problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 2 As shown in Figure 6.
In $\triangle A B C$, the three altitudes $A A_{1}, B B_{1}, C C_{1}$ intersect at point $H$. Prove: the perpendiculars from $A, B, C$ to $B_{1} C_{1}, C_{1} A_{1}, A_{1} B_{1}$ respectively must also intersect at one point. This point is precisely the circumcenter of $\triangle A B C$. | Prove: In Figure 6, there are two triangles $A B C$ and $A_{1} B_{1} C_{1}$ (the pedal triangle), where perpendiculars from $A_{1}, B_{1}, C_{1}$ to the sides $B C, C A, A B$ of $\triangle A B C$ meet at a point $H$. According to Theorem 2, perpendiculars from $A, B, C$ to $B_{1} C_{1}, C_{1} A_{1}, A_{1} B_{1}$ respec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,677 |
Example 5 In the plane of square $ABCD$, a line $l$ satisfies the following conditions: the distances from the four vertices of square $ABCD$ to line $l$ take only two values, one of which is three times the other. The number of such lines $l$ is ( ).
(A) 4 lines
(B) 8 lines
(C) 12 lines
(D) 16 lines | Solution: As shown in Figure 2.
(1) Four lines passing through the quartiles of one pair of opposite sides and parallel to the other pair of sides;
(2) Four lines passing through the points that externally divide one pair of opposite sides in the ratio $3: 1$ and parallel to the other pair of sides;
(3) Four lines pass... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,678 |
Example 3 Given that $A A_{1}, B B_{1}, C C_{1}$ are the altitudes of $\triangle A B C$ on sides $B C, C A, A B$ respectively. Let $A_{2}, B_{2}, C_{2}$ be the midpoints of $B_{1} C_{1}, C_{1} A_{1}, A_{1} B_{1}$ respectively. And let $A_{2} R, B_{2} S, C_{2} T$ be line segments perpendicular to $B C, C A, A B$ respect... | Proof:
As shown in Figure 7, from
$A_{1}, B_{1}, C_{1}$
draw the perpendiculars to
$BC, CA$,
$AB$ respectively, and they intersect at
H. According to Example 2,
it is known that, from
$A, B, C$ respectively draw the perpendiculars to $B_{1} C_{1}, C_{1} A_{1}, A_{1} B_{1}$, they intersect at point $O$, and point $O$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,679 |
Example 1: In a 10000-meter race on a 400-meter circular track at a school sports meet, two athletes, A and B, start running at the same time. B runs faster than A. At the 15th minute, A speeds up. At the 18th minute, A catches up with B and begins to overtake B. At the 23rd minute, A catches up with B again, and at 23... | Solution: According to the problem, draw the figure, as shown in Figure 2.
The broken line $O A B$ represents the movement of A, and the line segment $O C$ represents the movement of B. Since A catches up with B again at the 23rd minute, $E F=400$. And $B G=C K=10000$.
Since $\triangle D E F \sim \triangle D H B$, acc... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,680 |
Example 2: The distance between location A and location B is 70 kilometers. Two cars start from the two locations simultaneously and continuously travel back and forth between A and B. The first car starts from A, traveling at 30 kilometers per hour, and the second car starts from B, traveling at 40 kilometers per hour... | Solution: According to the problem, draw the graph, as shown in Figure 3.
The broken lines $O A B M$ and $D E F G$ represent the motion graphs of the first and second cars, respectively. Clearly, point $M$ indicates the second meeting of the first car, which departs from location A for the second time, with the second ... | 150 \text{ km and } 200 \text{ km} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,681 |
Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that... | Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up wit... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,682 |
Example 4 A mall has an escalator moving uniformly from the bottom to the top. Two people, A and B, are in a hurry to go upstairs. While riding the escalator, they both climb the stairs at a uniform speed. A reaches the top after climbing 55 steps, and B's climbing speed is twice that of A (the number of steps B climbs... | Solution: According to the problem, draw the graph, as shown in Figure 5.
$O C, O B, O A$ are the motion graphs of the escalator, person A riding the escalator, and person B riding the escalator, respectively.
Draw $A M \perp t$-axis, intersecting $O B$ and $O C$ at $E$ and $F$, respectively. Draw $B N \perp t$-axis, i... | 66 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,683 |
Proposition As shown in Figure 1, let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of $\triangle ABC$, respectively, and let $BC=a, CA$
$$
=b, AB=c, s=\frac{1}{2}
$$
$(a+b+c), \triangle AEF, \triangle BDF, \triangle CDE$ have areas denoted as $\triangle_{A}, \triangle_{B}, \triangle_{C}$, respectively. Then
$$
... | Proof: By the definition of the midpoints of the sides of a triangle, we know
$$
\begin{array}{l}
s=AB + AE = c + AE, \\
s=AC + AF = b + AF, \\
\therefore AE = s - c, AF = s - b. \\
\because \triangle_{A} = \frac{1}{2} AE \cdot AF \cdot \sin A \\
= \frac{1}{2}(s - b)(s - c) \sin A, \\
\therefore \frac{(s - b)(s - c)}{\... | 4 \sqrt{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,685 |
In $\triangle A B C$, $a 、 b 、 c$ are the lengths of its three sides, $m_{a} 、 h_{a}$ are the lengths of the median and altitude to side $B C$, and $\triangle$ is its area. Then we have
$$
\frac{m_{a}}{h_{a}} \leqslant \frac{a^{2}+b^{2}+c^{2}}{4 \sqrt{3} \triangle} \text {. }
$$ | Proof: From the triangle area formula $\triangle=\frac{1}{2} a h_{u}$, we get $h_{u}=\frac{2 \triangle}{a}$. Also, from the median length formula
$$
m_{a}=\frac{1}{2} \sqrt{2 b^{2}+2 c^{2}-a^{2}}
$$
we know that equation (1) is equivalent to
$$
\begin{array}{l}
3 a^{2}\left(2 b^{2}+2 c^{2}-a^{2}\right) \\
\leqslant\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,686 |
Proposition Let $t_{a}, t_{b}, t_{c}$ be the lengths of the angle bisectors of $\angle A, \angle B, \angle C$ in $\triangle ABC$. Then, $\frac{bc}{t_{a}^{2}}+\frac{ca}{t_{b}^{2}}+\frac{ab}{t_{c}^{2}} \geqslant 4$. | Proof: Let the semi-perimeter, circumradius, and inradius of $\triangle ABC$ be $s, R, r$, and let $a, b, c$ be the sides opposite to $\angle A, \angle B, \angle C$ respectively. Then we have:
$$
\begin{array}{l}
a b + b c + c a = s^{2} + 4 R r + r^{2}, \\
a^{2} + b^{2} + c^{2} = 2\left(s^{2} - 4 R r - r^{2}\right), \\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,687 |
Example 6 As shown in Figure 3, in the isosceles trapezoid $A B C D$, $A B / / D C, A B$ $=998, D C=1001, A D=$ 1999, point $P$ is on the line segment $A D$. The number of points $P$ that satisfy the condition $\angle B P C=90^{\circ}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) an integer not less than 3 | Solution: Clearly, the angle subtended by $BC$ at the midpoint $M$ of $AD$ is $90^{\circ}$;
Take a point $N$ on $AD$ such that $AN=998$, then $DN=$ 1001. Since $\triangle ABN$ and $\triangle DCN$ are both isosceles triangles, it is easy to see that the angle subtended by $BC$ at $N$ is also $90^{\circ}$.
