problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Five. (15 points) Take $n$ points inside $\triangle A B C$, and subdivide $\triangle A B C$ into several smaller triangles (any two small triangles either share a common vertex, or share a common edge, or have no common points, as shown in Figure 3). Now, color point $A$ red, point $B$ blue, and point $C$ black. The re... | Let the number of red-blue edges (i.e., edges with one endpoint being a red point and the other a blue point) inside $\triangle A B C$ be $k$. Suppose the number of characteristic triangles with three vertices of different colors (one red, one blue, and one black) is $p$, and the number of small triangles with vertices... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,968 |
1. The number of positive integer solutions to the indeterminate equation $2(x+y)=x y+7$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | -、1.B.
Assume $x \geqslant y$. From the condition, we have $(x-2)(y-2)=-3$. Then $\left\{\begin{array}{l}x-2=3, \\ y-2=-1\end{array}\right.$ or $\left\{\begin{array}{l}x-2=1, \\ y-2=-3 .\end{array}\right.$
Solving these, we get $x=5, y=1$ or $x=3, y=-1$ (discard).
Therefore, the original indeterminate equation has two ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,969 |
2. If $\alpha$ is an angle in the second quadrant, then $\pi-\frac{\alpha}{2}$ is ( ).
(A) an angle in the first quadrant
(B) an angle in the second quadrant
(C) an angle in the first or third quadrant
(D) an angle in the second or fourth quadrant | 2.D.
$$
\begin{array}{l}
\text { Given } 2 k \pi+\frac{\pi}{2}<\alpha<2 k \pi+\pi, \\
\Rightarrow k \pi+\frac{\pi}{4}<\frac{\alpha}{2}<k \pi+\frac{\pi}{2} \\
\Rightarrow -k \pi+\frac{\pi}{2}<\pi-\frac{\alpha}{2}<-k \pi+\frac{3 \pi}{4} .
\end{array}
$$
Thus, $\pi-\frac{\alpha}{2}$ is an angle in the second or fourth qu... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,970 |
3. Given the set $A=\left\{x \mid x^{2}-5 x+6<0\right\}$, $B=\left\{\left.x\right| x^{2}-a x+5=0\right\}$. If $A \cap B \neq \varnothing$, then the range of real number $a$ is ( ).
(A) $\left[2 \sqrt{5}, \frac{14}{3}\right)$
(B) $\left[2 \sqrt{5}, \frac{9}{2}\right)$
(C) $\left(\frac{9}{2}, \frac{14}{3}\right)$
(D) $\l... | 3.A.
Since $A=\{x \mid 2<x<3\}$, and $A \cap B \neq \varnothing$, the equation $x^{2}-a x+5=0$ has a solution in $(2,3)$. The problem is converted to finding the range of $a=\frac{x^{2}+5}{x}=x+\frac{5}{x}$ in $(2,3)$.
Since $x+\frac{5}{x}$ is monotonically decreasing on $(2, \sqrt{5}]$ and monotonically increasing o... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,971 |
4. $x$ is a real number, the maximum value of the function $f(x)=4 \cos ^{3} x-3 \cos ^{2} x$ $-6 \cos x+5$ is ( ).
(A) 7
(B) $\frac{27}{4}$
(C) $\frac{25}{4}$
(D) 5 | 4.B.
$$
\begin{array}{l}
f(x)=(\cos x-1)\left(4 \cos ^{2} x+\cos x-5\right) \\
=(4 \cos x+5)(\cos x-1)^{2} \\
=2\left(2 \cos x+\frac{5}{2}\right)(1-\cos x)(1-\cos x) \\
\leqslant 2\left[\frac{\left(2 \cos x+\frac{5}{2}\right)+(1-\cos x)+(1-\cos x)}{3}\right]^{3} \\
=2\left(\frac{3}{2}\right)^{3}=\frac{27}{4},
\end{arra... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,972 |
5. In any permutation $a_{1}, a_{2}, \cdots$, $a_{7}$ of $1,2, \cdots, 7$, the number of permutations where any two adjacent numbers are coprime is ( ) kinds.
(A) 288
(B) 576
(C) 864
(D) 1152 | 5.C.
First, let the numbers $1, 3, 5, 7$ be fully permuted, which gives $\mathrm{A}_{4}^{4}=24$ ways; then arrange the number 6, since the number 6 should not be adjacent to 3, in the already arranged sequence, there are 5 gaps between 1, 3, 5, 7 and at the beginning and end, excluding the two gaps around 3, there are... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,973 |
6. Three spheres with a radius of 1 are mutually externally tangent, and each sphere is also externally tangent to two other spheres with a radius of $r$. If these two spheres with a radius of $r$ are also externally tangent to each other, then the value of $r$ is $(\quad)$.
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D... | 6.D.
Let $O_{1}, O_{2}, O_{3}$ be the centers of three spheres with radius 1, and $I_{1}, I_{2}$ be the centers of two spheres with radius $r$. Then they form a solid figure (as shown in Figure 3),
$H$ is the centroid of $\triangle O_{1} O_{2} O_{3}$. Since $\triangle O_{1} O_{2} O_{3}$ is an equilateral triangle wit... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,974 |
7. The minimum value of the function $y=\sqrt{x^{2}+2 x+2}+\sqrt{x^{2}-2 x+2}$ is $\qquad$ . | $$
\begin{array}{l}
\text { II.7.2 } \sqrt{2} . \\
\begin{aligned}
y \geqslant 2 \sqrt[4]{\left(x^{2}+2 x+2\right)\left(x^{2}-2 x+2\right)} \\
\quad=2 \sqrt[4]{x^{4}+4} \geqslant 2 \sqrt{2} .
\end{aligned}
\end{array}
$$
The equality holds if and only if $x=0$, that is, when $x=0$, $y$ attains its minimum value of $2 ... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,975 |
For example, in $\triangle A B C$, $\angle A B C=60^{\circ}$, point $P$ is inside the triangle. Given $P A=10, P B=6, P C=7$. Find the maximum value of the area of $\triangle A B C$. | Solution: As shown in Figure 3, construct $\square ABCD$, and draw $PQ \perp BC$, connect $DQ, CQ, PD$. It is easy to see that
$$
\begin{array}{l}
DQ = PA = 10, \\
CQ = PB = 6.
\end{array}
$$
Let $PD = x$.
To find $\left(S_{\triangle ABC}\right)_{\text{max}}$, it is sufficient to find $\left(S_{\square ABCD}\right)_{\... | 36 + 22\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,976 |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=$ $9 \sqrt[3]{a_{n}}(n \geqslant 1)$. Then $\lim _{n \rightarrow \infty} a_{n}=$ $\qquad$ . | 8.27.
Given $\log _{3} a_{n+1}=2+\frac{1}{3} \log _{3} a_{n}$, let $b_{n}=\log _{3} a_{n}$, then
$$
\begin{array}{l}
b_{n+1}=\frac{1}{3} b_{n}+2, b_{1}=1 \\
\Rightarrow b_{n+1}-3=\frac{1}{3}\left(b_{n}-3\right) \\
\Rightarrow b_{n}-3=\left(b_{1}-3\right)\left(\frac{1}{3}\right)^{n-1}=-2 \times\left(\frac{1}{3}\right)^... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,977 |
9. If $(2 x-1)^{8}=a_{8} x^{8}+a_{7} x^{7}+\cdots+a_{1} x$ $+a_{0}$, then $a_{8}+a_{6}+a_{4}+a_{2}=$ $\qquad$ | 9.3280 .
Let $f(x)=(2 x-1)^{8}$, then
$$
\begin{array}{l}
a_{0}=f(0)=1, \\
a_{8}+a_{7}+\cdots+a_{0}=f(1)=1, \\
a_{8}-a_{7}+a_{6}-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0} \\
=f(-1)=(-3)^{8}=6561 .
\end{array}
$$
Therefore, $a_{8}+a_{6}+a_{4}+a_{2}$
$$
=\frac{1}{2}(f(1)+f(-1))-f(0)=3280 \text {. }
$$ | 3280 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,978 |
10. Let the three equations concerning $x$ be
$$
\begin{array}{l}
x^{2}+2 x \sin A_{1}+\sin A_{2}=0, \\
x^{2}+2 x \sin A_{2}+\sin A_{3}=0, \\
x^{2}+2 x \sin A_{3}+\sin A_{1}=0
\end{array}
$$
all have real roots, $\angle A_{1}, \angle A_{2}, \angle A_{3}$ are three interior angles of a convex $(4 n+2)$-sided polygon $A... | $10.4 \pi$.
Since the interior angles of a convex $4 n+2$-sided polygon are all multiples of $30^{\circ}$, its interior angles can only be $30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}$, and the values of their sines can only be $\frac{1}{2}, \frac{\sqrt{3}}{2}, 1$. Also, because the three equations all... | 4 \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,979 |
11. In $\triangle A B C$, $B C=12$, the height $h_{a}=8$ on side $B C$, $h_{b}$ and $h_{c}$ are the heights on sides $C A$ and $A B$ respectively. Then the maximum value of the product $h_{b} h_{c}$ is $\qquad$. | 11. $\frac{2304}{25}$.
As shown in Figure 4, let $B C=a, C A=b, A B=c$. From the area relationship, we have $b h_{b}=c h_{c}=a h_{a}=b c \sin \angle B A C$.
