problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Let $\left\{a_{n}\right\}$ be a positive geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and
$$
2^{10} S_{30}+S_{10}=\left(2^{10}+1\right) S_{20} \text {. }
$$
Then the common ratio of the sequence $\left\{a_{n}\right\}$ is ( ).
(A) $\frac{1}{8}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 1 | 5. C.
From the given information,
$$
\begin{array}{l}
S_{20}-S_{10}=2^{10}\left(S_{30}-S_{20}\right) \\
\Rightarrow \frac{S_{30}-S_{20}}{S_{20}-S_{10}}=2^{-10}=q^{10} .
\end{array}
$$
Since $a_{n}>0$, we have $q=\frac{1}{2}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,252 |
6. The three views of a geometric solid are shown in Figure 1. Its side view is an equilateral triangle with a side length of 1, and its top view is a rhombus formed by two equilateral triangles. Then the volume of this geometric solid is ( ).
Figure 1
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $\frac{1}{8}$ | 6. C.
The geometric body can be regarded as consisting of two triangular pyramids with a base that is an equilateral triangle of side length 1 and a height of $\frac{\sqrt{3}}{2}$. Therefore, $V=2 \times \frac{1}{3} \times \frac{\sqrt{3}}{4} \times 1^{2} \times \frac{\sqrt{3}}{2}=\frac{1}{4}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,253 |
11. As shown in Figure 2, in the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a square with side length $a$, $PD=a$, $PA=PC=\sqrt{2} a$. If a sphere is placed inside this quadrilateral pyramid, then the maximum radius of the sphere is ( ).
(A) $(\sqrt{2}-1) a$
(B) $\sqrt{2} a$
(C) $\left(1-\frac{\sqrt{2}}{2}\righ... | 11. C.
By the equal volume method, we have
$$
\begin{array}{l}
V_{\text {tetrahedron } P-A B C D}=\frac{1}{3} a^{2} \times a \\
=\frac{1}{3}\left(a^{2}+2 \times \frac{1}{2} a^{2}+2 \times \frac{1}{2} a \times \sqrt{2} a\right) r \\
\Rightarrow r=\left(1-\frac{\sqrt{2}}{2}\right) a .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,254 |
For a positive integer $n$, define the set
$$
S_{n}=\left\{\mathrm{C}_{n}^{n}, \mathrm{C}_{2 n}^{n}, \mathrm{C}_{3 n}^{n}, \cdots, \mathrm{C}_{n^{2}}^{n}\right\} \text {. }
$$
Prove: (1) There exist infinitely many composite numbers $n$ such that $S_{n}$ is not a complete residue system modulo $n$;
(2) There exist inf... | (1) Let $n=2 p$ ($p$ is an odd prime). Then the $p$-ary representation of $n$ is $\overline{(20)_{p}}$. When $k$ takes $\frac{p+1}{2}, p, 2p$, the last two digits of $k n$ in $p$-ary are $10, 00, 00$, respectively, and the second digit from the right is less than the digit 2 of $n$. Therefore, these three $\mathrm{C}_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,255 |
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then th... | 14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} \text {. }
$$
Since points $M, O, N$ are collinear, it follows that,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,256 |
15. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\frac{1}{2} \sin 2 y+a=0$. Then the value of $\cos (x+2 y)$ is $\qquad$ | 15. 1 .
Solving the system of equations and eliminating $a$ yields
$$
x=-2 y \Rightarrow \cos (x+2 y)=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,257 |
17. (10 points) Let the function
$$
f(x)=\log _{2}\left(a^{x}-b^{x}\right) \text {, }
$$
and $f(1)=1, f(2)=\log _{2} 12$.
(1) Find the values of $a$ and $b$;
(2) When $x \in[1,2]$, find the maximum value of $f(x)$. | Three, 17. (1) From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\log _{2}(a-b)=1, \\
\log _{2}\left(a^{2}-b^{2}\right)=\log _{2} 12
\end{array}\right. \\
\quad \Rightarrow\left\{\begin{array}{l}
a-b=2, \\
a^{2}-b^{2}=12
\end{array} \Rightarrow a=4, b=2 .\right.
\end{array}
$$
(2) From (1), we kno... | 2+\log _{2} 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,258 |
19. (12 points) Given vectors
$$
\begin{array}{l}
\boldsymbol{m}=(\sqrt{3} \sin \omega x, \cos \omega x), \\
\boldsymbol{n}=(\cos \omega x,-\cos \omega x),
\end{array}
$$
where $\omega>0$. The function $f(x)=\boldsymbol{m} \cdot \boldsymbol{n}$ has the smallest positive period of $\frac{\pi}{2}$.
(1) Find the value of... | 19. From the given, we have
$$
\begin{array}{l}
f(x)=\frac{\sqrt{3}}{2} \sin 2 \omega x-\cos ^{2} \omega x \\
=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1+\cos 2 \omega x}{2} \\
=\sin \left(2 \omega x-\frac{\pi}{6}\right)-\frac{1}{2} .
\end{array}
$$
Thus, $T=\frac{2 \pi}{2 \omega}=\frac{\pi}{2} \Rightarrow \omega=2$.
... | -1<k<\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,259 |
22. (12 points) Let the function $f(x)=\frac{x}{\ln x}-a x$.
(1) If the function $f(x)$ is monotonically decreasing on the interval $(1,+\infty)$, find the minimum value of the real number $a$;
(2) If there exist $x_{1}, x_{2} \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$, satisfying
$$
f\left(x_{1}\right) \leqslant f^{\... | 22. (1) From the given, we have $x>0, x \neq 1$.
Since $f(x)$ is a decreasing function on the interval $(1,+\infty)$, it follows that, on the interval $(1,+\infty)$, we always have
$$
\begin{array}{l}
f^{\prime}(x)=\frac{\ln x-1}{\ln ^{2} x}-a \leqslant 0 . \\
\text { Also, } f^{\prime}(x)=\frac{\ln x-1}{\ln ^{2} x}-... | a \geqslant \frac{1}{2}-\frac{1}{4 \mathrm{e}^{2}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,260 |
2. Polynomial
$$
p(x)=x^{3}-224 x^{2}+2016 x-d
$$
has three roots that form a geometric progression. Then the value of $d$ is $\qquad$ | 2. 729 .
Let the three roots of the polynomial be $a$, $b$, and $c$, and $b^{2}=a c$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b+c=224, \\
a b+b c+c a=2016, \\
a b c=d .
\end{array}\right.
$$
Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$.
Thus, $d=a b c=b^{3}=729$. | 729 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,261 |
5. In the tetrahedron $ABCD$,
\[
\begin{array}{l}
AB=1, BC=2 \sqrt{6}, CD=5, \\
DA=7, AC=5, BD=7 .
\end{array}
\]
Then its volume is $\qquad$ | 5. $\frac{\sqrt{66}}{2}$.
From $A B^{2}+B C^{2}=1^{2}+(2 \sqrt{6})^{2}=25=A C^{2}$, $B C^{2}+C D^{2}=(2 \sqrt{6})^{2}+5^{2}=49=B D^{2}$, we know that $A B \perp B C, B C \perp C D$.
As shown in Figure 2, construct rectangle $A B C E$.
Thus $V$ : volume of $D-A B C$
Since $A E \perp E C$,
$A E \perp C D$,
we know that... | \frac{\sqrt{66}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,262 |
6. The probability that the product of the points obtained from rolling a die three times is divisible by 6 is $\qquad$ | 6. $\frac{133}{216}$.
Let $P_{3}$ and $P_{6}$ represent the probabilities that the product is divisible by 3 and by 6, respectively.
Then $P_{3}=1-\left(\frac{4}{6}\right)^{3}=\frac{152}{216}$.
Let $P_{3}^{\prime}$ denote the probability that the product is divisible by 3 and is also odd.
