problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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2. Given real numbers $a, b, c, d, e, f$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
2 a+b+c+d+e+f=20, \\
a+2 b+c+d+e+f=40, \\
a+b+2 c+d+e+f=80, \\
a+b+c+2 d+e+f=160, \\
a+b+c+d+2 e+f=320, \\
a+b+c+d+e+2 f=640 .
\end{array}\right.
$$
Then the value of $f-e+d-c+b-a$ is $\qquad$ . | 2. 420 .
$$
\begin{array}{l}
\text { (6) - (5) + (4) - (3) + (2)-(1) gives } \\
f-e+d-c+b-a \\
=640-320+160-80+40-20=420 .
\end{array}
$$ | 420 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,215 |
3. As shown in Figure 2, in rhombus $A B C D$, the distances from vertex $A$ to sides $B C$ and $C D$ are $A E$ and $A F$, both equal to $5$, and $E F=6$. Therefore, the side length of rhombus $A B C D$ is $\qquad$ | 3. $\frac{125}{24}$.
As shown in Figure 2, connect $A C$ and $E F$, intersecting at point $G$. It is easy to prove that $E F \perp A C, E G=G F$.
In the right triangle $\triangle A E G$, since $A E=5, E G=\frac{1}{2} E F=3$, therefore, $A G=4$.
In the right triangle $\triangle A E C$, since $A E^{2}=A G \cdot A C, \a... | \frac{125}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,216 |
4. Given the quadratic function $y=x^{2}-x+a$ whose graph intersects the $x$-axis at two distinct points, the sum of the distances from these points to the origin does not exceed 5. Then the range of values for $a$ is $\qquad$ . | 4. $-6 \leqslant a0$, i.e., $a<\frac{1}{4}$, and $\left|x_{1}\right|+$ $\left|x_{2}\right| \leqslant 5$. Therefore,
$$
\begin{array}{l}
\left(\left|x_{1}\right|+\left|x_{2}\right|\right)^{2}=x_{1}^{2}+x_{2}^{2}+2\left|x_{1} x_{2}\right| \\
=\left(x_{1}+x_{2}\right)^{2}+2\left|x_{1} x_{2}\right|-2 x_{1} x_{2} \\
=1+2|a|... | -6 \leqslant a < \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,217 |
5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.
Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.
Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, plea... | 5. 5 .
From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$
Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$
Therefore, $n=2,8,222,668,2006$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,218 |
6. $[x]$ represents the greatest integer not greater than $x$, the equation
$$
[2 x]+[3 x]=8 x-\frac{7}{2}
$$
all real solutions are $\qquad$ . | 6. $\frac{13}{16}, \frac{17}{16}$.
From $2 x-1<[2 x] \leqslant 2 x, 3 x-1<[3 x] \leqslant 3 x$ and the given equation, we get
$$
\begin{array}{l}
5 x-2<8 x-\frac{7}{2} \leqslant 5 x \Leftrightarrow \frac{1}{2}<x \leqslant \frac{7}{6} \\
\Leftrightarrow \frac{1}{2}<8 x-\frac{7}{2} \leqslant \frac{35}{6} .
\end{array}
$... | \frac{13}{16}, \frac{17}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,219 |
Example 6 As shown in Figure 6, the two diagonals of trapezoid $ABCD$ intersect at point $K$. Two circles are drawn with the two non-parallel sides of the trapezoid as diameters. It is known that $K$ is outside these two circles. Prove: The lengths of the tangents drawn from point $K$ to these two circles are equal. | Explanation: From the structure of the figure, think of the cutting line theorem, i.e., through \( K A \cdot K M = K D \cdot K N \), to prove that "the squares of the tangents drawn from point \( K \) to the two circles are equal."
For this, think of \( \frac{K A}{K D} = \frac{K C}{K B} \), hoping to get \( \frac{K M}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,220 |
7. As shown in Figure 3, quadrilateral $ABCD$ is a right trapezoid $\left(\angle B=\angle C=90^{\circ}\right)$, and $AB=BC$. If there exists a point $M$ on side $BC$ such that $\triangle AMD$ is an equilateral triangle, then the value of $\frac{CD}{AB}$ is $\qquad$ | 7. $\sqrt{3}-1$.
Let $A B=B C=a, A M=M D=D A=x, C D=y$. Then
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-a^{2}}+\sqrt{x^{2}-y^{2}}=a, \\
(a-y)^{2}+a^{2}=x^{2} .
\end{array}\right.
$$
From equation (2), we know $x^{2}-a^{2}=(a-y)^{2}$, then
$$
x^{2}-y^{2}=2 a^{2}-2 a y.
$$
From Figure 3, it is clear that $x>y>0$.
Thus, $2... | \sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,221 |
8. As shown in Figure 4, the area of $\triangle A B C$ is $S$, and the perimeter is $p$. The three sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$ are outside $\triangle A B C$ and are at a distance of $h$ from the corresponding sides. Then the perimeter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\qquad$... | 8. $\left(1+\frac{p h}{2 S}\right) p,\left(1+\frac{p h}{2 S}\right)^{2} S$.
It is easy to see that the line $A A^{\prime}$ bisects both $\angle A^{\prime}$ and $\angle B A C$, so the incenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle A B C$ is the same. Let the inradii of $\triangle A^{\prime} B^{... | \left(1+\frac{p h}{2 S}\right) p,\left(1+\frac{p h}{2 S}\right)^{2} S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,222 |
9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$
Then the minimum value of $n$ is $\qquad$ | 9. 5 .
Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.
If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, t... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,223 |
10. Arrange the positive integers that can be expressed as the difference of squares of two positive integers in ascending order: $a_{1}, a_{2}, \cdots$, $a_{n}, \cdots$ (for example, $a_{1}=2^{2}-1^{2}=3, a_{2}=3^{2}-2^{2}=5, a_{3}$ $\left.=4^{2}-3^{2}=7, a_{4}=3^{2}-1^{2}=8, \cdots\right)$. Then, the value of $a_{1}+... | 10. 6999.
First, it is evident that even numbers that are not multiples of 4 cannot be the difference of squares of two integers. It is known that \(a_{1}=3, a_{2}=5, a_{3}=7\).
When \(k \geqslant 2\), we have
\[4 k=(k+1)^{2}-(k-1)^{2},\]
\[4 k+1=(2 k+1)^{2}-(2 k)^{2},\]
\[4 k+3=(2 k+2)^{2}-(2 k+1)^{2},\]
and \(4 k+(... | \frac{5 \pm \sqrt{7}}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,224 |
Three. (20 points) The system of equations about $x, y, z$
$$
\left\{\begin{array}{l}
3 x+2 y+z=a, \\
x y+2 y z+3 z x=6
\end{array}\right.
$$
has real solutions $(x, y, z)$. Find the minimum value of the positive real number $a$. | From the first equation, we get $3 x+2 y=a-z$, and from the second equation, we get
$$
x y=6-z(3 x+2 y)=6-z(a-z)=z^{2}-a z+6 \text {. }
$$
From $(3 x+2 y)^{2} \geqslant 4 \times 3 x \times 2 y$, we get
$$
(a-z)^{2} \geqslant 24\left(z^{2}-a z+6\right) \text {, }
$$
which simplifies to $23 z^{2}-22 a z+144-a^{2} \leqs... | \sqrt{23} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,225 |
Four. (20 points) Let $A$ be a given positive rational number.
(1) If $A$ is the area of a right-angled triangle with all three sides being rational numbers, prove that there must exist three positive rational numbers $x$, $y$, and $z$ such that $x^{2}-y^{2}=y^{2}-z^{2}=A$;
(2) If there exist three positive rational nu... | (1) Let $a, b, c$ be the lengths of the three sides of a right triangle, where $a, b, c$ are rational numbers, and $a^{2}+b^{2}=c^{2}, \frac{1}{2} a b=A$.
If $a=b$, then $2 a^{2}=c^{2}, \frac{c}{a}=\sqrt{2}$, which contradicts the assumption that $a, c$ are rational numbers. Hence, $a \neq b$.
Without loss of general... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,226 |
Let $H$ be the orthocenter of $\triangle ABC$, and let $D, E, F$ be three points on the circumcircle of $\triangle ABC$ such that $AD \parallel BE \parallel CF$. Let $S, T, U$ be the reflections of $D, E, F$ across the sides $BC, CA$, and $AB$, respectively. Prove that $S, T, U, H$ are concyclic. | First, we prove the following lemma.
Lemma: As shown in Figure 1, let \( O \) and \( H \) be the circumcenter and orthocenter of \( \triangle ABC \), respectively. Let \( P \) be any point on the circumcircle of \( \triangle ABC \), and let \( Q \) be the point symmetric to \( P \) with respect to the midpoint of \( B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,227 |
Given a positive integer $n$. Find the largest real number $c$, such that: if the sum of the reciprocals of a set of integers greater than 1 (which can be the same) is less than $c$, then it is always possible to divide this set of numbers into no more than $n$ groups, so that the sum of the reciprocals of the numbers ... | II. $c=\frac{n+1}{2}$.
