, \\
B C=b, C D=c, D A=d .
\end{array}
$$
Then the area $S$ of this quadrilateral is ( ).
(A) $\frac{1}{2}(a d+b c)$
(B) $\frac{1}{2}(a b+c d)$
(C) $\frac{1}{2}(a c+b ... | 8. C
On $\overparen{D C B}$, construct $\overparen{D C^{\prime}}=$
$\overparen{B C}$, it is easy to prove that $D C^{\prime}=b, B C^{\prime}$
$=c, \triangle A D C^{\prime}, \triangle A B C^{\prime}$
are right triangles, and
$\triangle B C D^{\prime}$ is equal in area to $\triangle D C^{\prime} B$, so we should choose ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,992 |
1. Let $x, y$ be real numbers, and $2 x^{2}+x y+2 y^{2}=2$. Then the range of values for $x^{2}-5 x y+y^{2}$ is $\qquad$ . | $$
\text { II. 1. }\left[-\frac{6}{5}, \frac{14}{3}\right]
$$
Let $x^{2}-5 x y+y^{2}=k$, then
$$
\begin{array}{l}
2 k=2 x^{2}+x y+2 y^{2}-11 x y=2-11 x y \\
x y=\frac{2-2 k}{11} \\
2(x+y)^{2}=2+3 x y=2+\frac{3(2-2 k)}{11} \\
\geqslant 0, \Rightarrow k \leqslant \frac{14}{3} \\
2(x-y)^{2}=2-5 x y=2-\frac{5(2-2 k)}{11} ... | \left[-\frac{6}{5}, \frac{14}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,993 |
3. Let the quadratic function $y=x^{2}+m x+n$ have its vertex on $y=-4$, and the graph intersects with the lines $y=x+2$, $y=1-\frac{1}{2} x$ each at one point within $0 \leqslant x \leqslant 2$. Then the range of values for $m$ is $\qquad$. | 3. $-8 \leqslant m \leqslant-2 \sqrt{6}$ or $4(\sqrt{2}-1) \leqslant$ $m \leqslant 2 \sqrt{5}$.
Since the vertex is on $y = -4$, we have $n = \frac{m^{2}}{4} - 4$. The quadratic function can be expressed as $y = (x + \frac{m}{2})^{2} - 4$. According to the problem, the vertex should be between points $P_{1}, P_{2}$ or... | -8 \leqslant m \leqslant -2 \sqrt{6} \text{ or } 4(\sqrt{2}-1) \leqslant m \leqslant 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,995 |
One, (20 points) The side lengths of squares $A B C D$ and $A E F G$ are $a$ and $b$ respectively, with $a > b$, and $A$ being the common vertex. $D C$ intersects $E F$ at $P$, and $A P \perp F C$.
Find $\angle E A D$. | Let $D P=x, P E=y$. Connect $D E$. Since $\angle D = \angle E=90^{\circ}, D, A, E, P$ are concyclic,
$$
\therefore \angle D A P=\angle D E P.
$$
Extend $A P$ to intersect $F C$ at $H$, then
$$
\begin{array}{l}
\angle A H C=90^{\circ}=\angle D. \\
\therefore \angle D A P=\angle D C H, \\
\angle D E P=\angle D C H, \tri... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,997 |
II. (20 points) Find all positive integer solutions $(x, y, z)$ of the indefinite
equation
$$
2 x^{2}+6 x y-12 x z+x-3 y+6 z-6=0
$$ | Let $u=y-2z$, then we have
$$
2 x^{2}+6 x u+x-3 u-6=0,
$$
which is equivalent to $(2 x-1)(x+3 u+1)=5$.
Thus, we should have
$$
\left\{\begin{array}{l}
2 x-1=5 \\
x+3 u+1=1
\end{array}\right.
$$
(1) or $\left\{\begin{array}{l}2 x-1=1, \\ x+3 u+1=5 .\end{array}\right.$
Solving (1), we get $\left\{\begin{array}{l}x=3, \... | x=3, y=2 t-1, z=t \text{ or } x=1, y=2 t+1, z=t | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,998 |
*Three. (20 points) 4 small balls with a radius of $r$ are placed in cylinder $A$, numbered from top to bottom as $1,2,3,4 . A$'s base radius is slightly larger than $r . B, C$ are cylinders identical to $A$. The balls in $A$ are moved unidirectionally to $C$ via $B$, meaning no balls can be moved from $C$ to $B$, or f... | Three, first consider the case of three small balls. At this time, there are a total of 5 arrangement methods, each method (from bottom to top) and the operation process is represented as follows:
(1)(2)(3) (1) $\rightarrow B \rightarrow C$, (2) $\rightarrow B \rightarrow C$, (3) $\rightarrow B \rightarrow$ C)
(1) (3) ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,999 |
Example 1. If in a Pythagorean triple, the difference between the hypotenuse and one of the legs is 1, then the form of the Pythagorean triple is
$$
2 a+1,2 a^{2}+2 a, 2 a^{2}+2 a+1 .
$$ | Proof: Let the hypotenuse be $c$, the other leg be $c-1$, and the base be $x$.
Since $(c, c-1)=1$, $x, c-1, c$ form a set of primitive Pythagorean triples. Also, since $c$ is odd, $c-1$ is even, thus $x$ is odd.
Let $x=2a+1$, then we have
$(2a+1)^{2}+(c-1)^{2}=c^{2}$.
Solving for $c$ gives $c=2a^{2}+2a+1$,
$$
c-1=2a^{2... | 2a+1, 2a^{2}+2a, 2a^{2}+2a+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,000 |
Example 2. Prove that for a Pythagorean triple $x, y, z$, it must be that $6 \mid xy$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | To prove that it is only necessary to prove the conclusion for the basic Pythagorean triples.
$x$ and $y$ must include one even number. By Theorem 2, we have
$$
\begin{aligned}
x & =2 m n, \quad y=m^{2}-n^{2} . \\
x y & =2 m n\left(m^{2}-n^{2}\right) \\
& =2 m n(m+n)(m-n) .
\end{aligned}
$$
When one of $m, n$ is a mul... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 707,001 |
Example 5. Let $S$ be an $n \times n$ square. Prove that regardless of the position of $S$ on the grid plane, it will not cover more than $(n+1)^{2}$ grid points. | Prove that if the four vertices of a square $S$ are all lattice points and parallel to the coordinate axes, then it exactly covers $(n+1)^{2}$ lattice points.
