problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
13. If $n^{2}+100$ can be divided by $n+10$, then, the maximum positive integer value of $n$ that satisfies the condition is $\qquad$ | 13. 190.
Since $n^{2}+100$ and $n^{2}-100$ are both divisible by $n+10$, therefore, 200 is divisible by $n+10$.
Hence, the largest positive integer that satisfies the condition is 190. | 190 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. The school offers four extracurricular interest classes in Chinese, Math, Foreign Language, and Natural Science for students to voluntarily sign up for. The number of students who want to participate in the Chinese, Math, Foreign Language, and Natural Science interest classes are 18, 20, 21, and 19, respectively. I... | Three, 15. The number of people not attending the Chinese interest class is 7, the number of people not attending the Math interest class is 5, the number of people not attending the Foreign Language interest class is 4, and the number of people not attending the Natural Science interest class is 6. Therefore, the maxi... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The line $l$ passes through the focus of the parabola $y^{2}=a(x+1)(a>0)$ and is perpendicular to the $x$-axis. If the segment cut off by $l$ on the parabola is 4, then $a=$ $\qquad$ . | $=.1 .4$.
Since the parabolas $y^{2}=a(x+1)$ and $y^{2}=a x$ have the same length of the focal chord with respect to their directrix, the general equation $y^{2}=a(x+1)$ can be replaced by the standard equation $y^{2}=a x$ for solving, and the value of $a$ remains unchanged. Using the formula for the length of the latu... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of right-angled triangles with integer side lengths and area (numerically) equal to the perimeter is $\qquad$ | 4.2.
Let the legs of a right-angled triangle be $a, b$, and $c = \sqrt{a^{2}+b^{2}} (a \leqslant b)$, then we have
$$
\frac{1}{2} a b = a + b + \sqrt{a^{2} + b^{2}}.
$$
Thus, $\frac{1}{2} a b - a - b = \sqrt{a^{2} + b^{2}}$.
Squaring both sides and simplifying, we get
$$
a b - 4 a - 4 b + 8 = 0.
$$
Then, $(a-4)(b-4)... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. A chord $AB$ is drawn through a focus $F$ of the ellipse $\frac{x^{2}}{6^{2}}+\frac{y^{2}}{2^{2}}=1$. If $|A F|=m,|B F|=n$, then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ | 5.3.
As shown in Figure 2, draw $A A_{1}$ perpendicular to the directrix at point $A_{1} . A E \perp x$-axis at point $E$.
$$
\begin{array}{l}
\text { Since } \frac{c}{a}=e=\frac{|F A|}{\left|A A_{1}\right|} \\
=\frac{|F A|}{|D F|+|F E|} \\
=\frac{m}{m \cos \alpha+\frac{b^{2}}{c}} .
\end{array}
$$
Therefore, $\frac{1... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ and a fixed point $P(t, 0)(t>0)$, a line $l$ with a slope of $\frac{1}{2}$ passes through point $P$ and intersects the ellipse $C$ at two distinct points $A$ and $B$. For any point $M$ on the ellipse, there exists $\theta \in[0,2 \pi]$, such that $O M=\co... | Let $M(x, y)$ and points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. The equation of line $AB$ is $x-2 y-t=0$.
From the cubic equation, we get
$2 x^{2}-2 t x+t^{2}-4=0$ and $8 y^{2}+4 t y+t^{2}-4=0$.
Thus, $x_{1} x_{2}=\frac{t^{2}-4}{2}$ and $y_{1} y_{2}=\frac{t^{2}-4}{8}$.
Given $O M=\cos \theta \cdot O ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (20 points) Let two positive real numbers $x, y, z$. Try to find the maximum value of the algebraic expression $\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}$.
| $$
\begin{array}{l}
\text { Six, because } \frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z} \\
=\frac{16 x+\frac{9 \sqrt{x \cdot 18 y}}{3}+\frac{3 \sqrt[3]{x \cdot 18 y \cdot 36 z}}{2}}{x+2 y+z} \\
\leqslant \frac{16 x+\frac{3(x+18 y)}{2}+\frac{x+18 y+36 z}{2}}{x+2 y+z}=18,
\end{array}
$$
Therefore, the equalit... | 18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Let $y=\frac{x}{1+x}=f(x)$, and $f(1)$ represents the value of $y$ when $x=1$, i.e., $f(1)=\frac{1}{1+1}=\frac{1}{2}$; $f\left(\frac{1}{2}\right)$ represents the value of $y$ when $x=\frac{1}{2}$, i.e., $f\left(\frac{1}{2}\right)=$ $\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3} ; \cdots \cdots$. Try to find
... | Explanation: Given $y=\frac{x}{1+x}=f(x)$, we have
$$
\begin{array}{l}
f\left(\frac{1}{n}\right)=\frac{\frac{1}{n}}{1+\frac{1}{n}}=\frac{1}{1+n} \\
f(n)=\frac{n}{1+n} \\
f(0)=\frac{0}{1+0}=0 .
\end{array}
$$
Thus, $f(n)+f\left(\frac{1}{n}\right)=\frac{1}{n+1}+\frac{n}{n+1}=1$.
Therefore, the original expression is
$$
... | 2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $A=\{1,2,3,4,5,6\}, f: A \rightarrow A$, the number of mappings $f$ that satisfy $f(f(x)) \neq x$ is $\qquad$ | 2.7360 .
First, there should be $f(x) \neq x$, and the number of mappings satisfying $f(x) \neq x$ is $5^{6}$.
If $x \rightarrow a, a \rightarrow x(a \neq x)$, then it is called a pair of cyclic correspondences.
If there is one pair of cyclic correspondences and $f(x) \neq x$, the number of such mappings is $C_{6}^{2}... | 7360 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\frac{2}{2007}\left[\left(1^{2}+3^{2}+5^{2}+\cdots+2005^{2}\right)-\right. \\
\left.\left(2^{2}+4^{2}+6^{2}+\cdots+2006^{2}\right)\right] .
\end{array}
$$ | Answer: -2006 . | -2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $P$ be a regular 2006-gon. If an end of a diagonal of $P$ divides the boundary of $P$ into two parts, each containing an odd number of sides of $P$, then the diagonal is called a "good edge". It is stipulated that each side of $P$ is a good edge.
Given 2003 non-intersecting diagonals inside $P$ that partition $... | 2. If an isosceles triangle has two good sides, it is briefly referred to as a "good triangle". Let $\triangle A B C$ be a good triangle, and $A B, B C$ be the good sides. Then, there are an odd number of edges between points $A$ and $B$; the same applies to $B$ and $C$. We say these edges belong to the good $\triangle... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the value of $2-2^{2}-2^{3}-\cdots-2^{2005}+2^{2000}$. | Explanation: Utilizing the characteristic $2^{n+1}-2^{n}=2^{n}$, rearrange the terms of the original expression in reverse order.
Therefore, the original expression
$$
\begin{array}{l}
=2^{2006}-2^{2005}-2^{2004}-\cdots-2^{3}-2^{2}+2 \\
=2^{2006}(2-1)-2^{2005}-\cdots-2^{3}-2^{2}+2 \\
=2^{2005}(2-1)-2^{2004}-\cdots-2^{3... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A construction company
has contracted two projects, each to be constructed by two different teams. According to the progress of the projects, the construction company can adjust the number of people in the two teams at any time. If 70 people are transferred from Team A to Team B, then the number of people in Team ... | 14. Let team A have $x$ people, then team B has $[2(x-70)-70]$ people, which means team B has $(2 x-210)$ people. Suppose $y$ people are transferred from team B to team A, making the number of people in team A three times that of team B. Then
$$
3(2 x-210-y)=x+y,
$$
which simplifies to $x=126+\frac{4}{5} y$.
