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3. The minimum value of the area of a right-angled triangle circumscribed around a circle with radius 1 is ( ).
(A) $3-2 \sqrt{2}$
(B) $3+2 \sqrt{2}$
(C) $6-4 \sqrt{2}$
(D) $6+4 \sqrt{2}$ | 3. (B).
Let the lengths of the two legs of a right triangle be $a$ and $b$, and the area be $\mathrm{S}$. Then we have
$$
\frac{1}{2}\left(a+b+\sqrt{a^{2}+b^{2}}\right) \times 1=\frac{1}{2} a b .
$$
(The area of the triangle is equal to the product of the semi-perimeter and the inradius.)
Simplifying, we get $a b-2 a-... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,149 |
4. A student solved a quadratic equation $a x^{2}+b x+c=0$ that has no real roots, but due to misreading the sign of a term, obtained the roots as $\frac{1 \pm \sqrt{31977}}{4}$. Then the value of $\frac{b+c}{a}$ is ( ).
(A) 1998
(B) 1999
(C) 1998 or 1999
(D) 1999 or 2000 | 4. (C).
The symbol that was misread can only be the quadratic term or the constant term, not the linear term. The equation solved by the student is
$-a x^{2}+b x+c=0$ or $a x^{2}+b x-c=0$.
By the relationship between roots and coefficients, we know
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ - \frac { b } { - a ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,150 |
Example 1 There are several points on a plane (the number of points is no less than 5), some of which are painted red, and the rest are painted blue. Suppose that no three points of the same color are collinear. Prove that there exists a triangle such that
(1) its three vertices are painted the same color;
(2) this tri... | Proof: For any points colored red or blue, there must be three points of the same color, and thus a triangle with three vertices of the same color must exist.
Consider a triangle with three vertices of the same color. If conclusion (2) does not hold, i.e., each side of every triangle with three vertices of the same co... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,151 |
Example 2 On the plane, there are $n$ points $(n \geqslant 2)$ that are not all collinear. Prove: there must exist a line that passes through exactly two of these points. | Proof: Consider each of the $n$ points. For each point, draw a straight line between every other pair of points (this line does not necessarily pass through only two of the points), and then draw a perpendicular from this point to each of the lines drawn. This gives several distances from the point to the lines. Since ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,152 |
Example 2 Given an isosceles trapezoid $A B C D$ where $A B / /$ $C D$, and the incircle of $\triangle B C D$ touches $C D$ at $E, F$ is a point on the angle bisector of $\angle D A C$, and $E F \perp C D$, the circumcircle of $\triangle A C F$ intersects $C D$ at $G$. Prove: $\triangle A F G$ is an isosceles triangle | Explanation: As shown in the figure, it is easy to guess that we need to prove \(FA = FG\), which means proving \(\angle FGA = \angle FAG\). However, from the fact that \(A, C, F, G\) are concyclic and \(AF\) bisects \(\angle CAD\), we know \(\angle FGD = \angle FAD\). We also need to prove \(\angle DGA = \angle DAG\).... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,153 |
13. The board of directors of Xinhua High-Tech Co., Ltd. has decided to invest 1.3 billion yuan in development projects this year. There are 6 projects to choose from (each project is either fully invested in or not invested in at all), with the required investment amounts and expected annual returns for each project a... | 13. $A, B, E$ | A, B, E | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,162 |
Example 3 Circle $\Gamma_{1}$ and circle $\Gamma_{2}$ intersect at points $M$ and $N$. Let $l$ be the common tangent of circles $\Gamma_{1}$ and $\Gamma_{2}$ that is closer to $M$, with $l$ tangent to $\Gamma_{1}$ at point $A$ and to $\Gamma_{2}$ at point $B$. Let the line through $M$ parallel to $l$ intersect $\Gamma_... | As shown in Figure 8, connect $M N$ and extend it to intersect $A B$ at $G$, then $G$ is the midpoint of $A B$ (a basic conclusion, which can be proven by the secant-tangent theorem). Since $P Q / / A B$, it follows that $M$ is the midpoint of $P Q$. (Basic Conclusion 6) Therefore, we need to prove that $E M \perp P Q$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,164 |
15. If the equation $\frac{2 k}{x-1}-\frac{x}{x^{2}-x}=\frac{k x+1}{x}$ has only one solution, find the value of $k$ and the solution of the equation. | Three, 15. The original equation simplifies to
$$
k x^{2}-3 k x+2 x-1=0 \text {. }
$$
When $k=0$, the original equation has a unique solution $x=\frac{1}{2}$.
When $k \neq 0$, for equation (1),
$$
\Delta=(3 k-2)^{2}+4 k=5 k^{2}+4(k-1)^{2}>0 \text {. }
$$
Therefore, there are always two distinct real roots. According ... | k=0, x=\frac{1}{2}; k=\frac{1}{2}, x=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,165 |
16. Given any four points in a plane, where no three points are collinear. Can we always select three points from these four to form a triangle such that this triangle has at least one interior angle not greater than $45^{\circ}$? Please prove your conclusion. | 16. Can.
Proof: (1) As shown in Figure 6, if four points $A, B, C, D$ form a convex quadrilateral, then there must be an interior angle $\leqslant 90^{\circ}$, let's assume it is $\angle A$. This is because, if we assume all four interior angles are greater than $90^{\circ}$, then
$$
\begin{aligned}
360^{\circ} & =\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,166 |
17. Paying taxes according to law is an obligation of every citizen. The "Individual Income Tax Law of the People's Republic of China" stipulates that citizens whose monthly salary and wage income does not exceed 800 yuan do not need to pay taxes; the part exceeding 800 yuan is the taxable income for the whole month, a... | 17. (1) In the citizen's income of 1350 yuan in October, the taxable part is 550 yuan. According to the tax rate table, he should pay $500 \times 5\% + 50 \times 10\% = 25 + 5 = 30$ (yuan).
(2) When $1300 < x \leqslant 2800$, 800 yuan is not taxable, and the taxable part is between 500 yuan and 2000 yuan. Among this, 5... | 30, y=(x-1300) \times 10\% + 25, 1600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,167 |
18. (1) As shown in Figure 4, in quadrilateral $ABCD$, $AB = AD$, $\angle BAD = 60^\circ$, and $\angle BCD = 120^\circ$. Prove: $BC + DC = AC$.
(2) As shown in Figure 5, in quadrilateral $ABCD$, $AB = BC$, $\angle ABC = 60^\circ$, and $P$ is a point inside quadrilateral $ABCD$ such that $\angle APD = 120^\circ$. Prove:... | 18. (1) As shown in Figure 8, extend $BC$ to $E$ such that $CE = CD$.
$$
\begin{array}{l}
\because \angle BCD = 120^{\circ}, \\
\therefore \angle DCE = 60^{\circ}. \\
\text{Also, } \because CD = CE,
\end{array}
$$
$\therefore \triangle CDE$ is an equilateral triangle.
Thus, $DE = CD = CE$,
$$
\begin{array}{c}
\angle CD... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,168 |
1. If the sum of positive integers $a$ and $b$ is $n$, then $n$ can be transformed into $a b$. Can this method be used several times to change 22 into 2001? | 1. Reverse calculation, $2001=3 \times 667$ from $3+667=670$; $670=10 \times 67$ from $10+67=77$; $77=7 \times 11$ from $7+11=18$. From any $n=1+(n-1)$, we can get $n-1=1 \times(n-1)$. Therefore, starting from 22, we can sequentially get 21, $20,19,18,77,670$ and 2001. | 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,169 |
2. In $\triangle A B C$, $D$, $E$, and $F$ are the midpoints of $B C$, $C A$, and $A B$ respectively. If one of $D E$, $E F$, $F D$ is longer than one of $A D$, $B E$, $C F$, prove that $\triangle A B C$ is an obtuse triangle. | 2. First assume $E F$ is longer than $A D$. As shown in Figure 1, $G$ is the intersection of $A D$ and $E F$, then
$$
\begin{aligned}
& \angle A G E, \\
& \angle G A F>\angle G F A .
