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
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class | __index_level_0__ int64 0 742k |
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Example 3 Find all integer pairs $(m, n)(m, n \geqslant 2)$, such that for any integer $x$ we have
$$
x^{n} \equiv x(\bmod m)
$$ | If there is a prime $p$ such that $p^{2} \mid m$, taking $x=p$ leads to a contradiction.
Therefore, $m=p_{1} p_{2} \cdots p_{k}$ (where $p_{i}$ are distinct primes).
Furthermore, $x^{n} \equiv x(\bmod m)$ is equivalent to the system of congruences
$x^{n} \equiv x\left(\bmod p_{i}\right) \quad(1 \leqslant i \leqslant k)... | m=p_{1} p_{2} \cdots p_{k}, n=1+u\left[p_{1}-1, p_{2}-1, \cdots, p_{k}-1\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,410 |
11. (20 points) In a non-obtuse $\triangle ABC$, prove: $\sin A + \sin B + \sin C > 2$.
| 11. Note that,
$$
\begin{array}{l}
\sin A+\sin B+\sin C-2 \\
= \sin A+\sin B+\sin (A+B)- \\
\left(\sin ^{2} A+\cos ^{2} A\right)-\left(\sin ^{2} B+\cos ^{2} B\right) \\
= \sin A(1-\sin A)+\sin B(1-\sin B)+ \\
\sin (A+B)-\left(\cos ^{2} A+\cos ^{2} B\right) \\
= \sin A(1-\sin A)+\sin B(1-\sin B)+ \\
\cos B(\sin A-\cos ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,411 |
12. (26 points) Question: Does there exist a sequence of positive integers $\left\{a_{n}\right\}$, satisfying $a_{2013}=2013$, and for each $k \in\{2,3, \cdots, 2013\}$, we have $\left|a_{k}-a_{k-1}\right|=20$ or 13; and the terms $a_{1}$, $a_{2}, \cdots, a_{2013}$ form a permutation of $1,2, \cdots, 2013$? Prove your ... | 12. Existence.
Since $20+13=33$, and 3312013, note that the "difference" operation has a "translation property", that is, if $\left|a_{k}-a_{k-1}\right|=20$ or 13, then for any integer $c$, we also have
$\left|\left(a_{k}+c\right)-\left(a_{k-1}+c\right)\right|=20$ or 13.
For this reason, first arrange the numbers in t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,412 |
1. Let $g(x)=\sum_{i=1}^{n} \mathrm{C}_{n}^{i} \frac{i x^{i}(1-x)^{n-i}}{n}$. Then $g(2014)=$ $\qquad$ | $$
-, 1.2014
$$
From $\frac{r}{n} \mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}$, we get
$$
g(x)=x \sum_{k=0}^{n-1} \mathrm{C}_{n-1}^{k} x^{k}(1-x)^{n-k}=x \text {. }
$$
Thus, $g(2014)=2014$. | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,413 |
2. Let the function be
$$
f(x)=\sqrt{\ln x+x^{2}-a}(a \in \mathbf{R}) .
$$
If there exists $b \in[1, \mathrm{e}]$, such that $f(f(b))=b$, then the range of the real number $a$ is . $\qquad$ | 2. $a \in [0,1]$.
Given that there exists $b \in [1, \mathrm{e}]$ such that $f(f(b))=b$, we get $f(b)=b$.
Thus, there exists $x \in [1, \mathrm{e}]$ such that
$$
\sqrt{\ln x + x^2 - a} = x,
$$
which implies $a = \ln x$ for $x \in [1, \mathbf{e}]$.
Therefore, $a \in [0,1]$. | a \in [0,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,414 |
3. Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=2,(n+1) x_{n+1}=x_{n}+n \text {. }
$$
Then the general term of the sequence $\left\{x_{n}\right\}$ is $x_{n}=$ $\qquad$ . | $\begin{array}{l}\text { 3. } 1+\frac{1}{n!} \text {. } \\ \text { From }(n+1) x_{n+1}=x_{n}+n \\ \Rightarrow x_{n+1}-1=\frac{x_{n}-1}{n+1} \\ \Rightarrow\left(x_{n+1}-1\right) \cdot(n+1)!=\left(x_{n}-1\right) \cdot n! \\ \Rightarrow\left\{\left(x_{n}-1\right) \cdot n!\right\} \text { is a constant sequence } \\ \Right... | x_{n}=1+\frac{1}{n!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,415 |
4. Simplify:
$$
\cos \frac{2 \pi}{2015}+\cos \frac{4 \pi}{2015}+\cdots+\cos \frac{2014 \pi}{2015}=
$$
$\qquad$ | $\begin{array}{l}\text { 4. }-\frac{1}{2} \text {. } \\ \cos \frac{2 \pi}{2015}+\cos \frac{4 \pi}{2015}+\cdots+\cos \frac{2014 \pi}{2015} \\ =\sum_{k=1}^{1007} \cos \frac{2 k \pi}{2015} \\ =\frac{1}{\sin \frac{\pi}{2015}} \sum_{k=1}^{1007}\left[\sin \frac{(2 k+1) \pi}{2015}-\sin \frac{(2 k-1) \pi}{2015}\right] \\ =\fra... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,416 |
5. As shown in Figure 1, in $\triangle A B C$, it is known that $O$ is the midpoint of $B C$, and points $M$ and $N$ are on sides $A B$ and $A C$ respectively, and
$$
\begin{array}{l}
A M=6, M B=4, A N=4, \\
N C=3, \angle M O N=90^{\circ} .
\end{array}
$$
Then $\cos A=$ | 5. $\frac{3}{8}$.
Let $\overrightarrow{A B}=a, \overrightarrow{A C}=b$. Then $|a|=10,|b|=7$.
Since $O$ is the midpoint of $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2} a+\frac{1}{2} b \text {. }
$$
From the problem, we know $\overrightarrow{A M}=\frac{3}{5} a, \overrightarrow{A N}=\frac{4}{7} b$.
$$
\begin{array... | \frac{3}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,417 |
6. Given $a, b, c \in \mathbf{R}_{+}$, and
$$
a+b+c=12, a b+b c+c a=45 \text{. }
$$
Then $\min \max \{a, b, c\}=$ $\qquad$ | 6. 5 .
Let $a=\max \{a, b, c\}$.
From $a+b+c=12$, we get $a \geqslant 4$.
$$
\begin{array}{l}
\text { By }(a-b)(a-c) \geqslant 0 \\
\Rightarrow a^{2}-a(12-a)+b c \geqslant 0 \\
\Rightarrow b c \geqslant 12 a-2 a^{2} . \\
\text { Also } 45=a b+b c+c a=b c+a(12-a) \\
\geqslant 12 a-2 a^{2}+a(12-a),
\end{array}
$$
Then ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,418 |
7. Given that $P$ is a point outside the plane $\alpha$ where $\triangle A B C$ lies, $P A \perp$ plane $\alpha, P B=P C=\sqrt{26}, \tan \angle P B C=\frac{3}{2}$. Then the maximum value of the distance from point $A$ to plane $P B C$ is $\qquad$ | 7. $\frac{3 \sqrt{2}}{2}$.
As shown in Figure 2, draw $P D \perp B C$ at point $D$, connect $A D$, and draw $A F \perp P D$ at point $F$.
Since $P A \perp$ plane $A B C, B C \perp P D$, we have
$B C \perp A D \Rightarrow B C \perp$ plane $P A D$
$\Rightarrow$ plane $P A D \perp$ plane $P B C$.
From $A F \perp P D$, we... | \frac{3 \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,419 |
8. The unit digit of $\left[\frac{10^{10000}}{10^{100}+9}\right]$ is | 8. 1 .
Notice that,
$$
\frac{10^{10000}}{10^{100}+9}=\frac{\left(10^{100}\right)^{100}-3^{200}}{10^{100}+9}+\frac{3^{200}}{10^{100}+9} \text {. }
$$
And $\left(10^{100}\right)^{100}-3^{200}=\left[\left(10^{100}\right)^{2}\right]^{50}-\left(9^{2}\right)^{50}$, so $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,420 |
Example 4 Find all prime triples $(p, q, r)$ such that
$$
p \text { | }\left(q^{r}+1\right), q\left|\left(r^{p}+1\right), r\right|\left(p^{q}+1\right) \text {. }
$$ | Given that $p, q, r$ are pairwise coprime, we know they are distinct.
(1) $p, q, r$ are all odd primes.
From $p \mid (q^r + 1)$, we get $\delta_p(q) = 2$ or $2r$.
If $\delta_p(q) = 2r$, then
$$
\begin{array}{l}
2r \mid (p-1) \Rightarrow p \equiv 1 \pmod{r} \\
\Rightarrow p^q + 1 \equiv 2 \pmod{r},
\end{array}
$$
which... | (2, 5, 3), (5, 3, 2), (3, 2, 5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,421 |
9. (16 points) Given $\odot O: x^{2}+y^{2}=4$, circle $M$ :
$$
(x-5 \cos \theta)^{2}+(y-5 \sin \theta)^{2}=1(\theta \in \mathbf{R}) \text {, }
$$
Through any point $P$ on circle $M$, draw two tangents $P E$ and $P F$ to $\odot O$, with the points of tangency being $E$ and $F$. Try to find the minimum value of $\overri... | 9. The center of circle $M$ is on the circle $x^{2}+y^{2}=25$.