Given $y>... | 130 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Place the numbers 1, 2, $3, \cdots, 9$ into the 9 circles in Figure 4, such that the sum of the numbers in the three circles on each side of $\triangle ABC$ and $\triangle DEF$ is 18.
(1) Provide one valid arrangement;
(2) How many different arrangements are there? Prove your conclusion. | 15. (1) Figure 9 gives one arrangement that meets the requirements.
(2) There are 6 different ways to fill in the numbers.
Let the sum of the three numbers in circles $A, B, C$ be $x$; the sum of the three numbers in circles $D, E, F$ be $y$; and the sum of the remaining three circles be $z$. Clearly, we have
$$
x+y+z... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. In the quadratic equation $x^{2}+m x+n=0$, the coefficients $m, n$ can take values from $1,2,3,4,5,6$. Then, among the different equations obtained, the number of equations with real roots is ( ).
(A) 20
(B) 19
(C) 16
(D) 10 | 7.B.
$\Delta=m^{2}-4 n \geqslant 0$, then $m^{2} \geqslant 4 n$. When $n=1$, $m=2,3,4,5,6$, a total of 5; when $n=2$, $m=3,4,5,6$, a total of 4; when $n=3$, $m=4,5,6$, a total of 3; when $n=4$, $m=4,5,6$, a total of 3; when $n=5$, $m=5,6$, a total of 2; when $n=6$, $m=5,6$, a total of 2. In total, 19. | 19 | Algebra | MCQ | Yes | Yes | cn_contest | false |
13. As shown in Figure 4, the area of isosceles right $\triangle ABC$ is $98$, and $D$ is a point on the hypotenuse $BC$ such that $BD: DC = 2: 5$. Then the area of the square $ADEF$ with side $AD$ is $\qquad$ | 13.116.
Draw $D G \perp A B, D H \perp A C$, with the feet of the perpendiculars at $G$ and $H$ respectively. It is easy to see that $A B=A C=14$.
Also, $B D: D C=G D: H D=2: 5$, so $D G=4, D H=10$. Therefore, $A D=\sqrt{4^{2}+10^{2}}$. Hence, $S_{\text {square ADEF }}=116$. | 116 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. (16 points) In an activity class, the teacher asked each student to make a rectangular box with a lid as shown in Figure 6. The length, width, and height of the rectangular box are $x \mathrm{~cm}, y \mathrm{~cm}$, and $z \mathrm{~cm}$, respectively. When Xiao Yang presented his box, he told the class: “The length,... | Three, 15. Since $x y=x z+3$, therefore, $x(y-z)=3$.
Also, $x, y, z$ are all positive integers, then
$$
\left\{\begin{array}{l}
x = 3, \\
y - z = 1
\end{array} \text { or } \left\{\begin{array}{l}
x=1, \\
y-z=3 .
\end{array}\right.\right.
$$
(1) When $\left\{\begin{array}{l}x=3, \\ y=z+1\end{array}\right.$, by $y z=x y... | 22 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. (18 points) Among 200 small balls numbered $1, 2, \cdots, 200$, any $k$ balls are drawn such that there must be two balls with numbers $m$ and $n$ satisfying
$$
\frac{2}{5} \leqslant \frac{n}{m} \leqslant \frac{5}{2} \text {. }
$$
Determine the minimum value of $k$ and explain the reasoning. | 17. Divide 200 balls numbered $1 \sim 200$ into 6 groups, such that the ratio of the numbers of any two balls in each group is no less than $\frac{2}{5}$ and no more than $\frac{5}{2}$.
Grouping as follows:
Group 1 $(1,2)$
Group 2 $\quad(3,4,5,6,7)$
Group 3 $(8,9,10, \cdots, 20)$
Group 4 $\quad(21,22,23, \cdots, 52)$
G... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. If a natural number $n$ allows the vertical addition
$$
n+(n+1)+(n+2)
$$
to not produce any carry, then $n$ is called a "continuous number". For example, 12 is a continuous number, because $12+13+14$ does not produce a carry; but 13 is not a continuous number. Then, the number of continuous numbers less than 1000 i... | 5.C.
There are $0,1,2$, a total of 3 single-digit numbers;
There are $\overline{a 0}, \overline{a 1}, \overline{a 2}, a=1,2,3$, a total of $3 \times 3=9$ two-digit numbers;
There are $\overline{a b 0}, \overline{a b 1}, \overline{a b 2}, a=1,2,3, b=0,1,2,3$, a total of $3 \times 4 \times 3=36$ three-digit numbers.
The... | 48 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. Let natural numbers $x, y$ satisfy
$$
x<y, x^{3}+19 y=y^{3}+19 x
$$
Then $x+y=$ $\qquad$ | 2.5.
From $x^{3}-y^{3}=19(x-y), x<y$, we get $x^{2}+x y+y^{2}=19$.
Thus, $3 x^{2}<x^{2}+x y+y^{2}=19$.
Since $x$ is a natural number, we have $x=2$.
Therefore, $y=3$, so $x+y=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The 2008 Olympic Games will be held in Beijing. Figure 3 is the Olympic five-ring logo, where $a, b, c, d, e, g, h, i$ are filled with different numbers from $1 \sim 9$. If the sum of the numbers in each circle is equal, then the maximum value of this sum is $\qquad$
| 3.14 .
Let the sum of the numbers in each circle be $N$, then
$$
(1+2+\cdots+9)+(b+d+f+h)=5 N .
$$
Therefore, $N=\frac{b+d+f+h}{5}+9$.
To maximize $N$, $b+d+f+h$ must be the largest possible value and a multiple of 5.
Since $b+d+f+h=6+7+8+9=30$ is the maximum, and $N=15$, at this point, no matter how you fill in the... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
(20 points) Read the material and answer the questions:
As shown in Figure 5, the shapes of rhombuses, rectangles, and squares differ. We refer to the degree of similarity between these shapes as "closeness."
When studying closeness,
the closeness of similar figures
should be equal.
(1) Let the two adjacent interior an... | (1)(i)20 (ii)0 (2)Unreasonable
In rectangles with side lengths of $a, b$ and $2a, 2b$, it is clear that these two rectangles are similar, but $|2a-2b| \neq |a-b|$.
The proximity of a rectangle can be defined as $\frac{a}{b}(a \leqslant b)$, then the closer $\frac{a}{b}$ is to 1, the closer the rectangle is to a square... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ consists of positive integers and is an increasing sequence, and satisfies $a_{n+2}=a_{n+1}+2 a_{n}\left(n \in \mathbf{N}_{+}\right)$. If $a_{5}$ $=52$, then $a_{7}=(\quad)$.
(A) 102
(B) 152
(C) 212
(D) Insufficient conditions, $a_{7}$ cannot be uniquely determined | 2.C.
$$
\begin{array}{l}
a_{3}=a_{2}+2 a_{1}, \\
a_{4}=a_{3}+2 a_{2}=3 a_{2}+2 a_{1}, \\
a_{5}=a_{4}+2 a_{3}=5 a_{2}+6 a_{1} .
