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$2 \cdot 46$ Find the smallest positive integer $n$, such that when $n$ different numbers $a_{1}, a_{2}, \cdots, a_{n}$ are chosen from the interval $(1,1000)$, there always exist two numbers $a_{i}, a_{j}, 1 \leqslant i, j \leqslant n$, satisfying the inequality
$$
0<a_{i}-a_{j}<1+3 \sqrt[3]{a_{i} a_{j}} .
$$ | ［Solution］Since
$$
\begin{aligned}
a_{i}-a_{j} & =\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)\left(\sqrt[3]{a_{i}^{2}}+\sqrt[3]{a_{i} a_{j}}+\sqrt[3]{a_{j}^{2}}\right) \\
& =\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)\left[\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)^{2}+3 \sqrt[3]{a_{i} a_{j}}\right],
\end{aligned}
... | 11 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
【Question 9】As shown in the figure, in rectangle $A B C D$, the area of $\triangle F D C$ is 4, and the area of $\triangle F D E$ is 2. Then the area of the shaded quadrilateral $A E F B$ is $\qquad$ . | Parse: Connect BE, by the trapezoid butterfly theorem, we know that $S_{\triangle B E F}=S_{\triangle C D F}=4$, so $S_{\triangle B C F}=4 \times 4 \div 2=8$ therefore $S_{\triangle B C D}=4+8=12, S_{A B C D}=12 \times 2=24$ so $S_{A E F B}=24-2-4-8=10$ | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If the positive integer $a$ makes the maximum value of the function $f(x)=x+\sqrt{13-2 a x}$ also a positive integer, then this maximum value equals | 1. 7 Detailed Explanation: $f(x)=y,(y-x)^{2}=13-2 a x$, then $x^{2}-(2 y-2 a) x+y^{2}-13=0, \Delta_{x} \geqslant 0 \Rightarrow 2 y \leqslant a+\frac{13}{a}$ $\in \mathbf{Z}^{+} \Rightarrow a=1, y_{\max }=7$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 67$ represents any permutation of the integers $1,2, \cdots, n$ using $a_{1}, a_{2}, \cdots, a_{n}$. Let $f(n)$ be the number of permutations that satisfy the conditions:
(1) $a_{1}=1$,
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1,2, \cdots, n-1$
Determine whether $f(1996)$ is divisible by 3. | [Solution] Clearly, $f(1)=f(2)=1, f(3)=2$.
Let $n \geqslant 4$. Since $a_{1}=1$, then $a_{2}=2$ or 3.
If $a_{2}=2$, the number of permutations of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ is equal to the number of permutations of $a_{2}-1, a_{3}-1, \cdots, a_{n}-1$. The latter is a permutation of $1,2, \cdots, n-1$ and sati... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 6 In a rectangle $A B C D$ with an area of 1 (including the boundary) there are 5 points, among which no three points are collinear. Find the minimum number of triangles, with these 5 points as vertices, whose area is not greater than $\frac{1}{4}$. | In rectangle $ABCD$, if the area of a triangle formed by any three points is no more than $\frac{1}{4}$, then these three points are called a "good triplet" or simply a "good group."
Let the midpoints of $AB$, $CD$, $BC$, and $AD$ be $E$, $F$, $H$, and $G$ respectively. The intersection of line segments $EF$ and $GH$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Among all the triangles formed by any 3 of the 8 vertices of a rectangular cuboid, the number of acute triangles is
(A) 0
(B) 6
(C) 8
(D) 24 | (C)
4. 【Analysis and Solution】The number of triangles formed by any 3 of the 8 vertices of a rectangular prism is $\mathrm{C}_{8}^{3}$, and these triangles can only be right triangles or acute triangles. There are 6 right triangles for each vertex of the rectangular prism when it serves as the right-angle vertex, so th... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. In the known sequence $1,4,8,10,16,19,21,25,30,43$, the number of subarrays whose sum is divisible by 11 is $\qquad$ . | 3. Since the problem is about divisibility by 11, we can first subtract multiples of 11 from each term to make the numbers smaller and easier to handle, resulting in the following sequence:
$$
1,4,-3,-1,5,-3,-1,3,-3,-1 .
$$
Let $S_{n}$ be the sum of the first $n$ terms, then
$$
\begin{array}{l}
s_{1}=1, s_{2}=5, s_{3}... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In $\triangle A B C$, $\angle C=100^{\circ}, \angle A=60^{\circ}$, take a point $P$ on the line $A C$ such that $\triangle P A B$ is an isosceles triangle, then the number of points $P$ that satisfy the condition is $\qquad$. | $2$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. If the equation with respect to $x$
$$
x^{2}+a x+b-3=0(a, b \in \mathbf{R})
$$
has real roots in the interval $[1,2]$, then the minimum value of $a^{2}+(b-4)^{2}$ is $\qquad$. | 8. 2 .
From the problem, we know that $b=-x^{2}-a x+3$.
Then $a^{2}+(b-4)^{2}$
\[
\begin{array}{l}
=a^{2}+\left(-x^{2}-a x-1\right)^{2} \\
=x^{2}(x+a)^{2}+2\left(x^{2}+a x\right)+a^{2}+1 \\
=\left(x^{2}+1\right)(x+a)^{2}+x^{2}+1 .
\end{array}
\]
Also, $x \in[1,2]$, so,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.48 The number of real roots of the equation $\sin x = \lg x$ is
(A) 1.
(B) 2.
(C) 3.
(D) Greater than 3.
(China High School Mathematics League, 1984) | [Solution] In the same coordinate system, draw the graphs of the functions $y_{1}=\sin x, y_{2}=\lg x$ (as shown in the figure) to determine the number of intersections between $y_{1}$ and $y_{2}$, which is the number of real roots of the equation $\sin x=\lg x$.
From the graph, there are three intersection points, mea... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
12.3 Given $i^{2}=-1$, the number of $n$ that makes $(n+i)^{4}$ an integer is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(31st American High School Mathematics Examination, 1980) | [Solution] Since $i^{2}=-1$, we have
$$
(n+i)^{4}=n^{4}-6 n^{2}+1+i\left(4 n^{3}-4 n\right),
$$
which is a real number if and only if $4 n^{3}-4 n=0$.
Solving this, we get $n=0,1,-1$. Therefore, there are only three values of $n$ that make $(n+i)^{4}$ a real number, and in these cases, $(n+i)^{4}=n^{4}-6 n^{2}+1$ is a... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. (5 points) Three people, A, B, and C, compare the number of candies they each have.
A says: “I have 13, 3 less than B, and 1 more than C.”
B says: “I am not the one with the least, C and I differ by 4, and A has 11.”
C says: “I have fewer than A, A has 10, and B has 2 more than A.”
If each person has one false stat... | 【Answer】Solution: Jia says "I have 13, 3 less than Yi" is in contradiction with Bing saying "Jia has 10, Yi has 2 more than Jia", Jia's statements are one right and one wrong, so Yi saying Jia has 11 must be wrong, the first two statements are correct; assuming Jia has 13, then Yi has 2 more than Jia, Yi would have 15,... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. Let $F(x)=|f(x) \cdot g(x)|$, where $f(x)=a x^{2}+b x+c, g(x)=c x^{2}+b x+a, x \in[-1,1]$. For any $a, b, c$ satisfying $|f(x)| \leqslant 1$. When $a, b, c$ vary, find the maximum value of $F(x)$. | 9. $|f(0)|=|c| \leqslant 1, |f(1)|=|a+b+c| \leqslant 1, |f(-1)|=|a-b+c| \leqslant 1$, so for any $x \in [-1, 1]$, $|a+bx+c| \leqslant 1$. And $|g(x)|=|cx^2+bx+a|=|c(x^2-1)+c+bx+a| \leqslant |c(x^2-1)|+|a+bx+c|$ $\leqslant 2$. Therefore, $F(x) \leqslant 2$. And when $f(x)=2x^2-1, g(x)=-x^2+2$, $F(0)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A person has a total of 23 coins of 5 cents, 10 cents, and 25 cents. Among them, the number of 10 cent coins is three more than the number of 5 cent coins. If the total value of the coins is 320 cents, then the number of 25 cent coins is ( ) more than the number of 5 cent coins.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 8. C.
