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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 4 As shown in Figure 2, in the right triangle $\triangle ABC$, it is known that $\angle BCA=90^{\circ}$, $CD$ is the altitude, the three sides of the right triangle $\triangle ABC$ are all integers, and $BD=11^{3}$. Find the ratio of the perimeters of the right triangles $\triangle BCD$ and $\triangle ACD$.
(20... | Let $B C=a, C A=b, A B=c$.
From Rt $\triangle B C D \backsim \mathrm{Rt} \triangle B A C$, we get
$$
\frac{B C}{B A}=\frac{B D}{B C} \text {, }
$$
which means $B C^{2}=B D \cdot B A$.
Therefore, $a^{2}=11^{3} c$.
Since $a^{2}$ is a perfect square and 11 is a prime number, let $c=11 k^{2}$ (where $k$ is a positive inte... | \frac{11}{60} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,847 |
1. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that the equation
$$
f([x] y)=f(x)[f(y)]
$$
holds for all $x, y \in \mathbf{R}$ (where $[z]$ denotes the greatest integer less than or equal to the real number $z$).
(French contribution) | 1. $f(x)=c$ (constant), where, $c=0$ or $1 \leqslant c<2$.
Let $x=0$ substitute into equation (1) to get
$$
f(0)=f(0)[f(y)]
$$
for all $y \in \mathbf{R}$.
Thus, there are two cases.
(1) When $f(0) \neq 0$, from equation (2) we know $[f(y)]=1$ for all $y \in \mathbf{R}$.
Therefore, equation (1) $\Leftrightarrow f([x] ... | f(x)=c \text{ (constant), where, } c=0 \text{ or } 1 \leqslant c<2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,848 |
2. Let the incenter of $\triangle A B C$ be $I$, and its circumcircle be $\Gamma$. The line $A I$ intersects the circle $\Gamma$ at another point $D$. Let $E$ be a point on the arc $\overparen{B D C}$, and $F$ be a point on the side $B C$ such that
$$
\angle B A F=\angle C A E<\frac{1}{2} \angle B A C .
$$
Let $G$ be ... | 2. As shown in Figure 1, let the line $A D$ intersect $B C$ at point $H$, the ray $D G$ intersect $A F$ at point $K$, and the ray $D G$ intersect the ray $E I$ at point $T$. Connect $C E$.
Notice that
$$
\begin{array}{l}
\angle D I C \\
=\angle I A C+\angle I C A \\
=\angle B C D+\angle I C B=\angle I C D .
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,849 |
3. Find all functions $g: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that for all $m, n \in \mathbf{N}_{+}, (g(m) + n)(m + g(n))$ is a perfect square.
(US Proposal) | 3. $g(n)=n+c$, where the constant $c$ is a non-negative integer.
First, the function $g(n)=n+c$ satisfies the problem (because at this time
$$
(g(m)+n)(m+g(n))=(n+m+c)^{2}
$$
is a perfect square).
First, we prove a lemma.
Lemma If a prime $p$ divides $g(k)-g(l)$, where $k$ and $l$ are positive integers, then $p \mid(... | g(n)=n+c | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,850 |
4. Let $P$ be a point inside $\triangle A B C$, and let the lines $A P$, $B P$, and $C P$ intersect the circumcircle $\Gamma$ of $\triangle A B C$ again at points $K$, $L$, and $M$, respectively. The tangent to circle $\Gamma$ at point $C$ intersects line $A B$ at point $S$. If $S C = S P$, prove that $M K = M L$. | 4. Suppose $C A>C B$. Then point $S$ is on ray $A B$. As shown in Figure 2, let line $S P$ intersect the circumcircle of $\triangle A B C$ at points $E$ and $F$.
By the problem statement and the power of a point theorem, we have
$$
\begin{array}{l}
S P^{2}=S C^{2}=S B \cdot S A . \\
\text { Then } \frac{S P}{S B}=\fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,851 |
5. There are six boxes $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$, and initially each box contains exactly 1 coin. Each time, one of the following two operations can be chosen to perform on them:
(1) Select a box $B_{j}$ $(1 \leqslant j \leqslant 5)$ that contains at least 1 coin, remove 1 coin from box $B_{j}$, and ad... | 5. The answer is affirmative.
Let $A=2010^{2010^{2010}}$.
If the number of coins in boxes $B_{i}, B_{i+1}, \cdots, B_{i+k}$ are $b_{i}, b_{i+1}, \cdots, b_{i+k}$, and after several operations, they become $b_{i}^{\prime}, b_{i+1}^{\prime}, \cdots, b_{i+k}^{\prime}$, we denote this as
$$
\left(b_{i}, b_{i+1}, \cdots, b... | (0,0,0,0,0, A) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,852 |
6. Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers. Suppose there exists a fixed positive integer $s$ such that for all $n>s$, we have $a_{n}=\max \left\{a_{k}+a_{n-k} \mid 1 \leqslant k \leqslant n-1\right\}$. Prove: There exist positive integers $l(l \leqslant s)$ and $N$, such that for all $n \geqs... | 6. From the problem, for each $n>s, a_{n}$ can be expressed as
$$
a_{n}=a_{j_{1}}+a_{j_{2}}\left(j_{1} 、 j_{2}s$, we can continue to express $a_{j 1}$ as the sum of two terms of the sequence. Continuing this process, we can express $a_{n}$ as
$$
a_{n}=a_{i_{1}}+a_{i_{2}}+\cdots+a_{i_{k}},
$$
where $1 \leqslant i_{1}, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,853 |
1. There are 2009 cards, each with one side golden and the other side black, arranged in a row on a long table. Initially, all cards have their golden side facing up. Two players stand on the same side of the table and take turns to perform the following operation: choose 50 adjacent cards, with the leftmost card havin... | 1. (1) Denote the cards with the black side up as 0, and the cards with the golden side up as 1.
For every arrangement of 2009 cards, reading from left to right corresponds to a non-negative integer in binary representation (the leading digit can be 0) in a one-to-one manner. Each operation reduces this number, so the... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,854 |
2. For any integer $n(n \geqslant 2)$, let $N(n)$ be the maximum number of triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2, \cdots, N(n))$, where $a_{i}, b_{i}, c_{i}$ are non-negative integers, and satisfy the following two conditions:
(1) For all $i, a_{i}+b_{i}+c_{i}=n$;
(2) If $i \neq j$, then $a_{i} \neq a_{j}, b_... | 2. Let the integer $n \geqslant 2, \left\{T_{1}, T_{2}, \cdots, T_{N}\right\}$ be any set of non-negative integer triples satisfying conditions (1) and (2).
Since the first component $a_{i}$ are all distinct, we have
$$
\sum_{i=1}^{N} a_{i} \geqslant \sum_{i=1}^{N}(i-1)=\frac{N(N-1)}{2} .
$$
Similarly,
$$
\sum_{2=1}^{... | \left[\frac{2 n}{3}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,855 |
3. Let $n$ be a positive integer. It is known that the sequence $c_{1}, c_{2}, \cdots, c_{n-1}$ satisfies that for each $i(i=1,2, \cdots, n-1)$, $c_{i}=0$ or 1. The sequences $a_{0}, a_{1}, \cdots, a_{n}$ and $b_{0}, b_{1}, \cdots, b_{n}$ satisfy $a_{0}=b_{0}=1, a_{1}=b_{1}=7$, and for all $i(i=1,2, \cdots, n-1)$,
$$
\... | 3. For a binary number of length $n$, $\omega = \sigma_{1} \sigma_{2} \cdots \sigma_{n}$, where $\sigma_{i} \in \{0,1\}, i=1,2, \cdots, n$ and $\sigma \in \{0,1\}$, set
$$
\omega \sigma = \sigma_{1} \sigma_{2} \cdots \sigma_{n} \sigma, \sigma \omega = \sigma \sigma_{1} \sigma_{2} \cdots \sigma_{n},
$$
and $\bar{\omega}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,856 |
Example 5 Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it does, please prove how many there are?
$(1994$, National Junior High School Mathematics League) | There exists a right-angled triangle that satisfies the conditions, and there is exactly one such triangle.
Let the hypotenuse of this right-angled triangle be \( c \), the two legs be \( a \) and \( b \), and the area be \( S \). Then,
\[
\left\{\begin{array}{l}
a \leqslant b < c < a + b, \\
a + b + c = 6, \\
a^{2} +... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,858 |
5. There are five identical empty buckets with a capacity of $2 \mathrm{~L}$ each, placed at the five vertices of a regular pentagon. Cinderella and her evil stepmother engage in the following rounds of competition: at the beginning of each round, the stepmother fetches $1 \mathrm{~L}$ of water from the river and distr... | 5. No, and Cinderella can ensure that the competition continues indefinitely.
Let the five buckets be $B_{0}, B_{1}, B_{2}, B_{3}, B_{4}$, where $B_{k} (k=0,1,2,3,4)$ is adjacent to $B_{k-1}$ and $B_{k+1}$, and all indices are taken modulo 5.
