problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 2 In the acute triangle $\triangle ABC$, $AB > AC$, $CD$ and $BE$ are the altitudes on sides $AB$ and $AC$ respectively, $DE$ intersects the extension of $BC$ at point $T$, a perpendicular line from point $D$ to $BC$ intersects $BE$ at point $F$, and a perpendicular line from point $E$ to $BC$ intersects $CD$ a... | The most profound impression given by the conditions of this problem is the abundance of perpendicular relationships: \(CD \perp AB\), \(BE \perp AC\), \(DF \perp BC\), \(EG \perp BC\). Therefore, it is a natural idea to prove the problem using the properties of right triangles or area relationships (the standard answe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,153 |
Proposition 2 In $\triangle A B C$, $A D 、 B E 、 C F$ intersect at point $O$, a line through $E$ parallel to $A D$ intersects $C F$ at $P$, and a line through $F$ parallel to $A D$ intersects $B E$ at $Q$. If $F E$ intersects the extension of $B C$ at $T$, then $Q 、 P 、 T$ are collinear.
Proposition 2 holds for any tr... | Interestingly, although this generalization proposition is more essential than ratio 2, finding the proof idea is easier because the auxiliary line $A D$ has already been drawn, and the parallel relationship is directly presented. The deep structure of the problem has become an external manifestation, so solving the pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,154 |
Example 1 Let $n$ be a positive integer greater than 1, and $a_{1}, a_{2}$, $\cdots, a_{n}$ be $n$ distinct integers. Prove that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
cannot be divisible by any positive degree and less than $n$ polynomial with integer coefficie... | To prove: It is only necessary to prove that the polynomial $f(x)$ cannot be factored into the product of two polynomials with integer coefficients and degrees greater than zero.
Assume $f(x)$ has a proper factor $g(x) \in \mathbf{Z}[x]$ (the set of all integer coefficient polynomials with the leading coefficient 1), ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,155 |
Similar to Example 1, we can ask: Is the polynomial
$$
\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
reducible? If so, what are the conditions? | We can obtain the following theorem.
Theorem 1 Polynomial
$$
\begin{aligned}
F(x)= & x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots \cdot \\
& \left(x-a_{n-1}\right)+1
\end{aligned}
$$
(Where $00$.
$$
\begin{array}{l}
\text { Thus, } \varphi\left(a_{i}\right)=\psi\left(a_{i}\right)=1, \\
\varphi(x)=x^{k}+\cdots, \psi... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,156 |
Example 6 The polynomial $x^{4}+m x^{3}+n x-16$ contains the factors $x-1$ and $x-2$. Then $m n=$ $\qquad$
(1989, Sichuan Province Junior High School Mathematics League) | Solution: Let $f(x)=x^{4}+m x^{3}+n x-16$. By the Factor Theorem, we have
$$
\left\{\begin{array} { l }
{ f ( 1 ) = 0 , } \\
{ f ( 2 ) = 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
m+n=15, \\
8 m+2 n=0 .
\end{array}\right.\right.
$$
Solving, we get $m=-5, n=20$.
Therefore, $m n=-100$. | -100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,157 |
As shown in Figure 2, the excircle $\odot O$ of $\triangle ABC$ touches sides $BC$, $AB$, and $AC$ at points $D$, $M$, and $N$ respectively. $DE$ is the diameter of $\odot O$, and $AE$ intersects $BC$ at $F$. Prove that $BF = CD$. | Proof: As shown in Figure 2, draw \( F O^{\prime} \perp B C \) intersecting \( A O \) at \( O^{\prime} \), and draw \( O^{\prime} G \perp A C \) and \( O^{\prime} H \perp A B \), with the feet of the perpendiculars being \( G \) and \( H \), respectively. Connect \( O N \).
Since \( D E \perp B C \) and \( \sigma^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,158 |
Given $a 、 b 、 c$ are real numbers, and
$$
a^{2}+b^{2}+c^{2}+2 a b=1, a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{8} \text {, }
$$
The roots of the quadratic equation $(a+b) x^{2}-(2 a+c) x-(a+b)=0$ are $\alpha 、 \beta$. Find the value of $2 \alpha^{3}+\beta^{-5}-\beta^{-1}$. | Solution: From the given, we have
$$
\left\{\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)+2 a b=1, \\
2 a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4} .
\end{array}\right.
$$
Thus, \(a^{2}+b^{2}+c^{2}\) and \(2 a b\) are the roots of the equation \(t^{2}-t+\frac{1}{4}=0\).
Since the roots of the equation \(t^{2}-t+\f... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,159 |
Given $\alpha$ is an acute angle. Prove:
$$
\frac{1}{\sin \alpha}+\frac{3 \sqrt{3}}{\cos \alpha} \geqslant 8 .
$$ | Prove: The original inequality is equivalent to
$$
\begin{array}{l}
\cos \alpha+3 \sqrt{3} \sin \alpha \geqslant 8 \sin \alpha \cdot \cos \alpha \\
\Leftrightarrow 3 \sqrt{3} \sin \alpha \geqslant(8 \sin \alpha-1) \cos \alpha .
\end{array}
$$
When $8 \sin \alpha-1 \leqslant 0$, it is easy to see from the above inequal... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,160 |
(Song Qing, Nanchang University Affiliated High School, 330029)
In a cyclic convex quadrilateral $ABCD$, $\angle A = 60^{\circ}$, $BC = CD = 1$. Extend $AB$ and $DC$ to intersect at point $E$, and extend $BC$ and $AD$ to intersect at point $F$. It is known that the perimeters of $\triangle BCE$ and $\triangle DCF$ are ... | Solution: As shown in Figure 3, it is easy to see that
$\angle E B C + \angle F D C = 180^{\circ}$.
It is also easy to see that $\angle E B C \neq 90^{\circ}$. Otherwise, if $\angle E B C = 90^{\circ}$, then the perimeter of $\triangle B C E$ would be $3 + \sqrt{3}$, which contradicts the given information.
Since $\an... | \frac{38}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,161 |
Example 7 When $m=$ $\qquad$, the polynomial
$$
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m
$$
can be factored into the product of two linear factors.
(1992, Zhengzhou City Junior High School Mathematics Competition) | Solution: First, factorize the quadratic term, we have
$$
12 x^{2}-10 x y+2 y^{2}=(3 x-y)(4 x-2 y) \text {. }
$$
Therefore, we can set
$$
\begin{array}{l}
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m \\
=(3 x-y+a)(4 x-2 y+b) \\
=(3 x-y)(4 x-2 y)+(4 a+3 b) x- \\
(2 a+b) y+a b .
\end{array}
$$
Comparing the coefficients of the ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,162 |
Example 8 Prove: $x^{3}+y^{3}+z^{3}-3 x y z$ has a factor $x+y+z$.
| Prove: Let $f(x)=x^{3}-3 y z x+\left(y^{3}+z^{3}\right)$.
When $x=-(y+z)$,
$$
\begin{array}{l}
f(-(y+z)) \\
=[-(y+z)]^{3}-3 y z[-(y+z)]+y^{3}+z^{3} \\
=0 .
\end{array}
$$
Therefore, $x^{3}+y^{3}+z^{3}-3 x y z$ has a factor $x+y+z$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,163 |
Example 9 Express $\frac{11 x^{2}-23 x}{(2 x-1)\left(x^{2}-9\right)}$ as partial fractions. | Solution: Let $N=\frac{11 x^{2}-23 x}{(2 x-1)(x+3)(x-3)}$
$$
\equiv \frac{A}{2 x-1}+\frac{B}{x+3}+\frac{C}{x-3} \text {. }
$$
Then $11 x^{2}-23 x$
$$
\begin{aligned}
\equiv & A(x+3)(x-3)+B(2 x-1)(x-3)+ \\
& C(2 x-1)(x+3) .
\end{aligned}
$$
Let $x=\frac{1}{2}$, substituting we get $A=1$;
Let $x=-3$, substituting we ge... | N=\frac{1}{2 x-1}+\frac{4}{x+3}+\frac{1}{x-3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,164 |
Example 10 Express $\frac{2 x^{2}+1}{x^{2}(x-1)^{2}}$ as partial fractions. | Solution: Let $N=\frac{2 x^{2}+1}{x^{2}(x-1)^{2}}$
$$
\equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{(x-1)^{2}} \text {. }
$$
Then $2 x^{2}+1$
$$
\begin{aligned}
\equiv & A x(x-1)^{2}+B(x-1)^{2}+ \\
& C x^{2}(x-1)+D x^{2} .
\end{aligned}
$$
Let $x=0$, we get $B=1$;
Let $x=1$, we get $D=3$.
Substitute and r... | \frac{2}{x}+\frac{1}{x^{2}}+\frac{-2}{x-1}+\frac{3}{(x-1)^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,165 |
Example 1 The lengths of the three sides of a triangle are $6, 8, 10$. Prove: there is only one line that bisects both the perimeter and the area of this triangle. | It is obvious that a triangle with side lengths of $6, 8, 10$ is a right-angled triangle, with a perimeter of 24 and an area of 24 as well.
