problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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2. In $\triangle A B C$, it is known that $A B=b^{2}-1, B C=$ $a^{2}, C A=2 a$, where $a$ and $b$ are both integers greater than 1. Then the value of $a-b$ is $\qquad$ | 2.0.
Since $a$ is an integer greater than 1, we have $a^{2} \geqslant 2 a$. By the triangle inequality, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
a^{2}+2 a>b^{2}-1, \\
2 a+b^{2}-1>a^{2}
\end{array}\right. \\
\Leftrightarrow(a-1)^{2}<b^{2}<(a+1)^{2} .
\end{array}
$$
Given $a, b \in \mathbf{N}_{+}$, hence $b=a... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,516 |
3. If a parallelogram can be divided into 92 equilateral triangles with a side length of 1, then all possible values of the perimeter of such a parallelogram are $\qquad$ . | 3. 94 and 50.
Let the lengths of two sides of a parallelogram be $a$ and $b$, where $a, b \in \mathbf{N}_{+}$, and $a \leqslant b$. Then
$$
\begin{array}{l}
a b \sin 60^{\circ}=92 \times \frac{\sqrt{3}}{4} \\
\Rightarrow a b=46 \\
\Rightarrow(a, b)=(1,46) \text { or }(2,23) .
\end{array}
$$
Therefore, all possible va... | 94 \text{ and } 50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,517 |
4. Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ . | 4.5.
Factorizing the left side of the equation, we get
$$
\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 \text {. }
$$
Since $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1} 、 x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
$$
\begin{array}{l}
\text { Also, } x_{1}+x_{2}=-1 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,518 |
5. As shown in Figure 1, in the right triangle $\triangle ABC$, $AC=1, BC=$ $2, P$ is a moving point on the hypotenuse $AB$. Draw $PE \perp BC, PF$ $\perp AC$, with the feet of the perpendiculars being $E$ and $F$, respectively, and connect $EF$. Then the minimum length of the line segment $EF$ is $\qquad$ | 5. $\frac{2}{\sqrt{5}}$.
From the problem, we know that the hypotenuse $A B=\sqrt{5}$.
Since $E F=C P$, therefore,
$E F$ is shortest $\Leftrightarrow C P$ is shortest $\Leftrightarrow C P \perp A B$. Hence, the minimum length of $E F$ is $\frac{1 \times 2}{\sqrt{5}}=\frac{2}{\sqrt{5}}$. | \frac{2}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,519 |
6. Let $a, b$ be the roots of the equation $x^{2}+68 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-86 x+1=0$. Then
$$
(a+c)(b+c)(a-d)(b-d)
$$
the value is $\qquad$. | 6. 2772 .
$$
\begin{array}{l}
\text { Given } a b=c d=1, a+b=-68, \\
c^{2}-8 b c+1=d^{2}-86 d+1=0, \\
\text { then }(a+c)(b+c)(a-d)(b-d) \\
=\left[a b+(a+b) c+c^{2}\right]\left[a b-(a+b) d+d^{2}\right] \\
=\left(1-68 c+c^{2}\right)\left(1+68 d+d^{2}\right) \\
=18 c \cdot 154 d=2772 .
\end{array}
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,520 |
Example 6 The function $y=|x+1|+|x+2|+|x+3|$. When $x=$ $\qquad$, $y$ has its minimum value, and the minimum value is $\qquad$
(2007, National Junior High School Mathematics Competition Zhejiang Regional Finals) | It is known that $|x+1|+|x+2|+|x+3|$ represents the sum of the distances (lengths of segments) from point $x$ to points $-1$, $-2$, and $-3$ on the number line. It is easy to see that when $x=-2$, this sum of distances is minimized, i.e., the value of $y$ is the smallest, at which point, $y_{\min }=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,521 |
7. In the Cartesian coordinate system, there are two points $P(-1,1)$ $Q(2,2)$, and the graph of the linear function $y=k x-1$ intersects the extension of line segment $P Q$ (the intersection point does not include $Q$). Then the range of the real number $k$ is $\qquad$. | 7. $\frac{1}{3}2$'s solution $\Leftrightarrow\left(k-\frac{1}{3}\right) x=\frac{7}{3}$ has a solution $x>2$
$$
\begin{array}{l}
\Leftrightarrow k-\frac{1}{3}>0, \text { and }-\frac{\frac{7}{3}}{k-\frac{1}{3}}>2 \\
\Leftrightarrow \frac{1}{3}<k<\frac{3}{2} .
\end{array}
$$ | \frac{1}{3}<k<\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,522 |
8. The number of all integer solutions $(x, y, z)$ for the equation $x y z=2009$ is $\qquad$ groups. | 8. 72 .
First consider the integer solutions for $00$, it is easy to know that there are
$$
2 \mathrm{~A}_{3}^{1}+2 \mathrm{~A}_{3}^{3}=18
$$
sets of positive integer solutions.
For each set of positive integer solutions, adding two negative signs can yield 3 sets of integer solutions with two negatives and one posit... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,523 |
9. As shown in Figure 2, in quadrilateral $ABCD$, $AB=BC=CD$, $\angle ABC=78^{\circ}$, $\angle BCD=162^{\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\angle AEB$ is | $9.21^{\circ}$.
As shown in Figure 5, draw $D O / / C B$, and
make $D O=C B$, connect
$O A$ and $O B$.
Since $B C=C D$,
thus, quadrilateral $B C D O$ is a rhombus. Therefore,
$$
\begin{array}{l}
A B=B C=B O . \\
\text { Since } \angle A B O=78^{\circ}-\left(180^{\circ}-162^{\circ}\right)=60^{\circ},
\end{array}
$$
The... | 21^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,524 |
10. As shown in Figure 3, in the right trapezoid $A B C D$, $\angle A B C$ $=\angle B C D=90^{\circ}, A B=$ $B C=10$, point $M$ is on side $B C$, such that $\triangle A D M$ is an equilateral triangle. Then the sum of the areas of $\triangle A B M$ and $\triangle D C M$ is $\qquad$ (require the exact value) | 10. $150(2-\sqrt{3})$.
As shown in Figure 6, draw $A E \perp C D$, intersecting the extension of $C D$ at point $E$. Then
$$
\begin{array}{l}
A E=B C \\
=A B=10 .
\end{array}
$$
Since quadrilateral $A B C E$ is a square, and $A D=A M$, we have
$$
\begin{array}{l}
\mathrm{Rt} \triangle A E D \cong \mathrm{Rt} \triangl... | 150(2-\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,525 |
II. (15 points) As shown in Figure 4, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, point $D$ is on side $CA$ such that
$$
\begin{array}{l}
CD=1, DA=3, \text { and } \\
\angle BDC=3 \angle BAC .
\end{array}
$$
Find the length of $BC$. | As shown in Figure 7, construct the angle bisector $AE$ of $\angle ABD$, and let $AE = x$.
Then $AE = x$,
$$
\begin{array}{l}
BC^2 = x^2 - (4 - x)^2 = 8x - 16, \\
BD^2 = BC^2 + 1^2 = 8x - 15, \\
AB^2 = BC^2 + 4^2 = 8x.
\end{array}
$$
From $\frac{BD}{BA} = \frac{DE}{EA}$, we get
$$
\frac{8x - 15}{8x} = \frac{(3 - x)^2}... | \frac{4\sqrt{11}}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,526 |
Three, (15 points) Find all four-digit numbers $\overline{a b c d}$ that satisfy the following condition:
$$
\overline{a b c d}=(\overline{a b}+\overline{c d})^{2} \text {, }
$$
where the digit $c$ can be 0. | Three, let $\overline{a b}=x, \overline{c d}=y$. Then
$$
x, y \in \mathbf{Z}, \text{ and } 10 \leqslant x \leqslant 99, 0 \leqslant y \leqslant 99.
$$
From the problem, we know
$$
(x+y)^{2}=100 x+y,
$$
which simplifies to $x^{2}-2(50-y) x+y^{2}-y=0$.
Since $x, y \in \mathbf{Z}$, we have
$$
\Delta=4(2500-99 y)
$$
is ... | 3025, 2025, 9801 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,527 |
Four, (15 points) Given a positive integer $n$ that satisfies the following condition: among any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$.
| Because $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor no greater than 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17,19,23$,
$29,31,37,41,43$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers greater than 1 and not exce... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,528 |
Five. (15 points) If two different real numbers $a$ and $b$ make $a^{2}+b$ and $a+b^{2}$ both rational numbers, then the pair $(a, b)$ is called "harmonious."
(1) Find a pair of irrational numbers $a$ and $b$ such that $(a, b)$ is harmonious;
(2) Prove that if $(a, b)$ is harmonious and $a+b$ is a rational number not e... | (1) Let $a=\frac{1+\sqrt{2}}{2}, b=\frac{1-\sqrt{2}}{2}$. Then $(a, b)$ is harmonious.
【Note】The answer is not unique.
(2) According to the problem, we have
$$
\left(a^{2}+b\right)-\left(b^{2}+a\right)=(a-b)(a+b-1)
$$
is a rational number, denoted as $q$.
