problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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1. Let $m$ be a positive integer. If there exists a positive integer $n$ such that $m$ can be expressed as the quotient of $n$ and the number of positive divisors of $n$ (including 1 and itself), then $m$ is called a "good number." Prove:
(1) $1, 2, \cdots, 17$ are all good numbers;
(2) 18 is not a good number.
(Li She... | 1. Let $d(n)$ be the number of positive divisors of the positive integer $n$.
(1) Since $p=\frac{8 p}{d(8 p)}$ for $p=3,5,7,11,13,17$, and
$$
\begin{array}{l}
1=\frac{2}{d(2)}, 2=\frac{8}{d(8)}, 4=\frac{36}{d(36)}, 6=\frac{72}{d(72)}, \\
8=\frac{96}{d(96)}, 9=\frac{108}{d(108)}, 10=\frac{180}{d(180)}, \\
12=\frac{240}{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,195 |
2. Let $\triangle A B C$ be an acute triangle, and let points $D$, $E$, $F$ lie on sides $B C$, $C A$, $A B$ respectively. The line segments $A D$, $B E$, $C F$ pass through the circumcenter $O$ of $\triangle A B C$. It is known that among the following six ratios
$$
\frac{B D}{D C}, \frac{C E}{E A}, \frac{A F}{F B}, \... | 2. From six ratios, take two, there are two types:
(1) involving the same side; (2) involving different sides.
(1) If the two ratios on the same side are both integers, let's assume they are $\frac{B D}{D C}$ and $\frac{C D}{D B}$. Since they are reciprocals and both are integers, they must both be 1, so $B D = D C$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,196 |
3. Let integer $n(n>3)$, non-negative real numbers $a_{1}, a_{2}$, $\cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=2$.
Find the minimum value of $\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}$.
(Provided by Zhu Huawei) | 3. From $a_{1}+a_{2}+\cdots+a_{n}=2$, we know the problem is equivalent to finding the maximum value of
$$
\begin{array}{l}
a_{1}-\frac{a_{1}}{a_{2}^{2}+1}+a_{2}-\frac{a_{2}}{a_{3}^{2}+1}+\cdots+a_{n}-\frac{a_{n}}{a_{1}^{2}+1} \\
=\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,197 |
4. In the plane, a set $S$ consists of $n(n \geqslant 3)$ points, and $P$ is a set of $m$ lines in this plane, such that $S$ is symmetric with respect to each line in $P$. Prove that $m \leqslant n$, and determine when equality holds.
（Provided by Bian Hongping） | 4. (1) Let the $n$ points in $S$ be $A_{1}, A_{2}, \cdots, A_{n}$.
Establish a Cartesian coordinate system, and let $A_{i}\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n)$.
It is easy to prove that $\sum_{i=1}^{n} B A_{i}=0 \Leftrightarrow B\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}, \frac{1}{n} \sum_{i=1}^{n} y_{i}\right)$.
T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,198 |
5. Let $D$ be a point inside $\triangle A B C$ such that $\angle D A C = \angle D C A = 30^{\circ}$ and $\angle D B A = 60^{\circ}$. Let $E$ be the midpoint of side $B C$, and $F$ be a point on side $A C$ such that $A F = 2 F C$. Prove that $D E \perp E F$.
(Ye Zhonghao provided) | 5. Proof 1: As shown in Figure 3, draw $D M \perp A C$ at point $M$, $F N \perp C D$ at point $N$, and connect $E M, E N$.
Let $C F=a, A F=2 a$, then
$$
C N=C F \cos 30^{\circ}=\frac{\sqrt{3} a}{2}=\frac{1}{2} C D,
$$
which means $N$ is the midpoint of $C D$.
Since $M$ is the midpoint of side $A C$, and $E$ is the mid... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,199 |
6. Given $a, b, c \geqslant 0, a+b+c=1$. Prove:
$$
\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3} \text {. }
$$
(Li Weiguo provided) | 6. Proof 1: Without loss of generality, assume $b \geqslant c$. Let
$$
\sqrt{b}=x+y, \sqrt{c}=x-y \text {. }
$$
Then $b-c=4 x y, a=1-2 x^{2}-2 y^{2}, x \leqslant \frac{1}{\sqrt{2}}$.
$$
\begin{array}{l}
\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c} \\
=\sqrt{1-2 x^{2}-2 y^{2}+4 x^{2} y^{2}}+2 x \\
\leqslant \sqrt{1-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,200 |
7. Given integers $a$, $b$, and $c$ whose absolute values do not exceed 10, the cubic polynomial $f(x)=x^{3}+a x^{2}+b x+c$ satisfies the condition $|f(2+\sqrt{3})|<0.0001$. Question: Is $2+\sqrt{3}$ necessarily a root of this polynomial?
（Provided by Zhang Jingzhong） | 7. Substituting $2+\sqrt{3}$, we get
$$
\begin{array}{l}
f(2+\sqrt{3})=(2+\sqrt{3})^{3}+a(2+\sqrt{3})^{2}+b(2+\sqrt{3})+c \\
= 8+12 \sqrt{3}+18+3 \sqrt{3}+4 a+4 \sqrt{3} a+ \\
3 a+2 b+\sqrt{3} b+c \\
=(26+7 a+2 b+c)+(15+4 a+b) \sqrt{3} .
\end{array}
$$
Let $7 a+2 b+c+26=m, 4 a+b+15=n$, then
$$
|m|\frac{1}{260},
\end{... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,201 |
3. In Rt $\triangle A B C$, $P$ is the midpoint of the hypotenuse $B C$, and $Q$, $R$ are points on $A B$, $A C$ respectively. Prove that the perimeter of $\triangle P Q R$ is greater than the length of $B C$.
untranslated text remains the same as the source text. | (提示: As shown in Figure 19, construct the symmetric points $M, N$ of point $P$ with respect to $AB, AC$, and connect $MQ, NR$. By symmetry, we know $PQ=MQ, PR=NR$. Connect $AP$, by symmetry we know $M, A, N$ are collinear, and $\angle MPN=90^{\circ}$. Therefore, $MN=2AP=BC$. Hence, $PQ+QR+RP=MQ+QR+RN>MN=BC$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,202 |
8. $n$ chess players participate in a chess tournament, with each pair of players competing in one match. The rules are: the winner gets 1 point, the loser gets 0 points, and in the case of a draw, both players get 0.5 points. If it is found after the tournament that among any $m$ players, there is one player who has w... | 8. First, prove two lemmas.
Lemma 1 When $n \geqslant m$, if the match results of $n$ players have property $P(m)$, then there must be a player who wins all others.
Proof of Lemma 1: When $n=m$, the proposition is obviously true.
Assume the proposition holds for $n$.
Then for $n+1$ players, choose any $n$ players. By ... | 2m-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,203 |
1. There are two four-digit numbers, their difference is 534, and the last four digits of their square numbers are the same. Then the larger four-digit number has ( ) possibilities.
(A) 1
(B) 2
(C) 3
(D) 4 | - 1.C.
Let the larger four-digit number be $x$, and the smaller four-digit number be $y$. Then,
$$
x-y=534 \text {, }
$$
and $x^{2}-y^{2}$ is divisible by 10000.
Since $x^{2}-y^{2}=(x+y)(x-y)=267 \times 2(x+y)$, then $x+y$ is divisible by 5000.
Let $x+y=5000 k\left(k \in \mathbf{N}_{+}\right)$.
Solving equations (1) ... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,204 |
2. In $\triangle A B C$, $D$ and $E$ are points on sides $A B$ and $B C$ respectively, and $\frac{B E}{E C}+\frac{B D}{D A}=2 \sqrt{2}, A E$ intersects $C D$ at $F$, and the extension of $B F$ intersects $A C$ at $G$. Then $\frac{B F}{F G}=(\quad)$.
(A) $\sqrt{2}$
(B) $2 \sqrt{2}$
(C) $4 \sqrt{2}$
(D) 8 | 2.B.
As shown in Figure 1, notice that
$$
\begin{array}{l}
\frac{B F}{F G}=\frac{S_{\triangle A B F}+S_{\triangle B F C}}{S_{\triangle A F C}} \\
=\frac{S_{\triangle B F F}}{S_{\triangle A F C}}+\frac{S_{\triangle B F C}}{S_{\triangle A F C}} . \\
\text { Also, } \frac{B E}{E C}=\frac{S_{\triangle A B F}}{S_{\triangle... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,205 |
3. Given that $x$ is a real number, and
$$
|3 x-1|+|4 x-1|+\cdots+|17 x-1|
$$
is a definite constant. Then this constant is ( ).
(A) 5
(B) 10
(C) 15
(D) 75 | 3.A.
