problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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For any integer $n>1$, let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{t}^{\alpha_{1}}$ be the standard factorization of $n$, define
$$
\omega(n)=t, \Omega(n)=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i} \text {. }
$$
Is it true that for any given positive integer $k$ and positive real numbers $\alpha, \beta$, there... | 4. The conclusion is affirmative.
Supplement the definitions $\omega(1)=\Omega(1)=0$, then for any positive integers $a, b$ we have
$$
\begin{array}{l}
\omega(a b) \leqslant \omega(a)+\omega(b), \\
\Omega(a b)=\Omega(a)+\Omega(b) .
\end{array}
$$
For any given positive integer $k$ and positive real numbers $\alpha, \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,306 |
5. Let the set $X=\{1,2, \cdots, 100\}$, and the function $f: X \rightarrow X$ satisfies the following conditions:
(1) For any $x \in X$, $f(x) \neq x$;
(2) For any 40-element subset $A$ of $X$, $A \cap f(A) \neq \varnothing$.
Find the smallest positive integer $k$ such that for any function $f$ satisfying the above c... | 5. First consider the function $f: X \rightarrow X$ defined as follows:
For $i=1,2, \cdots, 30, j=91,92, \cdots, 99$, define
$f(3 i-2)=3 i-1, f(3 i-1)=3 i$,
$f(3 i)=3 i-2, f(j)=100, f(100)=99$.
Clearly, the function $f$ satisfies condition (1).
For any 40-element subset $A$ of the set $X$, either there exists an integ... | 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,307 |
6. For non-empty sets of numbers $S, T$, define
$$
\begin{array}{l}
S+T=\{|s+t| \mid s \in S, t \in T\}, \\
2 S=\{2 s \mid s \in S\}.
\end{array}
$$
Let $n$ be a positive integer, and $A, B$ be non-empty subsets of $\{1,2, \cdots, n\}$. Prove: there exists a subset $D$ of $A+B$ such that
$$
D+D \subseteq 2(A+B) \text ... | 6. Let $S_{y}=\{(a, b) \mid a-b=y, a \in A, b \in B\}$.
Since $\sum_{y=1-n}^{n-1}\left|S_{y}\right|=|A||B|$, there exists $y_{0}\left(1-n \leqslant y_{0} \leqslant n-1\right)$, such that
$$
\left|S_{y_{0}}\right| \geqslant \frac{|A||B|}{2 n-1}>\frac{|A||B|}{2 n} \text {. }
$$
Let $D=\left\{a+b \mid(a, b) \in S_{y_{0}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,308 |
Example 4 Given that $P$ is a point inside $\triangle A B C$, $P D \perp B C$, $P E \perp C A, P F \perp A B$, and let the circumcircle of $\triangle D E F$ intersect $B C$, $C A$, and $A B$ at another point $D^{\prime}$, $E^{\prime}$, and $F^{\prime}$ respectively. Prove: The perpendiculars from points $D^{\prime}$, $... | Proof As shown in Figure 9, let the circumcenter of $\triangle D E F$ be $O$, and construct the point $P^{\prime}$ symmetric to point $P$ with respect to $O$. Connect $P^{\prime} D, P^{\prime} E, P^{\prime} F$. Draw $O D_{0} \perp B C$ at point $D_{0}$.
By the Perpendicular Chord Bisector Theorem, we know that $D_{0}$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,309 |
1. Given $a=\frac{1}{2+\sqrt{7}}, b=\frac{1}{2-\sqrt{7}}$. Then $a^{3}-a+b^{3}-b=$ $\qquad$ . | -1. $-\frac{64}{27}$.
$$
\begin{array}{l}
\text { Note that, } a+b=-\frac{4}{3}, a b=-\frac{1}{3} . \\
\text { Therefore, the original expression }=\left(a^{3}+b^{3}\right)-(a+b) \\
=(a+b)\left(a^{2}-a b+b^{2}-1\right) \\
=(a+b)\left[(a+b)^{2}-3 a b-1\right] \\
=\left(-\frac{4}{3}\right)\left(\frac{16}{9}+1-1\right)=-\... | -\frac{64}{27} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,310 |
3. In Rt $\triangle A B C$, it is known that $\angle B A C=90^{\circ}, A B=$ $6, A C=8$, points $E$ and $F$ are on side $A B$ such that $A E=2, B F=3$. Draw a line through point $E$ parallel to $A C$, intersecting side $B C$ at point $D$, connect $F D$ and extend it, intersecting the extension of side $A C$ at point $G... | 3. $\sqrt{265}$.
As shown in Figure 3.
From the problem, we know
$$
\begin{array}{l}
\triangle B E D \backsim \triangle B A C \\
\Rightarrow \frac{E D}{8}=\frac{4}{6} \\
\Rightarrow E D=\frac{16}{3} .
\end{array}
$$
In the right triangle $\triangle D E F$,
$$
\begin{array}{l}
F D=\sqrt{1^{2}+\left(\frac{16}{3}\right)... | \sqrt{265} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,312 |
4. For the convex pentagon $A B C D E$, the side lengths are sequentially $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. It is known that a quadratic trinomial in $x$ satisfies:
When $x=a_{1}$ and $x=a_{2}+a_{3}+a_{4}+a_{5}$, the value of the quadratic trinomial is 5;
When $x=a_{1}+a_{2}$, the value of the quadratic trinomial i... | 4. 0 .
Let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {, }
$$
and let the axis of symmetry of its graph be $x=x_{0}$.
By the problem, we know
$$
f\left(a_{1}\right)=f\left(a_{2}+a_{3}+a_{4}+a_{5}\right)=5,
$$
and $\square$
$$
\begin{array}{l}
a_{1} \neq a_{2}+a_{3}+a_{4}+a_{5} . \\
\text { Henc... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,313 |
5. A three-digit number that is divisible by 35 and whose digits sum to 15 is | 5. 735.
Since the sum of the digits is 15, the three-digit number is a multiple of 3. Also, since the three-digit number is a multiple of 35, this three-digit number is a multiple of $35 \times 3=105$.
Starting from 105, list the three-digit multiples of 105:
$$
105,210,315,420,525,630,735,840, 945 \text {, }
$$
Amon... | 735 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,314 |
$$
\begin{array}{l}
\text { 6. Given the quadratic equation in } x \text { is } \\
x^{2}+a x+(m+1)(m+2)=0
\end{array}
$$
For any real number $a$, the equation has real roots. Then the range of real number $m$ is . $\qquad$ | 6. $-2 \leqslant m \leqslant-1$.
From the given condition, we know $\Delta=a^{2}-4(m+1)(m+2) \geqslant 0$, which means
$$
a^{2} \geqslant 4(m+1)(m+2)
$$
holds for all real numbers $a$.
Obviously, the minimum value of $a^{2}$ is 0.
Therefore, the above condition is equivalent to
$$
(m+1)(m+2) \leqslant 0 \Rightarrow-2... | -2 \leqslant m \leqslant -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,315 |
7. Let the area of rectangle $A B C D$ be 2013, and point $E$ lies on side $C D$. Then the area of the triangle formed by the centroids of $\triangle A B E$, $\triangle B C E$, and $\triangle A D E$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output ... | 7. $\frac{671}{3}$.
As shown in Figure 4, let the centroids of $\triangle A B E$, $\triangle B C E$, and $\triangle A D E$ be $X$, $Y$, and $Z$ respectively.
By the properties of centroids and similar triangles, it is easy to know that the distances from points $Y$ and $Z$ to side $C D$ are both $\frac{1}{3} A D$, he... | \frac{671}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,316 |
8. In Rt $\triangle A B C$, the altitude $C D$ on the hypotenuse $A B$ is $3$, extend $D C$ to point $P$, such that $C P=2$, connect $A P$, draw $B F \perp A P$, intersecting $C D$ and $A P$ at points $E$ and $F$ respectively. Then the length of segment $D E$ is $\qquad$ | 8. $\frac{9}{5}$.
As shown in Figure 5, let $A D=a$.
In the right triangle $\triangle A B C$, given $C D=3, C D \perp A B$, we know $B D=\frac{9}{a}$.
Notice that, in the right triangles $\triangle A P D \sim \triangle E B D$,
then $\frac{D E}{B D}=\frac{A D}{P D} \Rightarrow D E$ $=\frac{9}{a} \times \frac{a}{5}=\fra... | \frac{9}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,317 |
9. (15 points) Given a non-zero real number $a$, solve the system of equations for $x$ and $y$
$$
\left\{\begin{array}{l}
x y-\frac{x}{y}=a, \\
x y-\frac{y}{x}=\frac{1}{a} .
\end{array}\right.
$$ | Sure, here is the translated text:
```
9. The original system of equations is
$$
\left\{\begin{array}{l}
x y-a=\frac{x}{y}, \\
x y-\frac{1}{a}=\frac{y}{x} .