Take this set of numbers as $a_{1}, a_{2}, \cdots, a_{4}$, if $a_{1}=a_{2}=\cdots=a_{s}=2$, then $c \leqslant \frac{n+1}{2}$.
We will prove the following by mathematical induction on $n$:
If $\sum_{i=1}^{s} \frac{1}{a_{i}}<\frac{n+1}{2}$, then $a_{1}, a_{2}, \cdots, a_{s}$ can be divided into no ... | c=\frac{n+1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,228 |
For a positive integer $M$, if there exist integers $a, b, c, d$ such that $M \leqslant a<b \leqslant c<d \leqslant M+49, a d=b c$, then $M$ is called a good number; otherwise, $M$ is called a bad number. Try to find the largest good number and the smallest bad number. | Three, the largest good number is 576, and the smallest bad number is 443.
Lemma If positive integers $a, b, c, d$ satisfy $a48^{2}$.
Thus, $u+v-2 \geqslant 49$, i.e., $u+v \geqslant 51$, a contradiction.
Therefore, 576 is the largest good number.
(2) When $1 \leqslant M \leqslant 288$, take an integer $n$ such that
$1... | 576, 443 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,229 |
Let $k(k \geqslant 3)$ be an odd number. Prove: there exists a non-integer coefficient integer-valued polynomial $f(x)$ of degree $k$, with the following properties:
(1) $f(0)=0, f(1)=1$;
(2) There are infinitely many positive integers $n$ such that, if the equation
$$
n=f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f... | First, prove a lemma.
Lemma There exists a $k$-degree integer-valued polynomial $f(x)$, with not all coefficients being integers, satisfying
$$
f(0)=0, f(1)=1 \text {, }
$$
and $f(x) \equiv\left\{\begin{array}{l}0\left(\bmod 2^{k}\right), x \text { is even; } \\ 1\left(\bmod 2^{k}\right), x \text { is odd. }\end{array... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,230 |
Example 7 As shown in Figure 7, the excircle $\odot O$ of $\triangle A B C$ touches sides $B C, A B, A C$ at $D, M, N$ respectively, $D E$ is the diameter of $\odot O$, and $A E$ intersects $B C$ at $F$. Prove: $B F=C D$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 7 As shown in Figure 7, the excircle $\odot O$ of $\triangle A B... | Explanation: This is a problem involving the length of a tangent, which can be facilitated by using an expression as a medium for substitution. A more intuitive approach is $C D = C N = A N - A C = p - b$ (where $p$ is the semiperimeter of $\triangle A B C$). To prove that $B F = p - b$, it is sufficient to prove that ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,231 |
Given positive integers $m, a, b, (a, b)=1$. $A$ is a non-empty subset of the set of positive integers, such that for any positive integer $n$, either $a n \in A$ or $b n \in A$. For all sets $A$ satisfying the above property, find the minimum value of $|A \cap \{1, 2, \cdots, m\}|$.
---
The problem is to find the mi... | (1) When $a=b=1$, we have $A \cap\{1,2, \cdots, m\}=\{1,2, \cdots, m\}$, $|A \cap\{1,2, \cdots, m\}|=m$.
(2) Without loss of generality, assume $a>b$. Let
$A_{1}=\left\{k \mid\right.$ if $a^{\alpha} \mid k, a^{\alpha+1} \nmid k$, then $\alpha$ is odd $\}$.
Now verify that $A_{1}$ satisfies the condition.
For any positi... | \sum_{i=1}^{\infty}(-1)^{i+1}\left[\frac{m}{a^{i}}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,232 |
Six, Given that $\triangle A B C$ covers the convex polygon $M$. Prove: there exists a triangle congruent to $\triangle A B C$ that can cover $M$, and one of its sides lies on a line that is parallel or coincident with a line on which a side of $M$ lies. | Six, first, let's assume that $M$ has three vertices on the sides of $\triangle A B C$ (as shown in Figure 3) or $M$ has one vertex coinciding with a vertex of $\triangle A B C$ (such as point $B$), and another vertex of $M$ is on the opposite side of point $B$ (such as
Assume the initial state is $\angle A C_{1} B_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,233 |
8.1. At noon 12:00, "Western Kid" and "Moscow Kid" start driving from two locations 90 km apart, heading towards each other at constant speeds. 2 hours later, the distance between them is again 90 km. An "uninformed person" claims: before meeting Moscow Kid, Western Kid and after meeting Western Kid, Moscow Kid togethe... | 8.1. Notice that, if after exactly $2 \mathrm{~h}$ their cars are again $90 \mathrm{~km}$ apart, then they must have met exactly after $1 \mathrm{~h}$. This indicates that the cowboy traveled for $1 \mathrm{~h}$ before meeting the Muscovite, and the Muscovite traveled for $1 \mathrm{~h}$ after meeting the cowboy. There... | proof | Other | proof | Yes | Yes | cn_contest | false | 718,234 |
8.2. Two people are playing a game, with a checker piece placed in the middle square of a $1 \times 2005$ strip of squares. They take turns moving the piece. First, Player A moves the piece 1 square in any direction, then Player B moves the piece 2 squares in any direction, then Player A moves 4 squares, then Player B ... | 8.2. Answer: Player A can win without relying on Player B's actions.
Notice that, Player A's movement distances are $1, 4, 16, \cdots, 1024$.
We will prove that: Player A can successfully move a distance of 1024 (this way, Player B loses, as he would need to move the piece 2048 spaces, which is impossible).
Since $1+... | Player A can win without relying on Player B's actions. | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,235 |
8.3. There are 19 cards. Can a non-zero digit be written on each card so that these 19 cards can be arranged to form a 19-digit number divisible by 11? | 8.3. Answer: Yes.
Write a 2 on each of 10 cards, and a 1 on each of the remaining cards. It is well known that a positive decimal integer is divisible by 11 if and only if the difference $S$ between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11. Under al... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,236 |
8.4. Let $\triangle A B C$ be an acute triangle, and the vertices $B, C$ have their symmetric points $B^{\prime}, C^{\prime}$ with respect to the lines $A C, A B$, respectively. Let $P$ be the other intersection point of the circumcircles of $\triangle A B B^{\prime}$ and $\triangle A C C^{\prime}$ except for point $A$... | 8. 4. As shown in Figure 1, let $\mathrm{MO}_{1}$ and $\mathrm{NO}_{2}$ be the perpendicular bisectors of sides $AB$ and $AC$, respectively, where $M$ and $N$ are the midpoints of sides $AB$ and $AC$, and points $O_{1}$ and $O_{2}$ lie on lines $AC$ and $AB$, respectively.
Clearly, the intersection point $O$ of $MO_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,237 |
8.5. Given that the sum of the digits of a positive integer $N$ is 100, and the sum of the digits of $5N$ is 50. Prove: $N$ is even. | 8.5. Let $S(A)$ denote the sum of the digits of the positive integer $A$. By observing the vertical addition of two positive integers $A$ and $B$, we can see that
$$
S(A+B) \leqslant S(A)+S(B),
$$
and the equality holds if and only if no carry occurs.
From the given conditions, we can deduce that no carry occurs in th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,238 |
8.6. In quadrilateral $ABCD$, $\angle A = \angle C$, the angle bisector of $\angle B$ intersects line $AD$ at point $P$, and the line through point $A$ perpendicular to $BP$ intersects line $BC$ at point $Q$. Prove: $PQ \parallel CD$. | 8.6. As shown in Figure 2, let the intersection of $B P$ and $A Q$ be $R$. It is easy to see that $B R$ is both the altitude and the angle bisector of $\triangle A B Q$, thus, this triangle is isosceles, with $A B = B Q$.
Hence, $\triangle B P A \cong \triangle B P Q$.
Then $\angle B Q P = \angle B A P$
$$
= \angle B C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,239 |
8.7. Find all positive integer pairs $(x, y)$ that satisfy the equations $x+y=a^{n}, x^{2}+y^{2}=a^{m}$ for some positive integers $a$, $n$, and $m$. | 8.7. Answer: $\left(2^{k}, 2^{k}\right), k$ is a non-negative integer.
From the problem, we have $a^{2 n}=x^{2}+2 x y+y^{2}=a^{m}+2 x y$. Therefore, $a^{2 n}>a^{m}$. Thus, $a^{2 n}$ is divisible by $a^{m}$.
Hence, $2 x y$ is divisible by $a^{m}=x^{2}+y^{2}$.
This leads to $2 x y \geqslant x^{2}+y^{2} \geqslant 2 x y$.... | (2^{k}, 2^{k}), k \text{ is a non-negative integer} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,240 |
8.8. There are 99 baskets, each containing apples and plums. Prove: It is possible to select 50 baskets such that they contain no less than half of all the apples, and no less than half of all the plums. | 8.8. Solution 1: Let $x_{i}$ represent the number of apples in the $i$-th basket.
Assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{99}$. We just need to divide the 2nd to 99th baskets into two groups, each with 49 baskets, such that the difference in the number of apples between the two groups is no more th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,241 |
Example 8 As shown in Figure 8, given that $A B$ is the diameter of $\odot O$, $B C$ is the tangent of $\odot O$, and $O C$ is parallel to the chord $A D$. A perpendicular line $D E$ is drawn from point $D$ to $A B$ at point $E$, and $A C$ is connected to intersect $D E$ at point $P$. Is $E P$ equal to $P D$? Prove you... | Explanation: Segment $E P$ is equal to $P D$.
