problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$ | 7.3.
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Thus, the desired value is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,716 |
8. If the inequality about $x$
$$
|x-a|+|x-12|<6
$$
does not hold for any real number $x$, then the range of the real number $a$ is $\qquad$ . | 8. $a \leqslant 6$ or $a \geqslant 18$. From the given, $12-a \geqslant 6$ or $a-12 \geqslant 6$. Solving, we get $a \leqslant 6$ or $a \geqslant 18$. | a \leqslant 6 \text{ or } a \geqslant 18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,717 |
11. Given that the perimeter of quadrilateral $\square A B C D$ is 28, a line through vertex $D$ is drawn perpendicular to sides $A B$ and $B C$, with the feet of the perpendiculars being $E$ and $F$ respectively. If $D E=3, D F=4$, find
(1) the lengths of sides $A B$ and $B C$;
(2) the length of $B E+B F$. | Three, 11. (1) According to the problem, $DE$ and $DF$ are the distances between the two pairs of opposite sides of $\square ABCD$.
Thus, $AB \cdot DE = BC \cdot DF$.
Let $AB = a, BC = b$.
Since $DE = 3, DF = 4$, we have $3a = 4b$.
Also, the perimeter of $\square ABCD$ is 28, so,
$a + b = \frac{1}{2} \times 28 = 14$.
F... | 2 + \sqrt{3} \text{ or } 14 + 7\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,719 |
13. As shown in Figure 4, given that $\odot O$ is the circumcircle of $\triangle A B C$, $\angle A=60^{\circ}$, and $I$ and $H$ are the incenter and orthocenter of $\triangle A B C$, respectively. Prove: $O I=H I$. | 13. As shown in Figure 7, connect $A O$ and extend it to intersect $\odot O$ at point $D$. Connect $A H$ and extend it to intersect $B C$ at point $E$. Connect $B D$.
Given $\angle A B D = \angle A E C = 90^{\circ}$,
$\angle A D B = \angle A C E$,
we know $\angle B A D = \angle C A E$.
Connect $A I$.
Since $A I$ bisec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,720 |
14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}, 9$ is a wise number.
(1) Try to determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arr... | 14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$
which means that all odd numbers greater than 1 are wise numbers.
When $k=2... | 2689 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,721 |
1. Harry and Terry simultaneously solve the calculation problem $8-(2+5)$. Harry arrives at the correct answer, while Terry, ignoring the parentheses, incorrectly solves it as $8-2+5$. If Harry's answer is $H$, and Terry's answer is $T$, then $H-T=(\quad)$.
(A) -10
(B) -6
(C) 0
(D) 6
(E) 10 | 1. A.
From the problem, we know
$$
\begin{array}{l}
H=8-(2+5)=8-7=1, \\
T=8-2+5=11 .
\end{array}
$$
Therefore, $H-T=-10$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,722 |
2. Paul owes Paula 35 cents, and his pocket contains only 5 cent, 10 cent, and 25 cent coins (in sufficient quantity), which he can use to pay Paula back. The difference between the maximum and minimum number of coins used in the payment is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 2. E.
The maximum number of coins in the change is seven, all being 5 cents; the minimum is 2 coins, one 10 cents and one 25 cents. Therefore, the difference between the maximum and minimum values is $7-2=5$. | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,723 |
3. The school assigned Isabella a task to finish reading a book in one week. In the first three days, she averaged 36 pages per day, and in the next three days, she averaged 44 pages per day, and on the last day, she read the final 10 pages. Therefore, the book has ( ) pages.
(A) 240
(B) 250
(C) 260
(D) 270
(E) 280 | 3. B.
$$
3 \times 36 + 3 \times 44 + 10 = 250 \text{.}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,724 |
5. Maggie's car can travel 32 miles on one gallon of gasoline. At the current price of $4 per gallon, how many miles can Maggie travel with $20 worth of gasoline?
(A) 64
(B) 128
(C) 160
(D) 320
(E) 640 | 5. C.
$$
20 \div 4 \times 32=160
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,725 |
8. During the middle school math club activities, 11 guests were invited to give specialized lectures. The club paid each guest the same amount, with the total amount being $\overline{1 A 2}$ dollars. What is the digit $A$ in the tens place of this three-digit number?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 8. D.
From the problem, we know $111 \overline{1 A 2}$. Therefore, $1+2-A=3-A \equiv 0(\bmod 11)$.
Note that, $0 \leqslant A \leqslant 9$. Then $-6 \leqslant 3-A \leqslant 0 \Rightarrow 3-A=0$ $\Rightarrow A=3$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,726 |
Example 6 As shown in Figure 5, quadrilateral $ABCD$ is inscribed in $\odot O$. Extend $AB$ and $DC$ to meet at point $P$, and extend $AD$ and $BC$ to meet at point $Q$. Draw two tangents from $Q$ to the circle, touching the circle at points $E$ and $F$. Prove that points $P$, $E$, and $F$ are collinear.
(1997, China M... | 【Analysis】Notice that, the triangle intersecting with the secant $P E F$ is only $\triangle Q E F$, and it is difficult to grasp that points $P, E, F$ are collinear. However, the triangles intersecting with the secant $P N M$ are $\triangle Q C D$ and $\triangle Q A B$. We can prove that points $P, M, N$ are collinear,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,727 |
12. A magazine prints six photos, which are of three celebrities and their baby photos, but the three baby photos are not labeled with whose they are, and readers have to choose themselves. Assuming each photo is equally likely. Then the probability that a reader randomly selects three photos, making the baby photos co... | 12. B.
Let event $A$ be the event of correctly selecting three baby photos. Then
$$
P(A)=\frac{1}{3!}=\frac{1}{6} .
$$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,728 |
13. If $m^{2}+n^{2}(m, n \in \mathbf{Z})$ is even, then which of the following statements is impossible.
(A) $m, n$ are both even
(B) $m, n$ are both odd
(C) $m+n$ is even
(D) $m+n$ is odd
(E) One of the above answers is impossible | 13. D.
From the problem, we know that $m^{2}$ and $n^{2}$ have the same parity.
When $m^{2}$ and $n^{2}$ are both even, $m$ and $n$ are both even, so $m+n$ is even;
When $m^{2}$ and $n^{2}$ are both odd, $m$ and $n$ are both odd, so $m+n$ is even.
Therefore, it is impossible for $m+n$ to be odd. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,729 |
15. As shown in Figure 3, the circumference of a circle with center $O$ is divided into 12 equal arcs, and the division points are labeled as $A, B, \cdots, L$. Then $\alpha+\beta=$ ( ).
(A) $75^{\circ}$
(B) $80^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$
(E) $150^{\circ}$ | 15. C.
Divide the circumference of $\odot O$ into 12 equal parts, with each arc having a degree measure of $\frac{360^{\circ}}{12}=30^{\circ}$.
Then $\alpha=\frac{1}{2} \angle G O E=30^{\circ}$,
$$
\beta=\frac{1}{2} \angle A O I=60^{\circ} \text {. }
$$
Thus, $\alpha+\beta=90^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,730 |
16. The basketball league has eight teams. Each season, each team plays two games (home and away) against each of the other teams in the league, and each team also plays four games against opponents outside the league. Therefore, in one season, the eight teams in the league play a total of ( ) games.
(A) 60
(B) 88
(C) ... | 16. B.
There are $\mathrm{C}_{8}^{2}=28$ pairs that can be formed from the eight teams in the league.
Since each pair of teams plays both a home and an away match, the eight teams in the league play a total of $28 \times 2=56$ matches.