Cinderella can ensure that at the start of each round, the following three ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,859 |
6. On a $999 \times 999$ grid, there is a lame bird that moves according to the following rules: from the center of any cell to the center of any adjacent cell (cells that share a common edge), each move must turn, meaning that any two consecutive moves must be perpendicular. A non-intersecting path of the lame bird re... | 6. $998^{2}-4=4\left(499^{2}-1\right)$ squares.
First, prove: This number is the upper bound of the number of squares the lame bird can pass through.
Let the center of the square in the $i$-th row and $j$-th column be denoted as $(i, j)$. Color the squares with four colors $A, B, C, D$ as follows:
If $(i, j) \equiv ... | 4(499^{2}-1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,860 |
8. For any integer $n(n \geqslant 2)$, calculate $h(n)$ using the following method (let the rightmost digit of $n$ be $r$):
(1) If $r=0$, then $h(n)$ is the number obtained by removing the rightmost digit 0 from $n$;
(2) If $1 \leqslant r \leqslant q$, split $n$ into two parts such that the right part $R$ satisfies tha... | 8. We can consider writing $n (n \geqslant 2)$ as a string of digits and as a number string to be the same, and extend the definition of $h$ to all non-empty strings of digits from 0 to 9.
For two functions $f$ and $g$, define $g \otimes f$ such that for all strings $x$, $(g \otimes f)(x) = g(f(x))$.
For a non-negativ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,861 |
1. In $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively, $B E$ and $C D$ intersect at point $G$, the circumcircle of $\triangle A B E$ and the circumcircle of $\triangle A C D$ intersect at point $P(P \neq A)$, the extension of $A G$ intersects the circumcircle of $\triangle A C D$ ... | 1. As shown in Figure 1, connect $PA$, $PB$, $PC$, $PD$, and $PE$.
Since $\angle BDP = \angle ACP$,
$\angle CEP = \angle ABP$,
thus, $\triangle DBP \backsim \triangle CEP$.
Therefore, $\frac{AB}{AC} = \frac{DB}{CE} = \frac{DP}{CP}$
$= \frac{\sin \angle DCP}{\sin \angle CDP} = \frac{\sin \angle BAP}{\sin \angle CAP}$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,862 |
$$
\begin{array}{l}
M=\left\{(x, y) \left\lvert\, y \geqslant \frac{1}{4} x^{2}\right.\right\}, \\
N=\left\{(x, y) \left\lvert\, y \leqslant-\frac{1}{4} x^{2}+x+7\right.\right\}, \\
\left.D_{r}\left(x_{0}, y_{0}\right)=|(x, y)|\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\right\} .
\end{array}
$$
F... | 2. Let $f(x)=\frac{1}{4} x^{2}$,
$g(x)=-\frac{1}{4} x^{2}+x+7=-\frac{1}{4}(x-2)^{2}+8$.
Then $M \cap N \neq \varnothing \Leftrightarrow f(x) \leqslant g(x)$ has real solutions $\Leftrightarrow x^{2}-2 x-14 \leqslant 0$ has real solutions.
Therefore, $M \cap N \neq \varnothing$.
In the Cartesian coordinate system, the v... | \sqrt{\frac{25-5 \sqrt{5}}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,863 |
4. A set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane, no three of which are collinear, is denoted as $D$. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side"... | 4. The minimum value of $n$ is 11.
When $n \geqslant 11$, regardless of how $l$ is chosen, there always exists a subset, let's assume it is $D_{1}$, such that $D_{1}$ contains at least six points.
Consider all the middle edges of the triangles in $D_{1}$, and color them red, then color the other edges blue.
By Ramse... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,864 |
5. Given that the incircle $\odot I$ of $\triangle ABC$ touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively, and $AI$, $BI$, $CI$ intersect the circumcircle $\odot O$ of $\triangle ABC$ at points $L$, $M$, and $N$ respectively. $LD$, $ME$, and $NF$ intersect $\odot O$ at points $P$, $Q$, and $R$... | 5. As shown in Figure 2, connect $P B, P C, P E, P F$.
Since $P D$ is the angle bisector of $\angle B P C$, we have,
$$
\begin{array}{l}
\frac{P B}{P C}=\frac{B D}{C D} \\
=\frac{B F}{C E} . \\
\text { Also, } \angle F B P \\
=\angle E C P,
\end{array}
$$
then $\triangle F B P \sim \triangle E C P$.
Thus, $\angle F P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,865 |
6. Let positive real numbers $a, b, c$ satisfy $a^{3}+b^{3}+c^{3}=3$. Prove:
$$
\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+\frac{1}{c^{2}+c+1} \geqslant 1 .
$$ | 6. Proof 1 Since $(a-1)^{2}(a+1) \geqslant 0$, we have $a^{3}+2 \geqslant a^{2}+a+1$.
Similarly, $b^{3}+2 \geqslant b^{2}+b+1$,
$$
c^{3}+2 \geqslant c^{2}+c+1 \text {. }
$$
Therefore, $\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+\frac{1}{c^{2}+c+1}$
$$
\geqslant \frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2} \text ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,866 |
Example 6 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathemat... | (1) As shown in Figure 3, let $A B=a, C D=b, A C=l$, and let the height from $A B$ in $\triangle A B C$ be $h_{1}$, and the height from $D C$ in $\triangle A D C$ be $h_{2}$.
Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D} \\
=S_{\triangle A B C}+S_{\triangle A D C} \\
=\frac{1}{2}\left(h_{1} a+h_{2} b\righ... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,867 |
8. A scientist invented a time machine, which looks like a circular subway track. Now (2010) is the first platform, the 2nd, 3rd, ..., 2009th platforms are 2011, 2012, ..., 4018, and the 2010th platform returns to the present (the departure platform). Later, the machine had a programming error, changing its operation r... | 8. (1) The time track has 2009 stations, and the positive integer powers of 2 not exceeding 2009 are
$2,4,8,16,32,64,128,256,512,1024$.
According to the rules,
(i) (1) $5 k+1(k=1,2,3)$
$\xrightarrow{3-k \text { times }} 16 \rightarrow 14=5 \times 2+4$;
(2) $5 k+1(k=4,5, \cdots, 51)$
$\xrightarrow{51-k \text { times }} ... | 812 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,868 |
10. 1. Distribute 40 pencils of four colors (10 pencils of each color) to 10 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$. | 10.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
Since there are 10 pencils of each color, and each student has 4 pencils, there must be a student who has at least two different colors of pencils.
Obviously, any other color of pencils must be o... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,869 |
10.3. Point $O$ is the circumcenter of non-isosceles acute $\triangle A B C$, $A D$ is the altitude of $\triangle A B C$, $M$ is the midpoint of side $B C$, and the lines through $O$ perpendicular to $A B$ and $A C$ intersect segment $A D$ at points $P$ and $Q$, respectively. $S$ is the circumcenter of $\triangle O P Q... | 10.3. Suppose $AB > AC$, and let $L$ be the midpoint of $AB$.
Notice that $\angle A O L = \frac{1}{2} \angle A O B = \angle A C B$.
$$
\begin{array}{l}
\text { Hence } \angle B A O = 90^{\circ} - \angle A O L \\
= 90^{\circ} - \angle A C B = \angle C A D .
\end{array}
$$
Since the corresponding sides of $\triangle O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,870 |
10.4. In a $100 \times 100$ grid, each cell contains a positive integer. If the sum of the numbers in the cells of a rectangle (composed of several cells) is a multiple of 17, then the rectangle is called "good". We can color the cells of some good rectangles in the grid, with each cell being colored at most once. It i... | 10.4. $d_{\max }=9744=100^{2}-16^{2}$.
First, we prove a lemma.
Lemma: In a $1 \times n$ grid, each cell contains a positive integer, then at least $n-16$ cells can be colored. Proof by induction on $n$.
When $n \leqslant 16$, the conclusion is obviously true.
Assume $n=k \geqslant 17$, and for all $n < k$, the conclu... | 9744 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,871 |
10.6. Given that $K$ is a point inside $\triangle A B C$ and lies on the bisector of $\angle B A C$, the line $C K$ intersects the circumcircle $\omega$ of $\triangle A B C$ at points $C$ and $M$. Circle $\Omega$ is tangent to line $C M$ at point $K$, and intersects segment $A B$ at points $A$ and $P$. Circles $\omega$... | 10.6. Let $R$ be the second intersection point of circle $\Omega$ and line segment $A C$. Then
$$
\angle M K P=\angle K A P=\angle K A R=\angle K P R .
$$
Thus, $P R \parallel C M$.
Hence $\angle C M Q=\angle C A Q=\angle R A Q=\angle R P Q$.
Therefore, line $M Q$ coincides with $P Q$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,872 |
10. 7. Given $n(n \geqslant 3)$ pairwise coprime positive integers. It is known that the product of any $n-1$ of these numbers divided by the remaining number yields the same remainder, denoted as $r$. Prove: $r \leqslant n-2$. | 10.7. If $r=0$, the conclusion is obvious.
Assume $r>0$.