Proof: As shown in Figure 8, let the intersection of lines $A D$ and $B C$ be point $F$. Since $A O=O B$ and $A F \parallel O C$, it follows that $B C=C F$. Since $B C$ is a tangent to circle $\odot O$, we have $B F \perp A B$. Given that $D E \perp A B$, it follows that $E... | E P = P D | Geometry | proof | Yes | Yes | cn_contest | false | 718,242 |
1. Simplify $\sqrt[3]{\frac{\sqrt{5}-1}{2}-\left(\frac{\sqrt{5}-1}{2}\right)^{2}}$ to get $(\quad)$.
(A) $\frac{\sqrt{5}-1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\sqrt{5}$
(D) $\sqrt[3]{5}$ | - 1.A.
Let $\frac{\sqrt{5}-1}{2}=a$, then $a^{2}+a-1=0$.
Thus, $a^{3}+a^{2}-a=0$, which means $a^{3}=a-a^{2}$.
Therefore, the original expression $=\sqrt[3]{a-a^{2}}=\sqrt[3]{a^{3}}=a$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,243 |
2. Given that $a$ is a natural number, the equation with respect to $x$
$$
2 x-a \sqrt{1-x}-a+4=0
$$
has at least one integer root. Then the number of possible values for $a$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2.C.
Obviously, $x \leqslant 1$.
Also, $2 x+4=a(\sqrt{1-x}+1) \geqslant 0$, so $x \geqslant-2$. When $x=1,0,-2$, $a$ is $6,2,0$ respectively; when $x=-1$, $a$ is not an integer. Therefore, there are 3 possible values for $a$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,244 |
3. As shown in Figure $1, A B$ is the diameter of semicircle $O$, $A C$ and $A D$ are both chords, and $\angle B A C = \angle C A D$. Therefore, which of the following equations is true? ( ).
(A) $A C^{2}=A B \cdot A D$
(B) $A B \cdot A D>A C^{2}$
(C) $A B+A D=2 A C$
(D) $A B+A D<2 A C$ | 3.D.
As shown in Figure 4, draw $O M \perp A D$ at point $M$, $O M$ intersects $A C$ at point $N$, and connect $O C$. It is easy to see that $O C / / A D$.
Therefore,
$$
\begin{array}{l}
\angle C O N=\angle C O M \\
=\angle O M A=90^{\circ}
\end{array}
$$
It is easy to see that $N C > O C =$
$$
\frac{1}{2} A B, A N >... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,245 |
4. Given the equation $x^{2}+2 p x-q=0$ has no real roots. Then the value of $p+q$ ( ).
(A) is less than $\frac{1}{4}$
(B) is equal to $\frac{1}{4}$
(C) is greater than $\frac{1}{4}$
(D) cannot be determined | 4. A.
From the problem, we have $4 p^{2}+4 q<0$. Therefore, $q<-p^{2}$. Then, $p+q<-p^{2}+p=-\left(p-\frac{1}{2}\right)^{2}+\frac{1}{4} \leqslant \frac{1}{4}$. Hence, $p+q<\frac{1}{4}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,246 |
5. Given that $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are non-negative real numbers, and
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=100 \text {, }
$$
$M$ is the maximum value of $x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}, x_{4}+x_{5}$. Then the minimum value $m$ of $M$ satisfies ( ).
(A) $20<m<30$
(B) $30<m<40$
(C) $40<m<45$
(D) $45<m<5... | 5.B.
According to the problem, we have
$$
M \geqslant x_{1}+x_{2}, M \geqslant x_{2}+x_{3}, M \geqslant x_{4}+x_{5} .
$$
Adding the three inequalities, we get $3 M \geqslant 100+x_{2} \geqslant 100$, which means $M \geqslant \frac{100}{3}$. When $x_{2}=x_{4}=0, x_{1}=x_{3}=x_{5}=\frac{100}{3}$, we have $M=\frac{100}{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,247 |
6. In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $E$. Given $\angle ABD=35^{\circ}, \angle ADB=20^{\circ}$, $\angle ACB=40^{\circ}, \angle ACD=70^{\circ}$. Then $\angle AEB=$ ).
(A) $60^{\circ}$
(B) $65^{\circ}$
(C) $70^{\circ}$
(D) $75^{\circ}$ | 6.D.
As shown in Figure 5, extend $AC$ to point $F$ such that $CF = CD$, and connect $FB$ and $FD$. It is easy to see that
$$
\begin{array}{l}
\angle AFD = \angle CFD \\
= \frac{1}{2} \angle ACD \\
= \frac{1}{2} \times 70^{\circ} = 35^{\circ} \\
= \angle ABD.
\end{array}
$$
Thus, points $A$, $B$, $F$, and $D$ are con... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,248 |
1. Given $A_{1}, A_{2}, \cdots, A_{n}$ are 11 stations sequentially on a straight highway, and
$$
\begin{array}{l}
A_{i} A_{i+2} \leqslant 12(i=1,2, \cdots, 9), \\
A_{i} A_{i+3} \geqslant 17(i=1,2, \cdots, 8) .
\end{array}
$$
If $A_{1} A_{11}=56 \mathrm{~km}$, then $A_{2} A_{7}=$ $\qquad$ $\mathrm{km}$. | Ni, 1.29.
According to the problem, we have $A_{1} A_{10} \geqslant 3 \times 17=51(\mathrm{~km})$, then
$$
A_{10} A_{11} \leqslant 5 \mathrm{~km} \text {; }
$$
Also, $A_{8} A_{11} \geqslant 17 \mathrm{~km}, A_{8} A_{10} \leqslant 12 \mathrm{~km}$, then $A_{10} A_{11} \geqslant 5 \mathrm{~km}$.
Therefore, $A_{10} A_{11... | 29 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,249 |
2. As shown in Figure 2, color the four small triangles using four colors, such that triangles sharing a common side are different colors. Therefore, there are $\qquad$ different ways to color them.
Translate the above text into English, please keep the original text's line breaks and format, and output the translatio... | 2.84.
When $\triangle A O D$ and $\triangle B O C$ are the same color, there are $4 \times 3 \times 3=36$ different coloring methods; when $\triangle A O D$ and $\triangle B O C$ are different colors, there are $4 \times 3 \times 2$ $\times 2=48$ different coloring methods.
Therefore, there are a total of $36+48=84$ d... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 718,250 |
3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$ | 3.8 .
Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$
Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$
which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \tim... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,251 |
4. An isosceles triangle has legs of length $a$ and a base of length $b(a$ $>b)$, and another isosceles triangle has legs of length $b$ and a base of length $a$. If the vertex angles of the two isosceles triangles are supplementary, then $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=$ | 4. $\sqrt{3}$.
Since the apex angles of the two isosceles triangles are supplementary, the base angles of the two isosceles triangles must be complementary.
Obviously, in Figure 6, $\triangle A C D$ is similar to the isosceles triangle with legs of length $a$ and base of length $b$.
Thus, $\frac{C D}{b}=\frac{b}{a}$,... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,252 |
Example 9 As shown in Figure 9, in trapezoid $A B C D$, $A D / / B C$, squares $A B G E$ and $D C H F$ are constructed on the two legs $A B$ and $C D$ respectively, and $E F$ is connected, with the midpoint of line segment $E F$ being $M$. Prove: $M A=M D$. | Explanation: Based on the characteristics of the figure (parallel lines, midpoints, squares, etc.), choose to use the "three lines coincide" method. For this, draw a perpendicular from point $M$ to $AD$, with the foot of the perpendicular being $N$.
To utilize “$M$ is the midpoint of $EF$”, draw perpendiculars from po... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,253 |
One, (20 points) Given $a>b>c, a+b+c=1$, $a^{2}+b^{2}+c^{2}=3$. Prove:
$$
-\frac{2}{3}<b+c<\frac{1}{2} \text {. }
$$ | From the given, we have
$$
b+c=1-a, (b+c)^{2}-2 b c=3-a^{2} \text{. }
$$
Thus, $b c=a^{2}-a-1$.
Therefore, the equation with roots $b$ and $c$ is
$$
x^{2}+(a-1) x+a^{2}-a-1=0 \text{. }
$$
Since $b>c$, we have
$$
\Delta=(a-1)^{2}-4\left(a^{2}-a-1\right)>0 \text{. }
$$
Simplifying, we get $3 a^{2}-2 a-5-\frac{2}{3}$.
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,254 |
II. (25 points) As shown in Figure 3, $A D, B E, C F$ are the three altitudes of acute $\triangle A B C$, and $H$ is the orthocenter. Take the midpoint $O$ of $A H$, and the ray $E O$ intersects $A B$ at point $P, F D$ intersects $B E$ at point $Q$. Prove:
$$
P Q \perp B C .
$$ | II. As shown in Figure 7, connect $DE$, $EF$, and $OF$. It is easy to see that points $A$, $F$, $H$, and $E$ are concyclic, and point $O$ is the center of this circle, i.e., point $O$ is the circumcenter of $\triangle AEF$. Therefore,
$$
\angle EOF = 2 \angle BAC.
$$
It is easy to prove that points $A$, $F$, $D$, $C$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,255 |
1. In a tetrahedron with four vertices of a rectangular prism as its vertices, the number of tetrahedra where all four faces are right-angled triangles is ( ).
