problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
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| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
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values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. In $\triangle A B C$, it is known that $\angle A=2 \angle B, C D$ is the angle bisector of $\angle C$, $A C=16, A D=8$. Then $B C=$
|
3. 24 .
From $\angle A=2 \angle B$, we know $B C>A C$.
As shown in Figure 5, take a point $E$ on side $B C$ such that $E C=A C$,
and connect $D E$.
Then $\triangle C E D \cong \triangle C A D$
$$
\begin{array}{l}
\Rightarrow E D=A D, \\
\angle C E D=\angle C A D . \\
\text { Hence } \angle B D E=\angle C E D-\angle D B E \\
= \angle A-\angle B=\angle B=\angle D B E
\end{array}
$$
Therefore, $B E=D E$.
Thus, $B C=B E+C E=A D+A C=8+16=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given prime numbers $p$ and $q$, such that $p^{3}-q^{5}=(p+q)^{2}$. Then $\frac{8\left(p^{2013}-p^{2010} q^{5}\right)}{p^{2011}-p^{2009} q^{2}}=$ $\qquad$.
|
4. 140.
If $p$ and $q$ have the same remainder when divided by 3, and since $p$ and $q$ are both prime numbers:
If the remainder is 0 when divided by 3, then $p=q=3$, at this time,
$$
p^{3}-q^{5}=0 \Rightarrow 3^{3}>q^{5} \Rightarrow q^{5}<27,
$$
such a prime number $q$ does not exist.
Therefore, it can only be $q=3$, and $p^{3}-243=(p+3)^{2}$, which means
$$
p\left(p^{2}-p-6\right)=252=2^{2} \times 3^{2} \times 7.
$$
Thus, $p$ must be one of the numbers 2, 3, 7.
Upon inspection, only $p=7$ satisfies the equation.
Therefore, $p=7, q=3$.
$$
\begin{array}{l}
\text { Hence } \frac{8\left(p^{2013}-p^{2010} q^{5}\right)}{p^{2011}-p^{2009} q^{2}}=\frac{8 p^{2010}\left(p^{3}-q^{5}\right)}{p^{2009}\left(p^{2}-q^{2}\right)} \\
=\frac{8 p(p+q)^{2}}{(p-q)(p+q)}=\frac{8 p(p+q)}{p-q} \\
=\frac{8 \times 7(7+3)}{4}=140 .
\end{array}
$$
|
140
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$
|
2.9.
From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. From the set $A=\{1,2, \cdots, 30\}$, select five different numbers such that these five numbers form an arithmetic sequence. The number of different arithmetic sequences obtained is $\qquad$ .
|
7. 196.
Let the five numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, with a common difference of $d$.
From $a_{5}-a_{1}=4d$, we know $a_{1} \equiv a_{5}(\bmod 4)$.
Divide the set $A=\{1,2, \cdots, 30\}$ into four categories based on the remainder modulo 4:
$$
\begin{array}{l}
B=\{1,5,9, \cdots, 29\}, \\
C=\{2,6,10, \cdots, 30\}, \\
D=\{3,7,11, \cdots, 27\}, \\
E=\{4,8,12, \cdots, 28\} .
\end{array}
$$
It is easy to see that sets $B$ and $C$ each have eight elements, and sets $D$ and $E$ each have seven elements.
The key to determining a five-term arithmetic sequence is to determine the first and last terms of the sequence, i.e., to select two elements from the sets $B, C, D, E$ in any order, hence we get
$$
A_{8}^{2}+A_{8}^{2}+A_{7}^{2}+A_{7}^{2}=196
$$
different arithmetic sequences.
|
196
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$.
|
2.1.
Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$.
Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the perpendiculars, respectively. Therefore,
$$
\begin{array}{l}
2|M F|-|M P|=|M A|-|M P| \\
\leqslant|M A|-|M B|=|A B|,
\end{array}
$$
where $|A B|$ is the distance from point $P$ to the right directrix, which equals 1.
Thus, when the y-coordinate of point $M$ is 1, the maximum value of $2|M F|-|M P|$ is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ .
|
3. 1 .
Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \in[1, \sqrt{2}])$.
From $t^{2}=x^{2}+y^{2}+2 x y=1+2 x y \Rightarrow x y=\frac{t^{2}-1}{2}$.
Thus, $x+y-x y=t-\frac{t^{2}-1}{2}=1-\frac{(t-1)^{2}}{2} \leqslant 1$, and
when $x=1, y=0$, the equality holds.
Therefore, the maximum value sought is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $x_{n}=\sum_{k=1}^{2013}\left(\cos \frac{k!\pi}{2013}\right)^{n}$. Then $\lim _{n \rightarrow \infty} x_{n}=$
|
4. 1953.
Notice that, $2013=3 \times 11 \times 61$.
Therefore, when $1 \leqslant k \leqslant 60$,
$$
\frac{k!}{2013} \notin \mathbf{Z}, \cos \frac{k!\pi}{2013} \in(-1,1).
$$
Hence $\lim _{n \rightarrow \infty} \cos ^{n} \frac{k!\pi}{2013}=0$.
And when $k \geqslant 61$, $\frac{k!}{2013}$ is an integer, and always an even number, at this time, $\cos \frac{k!\pi}{2013}=1$, accordingly, $\lim _{n \rightarrow \infty} \cos ^{n} \frac{k!\pi}{2013}=1$.
$$
\text { Therefore, } \lim _{n \rightarrow \infty} x_{n}=2013-60=1953 \text {. }
$$
|
1953
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the sum of all elements in a non-empty subset of $\{1,2, \cdots, 9\}$ is a multiple of 3, the subset is called a "Jin state subset". Then the number of such Jin state subsets is $\qquad$ .
|
5. 175.
If a proper subset is a peculiar subset, then its complement is also a peculiar subset. Therefore, we only need to consider peculiar subsets with fewer than or equal to 4 elements.
Thus, there are 3 peculiar subsets with only 1 element; there are $\mathrm{C}_{3}^{2}+3 \times 3=12$ peculiar subsets with exactly 2 elements; there are $3+3 \times 3 \times 3=30$ peculiar subsets with exactly 3 elements.
For 4-element subsets, since the remainders of the 4 elements when divided by 3 must have two that are the same, it is known that there are only
$$
(0,0,1,2),(1,1,1,0),(1,1,2,2),(2,2,2,0)
$$
these four possible combinations of remainders.
Thus, the number of peculiar subsets with 4 elements is
$$
\mathrm{C}_{3}^{2} \mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}+2 \mathrm{C}_{3}^{3} \mathrm{C}_{3}^{1}+\mathrm{C}_{3}^{2} \mathrm{C}_{3}^{2}=42 \text {. }
$$
Since the universal set is also a peculiar subset, the number of peculiar subsets is
$$
2(3+12+30+42)+1=175 .
$$
|
175
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
\begin{array}{c}
f(f(n))+f(n)=2 n+3 \text {, and } f(0)=1 . \\
\text { Then } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=
\end{array}
$$
|
3. 2772 .
Substituting $n=0$ into equation (1) yields
$$
f(1)+1=3 \Rightarrow f(1)=2 \text {. }
$$
Furthermore, let $n=1$.
From equation (1), we get $f(2)=3$.
Similarly, $f(n)=n+1(n=3,4, \cdots, 10)$.
$$
\text { Hence } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=2772 \text {. }
$$
|
2772
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the function
$$
f(x)=A \cos \left(\omega x+\frac{\pi}{4} \omega\right)(A>0)
$$
is decreasing on $\left(0, \frac{\pi}{8}\right)$. Then the maximum value of $\omega$ is
|
Ni,6.8.
Assume $\omega>0$.
To make $f(x)$ a decreasing function in $\left(0, \frac{\pi}{8}\right)$, combining the image of the cosine-type function, we must have
$$
\begin{array}{l}
\frac{T}{2} \geqslant \frac{\pi}{8} \Rightarrow \frac{\pi}{\omega} \geqslant \frac{\pi}{8} \Rightarrow \omega \leqslant 8 . \\
\text { When } \omega=8 \text {, } \\
f(x)=A \cos (8 x+2 \pi)(A>0),
\end{array}
$$
Obviously, in $\left(0, \frac{\pi}{8}\right)$, it is a decreasing function, which fits.
Therefore, the maximum value of $\omega$ is 8.
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$
then the minimum value of the function $g(f(x))$ is . $\qquad$
|
$-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$
Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$
It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x, y$ are two different real numbers, and
$$
x^{2}=2 x+1, y^{2}=2 y+1 \text {, }
$$
then $x^{6}+y^{6}=$ $\qquad$ .
|
4. 198 .
$$
\text { Let } S_{n}=x^{n}+y^{n} \text {. }
$$
From $x^{2}=2 x+1, y^{2}=2 y+1$, we get
$$
x^{n+2}=2 x^{n+1}+x^{n}, y^{n+2}=2 y^{n+1}+y^{n} \text {. }
$$
Thus, $S_{n+2}=2 S_{n+1}+S_{n}$.
Also, $S_{1}=2, S_{2}=6$, so
$$
S_{3}=14, S_{4}=34, S_{5}=82, S_{6}=198
$$
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are 20 distinct real numbers. If the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{10}\right|
\end{array}
$$
has a finite number of solutions, then this equation has at most $\qquad$ solutions.
|
7.9.
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right|- \\
& \left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{10}\right| .
\end{aligned}
$$
Thus, by the problem statement, $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{20}$ be the elements of the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{10}, b_{1}, b_{2}, \cdots, b_{10}\right\}
$$
arranged in increasing order, and in
$$
\left(-\infty, c_{1}\right],\left[c_{1}, c_{2}\right], \cdots,\left[c_{19}, c_{20}\right],\left[c_{20},+\infty\right)
$$
these 21 intervals, the function $f(x)$ is linear in each.
Notice that, in the interval $\left(-\infty, c_{1}\right]$,
$$
f(x)=a_{1}+a_{2}+\cdots+a_{10}-b_{1}-b_{2}-\cdots-b_{10}=m,
$$
and in the interval $\left[c_{20},+\infty\right)$, $f(x)=-m$.
Since the number of roots of the equation is finite, we have $m \neq 0$.
Moving along the number line from left to right. Initially, the coefficient of $x$ in $f(x)$ is 0. Each time we pass a $c_{i}\left(1 \leqslant i \leqslant 20, i \in \mathbf{Z}_{+}\right)$, the way one of the absolute values is removed changes, causing the coefficient of $x$ to change by $\pm 2$ (increase by 2 or decrease by 2). This indicates that the coefficient of $x$ is always even and does not change sign before it becomes 0. Therefore, the coefficient of $x$ in any two adjacent intervals is either both non-negative or both non-positive. Thus, $f(x)$ is either non-increasing or non-decreasing in such an interval union. As a result, if $f(x)=0$ has only a finite number of roots, then it has at most one root in each of the intervals $\left[c_{1}, c_{3}\right], \cdots,\left[c_{17}, c_{19}\right],\left[c_{19}, c_{20}\right]$. Furthermore, since $f\left(c_{1}\right)$ and $f\left(c_{20}\right)$ have different signs, and $f(x)$ changes sign at each root, the equation $f(x)=0$ has an odd number of roots. Therefore, it has at most nine roots.
On the other hand, it is not difficult to verify that if
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=5, a_{4}=8, a_{5}=9, \\
a_{6}=12, a_{7}=13, a_{8}=16, a_{9}=17, a_{10}=19.5, \\
b_{1}=2, b_{2}=3, b_{3}=6, b_{4}=7, b_{5}=10, \\
b_{6}=11, b_{7}=14, b_{8}=15, b_{9}=18, b_{10}=19,
\end{array}
$$
then the equation $f(x)=0$ has exactly nine roots.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If the remainder of $\underbrace{11 \cdots 1}_{n+1 \uparrow} 1$ divided by 3102 is 1, then the smallest positive integer $n$ is $\qquad$ .
|
8. 138 .
Notice that, $3102=2 \times 3 \times 11 \times 47$.
From $\underbrace{11 \cdots 1}_{n+1 \uparrow}=3102 k+1(k \in \mathbf{Z})$, we know $\underbrace{11 \cdots 10}_{n \uparrow}=3102 k$.
Thus, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by $2, 3, 11, 47$.
(1) For any positive integer $n$, obviously, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 2;
(2) $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 3 if and only if $31 n$;
(3) $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by 11 if and only if $2 \mid n$;
(4) Also, $\underbrace{11 \cdots 10}_{n \uparrow}=\frac{1}{9}\left(10^{n+1}-10\right)$,
$$
(9,47)=1,(10,47)=1 \text {, }
$$
then $47|\underbrace{11 \cdots 10}_{n \uparrow} \Leftrightarrow 47|\left(10^{n}-1\right)$.
By Fermat's Little Theorem, $10^{46} \equiv 1(\bmod 47)$.
Let $t$ be the smallest positive integer such that $10^{t} \equiv 1(\bmod 47)$. Then $t \mid 46$.
And $10 \equiv 10(\bmod 47)$,
$10^{2} \equiv 6(\bmod 47)$,
$10^{23} \equiv 46(\bmod 47)$,
thus $t=46$.
Therefore, $46|n \Leftrightarrow 47| \underbrace{11 \cdots 1}_{n \uparrow}$.
In summary, $n_{\min }=[2,3,46]=138$.
|
138
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
If $F_{a} γ F_{b} γ F_{c} γ F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$.
|
From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$
which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{b} γ F_{d-1} γ F_{d}$.
If $b \leqslant d-3$, then
$$
\left(F_{a}+F_{b}\right)+F_{d-1} \leqslant F_{d-2}+F_{d-1}=F_{d},
$$
which is a contradiction.
Therefore, $b=d-2$, at this point,
$$
F_{a}+\left(F_{d-2}+F_{d-1}\right)=F_{a}+F_{d}>F_{d} \text {. }
$$
Thus, the side lengths of the quadrilateral are $F_{a} γ F_{d-2} γ F_{d-1} γ F_{d}$.
Hence, $d-b=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Place 11 identical balls into six distinct boxes so that at most three boxes are empty. The number of ways to do this is $\qquad$.
|
10. 4212 .
If there are no empty boxes, there are $\mathrm{C}_{10}^{5}$ ways to place them; if there is one empty box, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}$ ways to place them; if there are two empty boxes, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}$ ways to place them; if there are three empty boxes, there are $\mathrm{C}_{6}^{3} \mathrm{C}_{10}^{2}$ ways to place them. Therefore, the total number of ways to place them with at most three empty boxes is $\mathrm{C}_{10}^{5}+\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}+\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}+\mathrm{C}_{6}^{3} \mathrm{C}_{10}^{2}=4212$
|
4212
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let the equation $x^{2}-m x-1=0$ have two real roots $\alpha, \beta (\alpha<\beta)$, and the function $f(x)=\frac{2 x-m}{x^{2}+1}$.
(1) Find the value of $\alpha f(\alpha)+\beta f(\beta)$;
(2) Determine the monotonicity of $f(x)$ in the interval $(\alpha, \beta)$, and provide a proof;
(3) If $\lambda, \mu$ are both positive real numbers, prove:
$$
\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|\alpha-\beta| .
$$
|
Three, 13. (1) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-m x-1=0$, we know
$$
\begin{array}{l}
\alpha+\beta=m, \alpha \beta=-1 . \\
\text { Then } f(\alpha)=\frac{2 \alpha-m}{\alpha^{2}+1}=\frac{2 \alpha-(\alpha+\beta)}{\alpha^{2}-\alpha \beta} \\
=\frac{\alpha-\beta}{\alpha(\alpha-\beta)}=\frac{1}{\alpha} \\
\Rightarrow \alpha f(\alpha)=1 .
\end{array}
$$
Similarly, $\beta f(\beta)=1$.
Thus, $\alpha f(\alpha)+\beta f(\beta)=2$.
(2) Note that,
$$
f^{\prime}(x)=-\frac{2\left(x^{2}-m x-1\right)}{\left(x^{2}+1\right)^{2}}=-\frac{2(x-\alpha)(x-\beta)}{\left(x^{2}+1\right)^{2}} \text {. }
$$
Therefore, when $x \in(\alpha, \beta)$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is monotonically increasing on $(\alpha, \beta)$.
