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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class | __index_level_0__ int64 0 742k |
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7. If real numbers $x, y$ satisfy
$$
\frac{x}{2^{10}+5^{3}}+\frac{y}{2^{10}+6^{3}}=1, \frac{x}{3^{10}+5^{3}}+\frac{y}{3^{10}+6^{3}}=1 \text {. }
$$
then $x+y=$ | $$
\text { II. } 7.2^{10}+3^{10}+5^{3}+6^{3} \text {. }
$$
According to the conditions, $2^{10}$ and $3^{10}$ are the two roots of the equation with respect to $t$:
$$
\frac{x}{t+5^{3}}+\frac{y}{t+6^{3}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(x+y-5^{3}-6^{3}\right) t+30^{3}-6^{3} x-5^... | 2^{10}+3^{10}+5^{3}+6^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,862 |
8. The vertex of the parabola is $O$, the focus is $F$, and $M$ is a moving point on the parabola, then the maximum value of $\frac{M O}{M F}$ is $\qquad$ | 8. $\frac{2 \sqrt{3}}{3}$.
Let the equation of the parabola be $y^{2}=2 p x$, then the vertex $O(0, 0)$ and the focus $F\left(\frac{\sqrt{p}}{2}, 0\right)$. Let $M(x, y)$, then
$$
\begin{array}{l}
\left(\frac{M O}{M F}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x-\frac{p}{2}\right)^{2}+y^{2}}=\frac{x^{2}+2 p x}{x^{2}+p x+\fr... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,863 |
9. Calculate: $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=$ | 9.4 .
$$
\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=\frac{2 \sin \left(30^{\circ}-10^{\circ}\right)}{\frac{1}{2} \sin 20^{\circ}}=4
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,864 |
10. Through a point $P$ on the line $l: y=x+9$, construct an ellipse with the shortest major axis, whose foci are $F_{1}(-3,0)$ and $F_{2}(3,0)$. Then the equation of the ellipse is $\qquad$. | 10. $\frac{x^{2}}{45}+\frac{y^{2}}{36}=1$.
As shown in Figure 4, let the point on line $l$ be $P(t, t+9)$, and take the symmetric point of $F_{1}(-3,0)$ about line $l$ as $Q(-9, 6)$. According to the definition of an ellipse, we have
$$
\begin{array}{l}
2 a=P F_{1}+P F_{2} \\
=P Q+P F_{2} \\
\geqslant Q F_{2} \\
=\sqr... | \frac{x^{2}}{45}+\frac{y^{2}}{36}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,865 |
11. To cut a rectangular prism into $k$ tetrahedra, the minimum value of $k$ is | 11.5 .
According to the equivalence, we only need to consider the cutting situation of a unit cube.
On the one hand, first, we need to show that 4 is not enough. If there were 4, since all faces of a tetrahedron are triangles and are not parallel to each other, the top face of the cube would have to be cut into at le... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,866 |
12. Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$ . | 12.133113.
We call this kind of number a "good number". There are 3 one-digit good numbers; 12 two-digit good numbers, which is $3 \times 4$; 48 three-digit good numbers, which is $3 \times 4^2$; $\cdots \cdots \cdot k$-digit good numbers have $3 \times 4^{k-1}$ numbers $(k=1$, $2, \cdots)$. Let $S_{n}=3 \sum_{k=1}^{n... | 133113 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,867 |
13. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{n+1}=\frac{n a_{n}}{(n+1)\left(n a_{n}+1\right)} . \\
\text { Let } x_{k}=a_{1}+a_{2}+\cdots+a_{k}, \\
y_{k}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}(k=1,2, \cdots) . \\
\text { Find } \sum_{k=1}^{n} x_{k} y_{k} .
... | Three, 13. The condition can be written as
$$
\frac{1}{(n+1) a_{n+1}}-\frac{1}{n a_{n}}=1 \text {. }
$$
Then $\frac{1}{n a_{n}}=\left[\frac{1}{n a_{n}}-\frac{1}{(n-1) a_{n-1}}\right]+$
$$
\begin{array}{l}
{\left[\frac{1}{(n-1) a_{n-1}}-\frac{1}{(n-2) a_{n-2}}\right]+\cdots+} \\
\left(\frac{1}{2 a_{2}}-\frac{1}{a_{1}}\... | \frac{n(n+1)\left(3 n^{2}+11 n+4\right)}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,869 |
14. As shown in Figure 2, the circumcenter of $\triangle A B C$ is $O, E$ is the midpoint of $A C$, and the line $O E$ intersects $A B$ at point $D, M, N$ are the circumcenter and incenter of $\triangle B C D$ respectively. If $A B=2 B C$, prove that $\triangle D M N$ is a right triangle. | 14. Proof 1: As shown in Figure 6, since points $O$ and $M$ are both on the perpendicular bisector of $BC$, let the line $OM$ intersect $BC$ at point $P$ and $\odot M$ at point $F$. Then $P$ is the midpoint of $BC$, and $F$ is the midpoint of $\overparen{BC}$.
Since $N$ is the incenter of $\triangle BCD$, points $D$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,870 |
15. If a four-digit number $n=\overline{a b c d}$ has digits $a, b, c, d$ such that any three of these digits can form the lengths of the sides of a triangle, then $n$ is called a "four-digit triangular number". Find the total number of four-digit triangular numbers. | 15. Call $(a, b, c, d)$ a digital group of $n$, then
$$
a, b, c, d \in M=\{1,2, \cdots, 9\} \text {. }
$$
(1) When the digital group contains only one value, i.e., $(a, a, a, a)(a=1,2, \cdots, 9)$, there are 9 values of $n$.
(2) When the digital group contains exactly two values $a, b(a>b)$.
(i) The digital group is of... | 1681 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,871 |
In $\square A B C D$, $C E \perp A B, C F \perp A D$, with feet of the perpendiculars at $E$ and $F$. Let $E F$ intersect the diagonal $B D$ at point $P$. If $A B: A D=2: 3$, find $P F: P E$. | Solution: As shown in the figure,
2, connect $A C$ intersecting
$B D$ at $O$, draw $E H / / B D$
intersecting $A C$ and
$A F$ at $G$ and $H$ respectively,
draw $O M \perp E F$ at $M$, connect $G M, M C, C P$.
Since $C E \perp A B, C F \perp A D$, therefore, quadrilateral $A E C F$ is a cyclic quadrilateral with $A C$ a... | 4:9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,872 |
The square of four positive integers sums to 2008, among which the unit digits of two numbers are 2, and the unit digits of the other two numbers are 0. Find these four numbers.
(Tian Yonghai, Suihua Education Institute, Heilongjiang Province, 152054) | Solution: According to the problem, we can set the 4 numbers to be $10x$, $10y$, $10a+2$, and $10b+2$, where $0 < x \leq y$, $0 \leq a \leq b$ ($x, y, a, b \in \mathbf{N}_{+}$).
From the given information, we have
$$
\begin{array}{l}
100 x^{2} + 100 y^{2} + (10 a + 2)^{2} + (10 b + 2)^{2} \\
= 2008.
\end{array}
$$
Sim... | 20, 40, 2, 2 \text{ or } 10, 20, 22, 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,873 |
Given 225 As shown in Figure 3, in $\triangle ABC$, $\angle B$ and $\angle C$ are acute angles. The inscribed square $DEFG$ has vertices $D$ and $E$ on side $BC$, and $G$ and $F$ on sides $AB$ and $AC$ respectively. The center of the square is $O$, and the line $AO$ extended intersects $BC$ at $K$. Let $BC = a$, the he... | Proof: (1) From $\triangle A G F \backsim \triangle A B C$
$$
\begin{array}{l}
\Rightarrow \frac{G F}{B C}=\frac{h-m}{h} \Rightarrow \frac{m}{a}=\frac{h-m}{h} \\
\Rightarrow \frac{1}{m}=\frac{1}{a}+\frac{1}{h} .
\end{array}
$$
(2) It is easy to know that $a=m(1+\cot B+\cot C)$.
Since $D F$ and $E G$ are the transversal... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,874 |
Given a sequence $\left\{a_{n}\right\}$ whose terms are all positive, and satisfies the relation $a_{n+1} \leqslant a_{n}^{k}-a_{n}^{k+1}\left(k \in \mathbf{N}_{+}\right)$. Prove that for all positive integers $n(n \geqslant 2)$, we have
$$
a_{n} \leqslant \frac{k^{k}}{(k+1)^{k+1}+(n-2) k^{k}} .
$$ | Proof: From $01 \text {. }$
Applying the $(k+1)$-variable mean inequality, we get
$$
\begin{array}{l}
a_{2} \leqslant a_{1}^{k}\left(1-a_{1}\right)=\frac{1}{k} a_{1} a_{1} \cdots a_{1}\left(k-k a_{1}\right) \\
\leqslant \frac{1}{k}\left[\frac{k a_{1}+\left(k-k a_{1}\right)}{k+1}\right]^{k+1}=\frac{k^{k}}{(k+1)^{k+1}},... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,875 |
Five. (20 points) Given that the distance from point $P$ to the fixed point $M(1,1)$ is $d_{1}$, and the distance from $P$ to the fixed line $l: x+y-2=0$ is $d_{2}$. When $d_{2}=\alpha d_{1}, \alpha \in(0,1]$, find the equation of the locus of point $P$.
---
The above text translated into English, preserving the orig... | Let point $P(x, y)$. According to the problem, we have
$$
\frac{|x+y-2|}{\sqrt{2}}=\alpha \sqrt{(x-1)^{2}+(y-1)^{2}} \text {, }
$$
which means
$$
\begin{array}{l}
|(x-1)+(y-1)| \\
=\sqrt{2} \alpha \sqrt{(x-1)^{2}+(y-1)^{2}} .
