problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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1. In an acute triangle $\triangle ABC$, $AB > AC$, $M$ is the midpoint of side $BC$, and $P$ is a point inside $\triangle AMC$ such that $\angle MAB = \angle PAC$. Let the circumcenters of $\triangle ABC$, $\triangle ABP$, and $\triangle ACP$ be $O$, $O_1$, and $O_2$ respectively. Prove that the line $AO$ bisects the ... | Proof 1 As shown in Figure 1, construct the circumcircles of $\triangle ABC$, $\triangle ABP$, and $\triangle ACP$. Extend $AP$ to intersect $\odot O$ at point $D$, and connect $BD$. Draw the tangent to $\odot O$ at point $A$, which intersects $\odot O_{1}$ and $\odot O_{2}$ at points $E$ and $F$, respectively. Connect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,407 |
4. Let $G=G(V ; E)$ be a simple graph, where $V$ is the set of vertices and $E$ is the set of edges, with $|V|=n$. A mapping $f: V \rightarrow \mathbf{Z}$ is called "good" if $f$ satisfies:
(1) $\sum_{v \in V} f(v)=|E|$;
(2) If any number of vertices are colored red, then there always exists a red vertex $v$ such that ... | 4. For a sorting $\tau=\left(v_{1}, v_{2}, \cdots, v_{n}\right)$ of vertices in $V$, define $f_{\mathrm{r}}: V \rightarrow \mathbf{Z}$ as follows: $f_{\mathrm{r}}(v)$ equals the number of vertices adjacent to $v$ that are ranked before $v$.
The following explains: $f_{\mathrm{r}}$ is a good mapping.
In the calculation ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,410 |
5. Given an integer $a_{1} \geqslant 2$, for integers $n \geqslant 2$, define $a_{n}$ as the smallest positive integer that is not coprime with $a_{n-1}$ and is not equal to $a_{1}, a_{2}, \cdots, a_{n-1}$. Prove: Every integer not less than 2 appears in the sequence $\left\{a_{n}\right\}$.
(Red-Hongbing) | 5. Prove the conclusion in three steps.
(1) The sequence $\left\{a_{n}\right\}$ contains infinitely many even numbers.
Assume that the sequence $\left\{a_{n}\right\}$ contains only finitely many even numbers. Then there must exist an integer $c$ such that all even numbers greater than $c$ do not appear. Thus, there mu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,411 |
1. Given that $a$ and $b$ are both unit vectors, and vector $c=a+2 b$ is perpendicular to $d=5 a-4 b$. Then the angle between $a$ and $b$ $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=$ $\qquad$ | $$
-1 \cdot \frac{\pi}{3} \text {. }
$$
Let the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=\theta$. Then, according to the condition for vectors to be perpendicular, we have
$$
\begin{array}{c}
0=c \cdot d=(a+2 b) \cdot(5 a-4 b) \\
=5 a^{2}+10 a \cdot b-4 a \cd... | \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,413 |
Example 6 As shown in Figure 6, in quadrilateral $A B C D$, $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively, $P$ is any point on the extension of diagonal $A C$,
$P F$ intersects $A D$ at point $M$,
$P E$ intersects $B C$ at point $N$,
$E F$ intersects $M N$ at point $K$.
Prove: $K$ is the midpoint ... | Take point $G$ on $F P$ such that $F G=M F$, and connect $G C, G N$. Take the midpoint $L$ of $A C$, and connect $F L, L E$. Then, by the converse of Corollary 1, we have
$$
G C / / F L, C N / / L E \text {. }
$$
At this point, by Corollary 3, we know $F E / / G N$, that is,
$$
F K / / G N \text {. }
$$
Since $F$ is ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,414 |
2. $\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}$ is equal to
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}\text { 2. } \frac{4 \sqrt{3}}{3} \text {. } \\ \frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} \\ =\frac{1}{\cos 70^{\circ}}-\frac{1}{\sqrt{3} \sin 70^{\circ}} \\ =\frac{\sqrt{3} \sin 70^{\circ}-\cos 70^{\circ}}{\sqrt{3} \sin 70^{\circ} \cdot \cos 70^{\circ}} \\ =\frac{\sin 70^{\circ} \... | null | Algebra | proof | Yes | Yes | cn_contest | false | 722,415 |
3. As shown in Figure 1,
A tangent line is drawn from a point $M$ outside the circle $\odot O$, touching $\odot O$ at point $B$. Line $M O$ intersects $\odot O$ at point $A$. Given that $M A=4$, $M B=4 \sqrt{3}$, and $N$ is the midpoint of arc $\overparen{A B}$. Then the area of the curvilinear triangle (shaded part) ... | $3.8-\frac{4 \pi}{3}$.
According to the conditions, extend $M O$ to intersect $\odot O$ at point $C$. Let the radius of $\odot O$ be $r$. Then $M C=4+2 r$.
By the secant-tangent theorem, we have
$$
M B^{2}=M A \cdot M C,
$$
which is $48=4(4+2 r)$.
Solving for $r$ gives $r=4$.
Therefore, $O C=O A=A M=4$.
Connect $O B$.... | 8-\frac{4 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,416 |
4. $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ The value is
The value of $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is | 4. 4 .
Let $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}=x$.
Cubing both sides and simplifying, we get
$$
x^{3}-6 x-40=0 \text {. }
$$
By observation, 4 is a root of the equation. Therefore,
$$
(x-4)\left(x^{2}+4 x+10\right)=0 \text {. }
$$
Since $\Delta=4^{2}-4 \times 10=-24<0$, the equation $x^{2}+4 x+10=0$ h... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,417 |
5. In the Cartesian coordinate system, regardless of the value of $m$, the parabola $y=m x^{2}+(2 m+1) x-(3 m+2)$ does not pass through the points on the line $y=-x+1$. The coordinates of all points that meet this condition are (write down the coordinates of all such points).
Translate the above text into English, ple... | 5. $(1,0),(-3,4),\left(\frac{3}{2},-\frac{1}{2}\right)$.
From $y=m x^{2}+(2 m+1) x-(3 m+2)$
$$
=m(x+3)(x-1)+(x-2)
$$
we can see that the parabola must pass through the points $A(1,-1)$ and $B(-3,-5)$.
Draw lines parallel to the $y$-axis through points $A$ and $B$ to intersect the line $y=-x+1$ at points $C(1,0)$ and ... | (1,0),(-3,4),\left(\frac{3}{2},-\frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,418 |
In a right-angled triangle $\triangle ABC$, the radius of the inscribed circle is $r$, and the length of the angle bisector of the right angle is $t$. Prove that the lengths of the two legs $a$ and $b$ of the right-angled triangle $\triangle ABC$ are the roots of the quadratic equation in $x$:
$$
(t-2 \sqrt{2} r) x^{2}... | In the right triangle $\triangle ABC$, $\angle C=90^{\circ}$,
$$
\begin{array}{c}
AB=c, AC \\
=b, CB=
\end{array}
$$
$a$, the center of the inscribed circle is $O$.
Connecting $OA$ and
$OB$, we have
$$
\begin{array}{l}
S_{\triangle ABC}=\frac{1}{2} ab. \\
\text { Also, } S_{\triangle ABC}=S_{\triangle ADC}+S_{\triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,419 |
Three, (15 points) Find the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that
(1) $f(1)=1$;
(2) For all $x, y \in \mathbf{N}_{+}$,
$$
f(x+y)=f(x)+f(y)+x y
$$
holds. | Three, let the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ satisfy the given conditions.
For positive integers $n, k$ we have
$$
f((k+1) n)=f(k n)+f(n)+k n^{2} \text {. }
$$
Let $k=1,2, \cdots, m-1$ and sum up to get
$$
\begin{array}{l}
f(m n)=m f(n)+[1+2+\cdots+(m-1)] n^{2} \\
=m f(n)+\frac{m(m-1)}{2} n^{... | f(m)=\frac{m(m+1)}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,420 |
Four. (15 points) As shown in Figure 2, in $\square A B C D$, the angle bisector of $\angle B A D$ intersects $B C$ at point $M$ and the extension of $D C$ at point $N$. The circumcircle $\odot O$ of $\triangle C M N$ intersects the circumcircle of $\triangle C B D$ at another point $K$. Prove:
(1) Point $O$ lies on th... | (1) As shown in Figure 4, from the given conditions, we have
$$
\begin{array}{l}
\angle B M A \\
=\angle M A D \\
=\angle B A M .
\end{array}
$$
Therefore,
$$
B A=B M \text{. }
$$
Similarly,
$$
M C=C N \text{. }
$$
Connect $O C$.
Then $O C$ bisects $\angle N C M$.
Connect $O B$, $O M$, and $O D$. Let $\angle B A D=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,421 |
Five. (15 points) Prove: In any given seven real numbers, there must exist two real numbers $x, y$, such that
$$
0 \leqslant \frac{x-y}{1+x y} \leqslant \frac{\sqrt{3}}{3} \text {. }
$$ | Let the seven given real numbers be $a_{i}$ $(i=1,2, \cdots, 7)$, and there exist seven real numbers $\theta_{i}$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan \theta_{i}=a_{i}$.