\end{aligned}
$$
$\because$ The sum of these four angles is $180^{\circ}$,
$$
\begin{aligned}
\therefore & \angle C A B \\
& =\angle G A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,170 |
3. A store sells 20 kilograms of cheese. The first 10 customers, after each purchase, the salesperson tells the other customers: “If the amount you buy is equal to the average of what the previous customers bought, is it possible?” If possible, after the next 10 customers buy, how much cheese will be left? | 3. It is possible.
For $i \leqslant k<10$, if the first $k$ customers buy cheese $\frac{\dot{k}}{10+k}$, then after the $k$-th customer buys, the average amount of cheese sold to each customer is $\frac{1}{10+k}$, and $\frac{10}{10+k}$ remains. Therefore, it is still enough to sell to 10 customers.
Notice that the amo... | 10 \mathrm{~kg} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,171 |
4. Five paper-made triangles are on the desk, all congruent to each other. Each triangle can be translated but not rotated.
(1) Can any one of them be completely covered by the other four?
(2) Prove that if they are equilateral triangles, then any one of them can be completely covered by the other four. | 4. (1) No. Suppose in $\triangle ABC$, point $B$ is very close to the midpoint of $CA$. Rotate this triangle around point $C$ by $72^{\circ}$, $144^{\circ}$, $216^{\circ}$, and $288^{\circ}$ to obtain four other triangles congruent to it. Since each triangle can only cover a very small part of the longer side of any ot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,172 |
5. On a $15 \times 15$ chessboard, there are 15 rooks placed such that none of them can attack each other. They each move one step like a knight. Prove that now there are two that can attack each other. | 5. Record the row and column numbers of each rook. Since they do not attack each other, the row numbers and column numbers are all different. Therefore, among these 30 numbers, there are 16 odd numbers and 14 even numbers. When a rook moves one knight's move, its row number changes by 1 and its column number changes by... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,173 |
1. $n$ is a non-negative integer, and $3n+1, 5n+1$ are both perfect squares. Then $7n+3$ is:
( A ) must be a prime number
(B) must be a composite number
(C) must be a perfect square
(D) none of the above | -、1.(D).
Let $n=0, 3n+1=5n+1=1^{2}, 7n+3=3$ be prime numbers. Take $n=16, 3n+1=7^{2}, 5n+1=9^{2}, 7n+3=115$ as composite numbers.
However, 3 and 115 are not perfect squares. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,174 |
For example, $\triangle ABC$ is an isosceles triangle, $AB = AC$. Suppose:
(i) $M$ is the midpoint of $BC$, $O$ is a point on line $AM$ such that $OB$ is perpendicular to $AB$;
(ii) $Q$ is any point on segment $BC$ different from $B$ and $C$;
(iii) $E$ is on line $AB$, $F$ is on line $AC$, such that $E$, $Q$, $F$ are d... | Poem interpretation: As shown in Figure 9, from the basic structure 4, we know that $Q E = Q F$ if and only if $B E = C F$. Since $O$ is on the axis of symmetry of $\triangle A B C$, we have
$$
\begin{array}{l}
O C = O B, \\
\angle O C F \\
= \angle O B E = 90^{\circ}.
\end{array}
$$
Therefore, $B E = C F$ if and only... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,175 |
3. The number of integer solution sets for the system of equations $\left\{\begin{array}{l}x^{2}+x y+x z=10 \\ y z+y^{2}=15\end{array}\right.$ is ( ).
(A) 2 sets
(B) 4 sets
(C) More than 5 sets
(D) None of the above answers | 3.(B).
Obviously, $y(y+z)=15$. Therefore, as long as $y+z \mid 15$, $y$ is an integer, then $z$ is also an integer. Given $x(x+y+z)=10$ ($x$ is an integer), after verification, it is found that only when $x=2, x=5, x=-2, x=-5$, $y+z \mid 15$. Hence, there are 4 pairs of $(y, z)$ corresponding to the four solutions of ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,177 |
4. The parabola $y=a x^{2}+b x+c(a>0)$ intersects the $y$-axis at $P$, and intersects the $x$-axis at different points $A, B, |O A|=\frac{|O B|}{2}=\frac{|O P|}{3}$. Then the product of all possible values of $b$ is ( ).
(A) $\frac{729}{16}$
(B) $-\frac{9}{4}$
(C) $-\frac{9}{2}$
(D) $\frac{81}{64}$ | 4.(A).
(1) When $A$ and $B$ are both on the left side of the $y$-axis,
$$
\because a>0, \Delta>0 \text {, }
$$
$\therefore$ point $P$ is above the $x$-axis.
Let $A(-k, 0)$, $B(-2k, 0)$, $P(0, 3k)$. Substituting these three points into $y = ax^2 + bx + c$, we get $b = \frac{9}{2}$.
(2) When $A$ and $B$ are both on the r... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,178 |
5. For any $\square ABC$, with the incenter $I$. Then, when $AB + AC \geqslant 2BC$, the relationship between the circumradius $R_{1}$ of $\square ABC$ and the circumradius $R_{2}$ of $\square IBC$ is ().
$$
\begin{array}{ll}
\text { ( A ) } R_{1} \geqslant R_{2} & \text { ( B ) } R_{1}>R_{2} \\
\text { ( C) } R_{1} \l... | 5.(A).
As shown in Figure 4, extend $A I$ to $D$ such that $D$ lies on the circumcircle of $\triangle ABC$. Connect $C D$ and $B D$. It can be proven through angle transformation that
$$
D I=D B=D C .
$$
Thus, $R_{2}=D C$.
Let $A D$ and $B C$ intersect at $M$.
From $\angle 1=\angle 2$ and $\angle A B C=\angle A D C$,... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,179 |
6. The minimum value of $\sqrt{x^{2}-6 x+10}+\sqrt{x^{2}+6 x+13}$ is ( ).
(A) 6
(B) $6+\frac{\sqrt{3}}{2}$
(C) $3 \sqrt{5}$
(D) $2 \sqrt{10}$ | 6.( C).
$x$ is a real number. Establish a Cartesian coordinate system, and take $A(x, 0), B(3, -1), C(-3, 2)$.
By the distance formula between two points, we get
$$
A B=\sqrt{(x-3)^{2}+1^{2}}=\sqrt{x^{2}-6 x+10} \text {. }
$$
And $A C=\sqrt{(x+3)^{2}+2^{2}}=\sqrt{x^{2}+6 x+13}$.
When $x=1$, we have $A B+A C=\sqrt{5}+2... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,180 |
1. The equation $(k-1) x^{2}+3 k x+(2 k+3)=0$ has roots greater than -5 and less than $-\frac{3}{2}$. Then the range of values for $k$ is | II. $1 . k=1$ or $\frac{11}{6}0$.
Let $f(x)=(k-1) x^{2}+3 k x+(2 k+3)$.
When $k>1$, $k-1>0$,
By the necessary and sufficient conditions for the distribution of roots of a quadratic equation
$$
\begin{array}{l}
f(-5)>0 \Rightarrow k>\frac{11}{6}, \\
f\left(-\frac{3}{2}\right)>0 \Rightarrow k<3 . \\
\text { Hence } \frac... | \frac{11}{6}<k<3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,181 |
3. Three-digit number $\overline{a b c}=a^{2}+1+(\overline{b c})^{2}$. Then $\overline{a b c}=$ | 3.726 .
From $100 a+\overline{b c}=a^{2}+(\overline{b c})^{2}+1$, we have $(\overline{b c}-1) \overline{b c}=100 a-a^{2}-1$.
