problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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4. As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, $A A_{1}=A B=A C$, and $M$ and $Q$ are the midpoints of $C C_{1}$ and $B C$ respectively. If for any point $P$ on the line segment $A_{1} B_{1}$, $P Q \perp A M$, then $\angle B A C$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{... | 4.D.
Let the midpoint of $A C$ be $R$. It is easy to know that $A_{1} R \perp A M$. Also, $P Q \perp A M$, so $A M \perp$ plane $A_{1} B_{1} Q R$. Therefore, $A M \perp Q R$, which means $A M \perp A B$. Also, $A B \perp A A_{1}$, so $A B \perp$ plane $A C C_{1} A_{1}$. Therefore, $A B \perp A C$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,693 |
5. As shown in Figure 2, three points $P$, $S$, and $R$ move at a uniform speed on the sides of $\triangle A B C$. When $t=0$, they start from $A$, $B$, and $C$ respectively, and when $t=1 \mathrm{~s}$, they simultaneously reach $B$, $C$, and $A$. Then, the fixed point in this motion process is the ( ) of $\triangle P ... | 5.D.
According to the problem, $\frac{A P}{A B}=\frac{B S}{B C}=\frac{C R}{C A}=\lambda$.
Let $G$ be the centroid of $\triangle P S R$, then
$$
\begin{array}{l}
A G=\frac{1}{3}(A P+A S+A R) \\
=\frac{1}{3}[\lambda A B+A B+\lambda B C+(1-\lambda) A C] \\
=\frac{1}{3}(A B+A C) .
\end{array}
$$
Therefore, $G$ is the cen... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,694 |
6. A bag contains 20 balls, of which 9 are white, 5 are red, and 6 are black. Now, 10 balls are randomly drawn from the bag, such that the number of white balls is no less than 3 and no more than 7, the number of red balls is no less than 2 and no more than 5, and the number of black balls is no more than 3. The number... | 6.A.
Notice that
$$
\begin{array}{l}
f(x)=\left(x^{3}+x^{4}+x^{5}+x^{6}+x^{7}\right) . \\
\quad\left(x^{2}+x^{3}+x^{4}+x^{5}\right)\left(1+x+x^{2}+x^{3}\right) \\
=\left(x^{3}+x^{4}+x^{5}+x^{6}+x^{7}\right)\left(x^{2}+2 x^{3}+3 x^{4}+\right. \\
\left.4 x^{5}+3 x^{6}+2 x^{7}+x^{8}\right)
\end{array}
$$
In this, the co... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,695 |
1. The number of positive integers $n$ that make $1^{2 \times 15}+2^{2005}+\cdots+n^{205}$ divisible by $n+2$ is | $=1.0$.
Let $S_{n}=1^{2008}+2^{2000}+\cdots+n^{2005}$, then
$$
S_{n}=n^{2008}+(n-1)^{2005}+\cdots+1^{2008} \text {. }
$$
By misalignment addition, we get
$$
\begin{aligned}
2 S_{n}= & 2+\left(2^{2005}+n^{2005}\right)+\left[3^{2005}+(n-1)^{2000}\right]+ \\
& \cdots+\left(n^{2005}+2^{2005}\right) .
\end{aligned}
$$
Fro... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,696 |
3. Let $\frac{\sin ^{4} x}{3}+\frac{\cos ^{4} x}{7}=\frac{1}{10}$. Then for a given positive integer $n, \frac{\sin ^{2 n} x}{3^{n-1}}+\frac{\cos ^{2 n} x}{7^{n-1}}=$ $\qquad$ | 3. $\frac{1}{10^{n-1}}$.
Let $\boldsymbol{\alpha}=\left(\frac{\sin ^{2} x}{\sqrt{3}}, \frac{\cos ^{2} x}{\sqrt{7}}\right), \boldsymbol{\beta}=(\sqrt{3}, \sqrt{7})$. Then $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=1$,
$|\alpha||\beta|=\sqrt{\frac{\sin ^{4} x}{3}+\frac{\cos ^{4} x}{7}} \cdot \sqrt{3+7}=1$.
Therefore,... | \frac{1}{10^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,698 |
4. As shown in Figure $3, M$ and $N$ are internal points on the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, and $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$. If points $B, M, N$ are collinear, then $\lambda$ $=$ | 4. $\frac{\sqrt{3}}{3}$.
Extend $E A$ and $C B$ to intersect at point $P$, and let the side length of the regular hexagon be 1.
It is easy to know that $P B=2, A$ is the midpoint of $E P$, $E A=A P=\sqrt{3}$.
From $A M=\lambda A C$, we get
$$
C M=(1-\lambda) C A \text {. }
$$
Also, $C P=3 C B, C A$ is the median of $... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,699 |
5. The system of equations $\left\{\begin{array}{l}x+x y+y=1, \\ x^{2}+x^{2} y^{2}+y^{2}=17\end{array}\right.$ has the real solution $(x, y)$ $=$ . | (Let $x+y=u, x y=v$. Answer: $x_{1}=\frac{3+\sqrt{17}}{2}$,
$$
\left.y_{1}=\frac{3-\sqrt{17}}{2} ; x_{2}=\frac{3-\sqrt{17}}{2}, y_{2}=\frac{3+\sqrt{17}}{2} .\right)
$$ | x_{1}=\frac{3+\sqrt{17}}{2}, y_{1}=\frac{3-\sqrt{17}}{2}; x_{2}=\frac{3-\sqrt{17}}{2}, y_{2}=\frac{3+\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,700 |
5. For all positive real numbers $a, b, c, d$,
$$
\left(\frac{a^{3}}{a^{3}+15 b c d}\right)^{\frac{1}{2}} \geqslant \frac{a^{x}}{a^{x}+b^{x}+c^{x}+d^{x}}
$$
the real number $x=$ | 5. $\frac{15}{8}$.
Rearranging the given inequality, we get
$$
\begin{array}{l}
\left(a^{x}+b^{x}+c^{x}+d^{x}\right)^{2}-a^{2 x} \geqslant 15 a^{2 x-3} b c d . \\
\text { The left side of equation (1) }=\left(b^{x}+c^{x}+d^{x}\right)\left(2 a^{x}+b^{x}+c^{x}+d^{x}\right) \\
\geqslant 15 \sqrt[3]{(b c d)^{x}} \cdot \sq... | \frac{15}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,701 |
6. Let $1 \leqslant r \leqslant n$. Then the arithmetic mean of the smallest numbers in all $r$-element subsets of the set $M=\{1,2, \cdots, n\}$ is $\qquad$ | 6. $\frac{n+1}{r+1}$.
The number of subsets of $M$ containing $r$ elements is $\mathrm{C}_{n}^{r}$, and the number of subsets of $M$ containing $r$ elements with the smallest number being the positive integer $k$ is $\mathrm{C}_{n-k}^{r-1}$. Therefore, the arithmetic mean of the smallest numbers in all subsets of $M$ ... | \frac{n+1}{r+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,702 |
Three. (20 points) Let real numbers $a, b \in [p, q] (q > p > 0)$. Prove that for any positive integer $m$, we have
$$
\begin{array}{l}
a^{-m} b^{m+2} + a^{m+2} b^{-m} \\
\leqslant \frac{a^{2} + b^{2}}{p^{2} + q^{2}} \left(p^{-m} q^{m+2} + p^{m+2} q^{-m}\right).
\end{array}
$$ | Let $c=\frac{q}{p}$, then $\frac{p}{q}=\frac{1}{c}$.
Since $a, b \in [p, q]$, we have $\frac{1}{c} \leqslant \frac{a}{b} \leqslant c$.
$$
\begin{array}{l}
\text { Hence }\left(\frac{a}{b}-\frac{1}{c}\right)\left(\frac{a}{b}-c\right) \leqslant 0, \text { i.e., } \\
\left(\frac{a}{b}\right)^{2}-\left(c+\frac{1}{c}\right)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,703 |
(1) Let $A$ and $B$ be constants. If for any $n \in \mathbf{N}$, we have $|A-B| \leqslant \frac{1}{n}$, prove that: $A=B$.
(2) Let $f(x)$ be a monotonic function defined on the interval $(0,+\infty)$, and for any $x, y \in (0,+\infty)$, we have
\[
\begin{array}{l}
f(x y)=f(x)+f(y), \\
f(a)=1 \quad(0<a \neq 1) .
\end{ar... | (1) Assume $A \neq B$.
According to the problem, $|A-B| \leqslant 1$.
Let $|A-B|=\frac{1}{m}$, where $m \geqslant 1$.
Take $n=[m]+1$, where $[m]$ represents the greatest integer not exceeding $m$, then $|A-B|=\frac{1}{m}>\frac{1}{n}$, which contradicts the given condition.
