problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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5. On the graph of the function $y=-\frac{a^{2}+1}{x}(a$ is a constant $)$, there are points $A\left(-1, y_{1}\right) 、 B\left(\frac{1}{4}, y_{2}\right) 、 C\left(\frac{1}{2}, y_{3}\right)$. The size relationship of the function values $y_{1} 、 y_{2} 、 y_{3}$ is ( ).
(A) $y_{1}<y_{2}<y_{3}$
(B) $y_{3}<y_{2}<y_{1}$
(C) $... | 5. C.
Since $-\left(a^{2}+1\right)<0$, in each quadrant, $y$ increases as $x$ increases. Therefore, $y_{1}<y_{2}$. Also, $\left(-1, y_{1}\right)$ is in the second quadrant, while $\left(\frac{1}{2}, y_{3}\right)$ is in the fourth quadrant, so $y_{3}<y_{1}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,293 |
Example 6 Proof: There exist positive integers $a_{i}(1 \leqslant i \leqslant 8)$, such that
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}}=2 .
\end{array}
$$ | Prove the construction of the identity:
$$
\begin{array}{l}
\sqrt{(2 i+1)^{2}}-\sqrt{(2 i+1)^{2}-1} \\
=2 i+1-2 \sqrt{i(i+1)} \\
=(\sqrt{i+1}-\sqrt{i})^{2} .
\end{array}
$$
Take \( a_{i}=(2 i+1)^{2} \) for \( i=1,2, \cdots, 8 \), then
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2... | 2 | Algebra | proof | Yes | Yes | cn_contest | false | 722,294 |
6. Given $a+b+c \neq 0$, and
$$
\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=p \text {. }
$$
Then the line $y=p x+p$ does not pass through the ( ) quadrant.
(A) first
(B) second
(C) third
(D) fourth | 6. D.
From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=c p, \\
a+c=a p, \\
c+a=b p
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2(a+b+c)=p(a+b+c) \\
a+b+c \neq 0
\end{array}\right. \\
\Rightarrow p=2 .
\end{array}
$$
Therefore, the line $y=p x+p$ cannot pass through the ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,295 |
7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ - | II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,296 |
8. Robots A and B simultaneously conduct a $100 \mathrm{~m}$ track test at a uniform speed, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. After calc... | 8. 1 .
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙 }}$, respectively. When 甲 is $1 \mathrm{~m}$ away from the finish line, the time spent is $t$. Then $v_{\text {乙 }} t=x-2$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}$.
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,297 |
10. There are three right-angled triangles containing a
$30^{\circ}$ angle, they are of different sizes,
but they have one side equal. Then, in the order of the areas of these three triangles from largest to smallest, the ratio of their hypotenuses is | 10. $2: \frac{2 \sqrt{3}}{3}: 1$.
Let the shorter leg of Triangle A equal the longer leg of Triangle B equal the hypotenuse of Triangle C. Then the hypotenuses of Triangles A, B, and C are $2$, $\frac{2 \sqrt{3}}{3}$, and $1$, respectively, i.e., the required ratio is $2: \frac{2 \sqrt{3}}{3}: 1$. | 2: \frac{2 \sqrt{3}}{3}: 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,299 |
11. (20 points) As shown in Figure 2, given that $\triangle ABC$ is an equilateral triangle, $E$ is a point on the extension of $AC$, and a point $D$ is chosen such that $\triangle CDE$ is an equilateral triangle. If $M$ and $N$ are the midpoints of segments $AD$ and $BE$ respectively, prove that $\triangle CMN$ is an ... | Three, 11. From $\triangle A C D \cong \triangle B C E$, we have $A D=B E, A M=B N$.
Also, $\triangle A M C \cong \triangle B N C$, so $C M=C N, \angle A C M=\angle B C N$. Therefore, $\angle N C M=\angle B C N-\angle B C M$, $\angle A C B=\angle A C M-\angle B C M$. Hence, $\angle N C M=\angle A C B=60^{\circ}$. Thus,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,300 |
13. (25 points) Given positive integers $x, y$ satisfy
$$
\frac{2}{x}+\frac{1}{y}=a\left(a \in \mathbf{N}_{+} \text {, and } x<y\right) \text {. }
$$ | 13. Given $x \geqslant 1, y \geqslant 2$, we have $\frac{2}{x}+\frac{1}{y} \leqslant 2+\frac{1}{2}$, i.e., $a \leqslant 2 \frac{1}{2}$.
Therefore, $a=1$ or 2.
When $a=1$, $\frac{2}{x}+\frac{1}{y}=1$.
If $x=1$, then $\frac{1}{y}=-1$, which contradicts that $y$ is a positive integer. Hence, $x \neq 1$.
If $x=2$, then $\... | x=2, y=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,302 |
Example 7 Prove:
$$
\frac{(x+a)(x+b)}{(c-a)(c-b)}+\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}=1 .
$$ | Prove the construction of the function
$$
\begin{aligned}
f(x)= & \frac{(x+a)(x+b)}{(c-a)(c-b)}+\frac{(x+b)(x+c)}{(a-b)(a-c)}+ \\
& \frac{(x+c)(x+a)}{(b-c)(b-a)}-1 .
\end{aligned}
$$
Then \( f(-a)=f(-b)=f(-c)=0 \).
Since \( a \neq b \neq c \), the quadratic function \( f(x) \) intersects the \( x \)-axis at three diff... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,305 |
3. In $\triangle A B C$, $A B=6, A C=8, \angle B A C=$ $90^{\circ}, A D, B E$ are the medians to sides $B C, A C$ respectively. Then the cosine of the angle between vectors $\overrightarrow{A D}$ and $\overrightarrow{B E}$ is equal to ( ).
(A) $\frac{\sqrt{13}}{65}$
(B) $\frac{\sqrt{3}}{2}$
(C) $-\frac{\sqrt{13}}{65}$
... | 3. C.
As shown in Figure 3, with $A$ as the origin and the two perpendicular sides as coordinate axes, we establish a rectangular coordinate system. Thus, $A(0,0)$, $B(6,0)$, $C(0,8)$, $D(3,4)$, and $E(0,4)$. Therefore,
$$
\begin{aligned}
\overrightarrow{A D} & =(3,4), \\
\overrightarrow{B E} & =(-6,4). \\
& \text { H... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,306 |
4. $\log _{2} \sin \frac{\pi}{12}+\log _{2} \sin \frac{\pi}{6}+\log _{2} \sin \frac{5 \pi}{12}=$ ( ).
(A) -3
(B) -1
(C) 1
(D) 3 | 4. A.
$$
\begin{array}{l}
\log _{2} \sin \frac{\pi}{12}+\log _{2} \sin \frac{\pi}{6}+\log _{2} \sin \frac{5 \pi}{12} \\
=\log _{2}\left(\sin \frac{\pi}{12} \cdot \frac{1}{2} \cdot \sin \frac{5 \pi}{12}\right) \\
=\log _{2}\left(\frac{1}{2} \sin \frac{\pi}{12} \cdot \cos \frac{\pi}{12}\right) \\
=\log _{2}\left(\frac{1}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,307 |
5. $a$ is a parameter, the function
$$
f(x)=(x+a) 3^{x-2+a^{2}}-(x-a) 3^{8-x-3 a}
$$
is an even function. Then the set of values that $a$ can take is ( ).
(A) $\{0,5\}$
(B) $\{-2,5\}$
(C) $\{-5,2\}$
(D) $\{1,2009\}$ | 5. C.
Given that the function $f(x)$ is an even function, we know that $f(-a)=f(a)$, which means $2 a \cdot 3^{8+a-3 a}=2 a \cdot 3^{a-2+a^{2}}$.
Thus, $a=0$ or $8+a-3 a=a-2+a^{2}$.
If $a=0$, then $f(x)=x\left(3^{x-2}-3^{8-x}\right)$ is not an even function. Therefore, $a \neq 0$.
If $8+a-3 a=a-2+a^{2}$, then $a=-5$ ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,308 |
For the function
$$
f(x)=\left\{\begin{array}{ll}
x^{2}+b x+c & x \geqslant 0 ; \\
-2, & x<0,
\end{array}\right.
$$
it is given that $f(4)=f(0), f(1)=-1$. Then the number of solutions to the equation $f(x)=x$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6. D.
$$
\begin{array}{l}
\text { Given } f(4)=f(0), f(1)=-1 \\
\Rightarrow 16+4 b+c=c, 1+b+c=-1 \\
\Rightarrow b=-4, c=2 . \\
\text { Therefore, } f(x)=\left\{\begin{array}{ll}
x^{2}-4 x+2, & x \geqslant 0 ; \\
-2, & x<0 .
\end{array}\right.
