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7. When $x=2, y=8$,
$$
\sqrt{x^{3}+x^{2} y+\frac{1}{4} x y^{2}}+\sqrt{y^{3}+x y^{2}+\frac{1}{4} x^{2} y}
$$
is equal to $\qquad$. | $\approx 7.24 \sqrt{2}$.
$$
\begin{array}{l}
\text { Original expression }=\sqrt{x\left(x^{2}+x y+\frac{1}{4} y^{2}\right)}+\sqrt{y\left(y^{2}+x y+\frac{1}{4} x^{2}\right)} \\
=\sqrt{x\left(x+\frac{1}{2} y\right)^{2}}+\sqrt{y\left(y+\frac{1}{2} x\right)^{2}} .
\end{array}
$$
When $x=2, y=8$, $x+\frac{1}{2} y>0, y+\fra... | 24 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,030 |
Example 10 As shown in Figure 14, villages $A$, $B$, and $C$ are located along an east-west oriented highway, with $AB=2 \text{ km}$, $BC=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle ADC=45^{\circ}$. Now the $\triangle ADC$ area is planned to be a development zone, exce... | Solution: As shown in Figure 15, construct the axisymmetric figure of Rt $\triangle A D B$ with respect to the line containing $D A$, which is Rt $\triangle A D B_{1}$. It is easy to know that
Rt $\triangle A D B \cong$
Rt $\triangle A D B_{1}$.
Construct the axisymmetric figure of Rt $\triangle C D E$ with respect to ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,031 |
8. As shown in Figure 5, in isosceles $\triangle A B C$, the base angle $\angle B=$ $15^{\circ}$, and the length of the leg $A B=10$. Then the area of this triangle is $\qquad$ | 8. 25 .
Draw the altitude $C D$ from $C$ to $B A$ extended, meeting at point $D$. In the right triangle $\triangle A C D$,
$$
\angle C A D=\angle B+\angle C=2 \angle B=30^{\circ} \text {. }
$$
Then $C D=\frac{1}{2} A C=\frac{1}{2} \times 10=5$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C D=\frac{1}{2} \ti... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,032 |
9. Given real numbers $a$, $b$, $x$, $y$ satisfy $a x+b y=3$, $a y-b x=5$. Then the value of $\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)$ is $\qquad$. | 9. 34 .
$$
\begin{array}{l}
\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \\
=a^{2} x^{2}+a^{2} y^{2}+b^{2} x^{2}+b^{2} y^{2} \\
=\left(a^{2} x^{2}+b^{2} y^{2}+2 a b x y\right)+\left(a^{2} y^{2}+b^{2} x^{2}-2 a b x y\right) \\
=(a x+b y)^{2}+(a y-b x)^{2} \\
=3^{2}+5^{2}=34 .
\end{array}
$$ | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,033 |
10. As shown in Figure 6, in two concentric circles with center 0, the chord $AB$ of the larger circle is tangent to the smaller circle at point $C$, and $AB=8 \text{ cm}$. Then the area of the shaded part in the figure is $\qquad$ | $10.16 \pi \mathrm{cm}^{2}$.
Connect $O A$ and $O C$. By the property of tangents, $O C \perp A B$. By the perpendicular diameter theorem, we get $A C=B C=4(\mathrm{~cm})$. In the right triangle $\triangle O A C$, $A C^{2}=O A^{2}-O C^{2}$.
Therefore, the area of the shaded part is
$$
\begin{array}{l}
S=\pi \cdot O A^{... | 16 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,034 |
11. Draw 4 cards from a deck of cards (excluding the joker and the big joker), and perform mixed operations of addition, subtraction, multiplication, division, and exponentiation based on the numbers on the cards (parentheses can be used, but each card cannot be reused), to make the result 24 or -24. Here, $\mathrm{A},... | 11. $(-13-3) \div(-4) \times 6 ;[3-(-13)] \times 6 \div$ $(-4)$. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,035 |
12. Observe the following equation:
$$
0 \times 0=0-0,1 \times \frac{1}{2}=1-\frac{1}{2}, \cdots \text {. }
$$
According to the pattern reflected in the equation, write two more equations that satisfy this pattern. $\qquad$ | 12. $2 \times \frac{2}{3}=2-\frac{2}{3} ; 3 \times \frac{3}{4}=3-\frac{3}{4}$.
If we use $x$ and $y$ to represent these two numbers, the pattern reflected in the equation can be expressed as $xy=x-y$. Therefore, we have
$$
y=\frac{x}{1+x} .
$$
Taking $x=2$, then $y=\frac{2}{3}$; taking $x=3$, then $y=\frac{3}{4}$. | 2 \times \frac{2}{3}=2-\frac{2}{3} ; 3 \times \frac{3}{4}=3-\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,036 |
13. Given $f(x)=\frac{x}{1+x}$. Find the value of the following expression:
$$
\begin{array}{l}
f\left(\frac{1}{2004}\right)+f\left(\frac{1}{2003}\right)+\cdots+f\left(\frac{1}{2}\right)+f(1)+ \\
f(0)+f(1)+f(2)+\cdots+f(2003)+f(2004) .
\end{array}
$$ | Three, 13. Since $f(x)=\frac{x}{1+x}$, then
$$
f\left(\frac{1}{n}\right)=\frac{\frac{1}{n}}{1+\frac{1}{n}}=\frac{1}{n+1}, f(n)=\frac{n}{1+n} \text {. }
$$
Therefore, $f\left(\frac{1}{n}\right)+f(n)=\frac{1}{n+1}+\frac{n}{n+1}=1$.
Thus, the original expression is
$$
\begin{array}{l}
=f\left(\frac{1}{2004}\right)+f(2004... | 2004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,037 |
14. As shown in Figure 7, in $\triangle A B C$, $A B=A C$, $\angle B A C=90^{\circ}$, $B D$ is a median, $A E \perp B D$, intersecting $B C$ at point $E$. Prove:
$$
B E=2 E C \text {. }
$$ | 14. As shown in Figure 8, extend $BA$ to point $F$ such that $AF = AD$, and connect $CF$. In $\triangle ABD$ and $\triangle ACF$, since $AB = AC$, $\angle BAD = \angle CAF = 90^{\circ}$, and $AD = AF$, we have:
$\triangle ABD \cong \triangle ACF$.
Therefore, $\angle ABD = \angle ACF$.
Since $\angle ABD + \angle ADB = 9... | BE = 2EC | Geometry | proof | Yes | Yes | cn_contest | false | 715,038 |
15. Given a quadratic equation in $x$
$$
\dot{x}^{2}-2 x-a^{2}-a=0 \quad (a>0) \text {. }
$$
(1) Prove: one root of this equation is greater than 2, and the other root is less than 2;
(2) If for $a=1,2, \cdots, 2004$, the two roots of the corresponding quadratic equations are $\alpha_{1}, \beta_{1}, \alpha_{2}, \beta_{... | 15. (1) Let the two roots of the equation be $x_{1}$ and $x_{2}$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=2, x_{1} x_{2}=-a^{2}-a . \\
\text { Then }\left(2-x_{1}\right)\left(2-x_{2}\right)=4-2\left(x_{1}+x_{2}\right)+x_{1} x_{2} \\
=4-2 \times 2-a^{2}-a=-a^{2}-a<0 .
... | -\frac{4008}{2005} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,039 |
1. Given $f(\bar{z}-\mathrm{i})=\bar{z}+2 z-2 \mathrm{i}-1$. Then $f\left(i^{3}\right)=(\quad)$.
(A) $2 \mathrm{i}-1$
(B) $-2 \mathrm{i}-1$
(C) $i-1$
(D) $-\mathrm{i}-1$ | -、1.B.
Let $\bar{z}-\mathrm{i}=\mathrm{i}^{3}=-\mathrm{i}$, so, $\bar{z}=0, z=0$. Therefore, $f\left(\mathrm{i}^{3}\right)=-2 \mathrm{i}-1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,040 |
Example 1 Suppose the equation $3 x^{2}-6(m-1) x+m^{2}+1$ $=0$ has 2 roots whose magnitudes sum to 2. Find the value of the real number $m$. | (1) When $\Delta=36(m-1)^{2}-12\left(m^{2}+1\right) \geqslant 0$, let the equation have 2 real roots $x_{1} 、 x_{2}$.
By Vieta's formulas, we have $x_{1} x_{2}=\frac{m^{2}+1}{3}>0$, so $x_{1}$ and $x_{2}$ have the same sign. Therefore,
$$
\left|x_{1}\right|+\left|x_{2}\right|=\left|x_{1}+x_{2}\right| .
$$
Again by Vi... | m=0 \text{ or } m=\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,042 |
3. Vector Set
$$
\begin{array}{l}
M=\{\boldsymbol{a} \mid \boldsymbol{a}=(-1,1)+x(1,2), x \in \mathbf{R}\}, \\
N=\{\boldsymbol{a} \mid \boldsymbol{a}=(1,-2)+x(2,3), x \in \mathbf{R}\} .
\end{array}
$$
Then $M \cap N=(\quad$.
(A) $\{(1,-2)\}$
(B) $\{(-13,-23)\}$
(C) $\{(-1,1)\}$
(D) $\{(-23,-13)\}$ | 3. B.
If $\boldsymbol{a}=\left(a_{1}, a_{2}\right) \in M \cap N$, then we have
$$
\begin{array}{l}
a_{1}=-1+x_{1}=1+2 x_{2}, \\
a_{2}=1+2 x_{1}=-2+3 x_{2} .
