problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9. 5. Proof: An arithmetic sequence with the first term 1 and a common difference of 729 must have infinitely many terms that are natural number powers of 10. | 9. 5. First, prove that for all natural numbers $n, 10^{81 n}-1$ is divisible by 729.
In fact, $10^{81 n}-1=\left(10^{81}\right)^{n}-1^{n}=\left(10^{81}-1\right) \cdot A$.
$$
\text { Then } \begin{array}{l}
\text { and } 10^{81}-1=\underbrace{99 \cdots 9}_{81 \uparrow} \\
=\underbrace{99 \cdots 9}_{9 \uparrow} \cdot 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,841 |
Example 4 Given $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$, and $y \leqslant|x|$. Find the range of values for $x$.
(1995, National Junior High School Competition (Ethnic Group))
Analysis: From $y \leqslant|x|$, we know that when $x \geqslant 0$, $y \leqslant x$; when $x<0$, $y \leqslant -x$. Therefore, these two ca... | Solution: When $x \geqslant 0$, we have $y \leqslant x$, i.e., $\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5} \leqslant x$, which also means $(x-2)(x-9) \leqslant 0$.
Solving this, we get $2 \leqslant x \leqslant 9$.
When $x<0$, we have $y \leqslant -x$, i.e., $\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5} \leqslant -x$, which... | 2 \leqslant x \leqslant 9 \text { or } -6 \leqslant x \leqslant -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,842 |
9. 6. Isosceles $\triangle ABC (AC = BC)$, $O$ is its circumcenter, $I$ is its incenter, point $D$ is on $BC$, $HD \perp BI$. Prove: $ID \parallel AC$.
保留了源文本的换行和格式。 | 9. 6. If it is known that the triangle is equilateral (point $O$ coincides with $I$), then the conclusion is obviously true.
If point $O$ is between point $I$ and $C$ (as shown in Figure 4). Draw the altitude $C E$ from the base, then
$$
\begin{array}{l}
\angle E I B=90^{\circ}-\frac{1}{2} \angle A B C, \\
\angle O D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,843 |
9. 7. On the table: there are two piles of coins with equal total weight, the number of coins in the first pile is $n$, the number of coins in the second pile is $m, S=\min \{n, m\}$. For any natural number $k$ not greater than $S$, in the order of coin weight from largest to smallest, the sum of the weights of the fir... | 9. 7. Let the weights of the first pile of $n$ coins be represented as $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, and the weights of the second pile of $m$ coins be represented as $y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{m}$.
Also, let $x_{1} \geqslant \cdots \geqslant x_{s} \geqslant x \geqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,844 |
9.8. Using a L-shaped piece formed by removing a 1x1 square from a 2x2 square to cover a rectangle with adjacent sides of 5 and 7 (the rectangle is divided into $5 \times 7$ unit squares), without overlapping but not exceeding the rectangle. Is it possible for each 1x1 unit square in the rectangle to be covered by the ... | 9.8. As shown in Figure 7, the squares of the rectangle are colored in black and white. In the black squares, the number -2 is recorded, and in the white squares, the number 1 is recorded. We find that the sum of the numbers in the squares covered by any "L-shape" is non-negative.
Assuming we can cover the squares of ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,845 |
10. 1. On the side $BC$ of the convex quadrilateral $ABCD$, take two points $E$ and $F$, with $E$ being closer to $B$ than $F$. If $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$, prove: $\angle FAC = \angle EDB$. | 10. 1. From $\angle E A F=\angle F D E$, we know that quadrilateral $A E F D$ is cyclic. Therefore, $\angle A E F+\angle F D A=180^{\circ}$. Also, $\angle B A E$ $=\angle C D F$, so we have
$$
\begin{array}{l}
\angle A D C+\angle A E C \\
=\angle F D A+\angle C D F+\angle A E F-\angle B A E \\
=180^{\circ} .
\end{array... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,846 |
10.3. $x, y$ are natural numbers. $k$ is a natural number of size 1. Find all natural numbers $n$ that satisfy: $3^{n}=x^{k}+y^{k}$, and provide a proof. | 10. 3. Answer: $n=2$.
At the time: Let $3^{n}=x^{k}+y^{k}$, where $x$ and $y$ are prime (assuming $x > y$), $k>1$, and $n$ is a natural number. Of course, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then $x^{k}$ and $y^{k}$ leave a remainder of 1 when divided by 3. Thus, the sum of $x^{k}$ and $y^{k}$ ... | 2 | Number Theory | proof | Yes | Yes | cn_contest | false | 708,848 |
10. 5. On each vertex of a cube, write a distinct natural number, and on each of its edges, write the greatest common divisor (GCD) of the two natural numbers at its ends. Is it possible for the sum of the numbers on the vertices to equal the sum of the numbers on the edges? Prove your conclusion. | 10. 5. Answer: No.
Proof: Let $a$ and $b$ be two natural numbers with $a > b$, and let the greatest common divisor of $a$ and $b$ be denoted as $(a, b)$. Then, we have $(a, b) \leqslant b \mathrm{H}(a, b) \leqslant \frac{a}{2}$. Therefore, when $a \neq b$, $(a, b) \leqslant \frac{a+b}{3}$.
Considering the 12 edges of... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,850 |
10. 6. In a platoon of a military camp, there are 3 sergeants and several soldiers. 3 sergeants take turns on duty in a sequential manner. The officer has issued the following orders:
1) Each day, the sergeant on duty must punish at least 1 soldier;
2) No soldier can be punished more than 2 times, nor can they be punis... | 10. 6. Answer: The 3rd duty soldier can do it.
Proof: We refer to each consecutive set of 3 duty days as a duty cycle. To avoid being confined, the 3rd duty soldier, at the end of each duty cycle, punishes: I: all soldiers who have been punished once (according to the third order, such soldiers always exist). Followin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,851 |
10.7. Given a convex polygon, none of whose sides are parallel, for each side, take the vertex farthest from the line containing that side - prove: all the angles subtended by the corresponding chords at the vertices are equal to $180^{\circ}$. | 10. 7. Let the vertex of the polygon farthest from the line containing side $a$ be denoted as $P_{0}$. Take any point $O$ in the plane, and all lines passing through $O$ and parallel to line $P Q Q$ (where point $Q$ is on side $a$) form two pairs of vertical angles. We call these two pairs of vertical angles the corres... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,852 |
Example 5 Draw a semicircle with line segment $AB$ as the diameter, with the center of the circle being $O$. $C$ is a point on the semicircle. If $OC^2 = AC \cdot BC$, then $\angle CAB=$ $\qquad$
(1995, National Junior High School Mathematics League) | Solution: (1) As shown in the figure, draw a perpendicular line from point $C$ to $AB$, intersecting $AB$ at $D$. By the area formula and the given information, we have $CD = \frac{1}{2} OC$, thus $\angle COB = 30^{\circ}$.
Since $\triangle AOC$ is an isosceles right triangle, $\angle COD$ is an exterior angle of $\tr... | 15^{\circ} \text{ or } 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,853 |
10.8. Ul: A clever student first writes 10 numbers on the blackboard, then asks another student to write 10 more numbers, making sure these 20 numbers are all distinct positive integers. The requirement is to find a quadratic polynomial: $x^{2}+p x+q$, where $p$ and $q$ are among the 20 numbers, such that the quadratic... | 10.8. Answer: Yes. For example, the numbers $-\frac{1}{4}, 1, 2, 5, 5^{2}, 5^{4}, 5^{6} \cdot 5^{-16} \cdot 5^{32}$.
} Theorem 1 (i) If $a>4$ and $a>b$, then the quadratic binomial $x^{2}+a x+b$ has two distinct real roots.
(2) If $a>0$, then at least one of the quadratic trinomials $x^{2}+a x + b$ and $x^{2}+b x+a$ ha... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,854 |
11. 2. Several people are walking uniformly along a straight line. It is known that during a certain period, the sum of the distances between all different pairs of people is continuously decreasing. Prove: It is always possible to find... a person, during this period, the sum of his distances to all other people is al... | 11. 2. Let there be $n$ pedestrians denoted by $P_{1}, P_{2}, \cdots, P_{n}$. We use the variable $V_{i j}$ to represent the approach speed of $P_{i}$ and $P_{j}$ (for any $1 \leqslant i, j \leqslant n$). This variable can be positive or negative (when walking in opposite directions). Over the entire observation period... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,856 |
11. 3. Proof: An $n$-sided pyramid with a base that is a regular $n$-gon $(n \geqslant 5)$ does not have a section that is a regular $(n+1)$-gon. | 1. 5. Suppose the regular $(n+1)$-gon $B_{1} B_{2} \cdots B_{n+1}$ is a section of the pyramid $S-A_{1} A_{2} \cdots A_{n}$, where the base $A_{1} A_{2} \cdots A_{n}$ is a regular $n$-gon. We will discuss three cases: $n=5, n=2 k-1(k>3)$, $n=2 k(k>2)$.
