problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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6. Let the complex number $z$ satisfy $|z|=1$. Then
$$
|(z+1)+\mathrm{i}(7-z)|
$$
cannot be ( ).
(A) $4 \sqrt{2}$
(B) $4 \sqrt{3}$
(C) $5 \sqrt{2}$
(D) $5 \sqrt{3}$ | 6. D.
Notice,
$$
\begin{array}{l}
|(z+1)+\mathrm{i}(7-z)|=|1-\mathrm{i}||z-3+4 \mathrm{i}| \\
=\sqrt{2}|z-(3-4 \mathrm{i})| .
\end{array}
$$
Since $z$ is on the unit circle, the range of the above expression is $[4 \sqrt{2}, 6 \sqrt{2}]$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,001 |
Example 6 Let $k$ be a given positive integer. Try to find all positive integers $a$, such that there exists a positive integer $n$, satisfying $n^{2} \mid\left(a^{n}-1\right)$ and $n$ has exactly $k$ distinct prime factors. | When $a=1$, let
$n=p_{1} p_{2} \cdots p_{k}\left(p_{1}<p_{2}<\cdots<p_{k}\right)$, then $n \times\left(2^{n}-1\right)$.
Proof by contradiction.
If $n \mid\left(2^{n}-1\right)$, then $n$ must be odd.
Let $q$ be the smallest prime factor of $n$. Then $q \mid\left(2^{n}-1\right)$ (since $q$ is odd).
By Fermat's Little The... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,002 |
1. Given functions
$$
f(x)=\sin x, g(x)=\sqrt{\pi^{2}-x^{2}}
$$
have the domain $[-\pi, \pi]$. The area of the region enclosed by their graphs is $\qquad$ . | $$
\text { II.1. } \frac{\pi^{3}}{2} \text {. }
$$
If the graph of $y=\sqrt{\pi^{2}-x^{2}}$ is completed to form a full circle, then by central symmetry, it is easy to see that the desired area is half of the area of the circle. | \frac{\pi^{3}}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 731,003 |
2. If $a$ is a positive real number, and
$$
f(x)=\log _{2}\left(a x+\sqrt{2 x^{2}+1}\right)
$$
is an odd function, then the solution set for $f(x)>\frac{3}{2}$ is | 2. $\left(\frac{7}{8},+\infty\right)$.
Given $f(x)+f(-x)=0$
$$
\begin{array}{l}
\Rightarrow\left(a x+\sqrt{2 x^{2}+1}\right)\left(-a x+\sqrt{2 x^{2}+1}\right)=1 \\
\Rightarrow\left(2-a^{2}\right) x^{2}=0 \Rightarrow a=\sqrt{2} .
\end{array}
$$
Since $\sqrt{2} x+\sqrt{2 x^{2}+1}$ is an increasing function on the inter... | \left(\frac{7}{8},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,004 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. For example, $[3]=3,[\sqrt{2}]=1,[-\pi]=-4$. Let $x$ be a positive real number. If $\left[\log _{2} x\right]$ is even, then $x$ is called a lucky number. Then the probability that a number randomly selected from the interval $(0,1)$ is a lucky ... | 3. $\frac{1}{3}$.
Notice that when $x \in (0,1)$, $\log _{2} x < 0$. Therefore, $\left[\log _{2} x\right]$ is even if and only if
$$
\begin{array}{l}
\log _{2} x \in \bigcup_{k=1}^{\infty}[-2 k, 1-2 k) \\
\Leftrightarrow x \in \bigcup_{k=1}^{\infty}\left[2^{-2 k}, 2^{1-2 k}\right) .
\end{array}
$$
The sum of the leng... | \frac{1}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,005 |
4. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=20$. Then the maximum value of $x y+8 x+y$ is $\qquad$ . | 4. 42 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x y+8 x+y)^{2} \\
\leqslant\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right) \\
=84 \times 21=42^{2} .
\end{array}
$$
Therefore, the maximum value sought is 42. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,006 |
5. Given a convex hexagon $A B C D E F$ with six equal side lengths, the interior angles $\angle A$, $\angle B$, and $\angle C$ are $134^{\circ}$, $106^{\circ}$, and $134^{\circ}$, respectively. Then the measure of the interior angle $\angle E$ is | 5. $134^{\circ}$.
Let's assume the side length is $1$, and the midpoints of $AC$ and $DF$ are $M$ and $N$, respectively, and the projection of point $A$ on $DF$ is $K$. Then
$$
\begin{array}{l}
\angle B A M=37^{\circ}, \angle M A F=97^{\circ}, \\
\angle A F K=83^{\circ} .
\end{array}
$$
Thus, $F K=\cos 83^{\circ}, K ... | 134^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,007 |
1. Given that $a, b, x, y$ are positive real numbers, satisfying:
$$
a+b+\frac{1}{a}+\frac{9}{b}=8, a x^{2}+b y^{2}=18 \text {. }
$$
Then the range of values for $a x+b y$ is . $\qquad$ | \begin{array}{l}\text { I. 1. }(3 \sqrt{2}, 6 \sqrt{2}] . \\ \text { From } 8=a+b+\frac{1}{a}+\frac{9}{b} \\ =\left(a+\frac{1}{a}\right)+\left(b+\frac{9}{b}\right) \\ \geqslant 2+2 \sqrt{9}=8 \\ \Rightarrow a=1, b=3 \\ \Rightarrow \frac{x^{2}}{18}+\frac{y^{2}}{6}=1 . \\ \text { Let } x=3 \sqrt{2} \cos \theta, y=\sqrt{6... | (3 \sqrt{2}, 6 \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,008 |
2. Let vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfy $|\boldsymbol{a}|=|\boldsymbol{b}|=1, \boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{2}$, and the angle between vectors $\boldsymbol{a}-\boldsymbol{c}$ and $\boldsymbol{c}-\boldsymbol{b}$ is $\frac{\pi}{3}$. Then the maximum value of $|\boldsymbol{c}... | 2. $\frac{2 \sqrt{3}}{3}$.
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. We discuss in two cases.
(1) As shown in Figure 2, points $O, A, C, B$ are concyclic, and point $C$ lies on the minor arc $\overparen{A B}$. At this time, $O C$ is the diameter of the circle, and
$$
|c|=|\overright... | \frac{2 \sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,009 |
3. In the Cartesian coordinate system $x O y$, two circles both pass through the point $(1,1)$, and are tangent to the line $y=\frac{4}{3} x$ and the $x$-axis. Then the sum of the radii of the two circles is . $\qquad$ | 3. $\frac{3}{2}$.
Let the equations of the two circles be
$$
\left(x-a_{i}\right)^{2}+\left(y-r_{i}\right)^{2}=r_{i}^{2}(i=1,2) \text {. }
$$
Since both circles pass through the point $P(1,1)$, we have
$$
\begin{array}{l}
\left(1-a_{i}\right)^{2}+\left(1-r_{i}\right)^{2}=r_{i}^{2} \\
\Rightarrow a_{i}^{2}-2 a_{i}-2 r... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,010 |
4. In $\triangle A B C$, the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. If $a^{2}+2\left(b^{2}+c^{2}\right)=2 \sqrt{2}$, then the maximum value of the area of $\triangle A B C$ is $\qquad$ | 4. $\frac{1}{4}$.
Let $M$ be the midpoint of $B C$.
By the median length formula, we have
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a A M \sin \angle A M C \\
\leqslant \frac{1}{2} a A M=\frac{a}{4} \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}} \\
=\frac{a}{4} \sqrt{2 \sqrt{2}-2 a^{2}} \\
=\frac{1}{4} \sqrt{\frac{2... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,011 |
5. In a regular tetrahedron $P-ABC$, points $E$ and $F$ are on edges $PB$ and $PC$, respectively. If $PE \neq PF$, and $AE = AF = 2$, $EF = \sqrt{3}$, then the volume of the tetrahedron $P-AEF$ is $\qquad$ | 5. $\frac{1}{8}$.
As shown in Figure 4.
By the problem, let's assume the height of the regular tetrahedron is $h$,
$$
\begin{array}{l}
P F=B E=x, \\
P E=F C=y .
\end{array}
$$
Then the edge length is $x+y$.
Thus, in $\triangle P E F$,
$$
x^{2}+y^{2}-x y=3 \text {; }
$$
In $\triangle P A F$, $x^{2}+y^{2}+x y=4$.
From... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,012 |
6. Given the inequality $\left|a x^{2}+b x+a\right| \leqslant x$ holds for $x \in$ $[1,2]$. Then the maximum value of $3 a+b$ is $\qquad$ | 6. 3 .
From the problem, we know that $\left|a\left(x+\frac{1}{x}\right)+b\right| \leqslant 1$.
Given $x \in[1,2]$, we have $t=x+\frac{1}{x} \in\left[2, \frac{5}{2}\right]$.
Thus, $|2 a+b| \leqslant 1$, and $\left|\frac{5}{2} a+b\right| \leqslant 1$.