Let there be $x$ coins of 5 cents. Then there are $x+3$ coins of 10 cents, and $23-(x+3)-x$ coins of 25 cents.
According to the problem,
$$
\begin{array}{l}
5 x+10(x+3)+25(23-x-(x+3))=320 \\
\Rightarrow x=6 .
\end{array}
$$
Therefore, the required number is $23-(x+3)-x-x=2$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 1 (1993 National High School League Question) The last two digits of $\left[\frac{10^{93}}{10^{31}+3}\right]$ are $\qquad$ (where [x] denotes the greatest integer not greater than $x$.) | Solution: Fill in 08. Reason: Since $10^{93}=\left(10^{31}\right)^{3}+3^{3}-3^{3}=\left(10^{3}+3\right)\left[\left(10^{31}\right)^{2}-10^{31} \cdot 3+\right.$ $\left.3^{2}\right]-3^{3}$, therefore
the original expression $=\left(10^{31}\right)^{2}-10^{31} \cdot 3+3^{2}-1=10^{31}\left[10^{31}-3\right]+8$. Hence, the la... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
45. 18 $k \star$ Find the smallest real number $\lambda$ such that the inequality
$$
5(a b c+a b d+a c d+b c d) \leqslant \lambda a b c d+12
$$
holds for any positive real numbers $a, b, c, d$ satisfying $a+b+c+d=4$. | When $a=b=c=d=1$, we get $\lambda \geqslant 8$. Below, we prove the inequality
$$
5(a b c+a b d+a c d+b c d) \leqslant 8 a b c d+12
$$
for any positive real numbers $a, b, c, d$ satisfying $a+b+c+d=4$.
Let $f(a, b, c, d) - 5(a b c + a a b d + a c d + b c d) - 8 a b c d$
$$
= a b(5 c + 5 d - 8 c d) + 5 c d(a + b).
$$
... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$11 \cdot 14 n$ is a non-negative integer. Given $f(0)=0, f(1)=1, f(n)=$ $f\left(\left[\frac{n}{2}\right]\right)+n-2\left[\frac{n}{2}\right]$. Determine $f(n)$. Find the maximum value of $f(n)$ for $0 \leqslant n \leqslant 1991$. (Here $[x]$ denotes the greatest integer not exceeding $x$)
(Japan Mathematical Olympiad, ... | \[ \text{Solution:} \] Express \( n \) as a binary number:
\[
n=\left(\overline{a_{k} a_{k-1} \cdots a_{2} a_{1}}\right)_{2} .
\]
Here \( a_{i}=0 \) or \( 1, i=1,2, \cdots, k \), then
\[
\begin{aligned}
& {\left[\frac{n}{2}\right]=\left(\overline{a_{k} a_{k-1} \cdots a_{2}}\right)_{2} . } \\
f(n) & \left.=f\left(\over... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 6 Determine all integers $n>3$, such that there exist $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in the plane, and real numbers $r_{1}$, $r_{2}, \cdots, r_{n}$ satisfying:
(1) No three points among $A_{1}, A_{2}, \cdots, A_{n}$ are collinear;
(2) For each triple $i, j, k(1 \leqslant j<k \leqslant n), \triangle A_... | Analysis Consider various possible cases of the convex hull, discuss and explore, guess that $n \geqslant 5$ is impossible. To illustrate this, it is sufficient to show that $n=5$ is not possible.
Solution Let the area of $\triangle A_{i} A_{j} A_{k}$ be denoted as $S_{i j k}$.
Lemma 1 If the vertices of a certain conv... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. As shown in the figure, points $P, Q, R$ are on the sides $AB$, $BC$, and $CA$ of $\triangle ABC$, respectively, and $BP=PQ=QR=RC=1$. The maximum area of $\triangle ABC$ is
(A) $\sqrt{3}$;
(B) 2;
(C) $\sqrt{5}$;
(D) 3. | 8. (B)
Solution First, if we denote $\triangle A P R, \triangle B Q P, \triangle C R Q, \triangle P Q R$ as I, II, III, N respectively, then $S_{1}, S_{1}, S_{\mathrm{N}}$ are all no greater than $\frac{1}{2} \times 1 \times 1=\frac{1}{2}$. Also, since $\angle P Q R=180^{\circ}-(\angle B+\angle C)$
$$
=\angle A \text ... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
6、Given $\left[\prod_{k=1}^{n} \frac{(2 k+1)^{4}+(2 k+1)^{2}+1}{(2 k)^{4}+(2 k)^{2}+1}\right]=44$. Then $n=$ $\qquad$, $[x]$ denotes the greatest integer not exceeding the real number $x$.
untranslated part:
已知 $\left[\prod_{k=1}^{n} \frac{(2 k+1)^{4}+(2 k+1)^{2}+1}{(2 k)^{4}+(2 k)^{2}+1}\right]=44$. 则 $n=$ $\qquad$ ... | 6. 5
Analysis: Given $n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$, and $n^{2}+n+1=(n+1)^{2}-(n+1)+1$, $n^{2}-n+1=(n-1)^{2}+(n-1)+1$, hence $\prod_{k=1}^{n} \frac{(2 k+1)^{4}+(2 k+1)^{2}+1}{(2 k)^{4}+(2 k)^{2}+1}=\frac{(2 n+1)^{2}+(2 n+1)+1}{3}$, solving for $n$ yields $n=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22. As shown in Figure 2, in rectangle $D E F A$, $A D=3, A F=$ $4, D C=C B=B A$. Then the area of the shaded part is ( ).
(A) 2
(B) $\frac{5}{2}$
(C) 3
(D) $\frac{7}{2}$
(E) 5 | 22. C.
Let $B E$ intersect $C F$ at point $O$.
In rectangle $D E F A$, we have
$$
\begin{array}{l}
A D / / E F \Rightarrow \triangle B C O \backsim \triangle E F O \\
\Rightarrow \frac{B O}{O E}=\frac{B C}{E F}=\frac{1}{3} \\
\Rightarrow S_{\triangle B C E}=\frac{1}{2} \times 1 \times 4=2, \\
S_{\triangle C O E}=\frac... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
N57 (46-4, Poland) The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1 \quad(n=1,2,3, \cdots) .
$$
Find all positive integers that are coprime with every term of this sequence. | We first prove the following conclusion: For any prime $p$ not less than 5, we have
$$
2^{p^{\cdots 2}}+3^{p^{\cdots 2}}+6^{p-2}-1 \equiv 0 \quad(\bmod p) .
$$
Since $p$ is a prime not less than 5, we have $(2, p)=1, (3, p)=1, (6, p)=1$. By Fermat's Little Theorem,
$$
2^{p-1} \equiv 1 \quad(\bmod p), 3^{p-1} \equiv 1 ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
20. Two isosceles triangles $T$ and $T'$ are not congruent, but they have equal perimeters and equal areas. The side lengths of triangle $T$ are $5$, $5$, and $8$, and the side lengths of triangle $T'$ are $a$, $a$, and $b$. Which of the following numbers is closest to $b$?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8 | 20. A.