Therefore, $A=B$.
(2) When $n$ is an integer, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,704 |
Five. (20 points) Through the center $O$ of the hyperbola $x^{2}-\frac{y^{2}}{4}=1$, draw two mutually perpendicular rays, intersecting the hyperbola at points $A$ and $B$. Try to find:
(1) The equation of the trajectory of the midpoint $P$ of chord $A B$;
(2) The distance from the center $O$ of the hyperbola to the li... | (1) Let $P(x, y)$, $A(x-m, y-n)$, and $B(x+m, y+n)$, then we have
$$
\begin{array}{l}
4(x-m)^{2}-(y-n)^{2}=4, \\
4(x+m)^{2}-(y+n)^{2}=4 .
\end{array}
$$
From $O A \perp O B$ we get
$$
x^{2}+y^{2}=m^{2}+n^{2} \text {. }
$$
(2) - (1) gives
$4 m x=n y$,
(2) + (1) gives
$$
4 x^{2}-y^{2}+4 m^{2}-n^{2}=4 \text {. }
$$
From... | \frac{4 \sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,705 |
Given positive integers $m, n (m \leqslant n)$, let $A=\{1,2, \cdots, n\}$. Find the number of mappings $f: A \rightarrow A$ that satisfy the following conditions:
(1) $f$ takes exactly $m$ values;
(2) If $k, l \in A, k \leqslant l$, then,
$$
f(f(k))=f(k) \leqslant f(l).
$$ | From condition (2), we know that $f$ is a non-decreasing function, and
$$
f(f(x))=f(x) \text {. }
$$
Let the range of $f$ be $\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$, and the array $\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}$ satisfies $x_{1}+x_{2}+\cdots+x_{m}=n$, and
$$
\begin{array}{l}
f(1)=f(2)=\cdots=f\left(x... | \mathrm{C}_{n+m-1}^{n-m} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,706 |
II. (50 points) Given a circle $\odot O_{1}$ with radius $r$ and a circle $\odot O$ with radius $R$ that are internally tangent at $D$. $\triangle ABC$ is inscribed in $\odot O$, and $AB, AC$ are tangent to $\odot O_{1}$ at $P, Q$ respectively. The intersection of $AO_{1}$ and $PQ$ is $M$. Prove that $M$ is the incente... | II. As shown in Figure 6, extend $A O_{1}$ to intersect $\odot O$ at $E$, connect $B E$ and $B M$, and extend $O_{1} O$ to intersect $\odot O$ at $K$. Clearly, $A O_{1}$ bisects $\angle B A C$.
From $O_{1} E \cdot O_{1} A = O_{1} D \cdot O_{1} K$, we get
$$
\begin{array}{l}
O_{1} E \cdot \frac{r}{\sin \frac{A}{2}} \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,707 |
Three, (50 points) Prove: any positive odd number $m$ can always divide an integer whose all digits are odd.
| Three, first prove the following lemma.
Lemma For any $k \in \mathbf{N}_{+}$, there always exists a $k$-digit number $n_{k}$ with all digits being odd, such that $5^{k} \mid n_{k}$.
Proof of the lemma: When $k=1$, it is obviously true.
Assume for $k$, $5^{k} \mid n_{k}$.
For $k+1$, when $5^{k+1} \mid n_{k}$, let $n_{k+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,708 |
As shown in Figure 2, in the square $ABCD$, with point $A$ as the center and $AB$ as the radius, draw arc $BD$ intersecting $AC$ at $E$. $\odot O_{1}$ is tangent to $AB$ and $AD$ and internally tangent to $\overparen{BD}$. $\odot O_{2}$ is tangent to $CB$ and $CD$ and externally tangent to $\overparen{BD}$. Draw the ta... | Proof: Given that points $O_{1}$ and $O_{2}$ lie on $AC$, and circles $\odot O_{1}$ and $\odot O_{2}$ are tangent to arc $\overparen{B D}$ at point $E$, thus,
$$
\begin{array}{l}
P E \perp A C, P E=E C . \\
\text { Let } A B=A D=A E=a,
\end{array}
$$
then $A C=\sqrt{2} a$.
Since $A O_{1}+O_{1} E=(\sqrt{2}+1) O_{1} E=a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,709 |
Given that $ABCD$ is a square, point $M$ (different from points $B, C$) is on side $BC$, and the perpendicular bisector $l$ of line segment $AM$ intersects $AB$ and $CD$ at points $E$ and $F$ respectively.
(1) Which is longer, $BE$ or $DF$? Please explain your reasoning.
(2) If $AB=1$, find the range of $|BE - DF|$ as ... | (1) As shown in Figure 3, let $A C$ and $B D$ intersect at point $O$, where $O$ is the center of the square $A B C D$. Line $l$ intersects $A M$ at the midpoint $N$ of $A M$. Connect $NO$, then $NO // B C$. Therefore, $\angle O N M = \angle N M B = \angle A M B B E' = D F' > D F$, which means $B E$ is longer than $D F$... | \left(0, \frac{1}{4}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,710 |
Example 1 In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $D$ is the midpoint of $A C$. Prove: $A B_{1} / /$ plane $D B C_{1}$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, please let me know! | Proof: Establish the spatial rectangular coordinate system as shown in Figure 1. Then,
$$
\begin{array}{l}
A(0,0,0) . \\
\text { Let } B_{1}(0, a, b) \text {. } \\
B(0, a, 0), \\
D\left(\frac{\sqrt{3} a}{4}, \frac{a}{4}, 0\right), \\
C_{1}\left(\frac{\sqrt{3} a}{2}, \frac{a}{2}, b\right) .
\end{array}
$$
Thus, $A B_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,711 |
Given $x, y, z \in \mathbf{R}_{+}, x+y+z=1$. Prove:
$$
\left(\frac{1}{x^{2}}-x\right)\left(\frac{1}{y^{2}}-y\right)\left(\frac{1}{z^{2}}-z\right) \geqslant\left(\frac{26}{3}\right)^{3} \text {. }
$$ | $$
\begin{array}{l}
\left(\frac{1}{x^{2}}-x\right)\left(\frac{1}{y^{2}}-y\right)=\frac{\left(1-x^{3}\right)\left(1-y^{3}\right)}{x^{2} y^{2}} \\
=(1-x)(1-y) \cdot \frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x^{2} y^{2}} \\
=(1-x-y+x y)\left(1+\frac{1}{x}+\frac{1}{x^{2}}\right)\left(1+\frac{1}{y}+\frac{1}{y^{2}}... | \left(\frac{1}{x^{2}}-x\right)\left(\frac{1}{y^{2}}-y\right)\left(\frac{1}{z^{2}}-z\right) \geqslant\left(\frac{26}{3}\right)^{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,712 |
Let $S=\{1,2, \cdots, n\}$. Find the smallest natural number $n$, such that when $S$ is arbitrarily divided into two subsets, there is always one subset that contains two different numbers $a$ and $b$, satisfying $(a+b) \mid a b$. | Solution: Two distinct numbers $a$ and $b$ that satisfy $(a+b) \mid a b$ are called a "good pair." Clearly, the answer to this problem is only related to these good pairs and has nothing to do with numbers that do not appear in any good pair. Therefore, we can list all the good pairs within a certain range, for example... | 40 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,713 |
Example 2: Prove that in the unit cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, the plane $A_{1} B C_{1} \parallel$ plane $D_{1} A C$. | Proof: Establish a spatial rectangular coordinate system as shown in Figure 2.
Then
$$
\begin{array}{l}
D(0,0,0) \text {, } \\
A_{1}(1,0,1) \text {, } \\
B(1,1,0) \text {, } \\
C(0,1,0) \text {, } \\
C_{1}(0,1,1) \text {, } \\
D_{1}(0,0,1) \text {, } \\
A(1,0,0) .
\end{array}
$$
Thus, \( A_{1} C_{1}=(-1,1,0), A_{1} \b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,714 |
Example 3 Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, M, N, P$ are the midpoints of edges $C C_{1}, B C, C D$ respectively. Prove: $A_{1} P \perp$ plane $D M N$.
---
The translation maintains the original text's line breaks and formatting. | Proof: Establish a spatial rectangular coordinate system as shown in Figure 3, and connect $D M$ and $D N$. Then
$$
\begin{array}{l}
D(0,0,0), \\
N\left(\frac{1}{2}, 1,0\right), \\
M\left(0,1, \frac{1}{2}\right) .
