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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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 13. When $n \in N$, is the integer part of $(3+\sqrt{7})^{n}$ odd or even? Prove your conclusion.
(1980, Beijing Spring Festival Contest Problem) | Solve for the integer part of $(3+\sqrt{7})^{n}$ being odd.
Let $(3+\sqrt{7})^{n}=a+\sqrt{7} b(a, b \in N)$, then we can construct
$$
(3-\sqrt{7})^{n}=a-\sqrt{7} b .
$$
Since $(3+\sqrt{7})^{n}+(3-\sqrt{7})^{n}=2 a \in N$, and $0<(3-\sqrt{7})^{n}<1$, the fractional part of $(3+\sqrt{7})^{n}$ is $1-(3-\sqrt{7})^{n}$.
$$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,218 |
Example 14. Prove: The first $n$ digits of the decimal part of $(5+\sqrt{26})^{n} (n \in N)$ are the same. (Hungarian Olympiad Problem)
---
The translation maintains the original format and line breaks as requested. | Proof Let $A=(5+\sqrt{26})^{n}$, construct
$$
B=(5-\sqrt{26})^{n},(n \in N)
$$
$$
\begin{array}{l}
\text { Then } A+B=2 \sum_{k=0}^{\left(\frac{n}{2}\right)} C_{n}^{2 k} 5^{n-2 k} \cdot 26^{k} \in N . \\
\text { Also, since } 251-10^{-n}$, thus the first $n$ digits after the decimal point are all 9;
(2) When $n$ is odd... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,219 |
Example 15. If $n \in N$, prove that $a_{n}=\sum_{k=0}^{n} 2^{3 k} C_{2 n+1}^{2 k+1}$ is not divisible by 5. (16th IMO Problem) | $$
\begin{array}{l}
(1+2 \sqrt{2})^{2 n+1}=\sum_{k=j}^{2 n} C_{2 n}^{k}(2 \sqrt{2})^{k} \\
=\sum_{k=0}^{n} C_{2 n+1}^{2 k+1} \cdot(2 \sqrt{2})^{2 k+1} \\
\quad+\sum_{k=0}^{n} C_{2 n+1}^{2 k} \cdot(2 \sqrt{2})^{2 k} \\
=2 \sqrt{2} \sum_{k=0}^{n} C_{2 n+1}^{2 k-1} \cdot 2^{3 k}+\sum_{k=0}^{n} C_{2 n+1}^{2 k} \cdot 2^{3 k... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,220 |
Example 1. (4th CMO) Let $x_{i} \in \mathbb{R}^{+}$ $(1 \leqslant i \leqslant n, n \geqslant 2), \sum x_{1}:=1$. Prove that:
$$
\sum \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum \sqrt{x_{i}}}{\sqrt{n-1}} .
$$
(In this article, $\sum$ denotes the sum from $i$ = 1 to $n$) | Prove $\begin{aligned} & \sum \frac{x_{i}}{\sqrt{1-x_{i}}} \\ = & \sum \frac{1}{\sqrt{1-x_{i}}}-\sum \sqrt{1-x_{i}} \\ \geqslant & \frac{n^{2}}{\sum \sqrt{1-x_{i}}}-\sum \sqrt{1-x_{i}} \\ \geqslant & \frac{1}{\sqrt{n} \sqrt{\sum\left(1-x_{i}\right)}} \\ & -\frac{\sqrt{n} \sqrt{\sum\left(1-x_{i}\right)}}{\sqrt{n-1}}=\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,221 |
Example 4. If $P$ is a point inside the equilateral $\triangle A B C$, and $P A, P B, P C$ are connected, with $P A^{2}=P B^{2}+P C^{2}$. Find $\angle B P C$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve As shown in the figure, rotate $\triangle A B P$ around vertex $A$ by $60^{\circ}$ to $\triangle A C P^{\prime}$, then $\triangle A P P^{\prime}$ is an equilateral triangle.
Therefore, $P P^{\prime}=A P$.
$$
\begin{array}{l}
\text { Also, } B P=C P^{\prime} \text {, } \\
\therefore P P^{\prime 2}=P C^{2}+P^{\prim... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,223 |
Example 5. (1984, National Competition Question) Let $a_{i} \in R^{+}$ $(1 \leqslant i \leqslant n)$. Prove: $\sum \frac{a_{i}^{2}}{a_{i+1}} \geqslant \sum a_{i}$. (Define $a_{n+1}$ $=a_{1}$) | $$
\begin{array}{l}
\text { Prove } \because \sum\left(\frac{a_{i}^{2}}{a_{i+1}}+a_{i+1}\right) \geqslant \sum 2 a_{i}, \\
\therefore \sum \frac{a_{i}^{2}}{a_{i+1}} \geqslant \sum 2 a_{i}-\sum a_{i+1}=\sum a_{i} .
\end{array}
$$
When pairing terms, it is also necessary to consider the possibility of equality in the in... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,225 |
Example 6. (1991, Asia Pacific Mathematical Competition) Let
$$
\begin{array}{l}
a_{i}, b_{i} \in R^{+}, \sum a_{i}=\sum b_{i}(1 \leqslant i \leqslant n) . \\
\text { Prove: } \sum \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} \sum a_{i} .
\end{array}
$$ | $$
\begin{array}{l}
\text { Prove } \sum\left(\frac{a_{i}^{2}}{a_{i}+b_{i}}+\frac{a_{i}+b_{i}}{4}\right) \geqslant \sum a_{i}, \\
\begin{array}{c}
\therefore \sum \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \sum a_{i}-\frac{1}{4} \sum a_{i}-\frac{1}{4} \sum b_{i} \\
=\frac{1}{2} \sum a_{i} .
\end{array}
\end{array}
$$
For... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,226 |
Example 8. (1994, Communication Competition Question) Find the maximum constant $k$, such that for any $x, y, z \in R^{+}$, we have
$$
\sum \frac{x}{\sqrt{y+z}} \geqslant k \sqrt{\sum x} .
$$ | Solution: First, let $x=y=z=1$, we get $k \leqslant \sqrt{\frac{3}{2}}$. Next, we prove $\sum \frac{x}{\sqrt{y+z}} \geqslant \sqrt{\frac{3}{2}} \sqrt{\sum x}$.
Assume without loss of generality that $x \geqslant y \geqslant z$, by Chebyshev's inequality, we have
$$
\begin{array}{l}
3 \sum_{x} \sqrt{y+z} \leqslant \sum ... | \sqrt{\frac{3}{2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,228 |
Example 1. (1993, National High School Mathematics Competition) Let $0<a<b$, and draw lines $l$ and $m$ through the fixed points $A(a, 0)$ and $B(b, 0)$, respectively, intersecting the parabola $y^{2}=x$ at four distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of $l$ a... | Let $l$ and $m$ intersect at $P\left(x^{\prime}, y^{\prime}\right)$, then the equations of $l$ and $m$ are:
$$
\begin{array}{l}
y^{\prime} x-\left(x^{\prime}-a\right) y+a y^{\prime}=0, \\
y^{\prime} x-\left(x^{\prime}-b\right) y+b y^{\prime}=0 .
\end{array}
$$
And $F: y^{2}-x=0$.
By Theorem 2, we get
$$
y^{\prime} \cd... | x^{\prime}=\frac{a+b}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,229 |
Example 2. (Hungarian Competition Problem) For the cyclic quadrilateral $ABCD$ inscribed in $\odot K$, the diagonals $AC$ and $BD$ intersect perpendicularly at $P$. The line through point $P$ and the midpoint $M$ of $AE$ intersects $CD$ at $M'$. Similarly, the points $N, N', G, G', U, H'$ are defined. Prove that the ei... | Prove that if $A C$ and $B D$ are used as coordinate axes to establish a Cartesian coordinate system, and the equation of $\odot K$ is $\left(x-x_{0}\right)^{2}+(y-\left.y_{0}\right)^{2}=r^{2}$, then $A\left(x_{1}, 0\right)$, $B\left(0, y_{1}\right)$, $C\left(x_{2}, 0\right)$, $D\left(0, y_{2}\right)$. Among them,
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,230 |
Example 1. (1989, Jilin Provincial Competition Question) Simplify
$$
\frac{a^{2}\left(\frac{1}{b}-\frac{1}{c}\right)+b^{2}\left(\frac{1}{c}-\frac{1}{a}\right)+c^{2}\left(\frac{1}{a}-\frac{1}{b}\right)}{a\left(\frac{1}{b}-\frac{1}{c}\right)+b\left(\frac{1}{c}-\frac{1}{a}\right)+c\left(\frac{1}{a}-\frac{1}{b}\right)} .
