problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the number of all real roots of the equation
$$
3 x^{2}-10[x]+3=0
$$
is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 8. D.
From $3 x^{2}-10[x]+3=0$
$$
\begin{array}{l}
\Rightarrow 10(x-1)<10[x]=3 x^{2}+3 \leqslant 10 x \\
\Rightarrow \frac{1}{3} \leqslant x \leqslant 3 .
\end{array}
$$
When $\frac{1}{3} \leqslant x<1$, the original equation has no solution;
When $1 \leqslant x<2$, the solution to the original equation is $x=\sqrt{\... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,736 |
9. If $a \in \mathbf{R}_{+}, b \in \mathbf{R}$, and
$$
\max \min _{x \in \mathbb{R}}\left\{2 x+4, a x^{2}+b, 5-3 x\right\}=2 \text {, }
$$
then $a+b=(\quad)$.
(A) -1
(B) 1
(C) 2
(D) 3 | 9. C.
Notice that the lines $y=2x+4$ and $y=5-3x$ pass through the points $(-1,2)$ and $(1,2)$, respectively, and these two points are symmetric with respect to the $y$-axis.
Therefore, when $b=0$, the parabola $y=ax^2$ passes through the points $(-1,2)$ and $(1,2)$, yielding $a=2$;
When $b \neq 0$, the parabola $y=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,737 |
10. Given the function
$$
f(x)=\cos (a \sin x)-\sin (b \cos x)
$$
has no zeros. Then the range of $a^{2}+b^{2}$ is ( ).
(A) $\left[0, \frac{\pi}{4}\right)$
(B) $\left[0, \frac{\pi^{2}}{2}\right)$
(C) $\left[0, \frac{\pi^{2}}{4}\right)$
(D) $\left[0, \frac{\pi}{2}\right)$ | 10. C.
From the given, we have
$$
\left\{\begin{array}{l}
2 k \pi + b \cos x = \frac{\pi}{2} - a \sin x, \\
(2 k + 1) \pi - b \cos x = \frac{\pi}{2} - a \sin x
\end{array} \quad (k \in \mathbf{Z})\right.
$$
has no solution.
Then $\sqrt{a^{2}+b^{2}} \sin (x+\varphi) = \frac{\pi}{2} + 2 k \pi$ or
$$
\sqrt{a^{2}+b^{2}} ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,738 |
11. Let real numbers $x, y$ satisfy the equation $(x+2)^{2}+y^{2}=1$. Then the maximum value of $\frac{y}{x}$ is $\qquad$ | $=11 \cdot \frac{\sqrt{3}}{3}$.
According to the problem, to maximize the value of $\frac{y}{x}$, it is necessary to draw a tangent from the origin to the circle
$$
(x+2)^{2}+y^{2}=1
$$
such that the slope of the tangent is maximized, at which point the maximum value is $\frac{\sqrt{3}}{3}$. | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,739 |
13. As shown in Figure 2, in the regular quadrilateral prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that the tangent value of the angle formed by $A B_{1}$ and the base $A_{1} B_{1} C_{1} D_{1}$ is $a$. Then the tangent value of the dihedral angle $A-B_{1} D_{1}-A_{1}$ is $\qquad$ | 13. $\sqrt{2} a$.
Let the side length of the base and the height of a regular quadrilateral prism be $x$ and $y$, respectively. Then $\frac{y}{x}=a$. Let the intersection point of the diagonals of the base be O. Then the plane angle of the dihedral angle $A-B_{1} D_{1}-A_{1}$ is $\angle A O A_{1}$.
Thus, $\tan \angle ... | \sqrt{2} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,741 |
14. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{aligned}
f(x+2) & =f(x)+2, \\
\text { then } \sum_{k=1}^{2014} f(k) & =
\end{aligned}
$$ | 14.2029105.
Notice that, $f(0)=0$.
By $f(x+2)=f(x)+2$, let $x=-1$. Then $f(1)=f(-1)+2 \Rightarrow f(1)=1$.
$$
\begin{array}{l}
\text { Also, } f(2 n)=\sum_{k=1}^{n}(f(2 k)-f(2 k-2))+f(0) \\
=2 n, \\
f(2 n-1)=\sum_{k=2}^{n}(f(2 k-1)-f(2 k-3))+f(1) \\
=2 n-1, \\
\text { Therefore, } \sum_{k=1}^{2014} f(k)=\sum_{k=1}^{20... | 2029105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,742 |
15. Let $P$ be a point on the ellipse $2 x^{2}+3 y^{2}=1$, $F_{1}$ and $F_{2}$ be the two foci of the ellipse, and $\angle F_{1} P F_{2}=\frac{\pi}{6}$. Then the area of $\triangle F_{1} P F_{2}$ is . $\qquad$ | 15. $\frac{2-\sqrt{3}}{3}$.
Let the lengths of the semi-major axis and semi-minor axis of the ellipse be $a$ and $b$, respectively. Then $a=\frac{\sqrt{2}}{2}, b=\frac{\sqrt{3}}{3} \Rightarrow c=\frac{\sqrt{6}}{6}$. Therefore, $\left|F_{1} F_{2}\right|=\frac{\sqrt{6}}{3},\left|P F_{1}\right|+\left|P F_{2}\right|=\sqrt... | \frac{2-\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,743 |
Example 1 Let $\left\{a_{1}, a_{2}, \cdots, a_{8}\right\}$ be an 8-element subset of the set $S=\{1,2, \cdots, 17\}$.
(1) Prove: there exists a positive integer $k$, such that the equation $a_{i}-a_{j}=k$ $(1 \leqslant i, j \leqslant 8)$ has at least three different solutions;
(2) Give a 7-element subset, such that for... | (1) Proof Without loss of generality, let $a_{1}<a_{2}<\cdots<a_{8}$. Denote
$$
\begin{array}{l}
d_{i}=a_{i+1}-a_{i}(i=1,2, \cdots, 7), \\
b_{i}=a_{i+2}-a_{i}(i=1,2, \cdots, 6) .
\end{array}
$$
If there does not exist a $k$ that satisfies the condition, then among these 13 numbers, $1,2, \cdots, 6$ each have at most t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,744 |
16. Let $f(x)$ be a function defined on $\mathbf{R}$, satisfying
$$
\left|f(x)+\cos ^{2} x\right| \leqslant \frac{3}{4},\left|f(x)-\sin ^{2} x\right| \leqslant \frac{1}{4} \text {. }
$$
Then the function $f(x)=$ . $\qquad$ | 16. $\sin ^{2} x-\frac{1}{4}$.
Notice,
$$
\begin{array}{l}
1=\sin ^{2} x+\cos ^{2} x \\
\leqslant\left|f(x)+\cos ^{2} x\right|+\left|f(x)-\sin ^{2} x\right| \\
\leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
Therefore, $\left|f(x)+\cos ^{2} x\right|=\frac{3}{4},\left|f(x)-\sin ^{2} x\right|=\frac{1}{4}$.
Thus, $... | \sin ^{2} x-\frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,745 |
17. A courier company undertakes courier services between 13 cities in a certain area. If each courier can take on the courier services for at most four cities, to ensure that there is at least one courier between every two cities, the courier company needs at least $\qquad$ couriers. | 17. 13.
From the problem, we know that there are $\mathrm{C}_{13}^{2}$ types of express delivery services between 13 cities. Each courier can handle at most $\mathrm{C}_{4}^{2}$ types of express delivery services between four cities. Therefore, at least $\frac{\mathrm{C}_{13}^{2}}{\mathrm{C}_{4}^{2}}=13$ couriers are ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,746 |
18. If $b, c \in \mathbf{R}$, the quadratic function $f(x)=x^{2}+b x+c$ has two distinct intersections with the $x$-axis in the interval $(0,1)$. Find the range of $c^{2}+(1+b) c$.
The text has been translated while preserving the original formatting and line breaks. | $$
\begin{array}{l}
f(x)=(x-r)(x-s) . \\
\text { Then } c=f(0)=r s, \\
1+b+c=f(1)=(1-r)(1-s) . \\
\text { Therefore } 0<c^{2}+(1+b) c=f(0) f(1) \\
=r s(1-r)(1-s)<\left(\frac{r+1-r}{2}\right)^{2}\left(\frac{s+1-s}{2}\right)^{2} \\
=\frac{1}{16} .