\end{array}
$$
Thus, $D N=\left(\frac{1}{2}, 1,0\right), D M=\left(0,1, \frac{1}{2}\right)$. Let the norm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,715 |
Example 4 Given a unit cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, $E$ is a moving point on $B C$, and $F$ is the midpoint of $A B$. Determine the position of point $E$ such that $C_{1} F \perp A_{1} E$. | Prove: Connect $A_{1} D, D E$, and establish a spatial rectangular coordinate system as shown in Figure 4. Then,
$$
\begin{array}{l}
D(0,0,0), \\
A_{1}(1,0,1), \\
E(a, 1,0), \\
F\left(1, \frac{1}{2}, 0\right), \\
C_{1}(0,1,1).
\end{array}
$$
Thus, $D A_{1}=(1,0,1), D E=(a, 1,0)$.
Let the normal vector of plane $A_{1} ... | a=\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,716 |
Example 5 Given a unit cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, where $E$ is the midpoint of $C C_{1}$. Prove: plane $A_{1} B D \perp$ plane $E B D$.
---
The translation maintains the original text's line breaks and formatting. | Proof 1: Establish a spatial rectangular coordinate system as shown in Figure 5. Then,
$$
\begin{array}{l}
D(0,0,0), \\
A_{1}(1,0,1), \\
B(1,1,0), \\
E\left(0,1, \frac{1}{2}\right).
\end{array}
$$
Thus, $D E=\left(0,1, \frac{1}{2}\right), D B=(1,1,0)$,
$$
D A_{1}=(1,0,1), D B=(1,1,0).
$$
Let the normal vector of plan... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,717 |
Example 6 In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=2, A A_{1}=A D=1$. Find the angle formed by $A B$ and the plane $A B_{1} C$. | Solution: Establish a spatial rectangular coordinate system as shown in Figure 7. Then
$$
\begin{array}{l}
D(0,0,0) \text {, } \\
A(1,0,0) \text {, } \\
C(0,2,0) \text {, } \\
B_{1}(1,2,1) .
\end{array}
$$
Thus, $A C=(-1,2,0), A B_{1}=(0,2,1)$.
Let the normal vector of plane $A B_{1} C$ be $\boldsymbol{n}=(x, y, z)$.
... | \arcsin \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,718 |
Example 7 In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the degree of the dihedral angle $A-B D_{1}-A_{1}$ is | Solution: Let the body be a unit cube. Establish a spatial rectangular coordinate system as shown in Figure 9, then
$$
\begin{array}{l}
D(0,0,0), \\
A(1,0,0), \\
D_{1}(0,0,1), \\
B(1,1,0), \\
A_{1}(1,0,1).
\end{array}
$$
Thus, $A D_{1}=(-1,0,1), A B=(0,1,0)$,
$$
A_{1} B=(0,1,-1), A_{1} D_{1}=(-1,0,0).
$$
Let the norm... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,719 |
Example 3 Let the three sides of $\triangle A B C$ be $a, b$, and $c$, and $\frac{a-b}{1+a b}+\frac{b-c}{1+b c}+\frac{c-a}{1+c a}=0$. Then the shape of $\triangle A B C$ must be a $\qquad$ triangle. | Solution: Remove the denominator from the original expression and set it as $f$, we get
$$
\begin{aligned}
f= & (a-b)(1+b c)(1+c a)+(b-c)(1+a b) . \\
& (1+c a)+(c-a)(1+b c)(1+a b) \\
= & a\left(b^{2}-c^{2}\right)+b\left(c^{2}-a^{2}\right)+c\left(a^{2}-b^{2}\right) \\
= & 0 .
\end{aligned}
$$
When $a=b$, $f=0$, by the ... | isosceles | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,720 |
Example 8 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dire... | Solution: As shown in Figure 10, in the rectangular coordinate system, we have
$$
\begin{array}{l}
D(0,0,0), \\
B_{1}(1,1,1), \\
E\left(1, \frac{1}{2}, 0\right), \\
E\left(\frac{1}{2}, 1,0\right) .
\end{array}
$$
Thus, $B_{1} E=\left(0,-\frac{1}{2},-1\right)$,
$$
\boldsymbol{B}_{1} \boldsymbol{F}=\left(-\frac{1}{2}, 0... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,721 |
Example 9 Given a rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$, where $A B=a, B C=b, C C_{1}=c$. Find the distance between the plane $A_{1} B D$ and the plane $B_{1} D_{1} C$. | Solution: Establish a spatial rectangular coordinate system as shown in Figure 11, then
$$
\begin{array}{l}
D(0,0,0) \\
A_{1}(b, 0, c) \\
B(b, a, 0) \text {, } \\
C(0, a, 0) .
\end{array}
$$
Thus, $D A_{1}=(b, 0, c), D B=(b, a, 0)$,
$$
D C=(0, a, 0) \text {. }
$$
Let the normal vector of plane $A_{1} B D$ be $\boldsy... | \frac{a b c}{\sqrt{a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,722 |
Example 10 Given a unit cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $M$ and $N$ are the midpoints of $B B_{1}$ and $B_{1} C_{1}$, respectively, and $P$ is the midpoint of line segment $M N$. Find the distance between $D P$ and $A C_{1}$. | Solution: Establish a spatial rectangular coordinate system as shown in Figure 12, then
$$
\begin{array}{l}
B_{1}(0,0,0), \\
A(0,1,1), \\
C_{1}(1,0,0), \\
D(1,1,1), \\
P\left(\frac{1}{4}, 0, \frac{1}{4}\right)
\end{array}
$$
Let the equation of the plane $\alpha$ passing through $D P$ and parallel to $A C_{1}$ be
$$
A... | \frac{\sqrt{86}}{86} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,723 |
Example 11 In the right quadrilateral prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the base is a right trapezoid $A B C D, C D \perp A D, A B=$ $2, A D=3, D C=6, A A_{1}=6, M, N$ are the midpoints of $C_{1} D_{1}, C C_{1}$ respectively. Find the distance from $M N$ to the plane $A D_{1} C$. | Solution: Establish a rectangular coordinate system as shown in Figure 13, then
$$
\begin{array}{l}
D(0,0,0), \\
A(3,0,0), \\
C(0,6,0), \\
D_{1}(0,0,6), (0,6,3).
\end{array}
$$
Thus, $A C=(-3,6,0), A D_{1}=(-3,0,6)$.
Let the normal vector of plane $A D_{1} C$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp ... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,724 |
1. Given a regular tetrahedron $ABCD$, the midpoints of $AB$, $BC$, and $CD$ are $E$, $F$, and $G$ respectively. Find the dihedral angle $C-FG-E$. | (Answer: $\pi-\arccos \frac{\sqrt{3}}{3$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
(Answer: $\pi-\arccos \frac{\sqrt{3}}{3}$. | \pi-\arccos \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,725 |
4. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the midpoints of edges $B B_{1}$ and $C C_{1}$ are $M$ and $N$, respectively. Find the angle formed by $A_{1} D$ and the plane $D_{1} M N$. | (Answer: $\arcsin \frac{\sqrt{10}}{5$.) | \arcsin \frac{\sqrt{10}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,726 |
5. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, there are points $M$, $N$, and $P$ on $C C_{1}$, $B C$, and $C D$ respectively, and $C M=C N$. To make $A_{1} P \perp$ plane $D M N$, determine the position of point $P$. | (Tip: $D P=C M$. ) | D P=C M | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,727 |
Example 1 A paper has a circle $\odot O$ with radius $R$ and a fixed point $A$ inside the circle, where $O A=a$. Fold the paper so that a point $A^{\prime}$ on the circumference coincides exactly with point $A$. Each such fold leaves a straight line crease. Find the set of points on all the crease lines when $A^{\prime... | Solution: Establish a rectangular coordinate system as shown in Figure 2. By the symmetry of the paper folding, point $A^{\prime}$ and point $A$ are symmetric with respect to the line $l$ of the fold, i.e., $l$ is the perpendicular bisector of segment $A A^{\prime}$.
Connecting $O A^{\prime}$ and intersecting $l$ at po... | \frac{\left(x-\frac{a}{2}\right)^{2}}{\frac{R^{2}}{4}}+\frac{y^{2}}{\frac{R^{2}}{4}-\frac{a^{2}}{4}} \geqslant 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,728 |
Example 2 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$. Fold $\triangle A C M$ along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Find the volume of the tetrahedron $A-B C M$. | Solution: The figures before and after folding are shown in Figure 3 and Figure 4, respectively. By comparing the two figures, we can see that in Figure 4, $A C=2, A M=2, B C=2 \sqrt{3}, \angle A C M=60^{\circ}, \angle B C M=30^{\circ}$.
Take the midpoint $D$ of $C M$, and connect $A D$.
