problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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Example 6 In $\triangle A B C$, it is known that $\frac{1}{2} A C>A B$. Prove: $\frac{1}{2} \angle A B C>\angle A C B$. | Proof: Since $A C>2 A B>A B$, therefore,
$\angle A B C>\angle A C B$.
As shown in Figure 6, construct $\angle A B D$
$=\angle A C B$, intersecting $A C$ at
D.
It suffices to prove
$\angle C B D>\angle A C B$.
Since $\triangle B A D \backsim$
$\triangle C A B$, therefore,
$$
\frac{B C}{B D}=\frac{A C}{A B}>2,
$$
i.e., ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,080 |
10.6. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Let $PQ$ and $RS$ be the external common tangents of these two circles (points $P$ and $R$ are on circle $\omega_{1}$, points $Q$ and $S$ are on circle $\omega_{2}$). Given that $RB \parallel PQ$, the ray $RB$ intersects circle $\omega_{2}$ ag... | 10.6. As shown in Figure 7, let the intersection point of line $AB$ and $PQ$ be $X$. Then
$$
\begin{array}{l}
X P^{2}=A X \cdot X B \\
=X Q^{2},
\end{array}
$$
which means $X$ is the midpoint of segment $PQ$.
It is easy to see that $AB \parallel PR$.
In fact, they are both perpendicular to the line connecting the cent... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,081 |
10.7. In a convex polyhedron, the degree of vertex $A$ (the number of edges emanating from a vertex is called the degree of that vertex) is 5, and the degree of all other vertices is 3. Each edge is colored blue, red, or purple. If the 3 edges emanating from any 3-degree vertex are exactly colored with 3 different colo... | 10.7. Observe any good coloring method. Note that, the total number of endpoints of edges of any color is even. In every good coloring method, since the number of edges of three different colors converging at each 3-degree vertex is 1 (thus, their parity is the same, and all are odd), the number of edges of three diffe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,082 |
11.1. Let $k>10$. Prove: In
$$
f(x)=\cos x \cdot \cos 2 x \cdot \cos 3 x \cdots \cdots \cos 2^{k} x
$$
one $\cos$ can be replaced by $\sin$, such that the resulting $f_{1}(x)$, for all real numbers $x$, satisfies $\left|f_{1}(x)\right| \leqslant \frac{3}{2^{k+1}}$. | 11.1. Since
$$
\begin{array}{l}
|\sin 3 x|=\left|3 \sin x-4 \sin ^{3} x\right| \\
=\left|3-4 \sin ^{2} x\right||\sin x| \leqslant 3|\sin x|,
\end{array}
$$
thus, if in $f(x)$ we replace $\cos 3 x$ with $\sin 3 x$, then the resulting $f_{1}(x)$ will satisfy
$$
\begin{array}{l}
\left|f_{1}(x)\right| \leqslant 3|\sin x||... | \frac{3}{2^{k+1}} | Algebra | proof | Yes | Yes | cn_contest | false | 719,083 |
11.2. Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $A_1$, $B_1$, and $C_1$ respectively. Connect $AA_1$, intersecting the incircle at point $Q$. Line $l$ is parallel to $BC$ and passes through point $A$. Lines $A_1C_1$ and $A_1B_1$ intersect line $l$ at points $P$ and $R$ respectively.... | 11.2. As shown in Figure 9. On one hand, since $BC$ is a tangent to the incircle, then
$$
\begin{array}{l}
\angle B A_{1} Q=\frac{1}{2} \overparen{A_{1} C_{1} Q} \\
=\angle A_{1} B_{1} Q . \\
\text { On the other hand, since }
\end{array}
$$
On the other hand, since the alternate interior angles formed by parallel lin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,084 |
11.4. In the infinite sequence $\left\{x_{n}\right\}$, the first term $x_{1}$ is a rational number greater than 1, and for any positive integer $n$, we have
$$
x_{n+1}=x_{n}+\frac{1}{\left[x_{n}\right]} \text {. }
$$
Prove: the sequence contains an integer term. | 11.4. If $\left[x_{n}\right]>\left[x_{n}-1\right]$, then $x_{n}$ is called a "special term". Clearly, there are infinitely many special terms (if $\left[x_{n}\right]=k$, then $\left.\left[x_{n+k}\right]>\left[x_{n}\right]\right)$.
Each special term's fractional part is expressed as a reduced fraction (if the special t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,085 |
11.6. Do there exist non-zero real numbers $a$, $b$, $c$, such that for any $n (n>3)$, there is a polynomial
$$
P_{n}(x)=x^{n}+\cdots+a x^{2}+b x+c,
$$
which has exactly $n$ integer roots (which may be the same)? | 11.6. Does not exist.
Assume there exist such non-zero real numbers $a$, $b$, $c$. By Vieta's formulas, they are all integers, and for any $n (n>3)$, either $c$ or $-c$ is the product of all the roots of $P_{n}(x)$, and thus, is the product of $n$ integers. For each such $P_{n}(x)$, consider all roots that are not $\p... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,086 |
11.7. Lisha chooses two skew edges of a given tetrahedron and constructs two spheres, each with one of these edges as its diameter. Can she always ensure that any point in the tetrahedron is contained in at least one of the spheres? | 11.7. Let the given tetrahedron be denoted as $A-BCD$. Find the pair of skew edges with the maximum sum of the squares of their lengths, and assume without loss of generality that these edges are $AB$ and $CD$.
Next, we prove that the spheres with diameters $AB$ and $CD$ cover every edge of the tetrahedron (we will on... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,087 |
Example 7 Given any four points in a plane, where no three points are collinear. Can we always select three points from these four to form a triangle such that this triangle has at least one interior angle not greater than $45^{\circ}$? Prove your conclusion. | Proof: Discuss according to the size of the interior angles.
(1) As shown in Figure 7, if quadrilateral $ABCD$ is a convex quadrilateral, then there must be an interior angle not greater than $90^{\circ}$. Without loss of generality, let it be $\angle A$. Thus,
$\angle A = \angle BAC + \angle CAD \leqslant 90^{\circ}$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,091 |
3. The minimum value of the fraction $\frac{5 x^{2}+30 x y+51 y^{2}}{x^{2}+6 x y+11 y^{2}}$ is ( ).
(A) -5
(B) -3
(C) 5
(D) 3 | 3. D.
$$
\begin{array}{l}
\text { Original expression }=5-\frac{4 y^{2}}{x^{2}+6 x y+11 y^{2}} \\
=5-\frac{4 y^{2}}{(x+3 y)^{2}+2 y^{2}} \\
=5-\frac{4}{\left(\frac{x}{y}+3\right)^{2}+2} .
\end{array}
$$
When $\frac{x}{y}=-3$, the original expression takes the minimum value 3. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,092 |
5. A set of 3 parallel lines on a plane intersects with another set of 5 parallel lines, the number of parallelograms that can be formed is ( ).
(A) 24
(B) 28
(C) 30
(D) 32 | 5.C.
Selecting any two lines from one group and any two lines from another group can form a parallelogram. The number of ways to choose any two lines from three parallel lines is 3, and the number of ways to choose any two lines from five parallel lines is 10. Therefore, the number of parallelograms is $3 \times 10=30... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,094 |
6. As shown in Figure 2, let $O$ be a point inside $\triangle ABC$. Connect $AO$, $BO$, and $CO$, and extend them to intersect $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Given that $S_{\triangle AOB}: S_{\triangle BOC}: S_{\triangle AOC}=$ 3:4:6. Then $\frac{OD}{AO} \cdot \frac{OE}{BO} \cdot \frac{... | 6. B.
Since $\frac{S_{\triangle A O B}}{S_{\triangle A B C}}=\frac{3}{13}, \frac{S_{\triangle B O C}}{S_{\triangle A B C}}=\frac{4}{13}, \frac{S_{\triangle A O C}}{S_{\triangle A B C}}=\frac{6}{13}$, then $\frac{O F}{C F}=\frac{3}{13}, \frac{O D}{A D}=\frac{4}{13}, \frac{O E}{B E}=\frac{6}{13}$. Therefore, $\frac{O F}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,095 |
1. Let $N=99999$. Then $N^{3}=$ | $$
\text { Two, 1.999970000 } 299999 \text {. }
$$
Since $N=10^{5}-1$, then
$$
\begin{array}{l}
N^{3}=\left(10^{5}-1\right)^{3}=10^{15}-3 \times 10^{10}+3 \times 10^{5}-1 \\
=10^{10}\left(10^{5}-3\right)+3 \times 10^{5}-1 \\
=999970000299999 .