Since each team plays four matches against opponents outside the league, the eight... | 88 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,731 |
19. A cube with an edge length of 3 inches is composed of 27 unit cubes, of which, 21 are red unit cubes and 6 are white unit cubes. If the surface of the large cube is to have as few white unit cubes as possible, then the area of the white part on the surface of the large cube as a fraction of the total surface area o... | 19. A
To minimize the white area, one white unit cube can be placed at the center of the larger cube, and the other five white unit cubes can be placed at the centers of any five faces of the larger cube. The surface area of a larger cube with an edge length of 3 is $6 \times 3^{2}$ $=54$, so the white area on the sur... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,732 |
23. Three girls from Euclid High School are on the school's softball team.
Shayla says: “All three of our uniform numbers are two-digit prime numbers. The sum of your two uniform numbers is today's date.”
Bao Fen says: “The sum of your two uniform numbers is my birthday, and it is earlier than today's date.”
Kaitlyn... | 23. A.
Notice that the maximum value of a date in a month is 31, and the smallest three two-digit prime numbers are $11, 13, 17$, their pairwise sums are $24, 28, 30$.
From the problem, we know that Katelyn's team jersey number is 11. | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 728,733 |
25. A closed section of highway, 1 mile long and 40 feet wide. Robert rides along the route shown in Figure 5 in a semicircular path. If his speed is 5 miles per hour, then the time he needs to ride this section of the highway is ( ) hours (Note: 1 mile = 5280 feet).
(A) $\frac{\pi}{11}$
( B) $\frac{\pi}{10}$
(C) $\fra... | 25. B.
The width of the road is 40 feet, which means each lane is 20 feet wide, so the diameter of the semicircle Robert rides is 40 feet.
The total length of the road is 1 mile, which is 5280 feet, so $\frac{5280}{40}=132$ semicircles, and the circumference of each semicircle is $\frac{1}{2} \times 2 \pi \times 20=2... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,734 |
1. “ $a=2, b=\sqrt{2}$ ” is the ( ) condition for “the curve $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a 、 b \in \mathbf{R}, a b \neq 0)$ passes through the point $(\sqrt{2}, 1)$ ”.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | - 1. A.
The sufficiency is obviously true.
When the curve $C$ passes through the point $(\sqrt{2}, 1)$, we have $\frac{2}{a^{2}}+\frac{1}{b^{2}}=1$.
Clearly, $a=-2, b=-\sqrt{2}$ also satisfy the above equation. Therefore, the necessity does not hold. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,735 |
2. Given a triangle with one angle greater than $120^{\circ}$ and side lengths $m$, $m+1$, and $m+2$. Then the range of the real number $m$ is ( ).
(A) $m>1$
(B) $m>3$
(C) $\frac{3}{2}<m<3$
(D) $1<m<\frac{3}{2}$ | 2. D.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array}{l}
m+(m+1)>m+2, \\
(m+2)^{2}>m^{2}+(m+1)^{2}+m(m+1)
\end{array}\right. \\
\Rightarrow 1<m<\frac{3}{2} .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,736 |
3. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $M$ is the midpoint of $B B_{1}$. Then the cosine value of the dihedral angle $M-C D_{1}-A$ is ( ).
(A) $\frac{\sqrt{3}}{6}$
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{6}}{3}$ | 3. C.
Taking $D$ as the origin, the lines $D A$, $D C$, and $D D_{1}$ as the $x$-axis, $y$-axis, and $z$-axis respectively to establish a spatial rectangular coordinate system. Then
$$
\begin{array}{l}
D(0,0,0), A(1,0,0), C(0,1,0), \\
D_{1}(0,0,1), M\left(1,1, \frac{1}{2}\right),
\end{array}
$$
and the normal vector ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,737 |
Example 2 Let $n>1$ be an integer, $(a, n)=1$. Prove: $a^{\varphi(n)} \equiv 1(\bmod n)$. | Proof: Let $a_{1}, a_{2}, \cdots, a_{\varphi(n)}$ form a reduced residue system modulo $n$.
By property 2, $a a_{1}, a a_{2}, \cdots, a a_{\varphi(n)}$ also form a reduced residue system modulo $n$.
According to property 4, we have
$$
\begin{array}{l}
a_{1} a_{2} \cdots a_{\varphi(n)} \equiv\left(a a_{1}\right)\left(a ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,738 |
5. Given that the three vertices of isosceles right $\triangle P Q R$ lie on the three sides of isosceles right $\triangle A B C$. Then the minimum value of $\frac{S_{\triangle P Q R}}{S_{\triangle A B C}}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) $\frac{1}{5}$ | 5. D.
Consider two cases.
(1) When the right-angle vertex of $\triangle P Q R$ is on the hypotenuse of $\triangle A B C$, as shown in Figure 2. Then
$$
\begin{array}{l}
P, C, Q, R \text { are concyclic } \\
\Rightarrow \angle A P R=\angle C Q R \\
=180^{\circ}-\angle B Q R \\
\Rightarrow \sin \angle A P R \\
=\sin \an... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,739 |
6. Given the general term of the sequence $\left\{a_{n}\right\}$
$$
a_{n}=\frac{n x}{(x+1)(2 x+1) \cdots(n x+1)}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
If $a_{1}+a_{2}+\cdots+a_{2015}<1$, then the value of the real number $x$ is ( ).
(A) $-\frac{3}{2}$
(B) $-\frac{5}{12}$
(C) $-\frac{9}{40}$
(D) $-\frac{11}{60... | 6. D.
From the problem, we know
$$
\begin{array}{l}
a_{n}=\frac{1}{(x+1)(2 x+1) \cdots[(n-1) x+1]}- \\
\quad \frac{1}{(x+1)(2 x+1) \cdots(n x+1)} . \\
\text { Then } \sum_{k=1}^{2015} a_{k} \\
=1-\frac{1}{(x+1)(2 x+1) \cdots(2015 x+1)}0 . \\
\text { Therefore, } x \in \bigcup_{k=1}^{100}\left(-\frac{1}{2 k-1},-\frac{1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,740 |
8. If the set
$$
\begin{aligned}
A= & \{(m, n) \mid(m+1)+(m+2)+\cdots+ \\
& \left.(m+n)=10^{2015}, m \in \mathbf{Z}, n \in \mathbf{Z}_{+}\right\},
\end{aligned}
$$
then the number of elements in set $A$ is ().
(A) 4030
(B) 4032
(C) $2015^{2}$
(D) $2016^{2}$ | 8. B.
From the given, we have
$$
n(n+2 m+1)=2^{2016} \times 5^{2015} \text {. }
$$
Thus, one of $n$ and $n+2 m+1$ is odd and the other is even.
Therefore, one of $n$ and $n+2 m+1$ is even, with the possibilities being $2^{2016}, 2^{2016} \times 5, 2^{2016} \times 5^{2}, \cdots, 2^{2016} \times 5^{2015}$, for a total ... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,741 |
10. If the sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms
$$
\begin{array}{l}
S_{n}=n^{3}-n^{2}\left(n \in \mathbf{Z}_{+}\right), \\
\text { then } \sum_{i=1}^{215} \frac{1}{a_{i}+8 i-2}=\text {. }
\end{array}
$$ | 10. $\frac{2015}{6048}$.