In $\triangle B C M$, draw $D E... | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,729 |
Example 3 As shown in Figure $5, H$ is the orthocenter of $\triangle A B C$, and $D$, $E$, $F$ are the midpoints of sides $B C$, $C A$, and $A B$, respectively. A circle centered at $H$ intersects $D E$ at points $P$ and $Q$, intersects $E F$ at points $R$ and $S$, and intersects $F D$ at points $T$ and $V$. Prove:
$$
... | Proof: In Figure 5, since the orthocenter $H$ is inside $\triangle ABC$, it is easy to see that $\triangle ABC$ is an acute triangle.
When $\triangle AEF$,
$\triangle BFD$, and $\triangle CED$
are folded along the midlines
$EF$, $FD$, and $DE$, respectively,
a tetrahedron $O-EFD$ (as shown
in Figure 6) is formed, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,730 |
Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$. | Solution: Let $x+y=u, xy=v$. From the given equation, we get $u+v=71, uv=880$.
By Vieta's formulas, $u$ and $v$ are the roots of the quadratic equation
$$
t^{2}-71t+880=0
$$
Solving this equation, we get $t=16$ or $t=55$.
Therefore, $\left\{\begin{array}{l}u=16, \\ v=55\end{array}\right.$ or $\left\{\begin{array}{l}u=... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,731 |
Example 4 In a plane, given six line segments $l_{1}, l_{2}, \cdots$, $l_{6}$, they are equal to the edges $A B, A C$, $A D, B C, B D, C D$ of the tetrahedron $A B C D$. Question: How to use a ruler and compass to construct a line segment equal to the altitude of the tetrahedron passing through vertex $A$? | Solution: As shown in Figure 7, $AO$ is an altitude of the tetrahedron. In the base plane $BCD$, draw $OE \perp BC$ at $E$, and $OF \perp CD$ at $F$. By the theorem of three perpendiculars, we get
$AE \perp BC, AF \perp CD$.
Now, unfold the lateral faces $ABC$ and $ACD$ onto the plane $BCD$ (as shown in Figure 8), then... | OA | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,732 |
Example 5 The lateral faces of the tetrahedron $ABCD$ are all acute triangles. Consider all closed broken lines $XYZTX$, where $X, Y, Z, T$ are interior points of the edges $AB, BC, CD, DA$ respectively. Prove:
(1) If $\angle DAB + \angle BCD \neq \angle ABC + \angle CDA$, then there is no shortest one among these clos... | Proof: When the tetrahedron \(ABCD\) is unfolded and laid flat (as shown in Figure 9, where \(C\) and \(C'\), \(D\) and \(D'\) are two points formed by unfolding from the same point),
\[
\begin{array}{l}
CD // C'D' \\
\Leftrightarrow \angle P = \angle Q. \\
\text{And } \angle P \\
= \angle BCD - \angle ABC, \\
\angle Q... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,733 |
Example 6 In all tetrahedra with six edge lengths of $2, 3, 3, 4, 5, 5$, what is the maximum volume? Prove your conclusion.
The above text is translated into English, preserving the original text's line breaks and format. | Solution: Since the difference between the two sides of a triangle is less than the third side, for any side triangle of a tetrahedron with the given edge lengths, if it contains an edge of length 2, then the lengths of the other two sides can only be the following four possibilities:
(1) 3,3; (2) 5,5; (3) 4,5; (4) 3,4... | \frac{8 \sqrt{2}}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 716,734 |
Example 7 Consider $\triangle A B C$ and $\triangle P Q R$ as shown in Figure 13. In $\triangle A B C$,
$$
\angle A D B=\angle B D C=\angle C D A=120^{\circ} \text {. }
$$
Prove: $x=u+v+w$. | Proof: First, as shown in Figure 14(a), construct a parallelogram on each side of $\triangle ABC$ with the sides of the triangle as diagonals, and let these three parallelograms share a common vertex $D$.
Second, cut along $AD$, $BD$, and $CD$ to separate the three parallelograms. The angles at the vertices of these p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,735 |
Example 1 For all positive real numbers $a, b, c, d$, prove:
$$
\begin{array}{l}
\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+ \\
\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3} .
\end{array}
$$ | Prove: Perform the linear transformation
$$
\left\{\begin{array}{l}
x=b+2 c+3 d, \\
y=c+2 d+3 a, \\
z=d+2 a+3 b, \\
w=a+2 b+3 c .
\end{array}\right.
$$
Consider \(a, b, c, d\) as variables, and solve the system of equations to get
$$
\left\{\begin{array}{l}
a=-\frac{5}{24} x+\frac{7}{24} y+\frac{1}{24} z+\frac{1}{24} ... | \frac{2}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,736 |
Example 2 If $x, y, z > 1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove:
$$
\sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} .
$$ | Proof: Let $\alpha, \beta, \gamma$ all be acute angles, then $\frac{1}{\cos ^{2} \alpha}>1, \frac{1}{\cos ^{2} \beta}>1, \frac{1}{\cos ^{2} \gamma}>1$.
Make the trigonometric substitution
$$
x=\frac{1}{\cos ^{2} \alpha}, y=\frac{1}{\cos ^{2} \beta}, z=\frac{1}{\cos ^{2} \gamma}.
$$
Then the condition simplifies to
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,737 |
Example 3 Let $a, b, c$ be positive numbers, and $abc=1$. Try to prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }
$$ | To prove Example 3, we first provide a generalization of Example 3.
Let \( m \in \mathbf{N}, m \geqslant 2, a, b, c \in \mathbf{R}_{+} \) and \( abc = 1 \). Try to prove:
\[
\frac{1}{a^{m}(b+c)} + \frac{1}{b^{m}(c+a)} + \frac{1}{c^{m}(a+b)} \geqslant \frac{3}{2}.
\]
Analysis: Polya once said, solving a problem is to t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,738 |
Question 1 Proof: The inequality
$$
\begin{array}{l}
\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+ \\
\frac{c^{2}}{(c+a)(c+b)} \geqslant \frac{3}{4}
\end{array}
$$
holds for all positive real numbers $a, b, c$. | Proof: From equation (1) we know
$$
\begin{array}{l}
\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+ \\
\frac{c^{2}}{(c+a)(c+b)} \\
\geqslant \frac{(a+b+c)^{2}}{(a+b)(a+c)+(b+c)(b+a)+(c+a)(c+b)} \\
=\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)} \\
=\frac{(a+b+c)^{2}}{(a+b+c)^{2}+(a b+b c+c a)} \\
\geqslant \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,739 |
For positive real numbers $a, b, c$ satisfying $a+b+c=1$, prove:
$$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) .
$$
No need to prove the equality case. | Proof: The inequality to be proved is equivalent to
$$
\begin{array}{l}
\frac{2 a+b+c}{b+c}+\frac{2 b+c+a}{c+a}+\frac{2 c+a+b}{a+b} \\
\leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) \\
\Leftrightarrow 3+2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \\
\leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,740 |
Question 3 If $0<x_{1}, x_{2}, \cdots, x_{n} \leqslant 1, n \geqslant 1$, prove:
$$
\begin{array}{l}
\frac{x_{1}}{1+(n-1) x_{1}}+\frac{x_{2}}{1+(n-1) x_{2}}+\cdots+ \\
\frac{x_{n}}{1+(n-1) x_{n}} \leqslant 1 .
\end{array}
$$ | Proof: Since $\frac{x_{i}}{1+(n-1) x_{i}}$
$$
\begin{array}{l}
=\frac{1}{n-1} \cdot \frac{(n-1) x_{i}+1-1}{1+(n-1) x_{i}} \\
=\frac{1}{n-1}-\frac{1}{n-1} \cdot \frac{1}{1+(n-1) x_{i}},
\end{array}
$$
Thus, equation (3)
$$
\begin{array}{l}
\Leftrightarrow \sum_{i=1}^{n} \frac{x_{i}}{1+(n-1) x_{i}} \\
=\frac{n}{n-1}-\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,741 |
Example 5 Given $x y z=1, x+y+z=2, x^{2}+$ $y^{2}+z^{2}=16$. Then $\frac{1}{x y+2 z}+\frac{1}{y z+2 x}+\frac{1}{z x+2 y}=$ | Solution: Squaring both sides of $x+y+z=2$ yields
$$
x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=4 \text {. }
$$
Substituting $x^{2}+y^{2}+z^{2}=16$ gives
$$
x y+y z+z x=-6 \text {. }
$$
From $x+y+z=2$, we get $z=2-x-y$. Therefore,
$$
\begin{array}{l}
\frac{1}{x y+2 z}=\frac{1}{x y-2 x-2 y+4} \\
=\frac{1}{(x-2)(y-2)} .