As shown in the figure, let $H$ be the largest polygon contained in the square $S$, with each vertex being a lattice point (if each vertex of $S$ is a lattice p... | (n+1)^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 707,002 |
5. A store first increases the price of color TVs by $40 \%$, then writes in the advertisement "Grand Sale, 20% off", as a result, each TV earns 270 yuan more than the original price. Therefore, the original price of each TV should be ( ) yuan.
(A) 2150
(B) 2200
(C) 2250
(D) 2300 | 5. C (Let the original price of each TV be $x$ yuan. According to the problem, we have $\frac{140 x}{100} \cdot \frac{80}{100}-x=$ 270.) | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,003 |
6. Let $P$ be any point on the hypotenuse $AC$ of an isosceles right $\triangle ABC$, $PE \perp AB$ at $E$, $PF \perp BC$ at $F$, $PG \perp EF$ at $G$, and take a point $D$ on the extension of $GP$ such that $PD = PB$. Then the relationship between $BC$ and $DC$ is ( ).
(A) equal but not perpendicular
(B) not equal but... | 6. $\mathrm{C}$ (Hint: Prove $\triangle D P C \cong \triangle B P C$ )
The text above has been translated into English, preserving the original formatting and line breaks. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,004 |
1. $x=\left(\frac{-2 a}{4+a}-\frac{\sqrt{|a|-3}+\sqrt{3-|a|}}{3-a} ; 1593\right.$ the intermediate digit of egg
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 二、1.6 (Hint: To make $x$ meaningful, it is necessary that (1) $|a|-3 \geqslant 0$;
(2) $3-|a| \geqslant 0$; (3) $4+a \neq 0$; (4) $3-a \neq 0$.) | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,005 |
3. In $\triangle A B C$, if $A B=A C, \angle B A C=$ $120^{\circ}, D$ is the midpoint of side $B C$, and $D E \perp A B$ at $E$. Then $A E: B D=$ | 3.
$1: 2 \sqrt{3}$ (Hint: Connect $A D$, then $A D \perp B C$. ) | 1: 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,007 |
4. In trapezoid $A B C D$, $A D / / B C, \angle B=$ $30^{\circ}, \angle C=60^{\circ}, E, M, F, N$ are the midpoints of $A B, B C$. $C D, D A$ respectively. Given $B C=7, M N=3$. Then $E F$ | 4. 4(Extend $B A, C D$ to meet at $H$, then $\triangle B H C$ is a right triangle.) | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,008 |
Three, (20 points) Solve the system of equations
$$
\left\{\begin{array}{l}
x(y+z-x)=39 \cdots 2 x^{2}, \\
y(z+x-y)=52-2 y^{2}, \\
(x+y-z)=78-2 z^{2} .
\end{array}\right.
$$ | $$
\begin{array}{l}
\text { Three, (1)+(2)+(3) gives } \\
(x+y+z)^{2}=169 .
\end{array}
$$
Taking the square root gives $x+y+z= \pm 13$,
Dividing (1), (2), (3) by (1) gives
$$
(x, y, z)=(3,4,6), \quad(-3,-4,-6) .
$$ | (x, y, z)=(3,4,6), \quad(-3,-4,-6) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,009 |
Four. (20 points) As shown in the figure, in the convex pentagon $A B C D E$, it is known that $S_{\triangle A B C}=1$, and $E C$ // $A B, A D$ // $B C, B E$ // $C D, C A$ // $D E, D B$ // $E A$. Try to find: the area of pentagon $A B C D E$.
---
The translation preserves the original text's line breaks and formattin... | Given, we have
\[
\begin{aligned}
S_{\triangle A C D} & =S_{\triangle C D E}=S_{\triangle D E A}=S_{\triangle E A B}=S_{\triangle A C B}=S_{\triangle A C F} \\
& =1 .
\end{aligned}
\]
Let \( S_{\triangle A E P} = x \). Then \( S_{\triangle D E} = 1 - x \).
Since \( A A E F \) and \( \triangle D E C \) have equal heigh... | \frac{5 + \sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,010 |
Five. (20 points) In a mountain bike race held in a city, two cyclists, A and B, start from point A to point B at the same time. Cyclist A runs $\frac{1}{3}$ of the time at speeds $v_{1}, v_{2}, v_{3}$ respectively; Cyclist B runs $\frac{1}{3}$ of the distance at speeds $v_{1}, v_{2}, v_{3}$ respectively. Who will reac... | Let the distance between $A$ and $B$ be $S$. The times taken by car A and car B to complete the journey are $t_{1}$ and $t_{2}$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{t_{1}}{3}\left(v_{1}+v_{2}+v_{3}\right)=S, \\
\frac{S}{3}\left(\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}... | t_{1} \leqslant t_{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,011 |
1. Given $a^{2 x}=\sqrt{2}+1(a>0)$, then the value of $\frac{a^{3 x}+a^{3 x}}{a^{x}+a^{x}}$ is $\qquad$ . | 一、1.2 $\sqrt{2}-1$ (Hint: Apply the sum of cubes formula)
untranslated part:
一、1.2 $\sqrt{2}-1$ (提示:应用立方和公式)
The translated part:
1.2 $\sqrt{2}-1$ (Hint: Apply the sum of cubes formula) | 2\sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,012 |
Example 6. Does there exist a closed broken line composed of an odd number of equal-length segments, such that its vertices are all lattice points? | Let $A_{i}\left(x_{i}, y_{i}\right)$ be the vertices of a closed polygonal line, where $x_{i}$ and $y_{i}(i=1,2, \cdots, n)$ are all integers. Let $\alpha_{i}=x_{i}-x_{i-1}, \beta_{i}=y_{i}-y_{i-1}$, where $x_{n+1}=x_{1}$, $y_{n+1}=y_{1}, i=1,2, \cdots, n$. By the given conditions, we have
$$
\begin{array}{l}
\alpha_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,013 |
2. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the two roots is $S_{1}$, the sum of the squares of the two roots is $S_{2}$, and the sum of the cubes of the two roots is $S_{3}$. Then the value of $a S_{1}+b S_{2}+c S_{3}$ is $\qquad$ | 2. 0
(Hint: Let the two roots of the equation be $x_{1}, x_{2}$. It is easy to see that $a x_{1}^{3}+b x_{1}^{2}+c x_{1}=$ $\left.0, a x_{2}^{3}+b x_{2}^{2}+c x_{2}=0.\right)$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,014 |
3. Given $x=1-\sqrt{3}$. Then $x^{5}-2 x^{4}-2 x^{3}$ $+x^{2}-2 x-1$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 1 (Hint: Original expression $=x^{3}\left(x^{2}-\right.$
$$
\left.2 x-2)+\left(x^{2}-2 x-2\right)+1\right)
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,015 |
5. Let there be $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying: $\left|x_{1}\right|$ $<1(i=1,2, \cdots, n)$, and $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|$ $=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right|$. Then the minimum value of $n$ is | 5. 20 (Hint:
From $\left|x_{i}\right|<1(i=1,2, \cdots, n)$, it is easy to know that $19 \leqslant\left|x_{1}\right|+\left|x_{2}\right|+\cdots$ $+\left|x_{n}\right|<n$. When $n=20$, we have
$$
x_{i}=\left\{\begin{array}{l}
0.95(i=1,3, \cdots, 19), \\
-0.95(i=2,4, \cdots, 20)
\end{array}\right.