Let $|P E|=|P F|=d$.
In the right triangle $\triangle P E O$, it is easy to see that
$4 \leqslant|P O| \leqslant 6,|O E|=2$.
Thus, $2 \sqrt{3} \leqslant d \leqslant 4 \sqrt{2}$.
Also, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,422 |
10. (20 points) Let the function
$$
f(x)=\log _{a}(1+a x)-\log _{a}(1-a x),
$$
where $a>0$, and $a \neq 1$.
(1) When $a>1$, if the inequality
$$
f(x) \geqslant \log _{a} 8 x
$$
holds for all $x$, find the range of values for $a$;
(2) If $f\left(x_{0}\right)=x_{0}-1$, prove: $\left|x_{0}\right|<1$. | $$
\begin{array}{l}
(1) \text{ Let } a x=t. \text{ Then } 01) \\
\Leftrightarrow \frac{8 t}{a} \leqslant \frac{1+t}{1-t} \\
\Leftrightarrow \frac{a}{8} \geqslant \frac{t(1-t)}{1+t}=-(t+1)-\frac{2}{t+1}+3 .
\end{array}
$$
Thus, $a \geqslant 24-16 \sqrt{2}$.
(2) From $\left\{\begin{array}{l}1-a x>0, \\ 1+a x>0, \\ a>0\e... | 24-16\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,423 |
11. (20 points) Let
$$
\begin{array}{l}
f(x)=a_{1} x^{2013}+a_{2} x^{2012}+\cdots+a_{2013} x+a_{2014} \\
=x^{13}\left(x^{10}+x^{2}+x\right)^{2000}, \\
b_{0}=1, b_{1}=2, b_{n+2}+b_{n}=b_{n+1}(n \in \mathbf{N}) .
\end{array}
$$
Find the value of $\sum_{i=1}^{2013} a_{i} b_{i}$. | 11. From the recurrence relation of $\left\{b_{n}\right\}$, we have
$$
\begin{array}{l}
b_{2}=1, b_{3}=-1, b_{4}=-2, b_{5}=-1, \\
b_{6}=1=b_{0}, b_{7}=2=b_{1} .
\end{array}
$$
Thus, $b_{6 k}=1, b_{6 k+1}=2, b_{6 k+2}=1, b_{6 k+3}=-1$, $b_{6 k+4}=-2, b_{6 k+5}=-1(k \in \mathbf{N})$.
Let $\lambda=\frac{1}{2}+\frac{\sqrt... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,424 |
One. (40 points) Let $P$ be a point on the circumcircle of $\triangle ABC$, and let $X$, $Y$, $Z$ be the projections of point $P$ onto sides $BC$, $CA$, $AB$ respectively. If point $P$ and $A$ are on opposite sides of side $BC$, prove:
$$
\frac{BC}{PX} = \frac{CA}{PY} + \frac{AB}{PZ}.
$$ | As shown in Figure 3, since $\angle B P C > \angle B P A$, a point $D$ can be taken on $B C$ such that $\angle C P D = \angle A P B$.
Thus, $\triangle P D C \sim \triangle P B A, \triangle P D X \sim \triangle P B Z$.
$$
\begin{array}{l}
\text { Then } \frac{C D}{A B} = \frac{P D}{P B} = \frac{P X}{P Z} \\
\Rightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,425 |
II. (40 points) Given real numbers $a, b, c \in [-1,1]$, and satisfying
$$
1 + 2abc \geq a^2 + b^2 + c^2 \text{.}
$$
Prove: For any positive integer $n$,
$$
1 + 2(abc)^n \geq a^{2n} + b^{2n} + c^{2n} .
$$ | It is known that at least two of $a, b, c$ are non-negative or non-positive. Therefore, at least one of $ab, bc, ca$ is non-negative. Without loss of generality, assume $ab \geqslant 0$.
Notice that $c \in [-1,1]$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\sum_{k=0}^{n-1}(a b)^{n-1-k} c^{k}\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,426 |
Three. (50 points) Let $h_{n}$ be the last non-zero digit of the decimal representation of $n!$. Prove: $0 . h_{1} h_{2} \cdots$ is an irrational number.
保留源文本的换行和格式，翻译结果如下：
Three. (50 points) Let $h_{n}$ be the last non-zero digit of the decimal representation of $n!$. Prove: $0 . h_{1} h_{2} \cdots$ is an irrationa... | Three, let $0 . h_{1} h_{2} \cdots$ be a rational number. Then there exist $N_{0}, T \in \mathbf{Z}_{+}$, such that for each $n \geqslant N_{0}$, we have $h_{n+T}=h_{n}$.
First, we prove: there exists $T_{1} \in \mathbf{Z}_{+}, T \mid T_{1}$, and the last non-zero digit of $T_{1}$ is 1.
In fact, let $T=2^{\alpha} \ti... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,427 |
The number 9 can be expressed as the sum of two consecutive positive integers $(9=4+5)$, and it can also be written as the sum of consecutive positive integers in exactly two different ways $(9=4+5=2+3+4)$. Question: Does there exist a positive integer that can be expressed as the sum of 2013 consecutive positive integ... | $$
\begin{array}{l}
s=3 \times 11^{18} \times 61^{52}. \\
\text { On the one hand, take } s=11^{17} \times 61^{51}-1006, \text { then } \\
s+(s+1)+\cdots+(s+2012) \\
=2013 n+\frac{2012 \times 2013}{2} \\
=3 \times 11^{18} \times 61^{52}=m,
\end{array}
$$
which means $m$ can be expressed as the sum of 2013 consecutive ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,429 |
In $\triangle A B C$, it is known that $A B=A C$, points $P$ and $Q$ are on sides $A B$ and $A C$ respectively, and $A P=P Q=Q C=\frac{\sqrt{3}}{3} B C$. Find the degree measure of $\angle B A C$. | Connect $C P$.
(1) If $\angle A P C$ is an acute angle, as shown in Figure 1, draw $A E \perp C P$, $A D \perp B C$, with the feet of the perpendiculars at $E$ and $D$ respectively.
From $A P=P Q=Q C$, we get $\angle P A Q=\angle P Q A, \angle Q C P=\angle Q P C$.
Also, $\angle P Q A=\angle Q P C+\angle Q C P$, so $\an... | 80^{\circ} \text{ or } 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,430 |
Given $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$. Prove:
$$
\sum_{i=1}^{n} \frac{1}{\left(1+\sqrt{x_{i}}\right)^{2}} \geqslant \frac{n^{2}}{2\left(2+\sum_{i=1}^{n} x_{i}\right)}
$$ | Prove that by Cauchy-Schwarz inequality,
$$
\begin{array}{l}
2\left(1+x_{i}\right)=\left(1^{2}+1^{2}\right)\left[1^{2}+\left(\sqrt{x_{i}}\right)^{2}\right] \\
\geqslant\left(1+\sqrt{x_{i}}\right)^{2} . \\
\text { Then } \sum_{i=1}^{n} \frac{1}{\left(1+\sqrt{x_{i}}\right)^{2}} \geqslant \sum_{i=1}^{n} \frac{1}{2\left(1+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,431 |
Example 5 Let $n \geqslant 2$ be a positive integer. Prove that:
$$
n \mid\left[1^{n-1}+2^{n-1}+\cdots+(n-1)^{n-1}+1\right]
$$
if and only if for each prime factor $p$ of $n$ we have
$$
p \left\lvert\,\left(\frac{n}{p}-1\right)\right. \text { and }(p-1) \left\lvert\,\left(\frac{n}{p}-1\right)\right. \text {. }
$$ | Prove that let $n=a p$. Then
$$
1^{n-1}+2^{n-1}+\cdots+(n-1)^{n-1} \equiv a \sum_{k=1}^{n-1} k^{n-1}(\bmod p).
$$
If $(p-1) \uparrow(n-1)$, take a primitive root $x$ modulo $p$, then $x^{n-1} \equiv 1(\bmod p)$.
Thus, $S=\sum_{k=1}^{n-1} k^{n-1} \equiv \sum_{k=1}^{n-1}(k x)^{n-1} \equiv x^{n-1} S(\bmod p)$.
This impli... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,432 |
378 (1) Does there exist $n(n \geqslant 2014)$ points on a circle with integer coordinates, such that the distance between any two points is an integer, and the area of any triangle formed by any three points is an integer?
(2) Does there exist $n(n \geqslant 2014)$ points on a circle, such that the distance between an... | Consider a circle $\odot O$ with the origin as its center and radius $r$. Let $P_{1}, P_{2}, \cdots, P_{n}$ be $n$ points on $\odot O$ arranged counterclockwise, and let
$$
P_{k}\left(r \cos 2 \theta_{k}, r \sin 2 \theta_{k}\right)(k=1,2, \cdots, n).
$$
Thus, the distance between any two points is
$$
\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,433 |
Example 6 Let $p$ be a prime. Prove: There exist infinitely many positive integers $n$, such that $p n+1$ is a prime.