Thus, $h_{b} h_{c}=\frac{a^{2} h_{a}^{2}}{b c}=a h_{a} \sin \angle B A C=96 \sin \angle B A C$.
Clearly, vertex $A$ lies on a line $l$ that is 8 units away from a... | \frac{2304}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,980 |
12. Given the ellipse $\frac{x^{2}}{4}+y^{2}=1$ and the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, where $F_{1} 、 F_{2}$ are the foci of the ellipse, and $P$ is an intersection point of the ellipse and the hyperbola. Then $\cos \angle F_{1} P F_{2}$ $=$ . $\qquad$ | 12. $-\frac{1}{3}$.
From the problem, we know that $F_{1}$ and $F_{2}$ are the common foci of the ellipse and the hyperbola, and $\left|F_{1} F_{2}\right|=2 \sqrt{3}$. According to the definitions of the ellipse and the hyperbola,
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=4, \quad || P F_{1}|-| P F_{2}||=2 \text {,... | -\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,981 |
13. As shown in Figure 1, take a point $P$ inside $\triangle A B C$ such that $\angle P B A = \angle P C A$. Draw $P D \perp A B$ at point $D$ and $P E \perp A C$ at point $E$. Prove that the perpendicular bisector of $D E$ must pass through the midpoint $M$ of $B C$. | Three, 13. As shown in Figure 5, let $L$ and $N$ be the midpoints of $PB$ and $PC$, respectively. Connect $MD$, $ME$, $ML$, $MN$, $DL$, and $EN$. Then,
$$
\begin{array}{l}
ML \perp \frac{1}{2} PC; \\
MN \perp \frac{1}{2} PB.
\end{array}
$$
$$
\text{Also, } \angle PDB = \angle PEC =
$$
$90^{\circ}$, thus,
$$
DL = \frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,982 |
14. Given the line $y=k x-1$ intersects the left branch of the hyperbola $x^{2}-$ $y^{2}=1$ at points $A$ and $B$. If another line $l$ passes through point $P(-2,0)$ and the midpoint $Q$ of segment $A B$, find the range of the y-intercept $b$ of line $l$. | 14. Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). From the problem, we establish the system of equations
\[
\left\{\begin{array}{l}
y=k x-1, \\
x^{2}-y^{2}=1
\end{array}\right.
\]
Eliminating \( y \) gives \(\left(1-k^{2}\right) x^{2}+2 k x-2=0\).
Given that the line intersects the left br... | (-\infty,-2) \cup(2+\sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,983 |
15. Given the sequence $\left\{a_{k}\right\}$ defined as follows:
$$
a_{k}=\frac{1}{2} \cdot \frac{3}{4} \cdots \cdots \cdot \frac{2 k-3}{2(k-1)} \cdot \frac{2 k-1}{2 k}(k=1,2, \cdots) \text {. }
$$
Prove: (1) $a_{n+1}<\frac{1}{\sqrt{2 n+3}}$;
(2) For any positive integer $n$, we have
$$
\sum_{k=1}^{n} a_{k}<\sqrt{2(n... | 15. (1) Let \( A = \frac{1}{2} \cdot \frac{3}{4} \cdots \cdots \cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2(n+1)} \),
\[
B = \frac{2}{3} \cdot \frac{4}{5} \cdots \cdots \frac{2n}{2n+1} \cdot \frac{2(n+1)}{2n+3} \text{. }
\]
Then \( A < B \). Therefore, \( A^2 < AB = \frac{1}{2n+3} \).
Thus, \( A < \frac{1}{\sqrt{2n+3}} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,984 |
16. Figure 2 is a street map composed of $3 \times 4$ unit squares, with lines representing roads. Person A starts from point $A(0,0)$ and walks to point $B(4,3)$ along the shortest path, while person B starts from point $B$ and walks to point $A$ along the shortest path. If they start at the same time and move at the ... | 16. Suppose A and B take $t=1$ to travel a unit length of a line segment.
As shown in Figure 6, it is not difficult to calculate that when A starts from point $A$ and $t=3$, the probabilities of reaching points $P_{1}(0,3)$, $P_{3}(1,2)$, $P_{5}(2,1)$, and $P_{7}(3,0)$ are
$$
\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \fr... | \frac{37}{256} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,985 |
1. In a season of football matches, the scoring rules are: 3 points for a win; 1 point for a draw; 0 points for a loss. A team has played 15 games and accumulated 33 points. If the order is not considered, the number of possible win, loss, and draw scenarios for the team is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | - 1.A.
Let the team win $x$ games, draw $y$ games, and lose $z$ games. Then $x$, $y$, and $z$ are non-negative integers, and satisfy
$$
\left\{\begin{array}{l}
x+y+z=15, \\
3 x+y=33 .
\end{array}\right.
$$
From equation (2), we get $y=3(11-x)$, substituting into equation (1) yields
$$
z=2(x-9) \text {. }
$$
Also, $0... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,986 |
2. As shown in Figure 1, if quadrilateral $A B C D$ is folded along $E F$, such that points $A$ and $B$ fall inside quadrilateral $E F C D$, there is a constant quantitative relationship between $\angle 1+\angle 2$ and $\angle A$, $\angle B$. Then, we have $(\quad$.
(A) $\angle 1+\angle 2=(\angle A+\angle B)-90^{\circ}... | 2. D.
As shown in Figure 4, let points $A$ and $B$ fall on points $A^{\prime}$ and $B^{\prime}$, respectively. Extend $E A$, $F B$, $E A^{\prime}$, and $F B^{\prime}$ to intersect at points $P$ and $P^{\prime}$, and connect $P P^{\prime}$. By symmetry, we have
$$
\begin{array}{l}
\angle E P P^{\prime}=\angle E P^{\pri... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,988 |
3. Given a quadratic equation $a x^{2}+b x+c=0$ $(a \neq 0, c \neq 0)$, the sum, difference, product, and quotient of its roots are $p$, $q$, $r$, and $s$, respectively. They have the following dependency relations:
(1) $p^{2}-q^{2}=4 r$,
(2) $\left(\frac{s+1}{p}\right)^{2}=\frac{s}{r}$,
(3) $\left(\frac{s-1}{q}\right)... | 3. D.
Let's assume the two roots of $a x^{2}+b x+c=0(a \neq 0, c \neq 0)$ are $x_{1}$ and $x_{2}$, then we have
$$
x_{1}+x_{2}=-\frac{b}{a}=p, x_{1} x_{2}=\frac{c}{a}=r .
$$
Thus, $|q|=\left|x_{1}-x_{2}\right|$
$$
=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{p^{2}-4 r} \text {, }
$$
which means $p^{2}-q^... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,989 |
4. Given that $a, b, c$ are distinct real numbers, and
$$
\begin{array}{l}
L=\frac{1}{(a-b)(a-c)}, M=\frac{1}{(b-a)(b-c)}, \\
N=\frac{1}{(c-a)(c-b)} .
\end{array}
$$
Then among the following conclusions:
(1) $a L+b M+c N=0$,
(2) $a^{2} L+b^{2} M+c^{2} N=1$,
(3) $a^{3} L+b^{3} M+c^{3} N=a+b+c$, the number of correct co... | 4.D.
Let $k=\frac{1}{(a-b)(b-c)(c-a)}$, then $L=k(c-b), M=k(a-c), N=k(b-a)$.
$$
\begin{array}{l}
\text { (1) } a L+b M+c N=k[a(c-b)+b(a-c)+c(b-a)] \\
=k \times 0=0 ;
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } a^{2} L+b^{2} M+c^{2} N=k\left[a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)\right] \\
=k \times \frac{1}{k}=1 ;
\end... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,990 |
6. If $a, b, c$ and $t_{a}, t_{b}, t_{c}$ represent the lengths of the three sides and the three angle bisectors of $\triangle ABC$, respectively, then among the following conclusions:
(1) $t_{a} t_{b} t_{c}>a b c$,
(2) $t_{a} t_{b} t_{c}=\sqrt{a b c}$,
(3) $t_{a} t_{b} t_{c}<a b c$,
(4) $t_{a} t_{b} t_{c}=a b c$, the ... | 6.D.
As shown in Figure 6, construct the circumcircle $\odot O$ of $\triangle ABC$, and let $AD$ be the angle bisector of $\angle BAC$. Extend $AD$ to intersect $\odot O$ at point $E$, and connect $BE$ and $CE$.
From $\triangle ABD \sim \triangle AEC$, we get $bc = t_a AE$.
Let $AE = x$, i.e., $bc = t_a x$.
Similarly,... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,992 |
1. Given $\frac{1949}{y^{2}}=\frac{2007}{x^{2}}$, and $\frac{1}{x}+\frac{1}{y}=1(x>$ $0, y>0)$. Then $\sqrt{1949 x+2007 y}=$ $\qquad$ | II. 1. $\sqrt{1949}+\sqrt{2007}$.
Since $\frac{1949}{y^{2}}=\frac{2007}{x^{2}}$, and $x>0, y>0$, therefore,
$\frac{y}{x}=\sqrt{\frac{1949}{2007}}$, which means $\frac{1}{x}=\sqrt{\frac{1949}{2007}} \cdot \frac{1}{y}$.