To prove that there is only one line that simultaneously bisects the perimeter and area of this triangle, we need to consider different scenarios of how this line intersects the s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,166 |
Example 2 Proof: There exists a unique triangle with three sides of lengths being three consecutive positive integers, and one of its three interior angles is twice another interior angle. | Explanation: Break down the problem into two parts:
(1) When one interior angle of a triangle is twice another interior angle, what is the relationship among the three sides?
(2) Under the premise of conclusion (1), what are the lengths of the three sides if they are consecutive positive integers?
(1) As shown in Figur... | 4, 5, 6 | Geometry | proof | Yes | Yes | cn_contest | false | 716,167 |
Example 8 Isosceles $\triangle ABC (AB = BC)$, point $O$ is its circumcenter, point $I$ is its incenter, point $D$ is on side $BC$, and $OD \perp CI$. Prove: $DI \parallel AB$.
(22nd Russian Mathematical Olympiad) | Proof: As shown in Figure 8, without loss of generality, let \( AC = 4 \) and \( \angle A = 2\alpha \) \((0^\circ < \alpha < 45^\circ)\). Then the equation of the perpendicular bisector of \( BC \) is
\[
\begin{array}{l}
y - \tan 2\alpha \\
= \cot 2\alpha \cdot (x - 1).
\end{array}
\]
Let \( x = 0 \), we get the circu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,168 |
5. The axial cross-section of a wine glass is part of a parabola, whose equation is $x^{2}=2 y(0 \leqslant y<15)$. If a glass ball with a radius of 3 is placed inside the cup, then the distance from the highest point of the ball to the bottom of the cup is $\qquad$ | 5.8 .
As shown in the figure, let the center of the sphere be at $(0, b)$.
Then we have
$$
x^{2}+(y-6)^{2}=9 \text {. }
$$
Therefore, the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2 y \\
x^{2}+(y-b)^{2}=9
\end{array}\right.
$$
has two solutions. For $y$,
$$
y^{2}+2(1-b) y+b^{2}-9=0 \text {. }
$$
By $\Delt... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,169 |
Three. (20 points) Given $p_{1}, p_{2}, \cdots, p_{2 n}$ are distinct integers, and $q_{1}, q_{2}, \cdots, q_{n-1}$ are distinct prime numbers $(n \geqslant 2, n \in \mathbf{N}), m$ is an integer root of the equation
$$
\begin{array}{l}
\left(x-p_{1}\right)\left(x-p_{2}\right) \cdots\left(x-p_{2 n}\right) \\
=\left(q_{... | Three, let $p_{1}, p_{2}, \cdots, p_{2n}$ be distinct integers, and $m$ be an integer. Therefore, $m-p_{1}, m-p_{2}, \cdots, m-p_{2n}$ are $2n$ distinct integers. According to the problem,
$$
\begin{array}{l}
\left(m-p_{1}\right)\left(m-p_{2}\right) \cdots\left(m-p_{2n}\right) \\
=\left(q_{1} q_{2} \cdots q_{n-1}\right... | p_{1}+p_{2}+\cdots+p_{2n}=2mn | Algebra | proof | Yes | Yes | cn_contest | false | 716,171 |
Four, (20 points) Given any 13 distinct real numbers with absolute values less than 1. Prove that there always exist two real numbers \( x \) and \( y \) such that
$$
\frac{\sqrt{6}+\sqrt{2}}{4}<x y+\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}<1
$$
holds. | Four, since $x, y$ are unequal real numbers with absolute values less than 1, we can set $x=\cos \alpha, y=\cos \beta, \alpha, \beta \in(0, \pi), \alpha \neq \beta$. Then
$$
\begin{array}{l}
x y+\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)} \\
=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta=\cos (\alpha-\b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,172 |
Five. (20 points) As shown in Figure 1, in the square $ABCD$ with side length 1, points $M_{1}$, $M_{2}, \cdots, M_{n} (n \geqslant 1)$ sequentially divide the broken line $BC + CD$ into $n+1$ equal parts. Try to find:
(1) $\sum_{k=1}^{n} \mathcal{\Lambda} M_{k}^{2}$;
(2) $\lim _{n \rightarrow \infty}\left(\frac{1}{n} ... | When $n$ is odd, point $M_{\frac{n+1}{2}}$ coincides with $C$, and the other division points are symmetric about the line $A C$.
When $k \leqslant \frac{n+1}{2}$, point $M_{k}\left(1, \frac{2 k}{n+1}\right)$ lies on side $B C$, and $A M_{4}^{2}=1+\frac{4 k^{2}}{(n+1)^{2}}$. At this time,
$$
\begin{array}{l}
\sum_{i=1}... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,173 |
In $\triangle A B C$, $\angle A C B=90^{\circ}$,
$C D$ is the altitude, points $O_{1} 、 O_{2}$ are the incenters of $\triangle A D C 、 \triangle C D B$ respectively, rays $D O_{1} 、 D O_{2}$ intersect $A C 、 B C$ at points $E 、 F$, line $O_{1} O_{2}$ intersects $A C 、 B C$ at points $M 、 N$. Prove: $\frac{1}{O_{1} \bar... | As shown in Figure 6, connect $\mathrm{CO}_{1}$. It is easy to see that
Rt $\triangle A D C \backsim$
Rt $\triangle C D B \backsim$
Rt $\triangle A C B$,
$D E$ and $D F$,
$D O_{1}$ and $D O_{2}$ are two pairs of opposite segments of $\triangle A D C$ and $\triangle C D B$, respectively. Using the properties of similar ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,174 |
II. (50 points) Given an odd function $f(x)$ defined on $(-\infty, 0) \cup (0, +\infty)$, which is decreasing on the interval $(0, +\infty)$, and $f(k)=0$ for some $k>0$. Another function is defined as
$$
\varphi(\alpha)=\sin ^{2} \alpha+(\\xi+1) \cos \alpha-\xi^{2}-\xi-k,
$$
where $\alpha \in [0, \pi]$.
If the set $A=... | Given $f(k)=0 \quad(k>0), f(x)$ is an odd function, so $f(-k)=-f(k)=0$.
Also, $f(x)$ is a decreasing function in $(0,+\infty)$. Therefore, $f(x)$ is also a decreasing function in $(-\infty, 0)$.
Thus, the solution set for $f(x)>0$ is
$\{x \mid |x| < 0\}$.
Then $B=\{\xi \mid f(\varphi(\alpha))>0, \forall \alpha \in[0, \... | \left\{\xi \mid \xi>1 \text { or } \xi<-\frac{5}{3}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,175 |
Three. (50 points) Prove that for any angles $\alpha, \beta, \gamma, \delta$, we have
$$
\begin{array}{l}
\sqrt{\sin ^{4} \alpha+\sin ^{4} \beta}+\sqrt{\cos ^{4} \beta+\sin ^{4} \gamma}+ \\
\sqrt{\cos ^{4} \gamma+\sin ^{4} \delta}+\sqrt{\cos ^{4} \alpha+\cos ^{4} \delta} \geqslant 2 \sqrt{2}
\end{array}
$$
holds. Equa... | Three, let $a=l \sin ^{2} \alpha, b=l \sin ^{2} \beta, c=l \sin ^{2} \gamma, d=$ $l \sin ^{2} \delta$. It is not difficult to obtain the algebraic form of the problem:
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}}+\sqrt{(l-b)^{2}+c^{2}}+\sqrt{(l-c)^{2}+d^{2}}+ \\
\sqrt{(l-a)^{2}+(l-d)^{2}} \geqslant 2 \sqrt{2} l .
\end{array}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,176 |
In $\triangle A B C$, $\angle A: \angle B: \angle C=2: 3: 4, A D \perp B C$ at $D, C E$ bisects $\angle C$ and intersects $A B$ at $E$ and $A D$ at $O$, the line $B O$ intersects $A C$ at $F$. Prove:
$$
\frac{A F}{F C}=\frac{\cos 40^{\circ}}{\cos 80^{\circ}} \text {. }
$$ | Proof: According to the problem,
$$
\begin{array}{l}
\angle A=40^{\circ}, \angle B=60^{\circ}, \\
\angle C=80^{\circ} .
\end{array}
$$
Since $A D \perp B C$ and $\angle A B C=60^{\circ}$, therefore, $B D=\frac{1}{2} A B$.
Given $\angle \mathrm{I}=\angle 2$, so $\frac{A E}{E B}=\frac{A C}{B C}$.
Since $A D$, $B F$, and... | \frac{\cos 40^{\circ}}{\cos 80^{\circ}} | Geometry | proof | Yes | Yes | cn_contest | false | 716,177 |
In $\triangle A B C$, $\angle B$ and $\angle C$ are both acute angles. From point $A$, a perpendicular line to $B C$ intersects $B C$ at point $K$. $Q$ and $P$ are points on $A C$ and $A B$ respectively, such that $\angle P K A = \angle Q K A$. Prove: $A K, B Q, C P$ are concurrent.