\end{array}
$$
Its graph is shown in Figure 4, with a total of 3 solutions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,309 |
1. The positive numbers $b_{1}, b_{2}, \cdots, b_{60}$ are arranged in sequence and satisfy $\frac{b_{2}}{b_{1}}=\frac{b_{3}}{b_{2}}=\cdots=\frac{b_{60}}{b_{59}}$.
Determine the value of $\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right)$. | Given that $\left\{b_{k}\right\}$ is a geometric sequence with a common ratio of $q$, then $b_{k}=b_{1} q^{k-1}$. Therefore, $b_{k} b_{61-k}=b_{1} b_{60}$. Hence,
$$
\begin{array}{l}
\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right) \\
=\log _{b_{1} b_{60}}\left(b_{1} b_{60}\right)^{30}=30 .
\end{array}
$$ | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,310 |
2. Let $\sin (\alpha+\beta)=\frac{4}{5}, \cos (\alpha-\beta)=\frac{3}{10}$, find the value of $(\sin \alpha-\cos \alpha)(\sin \beta-\cos \beta)$. | $\begin{array}{l}\text { 2. }(\sin \alpha-\cos \alpha)(\sin \beta-\cos \beta) \\ =(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta)- \\ \quad(\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta) \\ =\cos (\alpha-\beta)-\sin (\alpha+\beta)=-\frac{1}{2} .\end{array}$ | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,311 |
3. If the coordinates of the vertices of quadrilateral $A B C D$ are $A(1,2)$, $B(2,5)$, $C(7,3)$, and $D(5,1)$, find the area of quadrilateral $A B C D$. | 3. As shown in Figure 5, construct the circumscribed rectangle of quadrilateral $ABCD$, with its vertices at $M(1,1)$, $N(1,5)$, $P(7,5)$, and $T(7,1)$. Quadrilateral $ABCD$ is obtained by removing $\triangle AMD$, $\triangle ANB$, $\triangle BPC$, and $\triangle CTD$ from rectangle $MNPT$. It is easy to see that
$$
\b... | \frac{27}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,312 |
4. $[x]$ represents the greatest integer not exceeding the real number $x$. If
$$
\left[\log _{3} 6\right]+\left[\log _{3} 7\right]+\cdots+\left[\log _{3} n\right]=2009 \text {, }
$$
determine the value of the positive integer $n$. | 4. For $3^{k} \leqslant i \leqslant 3^{k+1}-1$, we have $\left[\log _{3} i\right]=k$. Therefore, $\sum_{i=3^{k}}^{3 k+1}\left[\log _{3} i\right]=2 \times 3^{k} k$.
Also, $3+\sum_{k=2}^{4} 2 \times 3^{k} k<2009<3+\sum_{k=2}^{5} 2 \times 3^{k} k$, hence $3^{5}<n<3^{6}-1$.
From $3+\sum_{k=2}^{4} 2 \times 3^{k} k+5\left(n... | 474 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,313 |
5. In $\triangle A B C$, $A B=B C>A C, A H$ and $A M$ are the altitude and median from vertex $A$ to side $B C$, respectively, and $\frac{S_{\triangle A M H}}{S_{\triangle A B C}}=\frac{3}{8}$. Determine the value of $\cos \angle B A C$. | 5. In $\triangle A B C$, draw $B D \perp A C$ at point $D$. Let $A B=A C=a$.
Since $\frac{S_{\triangle M M H}}{S_{\triangle A B C}}=\frac{M H}{B C}=\frac{3}{8}$, therefore,
$M H=\frac{3}{8} a$,
$H C=M C-M H=\frac{1}{2} a-\frac{3}{8} a=\frac{1}{8} a$.
Let $\angle B A C=\beta$. Then $\angle B C A=\beta$.
In Rt $\triangle... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,314 |
6. The graphs of the functions $f(x)=2 x^{2}-2 x-1$ and $g(x)=$ $-5 x^{2}+2 x+3$ intersect at two points. The equation of the line passing through these two points is $y=a x+b$. Find the value of $a-b$. | 6. The graphs of the functions $f(x)$ and $g(x)$ intersect at two points, which are the solutions to the system of equations $\left\{\begin{array}{l}y=2 x^{2}-2 x-1, \\ y=-5 x^{2}+2 x+3\end{array}\right.$. Therefore,
$$
\begin{array}{l}
7 y=5\left(2 x^{2}-2 x-1\right)+2\left(-5 x^{2}+2 x+3\right) \\
=-6 x+1 .
\end{arra... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,315 |
Example 8 Let $x, y, z$ be between 0 and 1. Prove that:
$$
x(1-y)+y(1-z)+z(1-x)<1 \text{. }
$$ | $$
\begin{array}{l}
f(x)=1-[x(1-y)+y(1-z)+z(1-x)] \\
=(y+z-1) x+(y z+1-y-z) .
\end{array}
$$
Since $00, \\
f(1)=(y+z-1)+(y z+1-y-z)=y z>0 .
\end{array}
$$
Since the graph of the linear function $f(x)$ is a straight line, therefore, when $00$ holds. Hence, the inequality to be proved is established. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,316 |
7. Given that the 6027-digit number $\frac{a b c a b c \cdots a b c}{2000 \uparrow a b c}$ is a multiple of 91. Find the sum of the minimum and maximum values of the three-digit number $\overline{a b c}$. | 7. From $91=7 \times 13, 1001=7 \times 11 \times 13$, we know 91 | 1001.
And $\overline{a b c a b c}=1001 \times \overline{a b c}$, thus, $91 \mid \overline{a b c a b c}$. $2009 \uparrow a b c$
$$
\overline{a_{2009 \uparrow a b c}^{a b c a b c}}=\underset{2 \times 1004 \uparrow a b c}{\overline{a b c a b c \cdots a b c... | 1092 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,317 |
8. As shown in Figure 2, in
rhombus $A B C D$, $\angle A B C$
$$
=120^{\circ}, B C=6 \sqrt{3} \text {, }
$$
$P$ is a moving point on the extension of $B C$
away from point $C$,
$A P$ intersects $C D$ at point $E$, and $B E$ is extended to intersect $D P$ at point
$Q$. When the moving point $P$ is at its initial positio... | 8. As shown in Figure 6, connect $B D$, and construct the circumcircle of $\triangle A B D$ intersecting $A P$ at point $F$. Connect $B F$, $D F$, $F C$, and $C Q$. It is easy to see that
$$
\begin{array}{l}
\angle D F B=\angle D F P=\angle B F P=120^{\circ}, \\
\angle B F E=\angle E C P=120^{\circ}.
\end{array}
$$
Th... | \frac{4 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,318 |
3. Given that $E$ is a point on the side $BC$ of square $ABCD$, such that $S_{\triangle ABE} 、 S_{\triangle AEC} 、 S_{\triangle ACD}$ form a geometric sequence. Then $\angle EAB=$ $\qquad$ (accurate to $1^{\prime \prime}$ ). | 3. $20^{\circ} 54^{\prime} 19^{\prime \prime}$ | 20^{\circ} 54^{\prime} 19^{\prime \prime} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,321 |
7. For integer $n>1$, let $x=1+\frac{1}{2}+\cdots+\frac{1}{n}$, $y=\lg 2+\lg 3+\cdots+\lg n$. Then the set of all integers $n$ that satisfy $[x]=[y]$ is $\qquad$ ( [a] denotes the greatest integer not exceeding the real number $a$). | 7. $\{5,6\}$ | \{5,6\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,325 |
8. A three-digit number has 3 digits, none of which are 0, and its square is a six-digit number that has exactly 3 digits as 0. Write down one such three-digit number: $\qquad$ | 8.448 or 548 or 949 | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,326 |
Example 9 Definition: In the plane, a set of points composed of $n$ points, if the perpendicular bisector of any two points in the set passes through at least one point in the set, then this set of points is called an "Zu Chongzhi point set" of $n$ points.
For example, in Figure 4 (a), the five vertices $A, B, C, D, E... | The three vertices and the midpoints of the three sides of an equilateral triangle can form a 6-point Zu Chongzhi point set, as shown in the 6 points $A, B, C, D, E, F$ in Figure 5 (a). The ten vertices of a regular pentagram on a flag form a 10-point Zu Chongzhi point set, as shown in the 10 points in Figure 5 (b). | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,327 |
II. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, there is an object $A$ flying at a speed of $0.3 \mathrm{~km} / \mathrm{s}$ parallel to the positive direction of the $x$-axis. When object $A$ is at point $N$, it is measured that $O N=10 \mathrm{~km}$, $\angle x O N=105^{\circ}$. At thi... | II. Suppose after $t \mathrm{~s}$, missile $B$ precisely hits the target at point $C$ (as shown in Figure 2),
and let $\angle x O B=\alpha$. Then
$$
\begin{array}{l}
N C=0.3 t, \\
O C=0.7 t, \\
\angle N O C=105^{\circ}-\alpha, \\
\angle O N C=75^{\circ} .