保留了原文的换行和格式。 | Prove that for any multiple $a$ of $p$,
$$
N=\frac{a^{p}-1}{a-1}=1+a+a^{2}+\cdots+a^{p-1}>1.
$$
Let $q$ be any prime factor of $N$. Then $a^{p} \equiv 1(\bmod q)$.
Thus, $\delta_{q}(a)=1$ or $p$.
If $a \equiv 1(\bmod q)$, then $N \equiv p(\bmod q)$.
This implies $p=q$, a contradiction (since $N$ and $p$ are coprime).
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,434 |
Example 7 Let $p$ be an odd prime. Prove: there exists a prime $q$, such that for any integer $n$ we have
$$
q \nmid\left(n^{p}-p\right) .
$$
[Note] There are infinitely many such $q$. | Proof Note
$$
\begin{array}{l}
N=\frac{p^{p}-1}{p-1}=1+p+p^{2}+\cdots+p^{p-1} \\
\equiv p+1\left(\bmod p^{2}\right) .
\end{array}
$$
There exists a prime factor $q$ of $N$ such that $p^{2} \nmid(q-1)$ (if every prime factor of $N$ leaves a remainder of 1 when divided by $p^{2}$, then $N$ also leaves a remainder of 1 w... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,435 |
1. Let prime $p \geqslant 5$. Prove: There exists $a \in\{1,2, \cdots, p-2\}$, such that
$$
a^{p-1} \not \equiv 1\left(\bmod p^{2}\right) \text {, and }(a+1)^{p-1} \neq 1\left(\bmod p^{2}\right) \text {. }
$$ | Let $S=\{1,2, \cdots, p-1\}$,
$$
A=\left\{a \in S \mid a^{p-1} \equiv 1\left(\bmod p^{2}\right)\right\} .
$$
For any $a \in S$, it is known that at least one of $a$ and $p-a$ belongs to the set $A$.
In particular, since $1 \notin A$, it follows that $p-1 \in A$.
Assume the set $A$ does not contain adjacent elements of... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,436 |
2. (1) Let $n \geqslant 4$ be a positive integer. Prove: there exists an integer $a$ such that
$$
1 \leqslant a \leqslant \frac{n}{4}+1 \text {, and } a^{n} \not \equiv a\left(\bmod n^{2}\right) \text {. }
$$
(2) Let $p$ be a prime, $p \equiv 1(\bmod 4)$. Prove: there exists a prime $q<\sqrt{p}$, such that
$$
q^{p-1} \... | (1) If $n$ is a composite number, let $p$ be the smallest prime factor of $n$. Then
$$
p \leqslant \sqrt{n} \leqslant \frac{n}{4}+1 .
$$
Since $p^{2} \mid n^{2}$, but $p^{2} \nmid\left(p^{n}-p\right)$, hence $a=p$ is the desired value.
Assume $n(n \geqslant 5)$ is a prime number.
If $2^{n} \equiv 2\left(\bmod n^{2}\ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,437 |
3. For each positive integer $n, L(n)$ denotes the number of integers $a$ satisfying $1 \leqslant a \leqslant n$ and $n!\left(a^{n}-1\right)$. Let $p_{1}, p_{2}, \cdots, p_{k}$ be all the distinct prime factors of $n$, and define
$$
T(n)=\left(p_{1}-1\right)\left(p_{2}-1\right) \cdots\left(p_{k}-1\right) .
$$
Prove: (... | For any prime power factor $p^{4}$ of $n$ (i.e., $p^{u} \mid n$ and $p^{u+1} \nmid n$), if $a^{n} \equiv 1\left(\bmod p^{u}\right)$, then for any integer $b$ we have
$$
\begin{array}{l}
(a+b p)^{n}-1=a^{n}-1+\sum_{i=1}^{n} \mathrm{C}_{n}^{i}(b p)^{i} a^{n-i} \\
=M p^{u}+\sum_{i=1}^{n-1} \mathrm{C}_{n}^{i}(b p)^{i} a^{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,438 |
Example 3 Let $n$ be a natural number greater than 2. Prove:
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<n-\frac{n-1}{\sqrt[n-1]{n}} .
$$ | 【Analysis and Proof】Transform the inequality into the form of "sum $>$ product", placing the power (product) to the right of the “ $>$ ” sign.
$$
\text { Then equation (1) } \Leftrightarrow n-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)>\frac{n-1}{\sqrt[n-1]{n}} \text {. }
$$
Combine appropriately to remo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,439 |
Example 1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ distinct positive integers, none of whose decimal representations contain the digit 9. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<30 .
$$ | Prove that if all positive integers in decimal representation without the digit 9 are arranged as $b_{1}, b_{2}, \cdots$, then
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots \\
\leqslant c+\frac{9}{10} c+\left(\frac{9}{10}\right)^{2} c+\left(\frac{9}{10... | 30 | Number Theory | proof | Yes | Yes | cn_contest | false | 727,442 |
Example 2 (1) Prove: for $n>2$ we have
$$
3-\frac{2}{(n-1)!}<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<3 .
$$
(2) For $n>2$ $\left(n \in \mathbf{Z}_{+}\right)$, we have
$$
b-\frac{c}{(n-2)!}<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<b .
$$ | (1) Proof: In $\frac{2}{2!}+\frac{7}{3!}+\cdots+\frac{n^{2}-2}{n!}$, adding one term $\frac{n+2}{n!}$, then
$$
\begin{array}{l}
\frac{2}{2!}+\frac{7}{3!}+\cdots+\frac{(n-1)^{2}-2}{(n-1)!}+\frac{n^{2}-2}{n!}+\frac{n+2}{n!} \\
=\frac{2}{2!}+\frac{7}{3!}+\cdots+\frac{(n-1)^{2}-2}{(n-1)!}+\frac{n+1}{(n-1)!} \\
=\frac{2}{2!... | a=5, b=9, c=4 | Inequalities | proof | Yes | Yes | cn_contest | false | 727,443 |
Example 3 Let $0<t_{1} \leqslant t_{2} \leqslant \cdots \leqslant t_{n}<1$. Prove:
$$
\left(1-t_{n}\right)^{2}\left[\frac{t_{1}}{\left(1-t_{1}^{2}\right)^{2}}+\frac{t_{2}^{2}}{\left(1-t_{2}^{3}\right)^{2}}+\cdots+\frac{t_{n}^{n}}{\left(1-t_{n}^{n+1}\right)^{2}}\right]
$$
$<1$. | Proof From the given conditions, we have
$$
\begin{array}{l}
\frac{t_{1}\left(1-t_{n}\right)^{2}}{\left(1-t_{1}^{2}\right)^{2}}=\frac{t_{1}\left(1-t_{n}\right)^{2}}{\left(1+t_{1}\right)^{2}\left(1-t_{1}\right)^{2}} \\
\leqslant \frac{t_{1}}{\left(1+t_{1}\right)^{2}}<\frac{t_{1}}{1+t_{1}}=1-\frac{1}{1+t_{1}}, \\
\frac{t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,444 |
Example 4 Let $\delta(x)$ be the greatest odd divisor of the positive integer $x$. For any positive integer $x$, prove:
$$
\left|\sum_{n=1}^{x} \frac{\delta(n)}{n}-\frac{2}{3} x\right|<1 .
$$ | Proof: Let $F(x)=\sum_{n=1}^{x} \frac{\delta(n)}{n}$.
Notice that, $\delta(2 m+1)=2 m+1$,
$$
\delta(2 m)=\delta(m)\left(m \in \mathbf{Z}_{+}\right) \text {. }
$$
Therefore, $F(1)=1$,
$$
\begin{array}{l}
F(2 x+1)=\sum_{n=1}^{2 x} \frac{\delta(n)}{n}+\frac{\delta(2 x+1)}{2 x+1}=F(2 x)+1, \\
F(2 x)=\sum_{n=1}^{\delta} \f... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,445 |
4. (Bernoulli's Inequality) Let $p$ be a positive number, and $t$ be a rational number greater than 1. Prove:
$$
(1+p)^{t}>1+t p
$$ | Given that $t>1$, we might as well set $t=\frac{m}{n}(m>n, m, n$ being natural numbers $)$.
Transform the inequality into the form of “product $\leqslant$ sum”
$$
\sqrt[m]{(1+t p)^{n}}<1+p \text {. }
$$
The root index $m$ suggests adding $m-n$ ones under the root, so we have
$$
\begin{array}{l}
\sqrt[m]{(1+t p)^{n}}=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,446 |
Example 1 Let $p$ be a prime number, and $m, n$ be non-negative integers. Find all triples $(p, m, n)$ that satisfy $p^{m}-n^{3}=8$. | Solve: From $p^{m}-n^{3}=8$, we know
$$
p^{m}=(n+2)\left(n^{2}-2 n+4\right) \text {. }
$$
Since $p$ is a prime, without loss of generality, assume
$$
p^{x}=n+2, p^{y}=n^{2}-2 n+4 \text {, }
$$
where $x, y \in \mathbf{Z}_{+}, x+y=m$.
