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
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class | __index_level_0__ int64 0 742k |
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6. Let $a>b>0$. Then the minimum value of $a^{3}+\frac{12}{b(a-b)}$ is $\qquad$ .
| 6.20
$$
\begin{array}{l}
a^{3}+\frac{12}{b(a-b)} \geqslant a^{3}+\frac{12}{\frac{(b+a-b)^{2}}{4}} \\
=a^{3}+\frac{48}{a^{2}}=\frac{a^{3}}{2}+\frac{a^{3}}{2}+\frac{16}{a^{2}}+\frac{16}{a^{2}}+\frac{16}{a^{2}} \geqslant 20 .
\end{array}
$$ | 20 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,800 |
Three, (20 points) (1) Given $a+\log _{2}(2 a+6)=11$ and $b+2^{b-1}=14$. Find the value of $a+b$.
(2) Given $f(x)=\frac{2}{2^{x-2}+1}$. Find
$$
f(-1)+f(0)+f(1)+f(2)+f(3)+f(4)+f(5)
$$
the value. | (1) Let $t=\log _{2}(2 a+6)$, then we have $t+2^{t-1}=14$. Since the function $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, we have $b$ $=t$. Therefore, $a+t=11$, which means $a+b=11$.
(2) Since $f(x)+f(4-x)=\frac{2}{2^{x-2}+1}+\frac{2}{2^{2-x}+1}$
$$
=\frac{2\left(2^{x-2}+1\right)}{2^{x-2}+1}=2 \text {, ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,801 |
Four. (20 points) As shown in Figure 1, the base of the pyramid $P-ABCD$ is a square, $PA \perp$ the base $ABCD$, $PA = AD = 2$, points $M$ and $N$ are on edges $PD$ and $PC$ respectively, and $PC \perp$ plane $AMN$.
(1) Prove that $AM \perp PD$;
(2) Find the size of the dihedral angle $P-AM-N$;
(3) Find the angle betw... | (1) Since quadrilateral $ABCD$ is a square, therefore, $CD \perp AD$.
Also, since $PA \perp$ the base $ABCD$, so, $PA \perp CD$. Hence, $CD \perp$ plane $PAD$.
Since $AM \subset$ plane $PAD$, then $CD \perp AM$.
And since $PC \perp$ plane $AMN$, we have $PC \perp AM$. Thus, $AM \perp$ plane $PCD$. Therefore, $AM \perp ... | \arccos \frac{\sqrt{3}}{3}, \arcsin \frac{\sqrt{3}}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 716,802 |
Five. (20 points) As shown in Figure 2, in the Cartesian coordinate system, a line segment $P Q$ of length 6 has one endpoint $P$ sliding on the ray $y=0(x \leqslant 0)$, and the other endpoint $Q$ sliding on the ray $x=0(y \leqslant 0)$. Point $M$ is on line segment $P Q$, and $\frac{P M}{M Q}=\frac{1}{2}$.
(1) Find t... | (1) Let $P\left(x_{1}, 0\right)$, $Q\left(0, y_{1}\right)$, and $M(x, y)$, where $x \leqslant 0$ and $y_{1} \leqslant 0$.
From the conditions, we have
$$
x_{1}^{2}+y_{1}^{2}=36 \text{. }
$$
Also, given $\lambda=\frac{P M}{M Q}=\frac{1}{2}$, we get
$$
x=\frac{x_{1}}{1+\frac{1}{2}}, y=\frac{\frac{1}{2} y_{1}}{1+\frac{1}... | 4 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,803 |
Six. (20 points) As shown in Figure 3, let
$$
P_{1}(1, \sqrt{3}), P_{2}(4,2 \sqrt{3}), \cdots, P_{n}\left(x_{n}, y_{n}\right)
$$
$\left(0<y_{1}<y_{2}<\cdots<y_{n}\right)$ be $n$ points on the curve $C: y^{2}=3 x(y \geqslant 0)$. Points $A_{i}(i=1,2, \cdots, n)$ are on the positive $x$-axis, satisfying that $\triangle A... | (1) Let point $A_{n}\left(a_{n}, 0\right)$. Using the graph, we can find that the x-coordinate of the point is the average of the x-coordinates of points $A_{n-1}$ and $A_{n}$, i.e., $x_{n}=\frac{a_{n-1}+a_{n}}{2}$. Therefore, the y-coordinate of point $P_{n}$ is $y_{n}=\sqrt{3\left(\frac{a_{n-1}+a_{n}}{2}\right)}$.
At... | 88 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,804 |
3. As shown in Figure 2, in $\odot O$, $MN$ is a chord, and the sides $AB$ and $GH$ of the squares $ABCD$ and $EFGH$ are also chords. Points $D, E, F, C$ are all on $MN$. If the distance from $O$ to $MN$ is $OS=h$, then $AB-HG$ equals ( ).
(A) $\frac{4}{3} h$
(B) $\frac{7}{4} h$
(C) $\frac{8}{5} h$
(D) $\frac{11}{6} h$ | 3.C.
As shown in Figure 7, it is easy to see that $S$ is the midpoint of $MN$, and also the midpoint of $DC$ and $EF$.
Obviously, points $H$, $S$, and $B$ are collinear.
Draw $OK \perp HB$ at $K$, then $BK=HK$.
Also, $\frac{SK}{OK}=\frac{HE}{ES}=2$, thus, $SK=\frac{2}{\sqrt{5}} h$.
From $2 SK=BS-HS$
$$
=\frac{\sqrt{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,807 |
Example 1 Given that $P Q R S$ is a cyclic quadrilateral, $\angle P S R=90^{\circ}$, a perpendicular is drawn from point $Q$ to $P R$ and $P S$, with the feet of the perpendiculars being points $H$ and $K$ respectively. Prove: $H K$ bisects $Q S$.
| Proof 1: As shown in Figure 1, let $H K$ intersect $Q S$ at point $T$, then
$$
\begin{array}{l}
\angle T S K \\
=90^{\circ}-\angle R S Q \\
=90^{\circ}-\angle R P Q \\
=\angle T K S .
\end{array}
$$
Therefore, $T S=T K$.
Also, $\angle T Q K=90^{\circ}-\angle T S K$
$$
=90^{\circ}-\angle T K S=\angle T K Q \text {, }
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,808 |
Example 2 Given an acute triangle $\triangle A B C$, the circle with $A B$ as its diameter intersects the altitude $C C^{\prime}$ and its extension at points $M, N$, and the circle with $A C$ as its diameter intersects the altitude $B B^{\prime}$ and its extension at points $P, Q$. Prove that $M, P, N, Q$ are concyclic... | Proof: As shown in Figure 3, since $AB$ and $AC$ are diameters of the two circles, the perpendicular bisectors of $MN$ and $PQ$ are $AB$ and $AC$, respectively. Therefore,
$$
\begin{array}{l}
AM=AN, \\
AP=AQ.
\end{array}
$$
Connect $AM, BM$.
In the right triangle $\triangle ABM$, $MC'$ is the altitude to the hypotenus... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,809 |
Example 3 Let $P$ be a point inside $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to intersect the opposite sides at points $A'$, $B'$, and $C'$, respectively. Then
$$
\frac{AP}{PA'} \cdot \frac{BP}{PB'} \cdot \frac{CP}{PC'} \geqslant 8 .
$$ | Proof: Let's denote
$$
\begin{array}{l}
S_{\triangle A P B}=S_{1}, S_{\triangle A N T}=S_{2}, S_{\triangle P A}=S_{3} \text{.} \\
\text{Then } \frac{A P}{P A^{\prime}}=\frac{S_{3}}{S_{\triangle P^{\prime} C}}=\frac{S_{1}}{S_{\triangle P A^{\prime} B}} \\
=\frac{S_{1}+S_{3}}{S_{2}} \geqslant \frac{2 \sqrt{S_{1} S_{3}}}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,810 |
II. (25 points) As shown in Figure 4, in the Cartesian coordinate system, the vertices of rectangle $O A B C$ are $A$ and $C$ on the $x$-axis and $y$-axis, respectively, and vertex $B$ is in the first quadrant. Points $E$ and $F$ are on $O A$ and $A B$, respectively. When $\triangle A E F$ is folded along $E F$, point ... | II. Since $y=a x^{2}-14 a x+49 a+4=a(x-7)^{2}+4$, therefore, $P(7,4)$.
As shown in Figure 9, we easily get
$$
\begin{array}{l}
A(14,0), \\
B(14,8), \\
C(0,8) .
\end{array}
$$
Let $A F^{\prime}=x(x$ be an integer), then
$$
\begin{array}{l}
D F=x, B F=8-x, \\
E D=E A=14-O E \\
=14-B D \\
=14-\sqrt{x^{2}-(8-x)^{2}}=14-4 ... | a \leqslant -\frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,811 |
Three, (25 points) Given a three-digit number $\overline{a b c}$ that satisfies $a+$ $b+c=a b c$. Find the sum of all such three-digit numbers.
| Three, from the problem, we get that $a$, $b$, and $c$ are all non-zero, and without loss of generality, assume $a \geqslant b \geqslant c$.
When $b c>3$, $a+b+c=a b c>3 a$, which means $b+c>2 a$. This contradicts $b+c \leqslant 2 a$.
When $b c=3$, then $b=3, c=1$. In this case, $a=2$, which contradicts $a \geqslant b... | 1332 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,812 |
1. Given
$$
A=\{n \in \mathbf{N} \mid 1 \leqslant n \leqslant 2006 \text { and }(n+4,30) \neq 1\} \text {. }
$$
then $\operatorname{card}(A)=(\quad)$.