Therefore, $3 a+b=2\left(\frac{5}{2} a+b\right)-(2 a+b) \leqslant 3$... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,013 |
7. In the sequence
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]
$$
there are $\qquad$ distinct integers ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.1515.
Let the $k$-th term of the known sequence be $\left[\frac{k^{2}}{2019}\right]$. Then, when $(k+1)^{2}-k^{2} \leqslant 2019$, i.e., $k \leqslant 1009$,
$$
\begin{array}{l}
\frac{(k+1)^{2}}{2019}=\frac{k^{2}}{2019}+\frac{2 k+1}{2019} \leqslant \frac{k^{2}}{2019}+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right... | 1515 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,014 |
8. On each face of a cube, randomly fill in one of the numbers 1, 2, $\cdots$, 6 (the numbers on different faces are distinct). Then number the eight vertices such that the number assigned to each vertex is the product of the numbers on the three adjacent faces. The maximum value of the sum of the numbers assigned to t... | 8. 343 .
Let the numbers on the six faces be $a, b, c, d, e, f$, and $(a, b), (c, d), (e, f)$ be the numbers on the opposite faces. Thus, the sum of the numbers at the eight vertices is
$$
\begin{array}{l}
(a+b)(c+d)(e+f) \\
\leqslant\left(\frac{(a+b)+(c+d)+(e+f)}{3}\right)^{3} \\
=7^{3}=343 .
\end{array}
$$
When $a=... | 343 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,015 |
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and $S_{n}=2 a_{n}-2$, the sequence $\left\{b_{n}\right\}$ satisfies
$$
S_{n} b_{n}=\frac{1}{2^{n}}\left(1+\sum_{k=1}^{n} \mathrm{C}_{n}^{k} a_{k}\right) .
$$
Prove: $\sum_{k=1}^{n} b_{k} \leqslant 3\left(1-\left(\... | When $n=1$,
from $S_{1}=2 a_{1}-2 \Rightarrow a_{1}=2$.
When $n \geqslant 2$,
from $S_{n}=2 a_{n}-2 \Rightarrow S_{n-1}=2 a_{n-1}-2$.
Subtracting the two equations gives
$$
\begin{array}{l}
a_{n}=2 a_{n}-2 a_{n-1} \Rightarrow a_{n}=2 a_{n-1} \\
\Rightarrow a_{n}=2^{n}, S_{n}=2^{n+1}-2 \\
\Rightarrow\left(2^{n+1}-2\righ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 731,016 |
10. (20 points) Given complex numbers $z_{1}, z_{2}, z_{3}$ with arguments $\alpha, \beta, \gamma$ respectively, and satisfying:
$$
\left|z_{1}\right|=1,\left|z_{2}\right|+\left|z_{3}\right|=2, z_{1}+z_{2}+z_{3}=0 .
$$
Find the range of $\cos (\alpha-\beta)+2 \cos (\beta-\gamma)+3 \cos (\gamma-\alpha)$. | 10. Since $z_{2}=-\left(z_{1}+z_{3}\right)$, we have
$$
\begin{array}{l}
\left|z_{2}\right|=\left|z_{1}+z_{3}\right|=2-\left|z_{3}\right| . \\
\text { By }\left|z_{3}\right|-\left|z_{1}\right| \leqslant\left|z_{1}+z_{3}\right| \leqslant\left|z_{1}\right|+\left|z_{3}\right| \\
\Rightarrow\left|z_{3}\right|-1 \leqslant 2... | [-4,0] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,017 |
11. (20 points) In the Cartesian coordinate system $x O y$, $A$ is a point on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, $M$ is a moving point on the line segment $O A$, and a line through $M$ intersects the ellipse at points $P$ and $Q$. If $\overrightarrow{P M}=2 \overrightarrow{M Q}$, find the maximum value of the area ... | 11. As shown in Figure 5.
Let \( P(2 \cos \alpha, \sin \alpha), Q(2 \cos \beta, \sin \beta) \).
Since \( \overrightarrow{P M}=2 \overrightarrow{M Q} \), we have
\[
x_{M}=\frac{2 \cos \alpha+4 \cos \beta}{3}, y_{M}=\frac{\sin \alpha+2 \sin \beta}{3} \text{. }
\]
Let \( \overrightarrow{O A}=u \overrightarrow{O M} (u>1)... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,018 |
Question 2 Let $R, S$ be two distinct points on a circle $\Gamma$, and $RS$ is not a diameter. Let $l$ be the tangent line to circle $\Gamma$ at point $R$. A point $T$ in the plane satisfies that $S$ is the midpoint of segment $RT$, and $J$ is a point on the minor arc $\overparen{RS}$ of circle $\Gamma$ such that the c... | Prove as shown in Figure 2, connect $A S, S K$, extend $K S$, and intersect $A T$ at point $P$, connect $P R$.
Given that $A, T, S, J$ and $R, K, S, J$ are each four points on a circle
$\Rightarrow \angle K R S = \angle K J S = \angle A T S$
$\Rightarrow A T \parallel R K \Rightarrow \angle R K S = \angle T P S$.
Also,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,019 |
One, (40 points) Let $n \in \mathbf{Z}, n \geqslant 2$, and non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Find
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)
$$
the maximum and minimum values. | Obviously, $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right) \geqslant 0$, when
$$
\left(x_{1}, x_{2}, x_{3}, \cdots, x_{n}\right)=(1,0,0, \cdots, 0)
$$
the equality holds.
Thus, the minimum value sought is 0.
$$
\begin{array}{l}
\text { Let } F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \le... | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,020 |
II. (40 points) As shown in Figure 1, quadrilateral $ABCD$ is inscribed in $\odot O, AC$ intersects $BD$ at point $M$, line $AD$ intersects $BC$ at point $N, \angle ANB=90^{\circ}, K$ is the point symmetric to $O$ with respect to $AB$. Prove: $OM \parallel KN$. | When $N A=N B$, the conclusion is obvious.
When $N A \neq N B$, let $A B$ intersect $C D$ at point $P$, and let $Q$ be the midpoint of $A B$. Then $Q$ is also the midpoint of $O K$, as shown in Figure 6.
By the properties of a cyclic quadrilateral, we have
$$
\begin{array}{l}
O M \perp N P, \\
N P^{2}=P A \cdot P B+N C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,021 |
Three. (50 points) A positive integer $N(N \geqslant 2)$ is called "good" if $N$ can be expressed as the product of an even number of primes (which can be the same). Question: Do there exist distinct positive integers $a_{1}, a_{2}, \cdots, a_{2018}$ such that the polynomial
$$
P(x)=\left(x+a_{1}\right)\left(x+a_{2}\ri... | Three, the conclusion is affirmative.
Positive integers that can be expressed as the product of an odd number of primes (which can be the same) are called "bad".
Let set $A$ represent all good positive integers, set $B$ represent all bad positive integers, and $C=\{n \mid n \geqslant 2, n \in \mathbf{Z}\}$.
By definit... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,022 |
Four. (50 points) A planet has 1000 cities $c_{1}$, $c_{2}, \cdots, c_{1000}$, and three airlines $X$, $Y$, and $Z$ provide flights between these cities. For any $1 \leqslant i<j \leqslant 1000$, exactly one airline operates a one-way flight from city $c_{i}$ to city $c_{j}$. Find the largest positive integer $n$, such... | Four, the maximum value of $n$ is 9.
For each city $c_{i}(i=1,2, \cdots, 1000)$, define a triplet of non-negative integers $\left(x_{i}, y_{i}, z_{i}\right)$ according to the following rules:
If there are no flights from company $X$ arriving at city $c_{i}$, set $x_{i}=0$; otherwise, there exists a largest positive in... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,023 |
2. Given that $G$ is the centroid of $\triangle A B C$ with an area of 24, and $D, E$ are the midpoints of sides $A B, B C$ respectively. Then the area of $\triangle D E G$ is $(\quad$.
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B.
$$
S_{\triangle D E G}=\frac{1}{4} S_{\triangle A C G}=\frac{1}{12} S_{\triangle A B C}=2 .
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,024 |
3. A super balance uses weights of $1, 3, 9, 27, 81, 243, 729$ grams, with three of each weight. To weigh 2019 grams of iron particles, the minimum number of weights needed is ( ) .
(A) 6
(B) 7
(C) 8
(D) 9 | 3. B.
From $3 \times 729-2 \times 81-9+3=2019$, we know the weights used are: 3 weights of 729 grams, 2 weights of 81 grams, 1 weight of 9 grams, and 1 weight of 3 grams, totaling 7 weights. Among them, the weights representing the subtracted grams are placed in the tray containing the iron particles. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 731,025 |
6. As shown in Figure 3, the incircle $\odot O$ of the right triangle $\triangle ABC$ touches the hypotenuse $AB$, the legs $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. Perpendiculars are drawn from point $D$ to $AC$ and $BC$, with the feet of the perpendiculars being $M$ and $N$. If the area of rectangle ... | 6. A.