Notice,
$$
a_{n}=\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n-1} a_{i}=3 n^{2}-5 n+2 \text {. }
$$
Also, $a_{1}=0$, then
$$
\begin{array}{c}
a_{n}=3 n^{2}-5 n+2\left(n \in \mathbf{Z}_{+}\right) . \\
\text {Therefore, } \sum_{i=1}^{2015} \frac{1}{a_{i}+8 i-2}=\sum_{i=1}^{2015} \frac{1}{3 i(i+1)} \\
=\fr... | \frac{2015}{6048} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,742 |
12. If $16^{\sin ^{2} x}+16^{\cos ^{2} x}=10$, then $\cos 4 x=$ | 12. $-\frac{1}{2}$.
Let $t=16^{\sin ^{2} x}(1 \leqslant t \leqslant 16)$. Then $16^{\cos ^{2} x}=16^{1-\sin ^{2} x}=\frac{16}{t}$.
Substituting into the original equation gives $t+\frac{16}{t}=10 \Rightarrow t=2$ or 8.
Solving yields $\sin ^{2} x=\frac{1}{4}$ or $\frac{3}{4}$.
Thus, $\cos 4 x=-\frac{1}{2}$. | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,743 |
13. Let the function
$$
f(x)=\min \left\{x^{2}-1, x+1,-x+1\right\},
$$
where $\min \{x, y, z\}$ denotes the minimum of $x$, $y$, and $z$. If $f(a+2)>f(a)$, then the range of real number $a$ is $\qquad$ | 13. $(-\infty,-2) \cup(-1,0)$.
When $a+2 \leqslant-1$, $a<a+2 \leqslant-1$, at this time, $f(a)<f(a+2)$;
When $-1<a+2<0$, $-3<a<-2$, at this time, $f(a) \leqslant f(-2)=-1<f(a+2)$;
When $0 \leqslant a+2 \leqslant 1$, $-2 \leqslant a<-1$, at this time, $f(a) \geqslant f(a+2)$;
When $1<a+2<2$, there is $-1<a<0$, at this... | (-\infty,-2) \cup(-1,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,744 |
14. Let the angle between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\frac{\pi}{3}$, the angle between vectors $\boldsymbol{c}-\boldsymbol{a}$ and $\boldsymbol{c}-\boldsymbol{b}$ be $\frac{2 \pi}{3}$, $|\boldsymbol{a}-\boldsymbol{b}|=5$, and $|\boldsymbol{c}-\boldsymbol{a}|=2 \sqrt{3}$. Then the maximum value of... | 14. 24 .
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. Then
$$
|\overrightarrow{A C}|=|c-a|=2 \sqrt{3},|\overrightarrow{A B}|=|a-b|=5 \text {. }
$$
Also, $\angle A O B=\frac{\pi}{3}, \angle A C B=\frac{2 \pi}{3}$, at this time, $O, A, C, B$ are concyclic.
By the Law of Sines, we get
$$... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,745 |
16. (16 points) Let $a, b \in \mathbf{R}$, and the function
$$
f(x)=a x^{2}+b(x+1)-2 \text {. }
$$
If for any real number $b$, the equation $f(x)=x$ has two distinct real roots, find the range of real numbers for $a$.
| Three, 16. From the problem, we know the equation
$$
a x^{2}+(b-1) x+b-2=0
$$
has two distinct real roots. Therefore,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a \neq 0, \\
\Delta=(b-1)^{2}-4 a(b-2)>0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a \neq 0, \\
b^{2}-2(1+2 a) b+8 a+1>0 .
\end{array}\right.
... | 0<a<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,746 |
18. (18 points) Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\left\{\begin{array}{l}
a_{1}>0, b_{1}>0, \\
a_{n+1}=a_{n}+\frac{1}{b_{n}}, \\
b_{n+1}=b_{n}+\frac{1}{a_{n}}
\end{array}\left(n \in \mathbf{Z}_{+}\right) .\right.
$$
Prove: $a_{50}+b_{50}>20$. | 18. From the problem, we have
$$
\begin{array}{l}
a_{n+1}^{2}+b_{n+1}^{2} \\
=a_{n}^{2}+b_{n}^{2}+\frac{1}{a_{n}^{2}}+\frac{1}{b_{n}^{2}}+2\left(\frac{a_{n}}{b_{n}}+\frac{b_{n}}{a_{n}}\right) .
\end{array}
$$
Then $a_{50}^{2}+b_{50}^{2}$
$$
\begin{array}{l}
=a_{1}^{2}+b_{1}^{2}+\sum_{i=1}^{49}\left(\frac{1}{a_{i}^{2}}... | a_{50}+b_{50}>20 | Algebra | proof | Yes | Yes | cn_contest | false | 728,747 |
1. Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, \cdots$ such that: $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$? | 1. Solution 1 exists. Let the sequence of all prime numbers in ascending order be
$2=p_{1}1$, so for $r, s \in \mathbf{Z}_{+}$,
$r=s$, we have $\left(a_{r}, a_{s}\right)>1$.
For positive integers $m1, \\
p_{4}\left|a_{2 m}, p_{4}\right| a_{2 n} \Rightarrow\left(a_{2 m}, a_{2 n}\right)>1,
\end{array}
$$
By $2 m1,
$$
A... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,748 |
3. A finite sequence of rational numbers is written on a blackboard. An operation refers to: first selecting any two numbers $a$ and $b$ from this sequence and erasing them, then writing down one of the following forms:
$$
\begin{array}{l}
a+b, a-b, b-a, a \times b, \\
\frac{a}{b}(b \neq 0), \frac{b}{a}(a \neq 0) .
\en... | 3. First, the numbers on the blackboard are initially $\frac{P_{i}(k)}{Q_{i}(k)}$, with $A(0)=1$ (considering 0 as $\frac{0}{1}$, hence for the form $(i=1,2, \cdots, n)$, where $P_{i}, Q_{i} \in \mathbf{Z}[k]$.
At this point, $P_{i}(k)=k+1, Q_{i}(k)=1$.
Notice,
\[
\frac{p(k)}{q(k)}+\frac{r(k)}{s(k)}=\frac{p(k) s(k)+q(k... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,750 |
4. In $\triangle A B C$, it is known that $D$ is the point where the incircle of $\triangle A B C$ touches side $B C$. Suppose points $J_{b} 、 J_{c}$ are the incenters of $\triangle A B D 、 \triangle A C D$, respectively. Prove: The circumcenter of $\triangle A J_{b} J_{c}$ lies on the angle bisector of $\angle B A C$. | 4. As shown in Figure 2, let the points of tangency of the incircle of $\triangle A B C$ with $A B$ and $A C$ be $F$ and $E$ respectively. Let the projections of $J_{b}$ and $J_{c}$ on side $A D$ be $K$ and $K^{\prime}$ respectively. $K$ is the point of tangency of the incircle of $\triangle A B D$ with side $A D$, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,751 |
Example 1 Allocate 24 volunteer slots to 3 schools. Then the number of allocation methods where each school gets at least one slot and the number of slots for each school is different is $\qquad$ kinds. ${ }^{[2]}$ | Let the quotas allocated to schools A, B, and C be $x$, $y$, and $z$ respectively.
First, without considering that $x, y, z$ are pairwise distinct.
From $x+y+z=24$, we get a total number of combinations $\mathrm{C}_{23}^{2}$.
Next, we separate out the number of positive integer solutions $(x, y, z)$ that are not pairwi... | 222 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,752 |
Example 2 Let $A, B, C, D$ be four non-coplanar points in space. With a probability of $\frac{1}{2}$, connect an edge between each pair of points, and whether any pair of points is connected by an edge is independent of each other. Find the probability that points $A, B$ can be connected by a (single edge or a sequence... | Solving: Each pair of points has 2 possibilities for whether they are connected by an edge, totaling $2^{6}=64$ possibilities.