Obviously, the left side is even,
so the right side $a$ must be odd to make both sides equal.
When $a=1$, $(\overline{b c}-1) \overline{b c}=98=2 \times 7^{2}$;
When $a=3$, $(\overline{b c}-1) \... | 726 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,183 |
One, (20 points) $k$ is the minimum value of $\sqrt{\frac{5 m^{2}-2 \sqrt{5} m+7}{m^{2}+1}}$ as $m$ takes all real values. Solve the system of equations for $x, y$ in the real numbers:
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-\frac{6}{x}}+\sqrt{y^{2}-y}=1+k, \\
y\left(x^{2} y-x^{2}+\frac{6}{x}-\frac{6 y}{x}\right)=\sqrt{... | Let $p=\frac{5 m^{2}-2 \sqrt{5} m+7}{m^{2}+1}$. Then
$$
\begin{array}{l}
(p-5) m^{2}+2 \sqrt{5} m+p-7=0 . \\
\Delta \geqslant 0 \Rightarrow p^{2}-12 p+30 \leqslant 0,
\end{array}
$$
Thus, $6-\sqrt{6} \leqslant p \leqslant 6+\sqrt{6}$.
When $m=\frac{\sqrt{5}}{\sqrt{6}-1}$, $p=6-\sqrt{6}$,
then $k=\sqrt{p_{\text {min }}... | (x, y) = (2, \sqrt{6}), (2, 1-\sqrt{6}), (\sqrt{6}, \frac{1 + \sqrt{5}}{2}), (\sqrt{6}, \frac{1 - \sqrt{5}}{2}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,185 |
Example 5 Isosceles $\triangle A B C, A B=B C, C D$ is its angle bisector, $O$ is its circumcenter, a perpendicular line to $C D$ through $O$ intersects $B C$ at $E$, and a line through $E$ parallel to $C D$ intersects $A B$ at $F$. Prove: $B E=F D$. | Explanation: As shown in Figure 10, extend $CD$ to intersect $\odot O$ at $N$, and draw line $EN$ to intersect $\odot O$ at $M$ and $AB$ at point $G$. Since $OE$ perpendicularly bisects chord $CN$, we have
$$
\begin{aligned}
\angle N & =\angle ECN \\
& =\angle NCA,
\end{aligned}
$$
which means $MN \parallel AC$.
Also,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,186 |
II. (25 points) As shown in Figure 2, in $\square ABC$, $AB > AC$, and $D$ is on $AB$, with $AC$ being the mean proportional between $AD$ and $AB$. There is a point $H$ on $DC$ such that $DH = \frac{1}{2} BD$. Extend $AH$ to $E$ such that $E$ lies on the circumcircle of $\square ABC$ and $\angle BED = 3 \angle DEH$.
P... | Given that
\[
\frac{A D}{A C}=\frac{A C}{A B},
\]
then
\[
\square A D C \backsim \square A C B.
\]
Thus,
\[
\begin{array}{l}
\angle A D C \\
=\angle A C B. \\
\because A, C, E, B
\end{array}
\]
are concyclic,
\[
\begin{array}{c}
\therefore \angle A E B= \\
\angle A C B=\angle A D C, \\
\therefore B, D, H, E
\end{array}... | AB^2 \cdot ED = \sqrt{3} AC^2 \cdot BE | Geometry | proof | Yes | Yes | cn_contest | false | 712,187 |
Three. (25 points) The roots of the equation $\left(m^{3}-2 m^{2}\right) x^{2}-$ $\left(m^{3}-3 m^{2}-4 m+8\right) x+12-4 m=0$ with respect to $x$ are all integers. Find the possible values of the real number $m$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the ... | When $m=0$, $-8 x+12=0$. $x$ is not an integer.
When $m=2$, $4 x+4=0 \Rightarrow x=-1$. Satisfied.
When $m \neq 0$ and $m \neq 2$, we have
\[
\begin{array}{l}
{\left[(m-2) x-(m-3) \right] m^{2} x+4=0 .} \\
x_{1}=\frac{m-3}{m-2}, \\
x_{2}=-\frac{4}{m^{2}} .
\end{array}
\]
From (1), we have $\left(x_{1}-1\right) m=2 x_{... | 1 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,188 |
1. Given the sets $E=\left\{x \left\lvert\, x=\cos \frac{n \pi}{3}\right., n \in \mathbf{Z}\right\}, F=$ $\left\{x \left\lvert\, x=\sin \frac{(2 m-3) \pi}{6}\right., m \in \mathbf{Z}\right\}$. The relationship between $E$ and $F$ is ( ).
(A) $E \subset F$
(B)E $=F$
( C ) $E \supset F$
(D) $E \cap F=\varnothing$ | -、1.(B).
Since the smallest positive period of the functions $x=\cos \frac{n \pi}{3}$ and $x=\sin \frac{(2 m-3) \pi}{6}$ is both 6, by taking $n, m=0, 1, 2, 3, 4, 5$, we get
$$
\begin{array}{l}
E=\left\{1, \frac{1}{2},-\frac{1}{2},-1\right\}, \\
F=\left\{-1,-\frac{1}{2}, \frac{1}{2}, 1\right\} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,189 |
2. Given $a=\frac{200}{157}$, and $b a^{b}=1$. Then the relationship between $A=\log _{b} \cos \frac{b}{a}$ and $B=\log _{b} \sin \frac{b}{a}$ is ().
(A) $A<B$
(B) $A>B$
(C) $A=B$
(D) Cannot be determined | 2.(A).
Given $a>1$, $a^{b}>0$, and $b a^{b}=1>0$, we get $b>0$.
Thus, $a^{b}>1$, and $0<\sin \frac{b}{a}$.
Therefore, $\log _{b} \cos \frac{b}{a}<\log _{b} \sin \frac{b}{a}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,190 |
3. Let the first term of an arithmetic sequence be $a(a \neq 0)$, and the second term be $b$. Then the necessary and sufficient condition for this sequence to have a term equal to 0 is ( ).
(A) $a-b$ is a positive integer (B) $a+b$ is a positive integer
(C) $\frac{b}{a-b}$ is a positive integer (D) $\frac{a}{a-b}$ is a... | 3.(D).
Since the common difference $d=b-a \neq 0$, otherwise it would be impossible for any term to be 0. Thus, let $a_{n}=a+(n-1)(b-a)=0$, we get $n-1=$ $\frac{a}{a-b}$, which implies that $\frac{a}{a-b}$ is a natural number.
Conversely, if $\frac{a}{a-b}$ is a natural number, then the $\frac{a}{a-b}+1$-th term is 0. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,191 |
4. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with the right focus at $F$, the right directrix at $l$, and a line intersecting the two branches of the hyperbola at points $P$ and $Q$, and intersecting $l$ at $R$. Then ( ).
( A) $\angle P F R>\angle Q F R$
( B) $\angle P F R<\angle Q F R$
... | 4. ( C ).
As shown in Figure 2, draw $P P^{\prime} \perp l$ and $Q Q^{\prime} \perp l$, with the feet of the perpendiculars being $P^{\prime}$ and $Q^{\prime}$, respectively. By the properties of similar triangles, we get
$$
\frac{|P R|}{|R Q|}=\frac{\left|P P^{\prime}\right|}{\left|Q Q^{\prime}\right|} \text {. }
$$
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,192 |
5. In the complex plane, the vertices $A, B, C$ of $\square A B C$ correspond to the complex numbers $3+2 \mathrm{i}, 3 \mathrm{i}, 2-\mathrm{i}$, respectively. The moving point $P$ corresponds to the complex number $z$. If the equation $|z|^{2}+\alpha z+\bar{\alpha} \bar{z}+\beta=0$ represents the circumcircle of $\sq... | 5.(C).