\end{array}
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\left(1-2 \alpha^{2}\rig... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,876 |
One, (50 points) As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle C = 90^{\circ}$, $CH \perp AB$, with $H$ as the foot of the perpendicular. $\odot O_{1}$ and $\odot O_{2}$ are the incircles of $\triangle AHC$ and $\triangle BHC$, respectively. The other external common tangent of the two circles in... | As shown in Figure 5, the altitude of a right triangle divides it into two similar right triangles, so $\triangle A H C \backsim \triangle C H B$. By the property of corresponding elements of similar triangles, we have $\frac{O_{1} H}{O_{2} H}=\frac{A C}{B C}$. Therefore, Rt $\triangle O_{1} O_{2} H \backsim$ Rt $\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,877 |
II. (50 points) Given the concentric circle $(x-\sqrt{2})^{2}+(y-\sqrt{3})^{2}=R^{2}(R>0)$ and the ellipse $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$ have common points. Find the minimum value of the radius $R$.
untranslated text retained the original format and line breaks. | $$
\begin{array}{l}
R^{2}=(\sqrt{3} \cos \theta-\sqrt{2})^{2}+(\sqrt{2} \sin \theta-\sqrt{3})^{2} \\
=3 \cos ^{2} \theta-2 \sqrt{6} \cos \theta+2+2 \sin ^{2} \theta-2 \sqrt{6} \sin \theta+3 \\
=7+\cos ^{2} \theta-2 \sqrt{6}(\cos \theta+\sin \theta) \\
=7+\frac{1+\cos 2 \theta}{2}-4 \sqrt{3} \sin \left(\theta+\frac{\pi}... | \sqrt{\frac{15-\sqrt{56 \sqrt{42}-167}}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,878 |
Three. (50 points) A subset of the set $S_{n}=\{1,2, \cdots, n\}$ is called a "good subset" if it does not contain two consecutive natural numbers. How many good subsets are there in $S_{n}$? | Let $S_{n}$ have $a_{n}$ good subsets.
When $n=1$, $S_{n}$ has subsets $\varnothing, \{1\}$, which are good subsets, $a_{1}=2$.
When $n=2$, $S_{n}$ has 4 subsets: $\varnothing, \{1\}$, $\{2\}, \{1,2\}$, except for $\{1,2\}$, all are good subsets, $a_{2}=3$.
For $n \geqslant 3$, let $M$ be a good subset of $S_{n}$.
If ... | a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,879 |
1. There are 2008 viewing stations set up along a circular marathon track, labeled as $c_{1}, c_{2}, \cdots, c_{2008}$ in a clockwise direction. These stations divide the track into 2008 segments. An athlete places mascots numbered $1, 2, \cdots, 2008$ at these stations in the following manner: He first places the 1st ... | - 1. C.
Let the mascot number placed at site $c_{k}$ be $x$, $x \in\{1,2, \cdots, 2008\}$. Then $29(x-1)+1 \equiv k(\bmod 2008)$.
When $k=2008$, we get
$29(x-1)+1 \equiv 0(\bmod 2008)$.
Let $29(x-1)+1=2008 y$, i.e., $x=\frac{2008 y-1}{29}+1=69 y+1+\frac{7 y-1}{29}$.
It is easy to see that when $y=-4+29 t, t \in \mathb... | 1732 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,880 |
2. Let $n \geqslant 2$ be a positive integer, and let $A_{n}$ denote the set of all proper fractions with denominator $n$. Then, the number of elements in the set $A_{2008} \cap\left(\bigcup_{k=1}^{2007} A_{k}\right)$ is ( ).
(A) 2007
(B) 1008
(C) 1007
(D) 1000 | 2.C.
Since $2008=2^{3} \times 251$ (251 is a prime number), among the numbers $1,2, \cdots, 2007$, the numbers that are not coprime with 2008 include all the even numbers, as well as $251,3 \times 251,5 \times 251,7 \times$ 251. In total, there are $1003+4=1007$ such numbers. For each such number $a$, the proper fract... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,881 |
3. The trajectory of the moving point in the complex plane
$$
Z\left(\frac{\cos \theta+\sin \theta}{1+\cos \theta \cdot \sin \theta}, \frac{\cos \theta \cdot \sin \theta}{1+\cos \theta \cdot \sin \theta}\right)(\theta \in \mathbf{R})
$$
is ( ).
(A) Circle
(B) Ellipse
(C) Hyperbola
(D) Parabola | $\begin{array}{l}\text { 3. A. } \\ =\left(\frac{\cos \theta+\sin \theta}{1+\cos \theta \cdot \sin \theta}\right)^{2}+\left(\frac{\cos \theta \cdot \sin \theta}{1+\cos \theta \cdot \sin \theta}\right)^{2} \\ =1\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 3. A. } \\ =\left(\frac{\cos \theta+\si... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,882 |
3. Rectangle $A B C D$ is a billiard table, with the length-to-width ratio $A B: B C=2: 1$. A ball is shot from the midpoint $P$ of $A B$ at an acute angle $\alpha$ to $A B$ towards side $B C$, and it sequentially bounces off sides $B C$, $C D$, $D A$, and $A B$ before hitting a ball $M$ at the center of the table. The... | 3. B.
As shown in Figure 2, construct the reflection of point $M$ about $AB$ to get point $M_{4}$, the reflection of $M_{4}$ about $AD$ to get point $M_{3}$, the reflection of $M_{3}$ about $CD$ to get point $M_{2}$, and the reflection of $M_{2}$ about $BC$ to get point $M_{1}$.
Let $P M_{1} \cap BC=P_{1}, P_{1} M_{2... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,883 |
4. For any given irrational numbers $a, b$ and real number $r$ $(r>0)$, the number of rational points on the circle $(x-a)^{2}+(y-b)^{2}=r^{2}$ is ( ).
(A) at most one
(B) at most two
(C) at least two, finite in number
(D) infinitely many | 4.B.
For a point $M(a, b)$, let $P(M, r)$ denote the number of rational points on the above circle.
First, construct a circle that meets the conditions and has at least two rational points.
For this, take points $A(0,0)$ and $B(2,2)$. The equation of the perpendicular bisector $l$ of segment $AB$ is $x+y=2$. On the ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,884 |
5. A circle $\odot O_{1}$ with radius $r$ is contained within a circle $\odot O$ with radius $R$. It is known that there exists a quadrilateral $A B C D$ that is externally tangent to $\odot O_{1}$ and inscribed in $\odot O$. Then the minimum value of $\frac{R}{r}$ is ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{... | 5.D.
Let the area of quadrilateral $ABCD$ be $S$, the semi-perimeter be $p$, $AB=a$, $BC=b$, $CD=c$, $DA=d$, and the areas of $\triangle ABC$, $\triangle BCD$, $\triangle CDA$, and $\triangle DAB$ be $S_1$, $S_2$, $S_3$, and $S_4$ respectively.
Since quadrilateral $ABCD$ has both an incircle and a circumcircle, then $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,885 |
6. In tetrahedron $ABCD$, $AB=2\sqrt{3}$, $BC=\sqrt{3}$, $AC=AD=BD=CD=3$, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. A beetle wants to travel from point $M$ to point $N$ along the surface of the tetrahedron. To minimize the distance traveled, which edge must it cross?
(A) $AC$
(B) $BC$
(C) $AD$
(D) $B... | 6. A.
From the side length conditions, it is easy to see that $\triangle A B C$ is a right triangle, with $\angle A C B=90^{\circ}, \angle A B C=60^{\circ}, \angle B A C=30^{\circ}$.
Let the shortest path length from point $M$, crossing edge $A C$ to reach point $M$ be denoted as $(A C)$, and similarly for other case... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,886 |
II. Fill-in-the-blank Questions (9 points each, total 54 points)
1. Remove all perfect squares and cubes from the natural numbers, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2008}=$ $\qquad$ | Two, 1.2062.
First, remove numbers of the form $n^{6}$. In the intervals $\left(1^{6}, 2^{6}\right),\left(2^{6}, 3^{6}\right),\left(3^{6}, 4^{6}\right), \cdots$, there are no more overlaps between square numbers and cube numbers.
Since $\sqrt{2^{6}}=2^{3}=8, \sqrt[3]{2^{6}}=4$, removing the interval endpoints, we know... | 2062 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,887 |
2. The vertices of a cube with edge length 1 are alternately colored red and blue, so that the two vertices on each edge are of different colors. Then, the volume of the common part of the tetrahedron formed by the red vertices and the tetrahedron formed by the blue vertices is $\qquad$ | 2. $\frac{1}{6}$.
As shown in Figure 5, note that
$$
S_{\triangle A_{1} B D}
$$
$$
=\frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{\sqrt{3}}{2} \text {. }
$$
Let $A C_{1}$ intersect the plane of $\triangle A_{1} B D$ at point $M$, and let $A M=h$.
According to the volume relationship of the tetrahedron $A_{1} A B D$, we have... | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,888 |
3. Let the even function $f(x)$ satisfy: $f(1)=2$, and when $x y$ $\neq 0$, $f\left(\sqrt{x^{2}+y^{2}}\right)=\frac{f(x) f(y)}{f(x)+f(y)}$. Then $f(5)$ $=$ | 3. $\frac{2}{25}$.
From the condition, we have $\frac{1}{f\left(\sqrt{x^{2}+y^{2}}\right)}=\frac{1}{f(x)}+\frac{1}{f(y)}$.
Since $f(-x)=f(x)$, we only need to consider the case where $x, y \in \mathbf{R}_{+}$.