From the problem, we know that the perimeter of triangle $T$ is 18, and the area is 12, i.e.,
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a+b=18, \\
\frac{1}{2} b \sqrt{a^{2}-\left(\frac{b}{2}\right)^{2}}=12
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2 a+b=18, \\
b \sqrt{4 a^{2}-b^{2}}=48
\end{a... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
【Question 5】
Given that $A$ is a positive integer and satisfies the following conditions: $\frac{5}{9}<\frac{9}{A}<1$. How many different values can $A$ take? | 【Analysis and Solution】
Since $\frac{5}{9}\frac{A}{9}>1$, that is, $1<\frac{A}{9}<\frac{9}{5}$;
thus $9<A<\frac{81}{5}$, which means $9<A<16 \frac{1}{5}$;
since $A$ is a positive integer;
therefore $A=10, 11, 12, 13, 14, 15, 16$; $A$ has $16-10+1=7$ different values. | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11. Given that the graph of the function $y=f(x)$ is symmetric to the graph of $y=2-\frac{1}{x}$ with respect to the line $y=x$. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1} = f\left(a_{n}\right)\left(n \in \mathbf{N}^{*}\right)$.
(1) If $a_{1}=3$, prove that there exists a positive integer $n_{0}$ such that ... | 11. Solution: (1) Since the graph of the function $y=f(x)$ is symmetric to the graph of $y=2-\frac{1}{x}$ about the line $y=x$, for any point $(x, y)$ on the graph of the function $y=f(x)$, the point $(y, x)$ is on the graph of the function $y=2-\frac{1}{x}$, i.e., $x=2-\frac{1}{y}$, so, $y=f(x)=\frac{1}{2-x}, a_{n+1}=... | +2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
24.3.11 For a competition with 4 multiple-choice questions, each with $A, B, C$ as options, for any three participants, there is at least one question where their answers are all different. How many people can participate in the competition at most? | Using the recursive method, let $f(n)$ represent the maximum number of participants when there are $n$ questions, then $f(1)=3$.
If a question makes the answers of three people different, we say this question can distinguish these three people. By the pigeonhole principle, for the first question, there must be an optio... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. $x, y \in \mathbf{R}$, and satisfy $\left\{\begin{array}{l}(x+1)^{\frac{3}{5}}+2023(x+1)=-2023, \\ (y+1)^{\frac{3}{5}}+2023(y+1)=2023\end{array}\right.$, then $x+y=$ | Since the function $f(x)=x^{\frac{3}{5}}+2023 x$ is an odd function, and it is monotonically increasing on $(0,+\infty)$, then according to the problem, $f(x+1)+f(y+1)=0 \Rightarrow x+1+y+1=0 \Rightarrow x+y=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate:
$$
\left(2^{2}-2\right)-\left(3^{2}-3\right)+\left(4^{2}-4\right)=(\quad) \text {. }
$$
(A) 1
(B) 2
(C) 5
(D) 8
(E) 12 | 1. D.
$$
\begin{array}{l}
\left(2^{2}-2\right)-\left(3^{2}-3\right)+\left(4^{2}-4\right) \\
=(4-2)-(9-3)+(16-4) \\
=2-6+12=8 .
\end{array}
$$ | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. $S_{n}$ represents the sum of the first $n$ terms of a geometric sequence with a common ratio $q \neq-1(q \in \mathbf{R})$, and let $M=$ $\left\{x \left\lvert\, x=\lim _{n \rightarrow \infty} \frac{S_{n}}{S_{2 n}}\right.\right\}$, the number of subsets of the set $M$ is ( ).
A. 2
B. 4
C. 7
D. 8 | 2. D.
Since $S_{2 n}=S_{n}+q^{n} S_{n}$, then $\frac{S_{n}}{S_{2 n}}=\frac{1}{1+q^{n}}$. When $|q|<1$, $\lim _{n \rightarrow+\infty} \frac{S_{n}}{S_{2 n}}=0$. Therefore, $M=\left\{0, \frac{1}{2}, 1\right\}$, so the number of subsets of $M$ is $2^{3}=8$ (subsets). | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. (10 points) Place the " $b$ " shaped polyominoes on an $8 \times 8$ grid: The polyominoes must align with the grid lines, and after placement, the total number of polyominoes in each row and column must be equal. How many such " $b$ " shaped polyominoes can be placed? (Note: The polyominoes can be rotated, but not f... | 【Answer】Solution: According to the analysis, as shown in the figure,
to make the blocks align with the grid lines, and after covering, the total number of blocks in each row and column of the grid must be equal, 7 more $b$-type blocks can be placed.
Therefore, the answer is: 7. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $ABCD$ be a rectangle with an area of 2, $P$ a point on side $CD$, and $Q$ the point where the incircle of $\triangle PAB$ touches side $AB$. The product $PA \cdot PB$ varies as point $P$ changes. When $PA \cdot PB$ is minimized,
(1) Prove: $AB \geqslant 2BC$;
(2) Find the value of $AQ \cdot BQ$.
| 3. (1) Since $S_{\triangle A P B}=\frac{1}{2} S_{A B C D}=1$, we have $\frac{1}{2} P A \cdot P B \cdot \sin \angle A P B=1$, so $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$,
with equality holding only when $\angle A P B=90^{\circ}$.
This indicates that point $P$ lies on the circle with $A B$ as its diameter,... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) The width of rectangle $O_{1} O_{2} B A$ is $A O_{1}=1$ cm. Circles $O_{1}$ and $O_{2}$ are drawn with centers at $O_{1}$ and $O_{2}$, respectively, and a radius of 1 cm, intersecting line segment $O_{1} O_{2}$ at points $C$ and $D$, as shown in the figure. What is the area of quadrilateral $A B C D$ in ... | 【Analysis】According to the problem, we can draw $D E \perp A B$ at $E$. It is easy to know that $A E=O_{1} D=O_{2} C$, and the areas of $\triangle A E D$ and $\triangle B C O_{2}$ are equal. Therefore, the area of the shaded part in the figure is equal to the area of the square $E B O_{2} D$. It is not difficult to fin... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$1 \cdot 47$ Let $x^{2}-x+a$ divide $x^{13}+x+90$, determine the positive integer value of $a$.
(24th Putnam Mathematical Competition, 1963) | [Solution] Let $x^{13}+x+90=\left(x^{2}-x+a\right) q(x)$, where $a$ is an integer, and $q(x)$ is a polynomial with integer coefficients.
$$
\begin{array}{c}
\text { Let } x=-1,0,1, \text { then } \\
\left\{\begin{array}{l}
(a+2) q(-1)=88, \\
a q(0)=90, \\
a q(1)=92 .
\end{array}\right.
\end{array}
$$
From (2),(3),
$$
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (6 points) Letters $a, b, c, d, e, f, g$ each represent one of the numbers from 1 to 7. If $a+b+c=c+d+e=c+f+g$, then the number of possible values for $c$ is $\qquad$.
| 【Solution】Solution: $a+b+c=c+d+e=c+f+g$, which means $a+b=d+e=f+g$, there can only be 3 cases:
(1) $1+7=2+6=3+5$, in this case $c=4$;
(2) $2+7=3+6=4+5$, in this case $c=1$;
(3) $1+6=2+5=3+4$, in this case $c=7$;
Therefore, the possible values of $c$ are $1, 4, 7$, a total of 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$28 \cdot 22$ Let $N$ be a number between 9 and 17. If the average of 6, 10, and $N$ is an even number, then it can only be
(A) 8.
(B) 10.
(C) 12.
(D) 14.
(E) 16.
(5th American Junior High School Mathematics Examination, 1989) | ［Solution］According to the problem, it is obvious that (let $\bar{x}$ be the average of the three numbers)
$$
\frac{9+6+10}{3}<\bar{x}<\frac{17+6+10}{3} \text {, }
$$
which means
$$
8 \frac{1}{3}<\bar{x}<11 \text {. }
$$
Thus, $\bar{x}=10$.
