problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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1. Given $n$ is a natural number, $n^{2}+4 n+2009$ can be expressed as the sum of the squares of four consecutive natural numbers. Then the sum of all $n$ that satisfy this condition is $\qquad$ . | 2.1.712.
Let $n^{2}+4 n+2009$
$$
=(m-1)^{2}+m^{2}+(m+1)^{2}+(m+2)^{2} \text {, }
$$
where $m$ is a positive integer. Then
$$
(n+2)^{2}+2005=(2 m+1)^{2}+5 \text {, }
$$
which implies $(2 m+1)^{2}-(n+2)^{2}=2000$.
Factoring, we get
$$
(2 m+n+3)(2 m-n-1)=2000 \text {. }
$$
Since $2 m+n+3$ and $2 m-n-1$ have the same pa... | 712 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,108 |
2. Given $k(k \geqslant 5)$ is a positive integer, the subset $M$ of $\{1,2, \cdots, 2 k-1\}$ satisfies: the sum of any number of elements in $M$ (at least one, or all) is not equal to $2 k$. Then $M$ can have at most $\qquad$ elements. | 2.k.
Divide $\{1,2, \cdots, 2 k-1\}$ into $k$ groups:
$$
\{1,2 k-1\},\{2,2 k-2\}, \cdots,\{k-1, k+1\},\{k\} \text {. }
$$
If $|M| \geqslant k+1$, then $M$ must contain two elements in the same group, whose sum is $2 k$, which is a contradiction.
Therefore, $|M| \leqslant k$.
Furthermore, take $M=\{k, k+1, \cdots, 2 k... | k | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,109 |
4. As shown in Figure 3, in the kite $ABCD$, $AB=AD$, $CB=CD$, the angle bisector of $\angle BDC$ intersects $BC$ at point $L$. Given that points $A$, $L$, $C$, and $D$ are concyclic. Then $2 \angle BAD + 3 \angle BCD$ = . $\qquad$ | $4.540^{\circ}$.
As shown in Figure 8, connect
$A L$ intersecting $B D$ at point $M$.
Let $A C$ and $B D$ intersect at point
$O$, and construct the symmetric point $L^{\prime}$ of $L$ with respect to $A C$. It is easy to see that $L^{\prime}$
lies on segment $C D$. Connect $A L^{\prime}$ and $M L^{\{\prime}$.
Since $A,... | 540^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,111 |
II. (25 points) As shown in Figure 4, given that the orthocenter of acute triangle $\triangle ABC$ is $H$, the midpoint of side $BC$ is $M$, the extension of $MH$ intersects the circumcircle $\odot O$ of $\triangle ABC$ at point $P$, and $OH \perp AM$. Prove: $AP$, $OH$, and $BC$ are concurrent. | II. From $O H \perp A M$, we know $A B \neq A C$.
Assume without loss of generality that $A B > A C$.
As shown in Figure 11, connect $O M$, and draw $A D \perp B C$ at point $D$. Then $H$ lies on $A D$.
By $A H=\frac{A B \cos \angle B A C}{\sin \angle A C B}=2 R \cos \angle B A C$,
$$
O M=R \cos \angle B A C,
$$
where... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,113 |
Three. (25 points) A city has built an urban subway to facilitate the travel of its citizens, with a total of 20 stations (including the starting and terminal stations). The subway company is preparing to submit a pricing scheme to the Price Bureau. There are two pricing schemes.
【Scheme One】Fare for each passenger = ... | Three, Plan Two.
Assume the number of people boarding at each station is $k$.
According to Plan One:
The average number of stations traveled by passengers boarding at the $i(i=1,2, \cdots, 19)$ station is
$$
\frac{1}{2}[1+(20-i)]=\frac{1}{2}(21-i) .
$$
Therefore, the ticket revenue is
$$
\begin{array}{l}
k \times 0.5 ... | 65k > \frac{209k}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,114 |
Example 1: Prove that $6411\left(2^{32}+1\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: Given $2^{9} \equiv 512 \equiv-129(\bmod 641)$, then
$$
\begin{array}{l}
2^{11} \equiv-516 \equiv 125(\bmod 641) \\
\Leftrightarrow 2^{13} \equiv 500 \equiv-141(\bmod 641) \\
\Leftrightarrow 2^{15} \equiv-564 \equiv 77(\bmod 641) \\
\Leftrightarrow 2^{18} \equiv 616 \equiv-25(\bmod 641) \\
\Leftrightarrow 2^{2} ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,115 |
Example 2 Starting from the last digit of $n$, every three digits are considered as one number, these numbers are sequentially denoted as $A_{1}, A_{2}, \cdots, A_{r}\left(A_{1}\right.$, $A_{2}, \cdots, A_{r-1}$ are all three-digit numbers, while $A_{r}$ may be less than three digits). Let
$$
\begin{array}{l}
S_{1}=A_{... | Proof: Let $n=\overline{a_{k} a_{k-1} \cdots a_{1} a_{0}}$. Then
$$
\begin{array}{l}
n=\overline{a_{2} a_{1} a_{0}}+a_{5} a_{4} a_{3} \cdot 10^{3}+\overline{a_{8} a_{7} a_{6}} \cdot 10^{6}+\cdots \\
=A_{1}+A_{2} \times 10^{3}+A_{3} \times 10^{6}+\cdots+A_{r} \times 10^{3 r-3} \\
\equiv A_{1}-A_{2}+A_{3}-A_{4}+\cdots+A_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,116 |
3. Find the remainder when $47^{37^{2}}$ is divided by 7.
Try to find the remainder of $47^{37^{2}}$ when divided by 7. | (Note that $47^{6} \equiv(-2)^{6} \equiv 2^{6} \equiv 8^{2} \equiv 1^{2}$ $\equiv 1(\bmod 7)$. Also, $37^{23} \equiv 1^{23} \equiv 1(\bmod 6)$, let $37^{23}$ $=6k+1$. Then $47^{37^{23}}=47^{k+1}=\left(47^{6}\right)^{k} \times 47 \equiv$ $\left.1^{k} \times 47 \equiv 47 \equiv 5(\bmod 7).\right)$ | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,117 |
7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.6.
Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$
and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,118 |
8. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c$ $=0$ form a geometric progression. Then $a^{3} c-b^{3}$ $=$ . $\qquad$ | 8.0.
Let the three roots be $d$, $d q$, and $d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a, \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b, \\
d^{3} q^{3}=-c .
\end{array}\right.
$$
Dividing (2) by (1) gives $d q=-\frac{b}{a}$.
Substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,119 |
9. (14 points) Given that $E$ is a point on the edge $BC$ of a regular tetrahedron $ABCD$ with edge length 2. The inradii of the tetrahedra $ABDE$ and $ACDE$ are $r_{1}$ and $r_{2}$, respectively. If $r_{1}^{-1}+r_{2}^{-1}=4 \sqrt{6}+4 \sqrt{3}$, then there are two points $E$ that satisfy the condition, denoted as $E_{... | Let $F$ be the midpoint of $BC$, and $EF = x$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
V_{\text {tetrahedron } 1 C D E}=\frac{\sqrt{2}}{3}(1+x), \\
S_{\triangle A B E}=S_{\triangle D B E}=\frac{\sqrt{3}}{2}(1-x), \\
S_{\triangle M C E}=S_{\triangle D C E}=\frac{\sqrt{3}}{2}(1+x),
\end{array}
$$
Then,
$... | \frac{2 \sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,120 |
10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$
Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{... | 10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$
Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,121 |
11. (15 points) Does there exist a sequence of positive integers $\left\{x_{n}\right\}$, satisfying
$$
x_{n+2}=\left(x_{n+1}, x_{n}\right)+2009(n=1,2, \cdots) \text {, }
$$
where $(u, v)$ denotes the greatest common divisor of $u$ and $v$, and the sequence contains at least $10^{2009}$ distinct numbers? | 11. Existence.
Below, we construct a sequence that satisfies the problem's requirements and has a sufficient number of distinct terms.
$$
\begin{array}{l}
\text { Let } a=2009 \text {. Take } \\
x_{1}=a+1, x_{2}=3 a+1, \\
x_{k+2}=\left(x_{k}-a\right)\left(x_{k+1}-a\right)(k=1,2, \cdots) .
\end{array}
$$
From $x_{k+3}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,122 |
One, (50 points) As shown in Figure 1, given that $\odot I, \odot O$ are the incircle and circumcircle of quadrilateral $ABCD$, respectively, $\odot I$ touches $AB, BC, CD, DA$ at points $E, F, G, H$. Construct $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}$ to touch $AB, BC, CD, DA$ at points $E, F, G, H$, respec... | (1) As shown in Figure 3. Clearly, $O, O_{1}, K, I, E, O_{1}$ are collinear in pairs, and $\frac{O K}{K O_{1}}=\frac{R}{r_{1}}, \frac{O_{1} E}{E I}=\frac{r_{1}}{r}\left(R, r, r_{1}\right.$ are the radii of $\odot O$, $\odot I, \odot O_{1}$, respectively).
