problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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135 As shown in Figure 3, $E$ is a point on the plane of $\triangle ABC$, satisfying $\angle EBA = \angle ECA$. $F$ and $J$ are the feet of the perpendiculars from $E$ to the internal and external angle bisectors of $\angle BAC$, respectively. $D$ is the midpoint of $BC$. Prove that $D$, $F$, and $J$ are collinear. | Proof: Extend $EF$ to intersect $AB$ at point $K$ and $AC$ at point $M$, and let $\alpha$
Figure 4
$$
=\frac{1}{?} \angle A \text{. Taking point } A
$$
as the origin, the internal angle bisector of $\angle A$ as the $x$-axis,
establish the rectangular coordinate system as shown in Figure 4. Since the external angle bi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,332 |
If the number of sides $n$ of a convex polygon $A_{1} A_{2} \cdots A_{n-1} A_{n}$ circumscribed around a circle is odd, and the side lengths $a_{i}=A_{i} A_{i+1}(i=1,2, \cdots, n$, $A_{n+1}=A_{1}$ ) are all rational numbers, prove that the lengths $b_{i}=A_{i} B_{i}, c_{i}=$ $B_{1} A_{t+1}(i=1,2, \cdots, n)$ of the two... | Proof: As shown in the figure
5. By the tangent length theorem,
we know $c_{i}=b_{i+1}$, so
$c_{i}=a_{i+1}-$
$c_{i+1}$. Therefore,
$$
\begin{aligned}
c_{1} & =a_{2}-c_{2} \\
& =a_{2}-\left(a_{3}-c_{3}\right) \\
& =a_{2}-a_{3}+c_{3} \\
& =a_{2}-a_{3}+\left(a_{4}-c_{4}\right)=a_{2}-a_{3}+a_{4}-c_{4} \\
& =\cdots=a_{2}-a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,333 |
4. Given positive integers $a$, $b$, $c$, define the function $f(x, y, z)=$ $a x+b y+c z$, where $x$, $y$, $z \in \mathbf{Z}$. Find the smallest positive integer value of $f(x, y, z)$. | (提示: Let $f_{\text {min }}=a x_{0}+b y_{0}+c z_{0}=d_{0}$. Set $(a, b, c)$ $=d$, obviously $d \mid d_{0}$, thus $d \leqslant d_{0}$. On the other hand, by Bézout's theorem, there exist integers $x, y, z$ such that $a x+b y+c z=d$. Since $d_{0}$ is the smallest, hence $d_{0} \leqslant d$. Therefore, $d_{0}=d$, i.e., $f_... | (a, b, c) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,334 |
5. If $p$ and $q$ are coprime, and $p, q \in \mathbf{N}$, then there exists a smallest $m = pq - p - q + 1$, such that for all $n \geqslant m, n$ can be written as $n = px + qy$ (where $x, y$ are non-negative integers). | (Tip: It is only necessary to prove that $pq-p-q$ cannot be expressed as $px+qy$. Proof by contradiction. If $pq-p-q=ap+bq(a, b$ non-negative), then $pq=(1+a)p+(1+b)q$, therefore, $p|(1+b), q|(1+a)$. Let $1+a=a'q, 1+b=b'p(a', b' \geqslant 1)$, we get $pq=(a'+b')pq$. But $a'+b'=1$, a contradiction.) | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,335 |
Example $3 P$ is any point inside rectangle $A B C D$. Prove: that a quadrilateral can be formed with $A P, B P, C P, D P$ as sides, and the two diagonals of this quadrilateral are equal to segments $A B$ and $B C$, and are perpendicular to each other. | Analysis: To form a quadrilateral with $AP$, $BP$, $CP$, and $DP$, they need to be translated to new positions so that their ends meet. For this, $\triangle APD$ is translated along $AB$ to the position of $\triangle BQC$. Then, $PA$, $PB$, $PC$, and $PD$ become the four sides of quadrilateral $PBQC$.
Proof: As shown ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,336 |
Example 1 The function $f(x)$ is continuous on $[0,1]$, $f(0)=$ $f(1)$, and for any different $x_{1}, x_{2} \in [0,1]$, we have $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\left|x_{1}-x_{2}\right|$.
Prove: $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\frac{1}{2}$.
(1983, National High School Mathemat... | Solution: Since $f(x)$ is continuous on $[0,1]$, $f(x)$ has a maximum value and a minimum value on $[0,1]$. Let's assume the maximum and minimum values are $M=f\left(t_{1}\right), m=f\left(t_{2}\right)$. $t_{1} 、 t_{2} \in[0,1]$.
(1) When $\left|t_{1}-t_{2}\right| \leqslant \frac{1}{2}$,
$$
\left|f\left(x_{2}\right)-f\... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 714,337 |
Example 2 Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n, n \geqslant 2)$ satisfies $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$. Prove:
$$
\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n} \text {. }
$$
(1989, National High School Mathematics Competition) | Proof: Let $S_{k}=x_{1}+x_{2}+\cdots+x_{k}$.
Since $\sum_{i=1}^{n} x_{i}=0, \sum_{i=1}^{n}\left|x_{i}\right|=1$, we have,
$$
S_{n}=0 \text {, and }\left|S_{i}\right| \leqslant \frac{1}{2}, i=1,2, \cdots, n-1 \text {. }
$$
Assume $S_{0}=0$, then
$$
\begin{array}{l}
x_{i}=S_{i}-S_{i-1} \quad(1 \leqslant i \leqslant n, i... | \frac{1}{2}-\frac{1}{2 n} | Inequalities | proof | Yes | Yes | cn_contest | false | 714,338 |
Example 3 Given that for real numbers $a, b$, the inequality $a \cos x + b \cos 3x > 1$ has no solution. Prove: $|b| \leqslant 1$.
(15th All-Soviet Union Mathematical Olympiad) | Prove: Taking $x=0, \pi, \frac{\pi}{3}, \frac{2 \pi}{3}$ and substituting into $a \cos x + b \cos 3 x \leqslant 1$ yields, $a+b \leqslant 1, -a-b \leqslant 1, \frac{a}{2}-b \leqslant 1, -\frac{a}{2}+b \leqslant 1$. Then
$$
-1 \leqslant a+b \leqslant 1 \text { and } -1 \leqslant \frac{a}{2}-b \leqslant 1,
$$
i.e., $-2 ... | |b| \leqslant 1 | Inequalities | proof | Yes | Yes | cn_contest | false | 714,339 |
Example 4 Let $f(x)$ and $g(x)$ be functions defined on $[0,1]$. Prove that there exist $x_{0}$ and $y_{0}$ such that $0 \leqslant x_{0} \leqslant 1$, $0 \leqslant y_{0} \leqslant 1$, and $\left|x_{0} y_{0}-f\left(x_{0}\right)-g\left(y_{0}\right)\right| \geqslant \frac{1}{4}$.
(20th American Mathematical Competition) | Proof: If for all $x_{0}, y_{0} \in [0,1]$,
$$
\left|x_{0} y_{0}-f\left(x_{0}\right)-g\left(y_{0}\right)\right| \geqslant \frac{1}{4}
$$
does not hold, we consider the special values at the endpoints, and get
$$
\begin{aligned}
1= & \mid[1 \times 1-f(1)-g(1)]-[1 \times 0- \\
& f(1)-g(0)]-[0 \times 1-f(0)- \\
& g(1)]+[... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,340 |
Example 5 Let $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. Prove that for any integer $k \geqslant 2$, there exist $n$ integers $a_{i},\left|a_{i}\right| \leqslant k-$ $1, i=1,2, \cdots, n$, such that
$$
\left|a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right| \leqslan... | Proof: By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|\right)^{2} \\
\leqslant\left(1^{2}+1^{2}+\cdots+1^{2}\right)\left(\left|x_{1}\right|^{2}+\left|x_{2}\right|^{2}+\cdots+\left|x_{n}\right|^{2}\right) \\
=n .
\end{array}
$$
Thus, $\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,341 |
Example 6 Given that $a, b, c$ are all real numbers, and $a > 100$. Prove: There are at most two integers $x$ such that
$$
\left|a x^{2}+b x+c\right| \leqslant 50 \text{. }
$$
(1991, Jiangsu Province Mathematics Summer Camp) | Proof: Suppose there are three different integers $x_{1}, x_{2}, x_{3}$, satisfying $\left|a x^{2}+b x+c\right| \leqslant 50$. Then by the pigeonhole principle, among $x_{1}, x_{2}, x_{3}$, there must be two that are both greater than (or both less than) $-\frac{b}{2 a}$. Without loss of generality, assume $x_{2}>x_{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,342 |
Example 7 Let $a, b, x, y \in \mathbf{R}$, and $x^{2}+y^{2}=1$. Prove: $\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \geqslant|a+b|$. | Proof: Constructing Complex Numbers.