Then $P_{3}^{\prime}=\left(\f... | \frac{133}{216} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,263 |
Example 2 Given a cyclic quadrilateral $ABCD$ with a point $P$ inside, satisfying $\angle DPA = \angle PBA + \angle PCD$, the projections of point $P$ onto the sides $AB$, $BC$, and $CD$ are $E$, $F$, and $G$ respectively. Prove: $\triangle APD \sim \triangle EFG$.
---
The translation maintains the original text's fo... | Proof 1: From $\angle D P A=\angle P B A+\angle P C D$, we know that the circumcircle of $\triangle A B P$ and the circumcircle of $\triangle D P C$ are tangent at point $P$.
By the Radical Axis Theorem, the common tangent line of the circumcircle of $\triangle A B P$ and the circumcircle of $\triangle D P C$ at the t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,264 |
One, (40 points) Let integer $n \geqslant 2$. Prove:
$$
\sum_{k=1}^{n-1} \frac{3^{k} \mathrm{C}_{n}^{k}}{8^{k} \mathrm{C}_{n-1}^{k}}<1 .
$$ | Let $x_{n}=\sum_{k=1}^{n-1}\left(\frac{3}{8}\right)^{k} \frac{\mathrm{C}_{n}^{k}}{\mathrm{C}_{n-1}^{k}}=\sum_{k=1}^{n-1}\left(\frac{3}{8}\right)^{k} \frac{n}{n-k}$. Then $x_{2}=\frac{3}{8} \times \frac{2}{1}=\frac{3}{4}<1$,
$$
x_{3}=\frac{3}{8} \times \frac{3}{2}+\left(\frac{3}{8}\right)^{2} \times \frac{3}{1}=\frac{63... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,265 |
Three. (50 points) As shown in Figure 1, the incircle of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. The midpoints of $BF$, $BD$, $CE$, and $CD$ are $M$, $N$, $P$, and $Q$ respectively. $MN$ and $PQ$ intersect at point $X$. Prove that the circumcircle of $\triangle XBC$ ... | Three, first prove a conclusion.
If points $B_{1}$ and $C_{1}$ are on the sides $TB$ and $TC$ of $\triangle TBC$ respectively, and $B_{1}C_{1} \parallel BC$, then the circumcircle of $\triangle TB_{1}C_{1}$ is tangent to the circumcircle of $\triangle TBC$.
Proof: As shown in Figure 4, we only need to consider the tan... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,266 |
469 Given that $M$ is a point inside the tetrahedron $ABCD$, and the areas of the faces opposite to vertices $A, B, C, D$ are $S_{A}, S_{B}, S_{C}, S_{D}$ respectively, and they satisfy
$$
S_{A} \overrightarrow{M A}+S_{B} \overrightarrow{M B}+S_{C} \overrightarrow{M C}+S_{D} \overrightarrow{M D}=\mathbf{0} \text {. }
$... | Proof As shown in Figure 1, let the distances from point $M$ to the planes opposite $A, B, C, D$ be $m_{A}, m_{B}, m_{C}, m_{D}$, respectively. Then the volume of the tetrahedron $M B C D$ is
$$
\begin{array}{l}
\left.\frac{1}{6} \mathrm{I}(\overrightarrow{M B} \times \overrightarrow{M D}) \cdot \overrightarrow{M C} \r... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,267 |
Given 470 Let $O$ be the circumcenter of acute $\triangle ABC$, and let $D$, $E$, and $F$ be the projections of points $A$, $B$, and $C$ onto sides $BC$, $CA$, and $AB$, respectively. Draw $DG \perp EF$, with the foot of the perpendicular being $G$. If $O$, $B$, $C$, and $G$ are concyclic, prove:
$$
\cos A \cdot \sin B... | Proof with auxiliary lines as shown in Figure 2.
Let $H$ be the orthocenter of $\triangle ABC$, and the sides of $\triangle ABC$ be $a$, $b$, and $c$.
Then $BD = c \cos \angle ABC$,
$CD = b \cos \angle BCA$,
$CE = a \cos \angle BCA$,
$BF = a \cos \angle ABC$.
Let $\angle BGD = \alpha$,
$\angle CGD = \beta$.
Since $DG \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,268 |
Given $\odot O_{1}$ and $\odot O_{2}$ are both internally tangent to $\odot O$, with points of tangency at $A$ and $B$, respectively. The chord $CD$ of $\odot O$ is the external common tangent of $\odot O_{1}$ and $\odot O_{2}$ (the length of arc $\overparen{AC}$ is less than the length of arc $\overparen{AD}$). Let th... | Proof of the lemma first.
Lemma As shown in Figure 3, $\odot O_{1}$ is internally tangent to $\odot O$ at point $A$, the chord $C D$ of $\odot O$ is tangent to $\odot O_{1}$ at point $F$, and $A F$ intersects $\odot O$ again at point $K$. Then $K$ is the midpoint of the arc $\overparen{C D}$ not containing point $A$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,269 |
Given $\triangle A B C$ is an acute triangle, $A D$, $B E$, and $C F$ are the three altitudes, $A D$ intersects $E F$ at point $S$, $M$ is the midpoint of side $B C$, $B S$ intersects $M E$ and $C S$ intersects $M F$ at points $P$ and $Q$ respectively. Prove: $P$, $A$, and $Q$ are collinear. | Proof As shown in Figure 5, let the orthocenter of $\triangle ABC$ be $H$, connect $DE$, and draw $PK \perp BC$ at point $K$.
Let $\angle CBS = \alpha, \angle CME = \beta, \angle CBH = \gamma$.
It is easy to see that $\beta = 2\gamma$.
Let the three interior angles of $\triangle ABC$ be $\angle A, \angle B, \angle C$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,270 |
Example 1 Given that $a, b, c, d$ are prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. [1] | Let
\[
\begin{array}{l}
a b c d=x+(x+1)+\cdots+(x+34) \\
=5 \times 7(x+17) .
\end{array}
\]
Assume \( a=5, b=7, c \leqslant d, c d=x+17 \).
Thus, \( d_{\text {min }}=5 \).
If \( d=5 \), then \( c_{\text {min }}=5 \), at this point,
\[
x=8, a+b+c+d=22 \text {; }
\]
If \( d=7 \), then \( c_{\text {min }}=3 \), at this ... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,271 |
2. Given positive integer $n=a b c<10000, a, b, c$ are all prime numbers, and $2 a+3 b=c, 4 a+c+1=4 b$. Find the value of $n$.
| From the given equations, we have
$$
b=6 a+1, c=20 a+3 \text{. }
$$
Then $a(6 a+1)(20 a+3)<10000$
$\Rightarrow 12 a^{3}<10000 \Rightarrow$ prime $a<5$.
Combining $b=6 a+1, c=20 a+3$ being primes, we can determine
$$
a=2, b=13, c=43, n=1118 \text{. }
$$ | 1118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,272 |
5. For any $\alpha, \beta \in\left(0, \frac{2 \pi}{3}\right)$, we have
$$
\begin{array}{l}
4 \cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+4 \cos ^{2} \beta- \\
3 \cos \alpha-3 \cos \beta-k<0 .
\end{array}
$$
Then the minimum value of $k$ is | 5.6.
Substitution $\left(x_{1}, x_{2}\right)=(2 \cos \alpha, 2 \cos \beta)$.
Then the given inequality becomes
$$
2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-2 k<0 \text {. }
$$
Since $x_{1}, x_{2} \in(-1,2)$, we have
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,273 |
11. (20 points) Let $a_{1} \in \mathbf{Z}_{+}$, and $a_{1} \leqslant 18$, define the sequence $\left\{a_{n}\right\}:$
$$
a_{n+1}=\left\{\begin{array}{ll}
2 a_{n}, & a_{n} \leqslant 18 ; \\
2 a_{n}-36, & a_{n}>18
\end{array}(n=1,2, \cdots) .\right.
$$
Find the maximum number of elements in the set $M=\left\{a_{n} \mid ... | 11. Given the positive integer $a_{1} \leqslant 18$ and the recursive formula for the sequence $\left\{a_{n}\right\}$, the following properties can be derived:
(1) $a_{n+1} \equiv 2 a_{n}(\bmod 36)$, i.e.,
$a_{n+1} \equiv 2 a_{n}(\bmod 4)$, and $a_{n+1} \equiv 2 a_{n}(\bmod 9)$;
(2) All terms are positive integers, and... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,274 |
II. (40 points) Prove:
$$
2^{4} \times 5^{7} \times 2003 \mid\left(2^{286}-1\right)\left(3^{500}-1\right)\left(1978^{100}-1\right) .
$$ | (1) Prove $5^{3} \mid \left(1978^{100}-1\right)$.