From the problem, let $O D=O E=O F=r$,
$$
\begin{array}{l}
A D=A F=m, B D=B E=n, C E=C F=r . \\
\text { Then } S=S_{\triangle A B C}=\frac{1}{2}(r+m)(r+n) \\
=\frac{1}{2}(r(r+m+n)+m n)=\frac{1}{2}(S+m n) \\
\Rightarrow S=m n .
\end{array}
$$
By $D M / / B C \Rightarrow \triangle A D M \backsim \triangle A B C$
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,026 |
Question 3 Given $n, k \in \mathbf{Z}_{+}, n>k$. Given real numbers $a_{1}, a_{2}, \cdots, a_{n} \in(k-1, k)$. Let positive real numbers $x_{1}, x_{2}, \cdots$, $x_{n}$ satisfy that for any set $I \subseteq\{1,2, \cdots, n\}$, $|I|=k$, there is $\sum_{i \in I} x_{i} \leqslant \sum_{i \in I} a_{i}$. Find the maximum val... | Solve the maximum value as $a_{1} a_{2} \cdots a_{n}$.
Let $U=\{1,2, \cdots, n\}, d_{i}=a_{i}-x_{i}$.
From the problem's condition, we know that for any $I \subseteq U,|I|=k$ we have
$$
\sum_{i \in I} d_{i} \geqslant 0 .
$$
Notice that,
$$
\begin{array}{l}
\prod_{i=1}^{n} x_{i}=\left(\prod_{i=1}^{n} a_{i}\right)\left(... | a_{1} a_{2} \cdots a_{n} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,027 |
2. A trading company purchases a certain type of fruit at a cost of 20 yuan per kilogram. After market research, it is found that the selling price $p$ (yuan per kilogram) of this fruit over the next 48 days is related to time $t$ (days) by the function
$$
p=\frac{1}{4} t+30(1 \leqslant t \leqslant 48, t \in \mathbf{Z}... | 2. 8. $75<n \leqslant 9.25$.
According to the problem, we have
$$
\begin{array}{l}
W=(p-20-n) y \\
=\left(\frac{1}{4} t+30-20-n\right)(120-2 t) \\
=-\frac{1}{2} t^{2}+2(n+5) t+120(10-n),
\end{array}
$$
The axis of symmetry is $t=2 n+10$.
From the problem, we know
$$
\begin{array}{l}
27.5<2 n+10 \leqslant 28.5 \\
\Rig... | 8.75 < n \leqslant 9.25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,028 |
4. Given four positive integers
$a, b, c, d$ satisfy:
$$
a^{2}=c(d+20), b^{2}=c(d-18) \text {. }
$$
Then the value of $d$ is $\qquad$ | 4. 180 .
Let $(a, b)=t, a=t a_{1}, b=t b_{1}$.
Then $\frac{d+20}{d-18}=\frac{c(d+20)}{c(d-18)}=\frac{a^{2}}{b^{2}}=\frac{a_{1}^{2}}{b_{1}^{2}}$ (simplest fraction).
Let $d+20=k a_{1}^{2}, d-18=k b_{1}^{2}$.
Eliminating $d$ yields
$$
\begin{array}{l}
k\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right)=2 \times 19 \\
\Rig... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,029 |
II. (25 points) As shown in Figure 5, circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. $AC$ is the diameter of $\odot O_{1}$, and $D$ is a point on arc $\overparen{BC}$. Lines $CD$ and $AD$ intersect $\odot O_{2}$ at points $E$ and $H$, respectively. The midpoint $F$ of $HE$ lies on line $BC$. P... | Extend $C F$, intersecting $\odot O_{2}$ at point $G$.
From $\angle A B C=\angle A B G=90^{\circ}$, we get $A G$ as the diameter of $\odot O_{2}$.
Thus, $A E \perp E G, A H \perp H G$.
Also, $A D \perp C D$, hence, $C D / / H G$.
It is easy to prove $\triangle C E F \cong \triangle G H F \Rightarrow C F=F G$.
Therefore... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,030 |
2. Planar Region
$$
\left\{(x, y) \mid \sqrt{1-x^{2}} \sqrt{1-y^{2}} \geqslant x y\right\}
$$
The area of the region is $\qquad$ . | 2. $2+\frac{\pi}{2}$.
From the problem, we know that $x \in[-1,1], y \in[-1,1]$. If $x y \leqslant 0$, then the inequality holds; if $x y>0$, then $1-x^{2}-y^{2} \geqslant 0$, i.e., $x^{2}+y^{2} \leqslant 1$. By drawing the graph, we can see that the area is $2+\frac{\pi}{2}$. | 2+\frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,031 |
3. Given that $a, b, c, d$ are positive integers, and $\log _{a} b=\frac{3}{2}, \log _{c} d=\frac{5}{4}, a-c=9$.
Then $a+b+c+d=$ $\qquad$ | 3. 198 .
Given $a=x^{2}, b=x^{3}, c=y^{4}, d=y^{5}$.
$$
\begin{array}{l}
\text { Given } a-c=x^{2}-y^{4}=9 \\
\Rightarrow\left(x+y^{2}\right)\left(x-y^{2}\right)=9 \\
\Rightarrow x+y^{2}=9, x-y^{2}=1 \\
\Rightarrow x=5, y^{2}=4 \\
\Rightarrow a=25, b=125, c=16, d=32 \\
\Rightarrow a+b+c+d=198 .
\end{array}
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,032 |
1. Let $c$ be a real number. If there exists $x \in [1,2]$, such that
$$
\max \left\{\left|x+\frac{c}{x}\right|,\left|x+\frac{c}{x}+2\right|\right\} \geqslant 5 \text{, }
$$
find the range of values for $c$.
(Yang Xiaoming, provided) | 1. Note that,
$$
\begin{array}{l}
\max \left\{\left|x+\frac{c}{x}\right|,\left|x+\frac{c}{x}+2\right|\right\} \\
=\left|x+\frac{c}{x}+1\right|+1 .
\end{array}
$$
The given condition can be transformed into: there exists $x \in[1,2]$, such that $\left|x+\frac{c}{x}+1\right| \geqslant 4$.
Let $g(x)=x+\frac{c}{x}+1(x \i... | c \in(-\infty,-6] \cup[2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,033 |
4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$ | 4.9.
From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$.
By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin ... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,034 |
5. Two players, A and B, are playing a table tennis match, with the agreement: the winner of each game gets 1 point, and the loser gets 0 points; the match stops when one player is 2 points ahead or after 6 games have been played. Suppose the probability of A winning each game is $\frac{3}{4}$, and the probability of B... | 5. $\frac{97}{32}$.
From the problem, we know that all possible values of $\xi$ are $2, 4, 6$.
Let each two games be a round. Then the probability that the match stops at the end of this round is
$$
\left(\frac{3}{4}\right)^{2}+\left(\frac{1}{4}\right)^{2}=\frac{5}{8} \text {. }
$$
If the match continues after this r... | \frac{97}{32} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,035 |
6. Construct two right triangles inscribed in the parabola $y=x^{2}$, with the point $M(1,1)$ on the parabola as the right-angle vertex: Rt $\triangle M A B$ and Rt $\triangle M C D$. Then the coordinates of the intersection point $E$ of line segments $A B$ and $C D$ are $\qquad$ | 6. $(-1,2)$.
For any point $P\left(x_{0}, y_{0}\right)$ on the parabola $x^{2}=2 p y$, draw perpendicular chords $P A$ and $P B$.
Then $A B$ always passes through the point $\left(-y_{0}, x_{0}+2 p\right)$.
Therefore, $E(-1,2)$. | (-1,2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,036 |
10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k 、 n \in \mathbf{Z}_... | 10. Let the indices $n$ that satisfy $S_{n}=0$ be arranged in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by m... | 1092 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,037 |
11. (20 points) In the Cartesian coordinate system $x O y$, the curve $C_{1}: y^{2}=4 x$,
curve $C_{2}:(x-4)^{2}+y^{2}=8$,
a line $l$ with an inclination angle of $45^{\circ}$ is drawn through a point $P$ on the curve $C_{1}$, intersecting the curve $C_{2}$ at two different points $Q$ and $R$. Find the range of $|P Q||... | 11. Let $P\left(t^{2}, 2 t\right)$. Then the line $l: y=x+2 t-t^{2}$.
Substituting into the equation of curve $C_{2}$ and rearranging, we get
$$
2 x^{2}-2\left(t^{2}-2 t+4\right) x+\left(t^{2}-2 t\right)^{2}+8=0 \text {. }
$$
Since $l$ intersects curve $C_{2}$ at two distinct points, we have
$$
\begin{array}{l}
\frac... | [4,8) \cup (8,200) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,038 |
One, (40 points) Find the smallest integer $c$, such that there exists a sequence of positive integers $\left\{a_{n}\right\}(n \geqslant 1)$ satisfying:
$$
a_{1}+a_{2}+\cdots+a_{n+1}<c a_{n}
$$
for all $n \geqslant 1$. | Given the problem, we have
$$
c>\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}}.
$$
For any \( n \geqslant 1 \), we have
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,039 |
In $\triangle A B C$, $A B=A C, O$ is the circumcenter, and $D E / / B C$ intersects $A B$ and $A C$ at points $D$ and $E$ respectively. $P$ is the circumcenter of $\triangle A D E$, and the circumcircle of $\triangle A P D$ intersects the circumcircle of $\triangle A O C$ at the second point $F$. Let $A F$ intersect $... | (1) As shown in Figure 2, connect $F D$ and $F C$.
