problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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One. (20 points) As shown in Figure 3, given that two circles are internally tangent at point $A$, the chord $A B$ of the larger circle intersects the smaller circle at point $D$, the chord $B C$ of the larger circle is tangent to the smaller circle at point $E$, and the extension of $A E$ intersects the larger circle ... | As shown in Figure 6, draw the common tangent line $MN$ through point $A$. Connect $ED$ and $CF$.
Then $\angle DEA = \angle BAN = \angle ECA$.
Since $BC$ intersects the smaller circle at point $E$
$$
\begin{array}{l}
\Rightarrow \angle CEA = \angle EDA \\
\Rightarrow \triangle CEA \backsim \triangle EDA \\
\Rightarrow ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,322 |
II. (25 points) Given the quadratic function
$$
y=a x^{2}+b x+c(a>0)
$$
the graph of which intersects the graph of the linear function $y=x$ at two points with abscissas $x_{1} 、 x_{2}$, and $0<x_{1}<x_{2}<\frac{1}{a}$.
(1) Try to express $b 、 c$ in terms of $a 、 x_{1} 、 x_{2}$;
(2) If $0<t<x_{1}$, when $x=t$, the val... | (1) From the given information,
$$
\begin{array}{l}
a x^{2}+b x+c=x \\
\Rightarrow a x^{2}+(b-1) x+c=0,
\end{array}
$$
its two roots are $x_{1}$ and $x_{2}$.
Thus, $x_{1}+x_{2}=-\frac{b-1}{a}, x_{1} x_{2}=\frac{c}{a}$.
Therefore, $b=-a\left(x_{1}+x_{2}\right)+1, c=a x_{1} x_{2}$.
(2) When $0<t<x_{1}<x_{2}<\frac{1}{a}$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,323 |
Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers. | Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Righta... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,324 |
2. A cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1 is cut by a plane passing through the body diagonal $A C_{1}$. Then the minimum value of the area of this section is $\qquad$ . | 2. $\frac{\sqrt{6}}{2}$.
As shown in Figure 1, it is easy to see that the section $A F C_{1} G$ is a parallelogram. Draw $F E \perp A C_{1}$, with the foot of the perpendicular at $E$. Then the area of the section $S=A C_{1} \cdot E F$.
Since $A C_{1}=\sqrt{3}$ is a constant, the area $S$ is minimized when $F E$ is m... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,326 |
Given a cyclic quadrilateral $A B C D$, the pairs of opposite sides $A B$ and $D C$, $A D$ and $B C$ intersect at points $E$ and $F$ respectively, and $M$ and $N$ are the midpoints of $A C$ and $B D$ respectively. Prove:
$$
\frac{2 M N}{E F}=\left|\frac{A C}{B D}-\frac{B D}{A C}\right| \text {. }
$$ | Proof As shown in Figure 1, construct parallelogram $BCDR$ with vertices $B, C, D$. $DR$ intersects $AB$ at point $P$, and $BR$ intersects $AD$ at point $Q$. Connect $AR, PQ, CR$.
Assume $AC \geqslant BD$.
By the Midline Theorem, we know $AR=2MN$.
Since $\angle EDP = \angle DCF = \angle EAD$, thus $\triangle EDA \backs... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,327 |
3. Given that the sum of the first $n(n>1)$ terms of an arithmetic sequence is 2013, the common difference is 2, and the first term is an integer. Then the sum of all possible values of $n$ is . $\qquad$ | 3. 2975 .
Let the first term of the sequence be $a_{1}$, and the common difference $d=2$. Then
$$
\begin{array}{l}
S_{n}=n a_{1}+n(n-1) \\
=n\left(a_{1}+n-1\right)=2013 .
\end{array}
$$
Also, $2013=3 \times 11 \times 61$, and $n$ is a divisor of 2013, so the sum of all possible values of $n$ is
$$
(1+3) \times(1+11) ... | 2975 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,328 |
4. Let $a, b, x \in \mathbf{N}_{+}$, and $a \leqslant b$. $A$ is the solution set of the inequality
$$
\lg b - \lg a < \lg x < \lg b + \lg a
$$
It is known that $|A|=50$. When $ab$ takes its maximum possible value,
$$
\sqrt{a+b}=
$$ | 4. 6 .
It is easy to know, $\frac{b}{a}<x<a b, a \neq 1$.
Therefore, $a \geqslant 2$,
$$
\begin{array}{l}
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 \\
\Rightarrow a b \leqslant 68 .
\end{array}
$$
Upon inspection, when and only when $a=2, b=34$, the equality holds.... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,329 |
5. The function defined on the domain $R$
$$
f(x)=|\lg | x-2||-1 \text{. }
$$
If $b<0$, then the equation concerning $x$
$$
f^{2}(x)+b f(x)=0
$$
has $\qquad$ distinct real roots. | 5. 8 .
From the problem, we know that the graph of $y=\lg x$ is first symmetrical about the $y$-axis, then the part below the $x$-axis is flipped up, followed by a rightward shift of 2 units and a downward shift of 1 unit to form the graph of $f(x)$. The zeros of $g(x)=f^{2}(x)+b f(x)$ are the roots of the equation. I... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,330 |
6. Let the points be
$$
A\left(a, a+\frac{1}{2}\right), B\left(a+1, a+\frac{3}{2}\right) \text {, }
$$
A moving point $P$ is such that its distance to point $M(1,0)$ is 1 unit greater than its distance to the $y$-axis, and its trajectory forms a curve $C$. The line segment $AB$ intersects curve $C$. Then the range of ... | 6. $\left[\frac{1}{2}-\sqrt{2}, \frac{3}{2}-\sqrt{2}\right] \cup\left[\frac{1}{2}+\sqrt{2}, \frac{3}{2}+\sqrt{2}\right]$. From the given conditions, the equation of curve $C$ is $y^{2}=4 x$.
To ensure that segment $A B$ intersects with curve $C$, it is sufficient for points $A$ and $B$ to be on opposite sides of the pa... | \left[\frac{1}{2}-\sqrt{2}, \frac{3}{2}-\sqrt{2}\right] \cup\left[\frac{1}{2}+\sqrt{2}, \frac{3}{2}+\sqrt{2}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,331 |
7. Given that $D$ is a point on side $A B$ of $\triangle A B C$, and $P$ is a point inside $\triangle A B C$, and it satisfies
$$
\begin{array}{l}
\overrightarrow{A D}=\frac{3}{4} \overrightarrow{A B}, \overrightarrow{A P}=\overrightarrow{A D}+\frac{2}{5} \overrightarrow{B C} . \\
\text { Then } \frac{S_{\triangle A P ... | $7 \cdot \frac{3}{10}$.
Notice,
$$
\begin{array}{l}
\overrightarrow{D P}=\frac{2}{5} \overrightarrow{B C} \Rightarrow D P / / B C \\
\Rightarrow \angle A D P=\angle B \\
\Rightarrow \frac{S_{\triangle M P D}}{S_{\triangle A B C}}=\frac{\frac{1}{2}|A D||D P| \sin \angle A D P}{\frac{1}{2}|A B||B C| \sin B} \\
\quad=\fra... | \frac{3}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,332 |
8. In a large cube composed of $2013 \times 2013 \times 2013$ unit cubes, if a line is drawn to pass through the centers of 2013 unit cubes, then such lines can be drawn $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Note: The translation is provided as requested, with the blank space ($\qquad$) preserved as in t... | 8. 12168589 .