\end{array}
$$
In $\triangle O N C$, by the Law of Sines we get... | 80.546^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,328 |
$$
\begin{array}{l}
\text { Three. (20 points) (1) Prove: } \\
\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=4 \sin ^{2} 2 x-3(x \in \mathbf{R}) ;
\end{array}
$$
(2) Find the value: $\prod_{t=0}^{2^{8}}\left(4 \sin ^{2} \frac{t \pi}{2^{9}}-3\right)$. | $$
\begin{array}{l}
=(1)\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=16 \sin ^{2} x \cdot \cos ^{2} x-12\left(\sin ^{2} x+\cos ^{2} x\right)+9 \\
=4 \sin ^{2} 2 x-3
\end{array}
$$
(2) Let $A_{k}=\prod_{i=0}^{2 k-1}\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)$.
When $k \geqslant 2$, by (1) we have
$$
... | -3 | Algebra | proof | Yes | Yes | cn_contest | false | 722,329 |
Four, (20 points) Two three-digit numbers written together form a six-digit number. If this six-digit number is exactly an integer multiple of the product of the original two three-digit numbers, find this six-digit number. | Let the six-digit number be abcdef.
By the problem, we can set $\overline{a b c d e f}=k \overline{a b c} \cdot \overline{d e f}$, that is,
$\overline{a b c} \cdot 1000+\overline{d e f}=k \overline{a b c} \cdot \overline{d e f}$.
Therefore, $\overline{a b c} \mid \overline{d e f}$.
Let $\overline{\operatorname{def}}=l ... | 143143 \text{ and } 167334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,330 |
One, (20 points) (1) Table 1 shows the data measured in a projectile experiment of an object.
Assuming the trajectory equation of the object is $y=a x^{2}$, find the value of the real number $a$ so that the sum of the squares of the distances from each measurement point $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3, 4)$ to ... | (1) From the problem, we know
$$
\begin{array}{l}
\sum_{i=1}^{4} P_{i} A_{i}^{2}=\sum_{i=1}^{4}\left(a x_{i}^{2}-y_{i}\right)^{2} \\
=a^{2} \sum_{i=1}^{4} x_{i}^{4}-2 a \sum_{i=1}^{4} x_{i}^{2} y_{i}+\sum_{i=1}^{4} y_{i}^{2} \\
=\sum_{i=1}^{4} x_{i}^{4}\left[a^{2}-\frac{2 \sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} ... | y=-2.49 x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,331 |
For a positive integer $n$, let $t_{n}=\frac{n(n+1)}{2}$. Writing down the last digits of $t_{1}=1, t_{2}=3, t_{3}=6, t_{4}=10, t_{5}=15 \cdots \cdots$ can form an infinite repeating decimal: $0.13605 \cdots$. Find the length of the repeating cycle of this decimal. | $$
\begin{array}{l}
t_{n+20}-t_{n}=\frac{(n+20)(n+20+1)}{2}-\frac{n(n+1)}{2} \\
=20 n+210=10(2 n+21)
\end{array}
$$
That is, the last digit of $t_{n+20}$ is the same as that of $t_{n}$.
Therefore, 20 is the length of the repeating cycle of this repeating decimal. This repeating decimal is
$$
0.13605186556815063100 .
$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,332 |
Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to ... | (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.
If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,333 |
$$
\begin{array}{l}
\text { 1. Let } A=\{x \mid x=2 n, n \in \mathbf{N}\} \\
B=\{x \mid x=3 n, n \in \mathbf{N}\}, \\
C=\left\{x \mid x=n^{2}, n \in \mathbf{N}\right\} .
\end{array}
$$
Then $A \cap B \cap C=$ | 1. $\left\{x \mid x=36 n^{2}, n \in \mathbf{N}\right\}$.
Notice that the numbers in $A \cap B \cap C$ are all perfect squares and can be divided by both 2 and 3, thus they can also be divided by 4 and 9. Therefore, $A \cap B \cap C=\left\{x \mid x=36 n^{2}, n \in \mathbf{N}\right\}$. | \left\{x \mid x=36 n^{2}, n \in \mathbf{N}\right\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,334 |
2. Draw two tangent lines to the circle $x^{2}+y^{2}=1$ through the point $(1,2)$. Then the area of the quadrilateral formed by these two tangent lines with the $x$-axis and $y$-axis is $\qquad$ | 2. $\frac{13}{8}$.
Obviously, one of the tangent points is $(1,0)$, and the corresponding tangent segment length is 2. Therefore, the quadrilateral formed is a trapezoid.
Let the other tangent point be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
\left(x_{0}-1\right)^{2}+\left(y_{0}-2\... | \frac{13}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,335 |
3. In the tetrahedron $P A B C$, it is known that
$$
\angle A P B=\angle B P C=\angle C P A=90^{\circ} \text {, }
$$
the sum of the lengths of all edges is $S$. Then the maximum volume of this tetrahedron is $\qquad$ . | 3. $\frac{S^{3}}{162(1+\sqrt{2})^{3}}$.
When the edges $A P=B P=C P$ of the tetrahedron $P A B C$ are equal, the volume of the tetrahedron is maximized.
Let $A P=B P=C P=x$. Then
$$
A B=B C=C A=\sqrt{2} x \text {. }
$$
From the given information,
$$
\begin{array}{l}
A P+B P+C P+A B+B C+C A \\
=3(1+\sqrt{2}) x=S .
\en... | \frac{S^{3}}{162(1+\sqrt{2})^{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,336 |
4. Let the function $y=f(x)$ satisfy for all $x \in \mathbf{R}$, $y=f(x) \geqslant 0$, and $f(x+1)=\sqrt{9-f^{2}(x)}$. It is known that when $x \in[0,1)$,
$$
f(x)=\left\{\begin{array}{ll}
2^{x}, & 0 \leqslant x \leqslant \frac{1}{2} ; \\
\lg (x+31), & \frac{1}{2} \leqslant x<1 .
\end{array}\right.
$$
Then $f(\sqrt{100... | 4. $\frac{3 \sqrt{3}}{2}$.
From the given information, we have
$$
f^{2}(x+2)=9-f^{2}(x+1)=f^{2}(x) \text {. }
$$
Also, $y=f(x) \geqslant 0$, so $f(x+2)=f(x)$. Therefore, $y=f(x)$ is a periodic function with a period of 2.
Since $10 \sqrt{10}-31 \in\left(\frac{1}{2}, 1\right)$, we have
$$
f(10 \sqrt{10}-31)=\lg 10 \sq... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,337 |
5. A die is rolled twice in succession, yielding the numbers $m$ and $n$. The probability that a right-angled triangle can be formed with the points $(0,0)$, $(1,-1)$, and $(m, n)$ as vertices is | 5. $\frac{5}{18}$.
Since $1 \leqslant m, n \leqslant 6$, there are a total of 36 possible values for $(m, n)$. The points $(m, n)$ that can form a right-angled triangle with the points $(0,0)$, $(1,-1)$ have the following 10 possible values:
$$
\begin{array}{l}
(1,1),(2,2),(3,3),(4,4),(5,5), \\
(6,6),(3,1),(4,2),(5,3)... | \frac{5}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,339 |
$$
\begin{array}{l}
\cos 10^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}+\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ} \\
=\quad .
\end{array}
$$ | 6. $\frac{1+\sqrt{3}}{8}$.
Original expression $=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ}+\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ}$ is easy to calculate
$$
\begin{array}{l}
8 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
=4\left(\cos 20^{\circ}-\cos 60^{\circ}\rig... | \frac{1+\sqrt{3}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,340 |
7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is | 7. 10 .
Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$ | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,341 |
8. Given positive integers whose digits are all non-zero and their sum is 7. Then the number of times the digit 3 appears in all these positive integers is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 8. 28 .
We only need to calculate the number of times 3 appears in each case.
For numbers with two 3s in the satisfying digits, there are 3 such numbers, with 3 appearing 6 times;
For numbers with one 3, the five-digit, four-digit, three-digit, and two-digit numbers that satisfy the condition are 5, $ \mathrm{~A}_{4}... | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 722,342 |
10. (14 points) Given positive real numbers $a, b, c, d$ satisfying $a+b+c+d=abcd$. Find the minimum value of $\sum a^{4}(bcd-1)$, where “$\sum$” denotes the cyclic sum.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated part:
10. (14 points) Given positive real numbers $a, b, c, d$ satisfying $a+b... | 10. Since $a, b, c, d > 0$, and $a + b + c + d = abcd$, we have $a(bcd - 1) = b + c + d$.