Prove: $A K, B Q, C P$ are concurre... | Prove: As shown in Figure 4, draw a line $l$ through point $A$ parallel to $BC$, extend $KQ$ and $KP$ to intersect line $l$ at $S$ and $R$,
Since $\angle P K A = \angle Q K A$, therefore,
$$
A R = A S \text{. }
$$
By $\triangle R A P \sim \triangle K B P$ and $\triangle S A Q \sim \triangle K C Q$, we have
$$
\begin{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,178 |
1. In a convex quadrilateral $ABCD$, the diagonals are perpendicular to each other. Through the midpoints of $AB$ and $AD$, draw perpendiculars to the opposite sides $CD$ and $CB$, respectively. Prove that the drawn lines and $AC$ are concurrent.
（15th All-Russian High School Mathematics Olympiad） | (提示: As shown in Figure 9, let $A(2 a, 0), \quad B(0,2 b)$, $C(2 c, 0), D(0,2 d)$, where $a, b>0, c, d<0$. Then $E(a, b), F(a, d)$. In the system of equations of the line passing through the intersection of $E G$ and $F H$, setting $\lambda=1, \mu=-1$, we get the equation $y=0$, which is the equation of the line $A C$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,179 |
For all positive real numbers $a, b$, prove:
$$
\sqrt{\frac{a}{3 a+b}}+\sqrt{\frac{b}{3 b+a}} \leqslant 1 .
$$ | Proof: Let the left side of the inequality be $M$. From the inequality $a^{2}+b^{2} \geqslant \frac{1}{2}(a+b)^{2}$, we get
$$
\begin{array}{l}
M \leqslant \sqrt{2\left(\frac{a}{3 a+b}+\frac{b}{3 b+a}\right)} \\
=\sqrt{2\left[\frac{1}{3}\left(1-\frac{b}{3 a+b}\right)+\frac{1}{3}\left(1-\frac{a}{3 b+a}\right)\right]} \\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,180 |
Given a unit square $ABCD$, $O$ is its center. Suppose point $P$ is on side $CD$, $AP$ intersects $OD$ at point $T$, $BP$ intersects $OC$ at point $R$, and $OP$ intersects $TR$ at point $K$.
(1) Try to find the maximum value of $S_{\triangle MPT}$ and the maximum value of $S_{\text{quadrilateral } RRP T}$;
(2) When poi... | Solution: (1) As shown in Figure 5, let $D P = x$, then $P C = 1 - x$.
Since $\angle A D T = \angle P D T = 45^{\circ}$, we have
$$
\frac{A T}{T P} = \frac{A D}{D P} = \frac{1}{x},
$$
which means $\frac{A P}{T P} = \frac{1 + x}{x}$.
Thus, we have
$$
\begin{array}{l}
S_{\triangle A P T} = \frac{T P}{A P} S_{\triangle ... | \frac{1}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,181 |
2. Let $M$ and $N$ be points on the sides $AC$ and $BC$ of $\triangle ABC$ respectively, and $\angle ACB=90^{\circ}$. Let $AN$ and $BM$ intersect at point $L$. Prove that the orthocenters of $\triangle AML$ and $\triangle BNL$ and point $C$ are collinear. (2002, Bulgarian Winter Mathematical Competition) | (提示: As shown in Figure 10, let $A(a, 0)$, $B(0, b)$, $M(m, 0)$, and $N(0, n)$. In the system of lines passing through the orthocenter of $\triangle A M L$, i.e., the line through point $A$ and perpendicular to $B M$, and the line through point $M$ and perpendicular to $A N$, setting $\lambda_{1}=1, \mu_{1}=-1$, we get... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,182 |
3. As shown in Figure 11, in quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$. Take a point $E$ on $CD$, and let $BE$ intersect $AC$ at point $F$. Extend $DF$ to intersect $BC$ at point $G$. Prove:
$$
\angle GAC = \angle EAC \text{.}
$$
(1999, National High School Mathematics Competition) | (Tip: Establish a rectangular coordinate system with point $A$ as the origin and $A C$ as the $y$-axis. Let $B(b, k b) 、 D(d,-k d) 、 C(0, c) 、 F(0, f)$, where $k$ is the slope of $A B$, and $b, c, f>0, d<0$. First, find the equation of the line system passing through point $G$
$$
\lambda[(k b-c) x-b y+b c]+\mu[(k d+f) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,183 |
4. The segments connecting point $K$ with the vertices $A$ and $D$ of rectangle $ABCD$ both intersect side $BC$. A perpendicular is drawn from point $B$ to $DK$, and a perpendicular is drawn from point $C$ to $AK$, and they intersect at point $M$. Prove: If $M \neq K$, then $MK \perp AJ$.
(17th All-Russian High School ... | (提示: 如图 12 , 建立直角坐标系. 设 $C(a, b)$ 、 $K(m, n),-ab$, 则有 $A(-a,-b)$, $B(-a, b), D(a,-b)$.
在过 $C F$ 与 $B E$ 的交点 $M$ 的直线系方程中, 令 $\mu=-\lambda=-1$, 得方程 $x=m$, 即为直线 $M K$的方程. )
(提示: As shown in Figure 12, establish a rectangular coordinate system. Let $C(a, b)$, $K(m, n), -ab$, then we have $A(-a,-b)$, $B(-a, b)$, $D(a,-b)$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,184 |
5. In $\triangle A B C$, $A M$ is the perpendicular from point $A$ to the angle bisector of $\angle C$, with $M$ being the foot of the perpendicular. $A N$ and $C L$ are the perpendiculars from points $A$ and $C$ to the angle bisector of $\angle B$, with $N$ and $L$ being the feet of the perpendiculars. The extension o... | (提示: 如图 13, 设 $A(0,2 a), \angle B=2 \alpha, \angle C=$ $2 \beta$. 则 $B(-2 a \cot 2 \alpha, 0) 、 C(2 a \cot 2 \beta, 0)$. 延长 $A M$ 到点 $M_{1} 、 A N$ 到点 $N_{1}$. 先证 $M N / / B C$, 且 $F$ 也是 $A C$ 的中点, 从而, 得
$F(a \cot 2 \beta, a)$. 再
由直线 $B F 、 C L$ 的
方程, 知过点 $E$ 的
直线系方程中, 令
$$
\lambda=-1, \mu=1 \text {, 可 }
$$
得
$$
\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,185 |
Example 1 Fill the $3 \times 3$ chessboard with the numbers $1,2, \cdots, 9$ so that the sums of the numbers in each row, each column, and both diagonals are equal. | Solution 1: Since a third-order magic square requires that the sums of each row, each column, and the two diagonals are equal, let's call this sum the "magic sum" and denote it as $S$. Note that the sum of the numbers in the three rows is the sum of all numbers from $1 \sim 9$, so $3S = 45$, which means $S = 15$.
Now ... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,186 |
Example 2 In a $3 \times 3$ chessboard, fill each square randomly with 1 or -1, and then perform the following operation: each square is replaced by the product of the numbers in all its adjacent (sharing a common edge) squares. Prove that, regardless of the initial numbers, all squares can be made to contain 1 after a... | Analysis: Due to the randomness of filling in the numbers 1 or -1 in each cell, we can consider using the letter $a_{i}$ to replace the numbers filled in. Algebraic analysis methods can be used to study the invariance in the neighborhood product transformation (i.e., the self-product of the numbers 1 or -1 is always eq... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,187 |
Example 3 As shown in Figure 4(a), it is a $3 \times 3$ grid, in which eight word blocks “希、望、杯、数、学、竞、赛、题” are placed, but in the wrong order. Question: Can the word blocks in the grid be moved to correct the order of the eight word blocks shown in Figure 4(a) to the order shown in Figure 4(b)? If yes, please write dow... | Solution: Let the eight characters “希、望、杯、数、学、竞、赛、题” be denoted as $1, 2, 3, 4, 5, 6, 7, 8$, respectively. Then, Figure 4(a) becomes Figure 5(a). Thus, adjusting the character blocks is equivalent to adjusting these numbers.
If the numbers in the $3 \times 3$ grid are arranged from left to right and from top to bottom... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,188 |
Example 4 Fill the numbers $1,2, \cdots, 8$ into the 8 squares surrounding the four sides of a $3 \times 3$ chessboard, so that the sum of the absolute values of the differences between adjacent numbers in these 8 squares is maximized. Find this maximum value. | Solution: Let the sum of the absolute values of the differences between adjacent numbers in the 8 squares be $M$. Note that rotating the 8 numbers in the outer ring of the $3 \times 3$ chessboard does not change the value of $M$. As shown in Figure 6, let the numbers in each square be denoted as $a_{1}, a_{2}, \cdots, ... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,189 |
Example 3 On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P$, $\triangle B C P$, $\triangle C D P$, and $\triangle D A P$ are equal, what condition must the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there b... | When solving this problem, it is generally considered that point $P$ is inside the quadrilateral $A B C D$.