(3) By $\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}-\alpha=\frac{\mu(\beta-\alpha)}{\lambda+\mu}>0$,
$$
\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}-\beta=\frac{\lambda(\alpha-\beta)}{\lambda+\mu}<0,
$$
we know $\alpha<\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}<\beta$.
Hence, by (2), $f(\alpha)<f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)<f(\beta)$.
Similarly, $f(\alpha)<f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)<f(\beta)$.
Thus, $\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|f(\alpha)-f(\beta)|$.
By (1), we know $f(\alpha)=\frac{1}{\alpha}, f(\beta)=\frac{1}{\beta}, \alpha \beta=-1$.
Then $|f(\alpha)-f(\beta)|=\left|\frac{1}{\alpha}-\frac{1}{\beta}\right|$
$$
=\left|\frac{\beta-\alpha}{\alpha \beta}\right|=|\alpha-\beta| \text {. }
$$
Thus, $\left|f\left(\frac{\lambda \alpha+\mu \beta}{\lambda+\mu}\right)-f\left(\frac{\mu \alpha+\lambda \beta}{\lambda+\mu}\right)\right|<|\alpha-\beta|$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4: There are 2013 balls placed around a circle, numbered $1, 2, \cdots, 2013$ in a clockwise direction. Starting from a certain ball $a$, and then taking every other ball in a clockwise direction, this process continues until only one ball remains on the circle. For what value of the number of ball $a$ will the last remaining ball be the 2013th ball?
|
First, consider a special case.
When the ball-picking procedure starts with the ball numbered 1, then when the person picking the balls reaches the next ball to be taken (numbered 3), it is essentially as if the ball numbered 2 has been placed at the farthest point in the direction of his movement. And when he takes the ball numbered 3 and moves to the next ball to be taken (numbered 5), it is again as if the ball numbered 4 has been placed at the farthest point in the direction of his movement. This process continues indefinitely.
Therefore, when the ball-picking procedure starts with the ball numbered 1, the problem is essentially the same as Example 3. In this case, since \(2013 = 2^{10} + 989\), the last remaining ball is numbered \(2 \times 989 = 1978\).
This is not the desired result, but it provides a clue.
If the ball-picking procedure is changed to start from the ball numbered \(x+1\), and the last remaining ball changes from 1978 to 2013, the goal is to find the number of the ball \(a\) by changing the initial position of the ball-picking.
It is easy to see that if the ball numbered \(x+k\) is renumbered as \(k^*\) in a clockwise direction, then the ball originally numbered \(k+1978\) becomes \(1978^*\). Therefore, when the ball-picking procedure starts from \(1^\circ\), the last remaining ball should be numbered \(1978^\circ\).
\[
\begin{array}{l}
\text{Let } 1978^* = 2013 \Rightarrow x + 1978 = 2013 \\
\Rightarrow x = 35 \Rightarrow x + 1 = 36.
\end{array}
\]
Therefore, when the ball \(a\) is numbered 36, the last remaining ball is exactly the ball numbered 2013.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n$!. If $M=1!\times 2!\times \cdots \times 10!$, then the number of perfect cubes among the positive divisors of $M$ is ( ).
(A) 468
(B) 684
(C) 846
(D) 648
|
2. A.
Notice that,
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 10! \\
=2^{9} \times 3^{8} \times 4^{7} \times 5^{6} \times 6^{5} \times 7^{4} \times 8^{3} \times 9^{2} \times 10 \\
=2^{9+2 \times 7+5+3 \times 3+1} \times 3^{8+5+2 \times 2} \times 5^{6+1} \times 7^{4} \\
=2^{38} \times 3^{17} \times 5^{7} \times 7^{4} .
\end{array}
$$
A perfect cube number \( n \) (where \( n \mid M \)) should have the form
$$
n=2^{3 x} \times 3^{3 y} \times 5^{3 z} \times 7^{3 u} (x, y, z, u \in \mathbf{N}),
$$
and \( 3 x \leqslant 38, 3 y \leqslant 17, 3 z \leqslant 7, 3 u \leqslant 4 \).
Therefore, such \( n \) has a total of \( 13 \times 6 \times 3 \times 2 = 468 \) numbers.
|
468
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of all positive integer solutions $(x, y, z)$ for the equation $x y+1000 z=2012$ is. $\qquad$
|
2. 18 .
First consider the positive integer solutions for $x \leqslant y$.
When $z=1$, $x y=1012=4 \times 253=2^{2} \times 11 \times 23$. Hence, $(x, y)=(1,1012),(2,506),(4,253)$,
$$
(11,92),(22,46),(23,44) \text {. }
$$
When $z=2$, $x y=12=2^{2} \times 3$. Hence, $(x, y)=(1,12),(2,6),(3,4)$.
Next, consider the positive integer solutions for $x>y$.
Similarly, there are 9 sets of positive integer solutions.
Therefore, there are a total of $2 \times 9=18$ sets of positive integer solutions.
|
18
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Xiao Li and Xiao Zhang are running at a constant speed on a circular track. They start from the same place at the same time. Xiao Li runs clockwise and completes a lap every 72 seconds, while Xiao Zhang runs counterclockwise and completes a lap every 80 seconds. At the start, Xiao Li has a relay baton, and each time they meet, the baton is passed from one to the other (the time taken for passing the baton is negligible). The relay baton returns to the starting point in the minimum of $\qquad$ seconds.
|
3. 720 .
The time taken from the start to the first meeting is
$$
\frac{1}{\frac{1}{72}+\frac{1}{80}}=\frac{720}{19}
$$
seconds, during which Xiao Li runs $\frac{10}{19}$ laps, and Xiao Zhang runs $\frac{9}{19}$ laps.
Divide the circular track into 19 equal parts, and number the points clockwise from the starting point as $0,1, \cdots, 18$. Then,
from $0 \sim \frac{720}{19}$ seconds, the baton moves from 0 to 10;
from $\frac{720}{19} \sim \frac{720}{19} \times 2$ seconds, the baton moves from 10 to 1;
from $\frac{720}{19} \times 2 \sim \frac{720}{19} \times 3$ seconds, the baton moves from 1 to 11;
generally, from $\frac{720}{19} \times 2 k \sim \frac{720}{19} \times(2 k+1)$ seconds, the baton moves from $k$ to $10+k$; from $\frac{720}{19} \times(2 k+1) \sim \frac{720}{19} \times(2 k+2)$ seconds, the baton moves from $10+k$ to $1+k$.
Therefore, when $\frac{720}{19} \times(2 \times 9+1)=720$ seconds, the baton returns to the starting point for the first time.
|
720
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: At each vertex of a regular 2009-gon, a non-negative integer not exceeding 100 is placed. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ such that it is always possible to make all the numbers at the vertices equal. ${ }^{[1]}$
(35th Russian Mathematical Olympiad)
|
Solve $k_{\min }=100400$.
Let the numbers at vertices $A_{1}, A_{2}, \cdots, A_{2009}$ be $a_{1}, a_{2}, \cdots, a_{2009}$ respectively.
First, take $a_{2}=a_{4}=a_{2008}=100, a_{1}=a_{3}=a_{2009}=0$.
For each operation, assign
$$
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) \text {. }
$$
If the adjacent vertex pair $\left(A_{1}, A_{2}\right)$ is operated once, then $S$ increases by 1;
If the adjacent vertex pair $\left(A_{1}, A_{2009}\right)$ is operated once, then $S$ decreases by 1;
If other adjacent vertex pairs $\left(A_{i}, A_{i+1}\right)(i=2,3 \cdots, 2008)$ are operated once, then $S$ remains unchanged.
Initially, $S=100400$; finally, $S=0$.
Thus, the adjacent vertex pair $\left(A_{1}, A_{2009}\right)$ must be operated at least 100400 times.
Therefore, $k \geqslant 100400$.
Next, we prove that for any initial state $\left(a_{1}, a_{2}, \cdots, a_{2009}\right)$ $\left(a_{j} \in \mathbf{N}, a_{j} \leqslant 100\right)$, all numbers can be made equal through at most 100400 operations.
In fact, for the adjacent vertex pair $\left(A_{i}, A_{i+1}\right)$, operate
$$
S_{i}=a_{i+2}+a_{i+4}+\cdots+a_{i+2008}
$$
times (with the convention $a_{j+2009}=a_{j}$), then $a_{i}$ will eventually become
$$
a_{i}+S_{i-1}+S_{i}=a_{1}+a_{2}+\cdots+a_{2009} \text {. }
$$
Thus, all numbers become equal, and the number of operations on the adjacent vertex pair $\left(A_{i}, A_{i+1}\right)$
$$
\begin{array}{l}
S_{i}=a_{i+2}+a_{i+4}+\cdots+a_{2008} \\
\leqslant 100 \times 1004=100400,
\end{array}
$$
which meets the problem's requirements.
|
100400
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.
---
Please note that the term "ζΈ©ι₯±ζ°" is translated as "sufficient number" to maintain the context and meaning in English. If you have a specific term you'd like to use, please let me know!
|
Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \cdots, b_{6}\right), A_{2}=\left(b_{7}, b_{8}, \cdots, b_{12}\right), \cdots, \\
A_{111}=\left(b_{661}, b_{662}, \cdots, b_{666}\right), \\
A_{112}=\left(b_{666}, b_{667}, \cdots, b_{671}\right) .
\end{array}
$$
Then the average value of the sums of these 112 groups is greater than
$$
\frac{1+2+\cdots+671}{112}=\frac{671 \times 672}{2 \times 112}=2013 \text {. }
$$
Therefore, there must be at least one group whose sum is greater than 2013. At this point, it is always possible to find a sequence of six consecutive numbers whose sum is greater than 2013.
Thus, 671 is a well-fed number.
Next, construct a permutation of $1,2, \cdots, 670$ such that the sum of any six consecutive numbers is less than or equal to 2013.
Construct a $112 \times 6$ number table: place $670,669, \cdots, 335$ sequentially in the left $112 \times 3$ number table (larger numbers on the left (top) side of smaller numbers), and place $1,2, \cdots, 334$ sequentially in the right $112 \times 3$ number table (smaller numbers on the left (top) side of larger numbers), resulting in the number table shown in Figure 8 (with two empty positions).
\begin{tabular}{cccccc}
670 & 669 & 668 & 1 & 2 & 3 \\
667 & 666 & 665 & 4 & 5 & 6 \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
340 & 339 & 338 & 331 & 332 & 333 \\
337 & 336 & 335 & 334 & &
\end{tabular}
Figure 8
Arrange the numbers in this table from left to right and from top to bottom in a single row, which will be the permutation that meets the condition (the sum of any six consecutive numbers is $2013-3k$ for $k=0,1,2,3$).
Thus, 670 is not a well-fed number.
Therefore, the smallest well-fed number is 671.
(Li Zhang, Tianjin Foreign Studies University Affiliated Foreign Languages School, 300230)
|
671
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. From $1,2, \cdots, 100$ choose three different numbers such that they cannot form the three sides of a triangle. The number of different ways to do this is.
|
2. 82075.
Let these three numbers be $i, j, k(1 \leqslant i<j<k \leqslant 100)$. Then $k \geqslant i+j$.
Therefore, the number of different ways to choose these three numbers is
$$
\begin{array}{l}
\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100}\left(\sum_{k=i+j}^{100} 1\right)\right]=\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100-i}(101-i-j)\right] \\
=\sum_{i=1}^{100}\left(\sum_{i=1}^{100-2 i} t\right)=\sum_{i=1}^{49} \frac{(100-2 i)(101-2 i)}{2} \\
=\sum_{k=1}^{49} k(2 k+1)(k=50-i) \\
=\frac{49 \times 50 \times(2 \times 49+1)}{3}+\frac{49 \times 50}{2} \\
=82075 .
\end{array}
$$
|
82075
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{1,2, \cdots, 6\}(i=0,1,2,3)$. If positive integers $m, n \in A$, and $m+n=2014(m>n)$, then the number of pairs of positive integers $(m, n)$ that satisfy the condition is $\qquad$.
|
3. 551.
Notice that, $2014=5 \times 7^{3}+6 \times 7^{2}+5$.
Given $m, n \in A$, let
$$
\begin{array}{l}
m=a \times 7^{3}+b \times 7^{2}+c \times 7+d, \\
n=a^{\prime} \times 7^{3}+b^{\prime} \times 7^{2}+d^{\prime},
\end{array}
$$
where $a, b, c, d, a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime} \in\{1,2, \cdots, 6\}$.
(1) When $d=1,2,3,4$, $d^{\prime}=5-d$. At this time,
$$
c=1,2, \cdots, 6, c^{\prime}=7-c.
$$
If $b=1,2,3,4$, then $b^{\prime}=5-b$. At this time,
$$
a=1,2,3,4, a^{\prime}=5-a.
$$
If $b=5$, then $b^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
If $b=6$, then $b^{\prime}=6$. At this time,
$$
a=1,2,3, a^{\prime}=4-a.
$$
Therefore, in this case, the number of $(m, n)$ that meet the conditions is
$$
4 \times 6(4 \times 4+3)=24 \times 19=456 \text{ (pairs). }
$$
(2) When $d=5$, $d^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
(3) When $d=6$, $d^{\prime}=12-d=6$. At this time,
$$
c=1,2, \cdots, 5, c^{\prime}=6-c.
$$
If $b=1,2,3,4$, then $b^{\prime}=5-b$. At this time,
$$
a=1,2,3,4, a^{\prime}=5-a.
$$
If $b=5$, then $b^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
If $b=6$, then $b^{\prime}=6$. At this time,
$$
a=1,2,3, a^{\prime}=4-a.
$$
Therefore, in this case, the number of $(m, n)$ that meet the conditions is
$$
5(4 \times 4+3)=95 \text{ (pairs). }
$$
In summary, there are $456+95=551$ pairs of $(m, n)$ that satisfy the conditions.
|
551
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { II. (40 points) Given } x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\} \\
(i=1,2, \cdots, 2013) \text {, let } \\
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2012} x_{2013} .
\end{array}
$$
Find the number of different integer values that $S$ can take.
|
Because $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, (\sqrt{2}-1)(\sqrt{2}+1)=1,
$$
Therefore, $x_{i} x_{i+1} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ instances of $3+2 \sqrt{2}$, $b$ instances of $3-2 \sqrt{2}$, and $c$ instances of 1. Then $a, b, c \in \mathbf{N}$, and $a+b+c=2012$.
$$
\begin{array}{l}
\text { Hence } S=\sum_{i=1}^{2012} x_{i} x_{i+1} \\
=(3+2 \sqrt{2}) a+(3-2 \sqrt{2}) b+c \\
=3 a+3 b+c+2 \sqrt{2}(a-b).
\end{array}
$$
If $S$ is an integer, then $a=b$. At this time,
$$
\begin{aligned}
S= & a \\
& +c=6 a+(2012-2 a) \\
& =4 a+2012=4(a+503)(0 \leqslant a \leqslant 1006).
\end{aligned}
$$
(1) When $a=b=1006$, $\left\{x_{i}\right\}$ contains at least 1007 instances of $\sqrt{2}-1$ and 1007 instances of $\sqrt{2}+1$, i.e., at least 2014 numbers, which is a contradiction.
When $a=b=1005$, $\left\{x_{i}\right\}$ contains at least 1006 instances of $\sqrt{2}-1$ and 1006 instances of $\sqrt{2}+1$.
(i) $\left\{x_{i}\right\}$ contains 1006 instances of $\sqrt{2}-1$ and 1007 instances of $\sqrt{2}+1$.
Since $S$ contains 1005 instances of $(\sqrt{2}-1)^{2}$, these 1006 instances of $\sqrt{2}-1$ are consecutive in $\left\{x_{i}\right\}$, i.e.,
$$
\begin{array}{l}
x_{i}=\sqrt{2}+1(i=1,2, \cdots, k), \\
x_{k+i}=\sqrt{2}-1(i=1,2, \cdots, 1006), \\
x_{k+1006+i}=\sqrt{2}+1(i=1,2, \cdots, 1007-k),
\end{array}
$$
where $k \in\{1,2, \cdots, 1006\}$.