Let $x=\sqrt{u}, y=\sqrt{v}$. Then
$\frac{1}{f(\sqrt{u+v})}=\frac{1}{f(\sqrt{u})}+\frac{1}{f(\sqrt{v})}$.
Let... | \frac{2}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,889 |
4. Calculate $3 \tan 10^{\circ}+4 \sqrt{3} \sin 10^{\circ}=$ | $\begin{array}{l}4 \cdot \sqrt{3} . \\ 3 \tan 10^{\circ}+4 \sqrt{3} \sin 10^{\circ}=\frac{3 \sin 10^{\circ}+2 \sqrt{3} \sin 20^{\circ}}{\cos 10^{\circ}} \\ =\frac{3 \sin 10^{\circ}+2 \sqrt{3} \sin \left(30^{\circ}-10^{\circ}\right)}{\cos 10^{\circ}} \\ =\frac{3 \sin 10^{\circ}+2 \sqrt{3}\left(\sin 30^{\circ} \cdot \cos... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,890 |
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then, in the plane, the area of the figure formed by all points satisfying $[x]^{2}+[y]^{2}=50$ is
保留源文本的换行和格式，直接输出翻译结果。 | 5.12.
First, consider the first quadrant.
From $50=1^{2}+7^{2}=7^{2}+1^{2}=5^{2}+5^{2}$, we get $([x],[y])=(1,7),(7,1),(5,5)$.
And from $[x]=1,[y]=7$, we get the unit square $1 \leqslant x<2,7 \leqslant y<8$, whose area is 1.
Similarly, from $([x],[y])=(7,1)$ and $(5,5)$, we also get one unit square each, and these t... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,891 |
6. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits be denoted as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". The sum of all coincidence numbers is $\qquad$ . | 6.531.
Let $n=\overline{a_{1} a_{2} \cdots a_{k}}\left(a_{1} \neq 0\right)$.
From $n-s(n)=p(n)$, we get
$$
\sum_{i=1}^{k-1} a_{i}\left(10^{k-i}-1\right)=a_{1} a_{2} \cdots a_{k} \text {, }
$$
which means $a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right)+m=0$,
where $m=a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,892 |
4. A school plans to renovate three teaching buildings during the summer vacation, and has asked three construction companies to bid on the repair costs for each building, as shown in Table 1 (unit: ten thousand yuan).
Table 1
\begin{tabular}{|l|c|c|c|}
\hline & Teaching Building 1 & Teaching Building 2 & Teaching Buil... | 4.C.
Take three numbers from different rows and columns in the table and find their sum:
$$
\begin{array}{ll}
13+19+21=53, & 13+22+15=50, \\
17+24+21=62, & 17+22+10=49, \\
20+24+15=59, & 20+19+10=49 .
\end{array}
$$
It can be seen that the minimum total bid is 490,000 yuan. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,893 |
$\begin{array}{l}\text { Three. (20 points) In an acute } \triangle A B C \text {, prove: } \\ \cos A+\cos B+\cos C \\ \geqslant 2(\cos A \cdot \cos B+\cos B \cdot \cos C+\cos C \cdot \cos A) .\end{array}$ | $$
\begin{array}{l}
\text { III. Since } \\
2 \leqslant \frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}, 2 \leqslant \frac{\sin B}{\sin C}+\frac{\sin C}{\sin B}, \\
2 \leqslant \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A},
\end{array}
$$
and
$$
\begin{array}{l}
\sin A \cdot \cos A+\sin B \cdot \cos B \\
=\frac{\sin 2 A+\si... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,894 |
Four, (20 points) Find all positive numbers $t=\frac{b^{2}}{a^{2}}$, such that on the right branch of the hyperbola $\frac{\dot{x}^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$, there always exists a focal chord $l$ that subtends a right angle at the origin. | Let the right focus of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{t a^{2}}=1$ be $F(c, 0)$, where $c=a \sqrt{1+t}$. Let the focal chord $l$ intersect the right branch of the hyperbola at points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$.
From $O M \cdot O N=0$, we get
$x_{1} x_{2}+y_{1} y_{2}=0$... | \frac{1+\sqrt{5}}{2} \leqslant t<2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,895 |
Five. (20 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=\left\{\begin{array}{l}
a_{n}-n, a_{n}>n ; \\
a_{n}+n, a_{n} \leqslant n .
\end{array}\right.
$$
(1) Find $a_{2008}$;
(2) Find the smallest positive integer $n$, such that $a_{n}=2008$. | Five, the initial values of the easy sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & $2 \mid 3$ & & & 6 & 7 & 8 & & 10 & 11 & 12 & & \\
\hline$a$ & 1 & 2 & 41 & & 10 & & 11 & & $12 \mid$ & 2 & 13 & 1 & 14 \\
\hline
\end{tabular}
Next, focus on the subscripts $n_{k}$ such ... | 5827 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,896 |
One, (50 points) As shown in Figure 1, in the acute triangle $\triangle ABC$ with unequal sides, the orthocenter is $H$, $D$ is the midpoint of side $BC$, the line $B H \cap A C=E$, $C H \cap A B=F$, $A H \cap B C=T$, the circumcircle of $\triangle B D E$ intersects the circumcircle of $\triangle C D F$ at $G$, and the... | As shown in Figure 7, connect $M B$ and $M C$. Since $B, C, E, F$ are concyclic with the center $D$, then
$$
D E=D F=D B=D C \text{. }
$$
Connect $G D, G E, G F$.
Since $B, D, E, G$ and $C, D, F, G$ are each concyclic, we have
$$
\begin{array}{l}
\angle D G E=\angle D B E=\angle T A C, \\
\angle D G F=\angle D C F=\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,897 |
In the Olympic volleyball qualifying tournament, there are $n$ teams participating, where each pair of teams plays one match, and each match must have a winner. If there are $k(3 \leqslant k \leqslant n)$ teams $A_{1}$, $A_{2}, \cdots, A_{k}$ such that: $A_{1}$ beats $A_{2}$, $A_{2}$ beats $A_{3}$, ..., $A_{k-1}$ beats... | With $A_{1}, A_{2}, \cdots, A_{k}$ as vertices.
If team $A_{i}$ beats $A_{j}$, then a directed edge is drawn between the two points: $A_{i} \rightarrow A_{j}$, thus forming an $n$-order tournament graph $G$. According to the conditions, the $n$ vertices of $G$ can be arranged into an $n$-order directed cycle, set as
$$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,898 |
Three, (50 points) If the three sides of a triangle are all rational numbers, and one of its interior angles is also a rational number, find all possible values of this interior angle.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result direc... | In $\triangle A B C$, let $B C=a, C A=b, A B=$ $c(a, b, c \in \mathbf{Q})$, and its internal angle
$$
\angle A=\frac{m^{0}}{n_{1}}=\frac{m}{n} 180^{\circ}.
$$
Since a similar transformation of the triangle does not change its internal angles, we can assume $a, b, c \in \mathbf{Z}_{+}$.
According to the cosine rule, $... | 60^{\circ}, 90^{\circ}, 120^{\circ} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,899 |
2. As shown in Figure 1, the two sides $A B C D$ and $A B E F$ of the triangular prism are both rectangles, and $P, Q$ are points on $A E$ and $B D$ respectively. If $P Q \parallel$ plane $B C E$, then, ( ).
(A) It must be that $B Q = D Q$ and $E P = P A$
(B) It must be that $E P = P A$, but it is possible that $B Q \n... | 2.D.
As shown in Figure 3, draw $PM \perp BE$ at point $M$, $QN \perp BC$ at point $N$, and connect $MN$. Since quadrilateral $ABEF$ is a rectangle, $PM \parallel AB$.
Similarly, $NQ \parallel AB$.
Therefore, $QN \parallel PM$.
Thus, points $P, Q, N, M$ are coplanar.
Since $PQ \parallel$ plane $BCE$, $PQ \subseteq$ pl... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,901 |
3. There is a dice-throwing game that can randomly display one of the numbers from 1 to 6. A chess piece is placed at the vertex of a square $ABCD$ with a side length of 1. Each time the dice is thrown, the chess piece moves once. The movement rule is to start from the current position and move a distance equal to the ... | 3. B.
For $X \in\{A, B, C, D\}$, let $P(X)$ denote the probability of moving from $A$ to point $X$ in one move. Therefore,
$$
P(A)+P(B)+P(C)+P(D)=1 .
$$
The distance between the points reached after two moves being greater than 1 is equivalent to the points being opposite vertices of the square. Thus, the required pr... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,902 |
4. Let $S_{n}$ be the sum of the first $n$ terms of a geometric sequence. If $x=S_{n}^{2}+S_{2 n}^{2}, y=S_{n}\left(S_{2 n}+S_{3 n}\right)$, then $x-y(\quad)$.
(A) is 0
(B) is a positive number
(C) is a negative number
(D) is sometimes positive and sometimes negative | 4. A.
Let the geometric sequence be $\left\{a_{n}\right\}$, with the common ratio $q$. Then
$$
\begin{array}{l}
S_{n}=a_{1}+a_{2}+\cdots+a_{n}, \\
S_{2 n}=S_{n}+a_{n+1}+a_{n+2}+\cdots+a_{2 n} \\
=S_{n}+q^{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
=S_{n}+q^{n} S_{n}=S_{n}\left(1+q^{n}\right), \\
S_{3 n}=S_{2 n}+a_{2 n... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,903 |
5. If the vertices of a regular tetrahedron $P Q M N$ are on the faces of a given tetrahedron $A B C D$, with one point on each face, then, ( ).
(A) When the tetrahedron $A B C D$ is a regular tetrahedron, there are infinitely many regular tetrahedrons $P Q M N$, otherwise, there is only one regular tetrahedron $P Q M ... | 5.D.