It is easy to know that $|A B|=|A C|$, and $A B \perp A C$, so $\square A B C$ is an isosceles right triangle. Its circumcircle equation is
$$
|z-(1+\mathrm{i})|=\sqrt{5} \text { , }
$$
which means $[z-(1+\mathrm{i})][\bar{z}-\overline{(1+\mathrm{i})}]=5$,
or $|z|^{2}-\overline{(1+\mathrm{i})} z-(1+\mathrm{i})... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,193 |
6. Three cylinders are tangent to each other on their sides, and their axes are pairwise perpendicular. If the radius of the base of each cylinder is 1, then the radius of the smallest sphere that is tangent to the sides of these three cylinders is ( ). Wanfang Data
(A) $\sqrt{2}-1$
(B) $\frac{\sqrt{2}-1}{2}$
(C) $\fra... | 6.(A).
Construct a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with an edge length of 2, such that the axes of three cylinders are the edges $A A_{1}$, $B C$, and $C_{1} D_{1}$ of the cube. It is easy to see that the side surfaces of these three cylinders are pairwise tangent, and their axes are also pairwise perpendicular... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,194 |
1. The maximum value of the function $y=\sin ^{4} x \cdot \cos x+\sin x \cdot \cos ^{4} x$ $\left(0<x<\frac{\pi}{2}\right)$ is $\qquad$ . | $$
\begin{array}{l}
\text { II. 1. } \frac{\sqrt{2}}{4} . \\
y=\sin x \cos x\left(\sin ^{3} x+\cos ^{3} x\right) \\
=\sin x \cos x(1-\sin x \cos x)(\sin x+\cos x) . \\
\because \sin x \cos x(1-\sin x \cos x) \\
\quad \leqslant\left[\frac{\sin x \cos x+(1-\sin x \cos x)}{2}\right]^{2}=\frac{1}{4},
\end{array}
$$
$\there... | \frac{\sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,195 |
2. Let $x, y \in \mathbf{R}$. If $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$ . | 2.4.
Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$
Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$
Substituting (1) into (2), we get $2 ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,196 |
Example 6 Let $\triangle A B C$ be an acute triangle, with circumcenter $O$ and circumradius $R, A O$ intersects the circumcircle of $\triangle B O C$ at point $A^{\prime}, B O$ intersects the circumcircle of $\triangle C O A$ at point $B^{\prime}$, and $C O$ intersects the circumcircle of $\triangle A O B$ at point $C... | Let the intersection of $A A^{\prime}$ and $B C$ be $M$. Since $O, B, A^{\prime}, C$ are concyclic, we have
$$
\begin{aligned}
\angle A^{\prime} & =\angle O B C, \\
& =\angle O C B .
\end{aligned}
$$
Thus, $\triangle O C M \sim \triangle O A^{\prime} C$,
which implies $O M \cdot O A^{\prime}$
$$
=O C^{2}=R^{2} \text {... | O A^{\prime} \cdot O B^{\prime} \cdot O C^{\prime}=8 R^{3} | Geometry | proof | Yes | Yes | cn_contest | false | 712,197 |
4. Let $a_{n}$ denote the integer closest to $\sqrt{n}(n \in \mathbf{N})$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{2001}}$ is $\qquad$ . | $4.88 \frac{7}{15}$.
Since $\sqrt{n}(n \in \mathbf{N})$ is either an integer or an irrational number,
$$
\begin{aligned}
\therefore a_{n}=k(k \in \mathbf{N}) & \Leftrightarrow k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2} \\
& \Leftrightarrow k^{2}-k+1 \leqslant n \leqslant k^{2}+k,
\end{aligned}
$$
That is, when $n$ takes the... | 88 \frac{7}{15} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,199 |
5. Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, where $O$ is the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\q... | 5.2.
Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter ).
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta}$.
Then $|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,200 |
6. Draw 63 lines on the coordinate plane: $y=b, y=\sqrt{3} x + 2b, y=-\sqrt{3} x+2b$, where $b=-10,-9,-8, \ldots, 8,9,10$. These lines divide the plane into several equilateral triangles. The number of equilateral triangles with side length $\frac{2}{\sqrt{3}}$ is $\qquad$ $-$ | 6.660.
The six outermost straight lines determine a regular hexagon with a side length of $\frac{20}{\sqrt{3}}$. The three straight lines passing through the origin $O$ divide this hexagon into six equilateral triangles with a side length of $\frac{20}{\sqrt{3}}$. Since the side length of each large triangle is 10 tim... | 660 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,201 |
Three, (20 points) Prove that there exist infinitely many such sequences, all of whose terms are distinct natural numbers, and the sum of the first $k$ terms of each sequence is divisible by $k$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translati... | Three, construct the sequence $\left\{a_{n}\right\}$, whose general term formula is $a_{n}=2 a n +$ $b(a, b \in \mathbf{N})$. The sequence $\left\{a_{n}\right\}$ is the sequence that meets the requirements.
First, the terms determined by $a_{n}=2 a n+b$ are all different natural numbers.
Second, $\left\{a_{n}\right\}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,202 |
Four. (20 points) For the tetrahedron $P-ABC$, the lengths of the three sides of the base are $a$, $b$, and $c$, and the lengths of the opposite edges are $a'$, $b'$, and $c'$. Prove: A new tetrahedron can be formed with $aa'$, $bb'$, and $c'$ as the lengths of the three sides of the base, and $b'c'$, $c'a'$, and $a'b'... | As shown in Figure 4, on the ray $PA$, take point $A'$ such that $PA' = b'c'$. In the plane of the side, construct $\angle PA'B' = \angle PBA$ and $\angle PA'C' = \angle PCA$, intersecting the rays $PB$ and $PC$ at $B'$ and $C'$ respectively, and connect $B'C'$. We will now prove that $P-A'B'C'$ is the tetrahedron that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,203 |
Five. (20 points) Given real numbers $a, b, c$ satisfy $b^{2}-a c<0$, and the curve represented by the equation $a(\lg x)^{2}+2 b(\lg x \lg y)+c(\lg y)^{2}=1$ passes through the point $\left(10, \frac{1}{10}\right)$. Prove that for any point $P(x, y)$ on this curve, the inequality
$$
-\frac{1}{\sqrt{a c-b^{2}}} \leqsla... | Given the curve passes through the point $\left(10, \frac{1}{10}\right)$, we have
$$
a-2 b+c=1 \text {. }
$$
Let $u=\lg x, v=\lg y$. Then the original equation becomes
$$
a u^{2}+2 b u v+c v^{2}=1 \text {. }
$$
Let $u+v=t$ then $v=t-u$, substituting into (2) we get
$$
(a-2 b+c) u^{2}+2(b-c) t u+c t^{2}-1=0 \text {. }... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,204 |
One, (50 points) As shown in Figure 1, in $\square A B C$, $B_{1}$ and $C_{1}$ are points on the extensions of $A B$ and $A C$ respectively, $D_{1}$ is the midpoint of $B_{1} C_{1}$, and $A D_{1}$ intersects the circumcircle of $\square A B C$ at $D$. Prove: $A B \cdot A B_{1} + A C \cdot A C_{1} = 2 A D \cdot A D_{1}$... | In $\square ABC$, connect $BD$, $CD$. Let $\angle BAD=\alpha$, $\angle CAD=\beta$, and the radius of the circumcircle of $\triangle ABC$ be $R$.
