problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Five, try to find two different natural numbers, whose arithmetic mean $A$ and geometric mean $G$ are both two-digit numbers. Among $A, G$, one can be obtained by swapping the units and tens digits of the other. | Five, Solution Let these two natural numbers be $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=2 A, \\
x_{1} x_{2}=G^{2} .
\end{array}\right.
$$
Thus, $x_{1}, x_{2}$ are the two roots of the equation $x^{2}-2 A x+G^{2}=0$. Therefore, $A \pm \sqrt{A^{2}-G^{2}}$ should be natural numbers, which means $\sqrt... | 98 \text{ and } 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,431 |
In Rt $\triangle A B C$,
$D 、 E 、 F$ are the midpoints of the three
sides, $H$ is the foot of the perpendicular from the right angle
to the hypotenuse $A B$, $G$ is the midpoint of $D H$, and let
$O$ be any point on $A B$. Prove that: $\angle E O F$ is maximized when it is $\angle E G F$. | Six (Figure omitted)
Prove that connecting $E F$, then $E F \parallel A B$. From the conditions, we can prove:
$$
\begin{array}{l}
D E=H F=\frac{1}{2} A B, \\
G E=G F .
\end{array}
$$
We want to prove $\angle E O F = \angle O F G$.
Extend $E G$ to $F^{\prime}$, such that $G F^{\prime}=G F$, and connect $O F^{\prime}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,432 |
2. The number of isosceles triangles with integer side lengths and a perimeter of 21 is ().
(A) 4
(B) 5
(C) 6
(D) 7 | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,434 |
3. The condition for the equation $\sqrt{\frac{x}{x-1}}=-\frac{\sqrt{x}}{\sqrt{x}-1}$ to hold is (.
(A) $\frac{x}{x-1} \geqslant 0$
(B) $x \geqslant 0$
(C) $x \geqslant 0$ and $x \neq 1$
(D) $x>1$ | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,435 |
4. The number of $x$ that satisfies $|x-2|+|x+1|=3$ is ().
(A) 0
(B) 1
(C) 2
(D) More than 2 | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,436 |
6. In $\triangle A B C$, $\angle A C B=90^{\circ}$, $C D$ and $C E$ are the altitude and the angle bisector of $\angle C$ on the base $A B$, respectively. If $\triangle C E D \backsim \triangle A B C$, then $\angle E C D$ equals ( ).
(A) $18^{\circ}$
(B) $20^{\circ}$
(C) $22.5^{\circ}$
(D) $30^{\circ}$ | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,438 |
3. In the acute triangle $\triangle A B C$, it is known that $\angle A=60^{\circ}$, $B D$ and $C E$ are the altitudes from $B$ and $C$ to the sides $A C$ and $A B$ of $\triangle A B C$, respectively. If $S_{\triangle A B C}=1$, then the area of quadrilateral $B C D E$ is $\qquad$ | 3. $\frac{3}{4}$ | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,443 |
In $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the angle bisector of $\angle C$, and $C A=3$, $C B=4$. Find the length of $C D$. | Three, Proof As shown in the figure, it is obvious that $AB=5$.
$\because CD$ bisects
$\angle C$,
$$
\therefore \frac{AD}{BD}=\frac{3}{4} \text{. }
$$
Then $AD=\frac{15}{7}$,
$$
BD=\frac{20}{7} \text{. }
$$
Draw $DE \perp BC$ at $E$, thus $DE \parallel AC$,
$$
\therefore \frac{BE}{CE}=\frac{BD}{AD}=\frac{4}{3} \text{... | \frac{12}{7} \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,445 |
Let $\alpha, \beta$ be the two real roots of the integer-coefficient equation $x^{2}+a x+b=0$, and $\alpha^{2}+\beta^{2}<4$. Try to find all possible integer pairs ( $a$, b).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{aligned}
\alpha^{2}+\beta^{2}: & =(\alpha+\beta)^{2}-2 \alpha \beta \\
& =a^{2}-2 b-2.
\end{aligned}
$$
Since the equation has two real roots, we have
$$
\Delta=a^{2}-4 b \geqslant 0.
$$
Thus, $2 b \leqslant a^{2}-2 b-2, b<2$ gives
$$
\begin{array}{c}
b=-1, b=0 \text { or } b=1. \\
\text { (1) } b=0 \text {,... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,446 |
Five, let $x^{2}+y^{2}=a^{2}, a^{2}<1$. Try to find the maximum and minimum values of $S=\sqrt{1-x^{2}}+\sqrt{1-y^{2}}$. | $$
\begin{aligned}
& \text { V. Solution } S=\sqrt{1-x^{2}}+\sqrt{1-y^{2}} \\
= & \sqrt{\left(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right)^{2}} \\
= & \sqrt{2-\left(x^{2}+y^{2}\right)+2 \sqrt{1-\left(x^{2}+y^{2}\right)+x^{2} y^{2}}} \\
= & \sqrt{2-a^{2}+2 \sqrt{1-a^{2}}+x^{2}\left(a^{2}-x^{2}\right)} \\
= & \sqrt{2-a^{2}+2 \sq... | S_{\text {max }}=1+\sqrt{1-a^{2}}, \quad S_{\text {min }}=\sqrt{4-2 a^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,447 |
2. Let $a, b$ be real numbers, and $a \neq 0$. For the quadratic function $y=a x^{2}+b x+c$,
(A) it has a maximum value of $\frac{4 a c-b^{2}}{4 a}$
(B) it does not necessarily have a maximum value
(C) its graph intersects the $x$-axis at two points.
(D) its graph is symmetric about the $y$-axis | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,449 |
3. If $a, b, \sqrt{a}+\sqrt{b}$ are all rational numbers, then the values of $\sqrt{a}, \sqrt{b}$ are ( ).
(A) both are rational numbers
(B) both are irrational numbers
(C) one is rational and the other is irrational
(D) any of the above three cases is possible | 3. A | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,451 |
4. Given $x-\frac{1}{x}=4$, then the value of $x+\frac{1}{x}$ is (
(A) $2 \sqrt{5}$
(B) $\pm 2 \sqrt{5}$
(C) $3 \sqrt{2}$
(D) $\pm 3 \sqrt{2}$. | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,452 |
5. Given that $a, b, c$ are all positive numbers, the equation $a x^{2}+b x$ $+c=0$ has real roots, then the equation $a c x^{2}+b^{2} x+a c=0$ ( . ).
(A) has two distinct positive roots
(B) has one positive root and one negative root
(C) may not have real roots
(D) has two distinct negative roots | 5. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,453 |
6. On the sides $AB, BC, CD$, and $DA$ of rectangle $ABCD$, points $E, F, G, H$ are taken respectively, such that $EFGH$ is a rectangle, then such a rectangle ( ).
(A) can only be made one
(B) can be made 4
(C) generally cannot be made
(D) can be made infinitely many | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,454 |
7. In $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$, it is known that $\frac{A B}{A^{\prime} B^{\prime}}=\frac{A C}{A^{\prime} C^{\prime}}$, and $\angle B=\angle B^{\prime}$, then the correct conclusion is ( ).