Let the internal division point $P$ on segment $O I$ satisfy $\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,123 |
Three. (50 points) Given $m n\left(m, n \in \mathbf{N}_{+}, n\right.$ is even) consecutive positive integers arranged in a sequence, denoted as $A=\left\{a_{i}\right\}_{i=1,2, \cdots, m}$. Rearrange the elements of $A$ to form a new sequence $\sigma(A)=\left\{b_{i}\right\}_{i=1,2, \cdots, m n}$. Define
$$
\begin{array}... | For any new permutation $\sigma(A)=\left\{b_{i}\right\}_{i=1,2, \cdots, m}$, we have $f(\sigma(A)) \geqslant \sum_{k=1}^{n} b_{n i+k} \geqslant g(\sigma(A))$,
where $i=0,1, \cdots, m-1$.
Adding these $m$ inequalities, we get
$m f(\sigma(A)) \geqslant \sum_{i=1}^{m} b_{i}=\sum_{i=1}^{m} a_{i} \geqslant m g(\sigma(A))$.
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,125 |
Initially 247 Suppose there are $n$ square cards of the same size, each inscribed with a positive integer. These $n$ positive integers are consecutive two-digit numbers, and the sum of these two-digit numbers equals the "composite number" of the first and last terms (such as $1+2+\cdots+5=15,4+5+\cdots$ $+29=429$). If ... | Solution: Clearly, $2009=41 \times 7^{2}$.
Given that the side lengths of the cards are integers greater than 1, and the sum of the areas of these cards is 2009, it follows that the area of each card is 49, and the total number of cards is $n=41$.
Let the numbers on the cards be
$$
\begin{array}{l}
a_{1}=10 a+b, a_{2}=... | 13,14, \cdots, 52,53 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,127 |
4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101. | (Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$
Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
) | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,128 |
As shown in Figure 2, in $\triangle ABC$, $AD$ is the internal angle bisector, $\angle ADC = 60^{\circ}$, point $E$ is on $AD$ such that $DE = DB$, and ray $CE$ intersects $AB$ at point $F$. Prove that $AF \cdot AB + CD \cdot CB = AC^2$. | Proof: As shown in Figure 2, extend $AC$ to $G$ such that $AG = AB$, and connect $DG$.
It is easy to see that $\triangle ADG \cong \triangle ADB$, thus $\angle G = \angle B$, $DG = DB$, and $\angle ADG = \angle ADB$. Since $DE = DB$, we have $DG = DE$.
Because $\angle ADG = \angle ADB$
$$
= 180^{\circ} - \angle ADC = 1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,129 |
For $a_{n}=\sqrt{\frac{n}{n+2}}-\frac{n}{n+1}$, let its sum of the first $n$ terms be $S_{n}$. Prove: for any positive integer $n$, we have
$$
0<S_{n}<1-\frac{\sqrt{2}}{2} .
$$ | Proof: First, we prove two lemmas.
Lemma 1 For positive integers $a_{1}, a_{2}, \cdots, a_{n}>0$, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \\
\geqslant \sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{n} a_{1}} .
\end{array}
$$
Proof of Lemma 1: Multiply both sides by 2, and use $a+b \geqslant 2 \sqr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,130 |
248 Given $P$ is a point inside $\triangle ABC$, $A_{1}$, $B_{1}$, $C_{1}$ are the midpoints of $\overparen{BPC}$, $\overparen{CPA}$, $\overparen{APB}$ respectively, and $P$, $A_{1}$, $B_{1}$, $C_{1}$ are distinct points. Prove: $P$, $A_{1}$, $B_{1}$, $C_{1}$ are concyclic. | Proof: As shown in Figure 3, connect the segments. Let $\angle B P C$, $\angle C P A$, and $\angle A P B$ be $\alpha$, $\beta$, and $\gamma$ respectively.
Since $P, \dot{A}_{1}$, $B_{1}$, and $C_{1}$ are not collinear, the segments $P A$, $P B$, and $P C$ are not equal. Assume $P A > P B > P C$. Then,
$$
\begin{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,131 |
6. Let $p$ and $8 p^{2}+1$ both be prime numbers. Prove that $8 p^{2}-p+2$ is also a prime number. | (Tip: If $p \equiv 0(\bmod 3)$, then $p=3$. In this case, $8 p^{2}-p+2=72-3+2=71$ is a prime number.
If $p \equiv \pm 1(\bmod 3), 8 p^{2}+1 \equiv 8+1 \equiv$ $0(\bmod 3)$, i.e., $3 \mid\left(8 p^{2}+1\right)$, which is a contradiction.) | 71 | Number Theory | proof | Yes | Yes | cn_contest | false | 721,133 |
7. Let $p$ be a prime number, and $q=4^{p}+p^{4}+4$ is also a prime number. Find the value of $p+q$.
The text above is translated into English, keeping the original text's line breaks and format. | (Tip: Mod 3 we get $q=4^{p}+p^{4}+4 \equiv 1^{p}+p^{4}+$ $1 \equiv p^{4}+2(\bmod 3)$. Also, $p^{2} \equiv 0,1(\bmod 3)$, so $p^{4}$ $\equiv 0,1(\bmod 3)$.
If $p^{4} \equiv 1(\bmod 3)$, then $q \equiv p^{4}+2 \equiv 1+2 \equiv$ $0(\bmod 3)$. But $q=4^{p}+p^{4}+4>3$, which contradicts the fact that $q$ is a prime number... | 152 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,134 |
Example 1 Let $a, b, c$ be positive numbers, and $ab + bc + ca = 3$. Prove:
$$
\begin{array}{l}
\frac{1}{1+a^{2}(b+c)}+\frac{1}{1+b^{2}(c+a)}+\frac{1}{1+c^{2}(a+b)} \\
\leqslant \frac{1}{abc}
\end{array}
$$ | Proof: According to the problem, by the AM-GM inequality, we have
$$
a b+b c+c a=3 \geqslant 3 \sqrt[3]{(a b c)^{2}},
$$
which implies $a b c \leqslant 1$.
$$
\begin{array}{l}
\text { Hence } \frac{1}{1+a^{2}(b+c)} \leqslant \frac{1}{a b c+a^{2}(b+c)} \\
=\frac{1}{a(a b+b c+c a)}=\frac{1}{3 a} .
\end{array}
$$
Simila... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,135 |
Example 2 Let $x, y, z$ be positive real numbers, and $x+y+z=3$. Prove:
$$
\frac{x^{3}}{y^{3}+8}+\frac{y^{3}}{z^{3}+8}+\frac{z^{3}}{x^{3}+8} \geqslant \frac{1}{9}+\frac{2}{27}(x y+y z+z x) .
$$ | Proof: By the AM-GM inequality,
$$
\begin{array}{l}
\frac{x^{3}}{y^{3}+8}+\frac{y+2}{27}+\frac{y^{2}-2 y+4}{27} \\
\geqslant 3 \sqrt[3]{\frac{x^{3}}{y^{3}+8} \cdot \frac{y+2}{27} \cdot \frac{y^{2}-2 y+4}{27}}=\frac{x}{3} .
\end{array}
$$
Similarly, $\frac{y^{3}}{z^{3}+8}+\frac{z+2}{27}+\frac{z^{2}-2 z+4}{27} \geqslant... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,136 |
Example 3 Let $x, y, z$ be non-negative numbers, and $x^{2}+y^{2}+z^{2}=3$. Prove:
$$
\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+\frac{z}{\sqrt{z^{2}+x+y}} \leqslant \sqrt{3} \text {. }
$$ | Prove: By the Cauchy-Schwarz inequality,
$$
3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+z)^{2} \text {. }
$$
Since \( x^{2}+y^{2}+z^{2}=3 \), we have
$$
x^{2}+y^{2}+z^{2} \geqslant x+y+z \text {. }
$$
By the Cauchy-Schwarz inequality,
$$
\left(x^{2}+y+z\right)(1+y+z) \geqslant(x+y+z)^{2} \text {. }
$$
Thus, it suf... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,137 |
Example 4 Let $a, b, c \in \left(\frac{1}{\sqrt{6}}, +\infty\right)$, and $a^{2} + b^{2} + c^{2} = 1$. Prove:
$$
\begin{array}{l}
\frac{1+a^{2}}{\sqrt{2 a^{2}+3 a b-c^{2}}}+\frac{1+b^{2}}{\sqrt{2 b^{2}+3 b c-a^{2}}}+\frac{1+c^{2}}{\sqrt{2 c^{2}+3 a b-b^{2}}} \\
\geqslant 2(a+b+c) .