Let $z_{1}=a x+b y \mathrm{i}, z_{2}=b x+a y \mathrm{i}$, then
$$
\begin{array}{l}
\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \\
=\left|z_{1}\right|+\left|z_{2}\right| \geqslant\left|z_{1}+z_{2}\right| \\
=|(a+b) x+(a+b) y \mathrm{i}| \\
=|a+b| \cdot|x+y \mathrm{i... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,343 |
Example 8 Given $f(x)=c_{0} z^{n}+c_{1} z^{n-1}+\cdots+$ $c_{n-1} z+c_{n}$ is an $n$-degree polynomial with complex coefficients. Prove:
There must exist a complex number $z_{0},\left|z_{0}\right| \leqslant 1$, and satisfying
$$
\left|f\left(z_{0}\right)\right| \geqslant\left|c_{0}\right|+\left|c_{n}\right| .
$$
(1994,... | Proof: For a given polynomial $f(z)$, the argument of the constant $c_{n}$ is determined (when $c_{n}=0$, the argument can be arbitrarily chosen), so, we can take a complex number $\mu$ with the same argument as $c_{n}$, such that $|\mu|=\left|c_{0}\right|+\left|c_{n}\right|$.
Construct the polynomial $f(z)-\mu$
$$
=c_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,344 |
Example 9 Let the function $f(x)$ satisfy $|f(m+n)-f(m)| \leqslant \frac{m}{n}$ for all rational numbers $m$ and $n$. Prove that for all positive integers $k$, we have
$$
\sum_{i=1}^{k}\left|f\left(2^{k}\right)-f\left(2^{i}\right)\right| \leqslant \frac{k(k-1)}{2} .
$$
(12th Korean Mathematical Olympiad) | Prove: When $k=1$, the left side $=$ the right side $=0$.
When $k=2$, the left side $=|f(4)-f(2)|$ $=|f(2+2)-f(2)| \leqslant 1=$ the right side.
Assume that when $k=n$, the conclusion holds, i.e.,
$$
\begin{array}{l}
\left|f\left(2^{n}\right)-f(2)\right|+\left|f\left(2^{n}\right)-f\left(2^{2}\right)\right|+\cdots+ \\
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,345 |
Example 1 Given two sets of real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\}$. If the mapping $f$ from $A$ to $B$ makes each element in $B$ have a preimage, and $f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant f\left(a_{100}\ri... | Solution: Let $b_{1}, b_{2}, \cdots, b_{50}$ be arranged in ascending order as $c_{1}<c_{2}<\cdots<c_{50}$ (since the elements of the set are distinct, there are no equal cases).
Divide the elements $a_{1}, a_{2}, \cdots, a_{100}$ of $A$ into 50 groups, each corresponding to the elements $c_{1}, c_{2}, \cdots, c_{50}$... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,346 |
Example 4 In hexagon $A B C D E F$, $A B / / E D$, $B C / / F E$, $C D / / A F$, and the difference between opposite sides $B C-F E$ $=D E-B A=F A-D C>0$. Prove: All interior angles of hexagon $A B C D E F$ are equal. | Analysis: The sum of the interior angles of a hexagon is $720^{\circ}$. To prove that all interior angles are equal, it is to prove that each interior angle is $120^{\circ}$. Therefore, the problem is to prove that each interior angle of the hexagon is $120^{\circ}$ under the condition that the opposite sides are paral... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,347 |
Example 2 Someone wants to go upstairs, and this person can go up 1 step or 2 steps at a time. If one floor has 18 steps, how many different ways can he go up one floor? | Solution 1: This person can climb one floor in a maximum of 18 steps and a minimum of 9 steps. Here, $a_{18}, a_{17}, a_{16}, \cdots, a_{9}$ represent the number of ways this person can climb the stairs in 18 steps, 17 steps, 16 steps, $\cdots, 9$ steps, respectively (for any two consecutive step counts, the latter tak... | 4181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,348 |
Example 3 Before the World Cup, a coach of a certain country, in order to examine the players $A_{1}, A_{2}, \cdots, A_{7}$, plans to have them all play in three training matches (each 90 minutes). Assuming that at any moment during the matches, only one of these players is on the field, and the total playing time (in ... | Let $A_{i}(i=1,2, \cdots, 7)$ be the total time in minutes that each player is on the field. According to the problem, we can set
$$
a_{i}=7 k_{i}(i=1,2,3,4), a_{i}=13 k_{i}(i=5,6,7),
$$
where $k_{i}(i=1,2, \cdots, 7) \in \mathbf{Z}_{+}$.
$$
\text { Let } \sum_{i=1}^{4} k_{i}=m, \sum_{i=5}^{7} k_{i}=n \text {, where }... | 42244 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,349 |
Example 4 consists of forming $n$-digit numbers using the digits $1,2,3$, and in this $n$-digit number, each of $1,2,3$ must appear at least once. How many such $n$-digit numbers are there? | Let $U$ be the set of $n$-digit numbers composed of $1, 2, 3$, $A_{1}$ be the set of $n$-digit numbers in $U$ that do not contain the digit 1, $A_{2}$ be the set of $n$-digit numbers in $U$ that do not contain the digit 2, and $A_{3}$ be the set of $n$-digit numbers in $U$ that do not contain the digit 3. Then,
$$
\beg... | 3^{n}-3 \times 2^{n}+3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,350 |
Example 5: A total of 240 students participating in a large group performance stand in a row facing the coach, and they report numbers in sequence from left to right as 1, 2, 3, 4, 5, ... The coach asks all students to remember their reported numbers and perform the following actions: First, all students who reported n... | Solution: (1) Let $U=\{1,2, \cdots, 240\}, A_{i}$ represent the set of all multiples of $i$ in $U$. Then
$|U|=240 ;\left|A_{3}\right|=\left[\frac{240}{3}\right]=80$,
$\left|A_{5}\right|=\left[\frac{240}{5}\right]=48,\left|A_{7}\right|=\left[\frac{240}{7}\right]=34$;
$\left|A_{15}\right|=\left[\frac{240}{15}\right]=16,\... | 118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,351 |
Example 1 If the parallel lines $E F$ and $M N$ intersect with the intersecting lines $A B$ and $C D$ to form the figure shown in Figure 1, then the number of consecutive interior angles obtained is ( ).
(A) 4
(B) 8
(C) 12
(D) 16 | The solution provided is (Figure 2), which decomposes the given figure into 8 basic "three-line eight-angle" shapes. Knowing that each basic shape has 2 consecutive interior angles, a total of 16 consecutive interior angles are obtained ${ }^{[1]}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,352 |
2003 CMO Problem 6:
Let $a, b, c, d$ be positive real numbers satisfying $ab + cd = 1$. Points $P_{i}(x_{i}, y_{i}) (i=1,2,3,4)$ are four points on the unit circle centered at the origin. Prove that:
$$
\begin{array}{l}
\left(a y_{1} + b y_{2} + c y_{3} + d y_{4}\right)^{2} + \left(a x_{4} + b x_{3} + c x_{2} + d x_{1}... | Proof: Let $\alpha=a y_{1}+b y_{2}+c y_{3}+d y_{4}$,
$$
\beta=a x_{4}+b x_{3}+c x_{2}+d x_{1} \text {. }
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\left(\sqrt{a d} y_{1}\right)^{2}+\left(\sqrt{b c} y_{2}\right)^{2}+\left(\sqrt{b c} y_{3}\right)^{2}+\left(\sqrt{a d} y_{4}\right)^{2}\right]... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,353 |
1. Given real numbers $a \neq b$, and satisfy
$$
(a+1)^{2}=3-3(a+1), 3(b+1)=3-(b+1)^{2} \text {. }
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13 | $-、 1 . B$.
Since $a, b$ are the roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Rearranging this equation, we get
$$
x^{2}+5 x+1=0 \text {. }
$$
Since $\Delta=25-4>0$, we have
$$
a+b=-5, a b=1 \text {. }
$$
Therefore, $a, b$ are both negative. Thus,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \s... | -23 | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,354 |
2. If the lengths of the two legs of a right triangle are $a, b$, and the length of the hypotenuse is $c$, and the altitude to the hypotenuse is $h$, then ( ).
(A) $a b=h^{2}$
(B) $a^{2}+b^{2}=2 h^{2}$
(C) $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{h^{2}}$
(D) $\frac{1}{a}+\frac{1}{b}=\frac{1}{h}$ | 2.C.
It is known that $ab > h^2$, $a^2 + b^2 > h^2 + h^2 = 2h^2$, $\frac{1}{2}(a+b)h > \frac{1}{2}ch = \frac{1}{2}ab$, therefore, (A), (B), and (D) are incorrect.
Furthermore, from $\frac{1}{2} \sqrt{a^2 + b^2} h = \frac{1}{2} ab$, we can derive $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{h^2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,355 |
3. The vertex of the parabola $y=a x^{2}+b x+c$ is $(4,-11)$, and the x-coordinates of its two intersections with the x-axis are one positive and one negative. Then, among $a$, $b$, and $c$, the positive number(s) is/are ( ).
(A) only $a$
(B) only $b$
(C) only $c$
(D) $a$ and $b$ | 3.A.
Given the vertex is $(4,-11)$, and the parabola intersects the $x$-axis at two points, we know $a > 0$. Let the $x$-coordinates of the two intersection points of the parabola with the $x$-axis be $x_{1}$ and $x_{2}$, i.e., $x_{1}$ and $x_{2}$ are the two roots of the equation $a x^{2}+b x+c=0$.