The circle w... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,689 |
2. For all positive real numbers $a, b, c$, prove:
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c u}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .
$$ | Proof 1 (Chen Yu): The original inequality is transformed into
$$
\frac{1}{\sqrt{1+\frac{8 b x}{a^{2}}}}+\frac{1}{\sqrt{1+\frac{8 a c}{b^{2}}}}+\frac{1}{\sqrt{1+\frac{8 a b}{c^{2}}}} \geqslant 1 \text {. }
$$
Let $\alpha=\frac{b c}{a^{2}}, \beta=\frac{a c}{b^{2}}, \gamma=\frac{a b}{c^{2}}$. Clearly, $\alpha, \beta, \g... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,690 |
5. In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. It is known that $\angle B A C^{\circ}=60^{\circ}$, and $A B+B P=A Q+B Q$. What are the possible values of the angles of $\triangle A B C$? | Solution 1 (Yang Xiao): Extend $AB$ to $B'$ such that $BB' = BP$, and take $C'$ on the ray $QC$ such that $QC' = QB$, as shown in Figure 4 or Figure 5.
If $C$ and $C'$ do not coincide, then connect $B'C'$, $B'P$, $C'P$, and $BC'$.
$$
\begin{array}{l}
\because AB + BP = AQ + QB, \\
\therefore AB + BB' = AQ + QC', \text{... | \angle BAC = 60^\circ, \angle ABC = 80^\circ, \angle ACB = 40^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,691 |
(50 points) In an acute triangle $\triangle ABC$, $AD \perp BC$, with $D$ as the foot of the perpendicular, $DE \perp AC$, with $E$ as the foot of the perpendicular. $DF \perp AB$, with $F$ as the foot of the perpendicular. $O$ is the circumcenter of $\triangle ABC$. Prove:
(1) $\triangle AEF \subset \triangle ABC$;
(2... | $$
\begin{array}{l}
\therefore \angle A E F=\angle A D F \\
=90^{\circ}-\angle B D F \\
=\angle B . \\
\text { Also } \angle A=\angle A, \\
\therefore \triangle A E F \backsim \triangle A B C .
\end{array}
$$
By the circumcenter of $\triangle A E F$ being the midpoint $O^{\prime}$ of $A D$,
$$
\begin{array}{l}
\theref... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,692 |
II. (50 points) Given the algebraic expression $-x^{3}+100 x^{2}+x$, the letter $x$ is only allowed to take values within the set of positive integers. When the value of this algebraic expression reaches its maximum, what is the value of $x$? Prove your conclusion. | Let $x=k$, the value of the algebraic expression is $a_{k}$.
$$
a_{k+1}-a_{k}=-3 k^{2}+197 k+100 \text {. }
$$
When $00$;
When $k \geqslant 67$, $a_{k+1}-a_{k}<0$.
$\therefore$ When $x=67$, the algebraic expression reaches its maximum value. | 67 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,693 |
Three, (50 points) (1) Prove that there exist non-zero integer pairs $(x, y)$ such that the algebraic expression $11 x^{2}+5 x y+37 y^{2}$ is a perfect square;
(2) Prove that there exist six non-zero integers $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$, where $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$, such that for... | (2) Let $x_{0}=t+1, y_{0}=2 t+1$, then
$$
11 x_{0}^{2}+5 x_{0} y_{0}+37 y_{0}^{2}=(13 t+n)^{2} \text {. }
$$
Thus, $(185-26 n) t=n^{2}-53$.
Let $\left\{\begin{array}{l}x=(185-26 n) x_{0}=n^{2}-26 n+132, \\ y=(185-26 n) y_{0}=2 n^{2}-26 n+79 .\end{array}\right.$
Then $11 x^{2}+5 x y+37 y^{2}$
$$
\begin{array}{l}
=[(185... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,694 |
Four. (15 points) Points
$D$ and $E$ are on
the sides $AC$
and $BC$ of
$\triangle ABC$, $\angle C$ is
a right angle, $DE \parallel AB$,
and $3 DE = 2 AB$,
$AE = 13, BD = 9$. Then, the length of $AB$ is | Four、 $\frac{15}{13} \sqrt{130}$ | \frac{15}{13} \sqrt{130} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,698 |
Nine, (15 points) satisfying the following two conditions:
(1) For all natural numbers $x, x^{2}-2001 x+n$ $\geqslant 0$
(2) There exists a natural number $x_{0}$, such that $x_{0}^{2}-2002 x_{0}+n$ $<0$ the number of positive integers $n$ equals $\qquad$ | $Nine, 1001$ | 1001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,704 |
2. As shown in Figure 1, given $A B=10, P$ is any point on line segment $A B$, and equilateral triangles $\triangle A P C$ and $\triangle B P D$ are constructed on the same side of $A B$ with $A P$ and $P B$ as sides, respectively. Then the minimum length of $C D$ is ( ).
(A) 4
(B) 5
(C) 6
(D) $5(\sqrt{5}-1)$ | 2.R
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,707 |
3. In a convex $n$-sided polygon, the maximum number of angles less than $108^{\circ}$ can be ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,708 |
4. The number of all integer solutions to the equation $\left(x^{2}+x-1\right)^{x+3}=1$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,709 |
Example $8 A$ and $B$ are two fixed points on a plane. Find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles right triangle. There are $\qquad$ such points $C$. | Solution: As shown in Figure 5, the vertices of the two isosceles right triangles with $AB$ as the hypotenuse are $C_{1}$ and $C_{2}$; the vertices of the four isosceles right triangles with $AB$ as the legs are $C_{3}, C_{4}, C_{5}, C_{6}$.
Therefore, there are a total of 6 points $C$ that meet the conditions. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,711 |
6. Person A and Person B start from the same location at the same time, heading in opposite directions, and after 1 hour they reach their respective destinations $A$ and $B$. If they still start from the original location and swap their destinations, then A arrives at $B$ 35 minutes after B reaches $A$. What is the rat... | 6. I))
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,712 |
7. Introduce a new operation * in the set of all real numbers, defined as follows:
(1) For any real numbers $a, b$, $a * b=(a+1) \cdot (b-1)$;
(2) For any real number $a$, $a^{* 2}=a * a$.
When $x=2$, the value of $\left[3 *\left(x^{* 2}\right)\right]-2 * x+1$ is ( ).
(A) 34
(B) 16
(C) 12
(D) 6 | 7.I))
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,713 |
11. If the length of the line segment joining the midpoints of one pair of opposite sides of a quadrilateral is $d$, and the lengths of the other pair of opposite sides are $a$ and $b$, then the relationship between $d$ and $\frac{a+b}{2}$ is $\qquad$ . | 11. $d \leqslant \frac{a+b}{2}$ | d \leqslant \frac{a+b}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,717 |
13. For the equation $k x^{2}-(k-1) x+1=0$ with respect to $x$, if it has rational roots, find the integer value of $k$. | Three, 13. (1) When $k=0$, $x=-1$, the equation has a rational root.
(2) When $k \neq 0$, since the equation has a rational root, if $k$ is an integer, then $\Delta=(k-1)^{2}-4 k=k^{2}-6 k+1$ must be a perfect square, i.e., there exists a non-negative integer $m$, such that
$$
k^{2}-6 k+1=m^{2} .
$$
Completing the squ... | 6 \text{ or } 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,719 |
14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of ... | 14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,720 |
15. There are 12 students standing in a circle, some of whom are holding flowers, with a total of 13 bouquets. They play a flower distribution game, following these rules: one of the students who has at least two bouquets of flowers takes two bouquets and gives one to each of the two adjacent students. Try to prove: du... | 15. Let's assume that at the beginning, fewer than 7 students hold flowers. We mark these 12 students as \(A_{1}, A_{2}, A_{3}, \cdots, A_{12}\) in a counterclockwise direction.
(1) During the flower distribution game, if any two adjacent students have one of them holding a flower, then after each subsequent distributi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,721 |
Example 9 In $\triangle A B C$, $\angle A>\angle B>\angle C$, $\angle A \neq 90^{\circ}$, draw a line that divides $\triangle A B C$ into two parts, and makes one of the parts similar to $\triangle A B C$. There are ( ) such non-parallel lines.
(A) 2
(B) 3
(C) 6
(D) 12 | Solution: (1) There are 3 lines parallel to each of the two sides;
(2) There are 3 anti-parallel lines to the three sides (i.e., the four points of the quadrilateral formed are concyclic).