(A) These two triangles must be similar
(B) These two triangles must not be similar
(C) ... | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,455 |
1. Given the radius of a circle is $r$, the side length of its inscribed regular pentagon is $a_{5}=r \sqrt{\frac{5-\sqrt{5}}{2}}$, then the side length of the inscribed regular decagon of the circle $a_{10}=$ $\qquad$ | $=1 \cdot \frac{1}{2}(\sqrt{5}-1) r$ | \frac{1}{2}(\sqrt{5}-1) r | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,457 |
2. The set of real solutions that satisfy the equation $\sqrt{x+3-2 \sqrt{x+2}}+$ $\sqrt{x+6-4 \sqrt{x+2}}=1$ is | 2. $-1 \leqslant x \leqslant 2$ | -1 \leqslant x \leqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,458 |
3. In trapezoid $A B C D$, $A B / / D C, \angle A=$ $90^{\circ}, A B=4, C D=3, B C=7, O$ is the midpoint of side $A D$, then the distance from $O$ to $B C$ is $\qquad$ . | 3. $2 \sqrt{3}$
3. Hint: Let $M$ be the midpoint of $B C$: then $O M=\frac{7}{2}$. Connect $O B, O C$. | 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,459 |
Example 4. Draw four lines $l_{1}, l_{2}, l_{3}, l_{4}$ through point $O$, and draw $A B / / l_{4}$ through any point $A$ on $l_{1}$, intersecting $l_{2}$ at $B$; draw $B C / / l_{1}$, intersecting $l_{3}$ at $C$; draw $C D / / l_{2}$, intersecting $l_{4}$; draw $D E / / l_{3}$, intersecting $l_{1}$ at $E$. Prove: $O E... | Connect $A D, B D$. As shown in the figure, from $A B / / l_{4}, C D$ $/ / l_{2}$ we know
$$
S_{\triangle A O D}=S_{\triangle B O D}=S_{\triangle B C C} .
$$
Extend $C D$ to intersect $\iota_{1}$ at $F$. Since $B C F O$ is a parallelogram,
$$
\therefore S_{\triangle A X C}=S_{\triangle C O F} .
$$
By a special case o... | O E \leqslant \frac{1}{4} A O | Geometry | proof | Yes | Yes | cn_contest | false | 706,461 |
One, to determine whether a positive integer can be divided by 7, the "tail-cutting method" can be used. For example, for 2527, cutting off the last digit 7 yields 252, then subtracting twice the cut-off last digit 7 from 252 gives 238. This is called one "tail-cutting". Performing another "tail-cutting" on 238 yields ... | Proof one: Let the positive integer be
$$
A=\overline{a_{n-1} a_{n-2} \cdots a_{1} a_{0}},
$$
i.e., $A=10^{n-1} a_{n-1}+10^{n-2} a_{n-2}+\cdots+10 a_{1}+a_{0}$. After one "cutting the tail", $A$ becomes
$$
\begin{aligned}
A^{\prime}= & \left(10^{n-2} a_{n-1}+10^{n-3} a_{n-2}+\cdots\right. \\
& \left.+10 a_{2}+a_{1}\ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,462 |
II. Given that a circle is divided into two arcs by two fixed points $A, B(A \neq$ $B$) on it. Try to determine the position of a moving point $P$ on the specified arc where $P A+P B$ reaches its maximum value? Prove your conclusion and find this maximum value. | When point $P$ is at the midpoint of a specified arc,
i.e., $P A = P B$, $P A + P B$ achieves its maximum value.
Since $A, B$ are two distinct fixed points and $P$ is on the specified arc $\overparen{A B}$, $\angle A P B$ is a constant (let it be $\alpha$).
$$
\text{Also, } 0 < \alpha < \pi \text{, }
$$
which means $... | \sqrt{\frac{2}{1 - \cos \alpha}} A B | Geometry | proof | Yes | Yes | cn_contest | false | 706,463 |
Three, in the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, $A_{1} A_{5} \parallel A_{3} A_{4}$. Let $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}$ be the midpoints of $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{5}, A_{5} A_{1}$, respectively. Connecting $M_{1} M_{2}, M_{2} M_{3}, M_{3} M_{4}, M_{4} M_{5}, M_{5} M_{1}$ in... | Three, Proof According to the problem statement.
Since $A_{1} A_{5} / / A_{3} A_{4}$,
Therefore, $S_{\triangle A_{1} A_{3} A_{4}}=S_{\triangle A_{3} A_{4} A_{5}}$.
Also, given that points $M_{1}, M_{2}, \cdots, M_{5}$ are the midpoints of $A_{1} A_{2}$, $A_{2} A_{3}, \cdots, A_{5} A_{1}$ respectively.
We know
$$
\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,464 |
1. Calculate $\frac{1}{1+\sqrt[4]{10}}+\frac{1}{1-\sqrt[4]{10}}+$ $\frac{2}{1+\sqrt{10}}$ The result obtained is $\qquad$ | - I. $-\frac{9}{4}$ | -\frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,465 |
3. If the equation $x^{2}+p x+q=0$ and the equation $x^{2}-\frac{1}{q} x+\frac{1}{q}=0$ both have equal real roots, and their roots are reciprocals of each other, then the value of $1992 p^{2}-12 q-6$ is . $\qquad$ | 3. 1983 | 1983 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,467 |
5. Let positive numbers $p, q, r$ satisfy $\frac{(p+q+r)^{2}}{p^{2}}=$ $\frac{2(p+q+r)^{2}}{q^{2}}=\frac{3(p+q+r)^{2}}{r^{2}}=k$, then the value of $k$ is $\qquad$. | 5. $6+2(\sqrt{2}+\sqrt{3}+\sqrt{6})$ | 6+2(\sqrt{2}+\sqrt{3}+\sqrt{6}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,469 |
Example 5. As shown in the figure, in $\triangle A B C$, $D, E$ are points on $B C, C A$ respectively, $B E, A D$ intersect at $H$, and $\frac{B D}{D C}=\frac{1}{3}, \frac{C E}{E A}=\frac{1}{2}, S_{\triangle A B C}=1$. Find the area of $\triangle A H E$. | In Theorem 4, let
$$
m=\frac{B D}{D C}=\frac{1}{3}, n=\frac{C E}{E A}, p=0
$$
(considering $A, F$ coincide), then
$$
\begin{aligned}
\frac{S_{\triangle A H F}}{S_{\triangle A B C}} & =\frac{1}{\left(1+\frac{1}{3}+\frac{1}{6}\right)\left(1+\frac{1}{2}\right)} \\
& =\frac{4}{9} . \\
\therefore S_{\triangle A H F} & =\fra... | \frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,472 |
2. A clothing vendor sells two sets of clothes at the same time, each set for 168 yuan. Based on the cost, one set yields a profit of $20 \%$, while the other incurs a loss of $20 \%$. Then, in this transaction, the vendor ( ).
(A) breaks even
(B) loses 37.2 yuan
(C) gains 14 yuan
(D) loses 14 yuan | 2. I))
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,474 |
3. In an office, six people $A, B, C, D, E, F$ take turns on night duty, with the order being $A \rightarrow B \rightarrow C \rightarrow D$ $\rightarrow E \rightarrow F \rightarrow A \rightarrow B \cdots$. In the first week, Sunday is covered by $A$, Monday by $B$, $\cdots$, Friday by $F$, and Saturday again by $A$, an... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,475 |
4. $P A$ is tangent to $\odot O$ at point $A, A B$ is a chord of $\odot O$. If the radius of $\odot O$ is $1, P A=1, A B=\sqrt{2}$, then the length of $P B$ is ( ).
(A) 1
(B) $\sqrt{5}$
(C) 1 or $\sqrt{5}$
(D) cannot be determined | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,476 |
5. In $\triangle A B C$, $\angle A>\angle B>\angle C, \angle A \neq 90^{\circ}$, draw a line to divide $\triangle A B C$ into two parts, and make one of the parts similar to $\triangle A B C$. The number of such non-parallel lines is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 5. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,477 |
6. There are 10 numbers: $1983^{3}+3 \times 1983^{2}+2 \times$ $1983,1984^{3}+3 \times 1984^{2}+2 \times 1984, \cdots, 1991^{3}$ $+3 \times 1991^{2}+2 \times 1991,1992^{3}+3 \times 1992^{2}+2$ $\times 1992$, among the following integers, the largest integer that can divide each of the 10 numbers is ( ).
(A) 2
(B) 3
(C)... | $6 . C$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,478 |
Three, given the quadratic function $y=a x^{2}+b x+c$ whose graph has a vertex $C$, and it intersects the $x$-axis at two distinct points $A$ and $B$.