\end{array}
$$ | 0 < c^2 + (1+b)c < \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,747 |
19. Given that $A$ is a moving point on the parabola $y^{2}=2 x$, and the fixed point $B(2,0)$, a circle $\odot C$ is constructed with $A B$ as its diameter. If the chord length of the circle $\odot C$ intercepted by the line $l$ : $x+k y-\frac{3}{2}=0$ is a constant, find this chord length and the value of the real nu... | 19. Let the moving point $A\left(x_{0}, y_{0}\right)\left(y_{0}^{2}=2 x_{0}\right)$ on the parabola. Then the equation of the circle with diameter $A B$ is
$$
\left(x-x_{0}\right)(x-2)+\left(y-y_{0}\right) y=0 \text {. }
$$
Let the line $l$ intersect the circle at points $P_{1}\left(x_{1}, y_{1}\right)$ and $P_{2}\lef... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,748 |
20. Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{1}=a, a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}(n \geqslant 1)
$$
Find all real numbers $a$ such that $0<a_{n}<1(n \geqslant 2)$. | 20. From the given conditions, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n-1}=\frac{a_{n}}{a_{n}-1} . \\
\text { Also, } a_{n+1}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}+a_{2}+\cdots+a_{n}-1} \\
=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}=\frac{a_{n}^{2}}{\left(a_{n}-\frac{1}{2}\right)^{2}+\frac{3}{4}}(n \geqslant 2) .
\... | a<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,749 |
21. Among the 100 integers from $1 \sim 100$, arbitrarily select three different numbers to form an ordered triplet $(x, y, z)$. Find the number of triplets that satisfy the equation $x+y=3z+10$. | (1) When $3 z+10 \leqslant 101$, i.e., $z \leqslant 30$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
S=\sum_{k=1}^{30}(3 k+9)=1665 \text{. }
$$
(2) When $3 z+10 \geqslant 102$, i.e., $31 \leqslant z \leqslant 63$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
\begin{aligned}
T & =\sum_{k=31}... | 3194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,750 |
22. Let positive real numbers $a$, $b$, $c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}+b^{2}=3, \\
a^{2}+c^{2}+a c=4, \\
b^{2}+c^{2}+\sqrt{3} b c=7 .
\end{array}\right.
$$
Find the values of $a$, $b$, and $c$. | 22. As shown in Figure 3, from point $O$, draw three line segments $O A$, $O B$, and $O C$ of lengths $a$, $b$, and $c$ respectively, such that $\angle A O B=90^{\circ}$ and $\angle A O C=120^{\circ}$. Then $\angle C O B=150^{\circ}$.
By the cosine rule, we have
$$
A B=\sqrt{a^{2}+b^{2}}=\sqrt{3}, A C=2, B C=\sqrt{7} \... | (a, b, c)=\left(\frac{6 \sqrt{37}}{37}, \frac{5 \sqrt{111}}{37}, \frac{8 \sqrt{37}}{37}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,751 |
2. Given that the first $n$ terms sum of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $S_{n}$ and $T_{n}$ respectively, and for all positive integers $n$, $\frac{a_{n}}{b_{n}}=\frac{2 n-1}{3 n+1}$. Then $\frac{S_{6}}{T_{5}}=$ $\qquad$ . | 2. $\frac{18}{25}$.
According to the problem, let
$$
a_{n}=k(2 n-1), b_{n}=k(3 n+1) \text {, }
$$
where, the constant $k \neq 0$.
Then $S_{6}=36 k, T_{5}=50 k$.
Thus, $\frac{S_{6}}{T_{5}}=\frac{36 k}{50 k}=\frac{18}{25}$. | \frac{18}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,753 |
3. The minimum value of the function $f(x)=\sqrt{2 x^{2}-x+3}+\sqrt{x^{2}-x}$ is $\qquad$ $ـ$ | 3. $\sqrt{3}$.
From $\left\{\begin{array}{l}2 x^{2}-x+3 \geqslant 0, \\ x^{2}-x \geqslant 0,\end{array}\right.$ we get the domain of the function $f(x)$ as $(-\infty, 0] \cup[1,+\infty)$.
The function $f(x)$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[1,+\infty)$, hence
$$
f(x)_{\te... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,754 |
Example 2 In a finite collection of subsets on the real number line, each subset is the union of two closed intervals, and any three subsets in the collection have a common point. Prove: There exists a point on the real number line that is a common point for at least half of the subsets in the collection. | Proof Let the given finite collection of subsets be
$$
\left\{F_{i} \mid 1 \leqslant i \leqslant n\right\} .
$$
Since $F_{i}$ is the union of two closed intervals, we set
$$
\begin{array}{l}
F_{i}=\left[a_{i}, b_{i}\right] \cup\left[c_{i}, d_{i}\right]\left(a_{i} \leqslant b_{i} \leqslant c_{i} \leqslant d_{i}\right) ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,755 |
4. Given that the arithmetic mean of $\sin \theta$ and $\cos \theta$ is $\sin \alpha$, and the geometric mean is $\sin \beta$. Then $\cos 2 \alpha-\frac{1}{2} \cos 2 \beta=$ $\qquad$ . | 4. 0 .
From the given, we have
$$
\begin{array}{l}
\sin \alpha=\frac{\sin \theta+\cos \theta}{2}, \sin ^{2} \beta=\sin \theta \cdot \cos \theta. \\
\text { Therefore, } \cos 2 \alpha-\frac{1}{2} \cos 2 \beta \\
=1-2 \sin ^{2} \alpha-\frac{1}{2}\left(1-2 \sin ^{2} \beta\right) \\
=\frac{1}{2}-2 \sin ^{2} \alpha+\sin ^{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,756 |
5. If the edge length of the cube $A_{1} A_{2} A_{3} A_{4}-B_{1} B_{2} B_{3} B_{4}$ is 1, then the number of elements in the set
$$
\left\{x \mid x=\overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}, i, j \in\{1,2,3,4\}\right\}
$$
is $\qquad$ | 5. 1 .
Solution 1 Note that,
$$
\begin{array}{l}
\overrightarrow{A_{1} B_{1}} \perp \overrightarrow{A_{i} A_{1}}, \overrightarrow{A_{1} B_{1}} \perp \overrightarrow{B_{1} B_{j}}(i, j \in\{2,3,4\}) \text {. } \\
\text { Hence } \overrightarrow{A_{1} B_{1}} \cdot \overrightarrow{A_{i} B_{j}}=\overrightarrow{A_{1} B_{1}}... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,757 |
6. A uniform cubic die is rolled twice, and the numbers that appear in succession are $a$ and $b$, respectively. The probability that the equation $x^{2}+a x+b=0$ has real roots is $\qquad$ (answer in simplest fraction form). | 6. $\frac{19}{36}$.
From the problem, since $a, b \in \{1,2, \cdots, 6\}$, the total number of basic events is $6^{2}=36$.
The necessary and sufficient condition for the equation $x^{2}+a x+b=0$ to have real roots is $\Delta=a^{2}-4 b \geqslant 0$.
For the values of $a^{2}, 4 b$ and the sign of $a^{2}-4 b$ as shown ... | \frac{19}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,758 |
7. If real numbers $a, b, c$ satisfy
$$
a+2 b+3 c=6, a^{2}+4 b^{2}+9 c^{2}=12 \text {, }
$$
then $a b c=$ . $\qquad$ | 7. $\frac{4}{3}$.
Solution 1 From the given, we have
$$
\begin{array}{l}
\left(a^{2}+4 b^{2}+9 c^{2}\right)-4(a+2 b+3 c)=12-4 \times 6 \\
\Rightarrow(a-2)^{2}+4(b-1)^{2}+(3 c-2)^{2}=0 \\
\Rightarrow a=2, b=1, c=\frac{2}{3} \\
\Rightarrow a b c=\frac{4}{3} .
\end{array}
$$
Solution 2 From the given, we have
$$
\begin{... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,759 |
8. In $\triangle A B C$, it is known that $B C=2 \sqrt{3}$, and the sum of the lengths of the medians on sides $A B$ and $A C$ is 6. Establish a Cartesian coordinate system with line $B C$ as the $x$-axis and the perpendicular bisector of side $B C$ as the $y$-axis. Then the equation of the locus of vertex $A$ is | 8. $\frac{x^{2}}{36}+\frac{y^{2}}{9}=1(y \neq 0)$.
Solution 1: Let the medians $C F$ and $B E$ of sides $A B$ and $A C$ intersect at point $G$. Then $G$ is the centroid of $\triangle A B C$.
$$
\begin{array}{l}
\text { By }|G B|+|G C|=\frac{2}{3}(|B E|+|C F|) \\
=\frac{2}{3} \times 6=4>2 \sqrt{3},
\end{array}
$$
we k... | \frac{x^{2}}{36}+\frac{y^{2}}{9}=1(y \neq 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,760 |
9. Given a tetrahedron $P-ABC$ with the base being an equilateral triangle of side length $4 \sqrt{3}$, $PA=3, PB=4, PC=5$. If $O$ is the center of $\triangle ABC$, then the length of $PO$ is $\qquad$ . | 9. $\sqrt{6}$.