These 6 lines all meet the requirements of the problem. Therefore, the answer is (C). | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,722 |
5. Fold a sheet of paper with a Cartesian coordinate system once so that point $A(0,2)$ coincides with point $B(4,0)$. If at this time point $C(7,3)$ coincides with point $D(m, n)$, then the value of $m+n$ is . $\qquad$ | 5. $\frac{34}{5}$ | \frac{34}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,727 |
10. If the inequality about $x$
$$
\frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7}{x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7}<0
$$
has a solution set that is the union of some intervals, and the sum of the lengths of these intervals is no less than 4, then the range of real number $a$ is $\qquad$ | $\begin{array}{l}10 .(-\infty, 1] \cup \\ {[3,+\infty)}\end{array}$ | (-\infty, 1] \cup [3,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,732 |
Example 10 In $\triangle ABC$, there is an interior angle of $36^{\circ}$. A line $AD$ through vertex $A$ divides this triangle into two isosceles triangles. Then the number of different shapes of $\triangle ABC$ that satisfy the above conditions (similar triangles are considered the same shape) is ( ).
(A) 2
(B) 3
(C)... | Solution: (1) When $\angle \mathrm{A}=$
$36^{\circ}$, let $\angle C=$
$\frac{1}{3} \angle A=12^{\circ}$. Then $\angle \mathrm{A}$
has a trisector that meets
the problem's requirements, as shown in Figure 6-1.
(2) When $\angle A \neq 36^{\circ}$, assume $\angle B=36^{\circ}$, then $\angle A D C=72^{\circ}, \angle C=72^{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,733 |
(16 points) Let $A, B, A, (1 \leqslant i \leqslant k)$ be sets.
(1) How many ordered pairs $(A, B)$ satisfy $A \cup B=\{a, b\}$? Why?
(2) How many ordered pairs $(A, B)$ satisfy $A \cup B=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$? Why?
(3) How many ordered tuples $\left(A_{1}, A_{2}, \cdots, A_{k}\right)$ satisfy $A_... | Ni, (1)9 $\quad(2) 3^{n} \quad(3)\left(2^{k}-1\right)^{n}$
(1) and (2) are special cases of (3), so we only need to prove (3).
To determine the number of ordered sets $\left(A_{1}, A_{2}, \cdots, A_{k}\right)$, we can divide it into $n$ steps.
In the first step, consider the possibilities for $a_{1}$ to belong to $A_... | \left(2^{k}-1\right)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,734 |
Three. (16 points) A publishing company has set the following prices for a best-selling book:
$$
C(n)=\left\{\begin{array}{l}
12 n, \text { if } 1 \leqslant n \leqslant 24 ; \\
11 n, \text { if } 25 \leqslant n \leqslant 48 ; \\
10 n, \text { if } n \geqslant 49 .
\end{array}\right.
$$
Here $n$ represents the number o... | (1) From $C(25)=275, C(24)=288, C(23)=$ 276, $C(22)=264$,
we have $C(25)<C(23)<C(24)$.
From $C(49)=490, C(48)=528, C(47)=517$,
$$
C(46)=506, C(45)=495, C(44)=484 \text {, }
$$
we have $C(49)<C(45)<C(46)<C(47)<C(48)$.
Therefore, there are 6 values of $n$ (i.e., $23,24,45,46,47,48$) where buying more than $n$ books cost... | 302 \text{ and } 384 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,735 |
Four, (18 points) The real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001 . \\
\text { Let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2 .
\end{array}
$$
$\cdots, 2$ 001. Find the maximum possible value of $\sum_{k=1}^{2000}\left|y... | For $k=1,2, \cdots, 2000$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
=\frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,736 |
1. If real numbers $a, b, c, d$ satisfy $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$, then the value of $\frac{a b+b c+c d+d a}{a^{2}+b^{2}+c^{2}+d^{2}}$ is ( ).
(A) 1 or 0
(B) -1 or 0
(C) 1 or -2
(D) 1 or -1 | -.1.D.
Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, we can get $k= \pm 1$.
When $k=-1$, $a=-b=c=-d$, the original expression $=-1$;
When $k=1$, the original expression $=1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,737 |
2. Real numbers $x, y$ satisfy the equation $x^{2}+2 y^{2}-2 x y+x-3 y+1=0$, then the maximum value of $y$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{3}{2}$
(C) $-\frac{3}{4}$
(D) does not exist | 2.B.
$$
\begin{array}{l}
x^{2}+(1-2 y) x+2 y^{2}-3 y+1=0 . \\
\Delta=(1-2 y)^{2}-4\left(2 y^{2}-3 y+1\right) \geqslant 0 .
\end{array}
$$
Then $4 y^{2}-8 y+3 \leqslant 0 \Rightarrow \frac{1}{2} \leqslant y \leqslant \frac{3}{2}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,738 |
3. As shown in Figure 1, $C$ is any point on line segment $A B$, $\triangle A C D$ and $\triangle B C E$ are equilateral triangles, $A F$ and $B G$ are tangents to the circle passing through points $D$, $C$, and $E$, with $F$ and $G$ being the points of tangency. Then ( ).
(A) $A F>B G$
(B) $A F=B G$
(C) $A F<B G$
(D) ... | 3. B.
Connect $A E$ and $B D$. Let $A E$ and $B D$ intersect at $P$, and $A E$, $B D$ intersect the circle at points $M$, $N$ respectively. It can be proven that $\triangle D C B \cong \triangle A C E$, then $A E = D B, \angle D P A = 60^{\circ}$.
Connect $M D$ and $N E$. It can be proven that $\triangle P D M$ and $... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,739 |
4. A point whose both x-coordinate and y-coordinate are integers is called an integer point. If the number of integer points (including the boundary) within the triangle formed by the line $y=-2 x+k$ (where $k$ is a positive integer) and the coordinate axes is 100, then $k$ equals ( ).
(A) 9
(B) 11
(C) 18
(D) 22 | 4.C.
(1) $k$ is an even number. Let $k=$ $2 m$. As shown in Figure 2, $x=0, y=$
$$
\begin{array}{l}
2 m ; x=1, y=-2 \times 1 \\
+2 m ; \cdots ; x=m-1, y= \\
-2(m-1)+2 m .
\end{array}
$$
There are $m+1$ integer points on the $x$-axis. Therefore, the total number of integer point coordinates is
$$
\begin{array}{l}
-2(1+... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,740 |
6. $\overline{a b}$ and $\overline{b a}$ represent two-digit integers, then the number of $\overline{a b}$ that satisfy the equation $(a+b) x^{2}-\overline{a b} x+\overline{b a}=0$ with both roots being integers is ( ).
(A)0
(B) 2
(C) 4
(D) 6 | 6. B.
Let the two integer roots be $x_{1}, x_{2}$, then
$$
x_{1}+x_{2}=\frac{10 a+b}{a+b}, x_{1} x_{2}=\frac{10 b+a}{a+b} \text {. }
$$
We have $x_{1}+x_{2}+x_{1} x_{2}=11$.
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=12=( \pm 1)( \pm 12) \\
=( \pm 2)( \pm 6)=( \pm 3)( \pm 4) .
\end{array}
$$
When $... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,742 |
1. Given $x-\frac{1}{x}=\sqrt{3}$, then $x^{5}+\frac{1}{x^{5}}=$ | \begin{array}{l}=1.19 \sqrt{7} \text { or }-19 \sqrt{7} . \\ x^{5}+\frac{1}{x^{5}}=x^{5}+x+\frac{1}{x^{5}}+\frac{1}{x}-\left(x+\frac{1}{x}\right) \\ =x^{3}\left(x^{2}+\frac{1}{x^{2}}\right)+\frac{1}{x^{3}}\left(x^{2}+\frac{1}{x^{2}}\right)-\left(x+\frac{1}{x}\right) \\ =\left(x+\frac{1}{x}\right)\left[\left(x^{2}+\frac... | \pm 19 \sqrt{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,743 |
Example 1 Suppose the equation $m x^{2}-(m-2) x+m-3=0$ has integer solutions, try to determine the value of the integer $m$, and find all integer solutions of the equation at this time. | Analysis: If $m=0$, then $2 x-3=0$, in this case the equation has no integer solutions;
When $m \neq 0$, consider $\Delta=-3 m^{2}+8 m+4$. Note that the coefficient of the quadratic term is negative, for the equation to have solutions, then
$-3 m^{2}+8 m+4 \geqslant 0$.