\end{array}
$$
Rearranging, we get
$$
\left\{\begin{array}{l}
x_{1}-2 x_{2}=2, \\
2 x_{1}-3 x_{2}=-3
\end{array}\right.
$$
Solving, we get $x_{1}=-12, x_{2}=-7$... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,043 |
4. If line $l$ is translated 5 units in the negative direction of the $x$-axis, and then 1 unit in the positive direction of the $y$-axis, and it returns to its original position, then the slope of line $l$ is ( ).
(A) $-\frac{1}{5}$
(B) -5
(C) $\frac{1}{5}$
(D) 5 | 4.A.
According to the problem, the line $l$ returns to its original position after being translated along the vector $a=(-5,1)$. Let the equation of the line $l$ be $y=a x+b$. After the translation, the equation still has the form $y^{\prime}=a x^{\prime}+b$, where
$$
\left\{\begin{array} { l }
{ y ^ { \prime } = y +... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,044 |
5. If $a, b$ satisfy $0<a<b<1$, then which of the following inequalities is correct? ( ).
(A) $a^{a}<b^{b}$
(B) $b^{a}<b^{b}$
(C) $a^{a}<b^{a}$
(D) $b^{b}<a^{a}$ | 5.C.
According to the monotonicity of power functions and exponential functions, we should have $b^{a}>a^{a}, b^{a}>b^{b}$.
The size relationship between $a^{a}$ and $b^{b}$ is uncertain. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,045 |
6. Let the right directrix of the hyperbola intersect the two asymptotes at points $A$ and $B$. The circle with diameter $AB$ passes through the right focus $F$. Then the eccentricity of the hyperbola is ( ).
(A) $\sqrt{2}$
(B) $2 \sqrt{2}$
(C) $\sqrt{3}$
(D) $2 \sqrt{3}$ | 6.A.
Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Then the equation of the right directrix is $x=\frac{a^{2}}{c}$, and the equations of the asymptotes are $y= \pm \frac{b}{a} x$.
From this, the coordinates of points $A$ and $B$ are $\left(\frac{a^{2}}{c}, \frac{a b}{c}\right)$ and ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,046 |
7. Given that the domain of the function $f(x)$ is $(a, b)$, and $b-a>2$. Then the domain of $F(x)=f(3 x-1)-f(3 x+1)$ is ( ).
(A) $\left(\frac{a-1}{3}, \frac{b+1}{3}\right)$
(B) $\left(\frac{a+1}{3}, \frac{b-1}{3}\right)$
(C) $\left(\frac{a-1}{3}, \frac{b-1}{3}\right)$
(D) $\left(\frac{a+1}{3}, \frac{b+1}{3}\right)$ | 7.B.
From the problem, we have $\left\{\begin{array}{l}a2$, then
$$
\frac{b-1}{3}-\frac{a+1}{3}=\frac{(b-a)-2}{3}>0,
$$
which means $\frac{a+1}{3}<\frac{b-1}{3}$.
Therefore, the solution to the inequality system is $\frac{a+1}{3}<x<\frac{b-1}{3}$.
Hence, the domain of $F(x)$ is $\left(\frac{a+1}{3}, \frac{b-1}{3}\rig... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,047 |
8. Given the function $f(x)=\frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$. If $g(x)=$ $f^{-1}(-x)$, then $g(x)$ is an increasing function on the interval $(\quad$.
(A) $(-1,+\infty)$
(B) $(-1,+\infty)$ is a decreasing function
(C) $(-\infty,-1)$ is an increasing function
(D) $(-\infty,-1)$ is a decreasing function | 8.C.
Let $y=\frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$. Solving for $x$ yields $x=\ln \frac{y+1}{y-1}$.
Therefore, $f^{-1}(x)=\ln \frac{x+1}{x-1}$.
Thus, $g(x)=f^{-1}(-x)=\ln \frac{-x+1}{-x-1}=\ln \frac{x-1}{1+x}$.
From $\frac{x-1}{1+x}>0$, we get $x1$.
Hence, the domain of $g(x)$ is $(-\infty,-1) \cup(1,+\infty)$.
Op... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,048 |
9. Given the equation in terms of $x$
$$
\sin ^{2} x-(2 a+1) \cos x-a^{2}=0
$$
has real solutions. Then the set of values for the real number $a$ is ( ).
(A) $\left[-\frac{5}{4}, 1-\sqrt{2}\right]$
(B) $\left[-\frac{5}{4}, 1+\sqrt{2}\right]$
(C) $[1-\sqrt{2}, 1+\sqrt{2}]$
(D) $\left[-\frac{3}{2}, 1-\sqrt{2}\right]$ | 9. B.
Transform the equation into
$$
\cos ^{2} x+(2 a+1) \cos x+a^{2}-1=0 \text {. }
$$
Let $t=\cos x$, then the equation becomes
$$
t^{2}+(2 a+1) t+a^{2}-1=0 \text {. }
$$
Let $f(t)=t^{2}+(2 a+1) t+a^{2}-1, t \in[-1,1]$.
By the problem, the real number $a$ should satisfy
$$
\left\{\begin{array}{l}
(2 a+1)^{2}-4\lef... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,049 |
10. As shown in Figure 2, in the tetrahedron $P-ABC$, $PA \perp$ the base $ABC$, $\angle ACB=90^{\circ}$, $AE \perp PB$ at $E$, $AF \perp PC$ at $F$. If $PA=AB=2$, $\angle BPC=\theta$, then when the area of $\triangle AEF$ is maximized, the value of $\tan \theta$ is ( ).
(A) 2
(B) $\frac{1}{2}$
(C) $\sqrt{2}$
(D) $\fra... | 10.D.
Since $P A \perp$ plane $A B C$, then $P A \perp B C$.
Also, $B C \perp A C$, so $B C \perp$ plane $P A C$.
Therefore, plane $P B C \perp$ plane $P A C$.
Since $A F \perp P C$, then $A F \perp$ plane $P B C$, hence $A F \perp E F$.
Also, $P A=A B=2, A E \perp P B$, so $A E=\sqrt{2}$.
In the right triangle $\tria... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,050 |
11. A person has 100,000 yuan and plans to invest in real estate or buy stocks. If a decision is made based on the profit table (Table 1), then the reasonable investment plan should be $\qquad$ .
Table 1
\begin{tabular}{|c|c|c|c|}
\hline & Probability of Profit & \begin{tabular}{c}
Buying Stocks \\
Profit
\end{tabular... | II. 11. Invest in Real Estate.
The expected profit from buying stocks is
$$
0.3 \times 10 + 0.5 \times 3 + 0.2 \times (-5) = 3.5 \text{ (ten thousand yuan) } \text{.}
$$
The expected profit from investing in real estate is
$$
0.3 \times 8 + 0.5 \times 4 + 0.2 \times (-4) = 3.6 \text{ (ten thousand yuan) } \text{.}
$$
... | Invest in Real Estate. | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,051 |
12. Given $\sin \alpha-\cos \alpha=\frac{1}{2}$. Then the value of $\sin ^{3} \alpha-$ $\cos ^{3} \alpha$ is $\qquad$. | $$
\begin{array}{l}
\text { 12. } \frac{11}{16} \\
\sin ^{3} \alpha-\cos ^{3} \alpha \\
=(\sin \alpha-\cos \alpha)\left(\sin ^{2} \alpha+\sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha\right) \\
=(\sin \alpha-\cos \alpha)(1+\sin \alpha \cdot \cos \alpha)
\end{array}
$$
Given $\sin \alpha-\cos \alpha=\frac{1}{2}$, and
$... | \frac{11}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,052 |
Example 2 For the quadratic equation in $x$
$$
x^{2}-(\tan \theta+\mathrm{i}) x+\mathrm{i}(2 \mathrm{i}-1)=0(\theta \in \mathbf{R}) \text {. }
$$
(1) If this equation has 1 real root, find the value of the acute angle $\theta$;
(2) For any real number $\theta$, prove: this equation cannot have a purely imaginary root. | (1) If this equation has 1 real root $a$, then by the condition of equality of complex numbers, we get
$$
\left\{\begin{array}{l}
a^{2}-a \tan \theta-2=0, \\
a+1=0 .
\end{array}\right.
$$
Solving this, we get $a=-1, \tan \theta=1$.
Since $0<\theta<\frac{\pi}{2}$, hence $\theta=\frac{\pi}{4}$.
(2) Proof by contradictio... | \theta=\frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,053 |
13. Place 5 small balls, red, yellow, blue, white, and black, into 5 boxes, red, yellow, blue, white, and black, respectively, with 1 ball in each box. The probability that the red ball is not in the red box and the yellow ball is not in the yellow box is $\qquad$ . | 13.0.65.
Placing 5 balls into 5 boxes has a total of \( N = 5! = 120 \) (ways).