(A) 8
(B) 16
(C) 24
(D) 48 | -、1.C.
It is known that there is a one-to-one correspondence between a tetrahedron satisfying the condition and a pair of skew lines among the edges of a rectangular parallelepiped. Since there are $\frac{6 \times 8}{2}=24$ pairs of skew lines among the edges of the rectangular parallelepiped, the number of tetrahedra ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,257 |
$$
\begin{aligned}
A= & \{z \in \mathbf{C} \mid(a+b \mathrm{i}) \bar{z}+(a-b \mathrm{i}) z+2 c \\
& =0, a, b, c \in \mathbf{R}, ab \neq 0\}, \\
B= & \{z \in \mathbf{C} \mid |z| = 1\} .
\end{aligned}
$$
If $A \cap B=\varnothing$, then with $|a|, |b|, |c|$ as the three sides ( ).
(A) must form an acute triangle
(B) must... | 2.D.
Let $z=x+y \mathrm{i} \in A$, then
$$
(a+b \mathrm{i}) \tilde{z}+(a-b \mathrm{i}) z+2 c=0,
$$
we get $(a+b \mathrm{i})(x-y \mathrm{i})+(a-b \mathrm{i})(x+y \mathrm{i})+2 c=0$.
Expanding and simplifying, we obtain $a x+b y+c=0$.
If $z=x+y \mathrm{i} \in B$, by $|z|=1$, we get $x^{2}+y^{2}=1$.
Thus, $A \cap B=\va... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,258 |
3. Given in the tetrahedron $S-ABC$, $SC \perp CB$, $SA \perp AB$, $CB \perp AB$, and the angles formed by $SA$, $SC$ with the plane of $\triangle ABC$ are equal. If $AC=6$, and the distance from $S$ to the plane $ABC$ is 4, then the distance between the skew lines $AC$ and $SB$ is ( ).
(A) $\frac{6 \sqrt{13}}{13}$
(B)... | 3. A.
As shown in Figure 3, draw $SD \perp$ plane $ABC$, with the foot of the perpendicular being $D$. Connect $AD$, $CD$, and $BD$, and let $AC$ intersect $BD$ at point $O$. Since $SC \perp CB$, according to the Three Perpendiculars Theorem, we have $CD \perp BC$. Similarly, $AB \perp AD$. Also, since $CB \perp AB$, ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,259 |
4. Given that $\triangle A B C$ is an equilateral triangle with side length 2, and points $M_{1}, M_{2}, \cdots, M_{n}$ are taken sequentially on segment $B C$, dividing $B C$ into $n+1$ equal parts. Then
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)=(\quad) \text {.... | 4. C.
As shown in Figure 4, draw $B D \perp A C$, with the foot of the perpendicular being $D$. Draw $M_{k} N_{k} \perp B D$ and $M_{k} P_{k} \perp A C$, with the feet of the perpendiculars being $N_{k}$ and $P_{k}$, respectively.
From $\frac{B M_{k}}{B C}=\frac{k}{n+1}$, we get
$$
\begin{array}{c}
M_{k} N_{k}=\frac{k... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,260 |
5. The number of pairs of positive integer solutions $(x, y)$ for the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2006}$ is ( ).
(A) 16
(B) 24
(C) 32
(D) 48 | 5.D.
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2006}$, we get
$x y-2006 x-2006 y-2006=0$.
Rearranging gives
$$
\begin{array}{l}
(x-2006)(y-2006)=2006 \times 2007 \\
=2 \times 3^{2} \times 17 \times 59 \times 223 .
\end{array}
$$
Thus, the positive integer solutions to the original equation are
$$
(1+1)(2+1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,261 |
6. If $a, b \in \mathbf{R}$, then
$$
\begin{aligned}
f(a, b)= & \sqrt{2 a^{2}-8 a+10}+\sqrt{b^{2}-6 b+10}+ \\
& \sqrt{2 a^{2}-2 a b+b^{2}}
\end{aligned}
$$
the minimum value is ( ).
(A) $2 \sqrt{10}$
(B) $1+\sqrt{2}+\sqrt{5}$
(C) $2 \sqrt{5}$
(D) 4 | 6.C.
$$
\begin{aligned}
f(a, b)= & \sqrt{(a-3)^{2}+(a-1)^{2}}+ \\
& \sqrt{(b-3)^{2}+1}+\sqrt{(a-b)^{2}+a^{2}}
\end{aligned}
$$
represents the perimeter of the triangle in the coordinate plane $x O y$ with vertices at points $P(3,1)$, $Q(a, a)$, and $R(b, 0)$. It is easy to see that the trajectories of points $Q$ and $... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,262 |
1. Given that the positive real number $x$ satisfies $\lg ^{2} 207 x<\lg ^{2000} x<$ $\lg ^{2088} x$. Then the range of values for $x$ is $\qquad$ | $$
\text { Ni.1. }\left(0, \frac{1}{10}\right) \text {. }
$$
Let $a=\lg x$, then $a \in \mathbf{R}, a^{2007}<1$.
Thus, $a<-1$, that is, $\lg x<-1$. Therefore, $0<x<\frac{1}{10}$. | \left(0, \frac{1}{10}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,263 |
1. As shown in Figure 10, in the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $N$. Point $M$ is on diagonal $BD$, and it satisfies $\angle BAM = \angle DAN$, $\angle BCM = \angle DCN$. Prove that $M$ is the midpoint of $BD$. | (提示: 如图 10, 设直
line $A M$ intersects the circumcircle of quadrilateral $A B C D$ at another point $P$. From $\angle B C P=\angle B A M=\angle D A N=\angle D B C$, we know that $C P / / D B$. Therefore, quadrilateral $B P C D$ is an isosceles trapezoid. It follows that $M$ is a point on the perpendicular bisector of $D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,264 |
2. The number of five-digit numbers where the sum of any two adjacent digits is divisible by 3 is $\qquad$ .
| 2.1254.
Classify the digits by their remainders, then the five-digit number $\overline{a b c d e}$ that meets the condition satisfies
$$
a \equiv c=e \equiv 2(\bmod 3), b \equiv d \equiv 1(\bmod 3),
$$
or $a=c=e=1(\bmod 3), b=d=2(\bmod 3)$,
or $a=b=c=d=e=0(\bmod 3)$.
If it is equation (1), then there are $3^{5}$ such... | 1254 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,265 |
3. Let the chord $P Q$ of the parabola $y^{2}=x$ be perpendicularly bisected by the line $l: y=k(x-1)+1(k \in \mathbf{Z})$. Then the length of the chord $P Q$ is $\qquad$ . | 3. $\sqrt{10}$.
Let the equation of line $P Q$ be $y=-\frac{1}{k} x+b$ (obviously $k \neq 0$, otherwise, $l$ cannot perpendicularly bisect $P Q$).
From $\left\{\begin{array}{l}y^{2}=x, \\ y=-\frac{1}{k} x+b,\end{array}\right.$ eliminating $x$ and rearranging gives
$$
y^{2}+k y-b k=0 \text {. }
$$
Since $P Q$ intersec... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,266 |
4. Given the function
$$
f(x)=2^{2 x+1} \cos ^{2} \frac{\theta}{2}-2 \sqrt{2} \times 6^{x}+9^{x}(\sin \theta+1)
$$
is always positive on $\mathbf{R}$. Then the range of the real number $\theta$ is
$\qquad$ | 4. $\left(2 k \pi, 2 k \pi+\frac{\pi}{2}\right)(k \in \mathbf{Z})$.
The given condition is equivalent to the inequality
$$
4^{x}(\cos \theta+1)+9^{x}(\sin \theta+1)>2 \sqrt{2} \times 6^{x}
$$
which holds.
Dividing both sides of the above inequality by $6^{x}$, we get
$$
\left(\frac{2}{3}\right)^{x}(\cos \theta+1)+\le... | \left(2 k \pi, 2 k \pi+\frac{\pi}{2}\right)(k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,267 |
5. Given non-negative real numbers $u, v, w$ satisfy $u+v+w=2$. Then the range of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is $\qquad$ - | 5. $[0,1]$.
Obviously, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2} \geqslant 0$.
If we take $u=2, v=w=0$, then the equality holds. Therefore, the minimum value of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is 0.
Assume without loss of generality that $u=\max \{u, v, w\}$.
Thus, $v w \leqslant u^{2}<2 u^{2}$. Therefore,
$$
\begin{... | [0,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,268 |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=6, a_{2}=20, a_{n}\left(a_{n}-8\right)=a_{n-1} a_{n+1}-12(n \geqslant 2) \text {. }
$$
Let $\{x\}=x-[x]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}=$ $\qquad$ | 6. $\frac{1}{2}$.
According to the recursive relation and initial value, it is easy to calculate $a_{3}=42$.
