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 |
|---|---|---|---|---|---|---|---|---|---|
II. (25 points) As shown in Figure 4, given that $\odot O$ is tangent to the sides $AB$ and $AC$ of $\triangle ABC$ at points $P$ and $Q$, respectively, and is tangent to the circumcircle of $\triangle ABC$ at point $T$. Let the midpoint of the chord $PQ$ be $I$. Prove that $IT$ bisects $\angle B T C$ | II. As shown in Figure 8, connect $O A$. From the given conditions, it is easy to see that $O A$ bisects $\angle B A C$, and $O A$ perpendicularly bisects $P Q$. Therefore, $O$, $I$, and $A$ are collinear.
Connect $O P$, $O T$, and $A T$, then
$$
O P^{2}=O I \cdot O A=O T^{2} \text {. }
$$
Thus,
$\triangle O T I \back... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,894 |
Three. (25 points) It is known that the unit digit of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{8}^{3}-x_{9}^{3}$ is 1, where $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from 2001, 2002, ..., 2009, and $8 x_{9}>$ $x_{1}+x_{2}+\cdots+x_{8}$. Find the value of $x_{9}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出... | Three, because $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from $2001, 2002, \cdots, 2009$, and the unit digits of $2001, 2002, \cdots, 2009$ are $1, 2, 3, 4, 5, 6, 7, 8, 9$. After cubing these unit digits, the resulting unit digits are $1, 8, 7, 4, 5, 6, 3, 2, 9$. Therefore, the unit digit of $x_{1}^{3} +... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,895 |
1. Given $a^{2}+b^{2}=1, b^{2}+c^{2}=2, c^{2}+a^{2}=$ 2. Then the minimum value of $a b+b c+c a$ is ( ).
(A) $\sqrt{3}-\frac{1}{2}$
(B) $-\sqrt{3}+\frac{1}{2}$
(C) $-\sqrt{3}-\frac{1}{2}$
(D) $\sqrt{3}+\frac{1}{2}$ | -、1.B.
From the problem, we have
$$
c= \pm \sqrt{\frac{3}{2}}, a= \pm \sqrt{\frac{1}{2}}, b= \pm \sqrt{\frac{1}{2}} .
$$
Notice that
$$
a b+b c+c a=\frac{(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2},
$$
so we only need to consider the minimum value of $|a+b+c|$.
To make $|a+b+c|$ as small as possible, we can take
$... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,896 |
2. A math test consists of 20 questions. The scoring rules are: 5 points for each correct answer, 0 points for no answer, and -2 points for each wrong answer. It is known that in this test, the total scores of Xiaoqiang and Xiaogang are equal, and the score is a prime number. Then the situation of Xiaoqiang and Xiaogan... | 2.D.
According to the problem, by enumerating the possibilities of answering 20 questions, 19 questions, ... correctly, we find:
(1) Xiaoqiang and Xiaogang may both answer 17 questions correctly, 1 question incorrectly, and leave the remaining 2 questions unanswered, scoring 83 points each;
(2) Xiaogang and Xiaoqiang ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,897 |
3. In $\triangle A B C$, $A D$ is the median on side $B C$, points $M$ and $N$ are on sides $A B$ and $A C$ respectively, and satisfy $\angle M D N = 90^{\circ}$. If $B M^{2} + C N^{2} = D M^{2} + D N^{2}$, then the relationship between $A D^{2}$ and $A B^{2} + A C^{2}$ is ( ).
(A) $A D^{2} > A B^{2} + A C^{2}$
(B) $A ... | 3. B.
As shown in Figure 5, draw a line through point $B$ parallel to $A C$ and let it intersect the extension of $N D$ at point $E$. Connect $M E$. Since $B D = D C$, we know $E D = D N$, thus $\triangle B E D \cong \triangle C N D$. Therefore, $B E = C N$. Clearly, $M D$ is the perpendicular bisector of $E N$, so
$$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,898 |
4. There are $n$ numbers, starting from the second number, each number is 3 more than the one immediately before it, i.e., $4, 7, \cdots, 3n + 1$, and the product of these numbers ends with exactly 32 zeros. Then the minimum value of $n$ is ( ).
(A) 125
(B) 126
(C) 127
(D) 128 | 4.D.
Since $(1+3 n) \div 5=2+3(n-3) \div 5$, therefore, among these $n$ numbers, only the 3rd, 8th, 13th, 18th, ... numbers are multiples of 5, and they are $5 \times 2, 5 \times 5, 5 \times 8, 5 \times 11, \cdots$. Among them, exactly 1 out of every 5 is a multiple of 25, and exactly 1 out of every 25 is a multiple o... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,899 |
5. Figure 1 is a simplified model of a traffic roundabout at a three-way intersection. During a certain peak period, the number of motor vehicles entering and exiting intersections $A, B$, and $C$ per unit time is shown in Figure 1. The $x_{1}, x_{2}$, and $x_{3}$ in the figure represent the number of motor vehicles pa... | 5.C.
According to the problem, we have $x_{1}=50+x_{3}-55=x_{3}-5$, so $x_{1}<x_{3}$. Similarly, $x_{1}<x_{2}, x_{3}<x_{2}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,900 |
2. When $x=\frac{\sqrt{29}-3}{2}$, the value of the algebraic expression
$$
x^{4}+5 x^{3}-3 x^{2}-8 x+9
$$
is $\qquad$. | $2.7 \sqrt{29}-32$.
Since $x=\frac{\sqrt{29}-3}{2}$ is a root of the equation $x^{2}+3 x-5=0$, then
$$
\begin{array}{l}
x^{4}+5 x^{3}-3 x^{2}-8 x+9 \\
=\left(x^{2}+3 x-5\right)\left(x^{2}+2 x-4\right)+14 x-11 \\
=14 x-11=14 \times \frac{\sqrt{29}-3}{2}-11=7 \sqrt{29}-32 .
\end{array}
$$ | 7 \sqrt{29}-32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,901 |
6. Given four distinct real numbers $x_{1}, x_{2}, x_{3}, x_{4} \left(x_{1}<x_{2}, x_{3}<x_{4}\right)$. Also, $a$ is a real number, the function $y_{1}=$ $x^{2}-4 x+a$ intersects the $x$-axis at points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$, and the function $y_{2}=x^{2}+a x-4$ intersects the $x$-axis at p... | 6. C.
$$
\begin{array}{l}
x_{1}<x_{2}<x_{3}<x_{4}, x_{1}<x_{3}<x_{2}<x_{4}, x_{1}<x_{3}<x_{4}<x_{2}, \\
x_{3}<x_{4}<x_{1}<x_{2}, x_{3}<x_{1}<x_{4}<x_{2}, x_{3}<x_{1}<x_{2}<x_{4} .
\end{array}
$$
Among the above 6 cases, the 3rd and 6th cases cannot occur (otherwise, the axes of symmetry of the two functions would be t... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,902 |
1. As shown in Figure $2, A D$
$/ / B C$, trapezoid $A B C D$
has an area of $180, E$ is
the midpoint of $A B, F$ is a point
on side $B C$, and $A F$
$/ / D C, A F$ intersects $E D$ and $B D$ at points $G$ and $H$. Let $\frac{B C}{A D}=$ $m(m \in \mathbf{N})$. If the area of $\triangle G H D$ is an integer, then the va... | 2, 1.2 or 5.
As shown in Figure 6, draw $B K / / A F$ intersecting $E D$ at point $K$, then $\triangle K E B \cong \triangle G E A$
$$
\begin{array}{l}
\text { Hence } \frac{G H}{A G}=\frac{G H}{B K} \\
=\frac{H D}{B D}=\frac{F C}{B C} \\
=\frac{A D}{B C}=\frac{1}{m} .
\end{array}
$$
Thus, we have $S_{\triangle A B D}... | 2, 1.2 \text{ or } 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,903 |
2. List the natural numbers $1, 2, \cdots, k^{2}$ in a square table (as shown in Table 1), then select any one number from the table, and subsequently remove the row and column containing that number. Then, perform the same operation on the remaining $(k-1)^{2}$ numbers in the square table, and continue this process $k... | 2. $\frac{1}{2} k\left(k^{2}+1\right)$.
Divide Table 1 into the following two number tables:
It is easy to see that each number in Table 1 equals the sum of the two numbers in the same position in the divided Table 2 and Table 3. Therefore, the sum of the $k$ selected numbers, which are neither in the same row nor in ... | \frac{1}{2} k\left(k^{2}+1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,904 |
3. As shown in Figure 3, let $A B$ and $C D$ be two perpendicular chords of a circle with center $O$ and radius $r$, dividing the circle into four parts (each part may degenerate to a point) in clockwise order denoted as $X, Y, Z, W$. Then the maximum value of $\frac{S_{X}+S_{Z}}{S_{Y}+S_{W}}$ (where $S_{U}$ represents... | 3. $\frac{\pi+2}{\pi-2}$.