Since the value of the original expression is a definite constant, removing the absolute value symbol should eliminate $x$.
$$
\begin{array}{l}
\text { and } 3+4+\cdots+12=13+14+\cdots+17 \text {, therefore, } \\
12 x-1 \leqslant 0 \text {, and } 13 x-1 \geqslant 0 \text {. } \\
\text { Solving, we get } \frac{1}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,206 |
4. Given that $\triangle A B C$ and $\triangle B C D$ are both equilateral triangles, and points $A$ and $D$ are on opposite sides of side $B C$. A line through $D$ intersects the extension of $A B$ at point $P$ and the extension of $A C$ at point $Q$. Connecting $P C$ and $Q B$ intersect at point $M$. Then $\angle B M... | 4. B. As shown in Figure 2, it is easy to see that $\triangle P B D \backsim \triangle D C Q$.
$$
\text { Therefore, } \frac{P B}{D C}=\frac{B D}{C Q} \text {. }
$$
Since $D C=B D=B C$, we have
$$
\frac{P B}{B C}=\frac{B C}{C Q} \text {. }
$$
Given $\angle P B C=\angle B C Q=$
$120^{\circ}$, we can conclude
$$
\trian... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,207 |
5. Given that $\alpha, \beta$ are the roots of the equation $x^{2}+x-1=0$. Then the value of $\alpha^{16}-987 \beta+411$ is ( ).
(A) 2006
(B) 2007
(C) 2008
(D) 2009 | 5.C.
From the given, we have $\alpha^{2}=1-\alpha$ and $\alpha+\beta=-1$. Therefore,
$$
\begin{array}{l}
\alpha^{4}=(1-\alpha)^{2}=1-2 \alpha+\alpha^{2}=2-3 \alpha, \\
\alpha^{8}=(2-3 \alpha)^{2}=4-12 \alpha+9 \alpha^{2}=13-21 \alpha, \\
\alpha^{16}=(13-21 \alpha)^{2}=169-546 \alpha+441 \alpha^{2}=610-987 \alpha .
\en... | 2008 | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,208 |
6. Given that $\odot O$ is the circumcircle of acute $\triangle A B C$, $N$ is the midpoint of $B C$, and point $M$ is on $O A$, and $\angle A B C=$ $4 \angle O N M, \angle A C B=6 \angle O N M$. Then the size relationship between $\angle O M N$ and $\angle O N M$ is ( ).
(A) $\angle O M N>\angle O N M$
(B) $\angle O M... | 6. B.
As shown in Figure 3, connect $O B$, extend $O N$ to intersect $\odot O$ at point $D$, and connect $A D$. Let $\angle O N M=x$, then
$$
\begin{array}{l}
\angle A B C=4 x, \\
\angle A C B=6 x . \\
\text { Hence } \angle A O B=2 \angle A C B \\
=12 x, \\
\angle B A C=180^{\circ}-10 x .
\end{array}
$$
By the conve... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,209 |
1. If the graph of the function with respect to $x$
$$
y=(a-3) x^{2}-(4 a-1) x+4 a
$$
intersects the coordinate axes at two points, then the value of $a$ is | $$
\text { II. 1.3, 0 or }-\frac{1}{40} \text {. }
$$
When $a-3=0$, i.e., $a=3$, the original function becomes $y=-11 x + 12$, and its graph intersects the coordinate axes at two points.
When $a-3 \neq 0$, i.e., $a \neq 3$, the original function is a quadratic function, which must intersect the $y$-axis at one point.... | a=3, 0 \text{ or } -\frac{1}{40} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,210 |
2. If $P$ is any point inside quadrilateral $A B C D$, and
$$
S_{\triangle P A B}+S_{\triangle P C D}=\frac{1}{2} S_{\text {quadrilateral } A B C D}
$$
always holds, then the shape of quadrilateral $A B C D$ is $\qquad$ | 2. Parallelogram.
As shown in Figure 4, draw $P P^{\prime} / / A B$, with point $P^{\prime}$ inside quadrilateral $A B C D$, and connect
$$
P^{\prime} A, P^{\prime} B, P^{\prime} C, P^{\prime} D \text {. }
$$
Then $S_{\triangle P R B}=S_{\triangle P A B}$.
From the given information,
Therefore, $S_{\triangle P C D}=... | Parallelogram | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,211 |
3. Given that $a$, $b$, and $c$ are positive numbers, and
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}>1 \text{. }
$$
Then, with $a$, $b$, and $c$ as sides, a triangle can be formed (fill in "can" or "cannot"). | 3. Can.
Given that $a$, $b$, and $c$ are positive numbers, removing the denominators from the given equation yields
$$
\begin{array}{l}
a\left(b^{2}+c^{2}-a^{2}\right)+b\left(a^{2}+c^{2}-b^{2}\right) \\
>2 a b c-c\left(a^{2}+b^{2}-c^{2}\right) .
\end{array}
$$
Factoring both sides of the above equation, we get
$$
\be... | Can | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,212 |
4. As shown in Figure 20, divide the three sides of any $\triangle ABC$ into four equal parts. The division points on sides $BC$, $CA$, and $AB$ are $A_{1}$, $A_{2}$, $A_{3}$, $B_{1}$, $B_{2}$, $B_{3}$, $C_{1}$, $C_{2}$, $C_{3}$, respectively. Let the perimeters of $\triangle ABC$ and $\triangle A_{1} B_{1} C_{1}$ be $... | (It is known that $C_{1} B_{3}=\frac{1}{4} B C$. In $\triangle B_{1} B_{3} C_{1}$, we have $C_{1} B_{3}+B_{3} B_{1}>B_{1} C_{1}$, i.e., $\frac{1}{4} B C+\frac{1}{2} C A>B_{1} C_{1}$. Similarly, $\frac{1}{4} C A+\frac{1}{2} A B>C_{1} A_{1}, \frac{1}{4} A B+\frac{1}{2} B C>A_{1} B_{1}$. Adding the three inequalities yiel... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,213 |
4. Quadrilateral $A B C D$ is a circumscribed quadrilateral of $\odot O$. Then the positional relationship between the incircle $\odot O_{1}$ of $\triangle A B C$ and the incircle $\odot O_{2}$ of $\triangle A C D$ is $\qquad$ | 4. Externally Tangent.
As shown in Figure 5, let $\odot O_{1}$ and $\odot O_{2}$ be tangent to $AC$ at points $E$ and $F$, respectively. Then,
$$
\begin{array}{l}
A E=\frac{1}{2}(A B+A C-B C), \\
A F=\frac{1}{2}(A D+A C-C D) .
\end{array}
$$
Since quadrilateral $ABCD$ is an ex-tangential quadrilateral of $\odot O$, w... | Externally Tangent | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,214 |
One, (20 points) Given that $a$, $b$, and $c$ are positive real numbers, and $abc=1$. Find the minimum value of
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}
$$ | When $a=b=c=1$, the original expression $=\frac{3}{4}$.
Below, we compare the original expression with $\frac{3}{4}$.
$$
\begin{array}{l}
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \\
\frac{c}{(c+1)(a+1)}-\frac{3}{4} \\
=\frac{a b+a c+b c+a+b+c-6}{4(a+1)(b+1)(c+1)} .
\end{array}
$$
Since $a b c=1$, and $a, b, c$ are p... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,215 |
II. (25 points) Is the statement "If two isosceles triangles have equal perimeters and equal areas, then they are congruent" true? If so, provide a proof; if not, give a counterexample. | II. The original proposition is a false proposition. Counterexamples are given as follows.
As shown in Figures 6 and 7, $\triangle ABC$ and $\triangle A'B'C'$ are both isosceles triangles, with $BC$ and $B'C'$ being their bases, and $AD$ and $A'D'$ being their altitudes.
Figure 6
Figure 7
Let $AD = 4k$, $A'D' = 2k$, $B... | not found | Geometry | proof | Yes | Yes | cn_contest | false | 719,216 |
Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.
保留源文本的换行和格式，直接输出翻译结果。 | Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$
From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$
Since the left side of equation (2) is a posi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,217 |
1. Given $\triangle A B C$ is an acute triangle,
$$
\begin{array}{l}
f(x)=-x^{2}+2 x+m, \\
p=\sin A+\sin B+\sin C, \\
q=\cos A+\cos B+\cos C .
\end{array}
$$
Then ( ).
(A) $f(p)>f(q)$
(B) $f(p)=f(q)$
(C) $f(p)<f(q)$
(D) Any of the above three cases is possible | $-、 1 . C$.
From $\triangle A B C$ being an acute triangle, we get
$$
\angle A+\angle B>\frac{\pi}{2} \Rightarrow \angle A>\frac{\pi}{2}-\angle B>0 \text {. }
$$
Thus, $\sin A>\cos B$.
Similarly, $\sin B>\cos C, \sin C>\cos A$.
Therefore, $p>q$.
Since $\cos A+\cos B+\cos C$
$$
=1+4 \sin \frac{A}{2} \cdot \sin \frac{B}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,218 |
3. Given that the surface area of a rectangular prism is $\frac{45}{2} \mathrm{~cm}^{2}$, and the total length of all edges is $24 \mathrm{~cm}$. Then, the maximum angle formed between the body diagonal of the rectangular prism and an edge is $($.