Therefore, the answer is $(B)$. | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Let point $O$ be inside a regular tetrahedron $A B C D$, and $G$ be the center of the regular tetrahedron. The line $O G$ intersects the faces of the tetrahedron at points $A_{1} \cdot B_{1}, C_{1} \cdot D_{1}$. Then the value of $\frac{A_{1} O}{A_{1} G}+\frac{B_{1} O}{B_{1} G}+\frac{C_{1} O}{C_{1} G}+\frac{D_{1} O}... | 2. D Let the distance from point $G$ to each face of the tetrahedron be $r$, and the distances from point $O$ to the planes $BCD$, $CDA$, $DAB$, $ABC$ be $h_{1}$,
$$
\begin{array}{c}
h_{2}, h_{3}, h_{4} \text {, then } \frac{V_{O-BCD}}{V_{G-BCD}}=\frac{h_{1}}{r}=\frac{A_{1} O}{A_{1} G}, \frac{V_{O}-CDA}{V_{G}-CDA}=\fra... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\sin y \cdot \cos y+a=0 .
\end{array}\right.
$$
Then $\cos (x+2 y)=$ . $\qquad$ | 2. From the given, $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$, let $f(t)=$ $t^{3}+\sin t$, then $f(x)=(-2 y)$. Since the function $f(t)=t^{3}+\sin t$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\therefore x=-2 y, x+2 y=0$, hence $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The first railway independently designed and constructed in our country, the "Jingzhang Railway," adopted a "human-shaped" design. Now, a section of the human-shaped road is to be built, as shown in the figure, where the length of $\mathrm{A} \rightarrow \mathrm{C}$ is 25 kilometers, the length of $\mathrm{C} \right... | $9$ | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given that $f(x)$ is an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
f(2+x)+f(2-x)=0 .
$$
If for $x \in[-1,0)$,
$$
f(x)=\log _{2}(1-x) \text {, }
$$
then $f(1)+f(2)+\cdots+f(2021)=$ $\qquad$ | - 1. -1 .
From $f(x)$ being an odd function defined on $\mathbf{R}$, we get
$$
f(0)=0, f(1)=-f(-1)=-1 \text {. }
$$
Taking $x=0$, we get
$$
f(2)=-f(2) \Rightarrow f(2)=0 \text {. }
$$
Let $y=2-x$, then
$$
\begin{array}{c}
f(y)=-f(2+(2-y)) \\
=-f(4-y)=f(y-4) .
\end{array}
$$
Thus, $f(x)$ is a periodic function with ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If the complex coefficient equation $(4+3 \mathrm{i}) x^{2}+m x+$ $4-3 \mathrm{i}=0$ has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$ | 4. 8 .
Let $\alpha$ be a real root of the original equation, $m=x+y \mathrm{i}$, then
$$
(4+3 \mathrm{i}) \alpha^{2}+(x+y \mathrm{i}) \alpha+4-3 \mathrm{i}=0,
$$
which means
$$
\left\{\begin{array}{l}
4 \alpha^{2}+\alpha x+4=0, \\
3 \alpha^{2}+\alpha y-3=0,
\end{array}\right.
$$
or equivalently
$$
\left\{\begin{arra... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2、To make the equation $\frac{1}{2} \times\left\{\frac{1}{3} \times\left[\frac{1}{4} \times(145-1)-\square\right]+4\right\}=7$ true, the number that should be filled in the box is | 【Analysis】The original expression becomes: $\frac{1}{3} \times\left[\frac{1}{4} \times 144-\square\right]+4=14$
$$
\begin{aligned}
\frac{1}{3} \times\left(\frac{1}{4} \times 144-\square\right) & =10 \\
\frac{1}{4} \times 144-\square & =30 \\
\square & =6
\end{aligned}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. Given the quadratic function $f(x)=4 x^{2}-4 a x+\left(a^{2}-2 a+2\right)$ has a minimum value of 2 on $0 \leqslant x \leqslant 1$, find the value of $a$.
| 18. $f(x)=4\left(x-\frac{a}{2}\right)^{2}-2 a+2$.
When $0 \leqslant \frac{a}{2} \leqslant 1$ i.e., $0 \leqslant a \leqslant 2$, $f_{\text{min}}(x)=-2 a+2=2$. This gives $a=0$.
When $\frac{a}{2}>1$, i.e., $a>2$, $f_{\text{min}}(x)=f(1)=4-4 a+a^{2}-2 a+2=2$. This yields $a=3-\sqrt{5}$ (discard) or $a=3+\sqrt{5}$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let the side length of a regular $n$-sided polygon be $a$, and the longest and shortest diagonals be $b$ and $c$ respectively. If $a=b-c$, then $n=$ $\qquad$ | 6. 9 .
Let the diameter of the circumcircle of a regular $n$-sided polygon be 1.
When $n=2k$,
$b=1, c=\sin \frac{\pi}{k}, a=\sin \frac{\pi}{2k}$.
Thus, $f(k)=\sin \frac{\pi}{k}+\sin \frac{\pi}{2k}=1$.
Since the function $f(k)$ is monotonically decreasing, and $f(4)>1>f(5)$,
there is no solution in this case.
When $n=2... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. BAIYUN Taxi charges a starting fee of 12 yuan for the first 3 kilometers (including 3 kilometers); from 3 kilometers to 7 kilometers (including 7 kilometers), it charges 2.6 yuan per kilometer; for distances greater than 7 kilometers, it charges 3.5 yuan per kilometer; now Yueyue needs to go to a place 30 kilometers... | 【Analysis】(1) When the ride distance does not exceed 3 kilometers, the charge is 12 yuan, and the average price per kilometer is at least 4 yuan (only when traveling 3 kilometers).
(2) When the ride distance does not exceed $3^{\sim} 10$ kilometers, due to $2.63.02$, the average price per kilometer will increase, and w... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. If real numbers $x, y$ satisfy $2^{x}+4 x+12=\log _{2}(y-1)^{3}+3 y+12=0$, then $x+y=$ | Answer: -2
Analysis: Let $s=x-2$, then $2^{x}+4 x+12=0 \Leftrightarrow 2^{s}=-s-5$; Let $t=y-1$, then $\log _{2}(y-1)^{3}+3 y+12=0 \Leftrightarrow \log _{2} t=-t-5$. Note that the graphs of the functions $y=2^{x}$ and $y=\log _{2} x$ are symmetric about the line $y=x$, and the graph of the function $y=-x-5$ is also sym... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (1984 National High School Competition) The number of real roots of the equation $\sin x=\lg x$ is $(\quad)$.
A. 1
B. 2
C. 3
D. Greater than 3 | 3. C. Reason: The number of solutions to the equation $\sin x=\lg x$ is the number of intersection points between the sine curve $\sin x$ and the logarithmic curve $\lg x$. First, determine the range of $x$. From the definition of $\lg x$, we know: $x>0$, and since $\sin x \leqslant 1$, it follows that $\lg x \leqslant... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Let the function $f(x)=x^{2}+a x+b$, for any $a, b \in \mathbf{R}$, there always exists $t \in[0,4]$, such that $|f(t)| \geqslant m$ holds, then the maximum value of the real number $m$ is $\qquad$. | Let $M=\max \left\{|f(0)|,|f(4)|,\left|f\left(-\frac{a}{2}\right)\right|\right\}=\max \left\{|b|,|16+4 a+b|,\left|b-\frac{a^{2}}{4}\right|\right\}$,
then $4 M \geqslant|b|+|16+4 a+b|+2\left|b-\frac{a^{2}}{4}\right| \geqslant\left|16+4 a+\frac{a^{2}}{2}\right|$ $=\frac{1}{2}\left|a^{2}+8 a+32\right|=\frac{1}{2}\left|(a+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the Cartesian coordinate system, with $(199,0)$ as the center and 199 as the radius, the number of integer points on the circumference of the circle is ( ) .