\end{array}
$$
The positive integer solutions to the equation $5 a_{2}+6 a_{1}=52$ are $a_{1}=2, a_{2}=8$ or $a_{1}=7, a_{2}=2$.
Since $a_{2}>a_{1}$, we have $a_{1}=2, a_{2}=8$.
Thus, $a_{7}=a... | 212 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 5 "Tossing a coin continuously until two consecutive heads appear" occurs in all possible ways on the $n(n \geqslant 2)$-th toss.
How many such ways are there? | Explanation: Let the number of methods that satisfy the conditions be $a_{n}$.
When $n=2$, there is only 1 case of "positive positive", $a_{2}=1$;
When $n=3$, there is only 1 case of "negative positive positive", $a_{3}=1$.
For $n \geqslant 4$, the number of methods $a_{n}$ where "positive positive" appears exactly a... | 233 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) Let real numbers $p, q, r$ satisfy: there exists $a$ which is one of $p, q, r$, and the other two are exactly the two real roots of the equation
$$
x^{2}+(a-3) x+a^{2}-3 a=0
$$
Find the minimum possible value of $p^{3}+q^{3}+r^{3}$. | Three, let $a=p$, then $q, r$ are exactly the two roots of the equation. By Vieta's formulas, we know $p+q+r=3$ and $qr=p^2-3p$.
From $\Delta=(p-3)^2-4(p^2-3p) \geqslant 0$, we know $-1 \leqslant p \leqslant 3$.
$$
\begin{array}{l}
\text { Then } p^2+q^2+r^2=p^2+(q+r)^2-2qr \\
=p^2+(3-p)^2-2(p^2-3p)=9 \text {. }
\end{a... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five, color the numbers in $S=\{0,1,2, \cdots, n\}$ with two colors arbitrarily. Find the smallest positive integer $n$, such that there must exist $x, y, z \in S$ of the same color, satisfying $x+y=2 z$. | Let $A=\{0,2,5,7\}$ and $B=\{1,3,4,6\}$. Then neither $A$ nor $B$ contains a 3-term arithmetic progression. By coloring the elements of $A$ red and the elements of $B$ yellow, we can see that $n>7$.
When $n \geqslant 8$, we will prove that there must be a color for which the numbers form a 3-term arithmetic progressio... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. A football team participates in the World Cup, and in a certain match, 11 players start the game, using a 4-4-2 formation (4 defenders, 4 midfielders, 2 forwards). There are 7 substitute players (2 defenders, 2 midfielders, 2 forwards, and 1 goalkeeper). The rules allow for a maximum of three substitutions. If the ... | 15. (1) Let the total number of substitutions in the match be $t$, where $t=0,1,2,3$.
When $t=0$, 11 players play the entire match, which is only 1 way.
When $t=1$, the substituted player can be a midfielder or a defender, resulting in $\mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} + \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} = 16... | 189 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The set of integer points on the plane
$$
S=\{(a, b) \mid 1 \leqslant a, b \leqslant 5(a, b \in \mathbf{Z})\},
$$
$T$ is a set of integer points on the plane, such that for any point $P$ in $S$, there exists a point $Q$ in $T$ different from $P$, such that the line segment $P Q$ contains no other integer points exce... | 5. The minimum number is 2.
First, we prove that $T$ cannot consist of only one point.
Otherwise, suppose
$$
T=\left\{Q\left(x_{0}, y_{0}\right)\right\}.
$$
In $S$, take a point $P\left(x_{1}, y_{1}\right)$ such that $\left(x_{1}, y_{1}\right) \neq \left(x_{0}, y_{0}\right)$, and $x_{1}$ has the same parity as $x_{0}... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set
$$
\begin{array}{l}
M=\{1,2, \cdots, 19\}, \\
A=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\} \subseteq M .
\end{array}
$$
Find the smallest $k$, such that for any $b \in M$, there exist $a_{i}, a_{j} \in A$, satisfying $b=a_{i}$ or $b=a_{i} \pm a_{j}\left(a_{i}, a_{j}\right.$ can be the same).
(Supplied b... | 6. By the hypothesis, in $A$, there are $k(k+1)$ possible combinations. Thus, $k(k+1) \geqslant 19$, i.e., $k \geqslant 4$.
When $k=4$, we have $k(k+1)=20$. Suppose $a_{1}<a_{2}<a_{3}<a_{4}$. Then $a_{4} \geqslant 10$.
(1) When $a_{4}=10$, we have $a_{3}=9$. At this time, $a_{2}=8$ or 7.
If $a_{2}=8$, then $20,10-9=1,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure $9, A B$ is the diameter of a semicircle, and $C$ is a point on the semicircular arc. One side $D G$ of the square $D E F G$ lies on the diameter $A B$,
and the other side $D E$ passes through the incenter $I$ of $\triangle A B C$, with point
$E$ on the semicircular arc. If the area of the squar... | (Given: $A D=\frac{b+c-a}{2}, B D=\frac{c+a-b}{2}$. From $D E^{2}$ $=A D \cdot D B$, we get $100=\frac{b+c-a}{2} \cdot \frac{c+a-b}{2}=\frac{a b}{2}$, hence $\left.S_{\triangle B B C}=100.\right)$ | 100 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $c(X)$ denote the number of subsets of set $X$. If $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ with different numbers of elements satisfy:
$$
\left(\left|A_{1}\right|+1\right)\left(\left|A_{2}\right|+1\right) \cdots\left(\left|A_{n}\right|+1\right)=2006
$$
and
$$
\begin{array}{l}
c\left(A_{1}\right)+c\left(A_{2}\rig... | 2.58.
Assume $\left|A_{1}\right|>\left|A_{2}\right|>\cdots>\left|A_{n}\right| \geqslant 0$.
When $n \geqslant 3$, by $c(X)=2^{|X|}$ and the condition, we know that $2^{\left|A_{1}\right|}+2^{\left|A_{2}\right|}+\cdots+2^{\left|A_{n}\right|}=2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{1} \cap A... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the ellipse $C_{1}$ and the hyperbola $C_{2}$ share foci $F_{1}(3,0), F_{2}(-3,0)$, and have coincident minor axes. Then the number of lattice points inside the region enclosed by the intersection points of $C_{1}$ and $C_{2}$ is $\qquad$
Translate the above text into English, please retain the original ... | 3.25.
From the problem, we can let
$$
C_{1}: \frac{x^{2}}{m+9}+\frac{y^{2}}{m}=1, C_{2}: \frac{x^{2}}{9-m}-\frac{y^{2}}{m}=1,
$$
where $0<m<9$.
Let the intersection point be $P\left(x_{0}, y_{0}\right)$. Then, since point $P$ lies on $C_{1}$ and $C_{2}$, we have $\frac{x_{0}^{2}}{m+9}+\frac{x_{0}^{2}}{9-m}=2$, which ... | 25 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Example 1 Given that $a, b$ are positive real numbers, the equations
$$
x^{2}+a x+2 b=0 \text{ and } x^{2}+2 b x+a=0
$$
both have real roots. Find the minimum value of $a^{2}+b^{2}$. | Explanation: Since the given equations all have real roots, we obtain the constraint conditions
$$
\left\{\begin{array}{l}
a^{2}-8 b \geqslant 0 \\
b^{2}-a \geqslant 0 \\
a>0, b>0
\end{array}\right.
$$
To find the minimum value of $a^{2}+b^{2}$, using algebraic methods is clearly challenging, so we consider a numerica... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the real number $a$ be such that the quadratic equation $5 x^{2}-5 a x+66 a-1715=0$ has two integer roots. Then all such $a$ are $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.870.
