problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. Given real numbers $x, y$ satisfy
$$
\left(x+\sqrt{x^{2}+2}\right)\left(y+\sqrt{y^{2}+4}\right)=16 .
$$
Then the value of $x \sqrt{y^{2}+4}+y \sqrt{x^{2}+2}$ is ( ).
(A) $\frac{27}{4}$.
(B) $\frac{29}{4}$
(C) $\frac{31}{4}$
(D) $\frac{33}{4}$ | 3. C.
From equation (1), we easily obtain
$$
\left(x-\sqrt{x^{2}+2}\right)\left(y-\sqrt{y^{2}+4}\right)=\frac{1}{2} \text {. }
$$
Expanding equations (1) and (2) respectively and then subtracting them, we get
$$
x \sqrt{y^{2}+4}+y \sqrt{x^{2}+2}=\frac{31}{4} .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,857 |
4. In an acute triangle $\triangle A B C$, the three altitudes $A D, B E$, and $C F$ intersect at point $H$. Then $H D: H E: H F=(\quad)$.
(A) $a: b: c$
(B) $\cos A: \cos B: \cos C$
(C) $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$
(D) $\frac{1}{\cos A}: \frac{1}{\cos B}: \frac{1}{\cos C}$ | 4. D.
As shown in Figure 2, it is easy to see that
$$
\begin{array}{l}
H D: H E \\
=\cos \angle C H D: \\
\quad \cos \angle C H E \\
=\cos B: \cos A .
\end{array}
$$
Similarly,
$$
H E: H F=\cos C: \cos B \text {. }
$$
From equations (1) and (2), we get
$$
H D: H E: H F=\frac{1}{\cos A}: \frac{1}{\cos B}: \frac{1}{\c... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,858 |
5. There is a sequence of numbers $a_{1}, a_{2}, \cdots, a_{m}$, and $S_{n}$ represents $a_{1}+a_{2}+\cdots+a_{n}$. If $a_{1}=1005$, and $S_{n}=n^{2} a_{n}$, then the value of $a_{2000}$ is ( ).
(A) $\frac{1}{2007}$
(B) $\frac{1}{2008}$
(C) $\frac{1}{2009}$
(D) $\frac{1}{2010}$ | 5. C.
From $S_{n}=n^{2} a_{n}$, we get $S_{n-1}=(n-1)^{2} a_{n-1}$.
Subtracting the two equations and rearranging, we obtain
$$
a_{n}=\frac{(n-1)^{2} a_{n-1}}{n^{2}-1}=\frac{n-1}{n+1} a_{n-1} \text {. }
$$
Therefore, $a_{n}=\frac{n-1}{n+1} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdot \frac{2}{4} \cdot \frac... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,859 |
6. In a convex quadrilateral $ABCD$, $AB \times CD$, $M$ is the midpoint of $AB$, $N$ is a point on $CD$, $AN$ intersects $DM$ at point $E$, and $CM$ intersects $BN$ at point $F$. If
$$
S_{\triangle ADE} + S_{\triangle BCF} = S_{\text{quadrilateral MENF}},
$$
then the value of $\frac{DN}{CD}$ is ( ).
(A) $\frac{1}{2}$... | 6. A.
From the given equation, we know that
$$
S_{\triangle A D N}+S_{\triangle B C N}=S_{\triangle M D C} \text {. }
$$
As shown in Figure 3, draw perpendiculars from points $A$, $B$, and $M$ to $CD$, intersecting $CD$ at points $A'$, $B'$, and $M'$, respectively. Then
$$
\begin{array}{l}
A A' / / M M' / / B B' . \\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,860 |
1. If real numbers $x, y$ satisfy $4 x^{2}+y^{2}=1$, then the minimum value of $\frac{4 x y}{2 x+y-1}$ is . $\qquad$ | $=.1 .1-\sqrt{2}$.
Notice
$$
\begin{array}{l}
\frac{4 x y}{2 x+y-1}=\frac{(2 x+y)^{2}-\left(4 x^{2}+y^{2}\right)}{2 x+y-1} \\
=\frac{(2 x+y)^{2}-1}{2 x+y-1}=2 x+y+1 .
\end{array}
$$
Let $2 x+y+1=k$. Then $y=k-2 x-1$.
Substitute into $4 x^{2}+y^{2}=1$ and rearrange to get
$$
8 x^{2}+4(1-k) x+\left(k^{2}-2 k\right)=... | 1-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,861 |
2. In $\triangle A B C$, $\angle A=60^{\circ}, \angle B=20^{\circ}$, extend $B C$ to $D$ such that $C D=A B$. Then $\angle C A D=$ $\qquad$ | 2. $80^{\circ}$.
As shown in Figure 4,
Take point $E$ on side $CD$ such that $BE = AB$, and connect $AE$. Then
$$
\begin{array}{l}
\angle AEC = \angle BAE = \frac{180^{\circ} - \angle B}{2} = 80^{\circ} \\
= \angle B + \angle BAC = \angle ACE.
\end{array}
$$
Therefore, $AC = AE$.
It is easy to see that $\triangle ABE... | 80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,862 |
3. Given that $x$, $y$, and $z$ are real numbers, and
$$
x^{2}+4 y^{2}+9 z^{2}=2009 \text{. }
$$
Then the maximum value of $x+2 y+3 z$ is $\qquad$ | $3.7 \sqrt{123}$.
By the AM-GM inequality, we have
$$
\frac{x+2 y+3 z}{3} \leqslant \sqrt{\frac{x^{2}+4 y^{2}+9 z^{2}}{3}}=\sqrt{\frac{2009}{3}} \text {. }
$$
Thus, $x+2 y+3 z \leqslant 7 \sqrt{123}$.
Equality holds if and only if $x=2 y=3 z=\sqrt{\frac{2009}{3}}$. | 7 \sqrt{123} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,863 |
4. In $\triangle A B C$, $A B=1, B C=2, \angle B-\angle C$ $=90^{\circ}$. Then the radius of the circumcircle of $\triangle A B C$ is $\qquad$ | 4. $\frac{\sqrt{3}+1}{2}$.
As shown in Figure 5, let $O$ be the center of the circumcircle of $\triangle ABC$, and construct the diameter $AD$, connecting $DB$ and $DC$. Then
$$
\begin{array}{l}
\angle ABD=90^{\circ}. \\
\text{By } \angle ABC \\
=90^{\circ}+\angle ACB, \\
\angle ABC=90^{\circ}+\angle DBC, \\
\text{the... | \frac{\sqrt{3}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,864 |
II. (25 points) In the equilateral $\triangle ABC$, points $D$ and $E$ are on $BC$ and $AC$ respectively, and $BD = \frac{1}{3} BC$, $CE = \frac{1}{3} AC$. Connect $AD$ and $BE$ to intersect at point $F$. Prove that $CF \perp AD$.
---
The translation maintains the original format and line breaks. | As shown in Figure 9, draw $AG \perp BC$ at point $G$, and connect $FG$. Then $BG = \frac{1}{2} BC$.
It is easy to see that
$\triangle ABD \cong \triangle BCE$.
Thus, $\angle BAD$
$$
= \angle CBE = \angle DBF.
$$
Therefore, $\triangle ABD \backsim \triangle BFD \Rightarrow \frac{BD}{FD} = \frac{AD}{BD}$, which means $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,867 |
Three. (25 points) Given that $a$ and $b$ are any real numbers, find the minimum value of the maximum number among $|2a+b|$, $|2a-b|$, and $|1-b|$. | Three, let the largest number among $|2a + b|$, $|2a - b|$, and $|1 - b|$ be $x$. Then,
\[ |2a + b| \leq x, \quad |2a - b| \leq x, \quad |1 - b| \leq x. \]
Thus,
\[ -x \leq 2a + b \leq x, \]
\[ -x \leq b - 2a \leq x, \]
\[ -x \leq 1 - b \leq x. \]
Adding (1) + (2) + 2 \times (3) gives:
\[ -4x \leq 2 \leq 4x. \]
Solving... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,868 |
1. Given three non-zero real numbers $a, b, c$, the set $A=$ $\left\{\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}\right\}$. Let $x$ be the sum of all elements in set $A$, and $y$ be the product of all elements in set $A$. If $x=2 y$, then the value of $x+y$ is $\qquad$ | $-1 .-6$.
From the problem, we have
$$
x=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}
$$
$$
\begin{aligned}
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-3, \\
y & =\frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} \\
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-1 .
\end{aligned}
$$
Then $x=y-2$.
Also, $x=2y$, solving this gives $y=-2$.
Thus... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,869 |
2. The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersects the negative half-axis of the $x$-axis at point $C$, and $A$ is a point on the ellipse in the first quadrant. The line $O A$ intersects the ellipse at another point $B$, and the left focus of the ellipse is $P$. Connecting $A P$ intersects $B ... | 2. $\frac{1}{4}$.
From the problem, we know $C(-a, 0), P(-c, 0)$.
Let $A\left(x_{0}, y_{0}\right)\left(y_{0} \neq 0\right)$.
Then $B\left(-x_{0},-y_{0}\right)$.
By $\overrightarrow{C D}=\frac{3}{2} \overrightarrow{D B}$, we have
$$
x_{D}=-\frac{2 a+3 x_{0}}{5}, y_{D}=-\frac{3 y_{0}}{5} \text {. }
$$
Then $k_{A P}=\fr... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,870 |
3. Given the recurrence relation $a_{n+1}=\left(\frac{a_{n}+2-\sqrt{2-a_{n}^{2}}}{2}\right)^{\frac{1}{2}}$ $(n \in \mathbf{N}), a_{0}=\frac{\sqrt{2}}{2}$. Then the general term formula $a_{n}=$ $\qquad$ | 3. $\sqrt{2} \cos \left(\frac{\pi}{4}+\frac{\pi}{12 \times 2^{n}}\right)$.