$\because D_{1}$ is the midpoint of the two sides,
$$
\begin{array}{r}
\therefore \boldsymbol{\square}_{AB_{1}D_{1}}=S_{\boldsymbol{\square}_{AC_{1}D_{1}}} \\
\quad=\frac{1}{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,205 |
II. (50 points) Given a real number
$m$ satisfies: there is one and only one square whose four vertices all lie on the curve $y=x^{3}+m x$. Try to find the area of this square. | Due to the curve $y=x^{3}+m x$ being symmetric about the origin, the center of the square must be the origin (otherwise, reflecting this square about the origin would yield another square with vertices on the curve).
Let one vertex of the square be $A(a, b)$, then $B(b, -a)$, $C(-a, -b)$, and $D(-b, a)$ are the other ... | 6 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,206 |
Three. (50 points) Let $f(x)=a x^{2}+b x+c$ be a quadratic trinomial with real coefficients. If for all integers $x$, $f(x)$ is a perfect square, prove: there exist a natural number $d$ and an integer $e$, such that
$$
f(x)=(d x+e)^{2} .
$$ | Three, since $f(0)$ is a perfect square, $c=e^{2}(e \in \mathbf{Z})$. To determine the convention, when $b \geqslant 0$, $e \geqslant 0$; when $b<0$.
Let the largest natural number square not exceeding $a$ be $d^{2}(d \in \mathbf{N})$, i.e., $d^{2} \leqslant a<0$, then the right side of the above equation is positive ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,207 |
Example 7 As shown in Figure 12,
$\odot O_{1}$ and $\odot O_{2}$ are tangent to the three sides of $\triangle A B C$, $E, F, G, H$ are the points of tangency, and the extensions of $E G$ and $F H$ intersect at point $P$. Prove: the line $P A$ is perpendicular to $B C$. | Explanation: As shown in Figure 12, draw $A D \perp B C$ at $D$, and extend $A D$ in the opposite direction to intersect $E G$ and $F H$ at points $P_{1}$ and $P_{2}$, respectively. We need to prove that $P_{1}$ and $P_{2}$ coincide, i.e.,
$$
\begin{array}{l}
P_{1} D=P_{2} D \\
\Leftrightarrow E D \cdot \tan \angle G E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,208 |
In $\triangle ABC$, $h_{a} 、 h_{b} 、 h_{c}, r_{a} 、 r_{b} 、 r_{c}$ represent the lengths of the three altitudes and the radii of the three excircles, respectively. Prove:
$$
h_{a}+h_{b}+h_{c} \leqslant r_{a}+r_{b}+r_{c} .
$$ | Proof: As shown in Figure 1, let $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$ be the excenters of $\triangle \mathrm{ABC}$. It is easy to see that $\mathrm{O}_{2}, \mathrm{~A}, \mathrm{O}_{3}, \mathrm{O}_{3}, \mathrm{~B}, O_{1}, O_{1}, C, O_{2}$ are collinear respectively.
From $\mathrm{O}_{1}, \mathrm{O}_{2}, \mat... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,209 |
For $101 n$ as an integer not less than 2, determine the largest constant $C(n)$ such that
$$
\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-x_{i}\right)^{2} \geqslant C(n) \cdot \min \left(x_{i+1}-x_{i}\right)^{2}
$$
holds for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}<x_{2}<\cdots<x_{n}$. | $$
\begin{array}{l}
\text { Solution: } \because x_{j}-x_{i}=\left(x_{j}-x_{j-1}\right)+\left(x_{j-1}-x_{j-2}\right) \\
+\cdots+\left(x_{i+1}-x_{i}\right) \\
\geqslant(j-i) \cdot \min \left(x_{i+1}-x_{i}\right),(j>i) \\
\therefore \frac{\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-x_{r}\right)^{2}}{\min \left(x_{i+1}-... | \frac{1}{12} n^{2}\left(n^{2}-1\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,211 |
$$
\begin{array}{l}
\log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\
=2 \log _{3}(5 x+2) \log _{5}(5 x+4)
\end{array}
$$
Solve the equation for \( x \):
$$
\begin{array}{l}
\log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\
=2 \log _{3}(5 x+2) \log _{5}(5 x+4)
\end{array}
$$ | First, we prove the lemma: If \( n \geqslant m > 1 \), then
\[
01, \\
\therefore \log _{n}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{m}\right), \\
\therefore \log _{n}(n+1)-1 \leqslant \log _{m}(m+1)-1, \\
\log _{n}(n+1) \leqslant \log _{m}(m+1) . \\
\fr... | x = \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,212 |
Example 8 A line $l$ does not intersect with a circle $\omega$ with center $O$, $E$ is a point on $l$, $O E \perp l, M$ is any point on $l$ different from $E$, from $M$ two tangents to circle $\omega$ are drawn, touching $\omega$ at $A$ and $B$, $C$ is a point on $M A$ such that $E C \perp M A, D$ is a point on $M B$ s... | Explanation: As shown in Figure 13, connect \( OB \), \( OM \), and \( AE \), and draw line \( AB \) to intersect the extension of \( DC \) at \( G \). Connect \( GE \). Since \( OA \perp MA \), \( OB \perp MB \), and \( OE \perp ME \), points \( A \), \( O \), \( B \), \( M \), and \( E \) are concyclic. Therefore, \(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,213 |
5. Prove: The indeterminate equation $x^{2}+y^{2}+z^{2}=2^{n} x y z$ has no positive integer solutions $x, y, z, n$ that are not all zero. | (Let $x_{1}$ be the smallest positive integer solution of $x^{2}+y^{2}+z^{2}=2^{n} x y z$, i.e., $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=2^{n} x_{1} y_{1} z_{1}$. Then $x_{1}, y_{1},$ and $z_{1}$ are all even numbers. Set $x_{1}=2 x_{2}, y_{1}=2 y_{2}, z_{1}=2 z_{2}$ and substitute into the equation, and we get another solution... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,214 |
2. If $a, b, c$ are three arbitrary integers, then, $\frac{a+b}{2}$, $\frac{b+c}{2}, \frac{c+a}{2}(\quad)$.
(A) None of them are integers
(B) At least two of them are integers
(C) At least one of them is an integer
(D) All of them are integers | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,216 |
Example 3 Given that $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers, $a_{1}$, $a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ are positive integers, let
$$
\begin{array}{l}
a=\frac{a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}}{a_{1}+a_{2}+\cdots+a_{n}}, \\
b=\frac{b_{1} x_{1}+b_{2} x_{2}+\cdots+b_{n} x_{n}}{b_{1... | Proof:
$$
\begin{aligned}
& |a-b|=\left|a-\frac{b_{1} x_{1}+b_{2} x_{2}+\cdots+b_{n} x_{n}}{b_{1}+b_{2}+\cdots+b_{n}}\right| \\
= & \frac{\mid b_{1}\left(a-x_{1}\right)+b_{2}\left(a-x_{2}\right)+\cdots+b_{n}\left(a-x_{n}\right)}{b_{1}+b_{2}+\cdots+b_{n}} \\
\leqslant & \frac{b_{1}\left|a-x_{1}\right|+b_{2}\left|a-x_{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,217 |
3. If $a, b$ are prime numbers, and $a^{2}-13 a+i n=0$, $b^{2}-13 b+m=6$, then the value of $\frac{b}{a}+\frac{a}{b}$ is ( ).