Let's assume the center of the circle falls in $Z$ as shown in Figure 7(a).
When the chord $AB$ moves upward, the shaded area in Figure 7(b) is greater than the unshaded area to its left, so $S_{X}+S_{Z}$ increases, while $S_{Y}+S_{W}$ decreases (note that the sum of the areas of $X$, $Y$, $Z... | \frac{\pi+2}{\pi-2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,905 |
4. A person rolls a die, adding up the numbers rolled each time, and stops if the total exceeds 20. Then, when he stops, the number he is most likely to have rolled is $\qquad$ | 4.21.
Consider the value obtained after the dice roll just before the one that exceeds 20 is $x$.
If $x=15$, then only rolling a 6 can result in 21;
If $x=16$, then rolling a 5 or 6 can result in 21 or 22, with each number having a probability of $\frac{1}{2}$;
If $x=17$, then rolling a 4, 5, or 6 can result in 21, 2... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,906 |
One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If eac... | (1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$
Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,907 |
II. (25 points) As shown in Figure 4, from a point $P$ outside a circle, draw two tangents $PA$ and $PB$ to the circle, with $A$ and $B$ being the points of tangency. Draw a secant line through point $P$ that intersects the circle at points $C$ and $D$. Draw a line through point $B$ parallel to $PA$ that intersects lin... | II. As shown in Figure 8, connect $B C$, $B A$, and $B D$. Therefore,
$$
\begin{array}{l}
\angle A B C=\angle P A C \\
=\angle E .
\end{array}
$$
Thus, $\triangle A B C \sim \triangle A E B$.
Hence, $\frac{B E}{B C}=\frac{A B}{A C}$, which means
$$
B E=\frac{A B \cdot B C}{A C} \text {. }
$$
Also, $\angle A B F=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,908 |
$$
\text { Three. (25 points) Let } 1 \leqslant a_{1}<a_{2}<\cdots<a_{n} \leqslant 21
$$
be $n$ arbitrary integers. If there are always 4 different numbers $a_{i}, a_{j}, a_{k}, a_{m}$ such that
$$
a_{i}+a_{m}=a_{j}+a_{k}(1 \leqslant i<j<k<m \leqslant n),
$$
then the order $n$ of the array $\left(a_{1}, a_{2}, \cdots... | (1) When $n=7$, $\{1,2,3,5,8,13,21\}$ does not meet the requirement, so $n=7$ is not a good number.
(2) It is only necessary to prove: for any 8 integers
$$
1 \leqslant a_{1}<a_{2}<\cdots<a_{8} \leqslant 21 \text {, }
$$
there are always 4 different numbers $a_{i}<a_{j}<a_{k}<a_{m}$ satisfying $a_{i}+$ $a_{m}=a_{j}+a_... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,909 |
1. Let $a_{1}, a_{2}, \cdots, a_{n}$ represent the digits from $0$ to $9$, and the $(n+1)$-digit number $\overline{a_{1} a_{2} \cdots a_{n} 2}$, when multiplied by 2, becomes $2 a_{1} a_{2} \cdots a_{n}$. Then the minimum value of $n$ is ( ).
(A) 15
(B) 16
(C) 17
(D) 18 | $-1 . C$.
Using vertical multiplication, as shown in Figure 1, work from right to left.
Figure $1^{\circ}$
Therefore, the minimum value of $n$ is 17. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,910 |
2. The clock points to a certain moment between 9 a.m. and 10 a.m., and at this moment, the hour hand 2 minutes earlier and the minute hand 2 minutes later are in a straight line (not considering the overlapping case). Then this moment is ( ).
(A) 9 a.m. $13 \frac{7}{11}$ minutes
(B) 9 a.m. 16 minutes
(C) 9 a.m. 12 min... | 2.D.
On the clock, there are 60 small divisions, starting from the mark 12, dividing the circumference into 60 equal parts. The minute hand moves 60 divisions per hour, while the hour hand moves 5 divisions per hour. The speed of the minute hand is 12 times that of the hour hand.
As shown in Figure 2, points $A$, $O$,... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 717,911 |
3. Given two sets of numbers, set $A$ is: $1,2, \cdots, 100$; set $B$ is: $1^{2}, 2^{2}, \cdots, 100^{2}$. For a number $x$ in set $A$, if there is a number $y$ in set $B$ such that $x+y$ is also a number in set $B$, then $x$ is called an "associated number". Therefore, the number of such associated numbers in set $A$ ... | 3.73.
Let $x+y=a^{2}, y=b^{2}$, then $1 \leqslant b<a \leqslant 100$.
And $x=a^{2}-b^{2}=(a+b)(a-b) \leqslant 100$, since $a+b$ and $a-b$ have the same parity, hence $a+b \geqslant(a-b)+2$.
(1) If $a-b=1$, then $a+b$ is odd, and $3 \leqslant a+b \leqslant$ 99. Thus, $a+b$ can take the values $3,5,7, \cdots, 99$, a tot... | 73 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,912 |
3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5 | 3. A.
As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$
Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$
Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$
Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Henc... | 9 | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,913 |
4. For each $x$, the function $y$ is the maximum of the functions
$$
y_{1}=2 x, y_{2}=x+3, y_{3}=-x+3
$$
Then the minimum value of the function $y$ is ( ).
(A)2
(B) 3
(C) 5
(D) 6 | 4. B.
As shown in Figure 4, the intersection point of the graphs of functions $y_{1}$ and $y_{2}$ is $(3,6)$, and the intersection point of the graphs of functions $y_{2}$ and $y_{3}$ is $(0,3)$.
From the problem and the graph, we have
$$
y=\left\{\begin{array}{ll}
-x+3, & x<0 ; \\
x+3, & 0 \leqslant x<3 \\
2 x, & x \... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,914 |
5. Given the quadratic function $y=a x^{2}+b x+c$ opens downwards, the vertex is located in the second quadrant, and it passes through the points $(1,0)$ and $(0,2)$. Then the range of values for $a$ is ( ).
(A) $a<0$
(B) $a<-1$
(C) $-2<a<0$
(D) $-1<a<0$ | 5.C.
From the given information, we have
$$
a0, a+b+c=0, c=2 \text {. }
$$
From $a+b+c=0$ and $c=2$, we get $b=-2-a$.
From $a-2$.
And $a0$, so the range of values for $a$ is $-2<a<0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,915 |
6. Let $G$ be the centroid of $\triangle ABC$, $r$ be the inradius of $\triangle ABC$, and the distances from point $G$ to sides $BC$, $CA$, and $AB$ be $GD$, $GE$, and $GF$ respectively. Let $s=\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}$. Then
(A) $s>\frac{3}{r}$
(B) $s=\frac{3}{r}$
(C) $s<\frac{3}{r}$
(D) Cannot be deter... | 6.B.
As shown in Figure 5, $A M$ is the median on $B C$, and $B G$, $C G$ are connected. Then
$$
\begin{array}{l}
S_{\triangle A C B}=S_{\triangle A C C} \\
=S_{\triangle B C C}=\frac{1}{3} S_{\triangle A B C} . \\
\text { Also, } S_{\triangle B C C}=\frac{1}{2} B C \cdot G D,
\end{array}
$$
Therefore,
$$
\frac{1}{G ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,916 |
1. Express the fraction $\frac{6}{n(n+1)(n+2)(n+3)}$ as the algebraic sum of four fractions with denominators $n$, $n+1$, $n+2$, and $n+3$, respectively. The result is $\qquad$. | \[
\begin{array}{l}
=1 \cdot \frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3} . \\
\frac{6}{n(n+1)(n+2)(n+3)} \\
=\frac{3}{n(n+3)}-\frac{3}{(n+1)(n+2)} \\
=\left(\frac{1}{n}-\frac{1}{n+3}\right)-\left(\frac{3}{n+1}-\frac{3}{n+2}\right) \\
=\frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3} .
\end{array}
\] | \frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,917 |
2. Given that $a$ and $b$ are positive numbers, and $2a + b = 2$. Then the minimum value of $\sqrt{4a^2 + 1} + \sqrt{b^2 + 4}$ is $\qquad$ | 2. $\sqrt{13}$.
As shown in Figure 6, construct a line segment $AB=2$, with points $C$ and $D$ on opposite sides of $AB$, and $CA \perp AB, DB \perp AB, CA=1, BD=2$, and $E$ is a point on $AB$.
Let $AE=2a, BE=b$. Connect $CE, DE, CD$.
