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
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class | __index_level_0__ int64 0 742k |
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Four. (20 points) It is known that the five-digit number $\overline{a b c d e}$ satisfies the following conditions:
(1) None of its digits are zero;
(2) It is a perfect square;
(3) The digit $a$ in the ten-thousands place is a perfect square, the two-digit number formed by the digits in the thousands and hundreds place... | Let $M^{2}=\overline{a b c d e}$, and $a=m^{2}$ (a single digit), $\overline{b c}=n^{2}$ (a two-digit number), $\overline{d e}=t^{2}$ (a two-digit number). Then
$$
M^{2}=m^{2} \times 10^{4}+n^{2} \times 10^{2}+t^{2} \text {. }
$$
From equation (1), we have
$$
\begin{array}{l}
M^{2}=\left(m \times 10^{2}+t\right)^{2} \... | 11664, 41616, 43681, 93636 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,115 |
Five. (20 points) Given a quadratic function whose graph opens upwards and does not pass through the origin $O$, with the vertex coordinates at $(1,-2)$, and intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$, and satisfies the relationship $\mid O C \mid^{2} = |O A| \cdot |O B|$.
(1) Find the... | Let the analytical expression of the quadratic function be
$$
y=a(x-1)^{2}-2=a x^{2}-2 a x+a-2,
$$
and $a>0, a \neq 2$, with the graph as shown in Figure 7.
The intersections of the graph with the $x$-axis and $y$-axis are $A\left(x_{1}, 0\right)$, $B\left(x_{2}, 0\right)$, and $C(0, a-2)$.
(1) From $|O C|^{2}=|O A| ... | \sqrt{2} \text{ or } (\sqrt{2}-1) \sqrt{2(\sqrt{2}-1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,116 |
Six. (20 points) As shown in Figure 3, in $\triangle A B C$, $D$ is a point on $A C$ such that $A D = D C + C B$. A perpendicular line from $D$ to $A C$ intersects the circumcircle at $M$. Prove that $M$ is the midpoint of the major arc $\overparen{A B}$.
保留源文本的换行和格式，直接输出翻译结果如下：
Six. (20 points) As shown in Figure 3,... | Six, as shown in Figure 8, extend $A C$ to $E$ such that $C E=B C$. Connect $M A$, $M B$, $M E$, and $B E$.
Since $A D=D C+B C$ $=D C+C E=D E$, and $M D \perp A E$, then
$$
M A=M E, \angle 1=\angle 2 \text {. }
$$
Also, $\angle 1=\angle 3$, so $\angle 2=\angle 3$.
Since $C E=B C$, we have
$$
\angle 4=\angle 5, \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,117 |
1. Let $a$ be a natural number, then the units digit of $a+a^{2}+a^{3}+\cdots+$ $a^{2002}$ cannot be ( ).
(A) 0
(B) 2
(C) 4
(D) 6 | -、1.C.
We only need to discuss $a$ being the ten digits $0,1,2, \cdots, 9$.
When $a=1$ or 6, the unit digit of $a+a^{2}+\cdots+a^{2000}$ is 2.
When $a \neq 1$ and $a \neq 6$, we have $a^{4 n}+a^{4 n+1}+a^{4 n+2}+$ $a^{4 n+3}$ (where $n$ is a positive integer) has a unit digit of 0. Therefore, the proposition is equiva... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,118 |
2. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle A=90^{\circ}, AB=$ $AC, \angle ABC$ is bisected by $BD$, and $DE \perp BC$ at $E$. If the perimeter of $\triangle CDE$ is 6, then the length of $BC$ is ( ).
(A) $6-3 \sqrt{2}$
(B) $12-6 \sqrt{2}$
(C) $6 \sqrt{2}$
(D) 6 | 2.D.
From $\angle C B D=\angle A B D, \angle D E B=\angle A=90^{\circ}, B D=$ $B D$, we know $\triangle B A D \cong \triangle B E D$. Therefore, $B E=A B=A C, D E=$ $A D$. Thus,
$$
\begin{array}{l}
B C=B E+E C=A C+E C \\
=A D+D C+E C=D E+D C+E C=6 .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,119 |
3. Insert a digit (including 0) into a two-digit number to form a three-digit number (for example, inserting the digit 5 into 39 results in 359). Some two-digit numbers, when a digit is inserted, become a three-digit number that is $k$ times (where $k$ is a positive integer) the original two-digit number. What is the m... | 3.D.
Let the two-digit number be $10a + b$, and the three-digit number after inserting a digit $m$ be $100a + 10m + b$ (where $a$, $m$, and $b$ are integers, and $0 < a < 10$, $0 \leqslant m, b < 10$). Then,
$$
k = \frac{100a + 10m + b}{10a + b} = 10 + \frac{10m - 9b}{10a + b}.
$$
Clearly, when $a$ and $b$ are as sma... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,120 |
4. If $a, c$ are both positive integers, and the equations $x^{2}+a x+$ $18=0$ and $x^{2}-a x+c=0$ have a common prime root. Then the value of $c$ is $(\quad$.
(A) -18
(B) -26
(C) -26 or -36
(D) 26 or 36 | 4.C.
Since $18=2 \times 3^{2}$, and the equation $x^{2}+a x+18=0$ has a prime root, this prime root can only be 2 or 3.
When the prime root is 2, we have $2^{2}+2 a+18=0$ and $2^{2}-$ $2 a+c=0$. Adding these two equations gives $c=-26$.
When the prime root is 3, we have $3^{2}+3 a+18=0$ and $3^{2}-$ $3 a+c=0$. Addin... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,121 |
5. As shown in Figure 2, with the sides $AB$ and $AD$ of the square $ABCD$ as diameters, semicircles are drawn outside the square. A line is drawn through $A$ intersecting the two semicircles at $E$ and $F$. If the area of the square $ABCD$ is $1997$, and $AE$ and $AF$ are integers, then the length of $EF$ is ( ).
(A) ... | 5.A.
Connect $B E, D F$, then $\angle B E A=\angle A F D=90^{\circ}$.
Also, $\angle E B A+\angle E A B=90^{\circ}, \angle E A B+\angle F A D=90^{\circ}$, hence $\angle E B A=\angle F A D$.
Since $A B=A D$, thus,
$\triangle B E A \cong \triangle A F D, D F=A E$.
Let $A E=x, A F=y$, then
$x^{2}+y^{2}=A D^{2}=1997$ (wher... | 63 | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,122 |
Example 1 As shown in Figure 2, from a point $C$ on the semicircle, a perpendicular line is drawn to the diameter $A B$, with the foot of the perpendicular being $D$. Construct $\odot O_{1}$ to be tangent to $\overparen{B C}$, $C D$, and $D B$ at points $E$, $F$, and $G$ respectively. Prove: $A C = A G$.
| Proof: Let the center of the semicircle be $O$, then $O$, $O_{1}$, and $E$ are collinear. Connect $O_{1} F$, we know $O_{1} F \perp C D, O_{1} F \parallel A B$. Connect $E F$, connect $A E$. Since $\angle F E O_{1}=\frac{1}{2} \angle F O_{1} O = \frac{1}{2} \angle E O B = \angle O E A$, we know that $E$, $F$, and $A$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,123 |
6. If the positive real numbers $a, b$ satisfy $a b=a+b+3$, then the minimum value of $a^{2}+b^{2}$ is ( ).
(A) -7
(B) 0
(C) 9
(D) 18 | 6. D.
Let $a+b=m$, then $ab=m+3$, and $m>0$. Therefore, $a, b$ are the roots of the equation $x^{2}-mx+m+3=0$, which gives us
$$
\Delta=m^{2}-4m-12=(m-6)(m+2) \geqslant 0 .
$$
Since $m>0$, we have $m-6 \geqslant 0, m \geqslant 6$. Thus,
$$
\begin{array}{l}
a^{2}+b^{2}=(a+b)^{2}-2ab \\
=m^{2}-2m-6=(m-1)^{2}-7 \geqslan... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,124 |
1. Let the polynomial $x^{3}-x-a$ and the polynomial $x^{2}+x$ $-a$ have a non-constant common factor, then $a=$ $\qquad$ . | II. 1.0 or 6.
Since $\left(x^{3}-x-a\right)-\left(x^{2}+x-a\right)=x(x+1)(x-2)$, the common factor of $x^{3}-x-a$ and $x^{2}+x-a$ must be one of $x$, $x+1$, or $x-2$.
When the common factor is $x$ or $x+1$, $a=0$;
When the common factor is $x-2$, $a=6$.
Therefore, $a=0$ or 6. | 0 \text{ or } 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,125 |
2. As shown in Figure 3, $A_{0} A_{1}$ is the diameter of a semicircle, $A_{0} A_{1}=2$, $A_{2}, A_{3}, \cdots, A_{k}$, $A_{k+1}, \cdots$, are points on the semicircle, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{... | 2. 11 .
As shown in Figure 6, connect
$$
A_{k} O, A_{k+1} O, A_{k} A_{1} \text {, }
$$
then
$$
\begin{array}{l}
\angle O A_{k} A_{k+1}= \\
\angle A_{1} A_{k} A_{k+1}+\angle A_{1} A_{k} O \\
=\angle A_{1} A_{k} A_{k+1}+\angle A_{k} A_{1} O=\left(\frac{k(k+1)}{2}\right)^{\circ} .
