problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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10. (14 points) Given the equation in terms of $x$
$$
x^{3} \sin \theta-(\sin \theta+2) x^{2}+6 x-4=0
$$
has 3 positive real roots. Find
$$
u=\frac{9 \sin ^{2} \theta-4 \sin \theta+3}{(1+\cos \theta)(2 \cos \theta-6 \sin \theta-3 \sin 2 \theta+2)}
$$
the minimum value. | 10. The original equation is $(x-1)\left(x^{2} \sin \theta-2 x+4\right)=0$.
Since the original equation has 3 positive real roots, the quadratic equation in $x$, $x^{2} \sin \theta-2 x+4=0$, has 2 positive real roots, i.e., $\square$
$$
\begin{array}{l}
\left\{\begin{array}{l}
\Delta=4-16 \sin \theta \geqslant 0, \\
\... | \frac{621}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,969 |
Example 1 In $\triangle ABC$, $\angle B$ is an acute angle, $AD$ is the altitude on side $BC$, and $CE$ is the altitude on side $AB$. When $\frac{2 BD}{BC}$ and $\frac{2 BE}{AB}$ are both integers, determine the shape of $\triangle ABC$ and prove your conclusion. | Analysis: To solve this type of problem, it is usually started by determining the range of integer values, utilizing the discreteness of integers for case-by-case discussion. For this, set \(\frac{2 B D}{B C}=a, \frac{2 B E}{A B}=b\), first determine the range of \(a b\) based on the given conditions, then discuss the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,970 |
Example 2 As shown in Figure 1, in rectangle $A B C D$, $D E=$ $B G$, and $\angle B E C=90^{\circ}$, $\frac{S_{\text {rectangle } A B C D}}{S_{\text {quadrilateral } E F C H}}=n, \frac{B C}{A B}=\lambda$.
Given that $n$ is a positive integer, $\lambda$ is a rational number. Prove: $\lambda$ is a positive integer. | Analysis: To solve this problem, we first need to determine the relationship between $n$ and $\lambda$, which can be obtained through a graph; then, according to the theory of prime numbers, we get that $\lambda$ is a positive integer.
Proof: It is easy to prove that quadrilateral $EFGH$ is a parallelogram.
Let $AB=a, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,971 |
Theorem 3 Any line that does not pass through the inversion center, its inverse is a circle passing through the inversion center, and vice versa. In particular, intersecting circles passing through the inversion center become intersecting lines that do not pass through the inversion center. | Proof: As shown in Figure 1, draw a perpendicular line $OC$ from $O$ to line $l$, with $C$ being the foot of the perpendicular, and $C'$ being the inverse point of $C$. Take any point $M$ on line $l$, and let $M'$ be the inverse point of $M$. Then,
$$
O M \cdot O M^{\prime}=O C \cdot O C^{\prime}=r^{2}
$$
Thus, $M, M'... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,972 |
Theorem 5 The angle between two lines or curves is invariant under inversion (the angle between two curves refers to the angle between their tangents). | Proof: We will prove a special case concerning the angle formed by a curve and a straight line.
As shown in Figure 3, let the inverse of curve $C$ be $C^{\prime}$, which intersects $O L$ at the inverse point $P^{\prime}$.
We will prove that the angle $x_{0}$ between the straight line $O L$ and the tangent to curve $C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,974 |
Example 1 Two circles are externally tangent at point $A$ and internally tangent to another circle $\odot T$ at points $B$ and $C$. Let $D$ be the midpoint of the chord of $\odot T$ cut by the common internal tangent of the smaller circles. Prove: When points $B$, $C$, and $D$ are not collinear, $A$ is the incenter of ... | Prove: With $A$ as the inversion center and $r$ (where $r$ is any real number) as the inversion radius, the inversion of Figure 4 results in Figure 5.
Among them, $\odot T_{1}$ and $\odot T_{2}$ become two parallel lines $T_{1}^{\prime}$ and $T_{2}^{\prime}$, $\odot T$ becomes a circle $\odot T^{\prime}$ tangent to th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,975 |
Example 2 In a segment, a pair of tangent circles is inscribed. For each pair of tangent circles, draw the common tangent line through their point of tangency. Prove: All the tangent lines pass through a single point.
| Proof: As shown in Figure 6, let $P$ be the point of tangency of the two circles $\odot O_{1}$ and $\odot O_{2}$. Perform an inversion with $P$ as the center of inversion. Thus, the two circles that are tangent at point $P$ are inverted into a pair of parallel lines $l_{1} \parallel l_{2}$. The chords and arcs that are... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,976 |
Example 3 As shown in Figure $8, Q$ is a point on the circle with diameter $AB$, $Q \neq A, B$, and the projection of $Q$ on $AB$ is $H$. The circle with center $Q$ and radius $QH$ intersects the circle with diameter $AB$ at points $C$ and $D$. Prove: $CD$ bisects the line segment $QH$.
(2006, Turkish National Team Sel... | Proof: Consider the inversion transformation with $Q$ as the inversion center and $\odot Q$ as the inversion circle. Then, $\odot O$ inverts to the line $CD$, and $AB$ inverts to a circle with $QH$ as its diameter and internally tangent to $\odot Q$ (as shown in Fig. 9).
Since $AB$ is the diameter of $\odot O$, $AB$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,977 |
Example 4 As shown in Figure 10, in isosceles right $\triangle ABC$, $\angle A=90^{\circ}, AB$ $=1, D$ is the midpoint of $BC$, and $E, F$ are two other points on side $BC$. $M$ is the other intersection point of the circumcircle of $\triangle ADE$ and the circumcircle of $\triangle ABF$, $N$ is the other intersection ... | Proof: With $A$ as the inversion center and $r=1$ as the inversion radius, perform an inversion transformation. Under the inversion transformation, denote the image of point $X$ as $X^{\prime}$.
By the properties of inversion,
$B, F, D, E, C$ are collinear
$\Leftrightarrow A, B^{\prime}, F^{\prime}, D^{\prime}, E^{\pri... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,978 |
Example 5 As shown in Figure 12, quadrilateral $ABCD$ is inscribed in $\odot O$, and diagonal $AC$ intersects $BD$ at $P$. Let the circumcenters of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, and $\triangle DAP$ be $O_{1}$, $O_{2}$, $O_{3}$, and $O_{4}$, respectively. Prove that $OP$, $O_{1}O_{3}$, and $O_{2}O_{... | Prove: Perform an inversion transformation with $P$ as the inversion center and the power of $P$ with respect to $\odot O$ as the inversion power. Then $\odot O$ inverts to itself, $\odot O_{i}(i=1,2,3,4)$ inverts to the lines of the sides of quadrilateral $A B C D$, and lines passing through point $P$ also invert to t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,979 |
Example 6 As shown in Figure $13, H$ is the orthocenter of $\triangle A B C$, and $P$ is any point inside $\triangle A B C$. Perpendiculars from $H$ to $P A, P B, P C$ are $H L, H M, H N$, intersecting the extensions of $B C, C A, B A$ at $X, Y, Z$. Prove that $X, Y, Z$ are collinear. | Proof: Let the altitudes of $\triangle ABC$ be $AD, BE, CF$, with feet at $D, E, F$. Then,
$$
HA \cdot HD = HB \cdot HE = HC \cdot HF.
$$
If $A, L, D, X$ are concyclic, then
$$
HL \cdot HX = HA \cdot HD.
$$
Similarly, $HM \cdot HY = HB \cdot HE$,
$$
HN \cdot HZ = HC \cdot HF.
$$
Construct an inversion with center $H... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,980 |
Example 7 As shown in Figure 14, take point $C$ on line segment $AB$, and use line segments $AC$, $BC$, and $AB$ as diameters to draw circles. Draw a line perpendicular to $AB$ through point $C$, forming curved triangles $\triangle ACD$ and $\triangle BCD$, with their inscribed circles being $\odot O_{1}$ and $\odot O_... | Proof: Let $A C=2 r_{1}, B C=2 r_{2}$. Then
$$
A B=2\left(r_{1}+r_{2}\right) \text {. }
$$
Below, we use inversion to calculate the radii of $\odot O_{1}$ and $\odot O_{2}$.
Construct an inversion with $C$ as the inversion center and $k=A C$.
$B C=4 r_{1} r_{2}$ as the inversion power (as shown in Figure 15), then $A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,981 |
Example 3 Let the inradius of $\triangle ABC$ be 1, and the lengths of the three sides be $BC=a, CA=b, AB=c$. If $a, b, c$ are all integers, prove that $\triangle ABC$ is a right triangle.
(3rd Northern Mathematical Olympiad Invitational Competition) | Analysis: How to convert geometric problems that are simple in conditions but rich in content into algebraic problems, and how to handle the algebraic problems after conversion is the difficulty in solving such problems. For this, inequality estimation can be used to continuously narrow down the range, and sometimes it... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,982 |
Example 1 Given an acute triangle $\triangle A B C$ with three interior angles satisfying $A>B>C$, let $\alpha$ represent the minimum of $A-B$, $B-C$, and $90^{\circ}-A$. Then the maximum value of $\alpha$ is $\qquad$ $(2005$, National Junior High School Mathematics League) | Given the conditions, we have
$$
\begin{aligned}
\alpha & =\frac{2 \alpha+\alpha+3 \alpha}{6} \\
& \leqslant \frac{2(A-B)+(B-C)+3\left(90^{\circ}-A\right)}{6} \\
& =\frac{270^{\circ}-(A+B+C)}{6} \\
& =15^{\circ} .