Since $\varphi\left(5^{3}\right)=5^{3}\left(1-\frac{1}{5}\right)=100$, and 5 is coprime with 1978, by Euler's theorem we have
$$
1978^{100} \equiv 1\left(\bmod 5^{3}\right) \Rightarrow 5^{3} \mid \left(1978^{100}-1\right).
$$
(2) Prove $10^{4} \mid \left(3^{500}-1\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,275 |
3. Let $S=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{980100}}$. Find the greatest positive integer $[S]$ that does not exceed the real number $S$. | Notice that, $\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}$.
Therefore, $\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$,
$2(\sqrt{n+1}-\sqrt{n})=\frac{2}{\sqrt{n+1+\sqrt{n}}}<\frac{1}{\sqrt{n}}$.
Thus, $1977<2(\sqrt{980101}-\sqrt{2})$
$$
<S<2(\sqrt{980100}-1)=1978 \text {. }
$$
Hence, $[S]=... | 1977 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,276 |
Four. (50 points) Let integer $n \geqslant 3$, and consider cards placed at $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ and point $O$. Perform the following operations:
Operation $A$: If the number of cards at some point $A_{i}$ is at least 3, then three cards can be taken out and one card can be placed at each of the th... | (1) First, adjust the number of cards at each of the $n$ points $A_{i}(i=1,2, \cdots, n)$ to be no less than $n+1$.
If the number of cards at some point $A_{i}$ is no less than 3, then perform operation $A$. Each such operation increases the number of cards at point $O$ by 1. After several operations $A$, operation $A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,277 |
473 Let positive real numbers $x, y, z$ satisfy $xyz=1$, $\alpha \geqslant 0$. Prove:
$$
\sum \frac{x^{\alpha+3}+y^{\alpha+3}}{x^{2}+xy+y^{2}} \geqslant 2 \text {. }
$$ | $$
\begin{array}{l}
\frac{x^{3}}{x^{2}+x y+y^{2}} \geqslant \frac{2}{3} x-\frac{1}{3} y \\
\Leftrightarrow 3 x^{3} \geqslant(2 x-y)\left(x^{2}+x y+y^{2}\right) \\
\Leftrightarrow x^{3}+y^{3} \geqslant x^{2} y+x y^{2} \\
\Leftrightarrow(x-y)^{2}(x+y) \geqslant 0 .
\end{array}
$$
The above inequality is obviously true.
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,278 |
In $\triangle A B C$, $\angle A=90^{\circ}, A B=$ $A C$, points $P, Q$ are on sides $A B, A C$ respectively, satisfying $P Q=$ $B P+C Q$. Let the circumcircle of $\triangle A P Q$ intersect side $B C$ at points $E, F$. Prove: $S_{\triangle A E F}=S_{\text {quadrilateral } P E F C}$. | Proof As shown in Figure 2, let
$$
\angle B A E=\alpha, \angle C A F=\beta, \angle E A F=\gamma .
$$
It is easy to see that $P Q$ is the diameter of the circumcircle of $\triangle A E F$.
By the Law of Sines, we have
$$
P E=P Q \sin \gamma, Q F=P Q \sin \beta \text {. }
$$
We can prove that $\triangle B P E \backsim ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,279 |
5. Let the positive integer $n$ be the sum of the reciprocals of four distinct positive integers $a$, $b$, $c$, $d$. Find the values of $n$ and $a+b+c+d$.
| Suppose $1 \leqslant a<b<c<d$.
Then $n=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$
$$
\leqslant 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \text {. }
$$
Thus, $n=1$ or 2.
Using the method of magnification and reduction, determine $(a, b, c, d)$ to be
$$
\begin{array}{l}
(2,3,7,42),(2,3,8,24),(2,3,9,18), \\
(2,3,10,15)... | 12, 24, 30, 31, 32, 37, 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,281 |
Example 2 If the set of integers $A \subseteq A+A$, where,
$$
A+A=\{a+b \mid a \in A, b \in A\},
$$
then $A$ is called "saturated". If all integers except 0 are the sum of all elements in some non-empty finite subset of the integer set $A$, then the set $A$ is called "free". Question: Does there exist an integer set t... | 【Analysis】Take the Fibonacci sequence
$$
F_{1}=1, F_{2}=1, F_{i+2}=F_{i+1}+F_{i}(i \geqslant 1) \text {. }
$$
Construct $A=\left\{F_{2 n} \mid n \in \mathbf{Z}_{+}\right\} \cup\left\{-F_{2 n+1} \mid n \in \mathbf{Z}_{+}\right\}$.
By $F_{2 n}=F_{2 n+2}+\left(-F_{2 n+1}\right)$,
$$
-F_{2 n+1}=\left(-F_{2 n+3}\right)+F_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,282 |
Example 5 For integer pairs $(a, b)(0<a<b<1000)$, a set $S \subseteq\{1,2, \cdots, 2003\}$ is called a "jump set" of the pair $(a, b)$: if for any element pair $\left(s_{1}, s_{2}\right)$, $s_{1} 、 s_{2} \in$ $S,\left|s_{1}-s_{2}\right| \notin\{a, b\}$.
Let $f(a, b)$ be the maximum number of elements in a jump set of ... | 【Analysis】This method is quite typical, the minimum value is obtained using the greedy idea, while the maximum value is derived using the pigeonhole principle. Unfortunately, during the exam, most candidates only answered one part correctly.
For the minimum value, take $a=1, b=2$, then
$$
\begin{array}{l}
\{1,2,3\},\{4... | 668 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,283 |
Example 6 Let $[x]$ denote the greatest integer not exceeding the real number $x$. Let the sets $A_{1}, A_{2}, \cdots, A_{n}$ all be three-element subsets of the set $\{1,2, \cdots, n\}$. Prove: In the set $\{1,2, \cdots, n\}$, there exist $\left[\frac{2 n}{5}\right]$ elements, which, when colored, ensure that each set... | 【Analysis】Looking at the problem in reverse, it can be proven: color $\left\lceil\frac{3 n}{5}\right\rceil$ elements so that each set $A_{i}$ contains at least one colored element.
To achieve the requirement with the minimum number of colorings, each time "greedily" color the element that appears the most frequently i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,284 |
Example 10 If a set of lines in the plane has no two lines parallel and no three lines concurrent, it is called "in general position". A set of lines in general position divides the plane into several regions, and the regions with finite area are called "bounded regions" of this set of lines. Prove: for sufficiently la... | 【Analysis】Each time "greedily" choose to color a line so that no blue area is generated. Suppose after coloring $k$ lines, no more can be colored, then the remaining $n-k$ lines, each line $l$ is at least the side of a finite area (called a "dangerous area") where all other sides are blue edges.
Below we estimate $n-k$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,286 |
3. Let $X$ be the set of open disks in the plane that can cover the set $E$. Prove: there exists a subset $\left\{D_{1}, D_{2}, \cdots, D_{n}\right\}$ of $X$ with mutually disjoint elements, such that $3 D_{1}, 3 D_{2}, \cdots, 3 D_{n}$ can cover the set $E$, where the disk $3 D$ denotes a disk with the same center as ... | Hint: Each time, take the disk with the largest radius that does not intersect with the existing disks. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,289 |
4. Let $S=\{1,2, \cdots, 1000000\}, A$ be a 101-element subset of $S$. Prove that there exist integers $t_{1}, t_{2}, \cdots, t_{100}$ in $S$ such that
$$
A_{j}=\left\{x+t_{j} \mid x \in A\right\}(j=1,2, \cdots, 100)
$$
are pairwise disjoint. | Hint Similar to Example 4, each time take the smallest $t_{i}$ such that adding it results in $A_{i}$ being disjoint from all previous sets. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,290 |
Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}$, $\cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ;
\end{array}
$$
$$
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. }
$$
Find t... | 【Detailed Estimation】It is known that, $a_{31} \leqslant 2015, a_{30} \leqslant 2014, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \leqslant 1984+i$.
