problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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3. Points with integer x-coordinates are called "sub-integer points". Lines are drawn through any two sub-integer points on the curve $y=\sqrt{9-x^{2}}$. The number of lines with an inclination angle greater than $30^{\circ}$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 15 | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,969 |
4. There is a regular tetrahedron and a regular square pyramid, all of whose edges are equal. After overlapping one side, the resulting geometric body is ( ).
(A) tetrahedron
(B) pentahedron
(C) hexahedron
(D) heptahedron | 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 | 721,970 |
5. Given $I$ is the incenter of $\triangle A B C$, $A C=2, B C$ $=3, A B=4$. If $\overrightarrow{A I}=x \overrightarrow{A B}+y \overrightarrow{A C}$, then the value of $x+y$ is $(\quad)$.
(A) $\frac{1}{3}$
(B) $\frac{2}{3}$
(C) $\frac{4}{9}$
(D) $\frac{5}{9}$ | 5. B.
In $\triangle ABC$, $I$ is the incenter, and $AI$ is extended to intersect $BC$ at point $D$. Then the ratio in which $D$ divides $BC$ is
$$
\lambda=\frac{AB}{AC}=\frac{4}{2}=2.
$$
Thus, $\overrightarrow{AD}=\frac{1}{3} \overrightarrow{AB}+\frac{2}{3} \overrightarrow{AC}$.
Given $BC=3$, then $BD=2$, $DC=1$.
In ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,971 |
6. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, \sqrt{\frac{1}{a_{n}^{2}}+4}=\frac{1}{a_{n+1}}$, and let $S_{n}=\sum_{i=1}^{n} a_{i}^{2}$. If $S_{2 n+1}-S_{n} \leqslant \frac{t}{30}$ holds for any $n$ $\left(n \in \mathbf{N}_{+}\right)$, then the minimum value of the positive integer $t$ is ( ).
(A) 10
(B) 9... | 6. A.
From the given $\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}=4$, we get $a_{n}^{2}=\frac{1}{4 n-3}$. Let $g(n)=S_{2 n+1}-S_{n}$. Then $g(n)-g(n+1)=a_{n+1}^{2}-a_{2 n+2}^{2}-a_{2 n+3}^{2}$
$$
=\frac{1}{4 n+1}-\frac{1}{8 n+5}-\frac{1}{8 n+9}>0,
$$
which means $g(n)$ is a decreasing function.
Therefore, $S_{2 n+1}-S_... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,972 |
Example 8 As shown in Figure 8, take a point $P$ inside $\triangle A B C$ such that $\angle P B A=\angle P C A$,
draw $P D \perp A B$ at point
$D, P E \perp A C$ at point
$E$. Prove: The perpendicular bisector of $D E$ must pass through
the midpoint $M$ of $B C$.
(2006, National
High School Mathematics League Sichuan ... | Prove as shown in Figure 8, let $L, N$ be the midpoints of $PB, PC$ respectively. Connect $MD, ME, ML, MN, DL, EN$. Then
$$
ML \Perp \frac{1}{2} PC, MN \Perp \frac{1}{2} PB \text{. }
$$
Since $\angle PDB = \angle PEC = 90^{\circ}$, we know
$$
DL = \frac{1}{2} PB, EN = \frac{1}{2} PC \text{. }
$$
Therefore, $DL = MN, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,974 |
8. The inequality $x^{2}+p x>4 x+p-3$ holds for all $0 \leqslant p \leqslant 4$. Then the range of real numbers $x$ is $\qquad$ . | 8. $x3$.
Hint: Convert the original inequality into a linear inequality about $p$ to get $p(x-1)+x^{2}-4 x+3>0$. Let $f(p)=p(x-1)+x^{2}-4 x+3$.
Then we only need $\left\{\begin{array}{l}f(0)>0, \\ f(4)>0 .\end{array}\right.$ | x<1 \text{ or } x>3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,975 |
9. Place 3 identical white balls, 4 identical red balls, and 5 identical yellow balls into three different boxes, allowing some boxes to contain balls of different colors. The total number of different ways to do this is (answer in numbers).
允许有的盒子中球的颜色不全的不同放法共有种 (要求用数字做答).
Allowing some boxes to contain balls of diff... | $\begin{array}{l}\text { 9. } 3150 \text {. } \\ \mathrm{C}_{5}^{2} \cdot \mathrm{C}_{6}^{2} \cdot \mathrm{C}_{7}^{2}=3150 \text {. }\end{array}$ | 3150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,976 |
10. If $0<x \leqslant 1, a=\left(\frac{\sin x}{x}\right)^{2}, b=\frac{\sin x}{x}$, $c=\frac{\sin x^{2}}{x^{2}}$. Then the size relationship of $a, b, c$ is $\qquad$ . | 10. $a<b \leqslant c$.
When $0<x \leqslant 1 \leqslant \frac{\pi}{2}$, $\sin x<x$. Therefore, $0<\frac{\sin x}{x}<1$.
Thus, $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x}{x}$, which means $a<b$.
Since $y=\frac{\sin x}{x}$ is a decreasing function on $(0,1]$, we have $\frac{\sin x}{x}<\frac{\sin x^{2}}{x^{2}}$, which... | a<b \leqslant c | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 721,977 |
11. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$ | 11. 9 .
Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, therefore, the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is
$$
\begin{array}{l}
1-(\sqrt{2}-\sqrt{3})^{2010} . \\
\text { Also } 0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300} \text {, then } \\
0.9<1-(\sqrt{2}-\sqrt{3})^... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,978 |
13. If $a, b, c \in(0,+\infty)$, prove:
$$
\frac{b+c}{2 a}+\frac{c+a}{2 b}+\frac{a+b}{2 c} \geqslant \frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b} .
$$ | Three, 13. From $(b+c)^{2}=b^{2}+c^{2}+2 b c \geqslant 4 b c$
$$
\Rightarrow \frac{b+c}{b c} \geqslant \frac{4}{b+c} \Rightarrow \frac{1}{b}+\frac{1}{c} \geqslant \frac{4}{b+c} \text {. }
$$
Since $a>0$, we have $\frac{a}{b}+\frac{a}{c} \geqslant \frac{4 a}{b+c}$.
Similarly, $\frac{b}{c}+\frac{b}{a} \geqslant \frac{4 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,980 |
14. If the function $f(x)$ defined on the set $A$ satisfies: for any $x_{1}, x_{2} \in A$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right],
$$
then the function $f(x)$ is called a concave function on $A$.
(1) Determine whether $f(x)=3 x^{2}+x$ is... | 14. (1) Given $f_{1}(x)=3 x_{1}^{2}+x_{1}$,
$$
\begin{array}{l}
f_{2}(x)=3 x_{2}^{2}+x_{2}, \\
f\left(\frac{x_{1}+x_{2}}{2}\right)=3\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\frac{x_{1}+x_{2}}{2},
\end{array}
$$
then $f\left(\frac{x_{1}+x_{2}}{2}\right)-\frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right]$
$$
... | m \geqslant 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,981 |
15. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}>0$, and
$$
a_{n+1}=\sqrt{\frac{3+a_{n}}{2}} \text {. }
$$
(1) Find the range of $a_{1}$ such that $a_{n+1}>a_{n}$
for any positive integer $n$;
(2) If $a_{1}=4$, let $b_{n}=\left|a_{n+1}-a_{n}\right|(n=1$, $2, \cdots)$, and let $S_{n}$ denote the sum of the f... | 15. (1) Research
$$
\begin{array}{l}
a_{n+1}-a_{n}=\sqrt{\frac{3+a_{n}}{2}}-\sqrt{\frac{3+a_{n-1}}{2}} \\
=\frac{a_{n}-a_{n-1}}{2\left(\sqrt{\frac{3+a_{n}}{2}}+\sqrt{\frac{3+a_{n-1}}{2}}\right)}(n \geqslant 2) .
\end{array}
$$
Notice that $\sqrt{\frac{3+a_{n}}{2}}+\sqrt{\frac{3+a_{n-1}}{2}}>0$.