A. 2
B. 4
C. 8
D. 16 | 3. B Let $(x, y)$ be an integer point on a known circle, then $(x-199)^{2}+y^{2}=199^{2}$.
Obviously, $(x, y)=(0,0),(199,199),(199,-199),(398,0)$ are four solutions to the equation.
When $y \neq 0$ and $y \neq \pm 199$, $y$ is coprime with 199 (since 199 is a prime number), at this time $199, y,|199-x|$ form a Pythago... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
1.54 In the Aba tribe, there are two letters. It is known that in this language, no word is the prefix of another word. Can the dictionary of this tribe's language contain: 3 words each with 4 letters, 10 words each with 5 letters, 30 words each with 6 letters, and 5 words each with 7 letters? | [Solution] With two different letters, a total of $2^{7}=$ 128 different 7-letter words can be formed. However, since no word can be the prefix of another word, for a 7-letter word, its first $4,5,6$ letters cannot be the same as those of any existing word. There are a total of $3 \times 8+10 \times 4+30 \times 2=124$ ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 8$ When $n$ is a positive integer, the sum of the decimal digits of $\left(10^{4 n^{2}+8}+1\right)^{2}$ is
(A) 4.
(B) $4 n$.
(C) $2+2 n$.
(D) $4 n^{2}$.
(E) $n^{2}+n+2$
(26th American High School Mathematics Examination, 1975) | [Solution] For any positive integer $k$, $10^{k}+1=100 \cdots 01$, and
$$
\begin{array}{r}
100 \cdots 01 \\
\times \quad 100 \cdots 01 \\
\hline 100 \cdots 01 \\
+\quad 100 \cdots 01 \\
\hline 100 \cdots 0200 \cdots 01
\end{array}
$$
$\therefore$ The sum of the digits is $1+2+1=4$. Therefore, the answer is $(A)$. | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Let $x_{i} \in\{1,2, \cdots, n\}, i=1,2, \cdots, n$ satisfy $\sum_{i=1}^{n} x_{i}=\frac{n(n+1)}{2}, x_{1} x_{2} \cdots x_{n}=n$!. Then, the largest positive integer $n$ for which $x_{1}, x_{2}, \cdots, x_{n}$ must be a permutation of $1,2, \cdots, n$ is $\qquad$. | 8. 8 .
When $n \geqslant 9$, the factors $8,6,3$ in $n!$ can be replaced by $9,4,4$, with $8+6+3=9+4+4$, $8 \times 6 \times 3=9 \times 4 \times 4$. Thus, the sum $x_{1}+x_{2}+\cdots+x_{n}$ and the product $x_{1} x_{2} \cdots x_{n}$ remain unchanged. When $n=8$, it is known that $x_{1}, x_{2}, \cdots, x_{8}$ is indeed ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$
$\qquad$ | 7. 3 .
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Therefore, the required value is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(13) Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy $a f(x)+b f(x-c)=1$ for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$. | 13 (1) From the given conditions, we know that $f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1$. Given $0 \leqslant x \leqslant \frac{\pi}{2}$, we have $\frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$, thus $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when $x=\frac{\p... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in Figure 3, the vertex of the parabola $y=x^{2}$ is the origin $O$, and two perpendicular lines are drawn through $O$ intersecting the parabola $y=x^{2}$ at points $P$ and $Q$, respectively. Connect $P Q$.
(1) When $P Q \parallel x$-axis, find the distance from point $O$ to $P Q$; (2) Find the maximum dis... | Three, 11. (1) When $P Q / / x$-axis, point $P$ and $Q$ are symmetric about the $y$-axis. At this time, $\triangle O P Q$ is an isosceles right triangle.
Assume point $Q(m, m)$.
Substitute into $y=x^{2}$, we get $m=1$.
Thus, $O M=m=1$.
Therefore, the distance from point $O$ to $P Q$ is 1.
(2) Assume point $P\left(a, a^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{7}\right\}$. Here $a_{i} \in \mathbf{Z}^{\prime}$, and let $n_{A}$ denote the number of triples $(x, y, z)$ such that: $x<y$, $x+y=z, x, y, z \in A$. Then the maximum possible value of $n_{A}$ is $\qquad$. | 8. 9 Let $a<b<c<d<e<f<g$. For any three numbers $x<y<z$, calculate how many times the middle number $y$ appears, thus, for $a, b, c, d, e, f, g$ there are at most $0,1,2,3,2,1,0$ times respectively. Constructing $A=\{1,2,3,4,5,6,7\}$ can achieve this. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the minimum value of the function $f(a, b)=(a-b)^{2}+\left(\sqrt{2-a^{2}}-\frac{9}{b}\right)^{2}$. | 3. From the problem, we know that $f(a, b)$ is the square of the distance from point $P\left(a, \sqrt{2-a^{2}}\right)$ on the circle $x^{2}+y^{2}=2$ to point $Q\left(b, \frac{9}{b}\right)$ on the hyperbola $x y=9$. Since $\sqrt{2-a^{2}} \geqslant 0$, point $P$ lies on the upper half of the circle $x^{2}+y^{2}=2(y \geqs... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.59 The sum of an infinite geometric series with common ratio $r$ (where $|r|<1$) is 15, and the sum of the squares of each term of this series is 45. The first term of this series is
(A) 12.
(B) 10.
(C) 5.
(D) 3.
(E) 2.
(21st American High School Mathematics Examination, 1970) | [Solution] Let the first term of the sequence be $a$, then we have
$$
\frac{a}{1-r}=15 \text {, and } \frac{a^{2}}{1-r^{2}}=45 \text {, }
$$
From $\frac{a^{2}}{1-r^{2}}=\frac{a}{1+r} \cdot \frac{a}{1-r}$, we get $\frac{a}{1+r}=3$. From $a=15(1-r)$ and $a=3(1+r)$, we can solve to get $a=5$. Therefore, the answer is $(C... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Let $P(x)=x^{5}-x^{2}+1$ have five roots $r_{1}$,
$$
\begin{array}{l}
r_{2}, \cdots, r_{5}, Q(x)=x^{2}+1 \text {. Then } \\
\quad Q\left(r_{1}\right) Q\left(r_{2}\right) Q\left(r_{3}\right) Q\left(r_{4}\right) Q\left(r_{5}\right) \\
\quad=
\end{array}
$$ | 7. 5.
From the problem, we know $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Let $R_{n}=\frac{1}{2}\left(a^{n}+b^{n}\right), a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, n=1,2, \cdots$ What is the units digit of $R_{12445}$?
(41st AHSME Problem) | Given that $a$ and $b$ are the roots of the equation $x^{2}=6 x-1$, according to Newton's formula, we have
$$
R_{n+2}=6 R_{n+1}-R_{n} \text {. }
$$
It is easy to see that $R_{1}=\frac{1}{2}(a+b)=3, R_{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)=17$. For convenience in writing, let $D_{n}$ represent the unit digit of $R_{n}... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There are ten cards, each with one of the ten digits $0-9$. They are arranged in a row to display the number 9072543681. If swapping the positions of two adjacent cards is considered one operation, then the minimum number of operations required to change the original number into one that is divisible by 99 is $\qqua... | 2.2
Analysis: A necessary and sufficient condition for a positive integer to be a multiple of 11 is that the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11 (counting from left to right). Since the number 9072543681 has a digit sum that ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (3 points) One of the heights of a triangle is 2 cm. If the height increases by 6 cm, with the base remaining unchanged, the area increases by 12 square cm. The area of the original triangle is $\qquad$ square cm. | 【Solution】Solution: Let the base of the triangle be $a$ cm
$$
\begin{aligned}
a \times(2+6) \div 2-2 a \div 2 & =12 \\
4 a-a & =12 \\
3 a & =12 \\
a & =4 ;
\end{aligned}
$$
The area of the original triangle is $4 \times 2 \div 2=4$ (square cm)
Answer: The area of the original triangle is 4 square cm.