Actually, the position of point $P$ relative to quadrilateral $A B C D$ has three possibilities:
(1) Point $P$ is inside the quadrilateral $A B C D$;
(2) Point $P$ is outside the quadrilateral $A B C D$;
(3) Poi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,190 |
Example 5 Fill the numbers $1,2, \cdots, 9$ into a $3 \times 3$ chessboard, such that the sum of the absolute values of the differences between adjacent (sharing a common edge) cells is maximized. Find this maximum value. | Solution: Let the sum of the absolute values of the differences between adjacent cells be $M$. Note that $M$ can be divided into two parts:
(1) The sum of the absolute values of the differences between adjacent numbers in the 8 surrounding cells, denoted as $M_{1}$;
(2) The sum of the 4 absolute values of the differenc... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,191 |
Example 6 Can 15 $1 \times 4$ rectangles and 1 $2 \times 2$ square cover an $8 \times 8$ chessboard. | Solution 1: Notice that
$$
15 \times(1 \times 4)+1 \times(2 \times 2)=64=8 \times 8 \text {, }
$$
i.e., 15 pieces of $1 \times 4$ rectangles and 1 piece of $2 \times 2$ square together provide the 64 squares required for an $8 \times 8$ chessboard.
Now consider the requirement for complete coverage.
As shown in Figure... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,192 |
Example 7 Given an $11 \times 12$ rectangular chessboard. Prove:
(1) It is possible to completely cover the $11 \times 12$ rectangular chessboard with 20 tiles of size $1 \times 6$ or $1 \times 7$;
(2) It is impossible to completely cover the $11 \times 12$ rectangular chessboard with 19 tiles of size $1 \times 6$ or $... | Proof: (1) $11 \times 12=(1 \times 6) a+(1 \times 7)(20-a)$,
i.e., $132=6 a+7(20-a)$.
Solving for $a$ gives $a=8$.
Thus, 8 horizontally placed $1 \times 6$ rectangles can form a $4 \times 12$ rectangle, and 12 vertically placed $1 \times 7$ rectangles can form a $7 \times 12$ rectangle. Using these two rectangles, we c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,193 |
The 15th question of the first test:
As shown in Figure 1, a tangent line is drawn through point $A(1,1)$ on the parabola $y=x^{2}$, intersecting the $x$-axis at point $D$ and the $y$-axis at point $B$. Point $C$ is on the parabola, and point $E$ is on line segment $A C$, satisfying $\frac{A E}{E C}=\lambda_{1}$. Point... | Solution: Since $y^{\prime}=\left.2 x\right|_{x=1}=2$, the equation of the tangent line passing through point $A$ is
$y-1=2(x-1)$, which is $y=2 x-1$.
Thus, $A(1,1), B(0,-1), D\left(-\frac{1}{2}, 0\right)$.
Therefore, $D$ is the midpoint of $A B$, and $C D$ is a median of $\triangle A B C$.
By the given conditions,
$\b... | y=\frac{1}{3}(3 x-1)^{2}\left(x \neq \frac{2}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,194 |
In $\triangle A B C$, let $A B>A C$, and draw the tangent line $l$ of the circumcircle of $\triangle A B C$ through point $A$. Also, construct a circle with center $A$ and radius $A C$ that intersects line segment $A B$ at point $D$, and intersects line $l$ at points $E$ and $F$. Prove: lines $D E$ and $D F$ pass throu... | Solution: As shown in Figure 2, establish a Cartesian coordinate system with $A$ as the origin and the line $l$ as the $x$-axis.
Let the radius of $\odot A$ be 1, and its equation is
$$
x^{2}+y^{2}=1 \text {. }
$$
Let the radius of the circumcircle of $\triangle A B C$ be $R$, and its equation is
$$
x^{2}+(y+R)^{2}=R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,195 |
Example 1 Given an acute triangle $\triangle A B C$ with three interior angles satisfying $A>B>C$. Let $\alpha$ denote the minimum of $A-B, B-C$, and $90^{\circ}-A$. Then the maximum value of $\alpha$ is $\qquad$ | Solution 1: Since $\alpha$ is the minimum of $90^{\circ}-A$, $A-B$, and $B-C$, we have
$$
\begin{array}{l}
90^{\circ}-A \geqslant \alpha, \\
90^{\circ}-B=\left(90^{\circ}-A\right)+(A-B) \geqslant 2 \alpha, \\
90^{\circ}-C=\left(90^{\circ}-A\right)+(A-B)+(B-C) \geqslant 3 \alpha .
\end{array}
$$
(1) + (2) + (3) gives
$$... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,196 |
Example 2 Given that the interior angles of quadrilateral $ABCD$ satisfy $A>B>C>D$, use $\alpha$ to denote the minimum of $180^{\circ}-A$, $A-B$, $B-C$, and $C-D$. Find the maximum value of $\alpha$.
保留源文本的换行和格式，直接输出翻译结果。 | $$
\begin{array}{l}
\text { Solution: } \alpha=\frac{4 \alpha+3 \alpha+2 \alpha+\alpha}{4+3+2+1} \\
\leqslant \frac{4\left(180^{\circ}-A\right)+3(A-B)+2(B-C)+(C-D)}{10} \\
=\frac{4 \times 180^{\circ}-(A+B+C+D)}{10}=36^{\circ} .
\end{array}
$$
This shows that $\alpha$ will not exceed $36^{\circ}$ at most.
On the other ... | 36^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,197 |
Example 3 Given the $n$ interior angles of an $n$-sided polygon, sorted in descending order as $A_{1}>A_{2}>\cdots>A_{n}(n \geqslant 4)$, let $\alpha$ denote the minimum of $180^{\circ}-A_{1}, A_{1}-A_{2}, \cdots, A_{n-1}-A_{n}$. Find the maximum value of $\alpha$.
---
The translation maintains the original text's li... | $$
\begin{array}{l}
\text { Solution: } \alpha=\frac{n \alpha+(n-1) \alpha+\cdots+\alpha}{n+(n-1)+\cdots+1} \\
\leqslant \frac{n\left(180^{\circ}-A_{1}\right)+(n-1)\left(A_{1}-A_{2}\right)+\cdots+\left(A_{n-1}-A_{n}\right)}{n+(n-1)+\cdots+1} \\
=\frac{2}{n(n+1)}\left[n \times 180^{\circ}-\left(A_{1}+A_{2}+\cdots+A_{n}\... | \frac{720^{\circ}}{n(n+1)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,198 |
Example 4 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\begin{array}{l}
A>a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=2 A .
\end{array}
$$
Let $\alpha$ be the minimum of $A-a_{1}, a_{1}-a_{2}, \cdots, a_{n-1}-a_{n}$. Find the maximum value of... | Solution: From the given, we have
$$
\begin{aligned}
\alpha & =\frac{n \alpha+(n-1) \alpha+\cdots+2 \alpha+\alpha}{n+(n-1)+\cdots+2+1} \\
& \leqslant \frac{n\left(A-a_{1}\right)+(n-1)\left(a_{1}-a_{2}\right)+\cdots+\left(a_{n-1}-a_{n}\right)}{n+(n-1)+\cdots+2+1} \\
& =\frac{n A-\left(a_{1}+a_{2}+\cdots+a_{n}\right)}{n+... | \frac{2(n-2) A}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,199 |
Example 5 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\begin{array}{l}
m>a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=A<n m<2 A+m .
