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6. A college student spent 37 days last summer studying advanced mathematics, and followed the following rules:
(i) study at least 1 hour each day;
(ii) study in whole hours, and no more than 12 hours per day;
(iii) the total study time does not exceed 60 hours.
Prove: there exists a consecutive period of days during w... | Proof: Let the student study $a_{i}$ hours on the $i$-th day, then the total study time for the first $i$ days is $A_{1}=a_{1}+a_{2}+\cdots+a_{i}$, where $1 \leqslant a_{1} \leqslant 12, i=1,2, \cdots, 37$. According to the problem, we have
$$
\begin{array}{l}
1 \leqslant A_{1}<A_{2}<\cdots<A_{37} \leqslant 60, \\
14 \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,645 |
8. A $9 \times 9$ grid of squares is colored in two colors, black and white, such that the number of black squares adjacent to each white square is greater than the number of white squares, and the number of white squares adjacent to each black square is greater than the number of black squares (squares sharing a commo... | Solution: A coloring scheme of a square matrix that satisfies the problem's conditions is called a good scheme. For a good coloring scheme, it is easy to prove the following conclusions:
(1) If two adjacent cells have the same color and are in different rows (or columns), then any two adjacent cells in these two rows (... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,647 |
9. There are $n$ students, and the teacher assigns $N$ problems to these students. Student $i$ gets $a_{i}$ problems, $1 \leqslant i \leqslant n$. If all $a_{i}$ are not equal, then the problems are reassigned using the following method: two students $i$ and $j$ together get $a_{i}+a_{j}$ problems. If this number is ev... | Solution: Rewrite the original proposition as follows:
There are $n$ non-negative integers that sum to $N$. In each step, choose any two numbers $a$ and $b$ with the same parity, and replace the original $a$ and $b$ with $\frac{a+b}{2}$. Is it possible after several steps to make all the numbers the same?
(1) Prove tha... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,648 |
10. $N$ teams participate in a hockey league, each team plays a match against every other team exactly once. It is known that among any 3 teams, not every pair of teams has a draw. Find the maximum number of draws.
(53rd Belarusian Mathematical Olympiad (Final C Category)) | Let $d_{i}$ be the number of draws for the $i$-th team, and let $k=\max _{1 \leqslant i \leqslant v} d_{1}$. Without loss of generality, assume $d_{v}=k$, and the $N$-th team draws with the $i$-th team $(1 \leqslant i \leqslant k)$.
By the given conditions, the $i$-th team and the $j$-th team $(1 \leqslant i, j \leqsl... | M = \left\{\begin{array}{ll} m^2, & \text{当 } N = 2m, \\ m(m+1), & \text{当 } N = 2m+1. \end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,649 |
11. There are $n$ teams participating in a football league, each team plays a match with every other team only once. It is stipulated that a win earns 3 points, a draw earns 1 point, and a loss earns 0 points. After the league ends, some teams may be disqualified, and thus their match results will also be canceled. The... | Solution: Since $f_{1}(T) \leqslant n-1(i=1,2, \cdots, n)$, we have $F(T) \leqslant n(n-1)$. Additionally, in the league $T_{0}$, all matches end in a draw, so for the $i$-th team to become the champion, it must disqualify the other $n-1$ teams, hence $f_{1}\left(T_{0}\right)=n-1$. Therefore,
$$
F\left(T_{0}\right)=n(n... | n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,650 |
12.12 People are sitting around a table in a circle. How many ways are there for 6 pairs of people to shake hands so that no arms cross?
(2001, British Mathematical Olympiad (Round 1)) | Solution: According to the problem, handshakes only occur on the table, not behind other people.
Let $N_{p}$ denote the total number of ways for $2p$ people sitting around a circular table to shake hands without crossing arms, and each person shakes hands only once. Consider each person as a point. Without loss of gen... | 132 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,651 |
Example 1 (Ball Partitioning Problem) There are $n$ balls, which are arbitrarily divided into two piles, and the product of the number of balls in these two piles is calculated; then one of the piles is arbitrarily divided into two piles, and the product of the number of balls in these two piles is calculated; and so o... | Explanation: This is a constant value problem, meaning the sum of the products of the number of balls in each pile is independent of the method of division and depends only on the number of balls $n$.
We can start with the simplest cases, making $n$ specific and simple. The simplest cases are when $n=1,2,3,4,5$.
For ... | f(n)=\frac{n(n-1)}{2} | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,652 |
Example 2 (Standard Segment Problem) On the two endpoints of line segment $A B$, one is marked in blue, and the other is marked in red. Insert $n$ points in the middle of the segment, and randomly mark these points with red or blue. This divides the original segment into $n+1$ non-overlapping smaller segments. Those sm... | Explanation: Start with simple cases. The simplest case is when no points are inserted between $A B$, at this time $A B$ is a standard line segment, i.e., there is only one standard line segment, and the number is obviously odd.
Next, if there is one point between $A B$, this point can be either red or blue, so the th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,653 |
Example 7 Let $O$ be the vertex of a parabola, $F$ the focus, and $P Q$ a chord passing through $F$. Given $|O F|=a,|P Q|=$ $b$. Find the area of $\triangle O P Q$.
(1991, National High School Mathematics Competition) | Analysis 1: To find the area of $\triangle ABC$, common formulas are $S=\frac{1}{2} a h_{a}$ or $S=\frac{1}{2} a b \sin C$. If the former is used to find the area of $\triangle OPQ$, the key is to find the distance from point $O$ to $PQ$, which requires establishing a Cartesian coordinate system and finding the equatio... | a \sqrt{a b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,654 |
15. Huáshì Supermarket offers discounts to customers, with the following rules:
(1) If a single shopping trip is less than 200 yuan, no discount is given;
(2) If a single shopping trip is between 200 yuan and 500 yuan, a 10% discount is given;
(3) If a single shopping trip exceeds 500 yuan, the first 500 yuan is given ... | Three, 15. The first payment of 198 yuan might be the actual price of the purchased items without any discount; it could also be the amount paid after a 10% discount. Therefore, we should discuss two scenarios.
(1) When 198 yuan is the amount paid for shopping without a discount, the original price of the purchased ite... | 712.4 \text{ yuan or } 730 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,655 |
16. When $m$ is an integer, does the equation $(2 m-1) x^{2}-$ $(2 m+1) x+1=0$ have rational roots? If so, find the value of $m$; if not, explain the reason. | 16. Since $m$ is an integer, $2 m-1 \neq 0$.
From $\Delta=(2 m+1)^{2}-4(2 m-1)=4 m(m-1)+5$,
and since $m(m-1)$ is a multiple of 2, we know that $\Delta$ can be expressed in the form $8 k+5\left(k=\frac{m(m-1)}{2}\right.$ is an integer), which means $\Delta$ is an odd number.
However, the square of an odd number $\lef... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,656 |
17. Given a wire of length $150 \mathrm{~cm}$, it is to be cut into $n(n>2)$ smaller segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, find the maximum value of $n$, and how many ways there are to cut the wire into $n$ segments that satisfy the co... | 17. Since the sum of $n$ segments is a fixed value of $150(\mathrm{~cm})$, to make $n$ as large as possible, the length of each segment must be as small as possible. Given that the length of each segment is no less than $1(\mathrm{~cm})$, and no three segments can form a triangle, the lengths of these segments can only... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,657 |
18. As shown in Figure $5, \odot O$ is the circumcircle of $\triangle A B C$, $\angle B A C=60^{\circ}$, and $H$ is the intersection of the altitudes $B D$ and $C E$ from sides $A C$ and $A B$ respectively. Take a point $M$ on $B D$ such that $B M = C H$.
(1) Prove that $\angle B O C = \angle B H C$;
(2) Prove that $\t... | 18. (1) From $\angle \mathrm{BAC}=60^{\circ}$, we know
$\angle B O C=2 \angle B A C=120^{\circ}$.
Since $\angle B H C=\angle D H E$
$$
=360^{\circ}-\left(90^{\circ}+90^{\circ}+\angle B A C\right)=120^{\circ} \text {, }
$$
Therefore, $\angle B O C=\angle B H C$.
(2) From $O B=O C$, we get $\angle O B C=\angle O C B$.