Since an $n$-sided pyramid has $n+1$ faces, the section intersects... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,857 |
11. 5 . Does there exist 3 natural numbers greater than 1, such that the square of each number minus 1 is divisible by any of the other two numbers? Prove your conclusion. | 11. 5. Answer: Does not exist.
Proof: Let $a \geqslant b \geqslant c$ be the numbers satisfying the problem's conditions.
Since $a^{2}-1$ is divisible by $b$, $a$ and $b$ are coprime. According to the conditions, $c^{2}-1$ is divisible by $a$ and $b$, and thus should also be divisible by their product $a b$. Therefore... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,859 |
11. 6. Isosceles $\triangle ABC, AB=BC, CD$ is its angle bisector, $O$ is its circumcenter. Through $O$ draw a perpendicular to $CD$ intersecting $BC$ at $E$, and through $E$ draw a line parallel to $CD$ intersecting $AB$ at $F$. Prove: $BE=FD$. | 11. 6. Let $O$ be the center of the circumcircle of $\triangle ABC$, and $K$ be the intersection of line $BO$ and $CD$ (see Figures 14 and 15).
Since the acute angle $\angle BOE$ and $\angle DCA$ have sides that are perpendicular to each other, we have $\angle BOE = \angle DCA$.
Thus,
$$
\angle BOE = \angle KCE
$$
($C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,860 |
11. 7. Is there a finite non-zero set of real numbers, $M$, such that for any natural number $n$, there always exists a polynomial of degree no less than $n$ with coefficients from $M$, all of whose roots are in $M$? Answer: Your conclusion.
Will your conclusion. | 11. 7. Answer: Does not exist.
Proof: If not, then there exists a set $M = \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$. Let $m = \min \left\{\left|a_{1}\right|, \left|a_{2}\right|, \cdots, \left|a_{n}\right|\right\}$, $M = \max \left\{\left|a_{1}\right|, \left|a_{2}\right|, \cdots, \left|a_{n}\right|\right\}$. By the ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,861 |
1. If real numbers $x, y$ satisfy $\left(\sqrt{x^{2}+1}+x\right)$
- $\left(\sqrt{y^{2}+1}+y\right)=1$, then the value of $x+y$ is ( ).
(A) -1
(B) 0
(C) 1
(D) cannot be determined | -1 (B).
Hint: Given that both sides of the equation are multiplied by $\sqrt{x^{2}+1}-x$,
$$
\sqrt{y^{2}+1}-y
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,863 |
Example 6 In $\triangle A B C$, $\angle A=50^{\circ}, H$ is the orthocenter of $\triangle A B C$, and $H$ does not coincide with $B$ or $C$. Then $\angle B H C=$ $\qquad$
(1994, Oriental Airlines Cup - Shanghai Junior High School Mathematics Competition) | Solution: (1) When $\triangle A B C$ is an acute triangle, as shown in the figure, $A D, B E$, and $C F$ are the three altitudes, then
$$
\begin{array}{c}
\angle E B C=90^{\circ}-\angle A C B, \\
\angle F C B=90^{\circ}-\angle A B C . \\
\text { Therefore } \begin{aligned}
\angle B H C & =180^{\circ}-\angle E B C-\angl... | 130^{\circ} \text{ and } 50^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,864 |
2. As shown in the figure, $AB$ is the diameter of $\odot O$ with a radius of 1, the perpendicular bisector of $OA$ intersects $AB$ at $C$, and intersects $\odot O$ at $P$ and $Q$. With $C$ as the center and $CP$ as the radius, a circle intersects $AB$ at $D$. Connect $QD$ and extend it to intersect $\odot O$ at $M$. T... | 2. (B).
Hint: It is easy to know that $\angle P Q M=45^{\circ}$. In the inscribed $\triangle P Q M$ of $\odot O$, by the Law of Sines, we get. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,865 |
3. Let $x, y$ be positive integers no greater than 9, the remainder of $x$ divided by 3 is $f(x)$, and the remainder of $y$ divided by 4 is $g(y)$. If $f(x)+2 g(y)=0$, then the value of $x+2 y$ is ( ).
(A) 11
(B) 22
(C) 25
(D) 27 | 3. (C).
According to the problem, we have
$$
f(x)=\left\{\begin{array}{l}
0, x=3,6,9, \\
1, x=1,4,7, \\
2, x=2,5,8 .
\end{array} \quad g(y)=\left\{\begin{array}{l}
0, y=4,8, \\
1, y=1,5,9, \\
2, y=2,6, \\
3, y=3,7 .
\end{array}\right.\right.
$$
Since $f(x)$ and $g(y)$ are both non-negative, to make $f(x) + 2g(y) = 0$... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,866 |
4. In isosceles $\triangle A B C$, $A B=A C=6, P$ is a point on side $B C$, and $P A=4$. Then $P B \cdot P C$ equals: ( ).
(A) 10
(B) 15
(C) 20
(D) 25 | 4. (C).
As shown in the figure, on the extensions of $AC$ and $BA$, intercept $AD = AE = AP = 4$, and connect $PD$ and $PE$. Then $\angle APD=$
$$
\begin{array}{l}
\angle ADP, \angle APE=\angle E, \angle BAP \\
=2 \angle E . \\
\begin{array}{l}
\because \angle APD= \angle APC-\angle CPD \\
=\angle B+\angle BAP-\angle... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,867 |
5. Let $m$ and $n$ be integers greater than 1. If the sum of the fractions $\frac{m^{2}-1}{n+1}$ and $\frac{n^{2}-1}{m+1}$ is an integer, then, ( ).
(A) $\frac{m^{2}-1}{n+1}$ and $\frac{n^{2}-1}{m+1}$ are both integers
(B) $\frac{m^{2}-1}{n+1}$ is an integer, $\frac{n^{2}-1}{m+1}$ is not an integer
(C) $\frac{m^{2}-1}{... | 5. (A).
Let $x=\frac{m^{2}-1}{n-1}, b=\frac{n^{2}-1}{m+1}$, then $ab=(m-1)(n-1)$ is an integer. Also, $a+b$ is an integer, so $a, b$ are the two roots of some integer-coefficient quadratic equation $x^{2}+p x+q=0$. Since $a, b$ are both rational numbers, the discriminant $\Delta=p^{2}-4 q$ of the equation is a perfect... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,868 |
6 . For the equation about $x$
$$
\begin{array}{l}
\left|x^{2}+b_{1} x+c_{1}\right|+\left|x^{2}+b_{2} x+c_{2}\right| \\
=\left|x^{2}+b_{3} x+c_{3}\right|
\end{array}
$$
the maximum number of real roots is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 6. (D).
Let $f_{i}(x)=x^{2}+b_{i} x+c_{i}(i=1,2,3)$, then the roots of the equation $\left|f_{1}(x)\right|+\left|f_{2}(x)\right|=\left|f_{3}(x)\right|$ are all the roots of the equation $f_{1}(x) \pm f_{2}(x)= \pm f_{3}(x)$. This combination represents four independent equations, and the coefficient of the quadratic t... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,869 |
1. If integers $a, b$ satisfy $\frac{2 \sqrt[3]{4}-b}{\sqrt[3]{4}-a \sqrt[3]{2}+1}$ $=\sqrt[3]{4}+1$, then $a=$ $\qquad$ ,$b=$ $\qquad$ | $$
\begin{array}{l}
\text { II. 1. } a=2, b=3 . \\
\text { Let } \sqrt[3]{2}=x \text {, then } \\
\frac{2 x^{2}-b}{x^{2}-a x+1}=x^{2}+1,
\end{array}
$$
which means
$$
\begin{array}{l}
x^{4}-a x^{3}-a x+b+1=0 . \\
\because x^{3}=2, \\
\therefore(2-a) x=2 a-b-1 .
\end{array}
$$
$\because x$ is an irrational number, and ... | a=2, b=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,870 |
2. Given an acute triangle $\triangle A B C$ with area $S$, perpendiculars are drawn from the midpoints of each side to the other two sides. Then, the area of the hexagon formed by these six perpendiculars (expressed in terms of $S$) is $\qquad$
Translate the above text into English, please retain the original text's ... | 2. $\frac{1}{2} S$.
As shown in the figure, the hexagon formed by six perpendicular lines is $D P E Q F R$. Let $O$ be the circumcenter of $\triangle A B C$. Since $\triangle A B C$ is an acute triangle, point $O$ is inside the medial triangle $\triangle D E F$. It is easy to prove that $O D P E, O E Q F, O F R D$ are... | \frac{1}{2} S | Number Theory | proof | Yes | Yes | cn_contest | false | 708,871 |
3. Given real numbers $a, b, c$ satisfy $\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$ $=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|$. If the quadratic function $y=a x^{2}+b x$ $+c$ intersects the $x$-axis at a fixed point, then the coordinates of this fixed point are $\qquad$. | 3. $(1,0)$.