Below, we calculate the number of ways points $A$ and $B$ are connected by a polyline.
(1) There is an edge $A B$, totaling $2^{5}=32$ ways.
(2) There is no edge $A B$, but there is an edge $C... | \frac{3}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,753 |
7. From the set $\{1,2, \cdots, 100\}$, randomly select elements $m$ and $n$ (which can be the same). Then the probability that the unit digit of $2^{m}+3^{n}$ is 3 is . $\qquad$ | 7. $\frac{3}{16}$.
Notice that, $2^{1}, 2^{2}, 2^{3}, 2^{4}, \cdots$ modulo 10 has a repeating cycle of $2, 4, 8, 6$, and $3^{1}, 3^{2}, 3^{3}, 3^{4}, \cdots$ modulo 10 has a repeating cycle of $3, 9, 7, 1$.
Since 41100, the last digit of $2^{m}$ is $2, 4, 8, 6$ with a probability of $\frac{1}{4}$ each, and the last ... | \frac{3}{16} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,755 |
8. The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$ | 8. 1141 .
Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$
satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{... | 1141 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,756 |
Sure, here is the translated text:
```
II. (40 points) Given positive real numbers $a, b, c, d$ satisfying $a+b+c+d=1$. Prove:
$$
\sum \frac{a}{a^{2}+1} \leqslant \frac{16}{17},
$$
where, “ $\sum$ " denotes the cyclic sum.
``` | From $a^{2}+\frac{1}{16} \geqslant 2 a \times \frac{1}{4}=\frac{a}{2}$, we get
$$
\frac{a}{a^{2}+1} \leqslant \frac{a}{\frac{a}{2}+\frac{15}{16}}=\frac{16 a}{8 a+15} \text {. }
$$
Thus, it suffices to prove $\sum \frac{16 a}{8 a+15} \leqslant \frac{16}{17}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,758 |
In convex quadrilateral $ABCD$, $\angle CBD = \angle CDB = \angle BAD$, and $P$ is a point inside quadrilateral $ABCD$. Through point $P$, draw lines parallel to sides $AB$ and $AD$, intersecting sides $BC$, $AD$, $AB$, and $CD$ at points $M$, $N$, $G$, and $H$ respectively. Prove that $M$, $N$, $G$, and $H$ are concyc... | Prove: As shown in Figure 2, draw a line through point $P$ parallel to $BD$ intersecting lines $AB$, $AD$, $CB$, and $CD$ at points $R$, $S$, $K$, and $L$ respectively. Connect $AP$, intersecting $BD$ at point $A'$, and connect $CP$, intersecting $BD$ at point $C'$.
Since $PN \parallel AB$ and $PG \parallel AD$, we hav... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,759 |
4. The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences. | Prompt: Categorized Count. Answer: 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,760 |
- Height 434 Proof: There exist infinitely many positive integers $n$ such that
(1) $n$ is a multiple of 2015;
(2) Removing one non-zero digit from $n$ results in a number that is still a multiple of 2015. | $$
\begin{array}{l}
\text { Proof: Let } n=a_{p} a_{p-1} \cdots a_{q+1} a_{q} a_{q-1} \cdots a_{1} a_{0} \\
=\overline{a_{p} a_{p-1} \cdots a_{q+1}} \times 10^{q+1}+a_{q} \times 10^{q}+\overline{a_{q-1} \cdots a_{1} a_{0}} \\
=10^{q+1} a+10^{q} b+c, \\
\end{array}
$$
where, $a \in \mathbf{Z}_{+}, 1 \leqslant b \leqsla... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,761 |
435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$. | (1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).
On the other hand, for any smaller parallelogram $F_1$ that is similar to the conve... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,762 |
Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, where $n \in \mathbf{Z}_{+}, n \geqslant 2$. Prove:
$$
\left[\prod_{i=1}^{n}\left(1+x_{i}\right)\right]\left(\sum_{i=1}^{n} \frac{1}{x_{i}}\right) \geqslant 2 n^{2} .
$$ | Prove that for any $i \in\{1,2, \cdots, n\}$, by the AM-GM inequality we have
$$
\begin{array}{l}
1+x_{i}=\frac{1}{n-1}+\frac{1}{n-1}+\cdots+\frac{1}{n-1}+x_{i} \\
\geqslant n \sqrt[n]{\left(\frac{1}{n-1}\right)^{n-1} x_{i}}, \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}} \geqslant \sqrt[n]{\frac{1}{x_{1} x_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,763 |
6. A password lock's password setting involves assigning one of the two numbers, 0 or 1, to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex either red or blue, such that for any two adjacent vertices, at least one of the number or color is the same. How many different pas... | Hint: Recursive method. Answer: $3^{n}+(-1)^{n}+2$. | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,765 |
Example 1 It is well known that using the first two positive integers 1, 2, a positive integer 21 can be arranged, which is a multiple of 7; using the first three positive integers $1,2,3$, a positive integer 231 can be arranged, which is also a multiple of 7. Prove: For each positive integer $n \geqslant 2$, the first... | 【Analysis】When $n=4$, the numbers $1, 2, 3, 4$ can be arranged as 4123.
When $n=5$, the numbers $1, 2, 3, 4, 5$ can be arranged as 51324.
However, listing them one by one cannot be generalized to a general case.
For this, we adopt the method of "waiting for the rabbit by the stump".
When $n=4$, we first give seven four... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,766 |
Example 5 On a plane, there are $n$ circular iron sheets, among which any two circles have a common point. Prove: At most 7 hooks are needed to catch them all (i.e., there exist seven points such that each circle contains at least one of these seven points).
Translate the above text into English, please keep the origi... | Proof: Let $\odot O$ be the circle with the smallest radius $r$ among the $n$ circles. Construct a circle with center $O$ and radius $\sqrt{3} r$, denoted as $\odot(O, \sqrt{3} r)$. Let the regular hexagon $A B C D E F$ be inscribed in $\odot(O, \sqrt{3} r)$.
Place 7 hooks at points $A, B, C, D, E, F, O$ respectively.... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,767 |
Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them. | From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$
Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the numb... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,768 |
Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation. | First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$
is odd.
For the general case, it is only necessary to adjust some " + " to " - ".
Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,769 |
Example: There are $4 n(n \geqslant 4)$ plates with a total of no less than 4 candies. Selecting any two plates, take one candy from each and place them into another plate, which is called one operation. Can all the candies be concentrated into one plate after a finite number of operations? Prove your conclusion. ${ }^... | First, we prove: it is possible to concentrate all the candies in two or three plates after a finite number of operations.
In fact, if the number of plates with candies is no less than three, take any three of them, denoted as $A, B, C$, and assume that $A, B, C$ contain $a, b, c (0 \leqslant c < b < a)$ candies respe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,770 |
Example 5 Express 2006 as the sum of five positive integers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. Let $S=\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}$.
(1) For what values of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is $S$ maximized?
(2) Suppose for any $1 \leqslant i, j \leqslant 5$, we have $\left|x_{i}-x_{j}\right| \leq... | (1) First, the values of $S$ are bounded, so there must be a maximum and a minimum value.
If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, and
$$
S=\sum_{1 \leqslant i x_{1} x_{2}.
\end{aligned}
$$
Rewrite $S$ as
$$
\begin{aligned}
S= & \sum_{1 \leqslant i0$.