Divide such straight lines into three categories, there are
$$
\begin{array}{l}
2013 \times 2013 \times 3+2013 \times 2 \times 3+4 \\
=12168589 \text { (lines). }
\end{array}
$$ | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,333 |
9. (16 points) Given that the three interior angles of $\triangle A B C$ satisfy
$$
\begin{array}{l}
\angle A+\angle C=2 \angle B, \\
\cos A+\cos C=-\frac{\sqrt{2} \cos A \cdot \cos C}{\cos B} .
\end{array}
$$
Find the value of $\cos \frac{A-C}{2}$. | Given that
$$
\angle B=60^{\circ}, \angle A+\angle C=120^{\circ} \text {. }
$$
Let $\frac{\angle A-\angle C}{2}=\alpha$. Then
$$
\angle A-\angle C=2 \alpha \text {. }
$$
Thus, $\angle A=60^{\circ}+\alpha, \angle C=60^{\circ}-\alpha$.
Therefore, $\cos A+\cos C$
$$
\begin{array}{l}
=\cos \left(60^{\circ}+\alpha\right)+... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,334 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$
where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$. | 10. Let $b_{n}=a_{n+1}-a_{n}$.
From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,335 |
11. (20 points) Given the ellipse $C$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
with left and right foci $F_{1}$ and $F_{2}$, and the right vertex $A$. $P$ is any point on the ellipse $C$, and the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is 3, and the minimum value is 2.... | 11. Since $P$ is any point on the ellipse, we have
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=2 a \text {, }
$$
and $a-c \leqslant\left|P F_{1}\right| \leqslant a+c$.
Thus, $y=\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$
$$
\begin{array}{l}
=\left|P F_{1}\right|\left|P F_{2}\right| \cos \angle F_{1} P F_... | \left(\frac{2}{7}, 0\right) | Geometry | proof | Yes | Yes | cn_contest | false | 726,336 |
One. (40 points) Let the midpoints of the three arcs $\overparen{A B}$, $\overparen{B C}$, and $\overparen{A C}$ on the circumcircle of $\triangle A B C$ be $P_{3}$, $P_{1}$, and $P_{2}$, respectively. Their reflections over $A B$, $B C$, and $A C$ are $R_{3}$, $R_{1}$, and $R_{2}$, respectively. If the lines connectin... | Let $\triangle ABC$ have side lengths $a, b, c$, $o = \frac{a+b+c}{2}$, and $I$ be the incenter of $\triangle ABC$.
First, we prove a lemma.
Lemma: The line connecting a vertex to the excenter is parallel and equal to ? times the line connecting the incenter to the midpoint of the corresponding side.
Proof: As shown i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,337 |
Sure, here is the translated text:
```
II. (40 points) Find all positive integer quadruples $(p, q, x, y)$, where $p<q$, and $(p, q)=1$, such that
$$
3^{x \cos \frac{8}{q} \pi}+1=2^{y} \text {. }
$$
``` | From the given, we have $3^{x \cos \frac{2}{q} \pi}=2^{y}-1$.
Thus, $x \cos \frac{p_{-\pi \in}}{q}$.
(1) If $x \cos \frac{p}{q} \pi=0$, then $y=1$.
(2) If $x \cos \frac{p}{q} \pi \neq 0$, then $x \cos \frac{p}{q} \pi \in \mathbf{N}_{+}$.
Therefore, $3^{x \cos \frac{p}{q^{\prime} \pi}}+1=2^{y} \equiv 1(\bmod 3)$
$\Righ... | (p, q, 0,1),(1,2, x, 1),(1,3,2,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,339 |
Three, (50 points) Let $n \in \mathbf{N}_{+}$. Prove: there exists $m \in \mathbf{N}_{+}$ such that the congruence equation $x^{2} \equiv 1(\bmod m)$ has at least $n$ solutions. | For any odd prime $p$, the congruence equation
$$
x^{2} \equiv 1(\bmod p)
$$
has exactly two distinct solutions:
$x=1$ or $p-1(\bmod p)$.
Take $s$ different odd primes, where $s$ is a positive integer satisfying $2^{s} \geqslant n$.
Let $m=p_{1} p_{2} \cdots p_{\mathrm{s}}$ be the product of these $s$ primes.
Consider... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,340 |
Four. (50 points) Given $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers, and $\left|x_{i}\right| \geqslant 1$, for $X=\{1,2, \cdots, n\}$, the subset $A=$ $\left\{i_{1}, i_{2}, \cdots, i_{t}\right\}$, define
$$
S(A)=x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{1}},
$$
where it is specified that $S(\varnothing)=0$. Question: From t... | Let's assume all $x_{i}>0$.
In fact, if there is some $x_{i}<0$, then replace $x_{i}$ with $-x_{i}$, and change the set $A$ to:
$A^{\prime}=A \cup\{i\}(i \notin A)$ or $A^{\prime}=A \backslash\{i\}(i \in A)$.
Thus, the "sum $S(A)$" becomes $S\left(A^{\prime}\right)=S(A)-x_{i}$, meaning all $2^{n}$ sums increase by $-x_... | \mathrm{C}_{n}^{\left[\frac{n}{2}\right]} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,341 |
Given $G$ is the centroid of $\triangle ABC$, $M$ is the midpoint of side $AC$, and $AC = 2\sqrt{3} GM$. $D$ is any point on the extension of $GA$, and $DM$ is connected. Point $E$ is taken on $DM$ such that $\angle AED = \angle CAG$. Line $CF \parallel AB$ intersects line $BE$ at point $F$, and $CD$ intersects $MF$ at... | Prove as shown in Figure 2, connect $CE$.
It is easy to know that points $B$, $G$, and $M$ are collinear, and $BM = 3GM$.
From the given, $AM = \sqrt{3} GM$.
Thus, $\frac{AM}{BM} = \frac{GM}{AM}$
$\Rightarrow \triangle ABM \backsim \triangle GAM$
$\Rightarrow \angle ABM = \angle GAM = \angle AED$.
Therefore, points $A$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,342 |
In trapezoid $A B C D$, it is known that $A D / /$ $B C, A D<B C, A C$ intersects $B D$ at point $O$, points $E$ and $F$ are on $A B$ and $C D$ respectively, satisfying $\angle A E D=\angle B E C$, $\angle D F A=\angle C F B$. Prove: $O E=O F$. | Proof As shown in Figure 3, extend $BA$ and $CD$ to intersect at point $P$, and extend $CE$ to point $X$ and $BF$ to point $Y$.
Let $AD=a, BC=b$.
Since $\angle PEX = \angle BEC = \angle AED$, we know that $EP$ is the external angle bisector of $\triangle ECD$, and
$$
\frac{ED}{EC} = \frac{PD}{PC} = \frac{AD}{BC} = \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,343 |
Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.
Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,344 |
Given $n$ is a positive integer. Prove:
$$
\sum_{i=1}^{n} \frac{i}{2^{i}} \geqslant n \sqrt{\frac{n}{2^{n+1}}} \text {. }
$$ | $\begin{array}{l}\text { Prove } \quad \sum_{i=1}^{n} \frac{i}{2^{i}}=\frac{1}{2}\left(\sum_{i=1}^{n} \frac{i}{2^{i}}+\sum_{i=1}^{n} \frac{i}{2^{i}}\right) \\ =\frac{1}{2}\left(\sum_{i=1}^{n} \frac{i}{2^{i}}+\sum_{i=1}^{n} \frac{n+1-i}{2^{n+1-i}}\right) \\ =\frac{1}{2} \sum_{i=1}^{n}\left(\frac{i}{2^{i}}+\frac{n+1-i}{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,345 |
Example 2 Given $z=(\sqrt{3}-3 \mathrm{i})^{n}$. If $z$ is a real number, then the smallest positive integer $n$ is ( ). ${ }^{[1]}$
(A) 3
(B) 4
(C) 5
(D) 6 | Solution 1 Notice that; when $n=1$, $z=\sqrt{3}-3 \mathrm{i}$ is not a real number;
when $n=2$, $z=(\sqrt{3}-3 \mathrm{i})^{2}=-6-6 \sqrt{3} \mathrm{i}$ is not a real number;
when $n=3$, $z=(\sqrt{3}-3 \mathrm{i})^{3}=-24 \sqrt{3}$ is a real number.
Thus, the smallest positive integer $n=3$.
Solution 2 Notice that;
$... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,346 |
Example 3 Given $|z|=1, k$ is a real number, $z$ is a complex number. Find the maximum value of $\left|z^{2}+k z+1\right|$. | Let $z=\cos q+\mathrm{i} \sin q$. Then
$$
\begin{array}{l}
z^{2}+k z+1 \\
=\cos 2 q+\mathrm{i} \sin 2 q+k(\cos q+\mathrm{i} \sin q)+1 \\
=(\cos 2 q+k \cos q+1)+\mathrm{i}(\sin 2 q+k \sin q) .
\end{array}
$$
Therefore, $|z^{2}+k z+1|$
$$
\begin{array}{l}
=\sqrt{(\cos 2 q+k \cos q+1)^{2}+(\sin 2 q+k \sin q)^{2}} \\
=\sq... | \left|z^{2}+k z+1\right|_{\max }=\left\{\begin{array}{ll}k+2, & k \geqslant 0 \text {; } \\ 2-k, & k<0 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,347 |
Example 4 Let $z$ be a complex number with modulus 2. Then the sum of the maximum and minimum values of $\left|z-\frac{1}{z}\right|$ is $\qquad$ [2] | Given $|z|=2$, we know
$$
\begin{array}{l}
|z+1|^{2}=(z+1)(\bar{z}+1) \\
=z \bar{z}+z+\bar{z}+1=5+2 \operatorname{Re} z, \\
|z-1|^{2}=(z-1)(\bar{z}-1) \\
=z \bar{z}-z-\bar{z}+1=5-2 \operatorname{Re} z .
\end{array}
$$
Therefore, $\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|$
$$
=\frac{|z+1||z-1|}{|z|}=\fr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,348 |
Example 5 Given
$$
A=\left\{z \mid z^{18}=1\right\} \text { and } B=\left\{\omega \mid \omega^{48}=1\right\}
$$
are sets of complex roots of unity,
$$
C=\{z w \mid z \in A, w \in B\}
$$
is also a set of complex roots of unity. How many distinct elements are there in the set $C$? ${ }^{[3]}$ | Notice that, $z=\cos \frac{2 k \pi}{18}+\mathrm{i} \sin \frac{2 k \pi}{18}(k \in \mathbf{Z})$ (18 distinct elements),
$\omega=\cos \frac{2 t \pi}{48}+\mathrm{i} \sin \frac{2 t \pi}{48}(t \in \mathbf{Z})$ (48 distinct elements),
$$
\begin{array}{l}
z \omega=\cos \frac{2 \pi(8 k+3 t)}{144}+\mathrm{i} \sin \frac{2 \pi(8 k... | 144 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,349 |
For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$ | Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mat... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,350 |
Example 7 Let three complex numbers
\[
\begin{array}{l}
z_{1}=\cos A+\mathrm{i} \sin A, \\
z_{2}=\cos B+\mathrm{i} \sin B, \\
z_{3}=\cos C+\mathrm{i} \sin C,
\end{array}
\]
and \( z_{1}+z_{2}+z_{3}=0 \).
Find \( \cos (A-B) \). | Given $z_{1}+z_{2}+z_{3}=0$, we have
$$
\begin{array}{l}
\cos A+\cos B+\cos C=0, \\
\sin A+\sin B+\sin C=0 . \\
\text { Therefore, } \cos A+\cos B=-\cos C, \\
\sin A+\sin B=-\sin C .
\end{array}
$$
Squaring both equations and then adding them, we get
$$
2+2 \cos (A-B)=1 \text { . }
$$
Thus, $\cos (A-B)=-\frac{1}{2}$. | \cos (A-B)=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,351 |
Example 8 As shown in Figure 1, in the acute $\triangle A B C$, it is known that $\angle A>60^{\circ}, H$ is the orthocenter of $\triangle A B C$, points $M$ and $N$ are on sides $A B$ and $A C$ respectively, $\angle H M B=\angle H N C=60^{\circ}, O$ is the circumcenter of $\triangle H M N$, point $D$ is on the same si... | Proof: Establish a complex plane with $O$ as the origin, and use the letters of each point to represent the complex number corresponding to that point.
Let $H=1, N=\mathrm{e}^{2 \alpha \mathrm{i}}, M=\mathrm{e}^{-2 \beta \mathrm{i}}$.
Then $|N H|=2 \sin \alpha,|M H|=2 \sin \beta$,
$\angle H C A=\angle H B A=\frac{5 \pi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,352 |
Example 3 As shown in Figure 1, divide $\triangle G R B$ into 25 small triangles, and color all these triangle vertices according to the following rules: vertex $G$ is colored green, vertex $R$ is colored red, vertex $B$ is colored blue, vertices on edge $G R$ are colored green or red, vertices on edge $R B$ are colore... | Prove that a triangle with three vertices colored in three different colors is called a "heterochromatic triangle".
Place a point in each of the 25 small triangles, with each point corresponding to a small triangle. Denote this graph as $H$.
(1) Merge a same-color segment $AB$ (where $A$ and $B$ are the same color) in... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,353 |
2. Find the smallest positive integer $n$, such that
$$
I=\left(\frac{1}{2}+\frac{1}{2 \sqrt{3}} \mathrm{i}\right)^{n}
$$
is a pure imaginary number, and find $I$. | Answer: $n=3, I=\frac{\sqrt{3}}{9} \mathrm{i}$. | n=3, I=\frac{\sqrt{3}}{9} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,355 |
3. Calculate:
$\arctan \frac{1}{3}+\arctan \frac{1}{5}+\arctan \frac{1}{7}+\arctan \frac{1}{8}$ $=$ . $\qquad$ | Due to $\arctan \frac{1}{k}=\arg (k+\mathrm{i})(k=3,5,7,8)$,
and $(3+\mathrm{i})(5+\mathrm{i})(7+\mathrm{i})(8+\mathrm{i})=650(1+\mathrm{i})$, its principal value of the argument is $\frac{\pi}{4}$. | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,356 |
Question 1 At different times during a day, the manager hands documents to the secretary for printing, each time placing a document on top of the stack of documents the secretary is to print. The secretary, when available, takes the document from the top of the stack to print, taking one at a time. If there are $n$ doc... | Solve: Record the secretary's printing order using a grid paper. The method is as follows.