$$
satisfying the condit... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,017 |
II. (30 points) Let $y=a x^{2}+b x+c$. It is known that when $x=1$, $y=0$; when $x=-1$, $y$ is an even number. Prove that the equation
$$
x^{2}-(a+b+c) x+b a+b c=0
$$
has two integer roots. | II. Transform the original equation into $(x-b)(x-a-c)=0$, then $x_{1}=b, x_{2}=a+c$.
Given $a+b+c=0$, and $a-b+c$ is even, so, $2i$ is even, $2(c+c)$ is even.
Limit 3, $a-c$ are integers? | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,018 |
Three. (30 points) Given the equation
$$
\left(a x+a^{2}-1\right)^{2}+\frac{x^{2}}{(x+a)^{2}}+2 a^{2}-1=0
$$
has real solutions. Find the range of real values for $a$. | Three, obviously, $a \neq 0$. Given that the equation can be transformed into
$$
\left(a^{2} i x+a\right)^{2}-2 a x+\frac{x^{2}}{(x+a)^{2}}=0 \text {, }
$$
which is $\left[a(x+a)-\frac{x}{x+a}\right)^{2}=0$.
Thus, $a(x+a)-\frac{x}{x+a}=0$, then
$$
a \cdot x^{2}+\left(2 a^{2}-1\right) x+a^{3}=0 \text {. }
$$
By the di... | -\frac{1}{2} \leqslant a \leqslant \frac{1}{2}, a \neq 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,019 |
Four, (30 points) In $\triangle A B C$, $\angle A=90^{\circ}, A D \perp B C$ at $D$, and the line parallel to $\angle B$ intersects $A D$ and $A C$ at $E$ and $F$ respectively. Prove:
(1) $A E=A F$;
(2) If $A C=\sqrt{2}$, then $\frac{E F}{C D}<A F \cdot C F$.
保留源文本的换行和格式，直接输出翻译结果。 | (1) In $\triangle B D E$ and $\triangle B A F$, since $\angle B D E = \angle B A F = 90^{\circ}$, and $\angle D B E = \angle A B F$, therefore, $\angle A F B = \angle D E B$.
Also, $\angle D E B = \angle F E A$,
thus $\angle A F E = \angle A E F$, which means $A E = A F$.
(2) Draw $A H \perp E F$ at $H$.
\[
\begin{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,020 |
1. If $0<p<15$, then the minimum value of the algebraic expression $|x-p|+|x-15|+|x-p-15|$ on $p \leqslant x \leqslant 15$ is ( ).
(A) 30
(B) 0
(C) 15
(D) an algebraic expression related to $p$ | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,021 |
3. The incorrect statement among the following is ( ).
(A) $\sqrt{-x^{3}}=-x \sqrt{-x}$
(B) The number of triangles with integer side lengths and a perimeter of 17 is 8
(C) $1-\sqrt{2}$ and $1+\sqrt{2}$ are negative reciprocals of each other
(D) $6 x^{2}-5 x+2$ can be factored over the real numbers | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,023 |
Example 1. Does there exist 1000000 consecutive integers, each of which has a repeated prime factor, i.e., is divisible by the square of some prime? (15th Putnam Mathematical Competition) | The number 1000000 in this problem is not the key; a stronger proposition can be proven:
There exist $n$ consecutive positive integers, each of which has a repeated prime factor.
In fact, if there exist $n$ consecutive positive integers
$$
m+1, m+2, \cdots, m+n,
$$
then the goal of the proof is: $m \equiv -i \pmod{p_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,024 |
4. In $\triangle A B C$, if $A B=2 B C, \angle B$ $=2 \angle A$, then $\triangle A B C$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle
(D) cannot be determined | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,025 |
5. Draw a line through any point (excluding the vertex) on the shortest side of a non-isosceles right triangle, intersecting the other side, dividing the original triangle into a quadrilateral and a smaller triangle. If the smaller triangle is similar to the original triangle, then the number of such lines is ( ).
(A) ... | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,026 |
6. The area enclosed by the graph of $|x-2|+|y-2|=2$ is ( ).
(A) $\sqrt{6}$
(B) $2 \sqrt{2}$
(C) 8
(D) $2 \sqrt{3}$ | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,027 |
8. There are 100 balls in the bag, including 28 red balls, 20 green balls, 12 yellow balls, 20 blue balls, 10 white balls, and 10 black balls. If you randomly draw balls from the bag, then the number of balls drawn from the bag must be at least ( ) balls.
(A) 100
(B) 75
(C) 68
(D) 77 | 8. B
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 | 707,029 |
3. Given $x-y=3, z=\sqrt{y^{2}-3 x^{2}}$. Then the maximum value of $z$ is $\qquad$ . | 3. $\frac{3}{2} \sqrt{6}$ | \frac{3}{2} \sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,032 |
4. Let $x, y$ be real numbers, and $x^{2}+x y+y^{2}=3$. Then, the range of $x^{2}-x y+y^{2}$ is $\qquad$ . | 4. $[1,9]$
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a form that does not require translation (it is a number and a mathematical interval notation), the output remains the same:... | [1,9] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,033 |
Three, (12 points) The equation $x^{2}+k x+4-k$ $=0$ has two integer roots. Find the value of $k$.
untranslated part:
关于 $x$ 的方程 $x^{2}+k x+4-k$ $=0$ 有两个整数根. 试求 $k$ 的值.
translated part:
The equation $x^{2}+k x+4-k$ $=0$ has two integer roots. Find the value of $k$. | Three, let $x_{1}, x_{2}$ be the roots of the equation, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} \cdot x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$, so
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$
By the problem statement,
$\left\{\begin{array}{l}x_{1}-... | k_{1}=-8 \text{ or } k_{2}=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,034 |
Example 2. Prove that for any positive integer $n$, there exist $n$ consecutive positive integers, none of which is a positive integer power of a prime. (30th IMO Problem) | In fact, if it can be proven that there exist $n$ consecutive positive integers, each of which can be divided by the product of at least two different primes, it is clear that these $n$ consecutive positive integers are not powers of prime numbers. Thus, Example 2 is merely a variation of Example 1.
