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6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column a... | 6.1.
Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.
Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,936 |
Example 2 There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs a total of 315 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs a total of 420 yuan. Question: How much would it cost to buy one piece each of A, B, and C? | Let the unit prices of A, B, and C be $x$, $y$, and $z$ yuan, respectively. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=315, \\
4 x+10 y+z=420 .
\end{array}\right.
$$
The problem actually only requires finding the value of $x+y+z$, without necessarily solving for $x$, $y$, and $z$ indi... | 105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,937 |
Example 3 In the interior of square $ABCD$, there are 1999 points. Using the four vertices of the square and the 1999 interior points as vertices, it is divided into some triangles. How many triangles can be formed in total? How many cuts are needed in total?
---
The translation maintains the original text's format a... | Solving from an overall perspective, first calculate the sum of the interior angles of all triangles.
In fact, the sum of the angles converging at a point inside a square is $360^{\circ}$, and the sum of the interior angles of the square is also $360^{\circ}$. Therefore, we have
$$
360^{\circ} \times 1999 + 360^{\circ... | 4000 \text{ triangles, } 5998 \text{ cuts} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,938 |
4. Place the integers $1,2, \cdots, 2 n$ in any order in $2 n$ positions labeled $1,2, \cdots, 2 n$, and then add the label of the position to each number. Prove: There will always be two numbers that are congruent modulo $2 n$. | Prompt: Use proof by contradiction. Assume that all remainders 0, 1, ..., 2n-1 appear. The sum of all integers and their position index numbers is
$$
\begin{array}{l}
S_{1}=2(1+2+\cdots+2 n) \\
=2 n(2 n+1) \equiv 0(\bmod 2 n),
\end{array}
$$
The sum of all remainders is
$$
\begin{array}{l}
S_{2}=0+1+\cdots+2 n-1 \\
=n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,939 |
1. Let
$$
\begin{aligned}
M= & \frac{\left(1-y^{2}\right)\left(1-z^{2}\right)}{y z}+\frac{\left(1-z^{2}\right)\left(1-x^{2}\right)}{z x}+ \\
& \frac{\left(1-x^{2}\right)\left(1-y^{2}\right)}{x y},
\end{aligned}
$$
where $x, y, z$ are distinct real numbers. If
$$
x+y+z=x y z \neq 0,
$$
then the value of $M$ is ( ).
(A... | - 1. C.
Notice
$$
\begin{array}{l}
\frac{\left(1-y^{2}\right)\left(1-z^{2}\right)}{y z}=\frac{1}{y z}-\left(\frac{y}{z}+\frac{z}{y}\right)+y z, \\
\frac{\left(1-z^{2}\right)\left(1-x^{2}\right)}{z x}=\frac{1}{z x}-\left(\frac{z}{x}+\frac{x}{z}\right)+z x, \\
\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)}{x y}=\frac{1... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,940 |
2. As shown in Figure 1, in the same plane, two congruent isosceles right triangles $\triangle A B C$ and $\triangle A F G$ are placed together, with $A$ as their common vertex, $\angle B A C=\angle A F G=90^{\circ}$. If $\triangle A B C$ is fixed and $\triangle A F G$ rotates around vertex $A$, the intersection points... | 2. B.
As shown in Figure 3, rotate $\triangle A C E$ $90^{\circ}$ clockwise around vertex $A$ to get $\triangle A B H$, and connect $H D$.
In $\triangle E A D$ and $\triangle H A D$, we have
$$
\begin{aligned}
A E & =A H, \\
A D & =A D, \\
\angle H A D & =\angle E A H-\angle F A G \\
=45^{\circ} & =\angle E A D .
\en... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,941 |
3. Given non-zero real numbers $a, b, c$ satisfy
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$
Then for positive integer $k$ such that
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$,
(2) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{... | 3. B.
From the given equation, we have
$$
\begin{array}{l}
(a b+b c+c a)(a+b+c)=a b \cdot c \\
\Rightarrow(a+b)(b+c) \cdot(c+a)=0 .
\end{array}
$$
Thus, $a+b=0$ or $b+c=0$ or $c+a=0$.
By symmetry, assume $a+b=0$. Then
$$
\begin{array}{l}
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{c^{2 k+1}} \\
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,942 |
4. In an acute triangle $\triangle A B C$, it is known that $A B=c, A C=b$ $(b>c), \angle B A C=60^{\circ}$, its orthocenter is $H$, circumcenter is $O, O H$ intersects $A B$ and $A C$ at points $D$ and $E$ respectively. Then the length of $D E$ is $($.
(A) $\frac{b+c}{2}$
(B) $\frac{b+c}{3}$
(C) $2(b-c)$
(D) $3(b-c)$ | 4. B.
As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
\angle B H C=\angle G H F \\
=180^{\circ}-\angle A \\
=120^{\circ} .
\end{array}
$$
Therefore, points $B, H, O, C$ are concyclic.
Hence, $\angle D H C=\angle O B C=30^{\circ}$.
Also, $\angle D C ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,943 |
5. Given a two-digit number $\overline{p q}$, such that the quadratic function $y=x^{2}+q x+p$ intersects the $x$-axis at two distinct points $A$ and $B$, with the vertex at $C$, and $S_{\triangle A B C} \leqslant 1$. Then the number of such two-digit numbers is
(A) 1
(B) 2
(C) 3
(D) 4 | 5. D.
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)\left(x_{1} \neq x_{2}\right)$. Then $x_{1}$ and $x_{2}$ are the two distinct real roots of the equation $x^{2}+q x+p=0$.
Thus, $x_{1}+x_{2}=-q, x_{1} x_{2}=p$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}\left|x_{1}-x_{2}\right|\left|y_{c}\right|$
$$
=\frac... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,944 |
1. Given $p$ is a prime number greater than 5, and $m$ is the smallest non-negative remainder when $\left(p^{2}+5 p+5\right)^{2}$ is divided by 120. Then the units digit of $2009^{m}$ is $\qquad$ | Notice that
$$
\begin{array}{l}
\left(p^{2}+5 p+5\right)^{2}=\left[\left(p^{2}+5 p+5\right)^{2}-1\right]+1 \\
=\left(p^{2}+5 p+6\right)\left(p^{2}+5 p+4\right)+1 \\
=(p+1)(p+2)(p+3)(p+4)+1 .
\end{array}
$$
Let $M=(p+1)(p+2)(p+3)(p+4)$.
Clearly, $5! \mid p M$.
Since $p$ is a prime number greater than 5, 120 divides $M$... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,946 |
2. For $\triangle A B C$, squares are constructed outward on its three sides $a, b, c$, with their areas denoted as $S_{a}, S_{b}, S_{c}$ respectively. If $a+b+c=18$, then the minimum value of $S_{a}+S_{b}+S_{c}$ is $\qquad$ | 2. 108.
$$
\begin{array}{l}
\text { Given }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0 \\
\Rightarrow 2\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2(a b+b c+c a) \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant \frac{18^{2}}{3}=108 . \\
\text { Therefore, }\left(... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,947 |
3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
then $x^{2... | 3. 36 .
It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$
Therefore,... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,948 |
4. Let $D$, $E$, and $F$ be points on the sides $BC$, $AB$, and $AC$ of $\triangle ABC$, respectively, such that $AE = AF$, $BE = BD$, and $CF = CD$. Given that $AB \cdot AC = 2BD \cdot DC$, $AB = 12$, and $AC = 5$. Then the inradius $r$ of $\triangle ABC$ is $\qquad$ | 4.2.
As shown in Figure 6, let
$$
\begin{array}{l}
A E=A F=x, \\
B E=B D=y, \\
C D=C F=z .
\end{array}
$$
Then $A B=x+y$,
$$
\begin{array}{l}
A C=x+z, \\
B C=y+z .
\end{array}
$$
From $A B \cdot A C=2 B D \cdot D C$
$$
\begin{array}{l}
\Rightarrow(x+y)(x+z)=2 y z \\
\Rightarrow x^{2}+x y+x z=y z \\
\text { Also } A ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,949 |
5. Draw all diagonals in a
(a) regular pentagon; (b) regular hexagon.