Rule: If the manager places another document on top of the stack of documents to be printed, draw a line segment with a slope of 1 to fill one grid, and call this operation $A$; if the secretary takes the top document from the s... | \mathrm{C}_{2 n}^{n}-\mathrm{C}_{2 n}^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,358 |
Question 2 A (chess) rook is located on an $m \times n$ chessboard. Two players take turns moving it horizontally or vertically, as many squares as they wish. The squares the rook passes over (including the ones it stops on) are colored, and the rook cannot pass over or stop on already colored squares. If a player cann... | (1) First, consider the scenario where the rook is located at one of the four corners of the chessboard.
Assume $n \geqslant m$.
If the first player always moves the rook as far as possible in the horizontal direction, forcing the second player to only make vertical moves, the first player can win. This is because, aft... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,359 |
1. In $\triangle A B C$, it is known that the height on side $B C$ is 3, and two lines $l_{1} 、 l_{2}$ parallel to $B C$ divide the area of $\triangle A B C$ into three equal parts. Then the distance between the lines $l_{1}$ and $l_{2}$ is | $-1 . \sqrt{6}-\sqrt{3}$.
As shown in Figure 2, let $l_{1}$ and $l_{2}$ intersect the altitude from $A$ to side $BC$ at points $E$ and $F$, respectively, intersect $AB$ at points $G$ and $J$, and intersect $AC$ at points $H$ and $K$.
By similar triangles, we have
$$
\frac{A E}{A D}=\sqrt{\frac{S_{\triangle 1 G H}}{S_{... | \sqrt{6}-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,360 |
2. When two dice are rolled simultaneously, if $P(a)$ represents the probability that the sum of the numbers facing up on the two dice is $a$, then
$$
\begin{aligned}
& P(1)+P(2)+P(3)+P(4)+P(5) \\
= &
\end{aligned}
$$ | 2. $\frac{5}{18}$.
Let the two dice be A and B. Throwing two dice has a total of $6 \times 6=36$ possible outcomes.
Obviously, $P(1)=0$.
The sum of the numbers facing up on the two dice being 2 only has one possibility: both A and B show 1, so $P(2)=\frac{1}{36}$.
The sum of the numbers facing up on the two dice bein... | \frac{5}{18} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,361 |
3. In the Cartesian coordinate system $x O y$, it is known that point $A(1,0)$, and point $B$ lies on the line $y=3x$. If $\triangle A O B$ is an isosceles triangle, then the coordinates of point $B$ are $\qquad$ | 3. $\left( \pm \frac{\sqrt{10}}{10}, \pm \frac{3 \sqrt{10}}{10}\right),\left(\frac{1}{5}, \frac{3}{5}\right),\left(\frac{1}{2}, \frac{3}{2}\right)$. Let $B(t, 3 t)(t \neq 0)$.
If $O A=O B$, then $\sqrt{t^{2}+(3 t)^{2}}=1$.
Thus, $t= \pm \frac{\sqrt{10}}{10}$.
Therefore, $B\left( \pm \frac{\sqrt{10}}{10}, \pm \frac{3 \s... | \left( \pm \frac{\sqrt{10}}{10}, \pm \frac{3 \sqrt{10}}{10}\right),\left(\frac{1}{5}, \frac{3}{5}\right),\left(\frac{1}{2}, \frac{3}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,362 |
4. In rectangle $A B C D$, it is known that $A B=5, B C=9$, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=C G=3, B F=D H=4, P$ is a point inside the rectangle. If the area of quadrilateral $A E P H$ is 15, then the area of quadrilateral $P F C G$ is $\qquad$ | 4. 11.
As shown in Figure 3, let the distances from $P$ to $AB$ and $AD$ be $a$ and $b$, respectively. Then the distances from $P$ to $BC$ and $CD$ are $5-b$ and $9-a$, respectively. Given that $S_{\text {quadrilateral } AEPH}=\frac{1}{2}(3a+5b)=15$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral PFCG }}=\frac{1}... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,363 |
Example 4 A country has $n$ cities, and between any two cities, there is at most one direct road, and these roads satisfy: after blocking all the roads connected to any one city, any two other cities in the country can still be connected through other roads.
Now, all the roads in the country are changed to one-way roa... | Solve $k_{\text {min }}=2 n-3$.
Using graph theory ideas.
Map cities to points, and roads between two cities to edges connecting two points, forming graph $G$.
Assume the length of each edge is 1. For any two points $X, Y$, let $d(X, Y)$ be the length of the shortest path from $X$ to $Y$.
For any two points $A, B$ in... | 2n-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,364 |
5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5.4.
If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$
Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (correspond... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,365 |
6. A moving point $P$ on a plane is at a distance of 3 from the line containing the segment $A B$ of length 10. When $P A \cdot P B$ reaches its minimum value, then $P A+P B=$ $\qquad$ | $6.4 \sqrt{10}$.
From $\frac{1}{2} P A \cdot P B \geqslant S_{\triangle P A B}=\frac{1}{2} \times 10 \times 3=15$, we know $P A \cdot P B \geqslant 30$, and the equality holds when $\angle A P B=90^{\circ}$. At this time,
$$
\begin{array}{l}
P A \cdot P B=30, \text{ and } P A^{2}+P B^{2}=10^{2} . \\
\text{Therefore, }... | 4 \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,366 |
7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$ | 7. 100 .
Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,367 |
8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer no... | 8. 2025078.
Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$
From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$
Ther... | 2025078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,368 |
9. (15 points) As shown in Figure 1, let $P$ be a point inside the square $ABCD$. Draw perpendiculars from $P$ to sides $BC$ and $CD$, with feet of the perpendiculars being $M$ and $N$ respectively. If $AP \perp MN$, prove that $AP = MN$ or $AP \perp BD$. | 9. As shown in Figure 4, extend $M P$ to intersect $A D$ at point $Q$. Let $A B=a, C M=x, C N=y$.
Then $A Q=a-x, P Q=a-y$.
Since $A P \perp M N$, therefore, Rt $\triangle P A Q \sim$ Rt $\triangle M N C$. Thus, $\frac{A Q}{P Q}=\frac{N C}{M C}$, which means $\frac{a-x}{a-y}=\frac{y}{x}$.
Hence, $(x-y)(a-x-y)=0$.
Theref... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,369 |
10. (15 points) Solve the system of equations
$$
\left\{\begin{array}{l}
a b+c+d=3 \\
b c+d+a=5 \\
c d+a+b=2 \\
d a+b+c=6
\end{array}\right.
$$ | $$
\begin{array}{l}
\text { 10. (1) +(2)-(3) - (4) gives } \\
(a+c)(b-d)+2(d-b)=0 \\
\Rightarrow(b-d)(a+c-2)=0 .
\end{array}
$$
If $b-d=0$, then $b=d$, which contradicts equations (1) and (4).
If $a+c-2=0$, then
$$
a+c=2 \text {. }
$$
By adding (3) and (4) and combining with equation (5), we get
$$
b+d=3 \text {. }
$... | (a, b, c, d)=(2,0,0,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,370 |
11. (20 points) Given a positive real number $a$, for any positive integer $n$, let
$$
f(n)=\left[\frac{n+\left[\frac{a}{n}\right]}{2}\right],
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
(1) If $f(5)=5$, find the range of values for $a$;
(2) Prove that $f(n)>\sqrt{a}-1$.