Proof: Take $2n$ d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,035 |
Four, (12 points) Given that $D$ is a point inside $\angle BAC$ of the equilateral $\triangle ABC$, and $DA=DB+DC$. Prove: $A$, $B, C, D$ are concyclic.
---
The translation maintains the original text's line breaks and format. | When the hour hand rotates $60^{\circ}$ to $\triangle A D^{\prime} C$, connect $D D^{\prime}$. Since $D$ is outside $\odot O$, $\angle D B C \neq \angle D A C$, so $A D^{\prime}$ does not coincide with $A D$. In $\triangle A D D^{\prime}$, $D A < D^{\prime} A + D D^{\prime} = D B + D C$. This is in contradiction with $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,036 |
Five. (12 points) Try to find the largest and smallest seven-digit numbers that can be divided by 165, formed by the 7 digits $0,1,2,3,4,5,6$ without repetition (require the reasoning process to be written out).
Find the largest and smallest seven-digit numbers that can be formed using the digits $0,1,2,3,4,5,6$ witho... | Five, since $165=3 \times 5 \times 11$, the number we are looking for must be a multiple of 3, 5, or 11. Also, $0+1+2+3+4+5+6=21$. Therefore, the sum of these seven digits is always a multiple of 3.
Let the sum of the four digits in the odd positions be $A$, and the sum of the three digits in the even positions be $B$... | 1042635, 6431205 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,037 |
1. For a convex $n$-sided polygon, the sum of $n-1$ interior angles, except one, is $1993^{\circ}$, then the value of $n$ is ).
(A) 12
(B) 13
(C) 14
(D) None of the above | 1. $\mathrm{C}\left(\right.$ Prompt: $\left.1993^{\circ}=11 \times 180^{\circ}+13^{\circ}\right)$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,038 |
2. Given $\sqrt{a}+\sqrt{b}=1$, and $\sqrt{a}=m+\frac{a-b}{2}, \sqrt{b}=n-\frac{a-\dot{b}}{2}$, where $m, n$ are rational numbers. Then ( ).
(A) $m \cdot n=\frac{1}{2}$
(B) $m^{2}+n^{2}=\frac{1}{2}$
(C) $m+n=\frac{1}{2}$
(D) $n-m=\frac{1}{2}$ | 2. B
(Hint: $\sqrt{a}=\frac{\sqrt{a}+\sqrt{b}}{2}+\frac{\sqrt{a}-\sqrt{b}}{2}=\frac{1}{2}$ $+\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{2(\sqrt{a}+\sqrt{b})}=\frac{1}{2}+\frac{a-b}{2}$, similarly $\sqrt{b}=\frac{1}{2}-\frac{a-b}{2}$.) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,039 |
3. The graph of the function $y=-x^{2}+p x+q$ intersects the $x$-axis at points $(a, 0),(b, 0)$. If $a>1>b$, then ( ).
(A) $p+q>1$
(B) $p-r q=1$
(C) $p+q<1$
(D) $f q$ ) $=0$ | 3. $\mathrm{A}$ (Hint: From the problem, $a, b$
are the roots of the equation $x^{2}-p x-q=0$. By Vieta's formulas $p+q=$ $(a+b)-a b=(a-1)(1-b)+1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,040 |
Long) and 1 public F: foundation code. A customer wants to buy two kilograms of candy. The salesperson places a 1-kilogram weight on the left pan and puts the candy on the right pan to balance it, then gives the candy to the customer. Then, the salesperson places the 1-kilogram weight on the right pan and puts more can... | 4. C (Hint: Set up the city
Bi Chang: Left Bi Chang $=a: b$, then the weight of the sugar fruit in the first weighing is $\frac{b}{a}$ kilograms; the weight of the fruit in the second weighing is $\frac{a}{b}$ kilograms. Since $a \neq b$, so $\frac{b}{a}+$ $\left.\frac{a}{b}=\frac{(a-b)^{2}}{a b}+2.\right)$ | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 707,041 |
5. As shown in the figure, in $\triangle A B C$, $D, E$ are the trisection points of $B C$, $M$ is the midpoint of $A C$, and $B M$ intersects $A D, A E$ at $G, H$. Then $B G: G H: H M$ equals ( ).
(A) $3: 2: 1$
(B) $4: 2: 1$
(C) $5: 4: 3$
(D) $5: 3: 2$ | 5. $\mathrm{D}$ (Hint: Draw $M N / / B C$ through $M$,
it can be obtained by the median theorem or Menelaus' theorem. Note: When the trisection points become 4, 5, or $n$ equal parts, what results can you get?) | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,042 |
6. Insert a one-digit number (including 0) into a two-digit number, and it becomes a three-digit number, for example, inserting 6 into 72 results in 762. Some two-digit numbers, when a one-digit number is inserted, become a three-digit number that is 9 times the original two-digit number. How many such two-digit number... | 6. B(Because "three-digit number" is 9 times "two-digit number", the sum of "three-digit number" and "two-digit number" should be 10 times "two-digit number". Therefore, the unit digit of this "two-digit number" can only be 0 and 5. Since the inserted digit should be less than the unit digit of the "two-digit number", ... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,043 |
* 7. An exam consists of 5 questions. The post-exam score statistics are as follows: $81 \%$ of the students got the first question right; $91 \%$ of the students got the second question right; $85 \%$ of the students got the third question right; $79 \%$ of the students got the fourth question right; $74 \%$ of the st... | 7. A (Hint:
Assume there are 100 students taking the exam, then the total number of wrong answers is $19 + 9 + 15 + 21 + 26 = 90$ (person-questions). For a student to be不合格 (unqualified), they must get three questions wrong, and $90 \div 3 = 30$, so at most 30 students can be unqualified. On the other hand, concentrat... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,044 |
1. Calculate $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots$
$$
+\frac{1}{1+2+3+\cdots+100}=
$$
$\qquad$ | 1. $\frac{200}{101}$ (Hint: From $\frac{1}{1+2+3+\cdots+n}$ $=\frac{2}{n(n+1)}$) | \frac{200}{101} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,045 |
Example 3. Let $m, n$ be natural numbers, satisfying: for any natural number $k, 11 k-1$ and $m$ and $11 k-1$ and $n$ have the same greatest common divisor. Prove: there exists an integer $l$, such that $m$ $=11^{t} n$.