Solution 1 As shown in Figure 6, connect $A O$ and extend it to intersect $B C$ at point $M$, then $M$ is the midpoint of side $B C$. Connect $P M$.
In $\triangle P B C$, by the median length formula, we get $P M=\frac{5}{\sqrt{2}}$.
In $\triangle P A M$, by the cosine rule, we get
$$
\cos \angle P A M=... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,761 |
10. Let $[m]$ denote the greatest integer not exceeding the real number $m$. Then the sum of all elements in the set $\left\{x \in \mathbf{R} \mid 9 x^{2}-30[x]+20=0\right\}$ is $\qquad$ | 10. $\sqrt{10}$.
$$
\begin{array}{l}
\text { Given } 9 x^{2}-30[x]+20=0 \\
\Rightarrow 30[x]=9 x^{2}+20>0 \\
\Rightarrow[x]>0 \Rightarrow x>0 . \\
\text { Also }[x] \leqslant x \Rightarrow[x]^{2} \leqslant x^{2}, \text { then } \\
9[x]^{2}-30[x]+20 \leqslant 0 \\
\Rightarrow(3[x]-5)^{2} \leqslant 5 .
\end{array}
$$
Th... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,762 |
One. (15 points) As shown in Figure 1, $A$ is a fixed point between two parallel lines $l_{1}$ and $l_{2}$, and the distances from point $A$ to lines $l_{1}$ and $l_{2}$ are $A M=1, A N=\sqrt{3}$. Let the other two vertices $C$ and $B$ of $\triangle A B C$ move on $l_{1}$ and $l_{2}$ respectively, and satisfy $A B<A C,... | (1) From $\frac{A B}{\cos B}=\frac{A C}{\cos C}$ and the Law of Sines, we get
$$
\frac{\sin C}{\cos B}=\frac{\sin B}{\cos C} \Rightarrow \sin 2 B=\sin 2 C \text {. }
$$
Since $A B<A C$, we have $\angle C<\angle B$. Therefore, $2 \angle B+2 \angle C=\pi \Rightarrow \angle B+\angle C=\frac{\pi}{2}$. Hence, $\triangle A ... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,763 |
II. (15 points) Given a sequence $\left\{a_{n}\right\}$ whose terms are all positive, the sum of its first $n$ terms is $S_{n}$, and for any $n \in \mathbf{Z}_{+}$, we have
$$
S_{n}^{2}-\left(n^{2}+n-1\right) S_{n}-\left(n^{2}+n\right)=0 \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\rig... | II. (1) From the problem, we know
$$
\left(S_{n}+1\right)\left(S_{n}-n^{2}-n\right)=0 \text {. }
$$
Since \( a_{n}>0 \), it follows that \( S_{n}>0 \left(n \in \mathbf{Z}_{+}\right) \).
Thus, \( S_{n}=n^{2}+n \).
When \( n \geqslant 2 \),
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1} \\
=n^{2}+n-\left[(n-1)^{2}+(n-1)\right]... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,764 |
Example 1 As shown in Figure 1, in $\square A B C D$, $\odot O$ passes through points $A, B, C$, and $E$ is a point on $B C$. Let the circumcircle of $\triangle A B E$ be $\odot P$, $P F \perp B C$, and $P F$ intersects $A B$ at point $F$. $C F$ intersects $\odot O$ at point $G$. Prove: $G, E, C, D$ are concyclic. | 【Analysis and Proof】As shown in Figure 1, extend $EF$ to intersect $\odot P$ at point $H$, and connect $HG$, $HA$.
Since $PF \perp BC$, by symmetry we know
$AH \parallel BE$
$\Rightarrow H, A, D$ are collinear $\Rightarrow EH=BA=CD$
$\Rightarrow$ Quadrilateral $HECD$ is an isosceles trapezoid
$\Rightarrow H, E, C, D$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,766 |
Example 2 As shown in Figure 2, the orthocenter of acute $\triangle ABC$ is $H$, $E$ is any point on segment $CH$, extend $CH$ to point $F$ such that $HF = CE$, draw $FD \perp BC$, $EG \perp BH$, with $D$ and $G$ being the feet of the perpendiculars, $M$ is the midpoint of segment $CF$, $O_{1}$ and $O_{2}$ are the circ... | 【Analysis and Proof】(1) As shown in Figure 2, let $E G$ intersect $D F$ at point $K$, and connect $A H$.
From $A C \perp B H, E K \perp B H, A H \perp B C, K F \perp B C$, we get $C A \parallel E K, A H \parallel K F, C H=E F$.
Therefore, $\triangle C A H \cong \triangle E K F, A H \cong K F$.
Thus, $A K \parallel H F$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,767 |
3. As shown in Figure 11, two circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $B$. A line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $C$ and $D$, respectively. Another line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $E$ and $F$, respe... | First, prove that $BA$, $CM$, and $FN$ are the angle bisectors of $\triangle BCF$, intersecting at the incenter $I$.
In circles $\Gamma_{1}$ and $\Gamma_{2}$, by the power of a point theorem, we have
$$
\begin{array}{l}
CI \cdot IM = AI \cdot IB, AI \cdot IB = NI \cdot IF, \\
NI \cdot IF = CI \cdot IM .
\end{array}
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,768 |
3. Given a sphere that is tangent to all six edges of a regular tetrahedron with edge length $a$. Then the volume of this sphere is $\qquad$ . | 3. $\frac{\sqrt{2}}{24} \pi a^{3}$.
Construct a cube with edge length $\frac{\sqrt{2}}{2} a$, and the radius of the sphere is $\frac{\sqrt{2}}{4} a$. | \frac{\sqrt{2}}{24} \pi a^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,769 |
4. Let the function be
$$
f(x)=k\left(x^{2}-x+1\right)-x^{3}(1-x)^{3} \text {. }
$$
If for any $x \in[0,1]$, we have $f(x) \geqslant 0$, then the minimum value of $k$ is $\qquad$ | 4. $\frac{1}{48}$.
For any $x \in [0,1]$, we have
$$
k \geqslant \frac{x^{3}(1-x)^{3}}{x^{2}-x+1} \text {. }
$$
Since $x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4}$, when $x=\frac{1}{2}$,
$$
\left(x^{2}-x+1\right)_{\min }=\frac{3}{4} \text {. }
$$
Notice that, $\sqrt{x(1-x)} \leqslant \... | \frac{1}{48} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,770 |
5. In the Cartesian coordinate plane, if a circle passing through the origin with radius $r$ is completely contained within the region $y \geqslant x^{4}$, then the maximum value of $r$ is $\qquad$ - . | 5. $\frac{3 \sqrt[3]{2}}{4}$.
By symmetry, to find the maximum value of $r$, we can set the equation of the circle as $x^{2}+(y-r)^{2}=r^{2}$.
From the problem, we know that for any $x \in[-r, r]$,
$$
r-\sqrt{r^{2}-x^{2}} \geqslant x^{4} \text {. }
$$
Let $x=r \cos \theta$.
We only need to discuss for $0 \leqslant \t... | \frac{3 \sqrt[3]{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,771 |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$
Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$ | 6.8.
From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$
Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,772 |
7. Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=6$. Then the maximum value of $x+\sqrt{x y}+\sqrt[3]{x y z}$ is $\qquad$ . | 7.8.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{1}{4} x+y+4 z \geqslant 3 \sqrt[3]{x y z}, \\
\frac{3}{4} x+3 y \geqslant 3 \sqrt{x y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
x+4 y+4 z \geqslant 3 \sqrt{x y}+3 \sqrt[3]{x y z} \\
\Rightarrow x+\sqrt{x y}+\sqrt[3]{x y z}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,773 |
8. Fill the four characters “马” “上” “成” “功” in a $5 \times 5$ grid, with at most one character in each small square. “马” “上” must be filled in from left to right, and “成” “功” must also be filled in from left to right. “马” “上” must be in the same row or in the same column from top to bottom, or “成” “功” must be in the sa... | 8.42100.
The problem is equivalent to:
Filling 2 $a$s and 2 $b$s in a $5 \times 5$ grid, with at most one letter in each small square, such that at least one pair of the same letters is in the same row or column.
First, consider the case where the same letters are neither in the same row nor in the same column.
The nu... | 42100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,774 |
9. (16 points) Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \subset \mathbf{Z}_{+}$. For all different subsets $B, C \subseteq A$, we have $\sum_{x \in B} x \neq \sum_{x \in C} x$. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<2 .
$$ | For the set $A=\left\{1,2,2^{2}, \cdots, 2^{n-1}\right\}$, it satisfies that for any $B \neq C, B 、 C \subseteq A, \sum_{x \in B} x \neq \sum_{x \in C} x$, and
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}2^{n_{0}-1}$.