From $a_{n}\left(a_{n}-8\right)=a_{n-1} a_{n+1}-12$,
we get
$$
a_{n+1}\left(a_{n+1}-8\right)=a_{n} a_{n+2}-12 \text {. }
$$
Subtracting (1) from (2) gives
$$
a_{n+1}^{2}-a_{n}^{2}-8\left(a_{n+1}-a_{n}\right)=a_... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,269 |
Three. (20 points) As shown in Figure 1, the center of the regular hexagon $A B C D E F$ is $O$. For any two points among $A, B, C, D, E, F, O$, a vector is drawn from one point to another. If two vectors are randomly selected, the absolute value of their dot product is taken as the random variable $\xi$. Try to find t... | Three, the number of vectors generated is $\mathrm{C}_{7}^{2}=21$, then we can take $\mathrm{C}_{21}^{2}=210$ pairs of vectors. Let the vectors taken be $a, b$. Since $\xi=|a \cdot b|=|a| \cdot|b| \cdot|\cos \langle a, b\rangle|$, therefore, the direction of the vectors can be ignored. Without loss of generality, let t... | \frac{11}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,270 |
Four. (20 points) Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with its upper vertex $B$ and right focus $F$, a line $l$ intersects the ellipse at points $M$ and $N$. Determine whether there exists a line $l$ such that point $F$ is the
(1) centroid; (2) orthocenter of $\triangle B M N$.
If it exists, find the equation o... | (1) Assume there exists a line $l$ that satisfies the condition, making point $F$ the centroid of $\triangle B M N$.
Let $M(\sqrt{2} \cos \alpha, \sin \alpha), N(\sqrt{2} \cos \beta, \sin \beta)$.
According to the problem, we have
$$
\left\{\begin{array}{l}
\sqrt{2} \cos \alpha+\sqrt{2} \cos \beta+0=3 \times 1, \\
\sin... | y=x-\frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,271 |
Five. (20 points) In an acute triangle $\triangle ABC$, it is known that
$$
\begin{array}{l}
\left(\sqrt{\frac{\sin B}{\sin A}}+\sqrt{\frac{\sin C}{\sin A}}\right) \cos A+ \\
\left(\sqrt{\frac{\sin C}{\sin B}}+\sqrt{\frac{\sin A}{\sin B}}\right) \cos B+ \\
\left(\sqrt{\frac{\sin A}{\sin C}}+\sqrt{\frac{\sin B}{\sin C}}... | Given that $\triangle A B C$ is an equilateral triangle.
First, we prove:
$$
\sqrt{\frac{\sin B}{\sin A}} \cos A+\sqrt{\frac{\sin A}{\sin B}} \cos B \geqslant \cos A+\cos B,
$$
with equality holding if and only if $\angle A=\angle B$.
$$
\begin{array}{l}
\sqrt{\frac{\sin B}{\sin A}} \cos A+\sqrt{\frac{\sin A}{\sin B}}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,272 |
One, (50 points) As shown in Figure 2, from a point $P$ outside the circle $\odot O$, draw two tangents and one secant to $\odot O$. The points of tangency are $A$ and $B$, and the secant intersects $\odot O$ at points $C$ and $D$. Draw $D E \perp P A$ and $D F \perp P B$, with the feet of the perpendiculars being $E$ ... | As shown in Figure 6, connect $E F$. Since $D E \perp P A$ and $D F \perp P B$, we know that points $P, E, D, F$ are concyclic. According to Ptolemy's theorem, we have
$$
\begin{array}{l}
D E \cdot P F + D F \cdot P E \\
= E F \cdot P D .
\end{array}
$$
Draw $O M \perp C D$, with the foot of the perpendicular at $M$. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,273 |
II. (50 points) Find the largest real number $m$ such that the inequality
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+m \leqslant \frac{1+x}{1+y}+\frac{1+y}{1+z}+\frac{1+z}{1+x}
$$
holds for any positive real numbers $x$, $y$, $z$ satisfying $x y z=x+y+z+2$. | Let $x=y=z=2$, we get $m \leqslant \frac{3}{2}$.
Below, we prove that the maximum value of $m$ is $\frac{3}{2}$.
Let $x=\frac{1}{a}-1, y=\frac{1}{b}-1, z=\frac{1}{c}-1$, then we have
$$
\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 \text {. }
$$
Re... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,274 |
Example 1 Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $S$, and the projections of $S$ onto sides $AB$ and $CD$ are $E$ and $F$ respectively. Prove that the perpendicular bisector of $EF$ bisects the segments $BC$ and $AD$.
(2003, German National Mathematics Competition (Second... | Proof: Auxiliary lines as shown in Figure 1. It is sufficient to prove that $E M = F M$, where $M$ is the midpoint of $A D$. Applying the median formula in $\triangle A E D$ and $\triangle A F D$, we have $A E^{2} + D E^{2}$
Figure 1 $= 2\left(A M^{2} + E M^{2}\right)$, $A F^{2} + D F^{2} = 2\left(A M^{2} + F M^{2}\ri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,275 |
2 Euler's Theorem
Theorem Let $M, N$ be the midpoints of the diagonals $A C, B D$ of quadrilateral $A B C D$. Then
$$
\begin{array}{l}
A B^{2}+B C^{2}+C D^{2}+D A^{2} \\
=A C^{2}+B D^{2}+4 M N^{2}
\end{array}
$$ | Proof: As shown in Figure 2, connect $B M$ and $D M$. By the median formula, we have
$$
\begin{array}{l}
A B^{2}+B C^{2} \\
=2\left(B M^{2}+A M^{2}\right), \\
D A^{2}+C D^{2} \\
=2\left(D M^{2}+A M^{2}\right).
\end{array}
$$
$$
\begin{array}{l}
\text { Also, } B M^{2}+D M^{2}=2\left(B N^{2}+M N^{2}\right), \\
4 A M^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,276 |
3. Prove: $t_{a}+t_{b}+t_{c} \leqslant \sqrt{3} p$. | (Hint: $t_{a}=\frac{2}{b+c} \sqrt{b c p(p-a)} \leqslant \sqrt{p(p-a)}$.
Similarly, $t_{b} \leqslant \sqrt{p(p-b)}, t_{c} \leqslant \sqrt{p(p-c)}$.
$$
\begin{array}{l}
t_{a}+t_{b}+t_{c} \\
\leqslant \sqrt{p(p-a)}-\sqrt{p(p-b)}+\sqrt{p(p-c)} \\
\leqslant \sqrt{(p+p+p)(p-a+p-b+p-c)}=\sqrt{3} p .)
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,277 |
3. As shown in Figure 2, in tetrahedron $ABCD$, $\triangle ADB$ is an isosceles right triangle, $\angle ADB=$ $90^{\circ}, AD=1$, and $\angle BDC=$ $\angle ADC=60^{\circ}$. Then the distance between the skew lines $AB$ and $CD$ is | 3. $\frac{1}{2}$.
$$
\begin{array}{l}
A B \cdot C D=(A D+D B) \cdot C D=A D \cdot C D+D B \cdot C D \\
=|A D| \cdot|C D| \cos \langle A D, C D\rangle+ \\
|D B| \cdot|C D| \cos \langle D B, C D\rangle \\
= C D \cos 60^{\circ}+C D \cos 120^{\circ}=0 .
\end{array}
$$
Thus, $\boldsymbol{A B} \perp \boldsymbol{C D}$. Ther... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,278 |
4. Non-zero sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$ satisfy
$$
a_{n+1}=a_{n}+3 b_{n}, b_{n+1}=a_{n}+b_{n} \text {. }
$$
Let $x_{n}=\frac{a_{n}}{b_{n}}$. When $\left|x_{2000}+x_{2000}\right|$ takes the minimum value, the value of $x_{2000}$ is $\qquad$ . | $4.3 \pm \sqrt{2}$.
From $x_{n+1}=\frac{a_{n+1}}{b_{n+1}}=\frac{a_{n}+3 b_{n}}{a_{n}+b_{n}}=\frac{x_{n}+3}{x_{n}+1}=1+\frac{2}{x_{n}+1}$,
we know that $\dot{x}_{2006}=1+\frac{2}{x_{200}+1}$.
Therefore, $\left|x_{2005}+x_{2006}\right|=\left|1+x_{200}+\frac{2}{1+x_{200}}\right| \geqslant 2 \sqrt{2}$.
The equality holds i... | 3 \mp \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,279 |
5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$ | 5.10.
From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$
It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$
If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,280 |
6. In $\triangle A B C$, the lengths of the three sides are $a, b, c$, and $b>\max \{a, c\}$. There are 3 non-zero real numbers $x_{0}, y_{0}, z_{0}$, satisfying that the line $a x+b y+c=0$ passes through the point $\left(\frac{z_{0}}{x_{0}}, \frac{2 y_{0}}{x_{0}}\right)$, and the point $\left(\frac{z_{0}}{y_{0}}, \fra... | 6. $\frac{5}{3}$.
According to the problem,
$$
\begin{array}{l}
a z_{0}+2 b y_{0}+c x_{0}=0, \\
4 z_{0}^{2}+x_{0}^{2}=4 y_{0}^{2} .