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{2012} x_{i} x_{i+1} \\
=(k-1)(\sqrt{2}+1)^{2}+1005(\sqrt{2}-1)^{2}+ \\
(1006-k)(\sqrt{2}+1)^{2}+2 \\
= 6032.
\end{array}
$$
(ii) $\left\{x_{i}\right\}$ contains 1007 instances of $\sqrt{2}-1$ and 1006 instances of $\sqrt{2}+1$.
Similarly,
$$
\begin{array}{l}
x_{i}=\sqrt{2}-1(i=1,2, \cdots, k), \\
x_{k+i}=\sqrt{2}+1(i=1,2, \cdots, 1006), \\
x_{k+1006+i}=\sqrt{2}-1(i=1,2, \cdots, 1007-k),
\end{array}
$$
where $k \in\{1,2, \cdots, 1006\}$.
Combining (i) and (ii), there are 2012 sequences $\left(x_{1}, x_{2}, \cdots, x_{2013}\right)$ that make $S$ reach the maximum value of 6032.
(2) Prove by mathematical induction: When
$$
a=b=t(t=1005,1004, \cdots, 1)
$$
there exists $\left\{x_{i}\right\}$ such that $S=4(a+503)$.
When $a=b=1005$, it has been proven in (1).
Assume that when $a=b=t(1<t \leqslant 1005)$, there exists $\left\{x_{i}\right\}$ such that $S=4(a+503)$.
Consider a segment of consecutive $\sqrt{2}+1$ (or $\sqrt{2}-1$) as a segment. Take one from each segment of $\sqrt{2}+1$ and $\sqrt{2}-1$ with a length greater than 1, and place them at the end of the sequence (if the original end is $\sqrt{2}-1$, place the taken $\sqrt{2}-1$ at the very end; if the original end is $\sqrt{2}+1$, place the taken $\sqrt{2}+1$ at the very end).
Thus, $a$ and $b$ each decrease by 1, and $c$ increases by 2. At this time, $S=4[(t-1)+503]$.
Therefore, when $a=b=t-1$, the conclusion holds.
In summary, the number of different integer values that $S$ can take is 1005.
|
1005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the set $X=\{1,2, \cdots, 100\}$, and the function $f: X \rightarrow X$ satisfies the following conditions:
(1) For any $x \in X$, $f(x) \neq x$;
(2) For any 40-element subset $A$ of $X$, $A \cap f(A) \neq \varnothing$.
Find the smallest positive integer $k$ such that for any function $f$ satisfying the above conditions, there exists a $k$-element subset $B$ of $X$ such that
$$
B \cup f(B)=X.
$$
[Note] For any subset $T$ of $X$, define
$f(T)=\{f(t) \mid t \in T\}$.
(Zhai Zhenhua)
|
5. First consider the function $f: X \rightarrow X$ defined as follows:
For $i=1,2, \cdots, 30, j=91,92, \cdots, 99$, define
$f(3 i-2)=3 i-1, f(3 i-1)=3 i$,
$f(3 i)=3 i-2, f(j)=100, f(100)=99$.
Clearly, the function $f$ satisfies condition (1).
For any 40-element subset $A$ of the set $X$, either there exists an integer $i(1 \leqslant i \leqslant 30)$ such that
$|A \cap\{3 i-2,3 i-1,3 i\}| \geqslant 2$,
in which case $A \cap f(A) \neq \varnothing$, or $91,92, \cdots, 100 \in A$, in which case $A \cap f(A) \neq \varnothing$ as well, i.e., the function $f$ also satisfies condition (2).
If a subset $B$ of the set $X$ satisfies $f(B) \cup B=X$, then
$|B \cap\{3 i-2,3 i-1,3 i\}| \geqslant 2(1 \leqslant i \leqslant 30)$,
and $91,92, \cdots, 98 \in B$, and at least one of $99$ and $100$ belongs to the set $B$, thus $|B| \geqslant 69$.
We need to prove: For any function $f$ that satisfies the conditions of the problem, there always exists a subset $B$ with no more than 69 elements such that
$f(B) \cup B=X$.
Consider all subsets $U \subseteq X$ that satisfy $U \cap f(U)=\varnothing$ (by condition (1), $\{x\} \cap f(\{x\})=\varnothing$, so such subsets $U$ exist). Among these subsets, choose a subset $U$ such that $|U|$ is maximized, and if there are multiple subsets $U$ that meet the requirement, then choose the one with the largest $|f(U)|$.
Let $V=f(U), W=X \backslash(U \cup V), U, V, W$ are pairwise disjoint, and their union is $X$.
By condition (2), $|U| \leqslant 39$.
Thus, $|V| \leqslant 39, |W| \geqslant 22$.
(i) For any $w \in W$, it must be that $f(w) \in U$.
Otherwise, let $U^{\prime}=U \cup\{w\}$.
Since $f(U)=V, f(w) \notin U$, and $f(w) \neq w$, it follows that $U^{\prime} \cap f\left(U^{\prime}\right)=\varnothing$, which contradicts the maximality of $|U|$.
(ii) If $w_{1}, w_{2} \in W\left(w_{1} \neq w_{2}\right)$, then $f\left(w_{1}\right) \neq f\left(w_{2}\right)$.
Otherwise, let $u=f\left(w_{1}\right)=f\left(w_{2}\right)$.
By conclusion (i), $u \in U$.
Let $U^{\prime}=(U \backslash\{u\}) \cup\left\{w_{1}, w_{2}\right\}$.
Since $f\left(U^{\prime}\right) \subseteq V \cup\{u\}, U^{\prime} \cap(V \cup\{u\})=\varnothing$, it follows that $U^{\prime} \cap f\left(U^{\prime}\right)=\varnothing$, which contradicts the maximality of $|U|$.
Let $W=\left\{w_{1}, w_{2}, \cdots, w_{m}\right\}$, and set
$u_{i}=f\left(w_{i}\right)(1 \leqslant i \leqslant m)$.
By (i) and (ii), $u_{1}, u_{2}, \cdots, u_{m}$ are distinct elements of the subset $U$.
(iii) For $1 \leqslant i|f(U)|$, which contradicts the maximality of $|f(U)|$.
Therefore, $f\left(u_{1}\right), f\left(u_{2}\right), \cdots, f\left(u_{m}\right)$ are distinct elements of the subset $V$.
Thus, $|V| \geqslant|W|$.
Since $|U| \leqslant 39$, it follows that $|V|+|W| \geqslant 61$.
Thus, $|V| \geqslant 31$.
Let $B=U \cup W$. Then
$|B| \leqslant 69$, and $f(B) \cup B \supseteq V \cup B=X$.
In conclusion, the smallest positive integer $k$ is 69.
|
69
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For the convex pentagon $A B C D E$, the side lengths are sequentially $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. It is known that a quadratic trinomial in $x$ satisfies:
When $x=a_{1}$ and $x=a_{2}+a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is 5;
When $x=a_{1}+a_{2}$, the value of the quadratic trinomial is $p$;
When $x=a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is $q$.
Then $p-q=$ $\qquad$
|
4. 0 .
Let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
and let the axis of symmetry of its graph be $x=x_{0}$.
By the problem, we know
$$
f\left(a_{1}\right)=f\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=5,
$$
and $\square$
$$
\begin{array}{l}
a_{1} \neq a_{2}+a_{3}+a_{4}+a_{5} . \\
\text { Hence } a_{1}+\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=2 x_{0} \\
\Rightarrow\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}+a_{5}\right)=2 x_{0} \\
\Rightarrow f\left(a_{1}+a_{2}\right)=f\left(a_{3}+a_{4}+a_{5}\right) \\
\Rightarrow p=q \Rightarrow p-q=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A three-digit number that is divisible by 35 and whose digits sum to 15 is
|
5. 735.
Since the sum of the digits is 15, the three-digit number is a multiple of 3. Also, since the three-digit number is a multiple of 35, this three-digit number is a multiple of $35 \times 3=105$.
Starting from 105, list the three-digit multiples of 105:
$$
105,210,315,420,525,630,735,840, 945 \text {, }
$$
Among them, only 735 has a digit sum of 15.
|
735
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-13) \text {. }
$$
Find all possible values of \( d \).
|
11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2013$.
Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant a_{3}$. Then
$a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=\frac{2013}{3}=671$
$\Rightarrow a_{2} a_{3}=\frac{2013}{a_{1}} \leqslant \frac{2013}{671}=3$.
If $a_{1}=671$, then $a_{2} a_{3}=3$. Thus, $a_{2}+a_{3} \leqslant 4$.
Hence $a_{1}+a_{2}+a_{3} \leqslant 671+4=675671$, at this point, $a_{2} a_{3}<3$.
Therefore, $a_{2} a_{3}=1, a_{1}=2013$, which contradicts $a_{1}+a_{2}+a_{3}=2013$.
When $n=5$, there exist $-1, -1, 1, 1, 2013$ satisfying
$(-1)+(-1)+1+1+2013$
$=(-1) \times(-1) \times 1 \times 1 \times 2013=2013$.
Thus, the minimum value of $n$ is 5.
12. From the given, we have $\frac{a^{2}}{b^{2}}=\frac{c(d+13)}{c(d-13)}=\frac{d+13}{d-13}$.
Let $(a, b)=t, a_{1}=\frac{a}{t}, b_{1}=\frac{b}{t}$. Then
$\frac{a_{1}^{2}}{b_{1}^{2}}=\frac{a^{2}}{b^{2}}=\frac{d+13}{d-13}$.
Since $\left(a_{1}, b_{1}\right)=1$, we know $\left(a_{1}^{2}, b_{1}^{2}\right)=1$, meaning the left side of equation (1) is a reduced fraction. Assume without loss of generality that
$d+13=k a_{1}^{2}, d-13=k b_{1}^{2}\left(k \in \mathbf{Z}_{+}\right)$.
Thus, $k\left(a_{1}^{2}-b_{1}^{2}\right)=26$, i.e., $a_{1}^{2}-b_{1}^{2}=\frac{26}{k}$ is a positive divisor of 26 and can be decomposed into the product of two unequal positive integers $a_{1}+b_{1}$ and $a_{1}-b_{1}$ with the same parity, hence $\frac{26}{k}=13$.
Then $a_{1}+b_{1}=13, a_{1}-b_{1}=1, k=2$. Thus, $a_{1}=7$.
Furthermore, $d=k a_{1}^{2}-13=2 \times 7^{2}-13=85$.
Upon verification, when $(a, b, c, d)=(14,12,2,85)$, it satisfies the conditions. Therefore, the value of $d$ is 85.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Write all positive integers in ascending order in a row. Then the 2013th digit from left to right is
|
$-1.7$
All single-digit numbers occupy 9 positions, all two-digit numbers occupy $2 \times 90=180$ positions, and next come the three-digit numbers in sequence. Since $2013-9-180=1824$, and $\frac{1824}{3}=608$, because $608+99=707$, the 2013th digit is the last digit of the three-digit number 707, which is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a, b, c$ are distinct positive integers. If the set
$$
\{a+b, b+c, c+a\}=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\} \text {, }
$$
where $n \in \mathbf{Z}_{+}$. Then the minimum value of $a^{2}+b^{2}+c^{2}$ is
|
6. 1297 .
From $n^{2}+(n+1)^{2}+(n+2)^{2}=2(a+b+c)$ being even, we know that among $n, n+1, n+2$, there are two odd numbers and one even number, which means $n$ is odd.
Obviously, $n>1$.
Without loss of generality, let $a<b<c$.
If $n=3$, then
$$
\begin{array}{l}
a+b=9, a+c=16, b+c=25 \\
\Rightarrow a+b+c=25 .
\end{array}
$$
In this case, $a=0$, which is a contradiction.
Therefore, $n \geqslant 5$.
When $n=5$, we have
$$
\begin{array}{l}
a+b=25, a+c=36, b+c=49 \\
\Rightarrow a=6, b=19, c=30 .
\end{array}
$$
In this case, $a^{2}+b^{2}+c^{2}=1297$.
|
1297
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The minimum value of the function $f(x)=\sum_{k=1}^{2013}|x-k|$ is
|
7.1013042.
Notice that, the median of $1,2, \cdots, 2013$ is 1007.
Therefore, when $x=1007$, the function reaches its minimum value
$$
\begin{array}{l}
f(1007)=2(1+2+\cdots+1006) \\
=1006 \times 1007=1013042
\end{array}
$$
|
1013042
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If three non-zero and distinct real numbers $a, b, c$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$, then $a, b, c$ are called "harmonic"; if they satisfy $a+c=2b$, then $a, b, c$ are called "arithmetic".
Given the set $M=\{x|| x | \leqslant 2013, x \in \mathbf{Z}\}$, the set $P$ is a three-element subset of set $M$, i.e., $P=\{a, b, c\} \subset M$. If the elements $a, b, c$ in set $P$ are both harmonic and arithmetic, then the set $P$ is called a "good set".
The number of different good sets is $\qquad$
|
3.1006.
If $a, b, c$ are both harmonic and arithmetic, then
$$
\left\{\begin{array}{l}
\frac{1}{a} + \frac{1}{b} = \frac{2}{c}, \\
a + c = 2b
\end{array} \Rightarrow \left\{\begin{array}{l}
a = -2b \\
c = 4b
\end{array}\right.\right.
$$
Thus, the good set is of the form $\{-2b, b, 4b\}(b \neq 0)$.
Since the good set is a three-element subset of set $M$, we have
$$
\begin{array}{l}
-2013 \leqslant 4b \leqslant 2013(b \in \mathbf{Z}, \text{ and } b \neq 0) \\
\Rightarrow -503 \leqslant b \leqslant 503(b \in \mathbf{Z}, \text{ and } b \neq 0).
\end{array}
$$
Therefore, $b$ can take 1006 values.
Thus, the number of good sets is 1006.
|
1006
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y$ satisfy $x y+1=4 x+y$, and $x>1$. Then the minimum value of $(x+1)(y+2)$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4.27.
From the problem, we know $y=\frac{4 x-1}{x-1}$.
$$
\begin{array}{l}
\text { Therefore, }(x+1)(y+2)=(x+1)\left(\frac{4 x-1}{x-1}+2\right) \\
\quad= \frac{3(x+1)(2 x-1)}{x-1} .
\end{array}
$$
Let $x-1=t>0$. Then
$$
\begin{array}{l}
(x+1)(y+2)=\frac{3(t+2)(2 t+1)}{t} \\
=6\left(t+\frac{1}{t}\right)+15 \geqslant 27,
\end{array}
$$
when and only when $t=\frac{1}{t}$, i.e., $t=1, x=2, y=7$, the equality holds.
Thus, the minimum value of $(x+1)(y+2)$ is 27.
|
27
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$
The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<\frac{1}{2}$, then
$$
\begin{array}{l}
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=0 \\
\Rightarrow \pi x=k \pi(k \in \mathbf{Z}) \\
\Rightarrow x=k(k \in \mathbf{Z}) .
\end{array}
$$
Combining $x \in[0,2 \pi]$, we know $x=0,1, \cdots, 6$.
Upon inspection, $x=0,2,4,6$ meet the requirements of the problem.
$$
\begin{array}{l}
\text { (2) If } \frac{1}{2} \leqslant\left\{\frac{x}{2}\right\}<1 \\
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=1 \\
\Rightarrow \pi x=2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \\
\Rightarrow x=2 k+\frac{1}{2}(k \in \mathbf{Z}) .
\end{array}
$$
In this case, $\left\{\frac{x}{2}\right\}=0$, which does not meet the requirements.
Therefore, the values of $x$ that satisfy the conditions are $0,2,4,6$, and their sum is 12.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and for any $x \in$ $\mathbf{R}$, we have
$$
f\left(f(x)-3^{x}\right)=4 .
$$
Then $f(2)=$
|
8. 10 .
From the problem, we know that $f(x)-3^{x}$ is a constant. Let's assume $f(x)-3^{x}=m$.
Then $f(m)=4, f(x)=3^{x}+m$.
Therefore, $3^{m}+m=4 \Rightarrow 3^{m}+m-4=0$.
It is easy to see that the equation $3^{m}+m-4=0$ has a unique solution $m=1$.