By the Pythagorean theorem, we have
$$
CE=\sqrt{4a^{2}+1}, DE=\sqrt{b^{2}+4}
$$
Si... | \sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,918 |
3. Given real numbers $a, b$ satisfy
$$
a^{2}=-1-5 a, 5 b=-1-b^{2} \text {. }
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is | 3. $-5 \pm \sqrt{21}$ or -23.
From the given,
$$
a^{2}+5 a+1=0, b^{2}+5 b+1=0 \text {. }
$$
When $a=b$, $a=\frac{-5 \pm \sqrt{21}}{2}$, then
$$
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=a+b=2 a=-5 \pm \sqrt{21} \text {. }
$$
When $a \neq b$, $a, b$ are the two real roots of the quadratic equation $x^{2}+5 x+1=0$. At... | -5 \pm \sqrt{21} \text{ or } -23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,919 |
4. Given that $A D$ is the internal angle bisector of $\triangle A B C$, and the extension of $A D$ intersects the circumcircle of $\triangle A B C$ at point $E$. Then the relationship between $A B \cdot A C - B D \cdot D C$ and $A D^{2}$ is $\qquad$ (fill in “equal” or “not equal”).
| 4. Equality.
As shown in Figure 7, connect $B E$.
Since $\angle B A E=\angle D A C$,
$$
\angle E=\angle C \text {, }
$$
we have $\triangle A B E \backsim \triangle A D C$.
Therefore, $\frac{A B}{A D}=\frac{A E}{A C}$, which means
$$
A B \cdot A C=A E \cdot A D \text {. }
$$
By the intersecting chords theorem, we get... | A B \cdot A C - B D \cdot D C = A D^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,920 |
One, (20 points) Find the positive integer solutions of the indeterminate equation
$$
29 a+30 b+31 c=2196
$$ | One, transform the original equation into
$$
\left\{\begin{array}{l}
29(a+b+c)+(b+2 c)=2196 \\
31(a+b+c)-(2 a+b)=2196 .
\end{array}\right.
$$
Since \(a, b, c\) are positive integers, from equation (1) we get
$$
\begin{array}{l}
29(a+b+c)=2196-(b+2 c) \\
\leqslant 2196-(1+2 \times 1)=2193 .
\end{array}
$$
Therefore, \... | 86 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,921 |
II. (25 points) In a convex quadrilateral $ABCD$, $E$ and $F$ are points on sides $AB$ and $AD$ respectively, such that $EF \parallel BD$. A line through point $E$ is drawn perpendicular to $CD$ at point $M$, and a line through point $F$ is drawn perpendicular to $BC$ at point $N$, and $EM$, $FN$, and $AC$ intersect at... | II. As shown in Figure 8. Given that $\angle A C D \neq 90^{\circ}$ and $\angle A C B \neq 90^{\circ}$, otherwise $E M / / A C$ or $F N / / A C$.
Draw $D Q \perp B C$ at point $Q$, and draw $B P \perp C D$ at point $P$. Let the line $B P$ intersect $A C$ at point $H$, and the line $D Q$ intersect $A C$ at point $H^{\p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,922 |
4. Given the three sides of $\triangle A B C$ are $A B=$ $2 \sqrt{a^{2}+576}, B C=\sqrt{a^{2}+14 a+625}, A C=$ $\sqrt{a^{2}-14 a+625}$, where $a>7$. Then the area of $\triangle A B C$ is $\qquad$ | 4.168.
Notice
$$
\begin{array}{l}
A B^{2}=(2 a)^{2}+48^{2}, \\
B C^{2}=(a+7)^{2}+24^{2}, \\
A C^{2}=(a-7)^{2}+24^{2} .
\end{array}
$$
As shown in Figure 6, with $A B$ as the hypotenuse, construct a right triangle $\triangle A B D$ on one side of $\triangle A B C$, such that
$$
\begin{array}{l}
B D=2 a, A D=48, \\
\an... | 168 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,923 |
Three, (25 points) In $\triangle A B C$, it is known that $\frac{b-c}{\cot \frac{A}{2}}+\frac{c-a}{\cot \frac{B}{2}}+\frac{a-b}{\cot \frac{C}{2}}=0$. Prove: $\frac{b-c}{\cot ^{2} \frac{A}{2}}+\frac{c-a}{\cot ^{2} \frac{B}{2}}+\frac{a-b}{\cot ^{2} \frac{C}{2}}=0$. | Three, as shown in Figure 9, let the incenter of $\triangle ABC$ be $I$ and the inradius be $r$, and the points of tangency with sides $BC$, $CA$, and $AB$ be $D$, $E$, and $F$ respectively; let the lengths of sides $BC$, $CA$, and $AB$ of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Connecting $AI$ and $IF$, we ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,924 |
1. If $a, b, c$ are three arbitrary integers, then $\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}$ are ( ).
(A) not integers
(B) at least one integer
(C) all integers
(D) at least two integers | -.1.C.
Notice that among three arbitrary integers $a, b, c$, at least two of them have the same parity. Without loss of generality, let these two numbers be $a, b$, then $\frac{a+b}{2}$ is an integer. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,925 |
3. The condition that can determine quadrilateral $ABCD$ is a rhombus is ( ).
(A) Diagonal $AC$ bisects diagonal $BD$, and $AC \perp$ $BD$
(B) Diagonal $AC$ bisects diagonal $BD$, and $\angle A=$ $\angle C$
(C) Diagonal $AC$ bisects diagonal $BD$, and bisects $\angle A, \angle C$
(D) Diagonal $AC$ bisects $\angle A, \a... | 3. D.
As shown in Figure 4, $AC$ bisects $BD$, $AC \perp BD$, and $AC$ also bisects $\angle A$ and $\angle C$, so we can rule out options (A) and (C). Option (B)’s condition can only lead to the conclusion that quadrilateral $ABCD$ is a parallelogram, hence we exclude option (B). | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,927 |
4. When $x-y=1$, the value of $x^{4}-x y^{3}-x^{3} y-3 x^{2} y$ $+3 x y^{2}+y^{4}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | 4. C.
When $x-y=1$,
$$
\begin{array}{l}
x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right) \\
=x^{2}+x y+y^{2}=(x-y)^{2}+3 x y=1+3 x y,
\end{array}
$$
Therefore, $x^{4}-x y^{3}-x^{3} y-3 x^{2} y+3 x y^{2}+y^{4}$
$$
\begin{array}{l}
=x\left(x^{3}-y^{3}\right)+y\left(y^{3}-x^{3}\right)+3 x y(y-x) \\
=(x-y)\left(x^{3}-y^{3}\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,928 |
5. Let $b$ take even numbers from 1 to 11, and $c$ take any positive integer. Then the number of quadratic equations $x^{2}-b x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 50
(C) 55
(C) 57
(D) 58 | 5.A.
When $b=2$, from $b^{2}-4 c=4-4 c>0$, we get $c<1$, so, $c=0$.
When $b=4$, from $b^{2}-4 c=16-4 c>0$, we get $c<4$, so, $c=0,1,2,3$.
When $b=6$, from $b^{2}-4 c=36-4 c>0$, we get $c<9$, so, $c=0,1, \cdots, 8$.
When $b=8$, from $b^{2}-4 c=64-4 c>0$, we get $c<16$, so, $c=0,1, \cdots, 15$.
When $b=10$, from $b^{2}-... | 50 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,929 |
6. Four people, A, B, C, and D, are playing a passing game. In the first pass, A passes the ball to one of the other three people. In the second pass, the person who receives the ball passes it to one of the other three people. This passing continues for a total of 4 times. What is the probability that the ball is pass... | 6.A.
Draw the tree diagram as shown in Figure 5.
From Figure 5, we know that the third pass cannot be made to A. There are 6 times when A can receive the ball, so there are $27-6=21$ times that meet the condition. The number of ways for 4 passes is $3^{4}$, thus the probability that meets the condition is $\frac{21}{3... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,930 |
1. Given $a+b+c=0, a>b>c$. Then the range of $\frac{c}{a}$ is $\qquad$ . | $$
\text{2.1. }-2b>c, \text{ get } a>0, cb>c, \text{ get}
$$
$$
1>\frac{b}{a}>\frac{c}{a} \text{. }
$$
Substitute equation (1) into equation (2) to get
$$
1>-1-\frac{c}{a}>\frac{c}{a} \text{. }
$$
Solving, we get $-2<\frac{c}{a}<-\frac{1}{2}$. | -2<\frac{c}{a}<-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,931 |
2. Calculate: $\frac{1^{2}}{1^{2}-100+5000}+\frac{3^{2}}{3^{2}-300+5000}+$ $\frac{5^{2}}{5^{2}-500+5000}+\cdots+\frac{99^{2}}{99^{2}-9900+5000}=$ $\qquad$ | 2.50.