Let these $n$ pairwise coprime positive integers be $a_{1}, a_{2}, \cdots, a_{n}$. Let
$$
P=a_{1} a_{2} \cdots a_{n}, P_{i}=\frac{P}{a_{i}}(i=1,2, \cdots, n) .
$$
Then $a_{i}>r$.
Consider $S=P_{1}+P_{2}+\cdots+P_{n}-r$, then $a_{i}|S \Rightarrow P| S$.
Notice t... | r \leqslant n-2 | Number Theory | proof | Yes | Yes | cn_contest | false | 722,873 |
10.8. In a certain country, some cities have direct two-way flights, and any two cities can be connected by a series of flights (referred to as a route). It is known that for any circle (a route where the start and end city are the same) consisting of an odd number of flights, removing all the flights in the circle wil... | 10.8. Consider a graph $G$, where the vertices correspond to cities, and two vertices are connected by an edge if and only if there is a flight between the corresponding cities. Thus, we obtain a connected graph, and removing all the edges of any odd cycle makes the graph disconnected.
We need to prove: The vertices o... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,874 |
1. The number of integer pairs $(m, n)$ that satisfy the equation $\frac{2011}{m}+\frac{1}{n}=1$ is ( ).
(A) 2
(B) 3
(C) 5
(D) 6 | -1. B.
The original equation can be transformed into
$$
\begin{array}{l}
(m-2011)(n-1)=2011 \text { (prime). } \\
\text { Also, } 2011=1 \times 2011=(-1) \times(-2011) \text {, } \\
m \neq 0,
\end{array}
$$
Then $m-2011$ can take three values $\pm 1,2011$, and accordingly $n-1$ also has three integer values. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,875 |
2. Person A and Person B are playing a card game. There are 40 cards in total, each card has a number from 1 to 10, and each number has four different suits. At the beginning, each person has 20 cards. Each person removes pairs of cards that differ by 5. In the end, Person B has two cards left, with the numbers 4 and $... | 2. D.
Only cards with numbers $x+5$ and $x$ (where $x$ is a positive integer not greater than 5) can be removed together, 4 can only be removed with 9, and 7 can only be removed with 2, so, $\{a, b\}=\{2,9\}$.
Thus, $|a-b|=7$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,876 |
3. The number of all real number pairs $(x, y)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{x+y}=y^{x-y} \\
x^{3} y=1
\end{array},\right.
$$
is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. D.
When $x=1$ or -1, from equation (2) we get $y= \pm 1$. Therefore, $(x, y)=(1,1),(-1,-1)$, both satisfy equation (1). When $x \neq \pm 1$, from equation (2) we get
$$
y=x^{-3} \text {. }
$$
Substituting equation (3) into equation (1) gives
$$
x^{x+y}=x^{3 y-3 x} \Rightarrow x+y=3 y-3 x \Rightarrow y=2 x \text {.... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,877 |
1. As shown in Figure 4, an equilateral triangle \( \triangle ABC \) with side length 26 is inscribed in a circle, and chord \( DE \parallel BC \) intersects \( AB \) and \( AC \) at points \( F \) and \( G \), respectively. If the length of \( AF \) is \( x \) and the length of \( DF \) is \( y \), and both \( x \) an... | Obviously, $D F=G E=y$.
Since $A F \cdot F B=D F \cdot F E$, therefore,
$$
x(26-x)=y(x+y) \text {. }
$$
Substituting the four options, only when $y=6$, $x$ meets the requirement. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,878 |
4. In two boxes, Jia and Yi, each contains the same number of pieces of jewelry. After transferring a piece of jewelry worth 50,000 yuan from Jia to Yi, the average value of the jewelry in Jia decreases by 10,000 yuan, and the average value of the jewelry in Yi increases by 10,000 yuan. Then the total value of the jewe... | 4. B.
Let the original number of jewels in box A be $n(n>1)$, with a total value of $x$ million yuan, and the total value in box B be $y$ million yuan. Then $\left\{\begin{array}{l}\frac{x-5}{n-1}+1=\frac{x}{n}, \\ \frac{y+5}{n+1}-1=\frac{y}{n}\end{array} \Rightarrow\left\{\begin{array}{l}x=-n^{2}+6 n, \\ y=-n^{2}+4 n... | 12 | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,879 |
6. As shown in Figure 3, given two circles $\odot O_{1}$ and $\odot O_{2}$ with radii 1 and 2 respectively, which are externally tangent to each other, line $A B$ is a common external tangent to the two circles, touching $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$ respectively. $A C$ and $B D$ are diameters o... | 6. C.
As shown in Figure 6, let the point of tangency of the two circles be $M$. Draw the common internal tangent $MN$ through $M$, intersecting $AB$ at point $N$. Connect $AM$, $BM$, and $CM$. It is easy to see that
$$
\begin{aligned}
& \angle AMB \\
= & \angle AMC=90^{\circ} .
\end{aligned}
$$
Therefore, points $B$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,881 |
1. Given three positive integers $a$, $b$, and $c$ whose squares sum to 2011, and the sum of their greatest common divisor and least common multiple is 388. Then the sum of the numbers $a$, $b$, and $c$ is $\qquad$ . | Let the greatest common divisor of $a$, $b$, and $c$ be $d$.
Then $a = d a_{1}$, $b = d b_{1}$, $c = d c_{1}$.
Assume without loss of generality that $a_{1} \geqslant b_{1} \geqslant c_{1}$, and the least common multiple of $a_{1}$, $b_{1}$, and $c_{1}$ is $m$.
From the problem, we have
$$
a_{1}^{2} + b_{1}^{2} + c_{1}... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,882 |
3. Given that the two angle bisectors $B D$ and $C E$ of $\triangle A B C$ intersect at point $I, I D=I E, \angle A B C=70^{\circ}$. Then the degree measure of $\angle A$ is $\qquad$ . | 3. $40^{\circ}$ or $60^{\circ}$.
When point $D^{\prime}$ does not coincide with $E^{\prime}$, as shown in Figure 8, on $BC$, take $BE^{\prime}=BE$, $CD^{\prime}=CD$, and connect $ID^{\prime}$, $IE^{\prime}$.
It is easy to prove
$\triangle BIE \cong \triangle BIE^{\prime}$,
$\triangle CID \cong \triangle CID^{\prime}$.... | 40^{\circ} \text{ or } 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,884 |
4. Given that $A$ and $B$ are two fixed points on a circle $\odot O$ with radius 2, $A B \neq 4$, and $P$ is a moving point on $\odot O$. When $P$ moves around $\odot O$ once, the path traced by the orthocenter of $\triangle A B P$ is $\qquad$ (add two special points so that the orthocenter $H$ of $\triangle A B P$ tra... | 4. $4 \pi$.
When point $P$ is on the major arc $\overparen{A B}$, and $\triangle A B P$ is an acute triangle, $\angle A H B=180^{\circ}-\angle P$;
When $\triangle A B P$ is an obtuse triangle,
$$
\angle A H B=\angle P \text {; }
$$
When $\triangle A B P$ is a right triangle, the orthocenter is $A$ or $B$.
When point ... | 4 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,885 |
One, (20 points) Integrity Store has mini greeting cards originally priced at 1 yuan each. For the New Year, the store is promoting this type of card with the following two discount methods.
Method One: In addition to a 10% discount, there is a "Buy 20 Get 3 Free" offer (i.e., for every 20 cards purchased, 3 cards are... | One, to buy 23 greeting cards using method one requires $1 \times 0.9 \times 20 = 18$ yuan, and using method two requires $1 \times 0.8 \times 23 = 18.4$ yuan. Clearly, buying 23 greeting cards using method one is cheaper.
If buying $m(00.8 m$, then use method two;
When $m=19$, $0.7 m + 1.9 = 0.8 m$, so either method... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,886 |
II. (25 points)
As shown in Figure 4, given that circle $\odot O$ is internally tangent to circle $\odot O^{\prime}$ at point $A$, chord $BC$ of $\odot O$ is tangent to $\odot O^{\prime}$ at point $D$, $AB$ and $AC$ intersect $\odot O^{\prime}$ at points $E$ and $F$ respectively, $AG$ and $EH$ are diameters of $\odot O... | (1) As shown in Figure 10, draw the external common tangent $PQ$ of the two circles through point $A$, extend $BM$ to intersect $\odot O$ at point $N$, and connect $AD, AH, AN$, and $EF$.
Then $\angle AEF = \angle QAC = \angle ABC \Rightarrow EF \parallel BC \Rightarrow \angle DFE = \angle CDF$.
It is easy to see that ... | \sqrt{3} | Geometry | proof | Yes | Yes | cn_contest | false | 722,887 |
Three. (25 points) Given positive integers $a, b, c, d$ satisfying $2009(a+2009 b) c=2011(a+2011 b) d$, $(2009, d)=1,(2011, c)=1,(c, d)=1$. When $a+b$ takes the minimum value, find the values of $a, b, c, d$.