In the tetrahedron $ABCD$, construct a sufficiently small regular tetrahedron $PQMN$ inside it, such that it can rotate freely within the tetrahedron, with the distances from each vertex of the regular tetrahedron $PQMN$ to the corresponding face being different. Through the vertex closest to each face, draw a pl... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,905 |
6. Let $X$ be a set of $n$ points in the plane, and for each point $A$ in $X$, there are exactly 3 points in $X$ that are at a distance of 1 from $A$. Then the minimum value of $n$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 6.C.
A line segment of length 1 is called a "good edge".
For any two points in $X$, if their distance is 1, then they are connected by a line segment. Thus, each point gives rise to 3 good edges. Therefore, $n$ points give rise to $3n$ good edges in total. Since each good edge has two endpoints and is counted twice, t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,906 |
1. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$, $F$, and $G$ are the midpoints of edges $B C$, $C C_{1}$, and $C D$ respectively. Then the angle formed by line $A_{1} G$ and plane $D E F$ is $\qquad$ | $=1.90^{\circ}$.
Since $E F \perp B_{1} C, E F \perp C D$, therefore, $E F \perp$ plane $B_{1} C D$. Thus, $E F \perp A_{1} G$.
In the square $A B C D$, $D G=C E$, so $A G \perp D E$. Since $A G$ is the projection of $A_{1} G$ in the plane $A B C D$, hence $D E \perp A_{1} G$.
Therefore, $A_{1} G \perp$ plane $D E F$. | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,907 |
2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ . | 2.1.
Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$
When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.
Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.
In fact, ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,908 |
3. Let $a, b$ be any two distinct positive integers. Then the minimum value of $\left|a b\left(a^{4}-b^{4}\right)\right|$ is $\qquad$ .
| 3.30.
$$
\begin{array}{l}
\text { Let } A=a b\left(a^{4}-b^{4}\right) \\
=a b(a-b)(a+b)\left(a^{2}+b^{2}\right) .
\end{array}
$$
First, prove: $2 \mid A$.
If $a$ and $b$ are both odd or both even, then $2 \mid (a-b)$, so $2 \mid A$;
If $a$ and $b$ are of different parity, then one of $a$ or $b$ is even, so $2 \mid ab$... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,909 |
4. If one focus of the ellipse is the orthocenter of the triangle formed by its three vertices, then the eccentricity of the ellipse $e=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. $\frac{\sqrt{5}-1}{2}$.
Let $F(c, 0)$ be a focus of the ellipse, which is the orthocenter of the triangle formed by the three vertices $A(a, 0)、 B(0, b)、 C(0,-b)$ of the ellipse, as shown in Figure 5. From $C F \perp A B$ we have
$$
\begin{array}{l}
\frac{b}{c}\left(-\frac{b}{a}\right)=-1 \\
\Rightarrow b^{2}=a c \... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 719,910 |
5. Let $n$ be a given natural number not less than 5. If
in the plane there are $n$ lines, among which exactly 3 lines are pairwise parallel, and in addition, no two lines are parallel, and no three lines are concurrent. Then the number of intersection points of these lines is $\qquad$ | 5. $\frac{n^{2}-n-6}{2}$.
First, remove 2 lines $l_{1}$ and $l_{2}$ from the 3 parallel lines, leaving $n-2$ lines, which gives
$$
\mathrm{C}_{n-2}^{2}=\frac{(n-2)(n-3)}{2}
$$
intersection points. Adding line $l_{1}$, it intersects with $n-3$ lines, resulting in $n-3$ intersection points. Similarly, adding line $l_{2... | \frac{n^{2}-n-6}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,911 |
6. Let the real number $x$ satisfy $\cos \left(x+30^{\circ}\right) \cdot \cos (x+$ $\left.45^{\circ}\right) \cdot \cos \left(x+105^{\circ}\right)+\cos ^{3} x=0$. Then, $\tan x$ $=$ . $\qquad$ | $$
\begin{array}{l}
6.2 \sqrt{3}-1 . \\
\text { Let } \angle A=30^{\circ}, \angle B=45^{\circ}, \angle C=105^{\circ} \text {. Then } \\
\angle A+\angle B+\angle C=180^{\circ} \text {. }
\end{array}
$$
Therefore, $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$.
$$
\begin{array}{l}
\text ... | 2\sqrt{3}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,912 |
Three, (20 points) Find the maximum value of the function $f(x)=\frac{\sqrt{x}}{x^{2}+1}$.
| Three, take $k>0$ ( $k$ is a parameter to be determined), then
$$
\begin{array}{l}
f(x)=\frac{\sqrt{x}}{x^{2}+1}=\frac{2 \sqrt{k x}}{2 \sqrt{k}\left(x^{2}+1\right)} \\
\leqslant \frac{x+k}{2 \sqrt{k}\left(x^{2}+1\right)} . \\
\text { Let } y=\frac{x+k}{x^{2}+1} . \text { Then } \\
y x^{2}-x+y-k=0 .
\end{array}
$$
Sin... | \frac{\sqrt[4]{27}}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,913 |
Given a positive integer $n$, solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+1=\frac{1}{x_{2}} \\
\ldots \ldots \\
x_{n-1}+1=\frac{1}{x_{n}} \\
x_{n}+1=\frac{1}{x_{1}}
\end{array}\right.
$$ | $$
x_{1}=x_{2}=\cdots=x_{n}=\frac{-1-\sqrt{5}}{2} \text { or } \frac{-1+\sqrt{5}}{2}
$$
are solutions to the system of equations.
Next, we prove: $x_{1}=x_{2}=\cdots=x_{n}$.
Subtracting the first two equations, we get
$$
x_{1}-x_{2}=\frac{1}{x_{2}}-\frac{1}{x_{3}}=\frac{x_{3}-x_{2}}{x_{2} x_{3}}.
$$
Similarly, $x_{i}... | x_{1}=x_{2}=\cdots=x_{n}=\frac{-1-\sqrt{5}}{2} \text { or } \frac{-1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,914 |
6. Assign 18 table tennis players numbered $1, 2, \cdots, 18$ to 9 tables for singles matches, with the rule that the sum of the numbers of the two players on each table must be a perfect square greater than 4. Let $\{7$ and 18 play against each other $\}$ be event $p$. Then $p$ is $(\cdots)$.
(A) impossible event
(B) ... | 6.D.
Since the sum of the two largest numbers is $18+17=35$ $<36$, the sum of the numbers of two players on the same table can only be 3 square numbers: $25, 16, 9$. Let the number of pairs with sums of $25, 16, 9$ be $x, y, z$ respectively $(x, y, z$ are non-negative integers $)$. According to the problem, we have
$$... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,915 |
Five. (20 points) Let the vertex of the parabola be $A$, and the focus be $F$. Draw a line $l$ through $F$ intersecting the parabola at points $P$ and $Q$. Lines $AP$ and $AQ$ intersect the directrix of the parabola at points $M$ and $N$, respectively. Under what condition does line $l$ ensure that the three lines $PN$... | When the line $l \perp A F$, by the symmetry of the parabola, it is known that the quadrilateral $P Q M N$ is a rectangle. At this time, the three lines $P N, Q M, A F$ are parallel to each other.
When the line $l$ is not perpendicular to $A F$, establish a rectangular coordinate system as shown in Figure 6. Let the e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,916 |
One, (50 points) As shown in Figure 2, let $H$ be the orthocenter of $\triangle A B C$, and $P$ be any point on the plane of $\triangle A B C$. Draw $H M \perp P B$ at point $M$, intersecting $A C$ at point $J$, and draw $H N \perp P C$ at point $N$, intersecting $A B$ at point $I$. Prove: $P H \perp I J$.
| As shown in Figure 7, let the three altitudes $AD$, $BE$, and $CF$ of $\triangle ABC$ intersect at point $H$. Then points $A$, $B$, $D$, and $E$ are concyclic. Hence,
$$
\begin{array}{l}
A H \cdot H D \\
=B H \cdot H E .
\end{array}
$$
Similarly,
$$
B H \cdot H E=C H \cdot H F \text {. }
$$
Let $A H \cdot H D=B H \cd... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,917 |
II. (50 points) Find all real triples $(a, b, c)$ such that for any integer $n$, we have $[n a] + [n b] = [n c]$. Here, $[x]$ denotes the greatest integer not exceeding the real number $x$.
保留源文本的换行和格式，直接输出翻译结果。 | Second, first prove: "For any integer $n$, $[n a]+[n b]=[n c]$" is equivalent to "$a, b$ at least one is an integer, and $c=a+b'$".
On one hand, if $a, b$ at least one is an integer, and $c=a+b$, then without loss of generality, assume $a$ is an integer. Then, for any integer $n, n a$ is an integer. Therefore, $n a=[n... | (m, t, m+t),(t, n, n+t),(m, n, m+n) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,918 |
Three, (50 points) Find the smallest positive integer $t$, such that for any convex $n$-gon $A_{1} A_{2} \cdots A_{n}$, as long as $n \geqslant t$, there must exist three points $A_{i} 、 A_{j} 、 A_{k}(1 \leqslant i<j<k \leqslant n)$, such that the area of $\triangle A_{i} A_{j} A_{k}$ is no more than $\frac{1}{n}$ of t... | Three, first prove a lemma.
Lemma For any convex hexagon $A_{1} A_{2} \cdots A_{6}$, there exists $1 \leqslant i\frac{S}{k+1}$, then
$S_{\text {pentagon } A_{1} A_{2} \cdots A_{k}}<S-\frac{S}{k+1}=\frac{k S}{k+1}$.