... | \sqrt{3} | Geometry | proof | Yes | Yes | cn_contest | false | 713,658 |
1. Person A first bought 3 sheep at an average price of $a$ dollars each at the market, and then bought 2 more sheep at an average price of $b$ dollars each. Later, he sold all the sheep to Person B at a price of $\frac{a+b}{2}$ dollars each, and found that he had lost money. The reason for the loss is ( ).
(A) $a=b$
(... | $-1 .(\mathrm{B})$.
$3 a+2 b>5 \times \frac{a+b}{2}$, solving yields $a>b$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,659 |
3. As shown in Figure 1, in the right trapezoid $A B C D$, the bases $A B=13, C D=8, A D \perp A B$ and $A D$ $=$ 12. Then the distance from $A$ to $B C$ is ( ).
(A) 12
(B) 13
(C) $\frac{12 \times 21}{13}$
(D) 10.5 | 3. (A).
Draw $C E \perp A B$ at $E$, then $A E C D$ is a rectangle.
Therefore, $A E=D C=8, C E=A D=12$.
Thus, $B E=A B-A E=5, B C=\sqrt{C E^{2}+B E^{2}}=$ 13, which means $\triangle A B C$ is an isosceles triangle, with the heights from the two equal sides being equal, so $A F$ $=C E=12$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,661 |
5. As shown in Figure 2, in $\triangle A B C$, $E F$ $/ / B C, S_{\triangle A E F}=S_{\triangle B C E}$. If $S_{\triangle B B C}$ $=1$, then $S_{\triangle C E F}$ equals ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{5}$
(C) $\sqrt{5}-2$
(D) $\sqrt{3}-\frac{3}{2}$ | 5. (C).
Let $S_{\triangle C E F}=x$.
Then $S_{\triangle A E F}=S_{\triangle B C E}=\frac{1-x}{2}, S_{\triangle A B C}=\frac{1+x}{2}$.
Thus, $\frac{A F}{A C}=\frac{S_{\triangle A E F}}{S_{\triangle M B C}}=\frac{1-x}{1+x}$.
But $E F \parallel B C,\left(\frac{A F}{A C}\right)^{2}=\frac{S_{\triangle A E F}}{S_{\triangle ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,663 |
6. If $x_{1} 、 x_{2}$ are the roots of the equation $x^{2}+x-3=0$, then the value of $x_{1}^{3}-4 x_{2}^{2}+19$ is ( ).
(A)6
(B) 4
(C) 2
(D) 0 | 6. (D).
From the given, we have $x_{1}^{2}+x_{1}-3=0$. Thus, $x_{1}^{3}=3 x_{1}-x_{1}^{2}$. Also, $x_{2}^{2}=3-x_{2}$, so,
$$
\begin{array}{l}
x_{1}^{3}-4 x_{2}^{2}+19=3 x_{1}-x_{1}^{2}-4\left(3-x_{2}\right)+19 \\
=-\left(x_{1}^{2}+x_{1}-3\right)+4\left(x_{1}+x_{2}\right)+4=0
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,664 |
1. Prove: If the triangle $\triangle A_{1} A_{2} A_{3}$ inscribed in the parabola $y^{2}=2 p x$ has two sides $A_{1} A_{2}$ and $A_{2} A_{3}$ both tangent to the parabola $x^{2}=2 q y$, then the third side $A_{1} A_{3}$ is also tangent to this parabola. | (Given: Let the coordinates of $A_{i}$ be $\left(\frac{y_{i}^{2}}{2 p}, y_{i}\right), i=1,2,3$. It is easy to find that the necessary and sufficient condition for the line $A_{1} A_{2}$ to be tangent to the parabola $x^{2}=2 q y$ is $2 p^{2} q+y_{1} y_{2}\left(y_{1}+y_{2}\right)=0$. (1) Similarly, the necessary and suf... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,665 |
2. In $\triangle A B C$, $A B=3, A C=4, B C=5$. Now fold it so that point $B$ coincides with point $C$. Then the length of the fold line is $\qquad$ | 2. $\frac{15}{8}$.
From the given information, $\triangle A B C$ is a right triangle with $\angle A=90^{\circ}$. As shown in Figure 4, the crease line $D E$ is the perpendicular bisector of segment $B C$. It is easy to see that $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C A B$. Therefore,
$$
\frac{D E... | \frac{15}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,667 |
3. For the system of equations $\left\{\begin{array}{l}x-y=2, \\ m x+y=3\end{array}\right.$, if the solution corresponds to a point in the first quadrant of the Cartesian plane, then the range of values for $m$ is | 3. $-10$, solving gives $m>-1$.
Thus, $y=x-2=\frac{3-2 m}{m+1}$. But $m+1>0$,
so, $3-2 m>0, m<\frac{3}{2}$. Therefore, $-1<m<\frac{3}{2}$. | -1<m<\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,668 |
4. For all positive integers $n$ greater than 2, the greatest common divisor of the number $n^{5}-5 n^{3}+4 n$ is
untranslated part:
untranslated part remains the same as it is a mathematical expression. | 4.120 .
$$
n^{5}-5 n^{3}+4 n=(n-2)(n-1) n(n+1)(n+2) \text {. }
$$
For any positive integer $n$ greater than 2, the number $n^{5}-5 n^{3}+4 n$ contains the common divisor $1 \times 2 \times 3 \times 4 \times 5=120$. Therefore, the greatest common divisor of these numbers is 120. | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,669 |
5. $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-2 m x+\left(m^{2}+2 m+3\right)=0$. Then the minimum value of $x_{1}^{2}+x_{2}^{2}$ is | $5.4 \frac{1}{2}$.
If the equation has real roots, then
$$
\Delta=4 m^{2}-4\left(m^{2}+2 m+3\right)=-8 m-12 \geqslant 0 \text {. }
$$
Thus, $m \leqslant-\frac{3}{2}$.
Also, from $x_{1}+x_{2}=2 m, x_{1} x_{2}=m^{2}+2 m+3$, we get $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=2\left[(m-1)^{2}-4\right]$... | 4 \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,670 |
6. If the equation $\frac{2 k}{x-1}-\frac{x}{x^{2}-x}=\frac{k x+1}{x}$ has only 1 solution, then $k=$ | 6. $k=0$ or $\frac{1}{2}$.
The original equation, after clearing the denominator and rearranging, becomes
$$
k x^{2}-(3 k-2) x-1=0 \text {. }
$$
If $k=0$, the only solution to equation (1) is $x=\frac{1}{2}$, which is also the only solution to the original equation.
If $k \neq 0$, then
$$
\Delta=(3 k-2)^{2}+4 k=9\lef... | k=0 \text{ or } \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,671 |
Three. (20 points) Given that the perimeter of right triangle $\triangle ABC$ is 14, and its area is 7. Try to find the lengths of its three sides.
untranslated part:
(20 分) 已知 Rt $\triangle A B C$ 的周长为 14 , 面积为 7. 试求它的三边长.
Here is the translated text with the format preserved:
Three. (20 points) Given that the per... | Three, let the three sides of $\triangle A B C$ be $a, b, c$, where $c$ is the hypotenuse. According to the problem, we have the system of equations:
$$
\left\{\begin{array}{l}
a+b+c=14, \\
\frac{1}{2} a b=7, \\
a^{2}+b^{2}=c^{2} .
\end{array}\right.
$$
From equation (1), we get $a+b=14-c$.
Thus, solving gives $c=6$.
... | 4-\sqrt{2}, 4+\sqrt{2}, 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,672 |
Five. (20 points) A scientific expedition team went to the upper reaches of a certain river to investigate an ecological area. After setting out, they advanced at a speed of $17 \mathrm{~km}$ per day, traveling upstream along the riverbank for several days before reaching their destination. They then spent several days... | Let the expedition team took $x$ days to reach the ecological area, $y$ days to return, and spent $z$ days on the investigation. Then,
$$
x+y+z=60,
$$
and $17 x-25 y=-1$ or $25 y-17 x=1$.
(Here $x, y$ are positive integers)
First, find a special solution $\left(x_{0}, y_{0}\right)$ of equation (1) (here $x_{0}, y_{0}$... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,674 |
2. Given: (1) The diameter $AB$ of a semicircle is $2r$;
(2) A line $l$ outside the semicircle is perpendicular to the extension of $BA$, with the foot of the perpendicular being $T$, and the length of $AT$ is $2a\left(2a<\frac{r}{2}\right)$;
(3) There are two distinct points $M$ and $N$ on the semicircle, and their di... | (Tip: Take $A$ as the pole and $AB$ as the polar axis to establish a polar coordinate system. Then the equation of the semicircle is $\rho=2 r \cos \theta$. Let $M\left(\rho_{M}, \theta_{1}\right) 、 N\left(\rho_{N}\right.$, $\theta_{2}$ ), we can find that $\cos \theta_{1}+\cos \theta_{2}=1$. Consequently, the conclusi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,676 |
2. The sum of two numbers is 6, and their difference is 8. The quadratic equation with these two numbers as roots is ( ).