Thus, $f(x)=3^{x}+1$,
and hence, $f(2)=3^{2}+1=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$).
|
9. 10 .
It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$
meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$
where $x_{1}<x_{2}<\cdots<x_{7}$ meets the requirements.
$$
\begin{array}{l}
\text { Then } x_{1}=1+1=2, x_{2} \leqslant 2+2=4, x_{3} \leqslant 8, \\
x_{4} \leqslant 16, x_{5} \leqslant 32, x_{6} \leqslant 64, x_{7} \leqslant 128 . \\
\text { By } x_{6}+x_{7} \leqslant 64+128=192<200 \text {, we know that } \\
200=x_{7}+x_{7} \Rightarrow x_{7}=100 ;
\end{array}
$$
Similarly, by $x_{5}+x_{6} \leqslant 32+64=96<100$, we know that
$$
x_{7}=100=x_{6}+x_{6} \Rightarrow x_{6}=50 \text {; }
$$
By $x_{4}+x_{5} \leqslant 16+32=48<50$, we know that
$$
x_{6}=50=x_{5}+x_{5} \Rightarrow x_{5}=25 \text {; }
$$
By $x_{3}+x_{4} \leqslant 8+16=24<25$, we know that
$$
x_{5}=25=x_{4}+x_{4} \Rightarrow x_{4}=\frac{25}{2} \text {, }
$$
which contradicts the fact that $x_{4}$ is an integer.
Thus, $|A|=9$ does not meet the requirements, i.e., $|A| \neq 9$.
Similarly, $|A| \leqslant 8$ does not meet the requirements.
In summary, the minimum value of $|A|$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a γ b γ c\right. \text { are positive integers not }
$$
exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$.
|
15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$
These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arithmetic sequence.
Let $x_{1}, x_{2}, \cdots, x_{7}$ be seven different elements in the set $P$ that form an arithmetic sequence, with a common difference of $d (d>0)$.
By the properties of the elements in the set $P$, we know that no element in the set $P$ is a multiple of 7.
Thus, by the pigeonhole principle, among the seven numbers $x_{1}, x_{2}, \cdots, x_{7}$, there exist two numbers that have the same remainder when divided by 7, and their difference is divisible by 7.
Without loss of generality, assume $x_{i}-x_{j} (i, j \in \{1,2, \cdots, 7\}, i<j)$ is divisible by 7. Then
$$
7 \mid (j-i) d \Rightarrow 7 \mid d \text{. }
$$
Let $d=7 m (m \in \mathbf{Z}_{+})$, and suppose
$$
x_{1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3},
$$
where $a_{1}, a_{2}, a_{3}$ are positive integers not exceeding 6.
Then $x_{i}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7(i-1) m$, where $i=2,3, \cdots, 7$.
$$
\begin{array}{l}
\text{By } x_{7} \leqslant 7^{3}+6 \times 7^{2}+6 \times 7+6, \\
x_{7} \geqslant 7^{3}+1 \times 7^{2}+1 \times 7+1+7 \times(7-1) m,
\end{array}
$$
we know $1 \leqslant m \leqslant 6$, i.e., the common difference $d$ can only be
$$
7 \times 1, 7 \times 2, \cdots, 7 \times 6 \text{. }
$$
Since $1 \leqslant m \leqslant 6$, and $(7, m)=1$, the remainders of $m, 2 m, \cdots, 6 m$ when divided by 7 are all different, and are one of 1, 2, $\cdots, 6$.
Therefore, there exists $k \in \{1,2, \cdots, 6\}$ such that $a_{2}+k m$ is divisible by 7.
Let $a_{2}+k m=7 t (t \in \mathbf{Z}_{+})$. Then
$$
\begin{array}{l}
x_{k+1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7 k m \\
=7^{3}+a_{1} \times 7^{2}+(a_{2}+k m) \times 7+a_{3} \\
=7^{3}+(a_{1}+t) \times 7^{2}+a_{3} .
\end{array}
$$
Thus, the coefficient of 7 (i.e., the second digit from the left) in the base-7 representation of $x_{k+1}$ is 0, which contradicts $x_{k+1} \in P$.
Therefore, any seven different numbers in the set $P$ cannot form an arithmetic sequence.
Hence, the maximum value of $n$ is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given a convex quadrilateral $ABCD$ whose four interior angles form an arithmetic sequence. In $\triangle ABD$ and $\triangle BCD$, if $\angle ABD = \angle CDB$, $\angle DAB = \angle CBD$, and the three interior angles of these two triangles also form an arithmetic sequence, then the maximum possible value of the sum of the two largest interior angles of quadrilateral $ABCD$ is ( ).
(A) 210
(B) 220
(C) 230
(D) 240
(E) 250
|
13. D.
Assume the four interior angles of the quadrilateral are
$$
\alpha, \alpha+d, \alpha+2d, \alpha+3d \text{.}
$$
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have
$$
\alpha+\alpha+3d=\alpha+d+\alpha+2d=180^{\circ} \text{.}
$$
Given that $\angle ABD = \angle CDB$ and $\angle DAB = \angle CBD$, we know that $\triangle ABD \sim \triangle BDC$.
Let $\angle ABD = \angle CDB = \alpha_1$,
$\angle DAB = \angle CBD = \alpha_2$,
$\angle ADB = \angle BCD = \alpha_3$.
Then the four interior angles of the quadrilateral are $\alpha_2, \alpha_1 + \alpha_2, \alpha_3, \alpha_1 + \alpha_3$.
Since the three interior angles of a triangle form an arithmetic sequence, one of $\alpha_1, \alpha_2, \alpha_3$ must be $60^{\circ}$.
We will consider three cases.
(1) $\alpha_1 = 60^{\circ}$.
Then the four interior angles of the quadrilateral are $\alpha_2, 60^{\circ} + \alpha_2, \alpha_3, 60^{\circ} + \alpha_3$. Since the sum of the interior angles of the quadrilateral is $360^{\circ}$, we have $\alpha_2 + \alpha_3 = 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
\alpha_1 + \alpha_2 + \alpha_1 + \alpha_3 = 120^{\circ} + \alpha_2 + \alpha_3 = 240^{\circ} \text{.}
$$
(2) $\alpha_2 = 60^{\circ}$.
Then $\alpha_1 + \alpha_3 = 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
\alpha_1 + \alpha_3 + \alpha_3 < 2(\alpha_1 + \alpha_3) = 240^{\circ} \text{.}
$$
(3) $\alpha_3 = 60^{\circ}$.
Then $\alpha_1 + \alpha_2 = 120^{\circ}$. Therefore, $\alpha_2 < 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
120^{\circ} + \alpha_2 < 240^{\circ} \text{.}
$$
In summary, the maximum possible value of the sum of the two largest interior angles of quadrilateral $ABCD$ is $240^{\circ}$.
|
240
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
19. In $\triangle A B C$, it is known that $A B=13, B C=14, C A=$ 15, points $D, E, F$ are on sides $B C, C A, D E$ respectively, and $A D \perp$ $B C, D E \perp A C, A F \perp B F$. If the length of segment $D F$ is $\frac{m}{n}$ $\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30
|
19. B.
As shown in Figure 3, with $D$ as the origin, the line $BC$ as the $x$-axis, and the line $AD$ as the $y$-axis, we establish a Cartesian coordinate system $x D y$.
It is easy to know that $p_{\triangle ABC}=13+14+15=42$,
$S_{\triangle ABC}=\sqrt{21(21-13) \times(21-14) \times(21-15)}=84$.
Thus, $AD=\frac{2 S_{\triangle ABC}}{BC}=12$.
Therefore, the coordinates of point $A(0,12), B(-5,0), C(9,0)$.
Since $DE \perp AC$, the equation of line $DE$ is
$$
y=\frac{3}{4} x \text {. }
$$
Because $AF \perp BF$, point $F$ lies on the circumcircle with $AB$ as the diameter.
Note that the equation of the circle with $AB$ as the diameter is
$$
\left(x+\frac{5}{2}\right)^{2}+(y-6)^{2}=\left(\frac{13}{2}\right)^{2} \text {. }
$$
Substituting equation (1) into equation (2) yields
$$
\begin{array}{l}
x^{2}+5 x+\frac{25}{4}+\frac{9}{16} x^{2}-9 x+36=\frac{169}{4} \\
\Rightarrow \frac{25}{16} x^{2}-4 x=0 \Rightarrow x_{1}=0, x_{2}=\frac{64}{25} .
\end{array}
$$
Therefore, point $F\left(\frac{64}{25}, \frac{48}{25}\right)$.
Thus, $DF=\sqrt{\left(\frac{64}{25}\right)^{2}+\left(\frac{48}{25}\right)^{2}}=\frac{16}{5}$.
Hence, $m+n=21$.
|
21
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $A=\{2,4, \cdots, 2014\}, B$ be any non-empty subset of $A$, and $a_{i} γ a_{j}$ be any two elements in set $B$. There is exactly one isosceles triangle with $a_{i} γ a_{j}$ as side lengths. Then the maximum number of elements in set $B$ is $\qquad$
|
7. 10 .
By symmetry, without loss of generality, assume $a_{i}<a_{j}$. Then there must exist an isosceles triangle with $a_{j}$ as the waist and $a_{i}$ as the base, and there is only one isosceles triangle with $a_{i}$ and $a_{j}$ as side lengths.
Thus, $a_{i}+a_{i} \leqslant a_{j} \Rightarrow a_{j} \geqslant 2 a_{i}$.
This indicates that, when the elements of set $B$ are arranged in ascending order, each subsequent element is at least twice the previous one.
Therefore, when the number of elements is maximized, the set
$$
B=\{2,4,8,16,32,64,128,256,512,1024\} \text {. }
$$
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad)
|
Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$
Since $D A$ and $D B$ are tangent to $\odot O$, we have
$$
\angle O A D=\angle O B D=90^{\circ} \text {. }
$$
Thus, $O, A, B, D$ are concyclic.
Since $F, A, B, D$ are concyclic, then $F, O, A, B, D$ are concyclic. Therefore,
$$
\angle O A B=\angle O F B=90^{\circ} \text {. }
$$
This contradicts $\angle O A D=90^{\circ}$.
Therefore, point $F$ coincides with $O$.
Next, consider the ratio of some segments
$$
\begin{aligned}
\frac{A M}{M B} & =\frac{S_{\triangle A C D}}{S_{\triangle B C D}}=\frac{\frac{1}{2} A C \cdot A D \sin \angle C A D}{\frac{1}{2} B C \cdot B D \sin \angle C B D} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cdot \frac{\sin \left(90^{\circ}+\angle C A F\right)}{1} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cos \angle C A F .
\end{aligned}
$$
Since $\angle C A F=\angle A C F=\angle B C A$, by the tangent length theorem, we know $A D=B D$.
Thus, $\frac{A M}{B M}=\frac{A C}{B C} \cos \angle B C A=\left(\frac{A C}{B C}\right)^{2}$.
Applying Menelaus' theorem to $\triangle B A E$ and the transversal $G M C$, we get
$$
\frac{B G}{G E} \cdot \frac{E C}{C A} \cdot \frac{A M}{M B}=1 \text {. }
$$
Therefore, $\frac{B G}{G E}=\frac{C A}{C E} \cdot \frac{M B}{M A}=\left(\frac{B C}{A C}\right)^{2} \frac{A C}{C E}=\frac{B C^{2}}{A C \cdot C E}$.
By the secant theorem and $F$ being the midpoint of $B C$, we have
$$
A C \cdot C E=C F \cdot C B=\frac{1}{2} B C^{2} \text {. }
$$
Therefore, $\frac{B G}{G E}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $a c+c^{2}=b^{2}-a^{2}$. If the longest side of $\triangle A B C$ is $\sqrt{7}$, and $\sin C=2 \sin A$, then the length of the shortest side of $\triangle A B C$ is $\qquad$.
|
6. 1 .
From $a c+c^{2}=b^{2}-a^{2}$
$\Rightarrow \cos B=-\frac{1}{2} \Rightarrow \angle B=\frac{2 \pi}{3}$.
Thus, the longest side is $b$.
Also, $\sin C=2 \sin A \Rightarrow c=2 a$.
Therefore, $a$ is the shortest side.
By the cosine rule,
$$
(\sqrt{7})^{2}=a^{2}+4 a^{2}-2 a \times 2 a \times\left(-\frac{1}{2}\right) \text {. }
$$
Solving, we get $a=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=$ $\qquad$ .
|
10.2013.
Obviously, when $k \geqslant 11$, $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=0$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{10}\left[\frac{2013+2^{k}}{2^{k+1}}\right] \\
= 1007+503+252+126+63+ \\
31+16+8+4+2+1 \\
= 2013 .
\end{array}
$$
|
2013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has one focus at $F_{1}(-\sqrt{3}, 0)$, and passes through the point $H\left(\sqrt{3}, \frac{1}{2}\right)$. Let the upper and lower vertices of the ellipse $E$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be any point on the ellipse different from $A_{1}$ and $A_{2}$. The lines $P A_{1}$ and $P A_{2}$ intersect the $x$-axis at points $M$ and $N$, respectively. If the line $O T$ is tangent to the circle $\odot G$ passing through points $M$ and $N$, with the point of tangency being $T$, prove that the length of segment $O T$ is a constant, and find this constant value.
|
12. From the problem, we have
$$
a^{2}-b^{2}=3, \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1 \text {. }
$$
Solving, we get $a^{2}=4, b^{2}=1$.
Thus, the equation of the ellipse $E$ is $\frac{x^{2}}{4}+y^{2}=1$.
From this, we know the points $A_{1}(0,1), A_{2}(0,-1)$. Let point $P\left(x_{0}, y_{0}\right)$.
Then $l_{P A_{1}}: y-1=\frac{y_{0}-1}{x_{0}} x$,
$$
l_{P A_{1}}: y+1=\frac{y_{0}+1}{x_{0}} x \text {. }
$$
Let $y=0$, we get
$$
x_{N}=\frac{-x_{0}}{y_{0}-1}, x_{M}=\frac{x_{0}}{y_{0}+1} \text {. }
$$
Thus, $|O M||O N|=\left|\frac{x_{0}}{y_{0}+1} \cdot \frac{-x_{0}}{y_{0}-1}\right|=\left|\frac{x_{0}^{2}}{y_{0}^{2}-1}\right|$.
Also, $\frac{x_{0}^{2}}{4}+y_{0}^{2}=1$, so, $x_{0}^{2}=4\left(1-y_{0}^{2}\right)$.
Therefore, $|O M||O N|=\left|\frac{x_{0}^{2}}{y_{0}^{2}-1}\right|=4$.
By the secant-tangent theorem, we have
$$
O T^{2}=|O M||O N|=4 \Rightarrow|O T|=2 .
$$
|
2
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant 6 \text {. }
$$
|
15. Notice,
$$
\begin{array}{l}
\frac{\sqrt{a^{2}+3 b c}}{a}=\frac{\sqrt{a^{2}+b c+b c+b c}}{a} \\
\geqslant \frac{\sqrt{4 \sqrt[4]{a^{2} b^{3} c^{3}}}}{a}=\frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a} .
\end{array}
$$
Similarly, $\frac{\sqrt{b^{2}+3 a c}}{b} \geqslant \frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}$,
$$
\begin{array}{l}
\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant \frac{2 \sqrt[8]{a^{3} b^{3} c^{2}}}{c} . \\
\text { Therefore, } \frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \\
\geqslant \frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a}+\frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}+\frac{2 \sqrt[8]{a^{3} b^{3} c^{2}}}{c} \\
\geqslant 2 \times 3 \sqrt[3]{\frac{\sqrt[8]{a^{8} b^{8} c^{8}}}{a b c}}=6 .
\end{array}
$$
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. If every prime factor of 2013 is a term in a certain arithmetic sequence $\left\{a_{n}\right\}$ of positive integers, then the maximum value of $a_{2013}$ is $\qquad$
|
$-, 1.4027$.
Notice that, $2013=3 \times 11 \times 61$.
If $3, 11, 61$ are all terms in a certain arithmetic sequence of positive integers, then the common difference $d$ should be a common divisor of $11-3=8$ and $61-3=58$. To maximize $a_{2013}$, the first term $a_{1}$ and the common difference $d$ should both be as large as possible. Thus, $a_{1}=3, d=2$.