Similarly, $b_{i} \leqslant 1984+i$.
Also, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, ... | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,291 |
Example As shown in Figure 1, three circles with centers at $A$, $B$, and $C$ are externally tangent to each other, and all are tangent to the line $l$. Let the radii of the three circles be $a$, $b$, and $c$ $(0<c<a<b)$. Prove:
$$
\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}
$$ | To prove that in fact, four circles on a plane that are pairwise tangent either all four circles are pairwise externally tangent, or three of the circles are pairwise externally tangent and simultaneously internally tangent to the fourth circle, i.e., the theorem's cases (1) and (2).
We will discuss this by classifica... | \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}} | Geometry | proof | Yes | Yes | cn_contest | false | 729,292 |
Example 4 Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2011^{3}}$.
Then the integer part of $4 S$ is ( ). ${ }^{[2]}$
(A) 4
(B) 5
(C) 6
(D) 7
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | When $k=2,3, \cdots, 2011$,
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {. }
$$
Then $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
... | 4 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,293 |
Example 1 Suppose in the gap between two mutually internally tangent circles, four internally tangent circles are constructed successively. If the four constructed circles, except for the first and last, are externally tangent to each other in sequence, then the radii $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$ of the four circ... | Proof As shown in Figure 3, let the two given circles be $\odot O$ and $\odot O^{\prime}$, with radii $a$ and $b$ respectively. The four circles constructed are $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$, and $\odot O_{4}$.
Then, for $\odot O$, $\odot O^{\prime}$, $\odot O_{1}$, and $\odot O_{2}$, using the formula (... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,294 |
Example 4 Let the semi-perimeter of $\triangle A B C$ be $p$, the inradius be $r$, and construct semicircles outside $\triangle A B C$ with $B C, C A, A B$ as diameters, respectively. Let the radius of the circle $\Gamma$ that is tangent to these three semicircles be $t$. Prove:
$$
\frac{p}{2}<t \leqslant \frac{p}{2}+\... | Proof As shown in Figure 6, let the center of circle $\Gamma$ be $O$, and $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ respectively. The circle $\Gamma$ is tangent to three semicircles at points $D'$, $E'$, $F'$, and the radii of these three semicircles are $d'$, $e'$, $f'$ respectively. Then $DD'$, $EE'$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,295 |
Example 5 In space, there are four spheres with radii of $2$, $2$, $3$, and $3$. Each sphere is externally tangent to the other three spheres. There is another smaller sphere that is externally tangent to these four spheres. Find the radius of the smaller sphere. | Let the radius of the required sphere be $r=\frac{1}{x}$. Then, by Descartes' theorem, we have
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+x\right)^{2} \\
=3\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{3^{2}}+x^{2}\right) \\
\Rightarrow\left(\frac{5}{3}+x\right)^{2}=3\lef... | \frac{6}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,296 |
Example 2 (Eckers' Theorem) Given three equilateral triangles $\triangle A_{1} B_{1} C_{1}$, $\triangle A_{2} B_{2} C_{2}$, and $\triangle A_{3} B_{3} C_{3}$ with vertices in reverse order. Let the centroids of $\triangle A_{1} A_{2} A_{3}$, $\triangle B_{1} B_{2} B_{3}$, and $\triangle C_{1} C_{2} C_{3}$ be $M_{1}$, $... | Proof and Diagram 2.
$$
\begin{array}{l}
\text { Let } \times\left(A_{1} B_{1}, A_{2} B_{2}\right)=\alpha, \\
\text { Let }\left(A_{2} B_{2}, A_{3} B_{3}\right)=\beta, \\
\text { Let }\left(A_{3} B_{3}, A_{1} B_{1}\right)=\gamma
\end{array}
$$
be the angles between corresponding sides of three equilateral triangles, w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,297 |
Example $\mathbf{3}$ (Feng - Van Aubel's Theorem) Construct four squares $S_{1}, S_{2}, S_{3}, S_{4}$ outwardly on the four sides of a closed broken line $A B C D$, with their centers being $O_{1}, O_{2}, O_{3}, O_{4}$ respectively. Prove:
$$
O_{1} \mathrm{O}_{3} \perp \mathrm{O}_{2} \mathrm{O}_{4}, \mathrm{O}_{1} \mat... | Prove the case for a convex quadrilateral $A B C D$ only.
As shown in Figure 3.
Notice,
$$
\begin{array}{l}
\left(O_{3}-O_{1}\right) \mathrm{i}=(D+F-A-E) \mathrm{i} \\
=(\overrightarrow{A D}+\overrightarrow{E B}+\overrightarrow{B C}+\overrightarrow{C F}) \mathrm{i} \\
=\overrightarrow{A H}+\overrightarrow{B A}+\overrig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,298 |
Example 5 Find all positive integers $n$, such that there exist positive integers $x_{1}, x_{2}, \cdots, x_{2012}$, satisfying
$x_{1}<x_{2}<\cdots<x_{2012}$,
and $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}=n$. | Given that
$$
x_{1} \geqslant 1, x_{2} \geqslant 2, \cdots, x_{2012} \geqslant 2012 .
$$
Thus, $n=\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}$
$$
\leqslant \frac{1}{1}+\frac{2}{2}+\cdots+\frac{2012}{2012}=2012 \text {. }
$$
When $n=1$, let
$$
\begin{array}{l}
x_{1}=2012, x_{2}=2 \times 2012, \cdots, ... | 1, 2, \cdots, 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,299 |
1. If the natural numbers $a$, $b$, $c$ satisfy $a > b$, and
$$
a+b=18, c-b=2 \text{, }
$$
then the maximum value of $b+c$ is ( ).
(A) 16
(B) 18
(C) 19
(D) 20 | 1. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,300 |
3. Person A and Person B start cycling from location $A$ at 7:00 in the morning, traveling along the same route at a constant speed to location $B$, which is 24 kilometers away from $A$. Person A rests for 5 minutes after every 4 kilometers, while Person B, traveling at 75% of Person A's speed, rests for 3 minutes afte... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,302 |
4. In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}$, $E$ is a point inside $\triangle ABC$, $EA=1$, $EB=\sqrt{2}$, $EC=\frac{\sqrt{2}}{2}$. Then the size of $\angle AEB$ is ( ).
(A) $90^{\circ}$
(B) $120^{\circ}$
(C) $135^{\circ}$
(D) $150^{\circ}$ | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,303 |
7. When two fair dice are rolled simultaneously, the probability that the sum of the numbers on the top faces is divisible by 3 is $\qquad$ | 7. $\frac{1}{3}$ | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,306 |
9. If the inequality about $x$
$$
|x+a| \geqslant|x-3|
$$
has a solution set that includes $x \geqslant a$, then the range of real number $a$ is $\qquad$ . | 9. $a \geqslant 1$ or $a=-3$ | a \geqslant 1 \text{ or } a = -3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,308 |
Example 6 For a positive integer $n$, denote $n!=1 \times 2 \times \cdots \times n$. Find all positive integer tuples $(a, b, c, d, e, f)$, such that
$$
a!=b!+c!+d!+e!+f!,
$$
and $a>b \geqslant c \geqslant d \geqslant e \geqslant f .{ }^{[1]}$ | Solve: From equation (1), we get
$$
a!>5(f!) \geqslant 5 \Rightarrow a \geqslant 3 \text {. }
$$
From $a>b \Rightarrow a!\geqslant a(b!)$.
