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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2. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the angles opposite to sides $a, b, c$ are $\angle A, \angle B, \angle C$ respectively, and $\sin A \cdot \sin B+\sin B \cdot \sin C+\cos 2 B=1$. Then $\angle B$ is $(\quad)$.
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{3}$
(C) $\frac{\... | 2. B.
Given that $a, b, c$ form a geometric sequence, we know $b^{2}=a c$.
From the given equation,
$$
\begin{array}{l}
\sin A \cdot \sin B+\sin B \cdot \sin C+1-2 \sin ^{2} B=1 . \\
\text { Since } \sin B \neq 0, \text { we have } \\
\sin A+\sin C=2 \sin B \Rightarrow a+c=2 b . \\
\text { By } \cos B=\frac{a^{2}+c^{2... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,367 |
4. Given the sequence $\left\{a_{n}\right\}$ composed of positive integers, and it is an increasing sequence, with
$$
a_{n+2}=a_{n+1}+2 a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
If $a_{5}=52$, then $a_{7}=(\quad)$.
(A) 102
(B) 152
(C) 212
(D) 272 | 4. C.
From the problem, we know that $a_{5}=5 a_{2}+6 a_{1}=52$.
Given that $a_{1}$ and $a_{2}$ are positive integers,
$$
6\left|\left(52-5 a_{2}\right) \Rightarrow 6\right|\left(a_{2}-2\right) \text {. }
$$
Let $a_{2}=6 k+2(k \in \mathbf{N})$.
Notice that, $52>5 a_{2}=30 k+10$.
Thus, $k=0$ or 1.
If $k=0$, then $a_{2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,368 |
5. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $A A_{1}=1, A D=2$. Then the distance between the skew lines $A_{1} D$ and $B_{1} D_{1}$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{2}{3}$
(D) $\frac{3}{2}$ | 5. C.
Let $A_{1} B$ and $A B_{1}$ intersect at point $O$, and draw $B_{1} M \perp D O$ at point $M$.
Since $A_{1} B \perp A B_{1}, A_{1} B \perp A D$, therefore, $A_{1} B \perp$ plane $A D C_{1} B_{1} \Rightarrow A_{1} B \perp B_{1} M$. Also, $B_{1} M \perp D O$, thus, $B_{1} M \perp$ plane $A D B$. Given $B_{1} D / /... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,369 |
6. As shown in Figure $1, M$ and $N$ are internal points on the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, respectively, and $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$. If points $B$, $M$, and $N$ are collinear, then $\lambda=$ ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{1}{3}$
(C) $\frac{\sqrt{2}}{2}$... | 6. A.
Set $B$ as the origin and the line $B C$ as the $x$-axis to establish a rectangular coordinate system.
Assume $B C=1$. Then
$$
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), C(1,0), E(1, \sqrt{3}) \text {. }
$$
Given $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$, using the section formula, we get
$$
M\left(\frac{3... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,370 |
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ . | 8.1.
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,371 |
12. Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $C$, and $AB$ be a chord of the ellipse passing through point $F_{2}$. In $\triangle F_{1} A B$,
$$
\left|F_{1} A\right|=3,|A B|=4,\left|B F_{1}\right|=5 .
$$
Then $\tan \angle F_{2} F_{1} B=$ $\qquad$ | 12. $\frac{1}{7}$.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Then, $F_{1}(-c, 0), F_{2}(c, 0), e=\frac{c}{a}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
\left|F_{1} A\right|=a+e x_{1}=3, \\
\left|F_{1} B\right|=a+e x_{2}=5, \\
|A B|=\le... | \frac{1}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,372 |
13. (20 points) Let real numbers $a$, $b$, $\lambda$ satisfy $0<a<b$, $0 \leqslant \lambda \leqslant 1$. Prove:
(1) $\ln a+1 \leqslant \frac{a \ln a-b \ln b}{a-b} \leqslant \ln b+1$;
(2) $[\lambda a+(1-\lambda) b] \ln [\lambda a+(1-\lambda) b]$
$\leqslant \lambda a \ln a+(1-\lambda) b \ln b$. | $$
\begin{array}{l}
f(x)=(x-b)(\ln x+1)-x \ln x-b \ln b . \\
\text { By } f^{\prime}(x)=\ln x+1+1-\frac{b}{x}-\ln x-1 \\
=1-\frac{b}{x}0 \text {. }
\end{array}
$$
Therefore, for any $00$, $g^{\prime}(1)<0$, then by the graph of the function we know
$$
g(x) \geqslant 0(0 \leqslant x \leqslant 1) .
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,373 |
14. (20 points) As shown in Figure 2, the circumcenter of the right triangle \( \triangle ABC \) is \( O \), with \( \angle C = 90^\circ \). A perpendicular line is drawn from point \( C \) to \( AB \), with the foot of the perpendicular being \( D \). A circle \( \odot O_1 \) is constructed to be tangent to arc \( \ov... | 14. Connect $B E, E F, F A, O O_{1}, O_{1} E, O_{1} F$. Then $O, O_{1}, E$ are collinear.
By $\angle E F O_{1}=\angle F E O_{1}=\frac{1}{2} \angle F O_{1} O$
$$
=\frac{1}{2} \angle O_{1} O B=\angle E A B,
$$
we know that $A, F, E$ are collinear.
Then $\triangle A F D \backsim \triangle A B E \Rightarrow A D \cdot A B=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,374 |
5. The range of real numbers $x$ that satisfy $\sqrt{1-x^{2}} \geqslant x$ is . $\qquad$ | 5. $\left[-1, \frac{\sqrt{2}}{2}\right]$.
Let $y=\sqrt{1-x^{2}}$, this is the upper half of the unit circle, which intersects the line $y=x$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. The portion of the semicircle to the left of the intersection point lies above the line $y=x$.
Therefore, $x \... | \left[-1, \frac{\sqrt{2}}{2}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,377 |
6. If real numbers $x, y, z \geqslant 0$, and
$$
x+y+z=30,3 x+y-z=50 \text {, }
$$
then the range of $T=5 x+4 y+2 z$ is | 6. $[120,130]$.
Notice,
$$
\begin{array}{l}
T=5 x+4 y+2 z \\
=(x+y+z)+(4 x+3 y+z) \\
=30+(4 x+3 y+z) .
\end{array}
$$
From $4 x+2 y=(x+y+z)+(3 x+y-z)=80$
$$
\Rightarrow T=110+(y+z) \text {. }
$$
Also, $20=(3 x+y-z)-(x+y+z)=2(x-z)$,
then $x-z=10$.
Since $x, z \geqslant 0$, we have $x \geqslant 10$.
Thus, from $x+y+z=... | [120,130] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,378 |
7. As shown in Figure 1, the edge length of the regular tetrahedron $ABCD$ is 2, and $A_{1}, B_{1}, C_{1}$ are the midpoints of edges $DA, DB, DC$ respectively. With $D$ as the center and a radius of 1, arcs $\overparen{A_{1} B_{1}}$ and $\overparen{B_{1} C_{1}}$ are drawn in the planes $DAB$ and $DBC$ respectively, an... | 7. $2 \sin 42^{\circ}$.
Make two expansions, then compare.
Note that, arc $\overparen{A}_{1} B_{1}$ is divided into five equal arcs by points $P_{1}, P_{2}, P_{3}, P_{4}$, each corresponding to a central angle of $12^{\circ}$; arc $\overparen{B_{1} C_{1}}$ is divided into five equal arcs by $Q_{1}, Q_{2}, Q_{3}, Q_{4}... | 2 \sin 42^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,379 |
10. (22 points) As shown in Figure 2, $H$ is the orthocenter of acute $\triangle ABC$. Take any point $E$ on segment $CH$, extend $CH$ to point $F$ such that $HF = CE$, draw $FD \perp BC$ at point $D$, draw $EG \perp BH$ at point $G$, let $M$ be the midpoint of segment $CF$, $O_{1}$ and $O_{2}$ are the circumcenters of... | 10. (1) As shown in Figure 3, let $EG$ intersect $DF$ at point $K$, and connect $AH$.
Since $AC \perp BH$, $EK \perp BH$, $AH \perp BC$, and $KF \perp BC$, we have $CA \parallel EK$ and $AH \parallel KF$.
Also, since $CH = EF$, it follows that $\triangle CAH \cong \triangle EKF$.