(A) $\frac{123}{22}$
(B) $\frac{125}{22}$ or 2
(C) $\frac{125}{22}$
(D) $\frac{123}{22}$ or 2 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,218 |
12. Given real numbers $a$ and $b$ satisfy $a^{2}+a b+b^{2}=1$, and $t=a b-a^{2}-b^{2}$. Then, the range of $t$ is $\qquad$ | 12. $-3 \leqslant t \leqslant -\frac{1}{3}$ | -3 \leqslant t \leqslant -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,227 |
Example 4 In an $n \times n$ square grid, non-negative integers are written. If the number at the intersection of a certain row and a certain column is 0, then the sum of the numbers in that row and that column is not less than $n$. Prove: The sum of all numbers in the table is not less than
$$
\frac{1}{2} n^{2} \text ... | Prove: When calculating the sum of all the numbers in each row and each column of an $n$-row and $n$-column grid, among these $2n$ sums, there must be a minimum. Suppose the sum of the numbers in a certain row is the smallest, and let this sum be $k$, then this row must have at least $n-k$ zeros.
Now consider the colu... | \frac{n^{2}}{2} | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,228 |
13. A student participates in military training and must shoot 10 times. In the 6th, 7th, 8th, and 9th shots, he scored 9.0, 8.4, 8.1, and 9.3 points, respectively. The average score of his first 9 shots is higher than the average score of his first 5 shots. If he wants the average score of 10 shots to exceed 8.8 point... | Three, 13. From the given information, the average score of the first 5 shots is less than $\frac{9.0+8.4+8.1+9.3}{4}=8.7$, the total score of the first 9 shots is at most $8.7 \times 9-0.1=78.2$. Therefore, the 10th shot must score at least $8.8 \times 10+0.1-78.2=9.9$ (rings). | 9.9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,229 |
14. As shown in Figure 3, given that point $P$ is an external point of $\odot O$, $P S$ and $P T$ are two tangents to $\odot O$, a secant line $P A B$ through point $P$ intersects $\odot O$ at points $A$ and $B$, and intersects $S T$ at point $C$. Prove:
$$
\frac{1}{P C}=\frac{1}{2}\left(\frac{1}{P A}+\frac{1}{P B}\rig... | 14. As shown in Figure 4, connect $P O$ intersecting $S T$ at point $D$, then $P O \perp S T$. Connect $S O$, and draw $O E \perp P B$, with the foot of the perpendicular at point $E$, then $E$ is the midpoint of $A B$. Therefore,
$$
P E=\frac{P A+P B}{2} \text {. }
$$
$\because C, E, O, D$ are concyclic,
$$
\therefore... | \frac{1}{P C}=\frac{1}{2}\left(\frac{1}{P A}+\frac{1}{P B}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 712,230 |
15. Given the equation about $x$
Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1}=\frac... | 15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,231 |
3. In a convex pentagon $A B C D E$, $\angle A=\angle B=120^{\circ}, E A$ $=A B=B C=2, C D=D E=4$. Then its area is ( ).
(A) $6 \sqrt{3}$
(B) $7 \cdot \sqrt{3}$
(C) $8 \sqrt{3}$
(D) $9 \sqrt{3}$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,234 |
5. An isosceles triangle with a rational perimeter, the height on the base is $\frac{1}{2}$ of the base length. Then the ( ) of the triangle.
(A) the legs and the height on the base are both rational numbers
(B) neither the legs nor the height on the base are rational numbers
(C) the legs are rational numbers, and the ... | 5.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,236 |
6. As shown in Figure 2, $\triangle A B C$ is an isosceles right triangle, $\angle C=90^{\circ}$. If $A D=$ $\frac{1}{3} A C, C E=\frac{1}{3} B C$, then the size relationship between $\angle 1$ and $\angle 2$ is ( ).
(A) $\angle 1>\angle 2$
(B) $\angle 1<, \angle 2$
(C) $\angle 1=\angle 2$
(D) Cannot be determined | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,237 |
Example 5 Proof: The equation
$$
x^{2}+y^{2}=3\left(z^{2}+u^{2}\right)
$$
has no positive integer solutions $(x, y, z, u)$. | Proof: Assume the equation has positive integer solutions $x, y, z, u$. Consider the sum of squares of $x$ and $y$, $x^{2} + y^{2}$. Since $x^{2} + y^{2}$ is a positive integer, there must be a smallest one among all $x^{2} + y^{2}$. We consider the set of positive integer solutions $(x, y, z, u)$ that makes $x^{2} + y... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,239 |
13. Given the parabola $y=x^{2}+p x+q$ has a point $M\left(x_{0}, y_{0}\right)$ located below the $x$-axis.
(1) Prove: this parabola intersects the $x$-axis at two points;
(2) Let the intersection points of this parabola with the $x$-axis be $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1}<x_{2}$. Prov... | Three, 13. (1) From the given information,
$$
\begin{aligned}
y_{0} & =x_{0}^{2}+p x_{0}+q \\
& =\left(x_{0}+\frac{p}{2}\right)^{2}-\frac{p^{2}-4 q}{4}\left(x_{0}+\frac{p}{2}\right)^{2} \geqslant 0. \text{ Given } p^{2}-4 q>0.
\end{aligned}
$$
Therefore, $x^{2}+p x+q=0$ has two distinct real roots, meaning the parabola... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,245 |
14. As shown in Figure 4, in any pentagon $A B C D E$, $M 、 N 、 P 、 Q$ are the midpoints of $A B 、 C D 、 B C$ 、 $D E$ respectively, and $K 、 L$ are the midpoints of $M N 、 P Q$ respectively. Prove that $K L \parallel A E$, and $K L = -\frac{1}{4} A E$ | 14. Connect $B E$, take the midpoint of it as $R$, and connect $M R$.
In $\triangle A B E$, since $M 、 R$ are the midpoints of $A B 、 B E$ respectively, then $M R \cong \frac{1}{2} A E$.
Connect $R N$, in quadrilateral $B C D E$,
$\because P 、 N, Q, R$ are the midpoints of each side respectively,
$\therefore$ quadrila... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,246 |
15. As shown in Figure 5, the line $y=-\frac{\sqrt{3}}{3} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. An isosceles right triangle $\triangle A B C$ with $\angle B A C=90^{\circ}$ is constructed in the first quadrant with segment $A B$ as one of its legs. If there is a point $P\left(a,... | 15. Let $y=0$ and $x=0$, we get the intersection points of the line $y=-\frac{\sqrt{3}}{3} x+1$ with the $x$-axis and $y$-axis as $A(\sqrt{3}, 0), B(0,1)$, i.e., $O A=\sqrt{3}, O B=1$.
Thus, $A B=2$.
Since $\triangle A B C$ is an isosceles right triangle,
$S_{\triangle A B C}=2$.
Also, since $S_{\triangle A B P}=S_{\tr... | \frac{\sqrt{3}-8}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,247 |
16. As shown in Figure 6, given point $P$ is moving on the arc $\overparen{A B}$ (excluding endpoints) of a sector $O A B$ with a radius of 6 and a central angle of $90^{\circ}$, $P H \perp O A$, with the foot of the perpendicular at $H$, and the centroid of $\triangle O P H$ is $G$.
(1) When point $P$ is moving on $\o... | 16. (1) Online destroy $G O$, $C$, $G H$ Ning, there is a line segment, which is $\mathrm{UH}$, whose length remains unchanged.
Extend $HQ$ to intersect $O$ at $E$, and extend $PG$ to intersect $AO$ at $D$.
$\because G$ is the centroid of $\triangle OPH$; and $\angle PHO=90^{\circ}$,
$$
\therefore GH=\frac{2}{3} HE=\fr... | \sqrt{6} \text{ or } 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,248 |
Example 6 There are $n$ boys and $m$ girls $(n, m \geqslant$ 2), each boy knows at least one girl, and each girl does not know all $n$ boys. Prove: Among them, there are two boys and two girls, where each boy knows exactly one of the girls, and each girl knows exactly one of the boys. | Proof: Let $a_{1}$ be the girl who knows the most boys.
By the problem's condition, each girl does not know all $n$ boys, so $a_{1}$ must not know at least one boy, let's say $a_{1}$ does not know boy $b_{1}$.