$... | Let $c-b=x, a-c=y, b-a=z$, then $x$
$$
\begin{array}{l}
+y+z=0 . \\
\text { Original expression }=\frac{\frac{1}{a}\left(c^{2}-b^{2}\right)+\frac{1}{b}\left(a^{2}-c^{2}\right)+\frac{1}{c}\left(b^{2}-a^{2}\right)}{\frac{1}{a}(c-b)+\frac{1}{b}(a-c)+\frac{1}{c}(b-a)} \\
-\frac{\frac{x}{a}(c+b)+\frac{y}{b}(a+c)+\frac{z}{c}... | a+b+c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,231 |
9. 1. Given that $n$ is a natural number, and $2 n+1$ and $3 n+1$ are both perfect squares. For this $n$, can $5 n+3$ be a prime number? | 9. 1. $5 n+3$ is not a prime number.
If $2 n+1=k^{2}, 3 n+1=m^{2}$, then
$$
\begin{array}{l}
5 n+3=4(2 n+1)-(3 n+1) \\
=4 k^{2}-m^{2}=(2 k+m)(2 k-m)
\end{array}
$$
is a composite number, because $2 k-m \neq 1$. Conversely, if $5 n+3=2 m+1$, then
$$
\begin{array}{l}
(m-1)^{2}=m^{2}-(2 m+1)+2 \\
=(3 n+1)-(5 n+3)+2=-2 n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,233 |
Example 5. As shown in the figure, on a semicircle with center $C$ and diameter $M N$, there are two distinct points $A$ and $B$. $P$ is on $C N$, and $\angle C A P=\angle C B P=10^{\circ}, \angle M C A=40^{\circ}$. Find $\angle B C N$. (34th American Competition Problem) | Solution: Clearly, $\angle C P A = 30^{\circ}$. Therefore, applying the Law of Sines in $\triangle A C P$ and $\triangle B C P$, we have:
$$
\begin{array}{l}
\frac{\sin 10^{\circ}}{C P}=\frac{\sin 30^{\circ}}{A C} \text { and } \frac{\sin 10^{\circ}}{C P}=\frac{\sin \angle C P B}{B C}. \\
\therefore \frac{\sin 30^{\cir... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,234 |
9. 2. Two line segments of length 1 intersect at point $O$, and $\angle A O C=60^{\circ}$. Prove: $A C+B D \geqslant 1$.
| 9.2. As shown in the figure, construct segment $C B_{1}$ such that $C B_{1} / / A B$ and $C B_{1}=A B$, then quadrilateral $A B B_{1} C$ is a parallelogram, hence $A C=B B_{1}$. By $B B_{1}+$ $B D \geqslant B_{1} D$, therefore, $A C+$ $B D \geqslant B_{1} D$. Also, $\triangle C B_{1} D$ is an equilateral triangle, so $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,235 |
9. 3. Let $f(x)$ be a quadratic trinomial. The trinomial $x^{2}$ - $f\left(\frac{1}{x}+1\right)$ or $(x-1)^{2} f\left(\frac{1}{x}-1\right)$ can replace $f(x)$. Can these operations transform the quadratic trinomial $x^{2}+4 x+3$ into the quadratic trinomial $x^{2}+10 x+3$? | 9.3. Not possible.
Let $f(x)=A x^{2}+B x+C$, then $\Delta=B^{2}-4 A C$. After performing the first operation, $f(x)$ becomes
$$
\begin{array}{l}
(A+B+C) x^{2}+(B+2 A) x+A, \\
\Delta=(B+2 A)^{2}-4(A+B+C) A=B^{2}-4 A C .
\end{array}
$$
While performing the second operation, we get
$$
\begin{array}{l}
C x^{2}+(B-2 C) x+... | not possible | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,236 |
9. 4. In a family photo album, there are 10 photos. Each photo has 3 people, with the man standing in the middle, his son standing on his left, and his brother on his right. If it is known that the 10 men standing in the middle are all different, then what is the minimum number of different people in these photos? | 9. 4. 16.
We refer to the 10 men standing in the middle of the photo as the main characters. We then classify all the men in the photo into levels: men who do not have a father in any photo are classified as level 0; when $k=0,1,2$, $\cdots$, men who have a father classified as level $k$ in some photo are classified a... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,237 |
9. 5. Integers $x, y, z$ satisfy the equation
$$
(x-y)(y-z)(z-x)=x+y+z \text {. }
$$
Prove: $x+y+z$ is divisible by 27. | 9. 5. As long as it is proven that the remainders of $x, y, z$ when divided by 3 are the same, then $x+y+z=(x-y)(y-z)(z-x)$ can be divisible by 27. If the remainders of $x, y, z$ when divided by 3 are different, then $(x-y)(y-z)(z-x)$ cannot be divisible by 3, while $x+y+z$ can be divisible by 3. Therefore, at least tw... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,238 |
9. 6. As shown in the figure. Inside the circle, there is a convex quadrilateral, the extensions of whose sides intersect the circumference at points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$, $D_{1}$, $D_{2}$. Prove: If $A_{1} B_{2}=B_{1} C_{2}=$ $C_{1} D_{2}=D_{1} A_{2}$, then the quadrilateral formed by t... | 9. 6. Let the angles between the lines $A_{1} A_{2}$ and $B_{1} B_{2}$, $C_{1} C_{2}$ and $D_{1} D_{2}$ be $\alpha, \beta$, respectively, and the arcs subtended by any equal chords $A_{1} B_{2}, B_{1} C_{2}$, $C_{1} D_{2}, D_{1} A_{2}$ are all equal to $\gamma$. Then, by the external angle theorem of a circle, we have
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,239 |
9.7. Place an even number of pawns on each row, each column, and each diagonal of a chessboard. Question: For this arrangement, what is the maximum number of pawns that can be placed? | 9. 7. 48.
Firstly, note that on a chessboard, there are 16 diagonals containing an odd number of squares, and these diagonals do not share any squares. Therefore, the number of pieces will not exceed \(64-16=48\). If a piece is placed in each square except those on the two main diagonals, the condition will be satisfi... | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,240 |
9. 8. On the blackboard, there are $n$ expressions of the form $* x^{2}+* x+* =0$ ($n$ is odd). Two people take turns playing a game, each step allowing the replacement of one “star” with a number not equal to 0. After $3 n$ steps, $n$ quadratic equations are obtained. The first person aims to make as many equations as... | 9. 8. $\frac{n+1}{2}$. process of the same kind.
Now, the first use of a number to replace the "star" in the expression is called breaking down the expression. The first player's first step should be to break down one of the expressions in their expression and fill in the number 1 before $x$. If the second player fill... | \frac{n+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,241 |
10. 1. Let the lengths of the three sides of a triangle all be prime numbers. Prove that the area of this triangle is not an integer. | 10.1. Let the sides of a triangle be $a, b, c$. By Heron's formula, we have
$$
16 S^{2}=p(p-2 a)(p-2 b)(p-2 c) .
$$
where $S$ is the area of the triangle and $p=a+b+c$ is the perimeter of the triangle.
Assume $S$ is an integer. Then from (1), we know that $p$ is even. Therefore, either $a, b, c$ are all even, or one o... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,242 |
10. 2. Draw two lines from the center of symmetry of two intersecting circles with equal radii, intersecting the circumferences of the two circles at 4 points not on the same straight line. Prove that these 4 points lie on the same circle. | $\overparen{B_{2} B_{3}}=\overparen{A_{1} A_{3}}$, since these two arcs are symmetric with respect to point O, thus, $\angle A_{3} A_{2} A_{1}=\angle B_{3} B_{1} B_{2}$. This means that the segment $A_{1} B_{2}$ can be seen from points $B_{1}$ and $A_{2}$ at the same angle. Therefore, points $A_{1}, A_{2}, B_{1}, B_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,243 |
10. 4. At a round table sit 30 people, each of whom is either a fool or a wise person. Ask all the people present the question: “Is the person to your right a wise person or a fool?” Wise people answer correctly, while fools may answer either correctly or incorrectly. It is known that the number of fools does not excee... | 10. 4. $F=8$.