---
The above text translated into English, preserving the original text's line breaks and format. | $$
\begin{array}{l}
2009|(a+2011 b) \Leftrightarrow 2009|(a+2 b) \\
\Leftrightarrow 2009 \text { divides } 1005 a+2010 b) \\
\Leftrightarrow 2009 \text { divides } 1005 a+b) .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
20111(a+2009 b) \Leftrightarrow 2011 \mid(a-2 b) \\
\Leftrightarrow 2011 \text { divides } (1005 ... | a=4019, b=1004, c=1007, d=1005 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,888 |
2. Among the triangles with side lengths being consecutive natural numbers and a perimeter not exceeding 100, the number of acute triangles is $\qquad$
$(1987$, National Junior High School Mathematics League) | 提示: Set the three sides of the triangle to be $n-1$, $n$, and $n+1$. When $n>24$, $(n-1)^{2}+n^{2}>(n+1)^{2}$, so the triangle is an acute triangle. Therefore, the number of acute triangles that meet the requirements is 29. | 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,889 |
1. The solution to the equation $\frac{1+3^{x}+3^{2 x}+\cdots+3^{2010 x}}{1+3^{-x}+3^{-2 x}+\cdots+3^{-2010 x}}=3$ is $x=$ | $-1 \cdot \frac{1}{2010}$.
The original equation is transformed into
$$
\begin{array}{l}
\frac{3^{2010 x}\left(3^{-2010 x}+3^{-2000 x}+\cdots+3^{-x}+1\right)}{1+3^{-x}+3^{-2 x}+\cdots+3^{-2010 x}}=3 \\
\Rightarrow 3^{2010 x}=3 \Rightarrow x=\frac{1}{2010} .
\end{array}
$$ | x=\frac{1}{2010} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,890 |
2. If real numbers $x, y$ satisfy $y^{2}=4 x$, then the range of $\frac{y}{x+1}$ is $\qquad$ . | 2. $[-1,1]$.
It is easy to know that the focus of the parabola is $(0,1)$, and the directrix is $x=-1$.
By the definition of a parabola, we have
$$
y^{2}=4 x \Leftrightarrow \sqrt{y^{2}+(x-1)^{2}}=|x+1| \text {, }
$$
Thus, $\left|\frac{y}{x+1}\right| \leqslant \frac{\sqrt{y^{2}+(x-1)^{2}}}{|x+1|}=1$.
When $x=1, y= \... | [-1,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,891 |
3. Set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}y=\sqrt{1-x}, \\ y=1-x^{2}\end{array}\right\}\right.\right.$ has the number of subsets as $\qquad$ | 3. 8 .
Notice that the number of elements in set $A$ is the number of solutions to the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2}
\end{array}\right.
$$
Substituting equation (1) into equation (2) and simplifying, we get
$$
x(x-1)\left(x^{2}+x-1\right)=0 \text {. }
$$
Solving this, we g... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,892 |
4. A tetrahedron with four right-angled triangular faces is called a
“tetrahedron with four right angles”. If in the tetrahedron $S-A B C$, $\angle S A B=\angle S A C=\angle S B C=90^{\circ}$, then the right angle of the other face is $\qquad$ | 4. $\angle A B C$.
From $\angle S A B=\angle S A C=90^{\circ}$, we know $S A \perp A B, S A \perp A C$.
Thus, $S A \perp B C$.
From $\angle S B C=90^{\circ}$, we know $S B \perp B C$.
Thus, $B C \perp$ plane $S A B \Rightarrow B C \perp A B$.
Therefore, $\angle A B C=90^{\circ}$. | \angle A B C=90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,893 |
5. Arrange the coefficients of the function $f(x)=\frac{a x+b}{c x+d}$ in their natural positions to form a 2x2 matrix, denoted as $A=$ $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$, where each number is called an element of the 2x2 matrix. Also, denote
$$
\begin{array}{l}
f(f(x))=\frac{a f(x)+b}{c f(x)+d} ... | 5. $a^{2} c+a c d+b c^{2}+c d^{2}$.
According to the rules of second-order matrix multiplication, we know that
$$
A^{3}=\left(\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right)
$$
the $a_{ij}$ should be the algebraic sum of the products of the elements in the $i$-th row of $A^{2}$ and the corresp... | a^{2} c+a c d+b c^{2}+c d^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,894 |
6. Let $A=\{1,3,5,7,9\}, B=\{2,4,6,8\}$. Now, by tossing a coin, numbers are drawn without replacement from $A$ and $B$ to form a nine-digit number. The rule is: if the coin shows heads, the smallest number from set $A$ is taken; if the coin shows tails, the smallest number from set $B$ is taken. When the numbers from ... | 6. $\frac{1}{256}$.
According to the rule, the sequence of heads and tails from tossing a coin is:
heads-tails-heads-tails-heads-tails-heads-tails.
Therefore, the probability of drawing the number sequence 123456789 is
$$
\left(\frac{1}{2}\right)^{8}=\frac{1}{256} \text {. }
$$ | \frac{1}{256} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,895 |
7. A company invested in a project in 2009, with both cash inputs and cash revenues every year. It is known that
(1) In 2009, the company invested 10 million yuan, and the investment will decrease by $20\%$ each subsequent year;
(2) In 2009, the company earned 5 million yuan, and the revenue will increase by $25\%$ eac... | 7.2013.
Let the total investment of the project over $n$ years from 2009 be $A_{n}$ million yuan, and the total revenue be $B_{n}$ million yuan.
From (1) we know
$$
\begin{array}{l}
A_{n}=\sum_{k=1}^{n} 1000(1-20 \%)^{k-1} \\
=\frac{1000\left[1-\left(\frac{4}{5}\right)^{n}\right]}{1-\frac{4}{5}}=5000\left[1-\left(\fra... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,896 |
9. (16 points) As shown in Figure 1, in the tetrahedron $O-ABC$,
$$
\begin{array}{l}
OA=OB=OC=1, \\
\overrightarrow{OA} \cdot \overrightarrow{OB}=\overrightarrow{OB} \cdot \overrightarrow{OC} \\
=\overrightarrow{OA} \cdot \overrightarrow{OC}=0,
\end{array}
$$
$D$ is the midpoint of $AC$, points $E$ and $F$ are on $OD$ ... | In fact, from the known information, we have
$$
\begin{array}{l}
\overrightarrow{E F}=\overrightarrow{O F}-\overrightarrow{O E} \\
=\left(\frac{2}{3} \overrightarrow{O A}+\frac{1}{3} \overrightarrow{O B}\right)-\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O C}) \\
=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,897 |
10. (20 points) On the x-axis of the Cartesian coordinate system $x O y$, take two fixed points $A_{1}(a, 0)$ and $A_{2}(-a, 0)$. On the y-axis, take two moving points $N_{1}\left(0, b_{1}\right)$ and $N_{2}\left(0, b_{2}\right)$, satisfying $b_{1} b_{2}=b^{2}$, where $a > b > 0$ (a and b are constants). Connect $A_{1}... | 10. (1) Let point $M(x, y)$.
From $A_{1}, M, N_{1}$ and $A_{2}, M, N_{2}$ being collinear, we get
$$
\begin{array}{l}
\frac{y-0}{x-a}=\frac{b_{1}-0}{0-a}(x \neq a), \\
\frac{y-0}{x+a}=\frac{b_{1}-0}{0+a}(x \neq-a) .
\end{array}
$$
Multiplying the two equations, we get
$$
\frac{y^{2}}{x^{2}-a^{2}}=\frac{b_{1} b_{2}}{-... | \frac{3}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 722,898 |
11. (20 points) Given the function
$$
f(x)=\frac{1}{3} x^{3}-(2 a+1) x^{2}+3 a(a+2) x+\frac{2}{3},
$$
where $a$ is a real number.
(1) When the graph of $y=f^{\prime}(x)$ has a unique common point with the $x$-axis in the interval $(0,6)$, find the range of the real number $a$;
(2) When $a=-1$, find the maximum and min... | 11. Given
$$
f^{\prime}(x)=x^{2}-2(2 a+1) x+3 a(a+2) \text {. }
$$
(1) The x-coordinates of the intersection points of the graph of the function $y=f^{\prime}(x)$ with the x-axis are the real roots of the quadratic equation
$$
x^{2}-2(2 a+1) x+3 a(a+2)=0
$$
which are
$$
x_{1}=a+2, x_{2}=3 a \text {. }
$$
We will disc... | 2 \leqslant a < 4 \text{ or } a = 1 \text{ or } -2 < a \leqslant 0 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 722,899 |
3. The sides of a right-angled triangle are all integers, and the numerical value of its area is equal to the numerical value of its perimeter. Try to determine the lengths of the three sides of this right-angled triangle.
$(2003$, Beijing Junior High School Mathematics Competition) | Let the two legs of a right triangle be $a$ and $b$ $(a \geqslant b)$, and the hypotenuse be $c$. Then
$$
\left\{\begin{array}{l}
a^{2}+b^{2}=c^{2}, \\
\frac{1}{2} a b=a+b+c .