The number of ways where the red ball is not in the red box and the yellow ball is not in the yellow box can be divided into two categories:
(1) The red ball is in the yellow box, in which case there are
$$
n_{1} = 4! = 24 \text{ (ways); }... | 0.65 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,054 |
14. Let $A_{1}, A_{2}$ be the two vertices on the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b$ $>0)$, and $P_{1} P_{2}$ be a chord perpendicular to the major axis. The intersection point of the lines $A_{1} P_{1}$ and $A_{2} P_{2}$ is $P$. Then the equation of the trajectory of point $P$ is... | 14. $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Let the coordinates of point $P_{1}$ be $\left(x_{0}, y_{0}\right)$, then $P_{2}\left(x_{0},-y_{0}\right)$, $A_{1}(-a, 0)$, $A_{2}(a, 0)$. The equation of the line $A_{1} P_{1}$ is
$$
y=\frac{y_{0}}{x_{0}+a}(x+a) .
$$
The equation of the line $A_{2} P_{2}$ is
$$
y=\frac... | \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,055 |
15. (12 points) Given $b>a>0$, and $a+b=1$. Consider the following four expressions:
$$
b, \frac{a+b}{2}, 2 a b, \frac{a^{4}-b^{4}}{a-b}
$$
Arrange them in descending order, and provide the corresponding proof. | Three, 15. The order of the four expressions from largest to smallest is
$$
b>\frac{a^{4}-b^{4}}{a-b}>\frac{a+b}{2}>2 a b \text {. }
$$
Given that $b>a>0$, and $a+b=1$, we have
$$
\begin{array}{l}
\frac{a^{4}-b^{4}}{a-b}=(a+b)\left(a^{2}+b^{2}\right)=a^{2}+b^{2}, \\
\frac{(a+b)^{2}}{2}=\frac{1}{2}=\frac{a+b}{2} .
\end... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,056 |
16. (12 points) In the plane quadrilateral $A B C D$,
$$
A B=a, B C=b, C D=c, D A=d,
$$
and $\boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b} \cdot \boldsymbol{c}=m, \boldsymbol{c} \cdot \boldsymbol{d}=\boldsymbol{d} \cdot \boldsymbol{a}=n$.
(1) When $m=n$, what kind of quadrilateral is $A B C D$? Prove your conclus... | 16. (1) When $m=n$, that is, $a \cdot b=b \cdot c=c \cdot d=d \cdot a$. From $a+b+c+d=0$, we get $b+d=-(a+c)$.
Since $a \cdot b+d \cdot a=b \cdot c+c \cdot d$, thus, $a \cdot(b+d)=c \cdot(b+d)$.
Then $a \cdot(a+c)=c \cdot(a+c)$, $a^{2}+a \cdot c=a \cdot c+c^{2}$.
Therefore, $a^{2}=c^{2}$, that is, $|a|^{2}=|c|^{2},|a|=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,057 |
17. (12 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and its first $n$ terms sum $S_{n}$ satisfies the relation
$$
3 t S_{n}-(2 t+3) S_{n-1}=3 t(t>0, n=2,3, \cdots) \text {. }
$$
(1) Prove that the sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) Let the common ratio of the sequence $\left\{a_... | 17. (1) From the given $3 S_{2}-(2 t+3) S_{1}=3 t$,
i.e., $3 t\left(a_{1}+a_{2}\right)-(2 t+3) a_{1}=3 t$.
Given $a_{1}=1$, solving for $a_{2}$ yields $a_{2}=\frac{2 t+3}{3 t}$.
Thus, $\frac{a_{2}}{a_{1}}=\frac{2 t+3}{3 t}$.
For $n \geqslant 2$, we have
$$
\begin{array}{l}
3 t S_{n+1}-(2 t+3) S_{n}=3 t, \\
3 t S_{n}-(... | -\frac{8}{9} n^{2}-\frac{4}{3} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,058 |
18. (15 points) As shown in Figure 3, the edge length of the cube $A B C D$ - $A_{1} B_{1} C_{1} D_{1}$ is $2, M, N, P$ are the midpoints of edges $C C_{1}, C B, C D$ respectively.
(1) Prove: $A_{1} P \perp$ plane $D M N$; (2) Find the volume of the tetrahedron $A_{1} D M N$. | 18. (1) As shown in Figure 3, connect $A P$, then
Rt $\triangle A D P \cong$ Rt $\triangle D C N$.
Thus, $\angle D A P = \angle C D N$,
$\angle D A P + \angle A D N = \angle C D N + \angle A D N = 90^{\circ}$.
Therefore, $A P \perp D N$.
Since $A_{1} A \perp$ plane $A B C D$, and $A P$ is the projection of $A_{1} P$ i... | \frac{4}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 715,059 |
19. (15 points) As shown in Figure 4, there is a pointer on a disk, initially pointing to the top of the disk. The pointer rotates clockwise around the center of the disk by an angle $\alpha$ each time, and $3.6^{\circ}<\alpha<180^{\circ}$. After 2,004 rotations, it returns to its initial position for the first time, p... | 19. Clearly, if
$3.6^{\circ}1$, then let
$n_{1}=\frac{n}{d}, n_{2}=\frac{2004}{d}$.
At this point, we have $\alpha=\frac{n_{1} \times 360^{\circ}}{n_{2}}$.
This means that after the pointer rotates $n_{2}$ times, each time rotating by $\alpha$, the pointer will rotate $n_{1}$ full circles and return to its initial posi... | 325 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,060 |
Find all integers $n$ such that
$$
n^{4}+6 n^{3}+11 n^{2}+3 n+31
$$
is a perfect square.
(Xu Wandang) | Let $A=n^{4}+6 n^{3}+11 n^{2}+3 n+31$. Then
$$
A=\left(n^{2}+3 n+1\right)^{2}-3(n-10)
$$
is a perfect square.
When $n>10$, we have $A<\left(n^{2}+3 n+1\right)^{2}$.
If $n \leqslant-3$ or $n \geqslant 0$, then $n^{2}+3 n \geqslant 0$.
Thus, $A \geqslant\left(n^{2}+3 n+2\right)^{2}$.
Therefore, $2 n^{2}+9 n-27 \leqslant... | n=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,061 |
II. Quadrilateral $ABCD$ is a convex quadrilateral, $I_{1}$ and $I_{2}$ are the incenters of $\triangle ABC$ and $\triangle DBC$, respectively. The line through points $I_{1}$ and $I_{2}$ intersects $AB$ and $DC$ at points $E$ and $F$, respectively. Extend $AB$ and $DC$ to meet at point $P$, and $PE = PF$. Prove that p... | II. As shown in Figure 1, connect $I_{1} B$, $I_{1} C$, $I_{2} B$, and $I_{2} C$.
Since $P E = P F$, we have $\angle P E F = \angle P F E$.
And $\angle P E F = \angle I_{2} I_{1} B - \angle E B I_{1} = \angle I_{2} I_{1} B - \angle I_{1} B C$,
$\angle P F E = \angle I_{1} I_{2} C - \angle F C I_{2} = \angle I_{1} I_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,062 |
Three, find all real numbers $k$ such that the inequality
$$
a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)
$$
holds for all $a, b, c, d \in[-1,+\infty)$.
(Xu Wanyi, problem contributor) | When $a=b=c=d=-1$, we have
$-3 \geqslant k \times(-4) \Rightarrow k \geqslant \frac{3}{4}$;
When $a=b=c=d=\frac{1}{2}$, we have
$4 \times \frac{1}{8}+1 \geqslant k \times\left(4 \times \frac{1}{2}\right) \Rightarrow k \leqslant \frac{3}{4}$.
Therefore, $k=\frac{3}{4}$.
Next, we prove the inequality
$$
a^{3}+b^{3}+c^{3}... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,063 |
Example 3 If the equation $z^{m}=1+\mathrm{i}(m \in \mathbf{N})$ has exactly 2 roots whose corresponding points in the complex plane fall precisely in the first quadrant, then the value of $m$ is ( ).
(A) $3,4,5,6$
(B) $4,5,6,7$
(C) $5,6,7,8$
(D) $6,7,8,9$ | Explanation: From the square root operation of complex numbers, we easily obtain
$$
\begin{array}{l}
\omega_{k}=\sqrt[2 m]{2}\left(\cos \frac{\frac{\pi}{4}+2 k \pi}{m}+\mathrm{i} \sin \frac{\frac{\pi}{4}+2 k \pi}{m}\right) \\
(k=0,1,2, \cdots, m-1) .
\end{array}
$$
Since the equation has and only has 2 roots in the fi... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,064 |
Let $n \in \mathbf{N}_{+}$, and let $d(n)$ denote the number of all positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $\{1, 2, \cdots, n\}$ that are coprime to $n$. Find all non-negative integers $c$ such that there exists a positive integer $n$ satisfying
$$
d(n) + \varphi(n) = n + c,
$$
and... | Let the set of all positive divisors of $n$ be $A$, and let the set of numbers in $1,2, \cdots, n$ that are coprime with $n$ be $B$. Since there is exactly one number $1 \in A \cap B$ in $1,2, \cdots, n$, we have
$d(n)+\varphi(n) \leqslant n+1$.
Thus, $c=0$ or 1.
(1) When $c=0$, from $d(n)+\varphi(n)=n$, we know that t... | c=0 \text{ or } 1; \text{ for } c=0, n=6, 8, 9; \text{ for } c=1, n=1, 4, \text{ or a prime number} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,065 |
Five, let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a_{2}=1$, and
$$
a_{n+2}=\frac{1}{a_{n+1}}+a_{n}, n=1,2, \cdots \text {. }
$$
Find $a_{2004}$.