\end{array}
$$ | 999970000299999 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,096 |
2. If $x-y=1, x^{3}-y^{3}=2$, then $x^{4}+y^{4}=$
$\qquad$ $x^{5}-y^{5}=$ $\qquad$ . | 2. $\frac{23}{9}, \frac{29}{9}$.
Since $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=2, x-y=1$, we have
$$
x^{2}+x y+y^{2}=2 \text{. }
$$
Also, $x^{2}-2 x y+y^{2}=1$, combining this with the above equation, we get
$$
x y=\frac{1}{3}, x^{2}+y^{2}=\frac{5}{3} \text{. }
$$
Therefore, $x^{4}+y^{4}=\left(x^{2}+y^{2}\rig... | \frac{23}{9}, \frac{29}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,097 |
3. Given Rt $\triangle A B C$ with the two legs $A C=5$, $B C=12, D$ is a point on $B C$. When $A D$ is the angle bisector of $\angle A$, then $C D=$ $\qquad$ | 3. $\frac{10}{3}$.
As shown in Figure 5, construct Rt $\triangle ABC$,
on side $AB$, mark $AE = AC$, draw $ED \perp AB$, intersecting $BC$ at point $D$. Then $AD$ is the angle bisector of $\angle A$.
Let $CD = x$. By $BE = 13 - 5 = 8$, $DE = x$, $BD = 12 - x$ and the Pythagorean theorem, we get $x^2 + 8^2 = (12 - x)^... | \frac{10}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,098 |
4. Two students, A and B, play chess. Winning a game earns 2 points, drawing a game earns 1 point each, and losing a game earns 0 points. They play three consecutive games, and the one with more points wins. What is the probability that A wins? | 4. $\frac{10}{27}$.
Each game has three possible outcomes: win, draw, or lose. For three games, there are $3 \times 3 \times 3=27$ possible outcomes. The scenarios where Player A wins are: winning all three games for 6 points, which has 1 scenario; winning two games and drawing one for 5 points, which has 3 scenarios;... | \frac{10}{27} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,099 |
Three, (18 points) Given that the line $l$ passes through the point $P(2,1)$, intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and $P A = P B$.
(1) Find the function expression of the line $l$;
(2) Let $\odot Q$ be the incircle of the right triangle $\triangle A O B$, which is tangent to $O A$, ... | (1) As shown in Figure 6, establish a coordinate system. Construct $\mathrm{Rt} \triangle A O B$ according to the problem. Draw $P H \perp O A$, with the foot of the perpendicular at $H$.
Since $P A = P B$, it follows that $O H = H A$.
Therefore, $A(4,0)$.
Let the function of line $l$ be $y = k x + b$.
Since points $... | x^{2} - 2 \sqrt{5} x + 4 = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,100 |
Example 1 Let $p, q, r$ be positive numbers, and satisfy $pqr=1$. Prove: for all $n \in \mathbf{N}_{+}$, we have
$$
\frac{1}{p^{n}+q^{n}+1}+\frac{1}{q^{n}+r^{n}+1}+\frac{1}{r^{n}+p^{n}+1} \leqslant 1 \text {. }
$$
(2004, Baltic Region Mathematical Competition) | Proof: By making the substitution $a=p^{n}, b=q^{n}, c=r^{n}$, then $a b c=1$, and the original inequality is equivalent to
$$
\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1 .
$$
By the rearrangement inequality
$$
\begin{array}{l}
a+b=a^{\frac{2}{3}} a^{\frac{1}{3}}+b^{\frac{2}{3}} b^{\frac{1}{3}} \geqsla... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,102 |
Five, (16 points) When $x \leqslant y \leqslant z$, find the positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{8}$. | First, given $x>1$, and $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}>0$. Therefore,
$$
\begin{array}{l}
\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x} \\
\Rightarrow \frac{1}{x}<\frac{7}{8} \leqslant \frac{3}{x} \\
\Rightarrow \frac{8}{7}<x \leqslant \frac{24}{7} .
\end{array}
$$
Thu... | (2,3,24) \text{ and } (2,4,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,103 |
Six. (16 points) As shown in Figure 3, given that $A D$ is the angle bisector of $\triangle A B C$, $A B < A C$, and a segment $C E = A B$ is cut off on $A C$. $M$ and $N$ are the midpoints of $B C$ and $A E$, respectively. Prove that $M N \parallel A D$.
untranslated part:
In the given problem, we need to prove that... | Six, as shown in Figure 7, connect $B E$. Let the midpoint of $B E$ be $F$, and connect $F N$ and $F M$.
Since $F N$ is the midline of $\triangle E A B$, we have
$$
F N=\frac{1}{2} A B,
$$
and $F N \parallel A B$.
Since $F M$ is the midline of $\triangle B C E$, we have
$$
F M=\frac{1}{2} C E \text {, and } F M \para... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,104 |
1. Given that the graph of the function $y=f^{-1}(x)$ passes through the point $(1, 0)$. Then the inverse function of $f\left(\frac{1}{3} x-2\right)$ must pass through the point ( ).
(A) $(1,6)$
(B) $(6,1)$
(C) $(0,6)$
(D) $(6,0)$ | -1.A.
Since the graph of $y=f(x)$ passes through the point $(0,1)$, therefore, the graph of $y=f\left(\frac{1}{3} x-2\right)$ passes through the point $(6,1)$, and its inverse function's graph passes through the point $(1,6)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,105 |
2. Given in rectangle $A B C D$, $A B=2, B C=3$. Then the modulus of $\boldsymbol{A B}+\boldsymbol{B} \boldsymbol{C}+\boldsymbol{A} \boldsymbol{C}$ equals ( ).
(A) 0
(B) 5
(C) $\sqrt{13}$
(D) $2 \sqrt{13}$ | 2.D.
Since $A B+B C+A C=2 A C$, and
$$
|A C|=\sqrt{A B^{2}+B C^{2}}=\sqrt{13} \text {, }
$$
Therefore, $|A B+B C+A C|=2 \sqrt{13}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,106 |
3. Given $\sin \alpha+\cos \alpha=\frac{7}{5}$, and $\tan \alpha>1$. Then $\boldsymbol{\operatorname { c o s }} \alpha=(\quad$. .
(A) $-\frac{3}{5}$
(B) $-\frac{4}{5}$
(C) $\frac{3}{5}$
(D) $\frac{4}{5}$ | 3.C.
Combining $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we easily get $\cos \alpha=\frac{3}{5}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,107 |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10 | 4.C.
Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,108 |
5. $\tan 70^{\circ} \cdot \cos 10^{\circ}\left(\sqrt{3} \tan 20^{\circ}-1\right)=$ ).
(A) 1
(B) 2
(C) -1
(D) -2 | $\begin{array}{l}\text { 5. C. } \\ \text { Original expression }=\frac{\cos 20^{\circ} \cdot \cos 10^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sqrt{3} \sin 20^{\circ}-\cos 20^{\circ}}{\cos 20^{\circ}} \\ =\frac{\cos 10^{\circ} \cdot 2 \sin \left(20^{\circ}-30^{\circ}\right)}{\sin 20^{\circ}} \\ =-\frac{2 \cos 10^{\circ} ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,109 |
7. Given $\boldsymbol{A} \boldsymbol{B}=(k, 1), \boldsymbol{A} \boldsymbol{C}=(2,3)$. Then the value of $k$ that makes $\triangle A B C$ a right triangle is ( ).
(A) $\frac{3}{2}$
(B) $1-\sqrt{2}$
(C) $1-\sqrt{3}$
(D) $-\sqrt{5}$ | 7.C.
If $\angle B A C=90^{\circ}$, then $A B \cdot A C=2 k+3=0$.
Solving for $k$ gives $k=-\frac{3}{2}$.
If $\angle A B C=90^{\circ}$, then
$$
C B \cdot A B=(A B-A C) \cdot A B=k^{2}-2 k-2=0 \text {. }
$$
Solving for $k$ gives $k=1 \pm \sqrt{3}$.
If $\angle A C B=90^{\circ}$, then
$$
C B \cdot A C=(A B-A C) \cdot A C... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,111 |
8. Given $x, y \in(-\sqrt{2}, \sqrt{2})$, and $x y=1$. Then the minimum value of $\frac{2}{2-x^{2}}+\frac{4}{4-y^{2}}$ is ( ).
(A) $\frac{20}{7}$
(B) $\frac{12}{7}$
(C) $\frac{16+4 \sqrt{2}}{7}$
(D) $\frac{16-4 \sqrt{2}}{7}$ | 8. C.