\end{array}
$$
Let $\alpha$ be the minimum of $m-a_{1}, a_{1}-a_{2}, \cdots, a_{n-1}-a_{n}$. Find the maximum ... | Solution: From the given, we have
$$
\begin{array}{l}
\alpha=\frac{n \alpha+(n-1) \alpha+\cdots+2 \alpha+\alpha}{n+(n-1)+\cdots+2+1} \\
\leqslant \frac{n\left(m-a_{1}\right)+(n-1)\left(a_{1}-a_{2}\right)+\cdots+\left(a_{n-1}-a_{n}\right)}{n+(n-1)+\cdots+2+1} \\
=\frac{2}{n(n+1)}\left[n m-\left(a_{1}+a_{2}+\cdots+a_{n}\... | \frac{2(n m-A)}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,200 |
Example 1 As shown in the figure, $\odot O$ intersects the sides $BC, CA, AB$ of $\triangle ABC$ at points $A_{1}$ and $A_{2}$, points $B_{1}$ and $B_{2}$, points $C_{1}$ and $C_{2}$, respectively. Given that the perpendiculars from points $A_{1}$, $B_{1}$, $C_{1}$ to $BC$, $CA$, $AB$ intersect at point $P$. Prove: The... | Proof: Taking point $O$ as the origin and $C_{2} B_{2}$ as the $x$-axis to establish a Cartesian coordinate system. Let the line passing through point $A_{2}$ and perpendicular to $B C$ be denoted as $'j$. The line passing through point $B_{2}$ and perpendicular to $A C$ intersects at point $Q$. Clearly, the two lines ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,201 |
1. Given an acute triangle $\triangle A B C$ with three interior angles satisfying $A>B>C$, let $\alpha$ denote the maximum of $90^{\circ}-A, A-B, B-C$. Find the minimum value of $\alpha$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result... | Answer: $15^{\circ}$ | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,202 |
2. Given a quadrilateral $ABCD$ whose four interior angles satisfy $A>B>C>D$, let $\alpha$ represent the minimum of $180^{\circ}-A$, $A-B$, $B-C$, $C-D$, and its maximum value is $\alpha_{0}$; let $\beta$ represent the maximum of $180^{\circ}-A$, $A-B$, $B-C$, $C-D$, and its minimum value is $\beta_{0}$. Try to compare... | Answer: $\alpha_{0}=\beta_{0}$ | \alpha_{0}=\beta_{0} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,203 |
In $\triangle A B C$, $\angle A C B=90^{\circ}, A C=$ $B C, D$ is a point on side $A B$, and the perpendicular bisector of line segment $C D$ intersects sides $A C$ and $B C$ at points $M$ and $N$, respectively. If $A D=a$, $B D=b(a, b$ are given positive numbers), try to find the lengths of $C M$ and $C N$ (expressed ... | Solution: As shown in Figure 1, let
$$
C M = D M = x,
$$
and draw
$$
D P \perp A M \text{ at point } P.
$$
Then,
$$
\begin{aligned}
D P & = \frac{a}{\sqrt{2}} = A P, \\
M P & = \left| \left( \frac{a + b}{\sqrt{2}} - x \right) - \frac{a}{\sqrt{2}} \right| = \left| \frac{b}{\sqrt{2}} - x \right|.
\end{aligned}
$$
From $... | \sqrt{2} - 1 \leqslant \frac{b}{a} \leqslant \sqrt{2} + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,204 |
Question 1: Do there exist positive integers $a, b$ such that the equation
$$
a^{3}+(a+b)^{2}+b=b^{3}+a+2
$$
holds? If they exist, find all values of $a, b$; if not, explain why. | Solution: From the original equation, we have
$$
\begin{array}{l}
a^{3}+(a+b)^{2}-a=b^{3}-b+2, \\
\text { hence } a(a+1)(a-1)+(a+b)^{2} \\
=b(b+1)(b-1)+2 .
\end{array}
$$
Given that \(a\) and \(b\) are positive integers, therefore,
$$
a(a+1)(a-1), b(b+1)(b-1)
$$
must be divisible by 3. However, the remainder of \((a+... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,205 |
Given real numbers $a, b, c, d$ are all distinct, and
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{a}=x \text {. }
$$
Try to find the value of $x$.
(2003, National Junior High School Mathematics League) | Given:
$$
\begin{array}{l}
\frac{1}{b}=x-a, \frac{1}{d}=x-c, b=x-\frac{1}{c}, d=x-\frac{1}{a} . \\
\text { Then } (x-a)\left(x-\frac{1}{c}\right)=1, (x-c)\left(x-\frac{1}{a}\right)=1 .
\end{array}
$$
Thus,
$$
c x^{2}-(a c+1) x+a-c=0,
$$
$$
a x^{2}-(a c+1) x+c-a=0 .
$$
Subtracting the two equations, we get
$$
(c-a) x... | x = \pm \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,206 |
Question 3 Given real numbers $a, b$ satisfy $a^{3}+b^{3}+3 a b=1$. Find $a+b$.
(2004, National Junior High School Mathematics League) | Solution: From the given, we have
$$
a^{3}+b^{3}+(-1)^{3}-3 a b(-1)=0 \text {. }
$$
By the formula
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right),
\end{array}
$$
if $a^{3}+b^{3}+c^{3}-3 a b c=0$, then $a+b+c=0$ or $a=b=c$.
Therefore, $a+b+(-1)=0$ or $a=b=-1$. Henc... | a+b=1 \text{ or } a+b=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,207 |
Question 4 If real numbers $a, b, c$ satisfy
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1 \text{, }
$$
find the value of $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}$.
(1999, Changsha Junior High School Mathematics Competition) | Solution: Construct the identity
$$
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 \text {. }
$$
Subtract the above identity from the given equation to get
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\en... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,208 |
Find the smallest positive integer $n$ such that there exists a positive integer $k$ for which $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds. | Solution: For the inequality $\frac{8}{15}1$, i.e., $m>8$.
Therefore, $n>56$.
When $n=7 m+p\left(m \in \mathbf{N}_{+}, p \in \mathbf{N}\right.$, and $1 \leqslant p$ $\leqslant 6$), we have
$$
\begin{array}{l}
6 m+\frac{6}{7} p0$, i.e., $m>p$.
Since $1 \leqslant p \leqslant 6$, to find the smallest $m$, when $p=1$, $m_... | 15 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,209 |
1. Among the 100 natural numbers from $1 \sim 100$, the percentage of prime numbers is ( ).
(A) $25 \%$
(B) $24 \%$
(C) $23 \%$
(D) $22 \%$ | - 1.A.
Among the 100 natural numbers from 1 to 100, there are prime numbers $2,3,5,7$, $11,13,17,19,23,29,31,37,41,43,47,53,59,61,67$, $71,73,79,83,89,97$ totaling 25, so the percentage of prime numbers among them is $25 \%$. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,210 |
2. The three sides of a triangle are all integers, and its perimeter equals 10. Then this triangle is ( ).
(A) a right triangle
(B) an obtuse triangle
(C) a triangle with exactly two equal sides
(D) a triangle with exactly one interior angle of $60^{\circ}$ | 2. C.
Decompose 10 into the sum of three positive integers, we have
$$
\begin{array}{l}
10=1+1+8=1+2+7=1+3+6=1+4+5 \\
=2+2+6=2+3+5=2+4+4=3+3+4
\end{array}
$$
There are a total of eight cases. According to the "sum of the lengths of any two sides of a triangle is greater than the length of the third side," only the gr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,211 |
Example 2 Given that $\triangle ABC$ is an acute triangle, and the circle $\odot M$ with diameter $AB$ intersects $AC$ and $BC$ at points $P$ and $Q$ respectively. Two tangents to $\odot M$ are drawn through points $A$ and $Q$, intersecting at point $R$; two tangents to $\odot M$ are drawn through points $B$ and $P$, i... | Proof: As shown in Figure 2, let
the equation of $\odot M$ be $x^{2}+y^{2}=1, \angle Q M X=2 \alpha$, $\angle P M X=2 \beta$. Connect $R M$ and $S M$, then $\angle S M B=\beta, \angle R M A=90^{\circ}-\alpha$. Therefore, we have $S(1, \tan \beta)$, $R(-1, \cot \alpha)$. Hence,
the equation of line $l_{R^{\prime}}$ is:
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,212 |
3. Given that $n$ is a positive integer, $S=1+2+\cdots+n$. Then the units digit of $S$ cannot be ( ).
(A) $0,1,2,3$
(B) $3,4,5,6$
(C) $3,5,6,8$
(D) $2,4,7,9$ | 3. D.
Given $S=\frac{n(n+1)}{2}$, and since $n$ and $n+1$ are two consecutive natural numbers, the units digit of $n(n+1)$ can only be $0, 2, 6$. Therefore, the units digit of $S$ can only be $0, 1, 3, 5, 6, 8$. Thus, the units digit of $S$ cannot be $2, 4, 7, 9$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,213 |
4. As shown in Figure 1, the diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect at point $O$. $S_{\triangle A O B}=4, S_{\triangle C O D}=9$. Then the minimum value of $S_{\text {quadrilateral } A B C D}$ is ( ).
(A) 22
(B) 25
(C) 28
(D) 32 | 4. B.
As shown in Figure 1, let $S_{\triangle M O D}=x, S_{\triangle B O C}=y$. Then $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 13+2 \sqrt{x y}$. From $\frac{x}{9}=\frac{4}{y}$, we have $x y=36$. Therefore, $S_{\text {quadrilateral } A B C D} \geqslant 13+2 \sqrt{x y}=13+12=25$. Hence, the minimum value of ... | 25 | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,214 |
5. If $|a-b|=1,|b+c|=1,|a+c|$ $=2$, then $|a+b+2 c|$ equals $(\quad)$.
(A) 3
(B) 2
(C) 1
(D) 0 | 5. A.
From $|a-b|=1$, we know $a-b=1$ or $a-b=-1$;
From $|b+c|=1$, we know $b+c=1$ or $b+c=-1$;
From $|a+c|=2$, we know $a+c=2$ or $a+c=-2$.
