problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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1. Let $M=x+\frac{y z-x^{2}}{x^{2}+y^{2}+z^{2}}$ (where $x, y, z$ are distinct real numbers). If $x$ is swapped with $y$ or $z$, the value of $M$ remains unchanged. When $x+y+z=1$, the value of $M$ is ( ).
(A) positive
(B) zero
(C) negative
(D) cannot be determined | $-1 . B$.
From the problem, we have
$$
\begin{array}{l}
M=x+\frac{y z-x^{2}}{x^{2}+y^{2}+z^{2}}=y+\frac{x z-y^{2}}{y^{2}+x^{2}+z^{2}} \\
=z+\frac{y x-z^{2}}{z^{2}+y^{2}+x^{2}} .
\end{array}
$$
Then $(x-y)\left(x^{2}+y^{2}+z^{2}\right)$
$$
\begin{array}{l}
=z(x-y)+\left(x^{2}-y^{2}\right), \\
(y-z)\left(x^{2}+y^{2}+z^{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,751 |
2. As shown in Figure 1, given three equal circles $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ with a common point $H$, points $A$, $B$, and $C$ are the other intersection points of these circles. Then $H$ is the ( ) of $\triangle A B C$.
(A) Circumcenter
(B) Incenter
(C) Orthocenter
(D) Centroid | 2.C.
Connect $A H, B H, C H$. Since $\odot \mathrm{O}_{1}, \odot \mathrm{O}_{2},$
$\odot O_{3}$ are three equal circles, then
$$
\begin{array}{l}
\angle A B H=\angle A C H, \angle B A H=\angle B C H, \\
\angle C B H=\angle C A H . \\
\text { Therefore, } \angle B A H+\angle A B C=\angle B A H+\angle A B H+\angle C B H... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,752 |
3. Given that $x_{1}, x_{2}$ are the roots of the equation $x^{2}+k x+p=0$, and $x_{3}, x_{4}$ are the roots of the equation $x^{2}+k x+q=0$. Then $\left(x_{1}-x_{3}\right)\left(x_{2}-x_{4}\right)\left(x_{1}-x_{4}\right)\left(x_{2}-x_{3}\right)=$ ( ).
(A) $(p+q)^{2}$
(B) $p^{2}+q^{2}$
(C) $(q-p)^{2}$
(D) $q^{2}-p^{2}$ | 3. C.
From the problem, we know $x_{1}+x_{2}=-k, x_{1} x_{2}=p$,
$$
\begin{array}{l}
x_{3}+x_{4}=-k, x_{3} x_{4}=q, \\
x_{1}^{2}+k x_{1}+p=0, x_{2}^{2}+k x_{2}+p=0 .
\end{array}
$$
Then $\left(x_{1}-x_{3}\right)\left(x_{2}-x_{4}\right)\left(x_{1}-x_{4}\right)\left(x_{2}-x_{3}\right)$ $=\left[x_{1}^{2}-\left(x_{3}+x_{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,753 |
4. Let $2^{a}=3, 2^{b}=6, 2^{c}=12$. Given the following four relationships among the real numbers $a, b, c$:
(1) $a+c=2b$,
(2) $a+b=2c-3$,
(3) $b+c=2a+3$, (4) $b^{2}-ac=1$.
The number of correct relationships is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4.D.
Notice that $2^{a}=3$,
$$
\begin{array}{l}
2^{b}=6=3 \times 2=2^{a} \times 2=2^{a+1}, \\
2^{c}=12=6 \times 2=2^{b} \times 2=2^{b+1} \\
=3 \times 4=2^{a} \times 2^{2}=2^{a+2} .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } b=a+1, c=b+1 \\
\Rightarrow a=b-1, c=b+1 \\
\Rightarrow a+c=2 b, b^{2}-a c=1 .
\end{arr... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,754 |
5. Given the function $y=-x^{2}+a x-\frac{a}{4}+\frac{1}{2}(0 \leqslant$ $x \leqslant 1$ ) has a maximum value of 2. Then the range of the real number $a$ is ( ).
(A) $0 \leqslant a \leqslant 2$
(B) $a2$
(C) $a=3$ or $a=-2$
(D) $a=-6$ or $a=\frac{10}{3}$ | 5.D.
$$
\begin{array}{l}
y=-x^{2}+a x-\frac{a}{4}+\frac{1}{2} \\
=-\left(x-\frac{a}{2}\right)^{2}+\frac{a^{2}-a+2}{4} .
\end{array}
$$
(1) If $x=\frac{a}{2}$ is to the left of $0 \leqslant x \leqslant 1$, then $\frac{a}{2}<0 \Rightarrow a<0$.
If $x=\frac{a}{2}$ is to the right of $0 \leqslant x \leqslant 1$, then $\fra... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,755 |
2. If the three sides of $\triangle A B C$ are $\sqrt{2}, \sqrt{3}, \sqrt{5}$, then its inradius $r=$ | 2. $\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2}$.
It is easy to know that $(\sqrt{2})^{2}+(\sqrt{3})^{2}=(\sqrt{5})^{2}$. Therefore, $\triangle A B C$ $S=\frac{\sqrt{6}}{2}$
Also, $r \cdot \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}=\frac{\sqrt{6}}{2}$, so
$r=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2}$. | \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,756 |
6. As shown in Figure 2, in rectangle $A B C D$, $A B=\sqrt{2} A D$. A semicircle is drawn outside the rectangle $A B C D$ with $A B$ as the diameter. $P$ is a moving point on the semicircle, and $P C, P D$ intersect $A B$ at points $E, F$ respectively. Then the triangle formed by segments $A E$, $B F$, and $A B$ is ( ... | 6. B.
As shown in Figure 4, draw $P H \perp C D$ at $H$, intersecting $A B$ at point $G$. Let $A D=$
1. Then $A B=\sqrt{2}$. Let
$$
\begin{array}{l}
A E=x, B F=y, \\
P G=h, A G=z .
\end{array}
$$
Since $\triangle P D C \backsim$
$\qquad$ $\triangle P F E$, we have
$$
\frac{P G}{P \tilde{H}}=\frac{E F}{C D} \text {. }... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,757 |
1. Given that $a, b, c$ are the three side lengths of a triangle. The following conclusions are given:
(1) Three segments with lengths $a^{2}, b^{2}, c^{2}$ can form a triangle;
(2) Three segments with lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form a triangle;
(3) Three segments with lengths $|a-b|, |b-c|, |c-a|$ can ... | $=1$.(1)(3)(4)(5).
When $a=3, b=4, c=5$, $a^{2}+b^{2}=c^{2}$, then the lengths $a^{2}, b^{2}, c^{2}$ cannot form a triangle. Therefore, conclusion (1) is incorrect.
Assume $a>b>c>0$, and satisfies $a^2=b^2+c^2$, then $\sqrt{b}+\sqrt{c}>\sqrt{a}$.
Similarly, $\sqrt{c}+\sqrt{a}>\sqrt{b}, \sqrt{a}+\sqrt{b}>\sqrt{c}$.
Hen... | (1)(3)(4)(5) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,758 |
2. Given $\triangle A B C$ with side lengths $a=17, b=18, c$ $=19$, a point $O$ inside $\triangle A B C$ is drawn perpendiculars to the three sides of $\triangle A B C$, intersecting at points $D$, $E$, and $F$, such that $B D+C E+A F=27$. Then $B D+B F$ $=$ $\qquad$ | 2.18.
As shown in Figure 5, let \( BD = x \), \( CE = y \), \( AF = z \). Then
\[
\begin{array}{l}
CD = 17 - x, \\
AE = 18 - y, \\
BF = 19 - z.
\end{array}
\]
Connect \( OA \), \( OB \), \( OC \).
In right triangles \( \triangle OBD \), \( \triangle OCE \), \( \triangle OAF \),
\[
\left\{\begin{array}{l}
x^{2} + OD^{... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,759 |
3. A positive integer $M$, when added to 36, 300, and 596 respectively, results in the squares of three different positive integers. The sum of the smallest and the largest of these three different positive integers is twice the middle one. Then this positive integer $M=$ | 3.925 .
According to the problem, let $M+36=(k-m)^{2}$,
$M+300=k^{2}$,
$M+596=(k+m)^{2}(m>0)$.
Then $(M+36)+(M+596)=2\left(k^{2}+m^{2}\right)$.
Therefore, $M+316=M+300+16=k^{2}+m^{2}$.