\end{array}
$$ | $$
\begin{array}{l}
\left(\sqrt{2 a^{2}+3 a b-c^{2}}+\sqrt{2 b^{2}+3 b c-a^{2}}+\sqrt{2 c^{2}+3 a c-b^{2}}\right) \\
\left(\frac{a^{2}}{\sqrt{2 a^{2}+3 a b-c^{2}}}+\frac{b^{2}}{\sqrt{2 b^{2}+3 b c-a^{2}}}+\frac{c^{2}}{\sqrt{2 c^{2}+3 a c-b^{2}}}\right) \\
\geqslant(a+b+c)^{2}, \\
\left(\sqrt{2 a^{2}+3 a b-c^{2}}+\sqrt{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,138 |
Example 5 Given that $a, b, c$ are all positive real numbers. Prove:
$$
(a+b)^{3}+4 c^{3} \geqslant 4\left(\sqrt{a^{3} b^{3}}+\sqrt{b^{3} c^{3}}+\sqrt{c^{3} a^{3}}\right) .
$$ | $$
\begin{array}{l}
(a+b)^{3}+4 c^{3} \\
=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}+4 c^{3} \\
=2\left(a^{2} b+a b^{2}\right)+\left(a^{2}+b^{2}\right)(a+b)+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} b^{3}}+\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right)^{2}+4 c^{3} \\
\geqslant 4 \sqrt{a^{3} \cdot b^{3}}+4 c^{\frac{3}{2}}\left(a^{\frac{3}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,139 |
Example 6 Let $x, y, z$ all be positive numbers, and $x+y+z \geqslant 1$. Prove:
$$
\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y} \geqslant \frac{\sqrt{3}}{2} .
$$ | Proof: By the AM-GM inequality, we have $x^{\frac{3}{2}}+y^{\frac{3}{2}}+y^{\frac{3}{2}} \geqslant 3 x^{\frac{1}{2}} y, x^{\frac{3}{2}}+z^{\frac{3}{2}}+z^{\frac{3}{2}} \geqslant 3 x^{\frac{1}{2}} z$. Adding these, we get $2\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^{\frac{3}{2}}\right) \geqslant 3 x^{\frac{1}{2}}(y+z)$.
T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,141 |
Example 7 Let $a, b, c$ be positive numbers. Prove that:
$$
\left(1+\frac{4 a}{b+c}\right)\left(1+\frac{4 b}{c+a}\right)\left(1+\frac{4 c}{a+b}\right)>25 .
$$ | Prove: Notice
$$
\begin{array}{l}
\left(1+\frac{4 a}{b+c}\right)\left(1+\frac{4 b}{c+a}\right)\left(1+\frac{4 c}{a+b}\right)>25 \\
\Leftrightarrow(b+c+4 a)(c+a+4 b)(a+b+4 c) \\
\quad>25(a+b)(b+c)(c+a) \\
\Leftrightarrow a^{3}+b^{3}+c^{3}+7 a b c \\
\quad>a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+a c^{2} .
\end{array}
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,142 |
Example 8 Let $x, y, z$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} . \\
a^{2}+b^{2}+c^{2}>2 \sqrt[3]{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}},
\end{array}
$$
i.e.,
$$
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)^{3}>8\left(a^{3} b^{3}+b^{... | Proof: Let $\frac{x y}{z}=a^{2}, \frac{y z}{x}=b^{2}, \frac{z x}{y}=c^{2}$. Since $x, y, z$ are positive real numbers, we have $x=c a, y=a b, z=b c$.
Thus, the original inequality transforms into
Translate the above text into English, preserving the original text's line breaks and format, and output the translation re... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,143 |
Example 9 Given that $a, b, c$ are positive numbers, and $a+b+c=1$. Prove:
$$
\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31} .
$$ | Proof: Notice
$$
\begin{array}{l}
\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31} \\
\Leftrightarrow \frac{9 a^{2}+9 a b c+9-31 a}{a^{2}+a b c+1}+ \\
\frac{9 b^{2}+9 a b c+9-31 b}{b^{2}+a b c+1}+ \\
\frac{9 c^{2}+9 a b c+9-31 c}{c^{2}+a b c+1} \geqslant 0 .
\e... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,144 |
Example 10 Given that $x, y, z$ are positive numbers, and $x+y+z = 1, k$ is a positive integer. Prove:
$$
\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \geqslant \frac{1}{7} .
$$ | Proof: Without loss of generality, let $x \geqslant y \geqslant z$. Then $x^{k} \geqslant y^{k} \geqslant z^{k}$.
By Chebyshev's inequality, we have
$$
\begin{array}{l}
3\left(x^{k+1}+y^{k+1}+z^{k+1}\right) \\
\geqslant(x+y+z)\left(x^{k}+y^{k}+z^{k}\right) .
\end{array}
$$
Since $x \geqslant y \geqslant z$, we have
$$... | \frac{1}{7} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,145 |
1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $n$ be a positive integer. Prove:
$$
\begin{array}{l}
\prod_{i=1}^{n}\left(1+x_{1}+x_{2}+\cdots+x_{i}\right) \\
\geqslant \sqrt{(n+1)^{n+1} x_{1} x_{2} \cdots x_{n}} .
\end{array}
$$ | (Tip: For the element $y_{1}=\frac{x_{1}}{1+x_{1}}$,
$$
\begin{array}{l}
y_{2}=\frac{x_{2}}{\left(1+x_{1}\right)\left(1+x_{1}+x_{2}\right)}, \\
y_{3}=\frac{x_{3}}{\left(1+x_{1}+x_{2}\right)\left(1+x_{1}+x_{2}+x_{3}\right)}, \\
\cdots \cdots \\
y_{n}=\frac{x_{n}}{\left(1+x_{1}+\cdots+x_{n-1}\right)\left(1+x_{1}+\cdots+x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,146 |
2. Given that $a, b, c$ are all positive numbers, and $a b + b c + c a = 1$. Prove:
$$
\sqrt{a^{3}+a}+\sqrt{b^{3}+b}+\sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c} .
$$ | (Hint: Use the condition $a b+b c+c a=1$ to transform the problem into proving
$$
\begin{array}{l}
\sqrt{a(a+b)(c+a)}+\sqrt{b(a+b)(b+c)}+ \\
\sqrt{c(c+a)(b+c)} \\
\geqslant 2 \sqrt{(a+b+c)(a b+b c+c a)},
\end{array}
$$
then apply the Cauchy-Schwarz inequality and Schur's inequality.) | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,147 |
3. Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geqslant \frac{3}{2} .
$$ | (Hint: First make the substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then use the Cauchy-Schwarz inequality and the AM-GM inequality.) | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,148 |
Example 4 Find the unit digit of the natural number $2^{100}+3^{101}+4^{100}$. | Solution: Notice that
$$
\begin{array}{l}
2^{100} \equiv 2^{1 \times 25} \equiv\left(2^{4}\right)^{25} \equiv 16^{25} \\
\equiv 6^{25} \equiv 6(\bmod 10), \\
3^{101} \equiv 3^{4 \times 25+1} \equiv\left(3^{4}\right)^{25} \times 3^{1} \\
\equiv 1^{25} \times 3^{1} \equiv 3(\bmod 10), \\
4^{102} \equiv 4^{2 \times 51} \e... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,151 |
Example 1 Let $a, b, c > 0$, and $ab + bc + ca = 1$. Prove: $\frac{b^{2}+c^{2}}{\sqrt{a^{2}+1}}+\frac{c^{2}+a^{2}}{\sqrt{b^{2}+1}}+\frac{a^{2}+b^{2}}{\sqrt{c^{2}+1}} \geqslant \sqrt{3}$. | Proof: Since $a b+b c+c a=1$, then,
$$
\begin{array}{l}
\frac{b^{2}+c^{2}}{\sqrt{a^{2}+1}}=\frac{b^{2}+c^{2}}{\sqrt{a^{2}+a b+b c+c a}} \\
=\frac{2\left(b^{2}+c^{2}\right)}{2 \sqrt{(a+b)(a+c)}} \\
\geqslant \frac{(b+c)^{2}}{(a+b)+(a+c)}=\frac{(b+c)^{2}}{2 a+b+c} \\
=\left[\frac{(b+c)^{2}}{2 a+b+c}+\frac{2 a+b+c}{4}\rig... | \sqrt{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,152 |
Example 2 Let $a, b, c > 0$, and $ab + bc + ca = 1$. Prove:
$$
\sqrt{a^{3}+a}+\sqrt{b^{3}+b}+\sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c} .
$$ | Proof: Notice
$$
\begin{array}{l}
a^{3}+a=a\left(a^{2}+1\right) \\
=a\left(a^{2}+a b+b c+c a\right) \\
=a(a+b)(a+c) .