Given $A B = A C$, $D E \parallel B C$,
$$
\Rightarrow A D = A E.
$$
$$
\text{Then } \angle P A E = \angle P A D = \angle P D A
$$
$\Rightarrow A E$ is the tangent to the circumcircle of $\triangle A P D$
$$
\Rightarrow \angle E A F = \angle A D F.
$$
Similarly, $\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,040 |
2. In the Cartesian coordinate system, if both the x-coordinate and y-coordinate of a point are rational numbers, then the point is called a "rational point"; otherwise, it is called an "irrational point". In the Cartesian coordinate system, draw any regular pentagon. Among its five vertices, which are more: rational p... | 2. In a Cartesian coordinate system, draw any regular pentagon, let its side length be $a$, and the diagonal length be $b$.
It is known that, $\frac{a}{b}=\frac{\sqrt{5}-1}{2}$.
Thus, $\frac{a^{2}}{b^{2}}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{3-\sqrt{5}}{2}$ is an irrational number.
Assume that among the five ver... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,041 |
Four. (50 points) Let $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a sequence of positive integers, satisfying: for each prime $p$, there are infinitely many terms in the sequence that are divisible by $p$. Prove that every positive rational number less than 1 can be written in the form:
$$
\frac{b_{1}}{a_{1}}+\frac{b_{2}}... | For any $n \in \mathbf{Z}_{+}, a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{Z}_{+}$, if $b_{i}(i=1,2, \cdots, n)$ runs through a complete residue system modulo $a_{i}$, then $A=$ $\sum_{i=1}^{n}\left(b_{i} \prod_{j=i+1}^{n} a_{j}\right)$ runs through a complete residue system modulo $\prod_{i=1}^{n} a_{i}$.
For any $\frac{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,042 |
Given non-negative real numbers $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=n$. Prove:
$$
\sum_{k=1}^{n} \sqrt{1+\sum_{i=1}^{k}(2 i+1) a_{k+1-i}^{2}} \leqslant \frac{n(n+3)}{2} .
$$ | To prove the conclusion:
$$
\sum_{i=1}^{n}(2 i-1) a_{n+1-i}^{2} \leqslant\left(\sum_{i=1}^{n} a_{i}\right)^{2} \text {. }
$$
We use induction on \( n \).
Indeed, when \( n=1 \), both sides of the inequality are equal. Assume the inequality holds for \( n \).
Now consider the case for \( n+1 \).
Notice that,
$$
\begin{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 731,043 |
Does there exist 2019 integer points such that the distance between any two points is irrational, and the area of any triangle formed by any three of these points is a multiple of 2019? | Consider the existence.
Take $\left(2 k \times 2019,(2 k \times 2019)^{2}\right)(k=0,1$, $\cdots, 2018)$.
Thus, the distance between points $\left(i, i^{2}\right)$ and $\left(j, j^{2}\right)$ is
$$
\begin{array}{l}
\sqrt{(i-j)^{2}+\left(i^{2}-j^{2}\right)^{2}} \\
=|i-j| \sqrt{1+(i+j)^{2}} .
\end{array}
$$
From $i+j<\s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,044 |
607 Given $E$ and $F$ are points on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BE$ and $CF$ intersect at point $P$, and $AP$ intersects $EF$ at point $Q$. A line through point $Q$ intersects $AB$, $BC$, and $CF$ at points $G$, $D$, and $M$, respectively. $CG$ and $DE$ intersect at point $N$. Prove that ... | To prove as shown in Figure 3, let line \(MN\) intersect \(AB\) and \(BC\) at points \(B_1\) and \(B_2\) respectively. To prove that points \(B\), \(M\), and \(N\) are collinear, we need to prove that
point \(B_1\) coincides with \(B_2\)
\[
\Leftrightarrow \frac{N B_1}{M B_1} = \frac{N B_2}{M B_2} \Leftrightarrow \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,045 |
For given positive integers $m, n$. Prove: there exists $k \in\{0,1, \cdots, n\}$, such that $(m+k) \nmid \mathrm{C}_{m+n-1}^{n}$. | $$
\begin{array}{l}
S_{n}=\sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} \frac{\mathrm{C}_{m+n-1}^{m}}{m+k} \\
=\frac{\mathrm{C}_{m+n-1}^{m}}{m} \sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} \frac{m}{m+k} \text {. } \\
\text { Let } a_{n}=\sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} \frac{m}{m+k} \text {. Then } \\
a_{n}=\sum_{k=0... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,046 |
Example 1 In $\triangle A B C$, $\angle A=108^{\circ}, A B=$ $A C$. Extend $A C$ to point $D$, and let $J$ be the midpoint of $B D$. Prove: When $A D=B C$, $A J \perp J C .{ }^{[1]}$ | Prove as shown in Figure 1, draw $DE \parallel BC$, and $DE = BC$, connect $EB$, $EA$, $EJ$.
Then we get $\square BCDE$.
Since $J$ is the midpoint of diagonal $BD$, therefore, $J$ is also the midpoint of diagonal $CE$.
$$
\begin{array}{l}
\text { By } \angle EDA = \angle BCA = \angle ABC = 36^{\circ}, \\
ED = BC = AD, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,047 |
5. As shown in Figure $9, D$ is a point inside $\triangle A B C, B D$ intersects $A C$ and $C D$ intersects $A B$ at points $E, F$ respectively, and $A F=B F=$ $C D, C E=D E$. Find the degree measure of $\angle B F C$. | Prompt: On $F C$, intercept $F G=A F$, and connect $A G$. Then $G C=D F, \angle E C D=\angle E D C=\angle F D B$.
For $\triangle A B E$ and the transversal $F D C$, applying Menelaus' theorem yields $A C=D B$.
Therefore, $\triangle A C G \cong \triangle B D F, A G=B F, \triangle A F G$ is an equilateral triangle, $\a... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,048 |
Example 1 Given that the incircle $\Gamma$ of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $A'$, $B'$, and $C'$ respectively, and points $K$, $L$ on the circle satisfy
$$
\begin{array}{l}
\angle A K B^{\prime}+\angle B K A^{\prime} \\
=\angle A L B^{\prime}+\angle B L A^{\prime}=180^{\circ} .
\end{array... | For a point $P$ on the minor arc $\overparen{A^{\prime} B^{\prime}}$ of circle $\Gamma$, we have $\angle A P B^{\prime}+\angle B P A^{\prime}<\angle A^{\prime} P B^{\prime}<180^{\circ}$.
From this, we know that points $K$ and $L$ are both on the major arc $A^{\prime} C^{\prime} B^{\prime}$.
By the given conditions,
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,049 |
Example 2 Let circle $\Gamma_{1}$ be the circumcircle of $\triangle ABC$, and let $D, E$ be the midpoints of arcs $\overparen{AB}$ (not containing point $C$) and $\overparen{AC}$ (not containing point $B$), respectively. The incircle of $\triangle ABC$ touches sides $AB$ and $AC$ at points $F$ and $G$, respectively. Li... | Proof As shown in Figure 11, let $I$ be the incenter of $\triangle ABC$, and line $DE$ intersects $AB$ and $AC$ at points $K$ and $L$ respectively.
Obviously, $B, I, E$ and $C, I, D$ are collinear respectively.
Since $\angle IBK = \angle CBI = \angle CBE$
$= \angle CDE = \angle IDK$,
we know that $B, I, K, D$ are conc... | AB = AC | Geometry | proof | Yes | Yes | cn_contest | false | 731,050 |
Example 3 Given that $M$ is any point on side $BC$ of $\triangle ABC$, circle $\Gamma$ is tangent to segments $AB$, $BM$ at points $T$, $K$ respectively, and is externally tangent to the circumcircle of $\triangle AMC$ at point $P$. Prove: If $TK \parallel AM$, then the circumcircles of $\triangle APT$ and $\triangle K... | Proof as shown in Figure 12.
From the problem, we know that the angle bisector of $\angle ABC$ is perpendicular to $TK$.
Since $TK \parallel AM$, the angle bisector of $\angle ABC$ is also perpendicular to $AM$.
Therefore, $BM = BA$.
Thus, the circumcenter of $\triangle AMC$ also lies on the angle bisector of $\angle A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,051 |
Example 4 In a cyclic quadrilateral $ABCD$, $AD = BD$, $M$ is the intersection of the two diagonals of the quadrilateral, $N$ is a point on diagonal $AC$ different from $M$, and $M, N, B,$ and the incenter $I$ of $\triangle BCM$ are concyclic. Prove:
$$
AN \cdot NC = CD \cdot BN .{ }^{[4]}
$$ | Proof as shown in Figure 13.