Thus, $AH \perp KF$. Therefore, $AK \p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,380 |
1. A certain electronic device contains three components, with probabilities of failure being $0.1$, $0.2$, and $0.3$, respectively. If one, two, or three components fail, the probabilities of the device malfunctioning are $0.25$, $0.6$, and $0.9$, respectively. Find the probability that the device malfunctions. | Let the generating function be
$$
\varphi(x)=(0.1 x+0.9)(0.2 x+0.8)(0.3 x+0.7) \text {, }
$$
where the coefficient of the
$x$ term corresponds to the “probability of one component failing”,
$x^{2}$ term corresponds to the “probability of two components failing”,
$x^{3}$ term corresponds to the “probability of three co... | 0.1601 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,381 |
1. Given that the circumradius of $\triangle A B C$ is $R$, and
$$
2 R\left(\sin ^{2} A-\sin ^{2} C\right)=(\sqrt{2} a-b) \sin B,
$$
where $a$ and $b$ are the sides opposite to $\angle A$ and $\angle B$ respectively. Then the size of $\angle C$ is | $-1.45^{\circ}$.
From the given equation and using the Law of Sines, we get
$$
\begin{array}{l}
a^{2}-c^{2}=(\sqrt{2} a-b) b \\
\Rightarrow a^{2}+b^{2}-c^{2}=\sqrt{2} a b \\
\Rightarrow \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{\sqrt{2}}{2} \\
\Rightarrow \angle C=45^{\circ} .
\end{array}
$$ | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,382 |
2. Let the set $A=\{x \mid 2 a+1 \leqslant x \leqslant 3 a+5\}$, $B=\{x \mid 3 \leqslant x \leqslant 33\}, A \subseteq A \cap B$.
Then the range of values for $a$ is | 2. $(-\infty,-4) \cup\left[1, \frac{28}{3}\right]$.
From $A \subseteq A \cap B$, we know $A \subseteq B$.
When $A=\varnothing$,
$2 a+1>3 a+5 \Rightarrow a<-4$.
When $A \neq \varnothing$, $a \geqslant-4$.
Thus $\left\{\begin{array}{l}2 a+1 \geqslant 3, \\ 3 a+5 \leqslant 33\end{array} \Rightarrow 1 \leqslant a \leqslan... | (-\infty,-4) \cup\left[1, \frac{28}{3}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,383 |
3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ . | 3.5.
Notice that,
$$
\begin{array}{l}
6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{9}+\cdots+\mathrm{C}_{11}^{10} 6-1 \\
=\mathrm{C}_{10}^{0} 6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\cdots+\mathrm{C}_{11}^{10} 6+\mathrm{C}_{11}^{11} 6^{0}-2 \\
=(6+1)^{11}-2=7^{11}-2 \\
\equiv(-1)^{11}-2 \equiv 5(\bmod 8) .
\end{... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,384 |
5. As shown in Figure 1, in the quadrilateral pyramid $P-A B C D$, $A D$
$/ / B C, \angle A B C=$
$90^{\circ}, P A \perp$ plane
$A B C D, P A=3, A D=2, A B=2 \sqrt{3}, B C=6$. Then the size of the dihedral angle $P-B D-A$ is | 5. $60^{\circ}$.
Draw $A F \perp B D$ at point $F$.
Since $P A \perp$ plane $A B C D$, we know $P F \perp B D$.
Thus, $\angle P F A$ is the size of the dihedral angle $P-B D-A$.
Because $\angle A B C=90^{\circ}, A D / / B C$, so, in the right triangle $\triangle A B D$, given $A B=2 \sqrt{3}, A D=2$, we know $B D=4, A... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,385 |
7. Given $\boldsymbol{a}$ and $\boldsymbol{b}$ are two mutually perpendicular unit vectors, and $c \cdot a=c \cdot b=1$. Then for any positive real number $t$, $\left|c+t a+\frac{1}{t} b\right|$ has the minimum value of $\qquad$.
Translate the above text into English, please retain the original text's line breaks and ... | $7.2 \sqrt{2}$.
Let $c=\lambda a+\mu b$. Then
$$
\begin{array}{l}
\boldsymbol{c} \cdot \boldsymbol{a}=\lambda|\boldsymbol{a}|^{2}+\mu \boldsymbol{a} \cdot \boldsymbol{b}=\lambda=1, \\
\boldsymbol{c} \cdot \boldsymbol{b}=\lambda \boldsymbol{a} \cdot \boldsymbol{b}+\mu|\boldsymbol{b}|^{2}=\mu=1 .
\end{array}
$$
Thus, $c... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,386 |
10. Let $f(x)$ be a function defined on the set of integers, satisfying the conditions
(1) $f(1)=1, f(2)=0$;
(2) For any $x, y$,
$$
f(x+y)=f(x) f(1-y)+f(1-x) f(y) \text {. }
$$
Then $f(2015)=$ $\qquad$ | 10. $\pm 1$.
In condition (2), let $x=0$, then
$$
f(y)=f(0) f(1-y)+f(1) f(y) \text {. }
$$
By $f(1)=1$, we know
$$
f(0) f(1-y)=0 \text {. }
$$
In the above equation, let $y=0$, then
$$
f(0) f(1)=0 \Rightarrow f(0)=0 \text {. }
$$
In condition (2), let $x=1,-1,2$ respectively, we get
$$
\begin{array}{l}
f(y+1)=f(1) ... | \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,387 |
12. (14 points) A flower shop purchases a certain number of roses from a farm at a price of 5 yuan per stem every day, and then sells them at a price of 10 yuan per stem. If they are not sold on the same day, the remaining roses are treated as garbage.
(1) If the flower shop purchases 16 stems of roses one day, find th... | 12. (1) When the daily demand $n \geqslant 16$, the profit $y=80$. When the daily demand $n<16$, the profit $y=10 n-80$. Therefore, the function expression of $y$ with respect to the positive integer $n$ is
$$
y=\left\{\begin{array}{ll}
10 n-80, & n<16 ; \\
80, & n \geqslant 16 .
\end{array}\right.
$$
(2) (i) The possi... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,388 |
Example 2 Let $n \geqslant 2$ be a positive integer. Find the maximum value of the constant $C(n)$ such that for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in(0,1)(i=1,2$, $\cdot \cdot, n)$, and
$$
\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant i<j \leqslant n)
$$
we ... | 【Analysis】From the equality of the given inequality, we can set
$$
x_{i}=\frac{1}{2}(i=1,2, \cdots, n) \text {. }
$$
Substituting into equation (1) yields
$$
\frac{n}{2} \geqslant C(n) \mathrm{C}_{n}^{2}\left(\frac{1}{2}+\frac{1}{2}\right) \Rightarrow C(n) \leqslant \frac{1}{n-1} \text {. }
$$
Next, we prove that: $C... | \frac{1}{n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,391 |
Question 2 Let $n$ be a positive even number. On the blackboard, there are $n$ real numbers. Each operation can arbitrarily erase two numbers $a, b$, and then write down the product of these two numbers $ab, ab$. Prove: Regardless of which real numbers are initially written on the blackboard, it is possible to make all... | The problem is very challenging. After proving that $n=2, n=4$ hold true using mathematical induction, the key is the operation for $n=6$.
The proof essentially generalizes the method for $n=6$ to the general case of $n=k$ (where $k$ is even).
Proof: Use mathematical induction to prove that the conclusion holds for a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,392 |
Question 3 Let $n(n>3)$ be a given integer. Find the largest integer $d$, such that for any set $S$ of $n$ integers, there exist four distinct non-empty subsets (which may intersect), the sum of the elements of each of which is divisible by $d$. | Solve for the largest integer $d=n-2$.
Let $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
For convenience, the sum of elements in set $A$ is denoted as $S(A)$.
When $d \geqslant n$, let
$a_{i} \equiv 1(\bmod n)(i=1,2, \cdots, n)$.
Then the sum of elements in set $S$ is $S(S) \equiv n(\bmod d)$.
Let $T \subseteq S$, an... | n-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,393 |
Question 4: Call an $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ "intermittently periodic" if there exist non-negative integers $i$ and positive integers $p$ such that $i+2 p \leqslant n$, and for all $j=1,2, \cdots, p$,
$$
a_{i+j}=a_{i+p+j}.
$$
Let $k$ be a given positive integer. Find the smallest positive i... | Solve for the smallest positive integer $n=2^{k}-1$.
Let $p$ be the period length of the array.
Suppose the period length of $a_{n+1}=i(i=1,2, \cdots, k)$ is $p_{i}$. If $i_{1} \neq i_{2} \in\{1,2, \cdots, k\}$, then $p_{i_{1}} \neq p_{i_{2}}$. Otherwise, $i_{1}=a_{n+1}=a_{n+1-p_{n_{1}}}=a_{n+1-p_{n}}=a_{n+1}^{\prime}=... | 2^{k}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,394 |
Problem 5: In an $n \times n$ grid, 101 cells are colored blue. It is known that there is a unique way to cut the grid along the grid lines into some rectangles, such that each rectangle contains exactly one blue cell. Find the minimum possible value of $n$.