(A) $x^{2}-6 x+7=0$
(B) $x^{2}-6 x-7=0$
(C) $x^{2}+6 x-8=0$
(D) $x^{2}-6 x+8=0$ | 2. (B)
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 | 713,677 |
3. If $a+\frac{1}{b}=1, b+\frac{2}{c}=1$, then, $c+\frac{2}{a}$ equals ().
(A) 1
(B) 2
(C) 3
(D) 4 | 3. (B).
When $a=0$, $b=1$ contradicts $b+\frac{2}{c}=1$.
Therefore, $a \neq 0$ and $b \neq 1$.
From $a=\frac{b-1}{b}$, we get $\frac{2}{a}=\frac{2b}{b-1}$;
From $\frac{2}{c}=1-b$, we get $c=\frac{2}{1-b}$.
Thus, $c+\frac{2}{a}=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,678 |
4. As shown in Figure $1, A$ is on $DE$, $F$ is on $AB$, and $AC=CE$, $\angle 1=\angle 2=\angle 3$. Then the length of $DE$ is equal to ( ).
(A) $DC$
(B) $BC$
(C) $AB$
(D) $AE+AC$ | 4. (C).
Since $\angle 1=\angle 3$, therefore, $\angle D=\angle B$.
Since $\angle 1=\angle 2$, hence $\angle B C A=\angle D C E$.
Also, $A C=C E$, so, $\triangle D C E \cong \triangle B C A$.
Thus, $D E=A B$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,679 |
5. As shown in Figure 2, $P$ is a point inside $\square A B C D$, and $S_{\triangle P A B}=5$, $S_{\triangle P A D}=2$. Then $S_{\triangle P A C}$ equals ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 5.(B).
$$
\begin{array}{l}
\text { Because } S_{\triangle P A B}+S_{\triangle P C D}=\frac{1}{2} S_{\triangle A B C D}=S_{\triangle D M C}, \\
\text { Therefore, } S_{\triangle P A C}=S_{\triangle D A C}-S_{\triangle A D P}-S_{\triangle D P C} \\
\quad=S_{\triangle P A B}-S_{\triangle P A D}=3 .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,680 |
$6.3 x^{2}-k x+4$ when divided by $3 x-1$ leaves a remainder of 3. Then, $k$ is ( ).
(A) -2
(B) 2
(C) 4
(D) 9 | 6. (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 | 713,681 |
7. In $\triangle A B C$, $\angle A B C=60^{\circ}, A D$ and $C E$ bisect $\angle B A C$ and $\angle A C B$ respectively. Then the relationship between the length of $A C$ and $A E+C D$ is ( ).
(A) $A C>A E+C D$
(B) $A C=A E+C D$
(C) $A C<A E+C D$
(D) Cannot be determined | 7.(B).
Let the intersection of $A D$ and $C E$ be $O$. Take a point $F$ on $A C$ such that $A F = A E$. Connect $O F$. Then,
$\triangle A E O \cong \triangle A F O$.
Furthermore, it can be proven that
$\triangle O D C \cong \triangle O F C$.
Therefore, $A C = A E + C D$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,682 |
8. There are $n$ integers, whose product is $n$, and whose sum is zero. Then $n$ ( ).
(A) must be even
(B) must be odd
(C) may be even or may be odd
(D) does not exist
Please retain the original text's line breaks and format, and output the translation result directly. | 8. (A).
$1, 2, -1, -2$ satisfy the condition, that is, $n$ can be 4.
If $n$ is odd, then $n$ numbers are all odd, the sum of an odd number of odd numbers cannot be zero. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,683 |
10. As shown in Figure 4, in quadrilateral $A B C D$, $A B=B C, \angle A B C$ $=\angle C D A=90^{\circ}, B E \perp A D$ at $E, S_{\text {quadrilateral } A B C D}=8$. Then the length of $B E$ is ( ).
(A) 2
(B) 3
(C) $\sqrt{3}$
(D) $2 \sqrt{2}$ | 10. (D).
Draw a perpendicular from $B$ to $CD$, intersecting the extension of $DC$ at point $F$.
It can be proven that $\triangle A B F \cong \triangle C B F$.
Then $B E^{2}=8, B E=2 \sqrt{2}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,685 |
4. Let the two branches of the hyperbola $xy=1$ be $C_{1}$ and $C_{2}$, and let the three vertices of the equilateral triangle $PQR$ lie on this hyperbola.
(1) Prove that $P$, $Q$, and $R$ cannot all lie on the same branch of the hyperbola;
(2) Suppose $P(-1,-1)$ is on $C_{2}$, and $Q$, $R$ are on $C_{1}$. Find the coo... | (Hint: (1) Grasp that the perpendicular bisectors of the three sides of an equilateral triangle all pass through the center of the triangle, and use proof by contradiction to derive a contradiction; (2) Note that the inclination angle of the line connecting point $P$ and the origin is $45^{\circ}$, thus, the inclinatio... | (2+\sqrt{3}, 2-\sqrt{3}),(2-\sqrt{3}, 2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,687 |
12. If an isosceles trapezoid can be divided into two isosceles triangles, it is called a golden trapezoid. The ratio of the four sides of a golden trapezoid is | 12. $1: 1: 1: \frac{\sqrt{5}+1}{2}$.
There is only one possibility, as shown in Figure 8.
Draw the angle bisector of $\angle B D C$, intersecting $B C$ at $E$. Then
$\triangle D E C \backsim \triangle B D C$, $B E=E D=C D$.
Let $B E=1, E C=x$, then $\frac{1}{x}=\frac{x+1}{1}$.
Since $x>0$, we have $x=\frac{\sqrt{5}-1}... | 1: 1: 1: \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,688 |
14. Mr. Huang's home phone number is an eight-digit number. The sum of the number formed by the first four digits and the number formed by the last four digits is 14405. The sum of the number formed by the first three digits and the number formed by the last five digits is 16970. Mr. Huang's home phone number is | 14. 82616144 .
Let the phone number be $100000 x+10000 y+z$, where $x$, $y$, and $z$ are all natural numbers, and $100 \leqslant x \leqslant 999, 0 \leqslant y \leqslant 9, 1000 \leqslant z \leqslant 9999$. Then
$$
\left\{\begin{array}{l}
10 x+y+z=14405, \\
x+10000 y+z=16970
\end{array} \Rightarrow 1111 y-x=285 .\righ... | 82616144 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,690 |
15. A bouncy ball falls from point $A$ to the ground, bounces up to point $B$, then falls to a platform $20 \mathrm{~cm}$ high, bounces up to point $C$, and finally falls to the ground. Each time it bounces, the height it reaches is $80 \%$ of the height from which it fell. It is known that point $A$ is $68 \mathrm{~cm... | Three, 15. Let the height of point $C$ above the ground be $x \mathrm{~cm}$, then $\frac{\frac{x-20}{80 \%}+20}{80 \%}-68=x$. Solving this, we get $x=132$. Answer:略.
Note: The word "略" at the end is not translated as it seems to be a placeholder or abbreviation in the original text, possibly indicating that the detail... | 132 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,691 |
16. As shown in Figure 6, if $\frac{A B}{U V}=\frac{C D}{V W}=\frac{E F}{W U}$, prove:
$$
\frac{B C}{X Y}=\frac{D E}{Y Z}=\frac{F A}{Z X}
$$ | 16. Through points $A$ and $B$, draw lines parallel to $UW$ and $WV$ respectively, and let their intersection be $P$. Connect $PE$ and $PD$. Then $\triangle ABP \backsim \triangle UVW$. Thus,
$$
\begin{array}{c}
\frac{AB}{UV}=\frac{BP}{VW} \\
=\frac{PA}{WU} .
\end{array}
$$
We get $CD = BP$, $EF = PA$.