Thus, we can obtain $2^{3}=8$ systems of three linear equations:
(1) $a-b=1, b+c=1, a+c=2$;
(2) $a-b=1, b+c=1, a+c=-2$;
(3) $a-b=1, b+c=-1, a+c=2$;
(4) $a-b=1, b+c=-1, a+c=-2$;
(... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,215 |
1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}... | $=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicir... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,216 |
2.2005 divided by a two-digit natural number, the maximum possible remainder is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.85 .
Starting from 2005, it is successively divided by $99, 98, 97, \cdots$, and the changes in the values of the remainders are observed:
$$
\begin{array}{l}
2005=99 \times 20+25=98 \times 20+45 \\
=97 \times 20+65=96 \times 20+85=95 \times 21+10 \\
=94 \times 21+31=93 \times 21+52=92 \times 21+73 \\
=91 \times 22+... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 716,217 |
3. Given $a^{2}+b c=14, b^{2}-2 b c=-6$. Then $3 a^{2}+4 b^{2}-5 b c=$ $\qquad$ | 3.18 .
$$
\begin{array}{l}
3 a^{2}+4 b^{2}-5 b c=3\left(a^{2}+b c\right)+4\left(b^{2}-2 b c\right) \\
=3 \times 14+4 \times(-6)=18
\end{array}
$$ | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,218 |
4. As shown in Figure 3, in the convex hexagon $A B C D E F$, $A D$, $B E$, and $C F$ intersect at point $O$. The area of each triangle formed by every three consecutive vertices is equal to 1. Then $S_{\text {hexagon } A B C D E F}=$ $\qquad$ | 4.6.
As shown in Figure 5, connect $B D$ and $C E$. Since
$$
S_{\triangle B C D}=S_{\triangle E C D}=1,
$$
therefore, $B E \parallel C D$.
Since $S_{\triangle M F}=S_{\triangle S A F}$
$$
=1 \text {, therefore, } B E \parallel A F \text {. }
$$
Thus, $B E \parallel A F \parallel C D$.
Similarly, $C F \parallel D E \... | 4.6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,219 |
5. There are 6 natural numbers that have the same remainder when divided by 12, and their product is 971425. Then the minimum value of the sum of these 6 natural numbers is $\qquad$ . | 5. 150.
Since 971425 leaves a remainder of 1 when divided by 12, and
$$
971425=5 \times 5 \times 7 \times 7 \times 13 \times 61,
$$
among the prime factors, there are two each that leave a remainder of 5, 7, and 1 when divided by 12. Since the product of two numbers that leave a remainder of 5 (or 7) when divided by ... | 150 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,220 |
Three, (15 points) Given non-zero real numbers $a, b, c$ satisfy $a+b+c=0$. Prove:
(1) $a^{3}+b^{3}+c^{3}=3 a b c$;
(2) $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9$. | Three, (1) From $a+b+c=0$, we get $a+b=-c$. Therefore, $(a+b)^{3}=-c^{3}$.
Thus, $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3}$.
Hence $a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c$.
$$
\begin{array}{l}
\text { (2) }\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b} \\
=1+\left(\frac{b-c}{a}+\frac... | 9 | Algebra | proof | Yes | Yes | cn_contest | false | 716,221 |
Four, (15 points) As shown in Figure 4, in $\triangle A B C$, $\angle B A C$ $=\angle B C A=44^{\circ}, M$
is a point inside $\triangle A B C$,
such that $\angle M C A=30^{\circ}$, $\angle M A C=16^{\circ}$. Find the measure of $\angle B M C$. | In $\triangle A B C$, from $\angle B A C=\angle B C A=44^{\circ}$, we get
$$
A B=B C, \angle A B C=92^{\circ} \text {. }
$$
As shown in Figure 6, draw $B D \perp$
$A C$ at point $D$, extend $C M$
to intersect $B D$ at point $O$, and connect
$O A$. Then we have
$$
\begin{array}{l}
\angle O A C=\angle M C A \\
=30^{\cir... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,222 |
Example 3 As shown in Figure 3, point $H$ is the orthocenter of $\triangle A B C$, and $P$ is any point. $H L$
$\perp P A$ intersects $P A$ and
$B C$ at points $L$ and $X$;
$H M \perp P B$ intersects
$P B$ and $C A$ at points $M$ and
$Y ; H N \perp P C$ intersects
$P C$ and $A B$ at points
$N$ and $Z$. Prove: $X, Y,$
$... | Proof: Establish a rectangular coordinate system with $H$ as the origin. Then $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) 、 C\left(x_{3}, y_{3}\right), P\left(x_{0}, y_{0}\right)$.
Since $H L \perp P A$, we have
the line $l_{m l}:\left(x_{0}-x_{1}\right) x+\left(y_{0}-y_{1}\right) y=0$.
Since $H$ is the or... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,223 |
Five. (10 points) A student wrote 17 natural numbers on the blackboard, with the unit digit of each number being one of the digits 0, 1, 2, 3, 4. Prove that from these 17 numbers, 5 numbers can be selected such that their sum is divisible by 5. | If the last digits of 17 numbers are $0,1,2,3,4$ each, we can select 5 numbers whose last digits are exactly $0,1,2,3,4$, respectively, then the last digit of the sum of these 5 numbers is 0, and their sum is divisible by 5.
If the last digits of 17 numbers are not each $0,1,2,3,4$, then there are at most 4 different ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,224 |
1. The maximum value of the real number $k$ for which the inequality
$$
\sqrt{x-3}+\sqrt{6-x} \geqslant k
$$
has solutions is ( ).
(A) $\sqrt{6}-\sqrt{3}$
(B) $\sqrt{3}$
(C) $\sqrt{6}+\sqrt{3}$
(D) $\sqrt{6}$ | -.1.D.
Let $y=\sqrt{x-3}+\sqrt{6-x}, 3 \leqslant x \leqslant 6$, then $y^{2} \leqslant 2[(x-3)+(6-x)]=6$.
Therefore, $0<y \leqslant \sqrt{6}$. Hence, the maximum value of the real number $k$ is $\sqrt{6}$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,225 |
2. Four points in space $A, B, C, D$ satisfy $|AB|=3$, $|BC|=7$, $|CD|=11$, $|DA|=9$. Then the value of $AC \cdot BD$ is ( ).
(A) only one
(B) two
(C) four
(D) infinitely many | 2.A.
Notice that $3^{2}+11^{2}=130=7^{2}+9^{2}$. Since $A B+B C+$ $C D+D A=0$, then
$$
\begin{array}{l}
D A^{2}=D A^{2}=(A \boldsymbol{B}+\boldsymbol{B C}+\boldsymbol{C D})^{2} \\
=A B^{2}-B C^{2}+C D^{2}+2(\boldsymbol{A B}+\boldsymbol{B C}) \cdot(B C+C D), \\
\text { i.e., } 2 A C \cdot B D=A D^{2}+B C^{2}-A B^{2}-C ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,226 |
3. $\triangle A B C$ is inscribed in a unit circle, and the angle bisectors of the three interior angles $A$, $B$, and $C$ are extended to intersect the circle at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\... | 3.A.
As shown in Figure 4, connect $B A_{1}$, then
$$
\begin{array}{l}
A A_{1}=2 \sin \left(B+\frac{A}{2}\right) \\
=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) .
\end{array}
$$
Therefore, $A A_{1} \cos \frac{A}{2}$
$$
\begin{array}{l}
=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) \cdot \cos \frac{A}{2} \\
=\cos \left... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,227 |
4. As shown in Figure $1, A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a cube. Any plane $\alpha$ is made perpendicular to the diagonal $A C^{\prime}$, such that $\alpha$ intersects each face of the cube. Let the area of the resulting cross-sectional polygon be $S$, and the perimeter be $l$. Then ( ).
(A) $S... | 4.B.
As shown in Figure 5, after cutting off two regular tetrahedra $A A^{\prime} B D$ and $C^{\prime}-D^{\prime} B^{\prime} C$ from a cube, a geometric body $V$ is obtained, with parallel planes $A^{\prime} B D$ and $D^{\prime} B^{\prime} C$ as its top and bottom bases. Each side face of $V$ is an isosceles right tri... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,228 |
5. The curve represented by the equation $\frac{x^{2}}{\sin \sqrt{2}-\sin \sqrt{3}}+\frac{y^{2}}{\cos \sqrt{2}-\cos \sqrt{3}}=1$ is ( ).
(A) an ellipse with foci on the $x$-axis
(B) a hyperbola with foci on the $x$-axis
(C) an ellipse with foci on the $y$-axis
(D) a hyperbola with foci on the $y$-axis | 5.C.
From $\sqrt{2}+\sqrt{3}>\pi$, we have $0\cos \left(\sqrt{3}-\frac{\pi}{2}\right)$,
which means $\sin \sqrt{2}>\sin \sqrt{3}$.