By the AM-GM inequality,
$$
\begin{array}{l}
abcd = a + b + c + d \geqslant 4 \sqrt[4]{abcd} \\
\Rightarrow abcd \geqslant (\sqrt[3]{4})^4,
\end{array}
$$
where equality holds if and only if $a = b = c = d = \sqrt[3]{4}$. Therefo... | 48 \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,344 |
12. (18 points) The inverse function of $f(x)$ is $y=\frac{x}{1+x}$, $g_{n}(x)+\frac{1}{f_{n}(x)}=0$, let $f_{1}(x)=f(x)$, and for $n>1\left(n \in \mathbf{N}_{+}\right)$, $f_{n}(x)=f_{n-1}\left(f_{n-1}(x)\right)$. Find the analytical expression for $g_{n}(x)\left(n \in \mathbf{N}_{+}\right)$. | (3) $\left\{\begin{array}{l}5-d+\frac{3}{q}=4, \\ 5+d+3 q=16\end{array} \Rightarrow 3 q+\frac{3}{q}=10\right.$ $\Rightarrow q=3$ or $\frac{1}{3} \Rightarrow d_{1,2}=1+\frac{3}{q_{1,2}}$.
Clearly, the larger $d$ is 10.
(4) $\left\{\begin{array}{l}5-d+\frac{3}{q}=8 \\ 5+d+3 q=8\end{array} \Rightarrow q+\frac{1}{q}=2\righ... | g_{n}(x)=2^{n-1}-\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,345 |
11. (18 points) Let $a_{1} , a_{2} , a_{3}$ form an arithmetic sequence, $a_{1}+a_{2}+a_{3}=15$; $b_{1} , b_{2} , b_{3}$ form a geometric sequence, $b_{1} b_{2} b_{3}$ $=27$. If $a_{1}+b_{1} , a_{2}+b_{2} , a_{3}+b_{3}$ are positive integers and form a geometric sequence, find the maximum value of $a_{3}$. | 11. Let $a_{1}=5-d, a_{2}=5, a_{3}=5+d, b_{1}=$ $\frac{3}{q}, b_{2}=3, b_{3}=3 q$. Then, from the conditions, $5-d+\frac{3}{q}, 5+$ $d+3 q$ are both positive integers, and
$$
\left(5-d+\frac{3}{q}\right)(5+d+3 q)=64 \text {. }
$$
Therefore, the possible values of $5-d+\frac{3}{q}$ and $5+d+3 q$ are essentially only fo... | 15+3\sqrt{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,346 |
1. A robot starts from the origin on a number line, moving in the positive direction of the number line, with a program of advancing 4 steps and then retreating 3 steps. Suppose the robot moves forward or backward 1 step per second, and each step is one unit length, $x_{n}$ represents the number corresponding to the ro... | -1. B.
It is known that the robot advances one unit length every $7 \mathrm{~s}$. From
$$
2007=7 \times 286+5,2011=7 \times 287+2 \text {, }
$$
we get
$$
\begin{array}{l}
x_{2007}=286 \times 1+4-1=289, \\
x_{2011}=287 \times 1+2=289 .
\end{array}
$$
Therefore, $x_{2007}-x_{2011}=0$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,347 |
2. There are the following four propositions:
(1) Through a point on a line, there is one and only one line perpendicular to this line;
(2) The equation $x \sqrt{(x+2)^{2}}+1=0$ has three different real solutions;
(3) If $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $\odot O_{1}$ has a radius of $5$,... | 2. A.
For a line in space, an infinite number of perpendicular lines can be drawn through a point on the line, so statement (1) is a false statement.
For statement (2), in the original equation, it is clear that $x<0$.
When $x<-2$, the original equation can be transformed into
$$
-x(x+2)+1=0 \text {. }
$$
Solving giv... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,348 |
1. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0
\end{array}
$$
$\left(a 、 b \in \mathbf{N}_{+}\right)$ have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^... | Given the known equations, we have $a \neq 1, b \neq 1$. Therefore, $a > 1, b > 1$, and $a \neq b$. Let $x_{0}$ be the common root of the two equations. It is easy to see that $x_{0} \neq 1$. By the definition of the roots of the equation, $a$ and $b$ are the two distinct real roots of the equation
$$
\left(1-x_{0}\rig... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,349 |
3. Given that $f(x)$ represents a fourth-degree polynomial in $x$, and $f(a)$ represents the value of $f(x)$ when $x=a$. If
$$
\begin{array}{l}
f(1)=f(2)=f(3)=0, \\
f(4)=6, f(5)=72,
\end{array}
$$
then the value of $f(6)$ is ( ).
(A) 200
(B) 300
(C) 400
(D) 600 | 3. B.
Since $f(1)=f(2)=f(3)=0$, the quartic polynomial $f(x)$ can be set as
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {. }
$$
Given $f(4)=6, f(5)=72$, we have
$$
\left\{\begin{array}{l}
6(4 a+b)=6, \\
24(5 a+b)=72 .
\end{array}\right.
$$
Solving these equations, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Th... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,350 |
4. In a convex pentagon $A B C D E$,
$$
A B=B C=C D=D E=E A,
$$
and $\angle C A D=\angle B A C+\angle E A D$. Then the degree measure of $\angle B A E$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 4. C.
As shown in Figure 2, $\triangle A B C$ is rotated counterclockwise around point $A$ by the degree of $\angle B A E$ to the position of $\triangle A E F$, and $D F$ is connected. Then
$$
\begin{array}{l}
B C=E F, \\
A C=A F, \\
\angle B A C=\angle E A F .
\end{array}
$$
Since $\angle C A D=\angle B A C+\angle E... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,351 |
5. Given that $a, b, c$ are the lengths of the three sides of a triangle, and real numbers $p, q$ satisfy $p+q=1$. Then the result of $p a^{2}+q b^{2}-p q c^{2}$ is $(\quad)$.
(A) Positive
(B) Zero
(C) Negative
(D) Any of the above is possible | 5. A.
Let $y=p a^{2}+q b^{2}-p q c^{2}$.
Substitute $q=1-p$ into the above equation:
$$
\begin{array}{l}
y=p a^{2}+(1-p) b^{2}-p(1-p) c^{2} \\
=c^{2} p^{2}+\left(a^{2}-b^{2}-c^{2}\right) p+b^{2} .
\end{array}
$$
Consider the above equation as a quadratic function of $p$, with the parabola opening upwards, then
$$
\be... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,352 |
6. The line $l:(2 m+1) x+(m+1) y=7 m+4$ is intercepted by the circle $\odot A$ with center $A(1,2)$ and radius 3, the shortest chord length is ( ).
(A) $\sqrt{15}$
(B) 4
(C) $\sqrt{17}$
(D) $3 \sqrt{2}$ | 6. B.
The equation of line $l$ is transformed into
$$
\begin{array}{l}
m(2 x+y-7)+(x+y-4)=0 . \\
\text { Let }\left\{\begin{array} { l }
{ 2 x + y - 7 = 0 , } \\
{ x + y - 4 = 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=3, \\
y=1 .
\end{array}\right.\right.
\end{array}
$$
Therefore, line $l... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,353 |
1. The probability of drawing at least 1 girl from 3 boys and 4 girls is $\qquad$ . | $$
=1 . \frac{34}{35} \text {. }
$$
Draw a tree diagram, from 7 people, 3 are drawn, there are a total of $7 \times 6 \times 5=210$ ways to draw, among which, the number of ways to draw 3 all being boys is 6, thus, the probability of drawing 3 all being boys is $6 \div 210=\frac{1}{35}$. Therefore, the probability of ... | \frac{34}{35} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,354 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$ (for example, $[3.1]=3,[-3.1]=-4$). Suppose the real number $x$ is not an integer, and $x+\frac{113}{x}=[x]+\frac{113}{[x]}$. Then the value of $x$ is | 3. $-10 \frac{3}{11}$.
Eliminating the denominator, we get
$$
x^{2}[x]+113[x]=x[x]^{2}+113 x \text {. }
$$
Rearranging and factoring, we get
$$
(x-[x])(x[x]-113)=0 \text {. }
$$
Since $x$ is not an integer, $x-[x] \neq 0$.
Therefore, $x[x]-113=0$.
Let $x=[x]+t(0<t<1)$. Substituting into equation (1), we get
$$
[x]^{... | -10 \frac{3}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,356 |
One. (20 points) Given that $a$ and $b$ are integers, the equation $a x^{2} + b x + 2 = 0$ has two distinct negative real roots greater than -1. Find the minimum value of $b$.
| Let the equation $a x^{2}+b x+2=0(a \neq 0)$ have two distinct negative real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}<0\right)$.
\end{array}\right.
$$
Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens up... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,358 |
II. (25 points) Given that $\triangle ABC$ is a scalene triangle, points $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$, respectively, and $OI \perp AI$. Prove that $AB + AC = 2BC$.