(2015, Bulgarian Mathematical Olympiad) | The minimum possible value of $n$ is 101.
First, prove the more general conclusion below.
Lemma Given an $n \times n$ grid $P$ with $m$ cells colored blue. A "good" partition of the grid $P$ is defined as: if the grid is divided along the grid lines into $m$ rectangles, and each rectangle contains exactly one blue cell... | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,395 |
Let $n$ be a positive integer. Prove:
$$
\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^{2}}\right) \cdots\left(1+\frac{1}{3^{n}}\right) \leqslant 2\left(1-\frac{1}{3^{n}}\right),
$$
with equality holding if and only if $n=1$. | Proof 1 Mathematical Induction.
When $n=1$,
the left side of equation (2) $=1+\frac{1}{3}=\frac{4}{3}=2\left(1-\frac{1}{3^{1}}\right)=$ the right side.
Obviously, inequality (2) holds.
Assume that when $n=k\left(k \geqslant 1, k \in \mathbf{Z}_{+}\right)$, inequality (2) holds, i.e., $\square$
$$
\begin{array}{l}
\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,398 |
Question 1 In the acute triangle $\triangle ABC$, $AB > BC$, $P$ and $Q$ are the midpoints of the minor arc $\overparen{AC}$ and the major arc $\overparen{AC}$ of its circumcircle $\odot O$, respectively. A perpendicular line is drawn from $Q$ to the segment $AB$, with the foot of the perpendicular being $M$. Prove: Th... | Prove: As shown in Figure 1, connect $A Q, B Q, C Q$, and draw $Q N \perp B C$ at point $N$.
It is easy to see that Rt $\triangle Q B M \cong$ Rt $\triangle Q B N$.
By $\angle Q A C=\angle Q B N=\angle Q B A=\angle Q C A$
$\Rightarrow A Q=C Q, Q N=Q M$
$\Rightarrow$ Rt $\triangle A Q M \cong$ Rt $\triangle C Q N$
$\Rig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,400 |
Question 2 In a cyclic quadrilateral $ABCD$, it is known that the midpoints of $AB, BC, CD, DA$ are $E, F, G, H$ respectively, and the orthocenters of $\triangle AHE, \triangle BEF, \triangle CFG, \triangle DGH$ are $W, X, Y, Z$ respectively. Prove: The area of quadrilateral $ABCD$ is equal to the area of quadrilateral... | Proof As shown in Figure 2, connect $A C, B D$. Let the orthocenters of $\triangle D A B$, $\triangle A B C$, $\triangle B C D$, and $\triangle C D A$ be $W_{1}$, $X_{1}$, $Y_{1}$, and $Z_{1}$, respectively. Let the circumcenter of the cyclic quadrilateral $A B C D$ be $O$. Connect $A W_{1}$, $B X_{1}$, $C Y_{1}$, $D Z... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,401 |
Given $\triangle A B C(A B>B C)$, the circumcircle of $\triangle A B C$ is circle $\Gamma, M$ and $N$ are points on sides $A B$ and $B C$ respectively, such that $A M=C N$. Line $M N$ intersects $A C$ at point $K, P$ is the incenter of $\triangle A M K$, and $Q$ is the excenter of $\triangle C N K$ that is tangent to s... | Proof: Let the midpoint of arc $\overparen{A B C}$ be $T$.
From the given, it is easy to know that $T A=T C$. Connect $T M$ and extend it to point $J$, such that $T J=T A$. Connect auxiliary lines as shown in Figure 3.
Since $\angle T A M=\angle T A B=\angle T C B=\angle T C N$,
$T A=T C, A M=C N$
$\Rightarrow \triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,402 |
Given an unequal-sided $\triangle ABC$ satisfying $\angle BCA = 90^{\circ}$, $D$ is the foot of the perpendicular from $C$ to side $AB$. On segment $CD$, take point $X$. Point $K$ lies on segment $AX$ such that $BK = BC$. Similarly, point $L$ lies on segment $BX$ such that $AL = AC$. Point $T$ on segment $AB$ satisfies... | Proof First, we prove a lemma.
Lemma As shown in Figure 1, with the same setup as the problem, and $H$ being the orthocenter of $\triangle A X B$. Then $\angle A L H=\angle B K H=90^{\circ}, H K=H L$.
Proof It is easy to prove that $H, C, D$ are collinear.
Let $A X$ intersect $B H$ at point $E$, and $B X$ intersect $A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,403 |
Question As shown in Figure 1, in the convex quadrilateral $A B C D$, $K$, $L$, $M$, and $N$ are points on the sides $A B$, $B C$, $C D$, and $D A$ respectively, satisfying
$$
\frac{A K}{K B}=\frac{D A}{B C}, \frac{B L}{L C}=\frac{A B}{C D}, \frac{C M}{M D}=\frac{B C}{D A}, \frac{D N}{N A}=\frac{C D}{A B} \text {, }
$$... | Prove that from the given information,
$$
\begin{array}{l}
\frac{A N}{N D}=\frac{A B}{C D}=\frac{B L}{L C} \\
\Rightarrow \frac{A N}{B L}=\frac{N D}{L C}=\frac{A N+N D}{B L+L C}=\frac{A D}{B C}=\frac{A K}{K B} \\
\Rightarrow \frac{A N}{A K}=\frac{B L}{B K} .
\end{array}
$$
Similarly, $\frac{B K}{B L}=\frac{C M}{C L}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,404 |
2. Let the real number $x$ satisfy $x^{3}=x+1$. If $x^{7}=a x^{2}+b x+c$, then the value of $a+b+c$ is ( ).
(A) 3
(B) 4
(C) 5
(D)6 | 2. C.
$$
\begin{array}{l}
x^{7}=x(x+1)^{2}=x^{3}+2 x^{2}+x=2 x^{2}+2 x+1 \\
\Rightarrow a+b+c=5 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,405 |
7. If $a$ is the positive root of the equation $x^{2}+3 x-2=0$, and $b$ is the root of the equation $x+\sqrt{x+1}=3$, then $a+b=$ $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. 2 .
Notice that, $b+1+\sqrt{b+1}=4$.
Let $t=\sqrt{b+1}-1$.
Then $(t+1)^{2}+(t+1)=4$, and $t>0$.
Thus, $a$ satisfies $(a+1)^{2}+(a+1)=4$.
Therefore, $a=t$.
Hence $b+1+a+1=4 \Rightarrow a+b=2$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,406 |
Example 4 Given a positive integer $n$. Find the largest constant $\lambda$, such that for all positive real numbers $x_{1}, x_{2}, \cdots, x_{2 n}$ satisfying
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+2\right)^{n} \geqslant \prod_{i=1}^{2 n} x_{i}
$$
we have
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n... | 【Analysis】When $x_{i}=2(1 \leqslant i \leqslant 2 n)$, the equality in formula (1) holds, at this time,
$$
\lambda \leqslant \frac{\frac{1}{2 n} \sum_{i=1}^{2 n}(2+1)^{n}}{2^{2 n}}=\frac{3^{n}}{2^{2 n}} \text {. }
$$
Below, we prove by contradiction: For all positive real numbers $x_{1}, x_{2}, \cdots, x_{2 n}$ that s... | \frac{3^{n}}{2^{2 n}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,407 |
8. As shown in Figure 2, point $A$ is on the positive $y$-axis, point $B$ is on the positive $x$-axis, $S_{\triangle A O B}=9$, segment $A B$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $C$ and $D$. If $C D=$ $\frac{1}{3} A B$, and $A C=B D$, then $k=$ . $\qquad$ | 8. 4 .
Let point $A\left(0, y_{A}\right), B\left(x_{B}, 0\right)$.
Given $C D=\frac{1}{3} A B, A C=B D$, we know that $C$ and $D$ are the trisection points of segment $A B$.
Thus, $x_{C}=\frac{1}{3} x_{B}, y_{C}=\frac{2}{3} y_{A}$.
Therefore, $k=x_{C} y_{C}=\frac{2}{9} x_{B} y_{A}=\frac{4}{9} S_{\triangle A O B}=4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,408 |
9. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then the solution to the equation
$$
[x]^{4}+\{x\}^{4}+x^{4}=2048
$$
is | 9. $-3-\sqrt{5}$.
$$
\begin{array}{l}
\text { Let } a=[x], b=\{x\} \text {. Then } \\
x=[x]+\{x\}=a+b . \\
\text { Hence } x^{4}+[x]^{4}+\{x\}^{4} \\
=(a+b)^{4}+a^{4}+b^{4} \\
=2\left(a^{4}+2 a^{3} b+3 a^{2} b^{2}+2 a b^{3}+b^{4}\right) \\
=2\left(a^{2}+a b+b^{2}\right)^{2} \\
\Rightarrow a^{2}+a b+b^{2}=32 .