Initially, mark the number 1 at each vertex and each intersection of the diagonals. In each step, you can change the sign of all numbers on a side or a diagonal. Can all the marked numbers be changed to -1 after several steps? | (a) cannot. The parity of the number of -1s on the pentagon boundary remains unchanged; (b) cannot. The product of the nine numbers colored black in Figure 2 remains unchanged. | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,950 |
One, (20 points) Given the quadratic equation in $x$
$$
\left(k^{2}-6 k+8\right) x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}-4=0
$$
has two integer roots. Find the value of the real number $k$.
| The original equation can be transformed into
$$
[(k-4) x+(k-2)][(k-2) x+(k+2)]=0 \text {. }
$$
Since $(k-4)(k-2) \neq 0$, we have
$$
\begin{array}{l}
x_{1}=-\frac{k-2}{k-4}=-1-\frac{2}{k-4}, \\
x_{2}=-\frac{k+2}{k-2}=-1-\frac{4}{k-2} .
\end{array}
$$
Thus, $k-4=-\frac{2}{x_{1}+1}$,
$$
k-2=-\frac{4}{x_{2}+1},
$$
whe... | k=6,3, \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,951 |
II. (25 points) As shown in Figure 2, given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P$ and $Q$. Through any point $M$ on $PQ$, draw a line intersecting $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $S$, $R$ and $D$, respectively. Then, draw lines through points $A$ and $D$ parallel to $PQ$, inters... | By the intersecting chords theorem, we have
$$
A M \cdot M S=P M \cdot M Q=R M \cdot M D \text {. }
$$
Then $\frac{A M}{M R}=\frac{M D}{M S}$
$$
\Rightarrow \frac{A R}{M R}=\frac{D S}{M S} \text {. }
$$
By $A B / / P Q / / C D$
$$
\begin{array}{l}
\Rightarrow \triangle A B R \backsim \triangle M P R \\
\Rightarrow \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,952 |
Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100? | Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$
Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$
where $k_{1} ... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,953 |
1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$
Then the minimum value of $a b c+\frac{9}{a b c}$ is | -1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$
Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$
is decreasing on $(0,3)$.
Si... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,954 |
2. Let $O$ be a point in the plane of acute $\triangle A B C$ (outside $\triangle A B C$), and let $C D \perp A B$ at point $D$. If
$$
\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c \text {, }
$$
then $\overrightarrow{O D}=$ $\qquad$ (express in terms of $a, b, c$). | 2. $a+\frac{(c-a) \cdot(b-a)}{|b-a|^{2}}(b-a)$.
From $\overrightarrow{A C} \cdot \overrightarrow{A B}=|\overrightarrow{A C}||\overrightarrow{A B}| \cos A$ $=|\overrightarrow{A B}||\overrightarrow{A D}|$,
we get $|\overrightarrow{A D}|=\frac{\overrightarrow{A C} \cdot \overrightarrow{A B}}{|\overrightarrow{A B}|}$.
Thu... | a+\frac{(c-a) \cdot(b-a)}{|b-a|^{2}}(b-a) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,955 |
3. The function
$$
f(x)=\left|\sin x+\frac{1}{2} \sin 2 x\right|(x \in \mathbf{R})
$$
has the range | 3. $\left[0, \frac{3 \sqrt{3}}{4}\right]$.
Obviously, $f(0)=0, f(x) \geqslant 0$.
Therefore, the minimum value of $f(x)$ is 0.
Next, we find the maximum value of $f(x)$.
Notice that
$$
\begin{array}{l}
f(x)=\left|\sin x+\frac{1}{2} \sin 2 x\right| \\
=\left|(1+\cos x) \sin x \right| \\
=\left|2 \cos ^{2} \frac{x}{2} \... | \left[0, \frac{3 \sqrt{3}}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,956 |
4. Given a finite arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a common difference of 2, the arithmetic mean of all its terms is 2011. If one term is removed, the arithmetic mean of the remaining terms is an integer. Then the number of ways to remove a term is $\qquad$. | 4.3.
According to the problem, we have
$$
\frac{1}{n}\left[n+\frac{n(n-1)}{2} \times 2\right]=2011 \text {. }
$$
Solving this, we get $n=2011$.
Thus, the sum of all terms in the sequence is $2011^{2}$.
Suppose that after removing the $k$-th term from the sequence, the arithmetic mean of the remaining terms is an inte... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,957 |
5. As shown in Figure 1, trapezoid $ABCD (AB \parallel CD \parallel y$-axis, $|AB| > |CD|)$ is inscribed in the ellipse.
$$
\begin{array}{r}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\
(a > b > 0), E
\end{array}
$$
is the intersection of diagonals $AC$ and $BD$. Let $|AB| = m, |CD| = n, |OE| = d$. Then the maximum va... | 5. $\frac{2 b}{a}$.
By symmetry, we know that point $E$ is on the $x$-axis.
Then point $E(d, 0)$.
Let $l_{B D}: x=k y+d$.
From $\left\{\begin{array}{l}x=k y+d, \\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\end{array}\right.$ eliminating $x$ yields
$$
\left(a^{2}+b^{2} k^{2}\right) y^{2}+2 d k b^{2} y+b^{2}\left(d^{2}-... | \frac{2 b}{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,958 |
6. A person practices shooting at a target. Initially, he is $100 \mathrm{~m}$ away from the target and performs the first shot. If this shot misses, he steps back $50 \mathrm{~m}$ to take the second shot, and continues this process. Before each shot, he steps back $50 \mathrm{~m}$ until he hits the target. It is known... | 6. $\frac{1}{2}$.
Let the event "hit on the $n$-th shot" be denoted as $A_{n}$, with its probability being $P\left(A_{n}\right)$. Then,
$$
P\left(A_{1}\right)=\frac{1}{4} \text {. }
$$
The distance from the target during the $n$-th shot is
$$
100+50(n-1)=50(n+1)(\text{m}) \text {, }
$$
thus $P\left(A_{n}\right)=\lef... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,959 |
7. As shown in Figure 2, in plane $\alpha$, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are on opposite sides of line $l$, with no common points with $l$, and are symmetric about line $l$. Now, if plane $\alpha$ is folded along line $l$ to form a right dihedral angle, then the six points $A, B, C, A_{1}, B_{1},... | 7.11.
Notice that after the fold, the three sets of four points
$$
\left(A, B, A_{1}, B_{1}\right) 、\left(B, C, B_{1}, C_{1}\right) 、\left(C, A, C_{1}, A_{1}\right)
$$
are all coplanar, therefore, these six points can determine
$$
C_{6}^{3}-3\left(C_{4}^{3}-1\right)=11 \text { (planes). }
$$ | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,960 |
6. Each vertex of a regular pentagon corresponds to an integer, such that the sum of these five integers is positive. If three consecutive vertices correspond to integers $x, y, z$, and the middle integer $y<0$, then the following transformation is performed: the integers $x, y, z$ are replaced by $x+y, -y, z+y$, respe... | Prompt: The answer to the question is affirmative.
For convenience, the circular sequence of five numbers can be written as a horizontal sequence $v, w, x, y, z$ (here $z$ and $v$ are adjacent).
Assume without loss of generality that $y < 0$. Then after the transformation, we get $v, w, x+y, -y, z+y$. This is a local ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,961 |
8. The complex sequence $z_{0}, z_{1}, \cdots$ satisfies
$$
\left|z_{0}\right|=1, z_{n+1}=\frac{i z_{n}}{\overline{z_{n}}} \text {. }
$$
If $z_{2011}=1$, then $z_{0}$ can have $\qquad$ possible values. | 8. $2^{2011}$.
Obviously, for any non-negative integer $n$, we have
$$
\left|z_{n}\right|=1 \text {. }
$$
Let $z_{n}=\mathrm{e}^{\mathrm{i} \theta_{n}}\left(\theta_{0} \in[0,2 \pi)\right)$. Then
$$
\begin{array}{l}
\mathrm{e}^{\mathrm{i} \theta_{n+1}}=\frac{\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{x}}{2}+\theta_{n}\... | 2^{2011} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,962 |
9. (16 points) Given $a \in \mathbf{R}_{+}$, the equation
$$
x^{2}-2 a x-2 a \ln x=0
$$
has a unique solution in $(0,+\infty)$. Find the value of $a$. | Let the function be
$f(x)=x^{2}-2 a x-2 a \ln x$.
Then $f^{\prime}(x)=2 x-2 a-\frac{2 a}{x}$
$$
=\frac{2}{x}\left(x^{2}-a x-a\right) \text {. }
$$
Let $f^{\prime}(x)=0$, i.e., $x^{2}-a x-a=0$.
Solving this, we get $x_{1}=\frac{a-\sqrt{a^{2}+4 a}}{2}<0$, and $f(x)$ is monotonically increasing.