(A) $\arccos \frac{1}{3}$
(B) $\arccos \frac{\sqrt{2}}{3}$
(C) $\arccos ... | 3. D.
Let the lengths of the three edges meeting at the same vertex be $a, b, c (a \geqslant b \geqslant c)$. Then
$$
\left\{\begin{array}{l}
2(a b + b c + c a) = \frac{45}{2} \\
4(a + b + c) = 24
\end{array} \Rightarrow \left\{\begin{array}{l}
a + b = 6 - c, \\
a b = c^{2} - 6 c + \frac{45}{4}.
\end{array}\right.\rig... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,220 |
4. Given that the circumcenter of $\triangle A B C$ is $O, B C>C A>A B$. Then $(\quad$.
(A) $\mathrm{OA} \cdot \mathrm{OB}>\mathrm{OA} \cdot \mathrm{OC}>\boldsymbol{O B} \cdot \mathrm{OC}$
(B) $\mathrm{OA} \cdot \mathrm{OB}>\mathrm{OB} \cdot \mathrm{OC}>\mathrm{OC} \cdot \mathrm{OA}$
(C) $\mathrm{OB} \cdot \mathrm{OC}>... | 4.A.
Let the circumradius of $\triangle A B C$ be $R$. Then
$$
\begin{array}{l}
O A \cdot O B=R^{2} \cos 2 C, \\
O B \cdot O C=R^{2} \cos 2 A, \\
O C \cdot O A=R^{2} \cos 2 B .
\end{array}
$$
Given $B C>C A>A B$, we have
$$
\sin A>\sin B>\sin C>0 \text {. }
$$
Thus, $1-2 \sin ^{2} A O A \cdot O C>O B \cdot O C$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,221 |
5. If $a>1, b>1$, then the minimum value of $\frac{a^{2}+b^{2}}{a+b-2}$ is ( ).
(A) 4
(B) 6
(C) 8
(D) does not exist | 5.A.
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{a+b-2} \geqslant \frac{2\left(\frac{a+b}{2}\right)^{2}}{a+b-2}=\frac{1}{2} \cdot \frac{(a+b-2+2)^{2}}{a+b-2} \\
=\frac{1}{2}\left[(a+b-2)+4+\frac{4}{a+b-2}\right] \\
\geqslant \frac{1}{2}\left[4+2 \sqrt{(a+b-2) \frac{4}{a+b-2}}\right]=4 .
\end{array}
$$
Equality holds if and... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,222 |
5. In a convex quadrilateral $A B C D$, $A B+A C+C D=16$. When the diagonals $A C$ and $B D$ are what values, will the area of quadrilateral $A B C D$ be maximized? What is the maximum area? | (Hint: Let $A B=x, A C=y$, then $C D=16-x-y$. And
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=S_{\triangle A B C}+S_{\triangle C D} \\
\leqslant \frac{1}{2} x y+\frac{1}{2} y(16-x-y) \\
=-\frac{1}{2}(y-8)^{2}+32 .
\end{array}
$$
Therefore, when $\angle B A C=\angle A C D=90^{\circ}, A C=8, B D=8 \sqrt{2}$,... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,224 |
1. Given $A=\{0,1,2,3,4,5,6,7\}, f: A \rightarrow$ $A$. If $i+j=7$, then $f(i) f(j)=i j$. Then, the number of mappings $f$ is $\qquad$ | For $0$ and $7$ elements, since $f(i) f(j)=0$, at least one of the images of $0$ and $7$ must be $0$. There are a total of $2 \times 8-1=15$ cases.
For the images of $1$ and $6$ elements, in this case, $f(i) f(j)=6=1 \times 6=2 \times 3$, so the images of $1$ and $6$ can be: $1, 6, 6, 1, 2, 3, 3, 2$, totaling 4 cases.... | 480 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,225 |
2. Given a sequence of positive terms $\left\{a_{n}\right\}$ satisfying
$$
a_{1}=\frac{3}{2}, a_{n+1}^{2}-a_{n}^{2}=\frac{1}{(n+2)^{2}}-\frac{1}{n^{2}} \text {. }
$$
Then the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{n}=$ $\qquad$ | $$
\begin{array}{l}
\text { 2. } \frac{n(n+2)}{n+1} . \\
a_{n}^{2}=\left(a_{n}^{2}-a_{n-1}^{2}\right)+\left(a_{n-1}^{2}-a_{n-2}^{2}\right)+\cdots+ \\
=\frac{\left(a_{2}^{2}-a_{1}^{2}\right)+a_{1}^{2}}{(n+1)^{2}}-\frac{1}{(n-1)^{2}}+\frac{1}{n^{2}}-\frac{1}{(n-2)^{2}}+\cdots+ \\
\frac{1}{3^{2}}-\frac{1}{1^{2}}+\left(\fr... | \frac{n(n+2)}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,226 |
3. A five-digit number consists of only the three different digits $1, 3, 5$. Then the number of such five-digit numbers is $\qquad$.
| 3.150.
Considering overall, there are
$$
3^{5}-C_{3}^{2}\left(2^{5}-2\right)-3=150 \text { (cases). }
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,227 |
4. Given that the area of $\triangle A B C$ is $S$, the three medians of $\triangle A B C$ form $\triangle A_{1} B_{1} C_{1}$, with area $S_{1}$, and the three medians of $\triangle A_{k} B_{k} C_{k}(k=1,2, \cdots)$ form $\triangle A_{k+1} B_{k+1} C_{k+1}$, with area $S_{k+1}$. Then
$$
S_{1}+S_{2}+\cdots+S_{n}+\cdots=
... | 4.3S.
As shown in Figure 2, the three medians of $\triangle ABC$ intersect at point $G$. Extend $AD$ to point $M$ such that $DM = DG$. Connect $BM$ and $CM$. Then quadrilateral $BMC G$ is a parallelogram. Therefore,
$$
\frac{BM}{CF} = \frac{GM}{AD} = \frac{BG}{BE} = \frac{2}{3}.
$$
Hence, $\triangle BGM$ is similar t... | 3S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,228 |
5. If $\theta \in\left(0, \frac{\pi}{2}\right), f(\theta)=\cos \theta+\theta \sin \theta$, then among the following judgments
(1) $f(\theta)>1$,(2) $f(\theta)1$; there also exists $\theta$, such that $f(\theta)<1$, the correct sequence number is $\qquad$ . | 5. (1).
Given $\theta \in\left(0, \frac{\pi}{2}\right)$, we know $\cos \theta>\cos ^{2} \theta, \theta>\sin \theta$. Therefore, $f(\theta)>\cos ^{2} \theta+\sin \theta \cdot \sin \theta=1$. | (1) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,229 |
6. Given $x, y \in \mathbf{R}$. If
$$
a(x+y) \leqslant 1+x^{2}+x y+y^{2}
$$
always holds, then the range of values for $a$ is $\qquad$ _. | 6. $[-\sqrt{3}, \sqrt{3}]$.
Notice that
$$
\begin{array}{l}
a(x+y) \leqslant 1+x^{2}+x y+y^{2} \\
=1+\frac{3}{4}(x+y)^{2}+\frac{1}{4}(x-y)^{2} .
\end{array}
$$
If $x+y=0$, it is clear that $a \in \mathbf{R}$.
If $x+y>0$, then
$$
a \leqslant \frac{1}{x+y}+\frac{3}{4}(x+y)+\frac{1}{4} \cdot \frac{(x-y)^{2}}{x+y}
$$
al... | [-\sqrt{3}, \sqrt{3}] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,230 |
Three, (20 points) When the faces of an $n(n \geqslant 3)$-sided pyramid are extended into planes, into how many parts can space be divided at most? | Three, let the $n$ side planes divide the space into $a_{n}$ parts. Since the $n$ planes intersect at one point, the $n$-th plane intersects with the previous $n-1$ planes, resulting in $n-1$ intersection lines. These $n-1$ intersection lines, intersecting at one point, divide the $n$-th plane into $2(n-1)$ parts, and ... | \frac{3}{2} n^{2}-\frac{1}{2} n+3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,231 |
Four. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{n+1}=a_{n}+\sqrt{a_{n}^{2}+1} . \\
\text { Prove: } a_{1}+a_{2}+\cdots+a_{n}>\frac{(n-1) \pi}{2} .
\end{array}
$$ | Given $a_{1}=1>0, a_{n+1}=a_{n}+\sqrt{a_{n}^{2}+1}$, it is easy to see that $a_{n}>0$.