Let the two integer roots be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have $a=x_{1}+x_{2}$, thus, $a$ is an integer. From the original equation, we get
$$
\begin{array}{l}
a=\frac{5 x^{2}-1715}{5 x-66}=x+13+\frac{x-857}{5 x-66} \in \mathbf{Z} \\
\L... | 870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. If from the set $S=\{1,2, \cdots, 20\}$, we take a three-element subset $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, such that it simultaneously satisfies: $a_{2}-a_{1} \geqslant 5,4 \leqslant a_{3}-a_{2} \leqslant 9$, then the number of all such subsets $A$ is $\qquad$ (answer with a specific number). | 6.251.
$a_{2}-a_{1} \geqslant 5 \Leftrightarrow\left(a_{2}-4\right)-a_{1} \geqslant 1$,
$4 \leqslant a_{3}-a_{2} \leqslant 9 \Leftrightarrow 1 \leqslant\left(a_{3}-7\right)-\left(a_{2}-4\right) \leqslant 6$.
Let $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-4, a_{3}^{\prime}=a_{3}-7$, then $a_{1}^{\prime}, a_{2}^{\prime}... | 251 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right) \cdot (x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $A=\{(x, y) \mid x>0, y>0\}$. | Explanation: If $x+y \leqslant 1$, then
$$
\begin{array}{l}
x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4} \text {. } \\
\text { Hence } f(x, y) \\
=6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y- \\
4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
=6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1) \text {. } \\
\left(x-\frac{1}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Try to find the smallest positive integer $m$, such that the following conditions are satisfied simultaneously:
(1) $\left[\frac{2}{1977} m^{2}\right] \geqslant m+2006$ ( $[x]$ denotes the greatest integer not exceeding $x$);
(2) $99^{m}$ leaves a remainder of 11 when divided by 190. | From condition (1), we know that $m > 1977$.
Otherwise, if $m \leqslant 1977$, then we have
$$
\begin{array}{l}
{\left[\frac{2}{1977} m^{2}\right] \leqslant \frac{2}{1977} m^{2} \leqslant \frac{2}{1977} \times 1977 m} \\
= 2m \cdot 1977.
$$
Let $m = 1977 + k$ ($k$ is a positive integer), and substitute it into the ineq... | 2004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Try to find the minimum value of the function
$$
f(x, y)=(2 y-1) \sin x+(1-y) \sin (1-y) x
$$ | It is known that for all $0 \leqslant x \leqslant \pi$, we have
$$
\sin x \geqslant 0, \sin (1-y) x \geqslant 0 \text {. }
$$
Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $a, b, c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary | Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By Vieta's formulas, we know $x_{1}x_{2} = \frac{c}{a} = 9$. Therefore,
\[ b^{2} \geqslant 4ac = 4 \times \frac{a}{c} \times c^{2} > 4 \times 9 \times 1^{2} = 36. \]
Thus, $b \geqslant 7$.
Also, $\frac{b}{a} = (-x_{1}) + (-x_{2}) \Rightarrow \frac{3}{2} b \... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10, |B|=9$, and set $A$ satisfies the following condition:
If $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$.
Let $A+B=\{a+b \mid a \in A, b \in B\}$.
Prove: $|A+B| \geqslant 50$ ( $|X|$ represents the number of elements in set $X$).
(2005, Nation... | Prove: Let $A=\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{9}\right\}, \\
C_{i}=\left\{a_{1}+b_{i}, a_{2}+b_{i}, \cdots, a_{10}+b_{i}\right\}
\end{array}
$$
where $a_{i}(i=1,2, \cdots, 10)$ are distinct.
Prove by contradiction.
$$
\left|C_{i} \cap C_{j}\right| \le... | 54 | Combinatorics | proof | Yes | Yes | cn_contest | false |
11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+\right.$ $\left.x^{2004}\right)=2006 x^{2005}$ is $\qquad$ | 11.1.
$$
\begin{array}{l}
\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 x^{2005} \\
\Leftrightarrow\left(x+\frac{1}{x^{2005}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 \\
\Leftrightarrow x+x^{3}+x^{5}+\cdots+x^{2005}+\frac{1}{x^{2008}}+\frac{1}{x^{2003}} \\
\quad+\cdots+\frac{1}{x}=... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Calculate:
\[
\begin{array}{l}
\sqrt{2005 \times 2006 \times 2007 \times 2008+1}-2006^{2} \\
=
\end{array}
\] | 12.2005 .
Let $n=2005$. Then
$$
\begin{array}{l}
\text { Original expression }=\sqrt{n(n+1)(n+2)(n+3)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n+1\right)^{2}}-(n+1)^{2}=n=2005 .
\end{array}
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given $x=2 \sqrt{2}+1$. Then the fraction
$$
\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=
$$
$\qquad$ | 13.2
Given $x=2 \sqrt{2}+1$, we know $x-1=2 \sqrt{2}$, then $x^{2}-2 x-7=0$. Therefore, $\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=\frac{\left(x^{2}-2 x-7\right)-2}{(x+2)\left(x^{2}-2 x-7\right)-1}=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
18. $A 、 n$ are natural numbers, and
$$
A=n^{2}+15 n+26
$$
is a perfect square. Then $n$ equals $\qquad$ | 18.23.
Let $A=n^{2}+15 n+26=(n+2)(n+13)$ be a perfect square, we can set $n+2=a^{2}, n+13=b^{2}(a, b$ are positive integers and $b$ $>a)$. Therefore, $11=b^{2}-a^{2}=(b+a)(b-a)$. Then we have $\left\{\begin{array}{l}b+a=11, \\ b-a=1 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=5, \\ b=6 .\end{ar... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. A rectangular prism has integer length, width, and height, and its volume is exactly $2006 \mathrm{~cm}^{3}$. After painting its surface red, it is then cut into small cubes with edge lengths of $1 \mathrm{~cm}$. If there are 178 small cubes with three red faces, then the number of small cubes with exactly two red ... | 19.1824 .
From the problem, we know that one of the edges of the cuboid must be $1 \mathrm{~cm}$, otherwise, only 8 small cubes would have three faces painted red. Let's assume the length, width, and height of the cuboid are $x \mathrm{~cm}, y \mathrm{~cm}, 1 \mathrm{~cm}$, respectively. Then we have
$$
\left\{\begin{... | 1824 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the first term and common difference of an arithmetic sequence be positive integers, the number of terms be a prime number no less than 3, and the sum of all terms be 2006. Then the number of such sequences is $\qquad$.
| Ni, 7.15.
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$, then
$$
n a+\frac{1}{2} n(n-1) d=2006,
$$
i.e., $n[2 a+(n-1) d]=2 \times 2006=2 \times 2 \times 17 \times 59$.
Since $2 a+(n-1) d \geqslant 2+n-1=n+1>n$, hence $n=17,59$.
(1) If $n=17$, then $a+... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
(x-11)^{5}+15(x-11)=5, \\
(y-4)^{5}+15(y-4)=-5 .
\end{array}\right.
$$
Then $x+y$ equals $\qquad$ . | 8. 15 .
Since the function $f(t)=t^{5}+15 t$ is strictly monotonically increasing in the real number domain, and $f(x-11)=f(4-y)$, then $x-11=4-y$. Therefore, $x+y=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. The 8 vertices of a cube can form
$\qquad$ non-equilateral triangles. | 11.48.