Divide $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ into six intervals of length $\frac{\pi}{6}$:
$$
\begin{array}{l}
\lef... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,422 |
1. Let $z$ be a complex number, $\alpha(z)$ denote the smallest positive integer $n$ such that $z^{n}=1$. Then for the imaginary unit $\mathrm{i}, \alpha$ (i) $=$ ( ).
(A) 8
(B) 6
(C) 4
(D) 2 | $-1, \mathrm{C}$.
Since $i^{1}=i, i^{2}=-1, i^{3}=-i, i^{4}=1$, therefore, the smallest positive integer $n$ that satisfies $z^{n}=1$ is $4$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,423 |
2. Given that the domain of the function $f(x)$ is $\mathbf{R}$. If $f(x+1)$ and $f(x-1)$ are both odd functions, then $(\quad)$.
(A) $f(x)$ is an even function
(B) $f(x)$ is an odd function
(C) $f(x+3)$ is an odd function
(D) $f(x+3)$ is an even function | 2. C.
Since $f(x+1)$ is an odd function, we have
$$
\begin{array}{l}
f(x+1)=-f(-x+1), \\
f(x)=-f(-x+2) .
\end{array}
$$
Similarly, since $f(x-1)$ is an odd function, we get
$$
f(x)=-f(-x-2) \text {. }
$$
From equations (1) and (2), we have
$$
f(-x+2)=f(-x-2) \text {. }
$$
Thus, $f(x+2)=f(x-2), f(x+4)=f(x)$.
Therefo... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,424 |
Example 7 Let the convex quadrilateral $ABDF$ have the extensions of its two pairs of opposite sides $AB$ and $FD$, $AF$ and $BD$ intersect at points $C$ and $E$ respectively. Prove that the midpoints $M$, $N$, $P$ of segments $AD$, $BF$, $CE$ are collinear. | Prove that, as shown in Figure 7, take the midpoints $Q$, $R$, and $S$ of $CD$, $BD$, and $BC$ respectively.
Thus, in
$\triangle ACD$,
$M$, $R$, and $Q$ are
collinear;
in $\triangle BCF$,
$S$, $R$, and $N$ are
collinear;
in $\triangle BCE$, $S$, $Q$, and $P$ are collinear.
At this point, since $MQ \parallel AB$, $NS \p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,425 |
3. A number $x$ is randomly chosen in the interval $[-1,1]$. Then the probability that the value of $\cos \frac{\pi x}{2}$ lies between 0 and $\frac{1}{2}$ is
(A) $\frac{2}{3}$
(B) $\frac{2}{\pi}$
(C) $\frac{1}{2}$
(D) $\frac{1}{3}$ | 3. D.
From $\cos \frac{\pi x}{2}=\frac{1}{2}$, we get $x= \pm \frac{2}{3}$.
From the graph of the function $y=\cos \frac{\pi x}{2}$, we know that the values of $x$ for which $\cos \frac{\pi x}{2}$ lies between 0 and $\frac{1}{2}$ fall within $\left[-1,-\frac{2}{3}\right]$ and $\left[\frac{2}{3}, 1\right]$.
Thus, the r... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,426 |
4. Let $f(x)$ be a function from $\mathbf{R} \rightarrow \mathbf{R}$, and for any real number, we have
$$
\begin{array}{l}
f\left(x^{2}+x\right)+2 f\left(x^{2}-3 x+2\right) \\
=9 x^{2}-15 x .
\end{array}
$$
Then the value of $f(50)$ is ( ).
(A) 72
(B) 73
(C) 144
(D) 146 | 4. D.
Substitute $1-x$ for $x$ in the given equation to get
$$
\begin{array}{l}
f\left(x^{2}-3 x+2\right)+2 f\left(x^{2}+x\right) \\
=9 x^{2}-3 x-6 .
\end{array}
$$
From the above equation and the original equation, eliminate $f\left(x^{2}-3 x+2\right)$ to get
$$
\begin{array}{l}
f\left(x^{2}+x\right)=3 x^{2}+3 x-4 \... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,427 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=0, \\
a_{n+1}=a_{n}+2+2 \sqrt{1+a_{n}}(n=1,2, \cdots) .
\end{array}
$$
Then $a_{2009}=(\quad$.
(A) 4036080
(B) 4036078
(C) 4036082
(D) 4036099 | 5. A.
From the given information, we have
$$
\begin{array}{l}
a_{n+1}+1=a_{n}+1+2 \sqrt{1+a_{n}}+1 \\
=\left(\sqrt{a_{n}+1}+1\right)^{2} .
\end{array}
$$
Since $a_{n+1}>0$, then
$$
\sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+1 \text { . }
$$
Therefore, the sequence $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,428 |
6. Given non-zero vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ satisfy
$$
\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right) \cdot \overrightarrow{B C}=0 \text {, }
$$
and $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|} \cdot \frac{\ov... | 6. D.
Since the line on which $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}$ lies passes through the incenter of $\triangle A B C$, it follows from
$$
\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,429 |
For $n \in \mathbf{N}_{+}$, calculate
$$
\begin{array}{l}
\mathrm{C}_{4 n+1}^{1}+\mathrm{C}_{4 n+1}^{5}+\cdots+\mathrm{C}_{4 n+1}^{4 n+1} \\
=
\end{array}
$$ | $$
\text { II.7. } 2^{4 n-1}+(-1)^{n} 2^{2 n-1} \text {. }
$$
When $n=1$, $\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{5}=2^{3}-2$;
When $n=2$, $\mathrm{C}_{9}^{1}+\mathrm{C}_{9}^{5}+\mathrm{C}_{9}^{9}=2^{7}+2^{3}$;
When $n=3$,
$$
\mathrm{C}_{13}^{1}+\mathrm{C}_{13}^{5}+\mathrm{C}_{13}^{9}+\mathrm{C}_{13}^{13}=2^{11}-2^{5} ;
$... | 2^{4 n-1}+(-1)^{n} 2^{2 n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,430 |
8. In $\triangle A B C$, $D$ is the midpoint of side $B C$. If $\overrightarrow{A D} \cdot \overrightarrow{A C}=0$, then the minimum value of $\tan C-\cot A$ is $\qquad$ | 8. $\sqrt{2}$.
Take the midpoint $E$ of $A B$. Then
$$
\begin{array}{l}
D E / / A C, \angle A D E=90^{\circ}, \\
\angle D A E=\angle A-90^{\circ} \text {. } \\
\text { Let } A C=b, A D=m . \text { Then } \\
\tan C=\frac{m}{b}, \tan \left(A-90^{\circ}\right)=\frac{b}{2 m}, \\
-\cot A=\frac{b}{2 m} .
\end{array}
$$
The... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,431 |
9. Calculate $\sqrt{\underbrace{44 \cdots 4}_{2 n \uparrow}+\underbrace{11 \cdots 1}_{n+1 \uparrow}-\underbrace{66 \cdots 6}_{n \uparrow}}$ $=$ . $\qquad$ | $$
\begin{array}{l}
\text { 9. } \underbrace{66 \cdots 6}_{n-1 \uparrow} . \\
\text { Let } \underbrace{11 \cdots 1}_{n \uparrow}=a \text {. Then } \\
10^{n}=\underbrace{99 \cdots 9}_{n \uparrow}+1=9 a+1, \\
\underbrace{44 \cdots 4}_{2 n \uparrow}=4 a \cdot 10^{n}+4 a=4 a(9 a+2), \\
\underbrace{11 \cdots 1}_{n+1 \uparr... | \underbrace{66 \cdots 6}_{n-1 \uparrow} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,432 |
10. Given the function
$$
f(x)=\left\{\begin{array}{ll}
(3-m) x-m, & x<1 ; \\
\log _{m} x, & x \geqslant 1
\end{array}\right.
$$
is monotonically increasing on the real number set $\mathbf{R}$. Then the range of the real number $m$ is. | 10. $\frac{3}{2} \leqslant m1, \\
3-m>0, \\
\log _{m} 1 \geqslant(3-m) \times 1-m .
\end{array}\right.
$
Solving gives $\frac{3}{2} \leqslant m<3$. | \frac{3}{2} \leqslant m < 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,433 |
11. The eccentricity of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1(b>0)$ is used as the radius, and the right focus is used as the center of a circle that is tangent to the asymptotes of the hyperbola. Then the eccentricity of the hyperbola is $\qquad$ | 11. $\frac{2 \sqrt{3}}{3}$.
It is known that $a=2, c=\sqrt{4+b^{2}}, e=\frac{\sqrt{4+b^{2}}}{2}$.
The equations of the asymptotes are
$$
\frac{x}{2} \pm \frac{y}{b}=0 \Rightarrow b x \pm 2 y=0 .
$$
The distance from the right focus $\left(\sqrt{4+b^{2}}, 0\right)$ to the asymptote is $=b$.