By the induction hypothesis, there must be $1 \leqslant i<j<r \leqslant n$, such that $S_{\triangle A_{i}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,919 |
2. Given the positive sequence $\left\{a_{n}\right\}$ :
$$
a_{1}=\alpha<\frac{\sqrt{2}}{2}, a_{n+1}^{2}=\frac{2 a_{n}^{2}}{1+2 a_{n}^{2}} \text {. }
$$
Then $\left\{a_{n}\right\}$ ).
(A) increasing and bounded above
(B) increasing and unbounded above
(C) decreasing and bounded below
(D) decreasing and unbounded below | 2. A.
$$
a_{1}0,
$$
so, $a_{n+1}>a_{n}$, the sequence $\left\{a_{n}\right\}$ is increasing. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,921 |
3. In the range $1 \sim 2000$, if a number is randomly selected, the probability that the integer can be divided by 6 but not by 4 is ).
(A) $\frac{1}{4}$
(B) $\frac{83}{1000}$
(C) $\frac{167}{2000}$
(D) $\frac{3}{4}$ | 3. C.
Let event $A$ be "the number is divisible by 6", and event $B$ be "the number is divisible by 4".
Since $333<\frac{2000}{6}<334$, then $P(A)=\frac{333}{2000}$.
The least common multiple of 6 and 4 is $12, 166<\frac{2000}{12}<167$, so there are exactly 166 numbers that are divisible by both 6 and 4, i.e., $P(A B)... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,922 |
4. Let $f(x)=-3 x^{3}+a x$, given that for all $x \in[0,1]$, it always holds that $|f(x)| \leqslant 1$. Then the range of the real number $a$ is ( ).
(A) $4 \leqslant a \leqslant 2 \sqrt[3]{6}$
(B) $\frac{3}{2} \sqrt[3]{6} \leqslant a \leqslant 4$
(C) $2 \leqslant a \leqslant 4$
(D) $2 \leqslant a \leqslant \frac{3}{2}... | 4.D.
From the problem, for $0 \leqslant x \leqslant 1$, we always have
$$
\begin{array}{l}
\left|a x-3 x^{3}\right| \leqslant 1 \Rightarrow-1 \leqslant a x-3 x^{3} \leqslant 1 \\
\Rightarrow 3 x^{3}-1 \leqslant a x \leqslant 3 x^{3}+1 .
\end{array}
$$
When $x=0$, $a$ can be any real number.
When $0<x \leqslant 1$, we... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,923 |
5. Let $P$ be a moving point on the edge $A B$ of the cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Then the minimum value of the dihedral angle formed by the plane $P D B^{\prime}$ and the plane $A D D^{\prime} A^{\prime}$ is ( ).
(A) $45^{\circ}$
(B) $\arctan \frac{\sqrt{2}}{2}$
(C) $\arctan \sqrt{2}$
(... | 5. B.
As shown in Figure 3, extend $A^{\prime} A$ and $B^{\prime} P$ to intersect at point $E$. By the properties of the cube, we know that $P A \perp$ plane $A E D$. Draw $A F \perp E D$ at $F$, and connect $P F$.
By the theorem of three perpendiculars, we know that $P F \perp E D$.
Therefore, $\angle P A F$ is the p... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,924 |
6. Let $P, A, B, C$ be four distinct points in space, and $\alpha \boldsymbol{P A}+\beta \boldsymbol{P B}+\gamma \boldsymbol{P C}=\mathbf{0}(\alpha, \beta, \gamma \in \mathbf{R})$.
Then $\alpha+\beta+\gamma=0$ and $\alpha \beta \gamma \neq 0$ is a ( ) for $A, B, C$ to be collinear.
(A) necessary and sufficient conditio... | 6.B.
Sufficiency.
From $\alpha+\beta+\gamma=0$, we get $\alpha=-\beta-\gamma$. Therefore,
$$
\begin{array}{l}
(-\beta-\gamma) \boldsymbol{P A}+\beta \boldsymbol{P} \boldsymbol{B}+\gamma \boldsymbol{P C}=\mathbf{0} \\
\Rightarrow \beta(\boldsymbol{P} \boldsymbol{B}-\boldsymbol{P A})+\gamma(\boldsymbol{P} \boldsymbol{C}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,925 |
1. Given $\alpha, \beta$ are acute angles, $\cos \alpha=\frac{\sqrt{5}}{5}, \cos \beta=$ $\frac{\sqrt{10}}{10}$. Then $\alpha+\beta=$ $\qquad$ . | $$
\text { 2. } \frac{3 \pi}{4} \text {. }
$$
From the given, $\sin \alpha=\frac{2 \sqrt{5}}{5}, \sin \beta=\frac{3 \sqrt{10}}{10}$, we get
$$
\begin{array}{l}
\cos (\alpha+\beta)=\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta \\
=\frac{\sqrt{5}}{5} \times \frac{\sqrt{10}}{10}-\frac{2 \sqrt{5}}{5} \times \f... | \frac{3 \pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,926 |
1. Given $x^{2}+y^{2} \leqslant 1$. Then the maximum value of the function $z=$ $\frac{\cos x+\cos y}{1+\cos x y}$ is $\qquad$ . | 2 1.1.
Assume $x \geqslant 0, y \geqslant 0$. When $0 \leqslant x \leqslant 1$, $0 \leqslant x y \leqslant y<1<\pi$, so $\cos x y \geqslant \cos y$.
Therefore, $\cos x+\cos y \leqslant 1+\cos x y$, where the equality holds when $x=y=0$.
Hence $z_{\max }=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,927 |
2. In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}$, points $D$, $E$, and $F$ are on sides $AB$, $BC$, and $CA$ respectively, $FD=FE=\frac{AC}{2}$, $\angle DFE=90^{\circ}$. Then $FC: CE: EF=$ $\qquad$ . | 2. $3: 4: 5$.
Let $A C=2 b, \angle A F D=\beta$. Then
$$
F D=b, \angle F E C=\beta \text {. }
$$
In $\triangle A F D$, by the Law of Sines, we get
$$
A F=\frac{b \sin \left(45^{\circ}+\beta\right)}{\sin 45^{\circ}}=b(\sin \beta+\cos \beta) \text {. }
$$
Also, $C E=b \cos \beta$, so,
$$
\begin{array}{l}
{[2 b-b(\sin \... | 3: 4: 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,928 |
3. Let $f=uv\left(u, v \in \mathbf{R}_{+}\right)$. It is known that for all positive triples $(a, b, c)$ satisfying $a+b+c=1$, the inequality
$$
u(ab+bc+ca)+vabc \leqslant \frac{7}{3}
$$
always holds. Then the maximum value of $f$ is . $\qquad$ | 3. $\frac{441}{4}$.
Given $a+b+c=1, a, b, c \in \mathbf{R}_{+}$, we have $a b+b c+c a \leqslant \frac{(a+b+c)^{2}}{3}=\frac{1}{3}$, and $a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3}=\frac{1}{27}$.
Thus, $u(a b+b c+c a)+v a b c \leqslant \frac{1}{3} u+\frac{1}{27} v$. Let $\frac{1}{3} u+\frac{1}{27} v \leqslant \fra... | \frac{441}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,929 |
4. Let the complex numbers be $z_{1}=-3-\sqrt{3} \mathrm{i}, z_{2}=\sqrt{3}+\mathrm{i}$, $z=\sqrt{3} \sin \theta+\mathrm{i}(\sqrt{3} \cos \theta+2)$.
Then the minimum value of $| z-z_{1}|+| z-z_{2} \mid$ is
$\qquad$. | $4.2+2 \sqrt{3}$.
$$
\begin{array}{l}
\left|z-z_{1}\right|+\left|z-z_{2}\right| \geqslant\left|z-z_{1}-\left(z-z_{2}\right)\right| \\
=\left|z_{1}-z_{2}\right|=2+2 \sqrt{3},
\end{array}
$$
The condition for equality to hold is
$$
\arg \left(z-z_{1}\right)=\arg \left(z_{2}-z\right),
$$
which means $\frac{\sqrt{3} \cos... | 2+2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,930 |
5. Given the asymptotes of a hyperbola are $x \pm 2 y=0$, and the minimum distance from a moving point $P$ on the hyperbola to point $A(5,0)$ is $\sqrt{6}$. Then the equation of the hyperbola is $\qquad$. | 5. $x^{2}-4 y^{2}=(5+\sqrt{6})^{2}$ and $4 y^{2}-x^{2}=4$.
Let the hyperbola equation be $x^{2}-4 y^{2}=\lambda(\lambda \neq 0)$, and the moving point $P(x, y)$. Then
$$
\begin{array}{l}
|P A|^{2}=(x-5)^{2}+y^{2} \\
=(x-5)^{2}+\frac{1}{4}\left(x^{2}-\lambda\right)=\frac{5}{4}(x-4)^{2}+5-\frac{\lambda}{4} .
\end{array}... | x^{2}-4 y^{2}=(5+\sqrt{6})^{2} \text{ and } 4 y^{2}-x^{2}=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,931 |
6. Given $0 \leqslant 6 x 、 3 y 、 2 z \leqslant 8$,
$$
\sqrt{12 x}+\sqrt{6 y}+\sqrt{4 z}=6 \text {. }
$$
Then the function
$$
f(x, y, z)=\frac{1}{1+x^{2}}+\frac{4}{4+y^{2}}+\frac{9}{9+z^{2}}
$$
has a maximum value of $\qquad$ . | 6. $\frac{27}{10}$.