Notice that
$$
\begin{array}{l}
\frac{k^{2}}{k^{2}-100 k+5000}+\frac{\left(100-k^{2}\right)}{(100-k)^{2}-100(100-k)+5000} \\
=\frac{k^{2}}{k^{2}-100 k+5000}+\frac{10000-200 k+k^{2}}{k^{2}-100 k+5000} \\
=\frac{2\left(k^{2}-100 k+5000\right)}{k^{2}-100 k+5000}=2 .
\end{array}
$$
Let $k=1,3,5, \cdots, 49$, and ad... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,932 |
Example 1 In $\triangle ABC$, $D$ is the midpoint of $AB$, points $E$ and $F$ are on the rays $CA$ and $CB$ respectively, and $DE = DF$. Perpendiculars are drawn from points $E$ and $F$ to $CA$ and $CB$ respectively, intersecting at point $P$. Prove that $\angle PAE = \angle PBF$.
---
The translation maintains the or... | As shown in Figure 1, take points \( A' \) and \( B' \) on \( CA \) and \( CB \) respectively, such that \( A' E = A E \) and \( B' F = B F \). Connect \( PA' \), \( PB' \), \( AB' \), and \( BA' \). It is easy to see that
\[
PA' = PA, \quad PB' = PB,
\]
and \( A' B = 2 DE = 2 DF = AB' \).
Thus, \( \triangle PA' B \co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,934 |
Example 2 In the convex quadrilateral $ABCD$, $P$ is inside $\triangle BCD$. It is known that $\angle PBC = \angle ABD, \angle PDC = \angle ADB$. Prove: $\angle APD$ and $\angle BPC$ are supplementary.
Translate the above text into English, please keep the original text's line breaks and format, and output the transla... | As shown in Figure 2, take point $G$ on $CD$ such that $\angle P G D = \angle A B D$, and connect $A G$, $B G$, and $P G$.
It is easy to see that $\triangle P G D \sim \triangle A B D$, thus, we can prove
$\triangle A P D \sim \triangle B G D$.
Therefore, $\angle A P D = \angle B G D$.
Also, $\angle P G D = \angle A B... | null | Geometry | proof | Yes | Yes | cn_contest | false | 717,935 |
Example 5 Given that there are $3 k$ points on the circumference of a circle, dividing the circumference into $3 k$ arcs, among which, there are $k$ arcs of lengths $1, 2, 3$ each. Prove that among these $3 k$ points, there must be two points that are antipodal (i.e., the line connecting the two points is a diameter of... | Explanation: Color the given $3k$ points all red, then add points on the circumference, dividing the entire circumference into arcs of length 1, and color all the newly added points blue. Clearly, there are also $3k$ blue points. Thus, the problem is reduced to proving: there is a pair of red points that are diametrica... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,936 |
Example 7 In a certain country, there are 1001 cities, and between any two cities, there is a one-way road connecting them. Each city has exactly 500 outgoing roads and 500 incoming roads. In this country, a region is demarcated, which includes 668 cities. Prove that from any city within this region, one can reach any ... | Explanation: Otherwise, there exist two cities $X$ and $Y$ in this region such that it is impossible to travel from city $X$ to city $Y$ along the roads within the region.
Divide all 668 cities in the region into two groups, $A$ and $B$. The set of all cities within the region that can be reached from city $X$ via the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,937 |
Example 8 Let $S=\{1,2, \cdots, 98\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which one number is coprime with the other 4 numbers, and in the other g... | Explanation: First, there are 49 even numbers in $S$, and the 49-element subset formed by them obviously does not meet the requirement. Therefore, the smallest natural number $n \geqslant 50$.
To prove that in any 50-element subset $T$ of $S$, there are 10 numbers that meet the requirement, we first prove a strengthen... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,938 |
Example 1 In a complete quadrilateral $A B C D E F$, $A B = A E$.
(1) If $B C = E F$, then $C D = D F$, conversely, if $C D = D F$, then $B C = E F$;
(2) If $B C = E F$ (or $C D = D F$), and $M$ is the Miquel point of the complete quadrilateral, then $M D \perp C F$, and the circumcenter $O_{1}$ of $\triangle A C F$ li... | Proof: Auxiliary lines as shown in Figure 1.
(1) By the property of a complete quadrilateral $1^{[1]}$ equation (2) (or applying Menelaus' theorem to $\triangle A C F$ and the transversal $B D E$), we have
$$
\frac{A B}{B C} \cdot \frac{C D}{D F} \cdot \frac{F E}{E A}=1.
$$
Since $A B=A E$, from the above equation we ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,942 |
Example 2 In a complete quadrilateral $A B C D E F$, $A C \perp B E, A E \perp C F$.
(1) If the projections of points $C$ and $E$ on the line containing diagonal $B F$ are $G$ and $H$ respectively, then $G B = F H$;
(2) If the extension of diagonal $A D$ intersects diagonal $C E$ at point $P$, and the circumcircle of $... | Proof: (1) As shown in Figure 3, from the given conditions, we know that points $C$, $E$, $F$, and $B$ are concyclic, and the midpoint $O$ of $CE$ is the center of the circle. Draw $OM \perp BF$ at point $M$. By the property of the perpendicular from the center to a chord, we have $BM = MF$.
Since $CG \parallel OM \par... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,943 |
Example 3 In $\triangle ABC$, $O$ is the incenter, and points $E, F$ are both on the larger side $BC$. Given that $BF=BA, CE=CA$. Prove: $\angle EOF=\angle B+\angle C$.
---
The translation maintains the original text's line breaks and format as requested. | Explanation: The auxiliary line is shown in Figure 3.
From the given information, it is easy to see that
$$
\begin{array}{l}
\triangle B O A \\
\cong \triangle B O F .
\end{array}
$$
Thus, $O A=O F$.
From the given information, it is easy to see that
$$
\triangle C O A \cong \triangle C O E \text {. }
$$
Thus, $O A=O... | \angle EOF=\angle ABC+\angle ACB | Geometry | proof | Yes | Yes | cn_contest | false | 717,945 |
Problem: Place the 2004 positive integers $1, 2, \cdots, 2004$ randomly on a circle. By counting the parity of all adjacent 3 numbers, it is known that there are 600 groups where all 3 numbers are odd, and 500 groups where exactly 2 numbers are odd. How many groups have exactly 1 odd number? How many groups have no odd... | Solution: These 2004 numbers are arbitrarily placed on a circle, with every 3 adjacent numbers forming a group, making a total of 2004 groups. Each number is part of 3 different groups. Let the number of groups with exactly 1 odd number be $x$, and the number of groups with no odd numbers be $y$. Considering the odd nu... | 206, 698 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,946 |
For a prime number $p(p \geqslant 3)$, define $F(p)=$ $\sum_{k=1}^{p-1} k^{r}$, where $r$ is a positive odd number not exceeding $p$; $f(p)=$ $\frac{1}{2}-\left\{\frac{F(p)}{p^{2}}\right\}$, where $\{x\}=x-[x]$ represents the fractional part of $x$. Find the value of $f(p)$. | Prove: When $p<r<2 p-1$, set
$$
r-1=(p-1)+r_{1} \text {, }
$$
then $1 \leqslant r_{1} \leqslant p-2$.
By Fermat's Little Theorem, $S_{r-1} \equiv S_{r_{1}}=0(\bmod p)$.
Therefore, $2 F(p) \equiv 0\left(\bmod p^{2}\right)$.
Since $(2, p)=1$, when $1<r<2 p-1$, $F(p) \equiv 0\left(\bmod p^{2}\right)$.
When $r=2 p-1$, by ... | f(p)=\left\{\begin{array}{ll}
\frac{1}{2 p}, & r=1 ; \\
\frac{1}{2}, & 1<r<2 p-1,2 \chi_{r} ; \\
-\frac{1}{2 p}, & r=2 p-1 .
\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,947 |
1. Given $\sqrt{x^{2}-4}+\sqrt{2 x+y}=0$. Then the value of $x-y$ is ( ).
(A) 2
(B) 6
(C) 2 or -2
(D) 6 or -6 | -、1.D.
Since $\sqrt{x^{2}-4} \geqslant 0, \sqrt{2 x+y} \geqslant 0$, therefore, it can only be that $\sqrt{x^{2}-4}=\sqrt{2 x+y}=0$.