If $a_{k}>2^{k-1}\left(k \geqslant n_{0}\right)$, then $a_{k} \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,775 |
10. (20 points) Given the functions
$$
f(x)=x^{2}-2 a x \text { and } g(x)=-x^{2}-1
$$
their graphs have two common tangents, and the perimeter of the quadrilateral formed by these four tangent points is 6. Find the value of the real number $a$. | 10. Let the functions $f(x)$ and $g(x)$ have a common tangent line passing through the points $\left(x_{1}, f\left(x_{1}\right)\right)$ and $\left(x_{2}, g\left(x_{2}\right)\right)$. Then the equation of the common tangent line is
$$
\begin{array}{l}
y=f\left(x_{1}\right)+f^{\prime}\left(x_{1}\right)\left(x-x_{1}\right... | a= \pm \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,776 |
11. (20 points) The ellipse $C_{1}$ and the hyperbola $C_{2}$ have common foci $( \pm c, 0)(c>0), C_{1}$ and $C_{2}$ have an eccentricity difference of no more than 1, and $C_{2}$ has an asymptote with a slope of no less than $\frac{4}{3}$. $C_{1}$ and $C_{2}$ intersect the positive x-axis at points $A$ and $B$, respec... | 11. Let the eccentricities of the ellipse $C_{1}$ and the hyperbola $C_{2}$ be $e_{1}$ and $e_{2}$, respectively.
Then the equations of the ellipse $C_{1}$ and the hyperbola $C_{2}$ are
$\left(1-e_{1}^{2}\right) x^{2}+y^{2}=\frac{1-e_{1}^{2}}{e_{1}^{2}} c^{2}$,
$\left(1-e_{2}^{2}\right) x^{2}+y^{2}=\frac{1-e_{2}^{2}}{e... | \frac{9 \sqrt{5}}{50} c^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,777 |
一、(40 points) Extend the sides $BC$ and $BA$ of $\triangle ABC$ to points $E$ and $F$ respectively, such that $AE = BE$ and $BF = CF$. Let $EA$ intersect $FC$ at point $D$. Let $H$ and $O$ be the orthocenter and circumcenter of $\triangle ABC$ respectively. Prove that the line $HO$ passes through point $D$. | Connect $O E, O F$.
By symmetry, $O E \perp A B, O F \perp B C$.
Thus, $O E / / C H, O F / / A H \Rightarrow \angle A H C=\angle F O E$.
$$
\begin{array}{l}
\text { Also, } \angle H A D=90^{\circ}-\angle A E B \\
=90^{\circ}-\angle B F C=\angle H C D .
\end{array}
$$
Applying the Law of Sines in $\triangle A H D$ and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,778 |
4. As shown in Figure $12, P A$ and $P B$ are tangent to $\odot O$ at points $A$ and $B$, respectively. $C$ is a point on the arc $\overparen{A B}$. Through $C$, draw $D E \perp P C$, intersecting the angle bisectors of $\angle A O C$ and $\angle B O C$ at points $D$ and $E$, respectively. Prove: $C D = C E$. | Let the line $D E$ intersect $O A$ at point $M$. Then $P, C, A$, and $M$ are concyclic.
It is easy to see that $\triangle P A C \backsim \triangle M O D \Rightarrow \frac{P C}{P A}=\frac{M D}{M O}$.
Therefore, since $O D$ bisects $\angle A O C \Rightarrow \frac{C D}{C O}=\frac{M D}{M O}=\frac{P C}{P B}$.
Since $P A=P B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,779 |
II. (40 points) Let $n(n \geqslant 3)$ be a given natural number, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i < j \leqslant n)$ as $m$. If $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, find the maximum value of $m$. | Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$.
On one hand,
$$
\begin{array}{l}
\sum_{1 \leqslant i < j \leqslant n}\left(a_{j}-a_{i}\right)^{2} \geqslant n \sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)^{2} \geqslant n(n-1) m^{2} .
\end{array}
$$
On the other hand, for $i < j$, we have
$$
\begin{ar... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,780 |
Three, (50 points) Prove: There exist infinitely many integer quadruples $(a, b, c, d)$ satisfying the following conditions:
(1) $a>c>0,(a, c)=1$;
(2) For any given positive integer $k$, there are exactly $k$ positive integers $n$ such that $(a n+b) \mid(c n+d)$. | When $k=1$, let $c=a-1, d=b+1(a>1$, $\left.a, d \in \mathbf{Z}_{+}\right)$, obviously, $a>c>0, (a, c)=1$.
When $n \geqslant 2$, $a n+b>(a-1) n+b+1>0$.
And when $n=1$, $(a n+b) \mid[(a-1) n+b+1]$.
Therefore, there are infinitely many integer tuples $(a, b, c, d)$.
When $k \geqslant 2$, let $a=2, b=1, c=1, d=\frac{p^{k}-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,781 |
For a given natural number $n \geqslant 2$, find the largest $N$ such that: no matter how the numbers $1,2, \cdots, n^{2}$ are filled into an $n \times n$ grid, there always exist two numbers in the same row or the same column whose difference is not less than $N$.
---
Please note that the mathematical symbols and ex... | The maximum value of $N$ is $\frac{(n-1)(n+2)}{2}$.
First, construct specific examples to prove:
$$
N \leqslant \frac{(n-1)(n+2)}{2} \text {. }
$$
(1) When $n$ is even, let $n=2 k$. As shown in Figure 2, divide the grid into four $k \times k$ regions $A, B, C, D$, and fill the numbers as follows.
The filling method for... | \frac{(n-1)(n+2)}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,782 |
As shown in Figure 2, given that the incircle of $\triangle ABC$ touches the sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively, and $AD$ intersects $EF$ at point $G$. Prove that $EB$ bisects $\angle CEF$ if and only if $FG = 4GE$. | Proof As shown in Figure 2, let the circle passing through points $B, E, F$ intersect $AC$ at another point $S$, and the extension of $AD$ intersects $BS$ at point $T$. Let the lengths of $AF, BD, CE$ be $x, y, z$ respectively.
By the symmetry of the figure,
$B F=S E, E F \parallel B S$.
Therefore, $\frac{F G}{G E}=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,783 |
In $\triangle A B C$, given that $I$ is the incenter, and the circumradius and inradius of $\triangle A B C$ are denoted as $R$ and $r$ respectively. Prove:
$$
A I^{2}+B I^{2}+C I^{2} \geqslant 6 R r .
$$ | Proof As shown in Figure 3, construct the circumcircle $\odot O$ of $\triangle ABC$, extend $AI, BI, CI$ to intersect $\odot O$ at points $A', B', C'$, respectively, and let the line $OI$ intersect $\odot O$ at points $X, Y$.
Let $AA'$ intersect $BC$ at point $D$, and connect $A'B$.
Let $BC = a, CA = b, AB = c$.
By th... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,784 |
Given $x_{i}(i=1,2, \cdots, n)$ are positive real numbers satisfying $\prod_{i=1}^{n} x_{i}=1, p \geqslant 0$. Prove:
$$
\prod_{i=1}^{n}\left(x_{i}+\sqrt{p}+\sqrt{p+1}-1\right) \geqslant(\sqrt{p}+\sqrt{p+1})^{n} .
$$ | Prove that when $p=0$, the inequality obviously holds.
Below, we prove the case when $p>0$.
Consider the function
$$
\begin{aligned}
f(x)= & x+\sqrt{p+1}+\sqrt{p-1}- \\
& (\sqrt{p+1}+\sqrt{p}) x^{\sqrt{p+1}-\sqrt{p}}(x>0) .
\end{aligned}
$$
Then $f^{\prime}(x)$
$$
\begin{array}{l}
=1-(\sqrt{p+1}+\sqrt{p})(\sqrt{p+1}-\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,785 |
390 If among $n$ elements exactly $m$ are the same, i.e., these $m$ elements are distinct from the remaining elements $(1 \leqslant m<n, n \geqslant 2)$. Let the number of ways to divide these $n$ elements into two groups be denoted by $H(n, m)$. Prove:
$$
H(n, m)=2^{n-m-1}(m+1)-1 .
$$ | Prove that when $n$ is even,
$$
\begin{array}{l}
H(n, m) \\
=\sum_{i=1}^{\left[\frac{n}{2}\right]} \sum_{j=\{(i)}^{-1} \mathrm{C}_{n-m}^{j}+\sum_{j=n(n, m)}^{\left[\frac{n-m}{2}\right]-1} \mathrm{C}_{n-m}^{j}+\lambda(n, m) \mathrm{C}_{n-m}^{\left[\frac{n-m}{2}\right]} \\
=(m+1)\left(\sum_{j=0}^{\left[\frac{n-m}{2}\righ... | 2^{n-m-1}(m+1)-1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,786 |
5. Given that the three vertices $A_{1}, B_{1}, C_{1}$ of $\triangle A_{1} B_{1} C_{1}$ are points on the lines of the sides $BC, CA, AB$ of $\triangle ABC$, and satisfy $\triangle A_{1} B_{1} C_{1} \backsim \triangle ABC$. Prove: The orthocenter $H$ of $\triangle A_{1} B_{1} C_{1}$ coincides with the circumcenter of $... | First, prove that points $A_{1} 、 B 、 H 、 C_{1}$ are concyclic, so $\angle H B A = \angle H A_{1} C_{1}$, and point $H$ is not inside $\triangle A_{1} C_{1} B$.