Since each boy knows at least one girl, we can assume that $b_{1}$ knows girl $a_{2}$.
Since $a_{2}$ knows n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,250 |
3. The output value of a factory in the second quarter increased by $x \%$ compared to the first quarter, and the output value in the third quarter increased by $x \%$ compared to the second quarter. Then the output value in the third quarter increased by ( ) compared to the first quarter.
(A) $2 x \%$
(B) $1+2 x \%$
(... | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,252 |
5. If $D$ is a point on side $AB$ of $\triangle ABC$, $\angle ADC = \angle BCA, AC=6, DB=5$, and the area of $\triangle ABC$ is $S$, then the area of $\triangle BCD$ is ( ).
(A) $\frac{3}{5} S$
(B) $\frac{4}{7} S$
(C) $\frac{5}{9} S$
(D) $\frac{6}{11} S$ | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,254 |
7. As shown in Figure 2, the two marked
wheels can rotate around the central axis, and the arrows above the wheels point to a number on each wheel. If the arrow above the left wheel points to the number $a$, and the arrow above the right wheel points to the number $b$, the total number of possible pairs $(a, b)$ is $n... | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,256 |
3. In trapezoid $A B C D$, $A B / / C D, A C 、 B D$ intersect at point $O$. If $A C=5, B D=12$, the midline length is $\frac{13}{2}$, the area of $\triangle A O B$ is $S_{1}$, and the area of $\triangle C O D$ is $S_{2}$, then $\sqrt{S_{1}}+\sqrt{S_{2}}$ $=$ . $\qquad$ | 3. $\sqrt{30}$ | \sqrt{30} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,260 |
1. Given 997 points in the plane, the midpoints of the line segments connecting every pair of points are colored red. Prove that there are at least 1991 red points. Can you find 997 points that result in exactly 1991 red points? | (遈示: Let $A B$ be the longest line connecting two points among 997 points. Point $A$ is connected to the other 996 points, and the midpoints of these lines are all within the circle centered at $A$ with a radius of $\frac{A B}{2}$. Point $B$ is connected to the other 996 points, and the midpoints of these lines are all... | 1991 | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,261 |
4. Given that the side lengths of rectangle $A$ are $a$ and $b$. If there is always another rectangle $B$ such that the ratio of the perimeter of rectangle $B$ to the perimeter of rectangle $A$ and the ratio of the area of rectangle $B$ to the area of rectangle $A$ are both equal to $k$. Then the minimum value of $k$ i... | 4. $\frac{4 a b}{(a+b)^{2}}$ | \frac{4 a b}{(a+b)^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,262 |
Four, (20 points) Given that $x$ and $y$ are real numbers, and satisfy
$$
\begin{array}{l}
x y + x + y = 17, \\
x^{2} y + x y^{2} = 66 .
\end{array}
$$
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$. | Given the conditions $x y + x + y = 17$ and $x y (x + y) = 66$, we know that $x y$ and $x + y$ are the two real roots of the equation
$$
t^{2} - 17 t + 66 = 0
$$
Solving this equation, we get
$$
t_{1} = 6, \quad t_{2} = 11.
$$
Thus, $x y = 6$ and $x + y = 11$; or $x y = 11$ and $x + y = 6$.
When $x y = 6$ and $x + y ... | 12499 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,264 |
Five. (20 points) Place the numbers 1, 2, 3, 4, 5, 6, 7, 8 at the 8 vertices of an octagon \(A B C D E F G H\), and let \(S_{1}, S_{2}, \cdots, S_{8}\) represent the sums of the numbers at the 8 groups of three adjacent vertices \((A, B, C), (B, C, D), \cdots, (H, A, B)\).
(1) Provide a placement such that \(S_{1}, S_... | (1) It is not difficult to verify that the filling method shown in Figure 7 satisfies $S_{1}, S_{2}, \cdots, S_{8}$ are all greater than or equal to 12;
(2) Obviously, each vertex appears in 3 groups out of the total 8 groups of 3 adjacent vertices, so we have
$$
\begin{array}{l}
S_{1}+S_{2}+\cdots+S_{8} \\
=(1+2+3+\cd... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,265 |
5. If the ratio of the lengths of the medians on the two legs of a right triangle is $m$, then the range of values for $m$ is $\qquad$ . | 5. $\frac{1}{2}<m<2$ | \frac{1}{2}<m<2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,270 |
6. If the equation $|1-x|=m x$ has a solution with respect to $x$, then the range of real number $m$ is
保留了原文的换行和格式。 | 6. $m<-1$ or $m \geqslant 0$ | m<-1 \text{ or } m \geqslant 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,271 |
3. Consider a $2 n \times 2 n$ square grid chessboard, and place a chess piece in any $3 n$ of its cells. Prove that it is possible to select $n$ rows and $n$ columns such that all $3 n$ chess pieces are within these $n$ rows and $n$ columns. | （Hint: Among the rows of pieces, there must be one with the most. Suppose there are $p_{1}$ pieces, then from the remaining $2 n-1$ rows, find the row with the most pieces, suppose there are $p_{2}$ pieces, and so on to find $n$ rows, totaling $p_{1}+\cdots+p_{n}$ pieces. Use proof by contradiction to show $p_{1}+\cdot... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,272 |
II. (16 points) Find all four-digit numbers that satisfy the following conditions: they are divisible by 111, and the quotient obtained is equal to the sum of the digits of the four-digit number. | $$
\overline{a b c d}=a \times 10^{3}+b \times 10^{2}+c \times 10+d
$$
satisfies the condition. Then
$$
\begin{array}{l}
\frac{a \times 10^{3}+b \times 10^{2}+c \times 10+d}{11 i} \\
=9 a+b+\frac{a-11 b+10 c+d}{11 i} . \\
\because-98 \leqslant a-11 b+10 c+d \leqslant 108, \text { and } \overline{a b c d} \text { is di... | 2997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,277 |
Three, (16 points) (1) In a $4 \times 4$ grid paper, some small squares are painted red, and then 2 rows and 2 columns are crossed out. If no matter how they are crossed out, at least one red small square remains uncrossed, how many small squares must be painted red at least? Prove your conclusion.
(2) If the “$4 \time... | Three, (1) At least 7 cells need to be colored.
If the number of colored cells $\leqslant 4$, then by appropriately crossing out 2 rows and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 5, then at least one row has 2 cells colored. By crossing out this row, the remaining colore... | 5 | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,278 |
Four (18 points) As shown in Figure 3, $ABCD$ is a square with a side length of 1, $U$ and $V$ are points on $AB$ and $CD$ respectively, $AV$ intersects with $DU$ at point $P$, and $BV$ intersects with $CU$ at point $Q$. Find the maximum area of quadrilateral $PUQV$. | $$
\begin{array}{l}
\because A U ̈ / / D V, \\
\therefore S_{\triangle U P V} \\
=S_{\triangle U D V}-S_{\triangle P D V} \\
=S_{\triangle A D V}-S_{\triangle P D V} \\
=S_{\triangle A S P}. \\
\end{array}
$$
Similarly, $S_{\triangle U Q V}=S_{\triangle B Q C}$.
Therefore, $S_{\text {quadrilateral PUQV }}=S_{\triangle... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,279 |
2. Given that $x$ is a positive integer. Then, the number of triangles that can be formed with sides $3$, $x$, and $10$ is ( ).