\end{arra... | -3-\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,409 |
10. In quadrilateral $A B C D$, $A B=B C=C D=$ $26, A D=30 \sqrt{3}, A C$ and $B D$ intersect at point $O, \angle A O B=$ $60^{\circ}$. Then $S_{\text {quadrilateral } A B C D}=$ $\qquad$ | 10. $506 \sqrt{3}$.
Let $O A=a, O B=b, O C=c, O D=d$. Then $S_{\triangle O A B}=\frac{1}{2} O A \cdot O B \sin 60^{\circ}=\frac{\sqrt{3}}{4} a b$,
$S_{\triangle O B C}=\frac{1}{2} O B \cdot O C \sin 120^{\circ}=\frac{\sqrt{3}}{4} b c$,
$S_{\triangle O C D}=\frac{1}{2} O C \cdot O D \sin 60^{\circ}=\frac{\sqrt{3}}{4} c... | 506 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,410 |
5. Given that the base $A B C$ of the regular tetrahedron $P-A B C$ is an equilateral triangle, and the center $O$ of the circumscribed sphere of the regular tetrahedron satisfies $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$. Then the cosine value of the dihedral angle $A-P B-C$ is $(\quad)$.
(A) ... | 5. C.
As shown in Figure 1.
From $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$, we know that $O$ is the circumcenter of $\triangle A B C$.
Let the side length of $\triangle A B C$ be $a$.
Then the height of this triangular pyramid $P O=O B=\frac{\sqrt{3}}{3} a$,
the length of the lateral edges $P ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,411 |
6. Let the prime $p$ satisfy that there exist positive integers $x, y$ such that $p-1=2 x^{2}, p^{2}-1=2 y^{2}$.
Then the number of primes $p$ that meet the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6. A.
Obviously, $p \neq 2$.
When the prime $p \geqslant 3$, obviously,
$$
p>y>x \text {, }
$$
and $p(p-1)=2(y+x)(y-x)$.
From $p \mid 2(y+x)(y-x)$, we know $p \mid(y+x)$.
Then $p \leqslant x+y<2 p \Rightarrow p=x+y$
$$
\begin{array}{l}
\Rightarrow p-1=2(y-x) \Rightarrow p=4 x-1 \\
\Rightarrow 4 x-1-1=2 x^{2} \Rightar... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,412 |
8. If $4^{a}=6^{b}=9^{c}$, then
$$
\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=
$$
$\qquad$ | 8. 0 .
Let $4^{a}=6^{b}=9^{c}=k$.
Then $a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,413 |
11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$ | 11.0.
As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,414 |
15. For any point \( P\left(x_{0}, y_{0}\right) \) on the right branch of the hyperbola \( x^{2}-\frac{y^{2}}{4}=1 \), draw a line \( l \) intersecting the two asymptotes at points \( A \) and \( B \). If \( P \) is the midpoint of \( AB \), prove:
(1) the line \( l \) intersects the hyperbola at only one point;
(2) th... | 15. (1) The equations of the two asymptotes of the hyperbola are
$$
y= \pm 2 x \text {. }
$$
When $y_{0}=0$, it is easy to get the equation of the line $l$ as $x=x_{0}$, at this time, the line $l$ intersects the hyperbola at only one point.
When $y_{0} \neq 0$, it is clear that the line $l$ has a slope, so we can set... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,415 |
16. Given that $a$ is a real constant, the function
$$
f(x)=\mathrm{e}^{-x} \sin x + a x \quad (x \in [0,2 \pi])
$$
(1) Let $g(x)$ be the derivative of $f(x)$, find the intervals of monotonicity of $g(x)$ within the interval $[0,2 \pi]$;
(2) If $f(x)$ has exactly one maximum and one minimum in the interval $(0,2 \pi)$,... | 16. (1) From $f(x)=\mathrm{e}^{-x} \sin x+a x$, we know $g(x)=f^{\prime}(x)=a-\mathrm{e}^{-x}(\sin x-\cos x)$. Then $g^{\prime}(x)=-2 \mathrm{e}^{-x} \cos x$.
From $g^{\prime}(x)=0$
$\Rightarrow x=\frac{\pi}{2}$ or $\frac{3 \pi}{2}$.
From $g^{\prime}(x)0$
$\Rightarrow \frac{\pi}{2}<x<\frac{3 \pi}{2}$.
From $g^{\prime}(... | -\mathrm{e}^{-2 \pi}<a<\mathrm{e}^{-\frac{\pi}{2}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,416 |
2. Given a positive integer $m$, and $n \geqslant m$, in an $m \times 2 n$ grid, what is the maximum number of dominoes $(1 \times 2$ or $2 \times 1$ small grids) that can be placed satisfying the following conditions:
(1) Each domino exactly covers two adjacent small grids;
(2) Any small grid is covered by at most one... | 2. The maximum value sought is $m n-\left[\frac{m}{2}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$), and for any alternating two rows, using $n$ pieces of $1 \times 2$ and $n-1$ pieces of $1 \times 2$ dominoes respectively ensures that the squares of the last row are exactly cover... | m n-\left[\frac{m}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,417 |
4. Let positive real numbers $x, y$ satisfy $x+y^{2016} \geqslant 1$. Prove: $x^{2016}+y>1-\frac{1}{100}$. | 4. (1) When $y>1-\frac{1}{100}$, it is obviously true.
(2) When $y \leqslant 1-\frac{1}{100}$, since
$$
x \geqslant 1-y^{2016} \geqslant 1-\left(1-\frac{1}{100}\right)^{2016} \text {, }
$$
Thus, by Bernoulli's inequality, we have
$$
x^{2016} \geqslant 1-2016\left(1-\frac{1}{100}\right)^{2016} \text {. }
$$
Therefore,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,418 |
5. Given a convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$, whose vertices lie on a circle $\Gamma$ with radius $R$. The diagonals $A_{1} B_{2}$, $A_{2} B_{3}$, and $A_{3} B_{1}$ intersect at a point $X$. For $i=1,2,3$, let $\Gamma_{i}$ be the circle that is tangent to the segments $X A_{i}$, $X B_{i}$, and the ar... | 5. (1) As shown in Figure 4, let $l_{1}$ be the tangent line of circle $\Gamma$, parallel to line $A_{2} B_{3}$ and on the same side of circle $\Gamma_{1}$ as line $A_{2} B_{3}$.
Similarly, define the tangent lines $l_{2}$ and $l_{3}$. The lines $l_{1}$ and $l_{2}$, $l_{2}$ and $l_{3}$, $l_{3}$ and $l_{1}$ intersect a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,419 |
6. A set of $n$ points in three-dimensional space, with no four points lying on the same plane, is divided into two subsets $A$ and $B$. An $AB$-tree is defined as a set of $n-1$ line segments, each with one endpoint in set $A$ and the other in set $B$, such that these segments do not form a cycle. An $AB$-tree can be ... | 6. Let $S$ be the set of these $n$ points, and the graph under consideration is a bipartite graph concerning sets $A$ and $B$, and it is a tree, which always maintains this property under the operation.
For a tree $T=(A \cup B, E)$, let $p_{T}(e)$ denote the number of points in set $A$ contained in the connected compo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,420 |
1. Let $n$ be a positive odd number, and $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers. Prove:
$$
\min _{i=1,2, \cdots, n}\left\{x_{i}^{2}+x_{i+1}^{2}\right\} \leqslant \max _{j=1,2, \cdots, n}\left\{2 x_{j} x_{j+1}\right\} \text {, }
$$
where, $x_{n+1}=x_{1}$. | 1. In the solution, all subscripts are understood modulo $n$.
Consider $n$ differences $x_{k+1}-x_{k}(k=1,2, \cdots, n)$.
Since $n$ is odd, there exists an index $j$ such that
$$
\left(x_{j+1}-x_{j}\right)\left(x_{j+2}-x_{j+1}\right) \geqslant 0 \text {. }
$$
Without loss of generality, assume that both factors on th... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,421 |
2. In a cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $X$. $C_1$, $D_1$, and $M$ are the midpoints of segments $CX$, $DX$, and $CD$ respectively. The line $AD_1$ intersects $BC_1$ at point $Y$, and the line $MY$ intersects $AC$ and $BD$ at different points $E$ and $F$. Prove: The line $XY$... | 2. As shown in Figure 1.
It is only necessary to prove $\angle E X Y=\angle E F X$ or $\angle A Y X+\angle X A Y=\angle B Y F+\angle X B Y$. From the fact that points $A, B, C, D$ are concyclic, we know $\triangle X A D \backsim \triangle X B C$.