Thus, $b=\frac{\sqrt{4+b^{2... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,434 |
12. The sum of all real roots of the equation $\left|x^{2}-3 x+2\right|+\left|x^{2}+2 x-3\right|=11$ is $\qquad$ | 12. $\frac{-19+5 \sqrt{97}}{20}$.
The original equation can be transformed into
$$
|x-1|(|x-2|+|x+3|)=11 \text {. }
$$
When $x \leqslant-3$, equation (1) becomes $2 x^{2}-x-12=0$, and both roots are greater than -3, so there is no solution in this case;
When $-3 < x \leqslant 2$, equation (1) becomes $-x^2 + 4x - 12 ... | \frac{-19+5 \sqrt{97}}{20} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,435 |
Example 8 Let the convex quadrilateral $ABDF$ have its two pairs of opposite sides $AB$ and $FD$ extended to meet at point $C$, and $AF$ and $BD$ extended to meet at point $E$. The line $AD$ intersects $BF$ and $CE$ at points $M$ and $N$ respectively. Prove that $\frac{AM}{AN}=\frac{DM}{DN}$. | Prove that as shown in Figure 8(a), if $B F \parallel C E$, then
$$
\frac{A M}{A N}=\frac{B F}{C E}=\frac{D M}{D N} \text {. }
$$
If $B F$ is replaced, as shown in Figure 8(b), let the line $B F$ intersect $C E$ at point $G$, and connect $A G$.
Draw a line $T P \parallel C G$ through point $D$, intersecting $A C$, $A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,436 |
13. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+\cdots+b_{n} x^{n},
\end{array}
$$
and $b_{1}+b_{2}+\cdots+b_{n}=1013$.
Then a possible value of the positive integer $n$ is | 13. 9 .
Let $x=0$, we get $b_{0}=n$.
Let $x=1$, we get
$$
2+2^{2}+\cdots+2^{n}=b_{0}+b_{1}+\cdots+b_{n},
$$
which is $2\left(2^{n}-1\right)=n+1013$.
Solving this, we get $n=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,437 |
14. Given a cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the centers of faces $A B C D$ and $A D D^{\prime} A^{\prime}$ are $M$ and $N$, respectively. Then the sine of the angle formed by the skew lines $M N$ and $B D^{\prime}$ is $\qquad$ | 14. $\frac{\sqrt{3}}{3}$.
Let the edge length of the cube be 1, and establish a spatial rectangular coordinate system with $\overrightarrow{D A} 、 \overrightarrow{D C} 、 \overrightarrow{D D^{\prime}}$ as the basis. Then $B(1,1,0)$. Therefore,
$$
D^{\prime}(0,0,1) 、 M\left(\frac{1}{2}, \frac{1}{2}, 0\right) 、 N\left(\f... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,438 |
15. (12 points) Given vectors
$$
\boldsymbol{a}=\left(x^{2}, x+1\right), \boldsymbol{b}=(1-x, t) \text {. }
$$
If the function $f(x)=\boldsymbol{a} \cdot \boldsymbol{b}$ is a monotonically increasing function on the interval $(-1,1)$, find the range of values for $t$. | Three, 15. By definition, we have
$$
\begin{array}{l}
f(x)=x^{2}(1-x)+t(x+1) \\
=-x^{3}+x^{2}+t x+t . \\
\text { Then } f^{\prime}(x)=-3 x^{2}+2 x+t .
\end{array}
$$
If $f(x)$ is a monotonically increasing function on the interval $(-1,1)$, then $f^{\prime}(x) \geqslant 0$ on $(-1,1)$.
Thus, $-3 x^{2}+2 x+t \geqslant... | t \geqslant 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,439 |
16. (12 points) As shown in Figure 1, a diagonal of the cyclic quadrilateral $ABCD$ divides a pair of opposite angles into four angles $\alpha_{1}$,
$\alpha_{2}$, $\alpha_{3}$, $\alpha_{4}$. Prove that:
$$
\begin{array}{l}
\sin \left(\alpha_{1}+\alpha_{2}\right) \cdot \\
\sin \left(\alpha_{2}+\alpha_{3}\right) \cdot \... | 16. By the Law of Sines, we have
$$
\begin{array}{l}
A C=2 R \sin \left(\alpha_{4}+\alpha_{1}\right)=2 R \sin \left(\alpha_{2}+\alpha_{3}\right), \\
B D=2 R \sin \left(\alpha_{1}+\alpha_{2}\right)=2 R \sin \left(\alpha_{3}+\alpha_{4}\right), \\
A B=2 R \sin \alpha_{3}, B C=2 R \sin \alpha_{2}, \\
C D=2 R \sin \alpha_{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,440 |
17. (12 points) If a sequence $\left\{a_{n}\right\}$ satisfies $a_{i-1} a_{i+1} \leqslant a_{i}^{2}$ for any three consecutive terms $a_{i-1}, a_{i}, a_{i+1}$, then the sequence is called a "logarithmically convex sequence". Suppose the positive sequence $a_{0}, a_{1}, \cdots, a_{n}$ is a logarithmically convex sequenc... | 17. Let $S=a_{1}+a_{2}+\cdots+a_{n-1}$. Then the desired inequality can be reduced to
$$
\begin{array}{l}
n^{2}\left(S+a_{0}+a_{n}\right) S \\
\geqslant\left(n^{2}-1\right)\left(S+a_{0}\right)\left(S+a_{n}\right),
\end{array}
$$
which is equivalent to $\left(S+a_{0}\right)\left(S+a_{n}\right) \geqslant n^{2} a_{0} a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,441 |
18. (14 points) A horizontal L-shaped corridor in a building is shown in Figure 2. The width of both corridors is $3 \mathrm{~m}$, and a piece of equipment with a rectangular cross-section needs to be moved horizontally into the L-shaped corridor. If the width of the rectangular cross-section of the equipment is $1 \ma... | 18. Establish a rectangular coordinate system with the lines $O B$ and $O A$ as the $x$-axis and $y$-axis, respectively. The problem then transforms into: first finding the minimum length of the segment $A B$ intercepted by the positive $x$-axis and the positive $y$-axis, which is a tangent to the circle with center $M... | 6 \sqrt{2}-2<7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,442 |
19. (14 points) Suppose there are 2009 people standing in a row, and they start counting from the first person, 1 to 3. Anyone who counts to 3 will leave the row, and the rest will move forward to form a new row. This process continues until only 3 people remain after the $p$-th round of counting. What were the initial... | 19. After the $p$-th round of counting, the first two people among the remaining 3 are obviously the first and second positions in the original queue.
Let the initial position of the third person be $a_{p+1}$. Then after the first round of counting, he stands at the $a_{p}$ position, $\cdots \cdots$ after the $p$-th r... | 1, 2, 1600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,443 |
1. A class of students learned at the award ceremony that the number of students (a student can submit multiple papers in one subject) who won awards in the subject paper competition is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline & Mathematics & Physics & Chemistry \\
\hline Provincial & 3 & 2 & 3 \\
\h... | - 1. B.
The class won a total of
$$
18+12+6+3+3+2=44 \text { (awards), }
$$
Those who won only two awards won a total of $13 \times 2=26$ awards.
Thus, the number of people who did not win two awards is 28 -
$$
13=15 \text { people. }
$$
Since everyone won at least one award, there are 44 $26-15=3$ awards left.
The... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,444 |
2. As shown in Figure 1, $A C=B C, D C=E C, \angle A C B=$ $\angle E C D=90^{\circ}$, and
$\angle E B D=42^{\circ}$. Then the degree measure of $\angle A E B$ is ( ).
(A) $128^{\circ}$
(B) $130^{\circ}$
(C) $132^{\circ}$
(D) $134^{\circ}$ | $$
\begin{array}{l}
\text { It is easy to see that } \triangle A C E \cong \triangle B C D . \\
\text { Therefore, } \angle C A E = \angle C B D . \\
\text { Then, } \angle A E B = \angle C A E + \angle E B C + 90^{\circ} \\
= \angle C B D + \angle E B C + 90^{\circ} \\
= \angle E B D + 90^{\circ} = 132^{\circ} .
\end{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,445 |
3. Arrange the numbers $\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, \cdots, 2 \sqrt{51}$ as shown in Figure 2.
\begin{tabular}{cccccc}
$\sqrt{2}$ & 2 & $\sqrt{6}$ & $2 \sqrt{2}$ & $\sqrt{10}$ & $2 \sqrt{3}$ \\
$\sqrt{14}$ & 4 & $3 \sqrt{2}$ & $2 \sqrt{5}$ & $\sqrt{22}$ & $2 \sqrt{6}$ \\
$\sqrt{26}$ & $2 \sqrt{7}$ & $... | 3. D.
Observing, we see that this sequence of numbers is $\sqrt{2 k}(k=1,2, \cdots, 102)$, where the largest rational number is $\sqrt{196}$, which is ranked 98th. Therefore, the largest rational number in this sequence is located in the 17th row and the 2nd column. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,446 |
1. In Rt $\triangle A B C$, the area is 120, and $\angle B A C=$ $90^{\circ}, A D$ is the median of the hypotenuse, a line $D E \perp$ $A B$ is drawn through point $D$, and $C E$ is connected to intersect $A D$ at point $F$. Then the area of $\triangle A F E$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24 | Draw $F G \perp A B$ at point $G$. Then $F G / / D E / / A C$,
we have $\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \frac{F G}{A C}=\frac{E G}{A E}$.