Thus, $f(x)$ attains its ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,963 |
10. (20 points) Given a positive integer $n(n \geqslant 3), a_{n}=$ $(\overline{1.00 \dot{0}})_{n}$ (i.e., $a_{n}$ equals the number represented in base $n$ as $\overline{1.00 \dot{2}}$). Try to find
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{a_{3}}+\frac{1}{a_{3} a_{4}}+\cdots+\frac{1}{a_{3} a_{4} \cdot... | 10. According to the problem, we have
$$
\begin{array}{l}
a_{n}=\overline{1 . \dot{0} \dot{2}}=1+\frac{2}{n^{3}}+\frac{2}{n^{6}}+\cdots=1+\frac{\frac{2}{n^{3}}}{1-\frac{1}{n^{3}}} \\
=\frac{1+\frac{1}{n^{3}}}{1-\frac{1}{n^{3}}}=\frac{n^{3}+1}{n^{3}-1}=\frac{n+1}{n-1} \cdot \frac{n^{2}-n+1}{n^{2}+n+1} .
\end{array}
$$
... | \frac{6}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,964 |
11. (20 points) On a line $l$ outside the ellipse $G$, take $n$ $\left(n \in \mathbf{N}_{+}, n \geqslant 2\right)$ different points $P_{1}, P_{2}, \cdots, P_{n}$. Through $P_{i}(1 \leqslant i \leqslant n)$, draw tangents $P A_{i}, P B_{i}$ to the ellipse $G$, with the points of tangency being $A_{i}$ and $B_{i}$. Denot... | 11. Let the ellipse $G: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$,
line $l: A x+B y+C=0$,
$$
P_{i}\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n) \text {. }
$$
Then the equation of the chord of tangents $l_{i}$ of point $P_{i}$ with respect to the ellipse $G$ is
$$
\frac{x_{i} x}{a^{2}}+\frac{y_{i} y}{b^{2}}=1
$$... | 2n | Geometry | proof | Yes | Yes | cn_contest | false | 723,965 |
One, (40 points) As shown in Figure $3, \triangle A B C$ is inscribed in circle $\Gamma$, $D$ is the midpoint of the minor arc $\overparen{B C}$, circle $\Gamma_{1}$ is tangent to circle $\Gamma$ at point $D$, and tangent to side $B C$ at point $E$. A tangent line $A P$ is drawn from point $A$ to circle $\Gamma_{1}$, w... | As shown in Figure 4, connect $A D, B D, C D, D E$. It is easy to see that $E$ is the midpoint of side $B C$, and $D E \perp B C$.
Draw $D F \perp A B, D G \perp A C$, with the feet of the perpendiculars being $F$ and $G$ respectively.
By the Simson line theorem, $F, E, G$ are collinear.
Since $D$ is the midpoint of a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,966 |
II. (40 points) Given positive real numbers $x, y, z$ satisfying
$$
(x+y+z)^{3}=32 x y z \text {. }
$$
Find the maximum and minimum values of $f(x, y, z)=\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}$. | Let's assume $x+y+z=4$. Then $xyz=2$.
Therefore, we only need to find the extremum of $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}$. At this point, we have
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4} \\
=\left(x^{2}+y^{2}+z^{2}\right)^{2}-2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \\
=[16-2(x y+y z+z x)]^{2}- \\
2(x y+y z+z x)^{2... | \frac{383-165 \sqrt{5}}{256} \text{ and } \frac{9}{128} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,967 |
Three. (50 points) Prove: There exist infinitely many sets of positive integers $(a, b, c, d)$, satisfying:
(1) $a>c>0$, and $(a, c)=1$;
(2) $a$ can take infinitely many values;
(3) There exist exactly $2^{2011}$ positive integers $n$ such that
$$
(a n+b) \mid(c n+d) .
$$ | Three, take any odd prime $p$, and let
$$
\begin{array}{l}
(a, b, c, d) \\
=\left(p-1, p-1, p-2,(p-2)+p^{2^{2011}}\right) .
\end{array}
$$
First, prove that this set of numbers satisfies the conditions in the problem.
Clearly, conditions (1) and (2) are satisfied.
Next, prove that this set of numbers satisfies conditi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,968 |
Four, (50 points) A company printed a batch of T-shirts, each T-shirt can have three different colors: red, yellow, and blue, and four different patterns. Now, this batch of T-shirts is to be distributed to $n$ new employees, with each employee receiving exactly 4 T-shirts with different patterns. Try to find the minim... | Four, the minimum value of $n$ is 19.
When $n=18$, the answer scenario shown in Table 1 does not meet the requirements.
[Note] In Table 1, (1), (2), (3), (4) are patterns, $A_{1}, A_{2}, \cdots, A_{18}$ are members, and $A$, $B$, $C$ represent red, yellow, and blue colors, respectively.
The proof below shows that when ... | 19 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,969 |
As shown in Figure 2, given that the internal angle bisector $A D$ of $\triangle A B C$ intersects $B C$ at point $D$, point $E$ is on $A B$ such that $A E=A C$, and point $F$ is on the extension of $A C$ such that $A F=A B$. The lines through points $E$ and $F$ perpendicular to $A B$ and $A C$ respectively intersect t... | Prove as shown in Figure 2, draw $A K \perp B C$ at point $K$, draw $P S / / B C$ intersecting $A K$ at point $S$, and connect $D E, A P$.
Since $A D$ is the angle bisector of $\triangle A B C$, and $A E = A C$, it is easy to prove that
$\triangle A D E \cong \triangle A D C$.
Thus, $D E = D C, \angle A E D = \angle A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,970 |
Do there exist six distinct positive integers such that: taking any two numbers to form a pair, subtracting the smaller number from the larger one in the same pair, and then adding up all the differences obtained from each pair, the sum is exactly 45, and the difference between the largest and smallest numbers is divis... | No solution exists.
Let's assume these six different positive integers are
$$
a_{1}, a_{2}, \cdots, a_{6}\left(a_{1}>a_{2}>\cdots>a_{6}\right) .
$$
According to the problem, any two of the six numbers can form a pair, resulting in 15 pairs. Thus,
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)+\cdots+\left(a_{1}-a_{6}\ri... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,971 |
Example 1 In $\triangle A B C$, it is known that $A B=A C=2$, and there are 100 different points $P_{1}, P_{2}, \cdots, P_{100}$ on side $B C$. Let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C(i=1,2, \cdots, 100)$.
Find the value of $m_{1}+m_{2}+\cdots+m_{100}$. | Solve As shown in Figure 2, since $\triangle A B C$ is an isosceles triangle, applying the property we get
$$
\begin{aligned}
& A P_{i}^{2} \\
= & A B^{2}-B P_{i} \cdot P_{i} C .
\end{aligned}
$$
Therefore, $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C=A B^{2}=4$.
Thus, $m_{1}+m_{2}+\cdots+m_{100}=400$. | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,972 |
Given $\triangle A B C$ with sides $B C=a, C A=$ $b, A B=c$, and point $Q$ inside the triangle. Let
$$
f=a Q A^{2}+b Q B^{2}+c Q C^{2} \text {. }
$$
(1) Find the minimum value of $f$;
(2) Determine the geometric position of point $Q$ when $f$ takes its minimum value. | (1) As shown in Figure 3, construct the symmetric points of point $Q$ with respect to $BC$, $CA$, and $AB$, denoted as $Q_{1}$, $Q_{2}$, and $Q_{3}$.
Then $S_{\triangle Q_{1}} + S_{\triangle Q_{Q}}$
$$
\begin{array}{l}
=\frac{1}{2} Q A^{2} \left(\sin 2 \alpha_{1} + \sin 2 \beta_{1}\right) \\
= Q A^{2} \sin \left(\alpha... | a b c | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,973 |
Given $x \in(0,2 \pi]$, and the six trigonometric values of $x$
$$
\sin x 、 \cos x 、 \tan x 、 \cot x 、 \sec x 、 \csc x
$$
are exactly the six edge lengths of a tetrahedron. Find the range of $x$. | Obviously, $x$ is an acute angle, and the six trigonometric function values are symmetric about $\frac{\pi}{4}$. Therefore, we only need to consider $x \in\left(0, \frac{\pi}{4}\right]$.
$$
\text { When } \frac{\pi}{6}<x \leqslant \frac{\pi}{4}
$$
At this time, it can be verified that the six trigonometric function va... | x \in\left(\frac{\pi}{6}, \frac{\pi}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,974 |
Example 2 As shown in Figure 3, given that $P T_{1}$ and $P T_{2}$ are secants of $\odot O$, intersecting $\odot O$ at points $S_{1}$ and $S_{2}$, respectively, and $P T_{1}=P T_{2}$. A line through point $P$ intersects $\odot O$ at points $Q$ and $R$ ($Q$ is between $R$ and $P$), and intersects $T_{1} T_{2}$ and $S_{1... | For the isosceles $\triangle P T_{1} T_{2}$, note that
$$
P S_{1} \cdot P T_{1}=P S_{2} \cdot P T_{2} \text {. }
$$
Thus, $P S_{1}=P S_{2}$.
