problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. As shown in Figure 2, $\triangle A B C$ is inscribed in $\odot O$, and $A B=A C$, diameter $A D$ intersects $B C$ at point $E, F$ is the midpoint of $O E$. If $B F \parallel$ $C D, B C=2 \sqrt{5}$, then $C D=$ $\qquad$ | 3. $\sqrt{6}$.
Since $A B=A C, A D$ is the diameter, so $C E=B E=\sqrt{5}$.
Also, because $B F / / C D$, we have $E F=D E$. And since $F$ is the midpoint of $O E$, then $O F=E F=D E$.
Let $D E=x$, then $O A=O D=3 x, A E=5 x$.
According to the intersecting chords theorem, we have $C E \cdot B E=D E \cdot A E$, which is... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,003 |
4. Zhang Hua wrote a five-digit number, which can be divided by 9 and 11. If the first, third, and fifth digits are removed, the resulting number is 35; if the first three digits are removed, the resulting number can be divided by 9; if the last three digits are removed, the resulting number can also be divided by 9. T... | 4.63954.
Let this five-digit number be $\overline{a 3 b 5 c}$, then $9 \mid \overline{5 c}$.
So, $9 \mid(5+c)$.
Since $5 \leqslant 5+c \leqslant 14$, thus $5+c=9, c=4$.
Similarly, $9 \mid \overline{a 3}$, then $9 \mid(a+3)$.
Since $4 \leqslant a+3 \leqslant 12$, so $a=6$.
Also, $91 \overline{a 3 b 5 c}$, then
$9 \mid(... | 63954 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,004 |
One, (20 points) In the same coordinate plane, the quadratic function $y=x^{2}-x+a$ intersects with the linear function $y=x$ at points $A$ and $B$. The function $y_{1}=y^{2}-y+a$ intersects with the linear function $y_{1}=x$ at the same points $A$ and $B$, and there are no other intersection points besides these two. ... | Given the quadratic function $y=x^{2}-x+a$ and the linear function $y=x$ intersect at points $A$ and $B$, we know that $x^{2}-x+a=x$ has two distinct real roots. Therefore, $\Delta=4-4a>0$, hence $a<1$.
(2) $x^{2}-x+a=-x$ and $x^{2}-x+a=x$ have the same real solutions, i.e., $-x=x$ solves to $x=0$. When $x=0$, substitu... | 0 \leqslant a < 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,005 |
II. (25 points) As shown in Figure 3, given $\triangle ABC$, extend $AB$ to point $E$ such that $BE = AC$; extend $BC$ to point $F$ such that $CF = AB$; extend $CA$ to point $D$ such that $AD = BC$. If the resulting $\triangle DEF$ is an equilateral triangle, prove that $\triangle ABC$ is also an equilateral triangle. | II. As shown in Figure 7, extend $EA$ to point $D'$ such that $AD' = AD$; extend $DC$ to point $F'$ such that $CF' = CF$; extend $FB$ to point $E'$ such that $BE' = BE$.
For convenience, let $BC = x$, $AC = y$, and $AB = z$. Then, $BE = BE' = y$, $BF = x + z = BD'$, and $\angle EBF = \angle E'BD'$.
Therefore, $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,006 |
Three iron cubes, each with an edge length of a positive integer, and the sum of their edge lengths is $38 \mathrm{~cm}$. After melting them, a cube with an edge length of $20 \mathrm{~cm}$ is obtained. What are the edge lengths of the original three iron cubes from largest to smallest? | Three cubes have edge lengths of $x \mathrm{~cm}, y \mathrm{~cm}$, and $z \mathrm{~cm}$, respectively, and $013^{3}$, then $z \geqslant 14$.
When $z=14$, if $y=14$, then $x=10$, but this does not satisfy equation (1), so $y<14$. Thus, $x \leqslant y \leqslant 13$.
Therefore, $x^{3}+y^{3}+z^{3} \leqslant 13^{3}+13^{3}+1... | 17 \mathrm{~cm}, 14 \mathrm{~cm}, 7 \mathrm{~cm} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,007 |
1. Let the plane angle of the dihedral angle formed by two adjacent lateral faces of a regular tetrahedron be $\theta$. Then the range of $\theta$ is ( ).
(A) $(0, \pi)$
(B) $\left(\frac{\pi}{6}, \pi\right)$
(C) $\left(\frac{\pi}{3}, \frac{2 \pi}{3}\right)$
(D) $\left(\frac{\pi}{3}, \pi\right)$ | - 1.D.
As shown in Figure 4, in the regular tetrahedron $V-$ $A B C$, let $A B=B C=C A=$ $a$, and the base angle of the lateral triangular faces be $\alpha$ $\left(0<\alpha<\frac{\pi}{2}\right)$. A plane perpendicular to $V C$ through $A B$ intersects $V C$ at point $D$. Thus, $A D$ and $B D$ are the altitudes on the ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,008 |
2. As shown in Figure 1, in $\triangle A B C$, point $P$ is a point on side $A B$, and it is given that
condition $M: \frac{A C}{A P}=\frac{B C}{C P}$;
condition $N: \triangle A B C \backsim$ $\triangle A C P$.
Then condition $M$ is condition $N$'s ( ).
(A) sufficient but not necessary condition
(B) necessary but not s... | 2.C.
Obviously, $N \Rightarrow M$. Now we prove: $M \Rightarrow N$.
In $\triangle A B C$ and $\triangle A C P$, using the Law of Sines, we have $\sin \angle A B C=\frac{A C}{B C} \sin A$, $\sin \angle A C P=\frac{A P}{C P} \sin A$.
But from the given condition, we get $\frac{A C}{B C}=\frac{A P}{C P}$, so, $\sin \ang... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,009 |
3. There are 10 multiple-choice questions, each with 4 options. A person randomly selects one option for each question as the answer. If the probability of getting exactly $k$ questions correct is the greatest, then the value of $k$ is $(\quad)$.
(A) 2
(B) 3
(C) 4
(D) 5 | 3.A.
The answer to each question is randomly and independently selected, and the probability of answering a question correctly is $\frac{1}{4}$, while the probability of answering incorrectly is $\frac{3}{4}$. This is an independent repeated trial, so the probability of answering exactly $k$ questions correctly is
$$
... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,010 |
Example 3 Let $M=\{1,2, \cdots, 20\}$. For any 9-element subset $S$ of $M$, the function $f(S)$ takes values in $M$. Prove that no matter how the function $f$ is defined, there always exists a 10-element subset $T$ of $M$ such that for all $k \in T$,
$$
f(T-\{k\}) \neq k .
$$
(1988, USA Mathematical Olympiad) | If a 10-element subset $T$ has the property: for any $k \in T$, there is $f(T-\{k\}) \neq k$, then $T$ is called a "good set". A 10-element subset that is not a good set is called a "bad set". Thus, the problem reduces to proving the existence of a good set.
If $T$ is a bad set, by definition there exists $k_{0} \in T... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,011 |
4. For the imaginary number $\mathrm{i}\left(\mathrm{i}^{2}=-1\right)$, consider the set $S=$ $\left\{\mathrm{i}, \mathrm{i}^{2}, \mathrm{i}^{3}, \mathrm{i}^{4}\right\}$. It is easy to see that the product of any two elements in $S$ is still in $S$. Now, we define the multiplicative identity $\theta$ in $S$: for any $a... | 4.D.
From $\mathrm{i}^{4}=1$, we know that for $k=1,2,3,4$, $\mathrm{i}^{4} \cdot \mathrm{i}^{k}=\mathrm{i}^{k} \cdot \mathrm{i}^{4}=\mathrm{i}^{k}$.
By definition, $\mathrm{i}^{4}$ is the identity element. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,012 |
5. The three sides of $\triangle A B C$ are $A B=a, B C$ $=b, C A=c$. If
$$
\left\{\begin{array}{l}
c=\sqrt{a^{2}-2}+\sqrt{b^{2}-2}, \\
a=\sqrt{b^{2}-3}+\sqrt{c^{2}-3}, \\
b=\sqrt{c^{2}-4}+\sqrt{a^{2}-4},
\end{array}\right.
$$
then the number of values less than 0 among $\boldsymbol{A B} \cdot \boldsymbol{B C}, \bolds... | 5.A.
As shown in Figure 5, construct $\mathrm{Rt} \triangle A B E$ with $a$ as the hypotenuse and $\sqrt{2}$ as one of the legs; construct $\mathrm{Rt} \triangle B C E$ with $b$ as the hypotenuse and $B E$ as one of the legs, such that $E C$ is on the extension of $A E$. Then
$$
\begin{array}{l}
C A=\sqrt{a^{2}-2}+\sq... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,013 |
6. The correct judgment about the positive integer solutions of the equation $1+2+\cdots+x=y^{2}$ is ( ).
(A) None
(B) 1
(C) More than 1 but finite
(D) Infinite | 6.D.
It is easy to know that $1+2+\cdots+8=\frac{8 \times 9}{2}=6^{2}$. Therefore, $x=8, y=6$ is a positive integer solution to the equation.
Assume $\left(x_{k}, y_{k}\right)$ is a positive integer solution to the equation, i.e.,
$$
1+2+\cdots+x_{k}=\frac{x_{k}\left(x_{k}+1\right)}{2}=y_{k}^{2} \text {, }
$$
then $1... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,014 |
1. As shown in Figure 2, the base of the regular triangular pyramid $A-BCD$ coincides with the side face $\triangle BCD$ of the regular tetrahedron $BCDE$. Connect $AE$. Then the angle between $AE$ and the plane $BCD$ is $\qquad$ | $=, 1.90^{\circ}$.