If $F=0$, then anyone sitting around the table can be pointed out. Now, let's assume $F \neq 0$. Group all the people sitting around the table: consecutive wise men form one group, and consecutive fools form another group, resulting in a total of $2k$ groups ($k$ groups of wise men, $k$ groups of fools).... | F=8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,244 |
Example 6. $\triangle A B C$ is an isosceles triangle, with vertex angle $A=$ $80^{\circ}, O$ is a point inside $\triangle A B C$, $\angle O B C=10^{\circ}$, $\angle O C B=30^{\circ}$. Find $\angle O A B$. (1992, Huanggang Region, Hubei Province Junior High School Mathematics Competition) | Let $N$ be the intersection of the angle bisector of $\angle A$ and the extension of $CO$ (as shown in the figure), then $BN$
$$
\begin{array}{l}
=CN. \\
\therefore \angle BNC=180^{\circ}-2 \angle NCB=120^{\circ}. \\
\because \angle ACB=\frac{1}{2}\left(180^{\circ}-\angle A\right)=50^{\circ}, \\
\therefore \angle ABN=5... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,245 |
10. 6. Is the following statement correct: Any two rectangles of equal area can be placed on a plane so that any horizontal line intersecting one of the rectangles also intersects the second rectangle, and the line intersects the two rectangles at segments of equal length. | 10. 6. Correct.
As shown in the figure, two rectangles $A_{1} B_{1} C_{1} D_{1}$ and $A_{2} B_{2} C_{2} D_{2}$ of equal area (just by overlapping their vertices $D_{1}$ and $D_{2}$, i.e., $D_{1}=D_{2}$, and placing the shorter side $D_{1} A_{1}$ of one rectangle on the longer side $D_{3} A_{2}$ of the other, while pla... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,246 |
10.7. A square wooden board is divided into $n^{2}$ unit squares by horizontal and vertical lines. Mark $n$ squares so that any rectangle with an area of at least $n$ and whose sides lie along the grid lines contains at least one marked square. Find the largest $n$ that satisfies this condition. | 10.7. 7 .
Obviously, if $n$ marked cells satisfy the conditions of the problem, then in each row and each column there is exactly one marked cell. Let $n \geqslant 3$ (obviously, $n=2$ is not the maximum), take the first row with a marked cell as $A$, the row adjacent to $A$ as $B$, and take either a row adjacent to $... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,247 |
10. 8. $\left\{a_{k}\right\}$ is a sequence of real numbers, and $\left\{a_{A}{ }_{1}\right\}$ is another sequence of real numbers, and $a^{\prime}{ }_{k}=\frac{a_{k}+a_{k+1}}{2}$, then $\left\{a^{\prime}{ }_{k}\right\}$ is called the average sequence of $\left\{a_{k}\right\}$. Consider the sequences $\left\{a_{k}\righ... | 10. 8. A sequence is called $m$-good if it and the first $m$ averages derived from it are all composed of integers. Prove using mathematical induction that if the sequence $\left\{x_{k}\right\}$ is good, then the sequence $\left\{x_{k}^{2}\right\}$ is $m$-good for any non-negative integer $m$. This will lead to the con... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,248 |
11. 2. On a plane, there are two right triangles, the medians to their hypotenuses are parallel. Prove: the angle between a leg of one triangle and a leg of the other triangle is less than half the angle between their hypotenuses. | 11. 2. Translate one of the right triangles so that their right-angle vertices $C$ and $C^{\prime}$ coincide, then apply a homothety centered at point $C$ to the same right triangle so that their medians coincide (as shown in the figure). Thus, the circle with center $E$ and radius $C E$ is the circumcircle of both tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,249 |
11. 3. Find all functions $f(x)$ that satisfy the following conditions: defined and positive for all positive $x$, and for all positive $x, y$ the equation $f\left(x^{y}\right)=f(x)^{f(y)}$ holds. | 11.3. $1, x$.
First, $f(x) \equiv 1$ satisfies the given conditions. Now let's find other solutions.
Assume for some $a>0, f(a) \neq 1$. Then from the equation
$$
f(a)^{f(x y)}=f\left(a^{x y}\right)=f\left(a^{x}\right)^{f(y)}=f(a)^{f(x) f(y)}
$$
we get $f(x y)=f(x) f(y)$
for any $x, y>0$. And from the equation
$$
\be... | 1, x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,250 |
11.4. Proof: There exists a natural number $n$, such that when an equilateral triangle with side length $n$ is divided by lines parallel to its sides into $n^{2}$ smaller equilateral triangles with side length 1, then among the vertices of these smaller triangles, one can select $1993 n$ vertices, such that no three of... | Incidentally, for any equilateral triangle with these points as vertices, all vertices either have different colors or the same color. It is easy to prove that if such a triangle is rotated $60^{\circ}$ around any of its vertices (without loss of generality, assume it has color $A$), then the vertex that remains in the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,251 |
11.7. In a tennis round-robin tournament, there are $n$ players participating in doubles matches (2 vs 2), and each player must face every other player as an opponent exactly once in the matches. Question: For what values of $n$ is such a round-robin tournament impossible? | 11. 7. $\quad n=8 k+1, k \in N$.
Assume that the round-robin tournament can take place. Then all opponents of a tennis player can be divided into several "pairs", so $n$ is odd. All players can form "pairs" that can be divided into several "two pairs" that can play a match, so the number of "pairs" that $n$ people can... | n=8k+1, k \in N | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,254 |
Example 7. $\triangle A B C$ is an isosceles triangle, with vertex angle $A=$ $80^{\circ}, O$ is a point inside $\triangle A B C$, $\angle O B C=10^{\circ}$, $\angle O C B=20^{\circ}$. Find $\angle O A C$. | Solve: As shown in the figure, construct the symmetric point $N$ of $O$ with respect to $AC$, and connect $AN, BN, CN$.
$$
\begin{array}{l}
\because \angle OCA = 30^{\circ}, \angle NCA = 30^{\circ}, \\
\therefore \angle OCN = 60^{\circ}.
\end{array}
$$
Therefore, $\triangle OCN$ is an equilateral triangle.
Thus, $ON =... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,256 |
Let $f$ be a function from $R \rightarrow R$, and
(1) for any $x, y \in R$,
$$
f(x)+f(y)+1 \geqslant f(x+y) \geqslant f(x)+f(y) ;
$$
(2) for any $x \in [0,1)$, $f(0) \geqslant f(x)$;
$$
\text{(3) } -f(-1)=f(1)=1 \text{.}
$$
Find all functions that satisfy the conditions. | First, prove that $f(0)=0$. From (1) and (2), we can deduce
$$
2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0),
$$
thus $f(0) \geqslant 0 \geqslant f(0)$, which means $f(0)=0$.
Next, prove that for any $x \in(0,1), f(x)=0$. From (2), for such an $x, f(x) \leqslant 0$. If $f(x)f(x)+f... | f(x)=[x] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,257 |
Given a non-degenerate $\triangle A B C$, let the circumcenter be $O$, the orthocenter be $H$, and the circumradius be $R$. Prove: $|O H|<3 R$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $G$ be the centroid of $\triangle A B C$, and $J$ be the center of the nine-point circle. $O$ and $J$ are harmonically conjugate with respect to $G$ and $H$, and $O, J, G, H$ all lie on the Euler (Euler) line of $\triangle A B C$. Considering the vectors starting from point $O$, we have
$$
\begin{aligned}
\overrigh... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,258 |
Three, let $n$ be an integer of the form $a^{2}+b^{2}$, where integers $a, b$ are coprime. If $p$ is a prime, $p<\sqrt{n}$, then $p \mid a b$. Determine all such $n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Three, first note that $n$ is a prime number, so $n=2$ or $n$ is a number of the form $4k+1$. Below we prove that $n \leq 5^2 = 25$.
Assume $n > 25$, then all primes not greater than 5 divide $ab$, and
$$
\begin{aligned}
n & =a^2 + b^2 \geq c^2 + d^2 \geq 2cd \\
& \geq 2(2 \times 3 \times 5) = 60 > 49 = 7^2.