保留了源文本的换行和格... | 11. (1) From $f(5)=5$, we know
$$
\begin{array}{l}
{\left[\frac{5+\left[\frac{a}{5}\right]}{2}\right]=5 \Leftrightarrow\left[\frac{a}{5}\right]=5 \text { or } 6} \\
\Leftrightarrow 5 \leqslant \frac{a}{5} < 7 \Leftrightarrow 25 \leqslant a < 35.
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } f(n)=\left[\frac{n... | f(n) > \sqrt{a} - 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,371 |
- 12. (20 points) Prove: In any 2013 distinct real numbers, there must exist two numbers $x, y$ such that
$$
2012|x-y||1-x y| \leqslant\left(1+x^{2}\right)\left(1+y^{2}\right) .
$$ | 12. First, for any real number $a$ we have
$$
-\left(1+a^{2}\right) \leqslant 2 a \leqslant 1+a^{2} \text {. }
$$
Therefore, $-\frac{1}{2} \leqslant \frac{a}{1+a^{2}} \leqslant \frac{1}{2}$.
If we write each of the 2013 distinct numbers $a$ in the form $\frac{a}{1+a^{2}}$,
and divide the segment from point $-\frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,372 |
1. It is known that Rochelle made 8 hamburgers with 3 pounds of meat. She still needs to prepare 24 hamburgers for the neighborhood picnic. Then she needs ( ) pounds of meat.
(A) 6
(B) $6 \frac{2}{3}$
(C) $7 \frac{1}{2}$
(D) 8
(E) 9 | 1.E.
$$
\frac{3}{8} \times 24=9
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,373 |
2. In a certain country, statisticians expect one baby to be born every 8 hours, and one person to die each day. Therefore, the annual increase in population in this country is approximately ( ) people.
(A) 600
(B) 700
(C) 800
(D) 900
(E) 1000 | 2. B.
From $\left(\frac{24}{8}-1\right) \times 365=730$, we know that the population increases by approximately 700 people each year. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,374 |
Example 5 A scientist stored the design blueprint of his time machine in a computer, setting the password to open the file as a permutation of $\{1,2, \cdots, 64\}$. He also designed a program that, when eight positive integers between $1 \sim 64$ are input each time, the computer will indicate the order (from left to ... | Let the password be denoted as $a_{1} a_{2} \cdots a_{n^{2}} (n=8)$.
First, write the numbers $1, 2, \cdots, n^{2}$ arbitrarily into an $n \times n$ grid (one number per cell). After the first $n$ operations, input each row of numbers once, and rearrange the numbers in each row from left to right according to the compu... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,375 |
4. Pete's family bought a 12-slice pizza. Pete ate one slice and shared another slice equally with his brother Paul. Then Pete ate ( ) of the whole pizza.
(A) $\frac{1}{24}$
(B) $\frac{1}{12}$
(C) $\frac{1}{8}$
(D) $\frac{1}{6}$
(E) $\frac{1}{4}$ | $\begin{array}{l}\text { 4. C. } \\ \frac{1.5}{12}=\frac{1}{8} \text {. }\end{array}$ | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 726,377 |
5. A rectangular photo is framed with a requirement that there be a 2-inch border around the photo. Upon measurement, the photo is found to be 8 inches tall and 10 inches wide. The area of the frame excluding the photo is ( ) square inches.
(A) 36
(B) 40
(C) 64
(D) 72
(E) 88 | 5. E.
As shown in Figure 4.
The area of the remaining part is
$$
2 \times 2 \times 4 + 2 \times 8 \times 2 + 10 \times 2 \times 2 = 88 \text{.}
$$ | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,378 |
6. Isabella is going to take four math tests scored out of 100. Her goal is to achieve an average score of 95. It is known that her scores for the first two tests were 97 and 91, respectively. After the third test, she believes she can reach her goal. What is the minimum score she needs to achieve on the third test?
(A... | 6. E:
Let's assume the score of the third exam is $x$. Then
$$
97+91+x \geqslant 95 \times 3 \Rightarrow x \geqslant 97 \text{. }
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,379 |
8. The mall advertisement reads "50% off sale on every item", and at checkout, if a coupon is used, the price is reduced by another 20%. Then, after using the coupon, the payment price is ( ) of the original.
(A) $10 \%$
(B) $33 \%$
(C) $40 \%$
(D) $60 \%$
(E) $70 \%$ | 8. C.
Let the original price be $x$, the current price be $x \times 50\% \times 80\%$.
$$
\text{Then } \frac{x \times 50\% \times 80\%}{x} \times 100\% = 40\% \text{.}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,381 |
9. The Fort Worth Zoo has several two-legged birds and four-legged mammals. During a visit to the zoo, Maggie counted a total of 200 heads and 522 legs among these animals. How many birds are there in the zoo? ( )
(A) 61
(B) 122
(C) 139
(D) 150
(E) 161 | 9. C.
Let the number of birds and mammals be $x$ and $y$ respectively. According to the problem,
$$
\left\{\begin{array} { l }
{ x + y = 2 0 0 , } \\
{ 2 x + 4 y = 5 2 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=139, \\
y=61 .
\end{array}\right.\right.
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,382 |
10. Among the four-digit numbers composed of the digits $2,0$ and $1,2$, those greater than 1000 are ( ) in number.
(A) 6
(B) 7
(C) 8
(D) 9
(E) 12 | 1. D.
The four-digit numbers that meet the criteria are:
$$
\begin{array}{l}
2012, 2021, 2102, 2120, 2201, 2202, \\
2210, 1022, 1202, 1220 .
\end{array}
$$ | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,383 |
The units digit of $12.13^{2012}$ is ( ).
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9 | 12. A.
From $13^{4}=1(\bmod 10)$
$$
\Rightarrow 13^{2012}=\left(13^{4}\right)^{503} \equiv 1(\bmod 10)
$$ | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,385 |
Example 1 Robot Crossing Ditch Problem.
On a straight road, a small ditch of width $d$ is dug at equal intervals, with the center distance between each adjacent two ditches being $a(a>d)$.
A robot moves forward along this straight road, with a step length of $b$, where $a$ and $b$ are incommensurable (i.e., $\frac{a}{... | Consider the straight path as a number line, and the robot's footprints as a sequence of points $\left\{x_{n}\right\}$ on the number line.
Imagine wrapping the number line around a circle with circumference $a$. Then all the small ditches $\left(\alpha_{k}, \beta_{k}\right)(k=1,2, \cdots)$ will coincide on the same ar... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,386 |
13. Jemore bought some pencils at the school bookstore, each costing more than 1 cent, and paid a total of $1.43; Selena also bought the same type of pencils and paid $1.87. Then Selena bought ( ) more pencils than Jemore.
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 | 13. C.
Notice that, $(143,187)=11$.
Then the price of a pencil is a divisor of 11.
Since the price is more than 1 cent, the price of each pencil is
11 cents, and the number of pencils Selena bought more than Jamir is
$$
\frac{187-143}{11}=4 \text {. }
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,387 |
14. In the annual football league of a certain middle school, it is stipulated that each team must play one match against every other team. If in the 2012 season all teams played a total of 21 matches. Then the number of participating teams is ( ).
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10 | 14. B.
There are $n$ teams participating in the league.
$$
\text { Then } \frac{n(n-1)}{2}=21 \Rightarrow n=7 \text {. }
$$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,388 |
15. Given an integer $m(m>2)$, when $m$ is divided by $3,4,5,6$, the remainder is 2 in each case. Then $m$ lies between ( ) .