Solving separately, we get $\left\{\begin{array}{l}x=2 \text { or }-2, \\ y=-2 x .\end{array}\right.$
Then $x-y=x-(-2 x)=3 x$, that is, $x-y$ is 6 or -6. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,949 |
2. A sample is $1,3,2,2, a, b, c$. It is known that the mode of this sample is 3, and the mean is 2. Then, the variance of this sample is ( ).
(A) 8
(B) 4
(C) $\frac{8}{7}$
(D) $\frac{4}{7}$ | 2.C.
Given the sample mean is 2, we have
$$
1+3+2+2+a+b+c=2 \times 7=14 \text {, }
$$
Thus, $a+b+c=6$.
Since the sample mode is 3, it means that at least two of $a, b, c$ are 3, and the other one is 0. Therefore, the sample variance is
$$
s^{2}=\frac{1}{7}(1+1+0+0+1+1+4)=\frac{8}{7} .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,950 |
3. A store sells two types of goods, A and B, both priced at 1800 yuan, where good A yields a profit of $20 \%$, and good B incurs a loss of $20 \%$. If one unit of each good is sold simultaneously, then, ( ).
(A) a total profit of 150 yuan
(B) a total loss of 150 yuan
(C) neither profit nor loss
(D) none of the above ... | 3.B.
According to the problem, the purchase price of product A is
$$
1800 \div(1+20 \%)=1500 \text { (yuan), }
$$
and the purchase price of product B is
$$
1800 \div(1-20 \%)=2250 \text { (yuan). }
$$
Therefore, the store's profit is
$$
1800 \times 2-(1500+2250)=-150 \text { (yuan), }
$$
which means the store incur... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,951 |
5. Given $\sqrt{25-x^{2}}-\sqrt{15-x^{2}}=2$. Then the value of $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 5.C.
Notice
$$
\begin{array}{l}
\left(\sqrt{25-x^{2}}-\sqrt{15-x^{2}}\right) \\
\left(\sqrt{25-x^{2}}+\sqrt{15-x^{2}}\right) \\
=\left(25-x^{2}\right)-\left(15-x^{2}\right)=10,
\end{array}
$$
Therefore, $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}=\frac{10}{2}=5$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,953 |
6. As shown in Figure $2, C E$ and $C F$ bisect $\angle A C B$ and $\angle A C D$ respectively, $A E \parallel C F$, $A F \parallel C E$, line $E F$ intersects $A B$ and $A C$ at points $M$ and $N$. If $B C = a, A C = b, A B = c$, and $c > a > b$, then the length of $M E$ is ().
(A) $\frac{c-a}{2}$
(B) $\frac{a-b}{2}$
... | 6.B.
Since $CE, CF$ bisect $\angle ACB, \angle ACD$ respectively, we have $\angle ECA = \frac{1}{2} \angle ACB, \angle ACF = \frac{1}{2} \angle ACD$. Therefore, $\angle ECF = \angle ECA + \angle ACF = \frac{1}{2} (\angle ACB + \angle ACD) = 90^\circ$. Since $AE \parallel CF, AF \parallel CE$, quadrilateral $AECF$ is a... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,954 |
7. As shown in Figure 3, the graph of the linear function $y=k x+b$ passes through the point $P(1,4)$, and intersects the positive x-axis and y-axis at points $A$ and $B$, respectively. $O$ is the origin. When the area of $\triangle A O B$ is minimized, the values of $k$ and $b$ are ( ).
(A) $-4,8$
(B) $-4,4$
(C) $-2,4... | 7.A.
Since point $P(1,4)$ lies on the graph of the linear function $y=k x+b$, the linear function can be expressed as $y=k x+(4-k)$.
Let $x=0$, we get $B(0,4-k)$.
Let $y=0$, we get $A\left(\frac{k-4}{k}, 0\right)$.
Connecting $P O$. Then
$$
\begin{array}{l}
S_{\triangle B O A}=S_{\triangle B O P}+S_{\triangle P O H} \... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,955 |
Example 4 In an acute triangle $\triangle ABC$, $AB$ is the largest side, $AC$ is the smallest side, $O$ is the circumcenter, and $H$ is the orthocenter. Prove:
$$
\angle OAH = \angle C - \angle B.
$$ | Explanation: As shown in Figure 4, draw the altitude $A D$, with the orthocenter $H$ on $A D$. Draw $O M \perp A B$ at point $M$. By the properties of the circumcenter of a triangle, we know $\angle A O M = \angle C$, so,
$$
\begin{array}{l}
\angle O A M = \angle C A D. \\
\text{Therefore, } \angle O A H = \angle B A C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,956 |
8. If the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ is satisfied by only one integer $k$, then the maximum value of the positive integer $n$ is ( ).
(A) 100
(B) 112
(C) 120
(D) 150 | 8. B.
From the given inequality, we have
$$
\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8} \text {, }
$$
then $\frac{6}{7}<\frac{k}{n}<\frac{7}{8}$,
so $\frac{6 n}{7}<k<\frac{7 n}{8}$.
Given that there is only one integer $k$ between $\frac{6 n}{7}$ and $\frac{7 n}{8}$, we have
$$
\frac{7 n}{8}-\frac{6 n}{7} \leqslant 2 \te... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,957 |
9. As shown in Figure 4, in $\triangle A B C$, $A B=3, A C=4$, $B C=5$, $\triangle A B D$, $\triangle A C E$, and $\triangle B C F$ are all equilateral triangles. Then the area of quadrilateral $A E F D$ is $\qquad$ | II, 9.6.
Since $B D=A B$,
$$
\begin{array}{l}
\angle D B F=\angle D B A-\angle F B A \\
=\angle F B C-\angle F B A=\angle A B C, \\
B F=B C,
\end{array}
$$
Therefore, $\triangle D B F \cong \triangle A B C$.
Similarly, $\triangle E F C \cong \triangle A B C$.
Thus, $A D=A B=E F, D F=A C=A E$.
Hence, quadrilateral $A E... | 9.6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,958 |
10. As shown in Figure 5, in trapezoid $A B C D$, $A D / / B C, C E$ is the bisector of $\angle B C D$, and $C E \perp A B$, with $E$ being the foot of the perpendicular, $B E=2 A E$. If the area of quadrilateral $A E C D$ is 1, then the area of trapezoid $A B C D$ is $\qquad$ . | $10.2 \frac{1}{7}$.
As shown in Figure 9, extend $B A$ and $C D$ to intersect at point $P$, then $C E \perp B P$. Therefore, $\triangle B C P$ is an isosceles triangle. Thus, $E P = B E = 2 A E, A P = A E$.
Hence, $A P: B P = 1: 4$.
Let $S_{\triangle B C P} = S$, then
$$
S_{\triangle P R D} = \left(\frac{1}{4}\right)^{... | 2 \frac{1}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,959 |
11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ . | 11.8.
From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$
Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,960 |
12. In the park, there are two rivers $O M$ and $O N$ converging at point $O$ (as shown in Figure 6, $\angle M O N=60^{\circ}$. On the peninsula formed by the two rivers, there is an ancient site $P$. It is planned to build a small bridge $Q$ and $R$ on each of the two rivers, and to construct three small roads to conn... | 12.300.
As shown in Figure 10, let the symmetric points of point $P$ with respect to $OM$ and $ON$ be $P_{1}$ and $P_{2}$, respectively. Connect $P_{1}P_{2}$, intersecting $OM$ and $ON$ at points $Q$ and $R$. By symmetry, we have
$$
\begin{array}{l}
PQ = P_{1}Q, \\
PR = P_{2}R, \\
PQ + QR + RP \\
= P_{1}Q + QR + RP_{2... | 300 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,961 |
13. Given integers $a$, $b$, and $c$ such that the equation
$$
(x+a)(x+b)+c(x-10)=(x-11)(x+1)
$$
holds for all $x$. Find the value of $c$. | Three, 13. Expanding the equation in the question, we get
$$
x^{2}+(a+b+c) x+a b-10 c=x^{2}-10 x-11 \text {. }
$$
The above equation holds for any $x$. Therefore,
$$
\left\{\begin{array}{l}
a+b+c=-10, \\
a b-10 c=-11 .
\end{array}\right.
$$
Eliminating the parameter $c$, we get
$$
10 a+10 b+a b=-111 \text {, }
$$
wh... | 20 \text{ or } 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,962 |
14. As shown in Figure 7, in Rt $\triangle A B C$, $\angle C=90^{\circ}, B C=2 A C$, $A D$ is the angle bisector of $\angle B A C$. Prove: $A B+2 B D=5 A C$. | 14. Proof 1: As shown in Figure 11, extend $AC$ to point $E$ such that $CE = 4AC$, and on $AE$, mark $AF = AB$, connect $BE$, $BF$, and extend $AD$ to intersect $BF$ at point $G$.