From the given $x... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,356 |
4. As shown in Figure 1, in $\triangle A B C$, $D E / / A B / /$ $F G$, and the ratio of the distances from $F G$ to $D E$ and $A B$ is $1: 2$. If the area of $\triangle A B C$ is 32, and the area of $\triangle C D E$ is 2, then the area of $\triangle C F G$ is ( ).
(A) 6
(B) 8
(C) 10
(D) 12 | 4.B.
From $D E / / A B / / F G$, we know,
$$
\frac{C D}{C A}=\sqrt{\frac{S_{\triangle C D E}}{S_{\triangle C B B}}}=\sqrt{\frac{2}{32}}=\frac{1}{4} \text {. }
$$
Also, $\frac{F D}{F A}=\frac{1}{2}$, so, $F D=D C$. Therefore,
$$
\frac{S_{\triangle C D E}}{S_{\triangle C F C}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,357 |
Example 5 Given that $D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $BD = CE$. Prove: $AB + AC > AD + AE$.
---
The translation maintains the original text's line breaks and format. | Analysis: To prove $A B+A C>A D+A E$, we can try to concentrate these four line segments.
For this purpose, translate $\triangle A C E$ to the position of $\triangle P D B$ (as shown in Figure 6), and the various relationships will naturally become apparent.
Proof: Draw a line through point $D$ parallel to $C A$ and a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,358 |
5. If $x$ and $y$ are non-zero real numbers, and satisfy $|x|+y=3$ and $|x| y+x^{3}=0$, then, $x+y$ equals ( ).
(A) 3
(B) $\sqrt{13}$
(C) $\frac{1-\sqrt{13}}{2}$
(D) $4-\sqrt{13}$ | 5.D.
Substitute $y=3-|x|$ into $|x| y+x^{3}=0$, we get $x^{3}-x^{2}+3|x|=0$.
(1) When $x>0$, $x^{3}-x^{2}+3 x=0$, the equation $x^{2}-x+3=0$ has no real roots.
(2) When $x<0$, $x^{3}-x^{2}-3 x=0$, solving gives $x=\frac{1-\sqrt{13}}{2}$.
Thus, $y=3-|x|=3+\frac{1-\sqrt{13}}{2}=\frac{7-\sqrt{13}}{2}$. Therefore, $x+y=4-... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,359 |
6. As shown in Figure 2, in $\triangle A B C$, $A B=A C, A D=A E$. $\angle B A D=60^{\circ}$. Then $\angle E D C=$ degrees. | II. 6. $30^{\circ}$.
Let $\angle C A D=2 \alpha$.
From $A B=A C$, we know
$$
\begin{array}{l}
\angle B=\frac{1}{2}\left(180^{\circ}-60^{\circ}-2 \alpha\right)=60^{\circ}-\alpha, \\
\angle A D B=180^{\circ}-\angle B-60^{\circ}=60^{\circ}+\alpha .
\end{array}
$$
From $A D=A E$, we know
$$
\angle A D E=90^{\circ}-\alpha ... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,360 |
8. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2$, $a x+b y=5$. Then the value of $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)$ is $\qquad$ . | 8. -5 .
Given $a+b=x+y=2$, we have
$$
(a+b)(x+y)=a x+b y+a y+b x=4 \text{. }
$$
Since $a x+b y=5$, then,
$$
\begin{array}{l}
a y+b x=-1 . \\
\text{ Thus, }\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right) \\
=(a y+b x)(a x+b y)=-5 .
\end{array}
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,362 |
9. As shown in Figure 4, in trapezoid $A B C D$, $A D / / B C$ $(B C > A D)$, $\angle D=$ $90^{\circ}$, $B C=C D=12$, $\angle A B E=45^{\circ}$. If $A E=$ 10, then the length of $C E$ is $\qquad$. | 9.4 or 6.
As shown in Figure 4, extend $DA$ to point $M$ such that $BM \perp BE$. Draw $BG \perp AM$ with $G$ as the foot of the perpendicular.
It is easy to see that quadrilateral $BCDG$ is a square.
Therefore, $BC = BG$.
Also, $\angle CBE = \angle GBM$, so
Rt $\triangle BEC \cong$ Rt $\triangle BMG$.
Thus, $BM = BE$... | 4 \text{ or } 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,363 |
10. Real numbers $x, y, z$ satisfy $x+y+z=5, xy+yz+zx=$ 3. Then the maximum value of $z$ is $\qquad$. | 10. $\frac{13}{3}$.
Since $x+y=5-z$, $xy=3-z(x+y)=3-z(5-z)=z^{2}-5z+3$, therefore, $x, y$ are the two real roots of the quadratic equation in $t$
$$
t^{2}-(5-z)t+z^{2}-5z+3=0
$$
Also,
$$
\begin{array}{l}
\text { and } \Delta=(5-z)^{2}-4\left(z^{2}-5z+3\right) \geqslant 0, \text { then } \\
3z^{2}-10z-13 \leqslant 0,... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,364 |
11. Through experimental research, experts have found that the attention index of junior high school students changes with the duration of the teacher's lecture. At the beginning of the lecture, students' interest surges, then there is a period during which their interest remains stable, followed by a period of dispers... | Three, 11. (1) When $0 \leqslant x \leqslant 10$, $y=-\frac{1}{5} x^{2}+\frac{24}{5} x+20$.
(2) When $20 \leqslant x \leqslant 40$, $y=-\frac{7}{5} x+76$.
When $0 \leqslant x \leqslant 10$, let $y=36$, solving gives $x=4, x=20$ (discard);
When $20 \leqslant x \leqslant 40$, let $y=36$, solving gives
$$
x=\frac{200}{7}... | 24 \frac{4}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,365 |
13. Given that $a, b$ are real numbers, the system of equations about $x, y$
$$
\left\{\begin{array}{l}
y=x^{3}-a x^{2}-b x, \\
y=a x+b
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the relationship that $a, b$ must satisfy. | 13. From the given, we have $y=x^{3}-x y$, which means $(x+1) y=x^{3}$.
It is clear that $x+1 \neq 0$, thus,
$$
y=\frac{x^{3}}{x+1}=x^{2}-x+1-\frac{1}{x+1} \text {. }
$$
Since $x, y$ are both integers, we have,
$$
x+1= \pm 1 \text {. }
$$
Therefore, $x=-2$ or $x=0$.
Thus, $y=8$ or $y=0$.
Hence, $x=-2, y=8$ or $x=0, y... | 2a - b + 8 = 0 \text{ or } b = 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,367 |
14. Let points $E$, $F$, $G$, $H$ be on the sides $AB$, $BC$, $CD$, $DA$ of a quadrilateral $ABCD$ with area 1, respectively, and
$$
\frac{A E}{E B}=\frac{B F}{F C}=\frac{C G}{G D}=\frac{D H}{H A}=k \text { ( } k \text { is a positive number). }
$$
Find the area of quadrilateral $E F G H$. | 14. As shown in Figure 7, connect $A C$. Given $\frac{C G}{G D}=\frac{D H}{H A}=k$, then
$$
\begin{array}{l}
\frac{D H}{D A}=\frac{k}{k+1}, \\
\frac{D P}{D A}=\frac{D G}{D C}=\frac{1}{k+1} .
\end{array}
$$
Therefore, $S_{\triangle D H C}=\frac{k}{(k+1)^{2}} \cdot S_{\triangle D A C}$.
Similarly, $S_{\triangle B E F}=\... | \frac{k^2 + 1}{(k + 1)^2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,368 |
Example 6 As shown in the figure,
$7, \triangle ABC$ is an equilateral
triangle, $\triangle A_{1} B_{1} C_{1}$
has sides $A_{1} B_{1}, B_{1} C_{1}$, and
$C_{1} A_{1}$ intersecting the sides of $\triangle ABC$
at $C_{2}$, $C_{3}$, $A_{2}$, $A_{3}$, $B_{2}$, and $B_{3}$. Given that $A_{2} C_{3}=C_{2} B_{3}=$ $B_{2} A_{3}... | Analysis: To prove $A_{1} B_{1} \perp A_{1} C_{1}$, it is only necessary to prove $\angle B_{1} A_{1} C_{1}=90^{\circ}$. Given that $\left(C_{2} C_{3}\right)^{2}+\left(B_{2} B_{3}\right)^{2}=\left(A_{2} A_{3}\right)^{2}$, this resembles the Pythagorean theorem, but $C_{2} C_{3}, B_{2} B_{3}, A_{2} A_{3}$ are not the th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,369 |
$1 . a$ is a non-zero real number, $x$ is a real number. Let proposition $M: x \in$ $\{-a, a\}$, let proposition $N: \sqrt{x^{2}}=a$ has a solution. Then $M$ is ( ) of $N$.
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient... | $-、 1 . B$.
Let $a$ be a positive number. When $x=-a$, $x \in \{-a, a\}$, M is true. But $\sqrt{(-a)^{2}}=a \neq -a$, N is not true. Therefore, M is not a sufficient condition for N.
If N is true, $\sqrt{x^{2}}=a$, it is clear that $a$ should be a non-negative number. But $a$ is not 0, so $a$ is a positive number. Th... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,370 |
3. Given that $f(x)$ is a function defined on the set of real numbers, and $f(x+5)=-f(x)$, when $x \in(5,10)$, $f(x)=\frac{1}{x}$. Then the value of $f(2003)$ is ( ).