From the given, we have $y=\frac{1}{x}$. Therefore,
$$
\begin{array}{l}
\frac{2}{2-x^{2}}+\frac{4}{4-y^{2}}=\frac{2}{2-x^{2}}+\frac{4}{4-\frac{1}{x^{2}}} \\
=\frac{-4 x^{4}+16 x^{2}-2}{-4 x^{4}+9 x^{2}-2}=1+\frac{7 x^{2}}{-4 x^{4}+9 x^{2}-2} \\
=1+\frac{7}{9-\left(4 x^{2}+\frac{2}{x^{2}}\right)} .
\end{array}
$$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,112 |
Example 2 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$. Prove:
$$
\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \\
\geqslant 0 .
\end{array}
$$ | $$
\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}-\frac{x^{5}-x^{2}}{x^{3}\left(x^{2}+y^{2}+z^{2}\right)} \\
=\frac{x^{2}\left(x^{3}-1\right)^{2}\left(y^{2}+z^{2}\right)}{x^{3}\left(x^{5}+y^{2}+z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)} \geqslant 0 . \\
\text { Therefore, } \sum \frac{x^{5}-x^{2}}{x^{5}+y^{2}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,113 |
9. The vertices of a quadrilateral lie on the sides of a square with side length 1. If the sum of the squares of the four sides is $t$, then the range of values for $t$ is ( ).
(A) $[1,2]$
(B) $[2,4]$
(C) $[1,3]$
(D) $[3,6]$ | 9.B.
As shown in Figure 1.
$$
\begin{array}{l}
t=a^{2}+b^{2}+c^{2}+d^{2} \\
=x^{2}+(1-n)^{2}+n^{2}+(1-m)^{2}+m^{2}+ \\
\quad(1-y)^{2}+y^{2}+(1-x)^{2} .
\end{array}
$$
Since $0 \leqslant x < 1$, we have
$$
\begin{array}{l}
x^{2}+(1-x)^{2} \\
=1-2 x(1-x) \leqslant 1, \\
x^{2}+(1-x)^{2} \\
=\frac{1+(2 x-1)^{2}}{2} \geqs... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,114 |
10. Chaoyang Electrical Factory and Hongxing Electrical Factory had equal output values in January 2005. The output value of Chaoyang Electrical Factory increased monthly by a constant amount, while the output value of Hongxing Electrical Factory increased monthly by a constant percentage. It is known that in January 2... | 10.A.
Let the output value of two factories in January 2005 be $a$, the monthly increase in output value of Chaoyang Electrical Factory be $d$, and the output value of Hongxing Electrical Factory be $q(q>1)$ times that of the previous month, then
$$
a+12 d=a q^{12} .
$$
The output values of the two factories in July ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,115 |
11. If the three medians $A D$, $B E$, $C F$ of $\triangle A B C$ intersect at point $M$, then $M A+M B+M C=$ | Ni, 11.0.
Diagram (omitted). Let the midpoint of $A B$ be $D$. By the parallelogram rule, we have
$$
\begin{array}{l}
M A+M B=2 M D=-M C \text {. } \\
\text { Therefore, } M A+M B+M C=0 \text {. } \\
\end{array}
$$ | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,116 |
12. For a real number $x$, it is defined that $[x]=n$ if and only if $n \leqslant x<n+1$ $\left(n \in \mathbf{N}_{+}\right)$. Then the solution set of the inequality
$$
4[x]^{2}-36[x]+45<0
$$
is . $\qquad$ | $12.2 \leqslant x<8$.
From $4[x]^{2}-36[x]+45<0$, we get $\frac{3}{2}<[x]<\frac{15}{2}$. Therefore, $2 \leqslant[x] \leqslant 7$. Hence, $2 \leqslant x<8$. | 2 \leqslant x<8 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,117 |
13. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1$ $\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\mathrm{S}_{2007}-2 \mathrm{~S}_{2006}+\mathrm{S}_{2000}=
$$
$\qquad$ | 13.3.
When $n$ is even, we have
$$
a_{1}+a_{2}=a_{3}+a_{4}=\cdots=a_{n-1}+a_{n}=1 .
$$
Thus, $S_{n}=\frac{n}{2}$.
When $n$ is odd, we have
$$
a_{1}=2, a_{2}+a_{3}=a_{4}+a_{5}=\cdots=a_{n-1}+a_{n}=1 \text {. }
$$
Thus, $S_{n}=2+\frac{n-1}{2}=\frac{n+3}{2}$.
Therefore, $S_{2007}-2 S_{2006}+S_{2005}$
$$
=1005-2 \times ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,118 |
14. In $\triangle A B C$, it is known that $A B=\sqrt{30}, A C=$ $\sqrt{6}, B C=\sqrt{15}$, point $D$ makes $A D$ bisect $B C$, and $\angle A D B=90^{\circ}$, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written in the form $\frac{m}{n}$, where $m, n$ are coprime positive integers. Then $m+n=$ | 14.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$. By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
Thus, $(\sqrt{30})^{2}-\left(\frac{x}{2}\right)^{2}$
$$
=\left(\frac{\sqrt{15}}{2}\right)^{2}-\left(\frac{x}{2}-\frac{\sqrt{57}}{2}\right)^{2} \text {. }
$$
Solving for $x$ gives $x=\frac{81}{\sqrt{57... | 65 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,119 |
15. (12 points) In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$. | Three, 15. From the given,
$\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B$.
Then, $\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C$ $=3 \sin B$.
Thus, $\sin A+\sin C+\sin (A+C)=3 \sin B$, which means $\sin A+\sin C=2 \sin B$.
Therefore, $2 \sin \frac{A+C}{2} \cdot \cos \frac{A... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,120 |
16. (12 points) Given the function
$$
f(x)=\frac{2 x^{2}+4 x+a}{x}, x \in[1,+\infty) \text {. }
$$
(1) When $a=2$, find the minimum value of the function $f(x)$;
(2) If for any $x \in[1,+\infty), f(x)>0$ always holds, try to find the range of the real number $a$.
| 16. (1) When $a=2$, $f(x)=2 x+\frac{2}{x}+4$.
Since $f(x)$ is an increasing function on $[1,+\infty)$, $f(x)$ has a minimum value of $f(1)=8$ on $[1,+\infty)$.
(2) On $[1,+\infty)$, $f(x)=\frac{2 x^{2}+4 x+a}{x}>0$ always holds if and only if $2 x^{2}+4 x+a>0$ always holds. Let
$$
g(x)=2 x^{2}+4 x+a=2(x+1)^{2}+a-2 \te... | a>-6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,121 |
17. (12 points) Given vectors
$$
\begin{array}{l}
\boldsymbol{a}=\left(\cos \frac{3}{2} x, \sin \frac{3}{2} x\right), \\
\boldsymbol{b}=\left(\cos \frac{x}{2},-\sin \frac{x}{2}\right), \text { and } x \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right] .
\end{array}
$$
(1) Find the range of $|a+b|$;
(2) Find the minimum val... | 17. (1) Since $x \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$, therefore, $-1 \leqslant \cos 2 x \leqslant 1$.
Also, $|\boldsymbol{a}+\boldsymbol{b}|=\sqrt{2+2 \cos 2 x}$, then $0 \leqslant|\boldsymbol{a}+\boldsymbol{b}| \leqslant 2$.
(2) Since $x \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$, therefore, $-1 \leq... | -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,122 |
18. (15 points) Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and its first $n$ terms sum $S_{n}$ satisfies $S_{n}=\frac{1}{4}\left(a_{n}-1\right)\left(a_{n}+3\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{S_{n}}$, try to find the first $n$ te... | 18. (1) From $a_{1}=S_{1}=\frac{1}{4}\left(a_{1}-1\right)\left(a_{1}+3\right)$ and $a_{n}>0$, we get $a_{1}=3$.
From $S_{n}=\frac{1}{4}\left(a_{n}-1\right)\left(a_{n}+3\right)$, we get $S_{n-1}=\frac{1}{4}\left(a_{n-1}-1\right)\left(a_{n-1}+3\right)$.
Thus, $a_{n}=\frac{1}{4}\left[\left(a_{n}^{2}-a_{n-1}^{2}\right)+2\... | \frac{3}{4}-\frac{2 n+3}{2(n+1)(n+2)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,123 |
Example 3 If $a_{i} \geqslant 0(1 \leqslant i \leqslant 4), \sum_{i=1}^{4} a_{i} a_{i+1}=$ 1, prove: $\sum_{i=1}^{4} \frac{a_{i}^{3}}{a_{i+1}+a_{i+2}+a_{i+3}} \geqslant \frac{1}{3}$, where, $a_{r+4}=a_{r}$.