Combining $a!=b!+c!+d!+e!+f!\leqslant 5 b!$, we get $a \leqslant 5$.
When $a=3$, it is not difficult to determine the positive integer tuple as $(3,2,1,1,1,1)$.
When $a=4$,
(1) If ... | (3,2,1,1,1,1),(5,4,4,4,4,4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,310 |
13. Given the non-isosceles acute triangle $\triangle ABC$ with orthocenter $H$, the midpoint of side $BC$ is $M$, and the extension of $AH$ intersects the circumcircle $\odot O$ of $\triangle ABC$ at point $D$. Connect $OM$.
(1) As shown in Figure 5, prove:
$$
OM=\frac{1}{2} AH \text {; }
$$
(2) Connect $AO$ and exten... | 13. (1) As shown in Figure 6, connect $B O$ and extend it to intersect $\odot O$ at point $B_{1}$, then connect $B_{1} C$.
Then $\angle B_{1} C B=90^{\circ}$, which means $B_{1} C \perp B C$.
In $\triangle B_{1} C B$, since $O$ and $M$ are the midpoints of $B B_{1}$ and $B C$ respectively, we have
$O M / / B_{1} C$, an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,311 |
2. In the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. For any positive integer $n$, can all rational points be colored using $n$ colors, with each point colored one color, such that on any line segment whose both endpoints are rational points, rational points of each... | 2. Sure.
Let $v(q)$ denote the power of 2 in the non-zero rational number $q$, i.e., if $q = \frac{a}{b}$ (where $a$ and $b$ are non-zero integers), then
$$
v(q) = v(a) - v(b),
$$
and let $v(0) = +\infty$.
Notice that for any rational numbers $x, y$, if
$$
v(x) < v(y),
$$
then $v(x + y) = v(x)$.
For any rational poi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,312 |
4. Let $a, b, b', c, m, q$ be positive integers, where $m>1, q>1, |b-b'| \geqslant a$. It is known that there exists a positive integer $M$ such that
$$
S_{q}(a n+b) \equiv S_{q}(a n+b') + c \pmod{m}
$$
holds for all integers $n \geqslant M$. Prove: Equation (1) holds for all positive integers $n$.
Here, $S_{q}(x)$ i... | 4. Let's assume $b > b'$. Then $b - b' \geqslant a$.
Thus, there exists a positive integer $r$ such that $b'ar - b'$, and $aq^t > b - ar$.
Therefore,
$$S_{q}(an + b) - S_{q}(an + b') = S_{q}(aq^t + b - ar) - S_{q}(aq^t - (ar - b'))$$
$$
= S_{q}(a) + S_{q}(b - ar) - S_{q}(aq^t - (ar - b')) \text{.}
$$
On one hand, ac... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,313 |
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in \mathbf{Z}_{+}\right)$ is $\qquad$ | -1.3 .
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x}=1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Therefore, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,314 |
2. Let $P$ be any point inside a regular tetrahedron $ABCD$ with edge length $\sqrt{2}$, and let the distances from point $P$ to the four faces be $d_{1}, d_{2}, d_{3}, d_{4}$. Then the minimum value of $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ is | 2. $\frac{1}{3}$.
From $V_{\text {tetrahedron } A B C D}$
$$
\begin{array}{l}
=V_{\text {trilateral pyramid } P-A B C}+V_{\text {trilateral pyramid } P-D B C}+ \\
V_{\text {trilateral pyramid } P-C D A}+V_{\text {trilateral pyramid } P-D A B} \\
\Rightarrow \frac{\sqrt{2}}{12}(\sqrt{2})^{3}=\frac{1}{3} \times \frac{\s... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,315 |
7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then
$$
g(x)=f(x)+f(1-x)
$$
the minimum value of $g(x)$ is . $\qquad$ | 7.8.
From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$
Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,316 |
8. Given the set
$$
\begin{array}{c}
A_{k}=\left\{2^{k-1}+\sum_{i=0}^{k-2} a_{i} 2^{a_{i}+i} \mid a_{i} \in\{0,1\},\right. \\
i=0,1, \cdots, k-2\} .
\end{array}
$$
Let $n_{k}$ denote the sum of all elements in the set $A_{k}$. Then $\sum_{k=1}^{2015} n_{k}=$ $\qquad$ | 8. $\frac{2^{4031}+1}{3}-2^{2015}$.
From binary, we know that regardless of how $a_{i}$ is chosen, the elements in the set $A_{k}$ are all distinct.
Considering that $a_{i}$ can take values of $0$ or $1$, when $a_{i}=1$,
$$
2^{a_{i}+i}=2^{i+1}(i=0,1, \cdots, k-2)
$$
appears $2^{k-2}$ times, then
$$
\begin{array}{l}
n... | \frac{2^{4031}+1}{3}-2^{2015} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,317 |
II. (16 points) Prove: Any positive integer $N$ can be expressed in the form $pq + uv$, where $u - v = 2(p - q)$, and $p, q, u, v \in \mathbf{Z}$. | If $N$ is even, let $N=2 m(m \in \mathbf{Z})$. Then $N=2 m=(2 m-3)(-1)+1 \times(4 m-3)$. Let $p=2 m-3, q=-1, u=1, v=4 m-3$. Therefore, $u-v=4 m-3-1=2(m-2)$ $=2[(2 m-3)-(-1)]=2(p-q)$.
If $N$ is odd, let $N=2 m+1(m \in \mathbf{Z})$. Then $N=2 m+1=(3 m+1)(1-m)+3 m \times m$. Let $p=3 m, q=m, u=3 m+1, v=1-m$. Clearly, $p,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,318 |
Five. (20 points) How many $n$-digit numbers can be formed from the digits $1,2, \cdots, 6$ such that $1$ and $6$ are adjacent? | Let the number of $n$-digit numbers we are looking for be $a_{n}$.
Let the number of $n$-digit numbers among $a_{n}$ that start with the digit 1 be $b_{n}$.
By swapping the positions of 1 and 6 in each of the $b_{n}$ numbers, we get $b_{n}$ numbers among the $a_{n}$ numbers that start with the digit 6.
The $a_{n}$ numb... | 6^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,320 |
1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$. | ,- 1.26 .
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers within this range are $9, 16, 25$. Let's assume $a<b<c$.
If $a+b+c=9$, then the possible values for $c$ are 6, 5, 4, in which case,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$
If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,321 |
2. If $\sin \alpha+\cos \alpha=\frac{1}{5}$, then $|\sin \alpha-\cos \alpha|=$ $\qquad$ | 2. $\frac{7}{5}$.
From the condition, we get
$$
\begin{array}{l}
2 \sin \alpha \cdot \cos \alpha=(\sin \alpha+\cos \alpha)^{2}-1=-\frac{24}{25} \\
\Rightarrow(\sin \alpha-\cos \alpha)^{2}=1-2 \sin \alpha \cdot \cos \alpha=\frac{49}{25} \\
\Rightarrow|\sin \alpha-\cos \alpha|=\frac{7}{5} .
\end{array}
$$ | \frac{7}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,322 |
3. The range of the function $y=\frac{\sqrt{1-x^{2}}}{2+x}$ is | 3. $\left[0, \frac{\sqrt{3}}{3}\right]$.
From the condition, we know $x \in[-1,1]$.
Let $x=\cos \alpha(\alpha \in[0, \pi])$. Then
$$
\begin{array}{l}
y=\frac{\sin \alpha}{2+\cos \alpha}(y \geqslant 0) \\
\Rightarrow 2 y=\sin \alpha-y \cos \alpha \\
=\sqrt{1+y^{2}} \sin (\alpha+\theta) \\
\leqslant \sqrt{1+y^{2}} \\
\... | \left[0, \frac{\sqrt{3}}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,323 |
4. The volume of a rectangular prism is 8 cubic centimeters, and the total surface area is 32 square centimeters. If the length, width, and height form a geometric sequence, then the sum of all the edges of this rectangular prism is $\qquad$ | 4.32 cm.