Given $A D=B D$, we can set
$\angle B A D=\angle D B A=\alpha$.
Then $\angle A C B=\angle A D B$
$$
=180^{\circ}-2 \alpha.
$$
Since points $M, N, B, I$ are concyclic, applying property 3(3) we know
$$
C N=C B.
$$
Connecting $B N$. Then $\angle C N B=\alpha$.
Thus, $\angle A N B=180^{\circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,052 |
Example 5 Given an acute triangle $\triangle ABC$ with the circumcircle $\odot O$ of radius $R, AB < AC < BC$, the incircle $\odot I$ of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. The circumcircles of $\triangle AEF$, $\triangle BDF$, and $\triangle CDE$ intersect $\odo... | Proof as shown in Figure 14.
(1) It is proved by Corollary 4(2).
(2) By Corollary 4(2), we know that points $D, E, A^{\prime}, B^{\prime}$ are concyclic, denoted as circle $\omega_{1}; E, F, B^{\prime}, C^{\prime}$ are concyclic, denoted as circle $\omega_{2}; D, F, A^{\prime}, C^{\prime}$ are concyclic, denoted as cir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,053 |
Example 6 As shown in Figure 15, in isosceles $\triangle ABC$, $AB=AC > BC$, and $D$ is a point inside $\triangle ABC$ such that $DA=DB+DC$. The perpendicular bisector of side $AB$ intersects the external angle bisector of $\angle ADB$ at point $P$, and the perpendicular bisector of side $AC$ intersects the external an... | Prove that since point $P$ lies on the perpendicular bisector of $AB$, we have $PA = PB$.
Point $P$ is also on the external angle bisector of $\angle ADB$. By Corollary 1, we know that points $A, B, D, P$ are concyclic.
Extend $DB$ to point $B_1$ such that $BB_1 = DC$.
Then $DA = DB_1$.
Since $\angle APB = \angle ADB =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,054 |
Question 1 In $\square A B C D$, $\angle D A C=90^{\circ}$, a perpendicular line from point $A$ to $D C$ is drawn, with the foot of the perpendicular being $H$. Point $P$ is on line $A C$ such that $P D$ is tangent to the circumcircle of $\triangle A B D$. Prove: $\angle P B A=\angle D B H$.
---
The translation maint... | Proof 1 As shown in Figure 1, extend $DP$, and let the second intersection point with the circumcircle of $\triangle APB$ be $J$. Connect $BJ$ and $AJ$.
Since $PD$ is tangent to the circumcircle of $\triangle ABD$ at point $D$, we have
$\angle PDA = \angle DBA$.
Also, in quadrilateral $ABCD$, $\angle DBA = \angle BDH$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,055 |
Question 2 Let $[x]$ denote the greatest integer not exceeding the real number $x$. Suppose $a, b$ are two distinct positive real numbers such that for any positive integer $n$, we have $[n a] \mid [n b]$. Prove: $a, b$ are both integers. ${ }^{[1]}$ | In [1], the proof is given using Wely's criterion. This paper provides an elementary proof.
Proof Let $\{x\}=x-[x]$.
Notice that, $x_{n}=\frac{[n b]}{[n a]}=\frac{n b-\{n b\}}{n a-\{n a\}} \in \mathbf{Z}_{+}$.
And $\{n a\} \in[0,1),\{n b\} \in[0,1)$, so when $n \rightarrow+\infty$, $x_{n} \rightarrow \frac{b}{a}$.
Sin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,056 |
2. If a set of three positive integers is composed of the side lengths of a right-angled triangle, it is called a "Pythagorean triplet set." For example, $\{6,8,10\}$ is a Pythagorean triplet set. Let $P$ and $Q$ be any two Pythagorean triplet sets. Prove: there exists an integer $m \geqslant 2$, and $m$ Pythagorean tr... | 2. First, prove a lemma.
Lemma For each integer $n(n \geqslant 3)$, there exists a Pythagorean triple containing $n$.
Proof By induction on $n$.
Since $\{3,4,5\}$ is a Pythagorean triple, the conclusion holds for $n=3,4,5$.
Assume $n \geqslant 6$, and the conclusion holds for $3 \leqslant k<n$.
If $n$ is even, let $n=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,057 |
4. Figure 2 is an ellipse with different lengths of major and minor axes.
(1) Prove: The smallest area circumscribed rhombus of this ellipse is unique;
(2) Please write down the process of constructing this rhombus using a ruler and compass.
(Yao Yijun provided the question) | 4. (1) Suppose in a coordinate system with the center of the ellipse as the origin and the symmetry axes as the coordinate axes, the equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
By the central symmetry of the ellipse, the points of tangency on the opposite sides of the c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,058 |
5. Given a positive integer $n$, fill each cell of an $n \times n$ grid with an integer. Perform the following operation on this grid: select a cell, and add 1 to each of the $2n-1$ cells in the selected cell's row and column. Find the maximum integer $N=N(n)$, such that no matter what the initial numbers in the grid a... | 5. Let the selected cell be in the $i$-th row and $j$-th column, and denote the operation in the problem as $M(i, j)$.
First, note that the result of a series of operations is independent of the order of the operations.
When $n$ is even, consider $2n-1$ operations $M(p, q)$ (where $p=i$ or $q=j$). These operations af... | N=n^{2}-n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,059 |
2. If the two foci of the hyperbola $C$ are exactly the two vertices of the ellipse $\Gamma$ : $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$, and the two vertices of the hyperbola $C$ are exactly the two foci of the ellipse $\Gamma$, then the equation of the hyperbola $C$ is $\qquad$ | 2. $\frac{x^{2}}{7}-\frac{y^{2}}{9}=1$.
According to the conditions, we know that the center of the hyperbola $C$ is at the origin, and its real axis of symmetry is the $x$-axis. Let its equation be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Then its vertices are $( \pm a, 0)$, and its foci are $( \pm c, 0)$.
The ve... | \frac{x^{2}}{7}-\frac{y^{2}}{9}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,062 |
3. The range of the function $y=\sqrt{7-x}+\sqrt{9+x}$ is the interval | 3. $[4,4 \sqrt{2}]$.
Obviously, the domain of the function is $x \in[-9,7]$, within this interval $y>0$.
Since $(7-x)+(9+x)=16$, that is
$$
\frac{7-x}{16}+\frac{9+x}{16}=1 \text {, }
$$
there exists $\alpha \in\left[0, \frac{\pi}{2}\right]$, such that
$$
\begin{array}{l}
\sqrt{\frac{7-x}{16}}=\sin \alpha, \sqrt{\frac... | [4,4 \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,063 |
4. If three angles $\alpha, \beta, \gamma$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $\qquad$ | 4. -3 .
According to the problem, $\alpha=\beta-\frac{\pi}{3}, \gamma=\beta+\frac{\pi}{3}$. Therefore, $\tan \alpha=\frac{\tan \beta-\sqrt{3}}{1+\sqrt{3} \tan \beta}, \tan \gamma=\frac{\tan \beta+\sqrt{3}}{1-\sqrt{3} \tan \beta}$.
Then, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,064 |
5. Let $x, y, z \in \mathbf{R}_{+}$, satisfying $x+y+z=x y z$. Then the function
$$
\begin{array}{l}
f(x, y, z) \\
=x^{2}(y z-1)+y^{2}(z x-1)+z^{2}(x y-1)
\end{array}
$$
has the minimum value of $\qquad$ | 5. 18 .
According to the conditions, we have
$$
y+z=x(y z-1) \Rightarrow y z-1=\frac{y+z}{x} \text{. }
$$
Similarly, $z x-1=\frac{z+x}{y}, x y-1=\frac{x+y}{z}$.
From $x y z=x+y+z \geqslant 3 \sqrt[3]{x y z} \Rightarrow x y z \geqslant 3 \sqrt{3}$, thus
$$
\begin{array}{l}
f(x, y, z)=2(x y+y z+z x) \\
\geqslant 2 \tim... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,065 |
6. The sequence of positive integers $\left\{a_{n}\right\}: a_{n}=3 n+2$ and $\left\{b_{n}\right\}$ $b_{n}=5 n+3(n \in \mathbf{N})$ have a common number of terms in $M=\{1,2, \cdots, 2018\}$ which is $\qquad$ | 6. 135.
It is known that 2018 is the largest common term of the two sequences within $M$. Excluding this common term, subtract 2018 from the remaining terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, to get
$$
\left\{\overline{a_{n}}\right\}=\{3,6,9, \cdots, 2016\},
$$
which are all multiples... | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,066 |
8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". Therefore, the sum of all coincidence numbers is | 8.531.
Let \( n = \overline{a_{1} a_{2} \cdots a_{k}} \left(a_{1} \neq 0\right) \).