(A) $-\frac{1}{8}$
(B) $\frac{1}{5}$
(C) $\frac{1}{3}$
(D) $-\frac{1}{3}$ | 3.A.
Since $f(x+10)=-f(x+5)=f(x)$, we have
$$
f(2003)=f(200 \times 10+3)=f(3)=-f(8)=-\frac{1}{8} .
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,372 |
4. The smallest real value of $x$ that satisfies the inequality $9^{x}-2 \times 3^{x}-3 \geqslant 0$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 3 | 4.C.
Let $3^{x}=t$. The original inequality can be transformed into $t^{2}-2 t-3 \geqslant 0$.
Solving this, we get $t \in(-\infty,-1] \cup[3,+\infty)$.
Since $t$ is positive, we have $3^{x}=t \geqslant 3$. Therefore, $x \geqslant 1$.
Thus, the smallest real value of $x$ that satisfies $9^{x}-2 \times 3^{x}-3 \geqslan... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,373 |
5. In $\triangle A B C$, it is known that $A B=5, A C \leqslant 3, B C \geqslant 7$. Then the minimum value of $\angle C A B$ is ( ).
(A) $\frac{\pi}{2}$
(B) $\frac{2 \pi}{3}$
(C) $\frac{3 \pi}{4}$
(D) $\frac{5 \pi}{6}$ | 5.B.
From the cosine rule, we have
$$
\begin{array}{l}
\cos \angle CAB = \frac{AC^2 + 5^2 - BC^2}{2 \times 5 \times AC} \leq \frac{3^2 + 25 - 7^2}{10 \times AC} \\
= -\frac{15}{10 \times AC} \leq -\frac{15}{10 \times 3} = -\frac{1}{2}.
\end{array}
$$
Since the cosine function is decreasing in the interval $(0, \pi)$,... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,374 |
6. If the real number $x$ satisfies $||x^{2}-4 x-5|-6|=a$ and there are exactly 6 values of $x$, then the range of the real number $a$ is ( ).
(A) $-6 \leqslant a \leqslant 0$
(B) $0<a \leqslant 3$
(C) $3<a<6$
(D) $6 \leqslant a<9$ | 6.C.
Sketch the graph of the function $y=|| x^{2}-4 x-5|-6|$, and observe the number of intersection points between the line $y=a$ and the graph to determine the range of the real number $a$ is $3<a<6$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,375 |
1. In square $A B C D$, $M$ is the midpoint of side $B C$, and $N$ is the midpoint of side $C D$. Then the value of $\sin \angle M A N$ is $\qquad$ | $$
\text { II. } 1 . \frac{3}{5} \text {. }
$$
As shown in Figure 4, let the side length of the square be $S_{\triangle A R M}=S_{\triangle N N N}=1, S_{\triangle C U N}=$ 0.5. Therefore, $S_{\angle M M N}=4-1-1-$ $0.5=1.5$. Also, $A M=A N=\sqrt{5}$, $S_{\triangle M M N}=\frac{1}{2} A M \cdot A N \sin \angle M A N$, s... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,376 |
2. Let $\min \{a, b, c\}$ be the minimum of $a, b, c$. If $x, y$ are any positive real numbers, then the maximum value of $M=\min \left\{x, \frac{1}{y}, y+\frac{1}{x}\right\}$ is $\qquad$. | 2. $\sqrt{2}$.
Given $-\frac{1}{y} \geqslant M, x \geqslant M, y+\frac{1}{x} \geqslant M$, then $y \leqslant \frac{1}{M}, \frac{1}{x} \leqslant \frac{1}{M}, M \leqslant y+\frac{1}{x} \leqslant \frac{2}{M}$.
Thus, $M^{2} \leqslant 2, M \leqslant \sqrt{2}$.
When $x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$, then $\frac{1}{y}=\sqr... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,377 |
3. Given the function $f(x)=\frac{2+x}{1+x}$. Let
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(1000)=m, \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{1000}\right)=n .
\end{array}
$$
Then the value of $m+n$ is $\qquad$ | 3. 2998.5
It is clear that $x \neq -1$. Since
$$
f(x)+f\left(\frac{1}{x}\right)=\frac{2+x}{1+x}+\frac{2+\frac{1}{x}}{1+\frac{1}{x}}=3, f(1)=\frac{3}{2} \text{, }
$$
therefore, $m+n=3 \times 999+\frac{3}{2}=2998.5$. | 2998.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,378 |
4. If $x_{1}$ and $x_{2}$ are the two distinct real roots of the equation $\sqrt{x+4}+\sqrt{9-3 x}=5$, then, the value of $x_{1}^{2}+x_{2}^{2}$ is $\qquad$ . | 4. $14 \frac{1}{16}$.
The original equation, when squared and simplified on both sides, yields
$$
\sqrt{x+4} \cdot \sqrt{9-3 x}=x+6 \text {, }
$$
Squaring both sides again gives
$$
9 x+36-3 x^{2}-12 x=x^{2}+12 x+36 \text {, }
$$
Combining terms results in $4 x^{2}+15 x=0$, or $x(4 x+15)=0$.
Solving yields $x_{1}=0, ... | 14 \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,379 |
Example 7 Consider $\triangle A B C$ and $\triangle P Q R$ as shown in Figure 9. In $\triangle A B C$, $\angle A D B=\angle B D C=\angle C D A=120^{\circ}$. Prove: $x=u+v+w$.
(3rd American High School Mathematics Examination) | Analysis: The problem states that in $\triangle ABC$, $BC=a, CA=b, AB=c, D$ is a point inside the shape, precisely satisfying $\angle ADB=\angle BDC=\angle CDA=120^{\circ}$, and $AD=u, BD=v, CD=w$. Prove that if there is a point $O$ inside an equilateral $\triangle PQR$ with side length $x$, such that $OP=a, OQ=b, OR=c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,380 |
5. In $\triangle A B C$, $\angle A B C=36^{\circ}, \angle A C B=42^{\circ}$, take a point $D$ on side $B C$ such that $B D$ is exactly equal to the radius of the circumcircle of $\triangle A B C$. Then $\angle D A C=$ $\qquad$ degrees. | 5. $54^{\circ}$.
Let $O$ be the circumcenter of $\triangle ABC$. It is easy to see that $\angle BAC = 102^{\circ}$ is an obtuse angle. Therefore, $O$ is outside $\triangle ABC$. Connect $OA$ and let it intersect side $BC$ at $D_1$. We will now prove that $D_1$ coincides with $D$.
From Figure 5, we can see that,
$$
\be... | 54^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,381 |
6. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=4 a_{n}+4 \sqrt{a_{n}}+1$, $n=1,2, \cdots$. Then the general term formula of the sequence is $\qquad$ . | $$
6.2^{2 n}-2^{n+1}+1
$$
From the given condition, we have $\sqrt{a_{n+1}}=2 \sqrt{a_{n}}+1$, so,
$$
\sqrt{a_{n+1}}+1=2\left(\sqrt{a_{n}}+1\right) \text {. }
$$
Therefore, $\left\{\sqrt{a_{n+1}}+1\right\}$ is a geometric sequence with the first term 2 and the common ratio 2. $\sqrt{a_{n}}+1=2^{n}$, which means $\sqr... | 2^{2 n}-2^{n+1}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,382 |
7. $n$ is a positive integer, $f(n)=\sin \frac{n \pi}{2}$. Then
$$
f(1991)+f(1992)+\cdots+f(2003)=
$$
$\qquad$ | 7. -1 .
It is easy to know that $f(1991)=-1, f(1992)=0, f(1993)=1$, $f(1994)=0, \cdots, f(2003)=-1$. Therefore, $f(1991)+f(1992)+\cdots+f(2003)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,383 |
8. If the three sides of a triangle form a geometric sequence, then the range of the common ratio $q$ is $\qquad$ . | 8. $\frac{\sqrt{5}-1}{2}$ $0, q \geqslant 1$. Then $a + aq > aq^2$, which means $q^2 - q - 1 < 0, 0 < q < 1$, and $q^2 + q - 1 > 0$. Solving this, we get
$$
\frac{\sqrt{5}-1}{2} < q < 1 \text{. }
$$
In summary, $\frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2}$. | \frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,384 |
1. Given that $x$ and $y$ are positive integers, and satisfy $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Then $x^{2} + y^{2} =$ $\qquad$ . | $-、 1.146$.
Let $a=x+y, b=x y, x y+x+y=a+b=71$, $x^{2} y+x y^{2}=x y(x+y)=a b=880$.
Therefore, $a, b$ are the two roots of $t^{2}-71 t+880=0$.
Solving, we get $a=x+y, b=x y$ equal to 16 and 55, respectively.
If $x+y=55, x y=16$, there are clearly no positive integer solutions. Therefore, only $x+y=16, x y=55$.
Thus, $x... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,385 |
3. The function defined on the set of positive integers is
$$
f(x)=\left\{\begin{array}{ll}
3 x-1, & x \text { is odd, } \\
\frac{x}{2}, & x \text { is even. }
\end{array}\right.
$$
Let $x_{1}=12, x_{n+1}=f\left(x_{n}\right), n \in \mathbf{N}$, then the number of elements in the set $\left\{x \mid x=x_{n}\right.$, $n ... | 3. 7 .