Let $a_{n}=\sqrt{2} \cos \theta_{n}\left(\theta_{n} \in\left(0, \frac{\pi}{2}\right)\right)$. Then $\theta_{0}=\frac{\pi}{3}$.
Substituting into the recursive formula, we get
$$
\begin{array}{l}
\cos \theta_{n+1}=\cos \left(\frac{\theta_{n}}{2... | \sqrt{2} \cos \left(\frac{\pi}{4}+\frac{\pi}{12 \times 2^{n}}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,871 |
5. Real numbers $x_{i} \geqslant 1(i=1,2, \cdots, 2009)$, and $\prod_{i=1}^{2009} x_{i}$ $=\mathrm{e}, \prod_{i=1}^{2009} x_{i}^{\ln x_{i}} \geqslant \mathrm{e}$. Then the value of $\sum_{i=1}^{2009} x_{i}$ is $\qquad$ | 5. $2008+$ e.
Since $\prod_{i=1}^{2009} x_{i}^{\ln x_{i}} \geqslant \mathrm{e}$, let $y_{i}=\ln x_{i}$.
Taking the natural logarithm on both sides, we get $\sum_{i=1}^{2009} y_{i}^{2} \geqslant 1$.
Also, from $\prod_{i=1}^{2009} x_{i}=\mathrm{e}$, we have $\sum_{i=1}^{2009} y_{i}=1$.
Substituting into the inequality, ... | 2008+\mathrm{e} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,873 |
6. In tetrahedron $ABCD$, $AD > AB$, $AD \perp AB$, $AD \perp AC$, $\angle BAC = \frac{\pi}{3}$. Let the areas of $\triangle ADB$, $\triangle ADC$, $\triangle ABC$, and $\triangle BCD$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively, and they satisfy $S_{1} + S_{2} = S_{3} + S_{4}$. Then the value of $\frac{S_{... | 6. $\frac{3}{2}$.
Let $A D=x, A B=y, A C=z$. Then
$$
\begin{array}{l}
B D=\sqrt{x^{2}+y^{2}}, C D=\sqrt{x^{2}+z^{2}}, \\
B C=\sqrt{y^{2}+z^{2}-y z} .
\end{array}
$$
And $\cos \angle B D C=\frac{B D^{2}+C D^{2}-B C^{2}}{2 B D \cdot C D}$
$$
=\frac{2 x^{2}+y z}{2 \sqrt{x^{2}+y^{2}} \cdot \sqrt{x^{2}+z^{2}}} \text {, }
... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,874 |
7. Given $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, the quadratic equation
$$
\left(\tan ^{2} \theta+\sec ^{2} \theta\right) x^{2}+2\left(\tan ^{2} \theta-\sin ^{2} \theta\right) x-\cos 2 \theta=0
$$
has a repeated root. Then the value of $\cos \theta$ is . $\qquad$ | 7. $\sqrt{\frac{3-\sqrt{5}}{2}}$.
Since the equation has a repeated root, $\Delta=0$, i.e., $\left(\tan ^{2} \theta-\sin ^{2} \theta\right)^{2}+\cos 2 \theta\left(\tan ^{2} \theta+\sec ^{2} \theta\right)=0$. Let $d=\cos ^{2} \theta$. Then $\left(\frac{1-d}{d}+d-1\right)^{2}+(2 d-1)\left(\frac{1-d}{d}+\frac{1}{d}\right... | \sqrt{\frac{3-\sqrt{5}}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,875 |
8. The total score of an exam consists of six 6-point questions, six 9-point questions, and twelve 5-point questions. Therefore, the number of different scores that can be formed by this test paper is $\qquad$ | 8. 136.
Since there are scores, and the scores can only be generated from the 150 positive integers from $1 \sim 150$, and the sum of the scores and the deducted points is also 150, we only need to consider the number of positive integers from $1 \sim 75$ that can be represented by $6x + 9y + 5z$ (where $x, y, z$ are ... | 136 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,876 |
1. Given that the graph of a certain linear function is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and passes through the point $(-1,-25)$. Then the number of integer points on the line segment $AB$ (including endpoints $A$ and $B$) ... | ( Hint: Use the method of exhaustion. The integer points are $(19,0)$, $(15,-5)$, $(11,-10)$, $(7,-15)$, $(3,-20)$, a total of 5.) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,877 |
1. (14 points) Let $n$ be a positive integer greater than 16, and let $A(n, k)$ and $B(n, k) (k \geqslant 3)$ be the number of $k$-term monotonic arithmetic sequences and geometric sequences with integer common ratios, respectively, in the set $\{1,2, \cdots, n\}$.
(1) Find the value of $A(2009,50)$;
(2) Prove: $n-4 \s... | 1. (1) First, find the general $A(n, \dot{k})$.
Let the set $\{1,2, \cdots, n\}$ have a monotonically increasing arithmetic sequence $\left\{a_{i}\right\}$, with common difference $d$, and $a_{k}=a_{1}+(k-1) d$. Then
$$
\begin{array}{l}
1+(k-1) d \leqslant a_{1}+(k-1) d \leqslant n \\
\Rightarrow d \leqslant \frac{n-1}... | 80360 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,878 |
2. (15 points) Given the hyperbola $S: x^{2}-4 y^{2}=8$, and a moving chord $A B$ on the right branch of $S$, with the midpoint of $A B$ being $C$.
(1) If the distance from $C$ to one of the asymptotes $l_{1}$ of the hyperbola that passes through the first and third quadrants is 1, and the other asymptote $l_{2}$ inter... | 2. (1) When $AB$ is not a perpendicular chord, let $l_{AB}: y=kx+b$. Combining with $S$ we get
$\left(1-4k^{2}\right)x^{2}-8kbx-4b^{2}-8=0$.
From $AB$ intersecting the hyperbola at two points, we know
$k \neq \pm \frac{1}{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we get $x_... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,879 |
One. (50 points) As shown in Figure 1, the excircle of $\triangle ABC$, $\odot O$, touches side $BC$ and the extensions of $AB$ and $AC$ at points $D$, $E$, and $F$ respectively. $BH \perp OC$ at point $H$, and $CG \perp OB$ at point $G$. Connect $GH$. Draw $DI \perp GH$, and let $BI$ and $CI$ intersect the extensions ... | First, prove a conclusion.
As shown in Figure 2, connect $E F$ intersecting $O B$ and $O C$ at points $G^{\prime}$ and $H^{\prime}$. Then $G^{\prime}=G, H^{\prime}=H$. That is, prove
$B H^{\prime} \perp O C, C G^{\prime} \perp O B$.
Connect $D G^{\prime}, D E, D F$.
It is easy to see that $\triangle B E G^{\prime} \con... | \frac{EN}{FM}=\frac{BD}{CD} | Geometry | proof | Yes | Yes | cn_contest | false | 721,881 |
(1) If $a, b, c$ are positive numbers, and $m$ is a non-negative number, then
$$
\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b} \geqslant \frac{a^{2}+m}{b c+m}+\frac{b^{2}+m}{c a+m}+\frac{c^{2}+m}{a b+m} \text {; }
$$
(2) If $a, b, c, d$ are positive numbers, then
$$
\begin{array}{l}
\frac{a^{2}}{a+b}+\frac{b^{2... | (1) It suffices to prove
$$
\sum\left(\frac{a^{2}}{b c}-\frac{a^{2}+m}{b c+m}\right) \geqslant 0,
$$
i.e., $\square$
$$
\sum \frac{m\left(a^{2}-b c\right)}{b c(b c+m)} \geqslant 0.
$$
Since $m \geqslant 0$, it suffices to prove
$$
\sum \frac{a^{2}-b c}{b c(b c+m)} \geqslant 0.
$$
By $b c \leqslant \frac{b^{2}+c^{2}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,882 |
Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.
| Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,883 |
Four. (50 points) Given integers $a, b$ are both non-zero, satisfying $a^{2}+b^{2}=p^{\alpha} q^{\beta}(\alpha \beta \in\{0,1,2\}, p, q$ are odd primes), and $p=m^{2}+n^{2}(m, n \in \mathbf{Z})$, $p-q$ is not a multiple of 4. If there exist integers $x, y$ such that
$$
\left(a^{2}+b^{2}\right) \mid\left(x^{2}+y^{2}\rig... | $$
\begin{array}{l}
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$
Let $a c+b d=x, a d-b c=y$ or $a c-b d=x$, $a d+b c=y$. Then,
$$
c=\frac{a x-b y}{a^{2}+b^{2}}, d=\frac{a y+b x}{a^{2}+b^{2}}
$$
or $c=\frac{a x+b y}{a^{2}+b^{2}}, d=\frac{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,884 |
Initially 261 the perimeter of an integer-sided triangle is 75, and squares are constructed on each side. The sum of the areas of the three squares is 2009. Find the difference between the longest and shortest sides of this triangle. | Solution: Given that the sum of three numbers is 75, their average is 25.
Let $2009=(25+a)^{2}+(25+b)^{2}+(25+c)^{2}$, where $a+b+c=0$. It is easy to see that
$a^{2}+b^{2}+c^{2}=134$.