\end{arr... | -\frac{4}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,742 |
Example 1 Let the non-zero sequence $\left\{a_{n}\right\}$ satisfy $a_{1} 、 a_{2}$, $\frac{a_{1}^{2}+a_{2}^{2}+b}{a_{1} a_{2}}$ are all integers, and $a_{n+2}=\frac{a_{n+1}^{2}+b}{a_{n}}$, where $b$ is a given integer. Prove: Every term of the sequence $\left\{a_{n}\right\}$ is an integer. | Proof: From the given, we have
$$
a_{n+2} a_{n}-a_{n+1}^{2}=b,
$$
Therefore, $q=1$, and
$$
\begin{aligned}
p & =-\frac{a_{3}+q a_{1}}{a_{2}}=-\frac{\frac{a_{2}^{2}+b}{a_{1}}+a_{1}}{a_{2}} \\
& =-\frac{a_{1}^{2}+a_{2}^{2}+b}{a_{1} a_{2}}
\end{aligned}
$$
is an integer.
By the proposition $(2) \Leftrightarrow(1)$, we k... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,743 |
Example 2 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=4, a_{1}=$ 22, and $a_{n}-6 a_{n-1}+a_{n-2}=0(n \geqslant 2)$. Prove: There exist two sequences of positive integers $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ such that
$$
a_{n}=\frac{y_{n}^{2}+7}{x_{n}-y_{n}} .
$$ | Proof: Since $a_{0}=4, a_{1}=22$, and
$$
a_{n}-6 a_{n-1}+a_{n-2}=0 \text {, }
$$
thus, $p=-6, q=1$, and $a_{2}=6 a_{1}-a_{0}=128$.
By the proposition $(1) \Leftrightarrow(3)$, we get
$$
a_{n}^{2}-6 a_{n} a_{n-1}+a_{n-1}^{2}=22^{2}-4 \times 128=-28 \text {. }
$$
Then, $a_{n}=\frac{a_{n}^{2}-2 a_{n} a_{n-1}+a_{n-1}^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,744 |
Example 3 Given the sequence $\left\{c_{n}\right\}$ satisfies
$$
\begin{array}{l}
c_{0}=1, c_{1}=0, c_{2}=2005, \\
c_{n+2}=-3 c_{n}-4 c_{n-1}+2008(n=1,2, \cdots) . \\
\text { Let } a_{n}=5\left(c_{n+2}-c_{n}\right)\left(502-c_{n-1}-c_{n-2}\right)+
\end{array}
$$
$4^{n} \times 2004 \times 501(n=1,2, \cdots)$. Is $a_{n}$... | Solve: Convert the linear non-homogeneous recurrence relation
$$
c_{n+2}=-3 c_{n}-4 c_{n-1}+2008
$$
into a linear homogeneous recurrence relation
$$
c_{n+2}-251=-3\left(c_{n}-251\right)-4\left(c_{n-1}-251\right) \text {, }
$$
where 251 is the root of the equation $x=-3 x-4 x+2008$. Let $d_{n}=c_{n}-251$. Then,
$$
\be... | a_{n}=501^{2} \times 2^{2}\left(t_{n}+2 t_{n-1}\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,745 |
Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0$,
$$
a_{n+1}=k a_{n}+\sqrt{\left(k^{2}-1\right) a_{n}^{2}+1}, n=0,1, \cdots \text {, }
$$
where $k$ is a given positive integer. Prove: every term of the sequence $\left\{a_{n}\right\}$ is an integer, and $2 k \mid a_{2 n}, n=0,1, \cdots$. | Proof: From the given, we have
$$
\left(a_{n+1}-k a_{n}\right)^{2}=\left(\sqrt{\left(k^{2}-1\right) a_{n}^{2}+1}\right)^{2} \text {, }
$$
which means $a_{n+1}^{2}-2 k a_{n+1} a_{n}+a_{n}^{2}=1$.
Thus, $p=-2 k, q=1$.
From the proposition $(3) \Leftrightarrow(1)$, we get
$$
a_{n+2}-2 k a_{n+1}+a_{n}=0 \text {. }
$$
Com... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,746 |
1. Given
(1) $a>0$;
(2) When $-1 \leqslant x \leqslant 1$, it satisfies
$$
\left|a x^{2}+b x+c\right| \leqslant 1 \text {; }
$$
(3) When $-1 \leqslant x \leqslant 1$, $a x+\dot{b}$ has a maximum value of 2. Find the constants $a, b, c$. | 1. From (1), we know that $y=a x^{2}+b x+c$ is a parabola opening upwards. From (1) and (3), we have
$$
a+b=2 \text{. }
$$
From (2), we have
$$
\begin{array}{l}
|a+b+c| \leqslant 1, \\
|c| \leqslant 1 .
\end{array}
$$
From (1) and (2), we have
$$
|2+c| \leqslant 1 \text{. }
$$
From (3) and (4), we get $c=-1$. Theref... | f(x)=2 x^{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,747 |
2. In $\triangle A B C$, it is known that $I$ is the incenter, $O$ is the circumcenter, $A B=5, B C=6, C A=4$. Prove: $O I \perp C I$. | 2. As shown in Figure 1, extend
$C I$ to intersect $A B$ at $F$, and connect $A I$, then
$$
\begin{array}{l}
\frac{B F}{A F}=\frac{6}{4}=\frac{3}{2}, \\
B F+A F=5 .
\end{array}
$$
From equations (1) and (2), we know
$$
A F=2 \text {. }
$$
Let $M$ be the midpoint of side $A C$, then $A M=2$.
Therefore, $A M=A F$.
Thu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,748 |
3. In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the... | 3. If a $9 \times 9$ grid is divided into 9 $3 \times 3$ grids, and each small cell in the same $3 \times 3$ grid is filled with the same number, and the numbers in any two different $3 \times 3$ grids are different, then each row and each column will have exactly three different numbers. Therefore, the maximum value o... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,749 |
1. Given $\frac{(2 x+z)^{2}}{(x+y)(-2 y+z)}=8$. Then $2 x+$ $4 y-z+6=$ $\qquad$ | 1. Hint: $(2 x+4 y-z)^{2}=0, 2 x+4 y-z+6=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,750 |
2. If $2 x^{2}+7 x y-15 y^{2}+a x+b y+3$ can be factored into the product of two linear polynomials with integer coefficients, where $a$ and $b$ are real numbers, then the minimum value of $a+b$ is $\qquad$ | $$
\begin{array}{c}
\text { If the original expression is }=(x+5 y)(2 x-3 y)+a x+b y+3, \\
\text { then }(a, b)=(-5,-12),(5,12),(-7,4),(7,-4). \\
\text { Therefore, }(a+b)_{\text {min }}=-17.
\end{array}
$$ | -17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,751 |
3. Given that $n$ is a positive integer, $1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}$ is the square of a rational expression $A$. Then, $A=$ $\qquad$ | 3. Hint: Original expression $=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^{2}$. Then $A= \pm \frac{n^{2}+n+1}{n^{2}+n}$. | \pm \frac{n^{2}+n+1}{n^{2}+n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,752 |
Example 6 Let $a, b, c$ be distinct real numbers. Prove that:
$$
\begin{array}{l}
\frac{a^{4}}{(a-b)(a-c)}+\frac{b^{4}}{(b-c)(b-a)}+ \\
\frac{c^{4}}{(c-a)(c-b)}>0 .
\end{array}
$$ | Solution: Let the left side of the inequality be $f$, and after finding a common denominator, we get
$$
f=\frac{-(b-c) a^{4}-(c-a) b^{4}-(a-b) c^{4}}{(a-b)(b-c)(c-a)} \text {. }
$$
Since the numerator is a 5th-degree homogeneous cyclic symmetric polynomial, and the denominator is a 3rd-degree homogeneous cyclic symmet... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,753 |
4. A computer user plans to purchase single-piece software and boxed disks, priced at 60 yuan and 70 yuan each, respectively, with a budget of no more than 500 yuan. According to the needs, the user must buy at least 3 pieces of software and at least 2 boxes of disks. How many different purchasing options are there? | 4. First buy 3 pieces of software and 2 boxes of disks. With the remaining 180 yuan, if you do not buy software, you can buy 0, 1, or 2 more boxes of disks; if you buy 1 more piece of software, you can buy 0 or 1 more box of disks; if you buy 2 more pieces of software, you cannot buy any more disks; if you buy 3 more p... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,754 |
5. Given the equation $6 x^{2}+2(m-13) x+12-m$ $=0$ has exactly one positive integer solution. Then the value of the integer $m$ is | 5. Since $\Delta=4(m-13)^{2}-24(12-m)$ is a perfect square, there exists a non-negative integer $y$ such that
$$
(m-13)^{2}-6(12-m)=y^{2} \text {, }
$$
i.e., $(m-10-y)(m-10+y)=3$.