(Supplied by Wu Weizhao) | $$
\begin{array}{l}
\text { From the given, } a_{n+2} a_{n+1}-a_{n+1} a_{n}=1, \text { so, } \left\{a_{n+1} a_{n}\right\} \text { is an arithmetic sequence with the first term 1 and common difference 1. } \\
\text { Therefore, } a_{n+1} a_{n}=n, n=1,2, \cdots . \\
\text { Thus, } a_{n+2}=\frac{n+1}{a_{n+1}}=\frac{n+1}{... | \frac{2003!!}{2002!!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,066 |
Six, color all the small squares of an $m \times n$ chessboard (consisting of $m$ rows and $n$ columns, $m \geqslant 3, n \geqslant 3$) with one of two colors, red or blue. If 2 adjacent (sharing a common edge) small squares are of different colors, then these 2 small squares are called a "standard pair." Let the numbe... | Solution 1: Divide all squares into three categories. The first category of squares are located at the four corners of the chessboard, the second category of squares are located on the boundary of the chessboard (excluding the four corners), and the remaining squares are the third category.
Fill all red squares with t... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,067 |
Seven, Given an acute triangle $ABC$ with unequal side lengths, the perimeter is $l$, and $P$ is a moving point inside it. The projections of point $P$ onto sides $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively. Prove that:
$$
2(A F + B D + C E) = l
$$
if and only if: point $P$ lies on the line connecting the ... | Let the sides of $\triangle ABC$ be $BC=a, CA=b, AB=c$, and assume $b \neq c$. As shown in Figure 2, establish a Cartesian coordinate system, and set $A(m, n)$, $B(0,0)$,
$$
C(a, 0), P(x, y).
$$
From $A F^{2}-B F^{2}$
$$
=A P^{2}-B P^{2},
$$
we get $A F^{2}-(c-A F)^{2}$
$$
=A P^{2}-B P^{2}.
$$
Thus, $2 c \cdot A F ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,068 |
6.2. A person has less than 1 Ruble (the currency unit of Russia, 1 Ruble = 100 Kopecks). He buys 5 pieces of candy, leaving 4 Kopecks; buys 6 pencils, leaving 3 Kopecks; buys 7 exercise books, leaving 1 Kopeck. Please answer: How much does each piece of candy cost? How much money does the person have in total? Try to ... | 6.2. The person has a total of 99 kopecks, and each piece of candy costs 19 kopecks.
We can reason as follows:
If we take out 9 kopecks from the person's money, the remaining amount can be divided by 5 and by 6, therefore, it can be divided by 30. Thus, the total amount of money can only be 39, 69, or 99 kopecks, and w... | 99 \text{ kopecks, 19 kopecks} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,071 |
6.3. Can the numbers $1,2, \cdots, 10$ be written in a row in some order so that the sum of every three consecutive numbers is not greater than 15? Explain your reasoning. | 6.3. Cannot.
The sum of all 10 numbers is 55. If the sum of any 3 consecutive numbers is not greater than 15, then the sum of the first 9 numbers is not greater than 45, so the 10th number is not less than 10, and thus, it must be 10. Similarly, the sum of the last 9 numbers is not greater than 45, so the 1st number i... | Cannot | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,072 |
6.4. Given three segments of lengths $1, 2, 3$, and divide the segment of length 3 into five segments arbitrarily. Prove: among these seven segments, it is possible to find three segments that can form a triangle. | 6.4. If one of the five segments has a length greater than 1, then it can form a triangle with the segments of lengths 1 and 2; if all five segments have lengths not greater than 1, then there must be two segments whose lengths sum to more than 1, and thus they can form a triangle with the segment of length 1. | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,073 |
Example 4 If the expansion of $\left(1+x+x^{2}\right)^{1000}$ is $a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2000} x^{2000}$,
then the value of $a_{0}+a_{3}+a_{6}+\cdots+a_{1998}$ is ( ).
(A) $3^{333}$
(B) $3^{666}$
(C) $3^{999}$
(D) $3^{2001}$
(2001, National High School Mathematics Competition) | Explanation: Let $f(x)=\left(1+x+x^{2}\right)^{1000}$
$$
\begin{array}{l}
=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2000} x^{2000}, \\
\omega=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3},
\end{array}
$$
$$
\begin{array}{l}
\text { then } a_{0}+a_{3}+a_{6}+\cdots+a_{1998} \\
=\frac{f(1)+f(\omega)+f\left(\omega^{2}\ri... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,075 |
7.2. There are 12 people in the room,
some of whom always lie, while the rest always tell the truth. One of them says: “There are no honest people here”; another says: “There is at most 1 honest person here”; the third person says: “There are at most 2 honest people here”; and so on, until the 12th person says: “There... | 7.2. There are 6 honest people in the room.
Obviously, the first few people are not honest. If the 6th person is telling the truth, then there are no more than 5 honest people in the room. On the other hand, from him onwards, the next 7 people are telling the truth, and thus, they should all be honest, which is imposs... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,076 |
7.3. There are three automatic coin exchange machines. Among them, the first machine can only exchange 1 coin for 2 other coins; the second machine can only exchange 1 coin for 4 other coins; the third machine can exchange 1 coin for 10 other coins. A person made a total of 12 exchanges, turning 1 coin into 81 coins. T... | 7.3. The first coin exchanger increases the number of coins by 1 each time, the second coin exchanger increases the number of coins by 3 each time, and the third coin exchanger increases the number of coins by 9 each time. Let the three coin exchangers be used $A$ times, $B$ times, and $C$ times, respectively, then we ... | A=B=2, C=8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,077 |
7.4. Take a point \( O \) inside a convex quadrilateral \( A B C D \). Prove: the inequality
\[
O A < A B, \quad O B < B C, \quad O C < C D, \quad O D < D A
\]
at least one of them holds. | 7.4. If $O A \geqslant A B$, then we have
$$
\angle O B A \geqslant \angle A O B \text {. }
$$
Adding such inequalities, we get
$$
\begin{array}{l}
\angle O B A+\angle O C B+\angle O D C+\angle O A D \\
\geqslant \angle A O B+\angle B O C+\angle C O D+\angle D O A=360^{\circ} .
\end{array}
$$
But this is impossible, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,078 |
8.3. Find the prime solutions of the equation $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ (where $[x]$ denotes the greatest integer not exceeding $x$). | 8.3. If \( p > \)
3, then \( p = 6k + 1 \) or \( p = 6k - 1 \). Substituting into the equation, we get \( q = 6k \) or \( q = 6k - 3 \). Since \( q \) is a prime number, \( q \neq 6k \); and the latter case is only possible when \( k = 1 \), hence \( p = 2, 3, 5 \). Substituting them into the equation, we get \( q = 1 ... | p = 3, q = 2; p = 5, q = 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,080 |
8.4. In a $6 \times 6$ grid, integers $1, 2, \cdots, 36$ have already been filled in a certain way. Prove: It is possible to delete one row and one column from the grid so that the sum of the remaining 25 numbers is even. | 8.4. Let the sum of the $i$-th row be denoted as $\alpha_{i}$, the sum of the $j$-th column as $\beta_{j}$, and the number in the cell at the intersection of the $i$-th row and the $j$-th column as $x_{i j}$. Since the sum of all numbers in the grid (denoted as $A$) is even, the $i$-th row and the $j$-th column can be ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,081 |
6.1. In every game of Chinese chess, the winner gets 1 point, a draw results in 0 points for both, and the loser loses 1 point. Several students participate in a Chinese chess tournament, with each pair of students playing one game. It is found that one student scores a total of 7 points, and another scores a total of ... | 6.1. Assuming there are no ties, then, the person who scored 20 points must have played an even number of games (the number of games he won minus the number of games he lost equals 20), while the person who scored 7 points must have played an odd number of games. But this is impossible. Because all participants played ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,082 |
6.2. The shape of a certain castle is a heptagon, and there is a bell tower at each of its vertices. Each side of the castle wall is guarded by the soldiers in the bell towers at the two endpoints of that side. How many soldiers are needed in total to ensure that each side of the wall is guarded by at least 7 soldiers? | 6.2. If there are no more than 3 guards on each tower, then no wall will be guarded by 7 guards. Observe any tower with at least 4 guards. At this point, the remaining 6 towers can be divided into 3 "adjacent pairs," with each "adjacent pair" corresponding to one wall, so each "adjacent pair" has at least 7 guards. The... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,083 |
6.3. Given a positive integer $n$. Prove: the sum of the digits of the number $n(n-1)$ and the sum of the digits of the number $(n+1)^{2}$ are not equal. | 6.3. The difference between the two given numbers is $3 n+1$, so, their remainders when divided by 3 are always different. And the remainder of an integer when divided by 3 is the remainder of the sum of its digits when divided by 3. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,084 |
6.4. A person collects coins. It is known that the diameter of all the coins he collects does not exceed $10 \mathrm{~cm}$. He sticks all his coins on a cardboard sheet measuring $30 \mathrm{~cm} \times 70 \mathrm{~cm}$. Prove: He can move them all to another cardboard sheet measuring $40 \mathrm{~cm} \times 60 \mathrm... | 6.4. As shown in Figure 6, a $30 \mathrm{~cm} \times 70 \mathrm{~cm}$ rectangle can be divided into two overlapping $30 \mathrm{~cm} \times$ $40 \mathrm{~cm}$ rectangles. Clearly, each coin can fit entirely within one of the $30 \mathrm{~cm} \times 40 \mathrm{~cm}$ rectangles. Two $30 \mathrm{~cm} \times$ $40 \mathrm{~... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,085 |
Example 5 Given the equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{i}, \overline{r_{i}}(i=1,2,3,4,5)$, where $\overline{r_{i}}$ is the conjugate of $r_{i}$. Find the value of $\sum_{i=1}^{5} \frac{1}{r_{i} \bar{r}_{i}}$.