Decomposing 134 into the sum of squares of three positive integers, we find the following four cases:
(1) $a^{2}+b^{2}+c^{2}=134=121+9+4... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,885 |
In $\triangle A B C$, $B C$ is the largest side, $A D$ is the altitude, and $B E$ is the angle bisector. Prove that $\frac{A B}{B C}=\frac{A D}{D C}$ if and only if $\angle A E B=45^{\circ}$. | Proof: Necessity.
As shown in Figure 2, draw $B K / / A D$ intersecting the ray $C A$ at point $K$, and draw $C F \perp B A$ at point $F$.
It is easy to see that
$$
\frac{K B}{B C}=\frac{A D}{D C}.
$$
Since $\frac{A B}{B C}=\frac{A D}{D C}$, it follows that $K B=A B$.
At this point, $\angle D A C=\angle K=\angle B A K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,886 |
$$
\begin{array}{l}
\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)\left(a^{2}+a b+b^{2}\right) \\
\geqslant \frac{1}{3}(b c+c a+a b)^{2}\left[2\left(a^{2}+b^{2}+c^{2}\right)+\right. \\
\quad(b c+c a+a b)] .
\end{array}
$$
Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\left(b^{2}+b c+c^{2}\righ... | Prove: Construct $\triangle A B C, O A=a$,
$$
\begin{array}{c}
O B=b, O C=c, \angle B O C \\
=\angle C O A=\angle A O B=
\end{array}
$$
$120^{\circ}$. Thus,
$$
\begin{array}{l}
B C=\sqrt{b^{2}+b c+c^{2}}, \\
C A=\sqrt{c^{2}+c a+a^{2}}, \\
A B=\sqrt{a^{2}+a b+b^{2}} .
\end{array}
$$
Let the circumradius of $\triangle A... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,887 |
Example 1 As shown in Figure 1, in $\triangle A B C$, $A B=A C$, $\angle B A C=90^{\circ}, B D$ is a median, $A E \perp B D$ intersects $B C$ at point $E$. Prove:
$$
B E=2 E C \text {. }
$$
(2004, Hebei Province
Junior High School Mathematics Innovation and Knowledge
Application Competition) | Proof As shown in Figure 1, extend $BA$ to point $F$, such that $AF=AD$, and connect $CF$.
Obviously, $\triangle ABD \cong \triangle ACF$
$$
\Rightarrow \angle ADB=\angle AFC \text{.}
$$
Notice that $\angle BAE=\angle ADB=\angle AFC$
$$
\Rightarrow AE \parallel FC \Rightarrow \frac{BE}{EC}=\frac{BA}{AF} \text{.}
$$
S... | BE=2EC | Geometry | proof | Yes | Yes | cn_contest | false | 721,888 |
Example 2 In an isosceles right $\triangle ABC$, $AC=BC=1$, $M$ is the midpoint of $BC$, $CE \perp AM$ at point $E$, intersects $AB$ at point $F$. Then $S_{\triangle BMF}=$ $\qquad$ . | From Figure 2, by the conclusion of Example 1, we have
$$
\begin{aligned}
A F & =2 B F \\
\Rightarrow & S_{\triangle B M F} \\
& =\frac{1}{3} S_{\triangle B M} .
\end{aligned}
$$
Since $M$ is the midpoint of $B C$, then
$$
S_{\triangle A B M}=\frac{1}{2} S_{\triangle A B C} .
$$
Therefore, $S_{\triangle B M F}=\frac{... | \frac{1}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,889 |
Example 11 As shown in Figure 11, 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:
$$
K L \parallel \frac{1}{4} A E \text {. }
$$ | Prove as shown in the figure:
11, Connect $B E$, take its midpoint $R$, and connect $M R$. Then $M R \parallel \frac{1}{2} A E$. Connect $R N$. Then $P, N, Q, R$ are the midpoints of the sides of quadrilateral $B C D E$. Therefore, quadrilateral $P N Q R$ is a parallelogram, and $R N, P Q$ bisect each other.
Since $L$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,890 |
1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$
Then $S_{\triangle A B C}: S_{\triangle O B C}=$ | $-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,891 |
3. Given two non-zero complex numbers $x$ and $y$ whose sum of cubes is zero. Then $\left(\frac{x}{x-y}\right)^{2000}+\left(\frac{y}{x-y}\right)^{2000}$ is $\qquad$ | 3. $2^{-1999}$ or -1.
Let $x=w y$. Then $w^{3}+1=0$.
Solving, we get $w=-1$ or $w^{2}-w+1=0$.
When $w=-1$,
$$
\begin{array}{l}
\text { the expression }=\left(\frac{w}{w-1}\right)^{2000}+\left(\frac{1}{w-1}\right)^{2000} \\
=2^{-2000}+2^{-2000}=2^{-1999} .
\end{array}
$$
When $w^{2}-w+1=0$,
$$
\begin{array}{l}
\text {... | 2^{-1999} \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,893 |
4. As shown in Figure 2, in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the dihedral angle $A-B D_{1}-A_{1}$ is $\alpha, A B: B C: C C_{1}=1: 1: 2$. Then $\tan \alpha=$ $\qquad$ | 4. $2 \sqrt{6}$.
As shown in Figure 5, connect $A C$, $A D_{1}$, $B D_{1}$, $A_{1} B$, and $C D_{1}$.
Since $A_{1} D_{1} / / B C$, it is known that points $A_{1}$, $B$, $C$, and $D_{1}$ are coplanar.
Draw $C E \perp B D_{1}$ at point $E$, intersecting $A_{1} B$ at point $F$, and connect $A E$.
By symmetry, we have
$... | 2 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,894 |
7. Given a quartic polynomial $f(x)$ whose four real roots form an arithmetic sequence with a common difference of 2. Then the difference between the largest and smallest roots of $f^{\prime}(x)$ is $\qquad$ | $7.2 \sqrt{5}$.
Let the four real roots of the quartic polynomial $f(x)$ be $a, a+2, a+4, a+6$. Then,
$$
f(x)=k(x-a)(x-a-2)(x-a-4)(x-a-6)
$$
$(k \neq 0)$.
Let $t=x-a-3$. Thus,
$$
\begin{array}{l}
f(x)=k(t+3)(t+1)(t-1)(t-3) \\
=k\left(t^{4}-10 t^{2}+9\right)=g(t) .
\end{array}
$$
Then, $g^{\prime}(t)=k\left(4 t^{3}-20 ... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,897 |
8. As shown in Figure 3, given that $M$ and $N$ are points on the sides $BC$ and $CD$ of the unit square $ABCD$ respectively, $\angle MAN=45^{\circ}, BE \perp$ $AN$ at point $E, DF \perp AM$ at point $F, BG \perp AM$ at point $G$, $DH \perp AN$ at point $H$. Then the area of quadrilateral $EFGH$ is | 8. $\frac{1}{4}$.
Let $\angle M A B=\alpha, \angle M A N=\beta=45^{\circ}, \angle N A D=\gamma$. Then
$$
\begin{array}{l}
S_{\triangle A E F}=\frac{1}{2} A E \cdot A F \sin \beta \\
=\frac{1}{2} A B \sin \gamma \cdot A D \sin \alpha \cdot \sin \beta \\
=\frac{1}{2} \sin \alpha \cdot \sin \beta \cdot \sin \gamma .
\end... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,898 |
9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this con... | 9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.
Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.
According to the principle that triangles w... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,899 |
10. (15 points) Given that $a$, $b$, and $c$ are the side lengths of $\triangle ABC$, and satisfy
$$
\sum \frac{1}{a} \sqrt{\frac{1}{b}+\frac{1}{c}} \geqslant \frac{3}{2} \sqrt{\prod\left(\frac{1}{a}+\frac{1}{b}\right)},
$$
where “$\sum$” and “$\prod$” denote cyclic sum and cyclic product, respectively, determine the ... | 10. Let $d=a+b, e=b+c, f=c+a$. Then, with $d, e, f$ as side lengths, a $\triangle D E F$ can be formed. Thus,
$\sin \frac{D}{2}=\sqrt{\frac{1-\cos D}{2}}=\sqrt{\frac{1-\frac{e^{2}+f^{2}-d^{2}}{2 e f}}{2}}$
$=\sqrt{\frac{(d-e+f)(d+e-f)}{4 e f}}=\sqrt{\frac{a b}{(b+c)(c+a)}}$.
Similarly, $\sin \frac{E}{2}=\sqrt{\frac{b c... | a=b=c | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,900 |
Example 12 In hexagon $A B C D E F$, $A B \parallel D E$, $B C \parallel E F$, $C D \parallel F A$, $A B+D E=B C+E F$, $A_{1}$, $B_{1}$, $D_{1}$, $E_{1}$ are the midpoints of sides $A B$, $B C$, $D E$, $E F$ respectively, and $A_{1} D_{1}=B_{1} E_{1}$. Prove: $\angle C D E=\angle A F E$.