(1) If the x-coordinate of point $C$ is 3, and the distance between points $A$ and $B$ is 8, find the roots of the equation $a x^{2}-(6 a-b) x+9 a-3 b+c=0$;
(2) If the d... | (1) Let point $B$ be to the left of point $A$. Then it is easy to know that the coordinates of point $A$ are (-1, 0), and the coordinates of point $B$ are (7, 0).
We can set the quadratic function as $y=a(x+1)(x-7)$.
Therefore, the coordinates of the vertex $C$ are $(3, -16a)$.
$$
\therefore-\frac{b}{2 a}=3, \fra... | x_{1}=2, x_{2}=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,480 |
Four, the perimeter of $\triangle A B C$ is $1992 \mathrm{~cm}$, a squirrel is located at point $P$ on side $A B$ (excluding points $A, B$). The squirrel first runs from point $P$ in a direction parallel to $B C$ to point $P_{1}$ on side $A C$, then immediately changes direction and runs in a direction parallel to $A B... | (1) When point $P$ is the midpoint of $AB$, the little squirrel can return to point $P$ by running according to the requirements of the problem.
When it returns to point $P$ for the first time, the distance it has run is obviously the perimeter of the medial triangle of $\triangle ABC$, which is $\frac{1992}{2}=996(\m... | 996 \text{ cm or } 1992 \text{ cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,481 |
Example 1. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ (Math Competition) | By analyzing, we compare $E_{1}$ with $x^{2}-x-1=0$ and find that $m, n$ are the two distinct real roots of the equation. By Vieta's formulas, we get
$$
\begin{aligned}
m+n= & 1, m n=-1 \\
m^{2}+n^{2} & =(m+n)^{2}-2 m n=3 \\
m^{3}+n^{3} & =(m+n)\left(m^{2}+n^{2}\right)-m n(m+n) \\
& =4 .
\end{aligned}
$$
By analogy wi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,483 |
Example 2. Solve the system of equations
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
y z+z t+t y=1, \\
z t+t x+x z=1, \\
t x+x y+y t=1 .
\end{array}\right.
$$
(3rd Junior High School Mathematics Competition, Jiangsu Province) | Analyzing the original equation, which has many elements and an unclear reduction strategy, we can first explore the system of equations by reducing the number of variables and making an analogy:
$$
\left\{\begin{array}{l}
x y + y z = 1, \\
x y + x z = 1, \\
x z + y z = 1
\end{array}\right.
$$
By subtracting the equat... | x = y = z = t = \pm \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,484 |
Example 2. The figure below is half of a Chinese chessboard. Prove:
(1) A knight cannot start from position $B$, jump to every point on the half chessboard exactly once;
(2) A rook starts from position $A$, moves one step at a time on this half chessboard, and after several steps, reaches position $B$. Then at least on... | (2) The chariot starts from -1, repeatedly passes through $+1, -1$, $\cdots$, and then moves to +1, so the sum of the numbers it passes through is 0. As mentioned earlier, the sum of all numbers on this half of the chessboard is -1, which means that the chariot has not traversed the entire half of the chessboard, or it... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,486 |
Example 3. Super chess is played on a $12 \times 12$ square board. Its knight moves from one corner of a $3 \times 4$ rectangle to the opposite corner. Is it possible for a knight to jump to each square of the board exactly once and then return to the starting square? (26th IMO Preliminary Question) | This kind of move is impossible.
Assume there exists a route that meets the requirements, and we label the chessboard alternately with +1 and -1. Clearly, a knight will move from a square labeled +1 to a square labeled -1 in one step, and from -1 to +1. Without loss of generality, let the knight start from a +1 square.... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,487 |
Example 4. A rectangular exhibition hall is divided into several rows and columns of rectangular exhibition rooms by walls intersecting vertically and horizontally. There are several square or round doors between every two adjacent rooms, and there are several doors connecting the hall to the outside only at the entran... | Prove that the exhibition room at the entrance and exit is marked as +1, all rooms adjacent to the +1 room are marked as -1, and all rooms adjacent to the -1 room are marked as +1. Thus, each room has one and only one corresponding +1 or -1, and the +1 and -1 of any two adjacent rooms are different.
A visitor enters t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,488 |
Example 5. There are $n$ boxes of matches arranged in a circle, and the following adjustment is made: if the sum of the number of matches in any consecutive 4 boxes is odd, then one match is taken from each of these boxes; otherwise, one match is added to each of these boxes. Moreover, after each set of 4 consecutive b... | Let $n$ boxes of matches be denoted as $a_{1}, a_{2}, \cdots, a_{n}$. If the number of matches in each $a_{k}$ is even, assign it a value of +1; otherwise, assign it a value of -1. Let the values assigned to $a_{1}, a_{2}, \cdots, a_{n}$ be $x_{1}, x_{2}, \cdots, x_{n}$. If the sum of the number of matches in $a_{i}, a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,489 |
Example 7. Suppose there are 11 teacups, all initially facing upwards. By randomly flipping them, a certain number of cups (flipped cups can be flipped again). Prove: no matter how many times they are flipped, it is impossible for all cup mouths to face downwards. | Define the cup mouth facing up as +1, and the cup mouth facing down as -1, and let $a_{k}$ be the product of these numbers after $k$ flips. Initial state:
$$
\underbrace{1,+1, \cdots,+1}_{n \uparrow+1}, a_{0}=(+1)^{11}=1 .
$$
After the first flip of $n_{1}$ cups (where $n_{1}$ is even),
it is easy to prove: generally... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 706,490 |
Example 8. At the endpoints of line segment $AB$, one is marked red and the other is marked blue. Insert $n$ points in the middle of the line segment, and randomly mark them with red and blue. This divides the original line segment into $n+1$ non-overlapping smaller segments, those whose endpoints have different colors... | Proof: Let the $n$ points inserted in the middle of line segment $AB$ be $A_{1}, A_{2}, \cdots A_{n}$. Suppose there are $k$ standard segments, and the points $A_{i}$ are associated with numbers $a_{i}$ as follows:
$$
\quad a_{i}=\left\{\begin{array}{l}
1, \text { if point } A_{i} \text { is red } \\
-1, \text { if poi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,491 |
As requested, here is the translation while preserving the original text's line breaks and format:
1. As shown in the figure, let the incircle of $\triangle A B C$ touch side $B C$ at $D$,
$N$ is the
midpoint of segment $A D$. Then
the line
$O N$ passes through
the midpoint of
$B C$ | Proof:
Draw the diameter $DE$ of $\odot O$, and connect $AE$. Draw $AH \perp BC$ at $H$, and draw $EI \perp AH$ at $I$. Let the line $ON$ intersect $BC$ at $M$. It is easy to prove that $\triangle OMD \sim \triangle AEI$, and thus $\frac{MD}{DH} = \frac{OD}{AI}$. Let $BC = a$, $CA = b$, $AB = c$, $p = \frac{1}{2}(a + b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,492 |
1. The number of ways to divide a square into four congruent shapes by drawing two straight lines is ( ).
(A) 1
(B) 2
(C) More than 2 but finite
(D) Infinite | 1. (D).
Draw any two perpendicular lines through the center of a square, which can divide the square into four figures. Each of these can be obtained by rotating the adjacent one by $90^{\circ}$, thus they are congruent. However, there are infinitely many ways to draw two perpendicular lines through a point. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,493 |
Example 3. Five circles are tangent to each other in sequence, and also tangent to lines $a$, $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, find the diameter of the circle in the middle. (34th American High School Mathematics Competition) | Analyzing the original problem involving the relationship between five circles, it is not easy to grasp. First, consider the case of three circles:
Let three circles $O_{1}, O_{2}, O_{3}$ be tangent to each other in sequence. If their radii are $r_{1}, r_{2}, r_{3}$ respectively, then by the similarity of triangles (a... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,495 |
3. Let the center of rhombus $A B C D$ be $O, E, F$ are two points inside the rhombus that are symmetric about $O$. Connect $C E, D F$, then the relationship between line segments $A B, C E, D F$ is ( ).