(A) 40 and 50
(B) 51 and 55
(C) 56 and 60
(D) 61 and 65
(E) 66 and 99 | $\begin{array}{l}\text { 15. D. } \\ {[3,4 ; 5,6]+2=62}\end{array}$
The translation is:
$\begin{array}{l}\text { 15. D. } \\ {[3,4 ; 5,6]+2=62}\end{array}$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,389 |
16. Using each of the ten digits $0 \sim 9$ exactly once to form two five-digit numbers. If the sum of the two numbers is to be maximized, then one of the numbers is ( ).
(A) 76531
(B) 86724
(C) 87431
(D) 96240
(E) 97403 | 16. C.
Obviously, the numbers in options B, D, and E are not arranged in descending order, so their sums are not the largest. Therefore, we only need to discuss options A and C. By forming the largest five-digit number with the remaining digits and adding it to the given number, we can verify that the sum with option ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,390 |
17. Divide a square with a side length of a positive integer into ten smaller squares, such that the side length of each smaller square is a positive integer, and at least eight of them have an area of 1. Then the minimum possible value of the side length of the original square is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7 | 17. B.
If the side length is 3, and the area is 9, it is impossible to divide it into ten smaller squares with integer side lengths; if the side length is 4, then it is possible (as shown in Figure 5). | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,391 |
18. There is a positive integer, which is neither a prime number nor a perfect square, and has no prime factors less than 50. Then the smallest such positive integer is ( ).
(A) 3127
(B) 3133
(C) 3137
(D) 3139
(E) 3149 | $\begin{array}{l}\text { 18. A. } \\ 53 \times 59=3127\end{array}$ | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,392 |
19. In a jar, there are red, green, and blue balls. It is known that there are 6 balls other than the red ones; 8 balls other than the green ones; and 4 balls other than the blue ones. Then the total number of balls in the jar is ( ).
(A) 6
(B) 8
(C) 9
(D) 10
(E) 12 | 19. C.
Let the number of red, green, and blue balls be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
y+z=6, \\
z+x=8, \\
x+y=4 .
\end{array}\right.
$$
Solving these equations, we get $x+y+z=9$. | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 726,393 |
21. Mara has a large white cube with a side length of 10 feet, and she has enough green paint to cover 300 square feet. Mara wants to paint a white square in the center of each face and use all the paint around the squares. Then the area of each white square is ( ) square feet.
(A) $5 \sqrt{2}$
(B) 10
(C) $10 \sqrt{2}$... | 21. D.
$$
\frac{10 \times 10 \times 6-300}{6}=\frac{300}{6}=50 \text {. }
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,395 |
22. Let set $R$ consist of nine different integers, six of which are $2,3,4,6,9, 14$. Then the number of possible medians of the elements in set $R$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8 | 22. C
From the given information, we know that the median of the nine numbers can only be selected from the seven numbers 3～9. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,396 |
23. Given that the perimeter of an equilateral triangle and a regular hexagon are the same. If the area of the equilateral triangle is 4, then the area of the regular hexagon is ( ).
(A) 4
(B) 5
(C) 6
(D) $4 \sqrt{3}$
(E) $6 \sqrt{3}$ | 23. C.
Let the side length of the equilateral triangle be $a$. Then the side length of the regular hexagon is $\frac{a}{2}$.
From the problem, we know
$$
\frac{\sqrt{3}}{4} a^{2}=4 \Rightarrow a^{2}=\frac{16}{\sqrt{3}} \text {. }
$$
Therefore, the area of the regular hexagon is
$$
\frac{3 \sqrt{3}}{2}\left(\frac{a}{2... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,398 |
24. As shown in Figure 2, a circle with a radius of 2 is divided into four equal arcs, which are then arranged to form a star shape. The ratio of the area of this shape to the area of the original circle is ( ).
(A) $\frac{4-\pi}{\pi}$
(B) $\frac{1}{\pi}$
(C) $\frac{\sqrt{2}}{\pi}$
(D) $\frac{\pi-1}{\pi}$
(E) $\frac{3}... | 24. A.
As shown in Figure 6.
Notice that, the area of the star-shaped figure is the difference between the area of the square with side length 4 and the area of the circle.
Therefore, the required ratio of the areas is
$$
\frac{4^{2}-4 \pi}{4 \pi}=\frac{4-\pi}{\pi} .
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,399 |
25. As shown in Figure 3, a square with an area of 4 is inscribed in a square with an area of 5, with each vertex of the smaller square touching the sides of the larger square, and each vertex divides the side into two segments of lengths $a$ and $b$. Then $ab=$ ( ).
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{1}{2}... | 25. C.
From the problem, we know
$$
\frac{1}{2} a b \times 4=5-4 \text {. }
$$
Solving for $a b$ gives $a b=\frac{1}{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,400 |
1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$
If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers. | 1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.
Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.
Below, we prove by contradiction: For any ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,401 |
2. Does there exist a pair of functions $(g, h)$ on $\mathbf{R} \rightarrow \mathbf{R}$ such that for any function $f: \mathbf{R} \rightarrow \mathbf{R}$, if for all $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(g(x))=g(f(x)), \\
f(h(x))=h(f(x)),
\end{array}
$$
then $f$ must be the identity function, i.e., $f(x) \... | 2. There exists such a pair of functions.
First, establish a bijection between $\mathbf{R}$ and the unit closed interval.
Thus, it is sufficient to find a pair of functions on the unit interval.
Here is a specific example: take positive real numbers $\alpha, \beta$, and define
$$
\begin{array}{l}
g(x)=\max \{x-\alpha,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,402 |
3. Given that quadrilateral $A B C D$ is inscribed in $\odot O$, line $A B$ intersects $C D$ at point $P$, $A D$ intersects $B C$ at point $Q$, and diagonals $A C$ and $B D$ intersect at point $R$. If $M$ is the midpoint of segment $P Q$, and $K$ is the intersection of segment $M R$ with $\odot O$, prove: $\odot O$ is ... | 3. Notice that, $P$, $Q$, $R$ are the poles (with respect to $\odot O$) of $QR$, $RP$, $PQ$.
Thus, $OP \perp QR$, $OQ \perp RP$, $OR \perp PQ$.
Therefore, $R$ is the orthocenter of $\triangle OPQ$.
If $MR \perp PQ$, then $M$, $R$, $O$ are collinear, and $\triangle PQR$ is symmetric about this line. The conclusion is ob... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,403 |
4. Let $P, P^{\prime}$ be two intersecting convex quadrilateral regions on a plane, and $O$ be a point in their intersection.
Assume that for any line passing through $O$, the segment it intercepts in region $P$ is longer than the segment it intercepts in region $P^{\prime}$. Question: Is it possible for the area rati... | 4. Possible.
For any $\varepsilon>0$, construct regions $P^{\prime}$ and $P$ such that the ratio of their areas is greater than $2-\varepsilon$.
Let $O$ be the center of the square $A B C D$, and let $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ be the reflections of $O$ about $A 、 B 、 C$ respectively.
Notice that $l$ is a... | 2-\frac{\varepsilon}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,404 |
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Given a positive number $k(k \geqslant 2)$, let $a_{1}=1$, and for any integer $n(n \geqslant 2)$, $a_{n}$ is the smallest solution greater than $a_{n-1}$ of the equation
$$
x=1+\sum_{i=1}^{n-1}\left[\sqrt[k]{\frac{x}{a_{i}}}\right]
$$
Prove: ... | 5. From the problem, we know that $a_{n}$ is the set of all positive integers that are non-prime $k$-th powers. Thus, the conclusion is proved.