For any non-negative integer $n$, we have
$$
\begin{array}{l}
n^{2}-2 n+4-(n+2)=n^{2}... | (p, m, n)=(2,3,0),(3,2,1),(2,4,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,448 |
Example 2 Proof: There do not exist integers $x, y, z$, satisfying
$$
2 x^{4}+2 x^{2} y^{2}+y^{4}=z^{2}(x \neq 0) .
$$ | Proof by contradiction.
Since $x \neq 0$, then $y \neq 0$.
Assume $(x, y)$ is an integer solution to the given equation.
Clearly, $(-x, y), (x, -y), (-x, -y)$ are also solutions to the original equation.
Therefore, without loss of generality, assume $x > 0, y > 0$ and $(x, y) = 1$, and assume $x$ is the smallest integ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,449 |
Example 4 Let $n$ be a natural number. Prove:
$$
\left(1+\frac{1}{n}\right)^{n}<\left(1+\frac{1}{n+1}\right)^{n+1} .
$$ | First, transform the inequality into the form of "product (L) $\leqslant$ sum", i.e.,
$$
\sqrt[n+1]{\left(1+\frac{1}{n}\right)^{n}}<1+\frac{1}{n+1} \text {. }
$$
Thinking of the root index $n+1$, we should have $n+1$ factors under the root, so we need to add a factor of 1. Thus,
$$
\begin{array}{l}
\sqrt[n+1]{\left(1+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,450 |
Example 3 Find all non-zero integer solutions $(a, b)$ of the indeterminate equation
$$
\left(a^{2}+b\right)\left(a+b^{2}\right)=(a-b)^{2}
$$
Solution (1) $a=b$.
Then $a^{2}+b=0$ or $a+b^{2}=0$.
Thus, $a=b=0$ (discard) or $a=b=-1$.
(2) $a \neq b$.
When $a=0$, we get $b=1$;
When $b=0$, we get $a=1$.
Now consider the ca... | (ii) $a>0>b$.
Let $b=-c$. Then $c>0$.
Thus, $\left(a^{2}-c\right)\left(a+c^{2}\right)=(a+c)^{2}$.
When $c=1$, $a^{2}-1=a+1$.
Therefore, $a=2, b=-1$.
When $c>1$, let $(a, c)=d, a=d a_{1}, c=d c_{1}$. Then $\left(a_{1}, c_{1}\right)=1$.
The original indeterminate equation transforms to
$$
a^{3}-a^{2}=c^{3}+c^{2}-a^{2} c... | (a, b) = (2, -1), (-1, -1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,451 |
Example 4 Find all non-negative integer solutions $(x, y, z, w)$ of the indeterminate equation
$$
2^{x} \times 3^{y}-5^{x} \times 7^{w}=1
$$
[6] | From $5^{2} \times 7^{\infty}+1$ being even, we know $x \geqslant 1$.
(1) When $y=0$, $2^{x}-5^{x} \times 7^{w}=1$.
If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, thus, $4 \mid x$.
Therefore, $31\left(2^{x}-1\right)$, which contradicts $2^{x}-5^{x} \times 7^{x}=1$.
If $z=0,2^{x}-7^{w}=1$, when $x=1,2,3$, after inspect... | (1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,452 |
Example 5 Proof: The system of equations
$$
\left\{\begin{array}{l}
x^{6}+x^{3}+x^{3} y+y=147^{157}, \\
x^{3}+x^{3} y+y^{2}+y+z^{9}=157^{147}
\end{array}\right.
$$
has no integer solutions. ${ }^{[7]}$ | Prove that assuming the system of equations has integer solutions.
(1) + (2) gives
$$
\left(x^{3}+y+1\right)^{2}+z^{9}=147^{157}+157^{147}+1 \text {. }
$$
Taking both sides modulo 19 and using Fermat's Little Theorem, we get
$$
\begin{array}{l}
147^{157}+157^{147}+1 \equiv 14^{157}+5^{147}+1 \\
=14^{18 \times 8+13}+5^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,453 |
Example 1 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=a_{2}=1, a_{n}=\sqrt{3} a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$
Find $a_{2013}$. | Let $x, y$ satisfy
$$
\begin{array}{l}
a_{n}-x a_{n-1}=y\left(a_{n-1}-x a_{n-2}\right) \\
\Rightarrow a_{n}=(x+y) a_{n-1}-x y a_{n-2} .
\end{array}
$$
Comparing with equation (1), we know that $x, y$ are the two roots of the equation
$$
t^{2}-m t+p=0
$$
and $x, y$ are both non-zero.
Clearly, $\left\{a_{n}-x a_{n-1}\r... | 1-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,454 |
Example 2 A bug crawls along a triangular iron ring, at each vertex the bug has an equal chance of crawling to one of the other two vertices. Then the probability that the bug is back at the starting point after $n$ moves is $\qquad$ [2] | 【Analysis】Suppose the bug starts from vertex $A$ of $\triangle A B C$.
Since there are two possible directions, clockwise or counterclockwise, at each vertex, there are $2^{n}$ possible ways of movement after $n$ steps.
Let the number of ways the bug can end up at points $A, B, C$ after $n$ moves be denoted as $A_{n},... | \frac{2^{n}+(-1)^{n} \times 2}{3 \times 2^{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,455 |
In $\triangle A B C$, $B C=a, C A=b, A B=c$, let the area of $\triangle A B C$ be $S$, and $\lambda>1$ be a real number. Prove:
$$
a^{2}+\frac{\lambda b^{2} c^{2}}{b^{2}+c^{2}} \geqslant 4 \sqrt{\lambda-1} S .
$$ | Proof: From the cosine theorem and the formula for the area of a triangle, we know equation (1):
$$
\begin{aligned}
\Leftrightarrow & b^{2}+c^{2}-2 b c \cos A+\frac{\lambda b^{2} c^{2}}{b^{2}+c^{2}} \\
& \geqslant 4 \sqrt{\lambda-1} \cdot \frac{1}{2} b c \sin A \\
\Leftrightarrow & \left(b^{2}+c^{2}\right)^{2}-2 b c\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,456 |
Let the function $f(x)$ satisfy $f(1)=1, f(4)=7$, and for any $a, b \in \mathbf{R}$, we have
$$
f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3} .
$$
Then $f(2014)=$ $\qquad$
(A) 4027
(B) 4028
(C) 4029
(D) 4030 | Solution 1 From $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$, we get
$f(2)=f\left(\frac{4+2 \times 1}{3}\right)=\frac{f(4)+2 f(1)}{3}=3$,
$f(3)=f\left(\frac{1+2 \times 4}{3}\right)=\frac{f(1)+2 f(4)}{3}=5$.
Thus, we conjecture that $f(n)=2 n-1\left(n \in \mathbf{Z}_{+}\right)$.
We prove this by mathematical in... | 4027 | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,457 |
Promotion 1 Let $s, t, u, v \in \mathbf{Q}, s \neq u, t \neq v, \lambda \in \mathbf{R}$, function $f(x)$ satisfies
$$
f(s)=t, f(u)=v,
$$
and for any $a, b \in \mathbf{R}$, there is
$$
f(\lambda a+(1-\lambda) b)=\lambda f(a)+(1-\lambda) f(b) .
$$
Find the value of $f(m) (m \in \mathbf{Q})$. | In fact, in the given conditions, let
$$
\begin{aligned}
a= & \frac{x}{\lambda}, b=0 ; a=0, b=\frac{y}{1-\lambda} . \\
& \text { Then } f(x)=\lambda f\left(\frac{x}{\lambda}\right)+(1-\lambda) f(0) ; \\
& f(y)=\lambda f(0)+(1-\lambda) f\left(\frac{y}{1-\lambda}\right) . \\
& \text { Hence } \lambda f\left(\frac{x}{\lam... | f(m)=\frac{t-v}{s-u} \cdot m+\frac{t u-s v}{u-s}(m \in \mathbf{Q}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,458 |
Example 1 For any positive integer $n$, non-negative numbers $x_{1}, x_{2}, \cdots$, $x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$. Prove:
$$
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \geqslant \frac{1}{2} \text {. }
$$ | Prove (1) When $n=1$, $1-x_{1} \geqslant \frac{1}{2}$.
(2) Assume when $n=k$,
$$
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{k}\right) \geqslant \frac{1}{2} \text {. }
$$
Then when $n=k+1$, let $x_{k}^{\prime}=x_{k}+x_{k+1}$.