Also, $\frac{1}{x}+\frac{1}{y}=1$, thus, $\left(1+\sqrt{\frac{1949}{2007}}\right) \frac{1}{y}=1$.
Henc... | \sqrt{1949}+\sqrt{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,993 |
2. In $\triangle A B C$, $I$ is the incenter, and $C A+A I=$ $B C$. If $\angle A=80^{\circ}$, then $\angle B=$ $\qquad$ | $2.40^{\circ}$.
As shown in Figure 7, connect $B I$ and $C I$, and take a point $D$ on $B C$ such that $C D = C A$, then connect $I D$.
Given $C A + A I = B C$, then
$D B = A I$.
It is easy to prove $\triangle A C I \cong \triangle D C I$
$$
\begin{array}{l}
\Rightarrow A I = I D = D B, \angle D I B = \angle D B I \\
... | 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,994 |
3. If the real number $a$ satisfies $a^{3}+a^{2}-3 a+2=\frac{3}{a}-$ $\frac{1}{a^{2}}-\frac{1}{a^{3}}$, then $a+\frac{1}{a}=$ $\qquad$ | 3.2 or -3.
Given the equation can be transformed into
$$
a^{3}+\frac{1}{a^{3}}+a^{2}+\frac{1}{a^{2}}+2-3\left(a+\frac{1}{a}\right)=0 \text {. }
$$
Let $a+\frac{1}{a}=x$, then $x^{2}=a^{2}+\frac{1}{a^{2}}+2$,
$$
x^{3}=a^{3}+3 a+\frac{3}{a}+\frac{1}{a^{3}}=a^{3}+\frac{1}{a^{3}}+3 x \text {. }
$$
Therefore, equation (1... | 2 \text{ or } -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,995 |
4. As shown in Figure 2, given $A C=2 B C, D A=$ $D B, D A \perp D B, C B \perp$ $A B, A B=1$. Then the area of the shaded part in the figure is $\qquad$ . | 4. $\frac{9-4 \sqrt{3}}{12}$.
Draw $E F \perp A B$, with the foot of the perpendicular at $F$. Given $B C=\frac{1}{2} A C$, we know $\angle C A B=30^{\circ}$. Therefore, $B C=A B \tan 30^{\circ}=\frac{\sqrt{3}}{3}$.
Also, $A F=E F \cot 30^{\circ}=\sqrt{3} E F$, and $B F=E F$, so,
$$
\begin{array}{l}
A F+B F=(\sqrt{3}+... | \frac{9-4 \sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,996 |
Let $n$ be a positive integer, and let $f(n)$ denote the number of positive integers not exceeding $\sqrt{n}$ (for example, $f(3)=1$, $f(9)=3$).
(1) Find $f(2007)$;
(2) Find the positive integer $n$ such that it satisfies
$$
f(1)+f(2)+\cdots+f(n)=2009 .
$$ | (1) From $44<\sqrt{2007}<45$, we have $f(2007)=44$.
(2) When $k=1,2,3$, $f(k)=1$, then
$$
f(1)+f(2)+f(3)=1 \times 3 \text {. }
$$
When $k=4,5, \cdots, 8$, $f(k)=2$, then
$$
f(4)+f(5)+\cdots+f(8)=2 \times 5 \text {. }
$$
When $k=9,10, \cdots, 15$, $f(k)=3$, then
$$
f(9)+f(10)+\cdots+f(15)=3 \times 7 \text {. }
$$
Whe... | 215 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,997 |
1. The sequence of real numbers $a_{0}, a_{1}, \cdots$ is defined as follows: for all non-negative integers $i, a_{i+1}=\left[a_{i}\right]\left\{a_{i}\right\}$, where $a_{0}$ is any real number, $\left[a_{i}\right]$ denotes the greatest integer not greater than $a_{i}$, and $\left\{a_{i}\right\}=a_{i}-\left[a_{i}\right... | 1. If $a_{0} \geqslant 0$, then $a_{i} \geqslant 0$. For $a_{i} \geqslant 1$, we have $\left[a_{i+1}\right] \leqslant a_{i+1}=\left[a_{i}\right]\left\{a_{i}\right\}i)$, such that $a_{j}a_{i+1}=\left[a_{i}\right]\left\{a_{i}\right\}>\left[a_{i}\right]$.
Thus, the sequence $\left\{\left[a_{i}\right]\right\}$ is non-decre... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,998 |
II. (25 points) As shown in Figure 3, from a fixed point $P$ outside $\odot O$, two tangents are drawn to $\odot O$, touching $\odot O$ at points $S$ and $T$. Take any point $C$ on the minor arc $\overparen{S T}$, and draw a tangent to $\odot O$ through point $C$, intersecting $P S$ and $P T$ at points $A$ and $B$ resp... | As shown in Figure 8, take the midpoint $M$ of $AB$, connect $ME$ and $MF$, and extend $AE$ to intersect $PT$ at point $G$, and extend $BF$ to intersect $PS$ at point $H$.
Since $\angle APE = \angle GPE$, $PE \perp AE$, therefore,
$$
\begin{array}{c}
PA = PG, AE = EG. \\
\text{Hence } ME = \frac{1}{2} BG \\
= \frac{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,999 |
Three. (25 points) A frog starts from point $(1,1)$ in a Cartesian coordinate system and can jump in the following two ways:
(1) It can jump from any point $(a, b)$ to $(2a, b)$ or $(a, 2b)$.
(2) For point $(a, b)$, if $a > b$, it can jump from $(a, b)$ to $(a-b, b)$; if $a < b$, it can jump from $(a, b)$ to $(a, b-a)$... | (i) It can reach point (1) $(48,40)$ and point (4) $(200,8)$.
The path from $(1,1)$ to $(48,40)$ is
$$
\begin{array}{l}
(1,1) \rightarrow(2,1) \rightarrow(4,1) \rightarrow(3,1) \rightarrow(3,2) \rightarrow(3,4) \\
\rightarrow(3,8) \rightarrow(3,5) \rightarrow(6,5) \rightarrow(12,5) \rightarrow(24,5) \\
\rightarrow(24,1... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,000 |
1. Arrange positive integers as shown in Figure 1. Then the sum of the numbers in the $n$-th row $S_{n}$ is ( ).
(A) $\frac{n^{2}(n+1)}{2}$
(B) $\frac{n\left(n^{2}+1\right)}{2}$
(C) $\frac{n(n+1)^{2}}{2}$
(D) $\frac{n^{3}+1}{2}$ | -、1.B.
The last natural number in the $n$th row, which is the $\frac{n(n+1)}{2}$th number, is
$$
\frac{n(n+1)}{2}=a_{1}+(n-1) \text {. }
$$
Thus, $a_{1}=\frac{n(n+1)}{2}-n+1=\frac{n(n-1)}{2}+1$ is the first term of the $n$th row. Therefore,
$$
S_{n}=\frac{n}{2}\left[\frac{n(n+1)}{2}+\frac{n(n-1)}{2}+1\right]=\frac{n\l... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,001 |
2. In an acute $\triangle A B C$, let
$$
y=\frac{\tan A+\tan B+\tan C}{\sin A+\sin B+\sin C} \text {. }
$$
Then ( ).
(A) $0<y<1$
(B) $y \geqslant 2$
(C) $y \geqslant 3$
(D) $1 \leqslant y \leqslant 2$ | 2. B.
$$
\begin{array}{l}
\tan A + \tan B + \tan C \\
= \frac{\sin (B+C)}{\cos A} + \frac{\sin (C+A)}{\cos B} + \frac{\sin (A+B)}{\cos C} \\
= \left(\frac{\cos C}{\cos B} + \frac{\cos B}{\cos C}\right) \sin A + \\
\left(\frac{\cos A}{\cos C} + \frac{\cos C}{\cos A}\right) \sin B + \\
\left(\frac{\cos B}{\cos A} + \fra... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,002 |
4. As shown in Figure 2, the base edge length of the regular triangular prism $A B C-A_{1} B_{1} C_{1}$ is $a$, and the side edge length is $\sqrt{2} a$. Then the angle between $A C_{1}$ and the side face $A B B_{1} A_{1}$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 4. A.
Take the midpoint $D$ of $A_{1} B_{1}$, and connect $A D$ and $C_{1} D$.
Since $C_{1} D \perp A_{1} B_{1}$, we have $C_{1} D \perp$ plane $A B B_{1} A_{1}$. Therefore, the angle formed by $A C_{1}$ and $A D$ is the required angle.
Given $C_{1} D=\frac{\sqrt{3}}{2} a$ and $A D=\frac{3}{2} a$, we have
$$
\tan \ang... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,004 |
5. Let the function $f(x)=\lg \frac{\sum_{i=1}^{n-1} i^{x}+n^{x} a}{n}$, where $a \in \mathbf{R}$, $n$ is a given positive integer, and $n \geqslant 2$. If the inequality $f(x)>(x-1) \lg n$ has a solution in the interval $[1,+\infty)$, then the range of the real number $a$ is ( ).
(A) $a \leqslant \frac{1-n}{2}$
(B) $a... | 5. B.