Therefore, the an... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. If the sum of the digits of the integer $3^{2012}$ is $a, a$'s sum of digits is $b, b$'s sum of digits is $c$, then $c=$ $\qquad$ . | $9$ | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 4 Given the function $f(x)=\log _{2}(x+1)$, the graph of $y=f(x)$ is shifted 1 unit to the left, and then the y-coordinates of all points on the graph are stretched to twice their original length (the x-coordinates remain unchanged), resulting in the graph of the function $y=g(x)$. Then the maximum value of the... | Given that $g(x)=2 \log _{2}(x+2)(x>-2)$, then
$$
\begin{aligned}
F(x) & =f(x)-g(x)=\log _{2}(x+1)-2 \log _{2}(x+2) \\
& =\log _{2} \frac{x+1}{(x+2)^{2}}=\log _{2} \frac{x+1}{x^{2}+4 x+4} \\
& =\log _{2} \frac{1}{\frac{x^{2}+4 x+4}{x+1}} \\
& =\log _{2} \frac{1}{x+1+\frac{1}{x+1}+2}(x>-1) .
\end{aligned}
$$
Since $x+1... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 31$ The maximum value of the function $y=-x-\frac{9}{x}+18(x>0)$ is
(A) 24 .
(B) 18 .
(C) 12 .
(D) 2 .
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics league, 1991) | [Solution] Let $y=-\left(x+\frac{9}{x}\right)+18, x>0$. When $x>0$, $x+\frac{9}{x} \geqslant 2 \sqrt{x \cdot \frac{9}{x}}=6$. Therefore, $y \leqslant-6+18=12$. That is, $y_{\text {max }}=12 \quad(x>0$ when $)$. Hence, the answer is $(C)$. | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. In the Cartesian coordinate system $x O y$, the moving point $A$ is on the circle $x^{2}+y^{2}=1$, and the coordinates of point $B$ are $(3,0)$. If point $C$ makes $\triangle A B C$ an equilateral triangle, then the maximum value of $|O C|$ is $\qquad$. | Given:
$$
z_{A}=\cos \theta+\mathrm{i} \sin \theta, \quad z_{B}=3,
$$
then
$$
\begin{array}{l}
z_{\overrightarrow{B A}}=(\cos \theta-3)+\mathrm{i} \sin \theta \\
\Rightarrow z_{\overrightarrow{B C}}=[(\cos \theta-3)+\mathrm{i} \sin \theta] \cdot\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right) \\
=\cos \left(\thet... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (6 points) Definition: $a \preccurlyeq b=\frac{\mathrm{a}-1}{\mathrm{~b}}$, then $2 \preccurlyeq(3 \preccurlyeq 4)=$ $\qquad$ | 【Answer】Solution: $3 \leqslant 4$
$$
\begin{array}{l}
=\frac{3-1}{4} \\
=\frac{1}{2} \\
2 \text { put (3 lead 4) } \\
=2 \pi\left(\frac{1}{2}\right) \\
=\frac{2-1}{\frac{1}{2}} \\
=2 \text {; } \\
\end{array}
$$
Therefore, the answer is: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Try to find all positive integers $n$, such that there exist at least two pairs of positive integers $(x, y)$, satisfying
$$
n=\frac{x^{2}+y}{x y+1} .
$$ | If $xy$, then
$$
\begin{aligned}
x y+1 \mid x^{2}+y & \Rightarrow x y+1 \mid x^{2} y+y^{2} \\
& \Rightarrow x y+1 \mid\left(x^{2} y+y^{2}\right)-x(x y+1) \\
& \Rightarrow x y+1 \mid y^{2}-x .
\end{aligned}
$$
But from $x y+1>x, x y+1>y^{2}+1>y^{2}$, we know $\left|y^{2}-x\right| \leqslant \max \left\{y^{2}, x\right\}<... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$
Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is . $\qquad$ | 6. 1 .
From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqs... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given real numbers $a, b, c, d$ satisfy $5^{a}=4,4^{b}=3,3^{c}=2,2^{d}=5$, then $(a b c d)^{2018}=$ | Rong's Problem 1.
Parse $5^{a}=4,4^{b}=3,3^{c}=2,2^{d}=5$ into logarithms, we have $a=\log _{5} 4=\frac{\ln 4}{\ln 5}, b=\frac{\ln 3}{\ln 4}, c=\frac{\ln 2}{\ln 3}, d=\frac{\ln 5}{\ln 2}$, so
$$
(a b c d)^{2018}=\left(\frac{\ln 4}{\ln 5} \times \frac{\ln 3}{\ln 4} \times \frac{\ln 2}{\ln 3} \times \frac{\ln 5}{\ln 2}\r... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
C1. The function $f$ is defined on the natural numbers $1,2,3, \ldots$ by $f(1)=1$ and
$$
f(n)=\left\{\begin{array}{ll}
f\left(\frac{n}{10}\right) & \text { if } 10 \mid n \\
f(n-1)+1 & \text { otherwise }
\end{array}\right.
$$
Note: The notation $b \mid a$ means integer number $a$ is divisible by integer number $b$.
... | Solution for (a):
$$
\begin{aligned}
f(2019) & =f(2018)+1=f(2017)+2=\ldots=f(2010)+9=f(201)+9 \\
& =f(200)+1+9 \\
& =f(20)+1+9 \\
& =f(2)+1+9 \\
& =f(1)+1+1+9 \\
& =1+1+1+9=12 .
\end{aligned}
$$
In general, we can see that if $n \equiv m(\bmod 10)$ then $f(n)=f(n-m)+m$, where $10 \mid(n-m)$. Therefore $f(n)$ gives the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 26 (Problem from the 30th Russian Mathematical Olympiad) Let $\left\{N_{1}, \cdots, N_{k}\right\}$ be an array of five-digit numbers (in decimal) such that any five-digit number with digits in non-decreasing order has at least one digit in the same position that is the same as a digit in one of the numbers $N_{... | The smallest value sought is 2. The reason is as follows:
Au
Prohibited
Lian
Mathematics
An array with the described property cannot consist of a single five-digit number. In fact, for any five-digit number \( N = \overline{a b c d e} \), there exists a digit \( g \) different from \( a, b, c, d, e \), thus, no digit o... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$ . | 1. 4
$$
\begin{aligned}
\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1} & =\frac{6 x^{2}+12 x+10}{x^{2}+2 x+2}=6-\frac{2}{x^{2}+2 x+2} \\
& =6-\frac{2}{(x+1)^{2}+1},
\end{aligned}
$$
$\therefore$ When $x=-1$, the formula takes the minimum value 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
II. (16 points) Let $x, y \in \mathbf{R}$, and $x^{2}+y^{2}=2$, $|x| \neq|y|$. Find the minimum value of $\frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}}$. | $$
\begin{array}{l}
\text { II. Since } x^{2}+y^{2}=2 \text {, then, } \\
(x+y)^{2}+(x-y)^{2}=4 . \\
\text { Therefore, } \frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}} \\
=\frac{1}{4}\left(\frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}}\right)\left((x+y)^{2}+(x-y)^{2}\right) \\
\geqslant \frac{1}{4}(1+1)^{2}=1 .