\end{array}\right.
$$
Multiplying the two equations gives
$$
\begin{array}{l}
(x y-a)\left(x y-\frac{1}{a}\right)=1 \\
\Rightarrow(x y)^{2}-\left(a+\frac{1}{a}\r... | (x, y) = \left(\frac{\sqrt{a^{2}+1}}{a}, \sqrt{a^{2}+1}\right) \text{ or } \left(-\frac{\sqrt{a^{2}+1}}{a}, -\sqrt{a^{2}+1}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,318 |
10. (15 points) As shown in Figure 2, given that the side length of square $ABCD$ is 1, a line passing through vertex $C$ intersects the rays $AB$ and $AD$ at points $P$ and $Q$ respectively. Find the maximum value of $\frac{1}{AP} +$ $\frac{1}{AQ} + \frac{1}{PQ}$. | 10. From $\triangle P B C \backsim \triangle P A Q \Rightarrow \frac{B C}{A Q}=\frac{B P}{A P}$ $\Rightarrow \frac{1}{A Q}=\frac{A P-1}{A P} \Rightarrow \frac{1}{A P}+\frac{1}{A Q}=1$.
Since $1=\frac{1}{A P}+\frac{1}{A Q} \geqslant 2 \sqrt{\frac{1}{A P} \cdot \frac{1}{A Q}}$, therefore, $A P \cdot A Q \geqslant 4$.
Als... | 1+\frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,319 |
Example 5 As shown in Figure 11, if $P$ is any point inside the equilateral $\triangle ABC$, and perpendiculars are drawn from points $A$, $B$, and $C$ to $PA$, $PB$, and $PC$ respectively, forming $\triangle DEF$. $O$ is the center of the equilateral $\triangle ABC$, and $Q$ is the point symmetric to $P$ with respect ... | Connect $P E$ and $P F$, and take their midpoints $E_{0}$ and $F_{0}$.
Since $A, E, C, P$ and $A, F, B, P$ are respectively concyclic, and $E_{0}, F_{0}$ are their circumcenters, by the perpendicular diameter theorem, we know that $E_{0}, F_{0}$ are respectively on the perpendicular bisectors of $A C$ and $A B$, i.e.,
... | 120^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 727,320 |
11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-1... | 11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,321 |
1. Write all positive integers in ascending order in a row. Then the 2013th digit from left to right is | $-1.7$
All single-digit numbers occupy 9 positions, all two-digit numbers occupy $2 \times 90=180$ positions, and next come the three-digit numbers in sequence. Since $2013-9-180=1824$, and $\frac{1824}{3}=608$, because $608+99=707$, the 2013th digit is the last digit of the three-digit number 707, which is 7. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,322 |
2. If the distances from the center of the ellipse to the focus, to the vertex of the major axis, and to the directrix can form a right triangle, then the eccentricity of the ellipse is $\qquad$
保留源文本的换行和格式，直接输出翻译结果。 | 2. $\sqrt{\frac{\sqrt{5}-1}{2}}$.
From $c^{2}+a^{2}=\left(\frac{a^{2}}{c}\right)^{2} \Rightarrow c^{2}=a b$.
$$
\begin{array}{l}
\text { Also } c^{2}+b^{2}=a^{2} \Rightarrow a b+b^{2}=a^{2} \\
\Rightarrow \frac{a}{b}=\frac{\sqrt{5}+1}{2} .
\end{array}
$$
Thus, $a^{2}=a b \cdot \frac{a}{b}=c^{2} \cdot \frac{\sqrt{5}+1... | \sqrt{\frac{\sqrt{5}-1}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,323 |
3. In a regular quadrilateral pyramid $P-ABCD$, it is known that $PA=5, AB$ $=6, M$ is the centroid of $\triangle PAD$. Then the volume of the tetrahedron $MPBC$ is $\qquad$ . | 3. $4 \sqrt{7}$.
As shown in Figure 2, let $PM$ intersect $AD$ at point $N$. Then $PN=\sqrt{PA^{2}-\left(\frac{AD}{2}\right)^{2}}=4$. Let $PH$ be the height of the pyramid $P-ABCD$. Then $PH=\sqrt{PN^{2}-\left(\frac{AB}{2}\right)^{2}}=\sqrt{7}$. Since $S_{\triangle H}$ shape $ABCD=36$, therefore, $S_{\triangle \triang... | 4 \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,324 |
4. The maximum value of the function $y=\frac{4-\sin x}{3-\cos x}$ is
保留了源文本的换行和格式。 | 4. $\frac{6+\sqrt{6}}{4}$.
Notice that, the function can be regarded as the slope of the line connecting the fixed point $(3,4)$ and the moving point $(\cos x, \sin x)$, and the trajectory of the moving point $(\cos x, \sin x)$ is a unit circle.
Let the equation of the line passing through the point $(3,4)$ be
$$
y-4=... | \frac{6+\sqrt{6}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,325 |
5. Let $[a]$ denote the greatest integer not exceeding the real number $a$, and let $\{a\}=a-[a]\left(a \in \mathbf{R}_{+}\right)$. If $a$, $[a]$, and $\{a\}$ form a geometric sequence in order, then $a=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output... | 5. $\frac{1+\sqrt{5}}{2}$.
Let $\{a\}=\theta(\theta \in[0,1)),[a]=k$.
By the problem, $\theta>0, k>1$, and $a\{a\}=[a]^{2}$
$$
\begin{array}{l}
\Rightarrow(\theta+k) \theta=k^{2} \\
\Rightarrow \theta^{2}+k \theta-k^{2}=0 \\
\Rightarrow \theta=\frac{-1+\sqrt{5}}{2} k \text { (negative value discarded). }
\end{array}
$... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,326 |
6. Given that $a, b, c$ are distinct positive integers. If the set
$$
\{a+b, b+c, c+a\}=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\} \text {, }
$$
where $n \in \mathbf{Z}_{+}$. Then the minimum value of $a^{2}+b^{2}+c^{2}$ is | 6. 1297 .
From $n^{2}+(n+1)^{2}+(n+2)^{2}=2(a+b+c)$ being even, we know that among $n, n+1, n+2$, there are two odd numbers and one even number, which means $n$ is odd.
Obviously, $n>1$.
Without loss of generality, let $a<b<c$.
If $n=3$, then
$$
\begin{array}{l}
a+b=9, a+c=16, b+c=25 \\
\Rightarrow a+b+c=25 .
\end{arr... | 1297 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,327 |
7. The minimum value of the function $f(x)=\sum_{k=1}^{2013}|x-k|$ is | 7.1013042.
Notice that, the median of $1,2, \cdots, 2013$ is 1007.
Therefore, when $x=1007$, the function reaches its minimum value
$$
\begin{array}{l}
f(1007)=2(1+2+\cdots+1006) \\
=1006 \times 1007=1013042
\end{array}
$$ | 1013042 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,328 |
8. Given $\sin \alpha+\sin \beta=\sin \gamma$, $\cos \alpha+\cos \beta=-\cos \gamma$.
Then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=$ $\qquad$ | 8. $\frac{3}{2}$.
Squaring the given condition, we get
$$
\begin{array}{l}
\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \cdot \sin \beta=\sin ^{2} \gamma, \\
\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cdot \cos \beta=\cos ^{2} \gamma .
\end{array}
$$
Adding the two equations yields $\cos (\alpha-\beta)=-\frac{... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,329 |
9. (16 points) As shown in Figure 1, given that through the three vertices $A, B, C$ of $\triangle ABC$, tangents are drawn to its circumcircle, intersecting the lines of the opposite sides at points $D, E, F$ respectively. Prove: $D, E, F$ are collinear. | 9. By the converse of Menelaus' theorem, it is sufficient to prove:
$$
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1 \text {. }
$$
By the angle between a chord and a tangent, we know
$$
\angle B A D=\angle A C B=\angle A B E, \angle A B C=\angle A C F \text {. }
$$
By $\triangle B A D \backsim \triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,330 |
Example 6 As shown in Figure 12, in the convex quadrilateral $ABCD$, it is known that the diagonal $AC$ is neither the angle bisector of $\angle A$ nor the angle bisector of $\angle C$. $P$ is a point inside the quadrilateral $ABCD$ that satisfies $\angle 1=\angle 2, \angle 3=\angle 4$. Prove that $PB=PD$ if and only i... | Prove as shown in Figure 12, draw perpendiculars from point $P$ to the four sides, with the feet of the perpendiculars being $E, F, G, H$.
From $\angle 1=\angle 2 \Rightarrow E H \perp A C$,
$\angle 3=\angle 4 \Rightarrow F G \perp A C$.
Thus, $E H \parallel F G$.
Therefore, quadrilateral $E F G H$ is a trapezoid.