Therefore, the maximum value of $a_{2013}$ is $3+2012 d=4027$.
|
4027
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, then the minimum value of $a+2b+3c$ is . $\qquad$
|
2. 36 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$
|
36
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the distances from the center of the ellipse to the focus, the endpoint of the major axis, the endpoint of the minor axis, and the directrix are all positive integers, then the minimum value of the sum of these four distances is $\qquad$ .
|
5.61.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distances from the center $O$ of the ellipse to the endpoints of the major axis, the endpoints of the minor axis, the foci, and the directrices are $a$, $b$, $c$, $d$ respectively, and satisfy
$$
c^{2}=a^{2}-b^{2}, d=\frac{a^{2}}{c} \text {. }
$$
Since $a$, $b$, $c$ form a Pythagorean triple, the Pythagorean triples satisfying $a \leqslant 20$ are
$$
\begin{aligned}
\{a, b, c\}= & \{3,4,5\},\{6,8,10\},\{9,12,15\}, \\
& \{12,16,20\},\{5,12,13\},\{8,15,17\},
\end{aligned}
$$
Among them, only $\frac{15^{2}}{9}=25$ and $\frac{20^{2}}{16}=25$.
Upon verification, when $(a, b, c, d)=(15,12,9,25)$, the value of $a+b+c+d$ is the smallest.
At this time, $a+b+c+d=61$.
|
61
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a composite number $k(1<k<100)$. If the sum of the digits of $k$ is a prime number, then the composite number $k$ is called a "pseudo-prime". The number of such pseudo-primes is . $\qquad$
|
7.23.
Let $S(k)$ denote the sum of the digits of $k$, and $M(p)$ denote the set of composite numbers with a pseudo-prime $p$.
When $k \leqslant 99$, $S(k) \leqslant 18$, so there are 7 prime numbers not exceeding 18, which are $2, 3, 5, 7, 11, 13, 17$.
The composite numbers with a pseudo-prime of 2 are $M(2)=\{20\}$.
Similarly, $M(3)=\{12,21,30\}$,
$$
\begin{array}{l}
M(5)=\{14,32,50\}, \\
M(7)=\{16,25,34,52,70\}, \\
M(11)=\{38,56,65,74,92\}, \\
M(13)=\{49,58,76,85,94\}, \\
M(17)=\{98\} .
\end{array}
$$
Thus, there are 23 pseudo-primes in total.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Make a full permutation of the elements in the set $\{1,2, \cdots, 8\}$, such that except for the number at the far left, for each number $n$ on the right, there is always a number to the left of $n$ whose absolute difference with $n$ is 1. The number of permutations that satisfy this condition is $\qquad$
|
8. 128 .
Suppose for a certain permutation that satisfies the conditions, the first element on the left is $k(1 \leqslant k \leqslant 8)$. Then, among the remaining seven numbers, the $8-k$ numbers greater than $k$, $k+1, k+2, \cdots, 8$, must be arranged in ascending order; and the $k-1$ numbers less than $k$, $1,2, \cdots, k-1$, must be arranged in descending order (not necessarily adjacent).
In fact, for any number $k+n$ greater than $k$, assume $k+n<8$.
If $k+n+1$ is placed to the left of $k+n$, then the other number $k+n+2$ that differs by 1 from $k+n+1$ must be placed to the left of $k+n+1$; similarly, the other number $k+n+3$ that differs by 1 from $k+n+2$ must be placed to the left of $k+n+2$; $\cdots$. Then the second number in the permutation cannot differ by 1 from $k$, which is a contradiction.
Therefore, $k+n+1$ must be placed to the right of $k+n$.
Similarly, the $k-1$ numbers less than $k$, $1,2, \cdots, k-1$, must be arranged in descending order.
Since when the first element on the left $k$ is determined, there are still seven positions on the right, and any $8-k$ of these positions can be chosen to fill in the numbers greater than $k$ (the remaining $k-1$ positions will fill in the numbers less than $k$), the number of ways to choose is $\mathrm{C}_{7}^{8-k}$; and once the positions are chosen, the method of filling in the numbers is uniquely determined.
Therefore, the total number of permutations is
$$
\sum_{k=1}^{8} \mathrm{C}_{7}^{8-k}=\sum_{k=0}^{7} \mathrm{C}_{7}^{j}=2^{7} .
$$
|
128
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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
|
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 $(a-5)(a-3) \geqslant 0$.
Thus, $a \geqslant 5$.
When $a=b=5, c=2$, the equality holds.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left(10^{10000}-9^{200}\right)$.
Since $\left(10^{100}+9\right) \mid \left[\left(10^{100}\right)^{2}-9^{2}\right]$, therefore, $\frac{10^{10000}-3^{200}}{10^{100}+9}$ is an integer.
By $\frac{3^{200}}{10^{100}+9}=\frac{9^{100}}{10^{100}+9}<1$, we know
$$
\begin{array}{l}
{\left[\frac{10^{10000}}{10^{100}+9}\right]=\frac{10^{10000}-3^{200}}{10^{100}+9}} \\
=\frac{10^{10000}-9^{100}}{10^{100}+9}=\frac{10^{10000}-81^{50}}{10^{100}+9},
\end{array}
$$
where the unit digit of the denominator is 9, and the unit digit of the numerator is 9, so the unit digit of the quotient is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $\overrightarrow{P E} \cdot \overrightarrow{P F}$.
|
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, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\overrightarrow{P F}| \cos \angle E P F$
$=d^{2} \cos \angle E P F$.
Let $\angle O P E=\theta$. Then,
$\tan \theta=\frac{|O E|}{|P E|}=\frac{2}{d}$.
Thus, $\cos \angle E P F=\cos 2 \theta$
$=\cos ^{2} \theta-\sin ^{2} \theta$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{d^{2}-4}{d^{2}+4}$.
Therefore, $\overrightarrow{P E} \cdot \overrightarrow{P F}=d^{2} \cdot \frac{d^{2}-4}{d^{2}+4}$
$=\left(d^{2}+4\right)+\frac{32}{d^{2}+4}-12$.
Let $d^{2}+4=x$. Then $x \in[16,36]$.
Using the function $f(x)=x+\frac{32}{x}$ which is monotonically increasing in the interval $[16,36]$, we know that when $d^{2}+4=16$,
$$
(\overrightarrow{P E} \cdot \overrightarrow{P F})_{\min }=6
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $\lambda=\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Then
$$
\lambda^{6}=1, \lambda^{10}=-\lambda \text{. }
$$
Therefore, $f(\lambda)=\lambda^{13}\left(\lambda^{10}+\lambda^{2}+\lambda\right)^{200}=\lambda^{13}\left(\lambda^{2}\right)^{200}$
$$
=\lambda^{413}=\frac{1}{2}-\frac{\sqrt{3}}{2}i.
$$
Thus, $\operatorname{Re} f(\lambda)=\frac{1}{2}$.
From $f(\lambda)=\sum_{i=1}^{2014} a_{i} \lambda^{2014-i}$, we know
$\operatorname{Re} f(\lambda)$
$$
=-a_{1}-\frac{1}{2} a_{2}+\frac{1}{2} a_{3}+a_{4}+\frac{1}{2} a_{5}-\frac{1}{2} a_{6} \cdots+\frac{1}{2} a_{2013}+a_{2014}.
$$
Since $a_{2014}=f(0)=0$, we have
$\operatorname{Re} f(\lambda)=-\frac{1}{2} \sum_{i=1}^{2013} a_{i} b_{i}=\frac{1}{2}$.
Thus, $\sum_{i=1}^{2013} a_{i} b_{i}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots \\
\leqslant c+\frac{9}{10} c+\left(\frac{9}{10}\right)^{2} c+\left(\frac{9}{10}\right)^{3} c+\cdots,
\end{array}
$$
where, $c=1+\frac{1}{2}+\cdots+\frac{1}{8}$.
$$
\begin{array}{l}
\text { Hence } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<10 c \\
=10\left(1+\frac{1}{2}+\cdots+\frac{1}{8}\right)<30 .
\end{array}
$$
|
30
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
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 induction.
(1) When $n=1,2,3$, it is clearly true.
(2) Assume that when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, the conclusion holds.
Then let $a=3 k+1, b=1$, we get
$f(3 k+1)=3 f(k+1)-2 f(1)=6 k+1$.
Similarly, let
$$
a=3 k+2, b=2 ; a=3 k+3, b=3 \text {. }
$$
Then $f(3 k+2)=6 k+3 ; f(3 k+3)=6 k+5$.
Therefore, when $n \in \mathbf{Z}_{+}$, we have $f(n)=2 n-1$.
Thus, $f(2014)=2 \times 2014-1=4027$.
Solution 2 In the given condition, let
$a=3 x, b=0 ; a=0, b=\frac{3}{2} y$.
Then $f(x)=\frac{f(3 x)+2 f(0)}{3}$;
$f(y)=\frac{f(0)+2 f\left(\frac{3}{2} y\right)}{3}$.
Hence $f(3 x)=3 f(x)-2 f(0)$;
$f\left(\frac{3}{2} y\right)=\frac{3 f(y)-f(0)}{2}$.
Again, in the given condition, let $a=3 x, b=\frac{3 y}{2}$, we get
$f(x+y)=\frac{f(3 x)+2 f\left(\frac{3 y}{2}\right)}{3}$
$=f(x)+f(y)-f(0)$,
which means $f(x+y)-f(0)=f(x)-f(0)+f(y)-f(0)$.
Let $g(x)=f(x)-f(0)$. Then
$g(x+y)=g(x)+g(y)$,
$g(1)=f(1)-f(0)=1-f(0)$.
It is easy to prove by mathematical induction that when $n \in \mathbf{Z}_{+}$, we have
$g(n)=n g(1)=n(1-f(0))$.
Thus, $g(4)=4(1-f(0))$.
Also, $g(4)=f(4)-f(0)=7-f(0)$, so
$4(1-f(0))=7-f(0)$.
Therefore, $f(0)=-1$.
Thus, $g(n)=n(1-f(0))=2 n$.
So, $f(n)=2 n-1\left(n \in \mathbf{Z}_{+}\right)$.
Hence $f(2014)=2 \times 2014-1=4027$.
From the above derivation, we know that the condition
$$
f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}
$$
is equivalent to the Cauchy equation
$$
g(x+y)=g(x)+g(y),
$$
where $g(x)=f(x)-f(0)$.
Next, we provide a method for solving the Cauchy equation in the field of rational numbers.
Lemma Let $f(x)$ be a function defined on the set of rational numbers $\mathbf{Q}$, and for any $x, y \in \mathbf{Q}$, we have
$f(x+y)=f(x)+f(y)$.
Then $f(x)=f(1) x(x \in \mathbf{Q})$.
Proof Let $x=y=0$. Then $f(0)=0$.
By the given condition, for any positive integer $n$ we have
$f(n+1)=f(n)+f(1)\left(n \in \mathbf{Z}_{+}\right)$.
Thus, $f(n)=f(1) n\left(n \in \mathbf{Z}_{+}\right)$
Also, let $x=n, y=-n\left(n \in \mathbf{Z}_{+}\right)$. Then
$f(0)=f(n)+f(-n)$.
Thus, $f(-n)=-f(1) n\left(n \in \mathbf{Z}_{+}\right)$.
Therefore, for any $n \in \mathbf{Z}$, we have $f(n)=f(1) n$.
For any rational number $\frac{q}{p}\left(q \in \mathbf{Z}, p \in \mathbf{Z}_{+}\right)$, we have
$f(1) q=f(q)$
$=\underbrace{f\left(\frac{q}{p}\right)+f\left(\frac{q}{p}\right)+\cdots+f\left(\frac{q}{p}\right)}_{p t}=p f\left(\frac{q}{p}\right)$.
Thus, $f\left(\frac{q}{p}\right)=f(1) \frac{q}{p}$.
In summary, for any $x \in \mathbf{Q}$, we have $f(x)=f(1) x$.
|
4027
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
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
|
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
|
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 the first term of 700 yuan and a common difference of -140 yuan. The expected prize money for participating in this competition is $\qquad$ yuan.
|
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, $E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7}$ $=500$ (yuan).
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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
|
14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$
Find the integer values of \( x \) for which \( A \) is a perfect square.
|
14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$
So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+x-1\right)^{2}$.
Let $A=y^{2}(y \in \mathbf{Z})$. Then
$$
\begin{array}{l}
|y|>\left|x^{2}+x-1\right| \\
\Rightarrow|y|-1 \geqslant\left|x^{2}+x-1\right| \\
\Rightarrow y^{2}-2|y|+1 \geqslant\left(x^{2}+x-1\right)^{2} \\
\Rightarrow-3(x-11)-2\left|x^{2}+x-1\right|+1 \geqslant 0 .
\end{array}
$$
Solving this inequality yields the integer values of $x$ as $\pm 3, \pm 2, \pm 1, 0, -4, -5$, but the corresponding $A$ for these values are not perfect squares.
In summary, the integer value of $x$ that makes $A$ a perfect square is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Seven balls of different colors are placed into three boxes numbered 1, 2, and 3. It is known that the number of balls in each box is not less than its number. The number of different ways to place the balls is $\qquad$
|
5.455.
(1) If the number of balls placed in boxes 1, 2, and 3 are 2, 2, and 3 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}=210$;
(2) If the number of balls placed in boxes 1, 2, and 3 are 1, 3, and 3 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{1} \mathrm{C}_{6}^{3}=140$;
(3) If the number of balls placed in boxes 1, 2, and 3 are 1, 2, and 4 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{1} \mathrm{C}_{6}^{2}=105$.
Thus, the total number of ways is $210+140+105=455$.
|
455
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
381 A chess piece starts from one corner of an $8 \times 8$ chessboard, and moving one square in the row direction or column direction is called a step. After several steps, it passes through every square without repetition and returns to the starting point.
(1) Prove: the number of steps in the row direction is different from the number of steps in the column direction;
(2) Question: what is the minimum difference between the number of steps in the row direction and the number of steps in the column direction?
|
(1) Construct an $8 \times 8$ grid, and connect every adjacent point with a line segment to form a graph. Assume the distance between adjacent points is 1, and call a square with an area of 1 and vertices on the grid points a "cell."
The original problem is equivalent to: Prove that in every Hamiltonian cycle of the graph that meets the requirements, the number of steps in the row direction is not equal to the number of steps in the column direction.
We can determine every Hamiltonian cycle that meets the conditions through the following operations.
First, draw a cycle as shown in Figure 3. Then, starting from the first operation, each time, select a cell inside the previous cycle such that exactly two of its four vertices are on the edge of the previous cycle. Remove the edge of the cell on the cycle and keep the other three edges, thus obtaining the next cycle. Continue this process until a Hamiltonian cycle that meets the conditions is formed.
Conversely, every Hamiltonian cycle that meets the conditions can be determined through the above operations.
Starting from the first operation, each operation will increase the perimeter of the cycle by 2. Since the perimeter of the cycle in Figure 3 is 28, and the perimeter of a Hamiltonian cycle that meets the conditions is 64, a total of 18 operations are required.
It is also easy to see that starting from the first operation, each operation will increase the number of pairs of adjacent points in the $6 \times 6$ grid in the middle of the $8 \times 8$ grid by 1 (no two pairs of points share a point). During each operation:
(i) If the two points are connected by a horizontal line, the operation increases the number of steps in the column direction by 2, while the number of steps in the row direction remains unchanged;
(ii) If the two points are connected by a vertical line, the operation increases the number of steps in the row direction by 2, while the number of steps in the column direction remains unchanged.
Thus, it is sufficient to prove the following proposition:
For a $6 \times 6$ grid, pair the 36 points such that each pair consists of two points that are adjacent in the row direction or the column direction. Then the number of pairs in the row direction and the number of pairs in the column direction (no two pairs share a point) must be different.
In fact, assume that there are 9 pairs in both the row direction and the column direction. Color the $6 \times 6$ grid as shown in Figure 4. It is easy to see that each pair in the column direction includes one black point and one white point, and each pair in the row direction includes two points of the same color. Thus, the 9 pairs in the column direction include 9 black points and 9 white points.