Each vertex leads to three edges whose endpoints can form an equilateral triangle, making a total of 8 equilateral triangles. Therefore, the answer is $\mathrm{C}_{8}^{3}-8=48$. | 48 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. There are 2006 distinct complex numbers, such that the product of any two of them (including self-multiplication) is one of these 2006 numbers. Find the sum of these 2006 numbers. | Three, 13. Let there be $n$ distinct complex numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$.
If one of them is 0, without loss of generality, let $\alpha_{n}=0$, then the set
$$
\left\{\alpha_{1}^{2}, \alpha_{1} \alpha_{2}, \cdots, \alpha_{1} \alpha_{n-1}\right\}=\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Find
$$
2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}
$$
the value. | 14. Solution 1: Since $k \mathrm{C}_{n}^{k}=n \mathrm{C}_{n-1}^{k-1}$, we have:
$$
\begin{array}{l}
\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-1}=n 2^{n-1} \\
\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1) \mathrm{C}_{n-1}^{k-1}+... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. In a mathematics competition, there are three types of questions: multiple-choice questions (6 questions), fill-in-the-blank questions (4 questions), and problem-solving questions (3 questions). Among them, each multiple-choice question and each fill-in-the-blank question is worth 7 points, and the problem-solving q... | G.D.
(1) The possible scores for multiple-choice and fill-in-the-blank questions are $0,7,14, \cdots, 70$, a total of 11 possibilities. The possible scores for problem-solving questions are $0,5,10, \cdots, 70$, a total of 15 possibilities. Therefore, there can be $11 \times 15=165$ results.
(2) The following 23 scores... | 117 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Arrange $n$ squares of different sizes without overlapping, so that the total area of the resulting figure is exactly 2,006. The minimum value of $n$ is $\qquad$ $\therefore$. | 3.3.
Let the side lengths of $n$ squares be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=2006$.
Since $x_{i}^{2} \equiv 0$ or $1(\bmod 4)$, and $2006 \equiv 2(\bmod 4)$, there must be at least two odd numbers among $x_{i}$.
If $n=2$, then $x_{1}$ and $x_{2}$ are both odd, let them be $2 p+... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given that $k$ and $a$ are positive integers, and $2004k + a$, $2004(k+1) + a$ are both perfect squares.
(1) How many ordered pairs of positive integers $(k, a)$ are there?
(2) Identify the minimum value of $a$ and explain your reasoning. | II. (1) Let $2004k + a = m^2$,
(1)
$$
2004(k+1) + a = n^2 \text{, }
$$
where $m, n$ are positive integers, then
$$
n^2 - m^2 = 2004 \text{. }
$$
Thus, $(n+m)(n-m) = 2004 = 2 \times 2 \times 3 \times 167$.
Noting that $m+n$ and $n-m$ have the same parity, we have
$$
\left\{\begin{array}{l}
n + m = 1002, \\
n - m = 2
\... | 137 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4.5 soccer teams are conducting a round-robin tournament (each pair of teams plays one match). It is known that Team A has played 3 matches, Team B has played more matches than Team A, Team C has played fewer matches than Team A, and Team D and Team E have played the same number of matches, but Team D and Team E have n... | 4.6.
Team B has played 4 matches. If Team C has only played 1 match, then Team C hasn't played against Team A, and Team A must have played against Teams D and E, so Teams D and E have each played 2 matches. Therefore, the total number of matches is
$$
(3+4+1+2+2) \div 2=6 \text {. }
$$
If Team C has played 2 matches,... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 3, $\triangle A B C$ is divided into six smaller triangles by three concurrent lines $A D$, $B E$, and $C F$. If the areas of $\triangle B P F$, $\triangle C P D$, and $\triangle A P E$ are all 1, find the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$. | Three, let the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$ be $x$, $y$, and $z$ respectively.
Since $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{B D}{D C}, \frac{S_{\triangle D A B}}{S_{\triangle A K C}}=\frac{B D}{D C}$,
then $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{S_{... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. A building has 4 elevators, each of which can stop at three floors (not necessarily consecutive floors, and not necessarily the lowest floor). For any two floors in the building, there is at least one elevator that can stop at both. How many floors can this building have at most? | 2. Let the building have $n$ floors, then the number of floor pairs is $\frac{n(n-1)}{2}$. Each elevator stops at 3 floors, which gives $\frac{3 \times 2}{2}=3$ floor pairs, so, $4 \times 3 \geqslant$ $\frac{n(n-1)}{2}$. Therefore, $n \leqslant 5$.
When $n=5$, the floors served by the four elevators are $(1,4,5),(2,4,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Divide the numbers $1,2, \cdots, 30$ into $k$ groups (each number can only appear in one group) such that the sum of any two different numbers in each group is not a perfect square. Find the minimum value of $k$.
Put the above text into English, please keep the original text's line breaks and format, and output the... | 3. First consider the numbers $6,19,30$.
Since $6+19=5^{2}, 6+30=6^{2}, 19+30=7^{2}$, these 3 numbers must be in 3 different groups. Therefore, $k \geqslant 3$.
Next, divide the 30 numbers $1,2, \cdots, 30$ into the following 3 groups:
$$
\begin{array}{l}
A_{1}=\{3,7,11,15,19,23,27,4,8,16,24\}, \\
A_{2}=\{1,5,9,13,17,... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Remove all positive integers that are divisible by 4 and those that leave a remainder of 1 when divided by 4, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$:
$$
2,3,6,7,10,11, \cdots \text {. }
$$
The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ i... | 8. It is known that $a_{2 n-1}=4 n-2, a_{2 n}=4 n-1(n=1,2$, $\cdots$ ). Therefore,
$$
\begin{aligned}
& S_{2 n}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\cdots+\left(a_{2 n-1}+a_{2 n}\right) \\
= & 5+13+21+\cdots+(8 n-3) \\
= & \frac{(5+8 n-3) n}{2}=(2 n)^{2}+n, \\
& S_{2 n-1}=S_{2 n}-a_{2 n}=4 n^{2}+n-(4 n-1)... | 2013021 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}=2^{2006}$, and for each positive integer $i, a_{i+1}$ is equal to the square of the sum of the digits of $a_{i}$. Find the value of $a_{2006}$. | 3. Since the remainders when $2, 2^{2}, 2^{3}, 2^{4}, 2^{5}, 2^{6}, \cdots$ are divided by 9 are $2, 4, 8, 7, 5, 1, \cdots$, therefore, the remainders when $2^{m+6}$ and $2^{m}$ are divided by 9 are equal.
But $2006 = 334 \times 6 + 2$, so the remainder when $a_{1}$ is divided by 9 is 4. Thus, the remainder when the s... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent, their radii are $112$ and $63$, respectively. The segment $A B$ is intercepted by their two external common tangents on their internal common tangent. Then, the length of $A B$ is $\qquad$ . | 4.168.
From the tangent theorem, we easily know that
$$
\begin{array}{l}
A B=\text { length of the external common tangent }=\sqrt{(112+63)^{2}-(112-63)^{2}} \\
=2 \sqrt{112 \times 63}=168 .
\end{array}
$$ | 168 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If 5 consecutive natural numbers are all composite, then this group of numbers is called a "twin 5 composite". So, among the natural numbers not exceeding 100, there are $\qquad$ groups of twin 5 composite. | Ni, 6.10.