\end{array}
$$
Since $A_{k} A_{k+1}60^... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,126 |
3. As shown in Figure 4, AM is the angle bisector of $\triangle A B C$, and $D, E$ are points on sides $A B, A C$ respectively. $D E$ intersects $A M$ at point $F$. If $A D=1, A E=2, B D$ $=3, E C=4$, then $\frac{A F}{A M}=$ $\qquad$ | 3. $\frac{5}{18}$.
Given that $A M$ bisects $\angle B A C$, we have
$$
\frac{D F}{F E}=\frac{A D}{A E}=\frac{1}{2}, \quad \frac{B M}{M C}=\frac{A B}{A C}=\frac{2}{3} \text {. }
$$
Therefore, $S_{\triangle N F F}=\frac{1}{3} S_{\triangle A D E}, S_{\triangle B B Y}=\frac{2}{5} S_{\triangle A B C}$, which means
$$
\beg... | \frac{5}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,127 |
One, (20 points) As shown in Figure 5, in the acute triangle $\triangle ABC$, $AB > AC, \angle A = 60^{\circ}$, and the two altitudes $BE$ and $CF$ intersect at point $H$. Prove:
$$
\begin{array}{l}
2 AC < \sqrt{3}(B H + \\
C H) < 2 AB .
\end{array}
$$ | As shown in Figure 7, extend $B E$ to $M$ such that $H M = H C$. Connect $A M$, $C M$, and $A H$. Then $\triangle C H M$ is an equilateral triangle, and $C M = H C$.
Since $A C \perp B E$, we have
$$
\begin{array}{l}
\angle H C E = \angle M C E \\
= 30^{\circ}.
\end{array}
$$
Also, $A C = A C$, so
$$
\triangle A H C \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,129 |
II. (25 points) Given a two-digit number, whose tens and units digits are $p$ and $q$ respectively, the quadratic function $y=x^{2}+q x+p$ intersects the $x$-axis at two distinct points $A$ and $B$, with the vertex at $C$, and $S_{\triangle A B C} \leqslant 1$.
(1) Find the range of $q^{2}-4 p$;
(2) Find all such two-d... | (1) Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)\left(x_{1} \neq x_{2}\right)$, then $x_{1}, x_{2}$ are the two distinct real roots of the equation $x^{2}+q x+p=0$. Therefore,
$$
x_{1}+x_{2}=-q, x_{1} x_{2}=p, q^{2}-4 p>0 \text{. }
$$
Also, $y_{c}=\frac{4 p-q^{2}}{4}\left(y_{c}\right.$ represents the y-coordina... | 23,65,34,86 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,130 |
Three. (25 points) Given that $a, b, c$ are real numbers. The functions are $y_{1} = a x^{2} + b x + c, y_{2} = a x + b (a > 0)$. When $-1 \leqslant x \leqslant 1$, it is given that $-1 \leqslant y_{1} \leqslant 1$ and $y_{2}$ has a maximum value of 2. Try to find the area of the figure formed by connecting in sequence... | Three, from $a>0$, we know that $y_{2}$ increases as $x$ increases.
Therefore, $a+b=2$.
When $x=0,1$, $-1 \leqslant c \leqslant 1, -1 \leqslant a+b+c \leqslant 1$, so $-1 \leqslant c=(a+b+c)-2 \leqslant 1-2 \leqslant-1$.
Thus, $c=-1$. Therefore, when $x=0$, $y_{1}=-1$ is the minimum value of $y_{1}=$ $a x^{2}+b x+c$ in... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,131 |
1. Given that $a, b, c$ are all positive numbers and none of them equals 1. If real numbers $x, y, z$ satisfy $a^{x}=b^{y}=c^{z}, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ $=0$, then the value of $a b c$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) 1
(C) 2
(D) 4 | -、1.B.
Let $a^{x}=b^{y}=c^{z}=t(t>0$, and $t \neq 1)$, then
$$
\frac{1}{x}=\log _{t} a, \frac{1}{y}=\log _{t} b, \frac{1}{z}=\log _{t} c \text {. }
$$
Since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\log _{6} a+\log _{6} b+\log _{2} c$ $=\log _{1}(a b c)=0$,
therefore, $a b c=1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,132 |
2. For any $x \in \mathbf{R}, f(x)=|\sin x|$, when $n \leqslant x$ $<n+1$ ( $n$ is an integer) then, $g(x)=x-n$. Among the 4 functions $f(x) 、 g(x) 、 f(x)+g(x) 、 f(x) g(x)$, the number of functions that are definitely periodic is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2.B.
It is easy to verify that the periods of $f(x)$ and $g(x)$ are $\pi$ and 1, respectively. Neither $f(x)+g(x)$ nor $f(x) g(x)$ are periodic functions.
In fact, if $f(x)+g(x)$ is a periodic function with period $T (T \neq 0)$, then for any $x \in \mathbf{R}$, we have
$$
f(x+T)+g(x+T)=f(x)+g(x) .
$$
In equation (1)... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,133 |
Example 1 A positive integer $n$ can be divided by 4 if and only if there exist $n$ integers, whose product is $n$, and whose sum is zero. | Proof: Sufficiency. Suppose there exist $n$ integers $a_{1}, a_{2}$, $\cdots, a_{n}$ satisfying
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=0, \\
a_{1} a_{2} \cdots a_{n}=n .
\end{array}
$$
If $n$ is odd, by equation (2), all $a_{i}(i=1$, $2, \cdots, n)$ are odd, and the sum of an odd number of odd numbers is odd, wh... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,134 |
Example 2 On a plane, there are $2 n$ points, no three of which are collinear. Among these points, $n$ points are colored red, and the other $n$ points are colored blue. Prove: There exists a line such that on each side of this line, the number of red points and blue points is the same.
untranslated text remains the ... | Proof: First, let's introduce a concept.
A convex polygon that contains a set of planar points $M$ and has the points of $M$ as its vertices is called the convex hull of the planar point set $M$, i.e., the smallest convex set containing $M$.
Suppose the convex hull of $2n$ known points is $P$, and let's assume $P$ is ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,135 |
Example 6 Consider two concentric circles in the same plane with radii $R$ and $r$ $(R>r)$. Let $P$ be a fixed point on the circumference of the smaller circle, $B$ be a moving point on the circumference of the larger circle, and the line $BP$ intersects the larger circle at another point $C$. The line $l$ passing thro... | Analysis:
(1) As shown in Figure 7, establish a Cartesian coordinate system. Let $\angle B P x = \alpha, A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3})$. Suppose $P B$ intersects the smaller circle at another point $A'$, with $A'(-x_{1}, -y_{1})$, then
$$
\begin{array}{l}
x_{1} = -r \cos 2 \alpha, y_{1} = -r \sin 2 ... | \left(x - \frac{r}{2}\right)^{2} + y^{2} = \frac{R^{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,136 |
6. As shown in Figure 2, a circle is inscribed in a square with side length 2, another square is inscribed in this circle, and then another circle is inscribed in the second square. This process continues. If $S_{n}$ represents the total area of the first $n$ circles, then $\lim S_{n}=(\quad)$.
(A) $2 \pi$
(B) $\frac{3... | 6.A.
The diameter of the 1st circle is equal to the side length of the square, $r_{1}=1$; the radius of the 2nd circle $r_{2}=\frac{\sqrt{2}}{2} r_{1}=\frac{\sqrt{2}}{2}$; generally, the radius of the $n$th circle $r_{n}=\frac{\sqrt{2}}{2} r_{n-1}$, which is a geometric sequence with the first term 1 and common ratio ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,137 |
1. Given $-2 \pi<\alpha<\beta<-\pi$. Then the range of $2 \alpha-\beta$ is $\qquad$ | 2. $(-3 \pi,-\pi)$.
From the given, we have $-4 \pi<2 \alpha<-2 \pi, \pi<-\beta<2 \pi$. Adding these, we get
$$
-3 \pi<2 \alpha-\beta<0 \text {. }
$$
But $\alpha<\beta$, so
$$
2 \alpha-\beta<2 \beta-\beta=\beta<-\pi \text {. }
$$
Combining (1) and (2), we get $-3 \pi<2 \alpha-\beta<-\pi$.
Analysis. A common mistake i... | -3 \pi<2 \alpha-\beta<-\pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,138 |
2. When point $P(x, y)$ is any point on the line $l$, point $Q(4 x+2 y, x+3 y)$ is also a point on this line. Then the equation of the line $l$ is $\qquad$ . | 2. $x+y=0$ or $x-2 y=0$.
Let points $P$ and $Q$ coincide (invariant points under transformation), we get
$$
\left\{\begin{array}{l}
x=4 x+2 y, \\
y=x+3 y .
\end{array}\right.
$$
Thus, the origin $(0,0)$ must lie on the line. According to this, by the collinearity of points $O, P, Q$, we get
$$
\frac{y}{x}=\frac{x+3 y... | x+y=0 \text{ or } x-2 y=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,139 |
3. Given $\lg a<0, \lg b<0, \lg c<0$, and $\lg (a+b+c)=0$. Then the maximum value of $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 0 .
Given that $a, b, c$ are positive numbers less than 1, and $a+b+c=1$. We have
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
$\geqslant 9$ (Cauchy-Schwarz inequality).