Solving this, we get $m=4$.
$$
\begin{array}{l}
\text { Also, }(M+596)-(M+36) \\
=(k+m)^{2}-(k-m)^{2}=4 k m,
\end{array}
$$
Thus, $k=... | 925 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,760 |
4. Given that the three sides of $\triangle A B C$ are exactly three consecutive positive integers, and its perimeter and area are $p_{1}$ and $S_{1}$, respectively. If the three sides of $\triangle A B C$ are each increased by 10, the new $\triangle A^{\prime} B^{\prime} C^{\prime}$ has a perimeter and area of $p_{2}$... | 4. $\frac{3}{5}$.
Let the three sides of $\triangle A B C$ be $a-1, a, a+1$, and the three sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $b-1, b, b+1$, where $b-a=10$.
From the triangle area formula, we have
$$
\begin{array}{l}
S_{1} S_{2}=\sqrt{\frac{3}{2} a \cdot \frac{1}{2} a\left(\frac{1}{4} a^{2}-1\rig... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,761 |
One, (20 points) For what value of $a$ does the equation
$$
\frac{1}{\sqrt{(a-1) x^{2}+2 x-a}}=\frac{1}{\sqrt{2(a-1) x-2 a+10}}
$$
have exactly one root? | (1) When $a=1$, the original equation has a unique solution $x=\frac{9}{2}$.
(2) When $a \neq 1$, from equation (1) we get
$$
(a-1) x^{2}+2(2-a) x+a-10=0 \text {. }
$$
Its discriminant is
$$
\Delta=4(2-a)^{2}-4(a-1)(a-10)=4(7 a-6) \text {. }
$$
If $\Delta=0$, i.e., $a=\frac{6}{7}$, then $x=\frac{a-2}{a-1}=8$.
Substit... | a=1 \text{ or } a=\frac{6}{7} \text{ or } a \geqslant \frac{15}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,762 |
II. (25 points) As shown in Figure 3, in isosceles $\triangle ABC$, $D$ is a point on the base $BC$, and $P$ is a point on $AD$. The circumcircles of $\triangle CPD$ and $\triangle BPD$ intersect $AC$ and $AB$ at points $E$ and $F$, respectively, and $I$ is the incenter of $\triangle ABC$. Prove that when $PD^2 = PE \c... | As shown in Figure 6, draw $P R \perp B C, P S \perp C A, P T \perp A B$, with the feet of the perpendiculars being $R, S, T$ respectively.
Connect $R S, R T$.
It is easy to see that $B, R, P, T$ and $C, R, P, S$ are concyclic. Therefore,
$$
\begin{array}{l}
\angle R P T=180^{\circ}-\angle A B C \\
=180^{\circ}-\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,763 |
Three, (25 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Find all positive numbers $x$ such that $[x]$ and $\{x\}$ satisfy the relation $[x]^{2}=x\{x\}$.
Translate the above text into English, please retain the original text's line breaks and format, and output... | Three, from the problem, we have $x=[x]+\{x\}$. Therefore, the original equation can be transformed into
$$
[x]^{2}=([x]+\{x\})\{x\} \text {. }
$$
(1) Suppose $\{x\}=0$, then $[x]=0$. Thus, $x=0$, which contradicts the given that $x$ is a positive number.
(2) Suppose $\{x\} \neq 0$, then $0<\{x\}<1$. From $[x]^{2}=([x]... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,764 |
2. If $a-2$ is a positive integer and a divisor of $3 a^{2}-2 a+10$, then the sum of all possible values of $a$ is $\qquad$ . | 2.51.
$$
\begin{array}{l}
\text { Given } \frac{3 a^{2}-2 a+10}{a-2} \\
=\frac{\left(3 a^{2}-6 a\right)+(4 a-8)+18}{a-2} \\
=3 a+4+\frac{18}{a-2}(a \neq 2),
\end{array}
$$
we know that $a-2$ must be a divisor of 18.
Thus, $a=3,4,5,8,11,20$.
Therefore, the sum of all possible values of $a$ is
$$
3+4+5+8+11+20=51 \text ... | 51 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,766 |
3. Given real numbers $a, b, c$ satisfy
$$
a+b+c=11 \text { and } \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{13}{17} \text {. }
$$
Then the value of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is $\qquad$ . | 3. $\frac{92}{17}$.
Since $a=11-(b+c)$, we have,
$$
\begin{array}{l}
\frac{a}{b+c}=\frac{11-(b+c)}{b+c}=\frac{11}{b+c}-1 . \\
\text { Therefore, } \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \\
=\frac{11}{a+b}-1+\frac{11}{b+c}-1+\frac{11}{c+a}-1 \\
=11\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3 \\
=11 \time... | \frac{92}{17} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,768 |
4. Given that $x$ is a real number. Then the maximum value of $\sqrt{2000-x}+$ $\sqrt{x-2000}$ is $\qquad$ . | 4.4.
$$
\begin{array}{l}
\text { Let } t_{1}=\sqrt{2008-x}+\sqrt{x-2000} \text {. Then } \\
t_{1}^{2}=8+2 \sqrt{(2008-x)(x-2000)} \\
\leqslant 8+8=16 .
\end{array}
$$
Therefore, $t \leqslant 4$, i.e., $t_{\max }=4$. At this point, $x=2004$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,769 |
5. A ten-digit number, whose digits can only be 2 or 3, and there are no two adjacent 3s. How many such ten-digit numbers are there?
保留源文本的换行和格式，直接输出翻译结果。 | 5.144.
Solution 1: Consider all positive integers with the given property (using a recursive method).
The number of ten-digit numbers starting with 2 is the same as the number of nine-digit numbers;
Ten-digit numbers starting with 3 must start with 32, and their number is the same as the number of eight-digit number... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,770 |
6. There are $n(n>3)$ integers on a circle with a sum of 94, each of which is equal to the absolute value of the difference between the two numbers that follow it (in a clockwise direction). Then all possible values of $n$ are | 6.141 .
Obviously, these $n$ integers are all non-negative.
Let $a$ be the largest one, and the next four numbers be $b, c, d, e$ respectively. Then $a=|b-c|$.
Thus, $b=a$ or $c=a$.
When $b=a$, $c=0$.
From $b=|c-d|$, we know $d=a$.
Also, from $c=|d-e|$, we get $d=e$.
Therefore, $n$ is a multiple of 3.
Assume $n=3m$. T... | 141 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,771 |
7. Given that $M$ is a four-digit perfect square. If the thousand's digit of $M$ is reduced by 3 and the unit's digit is increased by 3, another four-digit perfect square can be obtained. Then the value of $M$ is $\qquad$ | 7.4761 .
Let $M=\overline{a b c d}=A^{2}$ and $M$'s thousands digit decreases by 3 and the units digit increases by 3 to equal $B^{2}$. Then
$$
\left\{\begin{array}{l}
A^{2}=1000 a+100 b+10 c+d, \\
B^{2}=1000(a-3)+100 b+10 c+(d+3) .
\end{array}\right.
$$
Therefore, $A^{2}-B^{2}=2997$.
Hence $(A-B)(A+B)=3^{4} \times 3... | 4761 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,772 |
8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then ... | 8.12.
Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,773 |
9. Given the function $f(x)=a x^{2}-c(a, c$ are real numbers). If $-4 \leqslant f(1) \leqslant-1,-1 \leqslant f(2) \leqslant 2$, then the maximum value of $f(8)$ is $\qquad$ . | 9.122.
Since $f(1)=a-c, f(2)=4a-c$, therefore, $a=\frac{1}{3}(f(2)-f(1)), c=\frac{1}{3}(f(2)-4f(1))$.
Thus, $f(8)=64a-c=21f(2)-20f(1)$.
Hence $-1=21 \times(-1)-20 \times(-1) \leqslant f(8)$ $\leqslant 21 \times 2-20 \times(-4)=122$.
When $f(x)=2x^{2}-6$, it satisfies the given conditions, and $f(8)=122$. Therefore, t... | 122 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,774 |
10. Two plane mirrors are placed intersecting at an angle of $30^{\circ}$, with the intersection point at $V$. A horizontal beam of light is emitted from a source $S$, parallel to one of the mirrors $V W$, and reflects at point $A$ on the other mirror $U V$, then reflects again at point $B$ on mirror $V W$, and then re... | $10.2+\sqrt{3}$.
As shown in Figure 11, because $S A / / W V$, combined with the optical principle that the angle of reflection equals the angle of incidence, the sizes of the angles are marked in the figure.
Because $\angle B C V=90^{\circ}$, the light beam will return along the original path to point $S$, so, the to... | 2+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,775 |
11. The smallest positive integer $n$ for which $n^{2}-n+11$ has four prime factors (not necessarily distinct) is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 11.132.