By the induction hypothesis,
$$
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,459 |
Example 2 For any positive integer $n$, positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} x_{2} \cdots x_{n}=1$. Prove:
$$
\left(\sqrt{2}+x_{1}\right)\left(\sqrt{2}+x_{2}\right) \cdots\left(\sqrt{2}+x_{n}\right) \geqslant(\sqrt{2}+1)^{n} .
$$ | Proof 1 (1) When $n=1$,
$$
\sqrt{2}+x_{1}=\sqrt{2}+1 \geqslant(\sqrt{2}+1)^{1} \text {. }
$$
(2) Assume when $n=k$,
$$
\left(\sqrt{2}+x_{1}\right)\left(\sqrt{2}+x_{2}\right) \cdots\left(\sqrt{2}+x_{k}\right) \geqslant(\sqrt{2}+1)^{k} \text {. }
$$
Then when $n=k+1$, let $\left\{y_{k+1}\right\}=\left\{x_{k+1}\right\}$,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,460 |
Example 5 Let $a_{i}, b_{i} \in \mathbf{R}_{+}, \sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$. Prove:
$$
\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i} .
$$ | 【Analysis and Proof】Consider the local $\frac{a_{i}^{2}}{a_{i}+b_{i}}$.
To eliminate the denominator, pair it with a term $a_{i}+b_{i}$, so that,
$$
\sum_{i=1}^{n}\left[\frac{a_{i}^{2}}{a_{i}+b_{i}}+\left(a_{i}+b_{i}\right)\right] \geqslant \sum_{i=1}^{n} 2 a_{i} \text {. }
$$
However, this does not lead to inequality... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,461 |
2. At the fish market, a banner hangs that reads "Today's 50% off offer: one-pound packs of fish, half packs only 3 yuan." Then the regular full price per pound of fish is ( ) yuan.
(A)6
(B) 9
(C) 10
(D) 12
(E) 15 | 2. D.
From the problem, we know that half a pack of fish, which is one pound, is 3 yuan after a 50% discount. Therefore, the full pack of fish at 50% off is 6 yuan.
Thus, the regular price per pound of fish is 12 yuan. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,463 |
3. Calculate:
$$
4 \times(-1+2-3+4-\cdots+1000)=(\quad) \text {. }
$$
(A) -10
(B) 0
(C) 1
(D) 500
(E) 2000 | 3. E.
Notice,
$$
\begin{array}{l}
4 \times(-1+2-3+4-\cdots+1000) \\
=4[(-1+2)+(-3+4)+\cdots+(-999+1000)] \\
=4 \times(\underbrace{1+1+\cdots+1}_{500 \uparrow})=2000 .
\end{array}
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,464 |
4. Eight friends went to a restaurant for a meal, and the cost was to be shared equally among them. Because Judy forgot to bring money, the other seven friends each paid an extra 2.50 yuan. Therefore, the total cost of this meal was ( ) yuan.
(A) 120
(B) 128
(C) 140
(D) 144
(E) 160 | 4. C.
In fact, Judy should pay $2.50 \times 7 = 17.50$ yuan.
Since the meal cost is shared equally, each person pays 17.50 yuan.
Therefore, the total cost of the meal is $17.50 \times 8 = 140$ yuan. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,465 |
5. Given that Haman is a sixth-grade student, weighing 106 pounds, and his four sisters weigh 5 pounds, 5 pounds, 6 pounds, and 8 pounds, respectively. Then, among the mean and median of the weights of these five children, ( ) is larger, and the difference is ( ).
(A) Median, 60
(B) Median, 20
(C) Mean, 5
(D) Mean, 15
... | 5. E.
From the problem, we know the average of the five people is
$$
\frac{5+5+6+8+106}{5}=26 \text{; }
$$
The median of the five people is 6.
Thus, the average is larger, and the difference is $26-6=20$. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,466 |
6. As shown in Figure 1, each number in a box is determined by the product of the two numbers in the adjacent boxes of the upper layer (for example, $30=6 \times 5$). Then the number that should be filled in the unfilled box in the first row is ( ).
(A) $2 \quad$ (B) 3
$\begin{array}{lll}(\mathrm{C}) 4 & (\mathrm{D}) 5... | 6. C.
Let the unfilled number in the first row be $x$, and the unfilled number in the second row be $y$. According to the problem, we have
$$
\left\{\begin{array} { l }
{ 6 0 0 = 3 0 y , } \\
{ y = 5 x }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=4, \\
y=20 .
\end{array}\right.\right.
$$ | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,467 |
7. Trey and his mother were observing the passing trains by the railway. Starting from the locomotive, 6 carriages passed in the first 10 seconds, and the total time for the train to pass was 2 minutes and 45 seconds. Assuming the train was traveling at a constant speed, the train had approximately ( ) carriages.
(A) 6... | 7. C.
Notice that, 2 minutes 45 seconds $=165$ seconds. According to the problem, the number of carriages is $165 \times \frac{6}{10}=99$. Therefore, the train has approximately 100 carriages. | C | Other | MCQ | Yes | Yes | cn_contest | false | 727,468 |
8. When a fair coin is tossed three times in a row, the probability of getting at least two consecutive heads is ( ).
(A) $\frac{1}{8}$
(B) $\frac{1}{4}$
(C) $\frac{3}{8}$
(D) $\frac{1}{2}$
(E) $\frac{3}{4}$ | 8. C.
Tossing a uniformly distributed coin three times in a row, there are $2^{3}=8$ possible outcomes. Therefore, the scenarios with at least two consecutive heads are
(Head, Head, Head), (Head, Head, Tail), (Tail, Head, Head).
Thus, the probability of getting at least two consecutive heads is $\frac{3}{8}$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,469 |
9. The Hulk's distance for each jump is twice the distance of the previous jump. If the first jump is 1 meter, then the second jump is 2 meters, the third jump is 4 meters, and so on. When he first jumps more than 1 kilometer, it is his ( )th jump.
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13 | 9. C.
It is known that the distance of each jump of the Hulk forms a geometric sequence, i.e., $a_{n}=2^{n-1}\left(n \in \mathbf{Z}_{+}\right)$.
$$
\begin{array}{l}
\text { and } a_{10}=2^{9}=5121000,
\end{array}
$$
Therefore, when he first jumps over 1 kilometer, it is on the 11th jump. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,470 |
10.180 and 594's least common multiple to greatest common divisor ratio is ( ).
(A) 110
(B) 165
(C) 330
(D) 625
(E) 660 | 10. C.
Notice,
$$
\begin{array}{l}
\frac{[180,594]}{(180,594)}=\frac{2^{2} \times 3^{3} \times 5 \times 11}{2 \times 3^{2}} \\
=2 \times 3 \times 5 \times 11=330
\end{array}
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,471 |
Example 6 Let $a, b, c \in \mathbf{R}_{+}, abc=1$. Prove:
$$
\sum \frac{1}{a^{3}(b+c)} \geqslant \frac{3}{2},
$$
where, “ $\sum$ ” denotes the cyclic sum. | $$
\begin{array}{l}
\text { Thus, } \sum\left[\frac{1}{a^{3}(b+c)}+\frac{a(b+c)}{4}\right] \geqslant \sum \frac{1}{a} . \\
\text { Therefore, } \sum \frac{1}{a^{3}(b+c)} \geqslant \sum \frac{1}{a}-\sum \frac{a b}{2} \\
=\sum \frac{1}{a}-\sum \frac{1}{2 c}=\sum \frac{1}{2 a} \\
\geqslant \frac{1}{2} \times \frac{3}{\sqr... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,472 |
11. Ted's grandfather exercised on a treadmill for three days, running 2 miles each day. It is known that the speeds on the first, second, and third days were 5 miles/hour, 3 miles/hour, and 4 miles/hour, respectively. If his grandfather maintained a speed of 4 miles/hour throughout, he would have spent ( ) minutes les... | 11. D.
According to the problem, the required calculation is $\frac{2}{5}+\frac{2}{3}+\frac{2}{4}-\frac{6}{4}=\frac{1}{15}$ (hour) $=4$ (minutes). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,473 |
12. In 2013, there was a "special" sandal trading rule in the Winniebago County market: if you buy one pair of sandals, it is traded at the original price of 50 yuan; if you buy a second pair, it is traded at a 60% discount of the original price; if you buy a third pair, it is half price. It is known that Javier bought... | 12. B.
From the problem, what we need to find is
$$
150-(50+50 \times 60 \%+50 \times 50 \%)=45 \text { (yuan). }
$$
Thus, the savings amount to $\frac{45}{150} \times 100 \%=30 \%$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,474 |
13. When Clara calculated the total score of her test results in each subject, she unconsciously swapped the tens and units digits of a certain subject's score. Then the score that is most likely to be the difference between the incorrect total score and the correct total score is ( ).
(A) 45
(B) 46
(C) 47
(D) 48
(E) 4... | 13. A.
Let the correct total score be $10 a+b$. Then the incorrect total score is $10 b+a$.
Notice that, the difference between the two is
$$
10 a+b-(10 b+a)=9(a-b) \equiv 0(\bmod 9) \text {. }
$$
Since only option $\mathrm{A}$ fits, the most likely score difference is 45. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,475 |
14. Anbei has one green and one red jelly bean, while Beia has one green, one yellow, and two red jelly beans. If each of them randomly gives one of their jelly beans to the other, what is the probability that Anbei ends up with two jelly beans of the same color after the exchange?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(... | 14. C.