\end{align... | n = 2, 5, 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,259 |
Does there exist an infinite subset of the plane such that no three points are collinear and the distance between any two points is a rational number? | Four, the answer is yes. Below are two methods to construct such a set.
Method 1 There exists an angle $\alpha$ such that $\cos \alpha, \sin \alpha$ are both positive rational numbers (for example, $\sin \alpha=\frac{3}{5}$). Consider a circle with a rational radius $R$ and an arc of $2 \alpha$ radians. Clearly, $\fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,260 |
Given $A, B, C$ three columns of numbers. Column $A$ consists of decimal numbers of the form $10^{k}$, where $k \geqslant 1$ is an integer. Columns $B$ and $C$ are the numbers in column $A$ converted to binary and quinary (base-5) numbers, respectively:
\begin{tabular}{lll}
$A$ & $B$ & $C$ \\
10 & 1010 & 20 \\
100 & 11... | Five, Lemma (Betty's Theorem). Given two positive irrational numbers $x$ and $y$, such that $\frac{1}{x}+\frac{1}{y}=1$, then the set of positive integers can be written as the union of two disjoint sequences,
$$
\begin{array}{l}
{[x],[2 x],[3 x], \cdots ;} \\
{[y],[2 y],[3 y], \cdots .}
\end{array}
$$
Proof of the le... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,261 |
3. The product of the repeating decimals $0 . \overline{3}(=0.333 \cdots)$ and $0 . \overline{6}$ is ).
(A) $0 . \overline{18}$
(B) $0 . \overline{19}$
(C) $0 . \overline{2}$
(D) $0 . \overline{9}$
(E) None of the above answers is correct. | 3. (C) Since $0 . \overline{3}=\frac{1}{3}, 0 . \overline{6}=\frac{2}{3}$, the product is $\frac{2}{9}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,264 |
5. Danny has as many brothers as he has sisters, and Danny's older sister has exactly twice as many brothers as she has sisters. Then the number of Danny's parents' children is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7 | 5. (E) Let Dan have $x$ brothers and $x$ sisters. Then, his sister has $x+1$ brothers and $x-1$ sisters, and from $x+1=$ $2(x-1)$, we get $x=3$. | E | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 707,266 |
Example 8. In isosceles $\triangle A B C$, the vertex angle $B=20^{\circ}$. Points $D, E$ are taken on sides $B C, A B$ respectively, such that $\angle D A C=$ $60^{\circ}, \angle A C E=50^{\circ}$. Find $\angle A D E$. | Given the figure, let $F$ be the point symmetric to $D$ with respect to the angle bisector of $\angle B$, and $O$ be the intersection of $A D$ and $C F$. Connect $E O$.
To prove $\triangle E F D \cong \triangle E O D$, we only need to prove $F D=O D$ and $E F=E O$.
In $\triangle O F D$, all interior angles are $60^{\... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,267 |
7. Arrange 7 $3 \times 4$ rectangles without overlapping to form a larger rectangle. The minimum perimeter of this larger rectangle is ( ).
(A) 25
(B) 31
(C) 50
(D) 62
(E) None of the above answers is correct | 7. (E) There are three ways to arrange these 7 rectangles into a larger rectangle. Two obvious ways result in rectangles with perimeters of 50 and 62. The method we use is to arrange 4 rectangles into a $4 \times 12$ and 3 rectangles into a $3 \times 12$, thus forming a $7 \times 12$ rectangle, which has a perimeter of... | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,269 |
10. The age of a government substitute legislator is equal to his IQ. If he joins the cabinet, he will increase the average age of the cabinet members from 50 years to 51 years, and decrease the IQ from 114 to 111. Then, after he joins, the number of cabinet members is ( ).
(A) 12
(B) 15
(C) 20
(D) 30
(E) None of the a... | 10. (B) Let the age and IQ of the substitute member be $x$, and the number of cabinet members be $n$, then $x-51=n(51-50)$, and $111-x$ $=n(114-111)$, adding the two equations gives $60=4 n$. | 15 | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,272 |
11. Two three-digit numbers have the same digits in each place, but their order is exactly reversed. It is known that the product of these two three-digit numbers is 433755. Then, the tens digit of the three-digit number is ( ).
(A) 1
(B) 5
(C) 6
(D) 7
(E) None of the above answers is correct | 11. (C) The unit digit of a number is 5, and its hundred's digit is an odd number. If this odd number is 9, the product will exceed 450000; if it is less than or equal to 5, the product will be less than 360000. Therefore, the hundred's digit is 7. Let the ten's digit be $x$, then (5 $+7) x+3$ will yield the ten's digi... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,273 |
12. Starting from the number $a$, perform the following four operations:
Add 2, Subtract 2, Multiply by 2, Divide by 2, each operation only once, in any continuous order. For example, you can first multiply by 2, then subtract 2, etc. Then, the largest number that can be obtained at the end is ( ).
(A) $a$
(B) $a+1$
(... | 12. (C) You cannot perform addition and subtraction consecutively, nor can you perform multiplication and division consecutively, otherwise the answer will only be a mess. To make the expression reach its optimal value, addition should be arranged after division and before multiplication. Therefore, the best expression... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,274 |
13. $A B C D$ and $D E F G$ are two rectangles, and $A B=D E$ $=15, B C=E F=9 . E$ is on $A B$, and $E F$ intersects $B C$ at $H$. Then the area of $D E H C$ is ( ).
(A) 60
(B) 65
(C) 70
(D) 75
(E) None of the above answers is correct | 13. (D) By symmetry, $C$ should be on $F G$, and the area of $D E H C$ is twice the area of $\triangle D E H$. By the Pythagorean theorem, $E A=$ 12, so, $B E=3$. Then, by the similarity of $\triangle A D E$ and $\triangle B E H$, we get $H E=5$. Therefore, the area of $\triangle D E H$ is $\frac{75}{2}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,275 |
14. Starting from the number 3, repeat the operations of multiplying by 2 and subtracting 1, 30 times, the result obtained is ( ).
(A) $3 \times 2^{30}$
(B) $3 \times 2^{31}$
(C) $2^{31}+1$
(D) $2^{31}-1$
(E) None of the above answers are correct | 14. (C) From the problem, we have $3 \cdot 2^{30}-2^{29}-2^{28}-\cdots-2-1$ $=3 \cdot 2^{30}-\left(2^{30}-1\right)=2^{31}+1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,276 |
15. Around a $10 \sqrt{2}$ meters $\times 10$ meters rectangular building, there is a large, lush grassland. At a point $5 \sqrt{2}$ meters in front of the midpoint of the longer wall of the building, a wooden stake is driven into the ground. A 20-meter-long rope is then used to tie a goat to the stake. The area of the... | 15. (D) Let $A B C D$ be a building, and the stake $P$ is located outside the long side $A B$. Extend $P A$ to $E$ and $P B$ to $F$, it is easy to see that $A E=10=$ $B F$. The area where the goat can eat grass includes: (1) $\triangle P A B$. (2) A quarter circle with $P$ as the center, between the radii (length 20) $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,277 |
Example 9. In $\triangle A B C$, $\angle B A C=60^{\circ}$, $\angle A C B=45^{\circ}$. (1) Find the ratio of the three sides of this triangle $A B: B C: A C$; (2) Let $P$ be a point inside $\triangle A B C$, and $P A=\sqrt{6}+\sqrt{2}, P B=3 \sqrt{2}+\sqrt{6}$, $P C=3 \sqrt{2}+2 \sqrt{6}$. Find $\angle A P B, \angle B ... | (1) As shown in the figure, draw the altitude $B D$ on side $A C$, and let $A D=x$, then
$$
\begin{array}{l}
B D=\sqrt{3} x, A B=2 x \text {. } \\
\therefore D C=B D=\sqrt{3} x \text {, } \\
B C=\sqrt{6} x \text {, } \\
A C=A D+D C=(1+\sqrt{3}) x \text {. } \\
\end{array}
$$
Thus,
$$
A B: B C: A C=2: \sqrt{6}:(1+\sqrt... | 120^{\circ}, 150^{\circ}, 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,278 |
16. Let
$$
F(x)=\frac{1}{\sqrt{x+2 \sqrt{x-1}}}+\frac{1}{\sqrt{x-2 \sqrt{x-1}}} \text {, }
$$
then $F\left(\frac{3}{2}\right)$ is (
(A) 2
$\sqrt{3}$
(B) 3.5
(C) 4
(D) $3 \sqrt{2}$
(E) None of the above answers is correct
) | 16. (C) Method 1 Notice that $x \pm 2 \sqrt{x-1}=1 \pm 2 \sqrt{x-1}+(x-1)=(1 \pm \sqrt{x-1})^{2}$, thus
$$
\begin{array}{l}
F\left(\frac{3}{2}\right)=\frac{1}{1+\sqrt{\frac{3}{2}-1}}+\frac{1}{1-\sqrt{\frac{3}{2}-1}} \\
=\frac{\sqrt{2}}{\sqrt{2}+1}+\frac{\sqrt{2}}{\sqrt{2}-1} \\
=\sqrt{2}(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}+1... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,279 |