(12th American Invitational Mathematics Examination) | Explanation: The original equation can be transformed into
$$
\left(\frac{13 x-1}{x}\right)^{10}=-1 \text {. }
$$
Let the 10 complex roots of $y^{10}=-1$ be $\omega_{i}$ and $\overline{\omega_{i}}$ $(i=1,2,3,4,5)$, then we have $\omega_{i}=\frac{13 r_{i}-1}{r_{i}}$, which means
$$
\begin{array}{l}
\frac{1}{r_{i}}=13-\... | 850 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,086 |
6.5. There are 27 points distributed at equal intervals on a circle, each of which is colored either black or white. It is known that there are at least 2 points between any 2 black points. Prove: It is possible to find 3 white points that form the 3 vertices of an equilateral triangle. | 6.5. According to the coloring rule, there are at most 9 black points. We number the 27 points in sequence, and it is easy to see that they can form 9 equilateral triangles:
$$
(1,10,19),(2,11,20), \cdots,(9,18,27) \text {. }
$$
If there are no more than 8 black points, then there must be one equilateral triangle whos... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,087 |
6.6. Three magicians printed many different denominations of "money", each holding 100 rubles of "money". It is known that each of them can pay any amount from 1 to 25 rubles (including giving "change"). Prove: the money of the three magicians combined can pay any amount from 100 to 200 rubles (the denominations of the... | 6.6. Since the first two magicians together can pay various amounts of "goods money" from 1 to 50 rubles (including giving "change"), if we add the third magician's 100 rubles, then they can pay all different amounts of "goods money" from 100 to 150 rubles. By subtracting these amounts from 300, we can obtain all diffe... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,088 |
7.1. On January 1, in the Magic City's department store, the unit price (price per item) of 10 goods was 1 ruble each. Starting from January 1, the price of each item doubled or tripled daily compared to the previous day. It is known that on February 1, the unit prices of these 10 goods were all different. Prove: at th... | 7.1. Arrange the new prices in increasing order. It is clear that in every two adjacent prices, the latter price is at least 1.5 times the former, so the highest price is at least $1.5^{9}$ times the lowest price. And
$$
1.5^{9}>(2.25)^{4}>5^{2}=25 \text {. }
$$ | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,089 |
7.2. Does there exist a number of the form $1000 \cdots 001$ that can be divided by a number of the form $111 \cdots 11$? | 7.2. Only for 11
In fact, $1000 \cdots 001=99 \cdots 9+2$, and the remainder of $99 \cdots 9$ divided by $111 \cdots 11$ is a number of the form $99 \cdots 9$ with fewer digits than the divisor. Therefore, if $99 \cdots 9+2$ can be divided by $111 \cdots 11$, then $99 \cdots 9+2$ must equal $111 \cdots 11$. This is on... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,090 |
7.3. As shown in Figure 2, extend the sides of $\triangle ABC$ such that $BC_{1} = AC, AB_{1} = AB, CA_{1} = AB$. It is known that $\triangle A_{1}B_{1}C_{1}$ is an equilateral triangle. Prove: $\triangle ABC$ is also an equilateral triangle. | 7.3. Clearly, $\triangle A B_{1} C_{1} \cong \triangle C A_{1} B_{1}$, so,
$$
\angle C B_{1} A_{1}=\angle B_{1} C_{1} A, \angle C_{1} B_{1} A=\angle B_{1} A_{1} C \text {. }
$$
Since $\triangle A_{1} B_{1} C_{1}$ is an equilateral triangle, then $\angle C_{1} A_{1} B=$ $\angle C B_{1} A_{1}, \angle A_{1} C_{1} B=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,091 |
7.4. Mark 100 nodes (the intersections of the grid) on a grid paper. Prove: It is possible to find two points $A$ and $B$ such that the rectangle formed with them as vertices and the grid lines as edges (including the edges of the rectangle) contains at least 20 marked nodes. | 7.4. Find the leftmost, top, rightmost, and bottom 4 nodes $A, B, C, D$, so that all nodes fall within the rectangle formed by the grid lines passing through these 4 nodes. It is easy to see that the 5 rectangles with $(A, B), (B, C), (C, D), (D, A), (A, C)$ as pairs of opposite vertices cover all 100 nodes. Therefore,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,092 |
7.5. Solve the system of equations
$$
\left\{\begin{array}{l}
a^{2}+3 a+1=\frac{b+c}{2}, \\
b^{2}+3 b+1=\frac{a+c}{2}, \\
c^{2}+3 c+1=\frac{a+b}{2} .
\end{array}\right.
$$ | 7.5. Adding the three equations yields
$$
(a+1)^{2}+(b+1)^{2}+(c+1)^{2}=0 \text {. }
$$
Then \(a=b=c=-1\). This is the unique solution to the system of equations. | a=b=c=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,093 |
7.6. A person collects coins, and it is known that the diameter of all the coins he collects does not exceed $10 \mathrm{~cm}$. He sticks all his coins on a cardboard sheet measuring $30 \mathrm{~cm} \times 70 \mathrm{~cm}$. Recently, someone gave him a coin with a diameter of $25 \mathrm{~cm}$. Prove: Now he can stick... | 7.6. As shown in Figure 7, using two rectangles of $30 \mathrm{~cm} \times 55 \mathrm{~cm}$ and $30 \mathrm{~cm} \times 25 \mathrm{~cm}$ to cover a $30 \mathrm{~cm} \times 70 \mathrm{~cm}$ rectangle (with overlap), it is clear that each original coin is completely within one of the rectangles. Moving the two rectangles... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,094 |
7.7. A circular pond is divided into $n$ sectors, and $n+1$ frogs have landed in these sectors. Every second, two frogs that are in the same sector jump to the two adjacent sectors, one to the left and one to the right. Prove: There must be a moment when the number of sectors with frogs is at least half of the total nu... | 7.7. First, we prove that every sector will eventually have a frog jump into it. Suppose there is a sector $A$ where no frog ever jumps in. Starting from it, we number the sectors in a clockwise direction as $1, 2, \cdots, n$. We calculate the sum of the squares of the sector numbers where the frogs are located. Since ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,095 |
8.3. In a trapezoid, one of the diagonals has a length equal to the sum of the lengths of the two bases, and the angle between the two diagonals is $60^{\circ}$. Prove: the trapezoid is an isosceles trapezoid. | 8.3. As shown in Figure 8, on the line $AD$, intercept the segment $DE$, such that $DE=BC$. It is evident that $CE \parallel BD$. Since $AC=AD+BC$, therefore, $\triangle ACE$ is an isosceles triangle, and $\angle ACE=60^{\circ}$. Hence, $\triangle ACE$ is an equilateral triangle. Therefore, $CE=AC$, which means $BD=AC$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,096 |
Example 1 Let $x$ be any real number. Prove: $x, 2 x, \cdots$, $n x$ must contain a number, the absolute difference between which and some integer is no greater than $\frac{1}{n+1}$.
Analysis: If we can prove that the sum of the absolute differences between each number in $x, 2 x, \cdots, n x$ and some integer is no g... | Prove: If the absolute value of the difference between each $i x$ and some integer is greater than $\frac{1}{n+1}$, then
$$
i x-[i x]>\frac{1}{n+1},[i x]+1-i x>\frac{1}{n+1},
$$
i.e., there are $n$ values of $i x-[i x]$ in the interval $\left(\frac{1}{n+1}, \frac{n}{n+1}\right)$.
Consider the intervals
$$
\begin{array... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,097 |
Example 2 If 100 real numbers $a_{1}, a_{2}, \cdots a_{100}$, satisfy
$$
\left\{\begin{array}{l}
a_{1}-3 a_{2}+2 a_{3} \geqslant 0, \\
a_{2}-3 a_{3}+2 a_{4} \geqslant 0, \\
\cdots \cdots \\
a_{100}-3 a_{1}+2 a_{2} \geqslant 0 .
\end{array}\right.
$$
Prove: $a_{1}=a_{2}=\cdots=a_{100}$.
Analysis: If we directly solve t... | Proof: If $a_{1}, a_{2}, \cdots, a_{100}$ are not all equal, let $a_{k}$ be the largest among them, and assume that $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ contains terms smaller than $a_{k}$. Let $m$ be the largest number such that $a_{m}<a_{k}$, where $1 \leqslant m \leqslant k-1$. Then
$$
a_{m}<a_{m+1}=a_{m+2}=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,098 |
Example 3 Given non-negative numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=1$. Prove:
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+ \\
\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}
\end{array}
$$
has a minimum value, and find it. | Prove: The original expression is $\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$. Since the function $f(x) = \frac{1}{2-x}$ is increasing on $[0,1]$, for any $x \in [0,1]$, we have
$$
\left(x-\frac{1}{n}\right)\left(\frac{1}{2-x}-\frac{n}{2 n-1}\right) \geqslant 0 .
$$
Rearranging, we get
$$
\frac{2 n-1}{2 n} \cdot \frac{x}{2... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,099 |
2. Let $a=\sqrt{3}-\sqrt{2}, b=2-\sqrt{3}, c=$ $\sqrt{\sqrt{5}-2}$. Then the size relationship of $a, b, c$ is ( ).
(A) $c>a>b$
(B) $ab>c$
(D) $a>c>b$ | 2.A.
$$
a>b \Leftrightarrow 2 \sqrt{3}>2+\sqrt{2} \Leftrightarrow 12>6+4 \sqrt{2} \Leftrightarrow 3>2 \sqrt{2}
$$
$\Leftrightarrow 9>8$, holds;
$$
\begin{array}{l}
c>a \Leftrightarrow \sqrt{\sqrt{5}-2}>\sqrt{3}-\sqrt{2} \Leftrightarrow \sqrt{5}-2>5-2 \sqrt{6} \\
\Leftrightarrow 2 \sqrt{6}+\sqrt{5}>7 \Leftrightarrow 29+... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,100 |
3. $[x]$ represents the greatest integer not greater than $x$, then the number of all real solutions to the equation $x^{2}-8[x]+7=0$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 3.C.