Also, $00 \cdot \cos \sqrt{3}$ $0$, so the curve represented by the equation is an ellipse. And
$$
\begin{array}{l}
(\sin \sqrt{2}-\sin \sqrt{3})-(\cos \sqrt{2}-\cos \sqrt{3}) \\
=2 \sqrt{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,229 |
6. Let the set $T=\{0,1,2,3,4,5,6\}, M=$ $\left\{\left.\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}} \right\rvert\, a_{i} \in T, i=1,2,3,4\right\}$. If the elements of $M$ are arranged in descending order, then the 2005th number is $(\quad)$.
(A) $\frac{5}{7}+\frac{5}{7^{2}}+\frac{6}{7^{3}... | 6.C.
Use $\left[a_{1} a_{2} \cdots a_{k}\right]_{p}$ to denote a $k$-digit number in base $p$. Multiplying each number in set $M$ by $\tau^{4}$, we get
$$
\begin{aligned}
M^{\prime} & =\left\{a_{1} \times 7^{3}+a_{2} \times 7^{2}+a_{3} \times 7+a_{4} \mid a_{i} \in T, i=1,2,3,4\right\} \\
& =\left\{\left[a_{1} a_{2} a... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,230 |
7. Express the polynomial in $x$
$$
f(x)=1-x+x^{2}-x^{3}+\cdots-x^{19}+x^{20}
$$
as a polynomial in $y$
$$
g(y)=a_{0}+a_{1} y+\cdots+a_{20} y^{20},
$$
where $y=x-4$. Then $a_{0}+a_{1}+\cdots+a_{20}=$
$\qquad$ | $=.7 . \frac{5^{21}+1}{6}$.
From the problem, we know that the terms in the expression of $f(x)$ form a geometric sequence with the first term $1$ and the common ratio $-x$.
By the sum formula of a geometric sequence, we have
$$
f(x)=\frac{(-x)^{21}-1}{-x-1}=\frac{x^{21}+1}{x+1} \text {. }
$$
Let $x=y+4$, then $g(y)=\... | \frac{5^{21}+1}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,231 |
9. Let $\alpha, \beta, \gamma$ satisfy $0<\alpha<\beta<\gamma<2 \pi$. If for any $x \in \mathbf{R}, \cos (x+\alpha)+\cos (x+\beta)+$ $\cos (x+\gamma)=0$, then $\gamma-\alpha=$ $\qquad$ . | 9. $\frac{4 \pi}{3}$.
Let $f(x)=\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)$.
Given $x \in \mathbf{R}, f(x) \equiv 0$, we know
$$
f(-\alpha)=0, f(-\gamma)=0, f(-\beta)=0,
$$
which means
$$
\begin{array}{l}
\cos (\beta-\alpha)+\cos (\gamma-\alpha)=-1, \\
\cos (\alpha-\beta)+\cos (\gamma-\beta)=-1, \\
\cos (\alpha-\... | \frac{4 \pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,232 |
Example 4 Let $D$ be a point on side $BC$ of $\triangle ABC$, and point $P$ lies on segment $AD$. A line through point $D$ intersects segments $AB$, $PB$ at points $M$, $E$, and intersects the extensions of segments $AC$, $PC$ at points $F$, $N$. If $DE = DF$, prove that $DM = DN$.
(1st China Southeast Mathematical Oly... | Proof: As shown in Figure 4, let \( A(0, a), P(0, p) \), and let the equations of the lines be:
The line \( l_{A B} \):
\[
y = k_{1} x + a,
\]
The line \( l_{A C} \):
\[
y = k_{2} x + a,
\]
The line \( l_{P B} \):
\[
y = k_{3} x + p,
\]
The line \( l_{P C} \):
\[
y = k_{4} x + p,
\]
where \( k_{1}, k_{3} > 0 \) and \( ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,233 |
10. As shown in Figure 2, the volume of tetrahedron $D A B C$ is $\frac{1}{6}$, and it satisfies $\angle A C B=45^{\circ}$, $A D+B C+\frac{A C}{\sqrt{2}}=3$.
Then $C D=$ . $\qquad$ | 10. $\sqrt{3}$.
From $\frac{1}{3} A D \cdot\left(\frac{1}{2} B C \cdot A C \sin 45^{\circ}\right)$
$\geqslant V_{\text {tetrahedron DABC }}=\frac{1}{6}$,
that is, $A D \cdot B C \cdot \frac{A C}{\sqrt{2}} \geqslant 1$.
Also, $3=A D+B C+\frac{A C}{\sqrt{2}} \geqslant 3 \sqrt[3]{A D \cdot B C \cdot \frac{A C}{\sqrt{2}}}... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,234 |
11. If one side of the square
$A B C D$ lies on
the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$, then the minimum value of the area of the square is $\qquad$ | 11.80.
Let the side $AB$ of a square lie on the line $y=2x-17$, and the coordinates of the two vertices on the parabola be $C(x_1, y_1)$ and $D(x_2, y_2)$. Then the equation of the line $l$ on which $CD$ lies is $y=2x+b$. By solving the system of equations formed by the line $l$ and the parabola, we get $x^2=2x+b$. Th... | 11.80 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,235 |
12. If the sum of the digits of a natural number $a$ equals 7, then $a$ is called a "lucky number". Arrange all lucky numbers in ascending order as $a_{1}, a_{2}, a_{3}, \cdots$, if $a_{n}=$ 2005, then $a_{5 n}=$ $\qquad$ | 12.52000 .
It is known that the number of integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ that satisfy $x_{1} \geqslant 1$, $x_{i} \geqslant 0(i \geqslant 2)$ is $\mathrm{C}_{m+k-2}^{m-1}$. Taking $m=7$, we know that the number of $k$-digit lucky numbers is $P(k)=\mathrm{C}_{k+5}^{6}$.
Since 2005 is th... | 52000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,236 |
13. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=1, a_{n+1}=\frac{7 a_{n}+\sqrt{45 a_{n}^{2}-36}}{2}, n \in \mathbf{N} \text {. }
$$
Prove: (1) For any $n \in \mathbf{N}, a_{n}$ is a positive integer;
(2) For any $n \in \mathbf{N}, a_{n} a_{n+1}-1$ is a perfect square. | Three, 13. (1) From the given, we have $a_{1}=5$, and $\left\{a_{n}\right\}$ is strictly monotonically increasing. Transforming the given condition, we get
$$
2 a_{n+1}-7 a_{n}=\sqrt{45 a_{n}^{2}-36} \text {. }
$$
Squaring both sides and rearranging, we get
$$
a_{n+1}^{2}-7 a_{n} a_{n+1}+a_{n}^{2}+9=0 \text {. }
$$
F... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,237 |
One. (50 points) As shown in Figure 3, in $\triangle ABC$, let $AB > AC$. Draw the tangent line $l$ to the circumcircle of $\triangle ABC$ at point $A$.
Also, with point $A$ as the center and $AC$ as the radius, draw a circle that intersects line segment $AB$ at point $D$ and intersects line $l$ at points $E$ and $F$.... | (1) First, prove that $DE$ passes through the incenter of $\triangle ABC$.
As shown in Figure 7, connect $DE, DC$.
Draw the angle bisector of $\angle BAC$ to intersect $DE, DC$ at points $I, G$ respectively, and connect $IC$. Since $AD = AC$, we have
$AG \perp DC, ID = IC$.
Also, points $D, C, E$ lie on $\odot A$, so ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,238 |
II. (50 points) Let positive numbers $a, b, c, x, y, z$ satisfy $c y + b z = a$, $a z + c x = b$, $b x + a y = c$. Find the minimum value of the function
$$
f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}
$$ | From the conditions, we have
$$
b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0,
$$
which simplifies to $2 b c x+a^{2}-b^{2}-c^{2}=0$.
Solving for $x$ gives $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.
Similarly, $y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
Since $a, b, c, x, y, z$ are positive, from the a... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,239 |
1. As shown in Figure 1, point $P$ is on the circumcircle of $\triangle A B C$, line $C P$ and line $A B$ intersect at point $E$, line $B P$ and line $A C$ intersect at point $F$, the perpendicular bisector of side $A C$ intersects side $A B$ at point $J$, and the perpendicular bisector of side $A B$ intersects side $A... | 1. As shown in Figure 2, connect $B K$ and $C J$. Then
$$
\angle E=\angle A B P-\angle B P E .
$$
Since $A, B, P, C$ are concyclic, we know
$$
\angle B P E=\angle A .
$$
Thus, $\angle E=\angle A B P-\angle A$.
Also, since $K A=K B$, we have
$$
\angle A=\angle A B K \text {. }
$$
Therefore, $\angle E=\angle A B P-\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,241 |
Example 5 Let $H$ be any point on the altitude $C P$ of acute $\triangle A B C$, and $A H, B H$ intersect $B C, A C$ at points $M, N$ respectively.