Solving this, we get $\frac{4-2 \sqrt{7}}{3} \le... | m=1, \text{ solutions: } -2, 1; \, m=3, \text{ solution: } 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,744 |
Example 2 When $x$ is a rational number, why is the value of the algebraic expression $9 x^{2}+$ $23 x-2$ exactly the product of two consecutive even numbers.
Preserve the original text's line breaks and format, and output the translation result directly. | Analysis: Let two consecutive even numbers be $n, n+2$. The problem is transformed into: For what value of $n$ does the equation $9 x^{2}+23 x-2$ $=n(n+2)$ have rational roots.
The problem of rational roots is essentially also a problem of integer roots, requiring the discriminant of the equation to be a perfect squar... | -\frac{41}{9}, 2 \text { or } \frac{130}{9},-17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,745 |
Example 4 For any positive integer $n \geqslant 2$, prove: it is possible to select $2^{n}$ numbers from the set $M_{n}=\left\{1,2,3, \cdots, \frac{3^{n}+1}{2}\right\}$, such that no three of these numbers form an arithmetic progression. | Proof: Consider the set of all non-negative integers with no digit 2 in their ternary representation and with at most $n$ digits:
$$
\begin{aligned}
A= & \left\{x_{n-1} \cdot 3^{n-1}+x_{n-2} \cdot 3^{n-2}+\cdots+x_{1} \cdot 3\right. \\
& +x_{0} \mid x_{i}=0 \text { or } 1(i=0,1,2, \cdots, n-
\end{aligned}
$$
1)\}.
By ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,746 |
33. There is an arithmetic sequence with 100 terms and a common difference of 4, the sum of which is 20300. Then the first term of this sequence is ().
(A) 1
(B) 2
(C) 3
(D) 4
$(\mathrm{E}) *$ | 33. (E).
Let the first term be $a_{1}$. Then $a_{100}=a_{1}+99 \times 4$.
$$
\begin{array}{c}
\text { Also } S_{\mathrm{H \omega}}=100 \cdot \frac{a_{1}+a_{100}}{2}=20300 \\
\Rightarrow a_{1}+99 \times 2=203 \Rightarrow a_{1}=5 .
\end{array}
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,747 |
34. The sum of the infinite series $1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+\cdots$ is ).
(A) 1
(B) $\frac{9}{10}$
(C) $\frac{4}{5}$
(D) $\frac{7}{10}$
(E) * | 34. (C).
$$
\begin{array}{l}
1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+\cdots \\
=1+\left(-\frac{1}{4}\right)+\left(-\frac{1}{4}\right)^{2}+\left(-\frac{1}{4}\right)^{3}+\cdots \\
=\lim _{n \rightarrow \infty} \frac{1-\left(-\frac{1}{4}\right)^{n}}{1-\left(-\frac{1}{4}\right)}=\frac{4}{5} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,748 |
35. $K$ is a point on side $BC$ of equilateral $\triangle ABC$. If $\angle BAK = 15^{\circ}$, then the ratio of sides $\frac{AK}{AB}$ is ().
(A) $\frac{3 \sqrt{2}(3+\sqrt{3})}{2}$
(B) $\frac{\sqrt{2}(3+\sqrt{3})}{2}$
(C) $\frac{\sqrt{2}(3-\sqrt{3})}{2}$
(D) $\frac{3 \sqrt{2}(3-\sqrt{3})}{2}$
(E) * | 35. (C).
$$
\begin{array}{l}
\frac{A K}{A B}=\frac{\sin \angle A B K}{\sin \angle A K B}=\frac{\sin 60^{\circ}}{\sin 75^{\circ}}=\frac{\sin 60^{\circ}}{\cos 15^{\circ}} \\
=4 \cos 30^{\circ} \cdot \sin 15^{\circ}=2 \sqrt{3} \cdot \sqrt{\frac{1-\cos 30^{\circ}}{2}} \\
=\frac{\sqrt{2}}{2}(3-\sqrt{3}) .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,749 |
36. In quadrilateral $A B C D$, $A B=4, B C=1, C D=7$. If $\angle D A B=\angle B C D=120^{\circ}$. Then, $D A=(\quad)$.
(A) $2+3 \sqrt{5}$
(B) $2 \sqrt{3}+\sqrt{46-8 \sqrt{3}}$
(C) $2 \sqrt{3}+\sqrt{46+8 \sqrt{3}}$
(D) $-2+3 \sqrt{5}$
$(\mathrm{E}) *$ | 36. (D).
As shown in Figure 3, we have
$$
\begin{array}{l}
A B^{2}+A D^{2}+A B \cdot 4 I \\
=B C^{2}+C D^{2}+B C \cdot C D \text {. } \\
4^{2}+A D^{2}+4 \cdot D \\
=1^{2}+7^{2}+7 \text {. } \\
\end{array}
$$
That is,
then $A D^{2}+4 A D=41$.
Thus, $A D=-2+3 \sqrt{5}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,750 |
37. In $\angle ABC$, $AB=6, BC=5, CA=4$. If $\kappa$ is on $BC$ and $\frac{B K}{K C}=\frac{3}{2}$, then $A K=(\quad$.
(A) $2 \sqrt{3}$
(B) 4
(C) $3 \sqrt{2}$
(D) $2 \sqrt{\frac{21}{5}}$
(E) * | 37. (C).
Given $\frac{B K}{K C}=\frac{3}{2}, B C=5$. Therefore, $B D=3$.
$\because \cos B=\frac{A B^{2}+B C^{2}-A C^{2}}{2 \cdot A B \cdot B C}=\frac{A B^{2}+B K^{2}-A K^{2}}{2 \cdot A B \cdot B K}$.
$\therefore \frac{6^{2}+5^{2}-4^{2}}{2 \times 6 \times 5}=\frac{6^{2}+3^{2}-A K^{2}}{2 \times 6 \times 3}$. Hence, $A K... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,751 |
38. Equilateral $\angle A B C$ and equilateral $\triangle A B D$ are in two mutually perpendicular planes, then $\angle C A D=(\quad)$.
(A) $\cos ^{-1}\left(-\frac{1}{2}\right)$
(B) $\cos ^{-1}\left(\frac{1}{4}\right)$
(C) $\cos ^{-1}\left(-\frac{7}{16}\right)$
(D) $\frac{\pi}{2}$
(E) * | 38. (B).
As shown in Figure 4, take the midpoint $E$ of $AB$. According to the problem, we know that $CE \perp AB$, $DE \perp AB$, $CE \perp DE$, and $CE = DE = \frac{\sqrt{3}}{2} AB$.
Thus, $CD = \sqrt{2} CE = \frac{\sqrt{6}}{2} AB$.
Then, $\cos \angle CAI$
$$
=\frac{2 \cdot AB^{2} - CD^{2}}{2 \cdot AB^{2}} = 1 - \fr... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,752 |
39. $\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$ The value is ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{2}$
$(\mathrm{E}) *$ | 39. (B).
Let $\alpha=\tan ^{-1}\left(\frac{1}{2}\right) \cdot \beta=\tan ^{-1}\left(\frac{1}{3}\right)$. Then $\tan \alpha=\frac{1}{2}, \tan \beta=\frac{1}{3}, \alpha, \beta \in\left(0, \frac{\pi}{4}\right)$. $\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{\frac{1}{2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,753 |
40. The value of $x$ that satisfies $\cot ^{-1} 2=\cot ^{-1} x+\cot ^{-1} 7$ is ).
(A) 1
(B) 2
(C) 3
(D) 4
$(\mathrm{E}) *$ | 40. (C).