保留源文本的换行和格式，直接输出翻译结果如下：
II. (25 points) Given that $\triangle ABC$ is a scalene triangle, points $O$ and $I$ are the c... | As shown in Figure 6, extend $A I$ to intersect $\odot O$ at point $D$, and intersect $B C$ at point $E$. Connect $C I$ and $C D$. Then
$$
\begin{array}{l}
A I=I D, \\
\angle B=\angle D, \\
\angle B A D \\
=\angle B C D .
\end{array}
$$
Since $\angle D I C=\angle C A I+\angle A C I$
$$
\begin{array}{l}
=\frac{1}{2} \a... | AB + AC = 2BC | Geometry | proof | Yes | Yes | cn_contest | false | 722,359 |
Example 1 Quadrilateral $A B C D$ is a trapezoid, $E$ is a point on the upper base $A D$, and the extension of $C E$ intersects the extension of $B A$ at point $F$. A line through point $E$ parallel to $B A$ intersects the extension of $C D$ at point $M, B M$ intersects $A D$ at point $N$. Prove:
$$
\angle A F N=\angle... | Prove as shown in Figure 1, let $M N$ intersect $E F$ at point $P$.
Notice that $M E$ $/ / B F, N E / / B C$, then by Corollary 1, dividing (1) by (2) we get $\frac{P M}{P N}=\frac{P C}{P F}$.
Thus, by the converse of Corollary 1, we know $M C / / F N$.
Extend $F N$ and $M E$ to intersect at point $Q$. Then
$F Q / / M... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,360 |
Example 2 Let the intersection point of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ be $M$. Draw a line through $M$ parallel to $AD$ intersecting $AB$, $CD$ at points $E$, $F$ respectively, and intersecting the extension of $BC$ at point $O$. $P$ is a point on the circle with center $O$ and radius $O... | Prove as shown in the figure:
2, extend $A D$ and $B O$ to intersect at point $K$.
Notice that
$$
\begin{array}{l}
O M / / K A, \\
O E / / K A,
\end{array}
$$
Then by Corollary 2, we have
$$
\begin{array}{l}
\frac{O F}{O M}=\frac{K D}{K A}, \frac{O M}{O E}=\frac{K D}{K A} . \\
\text { Hence } \frac{O F}{O M}=\frac{O M... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,361 |
3. Let $P$ be a point outside $\odot O$, and draw two tangents from $P$ to $\odot O$, touching at points $A$ and $B$. Draw a line through $A$ parallel to $PB$ intersecting $\odot O$ at point $C$, connect $PC$ intersecting $\odot O$ at point $E$, connect $AE$, and extend $AE$ to intersect $PB$ at point $K$. Prove:
$$
PE... | Given $A C / / P B$, we have
$$
\angle K P E=\angle A C E=\angle K A P \text{. }
$$
Thus, $\triangle K P E \backsim \triangle K A P \Rightarrow \frac{K P}{K A}=\frac{K E}{K P}$
$$
\Rightarrow K P^{2}=K E \cdot K A \text{. }
$$
Since $K B^{2}=K E \cdot K A$, it follows that $P K=K B$.
Also, by $A C / / P B$, we have $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,362 |
11. (20 points) The ellipse $E$ with eccentricity $e=\sqrt{\frac{2}{3}}$ has its center at the origin $O$, and its foci on the $x$-axis. A line $l$ with slope $k (k \in \mathbf{R})$ passing through the point $C(-1,0)$ intersects the ellipse at points $A$ and $B$, and satisfies
$$
\overrightarrow{B A}=(\lambda+1) \overr... | 11. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then, from $e=\sqrt{\frac{2}{3}}$, we get $a^{2}=3 b^{2}$.
Thus, the equation of the ellipse becomes $x^{2}+3 y^{2}=3 b^{2}$.
Let $l: y=k(x+1)$, and the coordinates of the two intersection points with the ellipse be $A\left(x_{1... | \frac{x^{2}}{\frac{10}{7}}+\frac{y^{2}}{\frac{10}{21}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,363 |
一、(40 points) As shown in Figure 1, the three altitudes $A D, B E, C F$ of $\triangle A B C$ intersect at point $H, P$ is any point inside $\triangle A B C$. Prove: The circumcenters $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$ of $\triangle A P D, \triangle B P E, \triangle C P F$ are collinear. | Proof 1 As shown in Figure 3, draw line $PR \perp PA$, $PS \perp PB$, $PT \perp PC$, intersecting the lines of the three sides $BC$, $CA$, $AB$ of $\triangle ABC$ at points $R$, $S$, $T$ respectively, and connect $AR$, $BS$, $CT$. It is easy to know that the midpoints of these three segments are the circumcenters $O_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,364 |
Given a positive integer $n(n>1)$, for a positive integer $m$, the set $S_{m}=\{1,2, \cdots, m n\}$. The family of sets $\mathscr{T}$ satisfies the following conditions:
(1) Each set in $\mathscr{T}$ is an $m$-element subset of $S_{m}$;
(2) Any two sets in $\mathscr{T}$ have at most one common element;
(3) Each element... | Second, the maximum value of $m$ is $2n-1$.
First, estimate the upper bound of $m$.
On one hand, consider the set
$$
\begin{aligned}
U= & \left\{\left(i,\left\{T_{j}, T_{k}\right\}\right) \mid i \in S_{m}, T_{j} \neq T_{k},\right. \\
& \left.T_{j}, T_{k} \in \mathscr{T}, i \in T_{j} \cap T_{k}\right\} .
\end{aligned}
$... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,365 |
Three. (50 points) Given that $a_{1}, a_{2}, \cdots$ is a strictly increasing sequence of positive integers, and for any positive integers $m, n, \left(a_{m}, a_{n}\right) = a_{(m, n)}$ (where $(a, b)$ denotes the greatest common divisor of integers $a$ and $b$). If there exists a smallest positive integer $k$ such tha... | Three, first prove a lemma.
Lemma If positive integers $a, b, c$ satisfy $b^{2}=a c$, then $(a, b)^{2}=(a, c) a$.
Proof Let $d=(a, c), a=a_{0} d, c=c_{0} d$, $\left(a_{0}, c_{0}\right)=1$.
Then $b^{2}=a c=d^{2} a_{0} c_{0}$.
Hence $(a, b)^{2}=\left(a^{2}, b^{2}\right)=\left(a_{0}^{2} d^{2}, d^{2} a_{0} c_{0}\right)$
$=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,366 |
Four. (50 points) In an acute triangle $\triangle ABC$, prove:
(1) $\frac{\cos B \cdot \cos C}{\cos A}+\frac{\cos C \cdot \cos A}{\cos B}+\frac{\cos A \cdot \cos B}{\cos C} \geqslant \frac{3}{2}$, and determine the condition for equality;
$$
\text { (2) } \frac{\cos \frac{B}{2} \cdot \cos \frac{C}{2}}{\cos \frac{A}{2}}... | (1) Using the cosine theorem, the original inequality can be transformed into
$$
\sum \frac{\left(c^{2}+a^{2}-b^{2}\right)\left(a^{2}+b^{2}-c^{2}\right)}{2 a^{2}\left(b^{2}+c^{2}-a^{2}\right)} \geqslant \frac{3}{2},
$$
where, “ $\sum$ ” denotes the cyclic sum.
$$
\begin{array}{l}
\text { Let } u=\left(c^{2}+a^{2}-b^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,367 |
Given that there are only three positive integers between the fractions $\frac{112}{19}$ and $\frac{112+x}{19+x}$. Find the sum of all possible integer values of $x$. | Notice that $50$ or $x<-19$.
Therefore, the three positive integers between $\frac{112+x}{19+x}$ and $\frac{112}{19}$ are $3,4,5$.
From $2<\frac{112+x}{19+x}<3$, we get $\frac{55}{2}<x<74$.
Since $x$ is an integer greater than 0, thus, $x=28,29, \cdots, 73$.
Therefore, the sum of these numbers is
$$
\frac{(28+73) \time... | 2310 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,368 |
At first 272, from the eight points consisting of the vertices and midpoints of the sides of a square, how many isosceles triangles can be formed by selecting three points?
Will the above text be translated into English, please retain the original text's line breaks and format, and output the translation result direct... | Considering that there are no equilateral triangles that meet the requirements in this problem, we classify the isosceles triangles by their vertices.
(1) Using the vertices of the square as the vertices of the isosceles triangle. As shown in Figure 3, if $A$ is the vertex,
then $\triangle A G H$, $\triangle A B D$, an... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,369 |
As shown in Figure 4, in the acute triangle $\triangle ABC$, $D$ is the midpoint of $BC$, $BE \perp AC$ at point $E$, $CF \perp AB$ at point $F$, $H$ is the orthocenter, the extension of $DH$ intersects the extensions of $AC$ and $BA$ at points $P$ and $Q$, respectively. Point $G$ is on side $BH$ such that $GH = HE$, a... | Prove as shown in Figure 4, draw $D S \perp B E$ and $D T \perp C F$ at points $S$ and $T$ respectively.