From Table 1, it can be verified that for any positive integer $n, n^{2}-$ $n+11$ is not a multiple of $2,3,5,7$.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $\bmod 2$ & $\bmod 3$ & \multicolumn{3}{|c|}{$\bmod 5$} & \multicolumn{3}{|c|}{$\bmod 7$} \\
\hline$n$ & 0 & 1 & 0 & 1 & 2 & ... | 132 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,776 |
12. As shown in Figure 3, $\triangle A B C$ is an equilateral triangle, point $C$ is on the side $D E$ of rectangle $A B D E$, the inradius of $\triangle A B C$ is 1. Then the diameter of the circumcircle of rectangle $A B D E$ is $\qquad$ | 12. $\sqrt{21}$.
Let point $O$ be the center of the incircle of $\triangle ABC$, and $P$ be the point where the incircle touches side $AB$. Connect $OB$ and $OP$. Then $OB=2$, $BP=\sqrt{3}$, $AB=2\sqrt{3}$.
Thus, $PC=3$.
Since quadrilateral $ABDE$ is a rectangle and $CP \perp AB$, we have $AE=3$.
By the Pythagorean t... | \sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,777 |
2. In $\triangle A B C$, $\angle A B C=60^{\circ}$, points $O$ and $H$ are the circumcenter and orthocenter of $\triangle A B C$, respectively. Points $D$ and $E$ are on sides $B C$ and $A B$, respectively, such that $B D=B H, B E=B O$. Given that $B O=1$, find the area of $\triangle B D E$. | 2. As shown in Figure 12, draw the diameter $A F$ of the circumcircle of $\triangle A B C$, and connect $C F$, $B F$, and $C H$.
Since $B H \perp A C$ and $F C \perp A C$, we have $B H \parallel F C$.
Similarly, $C H \parallel F B$.
Therefore, quadrilateral $B H C F$ is a parallelogram.
Furthermore, since $F O = C O$ ... | \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,780 |
3. Given $t \in \mathbf{N}_{+}$. If $2^{t}$ can be expressed as $a^{b} \pm 1$ (where $a$ and $b$ are integers greater than 1), please find all possible values of $t$ that satisfy the above condition. | 3. Let $t$ be a positive integer such that $2^{t}=a^{b} \pm 1$. Clearly, $a$ is odd.
(1) If $b$ is odd, then
$$
2^{t}=(a \pm 1)\left(a^{b-1} \mp a^{b-2}+a^{b-3} \mp \cdots \mp a+1\right) \text {. }
$$
Since both $a$ and $b$ are odd, the sum or difference of an odd number of odd numbers is always odd, so
$$
a^{b-1} \mp... | t=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,781 |
1. (40 points) If the simplest fraction $\frac{p}{q}$ is written in decimal form as $0 . a b a b a b a b \cdots$ (non-negative integers $a$ and $b$ can be equal, but at least one is non-zero), then, among the fractions that meet the condition, how many different numerators are there? | 1. Decimal numbers in the form of 0.abababab $\cdots$ can all be written as $\frac{k}{99}$ $\left(k \in \mathbf{N}_{+}, k=1,2, \cdots, 98\right)$. Among the numbers $1,2, \cdots$, 99, those that are multiples of 3 or 11 are
$$
\left[\frac{99}{3}\right]+\left[\frac{99}{11}\right]-\left[\frac{99}{3 \times 11}\right]=39 \... | 63 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,782 |
3. (40 points) In the $8 \times 8$ grid shown in Figure 5, draw a closed polyline such that:
(1) This polyline is only parallel or perpendicular to the edges of the small squares, and does not intersect itself;
(2) This polyline passes through each small square at most once, and must pass through all the small squares ... | 3. According to the rules of the question, first determine the broken lines that each circled square must connect, resulting in Figure 15 (a);
Based on Figure 15 (a), then connect the lines that must be connected, resulting in Figure 15 (b);
Finally, based on the results of Figure 15 (b), continue to add the segments... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,784 |
4. (40 points) A rectangular box with dimensions $a_{1} \times b_{1} \times c_{1}$ can fit into another rectangular box with dimensions $a_{2} \times b_{2} \times c_{2}$ if and only if $a_{1} \leqslant a_{2} 、 b_{1} \leqslant b_{2} 、 c_{1} \leqslant c_{2}$. Therefore, among the rectangular boxes with dimensions $a \tim... | 4. Among the boxes that meet the conditions, there are 5 types of boxes that are cubes, and 20 types of boxes that are square-based but not cubic prisms, because they have five different heights and four different choices for the square side length.
In addition, there are 10 types of boxes where the length, width, and... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,785 |
6. (40 points) On the blackboard, the numbers 1 and 4 are originally written. For the numbers on the blackboard, an operation can be performed: each time, any two numbers can be chosen from the blackboard, and a new number \( c = ab + a + b \) can be added to the blackboard. After several operations, what is the smalle... | 6. From $c_{1}=a b+a+b=(a+1)(b+1)-1$, we have
$$
c_{1}+1=(a+1)(b+1) \text {. }
$$
Taking numbers $a$ and $c_{1}$, we get
$$
c_{2}+1=(a+1)\left(c_{1}+1\right)=(a+1)^{2}(b+1) \text {; }
$$
Taking numbers $b$ and $c_{1}$, we get
$$
c_{2}+1=(b+1)\left(c_{1}+1\right)=(a+1)(b+1)^{2} \text {; }
$$
$\qquad$
Suppose after sev... | 2047 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,787 |
7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.
保留源文本的换行和格式，直接输出翻译结果。 | 7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$
Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$
Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,788 |
Example 1 Arrange the positive odd numbers in Table 1 into five columns.
Table 1
\begin{tabular}{|l|r|r|r|r|r|}
\hline & Column 1 & Column 2 & Column 3 & Column 4 & Column 5 \\
\hline Row 1 & & 1 & 3 & 5 & 7 \\
\hline Row 2 & 15 & 13 & 11 & 9 & \\
\hline Row 3 & & 17 & 19 & 21 & 23 \\
\hline Row 4 & 31 & 29 & 27 & 25 &... | Notice that each row is arranged with 4 numbers, the 1st column is left empty in odd rows, and the 5th column is left empty in even rows. We just need to calculate which number 2007 is.
From $2 n-1=2007$, we get $n=1004$.
From $1004 \div 4=251$, we know that 2007 is in the 251st row. Since it can be divided evenly, 200... | 251st \text{ row and } 5th \text{ column} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,789 |
Example 2 Arrange all fractions of the form $\frac{m}{n}\left(m, n \in \mathbf{N}_{+}\right)$ in a row according to the following rules:
(1) If $m_{1} n_{1}<m_{2} n_{2}$, then $\frac{m_{1}}{n_{1}}$ is placed before $\frac{m_{2}}{n_{2}}$;
(2) If $m_{1} n_{1}=m_{2} n_{2}$ and $n_{1}<n_{2}$, then $\frac{m_{1}}{n_{1}}$ is ... | (1) For positive integers $m, n$, according to the rule, in $\frac{m}{n}$, $mn=1$, only $1 \times 1=1$; $mn=2$, only $1 \times 2=2$. Therefore, we can write down the sequence according to this rule:
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{1}{3}, \frac{4}{1}, \frac{2}{2}, \frac{1}{4}, \frac{5}{1}, \... | \frac{3}{2}, 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,790 |
3. A sequence of numbers, the first three numbers are $1, 9, 9$, and each subsequent number is the remainder of the sum of the three preceding numbers divided by 3. What is the 1999th number in this sequence? | (Tip: Apart from the first three numbers $1,9,9$, this sequence repeats every 13 terms (i.e., $1,1,2,1,1,1,0,2,0,2,1,0,0$). Since $1999-3=13 \times 153+7$, the 1999th number is the 7th number in the 154th cycle, which is exactly 0.) | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,791 |
$242 \triangle D E F$ is the triangle formed by the external angle bisectors of $\triangle A B C$, and the incircles of $\triangle B C D$, $\triangle C A E$, and $\triangle A B F$ touch the sides of $\triangle A B C$ at points $X$, $Y$, and $Z$, respectively. Prove that $D X$, $E Y$, and $F Z$ are concurrent. | Proof: As shown in Figure 3, let the circumradius of $\triangle ABC$ be $R$, $AB = c$, $BC = a$, $CA = b$. Let $p_A$, $p_B$, and $p_C$ represent the semiperimeters of $\triangle BCD$, $\triangle CAE$, and $\triangle ABF$, respectively.