\end{aligned}
$$
When equality holds in equation (1), we have
$$
A-B=B-C=90^{\circ}-A=\alpha=15^{\circ},
... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,983 |
Example 2 Let $a, b, c \in \mathbf{R}$ and $a+b+c=1$. Find the value of $\min \{\max \{a+b, b+c, c+a\}\}$.
(2001, Beijing High School Mathematics Competition) | Solution: Let $x=\max \{a+b, b+c, c+a\}$. Then
$$
\begin{array}{l}
x=\frac{x+x+x}{3} \\
\geqslant \frac{(a+b)+(b+c)+(c+a)}{3} \\
=\frac{2}{3}(a+b+c)=\frac{2}{3} .
\end{array}
$$
When $a+b=b+c=c+a=x=\frac{2}{3}$, we can solve for $a=b=c=\frac{1}{3}$ and $x=\frac{2}{3}$. Therefore,
$$
\min \{\max \{a+b, b+c, c+a\}\}=\fr... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,984 |
Example 3 For any $a>0, b>0$, find $\max \left\{\min \left\{\frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}\right\}$.
(2002, Beijing Middle School Mathematics Competition) | Solution 1: Let $x=\left\{\min \frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}>0$.
Then $x \leqslant \frac{1}{a} \Leftrightarrow a \leqslant \frac{1}{x}$,
$x \leqslant \frac{1}{b} \Leftrightarrow b \leqslant \frac{1}{x}$,
$x \leqslant a^{2}+b^{2}$
$\leqslant\left(\frac{1}{x}\right)^{2}+\left(\frac{1}{x}\right)^{2}=\frac{... | \sqrt[3]{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,985 |
Example 4 If $a, b$ are any positive real numbers, find $\max \left\{\min \left\{a, \frac{1}{b}, b+\frac{1}{a}\right\}\right\}$.
(2003, Beijing Middle School Mathematics Competition) | Solution: Let $x=\min \left\{a, \frac{1}{b}, b+\frac{1}{a}\right\}>0$. Then
$$
\begin{array}{l}
x \leqslant a \Leftrightarrow \frac{1}{a} \leqslant \frac{1}{x}, \\
x \leqslant \frac{1}{b} \Leftrightarrow b \leqslant \frac{1}{x}, \\
x \leqslant b+\frac{1}{a} \\
\leqslant \frac{1}{x}+\frac{1}{x}=\frac{2}{x} .
\end{array}... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,986 |
1. Given real numbers $x, y$ satisfy $\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3$. Then the value of $\frac{4}{x^{4}}+y^{4}$ is ( ).
(A) 7
(B) 5
(C) $\frac{7+\sqrt{13}}{2}$
(D) $\frac{1+\sqrt{13}}{2}$ | - 1.A.
Since $x^{2}>0, y^{2} \geqslant 0$, from the given conditions we have
$$
\begin{array}{l}
\frac{1}{x^{2}}=\frac{2+\sqrt{4+4 \times 4 \times 3}}{8}=\frac{1+\sqrt{13}}{4}, \\
y^{2}=\frac{-1+\sqrt{1+4 \times 3}}{2}=\frac{-1+\sqrt{13}}{2} . \\
\text { Therefore } \frac{4}{x^{4}}+y^{4}=\frac{2}{x^{2}}+3+3-y^{2} \\
=... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,987 |
2. A fair cube die, with faces numbered $1, 2, 3, 4, 5, 6$, is thrown twice. If the numbers on the top faces are $m$ and $n$ respectively, then the probability that the graph of the quadratic function $y=x^{2}+m x+n$ intersects the $x$-axis at two distinct points is ( ).
(A) $\frac{5}{12}$
(B) $\frac{4}{9}$
(C) $\frac{... | 2.C.
The total number of basic events is $6 \times 6=36$, which means we can get 36 quadratic functions. According to the problem,
$$
\Delta=m^{2}-4 n>0 \text {, i.e., } m^{2}>4 n \text {. }
$$
By enumeration, we know that there are 17 pairs of $m, n$ that satisfy the condition. Therefore, $P=\frac{17}{36}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,988 |
3. There are two concentric circles, with 4 different points on the circumference of the larger circle and 2 different points on the circumference of the smaller circle. The number of different lines that can be determined by these 6 points is at least ( ).
(A)6
(B) 8
(C) 10
(D) 12 | 3. B.
As shown in Figure 3, there are 4 different points $A, B, C, D$ on the circumference of a large circle, and the lines connecting them can determine 6 different lines; on the circumference of a small circle, there are two points $E, F$, at least one of which is not the intersection point of the diagonals $AC$ and... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,989 |
4. Given that $AB$ is a chord of the circle $\odot O$ with radius 1, and $AB = a < 1$. A regular $\triangle ABC$ is constructed inside $\odot O$ with $AB$ as one side. $D$ is a point on $\odot O$ different from point $A$, and $DB = AB = a$. The extension of $DC$ intersects $\odot O$ at point $E$. Then the length of $AE... | 4. B.
As shown in Figure 4, connect $O E$,
$O A$, and $O B$. Let $\angle D=\alpha$, then
$$
\begin{array}{l}
\angle E C A=120^{\circ}-\alpha \\
= \angle E A C . \\
\text { Also } \angle A B O \\
= \frac{1}{2} \angle A B D \\
= \frac{1}{2}\left(60^{\circ}+180^{\circ}-2 \alpha\right) \\
= 120^{\circ}-\alpha,
\end{arra... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,990 |
5. Arrange the numbers $1,2,3,4,5$ in a row, with the last number being odd, and such that the sum of any three consecutive numbers is divisible by the first of these three numbers. How many arrangements satisfy these conditions?
(A) 2
(B) 3
(C) 4
(D) 5 | 5.D.
Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be a permutation of $1,2,3,4,5$ that meets the requirements.
First, for $a_{1}, a_{2}, a_{3}, a_{4}$, there cannot be two consecutive even numbers, otherwise, all numbers after these two would be even, which contradicts the given conditions.
Second, if $a_{i}(1 \leqslant ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,991 |
6. For real numbers $u, v$, define an operation “ * ” as: $u * v = uv + v$. If the equation $x * (a * x) = -\frac{1}{4}$ has two distinct real roots, then the range of real numbers $a$ that satisfy the condition is . $\qquad$ | II. 6. $\left\{\begin{array}{l}
a>0 \text{ or } a<0 \\
a^2 - 1 > 0
\end{array}\right.$
Solving, we get $a>0$ or $a<-1$. | a>0 \text{ or } a<-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,992 |
7. Xiao Wang walks along the street at a uniform speed and finds that a No. 18 bus passes him from behind every 6 min, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals, then, the interval ... | 7.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, and the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, with the distance between two consecutive buses traveling in the same direction being $s \mathrm{~m}$.
From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,993 |
8. As shown in Figure 1, in $\triangle A B C$, $A B=7, A C$ $=11, M$ is the midpoint of $B C$, $A D$ is the angle bisector of $\angle B A C$, $M F / / A D$. Then the length of $F C$ is $\qquad$ | 8.9.
As shown in Figure 5, let $N$ be the midpoint of $AC$, and connect $MN$. Then $MN \parallel AB$.
Also, $MF \parallel AD$, so,
$\angle FMN = \angle BAD = \angle DAC = \angle MFN$.
Therefore, $FN = MN = \frac{1}{2} AB$.
Thus, $FC = FN + NC = \frac{1}{2} AB + \frac{1}{2} AC = 9$. | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,994 |
9. In $\triangle A B C$, $A B=7, B C=8, C A=9$, through the incenter $I$ of $\triangle A B C$ draw $D E / / B C$, intersecting $A B$ and $A C$ at points $D$ and $E$ respectively. Then the length of $D E$ is $\qquad$ | 9. $\frac{16}{3}$.
As shown in Figure 6, let the side lengths of $\triangle ABC$ be $a, b, c$, the radius of the incircle $\odot I$ be $r$, and the altitude from vertex $A$ to side $BC$ be $h_{a}$. Then,
$$
\frac{1}{2} a h_{a}=S_{\triangle A B C}=\frac{1}{2}(a+b+c) r .
$$
Therefore, $\frac{r}{h_{a}}=\frac{a}{a+b+c}$.... | \frac{16}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,995 |
10. All positive integer solution pairs $(x, y)$ of the equation $x^{2}+y^{2}=208(x-y)$ are $\qquad$ . | 10. $(48,32),(160,32)$.
Since 208 is a multiple of 4, the square of an even number is divisible by 4 with a remainder of 0, and the square of an odd number is divisible by 4 with a remainder of 1, so $x$ and $y$ must both be even.
Let $x=2a, y=2b$. Then
$$
a^{2}+b^{2}=104(a-b) \text {. }
$$
Similarly, $a$ and $b$ mus... | (48,32),(160,32) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,996 |
11.A. In the Cartesian coordinate system $x O y$, the graph of the linear function $y = kx + b (k \neq 0)$ intersects the positive half-axes of the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and makes the area of $\triangle O A B$ equal to $|O A| + |O B| + 3$.