Conn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,331 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=1+\frac{1}{n} \sum_{k=1}^{n} a_{k} \text {. }
$$
(1) Write out the first seven terms of the sequence;
(2) For any positive integer $n$, find the expression for $a_{n}$. | 10. (1) Calculate in sequence
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=\frac{5}{2}, a_{4}=\frac{17}{6}, \\
a_{5}=\frac{37}{12}, a_{6}=\frac{197}{60}, a_{7}=\frac{207}{60} .
\end{array}
$$
(2) Notice that,
$$
\begin{array}{l}
a_{n+1}=1+\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}, \\
a_{n}=1+\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n... | 2+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,332 |
11. (20 points) It is known that a box contains 100 red and 100 blue cards, each color of cards containing one card labeled with each of the numbers $1, 3, 3^2, \cdots, 3^{99}$. The total sum of the numbers on the cards of both colors is denoted as $s$.
For a given positive integer $n$, if it is possible to pick sever... | 11. Arrange the cards in the box in ascending order of their labels:
$$
1,1,3,3,3^{2}, 3^{2}, \cdots, 3^{99}, 3^{99},
$$
where items with the same value are of different colors. For each $k(1 \leqslant k \leqslant 100)$, the sum of the first $2k$ terms is less than $3^{k}$. Therefore, terms of the form $3^{n}$ must be ... | 2^{200}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,333 |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=32, a_{n+1}-a_{n}=2 n\left(n \in \mathbf{Z}_{+}\right)$. Then the minimum value of $\frac{a_{n}}{n}$ is $\qquad$. | $-1 . \frac{31}{3}$.
From the given, we know
$$
\begin{array}{l}
a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=32+\sum_{k=1}^{n-1} 2 k=n(n-1)+32 . \\
\text { Therefore, } \frac{a_{n}}{n}=n-1+\frac{32}{n} .
\end{array}
$$
When $n=5$, $\frac{a_{n}}{n}=\frac{52}{5}$;
When $n=6$, $\frac{a_{n}}{n}=\frac{31}{3}... | \frac{31}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,334 |
2. For the function $y=f(x)(x \in D)$, if for any $x_{1} \in D$, there exists a unique $x_{2} \in D$ such that
$$
\sqrt{f\left(x_{1}\right) f\left(x_{2}\right)}=M \text {, }
$$
then the function $f(x)$ is said to have a geometric mean of $M$ on $D$.
Given $f(x)=x^{3}-x^{2}+1(x \in[1,2])$. Then the geometric mean of th... | 2. $\sqrt{5}$.
Notice that, when $10.$
$$
Thus, $f(x)=x^{3}-x^{2}+1$ is an increasing function on the interval $[1,2]$, with its range being $[1,5]$.
Therefore, according to the definition of the geometric mean of the function $f(x)$, we have $M=\sqrt{5}$. | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,335 |
3. If three non-zero and distinct real numbers $a, b, c$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$, then $a, b, c$ are called "harmonic"; if they satisfy $a+c=2b$, then $a, b, c$ are called "arithmetic".
Given the set $M=\{x|| x | \leqslant 2013, x \in \mathbf{Z}\}$, the set $P$ is a three-element subset of set $M... | 3.1006.
If $a, b, c$ are both harmonic and arithmetic, then
$$
\left\{\begin{array}{l}
\frac{1}{a} + \frac{1}{b} = \frac{2}{c}, \\
a + c = 2b
\end{array} \Rightarrow \left\{\begin{array}{l}
a = -2b \\
c = 4b
\end{array}\right.\right.
$$
Thus, the good set is of the form $\{-2b, b, 4b\}(b \neq 0)$.
Since the good set ... | 1006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,336 |
4. Given real numbers $x, y$ satisfy $x y+1=4 x+y$, and $x>1$. Then the minimum value of $(x+1)(y+2)$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.27.
From the problem, we know $y=\frac{4 x-1}{x-1}$.
$$
\begin{array}{l}
\text { Therefore, }(x+1)(y+2)=(x+1)\left(\frac{4 x-1}{x-1}+2\right) \\
\quad= \frac{3(x+1)(2 x-1)}{x-1} .
\end{array}
$$
Let $x-1=t>0$. Then
$$
\begin{array}{l}
(x+1)(y+2)=\frac{3(t+2)(2 t+1)}{t} \\
=6\left(t+\frac{1}{t}\right)+15 \geqslant ... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 727,337 |
5. As shown in Figure 1, in the tetrahedron $A B C D$, it is known that $A B \perp$ plane $B C D, \triangle B C D$ is an equilateral triangle with a side length of 3. If $A B$ $=2$, then the surface area of the circumscribed sphere of the tetrahedron $A B C D$ is $\qquad$ | 5. $16 \pi$.
As shown in Figure 4, let the center of the equilateral $\triangle BCD$ be $O_{1}$, and the center of the circumscribed sphere of the tetrahedron $ABCD$ be $O$.
Then $O O_{1} \perp$ plane $BCD$, $O O_{1} / / AB$.
Thus, $B O_{1}=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 3=\sqrt{3}$.
Take the midpoint $E... | 16 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,338 |
6. Among the 10 vertices of a regular decagon, if any four points are chosen, the probability that the quadrilateral formed by these four points is a trapezoid is $\qquad$ | 6. $\frac{2}{7}$.
Let the regular decagon be $A_{1} A_{2} \cdots A_{10}$.
(1) With $A_{1} A_{2}$ as the base, there are 3 trapezoids: trapezoid $A_{1} A_{2} A_{3} A_{10}$, trapezoid $A_{1} A_{2} A_{4} A_{9}$, and trapezoid $A_{1} A_{2} A_{5} A_{8}$.
Similarly, there are 3 trapezoids each for $A_{2} A_{3}, A_{3} A_{4}... | \frac{2}{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,339 |
7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$
The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,340 |
8. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and for any $x \in$ $\mathbf{R}$, we have
$$
f\left(f(x)-3^{x}\right)=4 .
$$
Then $f(2)=$ | 8. 10 .
From the problem, we know that $f(x)-3^{x}$ is a constant. Let's assume $f(x)-3^{x}=m$.
Then $f(m)=4, f(x)=3^{x}+m$.
Therefore, $3^{m}+m=4 \Rightarrow 3^{m}+m-4=0$.
It is easy to see that the equation $3^{m}+m-4=0$ has a unique solution $m=1$.
Thus, $f(x)=3^{x}+1$,
and hence, $f(2)=3^{2}+1=10$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,341 |
9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$). | 9. 10 .
It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$
meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$
where $x_{1}<x_{2}<\cdots<x_{7}$ meets the require... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,343 |
10. Given the function
$$
f(x)=\left\{\begin{array}{ll}
x, & x \text { is irrational; } \\
\frac{q+1}{p}, & x=\frac{q}{p}\left(p, q \in \mathbf{Z}_{+},(p, q)=1, p>q\right) .
\end{array}\right.
$$
Then the maximum value of the function $f(x)$ in the interval $\left(\frac{7}{8}, \frac{8}{9}\right)$ is . $\qquad$ | 10. $\frac{16}{17}$.
(1) If $x$ is a rational number, and $x \in\left(\frac{7}{8}, \frac{8}{9}\right)$, let
$$
x=\frac{a}{a+\lambda} \in\left(\frac{7}{8}, \frac{8}{9}\right)\left(a 、 \lambda \in \mathbf{Z}_{+}\right) \text {. }
$$
From $\frac{7}{8}<\frac{a}{a+\lambda}<\frac{8}{9}$, we get
$$
\begin{array}{l}
7(a+\lamb... | \frac{16}{17} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,344 |
11. Arrange the sequence $\left\{a_{n}\right\}$, which consists of positive numbers, into a triangular array as shown in Figure 2 (the $n$-th row has $n$ numbers, and in the same row, the number with the smaller subscript is placed on the left). $b_{n}$ represents the number in the first column of the $n$-th row.
It i... | (1) Let the common ratio of $\left\{b_{n}\right\}$ be $q$.
According to the problem, $a_{12}$ is the number in the 5th row and 2nd column of the array; $a_{18}$ is the number in the 6th row and 3rd column of the array.
Then $b_{1}=1, b_{n}=q^{n-1}, a_{12}=q^{4}+d=17$, $a_{18}=q^{5}+2 d=34$.
Thus, $q=2, d=1, b_{n}=2^{n-... | 2^{62}+59 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,345 |
12. Given that $A$ and $B$ are two moving points on the parabola $C: y^{2}=4 x$, point $A$ is in the first quadrant, and point $B$ is in the fourth quadrant. Lines $l_{1}$ and $l_{2}$ pass through points $A$ and $B$ respectively and are tangent to the parabola $C$. $P$ is the intersection point of $l_{1}$ and $l_{2}$.
... | 12. (1) Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)\left(y_{1}>0>y_{2}\right)$. It is easy to know that the slope of line $l_{1}$ exists, let's assume it is $k_{1}$. Then $l_{1}: y-y_{1}=k_{1}\left(x-\frac{y_{1}^{2}}{4}\right)$.