From $x_{1}=12$ we get $x_{2}=6$, then $x_{3}=3, x_{4}=8, x_{5}=4$, $x_{6}=2, x_{7}=1, x_{8}=2, x_{9}=1, \cdots$, and thereafter it is a cycle of $2,1,2,1$, $\cdots$. Therefore, $x_{n}$ takes a total of 7 different values, i.e., the set $|x| x$ $\left.=x_{n}, n \in \mathbf{N}\right\}$ contains 7 elements. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,387 |
4. Given a geometric sequence $\left\{b_{n}\right\}$ with all terms being positive and an arithmetic sequence $\left\{a_{n}\right\}$, satisfying $b_{3}-b_{1}=9, b_{5}-b_{3}=36$, and $b_{1}=$ $a_{1}, b_{2}=a_{3}$. Let the sum of the first 6 terms of the geometric sequence be $G_{6}$, and the sum of the first 12 terms of... | 4. 324.
Let the common ratio be $q(q>0)$, then
$$
\left\{\begin{array}{l}
b_{1} q^{2}-b_{1}=9, \\
b_{1} q^{4}-b_{1} q^{2}=36,
\end{array}\right.
$$
which is equivalent to $\left\{\begin{array}{l}b_{1} q^{2}-b_{1}=9, \\ q^{2}\left(b_{1} q^{2}-b_{1}\right)=36 .\end{array}\right.$
Substituting (1) into (2) gives $q^{2}=... | 324 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,388 |
5. $M=\{-2,0,1\}, N=\{1,2,3,4,5\}$. Mapping $f: M$ $\rightarrow N$, such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number. Then the number of different mappings is $\qquad$ . | 5. 45 .
Given the mapping $f: M \rightarrow N$ where the independent variable is taken from $M=\{-2,0, 1\}$ and the values of $f(x)$ are taken from $N=\{1,2,3,4,5\}$. The problem is to find the number of different mappings such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number.
And $x+f(x)+x f(x)=x+x f(x)+1+f(x... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,389 |
```
(15 points) If $a, b, c$ are positive numbers, prove:
\[
\begin{array}{l}
\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \\
\geqslant \frac{a+b+c}{3} .
\end{array}
\]
``` | Because $a^{2}+a b+b^{2} \geqslant 3 a b$, we have
$$
\begin{array}{l}
\frac{a^{3}}{a^{2}+a b+b^{2}}=\frac{a^{3}+a^{2} b+a b^{2}-\left(a^{2} b+a b^{2}\right)}{a^{2}+a b+b^{2}} \\
=a-\frac{a b(a+b)}{a^{2}+a b+b^{2}} \geqslant a-\frac{a+b}{3} .
\end{array}
$$
Similarly, $\frac{b^{3}}{b^{2}+b c+c^{2}} \geqslant b-\frac{b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,390 |
Example 1 The relationship between the sets $M=\{u \mid u=12 m+8 n+4 l$, $m, n, l \in \mathbf{Z}\}$ and $N=\{u \mid u=20 p+16 q+12 r$, $p, q, r \in \mathbf{Z}\}$ is ( ).
(A) $M=N$
(B) $M \nsubseteq N, N \nsubseteq M$
(C) $M \subset N$
(D) $M \supset N$
(1989, National High School Mathematics Competition) | Explanation: Starting from the expressions of elements in two sets.
Since $12 m+8 n+4 l=4(3 m+2 n+l)$,
$$
20 p+16 q+12 r=4(5 p+4 q+3 r) \text {, }
$$
and $(3,2,1)=1,(5,4,3)=1$, by Bézout's theorem, $3 m+2 n+l$ and $5 p+4 q+3 r$ can both represent all integers. Therefore, $M=N=\{k \mid k=4 l, l \in \mathbf{Z}\}$.
Hence... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,391 |
Three. (15 points) As shown in Figure 2, a moving point $P$ is on the arc of a segment with $A B=1$ as the chord, and the segment contains a bow angle of $\frac{2 \pi}{3}$ (including the endpoints). Let $A P=x$, $B P=y$, and try to determine the maximum and minimum values of $k=3 x+2 y$.
---
The translation maintains... | In $\triangle A P B$, by the cosine rule we have
$$
x^{2}+y^{2}-2 x y \cos \frac{2 \pi}{3}=1 \text {, }
$$
which simplifies to $x^{2}+y^{2}+x y=1$.
Given $k=3 x+2 y$, we have
$$
y=\frac{k-3 x}{2} \text {. }
$$
Substituting (2) into (1) yields $7 x^{2}-4 k x+k^{2}-4=0$.
Since $x$ is a positive real number, this equati... | 2 \leqslant k \leqslant \frac{2 \sqrt{21}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,392 |
Four, (15 points) Given two circles with radii $R$ and $r$ that are externally tangent at point $P$, the distance from point $P$ to one of their external common tangents is equal to $d$. Prove: $\frac{1}{R}+\frac{1}{r}=\frac{2}{d}$.
| As shown in Figure 6, let the circle $\odot O$ with radius $R$ be externally tangent to the circle $\odot O_{1}$ with radius $r$ at point $P$. $AB$ is a common external tangent to the two circles, and $PC \perp AB$ at point $C$. Connect $OP$, $O_{1}P$. Then $O, P, O_{1}$ are collinear. Extend $BO_{1}$ to intersect $\od... | \frac{1}{R} + \frac{1}{r} = \frac{2}{d} | Geometry | proof | Yes | Yes | cn_contest | false | 714,393 |
Five. (15 points) Suppose there are $n$ distinct positive integers $a_{1}, a_{2}$, $\cdots, a_{n}$. Then among the integers of the form $t_{1} a_{1}+t_{2} a_{2}+\cdots+t_{n} a_{n}$ (where $t_{1}$ takes 1 or $-1, i=1,2, \cdots, n)$, there exist $\frac{n^{2}+n+2}{2}$ distinct integers that are either all odd or all even. | Let's assume $a_{1}<a_{2}<\cdots<a_{n}$, then $a=-a_{1}-a_{2}-\cdots-a_{n}$ is the smallest integer of the form $t_{1} a_{1}+t_{2} a_{2}+\cdots+t_{n} a_{n}$. $a+2 a_{1}=a_{1}-a_{2}-\cdots-a_{n}$ is also an integer of the form $t_{1} a_{1}+t_{2} a_{2}+\cdots+t_{n} a_{n}$. Generally, $a+2 a_{1}+2 a_{2}$ is also an intege... | \frac{n^{2}+n+2}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,394 |
3. Given that $\alpha$ and $\beta$ are both acute angles, $\cos (\alpha+\beta)=-\frac{4}{5}$. If we set $\sin \beta=x, \cos \alpha=y$, then the functional relationship between $y$ and $x$ is ( ).
(A) $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x(0<x<1)$
(B) $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x\left(\frac{4}{5}<x... | 3.B.
From the given conditions, we have $\sin (\alpha+\beta)=\frac{3}{5}, \cos \beta=$ $\sqrt{1-x^{2}}$. Therefore,
$$
\begin{aligned}
y & =\cos \alpha=\cos [(\alpha+\beta)-\beta] \\
& =\cos (\alpha+\beta) \cdot \cos \beta+\sin (\alpha+\beta) \cdot \sin \beta \\
& =-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x .
\end{alig... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,397 |
4. Given three distinct points $Z_{1}, Z_{2}, Z_{3}$ in the complex plane corresponding to the complex numbers $z_{1}, z_{2}, z_{3}$. If $\left(z_{2}-z_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}=0$, then $\triangle Z_{1} Z_{2} Z_{3}$ is ( ).
(A) an equilateral triangle
(B) an isosceles right triangle
(C) a right trian... | 4.B.
From the given equation, we get $z_{2}-z_{1}= \pm \mathrm{i}\left(z_{3}-z_{1}\right)$. Therefore, $Z_{1} Z_{2}=Z_{1} Z_{3}$, and $Z_{1} Z_{2} \perp Z_{1} Z_{3}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,398 |
5. If the ellipse $\frac{x^{2}}{m}+\frac{y^{2}}{p}=1$ and the hyperbola $\frac{x^{2}}{n}-\frac{y^{2}}{p}=1$ $(m, n, p>0)$ have common foci $F_{1}$ and $F_{2}$, and their intersection point is $Q$. Then, the area of $\triangle Q F_{1} F_{2}$ is ( ).
(A) $m+n$
(B) $\frac{m+n}{2}$
(C) $p$
(D) $\frac{p}{2}$ | 5.C.
According to the problem, we get $m-n=2p$, and $m+n=2(m-p)$.
From the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{m}+\frac{y^{2}}{p}=1, \\ \frac{x^{2}}{n}-\frac{y^{2}}{p}=1\end{array}\right.$
we solve to get $x^{2}=\frac{2 m n}{m+n}$, and then we find $y= \pm \frac{\sqrt{2} p}{\sqrt{m+n}}$.