Let $a_{n}=\tan \theta_{n}, \theta_{n} \in\left(0, \frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
a_{1}=\tan \theta_{1}=1, \theta_{1}=\frac{\pi}{4}, \\
a_{n+1}=\tan \theta_{n+1}=\tan \theta_{n}+\sqrt{\tan ^{2} \theta_{n}+1} \\
\qua... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,232 |
Five. (20 points) Let $M$ be the intersection point of the left directrix and the axis of symmetry of the ellipse $C$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
The left focus of the ellipse $C$ is $F$. Draw any line through $M$ with a non-zero slope that intersects the ellipse $C$ at points $P$ and $Q$. ... | As shown in Figure 3, according to the problem, let the equation of the line $P Q$ be
$$
y=k\left(x+\frac{a^{2}}{c}\right)
$$
$(k \neq 0)$.
Combining this with the equation of the ellipse $C$ and eliminating $y$, we get
$$
\begin{array}{l}
\left(b^{2}+a^{2} k^{2}\right) x^{2}+\frac{2 a^{4} k^{2}}{c} x+\frac{a^{6} k^{2}... | not found | Geometry | proof | Yes | Yes | cn_contest | false | 719,233 |
一、(50 points) As shown in Figure 1, in the acute triangle $\triangle ABC$, $O$ is the circumcenter, $OM \perp AB$ at point $M$, $AD$ and $BE$ are two altitudes, $ED = AE + BD$, $R$ and $r$ are the radii of the circumcircle and incircle of $\triangle ABC$ respectively. Prove that $OM$ is the arithmetic mean of $R$ and $... | As shown in Figure 4, draw $O N \perp$ $B C$ at point $N$, $O L \perp A C$ at point $L$, and connect $O A, O B, O C$.
Using the properties of the circumcenter, we have
$$
\begin{array}{l}
\angle B O N=\angle A, \\
O N=R \cos A, \\
\angle A O L=\angle B, \\
O L=R \cos B, \\
\angle A O M=\angle C, \\
O M=R \cos C .
\end{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,234 |
1. The number of real solutions $x$ that satisfy the equation
$$
\sqrt{x+5-4 \sqrt{x+1}}+\sqrt{x+10-6 \sqrt{x+1}}=1
$$
is ( ).
(A) 1
(B) 2
(C) 4
(D) infinitely many | $-1 . D$
Let $x_{0}=\sqrt{x+1}$, the equation can be rewritten as
$$
\left|x_{0}-2\right|+\left|x_{0}-3\right|=1 \text {. }
$$
Its geometric meaning is: on the number line, the sum of the distances from point $x_{0}$ to points 2 and 3 is 1.
Therefore, $x_{0}$ can take any real number satisfying $2 \leqslant x_{0} \le... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,235 |
2. As shown in Figure 1, in the unit square $A B C D$, a semicircle is drawn inside with $A B$ as the diameter. Tangents $C E$ and $D F$ (where $E$ and $F$ are the points of tangency) are drawn from points $C$ and $D$ to the semicircle. Then the length of segment $E F$ is ( ).
(A) $\frac{\sqrt{5}}{3}$
(B) $\frac{3}{5}$... | 2. B.
As shown in Figure 6, let the center of the semicircle be $G$, and connect $A E$, $A F$, $B E$, $B F$, and $G F$.
Since $A$, $D$, $F$, and $G$ are concyclic,
$\angle B G F = \angle A D F$.
Thus, $\triangle A D F \sim \triangle B G F$,
$$
\frac{B F}{A F} = \frac{B G}{A D} = \frac{1}{2}.
$$
Let $B F = x$, then $... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,236 |
One, (20 points) Given that $a_{i}, b_{i} (i=1,2,3)$ are real numbers, and $a_{1}^{2}-a_{2}^{2}-a_{3}^{2}$ and $b_{1}^{2}-b_{2}^{2}-b_{3}^{2}$ are at least one positive number. Prove: the quadratic equation in $x$
$$
\begin{array}{l}
x^{2}+2\left(a_{1} b_{1}-a_{2} b_{2}-a_{3} b_{3}\right) x+ \\
\left(a_{1}^{2}-a_{2}^{2... | Let's assume $a_{1}^{2}-a_{2}^{2}-a_{3}^{2}>0$, then $a_{1} \neq 0$.
Consider the quadratic equation
$\left(a_{1}^{2}-a_{2}^{2}-a_{3}^{2}\right) x^{2}+2\left(a_{1} b_{1}-a_{2} b_{2}-a_{3} b_{3}\right) x+$
$\left(b_{1}^{2}-b_{2}^{2}-b_{3}^{2}\right)=0$.
Let $f(x)=\left(a_{1}^{2}-a_{2}^{2}-a_{3}^{2}\right) x^{2}+2\left(a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,237 |
1. In a $4 \times 4$ grid, eighteen lines can be drawn, including eight lines (four horizontal and four vertical), and five diagonal lines from the top left to the bottom right and from the top right to the bottom left. These diagonal lines may pass through 2, 3, or 4 small squares. Ten chips are to be placed in the gr... | 2. As shown in Figure 11, the maximum score is 17.
The following proves that 18 points cannot be achieved.
Notice that, in Figure 11, there are always two opposite corner cells not covered by the same direction of five diagonal lines, and these two cells must be either both empty or both full. At the same time, there m... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,241 |
3. If a prime number remains prime after any permutation of its digits, it is called an "absolute prime" (such as $2,3,5,7,11,13(31), 17(71), 37(73), 79(97)$, $113(131,311), 199(919,991), 337(373,733)$, … are all absolute primes). Prove: The digits of an absolute prime cannot simultaneously contain the digits $1,3,7$ a... | 3. If an absolute prime number contains the digits 1, 3, 7, 9, then in its decimal representation, it cannot contain the digits 0, 2, 4, 5, 6, 8; otherwise, by appropriate rearrangement, this number can be divisible by 2 or 5.
Let \( N \) be an absolute prime number that contains the digits 1, 3, 7, 9.
\[
\begin{array... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,243 |
1. Figure 6 is an Olympic five-ring, with 9 small circles inside the rings. Please fill in the numbers $1 \sim 9$ into the small circles so that the sum of the numbers in each ring is 14. | The sum of these 9 numbers is 45, the sum of the numbers in the 5 circles is $5 \times 14=70$, and their difference $70-45=25$ is the sum of the numbers in the four small circles that fall in the overlapping parts of the circles.
The numbers in the outermost two circles can only be 9 and 5 on one side, and 6 and 8 on ... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 719,244 |
3. Figure 8 is a heptagram, it has a total of 14 intersection points. Please fill in the numbers $1,2, \cdots, 14$ at each intersection (one number per intersection) so that the sum of the four numbers on each line is equal (please provide one solution, no need to write out the reasoning process) | 3. Since each point has exactly two lines passing through it, for each correct filling method, if the sum of the four numbers on each line is $s$, then
$$
7 s=2(1+2+\cdots+14)=210.
$$
Solving for $s$ gives $s=30$.
It is easy to see that 14 must be collinear with either 1 or 2, and 13 must be collinear with one of 1, 2... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 719,246 |
4. Mary found that when a certain three-digit number is squared, the last three digits of the product are the same as the original three-digit number. What is the sum of all different three-digit numbers that satisfy this property? | 4. Let the three-digit number be $\overline{a b c}$, then
$$
\overline{a b c}^{2}=1000 k+\overline{a b c} \text {, }
$$
i.e., $\overline{a b c}(\overline{a b c}-1)=2^{3} \times 5^{3} k$.
Since $(\overline{a b c}, \overline{a b c}-1)=1$, therefore,
$2^{3} \mid \overline{a b c}$, and $5^{3} \mid(\overline{a b c}-1)$,
o... | 1001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,247 |
II. (25 points) As shown in Figure 4, from any point $M$ on the arc $\overparen{BC}$ of the circumcircle of $\triangle ABC$, draw $MD \perp BC$ at $D$. $P$ is a point on $AM$. Draw $PE \perp AC$, $PF \perp AB$, $PG \perp BC$, with $E$, $F$, and $G$ lying on $AC$, $AB$, and $AD$ respectively. Prove that $E$, $F$, and $G... | II. As shown in Figure 9, connect $GE$, $GF$, $MB$, and $MC$.
Since $PE \perp AC$,
$$
PF \perp AB, PG \perp BC,
$$
then
$$
\begin{array}{l}
\angle GPF \\
=\angle ABC, \\
\angle GPE=\angle ACB .
\end{array}
$$
By $PG \parallel MD$, we get
$$
\frac{PG}{MD}=\frac{AP}{AM} \text{. }
$$
Also, by $\triangle APE \sim \trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,248 |
5. Find all positive integers $m, n$ such that $m^{2}+1$ is a prime number, and
$$
10\left(m^{2}+1\right)=n^{2}+1 \text {. }
$$ | 5. From the given condition, we have
$$
9\left(m^{2}+1\right)=(n-m)(n+m) \text {. }
$$
Notice that $m^{2}+1$ is a prime number, and $m^{2}+1 \equiv 1$ or $2(\bmod 3)$, so $m^{2}+1$ is not a multiple of 3. Therefore,
$$
\left\{\begin{array}{l}
n-m=1,3,9, m^{2}+1, \\
n+m=9\left(m^{2}+1\right), 3\left(m^{2}+1\right), m^{... | (2,7),(4,13) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,249 |
7. There are two $3 \times 3$ grids, 1 and 2. Fill the numbers 1, 2, ..., 9 into grid 1 (one number per cell) in some order, and then fill grid 2 according to the following rule: the number in the cell at the intersection of the $i$-th row and $j$-th column of grid 2 is equal to the absolute value of the difference bet... | 7. Such an arrangement cannot be made.