Since $a+b-1 \neq 0$ and is a rational number, we have
$$
a-b=\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,529 |
1. Given the vector $\overrightarrow{O P}=\left(2 \cos \left(\frac{\pi}{2}+x\right),-1\right)$,
$$
\begin{array}{l}
\overrightarrow{O Q}=\left(-\sin \left(\frac{\pi}{2}-x\right), \cos 2 x\right), \\
f(x)=\overrightarrow{O P} \cdot \overrightarrow{O Q} .
\end{array}
$$
If $a$, $b$, and $c$ are the sides opposite to the... | -1. $\frac{15}{2}$.
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\overrightarrow{O P} \cdot \overrightarrow{O Q} \\
=-2 \cos \left(\frac{\pi}{2}+x\right) \cdot \sin \left(\frac{\pi}{2}-x\right)-\cos 2 x \\
=2 \sin x \cdot \cos x-\cos 2 x=\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right) \\
\Rightarrow \sqrt{2... | \frac{15}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,530 |
2. Let $a<-1$, and the variable $x$ satisfies $x^{2}+a x \leqslant-x$, and the minimum value of $x^{2}+a x$ is $-\frac{1}{2}$. Then $a=$ $\qquad$ . | 2. $-\frac{3}{2}$.
Given $a-(a+1)$, i.e., $-2<a<-1$, then $f(x)$ reaches its minimum value at $x=-(a+1)$, which is $f(-a-1)=a+1$.
Therefore, $a+1=-\frac{1}{2}$, which means $a=-\frac{3}{2}$.
If $-\frac{a}{2} \leqslant -(a+1)$, i.e., $a \leqslant -2$, then $f(x)$ reaches its minimum value at $x=-\frac{a}{2}$, which is ... | -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,531 |
For example, the minimum value of $7 y=2 x^{2}+4|x|-1$ is
$\qquad$
(2007, National Junior High School Mathematics Competition, Zhejiang Province Re-test) | Let $t=|x| \geqslant 0$. Then
$$
y=2 t^{2}+4 t-1=2(t+1)^{2}-3(t \geqslant 0) \text{.}
$$
It is easy to see that when $t=|x|=0$, i.e., $x=0$, $y$ is minimized at -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,532 |
3. Given five different real numbers, taking any two of them to find their sum results in ten sum numbers, where the smallest two sums are sequentially $32, 36, 37$, and the largest two sums are 48 and 51. Then the largest number among these five numbers is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 3. $\frac{55}{2}$.
Let the five numbers be $a, b, c, d, e (a<b<c<d<e)$. Then
$$
\begin{array}{l}
a+b=32, a+c=36, \\
c+e=48, d+e=51 .
\end{array}
$$
Below, we show that $b+c=37$.
Since $c-b=4, d-c=3, d-b=7$, we have $a+d=(a+b)+(d-b)=39$.
Thus, $b+c=37$.
Then $2a=(a+b)+(a+c)-(b+c)=31$. Therefore, $a=15.5, b=16.5, c=20.... | \frac{55}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,533 |
4. A semicircle with diameter $A B=2$, a perpendicular line to the plane of the circle is drawn through point $A$, and a point $S$ is taken on this perpendicular line such that $A S=A B$. $C$ is a moving point on the semicircle, and $M, N$ are the projections of point $A$ on $S B, S C$ respectively. When the volume of ... | 4. $\frac{\sqrt{3}}{2}$.
It is easy to know that $B C \perp$ plane $S A C$
$\Rightarrow B C \perp A N \Rightarrow A N \perp$ plane $S B C$
$\Rightarrow A N \perp S M \Rightarrow S M \perp$ plane $A M N$.
Thus, $V_{S-A M N}=\frac{1}{3} S M \cdot S_{\triangle M N M}$.
Given $S A=A B=2$, we get $A M=S M=\sqrt{2}$.
And $A... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,534 |
5. If the odd function $y=f(x)$ defined on $\mathbf{R}$ is symmetric about the line $x=1$, and when $0<x \leqslant 1$, $f(x)=\log _{3} x$, then the sum of all real roots of the equation $f(x)=-\frac{1}{3}+f(0)$ in the interval $(0,10)$ is . $\qquad$ | 5. 30 .
Given that the graph of the function $y=f(x)$ is symmetric about the line $x=1$, and $f(x)$ is an odd function, we have
$$
f(x+2)=f(-x)=-f(x) .
$$
Therefore, $f(x+4)=-f(x+2)=f(x)$, which means $f(x)$ is a periodic function with 4 as one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, ... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,535 |
6. In the Cartesian coordinate system $x O y$, the line $l_{1}$: $4 x+5 y=20$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. The line $l_{2}$ intersects the line segment $A B$ and $O A$ at points $C$ and $D$, respectively, and bisects the area of $\triangle A O B$. Then the minimum value of $... | 6. $5 \sqrt{41}-25$.
From the conditions, we know $O A=5, O B=4, A B=\sqrt{41}$. Let $\angle B A O=\theta$. Then $\cos \theta=\frac{5}{\sqrt{41}}$. Given $S_{\triangle A O B}=2 S_{\triangle A C D}$, we get $A C \cdot A D=\frac{1}{2} A B \cdot A O=\frac{5 \sqrt{41}}{2}$.
By the cosine rule, we have
$$
\begin{array}{l}
... | 5 \sqrt{41}-25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,536 |
7. If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$
is always a non-negative real number, then the maximum value of the real number $a$ is | 7. 1 .
From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0 .\end{array}\right.$
Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=1$, we have
$$
f(x)=x^{2}-2 x-2|x-2|+4,
$$
which is $f(x)=\left\{\begin{array}{ll}x^{2}, & x \leqslant 2 ; \\ x^{2}-4 x+8, &... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,537 |
8. The number of non-empty subsets of the set $\{1,2, \cdots, 2009\}$ whose elements sum to an odd number is $\qquad$. | 8. $2^{2008}$.
Solution 1 Let
$$
f(x)=(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{2009}\right) \text {. }
$$
Then the answer required in the problem is the sum of the coefficients of the odd powers of $x$ in the expansion of $f(x)$.
Thus, the answer is
$$
\frac{f(1)-f(-1)}{2}=\frac{2^{2009}-0}{2}=2^{2008} \text {. }
$... | 2^{2008} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,538 |
9. The real solution to the equation $x^{[x]}=\frac{9}{2}$ is $\qquad$ $([x]$ denotes the greatest integer not exceeding the real number $x$). | 9. $\frac{3 \sqrt{2}}{2}$.
Clearly, $x>0$.
If $x \geqslant 3$, then $[x] \geqslant 3$. Therefore,
$$
x^{[x]} \geqslant 3^{3}=27>\frac{9}{2} \text {; }
$$
If $0<x<2$, then $0 \leqslant[x]<2$. Therefore, $x^{[x]}<2^{2}=4<\frac{9}{2}$.
Thus, $2 \leqslant x<3 \Rightarrow[x]=2$
$$
\Rightarrow x^{2}=\frac{9}{2} \Rightarrow... | \frac{3 \sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,539 |
10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ . | 10. 6181.
For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$
there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \... | 6181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,540 |
11. Given $f(x)=\frac{a x+1}{3 x-1}$, and the equation $f(x)=-4 x+8$ has two distinct positive roots, one of which is three times the other. Let the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be $S_{n}$ and $T_{n}$ respectively, and $\frac{S_{n}}{T_{n}}=f(n)(n=1,2, \cd... | (1) From $f(x)=-4 x+8$, we get $\frac{a x+1}{3 x-1}=-4 x+8$.
Rearranging gives
$$
12 x^{2}-(28-a) x+9=0 \text {. }
$$
Let $x_{1} 、 x_{2}$ be the two roots of equation (1), and $x_{2}=3 x_{1}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=4 x_{1}=\frac{28-a}{12}, \\
x_{1} x_{2}=3 x_{1}^{2}=\frac{9}{12} .
\end{array} .\r... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,541 |
12. Given a parabola $C$ with its vertex at the origin and focus at $F(2,0)$, and a point $P(m, 0)(m \neq 0)$.
(1) Let the line $l_{1}$ passing through point $P$ with a slope of 1 intersect the parabola $C$ at points $A$ and $B$. If $m<0$, and the point $P$ is symmetric to point $Q$ with respect to the origin, find the... | 12. (1) From the given conditions, the parabola $C: y^{2}=8 x$, the line $l_{1}: y=x-m$, and the point $Q(-m, 0)$.
From $\left\{\begin{array}{l}y=x-m, \\ y^{2}=8 x,\end{array}\right.$ we get
$x^{2}-2(m+4) x+m^{2}=0$.
From the discriminant $\Delta>0$ of equation (1), we get $m>-2$.
Let $A\left(x_{1}, y_{1}\right)$ and $... | \frac{32 \sqrt{3}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,542 |
Example 8 Find the minimum value of the function
$$
f(x)=\max \left\{|x+1|,\left|x^{2}-5\right| \right\}
$$
and find the value of the independent variable $x$ when $f(x)$ takes the minimum value. | Solve As shown in Figure 1, in the same Cartesian coordinate system, draw the graphs of $f_{1}(x)=|x+1|$ and $f_{2}(x)=\left|x^{2}-5\right|$. The two graphs intersect at four points, $A$, $B$, $C$, and $D$, whose x-coordinates can be obtained by solving the equation $|x+1|=\left|x^{2}-5\right|$.