\end{array}
$$
Thus, the original inequality is equivalent to
$$
\begin{array}{l}
\sum \sqrt{a(a+b)(a+c)} \\
\geqslant 2 \sqrt{(a+b+c)(a b+b c+c a)} \\
\Leftrightarrow \sum a^{3}+2 \sum[\sqrt{a(a+b)(a+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,153 |
Example 3 Let $a, b, c > 0$, and $ab + bc + ca = 1$. Prove: $\sum \frac{1}{\left(1+a^{2}\right)\left(1+b^{2}\right)} \leqslant \frac{1}{8abc} \sum \frac{1}{a+b}$. | Prove: Notice
$$
\begin{array}{l}
\left(1+a^{2}\right)\left(1+b^{2}\right) \\
=\left(a b+b c+c a+a^{2}\right)\left(a b+b c+c a+b^{2}\right) \\
=(a+b)(a+c)(b+a)(b+c) .
\end{array}
$$
Therefore, $\sum \frac{1}{\left(1+a^{2}\right)\left(1+b^{2}\right)}$
$$
\begin{array}{l}
=\sum \frac{1}{(a+b)(a+c)(b+a)(b+c)} \\
=\frac{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,154 |
Example 4 Let $a, b, c \geqslant 0$, and $a b+b c+c a=1$.
$$
\begin{array}{l}
\text { Prove: } \frac{a^{3}}{1+b^{2}}+\frac{b^{3}}{1+c^{2}}+\frac{c^{3}}{1+a^{2}} \geqslant \frac{\sqrt{3}}{4} . \\
=\frac{1}{4} \sum a=\frac{1}{4} \sqrt{(a+b+c)^{2}} \\
\geqslant \frac{1}{4} \sqrt{3(a b+b c+c a)}=\frac{\sqrt{3}}{4} .
\end{a... | Proof: Notice that
$$
\begin{array}{l}
\frac{a^{3}}{1+b^{2}}=\frac{a^{3}}{a b+b c+c a+b^{2}}=\frac{a^{3}}{(b+a)(b+c)} \\
=\left[\frac{a^{3}}{(b+a)(b+c)}+\frac{b+a}{8}+\frac{b+c}{8}\right]-\frac{b}{4}-\frac{a}{8}-\frac{c}{8} \\
\geqslant 3 \sqrt[3]{\frac{a^{3}}{64}}-\frac{b}{4}-\frac{a}{8}-\frac{c}{8}=\frac{5 a}{8}-\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,155 |
Example 5 Let $a, b, c > 0$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a b c}$. Prove: $\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1} \leqslant \frac{3 \sqrt{3}}{4}$. | Proof: From $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a b c}$, we get $a b+b c+c a=1$.
By substituting $a b+b c+c a$ for 1 in equation (1), we have
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow \sum \frac{a}{a^{2}+a b+b c+c a} \leqslant \frac{3 \sqrt{3}}{4} \\
\Leftrightarrow \sum \frac{a}{(a+b)(a+c)} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,156 |
1. As shown in Figure 1, in the equilateral $\triangle ABC$, points $D$ and $E$ are on sides $BC$ and $CA$ respectively, such that $CD = AE$, $AD$ intersects $BE$ at point $P$, and $BQ \perp AD$ at point $Q$. Then $\frac{QP}{QB}=$ $\qquad$ | 1. $\frac{\sqrt{3}}{3}$.
It is easy to see that $\triangle A B E \cong \triangle C A D$. Then
$$
\begin{array}{l}
\angle A B E=\angle C A D \\
\angle D P B=\angle P A B+\angle A B E \\
=\angle P A B+\angle C A D=60^{\circ}
\end{array}
$$
Therefore, $\frac{Q P}{Q B}=\cot 60^{\circ}=\frac{\sqrt{3}}{3}$. | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,157 |
2. The inequality $x^{2}+|2 x-6| \geqslant a$ holds for all real numbers $x$. Then the maximum value of the real number $a$ is $\qquad$ | 2.5.
When $x \geqslant 3$,
left side $=x^{2}+2 x-6=(x+1)^{2}-7 \geqslant 9$;
When $x<3$,
left side $=x^{2}-2 x+6=(x-1)^{2}+5 \geqslant 5$.
Therefore, for the given inequality to hold for all real numbers $x$, the maximum value of $a$ should be 5. | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,158 |
3. Let $a_{n}$ denote the last digit of the number $n^{4}$. Then $a_{1}+a_{2}+\cdots+a_{2008}=$ $\qquad$ | 3.6632 .
Obviously, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, that is,
$$
1^{4}, 2^{4}, 3^{4}, 4^{4}, 5^{4}, 6^{4}, 7^{4}, 8^{4}, 9^{4}, 10^{4}
$$
the unit digits are
$$
\begin{array}{l}
1,6,1,6,5,6,1,6,1,0 \text {. } \\
\text { Then } a_{1}+a_{2}+\cdots+a_{10} \\
=1+6+1+6+5+6+1+6+1+0=33 \text {.... | 6632 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,159 |
4. In quadrilateral $A B C D$, $\angle A=60^{\circ}, A B=1$, point $E$ is on side $A B$ such that $A E: E B=2: 1, P$ is a moving point on diagonal $A C$. Then the minimum value of $P E+P B$ is $\qquad$ . | 4. $\frac{\sqrt{7}}{3}$.
Since quadrilateral $A B C D$ is a rhombus, point $B$ and $D$ are symmetric with respect to diagonal $A C$. Therefore,
$$
P E+P B=P E+P D \geqslant D E \text {. }
$$
Equality holds if and only if $P$ is the intersection of $D E$ and $A C$.
Thus, the minimum value of $P E+P B$ is the length o... | \frac{\sqrt{7}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,160 |
5. The solution to the equation $\frac{a x^{2}}{x-1}-2 a=a^{2}+1$ with respect to $x$ is $\qquad$ . | 5. When $a=0$, there is no solution; when $a \neq 0$, $x_{1}=a+1, x_{2}$ $=1+\frac{1}{a}$.
Notice that $\frac{a x^{2}}{x-1}-2 a=a^{2}+1$
$$
\Leftrightarrow\left\{\begin{array}{l}
x \neq 1, \\
a x^{2}-(a+1)^{2} x+(a+1)^{2}=0 .
\end{array}\right.
$$
When $a=0$, the left side of the original equation $=0$, the right side... | x_{1}=a+1, x_{2}=1+\frac{1}{a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,161 |
Example 5: There are $L$ squares arranged on a circle, one of which contains 1 chess piece. The chess piece moves counterclockwise around the circle: it moves 1 square the first time, 2 squares the second time, ... and $n$ squares the $n$-th time. Try to answer: which squares can the chess piece reach during its moveme... | Solution: Number the squares sequentially as $0,1,2,3,4$, and assume without loss of generality that the chess piece is in the square numbered 0. Then, the number of the square reached after the $n$th move is the remainder of $1+2+\cdots+n$ divided by 5.
For any positive integer $n$, let $n \equiv r(\bmod 5)(0 \leqsla... | 0,1,3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,162 |
6. As shown in Figure 2, let $P$ be a point inside the equilateral $\triangle ABC$ with side length 12. Draw perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $PD: PE: PF = 1: 2: 3$. Then, the area of quadrilateral $BDPF$ is | $$
6.11 \sqrt{3} \text {. }
$$
For the equilateral $\triangle ABC$ with a height of $12 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}$, then
$$
\begin{array}{l}
P D+P E+P F=6 \sqrt{3} . \\
\text { Also, } P D: P E: P F=1: 2: 3, \text { so, } \\
P D=\sqrt{3}, P E=2 \sqrt{3}, P F=3 \sqrt{3} .
\end{array}
$$
As shown in Figure 7... | 11 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,163 |
For the positive integer $n$, it is defined that $n!=1 \times 2 \times \cdots \times n$. Then, among all the divisors of the product $1! \times 2! \times \cdots \times 9!$, the number of divisors that are perfect squares is $\qquad$.
| 7.672.
Given $1!\times 2!\times \cdots \times 9!=2^{30} \times 3^{13} \times 5^{5} \times 7^{3}$, then the divisors of its perfect square must be in the form of $2^{2 a} \times 3^{2 b} \times$ $5^{2 c} \times 7^{2 d}$, where $a, b, c, d$ are non-negative integers, and $0 \leqslant a \leqslant 15,0 \leqslant b \leqslan... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,164 |
8. Given that $k$ is a positive integer not exceeding 2008, such that the equation $x^{2}-x-k=0$ has two integer roots. Then the sum of all such positive integers $k$ is $\qquad$ . | 8.30360
Since the roots of $x^{2}-x-k=0$ are $x=\frac{1 \pm \sqrt{4 k+1}}{2}$, therefore,
$x^{2}-x-k=0$ has two integer roots $\Leftrightarrow 4 k+1$ is a square of an odd number.