By the definition of internal and external angle bisectors, we have
$$
\begin{array... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,792 |
4. Arrange the odd numbers in the order shown in Figure 4, from top to bottom, referred to as rows. In the figure, the number 11 is the 2nd number from left to right in the 3rd row, and the number 29 is the 6th number in the 4th row. Therefore, 2003 is the
$\qquad$th row and the
$\qquad$th number. | (提示: 2003 is the 1002nd number when counting from top to bottom and from left to right in the number array, the first $n$ rows contain $1+3+\cdots+(2 n-1)=n^{2}$ numbers, and $31^{2}=961, 32^{2}=1024$, so 2003 is located in the 32nd row as the $1002-961=41$st number.) | 32\text{nd row, 41st number} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,793 |
5. Observe the array:
(1),(3,5),(7,9,11),(13,15,17,19), .....
In which group is 2003? | (Tip: Use the trial method. The first 45 groups have a total of $1+2+\cdots$ $+45=1035$ numbers. The last number of the 45th group is 2069, and the first number is 1981, so, 2003 is in the 45th group.) | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,794 |
Example 1 As shown in Figure 4, from a point $P$ outside a circle, two secants $P A B$ and $P C D$ are drawn, intersecting the circle at points $A, B, C,$ and $D$. Chords $A D$ and $B C$ intersect at point $Q$. A secant $P E F$ passing through point $Q$ intersects the circle at points $E$ and $F$. Prove:
$$
\frac{1}{P ... | Explanation: As shown in Figure 4, let $O$ be the center of the circle through which the secant line passes. According to the Power of a Point theorem, we can take a point $M$ on the ray $PF$ such that
$$
\begin{array}{l}
P A \cdot P B=P C \cdot P D \\
=P E \cdot P F=P Q \cdot P M .
\end{array}
$$
Connect $O B, O D, O... | \frac{1}{P E}+\frac{1}{P F}=\frac{2}{P Q} | Geometry | proof | Yes | Yes | cn_contest | false | 720,795 |
Example 2 In a convex quadrilateral $ABCD$, the diagonals intersect at point $P$, and the lines containing the two pairs of opposite sides intersect at points $Q$ and $R$. A line through $P$ intersects $AB$, $CD$, and $QR$ at points $M$, $N$, and $G$ respectively. Prove that $\frac{1}{MP} + \frac{1}{MG} = \frac{2}{MN}$... | Explanation: As shown in Figure 5, let $AC$ intersect $QR$ at point $S$. Then $S, P, C, A$ form a harmonic range. Therefore,
$$
\frac{SA}{AP} = \frac{SC}{CP}.
$$
By applying Menelaus' theorem to $\triangle PSG$ cut by line $QCN$ and to $\triangle PSG$ cut by line $AMQ$, we get
$$
\begin{array}{l}
\frac{PN}{NG} \cdot \... | \frac{1}{MP} + \frac{1}{MG} = \frac{2}{MN} | Geometry | proof | Yes | Yes | cn_contest | false | 720,796 |
For example, $3 A B$ is the diameter of $\odot O$, $C$ is a point on $A B$, a perpendicular line to $A B$ through point $C$ intersects $\odot O$ at point $D$, a tangent line to $\odot O$ through point $D$ intersects the extension of $A B$ at point $E, P$ is any point on $\odot O$. Prove: $\angle B P C=\angle B P E$.
-... | Explanation: As shown in Figure 6,
Connect $D A$, $D B$, and $A P$.
It is easy to prove that
$\angle A D B=90^{\circ}$,
$\angle B D E=\angle B A D$
$=\angle B D C$.
Figure 6
Thus, by Proposition 1, (1) and (4) $\Rightarrow(2)$ and (3), we know that $C$ and $E$ harmonically divide segment $A B$.
Also, $\angle A P B=90^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,797 |
Example 4. Let the two pairs of opposite sides of a convex quadrilateral $ABCD$ intersect at points $E$ and $F$, and the intersection of the two diagonals be point $P$. Draw $PO \perp EF$ at point $O$. Prove: $\angle BOC = \angle AOD$.
| As shown in Figure 7, extend $A C$ and $D B$ to intersect $E F$ at points $Q$ and $R$. If $B D$ is parallel to $E F$, then the point $R$ is at infinity.
$$
\angle P O A = \angle P O C, \angle P O B = \angle P O D .
$$
Therefore, $\angle B O C = \angle A O D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,798 |
Example 5 Let $O$ and $I$ be the circumcenter and incenter of $\triangle ABC$, respectively. The incircle of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Line $DF$ intersects $CA$ at point $P$, and line $DE$ intersects $AB$ at point $Q$. Points $M$ and $N$ are the midpoi... | As shown in Figure 8, it is easy to prove that \(\frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A}=1\).
Therefore, \(A D\), \(B E\), and \(C F\) are concurrent.
By Proposition 2, \(P\), \(E\), \(A\), and \(C\) form a harmonic range.
Since \(M\) is the midpoint of segment \(P E\), by conclusion (6) of the p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,799 |
Example 6 In an acute triangle $\triangle ABC$, the altitudes from vertices $A, B, C$ intersect the opposite sides at points $D, E, F$ respectively. A line through point $D$ parallel to $EF$ intersects $AC, AB$ at points $Q, R$ respectively, and $EF$ intersects $BC$ at point $P$. Prove that the circumcircle of $\triang... | As shown in Figure 9, take the midpoint $M$ of side $B C$.
From Proposition 2, we know that $P, D, B, C$ form a harmonic range.
Since $M$ is the midpoint of $B C$, we have:
$$
D M \cdot D P = D B \cdot D C.
$$
It is easy to prove that points $B, C, E, F$ are concyclic.
Since $R Q \parallel E F$, we have:
$$
\angle R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,800 |
1. Given $A$ is a point outside $\odot O$, a secant through $A$ intersects $\odot O$ at points $B$ and $C$, with point $B$ lying on the segment $AC$ internally. Two tangents are drawn from point $A$ to $\odot O$, touching the circle at points $S$ and $T$. Let $AC$ intersect $ST$ at point $P$. Prove:
$$
\frac{A P}{P C}=... | (Draw a perpendicular from point $O$ to $BC$ at point $H$. It is easy to prove that points $A, S, H, T$ are concyclic. Then $\triangle APT \sim \triangle ATH$, which means $AT^2 = AP \cdot AH = AB \cdot AC$. Since $H$ is the midpoint of $BC$, according to condition (4) in Judgment 1, $A, P, B, C$ form a harmonic range.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,801 |
Given a sequence of numbers:
$$
\begin{array}{l}
1, \frac{1}{2}, \frac{2}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{2}{3}, \frac{1}{3}, \\
\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{3}{4}, \frac{2}{4}, \frac{1}{4}, \cdots
\end{array}
$$
Question: (1) What is the position of $\frac{5}... | (1) Observing, we find that there is only 1 number with a denominator of 1, 3 numbers with a denominator of 2, 5 numbers with a denominator of 3, $\cdots \cdots$ and $2n-1$ numbers with a denominator of $n$. Therefore, the numbers with denominators from $1 \sim 7$ total
$$
(1+3+5+\cdots+13)=\frac{14 \times 7}{2}=49 \te... | 54 \text{ or } 60, \frac{4}{15} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,802 |
2. Given that the circumcenter of $\triangle A B C$ is $O, P$ is a point on the extension of $O A$, line $l$ is symmetric to $P B$ with respect to $B A$, line $h$ is symmetric to $P C$ with respect to $A C$, and $l$ intersects $h$ at point $Q$. If $P$ moves on the extension of $O A$, find the locus of $Q$. | (Let $l$ intersect $O A$ at point $R$, and let the other intersection of $O A$ with $\odot O$ be $D$. Then $\angle A B D=90^{\circ}$. Therefore, $B A$ and $B D$ are the internal and external angle bisectors of $\angle P B R$, respectively. Thus, $P, R, A, D$ form a harmonic range. Since $\angle A C D=90^{\circ}$, $C A$... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,803 |
3. In $\triangle A B C$, the circle passing through points $B$ and $C$ intersects sides $A C$ and $A B$ at another points $E$ and $F$ respectively. $B E$ and $C F$ intersect at point $P$, $A P$ intersects $B C$ at point $D$, and $M$ is the midpoint of side $B C$, with $D$ and $M$ not coinciding. Prove that $D$, $M$, $E... | (It is obvious that $E F$ is not parallel to $B C$. Let the intersection point of $E F$ and $B C$ be $Q$. Then $Q, D, C, B$ form a harmonic range. Since $M$ is the midpoint of $B C$, we have $Q M \cdot Q D = Q C \cdot Q B$. Using the secant theorem, we get $Q C \cdot Q B = Q E \cdot Q F$. Therefore, $Q M \cdot Q D = Q ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,804 |
4. In quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$. Take a point $E$ on $CD$, and let $BE$ intersect $AC$ at point $F$. Extend $DF$ to intersect $BC$ at point $G$. Prove:
$$
\angle GAC = \angle EAC.