2. Given the equation $(a+1) x^{2}+(|a+2|-\mid a-$ $10 \mid) x+a=5$ has two distinct real roots, then $a$ can be ( .
(A) 5
(B) 9
(C) 10
(D) 11 | 2. A
The original equation is $(a+1) x^{2}+(|a+2|-|a-10|) x+a-5=0$.
According to the problem, we have
$$
\Delta=(|a+2|-|a-10|)^{2}-4(a+1)(a-5)>0 \text {, }
$$
which simplifies to $6 a>a^{2}+|a+2| \cdot|a-10|$.
When $a$ takes the values $9,10,11$, none of them satisfy (*) equation;
When $a=5$, it satisfies (*) equatio... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,281 |
3. As shown in the figure, $G$ is the centroid of $\triangle A B C$. From the figure, it can be deduced that there are ( ) pairs of triangles with equal area.
(A) 15
(B) 18
(C) 30
(D) 33 | 3. D
If any triangle (regardless of size) in the figure can be regarded as a point on a line, then the solution can be divided into the following three steps:
(1) There are six triangles with an area of $\frac{1}{2} S_{\triangle A B C}$: $\triangle A B D, \triangle A D C, \triangle A F C, \triangle B F C, \triangle C ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,282 |
4. Given that $a, b, c$ are the three sides of a right triangle $\triangle A B C$, where $c$ is the hypotenuse. For $0<m<a+b-c$. If the sides $a, b, c$ are each shortened by the same length $m$, then the triangle formed by the segments $a-m, b-$ $m, c-m$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle... | 4. C
$$
\begin{array}{l}
\because(a-m)+(b-m)-(c-m) \\
\quad=a+b-c-m>0(\because m(c-m) .
\end{array}
$$
Similarly, $(a-m)+(c-m)>(\dot{b}-m)$,
$$
(b-m)+(c-m)>(a-m) .
$$
$\therefore$ A triangle with sides $a \cdots m, \dot{b}-m, c-m$ exists. Let such a triangle be $\triangle A^{\prime} B^{\prime} C^{\prime}$, with $c-m$ ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,283 |
5. Let $a, b, c$ be positive numbers, and $3^{a}=4^{b}=6^{c}$, then ( ).
(A) $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$
(B) $\frac{2}{c}=\frac{2}{a}+\frac{1}{b}$
(C) $\frac{1}{c}=\frac{2}{a}+\frac{2}{b}$
(D) $\frac{2}{c}=\frac{1}{a}+\frac{2}{b}$ | 5. B
Let $3^{a}=4^{b}=6^{c}=k$, then $a=\log _{k} 3, b=\log _{4} k, c=$ $\log _{6} k$, at this time
$$
\begin{array}{l}
\frac{2}{a}+\frac{1}{b}=2 \log _{k} 3+\log _{k} 4=\log _{k} 36, \\
\frac{2}{c}=2 \log _{k} 6=\log _{k} 36, \\
\therefore \frac{2}{c}=\frac{2}{a}+\frac{1}{b} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,284 |
6. Given the quadratic function $y=a x^{2}+b x+c(a \neq 0)$, where $a, b, c$ simultaneously satisfy the following conditions:
(1) $x-2$ is a factor of the left side of the equation $a x^{2}+b x-c=0$;
(2) The remainder when $a x^{2}+b x+c$ is divided by $x+2$ is -4;
(3) The difference between $c x^{2}+b x+c$ and -6 can ... | 6. D
According to the problem, using the root value method and the remainder theorem, we can obtain the system of equations
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 4 a + 2 b + c = 0 , } \\
{ 4 a - 2 b + c + 4 = 0 , } \\
{ a - b + c + 6 = 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
a=1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,285 |
3. $n$ is a natural number, $19 n+14$ and $10 n+3$ are both multiples of some natural number $d$ not equal to 1, then $d=$ $\qquad$ . | 3. 83.
Since $\frac{19 n+14}{10 n+3}=1+\frac{9 n+11}{10 n+3}$,
and $\frac{10 n+3}{9 n+11}=1+\frac{n-8}{9 n+11}$,
also $\frac{9 n+11}{n-8}=9+\frac{83}{n-8}$,
since 83 is a prime number, if 83 and $n-8$ are coprime, then $n=91$.
When $n=91$,
$$
\begin{array}{l}
19 n+14=19 \times 91+14=83 \times 21, \\
10 n+3=10 \times 9... | 83 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,288 |
4. As shown in the figure, $AB$ is the diameter of the semicircle $\odot O$, $CD \perp AB$. Let $\angle COD = \theta$, then $\frac{AD}{BD}$ - $\tan^2 \frac{\theta}{2}=$ $\qquad$ | 4. 1.
Connect $A C$, then $\angle C A D=\frac{1}{2} \angle \theta$, and $\operatorname{tg} \frac{\theta}{2}=\operatorname{tg} \angle C A D=\frac{C D}{A D}$.
According to the projection theorem, we know
$$
\begin{array}{l}
C D^{2}=A D \cdot B D . \\
\therefore \frac{A D}{B D} \operatorname{tg}^{2} \frac{\theta}{2}=\fra... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,290 |
One. (20 points) On a plane, there are two regular hexagons $A B C D E F$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ with equal side lengths, and the vertex $A^{\prime}$ of the regular hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ is at the center of the regula... | As shown in the figure, the area of the overlapping part $S_{A^{\prime} G B C H}$ is a constant. Connecting $A^{\prime} B$ and $A^{\prime} D$, since $A^{\prime}$ is the center of the regular hexagon $A B C D E F$, we have $A^{\prime} B=A^{\prime} D=A B$,
$$
\begin{array}{l}
\angle A^{\prime} B G=\angle A^{\prime} D H \... | \frac{\sqrt{3}}{2}(B C)^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 707,291 |
2. Let $a, d$ be non-negative real numbers, $b, c$ be positive real numbers, and $b+c \geqslant a+d$. Then the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\sqrt{2}-\frac{1}{2}$
(D) $\sqrt{2}$ | 2. C
Given $b+c \geqslant a+d$, then $b+c \geqslant \frac{1}{2}(a+b+c+d)$. Without loss of generality, assume $a+b \geqslant c+d$, then $\frac{1}{c+d} \geqslant \frac{1}{a+b}$. Therefore,
$$
\begin{array}{l}
\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
\geqslant \frac{1}{2} ... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,295 |
3. Let $a>b>0, A=\left(a^{\frac{2}{3}}+b^{\frac{3}{2}}\right)^{\frac{2}{3}}, B=\left(a^{\frac{2}{3}}$ $+b^{\frac{2}{3}}\right)^{\frac{3}{2}}, C=\left(a^{\frac{3}{2}}+a^{\frac{5}{6}} \cdot b^{\frac{2}{3}}\right)^{\frac{2}{3}}$. Then the size relationship of $A, B, C$ is ( ).