From the given, we have
$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2},
$$
which implies $\square$
$$
\begin{array}{l}
\frac{2}{a b}+\frac{2}{b c}+\frac{2}{c a}=0 . \\
\therefore a+b+c=0 . \\
\because y=a x^{2}+b x+c \text { intersects the } x \text... | (1,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,872 |
4. In the acute triangle $\triangle A B C$, the lengths of the three sides $a, b, c$ are all integers, and $a<b<c, a+b+c=20$. Then $\angle A$ $+\angle C=$ $\qquad$ . | 4. $120^{\circ}$.
$$
\because 20=(a+b)+c>c+c=2 c,
$$
there is $c<\frac{20}{3}$.
$$
\therefore \frac{20}{3}>c>0, \text{ then } \triangle A B C \text{ is an acute triangle.}
$$
$$
\begin{array}{l}
\because \cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}=\frac{1}{2}, \\
\therefore \angle B=60^{\circ},
\end{array}
$$
then $\angle... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,873 |
II. (25 points) In an acute triangle $\triangle ABC$, $M$ is the foot of the perpendicular from $A$ to $BC$, with $P$ and $Q$ being the feet of the perpendiculars from $B$ and $C$ to $AC$ and $AB$, respectively. If $AD \perp PQ$, and $AD$ intersects the circumcircle of $\triangle ABC$ at $D$, prove that $PQ=\frac{2 S_{... | Connect $C D$.
$$
\because M P \perp A B, M Q \perp A C,
$$
$\therefore A, P, M, Q$ are concyclic, and $A M$ is the diameter of this circle. By the Inscribed Angle Theorem,
$$
\begin{array}{l}
P Q=A M \cdot \sin 4. \\
\because A D \perp P, A C \perp M Q, \\
\therefore \angle 2=\angle 3 .
\end{array}
$$
Since $\angle 1... | PQ=\frac{2 S_{\triangle ABC}}{AD} | Geometry | proof | Yes | Yes | cn_contest | false | 708,876 |
Three, (25 points) On the first day of operation, the factory's production does not exceed 20 units, and the daily production increases thereafter, but the increase in production each time does not exceed 20 units. When the daily production reaches 1996 units, find the minimum value of the total production of the facto... | Three, suppose the production on the $n$-th day after the start of work reaches 1996 pieces, and the production on the first day is $a_{1}$ pieces, with the increase in production on the $i$-th day being $a_{i}$ pieces $(i=2,3, \cdots, n)$, then we have
$$
a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1996,0<a_{i} \leqslant 20 .
$$
... | 100600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,877 |
1. For every pair of real numbers $x, y$, the function $f$ satisfies the equation
$$
f(x+y)-f(x)-f(y)=1+xy,
$$
and $f(1)=1$. Then, the number of integers $n$ for which $f(n)=n (n \neq 1)$ is ( ) .
(A) 0
(B) 1
(C) 2
(L) 3 | -1 (B).
Let $x=1$, then
$$
\begin{array}{l}
f(1)+f(y)=f(y+1)-y-1, \\
f(y+1)-f(y)=y+2 .
\end{array}
$$
Let $y=1,2,3, \cdots, y-1$ and add all the equations to get
$$
f(y)=\frac{y^{2}+3 y-2}{2} \text {. }
$$
Thus, for all natural numbers $y$, equation (2) holds. If $f(y)=y$, we get $y^{2}+3 y-2=2 y, y=-2, y=1$. Therefo... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,878 |
3. In a sports competition, the Red and White teams each have 5 players participating. The scoring method is: players score points for their team based on their ranking in the competition, with the first place scoring 1 point, the second place scoring 2 points, and so on, and each player's score is unique. The team wit... | 3. (C).
Both teams scored a total of $1+2+\cdots+10=55$ points. Therefore, to win, a team needs to score at most $\left[\frac{55}{2}\right]=27$ points. Since each team must score at least $1+2+3+4+5=15$ points, the answer to this problem can only be 27 points. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,880 |
4. There are the following four propositions:
(1) A pyramid with a base that is a regular polygon and all other faces being isosceles triangles is a regular pyramid.
(2) A triangular pyramid with a base that is an equilateral triangle and all dihedral angles between adjacent lateral faces being equal is a regular trian... | 4. (D).
None of the four propositions are correct, and counterexamples are provided as follows:
(A) Counterexample: The base is an equilateral triangle with a side length of 3, the two side faces are isosceles triangles with side lengths of 3, 3, and 2, and one side face is an isosceles triangle with side lengths of 2... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,881 |
5. Let $f: N \rightarrow N$, and for all positive integers $n$, $f(n+1)>f(n), f(f(n))=3n$. Then the value of $f(1997)$ is ( ).
(A) 1997
(B) 1268
(C) 3804
(D) 5991 | 5. (C).
From the given conditions, we can prove: $f(1)=2, f(2)=3$,
$$
f(3 n)=f[f(f(n))]=3 f(n) .
$$
From (1) and $f(1)=2, f(2)=3$, we get
$$
f\left(3^{n}\right)=2 \cdot 3^{n}, f\left(2 \cdot 3^{n}\right)=3^{n+1} .
$$
Notice that there are $3^{n}-1$ natural numbers between $2 \cdot 3^{n}$ and $3^{n+1}$, and there are... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,882 |
6. The system of equations
$$
\left\{\begin{array}{l}
(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)=1+y^{\prime} \\
(1+y)\left(1+y^{2}\right)\left(1+y^{1}\right)=1+x^{7}
\end{array}\right.
$$
has ( ) solutions for $(x, y)$.
(A) 4
(B) 2
(C) 1
(D) 0 | 6. (B).
Given the equations can be transformed into
$$
\left\{\begin{array}{l}
1-x^{8}=(1-x)\left(1+y^{7}\right), \\
1-y^{8}=(1-y)\left(1+x^{7}\right) .
\end{array}\right.
$$
When $x=y$, it is clear that $x=1$. This can be solved to get
$$
(x, y)=(C, 0),(-1, \cdots 1) \text {. }
$$
When $x \neq y$,
if $x>0, y>0$, th... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,883 |
Example 8 Proof: Every natural number $n$ greater than 6 can be expressed as the sum of two natural numbers greater than 1 and coprime to each other. (1995, National Junior High School Mathematics League) | Solution: (1) If $n$ is odd, let $n=2k+1$, an integer of 2, then it can be written as $n=k+(k-1)$. It is evident that this representation meets the requirement.
(2) If $n$ is even, then we can set $n=4k$ or $4k+2$, where $k$ is a natural number greater than 1. Therefore, we should discuss two cases.
When $n=4k$, it ca... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,886 |
3. The solution to the equation $\sqrt{x^{2}-\frac{1}{2} x+1}+\sqrt{x^{2}-\frac{2}{3} x+1}$ $=1+\frac{\sqrt{30}}{6}$ is . $\qquad$ | 3. $\frac{13-2 \sqrt{30}}{7}$.
The original equation can be transformed into
$$
\begin{array}{l}
\sqrt{\left(x-\frac{1}{4}\right)^{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}}+\sqrt{\left(x-\frac{1}{3}\right)^{2}+\left(\frac{2 \sqrt{2}}{3}\right)^{2}} . \\
=1+\frac{\sqrt{30}}{6} .
\end{array}
$$
The left side of the equa... | \frac{13-2 \sqrt{30}}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,887 |
4. If the equation $x^{2}+(1-2 i) x+3 m-i=0(m$ $\in R$ ) has one real root and one complex root, then this complex root is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 4. $-\frac{1}{2}+2 i$.
Let the equation have a real root $\alpha$, then we have
$$
\left(\alpha^{2}+\alpha+3 m\right)+(-2 \alpha-1) i=0 .
$$
By the necessary and sufficient condition for the equality of complex numbers, we can solve to get $a=-\frac{1}{2}$. Let the other root be $\beta$, then $\alpha+\beta=2 i-1$, he... | -\frac{1}{2}+2 i | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 708,888 |
5. In the plane, there are four points $A, B, C, D$, where $A, B$ are fixed points, and $|A B|=\sqrt{3}, C, D$ are moving points, H. $|A D|=|D C|=|B C|=1$, let $S_{\triangle A B D}=S$ be the area of $\triangle A B D$, $S_{\triangle B C D}=T$ be the area of $\triangle B C D$. Then the range of $S^{2}+T^{2}$ is $\qquad$. | 5.
5. $\frac{2 \sqrt{3}-3}{4} \leqslant S^{2}+T^{2} \leqslant \frac{7}{8}$.
From $B D^{2}=1^{2}+\left(\sqrt{3}^{2}\right)-2$
$$
\begin{aligned}
& \cdot 1 \cdot \sqrt{3} \cos \alpha \\
= & 1^{2}+1^{2}-2 \cdot 1 \\
\cdot & 1 \cos \beta
\end{aligned}
$$
we get $\cos \beta=\sqrt{3} \cos \alpha-1$.
$$
\text { Also, } \beg... | \frac{2 \sqrt{3}-3}{4} \leqslant S^{2}+T^{2} \leqslant \frac{7}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,889 |
6. The smallest natural number $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$ $<a-1995 \frac{1}{3}$ hold for all natural numbers $n$ is $\qquad$. | 6. 1997.