Draw $A H \perp$ plane $B C D$ intersecting $\triangle B C D$ at point $H$. From $A B=A C=A D$, we get $H B=H C=H D$, so $H$ is the circumcenter of $\triangle B C D$. Then draw $E H_{1} \perp$ plane $B C D$ intersecting $\triangle B C D$ at point $H_{1}$. From $E B=E C=E D$, we get $H_{1} B=H_{1} C=... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,015 |
2. Given $a \geqslant b>0$, and $\sin \alpha$ is a root of the quadratic equation $a x^{2}+b x-b=0$. Then the maximum value of $\sin \alpha$ is $\qquad$ . | 2. $\frac{\sqrt{5}-1}{2}$.
Given that $\sin \alpha$ is a root of the equation, we have
$$
a \sin ^{2} \alpha+b \sin \alpha-b=0 \text {. }
$$
Obviously, $\sin \alpha \neq 0$, otherwise it would contradict $b>0$.
Also, from $a \geqslant b>0$, we get
$$
0=a \sin ^{2} \alpha+b \sin \alpha-b \geqslant b \sin ^{2} \alpha+b... | \frac{\sqrt{5}-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,016 |
4. For any $a, b \in \mathbf{R}$,
$$
\max \{|a+b|,|a-b|,|1-b|\}
$$
the minimum value is $\qquad$. | 4. $\frac{1}{2}$.
Let $x=\max \{|a+b|,|a-b|,|1-b|\} \geqslant 0$, then
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x \geqslant | a + b | , } \\
{ x \geqslant | a - b | , } \\
{ x \geqslant | 1 - b | }
\end{array} \Rightarrow \left\{\begin{array}{l}
-x \leqslant a+b \leqslant x, \\
-x \leqslant a-b \leqslant x, \... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,018 |
5. Draw the tangent line to the curve $y=3x-x^{3}$ through the point $A(2,-2)$. Then the equation of the tangent line is $\qquad$ . | 5. $y=-2$ or $9 x+y-16=0$.
Let the point of tangency be $P\left(x_{0}, y_{0}\right)$, then the equation of the tangent line at point $P$ is $\boldsymbol{y}-\boldsymbol{y}_{0}=k\left(x-x_{0}\right)$, where, $k=\left.y^{\prime}\right|_{x=x_{0}}=3-3 x_{0}^{2}$.
Since the tangent line passes through $A(2,-2)$, and point $... | y=-2 \text{ or } 9x+y-16=0 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 718,019 |
Example 4 Given two closed broken lines in the plane, each with an odd number of sides and such that the lines containing these sides are all distinct, and no three of these lines are concurrent. Prove: it is always possible to select one side from each broken line such that these two sides can serve as a pair of oppos... | Explanation: Consider one side of each of the two broken lines as a pair, called an "edge pair". Clearly, the positional relationship between the two edges in each edge pair has three different scenarios:
(1) The two edges intersect;
(2) One of the two edges, when extended, intersects the other edge (including endpoint... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,022 |
Four. (20 points) In a regular tetrahedron \(ABCD\), \(M, N, P, Q\) are the midpoints of edges \(AB, CD, BC, DA\) respectively. Connect \(MN, PQ\) intersecting at point \(O\), then connect \(OA, OB, OC, OD\). Prove:
(1) \(OA + OB + OC + OD = 0\);
(2) \(OA, OB, OC, OD\) have equal pairwise angles. | (1) As shown in Figure 7, it is easy to see that quadrilateral MPNQ is a parallelogram, with diagonals MN and PQ bisecting each other at point O. Therefore, we have
$$
O M + O N = 0.
$$
Since OM is the median of $\triangle AOB$, we have
$$
O A + O B = 2 O M.
$$
Similarly, $O C + O D = 2 O N$.
Adding these equations, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,023 |
Five. (20 points) Through the focus $F$ of the parabola $y^{2}=4 x$, draw a line $l$ intersecting the parabola at points $A$ and $B$.
(1) Prove that $\triangle A O B$ is not a right triangle.
(2) When the slope of $l$ is $\frac{1}{2}$, does there exist a point $C$ on the parabola such that $\triangle A B C$ is a right ... | (1) As shown in Figure 8, the focus of the parabola is $F(1, 0)$. All lines passing through point $F$ and intersecting the parabola at points $A$ and $B$ can be set as
$$
k y = x - 1,
$$
combined with the parabola $y^{2} = 4 x$. Eliminating $x$ yields
$$
y^{2} = 4 k y + 4,
$$
resulting in $y_{A} y_{B} = -4$.
$$
\begi... | C_{1}(1, -2), C_{2}(9, -6), C_{3}(14 - 6 \sqrt{5}, -6 + 2 \sqrt{5}), C_{4}(14 + 6 \sqrt{5}, -6 - 2 \sqrt{5}) | Geometry | proof | Yes | Yes | cn_contest | false | 718,024 |
一、(50 points) As shown in Figure 3, in the isosceles right triangle $\triangle A B C$, $D$ is the midpoint of the hypotenuse $A B$. A line $C C^{\prime} / / A B$ is drawn through the right-angle vertex $C$. Points $E$, $F$, and $M$ are on sides $A C$, $B C$, and $E F$ respectively, such that $\frac{C E}{A E}=$ $\frac{B... | From $A C = B C$ and $\frac{C E}{A E} = \frac{B F}{C F}$, we know $C E = B F$.
As shown in Figure 9, connect
$D F$, rotate $\triangle B D F$ counterclockwise by $90^{\circ}$ around point
$D$, then point $B$ moves to point $C$
and point $F$ moves to point $E$,
thus $\triangle C D E \cong$
$\triangle B D F$. Therefore, $... | M D = M H | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,025 |
II. (50 points) Let $M$ be a set composed of integers, satisfying:
(1) For all $a, b \in M, a \neq b$, it holds that $a+b \in M$;
(2) There exist $a, b \in M$ such that $a b < 0$.
Is it necessarily true that $M$ contains the element 0? Is $M$ necessarily an infinite set? Explain your reasoning. | Second, we first prove that $0 \in M$.
From the known condition (2), there exist $a, b \in M$ such that $ab < 0$.
If $a + b = 0$, then by $a \neq b, a \in M, b \in M$, we get
$0 = a + b \in M$.
If $a + b \neq 0$, then $a + b \neq a$ and $a + b \neq b$ (otherwise $ab = 0$ which contradicts $ab < 0$).
By condition (1), w... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,026 |
Three, (50 points) There are 7 points on a plane, and any three points form a scalene triangle with their pairwise connections. Prove: It is always possible to find 4 pairs of triangles, such that the common side of each pair is the longest side of one triangle and the shortest side of the other triangle. | Three, let the 7 points on the plane be denoted as $A_{1}, A_{2}, \cdots, A_{7}$. Since any three points form a non-equilateral triangle, each triangle has a longest side and a shortest side. Now, color the longest side of each triangle red, and the remaining sides blue, ensuring that each triangle has a red side.
We ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,027 |
Initially 193 In $\triangle A B C$, there are two points $D, E$ outside, and it satisfies that $\angle E A D$ is supplementary to $\angle C A B$, and $\angle E B D$ is supplementary to $\angle C B A$. Prove: $\angle A D E=\angle B D C$.
---
The translation maintains the original text's line breaks and format as reque... | Proof: As shown in Figure 2, draw $E X \perp D A$ at point $X$, $E Y \perp D B$ at point $Y$, and connect $X Y$.
From the given information, it is easy to see that
$$
\begin{array}{l}
\angle E A X=\angle C A B, \\
\angle E B Y=\angle C B A .
\end{array}
$$
Draw $C M \perp D A$ at point $M$, $C N \perp D B$ at point $N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,028 |
Initially 194 As shown in Figure $3, P$ is a point outside the equilateral
$\triangle A B C$, $P A$ intersects $B C$ at point $D$, and $P A=P B+P C$. Prove: $\frac{1}{P B}+\frac{1}{P C}=\frac{1}{P D}$. | Prove: Draw a line through point $B$ intersecting $AC$ at point $E$, such that $\angle PBE = 60^{\circ}$. On $BE$, mark $BF = BP$, and connect $AF$ and $PF$. Thus, $\triangle BPF$ is an equilateral triangle, so
$\angle ABE = \angle CBP$.
Therefore, $PF = PB$.
It is easy to prove that $\triangle ABF \cong \triangle CBP$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,029 |
Clothing 193 Let $a, b, c \in \mathbf{R}_{+}$, and satisfy
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+a b c=4 . \\
\text { Let } m=a^{3}+b^{3}+c^{3}, n=1+2 a b c, s=a+b+c .
\end{array}
$$
Try to compare the sizes of $m, n, s$, and find the maximum and minimum values of $m, n, s$ respectively. | Solution: (1) Obviously, when $a=b=c=1$, $a^{2}+b^{2}+c^{2}+a b c=4$, and $m=n=s=3$.
(2) It can be found that $(a, b, c)=\left(\frac{7}{4}, \frac{1}{2}, \frac{1}{2}\right)$ satisfies $a^{2}+b^{2}+c^{2}+a b c=4$.