Substituting the above equation into the parabo... | x=-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,346 |
13. As shown in Figure 3, in $\triangle ABC$, it is given that $\angle B=90^{\circ}$, and the incircle of the triangle touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Connect $AD$, intersecting the incircle at another point $P$, and connect $PC$, $PE$, $PF$, $FD$, and $ED$.
(1) Prove: $\fra... | 13. (1) From the given conditions,
$\angle A F P=\angle A D F$, and $\angle F A P=\angle F A D$.
Thus, $\triangle A F P \backsim \triangle A D F$
$$
\Rightarrow \frac{A P}{A F}=\frac{F P}{D F} \text {. }
$$
Similarly, $\frac{E P}{D E}=\frac{A P}{A E}$.
Since $A F=A E$, then
$\frac{E P}{D E}=\frac{A P}{A E}=\frac{A P}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,347 |
14. Given
$$
f(x)=2 \ln (x+1)+\frac{1}{x(x+1)}-1 \text {. }
$$
(1) Find the minimum value of $f(x)$ on the interval $[1,+\infty)$;
(2) For $n \in \mathbf{Z}_{+}, n \geqslant 2$, prove:
$$
\begin{array}{l}
\ln 1+\ln 2+\cdots+\ln n>\frac{(n-1)^{2}}{2 n}, \\
\ln ^{2} 1+\ln ^{2} 2+\cdots+\ln ^{2} n>\frac{(n-1)^{4}}{4 n^{3}... | 14. (1) Notice,
$$
\begin{array}{l}
f^{\prime}(x)=\frac{2}{x+1}-\frac{2 x+1}{x^{2}(x+1)^{2}} \\
=\frac{2 x^{3}+2 x^{2}-2 x-1}{x^{2}(x+1)^{2}} \\
=\frac{\left(2 x^{3}-1\right)+2 x(x-1)}{x^{2}(x+1)^{2}} .
\end{array}
$$
Thus, when $x \geqslant 1$, $f^{\prime}(x)>0$, meaning $f(x)$ is an increasing function on the interv... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 727,348 |
15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a 、 b 、 c\right. \text { are positive integers not }
$$
exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$. | 15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$
These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arit... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,349 |
8. Let the line $l_{1}: 3 x-2 y=1$, passing through point $A(-1,-2)$, line $l_{2}: y=1$ intersects $l_{1}$ at point $B$, and line $l_{3}$ has a positive slope, passing through point $A$, and intersects line $l_{2}$ at point $C$. If $S_{\triangle A B C}=3$, then the slope of line $l_{3}$ is ( ).
(A) $\frac{2}{3}$
(B) $\... | 8. B.
Since $l_{2}: y=1$, point $A(-1,-2)$, and $S_{\triangle A B C}=3$, therefore, $B C=2$.
From $\left\{\begin{array}{l}3 x-2 y=1, \\ y=1,\end{array}\right.$ we get point $B(1,1)$.
Thus, point $C(3,1)$ or $C(-1,1)$.
Since the slope of $l_{3}$ is positive, point $C(3,1)$.
Therefore, the slope of $l_{3}$ is $k=\frac{1... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,350 |
9. The sum of the prime factors of the arithmetic square root of the largest perfect square that divides 12! is ( ).
(A) 5
(B) 7
(C) 8
(D) 10
(E) 12 | 9. D.
Notice that, $12!=2^{10} \times 3^{5} \times 5^{2} \times 7 \times 11$.
Then the largest perfect square that can divide 12! has the square root of $2^{5} \times 3^{2} \times 5$.
Therefore, the sum of the prime factors is $2+3+5=10$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,351 |
11. Two bees $A$ and $B$ start flying from the same starting point at the same rate. Bee $A$ first flies 1 foot north, then 1 foot east, and then 1 foot up, and continues to fly in this manner; bee $B$ first flies 1 foot south, then 1 foot west, and continues to fly in this manner. When they are exactly 10 feet apart, ... | 11. A.
Taking the starting point of the two bees as the origin of coordinates, the southward direction as the positive direction of the $x$-axis, the eastward direction as the positive direction of the $y$-axis, and the upward direction as the positive direction of the $z$-axis, with 1 foot as the unit length, establi... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,352 |
Example 8 Given that $D$ is a moving point on the perpendicular bisector of side $BC$ of $\triangle ABC$, and $I_{1}, I_{2}$ are the incenters of $\triangle ABD$ and $\triangle ACD$ respectively. Prove: The circumcircle of $\triangle AI_{1}I_{2}$ must pass through another fixed point besides point $A$.
---
The transl... | 【Analysis】After exploration, it is known that the fixed point is the midpoint $E$ of the arc $\overparen{B A C}$ of the circumcircle of $\triangle A B C$. Therefore, it is only necessary to prove that quadrilateral $A E I_{1} I_{2}$ is a cyclic quadrilateral. For this, connect $A I_{2}$ and $E I_{2}$, and change to pro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,353 |
12. As shown in Figure 1, there are five cities $A, B, C, D, E$, connected by routes $\widetilde{A B}, \widetilde{A D}, \widetilde{A E}, \widetilde{B C}$, $\widetilde{B D}, \widetilde{C D}$, and $\widetilde{D E}$. Then, from city $A$ to $B$, there are ( ) different paths, requiring each path to be taken once, where eac... | 12. D.
Notice that, there is only one road passing through cities $C$ and $E$. Therefore, cities $C$ and $E$ can be merged into city $D$, making Graph 1 equivalent to Graph 2.
Discussing the two scenarios based on the first direction chosen from city $A$.
(1) $A \rightarrow D$.
$$
A \rightarrow D \rightarrow A \right... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,354 |
13. Given a convex quadrilateral $ABCD$ whose four interior angles form an arithmetic sequence. In $\triangle ABD$ and $\triangle BCD$, if $\angle ABD = \angle CDB$, $\angle DAB = \angle CBD$, and the three interior angles of these two triangles also form an arithmetic sequence, then the maximum possible value of the s... | 13. D.
Assume the four interior angles of the quadrilateral are
$$
\alpha, \alpha+d, \alpha+2d, \alpha+3d \text{.}
$$
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have
$$
\alpha+\alpha+3d=\alpha+d+\alpha+2d=180^{\circ} \text{.}
$$
Given that $\angle ABD = \angle CDB$ and $\angle DAB =... | 240 | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,355 |
14. Given two non-decreasing sequences of non-negative integers, with different first terms, and each subsequent term in the sequences is the sum of the two preceding terms. If the seventh term of both sequences is $N$, then the smallest possible value of $N$ is $(\quad)$.
(A) 55
(B) 89
(C) 104
(D) 144
(E) 273 | 14. C.
Let the first two terms of the first sequence be $x_{1}, x_{2}\left(x_{1}, x_{2} \in \mathrm{N}\right)$, and the first two terms of the second sequence be $y_{1}, y_{2}\left(y_{1}, y_{2} \in \mathrm{N}\right)$. Then the seventh terms of the two sequences are $5 x_{1}+8 x_{2}$ and $5 y_{1}+8 y_{2}$, respectively... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,356 |
16. Given that the perimeter of pentagon $A B C D E$ is 1, and all its interior angles are equal, extend each side outward to intersect at five points forming a star shape. Let $s$ be the perimeter of this star. Then the absolute value of the difference between the maximum and minimum possible values of $s$ is ( ).
(A)... | 16. A.
Consider the pentagram as composed of a pentagon and five triangles. Since the interior angles of the pentagon are all equal, each interior angle is
$$
\frac{(5-2) \times 180^{\circ}}{5}=108^{\circ} \text {. }
$$
Therefore, each triangle is an isosceles triangle with base angles of $72^{\circ}$.
Let the sides... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,357 |
17. Let $a, b, c \in \mathbf{R}$, and
$$
a+b+c=2, a^{2}+b^{2}+c^{2}=12 \text {. }
$$
Then the difference between the maximum and minimum values of $c$ is ( ).
(A) 2
(B) $\frac{10}{3}$
(C) 4
(D) $\frac{16}{3}$
(E) $\frac{20}{3}$ | 17. D.
From the problem, we have $\left\{\begin{array}{l}a+b=2-c, \\ a^{2}+b^{2}=12-c^{2} .\end{array}\right.$ Then,
$$
\begin{array}{l}
a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=\frac{(2-c)^{2}-\left(12-c^{2}\right)}{2} \\
=c^{2}-2 c-4 .
\end{array}
$$
Thus, the quadratic equation with roots $a, b$ is
$$
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,358 |
18. There are some coins on the table, and Barbara and Jenny take turns to take some. When Barbara takes coins, she can only take 2 or 4; when Jenny takes coins, she can only take 1 or 3. They decide who starts by flipping a coin; the one who takes the last coin wins. Assuming both use their best strategies, when there... | 18. B.