$(2008$, Beijing Middle School ... | Prove as shown in the figure:
12, construct $\square A B P F$, connect $D P$, take the midpoint $M$ of $D P$, then quadrilateral $B C D P$ is a trapezoid. Connect $B_{1} M, E_{1} M$. By the trapezoid midline theorem, we have
$$
\begin{array}{l}
B_{1} M / / C D / / B P / / A F, \\
M E_{1} / / D E / / F P / / A B,
\end{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,901 |
11. (15 points) Let $a_{0}, a_{1}, \cdots, a_{n}$ be real numbers, and $\lambda$ be a complex root of the real-coefficient equation $\sum_{i=0}^{n} a_{i} x^{i}=0$ with respect to $x$, and $|\lambda| \geqslant a_{n} \geqslant \cdots \geqslant a_{1} \geqslant a_{0} \geqslant 1$. Prove: $\lambda^{n+1}=1$. | 11. Since $\lambda$ is a root of the equation $\sum_{i=0}^{n} a_{i} x^{i}=0$, we have
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{i} \lambda^{i}=0 \Rightarrow(\lambda-1) \sum_{i=0}^{n} a_{i} \lambda^{i}=0 \\
\Rightarrow a_{n} \lambda^{n+1}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right) \lambda^{i}+a_{0},
\end{array}
$$
where $a_{... | \lambda^{n+1}=1 | Algebra | proof | Yes | Yes | cn_contest | false | 721,902 |
One, (50 points) Given that $P A$ and $P B$ are tangents to $\odot O$, with $A$ and $B$ as the points of tangency, $C$ is a point on the minor arc $\overparen{A B}$, and $A C > B C$. The extension of $P C$ intersects $A B$ and $\odot O$ at points $Q$ and $D$, respectively. $E$ is the midpoint of $A B$, and $C I$ bisect... | As shown in Figure 6, draw $B F \perp A C$ at point $F$, and connect $O B$, $O C$, $O D$, $O P$, $E F$, $A D$, and $B D$. Then
$$
P E \perp A B .
$$
By the secant-tangent theorem and the projection theorem, we have
$$
\begin{array}{l}
P C \cdot P D=P B^{2} \\
=P E \cdot P O .
\end{array}
$$
Therefore, points $C$, $D$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,903 |
Two, (50 points) The four-digit numbers $m$ and $n$ are reverse positive integers of each other, and $m+n=18 k+9\left(k \in \mathbf{N}_{+}\right), m$ and $n$ have 16 and 12 positive divisors (including 1 and themselves), respectively. The prime factors of $n$ are also prime factors of $m$, but $n$ has one fewer prime f... | Let $m=\overline{a b c d}, a d \neq 0$. Then $n=\overline{d c b a}$.
Given $m+n=9(2 k+1)$, then $9 \mid (m+n)$.
Thus, $9 \mid [(1000 a+100 b+10 c+d)+$
$$
\begin{array}{c}
(1000 d+100 c+10 b+a)], \\
9 \mid 2(a+b+c+d), 9 \mid (a+b+c+d).
\end{array}
$$
Therefore, $9 \mid m, 9 \mid n$.
Since $m+n$ is odd, it follows that ... | 1998 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,904 |
Three, (50 points) Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ such that for all positive numbers $x, y$, the following always holds:
$$
x y(f(x)+f(y)) \doteq(x+y) f(f(x) y) .
$$ | Three, substituting $x=y=t$ into equation (1) yields
$$
2 t^{2} f(t)=2 t f(f(t) t) \text {, }
$$
which simplifies to $t f(t)=f(t f(t))\left(\forall t \in \mathbf{R}_{+}\right)$.
If there exist $x_{1} \neq x_{2}$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$, then substituting $x=x_{1}, y=x_{2}$ into equation (1)... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,905 |
Four. (50 points) Does there exist 4098 sets
\[
\begin{aligned}
B_{i} & =\left\{a_{i 1}, a_{i 2}, \cdots, a_{i n}\right\}(i=1,2, \cdots, 4098, \\
a_{i j} \in A_{j} & =\{3 j-2,3 j-1,3 j\}, j=1,2, \cdots, 12)
\end{aligned}
\]
satisfying the following three conditions:
(1) \( B_{i} \cap B_{i+1}=\varnothing(i=1,2, \cdots,... | Let $(s, t)$ denote a set circle with $s$ sets $A_{1}, A_{2}, \cdots, A_{s}$, and $t$ sets $B_{1}, B_{2}, \cdots, B_{t}$ that satisfy the given conditions. Use $\left(B_{i}, p\right)$ to denote a set formed by all elements of $B_{i}$ and the $p$-th element of $A_{i+1}$.
If $B_{1}, B_{2}, \cdots, B_{t}$ satisfy the give... | 4098 | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,906 |
As shown in Figure 1, in $\odot O$, two chords $AC$ and $BD$ intersect at point $P$. Draw $PE \perp AB, PF \perp DC$, with the feet of the perpendiculars being $E$ and $F$ respectively, and $BF$ intersects $CE$ at point $Q$. Prove: $PQ \perp EF$. | Prove as shown in Figure 1, draw a perpendicular from point $E$ to $B F$ intersecting with the perpendicular from point $P$ to $E F$ at point $K$, and draw a perpendicular from point $F$ to $C E$ intersecting with the perpendicular from point $P$ to $E F$ at point $K'$. It is easy to see that
$\angle P E K + \angle B E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,907 |
In convex quadrilateral $A B C D$,
\[
\begin{array}{l}
\angle A B D=\angle C B D=45^{\circ}, \\
\angle A D B=30^{\circ}, \angle C D B=15^{\circ} .
\end{array}
\]
Find the ratio of the sides of quadrilateral $A B C D$. | Connect $A C$, take the incenter $I$ of $\triangle A B C$, connect $A I 、 C I$, and draw $A E \perp D C$ at point $E$.
From $\angle A B D=\angle C B D$ we know the incenter $I$ lies on $B D$.
According to the properties of the incenter of a triangle, we have
$$
\begin{array}{l}
\angle A I C=90^{\circ}+\frac{1}{2} \angl... | \sqrt{3}: 1:(1+\sqrt{3}): \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,908 |
263 Given that 2009 can be written as a three-digit number $x y z$ in base $k$. If $x+y+z=20+0+9$. Try to determine $x 、 y 、 z 、 k$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve: From the given conditions, we have
$$
\left\{\begin{array}{l}
x k^{2}+y k+z=2009, \\
x+y+z=29 .
\end{array}\right.
$$
(1) - (2) gives
$$
(k-1)[(k+1) x+y]=1980 \text {. }
$$
From the problem, we know $k>12\left(12^{3}=17282009\right)$.
Therefore, $k-1>11$.
Since $k-1<(k+1) x+y, 1980=44 \times 45$,
Thus, $k-1<44$... | (x, y, z, k)=(11,11,7,13),(7,13,9,16),(5,10,14,19),(4,11,14,21),(3,18,8,23),(2,2,25,31),(1,25,3,34) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,909 |
Given $k$ is an integer greater than 1, the function
$$
f_{k}(x)=k \sin x - \sin k x
$$
has a maximum value of $a_{k}$.
(1) Find the value of $a_{k}$;
(2) Prove: $a_{k+1} > a_{k} + \frac{4k-7}{4k+8}$.
Reference data: $\pi = 3.14159 \cdots, \pi^2 = 9.86960 \cdots$ | (1) Solving for the derivative of $f_{k}(x)$, we get
$f_{k}^{\prime}(x)=k \cos x - k \cos k x$.
Since $f_{k}(x)$ is a differentiable periodic function on $\mathbf{R}$, $f_{k}(x)$ has a maximum value. Therefore, the $x$ that maximizes $f_{k}(x)$ must be a root of $f_{k}^{\prime}(x)=0$.
Assume $x \in [0, 2\pi)$.
Setting ... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 721,910 |
1. As shown in Figure 13, in $\triangle ABC$, $D$ and $E$ are the midpoints of sides $BC$ and $AC$, respectively. $AD$ and $BE$ intersect at point $P$. If $\angle BPD = \angle C$, prove that the triangle formed by the three medians of $\triangle ABC$ is similar to $\triangle ABC$.
(2004, National Junior High School Mat... | As shown in Figure 13, extend $PD$ to point $F$ such that $PD = DF$. Connect $BF$, $FC$, $PC$, and $DE$. Then quadrilateral $BFCP$ is a parallelogram.
From $\angle BPD = \angle C$, we know that points $P$, $D$, $C$, and $E$ are concyclic.
Thus, $\angle PCF = \angle CPE = \angle CDE = \angle ABC$.
Also, $\angle PFC = \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,911 |
2. As shown in Figure 14, given that $A D$ is the angle bisector of $\triangle A B C$, $A B < A C$, and a segment $C E = A B$ is cut off on $A C$, with $M$ and $N$ being the midpoints of sides $B C$ and $A E$ respectively. Prove:
$$
M N \parallel A D \text {. }
$$ | As shown in Figure 14, connect $B E$, take the midpoint $F$ of $B E$, and connect $F N, F M$. It is easy to see that
$$
F N \Perp \frac{1}{2} A B, F M \Perp \frac{1}{2} C E .
$$
It is easy to prove $\angle D A C=\angle M N C \Rightarrow M N / / A D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,912 |
3. In the figure, $15, BD, CE$ are two altitudes of $\triangle ABC$, $F, G$ are the midpoints of sides $DE, BC$ respectively, $O$ is the circumcenter of $\triangle ABC$. Prove:
$$
AO \parallel FG \text{. }
$$
(2005, National High School Mathematics League Sichuan Preliminary) | As shown in Figure 15, extend $OA$ to intersect $DE$ at point $H$, and connect $OB$, $GD$, and $GE$. It is easy to see that points $B$, $C$, $E$, and $D$ are concyclic.
Thus, $\angle DEB = \angle DCB = \frac{1}{2} \angle AOB$.
It is also easy to see that $\angle EAH + \angle DEB = 90^{\circ}$.