For this, it is only necessary to take $b=c=\frac{1}{2}$, at this time,
$$
4=a^{2}+\frac{1}{4}+\frac{1}{4}+\... | m \geqslant 3 \geqslant s \geqslant n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,030 |
In Rt $\triangle ABC$, $\angle ACB=90^{\circ}$, $BC=a$, $CA=b$. Construct $CD$ perpendicular to $AB$, with the foot of the perpendicular at $D$. Let $G$ be the centroid of $\triangle ABC$, and draw a line through point $G$ that intersects $AC$, $CD$, and $BC$ at points $P$, $T$, and $Q$ respectively. If $PG=TQ$, find t... | Solution: As shown in Figure 4, let $M$ be the midpoint of $AB$, then
$$
\begin{array}{l}
\angle BCD = \angle CAM \\
= \angle ACM = \alpha.
\end{array}
$$
Let $\angle CPQ = \theta$, then
$$
\angle CQP = 90^\circ - \theta.
$$
Obviously, $CG = \frac{2}{3} CM = \frac{2}{3} \times \frac{1}{2} AB = \frac{1}{3} \sqrt{a^2 +... | CT = \frac{\sqrt{a^2 + b^2}}{3}, \quad \frac{1}{2} \leq \frac{b}{a} \leq 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,031 |
Example 1 In a convex quadrilateral $ABCD$, $\angle A - \angle B = \angle B - \angle C = \angle C - \angle D > 0$, and one of the four interior angles is $84^{\circ}$. Find the measures of the other angles. | Let $\angle A-\angle B=\angle B-\angle C=\angle C-\angle D=\alpha$. Then
$$
\begin{array}{l}
\angle C=\angle D+\alpha, \angle B=\angle D+2 \alpha, \\
\angle A=\angle D+3 \alpha . \\
\text { Also } \angle A+\angle B+\angle C+\angle D=360^{\circ} \text {, then } \\
\angle D+(\angle D+\alpha)+(\angle D+2 \alpha)+ \\
(\ang... | (1) \angle D=84^{\circ}, \alpha=4^{\circ}, \angle C=88^{\circ}, \angle B=92^{\circ}, \angle A=96^{\circ} \quad (2) \angle C=84^{\circ}, \alpha=12^{\circ}, \angle D=72^{\circ}, \angle B=96^{\circ}, \angle A | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,032 |
Example 2 In $\triangle A B C$, $\angle A=70^{\circ}$, point $I$ is the incenter. Given $A C+A I=B C$. Find the degree measure of $\angle B$.
---
The translation maintains the original text's format and line breaks. | Explanation: As shown in Figure 1, extend $C A$ to point $K$ such that $A K = A I$. Then
$$
\begin{array}{l}
C K = A C + A K \\
= A C + A I \\
= C B .
\end{array}
$$
$$
\text { Therefore, } \triangle C B I \cong \triangle C K I \text {. }
$$
Thus, $\angle C B I = \angle C K I$.
Since $A K = A I$, then $\angle C A I = ... | 35^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,033 |
5. As shown in Figure 7, in $\triangle A B C$, $A B=A C<B C$, points $D$ and $E$ are taken on the rays $A B$ and $C A$ respectively. It is known that $A D=B C=C E=D E$. Prove: $\angle B A C=100^{\circ}$. | (提示: Construct $\square B D F C$. It is easy to prove $\triangle E C F \cong \triangle D E A$. Then $E F=D A=B C=D F=D E$. Therefore, $\triangle D E F$ is an equilateral triangle. Let $\angle A B C=\alpha$, then $\angle A D F=\alpha, \angle A D E=180^{\circ}-4 \alpha$. So, $\left(180^{\circ}-4 \alpha\right)+\alpha=60^... | 100^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 718,034 |
22. (14 points) Given the sequence $\left\{x_{n}\right\}$,
$$
x_{1}=a, x_{n+1}=\frac{2 x_{n}}{1+x_{n}^{2}} \text {. }
$$
(1) Let $a=\tan \theta\left(0<\theta<\frac{\pi}{2}\right)$, if $x_{3}<\frac{4}{5}$, find the range of $\theta$;
(2) Define the function $f(x)$ on $(-1,1)$, for any $x, y \in(-1,1)$, there is
$$
f(x)-... | 22. (1) Since $x_{1}=a>0$, hence all $x_{n}>0$.
Also, $x_{n+1}=\frac{2}{\frac{1}{x_{n}}+x_{n}} \leqslant 1$, so, $x_{n} \in(0,1]$.
Since $x_{3}0$.
Solving, we get $x_{2}2$.
Also, $x_{2} \in(0,1]$, so $0<x_{2}<\frac{1}{2}$.
And $x_{2}=\frac{2 x_{1}}{1+x_{1}^{2}}=\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\sin 2 \theta$,... | 2^{n-2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,035 |
As shown in Figure $1, AB$ is the diameter of $\odot O$, the non-diameter chord $CD \perp AB, E$ is the midpoint of $OC$, connect $AE$ and extend it to intersect $\odot O$ at point $P$, connect $DP$ to intersect $BC$ at point $F$. Prove: $F$ is the midpoint of $BC$. | Connect $B D$. Since $A B$ is the diameter of $\odot O$, and $C D \perp A B$, therefore, $\angle A O C=\angle D B C$, that is, $\angle A O E=\angle D B F$.
Also, $\angle O A E=\angle B D F$, then
$\triangle A O E \backsim \triangle D B F$.
Thus, $\frac{A O}{D B}=\frac{O E}{B F}$.
Since $O A=O C, D B=B C$, then $\frac{O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,036 |
Let $p$ be a prime number greater than 2, and the sequence $\left\{a_{n}\right\}$ satisfies $n a_{n+1}=(n+1) a_{n}-\left(\frac{p}{2}\right)^{4}$.
Prove: When $a_{1}=5$, $16 \mid a_{81}$. | From the condition, we have $\frac{a_{n+1}}{n+1}=\frac{a_{n}}{n}-\frac{p^{4}}{16} \cdot \frac{1}{n(n+1)}$. Therefore, $\frac{a_{n}}{n}-\frac{p^{4}}{16 n}=\frac{a_{n-1}}{n-1}-\frac{p^{4}}{16(n-1)}=\cdots=\frac{a_{1}}{1}-\frac{p^{4}}{16}$,
which means $a_{n}=n a_{1}-\frac{1}{16}(n-1) p^{4}$.
When $a_{1}=5$, $a_{81}=81 \t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,037 |
Three, it is known that $A D$ is the altitude from vertex $A$ to side $B C$ of $\triangle A B C$, and $B C + A D = A B + A C$. Find the range of values for $\angle A$.
---
The original text has been translated into English, maintaining the original formatting and line breaks. | Three, let $A B=c, B C=a, C A=b, A D=h$. From the triangle area formula, we have $b c \sin A=a h$, then
$$
b c=\frac{a h}{\sin A} \text {. }
$$
From $B C+A D=A B+A C$, we get
$$
b+c=a+h .
$$
From the cosine rule, we have
$$
\begin{array}{l}
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b+c)^{2}-a^{2}-2 b c}{2 b c} \\... | \angle A \in\left[2 \operatorname{arccot} \frac{4}{3}, \frac{\pi}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,038 |
Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$
find the maximum and minimum values of $\Delta=a^{2}-4 b$. | Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$
or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,039 |
Given positive numbers $a, b, c$ satisfying $a+b+c=3$.
Prove:
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 .
\end{array}
$$ | $$
\begin{array}{l}
\text { 5. } \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}} \\
=\frac{1}{3}\left(1+\frac{2 a+6}{a^{2}-2 a+3}\right)=\frac{1}{3}\left[1+\frac{2 a+6}{(a-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{2 a+6}{2}\right)=\frac{1}{3}(4+a) .
\end{array}
$$
Similarly, $\frac{b^{... | 5 | Inequalities | proof | Yes | Yes | cn_contest | false | 718,040 |
Seven, can the positive integers $1,2, \cdots, 64$ be placed in the 64 squares of an $8 \times 8$ grid so that the sum of the numbers in any four squares forming a " $\square$ " shape (which can be rotated in any direction) is divisible by 5? | Seven, assuming the possibility.
Color the $8 \times 8$ grid in black and white, as shown in Figure 3.
Since $a+c+d+e$ and $a+b+c+e$ are both divisible by 5, their difference $d-b$ is also divisible by 5. Therefore, $d$ and $b$ have the same remainder $r_{1}\left(0 \leqslant r_{1} \leqslant 4\right)$ when divided by 5... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,041 |
Eight, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{k+1}=a_{k}+\frac{1}{2006} a_{k}^{2}, a_{0}=\frac{1}{2}, k \in \mathbf{N} \text {. }
$$
Prove: $1-\frac{1}{2008}<a_{2006}<1$. | From $a_{k+1}=a_{k}+\frac{1}{2006} a_{k}^{2}$, we get $a_{k+1}>a_{k}$, so the sequence $\left\{a_{n}\right\}$ is an increasing sequence. Then $a_{k+1}2-\frac{k}{2006}=\frac{4012-k}{2006}$.
For all $k(1 \leqslant k \leqslant 2006)$, we have
$$
a_{k}\frac{a_{k+1}}{2007}$.
Thus, $a_{k+1}=a_{k}+\frac{1}{2006} a_{k}^{2}>a_{... | 1-\frac{1}{2008}<a_{2006}<1 | Algebra | proof | Yes | Yes | cn_contest | false | 718,042 |
1. Given $a b c=1$, let
$$
\begin{aligned}
M & =\frac{1}{a+a b+1}+\frac{1}{b+b c+1}+\frac{1}{c+c a+1}, \\
N & =\frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c a+1} .
\end{aligned}
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determined | $-、 1 . B$.