(A) 1605
(B) 1537
(C) 1471
(D) 1404 | -1.C.
Since $30=2 \times 3 \times 5, (n+4,30) \neq 1$, therefore, $n+4=2k$ or $3k$ or $5k\left(k \in \mathbf{N}_{+}\right)$.
Also, $1 \leqslant n \leqslant 2006$, then $5 \leqslant n+4 \leqslant 2010$. Thus, there are $\left[\frac{2010}{2}\right]-2=1003$ numbers of the form $2k$, $\left[\frac{2010}{3}\right]-1=669$ nu... | 1471 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,813 |
3. The following 4 propositions are given:
(1) In $\triangle A B C$, $\sin A+\sin B-\sin C$ is always positive;
(2) In $\triangle A B C$, $\cos A+\cos B+\cos C$ is always positive;
(3) In $\triangle A B C$, $\cot A+\cot B+\cot C$ is always positive;
(4) In a non-right $\triangle A B C$, $\tan A+\tan B+\tan C$ is always... | 3. B.
(1) Correct. By the Law of Sines, we have
$$
\sin A+\sin B-\sin C=\frac{1}{2 R}(a+b-c)>0 \text {. }
$$
(2) Correct. When $\triangle A B C$ is an acute or right triangle, the algebraic expression is clearly positive. When $\triangle A B C$ is an obtuse triangle, assume without loss of generality that $\angle C$ is... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,815 |
4. Given a sequence of positive terms $\left\{a_{n}\right\}$ satisfying $a_{n}^{n}+a_{n}-1=$ $0\left(n \in \mathbf{N}_{+}\right)$. Below are 5 statements:
$$
\text { (1) } 0<a_{n} \leqslant \frac{n}{n+1} \text {; }
$$
(2) $\frac{n}{n+1} \leqslant a_{n}<1$;
(3) $\left\{a_{n}\right\}$ is an increasing sequence;
(4) $\lef... | 4. A.
Obviously, $0n a_{n}-(n-1)$
$\Rightarrow a_{n}^{n}+a_{n}-1>(n+1) a_{n}-n$.
Thus, $0<a_{n}<\frac{n}{n+1}$.
Also, $a_{1}=\frac{1}{2}$, so, $0<a_{n} \leqslant \frac{n}{n+1}\left(n \in \mathbf{N}_{+}\right)$.
If for some positive integer $n$, $a_{n} \geqslant a_{n+1}$, then $a_{n+1}^{n+1}=a_{n+1}^{n} a_{n+1}<a_{n+1}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,816 |
5. Given two ellipses $C_{1}$ and $C_{2}$, and $C_{1}$ is inside $C_{2}$. Let the equation of the inner ellipse $C_{1}$ be $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>$ $n>0)$, and the equation of the outer ellipse $C_{2}$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$. Draw a tangent line to $C_{1}$ at any ... | 5. D.
Let $M\left(x_{0}, y_{0}\right)$, then
$$
l_{P Q}: \frac{x_{0} x}{m^{2}}+\frac{y_{0} y}{n^{2}}=1 \text {. }
$$
Suppose $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right), R\left(x_{R}, y_{R}\right)$. The equations of the tangents at points $P 、 Q$ are respectively
$$
\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,817 |
6. Given point $P(x, y)$ lies on the curve $C$:
$$
(x \cos \theta + y \sin \theta)^{2} + x \sin \theta - y \cos \theta = 1
$$
Then the minimum value of $|O P|$ (where $O$ is the origin) is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1 | 6.C.
Let \( u = x \cos \theta + y \sin \theta, v = x \sin \theta - y \cos \theta \).
Then \( u^{2} + v = 1 \), and we have
\[
\begin{array}{l}
u^{2} + v^{2} = (x \cos \theta + y \sin \theta)^{2} + (x \sin \theta - y \cos \theta)^{2} \\
= x^{2} + y^{2} .
\end{array}
\]
\[
\begin{array}{l}
\text { Hence } |OP| = \sqrt{x... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,818 |
1. The range of the function $f(x)=\sin x+\cos x+\tan x+$ $\arcsin x+\arccos x+\arctan x$ is $\qquad$ . | $=、 1 \cdot\left[-\sin 1+\cos 1-\tan 1+\frac{\pi}{4}, \sin 1+\cos 1+\tan 1+\frac{3 \pi}{4}\right]$.
Obviously, $f(x)=\sin x+\cos x+\tan x+\arctan x+\frac{\pi}{2}$, $x \in[-1,1]$.
Below we prove: $g(x)=\cos x+\tan x$ is an increasing function on $[-1,1]$.
Let $x_{1} 、 x_{2} \in[-1,1]$, and $x_{1}0$, thus,
$$
\frac{\cos... | [-\sin 1+\cos 1-\tan 1+\frac{\pi}{4}, \sin 1+\cos 1+\tan 1+\frac{3 \pi}{4}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,819 |
2. Given $x, y \in \mathbf{R}_{+}, x^{n}+y^{n}=1(n>1, n$ $\in \mathbf{N}_{+}$). Then the maximum value of $x+4 y$ is | 2. $\left(1+4 i^{n-1}\right)^{\frac{n-1}{n}}$.
$$
\begin{array}{l}
\text { Since } x^{n}+\underbrace{a^{n}+a^{n}+\cdots+a^{n}}_{n-1 \uparrow} \geqslant n a^{n-1} x, \\
y^{n}+\underbrace{b^{n}+b^{n}+\cdots+b^{n}}_{n-1 \uparrow} \geqslant n b^{n-1} y,
\end{array}
$$
where $a, b \in \mathbf{R}_{+}$, thus
$$
\begin{array}... | \left(1+4^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,820 |
Example 4 with the same conditions as Example 3. Prove:
$$
S_{\triangle^{\prime} K C C} \leqslant \frac{1}{4} S_{\triangle A B C} .
$$ | Let $A B^{\prime}=b_{1}, B^{\prime} C=b_{2}, C A^{\prime}=a_{1}$, $A^{\prime} B=a_{2}, B C^{\prime}=c_{1}, C^{\prime} A=c_{2}$. Then
$$
\frac{S_{\triangle B C}}{S_{\triangle A H C}}=\frac{b_{1} c_{2}}{b c} \text {. }
$$
Similarly, we can find the other corresponding ratios. Thus,
$$
\begin{array}{l}
=1-\frac{b_{1} c_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,821 |
3. Find the sum: $\sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(\mathrm{C}_{n+1}^{k+1}+\mathrm{C}_{n+1}^{k+2}+\cdots+\mathrm{C}_{n+1}^{n+1}\right)$ $=$ | $3.2^{2 n}$.
$$
\begin{aligned}
P & =\sum_{k=0}^{n} C_{n}^{k}\left(C_{n+1}^{k+1}+C_{n+1}^{k+2}+\cdots+C_{n+1}^{n+1}\right) \\
& =\sum_{k=0}^{n} C_{n}^{k}\left[2^{n+1}-\left(C_{n+1}^{0}+C_{n+1}^{1}+\cdots+C_{n+1}^{k}\right)\right] \\
& =\sum_{k=0}^{n}\left(C_{n}^{k} \cdot 2^{n+1}\right)-\sum_{k=0}^{n} C_{n}^{k}\left(C_{... | 2^{2 n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,822 |
4. Given a tetrahedron $P-ABC$, $PA=a$, $PB=$ $b$, $PC=c$, the projection of $P$ on the base $ABC$ is $O$, which is the centroid of $\triangle ABC$, and $PO=h$. Then $V_{P-ABC}=$ $\qquad$ . | 4. $\sqrt{3 h^{4}-2 h^{2}\left(a^{2}+b^{2}+c^{2}\right)+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}} \cdot h$. As shown in Figure 1, according to the problem, we have
$$
\begin{array}{l}
A O=\sqrt{a^{2}-h^{2}}, \\
B O=\sqrt{b^{2}-h^{2}}, \\
C O=\sqrt{c^{2}-h^{2}} .
\end{array}
$$
Therefore, the lengths... | \sqrt{3 h^{4}-2 h^{2}\left(a^{2}+b^{2}+c^{2}\right)+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}} \cdot h | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,823 |
5. For any positive integer $k$, let $f_{1}(k)$ be the sum of the squares of the digits of $k$, and for $n \geqslant 2, f_{n}(k)=$ $f_{1}\left(f_{n-1}(k)\right)$. Then $f_{2006}(2006)=$ $\qquad$ | 5. 145 .
Notice that
$$
\begin{array}{l}
f_{1}(2006)=2^{2}+6^{2}=40, f_{2}(2006)=f_{1}(40)=16, \\
f_{3}(2006)=f_{1}(16)=37, f_{4}(2006)=f_{1}(37)=58, \\
f_{5}(2006)=f_{1}(58)=89, f_{6}(2006)=f_{1}(89)=145, \\
f_{7}(2006)=f_{1}(145)=42, f_{8}(2006)=f_{1}(42)=20, \\
f_{9}(2006)=f_{1}(20)=4, f_{10}(2006)=f_{1}(4)=16,
\en... | 145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,824 |
6. Given $A_{i}\left(t_{i}+\frac{1}{t_{i}}, t_{i}-\frac{1}{t_{i}}\right)(i=1,2,3$,
4) are four distinct points on a plane. If these four points are concyclic, then $t_{1} 、 t_{2} 、 t_{3} 、 t_{4}$ should satisfy $\qquad$ . | 6. $t_{1} t_{2} t_{3} t_{4}=1$.