(A) 2
(B) 3
(C) 5
(D) 7 | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,281 |
4. A batch of goods is divided into many packages of different sizes, with each package, including its packaging, weighing no more than 350 kilograms. The total weight of this batch of goods, including packaging, is 13500 kilograms. Now, a truck with a load capacity of 1500 kilograms is used to transport these goods. I... | (提示: First, load the first truck as much as possible, until it reaches 1500 kilograms, but adding one more package would exceed 1500 kilograms, and place this package aside. Then handle the second, third, ... trucks in the same way, until all 8 trucks are loaded. At this point, the total weight of the goods on the 8 tr... | 11 | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 712,283 |
5. As shown in Figure 1, in $\triangle A B C$, $\angle C=90^{\circ}$, $A D$ is the angle bisector of $\angle B A C$, and $B D: D C=2: 1$. Then $\angle B$ satisfies ( ).
(A) $0^{\circ}<\angle B<15^{\circ}$
(B) $\angle B=15^{\circ}$
(C) $15^{\circ}<\angle B<30^{\circ}$
(D) $\angle B=30^{\circ}$ | 5.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,285 |
6. The perimeter of a square is twice the circumference of a circle. Then the ratio of the area of the square to the area of the circle is ( ).
(A) $2 \pi$
(B) $\pi$
(C) 4
(D) 2 | 6.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,286 |
7. The product $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{1999^{2}}\right)(1-$ $\frac{1}{2000^{2}}$ ) equals ( ).
(A) $\frac{1999}{2000}$
(B) $\frac{2001}{2000}$
(C) $\frac{1999}{4000}$
(D) $\frac{2001}{4000}$ | 7.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,287 |
5. If the two roots of the equation $x^{2}-2 x+\frac{\sqrt{3}}{2}=0$ are $\alpha, \beta$, and they are also the roots of the equation $x^{4}+p x^{2}+q=0$, then $p=$ | $5 . \sqrt{3}-4$ | \sqrt{3}-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,293 |
Example 1 As shown in Figure $5, O$ and $I$ are the circumcenter and incenter of $\triangle A B C$, respectively. $A D$ is the altitude from $A$ to side $B C$, and $I$ lies on the segment $O D$. Prove that the circumradius of $\triangle A B C$ is equal to the radius of the excircle opposite to side $B C$. | Explanation: From the formula for the exradius, we have
$$
\begin{aligned}
r_{a} & =\frac{2 S_{\triangle A B C}}{b+c-a} \\
& =\frac{a \cdot A D}{b+c-a} .
\end{aligned}
$$
To prove \( R = r_{a} \), it suffices to prove
$$
\frac{R}{A D}=\frac{a}{b+c-a} \text {. }
$$
As shown in Figure 6, extend \( A I \) to intersect \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,294 |
8. As shown in Figure 4, in parallelogram $A B C D$, $A B=2, B C=$ $2 \sqrt{3}, A C=4$, a line $E F \perp A C$ is drawn through the midpoint $O$ of $A C$, intersecting $A D$ at $E$ and $B C$ at $F$. Then $E F=$ $\qquad$ | 8. $\frac{4 \sqrt{3}}{3}$ | \frac{4 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,297 |
Three, (16 points) Let the equation $x^{2}-|2 x-1|-4=0$. Find the sum of all roots that satisfy the equation.
| When $2 x-1>0$, i.e., $x>\frac{1}{2}$, the original equation becomes $x^{2}-2 x-3=0$.
Solving, we get $x_{1}=3, x_{2}=-1$ (discard).
When $2 x-1=0$, i.e., $x=\frac{1}{2}$, substituting into the original equation does not fit, discard.
When $2 x-1<0$, i.e., $x<\frac{1}{2}$, the original equation becomes $x^{2}+2 x-5=0$.... | 2-\sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,298 |
Four. (20 points) As shown in Figure 5, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are both equilateral triangles, and the midpoints of $B C$ and $B_{1} C_{1}$ are both $D$. Prove: $A A_{1} \perp O C_{1}$.
保留源文本的换行和格式，直接输出翻译结果如下：
Four. (20 points) As shown in Figure 5, $\triangle A B C$ and $\triangle A_{1} ... | As shown in Figure 6, connect $A D$, $A_{1} D$, extend $A A_{1}$ to intersect $D C$ at $O$, and intersect $C_{1} C$ at $E$.
In $\triangle A A_{1} D$ and $\triangle C C_{1} D$, we have
$\angle A D A_{1}=90^{\circ}-\angle A_{1} D C=\angle C D C_{1}$.
Also, since $\frac{A D}{D C}=\sqrt{3}$,
$\frac{D A_{1}}{D C_{1}}=\sqrt{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,299 |
Five. (20 points) At the New Year's Eve party, the school organized a knowledge competition on Chinese, mathematics, foreign language, the Olympics, and daily life common knowledge. The full score is set at 40 points, followed by 30 points, 20 points, 10 points, and 0 points, for a total of five scoring levels. Each gr... | Given that the total score of group $E$ is $E_{\text {total }} \geqslant 60$.
The total score of the five groups is
$5 \times(10+20+30+40)=500$ points.
If $E_{\text {total }}=70$, and each group's score is different, then
the total score of the five groups $\geqslant 70+80+90+100+180$
$$
=520 \geqslant 500 \text {, }
$... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,300 |
3. The natural number $n$ satisfies
$$
\left(n^{2}-2 n-2\right)^{n^{2}+47}=\left(n^{2}-2 n-2\right)^{16 n-16} \text {. }
$$
The number of such $n$ is ().
(A) 2
(E)1
(C) 3
(D) 4 | 3.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. The text you provided is already in English. If you meant to provide a different text in another language, p... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,303 |
4. In $\triangle A B C$, $\angle A B C=30^{\circ}$, side $A B=10$, side $A C$ can take one of the values 5, 7, 9, 11. The number of non-congruent triangles that satisfy these conditions is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,304 |
Example 1 A square grid paper with 8 rows and 8 columns, randomly color 32 of the cells black, and the remaining 32 cells white. The following operations are performed on the colored grid paper: each operation is to simultaneously change the color of all cells in any row or column. Can we get a grid paper with exactly ... | Solution: The requirement of the problem cannot be met.
Assume that a row (or column) contains $k$ black squares and $8-k$ white squares $(0 \leqslant k \leqslant 8)$.
If an operation is performed on this row (or column), the colors of the squares are changed, resulting in $8-k$ black squares and $k$ white squares.
T... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,305 |
In the following text, "倒 2 隹里" is an electronic display of English letters, where each operation can simultaneously change four letters in a row; the rule of change is that each English letter becomes its next letter according to the order of the English alphabet (i.e., $A$ becomes $B, B$ becomes $C, \cdots \cdots$, a... | Solution: According to the given operation rules, Table A cannot be transformed into Table B. The reason is as follows:
For convenience, we replace the letters in the table with their corresponding positions in the alphabet (i.e., $A$ is $1, B$ is $2, \cdots, Z$ is 26). Thus, Table A and Table B can be considered as t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 712,306 |
Example 1: 20 teams participate in the national football championship. To ensure that in any three teams that have already played, there are two teams that have already played against each other, what is the minimum number of matches that need to be played?
(3rd All-Union Mathematical Olympiad) | Solution: Suppose that in any three teams that have already played, there are two teams that have played against each other. Let team $A$ be the team with the fewest matches played, having played $k$ matches. Thus, the $k$ teams that have played against team $A$, as well as team $A$ itself, have all played no fewer tha... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,307 |
$\begin{array}{l}\text { II. ( } 50 \text { points) Let } a_{i} \in \mathbf{R}^{+}(i=1,2 \ldots, n) , \\ a_{1}+a_{2}+\ldots+a_{n}=1 \text {. Prove: } \\ \overline{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}} \\ \quad+\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}} \\ \quad+\ldots+\frac{a_{... | $\begin{array}{l}\text { II. Let } A=\frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}} \\ +\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}} \\ +\ldots+\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}} \text {, } \\ B=\frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}... | \frac{1}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,308 |
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