Let $1+4 k=(2 a+1)^{2}\left(a \in \mathbf{N}_{+}\right)$. Then $k=a(a+1)$.
Also, $k \leqslant 2008$, so $a \leqslant 44$.
Therefore, the su... | 30360 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,165 |
9. As shown in Figure 3, the center of the equilateral triangle $\triangle A_{1} B_{1} C_{1}$ with side length 1 is $O$. Rotate $\triangle A_{1} B_{1} C_{1}$ around the center $O$ to $\triangle A_{2} B_{2} C_{2}$ such that $A_{2} B_{2} \perp B_{1} C_{1}$. Then the area of the common part (i.e., the hexagon $A B C D E F... | 9. $\frac{3}{4}-\frac{\sqrt{3}}{4}$.
As shown in Figure 8, connect
$$
O B_{1}, O B_{2}, B_{1} B_{2} \text{. }
$$
Then
$$
\begin{array}{l}
O B_{1}=O B_{2}, \\
\angle O B_{1} B_{2} \\
=\angle O B_{2} B_{1} .
\end{array}
$$
Also, $\angle O B_{1} C=30^{\circ}=\angle O B_{2} C$, so,
$$
\angle C B_{1} B_{2}=\angle C B_{2}... | \frac{3}{4}-\frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,166 |
10. As shown in Figure 4, given $\angle B A D=\angle D A C=9^{\circ}, A D$ $\perp A E$, and $A B+A C=B E$. Then $\angle B=$ | $10.48^{\circ}$.
As shown in Figure 9, extend $B A$ to point $F$, such that
$$
A F=A C .
$$
From the given conditions, we know
$$
\begin{array}{l}
B F=B E, \\
\angle 3=90^{\circ}-\angle 1 \\
=90^{\circ}-\angle 2 \\
=\angle 4 .
\end{array}
$$
Thus, $\triangle E A F \cong \triangle E A C$, which implies $\angle A E F=\... | 48^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,167 |
II. (15 points) Inside a rectangle $ABCD$ (excluding the boundary), there is a point $P$, which is 1 unit away from vertex $A$ and sides $BC$, $CD$. Find the range of possible areas for rectangle $ABCD$. | II. As shown in Figure 10, $PE \perp BC$ at $E$, $PF \perp CD$ at $F$, then $PE = PF = 1$.
Let $BE = x$, $DF = y$. Then
$$
x^{2} + y^{2} = PA^{2} = 1,
$$
and $x + y > PA = 1$.
Thus, $S_{\text{quadrilateral } ABCD} = (1 + x)(1 + y)$
$$
= 1 + x + y + xy > 2,
$$
and when $x$ (or $y) \rightarrow 0$, $S \rightarrow 2$.
S... | 2 < S_{\text{quadrilateral } ABCD} \leqslant \frac{3}{2} + \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,168 |
Three, (15 points) Given real numbers $x, y$ satisfy the conditions:
$$
\left\{\begin{array}{l}
x+2 y>0, \\
x-2 y>0, \\
(x+2 y)(x-2 y)=4 .
\end{array}\right.
$$
Find the minimum value of $|x|-|y|$. | From the problem, we have
\[
\left\{
\begin{array}{l}
x > 2|y| \geqslant 0, \\
x^{2} - 4 y^{2} = 4.
\end{array}
\right.
\]
It is clear that \( x > 0 \), and by symmetry, we can assume \( y \geqslant 0 \).
Thus, \( |x| - |y| = x - y \).
Let \( u = x - y \). Then, from \( x > 2 y \geqslant 0 \), we know \( u > 0 \).
Subs... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,169 |
Four. (15 points) As shown in Figure 5, in a "V-shaped" groove $A B C D E F$, $\angle A, \angle B, \angle D, \angle E$ are all right angles. Given that $P M, P N$ are perpendicular to $A B, D E$ respectively, with the feet of the perpendiculars being $M, N$. Find the length of the broken line $M P N$ (accurate to 0.01)... | As shown in Figure 11, connect $F C$ intersecting $P N$ at point $Q$, then $Q$ is the midpoint of $F C$. Take the midpoint $R$ of $A B$, and connect $R Q$. Then $R Q$ is the midline of trapezoid $A B C F$. Draw $P S \perp R Q$, with the foot of the perpendicular at $S$, and draw $C G \perp E F$ and $C H \perp A F$, wit... | 15.50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,170 |
Five. (15 points) Find the largest positive integer $n$ that satisfies the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | Five, Proof: For any integers $x, k$, we have
$$
\left[\frac{x}{k}\right] \geqslant \frac{x-k+1}{k} \text {. }
$$
Fact .. By the division algorithm, there exist integers $q, r$ such that $x=k q+r(0 \leqslant r \leqslant k-1)$. Thus,
$$
\begin{array}{l}
{\left[\frac{x}{k}\right]=\left[\frac{k q+r}{k}\right]=q+\left[\fr... | 1715 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,171 |
Example 6 Let $a, b, c > 0, a + b + c = abc$. Prove:
$$
\sum \frac{1}{\sqrt{1+a^{2}}} \leqslant \frac{3}{2} .
$$ | Proof: Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$.
From $a+b+c=abc$, we get $xy+yz+zx=1$.
Thus, the original inequality is equivalent to
$$
\begin{array}{l}
\sum \frac{x}{\sqrt{x^{2}+1}} \leqslant \frac{3}{2} \\
\Leftrightarrow \sum \frac{2 x}{\sqrt{x^{2}+xy+yz+zx}} \leqslant 3 \\
\Leftrightarrow \sum \frac{2 x}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,172 |
1. (1) Can the set $\{1,2, \cdots, 96\}$ be represented as the union of its 32 three-element subsets, with the sum of the elements in each three-element subset being equal?
(2) Can the set $\{1,2, \cdots, 99\}$ be represented as the union of its 33 three-element subsets, with the sum of the elements in each two-element... | 1.(1) Cannot.
Because $32 \times(1+2+\cdots+96)=48 \times 97$.
(2) Can.
The sum of the elements of each three-element subset is
$$
\frac{1+2+\cdots+99}{33}=\frac{99 \times(99+1)}{33 \times 2}=150 \text {. }
$$
Divide $1,2, \cdots, 66$ into 33 groups, each containing two numbers, such that the sum of the two numbers ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,174 |
2. Given that the real-coefficient polynomial $\varphi(x)=a x^{3}+b x^{2}+c x+d$ has three positive roots, and $\varphi(0)<0$. Prove:
$$
2 b^{3}+9 a^{2} d-7 a b c \leqslant 0 \text {. }
$$ | 2. Let the real-coefficient polynomial $\varphi(x)=a x^{3}+b x^{2}+c x+d$ have three positive roots $x_{1}, x_{2}, x_{3}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a}, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=\frac{c}{a} .
\end{array}
$$
From $\varphi(0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,175 |
3. Find the smallest constant $a(a>1)$, such that for any point $P$ inside the square $ABCD$, there are always at least two of the triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, whose area ratio belongs to the interval $\left[a^{-1}, a\right]$.
(Li Weiguo) | 3. $a_{\min }=\frac{1+\sqrt{5}}{2}$.
First, prove: $a_{\text {min }} \leqslant \frac{1+\sqrt{5}}{2}$.
Let $\varphi=\frac{1+\sqrt{5}}{2}$. Assume the side length of the square is $\sqrt{2}$.
As shown in Figure 2, for a point $P$ inside the square $A B C D$, let $S_{1} 、 S_{2} 、 S_{3} 、 S_{4}$ represent the areas of $\t... | \frac{1+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,176 |
4. On the outside of the convex quadrilateral $A B C D$, construct: $\triangle A B Q, \triangle B C R, \triangle C D S, \triangle D A P$. Let the sum of the diagonals of quadrilateral $A B C D$ be $x$, and the sum of the lines connecting the midpoints of the opposite sides of quadrilateral $P Q R S$ be $y$. Find the ma... | 4. When quadrilateral $A B C D$ is a square, we get
$$
\frac{y}{x}=\frac{1+\sqrt{3}}{2} \text {. }
$$
The following is the proof: $\frac{y}{x} \leqslant \frac{1+\sqrt{3}}{2}$.