$f(x)>(x-1) \lg n$ has a solution in $[1,+\infty]$
$\Leftrightarrow \lg \frac{\sum_{i=1}^{n-1} i^{x}+n^{x} a}{n}>\lg n^{x-1}$ has a solution in $[1,+\infty)$
$\Leftrightarrow \sum_{i=1}^{n-1} i^{x}+n^{x} a>n^{x}$ has a solution in $[1,+\infty)$
$\Leftrightarrow a>1-\sum_{i=1}^{n-1}\left(\frac{i}{n}\right)^{x}$ ha... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,005 |
6. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=a_{2}=1, a_{3}=2, \\
a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}(n \geqslant 1),
\end{array}
$$
and $a_{n} a_{n+1} a_{n+2} \neq 1$ for all positive integers $n$. Then the value of $S_{100}=\sum_{k=1}^{100} a_{k}$ is $(\quad)$.... | 6. B.
For any $n \geqslant 1$, we have
$$
\begin{array}{l}
a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}, \\
a_{n+1} a_{n+2} a_{n+3} a_{n+4} \\
=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4} .
\end{array}
$$
Subtracting (2) from (1) gives
$$
\begin{array}{l}
a_{n+1} a_{n+2} a_{n+3}\left(a_{n}-a_{n+4}\right)=a_{n}-a_{n... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,006 |
1. When $n \in \mathbf{N}$, and $n>2$, the graphs of the functions $y=\log _{n-1} x$ and $y=\log _{n}(x+1)$ intersect at all points where | 2. $(n-1,1)$.
$$
\left\{\begin{array} { l }
{ y = \operatorname { log } _ { n - 1 } x , } \\
{ y = \operatorname { log } _ { n } ( x + 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=(n-1)^{y}, \\
x+1=n^{y} .
\end{array}\right.\right.
$$
Eliminating $x$ yields
$$
\begin{array}{l}
1+(n-1)^{y}=n^{y} \\
\Lef... | (n-1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,007 |
Example 1 Given that $B C$ is the longest side of $\triangle A B C$, and $O$ is any point inside $\triangle A B C$, the lines $O A, O B, O C$ intersect the opposite sides at points $A_{1}, B_{1}, C_{1}$. Prove:
$$
\begin{array}{l}
\text { (1) } O A_{1}+O B_{1}+O C_{1}<B C \text {; } \\
\text { (2) } O A_{1}+O B_{1}+O C... | Proof: (1) As shown in Figure 1, draw $OX \parallel AB$ and $OY \parallel AC$, intersecting $BC$ at points $X$ and $Y$ respectively. Then draw $XS \parallel CC_1$ and $YT \parallel BB_1$, intersecting $AB$ and $AC$ at points $S$ and $T$ respectively.
Since $\triangle OXY \sim \triangle ABC$, then $XY$ is the longest s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,009 |
Example 2 As shown in Figure 2, in $\triangle A B C$, $\angle B=$ $2 \angle C$. Prove:
$$
A C<2 A B \text {. }
$$ | Prove: Extend $C B$ to
$D$, such that $D B=A B$. Then we have
$$
\angle D=\angle B A D, \angle A B C=2 \angle D \text {. }
$$
By the given condition, $\angle A B C=2 \angle C$. Therefore,
$$
\angle D=\angle C \text {. }
$$
Thus, $A D=A C$.
In $\triangle A B C$, since $D B+A B>A D$, which means $2 A B>A D$, so, $A C<2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,010 |
Example 4 Prove: For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ $(n \geqslant 2)$, we have $\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}} \geqslant \frac{n}{n-1}$, where, $S=$ $\sum_{i=1}^{n} a_{i}$
(1976, British Mathematical Olympiad) | $\begin{array}{l}\text { Prove: } \sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}}=\sum_{i=1}^{n}\left(\frac{a_{i}}{S-a_{i}}+1\right)-n \\ =S\left(\frac{n}{S n-S}\right)^{2}\left[\sum_{i=1}^{n} \frac{\left(\frac{S n-S}{n}\right)^{2}}{S-a_{i}}\right]-n \\ \geqslant S\left(\frac{n}{S n-S}\right)^{2}\left\{\sum_{i=1}^{n}\left[2\left(... | \frac{n}{n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,011 |
4. As shown in Figure 4, given that the radii of circles $\odot O_{1}$ and $\odot O_{2}$, which intersect at points $A$ and $B$, are $5$ and $7$ respectively, and $O_{1} O_{2}=6$. A line through point $A$ intersects the two circles at points $C$ and $D$. $P$ and $O$ are the midpoints of line segments $CD$ and $O_{1} O_... | $4.2 \sqrt{7}$.
As shown in Figure 9, draw the diameter $A E$ of $\odot O_{1}$ through point $A$, connect $E B$ and extend it to intersect $\odot O_{2}$ at point $F$, connect $A O$ and extend it to intersect $E F$ at point $G$, connect $A B$, $C E$, $D F$, and $P G$. Let $A B$ intersect $O_{1} O_{2}$ at point $H$. Sinc... | 2 \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,012 |
One, (20 points) A certain musical instrument has 10 holes, sequentially labeled as hole 1, hole 2, $\cdots \cdots$, hole 10. When played, the sound quality index $D$ of the $n$-th hole is given by the relationship
$$
D=n^{2}+k n+90 .
$$
The lowest sound quality index of the instrument is $4 k+106$. Find the range of ... | For the parabola $D=n^{2}+k n+90$, the axis of symmetry is $n=-\frac{k}{2}$.
(1) When $-\frac{k}{2} \leqslant 1$, i.e., $k \geqslant-2$, we have
$$
n=1, D=4 k+106 \text {. }
$$
Thus, $1^{2}+k+90=4 k+106$.
Solving this, we get $k=-5$ (which does not meet the condition).
(2) When $-\frac{k}{2} \geqslant 10$, i.e., $k \l... | -9 \leqslant k \leqslant-7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,013 |
II. (25 points) As shown in Figure 5, $H$ is the orthocenter of $\triangle ABC$ inscribed in $\odot O$, $\angle A > 90^{\circ}$. The circle $\odot O_{1}$ with diameter $CH$ intersects $\odot O$ at another point $D$, and the extension of $HD$ intersects $AB$ at point $M$. Prove that $M$ is the midpoint of $AB$. | II. As shown in Figure 10, extend $DM$ to intersect $\odot O$ at point $E$, and connect $AH$, $AE$, $BE$, $BH$, $CD$, and $CE$.
Since $CH$ is the diameter of $\odot O_1$
$$
\Rightarrow \angle CDH = \angle CDE = 90^{\circ}
$$
$\Rightarrow CE$ is the diameter of $\odot O$
$$
\begin{array}{l}
\Rightarrow \angle CAE = \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,014 |
Three, (25 points) Given three natural numbers $a, b, c$ where at least $a$ is a prime number, and they satisfy
$$
\left\{\begin{array}{l}
(4 a+2 b-4 c)^{2}=443(2 a-442 b+884 c), \\
\sqrt{4 a+2 b-4 c+886}-\sqrt{42 b-2 a+2 c-443}=\sqrt{443} .
\end{array}\right.
$$
Find the value of $a b c$. | Let $x=\frac{4 a+2 b-4 c}{443}, y=\frac{2 a-442 b+884 c}{443}$. Then
$4 a+2 b-4 c=443 x$,
$2 a-442 b+884 c=443 y$.
Hence $x^{2}=y$.
From equation (1), we know that $(4 a+2 b-4 c)^{2}$ is divisible by 443. Since 443 is a prime number, 443 divides $(4 a+2 b-4 c)$. Therefore, $x$ is an integer.
From equation (5), we know ... | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,015 |
1. The maximum value of $y=\sin \left(\frac{\pi}{3}+x\right)-\sin 3 x$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) 1
(C) $\frac{16 \sqrt{3}}{25}$
(D) $\frac{8 \sqrt{3}}{9}$ | 1.D.
$$
\begin{array}{l}
\text { Let } \frac{\pi}{3}+x=t \text {, then } x=t-\frac{\pi}{3}, \\
\sin 3 x=\sin (3 t-\pi)=-\sin 3 t, \\
y=\sin \left(\frac{\pi}{3}+x\right)-\sin 3 x=\sin t+\sin 3 t \\
=4 \sin t\left(1-\sin ^{2} t\right) .
\end{array}
$$
Thus, $y^{2}=16 \sin ^{2} t\left(1-\sin ^{2} t\right)\left(1-\sin ^{2... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,016 |
2. As shown in Figure 1, the graph of the even function $f(x)$ is shaped like the letter $\mathrm{M}$, and the graph of the odd function $g(x)$ is shaped like the letter N. If the number of real roots of the equations
$$
\begin{array}{l}
f(f(x))=0, f(g(x))=0, \\
g(g(x))=0, g(f(x))=0
\end{array}
$$
are $a$, $b$, $c$, a... | 2. B.
Obviously, $f(x)=0$ has 3 roots $0, \pm p(1<p<2)$, $g(x)=0$ has 3 roots $0, \pm q(0<q<1)$.
The roots of the equation $f(f(x))=0$ are the roots of the equations $f(x)=0, \pm p$, among which, $f(x)=0$ has 3 roots, and $f(x)=\pm p(1<p<2)$ has no real roots, so $a=3$;
The roots of the equation $f(g(x))=0$ are the ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,017 |
3. Given a parallelepiped $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with volume $V$, and points $P$, $Q$, $R$ are the midpoints of edges $A^{\prime} B^{\prime}$, $C C^{\prime}$, $C D$ respectively. Then the volume of the tetrahedron $A-P Q R$ is ( ).