(1) Express $k$ in terms of $b$;
(2) Find t... | Three, 11.A.(1) Let $x=0$, we get $y=b(b>0)$; let $y=0$, we get $x=-\frac{b}{k}>0(k2)$.
(2) From (1), we know
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} b\left(-\frac{b}{k}\right)=\frac{b(b+3)}{b-2} \\
=\frac{(b-2)^{2}+7(b-2)+10}{b-2} \\
=b-2+\frac{10}{b-2}+7 \\
=\left(\sqrt{b-2}-\sqrt{\frac{10}{b-2}}\right)^... | 7+2 \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,997 |
11.B. Given the linear function $y_{1}=2 x$, and the quadratic function $y_{2}=x^{2}+1$. Does there exist a quadratic function $y_{3}=a x^{2}+b x+c$, whose graph passes through the point $(-5,2)$, and for any real number $x$, the function values $y_{1} 、 y_{2} 、 y_{3}$ satisfy $y_{1} \leqslant y_{3} \leqslant y_{2}$? I... | $11 . B$. There exists a quadratic function that satisfies the conditions.
Since $y_{1}-y_{2}=2 x-\left(x^{2}+1\right)$
$$
=-x^{2}+2 x-1=-(x-1)^{2} \leqslant 0 \text {, }
$$
Therefore, for any real number $x$, $y_{1} \leqslant y_{2}$ always holds.
Also, the graph of the quadratic function $y_{3}=a x^{2}+b x+c$ passes... | y_{3}=\frac{1}{3} x^{2}+\frac{4}{3} x+\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,998 |
12. A. Do there exist prime numbers $p$ and $q$ such that the quadratic equation $p x^{2}-q x+p=0$ has rational roots? | 12.A. Suppose the equation has a rational root. Then the discriminant is a perfect square. Let
$$
\Delta=q^{2}-4 p^{2}=n^{2},
$$
where $n$ is a non-negative integer. Then
$$
(q-n)(q+n)=4 p^{2} \text {. }
$$
Since $1 \leqslant q-n \leqslant q+n$, and $q-n$ and $q+n$ have the same parity, they must both be even. Theref... | p=2, q=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,999 |
12.B. Given that $a$ and $b$ are positive integers, the quadratic equation $x^{2}-2 a x+b=0$ has two real roots $x_{1}$ and $x_{2}$, and the quadratic equation $y^{2}+2 a y+b=0$ has two real roots $y_{1}$ and $y_{2}$. It is also given that $x_{1} y_{1}-x_{2} y_{2}=2008$. Find the minimum value of $b$. | 12.B. For the equation in $x$, $x^{2}-2 a x+b=0$, the roots are $a \pm \sqrt{a^{2}-b}$. For the equation in $y$, $y^{2}+2 a y+b=0$, the roots are $-a \pm \sqrt{a^{2}-b}$.
Let $\sqrt{a^{2}-b}=t$.
Then when $x_{1}=a+t, x_{2}=a-t, y_{1}=-a+t$, $y_{2}=-a-t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the c... | 62997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,000 |
13. A. Does there exist a triangle $\triangle A B C$ with side lengths that are exactly three consecutive positive integers, and one of its internal angles is twice another? Prove your conclusion. | 13. A. There exists a triangle that satisfies the conditions.
When the three sides of $\triangle ABC$ are $a=6, b=4, c=5$, $\angle A=2 \angle B$.
As shown in Figure 7, when $\angle A = 2 \angle B$, extend $BA$ to point $D$ such that $AD = AC = b$. Connect $CD$, then $\triangle ACD$ is an isosceles triangle.
Since $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,001 |
13. B. As shown in Figure 2, in $\triangle ABC$, the lengths of the three sides are $BC = a$, $CA = b$, $AB = c$, where $a$, $b$, and $c$ are all integers, and the greatest common divisor of $a$ and $b$ is 2. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find... | 13. B. As shown in Figure 8, extend $G I$, intersecting sides $B C$ and $C A$ at points $P$ and $Q$ respectively. Let the projections of the centroid $G$ on sides $B C$ and $C A$ be $E$ and $F$, and the inradius of $\triangle A B C$ be $r$. The lengths of the altitudes from $B C$ and $C A$ are $h_{a}$ and $h_{b}$ respe... | 35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,002 |
Example 5 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathemat... | Analysis: Note that the number of quadrilaterals $ABCD$ is determined by the number of triples of the lengths of $AB$, $CD$, and $AC$. Therefore, we can start by determining the shape of the quadrilateral $ABCD$.
Solution: (1) As shown in Figure 2, let
$$
AB = a, CD = b, AC = l
$$
$(a, b, l$ are all positive integers,... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,003 |
14.A. Choose $n$ numbers from $1,2, \cdots, 9$. Among them, there must be some numbers (at least one, or possibly all) whose sum is divisible by 10. Find the minimum value of $n$. | 14. A. When $n=4$, the numbers $1,3,5,8$ do not have any subset of numbers whose sum is divisible by 10.
When $n=5$, let $a_{1}, a_{2}, \cdots, a_{5}$ be five different numbers from $1,2, \cdots, 9$. If the sum of any subset of these numbers cannot be divisible by 10, then $a_{1}, a_{2}, \cdots, a_{5}$ cannot simultan... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,004 |
14. B. Given six distinct positive integers $a_{1}, a_{2}, \cdots, a_{6}\left(a_{1}<a_{2}<\cdots<a_{6}\right)$. From these six numbers, any three numbers are taken, denoted as $a_{i} 、 a_{j} 、 a_{k}$ $(i<j<k)$, and we define
$$
f(i, j, k)=\frac{1}{a_{i}}+\frac{2}{a_{j}}+\frac{3}{a_{k}} .
$$
Prove: There must exist thr... | 14.B. Among six positive integers, any three numbers can be chosen in 20 ways, thus determining 20 sets \((i, j, k)\). Since
\[
\begin{array}{l}
0 < f(i, j, k) = \frac{1}{a_{i}} + \frac{2}{a_{j}} + \frac{3}{a_{k}} \\
\leqslant \frac{1}{3} + \frac{2}{2} + \frac{3}{1} = \frac{13}{3},
\end{array}
\]
the points on the num... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,005 |
In $\triangle ABC$, $AB > AC$, its incircle touches side $BC$ at point $E$. Connect $AE$ and let it intersect the incircle at point $D$ (different from point $E$). Take a point $F$ on segment $AE$ different from point $E$ such that $CE = CF$. Connect $CF$ and extend it to intersect $BD$ at point $G$. Prove: $CF = FG$. ... | 1. Through point $D$, draw the tangent line $M N K$ of the incircle, intersecting $A B$, $A C$, and $B C$ at points $M$, $N$, and $K$ respectively.
From $\angle K D E = \angle A E K = \angle E F C$, we know $M K \parallel C G$.
By Newton's theorem, $B N$, $C M$, and $D E$ are concurrent. By Ceva's theorem, we have
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,006 |
$$
\begin{array}{l}
x_{1}=2, x_{2}=12, \\
x_{n+2}=6 x_{n+1}-x_{n}(n=1,2, \cdots) .
\end{array}
$$
Let $p$ be an odd prime, and $q$ be a prime factor of $x_{p}$. Prove: if $q \neq 2,3$, then $q \geqslant 2 p-1$. | II. It is known that
$$
x_{n}=\frac{1}{2 \sqrt{2}}\left[(3+2 \sqrt{2})^{n}-(3-2 \sqrt{2})^{n}\right](n \geqslant 1) \text {. }
$$
Let $a_{n}, b_{n} \in \mathbf{N}_{+}$.
Define $(3+2 \sqrt{2})^{n}=a_{n}+b_{n} \sqrt{2}$. Then
$$
(3-2 \sqrt{2})^{n}=a_{n}-b_{n} \sqrt{2} \text {. }
$$
It is known that $x_{n}=b_{n}, a_{n}^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,007 |
Prove: For any positive integer $n(n \geqslant 4)$, the subsets of the set $G_{n}=\{1,2, \cdots, n\}$ with at least 2 elements can be arranged in a sequence $P_{1}, P_{2}, \cdots, P_{2^{n}-n-1}$, such that $\left|P_{i} \cap P_{i+1}\right|=2\left(i=1,2, \cdots, 2^{n}-n-2\right)$. | First, when $n \geqslant 3$, use mathematical induction on $n$ to prove the following proposition:
For any positive integer $n(n \geqslant 3)$, the non-empty subsets of the set $G_{n}=\{1,2, \cdots, n\}$ can be arranged into a sequence $P_{1}, P_{2}, \cdots, P_{2^{n}-1}$, such that for any $i \in\lfloor 1$, $\left.2, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,009 |
Five, let $m$ and $n$ be given integers greater than 1, and $a_{ij}$ $(i=1,2, \cdots, n ; j=1,2, \cdots, m)$ be $mn$ non-negative real numbers, not all zero. Find
$$
f=\frac{n \sum_{i=1}^{n}\left(\sum_{j=1}^{m} a_{ij}\right)^{2}+m \sum_{j=1}^{m}\left(\sum_{i=1}^{n} a_{ij}\right)^{2}}{\left(\sum_{i=1}^{n} \sum_{j=1}^{m}... | Five, the maximum value of $f$ is 1.