As shown in Figure 3, let $P_{1} 、 Q_{1} 、 R_{1} 、 S_{1}$ be the midpoints of sides $D A$ 、 $A B 、 B C 、 C D$ respectively, and let $E 、 F 、 G ... | \frac{1+\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,177 |
5. As shown in Figure 1, given a convex quadrilateral $ABCD$ such that $AB = BC$, $AD = DC$, $E$ and $F$ are points on segments $AB$ and $AD$ respectively, satisfying that points $B$, $E$, $F$, $D$ are concyclic. Construct $\triangle DPE$ to be directly similar to $\triangle ADC$, and construct $\triangle BQF$ to be di... | 5. As shown in Figure 4, let the center of the circle passing through points $B, E, F, D$ be $O$. Connect $O B, O F, B D$.
In $\triangle B D F$, $O$ is the circumcenter, hence
$$
\begin{array}{l}
\angle B O F \\
=2 \angle B D A . \\
\text { Also, } \triangle A B D \sim \triangle C B D, \text{ then } \angle C D A=2 \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,178 |
6. Let the sequence of positive numbers $x_{1}, x_{2}, \cdots, x_{n}, \cdots$ satisfy
$$
\left(8 x_{2}-7 x_{1}\right) x_{1}^{7}=8
$$
and $x_{k+1} x_{k-1}-x_{k}^{2}=\frac{x_{k-1}^{8}-x_{k}^{8}}{\left(x_{k} x_{k-1}\right)^{7}}(k \geqslant 2)$.
Find the positive real number $a$, such that when $x_{1}>a$, the sequence is ... | 6. From $x_{k+1} x_{k-1}-x_{k}^{2}=\frac{x_{k-1}^{8}-x_{k}^{8}}{\left(x_{k} x_{k-1}\right)^{7}}$, we have
$\frac{x_{k+1}}{x_{k}}-\frac{x_{k}}{x_{k-1}}=\frac{1}{x_{k}^{8}}-\frac{1}{x_{k-1}^{8}}$,
which means $\frac{x_{k+1}}{x_{k}}-\frac{1}{x_{k}^{8}}=\frac{x_{k}}{x_{k-1}}-\frac{1}{x_{k-1}^{8}}=\cdots=\frac{x_{2}}{x_{1}}... | 8^{\frac{1}{8}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,179 |
7. Given a $2008 \times 2008$ chessboard, each small square has a different color. In each small square of the chessboard, fill in one of the 4 letters $C, G, M, O$. If in every $2 \times 2$ small chessboard there are the 4 letters $C, G, M, O$, then the chessboard is called a "harmonious chessboard". How many differen... | 7. There are $12 \times 2^{2088}-24$ different harmonious chessboards.
First, prove a conclusion.
Conclusion: In each harmonious chessboard, at least one of the following situations must occur:
(1) Each row is an alternating sequence of two letters;
(2) Each column is an alternating sequence of two letters.
In fact, a... | 12 \times 2^{2008} - 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,180 |
For a positive integer $n$, let
$$
f_{n}=\left[2^{n} \sqrt{2008}\right]+\left[2^{n} \sqrt{2009}\right] \text {. }
$$
Prove that the sequence $f_{1}, f_{2}, \cdots$ contains infinitely many odd numbers and infinitely many even numbers (where $[x]$ denotes the greatest integer less than or equal to the real number $x$).... | 8. Represent $\sqrt{2008}$ and $\sqrt{2009}$ in binary:
$\sqrt{2008}=101100 \cdot a_{1} a_{2} \cdots(2)$
and $\sqrt{2009}=\overline{101100 \cdot b_{1} b_{2} \cdots}(2)$.
First, prove: There are infinitely many even numbers in the sequence (proof by contradiction).
Assume there are only finitely many even numbers in the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,181 |
1. Given the set $S=\{1,2, \cdots, 3 n\}, n$ is a positive integer, $T$ is a subset of $S$, satisfying: for any $x, y, z \in T$ (where $x, y, z$ can be the same), we have $x+y+z \notin T$. Find the maximum number of elements in all such sets $T$. | 1. If we take $T_{0}=\{n+1, n+2, \cdots, 3 n\}$, then $\left|T_{0}\right|=2 n$, and the sum of any three numbers in $T_{0}$ is greater than $3 n$, i.e., not in $T_{0}$. Therefore, $\max |T| \geqslant 2 n$.
On the other hand, consider the sequence of three-element subsets
$A_{0}=\{n, 2 n, 3 n\}$,
$A_{k}=\{k, 2 n-k, 2 n+... | 2n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,182 |
2. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=2 a_{n}+n\left(1+2^{n}\right)(n=1,2, \cdots) \text {. }
$$
Try to find the general term $a_{n}$. (Wu Weizhao) | 2. Dividing both sides of the given recurrence relation by $2^{n+1}$ yields $\frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}+\frac{n}{2^{n+1}}+\frac{n}{2}$,
i.e., $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{n}{2^{n+1}}+\frac{n}{2}$.
Therefore, $\sum_{i=1}^{n}\left(\frac{a_{i+1}}{2^{i+1}}-\frac{a_{i}}{2^{i}}\right)=... | a_{n}=2^{n-2}\left(n^{2}-n+6\right)-n-1(n \geqslant 2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,183 |
Example 7 Let $p$ be a prime number no less than 5, and $2 p+1$ is also a prime number. Prove: $4 p+1$ is not a prime number. | Prove: Since $p(p>3)$ is a prime number, then,
$$
p \equiv \pm 1(\bmod 3).
$$
If $p \equiv 1(\bmod 3)$, then
$$
2 p+1 \equiv 2+1 \equiv 0(\bmod 3),
$$
which means $3 \mid (2 p+1)$.
But $2 p+1>3$, so $2 p+1$ is not a prime number, which is a contradiction.
Therefore, $p \equiv -1(\bmod 3)$.
Thus, $4 p+1 \equiv -4+1 \e... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,184 |
3. In $\triangle A B C$, $B C>A B, B D$ bisects $\angle A B C$ and intersects $A C$ at point $D, C P \perp B D$, with the foot of the perpendicular being $P, A Q \perp B P$, with the foot of the perpendicular being $Q, M$ is the midpoint of side $A C$, and $E$ is the midpoint of side $B C$. If the circumcircle $\odot O... | 3. As shown in Figure 3, extend $A Q$ to intersect $B C$ at point $N$, then $Q$ is the midpoint of $A N$. Also, $M$ is the midpoint of side $A C$, so $Q M / / B C$. Therefore,
$$
\angle P Q M=\angle P B C=\frac{1}{2} \angle A B C .
$$
Similarly, $\angle M P Q=\frac{1}{2} \angle A B C$.
Thus, $Q M=P M$.
Since $Q, H, P,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,185 |
4. Let positive integers $m, n \geqslant 2$, for any $n$-element integer set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, take each pair of different numbers $a_{i}, a_{j}(j>i)$, and compute the difference $a_{j}-a_{i}$. Arrange these $\mathrm{C}_{n}^{2}$ differences in ascending order to form a sequence, which is c... | For a given positive integer $m \geqslant 2$, if an integer $x$ when divided by $m$ leaves a remainder of $i(i \in\{0,1, \cdots, m-1\})$, then $x$ is said to belong to the residue class $K_{i}$ modulo $m$.
Let the elements of $A$ that belong to $K_{i}$ be $n_{i}(i=0, 1, \cdots, m-1)$, and the elements of the set $B=\{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,186 |
5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie) | 5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,187 |
6. As shown in Figure 1, the incircle $\odot I$ of $\triangle A B C$ touches $B C$ and $A C$ at points $M$ and $N$, respectively. $E$ and $F$ are the midpoints of sides $A B$ and $A C$, respectively. $D$ is the intersection of line $E F$ and $B I$. Prove that points $M$, $N$, and $D$ are collinear. | 6. As shown in Figure 4, connect $A D$. Then $\angle A D B=90^{\circ}$.
Connect $A I, D M$, and let $D M$ intersect $A C$ at point $G$.
Since $\angle A B I=\angle D B M$, we have $\frac{A B}{B D}=\frac{B I}{B M}$.
Thus, $\triangle A B I \backsim \triangle D B M$.
Therefore, $\angle D M B=\angle A I B=90^{\circ}+\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,188 |
7. Captain Jack and his pirates have plundered 6 treasure chests $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$, where $A_{i}$ $(i=1,2, \cdots, 6)$ contains $a_{i}$ gold coins (all $a_{i}$ are distinct). The pirates designed a layout for the chests (as shown in Figure 2) and assigned one of them to take turns with the capt... | 7. When the number of chests is 2, the captain has a winning strategy.
Lemma When the number of chests is 4, the captain has a winning strategy.
Proof of the lemma: When the number of chests is 4, there are two different ways they can be linked together, as shown in Figure 5.
Figure 5
Case 1: In the first round, the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,189 |
8. Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq$ $a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i... | And the $n$-digit waveform number $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfies $a_{1}>a_{2}$ is called a "B-type number". According to symmetry, when $n \geqslant 2$, the number of B-type numbers is also $g(n)$. Therefore, $f(n)=2 g(n)$.