(A) The sum of the two shorter line segments is greater than the longest line segment
(B) The sum of the two shorter... | 3. (A).
Take the symmetric point $G$ of $E$ with respect to $B D$, then $G$ is symmetric to $F$ with respect to $A C$, so $A G = C E, B G = D F$. In $\triangle G A B$, the sum of the two shorter sides must be greater than the longest side. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,496 |
4. A water pool has 5 inlet and outlet water pipes numbered (1), (2), ..., (5). It is known that the time it takes to fill the pool when certain pipes are opened is as shown in the table below. If all 5 pipes are opened simultaneously, ( ) hours will be needed to fill the pool.
(A.) $\frac{1}{5}$
(B) $\frac{1}{2}$
(C.)... | 4. (D).
Consider the sum of the fractions of the pool that can be filled in one hour by doubling the rate of all the pipes:
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=1 .
$$
This shows that the pool is exactly filled because the combined rate of the 5 pipes doubled for one hour | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 706,497 |
5. In $\triangle A B C$, $A D, B E$ are the altitudes on sides $B C, A C$, with $D, E$ being the feet of the perpendiculars. If $C E+C D=A B$, then $\angle C$ is ( ).
(A) acute angle
(B) right angle
(C) obtuse angle
(D) acute angle, right angle, obtuse angle are all possible | 5. (A).
If $\angle C=90^{\circ}$, then points $C, D, E$ coincide, which contradicts $C E+C D = A B$.
If $\angle C$ is an obtuse angle, then $D$ is on the extension of $B C$, and $E$ is on the extension of $A C$. By the property that in a right triangle the hypotenuse is the longest side, we have
$$
\begin{array}{l}
\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,498 |
6. In an oblique $\triangle A B C$, $a, b, c$ are the sides opposite to angles $A$, $B, C$ respectively, then the points $M(a, \sin A), N(b, \sin B)$, $P(c, \sin C)$ have the relationship $(\quad)$.
(A) The three points are collinear on $l$, and $l$ passes through the origin
(B) The three points are collinear on $l$, a... | 6. (A)
By the Law of Sines, we have $\left\{\begin{array}{l}c=2 R \sin A, \\ b=2 R \sin B, \\ c=2 R \sin C .\end{array}\right.$ This indicates that points $M, N, P$ all lie on the line $l: y=2 R x$, and $l$ passes through the origin. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,499 |
The legs of a right triangle are in the ratio $19: 3$, then the length of one leg is ( ).
(A) $\frac{1}{2}\left(1993^{2}+1\right)$
(B) $\frac{1}{2}\left(1993^{2}-1\right)$
(C) has more than one but a finite number of values
(D) has an infinite number of values | 7. (B).
Let the other leg be $a$, and the hypotenuse be $c$, then we have
$$
1993^{2}=c^{2}-a^{2}=(c+a)(c-a) .
$$
Since 1993 is a prime number, we have
$$
\left\{\begin{array}{l}
c+a=1993^{2}, \\
c-a=1 .
\end{array}\right.
$$
Thus, $a=\frac{1}{2}\left(1993^{2}-1\right)$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,500 |
8. Given that $a$ is a natural number, the number of integer solutions to the inequality
$$
a^{2} x^{4}-\left(a^{4}+1\right) x^{2}+a^{2}<0
$$
is ( ) .
(A) $2 a-2$
(B) $2 a-1$
(C) $2 a$
(D) $2 a+1$ | 8. (A).
The inequality is $a^{2}\left(x^{2}-a^{2}\right)\left(x^{2}-\frac{1}{a^{2}}\right)<0$,
which gives $\frac{1}{a^{2}}<x^{2}<a^{2}$.
Thus, $-a<x<-\frac{1}{a}$ or $\frac{1}{a}<x<a$.
The integer solutions are $-a+1,-a+2, \cdots,-1,1, \cdots, a-2, a-1$, a total of $2 a-2$ solutions. | 2a-2 | Inequalities | MCQ | Yes | Yes | cn_contest | false | 706,501 |
1. Let $x=\frac{1-\sqrt{5}}{2}$, then $\sqrt{\frac{1}{x}-\frac{1}{x^{3}}}-$ $\sqrt[3]{x^{4}-x^{2}}=$ $\qquad$ | $\begin{array}{l}\text { 1. } \sqrt{5} \text {. } \\ \text { From } x-\frac{1}{x}=\frac{1-\sqrt{5}}{2}-\left(-\frac{1+\sqrt{5}}{2}\right)=1 \text {, } \\ x+\frac{1}{x}=-\sqrt{5} \text {, we know that, } \\ \text { the original expression }=\sqrt{\frac{1}{x^{2}}\left(x-\frac{1}{x}\right)}-\sqrt[3]{x^{3}\left(x-\frac{1}{... | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,502 |
2. For the quadratic function $y=a x^{2}+b x(a b \neq 0)$, when $x$ takes $x_{1}, x_{2}\left(x_{1} \neq x_{2}\right)$, the function values are equal. Then, when $x$ takes $x_{1}+x_{2}$, the function value is | 2. Let $x$ take $x_{1}, x_{2}$, the function values are $-c$, then we have
$$
\left\{\begin{array}{l}
a x_{1}^{2}+b x_{1}=-c, \\
a x_{2}^{2}+b x_{2}=-c .
\end{array}\right.
$$
That is, $x_{1}, x_{2}$ are the two distinct roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$
and
$$
x_{1}+x_{2}=-\frac{b}{a} \text {. }
... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,503 |
3. In a right triangle $\triangle A O B$, the right-angle vertex $O$ is on the unit circle, and the hypotenuse is tangent to the unit circle. Extend $A O, B O$ to intersect the unit circle at $C, D$ respectively. Then the minimum value of the area of quadrilateral $A B C D$ is $\qquad$ | 3. $\frac{1}{2}(3+2 \sqrt{2})$.
Let the point of tangency be $E$,
then $O E=1$, thus
$A B$
$$
\begin{array}{l}
=O E \cdot A B \\
=A O \cdot B O \\
=\frac{A O^{2}+B O^{2}}{2} \\
-\frac{(A O-B O)^{2}}{2} \\
\leqslant \frac{A O^{2}+B O^{2}}{2} \\
=\frac{A B^{2}}{2}
\end{array}
$$
Thus, $A B \geqslant 2$.
When $A O=B O$,... | \frac{1}{2}(3+2 \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,504 |
4. Place the natural numbers $1,2,3,4, \cdots, 2 n$ in any order on a circle. It is found that there are $a$ groups of three consecutive numbers that are all odd, $b$ groups where exactly two are odd, $c$ groups where exactly one is odd, and $d$ groups where none are odd. Then $\frac{b-c}{a-d}=$ $\qquad$ . | 4. -3 .
Let the numbers on a circle be recorded in reverse order as $x_{1}, x_{2}$, $\cdots, x_{2 n}$, and satisfy
$$
x_{i}=\left\{\begin{array}{l}
-1, \text { when } x_{i} \text { represents an odd number, } \\
+1, \text { when } x_{i} \text { represents an even number. }
\end{array}\right.
$$
Then $x_{1}+x_{2}+\cdo... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,505 |
One, (20 points) If the two quadratic equations
$$
\begin{array}{l}
a^{2} x^{2}+a x-1=0, \\
x^{2}-a x-a^{2}=0
\end{array}
$$
have a common solution, find all possible values of $a$. | Let $x_{0}$ be the common solution of the two equations, then we have
$$
\begin{array}{l}
a^{2} x_{0}^{2} + a x_{0} - 1 = 0, \\
x_{0}^{2} - a x_{0} - a^{2} = 0 .
\end{array}
$$
Subtracting the two equations gives $\left(a^{2}-1\right)\left(x_{0}^{2}-1\right)=0$.