It is easy to know that the prime numbers not exceeding 100 are
$$
\begin{array}{l}
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47, \\
53,59,61,67,71,73,79,83,89,97 .
\end{array}
$$
There are 10 groups of twin 5 composites, namely
$$
\begin{array}{l}
24,25,26,27,28 ; 32,33,34,35,36 ; \\
48,49,50,51,52 ; 54,55,56,5... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. In $\triangle A B C$, $A C=B C, \angle A C B=90^{\circ}$, $D 、 E$ are points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$ | 7:36.
As shown in Figure 5, rotate $\triangle C E B$ $90^{\circ}$ clockwise to get $\triangle C E^{\prime} A$. Connect $E^{\prime} D$.
It is easy to see that $A E^{\prime}=B E=4$, $\angle E^{\prime} A D=90^{\circ}$, so
$$
\begin{array}{l}
E^{\prime} D=\sqrt{A E^{\prime 2}+A D^{2}} \\
=\sqrt{4^{2}+3^{2}}=5 .
\end{array... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. There are two sets of square paper pieces of the same size and the same number, one set black and one set white. Xiao Zhang first uses the white paper pieces to form a rectangle without any gaps in the middle, then uses the black paper pieces to surround the already formed white rectangle to create a larger rectang... | 10.350.
Let the first white rectangle be $a \times b$, then the first black rectangle is $(a+2)(b+2)$, the second white rectangle is $(a+4) \times (b+4), \cdots \cdots$, and the fifth black rectangle is $(a+18)(b+18)$.
Obviously, the difference in perimeter between each black rectangle and its inner white rectangle (... | 350 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $x$ and $y$ are real numbers, and satisfy
$$
\left(x+\sqrt{x^{2}+2008}\right)\left(y+\sqrt{y^{2}+2008}\right)=2008 \text {. }
$$
Then the value of $x^{2}-3 x y-4 y^{2}-6 x-6 y+2008$
is $\qquad$ | II, 1.2008.
From the given, we have
$$
x+\sqrt{x^{2}+2008}=\frac{2008}{y+\sqrt{y^{2}+2008}} .
$$
Rationalizing the denominator, we get
$$
x+\sqrt{x^{2}+2008}=\sqrt{y^{2}+2008}-y .
$$
Similarly, $y+\sqrt{y^{2}+2008}=\sqrt{x^{2}+2008}-x$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } x+y=0 \text {. } \\
\text ... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For real numbers $a, b, c$ satisfying
$$
a+b+c=0, \, abc=2 \text{.}
$$
then $u=|a|^{3}+|b|^{3}+|c|^{3}$ has the minimum value of
$\qquad$. | 2.10.
From the problem, we know that $a$, $b$, and $c$ must be one positive and two negative.
Without loss of generality, let $a>0$, $b<0$, and $c<0$.
Since $b+c=-a$ and $bc=\frac{2}{a}$, it follows that $b$ and $c$ are the two negative roots of the equation $x^{2}+a x+\frac{2}{a}=0$. Therefore, we have
$$
\Delta=a^{2... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given real numbers $a, b, c$ satisfy
$$
\left(-\frac{1}{a}+b\right)\left(\frac{1}{a}+c\right)+\frac{1}{4}(b-c)^{2}=0 \text {. }
$$
Then the value of the algebraic expression $a b+a c$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | 3. A.
The given equation can be rewritten as
$$
4(a b+1)(a c+1)+(a b-a c)^{2}=0,
$$
which simplifies to $(a b+a c)^{2}+4(a b+a c)+4=0$, or equivalently $[(a b+a c)+2]^{2}=0$.
Thus, $a b+a c=-2$. | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a-b+c=7, \\
a b+b c+b+c^{2}+16=0 .
\end{array}
$$
Then the value of $\left(a^{-1}-b^{-1}\right)^{a k c}(a+b+c)^{a+b+c}$ is $\qquad$ . | 2. -1 .
From equation (1), we get
$$
(-b)+(a+c+1)=8 \text {. }
$$
From equation (2), we get
$$
(a+c+1)(-b)=c^{2}+16 \text {. }
$$
Therefore, $a+c+1$ and $-b$ are the two roots of the equation
$$
x^{2}-8 x+c^{2}+16=0
$$
Thus, we have $\Delta=(-8)^{2}-4\left(c^{2}+16\right) \geqslant 0$.
Hence, $c^{2} \leqslant 0$.
I... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) There are three cylindrical containers $M$, $N$, and $P$, whose axial cross-sections are shown in figures 6(a), (b), and (c), respectively. The internal base areas are $S_{1} \mathrm{~cm}^{2}$, $S_{2} \mathrm{~cm}^{2}$, and $S_{3} \mathrm{~cm}^{2}$, and the internal heights are $h_{1} \mathrm{~cm}$, $h... | (1) From Figure 7, we know that filling containers $M$, $N$, and $P$ takes $60 \mathrm{~s}$. From Figure 8, we know that filling two of the three containers $M$, $N$, and $P$ takes $54 \mathrm{~s}$. Therefore, filling the top container in Figure 8 takes $60-54=6(\mathrm{~s})$.
Similarly, filling the top container in F... | 720 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the quadratic equation in $x$
$$
m^{2} x^{2}+(2 m+3) x+1=0
$$
there are two real roots whose product is 1; for the quadratic equation in $x$
$$
x^{2}+(2 a+m) x+2 a+1-m^{2}=0
$$
there is a real root that is greater than 0 and less than 4. Then the integer value of $a$ is . $\qquad$ | 2, -1.
According to the problem, we have $m^{2}=1, m= \pm 1$, and
$$
\Delta=(2 m+3)^{2}-4 m^{2}=12 m+9>0 \text {. }
$$
Thus, $m=1$.
Let $f(x)=x^{2}+(2 a+1) x+2 a$, then we should have
$$
f(0) f(4)=2 a(20+10 a)<0,
$$
which means $a(a+2)<0$.
Therefore, $-2<a<0$. Hence, $a=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In a certain basketball tournament, Xiao Ming played 10 games. In the 6th, 7th, 8th, and 9th games, he scored 23 points, 14 points, 11 points, and 20 points, respectively. His average score in the first 9 games was higher than his average score in the first 5 games. If the average score of the 10 games he played exc... | 3.29.
Let the average score of the first 5 games be $x$, then the average score of the first 9 games is
$$
\frac{5 x+23+14+11+20}{9}=\frac{5 x+68}{9} .
$$
According to the problem, $\frac{5 x+68}{9}>x$.
Solving this, we get $x<17$.
Therefore, the total score of the first 5 games is at most $5 \times 17-1=84$ points.
... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\left\{\begin{array}{l}
2006(x-y)+2000(y-z)+2008(z-x)=0, \\
2000^{2}(x-y)+2000^{2}(y-z)+2000^{2}(z-x)=2008 .
\end{array}\right.
$$
Then the value of $z-y$ is $\qquad$ . | $=、 1.2008$
Let $z-y=t$.
From the first equation, we get $z-2 x+y=0$.
Thus, $z-x=\frac{t}{2}$, and consequently, $x-y=\frac{t}{2}$.
Therefore, from the second equation, we can find that
$$
t=z-y=2008 .
$$ | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$
has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$ | $=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$
Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. A three-digit number $x y z$ (where $x$, $y$, and $z$ are distinct), rearranging its digits to form the largest and smallest possible three-digit numbers. If the difference between the largest and smallest three-digit numbers is equal to the original three-digit number, then this three-digit number is $\qquad$ | 3.495.