Therefore, $AH \perp DE$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,913 |
4. As shown in Figure 16, in rhombus $A B C D$, $\angle A=$ $100^{\circ}, M$ and $N$ are the midpoints of sides $A B$ and $B C$ respectively, $M P \perp C D$ at point $P$, then the degree measure of $\angle N P C$ is $\qquad$
(2007, Shandong Province Junior High School Mathematics Competition) | It is known that $\angle B=80^{\circ}$.
Notice that $B M=B N$. Then
$$
\angle B M N=\angle B N M=50^{\circ} \text {. }
$$
As shown in Figure 16, let the midpoint of $M P$ be $E$, and connect $N E$. It is easy to know that $N M=N P$. Then $\angle N M P=\angle N P M$. Therefore,
$$
\angle N P C=\angle N M B=50^{\circ} .... | 50^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,914 |
Example 1 Let $a, b$ be two distinct positive integers, and $\left(a+a b+b^{2}\right) \mid a b(a+b)$.
Prove: $|a-b|>\sqrt[3]{a b}$. | To solve this problem, we need to address two issues: First, the degrees of \(a^{2}+a b+b^{2}\) and \(a b(a+b)\) in the given conditions are different, so we can start by making them the same degree. Second, how to convert the divisibility relationship into an inequality, which requires applying some simple conclusions... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,915 |
Example 2: Let there be $n$ positive integers $a_{i}(1 \leqslant i \leqslant n)$ satisfying $2 \leqslant a_{1}<a_{2}<\cdots<a_{n}$, and let
$$
\begin{array}{l}
S=\sum_{k=1}^{n-1}\left[a_{k}, a_{k+1}\right]+\sum_{k=1}^{n} a_{k}, \\
t=\max _{1 \leq k \leq k \leq \pi}\left\{a_{k}-a_{k-1} .\right.
\end{array}
$$
with the ... | Explain the proof of the inequality with boundaries. Given $\left[a_{k}, a_{k+1}\right] \leqslant a_{k} a_{k+1}, a_{k}+1 \leqslant a_{k-1}$, therefore,
$$
\begin{array}{l}
S \leqslant \sum_{k=1}^{n-1} a_{k} a_{k+1}+\sum_{k=1}^{n} a_{k} \\
=\sum_{k=1}^{n-1} a_{k} a_{k+1}+\sum_{k=1}^{n-1} a_{k-1}+a_{1} \\
=\sum_{k=1}^{n-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,916 |
Example 3 Let $m, n \in \mathbf{N}_{+}, m>n$. Prove:
$$
[m, n]+[m+1, n+1]>\frac{2 m n}{\sqrt{m-n}} \text {. }
$$ | Let $m-n=k\left(k \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
{[m, n]+[m+1, n+1]} \\
=\frac{m n}{(m, n)}+\frac{(m+1)(n+1)}{(m+1, n+1)} \\
>\frac{m n}{(n+k, n)}+\frac{m n}{(n+k+1, n+1)} \\
=\frac{m n}{(k, n)}+\frac{m n}{(k, n+1)} .
\end{array}
$$
Let $(k, n)=d_{1},(k, n+1)=d_{2}$. Then $\left(d_{1}, d_{2}\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,917 |
Example 4 Arrange all positive divisors of 8128 that are less than itself in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$. Prove that:
$$
\sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)}<\frac{8127}{8128} .
$$ | By the Cauchy-Schwarz inequality, we have
$$
n \sum_{k=1}^{n} a_{k}^{2}=\sum_{k=1}^{n} 1^{2} \cdot \sum_{k=1}^{n} a_{k}^{2}>\left(\sum_{k=1}^{n} a_{k}\right)^{2} \text {. }
$$
Therefore, we can consider starting from determining the sum of all positive divisors of 8128 that are less than itself.
Given $8128=2^{6} \ti... | \frac{8127}{8128} | Number Theory | proof | Yes | Yes | cn_contest | false | 721,918 |
Example 3 As shown in Figure 3, in trapezoid $A B C D$, $\angle D=$ $90^{\circ}, M$ is the midpoint of $A B$. If $C M=6.5, B C+C D+$ $D A=17$, then the area of trapezoid $A B C D$ is ( ).
(A) 20
(B) 30
(C) 40
(D) 50
(2007, Shandong Province
Junior High School Mathematics Competition) | Solve As shown in Figure 3, extend $C M$ to intersect the extension of $D A$ at point $E$. Then $\triangle B C M \cong \triangle A E M$.
Therefore, $C E=2 C M=13, A E=B C$.
Let the area of trapezoid $A B C D$ be $S$. Then
$$
\left\{\begin{array}{l}
D E^{2}+C D^{2}=C E^{2}, \\
\frac{1}{2} D E \cdot C D=S .
\end{array}\r... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,919 |
Example 6 Let $\delta(n)$ denote the sum of all positive divisors of the positive integer $n$. Prove: there exist infinitely many $n \in \mathbf{N}_{+}$, such that for $k \in\{1,2, \cdots, n-1\}$, we have
$$
\frac{\delta(n)}{n}>\frac{\delta(k)}{k} \text {. }
$$
(24th IMO Shortlist Problem) | Assume the conclusion is not true, i.e., there are only finitely many $n \in \mathbf{N}_{+}$ such that when $k=1,2, \cdots, n-1$, $a_{n} > a_{k}$, where $a_{i}=\frac{\delta(i)}{i}$.
Let $N$ be the largest of these $i$. Then the maximum value of the sequence $\left\{a_{i}\right\}$ is $a_{N}$. This is because for each $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,921 |
Example 7 Let $p(n)$ denote the greatest odd divisor of the positive integer $n$. Prove that for all positive integers $n$,
$$
\left|\sum_{i=1}^{n} \frac{p(i)}{i}-\frac{2}{3} n\right|<1 \text {. }
$$
(32nd Annual William Lowell Putnam Mathematical Competition) | Notice that the largest odd divisor of an odd number is the number itself; an even number $2k$ and $k$ have the same odd divisors, so their largest odd divisors are equal, i.e.,
$$
\begin{array}{l}
p(2 k+1)=2 k+1, p(2 k)=p(k) . \\
\text { Let } S(n)=\sum_{i=1}^{n} \frac{p(i)}{i} \text {. Then } \\
S(2 n)=\sum_{k=1}^{n}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,922 |
2. Prove that for positive integers $k, m, n$, we have $[k, m][m, n][n, k] \geqslant [k, m, n]^{2}$.
(24th All-Russian Mathematical Olympiad) | Let $k=\prod_{i=1}^{t} p_{i}^{\alpha_{i}}, m=\prod_{i=1}^{t} p_{i}^{\beta_{i}}, n=\prod_{i=1}^{t} p_{i}^{\gamma_{i}}$, where $p_{i}$ are prime numbers, and $\alpha_{i}, \beta_{i}, \gamma_{i} (i=1,2, \cdots, t)$ are non-negative integers. Then
$$
\begin{array}{l}
{[k, m][m, n][n, k]} \\
=\prod_{i=1}^{i} p_{i}^{\max \lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,925 |
3. Let $n, k \in \mathbf{N}_{+}$, positive integers $a_{1}, a_{2}, \cdots, a_{k}$ satisfy $a_{1}<a_{2}<\cdots<a_{k} \leqslant n$, and for $1 \leqslant i<j \leqslant n$, we have $\left[a_{i}, a_{j}\right] \leqslant n$. Prove: $k \leqslant 2 \sqrt{n}+1$. | Let $t=[\sqrt{n}]$, and suppose $a_{i_{0}} \leqslant t<a_{i_{0}+1}$.
For any $i, j (i<j)$, we have
$$
n \geqslant\left[a_{i}, a_{j}\right]=\frac{a_{i} a_{j}}{\left(a_{i}, a_{j}\right)} \geqslant \frac{a_{i} a_{j}}{a_{j}-a_{i}},
$$
which implies $\frac{1}{a_{i}}-\frac{1}{a_{j}} \geqslant \frac{1}{n}$.
Then, $\frac{1}{a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,926 |
5. Let positive integers $a_{1}, a_{2}, \cdots, a_{n}$ be all no greater than $2 n$. Prove that when $n>4$, $\min _{1 \leqslant i<j \leqslant n}\left[a_{i}, a_{j}\right] \leqslant 6\left(\left[\frac{n}{2}\right]+1\right)$.
(1990, China National Training Team Selection Exam) | If $a_{1}, a_{2}, \cdots, a_{n}$ contains one number that is a multiple of another, then
$$
\min _{1 \leqslant i<j \leqslant n}\left[a_{i}, a_{j}\right] \leqslant 2 n \leqslant 6\left(\left[\frac{n}{2}\right]+1\right) .
$$
If $a_{1}, a_{2}, \cdots, a_{n}$ are such that none is a multiple of another, and when $a_{i} \l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,927 |
6. Let $q$ be a given real number, and $\frac{1+\sqrt{5}}{2}<q<2$. For any $n \in \mathbf{N}_{4}$, let the binary representation of $n$ be
$$
n=2^{k}+a_{k-1} \cdot 2^{k-1}+\cdots+a_{1} \cdot 2+a_{0},
$$
where $a, \in\{0,1 \mid(0 \leqslant i \leqslant k-1)$. Define the sequence $\left\{p_{n}\right\}: p_{n}=q^{k}+a_{k-1... | First, prove: when $n=(\underbrace{1010 \cdots 10}_{m \uparrow 10})_{2}$, i.e., it meets the requirement $\left(n=\frac{2\left(4^{m}-1\right)}{3}\right)$, at this time $n$ is even, denoted as $n=2 k$. In fact, for such $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
p_{2 k}=q^{2 m-1}+q^{2 m-3}+\cdots+q, \\
p_{2 k+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,928 |
Given positive real numbers $a, b, c$ satisfying $a+b+c = abc$. Prove:
$$
\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}+\frac{1}{\sqrt{1+c^{2}}} \leqslant \frac{3}{2},
$$
and determine when equality holds.