(A) $\frac{V}{12}$
(B) $\frac{V}{16}$
(C) $\frac{V}{24}$
... | 3. C.
Assume the section $P Q A$ intersects the plane $A B C D$ at $A E$ (point $E$ is on $C D$). Then, by the planes $A B B^{\prime} A^{\prime} \parallel$ plane $D C C^{\prime} D^{\prime}$, we have $Q E \parallel P A$. Therefore, draw a line through point $Q$ parallel to $P A$ intersecting $C D$ at point $E$, and ext... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,018 |
4. Let $H$ be a point on the plane of $\triangle A B C$, and let $a, b, c, h$ represent the vectors $O A, O B, O C, O H$ respectively. If $\boldsymbol{a} \cdot \boldsymbol{b}+\boldsymbol{c} \cdot \boldsymbol{h}=\boldsymbol{b} \cdot \boldsymbol{c}+\boldsymbol{a} \cdot \boldsymbol{h}=\boldsymbol{c} \cdot \boldsymbol{a}+\... | 4.D.
From $a \cdot b+c \cdot h=b \cdot c+a \cdot h$, we get $\boldsymbol{a} \cdot b+c \cdot h-b \cdot c-a \cdot h=0$, which is $(a-c) \cdot(b-h)=0$.
Therefore, $C A \cdot H B=0, H B \perp C A$.
Similarly, $\boldsymbol{H A} \perp \mathrm{BC}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,019 |
5. Let $n \in \mathbf{N}, r>0$. Then the number of complex roots of the equation
$$
x^{n+1}+r x^{n}-r^{n+1}=0
$$
with modulus $r$ is ( ).
(A) 0
(B) 1
(C) 2
(D) greater than 2 | 5.A.
The equation is transformed into
$$
x^{n}(x+r)=r^{n+1} \text {. }
$$
Since $|x|=r$, taking the modulus of both sides of equation (1) yields $|x+r|=r$. The complex roots of the original equation with modulus $r$ are the complex numbers corresponding to the intersection points of the two circles $|x|=r$ and $|x+r|... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,020 |
6. There are $p+q$ parts, among which $p$ are good and $q$ are defective. Now, they are checked one by one at random. The probability that all defective parts are found exactly when checking the $r$-th part ($q<p<r<p+q$) is ( ).
(A) $\frac{\mathrm{C}_{q}^{1} \mathrm{~A}_{r-1}^{q-1} \mathrm{~A}_{p}^{r-q}}{\mathrm{~A}_{p... | 6. D.
The problem is equivalent to:
Randomly selecting $r$ numbers from $1,2, \cdots, p$ and $-1,-2, \cdots,-q$ and arranging them in a row, find the probability that the resulting permutation satisfies the following conditions: either the permutation contains $q$ negative numbers, and the last number is negative (in ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,021 |
Example 5 Let $n \in \mathbf{N}_{+}, a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, and $\sum_{i=1}^{n} a_{i}=1, \sum_{i=1}^{n} b_{i}=1$. Find the minimum value of $\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}$.
(2004, French Team Selection Exam) | Solution: By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}\right) \\
\geqslant\left[\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}+b_{i}}} \sqrt{a_{i}+b_{i}}\right]^{2} \\
=\left(\sum_{i=1}^{n} a_{i}\right)^{2}=1 ... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,022 |
1. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=2008$, for $n>1$, $a_{1}+a_{2}+\cdots+a_{n}=(1+2+\cdots+n) a_{n}$. Then $a_{2007}=$ $\qquad$ | $\begin{array}{l}\text { 2 } 1 \cdot \frac{6}{2009} \cdot \\ \text { Given } a_{1}+a_{2}+\cdots+a_{n}=(1+2+\cdots+n) a_{n} \\ \Rightarrow a_{1}+a_{2}+\cdots+a_{n-1}=(1+2+\cdots+n-1) a_{n-1} \\ \Rightarrow a_{n}=(1+2+\cdots+n) a_{n}-(1+2+\cdots+n-1) a_{n-1} \\ \quad=\frac{n(n+1)}{2} a_{n}-\frac{n(n-1)}{2} a_{n-1} \\ \Ri... | \frac{6}{2009} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,023 |
3. In rectangle $A B C D$, $A B=4, A D=3$, fold the rectangle along diagonal $A C$ so that the projection of point $D$ on plane $A B C$, point $E$, falls on line $A B$. At this moment, the dihedral angle $B-A C-D$ is $\qquad$ | 3. $\arccos \frac{9}{16}$.
As shown in Figure 2, draw $E G \perp A C$
at point $G$, and $B H \perp A C$ at point $H$, and connect $D G$.
By the theorem of three perpendiculars, we get
$D G \perp A C$.
Thus, $\angle E G D$ is the plane angle of the dihedral angle $B-A C-D$.
Since $B H / / E G$, the size of $\angle E ... | \arccos \frac{9}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,025 |
4. Given a positive number $m$, if the intersection locus of the circle family $x^{2}+y^{2}=r^{2}$ and the line family $m x+m y=m+r$ is an ellipse, then the range of values for $m$ is $\qquad$ . | $4.0<m<\frac{\sqrt{2}}{2}$.
Let the intersection point be $P(x, y)$. Then
$$
x^{2}+y^{2}=r^{2}=(m x+m y-m)^{2} \text {. }
$$
Therefore, $\frac{\sqrt{x^{2}+y^{2}}}{|x+y-1|}=m$, which means
$$
\frac{\frac{\sqrt{x^{2}+y^{2}}}{|x+y-1|}}{\sqrt{2}}=\sqrt{2} m \text {. }
$$
Thus, the ratio of the distance from point $P$ to ... | 0<m<\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,026 |
5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ . | 5.1.
Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in th... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,027 |
6. Let $f(x)=a x-x^{3}\left(a \in \mathbf{R}, x \in\left[\frac{1}{2}, 1\right]\right)$ with its graph denoted as $C$. If any line with a slope no less than 1 intersects $C$ at most once, then the range of values for $a$ is $\qquad$ | 6. $a<\frac{7}{4}$.
Let $M\left(p, a p-p^{3}\right)$ and $N\left(q, a q-q^{3}\right)\left(p, q \in\left[\frac{1}{2}, 1\right]\right)$ be any two different points on the curve $C$, and the slope of the line $MN$ is $k$. According to the condition, for any $p \neq q, k<1$ always holds.
Since $k=a-\left(p^{2}+q^{2}+p q\... | a<\frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,028 |
Three, (20 points) Let $a, b$ be complex numbers, $0 \leqslant p \leqslant 1$. Prove: $|a+b|^{p} \leqslant|a|^{p}+|b|^{p}$. | For $p=0, p=1$, the inequality obviously holds.
For $0 < p < 1$, we have
$$
|a+b|^{p} \leqslant (|a|+|b|)^{p} \leqslant |a|^{p}+|b|^{p} \text{. }
$$
The inequality also holds.
Finally, if $a+b=0$, then
$$
|a|^{p}+|b|^{p} \geqslant 0=|a+b|^{p} \text{. }
$$
The inequality also holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,029 |
Four. (20 points) Let the line $y=\sqrt{3} x+b$ intersect the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$. The circle passing through $A$ and $B$ intersects the parabola $y^{2}=2 p x(p>0)$ at two other distinct points $C$ and $D$. Find the size of the angle between the lines $AB$ and $CD$.
| Let points $A, B, C, D$ lie on a circle with the equation
$$
x^{2}+y^{2}+d x+e y+f=0 \text {, }
$$
Then the coordinates of $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$ are the solutions to the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,030 |
Five. (20 points) Let $a_{1}=\frac{1}{2}$,
$$
a_{n+1}=\frac{a_{n}}{(1-\sqrt{2})^{n+1} a_{n}+\sqrt{2}+1}(n=1,2, \cdots) \text {. }
$$
Find $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}$. | Consider a more general sequence
$$
a_{1}=\frac{1}{p+q}, a_{n+1}=\frac{a_{n}}{p^{n+1} a_{n}+q} \text {. }
$$
Then $\frac{1}{a_{n+1}}=\frac{p^{n+1} a_{n}+q}{a_{n}}=p^{n+1}+\frac{q}{a_{n}}$.
So, $\frac{1}{a_{n+1}}-\frac{q}{a_{n}}=p^{n+1}$.
Therefore, $\left\{\frac{1}{a_{n+1}}-\frac{q}{a_{n}}\right\}$ is a geometric sequ... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,031 |
一、(50 points) Let $\triangle A B C$ be inscribed in the unit circle $\odot O$, and the center $O$ is inside $\triangle A B C$. If the projections of $O$ onto sides $B C$, $C A$, and $A B$ are points $D$, $E$, and $F$ respectively, find the maximum value of $O D+O E+O F$. | As shown in Figure 3, since $O D \perp B C$ and $O E \perp A C$, it follows that $D$ and $E$ are the midpoints of $B C$ and $A C$, respectively. Therefore, $D E = \frac{A B}{2}$, and points $O, D, C, E$ are concyclic.