Then $CD \xlong... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,692 |
4. In the group stage of the World Cup football tournament, each group of 4 teams plays a round-robin format, where the winning team gets 3 points, the losing team gets 0 points, and in the case of a draw, both teams get 1 point. After the group stage, the top 2 teams with the highest total points advance to the next r... | 4. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 713,696 |
5. Let $A, B, C, D$ be four distinct points arranged sequentially on a straight line. Two circles are drawn with diameters $AC$ and $BD$, respectively, intersecting at points $X$ and $Y$. The line $XY$ intersects $BC$ at $Z$. If $P$ is a point on the line $XY$ different from $Z$, the line $CP$ intersects the circle wit... | (Tip: Take the line $A C$ as the $x$-axis and the line $X Y$ as the $y$-axis to establish a Cartesian coordinate system. Prove that when point $P$ is determined, the intersection point $S$ of $A M$ with the $y$-axis and the intersection point $S^{\prime}$ of $B N$ with the $y$-axis are the same point.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,698 |
6. As shown in Figure 3, in trapezoid $A B C D$, $A D / / B C, A D=3, B C$ $=9, A B=6, C D=4$. If $E F / / B C$, and the perimeters of trapezoids $A E F D$ and $E B C F$ are equal, then the length of $E F$ is ().
(A) $\frac{45}{7}$
(B) $\frac{33}{5}$
(C) $\frac{39}{5}$
(D) $\frac{15}{2}$ | 6. (C)
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 | 713,699 |
7. As shown in Figure 4, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $AC=b$, $AB=c$. If $D$ and $E$ are points on $AB$ and the extension of $AB$ respectively, such that $BD=BC$, $CE \perp CD$, then the quadratic equation with the lengths of $AD$ and $AE$ as roots is ( ).
(A) $x^{2}-2 c x+b^{2}=0$
(B... | 7. (A) | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,700 |
13. As shown in Figure 6, Building A is $16 \mathrm{~m}$ tall, and Building B is located directly north of Building A. It is known that the angle between the sun's rays and the horizontal plane at noon on the winter solstice is $30^{\circ}$. At this time, find:
(1) If the distance between the two buildings is $20 \math... | Three, 13. (1) Suppose that when the sun is at its lowest in winter, the shadow of point $A$, the highest point of Building A, falls on point $C$ of Building B. Then, the length of $C D$ is the height of Building A's shadow on Building B.
Let $C E \perp A B$ at point $E$. In $\triangle A E C$,
$$
\angle A E C=90^{\circ... | 4.4 \text{ m}, 27.7 \text{ m} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,706 |
14. As shown in Figure 7, $\triangle ABC$ is an isosceles right triangle, $\angle C = 90^{\circ}, O$ is a point inside $\triangle ABC$, and the distances from point $O$ to the sides of $\triangle ABC$ are all equal to 1. $\triangle ABC$ is rotated $45^{\circ}$ clockwise around point $O$ to get $\triangle A_{1} B_{1} C_... | 14. (1) Connect $O C, O C_{1}$, intersecting $P Q, P N$ at points $D$ and $E$ respectively. According to the problem, $\angle C O C_{1}=45^{\circ}$. Since point $O$ is equidistant from the sides of $\triangle A B C$ and this distance is 1, $O C$ bisects $\angle A C B$.
Since $\angle A C B=90^{\circ}$, we have $\angle O... | 4 \sqrt{2}-2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,707 |
15. The teachers and students of a township primary school went to the county town for a visit. It was stipulated that the bus would depart from the county town and arrive at the school at 7:00 AM to pick up the visiting teachers and students and immediately head to the county town. However, the bus broke down on its w... | 15. Suppose it took \( x \) minutes to fix the car. As shown in Figure 8, let point \( A \) be the location of the county town, point \( C \) be the location of the school, and point \( B \) be the place where the teachers and students met the car on the way.
In the 30 minutes that the teachers and students were late ... | 38 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,708 |
In 2003, the 5th problem of the Chinese Mathematical Olympiad was as follows:
A company needs to hire a secretary, with a total of 10 people applying. The company manager decides to interview them in the order of their application, and the first 3 people will definitely not be hired. Starting from the 4th person, he wi... | Prove: Under the policy of the company manager, there is
(1) $A_{1}>A_{2}>\cdots>A_{8}=A_{9}=A_{10}$;
(2) The company has more than a $70 \%$ chance of hiring one of the top three most capable individuals; while there is no more than a $10 \%$ chance of hiring one of the bottom three least capable individuals.
This is... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,709 |
1. If $a$ and $b$ are both prime numbers, and $a^{2}+b=2003$, then the value of $a+b$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002 | $-、 1 .(\mathrm{C})$.
Since 2003 is an odd number, one of $a$ or $b$ must be the even prime number 2. If $b=2$, then $a^{2}=2001$, but 2001 is not a perfect square. Therefore, $a=2, b=1999$, and we have $a+b=2001$. | 2001 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,710 |
2. Let $a>0>b>c, a+b+c=1, M=$ $\frac{b+c}{a}, N=\frac{a+c}{b}, P=\frac{a+b}{c}$. Then the size relationship between $M, N, P$ is ( ).
(A) $M>N>P$
(B) $N>P>M$
(C) $P>M>N$
(D) $M>P>N$ | 2. (D).
Since $M=\frac{1}{a}-1, N=\frac{1}{b}-1$,
$$
P=\frac{1}{c}-1, N-P=\frac{c-b}{b c}<0, M>P$. Therefore, $M>P>N$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,711 |
3. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+c=$ $8, bc=a^{2}-12a+52$. Then the perimeter of $\triangle A B C$ is ( ).
(A) 10
(B) 14
(C) 16
(D) cannot be determined | 3. (B).
Since $(b-c)^{2}=64-4\left(a^{2}-12 a+52\right)$
$$
=-4(a-6)^{2} \geqslant 0 \text {, }
$$
Therefore, $a=6$, and $b=c$.
Also, $b+c=8$, then $b=c=4$. Thus, $a+b+c=14$. | 14 | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,712 |
4. Among the following 4 propositions:
(1) In a right-angled triangle, if two sides are 3 and 4, then the third side is 5;
(2) $x \sqrt{-\frac{1}{x}}=\sqrt{-x}$;
(3) A quadrilateral with equal and perpendicular diagonals is a square;
(4) If in quadrilateral $A B C D$, $A D / / B C$, and $A B + B C = A D + D C$, then qu... | 4. (B).
Only (4) is correct.
As shown in Figure 2, extend $C B$ to $E$ such that $B E=A B$. Extend $A D$ to $F$ such that $D F=D C$. Then quadrilateral $E C F A$ is a parallelogram. Therefore, $\angle E=\angle F$.
Given $\angle A B C=2 \angle E, \angle A D C=2 \angle F$,
we know $\angle A B C=\angle A D C$.
Since $A D... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,713 |
5. A four-digit number $\overline{a a b b}$ is a perfect square. Then the value of $a+b$ is ( ).
(A) 11
(B) 10
(C) 9
(D) 8 | 5.(A).
Since $\overline{a a b b}=1000 a+100 a+10 b+b$
$$
=11(100 a+b),
$$
By the problem, we can set $100 a+b=11 c^{2}$ (where $c$ is a positive integer),
Therefore, $101<100 a+b=11 c^{2}<999$,
That is, $9<c^{2}<90$.
Thus, $4 \leqslant c \leqslant 9$.
Upon inspection, when $c=8$, the condition is satisfied, at this t... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,714 |
6. If $\triangle A B C$ has exactly one solution when $\angle A B C=60^{\circ}, A C=12, B C=$ $k$, then the range of $k$ is ( ).
(A) $k=8 \sqrt{3}$
(B) $0<k \leqslant 12$
(C) $k \geqslant 12$
(D) $0<k \leqslant 12$ or $k=8 \sqrt{3}$ | 6.(D).
As shown in Figure 3, according to the problem, the circle with center $C$ and radius $k$ intersects the major arc $\mathscr{A C}$ at exactly one point $B$. It is easy to see that $0<k \leqslant 12$ or $k=$ $8 \sqrt{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,715 |
1. Given that $a$, $b$, and $c$ are real numbers, and the polynomial $x^{3}+$ $a x^{2}+b x+c$ can be divided by $x^{2}+3 x-4$. Then $2 a-2 b$ $-c=$ $\qquad$ . | $=、 1.14$.
From the problem, -4 and 1 are roots of the equation $x^{3}+a x^{2}+b x+c=0$, which means
$$
\begin{array}{l}
16 a-4 b+c=64, \\
a+b+c=-1 .