Since $m-10-y \leqslant m-10+y$, we have
$$
\left\{\begin{array} { l }
{ m - 1 0 - y = 1 , } \\
{ m - 1 0 + y = 3 }
\end{array} \text { or... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,755 |
6. In a square $ABCD$ with side length 1, points $M$, $N$, $O$, $P$ are on sides $AB$, $BC$, $CD$, $DA$ respectively. If $AM=BM$, $DP=3AP$, then the minimum value of $MN+NO+OP$ is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation resul... | 6. As shown in Figure 2, we have
$$
\begin{array}{l}
M N+N O+O P \\
=M N+N O_{1}+O_{1} P_{2} \\
\geqslant M P_{2},
\end{array}
$$
and the equality holds when points $N, O_{1}$ are both on the line segment $M P_{2}$. Then
$$
\begin{array}{l}
M P_{2}=\sqrt{M A_{1}^{2}+A_{1} P_{2}^{2}} \\
=\sqrt{\left(\frac{3}{2}\right)^... | \frac{\sqrt{85}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,756 |
7. Given $O$ is the circumcenter of $\triangle A B C$, $A D$ is the altitude on $B C$, $\angle C A B=66^{\circ}, \angle A B C=44^{\circ}$. Then, $\angle O A D=$ $\qquad$ . | 7. As shown in Figure 3, we have
$$
\begin{array}{l}
\angle O A D=90^{\circ}-\angle A E F \\
=90^{\circ}-(\angle A B C+\angle C B F)=90^{\circ}-\angle C A D-\angle A B C \\
=\angle A C B-\angle A B C=180^{\circ}-2 \angle A B C-\angle C A B=26^{\circ}
\end{array}
$$ | 26^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,757 |
8. Algebraic expression
$$
\begin{array}{l}
\sqrt{9 x^{2}+4}+\sqrt{9 x^{2}-12 x y+4 y^{2}+1}+ \\
\sqrt{4 y^{2}-16 y+20}
\end{array}
$$
When it reaches the minimum value, the values of $x$ and $y$ are respectively $\qquad$ | 8. As shown in Figure 4, we have
$$
\begin{array}{c}
\text { Original expression }=\sqrt{[0-(-2)]^{2}+(3 x-0)^{2}}+ \\
\sqrt{(1-0)^{2}+(2 y-3 x)^{2}}+ \\
\sqrt{(3-1)^{2}+(4-2 y)^{2}} \\
=A B+B C+C D \geqslant A D,
\end{array}
$$
where \( A(-2,0) , B(0,3 x) , C(1,2 y) , D(3,4) \), and when points \( B , C \) are on the... | x=\frac{8}{15}, y=\frac{6}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,758 |
10. The cost of a house consists of the cost of the above-ground part and the cost of the foundation part. For a house with an area of $N \mathrm{~m}^{2}$, the cost of the above-ground part is proportional to $N \sqrt{N}$, and the cost of the foundation part is proportional to $\sqrt{N}$. It is known that for a house w... | 10. Let the area of each house in square meters be $y$, and a total of $x$ identical houses are built, with the total cost being $S$. Then
$$
\left\{\begin{array}{l}
x y=80000, \\
S=(\alpha y \sqrt{y}+\beta \sqrt{y}) \cdot x, \\
\frac{\alpha \cdot 3600 \sqrt{3600}}{\beta \sqrt{3600}}=\frac{72}{100},
\end{array}\right.
... | 5000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,760 |
1. It is known that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
（Zhu Huawei provided the problem） | 1. Solution 1: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1$, $\alpha \beta=1$, the sum of the coefficients we are looking for is $S_{k}=\alpha^{k}+\beta^{k}$. From
$$
\begin{array}{l}
(\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
=\left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,761 |
2. As shown in Figure 1, from a point $P$ outside a circle, draw two tangents $P A$ and $P B$, with $A$ and $B$ being the points of tangency. Draw a secant line through point $P$ that intersects the circle at points $C$ and $D$. Draw a line through the point of tangency $B$ parallel to $P A$, intersecting lines $A C$ a... | 2. As shown in Figure 3, connect $B C$, $B A$, and $B D$, then
$$
\begin{array}{l}
\angle A B C=\angle P A C \\
=\angle E .
\end{array}
$$
Therefore, $\triangle A B C \backsim \triangle A E B$.
Thus, $\frac{B C}{B E}=\frac{A C}{A B}$, which means
$$
B E=\frac{A B \cdot B C}{A C} .
$$
Also, $\angle A B F=\angle P A B=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,762 |
3. Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.
(Tang Lihua) | 3. First, we have $n \geqslant 16$.
In fact, take the set $A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\}$, then $A_{0} \subseteq S, \left|A_{0}\right|=15, A_{0}$ contains any two numbers that are coprime, but there are no primes in it, which shows that $n \geqslant 16$.
Next, we prove: For any $A... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,763 |
Example 7 Let $a, b$ be the roots of the equation $x^{2}-3 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-4 x+2=0$. Given that $\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+$ $\frac{d}{a+b+c}=B$. Prove:
$$
\begin{array}{l}
\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+ \\
\frac{d^{2}}{a+b+c}=7 ... | Proof: By Vieta's formulas, we have
$$
a+b=3, a b=1 ; c+d=4, c d=2 \text {. }
$$
Then $a+b+c+d=3+4=7$.
Since $a^{2}+b^{2}=(a+b)^{2}-2 a b=7$,
$$
c^{2}+d^{2}=(c+d)^{2}-2 c d=12 \text {, }
$$
Therefore, $a^{2}+b^{2}+c^{2}+d^{2}=19$.
Thus, $\frac{a^{2}}{b+c+d}=\frac{a^{2}+7 a-7 a}{b+c+d}$
$$
=\frac{7 a-a(7-a)}{b+c+d}=\f... | 7 B-7 | Algebra | proof | Yes | Yes | cn_contest | false | 716,764 |
4. Given real numbers $x_{1}, x_{2}, \cdots, x_{n}(n>2)$ satisfy
$$
\left|\sum_{i=1}^{n} x_{i}\right|>1,\left|x_{i}\right| \leqslant 1(i=1,2, \cdots, n) \text {. }
$$
Prove: There exists a positive integer $k$, such that
$$
\left| \sum_{i=1}^{k} x_{i}-\sum_{i=k+1}^{n} x_{i} \right| \leqslant 1.
$$ (Leng Gangsong, prob... | 4. Let $g(0)=-\sum_{i=1}^{n} x_{i}$,
$$
\begin{array}{l}
g(k)=\sum_{i=1}^{k} x_{i}-\sum_{i=k+1}^{n} x_{i}(1 \leqslant k \leqslant n-1), \\
g(n)=\sum_{i=1}^{n} x_{i},
\end{array}
$$
Then $|g(1)-g(0)|=2\left|x_{1}\right| \leqslant 2$,
$$
\begin{array}{l}
|g(k+1)-g(k)| \\
=2\left|x_{k+1}\right| \leqslant 2, k=1,2, \cdots... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,765 |
5. As shown in Figure 2,
$\odot O_{1}, \odot O_{2}$ intersect at
$A$ and $B$. A line $DC$ through point $O_{1}$ intersects
$\odot O_{1}$ at point $D$ and
is tangent to $\odot O_{2}$ at point $C$,
$CA$ is tangent to $\odot O_{1}$ at point $A$, and the chord $AE$ of $\odot O_{1}$ is perpendicular to line $DC$. A line $AF... | 5. As shown in Figure 4, let $A E$ intersect $D C$ at point $H$, $A F$ intersect $B D$ at point $G$, and connect $A B$, $B C$, $B H$, $B E$, $C E$, $G H$.
By symmetry, side $C E$ is also a tangent to $\odot O_{1}$, and $H$ is the midpoint of $A E$.
Since $\angle H C B = \angle B A C$, and $\angle B A C = \angle B E H$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,766 |
6. In an isosceles right triangle \( \triangle ABC \), \( CA = CB = 1 \), and \( P \) is any point on the boundary of \( \triangle ABC \). Find the maximum value of \( PA \cdot PB \cdot PC \).