Let $\alpha=\cot ^{-1} 2 \cdot \beta=\cot ^{-1} 7$. Then $\tan \alpha=\frac{1}{2}, \tan \beta=\frac{1}{7}, \alpha, \beta \in\left(0, \frac{\pi}{4}\right)$.
$$
\begin{aligned}
\therefore \tan (\alpha-\beta) & =\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{\frac{1}{2}-\frac{1}{7}}{1+\frac{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,754 |
1. The number of integer solutions to the equation $\left(x^{2}-x-1\right)^{x+2}=1$ is ( ).
(A) 5
(B) 4
(C) 3
(D) 2 | -1. (B).
When $x^{2}-x-1=1$, $x_{1}=2, x_{2}=-1$; when $x+2=0$ and $x^{2}-x-1 \neq 0$, $x_{3}=-2$; when $x^{2}-x-1=-1$ and $x+2$ is even, $x=0, x=1$ (discard). Therefore, the integer solutions of the equation are
$$
x_{1,3}= \pm 2, x_{2}=-1, x_{4}=0 .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,755 |
Example 5 Proof: It is possible to color the positive integers $1,2, \cdots, 2000$ using 4 colors such that it does not contain an arithmetic sequence of 7 numbers of the same color.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direc... | Proof: The problem is equivalent to partitioning the set
$$
S=\{1,2, \cdots, 2000\}
$$
into 4 non-empty subsets $M_{1}, M_{2}, M_{3}, M_{4}$, such that
$$
M_{i} \cap M_{j}=\varnothing (i \neq j), M_{1} \cup M_{2} \cup M_{3} \cup M_{4}=S.
$$
Since $6 \times 7^{3} > 2000$, every number in $S$ can be represented as a 7-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,757 |
3. Given that the sum of 5 different positive odd numbers is 85. Then the range of the largest number $M$ among these five numbers is ( ).
(A) $23 \leqslant M \leqslant 67$
(B) $19 \leqslant M \leqslant 67$
(C) $21 \leqslant M \leqslant 69$
(D) $17 \leqslant M \leqslant 69$ | 3. (C).
Naturally, when taking 4 positive odd numbers as $1,3,5,7$, the maximum number reaches its maximum value of 69; when the maximum odd number is $x$, and the other four numbers are $x-8, x-6, x-4, x-2$, then $x-8+x-6+x-4+x-2+x=$ 85, $x=21$. Therefore, $21 \leqslant M \leqslant 69$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,758 |
5. On the hypotenuse $AB$ of the isosceles right $\triangle ABC$, take two points $M$ and $N$ such that $\angle MCN=45^{\circ}$. Let $AM=m, MN=x$, and $BN=n$. The shape of the triangle with side lengths $x, m, n$ is ( ).
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Varies with $x, m, n$ | 5.(B).
As shown in Figure 4.
Construct $C D \perp C M$, and
$C D = C M$, connect
$N D$ and $B D$. It is easy to prove
$\triangle C A M \cong \triangle C B D$.
Therefore, $\angle C B D = \angle A = 45^{\circ}, A M = B D = m$.
Since $\angle M C N = 45^{\circ}$,
we have $\angle N C D = \angle M C D - \angle M C N = 90^{\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,759 |
6. The graph of the quadratic function $y=-\frac{1}{2} x^{2}+\frac{1999}{2} x+1000$ passes through ( ) integer lattice points in the first quadrant (i.e., points with positive integer coordinates).
(A) 1000
(B) 1001
(C) 1999
(D) 2001 | 6. (C).
$y=-\frac{1}{2}(x-2000)(x+1)$ The range of the independent variable in the first quadrant is $0<x<2000$. When $x$ is odd, $x+1$ is even; when $x$ is even, $x-2000$ is even. Therefore, when $x=1,2, \cdots, 1999$, $y$ takes integer values.
Thus, the number of lattice points in the first quadrant is 1999. | 1999 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,760 |
1. Given non-zero real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}$ $=1$, and $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c$. $\left(\frac{1}{a}+\frac{1}{b}\right)=-3$. Then $a+b+c=$ $\qquad$ | $$
\text { Two, 1. } a+b+t=0, \pm 1 \text { . }
$$
When $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=-3$
clear the denominators, and factorize
$$
(a+b+c)(a b+b c+c a)=0,
$$
then $a+b+c=0$ or $a b+b c+c a=0$.
$$
\begin{array}{l}
\text { Also, } a^{2... | a+b+c=\pm 1, 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,761 |
3. Let $Q=\sqrt{5}-\sqrt{2}$, try to represent $Q$ with a cubic polynomial whose coefficients are rational numbers, then $\sqrt{5}=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { 3. }-\frac{1}{6} Q^{3}+\frac{17}{6} Q . \\
\because Q=\sqrt{5}-\sqrt{2}, \\
\therefore Q^{3}=11 \sqrt{5}-17 \sqrt{2} .
\end{array}
$$
(1) $\times 17-$ (2) gives
$$
17 Q-Q^{3}=6 \sqrt{5} \text {. Therefore, } \sqrt{5}=-\frac{1}{6} Q^{3}+\frac{17}{6} Q \text {. }
$$ | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,763 |
4. From a square iron sheet with a side length of 10 cm, circular pieces with a diameter of 1 cm can be cut out, at most $\qquad$ pieces.
| 4.106.
As shown in Figure 6, it is easy to see that $\angle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ is an equilateral triangle with a side length of $1 \mathrm{~cm}$, so the height $A O_{2}$ is $\frac{\sqrt{3}}{2} \mathrm{~cm}$. Assuming we can fit $k$ rows, then
$$
\frac{\sqrt{3}}{2}(k-1)+1 \leqslant 10 \text {... | 106 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,764 |
One. (20 points) Given the equations about $x$: $4 x^{2}-8 n x -3 n=2$ and $x^{2}-(n+3) x-2 n^{2}+2=0$. Does there exist a value of $n$ such that the square of the difference of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain th... | $$
\begin{aligned}
-\Delta_{1} & =(-8 n)^{2}-4 \times 4 \times(-3 n-2) \\
& =(8 n+3)^{2}+23>0 .
\end{aligned}
$$
Then $n$ is any real number, the first equation always has real roots.
Let the two roots of the first equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} . ... | n=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,765 |
II. (25 points) Several containers are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each container does not exceed 1 ton. To ensure that these containers can be transported in one go, how many trucks with a carrying capacity of 3 tons are needed at least? | First, note that the weight of each container does not exceed 1 ton, so the weight of containers that each vehicle can carry at one time will not be less than 2 tons; otherwise, another container can be added.
Let $n$ be the number of vehicles, and the weights of the containers they carry be $a_{1}, a_{2}, \cdots, a_{... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,766 |
Example 6 Let $S=\{x \mid x \geqslant 0$, in whose 4-ary representation only 0,1 are contained$\}, x$ is any non-negative real number not in $S$. Prove: there exists $y \in S$, such that $\frac{x+y}{2} \in S$.
Translate the above text into English, please keep the original text's line breaks and format, and output the... | Proof: To prove: There exist $y, z \in S$, such that
$$
x+y=2 z \text {. }
$$
Let $x=\sum_{i \leqslant m} x_{i} 4^{i}\left(i\right.$ can be negative), $x_{i} \in\{0,1, 2,3\}$. Take $t=\sum_{i \leqslant m+1}^{4} 4^{4}$, since $t$ has one more digit than $x$, the leading digit of $x + t$ will not carry over. Therefore, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,767 |
Three. (25 points) As shown in Figure 1, quadrilateral $ABCD$ is inscribed in $\odot O$, and $AB=AD$, $CB=CD$. Extend $AB$ and $DC$ to intersect at $E$, and the angle bisector of $\angle AED$ intersects $BC$ at $P$ and $AD$ at $K$. Extend $AD$ and $BC$ to intersect at $F$, and the angle bisector of $\angle BFA$ interse... | Three, as shown in Figure 7, connect $A C$,
$$
\begin{array}{l}
\because A B=A D, B C=C D, \\
A C=A C, \\
\therefore \angle A B C \cong \angle A D C,
\end{array}
$$
Therefore, $\triangle A B C = \triangle A D C$.
$$
\begin{array}{l}
\because \angle A B C + \angle A D C \\
\quad = 180^{\circ} .