Let the common ratio be $q$, and the length, width, and height be $q a, a, \frac{a}{q}$, respectively. Then $a^{3}=8 \Rightarrow a=2$.
$$
\begin{array}{l}
\text { Also } 2\left(q a \cdot a+a \cdot \frac{a}{q}+q a \cdot \frac{a}{q}\right)=32 \\
\Rightarrow q+\frac{1}{q}+1=4 .
\end{array}
$$
Therefore, the sum... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,324 |
10. (20 points) As shown in Figure 2, $A_{1}, B_{1}, C_{1}$ are points on the circumcircle $\odot O$ of $\triangle ABC$, satisfying $A A_{1} / / B B_{1} / / C C_{1}$. $P$ is any point on $\odot O$. If the line $P A_{1}$ intersects $BC$, $P B_{1}$ intersects $AC$, and $P C_{1}$ intersects $AB$ at points $D$, $E$, $F$ re... | 10. Notice that points $D$, $E$, and $F$ are on lines $BC$, $CA$, and $AB$, respectively.
$$
\begin{array}{l}
\text{Then } \angle A P E=\angle A P B_{1}=\angle B A A_{1}=\angle B P D, \\
\angle A P F=\pi-\angle A P C_{1}=\pi-\angle C P A_{1}=\angle C P D, \\
\angle B P F=\angle B A C_{1}=\angle C P B_{1}=\angle C P E .... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,325 |
11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
| 11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers in $1^{2}$, $2^{2}, \cdots, 10^{2}$ (allowing the same n... | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,326 |
2. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
|x-1|<x+2, \\
x<m
\end{array}\right.
$$
has two integer solutions. Then the range of the real number $m$ is ( ).
(A) $m<0$
(B) $0<m \leqslant 1$
(C) $1<m<2$
(D) $1<m \leqslant 2$ | 2. D.
It is easy to know,
$$
\begin{array}{l}
|x-1|-\frac{1}{2} \text{.}
\end{array}
$$
Therefore, the solution set of the original inequality system is $-\frac{1}{2}<x<m$. Considering there are two integer solutions, we know $1<m \leqslant 2$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 729,327 |
4. As shown in Figure 1, quadrilateral $ABCD$ is a rectangle, $AD=a$, $CD=b$. If points $E$, $F$, $G$ are on sides $BC$, $AD$, $DC$ respectively, and $\angle DAG = \angle CGE = \angle DGF$, then $\frac{EG + FG}{AG}$ equals ( ).
(A) $\frac{a}{b}$
(B) $\frac{b}{a}$
(C) $\frac{b}{a} + 1$
(D) $\frac{2b}{a}$ | 4. B.
Let $\angle D A G=\angle C G E=\angle D G F=\theta$. Then, by the given conditions,
$$
E G=\frac{G C}{\cos \theta}, F G=\frac{D G}{\cos \theta} .
$$
Thus, $E G+F G=\frac{G C+D G}{\cos \theta}=\frac{D C}{\cos \theta}$, and
$$
A G=\frac{A D}{\cos \theta} .
$$
Therefore, $\frac{E G+F G}{A G}=\frac{D C}{A D}=\frac... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,328 |
5. As shown in Figure $2, D, E, F$ are points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively, and $AD, BE, CF$ intersect at point $G$. If $S_{\triangle BDG}=8, S_{\triangle CDG}=6, S_{\triangle AEG}=14$, then $S_{\triangle ABC}$ equals ( ).
(A) 56
(B) 62
(C) 63
(D) Uncertain | 5. C.
$$
\begin{array}{l}
\text { Let } S_{\triangle C E G}=x . \\
\text { By } \frac{S_{\triangle A B G}}{S_{\triangle A E G}}=\frac{S_{\triangle B C G}}{S_{\triangle C E G}} \Rightarrow \frac{S_{\triangle A B G}}{14}=\frac{8+6}{x} \\
\Rightarrow S_{\triangle A B G}=\frac{196}{x} \text {. } \\
\text { By } \frac{S_{\t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,329 |
$$
\begin{aligned}
A & =|| x-y|+x+y-2 z|+|x-y|+x+y+2 z \\
& =4 \max \{x, y, z\},
\end{aligned}
$$
where, $\max \{x, y, z\}$ represents the maximum of $x$, $y$, and $z$. | Hint: Divide into three cases based on $x \geqslant y, x \geqslant z ; y \geqslant z, y \geqslant x ; z \geqslant x$, $z \geqslant y$ to simplify $A$, and obtain $A$ as $4 x, 4 y, 4 z$ respectively. | 4 \max \{x, y, z\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,330 |
4. If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$ | 4. 2 .
Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\Rightarrow 3^{n} \equiv 1(\bmod 8)
$$
$\Rightarrow n$ must be even.
Let $n=2+2 k(k \in \mathbf{N})$. Then
$$
2016+3^{n}=9\left(224+3^{2 k}\right) \text {. }
$$
So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,332 |
For any positive integer $n$, if among $2n$ distinct real numbers, any $n$ are selected and arranged in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$, and the remaining $n$ numbers are arranged in descending order as $b_{1}, b_{2}, \cdots, b_{n}$. Determine whether the value of
$$
\left|a_{1}-b_{1}\right|+\left|a_{2... | One, arrange these $2 n$ numbers in ascending order as $c_{1}, c_{2}, \cdots, c_{n}, \cdots, c_{2 n}$.
Below is the proof: For each $i(1 \leqslant i \leqslant n), a_{i} 、 b_{i}$ must be one less than or equal to $c_{n}$, and the other greater than $c_{n}$.
Otherwise, when $a_{i} 、 b_{i}$ are both less than or equal to... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,333 |
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x \text {, }
$$
find all positive integers $a$ that meet the requirement. | $$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times$ (2) $-x \times$ (1) gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,334 |
1. In the plane, $m$ points have no three points collinear, and their convex hull is an $n$-sided polygon. By appropriately connecting lines, a grid region composed of triangles can be obtained. Let the number of non-overlapping triangles be $f(m, n)$. Then $f(2016,30)=$ $\qquad$ | One, 1.4000.
Since no three points are collinear, we have
$$
f(m, n)=(n-2)+2(m-n)=2 m-n-2 \text {. }
$$
Therefore, $f(2016,30)=2 \times 2016-30-2=4000$. | 4000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,335 |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$ satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
According to the problem, for any $(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking $(x, y)=(1,0),(0,1)$, we get the fixed point $B(a, b)$ satisfying the necessary conditions $\left\{\begin{array}{l}a \leqslant 1, \\ b \leqslant 1,\end{array}\right.$, which implies $a+b \leqslant 2$.
Thus, $(a+b)_{\text ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,336 |
Example 1 Remove the big and small jokers from a deck of cards, and randomly draw five cards from the remaining 52 cards. The probability that at least two of the cards have the same number (or letter $K, Q, J, A$) is $\qquad$ (require calculating the numerical value of this probability, accurate to 0.01).
(2014, Natio... | 【Analysis】Let event $A$ be “drawing five cards randomly from 52 cards, where at least two cards have the same number (or letter $K, Q, J, A$)”. Then event $\bar{A}$ is “drawing five cards randomly from 52 cards, where each card has a different number (or letter $K, Q, J, A$)”.