\end{array}\right.
$$
Eliminating $c$ from the two equations, we get
$$
\begin{array}{l}
a b-4 a-4 b+8=0 \\
\Rightarrow(a-4)(b-4)=8 \\
\Rightar... | (5, 12, 13) \text{ or } (6, 8, 10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,900 |
One. (40 points) As shown in Figure 2, take any point $E$ on side $C D$ of the square $A B C D$. Draw a line through $E$ parallel to the diagonal $A C$, intersecting the diagonal $B D$ at point $G$, side $A D$ at point $H$, and the extension of side $B A$ at point $F$. Connect $B H$ to intersect $D F$ at point $M$. Pro... | (1) From the given, $F G // A C, A C \perp B D$.
Thus, $F G \perp B D$.
Also, $D A \perp F B$, so $H$ is the orthocenter of $\triangle D F B$.
Therefore, $B M$ through $H$ is also an altitude of the triangle, so $B M \perp D F$.
Consequently, $M, A, B, C, D$ are concyclic (or point $M$ lies on the circumcircle of squar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,901 |
8. For a positive integer $n$, let $S(n)$ denote the sum of the digits of $n$, and let $\varphi(n)$ denote the number of positive integers less than $n$ that are coprime to $n$. If $n$ is a three-digit number and satisfies $n=34 S(n)$, then the maximum value of $\varphi(n)$ is $\qquad$ | 8. 128 .
Let the three-digit number be
$$
n=100 a+10 b+c,
$$
where $a, b, c \in\{0,1, \cdots, 9\}, a>0$.
Then $100 a+10 b+c=34(a+b+c)$,
$$
8 b=22 a-11 c=11(2 a-c) .
$$
But $(8,11)=1$, so $11 \mid b \Rightarrow b=0$.
Thus, $2 a-c=0$.
When $a=1$, $c=2, n=102$;
When $a=2$, $c=4, n=204$;
When $a=3$, $c=6, n=306$;
When $... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,902 |
- 2. (40 points) Given $a>0, b>0$, and $\alpha, \beta$ are real numbers satisfying
$$
a^{2} \cos \alpha \cdot \cos \beta+b^{2} \sin \alpha \cdot \sin \beta=0
$$
Prove:
$$
\begin{array}{l}
\frac{\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)\left(a^{2} \cos ^{2} \beta+b^{2} \sin ^{2} \beta\right)}{a^{2}(\co... | Given $a^{2} \cos \alpha \cdot \cos \beta+b^{2} \sin \alpha \cdot \sin \beta=0$, we have
$$
\begin{array}{l}
a^{2} \frac{\cos (\alpha-\beta)+\cos (\alpha+\beta)}{2}+ \\
b^{2} \frac{\cos (\alpha-\beta)-\cos (\alpha+\beta)}{2}=0,
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
a^{2}\left(\cos ^{2} \frac{\alpha-\b... | \frac{a^{2} b^{2}}{a^{2}+b^{2}} | Algebra | proof | Yes | Yes | cn_contest | false | 722,903 |
$$
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\}: 1,3,5,7, \cdots \text {, }
$$
starting from the 5th term, $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$. Find:
$$
a_{2008}^{2}+a_{2009}^{2}+a_{2010}^{2}+a_{2011}^{2}
$$
Can it be divisible by 4? | Consider the sequence $\left\{a_{n}\right\}$ and its modulo 2 residue sequence $\left\{b_{n}\right\}$.
Given that $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$, we have $b_{n+4} \equiv b_{n}+b_{n+3}(\bmod 2)$.
By recursion, we get
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,904 |
As shown in Figure 3, from a point $P$ outside a circle, draw a tangent $PA$ and a secant $PB_{1}B_{2}$ to the circle, with the point of tangency at $A$ and the points of intersection at $B_{1}$ and $B_{2}$. Draw another secant $PCD$ through point $P$ intersecting the circle at points $C$ and $D$. Draw lines through $B... | Prove as shown in Figure 3, connect $B_{1} C$, $B_{1} A$, $B_{2} A$, $B_{2} D$.
Then $\angle A B_{1} C$
$$
=\angle P A C=\angle A E_{1} B_{1}.
$$
Since $\angle C A B_{1}=\angle B_{1} A E_{1}$, therefore, $\triangle A B_{1} C \backsim \triangle A E_{1} B_{1}$.
Thus, $\frac{B_{1} E_{1}}{B_{1} C}=\frac{A B_{1}}{A C}$, wh... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,906 |
Find all integer solution pairs $(x, y)$ for $y^{3}=8 x^{6}+2 x^{3} y-y^{2}$ when $x \in$ $[0,10]$. | Solve: From $y^{3}=8 x^{6}+2 x^{3} y-y^{2}$
$\Rightarrow 8 y^{3}=64 x^{6}+16 x^{3} y-8 y^{2}$
$\Rightarrow y^{2}(8 y+9)=\left(8 x^{3}+y\right)^{2}$
$\Rightarrow 8 y+9=\left(\frac{8 x^{3}}{y}+1\right)^{2}$.
From equation (1), we know $y \mid 8 x^{3}$, and $\frac{8 x^{3}}{y}$ is even.
Thus, $\left\{\begin{array}{l}8 x^{3... | (x, y)=(0,0),(0,-1),(1,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,907 |
279 Given positive real numbers $x, y, z$ satisfying $(\sqrt{3}+1) x y + 2 \sqrt{3} y z + (\sqrt{3}+1) z x = 1$.
(1) Find the minimum value of $x + y + z$;
(2) Find the minimum value of $\frac{\sqrt{3} x y}{z} + \frac{(8-4 \sqrt{3}) y z}{x} + \frac{\sqrt{3} z x}{y}$. | (1) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{(\sqrt{3}+1) x^{2}}{2}+(\sqrt{3}-1) y^{2} \geqslant 2 x y, \\
2 y^{2}+2 z^{2} \geqslant 4 y z, \\
(\sqrt{3}-1) z^{2}+\frac{(\sqrt{3}+1) x^{2}}{2} \geqslant 2 x z .
\end{array}
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
(\sqrt{3}+1)\... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,908 |
Height 280 As shown in Figure 4, from a point $P$ outside $\odot O$, two tangents to $\odot O$ are drawn, touching at points $A$ and $B$. A secant line through point $P$ intersects $\odot O$ at points $C$ and $D$ (with $PC < PD$). Let $AC$ and $DB$ intersect at point $E$, $CD$ and $AB$ intersect at point $F$, $EF$ and ... | To prove that point $K$ lies on the tangent line through point $C$.
As shown in Figure 4, draw the tangent line to $\odot O$ through $C$, intersecting $PB$ at point $K'$. We only need to prove that point $K'$ coincides with $K$, which requires proving that points $F$, $K'$, and $E$ are collinear.
By applying the conv... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,909 |
4. As shown in Figure 5, square $E F G H$ is inscribed in $\triangle A B C$, with $B C=\overline{a b}$ (this is a two-digit number), $E F=c$, and the height of the triangle $A D=d$.
It is known that $a$, $b$, $c$, and $d$ are exactly four consecutive positive integers in ascending order. Try to find the area of $\trian... | Given $\triangle A E F \backsim \triangle A B C$, we have
$$
\frac{E F}{B C}=\frac{A D-E F}{A D},
$$
which means $\frac{c}{10 a+b}=\frac{d-c}{d}$.
Also, $b=a+1, c=a+2, d=a+3$, substituting into the equation
(1) yields $a_{1}=1, a_{2}=5$.
Thus, $B C=12$ or $56 ; d=4$ or 8.
Therefore, $S_{\triangle A B C}=\frac{1}{2} B ... | 24 \text{ or } 224 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,910 |
Example 1 Find all positive integer pairs \((x, y)\) that satisfy the equation
\[
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2003 x}-\sqrt{2003 y}+\sqrt{2003 x y} \\
=2003
\end{array}
\]
[Analysis] Based on the characteristics of the problem, we can consider starting from factoring. | Given the equation can be transformed into
$$
\begin{array}{l}
\sqrt{y}(\sqrt{x})^{2}+\left[(\sqrt{y})^{2}+\sqrt{2003} \sqrt{y}-\sqrt{2003}\right] \sqrt{x}- \\
\sqrt{2003}(\sqrt{y}+\sqrt{2003})=0 .
\end{array}
$$
Thus, $(\sqrt{y} \sqrt{x}-\sqrt{2003})(\sqrt{x}+\sqrt{y}+\sqrt{2003})=0$.
It is clear that, $\sqrt{x}+\sqr... | (2003,1),(1,2003) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,911 |
Example 2 Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n... | Consider the general term. The $k$-th term is
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}} \\
=\frac{1}{\sqrt{k} \sqrt{k+1}(\sqrt{k}+\sqrt{k+1})} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k} \sqrt{k+1}} \\
=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots, n) .
\end{array}
$$
Thus, the original expression... | 39 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,912 |
2. Let $\sqrt{27-10 \sqrt{2}}=a+b$, where $a$ is a positive integer, and $b$ is between 0 and 1. Then $\frac{a+b}{a-b}=$ $\qquad$ . | Answer: $6 \sqrt{2}-7$. | 6 \sqrt{2}-7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,913 |
Initial 281 Given that there exist $k(k \in \mathbf{N}, k \geqslant 2)$ consecutive positive integers, the mean of their squares is a perfect square. Try to find the minimum value of $k$.
| Let the $k$ positive integers be $n+1, n+2, \cdots, n+k (n \in \mathbf{N})$. Then the mean of their squares is
$$
\begin{array}{l}
f(n, k)=\frac{1}{k} \sum_{i=1}^{k}(n+i)^{2} \\
=n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} .