(31st IMO Preliminary Problem) | Prove: Let $S=\sum_{i=1}^{4} a_{i}$, then
$$
\begin{array}{l}
\sum_{i=1}^{4}\left[\frac{a_{i}^{3}}{S-a_{i}}+\frac{a_{i}\left(S-a_{i}\right)}{9}\right] \geqslant \sum_{i=1}^{4} \frac{2}{3} a_{i}^{2} . \\
\text { Hence } \sum_{i=1}^{4} \frac{a_{i}^{3}}{S-a_{i}} \geqslant \frac{7}{9} \sum_{i=1}^{4} a_{i}^{2}-\frac{1}{9} S... | \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,124 |
19. (15 points) In mid-August 2006, the city of Zixing in Hunan Province encountered a once-in-a-century flood disaster. At point $O$ on the shore of Dongjiang Lake (consider the lake shore as a straight line), a rescue boat was parked. Due to the sudden breakage of the rope, the boat was blown away, with a direction m... | 19. As shown in Figure 3, suppose this person runs to point $A$ on the shore and then enters the water, catching up with the boat at point $B$. Let the boat's speed be $v$, the time it takes for the person to catch up with the boat be $t$, and the time the person spends chasing the boat on the shore be $k$ times $t$ $(... | 2 \sqrt{2} \mathrm{~km} / \mathrm{h} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,125 |
For real numbers $x, y$, the algebraic expression
$$
2 x^{2}+3 y^{2}-4 x y-5 x-6 y+7
$$
has the minimum value of ( ).
(A) $-28 \frac{7}{8}$
(B) $-27 \frac{5}{8}$
(C) $-26 \frac{3}{8}$
(D) $-25 \frac{1}{8}$ | - 1.C.
$$
\begin{array}{l}
2 x^{2}+3 y^{2}-4 x y-5 x-6 y+7 \\
=2 x^{2}-(4 y+5) x+3 y^{2}-6 y+7 \\
=2\left(x-\frac{4 y+5}{4}\right)^{2}+\left(y-\frac{11}{2}\right)^{2}-26 \frac{3}{8} .
\end{array}
$$
When $x-\frac{4 y+5}{4}=0, y-\frac{11}{2}=0$, i.e., $x=\frac{27}{4}, y=\frac{11}{2}$,
the expression (1) achieves its mi... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,126 |
2. Given that $a$, $b$, and $c$ are real numbers, and
$$
\sqrt[3]{(a+2008)^{8}+8}+\sqrt{|b+8|+1}=\sqrt{6 c-c^{2}} \text {. }
$$
Then the value of $\left(b+c^{2}\right)^{a}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 4 | 2.C.
It is easy to know that $(a+2008)^{8}+8 \geqslant 8$,
$$
|b+8|+1 \geqslant 1 \text {. }
$$
Then $\sqrt[3]{(a+2008)^{8}+8}+\sqrt{|b+8|+1} \geqslant \sqrt[3]{8}+\sqrt{1}=3$
$$
\Rightarrow \sqrt{6 c-c^{2}} \geqslant 3 \Rightarrow 6 c-c^{2} \geqslant 9 \Rightarrow(c-3)^{2} \leqslant 0 \text {. }
$$
Since $(c-3)^{2}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,127 |
3. In a right trapezoid $A B C D$, $A D / / B C$, $B C \perp C D$, $E$ is the midpoint of side $A B$, $B C=C D=C E$. Then the degree measure of $\angle B$ is ( ).
(A) $52.5^{\circ}$
(B) $62.5^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 3.D.
As shown in Figure 6, draw perpendiculars from points $A$ and $E$ to $BC$, with the feet of the perpendiculars being $G$ and $F$, respectively. Clearly, quadrilateral $AGCD$ is a rectangle.
Thus, $AG = CD$.
Also, $EF \parallel AG$, and $E$ is the midpoint of $AB$, so
$$
EF = \frac{1}{2} AG = \frac{1}{2} CD = \fra... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,128 |
5. Three people, A, B, and C, are playing a game. Each of them writes 100 different Chinese characters, and then they compare the characters written by each person. If a character is written by at least two people, it is deleted until there are no more identical characters. A, B, and C deleted 31, 27, and 39 characters... | 5.C.
Suppose there are $a$ Chinese characters that are the same for all three people, $b$ Chinese characters that are the same only for person A and B, $c$ Chinese characters that are the same only for person A and C, and $d$ Chinese characters that are the same only for person B and C. From the problem, we have:
$$
\... | 242 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,130 |
6. A "level-passing game" stipulates: On the $n$-th level, a die must be rolled $n$ times. If the sum of the points that appear in these $n$ rolls is greater than $\frac{3^{n}}{4}$, it counts as passing the level; otherwise, it does not count as passing. (Note: The die is a uniform cube with the numbers $1, 2, 3, 4, 5,... | 6.B.
To pass the first level, the score must be greater than $\frac{3}{4}$. Clearly, rolling a die once yields at least 1 point. Therefore, (1) is correct.
Since $4 \times 6=24, \frac{3^{4}}{4}=20 \frac{1}{4}, 24>20 \frac{1}{4}$, the fourth level can be passed. Therefore, (2) is correct.
Since $5 \times 6=30, \frac{... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,131 |
1. There are weights of $11 \mathrm{~g}$ and $17 \mathrm{~g}$ available in sufficient quantity. To weigh an object of mass $3 \mathrm{~g}$ on a balance, at least $\qquad$ such weights are needed. | Let the weights of 11 g and 17 g be used $x$ times and $y$ times, respectively. Then we have $11 x - 17 y = 3$ or $17 y - 11 x = 3$.
The solutions are $\left\{\begin{array}{l}x=8+17 t, \\ y=5+11 t\end{array}\right.$ or $\left\{\begin{array}{l}x=9+17 t, \\ y=6+11 t .\end{array}\right.$
where $t$ is an integer.
Since $x$... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,132 |
2. Given the lengths of the three altitudes of $\triangle A B C$ are $\frac{2}{n+6} 、 \frac{1}{3 n+12} 、 \frac{1}{8-2 n}$. Then the range of values for $n$ is | $$
\left\{
\begin{array}{l}
-\frac{14}{11}(16-4 n) S, \\
(6 n+24) S+(16-4 n) S>(n+6) S, \\
(n+6) S+(16-4 n)>(6 n+24) S .
\end{array}\right.
$$
Solving yields $-\frac{14}{11}<n<-\frac{2}{9}$. | -\frac{14}{11}<n<-\frac{2}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,133 |
3. Given that $a$ is a constant, the equation in terms of $x$
$$
x^{3}+\left(2 a-a^{2}\right) x-2 a^{2}=0
$$
has only one real solution in the real number range. Then the range of values for $a$ is $\qquad$ . | 3. $0 \leqslant a<8$.
The original equation is equivalent to
$$
\begin{array}{l}
\left(x^{3}-a^{2} x\right)+\left(2 a x-2 a^{2}\right)=0 \\
\Leftrightarrow x(x+a)(x-a)+2 a(x-a)=0 \\
\Leftrightarrow(x-a)\left(x^{2}+a x+2 a\right)=0 \\
\Leftrightarrow x=a \text { or } x^{2}+a x+2 a=0 .
\end{array}
$$
(1) If the equation... | 0 \leqslant a<8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,134 |
Example 8 In $\triangle A B C$, points $D, E, F$ are on sides $B C, C A, A B$ respectively. Prove:
$$
\min \left\{S_{\triangle A E F}, S_{\triangle B F D}, S_{\triangle C D E}\right\} \leqslant \frac{1}{4} S_{\triangle A B C} .
$$ | Proof: Let $\min \left\{S_{\triangle A E F}, S_{\triangle B F D}, S_{\triangle C D E}\right\}=S$. As shown in Figure 9, note that
$$
\begin{array}{l}
\frac{S_{\triangle A E F}}{S_{\triangle A B C}} . \\
\frac{S_{\triangle B F D}}{S_{\triangle A B C}} \cdot \\
\frac{S_{\triangle C D E}}{S_{\triangle A B C}} \\
=\frac{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,135 |
Example 9 In a regular $\triangle ABC$ with side length 1, points $M, N, P$ lie on sides $BC, CA, AB$ respectively, and $MB + CN + AP = 1$. Find the maximum area of $\triangle MNP$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direct... | Solution: As shown in Figure 10, let
$$
\begin{array}{l}
B M=x, \\
C N=y, \\
A P=z .