Consider two scenarios:
(1) The number of coins is 2013.
Notice that, according to the game rules, for any five coins, no matter who takes \( n \) coins, the other can take \( 5-n \) coins.
If Jenny goes first, she can take 3 coins, and then, no matter how many \( n \) (where \( n=2 \) or 4) coins Barbara tak... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,359 |
19. In $\triangle A B C$, it is known that $A B=13, B C=14, C A=$ 15, points $D, E, F$ are on sides $B C, C A, D E$ respectively, and $A D \perp$ $B C, D E \perp A C, A F \perp B F$. If the length of segment $D F$ is $\frac{m}{n}$ $\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24... | 19. B.
As shown in Figure 3, with $D$ as the origin, the line $BC$ as the $x$-axis, and the line $AD$ as the $y$-axis, we establish a Cartesian coordinate system $x D y$.
It is easy to know that $p_{\triangle ABC}=13+14+15=42$,
$S_{\triangle ABC}=\sqrt{21(21-13) \times(21-14) \times(21-15)}=84$.
Thus, $AD=\frac{2 S_{\... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,360 |
$$
\begin{array}{l}
\text { 20. Let } 135^{\circ}<\alpha<180^{\circ} \text {, and } \\
P\left(\cos \alpha, \cos ^{2} \alpha\right), Q\left(\cot \alpha, \cot ^{2} \alpha\right), \\
R\left(\sin \alpha, \sin ^{2} \alpha\right), S\left(\tan \alpha, \tan ^{2} \alpha\right)
\end{array}
$$
be the four vertices of a trapezoid... | 20. A.
Let $f, g, h, j$ represent the sine, cosine, tangent, and cotangent functions (in no particular order). Then the four vertices of the trapezoid are $\left(f, f^{2}\right),\left(g, g^{2}\right),\left(h, h^{2}\right),\left(j, j^{2}\right)$.
Since quadrilateral $P Q R S$ is a trapezoid, the two bases are parallel,... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,361 |
21. Consider a set of 30 parabolas defined as follows: all parabolas have the focus at $(0,0)$, and the directrices are of the form $y=a x+b$, where $a \in\{-2,-1,0,1,2\}, b \in$ $\{-3,-2,-1,1,2,3\}$. It is known that no three parabolas have a common intersection point. Then, the number of intersection points between a... | 21. C.
In a set, any two parabolas intersect at most at two points. If two parabolas have the same orientation (the foci are on the same side of the directrices, and the directrices are parallel), in this case, the two parabolas do not intersect.
Thus, the number of intersection points is
$$
2 \mathrm{C}_{30}^{2}-2 \t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,362 |
22. Let $m, n$ be integers greater than 1, and the equation in terms of $x$
$$
8 \log _{n} x \cdot \log _{m} x-7 \log _{n} x-6 \log _{m} x-2013=0
$$
has the smallest positive integer solution. Then $m+n=(\quad)$.
(A) 12
(B) 20
(C) 24
(D) 48
(E) 272 | 22. A.
By the change of base formula, we have
$$
8 \frac{1}{\lg n} \cdot \frac{1}{\lg m} \lg ^{2} x-\left(\frac{7}{\lg n}+\frac{6}{\lg m}\right) \lg x-2013=0 .
$$
Let $\lg x_{1}, \lg x_{2}$ be the two roots of the quadratic equation
$$
8 \frac{1}{\lg n} \cdot \frac{1}{\lg m} t^{2}-\left(\frac{7}{\lg n}+\frac{6}{\lg m... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,363 |
Example 1 As shown in Figure 1, given that $D$ is a point inside $\triangle A B C$, line $B D$ intersects side $A C$ and line $C D$ intersects side $A B$ at points $E$ and $F$ respectively, satisfying $A F=F B=C D, C E=D E$. Find $\angle B F C$. ${ }^{[1]}$
(2003, Japan Mathematical Olympiad (Final)) | As shown in Figure 1, let $G$ be a point on segment $CF$ such that $FG = CD$. Connect $AG$.
From $CE = DE \Rightarrow \angle ACD = \angle EDC = \angle FDB$.
Also, $FG = CD$, then
$$
GC = CF - FG = CF - CD = DF \text{. }
$$
Applying Menelaus' Theorem to $\triangle ABE$ and the transversal $CF$, we get
$$
\frac{BF}{FA} ... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,364 |
24. In $\triangle A B C$, it is known that $M$ is the midpoint of side $A C$, $C N$ bisects $\angle A C B$, and intersects side $A B$ at point $N$, $B M$ intersects $C N$ at point $X$, and $\triangle B X N$ is an equilateral triangle, $A C=2$. Then $B N^{2}=(\quad)$.
(A) $\frac{10-6 \sqrt{2}}{7}$
(B) $\frac{2}{9}$
(C) ... | 24. A.
By Menelaus' theorem, we have
$$
\frac{A N}{N B} \cdot \frac{B X}{X M} \cdot \frac{M C}{C A}=1 \Rightarrow A N=2 X M \text {. }
$$
Let $A N=2 a, N B=b$. Then $B X=X N=b, M X=a$.
In $\triangle A B M$, by the cosine rule, we get
$$
\begin{array}{l}
1=A M^{2}=A B^{2}+B M^{2}-2 A B \cdot B M \cos 60^{\circ} \\
=(2... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,365 |
1. Given $I$ is the incenter of $\triangle A B C$, $A C=2, B C=3$, $A B=4$. If $\overrightarrow{A I}=x \overrightarrow{A B}+y \overrightarrow{A C}$, then $x+y=$ $\qquad$ | $-1 . \frac{2}{3}$.
Connect $A I$ and extend it to intersect $B C$ at point $D$. Then $D$ divides the segment $B C$ in the ratio $\lambda=\frac{A B}{A C}=\frac{4}{2}=2$.
Thus, $\overrightarrow{A D}=\frac{1}{3} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A C}$.
Since $B C=3$, we have $B D=2, D C=1$.
In $\triangle A... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,367 |
2. Given the function $f(x)=x^{2}-2 x+3$. If when $1<x<2$, the solution set of the inequality $|f(x)-a| \geqslant 2$ is empty, then the range of the real number $a$ is $\qquad$ . | $2.1 \leqslant a \leqslant 4$.
From $|f(x)-a|<2$, we get
$$
\begin{array}{l}
x^{2}-2 x+1<a<x^{2}-2 x+5 . \\
\text { Therefore }\left(x^{2}-2 x+1\right)_{\max }<a<\left(x^{2}-2 x+5\right)_{\min } \\
\Rightarrow 1 \leqslant a \leqslant 4 .
\end{array}
$$ | 2.1 \leqslant a \leqslant 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,368 |
3. If $a, b, c$ form a geometric sequence, $\log _{c} a, \log _{b} c, \log _{a} b$ form an arithmetic sequence, then the common difference of this arithmetic sequence is $\qquad$ . | 3.0 or $\frac{3}{2}$.
From $a 、 b 、 c$ forming a geometric sequence $\Rightarrow b^{2}=a c$
$$
\Rightarrow \log _{b} a+\log _{b} c=2 \text {. }
$$
Let $x=\log _{b} a, y=\log _{b} c$. Then $x+y=2$.
From $\log _{c} a 、 \log _{b} c 、 \log _{a} b$ forming an arithmetic sequence
$$
\begin{array}{l}
\Rightarrow \log _{c} a... | \frac{3}{2} \text{ or } 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,369 |
4. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ with its two foci at $A(-1,0), B(1,0)$, a line $l$ passing through point $B$ intersects the right branch of the hyperbola at points $M$ and $N$, and $\triangle A M N$ is an isosceles right triangle with $N$ as the right-angle vertex. Then the le... | 4. $\frac{2 \sqrt{85+34 \sqrt{2}}}{17}$.
By the definition of a hyperbola, we have
$$
\begin{array}{l}
|A M|-|B M|=2 a, \\
|A N|-|B N|=2 a .
\end{array}
$$
From the given conditions, we get
$$
|A M|=\sqrt{2}|A N|=\sqrt{2}|M N|=\sqrt{2}(|B M|+|B N|) \text {. }
$$
Combining equations (1) and (2), we get
$$
|A M|=4 a,|... | \frac{2 \sqrt{85+34 \sqrt{2}}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,370 |
5. Given the function $f(x)=\mathrm{e}^{x}(\sin x+\cos x)$, where $x \in\left[-\frac{2011 \pi}{2}, \frac{2013 \pi}{2}\right]$. A tangent line is drawn to the graph of the function $f(x)$ through the point $M\left(\frac{\pi-1}{2}, 0\right)$. Let the x-coordinates of the points of tangency form the sequence $\left\{x_{n}... | 5. $1006 \pi$.
Let the coordinates of the tangent point be $\left(x_{0}, \mathrm{e}^{x_{0}}\left(\sin x_{0}+\cos x_{0}\right)\right)$.