Let $\dot{f}(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$, then $f(n+1)-f(n)=\frac{-1}{(2 n+3)(2 n+2)}<0$, which means $f(n)$ is a decreasing function when $n \geq 0$. Therefore, $f(1)$ is the maximum value. To ensure the inequality holds for all $n \in N$, the maximum value of $f(n)$ must be less t... | 1997 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,890 |
One, (25 points) Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $c$ is the semi-focal distance, and chord $AB$ passes through the right focus $F_{2}$. Find the maximum value of the area of $\triangle F_{1}AB$. | Given the points \(A\left(x_{1}, y_{1}\right)\), \(B\left(x_{2}, y_{2}\right)\), and the angle \(a\) between \(AB\) and the \(x\)-axis, and \(c^{2}=a^{2}-b^{2}\).
Since
\[
\begin{array}{l}
S_{\triangle F_{1} A B} \\
=S_{\triangle A F_{1} F_{2}}+S_{\triangle B F_{1} F_{2}} \\
=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y... | a b \text{ when } b \leqslant c \text{ and } \frac{2 b^{2} c}{a} \text{ when } b > c | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,891 |
II. (25 points) If $x_{i}>0, \sum_{i=1}^{n} x_{i}=1, x_{n+1}=$ $x_{1}, n>6$, prove:
$$
\prod_{i=1}^{n} \frac{1}{x_{i}+x_{i+1}}>n!
$$ | $$
\prod_{i=1}^{n}\left(x_{i}+x_{i+1}\right) \leqslant\left[\frac{\sum\left(x_{i}+x_{i+1}\right)}{n}\right]^{n}=\left(\frac{2}{n}\right)^{n} .
$$
Below we prove $\left(\frac{n}{2}\right)^{n}>n!\quad(n>6)$.
Using mathematical induction.
(1) When $n=7$,
$$
\begin{aligned}
71 \times 2^{7} & =3^{2} \times 2^{11} \times 5 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,892 |
Three. (35 points) Given that $\triangle A B C$ is an isosceles triangle. Point $M, A E \perp B M$ at $E, A F \perp C E$ at $F$. Prove: $A F \leqslant \frac{1}{3} A B$.
| Let $H$ be the midpoint of $BC$, and connect $AH, HM$. Let $\angle BAH=\alpha, \angle ABE=\beta, BD=1, DM=MC=t$.
Thus, $HM \parallel BD$.
From the fact that $A, D, M, E$ are concyclic and $A, D, H, C$ are concyclic, we get
\[
\begin{array}{l}
BD \cdot AB = BM \cdot BE, \\
BD \cdot AB = BH \cdot BC.
\end{array}
\]
Ther... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,893 |
Example 9 If $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=k$, then the graph of the line $y=k x+k$ must pass through ().
(A) the first, second, and third quadrants
(B) the first and third quadrants
(C) the second, third, and fourth quadrants
(D) none of the above is correct
(1995, Hebei Province Junior High School Mathem... | Solution: When $a+b+c \neq 0$, $\frac{2(a+b+c)}{a+b+c}=k$, $k=2$.
When $a+b+c=0$, $a+b=-c, b+c=-a, c+a=-b$, we get $k=-1$.
Therefore, when $a+b+c \neq 0$, the line $y=kx+k$ is such that when $a+b+c=0$, the line $y=kx+k$ becomes $y=-x-1$, at this time the graph passes through the second, third, and fourth quadrants.
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,895 |
Example 1 Find the positive integer root of the equation
$$
\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2}=\frac{133}{60}
$$
(1990, Shanghai Junior High School Mathematics Competition) | Solution: Since $n$ is a natural number, by the properties of natural numbers and
$$
\begin{array}{l}
\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2} \\
\therefore \frac{2+3+4}{n+2}<\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2} \\
<\frac{2+3+4}{n} .
\end{array}
$$
That is, $\frac{9}{n+2}<\frac{133}{60}<\frac{9}{n}$.
Solving this, ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,897 |
Solve the equation about real numbers $x$, $y$, $z$
$$
\begin{array}{l}
\left(8 z x^{2}-27 y^{2}+9 y z\right)^{2}+\left(3 y^{2}-y z+2 z^{2}\right. \\
-8 x)^{2}+9=6 x-x^{2} .
\end{array}
$$
(1991, Sichuan Province Junior High School Mathematics League) | The original equation can be transformed into
$$
\begin{array}{l}
\left(8 z x^{2}-27 y^{2}+9 y z\right)^{2}+\left(3 y^{2}-y z+2 z^{2}\right. \\
-8 x)^{2}+(x-3)^{2}=0 .
\end{array}
$$
By the property of non-negative numbers, we have
$$
\left\{\begin{array}{l}
8 z x^{2}-27 y^{2}+9 y z=0, \\
3 y^{2}-y z+2 z^{2}-8 x=0, \\... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,898 |
Example 11 Solve the equation
$$
2\left(x+\sqrt{x^{2}-1}\right)=(x-1+\sqrt{x+1})^{2} \text {. }
$$
(1987, Junior High School Mathematics Correspondence Competition in Some Provinces and Cities) | Solution: From the original equation, we know $x \geqslant 1$, so the original equation can be rewritten as
$$
\begin{array}{l}
(x+1)+2 \sqrt{x+1} \cdot \sqrt{x-1}+(x-1) \\
=(x-1+\sqrt{x+1})^{2} .
\end{array}
$$
Let $\sqrt{x+1}=m, \sqrt{x-1}=n(m>0$, $n \geqslant 0$ ), then the original equation can be transformed into... | x=1 \text{ or } x=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,899 |
5. The number of $n$ that makes each interior angle of a regular $n$-sided polygon an even number of degrees is ( ).
(A) 15
(B) 16
(C) 17
(D) 18 | 5. (B).
For a regular $n$-sided polygon, each interior angle is $\frac{(n-2) 180^{\circ}}{n}=180^{\circ}-\left(\frac{2^{3} \times 3^{2} \times 5}{n}\right)^{0} .2^{3} \times 3^{2} \times 5$ has $(1+3)(1+2)(1+1)=24$ divisors, but $n \geqslant 3$, so we should remove $1$ and $2$, which leaves 22. When $\frac{2^{3} \time... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,900 |
18. Arrange $mn$ numbers arbitrarily into a square matrix, with $m$ numbers in each row and $n$ numbers in each column. In each column, select the largest number, then from the selected $m$ numbers, choose the smallest number $a$; in each row, select the smallest number, then from the selected $n$ numbers, choose the l... | 6. (D).
If $a$ and $b$ are in the same row or column, according to the problem, we have $a \geqslant b$; if $a$ and $b$ are not in the same row or column, let the number at the intersection of the column containing $a$ and the row containing $b$ be $c$, then $a \geqslant c, c \geqslant b$, thus $a \geqslant b$. | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,901 |
2. Given positive integers $n$ and $a_{1}, a_{2}, \cdots, a_{n}$ such that $a_{1}+a_{2}+\cdots+a_{n}=1996$, and the product is maximized. Then $n=$ $\qquad$ | 2. 665 or 666.
Because
$$
2(a-2)>a(a \geqslant 5)
$$
$$
\begin{array}{l}
3 \times 3>2 \times 4 \\
3 \times 3>2 \times 2 \times 2 \\
3 \times 3 \times 2>4 \times 4 \\
a+1>a \times 1 \text { ( } a \text { is a positive integer) }
\end{array}
$$
Therefore, under the premise that $a_{1}+a_{2}+\cdots+a_{n}$ is constant, t... | 665 \text{ or } 666 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,903 |
4. Given that the roots of the equation $x^{2}-x-1=0$ are also roots of the equation $x^{6}-p x^{2}+q=0$. Then $p=$ $\qquad$ ,$q=$ $\qquad$ . | 4. $p=8, q=-21$.
Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-1=0$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=1, \\
x_{1} x_{2}=-1, \\
x_{1}^{6}-p x_{1}^{2}+q=0, \\
x_{2}^{6}-p x_{2}^{2}+q=0 .
\end{array}\right.
$$
From (1) and (2), we get $x_{1}^{2}+x_{2}^{2}=3, x_{1}^{4}+x_{2}^{4}=7$.