Therefore, $\triangle P S_{1} S_{2}$ is also an isosceles triangle.
Applying the property, we get
$$
P S^{2}=P S_{1}^{2}-S_{1} S \cdot S S_{2} \text {. }
$$
Since $S_{1} S_{2} \... | \frac{1}{P Q}+\frac{1}{P R}=\frac{1}{P S}+\frac{1}{P T} | Geometry | proof | Yes | Yes | cn_contest | false | 723,975 |
Example 3 Given that $D$, $E$, and $F$ are the midpoints of the sides $BC$, $CA$, and $AB$ of acute $\triangle ABC$, respectively. Points $P$, $Q$, and $R$ are taken on the extensions of $EF$, $FD$, and $DE$, respectively. If $AP = BQ = CR$, prove that the circumcenter of $\triangle PQR$ is the orthocenter of $\triangl... | Proof As shown in Figure 4, let the three altitudes of $\triangle ABC$ be $AL, BM, CN$, with the orthocenter $H$, and $EF$ intersects $AL$ at point $K$. Since $FE$ is the midline of $\triangle ABC$ and $AL \perp BC$, then $PE$ is the perpendicular bisector of segment $AL$: By applying the property, we get
$$
P H^{2}=A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,976 |
Example 4 Given that $E$ and $F$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$, and $CM$ and $BN$ are the altitudes from $C$ and $B$ to sides $AB$ and $AC$ respectively. Connect $EF$ and $MN$ intersecting at point $P$. Also, let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$ respective... | Prove that, as shown in Figure 5, connect $AO$ and $AH$. Let $\sigma_{1}$ and $H_{1}$ be the midpoints of $AO$ and $AH$, respectively. Then, points $A$, $M$, $H$, and $N$ are concyclic, and $H_{1}$ is the center of this circle.
Thus, point $H_{1}$ lies on the perpendicular bisector of segment $MN$.
Applying the propert... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,977 |
Example 5 As shown in Figure 6, $P A, P B$ are two tangent lines of $\odot O$, with points of tangency at $A, B$ respectively. A line through point $P$ intersects $\odot O$ at points $C, D$, and intersects chord $A B$ at point $Q$. Prove:
$$
P Q^{2}=P C \cdot P D-Q C \cdot Q D .{ }^{[2]}
$$ | Proof By the tangent-secant theorem, we have
$$
P A = P B \text{. }
$$
Applying the property, we get
$$
P Q^{2} = P B^{2} - B Q \cdot Q A \text{. }
$$
Noting that $P B^{2} = P C \cdot P D$,
$$
B Q \cdot Q A = Q C \cdot Q D \text{. }
$$
Thus, $P Q^{2} = P C \cdot P D - Q C \cdot Q D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,978 |
Example 6 Convex quadrilateral $A B C D$ is circumscribed around $\odot O$, the lines containing the two pairs of opposite sides intersect at points $E$ and $F$, and the diagonals intersect at point $G$. Prove: $O G \perp E F$. | Proof As shown in Figure 7, let $M, N, R, S$ be the points of tangency of $\odot O$ with quadrilateral $ABCD$. Then, by Newton's theorem, $AC, BD, MR, NS$ concur at point $G$.
By the tangent segment theorem, we know $EM = ER$.
Applying the property, we get
$$
EG^2 = EM^2 - MG \cdot GR.
$$
Similarly, $FG^2 = FS^2 - SG ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,979 |
Example 7 Given that $A B$ is a chord of $\odot O$, $M$ is the midpoint of arc $\overparen{A B}$, $C$ is any point outside $\odot O$, and from point $C$ two tangents $C S$ and $C T$ are drawn to $\odot O$. Connecting $M S$ and $M T$ intersect $A B$ at points $E$ and $F$ respectively. Perpendiculars from points $E$ and ... | Prove as shown in Figure 8, connect $O M, M X, X C$.
Then $O M \perp A B$.
Since $X E \perp A B$, we know $X E \parallel O M$.
Thus, $\angle X E S = \angle O M S = \angle X S E$, which means $X E = X S$.
By applying the property, we get
$$
X M^{2} = X E^{2} + M E \cdot M S \text{.}
$$
By the similarity of triangles, i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,980 |
Example 4 Find
$$
M=(x+1)(x+2) \cdots(x+n)
$$
the coefficient of $x^{n-2}$ in the expansion. | 【Analysis】The coefficient of $x^{n-2}$ in $M$ is the sum of the products of any two numbers taken from $1,2, \cdots, n$, i.e.,
$$
\begin{aligned}
A= & 1 \times 2+1 \times 3+\cdots+1 \times n+ \\
& 2 \times 3+2 \times 4+\cdots+2 \times n+ \\
& \cdots+(n-1) n .
\end{aligned}
$$
The above expression is quite complex and ... | \frac{1}{24}(n-1) n(n+1)(3 n+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,981 |
Example 8 Let $D$ be a point on side $BC$ of $\triangle ABC$, satisfying $\angle CAD = \angle CBA$. Circle $\odot O$ passes through points $B$ and $D$, and intersects line segments $AB$ and $AD$ at points $E$ and $F$, respectively. $BF$ and $DE$ intersect at point $G$. Connect $AO$ and $AG$, and let $M$ be the midpoint... | Prove as shown in Figure 9, take point $P$ on the extension of $A G$ such that $A G \cdot A P = A F \cdot A D$ (i.e., $G, P, D, F$ are concyclic). Then, from $A E \cdot A B = A F \cdot A D$, we know that $E, B, P, G$ are concyclic.
$$
\begin{array}{l}
\text{Thus, } \angle B P A = 180^{\circ} - \angle B E D \\
= 180^{\c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,982 |
1. Given points $A, B, C, D$ lie on the same circle, and $BC = DC = 4, AC$ intersects $BD$ at point $E, AE = 6$. If the lengths of segments $BE$ and $DE$ are both integers, find the length of $BD$. | Apply the property to $\triangle B C D$ to get
$$
C E^{2}=C D^{2}-B E \cdot E D=16-6 \cdot E C \text {. }
$$
Solving, we get $E C=2$.
From $B E \cdot E D=A E \cdot E C=12$, and
$$
B D<B C+C D=8 \text {, }
$$
we solve to get $B D=4+3=7$. | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,983 |
2. Given that quadrilateral $ABCD$ is inscribed in $\odot O$, the extensions of sides $AB$ and $DC$ intersect at point $P$, and the extensions of sides $AD$ and $BC$ intersect at point $Q$. Two tangents $QE$ and $QF$ are drawn from $Q$ to the circle, touching the circle at points $E$ and $F$ respectively. Prove: $P$, $... | Let the circumcircle of $\triangle Q C D$ intersect $P Q$ at point $G$, and the radius of $\odot O$ be $R$. From
$$
\angle P G C=\angle Q D C=\angle A B C \text {, }
$$
we know that points $B, C, G, P$ are concyclic.
Applying the property in $\triangle A O B$, we get
$$
\begin{array}{l}
P O^{2}=A O^{2}+P B \cdot P A=R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,984 |
3. Let quadrilateral $ABCD$ be inscribed in a circle, with diagonals $AC$ and $BD$ intersecting at point $G$, and lines $AB$ and $CD$ intersecting at point $P$. The circumcenters of $\triangle PAC$ and $\triangle PBD$ are $O_{1}$ and $O_{2}$, respectively. Prove:
$$
PG \perp O_{1} O_{2}
$$ | For $\triangle O_{1} A C$ and $\triangle O_{2} D B$, applying the properties we get
$$
\begin{array}{l}
O_{1} G^{2}=O_{1} C^{2}-C G \cdot G A, \\
O_{2} G^{2}=O_{2} D^{2}-D G \cdot G B .