Adding the two equations, we get
$$
\frac{3 F G}{A C}=\frac{A E}{A E}=1 \Rightarrow F G=\frac{1}{3} A C \text {. }
$$
Then $S_{\triangle A F E}=\frac{1}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,447 |
4. If a positive integer can be expressed as the difference of squares of two consecutive even numbers, it is called a "Morning Star Number" (such as $4=2^{2}-0^{2}, 12=4^{2}-2^{2}, 20=6^{2}-4^{2}$). Among the following statements about Morning Star Numbers, the correct ones are ( ) in number.
(1) 2010 is a Morning Sta... | 4. B.
Let two consecutive even numbers be $2 k+2$ and $2 k$. Then $(2 k+2)^{2}-(2 k)^{2}=4(2 k+1)$.
Therefore, Chen Yao numbers are multiples of 4, but not multiples of 8. Let any two positive odd numbers be $2 m+1, 2 n+1$. Then
$$
\begin{array}{l}
(2 m+1)^{2}-(2 n+1)^{2} \\
=4[m(m+1)-n(n+1)] .
\end{array}
$$
Thus, 8... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,448 |
5. As shown in Figure $3, A B$ is the diameter of the semicircle $\odot O$, $A C$ and $A D$ are both chords, and $\angle B A C = \angle C A D$. Which of the following equations is true? ( ).
(A) $A C^{2} = A B \cdot A D$
(B) $A B \cdot A D > A C^{2}$
(C) $A B + A D = 2 A C$
(D) $A B + A D < 2 A C$ | 5. D.
As shown in Figure 6, draw $O M \perp A D$ at point $M$, intersecting $A C$ at point $N$, and connect $O C$.
It is easy to see that $O C \parallel A D$
$$
\begin{array}{r}
\Rightarrow \angle C O N=\angle C O M \\
=\angle O M A=90^{\circ} .
\end{array}
$$
It is also easy to see that
$$
N C > O C = \frac{1}{2} A ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,449 |
6. A customs office received an intelligence report that a transnational drug trafficking group will hide drugs in a few boxes of a shipment entering the country, and the drug boxes are marked with a mysterious number:
$$
\overline{x 0 y z}=9 \overline{x y z} \text {. }
$$
Upon inspection, it was found that each box i... | 6. D.
Let the required four-digit number be $\overline{x 0 y z}$. Then
$$
\overline{x 0 y z}=9 \overline{x y z} \text {. }
$$
Thus, $1000 x+10 y+z=9(100 x+10 y+z)$.
Simplifying, we get $100 x=8(10 y+z)$.
From this, we know that the right side of the equation is divisible by 8.
Therefore, $x$ must be an even number: 2... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,450 |
1. Given that $a$ is an integer, the equation concerning $x$
$$
\frac{x^{2}}{x^{2}+1}-\frac{4|x|}{\sqrt{x^{2}+1}}=a-3
$$
has real roots. Then the possible values of $a$ are | 二、1.1,2,3.
From the original equation, we get
$$
\left(\frac{|x|}{\sqrt{x^{2}+1}}-2\right)^{2}=1+a \text {. }
$$
Also, from $0 \leqslant \frac{|x|}{\sqrt{x^{2}+1}}<1$, we know
$$
1<\left(\frac{|x|}{\sqrt{x^{2}+1}}-2\right)^{2} \leqslant 4,
$$
which implies $1<1+a \leqslant 4 \Rightarrow 0<a \leqslant 3$.
Therefore, t... | 1,2,3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,451 |
2. Let the three sides of the right triangle $\triangle ABC$ be $a$, $b$, and $c$, and $a < b < c$. If $\frac{b}{c+a} + \frac{a}{c+b} = \frac{13}{15}$, then $a: b: c=$ $\qquad$ | 2. 5: 12: 13.
Since $c^{2}-a^{2}=b^{2}, c^{2}-b^{2}=a^{2}$, we have
$$
\begin{array}{l}
\frac{13}{15}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{c-a}{b}+\frac{c-b}{a} \\
=\frac{c(a+b-c)}{a b}=\frac{2 c(a+b-c)}{(a+b)^{2}-\left(a^{2}+b^{2}\right)} \\
=\frac{2 c}{a+b+c}
\end{array}
$$
Thus, $13(a+b)=17 c$.
Squaring both sides an... | 5: 12: 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,452 |
3. If $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}=0$, then $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=$ $\qquad$ | 3.1 or $-\frac{3}{2}$.
Let $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=k$.
Then, by the given condition,
$$
\begin{array}{l}
a\left(k-\frac{b}{c+a-b}-\frac{c}{a+b-c}+\frac{a}{b+c-a}\right)+ \\
b\left(k-\frac{c}{a+b-c}-\frac{a}{b+c-a}+\frac{b}{c+a-b}\right)+ \\
c\left(k-\frac{a}{b+c-a}-\frac{b}{c+a-b}+\frac{c}{a+b... | 1 \text{ or } -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,453 |
4. As shown in Figure 4, given $\angle A O M=60^{\circ}$, there is a point $B$ on ray $O M$ such that the lengths of $A B$ and $O B$ are both integers, thus $B$ is called an "olympic point". If $O A=8$, then the number of olympic points $B$ in Figure 4 is $\qquad$ | 4. 4 .
As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$
which ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,454 |
One, (20 points) Let $a, b, c, d$ be real numbers satisfying the inequality
$$
(a+b+c)^{2} \geqslant 2\left(a^{2}+b^{2}+c^{2}\right)+4 d
$$
Prove: $a b+b c+c a \geqslant 3 d$. | From the given conditions, we have
$$
a^{2}-2(b+c) a+\left(b^{2}+c^{2}-2 b c+4 d\right) \leqslant 0 \text {. }
$$
Thus, there exists $r \geqslant 0$, such that
$$
a^{2}-2(b+c) a+\left(b^{2}+c^{2}-2 b c+4 d\right)+r=0 \text {. }
$$
Since $a$ is a real number, we have
$$
\begin{aligned}
\Delta & =4(b+c)^{2}-4\left(b^{2... | a b+b c+c a \geqslant 3 d | Inequalities | proof | Yes | Yes | cn_contest | false | 722,455 |
$\begin{array}{l}\quad \text { II. (25 points) As shown in Figure } 5, \text { in } \triangle A B C, \angle B A C \\ =90^{\circ}, \text { points } D \text { and } F \text { are on } \\ A C \text { and } B C \text { respectively, } B D \text { intersects } \\ A F \text { at point } E. \text { If } \\ \angle A D B=\angle... | As shown in Figure 8, extend $B A$ and $F D$ to intersect at point $K$, and draw $G H \parallel B D$ to intersect $C B$ and $F D$ at points $G$ and $H$ respectively. Then,
$$
\angle A D B = \angle D A H.
$$
Since $\angle C D F = \angle A D H$, it follows that $\angle D A H = \angle A D H$.
Thus, $A H = D H$.
Also, sin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,456 |
Three, (25 points) Six points in space (no three points are collinear) are connected pairwise, and these line segments are colored with red and blue. Among them, the line segments connected from point $A$ are all red. Find the number of triangles, with these six points as vertices, that have all three sides of the same... | Three, it is known that there are 20 triangles in total.
Let the number of monochromatic triangles be $x$. Then the number of bichromatic triangles is $20-x$. Each monochromatic triangle has three monochromatic angles, and each bichromatic triangle has one monochromatic angle.
Let the total number of monochromatic angl... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,457 |
2. Given that $D$ and $E$ are points on the sides $BC$ and $CA$ of $\triangle ABC$, respectively, and $BD=4, DC=1, AE=5, EC=2$. Connecting $AD$ and $BE$, they intersect at point $P$. Through $P$, draw $PQ \parallel CA$ and $PR \parallel CB$, which intersect side $AB$ at points $Q$ and $R$, respectively. The ratio of th... | Draw $E F / / A D$ intersecting $B C$ at point $F$. Then
$$
\begin{array}{l}
\frac{C F}{F D}=\frac{C E}{E A}=\frac{2}{5} \\
\Rightarrow F D=\frac{5}{7} . \\
\text { Also, } P Q / / C A, \text { so } \\
\frac{P Q}{E A}=\frac{B P}{B E}=\frac{B D}{B F}=\frac{28}{33} \Rightarrow P Q=\frac{140}{33} .
\end{array}
$$
Also, $... | \frac{400}{1089} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,458 |
1. Let $f(x)$ be an odd function defined on $(-\infty, 0) \cup(0$, $+\infty)$. When $x>0$, $x f^{\prime}(x)<0$ and $f(1)=0$. Then the solution set of the inequality $x f(x)<0$ is $\qquad$ | $-1 .(-\infty,-1) \cup(1,+\infty)$.