Let the equation of the circle be
$$
x^{2}+y^{2}+D x+E y+F=0 \text {. }
$$
If $P\left(t+\frac{1}{t}, t-\frac{1}{t}\right)$ lies on the circle, then
$$
\left(t+\frac{1}{t}\right)^{2}+\left(t-\frac{1}{t}\right)^{2}+D\left(t+\frac{1}{t}\right)+E\left(t-\frac{1}{t}\right)+P=0 \text {. }
$$... | t_{1} t_{2} t_{3} t_{4}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,825 |
Three. (20 points) Given $A=\{1,2,3\}, x_{i} \in A$, $i=1,2, \cdots, 2006$. Let
$$
\begin{array}{l}
a_{1}=x_{1}+x_{2}+\cdots+x_{2006}, \\
a_{2}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{2006}^{2}, \\
a_{n}=x_{1}^{n}+x_{2}^{n}+\cdots+x_{2000}^{n}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Try to express $a_{n}$ in terms of ... | Three, let $x_{i}(i=1,2, \cdots, 2006)$ take $1,2,3$ respectively $s, t, r$ times. Then
$$
\left\{\begin{array}{l}
s+t+r=2006, \\
a_{1}=s+2 t+3 r=2006+t+2 r, \\
a_{2}=s+4 t+9 r=2006+3 t+8 r .
\end{array}\right.
$$
Solving, we get $t=4 a_{1}-a_{2}-3 \times 2006$,
$$
r=2006-\frac{3 a_{1}-a_{2}}{2} \text {. }
$$
Thus, $... | a_{n} = \left(3^{n}-3 \times 2^{n}+3\right) 2006 + \left(2^{n+2}-\frac{1}{2} \times 3^{n+1}-\frac{5}{2}\right) a_{1} + \left(\frac{1}{2} \times 3^{n}-2^{n}+\frac{1}{2}\ | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,826 |
- Four, (20 points) Given that $P$ and $Q$ are two moving points on the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $O$ is the origin, the slope of line $O P$ is $k$, the product of the slopes of lines $O P$ and $O Q$ is $m$, and $p=|O P|^{2}+|O Q|^{2}$ is a constant independent of $k$.
(1) Find the ... | (1) The equation of the line $O P$ is $y=k x$, and when combined with the equation of the ellipse $C$, we get
$$
x^{2}=\frac{a^{2} b^{2}}{b^{2}+a^{2} k^{2}} \text {. }
$$
Then $|O P|^{2}=x^{2}+y^{2}=\left(1+k^{2}\right) x^{2}=\frac{a^{2} b^{2}\left(1+k^{2}\right)}{b^{2}+a^{2} k^{2}}$.
Similarly, we have
$$
\begin{arra... | 0<e_{2}-e_{1}<\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,827 |
Five. (20 points) Given a tetrahedron with four edges of length 1. Find the maximum volume of this tetrahedron.
| (1) If the edges of length $x, y$ are opposite edges, as shown in Figure 2.
Let $P A=P B=C A=C B=1$,
$A B=x, P C=y$.
Let the midpoint of $A B$ be $M$, and the midpoint of $P C$ be $N$. Connect
$P M, C M, M N$.
Then $P M \perp A B$,
$C M \perp A B$,
$P M=C M$
$=\sqrt{1-\frac{x^{2}}{4}}$.
Therefore, $M N \perp P C, A B ... | \frac{\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,828 |
一、(50 points) In $\triangle P M N$, $A$ and $B$ are points on sides $P M$ and $P N$ respectively, and satisfy $A M + A N = B M + B N$, $A N$ intersects $B M$ at point $Q$. Prove: Quadrilateral $P A Q B$ has an incircle. | As shown in Figure 4, on $MB$, intercept $ME = MA$, and on $NP$, intercept $NF = NA$. Connect $AE$, $EF$, and $FA$.
$$
\begin{array}{l}
\text{Then } BF = NF - NB \\
= NA - NB. \\
\text{Also } AM + AN \\
= BM + BN,
\end{array}
$$
Then $AN - BN$
$$
= BM - AM.
$$
Thus, $BF = BM - AM = BM - ME = BE$.
Draw the angle bisec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,829 |
II. (50 points) Given the set $X=\{2006,2006+$ $1,2006+2, \cdots, 2006+k\}$. Find the positive integer $k$, such that the set $X$ can be partitioned into three sets $A$, $B$, and $C$, satisfying:
(1) $A \cup B \cup C=X$,
$A \cap B=B \cap C=C \cap A=\varnothing$;
(2) The sum of the elements in sets $A$, $B$, and $C$ are... | (ii) When $m$ is even, i.e., $k=6n-1\left(n \in \mathbf{N}_{+}\right)$, the number of elements in set $X$ is a multiple of 6. It is only necessary to start from 2006, and in every 6 consecutive odd numbers, include the first two numbers in $A$, the middle two numbers in $B$, and the remaining two numbers in $C$. Thus, ... | k=3m-1\left(m \in \mathbf{N}_{+}, m>1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,830 |
Three, (50 points) Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
P=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}, \\
Q=\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{z+x}, \\
R=\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x} .
\end{array}
$$
Let $f=\max (P, Q, R)$, find $f_{\text {min }}$. | $$
\begin{array}{l}
\text { 3, } Q-R \\
=x\left(\frac{1}{z+x}-\frac{1}{y+z}\right)+y\left(\frac{1}{x+y}-\frac{1}{z+1}\right)+\left(\frac{1}{y+z}-\frac{1}{x+y}\right) \\
=\frac{x(y-x)}{(z+x)(y+z)}+\frac{y(z-y)}{(x+y)(z+x)}+\frac{z(x-z)}{(y+z)(x+y)} \\
=\frac{x\left(y^{2}-x^{2}\right)+y\left(z^{2}-y^{2}\right)+z\left(x^{... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,831 |
Example 5 Let $P$ be a point inside $\triangle ABC$, and let $D$, $E$, $F$ be the feet of the perpendiculars from $P$ to $BC$, $CA$, $AB$ respectively: Find all points $P$ that minimize $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$.
(22nd IMO) | Let the area of $\triangle ABC$ be $S$, then
$$
BC \cdot PD + CA \cdot PE + AB \cdot PF = 2S \text{.}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\left(\sqrt{\frac{BC}{PD}}\right)^{2}+\left(\sqrt{\frac{CA}{PE}}\right)^{2}+\left(\sqrt{\frac{AB}{PF}}\right)^{2}\right] .} \\
{\left[(\sqrt{BC \... | P \text{ is the incenter of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,832 |
In $\triangle A B C$, $A B>A C>B C$, points $M$ and $N$ are on sides $A B$ and $A C$ respectively, and satisfy $B M=C N=B C$. Prove: The sum of the distances from any point on line segment $M N$ to the three sides of $\triangle A B C$ is equal to the same value. | Proof: As shown in Figure 3, let the area of the convex quadrilateral $BCNM$ be $S$, $BM = CN = BC = a$. The distances from point $M$ to $BC$ and $AC$ are $MM_1$ and $MM_2$, respectively. The distances from point $N$ to $BC$ and $AB$ are $NN_1$ and $NN_2$, respectively. The distances from point $P$ on line segment $MN$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,833 |
Given $a, b, c$ are integers, and
$$
\left(a^{5}+b^{5}+c^{5}\right)+4(a+b+c)
$$
is a multiple of 120. Prove: $a^{3}+b^{3}+c^{3}$ is a multiple of 24. | Proof: Notice
$$
\begin{array}{l}
a^{5}-5 a^{3}+4 a=a\left(a^{4}-5 a^{2}+4\right) \\
=a\left(a^{2}-4\right)\left(a^{2}-1\right) \\
=(a-2)(a-1) a(a+1)(a+2)
\end{array}
$$
is the product of 5 consecutive integers, so, $a^{5}-5 a^{3}+4 a$ is a multiple of $1 \times 2 \times 3 \times 4 \times 5=120$.
Similarly, $b^{5}-5 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,834 |
In rectangle $ABCD$, $AB=1, BC=m, O$ is its center, $EO \perp$ plane $ABCD, EO=n$, and there exists a unique point $F$ on side $BC$ such that $EF \perp FD$. What conditions must $m, n$ satisfy for the angle between plane $DEF$ and plane $ABCD$ to be $60^{\circ}$? | Solution: As shown in Figure 4, let $FC = x$. Draw a perpendicular from point $O$ to $BC$, with the foot of the perpendicular being $G$. Then,
$$
\begin{array}{l}
GF = \frac{m}{2} - x, \\
OG = \frac{1}{2}, \\
EF^2 = EO^2 + OG^2 + GF^2 \\
= n^2 + \frac{1}{4} + \left(\frac{m}{2} - x\right)^2 \\
= x^2 - mx + \frac{m^2}{4}... | m = 2\sqrt{2}, n = \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,835 |
Given $a, b, c$ are positive numbers satisfying $a^{2}=b^{2}+c^{2}$. Prove:
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3} \\
\geqslant \frac{2 \sqrt{2}+1}{7}\left[a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\right] .
\end{array}
$$ | Proof: Note that
$$
\begin{array}{l}
b^{3}+b c^{2}+c^{3}+b^{2} c \\
\geqslant 2 \sqrt{ } b^{3} \cdot b c^{2}+2 \sqrt{c^{3} \cdot b^{2}} c=2 b^{2} c+2 b c^{2} .
\end{array}
$$
Then $b^{3}+c^{3} \geqslant b^{2} c+b c^{2}$.