Therefore, $a_{n+1}-a_{... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,982 |
16. As shown in Figure 1, in $\triangle ABC$, $AB=AC$, there is a circle that is internally tangent to the circumcircle of $\triangle ABC$, and is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively. Prove: The midpoint $O$ of segment $PQ$ is the incenter of $\triangle ABC$. | 16. Let the center of the small circle be $O_{1}$, and $\odot O_{1}$ is tangent to the circumcircle of $\triangle ABC$ at point $D$. Connect $AO_{1}$. Clearly, $AO_{1} \perp PQ$, and $\triangle ABC$ is an isosceles triangle. Therefore, $AO_{1}$ passes through the circumcenter of $\triangle ABC$, and point $D$ lies on t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,983 |
17. A set $S$ of points in space satisfies the property: the distances between any two points in $S$ are all different. Assume the coordinates $(x, y, z)$ of the points in $S$ are all integers, and $1 \leqslant x, y, z \leqslant n$. Prove: the number of elements in the set $S$ is less than
$$
\min \left\{(n+2) \sqrt{\f... | 17. Let $|S|=t$. Since the distance between any integer points satisfying $1 \leqslant x, y, z \leqslant n$ does not exceed the distance between $(1,1,1)$ and $(n, n, n)$, for any $\left(x_{1}, y_{1}, z_{1}\right)$, $\left(x_{2}, y_{2}, z_{2}\right) \in S$, we have
$$
\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\righ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,984 |
Example 9 Given that $AB$ is the diameter of semicircle $\odot O$, and $P$ is any point on diameter $AB$. With point $A$ as the center and $AP$ as the radius, draw $\odot A$, which intersects semicircle $\odot O$ at point $C$; with point $B$ as the center and $BP$ as the radius, draw $\odot B$, which intersects semicir... | Prove: As shown in Figure 9, connect $AC$, $AD$, $BC$, and $BD$, and draw perpendiculars from points $C$ and $D$ to $AB$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $CE \parallel DF$.
Since $AB$ is the diameter of $\odot O$, we have
$$
\angle ACB = \angle ADB = 90^{\circ}.
$$
In right tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,985 |
11. 1. In a country consisting of several cities, some of which are connected by roads, the following conditions are satisfied:
(1) All roads do not intersect;
(2) For any two cities, it is possible to travel from one city to another along the roads (possibly passing through other cities).
It is known that in each cit... | 11. 1. Consider the shortest route $l$ that passes through all cities. Let the starting and ending points of $l$ be $A$ and $B$, respectively, with a length of $N$. Then the numbers on the odometers in cities $A$ and $B$ are $N$. Let $C$ be any other city. Then city $C$ is on $l$. Therefore, the distance from city $C$ ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,986 |
11. 2. Let the sequence $a_{1}, a_{2}, \cdots$ satisfy $a_{1} \in(1,2)$, and $a_{k+1}=a_{k}+\frac{k}{a_{k}}(k=1,2, \cdots)$. Prove: There is at most one pair of positive integers $(i, j)(i<j)$, such that $a_{i}+a_{j}$ is an integer. | 11. 2. Let $b_{k}=a_{k}-k$. Then
$$
b_{k+1}=b_{k}-1+\frac{k}{k+b_{k}}=b_{k}\left(1-\frac{1}{k+b_{k}}\right) \text {. }
$$
From $b_{1}>0$, we get $b_{k}>0$, and $b_{k+1}<b_{k}$.
In particular, $b_{k} \leqslant b_{1}<1$.
Notice that $b_{2}=a_{1}+\frac{1}{a_{1}}-2$ is increasing with respect to $a_{1} \in(1,2)$.
Hence
$$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,987 |
11.3. In tetrahedron $ABCD$, all dihedral angles are not $90^{\circ}$, and the orthocenters of $\triangle ABC$, $\triangle ABD$, and $\triangle ACD$ are collinear. Prove: the center of the circumsphere of the tetrahedron and the midpoints of edges $AB$, $AC$, and $AD$ are coplanar. | 11. 3. Let $A B_{1}$, $A C_{1}$, and $A D_{1}$ be the altitudes of $\triangle A C D$, $\triangle A B D$, and $\triangle A B C$, respectively. Then the orthocenters of these three triangles lie on $A B_{1}$, $A C_{1}$, and $A D_{1}$, respectively, and do not coincide with vertex $A$.
Since they lie on a line $l$, $A B_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,988 |
11. 4. Let $S$ denote the set of all integer points $(x, y)$ in the plane that satisfy $x^{2}+y^{2} \leqslant 10^{10}$. Two players, A and B (starting with A), take turns to designate distinct points $A_{1}, A_{2}, \cdots$ in $S$ such that: $A_{i}(i=1,2, \cdots)$ is not symmetric to $A_{i+1}$ with respect to the origin... | 11.4. Player A has a winning strategy.
For a more general set $S$, prove the proposition: If $S$ is a finite set of points containing the origin and invariant under a $90^{\circ}$ rotation about the origin, then Player A has a winning strategy.
Let the set $S$ consist of $n$ points.
Prove the strengthened proposition ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,989 |
11.5. Given $1<a \leqslant b \leqslant c$. Prove:
$$
\begin{array}{l}
\log _{a} b+\log _{b} c+\log _{c} a \\
\leqslant \log _{b} a+\log _{c} b+\log _{a} c .
\end{array}
$$ | 11. 5. Let \( x = \log_{a} b, y = \log_{b} c \). Then the original inequality becomes
\[
x + y + \frac{1}{x y} \leqslant \frac{1}{x} + \frac{1}{y} + x y .
\]
After rearranging and combining terms, we get
\[
\frac{(x-1)(y-1)(x y-1)}{x y} \geqslant 0 \text{. }
\]
Since \( x, y \geqslant 1 \), the above inequality clear... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,990 |
11.6. A $10 \times 10$ chessboard has $k$ rooks. A square on the board that can be attacked by a rook is called "dangerous" (the square occupied by the rook itself is also considered dangerous). If removing any rook results in at least one dangerous square becoming safe, find the maximum possible value of $k$. | 11. 6. $k_{\max }=16$.
Consider a chessboard with $k$ rooks satisfying the problem's conditions. There are two cases to consider.
(1) Each row (column) has a rook.
In this case, all squares are dangerous. If there is a row (column) with at least two rooks, removing one of these rooks will still leave all squares dange... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,991 |
11. 7. In $\square A B C D$, $A_{1}$ and $C_{1}$ are points on sides $A B$ and $B C$ respectively. Segments $A C_{1}$ and $C A_{1}$ intersect at point $P$, and the second intersection point of the circumcircles of $\triangle A A_{1} P$ and $\triangle C C_{1} P$ is $Q$, which is located inside $\triangle A C D$. Prove:
... | 11. 7. Let $\omega_{A}$ and $\omega_{c}$ denote the circumcircles of $\triangle A A_{1} P$ and $\triangle C C_{1} P$, respectively. The rays $A Q$ and $C Q$ intersect the sides $C D$ and $A D$ at points $C_{2}$ and $A_{2}$, respectively.
From $A B / / C D$ and the fact that $A, A_{1}, P, Q$ are concyclic, we have
$$
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,992 |
1. Given the following four propositions:
(1) A non-rhombus parallelogram is divided into two pairs of congruent triangles by its diagonals, one pair being obtuse triangles and the other pair being acute triangles:
(2) If there are two points on a line that have equal (non-zero) distances to another line, then the two ... | - 1. A่.
Line segment $AB$ has two axes of symmetry, which are the perpendicular bisector of $AB$, the $y$-axis, and the $x$-axis on which $AB$ lies. Proposition (4) is a true proposition. The rest are false propositions.
For proposition (1), there is a counterexample as shown in Figure 4, in quadrilateral $\square A... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,993 |
2. A person travels from place A to place B, with regular buses running between A and B. The intervals between departures from both places are equal. He notices that a bus heading to A passes by every $6 \mathrm{~min}$, and a bus heading to B passes by every $12 \mathrm{~min}$. Then the interval between bus departures ... | 2. C.
Suppose the buses leave their respective starting stations every $t \mathrm{~min}$
$$
\left\{\begin{array} { l }
{ 6 ( v _ { \text { bus } } + v _ { \text { person } } ) = v _ { \text { bus } } t , } \\
{ 1 2 ( v _ { \text { bus } } - v _ { \text { person } } ) = v _ { \text { bus } } t }
\end{array} \Rightarro... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 721,994 |
3. As shown in Figure 1, the side length of square $A B C D$ is 2. Take any point $G$ on line segment $C D$ to construct square $C E F G$, and connect $B D$, $B F$, and $D F$. When point $G$ moves along line segment $C D$, the area of $\triangle B D F$ ( ).
(A) is greater than 2
(B) is equal to 2
(C) is less than 2
(D)... | 3. B.
Let the side length of square $C E F G$ be $b$. Then $S_{\triangle B E F}=\frac{(2+b) b}{2}=S_{\text {trapezoid } C E F D}$.