First, prove that $f \leqslant 1$.
This is equivalent to
$n \sum_{i=1}^{n}\left(\sum_{j=1}^{m} a_{i j}\right)^{2}+m \sum_{j=1}^{m}\left(\sum_{i=1}^{n} a_{i j}\right)^{2}$
$\leqslant\left(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i j}\right)^{2}+m n \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i j}^{2}$
... | \frac{m+n}{m n+\min \{m, n\}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,010 |
Six, find the largest constant $M(M>0)$, such that for any positive integer $n$, there exist positive real number sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ satisfying
$$
\begin{array}{l}
\text { (1) } \sum_{k=1}^{n} b_{k}=1, \\
2 b_{k} \geqslant b_{k-1}+b_{k+1}(k=2,3, \cdots, n-1) ; \\
\... | Lemma:
$$
\max _{1 \leqslant k \leqslant n}\left\{a_{k}\right\}\frac{n-k}{n-m} b_{m}, & m\frac{m}{2} b_{m}+\frac{n-m-1}{2} b_{m}=\frac{n-1}{2} b_{m} \text {. }
$$
This implies $b_{m}<\frac{2}{n-1}$. The lemma is proved.
Returning to the original problem.
Let $f_{0}=1$,
$$
f_{k}=1+\sum_{i=1}^{k} a_{i} b_{i}(k=1,2, \cdo... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,011 |
1. Given $7 \sin \alpha + 24 \cos \alpha = 25$. Then $\tan \alpha =$ ().
(A) $\frac{3}{4}$
(B) $\frac{4}{3}$
(C) $\frac{24}{7}$
(D) $\frac{7}{24}$ | - 1.D.
From $25^{2}=(7 \sin \alpha+24 \cos \alpha)^{2}+$ $(7 \cos \alpha-24 \sin \alpha)^{2}$,
we have $7 \cos \alpha-24 \sin \alpha=0$. Therefore, $\tan \alpha=\frac{7}{24}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,012 |
Example 6 As shown in Figure $3, \odot O$ has a diameter whose length is the largest integer root of the quadratic equation in $x$
$$
x^{2}+2(k-2) x+k
$$
$=0$ (where $k$ is an integer). $P$ is a point outside $\odot O$. A tangent $PA$ and a secant $PBC$ are drawn from point $P$ to $\odot O$, with $A$ being the point of... | Analysis: First, find the value of the diameter, then discuss the cases based on $P B$ not being a composite number.
Solution: Let the two integer roots of the equation $x^{2}+2(k-2) x+k=0$ be $x_{1}$ and $x_{2}$, with $x_{1}>x_{2}$. Then
$$
x_{1}+x_{2}=-2(k-2), x_{1} x_{2}=k \text {. }
$$
From the above two equation... | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,014 |
3. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $\tan A=\frac{1}{2}, \cos B=\frac{3 \sqrt{10}}{10}$. If the longest side of $\triangle A B C$ is 1, then the length of the shortest side is ( ).
(A) $\frac{2 \sqrt{5}}{5}$
(B) $\frac{3 \sqrt{5}}{5}$
(C) $\frac{... | 3. D.
Given $\cos B=\frac{3 \sqrt{10}}{10}$, we know that $\angle B$ is an acute angle. Thus, $\tan B=\frac{1}{3}$.
Therefore, $\tan C=\tan (\pi-A-B)$
$$
\begin{array}{l}
=-\tan (A+B) \\
=-\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=-1 .
\end{array}
$$
So, $\angle C=135^{\circ}$.
Thus, the side opposite $\angle C$ is... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,015 |
4. In a convex quadrilateral $ABCD$, $AB=\sqrt{3}, BC=CD=DA=1$. Let $S$ and $T$ be the areas of $\triangle ABD$ and $\triangle BCD$, respectively. Then the maximum value of $S^2 + T^2$ is ( ).
(A) $\frac{8}{7}$
(B) 1
(C) $\frac{7}{8}$
(D) 2 | 4.C.
As shown in Figure 2, let $B D = x, \angle D A B = \theta$, and construct $C E \perp B D$. Then $\sqrt{3}-1 < x < 2$, and $E$ is the midpoint of $B D$.
$$
\begin{array}{l}
x^{2}=1^{2}+(\sqrt{3})^{2}- \\
2 \times 1 \times \sqrt{3} \cos \theta, \\
C E^{2}= 1^{2}-\frac{x^{2}}{4} .
\end{array}
$$
Therefore, $S^{2}+T... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,016 |
5. In a right-angled triangle, the sines of the three interior angles form a geometric progression. Then the smallest angle of the triangle is equal to ( ).
(A) $\arcsin \frac{\sqrt{5}-1}{2}$
(B) $\arccos \frac{\sqrt{5}-1}{2}$
(C) $\arcsin \frac{\sqrt{5}+1}{4}$
(D) $\arccos \frac{\sqrt{5}+1}{4}$ | 5.A.
Let the three sides be $1$, $x$, and $x^{2}$ $(0<x<1)$. Then the sine of the smallest angle of the triangle is $x^{2}$.
From $1^{2}=x^{2}+\left(x^{2}\right)^{2}$, we solve to get $x^{2}=\frac{\sqrt{5}-1}{2}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,017 |
6. In a regular 2008-gon, the number of diagonals that are not parallel to any side is ( ).
(A) 2008
(B) $1004^{2}$
(C) $1004^{2}-1004$
(D) $1004^{2}-1003$ | 6. C.
For a regular $2n$-sided polygon $A_{1} A_{2} \cdots A_{2 n}$, the number of diagonals is
$$
\frac{1}{2} \times 2 n(2 n-3)=n(2 n-3) \text { (diagonals). }
$$
To calculate the number of diagonals parallel to a side $A_{1} A_{2}$:
Since $A_{1} A_{2} \parallel A_{n+1} A_{n+2}$, the endpoints of the diagonals paral... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,018 |
7. The minimum distance from a point on the parabola $y^{2}=2 x$ to the line $x+y+$ $1=0$ is $\qquad$ . | $$
=7 . \frac{\sqrt{2}}{4} \text {. }
$$
Let point $P\left(\frac{b^{2}}{2}, b\right)$ be the point where the distance to the line $x+y+1=0$ is minimized. Then
$$
d=\frac{\left|\frac{b^{2}}{2}+b+1\right|}{\sqrt{2}} \geqslant \frac{\sqrt{2}}{4} .
$$ | \frac{\sqrt{2}}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 720,019 |
8. Given the equation $\sin x+\sqrt{3} \cos x=M$ has two distinct solutions in $\frac{\pi}{4} \leqslant x$ $\leqslant \frac{5 \pi}{4}$. Then the range of $M$ is | 8. $\left(-2,-\frac{\sqrt{6}+\sqrt{2}}{2}\right]$.
The original expression can be transformed into $\sin \left(x+\frac{\pi}{3}\right)=\frac{M}{2}$. Using its graph and symmetry, we can determine that the equation $\sin x+\sqrt{3} \cos x=M$ has two different solutions in the interval $\frac{\pi}{4} \leqslant x \leqslan... | \left(-2,-\frac{\sqrt{6}+\sqrt{2}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,020 |
9. Let $O(0,0), A(1,0), B(0,1)$, and $P$ be a moving point on the line segment $AB$, $\boldsymbol{A P}=\lambda \boldsymbol{A B}$. If
$$
O P \cdot A B \geqslant P A \cdot P B,
$$
then the range of the real number $\lambda$ is $\qquad$ | 9. $\left[1-\frac{\sqrt{2}}{2}, 1\right]$.
Using the graph, we can set point $P(1-\lambda, \lambda)(0<\lambda<1)$. Then
$$
\begin{array}{l}
\boldsymbol{O P}=(1-\lambda, \lambda), \boldsymbol{A} \boldsymbol{B}=(-1,1), \\
\boldsymbol{P A}=(\lambda,-\lambda), \boldsymbol{P B}(\lambda-1,1-\lambda) .
\end{array}
$$
From $... | \left[1-\frac{\sqrt{2}}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,021 |
10. Given that $n$ is an integer, and the equation
$$
(n+1)^{2} x^{2}-5 n(n+1) x+\left(6 n^{2}-n-1\right)=0
$$
$(n \neq-1)$ has two integer roots. Then $n=$ $\qquad$ . | 10.0 or -2.
Decompose the equation into
$$
[(n+1) x-(3 n+1)][(n+1) x-(2 n-1)]=0 \text {. }
$$
The roots are obtained as
$$
\begin{array}{l}
x_{1}=\frac{3 n+1}{n+1}=3-\frac{2}{n+1}, \\
x_{2}=\frac{2 n-1}{n+1}=2-\frac{3}{n+1} .
\end{array}
$$
Therefore, for $x_{1}$ to be an integer, $n+1$ can only be $\pm 1$, $\pm 2$; ... | 0 \text{ or } -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,022 |
11. Given $O$ is the circumcenter of $\triangle A B C$, $|A B|=2$, $|A C|=1, \angle B A C=\frac{2 \pi}{3}$. Let $\boldsymbol{A B}=\boldsymbol{a}, \boldsymbol{A} \boldsymbol{C}=\boldsymbol{b}$. If $\boldsymbol{A} \boldsymbol{O}=\lambda_{1} \boldsymbol{a}+\lambda_{2} \boldsymbol{b}$, then $\lambda_{1}+\lambda_{2}=$ | 11. $\frac{13}{6}$.