保留源文本的换行和格式，翻译结果如下：
Example 3. Let $m, n$ be natural numbers, satisfying: for any natural number $... | To prove $m=11^{\prime} n$, is equivalent to proving $\frac{m}{n}=11^{l}$. This indicates that when $m$ has the form $11^{i} p$ and $n$ has the form $11^{j} q$, as long as $p=q$, the desired result can be obtained.
Proof: Let $m=11^{i} p, n=11^{\prime} q$, where $i, j$ are non-negative integers, and $p$ and $q$ are bo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,046 |
2. In a certain school, the number of students in the 7th, 8th, and 9th grades is the same. It is known that the number of boys in the 7th grade is the same as the number of girls in the 8th grade, and the boys in the 9th grade account for $\frac{3}{8}$ of the total number of boys in the school. Therefore, the proporti... | 2. $\frac{7}{15}$ | \frac{7}{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,047 |
5. As shown in the figure, in isosceles $\triangle ABC$, $AB=AC$, $\angle A=120^{\circ}$, point $D$ is on side $BC$, and $BD=1$, $DC=2$, then $AD=$ | 5. 1 (Hint: Draw $D N \perp A B$, and draw the median $A M$ of side $B C$. Thus, we can prove $\triangle A D N \cong \triangle A D M$, getting $A D=$ $2 D N=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,050 |
6. $M$ is a point inside the convex quadrilateral $A B C D$, and the points symmetric to $M$ with respect to the midpoints of the sides are $P, Q, R, S$. If the area of quadrilateral $A B C D$ is 1, then the area of quadrilateral $P Q R S$ is equal to | 6. 2 (Hint: Connect the midpoints of each pair of adjacent sides of quadrilateral $A B C D$, to get a parallelogram, whose area is easily known to be $\frac{1}{2}$. On the other hand, these four sides are precisely the midlines of four triangles that share $M$ as a common vertex and have the four sides of quadrilateral... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,051 |
*7. Investigate the stationery boxes of a class, and obtain the following results:
( I ) Students who have fountain pens do not have ballpoint pens;
(I) Students who have ballpoint pens also have brushes;
(I) Students who have pencils do not have brushes;
(IV) Students who do not have pencils have ballpoint pens.
Based... | 7. $(1) \times(2) \times(3) \vee(4) \vee \quad(5) \vee$ (Hint: If the original proposition is correct, its converse is not necessarily correct. So the second statement is not necessarily correct. From (I) and (N), we can derive that “students without pencils have brushes,” and combining this with (エ), we know that penc... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,052 |
Three, (10 points) Given the multiplication formulas:
$$
\begin{array}{l}
a^{5}+b^{5}=(a+b)\left(a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}\right) . \\
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) .
\end{array}
$$
Using or not using the above formulas, factorize the following expression:
$$
x^{8}... | $$
\begin{array}{l}
x^{8}+x^{6}+x^{4}+x^{2}+1 \\
=\frac{x^{10}-1}{x^{2}-1}=\frac{x^{5}-1}{x-1} \cdot \frac{x^{5}+1}{x+1} \\
=\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right) .
\end{array}
$$
The application of the formula is as follows:
$$
\begin{array}{l}
x^{8}+x^{6}+x^{4}+x^{2}+1 \\
=\frac{x^{10}... | \left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,053 |
* Four (12 points) Let $a_{1}, a_{2}, b_{1}, b_{2}$ be real numbers, $a_{1} \neq a_{2}$, and $\left(a_{1}+b_{1}\right)\left(a_{1}+b_{2}\right)=\left(a_{2}+b_{1}\right)\left(a_{2}+\right.$ $\left.b_{2}\right)=1$. Prove:
$$
\left(a_{1}+b_{1}\right)\left(a_{2}+b_{1}\right)=\left(a_{1}+b_{2}\right)\left(a_{2}+b_{2}\right)=... | Given that the quadratic equation $\left(x+b_{1}\right)\left(x+b_{2}\right) = 1$ has two real roots, these roots are $x=a_{1}, x=a_{2}$.
Obviously, the above equation is the same as $\left(x-a_{1}\right)\left(x-a_{2}\right)=0$, so we have
$\left(x+b_{1}\right)\left(x+b_{2}\right)-1=\left(x-a_{1}\right)\left(x-a_{2}\ri... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,054 |
Five, (12 points) As shown in the figure, in isosceles $\triangle ABC$, $AB = AC$, the vertex angle $A = 20^{\circ}$, and a point $D$ is taken on side $AB$ such that $AD = BC$. Find the measure of $\angle BDC$. | Five, Hint: Construct a regular $\triangle A C E$ outside $\triangle A B C$ with $A C$ as a side, and connect $D E$. Then $\triangle A B C \cong \triangle E A D$, so $\triangle E D C$ is an isosceles triangle. It is easy to find that $\angle B D C=30^{\circ}$. | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,055 |
*Six, (12 points) There are $n$ players
participating in a chess tournament. The scoring method is:
each game the winner gets 2 points, the loser
gets 0 points, and in a draw, both get 1 point. During the
tournament, the score sheet shows that the player with the highest score has $k$ points. Prove:
at this point, ther... | Six, let the sum of all players' scores at this time be $S$.
According to the problem, $S \leqslant k n$.
Assume that the player who has played the least number of games has played $m$ games. At this point, the total number of games played should be $\geqslant \frac{m n}{2}$.
Each game increases the total score by 2 po... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,056 |
Example 4. Does there exist 21 positive integers, each of which is divisible by at least one prime number not less than and not greater than 13? (15th United States of America Mathematical Olympiad) | Let the 21 consecutive positive integers be
$$
\begin{array}{l}
N-10, N-9, N-8, \cdots, N-1, N, \\
N+1, \cdots, N+10 .
\end{array}
$$
If $N=2 \times 3 \times 5 \times 7 k$, then
$$
N-10, \cdots, N-2, N, N+2, \cdots, N+10
$$
each of these is divisible by at least one of $2, 3, 5, 7$.
Next, we aim to make $N-1$ divisi... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,057 |
1. Let $P=\left(\frac{1}{3}\right)^{\frac{1}{5}}, Q=\left(\frac{1}{4}\right)^{\frac{1}{3}}, R=\left(\frac{1}{5}\right)^{\frac{1}{4}}$. Then the size relationship of $P, Q, R$ is ( ).