\end{array}
$$
Then $O_{2} P^{2}-O_{2} G^{2}$
$$
\begin{array}{l}
=O_{2} P^{2}-\left(O_{2} D^{2}-D G \cdot G B\right) \\
=D G \cdot G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,985 |
4. Let $M$ be a point inside $\odot O$, and $A_{i} B_{i} (i=1,2,3)$ be three chords of $\odot O$ passing through point $M$. The tangents to $\odot O$ at points $A_{i}$ and $B_{i}$ intersect at point $P_{i} (i=1,2,3)$. Prove that $P_{1}$, $P_{2}$, and $P_{3}$ are collinear. | For $\triangle P_{i} A_{i} B_{i}$, applying the property we get
$$
P_{i} M^{2}=P_{i} A_{i}^{2}-A_{i} M \cdot M B_{i} \text {. }
$$
Also, $P_{i} O^{2}=P_{i} A_{i}^{2}+A_{i} O^{2}$, thus, $P_{i} O^{2}-P_{i} M^{2}=A_{i} O^{2}+A_{i} M \cdot M B_{i}$ (a constant).
By the theorem of the locus of points with a fixed differen... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,986 |
5. Let $O$ and $I$ be the circumcenter and incenter of $\triangle ABC$, respectively. The incircle of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Line $FD$ intersects $CA$ at point $P$, and line $DE$ intersects $AB$ at point $Q$. Let $M$ and $N$ be the midpoints of segm... | For $\triangle A B C$ and the transversal $D F P$, applying Menelaus' theorem, we have
$$
\frac{B D}{D C} \cdot \frac{C P}{P A} \cdot \frac{A F}{B F}=1 \text {. }
$$
Noting that $B D=B F, C D=C E, A F=A E$, then
$$
\frac{A P}{A E}=\frac{C D}{C E} \text {. }
$$
Since $M$ is the midpoint of $P E$, we have
$$
\frac{P M+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,987 |
Question 1 Given that $f(x)$ is a non-negative function defined on $[0,1]$, and $f(1)=1$, for any $x, y, x+y \in [0,1]$ we have
$$
f(x+y) \geqslant f(x)+f(y) .
$$
Prove: $f(x) \leqslant 2 x(x \in[0,1])$. | Prove that for any $x, \Delta x, x+\Delta x \in [0,1]$, we have $f(x+\Delta x) - f(x) \geqslant f(\Delta x) \geqslant 0$.
Therefore, $f(x)$ is a non-decreasing function.
For any $x \in [0,1]$, there must exist $n \in \mathbf{N}_{+}$ such that $x \in \left[\frac{1}{2^n}, \frac{1}{2^{n-1}}\right]$.
We will use mathemati... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,988 |
Given $\left(x_{1}, y_{1}\right) 、\left(x_{2}, y_{2}\right) 、\left(x_{3}, y_{3}\right)$ are three points on the circle $x^{2}+y^{2}=1$, and satisfy
$$
x_{1}+x_{2}+x_{3}=0, y_{1}+y_{2}+y_{3}=0 \text {. }
$$
Prove: $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=\frac{3}{2}$. | Proof 1 Let $x_{i}=\cos \theta_{i}, y_{i}=\sin \theta_{i}(i=1,2,3)$. From $x_{1}+x_{2}+x_{3}=0, y_{1}+y_{2}+y_{3}=0, x_{i}^{2}+y_{i}^{2}=1$,
we get $\left(x_{2}+x_{3}\right)^{2}+\left(y_{2}+y_{3}\right)^{2}=1$.
Thus, $2\left(x_{2} x_{3}+y_{2} y_{3}\right)=-1$, which means
$\cos \left(\theta_{1}-\theta_{2}\right)=-\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,989 |
In $\triangle A B C$, $A B=A C, \angle C A B$ and $\angle A B C$'s internal angle bisectors intersect the sides $B C$ and $C A$ at points $D$ and $E$ respectively. Let $K$ be the incenter of $\triangle A D C$. If $\angle B E K=45^{\circ}$, find all possible values of $\angle C A B$. ${ }^{[1]}$ | As shown in Figure 1, let $A D$ and $B E$ intersect at point $I$, and the inradius of the incircle $\odot I$ of $\triangle A B C$ be $r$,
$$
\begin{array}{l}
\angle A B C \\
=2 \alpha\left(0<\alpha<45^{\circ}\right) .
\end{array}
$$
Since $A B=A C$, we have
$$
A D \perp B C, \angle A C B=2 \alpha \text {. }
$$
Therefo... | 60^{\circ} \text{ or } 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,990 |
Example 1 Let $a, b, c$ be the lengths of the three sides of a right-angled triangle, where $c$ is the length of the hypotenuse. Find the maximum value of $k$ such that $\frac{a^{3}+b^{3}+c^{3}}{a b c} \geqslant k$ holds. ${ }^{[2]}$ | Solve: Given $a^{2}+b^{2}=c^{2}$, let
$$
\begin{array}{l}
a=c \cos \theta, \\
b=c \sin \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right) .
\end{array}
$$
Then $f=\frac{a^{3}+b^{3}+c^{3}}{a b c}$
$$
=\frac{(\sin \theta+\cos \theta)^{3}-3 \sin \theta \cdot \cos \theta \cdot(\sin \theta+\cos \theta)+1}{\sin \the... | 2+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,991 |
Example 5 Can the numbers 1, 2, ..., 25 be filled into a $5 \times 5$ grid so that the sum of some numbers in each row equals the sum of the remaining numbers in the same row? | If each row is filled with numbers that meet the requirements, then the sum of the numbers in each row is even. Therefore, the total sum of all the numbers in the rows should be even. However, the sum of the numbers filled in is
$$
1+2+\cdots+25=\text { odd, }
$$
It is evident that a filling method that meets the requ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,992 |
Example 2 If $x$, $y$, $z$ are all real numbers, and $x^{2}+y^{2}+z^{2}$ $=1$, then the maximum value of $\sqrt{2} x y+y z$ is $\qquad$ | Solve: Given $x^{2}+y^{2}+z^{2}=1$, let
$$
\begin{array}{l}
x=\cos \theta \cdot \cos \varphi, \\
z=\cos \theta \cdot \sin \varphi, \\
y=\sin \theta \quad (\theta, \varphi \in [0, 2\pi)) . \\
\text{Then } \sqrt{2} x y + y z \\
=\sin \theta \cdot \cos \theta \cdot (\sqrt{2} \cos \varphi + \sin \varphi) \\
=\frac{1}{2} \s... | \frac{\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,993 |
Example 3 If $x, y, z$ are all positive real numbers, and $x^{2}+y^{2}+$ $z^{2}=1$, then the minimum value of $S=\frac{(z+1)^{2}}{2 x y z}$ is $\qquad$. | Given $x^{2}+y^{2}+z^{2}=1$, let
$$
\begin{array}{l}
x=\cos \theta \cdot \cos \varphi, \\
y=\cos \theta \cdot \sin \varphi \\
z=\sin \theta\left(\theta, \varphi \in\left(0, \frac{\pi}{2}\right)\right) .
\end{array}
$$
Then $S=\frac{(z+1)^{2}}{2 x y z}=\frac{(1+\sin \theta)^{2}}{\sin \theta \cdot \cos ^{2} \theta \cdot... | 3+2\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,994 |
1. Let $a$ be a prime number, and $7 a^{2}+8$ and $8 a^{2}+7$ are also prime numbers. Let $x=77 a+8, y=88 a+7$. Then, among the following scenarios, the one that must be true is ( ).
(A) $x, y$ are both prime numbers
(B) $x, y$ are both composite numbers
(C) One of $x, y$ is a prime number, and the other is a composite... | - 1. A.
When $a=3$,
$$
7 a^{2}+8=71 \text { and } 8 a^{2}+7=79
$$
are both prime numbers, and
$$
x=77 a+8=239, y=88 a+7=271
$$
are both prime numbers;
When the prime number $a$ is different from 3, then $a^{2}$ leaves a remainder of 1 when divided by 3.
Let $a^{2}=3 n+1$. Then
$$
\begin{array}{l}
7 a^{2}+8=21 n+15, ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,996 |
2. Simplify $\frac{\sqrt{3+2 \sqrt{2}}}{\sqrt{17+12 \sqrt{2}}}-\frac{\sqrt{3-2 \sqrt{2}}}{\sqrt{17-12 \sqrt{2}}}$, the result is ( ).
(A) $\sqrt{2}$
(B) $-\sqrt{2}$
(C) 2
(D) -2 | 2. D.
Notice that
$$
\begin{array}{l}
3+2 \sqrt{2}=(\sqrt{2}+1)^{2}, \\
3-2 \sqrt{2}=(\sqrt{2}-1)^{2}, \\
17+12 \sqrt{2}=(3+2 \sqrt{2})^{2}, \\
17-12 \sqrt{2}=(3-2 \sqrt{2})^{2} .