The equality holds when $a=b=c=\frac{1}{3}$.
Transforming the inequality (1), we get
$$
\begi... | 0 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,140 |
4. Given that the graph of a quadratic function is tangent to the lines containing the three sides of $\triangle A B C$. If the coordinates of the vertices of the triangle are $A\left(\frac{1}{4}, \frac{3}{2}\right), B\left(\frac{1}{2}, 1\right), C\left(\frac{3}{4}, 1\right)$, then the expression of the quadratic funct... | 4. $y=x^{2}-2 x+2$.
Let the quadratic function be (undetermined coefficient method)
$$
y=a x^{2}+b x+c \quad(a \neq 0) .
$$
By the two-point form, the equations of the three sides of the triangle can be found as
$$
\begin{array}{l}
A B: y=-2 x+2, \\
B C: y=1, \\
C A: y=-x+\frac{7}{4} .
\end{array}
$$
(1) and (2) comb... | y=x^{2}-2 x+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,141 |
5. On the altitude $A H$ of the regular tetrahedron $A B C D$, take the midpoint $M$, and connect $B M 、 C M$, then $\angle B M C=$ $\qquad$ . | 5. $90^{\circ}$.
Let the edge length of the regular tetrahedron $ABCD$ be $a$. As shown in Figure 10, connect $BH$ intersecting $CD$ at $E$, then $BE$ is the median of the equilateral $\triangle BCD$, so $BE = \frac{\sqrt{3}}{2} \cdot a$.
Furthermore, we have
$$
\begin{array}{l}
B H = \frac{2}{3} B E \\
= \frac{\sqrt... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,142 |
6. The sum of $n$ consecutive natural numbers starting from the positive integer $m$ is 2004, and $(m, n)>1$ (not coprime). Then the greatest common divisor $(m, n, 2004)=$
| 6. 12 .
According to the problem, $(m, n)>1$ and $m+(m+1)+\cdots+(m+n-1)=2004$, that is,
$$
\frac{(2 m+n-1) n}{2}=2004 \text {. }
$$
And $(2 m+n-1) n$
$$
=1 \times 4008=3 \times 1336=167 \times 24=501 \times 8 \text {, }
$$
where $2 m+n-1$ and $n$ are one odd and one even, and $2 m+n-1$ is greater than $n$. Therefor... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,143 |
Three. (20 points) Given the quadratic function $f(x)=a x^{2}+b x+c$.
(1) Prove: The roots of the equation $f(x)=x$ are also roots of the equation $f(f(x))=x$;
(2) Find the necessary and sufficient condition for the equation $f(f(x))=x$ to have 4 distinct real roots. | (1) Let $x_{i}$ be a root (fixed point) of the equation $f(x)=x$, then $f\left(x_{i}\right)=x_{i}$.
(1)
Substituting into the equation $f(f(x))=x$, we have the left side $=f\left(f\left(x_{i}\right)\right)=f\left(x_{i}\right)$ (by (1))
$$
=x_{i} \text{ (by (1)) }=\text{ right side. }
$$
This shows that $x_{i}$ is also... | (b-1)^{2}-4 a c>4 | Algebra | proof | Yes | Yes | cn_contest | false | 714,144 |
Four. (20 points) As shown in Figure 3, above the $O x$-axis is the upper part of the ellipse $\frac{x^{2}}{2}+y^{2}=1$, and below the $O x$-axis is the rectangle $A B C D$, where the length $A B$ of the rectangle equals the major axis of the ellipse, and the width of the rectangle equals the minor axis of the ellipse.... | It is known that $A(\sqrt{2}, 0)$, $B(-\sqrt{2}, 0)$, $C(-\sqrt{2},-2)$, $D(\sqrt{2},-2)$, and $P(\sqrt{2} \cos \theta, \sin \theta)(\theta \in(0, \pi))$. Then the equation of $P C$ is
$$
\frac{x+\sqrt{2}}{y+2}=\frac{\sqrt{2} \cos \theta+\sqrt{2}}{\sin \theta+2} \text {. }
$$
Let $y=0$, we get the x-coordinate of poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,145 |
Five. (20 points) The three sides of a triangle are positive integers in an arithmetic sequence, and the longest side is no greater than the positive integer $n$. The number of such triangles is denoted as $f(n)$ (congruent triangles are counted as one). Write down $f(1), f(2), f(3), f(4)$, and then find the formula fo... | When $n=1,2,3$, there are only equilateral triangles with a common difference of 0, $f(1)=1, f(2)=2, f(3)=3$.
When $n=4$, in addition to the equilateral triangles with a common difference of 0 and side lengths of $1,2,3,4$, there are also integer-sided triangles with a common difference of 1, $(2,3,4)$, giving $f(4)=5... | \frac{(n+1)(n+2)}{6}, \text{ if } 3 \nmid n; \frac{n(n+3)}{6}, \text{ if } 3 \mid n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,146 |
Example 7 In $\triangle ABC$, $AB=AC$, there is a circle that is internally tangent to the circumcircle of $\triangle ABC$, and is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively.
Prove that the midpoint of the line segment $PQ$ is the incenter of $\triangle ABC$.
(20th IMO) | Analysis: Let $O$ be the midpoint of $P Q$, then $O$ lies on the angle bisector $A D$ of $\angle B A C$. As shown in Figure 8, establish a rectangular coordinate system, set $O A=1, \angle B A O=\alpha$. Then the distances from $O$ to $A B$ and $A C$ are both $\sin \alpha$, so we only need to prove that the distance fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,147 |
One, (50 points) As shown in Figure 4, in the right triangles $\triangle OAB$ and $\triangle OCD$ sharing the vertex $O$, the legs $OB = OC$. If $AC \perp OD$, then $DB \perp OA$.
| As shown in Figure 11, extend $AC$ to intersect $OD$ at point $E$. In the right triangle $\triangle OCD$, since $AC \perp OD$, we have $OE \cdot OD = OC^2$.
Draw $BF \perp OA$ from point $B$ at point $F$. In the right triangle $\triangle OAB$, we have $OF \cdot OA = OB^2$.
Since $OB = OC$, it follows that
$OE \cdot O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,148 |
II. (50 points) From intuitive operation, if $\triangle A O B, \triangle B O C, \triangle C O A$ can be assembled into a triangular pyramid model $O-A B C$, then in $\triangle A B C$ there is the sum of any two sides greater than the third side. Based on this, make a general conjecture:
Proposition For positive number... | (1) When $\theta_{1}$ is all obtuse angles, it is a true proposition.
By $\cos \theta_{1}a, \\
\sqrt{b^{2}+c^{2}-2 b c \cos \theta_{2}}>c . \\
\text { Therefore, } \sqrt{a^{2}+b^{2}-2 a b \cos \theta_{1}}+\sqrt{b^{2}+c^{2}-2 b c \cos \theta_{2}} \\
>a+c=\sqrt{a^{2}+c^{2}+2 a c} \\
>\sqrt{a^{2}+c^{2}-2 a c \cos \theta_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,149 |
Example 1 In a non-isosceles acute triangle \( \triangle ABC \), the angle bisector of the acute angle formed by the altitudes \( AD \) and \( CF \) intersects sides \( AB \) and \( BC \) at points \( P \) and \( Q \), respectively. The angle bisector of \( \angle B \) intersects the segment connecting the orthocenter ... | Proof: As shown in Figure 1, extend $H M$ to $K$ such that $M K = H M$, and connect $A K$ and $C K$. Then $A H C K$ is a parallelogram.
Thus, $K C \perp B C$, $K A \perp A B$. Draw perpendiculars from $R$ to $A B$ and $B C$, with the feet of the perpendiculars being $P^{\prime}$ and $Q^{\prime}$, respectively. Then $R ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,152 |
Example 2 As shown in Figure 2, given that $B_{1}$ and $C_{1}$ are points on sides $AC$ and $AB$ of $\triangle ABC$, respectively, and $CC_{1}$ and $BB_{1}$ intersect at $D$. Prove: $\triangle ABD$ and $\triangle ACD$'s incircles are externally tangent if and only if quadrilateral $AB_{1}DC_{1}$ has an incircle. | Proof: When quadrilateral $A B_{1} D C_{1}$ has an incircle $\odot I$, let $\odot I$ touch $B B_{1}$ and $C C_{1}$ at points $B_{3}$ and $C_{3}$, respectively. Let the incircle of $\triangle A B D$, $\odot I_{1}$, touch $B B_{1}$ at point $B_{2}$, and the incircle of $\triangle A C D$, $\odot I_{2}$, touch $C C_{1}$ at... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,153 |
Example 3 Given $\triangle A B C$, construct a semicircle with the base $B C$ as the diameter, which intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Draw perpendiculars from $D$ and $E$ to $B C$, with feet of the perpendiculars being $F$ and $G$ respectively. The line segments $D G$ and $E F$ inter... | Prove: As shown in Figure 3, draw $A P \perp B C$ at $P$. It is sufficient to prove that $A, M, P$ are collinear, i.e., to prove
$$
\begin{array}{l}
\frac{F P}{P C} \cdot \frac{C A}{A E} \cdot \frac{E M}{M F} \\
=1 .