\end{array}
$$
Then $0 \leqslant x, y, z \leqslant 1$,
$$
\begin{array}{l}
x+y+z=1 . \\
\text { Hence } S_{\triangle M P P}+S_{\triangle B P M}+S_{\triangle C W V} \\
= \frac{1}{2}[x(1-z)+y(1-x)+z(1-y)] \sin 60^{\circ} \\
= \frac{\sq... | \frac{\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,136 |
6. As shown in Figure 21, in $\triangle ABC$, $AB=AC$, $D$ is the midpoint of $BC$, and $E$ is any point in $\triangle ABD$. Connect $AE$, $BE$, and $CE$. Prove: $\angle AEB > \angle AEC$ | (提示: 如图 21, 作点 $E$关于 $A D$ 的对称点 $E^{\prime}$, 联结 $A E^{\prime} 、 C E^{\prime} 、 E E^{\prime}$, 并延长 $E E^{\prime}$交 $A C$ 于点 $F$. 根据对称性得 $\triangle A B E \cong \triangle A C E^{\prime}$. 所以, $\angle A E B=\angle A E^{\prime} C$. 易知 $\angle A E^{\prime} C=\angle A E^{\prime} F+\angle C E^{\prime} F>\angle A E F+\angle C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,137 |
II. (50 points) Given that the center of the square $ABCD$ is at the origin of the coordinate system, and the four vertices of the square are on the graph of the function $f(x)=x^{3}+a x$. Find the range of the real number $a$.
Translate the above text into English, please retain the original text's line breaks and fo... | Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$. Then $B\left(-y_{0}, x_{0}\right) 、 C\left(-x_{0},-y_{0}\right) 、 D\left(y_{0},-x_{0}\right)$.
According to the problem, we have
$y_{0}=x_{0}^{3}+a x_{0}$,
$-x_{0}=y_{0}^{3}+a y_{0}$.
(1) $\times y_{0}-$ (2) $\times x_{0}$ gives
$x_{0}^{2}+y_{0}^{2}=x_{0} y... | a \in(-\infty,-2 \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,138 |
Three. (50 points) Given a set $A$ consisting of $3 n$ different positive integers. If the elements in set $A$ satisfy:
(1) The sum of any $n$ elements is no greater than $N$;
(2) The sum of any $n+1$ elements is greater than $N$.
Find the minimum value of $N$ (expressed in terms of $n$). | Let $A=\left\{a_{1}, a_{2}, \cdots, a_{3 n}\right\}$, where $a_{1}N .
$$
From equations (1) and (2), the above two inequalities can be written as
$$
\sum_{i=1}^{n}\left(a_{n+1}+n-1+i\right) \leqslant N, \sum_{j=1}^{n+1}\left(a_{n+1}+1-j\right)>N \text {. }
$$
Then $n a_{n+1}+\frac{n(3 n-1)}{2} \leqslant N$,
$$
\begin... | \frac{4 n^{3}+3 n^{2}+n}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,139 |
7. Given a convex hexagon $A B C D E F$ with side lengths at most 1. Prove: at least one of the diagonals $A D$, $B E$, $C F$ is no more than 2. | (提示: 如图 22 , 联结 $A C 、 C E 、 E A$. 在 $\triangle A E C$ 中,不妨设边 $C E$ 最大, 即 $C E \geqslant A C, C E \geqslant A E$. 对 $A 、 C$ 、 $D 、 E$ 四点用托勒密不等式, 有
$$
A D \cdot C E \leqslant A C \cdot E D+C D \cdot A E,
$$
$$
\text { Hence } A D \leqslant \frac{A C}{C E} \cdot D E+C D \cdot \frac{A E}{C E} \leqslant 1 \times 1+1 \times... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,140 |
8. As shown in Figure 23, in the convex quadrilateral $ABCD$, $M$ and $P$ are the midpoints of $BC$ and $CD$ respectively. It is known that $AM + AP = a$. Prove:
$S_{\text{quadrilateral } ABCD} < \frac{1}{2} a^{2}$. | (提示: 如图 23, 联结 $A C 、 M P$. 则
$S_{\triangle M P}+\frac{1}{4} S_{\triangle B D C}=S_{\text {四边开 } A N C P}=\frac{1}{2} S_{\text {四边开 } 1 B C D}$.
又 $S_{\triangle B D C}<S_{\text {四边送 } A B C D}$,
$S_{\triangle M P P} \leqslant \frac{1}{2} A M \cdot A P$
$\leqslant \frac{1}{2} \cdot \frac{(A M+A P)^{2}}{4}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,141 |
Example 3 Let $a, b, c$ be positive real numbers, and satisfy $abc = 1$. Prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }
$$ | Analysis: Directly eliminating the denominator to transform into an integral inequality will lead to higher-degree polynomials, thus encountering greater difficulties. Starting from the known $abc=1$, let's see if we can reduce the degree first.
Make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$, the ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,144 |
Example 4 Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$
10\left(a^{3}+b^{3}+c^{3}\right)-9\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 \text {. }
$$
(2005, China Western Mathematical Olympiad) | Prove: The original inequality is equivalent to (to distinguish from the coefficient letters, the variables below are $x, y, z$)
$$
\begin{array}{l}
f(x, y, z) \\
= 10\left(x^{3}+y^{3}+z^{3}\right)(x+y+z)^{2}- \\
9\left(x^{5}+y^{5}+z^{5}\right)-(x+y+z)^{5} \\
\geqslant 0 .
\end{array}
$$
Simple calculations yield
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,145 |
Example 6 Let positive real numbers $x, y, z$ satisfy $xyz \geqslant 1$. Prove:
$$
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0 \text {. }
$$
(46th IMO) | Prove: Note that
$$
\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} x y z}{x^{5}+\left(y^{2}+z^{2}\right) x y z} \\
=\frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z} . \\
\text { and } \frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z}=\frac{2\left(x^{4}-x^{2} y z\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,147 |
Example 7 Let $m_{a}, m_{b}, m_{c}$ be the lengths of the medians of $\triangle ABC$ corresponding to the sides, and $r_{a}, r_{b}, r_{c}$ be the radii of the excircles of $\triangle ABC$ corresponding to the sides. Prove:
$$
\frac{r_{a} r_{b}}{m_{a} m_{b}}+\frac{r_{b} r_{c}}{m_{b} m_{c}}+\frac{r_{c} r_{a}}{m_{c} m_{a}... | Proof: Note the familiar formula
$$
\begin{aligned}
2 p=a+b+c, \\
r_{a}=\sqrt{\frac{p(p-b)(p-c)}{p-a}}, \\
m_{a}=\frac{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}{2} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Then } \frac{r_{b} r_{c}}{m_{b} m_{c}}=\frac{4 p(p-a)}{\sqrt{\left(2 a^{2}+2 c^{2}-b^{2}\right)\left(2 a^{2}+2 b^{2}-c^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,148 |
For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$. | Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$
As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,149 |
In the acute triangle $\triangle ABC$, $AB < AC$, $AD$ is the altitude on side $BC$, $P$ is a point on segment $AD$, and through $P$ draw $PE \perp AC$, with the foot of the perpendicular being $E$, and draw $PF \perp AB$, with the foot of the perpendicular being $F$. $O_{1}$ and $O_{2}$ are the circumcenters of $\tria... | Proof: Sufficiency.
If $P$ is the orthocenter of $\triangle ABC$, since the circumcenters of $\triangle BDF$ and $\triangle CDE$ are the midpoints of $BP$ and $CP$, respectively, the nine-point circle of $\triangle ABC$ passes through $O_1$, $O_2$, $E$, and $F$, i.e., $O_1$, $O_2$, $E$, and $F$ are concyclic.
Necessity... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,151 |
Example 1 The sequence $\left\{u_{n}\right\}$ is defined as
$$
\begin{array}{l}
u_{1}=1, \\
u_{n+1}=\frac{1}{16}\left(1+4 u_{n}+\sqrt{1+24 u_{n}}\right) \quad(n \geqslant 1)
\end{array}
$$
Prove: The sequence $\left\{u_{n}\right\}$ contains only $u_{1}$ as an integer. | Proof: Let $a_{n}=\sqrt{1+24 u_{n}}(n \geqslant 1)$, then $u_{n}=\frac{a_{n}^{2}-1}{24}, u_{n+1}=\frac{a_{n+1}^{2}-1}{24}$.