\end{array}
$$
Therefore, the original expression
$$
\begin{array}{l}
=\frac{1}{\sqrt{3+2 \sqrt{2}}}-\frac{1}{\sqrt{3-2 \sqrt{2}}} \\
=\fr... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,997 |
$3.2^{2011}+3^{2011}$ 's last digit is ( ).
(A) 1
(B) 3
(C) 5
(D) 7 | 3. C.
The last digit of $2^{n}$ cycles in the order of $2, 4, 8, 6$, while the last digit of $3^{n}$ cycles in the order of $3, 9, 7, 1$.
Since 2011 is a number of the form $4k+3$, the last digit of $2^{2011}$ is 8. The last digit of $3^{2011}$ is 7, so the last digit of $2^{2011}+3^{2011}$ is 5: | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,998 |
4. The equation
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
has the solution(s) ( ).
(A) no solution
(B) exactly one solution
(C) exactly two solutions
(D) infinitely many solutions | 4. D.
Transform the equation into
$$
\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1 \text {. }
$$
Consider three cases.
(1) $\sqrt{x-1} \geqslant 3$.
Then equation (1) becomes
$$
(\sqrt{x-1}-2)+(\sqrt{x-1}-3)=1 \text {, }
$$
which implies $\sqrt{x-1}=3 \Rightarrow x=10$.
(2) $\sqrt{x-1} \leqslant 2$.
Then eq... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,999 |
5. As shown in Figure 1, a regular hexagon is divided into small equilateral triangles by three sets of parallel lines. The total number of equilateral triangles in the figure is ( ).
(A) 24
(B) 36
(C) 38
(D) 76 | 5. G.
Classification and Calculation: Let the side length of a regular hexagon be 2. Then there are 24 equilateral triangles with a side length of 1, 12 equilateral triangles with a side length of 2, and 2 equilateral triangles with a side length of 3. In total, there are 38 equilateral triangles. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,000 |
6. Let $a, b$ be integers, and the quadratic equation
$$
x^{2}+(2 a+b+3) x+\left(a^{2}+a b+6\right)=0
$$
has a double root $\alpha$, and the quadratic equation
$$
2 a x^{2}+(4 a-2 b-2) x+(2 a-2 b-1)=0
$$
has a double root $\beta$. Then the quadratic equation with integer coefficients and roots $\alpha, \beta$ is ( ).... | 6. A.
From the discriminants of the two equations being 0, we have
$$
\begin{array}{l}
(2 a+b+3)^{2}-4\left(a^{2}+a b+6\right)=0, \\
(4 a-2 b-2)^{2}-8 a(2 a-2 b-1)=0,
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
(b+3)^{2}=12(2-a), \\
(b+1)^{2}=2 a .
\end{array}
$$
Eliminating \(a\) gives \(7 b^{2}+18 b-9=0... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,001 |
1. Given the right triangle $\triangle A B C$ with side lengths of 3, 4, and 5. If its incircle is removed, the remaining area is $\qquad$ . | $=, 1.6-\pi$
$S_{\triangle M B C}=\frac{1}{2} \times 3 \times 4=6$.
Let the radius of the inscribed circle be $r$. Then
$$
S_{\triangle A B C}=\frac{1}{2} r(3+4+5)=6 r \Rightarrow r=1 \text {. }
$$
Therefore, the area of the inscribed circle is $\pi$.
Thus, the area of the remaining part is $6-\pi$. | 6-\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,002 |
Example 6 On the front of six pieces of paper, write down the integers $1,2, \cdots, 6$, shuffle them, then flip the pieces of paper over, and on their backs, write down the integers $1 \sim 6$ in any order. After that, calculate the absolute value of the difference between the numbers written on the front and back of ... | Proof: Let the numbers written on the front of the six cards be $a_{1}, a_{2}, \cdots, a_{6}$, and the numbers written on the back be $b_{1}, b_{2}, \cdots, b_{6}$. Then the absolute values of the differences between the numbers on the front and back of the six cards are $\left|a_{1}-b_{1}\right|, \left|a_{2}-b_{2}\rig... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,003 |
$\begin{array}{l}\text { 2. If } x^{3}+5 x^{2}-7 x-3 \\ =(x-4)^{3}+a(x-4)^{2}+b(x-4)+c \text {, } \\ \text { then }(a, b, c)=\end{array}$ | 2. $(17,81,113)$.
Notice that
$$
\begin{array}{l}
(x-4)^{3}+a(x-4)^{2}+b(x-4)+c \\
=x^{3}+(a-12) x^{2}+(b-8 a+48) x+ \\
\quad(16 a-4 b+c-64) .
\end{array}
$$
From $a-12=5, b-8 a+48=-7$,
$$
16 a-4 b+c-64=-3 \text {, }
$$
Solving these, we get $a=17, b=81, c=113$.
Therefore, $(a, b, c)=(17,81,113)$. | (17,81,113) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,004 |
3. As shown in Figure 2, the side length of square $ABCD$ is $1, E$ is a point outside side $CD$, satisfying $CE // BD$, $BE=BD$. Then $CE=$ | 3. $\frac{\sqrt{6}-\sqrt{2}}{2}$.
As shown in Figure 4, draw $E F \perp$
$B C$ at point $F$. Then
$$
E F=C F \text {. }
$$
Let $C F=x$. Then
$$
\begin{aligned}
2 & =B D^{2}=B E^{2} \\
& =B F^{2}+E F^{2} \\
& =(1+x)^{2}+x^{2} \\
\Rightarrow & 2 x^{2}+2 x-1=0
\end{aligned}
$$
$\Rightarrow x=\frac{\sqrt{3}-1}{2}$ (negat... | \frac{\sqrt{6}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,005 |
4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$... | 4.9.
For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.
For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$
are both multiples of 7, i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,006 |
One. (20 points) Determine for which positive integers $a$, the equation
$$
5 x^{2}-4(a+3) x+a^{2}-29=0
$$
has positive integer solutions? And find all positive integer solutions of the equation. | Rewrite the equation as
$$
(x-6)^{2}+(a-2 x)^{2}=65 \text {. }
$$
Since 65 can only be expressed as the sum of squares of two positive integers in two different ways: $65=1^{2}+8^{2}=4^{2}+7^{2}$, then
$$
\left\{\begin{array} { l }
{ | x - 6 | = 8 , } \\
{ | a - 2 x | = 1 }
\end{array} \text { or } \left\{\begin{arra... | x=14, a=29 \text { or } 27; x=13, a=22 \text { or } 30; x=5, a=2 \text { or } 18; x=7, a=6 \text { or } 22; x=2, a=11; x=10, a=13 \text { or } 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,007 |
II. (25 points) As shown in Figure 3, the circumcenter of acute triangle $\triangle ABC$ is $O$, and the circumradius is $R$. Extend $AO$, $BO$, and $CO$ to intersect sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Prove:
$$
\frac{1}{A D}+\frac{1}{B E}+\frac{1}{C F}=\frac{2}{R} .
$$ | Extend $A D$ to intersect $\odot O$ at point $M$.
Since $A D, B E, C F$ are concurrent at $O$, then
$$
\begin{array}{l}
\frac{O D}{A D}+\frac{O E}{B E}+\frac{O F}{C F} \\
=1 . \\
\text { and } \frac{O D}{A D}=\frac{R-D M}{2 R-D M} \\
=1-\frac{R}{2 R-D M}=1-\frac{R}{A D} .
\end{array}
$$
Similarly, $\frac{O E}{B E}=1-... | \frac{1}{A D}+\frac{1}{B E}+\frac{1}{C F}=\frac{2}{R} | Geometry | proof | Yes | Yes | cn_contest | false | 724,008 |
Three. (25 points) Let $k$ be a positive integer. Prove:
(1) If $k$ is the product of two consecutive positive integers, then $25 k+6$ is also the product of two consecutive positive integers;
(2) If $25 k+6$ is the product of two consecutive positive integers, then $k$ is also the product of two consecutive positive i... | (1) If $k$ is the product of two consecutive positive integers (let $k=n(n+1), n$ be a positive integer), then
$$
\begin{array}{l}
25 k+6=25 n(n+1)+6 \\
=25 n^{2}+25 n+6=(5 n+2)(5 n+3)
\end{array}
$$
is the product of two consecutive positive integers.