$$ | (Let $A C$ intersect $B D$ and $G E$ at points $H$ and $M$, and extend $G E$ to intersect $B D$ at point $N$. Then $B, D, H, N$ form a harmonic range. Since $A C$ bisects $\angle B A D$, we have $A H \perp A N$. Also, $G, E, M, N$ form a harmonic range, and $A M \perp A N$, so $A C$ bisects $\angle G A E$. Therefore, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,805 |
5. In $\triangle A B C$, $A B>A C$, its incircle touches side $B C$ at point $E$, and connects $A E$ intersecting the incircle at point $D$ (different from point $E$). Take a point $F$ on segment $A E$ different from $E$ such that $C E=C F$, and connect $C F$ and extend it to intersect $B D$ at point $G$. Prove: $C F=F... | (Draw the tangent line of the incircle through point $D$, intersecting $A B$, $A C$, and $B C$ at points $M$, $N$, and $K$ respectively. It is easy to prove that $M K \parallel C G$. By Newton's theorem, we know that $B N$, $C M$, and $A E$ are concurrent, so $B$, $C$, $E$, and $K$ form a harmonic range. Therefore, $B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,806 |
Example 1 Given $n(n \geqslant 2)$ lines in a plane, where no two are parallel and no three are concurrent, they divide the plane into several regions. Try to prove: it is possible to fill each region with a non-zero integer whose absolute value does not exceed $n$, such that the sum of the numbers in all regions on th... | When $n=2$, there is essentially only one construction (as shown in Figure 1, taking $k=1$ is the simplest).
When $n=3$, let the numbers filled in each region be $a, b, c, d, e, f, g$ (as shown in Figure 2). From $e+b+g=0=e+c+f$, we get $b+g=c+f$. Then, from $e+c+f=b+g+a+d$, we get $e=a+d$.
Similarly, $f=a+b, g=c+a$.
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,807 |
Example 3 MO brand football is made of several polygonal leather pieces sewn together with three different colored threads, and it satisfies the following conditions:
(1) Any side of a polygonal leather piece is exactly sewn to another side of the same length of another polygonal leather piece using the same color thre... | Analysis: Consider the tetrahedron \(ABCD\), and let the numbers at the four vertices be \(a, b, c, d\). Then,
\[abc = bcd = cda = dab = 1.\]
From \(abc = 1\), we get \(abcd = d\).
Similarly, \(a = abcd\), \(b = abcd\), \(c = abcd\).
Therefore, \(a = b = c = d\).
Substituting the above into equation (1), we get
\[a = b... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,809 |
Given $x 、 y 、 z \geqslant 0$, and $x+y+z=1$. Then
$$
0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} .
$$ | $$
x^{2}+y^{2}+z^{2}=1-2(x y+y z+z x) \text {, }
$$
we get
$$
\begin{array}{l}
\frac{13}{27} \leqslant x^{2}+y^{2}+z^{2}+4 x y z \\
=1-2(x y+y z+z x)+4 x y z,
\end{array}
$$
rearranging gives problem 2. | 0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 720,810 |
Question 4 Let $T$ be a triangle with a perimeter of 2, and let $x, y, z$ be the lengths of the sides of $T$. Then
$$
x y z+\frac{28}{27} \geqslant x y+y z+z x .
$$ | Question 1 $\rightarrow$ Question 4: From Question 1 and the formula
$$
k\left(x^{2}+y^{2}+z^{2}\right)+4 x y z \geqslant \frac{13}{27} k^{3},
$$
taking $k=2$ and
$$
k^{2}=2^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x) \text {, }
$$
we get $2\left[2^{2}-2(x y+y z+z x)\right]+4 x y z \geqslant \frac{13}{27} \time... | x y z+\frac{28}{27} \geqslant x y+y z+z x | Inequalities | proof | Yes | Yes | cn_contest | false | 720,812 |
Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th num... | Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,813 |
1. Given $\frac{a b}{a+b}=\frac{1}{15}, \frac{b c}{b+c}=\frac{1}{17}, \frac{c a}{c+a}=\frac{1}{16}$. Then the value of $\frac{a b c}{a b+b c+c a}$ is ( ).
(A) $\frac{1}{21}$
(B) $\frac{1}{22}$
(C) $\frac{1}{23}$
(D) $\frac{1}{24}$ | 1.D.
From the given, we have
$$
\frac{1}{a}+\frac{1}{b}=15, \frac{1}{b}+\frac{1}{c}=17, \frac{1}{c}+\frac{1}{a}=16 \text {. }
$$
Adding the three equations yields $2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=48$.
Therefore, the original expression $=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{24}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,816 |
2. As shown in Figure 1, the lengths of the sides of rectangle $A B C D$ are $A B=4$, $B C=8$. The rectangle is folded so that point $C$ coincides with point $A$. Then the length of the crease $E F$ is ( ).
(A) $3 \sqrt{2}$
(B) $2 \sqrt{5}$
(C) $\frac{15}{2}$
(D) $\frac{16}{3}$ | 2.B.
In Figure 1, connect $A C$ to intersect $E F$ at point $O$.
By the Pythagorean theorem, we know $A C=4 \sqrt{5}$.
When the rectangle is folded so that point $C$ coincides with $A$, we have $E F \perp A C$, then
$$
\begin{array}{l}
\text { Rt } \triangle E O C \backsim \mathrm{Rt} \triangle A B C \\
\Rightarrow \f... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,817 |
4. As shown in Figure 2, given that $P$ is a point inside a square $A B C D$ with side length $a$, and $\triangle P B C$ is an equilateral triangle. Then the circumradius of $\triangle P A D$ is ( ).
(A) $a$
(B) $\sqrt{2} a$
(C) $\frac{\sqrt{3}}{2} a$
(D) $\frac{1}{2} a$ | 4.A.
As shown in Figure 6, let the circumcenter of $\triangle P A D$ be $O$, and connect $O A$, $O P$, and $O D$.
Since $\triangle P A D$ and $\triangle O A D$ are both isosceles triangles, $O P$ is the perpendicular bisector of $A D$. Therefore, $O P \parallel A B$.
Given $\angle A B P=30^{\circ}$, then $\angle O P ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,819 |
5. As shown in Figure 3, the radius of the base of the cone $R=3 \mathrm{dm}$, the slant height $l=5 \mathrm{dm}$, $AB$ is the diameter of the base, $C$ is a point on the circumference of the base, $\angle COB=150^{\circ}$, $D$ is a point on $VB$, $VD$ $=\sqrt{7} \mathrm{dm}$. There is an ant that crawls from point $C$... | 5.B.
As shown in Figure 7, if the cone is cut along the generatrix $V C$ and then flattened onto the plane $V A B$, point $C$ will land on $C^{\prime}$. Therefore, the shortest path that the ant should take is the straight line $C^{\prime} D$.
According to the arc length formula of a sector, in the base sector $O B C... | 4 \sqrt{2} | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,820 |
6. If $a, b$ are integers, and the equation $a x^{2}+b x-$ $2008=0$ has two distinct prime roots, then the value of $3 a+b$ is ( ).
(A) 100
(B) 400
(C) 700
(D) 1000 | 6.D.
Let the roots of the equation be $x_{1}, x_{2}\left(x_{1}<x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} x_{2}=-\frac{2008}{a},
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
b=-a\left(x_{1}+x_{2}\right), \\
a x_{1} x_{2}=-2008=-2 \times 2 \times 2 \times 251.
\en... | 1000 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,821 |
1. Given real numbers $a>0, b>0$, satisfying $a+\sqrt{a}=2008, b^{2}+b=2008$. Then the value of $a+b$ is $\qquad$ | $=, 1.2008$
For the convenience of calculation, let $2008=m$.
From $a+\sqrt{a}-m=0$, we get
$$
\sqrt{a}=\frac{-1+\sqrt{1+4 m}}{2} \text{, }
$$
which means $a=\frac{1+2 m-\sqrt{1+4 m}}{2}$.
From $b^{2}+b-m=0$, we get
$$
b=\frac{-1+\sqrt{1+4 m}}{2} \text{. }
$$
Therefore, $a+b=m=2008$. | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,822 |
3. Three people, A, B, and C, play the "Rock, Paper, Scissors" game at the same time. The game rules are: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. The probability that A wins (ties are not counted) is $\qquad$ | 3. $\frac{1}{9}$.
A, B, and C each have 3 possible moves, so there are a total of $3 \times 3 \times 3=27$ possible outcomes. A wins (ties not considered) in only 3 scenarios: A throws rock, B and C throw scissors; A throws scissors, B and C throw paper; A throws paper, B and C throw rock. Therefore, the probability o... | \frac{1}{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,825 |
Three, (18 points) Given that the linear function $y=A x+B$ and the inverse proportion function $y=\frac{k}{x}$ intersect at points $M(2,3)$ and $N(-4, m)$.