Establish a rectangular coordinate system as shown in Figure 3. Then $A(0,0)$, $B(2, 0)$, and $C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
Obviously, the midpoint of $A C$ is $M\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$. Let $O(1, y)$.
Since $O M \perp A C$, we have $O M \cdot A C=0$, whic... | \frac{13}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,023 |
12. Let $a, b$ be positive real numbers. Then the minimum value of $\frac{a^{3}+b^{3}+4}{(a+1)(b+1)}$ is $\qquad$ . | $\geqslant \frac{3a+3b+3ab+3}{2(a+1)(b+1)}=\frac{3}{2}$. | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,024 |
1. As shown in Figure $4, \angle B C A$ $=90^{\circ}, C D$ is the altitude. It is known that the three sides of the right triangle $\triangle A B C$ are all integers, and $B D=11^{3}$. Find the ratio of the perimeters of the right triangles $\triangle B C D$ and $\triangle A C D$.
(2002, National Junior High School Mat... | (Let $B C=a, C A=b, A B=c$, we know $a^{2}=11^{3} c$. Let $c=11 k^{2}$. Then $a=11^{2} k, b=11 k \cdot$ $\sqrt{k^{2}-11^{2}}$. Let $k^{2}-11^{2}=m^{2}$. Get $k=61, m=$ 60 , the ratio of the perimeters is equal to the similarity ratio, i.e., $\frac{a}{b}=\frac{11}{60}$.) | \frac{11}{60} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,025 |
13. As shown in Figure 1, given the tetrahedron $O-ABC$ with lateral edges $OA$, $OB$, and $OC$ mutually perpendicular, and $OA=1$, $OB=OC=2$, $E$ is the midpoint of $OC$.
(1) Find the distance from point $O$ to plane $ABC$;
(2) Find the angle formed by the skew lines $BE$ and $AC$;
(3) Find the size of the dihedral an... | Three, 13. (1) As shown in Figure 4, with $O$ as the origin, $O B$, $O C$, and $O A$ as the $x$, $y$, and $z$ axes, respectively, we have:
$$
\begin{array}{l}
A(0,0,1) \text {, } \\
B(2,0,0) \text {, } \\
C(0,2,0) \\
E(0,1,0) .
\end{array}
$$
Let the normal vector of plane $A B C$ be $\boldsymbol{n}_{1}=(x, y, z)$.
Si... | \frac{\sqrt{6}}{3}, \arccos \frac{2}{5}, \arccos \frac{7 \sqrt{6}}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,026 |
14. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, with the sum of its first $n$ terms being $S_{n}$, and for all natural numbers $n$, we have
$$
S_{n}=\frac{3 n+1}{2}-\frac{n}{2} a_{n} \text {. }
$$
(1) Write down the first three terms of the sequence $\left\{a_{n}\right\}$;
(2) Find the general term fo... | 14. (1) According to the problem, $S_{n}=\frac{3 n+1}{2}-\frac{n}{2} a_{n}\left(a_{n}>0\right)$.
When $n=1$, $S_{1}=\frac{3+1}{2}-\frac{1}{2} a_{1}=a_{1}$, solving gives $a_{1}=\frac{4}{3}$.
When $n=2$, $S_{2}=\frac{6+1}{2}-\frac{2}{2} a_{2}=a_{1}+a_{2}$, solving gives $a_{2}=\frac{13}{12}$.
When $n=3$, $S_{3}=\frac... | S_{n}^{\prime}=\frac{n(n+1)(n+2)(n+3)}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,027 |
15. Given the parabola $x^{2}=4 y$ and a fixed point $P(0,8)$, $A$ and $B$ are two moving points on the parabola, and $\boldsymbol{A P}=\lambda P B(\lambda >0)$. Tangent lines are drawn through points $A$ and $B$ to the parabola, and their intersection point is $M$.
(1) Prove that the y-coordinate of point $M$ is a con... | 15. (1) Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. The parabola equation is $y=\frac{1}{4} x^{2}$, and differentiating gives $y^{\prime}=\frac{1}{2} x$. Therefore, the equations of the tangent lines at points $A$ and $B$ on the parabola are
$$
\begin{array}{c}
y=\frac{1}{2} x_{1}\left(x-x_{1}\ri... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,028 |
16. Given the even function
$$
\begin{aligned}
f(x)= & 5 \cos \theta \cdot \sin x-5 \sin (x-\theta)+ \\
& (4 \tan \theta-3) \sin x-5 \sin \theta
\end{aligned}
$$
has a minimum value of -6.
(1) Find the maximum value of $f(x)$ and the set of $x$ at which this maximum occurs.
(2) Let the function
$$
g(x)=\lambda f(\omeg... | 16. (1) Simplify to get
$$
f(x)=5 \cos x \cdot \sin \theta+(4 \tan \theta-3) \sin x-5 \sin \theta \text {. }
$$
Since $f(x)$ is an even function, we have
$(4 \tan \theta-3) \sin x=0$
for all $x \in \mathbf{R}$.
Thus, $4 \tan \theta-3=0, \tan \theta=\frac{3}{4}$.
Therefore, $f(x)=5 \cos x \cdot \sin \theta-5 \sin \thet... | \sqrt{3}+7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,029 |
1. Divide the numbers $1,2, \cdots, 20$ into two groups, Jia and Yi, such that the average of the numbers in group Jia is 1 more than the average of the numbers in group Yi. Then group Jia has ( ) numbers.
(A) 8
(B) 9
(C) 10
(D) 11 | $-1 . C$.
Let group A have $a$ numbers. Then group B has $20-a$ numbers. Let the average of the numbers in group B be $b$. Then the average of the numbers in group A is $b+1$. According to the problem,
$$
a(b+1)+b(20-a)=1+2+\cdots+20 \text {. }
$$
Simplifying the above equation gives $a=210-20 b$.
Since $0<a<20$, we h... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,030 |
2. In quadrilateral $ABCD$, $\angle A=\angle C=90^{\circ}$, $AB=AD$, $AE \perp BC$ at $E$. If the area of quadrilateral $ABCD$ is the maximum value of the binary function
$$
y=-x^{2}-2 k x-3 k^{2}-4 k+16
$$
then the length of $AE$ is ( ).
(A) 3
(B) $\sqrt{21}$
(C) $3 \sqrt{2}$
(D) Cannot be determined | 2.C.
As shown in Figure 1, draw $A F \perp C D$, with $F$ as the foot of the perpendicular.
In quadrilateral $A B C D$, since
$$
\begin{array}{l}
\angle B A D=\angle C \\
=90^{\circ},
\end{array}
$$
thus, $\angle B$
$$
\begin{array}{l}
=180^{\circ}-\angle A D C \\
=\angle A D F .
\end{array}
$$
It is easy to see th... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,031 |
3. $n$ is a positive integer, and it is defined that $n!=1 \times 2 \times \cdots \times n$ (for example, $1!=1,2!=1 \times 2$), then
$$
1!\times 1+2!\times 2+3!\times 3+\cdots+250!\times 250
$$
when divided by 2008, the remainder is ( ).
(A) 1
(B) 2
(C) 2007
(D) 2008 | 3.C.
Notice
$$
\begin{array}{l}
n! \times n = n! \times [(n+1)-1] \\
= n! \times (n+1) - n! = (n+1)! - n!.
\end{array}
$$
Then the original expression $=(2!-1!)+(3!-2!)+$
$$
\begin{aligned}
& (4!-3!)+\cdots+(251!-250!) \\
= & 251!-1=(251!-2008)+2007 .
\end{aligned}
$$
Since $2008 = 251 \times 8$, we have,
$$
2008 \t... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,032 |
4. In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}$. A line $l \parallel AB$ is drawn through $C$, and $F$ is a point on $l$ such that $AB = AF$. Then $\angle CFB=(\quad)$.
(A) $15^{\circ}$
(B) $75^{\circ}$
(C) $105^{\circ}$
(D) None of the above results is correct | 4.D.
As shown in Figure 2, when $A F$ intersects with line segment $B C$, draw $C D \perp A B$ at $D$, and draw $A E \perp l$ at $E$. Then
$$
\begin{array}{l}
A E=C D \\
=\frac{1}{2} A B=\frac{1}{2} A F .
\end{array}
$$
In the right triangle $\triangle A E F$, we have $\angle C F A=30^{\circ}$.
Therefore, $\angle F A... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,033 |
For each positive integer $n$, let $f(n)$ denote the last digit of $1+$ $2+\cdots+n$ (for example, $f(1)=1, f(2)=3$, $f(3)=6)$. Then
$$
f(1)+f(2)+\cdots+f(2008)
$$
has the tens digit ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5.C.
First, list 1 calculates the $f(n)$ for several positive integers:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & & \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 & & \\
\hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & & \\
\hline$f(... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,034 |
6. $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $CE$ and $DF$ are common tangents of the two intersecting circles, $C, D$ are on $\odot O_{1}$, $E, F$ are on $\odot O_{2}$, line $AB$ intersects $CE$ and $DF$ at $M$ and $N$ respectively. Then the result of $MN - CE$ is $(\quad)$.