Similarly, point $H$ is not inside $\triangle A_{1} B_{1} C$ or $\triangle A B_{1} C_{1}$.
Therefore, $H$ must be inside $\triangle A B C$.
Thus, $\angle H A B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,787 |
Example 2 Given that $P A$ and $P B$ are two tangents of $\odot O$, with points of tangency at $A$ and $B$ respectively, $M$ and $N$ are the midpoints of segments $A P$ and $A B$ respectively, the extension of $M N$ intersects $\odot O$ at point $C$, point $N$ is between $M$ and $C$, connecting $P C$ intersects $\odot ... | Proof As shown in Figure 1, let $C M$ intersect $\odot O$ at point $E$, and connect $P E, E O, O C, O P$.
By Property 1(2), we know that points $P, C, O, E$ are concyclic.
Obviously, points $P, N, O$ are collinear, and
$P N \cdot P O = P B^{2} = P D \cdot P C$.
Therefore, points $D, C, O, N$ are concyclic.
Thus, $\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,788 |
Example 3 Draw two tangents and a secant from a point outside a circle, with the points of tangency being $A, B$, and the secant intersecting the circle at points $C, D$, with $C$ between points $P, D$. Take a point $Q$ on the chord $C D$ such that $\angle D A Q = \angle P B C$. Prove: $\angle D B Q = \angle P A C .{ }... | Proof As shown in Figure 2.
By property $2(1)$, we know that quadrilateral $A C B D$ is a harmonic quadrilateral.
Thus, $A C \cdot B D=A D \cdot B C$.
By Ptolemy's theorem, we have
$$
2 A D \cdot B C=A B \cdot C D \text {. }
$$
By $\angle D A Q=\angle P B C=\angle C A B$,
$$
\angle A D Q=\angle A B C \text {, }
$$
we... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,789 |
Example 4 As shown in Figure 3, given that $P$ and $Q$ are the midpoints of the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$. If $\angle BPA = \angle DPA$, prove: $\angle AQB = \angle CQB .^{[4]}$ | Proof As shown in Figure 3, extend $BP$, intersecting the circle at point $E$.
Since $P$ is the midpoint of $AC$ and $\angle BPA = \angle DPA$, we know that $E$ and $D$ are symmetric with respect to the perpendicular bisector of $AC$.
$$
\begin{array}{l}
\text{Thus, } AE = CD, CE = AD. \\
\text{By } S_{\triangle ABE} =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,790 |
For example, $5 \odot O$ is the circumcircle of $\triangle A B C$, $A M 、 A T$ are the median and angle bisector respectively, the tangents to the circle through points $B 、 C$ intersect at point $P$, connect $A P$, intersecting $B C 、 \odot O$ at points $D 、 E$ respectively. Prove: $T$ is the incenter of $\triangle A ... | Proof As shown in Figure 4, let the line $O P$ intersect $\odot O$ at points $N$ and $L$.
Then point $M$ is on $O P$, and point $L$ is on line $A T$.
By property 2(3), we know that $M, P$ harmonically divide $N L$.
Since $N A \perp A L$, by the property of harmonic division of a segment, we know that $A L$ bisects $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,791 |
Example 8 Given two circles $\odot O_{1}$ and $\odot O_{2}$ with different radii intersecting at points $M$ and $N$, and $\odot O_{1}$ and $\odot O_{2}$ are internally tangent to $\odot O$ at points $S$ and $T$ respectively. Prove: $O M \perp M N$ if and only if $S$, $N$, and $T$ are collinear. | Proof As shown in Figure 7, let the tangents through points $S$ and $T$ intersect at point $P$. Then, by the Radical Center Theorem, we know that point $P$ lies on line $M N$. By Property 2(2), we have
$$
\begin{array}{l}
O M \perp M N \Leftrightarrow M \text { is the midpoint of chord } I J \text { of } \odot O \\
\Le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,794 |
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, $AC > BC$. Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively, and $CF \perp AB$ at point $F$. Let $P$ be a point on line $AB$ (point $P$ does not coincide with point $A$), such that $AF = PF$. Let $G$ be the midpoi... | 【Analysis and Proof】As shown in Figure 3, to prove that points $F, G, Z, Y$ are concyclic, we know from
$$
\angle Y G Z = \angle O G Z = 90^{\circ},
$$
that it suffices to prove $\angle O F X = 90^{\circ}$.
Draw $O E \perp A B$ at point $E$. Then $C H = 2 O E$.
From the given, $P B = P F - B F = A F - B F = 2 E F$.
On... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,795 |
5. Convex quadrilateral $A B C D$ is inscribed in a circle. The tangents to the circle at points $A$ and $C$ intersect at point $P$. If point $P$ does not lie on line $B D$, and $P A^{2}=P B \cdot P D$, prove: The intersection of $B D$ and $A C$ is the midpoint of $A C$. | Let $P B$ intersect the circle at point $E$, $P D$ intersect the circle at point $F$, the center of the circle be $O$, and $D B$ intersect $E F$ at point $M$.
By property 3, we know that $M$ lies on $A C$.
Furthermore, from $P E \cdot P B = P D \cdot P F = P A^{2} = P B \cdot P D$, we get $\triangle O P E \cong \triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,796 |
6. Two circles are externally tangent at point $A$, and internally tangent to another circle $\Gamma$ at points $B$ and $C$. Let $D$ be the midpoint of the chord of circle $\Gamma$ cut by the common internal tangent (through $A$) of the smaller circles. Prove: When points $B$, $C$, and $D$ are not collinear, $A$ is the... | Hint: Let the internal common tangent passing through point $A$ intersect circle $\Gamma$ at points $P$ and $Q$. By the Radical Center Theorem, the tangents through points $B$ and $C$ and line $PQ$ concur at point $K$.
By property $2(2)$, points $K$, $B$, $D$, and $C$ are concyclic.
Since $KB = KC$, point $A$ lies on t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,797 |
Example 1: A person places 2013 balls in sequence on a circle according to the following rules: First, place the first ball, then move counterclockwise around the circle, and whenever passing a previously placed ball, place a new ball in the gap; continue this process. The ball placed on the $k$-th time is numbered $k$... | Solve by the method of unraveling step by step, considering layers.
When the 1st and 2nd balls are placed, continue around one circle, then place the 3rd and 4th balls; after the second circle, place the 5th, 6th, 7th, and 8th balls; after the third circle, place the 9th, 10th, ..., 16th balls; ... after the k-th circ... | 1007, 252 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,798 |
Example 2 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$.
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Next, copy all the previous items $1,1,2$, and add the successor number 3 of 2, to get
$$
a_{4}=1, ... | From the construction method of the sequence $\left\{a_{n}\right\}$, it is easy to know that
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
In general, we have $a_{2^{n}-1}=n$, meaning the number $n$ first appears at the $2^{n}-1$ term, and if $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then
$... | 3952 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,799 |
Example 3 There are 800 points on a circle, numbered $1,2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ gaps in a cloc... | Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, such as the $2k$-th point, then according to the coloring rule, every red point colored afterward will also be an even-numbered point. In this case, if we rename the 2nd, 4th, ..., $2k$-th,... | 25 | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,800 |
Example 1 Given real numbers $x, y$ satisfy $x^{2}+3 x y+4 y^{2} \leqslant \frac{7}{2}$. Prove: $x+y \leqslant 2$. [1] | Proof Given conditions can be transformed into
$$
\left(x+\frac{3}{2} y\right)^{2}+\frac{7}{4} y^{2} \leqslant \frac{7}{2} \text {. }
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\lambda^{2}+\mu^{2}\right)\left[\left(x+\frac{3}{2} y\right)^{2}+\frac{7}{4} y^{2}\right] \\
\geqslant\left[\lambd... | x+y \leqslant 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 727,802 |
Example 2 Let $a$ and $b$ both be positive numbers, and $a+b=1$. Find
$$
E(a, b)=3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}
$$
the minimum value. | By the Cauchy-Schwarz inequality, we have
$$
\left(2 a^{2}+1\right)\left(\frac{1}{2}+\lambda^{2}\right) \geqslant(a+\lambda)^{2},
$$
where $\lambda$ is a positive constant to be determined.