(1) Find the expressions for the linear function $y=A x+B$ and the inverse proportion function $y=\frac{k}{x}$;
(2) Find the area of $\triangle M O N$. | Three, (1) Given that $M(2,3)$ is in the first quadrant, we know that $N(-4, m)$ is in the second or third quadrant. Draw the schematic diagram of the linear function $y=A x+B$ $(A>0)$ and the inverse proportion function $y=\frac{k}{x}(k>0)$ (as shown in Figure 10).
Since $M(2,3)$ and $N(-4, m)$ are both on the graph ... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,827 |
Four. (18 points) Given the quadratic function
$$
y=x^{2}-2 m x-2 m^{2}(m \neq 0)
$$
whose graph intersects the $x$-axis at points $A$ and $B$, and its vertex lies on the circle with $A B$ as its diameter.
(1) Prove: $A$ and $B$ are two distinct intersection points on the $x$-axis;
(2) Find the analytical expression o... | (1) Since $\Delta=4 m^{2}+8 m^{2}=12 m^{2}>0$, therefore, $A$ and $B$ are two distinct intersection points on the $x$-axis.
(2) Construct the parabola
$$
y=x^{2}-2 m x-2 m^{2}
$$
(see Figure 11). Let the intersection points with the $x$-axis be $A\left(x_{1}, 0\right)$ and
$$
B\left(x_{2}, 0\right)\left(x_{1}<x_{2}\rig... | \frac{2 \sqrt{6}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,828 |
Five. (15 points) As shown in Figure 5, given that $AB$ is a chord of $\odot O$, a line parallel to $AB$ is drawn through $O$ and intersects $\odot O$ at point $C$, and intersects the tangent line of $\odot O$ at point $B$ at point $D$. Prove:
$$
\angle ACB = \angle D \text{. }
$$ | Five, as shown in Figure 12, draw diameter $A M$, and connect $O B$, $B M$. Then $A B / / O D$, that is,
$$
\begin{array}{l}
\angle A B O \\
=\angle B O D .
\end{array}
$$
But $\angle A B O$
$$
=\angle B A O \text {, }
$$
then $\angle B A O=\angle B O D$.
Since $A M$ is the diameter and $B D$ is the tangent of the ci... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,829 |
5. Given that $\alpha, \beta$ satisfy $\alpha \lg \alpha=1004, \beta \cdot 10^{\beta}=1004$. Then $\alpha \beta$ equals ( ).
(A) $2 \sqrt{1004}$
(B) 1004
(C) $2 \sqrt{2008}$
(D) 2008 | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,834 |
Example 6 Arrange positive integers starting from 1 according to the pattern shown in Figure 2, where 2 is at the first turn, 3 is at the second turn, 5 is at the third turn, 7 is at the fourth turn, … then the number at the 2007th turn is $\qquad$ | To explore the relationship between the number of turns and the numbers at the turns, let $a_{n}$ represent the number corresponding to the $n$-th turn. If we observe in the order of natural numbers, then
$$
\begin{array}{l}
a_{1}=2, a_{2}=3, a_{3}=5, a_{4}=7, a_{5}=10, \\
a_{6}=13, a_{7}=17, a_{8}=21, \cdots \cdots
\e... | 1008017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,835 |
$$
\begin{array}{l}
6 \cdot \cos 0+\cos \frac{\pi}{7}+\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}+ \\
\cos \frac{4 \pi}{7}+\cos \frac{5 \pi}{7}+\cos \frac{6 \pi}{7}=(\quad) .
\end{array}
$$
(A) 4
(B) 3
(C) 2
(D) 1 | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,836 |
3. In the circle of Figure 2, chords $A B$ and $C D$ intersect perpendicularly at point $E$. If the lengths of segments $A E$, $E B$, and $E D$ are $2 \mathrm{~cm}$, $6 \mathrm{~cm}$, and $3 \mathrm{~cm}$ respectively, find the area of this circle. | 3. $\frac{65 \pi}{4}$ | \frac{65 \pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,839 |
11.2. Given that the weight ratio of any two out of $N$ weights belongs to $\left[\frac{4}{5}, \frac{5}{4}\right]$, and these $N$ weights can be divided into 10 groups of equal weight, as well as 11 groups of equal weight. Find the minimum possible value of $N$. | 11.2. $N_{\min }=50$.
An example for $N=50$.
In fact, 20 weights of $50 \mathrm{~g}$ and 30 weights of $40 \mathrm{~g}$ satisfy the requirement. These 50 weights can be divided into 10 equal groups, each consisting of 2 weights of $50 \mathrm{~g}$ and 3 weights of $40 \mathrm{~g}$. They can also be divided into 11 equa... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,845 |
Example 7 Arrange the squares of positive integers $1, 2, \cdots$ in a sequence:
$149162536496481100121 \cdots$.
The digit in the first position is 1, the digit in the fifth position is 6, the digit in the tenth position is 4, the digit in the 2008th position is $\qquad$ | To determine the digit at the 2008th position, we can only calculate by segmenting according to the number of digits in the squares.
$1^{2}$ to $3^{2}$, each result occupies 1 digit, totaling $1 \times 3=3$ digits;
$4^{2}$ to $9^{2}$, each result occupies 2 digits, totaling $2 \times 6=12$ digits;
$10^{2}$ to $31^{2}$,... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,846 |
11.3. Let $P$ be a set consisting of a finite number of primes. Prove: there exists a positive integer $x$, such that $x$ can be expressed as the sum of two positive integers raised to a prime power if and only if this prime belongs to $P$. | 11.3. First, prove the following lemma.
Lemma Let $p$ be a prime number, and $n$ be a positive integer. Then there exist positive integers $a, b$ such that $2^{n}=a^{p}+b^{p}$ if and only if $p \nmid (n-1)$.
Proof of the lemma: If $n-1=k p$, then
$$
2^{n}=\left(2^{k}\right)^{p}+\left(2^{k}\right)^{p} \text {. }
$$
Co... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,847 |
11.4. Given that each face of a tetrahedron can be covered by a unit disk. Prove: this tetrahedron can be covered by a sphere of radius $\frac{3}{2 \sqrt{2}}$. | 11.4. Note the following facts: the smallest circle covering a non-obtuse triangle is its circumcircle; the smallest circle covering an obtuse triangle is the circle with the largest side of the obtuse triangle as its diameter.
Let \( O \) and \( R \) be the center and radius of the circumsphere of a tetrahedron, resp... | R \leqslant \frac{3}{2\sqrt{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 720,848 |
11.5. Can the integers $51 \sim 150$ be placed in a $10 \times 10$ table such that for any two adjacent cells (i.e., they share a common edge) with numbers $a$ and $b$, at least one of the equations $x^{2}-a x+b=0$ and $x^{2}-b x+a=0$ has two integer roots. | 11.5. Not possible.
Assume there is a way to satisfy the requirement. Let $770$. This indicates that $1, a$ are its roots, and $b=1+a$.
If the equation $x^{2}-a x+b=0$ has two integer roots $x_{1}, x_{2}$, then $x_{1}+x_{2}=a, x_{1} x_{2}=b$;
If there are also $x_{1}, x_{2} \geqslant 2$, then
$$
b=x_{1}\left(a-x_{1}\... | Not possible. | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,849 |
11.6. A magician performs a guessing game with the audience to determine the area of a convex 2008-gon. A convex 2008-gon is drawn in a place where only the audience can see it. The magician specifies points on the polygon, either by naming vertices or by specifying points that divide a given side in a given ratio. Eac... | 11.6. Let $A_{1} A_{2} \cdots A_{2008}$ be a convex 2008-gon, $A=A_{2008}$. The diagonals from point $A$ divide the polygon into 2006 triangles:
$\triangle A A_{1} A_{2}, \triangle A A_{2} A_{3}, \cdots, \triangle A A_{2006} A_{2007}$.
Let $A M_{1}, A M_{2}, \cdots, A M_{2006}$ be the medians of these triangles.
We wi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,850 |
11.7. In a convex quadrilateral $A B C D$, $P$ and $Q$ are the intersections of the rays $B A$ and $C D$, and the rays $B C$ and $A D$, respectively. $H$ is the projection of point $D$ onto the line $P Q$. Prove: The quadrilateral $A B C D$ has an incircle if and only if the angle between the two tangents from $H$ to t... | 11.7. Let $\omega_{1}$ and $\omega_{2}$ be the incircles of $\triangle A D P$ and $\triangle C D Q$, respectively, with $I_{1}$ and $I_{2}$ being their centers, and $r_{1}$ and $r_{2}$ being their radii.