Remove the absolute value symbols from the numbers in Table 2, and the resulting table is denoted as
Table 3. Table 3. Then
\begin{tabular}{|l|l|l|}
\hline$c_{11}$ & $c_{12}$ & $c_{13}$ \\
\hline$c_{21}$ & $c_{22}$ & $c_{23}$ \\
\hline$c_{31}$ & $c_{32}$ & $c_{33}$ \\
\hline
\en... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 719,251 |
8. Let the diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$ be perpendicular to each other. Through the midpoint $M$ of $A B$, draw a perpendicular to $C D$, and through the midpoint $N$ of $A D$, draw a perpendicular to $C B$. Prove: The intersection of these two perpendiculars lies on the line $A C$.
| 8. As shown in Figure 18, take the midpoint $K$ of $AC$. Then
$MN \parallel BD, MK \parallel BC, NK \parallel CD$.
Thus, the two perpendicular lines and $AC$ are the lines containing the altitudes of $\triangle MNK$, so they are concurrent.
Therefore, the proposition is true. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,252 |
9. Arrange the $n$ positive integers from $1 \sim n (n>1)$ in a row so that the sum of any two adjacent numbers is a perfect square. What is the minimum value of $n$? | 9. The smallest positive integer $n$ is 15.
Since $n>1$, it includes 2. The smallest positive integer that, when added to 2, results in a perfect square is 7, represented as $2+7=9$ (the same notation applies below), so $n \geqslant 7$.
If $n=7$, we can get three non-adjacent segments: $(1,3,6),(2,7),(4,5)$.
Adding ... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,253 |
1. Given $a \neq 0$, and the equation in terms of $x$
$$
a x^{2}-b x-a+3=0
$$
has at most one solution. Determine whether the equation in terms of $x$
$$
(b-3) x^{2}+(a-2 b) x+3 a+3=0
$$
must have a solution, and prove your conclusion. | 1. From the problem, we know that the discriminant of equation (1) is
$$
\begin{array}{l}
\Delta_{1}=b^{2}+4 a(a-3) \leqslant 0 \\
\Rightarrow b^{2}+(2 a-3)^{2} \leqslant 9 \\
\Rightarrow\left\{\begin{array} { l }
{ - 3 \leqslant b \leqslant 3 , } \\
{ - 3 \leqslant 2 a - 3 \leqslant 3 }
\end{array} \Rightarrow \left\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,255 |
2. Given point $D$ is the midpoint of the base $BC$ of isosceles $\triangle ABC$, and $P$ is any point on the line segment $AB$. Let the perpendicular bisector of $BP$ intersect line $AD$ at point $E$, and $PC$ intersect $AD$ at point $F$. Prove that line $EP$ is the tangent to the circumcircle of $\triangle APF$. | 2. With $E$ as the center and $E B$ as the radius, draw a circle, then points $P$ and $C$ are both on the circumference of this circle. Connect $E C$. Then
$$
\begin{array}{l}
\angle P A E=90^{\circ}-\angle A B C \\
=90^{\circ}-\frac{1}{2} \angle P E C=\angle E P C,
\end{array}
$$
Therefore, $E P$ is the tangent to th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,256 |
3. Among the 2007 positive integers $1,2, \cdots, 2007$, what is the maximum number of integers that can be selected such that each selected number is coprime with 2007, and the sum of any three selected numbers is not a multiple of 7? | 3. When $1,2, \cdots, 2007$ are divided by 7, the remainders 1, 2, 3, 4, 5 each have $286+1=287$ numbers; the remainders 6, 0 each have 286 numbers.
Among $1,2, \cdots, 2007$, the numbers that are not coprime with 2007 are $3,2 \times 3,3 \times 3, \cdots, 669 \times 3$ and $223,2 \times 223,4 \times 223,5 \times 223,... | 386 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,257 |
1. Given in Rt $\triangle A B C$, $\angle C=90^{\circ}, \frac{A C}{B C}=$ $1+\frac{\sqrt{6}}{2}$. Then $\frac{A B}{A C}=$ .. $\qquad$ | $$
1.2 \sqrt{2}-\sqrt{3} \text{. }
$$
Given $\tan B=\frac{A C}{B C}=1+\frac{\sqrt{6}}{2}$, then
$$
\frac{A B}{A C}=\frac{1}{\sin B}=\sqrt{1+\cot ^{2} B}=2 \sqrt{2}-\sqrt{3} \text{. }
$$ | 2 \sqrt{2}-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,258 |
Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following... | Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,259 |
The expression $\frac{a^{2008}}{c^{2007}}+\frac{b^{2008}}{(2007-c)^{2007}}$ simplifies to
保留源文本的换行和格式，翻译结果如下：
The expression $\frac{a^{2008}}{c^{2007}}+\frac{b^{2008}}{(2007-c)^{2007}}$ simplifies to | $$
\begin{array}{l}
\text { Given } \frac{a^{2}}{c}+\frac{b^{2}}{2007-c}=\frac{1}{2007}=\frac{(a+b)^{2}}{c+(2007-c)} \\
\Rightarrow a^{2} \frac{2007-c}{c}+b^{2} \frac{c}{2007-c}=2 a b \\
\Rightarrow\left(\frac{a}{c}-\frac{b}{2007-c}\right)^{2}=0 \\
\Rightarrow \frac{a}{c}=\frac{b}{2007-c} .
\end{array}
$$
Also, $a+b=1... | \frac{1}{2007^{2007}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,260 |
3. The minimum value of the algebraic expression $113 \sqrt{x^{2}+3}-110 x$ is . $\qquad$ | $3.3 \sqrt{223}$.
Let $y=113 \sqrt{x^{2}+3}-110 x$, then $y^{2}+220 x y=3 \times 223 x^{2}+3 \times 113^{2}$,
which means $3 \times 223 x^{2}-220 y x+3 \times 113^{2}-y^{2}=0$.
Thus, $\Delta=(200 y)^{2}-4 \times 3 \times 223\left(3 \times 113^{2}-y^{2}\right)$
$$
=4 \times 113^{2}\left(y^{2}-3^{2} \times 223\right) \ge... | 3 \sqrt{223} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,261 |
5. Given in the Cartesian coordinate system $x O y$, the three vertices of $\triangle A B C$ are $A(2 \sqrt{2}, \sqrt{2}+\sqrt{6})$, $B(\sqrt{2}$, $\sqrt{2})$, and $C(5 \sqrt{2}, \sqrt{2})$. Then the coordinates of the intersection point of the altitude from $A$ to side $B C$ and the angle bisector of $\angle A B C$ ar... | 5. $\left(2 \sqrt{2}, \sqrt{2}+\frac{\sqrt{6}}{3}\right)$.
Let the altitude from $A$ to $BC$ and the angle bisector of $\angle ABC$ intersect at point $P(2 \sqrt{2}, \sqrt{2}+h)$. Then $\tan \angle ABC=\frac{\sqrt{6}}{\sqrt{2}}, \tan \angle PBC=\frac{h}{\sqrt{2}}$. Since $\angle ABC=2 \angle PBC$, we have $\frac{\sqrt... | \left(2 \sqrt{2}, \sqrt{2}+\frac{\sqrt{6}}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,263 |
6. It is known that a factory produced 10,000 units of a product in January, 12,000 units in February, and 13,000 units in March. In the $n$-th month, the factory produces $a b^{n}+c$ ten thousand units, where $a$, $b$, and $c$ are constants, and $n=1,2, \cdots, 12$. Then, how many ten thousand units will the factory p... | 6.1 .35 .
From the given conditions, we know
$$
a b+c=1, a b^{2}+c=1.2, a b^{3}+c=1.3 \text {. }
$$
Then $a b(b-1)=0.2, a b^{2}(b-1)=0.1$.
Thus $b=0.5, a=-0.8, c=1.4$.
Therefore, $a b^{4}+c=1.35$. | 1.35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,264 |
7. The equation
$$
3 x^{3}+2 \sqrt{2} x^{2}-(17-9 \sqrt{2}) x-(6-5 \sqrt{2})=0
$$
has solutions $x_{1}=$ $\qquad$ ,$x_{2}=$ $\qquad$ ,$x_{3}=$ $\qquad$ | $7 \cdot \frac{\sqrt{2}}{3}, \sqrt{2}-1,1-2 \sqrt{2}$.