Subtracting (4) ... | p=8, q=-21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,905 |
II. (25 points) Given $\triangle A B C$, where $B$ and $C$ are fixed points and $A$ is a moving point. Construct the diameters $A A_{1}$, $B B_{1}$, and $C C_{1}$ of the circumcircle of $\triangle A B C$. Let $M=S_{\triangle A_{1} B C}$, $N=S_{\triangle A_{2} B C}+S_{\triangle A B_{1} C}+S_{\triangle A D C_{1}}$. Deter... | (1) If $\triangle A B C$ is an obtuse triangle, it is obvious that $M>N$;
(2) If $\triangle A B C$ is a right triangle, it is obvious that $M>N$;
(3) If $\triangle A B C$ is an acute triangle, let $O$ be the circumcenter of $\triangle A B C$, and draw $O P_{1} \perp A B, O P_{2} \perp B C, O P_{3} \perp A C$, with $P_{... | M=N | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,907 |
Three, when arranging student dormitories, it is stipulated that two students live in one room. The school's preference order for various combinations of students from grades $A, B, C, D, E$ is
$$
A A, A B, A C, A D, A E, B B, B C, B D,
$$
$B E, C C, C D, C E, D D, D E, E E$.
That is, two students from grade $A$ living... | Three, Solution: According to the problem, we have
$$
a a>a b>a c>a d>a e>b b>b c>b d>b e>c c>c d
$$
$>c e>d d>d e>e e$, which means
$$
\left\{\begin{array}{l}
a>b>c>d>e, \\
a>\frac{b^{2}}{e}, \\
b>\frac{c^{2}}{e}, \\
c>\frac{d^{2}}{e} .
\end{array}\right.
$$
From (2), (3), and (4), we know that for a given $t$, to ma... | 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,908 |
1. Given the function $y=-\frac{x-a}{x-a-1}$, the graph of its inverse function is symmetric with respect to the point $(-1,3)$. Then the real number $a$ equals ( ).
(A) 2
(B) 3
(C) -2
(D) -4 | -1. (A).
From the problem, we know that the graph of the function $y=-\frac{x-a}{x-a-1}$ is centrally symmetric about the point (3, 1). By $y=-\frac{x-a}{x-a-1}=-1-\frac{1}{x-a-1}$, we get: $(x-a-1)(y+1)=-1$, which is a hyperbola centered at the point $(a+1,-1)$. Therefore, $a+1=3$, which means $a=2$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,909 |
Example 12 Solve the equation
$$
5 x^{2}+x-x \sqrt{5 x^{2}-1}-2=0 \text {. }
$$
(1979, National High School Mathematics Competition) | Solution: Let $\sqrt{5 x^{2}-1}=y$, then the original equation can be transformed into $y^{2}-x y+x-1=0$.
Solving, we get $y=1$ or $y=x-1$.
Thus, $\sqrt{5 x^{2}-1}=1$ or $\sqrt{5 x^{2}-1}=x-1$.
$\therefore x= \pm \frac{\sqrt{10}}{5}$ or $x=-1$ or $x=\frac{1}{2}$.
Upon verification, $x= \pm \frac{\sqrt{10}}{5}$ are the ... | x= \pm \frac{\sqrt{10}}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,910 |
2. We call an ellipse with an eccentricity equal to the golden ratio $\frac{\sqrt{5}}{2}-1$ a "beautiful ellipse". Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>i$ $>0$ ) be a beautiful ellipse, $F, A$ are its left focus and vertex, respectively, and $B$ is one endpoint of its minor axis. Then $\angle A B F$ equals ... | 2. (C).
From $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$, we get $a^{2}-a c-c^{2}=0$.
$$
\begin{aligned}
\because \quad & |A B|^{2}+|B F|^{2}-|F A|^{2} \\
& =\left(a^{2}+b^{2}\right)+a^{2}-(a+c)^{2} \\
& =2\left(a^{2}-a c-c^{2}\right)=0, \\
\therefore \quad & |A B|^{2}+|B F|^{2}=|F A|^{2}, \text{ hence } \angle A B F=90^{\circ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,911 |
3. Given that the lengths of the three sides of $\triangle A B C$ are $a, b, c$, and the complex numbers $z_{1}, z_{2}$ satisfy: $\left|z_{1}\right|=a,\left|z_{2}\right|=b,\left|z_{1}+z_{2}\right|=$ $c$, then the complex number $\frac{z_{1}}{z_{2}}$ is (
(A) a real number
(B) a pure imaginary number
(C) not a real numb... | 3. (C).
$$
\begin{array}{l}
\because \quad c^{2}=\left|z_{1}+z_{2}\right|=\left(z_{1}+z_{2}\right)\left(\overline{z_{1}}+\overline{z_{2}}\right) \\
=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2} \\
=a^{2}+b^{2}+\frac{z_{1}}{z_{2}} \cdot z_{2} \overline{z_{2}}+z_{1} \overli... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,912 |
4. The maximum value of the function $f(x)=\frac{\frac{1}{6} \cdot(-1)^{1+C_{2 x}^{x}} \cdot P_{x+2}^{5}}{1+C_{3}^{2}+C_{4}^{2}+\cdots+C_{x-1}^{2}}$ is ( ).
(A) 20
(B) 10
(C) -10
(D) -20 | 4. (D).
From $x-1 \geqslant 2$ and $x+2 \geqslant 5$, we get the domain of $f(x)$ as $\{x \mid x \geqslant 3, x \in N\}$.
$\because \quad C_{2 x}^{x}=\frac{(2 x)!}{x!\cdot x!}=\frac{(2 x) \cdot(2 x-1)!}{x \cdot(x-1)!\cdot x!}$ $=2 C_{2 x-1}^{x}$ is even,
$$
\begin{array}{l}
\therefore \quad \frac{1}{6} \cdot(-1)^{1+C_... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,913 |
6. Among all the subsets of the set $M=\{1,2,3, \cdots, 10\}$, there is a family of non-empty subsets such that the intersection of any two is not empty. What is the maximum number of such subsets?
(A) $2^{10}$
(B) $2^{9}$
(C) $10^{2}$
(D) $9^{2}$ | 6. (B).
Obviously, this family of subsets does not contain the empty set. According to the number of elements they contain, this family of subsets can be divided into 10 categories. Let's assume $A_{n_{i}}(i=1,2, \cdots, 10)$ is a subset containing $i$ elements, then the total number of subsets in this family that sat... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,915 |
1. In a Cartesian coordinate system, the two legs of a right triangle are parallel to the two coordinate axes, and the lines on which the two legs lie have the equations $y=3 x+1$ and $y=m x+2$, respectively. Then the value of the real number $m$ is $\qquad$. | II. $1 . \frac{3}{4}$ or 12 .
Let the lengths of the sides of the right triangle parallel to the $x$-axis and $y$-axis be $a, b$, respectively, then
$$
\left\{\begin{array}{l}
m=\frac{\frac{1}{2} b}{a}, \\
3=\frac{b}{\frac{1}{2} a} \text { or }
\end{array}\left\{\begin{array}{l}
m=\frac{b}{\frac{1}{2} a}, \\
3=\frac{\f... | \frac{3}{4} \text{ or } 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,916 |
2. Let $f(x)=\frac{a^{x}}{1+a^{x}}(a>0, a \neq 1), [m]$ denote the greatest integer not exceeding the real number $m$, then the range of the function $\left[f(x) -\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]$ is $\qquad$ . | 2. $\{-1,0\}$.
Let $g(x)=f(x)-\frac{1}{2}$, then
$$
\begin{aligned}
g(-x) & =\frac{a^{-x}}{1+a^{-x}}-\frac{1}{2}=\frac{1}{a^{x}+1}-\frac{1}{2} \\
& =\frac{\left(1+a^{x}\right)-a^{x}}{1+a^{x}}-\frac{1}{2} \\
& =\frac{1}{2}-\frac{a^{x}}{1+a^{x}} \\
& =-g(x) .
\end{aligned}
$$
$$
\begin{array}{ll}
\therefore \quad & {\le... | \{-1,0\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,917 |
3. Let $a, b, c$ be the three sides of a right triangle, with $c$ being the hypotenuse. The maximum value of $k$ such that the inequality $a^{2}(b+c)+b^{2}(c+a)$ $+c^{2}(a+b) \geqslant k a b c$ holds for all right triangles is $\qquad$. | $$
\begin{array}{l}
3.2+3 \sqrt{2} \\
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
=\left(a^{2}+b^{2}\right) c+a\left(b^{2}+\frac{c^{2}}{2}\right)+b\left(\frac{c^{2}}{2}+a^{2}\right) \\
+\frac{c}{2} \cdot c(a+b) \\
\geqslant 2 a b c+\sqrt{2} a b c+\sqrt{2} a b c+\frac{c}{2} \sqrt{a^{2}+b^{2}} \\
\cdot 2 \sqrt{a b} \\
\geqslant(... | 2+3\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,918 |
4. As shown in the figure, for the regular triangular prism $A B C-A_{1} B_{1} C_{1}$, all edges are of length 1. The section $B C_{1} D$ intersects the edge $A A_{1}$ at point $D$. Let the dihedral angles formed by this section with the base $A B C$ and the three side faces $A B B_{1} A_{1}$, $B C C_{1} B_{1}$, and $C... | 4. $\frac{3}{8} \sqrt{3}$.