Let $y=2 y_{1}, z=3 z_{1}$, then
$$
0 < x, y_{1}, z_{1} \leqslant \frac{4}{3}, \sqrt{\frac{x}{3}}+\sqrt{\frac{y_{1}}{3}}+\sqrt{\frac{z_{1}}{3}}=1,
$$
and $f(x, y, z)=\frac{1}{1+x^{2}}+\frac{1}{1+y_{1}^{2}}+\frac{1}{1+z_{1}^{2}}$.
By $1=\sqrt{\frac{x}{3}}+\sqrt{\frac{y_{1}}{3}}+\sqrt{\frac{z_{1}}{3... | \frac{27}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,932 |
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$:
$a_{1}=8, a_{2}=10, a_{n+1}+a_{n-1}<\frac{5}{2} a_{n}$.
Prove: $2^{n-1} a_{n}<4^{n}+4$.
保留源文本的换行和格式，直接输出翻译结果如下：
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$:
$a_{1}=8, a_{2}=10, a_{n+1}+a_{n-1}<\frac{5}{2} a_{n}$.
Prove: $2^{n-1} a_{n}<4^{n... | $$
\begin{array}{l}
a_{n+1}-2 a_{n}<\frac{1}{2}\left(a_{n}-2 a_{n-1}\right) . \\
\text { Let } b_{n}=a_{n+1}-2 a_{n} \text {, then } \\
b_{n}<\frac{1}{2} b_{n-1}<\left(\frac{1}{2}\right)^{2} b_{n-2}<\cdots<\left(\frac{1}{2}\right)^{n-1} b_{1} \\
=\left(\frac{1}{2}\right)^{n-1}\left(a_{2}-2 a_{1}\right)=-6 \times 2^{1-n... | 2^{n-1} a_{n}<4^{n}+4 | Inequalities | proof | Yes | Yes | cn_contest | false | 719,933 |
Four. (20 points) Given a regular tetrahedron $ABCD$ with edge length 2, a sphere $O$ is tangent to the faces $ABC$ and $DBC$ of the tetrahedron, with the points of tangency lying within (including the boundary) $\triangle ABC$ and $\triangle DBC$, and the sphere $O$ is also tangent to the edge $AD$.
(1) Prove: The cen... | (1) Let the midpoint of $AD$ be $E$, and connect $EB$, $EC$.
From the fact that $\triangle CAD$ and $\triangle BAD$ are both equilateral triangles, we know that $AD \perp EC$, $AD \perp EB$. Therefore, $AD \perp$ plane $BEC$, which means plane $BEC$ is the perpendicular bisector plane of $AD$. It is also easy to see th... | \frac{\sqrt{6}-\sqrt{2}}{2} \leqslant r \leqslant \frac{\sqrt{6}}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 719,934 |
Five. (20 points) As shown in Figure 1, $EF$ is a chord of the parabola $\Gamma: y^{2}=2px$. Tangents to $\Gamma$ at points $E$ and $F$ intersect at point $C$. Points $A$ and $B$ are on the rays $EC$ and $CF$ respectively, and $\frac{EC}{CA}=\frac{CF}{FB}=\lambda$.
(1) Prove that the line $AB$ is tangent to $\Gamma$;
(... | (1) Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then the equations of the tangents $E C$ and $C F$ are
$$
y_{1} y=p\left(x+x_{1}\right), y_{2} y=p\left(x+x_{2}\right).
$$
By solving the system of equations, we get
$$
C\left(\frac{y_{1} y_{2}}{2 p}, \frac{y_{1}+y_{2}}{2}\right).
$$
From $\frac{E C}{C... | 2 | Geometry | proof | Yes | Yes | cn_contest | false | 719,935 |
One, (50 points) As shown in Figure 2, in the obtuse triangle $\triangle ABC$,
$\angle B - \angle C = 90^{\circ}$,
$\angle C = \theta$, and the circumcircle
$\odot O$ has a radius of $R$.
$AD$ is a diameter of $\odot O$, and a tangent line is drawn through point $D$ intersecting the extension of $BC$ at $H$. A line par... | Let the line $O H$ intersect the extensions of $A C$ and $B A$ at $E^{\prime}$ and $F$ respectively. First, we prove: $O$ is the midpoint of segment $E^{\prime} F$.
As shown in Figure 5, let $\angle B A C=\alpha$, $\angle C B A=\beta$. Connect $O B$ and $O C$. Let
$$
\begin{array}{c}
\angle O A B=x, \\
\angle O A C=y,... | \frac{1}{2} R^{2} \sin ^{2} \theta \cdot \tan 2 \theta\left(1+\sin ^{2} 2 \theta\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,936 |
2. Given the sequence $\left\{a_{n}\right\}$, the first 4 terms are $a_{1}=2, a_{2}=3, a_{3}=5, a_{4}=8$. Please choose one of the following ways to answer:
(1) Write a recursive formula for the sequence $\qquad$ ;
(2) Write a general term formula for the sequence $\qquad$ . | 2. (1) $a_{n+2}=a_{n+1}+a_{n}, a_{n+1}-a_{n}=n$ etc.
$$
\begin{array}{l}
\text { (2) } a_{n}=\frac{n^{2}-n+4}{2} \\
a_{n}=\frac{n^{2}-n+4}{2}+(n-1)(n-2)(n-3)(n-4) b_{n} \\
a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] .
\end{array}
$$ | a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,937 |
```
Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 5)$. Find
\[
\begin{array}{l}
f= \frac{x_{1}+x_{3}}{x_{5}+2 x_{2}+3 x_{4}}+\frac{x_{2}+x_{4}}{x_{1}+2 x_{3}+3 x_{5}}+ \\
\frac{x_{3}+x_{5}}{x_{2}+2 x_{4}+3 x_{1}}+\frac{x_{4}+x_{1}}{x_{3}+2 x_{5}+3 x_{2}}+ \\
\frac{x_{5}+x_{2}}{x_{4}+2 x_{1}+3 x_{3}}
\end{array}
\]
the... | Let $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=1$, then $f(1,1,1,1,1)=\frac{5}{3}$.
The following proves: $f \geqslant \frac{5}{3}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
f=\sum_{i=1}^{5} \frac{x_{i}+x_{i+2}}{x_{i+4}+2 x_{i+6}+3 x_{i+8}} \\
\geqslant \frac{\left[\sum_{i=1}^{5}\left(x_{i}+x_{i+2}\right)\righ... | \frac{5}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,938 |
Three. (50 points) The most recent mathematics competition consisted of 6 problems, with each correct answer scoring 7 points and each incorrect (or unanswered) question scoring 0 points. After the competition, a certain team scored a total of 161 points, and it was found during the score tallying that: any two partici... | We find that by moving a point in the first column to another column in the same row, we can reduce the number of columns in Figure 6. For example, by making the move $(6,1) \rightarrow(6,2)$, we can simultaneously make the moves $(4,10) \rightarrow(6,3)$, $(3,9) \rightarrow(6,4)$, and $(5,9) \rightarrow(6,7)$, thus ob... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,939 |
1. If the set $M=\left\{x \left\lvert\, x^{2}-\frac{5}{4} x+a1\right., x \in \mathbf{R}\right\}
$
is an empty set, then the range of values for $a$ is ( ).
(A) $a \leqslant-\frac{3}{2}$
(B) $a \geqslant-\frac{3}{2}$
(C) $a \leqslant \frac{1}{4}$
(D) $a \geqslant \frac{1}{4}$ | $$
\begin{array}{l}
\text {-.1.D. } \\
N=\left\{x \left\lvert\, \frac{1}{2-x}>1\right., x \in \mathbf{R}\right\} \\
=\{x \mid 1<x<2, x \in \mathbf{R}\} .
\end{array}
$$
From $M \cap N=\varnothing$, we get that $x^{2}-\frac{5}{4} x+a<0$ has no solution in $1<x<2$, which means $x^{2}-\frac{5}{4} x+a \geqslant 0$ always ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,940 |
2. If $k$ is a given real number, such that the following system of equations about $a, b$
$$
\left\{\begin{array}{l}
a+3 b-1=0, \\
a^{2}+b^{2}-4 a-6 b+13-k=0
\end{array}\right.
$$
has real solutions, then the range of values for $k$ is ( ).
(A) $k \geqslant 10$
(B) $k \geqslant 12$
(C) $k \geqslant 15$
(D) $k \in \ma... | 2. A.
Notice that $a^{2}+b^{2}-4 a-6 b+13-k=0$, which is $(a-2)^{2}+(b-3)^{2}=k(k \geqslant 0)$.
Thus, the system of equations has real solutions
$\Leftrightarrow k \geqslant 0$ and $\sqrt{k}$ is greater than or equal to the distance from the center $(2,3)$ to the line $a+3 b-1=0$.
$$
\begin{array}{l}
\Leftrightarrow ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,941 |
3. As shown in Figure 1, in the pyramid $P-ABCD$, $AB$ $/ / CD, \angle ADC=90^{\circ}$, $PD \perp$ plane $ABCD$. If $PD=AD=AB=1$, $CD=2$, then the sine value $\sin \theta$ of the plane angle $\theta$ formed by the dihedral angle between plane $PAD$ and plane $PBC$ is ( ).
(A) $\frac{\sqrt{7}}{3}$
(B) $\frac{\sqrt{30}}{... | 3. B.
Given $A B / / C D$ and $\angle A D C=90^{\circ}$, we get $\angle D A B=$ $90^{\circ}$. Connect $B D$. By the Pythagorean theorem, we have $B D=\sqrt{2}, P B=$ $\sqrt{3}, P C=\sqrt{5}, B C=\sqrt{2}$ (obviously, $\angle P D B=90^{\circ}$ ).
Notice that $B C^{2}+P B^{2}=P C^{2}$, so $\angle P B C$ $=90^{\circ}$. ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,942 |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}$ denote the fractional part of $x$. Then the number of all real numbers $x$ that satisfy $2007 x + 2008\{x\} = 2009$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 4.C.