Also, $\l... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,399 |
6. Given a regular tetrahedron $ABCD$ with edge length 2, the sum of the areas of all sections obtained by planes equidistant from its four vertices is ( ).
(A) 4
(B) 3
(C) $\sqrt{3}$
(D) $3+\sqrt{3}$ | 6.D.
There are two types of such cross-sections:
(1) One side of the cross-section has one point, and the other side has three points, such as $\triangle B_{1} C_{1} D_{1}$ in Figure 4(a), where $B_{1}, C_{1}, D_{1}$ are the midpoints of edges $AB, AC, AD$ respectively. There are four such cross-sections, each with an... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,400 |
7. As shown in Figure 1, A, B, C, and D are four oil wells in an oil field, and it is planned to build three roads to connect these four oil wells (each road connects only two oil wells). Therefore, the number of different road construction plans is ( ) kinds.
(A) 12
(B) 14
(C) 16
(D) 18 | 7.C.
The number of roads connecting any two oil wells is $\mathrm{C}_{4}^{2}=6$. Choosing 3 out of these 6 roads gives $C_{6}^{3}=20$ ways. However, road construction plans that only connect three of the oil wells are not acceptable, and there are $\mathrm{C}_{4}^{\mathrm{i}}=4$ such plans. Therefore, the number of ro... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,401 |
Example 2 The minimum distance from the lattice points (points with integer coordinates) to the line $y=\frac{5}{3} x+\frac{4}{5}$ is ( ).
(A) $\frac{\sqrt{34}}{170}$
(B) $\frac{\sqrt{34}}{85}$
(C) $\frac{1}{20}$
(D) $\frac{1}{30}$
(2000, National High School Mathematics Competition) | Explanation: First, use the point-to-line distance formula to get the distance, then conduct the analysis.
Convert the given line to $25 x-15 y+12=0$. Let the distance from the integer point $\left(x_{0}, y_{0}\right)$ on the plane to the line be
$$
d=\frac{|25 x_{0}-15 y_{0}+12|}{5 \sqrt{34}} .
$$
Since $(25,15)=5$,... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,402 |
8. Given $O A=(4,3)$, the graph of the function $y=x^{2}+b x+c$ is translated by the vector $O A$ to obtain a graph that is tangent to the line $4 x+y-8=0$ at point $T(1,4)$. Then the original function is ( ).
(A) $y=x^{2}+2 x+1$
(B) $y=x^{2}+2 x$
(C) $y=x^{2}+2 x+2$
(D) $y=x^{2}+2 x-2$ | 8.D.
The derivative of the function $y=x^{2}+b x+c$ is $y^{\prime}=2 x+b$. If the point of tangency $T^{\prime}(x, y)$ is translated by the vector $O A=(4,3)$ to $T(1,4)$, then $x=1-4=-3, y=4-3=1$, which means $T^{\prime}(-3,1)$.
Given that the slope of the tangent line is -4, and the point of tangency $T^{\prime}(-3,... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,403 |
9. Let the sum of the first four terms of an arithmetic sequence be 26, and the sum of the last four terms be 110, and the sum of all terms of this sequence be 187. Then, the number of terms in this sequence is ( ).
(A) 11
(B) 22
(C) 8
(D) 16 | 9.A.
Let the sequence have $n$ terms: $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, we have
$$
\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4}=26, \\
a_{n-3}+a_{n-2}+a_{n-1}+a_{n}=110 .
\end{array}
$$
Since the arithmetic sequence satisfies
$$
a_{1}+a_{n}=a_{2}+a_{n-1}=a_{3}+a_{n-2}=a_{4}+a_{n-3} \text {, }
$$
w... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,404 |
10. Given in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the side face $A_{1} A D D_{1}$ is a square, $M$ is the midpoint of edge $C D$, and the angle between $A M$ and $C D_{1}$ is $\theta$. If $\sin \theta=\frac{\sqrt{78}}{9}$, then the value of $\frac{A A_{1}}{A B}$ is ( ).
(A) $\sqrt{2}$
(B) $\frac{2 \... | 10. A.
As shown in Figure 5, establish a coordinate system. Let
$D A=D D_{1}=a, D C=b$, then
$$
\begin{array}{l}
D_{1}(0,0, a), C(0, b, 0), \\
D_{1} C=(0, b,-a) ; A(a, 0,
\end{array}
$$
$0), M\left(0, \frac{b}{2}, 0\right), A M=$
$\left(-a, \frac{b}{2}, 0\right)$. Then
$$
\begin{array}{l}
\cos \theta=\frac{\boldsymbol... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,405 |
11. In a basketball game, with 0.1 seconds left, Team $A$ gets a chance for two free throws, and at that moment, Team $A$ is trailing 100:101. Statistical data on the free-throw shooter of the team shows: the first shot has a success rate of 0.6. If the first shot is successful, the success rate for the second shot is ... | Ni, 11. 101.36.
Let $\xi$ denote the team's final score, then
$$
\begin{array}{l}
P(\xi=100)=0.4 \times 0.3=0.12 . \\
P(\xi=101)=0.6 \times 0.2+0.4 \times 0.7=0.4 \\
P(\xi=102)=0.6 \times 0.8=0.48 . \\
\mathrm{E} \xi=100 \times 0.12+101 \times 0.4+102 \times 0.48=101.36 .
\end{array}
$$ | 101.36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,406 |
12. Let $t$ be a positive number, and the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term $t$ and common difference $2 t$, and its sum of the first $n$ terms is $S_{n}$. If $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}>$ $\sqrt{1-t}$, then the range of values for $t$ is $\qquad$ | 12. $00)$, then
$$
\begin{array}{c}
a_{n}=t+(n-1) 2 t=(2 n-1) t, \\
S_{n}=n t+\frac{n(n-1)}{2} \cdot 2 t=n^{2} t .
\end{array}
$$
Thus, $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{n \sqrt{t}}{(2 n-1) t}=\frac{1}{2 \sqrt{t}}$.
From $\frac{1}{2 \sqrt{t}}>\sqrt{1-t}$, we so... | 0<t \leqslant 1 \text{ and } t \neq \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,407 |
13. Given the sets $A=\left\{(x, y) \left\lvert\, \frac{y-3}{x-2}=a+1\right.\right\}, B=$ $\left\{(x, y) \mid\left(a^{2}-1\right) x+(a-1) y=15\right\}$. Then the possible values of $a$ that make $A \cap B=$ $\varnothing$ are . $\qquad$ | 13. $-4,-1,1, \frac{5}{2}$.
$A$ is $\{(x, y) \mid(a+1) x-y-2 a+1=0, x \neq 2\}$.
$A \cap B=\varnothing$ occurs only in one of the following three cases:
(1) When $a=-1$, since $A=\{(x, y) \mid y=3, x \neq 2\}$, $B=\left\{(x, y) \left\lvert\, y=-\frac{15}{2}\right.\right\}$, the lines $y=3$ and $y=-\frac{15}{2}$ are par... | -4, -1, 1, \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,408 |
14. Let the points on the curve $2 x^{2}+y^{2}=4 x+6$ that are farthest from and closest to the origin be $M$ and $N$, respectively. Then $|M N|=$ $\qquad$ . | 14. $\sqrt{15}$.
Given $y^{2}=-2 x^{2}+4 x+6 \geqslant 0$, we get $-1 \leqslant x \leqslant 3$.
Take any point $P(x, y)$ on the curve $2 x^{2}+y^{2}=4 x+6$, the distance from $P$ to the origin is
$$
\begin{array}{l}
d=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}-2 x^{2}+4 x+6} \\
=\sqrt{-(x-2)^{2}+10} .
\end{array}
$$
Since $-1 \l... | \sqrt{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,409 |
15. (12 points) $f(x)$ is a function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, it satisfies
$$
b[f(x+p)+f(x)]=a[1-f(x) f(x+p)].
$$
Here $a$, $b$, and $p$ are all non-zero constants. Prove that $f(x)$ is a periodic function, and find its period. | Three, 15. First, prove that for any $x \in \mathbf{R}$, $f(x) \neq -\frac{b}{a}$.
In fact, if $f(x) = -\frac{b}{a}$, then
$$
b f(x+p) - \frac{b^{2}}{a} = a + b f(x+p) \text{, }
$$
thus $a^{2} + b^{2} = 0$, which contradicts the fact that $a$ and $b$ are non-zero constants.
According to the given conditions, we have
$... | 2p | Algebra | proof | Yes | Yes | cn_contest | false | 714,410 |
17. (12 points) Ellipse $C$:
$A x^{2}+B y^{2}=1$ intersects with the line $l: x+2 y=7$ at points $P$ and $Q$. Point $R$ has coordinates $(2,5)$. If $\triangle P Q R$ is an isosceles right triangle, $\angle P R Q=90^{\circ}$, find the values of $A$ and $B$. | 17. Since $\triangle P Q R$ is an isosceles right triangle, $\angle P R Q=90^{\circ}$, the distance from point $R$ to $P Q$ is $\frac{|2+2 \times 5-7|}{\sqrt{1^{2}+2^{2}}}=\sqrt{5}$, so,
$$
|R P|=|R Q|=\sqrt{2} \times \sqrt{5}=\sqrt{10} \text {. }
$$
Thus, points $P$ and $Q$ are the intersections of the circle with ce... | A=\frac{3}{35}, B=\frac{2}{35} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,412 |
Example 3 Let $n \geqslant m \geqslant 1, n, m$ be integers. Prove: $\frac{(m, n)}{n} \cdot \mathrm{C}_{n}^{m}$ is an integer.