The total number of cases is $\mathrm{C}_... | 0.49 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,337 |
Example 2 Choose five different numbers from $1,2, \cdots, 20$, the probability that at least two of them are consecutive numbers is $\qquad$ [1]
(2013, National High School Mathematics Joint Competition) | 【Analysis】Suppose the five numbers taken out satisfy
$$
a_{1}<a_{2}<a_{3}<a_{4}<a_{5} .
$$
If these five numbers are not adjacent, then
$$
1 \leqslant a_{1}<a_{2}-1<a_{3}-2<a_{4}-3<a_{5}-4 \leqslant 16 \text {. }
$$
Thus, the number of ways to choose five non-adjacent numbers from $1,2, \cdots, 20$ is the same as the... | \frac{232}{323} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,338 |
5. A bag contains $n$ red balls and 5 white balls. It costs 1 yuan to draw once, each time drawing two balls from the bag. If the two balls are of different colors, it is considered a win, and the prize is 2 yuan. Then when $n=$ $\qquad$, the expected value is maximized. | Let the probability of winning once be $p_{n}$. Then
$$
p_{n}=\frac{\mathrm{C}_{n}^{1} \mathrm{C}_{5}^{1}}{\mathrm{C}_{n+5}^{2}}=\frac{10 n}{(n+5)(n+4)} .
$$
By the problem, if you win, you earn 1 yuan; if you don't win, you lose 1 yuan. Therefore, the mathematical expectation is
$$
E=p_{n}-\left(1-p_{n}\right)=2 p_{n... | 4 \text{ or } 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,339 |
6. Given the set $T=\{1,2, \cdots, 2010\}$, for each non-empty subset of $T$, calculate the reciprocal of the product of all its elements. Then the sum of all such reciprocals is $\qquad$ | 6. 2010 .
Notice that, in the expansion of the product
$$
(1+1)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)
$$
the $2^{2010}$ terms are all the reciprocals of positive integers, and each is precisely the reciprocal of the product of the numbers in one of the $2^{2010}$ subsets of set $T$. Removing t... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,340 |
8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line bre... | 8.2.
As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\begin{array}{l}
\tan \alpha=\tan \left(\beta-\frac{\pi}{4}\right) \\
=\frac{\tan \beta-1}{1+\tan \beta}=\frac{4}{3}, \\
\tan \angle x O B=\tan \left(\frac{\pi}{2}+\alpha\right) \\
... | 2 | Other | math-word-problem | Yes | Yes | cn_contest | false | 729,342 |
9. (16 points) If the complex number $z$ satisfies $4 z^{-2011}-3 \mathrm{i} z^{-2010}-3 \mathrm{i} z^{-1}-4=0$, find the range of $t=\overline{\left(\frac{3-4 \mathrm{i}}{z}\right)}+\overline{(3+4 \mathrm{i}) z}$.
Translate the above text into English, please retain the original text's line breaks and format, and out... | 2,9. Transform the known equation into
$$
\begin{array}{l}
4 z^{2011}+3 \mathrm{i} z^{2010}+3 \mathrm{i} z-4=0 \\
\Rightarrow z^{2010}(4 z+3 \mathrm{i})=4-3 \mathrm{i} z \\
\Rightarrow|z|^{2010}|4 z+3 \mathrm{i}|=|4-3 \mathrm{i} z| .
\end{array}
$$
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$. Then
$$
\begin{array}{l}
... | [-10,10] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,343 |
Example 2 Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{0}=b_{0}=1, \\
a_{n+1}=\alpha a_{n}+\beta b_{n}, b_{n+1}=\beta a_{n}+\gamma b_{n},
\end{array}
$$
where $\alpha, \gamma$ are positive integers, $\alpha<\gamma$, and $\alpha \gamma=\beta^{2}+1$. Prove that for a... | Prove the two lemmas first.
Lemma 1 If positive integers $m, n, k$ satisfy
$$
m n=k^{2}+1,
$$
then there exist $x_{1}, x_{2}, y_{1}, y_{2} \in \mathbf{N}$, such that the following three equations
$$
m=x_{1}^{2}+y_{1}^{2}, n=x_{2}^{2}+y_{2}^{2}, k=x_{1} x_{2}+y_{1} y_{2}
$$
hold simultaneously.
Proof of Lemma 1 Witho... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,345 |
Question 1 Given that $f(x)$ is a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, it always holds that
$$
f(x-f(y))=f(f(y))+x f(y)+f(x)-1 .
$$
Find $f(x)$, and calculate the value of $f(\sqrt{2014})$. ${ }^{[1]}$
$(2014$, Beijing High School Mathematics Competition (Grade 10)) | For any $y \in \mathbf{R}$, substituting $(f(y), y)$ into equation (1) yields
$$
\begin{array}{l}
f(0)=f(f(y))+f^{2}(y)+f(f(y))-1 \\
\Rightarrow f(f(y))=\frac{f(0)+1}{2}-\frac{f^{2}(y)}{2} .
\end{array}
$$
Substituting $(f(x), f(y))$ into equation (1) yields
$$
\begin{array}{l}
f(f(x)-f(y)) \\
=f(f(y))+f(x) f(y)+f(f(x... | f(x)=1-\frac{x^{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,346 |
Given $\triangle A B C$ is a determined acute triangle, $E, F$ are points on sides $A C, A B$ respectively, $M$ is the midpoint of $E F$, the perpendicular bisector of $E F$ intersects $B C$ at point $K$, the perpendicular bisector of $M K$ intersects $A C, A B$ at points $S, T$ respectively. If points $K, S, A, T$ are... | Prove that for any interesting point pair $(E, F)$, the corresponding $\triangle E F K$ is also interesting.
As shown in Figure 1, let $\triangle E F K$ be interesting. The circle passing through points $K, S, A, T$ is denoted as circle $\Gamma$. Extend $A M$ to intersect $T S$ and circle $\Gamma$ at points $N$ and $L... | \frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C} | Geometry | proof | Yes | Yes | cn_contest | false | 729,347 |
2. As shown in Figure $1, O$ is the center of the regular octagon $A B C D E F G H$, and $X$ is the midpoint of side $A B$. Then the area of the shaded part $S_{1}$ occupies the total area $S$ ( ).
(A) $\frac{11}{32}$
(B) $\frac{3}{8}$
(C) $\frac{13}{32}$
(D) $\frac{7}{16}$
(E) $\frac{15}{32}$ | 2. D.
From the given, $S_{\triangle O B X}=\frac{1}{2} S_{\triangle A O B}$.
Since $S_{\triangle A O B}=S_{\triangle B O C}=S_{\triangle C O D}=S_{\triangle D O E}$, therefore,
$$
\frac{S_{1}}{S}=\frac{3 S_{\triangle A O B}+\frac{1}{2} S_{\triangle A O B}}{8 S_{\triangle A O B}}=\frac{7}{16}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,348 |
Example 3: Each vertex of a regular nonagon is equally likely to be colored either red or blue. The probability that there exist three vertices of the same color forming an acute triangle is $\qquad$
(2013, National High School Mathematics League (B Volume)) | 【Analysis】If the convex polygon formed by vertices of the same color contains the center of the circumcircle of the regular nonagon, then there exist three vertices of this color such that the triangle they form also contains the center. Thus, this triangle is an acute triangle. Conversely, if the vertices of a certain... | \frac{247}{256} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,349 |
6. In $\triangle A B C$, if $A B=B C=29, A C=$ 42, then the area of $\triangle A B C$ is ( ).
(A) 100
(B) 420
(C) 500
(D) 609
(E) 701 | 6. B.
Draw $B D \perp A C$ at point $D$. Then
$$
\begin{array}{l}
S_{\triangle A B C}=2 S_{\triangle A B D}=2 \times \frac{1}{2} A D \cdot B D \\
=\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}} \times \frac{42}{2}=420 .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,350 |
7. Given that two boxes each contain three chips numbered $1$, $2$, and $3$, one chip is randomly drawn from each box, and the numbers on the two chips are then multiplied. What is the probability that the product is an even number? ( ).