Since $x=[x]+\{x\}$, by the given condition we have $2007[x]+4015\{x\}=2009$.
Thus, $\{x\}=\frac{2009-2007[x]}{4015}$.
Since $0 \leqslant\{x\}<1$, we have
$$
\begin{array}{l}
0 \leqslant \frac{2009-2007[x]}{4015}<1 \\
\Leftrightarrow\left\{\begin{array}{l}
2007[x] \leqslant 2009, \\
2009-2007[x]<4015
\end{array}\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,943 |
$$
\begin{array}{l}
\text { 5. Let } a_{1}=b_{1}=1, a_{n+1}=2 b_{n}+1, \\
b_{n+1}=2 a_{n}-1(n=1,2, \cdots) .
\end{array}
$$
If $d$ is a real constant such that for every positive
integer $k(k \geqslant 2)$, we have
$$
a_{2 k}+a_{2 k-1} \geqslant d\left(a_{2 k+2}+a_{2 k+1}\right),
$$
then the maximum value of $d$ is (... | 5.B.
Obviously, $a_{1}=1, b_{1}=1, a_{2}=2 b_{1}+1=3$, $a_{n+1}=2 b_{n}+1=2\left(2 a_{n-1}-1\right)+1=4 a_{n-1}-1(n \geqslant 2)$.
Let $d$ be an undetermined constant, then
$$
a_{n+1}+d=4 a_{n-1}+d-1=4\left(a_{n-1}+\frac{d-1}{4}\right) \text {. }
$$
Let $d=\frac{d-1}{4}$, we get $d=-\frac{1}{3}$.
Thus, $a_{n+1}-\frac... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,944 |
6. Among the 8 vertices, 12 midpoints of edges, 6 centers of faces, and 1 center of the cube, these 27 points, the number of groups of 8 points that lie on a common sphere is ( ).
(A) 4462
(B) 4584
(C) 4590
(D) 4602 | 6.B.
Place the cube in a spatial rectangular coordinate system, with the center of the cube as the origin, and the 8 vertices are $(2,2,2)$, $(-2,-2,-2)$, $(2,2,-2)$, $(2,-2,-2)$, where $(a, b, c)$ represents $(a, b, c)$ and its permutations.
The planes $z=0$, $z= \pm 2$ divide the 27 points into three layers. For ea... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,945 |
1. If $\theta$ is a real number, and $\frac{1+\sin \theta}{1+\cos \theta}<\frac{\cos \theta}{1-\sin \theta}$, then the range of $y=\sin \frac{\theta}{2}-\cos \frac{\theta}{2}$ is $\qquad$ (express the range as an inequality involving $y$). | $\begin{array}{l}\text { II. } 1 .(-\sqrt{2}, 0) \cup(0, \sqrt{2}) . \\ \frac{1+\sin \theta}{1+\cos \theta}>0\end{array}\right.\right. \\ \Leftrightarrow 2 k \pi-\frac{\pi}{2}<\theta<2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \\ \Leftrightarrow k \pi-\frac{\pi}{4}<\frac{\theta}{2}<k \pi+\frac{\pi}{4} . \\ \text { and } y=... | (-\sqrt{2}, 0) \cup (0, \sqrt{2}) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,946 |
2. Let $k$ be a complex number, and the quadratic equation in $x$, $x^{2}+k x-\frac{1}{2}=0$, has two complex roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{3}=k$, then $k=$ $\qquad$ . | 2.0 or $\frac{3}{2} \mathrm{i}$ or $-\frac{3}{2} \mathrm{i}$.
Since $x_{2}^{2}+k x_{2}-\frac{1}{2}=0$, we have
$$
x_{2}^{3}+k x_{2}^{2}-\frac{1}{2} x_{2}=0 \text {. }
$$
Thus, $x_{2}^{3}=-k x_{2}^{2}+\frac{1}{2} x_{2}$
$=-k\left(-k x_{2}+\frac{1}{2}\right)+\frac{1}{2} x_{2}$
$$
=\left(k^{2}+\frac{1}{2}\right) x_{2}-\... | 0 \text{ or } \pm \frac{3}{2} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,947 |
3. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-2 t-2\right]$, then the value of $t$ is $\qquad$ | 3. -1 .
The domain of an odd function is symmetric about the origin, and the right endpoint of the interval is not less than the left endpoint. Therefore,
$$
\left\{\begin{array} { l }
{ - t = t ^ { 2 } - 2 t - 2 > 0 , } \\
{ t \leqslant t ^ { 2 } - 2 t - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t^{2}-... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,948 |
3. Let the real constant $k$ be such that the equation $2 x^{2}+2 y^{2}-5 x y$ $+x+y+k=0$ represents two intersecting lines in the Cartesian coordinate system $x O y$, with the intersection point being $P$. If points $A$ and $B$ lie on these two lines respectively, and $|\boldsymbol{P A}|=|\boldsymbol{P B}|=1$, then $P... | 3. $\pm \frac{4}{5}$.
From the problem, we know that the quadratic polynomial in $x$ and $y$
$$
2 x^{2}+2 y^{2}-5 x y+x+y+k=0
$$
can be factored into the product of two linear factors.
Since $2 x^{2}+2 y^{2}-5 x y=(-2 x+y)(-x+2 y)$,
we have
$$
\begin{array}{l}
2 x^{2}+2 y^{2}-5 x y+x+y+k \\
=(-2 x+y+a)(-x+2 y+b),
\e... | \pm \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,949 |
5. The maximum value of the function $f(x)=(6-x)^{3}(x-1)^{\frac{2}{3}}(1 \leqslant x \leqslant 6)$ is
保留了源文本的换行和格式。 | $$
\begin{array}{l}
\text { 5. } \frac{3^{6} \times 5^{3}}{11^{4}} \sqrt[3]{1100} \\
(f(x))^{3}=(6-x)^{9}(x-1)^{2} \\
=\frac{1}{2^{9}}(12-2 x)^{9} \cdot \frac{1}{9^{2}}(9 x-9)^{2} \\
\leqslant \frac{1}{2^{9} \times 9^{2}}\left[\frac{9(12-2 x)+2(9 x-9)}{11}\right]^{11} \\
=\frac{9^{11} \times 2^{11}}{2^{9} \times 9^{2}}... | \frac{3^{6} \times 5^{3}}{11^{4}} \sqrt[3]{1100} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,951 |
6. The number of all distinct real solutions of the equation $2 x^{3}+6 x^{2}+12 x+8+x^{3} \sqrt{x^{2}+1}$ $+(x+2)^{3} \sqrt{(x+2)^{2}+1}=0$ is $\qquad$. | 6.1.
Rewrite the equation as
$$
\begin{array}{l}
x^{3}+x^{3} \sqrt{x^{2}+1}+(x+2)^{3}+ \\
(x+2)^{3} \sqrt{(x+2)^{2}+1}=0 . \\
\text { Let } f(x)=x^{3}+x^{3} \sqrt{x^{2}+1} \\
=x^{3}\left(1+\sqrt{x^{2}+1}\right) .
\end{array}
$$
Then equation (1) can be rewritten as
$$
f(x+2)=-f(x) \text {. }
$$
Clearly, $f(-x)=-f(x)... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,952 |
Four. (20 points) Let $k$ be a given non-zero real number, in the Cartesian coordinate system $x O y$, the equation of curve $C_{1}$ is $|x| \neq|y|$ and $\frac{x^{2}}{k-x}=\frac{y^{2}}{k+y}$, and point $A(k, k)$.
(1) Let $P$ be any point on $C_{1}$, try to find the equation of the locus $C_{2}$ of the midpoint $Q$ of ... | $$
\begin{array}{l}
\text { (1) } \frac{x^{2}}{k-x}=\frac{y^{2}}{k+y} \\
\Leftrightarrow x^{2}(k+y)=y^{2}(k-x)(x \neq k \text { and } y \neq-k) \\
\Leftrightarrow(x+y)(y-x) k=(x+y) x y \\
\Leftrightarrow(y-x) k=x y(\text { by }|x| \neq|y| \text {, we get } x+y \neq 0) \\
\Leftrightarrow y(k-x)=k x \\
\Leftrightarrow y=... | \theta=\arctan \frac{108}{145} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,954 |
One, (50 points) As shown in Figure $3, \odot O$ is the circumcircle of the square $ABCD$, point $P$ is on the minor arc $\overparen{AB}$ ($P$ does not coincide with $A$ or $B$), $DP$ intersects $AO$ and $AB$ at points $Q$ and $T$ respectively, the tangent to $\odot O$ at point $P$ intersects the extension of $DA$ at p... | (1) As shown in Figure 5, let the side length of the square $ABCD$ be 1, and denote $\angle ADP = \theta (0^{\circ} < \theta < 45^{\circ})$. Since $A G > A E$, it follows that $G$ and $E$ do not coincide, meaning $F, T, E$ cannot be collinear.
$$
A G > A E
$$
$$
\Rightarrow \frac{\sin \theta}{(1+\sqrt{2}) \cos \theta -... | \left(\frac{\sqrt{2}}{2}, 2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,955 |
(50 points)(1) Prove that there exist infinitely many pairs of positive integers $x, y$ such that $x^{2}-101 y^{2}=-1$.
(2) Given $0.30102<\lg 2<0.30103$,
$$
0.32220<\lg 2.1<0.32221 \text{. }
$$
Determine the first 6 digits before the decimal point and the first 131 digits after the decimal point of the number $(10+\s... | (1) Let $A_{2 n-1}=(\sqrt{101}-10)^{2 n-1}$, $B_{2 n-1}=(\sqrt{101}+10)^{2 n-1}(n=1,2, \cdots)$.