Let $B$ be the set of all positive integers that are non-prime $k$-th powers.
First, we prove: For any positive integer $c$, we have
$$
\sum_{\substack{b \in B \\ b \leq c}}\l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,405 |
6. Given that $1,2, \cdots, 2 n$ are placed at the vertices of a regular $2 n$-gon. A move is defined as selecting two numbers on the vertices of one edge of the $2 n$-gon and swapping them. Suppose after a finite number of moves, each pair of numbers has been swapped exactly once. Prove: there exists an edge that has ... | 6. Take any three numbers $i<j<k$, their order on the circumcircle of a regular $2n$-gon (in a clockwise direction) can only be $i, j, k$ or $i, k, j$.
If two of these numbers are swapped, the order changes. Thus, swapping these three numbers three times, the original clockwise order $i, j, k$ becomes $k, j, i$. Then,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,406 |
1. Given the set
$$
A=\{x \mid 5 x-a \leqslant 0, a \in \mathbf{N}\} \text {. }
$$
If $5 \in A \cap \mathbf{Z}$, then the minimum value of $a$ is | $$
-, 1.25
$$
From $A \left\lvert\,=\left(-\infty, \frac{a}{5}\right]\right.$, and $5 \in A \cap \mathbf{Z}$, we know
$$
\frac{a}{5} \geqslant 5 \Rightarrow a \geqslant 25 \text {. }
$$
Therefore, the minimum value of $a$ is 25. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,407 |
Example 3 refers to a number formed by the first few digits of a natural number $a$ as a "prefix" of $a$ (for example, 2, 20, 201, 2013 are all prefixes of the number 2013). Prove: For any given positive integer $M$, there exists a positive integer $n$, such that $M$ is a prefix of $2^{n}$. | Prove that when $M$ is exactly a non-negative integer power of 2, the conclusion is obvious.
For the following, assume $M$ is not a non-negative integer power of 2.
To prove: There exist positive integers $n, k$ such that $\left[\frac{2^{n}}{10^{k}}\right]=M$ (where $[x]$ denotes the greatest integer not exceeding the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,408 |
2. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
\begin{array}{l}
f(x+1)=f(-x), \\
f(x)=\left\{\begin{array}{ll}
1, & -1<x \leqslant 0 \\
-1, & 0<x \leqslant 1 .
\end{array}\right.
\end{array}
$$
Then $f(f(3.5))=$ $\qquad$ | 2. -1 .
From $f(x+1)=-f(x)$, we know $f(x+2)=f(x)$.
Then $f(3.5)=f(-0.5)=1$.
Therefore, $f(f(3.5))=f(1)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,409 |
3. In $\triangle A B C$, it is known that the angle between $\overrightarrow{A B}$ and $\overrightarrow{B C}$ is $150^{\circ},|\overrightarrow{A C}|=2$. Then the range of $|\overrightarrow{A B}|$ is | 3. $(0,4]$.
From the fact that the angle between $\overrightarrow{A B}$ and $\overrightarrow{B C}$ is $150^{\circ}$, we know $\angle B=30^{\circ}$.
By the Law of Sines, we have
$$
\frac{|\overrightarrow{A B}|}{\sin C}=\frac{|\overrightarrow{A C}|}{\sin 30^{\circ}}=4 \text {. }
$$
Thus, $|\overrightarrow{A B}|=4 \sin ... | (0,4] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,410 |
4. Given that the coordinates of points $M$ and $N$ satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x \geqslant 0, \\
y \geqslant 0, \\
x+2 y \leqslant 6, \\
3 x+y \leqslant 12 .
\end{array}\right.
$$
If $\boldsymbol{a}=(1,-1)$, then the range of $\overrightarrow{M N} \cdot \boldsymbol{a}$ is
$\qquad$ | 4. $[-7,7]$.
Let points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Then by the conditions we have
$$
\begin{array}{l}
0 \leqslant x_{1}, x_{2} \leqslant 4,0 \leqslant y_{1}, y_{2} \leqslant 3 . \\
\text { Hence } \overrightarrow{M N} \cdot a=(\overrightarrow{O N}-\overrightarrow{O M}) \cdot a \\
=\over... | [-7,7] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,411 |
5. Given the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
the graph always passes through a fixed point $A$. If point $A$ lies on the line
$$
\frac{x}{m}+\frac{y}{n}+1=0(m, n>0)
$$
then the minimum value of $3 m+n$ is | 5. 16 .
Note that the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
always passes through the fixed point $(-3,-1)$.
So point $A(-3,-1)$.
Then $-\frac{3}{m}-\frac{1}{n}+1=0 \Rightarrow 1=\frac{3}{m}+\frac{1}{n}$.
Thus, $3 m+n=(3 m+n)\left(\frac{3}{m}+\frac{1}{n}\right)$
$$
=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 \te... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,412 |
6. Given that $P$ is a moving point on the line $l$:
$$
k x+y+4=0(k>0)
$$
$P A$ and $P B$ are the two tangents from $P$ to the circle $C$:
$$
x^{2}+y^{2}-2 y=0
$$
with points of tangency $A$ and $B$ respectively. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$ | 6.2.
$$
\begin{array}{l}
\text { Given } S_{\text {quadrilateral } P A C B}=P A \cdot A C=P A \\
=\sqrt{C P^{2}-C A^{2}}=\sqrt{C P^{2}-1},
\end{array}
$$
we know that the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$. Also, $\sqrt{C P^{2}-1}=2$, then $C P=\sqrt{5}$. Using the point-to-line dist... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,413 |
7. In the tetrahedron $A-B C D$, it is known that the lateral edges $A B$, $A C$, and $A D$ are pairwise perpendicular, and the areas of $\triangle A B C$, $\triangle A C D$, and $\triangle A D B$ are $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, and $\frac{\sqrt{6}}{2}$, respectively. Then the volume of the circumscribe... | 7. $\sqrt{6} \pi$.
Let $A B=a, A C=b, A D=c$. Then
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a b=\frac{\sqrt{2}}{2}, \\
S_{\triangle A C D}=\frac{1}{2} b c=\frac{\sqrt{3}}{2}, \\
S_{\triangle D B}=\frac{1}{2} c a=\frac{\sqrt{6}}{2} .
\end{array}
$$
Multiplying and simplifying the above three equations yield... | \sqrt{6} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,414 |
8. Given the sequence $a_{n}$ satisfies
$$
\begin{array}{l}
a_{1}=m\left(m \in \mathbf{N}_{+}\right), \\
a_{n+1}=\left\{\begin{array}{ll}
\frac{a_{n}}{2}, & a_{n} \text { is even; } \\
3 a_{n}+1, & a_{n} \text { is odd. }
\end{array}\right.
\end{array}
$$
If $a_{4}=7$, then the possible values of $m$ are | 8.9 or 56.
(1) $m$ is an odd number.
From $a_{2}=3 a_{1}+1=3 m+1$ being even, we know $a_{3}=\frac{3 m+1}{2}$, $a_{4}=\frac{3 m+1}{4}=7$ or $a_{4}=\frac{3(3 m+1)}{2}+1=7$.
Solving gives $m=9$ or 1.
Upon verification, $m=9$.
(2) $m$ is an even number.