\end{array}
$$
From $D F / / A P$ we know
$$
\frac{F P}{B P}=\frac{D A}{A B} \text {. }
$$
Applying C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,154 |
Example 4 In an acute triangle $\triangle ABC$, $AD$ is the internal angle bisector of $\angle A$, point $D$ is on side $BC$, and through point $D$ perpendiculars $DE \perp AC$ and $DF \perp AB$ are drawn, with feet of the perpendiculars being $E$ and $F$ respectively. Connecting $BE$ and $CF$, they intersect at point ... | Proof: As shown in Figure 4, draw $A K \perp B C$ at $K$, then $\frac{B K}{K C}=$ $\frac{c \cos B}{b \cos C}$
First, we prove that
$A, H, K$ are collinear.
From $D F \perp$
$A B$ and $D E \perp A C$,
we have
$$
\begin{array}{l}
\frac{B F}{F A} \cdot \frac{A E}{E C}=\frac{D F \cot B}{D F \cot \angle D A F} \cdot \frac{D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,155 |
Example 5 Let $\odot O_{1}$ and $\odot O_{2}$ be externally tangent at $W$, and both are internally tangent to $\odot O$. Draw an external common tangent of $\odot O_{1}$ and $\odot O_{2}$ intersecting $\odot O$ at points $A$ and $B$, and draw an internal common tangent of $\odot O_{1}$ and $\odot O_{2}$ intersecting $... | Proof: As shown in Figure 5, let $AB$ be tangent to $\odot O_{1}$ and $\odot O_{2}$ at points $X$ and $Y$, respectively. $\odot O_{1}$ and $\odot O_{2}$ are tangent to $\odot O$ at points $P$ and $Q$, respectively. Extend $PX$ to intersect $\odot O$ at point $M$, and draw the common tangent $PT$ through point $P$. Then... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,156 |
Example 6 In a non-isosceles $\triangle ABC$, the incenter is $O$, and the points of tangency with sides $BC, CA, AB$ are $A_{1}, B_{1}, C_{1}$ respectively. $AA_{1}$ and $BB_{1}$ intersect $\odot O$ at $A_{2}$ and $B_{2}$ respectively. The angle bisector of $\angle C_{1} A_{1} B_{1}$ intersects $B_{1} C_{1}$ at point ... | Proof: (1) As shown in Figure 6, since \(A B_{1}\) and \(A C_{1}\) are tangents to \(\odot O\), we have
\[
\frac{A_{2} C_{1}}{A_{1} C_{1}} = \frac{A C_{1}}{A A_{1}} = \frac{A B_{1}}{A A_{1}} = \frac{A_{2} B_{1}}{A_{1} B_{1}}.
\]
Thus, \(\frac{A_{2} C_{1}}{A_{2} B_{1}} = \frac{A_{1} C_{1}}{A_{1} B_{1}}\).
Furthermore, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,157 |
Example 3 Try to find a positive integer $k$ other than 1, such that $k$ and $k^{4}$ can both be expressed as the sum of squares of two consecutive integers, and prove that such a $k$ is unique.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | $$
\begin{array}{l}
\text { Solution: Let } k=n^{2}+(n+1)^{2} . \\
\text { By }\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2}, \text { we have } \\
k^{2}=(2 n+1)^{2}+[2 n(n+1)]^{2}, \\
k^{4}=\left(4 n^{4}+8 n^{3}-4 n-1\right)^{2}+ \\
\quad\left(8 n^{3}+12 n^{2}+4 n\right)^{2} . \\
\text { Let }\l... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,158 |
Example 1 Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=$ 5. If $s=x^{2}+y^{2}$, then $\frac{1}{s_{\text {max }}}+\frac{1}{s_{\text {min }}}=$ $\qquad$ .
(1993. National High School Mathematics Competition) | Analysis: From $x^{2}+y^{2}=s$, we associate it with $\sin ^{2} \theta+\cos ^{2} \theta$ $=1$. Let's set $x=\sqrt{s} \cos \theta, y=\sqrt{s} \sin \theta$, and substitute into the given equation to get
$4 s \cos ^{2} \theta-5 s \cos \theta \cdot \sin \theta+4 s \sin ^{2} \theta=5$.
Solving this, we get $s=\frac{10}{8-5 ... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,159 |
Example 2 Given a positive integer $n$ and a positive number $M$. For all arithmetic sequences $a_{1}$, $a_{2}$, $a_{3}$, ..., satisfying the condition $a_{1}^{2}+a_{n+1}^{2} \leqslant M$, find the maximum value of $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$.
(1999, National High School Mathematics Competition) | Analysis: Noting that $a_{1}^{2}+a_{n+1}^{2} \leqslant M$, let $a_{1}=$ $\sqrt{k} \cos \theta, a_{n+1}=\sqrt{k} \sin \theta$, where $0 \leqslant k \leqslant M, 0 \leqslant$ $\theta<2 \pi$. Then
$$
\begin{aligned}
S & =\frac{n+1}{2}\left(a_{n+1}+a_{2 n+1}\right)=\frac{n+1}{2}\left(3 a_{n+1}-a_{1}\right) \\
& =\frac{n+1}... | \frac{(n+1) \sqrt{10 M}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,160 |
Example 3 The sequences $a_{0}, a_{1}, \cdots$ and $b_{0}, b_{1}, \cdots$ are defined as follows:
$$
\begin{array}{l}
a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, \\
n=0,1,2, \cdots ; \\
b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n=0,1,2, \cdots
\end{array}
$$
Prove that ... | Analysis: From the given conditions, we have $0$ $0, n=0,1,2, \cdots$. Let
$$
a_{n}=\sin \alpha_{n}, b_{n}=\tan \beta_{n}, \alpha_{n}, \beta_{n} \in\left(0, \frac{\pi}{2}\right).
$$
Then $a_{n+1}=\sin a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos a_{n}}=\sin \frac{a_{n}}{2}$.
Therefore, $\alpha_{n+1}=\frac{\alpha_{n}}{2}$, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,161 |
Example 4 Given any 7 real numbers, prove: there are 2 real numbers $x, y$, satisfying $0 \leqslant \frac{x-y}{1+x y}<\frac{\sqrt{3}}{3}$.
(16th Canadian Mathematical Olympiad) | Analysis: Based on the structural characteristics of $\frac{x-y}{1+x y}$, let these 7 real numbers be $\tan \theta_{i}, \theta_{i} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), i=1,2, \cdots, 7$. Divide $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ into 6 equal intervals. According to the pigeonhole principle, at least... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,162 |
Example 5 Let $a, b, c$ be positive real numbers, and $abc + a + c = b$. Try to determine the maximum value of $p = \frac{2}{a^2 + 1} - \frac{2}{b^2 + 1} + \frac{3}{c^2 + 1}$.
(1999, Vietnam Mathematical Olympiad) | Analysis: From the given conditions, we have $a+c=(1-a c) b$. Clearly, $1-a c \neq 0$. Therefore, $b=\frac{a+c}{1-a c}$. Let $a=\tan \alpha$, $b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$. Then
$$
\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \tan \gamma}=\tan (\a... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,163 |
Example 6 The range of the function $y=x+\sqrt{x^{2}}-3x+2$ is $\qquad$
(2001, National High School Mathematics Competition) | Analysis: From $x^{2}-3 x+2 \geqslant 0$, we get $x \leqslant 1$ or $x \geqslant 2$. Also, $x^{2}-3 x+2=\left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}$. Let $x-\frac{3}{2}=\frac{1}{2} \sec \theta, \theta \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]$, then $y=\frac{3}{2}+\frac{1}{2} \sec \theta+\frac... | \left[1, \frac{3}{2}\right) \cup [2, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,164 |
A paper has a circle $\odot O$ with radius $R$ and a fixed point $A$ inside the circle, where $OA=a$. Fold the paper so that a point $A^{\prime}$ on the circumference coincides exactly with point $A$. Each such fold leaves a straight line crease. Find the set of points on all such crease lines when $A^{\prime}$ takes a... | Solution 1: As shown in Figure 1, establish a Cartesian coordinate system, then we have $A(a, 0)$. Connect $O A^{\prime}$ to intersect the crease $l$ at point $M$, and connect $M A$. Then
$$
|M A| = |M A^{\prime}|
$$
$$
|O M| + |M A| = |O M| + |M A^{\prime}| = R.
$$
By the definition of an ellipse, the locus of point ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,165 |
In a convex quadrilateral $E F G H$, the vertices $E, F, G, H$ are on the sides $A B, B C, C D, D A$ of the convex quadrilateral $A B C D$, respectively, and satisfy $\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1$. The points $A, B, C, D$ are on the sides $H_{1} E_{1}, E_{1} F_{1},... | (1) As shown in Figure 1, if
$E F / / A C$, then
$$
\frac{B E}{E A}=\frac{B F}{F C} .
$$
Substituting the known conditions, we get $\frac{D H}{H A}=\frac{D G}{G C}$, so,
$$
H G / / A C \text {. }
$$
Thus, $E_{1} F_{1} / / A C / / H_{1} G_{1}$. Therefore, $\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda$.
(2) ... | \lambda | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,166 |
Given a positive integer $c$, let the sequence $x_{1}, x_{2}, \cdots$ satisfy $x_{1}=$ $c$, and $x_{n}=x_{n-1}+\left[\frac{2 x_{n-1}-(n+2)}{n}\right]+1, n=2,3, \cdots$, where $[x]$ denotes the greatest integer not exceeding $x$. Find the general term formula for the sequence $\left\{x_{n}\right\}$.