Thus, $a^{3}+(\sqrt{2} b)^{3} \geqslant a^{2} \cdot \sqrt{2} b+a(\sqrt{2} b)^{2}$, which means
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,836 |
Example 6 Let the three sides of $\triangle ABC$ be $a, b, c$, and the heights corresponding to these sides be $h_{a}, h_{b}, h_{c}$. Prove:
$$
\sqrt{3}(a+b+c) \geqslant 2\left(h_{a}+h_{b}+h_{c}\right) .
$$ | Prove: Let the area of $\triangle ABC$ be $S$.
$$
\begin{array}{l}
(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca \\
\geqslant 3ab+3bc+3ca \\
=\frac{3abc}{2S}(h_{a}+h_{b}+h_{c}) \\
=3 \times 2R(h_{a}+h_{b}+h_{c}).
\end{array}
$$
Since $a+b+c$
$$
=2R(\sin A+\sin B+\sin C) \leqslant 2R \cdot \frac{3\sqrt{3}}{2},
$$
then... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,837 |
Lemma In $\triangle A B C$, prove that:
$$
\begin{array}{l}
\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2} \\
\geqslant 2-8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2},
\end{array}
$$
The equality holds if and only if $\triangle A B C$ is an equilateral triangle. | Lemma Proof: Let $i, j, k$ be unit vectors in a plane, and $\boldsymbol{j}$ forms an angle of $\pi-\angle A$ with $k$, $k$ forms an angle of $\pi-\angle B$ with $i$, and $i$ forms an angle of $\pi-\angle C$ with $j$. Then
$$
\left(i \tan \frac{A}{2}+j \tan \frac{B}{2}+k \tan \frac{C}{2}\right)^{2} \geqslant 0 .
$$
Thu... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,838 |
Example 1 As shown in Figure 1, there is a point $B^{\prime}$ on line segment $A B$. Two semicircles are constructed with $A B$ and $A B^{\prime}$ as diameters. $M M^{\prime} \perp$ $A B, N N^{\prime} \perp A B$, points $M$, $M^{\prime}$, $N$, and $N^{\prime}$ are on the two semicircles, respectively. Prove: The circum... | Proof: Let the circumradii of $\triangle A M M^{\prime}$ and $\triangle A N N^{\prime}$ be $r_{1}$ and $r_{2}$, respectively. Suppose $M M^{\prime} \perp A B$ at point $E$.
According to Proposition 1, we have
$A M \cdot A M^{\prime}=A E \cdot 2 r_{1}$.
It is easy to prove that $A M^{2}=A E \cdot A B, A M^{\prime 2}=A E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,839 |
Example 2 Given $\triangle A B C$ is inscribed in $\odot O$, point $P$ is on $\overparen{B C}$, and $P X \perp B C, P Y \perp C A, P Z \perp A B, X, Y, Z$ are the three feet of the perpendiculars. Prove: $\frac{B C}{P X}=\frac{C A}{P Y}+\frac{A B}{P Z}$. | Proof: As shown in Figure 2, connect \( PA, PB, PC \). Let the radius of \(\odot O\) be \( R \).
By Proposition 1, we have
\[
\begin{array}{l}
PB \cdot PC = 2R \cdot PX, \\
PC \cdot PA = 2R \cdot PY, \\
PA \cdot PB = 2R \cdot PZ.
\end{array}
\]
Thus, \( PX = \frac{PB \cdot PC}{2R} \),
\[
PY = \frac{PC \cdot PA}{2R}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,840 |
Example 3 Given that $I$ is the incenter of $\triangle ABC$, the rays $AI$, $BI$, and $CI$ intersect the circumcircle $\odot O$ of $\triangle ABC$ at points $A'$, $B'$, and $C'$, respectively. Prove:
(1) $\frac{IB \cdot IC}{IA'} = \frac{IC \cdot IA}{IB'} = \frac{IA \cdot IB}{IC'}$;
(2) $\frac{IB' \cdot IC'}{IA} = \frac... | Proof: As shown in Figure 3, connect \(A' B\), \(A' C\), \(B' C'\), \(A B'\), \(A C'\), and draw \(ID \perp BC\) at point \(D\), and \(AI\) intersects \(B' C'\) at point \(K\).
Let the circumradius and inradius of \(\triangle ABC\) be \(R\) and \(r\), respectively, then \(ID = r\).
By the properties of the incenter of... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,841 |
Example 4 Given in $\triangle A B C$, $B E$ and $C F$ are two angle bisectors, and point $P$ is on $E F$. Prove: The longest distance from point $P$ to the sides of $\triangle A B C$ is equal to the sum of the other two distances. | Proof: As shown in Figure 5, draw \( P P_{1} \perp B C \) at point \( P_{1} \), \( P P_{2} \perp C A \) at point \( P_{2} \), \( P P_{3} \perp A B \) at point \( P_{3} \), \( E E_{1} \perp B C \) at point \( E_{1} \), \( F F_{1} \perp B C \) at point \( F_{1} \), \( E E_{2} \perp A B \) at point \( E_{2} \), \( F F_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,842 |
Example 5 As shown in Figure 6, in the acute triangle $\triangle ABC$, the distances from the circumcenter $O$ to $BC$, $CA$, and $AB$ are $\mathrm{OO}_{1}$, $\mathrm{OO}_{2}$, and $\mathrm{OO}_{3}$, respectively. The distances from the centroid $G$ to the sides are $G G_{1}$, $G G_{2}$, and $G G_{3}$, and the distance... | Proof: It is known that points $O, G, H$ are collinear, i.e., the Euler line of $\triangle ABC$, and $\frac{O G}{G H}=\frac{1}{2}$.
According to Proposition 2, from trapezoid $O O_{1} H_{1} H$ we get
$$
\begin{array}{l}
G G_{1}=\frac{H H_{1}+2 O O_{1}}{1+2} \\
=\frac{1}{3}\left(H H_{1}+2 O O_{1}\right) .
\end{array}
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,843 |
Example 3 In isosceles $\triangle A B C$, $P$ is any point on the base $B C$. A line through point $P$ parallel to the two legs intersects $A B$ and $A C$ at points $Q$ and $R$, respectively. Point $P^{\prime}$ is the reflection of point $P$ about line $Q R$. Prove: Point $P^{\prime}$ lies on the circumcircle of $\tria... | Analysis: This problem is to prove that points $A$, $P^{\prime}$, $B$, and $C$ are concyclic. Therefore, it is sufficient to prove that $\angle B P^{\prime} C = \angle B A C$.
Proof: As shown in Figure 4, connect $P^{\prime} B$, $P^{\prime} C$, $P^{\prime} Q$, and $P^{\prime} R$. Given that $P Q \parallel A C$, $P R \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,844 |
Example 6 As shown in Figure 7, in $\triangle A B C$, $D$ and $E$ are two points on side $B C$, and
$$
\begin{array}{l}
B D=D E=E C, G F \\
/ / B C(G \in A B, F \in
\end{array}
$$
$A C)$ and intersects $A D$ and $A E$
at points $P$ and $Q$, $B F$ intersects
$A D$ and $A E$ at points $M$ and $K$,
$C G$ intersects $A D$ ... | Prove: Let $D E=1$, then
$$
B D=E C=1, B E=D C=2 .
$$
It is easy to know that $G P=P Q=Q F$.
Let $P Q=a$, then
$$
G P=Q F=a, G Q=P F=2 a \text {. }
$$
From $G F / / B C$, we get
$\triangle Q K F \backsim \triangle E K \dot{B}, \triangle G P I \backsim \triangle C D I$, $\triangle Q N G \backsim \triangle E N C, \tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,845 |
For any given acute triangle $\triangle ABC$, there are two points $E, F$ on side $BC$ such that $\angle BAE = \angle CAF$. On sides $AB, AC$, take points $M, N$ respectively, so that points $A, M, F, N$ are concyclic. Extend $AE$ to intersect the circumcircle of $\triangle ABC$ at point $D$. Prove: $S_{\text{quadrilat... | Proof: As shown in Figure 1, connect $M N, D M, D N$. Let $\angle B A E=\angle C A F=\alpha$, $\angle E A F=\beta$, and the circumradius of $\triangle A B C$ be $R$. Then
$$
\begin{array}{l}
S_{\text {quadrilateral } M M D N} \\
=S_{\triangle M D D}+S_{\triangle A D N} \\
=\frac{1}{2} A D[A M \sin \alpha+A N \sin (\alp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,846 |
Given an acute triangle $\triangle ABC$, point $O$ is its circumcenter, and line $AO$ intersects side $BC$ at point $D$. Moving points $E$ and $F$ are located on sides $AB$ and $AC$, respectively, such that points $A$, $E$, $D$, and $F$ are concyclic. Prove that the length of the projection of segment $EF$ onto side $B... | For question 3: As shown in Figure 2, draw $A M \perp B C$ at $M$, extend $A M$ to intersect $\odot O$ at point $P$, connect $P E$ and $P F$, extend $A D$ to intersect $\odot O$ at point $Q$, and connect $B Q$.
Let the projection of $E F$ on $B C$ be $E_{0} F_{0}$, and the angle between $E F$ and $P A$ be $\alpha$. Th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,847 |
Let $\triangle A B C$ be an acute-angled triangle, with circumcenter $O$ and circumradius $R$. Let $A O$ intersect the circumcircle of $\triangle B O C$ at another point $A^{\prime}$, $B O$ intersect the circumcircle of $\triangle C O A$ at another point $B^{\prime}$, and $C O$ intersect the circumcircle of $\triangle ... | Text [1] replaces the circumcenter in the problem statement with the incenter, orthocenter (for acute triangles), and centroid, and proves conclusions similar to formula (1). This article makes a more general promotion of the problem.