The probability that both jelly beans in Abei Te's hand are green is $\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$,
The probability that both jelly beans in his hand are red is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$.
Therefore, the probability that the jelly beans are of the same color is
$$
\frac{1}{8}+\... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,476 |
15. Given
$$
3^{p}+3^{4}=90,2^{r}+44=76,5^{3}+6^{3}=1421 \text {. }
$$
Then $p r s=(\quad)$.
(A) 27
(B) 40
(C) 50
(D) 70
(E) 90 | 15. B. From the known information, we get $p=2, r=5, s=4$. Therefore, prs $=40$. | 40 | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,477 |
16. At Fibonacci Middle School, some students from grades six to eight participate in a community service. It is known that the ratio of the number of participants from grade eight to grade six, and from grade eight to grade seven are $5: 3$ and $8: 5$, respectively. Then the minimum number of people who participated i... | 16. E.
From the problem, we know that the number of eighth graders must be a multiple of both 5 and 8. Therefore, the minimum number of eighth graders is \([5,8]=40\). At this point, the number of sixth graders is \(40 \times \frac{3}{5}=24\) people, and the number of seventh graders is \(40 \times \frac{5}{8}=25\) pe... | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,478 |
17. If the sum of six consecutive positive integers is 2013, then the largest number among these six consecutive positive integers is ( ).
(A) 335
(B) 338
(C) 340
(D) 345
(E) 350 | 17. B.
Let the largest number among these six numbers be $n$. Then $n+(n-1)+\cdots+(n-5)=2013$.
Solving this, we get $n=338$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,479 |
18. As shown in Figure 2, Isabella is building a rectangular fort 12 feet long, 10 feet wide, and 5 feet high using cubic bricks with an edge length of 1 foot, and the floor and all four walls are 1 foot thick. Then the total number of bricks needed is ( ).
(A) 204
(B) 280
(C) 320
(D) 340
(E) 600 | 18. B.
The volume of the fort can be considered as the difference between the volumes of two solid rectangles, then
$$
\begin{array}{l}
V_{\text {outer }}=12 \times 10 \times 5=600, \\
V_{\text {inner }}=10 \times 8 \times 4=320 .
\end{array}
$$
Therefore, a total of $600-320=280$ blocks are needed. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,480 |
19. Bridget, Casey, and Hannah are discussing their scores on a math test. Hannah showed her score to Bridget and Casey, but they didn't show their scores to anyone. Casey said, "I am not the lowest score in the class." Then Bridget added, "I am not the highest score in the class." The order of the three girls' test sc... | 19. D.
From the problem, we know that Casey's statement indicates Casey's score is higher than Hannah's. Bridget's statement indicates her score is lower than Hannah's. Therefore, the order of scores from highest to lowest is Casey, Hannah, Bridget. | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,481 |
20. Given a rectangle $A B C D$ with a length of 2 and a width of 1 inscribed in a semicircle $\odot O$, and the length $B C$ of the rectangle lies on the diameter of the semicircle. Then the area of the semicircle $\odot 0$ is ( ).
(A) $\frac{\pi}{2}$
(B) $\frac{2 \pi}{3}$
(C) $\pi$
(D) $\frac{4 \pi}{3}$
(E) $\frac{5 ... | 20. C.
As shown in Figure 7.
Since rectangle $A B C D$ is inscribed in the semicircle $\odot O$, we have $O B=O C=\frac{1}{2} B C=1$.
Also, $C D=1$, so $O D=\sqrt{2}$.
Therefore, $S_{\text {semicircle }}=\frac{1}{2} \times(\sqrt{2})^{2} \pi=\pi$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,482 |
Example 7 Let $a, b, c > 0$. Prove:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\sqrt{\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}} \geqslant \frac{5}{2} .
$$ | 【Analysis and Proof】The left side of the inequality can be clearly divided into two parts:
$$
A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \text { and } B=\sqrt{\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}} \text {, }
$$
Both parts are symmetric expressions of $a, b, c$, but there is a structural difference ("internal differe... | \frac{5}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,483 |
21. Samantha's house is two blocks west and one block south of the southwest corner of City Park, and her school is two blocks east and two blocks north of the northeast corner of City Park. On her way to school, Samantha always rides her bike to the southwest corner of City Park, then cuts diagonally through the park ... | 21. E.
Samantha needs to walk a minimum of three blocks from her home to the southwest corner of the park: she walks one block east twice and one block north once, which gives her three possible routes; she needs to walk a minimum of four blocks from the northeast corner of the park to her school: she walks two blocks... | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,484 |
22. As shown in Figure 3, to build a grid using toothpicks that is 60 toothpicks long and 32 toothpicks wide, a total of ( ) toothpicks are needed.
(A) 1920
(B) 1952
(C) 1980
(D) 2013
(E) 3932 | 22. E.
The number of toothpicks needed for the first row of grids is
$$
32 \times 3+1=97 \text {. }
$$
The number of toothpicks needed for the second to sixtieth rows of grids is
$$
(32 \times 2+1) \times 59=3835 \text {. }
$$
Thus, the total number of toothpicks required is $97+3835=3932$. | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,485 |
23. As shown in Figure 4, in the right triangle $\triangle ABC$, it is known that $\angle ABC = 90^{\circ}$. Semi-circles are constructed outward on each side of $\triangle ABC$. The area of the semi-circle with diameter $AB$ is $8\pi$, and the arc length of the semi-circle with diameter $AC$ is $8.5\pi$. Then the radi... | 23. B.
From the problem, we know that $A B=8, A C=17$.
By the Pythagorean theorem, we have
$$
B C=\sqrt{A C^{2}-A B^{2}}=\sqrt{289-64}=15 \text {. }
$$
Thus, the radius of the semicircle with $B C$ as its diameter is 7.5. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,486 |
25. As shown in Figure 6, a small ball with a diameter of 4 inches rolls along a track starting from point $A$, and the track consists of three semicircles. It is known that the radii of the three semicircles are $R_{1}=100$ inches, $R_{2}=60$ inches, $R_{3}=80$ inches. If the ball remains in contact with the track and... | 25. A.
Note that, the diameter of the ball is 4 inches.
If it rolls on the inside track of a semicircle with radius $R$ inches, then the path of the ball's center is a semicircle with radius $R-2$ inches; if it rolls on the outside track of a semicircle with radius $R$ inches, then the path of the ball's center is a s... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,488 |
1. If a set $S$ satisfies that for any $a, b \in S$, $a \pm b \in S$, then the set $S$ is called a "closed set". Among the following sets, which one is not a closed set? ( ).
(A) Set of natural numbers $\mathbf{N}$
(B) Set of integers $\mathbf{Z}$
(C) Set of rational numbers $\mathbf{Q}$
(D) Set of real numbers $\mathb... | -,1.A.
By the definition of a closed set, we know that the set of integers $\mathbb{Z}$, the set of rational numbers $\mathbb{Q}$, and the set of real numbers $\mathbb{R}$ are all closed sets, but the set of natural numbers $\mathbb{N}$ is not a closed set. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,489 |
2. Let positive numbers $a, b$ satisfy $a<b$. If there are real numbers $x_{1}, y_{1}, x_{2}, y_{2}$ such that $x_{1}+y_{1}, x_{2}+y_{2}$ are the arithmetic mean and geometric mean of $a$ and $b$ respectively. Then the maximum value of $\frac{\sqrt{x_{1} y_{1}}}{\left(x_{2}+y_{2}\right)^{2}}$ is ( ).
(A) $\frac{\sqrt{a... | 2. D.
Given $x_{1}+y_{1}=\frac{a+b}{2}, x_{2} y_{2}=\sqrt{a b}(0<a<b)$, then
$$
\frac{\sqrt{x_{1} y_{1}}}{\left(x_{2}+y_{2}\right)^{2}} \leqslant \frac{\frac{x_{1}+y_{1}}{2}}{4 x_{2} y_{2}}=\frac{a+b}{16 \sqrt{a b}} .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,490 |
3. Given a line segment $AB$ of length 4, whose two endpoints lie on the parabola $y=x^{2}+x$. Then the shortest distance from the midpoint $P$ of the segment to the $x$-axis is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) 1
(D) $\frac{1}{2}$ | 3. B.
From the problem, we know that the equation of the parabola can be rewritten as $y=x^{2}-\frac{1}{4}$, and the equation of the directrix is $y=-\frac{1}{2}$.
To make the distance from point $P$ to the $x$-axis the shortest, it means that the sum of the distances from points $A$ and $B$ to the directrix is the s... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,491 |
4. As shown in Figure 1, given a quadrilateral pyramid $P-ABCD$ with a base that is a square of side length 1, $PA=1$, and $PA \perp$ the base $ABCD$. Then the distance from the midpoint $E$ of $PB$ to the plane $PCD$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{1}{4}$ | 4. C.