\end{array}
$$
From equation (1), we get $4 b^{2} y_{0}^{2}=\left(a z_{0}+c x_{0}\right)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(x_{0}^{2}+4 z_{0}^{2}\r... | \frac{5}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,281 |
Three, (20 points) Let three positive numbers $x, y, z$ satisfy $3^{x}=4^{y}=6^{x}$.
Prove: $5^{x-2}+25^{y-1}+5^{\frac{1-2 z}{z}}>1$.
| To prove $5^{x-2}+25^{y-1}+5^{\frac{1-2 z}{x}}>1$, it suffices to prove
$$
5^{x}+25^{y}+5^{\frac{1}{x}}>25 \text {. }
$$
Let $3^{x}=4^{y}=6^{z}=k(k>0)$, then
$$
x=\log _{3} k, y=\log _{4} k, z=\log _{6} k \text {. }
$$
Thus, $\frac{2}{z}=\log _{k} 36=\log _{k} 4+\log _{k} 9=\frac{1}{y}+\frac{2}{x}$.
Therefore, $\frac... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,282 |
Four. (20 points) Let $A\left(x_{0}, y_{0}\right)$ be a point on the ellipse $C_{1}$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
Points $B$, $C$, and $D$ are the reflections of point $A$ across the $y$-axis, the origin, and the $x$-axis, respectively. Point $E$ is a point on the ellipse $C_{1}$ such that $... | (1) From $A\left(x_{0}, y_{0}\right)$, we have
$$
B\left(-x_{0}, y_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(x_{0},-y_{0}\right).
$$
Then $l_{B D}: y=-\frac{y_{0}}{x_{0}} x$.
Let $E\left(x_{1}, y_{1}\right)$. Since points $A$ and $E$ are both on the ellipse $C_{1}$, we have
$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,283 |
$$
\begin{array}{l}
\text { Five. (20 points) Let } k=2\left(b^{2}+a^{2} d-a b c\right), \\
y_{0}=a^{2}, y_{1}=b^{2}, \\
y_{n+1}=\left(c^{2}-2 d\right) y_{n}-d^{2} y_{n-1}+k d^{n},
\end{array}
$$
where $a, b, c, d$ are all positive integers. Prove that for any $n \in \mathbf{N}, y_{n}$ is a perfect square. | Because $y_{0}=a^{2}, y_{1}=b^{2}$, so,
$$
\begin{array}{l}
y_{2}=\left(c^{2}-2 d\right) y_{1}-d^{2} y_{0}+k d \\
=\left(c^{2}-2 d\right) b^{2}-d^{2} a^{2}+2 d\left(b^{2}+a^{2} d-a b c\right) \\
=(b c-a d)^{2} .
\end{array}
$$
Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{n+1}=c x_{n}-d x_{n-1}\left(n \in \mat... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,284 |
(50 points) As shown in Figure 3, let quadrilateral $A E F G$ be an inscribed quadrilateral of $\odot O$. Extend $A E$ and $G F$ to intersect at point $B$, extend $E F$ and $A G$ to intersect at point $C$, and let $D$ be the midpoint of $B C$. Connect $A D$ to intersect the circle at point $H$.
(1) Prove: $A H \cdot B ... | 1. Extend $A D$ to point $P$ such that $A D = D P$.
From $C D = B D$, it follows that quadrilateral $A C P B$ is a parallelogram.
Therefore,
$$
\begin{array}{l}
\angle B F C = \angle G F E = 180^{\circ} - \angle G A E, \\
\angle G A E = \angle C P B. \\
\text { Thus, } \angle B F C + \angle C P B = 180^{\circ}.
\end{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,285 |
II. (50 points) Given $n(n \geqslant 2)$ real-coefficient quadratic functions $f_{i}(x)=x^{2}+a_{i} x+b_{i}(1 \leqslant i \leqslant n)$ whose discriminants $\Delta_{i}$ are all equal. If for any $1 \leqslant i<j \leqslant n$, the equation $f_{i}(x)+f_{j}(x)=0$ has two distinct real roots, prove that the equation $\sum_... | Let $k=\Delta_{i}=a_{i}^{2}-4 b_{i}$.
Since for any two equations $f_{i}(x)+f_{j}(x)=0$ there are two distinct real roots, then
$$
\begin{array}{l}
f_{i}(x)+f_{j}(x)=2 x^{2}+\left(a_{i}+a_{j}\right) x+\left(b_{i}+b_{j}\right) . \\
\text { By } \Delta_{i}=\left(a_{i}+a_{j}\right)^{2}-4 \times 2\left(b_{i}+b_{j}\right) \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,286 |
Three. (50 points) Given a positive integer $m$, the set
$$
M=\{1,2, \cdots, m\}, A=\left\{t_{1}, t_{2}, \cdots, t_{k}\right\},
$$
and $A \subseteq M$, where $t_{i}$ satisfies $2^{t_{i}}-t_{i}^{2}$ is a multiple of 7 $(1 \leqslant i \leqslant k, i \in \mathbf{N})$. Find $k_{\max }$. | Three, let $f(n)=2^{n}-n^{2}$. First, consider the $n$ that satisfies $7 \mid f(n)$. Note that $2^{3 c} \equiv 1(\bmod 7)(c \in \mathbf{N})$ and
$$
2^{3 c+1} \equiv 2(\bmod 7), 2^{3 c+2} \equiv 4(\bmod 7) \text {. }
$$
(1) When $n=7 a$,
$$
2^{7 a}-(7 a)^{2} \equiv 2^{7 a}=0(\bmod 7) \text {. }
$$
This is impossible.
(... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,287 |
4. Let the circumradius and inradius of $\triangle ABC$ be $R$ and $r$, respectively, and the circumcenter, incenter, and centroid be $O$, $I$, and $G$, respectively. Prove:
$$
R^{2}-O G^{2} \geqslant 4\left(r^{2}-I G^{2}\right) .
$$ | (Hint: From $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, we have
$$
\begin{array}{l}
r^{2}-I G^{2}=\frac{1}{36}\left[6(a b+b c+c a)-5\left(a^{2}+b^{2}+c^{2}\right)\right] \\
\left.\leqslant \frac{1}{36}\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4}\left(R^{2}-O G^{2}\right) .\right)
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,288 |
Find all integers $x$ such that $7 x^{5}+x^{3}-x$ +2 is divisible by 120.
The original text has been translated into English while preserving the original line breaks and format. | Solution: Let $f(x)=7 x^{5}+x^{3}-x+2$.
(1) $120=2^{3} \times 3 \times 5$, where $2^{3}, 3, 5$ are pairwise coprime.
(2) $3|f(x) \Leftrightarrow 3|\left(x^{5}+x^{3}-x-1\right)$.
Obviously, $3 \mid x(x-1)(x+1)=x^{3}-x$, so
$$
\begin{array}{l}
3|f(x) \Leftrightarrow 3|\left(x^{5}-1\right) \\
\Leftrightarrow 3 \mid\left[... | x=120k+34(k \in \mathbb{Z}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,289 |
In $\triangle A B C$, points $D, E, F$ are on sides $B C, C A, A B$ respectively, and $A D, B E, C F$ intersect at point $P$. It is known that $\triangle A B C$ is divided into six smaller triangles by $A D, B E, C F$, and the areas of three of these smaller triangles are $6, 4, 5$ respectively. Find the values of the ... | Solution: Notice that
$$
\begin{array}{l}
1=\frac{S_{\triangle A B P}}{S_{\triangle B D P}} \cdot \frac{S_{\triangle B D P}}{S_{\triangle B C P}} \cdot \frac{S_{\triangle B C P}}{S_{\triangle A B P}}=\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A} \\
=\frac{S_{\triangle A P}}{S_{\triangle C P D}} \cdot \fra... | x=15, y=20, z=10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,290 |
199 Given the sides of $\triangle A B C$ are $a, b, c$ which are all integers, divide each side into $n(n>1)$ equal segments. Let the sum of the squares of the distances from all these division points to their opposite vertices be denoted as $S$. Does there exist a positive integer $n(n>1)$ such that for any positive i... | Solution: As shown in Figure 4, let the $n$ equal division points on side $BC$ be $D_{1}$,
$$
D_{2}, \cdots, D_{n-1}, a_{1}=A D_{1}, a_{2}=A D_{2}, \cdots \cdots a_{n-1}=A D_{n-1};
$$
the $n$ equal division points on side $CA$ be $E_{1}, E_{2}, \cdots, E_{n-1}$,
$$
b_{1}=B E_{1}, b_{2}=B E_{2}, \cdots \cdots b_{n-1}=B... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,291 |
Example 1 Given $\triangle A B C$ is inscribed in a circle, $A B>A C$, and a tangent line $l$ is drawn through point $A$. A circle centered at $A$ with radius $A C$ intersects $l$ at $E$ and $F$, and intersects $A B$ at $D$. Prove that $D E$ and $D F$ pass through the incenter and an excenter of $\triangle A B C$ respe... | As shown in Figure 4, draw the angle bisector of $\angle B A C$ intersecting $D E$ and $D F$ at points $I$ and $K$. Connect $C E$ and $C I$.