\end{array}
$$
$$
\begi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,768 |
1.… a decreasing arithmetic sequence, the sum of the 5th powers of its first 7 terms is zero, and the sum of the 4th powers is 2002. Then the 7th term of this sequence is ( ).
(A) 2002
(B) -2002
(C) $3 \sqrt[4]{\frac{1001}{98}}$
(D) $-3 \sqrt[4]{\frac{1001}{98}}$ | $-.1 .(\mathrm{b})$
Let the arithmetic sequence be: $a_{t}$ ! with a common difference of $d(d<0)$.
From $u_{1}^{5}+u_{2}^{5}+\cdots+u_{7}^{5}=0$, we get
$$
\begin{array}{l}
a_{4}=0, a_{5}=-a_{3}=d, a_{n}=-a_{2}=2 d, \\
a_{7}=-a_{1}=3 d .
\end{array}
$$
From $a_{1}^{4}+a_{2}^{4}+\cdots+a_{7}^{4}=2002$, we get
$$
2 d^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,769 |
2. Given real numbers $x, y$ satisfy $x^{2}-x y+2 y^{2}=1$. Then the sum of the maximum and minimum values of $x^{2}+$ $2 y^{2}$ is equal to ( ).
(A) $\frac{8}{7}$
(B) $\frac{16}{7}$
(C) $\frac{8-2 \sqrt{2}}{7}$
(D) $\frac{8+2 \sqrt{2}}{7}$ | 2. (B).
From the given, we have $x^{2}+2 y^{2}=x y+1$.
$$
\begin{array}{l}
\text { Also, } \frac{x^{2}+2 y^{2}}{2 \sqrt{2}} \geqslant x y \geqslant-\frac{x^{2}+2 y^{2}}{2 \sqrt{2}}, \\
\therefore \frac{x^{2}+2 y^{2}}{2 \sqrt{2}}+1 \geqslant x^{2}+2 y^{2} \geqslant-\frac{x^{2}+2 y^{2}}{2 \sqrt{2}}+1 .
\end{array}
$$
S... | \frac{16}{7} | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,770 |
3. For any positive integer $n$, connect the origin $O$ with the point $A_{n}(n, n+3)$, and let $f(n)$ denote the number of all integer points on the line segment $O A_{n}$ except for the endpoints. Then the value of $f(1)+f(2)+f(3)+\cdots+f(2002)$ is ( ).
(A) 2002
(B) 2001
(C) 1334
(D) 667 | 3. (C).
The integer points $(x, y)$ on the line segment $O A_{n}$ satisfy $1=\frac{n+3}{n} \cdot x(0<x<n$, and $x \in \mathbf{N})$. When $n=3 k(k \in \mathbf{N})$,
$$
y=\frac{k+1}{k} \cdot x(0<x<3 k, x \in \mathbf{N}) .
$$
$\because k$ and $k+1$ are coprime,
$\therefore$ only when $x=k$ or $2 k$, $y \in \mathbf{N}$. T... | 1334 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,771 |
4. A sphere is circumscribed around the tetrahedron $ABCD$, and another sphere with radius 1 is tangent to the plane $ABC$ and the two spheres are internally tangent at point $D$. Given $AD = 3, \cos \angle BAC = \frac{4}{5}, \cos \angle BAD = \cos \angle CAD = \frac{1}{\sqrt{2}}$. The volume of the tetrahedron $ABCD$ ... | 4. (A).
First, prove that the height $D H$ of tetrahedron $A B C D$ is a diameter of a certain sphere.
As shown in Figure 2, let $D E \perp A B, D F \perp A C$, with the feet of the perpendiculars being $E$ and $F$, respectively, then
$$
\begin{array}{l}
A E=A F \\
=A D \cos \angle B A D=\frac{3}{\sqrt{2}} . \\
\cos \... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,772 |
5. Given complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$. And $z_{1}+z_{2}+z_{3}+z_{4}=0$. Then the quadrilateral formed by the points corresponding to these four complex numbers must be ( ).
(A) Rectangle
(B) Square
(C) Parallelogram
... | 5. (1).
Let the complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ correspond to points $A_{1}, A_{2}, A_{3}, A_{4}$, respectively. From $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$, we know that $A_{1} A_{2} A_{3} A_{4}$ is a quadrilateral inscribed in the unit circle.
From $z_{1}+z_{2}+... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,773 |
6. Let the area of an ellipse with a major axis of length $2a$ and a minor axis of length $2b$ be $\pi a b$. Given that the hypotenuse $BC=2$ of an isosceles right $\triangle ABC$. Then, the maximum value of the area of an ellipse that is tangent to each side of $\triangle ABC$ and whose major axis is parallel to the s... | 6.(B).
As shown in Figure 3, establish a rectangular coordinate system. Let the center of the inscribed ellipse of $\triangle ABC$ be $(0, b) (b>0)$, and the semi-major axis length be $a (0<a<1)$. Then the equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{(y-b)^{2}}{b^{2}}=1.
$$
The equation of line $AC$ is
$$
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,774 |
1. Given $\frac{\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha)}{\cos \alpha \cdot \cos \beta}=\frac{4}{9}$. Then the value of $\frac{\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha)}{\cos (\alpha+\beta+\gamma) \cdot \cos \gamma}$ is $\qquad$ . | $$
\begin{array}{l}
=1 \cdot \frac{4}{5} . \\
\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha) \\
=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta+2 \gamma)], \\
\cos \alpha \cdot \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)], \\
\cos (\alpha+\beta+\gamma) \cdot \cos \gamma \\
=\frac{1}{2}[\cos (\alp... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,775 |
2. From the 99 natural numbers $1,2,3, \cdots, 99$, the number of ways to choose two different numbers such that their sum is less than 99 is $\qquad$ ways. | 2.2352 .
Let the two selected numbers be $x$ and $y$, then $x+y$ has three cases:
$$
x+y=100, x+y<100, x+y>100 \text {. }
$$
Below, we only consider the first two scenarios:
(1) When $x+y=100$, the number of ways to select is 49;
(2) When $x+y<100$. This indicates that the number of ways to select when $x+y<100$ is e... | 2352 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,776 |
5. Let $a_{n}$ denote the number of $n$-digit decimals in the interval $[0.1)$ that do not contain the digit 9, and let $S_{n}$ denote the sum of these decimals. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{a_{n}}=$ $\qquad$ | 5. $\frac{4}{9}$.
Obviously, $a_{n}=9^{n}$, where the numbers with $1,2,3, \cdots, 8$ appearing in the $i$-th digit $(i=1,2, \cdots, n)$ each have $9^{n-1}$ occurrences. Then
$$
\begin{aligned}
S_{n}= & 9^{n-1}(1+2+3+\cdots+8) \\
& \times\left(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\cdots+\frac{1}{10^{n}}\righ... | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,780 |
6. A rectangular piece of land enclosed by fences has a length and width of $52 \mathrm{~m}$ and $24 \mathrm{~m}$, respectively. An agricultural science technician wants to divide this land into several congruent square test plots. The land must be fully divided, and the sides of the squares must be parallel to the bou... | 6.702 pieces.
Assume the land is divided into several squares with side length $x$, then there exist positive integers $m, n$, such that
$$
\frac{24}{x}=m \text{, and } \frac{52}{x}=n \text{. }
$$
$\therefore \frac{m}{n}=\frac{6}{13}$, i.e., $m=6 k, n=13 k(k \in \mathbf{N})$.