(Li Weiguo) | 6. (1) As shown in Figure 5, when $P \in A C$, we have
$$
P A \cdot P C \leqslant \frac{1}{4}, \quad P B \leqslant \sqrt{2}.
$$
Therefore, $P A \cdot P B \cdot P C \leqslant \frac{\sqrt{2}}{4}$,
where the equality does not hold (since the two equalities cannot hold simultaneously), i.e., $P A \cdot P B \cdot P C < \fr... | \frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,767 |
7. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$
10\left(a^{3}+b^{3}+c^{3}\right)-9\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 .
$$ | 7. Since $\sum a^{3}=1-3 \prod(a+b)$,
$$
\sum a^{5}=1-5 \prod(a+b)\left[\sum a^{2}+\sum a b\right],
$$
thus the original inequality
$$
\begin{aligned}
\Leftrightarrow & 10\left[1-3 \prod(a+b)\right]-9\left[1-5 \prod(a+b)\right. \\
& \left.\left(\sum a^{2}+\sum a b\right)\right] \geqslant 1 \\
\Leftrightarrow & 45 \pro... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,768 |
8. Let $n$ students be such that among any 3 of them, 2 know each other, and among any 4 of them, 2 do not know each other. Find the maximum value of $n$.
(Tang Lihua | 8. The maximum value of $n$ is 8.
When $n=8$, the example shown in Figure 7 satisfies the requirements, where $A_{1}, A_{2}, \cdots, A_{8}$ represent 8 students, and the line between $A_{i}$ and $A_{j}$ indicates that $A_{i}$ and $A_{j}$ know each other.
Suppose $n$ students meet the requirements of the problem, we w... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,769 |
1. Given non-constant sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy: $a_{n} 、 b_{n} 、 a_{n+1}$ form an arithmetic sequence, $b_{n} 、 a_{n+1} 、 b_{n+1}$ form a geometric sequence. Let $c_{n}=\sqrt{b_{n}}$. Then, among the following statements about the sequence $\left\{c_{n}\right\}$, the correct o... | $-1 . A$.
From the given, we know that $2 b_{n}=a_{n+1}+a_{n}, a_{n+1}^{2}=b_{n} b_{n+1}$, so $a_{n+1}=\sqrt{b_{n} b_{n+1}}$.
Therefore, $2 b_{n}=\sqrt{b_{n-1} b_{n}}+\sqrt{b_{n} b_{n+1}}$, which means
$$
2 \sqrt{b_{n}}=\sqrt{b_{n-1}}+\sqrt{b_{n+1}} \text {. }
$$
Hence $\left.\mid c_{n}\right\}$ is an arithmetic seque... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,770 |
2. In $\triangle A B C$, $a$, $b$, $c$ are the lengths of the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively. If $\cos A + \sin A - \frac{2}{\cos B + \sin B} = 0$, then the value of $\frac{a+b}{c}$ is ( ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2 | 2.B.
From $\cos A+\sin A-\frac{2}{\cos B+\sin B}=0$, we get
$$
\sqrt{2} \sin \left(A+\frac{\pi}{4}\right)-\frac{2}{\sqrt{2} \sin \left(B+\frac{\pi}{4}\right)}=0,
$$
which means $\sin \left(A+\frac{\pi}{4}\right) \cdot \sin \left(B+\frac{\pi}{4}\right)=1$.
By the boundedness of the sine function and the fact that $\an... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,771 |
3. The minimum value of the function $f(x)=9^{x}+9^{-x}-2\left(3^{x}+3^{-x}\right)$ is ( ).
(A) 1
(B) 2
(C) -3
(D) -2 | 3. D.
$$
\begin{array}{l}
f(x)=9^{x}+9^{-x}-2\left(3^{x}+3^{-x}\right) \\
=\left(3^{x}+3^{-x}\right)^{2}-2\left(3^{x}+3^{-x}\right)-2 . \\
\text { Let } t=3^{x}+3^{-x} \geqslant 2, \text { then } \\
y=t^{2}-2 t-2=(t-1)^{2}-3 .
\end{array}
$$
Therefore, the minimum value is -2. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,772 |
4. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the left focus is $F_{1}$, the vertices are $A_{1}$ and $A_{2}$, and $P$ is any point on the right branch of the hyperbola. Then the two circles with diameters $P F_{1}$ and $A_{1} A_{2}$ must ( ).
(A) intersect
(B) be internally tangent
(C) be externall... | 4.B.
Let the other focus of the hyperbola be $F_{2}$, and the midpoint of segment $P F_{1}$ be $C$. In $\triangle F_{1} F_{2} P$, $C$ is the midpoint of $P F_{1}$, and $O$ is the midpoint of $F_{1} F_{2}$. Therefore, we have
$$
O C=\frac{1}{2}\left|P F_{2}\right|=\frac{1}{2}\left(\left|P F_{1}\right|-\left|A_{1} A_{2}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,773 |
5. Let $A=\{1,2, \cdots, 10\}$. If the equation $x^{2}-b x- c=0$ satisfies $b, c \in A$, and the equation has at least one root $a \in A$, then the equation is called a "beautiful equation". The number of beautiful equations is ( ).
(A) 8
(B) 10
(C) 12
(D) 14 | 5.C.
From the problem, we know that the two roots of the equation are both integers and one is positive while the other is negative. When one root is -1, there are 9 beautiful equations that meet the requirements; when one root is -2, there are 3 beautiful equations that meet the requirements. Therefore, there are a t... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,774 |
Example 8 Find the real solutions of the system of equations $\left\{\begin{array}{l}x^{3}+x^{3} y^{3}+y^{3}=17, \\ x+x y+y=5\end{array}\right.$. | Let $x+y=u, xy=v$, then the original system of equations can be transformed into
$$
\left\{\begin{array}{l}
u^{3}+v^{3}-3uv=17, \\
u+v=5 .
\end{array}\right.
$$
(2) $)^{3}$ - (1) gives
$$
uv=6 \text{.}
$$
From equations (2) and (3), we get
$$
\left\{\begin{array}{l}
u=2, \\
v=3
\end{array} \text{ or } \left\{\begin{ar... | x_{1}=1, y_{1}=2 ; x_{2}=2, y_{2}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,775 |
6. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be any permutation of $1,2,3,4$, and $f$ be a one-to-one mapping from $\{1,2,3,4\}$ to $\{1,2,3,4\}$ such that $f(i) \neq i$. Consider the matrix
$$
A=\left[\begin{array}{cccc}
a_{1} & a_{2} & a_{3} & a_{4} \\
f\left(a_{1}\right) & f\left(a_{2}\right) & f\left(a_{3}\right) & f\left(a... | 6. C.
For a permutation of $a_{1}, a_{2}, a_{3}, a_{4}$, there can be 9 mappings satisfying $f(i) \neq i$. Since $a_{1}, a_{2}, a_{3}, a_{4}$ have a total of $A_{4}^{4}=24$ permutations, therefore, the number of tables satisfying the condition is $24 \times 9=216$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,776 |
10. Let $S=x^{2}+y^{2}-2(x+y)$, where 1, $y$ satisfy $\log _{2} x+\log _{2} y=1$. Then the minimum value of $S$ is $\qquad$ . | $$
10.4-4 \sqrt{2}
$$
From $\log _{2} x+\log _{2} y=1$, we get $x y=2$.
$$
\begin{array}{l}
\text { Also } S=x^{2}+y^{2}-2(x+y) \\
=(x+y)^{2}-2(x+y)-2 x y \\
=(x+y)^{2}-2(x+y)-4 \\
=[(x+y)-1]^{2}-5 \geqslant[2 \sqrt{x y}-1]^{2}-5 \\
=(2 \sqrt{2}-1)^{2}-5=4-4 \sqrt{2} .
\end{array}
$$ | 4-4 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,780 |
11. Let $\triangle A B C$ be inscribed in a circle $\odot O$ with radius $R$, and $A B=A C, A D$ be the altitude from $A$ to the base $B C$. Then the maximum value of $A D + B C$ is $\qquad$ . | 11. $R+\sqrt{5} R$.
If finally 3, let $\angle O B D=\alpha$,
then
$$
\begin{array}{l}
A D=R+R \sin \alpha, \\
\frac{1}{2} B C=B D=R \cos \alpha, \\
B C=2 R \cos \alpha . \\
\text { Therefore, } A D+B C \\
=R+R \sin \alpha+2 R \cos \alpha \\
=R+\sqrt{5} R \sin (\alpha+\varphi),
\end{array}
$$
where $\tan \varphi=2$.
T... | R+\sqrt{5} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,781 |
12. Let $r, s, t$ be integers, and the set $\left\{a \mid a=2^{r}+\right.$ $2^{s}+2^{t}, 0 \leqslant t<s<r \mid$ consists of numbers arranged in ascending order to form the sequence $\left\{a_{n}\right\}: 7,11,13,14, \cdots$. Then $a_{34}=$ $\qquad$ | 12.131.
Since $r, s, 1$ are integers $\mathrm{HL}, 0 \leqslant 1<s<r$, then $r$ takes the smallest value 2. At this time, the numbers that meet the condition are $C_{2}^{2}=1$.
When $r=3$, $s, 1$ can be chosen from $0,1,2$, the numbers that meet the condition are $C_{i}^{3}=3$.