Removing the absolute v... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,543 |
13. As shown in Figure $1, \odot O$ is tangent to line segment $AB$ at point $M$, and is tangent to the semicircle with $AB$ as its diameter at point $E$. $CD \perp AB$ at point $D$, $CD$ intersects the semicircle with $AB$ as its diameter at point $C$, and is tangent to $\odot O$ at point $F$. Connect $AC, CM$. Prove:... | 13. (1) As shown in Figure 2, let the midpoint of $AB$ be $P$. From the given conditions, $\odot P$ is internally tangent to $\odot O$ at point $E$, so $P, O, E$ are collinear. Connect $FO$.
Since $CD \perp AB$ and $CD$ is tangent to $\odot O$ at point $F$, we have $CD \perp OF$, $FO \parallel AP$, and $\angle EOF = \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,544 |
14. Let $x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\}(i=1,2, \cdots, 2010)$. Let $S=x_{1} x_{2}+x_{3} x_{4}+\cdots+x_{2009} x_{2010}$.
(1) Can $S$ be equal to 2010? Prove your conclusion;
(2) How many different integer values can $S$ take? | 14. (1) Since $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
\begin{array}{l}
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, \\
(\sqrt{2}-1)(\sqrt{2}+1)=1
\end{array}
$$
Therefore, $x_{2 i-1} x_{2 i} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ terms of $3+2 \sqrt{2}$, $b$ terms of $3-2 \sqrt{2}$, and $c$ terms of 1. Then ... | 503 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,545 |
15. Given positive real numbers $a, b, c$ satisfying $a+b+c \leqslant 3$, prove:
(1) $3>\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \geqslant \frac{3}{2}$;
(2) $\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)} \geqslant 2$. | 15. (1) Since $a, b, c$ are positive numbers, we have
$$
\begin{array}{l}
\frac{1}{a+1}<1, \frac{1}{b+1}<1, \frac{1}{c+1}<1 \\
\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}<3 .
\end{array}
$$
By the AM-GM inequality, we get
$$
\begin{array}{l}
\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right) . \\
{[(a+1)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,546 |
1. For a set $P$ consisting of a finite number of prime numbers, let $m(P)$ denote the maximum number of consecutive positive integers such that each of these integers is divisible by at least one element of $P$. Let $|P|$ denote the number of elements in the set $P$. Prove:
(1) $|P| \leqslant m(P)$, and the equality h... | 1. Let $11$.
(1) By the Chinese Remainder Theorem, there exists a positive integer $a$ such that $a \equiv -i \pmod{p_i} (i=1,2, \cdots, k)$, i.e., $p_i \mid (a+i)$.
Therefore, there exist $k$ consecutive positive integers $a+1, a+2, \cdots, a+k$ that satisfy the properties mentioned in the problem.
Thus, $m(P) \geqsl... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,547 |
2. For every positive integer $n$, find the largest constant $c_{n}$ with the following property: for any $n$ real-valued functions $f_{1}(x), f_{2}(x), \cdots$, $f_{n}(x)$ defined on the closed interval $[0,1]$, there exist real numbers $x_{1}, x_{2}, \cdots, x_{n}$, such that $0 \leqslant x_{i} \leqslant 1$ $(i=1,2, ... | 2. The desired maximum constant $c_{n}=\frac{n-1}{2 n}$.
On one hand, taking $x_{1}=x_{2}=\cdots=x_{n}=1$, we get
the left side of equation (1) $=\left|\sum_{i=1}^{n} f_{i}(1)-1\right|$;
taking $x_{1}=x_{2}=\cdots=x_{n}=0$, we get
the left side of equation (1) $=\left|\sum_{i=1}^{n} f_{i}(0)\right|$;
taking $x_{i}=0, ... | \frac{n-1}{2 n} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,548 |
4. Does there exist a two-variable polynomial $f\left(x_{1}, x_{2}\right)$ with integer coefficients and two points $A\left(a_{1}, a_{2}\right)$ and $B\left(b_{1}, b_{2}\right)$ on the plane that simultaneously satisfy all the following conditions?
(1) $A$ is an integer point (i.e., $a_{1}$ and $a_{2}$ are both integer... | 4. There exists a polynomial $f\left(x_{1}, x_{2}\right)$ and two points $A\left(a_{1}, a_{2}\right)$ and $B\left(b_{1}, b_{2}\right)$ on the plane.
$$
\begin{array}{l}
\text { Let } a_{1}=a_{2}=0, b_{1}=\frac{1}{3}, b_{2}=\frac{6029}{3}, \\
f\left(x_{1}, x_{2}\right)=\left(3 x_{1}-1\right)^{2}+2\left(6029 x_{1}-x_{2}\... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,550 |
5. Let $n$ be a given positive integer. If a set $K$ of integer points in the plane satisfies the following condition, it is called "connected": for any pair of points $R, S \in K$, there exists a positive integer $l$ and a sequence of points in $K$
$$
R=T_{0}, T_{1}, \cdots, T_{l}=S,
$$
where the distance between eac... | 5. The maximum possible value of $\mid \Delta(K)$ is $2 n^{2}+4 n+1$.
On one hand, take the set
$$
\begin{aligned}
K= & \{(0,0)\} \cup\{(0, i) \mid i=1,2, \cdots, n\} \cup \\
& \{(i, 0) \mid i=1,2, \cdots, n\},
\end{aligned}
$$
which is connected, and $|K|=2 n+1$. And
$$
\begin{aligned}
|\Delta(K)|= & \mid(0,0)\} \cu... | 2 n^{2}+4 n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,551 |
6. Given a polynomial $f$ with rational coefficients, its degree $d \geqslant 2$. Define the sequence of sets $f^{0}(Q), f^{1}(Q), \cdots$ as follows:
$$
f^{0}(Q)=Q, f^{n+1}(Q)=f\left(f^{n}(Q)\right)(n \geqslant 0)
$$
(for a given set $S$, $f(S)=\{f(x) \mid x \in S\}$ ).
Let $f^{\infty}(Q)=\bigcap_{n=0}^{\infty} f^{n}... | 6. Let $f(x)=\frac{1}{M}\left(a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}\right)$ $\left(M \in \mathbf{N}_{+}, a_{i} \in \mathbf{Z}, a_{d} \neq 0, d \geqslant 2\right)$.
First, we prove a lemma.
Lemma: Let $\frac{p}{q} \in \mathbf{Q}(p$ be a non-zero integer, $q$ be a positive integer, and $(p, q)=1)$. Then there exists a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,552 |
1. Real numbers $a, b$ satisfy
$$
|a+1|+|a+3|+|b+2|+|b-5|=9 \text {. }
$$
Let the maximum and minimum values of $ab + a + b$ be $m, n$ respectively. Then the value of $m+n$ is ( ).
(A) -12 (B) -10 (C) -14 (D) -8 | -1. A.
$$
\begin{array}{l}
\text { Since } 9=|a+1|+|a+3|+|b+2|+|b-5| \\
=(|a+1|+|-a-3|)+(|b+2|+|5-b|) \\
\geqslant|1-3|+|2+5|=9 .
\end{array}
$$
Thus $-3 \leqslant a \leqslant-1$, and $-2 \leqslant b \leqslant 5$
$$
\begin{array}{l}
\Rightarrow-2 \leqslant a+1 \leqslant 0, \text { and }-1 \leqslant b+1 \leqslant 6 \\
... | -12 | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,553 |
Example 9 On the blackboard, there are two positive integers, one is 2002, and the other is a number less than 2002. If the average of these two numbers $m$ is an integer, then the following operation can be performed: one of the numbers is erased and replaced by $m$. How many times can such an operation be performed a... | Without loss of generality, let these two positive integers be $a, b(a>b)$.
Consider the absolute value of the difference between the two numbers.
Initially, it is $|a-b|$. After the first operation, the absolute value of the difference between the two numbers is $\left|a-\frac{a+b}{2}\right|=\left|\frac{a-b}{2}\right|... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,554 |
2. For any natural numbers $m, n$ satisfying $m<\sqrt{11} n$, the inequality $11 n^{2} \geqslant m^{2}+k$ always holds. Then the maximum value of $k$ is ( ).
(A) 1
(B) 2
(C) 6
(D) 10 | 2. B.
Since $m0$.
Also, the remainder of $m^{2}$ divided by 11 can be $0,1,4,9,5$, 3, and $11 n^{2}$ can be divisible by 11, then the minimum value of $11 n^{2}-m^{2}$ is 2.
Since $11 n^{2}-m^{2} \geqslant k$, therefore, the maximum value of $k$ is 2. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 722,555 |
3. Given that the diagonals of $\square A B C D$ intersect at point $O$, $\angle A D C=40^{\circ}$, and $E$ is a point on side $B C$ such that $A D-A B=$ $2 B E$. Then the degree measure of $\angle B E O$ is ( ).
(A) $140^{\circ}$
(B) $150^{\circ}$
(C) $160^{\circ}$
(D) $165^{\circ}$ | 3. C.