\end{array}
$$
Since $f(n, k) \in \mathbf{Z}$, then $\frac{(k+1)(2 k+1)}{6} \in \mathbf{Z}$.
Thus, $2 ... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,914 |
As shown in Figure 2, in $\triangle ABC$, $AB=AC$, point $D$ is on side $BC$, and points $P$ and $Q$ are the circumcenters of $\triangle ABD$ and $\triangle ACD$, respectively. Prove:
$$
\begin{array}{l}
S_{\triangle PAC} + S_{\triangle QAB} \\
= S_{\triangle PBC} + S_{\triangle QBC}.
\end{array}
$$ | Proof As shown in Figure 2, construct $\square A B C E$, and connect $Q D, Q E$. It is easy to see that
$$
\begin{array}{l}
C E=A B=A C, \\
\angle C A E=\angle A C B=\angle A B C, \\
\angle A C E=\angle C A B . \\
\text { Since } \angle A Q C=\angle A Q D+\angle C Q D \\
=2(\angle A C D+\angle C A D) \\
=2 \angle A D B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,915 |
There are $281 n(n>3)$ football teams that have played $k$ matches, with each pair of teams playing at most one match, and there are no ties. It turns out that exactly one team remains undefeated. Additionally, it is found that for any two teams $A$ and $B$, there exists another team $C$ such that either $A$ beats $C$ ... | Solve $f(n)=2 n-2$.
Represent teams with points. If team $A$ beats team $B$, draw a line segment between $A$ and $B$ with an arrow pointing to $A$, forming a graph $G$.
Next, we prove: the minimum number of edges $k$ in a graph $G$ that meets the problem's requirements is $2 n-2$.
First, the graph in Figure 3 has exa... | 2n-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,916 |
282 As shown in Figure 5, $\triangle ABC$ is an acute triangle with unequal sides, $H$ and $O$ are its orthocenter and circumcenter, respectively. $P$ and $Q$ are the midpoints of sides $AB$ and $BC$, respectively. Ray $PH$ and $QH$ intersect $\odot O$ at points $M$ and $N$, respectively. Line $PQ$ intersects $MN$ at p... | Prove as shown in Figure 5, extend $AO$ to intersect $\odot O$ at point $A'$, and $CO$ to intersect $\odot O$ at point $C'$.
Since $O$ is the circumcenter of $\triangle ABC$, $AA'$ is the diameter of $\odot O$. Therefore, $A'C \perp AC$.
Also, $H$ is the orthocenter of $\triangle ABC$, so $BH \perp AC$.
Thus, $A'C \pa... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,917 |
3. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$
(Fourth Jiangsu Province Junior High School Mathematics Competition) | Given that $m$ and $n$ are the roots of the equation $x^{2}=x+1$, i.e., $x^{2}-x-1=0$.
Let $a_{k}=m^{k}+n^{k}$. Then
$$
a_{k}=a_{k-1}+a_{k-2}(k \geqslant 3) \text {. }
$$
By $a_{1}=m+n=1$,
$$
a_{2}=(m+n)^{2}-2 m n=3 \text {, }
$$
we know $a_{3}=a_{2}+a_{1}=4, a_{4}=a_{3}+a_{2}=7$.
Therefore, $m^{5}+n^{5}=a_{5}=a_{4}+... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,918 |
4. Given that $m$ and $n$ are rational numbers, and the equation
$$
x^{2}+m x+n=0
$$
has a root $\sqrt{5}-2$. Then $m+n=$ $\qquad$ .
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | $$
\begin{array}{l}
(\sqrt{5}-2)^{2}+m(\sqrt{5}-2)+n=0 \\
\Rightarrow(n-2 m+9)=(4-m) \sqrt{5} \\
\Rightarrow\left\{\begin{array} { l }
{ n - 2 m + 9 = 0 , } \\
{ 4 - m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=4, \\
n=-1
\end{array}\right.\right. \\
\Rightarrow m+n=3 \text {. } \\
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,919 |
5. Let $x$ and $y$ be positive integers such that
$$
\sqrt{x-116}+\sqrt{x+100}=y \text {. }
$$
Find the maximum value of $y$. | Prompt: By analogy with Example 6, we can prove that $\sqrt{x-116}$ and $\sqrt{x+100}$ are both natural numbers.
Let $\sqrt{x-116}=a, \sqrt{x+100}=b$. Then $b^{2}-a^{2}=216 \Rightarrow (b+a)(b-a)=216$.
Also, $b+a \equiv (b-a) \pmod{2}$, and since 216 is even, both $b-a$ and $b+a$ are even and positive.
Thus, $b-a \geqs... | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,920 |
6. Given $\sqrt{2009}=\sqrt{x}+\sqrt{y}$, and $0<x<y$. Find all integer pairs $(x, y)$ that satisfy the equation. | From $\sqrt{2009}=\sqrt{x}+\sqrt{y}$, we get
$$
\sqrt{\frac{x}{2009}}+\sqrt{\frac{y}{2009}}=1 \text {. }
$$
It is easy to prove that $\sqrt{\frac{x}{2009}}$ and $\sqrt{\frac{y}{2009}}$ are both rational numbers.
Let $\sqrt{\frac{x}{2009}}=\frac{q_{1}}{p_{1}}\left(p_{1} 、 q_{1}\right.$ are both positive integers, and $... | (x, y) = (41, 1476), (164, 1025), (369, 656) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,921 |
Example 1 Let the polynomial be
$$
f(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n} \text {. }
$$
Given any $n+1$ integers $x_{0}, x_{1}, \cdots, x_{n}$, and $x_{0}<x_{1}<\cdots<x_{n}$. Prove: there exists an $x_{i}$ such that
$$
\left|f\left(x_{i}\right)\right| \geqslant \frac{n!}{2^{n}} \text {. }
$$
(19th IMO Shortlist Proble... | 【Analysis】From the problem, we know that an $n$-degree polynomial can be completely determined by the given $n+1$ numbers.
Proof According to the Lagrange interpolation formula
$$
f(x)=\sum_{i=0}^{n} f\left(x_{i}\right) \prod_{\substack{j \neq x+i \\ 0 ; i \leqslant n}} \frac{x-x_{j}}{x_{i}-x_{j}},
$$
Assume the concl... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,922 |
Example 2 Given the function $f(x)=a x^{4}+b x^{3}+c x^{2}+d x$, satisfying:
(1) $a, b, c, d$ are all greater than 0;
(2) For any $x \in\{-2,-1,0,1,2\}$, $f(x)$ is an integer;
(3) $f(1)=1, f(5)=70$.
Determine whether, for each integer $x, f(x)$ is an integer?
(2008, National High School Mathematics League Shandong Prov... | By the Lagrange interpolation formula, we have
$$
\begin{array}{r}
f(x)=\sum_{i=-2}^{2} f(i) \prod_{\substack{j \neq i \neq i \\
-2 \in j \in 2}} \frac{x-j}{i-j} \\
=\frac{f(-2)}{4!} \sum_{j=-1}^{2}(x-j)-\frac{f(-1)(x+2)}{3!} \prod_{j=0}^{2}(x-j)+ \\
\frac{(x+2)(x+1)(x-1)(x-2)}{2!\times 2!} f(0)- \\
\frac{f(1)(x-2)}{3!... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,923 |
Example 4 Let $a, b, c$ be the lengths of the three sides of a non-isosceles $\triangle ABC$, and $S$ be its area. Prove:
$$
\begin{array}{l}
\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)} \\
>2 \times 3^{\frac{3}{4}} S^{\frac{1}{2}} .
\end{array}
$$ | Prove the construction of a quadratic polynomial
$$
f(x)=x^{3}-(x-a)(x-b)(x-c) \text {. }
$$
Then, by the Lagrange interpolation formula, we have
$$
\begin{array}{l}
\frac{(x-b)(x-c)}{(a-b)(a-c)} a^{3}+\frac{(x-a)(x-c)}{(b-a)(b-c)} b^{3}+\frac{(x-a)(x-b)}{(c-a)(c-b)} c^{3} \\
=x^{3}-(x-a)(x-b)(x-c) .
\end{array}
$$
B... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,925 |
Example 5 Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}... | Consider the function
$$
f(x)=\prod_{i=1}^{4}\left[x-(2 i-1)^{2}\right]-\prod_{i=1}^{4}\left[x-(2 i)^{2}\right] \text {. }
$$
Then $f\left(k^{2}\right)=\prod_{i=1}^{4}\left[k^{2}-(2 i-1)^{2}\right](k=2,4,6,8)$.