Therefore, the 9 pairs in the row direction also include 9 black points and 9 white points. But this is impossible.
In conclusion, the original proposition is proved.
(2) From (1), we know that the difference is at least 4 (since the number of steps in the row direction and the column direction are both even), and Figure 5 provides an example.
|
4
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary)
|
Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$
Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}>\frac{5}{16}+\frac{5}{16} \times \frac{1}{4}=\frac{25}{64}, \\
a_{3}>\frac{25}{64}+\frac{25}{64} \times \frac{1}{3}=\frac{25}{48}>\frac{1}{2}, \\
a_{4}>\frac{1}{2}+\frac{1}{4}=\frac{3}{4}, a_{5}=\frac{3}{4}+\frac{9}{16}>1,
\end{array}
$$
Thus, $a_{2012}>1$.
Therefore, $0<\frac{1}{a_{2012}}<1, \sum_{n=0}^{2011} \frac{1}{a_{n}+1} \in(3,4)$.
Hence, the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A_{1} A_{2} \cdots A_{101}$ be a regular 101-gon. Each vertex is colored either red or blue. Let $N$ be the number of obtuse triangles that satisfy the following conditions: the three vertices of the triangle are vertices of the 101-gon, the two acute vertices have the same color, and the color of the obtuse vertex is different from that of the acute vertices. Find:
(1) the maximum possible value of $N$;
(2) the number of different coloring methods that make $N$ achieve its maximum value (for two coloring methods, if the color on some $A_{i}$ is different, they are considered different coloring methods). (Tian Zhenhua, problem contributor)
|
2. Let $x_{i}=0$ or 1 represent $A_{i}$ being red or blue, respectively. For an obtuse triangle $\triangle A_{i-a} A_{i} A_{i+b}$, where vertex $A_{i}$ is the obtuse angle vertex, i.e., $a+b \leqslant 50$.
The coloring of the three vertices satisfies the condition if and only if
$$
\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right)=1 \text {, }
$$
otherwise, it equals 0, where the indices are taken modulo 101.
$$
\text { Hence } N=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right) \text {, }
$$
where, β $\sum_{(a, b)}$ β is the sum over all ordered pairs of positive integers $(a, b)$ such that $a+b \leqslant 50$, and there are
$$
49+48+\cdots+1=1225 \text { (pairs). }
$$
Substituting equation (1) into equation (2) gives
$$
\begin{array}{l}
N=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right) \\
=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}^{2}-x_{i} x_{i-a}-x_{i} x_{i+b}+x_{i-a} x_{i+b}\right) \\
=1225 \sum_{i=1}^{101} x_{i}^{2}+\sum_{i=1}^{101} \sum_{k=1}^{\infty}[k-1-2(50-k)] x_{i} x_{i+k} \\
=1225 n+\sum_{i=1}^{101} \sum_{k=1}^{\infty}(3 k-101) x_{i} x_{i+k},
\end{array}
$$
where $n$ is the number of blue vertices.
For any two vertices $A_{i} A_{j}(1 \leqslant i \leqslant j \leqslant 101)$, let
$$
d\left(A_{i}, A_{j}\right)=d\left(A_{j}, A_{i}\right)=\min \{j-i, 101-j+i\} .
$$
Let $B \subseteq\left\{A_{1}, A_{2}, \cdots, A_{101}\right\}$ be the set of all blue vertices. Then equation (3) can be written as
$$
N=1225 n-101 \mathrm{C}_{n}^{2}+3 \sum_{|P, Q| \leq B} d(P, Q),
$$
where $\{P, Q\}$ runs over all two-element subsets of $B$.
Assume $n$ is even; otherwise, change the color of each vertex, and the value of $N$ remains unchanged.
Let $n=2 t(0 \leqslant t \leqslant 50)$, and reindex all blue vertices in a clockwise direction starting from some point as $P_{1}, P_{2}, \cdots, P_{2 t}$. Hence,
$$
\begin{array}{l}
\sum_{|P, Q| \leq B} d(P, Q) \\
=\sum_{i=1}^{t} d\left(P_{i}, P_{i+t}\right)+\frac{1}{2} \sum_{i=1}^{t} \sum_{j=1}^{t-1}\left(d\left(P_{i}, P_{i+j}\right)+\right. \\
\left.d\left(P_{i+j}, P_{i+t}\right)+d\left(P_{i+t}, P_{i-j}\right)+d\left(P_{i-j}, P_{i}\right)\right) \\
\leqslant 50 t+\frac{101}{2} t(t-1),
\end{array}
$$
where the indices of $P_{i}$ are taken modulo $2 t$, and the inequality $d\left(P_{i}, P_{i+1}\right) \leqslant 50$ and
$$
\begin{array}{l}
d\left(P_{i}, P_{i+j}\right)+d\left(P_{i+j}, P_{i+t}\right)+ \\
d\left(P_{i+t}, P_{i-j}\right)+d\left(P_{i-j}, P_{i}\right) \leqslant 101
\end{array}
$$
are used.
Combining equations (4) and (5) gives
$$
\begin{array}{l}
N \leqslant 1225 n-101 \mathrm{C}_{n}^{2}+3\left[50 t+\frac{101}{2} t(t-1)\right] \\
=-\frac{101}{2} t^{2}+\frac{5099}{2} t .
\end{array}
$$
The right-hand side of the above equation achieves its maximum value when $t=25$.
Thus, $N \leqslant 32175$.
We need to consider the necessary and sufficient conditions for $N=32175$.
First, $t=25$, i.e., there are exactly 50 blue points.
Second, for $1 \leqslant i \leqslant t$, $d\left(P_{i}, P_{i+t}\right)=50$.
When $d\left(P_{i}, P_{i+1}\right)=50$, the equality in equation (6) must hold.
Therefore, the number of ways to choose 50 blue points such that $N$ achieves its maximum value is the number of ways to choose 25 diagonals from the longest 101 diagonals such that any two diagonals intersect internally (equivalent to having no common vertices).
Connect $A_{i}(i=1,2, \cdots, 101)$ with $A_{i+50}$ to form a graph $G$.
Note that 50 and 101 are coprime.
Thus, $1+50 n(0 \leqslant n \leqslant 100)$ forms a complete residue system modulo 101, i.e., graph $G$ is a cycle with exactly 101 edges.
Therefore, the number of ways to color 50 vertices blue such that $N$ achieves its maximum value, $S$, is the number of ways to select 25 edges from graph $G$ such that no two edges are adjacent in $G$.
Now, fix an edge $e$ in graph $G$. The number of ways to choose 25 edges including edge $e$ is $\mathrm{C}_{75}^{24}$, and the number of ways to choose 25 edges excluding edge $e$ is $\mathrm{C}_{76}^{25}$. Hence, the number of ways is
$$
S=\mathrm{C}_{75}^{24}+\mathrm{C}_{76}^{25} \text {. }
$$
Considering the case of 50 red points also has $S$ ways of coloring. Therefore, the number of coloring methods that make $N$ achieve its maximum value is $2 S$.
In summary, the maximum value of $N$ is 32175, and the corresponding number of coloring methods is $2 S=2\left(\mathrm{C}_{75}^{24}+\mathrm{C}_{76}^{25}\right)$.
|
32175
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (18 points) Given
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) .
$$
(1) If there exist positive numbers $a, b, c$ such that inequality (1) holds, prove: $k>5$;
(2) If there exist positive numbers $a, b, c$ such that inequality (1) holds, and any set of positive numbers $a, b, c$ that satisfy inequality (1) are the sides of some triangle, find the integer $k$ that satisfies the condition.
|
11. (1) Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, therefore, from inequality (1) we know
$$
k(a b+b c+c a)>5(a b+b c+c a) .
$$
Noting that $a, b, c$ are positive numbers.
Thus, $a b+b c+c a>0$.
Hence $k>5$.
(2) From (1) we know $k>5$.
Since $k$ is an integer, then $k \geqslant 6$.
Let $a=1, b=1, c=2$. It is easy to see that $a, b, c$ are not the lengths of the three sides of a triangle.
By the problem, we know that the inequality (1) cannot hold, i.e.,
$$
k(1+2+2) \leqslant 5(1+1+4) \Rightarrow k \leqslant 6 \text {. }
$$
Thus, $k=6$.
Next, we prove: $k=6$.
At this point, inequality (1) is
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$
Noting that $a=1, b=1, c=\frac{3}{2}$ satisfies inequality (2).
Assume $a, b, c$ satisfy inequality (2).
By symmetry, without loss of generality, assume $a \leqslant b \leqslant c$.
Transform inequality (2) into
$$
\begin{array}{l}
5 c^{2}-6 c(a+b)+5 a^{2}+5 b^{2}-6 a b<0 . \\
\text { Hence } \Delta=[6(a+b)]^{2}-20\left(5 a^{2}+5 b^{2}-6 a b\right) \\
=-64(a-b)^{2}+64 a b \\
\leqslant 64 a b \leqslant 16(a+b)^{2} .
\end{array}
$$
Thus, from inequality (3) we get
$$
\begin{aligned}
c & <\frac{6(a+b)+\sqrt{\Delta}}{10} \leqslant \frac{6(a+b)+4(a+b)}{10} \\
& =a+b .
\end{aligned}
$$
Therefore, $a, b, c$ are the lengths of the three sides of a triangle.
In summary, the integer $k=6$.
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
16. Given that one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$. Then the minimum value of the area of the square is $\qquad$ .
|
16. 80 .
From the problem, we know that the distance from a point $\left(x, x^{2}\right)$ on the parabola to the line $l$ is
$$
d=\frac{\left|2 x-x^{2}-17\right|}{\sqrt{5}}>0 .
$$
When $x=1$, $d$ reaches its minimum value $\frac{16}{\sqrt{5}}$.
To minimize the area of the square, and by the symmetry of the square, the distances from the two points on the parabola to $(1,1)$ should be equal. Thus, we can assume these two points are $\left(a, a^{2}\right),\left(2-a,(2-a)^{2}\right)(a<1)$.
Therefore, $\sqrt{[a-(2-a)]^{2}+\left[a^{2}-(2-a)^{2}\right]^{2}}$ $=\frac{\left|a^{2}-2 a-17\right|}{\sqrt{5}}$
$\Rightarrow a=-1$ or -7 .
Thus, the coordinates of the two points are $(-1,1),(3, a)$ or $(-7,49),(9,8)$.
Therefore, the minimum value of the required area is
$$
(3+1)^{2}+(9-1)^{2}=80 \text {. }
$$
|
80
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) In the sequence $\left\{x_{n}\right\}$,
$$
x_{n}=p^{n}+q^{n}, x_{1}=1, x_{3}=4 \text {. }
$$
(1) Prove: $x_{n+2}=x_{n+1}+x_{n}$;
(2) Determine the last digit of $x_{2011}$, but no proof is required.
|
11. (1) Notice,
$$
\begin{array}{l}
p+q=1, p^{3}+q^{3}=4, \\
p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right) .
\end{array}
$$
Thus, $p^{2}-p q+q^{2}=4, p^{2}+2 p q+q^{2}=1$.
$$
\begin{array}{l}
\text { Then } x_{2}=p^{2}+q^{2}=3 \\
\Rightarrow p^{2}+(1-p)^{2}=3 \\
\Rightarrow p^{2}=p+1, q^{2}=q+1 \text {. }
\end{array}
$$
Therefore, $x_{n+2}=p^{n+2}+q^{n+2}=p^{n} p^{2}+q^{n} q^{2}$
$$
\begin{array}{l}
=p^{n}(p+1)+q^{n}(q+1) \\
=p^{n+1}+q^{n+1}+p^{n}+q^{n}=x_{n+1}+x_{n} .
\end{array}
$$
(2) The first few terms of the sequence $\left\{x_{n}\right\}$ are:
$$
\begin{array}{l}
1,3,4,7,11,18,29,47,76,123,199,322, \\
521,843,1364,2207,3571,5778,9349, \\
15127,24476,39603,64079,103682, \\
\quad \ldots
\end{array}
$$
Observing, starting from the first term, every 11 terms, the last digit repeats, i.e., if
$$
x_{n} \equiv r_{n}(\bmod 10)(r \in\{0,1, \cdots, 9\}),
$$
then the sequence $\left\{r_{n}\right\}$ has a period of 12.
Since $2013=167 \times 12+9$, we have
$$
r_{2013}=r_{9}=6 .
$$
|
6
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that for any real number $x$ we have $a \cos x + b \cos 2x \geqslant -1$.
Then the maximum value of $a + b$ is $\qquad$
|
2. 2 .
Let $x=\frac{2 \pi}{3}$, then $a+b \leqslant 2$.
When $a=\frac{4}{3}, b=\frac{2}{3}$,
$$
\begin{array}{l}
a \cos x+b \cos 2 x=\frac{4}{3} \cos x+\frac{2}{3} \cos 2 x \\
=\frac{4}{3} \cos ^{2} x+\frac{4}{3} \cos x-\frac{2}{3} \\
=\frac{4}{3}\left(\cos x+\frac{1}{2}\right)^{2}-1 \\
\geqslant-1 .
\end{array}
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfies
$$
a_{n+2}=a_{n+1}-a_{n} \text {. }
$$
If the sum of the first 1000 terms of the sequence is 1000, then the sum of the first 2014 terms is $\qquad$ .
|
3. 1000 .
From $a_{n+2}=a_{n+1}-a_{n}$, we get
$$
a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n} \text {. }
$$
Therefore, for any positive integer $n$ we have
$$
a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0 \text {. }
$$
Also, $2014=1000(\bmod 6)$, so
$$
S_{2014}=S_{1000}=1000 \text {. }
$$
|
1000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$
real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$
|
4.4 .
Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$
Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $g(a-2)=-g(b-2)=g(2-b)$
$$
\Rightarrow a-2=2-b \Rightarrow a+b=4
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.
|
7.4.
Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$
Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant a+d+b^{2}+b\left(a d^{2}-2 a-2 d\right) .
\end{array}
$$
Thus, $a d^{2}-2 a-2 d \leqslant 0$.
If $d \geqslant 3$, then
$$
\begin{aligned}
a d^{2}- & 2 a-2 d \geqslant 3 a d-2 a-2 d \\
& =2 a d+a d-2 a-2 d \\
& \geqslant 2 d+3 a-2 a-2 d>0,
\end{aligned}
$$
which is a contradiction.
Therefore, $d \leqslant 2$.
(1) $d=2$.
From $a d^{2}-2 a-2 d \leqslant 0$, we get $a \leqslant 2$.
When $a=2$, from equation (1) we get $b^{2}-8 b+4=0$, in which case, $b$ has no integer solutions;
When $a=1$, from equation (1) we get
$$
b^{2}-4 b+3=0 \Rightarrow b=1 \text { or } 3 \text {, }
$$
In this case, the solutions to the original equation are
$$
(x, y)=(4,2) \text { or }(4,6) \text {. }
$$
(2) $d=1$.
From equation (1) we get
$$
\begin{array}{l}
2=(a-b-1)(b-1) \\
\Rightarrow(a-b-1, b-1)=(2,1) \text { or }(1,2) \\
\Rightarrow(a, b)=(5,2) \text { or }(5,3),
\end{array}
$$
In this case, the solutions to the original equation are
$$
(x, y)=(5,2) \text { or }(5,3) \text {. }
$$
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A middle school has 35 lights on each floor. To save electricity while ensuring the lighting needs of the corridors, the following requirements must be met:
(1) Two adjacent lights cannot be on at the same time;
(2) Any three consecutive lights cannot be off at the same time.
If you were to design different lighting methods, what is the maximum number of different lighting methods you can design?
$\qquad$ kinds of different lighting methods.
|
8. 31572 .
Notice that, $a_{n+3}=a_{n}+a_{n+1}, a_{3}=4, a_{4}=5$. Therefore, $a_{35}=31572$.
|
31572
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
a_{k}=\left[\frac{2014}{k}\right](k=1,2, \cdots, 100) \text {. }
$$
Then, among these 100 integers, the number of distinct integers is
|
6.69.