(Huang Yumin) | Obviously, when $n \geqslant 2$,
$$
x_{n}=x_{n-1}+\left[\frac{2\left(x_{n-1}-1\right)}{n}\right] \text {. }
$$
Let $a_{n}=x_{n}-1$, then
$$
\begin{array}{l}
a_{1}=c-1, \\
a_{n}=a_{n-1}+\left[\frac{2 a_{n-1}}{n}\right]=\left[\frac{n+2}{n} \cdot a_{n-1}\right], n=2,3, \cdots .
\end{array}
$$
Let $u_{n}=A \cdot \frac{(n... | x_{n}=\begin{cases}
\frac{c-1}{6}(n+1)(n+2)+1 & \text{if } c \equiv 1(\bmod 3) \\
\frac{c-2}{6}(n+1)(n+2)+n+1 & \text{if } c \equiv 2(\bmod 3) \\
\frac{c-3}{6}(n+1)(n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,167 |
Three, let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum... | Three, prove $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} \cdots A_{7}$ with vertices in $M$, and connect $A_{1} A_{5}$. By condition (2), there is at least one point in $M$ within the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_{5}$. Then, in the... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,168 |
Example 4 In the Cartesian coordinate system, if the coordinates of a point $\left(x_{0}, y_{0}\right)$, $x_{0}$ and $y_{0}$, are both integers, then the point is called an integer point. Try to prove that there does not exist a regular $n(n \geqslant 7)$-sided polygon in the coordinate plane such that all its vertices... | Proof: Suppose a regular $n$-sided polygon exists that satisfies the conditions, and the regular $n$-sided polygon with the shortest side length is $A_{1} A_{2} \cdots A_{n}$, then $A_{1}$, $A_{2}, \cdots, A_{n}$ are all lattice points.
In the coordinate plane, take any lattice point $O$, and draw $O B_{1}$ parallel a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,169 |
Given real number $a$ and positive integer $n$. Prove:
(1) There exists a unique real number sequence $x_{0}, x_{1}, \cdots, x_{n}, x_{n+1}$, satisfying
$$
\left\{\begin{array}{l}
x_{0}=x_{n+1}=0, \\
\frac{1}{2}\left(x_{i+1}+x_{i-1}\right)=x_{i}+x_{i}^{3}-a^{3}, i=1,2, \cdots, n .
\end{array}\right.
$$
(2) For the sequ... | (1) Existence. From $x_{i+1}=2 x_{i}+2 x_{i}^{3}-2 a^{3}-x_{i-1}$, $i=1,2, \cdots$ and $x_{0}=0$, we know that each $x_{1}$ is a real-coefficient polynomial of degree $3^{1-1}$ in $x_{1}$, thus, $x_{n+1}$ is a real-coefficient polynomial of degree $3^{n}$ in $x_{1}$. Since $3^{n}$ is an odd number, there exists a real ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,170 |
Given a positive integer $n(n \geqslant 2)$, let positive integers $a_{i}(i=1,2$, $\cdots, n$ ) satisfy $a_{1}<a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$.
Prove: For any real number $x$, we have
$$
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \times \frac{1}{... | When $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$, from $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$ we can get
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \frac{1}{a_{1}^{2}+x^{2}}\right)^{2} \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}|x|}\right)^{2} \\
=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,171 |
Six, Proof: Except for a finite number of positive integers, all other positive integers $n$ can be expressed as the sum of 2004 positive integers: $n=a_{1}+a_{2}+\cdots+a_{2004}$, and satisfy $1 \leqslant a_{1}<a_{2}<\cdots<a_{2004}, a_{i} \mid a_{i+1}$, $i=1,2, \cdots, 2003$.
(Chen Yonggao) | We prove a more general conclusion:
For any positive integer $r (r \geqslant 2)$, there always exists a positive integer $N(r)$, such that when $n \geqslant N(r)$, there exist positive integers $a_{1}, a_{2}, \cdots, a_{r}$, such that $n=$ $a_{1}+a_{2}+\cdots+a_{r}, 1 \leqslant a_{1}<a_{2}<\cdots<a_{r}, a_{i} \mid a_{i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,172 |
1. Given that $D$ is any point on side $AB$ of $\triangle ABC$. $E$ is any point on side $AC$, and connect $DE$. $F$ is any point on segment $DE$. Let $\frac{AD}{AB}=x, \frac{AE}{AC}=y, \frac{DF}{DE}=z$. Prove:
(1) $S_{\triangle BDF}=(1-x) y z S_{\triangle ABC}$,
$S_{\triangle CEF}=x(1-y)(1-z) S_{\triangle ABC}$;
(2) $... | 1. (1) As shown in Figure 2, we have
$$
\begin{array}{l}
S_{\triangle B D F}=z S_{\triangle B D E} \\
=z(1-x) S_{\triangle A B E} \\
=z(1-x) y S_{\triangle A B C}, \\
S_{\triangle C E F}=(1-z) S_{\triangle C D E} \\
=(1-z)(1-y) S_{\triangle A C D} \\
=(1-z)(1-y) x S_{\triangle A B C} .
\end{array}
$$
(2) From (1), we g... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,173 |
2. A class has 47 students, and the classroom has 6 rows, with 8 seats in each row. The seat located in the $i$th row and the $j$th column is denoted as $(i, j)$. For the new semester, the seats are to be rearranged. If a student's original seat is $(i, j)$ and the new seat is $(m, n)$, then the student's movement is d... | 2. Let the empty seat last semester be $\left(i_{0}, j_{0}\right)$, and the empty seat this semester be $\left(i_{1}, j_{1}\right)$.
Then
$$
\begin{aligned}
S= & {\left[\sum_{i=1}^{6} \sum_{j=1}^{8}(i+j)-\left(i_{0}+j_{0}\right)\right] } \\
& -\left[\sum_{i=1}^{6} \sum_{j=1}^{8}(i+j)-\left(i_{1}+j_{1}\right)\right] \\... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,174 |
3. As shown in Figure 1, $A B C D$ is a cyclic quadrilateral, $A C$ is the diameter of the circle, $B D \perp A C$, and the intersection of $A C$ and $B D$ is $E$. $F$ is on the extension of $D A$. Connect $B F$, and $G$ is on the extension of $B A$ such that $D G \parallel B F$. $H$ is on the extension of $G F$, and $... | 3. Connect $B H, E F, C G$. Since $\triangle B A F \backsim \triangle G A D$, then
$$
\frac{F A}{A B}=\frac{D A}{A G} \text {. }
$$
Since $\triangle A B E \backsim \triangle A C D$, then
$$
\frac{A B}{E A}=\frac{A C}{D A} \text {. }
$$
(1) $\times$ (2) gives $\frac{F A}{E A}=\frac{A C}{A G}$.
Since $\angle F A E=\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,175 |
4. (1) Prove: There exist five non-negative real numbers $a, b, c, d, e$ whose sum is 1, such that when they are placed arbitrarily on a circle, there are always two adjacent numbers whose product is not less than $\frac{1}{9}$;
(2) Prove: For any five non-negative real numbers $a, b, c, d, e$ whose sum is 1, they can ... | 4. (1) When $a=b=c=\frac{1}{3}, d=e=0$, placing $a, b, c, d, e$ arbitrarily on a circle, there will always be two $\frac{1}{3}$ that are adjacent, and their product is no less than $\frac{1}{9}$.
(2) Suppose $a \geqslant b \geqslant c \geqslant d \geqslant e \geqslant 0$, and place $a, b, c, d, e$ as shown in Figure 3.... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,176 |
5. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{1}=2, a_{n+1}=a_{n}^{2}-a_{n}+1, n=1,2, \cdots \text {. }
$$
Prove: $1-\frac{1}{2003^{203}}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200}}<1$. | 5. From the given, we have $a_{n+1}-1=a_{n}\left(a_{n}-1\right)$. Therefore,
$$
\begin{array}{l}
\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}} \\
=\left(\frac{1}{a_{1}-1}-\frac{1}{a_{2}-1}\right)+\left(\frac{1}{a_{2}-1}-\frac{1}{a_{3}-1}\right)+\cdo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,177 |
6. Given a positive integer $n(n \geqslant 2)$. Find the largest real number $\lambda$, such that the inequality $a_{n}^{2} \geqslant \lambda\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n}$ holds for any positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}<a_{2}<\cdots<a_{n}$. | 6. When $a_{i}=i, i=1,2, \cdots, n$,
$$
\lambda \leqslant(n-2) \div \frac{n-1}{2}=\frac{2 n-4}{n-1} \text {. }
$$
The following proves the inequality
$$
a_{n}^{2} \geqslant \frac{2 n-4}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n}
$$
for any integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $0<a_{1}<a_{2}<\cdo... | \frac{2 n-4}{n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,178 |
7. Let the three sides of $\triangle ABC$ be $AB=c$, $BC=a$, and $CA=b$, where $a$, $b$, and $c$ are distinct. Let $AD$, $BE$, and $CF$ be the angle bisectors of $\triangle ABC$, and suppose $DE=DF$. Prove:
(1) $\frac{a}{b+c}=\frac{b}{c+a}+\frac{c}{a+b}$;
(2) $\angle BAC > 90^{\circ}$. | 7. As shown in Figure 4.