(A) positive
(B) neg... | 6.A.
As shown in Figure 4, by the secant-tangent theorem, we have
$$
\begin{array}{l}
C M^{2}=M A \cdot M B \\
=E M^{2} .
\end{array}
$$
Thus, $C M=E M$.
Similarly, $D N=F N$.
Since $C E=D F$, we have
$$
C M=D N .
$$
Also, $D N^{2}=N B \cdot N A$, so
$M A \cdot M B=N B \cdot N A$
$\Rightarrow M A(M A+A B)=N B(N B+A ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,035 |
2. As shown in Figure 5, let the side lengths of $\triangle ABC$ be natural numbers, and the lengths of $AB$ and $AC$ are coprime.
The tangent to the circumcircle of $\triangle ABC$ at point $A$ intersects the extension of $BC$ at point $D$. Prove: The lengths of $AD$ and $CD$ are rational numbers, but not integers.
(1... | (Let $B C=a, A C=b, A B=c$. Then
$$
A D=\frac{a b c}{c^{2}-b^{2}}, C D=\frac{a b^{2}}{c^{2}-b^{2}} \text {. ) }
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,036 |
1. If $a, b$ are positive numbers, and
$$
a^{2009}+b^{2009}=a^{2007}+b^{2007} \text {, }
$$
then the maximum value of $a^{2}+b^{2}$ is $\qquad$ | Given $a, b$, without loss of generality, assume $a \geqslant b>0$.
Then $a^{2} \geqslant b^{2}, a^{2007} \geqslant b^{2007}$.
$$
\begin{array}{l}
\text { At this point, }\left(a^{2007}-b^{2007}\right)\left(a^{2}-b^{2}\right) \geqslant 0 \\
\Rightarrow a^{2009}+b^{2009} \geqslant a^{2} b^{2007}+a^{2007} b^{2} \\
\Right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,037 |
3. Given the line $y=\frac{1}{2} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Then the equation of the perpendicular bisector of line segment $A B$ is . $\qquad$ | 3. $y=-2 x-\frac{3}{2}$.
As shown in Figure 6, let the perpendicular bisector of line segment $AB$ intersect $AB$ at $C$ and the negative half of the $x$-axis at $D$. Connect $BD$.
Then $AD=BD$.
It is easy to see that $A(-2,0)$ and $B(0,1)$.
By the Midline Theorem of a triangle, we know $C\left(-1, \frac{1}{2}\right)$... | y=-2 x-\frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,039 |
4. Given points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$ respectively, and the extension of $DE$ intersects the extension of $BC$ at point $F$. Then $BD + BC$ $\qquad$ $CE + ED$ (fill in “$>$”, “<” or “$=$”).
| 4. $>$.
As shown in Figure 7, draw $D G / / B C$, and draw $C G / / A B$, intersecting at point $G$.
Then quadrilateral $B C G D$ is a parallelogram.
Thus, $D G = B C$,
$$
C G = B D \text{. }
$$
Let $C G$ intersect $E F$ at $H$.
In $\triangle D G H$,
$$
D G + G H > E D + E H ;
$$
In $\triangle E C H$,
$$
E H + C H ... | B C + B D > E D + C E | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,040 |
One. (20 points) Given $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Find the integer part of $x^{6}+x^{5}+2 x^{4}-4 x^{3}+3 x^{2}+4 x-4$.
| Given $x>0$.
If $\sqrt{1+x}>x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}>\sqrt{1+\sqrt{1+x}} \\
>\sqrt{1+x},
\end{array}
$$
which contradicts the assumption;
If $\sqrt{1+x}<x$, then
$$
\begin{array}{l}
x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}<\sqrt{1+\sqrt{1+x}} \\
<\sqrt{1+x},
\end{array}
$$
which also contradi... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,041 |
II. (25 points) In an acute triangle $\triangle ABC$, the three altitudes $AD$, $BE$, and $CF$ intersect at $H$. Connect $DF$ to intersect $BH$ at $P$, and draw $PQ \parallel AD$ to intersect $AB$ at $Q$. Prove that line $QE$ bisects segment $AH$.
保留源文本的换行和格式，直接输出翻译结果。 | As shown in Figure 8, connect $E F$. Let $Q E$ intersect $A H$ at point $O$.
From the fact that $A, F, H, E$ are concyclic, we have
$\angle F E H = \angle F A H$.
Since $P Q \parallel A D$, we get
$\angle F A H = \angle F Q P$.
Thus, $\angle F E H = \angle F Q P$.
Therefore, $F, P, E, Q$ are concyclic.
So, $\angle Q E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,042 |
Three. (25 points) Given that $a$, $b$, and $c$ are positive integers, and $b^{2}-4ac$ is a perfect square. Prove that $25a+5b+c$ is a composite number.
保留源文本的换行和格式，直接输出翻译结果如下：
```
Three. (25 points) Given that $a$, $b$, and $c$ are positive integers, and $b^{2}-4ac$ is a perfect square. Prove that $25a+5b+c$ is a co... | Three, for the quadratic equation $a x^{2}+b x+c=0$ with roots $x_{1}$ and $x_{2}$, we have
$$
x_{1}+x_{2}=-\frac{b}{a}0 \text {. }
$$
Therefore, $x_{1}<0, x_{2}<0$.
By the quadratic formula, we get
$$
x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \text {. }
$$
Given that $x_{1}$ and $x_{2}$ are both rational numbers... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,043 |
1. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the inequality $x^{2}+p x>4 x+p-3$ always holds. Then the range of $x$ is ( ).
(A) $x>3$
(B) $-13$ or $x<-1$ | -、1.D.
Transform the original inequality to get
$$
p(x-1)+\left(x^{2}-4 x+3\right)>0 \text {. }
$$
Let $f(p)=(x-1) p+\left(x^{2}-4 x+3\right)$. Then $f(p)$ is a linear function of $p$. From the original inequality, we have
$$
\left\{\begin{array} { l }
{ f ( 0 ) > 0 , } \\
{ f ( 4 ) > 0 }
\end{array} \text { that is ... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 720,044 |
2. Let
$$
f(n)=\frac{5+3 \sqrt{5}}{10}\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\frac{5-3 \sqrt{5}}{10}\left(\frac{1-\sqrt{5}}{2}\right)^{n} \text {. }
$$
Then $f(n+1)-f(n-1)=(\quad)$.
(A) $\frac{1}{2} f(n)$
(B) $2 f(n)+1$
(C) $f^{2}(n)$
(D) $f(n)$ | 2.D.
$$
\begin{array}{l}
f(n+1)-f(n-1) \\
=\frac{5+3 \sqrt{5}}{10}\left(\frac{1+\sqrt{5}}{2}\right)^{n}\left[\frac{1+\sqrt{5}}{2}-\left(\frac{1+\sqrt{5}}{2}\right)^{-1}\right]+ \\
\frac{5-3 \sqrt{5}}{10}\left(\frac{1-\sqrt{5}}{2}\right)^{n}\left[\frac{1-\sqrt{5}}{2}-\left(\frac{1-\sqrt{5}}{2}\right)^{-1}\right] \\
=\fr... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,045 |
3. Given $\sin (\alpha+2 \beta)=\frac{3}{5}, \sin (2 \alpha-3 \beta)=$ $\frac{4}{5}, \alpha \in\left[\frac{\pi}{12}, \frac{\pi}{4}\right], \beta \in\left[0, \frac{\pi}{12}\right]$. Then $\sin (8 \alpha-5 \beta)$ $=(\quad)$.
(A) $\frac{4}{25}$
(B) $-\frac{4}{125}$
(C) $-\frac{4}{5}$
(D) $\frac{4}{5}$ | 3.C.
Notice
$$
\begin{array}{l}
\sin (8 \alpha-5 \beta) \\
=\sin [2(\alpha+2 \beta)+3(2 \alpha-3 \beta)] \\
=\sin 2(\alpha+2 \beta) \cdot \cos 3(2 \alpha-3 \beta)+ \\
\quad \cos 2(\alpha+2 \beta) \cdot \sin 3(2 \alpha-3 \beta) .
\end{array}
$$
From the problem, we know that $\alpha+2 \beta \in\left[\frac{\pi}{12}, \f... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,046 |
3. As shown in Figure 6, square
$E F G H$ is inscribed in $\triangle A B C$.
Let $B C=\overline{a b}$ (this is a
two-digit number), $E F=c$,
the height of the triangle $A D=d$.
It is known that $a, b, c, d$ are exactly
four consecutive positive integers in ascending order. Find the area of $\triangle A B C$.
(1997, Anh... | (Given: $b=a+1, c=\frac{\overline{a b} \times d}{\overline{a b}+d}=$ $\frac{(10 a+a+1)(a+3)}{10 a+a+1+a+3}=a+2$, we get $a=1$ or $a=$ 5. Therefore, $S_{\triangle A B C}=24$ or 224.) | 24 \text{ or } 224 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,047 |
4. The five vertices of a regular quadrilateral pyramid $P-ABCD$ lie on the same sphere. If the side length of the base of the regular quadrilateral pyramid is 4, and the lateral edge length is $2 \sqrt{6}$, then the surface area of the sphere is ( ).