(A) $\frac{1}{9}$
(B) $\frac{2}{9}$
(C) $\frac{4}{9}$
(D) $\frac{1}{2}$
(E) $\frac... | 7. E.
$$
1-\frac{2}{3} \times \frac{2}{3}=\frac{5}{9} \text {. }
$$ | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,351 |
10. Among the four-digit integers from $1000 \sim 9999$, there are ( ) numbers whose digits are all different.
(A) 3024
(B) 4536
(C) 5040
(D) 6480
(E) 6261 | $\begin{array}{l}\text { 10. B. } \\ 9 \times 9 \times 8 \times 7=4536 \text {. }\end{array}$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,353 |
11. The car license plates in a certain town consist of four characters, the first of which must be a vowel $(A, E, I, O, U)$, the second and third must be different non-vowel letters, and the fourth must be a digit from $0$ to $9$. If a license plate is chosen at random from those that meet these conditions, the proba... | 11. B.
$$
\frac{1}{5} \times \frac{1}{21} \times \frac{1}{20} \times \frac{1}{10}=\frac{1}{21000} .
$$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,354 |
12. A cube has $(\quad)$ pairs of parallel edges.
(A) 6
(B) 12
(C) 18
(D) 24
(E) 36 | $\begin{array}{l}\text { 12. C. } \\ 6 \times 3=18 \text {. }\end{array}$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,355 |
Example 4 Mik participates in a ten-question multiple-choice test, getting one point for each correct answer, losing one point for each wrong answer, and getting zero points for not answering. His goal is to score at least 7 points, with 7 points being the passing score. Mik is now certain that his answers to the first... | 【Analysis】Mick got six questions right.
The probability of passing by doing one more question is $P_{1}=p$, the probability of passing by doing two more questions is $P_{2}=p^{2}$, the probability of passing by doing three more questions is
$$
P_{3}=p^{3}+\mathrm{C}_{3}^{2} p^{2}(1-p)=p^{2}(3-2 p) \text {, }
$$
The pr... | p=\frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,356 |
20. Jeff went to the store to buy socks, which are priced at 1 yuan, 3 yuan, and 4 yuan per pair. If Jeff paid a total of 24 yuan for 12 pairs of socks, and he bought at least one pair of each price, then how many pairs of 1-yuan socks did he buy?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8 | 20. D.
Let Jeff buy $x$ pairs of socks at 1 yuan per pair and $y$ pairs of socks at 3 yuan per pair. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+3 y+4(12-x-y)=24 \\
x, y \geqslant 1, \\
12-x-y \geqslant 1
\end{array}\right. \\
\Rightarrow 6 \frac{1}{2} \leqslant x \leqslant 7 \frac{2}{3} \text {. } \\
\end{arra... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,357 |
25. From a square with a side length of 5, cut out a small square with a side length of 1 from each of the four corners. In the remaining shape, a square with an area of ( ) can be inscribed.
(A) 9
(B) 12.5
(C) 15
(D) 15.5
(E) 17 | 25. C.
As shown in Figure 5, the required part is the square $A B C D$. By symmetry, we have
$$
\begin{array}{l}
S_{\text {square } A B C D}=S_{\text {square } A^{\prime} B^{\prime} C^{\prime} D^{\prime}}+4 S_{\triangle A A^{\prime} B^{\prime}} \\
=(5-1 \times 2)^{2}+4 \times \frac{1}{2} \times(5-1 \times 2) \\
=15 .
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,359 |
Example 5: A and B take turns rolling a fair die, with A starting first. The rules are: if A rolls a 1, A continues to roll; otherwise, B rolls. If B rolls a 3, B continues to roll; otherwise, A rolls. The two players always follow this rule. Then the probability that the $n$-th roll is made by A, $p_{n}=$ $\qquad$ [2]... | 【Analysis】The probability of A rolling a 1 and B rolling a 3 is $\frac{1}{6}$ each, and the probability of A not rolling a 1 and B not rolling a 3 is $\frac{5}{6}$ each.
Let the probability of A rolling on the $k$-th turn be $p_{k}$. Then the probability of B rolling is $1-p_{k}$.
Obviously, $p_{1}=1$.
If A rolls firs... | p_{n}=\frac{1}{2}+\frac{1}{2}\left(-\frac{2}{3}\right)^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,360 |
For integers $n \geqslant 3$, let
$$
f(n)=\log _{2} 3 \times \log _{3} 4 \times \cdots \times \log _{n-1} n \text {. }
$$
Then $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)=$ $\qquad$ | 2. 54
Notice,
$$
\begin{array}{l}
f(n)=\log _{2} 3 \times \frac{\log _{2} 4}{\log _{2} 3} \times \cdots \times \frac{\log _{2} n}{\log _{2}(n-1)} \\
=\log _{2} n .
\end{array}
$$
Thus, $f\left(2^{k}\right)=k$.
Therefore, $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)$
$$
=2+3+\cdots+10=54 \text ... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,361 |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of t... | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by ... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,362 |
5. Given $\cos \left(\theta+\frac{\pi}{4}\right)<3\left(\sin ^{5} \theta-\cos ^{5} \theta\right)$, where $\theta \in[-\pi, \pi]$. Then the range of $\theta$ is | 5. $\left[-\pi,-\frac{3 \pi}{4}\right) \cup\left(\frac{\pi}{4}, \pi\right]$.
Notice that,
$$
\begin{array}{l}
\cos \left(\theta+\frac{\pi}{4}\right)<3\left(\sin ^{5} \theta-\cos ^{5} \theta\right) \\
\Leftrightarrow \frac{\sqrt{2}}{2} \cos \theta+3 \cos ^{5} \theta<\frac{\sqrt{2}}{2} \sin \theta+3 \sin ^{5} \theta .
\... | \left[-\pi,-\frac{3 \pi}{4}\right) \cup\left(\frac{\pi}{4}, \pi\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,363 |
8. If real numbers $x, y$ satisfy
$$
x-3 \sqrt{x+1}=3 \sqrt{y+2}-y \text {, }
$$
then the maximum value of $x+y$ is $\qquad$. | $8.9+3 \sqrt{15}$.
Given the equation, i.e.,
$$
\begin{array}{l}
x+y=3 \sqrt{x+1}+3 \sqrt{y+2} . \\
\text { Let } u=\sqrt{x+1}, v=\sqrt{y+2} .
\end{array}
$$
Then $u \geqslant 0, v \geqslant 0, x+y=u^{2}+v^{2}-3$.
Thus, equation (1) can be transformed into
$$
\begin{array}{l}
u^{2}+v^{2}-3=3 u+3 v \\
\Leftrightarrow\l... | 9+3 \sqrt{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,364 |
12. (16 points) Let $n \in \mathbf{Z}_{+}, A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a sequence composed of the numbers 0 or 1, i.e.,
$$
a_{k}=0 \text { or } 1(1 \leqslant k \leqslant n) \text {. }
$$
(1) If $n \geqslant 3$, define another sequence $A^{\prime}=\left\{a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{n}... | 12. (1) Sequence $A=\{1,1, \cdots, 1\}$;
When $n$ is a multiple of 3, sequence $A$ also has the following three solutions:
$$
\begin{array}{l}
A=\{0,1,0,0,1,0, \cdots, 0,1,0\}, \\
A=\{0,0,1,0,0,1, \cdots, 0,0,1\}, \\
A=\{1,0,0,1,0,0, \cdots, 1,0,0\} .
\end{array}
$$
(2) Let $x_{n}, y_{n}, z_{n}, u_{n}$ represent the n... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,365 |
1. Let $a=\sum_{i=1}^{24}[\sqrt{i}]([x]$ denotes the greatest integer not exceeding the real number $x$), and the set $A=\left\{x\left|x \in \mathbf{Z}_{+}, x\right| a\right\}$. Then the number of elements in set $A$ is ( ).
(A) 4
(B) 6
(C) 8
(D) 12 | - 1. C.
Notice,
$$
\begin{aligned}
a= & \sum_{i=1}^{24}[\sqrt{i}] \\
= & ([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}])+ \\
& ([\sqrt{4}]+[\sqrt{5}]+\cdots+[\sqrt{8}])+ \\
& ([\sqrt{9}]+[\sqrt{10}]+\cdots+[\sqrt{15}])+ \\
& ([\sqrt{16}]+[\sqrt{17}]+\cdots+[\sqrt{24}]) \\
= & 1 \times 3+2 \times 5+3 \times 7+4 \times 9 \\
= & 70=2... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,366 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.