(A) $A<B<C$
(B) $B<C<A$
(C) $C<A<B$
(D) $A<C<... | 3. D
Given $A=a\left[1+\left(\frac{b}{a}\right)^{\frac{3}{2}}\right]^{\frac{2}{3}}, B=a\left[1+\left(\frac{b}{a}\right)^{\frac{2}{3}}\right]^{\frac{3}{2}}, C$ $=a\left[1+\left(\frac{b}{a}\right)^{\frac{2}{3}}\right]^{\frac{2}{3}}$, and $\left(\frac{b}{a}\right)^{\frac{3}{2}}<\left(\frac{b}{a}\right)^{\frac{2}{3}}$, th... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,296 |
4. In the tetrahedron $A-BCD$, $\angle BAC = \angle CAD = \angle DAB = 60^{\circ}, AB = 1, AC = 2, AD = 3$. Then the volume of the tetrahedron is ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{\sqrt{6}}{3}$
(D) Other | 4. A
Let $A C^{\prime}=A D^{\prime}=1$, then $A-B C^{\prime} D^{\prime}$ is a regular tetrahedron, with height $h=\frac{\sqrt{6}}{3}$. That is, the distance from point $B$ to plane $A C D$ is $h=\frac{\sqrt{6}}{3}$. Also, $S_{\triangle A C D}=\frac{3}{2} \sqrt{3}$, thus
$$
V_{A-B C D}=V_{B-A C D}=\frac{1}{3} S_{\trian... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,297 |
5. The number of different four-digit numbers that can be formed using the eight digits $1,1,2,2,3,3,4,4$ is ( ).
(A) 204
(B) 144
(C) 72
(D) 24 | 5. A
There are three cases:
(1) Four-digit numbers with no repeated digits: $P_{4}^{4}=24$ (ways).
(2) One pair of digits is the same: $C_{4}^{1} \cdot C_{3}^{2} \cdot C_{4}^{2} \cdot P_{2}^{2}=144$.
(3) Two pairs of digits are the same: $C_{4}^{2} \cdot C_{4}^{2}=36$. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,298 |
6. The integer solutions $(x, y)$ of the equation $\frac{1}{y}+\frac{1}{2(x+y)}=\frac{1}{y^{2}}$ are ( ).
(A) 0 sets
(B) 1 set
(C) finite sets (more than 1 set)
(D) infinitely many sets | 6. B
The equation is transformed into: $(x+2 y)^{2}=(x+y)[(x+y)+2]$.
(1) When the left side is 0, one of the factors on the right side must be 0. Solving yields $x=-4, y=2$ and $x=0, y=0$ (the extraneous solution is discarded).
(2) When the left side is not 0, it is a perfect square. However, on the right side, when $x... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,299 |
Example 1. In $\triangle A B C$, $B C=3, A C=4, A E$ and $B D$ are the medians to sides $B C$ and $A C$ respectively, and $A E \perp B D$. Then the length of $A B$ is ( ).
(A) $\sqrt{5}$
(B) $3 \sqrt{6}$
(C) $2 \sqrt{2}$
(D) $\sqrt{7}$
(E) cannot be determined | Given that $AE$ and $BD$ are medians of $\triangle ABC$, and $AE \perp BD$, according to property $3^{\circ}$, we know
$$
AC^{2} + BC^{2} = 5 AB^{2} \text{.}
$$
Therefore, $AB = \sqrt{\frac{AC^{2} + BC^{2}}{5}} = \sqrt{5}$. Hence, the correct choice is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,300 |
Example 1. Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$. Prove that: $yz + zx + xy - 2xyz \leq \frac{7}{27}$. (1984, IMO-1) | It is easy to see that when $x=y=z=\frac{1}{3}$, the equality in the inequality to be proven holds. By symmetry, we can assume $x \geqslant y \geqslant z$, thus, $x \geqslant \frac{1}{3} \geqslant z$. Let
$$
x^{\prime}=\frac{1}{3}, y^{\prime}=y, z^{\prime}=x+z-\frac{1}{3} .
$$
Then, $x^{\prime}+z^{\prime}=x+z, x \cdot... | \frac{7}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 707,302 |
In $\triangle ABC$, $\angle BAC=45^{\circ}$, $AD$, $BE$, and $CF$ are three altitudes. Prove that:
$$
DE + DF \leqslant BC.
$$ | Proof: As shown, connect $E F$. It is easy to prove that $B, C, E, F$ are concyclic.
$$
\left.\begin{array}{c}
\therefore \angle A E F=\angle A B C, \\
\text { Also } \angle E A F=\angle B A C
\end{array}\right\} \Rightarrow \triangle A E F \backsim \triangle A B C \Rightarrow
$$
Obviously, $\angle E D F=90^{\circ}$,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,303 |
21. On an $n \times n$ grid $(n>1)$, place $2n$ chess pieces, with each piece placed at the center of a cell. Prove: there must be four pieces that are located at the four vertices of a parallelogram. | Proof: For the convenience of narration, let's assume that the first $m$ rows of the chessboard each have at least two pieces, while the remaining $n-m$ rows each have at most one piece. Then the number of pieces in the first $m$ rows $\geqslant 2 n-(n-m)=n+m$.
Let the $i$-th row have $k_{i}$ pieces $(1 \leqslant i \l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,304 |
"Fragment numbers" (order unchanged), for example, the "fragment numbers" of 186 are $1,8,6,18$, 36, a total of 5. Find the $n$-digit natural numbers that satisfy the following conditions respectively.
(1) It is a perfect square, and all its "fragment numbers" are perfect squares;
(2) It is a prime number, and all its ... | Solution (1) All non-fog digits of $N$ are one-digit "fragment numbers" of $N$, which can only be $1,4,9$.
$N$ cannot contain the digit 0, otherwise there might be two-digit "fragment numbers" such as $10,40,90$ that are not perfect squares.
$N$ cannot contain the digit 1, otherwise there might be two-digit "fragment n... | 49, 23, 37, 53, 73, 373 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,305 |
Example 2. Let $A, B, C$ be the three interior angles of a triangle. Find the maximum value of $\sin A+\sin B+\sin C$.
untranslated text remains the same as requested, only the problem statement is translated. | From the trigonometric formula, we have
$$
\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}.
$$
It follows that, under the condition that $C$ is fixed, $\sin A+\sin B+\sin C$ achieves its maximum value when $A=B$. Therefore, it can be conjectured that the maximum value is obtained when $A=B=C$.
By symmetry, we c... | \frac{3 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,306 |
Example 3 For a quadratic trinomial $f(x)=a x^{2}+b x$ $+c$ with all coefficients being positive, and $a+b+c=1$. Prove that for any set of positive numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying $x_{1} x_{2} \cdots x_{n}=1$, we have
$$
f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right) \geqslant 1 .... | Obviously, $f(1)=1$. If $x_{1}=x_{2}=\cdots=x_{n}=1$, the equality in inequality (1) holds.
If $x_{1}, x_{2}, \cdots, x_{n}$ are not all equal, then there must be $x_{i}>1, x_{j}<1$.
\end{aligned}
$$
Note: In each term of the above expression, the factors in the two parentheses have opposite signs. It can be seen tha... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,307 |
Bi 4. For any real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the inequality $x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$. Find the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$. | When $x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}$ are all constants, from the relation
$$
\begin{array}{l}
\left(1-x_{n-1}\right)\left(1-x_{n}\right) \\
=1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n}
\end{array}
$$
it is evident that the larger $\left|x_{n-1}-x_{n}\right|$, the smaller the value of the above expression... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,308 |
Example 5. Given that $\theta_{1}, \theta_{2}, \cdots, \theta_{n}$ are all non-negative, and $\theta_{1}+\theta_{2}+\cdots+\theta_{n}=\pi$. Find the maximum value of $\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{n}$.