(A) $P<Q<R$
(B) $P<R<Q$
(C) $R<Q<P$
(D) $Q<R<P$ | 1. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,058 |
2. In the same plane, there are 1994 lines: $a_{1}, a_{2}$, $\cdots, a_{1994}$. If $a_{1} \perp a_{2}, a_{2} / / a_{3}, a_{3} \perp a_{4}, a_{4} / / a_{5}$, $\cdots$, then the positional relationship between $a_{1}$ and $a_{1994}$ is ( ).
(A) Coincident
(B) Parallel
(C) Perpendicular
(D) Intersecting but not perpendicu... | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,059 |
⒊ The minimum value of $x$ that satisfies the inequality $(a-3) x-3 x^{2} \geqslant a^{2}$ $-2 a$ is -1. Then the value of $a$ is ( ).
(A) -1 or 0
(B) -1 or 1
(C) 0 or 1
(D) 0 or -1 or 1 | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,060 |
5. As shown in the figure, in rhombus $A B C D$, $A B=4 a, E$ is on $B C$, $E C=2 a, \angle B A D=120^{\circ}, P$ is a point on $B D$. Then the minimum value of $P E+P C$ is ( ).
(A) $6 a$
(B) $5 a$
(C) $4 a$
(D) $2 \sqrt{3} a$ | 5. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,062 |
1. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+5+\sqrt{x-4}}+\frac{|x+y-z|}{4}=3 \text {, }
$$
then the last digit of $(5 x+3 y-3 z)^{1994}$ is | 1. 4 | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,064 |
* 2. Let $[x]$ denote the greatest integer not exceeding $x$. Then the solution to the equation $x^{2}-x[x]=[x]^{2}$ is | 2. $x_{1}=0, x_{2}=\frac{1+\sqrt{5}}{2}$ | x_{1}=0, x_{2}=\frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,065 |
4. There are four positive numbers, one of which is $\frac{1}{2}$. If any two numbers are taken from these four, among the remaining two numbers, there must be one number such that the sum of these three numbers is 1. Then these four numbers are | 4. $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{2}$ | \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,067 |
Example 5. Let $f(n) \in \mathbb{N}$ be the smallest number such that $\sum_{k=1}^{f(n)} k$ is divisible by $n$. Prove that $f(n) = 2n - 1$ if and only if $n = 2^m$ (where $m$ is a non-negative integer). (1976, New York State Mathematics Olympiad) | Proof (1) First, prove: when $n=2^{m}$, $f(n)=2 n-1$ is the smallest number such that $\sum_{k=1}^{f(n)} k$ is divisible by $n$. When $f(n)=2 n-1$, we have $\sum_{k=1}^{2 n-1} k=n(2 n-1)$. Therefore, $\sum_{k=1}^{2 n-1} k$ is divisible by $n$.
When $l \leqslant 2 n-2$, since $\sum_{k=1}^{l} k=\frac{l(l+1)}{2}$, one of... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,068 |
If $x, y$ are real numbers, and $1 \leqslant x^{2}+4 y^{2} \leqslant 2$. Find the range of $x^{2}-2 x y+4 y^{2}$. | Let $x^{2}+4 y^{2}=m$,
$$
x^{2}-2 x y+4 y^{2}=n.
$$
(1)-(2), we get $x \cdot 2 y=m-n$.
From (1),
$$
(x+2 y)^{2}=m+4 x y=3 m-2 n,
$$
thus $x+2 y= \pm \sqrt{3 m-2 n}$.
Therefore, $x, 2 y$ are the two real roots of the equation $t^{2} \pm \sqrt{3 m-2 n} t+(m-n)=0$.
By $\Delta=3 m-2 n-4(m-n) \geqslant 0$ and $3 m-2 n \geq... | \left[\frac{1}{2}, 3\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,069 |
In $\triangle A B C$, $A B=A C, O$ is the circumcenter of $\triangle A B C$, $D$ is the midpoint of $A B$, and $E$ is the centroid of $\triangle A C D$. Prove: $O E \perp C D$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $F$ be the midpoint of $AC$, then $E$ is on $DF$, and $\frac{DE}{DF} = \frac{2}{3}$.
Let $G$ be the midpoint of $BC$, $AG$ intersects $CD$ at $H$, then $H$ is the centroid of $\triangle ABC$.
Connect $FG$ intersecting $CD$ at $I$, then $I$ is the midpoint of $CD$.
$$
\begin{array}{c}
\therefore \quad \frac{DH}{DI... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,070 |
Three, how many five-digit numbers contain the digit 6 and are divisible by 3?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, among the 90000 five-digit numbers from 10000 to 90999, it is known that there are 30000 numbers divisible by 3.
Below, we will find the number of five-digit numbers that are divisible by 3 and do not contain the digit 6. We start by discussing the possibilities for each digit.
For the highest place, it cannot... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,071 |
1. In the polynomial of $x$ obtained by expanding $(\sqrt{3} x+\sqrt[3]{2})^{100}$, the number of terms with rational coefficients is ( ).
(A) 50
(B) 17
(C) 16
(D) 15 | -、1. The general term of the expansion of $\mathrm{B}(\sqrt{3} x+\sqrt[3]{2})^{100}$ is
$$
T_{r+1}=C_{100}^{r} 3^{50-\frac{r}{2}} \cdot 2^{\frac{r}{3}} x^{100-r} \text {. }
$$
Since $C_{100}^{\prime}$ is an integer, to make the coefficient of a certain term in the polynomial of $x$ a rational number, it is only necess... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,072 |
2. Given that $z$ satisfies $|z+5-12 i|=3$. Then the maximum value of $|zi|$ is ( ).
(A) 3
(B) 10
(C) 20
(D) 16 | 2. D
$|z+5-12 i|=3 \Rightarrow z$ corresponds to points on the circle with center $(-5,12)$ and radius 3, therefore,
$$
|z|_{\max }=\sqrt{5^{2}+12^{2}}+3=16 .
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,073 |
3. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ is the midpoint of $C D$, and $F$ is the midpoint of $A A_{1}$. The area of the section through $E$, $F$, and $B_{1}$ is ( ${ }^{\prime}$ ).