Subtracting the common part $S_{\text {trapezoid } C E F H}$, we get $S_{\triangle B C H}=S_{\triangle D H F}$. Therefore, $S_{\triangle B D F}=S_{\triangle B D H}+S_{\triangle D H F}=S_{\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,995 |
Example 10 As shown in Figure $10, M$ and $N$ are the midpoints of sides $AB$ and $CD$ of quadrilateral $ABCD$, respectively. $BN$ intersects $MC$ at point $P$, and $AN$ intersects $MD$ at point $Q$. Prove:
$S_{\text {quadrilateral } M Q N P}$
$$
=S_{\triangle B C P}+S_{\triangle A D Q} .
$$ | Notice that
$$
\begin{array}{l}
S_{\triangle A D Q}=\frac{1}{2} S_{\triangle A D B}-S_{\triangle A Q M}, \\
S_{\triangle B C P}=\frac{1}{2} S_{\triangle A C B}-S_{\triangle M B P} .
\end{array}
$$
Therefore, $S_{\triangle A D Q}+S_{\triangle B C P}$
$$
=\frac{1}{2} S_{\triangle A D B}+\frac{1}{2} S_{\triangle A C B}-S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,996 |
4. If the parabola $y=\frac{x^{2}}{2}-m x+m-1$ intersects the $x$-axis at integer points, then the axis of symmetry of the parabola is ( ).
(A) $x=1$
(B) $x=-1$
(C) $y=1$
(D) Cannot be determined | 4. A.
Let the parabola intersect the $x$-axis at integer points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$ $\left(x_{1}, x_{2} \in \mathbf{Z}, x_{1}<x_{2}\right)$. Then the equation $\frac{x^{2}}{2}-m x+m-1=0$ has integer roots $x_{1}$ and $x_{2}$, yielding the identity
$$
\frac{1}{2} x^{2}-m x+m-1=\frac{1}{2... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,997 |
5. As shown in Figure 2, two isosceles right triangles $A B P$ and $C D P$ are joined together to form a concave quadrilateral $A B C D$. When the triangle $C D P$ is rotated around point $P$ by an acute angle $\theta$, the relationship between $A C$ and $B D$ is ( ).
(A) $A C>B D$
(B) $A C=B D$
(C) $A C<B D$
(D) Canno... | 5. B.
In Figure 2, connect $A C$ and $B D$.
When the triangular board $C D P$ rotates around point $P$ by an acute angle $\theta$, from $\triangle A B P$ and $\triangle C D P$ being isosceles right triangles, we have
$$
\begin{array}{l}
A P=B P, C P=D P . \\
\text { By } \angle A P C=\angle B P D \\
\Rightarrow \trian... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,998 |
6. Four students each write a quadratic equation with real coefficients. The probability that exactly two of the four equations have no real roots is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{4}$
(C) $\frac{1}{5}$
(D) $\frac{3}{8}$ | 6. D.
Each equation either has or does not have real roots, giving two possibilities for each equation. For four equations, there are $2^{4}=16$ possible scenarios. Additionally, the first equation without real roots has four possible choices, and the second equation without real roots has two possible choices, but th... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,999 |
1. The following arrays are all composed of three numbers, they are:
$$
\begin{array}{l}
(1,2,3),(2,4,6),(3,8,11), \\
(4,16,20),(5,32,37), \cdots . \\
\left(a_{n}, b_{n}, c_{n}\right) .
\end{array}
$$
Please write an expression for $a_{n}, b_{n}, c_{n}$
$$
a_{n}=
$$
,$b_{n}=$ $\qquad$ ,$c_{n}=$ $\qquad$ | 2. $a_{n}=n, b_{n}=2^{n}, c_{n}=n+2^{n}$.
From $1,2,3,4,5, \cdots$ conjecture $a_{n}=n$;
From $2,4,8,16,32, \cdots$ conjecture $b_{n}=2^{n}$;
From each set of numbers being “the sum of the first two equals the third” conjecture $c_{n}=n+2^{n}$. | a_{n}=n, b_{n}=2^{n}, c_{n}=n+2^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,000 |
2. In $\triangle A B C$, $A B=13, A C=12, B C=$ 5 , take a point $P$ on $A C$. If through $P$ four lines can be drawn such that each line intercepts a triangle similar to $\triangle A B C$, then the maximum value of $C P$ is $\qquad$ | 2. $\frac{25}{12}$.
As shown in Figure 7, through a point $P$ on $AC$, we can always draw $PD \perp AB$, $PE \parallel BC$, $PG \parallel AB$, to get
$$
\text {Rt } \triangle APD \backsim \text{Rt} \triangle AEP
$$
$\backsim \text{Rt} \triangle PGC \backsim \text{Rt} \triangle ABC$.
The fourth line $PF$ should make
$$... | \frac{25}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,001 |
3. The first 24 digits of $\pi$ are
3. 14159265358979323846264 .
Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Then
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right) \\
\equiv \quad(\bmod 2) .
\end{array}
$$
$(\bmod 2)$. | 3. 0 .
In the first 24 digits of $\pi$, there are 13 odd numbers. When these 13 odd numbers are placed into 12 parentheses, by the pigeonhole principle, there must be two in the same parenthesis, making the difference in this parenthesis even. Thus, the product $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,002 |
4. The integer solutions of the equation $x^{3}+2 x^{2} y=2009$ are
$\qquad$ . | 4. $(1,1004),(-1,1005),(7,17),(-7,24)$.
From the factorization of 2009, we get
$$
\begin{array}{l}
x^{2}(x+2 y)=1^{2} \times 2009=7^{2} \times 41 \\
\Rightarrow\left\{\begin{array} { l }
{ x ^ { 2 } = 1 , } \\
{ x + 2 y = 2 0 0 9 }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ x = 1 , } \\
{ y = 1 0 0 4 , }
\... | (1,1004),(-1,1005),(7,17),(-7,24) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,003 |
一、(20 points) As shown in Figure 3, the diameter $C D$ of $\odot O$ is perpendicular to the chord $A B$ at point $M$. A chord $E F$ is drawn through $M$, with point $E$ inside arc $\overparen{B C}$. Connect $C F$ to intersect $A B$ at point $N$. When $A B$ is translated within $\odot O$, compare the lengths of $E M$ an... | Given that diameter $C D \perp A B$ at point $M$, $M$ is the midpoint of $A B$. When $A B$ is also a diameter, $B M, C M, E M$ are all radii of the circle. By the property "in a right triangle, the hypotenuse is greater than the legs," we have $C N > C M = E M$.
Next, we prove that when $A B$ is not a diameter, we also... | C N > E M | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,004 |
Given $a b c t \neq 0$, satisfying
$$
\left\{\begin{array}{l}
a=t b+c, \\
b=c\left(1+t+t^{2}\right) .
\end{array}\right.
$$
(1) Prove that $a$ is a root of the quadratic equation
$$
c x^{2}+c(b-2 c) x-(b-c)\left(b^{2}+c^{2}\right)=0
$$
When both real roots of the equation are $a$, find the value of $t$.
(2) When $a=15... | (1) It is known that $a, b, c, t$ are all non-zero, so we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
t=\frac{a-c}{b}, \\
b=c\left(1+t+t^{2}\right)
\end{array}\right. \\
\Rightarrow b=c\left[1+\frac{a-c}{b}+\left(\frac{a-c}{b}\right)^{2}\right] \\
\Rightarrow b^{3}=b^{2} c+b c(a-c)+c\left(a^{2}-2 a c+c^{2}\right) ... | t=-\frac{1}{2}, c=1, t=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,005 |
Three. (25 points) The year, month, and day of September 9, 2009, form the auspicious number 20090909, which means "long-lasting and never separating." It also happens to be a prime number that cannot be further decomposed. If numbers containing the prime factor 20090909 are defined as auspicious numbers, please prove ... | $$
\begin{aligned}
\equiv & \text { by } \frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{20090908} \\
= & \left(\frac{1}{1}+\frac{1}{20090908}\right)+\left(\frac{1}{2}+\frac{1}{20090907}\right)+ \\
& \cdots+\left(\frac{1}{10045454}+\frac{1}{10045455}\right) \\
= & \frac{20090909}{1 \times 20090908}+\frac{20090909}{2 \times 2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,006 |
Example 1 Arrange the four fractions $-\frac{1991}{1992} 、-\frac{91}{92} 、-\frac{1992}{1993} 、-\frac{92}{93}$ in ascending order: $\qquad$ | $$
\begin{array}{l}
-\frac{1991}{1992}+1=\frac{1}{1992}, \\
-\frac{91}{92}+1=\frac{1}{92}, \\
-\frac{1992}{1993}+1=\frac{1}{1993}, \\
-\frac{92}{93}+1=\frac{1}{93}. \\
\text { And } \frac{1}{1993}<\frac{1}{1992}<\frac{1}{93}<\frac{1}{92}, \text { so } \\
-\frac{1992}{1993}<-\frac{1991}{1992}<-\frac{92}{93}<-\frac{91}{9... | -\frac{1992}{1993}<-\frac{1991}{1992}<-\frac{92}{93}<-\frac{91}{92} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,007 |
Example 2 If $a=\frac{19951995}{19961996}, b=\frac{19961996}{19971997}$, $c=\frac{19971997}{19981998}$, then ().