Then the equation of the tangent line is
$$
y-\mathrm{e}^{x_{0}}\left(\sin x_{0}+\cos x_{0}\right)=2 \mathrm{e}^{x_{0}} \cos x_{0} \cdot\left(x-x_{0}\right) \text {. }
$$
Substituting... | 1006 \pi | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,371 |
6. As shown in Figure 1, given that line $l \perp$ plane $\alpha$, with the foot of the perpendicular being $O$, in the right triangle $\triangle ABC$, $BC=1, AC=2, AB=\sqrt{5}$. The right triangle moves freely in space under the following conditions: (1) $A \in l$, (2) $C \in \alpha$. Then the maximum distance between... | 6. $\sqrt{2}+1$.
It is easy to know that when the distance between points $B$ and $O$ is maximized, points $A$, $B$, $C$, and $O$ are coplanar.
Let $\angle A C O=\theta$. Then $O C=2 \cos \theta$.
In $\triangle B C O$,
$$
\begin{array}{l}
|O B|^{2}=|O C|^{2}+|B C|^{2}-2|O C||B C| \cos \left(90^{\circ}+\theta\right) \\... | \sqrt{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,372 |
7. Let $A=\{2,4, \cdots, 2014\}, B$ be any non-empty subset of $A$, and $a_{i} 、 a_{j}$ be any two elements in set $B$. There is exactly one isosceles triangle with $a_{i} 、 a_{j}$ as side lengths. Then the maximum number of elements in set $B$ is $\qquad$ | 7. 10 .
By symmetry, without loss of generality, assume $a_{i}<a_{j}$. Then there must exist an isosceles triangle with $a_{j}$ as the waist and $a_{i}$ as the base, and there is only one isosceles triangle with $a_{i}$ and $a_{j}$ as side lengths.
Thus, $a_{i}+a_{i} \leqslant a_{j} \Rightarrow a_{j} \geqslant 2 a_{i}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,373 |
8. Given that the quadratic equation with real coefficients $a x^{2}+b x+c=0$ has real roots. Then the maximum value of the positive real number $r$ such that
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant r a^{2}
$$
holds is . $\qquad$ | 8. $\frac{9}{8}$.
Let's assume $a=1$, and the two real roots of the equation $x^{2}+b x+c=0$ are $x_{1}$ and $x_{2}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
b=-x_{1}-x_{2}, c=x_{1} x_{2} . \\
\text { Then }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
=(1-b)^{2}+(b-c)^{2}+(c-1)^{2} \\
=\left(1+x_{1}+x_{2}\right)^{2}+\le... | \frac{9}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,374 |
Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. I... | Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$
Since $D A$ and $D B$ are tangent to $\odot O$... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,375 |
9. (16 points) When $x \geqslant 0$, find the minimum value $g(a)$ of the function
$$
f(x)=2 x+(|x-1|-a)^{2}
$$
| 9. Note that,
$$
\begin{array}{c}
f(x)=2 x+(|x-1|-a)^{2} \\
=x^{2}-2 a|x-1|+a^{2}+1 .
\end{array}
$$
(1) $a>1$.
(i) When $x \geqslant 1$,
$$
f(x)=x^{2}-2 a x+a^{2}+2 a+1
$$
is decreasing on $[1, a]$ and increasing on $[a,+\infty)$, thus,
$$
f(x)_{\min }=2 a+1 \text {. }
$$
(ii) When $0 \leqslant x \leqslant 1$,
$$
f(x... | g(a)=\left\{\begin{array}{ll}2 a+1, & a \geqslant 4 ; \\ a^{2}-2 a+1, & 0 \leqslant a<4 ; \\ 1-2 a, & -1<a<0 ; \\ a^{2}+2, & a \leqslant-1 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,376 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2 n}=a_{n}, a_{4 n-1}=0, a_{4 n+1}=1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Does there exist a positive integer $T$, such that for any $n \in$ $\mathbf{Z}_{+}$, we have $a_{n+T}=a_{n}$?
(2) Let $S=\frac{a_{1}}{10}+\frac{a_{2}... | 10. (1) Suppose there exists a positive integer $T$ such that for any $n \in \mathbf{Z}_{+}$, we have $a_{n+T}=a_{n}$. Then there exist infinitely many positive integers $T$ such that for any $n \in \mathbf{Z}_{+}$, we have $a_{n+T}=a_{n}$.
Let $T$ be the smallest positive integer among them.
If $T$ is odd, let $T=2 t-... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,377 |
11. (20 points) Let point $A(-2,0)$ and $\odot O: x^{2}+y^{2}=4$, $AB$ is the diameter of $\odot O$, from left to right $M$, $O$, $N$ are the four equal division points of $AB$, $P$ (different from $A$, $B$) is a moving point on $\odot O$, $PD \perp AB$ at point $D$, $\overrightarrow{PE}=\lambda \overrightarrow{ED}$, l... | 11. (1) It is easy to get the points $B(2,0), M(-1,0), N(1,0)$.
Let the points $P\left(x_{0}, y_{0}\right), C(x, y)$. Then $E\left(x_{0}, \frac{y_{0}}{1+\lambda}\right)$.
Since the line $P A$ intersects $B E$ at point $C$, we have $x_{0} \neq \pm 2$, $\frac{y}{x+2}=\frac{y_{0}}{x_{0}+2}$,
and $\frac{y}{x-2}=\frac{\fr... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,378 |
一、(40 points) As shown in Figure 2, given $\triangle ABC (AB \neq AC)$ with the incenter $I$, the excenter opposite to $B$ is $O$, the midpoint of $BC$ is $M$, and $MI$ intersects $AC$ at point $P$. Prove: $OP \parallel BC$. | As shown in Figure 3, connect $O B$ to intersect $A C$ and the circumcircle of $\triangle A B C$ at points $D, E$, respectively. Point $I$ is on $O B$, and $O F / / A C$ intersects the extension of $B C$ at point $F$.
Since $O C$ bisects $\angle A C F$, we have
$$
O F=C F \text {, and } \frac{B F}{F C}=\frac{B F}{O F}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,379 |
(1) Prove: $d \equiv 0(\bmod p)$ or $d \equiv 1(\bmod p)$;
(2) If $d$ is a prime different from $p$, then $x^{d}-1 \equiv$ $0(\bmod p)$ has exactly $d$ distinct solutions (i.e., incongruent modulo $p$). | (1) When $p=2$, the conclusion is obviously true. When $p \geqslant 3$, let's assume $d$ is a prime.
Notice that, $\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1$.
If $d \mid(x-1)$, then
$$
\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1 \equiv p \equiv 0(\bmod d) \text {. }
$$
Thus $d=p \Rightarrow d \equiv 0(\bmod p)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,380 |
Three. (50 points) Let $n \geqslant 4, M$ be a finite set of $n$ points in three-dimensional space, where no four points lie on the same plane. Color the points in set $M$ either white or black such that any sphere intersecting set $M$ in at least four points has the property that exactly half of these intersection poi... | Three, define $f: M \rightarrow\{-1,1\}$ as
$$
f(X)=\left\{\begin{array}{ll}
-1, & X \text { is white; } \\
1, & X \text { is black. }
\end{array}\right.
$$
From the given conditions, we know $\sum_{X \in S} f(X)=0$, where $S$ represents any sphere that passes through at least four points in the set $M$.
For any three... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,381 |
Four. (50 points) Let $\alpha$ be a real number, $1<\alpha<2$. Prove:
(1) $\alpha$ can be uniquely expressed as an infinite product
$$
\alpha=\prod_{i=1}^{+\infty}\left(1+\frac{1}{n_{i}}\right),
$$
where $n_{i}$ are positive integers, satisfying $n_{i}^{2} \leqslant n_{i+1}$;
(2) $\alpha$ is a rational number if and o... | (1) Construct the sequence $\left\{n_{i}\right\}$ and the ratio
$$
\theta_{k}=\frac{\alpha}{\prod_{i=1}^{k}\left(1+\frac{1}{n_{i}}\right)}\left(\theta_{k}>1\right)
$$
satisfying for all $k$.
Take $n_{k}$ as the smallest $n$ satisfying the following equation:
$$
1+\frac{1}{n}1$, we have
$$
\begin{array}{l}
\left(1+\fra... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,382 |
Given $G$ is the centroid of $\triangle ABC$, $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$ respectively, points $P$ and $Q$ lie on the ray $GD$, and satisfy $GP \cdot GD = GE^2$, $GQ \cdot GD = GF^2$. Connect $PF$ and $QE$.
Prove: (1) $\angle GPF = \angle GQE$;
$$
\text{(2) } PF = QE \text{.}
$$ | Prove (1) As shown in Figure 1, take points $R, T$ on rays $GF, GE$ respectively, such that
$$
GR \cdot GF = GP \cdot GD, \quad GT \cdot GE = GQ \cdot GD.
$$
Thus, points $R, F, P, D$ and points $T, E, Q, D$ are concyclic respectively.