(A) $\frac{55 \sqrt{5}}{192} a^{2}$
(B) $\frac{\sqrt{29}}{4} a^{2}$
(C) $\frac{11 \sqrt{29}}{48} a^{2}$... | 3. $C$ As shown in the figure, $B_{1} F=\frac{\sqrt{5}}{2} a, M N \underline{\mathbb{L}} B F, C M=\frac{1}{2}$. $A_{1} F=\frac{1}{4} a, C_{1} M=\frac{3}{4} a$, so, $M B_{1}=\frac{5}{4} a$. Connect $M F$, let $\angle M B_{1} F=\theta$,
$$
\begin{array}{l}
M F \\
\sqrt{(\sqrt{2} a)^{2}+\left(\frac{1}{4} a\right)^{2}} \\
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,074 |
* 4. The number of solutions to the equation $\sin ^{x} \alpha+\cos ^{x} \alpha=1\left(0<\alpha<\frac{\pi}{2}\right)$ is ( ).
(A) 0
(B) 1
(C) 2
(D) greater than 2 | 4. B Since $0<\sin \alpha, \cos \alpha<1$, the function $f(x)$ $=\sin ^{2} \alpha+\cos ^{x} \alpha$ is decreasing, so the original equation will not have more than one root. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,075 |
*5. Let $a$ be a positive integer, $a<100$, and $a^{3}+23$ is divisible by 24. Then, the number of such $a$ is ( ).
(A) 4
(B) 5
(C) 9
(D) 10 | 5. B Only consider $24 \mid\left(a^{3}-1\right)$, we have
$$
\begin{aligned}
a^{3}-1 & =(a-1)\left(a^{2}+a+1\right) \\
& =(a-1)[a(a+1)+1] .
\end{aligned}
$$
Since $a(a+1)+1$ is odd, for $24 \mid\left(a^{3}-1\right)$ to hold, it must be that $2^{3} \mid(a-1)$. If $a-1$ is not divisible by 3, then $3 \mid a(a+1)$, which... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,076 |
6. From $1,2, \cdots, 49$, choose six different numbers, among which at least two are adjacent. The total number of ways to choose is ( ).
(A) $C_{49}^{6}$
(B) $C_{44}^{6}$
(C) $C_{49}^{6}-C_{44}^{5}$
(D) $C_{19}^{6}-C_{44}^{6}$ | 6. D Let $a_{1}, a_{2}, \cdots, a_{6}$ be six different numbers taken from $1,2, \cdots, 49$. Without loss of generality, assume $a_{1}<a_{2}<\cdots<a_{6}$. Clearly, $a_{1} \leqslant a_{2}-1 \leqslant a_{3}-2 \leqslant a_{4}-3 \leqslant a_{5}-4 \leqslant a_{6}-5$, and the necessary and sufficient condition for $a_{1}, ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,077 |
1. Given $z_{1}, z_{2}$ correspond to points $P, Q$ on the complex plane, and $\left|z_{2}\right|=4,4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0$. Then the area of $\triangle O P Q$ formed by $P, Q$ and the origin $O$ is equal to | $$
\begin{array}{l}
\therefore\left|z_{1}\right|=\left|\frac{1 \pm \sqrt{3} i}{4}\right| \cdot\left|z_{2}\right|=2 . \text { and } \\
\arg \left(\frac{z_{1}}{z_{2}}\right)=\arg \frac{1 \pm \sqrt{3} i}{4}= \pm \frac{\pi}{3} . \therefore \angle P O Q=\frac{\pi}{3} . \\
\text { Therefore, } S_{\triangle O P Q}=\frac{1}{2}... | 2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,078 |
Example 6. Can a set $S$ containing 1990 natural numbers be found such that
(1) any two numbers in $S$ are coprime;
(2) the sum of any $k(\geqslant 2)$ numbers in $S$ is a composite number. (1990, China National Training Team Mock Exam) | Obviously, $S^{\prime}=\{5,4,21\}$ satisfies all the conditions of the problem. If we can generate another element based on this, such that after adding one element, the set of 4 elements also meets the conditions, and then use these 4 elements as the basis to generate the 5th element, and continue in this way, we can ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,079 |
2. Let the volume of tetrahedron $ABCD$ be $V$, $E$ be the midpoint of edge $AD$, and $F$ be on the extension of $AB$ such that $BF = AB$. The plane through points $C, E, F$ intersects $BD$ at $G$. Then the volume of tetrahedron $CDGE$ is $\qquad$ | 2. $\frac{V}{3}$ In $\triangle A F D$, since point $G$ is the intersection of median $B D$ and $E F$, $\therefore D G=\frac{2}{3} B D$.
$$
\begin{array}{l}
\because D E=\frac{1}{2} A D, \\
\therefore S_{\triangle D G E} \\
=\frac{1}{3} S_{\triangle A B D},
\end{array}
$$
Since $V_{C-D G E}$ and $V_{C-A B D}$ have the ... | \frac{V}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,080 |
3. Let $x, y, z>0$ and $x+y+z=1$. Then $\frac{1}{x}$ $+4+9$ has the minimum value of $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 36
$$
\begin{aligned}
\text { Given } & \left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z) \\
& \geqslant(1+2+3)^{2}=36 \\
\quad & \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant \frac{36}{x+y+z}=36 .
\end{aligned}
$$
Equality holds if and only if $x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}$. Therefore, the min... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,081 |
* 5. For a positive integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{1994}=$ $\qquad$. | 5. 7
$$
\begin{aligned}
1994 & =4 \times 498+2, \\
a_{1994} & =f\left(2^{4 \times 198+9}-1\right)=f\left(2^{4 \times 198} \times 8-1\right) \\
& =f\left(16^{448} \times 8-1\right)=f(6 \times 8-1)=7 .
\end{aligned}
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,083 |
* 6. Among $n(\geqslant 3)$ lines, exactly $p(\geqslant 2)$ are parallel, and no three lines among the $n$ lines intersect at the same point. Then the number of regions into which these $n$ lines divide the plane is | 6. $\frac{1}{2}\left(n^{2}-p^{2}+n+p+2\right) \quad p$ parallel lines divide the plane into $p+1$ regions. When adding $r$ of the remaining $n-p$ lines, let the plane be divided into $b_{p+r}$ regions. If another line is added, this line will intersect with the others at $p+r$ points, thus being divided into $p+r+1$ li... | \frac{1}{2}\left(n^{2}-p^{2}+n+p+2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,084 |
Given the equation of a circle as $x^{2}+y^{2}=4$. Try to find two points $A(s, t), B(m, n)$ on the coordinate plane, such that the following two conditions are satisfied:
(1) The ratio of the distance from any point on the circle to point $A$ and to point $B$ is a constant $k ;$.