Since $B E \perp A C$ and $D$ is the midpoint of $B C$, therefore, $S$ is the midpoint of $B E$, that is,
$$
\begin{aligned}
B S & =\frac{B E}{2}=\frac{B H+H E}{2} \\
\Rightarrow S H & =B H-B S=B H-\frac{B H+H E}{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,370 |
Given $a 、 b>1,0<\lambda \leqslant 1$. Prove:
$$
\frac{1}{a^{2}-\lambda}+\frac{1}{b^{2}-\lambda} \geqslant \frac{2}{a b-\lambda} \text {. }
$$ | Proof From the given conditions and the sum of terms of an infinite geometric series and the two-variable mean inequality, we have
$$
\begin{array}{l}
\frac{1}{a^{2}-\lambda}+\frac{1}{b^{2}-\lambda}=\frac{\frac{1}{a^{2}}}{1-\frac{\lambda}{a^{2}}}+\frac{\frac{1}{b^{2}}}{1-\frac{\lambda}{b^{2}}} \\
=\left(\frac{1}{a^{2}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,371 |
4. In $\triangle A B C$, construct $D E / / B C$ intersecting $A B$ at point $D$ and $A C$ at point $E$, and $D E=\frac{2}{3} B C, B E$ intersects $C D$ at point $O$, line $A O$ intersects $B C$ and $D E$ at points $M$ and $N$ respectively, $C N$ intersects $B E$ at point $F$, connect $F M$. Prove:
$$
F M=\frac{1}{4} A... | Given $D E / / B C$, we have
$$
\frac{D N}{B M}=\frac{N E}{M C}, \frac{D N}{M C}=\frac{N E}{B M} \text {. }
$$
Multiplying and dividing the above two equations respectively, we get
$$
D N=N E, B M=M C \text {. }
$$
Extend $C N$ to intersect $A B$ at point $K$.
Then, by $D E=\frac{2}{3} B C$, we know
$$
D N=N E=\frac{... | F M=\frac{1}{4} A B | Geometry | proof | Yes | Yes | cn_contest | false | 722,372 |
5. In $\triangle A B C$, $D$, $E$, $F$ are the midpoints of sides $A B$, $B C$, $A C$ respectively, $D M$, $D N$ are the angle bisectors of $\angle C D B$ and $\angle C D A$, intersecting $B C$ at point $M$ and $A C$ at point $N$, $M N$ intersects $C D$ at point $O$, the extensions of $E O$ and $F O$ intersect $A C$ an... | Connect $E F$. Then $E F // A B$.
Let $E F$ intersect $C D$ at point $G$. Then $E G = F G$.
Since $D M$ and $D N$ are angle bisectors, then
$\frac{C M}{B M} = \frac{C D}{B D} = \frac{C D}{A D} = \frac{C N}{A N}$.
Thus, $M N // A B // E F$.
At this point, $\angle O D M = \angle M D B = \angle D M O$.
Then $M O = D O$.
B... | C D = P Q | Geometry | proof | Yes | Yes | cn_contest | false | 722,373 |
Example 1 Let $k, l$ be two given positive integers. Prove: there are infinitely many positive integers $m(m \geqslant k)$, such that $\mathrm{C}_{m}^{k}$ is coprime with $l$. ${ }^{[1]}$ | Prove that when $l=1$, for any positive integer $m$, $\left(\mathrm{C}_{m}^{k}, l\right)=1$, the proposition is obviously true.
When $l \geqslant 2$, let $l=p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{u}^{t_{u}}\left(u, t_{1}, t_{2}, \cdots\right.$, $t_{u}$ are all positive integers, $p_{1}<p_{2}<\cdots<p_{u}$ are all prime... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,374 |
Example 2 Let $s$ and $t$ be non-negative integers satisfying $s+t=n$, and $p$ be a prime number. Write $n$ in the form
$$
n=n_{k} p^{k}+n_{k-1} p^{k-1}+\cdots+n_{1} p+n_{0} \text {, }
$$
where $0 \leqslant n_{i}<p(0 \leqslant i \leqslant k)$, and similarly for $s$ and $t$. Prove:
$$
\text { (1) } \frac{\left(s_{0}+t_... | Proof (1) By Theorem 1, the power of $p$ in $n!$ is
$$
\begin{array}{l}
\sum_{j=1}^{k}\left[\frac{n}{p^{j}}\right]=\sum_{j=1}^{k} \sum_{i=j}^{k} n_{i} p^{i-j}=\sum_{i=1}^{k} \sum_{j=1}^{i} n_{i} p^{i-j} \\
=\frac{1}{p-1} \sum_{i=1}^{k} n_{i}\left(p^{i}-1\right) \\
=\frac{1}{p-1}\left(n-\sum_{i=0}^{k} n_{i}\right) .
\en... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,375 |
Example 3 For all prime numbers $p$ and all positive integers $n(n \geqslant p)$, prove: $\mathrm{C}_{n}^{p}-\left[\frac{n}{p}\right]$ is divisible by $p$. | Proof Let $n=\left(a_{s} a_{s-1} \cdots a_{1} a_{0}\right)_{p}$, where,
$$
\begin{array}{l}
a_{i} \in\{0,1, \cdots, p-1\}, i=0,1, \cdots, s, \\
a_{s} \neq 0, p=(00 \cdots 010)_{p} .
\end{array}
$$
By Theorem 2, we have
$$
\begin{array}{l}
\mathrm{C}_{n}^{p} \equiv \mathrm{C}_{a_{s}}^{0} \mathrm{C}_{a_{s-1}}^{0} \cdots... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,376 |
For $n \geqslant k \geqslant 0$, let $a_{n} 、 b_{n}$ be the number of binomial coefficients $\mathrm{C}_{n}^{k}$ that are congruent to $1 、 2$ modulo 3, respectively. Prove: for all $n$, $a_{n}>b_{n}$. | Proof: Let $k=\left(c_{s} c_{s-1} \cdots c_{0}\right)_{3}$,
$$
n=\left(b_{s} b_{s-1} \cdots b_{0}\right)_{3} \text {, }
$$
where $c_{i}, b_{i} \in\{0,1,2\}, i=0,1, \cdots, s, b_{s} \neq 0$.
By Theorem 2, we have
$$
\mathrm{C}_{n}^{k} \equiv C_{b_{s}}^{c_{s}} \mathrm{C}_{b_{s-1}}^{c_{s-1}} \cdots \mathrm{C}_{b_{0}}^{c_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,377 |
Example 5 In the expansion of $(a+b)^{n}$, how many coefficients are odd? | Proof Let $n=\left(a_{t} a_{t-1} \cdots a_{0}\right)_{2}$,
$$
m=\left(b_{t} b_{t-1} \cdots b_{0}\right)_{2} \text {, }
$$
where, $a_{i} 、 b_{i} \in\{0,1\}, i=0,1, \cdots, t, a_{t} \neq 0$.
By Theorem 3, we know
$\mathrm{C}_{n}^{m}$ is odd
$$
\begin{aligned}
\Leftrightarrow & \left(a_{i}, b_{i}\right)=(0,0),(1,0),(1,1)... | 2^{S(n)} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,378 |
Example 6 Proof: For any prime $p$ and natural number $n$ $(n \geqslant p)$, we have $p \left\lvert\, \sum_{k=0}^{\left[\frac{n}{p}\right]}(-1)^{k} \mathrm{C}_{n}^{p k}\right.$. | Proof Let $k=\left(a_{i} a_{t-1} \cdots a_{0}\right)_{p}$,
$$
n=\left(b_{t} b_{t-1} \cdots b_{0}\right)_{p} \text {, }
$$
where, $a_{i} 、 b_{i} \in\{0,1, \cdots, p-1\}, i=0,1, \cdots, t$, $b_{t} \neq 0, q=\left(b_{t} b_{t-1} \cdots b_{1}\right)_{p}$.
Then $p k=\left(a_{t} a_{t-1} \cdots a_{0} 0\right)_{p}$,
$$
n=p q+b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,379 |
Example 7 Let $p$ be an odd prime. Prove:
$$
\sum_{j=0}^{n} C_{p}^{j} C_{p+j}^{j} \equiv 2^{p}+1\left(\bmod p^{2}\right) .
$$ | $$
\begin{array}{l}
\sum_{j=0}^{D} \mathrm{C}_{p}^{j} \mathrm{C}_{p+j}^{j} \equiv 2^{p}+1\left(\bmod p^{2}\right) \\
\Leftrightarrow \sum_{j=1}^{p} \mathrm{C}_{p}^{j} \mathrm{C}_{p+j}^{j} \equiv 2^{p}\left(\bmod p^{2}\right) \\
\Leftrightarrow \sum_{j=1}^{p-1} \frac{(p+j)!}{j!j!(p-j)!}+\frac{(2 p)!}{p!p!} \\
\quad \equ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,380 |
Example 3 As shown in Figure 3, in trapezoid $A B C D$, diagonal $A C$ is equal to side $B C$, $M$ is the midpoint of base $A B$, $L$ is a point on the extension of side $A D$, and $L M$ intersects $B D$ at point $N$. Prove: $\angle A C L=\angle B C N$.