Let $x=\sqrt{2} y$, substituting into the original equation we get
$$
6 \sqrt{2} y^{3}+4 \sqrt{2} y^{2}-17 \sqrt{2} y+18 y-6+5 \sqrt{2}=0 \text {. }
$$
It is easy to see that $y=\frac{1}{3}$ satisfies the condition. Therefore, $x_{1}=\frac{\sqrt{2}}{3}$. Thus,
$$
... | x_{1}=\frac{\sqrt{2}}{3}, x_{2}=\sqrt{2}-1, x_{3}=1-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,265 |
8. Given that the ratio of the square of the perimeter of rectangle $A B C D$ to its area is $k$. Then the ratio of the longer side to the shorter side of rectangle $A B C D$ is equal to | 8. $\frac{k-8}{8}+\frac{1}{8} \sqrt{k(k-16)}$.
Let the length and width of the rectangle be $a$ and $b$ ($a \geqslant b$). Then $\frac{4(a+b)^{2}}{a b}=k$, which means $4 a^{2}+(8-k) a b+4 b^{2}=0$. Let $t=\frac{a}{b}$, then $4 t^{2}+(8-k) t+4=0$. Solving for $t$ gives $t=\frac{1}{8}[(k-8)+\sqrt{k(k-16)}]$. | \frac{k-8}{8}+\frac{1}{8} \sqrt{k(k-16)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,266 |
9. Given a square piece of paper $A B C D$ with an area of $2007 \mathrm{~cm}^{2}$. Now fold the paper along a line segment (as shown in Figure 1), so that point $D$ lands on point $D^{\prime}$ on side $B C$, and point $A$ lands on point $A^{\prime}$, with $A^{\prime} D^{\prime}$ intersecting $A B$ at point $E$. Then t... | $9.6 \sqrt{223}$.
Let the side length of the square be $a=\sqrt{2007}, \angle D^{\prime} D C=\alpha$.
Then $\angle B D^{\prime} E=2 \alpha, C D^{\prime}=a \tan \alpha$, $B D^{\prime}=a(1-\tan \alpha)$.
Therefore, the perimeter of $\triangle B D^{\prime} E$ is
$$
\begin{array}{l}
a(1-\tan \alpha)(1+\tan 2 \alpha+\sec 2 ... | 6 \sqrt{223} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,267 |
10. If $x$ is an integer, $3<x<200$, and $x^{2}+(x+1)^{2}$ is a perfect square, then the value of the integer $x$ is $\qquad$. | 10.20 or 119.
Let $x^{2}+(x+1)^{2}=v^{2}$, then $(2 x+1)^{2}=2 v^{2}-1$. Let $u=2 x+1$, then $u^{2}-2 v^{2}=-1$. This is a Pell's equation, its fundamental solution is $\left(u_{0}, v_{0}\right)=(1,1)$.
All positive integer solutions can be obtained by
$$
u_{n}+v_{n} \sqrt{2}=\left(u_{0}+v_{0} \sqrt{2}\right)^{2 n+1}
... | 20 \text{ or } 119 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,268 |
1. If $a, b, c$ are 3 different positive integers, and $a b c=16$, then the maximum possible value of $a^{b}-b^{c}+c^{a}$ is ( ).
(A) 249
(B) 253
(C) 263
(D) 264 | $-1 . C$.
From the given, we easily know that $\{a, b, c\}=\{1,2,8\}$.
And when $b=1, c=2, a=8$, $a^{b}-b^{c}+c^{a}$ reaches the maximum value $8-1+2^{8}=263$. | 263 | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,269 |
1. As shown in Figure 1, in the right triangle $\triangle ABC$, $AB = 35$, and a square CDEF with side length 12 is inscribed in $\triangle ABC$. Then the perimeter of $\triangle ABC$ is ( ).
(A) 35
(B) 40
(C) 81
(D) 84 | -、1.D.
Let $B C=a, A C=b$. Then
$$
a^{2}+b^{2}=35^{2}=1225.
$$
Also, since Rt $\triangle A F E \backsim$ Rt $\triangle A C B$, we have $\frac{F E}{C B}=\frac{A F}{A C}$, which means $\frac{12}{a}=\frac{b-12}{b}$.
Thus, $12(a+b)=a b$.
From equations (1) and (2), we get
$$
(a+b)^{2}=1225+24(a+b).
$$
Solving this, we ge... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,270 |
2. It is known that the sum of the reciprocals of three consecutive positive integers equals $\frac{191}{504}$. Then the sum of these three numbers is ( ).
(A) 27
(B) 24
(C) 21
(D) 18 | 2.B.
Let these three consecutive positive integers be $n-1, n, n+1$. Then
$$
\frac{191}{504}=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}=\frac{3 n^{2}-1}{n\left(n^{2}-1\right)},
$$
which simplifies to $191 n^{3}-1512 n^{2}-191 n+504=0$.
Factoring gives $(n-8)\left(191 n^{2}+16 n-63\right)=0$.
When $n \geqslant 1$, $191 n... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,271 |
3. The number of positive simplest proper fractions with a denominator of 2007 is ( ).
(A) 675
(B) 1326
(C) 1329
(D) 1332 | 3. D.
Since $2007=3^{2} \times 223$, the number of positive integers less than 2007 and coprime with 2007 is
$$
\begin{array}{l}
2007-\left[\frac{2007}{3}\right]-\left[\frac{2007}{223}\right]+\left[\frac{2007}{3 \times 223}\right] \\
=1332 \text { (numbers). }
\end{array}
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,272 |
4. For a real number $x$, the symbol $[x]$ represents the greatest integer not greater than $x$, for example, $[3.14]=3,[-7.59]=$ -8. Then the number of integer solutions for the equation $\left[\frac{3 x+7}{7}\right]=4$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 4.B.
From $\left[\frac{3 x+7}{7}\right]=4$, we know $4 \leqslant \frac{3 x+7}{7}<5$, which gives $7 \leqslant$ $x<\frac{28}{3}$. Therefore, $x=7,8,9$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,273 |
5. As shown in Figure 1, given that the length of the rectangle is 8 and the width is 4, fold the rectangle along one of its diagonals and press it flat. Then the area of the overlapping part (the shaded triangle) is ( ).
(A) 10
(B) 12
(C) 14
(D) 16 | 5.A.
As shown in Figure 6, draw $E F \perp B D$ at $F$. It is easy to see that point $F$ is the midpoint of $B D$.
Also, points $A, B, F, E$ are concyclic, so
$$
D E \cdot D A=D F \cdot D B,
$$
which means $D E \cdot 8=2 \sqrt{5} \times 4 \sqrt{5} \Rightarrow D E=5$.
Thus, $S_{\triangle B D E}=\frac{1}{2} D E \cdot ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,274 |
1. Arrange positive integers starting from 1 in a number array according to the pattern shown in Figure 2, where 2 is at the 1st corner, 3 is at the 2nd corner, 5 is at the 3rd corner, 7 is at the 4th corner, $\cdots \cdots$. Then, the number at the 2007th corner is $\qquad$
Figure 2 | $$
\begin{array}{l}
a_{1}=2, a_{2 i}=a_{2 i-1}+i, \\
a_{2 i+1}=a_{2 i}+(i+1) .
\end{array}
$$
Since $2007=2 \times 1003+1$, we have,
$$
\begin{array}{l}
a_{2007}=1+2(1+2+\cdots+1003)+1004 \\
=1004^{2}+1=1008017 .
\end{array}
$$ | 1008017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,275 |
2. In a $3 \times 3$ grid, the numbers $1,2,3,4,5,6,7,8,9$ are filled in, with one number per cell. Now, the cells containing the maximum number in each row are colored red, and the cells containing the minimum number in each row are colored green. Let $M$ be the smallest number in the red cells, and $m$ be the largest... | 2.8
Obviously, $3 \leqslant m, M \leqslant 7$, and $m \neq M$. Therefore,
$$
M-m \in\{-4,-3,-2,-1,1,2,3,4\} \text {. }
$$
As shown in Figure 7, all 8 values can be obtained
\begin{tabular}{|l|l|l|}
\hline 7 & 9 & 8 \\
\hline 6 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 6 & 9 &... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,276 |
3. As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC$, and $P$, $Q$ are points on sides $AC$, $AB$ respectively, such that $AP = PQ = QB = BC$. Then $\angle PCQ =$ $\qquad$ | 3. $30^{\circ}$.
Let $\angle A=2 \alpha$, then
$$
\begin{array}{l}
\angle A B C=\angle A C B=90^{\circ}-\alpha, \\
\angle B C Q=\angle B Q C=45^{\circ}+\frac{\alpha}{2}, \\
\angle P C Q=\angle A C B-\angle B C Q=45^{\circ}-\frac{3 \alpha}{2} .