Equality holds if and only if $4 a^{2}=\frac{1}{\lambda^{2}}$, i.e., $\lambda^{2}=\frac{1}{4 a^{2}}$.
Thus, $\sqrt{1+2 a^{2}} \ge... | 5 \sqrt{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,803 |
Example 3 Let positive real numbers $a, b, c$ satisfy $2a^2 + b^2 = 9c^2$. Prove: $\frac{2c}{a} + \frac{c}{b} \geqslant \sqrt{3}$. | Given conditions can be transformed into
$$
\frac{2 a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=9 \text {. }
$$
By the Cauchy-Schwarz inequality, we have
$$
\left(\frac{2 a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}\right)\left(\lambda^{2}+\mu^{2}\right) \geqslant\left(\frac{\sqrt{2} \lambda a}{c}+\frac{\mu b}{c}\right)^{2},
$$
where $\l... | \sqrt{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,804 |
Example 4 Given positive real numbers $a, b, c, d$ satisfying $a\left(c^{2}-1\right)=b\left(b^{2}+c^{2}\right)$, and $d \leqslant 1$.
Prove:
$$
d\left(a \sqrt{1-d^{2}}+b^{2} \sqrt{1+d^{2}}\right) \leqslant \frac{(a+b) c}{2} .
$$ | Proof Let the parameter $\lambda>1$.
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
d\left(a \sqrt{1-d^{2}}+b^{2} \sqrt{1+d^{2}}\right) \\
\leqslant d \sqrt{\left(\frac{a^{2}}{\lambda}+b^{4}\right)\left[\lambda\left(1-d^{2}\right)+\left(1+d^{2}\right)\right]} \\
\leqslant \sqrt{\frac{1}{\lambda-1}\left(\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,805 |
Example 4 As shown in Figure 4, in $\triangle A B C$, $D, E, F$ are the midpoints of $B C, C A, A B$ respectively, $G$ is the centroid of $\triangle A B C$, the circumcircle $\odot O$ of $\triangle B C F$ intersects $B E$ at point $M$, the circumcircle $\odot O^{\prime}$ of $\triangle A B E$ intersects $A D$ at point $... | 【Analysis and Proof】According to the properties of the centroid, we have
$$
A G=2 D G, B G=2 E G, C G=2 F G \text {. }
$$
As shown in Figure 4, take $I$ and $J$ as the midpoints of $B G$ and $C G$ respectively.
Then $I J / / B C$
$$
\Rightarrow \angle G I J=\angle G B C=\angle G F M
$$
$\Rightarrow F, I, J, M$ are con... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,806 |
Example 5 Proof: For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}} \\
<4\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) .
\end{array}
$$ | Prove that the conclusion can be strengthened to
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}} \\
<2\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) .
\end{array}
$$
Let \( x_{1}, x_{2}, \cdots, x_{n} \) be undetermined positive real numbers.
By ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,807 |
Example 1 Color the points on the plane in red and blue, ensuring there are both red and blue points. For any given positive number $a$, prove:
(1) There exist two points of the same color on the plane, whose distance is $a$;
(2) There exist two points of different colors on the plane, whose distance is $a$. | 【Analysis and Proof】(1) It is very simple, a typical pigeonhole principle model with three elements (points), two pigeonholes (colors). However, it needs to satisfy another condition - the distance between two points is $a$.
To meet this condition, first construct a point set $A$, where the distance between any two po... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,808 |
Example 2: On an $n \times n$ chessboard, there are $n^{2}-n+1$ white pieces, with at most one piece per square. Each operation involves removing one white piece and placing one black piece (the black piece can be placed in any empty square), until all white pieces are removed and $n^{2}-n+1$ black pieces are placed. P... | 【Analysis and Proof】First, consider whether the number “ $n^{2}-n+1$ ” has any special significance.
There are always $n^{2}-n+1$ chess pieces on the board. Note that, $n^{2}-n+1$ involves too many objects. From the opposite perspective, it is equivalent to having $n-1$ empty cells, which implies that at least one row... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,809 |
Example 3 In a $4 n \times 4 n$ chessboard, each cell is filled with $0$, $1$, or $-1$. It is known that the absolute value of the sum of all numbers in the table does not exceed $16 n$. Prove: There exists an $n \times n$ sub-table, where the absolute value of the sum of all numbers does not exceed $n$.
Translate the... | 【Analysis and Proof】For any chessboard $A$, let $S(A)$ denote the sum of all numbers filled in $A$.
From the goal, we need to find an $n \times n$ sub-chessboard $A$ such that
$|S(A)| \leqslant n$.
Since it is not easy to find such a set directly, we can use proof by contradiction. Assume that for all $n \times n$ sub-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,810 |
Example 1 Given $-6 \leqslant x_{i} \leqslant 10(i=1,2, \cdots, 10)$, $\sum_{i=1}^{10} x_{i}=50$.
When $\sum_{i=1}^{10} x_{i}^{2}$ reaches its maximum value, among the ten numbers $x_{1}, x_{2}, \cdots, x_{10}$, the number of numbers equal to -6 is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | Notice that $f(x)=x^{2}$ is a convex function. According to the proposition, when $\sum_{i=1}^{10} x_{i}^{2}$ reaches its maximum value, at least nine of $x_{1}, x_{2}, \cdots, x_{10}$ are equal to -6 or 10.
Assume that there are $m$ values of -6 and $9-m$ values of 10. Then the remaining one is
$$
\begin{array}{l}
50... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,813 |
Example 2 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Try to find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason. | Given that $f(x)=x^{12}$ is a convex function, then $\sum_{i=1}^{197} x_{i}^{12}$ achieves its maximum value when at least 1996 of $x_{1}, x_{2}, \cdots, x_{1997}$ are equal to $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.
Assume that there are $t$ instances of $-\frac{1}{\sqrt{3}}$ and $1996-t$ instances of $\sqrt{3}$. Then t... | 189548 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,814 |
1. If $\frac{x}{2 y}=\frac{y}{x-y}$, then $\frac{x^{2}-2 x y-3 y^{2}}{x^{2}+4 x y+3 y^{2}}=(\quad)$.
(A) 0
(B) $-\frac{3}{2}$
(C) $-\frac{1}{5}$
(D) $-\frac{1}{5}$ or $-\frac{3}{2}$ | - 1. C.
From $\frac{x}{2 y}=\frac{y}{x-y}$, we get
$$
\begin{array}{l}
x^{2}-x y-2 y^{2}=0(x \neq y \neq 0) \\
\Rightarrow x=-y \text { or } x=2 y . \\
\text { Therefore, } \frac{x^{2}-2 x y-3 y^{2}}{x^{2}+4 x y+3 y^{2}}=\frac{(x-3 y)(x+y)}{(x+3 y)(x+y)} \\
=\frac{x-3 y}{x+3 y}=-\frac{1}{5}(x \neq-y) .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,815 |
2. Let $a=\sqrt{2}-1, b=\sqrt{5}-2, c=\sqrt{10}-3$. Then the size relationship of $a$, $b$, and $c$ is ( ).
(A) $a<b<c$
(B) $a<c<b$
(C) $b<c<a$
(D) $c<b<a$ | 2. D.
From the given, we know that $a-b=\sqrt{2}+1-\sqrt{5}$.
And $(\sqrt{2}+1)^{2}-5=2 \sqrt{2}-2>0$, thus, $a>b$.
Similarly, $b>c$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,816 |
Example 5 As shown in Figure 5, it is known that quadrilateral $ABCD$ is a cyclic quadrilateral, and its diagonals $AC$ and $BD$ are perpendicular to each other. Point $F$ is on side $BC$, line $EF \parallel AC$, and intersects $AB$ at point $E$, line $FG \parallel BD$, and intersects $CD$ at point $G$. Let the project... | Prove that connecting $G R$ intersects line $F Q$ at point $M$.
Below is the proof: Point $M$ lies on $A C$.
In fact, since $A C \perp B D$ and $F G \parallel B D$, we know $F G \perp A C$.
Combining $F Q \perp A D$ and the fact that points $A, B, C, D$ are concyclic, we get
$$
\begin{array}{l}
\angle Q F C=360^{\circ}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,817 |
3. Let $a$, $b$, and $c$ be positive integers, and $a \geqslant b \geqslant c$, satisfying $a+b+c=15$. Then the number of triangles with side lengths $a$, $b$, and $c$ is ( ) .
(A) 5
(B) 7
(C) 10
(D) 12 | 3. B.
The solution can be found using enumeration:
$$
\begin{array}{l}
(7,7,1),(7,6,2),(7,5,3),(7,4,4), \\
(6,6,3),(6,5,4),(5,5,5),
\end{array}
$$
There are 7 pairs in total. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,818 |
4. As shown in Figure 1, in quadrilateral $ABCD$, adjacent sides and angles are not equal, $E$ and $F$ are the midpoints of $BO$ and $DO$ respectively. Then, the number of pairs of congruent triangles in this figure is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 4. C.