Since $\angle I E C$
$$
\begin{array}{l}
=\angle D E C \\
=\frac{1}{2} \angle D A C \\
=\frac{1}{2} \angle B A C \\
=\angle I A C,
\end{array}
$$
Therefore, points $A, I, C, E$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,293 |
Example 2 In the acute triangle $\triangle ABC$, $AB \neq AC$, and $\cos B + \cos C = 1$. Points $E$ and $F$ are on the rays $AC$ and $AB$, respectively, and satisfy $\angle ABE = \angle ACF = 90^{\circ}$. Prove: $EF$ passes through one of the excenters of $\triangle ABC$. | Explanation: As shown in Figure 5, it is easy to see that points $F, E, C, B$ are concyclic. Therefore, we have
$$
\begin{array}{l}
\angle A E F=\angle A B C, \\
\angle A F E=\angle A C B .
\end{array}
$$
At this point,
$$
\begin{array}{l}
\cos \angle A E F+\cos \angle A F E \\
=\cos B+\cos C=1,
\end{array}
$$
which ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,294 |
Example 3 In $\square A B C D$, $M$ and $N$ are the excenters of $\triangle A B C$ and $\triangle A D C$ respectively. Prove:
$$
\angle A M C=\angle A N C .
$$ | As shown in Figure 6, let the incenter of $\triangle ABC$ and $\triangle ADC$ be $E$ and $F$ respectively, then $E$ lies on $AM$, and $F$ lies on $AN$. It is easy to see that
$$
\begin{array}{l}
\angle E C M=90^{\circ} \\
=\angle F C N .
\end{array}
$$
By the properties of the incenter, we have
$$
\begin{array}{l}
\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,295 |
Example 4 As shown in Figure 7, in $\triangle A B C$, $B C: C A: A B$ $=5: 4: 3$. The three excenters $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ are located inside $\angle A$, $\angle B$, and $\angle C$, respectively. Find the ratio of the sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$. | For simplicity, let
$BC=5, CA=4, AB=3$, then $\triangle ABC$ is a right triangle, $\angle BAC=90^{\circ}$.
By property 3 of the excenter, $B^{\prime}, A, C^{\prime}$, $C^{\prime}, B, A^{\prime}$, and $A^{\prime}, C, B^{\prime}$ are collinear, and $A^{\prime}A, B^{\prime}B, C^{\prime}C$ are the three altitudes of $\tri... | A^{\prime}B^{\prime}: A^{\prime}C^{\prime}: B^{\prime}C^{\prime}=\sqrt{9}: \sqrt{8}: \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,296 |
Example 5 In an acute triangle $\triangle ABC$, $AD$ is the altitude, $I$ and $O$ are the incenter and circumcenter respectively, and points $D$, $I$, and $O$ are collinear. Prove that the circumradius of $\triangle ABC$ is equal to the radius of the excircle tangent to side $BC$.
(1998, National High School Mathematic... | As shown in Figure 8, let the center of the excircle tangent to side $BC$ be $I_{A}$. Clearly, points $A$, $I$, and $I_{A}$ are collinear.
Draw $I E \perp A B$ at point $E$, and $I_{A} F \perp A B$ at point $F$. Let $I E = r$, $I_{A} F = r_{a}$, $BC = a$, $CA = b$, $AB = c$, and $p$ be the semiperimeter of $\triangle ... | r_{a} = R | Geometry | proof | Yes | Yes | cn_contest | false | 718,297 |
Example 6 As shown in Figure 9, in the convex quadrilateral $ABCD$, $AB = AC = BD$. Among its four interior angles, two are acute angles, with degree measures of $72^{\circ}$ and $66^{\circ}$. Find the degree measures of the other two interior angles. | Obviously, $\triangle A B C$ and $\triangle B A D$ are both isosceles triangles. Since the base angles of an isosceles triangle are acute, we know that $\angle B A D$ and $\angle A B C$ are both acute angles. Let's assume $\angle B A D = 72^{\circ}$ and $\angle A B C = 66^{\circ}$. At this point,
$$
\begin{array}{l}
\a... | 96^{\circ} \text{ and } 126^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,298 |
Example 7 As shown in Figure 10, in a plane, two lines $l_{1} / / l_{2}$, the distance between them is equal to $a$.
A square cardboard $A B C D$ with side length also equal to $a$ is now placed flat on the two parallel lines, such that $l_{1}$ intersects with $A B$ and $A D$, the intersection points being $E$ and $F; ... | As shown in Figure 10, connect $E H$ and $F G$ to get the intersection point $O$. Since point $H$ is equidistant from $A B$ and $l_{1}$, $E H$ bisects $\angle B E F$ and also bisects $\angle D H G$.
Since point $G$ is equidistant from $A D$ and $l_{1}$, $F G$ bisects $\angle D F E$ and also bisects $\angle B G H$.
Fr... | 2a | Geometry | proof | Yes | Yes | cn_contest | false | 718,299 |
Example 8 In $\triangle A B C$, there are points $A^{\prime}, B^{\prime}, C^{\prime}$ on sides $B C, C A, A B$ respectively, and they satisfy $\angle B A^{\prime} C^{\prime}=\angle C A^{\prime} B^{\prime}$, $\angle C B^{\prime} A^{\prime}=\angle A B^{\prime} C^{\prime}$, $\angle A C^{\prime} B^{\prime}=\angle B C^{\pri... | Explanation: As shown in Figure 11, extend \( C' A' \) and \( C' B' \) to \( X \) and \( Y \) respectively. Notice that
\[
\begin{array}{l}
\angle C A' B' \\
=\angle B A' C' \\
=\angle C A' X, \\
\angle C B' A' \\
=\angle A B' C' = \angle C B' Y.
\end{array}
\]
Therefore, \( C \) is an excenter of \( \triangle A' B' C... | \frac{3}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 718,300 |
Example 2: With point $C$ as the center of rotation, rotate $\triangle A B C$ to $\triangle A^{\prime} B^{\prime} C$. Let the midpoints of segments $B A^{\prime}$, $A C$, and $B^{\prime} C$ be $M$, $E$, and $F$ respectively. If $A C \neq B C$ and $E M = F M$, find the degree measure of $\angle E M F$.
(2005, Bulgarian ... | Proof: As shown in Figure 3, connect \( A A^{\prime} \) and \( B B^{\prime} \). Let
\[
\begin{array}{l}
B C = B^{\prime} C = a, \\
A C = A^{\prime} C = b, \\
A B = A^{\prime} B^{\prime} = c, \\
\angle A C B = \gamma.
\end{array}
\]
Since \(\triangle A A^{\prime} C \sim \triangle B B^{\prime} C\), we have
\[
\begin{arr... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,301 |
1. $\odot O_{1}$ intersects $\odot O_{2}$ at points $P, Q, \angle O_{1} P O_{2}<90^{\circ}$, the circle passing through $O_{1}, O_{2}, P$ intersects $\odot O_{1}, \odot O_{2}$ at points $A, B$. Prove: $Q$ is the excenter of $\triangle A B P$. | ( Hint: Try to prove that $A$, $O_{2}$, $Q$ and $B$, $O_{1}$, $Q$ are collinear respectively. ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,302 |
2. Given $AB, AC$ are tangent to $\odot O$ at points $B, C$, $OA$ intersects $BC$ at point $M$, and through $M$ another chord $EF$ of $\odot O$ is drawn. Prove: $\triangle ABC, \triangle AEF$ have a common excenter. | (提示: 设直线 $O A$ 交 $\odot O$ 于点 $X 、 Y, X$ 在 $\triangle A B C$的内部.易知 $A 、 B 、 O 、 C$ 四点共圆导致 $A 、 E 、 O 、 F$四点共圆. 先证 $X$ 为 $\triangle A B C 、 \triangle A E F$ 的公共内心, 再证 $Y$ 为两个三角形的公共旁心.)
(Hint: Let the line $O A$ intersect $\odot O$ at points $X, Y$, with $X$ inside $\triangle A B C$. It is easy to see that $A, B, O, C$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,303 |
3. Given that $\odot O$ is the incircle of the convex quadrilateral $ABCD$, extend $AB$ and $DC$ to meet at point $E$, and extend $AD$ and $BC$ to meet at point $F$. Prove:
$$
\angle BOE = \angle DOF.