As shown in Figure 3, take the midpoint $F$ of $BC$, and connect $OF$.
$$
\begin{array}{l}
\text { Then } BF=\frac{1}{2} BC, \\
OF=\frac{1}{2} AB, \\
OF \parallel AB .
\end{array}
$$
Therefore, $\angle OFC=\angle ABC=\angle ADC=40^{\circ}$.
Since $AD-AB=2BE$, we have,
$$
\begin{array}{l}
OF=\frac{1}{2} AB=\frac... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,556 |
4. Xiaohong calculated the sum of all consecutive five positive integers not greater than 2008, while Xiaoming calculated the sum of all consecutive six positive integers not greater than 2008. Then the number of the same sums calculated by Xiaohong and Xiaoming is ( ).
(A) 333
(B) 335
(C) 338
(D) 400 | 4. A.
The sum Xiao Hong is looking for can be expressed as
$$
1+2+3+4+5+5 m=5 m+15 \text {; }
$$
The sum Xiao Ming is looking for can be expressed as
$$
1+2+3+4+5+6+6 n=6 n+21 \text {, }
$$
where $m, n$ are natural numbers, $m \leqslant 2003, n \leqslant 2002$.
From the problem, we know
$$
\begin{array}{l}
5 m+15=6 ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,557 |
5. Given that $O$ is the circumcenter of acute $\triangle A B C$, points $D$ and $E$ are on sides $A B$ and $A C$ respectively, $\odot O_{1}$ passes through points $B$, $C$, $D$, and $E$, and the circumcircles of $\triangle A D E$, $\odot O_{2}$, and $\odot O_{1}$ both have a radius of 6. Then the distance between $O_{... | 5. D.
As shown in Figure 4, connect $B E$ and $D E$.
Since $\triangle A B C$ is an acute triangle, $\angle B A C$ and $\angle A B C$ are both acute angles.
Given that the radii of $\odot O_{1}$ and $\odot O_{2}$ are equal, and $D E$ is the common chord of the two circles, the minor arcs subtended by chord $D E$ in t... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,558 |
6. A box contains $m$ red balls, 9 white balls, and $n$ black balls, all of which are identical except for their color. If at least 17 balls must be drawn to ensure that there are 5 red balls, and at least 17 balls must be drawn to ensure that there are 8 balls of the same color. Then the value of the algebraic express... | 6. D.
From the problem, we know that $5+9+n=17$.
Solving for $n$ gives $n=3$.
If $m \geqslant 8$, then to draw eight balls of the same color, the minimum number of balls to be drawn is
$$
(8-1)+(8-1)+n+1=18>17,
$$
which is a contradiction.
Therefore, $m<8$.
Thus, $m+8+n=17 \Rightarrow m=6$.
Hence, $|m-n|+|m-5|=4$. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,559 |
1. Given real numbers $a$, $b$, $x$, $y$ satisfy $x y=2008^{2008}$,
$$
\frac{1}{1+2008^{a} x}+\frac{1}{1+2008^{b-2009} y}=1 \text {. }
$$
Then the value of $2008^{a+b}$ is | Ni, 1.2008.
Given the equation, after removing the denominator, we get
$$
\begin{array}{l}
1+2008^{b-2009} y+1+2008^{a} x \\
=\left(1+2008^{a} x\right)\left(1+2008^{b-2009} y\right) .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
2008^{a+b-2009} x y=1 \\
\Rightarrow 2008^{a+b-2009} \times 2008^{2008}=1 \\
\Ri... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,560 |
4. If the real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text {, }
$$
and $x-z=33$,
then the value of the algebraic expression $x-2 y+z$ is | 4. $\pm 31$.
Let $a=\sqrt[5]{x-y}, b=\sqrt[5]{y-z}$. Then
$$
\begin{array}{l}
a+b=3, \\
a^{5}+b^{5}=33 .
\end{array}
$$
(2) $\div$ (1) gives
$$
\begin{array}{l}
a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}=11 \\
\Rightarrow\left(a^{2}+b^{2}\right)^{2}-a b\left(a^{2}+b^{2}\right)-a^{2} b^{2}=11 \\
\Rightarrow\left[(a+b)^{2}... | \pm 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,563 |
One, (20 points) A TV station, to understand the viewing situation of its two characteristic programs $A, B, C$, surveyed 28 viewers and found: each viewer watched at least one of the programs; among the viewers who did not watch program $A$, the number of viewers who watched program $B$ was twice the number of viewers... | Let the number of viewers who only watch programs $A, B, C$ be $x, y, z$ respectively, the number of viewers who do not watch program $A$ but watch programs $B$ and $C$ be $m$, and the number of viewers who do not only watch program $A$ be $n$, as shown in Figure 7.
From the problem, we have
$$
\left\{\begin{array}{l}
... | 53.6\% | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,564 |
Example 10 Let $x_{1}, x_{2}, \cdots, x_{n}$ be any permutation of $1,2, \cdots, n$. Find
$$
u=\left|x_{1}-1\right|+\left|x_{2}-2\right|+\cdots+\left|x_{n}-n\right|
$$
the maximum value. | (1) If $n=2 k$ (where $k$ is a positive integer), then
$$
\begin{array}{l}
u_{\text {max }}=[n+n+\cdots+(k+1)+(k+1)]- \\
\quad(k+k+\cdots+1+1) \\
=2(1+2+\cdots+n)-4(1+2+\cdots+k) \\
=n(n+1)-2 k(k+1)=\frac{1}{2} n^{2} .
\end{array}
$$
(2) If $n=2 k+1$ (where $k$ is a positive integer), then
$$
\begin{array}{l}
u_{\text ... | \frac{1}{2} n^{2} \text{ for } n=2k, \frac{1}{2}\left(n^{2}-1\right) \text{ for } n=2k+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,565 |
II. (25 points) As shown in Figure 2, the large circle $\odot O_{1}$ intersects with the small circle $\odot O_{2}$ at points $M$ and $N$. Two chords $MA$ and $MB$ of $\odot O_{1}$ intersect $\odot O_{2}$ at points $C$ and $D$, respectively. Extend $AB$ to point $E$ and $CD$ to point $F$ such that $BE = AB$ and $DF = C... | (1) As shown in Figure 8, connect $A N$, $C N$, and $M N$.
$$
\begin{array}{c}
\text { Given } \angle M C N \\
=\angle M D N, \\
\text { then } \angle A C N \\
=\angle B D N . \\
\text { Also, } \angle C A N \\
=\angle D B N,
\end{array}
$$
Thus, $\triangle A C N \backsim \triangle B D N$
$$
\Rightarrow \frac{A N}{C N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,566 |
Three, (25 points) Find all positive integer triples $(a, b, c)$ that satisfy $a^{2}+2 b^{2}+3 c^{2}=2008$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, because $a^{2}+2 b^{2}+3 c^{2}=2008$, where $2 b^{2}$ and $2008$ are both even numbers, so $a^{2}+3 c^{2}$ is even, and more specifically, $a^{2}+c^{2}$ is even.
Therefore, $a$ and $c$ have the same parity.
If $a$ and $c$ are both odd, let $a=2 m-1, c=2 n-1$ (where $m, n$ are positive integers). Then
$$
\begin{a... | (14,30,2),(10,30,6),(22,6,22),(4,30,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,567 |
1. Function
$$
f(x)=27^{x}-3^{x+3}+1
$$
The minimum value of the function on the interval $[0,3]$ is $\qquad$ | - 1. -53.
Let $t=3^{x}(x \in[0,3])$. Then $f(x)=g(t)=t^{3}-27 t+1(t \in[0,27])$. And $g^{\prime}(t)=3 t^{2}-27=3(t-3)(t+3)$, so when $t \in[1,3]$, $g^{\prime}(t)0, g(t)$ is monotonically increasing.
Therefore, when $t=3$, $g(t)$ reaches its minimum value
$$
g(t)_{\min }=g(3)=-53,
$$
which means when $x=1$, $f(x)$ tak... | -53 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,568 |
2. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right] \text {, }
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
a_{2009}+a_{2010}=
$$ | 2. 2009.
Given $a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}$.
We will prove by mathematical induction:
$$
a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n \text {. }
$$
Obviously, when $n=1$, the conclusion holds.
Assume when $n=k$, the conclusion holds, i.e.,
$$
a_{k+2}-a_{k}=1, a_{k}+a_{k+1}=k \text {. }
$$
Then when ... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,569 |
3. If the set $A=\{x \mid x=6 n-1, n \in \mathbf{N}\}$,
$$
B=\{x \mid x=8 n+3, n \in \mathbf{N}\},
$$
then the number of elements in $A \cap B$ that are less than 2010 is $\qquad$ | 3. 84 .
According to the problem, if $x \in A$, then $x \equiv 5(\bmod 6)$; if $x \in B$, then $x \equiv 3(\bmod 8)$. Therefore, if $x \in A \cap B$, then $x \equiv 11(\bmod 24)$, i.e., $x=24 k+11(k \in \mathbf{N})$.
Thus, there are 84 elements that satisfy the condition. | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,570 |
4. If the equation
$$
n \sin x+(n+1) \cos x=n+2
$$
has two distinct real roots in $0<x<\pi$, then the minimum value of the positive integer $n$ is $\qquad$. | 4.4.