From \( n - s(n) = p(n) \), we get
\[
\begin{array}{l}
a_{1}\left(10^{k-1}-1\right) + a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \\
= a_{1} a_{2} \cdots a_{k},
\end{array}
\]
which simplifies to \( a_{1}\left(10^{k-1}-1-a_{... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,067 |
1. Let $S_{n} 、 T_{n}$ be the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$, respectively, and for any positive integer $n$, we have
$$
\frac{S_{n}}{T_{n}}=\frac{2 n+6}{n+1} \text {. }
$$
If $\frac{a_{m}}{b_{m}}$ is a prime number, then the value of the positive i... | - 1. A.
Given the conditions
$$
S_{n}=k n(2 n+6), T_{n}=k n(n+1) \text {. }
$$
When $m=1$, $\frac{a_{1}}{b_{1}}=\frac{S_{1}}{T_{1}}=\frac{8}{2}=4$, which does not satisfy the condition in the problem, so it is discarded.
When $m \geqslant 2$,
$$
\frac{a_{m}}{b_{m}}=\frac{S_{m}-S_{m-1}}{T_{m}-T_{m-1}}=\frac{4 m+4}{2 m... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 731,068 |
2. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse such that $\angle F_{1} P F_{2}=90^{\circ}$. If the area of $\triangle P F_{1} F_{2}$ is 2, then the value of $b$ is $(\quad)$.
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt... | 2. B.
Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n$. Then
$$
\begin{array}{l}
m+n=2 a, \\
m^{2}+n^{2}=4 c^{2}=4\left(a^{2}-b^{2}\right), \\
\frac{1}{2} m n=2 .
\end{array}
$$
From the above three equations, we get $b^{2}=2$, i.e., $b=\sqrt{2}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 731,069 |
6. For any positive integer $n$, define $Z(n)$ as the smallest positive integer $m$ such that $1+2+\cdots+m$ is a multiple of $n$. Regarding the following three propositions:
(1) If $p$ is an odd prime, then $Z(p)=p-1$;
(2) For any positive integer $a$, $Z\left(2^{a}\right)>2^{a}$;
(3) For any positive integer $a$, $Z\... | 6. D.
Notice that $1+2+\cdots+m=\frac{m(m+1)}{2}$.
In proposition (1), from $p \left\lvert\, \frac{m(m+1)}{2}\right.$, we know $2 p \mid m(m+1)$.
Since $p$ is an odd prime, then $p \mid m$ or $p \mid(m+1)$. Therefore, the minimum value of $m$ is $p-1$.
Thus, proposition (1) is correct.
In proposition (2), from $2^{a} ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 731,070 |
7. Given the function $f(x)=x+\frac{2}{x}$ on the interval $[1,4]$, the maximum value is $M$, and the minimum value is $m$. Then the value of $M-m$ is $\qquad$ | ニ.7.4.
Since $f(x)$ is monotonically decreasing on the interval $[1,3]$ and monotonically increasing on the interval $[3,4]$, the minimum value of $f(x)$ is $f(3)=6$.
Also, $f(1)=10, f(4)=\frac{25}{4}$, so the maximum value of $f(x)$ is $f(1)=10$.
Therefore, $M-m=10-6=4$. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 731,071 |
10. In the tetrahedron $P-ABC$, the three edges $PA$, $PB$, and $PC$ are pairwise perpendicular, and $PA=1$, $PB=PC=2$. If $Q$ is any point on the surface of the circumscribed sphere of the tetrahedron $P-ABC$, then the maximum distance from $Q$ to the plane $ABC$ is | 10. $\frac{3}{2}+\frac{\sqrt{6}}{6}$.
Notice that, the circumscribed sphere of the tetrahedron $P-ABC$ is the circumscribed sphere of the rectangular parallelepiped with $PA$, $PB$, and $PC$ as its length, width, and height, respectively. Its diameter is $2R=\sqrt{1^{2}+2^{2}+2^{2}}=3$.
Also, $\cos \angle BAC=\frac{1}... | \frac{3}{2}+\frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,072 |
11. Let the line $y=k x+b$ intersect the curve $y=x^{3}-x$ at three distinct points $A, B, C$, and $|A B|=|B C|=2$. Then the value of $k$ is | 11. 1.
Given that the curve is symmetric about the point $(0,0)$, and
$$
|A B|=|B C|=2 \text {, }
$$
we know that the line $y=k x+b$ must pass through the origin.
Thus, $b=0$.
Let $A(x, y)$. Then
$$
\begin{array}{l}
y=k x, y=x^{3}-x, \sqrt{x^{2}+y^{2}}=2 \\
\Rightarrow x=\sqrt{k+1}, y=k \sqrt{k+1} . \\
\text { Substi... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,073 |
13. Given the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$, the endpoints of the real axis are $A_{1}$ and $A_{2}$, and $P$ is a moving point on the hyperbola different from points $A_{1}$ and $A_{2}$. The lines $P A_{1}$ and $P A_{2}$ intersect the line $x=1$ at points $M_{1}$ and $M_{2}$, respectively. Prove: The ci... | Three, 13. Given, let $A_{1}(-2,0), A_{2}(2,0)$, and a moving point $P\left(x_{0}, y_{0}\right)$ on the hyperbola with $y_{0} \neq 0$. Thus,
$\frac{x_{0}^{2}}{4}-\frac{y_{0}^{2}}{3}=1$.
Also, $l_{P A_{1}}: y=\frac{y_{0}}{x_{0}+2}(x+2)$,
$l_{P A_{2}}: y=\frac{y_{0}}{x_{0}-2}(x-2)$,
then $M_{1}\left(1, \frac{3 y_{0}}{x_{... | \left(-\frac{1}{2}, 0\right),\left(\frac{5}{2}, 0\right) | Geometry | proof | Yes | Yes | cn_contest | false | 731,074 |
Example 6 As shown in Figure 6, in isosceles $\triangle ABC$, $AB = AC$, and $O$ is the midpoint of segment $AB$. Segment $OC$ intersects the circle $\odot O$ with diameter $AB$ at point $D$, and ray $BD$ intersects $AC$ at point $E$. If $AE = CD$, prove: $\angle BAC = 90^{\circ} .{ }^{[6]}$ | Prove that as shown in Figure 6, connect $A D$. Then
$$
\angle A D B=90^{\circ} \text {. }
$$
Apply Menelaus' theorem to $\triangle O A C$ and the transversal $B D E$ to get
$$
\frac{C D}{D O} \cdot \frac{O B}{B A} \cdot \frac{A E}{E C}=1 .
$$
Since $C D=A E, O B=O D, A B=A C$, we have $C D^{2}=C E \cdot C A \Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,075 |
16. Let the function $f(x)=p x-\frac{p}{x}-2 \ln x$.
(1) If $f(x)$ is a monotonically increasing function in its domain, find the range of the real number $p$;
(2) Let $g(x)=\frac{2 \mathrm{e}}{x}$, and $p>0$. If there exists at least one point $x_{0}$ in the interval $[1, \mathrm{e}]$ such that $f\left(x_{0}\right)>g\... | 16. (1) The domain of the function $f(x)$ is $\{x \mid x>0\}$. To make $f(x)$ a monotonically increasing function in its domain $(0,+\infty)$, it is sufficient that
$$
f^{\prime}(x)=p+\frac{p}{x^{2}}-\frac{2}{x} \geqslant 0
$$
holds for all $x$ in the interval $(0,+\infty)$, i.e.,
$$
p \geqslant\left(\frac{2 x}{x^{2}+... | p \in\left(\frac{4 \mathrm{e}}{\mathrm{e}^{2}-1},+\infty\right) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 731,077 |
1. Given $f(x)=\frac{10}{x+1}-\frac{\sqrt{x}}{3}$. Then the number of elements in the set $M=\left\{n \in \mathbf{Z} \mid f\left(n^{2}-1\right) \geqslant 0\right\}$ is $\qquad$. | ,- 1.6 .
From the problem, we know that $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and $f(9)=0$.
Then $f\left(n^{2}-1\right) \geqslant f(9) \Rightarrow 1 \leqslant n^{2} \leqslant 10$.
Thus, the number of elements in set $M$ is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,078 |
2. A box contains 9 good items and 3 defective items. Each time an item is taken, it is not replaced. What is the probability that 3 defective items have been taken before 2 good items are taken? $\qquad$ | $\begin{array}{l}\text { 2. } \frac{1}{55} . \\ P=\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10}+\frac{9}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{1}{9}+ \\ \frac{3}{12} \times \frac{9}{11} \times \frac{2}{10} \times \frac{1}{9}+\frac{3}{12} \times \frac{2}{11} \times \frac{9}{10} \times \frac{1}{... | \frac{1}{55} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,079 |
3. Let points $A$ and $B$ lie on the parabola $y^{2}=6 x$ and the circle $\odot C:(x-4)^{2}+y^{2}=1$, respectively. Then the range of $|A B|$ is | 3. $[\sqrt{15}-1,+\infty)$.
Since $|A B|_{\text {min }}=|A C|_{\text {min }}-1$, we only need to consider the range of $|A C|$.