Proposition Let $P$ be any point inside $\triangle ABC$, and the extensions of $AP$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,848 |
1. Among the positive integers $n$ less than 100, the possible values of $n$ that allow the fraction $\frac{1}{(3 n+32)(4 n+1)}$ to be expressed as a finite decimal in base 10 are $\qquad$ | 1.6 or 31.
From the problem, we know that $(3 n+32)(4 n+1)$ can only contain the prime factors 2 or 5. $4 n+1$ is odd, and $5 \leqslant 4 n+1<401$, so $4 n+1=5$ or 25 or 125. Therefore, $n=1$ or 6 or 31. When $n=1$, $3 n+32=35=5 \times 7$, which does not meet the condition. It is easy to verify that $n=6$ or 31 satisfi... | 6 \text{ or } 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,849 |
2. Arrange the digits $1,2,3,4,5,6,7,8,9$ in some order to form a nine-digit number abodefghi, and let
$$
A=\overline{a b c}+\overline{b c d}+\overline{c d e}+\overline{d e f}+\overline{e f g}+\overline{f g h}+\overline{g h i} .
$$
Then the maximum possible value of $A$ is | 2.4648 .
$A=111(c+d+e+f+g)+110 b+100 a+11 h+i$,
then when $c+d+e+f+g=9+8+7+6+5=35, b=4$, $a=3, h=2, i=1$, the maximum value of $A$ is 4648. | 4648 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,850 |
3. If a two-digit number $\bar{x}$ and a three-digit number $\overline{3 y z}$ have a product of 29,400, then, $x+y+z=$ $\qquad$ | 3.18.
According to the problem, we have
$$
73<\frac{29400}{400}<\overline{x 5} \leqslant \frac{29400}{300}=98 .
$$
Thus, $\overline{x 5}=75$ or 85 or 95.
Since $85 \times 29400, 95 \times 29400$, it can only be $\overline{x 5}=75$.
When $\overline{x 5}=75$, $\overline{3 y z}=\frac{29400}{75}=392$.
Therefore, $x=7, y=... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,851 |
5. As shown in Figure 1, the vertices of $\triangle O A B$ are $O(0,0)$, $A(2,1)$, and $B(10,1)$. The line $C D \perp x$-axis and bisects the area of $\triangle O A B$. If the coordinates of point $D$ are $(x, 0)$, then the value of $x$ is . $\qquad$ | $5.10-2 \sqrt{10}$.
Let $CD$ intersect $AB$ at point $E$, and intersect $OB$ at point $F$.
Since the equation of line $OB$ is $y=\frac{1}{10} x$, the coordinates of point $F$ are $\left(x, \frac{1}{10} x\right)$ (here $0<x<10$). Then
$$
\begin{array}{l}
EF=1-\frac{1}{10} x=\frac{10-x}{10}, \\
EB=10-x, AB=10-2=8 . \\
\t... | 10-2\sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,853 |
6. If the two quadratic equations $x^{2}+x+m=$ 0 and $m x^{2}+x+1=0$ each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$ . | 6. $\alpha=1$.
From the problem, we know $m \neq 0$, and $\Delta=1-4 m>0$, i.e., $m<\frac{1}{4}$ and $m \neq 0$; simultaneously, $\alpha^{2}+\alpha+m=0$, and $m \alpha^{2}+\alpha+1=0$.
Subtracting these equations gives $(1-m) \alpha^{2}+(m-1)=0$.
Since $m \neq 1$, we have $\alpha^{2}=1$. Solving this, we get $\alpha=... | \alpha=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,854 |
For example, $4 A B C D$ is a cyclic quadrilateral, $A C$ is the diameter of the circle, $B D \perp A C$, the intersection of $A C$ and $B D$ is $E$, point $F$ is on the extension of $D A$, connect $B F$, point $G$ is on the extension of $B A$ such that $D G / / B F$, point $H$ is on the extension of $G F$, and $C H \p... | Proof: As shown in Figure 5, connect $B H, E F, C G$. Since $\triangle B A F \sim \triangle G A D$, we have
$$
\frac{F A}{A B}=\frac{D A}{A G} \text{.}
$$
Also, since $\triangle A B E \sim \triangle A C D$, we have
$$
\frac{A B}{E A}=\frac{A C}{D A} \text{.}
$$
Multiplying (1) and (2) gives
$$
\frac{F A}{E A}=\frac{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,855 |
7. As shown in Figure 2, in trapezoid
$A B C D$, $A B / / D C$,
$D C=2 A B=2 A D$. If
$B D=6, B C=4$, then
$S_{A B C D}=\ldots \quad\left(S_{A B C D}\right.$ represents
the area of quadrilateral $A B C D$, the same below. | 7. 18
As shown in Figure 6, take the midpoint $M$ of $CD$, and connect $BM$, $AM$.
From the given conditions, it is easy to see that quadrilateral $ADMB$ is a rhombus, and quadrilateral $AMCB$ is a parallelogram. Therefore, $AM = BC = 4$.
Noting that $\triangle ADM \cong \triangle MAB \cong \triangle BMC$, we have $S... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,856 |
8. As shown in Figure 3, in $\square A B C D$, $M$ and $N$ are the midpoints of sides $B C$ and $D C$ respectively, $A N=1, A M=2$, and $\angle M A N=$ $60^{\circ}$. Then the length of $A B$ is $\qquad$. | 8. $\frac{2 \sqrt{13}}{3}$.
As shown in Figure 7, extend $A M$ to intersect the extension of $D C$ at point $F$. It is easy to prove that $\triangle A M B \cong \triangle F M C$, so $C F=A B$. Therefore, $N F=\frac{3}{2} A B$.
Draw a perpendicular from point $N$ to $A F$ at $H$, then $A H=\frac{1}{2} A N=\frac{1}{2},... | \frac{2 \sqrt{13}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,857 |
9. As shown in Figure 4, in $\triangle A B C$, points $E$ and $F$ are on sides $A B$ and $A C$ respectively, and $E F / / B C$. If $S_{\triangle A B C}=1, S_{\triangle A E F}=$ $2 S_{\triangle E B C}$, then $S_{\triangle C E F}=$ $\qquad$ | $9.3 \sqrt{3}-5$.
Let $\frac{A F}{F C}=t$. Then $S_{\triangle S B F}=\frac{C F}{F A} S_{\triangle A E F}=\frac{1}{t} S_{\triangle \triangle S F F}$.
Since $\triangle A E F \backsim \triangle A B C$, we have
$S_{\triangle \triangle E F}=\left(\frac{A F}{A C}\right)^{2} S_{\triangle \triangle B C}=\left(\frac{t}{1+t}\rig... | 3 \sqrt{3}-5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,858 |
10. Let $p$ be a prime number, and let the equation $x^{2}-p x-580 p=0$ have two integer roots. Then the value of $p$ is $\qquad$ . | 10.29.
Since $x^{2}-p x-580 p=0$ has two integer roots, we have
$$
\Delta=(-p)^{2}+4 \times 580 p=p(p+2320)
$$
is a perfect square.
From this, we know that $p 12320$.
Since $2320=2^{4} \times 5 \times 29$, we have $p=2$ or 5 or 29.
When $p=2$, $\Delta=2^{2}(1+1160)=2^{2} \times 1161=2^{2} \times 3^{2} \times 129$ is ... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,859 |
II. (20 points) Given a rectangle $A B C D$ with adjacent side lengths $a$ and $b$. Does there exist another rectangle $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ such that its perimeter and area are $\frac{1}{3}$ of the perimeter and area of rectangle $A B C D$? Prove your conclusion. | II. Let the lengths of the adjacent sides of rectangle $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be $m$ and $n$. According to the problem, we have
$$
m+n=\frac{1}{3}(a+b), \quad m n=\frac{1}{3} a b.
$$
Therefore, $m$ and $n$ are the two positive roots of the quadratic equation $x^{2}-\frac{1}{3}(a+b) x+\frac{1}{3}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,860 |
Three, (20 points) Given that $a$, $b$, and $c$ are all prime numbers greater than 3, and $2a + 5b = c$.
(1) Prove that there exists a positive integer $n > 1$ such that the sum $a + b + c$ of all three prime numbers $a$, $b$, and $c$ that satisfy the given condition is divisible by $n$;
(2) Find the maximum value of $... | (1) Since $c=2a+5b$, we have
$$
a+b+c=3a+6b=3(a+2b).
$$
Also, since $a, b, c$ are all prime numbers greater than 3, it follows that $3 \mid (a+b+c)$, i.e., there exists a positive integer $n>1$ (for example, $n=3$) such that $n! \mid (a+b+c)$.
(2) Since $a, b, c$ are all prime numbers greater than 3, $a, b, c$ are not... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false | 716,861 |
Four. (20 points) As shown in Figure 5, in the right triangle $\triangle ABC$, $CA > CB, \angle C = 90^{\circ}$, quadrilaterals $CDEF$ and $KLMN$ are two inscribed squares in $\triangle ABC$. It is known that $S_{\text{CDEF}} = 441, S_{KLMN} = 440$. Find the lengths of the three sides of $\triangle ABC$.
---
Translat... | Let the side length of square $CDEF$ be $x$, and the side length of square $KLMN$ be $y$. According to the problem, $x=21, y=2\sqrt{110}$.
Let $BC=a, CA=b, AB=c$, then $a^2 + b^2 = c^2$.