Since circles $\omega_{1}$ and $\omega_{2}$ are homothetic with respect to point $D$, we have
$$
\frac{r_{1}}{r_{2}}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,851 |
11.8.2n+3 players participate in a round-robin rapid chess tournament. The tournament proceeds and the requirements are as follows: no two games are held simultaneously; between any two games a player participates in, there are at least $n$ games. Prove: In such a tournament, there is a player who participates in both ... | 11.8. Label all the matches that all players participate in, in chronological order, as the 1st, 2nd, ..., $\mathrm{C}_{2 \mathrm{n}+3}^{2}$-th match. The difference in the match numbers of two consecutive matches that the same player participates in is called the "interval".
According to the tournament rules, all pla... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,852 |
1. Calculate $\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008}$ $=(\quad)$.
(A) 1
(B) -1
(C) 2008
(D) 2009 | -1.А.
Let $19492009=a, 2008=b$. Then
$$
\begin{array}{l}
\text { Original expression }=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b} \\
=\frac{b^{2}+2 b+1-b^{2}}{a^{2}-a^{2}+1+2 b}=\frac{2 b+1}{2 b+1}=1 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,853 |
2. When $m$ is a negative real number, the equation about $x$
$$
x^{2}-2 x+1+m\left(\frac{1}{x}-2\right)=0
$$
(A) has three distinct real roots
(B) has two distinct real roots
(C) has one real root
(D) has no real roots | 2.C.
The original equation can be transformed into $\frac{m}{x}=-(x-1)^{2}+2 m$. Let $y_{1}=\frac{m}{x}, y_{2}=-(x-1)^{2}+2 m$.
Draw the rough graphs of functions $y_{1}$ and $y_{2}$ (as shown in Figure 4). $y_{1}$ passes through the point $(1, m)$, and the vertex of $y_{2}$ is $(1,2 m)$. From the graph, it can be se... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,854 |
3. In Rt $\triangle A B C$, $\angle C=90^{\circ}, A C>B C$. $D$ is a point on side $A C$ ( $D$ does not coincide with $A$ or $C$), and satisfies $C D \cdot C A>C B^{2}$. A line $l$ is drawn through point $D$ to intersect $\triangle A B C$, such that the intersected triangle is similar to $\triangle A B C$. Then the num... | 3.C.
There are 3 lines $l$ that satisfy the conditions. As shown in Figure 5(a), among them, $l_{1} / / B C, l_{2} \perp A B, l_{3} / / A B$.
Figure 5
Note: If the line $l_{4}$ passing through point $D$ forms an acute angle $\alpha$ with $D C$ equal to $\angle B$, it does not intersect side $C B$, but intersects the e... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,855 |
4. If $\sqrt{m^{2}+1203}$ is an integer, then the sum of all positive integers $m$ that satisfy the condition is ( ).
(A)401
(B) 800
(C) 601
(D) 1203 | 4.B.
Let $\sqrt{m^{2}+1203}=n\left(n \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
n^{2}-m^{2}=1203 \\
\Rightarrow(n+m)(n-m)=401 \times 3=1203 \times 1 .
\end{array}
$$
Therefore, $\left\{\begin{array}{l}n+m=401, \\ n-m=3\end{array}\right.$ or $\left\{\begin{array}{l}n+m=1203, \\ n-m=1 .\end{array}\right.$
Sol... | 800 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,856 |
Example 8 List the positive integers that satisfy the condition "contain at least one digit 0 and are multiples of 4" in ascending order: $20,40,60,80,100,104, \cdots$ Then the 158th number in this sequence is ( ).
(A) 2000
(B) 2004
(C)2 0008
(D) 2012 | Explanation: In positive integers, the characteristic of being a multiple of 4 is that the last two digits are multiples of 4, with a total of 25 cases (including 7 cases with the digit 0: $00,04,08,20,40,60$, 80; and 18 cases without 0: $12,16,24,28,32$, $36,44,48,52,56,64,68,72,76,84,88,92$, 96).
The problem require... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,857 |
5. As shown in Figure $1, P$ is a point outside the square $A B C D, P A$ $=\sqrt{2}, P B=4$. Then the maximum length of $P D$ is ( ).
(A)5
(B) $4+\sqrt{2}$
(C) 6
(D) $4+\sqrt{3}$ | 5.C.
As shown in Figure 6, $\triangle P A D$ is rotated $90^{\circ}$ clockwise around point $A$ to get $\triangle P^{\prime} A B$. Therefore, $\triangle P^{\prime} A B \cong \triangle P A D$. Hence, $P^{\prime} B = P D$,
$$
\begin{array}{l}
P^{\prime} A = P A = \sqrt{2}, \\
\angle P A P^{\prime} = 90^{\circ}.
\end{arr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,858 |
6. Among the following statements, the correct option is ( ).
(A) A sample has only one mode.
(B) The median of a sample is always one of the numbers in the sample.
(C) The variance of a sample is always a positive number.
(D) Fans A, B, and C decide who gets the only ticket by drawing lots. They prepare three pieces o... | 6.D.
Option (A) is incorrect. For example, the mode of the sample $3,3,5,5,1,2$ has two numbers, 3 and 5.
Option (B) is incorrect. For example, the median of the sample $1,2,3,4$ is $\frac{2+3}{2}=2.5$, and 2.5 is not a data point in the sample.
Option (C) is incorrect. For example, in the sample $80,80,80,80,80$, t... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 720,859 |
1. Given the quadratic equation in $x$
$$
(n+1)(n+2) x^{2}+x-n(n+3)=0
$$
with two real roots $a_{n}$ and $b_{n}\left(n \in \mathbf{N}_{+}\right)$. Then the value of $a_{1} a_{2} \cdots a_{2000} \cdot b_{1} b_{2} \cdots b_{2000}$ is $\qquad$ | $$
=1 -\frac{1006}{3015} \text {. }
$$
According to Vieta's formulas, we have
$$
\begin{array}{l}
a_{n} b_{n}=-\frac{n(n+3)}{(n+1)(n+2)} . \\
\text { Therefore, } a_{1} a_{2} \cdots a_{2009} \cdot b_{1} b_{2} \cdots b_{2009} \\
=a_{1} b_{1} \cdot a_{2} b_{2} \cdots \cdot a_{2009} b_{2009} \\
=\left(-\frac{1 \times 4}{... | -\frac{1006}{3015} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,860 |
2. In quadrilateral $A B C D$, $\angle A: \angle B: \angle C$ : $\angle D=9: 12: 10: 17, A B=B C, C D=2$. Then the length of $B C$ is $\qquad$. | $2.1+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}$.
From the given information, we have
$$
\begin{array}{l}
\angle A=\frac{9}{9+12+10+17} \times 360^{\circ}=67.5^{\circ}, \\
\angle B=\frac{12}{9+12+10+17} \times 360^{\circ}=90^{\circ}, \\
\angle C=\frac{10}{9+12+10+17} \times 360^{\circ}=75^{\circ} .
\end{array}
$$
As shown ... | 2.1+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,861 |
3. The algebraic expression
$$
\sqrt{x^{2}+6 x+13}+\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}-4 y+5}
$$
reaches its minimum value when $x$ and $y$ are $\qquad$ | 3. $x=-1, y=1$.
Original algebraic expression
$$
\begin{array}{c}
=\sqrt{(x+3)^{2}+(0+2)^{2}}+ \\
\sqrt{(x-0)^{2}+(0-y)^{2}}+ \\
\sqrt{(0-1)^{2}+(y-2)^{2}} .
\end{array}
$$
As shown in Figure 9, construct the figure with $A(-3,-2) 、 B(x, 0)$ 、 $C(0, y) 、 D(1,2)$.
Then the original algebraic expression $=A B+B C+C D \... | x=-1, y=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,862 |
4. As shown in Figure $2, \odot O$ is internally tangent to $\odot O^{\prime}$ at point $P, \odot O$'s chord $A B$ is tangent to $\odot O^{\prime}$ at point $C$, and $A B / / O O^{\prime}$. If the area of the shaded part is $4 \pi$, then the length of $A B$ is $\qquad$ | 4.4.
As shown in Figure 10, connect $O^{\prime} C$ and $O A$. Draw $O D \perp A B$ at $D$. Then quadrilateral $O O^{\prime} C D$ is a rectangle.
Therefore, $O^{\prime} C=O D$.
Thus, $S_{\text {fill }}$
$=\pi O A^{2}-\pi O^{\prime} C^{2}$ $=\pi\left(O A^{2}-O D^{2}\right)=\pi A D^{2}=4 \pi$.
So, $A D=2$.