Substituting into the known recurrence relation, we get
$$
16 \times \frac{a_{n+1}^{2}-1}{24}=1+4 \times \frac{a_{n}^{2}-1}{24}+a_{n} \text {. }
$$
Simplifying, we get $4 a_{n+1}^{2}=a_{n}^{2}+6 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,152 |
Example 2 Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying
$$
\begin{array}{l}
\sqrt{a_{n} a_{n+1}+a_{n} a_{n+2}} \\
=3 \sqrt{a_{n} a_{n+1}+a_{n+1}^{2}}+2 \sqrt{a_{n} a_{n+1}},
\end{array}
$$
and $a_{1}=1, a_{2}=3$. Find the general term formula of the sequence $\left\{a_{n}\right\}$. | Solution: Dividing both sides of the given recurrence relation by $\sqrt{a_{n} a}$, we get
$$
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}=3 \sqrt{1+\frac{a_{n+1}}{a_{n}}}+2,
$$
which is equivalent to $\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}+1=3\left(\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1\right)$.
Let $b_{n}=\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1$.... | a_{n}=\left\{\begin{array}{ll}
1, & n=1 ; \\
\prod_{k=1}^{n-1}\left[\left(3^{k}-1\right)^{2}-1\right], & n>1 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,153 |
Example 3 Given the sequence $\left\{a_{n}\right\}$,
$$
a_{0}=1, a_{n}=\frac{\sqrt{1+a_{n-1}^{2}}-1}{a_{n-1}} \text {. }
$$
Prove: $a_{n}>\frac{\pi}{2^{n+2}}$. | Proof: Let $a_{n-1}=\tan \theta_{n-1}$. Then $a_{n}=\tan \theta_{n}$. Notice that
$$
a_{n}=\frac{\sqrt{1+a_{n-1}^{2}}-1}{a_{n-1}}=\frac{a_{n-1}}{\sqrt{1+a_{n-1}^{2}}+1} \text {. }
$$
Given $a_{0}=1>0$, we can get
$$
a_{2}>0, a_{3}>0, \cdots \cdots a_{n}>0 \text {. }
$$
Thus, $0<\theta_{n}<\frac{\pi}{2^{n+2}} \text {.... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,154 |
Example 4 Given $x_{1}=1, x_{n}=\sqrt[4]{a x_{n-1}}(a>0)$. Find the general term formula $x_{n}$ of the sequence. | Since $x_{1}=1>0$, we have
$$
x_{2}>0, x_{3}>0, \cdots \cdots x_{n}>0 \text {. }
$$
Taking the common logarithm of both sides of $x_{n}=\sqrt[4]{a x_{n-1}}$, we get
$$
\lg x_{n}=\frac{1}{4}\left(\lg x_{n-1}+\lg a\right) \text {. }
$$
Let $b_{n}=\lg x_{n}, b_{n-1}=\lg x_{n-1}$. Then
$$
b_{n}=\frac{1}{4}\left(b_{n-1}+\... | x_{n}=a^{\frac{1}{3}\left[1-\left(\frac{1}{4}\right)^{n-1}\right]} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,155 |
Example 5 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=10, a_{n+1}=\sqrt[n]{a_{n}}(n \geqslant 1)$. Find $a_{n}$. | Solution: Since $a_{1}=10>0$, we have
$$
a_{2}>0, a_{3}>0, \cdots \cdots a_{n+1}>0 .
$$
Taking the common logarithm of both sides of $a_{n+1}=\sqrt[n]{a_{n}}$ yields $\lg a_{n+1}=\frac{1}{n} \lg a_{n}$.
Let $b_{n}=\lg a_{n}$. Then $b_{n+1}=\lg a_{n+1}$.
Thus, $b_{n+1}=\frac{1}{n} b_{n}$, which means $\frac{b_{n+1}}{b_... | a_{n}=10^{(n-1)!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,156 |
Example 6 Let $x_{1}=1, x_{n+1}=\frac{x_{n}^{2}}{\sqrt{3 x_{n}^{4}+6 x_{n}^{2}+2}}$ $(n \geqslant 1)$. Find $x_{n}$. | Given the original recurrence relation, squaring both sides yields
$$
x_{n+1}^{2}=\frac{x_{n}^{4}}{3 x_{n}^{4}+6 x_{n}^{2}+2} \text {. }
$$
Let $y_{n}=x_{n}^{2}$. Then $y_{n+1}=x_{n+1}^{2}$.
Thus, $y_{n+1}=\frac{y_{n}^{2}}{3 y_{n}^{2}+6 y_{n}+2}$.
Therefore, $\frac{1}{y_{n+1}}=\frac{3 y_{n}^{2}+6 y_{n}+2}{y_{n}^{2}}$ ... | x_{n}=\frac{\sqrt{2}}{\sqrt{5^{2^{n-1}}-3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,157 |
Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition) | Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$
When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,158 |
Example 2-2 In $\triangle ABC$ with a fixed area $(S)$, its incircle touches sides $AB$, $AC$, and $BC$ at points $D$, $E$, and $F$, respectively. The following propositions are equivalent:
(1) $AC \cdot BC$ takes the minimum value;
(2) $\angle ACB=90^{\circ}$;
(3) $AD \cdot BD=S$. | Proof: According to the problem, by "the lengths of the two tangents from a point outside a circle to the circle are equal," we can set (as shown in Figure 2)
$$
\begin{array}{l}
x=B D=B F, \\
y=C E=C F, \\
z=A D=A E .
\end{array}
$$
Then the area of the triangle is
$$
\begin{array}{l}
S=\sqrt{p(p-a)(p-b)(p-c)} \\
=\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,159 |
Example 11 Find a point $P$ in a given plane such that the sum of its distances to the three vertices of $\triangle ABC$ is minimized (this point is called the Fermat point). | (1) Proof that point $P$ will not be outside $\triangle ABC$. As shown in Figure 12, divide the exterior of $\triangle ABC$ into 6 regions.
If point $P$ is in region I (as shown in Figure 13), then we have
$$
AB + AC \leq PB + PC < PA + PB + PC,
$$
which means the sum of the distances from point $A$ to the three verti... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,160 |
Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B ... | Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diamete... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,161 |
Question $P$ is any point on the incircle $\odot O$ of the equilateral $\triangle A_{1} A_{2} A_{3}$, and the distances from $P$ to $A_{1} A_{2}$, $A_{2} A_{3}$, and $A_{3} A_{1}$ are $d_{1}$, $d_{2}$, and $d_{3}$, respectively. Is $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}$ a constant when the position of point $P$ changes? Is $d... | Solution 1: When the position of point $P$ changes, $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}$ is a constant; $d_{1}^{4}+d_{2}^{4}+d_{3}^{4}$ is not a constant.
As shown in Figure 1, $\triangle A_{1} A_{2} A_{3}$ is inscribed in $\odot O$ at points $D$, $E$, and $F$. Suppose $P$ is any point on the minor arc $\overparen{D F}$, a... | d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = \frac{3}{8} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,162 |
2. The graphs of the functions $y=2x$ and $y=\frac{18}{x}$ intersect at points $A$ and $B$ (where $A$ is in the first quadrant). A perpendicular line $AC$ is drawn from $A$ to the $x$-axis, with the foot of the perpendicular being $C$. The area of $\triangle ABC$ is equal to ( ).
(A) 6
(B) 9
(C) 12
(D) 18 | 2.D.
Solve the system of equations $\left\{\begin{array}{l}y=2 x, \\ y=\frac{18}{x}\end{array}\right.$. The solutions are $A(3,6), B(-3,-6)$, so $C(3,0)$. Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 6 \times[3-(-3)]=18$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,163 |
3. In an isosceles right triangle $\triangle ABC$, $CA = CB = 3$, $E$ is a point on $BC$ such that $BE = 2$, and $P$ is a moving point on the hypotenuse $AB$. Then the minimum value of the sum of the lengths $PC + PE$ is ( ).
(A) $\sqrt{10}$
(B) $\sqrt{11}$
(C) $\sqrt{12}$
(D) $\sqrt{13}$ | 3.D.
As shown in Figure 2, construct the symmetric point $E_{1}$ of point $E$ with respect to $AB$, and connect $C E_{1}$ and $P E_{1}$. Then
$$
\begin{array}{l}
P C+P E \\
=P C+P E_{1} \geqslant C E_{1} \\
=\sqrt{B C^{2}+E_{1} B^{2}}=\sqrt{13} .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,164 |
4. Let $a=0.123456789101112 \cdots 998999$, where the digits are obtained by writing down the positive integers from 1 to 999 in sequence. Then the 2007th digit to the right of the decimal point is ( ).
(A) 0
(B) 4
(C) 5
(D) 6 | 4.C.