When $01$, $[a+b]>[a]$.
$$
\begin{array}{l}
\text { Also } \frac{2014}{k}-\frac{2014}{k+1}=\frac{2014}{k(k+1)} \\
\Rightarrow \frac{2014}{k}=\frac{2014}{k+1}+\frac{2014}{k(k+1)} .
\end{array}
$$
When $k \leqslant 44$, $\frac{2014}{k(k+1)}>1$;
When $k \geqslant 45$, $\frac{2014}{k(k+1)}<1$.
Therefore, when $k \leqslant 44$, $a_{1}, a_{2}, \cdots, a_{44}$ are all different,
$$
a_{44}=45, a_{45}=45, \cdots, a_{99}=20, a_{100}=20 \text {; }
$$
When $k \geqslant 45$, $a_{k}=a_{k+1}$ or $a_{k}=a_{k+1}+1$.
Thus, $a_{45}, a_{46}, \cdots, a_{100}$ take all the 25 integers from $20 \sim 44$.
Therefore, $a_{1}, a_{2}, \cdots, a_{100}$ contain a total of $44+25=69$ different integers.
|
69
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $A$ and $B$ be two different subsets of the set $\{a, b, c, d, e\}$, such that set $A$ is not a subset of set $B$, and $B$ is not a subset of set $A$. Then the number of different ordered pairs $(A, B)$ is
|
7.570.
Notice that, the set $\{a, b, c, d, e\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose the set $B$ contains $k(1 \leqslant k \leqslant 5)$ elements, then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ is
$$
\sum_{k=1}^{5}\left(2^{k}-1\right) \mathrm{C}_{5}^{k}=\sum_{k=0}^{5} 2^{k} \mathrm{C}_{5}^{k}-\sum_{k=0}^{5} \mathrm{C}_{5}^{k}=3^{5}-2^{5} .
$$
Similarly, the number of ordered pairs $(A, B)$ satisfying $B \subset A$ is $3^{5}-2^{5}$.
Therefore, the number of ordered pairs $(A, B)$ satisfying the condition is
$$
2^{5}\left(2^{5}-1\right)-2\left(3^{5}-2^{5}\right)=570 .
$$
|
570
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
385 A certain school's 2014 graduates numbered 2014 students. The school's six leaders must sign each student's graduation album. It is known that each leader must and can only use one of the three designated colored pens, and the color choices for the six leaders signing 2014 albums can be represented as a $6 \times 2014$ grid, called a color selection scheme. Find the smallest positive integer $n$, such that there exists a color selection scheme where in any $n$ albums, there are 8 albums where any two albums have at most four leaders' signatures in the same color.
|
Let 1, 2, 3 represent three colors, and the colors used by six school leaders for signing in each memorial album be represented by a six-element ordered array
$$
\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)
$$
This array is referred to as the sequence group $\left(x_{1}, x_{2}, \cdots, x_{6}\right)$. For $x_{2}, x_{3}, \cdots, x_{6} \in\{1,2,3\}$, the set
$$
\left\{\left(1, x_{2}, \cdots, x_{6}\right),\left(2, x_{2}, \cdots, x_{6}\right),\left(3, x_{2}, \cdots, x_{6}\right)\right\}
$$
is called a Type 1 set. Clearly, all sequence groups $\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)$ belong to 243 types.
Given $2014=8 \times 243+70$, we know that among 2014 memorial albums, there are nine albums of the same type. Select these nine albums;
Given $2005=8 \times 243+61$, we know that among the remaining 2005 albums, there are another nine albums of the same type. Select these nine albums;
Continuing this process, we can select seven times, obtaining a total of 63 albums. Among these, each set of eight albums includes at least two albums selected in the same round, which must be of the same type. This means that at least five school leaders signed with the same color in these two albums. This indicates that for any coloring scheme, there are 63 albums that do not meet the requirement. Therefore, $n \geqslant 64$.
On the other hand, consider the set
$$
S=\left\{\left(x_{1}, x_{2}, \cdots, x_{6}\right) \mid x_{i} \in\left\{1,2,3 \mid, \sum_{i=1}^{6} x_{i}=0(\bmod 3)\right\}\right. \text {. }
$$
The number of elements in set $S$ is 243, and any two different elements in set $S$ have at most four components the same. Arbitrarily remove 19 elements from set $S$, leaving 224 elements to form set $T$. Divide the 2014 memorial albums into 224 groups, with the first 223 groups each containing 9 albums, and the last group containing 7 albums. Ensure that each group corresponds to an element in set $T$ (i.e., the signatures in the same group are identical, while those in different groups are different). Clearly, any 64 albums must belong to at least eight groups, thus there exist eight albums corresponding to eight different elements in set $T$. Therefore, any two albums among these have at most four school leaders' signatures in the same color.
In summary, the minimum value of $n$ is 64.
οΌFang Tinggang, Chengdu No. 7 High School, Sichuan, 610041οΌ
|
64
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Real-coefficient polynomials $f_{i}(x)=a_{i} x^{2}+b_{i} x+c_{i}$ $\left(a_{i}>0, i=1,2, \cdots, 2011\right)$, and $\left\{a_{i}\right\} γ\left\{b_{i}\right\} γ\left\{c_{i}\right\}$ are all arithmetic sequences. If $F(x)=\sum_{i=1}^{204} f_{i}(x)$ has real roots, then at most how many polynomials in $\left\{f_{i}(x)\right\}$ have no real roots?
|
γAnalysisγFrom $\left\{a_{i}\right\}$ being an arithmetic sequence, we know
$$
\sum_{i=1}^{2011} a_{i}=(2 \times 1005+1) a_{1006}=2011 a_{1006} \text {. }
$$
Similarly, $\sum_{i=1}^{2011} b_{i}=2011 b_{1006}$,
$$
\sum_{i=1}^{2011} c_{i}=2011 c_{1006} \text {. }
$$
Therefore, $F(x)=\sum_{i=1}^{200} f_{i}(x)=2011 f_{1006}(x)$ has a real root, which means $f_{1006}(x)$ has a real root, denoted as $x_{0}$.
When $1 \leqslant k \leqslant 1005$, note that,
$$
f_{k}(x)+f_{2012-k}(x)=2 f_{1006}(x)
$$
has a real root $x_{0}$, i.e.,
$$
f_{k}\left(x_{0}\right)+f_{2012-k}\left(x_{0}\right)=0,
$$
then at least one of $f_{k}\left(x_{0}\right)$ and $f_{2012-k}\left(x_{0}\right)$ is less than or equal to zero.
Thus, at least one of $f_{k}\left(x_{0}\right)$ and $f_{2012-k}\left(x_{0}\right)$ has a real root.
Therefore, at least 1006 of $\left\{f_{i}(x)\right\}$ have real roots, meaning at most 1005 do not have real roots.
Next, we provide a specific example where 1005 of $\left\{f_{i}(x)\right\}$ do not have real roots:
For $f_{i}(x)=x^{2}+i-1006$,
when $1 \leqslant i \leqslant 1006$, $f_{i}(x)$ has real roots; when $1007 \leqslant i \leqslant 2012$, $f_{i}(x)$ does not have real roots.
γNoteγThis problem uses the extended form of the middle term formula
$$
2 a_{m}=a_{m-k}+a_{m+k} .
$$
|
1005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 1, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$ respectively. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then $m+n=$
|
3. 2 .
Solution 1 Notice that,
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} .
$$
Since points $M, O, N$ are collinear, we have,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
Solution 2 Since points $M, O, N$ are on the sides or their extensions of $\triangle A B C$, and the three points are collinear, by Menelaus' theorem we get
$$
\begin{array}{l}
\frac{A M}{M B} \cdot \frac{B O}{O C} \cdot \frac{C N}{N A}=1 . \\
\text { Hence } \frac{n-1}{1-m}=1 \Rightarrow m+n=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given
$$
\begin{array}{l}
A \cup B \cup C=\{a, b, c, d, e\}, A \cap B=\{a, b, c\}, \\
c \in A \cap B \cap C .
\end{array}
$$
Then the number of sets $\{A, B, C\}$ that satisfy the above conditions is. $\qquad$
|
7. 100 .
As shown in Figure 3, the set $A \cup B \cup C$ can be divided into seven mutually exclusive regions, denoted as $1, 2, \cdots, 7$.
It is known that element $c$ is in region 7, elements $a, b$ can appear in each of the regions 4, 7, with 4 possibilities; elements $d, e$ can appear in each of the regions $1, 2, 3, 5, 6$, with 5 possibilities each.
Thus, the number of sets $\{A, B, C\}$ that meet the conditions is $4 \times 5 \times 5=$ 100.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Use five different colors to color the six vertices of the triangular prism $A B C-D E F$, requiring each point to be colored with one color, and the two endpoints of each edge to be colored with different colors. Then the number of different coloring methods is $\qquad$.
|
9. 1920.
Transform the adjacency relationship of the vertices of a triangular prism into the regional relationship as shown in Figure 4, then the coloring methods for vertices $A$, $B$, and $C$ are $\mathrm{A}_{5}^{3}$ kinds.
For each determined coloring scheme of points $A$, $B$, and $C$, if we temporarily do not consider the requirement that the endpoints of each edge must be different colors, then vertices $D$, $E$, and $F$ also have $\mathrm{A}_{5}^{3}$ different coloring schemes. Let the set of all these coloring schemes be denoted as $P$, hence
$$
|P|=\mathrm{A}_{5}^{3}=60 \text{. }
$$
Let $P_{1}$, $P_{2}$, and $P_{3}$ represent the sets of all schemes in $P$ where points $A$ and $D$, $B$ and $E$, and $C$ and $F$ are the same color, respectively. Then
$$
\left|P_{i}\right|=\mathrm{A}_{4}^{2}=12, \left|P_{i} \cap P_{j}\right|=\mathrm{A}_{3}^{1}=3,
$$
where $1 \leqslant i<j \leqslant 3$,
$$
\left|P_{1} \cap P_{2} \cap P_{3}\right|=1 \text{. }
$$
By the principle of inclusion-exclusion, the total number of coloring schemes where the endpoints of any edge are different colors is
$$
\begin{array}{l}
\mathrm{A}_{5}^{3}\left(\mathrm{~A}_{5}^{3}-\left|P_{1} \cup P_{2} \cup P_{3}\right|\right) \\
=\mathrm{A}_{5}^{3}\left(\mathrm{A}_{5}^{3}-\sum_{i=1}^{3}\left|P_{i}\right|+\sum_{1<i<j<3}\left|P_{i} \cap P_{j}\right|-\left|P_{1} \cap P_{2} \cap P_{3}\right|\right) \\
=\mathrm{A}_{5}^{3}\left(\mathrm{A}_{5}^{3}-3 \mathrm{~A}_{4}^{2}+3 \mathrm{~A}_{3}^{1}-1\right) \\
=1920 .
\end{array}
$$
|
1920
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (15 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram.
|
12. Given that $\left|F_{1} F_{2}\right|=2$.
As shown in Figure 5, the inscribed quadrilateral $\square A B C D$ of the ellipse has a pair of opposite sides $B C$ and $A D$ passing through the foci $F_{1}$ and $F_{2}$, respectively.
Obviously, the center of symmetry of $\square A B C D$ is the origin.
Figure 5
Let the points be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$.
Then $S_{\square A B C D}=2 S_{\triangle A F, D}=\left|F_{1} F_{2}\right|\left|y_{1}-y_{4}\right|$ $=2\left|y_{1}-y_{4}\right|$.
If the slope of line $A D$ exists, let its equation be
$$
y=k(x-1) \text {. }
$$
Substituting into the ellipse equation and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \text {. }
$$
By Vieta's formulas, we have
$$
x_{1}+x_{4}=\frac{8 k^{2}}{3+4 k^{2}}, x_{1} x_{4}=\frac{4 k^{2}-12}{3+4 k^{2}} \text {. }
$$
Then $\left|y_{1}-y_{4}\right|^{2}=k^{2}\left|x_{1}-x_{4}\right|^{2}$
$$
\begin{array}{l}
=k^{2}\left[\left(x_{1}+x_{4}\right)^{2}-4 x_{1} x_{4}\right] \\
=\frac{k^{2}}{\left(3+4 k^{2}\right)^{2}}\left[64 k^{4}-16\left(3+4 k^{2}\right)\left(k^{2}-3\right)\right] \\
=9 \times \frac{16 k^{2}\left(k^{2}+1\right)}{\left(3+4 k^{2}\right)^{2}} .
\end{array}
$$
Thus, $S_{\triangle \triangle B C D}=2\left|y_{1}-y_{4}\right|$
$$
=6 \sqrt{\frac{16 k^{2}\left(1+k^{2}\right)}{\left(3+4 k^{2}\right)^{2}}}<6 \text {. }
$$
If the slope of $A D$ does not exist, i.e., $A D \perp x$ axis.
It is easy to get the points $A\left(1, \frac{3}{2}\right), D\left(1,-\frac{3}{2}\right)$.
Thus, $S_{\text {ΠΠBCD }}=2\left|y_{1}-y_{4}\right|=6$.
In summary, the maximum area of $\square A B C D$ is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Determine the least possible value of the largest term in an arithmetic sequence composed of seven distinct primes. ${ }^{[4]}$
(2005, British Mathematical Olympiad)
|
γAnalysisγLet these seven different prime numbers be $p_{i}(i=1,2$, $\cdots, 7)$, and
$$
\begin{array}{l}
p_{2}=p_{1}+d, p_{3}=p_{1}+2 d, p_{4}=p_{1}+3 d, \\
p_{5}=p_{1}+4 d, p_{6}=p_{1}+5 d, p_{7}=p_{1}+6 d,
\end{array}
$$
where the common difference $d \in \mathbf{Z}_{+}$.
Since $p_{2}=p_{1}+d$ and $p_{3}=p_{1}+2 d$ are two odd primes, the common difference $d$ is even, and $p_{1} \geqslant 3$.
If $3 \nmid d$, then $p_{2} γ p_{3}=p_{2}+d γ p_{4}=p_{2}+2 d$ are distinct modulo 3, one of them must be a multiple of 3, which contradicts that $p_{2}$, $p_{3}$, and $p_{4}$ are all primes greater than 3. Hence, $3 \mid d$, and $p_{1} \geqslant 5$.
If $5 \nmid d$, then
$$
\begin{array}{l}
p_{2} γ p_{3}=p_{2}+d γ p_{4}=p_{2}+2 d γ \\
p_{5}=p_{2}+3 d γ p_{6}=p_{2}+4 d
\end{array}
$$
these five numbers are distinct modulo 5, one of them must be a multiple of 5, which contradicts that they are all primes greater than 5. Therefore, $5 \mid d$, and $p_{1} \geqslant 7$.
In summary, $30 \mid d$, and $p_{1} \geqslant 7$.
If $p_{1}>7$, and $7 \nmid d$, then
$$
\begin{array}{l}
p_{1} γ p_{2}=p_{1}+d γ p_{3}=p_{1}+2 d γ p_{4}=p_{1}+3 d γ \\
p_{5}=p_{1}+4 d γ p_{6}=p_{1}+5 d γ p_{7}=p_{1}+6 d
\end{array}
$$
these seven numbers are distinct modulo 7, one of them must be a multiple of 7, which contradicts that they are all primes greater than 7.
If $p_{1}>7$, and $71 d$, then $2101 d$.
Thus, $p_{7}>7+210 \times 6=1267$.
Next, consider the case where $p_{1}=7, d=30 k$.
Notice that, $187=p_{1}+30 \times 6=11 \times 17$.
Therefore, $k \neq 1,2,3,6$.
Notice that, $247=p_{1}+30 \times 8=13 \times 19$.
Therefore, $k \neq 4$.
When $k=5$, the arithmetic sequence
$$
7,157,307,457,607,757,907
$$
are all primes, and $907<1267$.
In summary, the arithmetic sequence
$$
7,157,307,457,607,757,907
$$
is the solution.
γNoteγEnumerating and discussing different cases is a common method for solving number theory problems, and the common difference of an arithmetic sequence is often the subject of such case-by-case analysis.
|
907
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{aligned}
f(x+2) & =f(x)+2, \\
\text { then } \sum_{k=1}^{2014} f(k) & =
\end{aligned}
$$
|
14.2029105.
Notice that, $f(0)=0$.