By the Perpend... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,863 |
One, (20 points) Given the function
$$
y=k^{2} x^{2}+k\left(2 x-3 x^{2}\right)+2 x^{2}-2 x+1
$$
the graph does not pass through the fourth quadrant. Find the range of values for the constant $k$. | The original function can be transformed into
$$
y=(k-1)(k-2) x^{2}+2(k-1) x+1 \text {. }
$$
(1) When the function is a quadratic function, it can be known from the graph not passing through the fourth quadrant that the parabola must open upwards. There are two possibilities: one is that the parabola has no intersectio... | k \leqslant 1 \text{ or } k \geqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,864 |
II. (25 points) As shown in Figure 3, $P Q$ is tangent to $\odot O$ at point $Q$, and $Q$ is the midpoint of $\widehat{A B}$. Chord $A C = B C$, the extension of $C A$ intersects $P Q$ at point $D$, $O E \perp B C$ at point $E$, connect $D E$, and $F$ is the midpoint of $D E$. Given that $Q F = 2$, $A C = 6$. Find the ... | II. As shown in Figure 11, extend $PQ$ and $CB$ to intersect at $G$.
It is easy to see that $\overparen{QAC}$ is a semicircle. Connect $QC$, then $QC$ is the diameter of $\odot O$.
$$
\begin{array}{c}
\text { By } P Q \perp C Q, \\
\angle A C Q = \angle B C Q \\
\Rightarrow \triangle D C Q \cong \triangle G C Q \Right... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,865 |
Three, (25 points) Given that the difference between the product and the sum of two positive integers is equal to twice the sum of their greatest common divisor (GCD) and least common multiple (LCM). Find all such pairs of positive integers.
保留源文本的换行和格式，直接输出翻译结果。 | Let the required positive integers be $x, y$, and their greatest common divisor be $m\left(m \in \mathbf{N}_{+}\right)$. Then
$$
x=m a, y=m b\left(a, b \in \mathbf{N}_{+},(a, b)=1\right) \text {. }
$$
Thus, the least common multiple of $x, y$ is $m a b$.
According to the problem, we have
$$
m a \cdot m b-(m a+m b)=2(m... | 4,12 \text{ or } 6,6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,866 |
1. Given the complex number
$$
(x-2)+(y-2) \mathrm{i} \quad (x, y \in \mathbf{R})
$$
the modulus of which equals $x \cos 45^{\circ}+y \sin 45^{\circ}+11$. Then the locus of the moving point $P(x, y)$ is ( ).
(A) circle (B) ellipse (C) hyperbola (D) parabola | -、1.D.
From the given, we have
$$
\sqrt{(x-2)^{2}+(y-2)^{2}}=\frac{|x+y+\sqrt{2}|}{\sqrt{2}} \text {. }
$$
This indicates that the moving point $P(x, y)$ is equidistant from the fixed point $F(2,2)$ and the fixed line $x+y+\sqrt{2}=0$. According to the definition, the locus of the moving point $P(x, y)$ is a parabola. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,867 |
1. The digit at the 2007th position after the decimal point of the irrational number $0.2342343423434342343434342 \cdots$ is $\qquad$ . | (Observation: Note the position of the digit 2 after the decimal point. The $n$th 2 is at the $n^{2}$th position after the decimal point. Since $44^{2}<2007<45^{2}$, the digit at the 2007th position after the decimal point is between the 44th and 45th 2, with 44 instances of 34 in between. The digit 3 is in the odd pos... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,868 |
2. Given a point $P$ outside the plane of a convex quadrilateral $A B C D$, and $E 、 F 、 G 、 H 、 M 、 N$ are the midpoints of $A B$ 、 $P C 、 A D$ 、 $B C$ 、 $E F$ 、 $G H$ respectively. Then ( ).
(A) $P 、 D 、 M 、 N$ are coplanar, and $P D=4 M N$
(B) $P 、 D 、 M 、 N$ are not coplanar, and $P D=$ $4 M N$
(C) $P 、 D 、 M 、 N$ ... | 2.A.
As shown in Figure 3, take the midpoint $Q$ of $CD$, and connect the midpoints of the sides of quadrilateral $ABCD$ in sequence to form parallelogram $GEHQ$, then connect $QE$.
By the property that the diagonals of a parallelogram bisect each other, we know that $QE$ passes through the midpoint $N$ of $GH$, and ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,869 |
3. Rolling a die twice yields the numbers $m, n$, and the vector $\boldsymbol{a}=(m, n)$ is formed. Then the probability that the angle between $\boldsymbol{a}$ and the vector $\boldsymbol{b}=$ $(1,-1)$ is an angle within a right triangle is ( ).
(A) $\frac{5}{12}$
(B) $\frac{1}{2}$
(C) $\frac{7}{12}$
(D) $\frac{3}{4}$ | 3. C.
Solution 1: Since $m, n$ can take $1,2, \cdots, 6$, the vector $\boldsymbol{a}$ has $6 \times 6=36$ possible positions. When $\boldsymbol{a}$ and $\boldsymbol{b}=(1,-1)$ form an angle that is an interior angle of a right triangle,
$$
\begin{array}{l}
\cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\bolds... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,870 |
4. Given a differentiable function $f(x)(x \in \mathbf{R})$ that satisfies $f^{\prime}(x)>f(x)$. Then when $a>0$, the relationship between $f(a)$ and $\mathrm{e}^{a} f(0)$ is ( ).
(A) $f(a)\mathrm{e}^{a} f(0)$
(C) $f(a)=e^{a} f(0)$
(D) Cannot be determined, depends on $f(x)$ or $a$ | 4.B.
Let the function $F(x)=\frac{f(x)}{\mathrm{e}^{x}}(x \in \mathbf{R})$, then $F^{\prime}(x)=\frac{f^{\prime}(x)-f(x)}{\mathbf{e}^{x}}>0$.
Therefore, $F(x)$ is a monotonically increasing function on $\mathbf{R}$. Thus, for $a>0$, we have
$$
F(a)>F(0) \Rightarrow \frac{f(a)}{\mathrm{e}^{a}}>\frac{f(0)}{\mathrm{e}^{0... | B | Calculus | MCQ | Yes | Yes | cn_contest | false | 720,871 |
5. For any integer $n$, the remainder when the algebraic expression $n^{2}-n+4$ is divided by 9 will not be ( ).
(A) 0
(B) 1
(C) 6
(D) 7 | 5.A.
When $n=3$, the remainder can be 1;
When $n=4$, the remainder can be 7;
When $n=5$, the remainder can be 6.
The following proves that $n^{2}-n+4$ cannot be divisible by 9.
If not, then
$9 \mid\left[(n-2)^{2}+3 n\right]$.
Since $3 \mid 3 n$, it follows that $3 \mid(n-2)^{2}$.
But 3 is a prime number, so
$3 \mid (n... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,872 |
6. Each point on the circle with diameter $AB$ is colored one of red, yellow, or blue. It is known that $A$ and $B$ are colored red. Connecting points on the circle forms triangles. Given the following 4 conclusions:
(1) There must exist a right triangle with all three vertices colored red;
(2) There must exist a right... | 6.A.
It is known that connecting points on a circle forms a right-angled triangle if and only if the hypotenuse is the diameter. Now let's discuss the diameter.
If there are no red points on the circle other than points $A$ and $B$, then conclusion (1) does not hold; if there are no red points on the circle other tha... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,873 |
3. Given $M=\left\{x \left\lvert\, \frac{a x-6}{x^{2}-a}<0\right.\right\}$, and $\left\{\begin{array}{l}2 \in M, \\ 3 \notin M .\end{array}\right.$ Then the range of values for $\boldsymbol{a}$ is $\qquad$ . | 3. $[2,3) \cup(4,9]$.
From $2 \in M$, we have $\frac{2a-6}{2^2-a}<0$.
Solving this, we get $a \in (-\infty, 3) \cup (4, +\infty)$.
Also, from $3 \notin M$, we have $\frac{3a-6}{3^2-a} \geqslant 0$ or $a=9$.
Solving this, we get $a \in [2,9]$.
Combining these, we get $[2,3) \cup (4,9]$. | [2,3) \cup (4,9] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,876 |
4. Given a geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=-2 \text {. }
$$
Then the range of values for $a_{1}$ is $\qquad$ | 4. $(-4,-2) \cup(-2,0)$.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q$.
From the existence of the limit of partial sums, we have
$$
\left\{\begin{array}{l}
0<|q|<1, \\
\frac{a_{1}}{1-q}=-2 .
\end{array}\right.
$$
Then $0<\left|a_{1}+2\right|<2$.
Solving this, we get $a_{1} \in(-4,-2) \cup(-... | (-4,-2) \cup(-2,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,877 |
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