Let $F$ be the midpoint of side $PC$. Then $EF$ is the midline of $\triangle PBC$, and the distance from point $E$ to plane $PCD$ is half the distance from point $B$ to plane $PCD$. The distance from point $B$ to plane $PCD$ is equal to the distance from point $A$ to plane $PCD$, which is the altitude from side ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,492 |
5. Given that the angle between vectors $a$ and $b$ is $45^{\circ}$, and
$$
\begin{array}{l}
|a|=4, \\
\left(\frac{1}{2} a+b\right) \cdot(2 a-3 b)=12 .
\end{array}
$$
Then the projection of $\boldsymbol{b}$ onto $\boldsymbol{a}$ is equal to ( ).
(A) $\sqrt{2}$
(B) 4
(C) $4 \sqrt{2}$
(D) 1 | 5. D.
From the given information, we have
$$
\begin{array}{l}
a^{2}+\frac{1}{2} a \cdot b-3 b^{2}=12 \\
\Rightarrow|a|^{2}+\frac{1}{2}|a||b| \cos \theta-3|b|^{2}=12 \\
\Rightarrow 16+\frac{1}{2} \times 4|b| \cos 45^{\circ}-3|b|^{2}=12 .
\end{array}
$$
Solving, we get $|b|=\sqrt{2}$.
Thus, the projection of $\boldsymb... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,493 |
1. Let $a_{i} \geqslant 1(1 \leqslant i \leqslant n, n \geqslant 2)$. Prove:
$$
\begin{array}{l}
\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \\
\geqslant 2^{n} \frac{1+a_{1}+a_{2}+\cdots+a_{n}}{n+1} .
\end{array}
$$ | ```
Given $a_{i} \geqslant 1$, we can let $a_{i}=1+t_{i}$.
Then $t_{i} \geqslant 0(1 \leqslant i \leqslant n)$.
\[
\begin{array}{l}
\text { Left side }=2^{n}\left(1+\frac{t_{1}}{2}\right)\left(1+\frac{t_{2}}{2}\right) \cdots\left(1+\frac{t_{n}}{2}\right) \\
\geqslant 2^{n}\left(1+\frac{t_{1}}{2}+\frac{t_{2}}{2}+\cdots+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,494 |
6. Given that $x, y$ satisfy
$$
\left\{\begin{array}{l}
x+3 y-3 \leqslant 0, \\
x \geqslant 0, \\
y \geqslant 0 .
\end{array}\right.
$$
Then the range of $z=\frac{y+2}{x-1}$ is ( ).
(A) $[-2,1]$
(B) $(-\infty,-2] \cup[1,+\infty)$
(C) $[-1,2]$
(D) $(-\infty,-1] \cup[2,+\infty)$ | 6. B.
From the problem, we know that the system of inequalities represents the shaded area in Figure 3.
$$
z=\frac{y+2}{x-1} \text { can be seen as the slope of the line connecting point }(x, y) \text { and point } A(1,-2) \text { }
$$
It is easy to see that,
$$
k_{A O}=-2, k_{A B}=1 .
$$
Therefore, the range of $z$... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,495 |
7. Let sets $A, B, C$ satisfy
$$
A \cup B \cup C=\{1,2, \cdots, 9\} \text {. }
$$
Then the number of ordered triples $(A, B, C)$ is $\qquad$ . | Ni,7.7 ${ }^{9}$.
Place the numbers $1 \sim 9$ into three sets $A$, $B$, and $C$. First, consider the number 1 separately.
When the number 1 appears once, twice, or three times, the number of placement methods are $\mathrm{C}_{3}^{1}$, $\mathrm{C}_{3}^{2}$, and $\mathrm{C}_{3}^{3}$, respectively.
Therefore, the total n... | 7^9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,496 |
8. If the complex number $x$ satisfies $x+\frac{1}{x}=-1$, then $x^{2013}+\frac{1}{x^{2013}}=$ $\qquad$ . | 8. 2 .
Given that $x^{2}+x+1=0$.
Since the discriminant $\Delta=-3<0$, $x$ is a complex number.
Also, $x^{3}-1=(x-1)\left(x^{2}+x+1\right)=0$
$\Rightarrow x^{3}=1$.
Therefore, $x^{2013}+\frac{1}{x^{2013}}=\left(x^{3}\right)^{671}+\frac{1}{\left(x^{3}\right)^{671}}=1+1=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,497 |
9. Given the function
$$
f(x)=\frac{2 x}{a x+b}, f(1)=1, f\left(\frac{1}{2}\right)=\frac{2}{3} \text {. }
$$
Let $x_{1}=\frac{1}{2}, x_{n+1}=f\left(x_{n}\right)$. Then the general term formula for the sequence $\left\{x_{n}\right\}$ is $x_{n}=$ $\qquad$ . | 9. $\frac{2^{n-1}}{2^{n-1}+1}$.
From $f(1)=1, f\left(\frac{1}{2}\right)=\frac{2}{3}$, we get
$$
a=b=1 \Rightarrow f(x)=\frac{2 x}{x+1} \text {. }
$$
From $x_{1}=\frac{1}{2}, x_{2}=\frac{2}{3}, x_{3}=\frac{4}{5}, x_{4}=\frac{8}{9}$, we can conjecture
$$
x_{n}=\frac{2^{n-1}}{2^{n-1}+1} \text {. }
$$
Proof by mathemati... | \frac{2^{n-1}}{2^{n-1}+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,498 |
10. Let the function be
$$
f(x)=\cos 2 x+2 \sqrt{3} \sin x \cdot \cos x(x \in \mathbf{R}) \text {. }
$$
If there are ten distinct positive numbers $x_{i}$ satisfying
$$
f\left(x_{i}\right)=2, x_{i}<10 \pi(i=1,2, \cdots, 10),
$$
then $x_{1}+x_{2}+\cdots+x_{10}=$ . $\qquad$ | 10. $\frac{140 \pi}{3}$.
From the problem, we have $f(x)=2 \sin \left(2 x+\frac{\pi}{6}\right)$.
Also, $f\left(x_{i}\right)=2$, thus,
$$
\begin{array}{l}
\sin \left(2 x_{i}+\frac{\pi}{6}\right)=1 \\
\Rightarrow 2 x_{i}+\frac{\pi}{6}=2 k \pi+\frac{\pi}{2} \\
\Rightarrow x_{i}=k \pi+\frac{\pi}{6}(k \in \mathbf{Z}) .
\en... | \frac{140 \pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,499 |
11. Six candidates are sitting in the same row of seats with a passage on each side to take an exam. The order in which the candidates finish their papers is uncertain, and each person leaves their seat and exits the classroom immediately after finishing. Then the probability that at least one person will disturb other... | 11. $\frac{43}{45}$.
To avoid disturbing other students who are still taking the exam, one must leave only after the two students sitting next to them have left, meaning there are two choices each time, resulting in a total of $2^{5}$ possibilities. Therefore, the required probability is $1-\frac{2^{5}}{6!}=\frac{43}{... | \frac{43}{45} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,500 |
12. Write an arithmetic sequence consisting of three prime numbers with a common difference of 8. The sequence is $\qquad$ | 12.3,11,19.
Let the first prime be $p=3k, 3k+1,3k+2$ $\left(k \in \mathbf{Z}_{+}\right)$ and discuss each case.
(1) If $p=3k$, then the sequence is $3k, 3k+8,3k+16$. Since $p$ is a prime, $k=1$. Therefore, the sequence is $3,11,19$.
(2) If $p=3k+1$, then the sequence is $3k+1,3k+9$, $3k+17$. But $3k+9$ is not a prime,... | 3,11,19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,501 |
13. (16 points) As shown in Figure 2, given
semicircle $C: x^{2}+y^{2}=1(y \geqslant 0)$ and point $A(2,0)$, $B$ is any point on the semicircle, and an equilateral triangle $\triangle A B M$ is constructed with $A B$ as one side. When is the area of quadrilateral $O A M B$ maximized? And find this maximum value. | Three, 13. Let $\angle A O B=\theta(0<\theta<\pi)$.
Since $O B=1, O A=2$, we have
$$
A B^{2}=1^{2}+2^{2}-2 \times 1 \times 2 \cos \theta=5-4 \cos \theta \text {. }
$$
Therefore, $S_{\text {quadrilateral } O A M B}$
$$
\begin{array}{l}
=\frac{1}{2} \times 1 \times 2 \sin \theta+\frac{\sqrt{3}}{4}(5-4 \cos \theta) \\
=\... | 2+\frac{5 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,502 |
14. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=2, a_{2}+a_{4}=8,
$$
and for any $n \in \mathbf{Z}_{+}$, the function
$$
f(x)=\left(a_{n}-a_{n+1}+a_{n+2}\right) x+a_{n+1} \cos x-a_{n+2} \sin x
$$
satisfies $f^{\prime}\left(\frac{\pi}{2}\right)=0$.