(2) If $25 k+6$ is the product of two consecutive positive integer... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,009 |
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \text { and } \\
a_{n+2}=\left|a_{n+1}\right|-a_{n}
\end{array}\left(n \in \mathbf{N}_{+}\right) .
$$
Let $\left\{a_{n}\right\}$'s sum of the first $n$ terms be $S_{n}$. Then $S_{100}=$ | - 1. 89.
From the given, $a_{k+9}=a_{k}$.
Then $S_{100}=a_{1}+11\left(a_{1}+a_{2}+\cdots+a_{9}\right)=89$. | 89 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,010 |
2. In $\triangle A B C$, it is known that the angle bisector of $\angle B$ intersects $A C$ at point $K$. If $B C=2, C K=1, B K=\frac{3 \sqrt{2}}{2}$, then the area of $\triangle A B C$ is | 2. $\frac{15 \sqrt{7}}{16}$.
As shown in Figure 1, let $A C = b, A B = c$. Then, by the cosine rule, we have
$$
\begin{array}{l}
8 + 2 b^{2} - 2 c^{2} \\
= b .
\end{array}
$$
Also, $\frac{A K}{C K} = \frac{A B}{B C}$, so
$$
\frac{b-1}{1} = \frac{c}{2} \text{. }
$$
From equations (1) and (2), we know $b = \frac{5}{2}... | \frac{15 \sqrt{7}}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,011 |
3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . . | 3. 98.
Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.
Expanding and rearranging according to the... | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,012 |
4. Among the positive integers less than 20, each time three numbers are taken without repetition, so that their sum is divisible by 3. Then the number of different ways to do this is $\qquad$ . | 4.327.
Divide these 19 numbers into three categories based on the remainder when divided by 3:
$$
\begin{array}{l}
A_{1}: 3,6,9,12,15,18 ; \\
A_{2}: 2,5,8,11,14,17 ; \\
A_{3}: 1,4,7,10,13,16,19 .
\end{array}
$$
Thus, the number of ways to satisfy the conditions of the problem are only four scenarios.
(1) Choose any t... | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,013 |
Example 7 Given the number table
$$
\left(\begin{array}{cccc}
-1 & 2 & -3 & 4 \\
-1.2 & 0.5 & -3.9 & 9 \\
\pi & -12 & 4 & -2.5 \\
63 & 1.4 & 7 & -9
\end{array}\right) \text {. }
$$
Changing the sign of all numbers in any row or any column is called a "transformation". Question: Can all the numbers in the table be made... | Solution: Since each transformation changes the signs of four numbers in the table, and $(-1)^{4}=1$, the transformation will not change the sign of the product of all numbers in the table.
At the beginning, the product of the 16 numbers in the table is negative (overall), so no matter how many times the transformatio... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,014 |
5. If $x, y, z$ are all positive real numbers, and $x^{2}+y^{2}+z^{2}=1$, then the minimum value of $S=\frac{(z+1)^{2}}{2 x y z}$ is $\qquad$ . | $5.5+2 \sqrt{2}$.
Since $2 x y \leqslant x^{2}+y^{2}=1-z^{2}$, we have,
$$
\begin{aligned}
S & =\frac{(z+1)^{2}}{2 x y z} \geqslant \frac{(z+1)^{2}}{z\left(1-z^{2}\right)} \\
& =\frac{z+1}{(1-z) z}=\frac{z+1}{[2-(z+1)][(z+1)-1]} \\
& =\frac{1}{3-\left[(z+1)+\frac{2}{z+1}\right]} \\
& \geqslant \frac{1}{3-2 \sqrt{2}}=3... | 3+2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,015 |
6. Given the ellipse $\frac{x^{2}}{4}+y^{2}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, and $M$ as a point on the ellipse other than the endpoints of the major axis, $\angle F_{1} M F_{2}=2 \theta$, and the incenter of $\triangle M F_{1} F_{2}$ is $I$. Then $|M I| \cos \theta=$ $\qquad$ | $6.2-\sqrt{3}$.
Let the incircle of $\triangle M F_{1} F_{2}$ touch the sides $M F_{1}$ and $M F_{2}$ at points $D$ and $E$, respectively. Then
$$
M D=M E=\frac{1}{2}\left(M F_{1}+M F_{2}-F_{1} F_{2}\right) .
$$
Since $M D=2 M I \cos \theta$, we have
$$
M F_{1}+M F_{2}-F_{1} F_{2}=2 M I \cos \theta \text {. }
$$
In $... | 2-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,016 |
7. For all $x \in\left[-2, \frac{1}{2}\right]$, the inequality
$$
a x^{3}-x^{2}+x+1 \geqslant 0
$$
always holds. Then the range of the real number $a$ is $\qquad$ $\because$ | 7. $-10 \leqslant a \leqslant-1$.
Let $f(x)=a x^{3}-x^{2}+x+1$, the given condition is that $f(x) \geqslant 0$ for all $x \in\left[-2, \frac{1}{2}\right]$.
(1) When $x=0$, for all real numbers $a$, $f(x)=1>0$.
(2) When $x \in\left(0, \frac{1}{2}\right]$,
$$
f(x)=a x^{3}-x^{2}+x+1 \geqslant 0
$$
can be transformed int... | -10 \leqslant a \leqslant-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,017 |
8. Place 10 numbers on a given circle such that their total sum is 200, and the sum of any three consecutive numbers is not less than 58. Then the maximum value of the largest number among all sets of 10 numbers that satisfy the above requirements is $\qquad$ | 8. 26 .
Let the maximum number in all placements be $A$. Then
$$
A+3 \times 58 \leqslant 200 \Rightarrow A \leqslant 26 \text {. }
$$
In fact, $26,6,26,26,6,26,26,6,26,26$ satisfies. | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,018 |
9. (16 points) Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=\frac{1}{4}$,
and $a_{n+1}=\frac{(n-1) a_{n}}{n-a_{n}}(n=2,3, \cdots)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove that for all $n \in \mathbf{N}_{+}$, $\sum_{k=1}^{n} a_{k}^{2}<\frac{7}{6}$. | (1) From the given, for $n \geqslant 2$ we have
$$
\frac{1}{a_{n+1}}=\frac{n-a_{n}}{(n-1) a_{n}}=\frac{n}{(n-1) a_{n}}-\frac{1}{n-1} \text {. }
$$
Dividing both sides by $n$ and rearranging, we get
$$
\frac{1}{n a_{n+1}}-\frac{1}{(n-1) a_{n}}=-\left(\frac{1}{n-1}-\frac{1}{n}\right) \text {. }
$$
Then, $\sum_{k=2}^{n-... | \sum_{k=1}^{n} a_{k}^{2}<\frac{7}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,019 |
10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$
Find the integer value(s) of $x$ that make $P$ a perfect square. | 10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.
Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,020 |
11. (20 points) Given the line $y=x$ intersects the ellipse $C$ : $\frac{x^{2}}{16}+\frac{y^{2}}{11}=1$ at points $A$ and $B$, and a line $l$ passing through the right focus $F$ of the ellipse $C$ with an inclination angle of $\alpha$ intersects the chord $AB$ at point $P$, and intersects the ellipse $C$ at points $M$ ... | 11. (1) Let the inclination angle of the line $M N$ be $\alpha$, and let $\angle M F O=\theta$. Then
$$
\begin{array}{l}
\alpha+\theta=\pi, \\
|M N|=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \alpha}=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta} .
\end{array}
$$
The angle between $A B$ and $M N$ is $\frac{\pi}{4}+\theta$... | y=-\frac{1}{2} x+\frac{\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,021 |
1. Use interval notation to represent the domain of the function
$$
f(x)=\ln \left(\frac{1-x}{x+3}-1\right)
$$
as . $\qquad$ | $-1 .(-3,-1)$.
From $\frac{1-x}{x+3}-1>0$, we get
$$
\frac{-2-2 x}{x+3}>0 \Rightarrow-3<x<-1 \text {. }
$$
Therefore, the domain of $f(x)$ is $(-3,-1)$. | (-3,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,022 |
2. In $\triangle A B C$, if $\sin A+\cos A=-\frac{1}{3}$, then $\cos 2 A=$ . $\qquad$ | 2. $\frac{\sqrt{17}}{9}$.
From $\sin A+\cos A=-\frac{1}{3}$, we know
$$
1+\sin 2 A=\frac{1}{9} \Rightarrow \sin 2 A=-\frac{8}{9} \text {. }
$$
From $0<\angle A<\pi, \sin A+\cos A=-\frac{1}{3}$, we know
$$
\frac{3 \pi}{4}<\angle A<\pi \text {. }
$$
Therefore, $\cos 2 A=\sqrt{1-\left(-\frac{8}{9}\right)^{2}}=\frac{\sq... | \frac{\sqrt{17}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,023 |
3. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1}-2 a_{n}=2^{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the smallest positive integer $n$ for which $a_{n}>10$ holds is $\qquad$ . . | 3.3.