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{\sin \angle A F D}{\sin \angle F A D} \\
=\frac{A D}{F D}=\frac{A D}{E D}= \\
\frac{\sin \angle A E D}{\sin \angle D A E} \text {. } \\
\end{array}
$$
Thus, $\sin \angle A F D$
$$
=\sin \angle A E D \text {. }
$$
Therefore, $\angle A F D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,179 |
Example 5 Does there exist a natural number $n$, such that the sum of the digits of $n^{2}$ equals: (1) 2003; (2) 2002. | Solution: (1) Since the remainder of any natural number divided by 9 is equal to the remainder of the sum of its digits divided by 9, and $n^{2} \equiv$ $0,1,4,7(\bmod 9)$, while $2003 \equiv 5(\bmod 9)$, there does not exist such a natural number $n$ that the sum of the digits of $n^{2}$ equals 2003.
$$
\begin{array}{... | 2002 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,180 |
8. For any positive integer $n$, let the set of all positive divisors of $n$ be denoted as $S_{n}$. Prove: At most half of the elements in $S_{n}$ have a units digit of 3. | 8. Consider the following three cases:
(1) $n$ is divisible by 5. Let $d_{1}, d_{2}, \cdots, d_{m}$ be all the elements in $S_{n}$ whose unit digit is 3, then $S_{n}$ also includes $5 d_{1}, 5 d_{2}, \cdots, 5 d_{m}$, these $m$ elements whose unit digit is 5. Therefore, at most half of the elements in $S_{n}$ have a un... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,181 |
1. Place the numbers $1,2,3,4,5,6,7,8$ on the eight vertices of a cube such that the sum of any three numbers on each face is not less than 10. Find the minimum value of the sum of the four numbers on each face. | 1. Let the four numbers on a certain face be $a_{1}, a_{2}, a_{3}, a_{4}$, whose sum reaches the minimum value, and $a_{1}<a_{2}<a_{3}<a_{4}$. Since the sum of three different positive integers less than 5 is at most 9, it follows that $a_{4} \geqslant 6$. Therefore,
$$
a_{1}+a_{2}+a_{3}+a_{4} \geqslant 16 \text {. }
$... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,182 |
2. Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy the condition
$$
\begin{array}{c}
\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1 . \\
\text { Find the maximum value of }\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \text { . }
\end{array}
$$ | 2. When $n=1$, $\left(a_{2}\right.$ $\left.-a_{1}\right)^{2}=1$, hence $a_{2}-a_{1}=$ $\pm 1$. It is easy to see that the maximum value sought is 1.
When $n \geqslant 2$, let $x_{1}$
$$
\begin{array}{l}
=a_{1}, x_{i+1}=a_{1+1}-a_{1}, \\
i=1,2, \cdots, 2 n-1 \text {. Then } \\
\sum_{i=2}^{2 n} x_{i}^{2}=1, \text { and }... | \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,183 |
3. Let $n$ be a given positive integer. Find the smallest positive integer $u_{n}$, satisfying: for every positive integer $d$, the number of integers divisible by $d$ in any $u_{n}$ consecutive positive odd numbers is not less than the number of integers divisible by $d$ in the odd numbers $1,3,5, \cdots$, $2 n-1$. | 3. $u_{n}=2 n-1$.
(1) First, prove $u_{n} \geqslant 2 n-1$.
Since $u_{1} \geqslant 1$, assume $n \geqslant 2$. In the sequence $1,3, \cdots, 2 n-1$, the number of terms divisible by $2 n-1$ is 1. In the sequence $2(n+1)-1, 2(n+2)-1, \cdots, 2(n+2 n-2)-1$, the number of terms divisible by $2 n-1$ is 0. Therefore, $u_{n... | u_{n}=2 n-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,184 |
4. Prove: If the sum of the distances from any point $P$ inside a convex quadrilateral $ABCD$ to the sides $AB, BC, CD, DA$ is a constant, then $ABCD$ is a parallelogram.
| 4. Let the notation $d(P, l)$ represent the distance from point $P$ to line $l$. First, we prove a lemma.
Lemma: Suppose $\angle S A T=\alpha$ is a fixed angle, then the locus of a moving point $P$ inside $\angle S A T$ such that the sum of the distances from $P$ to the sides $A S$ and $A T$ is a constant $m$ is the l... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,185 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{0}=0, a_{n+1}=k a_{n}+$ $\sqrt{\left(k^{2}-1\right) a_{n}^{2}+1}, n=0,1,2, \cdots$, where $k$ is a given positive integer. Prove: Every term of the sequence $\left\{a_{n}\right\}$ is an integer, and $2 k \mid$ $a_{2 n}, n=0,1,2, \cdots$. | 5. From the given, we have
$$
a_{n+1}^{2}-2 k a_{n} a_{n+1}+a_{n}^{2}-1=0 .
$$
Therefore, $a_{n+2}^{2}-2 k a_{n+1} a_{n+2}+a_{n+1}^{2}-1=0$.
Subtracting the above two equations, we get
$$
a_{n+2}^{2}-a_{n}^{2}-2 k a_{n+1} a_{n+2}+2 k a_{n} a_{n+1}=0,
$$
which simplifies to $\left(a_{n+2}-a_{n}\right)\left(a_{n+2}+a_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,186 |
6. Convex quadrilateral $A B C D$ has an inscribed circle, which touches sides $A B, B C, C D, D A$ at points $A_{1}, B_{1}, C_{1}, D_{1}$, respectively. Connect $A_{1} B_{1}, B_{1} C_{1}, C_{1} D_{1}, D_{1} A_{1}$, and let points $E, F, G, H$ be the midpoints of $A_{1} B_{1}, B_{1} C_{1}, C_{1} D_{1}, D_{1} A_{1}$, re... | 6. As shown in Figure 4, let $I$ be the incenter of quadrilateral $ABCD$. Since $H$ is the midpoint of $D_1A_1$, and $AA_1$ and $AD_1$ are the tangents to $\odot 1$ through point $A$, it follows that $H$ lies on $AI$, and $AI \perp A_1D_1$.
Also, $ID_1 \perp AD_1$, so by the projection theorem, we have $IH \cdot IA = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,187 |
7. Let non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $\sum_{i=1}^{5} \frac{1}{1+x_{1}} = 1$. Prove: $\sum_{i=1}^{5} \frac{x_{1}}{4+x_{i}^{2}} \leqslant 1$. | 7. Let $y_{i}=\frac{1}{1+x_{i}}, i=1,2, \cdots, 5$, then $x_{i}=\frac{1-y_{i}}{y_{1}}$ $i=1,2, \cdots, 5$, and $\sum_{i=1}^{5} y_{i}=1$. Therefore,
$$
\begin{array}{l}
\sum_{i=1}^{5} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1 \Leftrightarrow \sum_{i=1}^{5} \frac{-y_{i}^{2}+y_{i}}{5 y_{i}^{2}-2 y_{i}+1} \leqslant 1 \\
\Leftr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,188 |
8.1650 students are arranged in 22 rows and 75 columns. It is known that among any two columns in the same row, the number of pairs of students of the same gender does not exceed 11. Prove: the number of boys does not exceed 928. | 8. Let the number of boys in the $i$-th row be $a_{i}$, then the number of girls is $75-a_{i}$. According to the problem, we have
$$
\sum_{i=1}^{2}\left(C_{a_{i}}^{2}+C_{n s-a_{i}}^{2}\right) \leqslant 11 \times C_{n s}^{2} .
$$
This is because, for any two columns, the number of pairs of students in the same row who ... | 928 | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,189 |
1. Let the function $f(x)=\log _{a} x(a>0, a \neq 1)$. If $f\left(x_{1} x_{2} \cdots x_{2003}\right)=8$.
Then the value of $f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{2003}^{2}\right)$ is ( ).
(A) 4
(B) 8
(C) 16
(D) $2 \log _{u} 8$ | \begin{array}{l}-、 1 . \mathrm{C} \\ f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{200 B}^{2}\right) \\ =\log _{a} x_{1}^{2}+\log _{a} x_{2}^{2}+\cdots+\log _{a} x_{200 B}^{2} \\ =2 \log _{a} x_{1} x_{2} \cdots x_{200 B}=2 f\left(x_{1} x_{2} \cdots x_{200 B}\right)=16 .\end{array}
Translates to:
\b... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,190 |
Example: $\odot O_{1}$ and $\odot O_{2}$ are contained within $\odot O$ and are tangent to $\odot O$ at two distinct points $M$ and $N$. $\odot O_{1}$ passes through point $O_{2}$, and the line through the two intersection points of $\odot O_{1}$ and $\odot O_{2}$ intersects $\odot O$ at points $A$ and $B$. Lines $M A$... | Analysis: As shown in Figure 1, let the radii of $\odot 0$, $\odot O_{1}$, and $\odot O_{2}$ be $r$, $r_{1}$, and $r_{2}$, respectively. Let $\angle O_{2} M O = \alpha$. Then the equation of $\odot O_{1}$ is
$$
\left(x - r_{1}\right)^{2} + y^{2} = r_{1}^{2}
$$
The equation of $\odot O$ is
$$
\left(x - r_{1} - r_{1} \co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,191 |
2. As shown in Figure $1, S-A B C$ is a triangular pyramid with three edges mutually perpendicular, and $O$ is a point within the base $A B C$. If $\angle O S A=\alpha, \angle O S B$ $=\beta, \angle O S C=\gamma$, then the range of $\tan \alpha \cdot \tan \beta \cdot \tan \gamma$ is ( ).