(A) $36 \pi$
(B) $12 \pi$
(C) $24 \pi$
(D) $48 \pi$ | 4. A.
As shown in Figure 3, let $O$ be the center of the circumscribed sphere, and $P H \perp$ the base $A B C D$ at $H$. Then
$$
\begin{array}{c}
O P=O C=R, \\
H C=2 \sqrt{2}, \\
O H=\sqrt{R^{2}-8}, \\
P H=R+\sqrt{R^{2}-8} . \\
P H^{2}+H C^{2}=P C^{2}, \text { thus, } \\
\left(R+\sqrt{R^{2}-8}\right)^{2}+(2 \sqrt{2})... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,048 |
5. If the function
$$
f(x)=\frac{1}{3} x^{3}-\frac{1}{2} a x^{2}+(a-1) x+1
$$
is decreasing in the interval $(1,4)$ and increasing in $(6,+\infty)$, then the range of the real number $a$ is ( ).
(A) $a \leqslant 5$
(B) $5 \leqslant a \leqslant 7$
(C) $a \geqslant 7$
(D) $a \leqslant 5$ or $a \geqslant 7$ | 5. B.
Notice that $f^{\prime}(x)=x^{2}-a x+(a-1)$ $=[x-(a-1)](x-1)$.
According to the problem, when $x \in(1,4)$, $f^{\prime}(x)0$.
Therefore, $4 \leqslant a-1,6 \geqslant a-1$.
Solving this, we get $5 \leqslant a \leqslant 7$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,049 |
6. Let $A N$, $B P$, and $C M$ be the medians of $\triangle A B C$ with unequal sides, $A N=3$, $B P=6$, and $S_{\triangle A B C}$ $=3 \sqrt{15}$. Then $C M=(\quad)$.
(A)6
(B) $3 \sqrt{3}$
(C) $3 \sqrt{6}$
(D) $6 \sqrt{3}$ | 6. C.
As shown in Figure 4, let the centroid of $\triangle ABC$ be $G$, and $CM = 3x$. Then $CG = 2x$, $GM = x$.
Extend $CM$ to point $Q$ such that $MQ = GM$. Connect $BQ$ and $AQ$. Then quadrilateral $AQB G$ is a parallelogram.
Thus, $AQ = BG = 4$.
Also, $S_{\triangle AQC} = S_{\triangle ABG} = \frac{1}{3} S_{\trian... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,050 |
1. Given $x_{i} \in \mathbf{R}_{+}(i=1,2,3,4)$, and $x_{1}+x_{2}+$ $x_{3}+x_{4}=\pi$. Then the minimum value of $\prod_{i=1}^{4}\left(\sin x_{i}+\frac{1}{\sin x_{i}}\right)$ is $\qquad$ _. | $=1 . \frac{81}{4}$.
Notice that
$$
\sin x_{1}+\frac{1}{2 \sin x_{1}}+\frac{1}{2 \sin x_{1}} \geqslant 3 \sqrt[3]{\frac{1}{4 \sin x_{1}}},
$$
which means $\sin x_{1}+\frac{1}{\sin x_{1}} \geqslant 3 \sqrt[3]{\frac{1}{4 \sin x_{1}}}$.
Thus, the original expression
$$
\geqslant 3^{4} \sqrt[3]{\frac{1}{4^{4} \sin x_{1} \... | \frac{81}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,051 |
2. As shown in Figure 1, given that $G$ is the centroid of $\triangle A B O$. If $P Q$ passes through point $G$, and
$$
\begin{array}{l}
O A=a, O B=b, \\
O P=m a, O Q=n b,
\end{array}
$$
then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ | 2.3.
From $O M=\frac{1}{2}(a+b)$, we know
$$
O G=\frac{2}{3} O M=\frac{1}{3}(a+b) .
$$
From the collinearity of points $P, G, Q$, we have $\boldsymbol{P G}=\lambda \boldsymbol{G} \boldsymbol{Q}$.
And $P G=O G-O P=\frac{1}{3}(a+b)-\dot{m} a$
$$
\begin{array}{l}
=\left(\frac{1}{3}-m\right) a+\frac{1}{3} b, \\
G Q=O Q-O... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,052 |
3. If $p$, $q$, $\frac{2 p-1}{q}$, $\frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$. Then $p+q=$ $\qquad$ . | 3.8 .
If $p=q$, then
$$
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \text {. }
$$
Given $p>1$, then $\frac{2 p-1}{q}$ is not an integer, which contradicts the problem statement. Therefore, $p \neq q$.
By symmetry, without loss of generality, assume $p>q$. Let
$$
\begin{array}{l}
\frac{2 p-1}{q}=m, \\
\frac{2 q-1}{p... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,053 |
4. In a square $A B C D$ with an area of 1, a point $P$ is chosen at random. Then the probability that the areas of $\triangle P A B$, $\triangle P B C$, $\triangle P C D$, and $\triangle P D A$ are all greater than $\frac{1}{6}$ is $\qquad$ . | 4. $\frac{1}{9}$.
As shown in Figure 5, establish a rectangular coordinate system with $A$ as the origin and $AB$ as the $x$-axis.
Let $P(x, y), 0 < x < 1, 0 < y < 1$. Then, the conditions are:
$$
\left\{\begin{array}{l}
\frac{1}{2} x > \frac{1}{6}, \\
\frac{1}{2} y > \frac{1}{6}, \\
\frac{1}{2}(1 - x) > \frac{1}{6},... | \frac{1}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,054 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{n+1}=a_{n}+2 n(n=1,2, \cdots) . \\
\text { Then } \sum_{k=1}^{2007} \frac{1}{a_{k+1}-1}=
\end{array}
$$ | 5. $\frac{2007}{2008}$.
It is easy to see that $a_{2}=3$.
When $n=1$,
$$
a_{2}-1=1+2-1=2=1 \times(1+1) \text {. }
$$
Assume $a_{k}-1=(k-1) k$.
When $n=k$,
$$
\begin{array}{l}
a_{k+1}-1=a_{k}+2 k-1 \\
=(k-1) k+2 k=k(k+1) .
\end{array}
$$
Thus, $\frac{1}{a_{k+1}-1}=\frac{1}{k(k+1)}$.
Then $\sum_{k=1}^{n} \frac{1}{a_{k... | \frac{2007}{2008} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,055 |
6. As shown in Figure 2, given the ellipse $\frac{x^{2}}{2}+y^{2}=1, A$ and $B$ are the intersection points of the ellipse with the $x$-axis, $D A$ $\perp A B, C B \perp A B$, and $|D A|$ $=3 \sqrt{2},|C B|=\sqrt{2}$. A moving point $P$ is on the arc $\overparen{A B}$ above the $x$-axis. Then the minimum value of $S_{\... | $6.4-\sqrt{6}$.
Given $C(\sqrt{2}, \sqrt{2}) 、 D(-\sqrt{2}, 3 \sqrt{2})$,
$$
|C D|=4, l_{C D}: y=-x+2 \sqrt{2} \text {. }
$$
Let $P(\sqrt{2} \cos \theta, \sin \theta)(0 \leqslant \theta \leqslant \pi)$.
Then the distance from point $P$ to line $C D$ is
$$
\begin{array}{l}
d=\frac{|\sqrt{2} \cos \theta+\sin \theta-2 \s... | 4-\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,056 |
Three, (20 points) Find the smallest real number $A$, such that for each quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds. | Three, let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {. }
$$
From the problem, we know
$$
|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1 .
$$
Notice that $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$,
$$
\begin{array}{l}
f(1)=a+b+c, \\
f... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,057 |
$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right), \\
x y+y z+z x=1
\end{array}\right.
$$
Find all real solutions to the system of equations. | Transform equation (1) into
$$
\frac{x}{5\left(1+x^{2}\right)}=\frac{y}{12\left(1+y^{2}\right)}=\frac{z}{13\left(1+z^{2}\right)} \text {. }
$$
From equation (3), it is known that $x$, $y$, and $z$ have the same sign.
From equation (2), we get $x=\frac{1-y z}{y+z}$.
Substituting the above equation into equation (1) yie... | \left(\frac{1}{5}, \frac{2}{3}, 1\right),\left(-\frac{1}{5},-\frac{2}{3},-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,059 |
Five. (20 points) In the Cartesian coordinate system, $O$ is the origin, $A$ and $B$ are points in the first quadrant, and $A$ lies on the line $y = (\tan \theta) x$, where $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right), |OA| = \frac{1}{\sqrt{2} - \cos \theta}, B$ is a point on the hyperbola $x^{2} - y^{2} = 1$ ... | As shown in Figure 6, draw the tangent line $B C$ of the hyperbola through point $B$, and $B C \parallel A O$.
Then $l_{B C}$:
$y=(\tan \theta) x+m$.
Substitute into the hyperbola equation $x^{2}-y^{2}=1$, and by $\Delta=0$ we get
$m^{2}=\tan ^{2} \theta-1$.