This contradicts the assumption that $S$ reaches its maximum value... | 402,402,402,400,400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,771 |
Given 2014 non-negative real numbers $a_{1}, a_{2}, \cdots, a_{2014}$ whose sum is 1. Prove: there exists a permutation $x_{1}, x_{2}, \cdots, x_{2014}$ of $a_{1}, a_{2}, \cdots, a_{2014}$, such that
$$
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2013} x_{2014}+x_{2014} x_{1} \leqslant \frac{1}{2014} \text {. }
$$ | This article provides a generalization of the problem and its concise proof.
Generalization Let $a_{i} \in \mathbf{R}(i=1,2, \cdots, n)$. Then there must exist a permutation $x_{1}, x_{2}, \cdots, x_{n}$ of $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1} \leqslant \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,772 |
Example 2 Given $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \\
\geqslant a b c+\sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)} .
\end{array}
$$ | Proof of the left side:
$$
\begin{array}{l}
=\frac{1}{2} \sqrt{\left[\sum_{\text {cyc }} b\left(a^{2}+b c\right)\right]\left[\sum_{\text {cyc }} c\left(a^{2}+b c\right)\right]} \\
\geqslant \frac{1}{2} \sum_{\text {cyc }} \sqrt{b c}\left(a^{2}+b c\right) \text { (Cauchy-Schwarz inequality) } \\
\geqslant \frac{3}{2} \s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,774 |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$ | To prove that since the inequality to be proved is homogeneous with respect to $a, b, c$, we may assume without loss of generality that $a+b+c=3$.
The original inequality is then $\sum_{\text {cyc }} \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant 8$.
Consider the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}(0<x<3)... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false | 728,775 |
Example 4 Let $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{x y}{z}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} .
$$ | Proof: Let $\frac{y z}{x}=a^{2}, \frac{z x}{y}=b^{2}, \frac{x y}{z}=c^{2}$.
Then $x=b c, y=c a, z=a b$.
$$
\begin{array}{l}
\text { Hence equation (1) } \Leftrightarrow \sum_{\text {cyc }} a^{2}>2 \sqrt[3]{\sum_{\text {cyc }} a^{3} b^{3}} \\
\Leftrightarrow\left(\sum_{\text {cyc }} a^{2}\right)^{3}>8 \sum_{\text {cyc }... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,776 |
Example 6 Given that $x, y, z$ are non-negative real numbers, and $x+y+z=1$. Prove:
$$
0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} \text {. }
$$ | Prove that by homogenizing equation (1) with the given conditions, we have
$$
\begin{array}{l}
0 \leqslant(x y+y z+z x)(x+y+z)-2 x y z \\
\leqslant \frac{7}{27}(x+y+z)^{3} .
\end{array}
$$
Let \( g_{1}=\sum_{\text {cyc }} x^{3}-\sum_{\text {sym }} x^{2} y+3 x y z \),
$$
g_{2}=\sum_{\text {sym }} x^{2} y-6 x y z, g_{3}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,777 |
Example 7 Proof: For positive real numbers $x, y, z$, we have
$$
\left(\sum_{\text {cyc }} x y\right)\left[\sum_{\text {cyc }} \frac{1}{(x+y)^{2}}\right] \geqslant \frac{9}{4} .
$$ | $$
\begin{array}{l}
\left(\sum_{\text {cyc }} x y\right)\left[\sum_{\text {cyc }}(x+y)^{2}(y+z)^{2}\right] \\
\geqslant \frac{9}{4} \prod(x+y)^{2} \\
\Leftrightarrow 4 \sum_{\text {sym }} x^{5} y-\sum_{\text {sym }} x^{4} y^{2}-3 \sum_{\text {sym }} x^{3} y^{3}+ \\
6 x^{2} y^{2} z^{2}+\sum_{\text {sym }} x^{4} y z-2 \s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,778 |
1. Given the inequality
$$
3(x-1)<2(x+a)-5
$$
has only three positive integer solutions. Then the range of $a$ is ( ).
(A) $\frac{5}{2} \leqslant a<3$
(B) $\frac{5}{2} \leqslant a \leqslant 3$
(C) $\frac{5}{2}<a \leqslant 3$
(D) $3<a<\frac{7}{2}$ | $-1 . C$.
From the given, we have $0<x<2 a-2$.
Combining the problem statement, we get
$$
3<2 a-2 \leqslant 4 \Rightarrow \frac{5}{2}<a \leqslant 3 .
$$ | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 728,779 |
Example 5 At different times of the day, the manager hands documents to the secretary for printing, each time placing the document on top of the stack of documents the secretary needs to print. The secretary takes one document from the top to print whenever there is time. If there are $n$ documents, and the manager han... | 【Analysis】For problems involving counting the return to state 0 after an equal number of "release" and "take out" operations, we can attempt to reduce the problem to the "line segment method."
Starting from the origin $O$, when a manager delivers a file, draw a line segment with a slope of 1 and a length of $\sqrt{2}$... | \frac{2}{n+1} \mathrm{C}_{2n-1}^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,781 |
5. Let $a, b, c$ be three numbers chosen from $1,2, \cdots, 5$ (no repetition allowed), and $ab+c$ is an odd number. Then the number of all possible values is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 5. D.
The 9 possible values that meet the requirements are
$$
5,7,9,11,13,17,19,21,23 .
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,782 |
7. The equation $x^{2}-a|x|+a^{2}-3=0$ has a unique real solution for $x$, then $a=$ $\qquad$ | 7. $-\sqrt{3}$.
From the problem, we know that if the original equation has only one real solution, then $x=0$.
Substituting $x=0$ into the original equation, we get $a= \pm \sqrt{3}$, where $a=\sqrt{3}$ does not meet the requirements of the problem, so it is discarded.
Thus, $a=-\sqrt{3}$. | -\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,784 |
8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$
Then $x+y-z=$ . $\qquad$ | 8. 0 .
Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$ | 0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,785 |
10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015. | 10.8 .
Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,786 |
12. Given that $x, n$ are positive integers, and $p$ is a prime number, satisfying $2 x^{3}+x^{2}+10 x+5=2 p^{n}$.
Find all possible values of $x+n+p$. | 12. From the given equation, we have
$$
\left(x^{2}+5\right)(2 x+1)=2 p^{n} \text {. }
$$
Obviously, $x^{2}+5>2 x+1$ (an odd number).
Combining with the problem, we know that $\frac{x^{2}+5}{2 x+1}$ is an integer, and
$$
\frac{x^{2}+5}{2 x+1}=\frac{1}{4}\left(2 x-1+\frac{21}{2 x+1}\right) \text {. }
$$
Thus, $x$ can ... | 6 \text{ or } 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,787 |
Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$. | 【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$
The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$
To compute such an "unordered sum", on... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,788 |
1. Calculate: $\left(2^{0}-1+5^{2}-0\right)^{-1} \times 5$ is ( ).
(A) -125
(B) -120
(C) 25
(D) $\frac{24}{5}$
(E) $\frac{1}{5}$ | 1. E.
$$
\left(2^{0}-1+5^{2}-0\right)^{-1} \times 5=\frac{1}{25} \times 5=\frac{1}{5} .
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,789 |
2. A box contains 25 tiles that are either triangular or square, with a total of 84 sides. How many square tiles are in the box?
(A) 3
(B) 5
(C) 7
(D) 9
(E) 11 | 2. D.
Let there be $a$ triangular tiles and $b$ square tiles in the box.