Given $a_{n+1}-2 a_{n}=2^{n+1}$, we know $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=1$.
Thus, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $a_{1}=2$, so $\frac{a_{n}}{2^{n}}=n \Rightarrow a_{n}=n \times 2^{n}$. Therefore, $\left\{a_{n}\right\}$... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,024 |
Example 8 In a $10 \times 10$ grid, natural numbers 1 to 100 are written; the first row from left to right contains $1 \sim 10$; the second row from left to right contains $11 \sim 20, \cdots \cdots$ and so on. Andrei attempts to divide the entire grid into $1 \times 2$ rectangles, and then calculate the product of the... | Solve: Call a $1 \times 2$ rectangle a "domino," and number the dominoes. Let the two numbers written in the $i$-th domino be $a_{i}, b_{i}$. Then,
$$
a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2} .
$$
Write such an expression for each domino. After summing, we find that the sum $S$ ... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,025 |
4. Let $f(x)$ be an odd function defined on $\mathbf{R}$, such that for any $x \in \mathbf{R}$, $f(x+2)=f(x)$, and when $x \in(0,1)$, $f(x)=x^{2}$. Then
$$
f\left(-\frac{3}{2}\right)+f(1)=
$$
$\qquad$ | 4. $\frac{1}{4}$.
From the problem, we know
$$
f\left(-\frac{3}{2}\right)=f\left(-\frac{3}{2}+2\right)=f\left(\frac{1}{2}\right)=\frac{1}{4} \text {. }
$$
Since $f(x)$ is an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have $f(x+2)=f(x)$, therefore,
$$
\begin{array}{l}
f(1)=f(-1)=-f(1), f(... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,026 |
5. As shown in Figure 1, in the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a square,
$\triangle PAB$ is an equilateral triangle, $O$ is the midpoint of side $AB$,
and $PO \perp$ plane $ABCD$. Then the cosine value of the dihedral angle $P-AC-D$ is $\qquad$ | 5. $-\frac{\sqrt{7}}{7}$.
Solution 1 As shown in Figure 4, draw $O E \perp A C$ at point $E$, and connect $P E$.
From $P O \perp$ plane $A B C D$, we know that $P E \perp A C, \angle P E O$ is the plane angle of the dihedral angle $P-A C-B$.
Let $A B=2$. Then
$$
P O=\sqrt{3}, O E=\frac{\sqrt{2}}{2}, P E=\frac{\sqrt{1... | -\frac{\sqrt{7}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,027 |
6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m 、 n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\q... | 6. 662 .
According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$
the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
... | 662 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,028 |
7. The minimum value of the function $f(x)=\sin ^{2 k} x+\cos ^{2 k} x\left(k \in \mathbf{N}_{+}\right)$ is . $\qquad$ | 7. $\left(\frac{1}{2}\right)^{k-1}$.
Let $\sin ^{2} x=t$. Then $0 \leqslant t \leqslant 1$, and
$$
f(x)=g(t)=t^{k}+(1-t)^{k} \text {. }
$$
Notice that
$$
g^{\prime}(t)=k t^{k-1}-k(1-t)^{k-1}\left(k \in \mathbf{N}_{+}\right) \text {. }
$$
Then when $0<t<\frac{1}{2}$, $g^{\prime}(t)<0$; when $\frac{1}{2}<t<1$, $g^{\pr... | \left(\frac{1}{2}\right)^{k-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,029 |
8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$). | 8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$
When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,030 |
9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positiv... | 9.3.
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
It is verified that $m=1, m=2$ do not meet t... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,031 |
10. As shown in Figure 2, let the area of the triangle formed by any three points chosen from the nine intersection points of the "田" shaped grid (consisting of four squares with side length 1) be $\xi$ (when the three points are collinear, $\xi=0$), then the mathematical expectation $E \xi=$ $\qquad$
Figure 2 | 10. $\frac{5}{6}$.
As shown in Figure 6, the number of ways to choose three points from nine points is $\mathrm{C}_{9}^{3}=84$.
When $\xi=0$, the three points are collinear, with 8 cases.
When $\xi=\frac{1}{2}$, there are 32 cases (12 cases each with $A B, B C, G H, H I$ as sides, totaling 12; 6 cases each with $D E... | \frac{5}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,032 |
11. For what value of the real number $a$ does the equation $a x=\ln x$ have no solution, one solution, or two solutions? | When $a \leqslant 0$, in the same coordinate system, draw the graphs of the functions $y=a x$ and $y=\ln x$. It is easy to see that the two graphs have exactly one intersection point, so the equation has exactly one solution.
When $a>0$, let $f(x)=a x-\ln x$. Then
$$
f^{\prime}(x)=a-\frac{1}{x}=\frac{a x-1}{x}(x>0) \te... | When\ a>\frac{1}{\mathrm{e}},\ the\ original\ equation\ has\ no\ solution;\ When\ a\ \leqslant\ 0\ or\ a=\frac{1}{\mathrm{e}},\ the\ original\ equation\ has\ one\ solution;\ When\ 0<a<\frac{1}{\mathrm{e}},\ the\ original\ equation\ has\ two\ solutions. | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 724,033 |
12. Given the function $f(x)=x|x-2a|$. Try to find the maximum value $g(a)$ of $f(x)$ on the interval $[0,1]$. | 12. When $x \in[0,1]$,
$$
\begin{array}{l}
f(x)=x|x-2 a|=\left|x^{2}-2 a x\right| \\
=\left|(x-a)^{2}-a^{2}\right| .
\end{array}
$$
Let $h(x)=(x-a)^{2}-a^{2}$.
(1) $a \leqslant 0$.
Then $h(x)$ is an increasing function on the interval $[0,1]$, and $h(x) \geqslant 0$, at this time, $g(a)=h(1)=1-2 a$.
(2) $a \geqslant ... | g(a)=\left\{\begin{array}{ll}
1-2 a, & a<\sqrt{2}-1 ; \\
a^{2}, & \sqrt{2}-1 \leqslant a<1 ; \\
2 a-1, & a \geqslant 1 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,034 |
13. As shown in Figure 3, in the acute triangle $\triangle ABC$, $AB = AC$, the angle bisector of $\angle ACB$ intersects $AB$ at point $D$. A perpendicular line to $CD$ through the circumcenter $O$ of $\triangle ABC$ intersects $AC$ at point $E$. A line parallel to $AB$ through $E$ intersects $CD$ at point $F$. Prove:... | 13. (1) As shown in Figure 7, connect $O C$, $O A$, and $O F$.
Since $A B=A C$ and $O$ is the circumcenter of $\triangle A B C$,
we have $O A$ bisects $\angle B A C$, and $O A=O C$.
Thus, $\angle O C A$
$$
\begin{aligned}
& =\angle O A C \\
& =\frac{1}{2} \angle B A C \\
= & \frac{1}{2}\left(180^{\circ}-2 \angle B\ri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,035 |
1. The family consists of four people. If Martha's scholarship is doubled, the family's total income will increase by $5 \%$; if Mom's salary is doubled, the total income will increase by $15 \%$; if Dad's salary is doubled, the total income will increase by $25 \%$. What percentage will the family's total income incre... | If each person's income doubles, the total family income will increase by $100 \%$. Of this $100 \%$, $5 \%$ belongs to Martha, $15 \%$ belongs to Mom, $25 \%$ belongs to Dad, and the remaining $55 \%$ is contributed by Grandpa. | 55\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,036 |
14. Given the hyperbola \( C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \) with an eccentricity of 2, a line \( l \) passing through the point \( P(0, m) (m>0) \) with a slope of 1 intersects the hyperbola \( C \) at points \( A \) and \( B \), and
\[
\overrightarrow{A P}=3 \overrightarrow{P B}, \overrightarr... | 14. (1) From the eccentricity of the hyperbola being 2, we know that $c=2a, b=\sqrt{3}a$.
Thus, the equation of the hyperbola becomes $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1$.
The equation of the line $l$ is $y=x+m$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1, \\ y=x+m,\end{array}\right... | x^{2}-\frac{y^{2}}{3}=1, M(-1,0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,037 |
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