When $x>0$, $x f^{\prime}(x)1$ or $10$.
Therefore, the solution set of the inequality $x f(x)<0$ is
$$
(-\infty,-1) \cup(1,+\infty) \text {. }
$$ | (-\infty,-1) \cup(1,+\infty) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 722,459 |
2. Given that $P(x, y)$ is a point on the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$. Then the minimum value of $3^{-x}+9^{y}$ is $\qquad$ . | 2. $\frac{2}{9}$.
According to the mean inequality, we have
$$
\begin{array}{l}
x-2 y \leqslant\left(\frac{x^{2}}{4}+1\right)+\left(y^{2}+1\right) \\
=2\left(\frac{x^{2}}{8}+\frac{y^{2}}{2}\right)+2=4 .
\end{array}
$$
On the other hand,
$$
\begin{array}{l}
3^{-x}+9^{y}=3^{-x}+3^{2 y} \\
\geqslant 2 \sqrt{3^{-(x-2 y)}... | \frac{2}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,460 |
3. If the six edges of a tetrahedron are all powers of 2, then in this tetrahedron, the lengths of the edges can have at most $\qquad$ values.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The $\qquad$ is kept as is, since it likely represents a space for an answer in the original text. | 3. 3 .
First, prove that if the three sides of a triangle are powers of 2, then the triangle must be isosceles.
Otherwise, let the three sides of the triangle be
$$
a=2^{u}, b=2^{v}, c=2^{w} (a>b>c) \text {. }
$$
Then $b+c \leqslant \frac{a}{2}+\frac{a}{2}=a$, which is a contradiction.
In the tetrahedron $ABCD$, if t... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,461 |
4. The range of the function $y^{\prime}=\frac{\sin 9 x}{\sin x}+\frac{\cos 9 x}{\cos x}$ is | 4. $\left[-\frac{5}{2}, 10\right)$.
It is clear that $\sin x \neq 0, \cos x \neq 0$. Thus, $x \neq \frac{k \pi}{2}(k \in \mathbb{Z})$. Therefore, $\cos 4 x \in[-1,1)$.
$$
\begin{array}{l}
\text { Hence } y=\frac{\sin 9 x}{\sin x}+\frac{\cos 9 x}{\cos x} \\
=2 \cos 8 x+\sin 8 x\left(\frac{\cos x}{\sin x}-\frac{\sin x}{... | \left[-\frac{5}{2}, 10\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,462 |
6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are | 6. 4 .
According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,464 |
7. In the right triangle $\triangle ABC$, $\angle C=90^{\circ}, AB=c$. Along the direction of vector $\overrightarrow{AB}$, points $M_{1}, M_{2}, \cdots, M_{n-1}$ divide the segment $AB$ into $n$ equal parts. Let $A=M_{0}, B=M_{n}$. Then
$$
\begin{array}{l}
\lim _{n \rightarrow+\infty} \frac{1}{n}\left(\overrightarrow{... | 7. $\frac{c^{2}}{3}$.
Let $C B=a, C A=b$. Then $a^{2}+b^{2}=c^{2}$. Hence
$$
\overrightarrow{C M}_{i}=\frac{n-i}{n} \overrightarrow{C A}+\frac{i}{n} \overrightarrow{C B} \text {. }
$$
By $\overrightarrow{C A} \cdot \overrightarrow{C B}=0$, we get
$$
\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_... | \frac{c^{2}}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 722,465 |
8. Planar Region
$$
\begin{aligned}
S= & \left\{(x, y) \mid x, y \in\left[0, \frac{\pi}{2}\right],\right. \\
& \left.\sin ^{2} x+\sin x \cdot \sin y+\sin ^{2} y \leqslant \frac{3}{4}\right\}
\end{aligned}
$$
The area of the region equals $\qquad$ | 8. $\frac{\pi^{2}}{6}$.
$$
\begin{array}{l}
\text { By } 2\left(\sin ^{2} x-\sin x \cdot \sin y+\sin ^{2} y\right) \\
= 2-2 \cos (x+y) \cdot \cos (x-y)+ \\
\cos (x+y)-\cos (x-y) \\
= \frac{3}{2}-2\left[\cos (x+y)+\frac{1}{2}\right] . \\
{\left[\cos (x-y)-\frac{1}{2}\right] } \\
\leqslant \frac{3}{2},
\end{array}
$$
w... | \frac{\pi^{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,466 |
9. (16 points) Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=a(a>0) \text {, }
$$
and for all positive integers $n$,
$$
x_{n+1} \geqslant(n+2) x_{n}-\sum_{i=1}^{n-1} i x_{i} \text {. }
$$
Prove: there exists a positive integer $n$, such that $x_{n}>2010$ !. | II. 9. First prove by mathematical induction: for all positive integers $n$,
$$
x_{n+1}>\sum_{i=1}^{n} i x_{i}>a \cdot n!\text {. }
$$
When $n=1$, it is obvious that $x_{2} \geqslant 3 x_{1}>x_{1}=a$.
Assume that the conclusion holds for $n \leqslant k$.
$$
\begin{array}{l}
\text { Then } x_{k+2} \geqslant(k+3) x_{k+1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,467 |
10. (20 points) Let $a, b, c, d$ be real numbers, and call $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$ a $2 \times 2$ number table, and refer to $a, b, c, d$ as the elements of the number table. If $a+b, c+d, a+c, b+d$ are all distinct, then the $2 \times 2$ number table is called a "good number table". I... | 10. Since each element in the number table $M$ has three different choices, there are a total of $3^{4}=81$ different number tables in $M$.
It is easy to see that, in a good number table, $a \neq d, b \neq c$.
Also, $a, b, c, d \in\{-1,0,1\}$, where two numbers are the same (let's assume $a=b$).
Clearly, $c \neq d$. Ot... | \frac{16}{81} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,468 |
Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 uni... | (1) Notice
$$
\begin{array}{l}
S=\mid-2-1+2-3+4-5+\cdots-9+101 \\
=1-2+1 \times 5 \mid=3 .
\end{array}
$$
Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times.
(2) In fact, after the electronic flea jumps four times, it returns to the ori... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,469 |
Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}<a_{2}<\cdots<a_{100}$) and the other in descending order (denoted as $b_{1}>b_{2}>\cdots>b_{100}$). Try to find
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\... | First, prove: For any term in the algebraic expression
$$
\left|a_{k}-b_{k}\right|(k=1,2, \cdots, 100)
$$
the larger number among $a_{k}$ and $b_{k}$ must be greater than 100, and the smaller number must not exceed 100.
(1) If $a_{k} \leqslant 100$ and $b_{k} \leqslant 100$, then by
$$
a_{1}b_{k+1}>\cdots>b_{100}
$$
w... | 10000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,470 |
1. Given
$$
|x+2|+|1-x|=9-|y-5|-|1+y| \text {. }
$$
Find the maximum and minimum values of $x+y$. | By the geometric meaning of absolute value, we have
$$
|x+2|+|x-1| \geqslant 3,|y-5|+|y+1| \geqslant 6 \text {. }
$$
Therefore, $|x+2|+|x-1|=3,|y-5|+|y+1|=6$.
Thus, $-2 \leqslant x \leqslant 1,-1 \leqslant y \leqslant 5$.
Hence, the maximum value of $x+y$ is 6, and the minimum value is -3. | 6, -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,471 |
9. (16 points) Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and point $A$ be on the $x$-axis and to the right of $F$. The circle with diameter $F A$ intersects the parabola at two distinct points $M$ and $N$ above the $x$-axis. Prove that $F M+F N=F A$.
---
The translation maintains the original text's fo... | Given $F\left(\frac{p}{2}, 0\right)$.
Let point $A(a, 0)$. Then $F A=a-\frac{1}{2} p$.
Therefore, the circle with diameter FA is
$$
\left(x-\frac{2 a+p}{4}\right)^{2}+y^{2}=\left(\frac{2 a-p}{4}\right)^{2} \text {. }
$$
Let $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$. Then $x_{1}$ and $x_{2}$ are the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,472 |
10. (20 points) Does there exist $\theta \in\left(0, \frac{\pi}{2}\right)$, such that $\sin \theta, \cos \theta, \tan \theta, \cot \theta$ can be arranged in an arithmetic sequence? Explain your reasoning. | 10. When $\theta \in\left(0, \frac{\pi}{2}\right)$, the graphs of the functions $y=\sin x$ and $y=\cos x$ are symmetric about the line $x=\frac{\pi}{4}$, and the graphs of the functions $y=\tan x$ and $y=\cot x$ are also symmetric about the line $x=\frac{\pi}{4}$. Moreover, when $\theta=\frac{\pi}{4}$, any permutation ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,473 |
11. (20 points) Let the function $f(x)=a x^{3}+b x^{2}+c x+d$ have an image $\Gamma$ with two extremum points $P$ and $Q$, where $P$ is the origin.