According to the problem, we have
$$
\left\{\begin{array}{l}
a+b=25, \\
3 a+4 b=84
\end{array} \Rightarrow(a, b)=(16,9)\right.
$$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,790 |
5. Mr. Patrick is the math teacher of 15 students. After a test, he found that the average score of the rest of the students, excluding Peyton, was 80 points, and the average score of the entire class, including Peyton, was 81 points. What was Peyton's score in this test? ( ) points.
(A)81
(B) 85
(C) 91
(D) 94
(E) 95 | 5. E.
$$
15 \times 81-14 \times 80=1215-1120=95 \text {. }
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,791 |
7. In the arithmetic sequence $13,16,19, \cdots, 70,73$, there are ( ) terms.
(A) 20
(B) 21
(C) 24
(D) 60
$(\mathrm{E}) 61$ | 7. B.
Notice that the first term of the arithmetic sequence is 13, the common difference is 3, and the last term is 73. Therefore, the number of terms is $\frac{73-13}{3}+1=21$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,792 |
10. For the letter arrangement abcd, there are ( ) different rearrangements such that no two letters that were adjacent in the original arrangement are adjacent in the new arrangement (for example, ab and ba should not appear in the new arrangement).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 10. C.
For different permutations of abcd, if a is in the first position, then the second position can only be c or d. Let's assume it is c. At this point, there is no letter to place in the third position. If we place b, then cb are adjacent; if we place d, then cd are adjacent, which is a contradiction. Therefore, a... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,793 |
Example 7 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 and 8. Find the number of different ways to select such a group of eight points. | 【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & ... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,794 |
14. The radius of the clock face is 20 cm, and it is externally tangent to another smaller disk with a radius of 10 cm at the 12 o'clock position. The smaller disk has a fixed pointer, which initially points vertically upward. When the smaller disk rolls clockwise along the outer edge of the larger clock face, always r... | 14. C.
As shown in Figure 3.
Let the angular displacement of the large circle be $\theta$. Then the angular displacement of the small circle is $360^{\circ}-\theta$.
Thus, $20 \theta=10\left(360^{\circ}-\theta\right) \Rightarrow \theta=120^{\circ}$.
Therefore, the two disks are externally tangent at the 4 o'clock posi... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,795 |
19. In an isosceles right $\triangle ABC$, it is known that $BC = AC$, $\angle C = 90^\circ$, and the area is $12.5$. The trisectors of $\angle ACB$ intersect the hypotenuse $AB$ at points $D$ and $E$. Then the area of $\triangle CDE$ is $(\quad)$.
(A) $\frac{5 \sqrt{2}}{3}$
(B) $\frac{50 \sqrt{3} - 75}{4}$
(C) $\frac{... | 19. D.
Given $A B=A C=5$.
As shown in Figure 5, draw $D F \perp A C$ at point $F$.
Since $\angle A C B=90^{\circ}$, and $C D$, $C E$ are the trisectors of $\angle A C B$, therefore, $\angle A C D=30^{\circ}$.
Then $A F=D F$, $C F=\sqrt{3} D F$,
$A C=(1+\sqrt{3}) D F=5$,
$D F=\frac{5}{\sqrt{3}+1}=\frac{5(\sqrt{3}-1)}{2... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,796 |
Example 8 Given a prime number $p$. Let $A=\left(a_{i j}\right)$ be a $p \times p$ matrix, satisfying
$$
\left\{a_{i j} \mid 1 \leqslant i, j \leqslant p\right\}=\left\{1,2, \cdots, p^{2}\right\} .
$$
We are allowed to perform the following operation on a matrix: first select a row or a column, and then add 1 or subtr... | 【Analysis】If $A$ is a good matrix, then after a finite number of operations, matrix $A$ can be transformed into a "zero matrix". Suppose all operations to transform matrix $A$ into a "zero matrix" involve subtracting $x_{i}$ from each number in the $i$-th row and subtracting $y_{j}$ from each number in the $j$-th colum... | 2(p!)^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,797 |
24. Given a quadrilateral $ABCD$ with all sides of positive integer lengths, perimeter $p$, $\angle B$ and $\angle C$ are both right angles, $AB=2$, and $CD=AD$. If $p<2015$, then the number of different possible values of the positive integer $p$ is ( ).
(A) 30
(B) 31
(C) 61
(D) 62
(E) 63 | 24. B.
As shown in Figure 7, let $B C=x$, $A D=C D=y$. Draw $A E \perp C D$ at point $E$.
Given $A B=2$,
$$
\angle B=\angle C=90^{\circ} \text {, }
$$
we know $A E=B C=x$.
In the right triangle $\triangle A D E$, by the Pythagorean theorem, we have
$$
y^{2}=(y-2)^{2}+x^{2} \Rightarrow x=2 \sqrt{y-1} \text {. }
$$
Th... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,798 |
1. In an acute triangle $\triangle ABC$, it is known that $CD \perp AB$ at point $D$, the angle bisector of $\angle ABC$ intersects $CD$ at point $E$, and intersects the circumcircle $\Gamma$ of $\triangle ADE$ at point $F$. If $\angle ADF=45^{\circ}$, prove that $CF$ is tangent to circle $\Gamma$. | 1. As shown in Fig. 1.
From $\angle C D F=90^{\circ}-45^{\circ}=45^{\circ}$, we know that line $D F$ bisects $\angle C D A$. Therefore, point $F$ lies on the perpendicular bisector of segment $A E$, and let this line intersect $A B$ at point $G$.
Let $\angle A B C=2 \beta$.
Since points $A, D, E, F$ are concyclic and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,799 |
2. Place $n^{2}$ dominoes of size $1 \times 2$ or $2 \times 1$ on a $2 n \times 2 n$ grid without overlapping, such that each $2 \times 2$ subgrid contains at least two uncovered cells, and these two cells are in the same row or column. Find the number of all possible arrangements that satisfy the condition. | 2. Divide the grid into $n^{2}$ sub-grids of size $2 \times 2$, each of which can be covered by at most two dominoes. Since the dominoes exactly cover $2 n^{2}$ cells, each sub-grid must have exactly two cells covered, and these two cells must be in the same row or column.
Next, we show that these two cells are covere... | \left(\mathrm{C}_{2 n}^{n}\right)^{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,800 |
4. Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, \cdots$,
$a_{n}, \cdots$ satisfying: for any positive integer $n$,
$$
a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} \text { ? }
$$ | 4. Suppose there exists a sequence of positive integers $\left\{a_{n}\right\}$ that satisfies the given conditions.
$$
\text { Let } b_{n}=a_{n+1}-a_{n}(n \geqslant 2) \text {. }
$$
By definition, for every
$$
b_{n}=\sqrt{a_{n}+a_{n-1}}(n \geqslant 2),
$$
we have $b_{n+1}^{2}-b_{n}^{2}=\left(a_{n+1}+a_{n}\right)-\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,801 |
1. Fill two $a$s and two $b$s into the 16 cells shown in Figure 3, with at most one letter per cell. If the same letters must not be in the same row or column, find the number of different ways to fill the cells. | Prompt: Categorized Count. Answer: 3960. | 3960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,802 |
2. In quadrilateral $ABCD$, $AC$ bisects $\angle BCD$, $\angle CAD=20^{\circ}, \angle CBD=40^{\circ}$. Then $\angle ABD=(\quad)$.