From the given, we have $\frac{1}{n}=-1-\frac{-1-\sin x}{2-\cos x}$.
And $\frac{-1-\sin x}{2-\cos x}$ represents the slope $k$ of the line connecting a moving point $P(\cos x, \sin x)$ on the upper half of the unit circle (excluding endpoints) and a fixed point $Q(2,-1)$.
To satisfy the problem, the line $PQ$ mu... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,571 |
5. If $a>b>c, a+b+c=0$, and $x_{1}, x_{2}$ are the two real roots of $a x^{2}+b x+c=0$. Then the range of $\left|x_{1}^{2}-x_{2}^{2}\right|$ is $\qquad$ | 5. $[0,3)$.
From $a+b+c=0$, we know that the equation $a x^{2}+b x+c=0$ has a real root of 1. Without loss of generality, let $x_{1}=1$.
Then, by Vieta's formulas, we know $x_{2}=\frac{c}{a}$.
Given $a>b>c, a+b+c=0$, it follows that $a>0, c-a-c>c$.
Thus, $-2<\frac{c}{a}<-\frac{1}{2}$.
Therefore, $\frac{1}{4}<x_{2}^{2}... | [0,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,572 |
6. In the tetrahedron $O-ABC$, if the three edges meeting at point $O$ are pairwise perpendicular and have a length of $\sqrt{3}$, then the total length of the curve formed by the points on the surface of the tetrahedron that are a distance of 2 from point $A$ is $\qquad$ . | 6. $\frac{3 \pi}{2}$.
As shown in Figure 2, points $M$ and $N$ are on edges $AB$ and $AC$, respectively, and $AM = AN = 2$; points $E$ and $F$ are on edges $OB$ and $OC$, respectively, and $OE = OF = 1$.
Thus, $AE = AF = 2$.
Therefore, the curves formed by the points that meet the conditions are:
(1) In the plane $OBC... | \frac{3 \pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,573 |
7. There are $n$ ellipses centered at the origin with the coordinate axes as axes of symmetry, and the directrices of all ellipses are $x=1$. If the eccentricity of the $k$-th $(k=1,2, \cdots, n)$ ellipse is $e_{k}=2^{-k}$, then the sum of the major axes of these $n$ ellipses is $\qquad$ | $7.2-2^{1-n}$.
Let the semi-major axis of the $k$-th ellipse be $a_{k}$, and the focal distance be $c_{k}$. Then, by the problem statement, we have
$$
\frac{a_{k}^{2}}{c_{k}}=1, e_{k}=\frac{c_{k}}{a_{k}}=2^{-k} \text {. }
$$
Thus, $a_{k}=2^{-k}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{n}$
$$
=2^{-1}+2^{-2}+\cdots+2^{-n}=1-... | 2-2^{1-n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,574 |
8. A school is holding a basketball shooting contest, with a total of 64 participants. It is known that each participant has a shooting accuracy of $\frac{3}{4}$ per shot. The rules are: only those who hit two consecutive shots will be admitted, and once admitted, they stop shooting; otherwise, they continue shooting u... | 8. $\frac{189}{4}$.
The probability of each participant being admitted is
$$
\begin{aligned}
p & =\frac{3}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} \\
& =\frac{189}{256} .
\end{aligned}
$$
Therefore, the number of admi... | \frac{189}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,575 |
3. A certain school has 50 students, with an equal number of boys and girls. They sit in a circle for a campfire party. Prove that there must be a student who has two girls sitting next to him. | Prompt: Alternate 50 seats with black and white colors. Assume that no matter how they are seated, it is impossible to find a student who has girls on both sides. Then, among the 25 black-marked seats, there can be at most 12 girls, otherwise, there will be at least two girls with a student in between. Similarly, the 2... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,576 |
3. The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2009|x|+2010$ intersects the x-axis is $\qquad$ . | 3. 0 .
The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2009|x|+2010=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,577 |
4. In a convex quadrilateral $ABCD$, $P$ and $Q$ are the centroids of $\triangle ABC$ and $\triangle ADC$, respectively, and $PQ \perp AC$. If the lengths of three of the four sides are $5$, $6$, and $7$, then the length of the fourth side is . $\qquad$ | 4. $2 \sqrt{3}$ or $2 \sqrt{15}$ or $\sqrt{38}$.
As shown in Figure 5, take the midpoint $M$ of $A C$. Then $B, P, M$ and $D, Q, M$ are collinear respectively.
Since $\frac{B P}{P M}=2$
$$
=\frac{D Q}{Q M},
$$
therefore, $P Q \parallel B D$.
Also, $P Q \perp A C$, so $B D \perp A C$.
Thus, $A B^{2}+C D^{2}=B C^{2}+D ... | 2 \sqrt{3} \text{ or } 2 \sqrt{15} \text{ or } \sqrt{38} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,578 |
One, (20 points) Let
$$
a>b>0, a+b \geqslant \frac{5}{9}, a b \leqslant \frac{4}{81} \text {. }
$$
Prove: $a \geqslant \frac{4}{9}$. | $$
\begin{array}{l}
\text { I. It is easy to see that }\left(a-\frac{4}{9}\right)\left(b-\frac{4}{9}\right) \\
=a b-\frac{4}{9}(a+b)+\frac{16}{81} \\
\leqslant \frac{4}{81}-\frac{4}{9} \times \frac{5}{9}+\frac{16}{81}=0 .
\end{array}
$$
Since $a-\frac{4}{9}>b-\frac{4}{9}$, therefore, $a-\frac{4}{9} \geqslant 0$, which... | a \geqslant \frac{4}{9} | Inequalities | proof | Yes | Yes | cn_contest | false | 722,579 |
As shown in Figure 1, the equilateral $\triangle ABC$ is inscribed in a circle, $M$ is the midpoint of side $AB$, and point $P$ lies on the arc $\overparen{AB}$, such that $PM \parallel BC$. Prove:
$$
\frac{PA + PB}{PA - PB} = \frac{PA^3}{PB^3}
$$ | As shown in Figure 6, extend $PM$, intersecting $AC$ and the circumcircle of $\triangle ABC$ at points $N$ and $Q$ respectively. Connect $PC$, $AQ$, and $CQ$. It is easy to see that $N$ is the midpoint of side $AC$.
$$
\begin{array}{l}
\text{Given } PM \parallel BC \\
\Rightarrow \overparen{CQ} = \overparen{BP}, \\
\o... | \frac{PA + PB}{PA - PB} = \frac{PA^3}{PB^3} | Geometry | proof | Yes | Yes | cn_contest | false | 722,580 |
Three. (25 points) Given the function
$$
y=(a+2) x^{2}-2\left(a^{2}-1\right) x+1 \text {, }
$$
where the independent variable $x$ is a positive integer, and $a$ is also a positive integer. Question: For what value of $x$ is the function value minimized? | Three, the function is organized as
$$
y=(a+2)\left(x-\frac{a^{2}-1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2} \text {. }
$$
Then its axis of symmetry is
$$
x=\frac{a^{2}-1}{a+2}=(a-2)+\frac{3}{a+2} \text {. }
$$
Since $a$ is a positive integer, we have
$$
\begin{array}{l}
0<1$ when, $a-2<0$, i.e., $a=1$... | x=\left\{\begin{array}{ll}
1, & a=1 ; \\
a-1, & a=2 \text { or } 3 ; \\
2 \text { or } 3, & a=4 ; \\
a-2, & a>4 \text { and } a \text { is an integer }
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,581 |
1. Given a positive integer $n$ such that the last two digits of $3^{n}$ form a two-digit prime number. Then the sum of all $n$ that satisfy this condition and do not exceed 2010 is $\qquad$ . | $-, 1.909128$.
Considering $3^{k}(k=1,2, \cdots)$ modulo 100, the remainders are sequentially $3,9,27,81,43,29,87,61,83,49,47,41$, $23,69,7,21,63,89,67,1 ; 3,9, \cdots$,
where $43,29,61,83,47,41,23,89,67$ are two-digit primes, corresponding to
$$
k \equiv 5,6,8,9,11,12,13,18,19(\bmod 20) .
$$
Thus, the sum of all $n$ ... | 909128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,582 |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
[\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}]
$$
is equal to
(there are a total of 2009 square roots). | 2. 45 .
Let $A=\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}$.
Then $A>\sqrt{2010+\sqrt{2009}}$
$$
\begin{array}{l}
>\sqrt{2010+44}>45, \\
A<\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{2009}}}} \\
<\sqrt{2010+46}<46 .
\end{array}
$$
Thus, $[A]=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,583 |
3. For a positive integer $k$, the equation
$$
\left(a^{2}-k\right)\left(b^{2}-k\right)=c^{2}-k
$$
has $\qquad$ integer solution sets $(a, b, c)$. | 3. Infinitely many.
Let $a=b+1, c=ab-k$. Then
$$
\begin{array}{l}
c^{2}-k=a^{2} b^{2}-2 k a b+k^{2}-k, \\
\left(a^{2}-k\right)\left(b^{2}-k\right)=a^{2} b^{2}-k\left(a^{2}+b^{2}\right)+k^{2} .