Notice,
$$
\begin{array}{l}
|A C|^{2}=(x-4)^{2}+y^{2} \\
=(x-4)^{2}+6 x \\
=x^{2}-2 x+16 \\
=(x-1)^{2}+15 .
\end{array}
$$
Also, $x \geqslant 0$, so $|A C|_{\min }=\sqrt{15}$.
Therefore, the... | [\sqrt{15}-1,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,080 |
5. Given an arithmetic sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}<0, a_{100} \geqslant 74, a_{200}<200 \text {, }
$$
and the number of terms of this sequence in the interval $\left(\frac{1}{2}, 8\right)$ is 2 less than the number of terms in the interval $\left[14, \frac{43}{2}\right]$. Then the general term ... | 5. $a_{n}=\frac{3}{4} n-1$.
From the problem, we know that $\frac{15}{2}$ and $6$ are both multiples of the common difference $d$. Let $d=\frac{3}{2 m}\left(m \in \mathbf{Z}_{+}\right)$.
From the given conditions, we solve to get
$$
\begin{array}{l}
\frac{74}{99}a_{1} \geqslant 74-99 d=-\frac{1}{4} .
\end{array}
$$
S... | a_{n}=\frac{3}{4} n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,081 |
6. Given $0 \leqslant y \leqslant x \leqslant \frac{\pi}{2}$, and
$$
4 \cos ^{2} y+4 \cos x \cdot \sin y-4 \cos ^{2} x \leqslant 1 \text {. }
$$
then the range of $x+y$ is | 6. $\left[0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, \pi\right]$.
Notice,
$$
\begin{array}{l}
4 \cos ^{2} y+4 \cos x \cdot \sin y-4 \cos ^{2} x-1 \\
= 2(\cos 2 y-\cos 2 x)+4 \cos x \cdot \sin y-1 \\
= 4 \sin (x+y) \cdot \sin (x-y)+2 \sin (x+y)- \\
2 \sin (x-y)-1 \\
=(2 \sin (x+y)-1)(2 \sin (x-y)+1) .
\end{arr... | \left[0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, \pi\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 731,082 |
1. As shown in Figure 7, in isosceles $\triangle ABC$, $\angle BAC$ is an obtuse angle. Extend side $AB$ to point $D$, and extend side $CA$ to point $E$, connect $DE$, such that $AD = BC = CE = DE$. Prove:
$$
\angle BAC = 100^{\circ} \text{.}
$$ | Prompt As shown in Figure 7, translate $BC$ to $DF$, connect $CF$, $EF$, to get quadrilateral $BDFC$.
It can be proven that $\triangle CFE \cong \triangle AED$, and $\triangle DEF$ is an equilateral triangle.
Let $\angle BAC = \alpha$. Then
$$
\angle ADF = 90^{\circ} - \frac{1}{2} \alpha, \angle ADE = 2 \alpha - 180^{... | 100^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 731,083 |
8. Given $a_{k}$ as the number of integer terms in $\log _{2} k, \log _{3} k, \cdots, \log _{2018} k$. Then $\sum_{k=1}^{2018} a_{k}=$ $\qquad$ | 8.4102 .
Let $b_{m}$ be the number of integer terms in $\log _{m} 1, \log _{m} 2, \cdots, \log _{m} 2018$.
Then, $\sum_{k=1}^{2018} a_{k}=\sum_{m=2}^{2018} b_{m}$.
Notice that, $\log _{m} t$ is an integer if and only if $t$ is a power of $m$.
\[
\begin{array}{l}
\text { Then } b_{2}=11, b_{3}=7, b_{4}=6, b_{5}=b_{6}=5... | 4102 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,084 |
9. (16 points) Let
$$
\begin{array}{l}
P(z)=z^{4}-(6 \mathrm{i}+6) z^{3}+24 \mathrm{iz}^{2}- \\
(18 \mathrm{i}-18) z-13 .
\end{array}
$$
Find the area of the convex quadrilateral formed by the four points in the complex plane corresponding to the four roots of $P(z)=0$. | II. 9. Notice that, $P(1)=0$.
Then $P(z)=(z-1)\left(z^{3}-(6 \mathrm{i}+6) z^{2}+24 \mathrm{iz}^{2}+\right.$ $(18 \mathrm{i}-5) z+13)$.
Let $Q(z)=z^{3}-(6 \mathrm{i}+5) z^{2}+(18 \mathrm{i}-5) z+13$.
Then $Q(\mathrm{i})=0$.
Hence $Q(z)=(z-\mathrm{i})\left(z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}\right)$.
Using the quadr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,085 |
II. (40 points) Given an $m$-row $n$-column grid in which some of the small squares are colored black, satisfying that each row and each column contains at least one black square, and the number of black squares in the row of each black square is no less than the number of black squares in its column. Prove: $m \leqsla... | For $i=1,2, \cdots, m$, let the $i$-th row have $x_{i}$ $\left(x_{i} \geqslant 1, x_{i} \in \mathbf{Z}\right)$ blackened cells; for $j=1,2, \cdots, n$, let the $j$-th column have $y_{j}\left(y_{j} \geqslant 1, y_{j} \in \mathbf{Z}\right)$ blackened cells.
For any $i=1,2, \cdots, m ; j=1,2, \cdots, n$, let
$a_{i j}=\lef... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 731,086 |
Four, (50 points) Let $k \in \mathbf{N}, k \geqslant 2, N=p_{1} p_{2} \cdots p_{k}$ $\left(p_{1}, p_{2}, \cdots, p_{k}\right.$ be distinct primes), $A$ be the set of all non-negative integers less than $N$ and not coprime with $N$, denoted as $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$;
Set $R=\left\{r_{i} \mid r_{i... | Assume $A=R$, consider a prime factor $p$ of $N$.
Let there be $t$ numbers in set $A$ divisible by $p$. Then there are also $t$ numbers in set $B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ divisible by $p$. Assume $a_{i_{1}}, a_{i_{2}}, \cdots, a_{i_{1}}, b_{j_{1}}, b_{j_{2}}, \cdots, b_{j_{1}}$ are all divisible by $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,087 |
1. Given that the three sides of a triangle are consecutive natural numbers. If the largest angle is twice the smallest angle, then the perimeter of the triangle is $\qquad$ . | -,1.15.
Assuming the three sides of a triangle are $n-1$, $n$, and $n+1$, with the largest angle being $2 \theta$ and the smallest angle being $\theta$.
Then, by the Law of Sines, we have
$$
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \cos \theta=\frac{n+1}{2(n-1)} \text {. }
$$
By the Law of Cosines... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,088 |
2. The sequence $a_{0}, a_{1}, \cdots$ satisfies:
$$
a_{0}=\sqrt{5}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}},
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $a_{2019}=$ $\qquad$ | $2.8076+\sqrt{5}$.
By calculating the first few terms, we find
$$
a_{n}=4 n+\sqrt{5}(n \in \mathbf{N}) \text {. }
$$
We will prove this by mathematical induction.
Check $a_{0}=\sqrt{5}, a_{1}=4+\sqrt{5}$.
Assume that when $n=k$, $a_{k}=4 k+\sqrt{5}$.
Then when $n=k+1$,
$$
\begin{array}{l}
a_{k+1}=\left[a_{k}\right]+\f... | 8076+\sqrt{5} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,089 |
3. A coin-flipping game, starting from the number $n$, if the coin lands heads up, subtract 1, if the coin lands tails up, subtract 2. Let $E_{n}$ be the expected number of coin flips before the number becomes zero or negative. If $\lim _{n \rightarrow \infty}\left(E_{n}-a n-b\right)=0$, then the pair $(a, b)=$ $\qquad... | 3. $\left(\frac{2}{3}, \frac{2}{9}\right)$.
From the problem, we have the recurrence relation
$E_{n}=\frac{1}{2}\left(E_{n-1}+1\right)+\frac{1}{2}\left(E_{n-2}+1\right)$
$\Rightarrow E_{n}=1+\frac{1}{2}\left(E_{n-1}+E_{n-2}\right)(n \geqslant 2)$.
Let $F_{n}=E_{n}-\frac{2}{3} n$. Then
$$
F_{n}=\frac{1}{2}\left(F_{n-1}+... | \left(\frac{2}{3}, \frac{2}{9}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,091 |
6. Given a regular tetrahedron $ABCD$ with edge length 1, $M$ is the midpoint of $AC$, and $P$ lies on the segment $DM$. Then the minimum value of $AP + BP$ is $\qquad$ | 6. $\sqrt{1+\frac{\sqrt{6}}{3}}$.
Let $\angle B D M=\theta$, in $\triangle B D M$,
$$
\begin{array}{l}
B D=1, B M=M D=\frac{\sqrt{3}}{2} \\
\Rightarrow \cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3} .
\end{array}
$$
Rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C... | \sqrt{1+\frac{\sqrt{6}}{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 731,092 |
8. The integer sequence $\left\{a_{i, j}\right\}(i, j \in \mathbf{N})$, where,
$$
\begin{array}{l}
a_{1, n}=n^{n}\left(n \in \mathbf{Z}_{+}\right), \\
a_{i, j}=a_{i-1, j}+a_{i-1, j+1}(i, j \geqslant 1) .