By Ptolemy's theorem, we have
$$
O D \cdot \frac{A C}{2} + O E \cdot \frac{B C}{2} = O C \cdot \frac{A... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,032 |
Example 6 Let $a, b, c \geqslant 0, ab + bc + ca = \frac{1}{3}$. Prove: $\square$
$$
\frac{1}{a^{2} - bc + 1} + \frac{1}{b^{2} - ca + 1} + \frac{1}{c^{2} - ab + 1} \leqslant 3.
$$
(2005, China National Training Team Test) | Proof: Let $M=a+b+c, N=ab+bc+ca$. Then the original inequality is equivalent to
$$
\begin{array}{l}
\sum \frac{1}{a^{2}-bc+N+2N} \leqslant \frac{1}{N} \\
\Leftrightarrow \sum \frac{N}{aM+2N} \leqslant 1 \\
\Leftrightarrow \sum\left(\frac{-N}{aM+2N}+\frac{1}{2}\right) \geqslant -1+\frac{3}{2} \\
\Leftrightarrow \sum \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,033 |
II. (50 points) Given a finite arithmetic sequence of positive integers $a_{1}, a_{2}, \cdots, a_{k}$ with a common difference greater than 0, where $a_{1}+a_{k}$ is a prime number. Two players, A and B, take turns removing stones from a pile of $n$ stones, with the rule that each player can take $a_{i}(1 \leqslant i \... | (1) When $r=0$, the strategy for player B is: If player A takes $a_{i}$ stones, then by the property of arithmetic sequences, there exists $a_{j}$ such that $a_{i}+a_{j}=a_{1}+a_{k}$. Player B takes $a_{j}$ stones. This way, B ensures that the number of stones left after his turn is always congruent to 0 modulo $a_{1}+... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,034 |
Three. (50 points) Let $f_{1}(x)=x+[\sqrt[k]{x}]$, where $k$ is a positive integer greater than 1, and
$$
f_{n}(x)=f_{1}\left(f_{n-1}(x)\right)(n \geqslant 2) \text {. }
$$
Prove: For every fixed positive integer $m$, the sequence $\left\{f_{n}(m)\right\}(n=1,2, \cdots)$ contains at least one integer that is a $k$-th ... | For any positive integer $m$, let's assume
$$
A^{k} \leqslant ma_{1}$, such that
$$
f_{a_{2}-1}(m)<(A+2)^{k},
$$
and $f_{a_{2}}(m) \geqslant(A+2)^{k}$, i.e.,
$$
(A+2)^{k} \leqslant f_{a_{2}}(m)<(A+1)+(A+2)^{k} .
$$
Let $f_{a_{2}}(m)=(A+2)^{k}+q(q \in \mathbf{N})$.
If $q=0$, then the proposition is proved.
If $q \neq ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,035 |
Initially 211, try to find the range of real number $m$ that satisfies the existence of two real numbers $a, b$, such that $a \leqslant 0, b \neq 0$, and
$$
\frac{a-b+\sqrt{a b}}{m a+3 b+\sqrt{a b}}=\frac{1}{m} \text {. }
$$ | Solution: First, it must be that $m \neq 0$.
Below, we discuss the cases for $a$.
(1) If $a=0$, then $\frac{-b}{3 b}=\frac{1}{m}(m \neq 0)$.
Since $b \neq 0$, we have $-\frac{1}{3}=\frac{1}{m}$, which means $m=-3$.
(2) If $a \neq 0$, then from $a \leqslant 0$ we get $a0$.
Thus, the given equation becomes $\frac{a-t a-... | -3 \leqslant m < -5 + 2\sqrt{5} \text{ or } -5 + 2\sqrt{5} < m < 0 \text{ or } 0 < m < 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,036 |
Initially 212 It is known that there is a four-digit number, and the sum of this four-digit number and the sum of its digits is 2008. Find this four-digit number and explain the reason.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result dire... | Solution: Let the four-digit number be $\overline{a b c d}$. According to the problem,
$$
1000 a+100 b+10 c+d+a+b+c+d=2008 \text {, }
$$
which simplifies to $1001 a+101 b+11 c+2 d=2008$.
Obviously, when $a \geqslant 3$, $1001 a>3000>2008$, so $a \geqslant 3$ is impossible. Therefore, $a=1$ or $a=2$.
(1) When $a=1$, su... | 1985 \text{ or } 2003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,037 |
Given $a, b, c$ are positive numbers, and $a+b+c=1$. Prove:
$$
\left(1+\frac{a^{3}}{1+a}\right)^{a}\left(1+\frac{b^{3}}{1+b}\right)^{b}\left(1+\frac{c^{3}}{1+c}\right)^{c} \geqslant \frac{37}{36} .
$$ | Prove: First, prove
$$
G(x)=\ln \left(1+\frac{x^{3}}{1+x}\right)-\frac{33}{148}\left(x-\frac{1}{3}\right)-\ln \left(\frac{37}{36}\right) \geqslant 0.
$$
Notice that
$$
\begin{array}{l}
G^{\prime}(x)=\frac{2 x^{3}+3 x^{2}}{\left(1+x+x^{3}\right)(1+x)}-\frac{33}{148} \\
=\frac{(3 x-1)\left(-11 x^{3}+84 x^{2}+165 x+33\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,038 |
212 Proof: There exist infinitely many positive integers $N$, where the unit digit of $N$ is not $0$, $N=n^{2}=\overline{A_{1} A_{2} A_{3} A_{4}}$, and $A_{1}, A_{2}, A_{3}, A_{4}$ these 4 numbers are all perfect squares with their first digit not being 0. | Proof: To find a sequence of square numbers that meet certain conditions, first find a number composed of two square numbers (it is not required that this number itself is a perfect square), and this number must be even (it must be a multiple of 4), such as 436, which is composed of 4 and 36, both perfect squares.
Belo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,039 |
Example 7 Let positive numbers $a, b, c$ satisfy $abc=1, n \in \mathbf{N}_{+}$. Prove:
$$
\frac{c^{n}}{a+b}+\frac{b^{n}}{c+a}+\frac{a^{n}}{b+c} \geqslant \frac{3}{2} .
$$
(2004, German Team Selection Exam) | Proof: Without loss of generality, let $a \geqslant b \geqslant c$. Then
$$
a^{n-1} \geqslant b^{n-1} \geqslant c^{n-1} \text {, }
$$
and $\frac{a}{b+c} \geqslant \frac{b}{c+a} \geqslant \frac{c}{a+b}$.
By the rearrangement inequality, we have
$$
\frac{c^{n}}{a+b}+\frac{b^{n}}{c+a}+\frac{a^{n}}{b+c}
$$
$\geqslant \fra... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,040 |
Example 8 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ all be positive numbers, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}} .
$$
(4th National High School Mathematics Winter Camp) | Proof: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$. Clearly, we have
$$
\frac{1}{\sqrt{1-x_{1}}} \leqslant \frac{1}{\sqrt{1-x_{2}}} \leqslant \cdots \leqslant \frac{1}{\sqrt{1-x_{n}}} .
$$
Using Chebyshev's inequality and applying the power mean inequality in the direction ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,041 |
Example 10 Let $a, b, c, d$ be positive real numbers, and satisfy $abcd=1$. Prove:
$$
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1 .
$$ | $$
\begin{array}{l}
\text { In fact, equation (1) is equivalent to } \\
(1+a b)\left[(1+a)^{2}+(1+b)^{2}\right] \\
\geqslant(a+b+a b+1)^{2} \\
\Leftrightarrow(1+a b)\left[2+2(a+b)+a^{2}+b^{2}\right] \\
\geqslant(a+b+a b+1)^{2} \\
\Leftrightarrow 2(1+a b)+2(1+a b)(a+b)+ \\
a b\left(a^{2}+b^{2}\right)+a^{2}+b^{2} \\
\geq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,043 |
1. Let positive numbers $a, b, c$ satisfy $abc=1$. Prove:
$$
\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} \geqslant 1 \text{. }
$$ | $$
\begin{array}{l}
\text { (Hint: } \frac{1}{1+2 a} \geqslant \frac{a^{-\frac{2}{3}}}{a^{-\frac{2}{3}}+b^{-\frac{2}{3}}+c^{-\frac{2}{3}}} \\
\Leftrightarrow a^{-\frac{2}{3}}+b^{-\frac{2}{3}}+c^{-\frac{2}{3}} \geqslant a^{-\frac{2}{3}}+2 a^{\frac{1}{3}} \\
\Leftrightarrow b^{-\frac{2}{3}}+c^{-\frac{2}{3}} \geqslant 2(b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,045 |
2. Prove that for all real numbers $y_{1}, y_{2}, x_{1}, x_{2}$, and $z_{1}, z_{2}$ satisfying
$$
x_{1}>0, x_{2}>0, x_{1} y_{1}-z_{1}^{2}>0, x_{2} y_{2}-z_{2}^{2}>0
$$
the inequality
$$
\begin{array}{l}
\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}} \\
\leqslant \frac{1}{x_{1} ... | (Let $x_{1} y_{1}-z_{1}^{2}=a>0, x_{2} y_{2}-z_{2}^{2}=b>0$. Then $x_{1} y_{1}=z_{1}^{2}+a, x_{2} y_{2}=b+z_{2}^{2}>0$. Thus,
$$
\begin{array}{l}
\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2} \\
=(\sqrt{a}+\sqrt{b})^{2}+\left(\sqrt{\frac{x_{1}}{x_{2}} b}-\sqrt{\frac{x_{2}}{x_{1}} a}\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,046 |
3. Let $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1$. Prove:
$$
\sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \geqslant \frac{\left(n^{2}+1\right)^{2}}{n} .