Let $[x]=n$ (where $n$ is an integer), then $n \leqslant x4$. Therefore, $n=1,5,6,7$, and accordingly $x=1, \sqrt{33}, \sqrt{41}, 7$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,101 |
4. As shown in Figure 1, in trapezoid $A B C D$, $A B$ //
$D C, A B=B C=\frac{3}{2}$,
$A D=A C=\sqrt{3}$, $E$ and $F$
are the midpoints of $C D$ and $A B$ respectively. Then the length of $E F$ is ( ).
(A) $\sqrt{2}$
(B) $2 \sqrt{2}$
(C) $\frac{\sqrt{41}}{4}$
(D) $\frac{\sqrt{41}}{5}$ | 4. C.
As shown in Figure 6, connect $A E$. Since $A B / / C D, A B=B C$, $A C=A D$, we have,
$$
\begin{array}{l}
\angle B C A=\angle B A C \\
=\angle A C D=\angle A D C .
\end{array}
$$
Therefore, $\triangle A D C \backsim \triangle B A C$. Hence, $\frac{D C}{A C}=\frac{A C}{B C}$.
Thus, $D C=2, E C=1$.
Since $D E=E ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,102 |
5. Let $k$ be a real number, and $\alpha, \beta$ be the roots of the equation $x^{2} + k x - 1 = 0$. If $(|\alpha| - \beta)(|\beta| - \alpha) \geqslant 1$, then the range of $k$ is ( ).
(A) $k \geqslant \sqrt{\sqrt{5}-2}$
(B) $k \leqslant \sqrt{\sqrt{5}-2}$
(C) $k \leqslant -\sqrt{\sqrt{5}-2}$ or $k \geqslant \sqrt{\sq... | 5.A.
Since $\Delta=k^{2}+4>0$, therefore, $\alpha, \beta$ are two distinct real roots. By Vieta's formulas, we have
$$
\alpha+\beta=-k, \alpha \beta=-1 .
$$
Since the left side of the inequality in the problem is symmetric with respect to $\alpha, \beta$, we can assume without loss of generality that $\alpha < \beta$... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,103 |
6. As shown in Figure $2, \odot O$ is the circumcircle of square $ABCD$, with $O$ as the center. Point $P$ is on the minor arc $AB$, and $DP$ intersects $AO$ at point $Q$. If $PQ=QO$, then $\frac{QC}{AQ}$ equals ( ).
(A) $2 \sqrt{3}-1$
(B) $\sqrt{3}+2$
(C) $\sqrt{3}+\sqrt{2}$
(D) $2 \sqrt{3}$ | 6. B.
As shown in Figure 7, let's assume $A O=$ $O C=1$, then $A D=C D=\sqrt{2}$.
Let $Q O=Q P=x$, then
$$
\begin{array}{l}
A Q=1-x, \\
Q C=1+x .
\end{array}
$$
By the intersecting chords theorem, we get
$$
\begin{array}{l}
D Q=\frac{A Q \cdot Q C}{Q P} \\
=\frac{1-x^{2}}{x} .
\end{array}
$$
Construct $Q H \perp A D... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,104 |
1. As shown in Figure 3, in trapezoid $A B C D$, $A B / / C D$, $A B=a, C D=b(a$ $<b), \angle B C D=$ $60^{\circ}, \angle A D C=45^{\circ}$. The area of trapezoid $A B C D$ is equal to $\qquad$ (express the area in terms of $a$ and $b$). | $$
\text { II. 1. } \frac{3-\sqrt{3}}{4}\left(b^{2}-a^{2}\right) \text {. }
$$
As shown in Figure 8, construct $A E \perp$
$$
D C, B F \perp D C, E \text{ and } F
$$
as the feet of the perpendiculars. Then
$$
\begin{array}{l}
E F=A B=a . \\
\text { Let } A E=B F=h,
\end{array}
$$
then $D E=A E=h$.
$$
C F=\frac{B F}{... | \frac{3-\sqrt{3}}{4}\left(b^{2}-a^{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,105 |
2. Let $k$ be a real number, and the quadratic equation $x^{2}+k x+k+1=0$ has two real roots $x_{1}$ and $x_{2}$. If $x_{1}+2 x_{2}^{2}=k$, then $k$ equals $\qquad$ . | 2.5.
From $\Delta \geqslant 0$ we get
$$
k^{2}-4 k-4 \geqslant 0 \text {. }
$$
Since $x_{2}^{2}=-k x_{2}-k-1$, then,
$$
\begin{array}{l}
x_{1}+2 x_{2}^{2}=k \Rightarrow x_{1}-2 k x_{2}=3 k+2 \\
\Rightarrow\left(x_{1}+x_{2}\right)-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-k-(2 k+1) x_{2}=3 k+2 \\
\Rightarrow-(2 k+1) x_{2}=2(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,106 |
3. The number of integers $x$ that make $x^{2}+x-2$ a perfect square is $\qquad$.
Make the above text in English, please keep the original text's line breaks and format, and output the translation result directly. | 3.4 .
Let $x^{2}+x-2=y^{2}, y \in \mathbf{N}$, then $4 x^{2}+4 x-8=(2 y)^{2}$,
i.e., $(2 x-2 y+1)(2 x+2 y+1)=3^{2}$.
Therefore, we have $\left\{\begin{array}{l}2 x-2 y+1=1, \\ 2 x+2 y+1=9 ;\end{array}\left\{\begin{array}{l}2 x-2 y+1=3, \\ 2 x+2 y+1=3 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}2 x-2 y+1=-3, \\... | 4 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,107 |
4. As shown in Figure 4, in $\triangle A B C$, $B C=a, C A=b, A B=$ $\sqrt{a^{2}+b^{2}}, C D \perp$ $A B$. If the incenter of $\triangle A C D$ and $\triangle B C D$ are $I_{1}$ and $I_{2}$ respectively, then $I_{1} I_{2}=$ $\qquad$ (express the answer in terms of $a$ and $b$). | 4. $\frac{\sqrt{2}}{2}\left(a+b-\sqrt{a^{2}+b^{2}}\right)$.
Let the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ be $r$, $r_{1}$, and $r_{2}$, respectively, then $r=\frac{a+b-\sqrt{a^{2}+b^{2}}}{2}$.
It is easy to see that $\angle I_{2} D I_{1}=90^{\circ}$.
By the similarity of right triangle... | \frac{\sqrt{2}}{2}\left(a+b-\sqrt{a^{2}+b^{2}}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,108 |
一、(20 points) Find all real numbers $k$ such that the quadratic equation in $x$
$$
k x^{2}-2(3 k-1) x+9 k-1=0
$$
has two integer roots. | Let the equation be
$$
k x^{2}-2(3 k-1) x+9 k-1=0
$$
with two integer roots $x_{1}, x_{2}$, and $x_{1} \leqslant x_{2}$. Then
$$
x_{1}+x_{2}=6-\frac{2}{k}, x_{1} x_{2}=9-\frac{1}{k} \text {. }
$$
Eliminating $\frac{1}{k}$ from the above two equations, we get
$$
2 x_{1} x_{2}-x_{1}-x_{2}=12 \text {, }
$$
which simpli... | k=-\frac{1}{4}, \frac{1}{9}, \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,109 |
Example 4 If $a, b, c$ are the lengths of the three sides of a triangle, $2 p = a + b + c, n \in \mathbf{N}_{+}$, then
$$
\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geqslant\left(\frac{2}{3}\right)^{n-2} \cdot p^{n-1} .
$$
(28th IMO Preliminary Problem) | Prove: Since the function $f(x)=\frac{x^{n-1}}{2 p-x}$ is an increasing function on $(0,2 p)$, for any $x \in(0,2 p)$, we have
$$
\left(x-\frac{2 p}{3}\right)\left[f(x)-f\left(\frac{2 p}{3}\right)\right] \geqslant 0 \text {. }
$$
Rearranging, we get
$$
\begin{array}{l}
\frac{x^{n}}{2 p-x} \\
\geqslant \frac{2 p x^{n-1... | \left(\frac{2}{3}\right)^{n-2} \cdot p^{n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,110 |
II. (25 points) As shown in Figure 5, in $\triangle A B C$, $\angle C=$ $90^{\circ}$, $D$ is a point on side $B C$, and $\angle A D C=45^{\circ}$. Draw $D E \perp A B$ at point $E$, and $A E: B E=10: 3$. If $D E=\sqrt{2}$, try to find the length of the angle bisector $C F$ of $\angle C$. | Let $B E=3 x, A E=10 x$. By the Pythagorean theorem, it is easy to get
$$
\begin{array}{l}
B D=\sqrt{2+9 x^{2}}, A D=\sqrt{2+100 x^{2}}, \\
C D=A C=\frac{A D}{\sqrt{2}}=\sqrt{1+50 x^{2}} .