Let the three-digit number $\overline{y z}$, after rearrangement, form the largest three-digit number $\overline{a b c}(a>b>c)$, then the smallest three-digit number is $\overline{c b a}$.
Since $1 \leqslant a \leqslant 9,1 \leqslant b \leqslant 9,1 \leqslant c \leqslant 9$, and
$$
\begin{array}{l}
\overline{a ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) It is known that the unit digit of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{8}^{3}-x_{9}^{3}$ is 1, where $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from 2001, 2002, ..., 2009, and $8 x_{9}>$ $x_{1}+x_{2}+\cdots+x_{8}$. Find the value of $x_{9}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出... | Three, because $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from $2001, 2002, \cdots, 2009$, and the unit digits of $2001, 2002, \cdots, 2009$ are $1, 2, 3, 4, 5, 6, 7, 8, 9$. After cubing these unit digits, the resulting unit digits are $1, 8, 7, 4, 5, 6, 3, 2, 9$. Therefore, the unit digit of $x_{1}^{3} +... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A person rolls a die, adding up the numbers rolled each time, and stops if the total exceeds 20. Then, when he stops, the number he is most likely to have rolled is $\qquad$ | 4.21.
Consider the value obtained after the dice roll just before the one that exceeds 20 is $x$.
If $x=15$, then only rolling a 6 can result in 21;
If $x=16$, then rolling a 5 or 6 can result in 21 or 22, with each number having a probability of $\frac{1}{2}$;
If $x=17$, then rolling a 4, 5, or 6 can result in 21, 2... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If eac... | (1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$
Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given two sets of numbers, set $A$ is: $1,2, \cdots, 100$; set $B$ is: $1^{2}, 2^{2}, \cdots, 100^{2}$. For a number $x$ in set $A$, if there is a number $y$ in set $B$ such that $x+y$ is also a number in set $B$, then $x$ is called an "associated number". Therefore, the number of such associated numbers in set $A$ ... | 3.73.
Let $x+y=a^{2}, y=b^{2}$, then $1 \leqslant b<a \leqslant 100$.
And $x=a^{2}-b^{2}=(a+b)(a-b) \leqslant 100$, since $a+b$ and $a-b$ have the same parity, hence $a+b \geqslant(a-b)+2$.
(1) If $a-b=1$, then $a+b$ is odd, and $3 \leqslant a+b \leqslant$ 99. Thus, $a+b$ can take the values $3,5,7, \cdots, 99$, a tot... | 73 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5 | 3. A.
As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$
Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$
Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$
Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Henc... | 9 | Geometry | MCQ | Yes | Yes | cn_contest | false |
One, (20 points) Find the positive integer solutions of the indeterminate equation
$$
29 a+30 b+31 c=2196
$$ | One, transform the original equation into
$$
\left\{\begin{array}{l}
29(a+b+c)+(b+2 c)=2196 \\
31(a+b+c)-(2 a+b)=2196 .
\end{array}\right.
$$
Since \(a, b, c\) are positive integers, from equation (1) we get
$$
\begin{array}{l}
29(a+b+c)=2196-(b+2 c) \\
\leqslant 2196-(1+2 \times 1)=2193 .
\end{array}
$$
Therefore, \... | 86 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the three sides of $\triangle A B C$ are $A B=$ $2 \sqrt{a^{2}+576}, B C=\sqrt{a^{2}+14 a+625}, A C=$ $\sqrt{a^{2}-14 a+625}$, where $a>7$. Then the area of $\triangle A B C$ is $\qquad$ | 4.168.
Notice
$$
\begin{array}{l}
A B^{2}=(2 a)^{2}+48^{2}, \\
B C^{2}=(a+7)^{2}+24^{2}, \\
A C^{2}=(a-7)^{2}+24^{2} .
\end{array}
$$
As shown in Figure 6, with $A B$ as the hypotenuse, construct a right triangle $\triangle A B D$ on one side of $\triangle A B C$, such that
$$
\begin{array}{l}
B D=2 a, A D=48, \\
\an... | 168 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $b$ take even numbers from 1 to 11, and $c$ take any positive integer. Then the number of quadratic equations $x^{2}-b x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 50
(C) 55
(C) 57
(D) 58 | 5.A.
When $b=2$, from $b^{2}-4 c=4-4 c>0$, we get $c<1$, so, $c=0$.
When $b=4$, from $b^{2}-4 c=16-4 c>0$, we get $c<4$, so, $c=0,1,2,3$.
When $b=6$, from $b^{2}-4 c=36-4 c>0$, we get $c<9$, so, $c=0,1, \cdots, 8$.
When $b=8$, from $b^{2}-4 c=64-4 c>0$, we get $c<16$, so, $c=0,1, \cdots, 15$.
When $b=10$, from $b^{2}-... | 50 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Calculate: $\frac{1^{2}}{1^{2}-100+5000}+\frac{3^{2}}{3^{2}-300+5000}+$ $\frac{5^{2}}{5^{2}-500+5000}+\cdots+\frac{99^{2}}{99^{2}-9900+5000}=$ $\qquad$ | 2.50.
Notice that
$$
\begin{array}{l}
\frac{k^{2}}{k^{2}-100 k+5000}+\frac{\left(100-k^{2}\right)}{(100-k)^{2}-100(100-k)+5000} \\
=\frac{k^{2}}{k^{2}-100 k+5000}+\frac{10000-200 k+k^{2}}{k^{2}-100 k+5000} \\
=\frac{2\left(k^{2}-100 k+5000\right)}{k^{2}-100 k+5000}=2 .
\end{array}
$$
Let $k=1,3,5, \cdots, 49$, and ad... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Let $S=\{1,2, \cdots, 98\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which one number is coprime with the other 4 numbers, and in the other g... | Explanation: First, there are 49 even numbers in $S$, and the 49-element subset formed by them obviously does not meet the requirement. Therefore, the smallest natural number $n \geqslant 50$.
To prove that in any 50-element subset $T$ of $S$, there are 10 numbers that meet the requirement, we first prove a strengthen... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ . | 11.8.
From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$
Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. In the park, there are two rivers $O M$ and $O N$ converging at point $O$ (as shown in Figure 6, $\angle M O N=60^{\circ}$. On the peninsula formed by the two rivers, there is an ancient site $P$. It is planned to build a small bridge $Q$ and $R$ on each of the two rivers, and to construct three small roads to conn... | 12.300.
As shown in Figure 10, let the symmetric points of point $P$ with respect to $OM$ and $ON$ be $P_{1}$ and $P_{2}$, respectively. Connect $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $Q$ and $R$. By symmetry, we have
$$
\begin{array}{l}
PQ = P_{1}Q, \\
PR = P_{2}R, \\
PQ + QR + RP \\
= P_{1}Q + QR + RP_{2... | 300 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given $n$ four-element sets $A_{1}, A_{2}, \cdots, A_{n}$, any two of which have exactly one common element, and
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=n .
$$
Find the maximum value of $n$. Here $\operatorname{Card} A$ is the number of elements in set $A$. | Consider any element $a \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}$.
If each $A_{i}$ contains $a$, then by the given condition, the other elements in each $A_{i}$ are all different. Hence,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=3 n+1>n,
$$
which contradicts the given condition.
The... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Choose several colors from 6 different given colors to color a cube, with each face colored with one color and ensuring that any two adjacent faces sharing a common edge are of different colors. How many different coloring schemes are there (if one of the two colored cubes can be rotated to match the coloring... | Explanation: Clearly, the maximum number of colors used is 6, and the minimum is 3. Below, we use the grouping method to count.
(1) When using 6 colors, each color is used to paint one face. According to the problem, the first color can be placed on top by flipping. At this point, the bottom is one of the remaining 5 ... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Six, find the smallest real number $m$, such that for any positive real numbers $a, b, c$ satisfying $a + b + c = 1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(Xiong Bin) | Six, Solution 1: When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$
for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since for $0<x<1$, we have
$$
\begin{array}{l}
27 x^{3} \geqslant 6 x^{2}+5... | 27 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$
then $a^{2007}+b^{2007}=$ $\qquad$ | Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Zhang Hua wrote a five-digit number, which can be divided by 9 and 11. If the first, third, and fifth digits are removed, the resulting number is 35; if the first three digits are removed, the resulting number can be divided by 9; if the last three digits are removed, the resulting number can also be divided by 9. T... | 4.63954.
Let this five-digit number be $\overline{a 3 b 5 c}$, then $9 \mid \overline{5 c}$.
So, $9 \mid(5+c)$.
Since $5 \leqslant 5+c \leqslant 14$, thus $5+c=9, c=4$.
Similarly, $9 \mid \overline{a 3}$, then $9 \mid(a+3)$.
Since $4 \leqslant a+3 \leqslant 12$, so $a=6$.
Also, $91 \overline{a 3 b 5 c}$, then
$9 \mid(... | 63954 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$
find the maximum and minimum values of $\Delta=a^{2}-4 b$. | Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$
or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given positive numbers $a, b, c$ satisfying $a+b+c=3$.
Prove:
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 .
\end{array}
$$ | $$
\begin{array}{l}
\text { 5. } \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}} \\
=\frac{1}{3}\left(1+\frac{2 a+6}{a^{2}-2 a+3}\right)=\frac{1}{3}\left[1+\frac{2 a+6}{(a-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{2 a+6}{2}\right)=\frac{1}{3}(4+a) .
\end{array}
$$
Similarly, $\frac{b^{... | 5 | Inequalities | proof | Yes | Yes | cn_contest | false |
3. The product of all integers $m$ that make $m^{2}+m+7$ a perfect square is ( ).
(A) 84
(B) 86
(C) 88
(D) 90 | 3. A.
Let $m^{2}+m+7=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $m^{2}+m+7-k^{2}=0$.
Solving for $m$, we get $m=\frac{-1 \pm \sqrt{4 k^{2}-27}}{2}$.
Since $m$ is an integer, we should have $4 k^{2}-27=n^{2}\left(n \in \mathbf{N}_{+}\right)$, which means $(2 k+n)(2 k-n)=27$.
Solving, we get $\left\{\begin{array}{l}... | 84 | Algebra | MCQ | Yes | Yes | cn_contest | false |
One, (20 points) Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Try to find the minimum value of $S=7x+5y$. | Let $x+2 y=5 A, x+y=3 B, A, B$ be integers.
Since $x, y \geqslant 0$, it follows that $A, B \geqslant 0$.
It is also easy to see that $x=6 B-5 A, y=5 A-3 B$, thus,
$$
\begin{array}{l}
2 x+y=9 B-5 A \geqslant 99, \\
S=7 x+5 y=27 B-10 A .
\end{array}
$$
Therefore, the problem becomes: For integers $A, B \geqslant 0, 6 B... | 366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle BAC=90^{\circ}$, $AB=AC$, points $D_{1}$ and $D_{2}$ are on $AC$, and satisfy $AD_{1}=CD_{2}$, $AE_{1} \perp BD_{1}$, $AE_{2} \perp BD_{2}$, intersecting $BC$ at points $E_{1}$ and $E_{2}$, respectively. Prove that: $\frac{CE_{2}}{BE_{2}}+\frac{CE_{1}}{B... | $$
\begin{array}{l}
\frac{C E_{1}}{B E_{1}}=\frac{A F}{A B}, \\
\angle E_{1} A C=\angle F C A . \\
\text { Also, } \angle A B D_{1}=90^{\circ}-\angle B A E_{1}=\angle E_{1} A C, \text { then } \\
\angle A B D_{1}=\angle A C F .
\end{array}
$$
Since $\angle B A D_{1}=\angle C A F=90^{\circ}, A B=A C$, therefore, $\tria... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2, the three edges of the cube are $AB$, $BC$, and $CD$, and $AD$ is the body diagonal. Points $P$, $Q$, and $R$ are on $AB$, $BC$, and $CD$ respectively, with $AP=5$, $PB=15$, $BQ=15$, and $CR=10$. What is the area of the polygon formed by the intersection of the plane $PQR$ extended in al... | Explanation: Since $B P = B Q$, therefore, $P Q \parallel A C$. Thus, the line through point $R$ and parallel to $P Q$ intersects $A F$ at point $U$, and $A U = C R$.
Since the line through point $R$ and parallel to $P Q$ lies in the plane $P Q R$, $U$ is a vertex of the polygon formed by the intersection. Also, the m... | 525 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that the length, width, and height of a rectangular prism are all integers, and the volume equals the surface area. Then the maximum value of its volume is
| 2.882.
Lemma When the sum of the reciprocals of two integers is a constant, the larger their difference, the greater their product.
Let the length, width, and height be $a$, $b$, and $c$, respectively, then
$$
a b c=2(b c+c a+a b) \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2} .
$$
Assume without los... | 882 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding $x$. Then the last two digits of $\left[\frac{10^{2006}}{10^{99}-3}\right]$ are $\qquad$ . | 3.23.
From $10^{2000}=10^{59 \times 34}-3^{34}+3^{34}$, we know
$$
\left[\frac{10^{2006}}{10^{99}-3}\right]=\frac{10^{99 \times 34}-3^{34}}{10^{59}-3}=\sum_{k=0}^{33} 10^{59(33-k)} \times 3^{k} \text {. }
$$
Therefore, the last two digits we are looking for are the remainder of $3^{33}$ modulo 100.
Also, $3^{32}=(10-... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5: Through the vertex $A$ of the regular tetrahedron $ABCD$, make a section in the shape of an isosceles triangle, and let the angle between the section and the base $BCD$ be $75^{\circ}$: The number of such sections that can be made is $\qquad$. | Explanation: Let the edge length of a regular tetrahedron be 1. Draw $AO \perp$ plane $BCD$ at point $O$, then $AO=\frac{\sqrt{6}}{3}$. With $O$ as the center and $\frac{\sqrt{6}}{3} \cot 75^{\circ}$ as the radius, draw a circle on plane $BCD$. It is easy to see that this circle is inside $\triangle BCD$, and the inter... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If the quadratic function $f(x)=a x^{2}+b x+c$ has values whose absolute values do not exceed 1 on $[0,1]$, what is the maximum value of $|a|+$ $|b|+|c|$? | Solution 1: For $x \in [0,1]$, we have $|f(x)| \leqslant 1$, thus
$$
\begin{array}{l}
|f(0)| \leqslant 1, \left|f\left(\frac{1}{2}\right)\right| \leqslant 1, |f(1)| \leqslant 1. \\
\text { Also } f(0)=c, \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c, \\
f(1)=a+b+c,
\end{array}
$$
Solving simultaneously, w... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.