$$ | $\begin{array}{l}\text { (Hint: } \sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \\ =\frac{n^{2}+1}{n} \sum_{i=1}^{n}\left[\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \frac{n}{n^{2}+1}\right] \\ \geqslant \frac{n^{2}+1}{n} \sum_{i=1}^{n}\left[2\left(a_{i}+\frac{1}{a_{i}}\right)-\frac{n^{2}+1}{n}\right] \\ =\frac{n^{2}... | \frac{\left(n^{2}+1\right)^{2}}{n} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,048 |
4. Let positive numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) .
$$ | (Tip: The original inequality is equivalent to
$$
\begin{array}{l}
\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \geqslant \frac{3}{2}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
\Leftrightarrow \frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)} \geqslant \frac{3}{2} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,049 |
5. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the side lengths of a quadrilateral with perimeter $2 S$. Prove:
$$
\sum_{i=1}^{4} \frac{1}{a_{i}+S} \leqslant \frac{2}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\sqrt{\left(S-a_{i}\right)\left(S-a_{j}\right)}} .
$$ | ( Hint: Just prove
$$
\sum_{i=1}^{4} \frac{1}{a_{i}+S} \leqslant \frac{4}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{2 S-a_{i}-a_{j}} .
$$
Let $a_{1}=a, a_{2}=b, a_{3}=c, a_{4}=d$, the above is equivalent to
$$
\begin{array}{l}
\frac{2}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{a+d}+\frac{1}{b+c}+\frac{1}{b+d}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,050 |
Question: Let $f(x)=\sin (\omega x+r \pi)$, where the constant $r \in$ Q. If $\omega \in \mathbf{Q}$ such that for any $n \in \mathbf{Z}$, $f(x)$ attains at least one maximum and one minimum value in the interval $[n, n+1]$, then what condition should $\omega$ satisfy? | Lemma Let $a \in \mathbf{R}, a \neq 0$, for any $M \geqslant 1$, there exists $m_{0} \in \mathbf{N}_{+}, m_{0} \leqslant[M], n_{0} \in \mathbf{Z}$, such that
$$
\left|m_{0} a-n_{0}\right|0$, let $m_{0}=j-i, n_{0}=[j a]-[i a]$;
if $j-i1$, let $T=1+d$, then $d>0$.
Without loss of generality, assume $\omega>0$.
Notice th... | |\omega| \geqslant 2 \pi | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,052 |
1. There are $n(n \geqslant 2)$ lamps $L_{1}, L_{2}, \cdots, L_{n}$, which are either on or off. We change the state of some lamps simultaneously every second according to the following method: if $L_{i}(i=1,2, \cdots, n)$ and its adjacent lamps (when $i=1$ or $i=n$, only one lamp is adjacent, otherwise two lamps are a... | 1. (1) Let $n=2^{k}$.
Let $A_{k}$ be a $2^{k} \times 2^{k}$ matrix, where each element is either 0 or 1 (0 indicates the light is off, 1 indicates the light is on). Each row represents the state of $n$ lights at a certain moment. The first row is the initial state $(1, \underbrace{0,0, \cdots, 0}_{2^{k}-1 \uparrow})$,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,053 |
3. Let $S$ be a finite set of points in the plane, no three of which are collinear. For each convex polygon $P$ with vertices in $S$, let $a(P)$ be the number of vertices of $P$, and $b(P)$ be the number of points in $S$ that are outside $P$. Prove: For any real number $x, \sum_{P} x^{a(P)}(1-x)^{b(P)}=1$, where $P$ ra... | 3. Let $S$ contain $n$ points, and for each convex polygon $P$ with vertices in $S$, let $c(P)$ be the number of points in $P$ that belong to $S$. Thus, $a(P)+b(P)+c(P)=n$.
Let $1-x=y$, then we have
$$
\begin{array}{l}
\sum_{P} x^{a(P)}(1-x)^{b(P)}=\sum_{P} x^{a(P)} y^{b(P)} \\
=\sum_{P} x^{a(P)} y^{b(P)}(x+y)^{c(P)} \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,054 |
6. There is an upward-pointing equilateral triangular cardboard $T$ with side length $n$, which contains $n$ upward-pointing unit equilateral triangular holes. A diamond is a rhombus with internal angles of $60^{\circ}$ and $120^{\circ}$ and side length 1. Prove: $T$ can be tiled by diamonds if and only if each upward-... | 6. Necessity. $T^{\prime}$ is an equilateral triangle of side length $k$ in $T$, with $h$ holes inside. Then each diamond can cover one or two unit equilateral triangles in $T^{\prime}$. Let the diamond-covered $T^{\prime}$ be $R$.
Since the boundary of $T$ consists of upward-pointing unit equilateral triangles, $R$ i... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,057 |
Example 4 Given a convex hexagon $A B C D E F$ with all side lengths equal to 1. Prove: at least one of the diagonals $A D, B E, C F$ does not exceed 2. | Proof: As shown in Figure 4, since
$$
\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 720^{\circ},
$$
we can assume without loss of generality that
$$
\angle A + \angle F \leqslant \frac{720^{\circ}}{3} = 240^{\circ}.
$$
Construct rhombus $ABGF$, then
$$
\angle GFE \leqslant 60^{\circ}, \quad FG = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,058 |
8.1. Given real numbers $a$, $b$, $c$. Prove: at least one of the following three equations
$$
\begin{array}{l}
x^{2}+(a-b) x+(b-c)=0, \\
x^{2}+(b-c) x+(c-a)=0, \\
x^{2}+(c-a) x+(a-b)=0,
\end{array}
$$
has a solution. | 8.1. Since $(b-c)+(c-a)+(a-b)=0$, at least one of the three addends is non-positive. Without loss of generality, let's assume $b-c \leqslant 0$. Then, the discriminant of the first equation is
$$
\Delta=(a-b)^{2}-4(b-c) \geqslant 0 .
$$
Thus, the equation has real roots. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,060 |
8.2. Write the 100 natural numbers from 1 to 100 in a $10 \times 10$ grid, with one number in each cell. In each operation, you can swap the positions of any two numbers. Prove that it is possible to ensure that the sum of any two numbers in cells that share a common edge is a composite number after just 35 operations. | 8.2. A vertical straight line $m$ divides the grid into two halves. In one of the halves, there are no more than 25 even numbers, let's assume it is the right half. Thus, in the left half, there is an equal number of odd numbers. By swapping the even numbers in the right half with the odd numbers in the left half one b... | 35 | Number Theory | proof | Yes | Yes | cn_contest | false | 719,061 |
8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$. | 8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.
From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,062 |
8.4. A magician and his assistant perform the following act: First, the assistant draws a circle on the blackboard, and the audience marks 2007 distinct points on the circle. Then, the assistant erases one of these points. After that, the magician comes on stage, observes the circle on the blackboard, and points out th... | 8.4. Give an example of one of the conventions.
Consider the 2007 segments of the circumference divided by 2007 points. Suppose $\overparen{A B}$ is the longest segment (if there are multiple longest segments, choose any one), and assume $\overparen{A B}$ is in the clockwise direction from point $A$ (i.e., in the coun... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,063 |
8.5. The distance from Maikop to Belorechensk is $24 \mathrm{~km}$. Three friends need to travel between the two places: two of them from Maikop to Belorechensk, and the third from Belorechensk to Maikop. The three friends have only one bicycle. Initially, the bicycle is in Maikop. Each person can walk (walking speed i... | 8.5. At first, A and B start from Maykop towards Belaya Rechenskaya, A on a bicycle, and B on foot; while C starts from Belaya Rechenskaya towards Maykop. After $\frac{24}{18+6}=1 \mathrm{~h}$, A and C meet, and A hands over the bicycle to C. At this point, B is $6 \mathrm{~km}$ away from Maykop. He can stop to rest an... | 2 \mathrm{~h} 40 \mathrm{~min} | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 719,064 |
8.6. In $\triangle A B C$, a line passing through the incenter $I$ intersects sides $A B$ and $B C$ at points $M$ and $N$, respectively. It is given that $\triangle B M N$ is an acute triangle. Take points $K$ and $L$ on side $A C$ such that $\angle I L A=\angle I M B$ and $\angle I K C=\angle I N B$. Prove: $A M+K L+C... | 8.6. As shown in Figure 4, perpendiculars $I C_{1}, I A_{1}, I B_{1}$ are drawn from the incenter $I$ to the sides $A B, B C, C A$ respectively. Clearly, these perpendiculars are equal in length. Moreover, by the properties of tangents, we know that $A C_{1}=A B_{1}, C A_{1}=C B_{1}$. Therefore,
Rt $\triangle I K B_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,065 |
8.7. For a natural number $n(n>3)$, we use “$n$ ?" to denote the product of all prime numbers less than $n$. Solve the equation
$$
n ?=2 n+16
$$ | 8.7. The given equation is
$$
n ?-32=2(n-8) \text {. }
$$
Since $n$ ? cannot be divisible by 4, it follows from equation (1) that $n-8$ is odd.