Next, we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit A-... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,190 |
1. Given the quadratic function $f(x)=x^{2}-3 x+2$. Then the number of distinct real roots of the equation $f(f(x))=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | -1.D.
Since $f(f(x))=x^{4}-6 x^{3}+10 x^{2}-3 x$, then $x(x-3)\left(x^{2}-3 x+1\right)=0$.
Solving, we get $x_{1}=0, x_{2}=3, x_{3,4}=\frac{3 \pm \sqrt{5}}{2}$.
Therefore, the original equation has 4 distinct real roots. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,191 |
2. For the parabola $y=a x^{2}+b x+1$, the parameters $a$ and $b$ satisfy $8 a^{2}+4 a b=b^{3}$. Then, as $a$ and $b$ vary, the vertex of the parabola must lie on ( ).
(A) parabola
(B) hyperbola
(C) circle or ellipse
(D) line | 2.B.
The vertex coordinates of the parabola $y=a x^{2}+b x+1$ are $\left(-\frac{b}{2 a}, \frac{4 a-b^{2}}{4 a}\right)$. Let $x=-\frac{b}{2 a}, y=\frac{4 a-b^{2}}{4 a}$. Then $\frac{b}{a}=-2 x, y=1-\frac{b^{2}}{4 a}=1+\frac{b x}{2}$.
Since $a \neq 0$, the conditions that $a$ and $b$ satisfy are equivalent to $8+4 \cdo... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,192 |
3. As shown in Figure 1, given that the midpoints of the three sides $BC$, $CA$, and $AB$ of $\triangle ABC$ are $L$, $M$, and $N$ respectively, and points $D$ and $E$ are on $BC$ and $AB$ respectively, satisfying that $AD$ and $CE$ both bisect the perimeter of $\triangle ABC$. Points $P$ and $Q$ are the symmetric poin... | 3. A.
Let the three sides and the semi-perimeter of $\triangle A B C$ be $a$, $b$, $c$, and $p$ respectively. Then,
$$
B Q=A E=p-b, B P=C D=p-b .
$$
Thus, $B Q=B P$.
Since $L M / / A B$, we have,
$$
L F=L P=B P-B L=p-b-\frac{a}{2}=\frac{c-b}{2} \text {. }
$$
Therefore, $F M=L M-L F=\frac{b}{2}=A M$,
$$
\angle M A F=... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,193 |
4. If the roots of the equation $a^{x}+2 x-4=0(a>0, a \neq 1)$ are $u_{1}, u_{2}, \cdots, u_{k}\left(k \in \mathbf{N}_{+}\right)$, and the roots of the equation
$$
\log _{a} 2 x+x-2=0(a>0, a \neq 1)
$$
are $v_{1}, v_{2}, \cdots, v_{l}\left(l \in \mathbf{N}_{+}\right)$. Then
$$
\frac{u_{1}+u_{2}+\cdots+u_{k}+v_{1}+v_{2... | 4.C.
The equation $a^{x}+2 x-4=0 \Leftrightarrow \frac{a^{x}}{2}=2-x$, its roots are the x-coordinates of the intersection points of $y=\frac{a^{x}}{2}$ and $y=2-x$.
The equation $\log _{a} 2 x+x-2=0 \Leftrightarrow \log _{a} 2 x=2-x$, its roots are the x-coordinates of the intersection points of $y=\log _{a} 2 x$ an... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,194 |
Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.
保留源文本的换行和格式，直接输出翻译结果。 | When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$
But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,195 |
5. Consider the set $S=\{1,2, \cdots, 10\}$ and all its non-empty subsets. If the number of even numbers in a non-empty subset is no less than the number of odd numbers, then this subset is called a "good subset". Therefore, the number of good subsets is ( ) .
(A) 631
(B) 633
(C) 635
(D) 637 | 5.D.
Suppose a good subset contains $i(i=1,2,3,4,5)$ even numbers. Then the number of odd numbers can be $j(j=0,1, \cdots, i)$.
Therefore, the number of good subsets is
$$
\begin{array}{l}
\sum_{i=1}^{5}\left(C_{5}^{i} \sum_{j=0}^{i} C_{5}^{j}\right) \\
= C_{5}^{1}\left(C_{5}^{0}+C_{5}^{1}\right)+C_{5}^{2}\left(C_{5}... | 637 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,196 |
6. Let the number of distinct positive integer solutions $(x, y, z)$ to the indeterminate equation $x^{2}+y^{2}+z^{2}-x y z+10=0$ that satisfy $x, y, z > 2008$ be $k$. Then $k$ satisfies ( ).
(A) $k=0$
(B) $1 \leqslant k \leqslant 2008$
(C) $k>2008$, but $k$ is a finite number
(D) $k$ is infinite | 6.D.
$\left(x_{0}, y_{0}, z_{0}\right)=(3,4,5)$ is a particular solution to the original indeterminate equation.
For any positive integer solution $\left(x_{1}, y, z\right)$ of the original indeterminate equation, assuming $x_{1} \leqslant y \leqslant z$, and $x_{1}z^{2} \text {. }$
Therefore, $x_{2}$ is a positive ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,197 |
7. The function $f(x)=\frac{\sin \left(x+45^{\circ}\right)}{\sin \left(x+60^{\circ}\right)}, x \in\left[0^{\circ}\right.$, $\left.90^{\circ}\right]$. Then the product of the maximum and minimum values of $f(x)$ is | $=7 . \frac{2 \sqrt{3}}{3}$.
Since $f^{\prime}(x)=\frac{\sin 15^{\circ}}{\sin ^{2}\left(x+60^{\circ}\right)}>0$, therefore, $f(x)$ is monotonically increasing. Thus, the maximum value is $f\left(90^{\circ}\right)=\sqrt{2}$, and the minimum value is $f\left(0^{\circ}\right)=\frac{\sqrt{2}}{\sqrt{3}}$, their product is $... | \frac{2 \sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,198 |
8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ . | 8. -1 .
Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$
Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with c... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,199 |
9. Consider the 25 grid points in a $4 \times 4$ square grid. The number of different lines passing through at least 3 grid points is $\qquad$ .
| 9.32.
There are 10 horizontal and vertical lines, 10 lines parallel to the two diagonals, and 12 other lines that meet the conditions. In total, there are 32 lines. | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,200 |
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the value of $\sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right]$ is $\qquad$ . | 10.2015028.
For $k=1,2, \cdots, 2008$, since $\frac{2008 k}{2009}$ is not an integer, so,
$$
\begin{array}{l}
{\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right]} \\
=\left[\frac{2008 k}{2009}\right]+\left[2008-\frac{2008 k}{2009}\right]=2007 . \\
\text { Therefore, } \sum_{k=1}^{2008}\left[\frac{... | 2015028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,201 |
11. Given the rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ satisfies $A A_{1}=2, A D=3, A B=251$, the plane $A_{1} B D$ intersects $C C_{1} 、 C_{1} B_{1} 、 C_{1} D_{1}$ at points $L 、 M 、 N$. Then the volume of the tetrahedron $C_{1} L M N$ is $\qquad$ . | 11.2008.
Since $B D / / M N, A_{1} D / / L M, A_{1} B / / L N$, therefore, $A_{1} 、 B 、 D$ are the midpoints of $M N 、 L M 、 L N$ respectively. Thus,
$$
\begin{array}{l}
C_{1} L=2 C C_{1}=4, C_{1} M=2 C_{1} B_{1}=6, \\
C_{1} N=2 C_{1} D_{1}=502 .