Since $A D_{1}, B C_{1}$ are the corresponding medians of the similar tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,422 |
5. Let $k, n$ be integers, $k \geqslant 2, k \leqslant n \leqslant 2 k-1$. Some $1 \times k$ or $k \times 1$ rectangular pieces of paper are placed on an $n \times n$ grid, such that each piece exactly covers $k$ cells, and no two pieces overlap, until no more pieces can be placed. For any such $k, n$, find the minimum... | 5. When $n=k$, the required minimum value is $n$;
When $k<n<2k$, the required minimum value is $\min \{n, 2n-2k+2\}$.
When $n=k$, the conclusion is obvious.
Assume $k<n<2k$ below.
If $k<n<2k-1$, then
$$
\min \{n, 2n-2k+2\}=2n-2k+2 \text{. }
$$
At this time, $2n-2k+2$ pieces of paper can be placed in the grid as follo... | \min \{n, 2n-2k+2\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,423 |
6. Let
$S=\left\{n \in \mathbf{Z}_{+}\right.$。There exists $n^{2}+1 \leqslant d \leqslant n^{2}+2 n$ such that $\left.d \mid n^{4}\right\}$.
Prove: For $m \in \mathbf{Z}$, the set $S$ contains infinitely many elements of the form $7 m$, $7 m+1$, $7 m+2$, $7 m+5$, $7 m+6$, and $S$ contains no elements of the form $7 m+... | 6. First, prove a lemma.
Lemma $n \in S$ if and only if at least one of $2 n^{2}+1$ and $12 n^{2}+9$ is a perfect square.
Proof Let $d=n^{2}+m(1 \leqslant m \leqslant 2 n)$ be a divisor of $n^{4}$.
Then $n^{2}=d-m \Rightarrow d \left\lvert\, m^{2} \Rightarrow \frac{m^{2}}{d} \in \mathbf{Z}_{+}\right.$.
By $n^{2}<d<(n+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,424 |
$\begin{array}{l}\text { 2. } \sqrt{3} \tan 25^{\circ}+\frac{\tan 25^{\circ}}{\tan 185^{\circ}}+ \\ \frac{\tan 25^{\circ} \cdot \cos 10^{\circ}-\tan 25^{\circ} \cdot \sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}} \\ =\end{array}$ | 2. $2 \sqrt{3}+3$.
Original expression
$$
\begin{array}{l}
=\tan 25^{\circ}\left(\tan 60^{\circ}+\cot 185^{\circ}+\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}\right) \\
=\tan 25^{\circ}\left(\tan 60^{\circ}+\tan 85^{\circ}+\tan 35^{\circ}\right) \\
=\tan 25^{\circ} \cdot \tan 60^{\circ} \cdo... | 2 \sqrt{3}+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,425 |
3. Given
$$
5 x+16 y+33 z \geqslant 136\left(x 、 y 、 z \in \mathbf{R}_{+}\right) \text {. }
$$
then the minimum value of $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$ is | 3. 50 .
Notice that,
$$
\begin{array}{l}
x^{3}+x^{2}-5 x+3=(x+3)(x-1)^{2} \geqslant 0 \\
\Rightarrow x^{3}+x^{2} \geqslant 5 x-3, \\
y^{3}+y^{2}-16 y+20=(y+5)(y-2)^{2} \geqslant 0 \\
\Rightarrow y^{3}+y^{2} \geqslant 16 y-20, \\
z^{3}+z^{2}-33 z+63=(z+7)(z-3)^{2} \geqslant 0 \\
\Rightarrow z^{3}+z^{2} \geqslant 33 z-6... | 50 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,426 |
4. In the right prism $A B C-A_{1} B_{1} C_{1}$, the side length of the square $A A_{1} C_{1} C$ is 4, the plane $A B C \perp$ plane $A A_{1} C_{1} C, A B=$ $3, B C=5$. If there is a point $D$ on the line segment $B C_{1}$ such that $A D \perp A_{1} B$, then the value of $\frac{B D}{B C_{1}}$ is | 4. $\frac{9}{25}$.
Establish a spatial rectangular coordinate system with $\overrightarrow{A C}$ as the $x$-axis, $\overrightarrow{A B}$ as the $y$-axis, and $\overrightarrow{A A_{1}}$ as the $z$-axis. Let $D(x, y, z)$ be a point on the line segment $B C_{1}$, and $\overrightarrow{B D}=\lambda \overrightarrow{B C_{1}}... | \frac{9}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,427 |
Example 8 Let $n(n \geqslant 2)$ be a fixed integer. Determine the smallest constant $c$ such that
$$
\sum_{1<i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$
holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$, and determine the necessa... | 【Analysis】Obviously, when $x_{1}=x_{2}=\cdots=x_{n}=0$, inequality (1) always holds.
If $x_{i}(i=1,2, \cdots, n)$ are not all 0, by homogeneity, we can assume $\sum_{i=1}^{n} x_{i}=1$.
When $n=2$, $x_{1}+x_{2}=1$,
the left side of equation (1)
$$
\begin{array}{l}
=x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}\right)=x_{1} x_{2... | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,428 |
6. Given that all positive integers are in $n$ sets, satisfying that when $|i-j|$ is a prime number, $i$ and $j$ belong to two different sets. Then the minimum value of $n$ is $\qquad$ | 6. 4 .
It is easy to see that $n \geqslant 4$ (2, 4, 7, 9 must be in four different sets).
Also, when $n=4$, the sets
$$
A_{i}=\left\{m \in \mathbf{Z}_{+} \mid m \equiv i(\bmod 4)\right\}(i=0,1,2,3)
$$
satisfy the condition.
Therefore, the minimum value of $n$ is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,429 |
7. Given that the ellipse $C$ passes through the point $M(1,2)$, with two foci at $(0, \pm \sqrt{6})$, and $O$ is the origin, a line $l$ parallel to $OM$ intersects the ellipse $C$ at points $A$ and $B$. Then the maximum value of the area of $\triangle OAB$ is $\qquad$ | 7. 2 .
According to the problem, let $l_{A B}: y=2 x+m$.
The ellipse equation is $\frac{y^{2}}{8}+\frac{x^{2}}{2}=1$. By solving the system and eliminating $y$, we get $16 x^{2}+8 m x+2\left(m^{2}-8\right)=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\fra... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,430 |
8. Given positive integers $a, b, c, x, y, z$ satisfying $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$,
and $\left\{\begin{array}{l}2 a+b+4 c=4 x y z, \\ 2 x+y+4 z=4 a b c .\end{array}\right.$
Then the number of six-tuples $(a, b, c, x, y, z)$ that satisfy the conditions is $\qquad$ gro... | 8. 0 .
Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,431 |
9. (16 points) Given the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ satisfy
$$
\begin{array}{l}
x_{1}=1, x_{n+1}=x_{n} \sqrt{\frac{1}{2\left(\sqrt{1-x_{n}^{2}}+1\right)}}, \\
y_{1}=0, y_{n+1}=\sqrt{\frac{y_{n}+1}{2}} .
\end{array}
$$
Prove: (1) $x_{n} x_{n+1}+y_{n} y_{n+1}=y_{n+1}$;
(2) If $a_{n}=x_{n... | (1) Let $x_{1}=1=\sin \frac{\pi}{2}, y_{1}=0=\cos \frac{\pi}{2}$, then $x_{2}=x_{1} \sqrt{\frac{1}{2\left(\sqrt{1-x_{1}^{2}}+1\right)}}=\sin \frac{\pi}{4}$, $y_{2}=\cos \frac{\pi}{4}$.
By mathematical induction, we can prove that
$$
x_{n}=\sin \frac{\pi}{2^{n}}, y_{n}=\cos \frac{\pi}{2^{n}} \text {. }
$$
Thus, $x_{n} ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,432 |
Three. (50 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Suppose $2n$ distinct odd primes $p_{1}, p_{2}, \cdots, p_{2 n}$ satisfy
$$
4\left(\sum_{k=1}^{1006}(-1)^{\left[\frac{2016 k}{1000}\right]}\left\{\frac{2016 k}{1007}\right\}\right)+\frac{2}{1007}
$$
is a non-... | Three, first prove two lemmas.
Lemma 1 If $(m, n)=1$, and $n$ is odd, $m$ is even, then
$$
\sum_{k=1}^{n-1}(-1)^{\left[\frac{m k}{n}\right]}\left\{\frac{m k}{n}\right\}+\frac{1}{2 n}=\frac{1}{2} \text {. }
$$
Proof of Lemma 1 Since $(m, n)=1$, thus, $m k$ takes all values $1,2, \cdots, n-1$ modulo $n$.