(2000, Putnam Mathematical Competition) | Explanation: The key to this problem is how to represent $(m, n)$.
By Bézout's identity, there exist $r, t \in \mathbf{Z}$, such that $m r + n t = (m, n)$. Therefore,
\[
\begin{array}{l}
\frac{(m, n)}{n} \cdot \mathrm{C}_{n}^{m} = \frac{m r + n t}{n} \cdot \mathrm{C}_{n}^{m} = r \frac{m}{n} \cdot \mathrm{C}_{n}^{m} + t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,413 |
18. (15 points) Let the set $M=\{n \in \mathbf{Z} \mid 0 \leqslant n \leqslant 11\}$, and the set $F=\{(a, b, c, d) \mid a, b, c, d \in M\}$. The mapping $f: F \rightarrow \mathbf{Z}$ is defined such that $(a, b, c, d)$ corresponds to $a b - c d$. If $(u, v, x, y)$ corresponds to 39, and $(u, y, x, v)$ corresponds to 6... | 18. According to the problem, we have
$$
\left\{\begin{array}{l}
u v - x y = 39, \\
u y - x v = 66 .
\end{array}\right.
$$
Adding (2) and (1) and simplifying, we get
$$
(u - x)(v + y) = 105 = 3 \times 5 \times 7,
$$
Subtracting (1) from (2) and simplifying, we get
$$
(u + x)(y - v) = 27 = 3^{3}.
$$
Since \( x, y, u, v... | x = 1, y = 9, u = 8, v = 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,414 |
19. (15 points) As shown in Figure 3, given a cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 2, point $E$ is the midpoint of edge $C D$. Find the distance between the skew lines $A_{1} C_{1}$ and $B_{1} E$. | 19. As shown in Figure 7, establish a spatial rectangular coordinate system $D-xyz$,
then $A_{1}(2,0,2), C_{1}(0,2,2), B_{1}(2,2,2), E(0,1,0)$.
$$
\begin{array}{l}
A_{1} C_{1}=(-2,2,0), \\
B_{1} E=(-2,-1,-2), \\
A_{1} B_{1}=(0,2,0) .
\end{array}
$$
Let $PQ$ be the common perpendicular segment of the skew lines $A_{1} ... | \frac{4 \sqrt{17}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,415 |
3. If the inequality $\sqrt{a}+\sqrt{b} \leqslant m \sqrt[4]{a^{2}+b^{2}}$ holds for all positive real numbers $a, b$, then the minimum value of $m$ is ( ).
(A) 2
(B) $\sqrt{2}$
(C) $2^{\frac{3}{4}}$
(D) 4 | 3.C.
Since $\sqrt{a}+\sqrt{b} \leqslant \sqrt{2(a+b)} \leqslant \sqrt{2 \sqrt{\frac{a^{2}+b^{2}}{2}}}$. The equality holds if and only if $a=b$.
Therefore, $m \geqslant 2^{\frac{3}{4}}$, which means the minimum value of $m$ is $2^{\frac{3}{4}}$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,418 |
4. The number of intersection points of the curves $2 x^{2}-x y-y^{2}-x-2 y-1=0$ and $3 x^{2}-$ $4 x y+y^{2}-3 x+y=0$ is ( ).
(A) 2
(B) 3
(C) 4
(D) infinitely many | 4.D.
$2 x^{2}-x y-y^{2}-x-2 y-1=0$, i.e.,
$2 x+y+1=0$ or $x-y-1=0$.
$3 x^{2}-4 x y+y^{2}-3 x+y=0$, i.e.,
$3 x-y=0$ or $x-y-1=0$.
Therefore, both curves can degenerate into two lines, and one pair of lines coincide. Hence, the two curves have infinitely many common points. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,419 |
5. Let $0<a<1$. If $x_{1}=a, x_{2}=a^{x_{1}}, x_{3}=a^{x_{2}}, x_{4}$ $=a^{x_{3}}, \cdots, x_{n}=a^{x_{n}-1}, \cdots$, then the sequence $\left\{x_{n}\right\}(\quad$.
(A) increasing
(B) odd terms increase, even terms decrease
(C) decreasing
(D) even terms increase, odd terms decrease | 5.B.
(1) $x_{1}=a^{1}, x_{2}=a^{x_{1}}$, since $1>a$, then $a^{1}a^{x_{1}}$. Therefore, $a^{1}a^{0}=1$, so $x_{2}>x_{4}$.
By analogy, we get $x_{2}>x_{4}>x_{6} \cdots$.
Thus, the sequence $\left\{x_{n}\right\}$ has increasing odd terms and decreasing even terms. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,420 |
6. In a cube $C$ with side length 1, construct a large inscribed sphere $O_{1}$, and then in one of the corners of $C$, construct a small sphere $O_{2}$, such that it is externally tangent to the large sphere and simultaneously tangent to three faces of the cube. Then the surface area of sphere $O_{2}$ is ( ).
(A) $(7-... | 6.A.
As shown in Figure 4, let the radius of sphere $\mathrm{O}_{2}$ be $r$, and assume sphere $O_{2}$ is located within $\angle D^{\prime}$. Then $O_{1}$ and $O_{2}$ lie on the diagonal $B D^{\prime}$. Let $\angle A D^{\prime} B=\theta$, then $\sin \theta=\frac{1}{\sqrt{3}}$. In $\triangle D^{\prime} E O_{2}$, $D^{\p... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,421 |
8. In the expansion of $\left(4 x^{2}-2 x-5\right)\left(1+\frac{1}{x^{2}}\right)^{5}$, the constant term is $\qquad$ | 8. 15 .
$$
\text { Original expression }=\left(4 x^{2}-2 x-5\right)\left(1+\frac{5}{x^{2}}+\frac{10}{x^{4}}+\cdots\right) \text {. }
$$
Therefore, the constant term in the expanded form is $(-5) \times 1+4 \times 5=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,423 |
Example 1 The sum of three integers $a$, $b$, and $c$ is a multiple of 6. Then, the remainder when their sum of cubes is divided by 6 is ( ).
(A) 0
(B) 2
(C) 3
(D) uncertain
(10th "Hope Cup" National Mathematics Invitational Competition (Grade 8)) | Explanation: Given that the sum of three numbers $a$, $b$, and $c$ is a multiple of 6, it is not straightforward to directly determine the remainder when $a^{3}+b^{3}+c^{3}$ is divided by 6. However, if we consider the difference between the two, we can construct:
$$
\begin{array}{l}
\left(a^{3}+b^{3}+c^{3}\right)-(a+b... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,424 |
Example 2 Given
$$
\left(x+\sqrt{x^{2}+2000}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 \text {. }
$$
Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$
(2002, Junior High School Mathematics Competition, Jiangsu Province) | Explain: Construct the rationalizing factors $x-\sqrt{x^{2}+2002}$ and $y-\sqrt{y^{2}+2002}$ for $x+\sqrt{x^{2}+2002}$ and $y+\sqrt{y^{2}+2002}$. Because
$$
\begin{array}{l}
\left(x+\sqrt{x^{2}+2002}\right)\left(x-\sqrt{x^{2}+2002}\right) . \\
\quad\left(y+\sqrt{y^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=(-... | 58 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,425 |
Example 4 Given that $a, b, c$ are real numbers, the functions are
$$
f(x)=a x^{2}+b x+c, g(x)=a x+b,
$$
when $-1 \leqslant x \leqslant 1$, $|f(x)| \leqslant 1$.
(1) Prove: $|c| \leqslant 1$;
(2) Prove: When $-1 \leqslant x \leqslant 1$, $|g(x)| \leqslant 2$;
(3) Suppose $a>0$, when $-1 \leqslant x \leqslant 1$, the m... | (1) Proof: From the condition, taking $x=0$, we get
$|c|=|f(0)| \leqslant 1$, i.e., $|c| \leqslant 1$.
(2) Proof: When $a>0$, $g(x)=a x+b$ is an increasing function on
$[-1,1]$, then we have
$$
g(-1) \leqslant g(x) \leqslant g(1) .
$$
Since $|f(x)| \leqslant 1(-1 \leqslant x \leqslant 1),|c| \leqslant 1$, then
$$
\beg... | f(x)=2 x^{2}-1 | Algebra | proof | Yes | Yes | cn_contest | false | 714,426 |
13. As shown in Figure 4, in Rt $\triangle ABC$, $\angle ACB = 90^{\circ}$, $CD$ is the angle bisector, $DE \parallel BC$ intersects $AC$ at point $E$, $DF \parallel AC$ intersects $BC$ at point $F$. Prove:
(1) Quadrilateral CEDF is a square; (2) $CD^2 = 2 AE \cdot BF$. | Three, 13. (1) Since $\angle A C B=90^{\circ}, D E / / B C, D F / /$ $A C$, therefore,
$D E \perp A C, D F \perp B C$.
Thus, $\angle E C F=\angle D E C=\angle D F C=90^{\circ}$.