(1) When point $Q(1,2)$, find the analytical expression of $f(x)$;
(2) When point $Q$ is on the circle $C:(x-2)^{2}+(y-3)^{2}=1$, find the maximum value of the slope of the... | 11. Since $f(x)=a x^{3}+b x^{2}+c x+d$, we have
$$
f^{\prime}(x)=3 a x^{2}+2 b x+c \text {. }
$$
Because the graph $\Gamma$ has an extremum point $P$ at the origin, we have
$$
f^{\prime}(0)=0 \text {, and } f(0)=0 \text {. }
$$
Thus, $c=d=0$.
(1) When point $Q(1,2)$, from $f^{\prime}(1)=0$ and $f(1)=2$, we get
$$
3 a... | 3+\sqrt{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 722,474 |
One, (40 points) As shown in Figure 1, let the projections of vertex $D$ of the cyclic quadrilateral $ABCD$ onto lines $AB$, $BC$, and $CA$ be $P$, $Q$, and $R$, respectively, and let the angle bisectors of $\angle ABC$ and $\angle ADC$ intersect at point $E$. Prove that point $E$ lies on $AC$ if and only if $PR = QR$. | By Simson's theorem, we know that points $P$, $Q$, and $R$ are collinear. From the problem, it is easy to see that points $C$, $Q$, $D$, and $R$ are concyclic. Therefore, $\angle DCA = \angle DQR = \angle DQP$.
Similarly, since points $A$, $P$, $R$, and $D$ are concyclic, we have $\angle DAC = \angle DPR = \angle DPQ$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,475 |
II. (40 points) Given the perimeter of $\triangle A_{i} B_{i} C_{i}$ $(i=1,2)$ is 1, with the lengths of the three sides being $a_{i} 、 b_{i} 、 c_{i}$. Let $p_{i}=a_{i}^{2}+b_{i}^{2}+c_{i}^{2}+4 a_{i} b_{i} c_{i}(i=1,2)$.
Prove: $\left|p_{1}-p_{2}\right|<\frac{1}{54}$. | Given $a_{i}+b_{i}+c_{i}=1$.
Assume $a_{i} \leqslant b_{i} \leqslant c_{i}$. Then, by the triangle inequality, we have
$$
0<a_{i} \leqslant b_{i} \leqslant c_{i}<\frac{1}{2} .
$$
It is not difficult to obtain
$$
\begin{array}{l}
0<\left(1-2 a_{i}\right)\left(1-2 b_{i}\right)\left(1-2 c_{i}\right) \\
\leqslant\left(\f... | \left|p_{1}-p_{2}\right|<\frac{1}{54} | Inequalities | proof | Yes | Yes | cn_contest | false | 722,476 |
Three, (50 points) Do there exist distinct prime numbers $p, q, r, s$ such that their sum is 640, and $p^{2}+q s$ and $p^{2}+q r$ are both perfect squares? If they exist, find the values of $p, q, r, s$; if not, explain the reason.
---
The above text has been translated into English, preserving the original text's li... | Given $p+q+r+s=640$, and $p, q, r, s$ are distinct prime numbers, we know that $p, q, r, s$ are all odd numbers.
Let
\[
\left\{\begin{array}{l}
p^{2}+q s=m^{2}, \\
p^{2}+q r=n^{2}
\end{array}\right.
\]
Without loss of generality, assume $s1$, then from $m-p<n-p<n+p$, we get
\[
\begin{array}{l}
m+p=q=n-p. \\
\text{Thus... | p=167, q=67, r=401, s=5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,477 |
Four, (50 points) For $n$ distinct positive integers, among any six numbers, there are at least two numbers such that one can divide the other. Find the minimum value of $n$ such that among these $n$ numbers, there must exist six numbers where one can be divided by the other five.
| The smallest positive integer $n=26$.
The proof is divided into two steps.
【Step 1】When $n \leqslant 25$, the condition is not satisfied.
Construct the following 25 positive integers:
(1) $2^{5}, 2^{4}, 2^{3}, 2^{2}, 2^{1}$;
(2) $3^{5}, 3^{4}, 3^{3}, 3^{2}, 3^{1}$;
(3) $5^{5}, 5^{4}, 5^{3}, 5^{2}, 5^{1}$;
(4) $7^{5}, 7... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,478 |
2. Divide the 100 natural numbers $1, 2, \cdots, 100$ into 50 groups, each containing two numbers. Now, substitute the two numbers in each group (denoted as $a$ and $b$) into $\frac{1}{2}(|a-b|+a+b)$ for calculation, and obtain 50 values. Find the maximum value of the sum of these 50 values. | Since the 100 numbers from $1 \sim 100$ are all different, in each pair of numbers, there must be a larger number. Therefore, the result of the calculation is the larger of the two numbers. Hence, the maximum sum of these 50 values is
$$
51+52+\cdots+100=3775
$$ | 3775 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,479 |
3. On the number line, points with coordinates $1, 2, \cdots, 2006$ are called "marked points". A frog starts from point 1, and after 2006 jumps, it visits all the marked points and returns to the starting point. What is the maximum length of the path the frog can jump? Explain your reasoning.
(2006, Shandong Province ... | Let the points the frog reaches in sequence be $x_{1}, x_{2}, \cdots, x_{2006}\left(x_{1}=1\right)$. Then the total length of the jumps is
$$
S=\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{2006}-x_{1}\right| \text {. }
$$
We call $\{1,2, \cdots, 1003\}$ the upper half, and $\{1004, 1005, \cdots, 2... | 2 \times 1003^2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,480 |
4. Find the value of the real number $a$ such that the graph of the function
$$
f(x)=(x+a)(|x-a+1|+|x-3|)-2 x+4 a
$$
is centrally symmetric. | When $x \leqslant \min \{a-1,3\}$,
$$
f(x)=-2 x^{2}-a x+a^{2}+6 a \text {; }
$$
When $x \geqslant \max \{a-1,3\}$,
$$
f(x)=2 x^{2}+(a-4) x-a^{2}+2 a \text {. }
$$
It can be seen that the graph of the function $f(x)$ consists of two parabolic arcs on the left and right sides, and a line segment in the middle. The grap... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,481 |
Example 2 Consider a permutation $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ of $\{1,2, \cdots, n\}$. If
\[
\begin{array}{l}
a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, \\
a_{1}+a_{2}+\cdots+a_{n}
\end{array}
\]
contains at least one perfect square, then the permutation is called a "quadratic permutation."
Find al... | Let $a_{i}=i(1 \leqslant i \leqslant n), b_{k}=\sum_{i=1}^{k} a_{i}$.
If $b_{k}$ is a perfect square, set $b_{k}=m^{2}$, that is,
$$
\begin{array}{l}
\frac{k(k+1)}{2}=m^{2} . \\
\text { Then } \frac{k}{2}<m, \text { and } \\
b_{k+1}=\frac{(k+1)(k+2)}{2} \\
=m^{2}+k+1<(m+1)^{2} .
\end{array}
$$
Therefore, $b_{k+1}$ is ... | n=\frac{(3+2 \sqrt{2})^{k}+(3-2 \sqrt{2})^{k}-2}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,483 |
Example 3 Consider the system of equations
$$
\left\{\begin{array}{l}
x+y=z+u, \\
2 x y=z u .
\end{array}\right.
$$
Find the maximum real constant $m$ such that for any positive integer solution $(x, y, z, u)$ of the system, when $x \geqslant y$, we have $m \leqslant \frac{x}{y}$.
(42nd IMO Shortlist) | Solve: Subtracting four times the second equation from the square of the first equation, we get
$$
x^{2}-6 x y+y^{2}=(z-u)^{2} \text {, }
$$
which is $\left(\frac{x}{y}\right)^{2}-6\left(\frac{x}{y}\right)+1=\left(\frac{z-u}{y}\right)^{2}$.
Let $\frac{x}{y}=w, f(w)=w^{2}-6 w+1$.
When $w=3 \pm 2 \sqrt{2}$, $f(w)=0$.
Si... | 3+2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,484 |
Example 4 Let $x_{0}+\sqrt{2003} y_{0}$ be the fundamental solution of the equation
$$
x^{2}-2003 y^{2}=1
$$
Find the solution $(x, y)$ of the equation such that $x, y>0$ and all prime factors of $x$ divide $x_{0}$.
(2003, China National Training Team Test) | We prove that apart from the fundamental solution, the equation has no other integer solutions that meet the conditions of the problem. Therefore, $\left(x_{0}, y_{0}\right)$ is the only solution that meets the conditions of the equation.
Since 2003 is not a perfect square,
$$
x^{2}-2003 y^{2}=1
$$
is a Pell's equati... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,485 |
Example 5 Find the integer solutions of the equation $x^{2}-15 y^{2}=61$.