Let $m k=n t+r$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,433 |
Question 1 A family of sets $\mathscr{F}$ is called "perfect": if for any three sets $X_{1}, X_{2}, X_{3} \in \mathscr{F}$, the sets $\left(X_{1} \backslash X_{2}\right) \cap X_{3}$ and $\left(X_{2} \backslash X_{1}\right) \cap X_{3}$ contain at least one empty set. Prove: if $\mathscr{F}$ is a perfect subfamily of a f... | Let $\mathscr{F}$ be a perfect subset family of $U$.
We will prove by induction on $|U|$ that: $|\mathscr{F}| \leqslant |U| + 1$.
If $|U| = 0$, i.e., $U = \varnothing$, then $|\mathscr{F}| \leqslant 1$ holds.
Assume the conclusion holds for $|U| = k-1$ ($k \geqslant 1$).
Consider the case $|U| = k$.
Take the element $A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,434 |
Example 1 The real solutions of the equation $[2 x]+[3 x]=8 x-\frac{1}{2}$ are $\qquad$
(2006, "Xinzhi Cup" Shanghai Junior High School Mathematics Competition) | Solve: From $[x] \leqslant x<[x]+1$, we get
$$
\begin{array}{l}
2 x-1<[2 x] \leqslant 2 x, \\
3 x-1<[3 x] \leqslant 3 x .
\end{array}
$$
Combining with the original equation, we get
$$
\begin{array}{l}
5 x-2<8 x-\frac{7}{2} \leqslant 5 x \\
\Rightarrow \frac{1}{2}<x \leqslant \frac{7}{6} \Rightarrow \frac{1}{2}<8 x-\f... | x=\frac{13}{16}, \frac{17}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,435 |
Example 2 Given $0<a<1$, and satisfies
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]=$ $\qquad$ | Solve:
$$
\begin{array}{l}
0<a<1 \\
\Rightarrow 0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2 \\
\Rightarrow\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right] \text { is either }
\end{array}
$$
0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 ... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,436 |
Example 2 Proof: The equation $x^{2}-2 y^{2}=1$ has infinitely many positive integer solutions $(x, y)$.
Translating the provided text into English while preserving the original line breaks and formatting, the result is as follows:
Example 2 Proof: The equation $x^{2}-2 y^{2}=1$ has infinitely many positive intege... | Prove that using inductive construction, the original equation has infinitely many positive integer solutions.
The first solution is taken as $x_{1}=3, y_{1}=2$.
Assume for $n \geqslant 1$, a solution $\left(x_{n}, y_{n}\right)$ has been obtained, i.e.,
$$
x_{n}^{2}-2 y_{n}^{2}=1 \text {. }
$$
Next, construct $x_{n+1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,437 |
8. There is a fair coin, and now a continuous coin-tossing game is being played, with the following rules: During the tossing process, at any time, if an odd number of consecutive heads is followed by a tail, the game stops; otherwise, the game continues. The game is played for a maximum of 10 tosses. What is the expec... | 8. $4 \frac{135}{256}$.
Let the mathematical expectation of the number of tosses in the game with a maximum of $n$ tosses be $E_{n}$. It is easy to see that $E_{1}=1, E_{2}=2$.
Consider the case when $n \geqslant 3$.
When the maximum number of tosses is $n$, we can discuss the following three scenarios.
(1) The first ... | 4 \frac{135}{256} | Other | math-word-problem | Yes | Yes | cn_contest | false | 729,439 |
10. (20 points) Define the sequence $\left\{a_{n}\right\}$:
$$
\begin{array}{l}
a_{0}=1, a_{1}=2.1, \\
a_{n+1} a_{n-1}^{2}+2 a_{n-1}^{2} a_{n}=a_{n}^{3}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Prove: For any non-negative integer $k$, we have
$$
\sum_{i=0}^{k} \frac{1}{a_{i}}<2.016
$$ | 10. The recursive formula can be written as
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=\left(\frac{a_{n}}{a_{n-1}}\right)^{2}-2\left(n \in \mathbf{N}_{+}\right) . \\
\text {Let } b_{n}=\frac{a_{n+1}}{a_{n}}(n \in \mathbf{N}) . \text { Then } \\
b_{n}=b_{n-1}^{2}-2 .
\end{array}
$$
More generally, let $a_{1}=a>2$. Then
... | 2.016 | Algebra | proof | Yes | Yes | cn_contest | false | 729,440 |
One, (40 points) Given $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers. Try to find
$$
E\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}+\sum_{i=1}^{n} x
$$
the minimum value. | Notice,
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1} \\
=\frac{1}{2}\left[x_{1}^{2}+\sum_{i=1}^{n-1}\left(x_{i}+x_{i+1}\right)^{2}+x_{n}^{2}\right] .
\end{array}
$$
Discuss in two cases.
(1) When $n=2 k$, by
$$
\begin{array}{l}
{\left[1+\left(\frac{k+1}{k}\right)^{2}+1+\cdots+\left(\frac... | -\frac{n(n+2)}{8(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,441 |
Four. (50 points) Let $A$ and $B$ be two point sets on a plane, satisfying $A \cap B=\varnothing,|A|=|B|=n$, and no three points are collinear. Connect several line segments between sets $A$ and $B$, with each line segment having one endpoint in set $A$ and the other endpoint in set $B$, and at most one line segment be... | Let set $A$ have $s$ points that do not lead to $l$ edges, and $n-s$ points that exactly lead to $l$ edges. Let set $B$ have $t$ points that do not lead to $l$ edges, and $n-t$ points that exactly lead to $l$ edges.
Due to symmetry, we can assume $s \leq t$.
Let $A = A_{1} \cup A_{2}$, where $A_{1}$ is the set of all p... | |S| \leq \begin{cases}
(n-l) 2l, & \left[\frac{n-l}{2}\right] > l; \\
\left(n - \left[\frac{n-l}{2}\right]\right)\left(\left[\frac{n-l}{2}\right] + l\right), & \left[\frac{n-l}{2}\right] \leq l.
\end{cases} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,442 |
Example 3 Let set $A$ be a finite set, and
$$
\begin{array}{l}
|A| \geqslant 2, \\
Q(A)=\left\{\left.\frac{a-b}{c-d} \right\rvert\, a, b, c, d \in A, c \neq d\right\} .
\end{array}
$$
Find the smallest real number $\lambda$, such that $|Q(A)| \leqslant \lambda|A|^{4}$ holds for all sets $A$.
(2015, National Mathematic... | Solve $\lambda$ with the minimum value being $\frac{1}{2}$.
Define $D(A)=\{a-b \mid a, b \in A, a \geqslant b\}$.
Thus, $|D(A)| \leqslant 1+\frac{1}{2}|A|(|A|-1)$.
Since the set $Q(A)$ can be expressed as
$$
\begin{aligned}
Q(A)= & \{0, \pm 1\} \cup\left\{\left. \pm \frac{x}{y} \right\rvert\, x, y \in D(A),\right. \\
&... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,443 |
Example 7 Let $n$ be a positive integer,
$$
S_{n}=\left\{\left(a_{1}, a_{2}, \cdots, a_{2^{n}}\right) \mid a_{i}=0,1\right\} \text {. }
$$
For $a, b \in S_{n}$,
$$
a=\left(a_{1}, a_{2}, \cdots, a_{2^{n}}\right), b=\left(b_{1}, b_{2}, \cdots, b_{2^{n}}\right) \text {, }
$$
define $d(a, b)=\sum_{i=1}^{2^{n}}\left|a_{i}... | Let set $A$ contain $k$ elements whose first component is 1, denoted as $e_{1}, e_{2}, \cdots, e_{k}$. Let
$$
T=\sum_{1 \leqslant i<j \leqslant k} d\left(e_{i}, e_{j}\right) .
$$
Then $T \geqslant \mathrm{C}_{k}^{2} \cdot 2^{n-1}$.
For each component of elements $e_{1}, e_{2}, \cdots, e_{k}$, let there be $x_{i}$ ones... | 2^{n+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,445 |
Example 2 Let $a, b, c \geqslant 0$. Prove:
$$
\frac{3}{2} \sum a \geqslant \sum \sqrt{a^{2}+b c} \text {. }
$$ | Prove that
$$
\begin{array}{l}
\sum \frac{a^{2}+b c}{(a+b+c)^{2}}+\frac{2 \sqrt{\prod\left(a^{2}+b c\right)}}{(a+b+c)^{3}} \leqslant 1 \\
\Leftrightarrow 2 \sqrt{\prod\left(a^{2}+b c\right)} \leqslant\left(\sum a\right)\left(\sum a b\right) \\
\Leftrightarrow \prod(a-b)^{2}+6 a b c \sum a(b-c)^{2}+ \\
\quad 49 a^{2} b^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,446 |
Example 4 Let $x, y \in \mathbf{R}$. Given
$$
\left(\sqrt{x^{2}+1}-x+1\right)\left(\sqrt{y^{2}+1}-y+1\right)=2 \text{. }
$$
Find the value of $xy$. | Solve equation (1)
$$
\begin{array}{l}
\Leftrightarrow \sqrt{x^{2}+1}-x+1=\frac{2}{\sqrt{y^{2}+1}-y+1} \\
\Leftrightarrow \sqrt{x^{2}+1}-x+1=\frac{\sqrt{y^{2}+1}+y-1}{y} \\
\Rightarrow \sqrt{x^{2}+1}-x+1=\sqrt{\left(\frac{1}{y}\right)^{2}+1}-\frac{1}{y}+1 .