(A) $a<b<c$
(B) $b<c<a$
(C) $c<b<a$
(D) $a<c<b$ | Solve: Subtract 1 from each fraction to get
$$
\begin{array}{l}
a-1=-\frac{10001}{19961996}, \\
b-1=-\frac{10001}{19971997}, \\
c-1=-\frac{10001}{19981998} . \\
\text { Since } a-1<b-1<c-1 \\
\Rightarrow a<b<c .
\end{array}
$$
Therefore, the answer is (A). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,008 |
Example 11 Rationalize the denominator and simplify $\frac{2 \sqrt{6}}{\sqrt{3}+\sqrt{2}-\sqrt{5}}$. | $\begin{array}{l}\text { Solve the original expression }=\frac{(3+2 \sqrt{6}+2)-5}{\sqrt{3}+\sqrt{2}-\sqrt{5}} \\ =\frac{(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{5})^{2}}{\sqrt{3}+\sqrt{2}-\sqrt{5}} \\ =\sqrt{3}+\sqrt{2}+\sqrt{5} .\end{array}$ | \sqrt{3}+\sqrt{2}+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,009 |
1. Let the real number
$$
x=\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}} \text {. }
$$
Then $\sqrt{2009 x}=$ $\qquad$ | $$
\begin{array}{l}
\text { Two } \\
x^{3}=(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}})^{3} \\
=40+6(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}) \\
=40+6 x,
\end{array}
$$
From $x^{3}=(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}})^{3}$
That is,
$$
\begin{array}{l}
x^{3}-6 x-40=0 \\
\Rightarrow(x-4... | 14 \sqrt{41} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,010 |
3. If a positive integer is written on each face of a cube, and then a number is written at each vertex, which is equal to the product of the two integers on the faces passing through that vertex, then, when the sum of the numbers at the vertices of the cube is 290, the sum of the numbers on the faces of the cube is | 3. 36 .
Let the numbers on each face of the cube be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$, $x_{6}$, and the numbers written at the vertices be
$$
\begin{array}{l}
x_{1} x_{2} x_{5}, x_{2} x_{3} x_{5}, x_{3} x_{4} x_{5}, x_{4} x_{1} x_{5}, x_{1} x_{2} x_{6}, \\
x_{2} x_{3} x_{6}, x_{3} x_{4} x_{6}, x_{4} x_{1} x_... | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,012 |
4. As shown in Figure $3, \odot O$ has a chord $AB \perp CD$, with the foot of the perpendicular being $G$. A square $CGFE$ is constructed with $CG$ as one side, and point $E$ lies on $\odot O$. If the radius of $\odot O$ is 10, and the distance from the center to the chord $AB$ is 5, then the side length of the square... | 4. $2 \sqrt{19}-4$.
As shown in Figure 13, draw $O M \perp C E$ at point $M$, intersecting $A B$ at point $H$, draw $O N \perp C D$ at point $N$, and connect $O B$ and $O D$. It is easy to see that
$$
\begin{array}{l}
O H=5, \\
O B=O D=10, \\
B H=5 \sqrt{3} .
\end{array}
$$
Assume the side length of the square $C G F... | 2 \sqrt{19}-4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,013 |
One. (20 points) Given an isosceles triangle with a vertex angle less than $60^{\circ}$, the lengths of its three sides are all positive integers. Construct a square outward on each side, such that the sum of the areas of the three squares is 2009. Find the perimeter of this isosceles triangle. | Let the length of the legs of the isosceles triangle be $a$, and the length of the base be $b$. According to the problem, we have
$$
2 a^{2}+b^{2}=2009 \text{. }
$$
Therefore, $a^{2}=\frac{2009-b^{2}}{2}\frac{2009}{3}>669$.
Thus, $669<a^{2}<1005 \Rightarrow 25<a<32$.
Since $a$ and $b$ are positive integers, it is veri... | 77 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,014 |
II. (25 points) As shown in Figure 4, given that $\odot O_{1}$ and $\odot O_{2}$ are separated, from point $O_{1}$, draw tangents $O_{1} A$ and $O_{1} B$ ($A$, $B$ are the points of tangency) intersecting $\odot O_{1}$ at points $E$ and $F$; from point $O_{2}$, draw tangents $O_{2} C$ and $O_{2} D$ ($C$, $D$ are the po... | Proof 1 As shown in Figure 14, connect $O_{1} C, O_{2} A, C E, A G, A C$.
Since $O_{1} A$ is tangent to $\odot O_{2}$ at point $A$ and $O_{2} C$ is tangent to $\odot O_{1}$ at point $C$, we have
$\angle O_{1} A O_{2} = \angle O_{1} C O_{2} = 90^{\circ}$.
Thus, $O_{1}, O_{2}, A, C$ are concyclic.
Therefore, $\angle A O_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,015 |
Three. (25 points) Given distinct positive real numbers $x, y, z$ satisfying
$$
\begin{array}{l}
(x+y)^{2}-9=x y, \\
(y+z)^{2}-5=y z, \\
(z+x)^{2}-4=z x .
\end{array}
$$
Prove: (1) $\frac{16}{(x-z)^{2}}=\frac{1}{(y-x)^{2}}=\frac{25}{(y-z)^{2}}=(x+y+z)^{2}$;
(2) $x+y+z=\sqrt{9+\sqrt{60}}$. | (1) From equations (1), (2), and (3), we have
$$
\begin{array}{l}
x^{2}+y^{2}+x y=9, \\
y^{2}+z^{2}+y z=5, \\
z^{2}+x^{2}+z x=4 .
\end{array}
$$
Adding (4), (5), and (6) gives
$$
2(x+y+z)^{2}-3(x y+y z+z x)=18 \text {. }
$$
Subtracting (5) from (4) gives
$$
\begin{array}{l}
x^{2}+x y-z^{2}-y z=4 \\
\Rightarrow(x+y+z)(x... | x+y+z=\sqrt{9+\sqrt{60}} | Algebra | proof | Yes | Yes | cn_contest | false | 722,016 |
1. A deck of playing cards, excluding the big and small jokers, has a total of 52 cards. After shuffling, four people take turns to draw 13 cards each. The probability that the two red $\mathrm{A}$ s (i.e., the Heart A and the Diamond A) are in the same person's hand is $\qquad$ . | $-1 . \frac{4}{17}$.
Notice that, after the cards are shuffled, each person's cards are determined, i.e., the 52 cards are arranged in 52 positions. Let the four groups of card numbers be:
$$
\begin{array}{l}
1,5,9,13, \cdots, 49 ; 2,6,10,14, \cdots, 50 ; \\
3,7,11,15, \cdots, 51 ; 4,8,12,16, \cdots, 52 .
\end{array}
$... | \frac{4}{17} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,017 |
2. Given $\odot C:(x-1)^{2}+(y-2)^{2}=25$ and the line $l:(2 a+1) x+(a+1) y=4 a+4 b(a, b \in \mathbf{R})$. If for any real number $a$, the line $l$ always intersects with $\odot C$, then the range of values for $b$ is $\qquad$. | 2. $\left[\frac{3-\sqrt{5}}{4}, \frac{3+\sqrt{5}}{4}\right]$.
The center of the circle $C(1,2)$, radius $r=5$.
By the distance formula from a point to a line, the distance from the center $C$ to the line $l$ is
$$
\begin{aligned}
d & =\frac{|(2 a+1)+2(a+1)-4 a-4 b|}{\sqrt{(2 a+1)^{2}+(a+1)^{2}}} \\
& =\frac{|4 b-3|}{\... | \left[\frac{3-\sqrt{5}}{4}, \frac{3+\sqrt{5}}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,018 |
3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$ | 3. 9 .
Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$
which means $3^{a} a-3^{\circ} \... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,019 |
Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the give... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,020 |
4. The solution set of the inequality about $x$
$$
2 x^{2}-4 x \sin x+1 \leqslant \cos 2 x
$$
is $\qquad$ . | 4. $\{0\}$.
The original inequality can be transformed into
$$
(x-\sin x)^{2} \leqslant 0 \Rightarrow x-\sin x=0 \text {. }
$$
Construct the function $f(x)=x-\sin x$.