Note that when the value of $k$ is as larg... | 702 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,781 |
Three. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}$. And for any positive integer $n$, we have
$$
a_{n+2}=2(n+2) a_{n+1}-(n+2)(n+1) a_{n}+\frac{n^{2}+3 n+1}{n+3} \text {. }
$$
Try to find the general term formula of the sequence $a_{n}$. | $$
\begin{array}{l}
\frac{a_{n+2}}{(n+2)!}=\frac{2 a_{n+1}}{(n+1)!}-\frac{a_{n}}{n!}+\frac{n^{2}+3 n+1}{(n+3)!} . \\
\therefore \frac{a_{n+2}}{(n+2)!}-\frac{1}{(n+3)!} \\
=2\left[\frac{a_{n+1}}{(n+1)!}-\frac{1}{(n+2)!}\right]-\left[\frac{a_{n}}{n!}-\frac{1}{(n+1)!}\right] .
\end{array}
$$
Therefore, $\left\{\frac{a_{n... | a_{n}=\frac{1}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,782 |
Four. (20 points) The hyperbola with $BC$ as its real axis intersects the extensions of the other two sides $AB$ and $AC$ of the triangle at $E$ and $F$. Tangents to the hyperbola at points $E$ and $F$ intersect at point $P$. Prove that $AP \perp BC$. | Four, taking the line $BC$ as the $x$-axis and the perpendicular bisector of $BC$ as the $y$-axis to establish a rectangular coordinate system, as shown in Figure 4. Suppose the equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>$
$0)$, then $B(-a, 0)$. $C(a, 0)$.
Using the parametric equat... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,783 |
Five. (20 points) Let $a, b, c$ be distinct real numbers. Prove that $(a-b)^{2}, (b-c)^{2}, (c-a)^{2}$ contain at least one that is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$. | Let's assume $a < b < c$, and denote
$$
m^{2}=\min \left\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\right\} \text {. }
$$
Then $b-a \geqslant|m|, c-b \geqslant|m|$.
$$
\begin{array}{l}
\text { Thus, } c-a=(c-b)+(b-a) \geqslant 2|m| \text {. } \\
\because 6 m^{2} \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
=2\left(a^{2}+b^{2}+c^{2}\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,784 |
一、(50 points) As shown in Figure 1, in $\angle ABC$, $AB > AC$, and $AD$ is the angle bisector of $\angle BAC$. Point $E$ is inside $\triangle ABC$, and $EC \perp AD$, $ED \parallel AC$. Prove that ray $AE$ bisects side $BC$. | As shown in Figure 5, ray $AE$ intersects $BC$ at $M$, and the extensions of $CE$ and $DE$ intersect $AB$ at $F$ and $G$. Let $BC = a$, $CA = b$, and $AB = c$. It is easy to see that $AF = AC = b$ and $FB = c - b$. $\triangle FBC$ is intersected by line $GED$, and by Menelaus' theorem, we have
$$
\frac{CE}{EF} \cdot \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,785 |
II. (50 points) Let $F_{n}=x^{n} \sin (n A)+y^{n} \sin (n B)+$ $z^{n} \sin (n C)$, where $x, y, z$ and $A, B, C$ are real numbers, and $A+B$ $+C=k \pi(k \in \mathbf{Z})$. Prove: If $F_{1}=F_{2}=0$, then for all positive integers $n$, we have $F_{n}=0$. | Let $\alpha=x(\cos A+\mathrm{i} \sin A), \beta=y(\cos B+\mathrm{i} \sin B)$. $\gamma=z(\cos C+\mathrm{i} \sin C)$, consider the equation with roots $\alpha, \beta, \gamma$
$$
t^{3}-a t^{2}+b t-c=0,
$$
where, $a=\alpha+\beta+\gamma=x \cos A+y \cos B+z \cos C+\mathrm{i} F_{1}$
$$
\begin{aligned}
& =x \cos A+y \cos B+z \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,786 |
Three. (50 points) Let $n \geqslant 2$ be a fixed integer. Define the remainder of any integer coordinate point $(i, j)$ modulo $n$ as the remainder of $i+j$ modulo $n$. Find all integer pairs $(a, b)$ such that the rectangle with vertices $(0,0), (a, 0)$, $(a, b), (0, b)$ has the following properties:
(i) The integer ... | On the boundary of a rectangle, there are $2(a+b)$ integer points. Thus, $n \mid 2(a+b)$.
Inside the rectangle, there are $(a-1)(b-1)$ integer points, so we have $n!(a-1)(b-1)$.
When $n=2$. To make the number of points that are divisible by 2 with a remainder of 0.1 the same, we must have $21(a-1)(b-1)$.
Thus, at least... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,787 |
Example 8 Proof: The sequence $\left\{2^{n}-3\right\}(n=2,3, \cdots)$ contains an infinite subsequence, in which any two terms are coprime. | Proof 1: Inductive construction of the required infinite subsequence $\left\{2^{n_{k}}-3\right\}(k=1,2, \cdots)$.
Let $n_{1}=3$, and inductively assume that $n_{1}, n_{2}, \cdots, n_{k}$ have been chosen such that the $k$ numbers $2^{n_{j}}-3(j=1,2, \cdots, k)$ are pairwise coprime. Let the complete set of prime divis... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,788 |
Example $1 \quad X=\{1,2,3, \cdots, 2 n+1\} . A$ is a subset of $X$, with the property: the sum of any two numbers in $A$ is not in $A$, find $\max |A|$. | Let's first consider a specific $A$. In the problem, $A$ is a set of numbers with a certain property, but the sum of any two numbers in $A$ does not have this property.
When giving examples, start with the familiar. The most common property of integers is parity, and the sum of any two odd numbers is not odd. Therefor... | n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,789 |
For example, $2 X=\{1,2,3, \cdots, n\}, A, B, C$ are partitions of $X$, i.e., $A \cup B \cup C=X$, and the intersections of $A, B, C$ are pairwise empty. If one element is taken from each of $A, B, C$, then the sum of any two is not equal to the third. Find $\max \min (|A|,|B|,|C|)$. | This problem is a bit more challenging than the previous one, but we should still start with specific examples. Let's still consider the parity. If \( A \) consists of the odd numbers in \( X \), and \( B \cup C \) consists of the even numbers in \( X \), then they meet the requirements of the problem. In this case,
\[... | \left[\frac{n}{4}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,790 |
For example, 3100 weights with masses of $1,2, \cdots, 100$ grams are placed on both sides of a balance, and it is exactly balanced. Prove: it is always possible to remove 2 weights from each side, and the balance will still be maintained.
untranslated text remains the same as requested. However, if you need any furt... | Starting from the specific, let's assume 1 (i.e., a 1-gram weight) is on the left side. Further, let $1, 2, \cdots, k$ be on the left side, and $k+1$ be on the right side. At this point, there are two scenarios:
(i) $k+2$ is on the right side.
If $k+3$ is on the left side, then, removing $k+3$ and $k$ from the left sid... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,791 |
Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$ | The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$
Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$
which simplifies to $a+b... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,792 |
Example 5 Find the sum of all numbers in the following square matrix
\begin{tabular}{ccccc}
1901 & 1902 & $\cdots$ & 1949 & 1950 \\
1902 & 1903 & $\cdots$ & 1950 & 1951 \\
$\cdots$ & & & & \\
1950 & 1951 & $\cdots$ & 1998 & 1999
\end{tabular} | Many people first use formula (5) to find the sum of the first row:
$$
S_{1}=\frac{1901+1950}{2} \times 50=96275 \text {; }
$$
Then they find the sums of the 2nd, 3rd, ..., 50th rows; finally, they add these sums together. If one notices that each number in the second row is 1 greater than the corresponding number in ... | 4875000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,793 |
Example 3 Find all real numbers $p$ such that the cubic equation $5 x^{3}$ $-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers. | Analysis: It can be observed that 1 is a solution to the equation. The equation can be transformed into
$$
(x-1)\left(5 x^{2}-5 p x+66 p-1\right)=0 \text {. }
$$
The problem is then reduced to: Find all real numbers $p$ such that the equation
$$
5 x^{2}-5 p x+66 p-1=0
$$
has natural number solutions.