If $a^{2}-1=0$, then the original equations become
$$
x^... | \frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,507 |
c Lu
$$
\begin{array}{l}
30-6(a+b+c)+(a b+b c+c a)=0, \\
81-9(a+b+c)+(a b+b c+c a)=0 .
\end{array}
$$
Find $a, b, c$. | Three, it is known that 6 and 9 are the roots of the quadratic equation
$$
x^{2}-(a+b+c) x+(a b+b c+c a)=0
$$
Thus, we have
$$
\left\{\begin{array}{l}
a+b+c=6+9=15, \\
a b+b c+c a=6 \times 9=54 .
\end{array}\right.
$$
Since $a>b>c$, it follows that
$$
15=a+b+c>3 c,
$$
Solving this, we get $c<5$. If $c \geq 5$, then ... | a=10, b=4, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,509 |
1. The graph of the function $y=\arccos x(-1 \leqslant x \leqslant 1)$ is symmetric about the $y$-axis and is denoted as $c_{1}$, and the graph of $c_{1}$ symmetric about the line $y=x$ is denoted as $c_{2}$. Then the equation of $c_{2}$ is ( . ).
- (A) $y=\cos x(0 \leqslant x \leqslant \pi)$
(B) $y=\arcsin x(-1 \leqsl... | i. (C).
From the problem, we know that the equation of $c_{1}$ is $y=\arccos (-x)(-1 \leqslant x \leqslant 1)$, and $c_{2}$ is the inverse function of $c_{1}$ (since $c_{1}$ and $c_{2}$ are symmetric with respect to the line $y=x$), so the equation of $c_{2}$ is $y=-\cos x(0 \leqslant x \leqslant \pi)$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,510 |
2. The number of positive integers $k$ for which the equation $a^{2}+b^{2}=k \cdot a b$ has positive integer solutions $(a, b)$ is ( ).
(A) 0
(B) 1
(C) more than 1, but only finitely many
(D) infinitely many | 2. (B).
Let the greatest common divisor of $a$ and $b$ be $d$, $a=d \cdot a_{1}$, $b=d \cdot b_{1}$ ($a_{1}$ and $b_{1}$ are coprime), substituting into $a^{2}+b^{2}=k \cdot a b$ yields $a_{1}^{2}+b_{1}^{2}=k \cdot a_{1} b_{1}$. Therefore, $a_{1} \mid b_{1}$ and $b_{1} \mid a_{1}$, hence $a_{1}=b_{1}$, at this point $... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,511 |
3. As shown in the figure, in the vertical coordinate plane $x o y$, line $l$ passes through the origin $O$, and $l$ is in the first and the I quadrant, with the angle between $l$ and the $x$-axis being $\alpha\left(0^{\circ}<\alpha<90^{\circ}\right)$. There is a point mass (neglecting the size of the point mass) at po... | 3. (C).
For a circle, let $O A=a$, and the acceleration due to gravity be $g$. Clearly, $\angle A O B=90^{\circ}-\alpha$. By the sine rule, we have
$$
\begin{array}{l}
\frac{A B}{\sin \left(90^{\circ}-a\right)}= \\
\frac{a}{\sin \left(180^{\circ}-90^{\circ}+a-\theta\right)},
\end{array}
$$
which means $A B=\frac{a \c... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,512 |
4. Let $a_{n}=4^{2 n+1}$
$$
+3^{n+2}(n=1,2,3 \text {. }
$$
$\cdots)$, $p$ is the smallest prime that can divide infinitely many terms of
$a_{1}, a_{2}, a_{3}, \cdots$, $q$ is the smallest prime that can divide every term of $a_{1}, a_{2}, a_{3}, \cdots$, then $p$ - $q$ is ( ).
(A) $5 \times 7$
(B) $7 \times 13$
(C) $13... | 4. (D).
Obviously, $p$ is not equal to 2 and 3. Let $n+2=4k (k \in \mathbb{N})$, then
$$
\begin{aligned}
a_{n} & =4 \times 16^{n}+3^{n+2}=4+3^{4 k} \\
& =4+9^{2 k} \equiv 4+(-1)^{2 k} \equiv 0(\bmod 5) .
\end{aligned}
$$
Therefore, $p=5$. This eliminates (B) and (C). Since $a_{1}=7 \times$
13, $a_{2}=4^{5}+3^{4}=4 \t... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,513 |
6. As shown in the figure, a semicircle with radius
1 has its center at $O$ and diameter $A B$. A moving point $C$ is on the entire
circumference. Extend $A C$ to point $D$ such that $\frac{C D}{C B}=k(k>0, k$ is a constant). Then the length of the curve traced by point $D$ is ().
(A) $(1+k) \cdot \pi$
(B) $\pi \cdot ... | 6. (D).
As shown in the figure, let $\angle C A B=\theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then $\angle C O B=2 \theta, C B=2 \sin \theta, C D=2 k \sin \theta, C$ point's coordinates are $(\cos 2 \theta, \sin 2 \theta)$. Draw $C E / / A B$, then $\angle D C E = \angle C A B = \theta$. Let the co... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,515 |
1. The number of solutions to the equation $\sin |x|=|\cos x|$ in the interval $[-10 \pi, 10 \pi]$ is $\qquad$ | 1. E0.
Since $\sin |x|$ and $|\cos x|$ are both even functions, we only need to consider the number of solutions within $[0,10 \pi]$. In this case, the equation becomes $\sin x=|\cos x|$. When $0 \leqslant x \leqslant \frac{\pi}{2}$, the equation becomes $\sin x=\cos x$, which has only one solution $x=\frac{\pi}{4}$; ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,516 |
Example 5. Given two squares with side length $a$, one of the vertices of one square coincides with the center of the other square, and it rotates around the center $O$. Prove: Regardless of how it rotates, the area of the overlapping part of the two squares is a constant. | $$
\begin{array}{l}
\text{Since } ABCD \text{ is a square and } O \text{ is the center,} \\
\therefore OA = OB, \angle OAM = \angle OBN = 45^{\circ}. \\
\text{Also, } \angle AOM = \angle AOB - \angle MOB \\
= \angle MON - \angle MOB = \angle BON, \\
\therefore \triangle AOM \cong \triangle BON. \\
\therefore S_{OMBN} =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,517 |
2. $[x]$ represents the greatest integer not exceeding $x$, then
$$
\sum_{k=1}^{99}[\sqrt{k}]=
$$
$\qquad$ | 2. 615 .
Obviously, when $i^{2} \leqslant k \leqslant(i+1)^{2}-1=i^{2}+2 i(i=1,2,3$ $\cdots, 9)$, we have $[\sqrt{k}]=i$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{9}[\sqrt{k}]=\sum_{i=1}^{9} i(2 i+1) \\
=2 \cdot \sum_{i=1}^{9} i^{2}+\sum_{i=1}^{9} i \\
=2 \times \frac{9 \times 10 \times 19}{6}+\frac{9 \times 10}{2}=... | 615 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,518 |
3. Two spheres with a radius of 1 are externally tangent to each other, and both are internally tangent to a cylindrical surface with a radius of 1. Another smaller sphere is externally tangent to both of these spheres and internally tangent to the cylindrical surface. A plane passing through the center of the smaller ... | 3. $\frac{4}{5}$.
As shown in the figure, let the two spheres $O_{1}$ and $O_{2}$ with radius 1 be externally tangent at point $S$, and the smaller sphere $O_{3}$ be internally tangent to the cylindrical surface at point $T$. The radius of the smaller sphere $O_{3}$ is $r$. By the Pythagorean theorem, we have
$$
1=T S... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,519 |
5. For sets $A, B, C$ (not necessarily distinct), the union $A \cup B \cup C=\{1,2,3,4,5,6,7,8,9,10\}$, then under this condition, the number of possible triples $(A, B, C)$ is (write the answer in the form $a^{b}$). | 5. $7^{10}$.