\end{array}
$$
Therefore, the volume of the tetrahedron $C_{1} L M N$ i... | 2008 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,202 |
12. As shown in Figure 2, given a circle $\odot O$ with radius $R$ and a line $l$ outside $\odot O$, the projection of $O$ on $l$ is $H$, $O H=d$, and $O H$ intersects $\odot O$ at points $C$ and $D$, with $C H > D H$. Let $P$ and $Q$ be points on $l$, both on the same side of $H$, and $P H = a$, $Q H = b (a > b)$. In ... | $$
12 \cdot 2 a^{2}+2 b^{2}+4 d^{2}+4 R^{2} \text {. }
$$
Let the midpoint of $A_{i} B_{i}$ be $M_{i}$, and the midpoint of $P Q$ be $N$. By the median formula, we have
$$
\begin{array}{l}
P A_{i}^{2}+P B_{i}^{2}+Q A_{i}^{2}+Q B_{i}^{2} \\
=P Q^{2}+A_{i} B_{i}^{2}+4 M_{i} N^{2} \\
=(a-b)^{2}+A_{i} B_{i}^{2}+4 M_{i} N^... | 2 a^{2}+2 b^{2}+4 d^{2}+4 R^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,203 |
13. Given an acute triangle $\triangle A B C$ with midpoints of sides $B C$, $C A$, and $A B$ being $D$, $E$, and $F$ respectively, take points $P$, $Q$, and $R$ on the extensions of $E F$, $F D$, and $D E$ respectively. If $A P = B Q = C R$, prove that the circumcenter of $\triangle P Q R$ is the orthocenter of $\tria... | Three, 13. As shown in Figure 3, let the three altitudes of $\triangle ABC$ be $AL$, $BM$, and $CN$, with the orthocenter being $H$. $EF$ intersects $AL$ at point $K$. Then,
$$
\begin{array}{l}
A P^{2} \\
=P K^{2}+A K^{2} \\
=P H^{2}-K H^{2}+A K^{2} \\
=P H^{2}+(A K+K H)(A K-K H) \\
=P H^{2}+A H \cdot H L \text {. } \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,204 |
14. Given the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies:
$$
a_{1}=1, a_{2}=1, a_{n+1}=\frac{n^{2} a_{n}^{2}+5}{\left(n^{2}-1\right) a_{n-1}}(n \geqslant 2) \text {. }
$$
Find the general term formula for $a_{n}$. | 14. From $\left(n^{2}-1\right) a_{n-1} a_{n+1}=n^{2} a_{n}^{2}+5$,
$$
\left[(n-1)^{2}-1\right] a_{n-2} a_{n}=(n-1)^{2} a_{n-1}^{2}+5 \text {, }
$$
subtracting the two equations gives
$$
\begin{array}{l}
\left(n^{2}-1\right) a_{n-1} a_{n+1}-n(n-2) a_{n-2} a_{n} \\
=n^{2} a_{n}^{2}-(n-1)^{2} a_{n-1}^{2},
\end{array}
$$
... | a_{n}=\frac{1}{n}\left[\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,205 |
1. Prove: The sum of the cubes of three consecutive positive integers is a multiple of 9. | (Tip: Let $S=(n-1)^{3}+n^{3}+(n+1)^{3}$ $=3 n\left(n^{2}+2\right)$.
If $n \equiv 0(\bmod 3)$, then $3 \mid n\left(n^{2}+2\right)$.
If $n \equiv \pm 1(\bmod 3)$, then
$$
n\left(n^{2}+2\right) \equiv \pm 1 \times(1+2) \equiv 0(\bmod 3) \text {, }
$$
i.e., $3 \mid n\left(n^{2}+2\right)$. ) | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,206 |
15. There are 10 players $A_{1}, A_{2}, \cdots, A_{10}$, whose points are $9,8,7,6,5,4,3,2,1,0$, and their rankings are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th. Now a round-robin tournament (i.e., each pair of players plays exactly one match) is held, and each match must have a winner. If the player with a hi... | 15. The minimum cumulative score of the new champion is 12.
If the new champion's score does not exceed 11 points, then
$A_{1}$ can win at most 2 games; $A_{2}$ can win at most 3 games;
$A_{3}$ can win at most 4 games; $A_{4}$ can win at most 5 games.
$A_{5}$ can increase by at most 6 points, but there are only 5 play... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,207 |
1. Given real numbers $a, b$ satisfy $3^{a}=2007, 223^{b}=$ 2007. Then the value of $\frac{1}{a}+\frac{1}{2 b}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) 1 | - 1.A.
From $3^{a}=2007 \Rightarrow 9^{\frac{1}{2} a b}=2007^{b}$.
Also, $223^{b}=2007 \Rightarrow 223^{\frac{1}{2} a b}=2007^{\frac{a}{2}}$.
From the above two equations, we have
$$
\begin{array}{l}
\frac{1}{2} a b=\frac{a}{2}+b \Rightarrow a b=a+2 b \\
\Rightarrow \frac{1}{a}+\frac{1}{2 b}=\frac{2 b+a}{2 a b}=\frac{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,208 |
2. As shown in Figure $1, D$ and $E$ are on the extensions of sides $B A$ and $B C$ of equilateral $\triangle A B C$, respectively, with $A D=A C+$ $C E, \angle B D C=10^{\circ}$. Then the degree measure of $\angle C D E$ is ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $45^{\circ}$
(D) $35^{\circ}$ | 2. B.
As shown in Figure 4, extend $B E$ to $F$ such that $E F = B C$. Connect $D F$. According to the problem, $\triangle B D F$ is an equilateral triangle. Therefore,
$$
\begin{array}{l}
\angle B D F = \angle F = 60^{\circ}, \\
B D = D F, \\
\angle B = \angle F .
\end{array}
$$
Thus, $\triangle B C D \cong \triangl... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,209 |
3. Given the parabola $y=a x^{2}+b x+c$ passes through two points $A(-1,2)$ and $B(-2,3)$, and does not pass through the first quadrant. If $s=a+b-c$, then the range of $s$ is ( ).
(A) $s \leqslant-3$
(B) $s<2$
(C) $s \leqslant 2$
(D) $s<-3$. | 3.A.
From the problem, we have $\left\{\begin{array}{l}a-b+c=2, \\ 4 a-2 b+c=3 \text {. }\end{array}\right.$
Solving, we get $a=\frac{b}{3}+\frac{1}{3}, c=\frac{2}{3} b+\frac{5}{3}$.
Since the parabola does not pass through the first quadrant, we know $a<0, c \leqslant 0$, i.e., $\square$
$$
\frac{b}{3}+\frac{1}{3}<0,... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,210 |
4. A netizen's $QQ$ number is $m, m$ when divided by 10000, the sum of the quotient and the remainder is $18889, m$ when divided by 100000, the sum of the quotient and the remainder is 58741. Then the hundred's digit of $m$ is $(\quad$.
(A) 4
(B) 5
(C) 6
(D) 7 | 4.C.
Let the remainder when $m$ is divided by 10000 be $a$, and the quotient when divided by 10 be $b$, with the remainder being $c$. According to the problem, we have
$$
\left\{\begin{array}{l}
10 b+c+a=18889, \\
10000 c+a+b=58741 .
\end{array}\right.
$$
Multiplying (2) by 10 and subtracting (1) gives
$$
99999 c+9 a=... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,211 |
5. Given a point $P$ in a plane, the distances from $P$ to the three sides of an equilateral $\triangle ABC$ on the lines $l_{1}, l_{2}, l_{3}$ are $6, 9, 12$ respectively. The number of possible values for the side length of $\triangle ABC$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5.D.
Let the height of the equilateral $\triangle ABC$ be $h$, and the distances from point $P$ to the three sides of $\triangle ABC$ be $h_{1}, h_{2}, h_{3}$.
As shown in Figure 5, when $P \in \mathrm{I}$, $h=h_{1}+h_{2}+h_{3}=27$; when
$P \in$ II, $h=h_{1}+h_{2}-$
$h_{3}$, there are 3 possible values; when $P$
$\in... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,212 |
6. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $s$ and the minimum value is $t$. Then the value of $s-t$ is ( ).
(A) 12
(B) 14
(C) 11
(D) 13 | 6.C.
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10} \Rightarrow(n+\sqrt{10})^{2}-3 n^{2}=p$
$$
\Rightarrow 2 n^{2}-2 \sqrt{10} n+(p-10)=0 \text {. }
$$
Since $n$ is a real number, then
$$
(-2 \sqrt{10})^{2}-4 \times 2(p-10) \geqslant 0 \text {. }
$$
Solving this, we get $p \leqslant 15$. Th... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,213 |
1. If a positive integer cannot be expressed as the difference of squares of two positive integers, then this positive integer is called a "non-wise number". If these non-wise numbers are arranged in ascending order, then the 2009th non-wise number is $\qquad$ | ニ、1.8026.
1 cannot be expressed as the difference of squares of two positive integers, so 1 is the 1st non-wise number.
Odd numbers greater than 1, $2 n+1\left(n \in N_{+}\right)$, can be expressed as $(n+1)^{2}-n^{2}$.
Even numbers divisible by 4, $4 m$, can be expressed as $(m+1)^{2}-(m-1)^{2}(m>1)$. However, 4 can... | 8026 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,214 |
2. Given that line $l$ passes through vertex $A, B$ of square $A B C D$, $B E \perp l$ at point $E, D F \perp l$ at point $F, B E=3, D F$ $=4$. Then the length of $E F$ is $\qquad$ . | 2.1 or 7.
As shown in Figure 6, when $l$ is outside the square $A B C D$, it is easy to see that
$$
\begin{array}{l}
\triangle A B E \cong \triangle D A F \\
\Rightarrow A E=D F=4, A F=B E=3 \\
\Rightarrow E F=A F+A E=7 .
\end{array}
$$
As shown in Figure 7, when $l$ passes through the interior of the square $A B C D... | 1 \text{ or } 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,215 |
3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ . | 3.2 .
Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,216 |
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