(1998, Korean Mathematical Olympiad) | Let $a=\tan A, b=\tan B, c=\tan C$. Then the inequality to be proven can be transformed into
$$
\cos A+\cos B+\cos C \leqslant \frac{3}{2} \text {. }
$$
The given equation can be transformed into
$$
\tan A+\tan B+\tan C=\tan A \cdot \tan B \cdot \tan C
$$
or $-\tan A=\frac{\tan B+\tan C}{1-\tan B \cdot \tan C}=\tan (... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,929 |
Example 4 As shown in Figure 4, in rhombus $A B C D$, $\angle A B C$ $=120^{\circ}, F$ is the midpoint of $D C$, and the extension of $A F$ intersects the extension of $B C$ at point $E$. Then the degree measure of the acute angle formed by line $B F$ and $D E$ is ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $50^{\circ}$... | As shown in Figure 4, let $B F$ intersect $D E$ at point $M$, and connect $B D$. Then $\triangle B C D$ is an equilateral triangle.
From $F$ being the midpoint of $C D$, we know $\angle M B C=30^{\circ}$.
Since $A D / / C E$, then $\triangle A D F$ and $\triangle E C F$ are centrally symmetric about point
$F$. Therefor... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,930 |
Question 6 Let $\triangle A B C$ be an acute-angled triangle. Prove that:
$$
\begin{array}{l}
\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2}+ \\
8 \cos A \cdot \cos B \cdot \cos C \geqslant 4 . \\
\text { (MOSP2000) }
\end{array}
$$ | Proof: Write the inequality to be proved in terms of $\cos ^{2} A$, $\cos ^{2} B$, and $\cos ^{2} C$. From the identity (3), we have
$$
\begin{array}{l}
4-8 \cos A \cdot \cos B \cdot \cos C \\
=4\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right) .
\end{array}
$$
Thus, it suffices to prove
$$
\begin{array}{l}
\quad\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,931 |
Question 7 Let $\alpha_{i}>0, \beta_{i}>0(1 \leqslant i \leqslant n, n>1)$, and $\sum_{i=1}^{n} \alpha_{i}=\sum_{i=1}^{n} \beta_{i}=\pi$. Then $\sum_{i=1}^{n} \frac{\cos \beta_{i}}{\sin \alpha_{i}} \leqslant \sum_{i=1}^{n} \cot \alpha_{i}$.
(29th IMO Preliminary Question) | Prove that when $n=2$,
$$
\begin{array}{l}
\frac{\cos \beta_{1}}{\sin \alpha_{1}}+\frac{\cos \beta_{2}}{\sin \alpha_{2}}=\frac{\cos \beta_{1}}{\sin \alpha_{1}}-\frac{\cos \beta_{1}}{\sin \alpha_{1}} \\
=0=\cot \alpha_{1}+\cot \alpha_{2} .
\end{array}
$$
When $n=3$, it is to be proved: Given the internal angles of two ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,932 |
Question 8 Let $a, b, c$ be given positive real numbers. Find all positive real numbers $x, y, z$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
x+y+z=a+b+c, \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .
\end{array}\right.
$$
(36th IMO Shortlist) | First, we prove the following lemma.
Lemma: The equation
$$
x^{2}+y^{2}+z^{2}+x y z=4
$$
has positive real solutions \(x, y, z\) if and only if there exists an acute triangle \(\triangle ABC\) such that \(x=2 \cos A, y=2 \cos B, z=2 \cos C\).
Proof of the lemma: First, by the identity (3), all triples \((2 \cos A, 2 ... | x = \frac{b + c}{2}, y = \frac{c + a}{2}, z = \frac{a + b}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,933 |
Given $x, y, z$ are positive real numbers, $xy + yz + zx + xyz = 4$. Prove: $x + y + z \geqslant xy + yz + zx$.
(1998, Indian Mathematical Olympiad) | Prove that the known equation can be transformed into
$$
(\sqrt{x y})^{2}+(\sqrt{y z})^{2}+(\sqrt{z x})^{2}+\sqrt{x y} \cdot \sqrt{y z} \cdot \sqrt{z x}=4 \text {. }
$$
From Lemma 8, we know that there exists an acute triangle $\triangle ABC$ such that
$$
\begin{array}{l}
\sqrt{y z}=2 \cos A, \sqrt{z x}=2 \cos B, \\
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,934 |
Question 11 Given $\alpha+\beta+\gamma=180^{\circ}(\alpha, \beta, \gamma \geqslant$ $0)$. Find the maximum value of $3 \cos \alpha+4 \cos \beta+5 \cos \gamma$. | Solve: In inequality (1), let
$$
2 y z=3,2 z x=4,2 x y=5 \text {, }
$$
i.e., $x=\sqrt{\frac{4 \times 5}{2 \times 3}}, y=\sqrt{\frac{3 \times 5}{2 \times 4}}, z=\sqrt{\frac{3 \times 4}{2 \times 5}}$,
and $\alpha=\angle A, \beta=\angle B, \gamma=\angle C$. Then
$3 \cos \alpha+4 \cos \beta+5 \cos \gamma \leqslant \frac{7... | \frac{769}{120} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,935 |
Example 1 Prove:
$$
\frac{n}{2}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^{n}-1}<n(n \geqslant 2) .
$$ | Proof Let $x_{n}=n$,
$$
y_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^{n}-1} \text {. }
$$
Then $x_{2}=2, y_{2}=1+\frac{1}{2}+\frac{1}{3}$. Hence $x_{2}>y_{2}$.
Also, $x_{k}-x_{k-1}=1$,
$$
y_{k}-y_{k-1}=\frac{1}{2^{k-1}}+\frac{1}{2^{k-1}+1}+\cdots+\frac{1}{2^{k}-1} \text {, }
$$
Thus, it suffices to prove
$$
1>\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,936 |
Example 2 Given that $n, k$ are positive integers. Prove:
$$
\frac{n^{k+1}}{k+1}<\sum_{i=1}^{n} i^{k}<\left(1+\frac{1}{n}\right)^{k+1} \frac{n^{k+1}}{k+1} \text {. }
$$ | Let $x_{n}=\sum_{i=1}^{n} i^{k}, y_{n}=\frac{n^{k+1}}{k+1}$, $z_{n}=\left(1+\frac{1}{n}\right)^{k+1} \frac{n^{k+1}}{k+1}=\frac{(1+n)^{k+1}}{k+1}$.
Notice that $x_{1}=1, y_{1}=\frac{1}{k+1}$. Thus, $x_{1}>y_{1}$.
Also, $x_{p}-x_{p-1}=p^{k}$,
$$
y_{p}-y_{p-1}=\frac{p^{k+1}-(p-1)^{k+1}}{k+1} \text {, }
$$
Then, by Theore... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,937 |
Example 3 Given $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{N}_{+}\right)$. Prove that for all $n \geqslant 2$ we have
$$
a_{n}^{2}>2 \sum_{i=2}^{n} \frac{a_{i}}{i} \text {. }
$$ | Prove that let $x_{n}=a_{n}^{2}, y_{n}=2 \sum_{i=2}^{n} \frac{a_{i}}{i}$. Then
$$
x_{2}=a_{2}^{2}=\frac{9}{4}, y_{2}=2\left(\frac{\frac{3}{2}}{2}\right)=\frac{3}{2} \text {. }
$$
Thus, $x_{2}>y_{2}$.
$$
\begin{array}{l}
\text { Also, } x_{k}-x_{k-1}=a_{k}^{2}-a_{k-1}^{2} \\
=\left(a_{k}+a_{k-1}\right)\left(a_{k}-a_{k-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,938 |
Example 4 Let $x_{1}, x_{2}, \cdots, x_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ be any two sets of real numbers satisfying the conditions:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\sum_{i=1}^{n}\left|x_{i}\right|=1$;
(3) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, a_{1}>a_{n}$
For any $n \geqslant 2$, try to find... | When $n=2$, $x_{1}=\frac{1}{2}, x_{2}=-\frac{1}{2}$ or $x_{1}=-\frac{1}{2}, x_{2}=\frac{1}{2}$.
Thus, the original inequality becomes
$$
\frac{1}{2}\left|a_{1}-a_{2}\right|=\frac{1}{2}\left(a_{1}-a_{2}\right) \leqslant A\left(a_{1}-a_{2}\right) \text {. }
$$
Therefore, $\frac{1}{2} \leqslant A$.