(A) $[2 \sqrt{2},+\infty)$
(B) ... | 2.A.
As shown in Figure 6, through $O$, planes parallel to $S A$, $S B$, and $S C$ intersect $S A$, $S B$, and $S C$ at $D$, $E$, and $F$, respectively, thus forming a rectangular parallelepiped $M F S E-O N D P$ with $S O$ as the diagonal. According to the problem,
$$
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,192 |
3. A water tank is equipped with 9 inlet and outlet pipes numbered $1,2, \cdots, 9$. Some are only for inlet, and some are only for outlet. It is known that the pipe numbers opened and the time required to fill the tank are as shown in Table 1. If all 9 pipes are opened at the same time, the time required to fill the t... | 3.B.
Let it take $t$ hours to fill the pool, then
i.e., $\left[\frac{\frac{1}{2}\left(1-\frac{1}{16}\right)}{1-\frac{1}{2}}+\frac{\frac{1}{31}\left(1-\frac{1}{32}\right)}{1-\frac{1}{2}}\right] t=2$.
Solving for $t$ gives $t=2$. | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,193 |
4. If a circle with a diameter defined by a chord passing through a focus of a conic section has no common points with the corresponding directrix, then the conic section is ( ).
(A) Hyperbola
(B) Ellipse
(C) Parabola
(D) Ellipse or Hyperbola | 4.B.
As shown in Figure 7, F is the focus, and the corresponding directrix is $l$. AB is the focal chord, M is the midpoint of AB, and the projections of A, M, B on $l$ are G, N, H, respectively. Let $e$ be its eccentricity. According to the definition of conic sections, we have
$$
\begin{aligned}
A F & =e \cdot A G, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,194 |
5. There are 10 different balls, including 2 red balls, 5 yellow balls, and 3 white balls. If getting 1 red ball scores 5 points, getting 1 yellow ball scores 1 point, and getting 1 white ball scores 2 points, then the number of ways to draw 5 balls such that the total score is greater than 10 points and less than 15 p... | 5.C.
There are four scenarios for drawing different balls: red-red-white-yellow-yellow, red-red-yellow-yellow-yellow, red-white-white-white-yellow, red-white-white-yellow-yellow. The number of different ways to draw them is
$$
C_{3}^{1} C_{5}^{2} + C_{5}^{3} + C_{2}^{1} C_{3}^{3} C_{5}^{1} + C_{2}^{1} C_{3}^{2} C_{5}^... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,195 |
6. Natural numbers are arranged according to the following pattern:
$$
\begin{array}{rrrrr}
1-2 & 5 & 10 & 17 \\
4-3 & 1 & 1 & 1 \\
4 & 11 & 18 \\
9-8 & 1 & 1 & 1 \\
9-7 & 12 & 19 \\
16-15-14 & -13 & 20 \\
25-24-23-22-21
\end{array}
$$
Then the number in the 2002nd row from the top and the 2003rd column from the left ... | 6.D.
From the observation, we can deduce the arrangement characteristics of this natural number table:
(1) Every number in the 1st column is a perfect square, and it is exactly equal to the square of its row number, i.e., the 1st number in the $n$-th row is $n^{2}$;
(2) The $n$-th number in the 1st row is $(n-1)^{2}+1... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,196 |
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2008}+2002(y-2)=1 .
\end{array}\right.
$$
Then $x+y=$ | 7. 3
Construct the function $f(t)=t^{2008}+2002 t$. It is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and it is also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y, x+y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,197 |
8. The acute angle that satisfies $2 \sin ^{2} x+\sin x-\sin 2 x=3 \cos x$ is = . $\qquad$ | 8. $\frac{\pi}{3}$.
Since $x$ is an acute angle, then $\cos x \neq 0$. Dividing both sides of the given equation by $\cos x$ yields $2 \sin x \cdot \tan x + \tan x - 2 \sin x = 3$, which simplifies to
$$
(2 \sin x + 1)(\tan x - 1) = 2 \text{.}
$$
Since the function $f(x) = (2 \sin x + 1)(\tan x - 1)$ is strictly incr... | \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,198 |
9. Let $\min \{a, b\}$ be the minimum of the two numbers $a$ and $b$. When positive numbers $x$ and $y$ vary, $t=\min \left\{x, \frac{y}{x^{2}+y^{2}}\right\}$ also varies. Then the maximum value of $t$ is | 9. $\frac{\sqrt{2}}{2}$.
$t^{2} \leqslant x \cdot \frac{y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2}$, so $t \leqslant \frac{\sqrt{2}}{2}$, with equality if and only if $x=y=\frac{\sqrt{2}}{2}$. | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,199 |
10. Given that $n$ is a natural number, the polynomial $P(x)=\sum_{h=0}^{n} \mathrm{C}_{n}^{h} x^{n-h} \cdot(x-1)^{h}$ can be expanded into an ascending power series of $x$ as $a_{0}+a_{1} x+a_{2} x^{2}+$ $\cdots+a_{n} x^{n}$. Then $\left|a_{0}\right|+\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=$ $\... | 10. $3^{n}$.
$$
\begin{array}{l}
P(x)=(x+x-1)^{n}=(2 x-1)^{n} \\
=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}(2 x)^{k}(-1)^{n-k},
\end{array}
$$
Therefore, $\sum_{k=0}^{n}\left|a_{k}\right|=\sum_{k=0}^{n} C_{n}^{k}(2)^{k}=3^{n}$. | 3^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,200 |
Example 2 A circle $\odot O$ is tangent to two parallel lines $l_{1}$ and $l_{2}$, a second circle $\odot O_{1}$ is tangent to $l_{1}$ at point $A$, and externally tangent to $\odot O$ at point $C$; a third circle $\odot O_{2}$ is tangent to $l_{2}$ at point $B$, externally tangent to $\odot O$ at point $D$, and extern... | Analysis: As shown in Figure 2, let $\angle O_{1} O x=\alpha, \angle O_{2} O x=\beta$, and the radii of the three circles be $R, r_{1}, r_{2}$, respectively. Then,
$$
\begin{array}{l}
O_{1}\left(\left(R+r_{1}\right) \cos \alpha,\left(R+r_{1}\right) \sin \alpha\right), \\
A\left(\left(R+r_{1}\right) \cos \alpha, R\right... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,202 |
12. Given that $a, b$ are distinct positive numbers, insert two sets of numbers: $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$, such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence. Then among the following inequal... | 12.(1)、(4).
It is easy to know that $\frac{1}{n} \sum_{k=1}^{n} x_{n}=\frac{a+b}{2}=\sqrt{a b}+\frac{(\sqrt{a}-\sqrt{b})^{2}}{2}>$ $\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, so (1) holds, and (2) does not hold.
It is also easy to know that $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}=\sqrt{a b}=\left(\frac{\sq... | 12.(1)、(4) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,203 |
13. Given the curve $x y^{2}=1(y>0)$, the slope $k$ of the tangent line at point $\left(x_{0}, y_{0}\right)$ on the curve is $k=-\frac{1}{2 \sqrt{x_{0}^{3}}}$. A line parallel to the $y$-axis is drawn through point $P_{1}(1,0)$, intersecting the curve at $Q_{1}$. A tangent line is drawn through $Q_{1}$, intersecting th... | 13. Let $Q_{n}\left(x_{n}, y_{n}\right)$, then the slope $k$ of the line $Q_{n} P_{n+1}$ is $-\frac{1}{2 \sqrt{x_{n}^{3}}}$. Therefore, the equation of $Q_{n} P_{n+1}$ is
$$
y-y_{n}=-\frac{1}{2 \sqrt{x_{n}^{3}}}\left(x-x_{n}\right) .
$$
Let $y=0$ to get $x_{n+1}=x_{n}+2 y_{n} \sqrt{x_{n}^{3}}$. Since $x_{n} y_{n}^{2}=... | \frac{1}{3^{1001}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 714,204 |
14. Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function
$$
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
$$
and provide a proof. | 14. Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1} \backslash t_{2} \in(0,1)$ and $t_{1}<t_{2}$, we have
$$
\left(t_{1}-t_{2}\ri... | 0 | Algebra | proof | Yes | Yes | cn_contest | false | 714,205 |
15. As shown in Figure 4, in tetrahedron $ABCD$, $AB \perp BC$, $BC \perp CD$, $CD \perp AB$.
(1) Identify the side face that is perpendicular to face $BCD$, and prove it;
(2) If $AB = BC = 1$,
$CD = x$, the dihedral angle $C-AD$
$-B$ has a plane angle $\alpha$, $\sin \alpha$
$= f(x)$, find the expression for $f(x)$ an... | 15. (1) From $A B \perp B C$ and $A B \perp C D$, we know $A B \perp$ plane $B C D$, hence plane $A B C \perp$ plane $B C D$, and plane $A B D \perp$ plane $B C D$.