Since $f^{\prime}(x)=1-\cos x \geqslant 0$, the function $f(x)=x-\sin x$ is monotonically increasing on $(-\infty,+\infty)$, and it is easy to see that... | \{0\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,021 |
5. Let the increasing function $f(x)=k x-2$. It is stipulated: given an initial value $a$ of a real number $x$, assign $x_{1}=f(a)$, if $x_{1} \leqslant$ 2012, then assign $x_{2}=f\left(x_{1}\right)$, and so on; if $x_{n-1} \leqslant 2012$, then assign $x_{n}=f\left(x_{n-1}\right)$, otherwise stop the assignment; if $x... | 5. $\left(\frac{1007}{1006},+\infty\right)$.
Obviously, $k>0, x_{1}=k a-2$.
When $n \geqslant 2$, we have $x_{n}=k x_{n-1}-2$.
(1) If $k=1$, then $x_{n}-x_{n-1}=-2$, so $x_{n}=x_{1}-(n-1) \cdot 2 = k a - 2 - (n-1) \cdot 2$.
(2) If $k>0$, and $k \neq 1$, let
$$
\left(x_{n}-\lambda\right)=k\left(x_{n-1}-\lambda\right) \... | \left(\frac{1007}{1006},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,022 |
6. Given $[x]$ represents the greatest integer not exceeding the real number $x$. If $P=\sum_{i=1}^{2010}\left[\frac{6^{i}}{7}\right]$, then the remainder when $P$ is divided by 35 is | 6. 20 .
First, consider $S=\frac{6}{7}+\frac{6^{2}}{7}+\cdots+\frac{6^{2010}}{7}$.
In equation (1), no term is an integer, but the sum of any two adjacent terms is an integer (since $\frac{6^{k}}{7}+\frac{6^{k+1}}{7}=6^{k}$ $(k \in \mathbf{Z})$ is an integer).
If the sum of two non-integer numbers is an integer, then... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,023 |
7. The range of the function $f(x)=x(\sqrt{1+x}+\sqrt{1-x})$ is $\qquad$ | 7. $\left[-\frac{8 \sqrt{3}}{9}, \frac{8 \sqrt{3}}{9}\right]$.
Obviously, the domain of this function is $[-1,1]$.
$$
\begin{array}{l}
\text { Also, } f(-x)=-x(\sqrt{1-x}+\sqrt{1+x}) \\
=-x(\sqrt{1+x}+\sqrt{1-x})=-f(x),
\end{array}
$$
Thus, the function $f(x)$ is an odd function.
Let $\sqrt{1-x^{2}}=t$. Then
$$
\begi... | \left[-\frac{8 \sqrt{3}}{9}, \frac{8 \sqrt{3}}{9}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,024 |
8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$ | 8. 0 .
Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,025 |
9. (14 points) Given $t \in \mathbf{N}_{+}, t \neq 0,1(\bmod 9)$. Arrange all $t$-digit numbers from $\underbrace{11 \cdots 1}_{1 \uparrow}$ to $\underbrace{99 \cdots 9}_{t}$ in any order into a single number, denoted as $A$. Prove that $A$ is not a perfect square. | 9. It is known that the number of $t$-digit numbers is
$$
m=\underbrace{88 \cdots 89}_{(t-1) \uparrow} \text { (numbers). }
$$
Let the $m$ $t$-digit numbers in $A$ be $a_{m}$, $a_{m-1}, \cdots, a_{1}$, i.e.,
$$
A=\sum_{i=1}^{m} a_{i} \times 10^{t(i-1)} .
$$
Thus, $A \equiv \sum_{i=1}^{m} a_{i}(\bmod 9)$.
Let $t \equi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,026 |
10. (15 points) Given that $P$ is a point on the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ with a focal distance of $4 \sqrt{2}$, a line passing through $P$ intersects the two asymptotes of the hyperbola $C$ at points $P_{1}$ and $P_{2}$, and $3 \overrightarrow{O P}=\overrightarrow{O P_{1}}+2 \ov... | 10. (1) As shown in Figure 2, let
$P\left(x_{0}, y_{0}\right)$.
From $3 \overrightarrow{O P}$
$=\overrightarrow{O P_{1}}+$
$$
2 \overrightarrow{\mathrm{OP}_{2}},
$$
we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
3 x_{0}=x_{1}+2 x_{2}, \\
3 y_{0}=y_{1}+2 y_{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}... | x^{2}-y^{2}=4 | Algebra | proof | Yes | Yes | cn_contest | false | 722,027 |
11. (15 points) Given seven different points on a circle, vectors are drawn from any one point to another (for points $A$ and $B$, if vector $\overrightarrow{A B}$ is drawn, then vector $\overrightarrow{B A}$ is not drawn). If the four sides of a convex quadrilateral determined by any four points are four consecutive v... | 11. Let the seven points on the circumference be $P_{1}, P_{2}, \cdots, P_{7}$. The number of vectors starting from point $P_{i} (i=1,2, \cdots, 7)$ is $x_{i} (i=1,2, \cdots, 7)$, then $0 \leqslant x_{i} \leqslant 6$, and
$$
\sum_{i=1}^{7} x_{i}=\mathrm{C}_{7}^{2}=21 \text {. }
$$
First, find the minimum number of "no... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,028 |
One, (50 points) As shown in Figure 1, there is a point $A$ inside the circle $\odot O$ with diameter $B C$, where $A B \neq A C$. The extensions of $B A$ and $C A$ intersect $\odot O$ at points $M$ and $N$ respectively, and the angle bisectors of $\angle M A N$ and $\angle M O N$ intersect at point $R$. Prove:
The cir... | As shown in Figure 6, when $AB > AC$, since $OM = ON$, $\angle ROM = \angle RON$, and $OR = OR$, we have:
$\triangle ROM \cong \triangle RON$
$\Rightarrow RM = RN, \angle RMN = \angle RNM$.
Let the line $AR$ intersect $BC$ at point $P$, and draw $RD \perp AB$ intersecting the extension of $BA$ at point $D$, and draw $R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,029 |
II. (50 points) Let real numbers $a, b, c, \lambda$ satisfy $a \geqslant \lambda > 0$, $b \geqslant \lambda, c \geqslant \lambda$. Prove:
$$
\frac{a}{\sqrt{\lambda b-\lambda^{2}}+c}+\frac{b}{\sqrt{\lambda c-\lambda^{2}}+a}+\frac{c}{\sqrt{\lambda a-\lambda^{2}}+b} \geqslant 2 \text {. }
$$ | Given $a \geqslant \lambda > 0$, by the AM-GM inequality, we have
$$
\begin{array}{l}
\sqrt{\lambda a - \lambda^{2}} = \sqrt{\lambda(a - \lambda)} \\
\leqslant \frac{\lambda + (a - \lambda)}{2} = \frac{a}{2}.
\end{array}
$$
Similarly, $\sqrt{\lambda b - \lambda^{2}} \leqslant \frac{b}{2}, \sqrt{\lambda c - \lambda^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,030 |
1. Given $a=81^{31}, b=27^{41}, c=9^{61}$. Then the size relationship of $a$, $b$, and $c$ is ( ).
(A) $a>b>c$
(B) $a>c>b$
(C) $ac>a$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Hint: Transform $a$, $b$, and $c$ into constants with a base of 3. Answer: (A). | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,031 |
Three, (50 points) Try to find the smallest positive integer $M$, such that the sum of all positive divisors of $M$ is 4896. | Let $M=\prod_{i=1}^{n} p_{i}^{\alpha_{i}}\left(p_{i}\right.$ be a prime, $\alpha_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, n)$. Denote
$$
f(p, \alpha)=\sum_{k=0}^{\alpha} p^{k}\left(p\right.$ be a prime, $\left.\alpha \in \mathbf{N}_{+}\right)$.
From the problem, we know
$$
\prod_{i=1}^{n} f\left(p_{i}, \alpha_{i}\right)=... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,032 |
Four. (50 points) Let $n$ be a positive integer, and let $S_{n}$ be a subset of the set $A_{n}=\left\{m \mid m \in \mathbf{N}_{+}\right.$, and $\left.m \leqslant n\right\}$. Moreover, the difference between any two numbers in $S_{n}$ is not equal to 4 or 7. If the maximum number of elements in $S_{n}$ is denoted by $M_... | It is known that the difference between any two numbers among $1,4,6,7,9$ is not 4 or 7. Adding 11 to each of these numbers gives $12, 15, 17, 18, 20$, which clearly also have the same property, and the difference between any of these numbers and any of the first five numbers is also not 4 or 7. By this reasoning, for ... | 922503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,033 |
In $\triangle A B C$, points $M$ and $N$ are on sides $A B$ and $A C$ respectively, with $A M=A N$. Points $D$ and $E$ are the midpoints of $C M$ and $B N$ respectively, and $B D=C E$. Prove: $A B=A C$. | Prove that, as shown in Figure 2, draw \( M Q \perp A N \) at point \( Q \), and \( N P \perp A M \) at point \( P \). Then
\[
A Q = A P, \quad M Q = N P.