Next, we prove that fo... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,939 |
Example 5 Given two sets of numbers $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}$, $y_{2}, \cdots, y_{n}$, and they satisfy
$$
\begin{array}{l}
\text { (1) } x_{1}>x_{2}>\cdots>x_{n}>0, \\
y_{1}>y_{2}>\cdots>y_{n}>0 \text {; } \\
\text { (2) } x_{1}>y_{1}, x_{1}+x_{2}>y_{1}+y_{2}, \\
\cdots \cdots \\
x_{1}+x_{2}+\cdots+x_{... | Proof: Let $A=\sum_{i=1}^{n} x_{i}^{l}=\sum_{i=1}^{n} x_{i} x_{i}^{l-1}$,
$$
S_{k}=\sum_{j=1}^{k} x_{j}, S_{k}^{\prime}=\sum_{j=1}^{k} y_{j}, b_{i}=x_{i}^{l-1} \text {. }
$$
From the given conditions, we have
$$
S_{k}>S_{k}^{\prime}(k=1,2, \cdots, n) \text {. }
$$
Applying Theorem 2, we get
$$
\begin{array}{c}
A=\sum... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,940 |
Example 5 As shown in Figure 5, in $\triangle A B C$, $A C=7, B C=$ $4, D$ is the midpoint of $A B$, $E$ is a point on side $A C$, and $\angle A E D=$ $90^{\circ}+\frac{1}{2} \angle C$. Find the length of $C E$. | As shown in Figure 5, draw $B F / / D E$ intersecting $A C$ at point $F$, and draw the angle bisector of $\angle A C B$ intersecting $A B$ at point $G$, and intersecting $B F$ at point $H$. Then $\angle A E D=\angle A F B=\angle C H F+\frac{1}{2} \angle C$.
From $\angle A E D=90^{\circ}+\frac{1}{2} \angle C$, we get
$$... | 5.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,941 |
In $\triangle A B C$, the angle bisector of $\angle A$ intersects $B C$ at point $D$, and the projections of $D$ on $A B$ and $A C$ are $E$ and $F$, respectively. Let the length of $E F$ be $l_{A}$, and similarly define $l_{B}$ and $l_{C}$. If the perimeter of $\triangle A B C$ is $l$, prove:
$$
\frac{l_{A} l_{B} l_{C}... | First, if the three sides $BC$, $CA$, $AB$ of $\triangle ABC$ have lengths $a$, $b$, $c$ respectively, then $l=a+b+c$.
Inspired by equation (1), the following problem is proposed:
Can inequality (1) be strengthened to
$$
l_{A}+l_{B}+l_{C} \leqslant \frac{3}{4}(a+b+c) \text {, }
$$
or $l_{A} l_{B} l_{c} \leqslant \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,942 |
Question Polynomial
$$
(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}} \cdots\left(1-z^{32}\right)^{b_{32}}\left(b_{i} \in \mathbf{N}_{+}\right)
$$
has the following property: when expanded, if terms with powers of $z$ higher than 32 are ignored, what remains is $1-2 z$. Find $b_{32}$.
(17th United States of America Mathema... | The solution to this problem is relatively straightforward:
$$
\begin{array}{l}
(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}} \cdots\left(1-z^{32}\right)^{b_{32}} \\
\equiv 1-2 z\left(\bmod z^{33}\right) .
\end{array}
$$
By comparing the coefficients of $z$, we get $b_{1}=2$.
Let the polynomial on the left side of the abo... | 2^{27}-2^{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,943 |
1. When $m=-\frac{1}{6}$, the value of the algebraic expression
$$
\frac{21-5 m}{m^{2}-9}-\frac{m}{m^{2}-9} \div \frac{m}{m+3}-\frac{m-3}{m+3}
$$
is ( ).
(A) -1
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) 1 | 1. A.
When $m \neq \pm 3$, the given algebraic expression is meaningful. Then
$$
\begin{array}{l}
\frac{21-5 m}{m^{2}-9}-\frac{m}{m^{2}-9} \div \frac{m}{m+3}-\frac{m-3}{m+3} \\
=\frac{21-5 m}{m^{2}-9}-\frac{m+3}{m^{2}-9}-\frac{m-3}{m+3} \\
=\frac{21-5 m-m-3-m^{2}+6 m-9}{m^{2}-9} \\
=\frac{9-m^{2}}{m^{2}-9}=-1 .
\end{a... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,944 |
2. Given that the longest diagonal of a regular octagon is equal to $a$, and the shortest diagonal is equal to $b$. Then the area of this regular octagon is ( ).
(A) $a^{2}+b^{2}$
(B) $a^{2}-b^{2}$
(C) $a+b$
(D) $a b$ | 2. D.
As shown in Figure 3, in the regular octagon, the longest diagonal is
$$
\begin{array}{c}
A E=B F=C G \\
=D H=a,
\end{array}
$$
the shortest diagonal is
$$
\begin{array}{l}
A C=B D=C E \\
=D F=E G=F H \\
=G A=H B=b .
\end{array}
$$
By cutting and rearranging as shown in Figure 3, we get
$$
S_{\text {regular oc... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,945 |
3. Calculate $\frac{1}{6 \times 11}+\frac{1}{11 \times 16}+\frac{1}{16 \times 21}+$ $\frac{1}{21 \times 26}+\frac{1}{26 \times 31}+\frac{1}{31 \times 36}$ The value is ( ).
(A) $\frac{1}{18}$
(B) $\frac{1}{36}$
(C) $\frac{1}{33}$
(D) $\frac{1}{66}$ | 3. B.
$$
\begin{array}{l}
\frac{1}{6 \times 11}+\frac{1}{11 \times 16}+\frac{1}{16 \times 21}+ \\
\frac{1}{21 \times 26}+\frac{1}{26 \times 31}+\frac{1}{31 \times 36} \\
= \frac{1}{5}\left(\frac{1}{6}-\frac{1}{11}\right)+\frac{1}{5}\left(\frac{1}{11}-\frac{1}{16}\right)+ \\
\frac{1}{5}\left(\frac{1}{16}-\frac{1}{21}\r... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,946 |
4. Given that $n$ is a positive integer, denote $1 \times 2 \times \cdots \times n=n!$ (for example, $1!=1,4!=1 \times 2 \times 3 \times 4=24$). If $M$ $=1!\times 2!\times \cdots \times 9!$, then the number of divisors of $M$ that are perfect squares is $(\quad)$.
(A) 504
(B) 672
(C) 864
(D) 936 | 4. B.
Notice
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 9! \\
=2^{8} \times 3^{7} \times 4^{6} \times 5^{5} \times 6^{4} \times 7^{3} \times 8^{2} \times 9 \\
=2^{30} \times 3^{13} \times 5^{5} \times 7^{3} .
\end{array}
$$
Since a perfect square number $n$ has the form
$$
n=2^{2 x} \times 3^{2 y} \times... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,947 |
5. Express 2009 as the difference of squares of two integers. Then the number of different ways to do this is ( ).
(A) 16
(B) 14
(C) 12
(D) 10 | 5. C.
Let $x^{2}-y^{2}=2009$, that is,
$$
(x+y)(x-y)=2009=7^{2} \times 41 \text {. }
$$
Then 2009 has 6 positive divisors, which are $1$, $7$, $41$,
$49$, $287$, and 2009.
Therefore, the corresponding system of equations is
$$
\left\{\begin{aligned}
x+y= & -1,-7,-41,-49,-287, \\
& -2009,1,7,41,49,287,2009 ; \\
x-y= &... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,948 |
1. Calculate $\frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9$ The value equals $\qquad$ . | $\begin{array}{l} \text { II.1.3 481. } \\ \frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9 \\ = \frac{(45.1-13.9)\left(45.1^{2}+45.1 \times 13.9+13.9^{2}\right)}{45.1-13.9}+ \\ 45.1 \times 13.9 \\ = 45.1^{2}+2 \times 45.1 \times 13.9+13.9^{2} \\ =(45.1+13.9)^{2}=59^{2}=3481 .\end{array}$ | 3481 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,949 |
2. Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]$ equals $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$). | 2. 6 .
Given $0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$, then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]=0$ or 1. From the problem, we know that 18 of them are equal to 1. Therefore,
$$
\begin{array}{l}
{\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,950 |
3. As shown in Figure 1, in $\square A B C D$, $A D=a, C D=b$, draw the heights $h_{a}, h_{b}$ from point $B$ to sides $A D$ and $C D$ respectively. Given $h_{a} \geqslant a, h_{b} \geqslant b$, and the diagonal $A C=20$. Then the area of $\square A B C D$ is $\qquad$ | 3. 200 .
Given $h_{a} \geqslant a, h_{b} \geqslant b$, in addition, $a \geqslant h_{b}, b \geqslant h_{a}$, then $h_{a} \geqslant a \geqslant h_{b} \geqslant b \geqslant h_{a}$, i.e., $h_{a}=a=h_{b}=b$, which means that quadrilateral ABCD is a square.
Since the diagonal of this square $AC=20$, the area of $\square ABC... | 200 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,951 |
Example 6 As shown in Figure 6, in quadrilateral $ABCD$, $E$ and $F$ are the midpoints of $AB$ and $CD$ respectively, $P$ is any point on the extension of diagonal $AC$, $PF$ intersects $AD$ and $PE$ intersects $BC$ at points $M$ and $N$ respectively, and $EF$ intersects $MN$ at point $K$.
Prove: $K$ is the midpoint ... | Prove as shown in Figure 6, take a point $G$ on $PF$ such that $GF = FM$, and connect $GC, GN$. Then $CG \parallel DM$.
Let the midpoint of $CA$ be $L$, and connect $LE, LF$. Then $LE$ and $LF$ are the midlines of $\triangle ABC$ and $\triangle ACD$, respectively, so we have
$LF \parallel AD, LE \parallel BC, \angle D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,952 |
4. Given $\triangle A B C, \angle A, \angle B, \angle C$ have exterior angles in the ratio $\alpha: \beta: \gamma(\alpha, \beta, \gamma$ are all positive numbers $)$. Then $\angle A: \angle B: \angle C$ equals $\qquad$ (express the ratio using $\alpha, \beta, \gamma$). | 4. $(\beta+\gamma-\alpha):(\gamma+\alpha-\beta):(\alpha+\beta-\gamma)$.