From $a b c=1$, we get
$$
\begin{array}{l}
\frac{1}{b+b c+1}=\frac{a}{a b+a b c+a}=\frac{a}{a+a b+1}, \\
\frac{1}{c+c a+1}=\frac{a b}{a b c+a^{2} b c+a b}=\frac{a b}{a+a b+1} .
\end{array}
$$
Therefore, $M=1$.
Similarly, $N=1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,043 |
2. Paving the ground with three types of regular polygon tiles, with vertices fitting together perfectly, and fully covering the ground. Let the number of sides of the regular polygon tiles be $x$, $y$, and $z$. Then,
(A) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
(B) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$
(C) ... | 2. B.
Let the number of sides of the tiles be $x, y, z$, and the internal angles of each tile be $\alpha, \beta, \gamma$. Then
$$
\begin{array}{l}
(x-2) 180^{\circ}=\alpha x, (y-2) 180^{\circ}=\beta y, \\
(z-2) 180^{\circ}=\gamma z . \\
\text { Hence } \frac{x-2}{x}+\frac{y-2}{y}+\frac{z-2}{z}=\frac{\alpha+\beta+\gamm... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,044 |
Example 1: Through the midpoints $E$ and $F$ of the edges $AB$ and $BC$ of the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, a section is made such that the angle between the section and the base is $45^{\circ}$. The shape of this section is ( ).
(A) Triangle or pentagon
(B) Triangle or hexagon
(C) Hexagon
(D) Triangle or quadrila... | Explanation: As shown in Figure 1, it is clear that a section passing through points $E$ and $F$ must intersect the edge $B B_{1}$, and this section is a triangle.
Let the angle between the section passing through point $D_{1}$ and the base be $\alpha$. It is easy to find that
$$
\begin{array}{l}
\tan \alpha=\tan \ang... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,045 |
3. The product of all integers $m$ that make $m^{2}+m+7$ a perfect square is ( ).
(A) 84
(B) 86
(C) 88
(D) 90 | 3. A.
Let $m^{2}+m+7=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $m^{2}+m+7-k^{2}=0$.
Solving for $m$, we get $m=\frac{-1 \pm \sqrt{4 k^{2}-27}}{2}$.
Since $m$ is an integer, we should have $4 k^{2}-27=n^{2}\left(n \in \mathbf{N}_{+}\right)$, which means $(2 k+n)(2 k-n)=27$.
Solving, we get $\left\{\begin{array}{l}... | 84 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,046 |
4. As shown in Figure 1, in $\triangle A B C$, $\angle C=90^{\circ}$, $A C=2$, $B C=1$, points $A$ and $C$ are on the $x$-axis and $y$-axis, respectively. When point $A$ moves on the $x$-axis, point $C$ moves on the $y$-axis accordingly. During the movement, the maximum distance from point $B$ to the origin $O$ is ( ).... | 4.B.
As shown in Figure 6, take the midpoint $D$ of $AC$, and connect $OD$, $BD$, $OB$. It is easy to see that $OD=1$, $BD=\sqrt{2}$. Since the shortest distance between two points is a straight line, i.e., $OB \leqslant OD+BD=1+\sqrt{2}$, thus the maximum distance from point $B$ to the origin $O$ is $\sqrt{2}+1$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,047 |
5. As shown in Figure 2, the side length of square $A B C D$ is 2. Equilateral triangles $\triangle A B E$, $\triangle B C F$, $\triangle C D G$, and $\triangle D A H$ are constructed outward from each side. Then the perimeter of quadrilateral $A F G D$ is ( ).
(A) $4+2 \sqrt{6}+2 \sqrt{2}$
(B) $2+2 \sqrt{6}+2 \sqrt{2}... | 5. A.
As shown in Figure 7, connect $A F$, $A G$, and $G F$.
In the isosceles $\triangle A B F$, the vertex angle is $90^{\circ}+60^{\circ}=150^{\circ}$. In the isosceles $\triangle F C G$, the vertex angle is $360^{\circ}-90^{\circ}-2 \times 60^{\circ}=150^{\circ}$.
Since $A B=B C=F C$, we have $\triangle A B F \cong... | 4+2 \sqrt{6}+2 \sqrt{2} | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,048 |
6. A bookstore sells the "Junior High School Math Competition" series, which includes one volume for each of the 7th, 8th, and 9th grades. Xiao Zhao, Xiao Li, and Xiao Chen all go to buy this series. Each person can buy one or none of each volume, and together they find that each volume has been purchased, and each vol... | 6.C.
For example, for the textbooks for seventh grade, each person has 2 choices: to buy or not to buy. Therefore, 3 people have $2 \times 2 \times 2=8$ choices. However, the 2 choices where all 3 people either do not buy or all buy are not allowed, so there are actually 6 choices. Thus, there are a total of different... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,049 |
1. When $n=1,2, \cdots, 2006$, the sum of the lengths of the segments cut off by the x-axis for all quadratic functions $y=n(n+1) x^{2}-(2 n+1) x+1$ is $\qquad$ . | 2. 1. $\frac{2006}{2007}$.
Let $y=0$, we get $n(n+1) x^{2}-(2 n+1) x+1=0$.
Transform it into $(n x-1)[(n+1) x-1]=0$, we get
$x_{1}=\frac{1}{n+1}, x_{2}=\frac{1}{n}$.
Thus, $\left|x_{2}-x_{1}\right|=\frac{1}{n}-\frac{1}{n+1}$.
By taking $n=1,2, \cdots, 2006$, we get the sum of the lengths of the segments intercepted on ... | \frac{2006}{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,050 |
2. Given that quadrilateral $A B C D$ is a square, $P$ is a point on side $B C$, and line $D P$ intersects the extension of $A B$ at point $Q$. If $D P^{2}-B P^{2}=B P \cdot B Q$, then $\angle C D P=$ $\qquad$ . | 2.22.5
Let $\theta=\angle C D P=$ $\angle P Q B$. As shown in Figure 8, take a point $T$ on side $A B$ such that $B T=B P$, and connect $D T$ and $D B$.
By symmetry, it is easy to see that
$$
\begin{array}{l}
D T=D P, \\
\angle B D T=\angle B D P, \angle A D T=\angle C D P=\theta .
\end{array}
$$
From the given condi... | 22.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,051 |
3. Let $x$, $y$, $z$ be real numbers, and
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$
Then $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=$ | 3.1 and -3.
Let $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=k$. Then
$$
\begin{array}{l}
\frac{x}{y+z}=k-\frac{y}{z+x}-\frac{z}{x+y}, \\
\frac{y}{z+x}=k-\frac{z}{x+y}-\frac{x}{y+z}, \\
\frac{z}{x+y}=k-\frac{x}{y+z}-\frac{y}{z+x} .
\end{array}
$$
Substituting the above three equations into
$$
\frac{x^{2}}{y+z}+\frac{y^... | 1 \text{ and } -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,052 |
4. As shown in Figure 3, $A B$ and $C D$ are two chords of $\odot O$, $\angle A O B$ and $\angle O C D$ are supplementary, $\angle O A B$ and $\angle C O D$ are equal. Then the value of $\frac{A B}{C D}$ is $\qquad$ | 4. $\frac{3+\sqrt{5}}{2}$.
It is easy to get $\angle A O B=108^{\circ}, \angle O C D=72^{\circ}$, $\angle A=\angle B=\angle C O D=36^{\circ}$. As shown in Figure 9, take $B M=O B$. Then $\triangle B O M \cong \triangle O C D$. Thus, $C D=O M, A M=O M$. Also, $\triangle M A O \sim \triangle O A B$, so $O A^{2}=A B \cdo... | \frac{3+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,053 |
One, (20 points) Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Try to find the minimum value of $S=7x+5y$. | Let $x+2 y=5 A, x+y=3 B, A, B$ be integers.
Since $x, y \geqslant 0$, it follows that $A, B \geqslant 0$.
It is also easy to see that $x=6 B-5 A, y=5 A-3 B$, thus,
$$
\begin{array}{l}
2 x+y=9 B-5 A \geqslant 99, \\
S=7 x+5 y=27 B-10 A .
\end{array}
$$
Therefore, the problem becomes: For integers $A, B \geqslant 0, 6 B... | 366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,054 |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle BAC=90^{\circ}$, $AB=AC$, points $D_{1}$ and $D_{2}$ are on $AC$, and satisfy $AD_{1}=CD_{2}$, $AE_{1} \perp BD_{1}$, $AE_{2} \perp BD_{2}$, intersecting $BC$ at points $E_{1}$ and $E_{2}$, respectively. Prove that: $\frac{CE_{2}}{BE_{2}}+\frac{CE_{1}}{B... | $$
\begin{array}{l}
\frac{C E_{1}}{B E_{1}}=\frac{A F}{A B}, \\
\angle E_{1} A C=\angle F C A . \\
\text { Also, } \angle A B D_{1}=90^{\circ}-\angle B A E_{1}=\angle E_{1} A C, \text { then } \\
\angle A B D_{1}=\angle A C F .
\end{array}
$$
Since $\angle B A D_{1}=\angle C A F=90^{\circ}, A B=A C$, therefore, $\tria... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 718,055 |
Example 2 As shown in Figure 2, the three edges of the cube are $AB$, $BC$, and $CD$, and $AD$ is the body diagonal. Points $P$, $Q$, and $R$ are on $AB$, $BC$, and $CD$ respectively, with $AP=5$, $PB=15$, $BQ=15$, and $CR=10$. What is the area of the polygon formed by the intersection of the plane $PQR$ extended in al... | Explanation: Since $B P = B Q$, therefore, $P Q \parallel A C$. Thus, the line through point $R$ and parallel to $P Q$ intersects $A F$ at point $U$, and $A U = C R$.