\end{array}
$$
From $(A P \cdot 2 \cos 2 \alpha+B Q) 2 \sin \alpha=B C$, w... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,277 |
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}\left(1+\frac{1}{a}\right)+\frac{1}{c}\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)+ \\
\frac{1}{d}\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)- \\
\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\left(1+\f... | $\begin{array}{l}\text { 4. }-1 \text {. } \\ \frac{1}{a}+\frac{1}{b}\left(1+\frac{1}{a}\right)+\frac{1}{c}\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)+ \\ \frac{1}{d}\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \\ =-1+\left(1+\frac{1}{a}\right)+\frac{1}{b}\left(1+\frac{1}{a}\r... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,278 |
Three, (15 points) Given that $a, b, c$ are real numbers. If
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \frac{c^{2}+a^{2}-b^{2}}{2 a c}, \frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
sum up to exactly 1, prove: two of these fractions are 1, and one is -1. | $$
\begin{aligned}
& \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 a c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}=1, \\
& \left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}-1\right)+\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}-1\right)+ \\
& \left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1\right)=0 \\
\Rightarrow & \left(\frac{b^{2}+c^{2}-a^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,280 |
2. Let $n=9+99+\cdots+\underbrace{99 \cdots 9}_{99 \uparrow}$. Then in the decimal representation of $n$, the digit 1 appears ( ) times.
(A) 50
(B) 90
(C) 99
(D) 100 | 2.C.
Since $n=(10-1)+(100-1)+\cdots+(1 \underbrace{00 \cdots 0}_{99 \uparrow}-1)$ $=\underbrace{11 \cdots 10}_{99}-99=\underbrace{11 \cdots 1}_{91 \uparrow} 011$,
therefore, in the decimal representation of $n$, the digit 1 appears $97+2=99$ (times). | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,281 |
Four, (10 points) As shown in Figure 5, in $\triangle A B C$, $\angle A B C=46^{\circ}, D$ is a point on side $B C$, $D C=$ $A B, \angle D A B=21^{\circ}$. Determine the degree measure of $\angle C A D$.
| As shown in Figure 8, construct the axisymmetric figure $\triangle A E D$ of $\triangle A B D$ with respect to $A D$, then
$$
\begin{array}{l}
\angle E A D=21^{\circ}, \\
A E=A B .
\end{array}
$$
Therefore, $D E=B D$.
It is easy to see that
$$
\angle A D C=21^{\circ}+46^{\circ}=67^{\circ} \text {. }
$$
Thus, $\angle ... | 67^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,282 |
Five. (15 points) If for any $n$ consecutive positive integers, there always exists a number whose sum of digits is a multiple of 8. Determine the minimum value of $n$. And explain the reason.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translati... | Five, first prove that when $n \leqslant 14$, the property in the problem does not hold.
When $n=14$, for
$9999993, 9999994, \cdots, 10000006$
these 14 consecutive integers, the sum of the digits of any number cannot be divisible by 8.
Therefore, when $n \leqslant 14$, the property in the problem does not hold.
Thus, t... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,283 |
1. The original price of a commodity is $a$ yuan. During the Spring Festival promotion, it is reduced by $20 \%$. If the price is to be restored to the original price after the festival, the current selling price should be increased by ( ).
(A) $15 \%$
(B) $20 \%$
(C) $25 \%$
(D) $30 \%$ | - 1.C.
Let the price increase be $x \%$. According to the problem, the equation is $a(1-20 \%)(1+x \%)=a$. Solving this, we get $x=25$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,284 |
2. Given real numbers $a, b, c$ satisfy $a+b+c \neq 0$, and $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k$. Then the line $y=k x$ -3 must pass through ( ).
(A) the first, second, and third quadrants
(B) the first, second, and fourth quadrants
(C) the second, third, and fourth quadrants
(D) the first, third, and fourth q... | 2.D.
By the property of ratios, we have
$$
k=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}.
$$
Thus, the line is $y=\frac{1}{2} x-3$.
Therefore, the line passes through the first, third, and fourth quadrants. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,285 |
3. The number of integers $x$ that satisfy $\frac{1}{\sqrt{3}-\sqrt{2}}<x<\frac{2}{\sqrt{6}-\sqrt{5}}$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 3.C.
$$
\begin{array}{l}
3.1=1.7+1.4<\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+\sqrt{2}<1.8+1.5=3.3, \\
9.2=2(2.4+2.2)<\frac{2}{\sqrt{6}-\sqrt{5}} \\
=2(\sqrt{6}+\sqrt{5})<2(2.5+2.3)=9.6 .
\end{array}
$$
Therefore, the integers $x$ that satisfy the conditions are $4,5,6,7,8,9$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 719,286 |
4. $a, b, c$ are not all 0, satisfying $a+b+c=0$, $a^{3}+b^{3}+c^{3}=0$, the positive integer $n$ that makes $a^{n}+b^{n}+c^{n}=0$ always true is called a "good number". Then the number of good numbers among the positive integers not exceeding 2007 is ( ).
(A) 2
(B) 1004
(C) 2006
(D) 2007 | 4.B.
$$
\begin{array}{l}
\text { Given } a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=0,
\end{array}
$$
we know that $0=a^{3}+b^{3}+c^{3}=3 a b c$.
Thus, at least one of $a b c$ is 0.
From the condition, we know that only one of $a b c$ is 0, and the other two are opposite number... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,287 |
5. As shown in Figure 1, in $\triangle A B C$, $C B>C A, \angle B A C$ $=80^{\circ}, D$ is a point on $A B$ such that $C B-C A=B D$, $I$ is the intersection of the three angle bisectors of $\triangle A B C$. Then $\angle I D A=$ ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $50^{\circ}$
(D) $60^{\circ}$ | 5. B.
As shown in Figure 7, take point $A_{1}$ on $CB$ such that $CA_{1} = CA$. Connect $IA_{1}$, $IA$, $A_{1}D$, $CI$, and $BI$. Then $BD = BA_{1}$.
Figure 7
It is easy to prove that $\triangle ACI \cong \triangle A_{1}CI$,
$\triangle A_{1}BI \cong \triangle DBI$.
Therefore, $AI = A_{1}I = DI$.
Thus, $\angle IDA = \... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,288 |
6. As shown in Figure 2, in the regular pentagon $A B C D E$, the diagonals intersect at points $A_{1} 、 B_{1} 、 C_{1} 、 D_{1} 、 E_{1}$. All congruent triangles are considered as one class, called a "congruence class" (for example, $\triangle A B C 、 \triangle B C D$ and $\triangle C D E$ all belong to the same congrue... | 6. C.
In Figure 2, there are no triangles with all three vertices being from $A_{1}, B_{1}, C_{1}, D_{1}, E_{1}$.
(1) The congruence classes of triangles with all vertices taken from $A, B, C, D, E$ are 2: the $\triangle ABC$ congruence class and the $\triangle ABD$ congruence class;
(2) The congruence classes of tria... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,289 |
1. As shown in Figure 3, $O$ is a point inside $\triangle A B C$, extending $A O$ intersects $B C$ at point $D$, extending $B O$ intersects $A C$ at point $E$, the areas of $\triangle B O D$, $\triangle A O E$, and $\triangle A O B$ are 7, $14$, and $21$ respectively. Then $\frac{B D}{C D}=$ $\qquad$ . | $$
=1 \cdot \frac{7}{8} \text {. }
$$
As shown in Figure 8, connect $C O$, and let the areas of $\triangle O C D$ and $\triangle O C E$ be $x$ and $y$ respectively. From the given conditions, we have
$$
\left\{\begin{array}{l}
\frac{14 + y}{21} = \frac{x}{7}, \\
\frac{7 + x}{21} = \frac{y}{14}.
\end{array} \text { Sol... | \frac{7}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,290 |
2. As shown in Figure 4, in the convex quadrilateral $ABCD$, $AB=1$, $BC=2$, $CD=\sqrt{3}$, $\angle ABC=120^{\circ}$, $\angle BCD=90^{\circ}$.
Then $AD=$ $\qquad$ | 2. $\sqrt{7}$.
As shown in Figure 9, take the midpoint $E$ of $BC$, and connect $ED$, $EA$. Draw $BF \perp EA$ at $F$.
By the Pythagorean theorem, we know $DE=2$, so $\angle CED=60^{\circ}$.
Since $BA=BE$, we have $\angle BEA=30^{\circ}$.
Therefore, $\angle AED=90^{\circ}$.
Also, $EA=2EF=2\sqrt{1-\frac{1}{4}}=\sqrt{3}... | \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,291 |
3. Given $f(x)=x^{2}+6 a x-a, y=f(x)$, the graph of which intersects the $x$-axis at two distinct points $\left(x_{1}, 0\right),\left(x_{2}\right.$, $0)$, and
$$
\begin{array}{l}
\frac{a}{\left(1+x_{1}\right)\left(1+x_{2}\right)}-\frac{3}{\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)} \\
=8 a-3 .
\end{array}
$$
The... | 3. D.