1 to 9 includes 9 digits; $10 \sim 99$ includes $2 \times 90=180$ digits; $100 \sim 999$ includes $3 \times 900=2700$ digits.
Since $2007-9-180=1818$, and $\frac{1818}{3}=606$, the 2007th digit is the 3rd digit of the 606th three-digit number 705, which is 5. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,165 |
5. Let $n$ be a certain positive integer. When substituting $n$ into the algebraic expression $n^{5}-n$ to calculate its value, four students came up with the following four results, of which only one is correct. Then this correct result is ( ).
()) 777 ()) 777 ()) 1776 ()) 7779 | 5.A.
Since $I=n^{5}-n=(n-1) n(n+1)\left(n^{2}+1\right)$, it is clear that $I$ is a multiple of 2, thus eliminating options (B) and (D). When $n=5k, 5k+1, 5k-1$, $I$ is a multiple of 5; when $n=5k-2, 5k+2$, $n^{2}+1$ is a multiple of 5. Therefore, $I$ is a multiple of 5. Hence, regardless of what positive integer $n$ i... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,166 |
6. In the convex quadrilateral $A B C D$, $\angle B=\angle D=$ $90^{\circ}, \angle C=120^{\circ}, A B=3, B C=\sqrt{3}$. Then $A D=$ ( ).
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) $3 \sqrt{3}$ | 6. B.
As shown in Figure 3, extend $BC$ and $AD$ to intersect at point $E$. Since $\angle BCD=120^{\circ}$, we have
$$
\begin{array}{l}
\angle ECD=60^{\circ} \Rightarrow \angle E=30^{\circ} \\
\Rightarrow EA=2AB=6 \\
\Rightarrow BE=\sqrt{6^{2}-3^{2}}=3\sqrt{3}
\end{array}
$$
Also, $BC=\sqrt{3} \Rightarrow CE=2\sqrt{3... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,167 |
1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ . | Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$
Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,168 |
Example 12 In a convex quadrilateral $ABCD$, we have
$$
AB \cdot CD + AD \cdot BC \geqslant AC \cdot BD \text{.}
$$
Equality holds if and only if the quadrilateral $ABCD$ is cyclic (this inequality is known as Ptolemy's inequality). | Proof: As shown in Figure 17, take a point \( M \) on side \( AB \) or its extension, and take a point \( N \) on side \( AD \) or its extension, such that
\[
\begin{array}{l}
AB \cdot AM = AC^2 \\
= AD \cdot AN.
\end{array}
\]
Connect \( MC \), \( NC \), and \( MN \).
Then, \(\triangle ABC \sim \triangle ACM\).
Thus,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,170 |
3. In Rt $\triangle A B C$, $C B=3, C A=4, M$ is a moving point on the hypotenuse $A B$, and through $M$ draw $M D \perp A C$ at point $D$, and through $M$ draw $M E \perp C B$ at point $E$. Then the minimum value of line segment $D E$ is | 3. $\frac{12}{5}$.
As shown in Figure 4, connect $C M$. From the given conditions, quadrilateral $C D M E$ is a rectangle, so $D E=C M$. Clearly, when $C M \perp A B$, $C M$ is the shortest. At this time, $C M=\frac{12}{5}$. | \frac{12}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,171 |
4. Given an isosceles triangle $\triangle A B C$ with side lengths $a$, $b$, and $c$ all being integers, and satisfying $a+b c+b+c a=24$. Then the number of such triangles is $\qquad$. | 4.3.
$$
\begin{array}{l}
\text { Since } a+b c+b+a a \\
=(a+b)(c+1)=24=12 \times 2=8 \times 3=6 \times 4, \text { and }
\end{array}
$$
$\triangle A B C$ is an isosceles triangle, so the length of the base can only be $c$.
Thus, there are 3 triangles that satisfy the conditions:
$$
c=1, a=b=6 ; c=2, a=b=4 ; c=3, a=b=3 \... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,172 |
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,173 |
Four, (25 points) Given that $x, y, z$ are all non-negative real numbers, satisfying $x+y-z=1, x+2 y+3 z=4$, let $w=3 x +2 y+z$. Find the maximum and minimum values of $w$.
| Solve the system $\left\{\begin{array}{l}x+y=1+z, \\ x+2 y=4-3 z .\end{array}\right.$ to get $\left\{\begin{array}{l}x=5 z-2, \\ y=3-4 z .\end{array}\right.$ From $\left\{\begin{array}{l}x=5 z-2 \geqslant 0, \\ y=3-4 z \geqslant 0\end{array} \Rightarrow \frac{2}{5} \leqslant z \leqslant \frac{3}{4}\right.$. Then $w=3 x... | \frac{16}{5} \text{ and } 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,174 |
Five. (25 points) As shown in Figure 1, in square $ABCD$, points $E$ and $F$ are on sides $BC$ and $CD$ respectively, satisfying $EF = BE + DF$. Lines $AE$ and $AF$ intersect diagonal $BD$ at points $M$ and $N$ respectively. Prove:
(1) $\angle EAF = 45^{\circ}$;
(2) $MN^2 = BM^2 + DN^2$ | (1) As shown in Figure 5, extend $C D$ to point $E_{1}$ such that $B E = D E_{1}$. Then
$$
\begin{array}{l}
\triangle A B E \cong \triangle A D E_{1} \\
\Rightarrow \angle B A E = \angle D A E_{1} \\
\Rightarrow A E = A E_{1} \\
\Rightarrow \angle E A E_{1} = 90^{\circ} .
\end{array}
$$
In $\triangle A E F$ and $\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,175 |
1. In a regular quadrilateral pyramid $P-A B C D$, $\angle A P C=$ $60^{\circ}$. Then the cosine value of the dihedral angle $A-P B-C$ is $(\quad)$.
(A) $\frac{1}{7}$
(B) $-\frac{1}{7}$
(C) $\frac{1}{2}$
(D) $-\frac{1}{2}$ | -、1.B.
As shown in Figure 4, within the side $PAB$, construct $AM \perp PB$, with the foot of the perpendicular being $M$. Connect $CM$ and $AC$, then $\angle AMC$ is the plane angle of the dihedral angle $A-PB-C$. Suppose $AB=2$, then $PA=AC=2\sqrt{2}$, and the slant height is $\sqrt{7}$.
Thus, $2 \times \sqrt{7} = A... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,176 |
2. Let the real number $a$ be such that the inequality
$$
|2 x-a|+|3 x-2 a| \geqslant a^{2}
$$
holds for all real numbers $x$. Then the set of all $a$ that satisfy this condition is ( ).
(A) $\left[-\frac{1}{3}, \frac{1}{3}\right]$
(B) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(C) $\left[-\frac{1}{4}, \frac{1}{3}\right... | 2.A.
For $k \in \mathbf{R}$, let $x=\frac{1}{2} k a$. Then the original inequality becomes
$$
|a| \cdot|k-1|+\frac{3}{2}|a| \cdot\left|k-\frac{4}{3}\right| \geqslant|a|^{2} \text {. }
$$
From this, it is easy to see that the original inequality is equivalent to
$$
|a| \leqslant|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\rig... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 719,177 |
3. Nine small balls numbered $1,2, \cdots, 9$ are placed in a bag. These balls are identical except for their numbers. Person A draws a ball from the bag, with the number being $a$, and then returns it. Person B then draws a ball from the bag, with the number being $b$. The probability of the event that the inequality ... | 3. D.
Person A and Person B each draw a small ball, with 9 different outcomes for each, so the total number of basic events is $9^{2}=81$.
From the inequality $a-2 b+10>0$, we get $2 b<a+10$.
Thus, when $b=1,2,3,4,5$, in each case $a$ can take every value from 1, $2, \cdots, 9$, making the inequality true, which resul... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 719,178 |
4. Let the function $f(x)=3 \sin x+2 \cos x+1$. If real numbers $a, b, c$ make $a f(x)+b f(x-c)=1$ hold for any real number $x$, then the value of $\frac{b \cos c}{a}$ is ( ).
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) -1
(D) 1 | 4.C.
From the given, we have
$$
\begin{array}{l}
f(x)=\sqrt{13} \sin (x+\varphi)+1, \\
f(x-c)=\sqrt{13} \sin (x+\varphi-c)+1,
\end{array}
$$
where $0<\varphi<\frac{\pi}{2}$ and $\tan \varphi=\frac{2}{3}$.