Let the exterior angles of $\angle A$, $\angle B$, and $\angle C$ be $\alpha x$, $\beta x$, and $\gamma x$ respectively. From the problem, we have
$$
\begin{array}{l}
\alpha x+\beta x+\gamma x=360^{\circ} \\
\Rightarrow x=\frac{360^{\circ}}{\alph... | (\beta+\gamma-\alpha):(\gamma+\alpha-\beta):(\alpha+\beta-\gamma) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,953 |
5. When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=
$$
$\qquad$ . | 5. 2 .
Notice that
$$
\begin{array}{l}
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
= \sqrt{(x-1)+2 \sqrt{x-1}+1}+ \\
\sqrt{(x-1)-2 \sqrt{x-1}+1} \\
= \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}} \\
=|\sqrt{x-1}+1|+|\sqrt{x-1}-1| .
\end{array}
$$
Since $1 \leqslant x \leqslant 2$, we have $\sqrt{x-1}-1 \leq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,954 |
Three, (10 points) Let $a+b+c=0, a^{2}+b^{2}+c^{2}=1$.
(1) Find the value of $ab+bc+ca$;
(2) Find the value of $a^{4}+b^{4}+c^{4}$. | Three, (1) Since $a+b+c=0$, then,
$$
\begin{array}{l}
(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=0 .
\end{array}
$$
Given $a^{2}+b^{2}+c^{2}=1$, then $a b+b c+c a=-\frac{1}{2}$.
$$
\begin{array}{l}
\text { (2) } a b+b c+c a=-\frac{1}{2} \\
\Rightarrow(a b+b c+c a)^{2}=\frac{1}{4} \\
\Rightarrow a^{2} b^{2}+b^... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,955 |
Four, (15 points) As shown in Figure 2,
In the hexagon $A B C D E F$,
$$
\begin{array}{l}
A B=B C=C D \\
=D E=E F=F A,
\end{array}
$$
and
$$
\begin{array}{l}
\angle A+\angle C+\angle E \\
=\angle B+\angle D+\angle F .
\end{array}
$$
Prove:
$$
\angle A=\angle D, \angle B=\angle E, \angle C=\angle F \text {. }
$$ | As shown in Figure 4, connect $A E, E C, C A$.
Since the sum of the interior angles of a hexagon is $720^{\circ}$, and $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$, we have:
$$
\begin{aligned}
& \angle B A F + \\
& \angle B C D + \angle D E F \\
= & \angle A B C + \angle C D E + \angle E F A = 360... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,956 |
Five, (15 points) (1) Prove: A natural number composed of 2009 ones and any number of zeros is not a perfect square;
(2) Explain that there exists a natural number of the form $\underbrace{11 \cdots 1}_{2009 \uparrow} * \cdots * *$ (where * represents Arabic numerals) whose leftmost 2009 digits are all 1 and is a perfe... | (1) Any natural number can be expressed in the form $3 k+r$ $(r=0,1,2)$, and
$$
(3 k+r)^{2}=9 k^{2}+6 k r+r^{2}\left(r^{2}=0,1,4\right),
$$
which means $r^{2}$, when divided by 3, leaves a remainder of 0 or 1. This implies that the square of an integer, when divided by 3, leaves a remainder of 0 or 1, meaning that a n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,957 |
1. Let $M$ be a set obtained by removing a finite number of elements from the set of real numbers $\mathbf{R}$. Prove: for any positive integer $n$, there exists an $n$-degree polynomial $f(x)$ such that all coefficients and the $n$ real roots of $f(x)$ belong to $M$. (Feng Zhigang provided) | 1. Let $a$ be the element with the largest absolute value in the set $T=\{x \in \mathbf{R} \mid x \notin M\}$, and take a real number $k>\max \{|a|, 1\}$.
For any positive integer $n$, consider the $n$-degree polynomial $f(x)=k(x+k)^{n}$. The coefficient of its $m$-th term is
$$
k \mathrm{C}_{n}^{m} \cdot k^{n-m} \geq... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,958 |
2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided) | 2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,959 |
3. Let $H$ be the orthocenter of acute $\triangle A B C$, and $D$ be the midpoint of side $B C$. A line through point $H$ intersects sides $A B$ and $A C$ at points $F$ and $E$, respectively, such that $A E=A F$. Ray $D H$ intersects the circumcircle of $\triangle A B C$ at point $P$. Prove that points $P, A, E, F$ are... | 3. As shown in Figure 2, extend $H D$ to point $M$ such that $H D = D M$, and connect $B M, C M, B H, C H$.
Since $D$ is the midpoint of side $B C$, quadrilateral $B H C M$ is a parallelogram. Therefore,
$$
\angle B M C = \angle B H C = 180^{\circ} - \angle B A C,
$$
which means $\angle B M C + \angle B A C = 180^{\c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,960 |
4. Prove that for any given positive integer $k$, there exist infinitely many positive integers $n$ such that
$$
2^{n}+3^{n}-1, 2^{n}+3^{n}-2, \cdots, 2^{n}+3^{n}-k
$$
are all composite.
(Chen Yonggao provided) | For any given positive integer $k$, take a sufficiently large positive integer $m$ such that $2^{m}+3^{m}-k>1$.
Consider the $k$ positive integers greater than 1: $2^{m}+3^{m}-1, 2^{m}+3^{m}-2, \cdots, 2^{m}+3^{m}-k$. Take a prime factor $p_{1}, p_{2}, \cdots, p_{k}$ of each number, and let
$n_{t}=m+t \prod_{i=1}^{k}\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,961 |
5. Let the sequence $\left\{x_{n}\right\}$ satisfy $x_{1} \in\{5,7\}$ and for $k \geqslant 1$, $x_{k+1} \in\left\{5^{x_{k}}, 7^{x_{k}}\right\}$. Determine all possible values of the last two digits of $x_{2009}$.
（Liang Yingde provided） | 5. Let $n=2009$. Then
(1) If $x_{n}=7^{5^{x_{n}-2}}$, then $x_{n} \equiv 7(\bmod 100)$.
In fact, since $5^{x_{n-2}} \equiv 1(\bmod 4)$, so,
$$
\begin{array}{l}
5^{x_{n-2}}=4 k+1\left(k \in \mathbf{N}_{+}\right) \text {. } \\
\text { Hence } x_{n}=7^{5^{x_{n-2}}}=7^{4 k+1}=(50-1)^{2 k} \times 7 \\
\equiv 7(\bmod 100) \t... | 07, 25, 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,962 |
Example 7 As shown in Figure 7, in $\triangle A B C$, $A B=A C, A D \perp B C$, with the foot of the perpendicular being $D, E$ and $G$ are the midpoints of $A D$ and $A C$ respectively, $D F \perp B E$, with the foot of the perpendicular being $F$. Prove: $F G=D G$.
(2006, National Junior High School Mathematics Leagu... | Note that $D G$ is the median on the hypotenuse of the right triangle $\triangle A D C$. Therefore, to prove $F G = D G$, it is sufficient to show that $F G$ is also the median on the hypotenuse of a right triangle. Hence, we start by proving that $\triangle A F C$ is a right triangle.
As shown in Figure 7, connect $A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,963 |
6. As shown in Figure 1, let $D$ be a point on side $BC$ of acute $\triangle ABC$. The circle with diameter $BD$ intersects lines $AB$ and $AD$ at points $X$ and $P$ (distinct from points $B$ and $D$), respectively. The circle with diameter $CD$ intersects lines $AC$ and $AD$ at points $Y$ and $Q$ (distinct from points... | 6. As shown in Figure 3, connect $X Y$, $D X$, $B P$, and $D Y$.
From the given information, we have
$B, P, D, X$ and
$C, Y, Q, D$ are
respectively
concyclic. Therefore,
$$
\begin{array}{l}
\angle A X M \\
=\angle B X P \\
=\angle B D P \\
=\angle Q D C \\
=\angle A Y N .
\end{array}
$$
Thus, Rt $\triangle A M X \back... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,964 |
7. There are $n (n>12)$ people participating in a mathematics invitational competition. The test consists of fifteen fill-in-the-blank questions, with each correct answer worth 1 point and no answer or a wrong answer worth 0 points. Analyzing every possible score situation, it is found that as long as the sum of the sc... | 7. The minimum possible value of $n$ is 911.
(1) First, prove: 911 satisfies the condition.
If each student answers at least three questions correctly, since the number of different ways a student can answer three questions correctly is $\mathrm{C}_{15}^{3}=455$, then if there are 911 students participating, by the pig... | 911 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,965 |
8. Real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=0,
$$
and $2 a_{k} \leqslant a_{k-1}+a_{k+1}(k=2,3, \cdots, n-1)$.
Find the smallest $\lambda(n)$, such that for all $k \in\{1$, $2, \cdots, n\}$, we have
$$
\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \left\{\left|... | 8. $\lambda(n)_{\min }=\frac{n+1}{n-1}$.
First, take $a_{1}=1, a_{2}=-\frac{n+1}{n-1}$,
$$
a_{k}=-\frac{n+1}{n-1}+\frac{2 n(k-2)}{(n-1)(n-2)}(k=3,4, \cdots, n) \text {. }
$$
Then it satisfies $a_{1}+a_{2}+\cdots+a_{n}=0$ and
$$
2 a_{k} \leqslant a_{k-1}+a_{k+1}(k=2,3, \cdots, n-1) \text {. }
$$
At this point, $\lamb... | \frac{n+1}{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,966 |
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