(A) $50^{\circ}$
(B) $60^{\circ}$
(C) $70^{\circ}$
(D) $80^{\circ}$ | $$
\begin{array}{l}
\angle C A E = \angle C B E \\
= 40^{\circ}, \\
\angle D A E \\
= \angle C A E - \angle C A D \\
= 20^{\circ} = \angle C A D, \\
\angle A E B = \angle A C B = \angle A C D . \\
\text { Also, } A D = A D, \text { so } \\
\triangle A D E \cong \triangle A D C \\
\Rightarrow A E = A C \Rightarrow \ove... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,803 |
5. There are 53 books of mathematics and physics on the bookshelf, where no two physics books are placed next to each other, but each mathematics book is adjacent to another mathematics book. Given the following four statements:
(1) There are at least 35 mathematics books;
(2) There are at most 18 physics books;
(3) Th... | 5. C.
The scenario with the fewest mathematics books $(S)$ occurs when the number of physics books (W) is at its maximum, as shown below:
$$
\underbrace{\text { WSSWS...WSSSW, }}_{\text {17 groups }}
$$
Three mathematics books SSS can also appear simultaneously in other positions, as indicated by arrangements (1), (2... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,804 |
6. As shown in Figure 2, circles $\odot O_{1}$ and $\odot O_{2}$ are externally separated, and their line of centers $O_{1} O_{2}$ intersects $\odot O_{1}$ at points $A$ and $B$, and intersects $\odot O_{2}$ at points $C$ and $D$. Circle $\odot O_{3}$ is internally tangent to $\odot O_{1}$ at point $B$, and circle $\od... | 6. B.
Let the radius of $\odot O_{i}(i=1,2,3,4)$ be $r_{i}, BC=m$.
Obviously, $\triangle A E O_{3} \sim \triangle A F O_{2}$.
Thus, $\frac{O_{3} E}{O_{2} F}=\frac{A O_{3}}{A O_{2}} \Rightarrow \frac{r_{3}}{r_{2}}=\frac{2 r_{1}-r_{3}}{2 r_{1}+m+r_{2}}$
$\Rightarrow r_{3}=\frac{2 r_{1} r_{2}}{2 r_{1}+m+2 r_{2}}$.
Simila... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,805 |
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
then the value of $x+\frac{1}{x}$ is | 2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,806 |
4. As shown in Figure 4, given that the two medians $B D$ and $C E$ of $\triangle A B C$ intersect at point $G$, and points $A, D, G, E$ are concyclic, $B C=6$. Then the length of $A G$ is $\qquad$. | 4. $2 \sqrt{3}$.
As shown in Figure 10, extend $A G$ to intersect $B C$ at point $F$, and connect $D E$, which intersects $A G$ at point $M$.
Obviously, $G$ is the centroid of $\triangle A B C$, and $A G=2 G F$.
Also, $D E \Perp \frac{1}{2} B C$, and $M$ is the midpoint of $A F$, thus
$$
\begin{array}{l}
E M=\frac{1}{... | 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,808 |
2. The maximum value of the function $f(x)=\frac{\sqrt{x}}{x^{2}+x+4}(x>0)$ is $\qquad$ . | 2. $\frac{1}{6}$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
x^{2}+x+1+1+1+1 \\
\geqslant 6 \sqrt[6]{x^{2} x \times 1 \times 1 \times 1 \times 1} \\
\Rightarrow x^{2}+x+4 \geqslant 6 \sqrt{x} \Rightarrow f(x) \leqslant \frac{1}{6} .
\end{array}
$$
When $x=1$, the equality holds.
Thus, the maximum value of $... | \frac{1}{6} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,809 |
5. Given real numbers $x, y$ satisfy
$$
\left(2015+x^{2}\right)\left(2015+y^{2}\right)=2^{22} \text {. }
$$
Then the maximum value of $x+y$ is | 5. $2 \sqrt{33}$.
Notice that, $2^{22}=2048^{2}=(2015+33)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[(2015-33)+33+x^{2}\right]\left[(2015-33)+y^{2}+33\right]} \\
\geqslant[(2015-33)+\sqrt{33} y+x \sqrt{33}]^{2} \\
\Rightarrow(2015+33)^{2} \\
\quad \geqslant[(2015-33)+\sqrt{33}(x+y)]^{2}... | 2 \sqrt{33} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,811 |
6. Let $M$ be the midpoint of the height $D D_{1}$ of the regular tetrahedron $A B C D$. Then the dihedral angle $A-M B-C$ in radians is $\qquad$ . | 6. $\frac{\pi}{2}$.
Let the edge length $AB=a$. It is easy to get the height $D D_{1}=\frac{\sqrt{6}}{3} a$.
Thus, $M A=\sqrt{M D_{1}^{2}+A D_{1}^{2}}$
$$
=\sqrt{\left(\frac{D D_{1}}{2}\right)^{2}+\left(\frac{A B}{\sqrt{3}}\right)^{2}}=\frac{\sqrt{2}}{2} a \text {. }
$$
Notice that,
$$
\begin{aligned}
M B & =M A \Rig... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,812 |
Example 1 Let $p$ be a prime number, and the number of distinct positive divisors of $p^{2}+71$ does not exceed 10. Find the value of $p$.
(2006, National High School Mathematics League, Jiangsu Province Preliminary Contest) | When $p=2,3$, it is verified that they satisfy the problem statement.
When $p>3$,
$$
p^{2}+71=(p+1)(p-1)+72 \text { . }
$$
If $p=3 k+1$ (where $k$ is even), then
$$
p^{2}+71=3 k(3 k+2)+72 \Rightarrow 24!\left(p^{2}+71\right) \text { . }
$$
Let $p^{2}+71=24 m=2^{3} \times 3 m(m \geqslant 4)$.
When $m$ contains a prime... | 2 \text{ or } 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,813 |
Example 2 Find all positive integers $n$, such that the product of all positive divisors of $n$ is $n^{3}$.
[1]
(2013, Croatian Mathematical Competition) | Let $1\alpha_{2}>\cdots>\alpha_{k}$.
Then $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=6$
$\Rightarrow k=1, \alpha_{1}=5$ or $k=2, \alpha_{1}=2, \alpha_{2}=1$.
Thus, $n=p^{5}$ or $n=p^{2} q(p, q$ being primes $)$ are all the solutions. | n=p^{5} \text{ or } n=p^{2} q | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,814 |
One. (40 points) As shown in Figure 4, given that $O$ is a point inside the acute triangle $\triangle ABC$, the projections of $O$ on $BC$, $CA$, and $AB$ are $A_1$, $B_1$, and $C_1$ respectively. Lines through points $A$ and $B$ perpendicular to $B_1C_1$ and $A_1C_1$ intersect at point $P$, and the projection of $P$ o... | 1. Construct $C P^{\prime} \perp A_{1} B_{1}$ at point $P^{\prime}$.
Notice,
$$
\begin{array}{l}
\angle P A B=\angle O C_{1} B_{1}, \angle P A C=\angle O B_{1} C_{1}, \\
\angle P B A=\angle O C_{1} A_{1}, \angle P B C=\angle O A_{1} C_{1}, \\
\angle P^{\prime} C B=\angle B_{1} A_{1} O, \angle P^{\prime} C A=\angle A_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,815 |
II. (40 points) Given positive real numbers $a, b, c$ satisfying $a+b+c < abc$.
Prove: $\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}+\frac{1}{\sqrt{1+c^{2}}}<\frac{3}{2}$. | Let $a+b+c=k^{2} a b c(0<k<1)$.
Then $k(a+b+c)=k^{3} a b c$.