According to the figure, there are 7 pairs of congruent triangles. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,819 |
5. If integers $a, b$ cannot be divided by 5, but $a+b$ can be divided by 5, then which of the following expressions cannot be divided by 5?
(A) $2a+3b$
(B) $2a-3b$
(C) $3a+8b$
(D) $3a-7b$ | 5. A.
Verify one by one
$$
\begin{array}{l}
2 a-3 b=2(a+b)-5 b, \\
3 a+8 b=3(a+b)+5 b, \\
3 a-7 b=3(a+b)-10 b,
\end{array}
$$
can all be divided by 5.
But $2 a+3 b=2(a+b)+b$, cannot be divided by 5. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,820 |
6. There is a certain amount of saltwater with a concentration of $3 \%$. After adding $a$ grams of pure water, the concentration of the saltwater becomes $2 \%$. After adding another $a$ grams of pure water, the concentration of the saltwater at this point is ( ).
(A) $1 \%$
(B) $1.25 \%$
(C) $1.5 \%$
(D) $1.75 \%$ | 6. C.
Let's assume that in a 3% salt solution, there are 97 grams of water and 3 grams of salt. Therefore, adding 50 grams of water can turn it into a 2% salt solution, and then adding another 50 grams of water can turn it into a 1.5% salt solution. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,821 |
1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$
then $a b c=$ . $\qquad$ | $$
\text { Two, 1. }-9 \text {. }
$$
From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$
Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,822 |
3. If $a, b$ are positive integers, and satisfy $5a+7b=50$, then $ab=$ . $\qquad$ | 3. 15 .
From the problem, we know
$$
a=\frac{50-7 b}{5}=10-b-\frac{2 b}{5} \text { . }
$$
Therefore, $a=3, b=5$.
Thus, $a b=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,824 |
4. Given $S_{\triangle M B C}=4, 3 A B=2 B C$, draw the perpendicular from point $C$ to the angle bisector $B E$ of $\angle A B C$, and let the foot of the perpendicular be $D$. Then $S_{\triangle B D C}=$ $\qquad$ | 4.3.
Construct the reflection of point $C$ about $BD$ as point $F$, thus, $BC=BF$. Connect $FA$ and $FD$, then point $F$ lies on the extension of $BA$, and points $C$, $D$, and $F$ are collinear.
$$
\text{Therefore, } \frac{S_{\triangle ABC}}{S_{\triangle FBC}}=\frac{AB}{FB}=\frac{AB}{BC}=\frac{2}{3} \text{.}
$$
Sinc... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,825 |
Three. (20 points) As shown in Figure 3, in the convex quadrilateral $ABCD$, it is known that $\angle ABC + \angle ADC = 180^{\circ}$, $AC$ bisects $\angle BAD$, and a perpendicular line from point $C$ to $AB$ intersects $AB$ at point $E$. Prove:
$$
AE = \frac{1}{2}(AB + AD).
$$ | Three, as shown in Figure 4, construct the symmetric points $E'$ and $B'$ of $B$ and $E$ with respect to $AC$, and connect $E B'$.
Since $AC$ is the angle bisector of $\angle BAC$, the points $E'$ and $B'$ lie on the extension of $AD$, and $\triangle CBE \cong \triangle CB'E'$. Therefore, $\angle CB'D = \angle ABC = 1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,826 |
Four. (25 points) Given three linear functions
$$
y=x+1, y=1-x, y=\frac{1}{2} x+b \text {. }
$$
(1) If the graphs of these three functions can form a triangle, find the range of the real number $b$;
(2) If the area of the triangle formed by the graphs of these three functions is $\frac{4}{3}$, find the value of the rea... | (1) Obviously, the line $l_{1}: y=x+1$ intersects with $l_{2}: y=1-x$ at point $P(0,1)$.
Only when the line $l_{3}: y=\frac{1}{2} x+b$ passes through point $P(0,1)$, the three lines cannot form a triangle, at this time, $b=1$.
Therefore, when $b \neq 1$, the three lines can form a triangle.
Thus, the intersection poin... | b=0 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,827 |
Example 6 Given that quadrilateral $A B C D$ is inscribed in $\odot O$, the opposite sides $A B$ and $D C$ intersect at point $P$, and $A D$ and $B C$ intersect at point $Q$. A circle passing through points $A$ and $B$ is tangent to line $P Q$ at points $M$ and $N$. Prove: The intersection of lines $M B$ and $N D$ lies... | Prove as shown in Figure 6, take point $K$ on $PQ$ such that points $P, K, C, B$ are concyclic.
Then $\angle PKC = \angle ABC = \angle CDQ$.
Thus, points $C, K, D, Q$ are concyclic.
By the secant theorem, we have
\[
\begin{array}{l}
PB \cdot PA + QD \cdot QA = PC \cdot PD + QC \cdot QB \\
= PK \cdot PQ + QK \cdot QP = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,828 |
Five. (25 points) Given positive integers $a$, $b$, $c$ satisfying $a<b<c$, and $\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=1$. Find the values of $a$, $b$, and $c$.
untranslated part:
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。
---
Five. (25 points) Given positive integers $a$, $b$, $c$ satisfying $a<b<c$, and $\frac{1}{a-... | Five, because $a\frac{1}{b-1}>\frac{1}{c-1} \text {. }$
Thus, $\frac{3}{a-1}>1 \Rightarrow a=2 \text {. }$
Therefore, $a=3$.
Similarly, $\frac{2}{b-1}>\frac{1}{2} \Rightarrow b<5$.
Since $a<b$, then $b=4$.
Substituting into $\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=1$, we get $c=7$.
In summary, $a=3, b=4, c=7$. | a=3, b=4, c=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,829 |
2. In $\triangle A B C$, it is known that $\angle A=120^{\circ}, A B=5, B C$ =7. Then $\frac{\sin B}{\sin C}=(\quad)$.
(A) $\frac{8}{5}$
(B) $\frac{5}{8}$
(C) $\frac{5}{3}$
(D) $\frac{3}{5}$ | 2. D.
By the Law of Sines, we have
$$
\begin{array}{l}
\sin C=\frac{c}{a} \sin A=\frac{5}{7} \times \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{14} . \\
\text { Therefore, } \cot C=\frac{11}{5 \sqrt{3}} . \\
\text { Hence, } \frac{\sin B}{\sin C}=\frac{\sin (A+C)}{\sin C} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\s... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,831 |
3. The function that is both odd and monotonically decreasing on the interval $[-1,1]$ is ( ).
(A) $f(x)=\sin x$
(B) $f(x)=-|x+1|$
(C) $f(x)=\ln \frac{2-x}{2+x}$
(D) $f(x)=\frac{1}{2}\left(a^{x}+a^{-x}\right)$ | 3. C.
If $f(x)=\ln \frac{2-x}{2+x}$, then
$$
f(-x)=\ln \frac{2+x}{2-x}=-\ln \frac{2-x}{2+x}=-f(x) \text {. }
$$
Therefore, $f(x)=\ln \frac{2-x}{2+x}$ is an odd function.
Also, $t=\frac{2-x}{2+x}=-1+\frac{4}{2+x}$ is a monotonically decreasing function on the interval $[-1,1]$, and $y=\ln t$ is a monotonically increasi... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,832 |
4. A folk song competition is held in a certain place, with a pair of singers from each of six provinces participating in the final. Now, four winners need to be selected. The probability that exactly two of the four selected are from the same province is ).
(A) $\frac{16}{33}$
(B) $\frac{33}{128}$
(C) $\frac{32}{33}$
... | 4. A.
The probability that exactly two of the four selected contestants are from the same province is
$$
P=\frac{\mathrm{C}_{6}^{1} \mathrm{C}_{5}^{2} \times 2 \times 2}{\mathrm{C}_{12}^{4}}=\frac{16}{33} .
$$ | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,833 |
5. If the sequence $\left\{a_{n}\right\}: a_{1}, a_{2}, \cdots, a_{5}$ satisfies
$$
0 \leqslant a_{1}<a_{2}<a_{3}<a_{4}<a_{5},
$$
and for any $i, j(1 \leqslant i \leqslant j \leqslant 5)$, $a_{j}-a_{i}$ is in the sequence.
(1) $a_{1}=0$;
(2) $a_{5}=4 a_{2}$;
(3) $\left\{a_{n}\right\}$ is an arithmetic sequence;
(4) Th... | 5. D.
Since $a_{1}-a_{1}=0 \in \left\{a_{n}\right\}$, we have $a_{1}=0$. Therefore, statement (1) is correct.