\end{array}
$$
Since $a^{2}+b^{2}=(b+1)^{2}+b^{2}$
$$
=2 b(b+1)+1=2 a b+1 \text {, }
$$
Therefore, $\left(a^{2}-k\right)\lef... | Infinitely many | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,584 |
4. Given
$$
A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n}(n \in \mathbf{N})
$$
can be expressed as the product of $k(k \in \mathbf{N}, k>1)$ consecutive integers. Then $n+k=$ $\qquad$ . | 4. 2 .
If $k \geqslant 4$, then $81 A$.
$$
\begin{array}{l}
\text { But } A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n} \\
\equiv 1+4-3^{7 n} \equiv 5-3^{n} \not \equiv 0(\bmod 8),
\end{array}
$$
Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,585 |
5. Let
$$
\begin{array}{l}
f(x)=\left(x+\frac{1-\sqrt{3} \mathrm{i}}{2}\right)^{2010} \\
=\sum_{k=0}^{2010} a_{k} x^{k}+\mathrm{i} \sum_{k=0}^{2010} b_{k} x^{k},
\end{array}
$$
where, $a_{k}, b_{k} \in \mathbf{R}, k=0,1, \cdots, 2010$. Then
$$
\sum_{k=0}^{670}\left(a_{3 k}+b_{3 k}\right)=
$$
$\qquad$ | 5. $-2 \times 3^{1004}$.
Notice that
$$
\begin{array}{l}
f(x)=\left(x+\mathrm{e}^{-\frac{\pi}{3} \mathrm{i}}\right)^{2010}=\left(1+\mathrm{e}^{\frac{\pi}{3} \mathrm{i}} x\right)^{2010} \\
\sum_{k=0}^{2010} \mathrm{C}_{2010}^{k} x^{k} \cdot \mathrm{e}^{-\frac{\pi}{3} \mathrm{i}},
\end{array}
$$
So $\sum_{k=0}^{670} b_... | -2 \times 3^{1004} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,586 |
6. A photographer took some photos of eight people at a party, with any two people (there are 28 possible combinations) appearing in exactly one photo. Each photo can be a duo or a trio. How many photos did the photographer take at least?
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。 | 6. 12 .
Let the number of group photos with three people be $x$, and the number of group photos with two people be $y$. Then $3 x+y=28$.
Thus, when $x$ is maximized, the total number of photos taken is minimized.
When $x>8$, by $\frac{3 x}{8}>3$ we know that there is a person $A$ who appears 4 times, and in the 4 grou... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,587 |
7. A small sphere with radius $r$ and a large sphere with radius $R$ are freely moving in all directions inside a regular tetrahedron container with internal edge length $l$. If
$$
r=1, R=2, l=\frac{14 \sqrt{6}}{3},
$$
then the area of the container's inner wall that the small sphere can never touch is . $\qquad$ | 7. $88 \sqrt{3}+\frac{128 \pi}{9}$.
Let the edge length of a regular tetrahedron be $l_{0}$ and the radius of its inscribed sphere be $r_{0}$. It is known that $l_{0}=2 \sqrt{6} r_{0}$, and the distance from the points where the inscribed sphere touches the faces to the edges is $\sqrt{2} r_{0}$.
First, consider the ... | 88 \sqrt{3}+\frac{128 \pi}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,588 |
9. (16 points) Given a moving point $P$ on the parabola $y^{2}=4 x$, and the focus $F(1,0)$. Find the maximum value of the inradius $r$ of $\triangle O P F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | By symmetry, without loss of generality, let
$$
P\left(\frac{y^{2}}{4}, y\right)(y>0) \text {. }
$$
It is easy to see that $S_{\triangle O P F}=\frac{y}{2}$, and
$$
O F=1, P F=\frac{y^{2}}{4}+1, O P=\frac{y \sqrt{y^{2}+16}}{4} \text {. }
$$
Thus, $\frac{1}{r}=\frac{O F+P F+O P}{2 S_{\triangle O P F}}$
$$
=\frac{2}{y}... | \frac{2 \sqrt{3}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,590 |
10. (20 points) Given
$$
A=\{1,2, \cdots, 2014\}
$$
is a non-empty subset, satisfying that the sum of its elements is a multiple of 5. Find the number of such subsets. | 10. Let $A_{j}=\{x \in A \mid x \equiv j(\bmod 5)\}(j=0, 1, \cdots, 4)$. Then
$$
\left|A_{0}\right|=402,\left|A_{j}\right|=403(j=1,2,3,4).
$$
Let $T$ be a non-empty subset that satisfies the condition, and
$$
T \cap A_{j}=T_{j},\left|T_{j}\right|=t_{j}(j=0,1, \cdots, 4).
$$
Then $\sum_{j=0}^{4} j t_{j} \equiv 0(\bmod... | \frac{1}{5}\left(2^{2014}+2^{404}\right)-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,591 |
11. (20 points) In the Cartesian coordinate system, there are three frogs $A$, $B$, and $C$, with their initial positions being
$$
\begin{array}{l}
A_{0}(4,6), B_{0}(-\sqrt{5}-2,-\sqrt{5}-3), \\
C_{0}(6+\sqrt{5}, 9+\sqrt{5}).
\end{array}
$$
First, $A$ jumps to its symmetric point with respect to $B$, then $B$ jumps to... | 11. Let the three sides of $\triangle A_{n} B_{n} C_{n}$ be $a_{n}, b_{n}, c_{n}$. Then, from the problem, we have
$$
\left\{\begin{array}{l}
A_{n+1}+A_{n}=2 B_{n}, \\
B_{n+1}+B_{n}=2 C_{n}, \\
C_{n+1}+C_{n}=2 A_{n+1}.
\end{array}\right.
$$
From equation (1), we get
$$
B_{n}=\frac{1}{2}\left(A_{n+1}+A_{n}\right).
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,592 |
I. (40 points) Let $P$ be any point on the circumcircle $\odot O$ of acute $\triangle ABC$, and let the Simson line of $P$ with respect to $\triangle ABC$ be $l_{P}$. Let $P'$ be the antipode of $P$ on $\odot O$, and let the intersection of $l_{P}$ and $l_{P'}$ be $P''$. Prove: For any two points $P, Q$ on $\odot O$, $... | Lemma 1 In $\triangle ABC$, any two Simson lines $l_{P_{1}}$ and $l_{P_{2}}$ are not parallel.
Proof Otherwise, suppose $l_{P_{1}} \parallel l_{P_{2}}$ intersect the lines $AB$ and $AC$ at points $C_{1}, C_{2}, B_{1}, B_{2}$ respectively.
From $\triangle AB_{1}C_{1}$ and $\triangle AB_{2}C_{2}$ being similar, their c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,593 |
II. (40 points) As shown in Figure 2, there are three types of paper pieces (which can be flipped).
Prove: (1) When $n \equiv 2(\bmod 4)$, an $n \times n$ board cannot be divided into several I-type and II-type paper pieces;
(2) When $n$ is an even number greater than 2, an $n \times n$ board can be divided into sever... | (1) Prove by contradiction.
Assume that an $n \times n$ cardboard can be divided into $u$ I-type and $v$ II-type pieces.
Label the squares of the $n \times n$ cardboard from the top left to the bottom right as shown in Figure 6.
Next, denote the cells filled with 1, 2, 3, 4 as 1, 2, 3, 4, respectively. Each I-type pi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,594 |
Three, (50 points) Given $m \in \mathbf{N}_{+}$, let $f(a)$ denote the sum of the digits of the positive integer $a$. Try to find the minimum value of the positive integer $t$, such that among any $t$ consecutive positive integers, there is always a number $c$ satisfying $m \mid f(c)$.
---
Note: The translation prese... | Let $k=\left[\frac{m-1}{9}\right], r=m-9 k$. Then,
when $10 \times(a+1)$, $f(a+1)=f(a)+1$;
when $10^{i} \|(a+1)\left(i \in \mathbf{N}_{+}\right)$,
$$
f(a+1)=f(a)+1-9 i=f\left(a+1-10^{i}\right)+1 \text {. }
$$
First, we prove: In any $t=2 r \cdot 10^{k}-1$ consecutive positive integers, there is always a number $c$ suc... | t_{\min } \leqslant 2 r \cdot 10^{k}-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,595 |
Four. (50 points) Select 1005 numbers $a_{1}, a_{2}, \cdots, a_{1005}$ from 1 to 2010 such that their total sum is 1006779, and the sum of any two of these 1005 numbers is not equal to 2011.
(1) Prove that $\sum_{i=1}^{1005} a_{i}^{2}-4022 \sum_{i=1}^{1005} a_{i}^{3}+\sum_{i=1}^{1005} a_{i}^{4}$ is a constant;
(2) When... | Divide $\{1,2, \cdots, 2010\}$ into 1005 groups:
$$
A_{i}=\{i, 2011-i\}(i=1,2, \cdots, 1005) .
$$
Since the sum of any two numbers in $\left\{a_{i}\right\}$ is not equal to 2011, exactly one number is taken from each group. First, take the even numbers from each group to form
$\left\{b_{i}\right\}$, where $b_{i} \in A... | 44253 | Algebra | proof | Yes | Yes | cn_contest | false | 722,596 |
Initial 275 Given $\qquad$
$$
\begin{array}{l}
23 \times \overline{a b c}=4 \times \overline{p q r}, 16 \times \overline{r p q}=5 \times \overline{x y z}, \\
22 \times \overline{z x y}=7 \times c a b,
\end{array}
$$
where $\overline{a b c} 、 \overline{p r} 、 \overline{x y z}$ represent three-digit numbers, and the sam... | Let the three equations in the problem be denoted as (1), (2), and (3).