By the Lagrange interpolation formula, we have
$$
f(x)=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\sub... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,926 |
Example 3 Given that $a$ is a non-negative integer, the equation $2 x-a \sqrt{1-x}-a+4=0$ has at least one integer root. Then the number of possible values of $a$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1
(2008, National Junior High School Mathematics League Sichuan Preliminary) | 【Analysis】From the problem, we know that the equation about $x$
$$
2 x-a \sqrt{1-x}-a+4=0
$$
has an integer root (let it be $x_{0}$). Then
$$
2 x_{0}-a \sqrt{1-x_{0}}-a+4=0 .
$$
Expressing $x_{0}$ in terms of $a$ is relatively difficult, so consider parameter swapping, expressing $a$ in terms of $x_{0}$. Thus,
$$
\be... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,927 |
Example 1 Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{1}=b_{1}$, and for any $n \in \mathbf{N}_{\text {, we have }}$
$$
a_{n}+b_{n}=1, \frac{a_{n+1}}{a_{n}}=\frac{b_{n}}{1-a_{n}^{2}} \text {. }
$$
(1) Determine whether the sequence $\left\{\frac{1}{a_{n}}\right\}$ is an arithmetic ... | (1) Solution From the given condition, we have
$$
\frac{a_{n+1}}{a_{n}}=\frac{1}{1+a_{n}} \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_{n}}+1 \Rightarrow a_{n}=\frac{1}{n+1} \text {. }
$$
(2) Proof From (1), we get
$$
a_{n}=\frac{1}{n+1}, b_{n}=1-a_{n}=\frac{n}{n+1} \text {. }
$$
Thus, the inequality to be proved is equiv... | \left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1}{n}\right)^{n} | Algebra | proof | Yes | Yes | cn_contest | false | 722,928 |
Example 2 Let real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$
\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4} \text {. }
$$
(2007, China Western Mathematical Olympiad) | Proof: Let $f(x)=\frac{1}{5 x^{2}-4 x+11}$. Then
$$
f^{\prime}(x)=\frac{-2(5 x-2)}{\left(5 x^{2}-4 x+11\right)^{2}} \text {. }
$$
Thus, $f(x)$ is monotonically increasing on $\left(-\infty, \frac{2}{5}\right)$ and monotonically decreasing on $\left[\frac{2}{5},+\infty\right)$.
Since $f^{\prime}(1)=-\frac{1}{24}$, the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,929 |
Example 3 For non-empty sets $A$ and $B$, define the operation:
$A \oplus B=\{x \mid x \in A \cup B$, and $x \notin A \cap B\}$.
Given two open intervals $M=(a, b), N=(c, d)$, where $a, b, c, d$ satisfy
$$
a+b<c+d, ab=cd<0 .
$$
Then $M \oplus N=(\quad)$.
(A) $(a, b) \cup(c, d)$
(B) $(a, c) \cup(b, d)$
(C) $(a, d) \cap... | Slightly explained: Let $a b=c d=t(t<0)$. Then
$$
a<0<b, c<0<d \text {. }
$$
Construct the functions
$$
\begin{array}{l}
f(x)=x^{2}-(a+b) x+t, \\
g(x)=x^{2}-(c+d) x+t .
\end{array}
$$
Then $a, b$ are the two roots of the equation $f(x)=0$, and $c, d$ are the two roots of the equation $g(x)=0$.
$$
\begin{array}{l}
\te... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,930 |
Example 4 Find the smallest positive integer $n$, such that for any sequence of $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=2007$, there must be a sum of some consecutive terms equal to 30.
(Fourth China Southeast Mathematical Olympiad) | First, construct an integer sequence $a_{1}, a_{2}, \cdots, a_{1017}$ with 1017 terms, such that no consecutive terms sum to 30. For this, take
$$
a_{1}=a_{2}=\cdots=a_{29}=1, a_{30}=31,
$$
and $a_{30 m+i}=a_{i}(i=1,2, \cdots, 30, m \in \mathbf{N})$.
Thus, $\left\{a_{k}\right\}$ is:
$$
\begin{array}{l}
1,1, \cdots, 1,... | 1018 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,931 |
Example 5 Given the function $f(x)=\mathrm{e}^{x}-x$ (where $\mathrm{e}$ is the base of the natural logarithm).
(1) Find the minimum value of the function $f(x)$;
(2) If $n \in \mathbf{N}_{+}$, prove:
$$
\left(\frac{1}{n}\right)^{n}+\left(\frac{2}{n}\right)^{n}+\cdots+\left(\frac{n-1}{n}\right)^{n}+\left(\frac{n}{n}\ri... | (1) From $f^{\prime}(x)=\mathrm{e}^{x}-1$, we know $f(x) \geqslant 1$.
(2) Proof From (1), we know that for any real number $x$, we have $\mathrm{e}^{x}-x \geqslant 1$, that is,
$$
1+x \leqslant \mathrm{e}^{x} \text {. }
$$
Construct the sequence $\left\{x_{i} \mid i=1,2, \cdots, n\right\}$, where,
$$
\begin{array}{l}... | \frac{\mathrm{e}}{\mathrm{e}-1} | Calculus | proof | Yes | Yes | cn_contest | false | 722,932 |
$$
\text { 1. Given } \sqrt[3]{a}+a=\sqrt{6}, b^{3}+b=\sqrt{6} \text {. Then } a+b
$$ | Let $f(x)=x^{3}+x$.
Since $f^{\prime}(x)=3 x^{2}+1>0$, $f(x)$ is monotonically increasing on $(-\infty,+\infty)$.
Given that $f(\sqrt[3]{a})=f(b)$.
Therefore, $\sqrt[3]{a}=b \Rightarrow a=b^{3}$.
Thus, $a+b=b^{3}+b=\sqrt{6}$. | a+b=\sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,933 |
2. If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{2^{10}+5^{3}}+\frac{y}{2^{10}+6^{3}}=1, \\
\frac{x}{3^{10}+5^{3}}+\frac{y}{3^{10}+6^{3}}=1,
\end{array}
$$
then $x+y=$ $\qquad$ (the result should be written in the form of a power). | From the given, we know that $2^{10}$ and $3^{10}$ are the two roots of the equation $\frac{x}{t+5^{3}}+\frac{y}{t+6^{3}}=1$ with respect to $t$, which means they are the roots of
$$
t^{2}-\left(x+y-5^{3}-6^{3}\right) t+\left(5^{3} \times 6^{3}-5^{3} y-6^{3} x\right)=0
$$
Therefore, $2^{10}+3^{10}=x+y-5^{3}-6^{3}$.
$$... | 2^{10}+3^{10}+5^{3}+6^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,934 |
If the function $f: \mathbf{N} \times \mathbf{Z} \rightarrow \mathbf{Z}$ satisfies the following conditions:
(1) $f(0,0)=1, f(0,1)=1$;
(2) For any $k \notin\{0,1\}, f(0, k)=0$;
(3) For each $n \geqslant 1$ and integer $k$, $f(n, k)=f(n-1, k)+f(n-1, k-2 n)$. $(2009$, Turkish Mathematical Olympiad) | Solution 1 Below we prove a lemma using mathematical induction.
Lemma $f(n, k)$ is the coefficient of $x^{k}$ in $(1+x) \prod_{i=1}^{n}\left(1+x^{2 i}\right)$.
Proof When $n=0$, conditions (1) and (2) are obviously satisfied.
Assume that the conclusion holds for $n=m$.
When $n=m+1$, the coefficient of $x^{k}$ in $(1+... | 2^{2008} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,935 |
3. Given that the incircle of $\triangle A B C$ touches sides $A B$ and $A C$ at points $Z$ and $Y$, respectively, and $B Y$ intersects $C Z$ at point $G$. Points $R$ and $S$ satisfy that quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms. Prove: $G R = G S$. | 3. As shown in Figure 1, let the incircle and the excircle of $\triangle ABC$ opposite to $\angle A$ be circles $\Gamma$ and $\Gamma_{a}$, respectively. Circles $\Gamma$ and $\Gamma_{a}$ touch side $BC$ at points $X$ and $T$, respectively. Circle $\Gamma_{a}$ touches lines $AB$ and $AC$ at points $P$ and $Q$, respectiv... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,936 |
4. Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$, line $AD$ and $BC$ intersecting at point $F$, and $G$, $H$ being the midpoints of sides $AB$ and $CD$ respectively. Prove: $EF$ is tangent to the circle passing through points $E$, $G$, and $H$ at point $E$. | 4. As shown in Figure 2.
Since $\angle B A D = \angle F C D$, therefore,
$\triangle F A B \backsim \triangle F C D$.
Let the transformation $\mathscr{T}$ be the reflection about the angle bisector of $\angle D F C$ followed by a homothety with center $F$ and ratio $\frac{F A}{F C}$. Then $\mathscr{T}$ transforms $F$ to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,937 |
Example 4 Given that $a$ is an integer, the equation concerning $x$
$$
\frac{x^{2}}{x^{2}+1}-\frac{4|x|}{\sqrt{x^{2}+1}}+2-a=0
$$
has real roots. Then the possible values of $a$ are $\qquad$
(2008, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】Take any real root $x_{0}$ of the equation. Then
$$
\frac{x_{0}^{2}}{x_{0}^{2}+1}-\frac{4\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}+2=a \text {. }
$$
Notice that $\frac{x_{0}^{2}}{x_{0}^{2}+1}=\left(\frac{\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}\right)^{2}$.