By Vieta's formu... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,794 |
Example 6 Set $S=\{x \mid x$ is a 9-digit number in decimal, digits are $1,2,3\}$. Mapping $f: S \longrightarrow\{1,2,3\}$. And for any pair of $x$ and $y$ in $S$ that have different digits in the same position, $f(x) \neq f(y)$. Find $f$ and its number. | First, we should clarify the problem. $S$ can be regarded as the set of 9-tuple ordered groups $\left(x_{1}, x_{2}, \cdots, x_{9}\right)$, where $x_{1} \in \{1, 2, 3\}, 1 \leqslant i \leqslant 9$. Let $S_{i}$ be the set of elements in $S$ that map to $i$ $(i=1,2,3)$. Then $S_{1}, S_{2}, S_{3}$ form a partition of $S$, ... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,795 |
Example 2: When A was B's current age, B was 10 years old; when B was A's current age, A was 25 years old. Who is older, A or B? How many years older? | Solution: Let the age difference between A and B be $k$ years, which is an undetermined constant. When $k>0$, A is older than B; when $k<0$, A is younger than B. Then, A's age $y$ and B's age $x$ have a linear relationship:
$$
y=x+k \text {. }
$$
After designing this dynamic process, the given conditions become 3 "mom... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,797 |
Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$. | Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$
which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$
where all th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,798 |
Example 5 Find the range of real number $a$ such that for any real number $x$ and any $\theta \in\left[0, \frac{\pi}{2}\right]$, the following inequality always holds:
$$
\begin{array}{l}
(x+3+2 \sin \theta \cdot \cos \theta)^{2}+(x+a \sin \theta+ \\
a \cos \theta)^{2} \geqslant \frac{1}{8} .
\end{array}
$$
(1996, Nati... | Solution 1: Conventional Approach.
Rearrange the inequality with $x$ as the variable:
$$
\begin{array}{l}
16 x^{2}+16[3+2 \sin \theta \cdot \cos \theta+a(\sin \theta \\
+\cos \theta)] x+8(3+2 \sin \theta \cdot \cos \theta)^{2} \\
+8 a^{2}(\sin \theta+\cos \theta)^{2}-1 \geqslant 0 .
\end{array}
$$
Let the left side of... | a \in(-\infty, \sqrt{6}] \cup\left[\frac{7}{2},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,800 |
Example 6 Given real numbers $x, y$ satisfy $(3 x+y)^{5}+x^{5}$ $+4 x+y=0$. Find the value of $4 x+y$. | Given that it can be transformed as
$$
(3 x+y)^{5}+(3 x+y)=-\left(x^{5}+x\right) \text {. }
$$
Let $f(t)=t^{5}+t$ be a monotonic odd function. Equation (1) indicates that the two function values are equal, i.e.,
$$
f(3 x+y)=-f(x)=f(-x) .
$$
Therefore, the independent variables are equal, which gives
$$
3 x+y=-x,
$$
... | 4 x+y=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,801 |
Example 7 Prove the identity
\[
\begin{array}{l}
(b-c)(d-b)(d-c)+(c-a)(d-c) \\
\cdot(d-a)+(a-b)(d-a)(d-b) \\
=-(b-c)(c-a)(a-b) .
\end{array}
\] | Prove: When $a, b, c$ have at least two equal, the original expression can be directly verified as an identity. When $a, b, c$ are all different, consider $d$ as a variable, then the left side of the original expression is a polynomial in $d$ of degree no more than two.
$$
\begin{aligned}
f(d)= & (b-c)(d-b)(d-c)+(c-a) ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,802 |
Example 8 There are 5 medicine boxes, every 2 boxes contain one same medicine, each medicine appears in exactly 2 boxes, how many kinds of medicines are there? | Solution: Represent the medicine boxes as 5 points. When a medicine box contains the same medicine, draw a line segment between the corresponding points.
Since every 2 medicine boxes have one kind of the same medicine, a line should be drawn between every two points. Also, because each kind of medicine appears in exac... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,803 |
1 Addition and Subtraction
Addition and subtraction is not only an important method when looking for ideal approximate fractions. It is also a clever method to solve the aforementioned problems. Then, what is the addition and subtraction of two fractions? For example, given the fractions $\frac{a}{b}$ and $\frac{c}{d}$... | Prove: Suppose the fraction $\frac{e}{f}$ satisfies $\frac{a}{b}<\frac{e}{f}<\frac{c}{d}$, then we must have
$$
\left\{\begin{array}{l}
b e-a f \geqslant 1 . \\
c f-d e \geqslant 1 .
\end{array}\right.
$$
By multiplying the first equation by $d$ and the second by $b$ and then adding them, we get
$$
f(b c-a d) \geqslan... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,804 |
Example 4 Let $m$ be an integer, and $4<m<40$, and the equation $x^{2}-2(2 m-3) x+4 m^{2}-14 m+8=0$ has two integer roots. Find the value of $m$ and the roots of the equation. | Analysis: Considering the discriminant $\Delta=4(2 m+1)$, since it is a linear expression in terms of $m$, the methods used in Example 1 and Example 2 are not applicable.
Given $4<m<40$, we know that
$9<2 m+1<81$.
To make the discriminant a perfect square, $2 m+1$ can only be $25$ or $2 m+1=49$.
When $2 m+1=25$, $m=12... | m=12, \text{ roots } 16 \text{ and } 26; \, m=24, \text{ roots } 38 \text{ and } 52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,805 |
Example 1 Find the smallest positive integer $n$ such that for this $n$, there is a unique positive integer $k$ satisfying
$$
\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13} \text {. }
$$
(Fifth American Mathematical Invitational) | Solution: Since the fractions $\frac{8}{15}$ and $\frac{7}{13}$ satisfy $15 \times 7 - 8 \times 13 = 1$, it must be that $\frac{n}{n+k} = \frac{8+7}{15+13} = \frac{15}{28}$, thus we get $n=15, k=13$. | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,806 |
Example 2 Let $a, b$ be positive integers and satisfy
$$
\frac{2}{3}<\frac{a}{b}<\frac{5}{7} \text {. }
$$
When $b$ is the minimum value, $a+b=$ $\qquad$
(Fifth "Hope Cup" National Mathematics Invitational Competition Training Question) | Solution: According to the theorem, we have
$$
\frac{a}{b}=\frac{2+5}{3+7}=\frac{7}{10} \text {. }
$$
Thus, $a=7, b=10$. Therefore, $a+b=17$.
From the above theorem, it can be seen that although addition and subtraction is a rather clever and simple method, it is not suitable for the case where $b c-a d>1$. For exampl... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,807 |
Example 5 Find the fraction $\frac{a}{b}$ with the smallest denominator such that
$$
\frac{2}{5}<\frac{a}{b}<\frac{7}{9} .
$$ | Solution: $\frac{2}{5}<\frac{a}{b}<\frac{7}{9} \Leftrightarrow \frac{9}{7}<\frac{b}{a}<\frac{5}{2}$
$$
\Leftrightarrow \frac{2}{7}<\frac{b-a}{a}<1 \frac{1}{2} \text {. }
$$
By observation, the rational number with the smallest denominator between $\frac{2}{7}$ and $1 \frac{1}{2}$ is clearly 1, so we only need to set
$... | \frac{1}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,808 |
Example 1 Let the terms of the sequence $\left.a_{n}\right\}$ be
$$
1,2,2,3,3,3, \cdots, \underbrace{n, n, \cdots, n}_{n \uparrow}, \cdots .
$$
Find $a_{n}$ and $S_{n}$. | Solution: This is an interesting sequence, where the number of terms in each segment increases (in an arithmetic sequence).
Let the general term formula of the sequence $\{a_{n}\}$ be $a_{n}=m$ (where $m$ is the sequence number of each segment in the segmented sequence). By the sum formula of an arithmetic sequence, t... | a_{n}=\left[\frac{1+\sqrt{8 n-7}}{2}\right], \quad S_{n}=m n-\frac{1}{6} m(m+1)(m-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,809 |
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