Since $0=a_{1}<a_{3}-a_{2}<a_{4}-a_{2}<a_{5}-a_{2}<a_{5}$, and $a_{3}-a_{2}, a_{4}-a_{2}, a_{5}-a_{2} \in \left\{a_{n}\right\}$, we have
$$
\begin{array}{l}
a_{3}-a_{2}=a_{2}, a_{4}-a_{2}=a_{3}, a_{5}-a_{2}=a... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,834 |
6. Given the function $y=f(x)$ is defined on $\mathbf{R}$, with a period of 3, and Figure 1 shows the graph of the function in the interval $[-2,1]$. Then $\frac{f(2014)}{f(5) f(15)}=$ $\qquad$ . | $$
\text { II,6. }-2 \text {. }
$$
From the problem statement and combining with the graph, we know that
$$
\frac{f(2014)}{f(5) f(15)}=\frac{f(1)}{f(-1) f(0)}=\frac{2}{(-1) \times 1}=-2 \text {. }
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,835 |
7. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
|\overrightarrow{A B}|=\sqrt{3},|\overrightarrow{B C}|=1, \\
|\overrightarrow{A C}| \cos B=|\overrightarrow{B C}| \cos A \text {. } \\
\text { Then } \overrightarrow{A C} \cdot \overrightarrow{A B}=
\end{array}
$$ | 7.2.
Let the three sides of $\triangle A B C$ be $a, b, c$. From the given condition and using the Law of Sines, we have
$$
\begin{array}{l}
\sin B \cdot \cos B=\sin A \cdot \cos A \\
\Rightarrow \sin 2 B=\sin 2 A \\
\Rightarrow \angle B=\angle A \text { or } \angle B+\angle A=90^{\circ} .
\end{array}
$$
We will disc... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,836 |
8. Figure 2 is the three views of a geometric solid. Then the volume of the solid is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 8. $\frac{4}{3}$.
From the three views, we can determine that the geometric body is a triangular pyramid $P-ABC$, as shown in Figure 3.
From the top view, we know that plane $PAB \perp$ plane $ABC$, $AC = BC$, $OC = 2$, and $OA = OB = 1$.
From the left view, we know that $OP \perp AB$, and thus, $OP \perp$ plane $AB... | \frac{4}{3} | Other | math-word-problem | Yes | Yes | cn_contest | false | 727,837 |
9. Given four points $O, A, B, C$ on a plane, satisfying
$$
O A=4, O B=3, O C=2, \overrightarrow{O B} \cdot \overrightarrow{O C}=3 \text {. }
$$
Then the maximum area of $\triangle A B C$ is $\qquad$ | $9.2 \sqrt{7}+\frac{3 \sqrt{3}}{2}$.
From the problem, we know
$$
B C=\sqrt{O B^{2}+O C^{2}-2 \overrightarrow{O B} \cdot \overrightarrow{O C}}=\sqrt{7} \text {, }
$$
and the angle between $O B$ and $O C$ is $60^{\circ}$.
Consider three circles $\Gamma_{1} 、 \Gamma_{2} 、 \Gamma_{3}$ centered at the origin $O$ with radi... | 2 \sqrt{7}+\frac{3 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,838 |
Example 7 Given a convex quadrilateral $A B C D$ inscribed in $\odot O$, the diagonals $A C$ and $B D$ intersect at point $P$, and through $P$ perpendiculars are drawn to lines $A B, B C, C D, D A$, with the feet of the perpendiculars being $E, F, G, H$ respectively. Prove that $E H$, $B D$, and $F G$ are concurrent or... | 【Analysis and Proof】As shown in Figure 7, connect $E F, G H$.
From $P E \perp A B, P F \perp B C$, we know that points $P, E, B, F$ are concyclic.
Similarly, points $P, E, A, H$ are concyclic.
Thus, $\angle P E H = \angle P A H = \angle D B C = \angle P E F$.
This indicates that $P E$ bisects $\angle H E F$.
Similarly,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,839 |
10. The system of equations
$$
\left\{\begin{array}{l}
a+b+c+d=-2, \\
a b+a c+a d+b c+b d+c d=-3, \\
b c d+a c d+a b d+a b c=4, \\
a b c d=3
\end{array}\right.
$$
has a set of real solutions $(a, b, c, d)=$ . $\qquad$ | 10. $\left(\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}\right)$.
By Vieta's formulas, $a, b, c, d$ are exactly the four real roots of the equation
$$
x^{4}+2 x^{3}-3 x^{2}-4 x+3=0
$$
Notice that,
$$
\begin{array}{l}
x^{4}+2 x^{3}-3 x^{2}-4 x+3=0 \\
\Rightarrow\left(x^{2... | \left(\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,840 |
11. (15 points) Let the sets be
$$
\begin{array}{l}
A=\left\{x \mid x^{2}+3 x+2 \leqslant 0\right\}, \\
B=\left\{x \mid x^{2}+a x+b \leqslant 0\right\} .
\end{array}
$$
(1) If $\left(\complement_{\mathrm{R}} A\right) \cap B=\{x \mid-1<x \leqslant 2\}$,
$$
\left(\complement_{\mathbf{R}} A\right) \cup B=\mathbf{R} \text ... | Three, 11. From the given, we have
$$
A=[-2,-1], \complement_{R} A=(-\infty,-2) \cup(-1,+\infty) \text {. }
$$
Let $f(x)=x^{2}+a x+b$.
(1) From $\left(\complement_{\mathrm{R}} A\right) \cap B=\{x \mid-10$, then
$$
\left\{\begin{array}{l}
\Delta>0, \\
f(-1) \geqslant 0, \\
f(-2) \geqslant 0, \\
-2 \leqslant-\frac{1}{a}... | a \in [-2, 2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,841 |
12. (25 points) Function
$$
f(x)=2(2 \cos x+1) \sin ^{2} x+\cos 3 x(x \in \mathbf{R}) \text {. }
$$
Find the maximum value of the function $f(x)$. | 12. Notice,
$$
\begin{array}{l}
f(x)=2 \sin 2 x \cdot \sin x+2 \sin ^{2} x+\cos (2 x+x) \\
= 2 \sin 2 x \cdot \sin x+2 \sin ^{2} x+\cos 2 x \cdot \cos x- \\
\sin 2 x \cdot \sin x \\
= \sin 2 x \cdot \sin x+\cos 2 x \cdot \cos x+2 \sin ^{2} x \\
= \cos (2 x-x)+2 \sin ^{2} x=2 \sin ^{2} x+\cos x \\
=-2 \cos ^{2} x+\cos ... | \frac{17}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,842 |
13. (25 points) The equation of line $m$ is $y=k x+1, A$ and $B$ are two points on line $m$, whose x-coordinates are exactly the two different negative real roots of the quadratic equation in $x$
$$
\left(1-k^{2}\right) x^{2}-2 k x-2=0
$$
The line $l$ passes through point $P(-2,0)$ and the midpoint of segment $A B$, $... | 13. Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=\frac{2 k}{1-k^{2}} \text {. }
$$
Let the midpoint of line segment $A B$ be $M$. Then
$$
\begin{array}{l}
x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{k}{1-k^{2}}, \\
y_{M}=k x_{M}+1=\frac{1}{1-k^{2}} .
\end{array}
$$
Let the line $l$ in... | 4+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,843 |
14. (25 points) If there exist sets $A$ and $B$ such that
$$
A \cap B=\varnothing, A \cup B=\mathbf{Z}_{+},
$$
then $(A, B)$ is called a bipartition of $\mathbf{Z}_{+}$.
$$
\begin{array}{l}
\text { (1) Let } A=\left\{x \mid x=3 k\left(k \in \mathbf{Z}_{+}\right)\right\}, \\
B=\left\{x \mid x=3 k \pm 1\left(k \in \math... | 14. (1) Since $1 \notin A, 1 \notin B$, therefore, $A \cup B \neq \mathbf{Z}_{+}$.
Hence, $(A, B)$ is not a bipartition of $\mathbf{Z}_{+}$.
(2) It can be found.
In the set of positive integers $\mathbf{Z}_{+}$, an infinite geometric sequence can be uniquely represented by a pair of positive integers $(a, q)$, where $a... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,844 |
6. Ed and Ann have lunch together, Ed and Ann order medium and large lemonade drinks, respectively. It is known that the large size has a capacity 50% more than the medium size. After both have drunk $\frac{3}{4}$ of their respective drinks, Ann gives the remaining $\frac{1}{3}$ plus 2 ounces to Ed. After lunch, they f... | 6. D.
Let the size of the medium lemon drink be $x$ ounces. According to the problem, we have
$$
x+\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x+2=\frac{3}{2} x-\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x-2 \text {. }
$$
Solving for $x$, we get $x=16$.
Therefore, the total amount they drank is $16+\frac... | 40 | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,845 |
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