\end{array}
$$
Then the unit digit of the value taken by $a_{128,1}$ is | 8. 4 .
By the recursive relation, we have
$$
\begin{array}{l}
a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}, \\
a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2} .
\end{array}
$$
Accordingly, by induction, we get
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 731,093 |
11. (20 points) Does there exist 2018 different complex number sets $\left(z_{1}, z_{2}, \cdots, z_{2019}\right)$, satisfying $\left|z_{1}\right|=\left|z_{2}\right|=1$, $\left|z_{2019}\right|=\sqrt{1010}$, and
$$
\begin{array}{l}
\left|2 z_{i+2}-\left(z_{i+1}+z_{i}\right)\right|=\left|z_{i}-z_{i+1}\right|, \\
\left|2 z... | 11. Let $z_{k}$ correspond to the point $Z_{k}$ in the complex plane.
As shown in Figure 2, take a set of $z_{1}, z_{2}$, and construct the coordinate system with the perpendicular bisector of $Z_{1} Z_{2}$ as the $x$-axis, $\left|O Z_{1}\right|=1$, and let the angle between $\overrightarrow{O Z_{1}}$ and the positive... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,094 |
3. In an acute triangle $\triangle ABC$, the largest altitude $AH$ equals the median $BM$. Prove: $\angle ABC \leqslant 60^{\circ}$. | Draw $M K \perp B C$ through point $M$ at point $K$. Then $M K=\frac{1}{2} A H=\frac{1}{2} B M, \angle M B K=30^{\circ}$.
Draw $C P \perp A B$ at point $P$, and draw $M Q \perp A B$ through point $M$ at point $Q$.
Thus, $B 、 K 、 M 、 Q$ are concyclic.
Also, $M Q=\frac{1}{2} C P \leqslant \frac{1}{2} A H=M K$, so
$\angl... | \angle ABC \leqslant 60^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 731,095 |
II. (40 points) As shown in Figure 1, the circumcenter of acute $\triangle ABC$ is $O$. There is a point $D$ on line segment $BC$. A circle $\Gamma$ is constructed with $OD$ as its diameter. Circles $\Gamma_{1}$ and $\Gamma_{2}$ are the circumcircles of $\triangle ABD$ and $\triangle ACD$, respectively. Circles $\Gamma... | II. Connecting auxiliary lines as shown in Figure 3. It is easy to see that,
$$
\begin{array}{l}
\angle A F O=\angle A F D-\angle O F D \\
=180^{\circ}-\angle A B C-90^{\circ} \\
=90^{\circ}-\angle A B C=\angle A C O
\end{array}
$$
$\Rightarrow A, O, F, C$ are concyclic
$$
\Rightarrow \angle D F C=180^{\circ}-\angle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,096 |
The number of peaches on each peach tree forms an array $(1,2, \cdots, 63)$. Each worker picks once per round. If they take turns picking in the order of Jia and Yi, does Yi have a winning strategy? If so, please write it down; if not, please explain the reason.
Four, (50 points) A peach orchard has $n\left(n \in \mat... | Four, Party B has a winning strategy.
First, represent all elements in the set $S=\{1,2, \cdots, 63\}$ as six-bit binary numbers (e.g., $6=(000110)_{2}$, $\left.63=(111111)_{2}\right)$. It is easy to see that the number 1 appears 32 times in each bit position across these 63 binary numbers.
If $n$ binary numbers have ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 731,098 |
4. As shown in Figure 8, in the hexagon $A B C D E F$,
$$
\begin{aligned}
& A B=B C=C D=D E=E F=F A, \\
\text { and } & \angle A+\angle C+\angle E=\angle B+\angle D+\angle F
\end{aligned}
$$
$$
\text { Prove: } \angle A=\angle D, \angle B=\angle E, \angle C=\angle F \text {. }
$$ | ```
Translate $\triangle A B C$ to $\triangle P F E$, and connect $A P$. Then
\[
\begin{array}{l}
P E=A C, P F=A B=D E, \angle P F E=\angle A B C, \\
\angle A F P=360^{\circ}-\angle A F E-\angle P F E=\angle D . \\
\text { Therefore, } \triangle A F P \cong \triangle C D E, A P=C E .
\end{array}
\]
\[
\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,099 |
Example 1: The leader of an IMO team chooses positive integers $n, k (n > k)$, and informs the deputy leader and the contestants. Then, the leader secretly tells the deputy leader a binary number string of $n$ digits. As a result, the deputy leader writes down all binary number strings of $n$ digits that differ from th... | 【Analysis】The original solution adopts a horizontal consideration, that is, comparing each $n$-bit binary string with the leader's number string. In fact, a vertical thinking approach can also be used, which involves considering the number of 0s and 1s in each position of the $\mathrm{C}_{n}^{k}$ $n$-bit binary strings... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 731,100 |
Example 2 Let $n>1$ be an integer, and $k$ be the number of distinct prime factors of $n$. Prove: there exists an integer $a\left(1<a<\frac{n}{k}+1\right)$, such that $n \mid\left(a^{2}-a\right)$.
[2]
(2011, China National Training Team Selection Exam) | Proof: Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $p_{i}$ are distinct primes, $\alpha_{i} \in \mathbf{Z}_{+}$.
$$
\begin{array}{l}
\text { Let } N=\{1,2, \cdots, n\} \text {. Then } \\
n!\left(x^{2}-x\right) \\
\Leftrightarrow x(x-1) \equiv 0(\bmod n) \\
\Leftrightarrow x \equiv \beta_{i}\left(\bmod p_{i}^{\alp... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,101 |
The points $A_{1}, A_{2}, A_{3}, A_{4}$ and $B_{1}, B_{2}, B_{3}, B_{4}$ satisfy
$$
A_{1} A_{2}: A_{2} A_{3}: A_{3} A_{4}=B_{1} B_{2}: B_{2} B_{3}: B_{3} B_{4}
$$
$(A_{1} A_{2}$ represents the quantity of the directed segment $\overline{A_{1} A_{2}}$, not the length, the same applies below), $O, P, Q$ are any three poi... | For question 2, first take any two points \( S \) and \( T \) on the line \( A_1 A_2 \); then take two points \( U \) and \( V \) on the line \( B_1 B_2 \) such that
\[
U B_1 : B_1 B_2 : B_2 V = S A_1 : A_1 A_2 : A_2 T.
\]
Clearly, such \( U \) and \( V \) are uniquely determined.
From equation (1), we get
\[
S A_i : ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 731,102 |
Example 1 On a plane, there are $2 n(n>1, n \in \mathbf{N})$ points with no three points collinear. Line segments are drawn between any two points, and any $n^{2}+1$ of these line segments are colored red. Prove: there are at least $n$ triangles with all three sides red.
(2017, National High School Mathematics League J... | 【Analysis】By slightly modifying the problem, let's only retain the red segments and delete the rest, i.e., to prove: if there are $n^{2}+1$ edges connecting $2n$ points (no three points are collinear), then there exist $n$ triangles.
Proof Let these $2n$ points be $v_{0}, v_{1}, \cdots, v_{2n-1}$. Then
$$
\sum_{i=0}^{2... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 731,103 |
Example 2 For a set $P$ consisting of a finite number of prime numbers, let $m(P)$ denote the maximum number of consecutive positive integers, each of which is divisible by at least one element of $P$. Let $|P|$ denote the number of elements in the set $P$. Prove:
(1) $|P| \leqslant m(P)$, equality holds if and only if... | Let the elements of $P$ be denoted as $1k$. Then for $i=1,2, \cdots, k$, in any $k+1$ consecutive positive integers, there is at most one multiple of $p_{i}$. Therefore, among them, at most $k$ positive integers can be divisible by some number in $P$, which does not satisfy the property in the question. At this point,
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 731,104 |
Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=a_{2}=a_{3}=1, a_{4}=-1, a_{5}=0,
$$
and for any $n \in \mathbf{Z}_{+}$, we have
$$
a_{n+5}=3 a_{n+4}-4 a_{n+3}+4 a_{n+2}-3 a_{n+1}+a_{n} \text {. }
$$
Find the general term formula for the sequence $\left\{a_{n}\right\}$. | This is a fifth-order linear recurrence sequence, and the corresponding characteristic equation is
$$
x^{5}-3 x^{4}+4 x^{3}-4 x^{2}+3 x-1=0 \text {. }
$$
The five roots can be found to be $i, -i, 1, 1, 1$, where 1 is a triple root.
By theorem, we can assume
$$
a_{n}=C_{1} i^{n}+C_{2}(-\mathrm{i})^{n}+\left(C_{3} n^{2}... | a_{n}= \frac{-3+7 \mathrm{i}}{8} \mathrm{i}^{n}+\frac{-3-7 \mathrm{i}}{8}(-\mathrm{i})^{n}+ \frac{3 n^{2}-19 n+27}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 731,105 |
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