Notice that
$$
a x + b x = 2\left(S_{\triangle CKB} + S_{\triangle CM}\right) = 2 S_{\triangle ABC} = a b,
$$
Thus, $x = \frac{a b}{... | a = 231 - 63 \sqrt{11}, b = 231 + 63 \sqrt{11}, c = 42 \sqrt{110} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,862 |
1. Define the operation symbol “ * ” as: $a * b=$ $\frac{a+b}{a b}$ (where $a$ and $b$ are not 0). There are two conclusions below:
(1) The operation “ * ” satisfies the commutative law;
(2) The operation “ * ” satisfies the associative law.
Among them ( ).
(A) Only (1) is correct
(B) Only (2) is correct
(C) Both (1) a... | 1.A.
From $a * b=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$, it is easy to see that the operation “*” satisfies the commutative law.
$$
\begin{array}{l}
\text { and } a *(b+c)=\frac{1}{a}+\frac{1}{b+c}, \\
a * b+a * c=\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c},
\end{array}
$$
By the arbitrariness of $a$, we can see... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,863 |
2. Below are 4 sets of positive integers:
(1) multiples of 3 from 1 to 101;
(2) multiples of 4 from 1 to 101;
(3) multiples of 5 from 1 to 101;
(4) multiples of 6 from 1 to 101.
Among them, the set with the largest average is ( ).
(A)(1)
(B)(2)
(C)(3)
(D)(4) | 2.C.
It is easy to know that the average value of multiples of 3 in 1 101 is $\frac{3+99}{2}=51$; the average value of multiples of 4 is $\frac{4+100}{2}=52$; the average value of multiples of 5 is $\frac{5+100}{2}=52.5$; the average value of multiples of 6 is $\frac{6+96}{2}=51$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,864 |
3. There are 3 conclusions below:
(1) There exist two different irrational numbers, the difference of which is an integer;
(2) There exist two different irrational numbers, the product of which is an integer;
(3) There exist two different non-integer rational numbers, the sum and quotient of which are integers.
Among t... | 3. D.
Example: $\sqrt{5}+1, \sqrt{5}-1$ satisfy conclusions (1) and (2); $\frac{5}{3}, \frac{1}{3}$ satisfy conclusion (3). | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,865 |
Example 5 In $\triangle A B C$, take points $Q, P$ on sides $A B, A C$ respectively, such that $\angle P B C=\angle Q C B=\frac{1}{2} \angle A$. Prove: $B O=C P$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 5 In $\triangle A B C$, take points $Q, P$ on sides $A B, A C$ respectively, such that $\angle P B C=\angle Q C B=\frac{1}{... | Proof: From the given,
$$
\angle P B C=\angle Q C B=\frac{1}{2} \angle A \text {. }
$$
Then $\angle B Q C+\angle C P B$
$$
\begin{array}{c}
=\left(\angle A+\angle C-\frac{1}{2} \angle A\right)+\left(\angle A+\angle B-\frac{1}{2} \angle A\right) \\
=\angle A+\angle B+\angle C=180^{\circ} .
\end{array}
$$
As shown in F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,866 |
4. If the lengths of two sides of $\triangle A B C$ are $a$ and $b$, then the area of $\triangle A B C$ cannot be equal to ( ).
(A) $\frac{1}{4}\left(a^{2}+b^{2}\right)$
(B) $\frac{1}{2}\left(a^{2}+b^{2}\right)$
(C) $\frac{1}{8}(a+b)^{2}$
(D) $\frac{1}{4} a b$ | 4.B.
$S_{\triangle A B C}=\frac{1}{2} a b \sin C \leqslant \frac{1}{2} a b$
$$
\leqslant \frac{1}{4}\left(a^{2}+b^{2}\right)<\frac{1}{2}\left(a^{2}+b^{2}\right) .
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,867 |
5. If $m$ and $n$ are odd numbers, the quadratic equation $x^{2}+$ $m x+n=0$ has two real roots, then the nature of these real roots is ( ).
(A) There are both odd and even roots
(B) There are neither odd nor even roots
(C) There are even roots, but no odd roots
(D) There are odd roots, but no even roots | 5.B.
If $x^{2}+m x+n=0 (m, n$ are odd numbers) has integer roots, then both roots are integers and are odd (since $n$ is odd). Therefore, their sum is even, which contradicts the fact that $m$ is odd. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,868 |
7. As shown in Figure 2, the system of inequalities representing the shaded region is ( ).
(A) $\left\{\begin{array}{l}2 x+y \geqslant 5, \\ 3 x+4 y \geqslant 9 \\ y \geqslant 0\end{array}\right.$,
(B) $\left\{\begin{array}{l}2 x+y \leqslant 5, \\ 3 x+4 y \leqslant 9, \\ y \geqslant 0\end{array}\right.$
(C) $\left\{\be... | 7.D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,870 |
8. There are six measuring cups $A, B, C, D, E, F$, with volumes of $16 \mathrm{~mL}, 18 \mathrm{~mL}, 22 \mathrm{~mL}, 23 \mathrm{~mL}, 24 \mathrm{~mL}$, and $34 \mathrm{~mL}$. Some of the cups are filled with alcohol, some are filled with distilled water, and one cup is empty. The volume of alcohol is twice the volum... | 8.D.
Since $V_{\text {empty }}=2 V_{\text {water }}$, the total volume of the graduated cylinders filled with liquid can be divided by 3. Also, $16+18+22+23+24+34$ leaves a remainder of 2 when divided by 3, so the empty graduated cylinder's volume is 23, which means $D$ is the empty graduated cylinder. Therefore,
$$
V... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 716,871 |
9. Let $a$, $b$, $c$ be the lengths of the sides of $\triangle ABC$, simplify
$$
\sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}}
$$
The result is $\qquad$. | $\begin{array}{l}=9 . a+b+c . \\ \sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}} \\ =(b+c-a)+(c+a-b)+(a+b-c) \\ =a+b+c .\end{array}$
The translation is as follows:
$\begin{array}{l}=9 . a+b+c . \\ \sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}} \\ =(b+c-a)+(c+a-b)+(a+b-c) \\ =a+b+c .\end{array}$ | a+b+c | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,872 |
$\begin{array}{l}\text { 10. As shown in Figure } 3, D C \\ / / A B, \angle B A E= \\ \angle B C D, A E \perp D E, \\ \angle D=130^{\circ} \text {. Then } \angle B \\ =\end{array}$ | $10.40^{\circ}$.
Extend $A E$ and $C D$ to intersect at point $F$. Given $\angle B A E=\angle B C D$ and $D C / / A B$, it is easy to see that quadrilateral $A B C F$ is a parallelogram.
$$
\text { Also, } \angle F=\angle C D E-\angle D E F=130^{\circ}-90^{\circ}=40^{\circ} \text {, }
$$
Therefore, $\angle B=\angle F=... | 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,873 |
11. When seven dice are rolled simultaneously, the probability that the sum of the numbers on the seven faces is 10 is equal to the probability that the sum of the numbers on the seven faces is $a(a \neq 10)$. Then, $a=$ $\qquad$ | 11.39.
$$
\text { Given }\left(x_{1}, x_{2}, \cdots, x_{7}\right) \rightarrow\left(7-x_{1}, 7-x_{2}, \cdots, 7-x_{7}\right)
$$
we know that the probability of the sum of the points being 10 is the same as the probability of the sum of the points being $49-10$ $=39$. | 39 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,874 |
12. The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.
| 12.4.
From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.
Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,875 |
13. As shown in Figure 4, in the Cartesian coordinate system, there are four points $A(-2,4) 、 B(-2,0) 、 C(2,-3) 、 D(2,0)$. Let $P$ be a point on the $x$-axis, and the triangle formed by $P A 、 P B 、 A B$ is similar to the triangle formed by $P C 、 P D 、 C D$. Write down the coordinates of all points $P$ that satisfy t... | $$
\frac{P B}{A B}=\frac{P D}{C D} \text { or } \frac{P B}{A B}=\frac{C D}{P D} ,
$$
i.e., $\frac{|x+2|}{4}=\frac{|x-2|}{3}$ or $\frac{|x+2|}{4}=\frac{3}{|x-2|}$.
Solving, we get $x=-4, \frac{2}{7}, 14,4$. | x=-4, \frac{2}{7}, 14, 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,876 |
Example 6 In trapezoid $A B C D$, $A D / / B C, B C=$ $B D=1, A B=A C, C D<1, \angle B A C+\angle B D C$ $=180^{\circ}$ Find the value of $C D$. | Solution: As shown in Figure 7, construct: point $D$ as the symmetric point of $\mathrm{K} B C$, and connect $A E$, $B E$, and $C E$. Let $A E^{1} \cdot \mathrm{j}$ $B C$ intersect at point $F$.
If $A I) B C$, then points $A$ and $E$ are equidistant from $B C$, so,
$$
A F = F E.
$$
Let $C I) = C E = x$, $A F' = F E =... | \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,877 |
15. As shown in Figure 5, there is an iron tower $A B$ at the top of a slope. Under the sunlight, the tower's shadow $D E$ is cast on the slope. It is known that the base width of the tower $C D=14 \mathrm{~m}$, the length of the tower's shadow $D E=36 \mathrm{~m}$, and both Xiao Ming and Xiao Hua are $1.6 \mathrm{~m}$... | 15.20.
Draw a perpendicular from point $D$ to $CD$ intersecting $AE$ at point $F$, and draw a perpendicular from point $F$ to $AB$, with the foot of the perpendicular being $G$. It is easy to see that
$$
\begin{array}{l}
D F=\frac{1.6}{4} D E=14.4(\mathrm{~m}), \\
A G=\frac{1.6}{2} F G=\frac{1.6}{2} B D=5.6(\mathrm{~m... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,879 |
16. Let the product of all distinct positive divisors of 2005 be $a$, and the product of all distinct positive divisors of $a$ be $b$. Then $b=$ $\qquad$ | 16.2 0059.