Assume $n>9$, then $n-8$ has an odd prime factor $p$.
Also, $p2 \times 9+16$.
When $n=7$, it is clearly a root of the equation:
However, when $n=5$, we have $n ?=6<16$.
Theref... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,066 |
8.8. In a $10 \times 10$ grid, natural numbers 1 to 100 are written: the first row from left to right contains $1 \sim 10$; the second row from left to right contains $11 \sim 20$; and so on. Andrei tries to divide the entire grid into $1 \times 2$ rectangles, calculate the product of the two numbers in each rectangle,... | 8.8 . A rectangle of size $1 \times 2$ is called a "domino". Number the dominoes. Let the two numbers written in the $i$-th domino be $a_{i}$ and $b_{i}$, then
$$
a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2} .
$$
Write such an expression for each domino. After summing, it is found t... | S \text{ is the smallest when all dominoes are placed vertically.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,067 |
9.1. Let $f(x)$ and $g(x)$ be quadratic trinomials with leading coefficients of 1. If the equations
$$
f(g(x))=0 \text { and } g(f(x))=0
$$
have no real roots, prove that at least one of the equations $f(f(x))=0$ and $g(g(x))=0$ has no real roots. | 9.1. Suppose one of the equations $f(x)=0$ and $g(x)=0$ has no real roots (for example, $f(x)=0$ has no real roots). Then for any real number $x$, we have $f(x)>0$. Therefore, for any real number $x$, we have $f(f(x))>0$. In this case, the proposition is already established.
Now assume that both equations have real ro... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,068 |
Example 5 There are three villages $A$, $B$, and $C$ forming a triangle (as shown in Figure 5). The ratio of the number of primary school students in villages $A$, $B$, and $C$ is $1: 2: 3$. A primary school needs to be established. Where should the school be located to minimize the total distance $S$ traveled by the s... | Solution: Let the primary school be located at point $P$, and the number of students from villages $A$, $B$, and $C$ be $a$, $2a$, and $3a$ respectively. Then,
$$
\begin{array}{l}
S=a P A+2 a P B+3 a P C \\
=a(P A+P C)+2 a(P B+P C) \\
\geqslant a A C+2 a B C .
\end{array}
$$
Equality holds if and only if $P=C$.
Theref... | P=C | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,069 |
9.2. On the blackboard, there are 100 fractions. Among their numerators, the natural numbers 1 to 100 each appear exactly once; among their denominators, the natural numbers 1 to 100 also each appear exactly once. If the sum of these 100 fractions can be simplified to a fraction with a denominator of 2, prove that it i... | 9.2. Suppose at the beginning, one of the fractions involved in the summation is $\frac{a}{2}$.
Next, we prove: among these fractions involved in the summation, there must be a fraction $\frac{b}{c}$, where $c$ is odd, and $b$ has a different parity from $a$.
In fact, among these fractions involved in the summation, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,070 |
9.3. Consider a regular $(2n+1)$-gon ($n>1$). Two players play a game according to the following rules: They take turns drawing diagonals inside the regular polygon, each time drawing one new (previously undrawn) diagonal that intersects exactly an even number of already drawn diagonals (the intersection points are ins... | 9.3. Let the person who starts first be called A, and the one who starts later be called B. If $n$ is odd, then B has a winning strategy; if $n$ is even, then A has a winning strategy.
Since the number of sides of a regular polygon is odd, for any diagonal, there will be an odd number of vertices on one side and an ev... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,071 |
9.4. In $\triangle A B C$, $B B_{1}$ is the angle bisector. Through point $B_{1}$, draw a perpendicular to $B C$ intersecting the circumcircle of $\triangle A B C$ at point $K$ on the arc $\overparen{B C}$. Through point $B$, draw a perpendicular to $A K$ intersecting $A C$ at point $L$. Prove that points $K$, $L$, and... | 9.4. As shown in Figure 5, perpendiculars are drawn from points $B_{1}$ and $B$ to $BC$ and $AK$, with the feet of the perpendiculars being $S$ and $T$ respectively.
In $\triangle A L T$ and $\triangle B S K$, we have
$$
\begin{array}{l}
\angle S B K=\angle L A T \\
=\alpha .
\end{array}
$$
Therefore, $\angle B_{1} L... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,072 |
9.5. On each vertex of a convex 100-gon, two different numbers are written. Prove: it is possible to erase one number from each vertex so that the remaining numbers on any two adjacent vertices are different.
Translating the text into English while preserving the original formatting and line breaks, the result is as... | 9.5. If the same pair of numbers $a$ and $b$ are written at each vertex, then it is sufficient to leave $a$ at all even-numbered vertices and $b$ at all odd-numbered vertices to meet the requirement.
Now assume this is not the case, i.e., we can find two adjacent vertices $A$ and $B$ with two different pairs of number... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,073 |
9.6. In the acute triangle $\triangle ABC$, points $M$ and $N$ are the midpoints of sides $AB$ and $BC$, respectively, and $H$ is the foot of the perpendicular from vertex $B$. The circumcircles of $\triangle AHN$ and $\triangle CHM$ intersect at point $P (P \neq H)$. Prove that the line $PH$ passes through the midpoin... | 9.6. As shown in Figure 6, let the circumcircles of $\triangle A H N$ and $\triangle C H M$ be denoted as $\omega_{1}$ and $\omega_{2}$, respectively. Suppose line $M N$ intersects circles $\omega_{1}$ and $\omega_{2}$ at another point $D$ and $E$, respectively, and intersects $P H$ at point $S$.
Since $H N$ is the me... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,074 |
9.8. Ji Ma calculated the reciprocal of the factorial of each integer from $80 \sim 99$, and printed the resulting decimal fractions on 20 infinitely long strips of paper (for example, Sasha cut a segment from one of the strips, which had exactly $n$ digits without a decimal point. If Sasha does not want Ji Ma to guess... | 9.8. The maximum value of $n$ is 155.
Assuming that on the slips of paper, $\frac{1}{k!}$ and $\frac{1}{l!}$ (where $k < l$) are written, the first 156 digits of the decimal expansion of $\frac{1}{k!} - \frac{1}{l!}$ are greater than $\frac{1}{10^{16}}$. Therefore, $n < 156$.
Thus, for any segment of 156 digits, we c... | 155 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,075 |
Composed of unit squares, use $2 \times 1$ rectangles to cover the surface of a cube along the grid lines without overlapping and without gaps (some $2 \times 1$ rectangles will definitely "span" two faces). Prove: the number of $2 \times 1$ rectangles that span two faces must be odd. | 10.1. A $2 \times 1$ rectangle is called a "domino".
A $9 \times 9 \times 9$ cube is colored on each of its faces according to the rules of a chessboard, such that the squares at each corner are black. This results in each face having 41 black squares and 40 white squares, and each domino that spans two faces consists... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,076 |
$$
\begin{array}{l}
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n} . \\
\text { Let } m=\min \left\{a_{0}, a_{0}+a_{1}, \cdots, a_{0}+a_{1}+\cdots+a_{n}\right\} .
\end{array}
$$
Prove: When $x \geqslant 1$, we have $P(x) \geqslant m x^{n}$. | 10.2. Since for each $k$ we have
$$
\begin{array}{l}
a_{0}+a_{1}+\cdots+a_{k} \geqslant m \\
\Rightarrow -m+a_{0}+a_{1}+\cdots+a_{k} \geqslant 0 .
\end{array}
$$
Thus, as long as $x \geqslant 1$, we have
$$
\begin{array}{l}
P(x)-m x^{n} \\
=\sum_{i=0}^{n-1}\left(-m+\sum_{k=0}^{i} a_{k}\right)\left(x^{n-i}-x^{n-i-1}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,077 |
10.4. A magician and his assistant perform the following act: First, the assistant asks the audience to write down $N$ numbers in a row on a blackboard, then the assistant covers up two adjacent numbers. After that, the magician comes on stage and guesses the two adjacent covered numbers (including their order). To ens... | 10.4. $N=101$.
For convenience, a sequence of $m$ digits is called an “$m$-digit number”. Suppose for some value of $N$, the magician can guess the result, so the magician can restore any two-digit number to the original $N$-digit number (the number of $N$-digit numbers that can be restored is denoted as $k_{1}$). Thi... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 719,078 |
10.5. Given $n(n>2)$ vectors. If the length of one of the vectors is not less than the length of the sum of the remaining vectors, then the vector is called "long". If all $n$ vectors are long, prove: their sum is zero. | 10.5. Let the given $n$ vectors be denoted as $a_{1}, a_{2}, \cdots, a_{n}$, and their sum as $\sigma$.
By the problem statement, for each $k$, we have $\left|a_{k}\right| \geqslant\left|\boldsymbol{\sigma}-a_{k}\right|$, which means $a_{k} \cdot a_{k} \geqslant \sigma \cdot \sigma-2 \sigma \cdot a_{k}+a_{k} \cdot a_{k... | \boldsymbol{\sigma}=\mathbf{0} | Algebra | proof | Yes | Yes | cn_contest | false | 719,079 |
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