Thus, $a f(x)+b f(x-c)=1$ can be rewritten as
$$
\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b \sin (x+\varphi-c)+a+b=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,179 |
1. In $\triangle A B C$, if $\angle B=n \angle C(n$ is a positive integer not less than 2), then $A C \leqslant n A B$. | (提示: As shown in Figure 18, on the circumcircle of $\triangle A B C$, divide the arc $\overparen{A C}$ into $n$ equal parts, and connect the adjacent points to get $n$ equal chords, each of which is equal to $A B$. Since the length of the broken line $A A_{1} A_{2} \cdots A_{n-1} C$ is greater than $A C$, therefore, $A... | A C \leqslant n A B | Geometry | proof | Yes | Yes | cn_contest | false | 719,181 |
6. Given that $A$ and $B$ are two subsets of the set $\{1,2, \cdots, 100\}$, satisfying that $A$ and $B$ have the same number of elements, and $A \cap B$ is an empty set. If $n \in A$, then $2n+2 \in B$, the maximum number of elements in the set $A \cup B$ is ( ).
(A) 62
(B) 66
(C) 68
(D) 74 | 6. B.
First, we need to prove that $|A \cup B| \leqslant 66$, which only requires proving that $|A| \leqslant 33$. For this, we only need to prove: if $A$ is any 34-element subset of $\{1,2, \cdots, 49\}$, then there must exist $n \in A$ such that $2 n+2 \in A$. The proof is as follows:
Divide $\{1,2, \cdots, 49\}$ in... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,182 |
7. In the Cartesian coordinate system, there are four fixed points $A(-3,0)$, $B(1,-1)$, $C(0,3)$, $D(-1,3)$, and a moving point $P$. Then the minimum value of $|P A|+|P B|+|P C|+|P D|$ is $\qquad$. | $$
\text { II. } 7.3 \sqrt{2}+2 \sqrt{5} \text {. }
$$
As shown in Figure 5, let $A C$ and $B D$ intersect at point $F$, then
$$
\begin{array}{l}
|P A|+|P C| \\
\geqslant|A C|=|F A|+|F C|, \\
|P B|+|P D| \\
\geqslant|B D|=|F B|+|F D| .
\end{array}
$$
Therefore, when the moving point $P$ coincides with $F$, $|P A|+|P ... | 3 \sqrt{2}+2 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,183 |
8. In $\triangle A B C$ and $\triangle A E F$, $B$ is the midpoint of $E F$, $A B=E F=1, B C=6, C A=\sqrt{33}$. If $A B \cdot A E + A C \cdot A F=2$, then the cosine value of the angle between $E F$ and $B C$ is $\qquad$ | 8. $\frac{2}{3}$.
Since $A B \cdot A E+A C \cdot A F=2$, we have $A B \cdot(A B+B E)+A C \cdot(A B+B F)=2$, which means
$$
A B^{2}+A B \cdot B E+A C \cdot A B+A C \cdot B F=2 \text {. }
$$
Since $A B^{2}=1, B E=-B F$,
$$
\boldsymbol{A} \cdot \boldsymbol{A B}=\sqrt{33} \times 1 \times \frac{33+1-36}{2 \times \sqrt{33}... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,184 |
9. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, a sphere is constructed with vertex $A$ as the center and $\frac{2 \sqrt{3}}{3}$ as the radius. The length of the curve formed by the intersection of the sphere's surface with the cube's surface is equal to | 9. $\frac{5 \sqrt{3} \pi}{6}$.
As shown in Figure 6, the sphere intersects with all six faces of the cube, resulting in two types of intersection lines: one type on the three faces containing vertex $A$; the other type on the three faces not containing vertex $A$.
On the face $A A_{1} B_{1} B$, the intersection line ... | \frac{5 \sqrt{3} \pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,185 |
11. Given the function
$$
f(x)=\frac{\sin \pi x-\cos \pi x+2}{\sqrt{x}}\left(\frac{1}{4} \leqslant x \leqslant \frac{5}{4}\right) \text {. }
$$
Then the minimum value of $f(x)$ is $\qquad$ . | 11. $\frac{4 \sqrt{5}}{5}$.
It is known that $f(x)=\frac{\sqrt{2} \sin \left(\pi x-\frac{\pi}{4}\right)+2}{\sqrt{x}}\left(\frac{1}{4} \leqslant x \leqslant \frac{5}{4}\right)$.
Let $g(x)=\sqrt{2} \sin \left(\pi x-\frac{\pi}{4}\right)\left(\frac{1}{4} \leqslant x \leqslant \frac{5}{4}\right)$, then $g(x) \geqslant 0, g... | \frac{4 \sqrt{5}}{5} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,186 |
13. Let $a_{n}=\sum_{k=1}^{n} \frac{1}{k(n+1-k)}$. Prove that for any positive integer $n \geqslant 2$, $a_{n+1}<a_{n}$. | Three, 13. Since
$$
\frac{1}{k(n+1-k)}=\frac{1}{n+1}\left(\frac{1}{k}+\frac{1}{n+1-k}\right),
$$
then $a_{n}=\frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k}$.
Thus, for any positive integer $n \geqslant 2$, we have
$$
\begin{array}{l}
\frac{1}{2}\left(a_{n}-a_{n+1}\right)=\frac{1}{n+1} \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{n+2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,188 |
14. Given a line $l$ passing through the point $(0,1)$ intersects the curve $C$: $y=x+\frac{1}{x}(x>0)$ at two distinct points $M$ and $N$. Find the locus of the intersection of the tangents to the curve $C$ at points $M$ and $N$. | 14. Let points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$, and the tangents to curve $C$ at points $M$ and $N$ be $l_{1}$ and $l_{2}$, respectively, with their intersection point being $P\left(x_{P}, y_{P}\right)$. If the slope of line $l$ is $k$, then the equation of $l$ is $y=k x+1$.
From the syste... | (2,2) \text{ to } \left(2, \frac{5}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,189 |
15. Let the function $f(x)$ satisfy $f(x+2\pi)=f(x)$ for all real numbers $x$. Prove: There exist 4 functions $f_{i}(x)$ $(i=1,2,3,4)$ such that
(1) For $i=1,2,3,4$, $f_{i}(x)$ is an even function, and for any real number $x$, $f_{i}(x+\pi)=f_{i}(x)$;
(2) For any real number $x$, we have
$$
\begin{aligned}
f(x)= & f_{1... | 15. Let \( g(x) = \frac{f(x) + f(-x)}{2} \),
\[
h(x) = \frac{f(x) - f(-x)}{2} \text{.}
\]
Then \( f(x) = g(x) + h(x) \), and \( g(x) \) is an even function, \( h(x) \) is an odd function.
For any \( x \in \mathbf{R} \),
\[
\begin{array}{l}
g(x + 2\pi) = g(x), \quad h(x + 2\pi) = h(x). \\
\text{Let } f_{1}(x) = \frac{g... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,190 |
In acute $\triangle ABC$, $AB < AC$, $AD$ is the altitude on side $BC$, $P$ is a point on segment $AD$, $PE \perp AC$, with the foot of the perpendicular being $E$, and $PF \perp AB$, with the foot of the perpendicular being $F$. $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle BDF$ and $\triangle CDE$, respecti... | I. As shown in Figure 7, connect
$B P, C P, O_{1} O_{2}, E O_{2},$
$E F, F O_{1}$.
Since $P D \perp B C, P F$
$\perp A B$, therefore, $B, D, P, F$
are concyclic, and $B P$ is the
diameter of this circle.
Also, $O_{1}$ is the circumcenter of $\triangle B D F$, so $O_{1}$ is the midpoint of $B P$.
Similarly, $C, D, P, E$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,192 |
II. (50 points)
As shown in Figure 3, in a $7 \times 8$
rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected." Now, some of the 56 chess pieces are to be removed ... | Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.
Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,193 |
Three. (50 points) Let the set $P=\{1,2,3,4,5\}$. For any $k \in P$ and positive integer $m$, denote
$$
f(m, k)=\sum_{i=1}^{5}\left[m \sqrt{\frac{k+1}{i+1}}\right],
$$
where $[a]$ represents the greatest integer not exceeding $a$.
Prove that for any positive integer $n$, there exist $k \in P$ and a positive integer $m... | Three, define the set $A=\left\{m \sqrt{k+1} \mid m \in \mathbf{N}_{+}, k \in P\right\}$.
Since for any $k, i \in P$, and $k \neq i$, $\frac{\sqrt{k+1}}{\sqrt{i+1}}$ is an irrational number, then for any $k_{1}, k_{2} \in P$ and positive integers $m_{1}, m_{2}$,
$$
m_{1} \sqrt{k_{1}+1}=m_{2} \sqrt{k_{2}+1} \Leftrightar... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,194 |
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