Since $AF = AB$ and $AG$ bisects $\angle BAC$, we have:
$$
\begin{array}{l}
AG \perp BF, \text{ and } BG = GF. \\
\text{Also, } AC \perp BC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,963 |
15. On the number line, points with coordinates
$1,2, \cdots, 2006$ are called marked points. A frog starts from point 1, and after 2006 jumps, it visits all the marked points and returns to the starting point. What is the maximum length of the total path the frog has jumped? Explain your reasoning. | 15. Let the points the frog reaches in sequence be $X_{1}, X_{2}, \cdots, X_{2006}$, with $X_{1}=1$. The total length of the path jumped over is
$$
S=\left|X_{1}-X_{2}\right|+\left|X_{2}-X_{3}\right|+\cdots+\left|X_{2006}-X_{1}\right| \text {. }
$$
For each point $X_{i}$, there is one entry and one exit, so $X_{i}$ ap... | 2 \times 1003^2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,964 |
1. If $-2 \leqslant|x+1|-|a x-1| \leqslant 2$ holds for all $x \in$ $\mathbf{R}$, then the number of real numbers $a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) infinitely many | -、1.C.
The original inequality always holds only when $a= \pm 1$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,965 |
2. Given that $\triangle A B C$ is inscribed in a unit circle. Then the three segments of lengths $\sin A, \sin B, \sin C$ ( ).
(A) can form a triangle, whose area is greater than $\frac{1}{2}$ of the area of $\triangle A B C$
(B) can form a triangle, whose area is equal to $\frac{1}{2}$ of the area of $\triangle A B C... | 2.C.
By the Law of Sines, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2$, so the triangle formed by $\sin A, \sin B, \sin C$ is similar to $\triangle ABC$, and its area is $\frac{1}{4}$ of the area of $\triangle ABC$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,966 |
Example 5 In $\square A B C D$, the angle bisector of $\angle B A D$ intersects $B C$ and $D C$ at points $F$ and $E$, respectively. $O$ is the circumcenter of $\triangle C E F$. Prove: $\angle A B C=2 \angle O B D$.
(2006, National Junior High School Mathematics League) | Explanation: As shown in Figure 5, connect $O F$, $O C$, and $O D$. It is easy to see that $A B = B F$ and $C E = C F$. Therefore, the perpendicular bisector of the base $E F$ of the isosceles $\triangle C E F$ must pass through the circumcenter $O$ and the vertex $C$. At this point, $\angle O C E = \angle O C F = \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,967 |
4. Let $a_{n}=n^{2}+n+2(n=1,2, \cdots)$. Then in the sequence $\left\{a_{n}\right\}$ ).
(A) there are infinitely many prime numbers
(B) there are only finitely many prime numbers
(C) there are infinitely many square numbers
(D) there are only finitely many square numbers | 4.D.
Since $a_{n}=n^{2}+n+2=n(n+1)+2$, we have $2 \mid a_{n}$, and $a_{n}>2$. Therefore, $a_{n}$ must be a composite number. This eliminates options (A) and (B). Furthermore, since when $n \geqslant 2$, $n^{2}<n^{2}+n+2<n^{2}+2 n+1=(n+1)^{2}$, it follows that $a_{1}$ is the only perfect square when $n=1$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,969 |
5. If $a, b, c \in \mathbf{N}$, and
$$
29 a+30 b+31 c=366 \text{. }
$$
then $a+b+c=(\quad)$.
(A) 10
(B) 12
(C) 14
(D) 16 | 5.B.
From $29a + 30b + 31c = 366 (a, b, c \in \mathbf{N})$, we get $29(a + b + c) \leqslant 366 \leqslant 31(a + b + c)$.
Therefore, $11 < a + b + c < 13$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,970 |
6. Xiao Ming and Xiao Hua are playing a game, with the scoring rules as follows: at the beginning, each person's score is 1 point; after each round, they multiply their score by 3. After the game ends, the difference between Xiao Ming's score and Xiao Hua's score is exactly a positive integer multiple of 675. Then, how... | 6.B.
Let Xiaoming jump $m$ times, and Xiaohua jump $n$ times. From the problem, we have $3^{m}-3^{n}=675 k\left(k \in \mathbf{N}_{+}\right)$,
which means $3^{n}\left(3^{m-n}-1\right)=3^{3} \times 5^{2} \times k$.
Therefore, $3^{m-n}-1$ is a multiple of 25. Then $\min (m-n)=20$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,971 |
7. The solution set of the inequality $(x-2) \sqrt{x^{2}-2 x-3} \geqslant 0$ is $\qquad$ . | From $x^{2}-2 x-3=(x+1)(x-3) \geqslant 0$, we get $x \geqslant 3$ or $x \leqslant-1$.
$x \geqslant 3$ satisfies the inequality.
When $x \leqslant-1$, only $x=-1$ satisfies the inequality. | x \geqslant 3 \text{ or } x = -1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,972 |
8. If the line $y=x$ is tangent to the curve $y=x^{3}-3 x^{2}+p x$, then the value of the real number $p$ is $\qquad$ | 8.1 or $\frac{13}{4}$.
Let the point of tangency be $(m, m)$, then by the point of tangency on the curve we have
$$
m=m^{3}-3 m^{2}+p m \text {. }
$$
Thus, $m$ satisfies
$$
m=0 \text {, }
$$
or $m^{2}-3 m+p=1$.
The slope of the tangent line $y=x$ is 1, and the slope at the point of tangency on the curve is also 1, h... | 1 \text{ or } \frac{13}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,973 |
9. Given the function
$$
f(x)=\left\{\begin{array}{ll}
3 x-1 & x \leqslant 1 ; \\
\frac{2 x+3}{x-1} & x>1
\end{array}\right.
$$
If the graph of the function $y=g(x)$ is symmetric to the graph of the function $y=f^{-1}(x+1)$ about the line $y=x$, then the value of $g(11)$ is | 9. $\frac{3}{2}$.
The graph of $y=f^{-1}(x)$ is shifted one unit to the left to get the graph of $y=f^{-1}(x+1)$. The graph of $y=f(x)$ is symmetrical to the graph of $y=f^{-1}(x)$ about the line $y=x$, and the graph of $y=f^{-1}(x+1)$ is symmetrical to the graph of $y=g(x)$ about the line $y=x$ as well. Therefore, the... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,974 |
10. In tetrahedron $ABCD$, $AB=CD=6$, $AC=$ $AD=BC=BD=5$. Then the radius $r$ of the inscribed sphere is $\qquad$. | 10. $\frac{3 \sqrt{7}}{8}$.
Let the midpoint of $C D$ be $E$. In $\triangle A B E$,
$$
\begin{array}{l}
A E=B E=\sqrt{5^{2}-3^{2}}=4, \\
E H=\sqrt{4^{2}-3^{2}}=\sqrt{7} .
\end{array}
$$
Therefore, the height $h$ on $B E$ is $h=\frac{6 \sqrt{7}}{4}=\frac{3 \sqrt{7}}{2}$, and $h$ is also the height of the tetrahedron $... | \frac{3 \sqrt{7}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,975 |
11. Given $\alpha, \beta \in\left[0, \frac{\pi}{4}\right]$. Then the maximum value of $\sin (\alpha-\beta)+$ $2 \sin (\alpha+\beta)$ is $\qquad$ . | 11. $\sqrt{5}$.
$$
\begin{array}{l}
\text { Since } \sin (\alpha-\beta)+2 \sin (\alpha+\beta) \\
=3 \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta,
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\text { the above expression } \leqslant \sqrt{(3 \sin \alpha)^{2}+\cos ^{2} \alpha... | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,976 |
12. After removing one element from the set $\{1!, 2!, \cdots, 24!\}$, the product of the remaining elements is exactly a perfect square. | 12.12!.
From $(2 k)!=(2 k)(2 k-1)!$, the product can be transformed into
$$
\begin{array}{l}
24 \times(23!)^{2} \times 22 \times(21!)^{2} \times \cdots \times 4 \times(3!)^{2} \times 2 \\
=\left[2^{6} \times(23!) \times(21!) \times \cdots \times(3!)\right]^{2} \times(12!) .
\end{array}
$$
Therefore, the factor 12 ! i... | 12! | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,977 |
Example 6 In $\triangle ABC$, $\angle ABC = 2 \angle ACB$, and a point $P$ inside the triangle satisfies $AB = AP$, $PB = PC$. Prove that $AP$ is the trisector of $\angle BAC$.
(1994, Hong Kong Mathematical Olympiad) | As shown in Figure 6, with the perpendicular bisector of side $BC$ as the axis, perform the axial symmetry of $\triangle ABP$ to get $\triangle DCP$, and connect $AD$, $CD$, and $PD$.