Translate the above text into English, please keep the original t... | Proof As shown in Figure 3, let $LC$ intersect $AB$ at point $E$, extend $CN$ to intersect $AB$ at point $F$, and extend $LN$ and $DC$ to intersect at point $G$.
Notice that $AM$ // $DG, BM / / DG$, then by Corollary 2, we have $\frac{AE}{AM}=\frac{DC}{DG}, \frac{BF}{BM}=\frac{DC}{DG}$. Therefore, $\frac{AE}{AM}=\frac... | null | Geometry | proof | Yes | Yes | cn_contest | false | 722,381 |
1. How many pairs $(n, r)$ are there in the array satisfying $0 \leqslant r \leqslant n \leqslant 63$ for which the binomial coefficient $\mathrm{C}_{n}^{r}$ is even (assuming $\left.\mathrm{C}_{0}^{0}=1\right) ?$ | From Example 5, we know that the number of odd numbers in $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ is $2^{S(n)},$ where $S(n)$ is the sum of the binary digits of $n$.
Since $63=2^{6}-1=(111111)_{2}$, when $0 \leqslant n \leqslant 63$, we have $0 \leqslant S(n) \leqslant 6$.
Classifying and... | 1351 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,382 |
3. Prove: For any natural number $n(n \geqslant k), \mathrm{C}_{n}^{k}$, $\mathrm{C}_{n+1}^{k}, \cdots, \mathrm{C}_{n+k}^{k}$, the greatest common divisor equals 1. | Let $\left(\mathrm{C}_{n}^{k}, \mathrm{C}_{n+1}^{k}, \cdots, \mathrm{C}_{n+k}^{k}\right)=d$, and $p \mid d$ (where $p$ is a prime number). Then $p \mid \mathrm{C}_{n}^{k}$.
By Theorem 3, performing the $p$-adic subtraction of $n$ and $k$ must result in a borrow.
Now, let $n=\left(c_{s} c_{s-1} \cdots c_{0}\right)_{p}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,384 |
4. Prove: A natural number $n$ greater than 1 is a prime number if and only if: for every suitable natural number $k(1 \leqslant k \leqslant n-1)$, the binomial coefficient $\mathrm{C}_{n}^{k}=\frac{n!}{k!(n-k!)}$ is divisible by $n$. | Prompt: By Theorem 3, the necessity holds, so we only need to prove the sufficiency.
If $n$ is not a prime number, then $n$ must have a prime factor $p < n$.
Let $n = p^s n_1 (s \geqslant 1)$, where $p$ does not divide $n_1$. Take $k = p$, then
$$
\begin{array}{l}
\mathrm{C}_{n}^{p} = \frac{n(n-1)(n-2) \cdots [n-(p-1)]... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,385 |
5. Prove: $\mathrm{C}_{n}^{m}(0 \leqslant m \leqslant n)$ are not divisible by the prime $p$ if and only if $n$ is of the form $s p^{k}-1(1 \leqslant s \leqslant p)$. | When $n=s p^{k}-1(1 \leqslant s \leqslant p)$,
$$
\begin{array}{l}
n=\left(a_{k} a_{k-1} \cdots a_{0}\right)_{p}, \\
a_{k}=s-1 \in\{0,1, \cdots, p-1\}, \\
a_{k-1}=a_{k-2}=\cdots=a_{0}=p-1 .
\end{array}
$$
Therefore, performing the $p$-ary subtraction of $n$ and $m$ will not result in borrowing.
By Theorem 3, $\mathrm{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,386 |
Example 1 As shown in Figure 1, divide each side of the equilateral $\triangle ABC$ into $n$ equal parts,
draw lines parallel to the other two sides through the division points, the resulting figure is a triangular grid array, briefly called a "triangular grid". Then, how many parallelograms are there in the triangular... | In the triangular grid, there are three sets of parallel lines, denoted as $W_{a}$, $W_{b}$, and $W_{c}$, parallel to $BC$, $CA$, and $AB$ respectively. Let $BC = CA = AB = n$. Extend $AB$ to point $B'$ and $AC$ to point $C'$ such that $BB' = CC' = 1$, and connect $B'C'$. Extend the diagonal grid lines to intersect wit... | 3 \mathrm{C}_{n+2}^{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,387 |
Example 2 Divide each side of the equilateral $\triangle A B C$ into $n$ equal parts, and draw lines parallel to the other two sides through the division points, resulting in a triangular grid array, or simply a triangular grid. How many rhombuses are there in the triangular grid?
untranslated text:
将上面的文本翻译成英文，请保留源文... | Let $n=2 m+\varepsilon(\varepsilon \in\{0,1\})$.
Using the method from Example 1, first consider the A-type rhombi formed by the intersection of $W_{b}$ and $W_{c}$.
As shown in Figure 1, take any A-type rhombus $D E I J$ with side length $k(k \leqslant m)$, where $D E, I J \in W_{b}$ and $E I, D J \in W_{c}$. Extend ... | g(n)=\frac{m}{2}\left[4 m^{2}+(6 \varepsilon+3) m+\left(3 \varepsilon^{2}+3 \varepsilon-1\right)\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,388 |
Example 3 Divide each side of the equilateral $\triangle A B C$ into $n$ equal parts, and draw lines parallel to the other two sides through the division points, resulting in a triangular grid array, or simply a triangular grid. How many triangles are there in the triangular grid?
Translate the above text into English... | As shown in Figure 1, the triangles (all equilateral) can be divided into two categories: those pointing upwards (such as $\triangle D J E$) and those pointing downwards (such as $\triangle J E I$). A downward-pointing triangle can be flipped along its top edge to form an upward-pointing triangle, and two such triangle... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,389 |
Example 4 As shown in Figure 2, given a triangular array of $\frac{n(n+1)}{2}$ points, let the number of all equilateral triangles formed by three points in the array be $f(n)$. Find $f(n)$
Figure 2 expression. ${ }^{\text {[3] }}$ | Let $f(i)(i=1,2, \cdots, n)$ denote the number of all equilateral triangles formed by the points in the first $i$ rows. It is known that
$$
f(1)=0, f(2)=1, f(3)=5 \text {. }
$$
Next, we seek the relationship between $f(n)$ and $f(n-1)$.
Clearly, $f(n)$ equals $f(n-1)$ plus the number of all equilateral triangles that ... | \frac{(n-1) n(n+1)(n+2)}{24} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,390 |
1. If $x^{2}-12 x+1=0$, then the unit digit of $x^{4}+x^{-4}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | - 1. B.
From $x^{2}-12 x+1=0$, we know $x \neq 0$.
Then $x+\frac{1}{x}=12 \Rightarrow x^{2}+x^{-2}=12^{2}-2=142$
$$
\Rightarrow x^{4}+x^{-4}=142^{2}-2 \text { . }
$$
Thus, the unit digit of $x^{4}+x^{-4}$ is $4-2=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,391 |
Example 4 In the acute triangle $\triangle ABC$, $AB > AC$, $CD$ and $BE$ are the altitudes on sides $AB$ and $AC$ respectively, $DE$ intersects the extension of $BC$ at point $T$, a perpendicular line from point $D$ to $BC$ intersects $BE$ at point $F$, and a perpendicular line from point $E$ to $BC$ intersects $CD$ a... | Prove that in Figure 4, let $CD$ and $BE$ intersect at point $H$, connect $AH$ and extend it to intersect $BC$ at point $K$, extend $DF$ to intersect $BC$ at point $M$, and extend $EG$ to intersect $BC$ at point $N$. Then
$$
DM \parallel AK \parallel EN \text{. }
$$
Since the parallel lines $DM, AK$ intercept the line... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,392 |
2. Given the quadratic function $y=(x-a)(x-b)-\frac{1}{2}$ $(a<b)$, and $x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)$ are the two roots of the equation
$$
(x-a)(x-b)-\frac{1}{2}=0
$$
The size relationship of the real numbers $a 、 b 、 x_{1} 、 x_{2}$ is ( ).
(A) $a<x_{1}<b<x_{2}$
(B) $a<x_{1}<x_{2}<b$
(C) $x_{1}<a<x_{2}<b$
(D... | 2. D.
Let $y^{\prime}=(x-a)(x-b)$, its graph is the parabola (1) in Figure 4, and its intersections with the $x$-axis have the abscissas $a$ and $b$.