(1) Prove: $\angle N P C=\angle M P C$;
(2) Let $O$ be the intersection of $M N$ and $C P$, and an arbitrary line through point $O$ intersects the sides of quadrilateral $C... | Proof: (1) As shown in Figure 5, establish a Cartesian coordinate system, and let \( A(a, 0) \), \( B(b, 0) \), \( C(0, c) \), \( H(0, h) \), and \( b, c, h > 0 \), \( a < 0 \). Then,
the line \( l_{A C} \):
$$
\frac{x}{a} + \frac{y}{c} = 1;
$$
the line \( l_{B H} \):
$$
\frac{x}{b} + \frac{y}{h} = 1.
$$
Thus, the fam... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,242 |
2. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right), \\
x y+y z+z x=1
\end{array}\right.
$$
(Supplied by Zhu Huawei) | 2. From $x+\frac{1}{x}=\frac{x^{2}+x y+y z+z x}{x}$
$$
=\frac{(x+y)(x+z)}{x},
$$
the first equation can be transformed into
$$
\begin{array}{l}
5 y z(x+y)(x+z)=12 x z(y+x)(y+z) \\
=13 x y(z+x)(z+y) . \\
\text { Let }\left\{\begin{array}{l}
x(y+z)=a, \\
y(z+x)=b, \\
z(x+y)=c .
\end{array}\right.
\end{array}
$$
then $a... | \left(\frac{1}{5}, \frac{2}{3}, 1\right) \text { and }\left(-\frac{1}{5},-\frac{2}{3},-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,243 |
4. Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets of integers $A_{1}$, $A_{2}$, \cdots, $A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z}$, and for any two numbers $b>c$ in each $A_{i}$, we have
$$
b-c \geqslant a^{i} .
$$
(Y... | 4. If $0 < a < 2$, let $A_{i}=\left\{2^{i-1} \mathrm{~m} \mid \mathrm{m}\right.$ is odd $\}, i=1,2, \cdots, n-1$, $A_{n}=\left\{2^{n-1}\right.$ multiples $\}$.
Then this partition satisfies the requirements.
If $a \geqslant 2$, suppose $A_{1}, A_{2}, \cdots, A_{n}$ satisfy the requirements.
Let $M=\left\{1,2, \cdots, 2... | a < 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,245 |
5. Let positive real numbers $x, y$ satisfy $x^{3}+y^{3}=x-y$. Prove: $x^{2}+4 y^{2}<1$.
(Xiong Bin) | 5. By the AM-GM inequality,
$5 y^{3}+x^{2} y \geqslant 2 \sqrt{5 x^{2} y^{4}}>4 x y^{2}$.
Therefore, $\left(x^{2}+4 y^{2}\right)(x-y)<x^{3}+y^{3}$. Hence $x^{2}+4 y^{2}<\frac{x^{3}+y^{3}}{x-y}=1$. | x^{2}+4 y^{2}<1 | Inequalities | proof | Yes | Yes | cn_contest | false | 716,246 |
6. Let $n(n \geqslant 3)$ be a positive integer. If there are $n$ lattice points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane such that: when $\left|P_{i} P_{j}\right|$ is a rational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both irrational; when $\left|P_{i} P_{... | 6. We assert that the smallest good number is 5, and 2005 is a good number.
In a triplet $\left(P_{i}, P_{j}, P_{k}\right)$, if $\left|P_{i} P_{j}\right|$ is a rational number (or irrational number), and $\left|P_{i} P_{k}\right|, \left|P_{j} P_{k}\right|$ are irrational numbers (or rational numbers), then $\left(P_{i... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,247 |
7. Let $m, n$ be integers, $m>n \geqslant 2, S=\{1,2$, $\cdots, m\}, T=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a subset of $S$. It is known that any two numbers in $T$ cannot simultaneously divide any number in $S$. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{m+n}{m} .
$$ | 7. Construct $T_{i}=\left\{b \in S\left|a_{i}\right| b\right\}, i=1,2, \cdots, n$. Then
$$
\left|T_{i}\right|=\left[\frac{m}{a_{i}}\right] \text {. }
$$
Since any two numbers in $T$ cannot simultaneously divide a number in $S$, when $i \neq j$, we have $T_{i} \cap T_{j}=\varnothing$. Thus,
$$
\sum_{i=1}^{n}\left|T_{i}... | \sum_{i=1}^{n} \frac{1}{a_{i}}<\frac{m+n}{m} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,248 |
8. Given real numbers $a, b (a>b>0)$, place a rectangle with length $a$ and width $b$ inside a square (including the boundary). What is the minimum length of the side of the square? | 8. Let the rectangle be $ABCD, AB=a, BC=b$, with the center at point $O$. Establish a Cartesian coordinate system with $O$ as the origin, where the $x$-axis and $y$-axis are parallel to the sides of the rectangle. We will discuss two scenarios.
(1) The line segment $BC$ does not intersect the coordinate axes. As shown ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,249 |
1. [ $x$ ] represents the greatest integer not greater than $x$, and $\{x\}$ $=x-[x]$. Then, among the following relations, the incorrect one is ( ).
(A) $[\sqrt{45}]=6$
(B) $\{\sqrt{45}\}=\sqrt{45}-6$
(C) $[-\sqrt{45}]=-6$
(D) $\{-\sqrt{45}\}=7-\sqrt{45}$ | - 1.C.
Since $-7<-\sqrt{45}<-6$, therefore,
$$
[-\sqrt{45}]=-7 .
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,250 |
2. Arrange the positive even numbers in 5 columns as shown in Table 1.
Table 1
\begin{tabular}{|l|c|c|c|c|c|}
\hline & Column 1 & Column 2 & Column 3 & Column 4 & Column 5 \\
\hline Row 1 & & 2 & 4 & 6 & 8 \\
\hline Row 2 & 16 & 14 & 12 & 10 & \\
\hline Row 3 & & 18 & 20 & 22 & 24 \\
\hline Row 4 & 32 & 30 & 28 & 26 & ... | 2. D.
From Table 1, we know that the numbers in the 1st column are of the form $16 k$ (where $k$ is an integer), the numbers in the 2nd column are of the form $16 k \pm 2$, the numbers in the 3rd column are of the form $16 k \pm 4$, the numbers in the 4th column are of the form $16 k \pm 6$, and the numbers in the 5th... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,251 |
3. Given the quadratic function $y=x^{2}+b x-3(-2 \leqslant b$ $\leqslant 2$ ), as $b$ changes from -2 to 2, the corresponding parabola also moves. Among the following descriptions about the direction of the parabola's movement, the correct one is ( ).
(A) First moves to the upper left, then to the lower left
(B) First... | 3. A.
The movement of the parabola's position is consistent with the movement of the vertex $\left(-\frac{b}{2},-\frac{b^{2}+12}{4}\right)$. As $b$ changes from -2 to 2, $-\frac{b}{2}$ continuously decreases, so it always moves to the left; while $-\frac{b^{2}+12}{4}$ first increases and then decreases, so it first mo... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,252 |
Example 6 A line $l$ does not intersect circle $\odot O$, $E$ is a point on line $l$, $O E \perp l, M$ is any point on line $l$ different from $E$, from point $M$ two tangents to $\odot O$ are drawn, touching $\odot O$ at points $A$ and $B$, $C$ is a point on $M A$ such that $E C \perp M A$, $D$ is a point on $M B$ suc... | Proof: As shown in Figure 6, let the equation of circle $\odot O$ be $x^{2}+(y-a)^{2}=r^{2}$, and let $M(s, 0)$, with the slopes of $M A$ and $M B$ being $k_{1}$ and $k_{2}$, respectively. Then the equation of the line system passing through point $C$ is
$$
\begin{array}{c}
\lambda_{1}\left(k_{1} \dot{x}-y-\right. \\
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,253 |
4. In $\triangle A B C$, $A B=A C, \angle A=36^{\circ}$, $P$ is a point in the plane of $\triangle A B C$, and $\triangle P A B$, $\triangle P B C$, $\triangle P A C$ are all isosceles triangles. Then the number of points $P$ that satisfy this condition is ( ) .
(A) 10
(B) 8
(C) 6
(D) 4 | 4. B.
As shown in Figure 2, when $B P = C P$, point $P$ lies on the perpendicular bisector of $B C$, and there are 4 points that meet the requirement;
When $B C = B P$, point $P$ lies on the circle $\odot B$ with $B C$ as the radius, and there are 2 points that meet the requirement;
Similarly, when $B C = C P$, ther... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,254 |
5. Reversing the digits of 1746 gives 6471, the new number is 4725 more than the original number. Then, among all four-digit numbers, the number of numbers that satisfy the condition that the new number is 4725 more than the original number is ( ) .
(A) 16
(B) 20
(C) 24
(D) 28 | 5. D.
Let the new number be $\overline{a b c d}$, then the original number is $\overline{d c b a}$. From the given condition, we have $\overline{a b c d}-\overline{d c b a}=999(a-d)+90(b-c)=4725$, which simplifies to $111(a-d)+10(b-c)=525$.
Clearly, 5 divides $(a-d)$.
Noting that $0<a-d<9$, it can only take $a-d=5$, yi... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,255 |
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