It is easy to see that quadrilateral $ABCD$ is an isosceles trapezoid, so points $A$, $B$, $C$, and $D$ are concyclic.
Since $\angle DC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,978 |
$$
\begin{array}{l}
\text { 13. Given the function } f(x)=-2 x+4 \text {, let } \\
S_{n}=f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)+f(1) \\
\left(n \in \mathbf{N}_{+}\right) \text {. }
\end{array}
$$
If the inequality $\frac{a^{n}}{S_{n}}<\frac{a^{n+1}}{S_{n+1}}$ always hold... | Three, 13. Since $S_{n}=-2\left(\frac{1}{n}+\cdots+\frac{n-1}{n}+1\right)+4 n=$ $3 n-1$, from $\frac{a^{n}}{S_{n}}0$, then $a^{n}0$, which is a contradiction. Therefore, when $a0$, since $a^{n}>0$, from equation (1),
$$
a>\frac{3 n+2}{3 n-1}=1+\frac{3}{3 n-1}
$$
decreases as $n$ increases.
When $n=1$, $1+\frac{3}{3 n-... | a>\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,979 |
14. Given point $A(a, b)$, parabola
$$
C: y^{2}=2 p x(a \neq 0, b \neq 0, a \neq 2 p) \text {. }
$$
Draw a line $l$ through point $A$, intersecting parabola $C$ at points $P$ and $Q$. If the circle with diameter $PQ$ passes through the vertex of parabola $C$, find the equation of line $l$. | 14. If the line $l$ passes through the origin, it obviously meets the requirement, and the equation is $y=\frac{b}{a} x$.
If the line $l$ does not pass through the origin, let its equation be
$$
x=m(y-b)+a \text {. }
$$
Let $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right)$, then
$$
O P \perp O Q \Leftrightarro... | b x-(a-2 p) y-2 b p=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,980 |
15. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, E$ and $F$ are moving points on edges $A B$ and $B C$ respectively, and $A E=B F$. Find the minimum value of the angle formed by lines $A_{1} E$ and $C_{1} F$ (express the result using inverse trigonometric functions). | 15. Solution 1: As shown in Figure 1, extend $DC$ to point $G$ such that $CG = AE$, and connect $C_1G$ and $FG$. From the problem, we know that $A_1E \parallel C_1G$, and the angle formed by $A_1E$ and $C_1F$ is equal to $\angle FC_1G$.
Let $AE = CG = x (0 \leqslant x \leqslant 1)$, then we have
$$
\begin{array}{l}
CF ... | \arccos \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,981 |
One, (50 points) Given that quadrilateral $ABCD$ is a cyclic quadrilateral, lines $AC$ and $BD$ intersect at point $P$, and $\frac{AB}{AD} = \frac{CB}{CD}$. Let $E$ be the midpoint of $AC$. Prove that $\frac{BE}{ED} = \frac{BP}{PD}$.
---
The translation is provided as requested, maintaining the original format and li... | By Ptolemy's theorem, we have
$$
A B \cdot C D + A D \cdot B C = A C \cdot B D \text{.}
$$
Since \( A B \cdot C D = A D \cdot B C \) and \( A E = E C \), we have
$$
2 A B \cdot C D = 2 A E \cdot B D = 2 E C \cdot B D \text{,}
$$
which means \( A B \cdot C D = A E \cdot B D = E C \cdot B D \).
In \(\triangle C E D\) a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,982 |
(50 points) Let $a, b, c$ be positive numbers, and let $d$ be the minimum of $(a-b)^{2}, (b-c)^{2}, (c-a)^{2}$.
(1) Prove that there exists $\lambda(0<\lambda<1)$, such that
$$
d \leqslant \lambda\left(a^{2}+b^{2}+c^{2}\right) \text {; }
$$
(2) Find the smallest positive number $\lambda$ for which the inequality (1) ho... | (1) By the definition of $d$, we have
$$
d \leqslant(a-b)^{2}, d \leqslant(b-c)^{2}, d \leqslant(c-a)^{2}.
$$
Adding these three inequalities, we get
$$
\begin{array}{l}
3 d \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
=2\left(a^{2}+b^{2}+c^{2}\right)-2 a b-2 b c-2 c a \\
0, \text{ and } d=(b-c)^{2}.
\end{array}
$$
Thus, ... | \frac{1}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,983 |
Three. (50 points) Given $n$ four-element sets $A_{1}, A_{2}, \cdots, A_{n}$, any two of which have exactly one common element, and
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=n .
$$
Find the maximum value of $n$. Here $\operatorname{Card} A$ is the number of elements in set $A$. | Consider any element $a \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}$.
If each $A_{i}$ contains $a$, then by the given condition, the other elements in each $A_{i}$ are all different. Hence,
$$
\operatorname{Card}\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=3 n+1>n,
$$
which contradicts the given condition.
The... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,984 |
Let $a>b>0$,
$$
f(x)=\frac{2(a+b) x+2 a b}{4 x+a+b} \text {. }
$$
Prove: There exists a unique positive number $x$, such that
$f(x)=\left(\frac{a^{\frac{1}{3}}+b^{\frac{1}{3}}}{2}\right)^{3}$.
(Li Shenghong) | Solution 1: Let $t=\left(\frac{a^{\frac{1}{3}}+b^{\frac{1}{3}}}{2}\right)^{3}$.
From $t=\frac{2(a+b) x+2 a b}{4 x+a+b}$, we get
$$
[2(a+b)-4 t] x=t(a+b)-2 a b \text {. }
$$
To prove that equation (1) has a unique positive solution $x$, it is only necessary to prove
$$
2(a+b)-4 t>0 \text { and } t(a+b)-2 a b>0 \text {,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,985 |
$$
\begin{array}{r}
\text { 2. As shown in Figure 1, in } \triangle A B C \\
\text { where } \angle A B C=90^{\circ}, D \text{ and } G
\end{array}
$$
2. As shown in Figure 1, in $\triangle A B C$
are two points on side $C A$, connect
$B D, B G$. Draw perpendiculars from
points $A, G$ to $B D$, with feet of
perpendicu... | II. As shown in Figure 4, construct the circumcircle $W$ of the right triangle $\triangle ABC$, and extend $BD$ and $AE$ to intersect the circle $\mathbb{W}$ at points $K$ and $J$, respectively. Connect $BJ$, $CJ$, $KJ$, $FJ$, and $CK$. It is easy to see that
$\angle BAJ = \angle KBC$.
Thus, $BJ = KC$.
Therefore, quadr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,986 |
Three, a deck of cards consists of 52 cards, among which, "Diamonds", "Clubs", "Hearts", and "Spades" each have 13 cards, numbered sequentially as $2,3, \cdots, 10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$. Cards of the same suit and adjacent numbers are called "straight flush" cards, and $\mathrm{A}$ can be co... | $$
\begin{array}{l}
\text { Let } n \geqslant 3 \text {, from } \\
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right) \text {, } \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right), D=\left(d_{1}, d_{2}, \cdots, d_{n}\right) \\
\end{array}
$$
Select $n$ terms from these four sequences, sa... | 3^{13}-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,987 |
For any positive integer $n$, let $a_{n}$ be the real root of the equation $x^{3} + \frac{x}{n} = 1$. Prove:
(1) $a_{n+1} > a_{n}$;
(2) $\sum_{i=1}^{n} \frac{1}{(i+1)^{2} a_{i}} < a_{n}$.
(Ben Shenghong, problem contributor) | (1) From $a_{n}^{3}+\frac{a_{n}}{n}=1$, we get $0 < a_{n} < 1$, so,
$$
a_{n+1}-a_{n}>0,
$$
which means $a_{n+1}>a_{n}$.
(2) Since $a_{n}\left(a_{n}^{2}+\frac{1}{n}\right)=1$, we have,
$$
a_{n}=\frac{1}{a_{n}^{2}+\frac{1}{n}}>\frac{1}{1+\frac{1}{n}}=\frac{n}{n+1} \text {. }
$$
Thus, $\frac{1}{(n+1)^{2} a_{n}}<\frac{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,988 |
Example 1 Choose several colors from 6 different given colors to color a cube, with each face colored with one color and ensuring that any two adjacent faces sharing a common edge are of different colors. How many different coloring schemes are there (if one of the two colored cubes can be rotated to match the coloring... | Explanation: Clearly, the maximum number of colors used is 6, and the minimum is 3. Below, we use the grouping method to count.