保持源文本的换行和格式。 | First, find the integer solution suitable for
$$
l^{2} \equiv 15(\bmod 61)\left(0 \leqslant l \leqslant \frac{61}{2}\right)
$$
which means solving the equation
$$
l^{2}=15+61 h\left(l^{2} \leqslant 900\right)
$$
Since $0 \leqslant h \leqslant\left[\frac{900}{61}\right]=14$, we test $h=0,1, \cdots, 14$ one by one, and... | x+\sqrt{15} y= \pm(4+\sqrt{15})^{n}(14 \pm 3 \sqrt{15}) \text{ and } x+\sqrt{15} y= \pm(4+\sqrt{15})^{n}(11 \pm 2 \sqrt{15}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,486 |
1. If $x$ is an integer, $3<x<200$, and $x^{2}+(x+1)^{2}$ is a perfect square, find the integer value of $x$.
$(2007$, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Let $x^{2}+(x+1)^{2}=v^{2}$. Then
$$
(2 x+1)^{2}+1=2 v^{2} \text {. }
$$
Let $u=2 x+1$. Then
$$
u^{2}-2 v^{2}=-1 \text {. }
$$
This is a Pell's equation.
It is easy to see that $\left(u_{0}, v_{0}\right)=(1,1)$ is a solution to equation (1), and the solution formula for Pell's equation (1) is
$$
u_{n}+\sqrt{2} v_{n}=... | 20, 119 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,487 |
Example 3 Proof:
\[
\begin{aligned}
A & =|| x-y|+x+y-2 z|+|x-y|+x+y+2 z \\
& =4 \max \{x, y, z\},
\end{aligned}
\]
where, \(\max \{x, y, z\}\) represents the maximum of the three numbers \(x, y, z\). | Prove (1) When $x \geqslant y, x \geqslant z$,
$$
\begin{array}{l}
A=|x-y+x+y-2 z|+x-y+x+y+2 z \\
=2 x-2 z+2 x+2 z=4 x ;
\end{array}
$$
(2) When $y \geqslant z, y \geqslant x$,
$$
\begin{array}{l}
A=|y-x+x+y-2 z|+y-x+x+y+2 z \\
=2 y-2 z+2 y+2 z=4 y ;
\end{array}
$$
(3) When $z \geqslant x, z \geqslant y$,
$$
\begin{arr... | A=4 \max \{x, y, z\} | Algebra | proof | Yes | Yes | cn_contest | false | 722,488 |
3. Prove: The equation $x^{3}+y^{3}+z^{3}+t^{3}=1999$ has infinitely many integer solutions. | First, there is a set of solutions:
$$
10^{3}+10^{3}+0^{3}+(-1)^{3}=1999 \text {. }
$$
Next, we seek integer solutions of the form
$$
x=10+k, y=10-k, z=m, t=-1-m
$$
$(k, m \in \mathbf{Z})$.
Substituting into the original equation, we get
$$
\begin{array}{l}
(10+k)^{3}+(10-k)^{3}+m^{3}+(-1-m)^{3} \\
=1999 .
\end{array}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,489 |
4. Try to find a set of positive integer solutions for the equation $x^{2}-51 y^{2}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\text { Hint: } \begin{aligned}
x^{2} & =51 y^{2}+1 \\
& =49 y^{2}+14 y+1+2 y^{2}-14 y \\
& =(7 y+1)^{2}+2 y(y-7) .
\end{aligned}
$$
Let $y=7$. Then $x=50$. | x=50, y= | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,490 |
5. Prove: There exist infinitely many pairs of positive integers $(k, n)$ such that
$$
1+2+\cdots+k=(k+1)+(k+2)+\cdots+n \text {. }
$$
(26th IMO Shortlist Problem) | Given the equation can be transformed into
$$
2(2 k+1)^{2}=(2 n+1)^{2}+1 \text {. }
$$
Let $x=2 n+1, y=2 k+1$.
We get the Pell's equation $x^{2}-2 y^{2}=-1$, with the fundamental solution $(x, y)=(1,1)$.
Therefore, the original equation has infinitely many solutions. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,491 |
6. Prove: There are infinitely many positive integers $n$ such that the average $\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}$ is a perfect square. The first such number is of course 1. Write down the next two such positive integers after 1.
(31st IMO Shortlist) | $$
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=\frac{2 n^{2}+3 n+1}{6} \text { . }
$$
Let $\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2}$. Then
$$
2 n^{2}+3 n+1=6 m^{2} \text {, }
$$
which is $(4 n+3)^{2}-3(4 m)^{2}=1$.
Let $x=4 n+3, y=4 m$, we get the Pell's equation
$$
x^{2}-3 y^{2}=1 \text { . }
$$
This problem can be seen as ... | 337, 65521 | Number Theory | proof | Yes | Yes | cn_contest | false | 722,492 |
Example 1 Given that $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ are positive real numbers satisfying $\sum_{i=1}^{n} x_{i}^{m+1}=1$ (where $m$ is a positive integer). Prove:
$$
\sum_{i=1}^{n} x_{i}^{m} \geqslant \frac{2(m+2)^{\frac{m+2}{m+1}}}{m+1} \sum_{1 \leq i, j \leq n} x_{i}^{m+1} x_{j}^{m+1} \text {. }
$$ | Prove that
$$
\begin{array}{l}
\underbrace{x_{i}^{m}+x_{i}^{m}+\cdots+x_{i}^{m}}_{m+1 \uparrow}+(m+2)^{\frac{m+2}{m+1} x_{i}^{2 m+2}} \\
\geqslant(m+2) \sqrt[m+2]{x_{i}^{m} x_{i}^{m} \cdots x_{i}^{m}}(m+2)^{\frac{m+2}{m+2} x_{i}^{2 m+2}} \\
=(m+2)^{\frac{m+2}{m+1} x_{i}^{m+1}} \text {. } \\
\end{array}
$$
Therefore, \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,493 |
Example 2 ( $n$-variable irrational fractional inequality) If $x_{i}>0$ $(i=1,2, \cdots, n)$, prove:
$$
\sum_{i=1}^{n}\left(\frac{x_{i}}{\sum_{j=1}^{n} x_{j}-x_{i}}\right)^{\frac{n-1}{n}} \geqslant \frac{n \sqrt[n]{n-1}}{n-1}(n \geqslant 2) .
$$ | $$
\begin{array}{l}
\text { Prove }\left(\frac{x_{i}}{\sum_{j=1}^{n} x_{j}-x_{i}}\right)^{\frac{n-1}{n}} \\
=\frac{\sqrt[n]{n-1} x_{i}}{\sqrt[n]{(n-1) x_{i}\left(\sum_{j=1}^{n} x_{j}-x_{i}\right)^{n-1}}} \\
\geqslant \frac{n \sqrt[n]{n-1} x_{i}}{(n-1) x_{i}+(n-1)\left(\sum_{j=1}^{n} x_{j}-x_{i}\right)} \\
\quad=\frac{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,494 |
Example 3 ($n$-variable fractional inequality) Given $x_{1}, x_{2}$, $\cdots, x_{n} \in \mathbf{R}_{+}$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}} \geqslant \frac{A^{m-1}}{n^{m-2}(n-1)},
$$
where, $A=x_{1}+x_{2}+\cdots+x_{n}, m \geqslant 2$. | Proof
\[
\sum_{i=1}^{n} \frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}}+\frac{A^{m-1}}{(n-1) n^{m-2}}
\]
\[
\begin{array}{l}
=\sum_{i=1}^{n}\left[\frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}}+\frac{\left(\frac{A}{n}\right)^{m-2}\left(\sum_{j=1}^{n} x_{j}-x_{i}\right)}{(n-1)^{2}}\right] \\
\geqslant 2\left(\frac{A}{n}\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,495 |
As shown in Figure 1, in $\odot O$, chord $CD$ is perpendicular to diameter $AB$, $M$ is the midpoint of $OC$, the extension of $AM$ intersects $\odot O$ at point $E$, and $DE$ intersects $BC$ at point $N$. Prove: $BN = CN$.
(2009, National Junior High School Mathematics Competition Huanggang City Selection) | Proof 1: Using the corollary of the theorem of parallel lines dividing segments proportionally.
As shown in Figure 1, connect $M N$ and $C E$.
Since chord $C D$ is perpendicular to diameter $A B$,
$$
\begin{array}{l}
\Rightarrow \overparen{A C}=\overparen{A D} \Rightarrow \angle M E N=\angle B=\angle M C N \\
\Rightarr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,496 |
Question As shown in Figure 1, in the convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Draw any two lines through point $O$ that intersect sides $AD$, $BC$, $AB$, and $CD$ at points $E$, $F$, $G$, and $H$, respectively. Lines $GF$ and $EH$ intersect $BD$ at points $I$ and $J$. Prove:
$$
... | Let $\angle A O B=\angle C O D=\theta, \angle B O G=\angle D O H=\alpha, \angle B O F=\angle D O E=\omega$.
Then by the Angle Subtended Theorem, we have
$$
\begin{array}{l}
\frac{\sin \theta}{O G}=\frac{\sin \alpha}{O A}+\frac{\sin (\theta-\alpha)}{O B}, \\
\frac{\sin (\pi-\theta)}{O F}=\frac{\sin \omega}{O C}+\frac{\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,497 |
Let $a, b, x, y \in \mathbf{R}$, satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$.