Since the line through point $R$ and parallel to $P Q$ lies in the plane $P Q R$, $U$ is a vertex of the polygon formed by the intersection. Also, the m... | 525 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,056 |
1. The mapping $f$ from the set $A$ of planar vectors to $A$ is defined by $f(x)=x-2(x \cdot a) a$, where $a$ is a constant vector. If the mapping $f$ satisfies $f(x) \cdot f(y)=x \cdot y$ for any $x, y \in A$, then the coordinates of $a$ could be ( ).
(A) $\left(\frac{\sqrt{5}}{2},-\frac{1}{2}\right)$
(B) $\left(\frac... | $$
\begin{array}{l}
\text {-,1.D. } \\
f(x) \cdot f(y)=[x-2(x \cdot a) a] \cdot[y-2(y \cdot a) a] \\
=x \cdot y+4\left(a^{2}-1\right)(x \cdot a)(y \cdot a)
\end{array}
$$
Therefore, from the condition, we should get $a^{2}-1=0$, i.e., $|a|=1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,058 |
2. Let $x>3, M=3 \sqrt{x-1}+\sqrt{x-3}, N=$ $\sqrt{x}+3 \sqrt{x-2}$. Then the size relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M<N$
(C) $M=N$
(D) Any of the three situations is possible | 2.B.
Given $x>3$, we have
$$
\begin{array}{l}
\sqrt{x}-\sqrt{x-1}=\frac{1}{\sqrt{x}+\sqrt{x-1}} \\
<\frac{1}{\sqrt{x-2}+\sqrt{x-3}} \\
=\sqrt{x-2}-\sqrt{x-3} .
\end{array}
$$
Thus, $\sqrt{x}+\sqrt{x-3}<\sqrt{x-1}+\sqrt{x-2}$,
which implies $\frac{3}{\sqrt{x}-\sqrt{x-3}}<\frac{1}{\sqrt{x-1}-\sqrt{x-2}}$, or equivalent... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,059 |
3. Given the function $f(x)=\log _{a}\left(a x^{2}-x+\frac{1}{2}\right)$ is always positive on $[1,2]$. Then the range of the real number $a$ is ( ).
(A) $\left(\frac{1}{2},+\infty\right)$
(B) $\left(\frac{1}{2}, \frac{5}{8}\right) \cup(1,+\infty)$
(C) $\left(\frac{1}{2}, 1\right) \cup\left(\frac{3}{2},+\infty\right)$
... | 3.D.
When $a>1$, $a x^{2}-x+\frac{1}{2}>1 \Leftrightarrow a>\frac{1}{2}\left(\frac{1}{x}\right)^{2}+\frac{1}{x}$.
Given $\frac{1}{x} \in\left[\frac{1}{2}, 1\right]$, we know $a>\frac{3}{2}$;
When $0<a<1$,
$$
\begin{array}{l}
0<a x^{2}-x+\frac{1}{2}<1 \\
\Leftrightarrow-\frac{1}{2}\left(\frac{1}{x}\right)^{2}+\frac{1}{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,060 |
4. As shown in Figure 1, in $\triangle B C D$, $\angle B C D=90^{\circ}, B C=C D$ $=1, A B \perp$ plane $B C D, \angle A D B$ $=60^{\circ}$. Points $E$ and $F$ are on $A C$ and $A D$ respectively, such that plane $B E F \perp$ plane $A C D$, and $E F / / C D$. Then the sine value of the dihedral angle formed by plane $... | 4.B.
From the given, we know that the edge of the dihedral angle passes through point $B$ and is parallel to $CD$. Since $CD \perp$ plane $ABC$, $\angle CBE$ is the plane angle of the dihedral angle. Also, since $AE \perp EF$, it follows that $AE \perp$ plane $BEF$. Therefore, $AE \perp BE$. Thus, $\angle CBE = \angle... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,061 |
5. The number of different shapes of ellipses with eccentricity $e=\log _{p} q$ (where $p, q$ are positive integers not exceeding 9) is ( ).
(A) 26
(B) 27
(C) 28
(D) 30 | 5.A.
Given $0<e<1$, we know $2 \leqslant q<p \leqslant 9$.
For $p=3,4,5,6,7,8,9$, the corresponding $e$ values are $1,2,3$, $4,5,6,7$ respectively, totaling 28.
However, $\log _{3} 2=\log _{9} 4, \log _{4} 2=\log _{9} 3$, so there are 26 distinct $e$ values.
Editor's note: Similar ellipses have the same shape. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,062 |
6. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{1}{4}, a_{2}=\frac{1}{5}$, and
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}=n a_{1} a_{n+1}
$$
for any positive integer $n$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{97}}$ is $(\quad)$.
(A) 5032
(B) 5044
(C) 5048
(D) 5050 | 6.B.
Given $a_{1} a_{2}+a_{2} a_{3}=2 a_{1} a_{3}$ and $a_{1}=\frac{1}{4}, a_{2}=\frac{1}{5}$, we get $a_{3}=\frac{1}{6}$.
Given $a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}=3 a_{1} a_{4}$, we can get $a_{4}=\frac{1}{7}$.
Furthermore, we can prove by mathematical induction that $a_{n}=\frac{1}{n+3}$.
$$
\begin{array}{l}
\tex... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,063 |
1. Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$, and the complex number
$$
\begin{array}{l}
z=\frac{\sqrt{65}}{5} \sin \frac{A+B}{2}+\mathrm{i} \cos \frac{A-B}{2}, \\
|z|=\frac{3 \sqrt{5}}{5} .
\end{array}
$$
Then the maximum value of $\angle C$ is $\qquad$ | $=1 . \pi-\arctan \frac{12}{5}$.
From $|z|^{2}=\frac{9}{5}$, we know
$$
\frac{13}{5} \sin ^{2} \frac{A+B}{2}+\cos ^{2} \frac{A-B}{2}=\frac{9}{5} \text {, }
$$
we get $13 \cos (A+B)=5 \cos (A-B)$.
Further, we have $\tan A \cdot \tan B=\frac{4}{9}$.
Thus, $\angle A$ and $\angle B$ are both acute angles, and
$$
\tan (A+... | \pi-\arctan \frac{12}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,064 |
2. Given that the length, width, and height of a rectangular prism are all integers, and the volume equals the surface area. Then the maximum value of its volume is
| 2.882.
Lemma When the sum of the reciprocals of two integers is a constant, the larger their difference, the greater their product.
Let the length, width, and height be $a$, $b$, and $c$, respectively, then
$$
a b c=2(b c+c a+a b) \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2} .
$$
Assume without los... | 882 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,065 |
3. Let $[x]$ denote the greatest integer not exceeding $x$. Then the last two digits of $\left[\frac{10^{2006}}{10^{99}-3}\right]$ are $\qquad$ . | 3.23.
From $10^{2000}=10^{59 \times 34}-3^{34}+3^{34}$, we know
$$
\left[\frac{10^{2006}}{10^{99}-3}\right]=\frac{10^{99 \times 34}-3^{34}}{10^{59}-3}=\sum_{k=0}^{33} 10^{59(33-k)} \times 3^{k} \text {. }
$$
Therefore, the last two digits we are looking for are the remainder of $3^{33}$ modulo 100.
Also, $3^{32}=(10-... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,066 |
Example 3 A closed figure formed by the intersection of a plane with the surface of a cube is called the "section figure" of the cube. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $E$ is the midpoint of $A B$, and $F$ is the midpoint of $C C_{1}$. Then the perimeter of the section figure passing thro... | Solution 1: As shown in Figure 3, extend $D_{1} F$ and $DC$ to meet at point $P$, draw line $EP$ to intersect $BC$ at point $N$ and the extension of $DA$ at point $S$, connect $D_{1} S$ to intersect $A_{1} A$ at point $M$, then the pentagon $D_{1} M E N F$ is the cross-sectional figure.
By the property of corresponding... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,067 |
4. In the tetrahedron $O A B C$, $O A \perp$ plane $A B C, A B \perp$ $A C$, point $P$ satisfies $O P=l O A+m O B+n O C$, where $l, m, n$ are positive numbers and $l+m+n=1$. If the line $O P$ is formed by points that are equidistant from the planes $O B C$, $O C A$, and $O A B$, then the cosine value of the dihedral an... | 4. $\frac{n}{l}$.
From the conditions, we know that point $P$ is inside $\triangle A B C$ and the area ratios of $\triangle P B C$, $\triangle P C A$, and $\triangle P A B$ are $l: m: n$. Furthermore, from the composition of $O P$, we know that the area ratios of $\triangle O B C$, $\triangle O C A$, and $\triangle O ... | \frac{n}{l} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,068 |
3. The number of trapezoids formed by the $4 n+2$ vertices of a regular $4 n+2$-sided polygon is $\qquad$ . | $$
5.2 n(2 n-1)(2 n+1) .
$$
A regular $4 n+2$-sided polygon has $2 n+1$ diameters. All trapezoids can be divided into two categories: one with bases parallel to a certain diameter, and there are $(2 n+1)\left(\mathrm{C}_{2 n+1}^{2}-n\right)$ such trapezoids; the other with bases perpendicular to a certain diameter, an... | 2 n(2 n-1)(2 n+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,069 |
6. For $x \in \mathbf{R}, n \in \mathbf{N}_{+}$, define
$$
\mathrm{C}_{x}^{n}=\frac{x(x-1) \cdots(x-n+1)}{n!} .
$$
Let $P(x)$ be a polynomial of degree 6 that satisfies
$$
P(0)=1, P(k)=2^{k-1}(k=1,2, \cdots, 6) \text {. }
$$
Express $P(x)=$
$\qquad$
using $\mathrm{C}_{x}^{k}(k=1,2, \cdots, 6)$. | $6.1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6}$.