(2) Draw $C E \perp B D$, with the foot of the perpendicular at $E$, then $C E \perp$ plane $A B D$. Draw $E F \perp A D$, with the foot of the perpendicul... | f(x)=\frac{1}{\sqrt{2}} \sqrt{1+\frac{1}{1+x^{2}}}\left(x \in \mathbf{R}_{+}\right), \left(\frac{\pi}{4}, \frac{\pi}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,206 |
16. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for $x>1$, and for any real numbers $x, y \in \mathbf{R}$, $f(x+y) = f(x) f(y)$ holds. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=f(0)$, and
$$
f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)} \quad(n \in \mathbf{N}) .
$$
(1) Find the value... | 16. (1) Let $x=-1, y=0$, we get
$f(-1)=f(-1) f(0), f(0)=1$. Hence $a_{1}=f(0)=1$.
When $x>0$, $-x<0, f\left(x_{2}-x_{1}\right)<1$.
\end{array}
$$
Therefore, $f\left(x_{1}\right)>f\left(x_{2}\right)$, the function $y=f(x)$ is a monotonically decreasing function on $\mathbf{R}$.
$$
\begin{array}{l}
\text { By } f\left(a... | a_{2000} = 3999, k = \frac{2}{3} \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,207 |
17. Company A and Company B, both had a market share of $A$ in 1996. According to market analysis and predictions, Company A's market share in the $n$-th year (1996 being the 1st year) is $a_{n}=\frac{A}{40}\left(n^{2}-n+40\right)$, and Company B's market share has increased year by year since 1996, following the patte... | Figure 5
17. (1) Let the market share of Company B in the $n$-th year be $b_{n}$. From the analysis of Figure 5, we get
$$
b_{n}=A+\frac{A}{2}+\frac{A}{4}+\cdots+\frac{A}{2^{n-1}}=\left(2-\frac{1}{2^{n-1}}\right) A .
$$
(2) According to the problem, 2015 is the 20th year, then
$$
\begin{array}{l}
a_{20}=\frac{A}{40}\le... | b_{20}<a_{20} 20 \% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,208 |
18. Find the number of right-angled triangles on the plane that satisfy the following conditions:
(1) The three vertices of the triangle are integer points, with the origin as the right-angle vertex;
(2) The coordinates of the incenter $M$ are $(96 p, 672 p)$, where $p$ is a prime number. | 18. Let the right triangle $\triangle OAB$ shown in Figure 8 satisfy the given conditions, then the slope of the line $OM$ is $\tan \alpha = 7$;
the slope of the line $OA$ is $\tan \left(\alpha - 45^{\circ}\right) = \frac{\tan \alpha - 1}{1 + \tan \alpha} = \frac{3}{4}$;
the slope of the line $OB$ is $-\frac{4}{3}$.
T... | 108, 42, 60 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,209 |
1. Given the function
$$
f(x)=\sqrt{\sin ^{4} x+4 \cos ^{2} x}-\sqrt{\cos ^{4} x+4 \sin ^{2} x} .
$$
Then $f(x)$ can be simplified to (.
(A) $\cos 2 x$
(B) $\sin 2 x$
(C) $\cos x-\sin x$
(D) $\cos \frac{x}{2}-\sin \frac{x}{2}$ | $-、 1 . A$
Since $\sin ^{4} x+4 \cos ^{2} x=\left(\sin ^{2} x-2\right)^{2}$, therefore,
$$
\sqrt{\sin ^{4} x+4 \cos ^{2} x}=2-\sin ^{2} x \text {. }
$$
Similarly, $\sqrt{\cos ^{4} x+4 \sin ^{2} x}=2-\cos ^{2} x$.
Thus, $f(x)=\cos ^{2} x-\sin ^{2} x=\cos 2 x$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,210 |
2. Given positive numbers $a_{1}, a_{2}, \cdots, a_{7}$ form a geometric sequence. If the sum of the first 5 terms is $7 \sqrt{2}+6$, and the sum of the last 5 terms is $14 \sqrt{2}+12$, then $a_{6}$ equals ( )
(A) 4
(B) $4 \sqrt{2}$
(C) 8
(D) $8 \sqrt{2}$ | 2.C.
Because $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$
$$
\begin{array}{l}
=a_{1}\left(1+q+q^{2}+q^{3}+q^{4}\right)=7 \sqrt{2}+6, \\
a_{3}+a_{4}+a_{5}+a_{6}+a_{7} \\
=a_{1} q^{2}\left(1+q+q^{2}+q^{3}+q^{4}\right)=14 \sqrt{2}+12,
\end{array}
$$
Dividing the second equation by the first, we get $q^{2}=2$.
Since $a_{1}, a_{2}, \c... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,211 |
3. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $E$ is the midpoint of $B C$, and $D$ is a moving point on $A A_{1}$, with $\frac{A D}{D A_{1}}=m$. If $A E / /$ plane $D B_{1} C$, then the value of $m$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) $\frac{2}{3}$
(D) 1 | 3.D.
Pass through $A A_{1}$ and $A E$ to make plane $A_{1} A E F$ intersect with plane $B C C_{1} B_{1}$ at $E F$. Let $E F$ intersect $B_{1} C$ at point $O$, and connect $D O$. Since $A A_{1} / /$ plane $B C C_{1} B_{1}, A E / /$ plane $D B_{1} C$, therefore,
$$
A A_{1} / / E F, A E / / D O \text {. }
$$
Thus, quadr... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,212 |
Example 3 Given two intersecting circles $\odot O_{1}$ and $\odot O_{2}$ in the plane, point $A$ is one of the intersection points. Two moving points $M_{1}$ and $M_{2}$ start from point $A$ simultaneously, moving at constant speeds along $\odot O_{1}$ and $\odot O_{2}$ in the same direction, and each returns to point ... | Analysis: As shown in Figure 4, let $O A=1, \angle A O_{1} O_{2}=\alpha$,
$$
\angle A O_{2} O_{1}=
$$
$\beta$, angular velocity is
$\omega, P\left(x_{0}\right.$,
$\left.y_{0}\right)$, then
$$
\begin{array}{l}
r_{1}=\csc \alpha, \\
r_{2}=\csc \beta .
\end{array}
$$
Therefore, after time $t$, the coordinates of $M_{1} 、... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,213 |
4. There are 20 cards each inscribed with the numbers $1,2, \cdots, 19,20$. They are placed in a box, and 4 people each draw one card. The two people who draw the smaller numbers are in one group, and the two people who draw the larger numbers are in another group. If two of them draw 5 and 14, the probability that the... | 4.D.
Since two people have drawn the cards 5 and 14, the other two need to draw from the remaining 18 cards, which results in $18 \times 17$ scenarios. If the two who drew 5 and 14 are in the same group, there are two cases:
(1) 5 and 14 are the smaller numbers, and the other two need to draw from the 6 cards numbered... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,214 |
5. Let the quadratic function $f(x)=a x^{2}+b x+c$ (where $a, b, c$ are integers), and 4 students calculate the function values. Student A gets: $f(7)=$ -1; Student B gets: $f(1)=3$; Student C gets: $f(4)=-4$; Student D gets: $f(2)=4$. Among them, only 1 student made a mistake, then the student who made the mistake is ... | 5.B.
Since $f(m)-f(n)=(m-n)(a m+a n+b)$, then $(m-n) \mid(f(m)-f(n))$.
Verification: $(7-1) \mid(-1-3),(7-4) \mid(-1+4)$,
$$
\begin{array}{l}
(7-2) \mid(-1-4),(1-4) \mid(3+4), \\
(1-2) \mid(3-4),(4-2) \mid(-4-4) .
\end{array}
$$
Therefore, B made a calculation error. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,215 |
6. Choose two different prime numbers between 4 and 18, then subtract their sum from their product, the resulting number can be ( ).
(A) 21
(B) 60
(C) 119
(D) 231 | 6.C.
Let the two chosen prime numbers be $x$ and $y$, and the resulting number be $T$, then
$$
T=xy-(x+y).
$$
Thus,
$$
T+1=xy-x-y+1=(x-1)(y-1).
$$
Since $x$ and $y$ are both primes greater than 4, $x$ and $y$ are both odd, and $(x-1)$ and $(y-1)$ are both even. Therefore, $4 \mid (T+1)$.
However, $4 \times 22, 4 \nm... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,216 |
7. The monotonic increasing interval of the function $f(x)=\log _{\frac{1}{3}}\left(x^{2}-5 x+6\right)$ is $\qquad$ . | 二、7. $(-\infty, 2)$.
Because the domain of the function $f(x)=\log _{\frac{1}{3}}\left(x^{2}-5 x+6\right)$ is $x^{2}-5 x+6>0$, which means $x<2$ or $x>3$. Also, $x^{2}-5 x+6=\left(x-\frac{5}{2}\right)^{2}-\frac{1}{4}$, its decreasing interval is $\left(-\infty, \frac{5}{2}\right]$, so the monotonic increasing interval ... | (-\infty, 2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,217 |
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=$ $b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is $\qquad$ | 8. 2 .
Let the inradius of $\triangle ABC$ be $r$. Since $a \cos A = b \cos B$, by the Law of Sines, we get $b \sin A = a \sin B$. Therefore, $\sin 2A = \sin 2B$.
Since $A \neq B$, then $A + B = 90^{\circ}$. Thus, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}, a^2 + b^2 = c^2$.
Also, since $c = 10$, and ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,218 |
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