\end{array}
$$
Let $f(x)=\sqrt{x^{2}+1}-x+1$.
Since $f^{\prim... | xy=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,447 |
Example 3 The number of solutions to the equation $x^{2}-2[x]-3=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(2009, National Junior High School Mathematics Competition) | Solve: From $x^{2}-2[x]-3=0 \Rightarrow x^{2}-3=2[x]$. And $[x] \leqslant x$, thus,
$$
\begin{array}{l}
x^{2}-3 \leqslant 2 x \\
\Rightarrow x^{2}-2 x-3 \leqslant 0 \\
\Rightarrow-1 \leqslant x \leqslant 3
\end{array}
$$
$\Rightarrow$ The possible values of $[x]$ are $-1,0,1,2,3$.
When $[x]=-1$, $x^{2}=1 \Rightarrow x=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,448 |
$$
|2-x|+(\sqrt{x-2})^{2}-\sqrt{4 x^{2}-4 x+1}
$$
The result obtained is ( ).
(A) $1-2 x$
(B) $5-4 x$
(C) 3
(D) -3 | 1. D.
From the condition, we know $x-2 \geqslant 0 \Rightarrow x \geqslant 2$. Therefore, the original expression $=(x-2)+(x-2)-(2 x-1)=-3$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,449 |
2. If the real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
x+2 y+3 z=0, \\
2016 x+2015 y+2014 z=2017,
\end{array}
$$
then $x+y+z=$ ( ).
(A) -1
(B) 0
(C) 1
(D) 2016 | 2. C.
From equations (1) and (2), we get
$$
\begin{array}{l}
2017=2017(x+y+z) \\
\Rightarrow x+y+z=1 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,450 |
4. If $k$ is an integer such that the equation in $x$
$$
k x^{2}-(2 k+3) x+3=0
$$
has rational roots, then $k$ is called a "good number". Then, the number of good numbers $k$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 4. C.
If $k=0$, the original equation becomes
$$
-3 x+3=0 \Rightarrow x=1 \text {. }
$$
Thus, $k=0$ is a good number.
If $k \neq 0$, then the discriminant of the original equation is
$$
\Delta=(2 k+3)^{2}-12 k=4 k^{2}+9 \text {, }
$$
a perfect square.
Let $4 k^{2}+9=m^{2}$ (where $m$ is a positive integer)
$$
\Right... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,451 |
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$ | $=, 7.1$.
Connect $O A, D E$.
$$
\begin{array}{l}
\text { Since } \angle B O C=2 \angle B A C=120^{\circ} \\
\Rightarrow \angle C O E=60^{\circ}=\angle D A E \\
\Rightarrow A, D, O, E \text { are concyclic } \\
\Rightarrow \angle D E B=\angle D A O=\angle D B E \\
\Rightarrow D B=D E .
\end{array}
$$
Similarly, $C E=D... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,452 |
For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$, and $\{x\}$ represents the fractional part of $x$. Then
$$
\begin{array}{l}
\left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} \\
=
\end{array}
$$ | $-, 1.1007$.
Notice that,
$$
\frac{2014^{n}}{2015}=\frac{(2015-1)^{n}}{2015}=M+\frac{(-1)^{n}}{2015} \text {, }
$$
where $M$ is an integer.
Then $\left\{\frac{2014^{n}}{2015}\right\}=\left\{\begin{array}{ll}\frac{1}{2015}, & n \text { is even; } \\ \frac{2014}{2015}, & n \text { is odd. }\end{array}\right.$
Therefore,... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,453 |
2. Given $a_{1}, a_{2}, \cdots, a_{9}$ as any permutation of $1,2, \cdots, 9$. Then the minimum value of $a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9}$ is $\qquad$ | 2. 214.
Since $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 9)$, by the AM-GM inequality, we have
$$
\begin{array}{l}
a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9} \\
\geqslant 3 \sqrt[3]{a_{1} a_{2} \cdots a_{9}}=3 \sqrt[3]{9!} \\
=3 \sqrt[3]{(2 \times 5 \times 7) \times(1 \times 8 \times 9) \times(3 \times 4 \tim... | 214 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,454 |
3. Given $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are positive integers. The set of sums obtained by taking any four of these numbers is $\{44, 45, 46, 47\}$. Then these five numbers are $\qquad$ | 3. $10,11,11,12,13$.
From these five positive integers, if we take any four numbers and find their sum, we can get five sum values. Since there are only four different sum values in this problem, it indicates that two of the sum values must be equal.
The sum of these five sum values is
$4\left(x_{1}+x_{2}+x_{3}+x_{4}... | 10, 11, 11, 12, 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,455 |
4. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+2015(x-1)=-1, \\
(y-1)^{3}+2015(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ | 4. 2 .
Notice that the function $f(z)=z^{3}+2015 z$ is a monotonically increasing function on $(-\infty,+\infty)$.
From the given condition, we have $f(x-1)=f(1-y)$.
Therefore, $x-1=1-y \Rightarrow x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,456 |
8. Given a regular tetrahedron $P-ABC$ with a base edge length of 1, and the distance from point $P$ to the base $ABC$ is $\sqrt{2}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$ . | 8. $\frac{\sqrt{2}}{6}$.
As shown in Figure 1, draw $P O_{1} \perp$ plane $A B C$ at point $O_{1}$. Take the midpoint $D$ of $A B$, and connect $C D$.
Then $O_{1} D \perp A B$.
Connect $P D$.
Then $P D \perp A B$.
According to the problem, the incenter of the regular tetrahedron $P-A B C$ lies on the line segment $P O... | \frac{\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,458 |
11. (20 points) Nine people attend a party, and it is known that every pair of them has drunk with exactly the same one person. Prove: there is exactly one person who has drunk with everyone, and each of the others has drunk with exactly two people.
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。
---
11. (20 points) Nine people ... | 11. Let $u_{1}, u_{2}, \cdots, u_{9}$ represent the nine people attending the party. Construct a graph $G=(V, E): V=\left\{u_{1}, u_{2}, \cdots, u_{9}\right\}$, where $u_{i}$ and $u_{j}$ are connected by an edge (denoted as $u_{i} u_{j} \in E$) if and only if $u_{i}$ and $u_{j}$ have had a drink together.
Let $N\left(... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,459 |
1. Let the sets be
$$
\begin{array}{l}
M=\left\{y \left\lvert\, y=x^{\frac{1}{2}}\right., x \in[1,4]\right\}, \\
N=\left\{x \mid y=\log _{2}(1-x)\right\} .
\end{array}
$$
Then $M \cap \complement_{\mathbf{R}} N=(\quad)$.
(A) $\{x \mid 1 \leqslant x \leqslant 2\}$
(B) $\{x \mid 1 \leqslant x \leqslant 4\}$
(C) $\{x \mi... | $-1 . \mathrm{A}$.
It is known that $M=[1,2], N=(-\infty, 1)$.
Therefore, $M \cap \complement_{\mathbb{R}} N=\{x \mid 1 \leqslant x \leqslant 2\}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,460 |
2. An interval containing a zero of the function $f(x)=\mathrm{e}^{x}+2 x-3$ is ().
(A) $(-1,0)$
(B) $\left(0, \frac{1}{2}\right)$
(C) $\left(\frac{1}{2}, 1\right)$
(D) $\left(1, \frac{3}{2}\right)$ | 2. C.
Knowing the following:
$$
\begin{array}{l}
f(-1)=\mathrm{e}^{-1}-50 \\
f\left(\frac{3}{2}\right)=\mathrm{e}^{\frac{3}{2}}>0
\end{array}
$$
Therefore, the solution is $x \in\left(\frac{1}{2}, 1\right)$. | C | Calculus | MCQ | Yes | Yes | cn_contest | false | 729,461 |
8. Let real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+1 \geqslant 0 \\
y+1 \geqslant 0 \\
x+y+1 \leqslant 0 .
\end{array}\right.
$$
Then the maximum value of $2 x-y$ is $\qquad$ | 8. 1 .