From $\Delta=36 a^{2}+4 a>0$, we get $a>0$ or $a<-\frac{1}{9}$. By the problem, we can set
$$
\begin{array}{l}
f(x)=x^{2}+6 a x-a=\left(x-x_{1}\right)\left(x-x_{2}\right) . \\
\text { Then }\left(1+x_{1}\right)\left(1+x_{2}\right)=f(-1)=1-7 a, \\
\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right) \\
=f(1-6 a)=1-7 ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,292 |
3. The number of positive integers $m$ that make $m^{2}+m+7$ a perfect square is $\qquad$ . | 3.2.
It has been verified that when $m=1$, $m^{2}+m+7=9$ is a perfect square;
when $m=2,3,4,5$, $m^{2}+m+7$ are not perfect squares;
when $m=6$, $m^{2}+m+7=49$ is a perfect square.
When $m>6$, $m^{2}+m+7$ are not perfect squares.
Therefore, there are only 2 positive integers $m$ that meet the condition. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,293 |
4. As shown in Figure 5, positive real numbers $a_{1}, a_{2}, \cdots, a_{8}$ are marked at the corresponding vertices of a cube, such that the number at each vertex is the average of the three numbers marked at the adjacent vertices. Let
$$
\begin{array}{l}
M=\left(a_{1}+2 a_{2}+3 a_{3}+4 a_{4}\right) . \\
\quad\left(5... | 4. $\frac{65}{2}$.
Assume that among the numbers marked in Figure 5, $a_{1}$ is the largest, then
$$
a_{1} \geqslant a_{2}, a_{1} \geqslant a_{4}, a_{1} \geqslant a_{5} .
$$
From $a_{1}=\frac{a_{2}+a_{4}+a_{5}}{3}$, we know
$$
\frac{a_{2}+a_{4}+a_{5}}{3} \leqslant a_{1} \text {. }
$$
The equality holds if and only i... | \frac{65}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,294 |
Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$
Determine the maximum value of the real number $m$. | Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.
And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,295 |
Four. (25 points) As shown in Figure 6, in the convex quadrilateral $ABCD$, $E$ is a point inside $\triangle ACD$ such that $AC = AD = 2AB = 2AE$, $\angle BAE + \angle BCE = 90^{\circ}$, and $\angle BAC = \angle EAD$. Prove:
(1) $\angle CED = 90^{\circ}$;
(2) $BC^2 + CE^2 = 4BE^2$.
保留源文本的换行和格式，直接输出翻译结果。 | (1) As shown in Figure 6.
Since $B A=E A, \angle B A C=\angle E A D, C A=$ $D A$, therefore,
$$
\begin{array}{l}
\triangle B A C \cong \triangle E A D \\
\Rightarrow \angle A E D=\angle A B C, E D=B C . \\
\text { Also, } \angle A E D+\angle D E C+\angle C E A=360^{\circ}, \\
\angle A B C+\angle B C E+\angle C E A+\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,296 |
Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition. | Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$
or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,297 |
In the pentagram $A B C D E$, the intersection points of each line segment are shown in Figure 3, $A G=G F=F D, B G=$ $3 E G, F C=2 F E$. If the area of pentagon $F G H I J$ is 91, try to find the area of the pentagram $A B C D E$. | Solution: As shown in Figure 3, connect $F H, F I, G J, H J, A E,$ and $B C$.
From $\frac{S_{\triangle A E C}}{S_{\triangle A C F}} \cdot \frac{S_{\triangle C F A}}{S_{\triangle C A G}}$.
$\frac{S_{\triangle C A G}}{S_{\triangle C A E}}=1$, we get
$$
\frac{E C}{C F} \cdot \frac{F A}{A G} \cdot \frac{G H}{H E}=1,
$$
wh... | 349 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,298 |
Find all triples of integers $a, b, c$ such that
$$
c-(b+1)=(b+1)-a \text{, and } \frac{c-1}{b}=\frac{b}{a+1} \text{. }
$$ | Given the problem, we have
$$
\left\{\begin{array}{l}
a+c=2 b+2, \\
(a+1)(c-1)=b^{2}
\end{array}(a+1 \neq 0, b \neq 0),\right.
$$
which can be rewritten as
$$
\left\{\begin{array}{l}(a+1)+(c-1)=2 b+2, \\ (a+1)(c-1)=b^{2}\end{array}(a+1 \neq 0, b \neq 0)\right.
$$
Thus, \(a+1\) and \(c-1\) are the two non-zero integer ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,299 |
Given 215 As shown in Figure $4, \odot O_{1}, \odot O_{2}$ and $\odot O_{3}$ intersect at point $P, \odot O_{1}$ and $\odot O_{2}$ intersect at point $A$, a line passing through point $A$ intersects $\odot O_{1}, \odot O_{2}$ at points $B, C$, the extension of $A P$ intersects $\odot O_{3}$ at point $D, D E / / B C$ in... | Proof: As shown in Figure 4, let $\odot O_{3}$ intersect $\odot O_{1}$ and $\odot O_{2}$ at points $F$ and $H$, respectively. Connect $E F, B F, E H, C H, D F, D H, F H, P F, P H$. Then,
$$
\begin{array}{l}
\angle P F E + \angle P H E = 180^{\circ}, \\
\angle P H E = \angle P A C = \angle P F B \\
= 180^{\circ} - \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,300 |
Let $f(n)$ denote the number of distinct factorizations of a positive integer $n$ into a product of factors greater than 1 (the order of factors does not matter), such as $f(12)=3$. This is because $12=4 \times 3=2 \times 6=3 \times 2 \times 2$. Specifically, define $f(1)=1$. If $n=p_{1}^{m} p_{2}^{l}$, prove that:
$$
... | Proof: (1) When $l=0$, $n=p_{1}^{m}$.
If $P(m)$ represents the number of additive partitions of a positive integer $m$ (i.e., the number of ways to partition $m$ into the sum of several addends), then we have
$f(n)=P(m)$.
For $P(m)$, establish the recursive inequality
$P(m) \leqslant P(m-1)+P(m-2) \quad (m \geqslant 3)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,301 |
4. If the inequality $a x^{2}+7 x-1>2 x+5$ holds for -1 $\leqslant a \leqslant 1$, then the range of $x$ is ( ).
(A) $2 \leqslant x \leqslant 3$
(B) $2<x<3$
(C) $-1 \leqslant x \leqslant 1$
(D) $-1<x<1$ | 4. B.
From the problem, we know that the inequality $a x^{2}+7 x-1>2 x+5$ holds for $-1 \leqslant a \leqslant 1$, which means the inequality about $a$, $x^{2} a+5 x-6>0$, holds for $-1 \leqslant a \leqslant 1$.
Let $g(a)=x^{2} a+5 x-6$. Then
$g(-1)=-x^{2}+5 x-6>0$,
$g(1)=x^{2}+5 x-6>0$.
Solving these, we get $2<x<3$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 719,302 |
5. In Rt $\triangle A B C$, $\angle B=60^{\circ}, \angle C=$ $90^{\circ}, A B=1$, equilateral triangles $\triangle A B R$, $\triangle B C P$, and $\triangle C A Q$ are constructed outward on $A B$, $B C$, and $C A$ respectively. Line $Q R$ intersects $A B$ at point $T$. Then the area of $\triangle P R T$ is equal to ( ... | 5. A.
As shown in Figure 4, connect $P Q$. From the given conditions, we have
$$
\begin{array}{l}
B C=\frac{1}{2}, A C=\frac{\sqrt{3}}{2}, \angle Q A T=90^{\circ}, \\
\angle Q C P=150^{\circ}, P, B, R \text { are collinear. }
\end{array}
$$
Since $S_{\triangle A Q T}$
$$
\begin{array}{l}
=\frac{1}{2} A T \cdot A Q \\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,303 |
6. On a $3 \times 5$ chessboard, a piece can move one square at a time either horizontally or vertically, but not diagonally. Starting from some undetermined squares, the piece must pass through each small square exactly once, but it does not need to return to the starting square. Among these 15 small squares, ( ) can ... | 6. B.
As shown in Figure 5, a $3 \times 5$ chessboard is colored in black and white.
In Figure 5, there are 8 black small squares and 7 white small squares. Since the chess piece alternates between black and white with each move, the 7 white squares cannot be the starting point.
On the other hand, any of the 8 black... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,304 |
1. Square $A B C D$ has a side length of $5, E$ is a point on side $B C$ such that $B E=3, P$ is a point on the diagonal $B D$ such that $P E+P C$ is minimized. Then $P B=$ $\qquad$ | = $1 \cdot \frac{15 \sqrt{2}}{8}$.
Since $P E+P C=P E+P A$, therefore, when $A$, $P$, and $E$ are collinear, $P E+P A$ is minimized.
As shown in Figure 7, establish a rectangular coordinate system, with $B$ as the origin and $B A$ as the $x$-axis. Then
$$
\begin{array}{l}
l_{B D}: y=x, \\
l_{A E}: 3 x+5 y=15 .
\end{ar... | \frac{15 \sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,305 |
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