\end{array}
$$
(1) $\times 6$ - (2) gives $2 a-2 b-c=14$. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,716 |
2. If positive integers $a$, $b$, and $c$ satisfy $a b + b c = 518$, $a b - a c = 360$, then the maximum value of $a b c$ is $\qquad$ . | 2. 1008 .
From $ab+bc=518, ab-ac=360$, subtracting the two equations, we get $c(a+b)=2 \times 79$.
Upon verification, taking $c=2, a+b=79$, then $ab=518-bc=518-2b$.
Upon inspection, when $b=7$, $ab$ reaches its maximum value of 504. Therefore, the maximum value of $abc$ is 1008. | 1008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,717 |
3. If $a b c=1, \frac{x}{1+a+a b}+\frac{x}{1+b+b c}+$ $\frac{x}{1+c+a c}=2003$, then $x=$ $\qquad$ | 3. 2003 .
Given $a b c=1$, we have
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}=1,
$$
thus $x=2003$. | 2003 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,718 |
4. As shown in Figure $1, A B$ is the diameter of semicircle $O$, quadrilaterals $C D M N$ and $D E F G$ are both squares, where $C, D, E$ are on $A B$, and $F, N$ are on the semicircle. If $A B=10$, then the sum of the areas of square $C D M N$ and square $D E F G$ is $\qquad$
Translate the above text into English, p... | 4.25 .
As shown in Figure 4, connect $O N, O F$.
Let $C N=x, E F=y$, then
$$
\begin{array}{l}
x^{2}+(x+D O)^{2}=25, \\
y^{2}+(y-D O)^{2}=25 .
\end{array}
$$
(1) - (2), we get
$$
\left(x^{2}-y^{2}\right)+D O \cdot(x+y)=0
$$
Since $x+y>0$, therefore, $D O=y-x$. Hence $x^{2}+y^{2}=25$. | 4.25 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,719 |
Example 7 If $\alpha, \beta$ are two fixed points of the quadratic function $F(x)=A x^{2}+B x+C$, then $\alpha, \beta$ are also two fixed points of the quartic function $F(F(x))=A\left(A x^{2}+B x+C\right)^{2}+B\left(A x^{2}+B x+\right.$ $C)+C$. | Prove: From $A \alpha^{2}+B \alpha+C=\alpha, A \beta^{2}+B \beta+C=\beta$ eliminating $B, C$, we can get
$$
F(x)=x+A(x-\alpha)(x-\beta) .
$$
Then $F(F(x))-x$
$$
\begin{aligned}
= & F(x)+A[F(x)-\alpha][F(x)-\beta]-x \\
= & A(x-\alpha)(x-\beta)+A[x+A(x-\alpha) . \\
& (x-\beta)-\alpha][x+A(x-\alpha) . \\
& (x-\beta)-\bet... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,720 |
One. (20 points) Given that $A D$ is the angle bisector of $\triangle A B C$, $I$ is a point on segment $A D$, and $\angle B I C=90^{\circ}+$ $\frac{1}{2} \angle B A C$. Prove: $I$ is the incenter of $\triangle A B C$. | As shown in Figure 5, construct the circumcircle $\odot O$ of $\triangle A B C$. Let the extension of $A D$ intersect $\odot O$ at $M$, and the extension of $B I$ intersect $\odot O$ at $E$. Connect $M E, C E, C M, I C$ and let $I C$ intersect $M E$ at point $\boldsymbol{F}$. Then
$$
\begin{array}{c}
\angle M E C=\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,721 |
II. (25 points) A steamboat tows several small boats, each fully loaded with equal weight of goods, to transport goods between two ports. It is known that when towing 4 small boats each time, it can make 16 round trips in a day; when towing 7 small boats each time, it can make 10 round trips in a day. The number of rou... | Let's assume that $x$ boats are towed each time every day, and the round trips are made $y$ times a day, with each boat having a carrying capacity of $M$. If $y=k x+b$, then
$$
\left\{\begin{array}{l}
4 k+b=16 \\
7 k+b=10 .
\end{array}\right.
$$
Solving these equations, we get $k=-2, b=24$, which means $y=-2 x+24$.
Th... | x=6, y=12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,722 |
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
``` | Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, and only let $c$ vary, where $c$ can take any integer from 1 to 9.
Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that
when $c=1$, $P$ reaches its maximum value, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,723 |
$$
\begin{array}{l}
P=\left\{n \left\lvert\, \frac{3^{n}+4^{n}}{5} \in \mathbf{N}\right., n \in \mathbf{N}\right\}, \\
Q=\left\{m \mid m=(2 k-1)^{2}+1, k \in \mathbf{N}\right\} .
\end{array}
$$
Then the relationship between $P$ and $Q$ is $($.
(A) $P=Q$
(B) $P \varsubsetneqq Q$
(C) $Q \subsetneq P$
(D) $P \not Q$ and ... | $-1 .(\mathrm{C})$.
List the unit digits of $3^{n}$ and $4^{n}$, it is easy to see that when $n=2,6,10,14$, $18, \cdots$, $\frac{3^{n}+4^{n}}{5} \in \mathbb{N}$.
Therefore, $P=\left\{n \mid n=4 m^{\prime}-2, m^{\prime} \in \mathbb{N}_{+}\right\}$.
On the other hand, for any $m \in Q, m=4\left(k^{2}-k+1\right)-2$ $\in P... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,724 |
2. As shown in Figure 1, given the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, points $M$ and $N$ are on $A B_{1}$ and $B C_{1}$ respectively, and $A M=B N$. Then,
(1) $A A_{1} \perp M N$;
(2) $A_{1} C_{1} / / M N$;
(3) $M N / /$ plane $A_{1} B_{1} C_{1} D_{1}$;
(4) $M N$ and $A_{1} C_{1}$ are skew lines.
Among the above 4... | 2. (B).
As shown in Figure 3, draw $M P \perp A B$ at $P$, and draw $N Q \perp B C$ at $Q$, then connect $P Q$. Then $P M \parallel Q N$, and $\triangle A M P$ and $\triangle B N Q$ are both isosceles right triangles. Also, $A M = B N$, so $P M = Q N$. It is easy to see that $P M \perp$ plane $A B C D$, hence $P M \pe... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,725 |
3. Let $S_{n}$ and $a_{n}$ represent the sum and the number of $n$-digit decimals in the interval $[0,1)$ that do not contain the digit 9, respectively. Then the value of $\lim _{n \rightarrow \infty} \frac{a_{n}}{S_{n}}$ is $(\quad)$.
(A) $\frac{3}{4}$
(B) $\frac{5}{4}$
(C) $\frac{7}{4}$
(D) $\frac{9}{4}$ | 3. (D).
It is easy to know that $a_{n}=9^{n}$, where in the $i$-th digit $(i=1,2, \cdots, n)$ of the decimal, the numbers with $0,1,2, \cdots, 8$ each appear $9^{n-1}$ times. Then
$$
\begin{aligned}
S_{n}= & 9^{n-1}(1+2+\cdots+8) \times\left(\frac{1}{10}+\frac{1}{10^{2}}+\cdots+\frac{1}{10^{n}}\right) \\
= & 4 \times ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,726 |
5. For all real numbers $x$, the values of all quadratic functions $f(x) = ax^2 + bx + c$ $(a < b)$ are non-negative real numbers. Then the maximum value of $\frac{b-a}{a+b+c}$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) 3
(D) 2 | 5. (A).
From the problem, we know that $a>0$ and $b^{2}-4 a c \leqslant 0$, which means $c \geqslant \frac{b^{2}}{4 a}$.
Let $b-a=t \quad(t \geqslant 0)$, then $b=a+t$. Therefore,
$$
\begin{array}{l}
M= \frac{b-a}{a+b+c} \leqslant \frac{t}{a+(a+t)+\frac{(a+t)^{2}}{4 a}} \\
=\frac{4 a t}{(3 a+t)^{2}} \leqslant \frac{4... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,728 |
6. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, one of its foci is $F_{1}$, the vertices are $A_{1}$ and $A_{2}$, and $P$ is any point on the hyperbola. Then the two circles with diameters $P F_{1}$ and $A_{1} A_{2}$, respectively, must ( ).
(A) intersect
(B) be tangent
(C) be separate
(D) any of the ... | 6. (B).