Let $A D=x$, then $S_{\triangle B C_{1} D}=\frac{1}{4} \sqrt{4 x^{2}-4 x+7}$.
Thus, the projection areas of the section $B C_{1} D$ on the base $A B C$ and the three side faces $A B B_{1} A_{1}, B C C_{1} B_{1}, C A A_{1} C_{1}$ are $\frac{\sqrt{3}}{4}, \frac{1}{4}(2-x), \frac{1}{4}(2 x-1), ... | \frac{3}{8} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,919 |
5. Given $f(x)$ is a function defined on the set of real numbers, and $f(x+2)[1-f(x)]=1+f(x)$. If $f(1)=$ $2+\sqrt{3}$, then the value of $f(1949)$ is $\qquad$ | $$
\begin{array}{l}
\text { 5. } \sqrt{3}-2 \text {. } \\
\text { From } f(x+2)[1-f(x)]=1+f(x) \text {, we get } f(x+2) \\
=\frac{1+f(x)}{1-f(x)} \text {. } \\
\therefore \quad f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} \\
=-\frac{1}{f(x)} \text {. } \\
\therefore \quad f(x... | \sqrt{3}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,920 |
Example 13 Find the sum of all negative roots of the equation $x^{3}-\frac{3}{2} \sqrt[3]{6} x^{2}+3=0$.
(2nd Hope Cup Mathematics Competition) | Solution: The original equation can be transformed into
$$
2 x^{3}-3 \sqrt[3]{6} x^{2}+6=0 \text {. }
$$
Let $\sqrt[3]{6}=y$, then the original equation can be transformed into
$$
2 x^{3}-3 x^{2} y+y^{3}=0 \text {, }
$$
which is $(x-y)^{2}(2 x+y)=0$.
Thus, $x=y$ or $x=-\frac{1}{2} y$,
i.e., $x=\sqrt[3]{6}$ or $x=-\fr... | -\frac{1}{2} \sqrt[3]{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,921 |
6. Let $x_{1}$ be the greatest negative root of the equation $\sqrt{3} \sin x-3 \cos x=2 a-1$, and $x_{2}$ be the smallest positive root of the equation $2 \cos ^{2} x-2 \sin ^{2} x=a$. Then, the range of real numbers $a$ for which the inequality $\left|x_{1}\right| \leqslant x_{2}$ holds is . $\qquad$ | 6. $\frac{1}{2}-\sqrt{3} \leqslant a \leqslant-1$ or $a=2$.
The equation $\sqrt{3} \sin x-3 \cos x=2 a-1$ is equivalent to
$\sin \left(x-\frac{\pi}{3}\right)=\frac{2 a-1}{2 \sqrt{3}}$.
From $-1 \leqslant \frac{2 a-1}{2 \sqrt{3}} \leqslant 1$, we get
$$
\frac{1}{2}-\sqrt{3} \leqslant a \leqslant \frac{1}{2}+\sqrt{3} \t... | \frac{1}{2}-\sqrt{3} \leqslant a \leqslant-1 \text{ or } a=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,922 |
I. (25 points) The Xiguang Factory glasses workshop has received a batch of tasks, requiring the processing of 6000 Type A parts and 2000 Type B parts. This workshop has 214 workers, and each of them can process 3 Type B parts in the time it takes to process 5 Type A parts. These workers are to be divided into two grou... | One, Solution: Let the number of people in the group that includes those with the surname $A$ be $x$, and in a unit of time, the number of $\mathrm{C} . \mathrm{A}$ parts processed by one person is $5 k$, then the number of the other parts is $3 k$.
The time required to process type $A$ parts is $t_{A}(x)=\frac{6000}{5... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,923 |
II. (25 points) Given a quadrilateral whose sides are all integers, and the length of any one side can divide the sum of the lengths of the other three sides. Prove: This quadrilateral must have two equal sides. | II. Proof: (By contradiction) Assume that any two sides of the quadrilateral are not equal. Let $a_{1}>a_{2}>a_{3}>a_{4}$ represent the lengths of the four sides, and $p=a_{1}+a_{2}+a_{3}+a_{4}$ represent the perimeter of the quadrilateral.
$$
\begin{array}{l}
\because \quad a_{i} \mid p-a_{i}, \\
\therefore \quad a_{i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,924 |
Three. (35 points) The real number sequence $a_{1}, a_{2} \cdots, a_{1997}$ satisfies:
$$
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1996}-a_{1997}\right|=
$$
1997. If the sequence $\left\{b_{n}\right\}$ satisfies:
$$
b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, 1997),
$$
find the max... | $$
\begin{array}{l}
\text { III. Solution: From } \left.\left|b_{k}-b_{k+1}\right|=\frac{1}{k(k+1)} \right\rvert\,(k+1)\left(a_{1}+a_{2}\right. \\
\left.+\cdots+a_{k}\right)-k\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \mid \\
=\frac{1}{k(k+1)}\left|a_{1}+a_{2}+\cdots+a_{k}-k a_{k+1}\right| \\
\left.=\frac{1}{k(k+1)} \rig... | 1996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,925 |
Four. (35 points) Given two heptagonal pyramids, they share a common base $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$, with vertices $P_{1}$ and $P_{2}$ on either side of the base. Now, each of the following segments is colored either red or blue: $P_{1} P_{2}$, all diagonals on the base, and all lateral edges. Prove t... | Four, first prove a lemma:
Lemma: If among the 7 edges $\underbrace{}_{i} A_{j}(i \in$
$\{1,2\}, j \in\{1,2, \cdots$,
7\}) there are 5 edges of the same color,
then the proposition holds.
Proof: Under the given conditions, there must be three non-adjacent edges $P_{i} A_{r}, P_{i} A_{s}, P_{i} A_{t}$ on the same sid... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,926 |
Initial 49. Three circles with radii of $1, 2, 3$ are externally tangent to each other.
(1) Find the radius of the circle that is externally tangent to all three circles.
(Nanchang Liaoning Normal University High School Grade 1, Class 1, 116023)
(2) Find the radius of the circle that is internally tangent to all three ... | Solution: The side lengths of the triangle formed by the centers of the three circles are $3$, $4$, and $5$, so this triangle is a right triangle. Let the coordinates of its vertices be $(0,0)$, $(0,3)$, and $(4,0)$.
(1) Let the coordinates of the center of the required circle be $(u, v)$, and the radius be $x$. Then, ... | \frac{6}{23} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,927 |
Given $\triangle ABC$, with $I$ as its incenter, the internal angle bisectors of $\angle A$, $\angle B$, and $\angle C$ intersect their opposite sides at $A'$, $B'$, and $C'$, respectively. Prove:
$$
\begin{aligned}
\frac{5}{4}< & \frac{A I \cdot B I}{A A' \cdot B B'}+\frac{B I \cdot C I}{B B' \cdot C C'} \\
& +\frac{C... | Proof: Let $BC = a$, $CA = b$, $AB = c$, and assume $a + b + c = 1$. Then it is easy to see that
$$
\begin{array}{l}
\frac{AI}{AA'} = b + c, \quad \frac{BI}{BB'} = c + a, \\
\frac{CI}{CC'} = a + b, \quad \frac{IA'}{AA'} = a, \\
\frac{IB'}{BB'} = b, \quad \frac{IC'}{CC'} = c.
\end{array}
$$
Therefore, we only need to p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,928 |
49. There are two watches of different sizes, both of which keep accurate time. On their faces, there exists a fixed point $M$, such that the triangle formed by point $M$ and the tips of the second hands of the two watches is similar at any given moment.
(Ye Tingyu, Ren Yansong Middle School, Ruian City, Zhejiang, 3252... | Proof: Construct a complex plane as shown, let the complex numbers corresponding to points $P, Q$ be:
$$
\begin{array}{l}
p=r_{1} e^{i\left(b_{1} \mid w \tau\right)}, \\
q=a+r_{2} e^{j\left(\theta_{2}+w\right)}
\end{array}
$$
(Where $t$ is a parameter)
Assume that a point $M$ exists, whose corresponding complex number ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,929 |
Given $\triangle ABC$, $R$ and $r$ are the radii of its circumcircle and incircle, respectively, and $\angle A \geqslant \angle B \geqslant \angle C$. Prove:
$$
\cos (A-C) \leqslant \frac{1}{2}+\frac{r}{R} \leqslant 1 .
$$
(Zhang Gaili, Zhoushan No.1 High School, Zhejiang, 316200) | Prove: Substituting $\cos A+\cos B+\cos C=1+\frac{r}{R}$ into (I), we get
$$
\cos A+\cos B+\cos C \geqslant \frac{1}{2}+\cos (A-C) .
$$
Applying the sum-to-product trigonometric formulas, it transforms to
$2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}+1-\sin ^{2} \frac{B}{2}$
$$
\geqslant 2 \cos ^{2} \frac{A-C}{2}-\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,930 |
Example 14 Solve the equation
\[
\begin{array}{l}
(x+1)(x+2)(x+3)(x+4) \\
=(x+1)^{2}+(x+2)^{2}+(x+3)^{2}+(x+4)^{2} .