From $P(0)=1$, we know there exists a polynomial $Q_{1}(x)$ such that
$$
P(x)=1+x Q_{1}(x) \text {. }
$$
Thus, $1=P(1)=1+Q_{1}(1)$, which implies $Q_{1}(1)=0$.
There also exists a polynomial $Q_{2}(x)$ such that
$$
Q_{1}(x)=(x-1) Q_{2}(x) \text {, }
$$
i.e., $P(x)=1+x(x-1) Q_{2}(x... | 1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,070 |
Three, (20 points) A town has three teahouses, $A$, $B$, and $C$. The new mayor drinks tea at only one of these three places each day. It is known that on the first day, the probability of him going to each place is $\frac{1}{3}$. If he goes to $A$ on a certain day, the probabilities of him going to $A$, $B$, and $C$ t... | Three, let the probabilities of the mayor going to places $A$, $B$, and $C$ on the $n$-th day be $a_{n}$, $b_{n}$, and $c_{n}$, respectively. Then $a_{n}+b_{n}+c_{n}=1$, and
$$
\left\{\begin{aligned}
a_{n+1} & =\frac{1}{2} a_{n}+\frac{1}{4} b_{n}+\frac{1}{4}\left(1-a_{n}-b_{n}\right) \\
& =\frac{1}{4} a_{n}+\frac{1}{4}... | a_{n}=\frac{1}{3}, b_{n}=\frac{10}{27}-\frac{1}{27}\left(\frac{1}{4}\right)^{n-1}, c_{n}=\frac{8}{27}+\frac{1}{27}\left(\frac{1}{4}\right)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,071 |
Four, (20 points) Given $f(x)=2 x-\frac{2}{x^{2}}+\frac{a}{x}$, where the constant $a \in(0,4]$. Find all real numbers $k$, such that for any $x_{1} 、 x_{2} \in \mathbf{R}_{+}$, it always holds that
$$
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geqslant k\left|x_{1}-x_{2}\right| .
$$ | When $x_{1}=x_{2}$, $k$ is arbitrary.
When $x_{1} \neq x_{2}$, the inequality becomes $\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \geqslant k$.
$$
\begin{array}{l}
\text { Since } \frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}} \\
=\frac{1}{x_{1}-x_{2}}\left[2\left(x_{1}-x_{2}\... | \left(-\infty, 2-\frac{a^{3}}{108}\right] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 718,072 |
Five. (20 points) The center of the ellipse $\Gamma$ is at the origin $O$, with foci on the $x$-axis, and the eccentricity $e=\sqrt{\frac{2}{3}}$. The line $l$ intersects the ellipse $\Gamma$ at points $A$ and $B$, satisfying $CA=2BC$, where the fixed point $C(-1,0)$. When $\triangle OAB$ achieves its maximum value, fi... | Let the equation of $\Gamma$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
From $b^{2}=\left(1-e^{2}\right) a^{2}=\frac{1}{3} a^{2}$, we know that the equation of $\Gamma$ can be transformed into $x^{2}+3 y^{2}=a^{2}$.
From $C A=2 B C$, we know that $l$ is not parallel to the coordinate axes. We can set $l: y=... | x^{2}+3 y^{2}=5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,073 |
One, (50 points) In $\triangle A B C$, $A B=A C, O$ is the midpoint of $B C$, a circle with $O$ as the center is tangent to $A B$ and $A C$ at points $E$ and $F$ respectively. On the arc $\overparen{E F}$ of $\odot O$, take any point $D$, draw the tangent line through $D$ intersecting $A B$ and $A C$ at points $P$ and ... | As shown in Figure 2, connect $O A, O D, O E, O F, O P,$ and $O Q$. To prove $Q R / / A B$, we need to prove that
$$
\frac{E R}{R F}=\frac{A Q}{Q F}.
$$
By Menelaus' theorem, we have
$$
\frac{E R}{R F}=\frac{E P}{P A} \cdot \frac{A C}{C F}.
$$
It is also easy to see that $\triangle A O C \sim \triangle O F C$, hence
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,074 |
II. (50 points) Let $x, y, z$ be non-negative real numbers satisfying $x^{2}+y^{2}+z^{2}=1$. Prove that:
$$
x+y+z-2 x y z \leqslant \sqrt{2} \text {. }
$$ | Let $x+y+z-2 x y z=f$.
Take $x=\cos \theta \cdot \cos \varphi, y=\cos \theta \cdot \sin \varphi, z=\sin \theta$,
where $\theta, \varphi \in \left[0, \frac{\pi}{2}\right]$.
Let $\cos \varphi + \sin \varphi = t$, then $t \in [1, \sqrt{2}]$, and
\[
\begin{aligned}
f &= \cos \theta (\sin \varphi + \cos \varphi) + \sin \the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,075 |
As shown in Figure 3, from a point \( P \) outside the circle \( \odot O \), two tangents to \( \odot O \) are drawn, touching the circle at points \( A \) and \( B \). Through point \( A \), a line parallel to \( PB \) is drawn, intersecting \( \odot O \) at point \( C \). Line \( PC \) intersects \( \odot O \) again ... | Proof: Since $A C / / P B$, therefore, $\angle E P D = \angle D C A$.
Also, $\angle D C A = \angle D A P$, thus $\angle E P D = \angle E A P$.
Hence, $\triangle E P D \backsim \triangle E A P$.
Therefore, $\frac{E P}{E D} = \frac{E A}{E P}$, which means $E P^{2} = E D \cdot E A$.
By the secant-tangent theorem, $E B^{2... | BC^2 = 2 AC \cdot BE | Geometry | proof | Yes | Yes | cn_contest | false | 718,076 |
Example 4 Suppose the base of the pyramid $P-ABCD$ is not a parallelogram, and a plane $\alpha$ cuts this pyramid such that the section is a parallelogram. Then the number of such planes $\alpha$ ( ).
(A) does not exist
(B) is only one
(C) is exactly two
(D) is infinitely many | Explanation: As shown in Figure 5, extend \( BA \) and \( CD \) to intersect at point \( M \), and connect \( PM \). Then \( PM \) is the intersection line of the lateral faces \( PAB \) and \( PCD \). Similarly, \( PN \) is the intersection line of the lateral faces \( PAD \) and \( PBC \). Let the plane determined by... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,077 |
Initially 196, try to find all integers $k$ such that the quadratic equation in $x$, $x^{2}-2(3 k-1) x+9 k-1=0$, has at least one rational root with a denominator of 4 (the denominator and numerator do not have to be coprime). | Solution: Let $x=\frac{y}{4}(y \in \mathbb{Z})$, substituting into the given equation yields
$$
\frac{y^{2}}{16}-2(3 k-1) \frac{y}{4}+9 k-1=0 \text{. }
$$
Therefore, the quadratic equation in $y$ is
$$
y^{2}-8(3 k-1) y+16(9 k-1)=0
$$
which has at least one integer root.
From the above equation, we get
$$
k=\frac{y^{2... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,078 |
Given $\square A B C D, E G 、 F H$ are two line segments between the opposite sides, their intersection point $O$ is not the center of $\square A B C D$, and satisfies $\angle A E G=\angle A H F$. Then the sufficient and necessary condition for $A C 、 E F 、 H G$ to be concurrent is:
(1) either $E G 、 F H$ are parallel ... | Proof: Sufficiency.
(1) As shown in Figure 4, $EG$ and $FH$ are parallel to one pair of sides of the parallelogram. Let the line $FE$ intersect $CA$ at point $P$, and the line $GH$ intersect $CA$ at point $P'$. In $\triangle ABC$ and $\triangle ADC$, by Menelaus' theorem, we have
\[
\frac{AE}{EB} \cdot \frac{BF}{FC} \c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,079 |
Given positive numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$.
(1) Find the minimum value of $\frac{x^{2} y}{z^{2}}+\frac{y^{2} z}{x^{2}}+\frac{z^{2} x}{y^{2}}$;
(2) Find the minimum value of $\frac{x^{2} y^{3}}{z^{4}}+\frac{y^{2} z^{3}}{x^{4}}+\frac{z^{2} x^{3}}{y^{4}}$. | Solution: (1) Let $s=\frac{x^{2} y}{z^{2}}+\frac{y^{2} z}{x^{2}}+\frac{z^{2} x}{y^{2}}$, then
$$
\begin{array}{l}
s^{2}=\frac{x^{4} y^{2}}{z^{4}}+\frac{y^{4} z^{2}}{x^{4}}+\frac{z^{4} x^{2}}{y^{4}}+2\left(\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^{3}}{x}\right) \\
\geqslant 3\left(\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,080 |
Example 5: Through the vertex $A$ of the regular tetrahedron $ABCD$, make a section in the shape of an isosceles triangle, and let the angle between the section and the base $BCD$ be $75^{\circ}$: The number of such sections that can be made is $\qquad$. | Explanation: Let the edge length of a regular tetrahedron be 1. Draw $AO \perp$ plane $BCD$ at point $O$, then $AO=\frac{\sqrt{6}}{3}$. With $O$ as the center and $\frac{\sqrt{6}}{3} \cot 75^{\circ}$ as the radius, draw a circle on plane $BCD$. It is easy to see that this circle is inside $\triangle BCD$, and the inter... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,081 |
Example 6 As shown in Figure 7, given a cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$. Any plane $\alpha$ is perpendicular to the diagonal $A C_{1}$, such that plane $\alpha$ intersects each face of the cube. Let the area of the resulting cross-sectional polygon be $S$, and the perimeter be $l$. Then ().