\end{array}
$$
From $S_{\triangle A B D}=\frac{1}{2} A B \cdot D E=\frac{1}{2} B D \cdot A C$, we get
$$
\frac{1}{2} \times 13 x \... | \frac{6}{5} \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,111 |
Three. (25 points) Find all integers $n$ such that $2 n^{2}+1$ divides $n^{3}-5 n+5$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Three. (25 points) Find all integers $n$ such that $2 n^{2}+1$ divides $n^{3}-5 n+5$. | Three, since $2 n^{2}+1$ is an odd number, " $2 n^{2}+1$ divides $n^{3}-5 n+5$ " is equivalent to " $2 n^{2}+1$ divides $2 n^{3}-10 n+10$ ". Let
$$
2 n^{3}-10 n+10=\left(2 n^{2}+1\right)(n+A),
$$
where $A$ is an undetermined integer.
After expanding and rearranging, we get
$$
\begin{array}{l}
2 A n^{2}+11 n+A-10=0(A \n... | 0 \text{ or } -1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,112 |
2. Given that the side length of rhombus $A B C D$ is $a, \angle A=$ $\frac{\pi}{3}$. Fold the rhombus $A B C D$ along the diagonal to form a dihedral angle $\theta$. If $\theta \in\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right]$, then the maximum distance between the skew lines $A C$ and $B D$ is ( ).
(A) $\frac{3}{2} a$
... | 2.D.
(1) As shown in Figure 3, fold along $AC$ to form a dihedral angle, and let $O$ be the midpoint of $AC$. Since $AC \perp OB$ and $AC \perp OD$, we have $AC \perp$ plane $BOD$. Therefore, $\angle BOD = \theta$. Draw $OE \perp BD$ at $E$, then $OE$ is the distance between $AC$ and $BD$.
From $\angle BAC = \frac{1}{2... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,114 |
3. Among the subsets of the set $\{1,2, \cdots, 100\}$, some subsets satisfy the condition: no number is twice another number. The maximum number of elements in such a subset is ( ).
(A) 67
(B) 68
(C) 69
(D) 70 | 3. A.
Let $A_{1}=\{51,52, \cdots, 100\}, A_{2}=\{26,27, \cdots, 50\}$, $A_{3}=\{13,14, \cdots, 25\}, A_{4}=\{7,8, \cdots, 12\}$,
$$
A_{5}=\{4,5,6\}, A_{6}=\{2,3\}, A_{7}=\{1\} \text {. }
$$
Then $A_{1} \cup A_{3} \cup A_{5} \cup A_{7}$ contains $50+13+3+1=67$ elements, and none of these elements is twice another.
If... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,115 |
4. Let $a>0, a \neq 1$,
$$
f(x)=\log _{a}\left(x+\sqrt{x^{2}-1}\right)(x \geqslant 1) \text {. }
$$
If $f^{-1}(n)<\frac{3^{n}+3^{-n}}{2}$, where $n$ is a positive integer, then the range of $a$ is ( ).
(A) $\left(0, \frac{1}{3}\right)$
(B) $\left(\frac{1}{3}, 1\right) \cup(1,3)$
(C) $(1,3)$
(D) $(1,+\infty)$ | 4.C.
It is easy to get $f^{-1}(x)=\frac{1}{2}\left(a^{x}+a^{-x}\right)$.
When $0<1$, we get $f(x) \geqslant 0$.
In $f^{-1}(n)$, $n>0$, so $a>1$.
$$
\begin{array}{l}
\text { By } \frac{a^{n}+a^{-n}}{2}<\frac{3^{n}+3^{-n}}{2} \\
\Rightarrow\left(3^{n}-a^{n}\right)\left(1-3^{n} a^{n}\right)<0 \\
\Rightarrow\left(\frac{1}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,116 |
5. The sum of the coefficients of the integer power terms of $x$ in the expansion of $(\sqrt{x}+2)^{2 n+1}$ is ( ).
(A) $\frac{1}{2}\left(3^{2 n+1}+1\right)$
(B) $\frac{1}{2}\left(2^{2 n+1}+1\right)$
(C) $\frac{1}{3}\left(3^{2 n+1}+1\right)$
(D) $\frac{1}{3}\left(2^{2 n+1}+1\right)$ | 5. A.
Let $A=(\sqrt{x}+2)^{2 n+1}, B=(-\sqrt{x}+2)^{2 n+1}$.
By the binomial theorem, the sum of the integer power terms of $x$ in $A$ and $B$ is the same, denoted as $f(x)$, and the sum of the non-integer power terms are opposite numbers, which cancel each other out when added. Therefore, we have
$$
2 f(x)=(\sqrt{x}+... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,117 |
6. If the parabola satisfies the following conditions:
(1) the directrix is the $y$-axis;
(2) the vertex is on the $x$-axis;
(3) the minimum distance from point $A(3,0)$ to a moving point $P$ on the parabola is 2.
Then the number of such parabolas is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6.D.
Suppose there exists such a parabola, with vertex $B$ on the $x$-axis. Let $B(a, 0)$, then the equation of the parabola with the $y$-axis as the directrix is
$$
y^{2}=4 a(x-a) \text {. }
$$
From condition (3), we know that $a>0$. Let $P\left(\frac{m^{2}}{4 a}+a, m\right)$, thus,
$$
\begin{array}{l}
|A P|^{2}=\le... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,118 |
1. Let $x^{2}+y^{2} \leqslant 2$. Then the maximum value of $\left|x^{2}-2 x y-y^{2}\right|$ is $\qquad$ . | $$
\text { Two } 1.2 \sqrt{2}
$$
Let $x=r \cos \theta, \boldsymbol{y}=r \sin \theta, 0 \leqslant \theta<2 \pi$.
From the problem, we know $|r| \leqslant \sqrt{2}$. Therefore,
$$
\begin{array}{l}
\left|x^{2}-2 x y-y^{2}\right| \\
=\left|r^{2} \cos ^{2} \theta-2 r^{2} \sin \theta \cdot \cos \theta-r^{2} \sin ^{2} \theta... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,119 |
Example 5 Let $a_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, n, k \in \mathbf{N}_{+}$. Prove:
$$
\begin{array}{l}
\left(\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}}\right)^{k}+\left(\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}}\right)^{k}+ \\
\cdots+\left(\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}}\right)^{k} \\
\geqslant \frac{n}{(n-1)^{k... | Proof: Let $S=\sum_{i=1}^{n} a_{i}$, then the problem is reduced to proving the inequality
$$
\sum_{i=1}^{n}\left(\frac{a_{i}}{S-a_{i}}\right)^{k} \geqslant \frac{n}{(n-1)^{k}} .
$$
Since the function $f(x)=\frac{x^{k-1}}{(S-x)^{k}}$ is increasing on $(0, S)$, for any $x \in(0, S)$, we have
$$
\left(x-\frac{S}{n}\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,121 |
3. Let $z_{n}=\left(\frac{1-\mathrm{i}}{2}\right)^{n}, n \in \mathbf{N}_{+}$, and
$$
S_{n}=\sum_{k=1}^{n}\left|z_{k+1}-z_{k}\right| \text {. }
$$
Then $\lim _{n \rightarrow \infty} S_{n}=$ $\qquad$ | $3.1+\frac{\sqrt{2}}{2}$.
Since $z_{k}-z_{k-1}=\left(\frac{1-\mathrm{i}}{2}\right)^{k}-\left(\frac{1-\mathrm{i}}{2}\right)^{k-1}$ $=\left(\frac{1-\mathrm{i}}{2}\right)^{k-1}\left(\frac{-1-\mathrm{i}}{2}\right)$,
thus $\left|z_{k}-z_{k-1}\right|=\left(\frac{\sqrt{2}}{2}\right)^{k}, \cdots$,
$\left|z_{2}-z_{1}\right|=\le... | 1+\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,122 |
4. Take any 2005 non-collinear points inside $\triangle A B C$ and add the 3 vertices of $\triangle A B C$, making a total of 2008 points as vertices. Connect these points to form non-overlapping small triangles. The total number of small triangles that can be formed is $\qquad$
Translate the above text into English, ... | 4.4011 .
In $\triangle ABC$, suppose there are $m$ points that meet the conditions, and there are $a_{m}$ small triangles that meet the conditions. Now, add a point $F$, which must fall into a certain small $\triangle PQR$, as shown in Figure 4. Suppose at this point there are $a_{m+1}$ small triangles that meet the c... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,123 |
5. Let the function $f(x)=a+\sqrt{-x^{2}-4 x}, g(x)$ $=\frac{4}{3} x+1$. If $f(x) \leqslant g(x)$ always holds, then the range of the real number $a$ is $\qquad$. | 5. $a \leqslant-5$.
From the problem, we have
$$
\begin{array}{l}
\sqrt{-x^{2}-4 x} \leqslant \frac{4}{3} x+1-a . \\
\text { Let } y_{1}=\sqrt{-x^{2}-4 x}, \\
y_{2}=\frac{4}{3} x+1-a .
\end{array}
$$
Then equation (2) represents the upper half of the circle $(x+2)^{2}+y_{1}^{2}=4$ (including the endpoints), and equat... | a \leqslant -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,124 |
6. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$, and $Q, R, S, P$ are the midpoints of $PA, QB, RC, SD$ respectively. If $V_{P-ABC}$ represents the volume of the tetrahedron $P-ABC$, and similarly for the others, then
$$
V_{P-ABC}: V_{P-BCD}: V_{P-CDA}: V_{P-DAB}=
$$
$\qquad$ | $6.8: 1: 2: 4$.