Then $\angle GPF = \angle CRD, \angle GQE = \angle BTD$.
Draw $RS \perp CD$. It is ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,383 |
374 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=1, a_{n+2}=a_{n+1}+a_{n} \text {. }
$$
Prove: $T=\sum_{i=1}^{+\infty} \frac{1}{\left(a_{i}^{2}\right)!}$ cannot be written in the form $\frac{p+q \sqrt{r}}{s}$, where $p, q, r, s \in \mathbf{Z}_{+}$, and $r$ is not a perfect square. | Prove a lemma first.
Lemma If $\sqrt{r}(r \in \mathbf{N})$ is an irrational number, and the positive fraction $\frac{x}{y}\frac{1}{2 r y^{2}}$.
Proof: Equation (1) $\Leftrightarrow \sqrt{r}>\frac{1}{2 r y^{2}}+\frac{x}{y}$
$\Leftrightarrow r>\frac{x^{2}}{y^{2}}+\frac{x}{r y^{3}}+\frac{1}{4 r^{2} y^{4}}$
$\Leftrightarro... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,384 |
Define the function $\gamma(n)$ as the sum of all positive integers less than $n$ and coprime to $n$, where $n \in \mathbf{Z}_{+}, n>1$. Prove: $\gamma(n)$ is odd if and only if $n=2$ or $p^{k}$, where $k \in \mathbf{Z}_{+}$, $p$ is a prime of the form $4a+3$. | Proof: First, we prove a lemma.
Lemma If $(m, n)=1$, then
$$
\gamma(m n)=2 \gamma(m) \gamma(n) \text {. }
$$
Proof: If $a$ is any positive integer less than $n$ and coprime to $n$, then $n-a$ must also be a positive integer less than $n$ and coprime to $n$. This shows that the average of all positive integers less tha... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,385 |
Example 1 (Bernoulli's Inequality) Let $n \geqslant 2$, and non-zero real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are all greater than -1 and have the same sign. Then
$$
\begin{array}{l}
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \\
>1+x_{1}+x_{2}+\cdots+x_{n} .
\end{array}
$$ | 【Analysis and Proof】Use mathematical induction on $n$.
When $n=2$, using the method of discarding terms, we have
$$
\begin{array}{l}
\left(1+x_{1}\right)\left(1+x_{2}\right) \\
=1+x_{1}+x_{2}+x_{1} x_{2} \\
>1+x_{1}+x_{2} .
\end{array}
$$
The conclusion holds.
Assume that the conclusion holds for $n=k$, i.e.,
$$
\begi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,386 |
Example 2 Let $n \in \mathbf{Z}_{+}, k \geqslant 2 n-1$. Prove that in a triangle,
$$
\sum \frac{a^{n}}{k\left(b^{n}+c^{n}\right)-a^{n}} \geqslant \frac{3}{2 k-1},
$$
where, “ $\sum$ ” denotes the cyclic sum. | 【Analysis and Proof】Introducing the parameter $\lambda$, we have
$$
\frac{a^{n}}{k\left(b^{n}+c^{n}\right)-a^{n}}+\lambda=\frac{(1-\lambda) a^{n}+k \lambda\left(b^{n}+c^{n}\right)}{k\left(b^{n}+c^{n}\right)-a^{n}} \text {. }
$$
To make the numerator of the above expression symmetric, let $1-\lambda=k \lambda$, yieldin... | \frac{3}{2 k-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,387 |
Example 1 Given that $p$ is a prime number and $k$ is a positive integer. Try to find the remainder $S=1^{k}+2^{k}+\cdots+(p-1)^{k}(\bmod p)$. | If $(p-1) \mid k$, then when $(i, p)=1$,
$$
i^{k} \equiv 1(\bmod p) \text {. }
$$
Thus, $S \equiv p-1 \equiv-1(\bmod p)$.
If $(p-1) \nmid k$, let $a$ be a primitive root modulo $p$. Then
$$
p \nmid\left(a^{k}-1\right) \text {, }
$$
and $a i(1 \leqslant i \leqslant p-1)$ is also a reduced residue system modulo $p$.
He... | S \equiv \begin{cases} -1 & \text{if } (p-1) \mid k \\ 0 & \text{if } (p-1) \nmid k \end{cases} (\bmod p) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,388 |
6. In $\triangle A B C$, it is known that $a, b, c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $a c+c^{2}=b^{2}-a^{2}$. If the longest side of $\triangle A B C$ is $\sqrt{7}$, and $\sin C=2 \sin A$, then the length of the shortest side of $\triangle A B C$ is $\qquad$. | 6. 1 .
From $a c+c^{2}=b^{2}-a^{2}$
$\Rightarrow \cos B=-\frac{1}{2} \Rightarrow \angle B=\frac{2 \pi}{3}$.
Thus, the longest side is $b$.
Also, $\sin C=2 \sin A \Rightarrow c=2 a$.
Therefore, $a$ is the shortest side.
By the cosine rule,
$$
(\sqrt{7})^{2}=a^{2}+4 a^{2}-2 a \times 2 a \times\left(-\frac{1}{2}\right) \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,389 |
7. Cut a 7-meter long iron wire into two segments (it can also be directly cut into two segments), so that the difference in length between these two segments does not exceed 1 meter. If these two segments are used to form two circles respectively, then the maximum value of the sum of the areas of these two circles is ... | 7. $\frac{25}{4 \pi}$.
Let the lengths of the two segments be $x$ meters and $y$ meters, respectively. Then $x$ and $y$ satisfy the following relationships:
$$
\left\{\begin{array}{l}
x>0, \\
y>0, \\
x+y \leqslant 7, \\
|x-y| \leqslant 1 .
\end{array}\right.
$$
The planar region is the shaded part shown in Figure 4. ... | \frac{25}{4 \pi} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,390 |
8. As shown in Figure 3, given a regular tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ mutually perpendicular, and each of length $2, E, F$ are the midpoints of edges $AB, AC$ respectively, $H$ is the midpoint of line segment $EF$, and a plane is constructed through $EF$ intersecting the lateral edges $OA, ... | 8. $\sqrt{5}$.
As shown in Figure 5, draw $O N \perp A_{1} B_{1}$ at point $N$, and connect $C_{1} N$.
Since $O C_{1} \perp$ plane $O A_{1} B_{1}$, by the theorem of three perpendiculars, we know $C_{1} N \perp A_{1} B_{1}$:
Therefore, $\angle O N C_{1}$ is the plane angle of the dihedral angle $O-A_{1} B_{1}-C_{1}$. ... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,391 |
9. Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and $A$ and $B$ are two moving points on the parabola, satisfying $\angle A F B=120^{\circ}$. A perpendicular line $M N$ is drawn from the midpoint $M$ of chord $A B$ to the directrix of the parabola, with the foot of the perpendicular being $N$. Then the maximum... | 9. $\frac{\sqrt{3}}{3}$.
According to the problem and the definition of a parabola, we have
$$
\begin{array}{l}
|A F|=\left|A A_{1}\right|,|B F|=\left|B B_{1}\right| \\
\Rightarrow|A F|+|B F|=\left|A A_{1}\right|+\left|B B_{1}\right|=2|M N| \\
\Rightarrow \frac{|M N|}{|A B|}=\frac{|A F|+|B F|}{2|A B|} .
\end{array}
$$... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,392 |
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=$ $\qquad$ . | 10.2013.
Obviously, when $k \geqslant 11$, $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=0$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{10}\left[\frac{2013+2^{k}}{2^{k+1}}\right] \\
= 1007+503+252+126+63+ \\
31+16+8+4+2+1 \\
= 2013 .
\end{array... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,393 |
11. Let $f(x)$ be an increasing function defined on $[1,+\infty)$, and the inequality
$$
f\left(k-\cos ^{2} x\right) \leqslant f\left(k^{2}+\sin x\right)
$$
holds for all $x$. Find the range of real numbers $k$. | $$
\begin{array}{l}
\left\{\begin{array}{l}
k-\cos ^{2} x \geqslant 1, \\
k-\cos ^{2} x \leqslant k^{2}+\sin x
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
k \geqslant 1+\cos ^{2} x, \\
k^{2}-k \geqslant \sin ^{2} x-\sin x-1 .
\end{array}\right.
\end{array}
$$
From equation (1), we get $k \geqslant 2$.
Sin... | k \geqslant 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,394 |
12. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has one focus at $F_{1}(-\sqrt{3}, 0)$, and passes through the point $H\left(\sqrt{3}, \frac{1}{2}\right)$. Let the upper and lower vertices of the ellipse $E$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be any point on the ellipse differ... | 12. From the problem, we have
$$
a^{2}-b^{2}=3, \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1 \text {. }
$$
Solving, we get $a^{2}=4, b^{2}=1$.