From equation (1), it is easy to see that $a=1$;
From equation (2), it is easy to see that $1 \leqslant r \leqslant 3$;
From equation (3), it is easy to see that $1 \leqslant z \leqslant 3$.
Furthermore, from equation (1), we know $2 \mid c$, from ... | \begin{array}{l}
23 \times 148=4 \times 851, \\
16 \times 185=5 \times 592, \\
22 \times 259=7 \times 814
\end{array} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,597 |
Given circles $\odot O_{1}$ and $\odot O_{2}$ with radii $r$ and $R$ respectively, they are externally tangent at point $P$. A square $ABCD$ is inscribed in $\odot O_{1}$, and point $P$ lies on the minor arc $\overparen{AB}$. Tangents $AA_{1}$, $BB_{1}$, $CC_{1}$, and $DD_{1}$ are drawn from points $A$, $B$, $C$, and $... | Connect $A P$ and extend it to intersect $\odot \mathrm{O}_{2}$ at point $G$, connect $O_{1} O_{2}$ and extend it in both directions to intersect $\odot O_{1}$ and $\odot O_{2}$ at points $E$ and $F$ respectively. Then point $P$ is on the line segment $E F$, connect $A E$ and $G F$.
Since $E P$ and $F P$ are diameters ... | \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,598 |
Given positive numbers $a, b$ satisfying $a+b<1$. Prove:
$$
\frac{(a-1)^{2}+b(2 a-b)}{(b-1)^{2}+a(2 b-a)} \geqslant \min \left\{\frac{b}{a}, \frac{a}{b}\right\} .
$$ | Proof Given that
$$
\begin{array}{l}
(b-1)^{2}+a(2 b-a) \\
=b^{2}+2(a-1) b-a^{2}+1 \\
=(b+a-1)^{2}+2 a(1-a)>0 .
\end{array}
$$
(1) When $a \geqslant b>0$, $\min \left\{\frac{b}{a}, \frac{a}{b}\right\}=\frac{b}{a}$.
Thus, it suffices to prove
$$
\begin{array}{l}
\frac{(a-1)^{2}+b(2 a-b)}{(b-1)^{2}+a(2 b-a)} \geqslant \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,599 |
As shown in Figure 3, in the convex quadrilateral $ABCD$, the opposite sides $AB$ and $DC$ are not parallel. The diagonals $AC$ and $BD$ intersect at point $P$, $PE \perp AB$ at point $E$, $PF \perp DC$ at point $F$, and points $E$ and $F$ lie on sides $AB$ and $DC$ respectively. Points $M$ and $N$ are the midpoints of... | Proof First, we prove a lemma.
Lemma As shown in Figure 3, in a convex quadrilateral $ABCD$, the opposite sides $AB$ and $DC$ are not parallel. Points $E$ and $F$ are on sides $AB$ and $DC$ respectively, and $M$ and $N$ are the midpoints of sides $BC$ and $AD$ respectively. Then the line $MN$ bisects $EF$ if and only i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,600 |
Example 1 Let $n$ be a positive integer. Prove: if the sum of all positive divisors of $n$ is a power of 2, then the number of these positive divisors is also a power of 2. ${ }^{[1]}$
(2009, Middle European Mathematical Olympiad) | Proof: Let $n=p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{k}^{L_{k}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are distinct primes, and $s_{i} \in \mathbf{N}_{+}(i=1,2, \cdots, k)$. Then the sum of all positive divisors of $n$ can be expressed as
$$
\begin{array}{l}
\left(1+p_{1}+\cdots+p_{1}^{s_{1}}\right)\left(1+p_{2}+\cdots+p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,601 |
Example 2 Suppose a positive integer satisfies the following property: the sum of all its positive divisors that are not congruent to 2 modulo 4 is equal to 1000. Find all positive integers that satisfy the above property. ${ }^{\text {[2] }}$
$(2008$, Japan Mathematical Olympiad) | For a positive integer $n$, let $S(n)$ be the sum of all positive divisors of $n$ that are not congruent to 2 modulo 4. Suppose the prime factorization of $n$ is
$$
2^{m} p_{1}^{m_{1}} p_{2}^{m_{2}} \cdots p_{k}^{m_{k}}\left(m, m_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, k\right) .
$$
Since an integer is congruent to 2 m... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,602 |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its divisors 1, $2,17,34,59,118,1003,2006$ is 3240. Find the smallest good number. ${ }^{[3]}$
(... | \left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)=8.
Therefore, when $k=1$, $\alpha_{1}=7$;
when $k=2$, $\alpha_{1}=1, \alpha_{2}=3$ or $\alpha_{1}=3, \alpha_{2}=1$;
when $k=3$, $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$;
when $k \geqslant 4$, there is no solution.
(1) If $n=p^{7}$ ( $p$ is a... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,603 |
Example 5 Find all positive integers $n$, such that $n$ is composite, and all its divisors greater than 1 can be placed on a circle so that any two adjacent divisors are not coprime. ${ }^{[6]}$
(2005, USA Mathematical Olympiad) | If $n=p q$ (where $p$ and $q$ are different primes), then its factors greater than 1, $p$, $q$, and $pq$, will always be adjacent when placed on a circle, which does not meet the requirement.
If $n=p^{m}$ (where $p$ is a prime and $\left.m \geqslant 2\right)$, then no matter how the factors of $n$ greater than 1 are p... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,605 |
Example 6 (1) If a positive integer $k(k \geqslant 3)$ satisfies: there are $k$ positive integers such that any two are not coprime, and any three are coprime, find all possible values of $k$;
(2) Does there exist an infinite set of positive integers that satisfies the conditions in (1)? ? ${ }^{[7]}$
(2003, Belarusian... | (1) Since there are infinitely many prime numbers, denoted as $p_{1}$, $p_{2}, \cdots$. To satisfy the condition that any two are not coprime and any three are coprime, we can construct $a_{1}, a_{2}, \cdots, a_{k}$ as follows:
$$
\begin{array}{l}
a_{1}=p_{1} p_{2} \cdots p_{k-1}, \\
a_{2}=p_{1} p_{k} p_{k+1} \cdots p_... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,606 |
1. Given that $N$ is a positive integer, and there are exactly 2005 ordered pairs of positive integers $(x, y)$ satisfying $\frac{1}{x}+\frac{1}{y}=\frac{1}{N}$. Prove: $N$ is a perfect square. ${ }^{[8]}$
(2005, British Mathematical Olympiad) | Note that $x, y > N$. Otherwise, $\frac{1}{x}$ or $\frac{1}{y}$ will be greater than $\frac{1}{N}$. Then
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{N} \Rightarrow y=\frac{N^{2}}{x-N}+N \text{. }
$$
Thus, each positive divisor $d$ of $N^{2}$ uniquely corresponds to a solution
$$
\begin{array}{l}
(x, y)=\left(d+N, \frac{N^{2}}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,607 |
2. Find a natural number $N$, such that it is divisible by 5 and 49, and including 1 and $N$, it has a total of 10 divisors. | Let $N=2^{a_{1}} 3^{a_{2}} 5^{a_{3}} 7^{a_{4}} \cdots p_{n}^{a_{n}}\left(a_{i} \in \mathbf{N}\right.$, $i=1,2, \cdots, n)$. Then
$$
\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{n}+1\right)=10 \text {. }
$$
Since $5\left|N, 7^{2}\right| N$, then $a_{3}+1 \geqslant 2, a_{4}+1 \geqslant 3$. Therefore, $a_{1}, ... | 5 \times 7^{4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,608 |
4. Let $b, n$ be integers greater than 1. If for every positive integer $k$ greater than 1, there exists an integer $a_{k}$ such that $k!\left(b-a_{k}^{n}\right)$, prove: there exists an integer $A$ such that
$$
b=A^{n} \text {. }{ }^{\text {9] }}
$$
(48th IMO Shortlist Problem) | Let the prime factorization of $b$ be
$$
b=p_{1}^{a_{1}} p_{2}^{\alpha_{2}} \cdots p_{s}^{\alpha_{1}},
$$
where $p_{1}, p_{2}, \cdots, p_{s}$ are distinct primes, and $\boldsymbol{\alpha}_{i} \in$ $\mathbf{N}_{+}(i=1,2, \cdots, s)$.
We will prove that all exponents $\alpha_{1}$ are divisible by $n$.
Let $A=p_{1}^{\fra... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,610 |
Example 1 In the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=2, \\
a_{n+1}=(\sqrt{2}-1)\left(a_{n}+2\right)(n=1,2, \cdots) .
\end{array}
$$
Find the general term formula of $\left\{a_{n}\right\}$.
(2007, National College Entrance Examination I Paper (Science)) | From the given, we know
$$
a_{n+1}=(\sqrt{2}-1) a_{n}+2(\sqrt{2}-1),
$$
which means $a_{n+1}-\sqrt{2}=(\sqrt{2}-1)\left(a_{n}-\sqrt{2}\right)$.