(1985, IMO Candidate Problem 12) | First, consider the case where $\theta_{1}+\theta_{2}$ is a constant. In this case,
$$
\begin{array}{l}
\sin ^{2} \theta_{1}+\sin ^{2} \theta_{2} \\
=\left(\sin \theta_{1}+\sin \theta_{2}\right)^{2}-2 \sin \theta_{1} \sin \theta_{2} \\
= 4 \sin ^{2} \frac{\theta_{1}+\theta_{2}}{2} \cos ^{2} \frac{\theta_{1}-\theta_{2}... | \frac{9}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,309 |
Example 6. Mark $4 n$ points on a circle and alternately color them red and blue. Pair all blue points into $n$ pairs, and connect each pair with a blue line; do the same for the $2 n$ red points. Prove that there are at least $n$ pairs of red and blue line segments that intersect (i.e., in these $n$ pairs, each one ha... | Proof: Let the blue points be labeled as $1, 3, \cdots, 4n$ 1, and the red points as $2, 4, \cdots, 4n$. Connect a blue line between $2j-1$ and $4n-(2j-1)$, and a red line between $2j$ and $4n-2j$ (as shown in the figure, where solid lines represent blue lines and dashed lines represent red lines). Clearly, there are e... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,310 |
Example 7. Let $s_{1}, x_{2}, c_{3}, x_{4}$ all be positive real numbers, and $x_{1} + x_{2} + x_{3} - x_{4} = \pi$. Find the minimum value of the expression
\[
\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)
- \left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x... | Let $x_{1}+x_{2}$ be a constant. Since
$$
\begin{array}{l}
\sin x_{1} \sin x_{2} \\
=\frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right],
\end{array}
$$
we know that the value of $\sin x_{1} \sin x_{2}$ increases as $\left|x_{1}-x_{2}\right|$ decreases. Let the expression in question be... | 81 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,311 |
Example 8. Given non-negative real numbers $p, q, r$ satisfying the condition $p^{2}+q^{2}+r^{2}=2$. Prove that $p+q+r-p q r \leqslant 2$. | $$
\begin{array}{l}
p+q+r-p q r \\
=p+q-\frac{1}{2}(p+q)^{2} r+\frac{1}{2}\left(p^{2}+q^{2}\right) r+r \\
=-\frac{r}{2}\left(p+q-\frac{1}{r}\right)^{2}+\frac{1}{2 r}+2 r-\frac{r^{3}}{2} \\
\leqslant \frac{1}{2 r}-2 r-\frac{r^{3}}{2} \\
=\frac{1}{2 r}\left(-r^{4}+4 r^{2}+1-4 r\right)+2 \\
=-\frac{1}{2 r}(r-1)^{2}\left(r... | p+q+r-p q r \leqslant 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 707,312 |
Example 1. On the three sides of $\triangle A B C$, there are $l, m, n$ points respectively. Lines are drawn from each vertex to the points on its opposite side. If no three of these lines intersect at the same point, how many regions do they divide $\triangle A B C$ into? | This problem is relatively simple. First, draw $l$ lines from $A$, obtaining $l+1$ regions. Consider any line drawn from $B$, which is divided into $l+1$ segments by the $l$ lines drawn from $A$, thus increasing the number of regions by $l+1$. If $m$ lines are drawn from $B$, the number of regions increases by $m(l+1)$... | l m+m n+n l+l+m+n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,313 |
Example 2. On the sides $AB$, $CD$ of square $ABCD$, there are $m$, $n$ points respectively. Lines are drawn between every two points, with no three lines intersecting at the same point. How many regions do these lines divide the square into? | This problem can also be solved following the approach of Example 1, but the process is more complicated and requires certain skills. As shown in the figure, let the $m$ points on side $AB$ from top to bottom be $A_{1}, A_{2}, \cdots, A_{m}$, and the $n$ points on side $CD$ from bottom to top be $B_{0}, B_{1}, \cdots, ... | C_{m}^{2} C_{n}^{2} + m n + 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,314 |
Point, point $E, F$ are the centroids of $\triangle A B D$ and $\triangle A C D$ respectively, connecting $E, F$ intersects $A D$ at point $G$. What is the value of $\frac{D G}{G A}$? (1991-1992 Guangzhou, Luoyang, Fuzhou, Wuhan, Chongqing Junior High School League) | Connect $B E, C F$, and extend them to intersect at $M$. According to property $5^{\circ}$, $M$ is the midpoint of $A D$.
$\because E, F$ are the centroids of
$\triangle A B D$ and $\triangle A C D$,
$$
\therefore \frac{M E}{M B}=\frac{M F}{M C}=\frac{1}{3} \text {. }
$$
Thus, $E F / / B C, E G / / B D$.
Therefore, $\... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,315 |
Example 1. On a plane, there are ? straight lines, any two of which intersect, and no three lines intersect at the same point. How many regions will these lines divide the plane into? | Solution: Obviously, the number of intersection points of $n$ lines is $p_{n}=C_{n}^{2}$, so the number of regions they divide the plane into is
$$
S_{n}=C_{n}^{2}+n+1=\frac{1}{2}\left(n^{2}+n+2\right) .
$$ | \frac{1}{2}\left(n^{2}+n+2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,316 |
Example 2. On a plane, there are $n$ lines, among which $k$ lines are pairwise parallel, and $t$ lines pass through the same point $(k+t<n)$, in addition, any two intersect, and any 3 do not share a point. How many regions do they divide the plane into: | The number of intersection points is given by $p_{n}=C_{n}^{2}-C_{k}^{2}-C_{t}^{2}+(t-1)$. According to Theorem 1, the number of regions they divide the plane into is
$$
S_{n}=C_{n}^{2}-C_{1}^{2}-C_{t}^{2}+n+t.
$$
The figure below gives an example F for $n=8, k=4, t=3$. In this case, $S_{8}=C_{8}^{2}-C_{4}^{2}-C_{3}^{... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,317 |
Example 3. On the sides $AB, BC, CD, DA$ of rectangle $ABCD$, there are $n, p, n, q$ points respectively, and lines are drawn connecting every two points on opposite sides. If no three lines intersect at a point inside the rectangle, how many regions do these lines divide the rectangle $ABCD$ into? | Consider the intersection points of these lines within the rectangle, they can be divided into 3 categories: one category is the intersection points of lines connecting points on the edges $A B$ and $C D$. Each such intersection point corresponds to two points on $A B$ and two points on $C D$, and vice versa. There are... | S = C_{m}^{2} C_{n}^{2} + C_{p}^{2} C_{q}^{2} + m n p q + m n + p q + 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,318 |
Example 4. Connecting every pair of $n$ points on a circle, with no three lines intersecting at the same point inside the circle, how many regions do they divide the circle into?
| Consider the intersection points of these lines inside the circle. Each such intersection point corresponds to four points on the circumference, and vice versa, so the number of intersection points is $C_{n}^{4}$. It is also clear that the number of lines is $C_{n}^{2}$, so the number of regions they divide the circle ... | S=C_{n}^{4}+C_{n}^{2}+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,319 |
Example 5. $n$ circles on a plane, any two of which intersect at two points, and no three circles pass through the same point, divide the plane into how many regions?
---
The translation maintains the original text's format and line breaks as requested. | Solve for the number of intersection points $p=2 C_{n}^{2}$, thus the number of regions the plane is divided into is $S=2 C_{n}^{2}+2=n^{2}-n+2$. | n^{2}-n+2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,320 |
Example б. On a plane, there are $n$ ellipses, any two of which intersect at 4 points, and no three ellipses pass through the same point. How many regions does this divide the plane into? | Solve the number of regions $S=4 C_{n}^{2}+2=2 n^{2}$ $-2 n+2$ | 2 n^{2} - 2 n + 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,321 |
Example 1. For what integer values of $x$ is $f(x)=x^{2}+5 x-20$ a perfect square? | Solution: Since $\Delta=5^{2}-4 \times(-20)=3 \times 5 \times$ 7, $\Delta$ has the following eight decompositions $p_{2} p_{1}\left(p_{1} \geqslant p_{2}\right): 7$ $\times 15,5 \times 21,3 \times 35,1 \times 105,(-7) \times(-15)$, $(-5) \times(-21),(-3) \times(-35),(-1) \times$ (-105). According to the conclusion 1 of... | x=3,4,7,24,-8,-9,-12,-29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,322 |
Consider a biased coin, the probability of landing heads up is $\frac{2}{3}$. If this coin is flipped 50 times, then the probability that the number of heads is even is (․).