$(1990$, American Mathematical Invitational) | Solution 1:
From
$a x^{3}+b y^{3}$
we get
$$
=\left(a x^{2}+b y^{2}\right)(x+y)-(a x+b y) x y,
$$
thus
$$
\begin{array}{l}
16=7(x+y)-3 x y . \\
\text { From } a x^{4}+b y^{4} \\
=\left(a x^{3}+b y^{3}\right)(x+y)-\left(a x^{2}+b y^{2}\right) x y,
\end{array}
$$
From $a x^{4}+b y^{4}$
we get
$$
42=16(x+y)-7 x y \text {... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,498 |
Example 4 If real numbers $a, b$ satisfy the conditions
$$
a^{2}+b^{2}=1,|1-2 a+b|+2 a+1=b^{2}-a^{2} \text {, }
$$
then $a+b=$ $\qquad$
(2009, National Junior High School Mathematics Joint Competition) | Given $a^{2}+b^{2}=1$, we have
$$
b^{2}=1-a^{2} \text {, and }-1 \leqslant a \leqslant 1,-1 \leqslant b \leqslant 1 \text {. }
$$
From $|1-2 a+b|+2 a+1=b^{2}-a^{2}$, we get
$$
\begin{array}{l}
|1-2 a+b|=b^{2}-a^{2}-2 a-1 \\
=\left(1-a^{2}\right)-a^{2}-2 a-1=-2 a^{2}-2 a .
\end{array}
$$
Thus, $-2 a^{2}-2 a \geqslant ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,499 |
1. Calculate
$$
\frac{2010^{2}-2009^{2}}{2010^{2}-2009 \times 2011+2 \times 2009}
$$
The value is ( ).
(A) 1
(B) -1
(C) 2009
(D) 2010 | - 1. A.
$$
\begin{array}{l}
\text { Original expression }=\frac{2010^{2}-2009^{2}}{2010^{2}-2009(2011-2)} \\
=\frac{2010^{2}-2009^{2}}{2010^{2}-2009^{2}}=1 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,500 |
3. As shown in Figure 2, in trapezoid $A B C D$, $A B / / D C, A B$ $\perp B C, E$ is the midpoint of $A D$, $A B+B C+C D=6, B E$ $=\sqrt{5}$. Then the area of trapezoid $A B C D$ is ( ).
(A) 13
(B) 8
(C) $\frac{13}{2}$
(D) 4 | 3. D.
As shown in Figure 10, draw $E F / / A B$ intersecting $B C$ at point $F$. Then
$$
\begin{array}{c}
B F=\frac{1}{2} B C, \\
E F=\frac{1}{2}(A B+C D)=\frac{1}{2}(6-B C) .
\end{array}
$$
Since $A B \perp B C$, it follows that $E F \perp B C$.
Therefore, in the right triangle $\triangle B F E$, we have
$$
\begin{a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,502 |
4. The graph of a certain linear function is parallel to the line $y=\frac{1}{2} x+3$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through point $C(-2,-4)$. Then, on the line segment $AB$ (including points $A$ and $B$), the number of points with both integer coordinates is (... | 4. B.
According to the problem, let the analytical expression of the linear function be
$$
y=\frac{1}{2} x+b \text {. }
$$
Since point $C(-2,-4)$ lies on the graph of this function, we have
$$
-4=\frac{1}{2} \times(-2)+b \Rightarrow b=-3 \text {. }
$$
Therefore, $y=\frac{1}{2} x-3$.
We can find points $A(6,0)$ and $... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,503 |
5. As shown in Figure 3, a circular paper piece with radius $r$ moves arbitrarily within an equilateral triangle with side length $a(a \geqslant 2 \sqrt{3} r)$. Then, within this equilateral triangle, the area of the part that the circular paper piece "cannot touch" is ( ).
(A) $\frac{\pi}{3} r^{2}$
(B) $\frac{(3 \sqrt... | 5. C.
As shown in Figure 11, when the circular paper piece moves to a position where it is tangent to the two sides of $\angle A$, draw perpendiculars from the center $O_{1}$ of the circular paper piece to the two sides, with the feet of the perpendiculars being $D$ and $E$, respectively, and connect $A O_{1}$. Then, ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,504 |
6. As shown in Figure 4, there are five playing cards with point values of $2,3,7,8,9$ respectively. If two cards are drawn at random, the probability that the product of their point values is an even number is $\qquad$ . | 2.6. $\frac{7}{10}$.
According to the problem, when the order of drawing cards is not considered, a tree diagram as shown in Figure 12 can be drawn.
From Figure 12, it can be seen that, when drawing two cards from five, there are a total of 10 possible draws, among which, the product of the numbers drawn is even in 7 ... | \frac{7}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,505 |
8. As shown in Figure 6, in $\triangle A B C$, the median $C M$ and the altitude $C D$ trisect $\angle A C B$. Then the measure of $\angle B$ is $\qquad$ | 8. $30^{\circ}$.
According to the problem, we have
$$
\begin{array}{l}
C D \perp A B, A M=M B, \\
\angle A C D=\angle M C D=\angle B C M .
\end{array}
$$
Since $\angle A C D=\angle M C D, C D \perp A M$, therefore,
$$
A D=D M=\frac{1}{2} A M \text {. }
$$
As shown in Figure 13, draw $M N \perp B C$ at point $N$.
Sin... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,507 |
9. There are $n$ consecutive natural numbers $1,2, \cdots, n$, after removing one of the numbers $x$, the average of the remaining numbers is 16. Then the values of $n$ and $x$ that satisfy the condition are $\qquad$ | 9. $n=30, x=1$ or $n=31, x=16$ or $n=$ $32, x=32$.
The sum of $n$ consecutive natural numbers is
$$
S_{n}=\frac{n(n+1)}{2} \text {. }
$$
If $x=n$, then the average of the remaining numbers is
$$
\frac{S_{n}-n}{n-1}=\frac{n}{2} \text {; }
$$
If $x=1$, then the average of the remaining numbers is
$$
\frac{S_{n}-1}{n-1}... | n=30, x=1 \text{ or } n=31, x=16 \text{ or } n=32, x=32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,508 |
10. Mother's Day is here, and Xiao Hong, Xiao Li, and Xiao Ying go to a flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the above flowers. Then she... | 10. 20 .
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $z$ yuan, respectively. Then
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text {. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text {. }
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,509 |
Example 5 Solve the inequality
$$
|| x-2|-6|+|3 x+2| \geqslant 8 \text {. }
$$ | Solve for the roots from $x-2=0, 3x+2=0, |x-2|-6=0$, we get
$$
x=2, x=-\frac{2}{3}, x=8, x=-4.
$$
(1) When $x \geqslant 8$, $4x-6 \geqslant 8$, solving gives $x \geqslant \frac{7}{2}$, thus $x \geqslant 8$;
(2) When $2 \leqslant x<8$, $2x+10 \geqslant 8$, solving gives $x \geqslant -1$, thus $2 \leqslant x<8$;
(3) When... | x \geqslant \frac{1}{2} \text{ or } x \leqslant -3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,510 |
12. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. | 12. By the Pythagorean theorem, we have
$$
c^{2}=a^{2}+b^{2} \text {. }
$$
Also, $c=\frac{1}{3} a b-(a+b)$, substituting into equation (1) gives
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,512 |
13. As shown in Figure 8, in $\triangle A B C$, $\angle A B C=45^{\circ}$, point $D$ is on side $B C$, $\angle A D C=60^{\circ}$, and $B D=\frac{1}{2} C D$. $\triangle A C D$ is reflected over line $A D$ to get $\triangle A C^{\prime} D$, and $B C^{\prime}$ is connected.
(1) Prove:
$$
B C^{\prime} \perp B C \text {; }
... | 13. (1) Since $\triangle A C^{\prime} D$ is obtained by reflecting $\triangle A C D$ over $A D$, we know that
$$
C^{\prime} D=C D, \angle A D C^{\prime}=\angle A D C \text {. }
$$
Since $B D=\frac{1}{2} C D, \angle A D C=60^{\circ}$, we have
$$
\begin{array}{l}
B D=\frac{1}{2} C^{\prime} D, \\
\angle B D C^{\prime}=18... | 75^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 722,513 |
14. (1) As shown in Figure 9 (a), within the square $ABCD$, two moving circles $\odot O_{1}$ and $\odot O_{2}$ are externally tangent to each other, and $\odot O_{1}$ is tangent to sides $AB$ and $AD$, while $\odot O_{2}$ is tangent to sides $BC$ and $CD$. If the side length of the square $ABCD$ is $1$, and the radii o... | 14. (1) (i) As shown in Figure 15, in the square $ABCD$, connect $AC$. Clearly, points $O_1$ and $O_2$ lie on $AC$, and
$$
\begin{array}{l}
AO_1 = \sqrt{2} r_1, \\
O_1O_2 = r_1 + r_2, \\
CO_2 = \sqrt{2} r_2
\end{array}
$$
From $AC = AO_1 + O_1O_2 + CO_2 = \sqrt{2}$, we get
$$
\sqrt{2} r_1 + r_1 + r_2 + \sqrt{2} r_2 = ... | \left(\frac{37}{8} - \frac{5\sqrt{3}}{2}\right) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,514 |
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