The parabola $y=(x-a)(x-b)-\frac{1}{2}$ is obtained by translating parabola (1) downward by $\frac{1}{2}$ unit (as shown in parabola (2) in Figure 4), and its intersecti... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,393 |
3. Given a right triangle with a perimeter of 14, and the median to the hypotenuse is 3. Then the area of the right triangle is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 3. C.
Since the median to the hypotenuse is 3, the hypotenuse is 6. Let the lengths of the two legs be $a$ and $b$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=8, \\
a^{2}+b^{2}=36
\end{array}\right. \\
\Rightarrow 2 a b=(a+b)^{2}-\left(a^{2}+b^{2}\right)=28 .
\end{array}
$$
Thus, $S=\frac{1}{2} a b=7$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,394 |
4. The equation $|2 x-1|-a=0$ has exactly two positive solutions. Then the range of values for $a$ is ( ).
(A) $-1<a<0$
(B) $-1<a<1$
(C) $0<a<1$
(D) $\frac{1}{2}<a<1$ | 4. C.
Since the equation $|2 x-1|-a=0$ has exactly two positive solutions, we have
$$
\left\{\begin{array}{l}
a>0, \\
x=\frac{a+1}{2}>0, \\
x=\frac{1-a}{2}>0 .
\end{array}\right.
$$
Solving this, we get $0<a<1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,395 |
5. When three non-negative real numbers $x, y, z$ satisfy the equations $x+3 y+2 z=3$ and $3 x+3 y+z=4$, the minimum and maximum values of $M=3 x-2 y+4 z$ are ( ).
(A) $-\frac{1}{7}, 6$
(B) $-\frac{1}{6}, 7$
(C) $\frac{1}{5}, 8$
(D) $-\frac{1}{8}, 5$ | 5. B.
From $\left\{\begin{array}{l}x+3 y+2 z=3, \\ 3 x+3 y+z=4,\end{array}\right.$ we get
$$
\left\{\begin{array}{l}
y=\frac{5}{3}(1-x), \\
z=2 x-1 .
\end{array}\right.
$$
Substituting into the expression for $M$ yields
$$
\begin{aligned}
M=3 & -2 y+4 z \\
& =3 x-\frac{10}{3}(1-x)+4(2 x-1) \\
& =\frac{43}{3} x-\frac{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,396 |
6. As shown in Figure 1, the side lengths of the equilateral triangle and the square are both $a$. In the plane of the figure, $\triangle P A D$ is rotated counterclockwise around point $A$ until $A P$ coincides with $A B$. This process is then continued by rotating the triangle around points $B$, $C$, and $D$ respecti... | 6. C.
As shown in Figure 5, the path that point $P$ travels is composed of three circular arcs with a radius of $a$ and central angles of $210^{\circ}$, $210^{\circ}$, and $150^{\circ}$. Therefore, the total length is
$$
2 \pi a\left(\frac{210^{\circ}}{360^{\circ}} \times 2+\frac{150^{\circ}}{360^{\circ}}\right)=\frac... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,397 |
1. $y=\frac{6 x^{2}+4 x+3}{3 x^{2}+2 x+1}$ The maximum value is $\qquad$ | 2.1. $\frac{7}{2}$.
Notice that $y=2+\frac{1}{3 x^{2}+2 x+1}$.
And $y=3 x^{2}+2 x+1$ has a minimum value of $\frac{12-4}{12}=\frac{2}{3}$. Therefore, the maximum value sought is $\frac{7}{2}$. | \frac{7}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,398 |
2. Given the equation in terms of $x$
$$
a x^{2}=k x+b(a b k \neq 0)
$$
with roots $x_{1} 、 x_{2}$, and
$$
k x+b=0(k b \neq 0)
$$
with root $x_{3}$. The quantitative relationship satisfied by $x_{1} 、 x_{2} 、 x_{3}$ is $\qquad$ | 2. $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$.
From $a x^{2}-k x-b=0$, we get
$x_{1}+x_{2}=\frac{k}{a}, x_{1} x_{2}=-\frac{b}{a}$.
Also, since $x_{3}=-\frac{b}{k}$, we have
$x_{3}\left(x_{1}+x_{2}\right)=-\frac{b}{a}$.
Thus, $x_{3}\left(x_{1}+x_{2}\right)=x_{1} x_{2}$, which means
$$
\frac{1}{x_{1}}+\frac{1}{x_... | \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,399 |
3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is | 3. 12 .
Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\tria... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,400 |
4. Each bag contains 5 small balls that are identical in shape, size, and material, numbered $1 \sim 5$. Without seeing the balls, one ball is randomly drawn from the bag, the number is noted, and the ball is then returned. Another ball is then drawn from the bag. The probability that the number on the ball drawn the s... | 4. $\frac{2}{5}$.
All possible outcomes of drawing the balls twice are $5 \times 5=25$, among which, the number of cases where the number on the ball drawn the second time is less than the number on the ball drawn the first time is
$$
4+3+2+1+0=10 \text { (cases). }
$$
Therefore, the required probability is $\frac{10... | \frac{2}{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,401 |
Three, (16 points) Given the inverse proportion function $y=\frac{6}{x}$ and the linear function $y=k x-5$ both pass through the point $P(m, 3)$.
(1) Find the value of $k$;
(2) Draw a line parallel to the $y$-axis, intersecting the graph of the linear function at point $A$ and the graph of the inverse proportion functi... | (1) Substituting the coordinates of point $P(m, 3)$ into the two functions, we get
$$
\left\{\begin{array}{l}
3=\frac{6}{m}, \\
3=k m-5 .
\end{array}\right.
$$
Solving, we get $m=2, k=4$.
(2) As shown in Figure 6, draw $P E \perp A B$. To make $P A=P B$, we only need $B E=E A$.
Since the x-coordinates of $A$ and $B$ a... | a=2 \text{ or } a= \pm \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,402 |
Example 5 As shown in Figure 5, in $\triangle A B C$, $A B<A C, I$ is the incenter, $M$ is the midpoint of side
$B C$
, $P$
is a point on side $B C$,
and $A P \parallel$
$I M, Q$ is a point on side $A P$.
If
quadrilateral $I M P Q$ is a parallelogram, prove: $\triangle M P Q$ is a right triangle. | Let $B C=a, A C=b, A B=c$, extend $I Q$ to intersect $A B$ and $A C$ at points $N$ and $K$ respectively, and connect $B I$.
Since $B I$ bisects $\angle B$, we have $I N=B N$.
Similarly, $I K=K C$.
Let the incircle $\odot I$ of $\triangle A B C$ touch $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,403 |
Four. (16 points) Given the quadratic function
$$
y=x^{2}-2(m-1) x+m^{2}-2 \text {. }
$$
(1) Prove: Regardless of the value of $m$, the vertex of the quadratic function's graph always lies on the same straight line, and find the function expression of this straight line;
(2) If the quadratic function's graph intercepts... | (1) The vertex coordinates of the quadratic function's graph are $(m-1,2 m-3)$. Eliminating $m$ yields the function relationship as
$$
y=2 x-1 \text {. }
$$
Therefore, regardless of the value of $m$, the vertex of the quadratic function is always on this line.
(2) Suppose the graph of the quadratic function intersects... | y=x^{2}+3 x-\frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,404 |
Five, (17 points) Find the positive integer solutions of the equation
$$
x^{2}+6 x y-7 y^{2}=2009
$$ | Five, factoring on the left side, we get
$$
(x-y)(x+7 y)=2009 \text{. }
$$
Since $2009=7 \times 7 \times 41$, we have:
When $x-y=1,7,41,49,287,2009$, correspondingly, $x+7 y=2009,287,49,41,7,1$.
Also, since $x, y$ are positive integers, then
$$
x-y<x+7 y \text{. }
$$
Therefore, among the six relationships above, only... | (252,251),(42,35),(42,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,405 |
Six. (17 points) As shown in Figure 3, given that $AB$ is the diameter of $\odot O$, $AC$ is tangent to $\odot O$ at point $A$, connect $CO$ and extend it to intersect $\odot O$ at points $D$ and $E$, connect $BD$ and extend it to intersect side $AC$ at point $F$.
(1) Prove: $AD \cdot AC = DC \cdot EA$;
(2) If $AC = nA... | (1) As shown in Figure 7, connect $A D$ and $A E$.
Since $\angle D A C = \angle D E A$, we have
$$
\begin{array}{l}
\triangle A D C \backsim \triangle E A C \Rightarrow \frac{A D}{D C} = \frac{E A}{A C} \\
\Rightarrow A D \cdot A C = D C \cdot E A .
\end{array}
$$
(2) Since $\angle C D F = \angle 1 = \angle 2 = \angle ... | \frac{\sqrt{1 + 4 n^{2}} - 1}{2 n} | Geometry | proof | Yes | Yes | cn_contest | false | 722,406 |
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