Similarly, when $r=4$, the numbers that... | 131 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,782 |
13. As shown in Figure 2, $B D$ and $C E$ are two altitudes of $\triangle A B C$, $F$ and $G$ are the midpoints of $D E$ and $B C$ respectively, and $O$ is the circumcenter of $\triangle A B C$. Prove:
$$
A O \parallel F G \text {. }
$$ | Three, 13. Given 4, and
$$
\begin{array}{l}
\angle B D C=\angle B E C \\
=90^{\circ}, \\
B G=G C,
\end{array}
$$
then $D G=\frac{1}{2} B C$
$$
=E G \text {. }
$$
Also, $D F=F E$, so $G F \perp D E$.
Extend $O A$ to intersect $D E$ at point $H$, and connect $O B$.
Since $\angle B D C=\angle B E C=90^{\circ}$, points $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,783 |
14. Given a square $A B C D$ with two vertices $A$ and $B$ on the parabola $y=x^{2}$, and points $C$ and $D$ on the line $y=x-4$. Find the side length $d$ of the square. | 14. Let $A\left(t_{1}, t_{1}^{2}\right) 、 B\left(t_{2}, t_{2}^{2}\right)$, and $t_{1} \neq t_{2}$.
If $A B / / D C$, then $1=\frac{t_{2}^{2}-t_{1}^{2}}{t_{2}-t_{1}}$, which means $t_{1}+t_{2}=1$.
- Therefore,
$$
\begin{array}{l}
d^{2}=|A B|^{2}=\left(t_{1}-t_{2}\right)^{2}+\left(t_{1}^{2}-t_{2}^{2}\right)^{2} \\
=\lef... | 3 \sqrt{2} \text{ or } 5 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,784 |
15. Let real numbers $a, b$ satisfy $a=x_{1}+x_{2}+x_{3}=$ $x_{1} x_{2} x_{3}, a b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$, where $x_{1}, x_{2}, x_{3}>0$. Find the maximum value of $P=\frac{a^{2}+6 b+1}{a^{2}+a}$. | So, $a \geqslant 3 \sqrt{3}$.
Also, $3\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right) \leqslant\left(x_{1}+x_{2}+x_{3}\right)^{2}$, then
$3 a b \leqslant a^{2}$, which means $3 b \leqslant a$.
Thus, $P=\frac{a^{2}+6 b+1}{a^{2}+a} \leqslant \frac{a^{2}+2 a+1}{a^{2}+a}$
$$
=1+\frac{1}{a} \leqslant 1+\frac{1}{3 \sqrt{3}}... | \frac{9+\sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,785 |
1. Given $x+y=3, x^{2}+y^{2}-x y=4$. Then $x^{4}+y^{4}+$ $x^{3} y+x y^{3}$ is $\qquad$ | (Hint: Let $x+y=u, x y=v$. Answer: 36 .) | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,786 |
16. In a class, 20 students each took a Chinese and a Math test, scored on a 10-point scale, meaning the scores are integers from 0 to 10. The test results are:
(1) No one scored 0;
(2) No two students have the same scores in both Chinese and Math.
We say “student A performs better than student B” if “student A’s scor... | 16. If student $A$ performs better than $B$, it is denoted as $A>B$.
The original problem is equivalent to proving: there exist three students $A$, $B$, and $C$ such that $A>B>C$.
Let $\left(a_{i}, b_{i}\right)$ represent the Chinese and Math scores of the $i$-th student $(i=1,2, \cdots, 20)$, then,
$\left(a_{i}, b_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,787 |
1. In the parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$, $M$ is the intersection of $A_{1} C_{1}$ and $B_{1} D_{1}$. If $\boldsymbol{A B}=\boldsymbol{a}, \boldsymbol{A D}=\boldsymbol{b}, \boldsymbol{A} \boldsymbol{A}_{1}=\boldsymbol{c}$, then among the following vectors, the vector equal to $\boldsymbol{B} \boldsymb... | $$
-\mathbf{1 . A} .
$$
It is easy to calculate $B M=-\frac{1}{2} a+\frac{1}{2} b+c$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,788 |
3. Given that point $P$ moves on the line $y=x+2$, points $A(2,2)$ and $B(6,6)$ satisfy that $\angle APB$ reaches its maximum value. Then the coordinates of point $P$ are ( ).
(A) $(0,2)$
(B) $(1,3)$
(C) $(2,4)$
(D) $(3,5)$ | 3.D.
From the image, we can see that the point of tangency $P$ between the circle passing through $A$ and $B$ and the line is the point $P$, so the coordinates of point $P$ are $(3,5)$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,790 |
4. Given $1+2 \times 3+3 \times 3^{2}+\cdots+n \times 3^{n-1}$ $=3^{n}(n a-b)+c$, for all $n \in \mathbf{N}_{+}$. Then, the values of $a, b, c$ are (.
(A) $a=0, b=c=\frac{1}{4}$
(B) $a=b=c=\frac{1}{4}$
(C) $a=\frac{1}{2}, b=c=\frac{1}{4}$
(D) There do not exist such $a, b, c$ | 4.C.
It is known that $S_{n}=3 S_{n-1}+\frac{3^{n}-1}{2}$.
Also, $S_{n}=S_{n-1}+n \cdot 3^{n-1}$, so $S_{n}=3^{n}\left(\frac{n}{2}-\frac{1}{4}\right)+\frac{1}{4}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,791 |
5. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and $A(0,-1)$, $B(3,1)$ are two points on its graph. Then, the complement of the solution set of $|f(x+1)|<1$ is ( ).
(A) $(3,+\infty)$
(B) $[2,+\infty$ )
(C) $(-1,2)$
(D) $(-\infty,-1] \cup[2,+\infty)$ | 5.D.
Using $0<x+1<3$ we get the solution set of the original inequality as $(-1,2)$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,792 |
6. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=3$, and $a_{n+1}=a_{n}^{2}-$ $(3 n-1) a_{n}+3$. Then the sum of the first 11 terms of the sequence $\left\{a_{n}\right\}$, $S_{11}=(\quad)$.
(A) 198
(B) 55
(C) 204
(D) 188 | 6. A.
Substituting $a_{1}=3$ yields $a_{2}=6$, substituting $a_{n}=3 n$ yields $a_{n+1}=3(n+1)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,793 |
1. Given that $O$ is a point inside $\triangle A B C$, and satisfies
$$
O A \cdot O B=O B \cdot O C=O C \cdot O A \text {. }
$$
Then point $O$ is the $\triangle A B C$'s $\qquad$ . | ニ、1. Orthocenter.
From $O A \cdot O B=O B \cdot O C$ we can get $O B \cdot A C=0$, so, point $O$ lies on the altitude of side $A C$. Similarly, we can prove that point $O$ also lies on the altitudes of the other sides. Therefore, point $O$ is the orthocenter of $\triangle A B C$. | Orthocenter | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,794 |
2. If the focal distance of the conic section $\frac{x^{2}}{k-2}+\frac{y^{2}}{k+5}=1$ is independent of the real number $k$, then its foci coordinates are $\qquad$ | 2. $(0, \pm \sqrt{7})$.
Since $(k+5)-(k-2)=7$ is a constant, it represents an ellipse with foci on the $y$-axis, and $c=\sqrt{7}$.
Therefore, the coordinates of its foci should be $(0, \pm \sqrt{7})$. | (0, \pm \sqrt{7}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,795 |
3. Let points $A(1,1,0)、B(1,0,1)、C(0,1,1)$. Then the shape of $\triangle A B C$ is $\qquad$ | 3. Equilateral triangle.
It is easy to prove that $|A B|=|B C|=|C A|$. | Equilateral triangle | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,796 |
2. Factorize:
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)-a^{3}-b^{3}-c^{3}-2 a b c .
$$ | ( Hint: When $a+b=c$, the original expression $=0$. Answer: $(a+b-c)(b+c-a)(c+a-b)$. | (a+b-c)(b+c-a)(c+a-b) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,797 |
4. Let the line $x \cos \theta+y \sin \theta=2, \theta \in$ $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$ have a common point. Then the range of $\theta$ is $\qquad$ | 4. $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$.
Since $2^{2}=(x \cos \theta+y \sin \theta)^{2}$
$$
\begin{array}{l}
=\left(\frac{x}{\sqrt{6}} \times \sqrt{6} \cos \theta+\frac{y}{\sqrt{2}} \times \sqrt{2} \sin \theta\right)^{2} \\
\leqslant\left(\frac{x^{2}}{6}+\frac{y^{2}}{2}\right)\left(6 \cos ^{2} \theta+2 \sin ^{... | \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,798 |
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