\]
Assume \( A B > A C \), then
\[
B P > C Q.
\]
Thus, \( B N > C M \).
Let \( O \) be the midpoint of \( B C \), and connect \( O D \) and \( O E \). Then
\[
\beg... | A B = A C | Geometry | proof | Yes | Yes | cn_contest | false | 722,034 |
In $\triangle ABC$, the three sides are all integers, $AB > BC > CA, AB = 2AC, \angle BAC$'s bisector intersects $BC$ at point $D$. Three squares are constructed with the sides of $\triangle ABC$ as one side each, and the sum of the areas of the three squares is 2009. Prove: The lengths of $BD$ and $DC$ are both intege... | Proof: Let $AC = x, BC = y$. Then $AB = 2x$.
Obviously, $x < y < 2x$.
According to the problem, we have $(2x)^2 + x^2 + y^2 = 2009$, which simplifies to $y^2 = 2009 - 5x^2$.
From $x < y < 2x$, we know
$$
\begin{array}{l}
x^2 < y^2 = 2009 - 5x^2 < 4x^2. \\
\text{Then } \frac{2009}{9} < x^2 < \frac{2009}{6} \Rightarrow 2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,035 |
265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$
always holds. Try to find the maximum value of $k$. | Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$
Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,036 |
Given $a, b, c > 0$, and $a + b + c = 2S$. Prove that for any integer $n$, we have
$$
\sum \frac{a^{n}}{b+c} \geqslant\left(\frac{2}{3}\right)^{n-2} S^{n-1},
$$
where, “$\sum$” denotes the cyclic sum. | Prove that since the inequality is symmetric with respect to $a, b, c$, we can assume without loss of generality that $a \geqslant b \geqslant c$.
First, prove that when $n$ is a positive integer, the inequality holds. In this case, we have $a^{n} \geqslant b^{n} \geqslant c^{n}$.
When $n=1$, $\sum \frac{a}{b+c} \geqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,037 |
2. $\frac{1}{2002}+\frac{1}{3.003}-\frac{1}{4004}+\frac{1}{6006}-\frac{1}{8008}$ $=(\quad)$.
(A) $\frac{1}{6006}$
(B) $-\frac{3}{7007}$
(C) $\frac{5}{8008}$
(D) $-\frac{7}{9009}$ | Prompt: Transform the denominator of each fraction to
$$
\begin{array}{l}
2 \times 1001, 3 \times 1001, 4 \times 1001, \\
6 \times 1001, 8 \times 1001 .
\end{array}
$$
Answer: $\frac{5}{8008}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,038 |
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2005}\right)\left(1+\frac{1}{2}+\cdots+\frac{1}{2004}\right)- \\
\left(1+\frac{1}{2}+\cdots+\frac{1}{2005}\right)\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2004}\right)
\end{array}
$$ | Prompt: Let $a=\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2005}, b=$ $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2004}$. Answer: $\frac{1}{2005}$. | \frac{1}{2005} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,039 |
4. Calculate: $\frac{1}{11 \times 13 \times 15}+\frac{1}{13 \times 15 \times 17}+\cdots+$ $\frac{1}{29 \times 31 \times 33}=$ $\qquad$ | Hint: Original expression $\times 4$
$$
\begin{aligned}
= & \left(\frac{1}{11 \times 13}-\frac{1}{13 \times 15}\right)+\left(\frac{1}{13 \times 15}-\frac{1}{15 \times 17}\right)+ \\
& \cdots+\left(\frac{1}{29 \times 31}-\frac{1}{31 \times 33}\right) \\
= & \frac{1}{11 \times 13}-\frac{1}{31 \times 33}=\frac{80}{13299} ... | \frac{20}{13299} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,040 |
5. Simplify: $\frac{\sqrt{3}+\sqrt{5}}{3-\sqrt{6}-\sqrt{10}+\sqrt{15}}$. | ```
Divide the denominator into two groups to get
\[
\begin{array}{l}
(3+\sqrt{15})-(\sqrt{6}+\sqrt{10}) \\
=(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{2}) \text {. Answer: } \sqrt{3}+\sqrt{2} .
\end{array}
\]
``` | \sqrt{3}+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,041 |
Example 1 Let the two pairs of opposite sides of the convex quadrilateral $ABCD$ intersect at points $E$ and $F$, and the two diagonals intersect at point $P$. Draw $PO \perp EF$ at point $O$. Prove that:
$$
\angle BOC = \angle AOD.
$$ | Discuss in two cases.
(1) When $B D / / E F$, this problem is easier to handle, and is not discussed here.
(2) When $B D \times E F$, let's assume $D B$ and $F E$ intersect at point $Q$ (as shown in Figure 2).
Extend $A C$ to intersect $E F$ at point $K$, and let $\angle B A D = \alpha, \angle B C D = \beta$. It is eas... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,042 |
Example 2 In $\triangle A B C$, $A B>A C, A M 、 A N$ are the bisectors of $\angle B A C$ and its exterior angle, respectively. A circle $\odot O$ is constructed with $M N$ as its diameter, and point $P$ is inside $\triangle A B C$ and on $\odot O$. Prove:
$$
\angle A P C-\angle A B C=\angle A P B-\angle A C B \text {. ... | As shown in Figure 3, extend $B P$ to point $X$, and connect $P N$. It is easy to see that $\odot O$ is the Apollonian circle with respect to segment $B C$.
Notice that point
$P$ lies on the Apollonian circle,
by the property of the Apollonian circle, we have
$$
\frac{B N}{N C}=\frac{P B}{P C}.
$$
Thus, $P N$ must bis... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,043 |
Example 3 Given an acute triangle $\triangle P B C, P B \neq P C$. Let $A$ and $D$ be points on $P B$ and $P C$ respectively, and let $A C$ intersect $B D$ at point $O$. Draw $O E \perp A B, O F \perp C D$, with $E$ and $F$ being the feet of the perpendiculars, and let the midpoints of segments $B C$ and $A D$ be $M$ a... | Prove: The perpendicular bisector of $EF$ simultaneously bisects sides $BC$ and $AD$, i.e., $ME = MF, NE = NF$.
This naturally leads to $ME \cdot NF = MF \cdot NE$.
The conclusion of part (2) is: $A, B, C, D$ do not necessarily lie on the same circle.
In fact, when $AD \parallel BC$ (it is easy to prove that $A, B, C,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,044 |
Example 3 If $x, y$ are solutions to the system of equations
$$
\left\{\begin{array}{l}
1995 x+1997 y=5989 \\
1997 x+1995 y=5987
\end{array}\right. \text {, }
$$
then $\frac{x^{3} y^{2}}{x^{2}+y^{2}}=$ . $\qquad$ | Solving the given system of equations, we get
$$
\left\{\begin{array}{l}
1995 x+1997 y=1995 \times 1+1997 \times 2, \\
1997 x+1995 y=1997 \times 1+1995 \times 2
\end{array}\right.
$$
Thus, $x=1, y=2$.
Therefore, $\frac{x^{3} y^{2}}{x^{2}+y^{2}}=\frac{1^{3} \times 2^{2}}{1^{2}+2^{2}}=\frac{4}{5}$. | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,045 |
Example 4 In $\triangle A B C$, $A B=A C, A M$ is the altitude, a line $l \perp A M$ is drawn through point $A$, and a line $D N \perp l$ is drawn through a point $D$ on $B C$ at point $N$. Points $E$ and $F$ are on sides $A B$ and $A C$ respectively, such that $\angle B D E=\angle C D F$. Prove that $M N$ bisects $E F... | As shown in Figure 5, draw $E E^{\prime} / / F F^{\prime} / / B C$, with points $E^{\prime}$ and $F^{\prime}$ both on $M N$.
Let $E F$ intersect $M N$ at point $O$, extend $M N$ to intersect $B A$ at point $K$, and let $M K$ intersect $A C$ at point $G$. Connect $D G$ and $D K$.
Notice that $B M = M C$,
$\triangle B D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,046 |
Example 5 Let points $A, B, C, D$ be on the same straight line in sequence, $AB=6, BC=3, CD=2$. Given that point $P$ is outside the line $AD$, and satisfies $\angle APB=\angle BPC=\angle CPD$. Determine the geometric position of point $P$. | As shown in Figure 6, first construct the Apollonius circle $\omega_{1}$ of segment $A C$ with respect to $6: 3$ (or $2: 1$), then construct the Apollonius circle $\omega_{2}$ of segment $B D$ with respect to $3: 2$.
Thus, the intersection point $P$ of circles $\omega_{1}$ and $\omega_{2}$ is the desired point.