From $2005=5 \times 401$, we know that any positive divisor of 2005 is the product of one of $1, 5$ and one of $1, 401$, so $a=5^{2} \times 401^{2}$. Furthermore, any positive divisor of $a$ is the product of one of $1, 5, 5^{2}$ and one of $1, 401, 401^{2}$, hence
$$
b=\left(5 \times 5^{2}\right)^{3} \time... | 2005^9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,880 |
18. Given Rt $\triangle A B C$ and Rt $\triangle A D C$ share the hypotenuse $A C, M$ and $N$ are the midpoints of $A C$ and $B D$ respectively, and $M$ and $N$ do not coincide.
(1) Is line segment $M N$ perpendicular to $B D$? Please explain your reasoning.
(2) If $\angle B A C=30^{\circ}, \angle C A D=45^{\circ}, A C... | 18. (1) As shown in Figures 7 and 8, connect $M B$ and $M D$.
Since $M$ is the midpoint of $A C$, $\angle A B C = \angle A D C = 90^{\circ}$, then $M B = M D = \frac{1}{2} A C$.
Also, $N$ is the midpoint of $B D$, by the property of isosceles triangles, $M N \perp B D$.
(2) Draw $C H \perp B D$, with the foot of the p... | \sqrt{2 - \sqrt{3}} \text{ or } \sqrt{2 + \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,882 |
19. Given that $x, y$ are positive integers, and satisfy
$$
x y-(x+y)=2 p+q \text{, }
$$
where $p, q$ are the greatest common divisor and the least common multiple of $x$ and $y$, respectively. Find all such pairs $(x, y)(x \geqslant y)$. | 19. From the problem, let $x=a p, y=b p(a, b$ be positive integers and $(a, b)=1, a \geqslant b)$. Then, $q=a b p$. At this point, the equation in the problem is $a p \cdot b p-(a p+b p)=2 p+a b p$.
By $p>0$, the above equation simplifies to
$(p-1) a b=a+b+2$.
Equation (1) shows that $0<p-1=\frac{1}{a}+\frac{1}{b}+\fra... | (x, y) = (5, 5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,883 |
20. 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 number 1 or... | 20. (1) Divide the sequence of numbers into $n$ groups:
These $n$ groups have a total of $1+2+\cdots+n=\frac{1}{2} n(n+1)$ numbers.
When $n=62$, there are 1953 numbers;
When $n=63$, there are 2016 numbers.
It is clear that the 2005th number is in the 63rd group, and it is not the last number in that group, so the 2005... | 7789435 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,884 |
1. Let the function $f(x)$ have the domain $\mathbf{R}$, and for any real number $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), f(\tan x)=\sin 2 x$. Then the maximum value of $f(2 \sin x)$ is $(\quad)$.
(A) 0
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{2}}{2}$
(D) 1 | -,1.D.
From $f(\tan x)=\frac{2 \tan x}{1+\tan ^{2} x}$, we know $f(x)=\frac{2 x}{1+x^{2}}$. Therefore, $f(2 \sin x)=\frac{4 \sin x}{1+4 \sin ^{2} x} \leq 1$.
When $\sin x=\frac{1}{2}$, the equality holds.
Thus, the maximum value of $f(2 \sin x)$ is 1. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,885 |
2. The real number sequence $\left\{a_{n}\right\}$ is defined as, $a_{1}=1, a_{9}=7$,
$$
a_{n+1}=\frac{a_{n}^{2}-a_{n-1}+2 a_{n}}{a_{n-1}+1}, n=2,3, \cdots
$$
Then the value of $a_{5}$ is ( ).
(A) 3
(B) -4
(C) 3 or -4
(D) 8 | 2. A.
Let $b_{n}=a_{n}+1$, then
$$
b_{n+1}-1=\frac{b_{n}^{2}-b_{n-1}}{b_{n-1}}=\frac{b_{n}^{2}}{b_{n-1}}-1 .
$$
Therefore, $b_{n}^{2}=b_{n+1} b_{n-1}$. Hence $b_{5}^{2}=b_{1} b_{9}=16$.
Thus, $b_{5}=4$ (discard -4, as $b_{1} 、 b_{5} 、 b_{9}$ have the same sign). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,886 |
3. For a regular tetrahedron $ABCD$ with edge length $1$, $E$ is a point inside $\triangle ABC$. The sum of the distances from point $E$ to the sides $AB$, $BC$, and $CA$ is $x$, and the sum of the distances from point $E$ to the planes $DAB$, $DBC$, and $DCA$ is $y$. Then $x^{2}+y^{2}$ equals ( ).
(A) 1
(B) $\frac{\sq... | 3.D.
Point $E$ to the distances of sides $A B, B C, C A$ sum up to the height of $\triangle A B C$, which is $\frac{\sqrt{3}}{2}$, hence $x=\frac{\sqrt{3}}{2}$.
$$
\begin{array}{l}
\frac{1}{3} S_{\triangle A B C} \cdot h \\
=\frac{1}{3} S_{\triangle \operatorname{MAB}} \cdot y_{1}+\frac{1}{3} S_{\triangle H C C} \cdot... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,887 |
Example 7 In: acute $\triangle A B C$, $A B \neq A C, A D$ is known, $B, C, E, F$ are points in between. Question: Is point $H$ necessarily the orthocenter of $\triangle A B C$? Answer: Prove your argument. | Solution: The case is affirmative.
If 8, draw the perpendicular bisector of $A D$: take a point $G$, then we have
$$
\begin{array}{l}
A I I \cdot A G=A F \cdot A B \\
=A E \cdot A C .
\end{array}
$$
(1) If $G, D$ coincide,
then
$$
\begin{array}{l}
\angle A F H=\angle A G B . \\
\angle A E H=\angle A G C .
\end{array}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,888 |
4. The sequence $x_{1}, x_{2}, \cdots, x_{100}$ satisfies the following conditions: For $k=1,2, \cdots, 100, x_{k}$ is less than the sum of the other 99 numbers by $k$. Given that $x_{S 0}=\frac{m}{n}, m, n$ are coprime positive integers. Then $m+n$ equals $(\quad)$.
(A) 50
(B) 100
(C) 165
(D) 173 | 4.D.
Let $S=x_{1}+x_{2}+\cdots+x_{100}$, then $x_{k}=\left(S-x_{k}\right)-k$, which means $k+2 x_{k}=S$.
Summing over $k$ we get
$$
(1+2+\cdots+100)+2 S=100 S \text {. }
$$
Therefore, $S=\frac{2525}{49}$. Thus, $x_{50}=\frac{S-50}{2}=\frac{75}{98}$.
Hence, $m+n=173$. | 173 | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,889 |
5. If $\sin x+\sin y=\frac{\sqrt{2}}{2}, \cos x+\cos y=$ $\frac{\sqrt{6}}{2}$, then $\sin (x+y)$ equals $(\quad)$.
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{6}}{2}$
(D) 1 | 5. B,
Square the two expressions and add them, we get
$$
\begin{array}{l}
\left(\sin ^{2} x+\cos ^{2} x\right)+\left(\sin ^{2} y+\cos ^{2} y\right)+ \\
2(\cos x \cdot \cos y+\sin x \cdot \sin y)=2 .
\end{array}
$$
Therefore, $\cos (x-y)=0$.
Multiply the two expressions, we get
$\left(\sin x^{\prime} \cos y+\sin y^{\p... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,890 |
6. $P$ is a moving point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ in the first quadrant. Tangents $P A$ and $P B$ are drawn from point $P$ to the circle $x^{2}+y^{2}=9$, touching the circle at points $A$ and $B$ respectively. The line $A B$ intersects the $x$-axis and $y$-axis at points $M$ and $N$ respectiv... | 6.C.
Let $P(4 \cos \theta, 3 \sin \theta), \theta \in\left(0, \frac{\pi}{2}\right)$, then the equation of the line $A B$ is
$4 x \cos \theta+3 y \sin \theta=9$.
Thus, $|O M|=\frac{9}{4 \cos \theta},|O N|=\frac{3}{\sin \theta}$,
$S_{\triangle M O N}=\frac{1}{2}|O M| \cdot|O N|=\frac{27}{4 \sin 2 \theta} \geqslant \frac... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,891 |
7. Real numbers $x, y, z$ satisfy
$$
x^{2}+2 y=7, y^{2}+4 z=-7, z^{2}+6 x=-14 \text {. }
$$
Then $x^{2}+y^{2}+z^{2}$ equals $\qquad$ | $$
\begin{array}{l}
\left(x^{2}+6 x\right)+\left(y^{2}+2 y\right)+\left(z^{2}+4 z\right)=-14 . \\
\text { Therefore, }(x+3)^{2}+(y+1)^{2}+(z+2)^{2}=0 .
\end{array}
$$
So, $x=-3, y=-1, z=-2$. | x=-3, y=-1, z=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,892 |
8. Let $S$ be a non-empty subset of the set $\{1,2, \cdots, 15\}$. If a positive integer $n$ satisfies: $n \in S, n+|S| \in S$, then $n$ is called a "model number" of the subset $S$, where $|S|$ denotes the number of elements in the set $S$. For all non-empty subsets $S$ of the set $\{1,2, \cdots, 15\}$, the sum of the... | $8.13 \times 2^{12}$.
Just find, for each $n$, how many subsets there are such that $n$ is a model number, and then sum over $\mathrm{F} n$.
If $n$ is a model number of $S$, and $S$ contains $k$ elements, then $n \in S, n+k \in S$. Therefore, $k \geqslant 2$.