\end{array}
\] | Solution: Let $y=\frac{1}{4}(x+1+x+2+x+3+x+4)=x+\frac{5}{2}$, the original equation can be transformed into
$$
\begin{array}{l}
\left(y-\frac{3}{2}\right)\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right)\left(y+\frac{3}{2}\right) \\
=\left(y-\frac{3}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\r... | y= \pm \frac{1}{2} \sqrt{13+4 \sqrt{15}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,931 |
Example 16 Solve the equation
$$
(2 x+1)^{7}+x^{7}+3 x+1=0 .
$$ | Solution: The original equation can be transformed into
$$
(2 x+1)^{7}+(2 x+1)=-\left(x^{7}+x\right) \text {. }
$$
Construct the function $f(x)=x^{7}+x$, the original equation can be rewritten as
$$
f(2 x+1)=-f(x) \text {. }
$$
Since $f(x)=x^{7}+x$ is an odd function, we have
$$
f(2 x+1)=f(-x) \text {. }
$$
Also, $f... | x=-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,933 |
Example 17 If $a \geqslant b>c$ and $a<b+c$, solve the equation
$$
b \sqrt{x^{2}-c^{2}}+c \sqrt{x^{2}-b^{2}}=a x \text {. }
$$
(1993, Nanjing City Junior High School Mathematics Competition) | Solution: According to the problem, we can construct $\triangle ABC$ with $a, b$, and $c$ as sides, such that $BC = a, CA = b, AB = c$, and $a \geqslant b > c$. Draw perpendiculars to $AC$ and $BA$, intersecting at point $D$. Thus, $ABDC$ is inscribed in a circle with diameter $AD$. By Ptolemy's theorem, we have
$$
AC ... | x = \frac{abc}{2 \sqrt{s(s-a)(s-b)(s-c)}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,934 |
Example 18 The solution to the equation $\left(1+x^{2}\right)^{2}=4 x\left(1-x^{2}\right)$ is $\qquad$
(2nd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Solution: Let $x=\operatorname{tg}^{2} \theta\left(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)$, the original equation becomes
$$
\left(1+\operatorname{tg}^{2} \theta\right)^{2}=4 \operatorname{tg} \theta\left(1-\operatorname{tg}^{2} \theta\right) .
$$
Using $\operatorname{tg} \theta=\frac{\sin \theta}{\cos \theta}$ an... | x=\tan^2\left(\frac{k \pi}{2}+\frac{\pi}{8}\right) \text{ (} k=0,1,2,3 \text{)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,935 |
Example 1 Does there exist a sequence of natural numbers such that each natural number appears exactly once, and for any $k=1,2,3, \cdots$, the sum of the first $k$ terms of the sequence is divisible by $k$?
(21st All-Russian High School Mathematics Competition) | Solution: Construct the sequence $\left\{a_{n}\right\}$ as follows: take $a_{1}=1$. Suppose the first $n$ terms satisfying the condition have been chosen, which are $a_{1}, a_{2}$, $\cdots, a_{n}$. Let $m$ be the smallest natural number outside these terms, and $M$ be the largest natural number among these terms. Denot... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,936 |
Example 2 Proof: For each $n \in N(n \geqslant 2)$, there exists a permutation $p_{1}, p_{2}, p_{3}, \cdots, p_{n}$ of $1,2,3, \cdots, n$ such that for $k=1,2,3, \cdots, n-1$, the number $p_{k+1}$ divides $p_{1}+p_{2}+\cdots+p_{k}$.
(1991, German Mathematical Olympiad) | Proof: By construction.
When $n=2m$,
$$
m+1,1, m+2,2, \cdots, m+m, m
$$
is a permutation of $1,2,3, \cdots, 2m$, and for $l(1 \leqslant l \leqslant m, l \in \mathbb{N})$, we have
$$
\begin{array}{l}
(m+1)+1+(m+2)+2+\cdots \\
+(m+l-1)+(l-1)+(m+l) \\
=l m+2[1+2+\cdots+(l-1)]+l \\
=l m+l(l-1)+l=l(m+l)
\end{array}
$$
is... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,937 |
Example 3 Solve the equation
$$
\frac{1}{6 x-5}-\frac{1}{5 x+4}=\frac{1}{5 x-4}-\frac{1}{4 x+5} .
$$ | Solution: By differentiating both sides of the equation, we get
$$
\frac{9-x}{30 x^{2}-x-20}=\frac{9-x}{20 x^{2}+9 x-20}
$$
By the property of fractions, when two fractions are equal, if the numerators are equal, then the fractions are zero or the denominators are equal.
$$
\therefore 9-x=0
$$
or $30 x^{2}-x-20=20 x^... | x_{1}=9, x_{2}=0, x_{3}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,938 |
Example 3 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad) | Proof: First, we prove that it is impossible for all 98 sums to be odd. We use proof by contradiction.
If all the sums are odd, then the arrangement of these 100 natural numbers can only be one of the following four cases:
(1) odd odd odd odd odd odd ...; (2) odd even even odd even even ...;
(3) even odd even even odd... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,939 |
Example 4 Arrange the natural numbers $1,2,3, \cdots, n$ in any order: $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$. If $a_{1}=m$, then reverse the order of the first $m$ numbers in this arrangement. Prove that after a finite number of transformations, 1 can be placed at the beginning. | Prove: When $n=1$, the proposition holds.
Assume when $n=k$, the proposition holds.
When $n=k+1$, let the permutation of $1, 2, \cdots, k, k+1$ be
$a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}$.
(1) If $a_{k+1}=k+1$, then the first $k$ numbers of this permutation consist of $1, 2, \cdots, k$. According to the induction hypoth... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,940 |
Example 5 On a bookshelf, the 10-volume collections of Ilif and Petrov are placed in a random order. The librarian can take any volume from the shelf and place it in the 5th position (counting from the left) at each step. Prove: The librarian can, through a series of such operations, arrange the collection in increasin... | Proof: First, perform the above operation on the 1st volume, then sequentially perform the operation on all volumes to the left of the 1st volume until the 1st volume is in the 1st position. Next, perform the operation on the 10th volume, then sequentially perform the operation on all volumes to the right of the 10th v... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,941 |
Example 6 Try to write the first $n$ natural numbers on a circle so that the difference between any two adjacent numbers does not exceed 2. Then prove: this method is unique, and for this, it is enough to pay attention to the number closest to each number.
$(1928$, Hungarian Mathematical Competition) | Solution: On both sides of 1, only 2 and 3 can be placed (Figure 1 and 2). On the other side of number 2, no number other than 4 can be placed, and on the other side of 3, only 5 can be placed, and so on. Therefore, on one side of 1, only even numbers $2,4,6, \cdots$ can be placed in sequence; on the other side of 1, o... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,942 |
Example 7 Let $D=\{1,2,3, \cdots, 10\}, f(x)$ be a one-to-one mapping from $D$ to $D$. Let $f_{n+1}(x)=f\left[f_{n}(x)\right]$, $n \in N ; f_{1}(x)=f(x)$. Try to find a permutation $x_{1}, x_{2}, x_{3}, \cdots, x_{10}$ of $D$, such that the following equation holds:
$$
\sum_{i=1}^{10} x_{i} f_{2520}(i)=220 \text {. }
$... | Solution: Since $f(x)$ is a one-to-one mapping from $D$ to $D$, for $i=1,2, \cdots, 10$, by $\left\{i, f_{1}(i), f_{2}(i), \cdots\right.$, $\left.f_{10}(i)\right\} \subseteq D$, according to the pigeonhole principle, there exists $r_{i}\left(1 \leqslant r_{i} \leqslant\right.$ $10)$, such that $f_{n_{i}}(i)=i$.
It is ... | 10, 9, 8, \cdots, 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,943 |
For example, $8 p$ is a permutation of the set $S_{n}=\{1,2, \cdots, n\}$: an element $j \in S_{n}$, if it satisfies $p(j)=j$, then it is called a fixed point of $p$. Let $f_{n}$ be the number of permutations of $S_{n}$ with no fixed points, and $g_{n}$ be the number of permutations with exactly one fixed point. Prove:... | Proof: Using the principle of inclusion-exclusion, we can obtain the derangement formula for $n$ elements:
$$
\begin{aligned}
F(n)= & n!\left[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots\right. \\
& \left.+\frac{(-1)^{n}}{n!}\right] .
\end{aligned}
$$
It is known that $f_{n}=F(n), g_{n}=C_{n}^{1} F(n-1)$.
Thus, $f_{n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,944 |
Example 9 The permutation of integers $1,2, \cdots, n$ satisfies: each number is either greater than all the numbers before it, or less than all the numbers before it. How many such permutations are there?