Let $\frac{\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}=t$.... | 0, 1, 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,938 |
5. Let $P$ be a convex polygon symmetric with respect to point $O$. Prove: There exists a parallelogram $R$ such that $P \subset R$, and $\frac{|R|}{|P|} \leqslant \sqrt{2}(|R|, |P|$ represent the areas of $R$ and $P$, respectively $)$. | 5. Construct parallelograms $R_{1}$ and $R_{3}$ containing $P$ such that
$$
\left|R_{1}\right| \leqslant \sqrt{2}|P| \text { and }\left|R_{3}\right| \leqslant \sqrt{2}|P|
$$
at least one of them holds.
Choose two points $A$ and $B$ in $P$ such that the area of $\triangle O A B$ is maximized.
As shown in Figure 3, let... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,939 |
6. Let the sides $AD$ and $BC$ of quadrilateral $ABCD$ intersect at point $P$, and $AB$ and $CD$ are not parallel. The circumcenters of $\triangle ABP$ and $\triangle DCP$ are $O_{1}$ and $O_{2}$, respectively, and the orthocenters are $H_{1}$ and $H_{2}$, respectively. $E_{1}$ and $E_{2}$ are the midpoints of segments... | 6. As shown in Figure 4, let the perpendicular from $E_{1}$ to $C D$ intersect $P H_{1}$ at point $X$, and the perpendicular from $E_{2}$ to $A B$ intersect $P H_{2}$ at point $Y$. Let the angle between lines $A B$ and $D C$ be $\varphi$, and let $M$ and $N$ be the midpoints of $P H_{1}$ and $P H_{2}$, respectively.
Le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,940 |
7. Given that the incenter of $\triangle A B C$ is $I$, and $X, Y, Z$ are the incenters of $\triangle B I C$, $\triangle C I A$, and $\triangle A I B$ respectively. Prove: if $\triangle X Y Z$ is an equilateral triangle, then $\triangle A B C$ is also an equilateral triangle. | 7. As shown in Figure 5, since $A Z, A I, A Y$ trisect $\angle B A C$, we can set $\angle B A C=4 \alpha$.
Similarly, let $\angle A B C=4 \beta, \angle B C A=4 \gamma$.
Then $\alpha+\beta+\gamma=45^{\circ}$, and $0^{\circ}<\alpha, \beta, \gamma<45^{\circ}$.
Since $\angle B I C=90^{\circ}+2 \alpha$, it follows that
$\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,941 |
8. Let quadrilateral $A B C D$ have an incircle, and let line $g$ through point $A$ intersect segments $B C$ and $D C$ at points $M$ and $N$, respectively. Let $I_{1}$, $I_{2}$, and $I_{3}$ be the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Prove that the orthocenter of $\tri... | 8. As shown in Figure 6, let the incircles of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$ be circles $\Gamma_{1}$, $\Gamma_{2}$, and $\Gamma_{3}$, respectively.
First, prove that: if a tangent line $h$ is drawn from point $C$ to circle $\Gamma_{1}$, different from $C B$, then $h$ is also tangent to cir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,942 |
1. Given that $a$, $b$, and $c$ are integers, and satisfy $(a-b)^{10}+(a-c)^{10}=1$.
Then $|a-b|+|b-c|+|c-a|=(\quad)$.
(A)1
(B) 2
(C) 3
(D) 4 | $-1 . B$
From the problem, we know that $a-b$ and $a-c$ are both integers.
According to the given equation, $|a-b|$ and $|a-c|$ are one 0 and one 1.
Therefore, $|a-b|+|b-c|+|c-a|$ $=1+|(a-b)-(a-c)|=2$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,943 |
2. Given real numbers $a, b, c$ satisfy the equations
$$
2 \sqrt{a}+3|b|=6, \quad 4 \sqrt{a}-9|b|=6 c \text {. }
$$
Then the maximum possible value of $c$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. C.
From the two known equations, we get
$$
\sqrt{a}=\frac{3}{5}(c+3),|b|=\frac{2}{5}(2-c) \text {. }
$$
Since $|b| \geqslant 0$, therefore, $c \leqslant 2$.
When $c=2$, we get $a=9, b=0$, which satisfies the known equations.
Therefore, the maximum value that $c$ can take is 2. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,944 |
3. If $a, b$ are two positive numbers, and $\frac{a-1}{b}+\frac{b-1}{a}+1=0$, then ( ).
(A) $0<a+b \leqslant \frac{1}{3}$
(B) $\frac{1}{3}<a+b \leqslant 1$
(C) $1<a+b \leqslant \frac{4}{3}$
(D) $\frac{4}{3}<a+b \leqslant 2$ | 3. C.
From the problem, we have $a^{2}+a b+b^{2}=a+b$.
Then $a b=(a+b)^{2}-(a+b)$
$$
=(a+b)(a+b-1) \text {. }
$$
Since $a, b$ are two positive numbers, then $a b>0, a+b>0$.
Therefore, $a+b-1>0 \Rightarrow a+b>1$.
By $a b \leqslant \frac{(a+b)^{2}}{4}$, and combining with equation (1), we get
$$
\frac{a+b}{4} \geqslan... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,945 |
4. Given that the two roots of the equation $x^{2}-3 x-1=0$ are also roots of the equation $x^{4}+a x^{2}+b x+c=0$. Then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0 | 4. A.
Let $m$ be a root of the equation $x^{2}-3 x-1=0$. Then $m^{2}=3 m+1$.
Given that $m^{4}+a m^{2}+b m+c=0$.
Substituting equation (1) into the above equation and simplifying, we get
$$
(9+a) m^{2}+(6+b) m+c+1=0 \text {. }
$$
Thus, the roots of the equation $x^{2}-3 x-1=0$ are also the roots of the equation $(9+a... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,946 |
5. In $\triangle A B C$, it is known that $\angle C A B=60^{\circ}, D, E$ are points on sides $A B, A C$ respectively, and $\angle A E D=60^{\circ}$, $E D+D B=C E, \angle C D B=2 \angle C D E$. Then $\angle D C B$ $=(\quad)$.
(A) $15^{\circ}$
(B) $20^{\circ}$
(C) $25^{\circ}$
(D) $30^{\circ}$ | 5. B.
As shown in Figure 2, extend $A B$ to point $F$ such that $B F = A D$, and connect $C F$. It is easy to see that $\triangle A D E$ is an equilateral triangle. Then
$$
\begin{array}{l}
\angle E D B = 120^{\circ}. \\
\text{Also, } C E = E D + D B \\
= A D + D B \\
= D B + B F \\
= D F,
\end{array}
$$
Thus, $\tria... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,947 |
For a natural number $n$, let the sum of its digits be denoted as $a_{n}$, for example,
$$
\begin{array}{l}
a_{2009}=2+0+0+9=11, \\
a_{2010}=2+0+1+0=3 .
\end{array}
$$
Then $a_{1}+a_{2}+\cdots+a_{2010}=$ ( ).
(A) 28062
(B) 28065
(C) 28067
(D) 28068 | 6. D.
Consider all natural numbers from 1 to 2010 as four-digit numbers (if $n$ is less than four digits, add 0s at the beginning to make it four digits, which does not change the value of $a_{n}$).
Notice that, the number of times 1 appears in the thousands, hundreds, tens, and units place are $10^{3}$, $2 \times 10... | 28068 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,948 |
Example 5 The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Given that $n$ is a positive integer, we have
$$
\begin{array}{l}
0\frac{1}{1005} \\
\quad \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} .
\end{array}
$$
Let $1004=a$. Then
$$
\begin{array}{l}
\frac{n}{n+2 a+1}<\frac{a^{2}}{(a+1)^{2}} \\
\Rightarrow\left[(a+1)^{2}-a^{2}\right] n<a^{2}(2 a+1) \\
\Rightarrow(2 a... | 1008015 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,949 |
1. Given real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=19, \\
x+y=1 .
\end{array}\right.
$$
then $x^{2}+y^{2}=$ $\qquad$ | II, 1.13.
From $x^{3}+y^{3}=19$, we get $(x+y)\left[(x+y)^{2}-3 x y\right]=19$. Substituting $x+y=1$ yields $x y=-6$. Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=13$. | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,950 |
2. The graph of the quadratic function $y=x^{2}+b x+c$ intersects the positive half-axis of the $x$-axis at points $A$ and $B$, and intersects the positive half-axis of the $y$-axis at point $C$. It is known that $A B=\sqrt{3} A C, \angle C A O=30^{\circ}$. Then $c=$ $\qquad$. | 2. $\frac{1}{9}$.
From the problem, we know $C(0, c), O C=c$.
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$.
Then $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+b x+c=0$.
By the relationship between roots and coefficients, we have $x_{1} x_{2}=c$.
Also, $\angle C A O=30^{\circ}$, so
$A C=2 c, A B... | \frac{1}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,951 |
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