Let $P$ be a point on the hyperbola, $F_{1}$ be the left focus, and take the midpoint $M$ of $P F_{1}$. Then $|O M|$ is the distance between the centers of the circles, $\frac{\left|P F_{1}\right|}{2}$ and $a$ are the radii of the two circles. In $\triangle P F_{1} F_{2}$, the midline $|O M|=\frac{1}{2}\left|P... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,729 |
1. Given the complex number $z_{1}=2+\mathrm{i}, 2 z_{2}=$ $\frac{z_{1}+\mathrm{i}}{(2 \mathrm{i}+1)-z_{1}}$. If the three interior angles $\angle A$, $\angle B$, and $\angle C$ of $\triangle A B C$ form an arithmetic sequence, and $u=\cos A+$ $2 \mathrm{i} \cos ^{2} \frac{C}{2}$, then the range of $\left|u+z_{2}\right... | II. $1 \cdot\left[\frac{\sqrt{2}}{2}, \frac{\sqrt{5}}{2}\right)$.
From the problem, we know that $z_{2}=-\mathrm{i}, \angle B=60^{\circ}, \angle A+\angle C=120^{\circ}$, then
$$
u+z_{2}=\cos A+\left(2 \cos ^{2} \frac{C}{2}-1\right) \mathrm{i}=\cos A+\mathrm{i} \cos C .
$$
Therefore, $\left|u+z_{2}\right|^{2}=\cos ^{2}... | \left[\frac{\sqrt{2}}{2}, \frac{\sqrt{5}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,730 |
Solve the equation $x^{4}-6 x^{3}+10 x^{2}-4 x=0$.
List 8 | Solution: The original equation can be transformed into
$$
\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2-x=0 \text {. }
$$
Let $F(x)=x^{2}-3 x+2$, then the two fixed points of $F(x)$ are $2+\sqrt{2}$ and $2-\sqrt{2}$ (by the definition of a function's fixed point, we get $x^{2}-3 x+2=x$, i.e., $x^{2}-4 x+2=... | 2+\sqrt{2}, 2-\sqrt{2}, 0, 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,731 |
2. Point $P(a, b)$ is in the first quadrant, and a line $l$ is drawn through point $P$, intersecting the positive $x$ and $y$ axes at points $A$ and $B$ respectively. Then, when $P A^{2}+P B^{2}$ takes the minimum value, the slope of line $l$ is $\qquad$ . | 2. $-\frac{\sqrt{a b}}{a}$.
As shown in Figure 5, construct $P C \perp x$-axis at $C$, and $P D \perp y$-axis at $D$, then $P D=a, P C=b$.
Let $\frac{P B}{P A}=\lambda \quad(\lambda>0)$.
From $\triangle P A C \backsim \triangle B P D$, we get
$$
\frac{P B}{P A}=\frac{B D}{P C}=\frac{P D}{A C}=\lambda \text {. }
$$
Th... | -\frac{\sqrt{a b}}{a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,732 |
3. Let $x=\sin \alpha+\cos \alpha$, and $\sin ^{3} \alpha+\cos ^{3} \alpha>$ 0. Then the range of $x$ is $\qquad$ | 3. $(0, \sqrt{2}]$.
Since $x=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)$, then
$$
\begin{array}{l}
\sin \alpha \cdot \cos \alpha=\frac{x^{2}-1}{2}, \text { and }-\sqrt{2} \leqslant x \leqslant \sqrt{2} . \\
\text { Also } \sin ^{3} \alpha+\cos ^{3} \alpha>0 \\
\Leftrightarrow(\sin \alpha+\... | (0, \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,733 |
4. In a regular tetrahedron $ABCD$, points $M$ and $P$ are the midpoints of $AD$ and $CD$ respectively, and points $N$ and $Q$ are the centers of $\triangle BCD$ and $\triangle ABC$ respectively. Then the cosine of the angle between the lines $MN$ and $PQ$ is | 4. $\frac{1}{18}$.
As shown in Figure 6, let the edge length of the regular tetrahedron $ABCD$ be 1, and let $M$ and $P$ be the midpoints of $AD$ and $CD$ respectively. $N$ and $Q$ are the centers of $\triangle BCD$ and $\triangle ABC$ respectively. Take the midpoint $F$ of $BC$, and connect $AF$ and $DF$. Then $AF = ... | \frac{1}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,734 |
5. In the expansion of $(\sqrt{x}+2)^{2 n+1}$, the sum of the coefficients of the terms where the exponent of $x$ is an integer is | $$
\text { 5. } \frac{1}{2}\left(3^{2 n+1}+1\right) \text {. }
$$
The $(r+1)$-th term in the expansion is
$$
T_{r+1}=\mathrm{C}_{2 n+1}^{r} x^{\frac{2 n+1-r}{2}} \cdot 2^{r} \text {. }
$$
Since the exponent of $x$ is an integer, $r$ must be odd. Let
$$
S=\mathrm{C}_{2 n+1}^{1} \cdot 2+\mathrm{C}_{2 n+1}^{3} \cdot 2^{... | \frac{1}{2}\left(3^{2 n+1}+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,735 |
6. For sets $A$, $B$, and $C$ (not necessarily distinct), the union $A \cup B \cup C=\{1,2,3, \cdots, n\}$. The number of ordered triples of sets $(A, B, C)$ that satisfy this condition is
| $6.7^{7}$.
As shown in Figure 7, $A \cup B \cup C$ is divided into 7 non-overlapping parts $X, Y, Z, W, U, V, T$. In this way, each of the elements $1,2,3, \cdots, n$ in $A \cup B \cup C$ has 7 allocation methods, so the total number of allocation schemes for these elements is $7^{n}$. Any 2 different allocation scheme... | 7^n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,736 |
Three. (20 points) Given the ellipse $\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$, draw two perpendicular chords $A C$ and $B D$ through the center $O$. Let the eccentric angles of points $A$ and $B$ be $\theta_{1}$ and $\theta_{2}$, respectively. Find the range of $\left|\cos \left(\theta_{1}-\theta_{2}\right)\r... | When $A C, B D$ coincide with the coordinate axes,
$$
\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|=0 \text {; }
$$
When $A C, B D$ do not coincide with the coordinate axes, let $\angle x O A=\varphi_{1}$, $\angle x O B=\varphi_{2}$, then $\varphi_{1}=\varphi_{2} \pm \frac{\pi}{2}+2 k \pi(k \in \mathbf{Z})$.
Th... | 0 \leqslant\left|\cos \left(\theta_{1}-\theta_{2}\right)\right| \leqslant \frac{a^{2}-b^{2}}{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,737 |
Four. (20 points) Let the domain of the function $f(x)$ be $\mathbf{R}$. For any real numbers $\alpha, \beta$, we have
$$
f(\alpha) + f(\beta) = 2 f\left(\frac{\alpha + \beta}{2}\right) f\left(\frac{\alpha - \beta}{2}\right),
$$
and $f\left(\frac{\pi}{3}\right) = \frac{1}{2}, f\left(\frac{\pi}{2}\right) = 0$.
(1) Prov... | (1) Since $f\left(\frac{\pi}{3}\right)+f\left(\frac{\pi}{3}\right)=2 f\left(\frac{\pi}{3}\right) f(0)$, and $f\left(\frac{\pi}{3}\right)=\frac{1}{2}$, then $f(0)=1$.
Also, $f(x)+f(-x)=2 f(0) f(x)$,
$f(x)+f(\pi-x)=2 f\left(\frac{\pi}{2}\right) f\left(x-\frac{\pi}{2}\right)$, and
$f\left(\frac{\pi}{2}\right)=0$,
Therefor... | \frac{27}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 713,738 |
一、(50 points) As shown in Figure 2, $\odot O$ is tangent to side $AB$ of $\triangle ABC$ at point $D$, and tangent to side $AC$ at point $C$. $M$ is a point on side $BC$, and $AM$ intersects $DC$ at point $N$. Prove that $M$ is the midpoint of $BC$ if and only if $ON \perp BC$. | I. Sufficiency. Draw $EF // BC$ through point $N$, intersecting $AB$ and $AC$ at $E$ and $F$ respectively, and connect $OC$, $OD$, $OE$, and $OF$. Since $ON \perp BC$, then $ON \perp EF$. Also, $OC \perp AC$, so points $N$, $O$, $C$, and $F$ are concyclic. Therefore,
$\angle NFO = \angle NCO$.