Also, $C D$ is the angle bisector, so, $D E=D F$.
Therefore, quadrilateral $C E D F$ is a square.
(2) Since $D E / / B C$, therefore, $\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,429 |
14. Let the larger root of the equation $2002^{2} x^{2}-2003 \times 2001 x-1=0$ be $r$, and the smaller root of the equation $2001 x^{2}-2002 x+1=0$ be $s$. Find the value of $r-s$. | 14. Since $2002^{2} x^{2}-2003 \times 2001 x-1=0$, the two roots are
$$
x_{1}=1, x_{2}=-\frac{1}{2002^{2}} \text {. }
$$
Therefore, $r=x_{1}=1$.
Similarly, the two roots of the equation $2001 x^{2}-2002 x+1=0$ are $x_{1}=1, x_{2}=\frac{1}{2001}$.
Thus, $s=x_{2}=\frac{1}{2001}$.
Hence, $r-s=1-\frac{1}{2001}=\frac{2000}... | \frac{2000}{2001} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,430 |
15. In an $18 \times 18$ grid, each cell is filled with a distinct positive integer. Prove that no matter how the numbers are filled, there will be at least two pairs of adjacent cells (two cells sharing a common edge are called a pair of adjacent cells) such that the difference between the numbers in each pair of adja... | 15. Let $a$ and $b$ be the smallest and largest numbers among these 324 positive integers, respectively. Since these numbers are all distinct, $b-a \geqslant 323$.
(1) The cells containing $a$ and $b$ are neither in the same row nor in the same column.
Starting from the cell containing $a$, one can reach the cell conta... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,431 |
2. As shown in Figure 1, $AB$ is the diameter of $\odot O$, $C$ is a moving point on $AB$ (point $C$ does not coincide with $A$ or $B$), $CD \perp AB$, $AD$ and $CD$ intersect $\odot O$ at $E$ and $F$ respectively. Then the one that is definitely equal to $AB \cdot AC$ is ( ).
(A) $AE \cdot AD$
(B) $AE \cdot ED$
(C) $C... | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,433 |
3. In $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$, it is known that $A B<A^{\prime} B^{\prime}, B C$ $<B^{\prime} C^{\prime}, C A<C^{\prime} A^{\prime}$. Among the following conclusions:
(1) The altitude from $A B$ in $\triangle A B C$ is less than the altitude from $A^{\prime} B^{\prime}$ in $\t... | 3.A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,434 |
4. Let $s=\frac{1}{(1+x)^{2}}+\frac{1}{(1-x)^{2}}$. Then, the relationship between $s$ and 2 is ( ).
(A) $s=2$
(B) $s<2$
(C) $s>2$
(D) The relationship between $s$ and 2 depends on the value of $x$ | 4.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,435 |
Example 5 Given that when $x \in[0,1]$, the inequality
$$
x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta>0
$$
always holds. Try to find the range of $\theta$. | Solution: If for all $x \in [0,1]$, we have
$$
f(x) = x^{2} \cos \theta - x(1-x) + (1-x)^{2} \sin \theta > 0 \text{.}
$$
Then $\cos \theta = f(1) > 0$, $\sin \theta = f(0) > 0$.
Let $x_{0} = \frac{\sqrt{\sin \theta}}{\sqrt{\cos \theta} + \sqrt{\sin \theta}} \in (0,1)$, then
$$
\sqrt{\cos \theta} x_{0} - \sqrt{\sin \th... | 2 k \pi + \frac{\pi}{12} < \theta < 2 k \pi + \frac{5 \pi}{12}, \quad k \in \mathbf{Z} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,437 |
10. In Rt $\triangle A B C$, $\angle A=90^{\circ}, A B=5 \text{ cm}, A C=$ $4 \text{ cm}$, then the length of the angle bisector $A D$ of $\angle A$ is $\qquad$ $\text{cm}$. | 10. $\frac{20}{9} \sqrt{2}$ | \frac{20}{9} \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,442 |
12. As shown in Figure $4, \odot C$ passes through the origin and intersects the coordinate axes at points $A$ and $D$ respectively. It is known that $\angle O B A=30^{\circ}$, and the coordinates of point $D$ are $(0,2)$. Then the coordinates of points $A$ and $C$ are $\qquad$
$\qquad$ | 12. $\left(-\frac{2 \sqrt{3}}{3}, 0\right) \quad\left(-\frac{\sqrt{3}}{3}, 1\right)$ | \left(-\frac{2 \sqrt{3}}{3}, 0\right) \quad\left(-\frac{\sqrt{3}}{3}, 1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,444 |
15. Class (8) of the third year of junior high school still has a class fund of $m$ (where $m$ is a positive integer less than 400) yuan, and plans to buy 1 photo album for each student. A wholesale and retail stationery store stipulates: if 50 or more photo albums are purchased, they can be sold at the wholesale price... | Three, 15. Let the class have $x$ students, and the retail price of each photo album be $y$ yuan. According to the problem, we have
$$
x y=(x+12)(y-2),
$$
and the integer $x$ satisfies $38<x<50$.
From (1), we get $12 y-2 x-24=0, y=\frac{x}{6}+2$. Therefore,
$$
x y=\frac{x^{2}}{6}+2 x \text {. }
$$
From (3) and $x y=m... | x=42, y=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,447 |
1. Given the functions $f(x)=x^{2}+2 b x+1$ and $g(x)=2 a(x+b)$, where $x$, $a$, and $b$ are real numbers, such that the graphs of $y=f(x)$ and $y=g(x)$ do not intersect in the $x O y$ plane, the set of real number pairs $(a, b)$ that satisfy this condition forms a point set $A$. Then, the area of the figure $T$ repres... | (Due to the pair of numbers $(a, b)$ making the equation $x^{2}+2 b x+1=$ $2 a(x+b)$ have no real roots, thus, $\Delta<0$, which means $a^{2}+b^{2}<1$ represents the area inside the circle. The area of figure $T$ is $\pi$.) | \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,448 |
16. Given the equation about $x$, $x^{2}+4 x+3 k-1=0$, the sum of the squares of its two real roots is not less than the product of these two roots, and for the inverse proportion function $y=\frac{1+5 k}{x}$, the graph's two branches in their respective quadrants show that the ordinate $y$ decreases as the abscissa $x... | 16. From the problem, the equation
$$
x^{2}+4 x+3 k-1=0
$$
has real roots, so $\Delta=4^{2}-4(3 k-1) \geqslant 0$.
Solving this, we get $k \leqslant \frac{5}{3}$.
Let $x_{1} 、 x_{2}$ be the roots of (1), by the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-4, x_{1} x_{2}=3 k-1 \text {. }
$$
Sin... | 0, 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,449 |
17. Find the sum of the reciprocals of all positive divisors of 360. | 17. Let the sum of all positive divisors of a positive integer $a$ be $b$. Let $d_{1}$, $d_{2}, \cdots, d_{n}$ be the positive divisors of $a$ arranged in ascending order, so $d_{1}=1, d_{n}=a$. Since the least common multiple of the denominators in $S=\sum_{i=1}^{n} \frac{1}{d_{i}}$ is $d_{n}=a$, we have
$$
S=\sum_{i=... | \frac{13}{4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,450 |
18. As shown in Figure 5, in $\triangle ABC$, $BC=6$, $AC=4\sqrt{2}$, $\angle C=45^{\circ}$, $P$ is a moving point on side $BC$. A line through $P$ parallel to $AB$ intersects $AC$ at point $D$. Connect $AP$. The areas of $\triangle ABP$, $\triangle APD$, and $\triangle CDP$ are denoted as $S_{1}$, $S_{2}$, and $S_{3}$... | 18. (1) From the problem, we know that $B P=x, 0<x<6$, and $\frac{A D}{B P}=\frac{A C}{B C}$,
so $A D=\frac{A C}{B C} \cdot B P=\frac{4 \sqrt{2}}{6} x=\frac{2 \sqrt{2}}{3} x$.
$S_{2}=S_{\triangle A P D}=\frac{1}{2} A D \cdot P M$
$=\frac{1}{2} \times \frac{2 \sqrt{2}}{3} x \times \frac{\sqrt{2}}{2}(6-x)=2 x-\frac{1}{3... | S_{1}=S_{2}+S_{3}, S_{2}=S_{3}, S_{1}=2 S_{2}, S_{1}=2 S_{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,451 |
1. Given that $a, b, c, d$ are four distinct real numbers, and $(b+d)(b+a)=1,(c+d)(c+a)=1$. Then the value of $(b+d)(c+d)$ is $(\quad)$.
(A) 1
(B) -1
(C) 2
(D) 0 | $$
-、 1 . B
$$
From the conditions, we have
$$
b^{2}+a b+b d+a d=1, c^{2}+a c+c d+a d=1 \text {. }
$$
Subtracting the two equations yields $b^{2}-c^{2}+a(b-c)+d(b-c)=0$.
Thus, $(b-c)(b+c+a+d)=0$.
Since $b \neq c$, i.e., $b-c \neq 0$, then
$$
\begin{array}{l}
b+c+a+d=0, c+d=-(b+a) . \\
\text { Therefore, }(b+d)(c+d)=-... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,452 |
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