(A) $25\left(\frac{2}{3}\right)^{50}$
(B) $\frac{1}{2}\left(i-\frac{i}{3^{50}}\right)$
(C) $\frac{1}{2}$
(I) $\frac{3}{2}\left(1+\frac{1}{3^{50}}\... | The required probability is the sum of the even powers of $\frac{2}{3}$ in the expansion of $\left(\frac{2}{3}+\frac{1}{3}\right)^{50}$. By the above corollary, this is known to be $\frac{1}{2}\left(1+\frac{1}{3^{50}}\right)$. | \frac{1}{2}\left(1+\frac{1}{3^{50}}\right) | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,325 |
Example 4. In $\triangle ABC$, $G$ is the centroid, and $P$ is a point inside the triangle. The line $PG$ intersects the lines $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively. Prove that: $\frac{A'P}{A'G} + \frac{B'P}{B'G} + \frac{C'P}{C'G} = 3$
(1991, Huanggang Region Junior High School Competition) | Prove that by connecting $B G, G C, P B, P C$, and drawing $G G^{\prime} \perp B C$ at $G^{\prime}$, $P P^{\prime} \perp B C$ at $P^{\prime}$, then
$$
\begin{array}{c}
P P^{\prime} / / C G^{\prime}, \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P^{2}}{A^{\prime} G} . \\
\text { Also, } \frac{S_{\triangle P B C}}{S... | 3 | Geometry | proof | Yes | Yes | cn_contest | false | 707,326 |
$\mathrm{IMO10.3}$ problem and its extension from [1], consider the system ( $n$ $\geqslant m$,
$$
\text { (*) }\left\{\begin{array}{l}
f_{1}\left(x_{1}\right)+f_{2}\left(x_{2}\right)+\cdots+f_{m}\left(x_{m}\right)=0, \\
f_{1}\left(x_{2}\right)+f_{2}\left(x_{3}\right)+\cdots+f_{m}\left(x_{m+1}\right)=0, \\
\cdots \cdot... | To prove: If $\xi$ is a zero of $S(x)$, then $(\xi, \xi, \cdots, \xi)$ is a solution of (*), hence iii is correct. Adding all the equations in (*) yields:
$$
\begin{array}{l}
\sum_{i=1}^{n}\left[f_{1}\left(x_{i}\right)+\cdots+f_{m}\left(x_{i}\right)\right] \\
=\sum_{i=1}^{n} S\left(x_{i}\right)=0 .
\end{array}
$$
When... | (3, \cdots, 3) | Algebra | proof | Yes | Yes | cn_contest | false | 707,327 |
Proposition: Let line segments $A B$ and $C D$ be non-collinear, and $P$ be a fixed point on the same plane. Then, the locus of point $P$ that makes $S_{\triangle P A B} + S_{\triangle P C D}$ a constant is a parallelogram. | Let $AB = a, CD = b$, and take the intersection point $O$ of lines $AB$ and $CD$ as the origin, with $AB$ as the $x$-axis, to establish a coordinate system (as shown in the figure). The inclination angle of line $CD$ is $\theta (0 < \theta < \pi)$, so its equation is $x \sin \theta - y \cos \theta = 0$. It is easy to s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,328 |
Let the volume of a tetrahedron be $V, S_{i}, i=1, \cdots, 4, a_{j}$ be its areas and edges, respectively, then (the summation limits are omitted below):
(1) $\sum S_{i}^{2} \geqslant 9 \sqrt[3]{3} V^{\frac{4}{3}}+\frac{1}{2} Q$.
(Regular tetrahedron)
(2) $\sum a_{j}^{2} \geqslant 12 \sqrt[3]{9} V^{\frac{2}{3}}+\frac{1... | Prove (1) $\sum S_{i}^{2}-\frac{1}{2} \alpha$
$$
\begin{array}{l}
=\sum S_{i}^{2}-\left(\sum S_{i}^{2}-\sum S_{i} S_{i+1}\right) \\
=\sum S_{i} S_{i+1} \geqslant 4\left(\prod S_{i}\right)^{\frac{1}{2}} \\
\geqslant 4 \cdot \frac{9 \sqrt[3]{3}}{2^{2}} V^{\frac{4}{3}} . \text { Q.E.D. }
\end{array}
$$
The last step uses... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,329 |
Example 1. Find a point $M$ on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ such that the sum of its distances to the fixed points $A(\sqrt{7}, 0), B(1,2)$, i.e., $|M A|+|M B|$, is maximized or minimized. | Solve $A(\sqrt{7}, 0)$ and $A^{\prime}(-\sqrt{7}, 0)$ as two foci (as shown in Figure 1). Connect $A^{\prime} B$ intersecting the ellipse at $M_{1}$ and $M^{\prime}, \quad$ then $\left|A M_{1}\right|+$ $\left|B M_{1}\right|=8-$ $\sqrt{12-2 \sqrt{7}}$ is the minimum value, while $\left|A M^{\prime}\right|+\left|B M^{\pr... | 8 \pm \sqrt{12-2 \sqrt{7}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,330 |
Example 2. Find a point $M$ on the hyperbola $\frac{x^{2}}{3} -\frac{y^{2}}{1}=1$ such that the sum of its distances to points $A(2,0)$ and $B$ $(8,2 \sqrt{2})$ is minimized. | Solve: As shown in Figure 2, $A(2,0), A^{\prime}(-2,0)$ are the foci. Connect $A^{\prime} B$, intersecting the hyperbola (right branch) at $M$, then $M(3$, $\sqrt{2}$ ) is the point required, and the minimum value is $4 \sqrt{3}$. | 4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,331 |
1. Express 15 as the sum of two or more consecutive natural numbers, the number of different ways to do this (regardless of the order of the addends) is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,332 |
2. The marked price of a certain commodity is 13200 yuan. If it is sold at ?% discount, a profit of $1.0 \%$ (for raw goods) can still be made.
(A) 10800
(B) 10560
(C) 10692
(D) 11880 | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,333 |
3. Given the following shapes: (1) equilateral triangle; (2) rectangle; (3) rhombus; (1) circle. The shape(s) that have exactly two lines of symmetry are ( ).
(A)(1)(2) (B)(1)(2)(3)
(C)(2)(3) (D) (2)(3)(4) | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,334 |
4. As shown in the figure, given that $M, N$, $K$ are three points on the circle $\odot O$ with radius $R$, $\angle M O N=120^{\circ}$, and $K$ is the midpoint of $\overparen{M N}$, then the area of the shaded part is ( ).
(A) $\frac{1}{3} \pi R^{2}$
(B) $\frac{1}{4} \pi R^{2}$
(C) $\frac{5}{24} \pi R^{2}$
(D) $\frac{1... | 4. 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 | 707,335 |
5. A family of three (father, mother, and son) is preparing to travel. Travel Agency A says: “If the father buys a full-price ticket, the others can enjoy a 70% discount.” Travel Agency B says, “For family travel, group ticket pricing applies, which is charged at $\frac{4}{5}$ of the original price.” If the prices of t... | 5. C
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 | 707,336 |
Example 5. As shown in the figure, $P$ is a point inside an equilateral triangle with side length 1, then at least one of $P A, P B, P C$ has a length not exceeding $\frac{\sqrt{3}}{3}$. (4th Junior High School Competition in Jiangsu Province) | Proof: Let $G$ be the centroid of $\triangle ABC$, then $AG = BG = CG = \frac{\sqrt{3}}{3}$. Circles are drawn with $A, B, C$ as centers, passing through point $G$, dividing $\triangle ABC$ into six regions as shown in the figure.
(1) When $P$ is in region I, it is obvious that $PA \leqslant \frac{\sqrt{3}}{3}$. Simila... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,337 |
6. Given $a_{1}>a_{2}>a_{3}, b_{1}>b_{2}>b_{3}$. Among the following expressions, the one with the largest value is ()
(A) $a_{2} b_{3}+a_{2} b_{1}+a_{3} b_{1}$
(B) $a_{1} \dot{b}_{1}+a_{2} b_{2}+a_{2} b_{3}$
(C) $a_{1} b_{2}+a_{2} b_{3}+a_{3} b_{1}$
(D) $a_{2} b_{1}+a_{3} b_{2}+a_{1} b_{3}$ | 6. 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 | 707,338 |
7. In the plane, there are three fixed non-collinear points $A, B, C$, and a line $l$ satisfies the condition: the distances from $A, B$ to $l$ are equal, and each is half the distance from $C$ to $l$. Then the number of such lines $l$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 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 | 707,339 |
8. Cut a 100 cm long wire into $n$ small segments $(n \geqslant 3)$, each segment being no less than 10 cm. If for any cutting method, there always exist 3 small segments that can form a triangle, then the minimum value of $n$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 8. C
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 | 707,340 |
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