\end{array}
$$
When... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there exists a row where the cells intersecting with these two columns are both filled... | 8. 5 .
Construct the application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be bought by $x_{i}$ peo... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the curve $C: x^{2}-y^{2}=2(x>0)$, then the minimum value of $f=\overrightarrow{O A} \cdot \overrightarrow{O B}$ is $\qquad$ . | 7.2 Detailed Explanation: From the problem, we know that $x_{1}^{2}-y_{1}^{2}=2, x_{2}^{2}-y_{2}^{2}=2$, where $x_{1}, x_{2}>0$.
$$
\begin{array}{l}
\because f=x_{1} x_{2}+y_{1} y_{2}=\frac{\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}-\left(y_{1}-y_{2}\right)^{2}}{4} \\
\quad=\... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
51 Let the set $S=\{100,101,102, \cdots, 999,1000\}, A=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n-1}, a_{n}\right.$ $\mid a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $\left.\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\cdots=\frac{a_{n}}{a_{n-1}}=q>1\right\}$. Try to find the maximum possible number of elements in ... | 51 Let $a_{1}, a_{n} \in S$, then $q^{n-1}=\frac{a_{n}}{a_{1}} \in \mathbf{Q}$. If $t$ is the smallest positive integer such that $q^{t} \in \mathbf{Q}$, then for any positive integer $k$ such that $q^{k} \in \mathbf{Q}$, by the division algorithm $k=c t+d$ (where $c, d$ are non-negative integers and $d<t$), thus $q^{d... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(2) For the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ with foci $F_{1}, F_{2}$, if a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$. | (2) 9 Hint: It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$, so $\left(P F_{1}+\right.$ $\left.P F_{2}\right)^{2}=10^{2}$, in the right-angled $\triangle P F_{1} F_{2}$, $P F_{1}^{2}+P F_{2}^{2}=8^{2}$, from the above two equations, we get $S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9$. | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Let real numbers $x, y$ satisfy:
$$
x^{3}=3 y^{2} x+5-\sqrt{7}, y^{3}=3 x^{2} y+5+\sqrt{7} \text {. }
$$
Then the value of $x^{2}+y^{2}$ is ( ).
(A) $25-8 \sqrt{7}$
(B) $2 \sqrt{7}$
(C) $12 \sqrt{7}-25$
(D) 4 | 5. D.
Rearranging terms, we get
$$
x^{3}-3 x y^{2}=5-\sqrt{7}, \quad y^{3}-3 x^{2} y=5+\sqrt{7} \text {. }
$$
Squaring and adding the above two equations, and simplifying, we get
$$
\begin{array}{l}
x^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}+y^{6}=64 \\
\Rightarrow\left(x^{2}+y^{2}\right)^{3}=64 \Rightarrow x^{2}+y^{2}=4 .
\e... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. A and B are partners in a business, and together they made a profit of $a^{2}$ yuan (where $a$ is a two-digit natural number). When dividing the money, A takes 100 yuan first, then B takes 100 yuan, followed by A taking another 100 yuan, B taking another 100 yuan, and so on, until the last amount taken is less than... | $4$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. Given that $a, b, c$ are positive real numbers, and $a+b+c=1$, find the minimum value of $u=\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}$. | 12. Consider the function $f(x)=\frac{x}{1+x^{2}}, x \in(0,1)$. It is easy to prove that for $x_{1}, x_{2} \in(0,1)$, we have $\left(x_{1}-x_{2}\right) \cdot\left[f\left(x_{1}\right) -f\left(x_{2}\right)\right] \geqslant 0$. Therefore, we have $\left(a-\frac{1}{3}\right)\left[f(a)-f\left(\frac{1}{3}\right)\right] \geqs... | 0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. 2008006 has ( ) prime factors.
(A) 4
(B) 5
(C) 6
(D) 7 | 2、Choose C
Solution: Since $20082006=2006 \times 1000+2006=2006 \times 1001=(2 \times 17 \times 59) \times$
$$
(7 \times 11 \times 13)
$$ | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
6.75 Given that $x, y, z$ are positive numbers, and satisfy the equation
$$
x y z(x+y+z)=1 \text{, }
$$
find the minimum value of the expression $(x+y)(y+z)$. | [Solution] From the given conditions and the arithmetic-geometric mean inequality, we have
$$
\begin{aligned}
(x+y)(y+z) & =(x+y+z) y+x z \\
& =\frac{1}{x y z} \cdot y+x z \\
& =\frac{1}{x z}+x z \\
& \geqslant 2 .
\end{aligned}
$$
On the other hand, when $x=1, y=\sqrt{2}-1, z=1$, the given equation is satisfied, and
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let the set $M=\{1,99,-1,0,25,-36,-91,19,-2,11\}$, and denote all non-empty subsets of $M$ as $M_{i}, i=1,2, \cdots$, 2013. The product of all elements in each $M_{i}$ is $m_{i}$. Then $\sum_{i=1}^{2013} m_{i}=$ $\qquad$ . | $$
\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }
$$
Solve:
$$
\sum_{i=1}^{2013} m_{i}=(1+1)(1+99)(1-1)(1+0) \cdots(1+11)-1=-1 \text {. }
$$ | -1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is $\qquad$. | 9.4 .
Let the circumcenter of $\triangle A B C$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $M N=4$, so $\triangle O M N$ is an isosceles right triangle, and
$$
\begin{array}{l}
\overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
\te... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) A poplar tree, a willow tree, a locust tree, a birch tree, and a sycamore tree are planted in a row, with the distance between adjacent trees being 1 meter. The distance between the poplar tree and the willow tree, and the locust tree is the same, and the distance between the birch tree and the poplar tre... | 【Answer】Solution: The distance between the poplar tree and the willow tree, as well as the locust tree, is equal, so the positions of all three trees could be: Willow Poplar Locust, Willow Poplar Locust $\square$, $\square$ Willow Poplar Locust $\square$, Willow Poplar Locust, where $\square$ represents an unknown posi... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
19. (3 points) As shown in the figure, an isosceles triangle $A B C$ is folded along $E F$, with vertex $A$ coinciding with the midpoint $D$ of the base, if the perimeter of $\triangle$ $A B C$ is 16 cm, and the perimeter of quadrilateral $B C E F$ is 10 cm, then $B C=$ $\qquad$ cm. | 【Solution】Solution: The perimeter of $\triangle A B C$ is 16 cm, so the perimeter of $\triangle A E F$ is: $16 \div 2=8$ (cm), the sum of the perimeters of $\triangle A E F$ and quadrilateral $B C E F$ is: $8+10=18$ (cm), therefore $B C=18-16=2$ (cm), answer: $B C=2$ cm. Hence the answer is: 2 . | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. If $n \in \mathbf{N}^{*}$, then $\lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}\right)=$ $\qquad$ (Contributed by Jian Weifeng) | 7. 1 .
By the induced formula, we know
$$
\sin ^{2}\left(\pi \sqrt{n^{2}+n}\right)=\sin ^{2}\left(\left(\sqrt{n^{2}+n}-n\right) \pi\right)=\sin ^{2}\left(\frac{n}{\sqrt{n^{2}+n}+n} \pi\right) .
$$
Therefore,
$$
\begin{aligned}
\lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}\right) & =\lim _{n \rightarr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
10. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}, x_{1999}-x_{601}=$ | 10. 0 .