Also, $k \leqslant 15-n$, and the other $k-2$ elements of... | 13 \times 2^{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,893 |
$$
\begin{array}{l}
\text { 9. For } \frac{1}{2} \leqslant x \leqslant 1 \text {, when } \\
(1+x)^{5}(1-x)(1-2 x)^{2}
\end{array}
$$
reaches its maximum value, $x=$ $\qquad$ . | 9. 8
The maximum value of the expression $[\alpha(1+x)]^{5}[\beta(1-x)][\gamma(2 x-1)]^{2}$, where $\alpha, \beta, \gamma$ are positive integers satisfying $5 \alpha-\beta+4 \gamma=0$, $\alpha(1+x)=\beta(1-x)=\gamma(2 x-1)$. It is given that
$$
\begin{array}{l}
\beta-\alpha \\
\beta+\alpha
\end{array}=\frac{\beta+\gam... | x=\frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,894 |
10. The function $f(x)$ satisfies: for any real numbers $x, y$, we have
$$
\frac{f(x) f(y)-f(x y)}{3}=x+y+2 .
$$
Then the value of $f(36)$ is | 10.39.
Let $x=y=0$, then $\frac{f^{2}(0)-f(0)}{3}=2$.
Therefore, $f(0)=-2$ or $f(0)=3$.
If $f(0)=-2$, let $y=0$, then $\frac{f(x) f(0)-f(0)}{3}=x+2$. So, $f(x)=-\frac{3}{2} x-2$. Substituting into the original equation shows it does not satisfy the condition.
If $f(0)=3$, solving yields $f(x)=x+3$, substituting for v... | 39 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,895 |
11. The volume of the regular tetrahedron $ABCD$ is $1, O$ is its center. The regular tetrahedron $A'B'C'D'$ is symmetric to the regular tetrahedron $ABCD$ with respect to point $O$. Then the volume of the common part of these two regular tetrahedrons is $\qquad$ . | 11. ${ }_{2}^{1}$.
If we place the tetrahedron $A-BCD$ on a horizontal plane, it is easy to see that the distance from the center to point $A$ is $\frac{3}{4}$ of the distance from point $A$ to the base. Therefore, the reflective symmetry plane is a horizontal plane equidistant from point $A$ and the base. Thus, it cu... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,896 |
12. On the hyperbola $x y=1$, the point with the abscissa $\frac{n}{n+1}$ is $A_{n}$, and the point with the abscissa $\frac{n+1}{n}$ is $B_{n}\left(n \in \mathbf{N}_{+}\right)$. The point with coordinates $(1,1)$ is denoted as $M, P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$. Then $... | $12.200 \frac{50}{101}$. $|A_{n} M|=|B_{n} M|$, and $k_{A_{n} B_{n}}=-1$. Therefore, $\triangle M A_{n} B_{n}$ is an isosceles triangle with $A_{n} B_{n}$ as the base, and the slope of the base is -1. Since point $M$ lies on the line $y=x$, the equation of the perpendicular bisector of the base is $y=x$. This gives $x_... | 200 \frac{50}{101} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,897 |
13. As shown in Figure 1, given that the incenter of $\triangle ABC$ is $I, AC \neq BC$, the incircle is tangent to sides $AB, BC, CA$ at points $D, E, F$ respectively, and $S = CI \cap EF$. Connect $CD$ to intersect the incircle again at point $M$, and the tangent line through point $M$ intersects the extension of $AB... | 13. (1) In the right triangle $\triangle CFI$, by the projection theorem, we have $FI^2 = SI \cdot CI = DI^2$.
Therefore, $\frac{DI}{SI} = \frac{CI}{DI}$.
And $\angle CID = \angle DIS$, so $\triangle CDI \sim \triangle DSI$.
(2) As shown in Figure 2, connect
$IM$, $IF$. Since $D$, $I$,
$M$, $G$ are concyclic, and
from ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,898 |
Example 8 Given that the centroid $C$ of $\triangle A B C$ has a symmetric point $G^{\prime}$ with respect to side $B C$. Prove: $A 、 B 、 G^{\prime} 、 C$ are concyclic if and only if $A B^{2}+A C^{2}=2 B C^{2}$. | Proof: As shown in Figure 9, let $AD$, $BE$, and $CF$ be the medians of $\triangle ABC$. Symmetric to $BC$, then $\angle BGC = \angle BG'C$.
(1) If $A$, $B$, $G'$, and $C$ are concyclic, then
$\angle BG'C + \angle BAC = 180^\circ$.
Also, $\angle ECF = \angle BGC = \angle BC'C$, so,
$\angle ECF + \angle EAF = 180^\circ$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,899 |
14. Let $a$, $b$, $c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$. | 14. Let the two real roots of the equation be $x_{1}, x_{2}$. By Vieta's formulas, we know that $x_{1}, x_{2}$ are both negative.
From $\frac{c}{a}=x_{1} x_{2}9$. Therefore, $b^{2} \geqslant 4 a c=4 \times \frac{a}{c} \times c^{2}>4 \times 9 \times 1^{2}=36$.
Solving this, we get $b>6$. Hence, $b \geqslant 7$.
Also, $\... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,900 |
15. Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10,|B|=9$, and set $A$ satisfies the following condition: if $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$. Let $A+B=\{a+b \mid a \in A$, $b \in B\}$. Prove: $|A+B| \geqslant 50(|X|$ represents the number of elements in set $X$). | 15. Consider the general case.
Let $|A|=m,|B|=n, A+B=\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$.
For any $1 \leqslant i \leqslant k$, let $s_{i}$ have $f(i)$ ways to be expressed in the form $a+b$, where $a \in A, b \in B$, i.e.,
$$
s_{i}=a_{i 1}+b_{i 1}=a_{i 2}+b_{i 2}=\cdots=a_{i f(i)}+b_{i f(i)} .
$$
Then it is c... | 50 | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,901 |
1. In the arithmetic sequence $\left\{a_{n}\right\}$, $S_{n}$ is the sum of the first $n$ terms of $a_{n}$, $S_{4}=1, S_{8}=4$. Then
$$
a_{2012}+a_{2003}+a_{2004}+a_{2005}
$$
is ( ).
(A) $\frac{2003}{2}$
(B) $\frac{2005}{2}$
(C) $\frac{2007}{2}$
(D) $\frac{2001}{2}$ | $$
\begin{array}{l}
-1 . \mathrm{A} . \\
S_{8}-2 S_{4} \\
=\left(a_{8}-a_{4}\right)+\left(a_{1}-a_{3}\right)+\left(a_{6}-a_{2}\right)+\left(a_{5}-a_{1}\right) \\
=16 d=2 .
\end{array}
$$
$$
\begin{array}{l}
\text { and } a_{2002}+a_{2003}+a_{2004}+a_{2005}-S_{4} \\
=4 \times 2001 d=\frac{2001}{2} .
\end{array}
$$
Ther... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,902 |
2. The function $f(x)$ is defined on $\mathbf{R}$ as an odd function, and for any $x \in \mathbf{R}$, we have
$$
f(x+10)=f(x)+f(5-x) .
$$
If $f(5)=0$, then the value of $f(2005)$ is $(\quad)$.
(A) 2000 (B) 2005 (C) 2008 (D) 0 | 2.D.
From the problem, we have $f((5-x)+10)=f(5-x)+f(x)$.
Therefore, $f(x+10)=f(15-x)=-f(x-15)$.
Thus, $f(x)=-f(x-25)=f(x-50)$.
Hence, $f(x)$ is a periodic function with a period of 50.
Therefore, $f(2005)=f(50 \times 40+5)=f(5)=0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,903 |
4. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $A_{1} B \perp$ $B_{1} C$. Then the angle formed by $A_{1} B$ and the plane $B B_{1} C_{1} C$ is equal to ( ).
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $\arctan \sqrt{2}$
(D) $90^{\circ}$ | 4.A.
As shown in Figure 1, let $M$ be the midpoint of $AB$, and connect $CM$ and $B_{1}M$. Clearly, $CM \perp$ plane $AA_{1}B_{1}B$.
By the theorem of three perpendiculars, $B_{1}M \perp A_{1}B$. Therefore,
$\triangle B_{1}BM \sim \triangle A_{1}B_{1}B$.
Thus, $A_{1}B_{1}: BB_{1} = \sqrt{2}: 1$.
Let $N$ be the midpo... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,905 |
5. The following numbers are 4 of the possible sums of the numbers facing up when three dice are rolled simultaneously. Among them, ( ) has the highest probability of appearing in one roll.
(A) 7
(B) 8
(C) 9
(D) 10 | 5.D.
The probability of getting a sum of 7 is $\frac{15}{6^{3}}$; the probability of getting a sum of 8 is $\frac{21}{6^{3}}$; the probability of getting a sum of 9 is $\frac{25}{6^{3}}$; the probability of getting a sum of 10 is $\frac{27}{6^{3}}$. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,906 |
6. Use 4 different colors to paint the faces of a regular tetrahedron, with each face painted one color, and no face left unpainted. There are () different ways to do this.
(A) 48
(B) 36
(C) 42
(D) 47 | 6. B.
Using one color has $\mathrm{C}_{4}^{1}=4$ different ways of coloring; using two colors has $\mathrm{C}_{4}^{2} \times 3=18$ different ways of coloring; using three colors has $\mathrm{C}_{4}^{3} \cdot \mathrm{C}_{3}^{1}$ $=12$ different ways of coloring; using four colors has $\mathrm{C}_{4}^{4} \cdot \mathrm{C... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,907 |
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