As shown in the figure, $A \cup B \cup C$ is divided into seven non-overlapping parts $X, Y, Z, W$, $U, V, T$ (where $Y = A \cap B \cap C, T = A - B - C, X = A \cap B - C$, etc.). Thus, each of the elements $1, 2, \cdots, 9, 10$ has 7 possible allocation places in $A \cup B \cup C$, so the total number of... | 7^{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,521 |
6. A moving point $P(x, y)$ starts from the origin $O$ and moves along the parabola $y=x^{2}$ to point $A\left(\frac{3}{2}, \frac{9}{4}\right)$, then during the movement, when $P O+P A$ is maximized, the x-coordinate of point $P$ is $\qquad$ | 6. As shown in the figure
With $O$ and $A$ as two
foci, draw an ellipse that
is tangent to the $\overparen{O A}$ arc (parabolic
arc segment) at $F$. Point ( $\overparen{C .4}$ arc
segment is within the
ellipse or on its boundary
. If there are several tangent
points, choose any one of them). By the definition of an el... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 706,522 |
Example 6. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \quad (1) \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 x\right)=0
\end{array}
$$
(Where $a, b$ are positive integers) have a common root. Find the value of $\frac{... | Analyzing, since the two equations have a common root (known number), let's denote it as $x_{0}$. Clearly, $x_{0} \neq 1$ (otherwise $a=b$), so the equation in terms of the unknown $x$ can be transformed into equations in terms of $a$ and $b$:
$$
\left(1-x_{0}\right) a^{2}+\left(x_{0}^{2}+2\right) a-\left(x_{0}^{2}+2 x... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,525 |
Example 7. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the range of $x$ that makes the inequality $x^{2}+p x>4 x+p-3$ true is $\qquad$ . (The 7th Junior High School Mathematics Competition in Jiangsu Province) | Analysis: If we solve by conventionally regarding $x$ as the main variable, we need to classify and discuss. By changing the perspective to regard $p$ as the main variable, the original problem can be reduced to finding the range of $x$ for which the linear function $f(p) = (x-1)p + x^2 - 4x + 3 > 0$ holds in the inter... | x > 3 \text{ or } x < -1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,526 |
Example 9. If $x \neq 0$, then
$$
\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}
$$
The maximum value is $\qquad$. $(1992$, National Junior High School Competition) | Analyzing this type of problem generally involves rationalizing the numerator to achieve reduction. Noting that the maximum value is obtained when $x>0$, the original expression can be transformed into:
$$
\begin{array}{l}
\sqrt{\left(x-\frac{1}{x}-0\right)^{2}+(0-\sqrt{3})^{2}} \\
-\sqrt{\left(x-\frac{1}{x}-0\right)^{... | \sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,528 |
Example 10. Given that $x, y, z$ are all positive real numbers. Prove:
$$
\begin{array}{l}
\sqrt{x^{2}+x y+y^{2}}+\sqrt{x^{2}+x z+z^{2}} \\
>\sqrt{y^{2}+y z+z^{2}} .
\end{array}
$$ | Prove the construction of $\triangle ABC$ (as shown in the figure), such that
$$
\angle A O B=\angle B O C=\angle C O A=120^{\circ} \text {, }
$$
and let $A O=x, B O=y, C O=z$.
By the Law of Cosines, we have
$A B=$
$$
\begin{array}{c}
\sqrt{x^{2}+x y+y^{2}}, \\
A C=
\end{array}
$$
$$
\begin{array}{c}
\sqrt{x^{2}+x z+z... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,529 |
1. As shown in Figure 1, let the area of $\triangle A B C$ be $1, A F=\frac{1}{m} A B, B D=\frac{1}{n} B C, C E=\frac{1}{p} C A$. Then the area of $\triangle D E F$ is . (1984, Xi'an City Junior High School Competition) | From the conditions, we have
$$
\frac{A F}{F B}=\frac{1}{m-1}, \frac{B D}{D C}=\frac{1}{n-1}, \frac{C E}{E A}=\frac{1}{p-1} .
$$
Substituting $\frac{1}{m-1}, \frac{1}{n-1}, \frac{1}{p-1}$ for $m, n, l$ in the theorem, respectively, and simplifying, we get
$$
\begin{aligned}
S_{\triangle D E F}= & \frac{1}{m n p}(m n p... | \frac{1}{m n p}(m n p-m n-n p-p m+m+n+p) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,530 |
Example 2. The area of equilateral $\triangle ABC$ is $36, P, Q, R$ are points on $BC, AB, CA$ respectively, and $BP=\frac{1}{3} BC, AQ=QB, PR \perp$ $AC$, then the area of $\triangle PQR$ is ( ).
(A) 10
(B) 12
(C) 14
(D) 15
(1988, 4th National Partial Provinces, Municipal Junior High School Correspondence Competition) | Yi As shown in Figure 3, from the conditions, it is easy to know that $\frac{A Q}{Q B}=1, \frac{B P}{P C}=$ $\frac{1}{2}, \frac{C R}{R A}=\frac{1}{2}$. Given $m=1$, $n=\frac{1}{2}$, $l=\frac{1}{2}$, we can find that $S_{\triangle P Q R}=10$. Therefore, the correct choice is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,531 |
Example 1. When $n=6$, find the value of $P(6)$.
| Solve: Make the partition table for 6
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l}
\begin{tabular}{l}
$k-$ \\
partition \\
\hline
\end{tabular}
\end{tabular} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \begin{tabular}{l}
partition \\
ways
\end{tabular} & \begin{tabular}{l}
6
\end{tabular} & \begin{tabular}{l}
\be... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,532 |
Example 2. If the integer $n \geqslant 12$, then $n$ can be written as the sum of composite numbers. | When discussing integer problems, a classification method is often effective. By dividing integers into three categories, every $n$ greater than 11 can be written as $n = 3q + r$ (where $q \geq 4, r=0,1,2$). When $r=0$, $n=3q=3$ - $(q-2)+6$ is the sum of two composite numbers; similarly, when $r=1$, $n$ $=3(q-1)+4$, an... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,533 |
Example 3. If the natural number $n$ is not in the form of a multiple of 2, then it can be expressed as the sum of two or more consecutive natural numbers.
Analysis Each natural number can be expressed in the form $2^{p}(2 q+1)$, where $p$ and $q$ are non-negative integers, which means it is the sum of $2 q+1$ terms o... | Let $n=2^{p} \cdot(2 q+1)$, where the integers $p$ and $q$ satisfy $p \geqslant 0$ and $q \geqslant 1$. If $q<2^{p}$, then
$$
n=\left(2^{p}-q\right) +\left(2^{p}-q+1\right)+\cdots+\left(2^{p}-1\right)+2^{p}+\left(2^{p}+1\right)+\cdots+ \left(2^{p}+q\right);
$$
If $q \geqslant 2^{p}$, then
$$
n=\left(q-2^{p}+1\right)+... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,534 |
Example 4. The sequence $1,2,3,5,8, \cdots$, starting from the third term, each term is equal to the sum of the two preceding ones. This sequence is called the Fibonacci sequence, and each term in it is called a Fibonacci number. Prove that any natural number $n$ can be expressed as the sum of several different Fibonac... | Solve by mathematical induction on $n$. When $n=1$, the conclusion is obviously true. Because 1 itself is a Fibonacci number.
Assume the conclusion holds for $n < k$. Let $m$ be the largest Fibonacci number less than $k$. Then $k - m$ can be expressed as the sum of Fibonacci numbers $m_{i} (i=1,2, \cdots, s)$, otherwi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,535 |
Example 5. The partition number $P(n, k)$ of a natural number $n$ satisfies the recurrence relation
$$
P(n+k, k)=P(n, 1)+P(n, 2)+\cdots+P(n, k) \text {, }
$$
where $P(n, 1)=P(n, n)=1$, and $1 \leqslant k \leqslant n$. | According to the definition, the right side of the recurrence relation refers to the total number of partitions of $n$ with length not exceeding $k$, while the left side refers to the total number of $k$-partitions of $n+k$. The former partitions form the set $X$, and the latter partitions form the set $Y$. Define a ma... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,536 |
Example 6. Prove that $P(n, k)=P(n-k)$, where $\frac{n}{2} \leqslant k \leqslant n$.