Then $A_{2 n-1} B_{2 n-1}=1$.
Also, let $B_{2 n-1}=x_{2 n-1}+y_{2 n-1} \sqrt{101}$, where
$x_{2 n-1}, y_{2 n-1} \in \mathbf{N}_{+}$. By the binomial theorem,
$A_{2 n-1}=-x_{2 n-1}+y_{2 n-1} \sqrt{101}$.
Substituting $A_{2 n... | 402020 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,956 |
Three. (50 points) Question: Can 2008 be expressed in the form $a_{1}^{x_{1}}+a_{2}^{x_{2}}+\cdots+a_{m}^{x_{m}}-b_{1}^{y_{1}}-b_{2}^{y_{2}}-\cdots-b_{n}^{y_{n}}$? If so, are there infinitely many such representations? Here, $m, n$ are both positive integers greater than 100 and less than 170, and $m-n \geqslant 50 ; a... | Three, the representation that satisfies the requirements of the problem exists and there are infinitely many. The reasons are as follows.
For convenience, let's take \( x_{1}=x_{2}=\cdots=x_{m}=y_{1}=y_{2}=\cdots=y_{n}=2 \), and consider the sum
\[
\begin{array}{l}
S_{n}=1^{2}+2^{2}+\cdots+n^{2}-(n+1)^{2}-(n+2)^{2}-\... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,957 |
Five. (20 points) Let $N=\{0,1,2, \cdots\}$, and given a $k \in \mathbf{N}$. Try to find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for every $n \in \mathbf{N}$, we have
$$
f(n)<f(n+1) \text {, and } f(f(n))=n+2 k \text {. }
$$ | Five, first prove the following lemma.
Lemma If the function $f: \mathbf{N} \rightarrow \mathbf{N}$ satisfies $f(n)<f(n+1)$ for every $n \in \mathbf{N}$, then for any $m, n \in \mathbf{N}, m \geqslant n$, we must have
$$
f(m)-f(n) \geqslant m-n \text {, and } f(m) \geqslant m.
$$
Proof of the lemma: Since $f(n)<f(n+1)... | f(n)=n+k | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,958 |
4. Given a circle centered at the origin with radius $R(R>0)$ intersects the line segment connecting $M(1,1)$ and $N\left(\frac{7}{4}, 0\right)$. Then the range of values for $R$ is $\qquad$. | 4. $\left[\frac{7}{5}, \frac{7}{4}\right]$.
Let the equation of $M N$ be $4 x+3 y-7=0$. Then the line passing through the origin and perpendicular to $M N$ is $3 x-4 y=0$. We can find the foot of the perpendicular from the origin to $M N$ to be $H\left(\frac{28}{25}, \frac{21}{25}\right)$, and
$$
O H=\frac{|4 \times 0... | \left[\frac{7}{5}, \frac{7}{4}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,959 |
1. Equation
$$
\begin{array}{l}
\sqrt{x_{1}-1}+2 \sqrt{x_{2}-4}+3 \sqrt{x_{3}-9} \\
=\frac{1}{2}\left(x_{1}+x_{2}+x_{3}\right)
\end{array}
$$
The real solution $\left(x_{1}, x_{2}, x_{3}\right)=$ | -1. $(2,8,18)$.
Multiplying both sides of the equation by 2 and rearranging, we get
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}-2 \sqrt{x_{1}-1}-4 \sqrt{x_{2}-4}- \\
6 \sqrt{x_{3}-9}=0 .
\end{array}
$$
Completing the square, we have
$$
\begin{array}{l}
\left(\sqrt{x_{1}-1}-1\right)^{2}+\left(\sqrt{x_{2}-4}-2\right)^{2}+ \\
... | (2,8,18) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,960 |
3. The complex sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=0, a_{n}=a_{n-1}^{2}+\mathrm{i}(n \geqslant 2) .
$$
Then the sum of its first 2007 terms is $\qquad$ | 3. $-1003+2$ i.
From the given, we have $a_{1}=0, a_{2}=\mathrm{i}, a_{3}=-1+\mathrm{i}$,
$$
a_{4}=-\mathrm{i}, a_{5}=-1+\mathrm{i}, a_{6}=-\mathrm{i}, \cdots \cdots
$$
It can be seen that, when $n \geqslant 3$,
$$
a_{n}=\left\{\begin{array}{ll}
-1+\mathrm{i}, & n \text { is odd; } \\
-\mathrm{i}, & n \text { is even... | -1003+2\mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,962 |
4. Given that $\alpha-l-\beta$ is a dihedral angle of size $45^{\circ}$, $C$ is a fixed point inside the dihedral angle, and the distances from $C$ to the half-planes $\alpha$ and $\beta$ are $\sqrt{2}$ and 6, respectively. $A$ and $B$ are moving points in the half-planes $\alpha$ and $\beta$, respectively. Then the mi... | 4. $10 \sqrt{2}$.
As shown in Figure 4, construct the symmetric points $P$ and $Q$ of point $C$ with respect to planes $\alpha$ and $\beta$, respectively. It is easy to prove that when $A$ and $B$ are the intersection points of line $P Q$ with planes $\alpha$ and $\beta$, respectively, the perimeter of $\triangle A B ... | 10 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,963 |
5. Given the correspondence of points in a Cartesian coordinate system
rule $f: P(m, n) \rightarrow P^{\prime}(\sqrt{m}, \sqrt{n})(m \geqslant 0, n \geqslant$ 0 ). If a segment of a curve corresponds to an arc of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(x \geqslant 0, y \geqslant 0)$ under the corresponde... | 5. $y=b^{2}\left(1-\frac{x}{a^{2}}\right)\left(0 \leqslant x \leqslant a^{2}\right)$.
Let the curve equation be $y=f(x)(s \leqslant x \leqslant t)$, then the point $P(x, f(x))$ on the curve corresponds to the point $P^{\prime}(\sqrt{x}, \sqrt{f(x)})$ on an arc of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(... | y=b^{2}\left(1-\frac{x}{a^{2}}\right)\left(0 \leqslant x \leqslant a^{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,964 |
6. Given $f(n)=\cos \frac{n \pi}{4}$. Calculate:
$$
f(1) f(3) \cdots f(2 n-1)=
$$ | 6. $\left\{\begin{array}{ll}\left(-\frac{1}{2}\right)^{\frac{n}{2}}, & n \text { is even; } \\ \left(\frac{1}{2}\right)^{\frac{n}{2}}, & n \text { is odd. }\end{array}\right.$
When $k \in \mathbf{Z}_{+}$,
$$
\begin{array}{l}
f(2 k-1) f(2 k+1) \\
=\cos \frac{(2 k-1) \pi}{4} \cdot \cos \frac{(2 k+1) \pi}{4} \\
=\frac{1}... | \left\{\begin{array}{ll}\left(-\frac{1}{2}\right)^{\frac{n}{2}}, & n \text { is even; } \\ \left(\frac{1}{2}\right)^{\frac{n}{2}}, & n \text { is odd. }\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,965 |
7. Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=0, x_{2}=1, x_{n}=\frac{x_{n-1}+x_{n-2}}{2}(n \geqslant 3)
$$
then the general term formula of the sequence $\left\{x_{n}\right\}$ is $x_{n}=$ $\qquad$ | 7. $\frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{n-1}$.
From $x_{n}=\frac{x_{n-1}+x_{n-2}}{2}$ we get
$x_{n}+\frac{1}{2} x_{n-1}=x_{n-1}+\frac{1}{2} x_{n-2}$.
Also, $x_{2}+\frac{1}{2} x_{1}=1$, so the sequence $\left\{x_{n}+\frac{1}{2} x_{n-1}\right\}$
is a constant sequence, with each term being 1, i.e.,
$$
\beg... | \frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,966 |
8. Given $\odot M:(x-1)^{2}+(y-3)^{2}=4$, there exists a secant line $P B A$ of $\odot M$ through the point $P(a, 0)$ on the $x$-axis, such that $P B=P A$. Then the range of the $x$-coordinate $a$ of point $P$ is $\qquad$ | 8. $1-3 \sqrt{3} \leqslant a \leqslant 1+3 \sqrt{3}$.
The center of the circle $M(1,3)$, diameter $d=4$.
As shown in Figure 5, draw a secant through point $P$ and
$M$. By the secant theorem, we have
$$
\begin{array}{l}
\left(P M+\frac{d}{2}\right) . \\
\left(P M-\frac{d}{2}\right) \\
= P B \cdot P A . \\
\text { Hence... | 1-3 \sqrt{3} \leqslant a \leqslant 1+3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,967 |
9. (14 points) For any positive integer $n$, let $S(n)$ denote the number of positive integer pairs $(x, y)$ that satisfy the indeterminate equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$. (For example, the positive integer pairs that satisfy $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ are $(6,3),(4,4),(3,6)$, so $S(2)=3$). F... | From $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}\left(x, y, n \in \mathbf{Z}_{+}\right)$, we know $x>n, y>n$.
Let $x=n+a, y=n+b\left(a, b \in \mathbf{Z}_{+}\right)$. Then $\frac{1}{n+a}+\frac{1}{n+b}=\frac{1}{n} \Leftrightarrow n^{2}=a b$.
Therefore, $S(n)$ equals the number of positive integer pairs $(a, b)$. Thus, $S(n)$ eq... | n=p_{1}^{1003} \text{ or } n=p_{1} p_{2}^{334} \text{ or } n=p_{1}^{4} p_{2}^{111} \text{ or } n=p_{1} p_{2} p_{3}^{111} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,968 |
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