Similarly, since points $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,739 |
II. (50 points) Find all positive integers $n$ that can be expressed as $n=$ $\frac{(a+b+c)^{2}}{a b c}\left(a, b, c \in \mathbf{Z}_{+}\right)$. | Let's assume $a \leqslant b \leqslant c$, and $(a, b, c)$ is the solution that minimizes $c$. From $n=\frac{(a+b+c)^{2}}{a b c}$, we get
$$
c^{2}+2(a+b) c+(a+b)^{2}=n a b c,
$$
which simplifies to $c^{2}-[n a b-2(a+b)] c+(a+b)^{2}=0$.
Let the other root of equation (1) be $c^{\prime}$, then
$$
\begin{array}{l}
c+c^{\p... | 1,2,3,4,5,6,8,9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,740 |
Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successf... | Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.
Simil... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,741 |
Example 9 Let $S$ be the unit circle in the complex plane (i.e., the set of complex numbers with modulus 1), and let $f$ be a mapping from $S$ to $S$. For any $z \in S$, define
$$
\begin{array}{l}
f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \cdots, \\
f^{(k)}(z)=\underbrace{f(\cdots(f(z)) \cdots) .}_{k \uparrow}
\end{array}
$... | When $f^{(n)}(C)=C$, it is known that $C$ is an $n$-fixed point of $f$. Clearly, an $n$-periodic point must be an $n$-fixed point. Conversely, if $C$ is an $n$-fixed point, then there must be a smallest positive integer $m$ such that $C$ is an $m$-periodic point, and at this time, $m \mid n$. Otherwise, let $n=k m+r(0<... | m^{1989}-m^{663}-m^{153}-m^{117}+m^{51}+m^{39}+m^{9}-m^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,742 |
As shown in Figure 1, in $\triangle A B C$, $\angle C=90^{\circ}$. Extend $A C$ to $A^{\prime}$ and extend $B C$ to $B^{\prime}$ such that $A A^{\prime}=B B^{\prime}=A B$. Prove: The perpendicular line $O E$ drawn from the midpoint $O$ of $A B$ must pass through the incenter of $\triangle A B C$.
Prove that the perpen... | Prove: As shown in Figure 2, construct the angle bisector $CD$ of $\angle ACB$ intersecting $OE$ at point $I$. Prove that point $I$ is the incenter of $\triangle ABC$.
Connect $AI$, draw $OS \perp BC$ at $S$, and let line $OE$ intersect $BC$ at $K$ and intersect ray $AA'$ at $M$.
Let $BC=a$, $CA=b$, $AB=c$.
Then $CB'=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,743 |
Find all pairs of positive integers $(a, b)$ such that $a b - a^{2} + b + 1$ divides $a b + 1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Since $a b-a^{2}+b+1=(a+1)(b-a+1)$, $a b+1=b(a+1)-(b-1)$, from the given divisibility conditions, we can get
$(a+1) \mid(b-1)$.
Let $b-1=k(a+1)$ (where $k$ is a non-negative integer), i.e., $b=k a+k+1$, substituting into the given divisibility condition, we get
$$
(a+1)[(k-1) a+k+2] \mid(k a+1)(a+1) .
$$
Thu... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,744 |
Clothing 125 If the center of the square $A_{1} A_{2} A_{3} A_{4}$ is $O, \odot O$ is inside the square $A_{1} A_{2} A_{3} A_{4}$, $P$ is any point on $\odot O$, and the distances from $P$ to $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{1}$ are $d_{1}, d_{2}, d_{3},$ and $d_{4}$, respectively, then $d_{1}^{3}+d_{2}... | Prove: (1) If $\odot O$ is exactly the incircle of the square $A_{1} A_{2} A_{3} A_{4}$, we prove that $d_{1}+d_{2}+d_{3}+d_{4}, d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}, d_{1}^{3}+d_{2}^{3}+d_{2}^{3}+d_{a}^{3}$ are constants.
As shown in Figure 3, let the incircle touch $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_... | 10 r^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,745 |
Let $n$ be a positive integer, $f(n)=\sum_{i=1}^{2 n+1} \frac{1}{n+i}$. Prove that
$$
\frac{13}{12} \leqslant f(n)<\frac{11}{10} .
$$ | Prove: First consider the monotonicity of $f(n)$.
$$
\begin{array}{l}
f(n+1)-f(n) \\
=\frac{1}{3 n+2}+\frac{1}{3 n+3}+\frac{1}{3 n+4}-\frac{1}{n+1} \\
=\frac{2}{(3 n+2)(3 n+3)(3 n+4)}>0 .
\end{array}
$$
Therefore, $f(n)$ is monotonically increasing with respect to $n$.
Hence, $f(n) \geqslant f(1)=\frac{1}{2}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,746 |
Example 10 Given a sequence $\left\{a_{n}\right\}: a_{1}, a_{2}, \cdots$, where $a_{1}$ is a natural number, and $a_{n+1}=\left[1.5 a_{n}\right]+1([x]$ represents the greatest integer not exceeding the real number $x)$. Can we choose the value of $a_{1}$ so that the first $10^{5}$ terms of the sequence $\left\{a_{n}\ri... | Solution: If $1 \leqslant n \leqslant 10^{5}$, $a_{n}$ is even, then $\left[1.5 a_{n}\right]=1.5 a_{n}$.
Thus, $a_{n+1}=1.5 a_{n}+1\left(1 \leqslant n \leqslant 10^{5}\right)$.
Since the fixed point of $f(x)=1.5 x+1$ is $x_{0}=$ -2, by Theorem 1, we have
$$
a_{n}=\left(\frac{3}{2}\right)^{n-1}\left(a_{1}+2\right)-2\lef... | a_{1}=2^{100000} \cdot N-2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,747 |
Example 3 (Identity Problem) Given the identity
$$
\begin{array}{l}
(2 x-1)^{8}-(a x+b)^{8}=\left(x^{2}+c x+d\right)^{4} \\
(a>0) .
\end{array}
$$
Find the values of the real numbers $a, b, c, d$. | Explanation: Since the equation holds for all real numbers $x$,
we can take some special values of $x$ and perform calculations.
Taking $x=0$, equation (1) becomes
$1-b^{8}=d^{4}$.
Taking $x=\frac{1}{2}$, equation (1) becomes
$$
-\left(\frac{a}{2}+b\right)^{8}=\left(\frac{1}{4}+\frac{c}{2}+d\right)^{4} \geqslant 0 .
$$... | a=\sqrt[8]{255}, b=-\frac{\sqrt[8]{255}}{2}, c=-1, d=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,748 |
Example 11 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and $a_{n}=$ $1+\frac{1}{a_{n-1}}, n=2,3, \cdots$. Find $a_{n}$. | Given: $f(x)=1+\frac{1}{x}$ has two distinct fixed points
$$
x_{1}=\frac{1}{2}(1+\sqrt{5}), \quad x_{2}=\frac{1}{2}(1-\sqrt{5})
$$
(From the definition of a fixed point of a function, we have $1+\frac{1}{x}=x$, solving which gives these results). By the conclusion of Theorem 2(1), we have
$$
\begin{array}{l}
\frac{a_{n... | a_{n}=\frac{1}{2}\left[\frac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}\right], n \in \mathbf{N} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,749 |
Example 12 Let the sequence $\left\{a_{n}\right\}$ satisfy $\left(2-a_{n}\right) a_{n+1}=$ $1, n \in \mathbf{N}$. Prove that: $\lim _{n \rightarrow \infty} a_{n}=1$.