Analysis According to the definition of partitions, $P(n, k)$ represents the total number of $k$-partitions of $n$, while $P(n-k)$ represents the total number of all partitions of $n-k$. Note that each partition of $n-k$
$$
n-k=n_{1}+n... | Let the set of $k$-partitions of $n$ be denoted by $A$, and the set of partitions of $n-k$ be denoted by $B$. For $n=n_{1}+n_{2}+\cdots+n_{k} \in A$, where $n_{1} \geqslant n_{2} \geqslant \cdots \geqslant n_{k} \geqslant 1$, we can obtain the partition
$$
n-k=\left(n_{1}-1\right)+\left(n_{2}-1\right)+\cdots+\left(n_{k... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,537 |
Example 7. If all parts in a partition are distinct, it is called a distinct partition. The total number of distinct partitions of a natural number $n$ is denoted by $D(n)$; and if all parts are odd, it is called an odd partition. The total number of odd partitions of a natural number $n$ is denoted by $O(n)$. Try to p... | Proof: Let the odd number $2k+1$ appear $p$ times in an odd partition of $n$, and write $p$ as a sum of powers of 2:
$$
p=2^{i_{1}}+2^{i_{2}}+\cdots+2^{i_{1}},
$$
where $i_{1}>i_{2}>\cdots>i_{2} \geqslant 0$. Then,
$$
\begin{aligned}
(2 k+1) p= & (2 k+1) \cdot 2^{i_{1}}+(2 k+1) \cdot 2^{i} 2 \\
& +\cdots+(2 k+1) \cdot ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,538 |
Example 8. Let the number of partitions of a natural number $n$ into sums of powers of 2 be denoted by $B(n)$. Prove:
(i) $B(2 n+1)=B(2 n)$;
(ii) $B(2 n)=B(2 n-1)+B(n)$;
(iii) For $n \geqslant 2$, $B(n)$ is even. | Let $2 n+1=2^{i_{1}}+2^{i_{2}}+\cdots+2^{i}+1$ be a partition of $2 n+1$, where $i_{1} \geqslant i_{2} \geqslant \cdots \geqslant i_{r} \geqslant 0$. Then, we have
$$
2 n=2^{i_{1}}+2^{i_{2}}+\cdots+2^{i} \text {. }
$$
This mapping is a bijection, hence $B(2 n+1)=B(2 n)$.
Notice that the set of partitions of $2 n$ into... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,539 |
2. Let $n$ be a natural number, and let $c(n)$ denote the number of ways to write $n$ as a sum of powers of 2, where each power of 2 can appear at most three times. For example, $c(8)=5$:
$$
\begin{array}{l}
8=8=4+4=4+2+2=4+2+1+1=2+2+2+1 \\
+1 .
\end{array}
$$
Find $c(n)$ (44th Putnam Mathematical Competition) | 2. Let $2 n=e_{0}+2 e_{1}+4 e_{2}+\cdots+2^{*} \cdot e_{k}$ be a partition of $2 n$, where $e_{0} \in\{0,2\}, e_{i} \in\{0,1,2,3\}(i=1,2,3$, $\cdots, k)$. If $e_{0}=0$, then $e_{1}+2 e_{2}+\cdots+2^{k-1} \cdot e_{k}$ is a partition of $n$. If $e_{0}=2$, then $e_{1}+2 e_{2}+\cdots+2^{k-1} \cdot e_{k}$ is a partition of ... | C(n)=\left[1+\frac{n}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,541 |
Example 3. Let the sides of $\triangle ABC$ be $BC=a, CA=b, AB=c$, and $AD, BE, CF$ be the angle bisectors of $\triangle ABC$. The areas of $\triangle DEF$ and $\triangle ABC$ are $S_{\triangle DEF}$ and $S_{\triangle ABC}$, respectively.
Prove: $\frac{S_{\triangle DEF}}{S_{\triangle ABC}}=\frac{2 a b c}{(a+b)(b+c)(c+a... | Proof According to the property of the internal angle bisector of a triangle, we have $\frac{A F}{F B}=\frac{b}{a}$,
$$
\frac{B D}{D C}=\frac{c}{b}, \frac{C E}{E A}=\frac{a}{c} \text {. }
$$
Therefore, according to the theorem, we have
$$
\begin{aligned}
& =\frac{\frac{S_{\triangle D F F}}{a} \cdot \frac{c}{b} \cdot \... | \frac{2 a b c}{(a+b)(b+c)(c+a)} | Geometry | proof | Yes | Yes | cn_contest | false | 706,542 |
～Let the positive integer $n$ be expressed as the sum of several 1s and 2s (considering the order of the addends) in $a(n)$ different ways. For example, 4 $=1+1+2=1+2+1=2+1+1=2+2=1+1+1+$ 1, so $a(4)=5$. Let $n$ be expressed as the sum of integers greater than 1 (also considering the order of the addends) in $b(n)$ diff... | 3. Establish a bijection between the first set of partitions $A$ and the second set of partitions $B$. Let $n=a_{1}+a_{2}+\cdots+a_{m} \in A$, where $a_{i_{1}}+a_{i_{2}}+\cdots+a_{i_{k}}=2, 1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant m$,
and the rest of the $a$'s are equal to 1, then
$$
\begin{array}{l}
a_{1}+a_{2}+... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,543 |
Example 1. As shown in the figure, given that the angle formed by the slant edge of a regular quadrilateral pyramid and the base is $\alpha$, and the dihedral angle between two adjacent side faces is $\beta$.
Prove: $\cos \beta$
$$
=\frac{\cos ^{2} \alpha}{\cos ^{2} \alpha-2} \text {. }
$$
(1983, National College Entra... | Proof: Let $S C=a, S O \perp$ plane $A C$ at $O$, then $S O=a - \sin \alpha, A C=2 a \cos \alpha$. In $\triangle A E C$, $\angle A E C$ is the plane angle of the dihedral angle $C-S D-A$. $O$ is the midpoint of $A C$, $A E=E C$, so $O E \perp A C$, thus $O E=\frac{a \cdot \cos \alpha}{\operatorname{tg} \frac{\beta}{2}}... | \cos \beta=\frac{\cos ^{2} \alpha}{\cos ^{2} \alpha-2} | Geometry | proof | Yes | Yes | cn_contest | false | 706,545 |
Example 2. Let the side lengths of the tetrahedron $P-ABC$ be $a, b, c$, and they are mutually perpendicular.
(1) Draw the height $h$ from vertex $P$ to the base $ABC$.