$$ | (Let the incenter of $\triangle B C E$ and $\triangle C D F$ be $I_{1}$ and $I_{2}$ respectively, with $I_{1}$ on $O E$ and $I_{2}$ on $O F$. Use the property 1 of the excenter to prove that $O$, $B$, $I_{1}$, $C$ and $O$, $C$, $I_{2}$, $D$ are concyclic respectively.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,304 |
4. Given that the three excenters of $\triangle A B C$ are $X, Y, Z$ (located inside $\angle A, \angle B, \angle C$ respectively). Let $X X^{\prime} \perp B C, Y Y^{\{\prime} \perp C A, Z Z^{\prime} \perp A B$. Prove: $X X^{\prime}, Y Y^{\prime}, Z Z^{\prime}$ are concurrent. | (Tip: Using the property 3 of the excenter, points $A, B, C$ are on the three sides of $\triangle XYZ$, $XA, YB, ZC$ are the three altitudes. It is easy to prove $\angle X'XY = \angle ZXA$. Draw $YK \perp XY$, point $K$ is on $XX'$. Try to prove that $X, Y, K, Z$ are concyclic, and $XK$ is the diameter of the circumcir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,305 |
5. Given $\triangle A B C$, point $D$ is on side $B C$, $O 、 O_{1} 、 O_{2}$ are the excenters of $\triangle A B C 、 \triangle A B D 、 \triangle A C D$ with respect to $\angle A$, and $O E$ $\perp B C$ at point $E$. Prove: $E O_{1} \perp E O_{2}$. | (Draw $O_{1} M \perp B C$ at point $M, O_{2} N \perp B C$ at point $N$. Using the property 2 of the excenter, prove: $M D=E N, M E=D N$. Let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $r_{1}$ and $r_{2}$, respectively. It is easy to prove that $D M \cdot D N=$ $r_{1} r_{2}$, then prove $E M \cdot E N=r_{1} r_{2}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,306 |
1 Learn to "set limits on group formation"
Example 1 Given $x_{1}, x_{2}, x_{3} > 0$, and $x_{1} + x_{2} + x_{3} = 1$. Prove: $\sum_{i=1}^{3} \sqrt[3]{\frac{1-x_{i}}{1+x_{i}}} > 2$, and 2 is the best. | Proof: Let $S(x)=\sqrt[3]{\frac{1-x}{1+x}}=\left(\frac{2}{1+x}-1\right)^{\frac{1}{3}}$ $(0<x<1)$. We need to prove that $S(x) > 2x - 1$.
Since $0 < x < 1$, we have $0 < \frac{2}{1+x} - 1 < 1$. Therefore, $\left(\frac{2}{1+x} - 1\right)^{\frac{1}{3}} > 2x - 1$.
To prove this, we can consider the function $f(x) = \left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,307 |
2 Proficient in two-handed adjustments
Example 2 Given $x, y, z > 0$, and $x+y+z=1$. Prove: $\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1} \leqslant \frac{3 \sqrt{3}}{7}$. | Proof: Let $S(x)=\frac{\sqrt{x}}{4 x+1}(0<x \leqslant b, b>0)$, and $\frac{1}{2}b+\frac{1}{4}>\frac{3}{4} \Rightarrow 0<b<1$. If $x, y, z \in(0, b]$, the proof is complete. Otherwise, there are two cases:
(1) If $x, y \in(0, b], z \in(b, 1)$, then $x+y+z \leqslant \frac{3}{4} \Rightarrow x+y \leqslant \frac{3}{4}-z<\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,308 |
3 Use the definition of intervals flexibly
Example 3 Let $A, B, C$ be the three interior angles of an acute $\triangle ABC$. Prove that: $\sum \frac{\cos ^{2} A}{1+\cos A} \geqslant \frac{1}{2}$. | Proof: Let \( S(A) = \frac{\cos^2 A}{1 + \cos A} \)
\[
= \frac{1}{1 + \cos A} + \cos A - 1 \text{, }
\]
where \( 0 < A < \frac{\pi}{2} \). Then \( S''(A) \) has a unique zero point \( \varphi \) satisfying \( \frac{\pi}{4} < \varphi < \frac{\pi}{2} \) (incorrect, should be \( \angle B \) increases to \( \varphi \) fir... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,309 |
Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-negative real numbers, not all zero, and $r_{1}, r_{2}, \cdots, r_{n}$ be non-negative real numbers such that the inequality
$$
\sum_{k=1}^{n} r_{k}\left(x_{k}-a_{k}\right) \leqslant\left(\sum_{k=1}^{n} x_{k}^{2}\right)^{\frac{1}{2}}-\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{... | Solution: Let $x_{1}=t \neq a_{1}, x_{2}=a_{2}, \cdots, x_{n}=a_{n}$.
From equation (1), we have
$r_{1}\left(t-a_{1}\right)$
$\leqslant\left(t^{2}+\sum_{k=2}^{n} a_{k}^{2}\right)^{\frac{1}{2}}-\left(a_{1}^{2}+\sum_{k=2}^{n} a_{k}^{2}\right)^{\frac{1}{2}}$.
Let $b^{2}=\sum_{k=2}^{n} a_{k}^{2}, f(t)=\left(t^{2}+b^{2}\rig... | r_{i}=a_{i}\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{-\frac{1}{2}}(i=1,2, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,310 |
Example 5 Prove: In $\triangle A B C$,
$$
\begin{array}{l}
\cot A+\cot B+\cot C \\
\geqslant \frac{1}{3}\left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right)
\end{array}
$$ | Prove: Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, we need to prove
$$
\frac{1-x^{2}}{2 x}+\frac{1-y^{2}}{2 y}+\frac{1-z^{2}}{2 z}
$$
$$
\geqslant \frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)
$$
$$
\Leftrightarrow \frac{1}{6}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geqslant... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,311 |
3 Carnot's Theorem
Theorem If $G$ is the centroid of $\triangle A B C$, and $P$ is any point on the plane of $\triangle A B C$. Then
$$
\begin{array}{l}
P A^{2}+P B^{2}+P C^{2} \\
=G A^{2}+G B^{2}+G C^{2}+3 P G^{2} \\
=\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)+3 P G^{2} .
\end{array}
$$
where the last equality is know... | Proof: Let the midpoint of $BC$ be $M$, and connect $AM$, $PM$. Let the trisection points of $AM$ be $N$, $G$. Then $G$ is the centroid of $\triangle ABC$. By the median formula, we have
$$
\begin{array}{l}
P B^{2}+P C^{2}=2\left(P M^{2}+B M^{2}\right), \\
P A^{2}+P G^{2}=2\left(P N^{2}+N G^{2}\right), \\
P M^{2}+P N^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,312 |
Example 6 (Finsler-Hadwiger Inequality) The side lengths of $\triangle ABC$ are $a, b, c$, and the area is $S$. Prove:
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2} \\
\geqslant 4 \sqrt{3} S+(a-b)^{2}+(b-c)^{2}+(a-c)^{2}
\end{array}
$$ | Prove: After organizing, it is to prove
$2(a b+a c+b c) \geqslant a^{2}+b^{2}+c^{2}+4 \sqrt{3} S$.
Also, $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$
and $S=2 R^{2} \sin A \cdot \sin B \cdot \sin C$, it is to prove
$$
\begin{array}{l}
2(\sin A \cdot \sin B+\sin B \cdot \sin C+\sin A \cdot \sin C) \\
\geqslant \sin ^{2} A... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,313 |
Example 7 In a non-obtuse $\triangle A B C$, prove:
$$
\begin{array}{l}
\cot ^{3} A+\cot ^{3} B+\cot ^{3} C+6 \cot A \cdot \cot B \cdot \cot C \\
\geqslant \cot A+\cot B+\cot C .
\end{array}
$$ | Proof: Let \( x = \cot A, y = \cot B, z = \cot C \), i.e., we need to prove \( x^{3} + y^{3} + z^{3} + 6xyz \geq x + y + z \).
Without loss of generality, assume \( x \geq y \geq z \), then
\[
\begin{array}{l}
x^{3} + y^{3} + z^{3} + 6xyz - (x + y + z) \\
= x^{3} + y^{3} + z^{3} + 6xyz - \\
(x + y + z)(xy + yz + zx) \\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,314 |
Example 8 In $\triangle A B C$, prove:
$$
\begin{array}{l}
\left(\sin \frac{B}{2}+\sin \frac{C}{2}\right)^{2}+\left(\sin \frac{C}{2}+\sin \frac{A}{2}\right)^{2}+ \\
\left(\sin \frac{A}{2}+\sin \frac{B}{2}\right)^{2} \leqslant 3
\end{array}
$$ | Prove: Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, i.e., to prove
$$
\left(\frac{y}{\sqrt{1+y^{2}}}+\frac{z}{\sqrt{1+z^{2}}}\right)^{2}+\left(\frac{z}{\sqrt{1+z^{2}}}+\frac{x}{\sqrt{1+x^{2}}}\right)^{2}+
$$
$$
\left(\frac{x}{\sqrt{1+x^{2}}}+\frac{y}{\sqrt{1+y^{2}}}\right)^{2} \leqslant 3 \text {. ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,315 |
For example, let the inradius and circumradius of $\triangle ABC$ be $r$ and $R$ respectively. Prove:
$$
\begin{array}{l}
\sin \frac{A}{2} \cdot \sin \frac{B}{2}+\sin \frac{B}{2} \cdot \sin \frac{C}{2}+\sin \frac{C}{2} \cdot \sin \frac{A}{2} \\
\leqslant \frac{5}{8}+\frac{r}{4 R} .
\end{array}
$$ | Proof: Let \( x = \tan \frac{A}{2}, y = \tan \frac{B}{2}, z = \tan \frac{C}{2} \).
From \( \frac{r}{R} = \frac{2 \sin A \cdot \sin B \cdot \sin C}{\sin A + \sin B + \sin C} \), we have
\[
\begin{array}{l}
\frac{r}{R} = \frac{16 x y z}{\left(1 + x^{2}\right)\left(1 + y^{2}\right)\left(1 + z^{2}\right)} \div \\
\left(\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,316 |
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