As shown in Figure 5, let the distances from points $A, P$ to the plane $BCD$ be $h_{1}, h_{P}$, and similarly for the others. Thus, we have
$$
\begin{array}{c}
h_{Q}=\frac{1}{2}\left(h_{A}+h_{P}\right), \\
h_{R}=\frac{1}{2} h_{Q} \\
=\frac{1}{4}\left(h_{1}+h_{P}\right), \\
h_{S}=\frac{1}{2} h_{R}=\frac... | 8: 1: 2: 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,125 |
Three. (20 points) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>$ $0, b>0)$, $A$ and $B$ are two points on the same branch of the hyperbola. The perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P$. Find the range of the $x$-coordinate of point $P$. | Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $P\left(x_{0}, 0\right)$, with the midpoint of $AB$ being $M\left(x^{\prime}, y^{\prime}\right)$. Then,
$$
\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1, \frac{x_{2}^{2}}{a^{2}}-\frac{y_{2}^{2}}{b^{2}}=1 \text {. }
$$
Subtracting the two equations... | x_{0}>\frac{a^{2}+b^{2}}{a} \text{ or } x_{0}<-\frac{a^{2}+b^{2}}{a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,126 |
Four. (20 points) Let $a, b, c, m$ satisfy the condition
$$
\frac{a}{m+2}+\frac{b}{m+1}+\frac{c}{m}=0,
$$
and $a \geqslant 0, m>0$. Prove that the equation $a x^{2}+b x+c=0$ has a root $x_{0} \in(0,1)$. | (1) When $a=0$, if $b \neq 0$, then $x_{0}=-\frac{c}{b}=$ $\frac{m}{m+1} \in(0,1)$; if $b=0$, then $c=0$. In this case, any real number is a root of the equation, so there exists a root $x_{0} \in(0,1)$.
(2) When $a>0$, let $f(x)=a x^{2}+b x+c$, then
$$
\begin{array}{l}
f\left(\frac{m}{m+1}\right)=a\left(\frac{m}{m+1}\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,127 |
Five, (20 points) Find the smallest natural number $k$, such that for any $x \in [0,1]$ and $n \in \mathbf{N}_{+}$, the inequality
$$
x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}
$$
always holds. | Given $x \in [0,1]$, by the AM-GM inequality, we have
$$
\begin{array}{l}
x^{k}(1-x)^{n} \\
=\underbrace{x \cdots \cdots x}_{k \uparrow} x \cdot \underbrace{(1-x) \cdot(1-x) \cdots \cdots(1-x)}_{n \uparrow} \\
=\left(\frac{n}{k}\right)^{n} \underbrace{x \cdots \cdots x}_{k \uparrow} \cdot \underbrace{\frac{k(1-x)}{n} \... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,128 |
一、(50 points) As shown in Figure 2, in $\triangle ABC$, $AB < AC$, and $\odot O$ is the circumcircle of $\triangle ABC$. A tangent line to $\odot O$ at point $A$ intersects the extension of $BC$ at point $D$. Perpendiculars from points $B$ and $C$ to $BC$ intersect the perpendicular bisectors of $AB$ and $AC$ at points... | As shown in Figure 2, since $AD$ is the tangent of $\odot O$, we have $\angle DAB = \angle ACB$.
Thus, $\triangle DAB \backsim \triangle DCA$.
Therefore, $\frac{DB}{DA} = \frac{AB}{AC}$, $\frac{DA}{DC} = \frac{AB}{AC}$.
Multiplying the two equations, we get $\frac{DB}{DC} = \left(\frac{AB}{AC}\right)^2$.
Let $M$ and $N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,129 |
Sure, here is the translated text:
```
II. (50 points) The sequence $\left\{x_{n}\right\}$ satisfies:
$$
\begin{array}{l}
x_{1} \in(0,1), \\
x_{n+1}=\left\{\begin{array}{ll}
\frac{1}{x_{n}}-\left[\frac{1}{x_{n}}\right], & x_{n} \neq 0, \\
0, & x_{n}=0
\end{array}\left(n \in \mathbf{N}_{+}\right) .\right.
\end{array}
$... | ```
Let $\{x\}=x-[x]$.
When $n=1$, by $x_{1} \in(0,1)$, we get $x_{1}\frac{f_{k+2}}{2 f_{k+2}}=\frac{1}{2}$, \\
Thus, $x_{1}+x_{2}=x_{1}+\frac{1}{x_{1}}-1=f\left(x_{1}\right)-1 \\
<f\left(\frac{f_{k+2}}{f_{k+3}}\right)-1=\frac{f_{k+2}}{f_{k+3}}+\frac{f_{k+3}}{f_{k+2}}-1=\frac{f_{k+2}}{f_{k+3}}+\frac{f_{k+1}}{f_{k+2}}$.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,130 |
Three, (50 points) Let $p(p \geqslant 5)$ be a prime number. Does there exist an integer $n$ in the interval $[1, p-2]$ such that $n^{p-1}-1$ and $(n+1)^{p-1}-1$ are both not divisible by $p^{2}$? Please prove your conclusion. | Let $S=\{1,2, \cdots, p-1\}$, $A=\left\{n \mid n \in S\right.$, and $\left.n^{p-1} \equiv 1\left(\bmod p^{2}\right)\right\}$.
We will prove that $|A| \geqslant \frac{p-1}{2}$.
In fact, if $1 \leqslant n \leqslant p-1$, by the binomial theorem, we have $(p-n)^{p-1}-n^{p-1}=-(p-1) n^{p-2} p \neq 0\left(\bmod p^{2}\right)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,131 |
Example 6 Let $x_{i} \in \mathbf{R}_{+}, 1 \leqslant i \leqslant n, n \geqslant 2, \sum_{i=1}^{n} x_{i}$ =1. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{1}{\sqrt{n-1}} \sum_{i=1}^{n} \sqrt{x_{i}} .
$$
(4th National High School Mathematics Winter Camp) | Prove: Since the function $f(x)=\frac{1}{\sqrt{1-x}}$ is increasing on $(0,1)$, for any $x \in(0,1)$, we always have $\left(x-\frac{1}{n}\right)\left[f(x)-f\left(\frac{1}{n}\right)\right] \geqslant 0$, i.e., $\frac{x}{\sqrt{1-x}} \geqslant \frac{1}{n \sqrt{1-x}}+\sqrt{\frac{n}{n-1}}\left(x-\frac{1}{n}\right)$.
By $x_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,132 |
In $\triangle ABC$, $AB=AC$, $\odot O$ is its circumcircle, $BN$ bisects $\angle ABC$, and $N$ is on $\odot O$. Points $E$ and $F$ are on sides $AB$ and $AC$ respectively, satisfying $EO \perp BN$, $EF \perp EO$. Prove: $AE^2 = BE \cdot AF$.
---
The translation maintains the original text's line breaks and format as ... | Proof: As shown in Figure 2, let line $EN$ intersect $\odot O$ at $M$ and intersect $AC$ at $K$, and let $BN$ intersect $AC$ at $D$.
Since $EO \perp BN$ and $EO$ is the perpendicular bisector of $BN$, we have
$$
EN = EB \text{.}
$$
From $ME \cdot EN = AE \cdot EB$, we get
$$
ME = AE \text{.}
$$
It is easy to see that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,133 |
Find all rational numbers $r$ such that
$$
\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}} \text{ and } \frac{r^{3}-3 r^{2}+4 r-1}{r-2 r^{2}}
$$
are both integers. | Solution: From $\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}}-\frac{r^{3}-3 r^{2}+4 r-1}{r-2 r^{2}}$
$$
=\frac{8 r^{2}-8 r+2}{r(1-2 r)}=-4+\frac{2}{r} \in \mathbf{Z} \text {, }
$$
we have $r=\frac{2}{k}, k \in \mathbf{Z}, k \neq 0,4$.
Substituting $r=\frac{2}{k}$ into $\frac{r^{3}+5 r^{2}-4 r+1}{r-2 r^{2}} \in \mathbf{Z}$, we... | r= \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,134 |
Height 147 Find the maximum and minimum three-digit prime number $n$ that satisfies $\sum_{k=0}^{n} k^{2} \mathrm{C}_{n}^{k} \mid \sum_{k=0}^{n} k^{4} \mathrm{C}_{n}^{k}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $\mathrm{C}_{n}^{k}=\frac{n!}{(n-k)!k!}$, we have
$$
n \mathrm{C}_{n}^{k}-k \mathrm{C}_{n}^{k}=(k+1) \mathrm{C}_{n}^{k+1} \text {. }
$$
By summing both sides of equation (1) for $k=0,1, \cdots, n-1$ and making appropriate adjustments, and similarly multiplying both sides of equation (1) by $k, k^{2}, k^{3}, \cdo... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,135 |
Given positive numbers $a, b (a < b)$ and a positive integer $n$, and a sequence of numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in [a, b], i=1,2, \cdots, n$. Find
$$
f=\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}
$$
the ... | Solution: (1) First, prove an inequality.
Let $b_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, m$, then
$$
\begin{array}{l}
\left(1+b_{1}\right)\left(1+b_{2}\right) \cdots\left(1+b_{m}\right) \\
\geqslant\left(1+\sqrt[m]{b_{1} b_{2} \cdots b_{m}}\right)^{m} .
\end{array}
$$
Consider $f(x)=\lg \left(1+10^{x}\right)(x \in \mat... | f \leqslant\left(a^{\frac{1}{n+1}}+b^{\frac{1}{n+1}}\right)^{-(n+1)} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,136 |
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