Let $k a=\tan A, k b=\tan B, k c=\tan C$, where,
$$
\angle A, \angle B, \angle C \in\left(0, \frac{\pi}{2}\right) \text {. }
$$
Then $\tan A+\tan B+\tan C$
$$
=\tan A \cdot \tan B \cdot \tan C \text {. }
$$
It is easy to see that $\angle A+\angle B+\angle C... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,816 |
In the spatial rectangular coordinate system, if the three coordinates of a point are all integers, then the point is called a "lattice point". If two lattice points have two corresponding coordinates equal, and the remaining coordinate differs by exactly 1, then these two lattice points are called "adjacent". Initiall... | Three, the answer is negative.
Assume that after a finite number of operations, it can be made such that the coordinates $(x, y, z)$ of all chess pieces satisfy $|x|+|y|+|z| \geqslant 6$, and let the set of coordinates of all chess pieces at this time be $T$.
$$
\text { Let } S=\sum_{(x, y, z) \in T} \frac{1}{3^{|x|+|y... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,817 |
Four. (50 points) Given a prime number $p$ and a positive integer $k$ $(k<p)$. Find all positive integers $n(n \geqslant p-1)$, such that
$$
\mathrm{C}_{n}^{i}, \mathrm{C}_{n}^{i+1}, \cdots, \mathrm{C}_{n}^{i+p-1}(i=1,2, \cdots, n-p+1)
$$
contain exactly $k$ numbers divisible by $p$. | Four, first prove a lemma.
Lemma For any positive integers $m, n (m < n)$, let $n = (a_1 a_2 \cdots a_t)_p, m = (b_1 b_2 \cdots b_t)_p$ (allowing $b_1 = b_2 = \cdots = b_i = 0$). Then $p \nmid \binom{n}{m}$ if and only if for all $i = 1, 2, \cdots, t$, we have $b_i \leq a_i$.
Proof Use $v_p\left(\binom{n}{m}\right)$ t... | n = (q+1)p^t - k - 1 \left(1 \leq q \leq p-1, t \in \mathbb{Z}_+\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,818 |
Given a point $P$ inside circle $\odot O$, three chords $A B$, $C D$, and $E F$ pass through $P$. If a line through $P$ and parallel to $B F$ intersects the rays $E A$, $E C$, and $D F$ at points $Q$, $M$, and $N$ respectively, prove:
$$
\frac{1}{P Q}=\frac{1}{P M}+\frac{1}{P N} .
$$ | Proof As shown in Figure 1, let line $P Q$ intersect $\odot O$ at points $R$ and $S$, and connect $R C$, $R E$, $S C$, and $S E$.
Since $\triangle C R E$ and $\triangle C S E$ share the common base $C E$, $\angle C R E = \angle C S E$, and the line connecting vertices $R$ and $S$ intersects $C E$ at point $M$, it is e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,819 |
As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB > AC$, $BD$ and $CE$ are the angle bisectors of $\triangle ABC$, and $I$ is the incenter of $\triangle ABC$. The circumcircle of $\triangle AEC$ intersects the circumcircle of $\triangle BIC$ at point $M$ (other than point $C$), and the circumcircle of $\... | Prove as shown in Figure 2, connect $A M, A N, I M, I N, B M, C M, B N, C N$.
From the given information,
$M, B, C, I$ are concyclic
$$
\begin{aligned}
\Rightarrow & \angle B M C=\angle B I C \\
& =\angle B A C+\frac{\angle A B C+\angle B C A}{2} \\
& =90^{\circ}+\frac{\angle B A C}{2},
\end{aligned}
$$
$A, E, M, C$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,820 |
For positive integers $m, n$, with $m<n$. Prove that for any $x \in\left(0, \frac{\pi}{2}\right)$, we have
$$
2\left|\sin ^{n} x-\cos ^{n} x\right| \leqslant 3\left|\sin ^{m} x-\cos ^{m} x\right| .
$$ | Prove that it is sufficient to prove for $x \in\left(0, \frac{\pi}{4}\right)$:
$$
\text { Let } f_{n}=\cos ^{n} x-\sin ^{n} x \text {. }
$$
It can be verified:
$$
\begin{array}{l}
f_{2} \leqslant \sqrt{2} f_{1}, f_{3} \leqslant \frac{3}{2} f_{2}, f_{3} \leqslant \frac{3}{2} f_{1}, \\
f_{2}=f_{4} \leqslant f_{3} .
\end... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,821 |
Given an integer $n \geqslant 3$. Find the maximum positive number $\lambda$, such that if positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}<\lambda\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{3} \text {, }
$$
then $a_{1}, a_{2}, a_{3}$ can definitely be the lengths of the sides ... | Solve $\lambda_{\max }=\frac{10}{[8+\sqrt{10}(n-3)]^{2}}$.
(1) If $\lambda>\frac{10}{[8+\sqrt{10}(n-3)]^{2}}$, take
$$
\begin{array}{l}
a_{1}=4, a_{2}=a_{3}=2, \\
a_{4}=a_{5}=\cdots=a_{n}=\sqrt{10} .
\end{array}
$$
Then the inequality holds, but $a_{1} 、 a_{2} 、 a_{3}$ cannot form a triangle.
(2) If $\lambda=\frac{8}{... | \lambda_{\max }=\frac{10}{[8+\sqrt{10}(n-3)]^{2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,822 |
Example 2 As shown in Figure 2, given three fixed points of a line segment in sequence as $A, B, C$, and $\odot O$ is a circle passing through points $A, C$ with its center not on $AC$. The tangents to $\odot O$ at points $A, C$ intersect at point $P$, and $PB$ intersects $\odot O$ at point $Q$. Prove: The bisector of ... | Proof: Let the angle bisector of $\angle A Q C$ intersect $A C$ at point $R$ and $\odot O$ at point $S$, where $S$ and $Q$ are distinct points. From the fact that $\triangle P A C$ is an isosceles triangle, we have
$$
\frac{A B}{B C}=\frac{\sin \angle A P B}{\sin \angle C P B} \text {. }
$$
Similarly, in $\triangle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,823 |
Example 3 As shown in Figure 3, given $\triangle A B C$ and three points $M$, $N$, $R$ outside the triangle, satisfying
$$
\begin{array}{l}
\angle B A R \\
=\angle C A N=\alpha, \\
\angle C B M \\
=\angle A B R=\beta, \\
\angle A C N \\
=\angle B C M=\gamma .
\end{array}
$$
Prove: $A M$, $B N$, $C R$ are concurrent. | Proof: Let $A M$ intersect $B C$ at point $M^{\prime}$, $B N$ intersect $A C$ at point $N^{\prime}$, and $C R$ intersect $A B$ at point $R^{\prime}$. The three internal angles of $\triangle A B C$ are denoted as $\angle A$, $\angle B$, and $\angle C$.
From the given information, we have
$$
\begin{array}{l}
\frac{B M^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,824 |
Example 5 As shown in Figure 5, it is known that quadrilateral $A B C D$ is inscribed in a circle, the extensions of $A B$ and $D C$ intersect at point $E$, the extensions of $A D$ and $B C$ intersect at point $F$, $P$ is any point on the circle, and $P E$ and $P F$ intersect the circle again at points $R$ and $S$ resp... | Prove: As shown in Figure 5, connect $PD$, $AS$, $RC$, $BR$, $AP$, and $SD$.
From $\triangle EBR \sim \triangle EPA$, $\triangle FDS \sim \triangle FPA$
$$
\Rightarrow \frac{BR}{PA}=\frac{EB}{EP}, \frac{PA}{DS}=\frac{FP}{FD} \text{. }
$$
Multiplying the two equations gives
$$
\frac{BR}{DS}=\frac{EB}{EP} \cdot \frac{FP... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,825 |
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