(Putnam Mathematical Competition, 43rd edition) | Prove: $f(x)=\frac{1}{2-x}$ has two identical fixed points $x_{1}=x_{2}=1$ (by the definition of a fixed point of a function, $\frac{1}{2-x}=x$, solving gives), by the conclusion of Theorem 2(2), we have
$$
\frac{1}{a_{n+1}-1}=\frac{-2}{2}+\frac{1}{a_{n}-1} \text {. }
$$
Therefore, the sequence $\left\{\frac{1}{a_{n}-... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,750 |
Example 13 In the sequence $\left\{a_{n}\right\}$, $a_{1}=a>2$, and $a_{n+1}$ $=\frac{a_{n}^{2}}{2\left(a_{n}-1\right)}, n=1,2, \cdots$. Find $a_{n}$. | Let $a_{n}=b_{n}+1$, then we have
$$
b_{n+1}=\frac{b_{n}^{2}+1}{2 b_{n}}=\frac{\left(b_{n}+1\right)^{2}+\left(b_{n}-1\right)^{2}}{\left(b_{n}+1\right)^{2}-\left(b_{n}-1\right)^{2}} \text {. }
$$
Using the conclusion of Theorem 3, we get
$$
b_{n}=\frac{\left(b_{1}+1\right)^{2^{n-1}}+\left(b_{1}-1\right)^{2^{n-1}}}{\lef... | \frac{2}{1-\left(1-\frac{2}{a}\right)^{2^{n-1}}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,751 |
Let $F=\max _{1<x<3}\left|x^{3}-a x^{2}-b x-c\right|$, when $a$, $b$, $c$ take all real numbers, find the minimum value of $F$. | Below is another simplified solution:
Let $f(x)=x^{3}-a x^{2}-b x-c$, and take 4 points symmetrically in the interval $[1,3]$: $x=1, \frac{3}{2}, \frac{5}{2}, 3$. It is easy to verify that
$$
-\frac{2}{3} f(1)+\frac{4}{3} f\left(\frac{3}{2}\right)-\frac{4}{3} f\left(\frac{5}{2}\right)+\frac{2}{3} f(3)=1
$$
holds.
$$
\b... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,752 |
Proposition Let $D$ and $E$ be any two points on the side $BC$ of $\triangle ABC$ that do not coincide with the vertices $B$ and $C$, respectively. The inradii of $\triangle ABD$, $\triangle ACE$, $\triangle ABE$, $\triangle ACD$, and $\triangle ADE$ are denoted as $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, and $r_{5}$, respe... | Lemma ${ }^{[1]}$ Given $\triangle A B C$, the altitude on side $B C$ is $h, N$ is a point on side $B C$, and the inradii of $\triangle A B N$ and $\triangle A N C$ are $r_{1}$ and $r_{2}$, respectively. Then the inradius $r$ of $\triangle A B C$ satisfies
$$
r=r_{1}+\dot{r}_{2}-\frac{2 r_{1} r_{2}}{h} .
$$
Propositio... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,753 |
In 1967, V.O. Cordon established the inequality between the sides and altitudes of a triangle:
$$
\sum \frac{a^{2}}{h_{b}^{2}+h_{c}^{2}} \geqslant 2 .^{[1]}
$$
Reference [2] strengthened inequality (1) to:
$$
\sum \frac{a^{2}}{t_{b}^{2}+t_{c}^{2}} \geqslant 2
$$
$\left(t_{a} 、 t_{b} 、 t_{c}\right.$ are the lengths of ... | This article will prove that $\sum \frac{a^{2}}{m_{b}^{2}+m_{c}^{2}} \leqslant 2\left(m_{a}, m_{b}, m_{c}\right.$ are the lengths of the medians of a triangle), with equality if and only if $\triangle A B C$ is an equilateral triangle.
$$
\begin{array}{l}
\text { Proof: } \sum \frac{a^{2}}{m_{b}^{2}+m_{c}^{2}}=\sum \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,754 |
2. Among all the interior angles of a convex decagon, the maximum number of acute angles is ( ).
(A) 0
(B) 1
(C) 3
(D) 5 | 2. (C).
Since the sum of the exterior angles of any convex polygon is $360^{\circ}$, the number of obtuse exterior angles is at most 3, which means the number of acute interior angles is at most 3. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,756 |
3. If the function $y=k x(k>0)$ intersects with the function $y=\frac{1}{x}$ at points $A$ and $C$, and $A B$ is perpendicular to the $x$-axis at point $B$. Then the area of $\triangle A B C$ is ( ).
(A) 1
(B) 2
(C) $k$
(D) $k^{2}$ | 3. (A).
Let $A(x, y)$, then $x y=1$, so $S_{\triangle A B O}=\frac{1}{2} x y=\frac{1}{2}$. Since $\triangle A B O$ and $\triangle C B O$ have the same base and height, therefore, $S_{\triangle A B C}=2 \times S_{\triangle A B O}=1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,757 |
4. The number of positive integer pairs $(x, y)$ that satisfy the equation $x \sqrt{y}+\sqrt{x} y-\sqrt{2003 x}-\sqrt{2003 y}+$ $\sqrt{2003 x y}=200 B$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4.(B).
From the given equation, we have
$$
\begin{array}{l}
(\sqrt{x y}-\sqrt{2003})(\sqrt{x}+\sqrt{y}+\sqrt{2003})=0 \\
\text { and } \sqrt{x}+\sqrt{y}+\sqrt{2003}>0,
\end{array}
$$
Therefore, $\sqrt{x y}-\sqrt{2003}=0$. Hence $x y=2003$.
Since 2003 is a prime number, it must be the case that
$$
\left\{\begin{array}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,758 |
Example 4 (Chess Piece Movement Problem) On a coordinate plane, there are 4 chess pieces, the centers of whose coordinates are all integers. Each chess piece can move according to a vector whose start and end points are the centers of any two other chess pieces (a vector is a directed line segment). Prove: Any two ches... | Explanation: In this problem, the chess pieces can move in any direction. We still choose a special case to consider.
So, what is the special case? Of course, it is when several chess pieces are on the same straight line.
First, we consider the case where 3 chess pieces are on the same straight line and have integer ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 713,759 |
5. Let the area of $\triangle A B C$ be $1, D$ be a point on side $A B$, and $\frac{A D}{A B}=\frac{1}{3}$. If a point $E$ is taken on side $A C$ such that the area of quadrilateral $D E C B$ is $\frac{3}{4}$, then the value of $\frac{C E}{E A}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) $\frac{1... | 5. (B).
As shown in Figure 3, connect $B E$,
$$
S_{\triangle A D E}=1-\frac{3}{4}=\frac{1}{4} \text {. }
$$
Let $\frac{C E}{A C}=x$, then
$$
S_{\triangle A B E}=1-x \text {. }
$$
$$
S_{\triangle A D E}=\frac{1-x}{3}=\frac{1}{4}, x=\frac{1}{4} \text {. Therefore, } \frac{C E}{E A}=\frac{1}{3} \text {. }
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,760 |
6. As shown in Figure 1, in $\square A B C D$, the circle passing through points $A, B, C$ intersects $A D$ at point $E$, and is tangent to $C D$. If $A B=4, B E=5$, then the length of $D E$ is ( ).
(A) 3
(B) 4
(C) $\frac{15}{4}$
(D) $\frac{16}{5}$
Figure 1 | 6. (D).
As shown in Figure 4, connect $AC$ and $CE$.
Since $AE // BC$, quadrilateral $ABCE$ is an isosceles trapezoid.
Therefore, $AC = BE = 5$.
Also, because $DC // AB$ and $DC$ is tangent to the circle,
we have $\angle BAC = \angle ACD = \angle ABC$.
Thus, $AC = BC = AD = 5$, and $DC = AB = 4$.
Since $DC^2 = AD \cdo... | \frac{16}{5} | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,761 |
1. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c$ $=$ $\qquad$ . | -1 .
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1} 、 x_{2}$ must have opposite signs. Thus, $x_{1} x_{2}=\frac{c}{a}<0$.
By the projection theorem, we know $|O C|^{2}=|A O| \cdot|B O|$, i.e., $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,762 |
2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ . | 2.4 .
From the problem, we have
$$
\left\{\begin{array}{l}
3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\
3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 .
\end{array}\right.
$$
Solving this, we get \(3 \frac{8}{21}<m<4 \frac{13}{45}\). Therefore, \(m=4\). | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,763 |
3. As shown in Figure $2, A A^{\prime} 、 B B^{\prime}$ are the bisectors of $\angle E A B 、 \angle D B C$ respectively. If $A A^{\prime}=B B^{\prime}$
$=A B$, then the degree measure of $\angle B A C$
is $\qquad$ . | 3. 12 ${ }^{\circ}$.
Let the degree of $\angle B A C$ be $x$.
Since $A B=B B^{\prime}$, thus $\angle B^{\prime} B D=2 x, \angle C B D=4 x$.
Also, since $A B=A A^{\prime}$, then
$\angle A A^{\prime} B=\angle A B A^{\prime}=\angle C B D=4 x$.
Since $\angle A^{\prime} A B=\frac{1}{2}\left(180^{\circ}-x\right)$,
thus $\fr... | 12^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,764 |
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