(2) $s>m, t>n$, and $m, n$ are both na... | Let any point on the circle be $P(x, y)$, then
$$
\frac{\sqrt{(s-x)^{2}+(1-y)^{2}}}{\sqrt{(m-x)^{2}+(n-y)^{2}}}=k .
$$
Taking $P_{1}(2,0), P_{2}(-2,0)$, we get
$$
\left.\begin{array}{l}
\frac{(s-2)^{2}+t^{2}}{(m-2)^{2}+n^{2}}=k^{2} \\
\frac{(s+2)^{2}+t^{2}}{(m+2)^{2}+n^{2}}=k^{2}
\end{array}\right\} \Rightarrow S=k^{2... | (2,2),(1,1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,085 |
*Two, find the real-coefficient polynomial $f(x)$ that satisfies the following conditions:
(1) For any real number $a$, we have
$$
f(a+1)=f(a)+f(1) \text {; }
$$
(2) There exists a real number $k_{1} \neq 0$, such that $f\left(k_{1}\right)=k_{2}$, $f\left(k_{2}\right)=k_{3}, \cdots, f\left(k_{n-1}\right)=k_{n}, f\left(... | II. Taking $a=0$, from (1) we get $f(1)=f(0)+f(1)$. Thus, $f(0)=0$, and $f(x)$ does not contain a constant term. We denote
$$
f(x)=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}
$$
From condition (1), we know
$$
\begin{array}{l}
a_{1}(1+x)+a_{2}(1+x)^{2}+\cdots+a_{n}(1+x)^{n} \\
=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+a_{1}+a... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,086 |
*Three, the sum of all divisors of a positive integer $n$ is denoted by $f(n)$, (for example, $f(4)=1+2+4=7$). Try to answer the following questions:
(1) Prove: If $m$ and $n$ are coprime, then
$$
f(m n)=f(m) f(n) ;
$$
(2) When $a$ is a divisor of $n$ ($a<n$), and $f(n)=n+a$. Try to prove that $n$ is a prime number. Fu... | (1) Let the divisors of $m$ be $1=m_{1}, m_{2}, \cdots, m_{p}=m$, and the divisors of $n$ be $1=n_{1}, n_{2}, \cdots, n_{q}=n$. Since $m, n$ are coprime, the divisors of $m n$ are $m_{i} n_{j}(i=1,2, \cdots, p ; j=1,2, \cdots, q)$. Therefore,
$$
\begin{aligned}
f(m, n)= & m_{1} n_{1}+\cdots+m_{1} n_{q}+m_{2} n_{1}+\cdo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,087 |
* Four, the sequence $1,1,3,3,3^{2}, 3^{2}, \cdots, 3^{1992}, 3^{1992}$ consists of two 1s, two 3s, two $3^{2}$s, ..., two $3^{1992}$s arranged in ascending order. The sum of the terms in the sequence is denoted as $S$. For a given natural number $n$, if it is possible to select some terms from different positions in t... | For each $k(1 \leqslant k \leqslant 1992)$, it is easy to see that the sum of the first $2k$ terms of the sequence is less than $3^k$, so terms of the form $3^k$ must be chosen from two $3^k$ terms. Thus, selecting one $3^k$ has two methods, and selecting two $3^k$ has only one method.
For each number $n$ in the set $... | 4^{1993}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,088 |
Example 2. Given a real number $c>1$. Prove: the two numbers $\sqrt{c}-$ $\sqrt{c-1}$ and $\sqrt{c+1}-\sqrt{c}$ are always one larger and the other smaller. (Canadian Mathematical Competition) | Solve: From $(\sqrt{c})^{2}-$ $(\sqrt{c-1})^{2}=(\sqrt{c+1})^{2}$ $-(\sqrt{c})^{2}=1$, we are inspired to construct a right triangle $\triangle ABC$ (as shown in the figure), such that $BC=1, AB=\sqrt{c+1}, CD=\sqrt{c-1}$, then $AC$ $=BD=\sqrt{c}$. By $AD+BD>AB$, we obtain the desired result. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,089 |
The following problem is from the third United States of America Mathematical Olympiad (1974) ${ }^{(1)}$ :
In the triangles $\triangle A B C$ and $\triangle P Q R$ shown in the figure below, the lengths of the segments are as indicated, and in $\triangle A B C$, $\angle A D B=$ $\angle B D C=\angle C D A=120^{\circ}$... | Proof 1: Let the side length of $\triangle PQR$ be $u+v+w$. On the base passing through $PQ$, intercept $QD_1 = u, PD_2 = v$. Draw lines parallel to one side of the equilateral triangle through $D_1$ and $D_2$: $D_1E_1 \parallel QR, D_2E_2 \parallel PR$. If $D_1E_1$ intersects $D_2E_2$ at $M$, connect $MP, MQ, MR$. It ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,090 |
Example 3. Let $\approx \in C$. Solve the equation:
$$
\frac{1}{2}(z-1)=\frac{\sqrt{3}}{2}(1+z) i \text {. }
$$ | Solve the equation $(1-\sqrt{3} i) z=1+\sqrt{3} i$, knowing $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$, and $\omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i$. Then the equation can be transformed into $\omega z=\omega^{2}$, thus obtaining the solution $z=\omega$. | z=\omega | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,091 |
Example 4. If $\operatorname{tg}(A+B)=3 \operatorname{tg} A$, prove: $\sin 2(A+B)+\sin 2 A=2 \sin 2 B$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | To prove the identity and given the conditional equation with different function names and unevenly distributed angles, to find the sum, start from satisfying the entire condition:
$$
\begin{array}{r}
\operatorname{tg}(A+B)=3 \operatorname{tg} A \text { can be transformed as } \\
\frac{\sin (A+B) \cos A}{\cos (A+B) \si... | null | Algebra | proof | Yes | Yes | cn_contest | false | 707,092 |
Initially 17. M $1,2,3,4, \cdots, 1994$ are 1994 numbers from which $k$ numbers are arbitrarily selected, such that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
In the 1994 numbers $1,2,3,4, \cdots, 1994$, arbitrarily select $k$ numbers, so ... | Solve for the maximum value of $k$ being 11.
(1) From $A=\{1,2,3,4, \cdots, 1994\}$, select $\{1,2,4,8,16,32,64,128,256,512,1024\}$, a total of 11 numbers, using the recursive formula: $a_{1}=1, a_{n+1}=2 a_{n}$. Any two numbers $a_{i}, a_{j}$ $(i<j)$ in this set satisfy $a_{i}<a_{j}$, and $2 a_{i} \leqslant a_{j}$. Th... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,093 |
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