(1) When using 6 colors, each color is used to paint one face. According to the problem, the first color can be placed on top by flipping. At this point, the bottom is one of the remaining 5 ... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,989 |
Five, As shown in Figure 2, in $\triangle A B C$, $\angle A=60^{\circ}$, the incircle $\odot I$ of $\triangle A B C$ touches sides $A B$ and $A C$ at points $D$ and $E$, respectively. Line $D E$ intersects lines $B I$ and $C I$ at points $F$ and $G$, respectively. Prove:
$$
F G=\frac{1}{2} B C .
$$ | Proof 1: As shown in Figure 6, connect $C F$, $B G$, $I D$, $I E$, and $A I$. Then, $A$, $D$, $I$, and $E$ are concyclic. Therefore, $\angle I D E = \frac{1}{2} \angle A$.
Hence, $\angle B D F$
$$
= 90^{\circ} + \frac{1}{2} \angle A.
$$
Also, $\angle B I C = 180^{\circ} - \frac{1}{2}(\angle B + \angle C) = 90^{\circ} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,990 |
Six, find the smallest real number $m$, such that for any positive real numbers $a, b, c$ satisfying $a + b + c = 1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(Xiong Bin) | Six, Solution 1: When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$
for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since for $0<x<1$, we have
$$
\begin{array}{l}
27 x^{3} \geqslant 6 x^{2}+5... | 27 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,991 |
(1) Find the number of positive integer solutions $(m, n, r)$ for the indeterminate equation
$$
m n + n r + m r = 2(m + n + r)
$$
(2) For a given integer $k (k > 1)$, prove that the indeterminate equation $m n + n r + m r = k(m + n + r)$ has at least $3k + 1$ sets of positive integer solutions $(m, n, r)$.
(Yuan Hanhu... | (1) If $m, n, r \geqslant 2$, from $m n \geqslant 2 m, n r \geqslant 2 n, m r \geqslant 2 r$, we get $m n+n r+m r \geqslant 2(m+n+r)$.
Thus, all the inequalities above hold with equality.
Therefore, $m=n=r=2$.
If $1 \in\{m, n, r\}$, without loss of generality, let $m=1$, then $n r+n+r=2(1+n+r)$.
Thus, $(n-1)(r-1)=3$.
T... | 3k + 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,992 |
For a circle with a circumference of $n\left(n \in \mathbf{N}_{+}\right)$, the smallest positive integer $p_{n}$ that satisfies the following conditions is called the "circle partition number": If there are $p_{n}$ points $A_{1}, A_{2}, \cdots, A_{P_{n}}$ on the circumference, for each integer $m$ in $1,2, \cdots, n-1$... | Let $p_{n}=k$.
Since among $k$ points, every two points can form a major arc and a minor arc, there can be at most $k(k-1)$ arc lengths.
When $k(k-1) \geqslant 20$, then $k \geqslant 5$;
When $k(k-1) \geqslant 30$, then $k \geqslant 6$.
On the other hand, when $k=5$, a partition graph can be provided (as shown in Figur... | p_{21}=5, \; T_{21}=(1,3,10,2,5); \; p_{31}=6, \; T_{31}=(1,2,7,4,12,5),(1,2,5,4,6,13),(1,3,2,7,8,10),(1,3,6,2,5,14),(1, | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,993 |
1.111 $111222222-333333=x^{2}$ and $x>$ 0. Then $x=(\quad)$.
(A) 332233
(B) 223377
(C)333 333
(D) 331177 | $\begin{array}{l}\text {-1.C. } \\ 111111222222-333333 \\ =111111 \times(10^{6}+2)-111111 \times 3 \\ =111111 \times(10^{6}-1)=\frac{999999}{9}(10^{6}-1) \\ =\frac{10^{6}-1}{9}(10^{6}-1)=\left(\frac{10^{6}-1}{3}\right)^{2}=333333^{2} .\end{array}$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,994 |
2. Let $0<a, b, c, d<1$. Then the relationship between $a(1-b)+b(1-c)+c(1-d)+d(1-a)$ and 2 is ( ).
(A) greater than
(B) less than
(C) equal to
(D) cannot be determined | 2.B.
As shown in Figure 4, for a square
with side length 1, then
$$
\begin{array}{c}
\frac{1}{2}[a(1-b)+ \\
b(1-c)+ \\
c(1-d)+ \\
d(1-a)] \\
=S_{1}+S_{2}+S_{3}+S_{4} \\
<S_{\text {square }}(B C D=1 .
\end{array}
$$
Therefore, $a(1-b)+b(1-c)+c(1-d)+d(1-a)<2$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,995 |
3. As shown in Figure 1, in $\triangle A B C$, $A B=A C$, points $D, E$ are on sides $A B, A C$ respectively, $D M$ bisects $\angle B D E$, and $E N$ bisects $\angle D E C$. If $\angle D M N=$ $110^{\circ}$, then $\angle D E A=$ ( ).
(A) $40^{\circ}$
(B) $50^{\circ}$
(C) $60^{\circ}$
(D) $70^{\circ}$ | 3.A.
Since $AB = AC$, therefore, $\angle B = \angle C$.
$$
\begin{array}{l}
\text{Also, } \angle B = \angle DMN - \angle BDM = \angle DMN - \angle MDE, \\
\angle C = \angle MNE - \angle NEC = \angle MNE - \angle NED,
\end{array}
$$
Thus, $\angle DMN - \angle MDE = \angle MNE - \angle NED$,
which means $\angle DMN + \... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,996 |
4. Given $x+\frac{2}{x}=1$. Then $\frac{x^{2}}{x^{4}+5 x^{2}+4}=$
(A) 1
(B) $\frac{1}{4}$
(C) $\frac{1}{3}$
(D) $\frac{1}{2}$ | 4.D.
$$
\frac{x^{4}+5 x^{2}+4}{x^{2}}=x^{2}+5+\frac{4}{x^{2}}=\left(x+\frac{2}{x}\right)^{2}+1=2 \text{. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,997 |
5. Let $a, b, c, d$ be distinct real numbers, and
$$
\begin{array}{l}
\left(a^{2}-c^{2}\right)\left(a^{2}-d^{2}\right)=1, \\
\left(b^{2}-c^{2}\right)\left(b^{2}-d^{2}\right)=1 .
\end{array}
$$
Then $a^{2} b^{2}-c^{2} d^{2}=(\quad)$.
(A)0
(B) -1
(C) 1
(D) Cannot be determined | 5. B.
$a^{2}, b^{2}$ are the roots of the equation $\left(x-c^{2}\right)\left(x-d^{2}\right)=1$. Expanding it, we get $x^{2}-\left(c^{2}+d^{2}\right) x+c^{2} d^{2}-1=0$. By the relationship between roots and coefficients, we have $a^{2} b^{2}=c^{2} d^{2}-1$. Therefore, $a^{2} b^{2}-c^{2} d^{2}=-1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,998 |
6. If $|x|+|y|=1$, then the range of values for $S=y-2 x$ is ( ).
(A) $-4 \leqslant S \leqslant 4$
(B) $-3 \leqslant S \leqslant 3$
(C) $-2 \leqslant S \leqslant 2$
(D) $-1 \leqslant S \leqslant 1$ | 6.C.
As shown in Figure $5, |x|+|y|=1$
the graph is a square $ABCD, S$ reaches its minimum value at point $A(1,0)$, and reaches its maximum value at point $C(-1,0)$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,999 |
Example 2: On a $12 \times 12$ super chessboard, there is a "super knight" that moves by jumping from one corner of a $3 \times 4$ rectangular block to the opposite corner. Can this super knight start from a certain square, visit each square exactly once, and then return to the starting point?
(26th IMO Candidate Probl... | (1) Notice that, the squares on an international chessboard are originally divided into two groups, black and white. According to the rules, the super knight must jump from one color to another with each move, meaning that the color of the square the super knight is on changes with each step (invariant).
(2) Divide the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,000 |
1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$
then $a^{2007}+b^{2007}=$ $\qquad$ | Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,001 |
2. The side length of square $ABCD$ is equal to 7. Points $E$, $F$, $G$, and $H$ are taken on sides $AB$, $BC$, $CD$, and $DA$ respectively. If quadrilateral $EFGH$ is a rhombus with a side length of 5, then the lengths of the diagonals $EG$ and $FH$ of the rhombus are $\qquad$ | 2. $5 \sqrt{2}, 5 \sqrt{2}$.
As shown in Figure 6, since quadrilateral $E F G H$ is a rhombus, $F H$ and $E G$ are perpendicular to each other. Therefore, it can be proven that $F H=E G$. Hence, quadrilateral $E F G H$ is a square. Thus,
$\triangle A E H \cong \triangle B F E$ $\cong \triangle C G F \cong \triangle D ... | 5 \sqrt{2}, 5 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,002 |
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