(A) $S$ is a consta... | Consider the special case first.
Assume the edge length of the cube is 1. As shown in Figure 7, take points $E$, $F$, $G$, $H$, $I$, and $J$ as the midpoints of the six edges. Clearly, the regular hexagon $E F G H I J$ is a valid cross-section, with its perimeter $l_{1}=3 \sqrt{2}$ and area $S_{1}=\frac{3 \sqrt{3}}{4}$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,082 |
Example 7 As shown in Figure 8, each face of the tetrahedron $ABCD$ is an acute triangle, and $AB=CD$ $=a, AC=BD=b, AD=$ $BC=c$, plane $\pi$ intersects edges $AB$, $BC$, $CD$, $DA$ at points $P$, $Q$, $R$, $S$ respectively. The minimum value of the perimeter of quadrilateral $PQRS$ is ( ).
(A) $2 a$
(B) $2 b$
(C) $2 c$... | Explanation: As shown in Figure 9, the lateral faces of the tetrahedron are unfolded into a plane figure. Since all faces of the tetrahedron are acute triangles, and
$$
\begin{array}{l}
A B=C D, A C=B D, A D=B C, \\
\text { thus, in the unfolded plane figure, } \\
A D \Perp B C \\
\text { and } A^{\prime} D^{\prime} \... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,083 |
Example 8 In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=a, B C=b, C C_{1}=c(a>b>c)$. Let the area of the section passing through $B D_{1}$ be $S$. Find the minimum value of $S$, and indicate the position of the section when $S$ is at its minimum (i.e., indicate the position of the intersection points... | (1) The sections $A B C_{1} D_{1}$, $B C D_{1} A_{1}$, and $D B B_{1} D_{1}$ are all rectangles, with their areas denoted as $S_{1}$, $S_{2}$, and $S_{3}$, respectively, then
$$
\begin{array}{l}
S_{1}=a \sqrt{b^{2}+c^{2}}, S_{2}=b \sqrt{c^{2}+a^{2}}, \\
S_{3}=c \sqrt{a^{2}+b^{2}} .
\end{array}
$$
Since $a>b>c$, it is ... | \frac{b c}{\sqrt{b^{2}+c^{2}}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,084 |
1. The plane of a cube cannot be:
(1) an obtuse triangle; (2) a right triangle; (3) a rhombus; (4) a regular pentagon; (5) a regular hexagon.
The correct option is ( ).
(A)(1)(2)(5)
(B)(1)(4)
(C)(2)(3)(4)
(D)(3)(4)(5) | (The cross-section of a cube can be an acute triangle, isosceles triangle, equilateral triangle, but it cannot be an obtuse triangle or a right triangle; for quadrilaterals, it can be an isosceles trapezoid, parallelogram, rhombus, rectangle, but it cannot be a right trapezoid; for pentagons, it can be any pentagon, bu... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,085 |
Example 3 The radius of the circle is both the mean proportional between two chords and equal to the difference of these two chords. Find the degree measures of the central angles subtended by the two chords.
---
The translation maintains the original text's line breaks and format. | Explanation: As shown in Figure 2, $AB$ and $CD$ are two chords of $\odot O$, with $AB > CD$. Take $AM = CD$, then $BM = OB'$.
From the given information,
$$
\begin{array}{l}
OA^2 = AB \cdot CD \\
= AB \cdot AM,
\end{array}
$$
we get $\frac{OA}{AB} = \frac{AM}{OA}$.
Therefore, $\triangle OAB \sim \triangle MAO$.
Thus,... | 108^\circ, 36^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,086 |
2. Given a regular tetrahedron $ABCD$ with edge length 2, the sum of the areas of the sections obtained by all planes equidistant from its four vertices is $(\quad)$.
(A) 4
(B) 3
(C) $\sqrt{3}$
(D) $3+\sqrt{3}$ | (Tip: There are two types of cross-sections: (1) one side of the cross-section has 1 point, and the other side has 3 points, there are 4 such cross-sections; (2) each side of the cross-section has 2 points, there are 3 such cross-sections., Answer: (D).) | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,087 |
3. The cross-sectional area of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ through the diagonal $B D_{1}$ is $S$, and $S_{1}$ and $S_{2}$ are the maximum and minimum values of $S$, respectively. Then $\frac{S_{1}}{S_{2}}$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{2 \sqrt{3}}{3}$
(D) $\frac{2 \... | (Answer: (C).) | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,088 |
4. Prove: The area of any section passing through the center of a cube is not less than the area of one of the cube's faces.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The note above is a repetition of the instruc... | (It is known that the cross-section is a quadrilateral or a hexagon. If the cross-section is a quadrilateral, then it does not intersect with two opposite faces of the cube, and the projection of the cross-section on these two faces is the entire face. If the cross-section is a hexagon, by examining the unfolded side v... | null | Geometry | proof | Yes | Yes | cn_contest | false | 718,089 |
5. As shown in Figure 11, the base of the pyramid $S-ABCD$ is a rectangle $ABCD$ with center $O$, where $AB=4$, $AD=12$, $SA=3$, $SB=5$, and $SO=7$. A section of the pyramid is made through the vertex $S$, the center $O$ of the base, and a point $N$ on the edge $BC$. What is the value of $BN$ when the area of the resul... | (Tip: From the conditions, it is easy to know that $S A \perp$ plane $A B C D$. Also, $O M = O N$, so $S_{\triangle \operatorname{seN}} = 2 S_{\triangle \text{ Swo }}$. It is easy to know that when the distance from point $M$ to $S O$ is the distance between the skew lines $A B$ and $S O$, $S_{\triangle \operatorname{s... | BN = 7 \frac{11}{13}, \text{ minimum area } = \frac{42 \sqrt{13}}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,090 |
Example 1 Let $a_{i} \in \mathbf{R}_{+}, \alpha \geqslant 1, \beta>0, n \in \mathbf{N}_{+}$ and $n \geqslant 3$. The sequence $\left\{\lambda_{k}\right\}(k=1,2, \cdots, n)$ is a positive arithmetic sequence, $a_{n+i}=a_{i}(i=1,2, \cdots, n)$. Also, $s=\sum_{i=1}^{n} a_{i}$. Prove:
$$
\begin{array}{l}
\sum_{i=1}^{n} \fr... | Proof: Let the left side of (2) be $A$, from (1) we get
$$
\begin{array}{l}
A=\sum_{i=1}^{n} \frac{a_{i}^{a_{i}+\beta}}{\left(\lambda_{1} a_{i} a_{i+1}+\lambda_{2} a_{i} a_{i+2}+\cdots+\lambda_{n-1} a_{i} a_{i+n-1}\right)^{\beta}} \\
\geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{\alpha+\beta}}{n^{\alpha-1}\left[\s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,091 |
Example 2 Let $a, b, c$ be positive real numbers, $s=a+b+c, \alpha \geqslant 1, \beta \geqslant 0, \lambda \geqslant 0$. Then
(1) When $0 \leqslant \lambda \leqslant 8$, we have
$$
\begin{array}{l}
\frac{a^{\alpha}}{\left(a^{2}+\lambda b c\right)^{\beta}}+\frac{b^{\alpha}}{\left(b^{2}+\lambda c a\right)^{\beta}}+\frac{... | Proof: Let the left side of equation (3) be $A$, and $\sum$ represents the cyclic sum (the same applies below). From equation (1), we have
$$
\begin{array}{l}
A=\sum \frac{a^{a}}{\left(a^{2}+\lambda b c\right)^{\beta}}=\sum \frac{a^{\alpha+\beta}}{\left(a^{3}+\lambda a b c\right)^{\beta}} \\
\geqslant \frac{\left(\sum ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,092 |
Example 3 Let $x_{i}>0(i=1,2, \cdots, n), \alpha \geqslant 1$, $\beta>0, \alpha \geqslant(n-1) \beta$, and $x_{1} x_{2} \cdots x_{n}=1, n \geqslant 2$. Then
$$
\begin{array}{l}
\frac{x_{1}^{\alpha}}{\left(1+x_{2}\right)^{\beta}\left(1+x_{3}\right)^{\beta} \cdots\left(1+x_{n}\right)^{\beta}}+ \\
\frac{x_{2}^{\alpha}}{\l... | Proof: Without loss of generality, let $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, then
$$
\begin{array}{l}
x_{1}^{a} \geqslant x_{2}^{a} \geqslant \cdots \geqslant x_{n}^{a} \text {, } \\
\frac{1}{\left(1+x_{2}\right)^{\beta}\left(1+x_{3}\right)^{\beta} \cdots\left(1+x_{n}\right)^{\beta}} \\
\geqslant \f... | \frac{n}{2^{(n-1) \beta}} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,093 |
Example 4 Let $a, b, c$ be positive numbers, $p-2q \geqslant 1, q>0$, and $abc=1$. Then
$$
\sum \frac{1}{a^{p}(b+c)^{q}} \geqslant \frac{3}{2^{q}} \text {. }
$$ | Proof: From equation (1) we have
$$
\begin{array}{l}
\sum \frac{1}{a^{p}(b+c)^{q}}=\sum \frac{\left(\frac{1}{a}\right)^{p-q}}{\left(\frac{1}{b}+\frac{1}{c}\right)^{q}} \\
\geqslant \frac{\left(\sum \frac{1}{a}\right)^{p-q}}{3^{p-2 q-1}\left[\sum\left(\frac{1}{b}+\frac{1}{c}\right)\right]^{q}} \\
=\frac{\left(\sum \frac... | \frac{3}{2^{q}} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,094 |
Example 5 Let positive numbers $a, b, c, x, y, z$ satisfy $c y + b z = a, a z + c x = b, b x + a y = c$.