(1) Find the general term formula of the sequ... | 14. (1) From the given, we have
$$
f^{\prime}(x)=a_{n}-a_{n+1}+a_{n+2}-a_{n+1} \sin x-a_{n+2} \cos x \text {. }
$$
Also, \( f^{\prime}\left(\frac{\pi}{2}\right)=0 \), so
$$
a_{n}-a_{n+1}+a_{n+2}-a_{n+1}=0 \text {. }
$$
Thus, \( a_{n+1}-a_{n}=a_{n+2}-a_{n+1} \).
Therefore, \( \left\{a_{n}\right\} \) is an arithmetic s... | b_{1}+b_{2}+\cdots+b_{2012}<2013 \times 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,503 |
15. (20 points) Given the functions
$$
\begin{array}{l}
f(x)=\ln (1+x), \\
g(x)=\frac{x(1+\lambda x)}{1+x}(\lambda>0),
\end{array}
$$
and for any $x \geqslant 0$, $f(x) \leqslant g(x)$.
(1) Find the range of $\lambda$;
(2) If $a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$, prove:
$$
a_{2 n}-a_{n}+\frac{1}{4 n}>\... | 15. (1) Let
$$
\begin{array}{l}
H(x)=f(x)-g(x) \\
=\ln (1+x)-\frac{x(1+\lambda x)}{1+x} .
\end{array}
$$
Then $H(0)=0$,
$$
H^{\prime}(x)=\frac{(1-2 \lambda) x-\lambda x^{2}}{(1+x)^{2}} \text {. }
$$
(i) $\lambda < \frac{1}{2}$.
Thus, when $x>0$, $H^{\prime}(x)>0$. Therefore, $H(x)$ is increasing on $(0, +\infty)$. Sin... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 727,504 |
2. Let $a_{n}=n$. Prove: $\sum_{k=1}^{n} \frac{\sqrt{k}}{a_{k}^{2}}<3$. | $\begin{array}{l}\text { Hint: } \sum_{k=1}^{n} \frac{\sqrt{k}}{a_{k}^{2}}=\sum_{k=1}^{n} \frac{1}{\sqrt{k^{3}}} \\ <1+\sum_{k=2}^{n} \frac{\sqrt{k+1}-\sqrt{k-1}}{\sqrt{(k-1)(k+1)}} \\ =1+\sum_{k=2}^{n}\left(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k+1}}\right) \\ =1+1+\frac{\sqrt{2}}{2}-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,505 |
16. (20 points) In communication engineering, an $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is often used to represent information, where $a_{i}=0$ or $1(i, n \in \mathbf{Z}_{+})$. Let
$$
\begin{array}{l}
w=(\underbrace{0,0, \cdots}_{n \uparrow 0}), \\
u=\left(a_{1}, a_{2}, \cdots, a_{n}\right), \\
v=\left(b_... | 16. Let $u$ and $v$ have $m$ positions where both are 0, and $s$ positions where both are 1. Thus, the number of positions where one is 1 and the other is 0 is $n-m-s$ (where $n \geqslant m+s$).
Therefore, $d(u, w)+d(v, w)=2 s+(n-m-s)$
$=d(u, v)+2 s \geqslant d(u, v)$. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,506 |
$$
\begin{array}{l}
A=\left\{1,10, \frac{1}{10}\right\}, \\
B=\{y \mid y=\lg x, x \in A\} .
\end{array}
$$
Then $A \cap B=$ | $-1 .\{1\}$.
From the problem, we know
$$
\begin{aligned}
B & =\left\{y \mid y=\lg 1, y=\lg 10, y=\lg \frac{1}{10}\right\} \\
& =\{0,1,-1\} .
\end{aligned}
$$
Therefore, $A \cap B=\{1\}$. | \{1\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,507 |
2. Given the complex number $z$ satisfies $z+|z|=2+\mathrm{i}$. Then $z=$
$\qquad$ | 2. $\frac{3}{4}+\mathrm{i}$.
Let $z=x+\mathrm{i}(x \in \mathbf{R})$. Then
$$
\begin{array}{l}
x+\mathrm{i}+\sqrt{x^{2}+1}=2+\mathrm{i} \\
\Rightarrow x+\sqrt{x^{2}+1}=2 \Rightarrow x=\frac{3}{4} \\
\Rightarrow z=\frac{3}{4}+\mathrm{i}
\end{array}
$$ | \frac{3}{4}+\mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,508 |
3. A science and technology innovation competition sets first, second, and third prizes (all participants will receive an award), and the probabilities of winning the corresponding prizes form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prizes form an arithmetic sequence with... | 3. 500 .
Let the prize money obtained be $\xi$ yuan. Then $\xi=700,560,420$. From the problem, we know
$$
\begin{array}{l}
P(\xi=700)=a, P(\xi=560)=2 a, \\
P(\xi=420)=4 a .
\end{array}
$$
From $7 a=1$, we get $a=\frac{1}{7}$.
Therefore, $E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7}$ $=50... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,509 |
4. $\cos ^{2} 75^{\circ}+\cos ^{2} 15^{\circ}+\cos 75^{\circ} \cdot \cos 15^{\circ}=$ | $\begin{array}{l}\text { 4. } \frac{5}{4} . \\ \cos ^{2} 75^{\circ}+\cos ^{2} 15^{\circ}+\cos 75^{\circ} \cdot \cos 15^{\circ} \\ =\cos ^{2} 75^{\circ}+\sin ^{2} 75^{\circ}+\sin 15^{\circ} \cdot \cos 15^{\circ} \\ =1+\frac{1}{2} \sin 30^{\circ}=\frac{5}{4} .\end{array}$ | \frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,510 |
6. Given the equation of $\odot O$ is $x^{2}+y^{2}=1$, and point $A(1,0)$. Take a moving point $B$ on $\odot O$, and let point $P$ satisfy $\overrightarrow{A P}=\lambda \overrightarrow{O B}(\lambda \in \mathbf{R})$ and $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$. Then the equation of the trajectory of point $P$... | 6. $y^{2}=2 x-1$.
Let point $P(x, y)$. Then by $\overrightarrow{A P}=\lambda \overrightarrow{O B}(\lambda \in \mathbf{R})$, we get $B(k(x-1), k y)\left(\lambda=\frac{1}{k}\right)$.
Substituting the coordinate formula into $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$, we get
$$
k=\frac{x}{(x-1)^{2}+y^{2}} \text ... | y^{2}=2 x-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,512 |
7. Given $l_{1}, l_{2}, \cdots, l_{100}$ are 100 distinct and coplanar lines. If the lines numbered $4 k\left(k \in \mathbf{Z}_{+}\right)$ are parallel to each other, and the lines numbered $4 k-1$ all pass through point $A$, then the maximum number of intersection points of these 100 lines is $\qquad$ . | 7.4351 .
According to the problem, the number of combinations of any two lines out of 100 lines is
$$
\mathrm{C}_{100}^{2}-\mathrm{C}_{25}^{2}-\mathrm{C}_{25}^{2}+1=4351 .
$$ | 4351 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,513 |
8. Given a regular tetrahedron $A_{1} A_{2} A_{3} A_{4}$, four parallel planes $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ are drawn through its four vertices. If the distance between each pair of adjacent planes is 1, then the volume of the tetrahedron is $\qquad$. | 8. $\frac{5 \sqrt{5}}{3}$.
As shown in Figure 3, a regular tetrahedron is complemented to form a cube, $E_{1}$ and $F_{1}$ are the midpoints of $A_{1} B_{1}$ and $C_{1} D_{1}$, respectively. The planes $E E_{1} D_{1} D$ and $B B_{1} F_{1} F$ are two parallel planes, and their distance is 1.
As shown in Figure 4, let t... | \frac{5 \sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,514 |
9. (14 points) Let $\triangle A B C$ have $\angle A, \angle B, \angle C$ opposite sides $a, b, c$ respectively, and $2 a \cos C=2 b-c$. Find
(1) the size of $\angle A$;
(2) the range of $b+c$ when $a=1$.
保留源文本的换行和格式，直接输出翻译结果。 | 9. (1) From $2 a \cos C=2 b-c$, we get
$$
\begin{array}{l}
\sin A \cdot \cos C+\frac{1}{2} \sin C=\sin B \\
=\sin (A+C)=\sin A \cdot \cos C+\cos A \cdot \sin C \\
\text { then } \frac{1}{2} \sin C=\cos A \cdot \sin C \text {. }
\end{array}
$$
Since $\sin C \neq 0$, thus, $\cos A=\frac{1}{2}$.
Also, $0a=1$, hence the r... | \cos A=\frac{1}{2}, (1,2] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,515 |
3. Let $0 \leqslant a, b, c \leqslant 1$. Prove:
$$
\sum \frac{a}{b+c+1}+(1-a)(1-b)(1-c) \leqslant 1,
$$
where, “ $\sum$ ” denotes the cyclic sum. | Suppose $0<a \leqslant b \leqslant c \leqslant 1$. Then
$$
a+b+1 \leqslant a+c+1 \leqslant b+c+1 \text {. }
$$
Thus, by replacing the denominators in the inequality with $a+b+1$, we get the left side of the inequality $\leqslant \frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c)$.
Using the method of analysis to prove:
$$
\frac{a+b+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,516 |
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