Prove: $\frac{1}{h^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$,
(2) Let $M$ be a point on $\triangle ABC$, and the distances from this point ... | Proof (1) As shown in the figure,
Let the areas of $\triangle P A B, \triangle P B C, \triangle P A C$ be $S_{1}, S_{2}, S_{3}$, and the areas of their projections on the base $\triangle O A B, \triangle O B C, \triangle O C A$ be $S^{\prime}{ }_{1}, S^{\prime}{ }_{2}, S^{\prime}{ }_{3}$, respectively. The area of $\tr... | \frac{1}{27} | Geometry | proof | Yes | Yes | cn_contest | false | 706,546 |
Example 3. In a trihedral angle with dihedral angles $\angle A O B, \angle B O C, \angle C O A$ being equal, take a point $S$ such that its distances to the edges of the trihedral angle are equal. Through $S$, draw a plane intersecting the edges $O A, O B, O C$ at points $M, N, P$. Prove: $\frac{1}{O M}+\frac{1}{O N}+\... | Solve As shown in the figure, let
$\angle A O B=\angle B O C=$ $\angle C O A=\theta, O A$ and the angle between plane $O B C$ is $a$ (a constant). Since $S$ is a fixed point and is equidistant from the three edges of the trihedral angle, it is evident that the distances from point $S$ to the three planes are also equal... | \frac{\sin a}{d} | Geometry | proof | Yes | Yes | cn_contest | false | 706,547 |
Example 3. (1981, College Entrance Examination Question) Given the hyperbola
$$
x^{2}-\frac{y^{2}}{2}=1 \text {. }
$$
(1) A line $l$ passing through point $A(2,1)$ intersects the given hyperbola at points $P_{1}, P_{2}$. Find the equation of the locus of the midpoint $P$ of $P_{1}, P_{2}$, and identify what kind of cur... | (1) Let $f(x, y)=x^{2}-\frac{y^{2}}{2}-1$, and let the midpoint of the chord $P_{1} P_{2}$ passing through $A(2,1)$ be $P\left(x_{0}, y_{0}\right)$. Then the equation of $P_{1} P_{2}$ is
$$
f(x, y)-f\left(2 x_{0}-x, 2 y_{0}-y\right)=0,
$$
which simplifies to $-2 x_{0}^{2}+2 x_{0} x+y_{0}^{2}-y_{0} y=0$.
Since $A(2,1)$... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,549 |
Let $a_{i} \in R^{+}(i=1,2, \cdots, k), m, n \in N$, then $\frac{1}{k} \sum_{i=1}^{k} a_{i}^{+^{n}} \geqslant\left(\frac{1}{k} \sum_{i=1}^{k} a_{i}^{n}\right)\left(\frac{1}{k} \sum_{i=1}^{k} a_{i}^{n}\right)$. The equality holds if and only if $a_{1}=a_{2}=\cdots=a_{k}$. | Prove that (1) is equivalent to
$$
k \sum_{i=1}^{k} a_{i}^{n+n}-\left(\sum_{i=1}^{k} a_{i}^{n}\right)\left(\sum_{i=1}^{1} a_{i}^{*}\right) \geqslant 0, .
$$
and is also equivalent to
Since $a_{i}^{-}-a_{j}^{n}=a_{i}^{*}-a_{i}^{*}$ is always of the same sign, the final inequality holds.
Example 1. (Polish Competition ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,550 |
Theorem: If $f(x)$ is a monotonic odd function, then the equation $f(a x+b)+f(x)=0$ has the same solution as the equation $(a x+b)+x=0$.
| Prove that from $f(-x)=-f(x)$, the equation $f(a x + b) + f(x) = 0$ can be transformed into $f(a x + b) = f(-x)$. Since $f(x)$ is a monotonic function, $f$ is a one-to-one mapping, hence the necessary and sufficient condition for $f(a x + b) = f(-x)$ to hold is $a x + b = -x$. Proof completed.
---
Translate the above... | null | Algebra | proof | Yes | Yes | cn_contest | false | 706,552 |
Example 1. Solve the equation
$$
(x+6)^{1991}+x^{1991}+2 x+6=0 .
$$ | $\because$ Solving $f(x)=x^{1991}+x$ yields an increasing odd function. Therefore,
$(x+6)+x=0$, the original equation has a unique real solution $x=-3$. | x=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,554 |
2. The two roots of the equation $x^{2}+p x+q=0$ are non-zero integers, and $p+q=198$, then $p=$ $\qquad$ . | 2. -202
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | -202 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,557 |
7. The sum of all positive numbers $a$ that make the roots of the equation $a^{2} x^{2}+a x+1-7 a^{2}=0$ integers is $\qquad$ . | 7. $\frac{11}{6}$ | \frac{11}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,562 |
8. For natural numbers $x, y$, $(x, y)$ is called a pair. It is stipulated that the pair $(x, y)$ is the same as $(y, x)$. If two pairs satisfy the following condition: the sum of the two numbers in one pair equals the product of the two numbers in the other pair, then these two pairs are called "companion pairs". Exce... | $\begin{array}{l}8 .(2,3) \text { and } \\ (1,5)\end{array}$ | (2,3) \text{ and } (1,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,563 |
Column 5. As shown in Figure 1, in $\triangle ABC$, point $F$ is the midpoint of side $AB$, and $D, E$ are points on $BC, CA$ respectively. Prove: The area of $\triangle DEF$ does not exceed the sum of the areas of $\triangle AFE$ and $\triangle BFD$.
(1983, 17th All-Soviet Union Mathematical Olympiad for Middle School... | Given that $A F: F B=1$. Let $B D: D C=n$, $C E: E A=l$. By the theorem and its corollary, we have
$$
\begin{array}{l}
S_{\triangle D E F}=\frac{n l+1}{2(n+1)(l+1)} \cdot S_{\triangle A B C}, \\
S_{\triangle A F E}+S_{\triangle B F D} \\
=\frac{1}{2(l+1)} \cdot S_{\triangle A B C}+\frac{n}{2(n+1)} \cdot S_{\triangle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,564 |
II. (16 points) Find all three-digit numbers such that the last three digits of their square are the original two-digit number.
The above text has been translated into English, retaining the original text's line breaks and format. | Let the three-digit number $m$ satisfy the condition, then $m^{2}=k \times 1000+$ $m$, where $k$ is a natural number. Therefore,
$$
m(m-1)=k \times 1000=k \times 2^{3} \times 5^{3} .
$$
$\because m, m-1$ are two coprime natural numbers, and one is odd while the other is even, hence the odd number among $m, m-1$ is a mu... | 625 \text{ or } 376 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,567 |
Three. (This question is worth 16 points) Let $n$ be a natural number greater than 1. From an $n \times n$ square, remove a $1 \times 1$ square from one of its corners, and divide this shape into $k$ triangles of equal area. Find the minimum value of $k$.
Translate the above text into English, preserving the original ... | Three, as shown in the figure, among the triangles of equal area that have common points with the broken line $ABC$, there must be a triangle whose one side is a part of $AB$ (or $AC$). The length of this side is no more than 1, and the height on this side is no more than $n-1$. Therefore, the area of this triangle doe... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,568 |
Four. (This question is worth 18 points) Let $\triangle A B C$ be an equilateral triangle with side length 6. Draw a line $l$ through vertex $A$, and let the distances from vertices $B$ and $C$ to $l$ be denoted as $d_{1}, d_{2}$. Find the maximum value of $d_{1} + d_{2}$.
保留源文本的换行和格式，直接输出翻译结果。 | Extend $B A$ to $B^{\prime}$, such that $A B^{\prime}=A B$, and connect $B^{\prime} C$. Then a line $l$ passing through vertex $A$ either intersects $B C$ or intersects $B^{\prime} C$.
(1) If $l$ intersects $B C$ at $D$. Then
$$
\begin{array}{l}
\frac{1}{2}\left(d_{1}+d_{2}\right) \cdot A D \\
=S_{\triangle A B D}+S_{\... | 6 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,569 |
2. For real numbers $x, y$, define a new operation $x: y=a: c+b y+$ $c x y$. Here $a, b, c$ are constants, and the usual operations are addition and multiplication, with $1 \times 2=: 3,2 \times 3=4$, and there exists a non-zero constant $d$, such that for any $x$, we have $x$ * $d=x$. Then the value of $d$ is ( ).
(A)... | 2. B.
Solution $1 * 2=3 \Rightarrow a+2 b+2 c=3$,
$2 * 3=4 \Rightarrow 2 a+3 b+6 c=4$,
$x * d=x \Rightarrow a x+b d+c d x=x$.
Let $x=0$, then we have $b d=0, d \neq 0, b=0$.
$\Rightarrow\left\{\begin{array}{l}a+2 c=3, \\ a+3 c=2\end{array} \Rightarrow a=5, c=-1\right.$.
Let $x=1$. Substituting into (3) we get
$$
5+(-1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,571 |
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