Also, let $\alpha, \beta$ be positive numbers, and $\alpha \geqslant 2 \beta, \beta \geqslant 1$. Find the minimum value of the function
$$
f(x, y, z)=\frac{x^{\alpha}}{(1+x)^{\beta}}+\frac{y^{\alpha}}{(1+y)^{\beta}... | Given the conditions $c y+b z=a, a z+c x=b$, $b x+a y=c$, we easily obtain
$$
\begin{array}{c}
x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \\
y=\frac{c^{2}+a^{2}-b^{2}}{2 c a}, \\
z=\frac{a^{2}+b^{2}-c^{2}}{2 a b} .
\end{array}
$$
From equation (1) and the AM-GM inequality, we have
$$
\begin{array}{l}
f(x, y, z)=\sum \frac{x^{... | \frac{1}{3^{\beta-1} \times 2^{\alpha-\beta}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,095 |
Example 4 Two isosceles triangles have supplementary vertex angles, one triangle has side lengths of $a, a, b (a > b)$, and the other triangle has side lengths of $b, b, a$. Find the degree measures of their interior angles.
---
The translation maintains the original text's format and line breaks as requested. | Explanation: Two isosceles triangles are combined to form the shape shown in Figure 3, where,
$$
\begin{array}{l}
A B= \\
A C=C D=a, B C= \\
B D=b .
\end{array}
$$
Since $\angle B A C + \angle D B C = 180^{\circ}$, the base angles must be complementary, i.e.,
$$
\angle A C B + \angle B C D = 90^{\circ}.
$$
Draw $B E ... | \angle B A C = 30^{\circ}, \angle A B C = \angle A C B = 75^{\circ}, \angle B D C = \angle B C D = 15^{\circ}, \angle D B C = 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,097 |
Example 5 In a complete quadrilateral $A B C D E F$, the diagonals $A D$ and $B F$ intersect at point $G$. If the circle passing through points $D$, $F$, and $G$ is tangent to sides $A E$ and $B E$ at points $F$ and $D$ respectively, then line $C G$ is a tangent to the circumcircle of $\triangle D F G$. | Proof: As shown in Figure 11, let the tangent line of the circumcircle of $\triangle D F G$ passing through point $G$ intersect line $F D$ at point $C^{\prime}$.
The line $D G$ intersects the tangent line of the circumcircle of $\triangle D F G$ passing through point $F$ at point $A$, and the line $F G$ intersects the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,098 |
Example 6 In the complete quadrilateral CFBEGA, the line containing diagonal $CE$ intersects the circumcircle of $\triangle ABC$ at point $D$. The circle passing through point $D$ and tangent to $FG$ at point $E$ intersects $AB$ at point $M$. Given $\frac{AM}{AB}=t$. Find $\frac{GE}{EF}$ (expressed in terms of $t$). | Solution: As shown in Figure 12, connect $A D$, $M D$, and $B D$. We have
$$
\begin{array}{l}
\angle D M A \\
=\angle D E G \\
=\angle F E C, \\
\angle F G E \\
=\angle M A D,
\end{array}
$$
Thus, $\triangle E F C \backsim \triangle M D A$.
Therefore, $\frac{E F}{M D}=\frac{C E}{A M}$, which means
$$
E F \cdot A M=M D... | \frac{t}{1-t} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,099 |
Example 7 In the complete quadrilateral $B X A P C R$, $\odot O_{1}$ is tangent to $A B$ at point $A$ and to $X C$ at point $P$; $\odot O_{2}$ passes through points $C$ and $P$, and is tangent to $A B$ at point $B$. $\odot O_{1}$ and $\odot O_{2}$ intersect at point $P$ and another point $Q$. Prove: the circumcircle of... | Proof: As shown in Figure 13, connect $A Q$ and $B Q$. Since
$$
\begin{array}{l}
\angle B P R \\
=\angle P B A + \angle B A P \\
=\angle B C X + \angle A P X \\
=\angle B C X + \angle C P R \\
=\angle B R P,
\end{array}
$$
thus, $B P = B R$.
By the converse of the tangent-chord angle theorem, it suffices to prove
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,100 |
Example 8 In a complete quadrilateral $A B C D E F$, $\odot O$ is internally tangent to the sides $A B, B D, D F, F A$ of quadrilateral $A B D F$ at points $P, Q, R, S$. Prove:
(1) $A D, B F, P R, Q S$ are concurrent;
(2) $A C - C D = A E - D E$.
保留了原文的换行和格式，以下是翻译结果：
```
Example 8 In a complete quadrilateral $A B C D ... | Proof: (1) As shown in Figure 14, let $B F$ intersect $Q S$ at point $M$, and $B F$ intersect $P R$ at point $M^{\prime}$.
We will prove that point $M$ coincides with $M^{\prime}$.
Applying Menelaus' theorem to $\triangle B E F$ and the transversal $Q M S$, and to $\triangle B C F$ and the transversal $P M^{\prime} R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,101 |
Example 1 As shown in Figure 1, the lengths of the sides $AB$, $BC$, $CD$, and $DA$ of quadrilateral $ABCD$ are 1, 9, 8, and 6, respectively. For the following statements:
(1) Quadrilateral $ABCD$ is circumscribed around a circle;
(2) Quadrilateral $ABCD$ is not inscribed in a circle;
(3) The diagonals are not perpendi... | Solution 1: If $\angle A D C=90^{\circ}$, then $A C=10$; if $\angle A D C>90^{\circ}$, then $A C>10$. From (4), we get $A C \geqslant 10$.
But in $\triangle A B C$, $A C A C^{2}$.
Therefore, $\angle A D C$ is an acute angle, which means proposition (4) is false, thus negating options (A) and (D). The options (B) and (... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,102 |
Example 2 If the quadratic function $f(x)=a x^{2}+b x+c$ has values whose absolute values do not exceed 1 on $[0,1]$, what is the maximum value of $|a|+$ $|b|+|c|$? | Solution 1: For $x \in [0,1]$, we have $|f(x)| \leqslant 1$, thus
$$
\begin{array}{l}
|f(0)| \leqslant 1, \left|f\left(\frac{1}{2}\right)\right| \leqslant 1, |f(1)| \leqslant 1. \\
\text { Also } f(0)=c, \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c, \\
f(1)=a+b+c,
\end{array}
$$
Solving simultaneously, w... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,103 |
Example 3 As shown in Figure 2, the three altitudes of acute $\triangle A B C$ intersect at point $H$. How many triangles are there in Figure 2?
The text above is translated into English, preserving the original text's line breaks and format. | Solution 1: Let the set of triangles with vertex $A$ be $P_{1}$. Then, the triangles with vertices on $A B$ and $A D$ are 3, the triangles with vertices on $A D$ and $A C$ are 3, and the triangles with vertices on $A B$ and $A C$ are 3, giving $\left|P_{1}\right|=9$.
Similarly, the sets of triangles with vertices at $... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,104 |
Example 4: Team A and Team B each send out 7 members to participate in a Go chess tournament according to a pre-arranged order. Both sides start with the No. 1 member competing, the loser is eliminated, and the winner then competes with the No. 2 member of the losing side, …, until all members of one side are eliminate... | Let the members of Team A be denoted as $A_{1}, A_{2}, \cdots, A_{7}$, and the members of Team B as $B_{1}, B_{2}, \cdots, B_{7}$, with the subscripts indicating their predetermined order of appearance. A match process corresponds to the "ordered" permutation of these 14 elements, where the order from left to right rep... | \mathrm{C}_{14}^{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,105 |
Let $p, q \in \mathbf{R}_{+}, x \in\left(0, \frac{\pi}{2}\right)$. Find the minimum value of the function $f(x)=\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}}$. | Solution: Let $\alpha=\frac{5}{4}, \beta=5$, then, $\frac{1}{\alpha}+\frac{1}{\beta}=1$.
By Hölder's inequality, we have
$$
\begin{array}{l}
p^{\frac{4}{5}}+q^{\frac{4}{5}} \\
=\frac{p^{\frac{4}{5}}}{(\sin x)^{\frac{2}{5}}}(\sin x)^{\frac{2}{5}}+\frac{q^{\frac{4}{5}}}{(\cos x)^{\frac{2}{5}}}(\cos x)^{\frac{2}{5}} \\
\l... | \left(p^{\frac{4}{5}}+q^{\frac{4}{5}}\right)^{\frac{5}{4}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,106 |
In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, the incircle $\odot O$ of $\triangle ABC$ is tangent to $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $AD$, intersecting the incircle $\odot O$ at point $P$, and connect $BP$ and $CP$. If $\angle BPC=90^{\circ}$, prove:
$$
AE + AP... | Proof: As shown in Figure 1, let \( A E = A F = x, B D = B F = u, C D = C E = v, P A = m, P B = h, P C = l, P D = n \).
According to formulas (1), the secant-tangent theorem, and the Pythagorean theorem, we have
\[
\begin{array}{l}
x^{2} = m(m+n), \\
h^{2} = \frac{n(x+u)^{2} + m u^{2}}{m+n} - m n, \\
l^{2} = \frac{n(x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,107 |
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