By $f(x+2)=f(x)+2$, let $x=-1$. Then $f(1)=f(-1)+2 \Rightarrow f(1)=1$.
$$
\begin{array}{l}
\text { Also, } f(2 n)=\sum_{k=1}^{n}(f(2 k)-f(2 k-2))+f(0) \\
=2 n, \\
f(2 n-1)=\sum_{k=2}^{n}(f(2 k-1)-f(2 k-3))+f(1) \\
=2 n-1, \\
\text { Therefore, } \sum_{k=1}^{2014} f(k)=\sum_{k=1}^{2014} k=2029105 .
\end{array}
$$
|
2029105
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. A courier company undertakes courier services between 13 cities in a certain area. If each courier can take on the courier services for at most four cities, to ensure that there is at least one courier between every two cities, the courier company needs at least $\qquad$ couriers.
|
17. 13.
From the problem, we know that there are $\mathrm{C}_{13}^{2}$ types of express delivery services between 13 cities. Each courier can handle at most $\mathrm{C}_{4}^{2}$ types of express delivery services between four cities. Therefore, at least $\frac{\mathrm{C}_{13}^{2}}{\mathrm{C}_{4}^{2}}=13$ couriers are needed.
Below is a construction showing that 13 couriers can meet the requirements:
$$
\begin{array}{l}
1,2,3,4 ; 1,5,6,7 ; 1,8,9,10 ; 1,11,12,13 ; \\
2,5,8,11 ; 2,6,9,12 ; 2,7,10,13 ; \\
3,5,9,13 ; 3,6,10,11 ; 3,7,8,12 ; \\
4,5,10,12 ; 4,6,8,13 ; 4,7,2,11 .
\end{array}
$$
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
21. Among the 100 integers from $1 \sim 100$, arbitrarily select three different numbers to form an ordered triplet $(x, y, z)$. Find the number of triplets that satisfy the equation $x+y=3z+10$.
|
(1) When $3 z+10 \leqslant 101$, i.e., $z \leqslant 30$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
S=\sum_{k=1}^{30}(3 k+9)=1665 \text{. }
$$
(2) When $3 z+10 \geqslant 102$, i.e., $31 \leqslant z \leqslant 63$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
\begin{aligned}
T & =\sum_{k=31}^{63}(191-3 k) \\
& =\sum_{k=1}^{33}[191-3(k+30)]=1650 .
\end{aligned}
$$
Now consider the case where $x, y, z$ are equal.
First, $x, y, z$ cannot all be equal.
If $x=y$, then the number of ternary tuples satisfying $x+y=3 z+10$ is $A=31$.
If $x=z$, then the number of ternary tuples satisfying $x+y=3 z+10$ is $B=45$.
If $y=z$, then the number of ternary tuples satisfying $x+y=3 z+10$ is $C=45$.
In summary, the number of ternary tuples is
$$
S+T-A-B-C=3194 .
$$
|
3194
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ .
|
4. 0 .
From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{2} \beta \\
=\frac{1}{2}-2\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}+\sin \theta \cdot \cos \theta \\
=\frac{1}{2}-\frac{1+2 \sin \theta \cdot \cos \theta}{2}+\sin \theta \cdot \cos \theta=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the edge length of the cube $A_{1} A_{2} A_{3} A_{4}-B_{1} B_{2} B_{3} B_{4}$ is 1, then the number of elements in the set
$$
\left\{x \mid x=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}, i, j \in\{1,2,3,4\}\right\}
$$
is $\qquad$
|
5. 1 .
Solution 1 Note that,
$$
\begin{array}{l}
\overrightarrow{A_{1} B_{1}} \perp \overrightarrow{A_{i} A_{1}}, \overrightarrow{A_{1} B_{1}} \perp \overrightarrow{B_{1} B_{j}}(i, j \in\{2,3,4\}) \text {. } \\
\text { Hence } \overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=\overrightarrow{A_{1} B_{1}} \cdot\left(\overrightarrow{A_{i} A_{1}}+\overrightarrow{A_{1} B_{1}}+\overrightarrow{B_{1} B_{j}}\right) \\
=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} A_{1}}+{\overrightarrow{A_{1} B_{1}}}^{2}+\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{B_{1} B_{j}}={\overrightarrow{A_{1} B_{1}}}^{2}=1 \text {. } \\
\end{array}
$$
Solution 2 As shown in Figure 4, for any $i, j \in\{1,2,3,4\}$, the projection of $\overrightarrow{A_{i} B_{j}}$ onto $\overrightarrow{A_{1} B_{1}}$ is $\overrightarrow{A_{1} B_{1}}$, hence
$$
\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$
Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$
|
6.8.
From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$
Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times 2013=4028>3969=63^{2} \text {. }
$$
Also, $a_{1}=\sqrt{2}, a_{n+1}=a_{n}+\frac{1}{a_{n}}>2\left(n \in \mathbf{Z}_{+}\right)$, so
$$
\begin{array}{l}
a_{2014}^{2}=2+2 \times 2013+\frac{1}{2}+\sum_{i=2}^{2013} \frac{1}{a_{i}^{2}} \\
<4028+\frac{1}{2}+\frac{1}{4} \times 2012=4531.5 \\
<4624=68^{2} .
\end{array}
$$
Therefore, $9 \approx \sqrt{63}<\sqrt{a_{2014}}<\sqrt{68} \approx 8.3$.
Thus, the natural number closest to $\sqrt{a_{2014}}$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=6$. Then the maximum value of $x+\sqrt{x y}+\sqrt[3]{x y z}$ is $\qquad$ .
|
7.8.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{1}{4} x+y+4 z \geqslant 3 \sqrt[3]{x y z}, \\
\frac{3}{4} x+3 y \geqslant 3 \sqrt{x y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
x+4 y+4 z \geqslant 3 \sqrt{x y}+3 \sqrt[3]{x y z} \\
\Rightarrow x+\sqrt{x y}+\sqrt[3]{x y z} \leqslant \frac{4}{3}(x+y+z)=8 .
\end{array}
$$
When $x: y: z=16: 4: 1$, the equality holds. Therefore, the maximum value is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Fill the four characters β马β βδΈβ βζβ βεβ in a $5 \times 5$ grid, with at most one character in each small square. β马β βδΈβ must be filled in from left to right, and βζβ βεβ must also be filled in from left to right. β马β βδΈβ must be in the same row or in the same column from top to bottom, or βζβ βεβ must be in the same row or in the same column from top to bottom. The number of different ways to fill the grid is $\qquad$ (answer with a number).
|
8.42100.
The problem is equivalent to:
Filling 2 $a$s and 2 $b$s in a $5 \times 5$ grid, with at most one letter in each small square, such that at least one pair of the same letters is in the same row or column.
First, consider the case where the same letters are neither in the same row nor in the same column.
The number of ways to place 2 $a$s such that they are neither in the same row nor in the same column is $\mathrm{C}_{5}^{2} \mathrm{~A}_{5}^{2} = 200$. Similarly, the number of ways to place 2 $b$s such that they are neither in the same row nor in the same column is also $\mathrm{C}_{5}^{2} \mathrm{~A}_{5}^{2} = 200$.
When both conditions are satisfied, there are two scenarios that do not meet the requirement:
(1) The 2 $a$s are in the same squares as the 2 $b$s, which has 200 ways;
(2) One of the 2 $a$s is in the same square as one of the 2 $b$s, which has $\mathrm{C}_{25}^{1} \mathrm{~A}_{16}^{2} = 25 \times 240 = 6000$ ways.
Therefore, the number of ways where the same letters are neither in the same row nor in the same column is $40000 - 200 - 6000 = 33800$.
Thus, the number of ways that meet the problem's conditions is
$$
\frac{A_{25}^{4}}{2! \times 2!} - 33800 = 75900 - 33800 = 42100 \text{. }
$$
|
42100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$.
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Next, copy all the previous items $1,1,2$, and add the successor number 3 of 2, to get
$$
a_{4}=1, a_{5}=1, a_{6}=2, a_{7}=3 ;
$$
Next, copy all the previous items $1,1,2,1,1,2,3$, and add the successor number 4 of 3, to get the first 15 items as
$$
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4 \text {; }
$$
Continue in this manner.
Try to find $a_{2000}$ and the sum of the first 2000 terms of the sequence $S_{2000}$.
|
From the construction method of the sequence $\left\{a_{n}\right\}$, it is easy to know that
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
In general, we have $a_{2^{n}-1}=n$, meaning the number $n$ first appears at the $2^{n}-1$ term, and if $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then
$$
\begin{array}{l}
a_{m}=a_{k} . \\
\text { Since } 2000=2^{10}-1+977, \\
977=2^{9}-1+466, \\
466=2^{8}-1+211, \\
211=2^{7}-1+84, \\
84=2^{6}-1+21, \\
21=2^{4}-1+6, \\
6=2^{2}-1+3,
\end{array}
$$
Thus, $a_{2000}=a_{977}=a_{466}=a_{211}=a_{84}$
$$
=a_{21}=a_{6}=a_{3}=2 \text {. }
$$
To find $S_{2000}$, we first calculate $S_{2^{n-1}}$.
By the construction method of the sequence $\left\{a_{n}\right\}$, we know that in the first $2^{n}-1$ terms, there are
1 $n$, 2 $(n-1)$, $2^{2}$ $(n-2)$, ...,
$2^{k}$ $(n-k)$, ..., $2^{n-1}$ 1.
Thus, $S_{2^{n}-1}=n+2(n-1)+2^{2}(n-2)+\cdots+$
$$
2^{n-2} \times 2+2^{n-1} \times 1 \text {. }
$$
Therefore, $2 S_{2^{n}-1}=2 n+2^{2}(n-1)+2^{3}(n-2)+\cdots+$
$$
2^{n-1} \times 2+2^{n} \text {. }
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
S_{2^{n-1}}=-n+\left(2+2^{2}+\cdots+2^{n-1}+2^{n}\right) \\
=2^{n+1}-(n+2) .
\end{array}
$$
When $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, we have
$$
\begin{array}{l}
S_{m}=S_{2^{n}-1}+\left[a_{\left(2^{n}-1\right)+1}+a_{\left(2^{n}-1\right)+2}+\cdots+a_{\left(2^{n}-1\right)+k}\right] \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$
Therefore, $S_{2000}=S_{2^{10}-1}+S_{977}, S_{977}=S_{2^{9}-1}+S_{466}$,
$$
\begin{array}{l}
S_{466}=S_{2^{8}-1}+S_{211}, S_{211}=S_{2^{7}-1}+S_{84}, \\
S_{84}=S_{2^{6}-1}+S_{21}, S_{21}=S_{2^{4}-1}+S_{6}, S_{6}=8 .
\end{array}
$$
From equation (1), we get
$$
\begin{array}{l}
S_{2000}=\left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
\\
\left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+8 \\
=3952 .
\end{array}
$$
|
3952
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 There are 800 points on a circle, numbered $1,2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ gaps in a clockwise direction and color the point reached red. Continue this process until no new red points appear on the circumference.
How many red points can be obtained at most on the circumference? Prove your conclusion.
|
Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, such as the $2k$-th point, then according to the coloring rule, every red point colored afterward will also be an even-numbered point. In this case, if we rename the 2nd, 4th, ..., $2k$-th, ..., $2n$-th points as $1^*$, $2^*$, ..., $k^*$, ..., $n^*$, it is not difficult to see that the number of red points that can be colored starting from the $2k$-th point on a circle with $2n$ points is the same as the number of red points that can be colored starting from the $k^*$-th point on a circle with $n$ points, and they correspond one-to-one.
(2) On a circle with $2n$ points, if the first red point is an odd-numbered point, then the second point colored must be an even-numbered point, such as the $2k$-th point. According to the discussion in (1), this also corresponds to a coloring method starting from the $k$-th point on a circle with $n$ points, and the number of red points colored is exactly one more than the number of red points obtained by this coloring method on a circle with $n$ points.
(3) Let the maximum number of red points that can be colored on a circle with $n$ points be $f(n)$, and suppose that the number of red points colored starting from the $k$-th point reaches this maximum value. Then, on a circle with $2n$ points, the number of red points colored starting from the $k$-th point should reach the maximum value $f(n)+1$ (since the second red point is the $2k$-th point).
Thus, $f(2n) = f(n) + 1$.
If $n = 2^k m$, from the above formula we get
$f(2^k m) = f(m) + k$.
Since $800 = 2^5 \times 25$, we have
$f(800) = f(25) + 5$.
Next, we calculate $f(25)$.
We only need to note that on a circle with 25 points, if the first red point is a multiple of 5, then every red point colored afterward will also be a multiple of 5. Therefore, starting from such a point, at most five red points can be colored.
If the first red point is not a multiple of 5, then every red point colored afterward will also not be a multiple of 5. Therefore, starting from such a point, at most 20 red points can be colored.
On the other hand, starting from any such point, 20 red points can be colored. For example, starting from the 1st point, we can color in sequence
$$
1, 2, 4, 8, 16, 7, 14, 3, 6, 12, 24, 23, 21, 17, 9,
$$
$18, 11, 22, 19, 13$.
Therefore, $f(25) = 20$.
Hence, from equation (1) we get
$$
f(800) = 20 + 5 = 25,
$$
which means at most 25 red points can be colored.
|
25
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Try to find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason.
|
Given that $f(x)=x^{12}$ is a convex function, then $\sum_{i=1}^{197} x_{i}^{12}$ achieves its maximum value when at least 1996 of $x_{1}, x_{2}, \cdots, x_{1997}$ are equal to $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.
Assume that there are $t$ instances of $-\frac{1}{\sqrt{3}}$ and $1996-t$ instances of $\sqrt{3}$. Then the remaining one is
$$
\begin{array}{l}
-318 \sqrt{3}-\left[-\frac{1}{\sqrt{3}} t+(1996-t) \sqrt{3}\right] \\
=\frac{4}{\sqrt{3}} t-2314 \sqrt{3} .
\end{array}
$$
From the given information,
$$
\begin{array}{l}
-\frac{1}{\sqrt{3}} \leqslant \frac{4}{\sqrt{3}} t-2314 \sqrt{3} \leqslant \sqrt{3} \\
\Rightarrow 1735.25 \leqslant t \leqslant 1736.25 .
\end{array}
$$
Noting that $t \in \mathbf{N}$, hence $t=1736$.
Thus, $1996-t=260$.
Therefore, according to the proposition, when $x_{1}, x_{2}, \cdots, x_{1997}$ have 1736 instances of $-\frac{1}{\sqrt{3}}$, 260 instances of $\sqrt{3}$, and one instance of $\frac{2 \sqrt{3}}{3}$, $\sum_{i=1}^{1997} x_{i}^{12}$ achieves its maximum value, which is
$$
\begin{array}{l}
\left(-\frac{1}{\sqrt{3}}\right)^{12} \times 1736+(\sqrt{3})^{12} \times 260+\left(\frac{2 \sqrt{3}}{3}\right)^{12} \\
=189548 .
\end{array}
$$
|
189548
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$
then $a b c=$ . $\qquad$
|
$$
\text { Two, 1. }-9 \text {. }
$$
From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$
Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $a, b$ are positive integers, and satisfy $5a+7b=50$, then $ab=$ . $\qquad$
|
3. 15 .
From the problem, we know
$$
a=\frac{50-7 b}{5}=10-b-\frac{2 b}{5} \text { . }
$$
Therefore, $a=3, b=5$.
Thus, $a b=15$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$
|
4.3.
Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$
Since $S_{\triangle FBC}=2 S_{\triangle DBC}$, we have $\frac{S_{\triangle ABC}}{S_{\triangle DBC}}=\frac{4}{3}$. Thus, $S_{\triangle BDC}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the function $y=f(x)$ is defined on $\mathbf{R}$, with a period of 3, and Figure 1 shows the graph of the function in the interval $[-2,1]$. Then $\frac{f(2014)}{f(5) f(15)}=$ $\qquad$ .
|
$$
\text { II,6. }-2 \text {. }
$$
From the problem statement and combining with the graph, we know that
$$
\frac{f(2014)}{f(5) f(15)}=\frac{f(1)}{f(-1) f(0)}=\frac{2}{(-1) \times 1}=-2 \text {. }
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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