(i) When $m=4 m_{0}$,
$$
\begin{array}{l}
a_{2}=\frac{a_{1}}{2}=2 m_... | 9 \text{ or } 56 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,415 |
9. (14 points) In $\triangle ABC$, it is known that the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, the circumradius of $\triangle ABC$ is $R=\sqrt{3}$, and it satisfies
$\tan B + \tan C = \frac{2 \sin A}{\cos C}$.
Find: (1) $\angle B$, $b$;
(2) the maximum area of $\tri... | (1) From the given,
$$
\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}=\frac{2 \sin A}{\cos C},
$$
which means $\sin B \cdot \cos C+\sin C \cdot \cos B=2 \sin A \cdot \cos B$.
Thus, $\sin (B+C)=2 \sin A \cdot \cos B$.
Since $\angle A+\angle B+\angle C=\pi$, we have
$\sin (B+C)=\sin A$.
Given $\sin A \neq 0$, then $\cos B=... | \frac{9 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,416 |
10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the sh... | 10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles.
As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position o... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,417 |
11. (14 points) Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms as $S_{n}$, and satisfy
$$
S_{2}=3,2 S_{n}=n+n a_{n}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Find: (1) The general term formula of the sequence $\left\{a_{n}\right\}$;
(2) The sum of the first $2 n$ terms $T_{2 n}$ of th... | 11. (1) From the problem, we know that when $n=1$, $a_{1}=1$.
When $n \geqslant 2$,
$$
\begin{array}{l}
2 S_{n}=n+n a_{n}, \\
2 S_{n-1}=n-1+(n-1) a_{n-1} .
\end{array}
$$
Subtracting the above two equations gives
$$
2 a_{n}=1+n a_{n}-(n-1) a_{n-1} \text {. }
$$
Then $2 a_{n+1}=1+(n+1) a_{n+1}-n a_{n}$.
Subtracting (... | 2^{2 n+1}+n^{2}+2 n-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,418 |
Example 5 Let $S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{N}_{+}\right)$. Prove: For any real numbers $a, b$ satisfying $0 \leqslant a<b \leqslant 1$, the sequence $\left\{S_{n}-\left[S_{n}\right]\right\}$ has infinitely many terms belonging to $(a, b)$ ( $[x]$ denotes the greatest integer not exceeding ... | Note that,
$(0,+\infty)=\bigcup_{k=0}^{\infty}(k, k+1]$.
Let $a_{k}=k+a, b_{k}=k+b(k=0,1, \cdots)$.
Then $b_{k}-a_{k}=b-a$.
In the small intervals $\left(a_{0}, b_{0}\right),\left(a_{1}, b_{1}\right), \cdots$, the distance between the centers of any two adjacent intervals is 1.
Imagine a robot starting from the origin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,419 |
13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$
and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis... | 13. (1) Let $A(a, 0), C(c, 0)$.
From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$
where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,421 |
14. (15 points) Given positive real numbers $x, y$ satisfy
$$
\begin{array}{l}
a=x+y, \\
b=\sqrt{x^{2}+7 x y+y^{2}} .
\end{array}
$$
(1) When $y=1$, find the range of $\frac{b}{a}$;
(2) If $c^{2}=k x y$, and for any positive numbers $x, y$, the segments with lengths $a, b, c$ can always form a triangle, find the range ... | 14. (1) Notice,
$$
\begin{array}{l}
\frac{b}{a}=\frac{\sqrt{x^{2}+7 x+1}}{x+1} \\
=\sqrt{\frac{x^{2}+7 x+1}{x^{2}+2 x+1}} \\
=\sqrt{1+\frac{5 x}{x^{2}+2 x+1}} \\
=\sqrt{1+\frac{5}{x+2+\frac{1}{x}}} .
\end{array}
$$
Given $x+2+\frac{1}{x} \geqslant 4$, we know
$$
0\sqrt{x^{2}+7 x y+y^{2}}-(x+y),\end{array}\right.$ whic... | (1,25) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,422 |
I. Fill in the Blanks (8 points each, total 64 points)
1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$. | $-, 1.327$
$$
C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327
$$ | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,423 |
2. If a line segment of length 3 cm is randomly divided into three segments, the probability that these segments can form a triangle is $\qquad$ . | 2. $\frac{1}{4}$.
Let the lengths of two segments be $x$ and $y$. Then the length of the third segment is $3-x-y$. Represent the segments using the point $(x, y)$.
Thus, the region formed is
$$
\Omega=\{(x, y) \mid 03-x-y, \\
& x+3-x-y>y, y+3-x-y>x\},
\end{aligned}
$$
The area $S_{A}=\frac{9}{8}$.
Therefore, the prob... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,424 |
3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$ | 3. 1 .
From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\tri... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,425 |
4. If the acute angle $\alpha$ satisfies
$$
\frac{1}{\sqrt{\tan \frac{\alpha}{2}}}=\sqrt{2 \sqrt{3}} \cdot \sqrt{\tan 10^{\circ}}+\sqrt{\tan \frac{\alpha}{2}},
$$
then $\alpha=$ $\qquad$ | 4. $50^{\circ}$.
The original expression can be transformed as follows:
$$
\begin{array}{l}
\frac{1}{\sqrt{\tan \frac{\alpha}{2}}}-\sqrt{\tan \frac{\alpha}{2}}=\sqrt{2 \sqrt{3}} \cdot \sqrt{\tan 10^{\circ}} \\
\Rightarrow \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}+\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alph... | 50^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,426 |
5. Given the function
$$
f(x)=\left\{\begin{array}{ll}
\left|\log _{2} x\right|, & 04 .
\end{array}\right.
$$
If $a, b, c, d$ are all different, and
$$
f(a)=f(b)=f(c)=f(d),
$$
then the range of $a b c d$ is $\qquad$ | 5. $(32,35)$.
Assume without loss of generality that $a4$ and $f(c)=f(d)$, then we have $c+d=12$, and $4<c<5$.
Thus, $a b c d=c d=c(12-c)$
$$
=-(c-6)^{2}+36 \text {. }
$$
When $4<c<5$, $32<a b c d<35$. | (32,35) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,427 |
6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$ | 6. 54.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$
... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,428 |
7. An ant starts from vertex $A$ of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, and travels along the surface to reach vertex $C_{1}$. The shortest distance is 6. Then the maximum volume of the rectangular prism is $\qquad$ . | 7. 12 $\sqrt{3}$.
Let the lengths of the three edges starting from the same vertex of a rectangular prism be $a, b, c (a \leqslant b \leqslant c)$. Then the shortest distance from point $A$ to vertex $C_{1}$ along the surface of the rectangular prism is
$$
\begin{array}{l}
\sqrt{(a+b)^{2}+c^{2}}=6 \\
\Rightarrow(a+b)^... | 12 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,429 |
Example 6 Blichfeldt's Lemma.
Suppose there is a plane region $R$, whose area is greater than $n$ $\left(n \in \mathbf{N}_{+}\right)$. Then, regardless of the original position and shape of $R$, it can always be moved to a new position by translation alone, such that it covers at least $n+1$ lattice points (for example... | Prove that the lines perpendicular to the coordinate axes drawn through each grid point on the longitudinal and transverse coordinate axes are called "grid lines", and the regions divided by these grid lines are all basic grid squares.
First, dye the region $R$ red, and do not dye the rest of the plane. Then, cut the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,430 |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$ | 8. 18084.
When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence ... | 18084 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,431 |
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