By symm... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,047 |
Example 6 As shown in Figure 7, $ABCD$ and $A'B'C'D'$ are maps of the same region of a country drawn to different scales.
When they are overlapped,
Figure 7
prove that there is only one point $O$ on the small map such that the point $O'$ directly below it on the large map represents the same location in the count... | For the convenience of calculation, the figures $ABCD$ and $A'B'C'D'$ given in the problem are both squares, and the letters of the two squares are arranged counterclockwise.
It should be emphasized that the essence of this problem is to find a point $O$ inside the smaller square $A'B'C'D'$, such that $\triangle OA'B'... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,048 |
1. In a convex quadrilateral $ABCD$, $AC \perp BD$, with the foot of the perpendicular being $K$. Extend $AB$ and $DC$ to meet at point $E$, and extend $BC$ and $AD$ to meet at point $F$. Prove that $\angle BKE = \angle DKF$. | If $B D \times X=E F$, extend $B D$ and $E F$ to intersect at point $Y$, and extend $A C$ to intersect $E F$ at point $X$. By Ceva's Theorem and Menelaus' Theorem, it is easy to get $\frac{E X}{X F}=\frac{E Y}{Y F}$. Note that $A C \perp B D$, i.e., $K X \perp K Y$, at this time, the circle with $X Y$ as its diameter i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,049 |
2. In $\triangle A B C$, $A B=4, C A: C B=5: 3$. Try to find $\left(S_{\triangle A B C}\right)_{\max }$. | Construct the Apollonian circle $\odot O$ of $AB$ with the ratio $5:3$, and find the point $C$ on $\odot O$ that is farthest from $AB$. It is easy to see that the diameter $MN=7.5$ of $\odot O$. Draw the radius $OC \perp MN$ through $O$. Then $OC=3.75$. Therefore, $\left(S_{\triangle ABC}\right)_{\max }=7.5$. | 7.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,050 |
3. $\odot O_{1} 、 \odot O_{2}$ are unequal and externally separated. There is a point $P$, which subtends equal angles for $\odot O_{1}$ and $\odot O_{2}$. Determine the geometric position of point $P$. | Draw the inner and outer common tangents of $\odot O_{1}$ and $\odot O_{2}$, which intersect the line segment $O_{1} O_{2}$ at points $A$ and $B$. The circle $\omega$ with diameter $A B$ is the Apollonian circle of the line segment $O_{1} O_{2}$ with respect to $r_{1}: r_{2}$ (where $r_{1}$ and $r_{2}$ are the radii of... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,051 |
4. Through a point $P$ outside $\odot O$, draw $PA$ tangent to $\odot O$ at point $A$, $PO$ intersects $\odot O$ at points $M$ and $N$, $AB \perp MN$ at point $B$, points $E$ and $F$ are both on $\odot O$. Prove: $\frac{PE}{BE}=\frac{PF}{BF}$. | Connect $A M$ and $A N$. It is easy to prove that $A M$ bisects $\angle P A B$, and $A N$ bisects the exterior angle of $\angle P A B$. Therefore, $\frac{P M}{M B}=\frac{P A}{A B}=\frac{P N}{N B}$, which means $\odot O$ is the Apollonian circle of segment $P B$.
Since points $E$ and $F$ are both on $\odot O$, we have,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,052 |
5. In a convex quadrilateral $ABCD$, $AC$ intersects $BD$ at point $O$, and $M$, $N$ are the midpoints of $AB$, $DC$ respectively, and points $M$, $O$, $N$ are collinear. Draw $OE \perp AD$ at point $E$, and $OF \perp BC$ at point $F$. Prove: $ME \cdot NF = MF \cdot NE$. | It is known that $S_{\triangle O D N}=S_{\triangle O C N}, S_{\triangle O B M}=S_{\triangle O A M}$.
$$
\begin{array}{l}
\text { Then } \frac{O D \cdot O N}{O B \cdot O M}=\frac{O C \cdot O N}{O A \cdot O M} \Rightarrow \frac{O D}{O B}=\frac{O C}{O A} \\
\Rightarrow \triangle O C D \backsim \triangle O A B \Rightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,053 |
Given real numbers $a, b \neq 0$, let
$$
x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{a b}{|a b|} \text {. }
$$
Then the sum of the maximum and minimum values of $x$ is $\qquad$ [1] | Question 1 Original Solution ${ }^{[1]}$ According to the number of negative numbers in $a$ and $b$, there are three cases:
(1) If $a$ and $b$ are both positive, then $x=3$;
(2) If $a$ and $b$ are one positive and one negative, then $x=-1$;
(3) If $a$ and $b$ are both negative, then $x=-1$.
In summary, the maximum valu... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,054 |
Question 2 Given that $a, b, c$ are non-zero real numbers, and
$$
M=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b}{|a b|}+\frac{b c}{|b c|}+\frac{c a}{|c a|}+\frac{a b c}{|a b c|} \text {. }
$$
Find the value of $M$. | Solution ${ }^{[2]}$ Considering the "equal status" of $a, b, c$ in $M$, the value of $\frac{a b}{|a b|}+\frac{b c}{|b c|}+\frac{c a}{|c a|}$ depends on the sign of the product of two numbers, and the value of $\frac{a b c}{|a b c|}$ depends on the sign of the product of three numbers. Therefore, we can classify accord... | -1 \text{ or } 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,055 |
Example 4 Let $a=\frac{1994}{1995}, b=\frac{1993}{1994}$. Try to compare the sizes of $a$ and $b$. | Notice that
$$
\begin{array}{l}
\frac{1994}{1995}=\frac{1994^{2}}{1995 \times 1994}, \\
\frac{1993}{1994}=\frac{1993 \times 1995}{1995 \times 1994}=\frac{1994^{2}-1}{1995 \times 1994} .
\end{array}
$$
Therefore, $\frac{1994}{1995}>\frac{1993}{1994}$, i.e., $a>b$. | a > b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,056 |
Example 1 Let
$$
\begin{array}{l}
A=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{d}{|d|}-\frac{a b}{|a b|}-\frac{a c}{|a c|}- \\
\frac{a d}{|a d|}-\frac{b c}{|b c|}-\frac{b d}{|b d|}-\frac{c d}{|c d|}+\frac{a b c}{|a b c|}+ \\
\frac{a b d}{|a b d|}+\frac{a c d}{|a c d|}+\frac{b c d}{|b c d|}-\frac{a b c d}{|a b c d|... | According to Proposition 2, we have
$$
\begin{aligned}
A & =1-\left(1-\frac{a}{|a|}\right)\left(1-\frac{b}{|b|}\right)\left(1-\frac{c}{|c|}\right)\left(1-\frac{d}{|d|}\right) \\
& =\left\{\begin{array}{ll}
1, & \text{if at least one of } a, b, c, d \text{ is positive; } \\
-15, & \text{if all of } a, b, c, d \text{ are... | 1 \text{ or } -15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,057 |
Example 2 Let
$$
\begin{aligned}
B= & \frac{a}{|a|}-\frac{b}{|b|}-\frac{c}{|c|}+\frac{d}{|d|}+\frac{a b}{|a b|}+\frac{a c}{|a c|}- \\
& \frac{a d}{|a d|}-\frac{b c}{|b c|}+\frac{b d}{|b d|}+\frac{c d}{|c d|}+\frac{a b c}{|a b c|}- \\
& \frac{a b d}{|a b d|}-\frac{a c d}{|a c d|}+\frac{b c d}{|b c d|}-\frac{a b c d}{|a ... | $$
\begin{array}{l}
B=1-\left(1-\frac{a}{|a|}\right)\left(1+\frac{b}{|b|}\right)\left(1+\frac{c}{|c|}\right)\left(1-\frac{d}{|d|}\right) \\
=\left\{\begin{array}{ll}
1, & a, d(b, c) \text { at least one is positive (negative) ; } \\
-15, & a, d \text { are both negative, and } b, c \text { are both positive. }
\end{arr... | 1 \text{ or } -15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,058 |
Example 3 Given $a b c<0$, let
$$
P=\frac{a}{|a|}+\frac{|b|}{b}+\frac{c}{|c|}+\frac{|a b|}{a b}+\frac{a c}{|a c|}+\frac{|b c|}{b c} \text {. }
$$
Find the value of $a P^{3}+b P^{2}+c P+2009$. | Given $a b c<0$, we get $\frac{a b c}{|a b c|}=-1$, and at least one of $a, b, c$ is negative. According to the corollary, we can change the position of the absolute value symbol, then
$$
\begin{aligned}
P & =\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b}{|a b|}+\frac{a c}{|a c|}+\frac{b c}{|b c|} \\
& =\left(1+\... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,059 |
Given that $A_{1}, B_{1}, C_{1}$ are points on the sides $BC, CA, AB$ of $\triangle ABC$ respectively, the circumcircles of $\triangle AB_{1}C_{1}, \triangle BC_{1}A_{1},$ and $\triangle CA_{1}B_{1}$ intersect the circumcircle of $\triangle ABC$ at points $A_{2}, B_{2}, C_{2}\left(A_{2} \neq A, B_{2} \neq B, C_{2} \neq... | Prove as shown in Figure 1, connect $A_{2} B$, $A_{2} C$, $A_{2} B_{1}$, and $A_{2} C_{1}$.