Also, $l_{A O}: y=(\tan \theta) x$,
$S_{\triangle O A B}=\fra... | \frac{\sqrt{6}}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,060 |
一、(50 points) Given that $F$ and $G$ are the projections of a point $M$ inside the acute triangle $\triangle ABC$ onto the sides $BC$ and $AC$, respectively. Prove:
$$
A B - F G \geqslant \frac{M F \cdot A G + M G \cdot B F}{C M},
$$
and determine the condition for equality. | As shown in Figure 7, draw perpendiculars from points $A$ and $B$ to $GF$, with $P$ and $Q$ being the feet of the perpendiculars.
Then,
$$
\begin{array}{l}
A B \geqslant P Q \\
= P G + G F + F Q.
\end{array}
$$
Since points $C, F, M, G$ are concyclic, we have
$$
\angle C M F = \angle C G F = \angle A G P.
$$
Therefor... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,061 |
II. (50 points) Find all positive integers $x, y, z$ such that $\sqrt{\frac{2009}{x+y}}+\sqrt{\frac{2009}{y+z}}+\sqrt{\frac{2009}{z+x}}$ is an integer. | II. First, prove a lemma.
Lemma If $p, q, r$ are rational numbers, and $S=\sqrt{p}+\sqrt{q}+\sqrt{r}$ is also a rational number, then $\sqrt{p}, \sqrt{q}, \sqrt{r}$ must all be rational numbers.
Proof of the lemma: Notice that
$(\sqrt{p}+\sqrt{q})^{2}=(S-\sqrt{r})^{2}$
$\Rightarrow 2 \sqrt{p q}=S^{2}+r-p-q-2 S \sqrt{r}... | x=4018, y=28126, z=4018 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,062 |
Three. (50 points) Let non-negative real numbers $x_{i}(i=1,2, \cdots, n)$ for any $n>2$, have
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1} \\
\geqslant x_{1}^{a} x_{2}^{b} x_{3}^{a}+x_{2}^{a} x_{3}^{b} x_{4}^{a}+\cdots+x_{n}^{a} x_{1}^{b} x_{2}^{a} .
\end{array}
$$
Find the values of t... | When $n=4$, equation (1) becomes
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1} \\
\geqslant x_{1}^{a} x_{2}^{b} x_{3}^{a}+x_{2}^{a} x_{3}^{b} x_{4}^{a}+x_{3}^{a} x_{4}^{b} x_{1}^{a}+x_{4}^{a} x_{1}^{b} x_{2}^{a} .
\end{array}
$$
Let $x_{1}=x_{2}=x_{3}=x_{4}=x \neq 0$, then from equation (2) we ge... | \left(\frac{1}{2}, 1\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,063 |
As shown in Figure 3, in $\triangle ABC (AB > AC)$, point $D_{1}$ is on side $BC$. A circle is drawn with $AD_{1}$ as its diameter, intersecting $AB$ at point $M$ and the extension of $AC$ at point $N$. Connect $MN$, and draw $AP \perp MN$ at point $P$, intersecting $BC$ at point $D_{2}$. $AE$ is the bisector of the ex... | Prove: As shown in Figure 3, connect $M D_{1}$.
Since $A D_{1}$ is the diameter of the circle, therefore,
$\angle A M D_{1}=90^{\circ}$.
It is easy to see that $\angle A P N=90^{\circ}$, hence
$\angle A M D_{1}=\angle A P N$.
Also, $\angle A D_{1} M=\angle A N P$, then
Rt $\triangle A M D_{1} \backsim \operatorname{Rt}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,064 |
For all positive integers $x, k$, satisfying $\frac{24 k}{x^{3}-x-2}=x$. Prove: $x$ is a multiple of 6.
The text is translated as requested, maintaining the original line breaks and format. | Prove: Since four consecutive natural numbers must include two even numbers, one of which is a multiple of 4, and because four consecutive natural numbers must include at least one number that is a multiple of 3, and since $2 \times 3 \times 4 = 24$, four consecutive natural numbers must be a multiple of 24.
Notice tha... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,065 |
Prove that there exist infinitely many positive integers $A$ that satisfy the following conditions:
(1) The digits of $A$ do not contain the digit 0;
(2) $A$ is a perfect square;
(3) The sum of the digits of $A$ is also a perfect square. | First, prove that the infinite sequence composed of the digits 1, 5, and 6:
\[ 1156, 111556, 11115556, 1111155556, \ldots \]
Each term is the square of:
\[ 34, 334, 3334, 33334, \ldots \]
That is, prove:
\[ \begin{array}{l}
=\underbrace{33 \cdots 3^{2}}_{n+1 \uparrow}+2 \times \underbrace{33 \cdots 3}_{n+1 \uparrow}+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,066 |
For a cyclic convex quadrilateral $ABCD$ with unequal sides, the circle center $O$ is inside the quadrilateral, and $AC, BD$ intersect at point $P$.
(1) Prove: The circumcircles of $\triangle OAB$, $\triangle OCD$, $\triangle PBC$, and $\triangle PDA$ intersect at one point;
(2) If the four circles in (1) intersect at ... | Proof: As shown in Figure 4, from \( A D \neq B C \), we know that \( A B \) is not parallel to \( C D \), hence the circumcircles of \( \triangle O A B \) and \( \triangle O C D \) do not touch. Let their intersection point be \( M \), then \( A, B, O, M \) and \( D, C, O, M \) are respectively concyclic.
Connecting \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,067 |
Example 1 As shown in Figure 1, in square $A B C D$, $E$ is a fixed point on $B C$, and $B E=10, E C=14, P$ is a moving point on $B D$. Then the minimum value of $P E+P C$ is $\qquad$
(2006, Zhejiang Province Junior High School Mathematics Competition) | Analysis: Considering the minimum value of the sum of two line segments, it is relatively easy to think of using the property "the shortest distance between two points is a straight line." This requires transforming one of the line segments to the other side. Therefore, by the axial symmetry of the square (as shown in ... | 26 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,068 |
Example 2 As shown in Figure 2, the side lengths of square $ABCD$ and square $CGEF$ are $2$ and $3$, respectively, and points $B$, $C$, and $G$ are collinear. $M$ is the midpoint of line segment $AE$, and $MF$ is connected. Then the length of $MF$ is $\qquad$
(2006, Zhejiang Province Junior High School Mathematics Comp... | Analysis 1: As shown in Figure 2, since $M$ is the midpoint of $A E$, it leads to the idea of a midline. The other midpoint that comes to mind is the center of symmetry of the square. Therefore, connect $F G$ and $C E$ to intersect at point $O$, then connect $A C$. It is easy to see that $F G \parallel A C$; connect $M... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,069 |
3. As shown in Figure 16, in the right trapezoid $A B C D$, $A B=B C=$ $4, M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}: S_{\triangle A B M}=$ | (Hint: Complete the trapezoid into a square. Then it is easy to know that Rt $\triangle A B M \cong \mathrm{Rt} \triangle A E D$. Therefore, $E D=B M=x$, i.e., $C D=C M=4-x$. In $\triangle C D M$, $4^{2}+x^{2}=$ $2(4-x)^{2}$, then $x=8 \pm 4 \sqrt{3}(x=8+4 \sqrt{3}$ is discarded $)$. Hence $\left.S_{\triangle C D M}: S... | 1: 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,070 |
As shown in Figure 3, from a point $P$ outside circle $\odot O$, two tangents to $\odot O$ are drawn, touching the circle at points $A$ and $B$. Another secant line through $P$ intersects $\odot O$ at points $C$ and $D$ (with $P C < P D$). Line $B C$ is extended to intersect $P A$ at point $E$, and a tangent from $C$ t... | Proof: As shown in Figure 3, let $FB$ intersect $\odot O$ at $K_{1}$. Connect $CK_{1}$, $K_{1}A$, and $AB$.
It is easy to see that $\triangle FK_{1}C \sim \triangle FCB$,
$\triangle FK_{1}A \sim \triangle FAB$.
Thus, $\frac{CK_{1}}{BC} = \frac{FC}{FB} = \frac{FA}{FB} = \frac{K_{1}A}{AB}$.
Therefore, $\frac{CK_{1}}{K_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,071 |
As 230 As shown in Figure 4, for the cyclic quadrilateral $ABCD$, the two pairs of opposite sides $AB$ and $DC$,
$AD$ and $BC$ intersect at points
$E$ and $F$, respectively. Tangents to the circle are drawn from points $B$ and $D$, and
they intersect at point $P$. Prove that:
$E$, $P$, and $F$ are collinear. | Proof: As shown in Figure 4, connect $B D, P E, P F$.
Since $A, B, C,$ and $D$ are concyclic, we have
$$
\begin{array}{l}
\angle A B C + \angle A D C = 180^{\circ}. \\
\text { Also, } \angle A E D + \angle A + \angle A D E = 180^{\circ}, \\
\angle A F B + \angle A + \angle A B F = 180^{\circ},
\end{array}
$$
Thus,
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,072 |
229 Given $a, b, c \in \mathbf{R}_{+}, abc=1$. Prove:
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c} \geqslant 4 .
$$ | Proof: By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$. Then
$$
\begin{array}{l}
a \geqslant 1, c \leqslant 1 . \\
\text { Let } f(a, b, c)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c} . \\
f(a, b, c)-f(a, \sqrt{b c}, \sqrt{b c}) \\
=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,073 |
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