(21st Canadian High School Mathematics Competition) | Let the number of permutations we are looking for be $a_{n}$.
Obviously, $a_{1}=1$.
For $n \geqslant 2$, consider the largest number $n$. If $n$ is placed in the $i$-th position, then the $n-i$ numbers after it are completely determined, i.e., they can only be $n-i, n-i-1, \cdots, 1$; and the $i-1$ numbers before it ha... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,945 |
Example 10 Let $n \in Z^{+}$. We say a permutation $\left(x_{1}, x_{2}, x_{3}, \cdots, x_{2 n}\right)$ of $1,2,3, \cdots, 2 n$ has property $p$, if there is at least one $i(1 \leqslant i \leqslant 2 n-1, i \in N)$ such that $\left|x_{i}-x_{i+1}\right|=n$. Prove that for any $n$, the permutations with property $p$ are m... | Proof: Let $A=\left\{\left(x_{1}, x_{2}, \cdots, x_{2 n}\right) \mid\left(x_{1}, x_{2}\right.\right.$, $\left.\cdots, x_{2 n}\right)$ has property $\left.p\right\}, B=\left\{\left(y_{1}, y_{2}, \cdots, y_{2 n}^{\prime}\right)\right.$ $\mid\left(y_{1}, y_{2}, \cdots, y_{2 n}\right)$ does not have property $\left.p\right... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,946 |
Example 1 Let $m, n, p$ be positive real numbers, and $m^{2}+n^{2}-p^{2}=0$. Find the minimum value of $\frac{p}{m+n}$.
(1988, Luo Jij Province Mathematics Competition) | Proof: According to the geometric meaning of the condition, using the Pythagorean theorem, starting from a right-angled triangle, construct a right-angled trapezoid \(CDEA\) as shown in the figure. Here, \(CD\) and \(AE\) are the legs, with \(CD \leqslant AE\), i.e., \(m+n \leqslant \sqrt{2} p\). When \(m=n\), the righ... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,947 |
Example 2 Let $a, b \in \mathbb{R}^{+}$, and $|\alpha| > a$. Then, regardless of the value of $\theta$, we always have
$$
\begin{array}{l}
\frac{\alpha \beta - \sqrt{a^{2} \beta^{2} + b^{2} \alpha^{2} - a^{2} b^{2}}}{\alpha^{2} - a^{2}} \\
\leqslant \frac{\beta + b \sin \theta}{\alpha + a \cos \theta} \leqslant \frac{\... | Prove: Let $k=\frac{\beta+b \sin \theta}{\alpha+a \cos \theta}$ represent the slope of the line connecting point $A(\alpha, \beta)$ and point $P(-a \cos \theta, b \sin \theta)$.
Point $P$ lies on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Since $|\alpha|>a$, point $(\alpha, \beta)$ is to the right of the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,948 |
Example 4 Solve the equation
\[
\begin{array}{l}
\sqrt{x^{2}+5 x-14}+\sqrt{x+7}+\sqrt{2-x}+x-5 \\
=0 .
\end{array}
\] | Solution: By the properties of quadratic radicals, we get
$$
\left\{\begin{array}{l}
x^{2}+5 x-14 \geqslant 0, \\
x+7 \geqslant 0, \\
2-x \geqslant 0 .
\end{array}\right.
$$
Solving, we get $-7 \leqslant x \leqslant-7$ or $2 \leqslant x \leqslant 2$.
That is, $x=-7$ or $x=2$.
Upon verification, $x=2$ is a root of the ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,949 |
Example 3 Given $a, b, c \in (-1,1)$.
Prove: $abc + 2 > a + b + c$.
untranslated text remains unchanged. | Proof: Treating $a$ as a variable and $b, c$ as constants, we denote
$$
\begin{aligned}
f(a) & =a i x-2-(a+b+c) \\
& =(b c-1) a+2-b-c, \\
& a \in(-1,1) .
\end{aligned}
$$
Now, we only need to prove $f(a)>0$ based on the properties of a linear function.
$$
\begin{array}{l}
\because b, c \in(-1,1), b cf(1)=1-b-c+b c \\
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,950 |
Example 4 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}$ are real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1, 0<\lambda_{1} \leqslant \lambda_{2} \leqslant \cdots \leqslant \lambda_{n}$. Prove:
$$
\left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right)\left(\s... | Prove: The conclusion of the problem has a discriminant structure. Construct the corresponding $\cdots$-degree function, let
$$
\begin{aligned}
f(x)= & \left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right) x^{2}-\left(\frac{\lambda_{1}+\lambda_{n}}{\sqrt{\lambda_{1} \lambda_{n}}}\right) x \\
& +\left(\sum_{i=1}^{n} a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,951 |
Example 5 Let real numbers $a, b, c$ satisfy:
$$
\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0 .
\end{array}\right.
$$
Find the range of values for $a$.
(1986, National High School Competition) | Proof: From (1) we get $b c=a^{2}-8 a+7$.
(2) - (1) gives $(b+c)^{2}=(a-1)^{2}$.
Thus $b+c= \pm(a-1)$.
By the inverse of Vieta's formulas, from (3) and (4) we get
$b, c$ are the roots of the equation $x^{2} \mp(a-1) x+\left(a^{2}-8 a+7\right)$ $=0$.
$$
\begin{array}{l}
\because b, c \in R, \\
\therefore \Delta=[\mp(a-... | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,952 |
Example 6 Prove: $\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right)>\frac{\sqrt{2 n+1}}{2}$. Here, $n \in N$, and $n \geqslant 2$. | Prove: Construct the sequence $\left\{T_{n}\right\}: 1+\frac{1}{3},\left(1+\frac{1}{3}\right)$
- $\left(1+\frac{1}{5}\right), \cdots,\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots(1+$ $\left.\frac{1}{2 n+1}\right), \cdots$, then
$$
\begin{array}{c}
\frac{T_{n}}{T_{n-1}}=1+\frac{1}{2 n+1}=\sqrt{\frac{(2 n+2... | \left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right)>\frac{\sqrt{2 n+1}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,953 |
Example 7 Proof: For any numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, whose sum is 1, the inequality $\frac{a_{1}{ }^{2}}{a_{1}+a_{2}}+\frac{a_{2}{ }^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}{ }^{2}}{a_{n-1}+a_{n}}$ $+\frac{a_{n}{ }^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$ holds.
(24th All-Soviet Union Middle School Mathemati... | Proof: Let the left side of the inequality be denoted as $A$, and construct the symmetric form of $A$, let
$$
\begin{array}{l}
B=\frac{a_{2}^{2}}{a_{1}+a_{2}}+\frac{a_{3}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n-1}+a_{n}} \\
+\frac{a_{1}^{2}}{a_{n}+a_{1}}, \text{ then } \\
A-B \\
\quad=\left(a_{1}-a_{2}\right)+\l... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,954 |
Example 9 For a finite set $A$, function $f: N \rightarrow A$ has only the following property: if $i, j \in N, |H| i-j |$ is a prime number, then $f(i) \neq f(j)$. How many elements does set $A$ have at least? | Solution: Let $|A|$ denote the number of elements in the finite set $A$, and estimate the lower bound of $|A|$.
Since the absolute value of the difference between any two numbers among $1, 3, 6, 8$ is a prime number, by the problem's condition: $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefor... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,956 |
As is known, when $P$ is a point inside $\triangle ABC$ and satisfies $\angle PAB = \angle PBC = \angle PCA = \theta$, point $P$ is called the Bubka point of $\triangle ABC$, and $\theta$ is called the Bubka angle of $\triangle ABC$. It is also known that $\csc^2 \theta = \csc^2 A + \csc^2 B + \csc^2 C$.
We have disco... | Proof: (1) As shown in the figure, let $A, B, C, D$ represent $\angle DAB, \angle ABC, \angle BCD, \angle CDA$ respectively, and let $\alpha, \beta, \gamma, \omega$ represent $\angle BPC, \angle CPD, \angle DPA, \angle APB$ respectively.
$$
\begin{array}{l}
\because \quad \angle DAP + \theta = 180^{\circ} - \gamma, \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,957 |
A Simple Proof of Weitzenböck's Inequality
Zhang Yanwei
(Department of Education, Shuyang County, Jiangsu Province, 223600)
Let the area of $\triangle ABC$ be $\triangle$, the lengths of the three sides be $a$, $b$, and $c$, the lengths of the medians to these sides be $m_{a}$, $m_{b}$, and $m_{c}$, and the altitudes ... | $$
\begin{array}{l}
m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right) \\
\text { Also, } \dot{h}_{a}=\frac{2 \Delta}{a}, h_{b}=\frac{2 \Delta}{b}, h_{c}=\frac{2 \Delta}{c}, \\
\therefore h_{a}^{2}+h_{b}^{2}+h_{c}^{2}=4 \Delta^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) . \\
\be... | a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} \Delta | Inequalities | proof | Yes | Yes | cn_contest | false | 708,958 |
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