$x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$. Let $x_{n}=\tan \alpha_{n}$, then $x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right)$, so $x_{n+6}=x_{n}$, which means $\left\{x_{n}\right\}$ is a periodic sequence. $1999=333 \times 6+1,601=6 \times 100+1$, thus $x_{1999}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. If the three interior angles $A, B, C$ of $\triangle A B C$ satisfy $\cos A=\sin B=2 \tan \frac{C}{2}$, then the value of $\sin A+\cos A+2 \tan A$ is $\qquad$. | Obviously, $B$ is an obtuse angle, then $\cos A=\cos \left(B-90^{\circ}\right) \Rightarrow B-A=90^{\circ} \Rightarrow 2 A+C=90^{\circ}$, thus $\cos A=2 \tan \left(45^{\circ}-A\right)=\frac{2(1-\tan A)}{1+\tan A} \Rightarrow 2-2 \tan A=\cos A+\sin A$ $\Rightarrow \sin A+\cos A+2 \tan A=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 32$ A student observed the weather for $x$ days during the summer vacation, and the results were:
(1) There were 7 sunny mornings;
(2) There were 5 sunny afternoons;
(3) It rained 8 times;
(4) On the days when it rained in the afternoon, the morning was sunny.
Then $x$ equals
(A) 8.
(B) 9.
(C) 10.
(D) 11.
(Ch... | ［Solution］From (4), we know that there is no weather where it rains both in the morning and in the afternoon. Therefore, there can only be the following three scenarios:
Morning clear, Afternoon clear;
Morning clear, Afternoon rain;
Morning rain, Afternoon clear.
7 mornings clear, 5 afternoons clear
(a total of 8 rainy... | 10 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
1 Call a year ultra-even if all of its digits are even. Thus $2000,2002,2004,2006$, and 2008 are all ultra-even years. They are all 2 years apart, which is the shortest possible gap. 2009 is not an ultra-even year because of the 9 , and 2010 is not an ultra-even year because of the 1 .
(a) In the years between the year... | Solution:
(a) The longest possible gap has length 1112. There are four valid examples of two ultra-even years that far apart with no ultra-even years between them: 888 to 2000,2888 to 4000,4888 to 6000 or 6888 to 8000 . ( 8888 to 10000 is not acceptable as 10000 is not ultra-even).
To see that no longer gap is possible... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. 15 Choose any 1962-digit number that is divisible by 9, and let the sum of its digits be $a$, the sum of the digits of $a$ be $b$, and the sum of the digits of $b$ be $c$. What is $c$? | [Solution] Since any number and the sum of its digits have the same remainder when divided by 9,
thus $c \geqslant 9$, and $9 \mid c$.
On the other hand, $a \leqslant 1962 \times 9 < 19999$,
therefore $b < 1 + 4 \times 9 = 37$,
so $c \leqslant 11$. Hence, $c = 9$. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.46 On an $8 \times 8$ chessboard, some squares are marked with stars, such that
(1) no two squares marked with stars share a common edge or vertex;
(2) every unmarked square shares a common edge or vertex with at least one marked square.
What is the minimum number of squares that need to be marked with stars? Explai... | [Solution] The right figure is marked with 9 asterisks, which obviously meets the requirements of the problem. And in the figure, 9 checkmarks are marked. No matter which square on the chessboard is marked with an asterisk, it can share at most one side or one vertex with one of these 9 squares. Therefore, to meet the ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (This question is worth 14 points) Find the smallest positive integer $n$, such that for any positive integer $k \geq n$, in the set $M=\{1,2, \cdots, k\}$ of the first $k$ positive integers, for any $x \in M$, there always exists another number $y \in M(y \neq x)$, such that $x+y$ is a perfect square. | It is easy to see that when $n \leq 6$, in $M=\{1,2,3,4,5,6\}$, the number 2 added to any other number is never a square number.
The following proof shows that the minimum value of $n$ is 7.
If positive integers $x, y (x \neq y)$ satisfy: $x+y=$ a square number, then $\{x, y\}$ is called a "square pair". Clearly, in $M... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. As shown in the figure, points $A$, $B$, $C$, and $D$ lie on the same circle, and $BC = CD = 4$, $AE = 6$. The lengths of segments $BE$ and $DE$ are both positive integers. Then the length of $BD$ is $\qquad$ | 4. 7
Solution: Let $E C=x, B E=y$, $E D=z$, from $\triangle D C E \sim \triangle A C D$,
we get $\frac{C D}{C A}=\frac{E C}{D C}$, that is $\frac{4}{6+x}=\frac{x}{4}$.
Solving this, we get $x=2 \cdot(x=-8$ is not feasible)
Also, from $A E \cdot E C=B E \cdot E D$
we get $y z=6 \times 2=12$.
However, in $\triangle B... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15. The quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfies the following conditions:
(1) $f(-1)=0$;
(2) For $x \in \mathbf{R}$, $f(x) \geqslant x$;
(3) For $x \in(0,2)$, $f(x) \leqslant \frac{(x+1)^{2}}{4}$.
If $x_{1}, x_{2}, x_{3} \in(0,2)$, and $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1... | 15. Since $f(x) \geqslant x$, we have $f(1) \geqslant 1$; also, since $f(x) \leqslant \frac{(x+1)^{2}}{4}$, we have $f(1) \leqslant \frac{(1+1)^{2}}{4}=1$, thus $f(1)=1$. Given $f(-1)=0$, we have
$$
\left\{\begin{array}{l}
a+b+c=1 \\
a-b+c=0
\end{array}\right.
$$
Solving these, we get $b=\frac{1}{2}, a+c=\frac{1}{2}$.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(8) The solution to the equation $3 \cdot 16^{x}+2 \cdot 81^{x}=5 \cdot 36^{x}$ is | $80, \frac{1}{2}$ Hint: The original equation can be transformed into $3 \cdot\left(4^{x}\right)^{2}-5 \cdot 4^{x} \cdot 9^{x}+2 \cdot\left(9^{x}\right)^{2}=$ 0 . Dividing both sides by $\left(9^{x}\right)^{2}$, we get
$$
3\left[\left(\frac{4}{9}\right)^{x}\right]^{2}-5 \cdot\left(\frac{4}{9}\right)^{x}+2=0 .
$$
Let $... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2 As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $\angle B A C=100^{\circ}$, point $D$ is on the extension of side $A B$, and $A D=B C$. Find the measure of $\angle B C D$. ${ }^{[2]}$ | Solve as shown in Figure 2, construct an equilateral $\triangle A C E$, and connect $D E$.
Then $C E=C A=A E, \angle E A C=60^{\circ}$.
Thus, $\angle D A E=\angle B A C-\angle E A C$
$=40^{\circ}=\angle A B C$.
Also, $A E=A C=A B, A D=B C$, then
$\triangle D A E \cong \triangle C B A$.
Therefore, $D E=A C=C E$,
$\angle... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 8 Let $f(a, b, c)=\frac{1}{\sqrt{1+2 a}}+\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}$, where $a, b, c>0$ and $abc=1$, find the minimum value of the constant $\lambda$ such that $f(a, b, c)<\lambda$ always holds. | This is not a simple problem. We guess the minimum value of $\lambda$ is 2, that is to prove $\frac{1}{\sqrt{1+2 a}}+$ $\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}\frac{\sqrt{a b}}{\sqrt{2+a b}}$,
Since $\frac{a}{1+2 a}+\frac{b}{1+2 b} \geqslant \frac{2 \sqrt{a b}}{\sqrt{(1+2 a)(1+2 b)}}$, and $\frac{2 \sqrt{a b}}{\s... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
76. Fill a 5x5 table with 5 ones, 5 twos, 5 threes, 5 fours, and 5 fives (one number per cell), such that the absolute difference between any two numbers in the same column does not exceed 2. Consider the sum of the numbers in each column, and let the minimum of these 5 sums be \( M \). Then the maximum value of \( M \... | answer: 10 | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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