When $x=0, y=-1$, $2 x-y$ takes the maximum value 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,463 |
12. Given four points $O, A, B, C$ on a plane, satisfying $OA=4, OB=3, OC=2, \overrightarrow{OB} \cdot \overrightarrow{OC}=3$. Then the maximum value of $S_{\triangle ABC}$ is $\qquad$ | 12. $2 \sqrt{7}+\frac{3 \sqrt{3}}{2}$.
From the given information,
$$
B C=\sqrt{O B^{2}+O C^{2}-2 \overrightarrow{O B} \cdot \overrightarrow{O C}}=\sqrt{7} \text {, }
$$
and the angle between $O B$ and $O C$ is $60^{\circ}$.
Consider a circle centered at the origin $O$ with radii 2, 3, and 4, denoted as $\odot O_{1}$... | 2 \sqrt{7}+\frac{3 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,464 |
2. Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$.
Then the number of such $n$ is $\qquad$. | From $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$, knowing that equality holds, we find that $n$ is a common multiple of $3$ and $6$, i.e., a multiple of $6$. Therefore, the number of such $n$ is $\left[\frac{2006}{6}\right]=334$. | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,465 |
2. The solution to the equation $10^{x} \times 100^{2 x}=1000^{5}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 2. C.
Taking the common logarithm on both sides, we get
$$
\begin{array}{l}
x \lg 10+2 x \lg 100=5 \lg 1000 \\
\Rightarrow x+4 x=5 \times 3 \Rightarrow x=3 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,466 |
3. Ben and David go to buy bagels. For every 1 yuan Ben spends, David can save 25 fen. This way, Ben spends 12.50 yuan more than David. Then the total amount spent by both is ( ) yuan.
(A) 37.50
(B) 50.00
( C) 87.50
(D) 90.00
(E) 92.50 | 3. C.
From the information that Ben spent 12.50 yuan more than David, we know Ben spent $12.50 \div 0.25=50.00$ yuan, and David spent $50.00-12.50=37.50$ yuan. Therefore, $50.00+37.50=87.50$ yuan. | 87.50 | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,467 |
6. Meena lists the numbers from 1 to 30, and Emilie copies these numbers and changes all the digit 2s to digit 1s. Both then find the sum of the numbers they have written. How much greater is Meena's sum than Emilie's?
(A) 13
(B) 26
(C) 102
(D) 103
(E) 110 | 6. D.
For numbers with 2 in the tens place, changing 2 to 1 reduces the total sum by 10; for numbers with 2 in the units place, changing 2 to 1 reduces the total sum by 1.
Numbers with 2 in the tens place are $20, 21, \cdots, 29$, a total of 10; numbers with 2 in the units place are $2, 12, 22$, a total of 3. Therefo... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,468 |
7. Given seven numbers $60, 100, x, 40, 50, 200, 90$ with an average and median both equal to $x$. Then $x$ is ( ).
(A) 50
(B) 60
(C) 75
(D) 90
(E) 100 | 7. D.
Since the average of the seven numbers is $x$, then
$$
\begin{array}{l}
\frac{60+100+x+40+50+200+90}{7}=x \\
\Rightarrow 7 x=540+x \Rightarrow x=90 .
\end{array}
$$
Substituting $x=90$ we get
$$
40, 50, 60, 90, 90, 100, 200 \text {. }
$$
At this point, 90 is also the median, so $x=90$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,469 |
3. Given that the real number $x$ satisfies $x^{3}+\frac{1}{x^{3}}=18$. Then $\{x\}+\left\{\frac{1}{x}\right\}=(\quad)$.
(A) $\frac{1}{2}$
(B) $3-\sqrt{5}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) 1
$(2014$, National Junior High School Mathematics League) | Notice,
$$
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=18
$$
Let $x+\frac{1}{x}=a$. Then
$$
a\left(a^{2}-3\right)=18 \Rightarrow a=3 \text {. }
$$
Solving gives $x=\frac{3 \pm \sqrt{5}}{2}$.
Clearly, $\{x\}+\left\{\frac{1}{x}\right\}=1$. Therefore, the answer is (D). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,470 |
8. The cunning rabbit and the foolish fox agreed: if the fox crosses the bridge in front of the rabbit's house each time, the rabbit will give the fox money to double the fox's money, but each time the fox crosses the bridge, it has to pay the rabbit 40 cents as a toll. Hearing that his money would double every time he... | 8. C.
Let $f(x)=2 x-40$. Then
$$
\begin{array}{l}
f(f(f(x)))=0 \\
\Rightarrow 2[2(2 x-40)-40]-40=0 \\
\Rightarrow x=35 .
\end{array}
$$ | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 729,471 |
9. Arrange 2016 coins into a triangle. Place 1 coin in the 1st row, 2 coins in the 2nd row, ….... place $N$ coins in the $N$th row. Then the sum of the digits of $N$ is $(\quad)$.
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10 | 9. D.
From the problem, we know
$$
\begin{array}{l}
1+2+\cdots+N=\frac{N(N+1)}{2}=2016 \\
\Rightarrow N=63 \Rightarrow 6+3=9 .
\end{array}
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,472 |
10. As shown in Figure 1, a carpet has three different colors, and the areas of the three differently colored regions form an arithmetic sequence. The width of the smallest rectangle in the middle is 1 foot, and the surrounding width of the other two shaded areas is also 1 foot. Then the length of the smallest rectangl... | 10. B.
Let the length of the central small rectangle be $x$ feet. Then the area of the small rectangle is $x \cdot 1$, the area of the middle region is
$$
(x+2) \times 3 - x = 2x + 6 \text{, }
$$
and the area of the outermost region is
$$
(x+4) \times 5 - (x+2) \times 3 = 2x + 14 \text{. }
$$
Since the areas of the ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,473 |
13. Five friends go to the cinema to watch a movie, and their seats are in a row from left to right, numbered 1 to 5. During the movie, Ada goes to the lobby to buy some popcorn. When she returns, she finds that Bea has moved two seats to the right, Cissy has moved one seat to the left, Dee and Idi have swapped seats, ... | 13. B.
Let $(A, B, C, D, E)$ be the initial seat numbers of five people. Then
$$
A+B+C+D+E=1+2+3+4+5 \text {. }
$$
Assuming the positive direction is to the right, if Ada moves $x$ seats, then
$$
\begin{array}{l}
A+x+B+2+C-1+E+D \\
=1+2+3+4+5 \\
\Rightarrow x=-1 .
\end{array}
$$
Thus, after moving 1 seat to the left... | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 729,475 |
14. There are $(\quad)$ different ways to write 2016 as the sum of a natural number multiple of 2 and a natural number multiple of 3 (for example, $2016=1008 \times 2+0 \times 3=402 \times 2+404 \times 3$).
(A) 236
(B) 336
(C) 337
(D) 403
(E) 672 | 14. C.
Let $2016=2x+3y(x, y \in \mathbf{N})$. Then $y=\frac{2016-2x}{3}=672-\frac{2x}{3}$.
Since $y \geqslant 0$, we have $0 \leqslant x \leqslant 1008(x \in \mathbf{N})$.
Also, since $y \in \mathbf{N}$, it follows that $2x \equiv 0(\bmod 3)$.
Given $(2,3)=1$, we know $3 \mid x$.
Therefore, there are $\left[\frac{1008... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,476 |
15. As shown in Figure 3, seven small circular cookies with a radius of 1 inch are cut from a circular dough. Adjacent small cookies are externally tangent to each other, and all small cookies are internally tangent to the dough. The remaining dough is reshaped into a circular cookie of the same thickness. The radius o... | 15. A.
From the problem, we know that the radius of the large dough circle is 3 inches. Therefore, the radius of the cookie made from the leftover dough is
$$
\sqrt{\frac{3^{2} \pi - 7 \pi \times 1^{2}}{\pi}} = \sqrt{2} .
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,477 |
16. Given the coordinates of the three vertices of $\triangle A B C$ are $A(0,2), B(-3,2), C(-3,0)$, construct $\triangle A^{\prime} B^{\prime} C^{\prime}$, which is the reflection of $\triangle A B C$ over the $x$-axis, and then rotate $\triangle A^{\prime} B^{\prime} C^{\prime}$ counterclockwise by $90^{\circ}$ aroun... | 16. D.
Consider any point $P(x, y)$ on the coordinate plane. The point symmetric to $P$ with respect to the $x$-axis is $P^{\prime}(x,-y)$. Then, rotating point $P^{\prime}$ counterclockwise by $90^{\circ}$ around the origin $O$, we get $P^{\prime \prime}(y, x)$. Therefore, $P$ and $P^{\prime \prime}$ are symmetric wi... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,478 |
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