Thus, the equation of the ellipse $E$ is $\frac{x^{2}}{4}+y^{2}=1$.
From this, we know the points $A_{1}(0,1), A_{2}(0,-1)$. Let point $P\left(x_{0}, y_{0}\right)$.
Then $l_{P A_{1}}: y... | 2 | Geometry | proof | Yes | Yes | cn_contest | false | 727,395 |
13. Given the functions $f(x)=a x(a \in \mathbf{R}), g(x)=\ln x$.
(1) If the function $F(x)=f(x)-g(x)$ has an extremum of 1, find the value of $a$;
(2) If the function $G(x)=f(\sin (1-x))+g(x)$ is increasing on the interval $(0,1)$, find the range of values for $a$;
(3) Prove: $\sum_{k=1}^{n} \sin \frac{1}{(k+1)^{2}}<\... | 13. (1) From the problem, we know
$$
F(x)=a x-\ln x(x>0) \text {. }
$$
Thus, $F^{\prime}(x)=a-\frac{1}{x}(x>0)$.
(i) When $a \leqslant 0$, $F(x)$ is monotonically decreasing on $(0,+\infty)$, with no extremum.
(ii) When $a>0$, $F^{\prime}(x)=0 \Rightarrow x=\frac{1}{a}$. Therefore, $F(x)$ is monotonically decreasing o... | \sum_{k=1}^{n} \sin \frac{1}{(k+1)^{2}}<\ln 2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,396 |
14. Given that $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, and $4 S_{n}=3 a_{n}+2^{n+1}(n \in \mathbf{N})$.
Find: (1) the relationship between $a_{n}$ and $a_{n-1}$;
(2) all values of $a_{0}$ that make the sequence $a_{0}, a_{1}, \cdots$ increasing. | 14. (1) From the problem, we have
$$
4 S_{n-1}=3 a_{n-1}+2^{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Subtracting the given equation from the above equation, we get
$$
\begin{array}{l}
4 a_{n}=3 a_{n}-3 a_{n-1}+2^{n} \\
\Rightarrow a_{n}=2^{n}-3 a_{n-1}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
(2) Fro... | a_{0}=\frac{2}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,397 |
15. Given that $a, b, c$ are positive real numbers. Prove:
$$
\frac{\sqrt{a^{2}+3 b c}}{a}+\frac{\sqrt{b^{2}+3 a c}}{b}+\frac{\sqrt{c^{2}+3 a b}}{c} \geqslant 6 \text {. }
$$ | 15. Notice,
$$
\begin{array}{l}
\frac{\sqrt{a^{2}+3 b c}}{a}=\frac{\sqrt{a^{2}+b c+b c+b c}}{a} \\
\geqslant \frac{\sqrt{4 \sqrt[4]{a^{2} b^{3} c^{3}}}}{a}=\frac{2 \sqrt[8]{a^{2} b^{3} c^{3}}}{a} .
\end{array}
$$
Similarly, $\frac{\sqrt{b^{2}+3 a c}}{b} \geqslant \frac{2 \sqrt[8]{a^{3} b^{2} c^{3}}}{b}$,
$$
\begin{arr... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false | 727,398 |
1. If every prime factor of 2013 is a term in a certain arithmetic sequence $\left\{a_{n}\right\}$ of positive integers, then the maximum value of $a_{2013}$ is $\qquad$ | $-, 1.4027$.
Notice that, $2013=3 \times 11 \times 61$.
If $3, 11, 61$ are all terms in a certain arithmetic sequence of positive integers, then the common difference $d$ should be a common divisor of $11-3=8$ and $61-3=58$. To maximize $a_{2013}$, the first term $a_{1}$ and the common difference $d$ should both be as ... | 4027 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,400 |
2. If $a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, then the minimum value of $a+2b+3c$ is . $\qquad$ | 2. 36 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$ | 36 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,401 |
3. If $S_{n}=n!\left[\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}-1\right]$, then
$$
S_{2013}=
$$ | 3. $-\frac{1}{2014}$.
From $\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$, we know
$$
\begin{array}{l}
\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!} \\
=\frac{1-1}{1!}+\frac{2-1}{2!}+\frac{3-1}{3!}+\cdots+\frac{(n+1)-1}{(n+1)!} \\
=1-\frac{1}{(n+1)!} .
\end{array}
$$
Therefore, $S_{n}=n!\... | -\frac{1}{2014} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,402 |
4. If the surface area of a cube $X$ is equal to that of a regular tetrahedron $Y$, then the ratio of their volumes $\frac{V_{X}}{V_{Y}}=$ $\qquad$ . | 4. $\sqrt[4]{3}$.
Let the surface area be 12.
Then the area of each face of the cube is 2, and its side length is $\sqrt{2}$. Thus, $V_{x}=2^{\frac{3}{2}}$.
Given that the area of each face of the regular tetrahedron is 3, let its side length be a. Then
$$
\frac{\sqrt{3}}{4} a^{2}=3 \Rightarrow a=2 \times 3^{\frac{1}... | \sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,403 |
5. If the distances from the center of the ellipse to the focus, the endpoint of the major axis, the endpoint of the minor axis, and the directrix are all positive integers, then the minimum value of the sum of these four distances is $\qquad$ .
| 5.61.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distances from the center $O$ of the ellipse to the endpoints of the major axis, the endpoints of the minor axis, the foci, and the directrices are $a$, $b$, $c$, $d$ respectively, and satisfy
$$
c^{2}=a^{2}-b^{2}, d=\frac... | 61 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,404 |
6. The range of the function $f(x)=\sqrt{3 x-6}+\sqrt{3-x}$ is
$\qquad$ | 6. $[1,2]$.
Notice that the domain of $f(x)=\sqrt{3(x-2)}+\sqrt{3-x}$ is $[2,3]$. Therefore, let $x=2+\sin ^{2} \alpha\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$.
$$
\begin{array}{l}
\text { Then } f(x)=\sqrt{3 \sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \alpha} \\
=\sqrt{3} \sin \alpha+\cos \alpha=2 \sin \left(\... | [1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,405 |
7. Given a composite number $k(1<k<100)$. If the sum of the digits of $k$ is a prime number, then the composite number $k$ is called a "pseudo-prime". The number of such pseudo-primes is . $\qquad$ | 7.23.
Let $S(k)$ denote the sum of the digits of $k$, and $M(p)$ denote the set of composite numbers with a pseudo-prime $p$.
When $k \leqslant 99$, $S(k) \leqslant 18$, so there are 7 prime numbers not exceeding 18, which are $2, 3, 5, 7, 11, 13, 17$.
The composite numbers with a pseudo-prime of 2 are $M(2)=\{20\}$.... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,406 |
8. Make a full permutation of the elements in the set $\{1,2, \cdots, 8\}$, such that except for the number at the far left, for each number $n$ on the right, there is always a number to the left of $n$ whose absolute difference with $n$ is 1. The number of permutations that satisfy this condition is $\qquad$ | 8. 128 .
Suppose for a certain permutation that satisfies the conditions, the first element on the left is $k(1 \leqslant k \leqslant 8)$. Then, among the remaining seven numbers, the $8-k$ numbers greater than $k$, $k+1, k+2, \cdots, 8$, must be arranged in ascending order; and the $k-1$ numbers less than $k$, $1,2, ... | 128 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,407 |
9. (20 points) Let the parabola $y^{2}=2 p x(p>0)$ intersect the line $x+y=1$ at points $A$ and $B$. If $O A \perp O B$, find the equation of the parabola and the area of $\triangle O A B$. | Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $y^{2}=2 p x$ and $x+y=1$, we get
$$
y^{2}+2 p y-2 p=0 \text {. }
$$
Thus, $\dot{x}_{1}=1+p-\sqrt{p^{2}+2 p}$,
$$
\begin{array}{l}
y_{1}=-p+\sqrt{p^{2}+2 p}, \\
x_{2}=1+p+\sqrt{p^{2}+2 p} \\
y_{2}=-p-\sqrt{p^{2}+2 p} .
\end{array}
$$
Since $O A... | \frac{\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,408 |
10. (20 points) As shown in Figure 1, in quadrilateral $ABCD$, it is known that $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $P$ is a point on diagonal $BD$. Lines $EP$ and $PF$ intersect the extensions of $AB$ and $DC$ at points $M$ and $N$, respectively. Prove: segment $MN$ is bisected by l... | 10. As shown in Figure 2, let $E F$ and $M N$ intersect at point $G$.
Applying Menelaus' theorem to line $E F$ and $\triangle P M N$, we get $\frac{N G}{G M} \cdot \frac{M E}{E P} \cdot \frac{P F}{F N}=1$.
To prove that $G$ is the midpoint of segment $M N$, it suffices to prove
$$
\frac{P F}{N F}=\frac{P E}{M E} \text... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,409 |
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