From the fact that points $A$, $A_{2}$, $B$, and $C$ are concyclic, we know $\angle A B A_{2} = \angle A C A_{2}$.
From the fact that points $A$, $A_{2}$, $C_{1}$, and $B_{1}$ are concyclic, we know $\angle A C_{1} A_{2} = \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,060 |
Question 1 Let $n$ be a positive integer, $D_{n}$ be the set of all positive divisors of $2^{n} 3^{n} 5^{n}$, $S \subseteq D_{n}$, and any number in $S$ cannot divide another number in $S$. Find the maximum value of $|S|$. ${ }^{[1]}$ | The answer to this question is $\left[\frac{3(n+1)^{2}+1}{4}\right]$, where $[x]$ represents the greatest integer not exceeding the real number $x$.
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。
The answer to this question is $\left[\frac{3(n+1)^{2}+1}{4}\right]$, where $[x]$ represents the greatest integer not exceeding the re... | \left[\frac{3(n+1)^{2}+1}{4}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,061 |
Question 2 Let $n$ be a positive integer. Find the minimum value of $q=f(n)$ for which the following conclusion always holds: If $q$ students all participate in three mock exams of a mathematics competition, and each person's score in each exam is an integer from 0 to $n$, then there must be two students $A$ and $B$, s... | The answer to question 2 is $\left[\frac{3(n+1)^{2}+1}{4}\right]+1$ | \left[\frac{3(n+1)^{2}+1}{4}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,062 |
Question 4 Let $m, n, p$ be given positive integers, and $m \leqslant n \leqslant p \leqslant m+n$. Find the minimum value of $q=f(m, n, p)$ such that the following conclusion always holds: If $q$ students all participate in three mock exams of a mathematics competition, and the score of each student in the first exam ... | $$
\begin{array}{l}
\text { The solution to problem 4, the minimum value of } q \text { is } \\
q_{0}=(m+1)(n+1)-\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+2}{2}\right]+1 . \\
\text { First, we prove that when } q=q_{0}-1 \text {, it does not guarantee the existence of } \\
\text { two students satisfying the requ... | q_{0}=(m+1)(n+1)-\left[\frac{m+n-p+1}{2}\right]\left[\frac{m+n-p+2}{2}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,063 |
1. Given positive real numbers $a$ and $b$ satisfy $ab = a + b$. Then $\frac{a}{b} + \frac{b}{a} - ab = (\quad)$.
(A) -2
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) 2 | $\begin{array}{l}\text { - 1. A. } \\ \frac{a}{b}+\frac{b}{a}-a b=\frac{a^{2}+b^{2}}{a b}-a b \\ =\frac{(a+b)^{2}-2 a b}{a b}-a b \\ =a b-2-a b=-2 .\end{array}$
The translation is as follows:
$\begin{array}{l}\text { - 1. A. } \\ \frac{a}{b}+\frac{b}{a}-a b=\frac{a^{2}+b^{2}}{a b}-a b \\ =\frac{(a+b)^{2}-2 a b}{a b}-... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,064 |
2. Given that $E$ is any point on side $C D$ of square $A B C D$, and $E F \perp A C$ at point $F$. Extend $B F$ to intersect line $A E$ at point $G$. Then $\angle B G C=(\quad)$.
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 2. B.
As shown in Figure 2, connect $D F$.
Since quadrilateral $A B C D$ is a square,
$$
\begin{array}{l}
\Rightarrow \angle D C F=\angle B C F=45^{\circ}, D C=B C \\
\Rightarrow \triangle D F C \cong \triangle B F C \Rightarrow \angle F B C=\angle F D C . \\
\text { Also, } \angle A D E=90^{\circ}, \angle E F A=90^{\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,065 |
3. Given that $B D$ is the median of $\triangle A B C$, $A C=6$, and $\angle A D B=45^{\circ}, \angle C=30^{\circ}$. Then $A B=(\quad)$.
(A) $\sqrt{6}$
(B) $2 \sqrt{3}$
(C) $3 \sqrt{2}$
(D) 6 | 3. C.
As shown in Figure 3, draw a perpendicular from point $B$ to $AC$, intersecting the extension of $CA$ at point $H$. Then
$$
\begin{array}{l}
H D=H B, H C=\sqrt{3} H B \\
\Rightarrow H C-H D=(\sqrt{3}-1) H B=3 \\
\Rightarrow H B=\frac{3}{2}(\sqrt{3}+1), H A=\frac{3}{2}(\sqrt{3}-1) \\
\Rightarrow A B=\sqrt{H A^{2}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,066 |
Example 5 Let $a=\frac{20052005}{20062006}, b=\frac{20062006}{20072007}$, $c=\frac{20072007}{20082008}$.
Try to compare the sizes of $a$, $b$, and $c$. | (2007, Nanning Junior High School Mathematics Competition)
Solve by dividing 1 by $a$ we get
$$
\begin{array}{l}
\frac{1}{a}=\frac{20062006}{20052005} \\
=\frac{2006 \times 10001}{2005 \times 10001} \\
=\frac{2006}{2005}=1+\frac{1}{2005} .
\end{array}
$$
Similarly, $\frac{1}{b}=1+\frac{1}{2006}, \frac{1}{c}=1+\frac{1}... | a<b<c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,067 |
4. If the three roots of the equation $x^{3}-5 x^{2}+(4+k) x-k=0$ can be the three sides of an isosceles triangle, then the value of the real number $k$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. B.
Obviously, $x=1$ is a root of the equation. Then
$$
(x-1)\left(x^{2}-4 x+k\right)=0 \text {. }
$$
According to the condition, $x=1$ is a root of $x^{2}-4 x+k=0$, or $x^{2}-4 x+k=0$ has two equal roots.
Thus, $k=3$ or 4.
When $k=3$, the three roots of the equation are $1, 1, 3$, which cannot be the three sides o... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,068 |
5. In Rt $\triangle A B C$, $C D$ is the altitude on the hypotenuse $A B$, $A C=3, B C=4$, and $r, r_{1}, r_{2}$ represent the radii of the inscribed circles of $\triangle A B C, \triangle A C D, \triangle B C D$ respectively. Then $r+r_{1}+r_{2}=(\quad)$.
(A) $\frac{12}{5}$
(B) 3
(C) 4
(D) 5 | 5. A.
As shown in Figure 4, by the Pythagorean theorem,
$$
A B=5 \text {, }
$$
By the area relationship,
$$
C D=\frac{12}{5} .
$$
The inradius of the right triangle $\triangle A B C$ is $r=\frac{1}{2}(C A+C B-A B)$.
Similarly, $r_{1}=\frac{1}{2}(D C+D A-C A)$,
$$
r_{2}=\frac{1}{2}(D C+D B-C B) \text {. }
$$
Thus, $... | \frac{12}{5} | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,069 |
6. If $x, y$ are positive integers, then the number of positive integer solutions $(x, y)$ for the equation $15 \times 2^{x}+1=$ $y^{2}$ is $(\quad)$.
(A) 0
(B) 1
(C) 2
(D) 4 | 6. C.
Obviously, $y$ is an odd number greater than 1.
Let $y=2k+1\left(k \in \mathbf{N}_{+}\right)$. Then $15 \times 2^{x-2}=k(k+1)$.
Note that $k$ and $k+1$ are one odd and one even.
(1) When $k$ is odd, $k=3,5,15$. By calculation, when $k=3$, there is no solution; when $k=5$, $x=3, y=11$; when $k=15$, $x=6, y=31$.
(... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,070 |
1. Given real numbers $a, b, c$ satisfy
$$
\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}, \frac{c a}{c+a}=\frac{1}{5} \text {. }
$$
then $a b+b c+c a=$ $\qquad$ | $$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{c}+\frac{1}{a}=5 \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6 \\
\Rightarrow \frac{1}{c}=3, \frac{1}{a}=2, \frac{1}{b}=1 \\
\Rightarrow a=\frac{1}{2}, b=1, c=\frac{1}{3} \\
\Rightarrow a b+b c+c a=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,071 |
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