Therefore, the sequence $\left\{a_{n}-\sqrt{2}\right\}$ is a geometric sequence with the first term $2-\sqrt{2}$ and the common ratio $\sqrt{2}-1$.
Thus, $a_{n}-\sqrt{2}=\sqr... | a_{n}=\sqrt{2}\left[(\sqrt{2}-1)^{n}+1\right](n=1,2, \cdots) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,611 |
Example 2 In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n+1}=4 a_{n}-3 n+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Find the general term formula of the sequence $\left\{a_{n}\right\}$. | Given $a_{n+1}=4 a_{n}-3 n+1$, let
$$
\begin{array}{l}
a_{n+1}-[a(n+1)+b] \\
=4\left[a_{n}-(a n+b)\right] .
\end{array}
$$
Then $a(n+1)+b-4(a n+b)=-3 n+1$.
We get $a=1, b=0$.
Thus, $a_{n+1}-(n+1)=4\left(a_{n}-n\right)$.
Also, $a_{1}-1=1 \neq 0$, so the sequence $\left\{a_{n}-n\right\}$ is a geometric sequence with the... | a_{n}=4^{n-1}+n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,612 |
Example 3 In the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=2, \\
a_{n+1}=\lambda a_{n}+\lambda^{n+1}+(2-\lambda) 2^{n}\left(n \in \mathbf{N}_{+}, \lambda>0\right) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Find the sum of the first $n$ terms $S_{n}$ o... | (1) From the problem, we have
$$
\frac{a_{n+1}}{\lambda^{n+1}}-\left(\frac{2}{\lambda}\right)^{n+1}=\frac{a_{n}}{\lambda^{n}}-\left(\frac{2}{\lambda}\right)^{n}+1 \text {. }
$$
Therefore, $\left\{\frac{a_{n}}{\lambda^{n}}-\left(\frac{2}{\lambda}\right)^{2}\right\}$ is an arithmetic sequence with the first term 0 and c... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,613 |
Example 4 Let $a>0$. As shown in Figure 1, given the line $l: y=a x$ and the curve $C: y=x^{2}$, the x-coordinate of point $Q_{1}$ on curve $C$ is $a_{1}\left(0<a_{1}<a\right)$. From point $Q_{n}(n \geqslant 1)$ on curve $C$, draw a line parallel to the $x$-axis, intersecting line $l$ at point $P_{n+1}(n=1,2, \cdots)$.... | Notice
$$
\begin{array}{l}
Q_{n}\left(a_{n}, a_{n}^{2}\right), P_{n+1}\left(\frac{a_{n}^{2}}{a}, a_{n}^{2}\right), \\
Q_{n+1}\left(a_{n+1}, a_{n+1}^{2}\right)=\left(\frac{a_{n}^{2}}{a}, \frac{a_{n}^{4}}{a^{2}}\right) .
\end{array}
$$
Thus, $a_{n+1}=\frac{a_{n}^{2}}{a}$.
Taking the common logarithm, we get $\lg a_{n+1}... | a_{n}=a\left(\frac{a_{1}}{a}\right)^{2 n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,614 |
Example 5 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{3}{2}$, and $a_{n}=\frac{3 n a_{n-1}}{2 a_{n-1}+n-1}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: For all positive integers $n$, the inequality $a_{1} a_{2... | (1) Taking the reciprocal of the known recurrence relation, we get
$$
\frac{n}{a_{n}}=\frac{1}{3} \cdot \frac{n-1}{a_{n-1}}+\frac{2}{3} \text {. }
$$
Let $b_{n}=\frac{n}{a_{n}}$. Then $b_{n}=\frac{1}{3} b_{n-1}+\frac{2}{3}$.
Thus, $b_{n}-1=\frac{1}{3}\left(b_{n-1}-1\right)$.
Therefore, the sequence $\left\{b_{n}-1\rig... | a_{n}=\frac{3^{n} n}{3^{n}-1} | Algebra | proof | Yes | Yes | cn_contest | false | 722,615 |
Example 6 Given the function $f(x)=x^{2}+x-1, \alpha 、 \beta$ $(\alpha>\beta)$ are the two roots of the equation $f(x)=0$, and $f^{\prime}(x)$ is the derivative function.
Let $a_{1}=1, a_{n+1}=a_{n}-\frac{f\left(a_{n}\right)}{f^{\prime}\left(a_{n}\right)}(n=1,2, \cdots)$.
(1) Find the values of $\alpha 、 \beta$;
(2) It... | (1) From $x^{2}+x-1=0$, we get $x=\frac{-1 \pm \sqrt{5}}{2}$. Therefore, $\alpha=\frac{-1+\sqrt{5}}{2}, \beta=\frac{-1-\sqrt{5}}{2}$.
(2) Note that
$$
\begin{array}{l}
f^{\prime}(x)=2 x+1, \\
a_{n+1}=a_{n}-\frac{a_{n}^{2}+a_{n}-1}{2 a_{n}+1}=\frac{a_{n}^{2}+1}{2 a_{n}+1}, \\
\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=\frac{\... | 4\left(2^{n}-1\right) \ln \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,616 |
Example 8 Given a sequence $\left\{a_{n}\right\}$ with all terms being positive, the sum of the first $n$ terms $S_{n}$ satisfies $S_{1}>1$, and
$$
6 S_{n}=\left(a_{n}+1\right)\left(a_{n}+2\right)\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula for $\left\{a_{n}\right\}$;
(2) Let the se... | (1) The problem gives the relationship between $S_{n}$ and $a_{n}$, which needs to be transformed into a relationship between $a_{n-1}$ and $a_{n}$. From
$$
a_{1}=S_{1}=\frac{1}{6}\left(a_{1}+1\right)\left(a_{1}+2\right) \text {, }
$$
we solve to get $a_{1}=1$ (discard) or 2.
$$
\begin{array}{l}
\text { Also, } a_{n+... | a_{n}=3n-1 | Algebra | proof | Yes | Yes | cn_contest | false | 722,617 |
Example 1 Given that $P A$ and $P B$ are two tangents drawn from a point $P$ outside the circle $\odot O$, $M$ and $N$ are the midpoints of segments $A P$ and $A B$ respectively. Extend $M N$ to intersect $\odot O$ at point $C$, with point $N$ between $M$ and $C$. $P C$ intersects $\odot O$ at point $D$, and extend $N ... | Prove: As shown in Figure 1, since $P A$ and $P B$ are tangents to $\odot O$, $A B$ is the polar line of point $P$ with respect to $\odot O$.
Let $P C$ intersect $A B$ at point $E$. Therefore, by property 3 of polarity, $(P, E, D, C)$ is a harmonic range of points.
Since $M N \parallel P B$, with $N$ as the center of... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,618 |
Example 2 Given a point $X$ outside the circle $\odot O$, two tangents are drawn from $X$ to $\odot O$, touching the circle at points $A$ and $B$. A line through $X$ intersects $\odot O$ at points $C$ and $D$, such that $C A \perp B D$. If $C A$ and $B D$ intersect at point $F$, $C D$ and $A B$ intersect at point $G$, ... | Proof: By the polarity property 3, $(X, G, D, C)$ is a harmonic range, and since $C A \perp B D$, it follows from Lemma 3 that
$$
\angle G F D=\angle D F X.
$$
As shown in Figure 2, let the circumcircle of
$\triangle G F X$
intersect $B F$ at point $H^{\prime}$.
Then $G H^{\prime}=X H^{\prime}$, meaning
point $H^{\pri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,619 |
Example 3 Let $E$ and $F$ be the points of tangency of the incircle $\odot I$ of $\triangle ABC$ with sides $AC$ and $AB$, respectively. Let $M$ be the midpoint of $BC$, and let $AM$ intersect $EF$ at point $N$. The circle $\odot M$ with diameter $BC$ intersects $BI$ and $CI$ at points $X$ and $Y$, respectively. Prove:... | Prove: As shown in Figure 3, draw a line through point $A$ parallel to $BC$ intersecting $EF$ at point $Z$, and $BC$ intersects $EF$ at point $T$.
First, prove that points $X, Y$ lie on line $EF$.
Assume $BI$ intersects $EF$ at point $X'$.
Since $\triangle BFI \cong \triangle BDI$, symmetrically we get
$\angle FXB = \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,620 |
For example, $4 \odot I$ is the incircle of $\triangle A B C$, and $D, E, F$ are the points of tangency of $\odot I$ with sides $B C, C A, A B$, respectively. Let $\odot I$ intersect $A D$ at point $M$, and $N$ be the intersection of the circumcircle of $\triangle C D M$ with $D F$. Let $C N$ intersect $A B$ at point $... | Prove that, as shown in Figure 4, let $E F$ intersect $C G$ at point $X$, and $E F$ intersect $B C$ at point $T$.
Applying Menelaus' theorem to $\triangle B C G$ and the transversal $D N F$, we get
$$
\frac{B F}{F G} \cdot \frac{G N}{N C} \cdot \frac{C D}{D B}=1 .
$$
Thus, $C D=3 G F \Leftrightarrow C N=3 N G$.
By Le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,621 |
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