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Example 5 In a regular pentagon $A B C D E$, there is a point $P$ inside. It is known that $\angle A B P=6^{\circ}, \angle A E P=12^{\circ}$. Find the degree measure of $\angle P A C$. | Explanation: As shown in Figure 4, construct the equilateral triangle $\triangle AEF$, and connect $BF$. It is easy to know that the interior angle of the regular pentagon $ABCDE$ is $108^{\circ}$. Therefore,
$$
\begin{array}{l}
\angle BAF = 108^{\circ} + 60^{\circ} \\
= 168^{\circ}.
\end{array}
$$
Since $AF = AE = AB... | 12^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,108 |
1. Simplify $\sqrt{9 x^{2}-6 x+1}-(\sqrt{3 x-5})^{2}$, the result is ( ).
(A) $6 x-6$
(B) $-6 x+6$
(C) -4
(D) 4 | $$
-1 . D \text {. }
$$
From $\sqrt{3 x-5}$, we get $3 x-5 \geqslant 0$. Then
$$
\begin{array}{l}
\sqrt{9 x^{2}-6 x+1}-(\sqrt{3 x-5})^{2} \\
=|3 x-1|-|3 x-5| \\
=(3 x-1)-(3 x-5)=4 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,110 |
2. The smallest integer $k$ that makes the quadratic equation in $x$
$$
2 x(k x-4)-x^{2}+6=0
$$
have no real roots is ( ).
(A) -1
(B) 2
(C) 3
(D) 4 | 2. B.
Organize the quadratic equation $2 x(k x-4)-x^{2}+6=0$, to get $(2 k-1) x^{2}-8 x+6=0$.
Since the equation has no real roots, then $\Delta=64-24(2 k-1) \leqslant 0$.
Solving for $k$ yields $k \geqslant 1 \frac{5}{6}$.
Therefore, the smallest integer $k$ is 2. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,111 |
3. On the sides $AB, BC, CD, DA$ of the square $ABCD$, points $E, F, G, H$ are taken arbitrarily. Among the quadrilaterals $EFGH$ obtained, the number of squares is ( ).
(A) 1
(B) 2
(C) 4
(D) infinitely many | 3.D.
As shown in Figure 5, as long as $A E=B F=C G=D H$, quadrilateral $E F G H$ is a square.
Figure 5
Figure 6 | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,112 |
4. As shown in Figure 1, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are perpendicular to each other. If $AB=3, BC$ $=4, CD=5$, then the length of $AD$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{3}$
(D) $4 \sqrt{2}$ | 4.A.
As shown in Figure 6, let the four segments into which the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are divided be $a, b, c, d$. By the Pythagorean theorem, we have
$$
a^{2}+b^{2}=3^{2}, b^{2}+c^{2}=4^{2}, c^{2}+d^{2}=5^{2} \text {. }
$$
Then $a^{2}+d^{2}=18=AD^{2}$. Therefore, $AD=3 \sqrt{2}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,113 |
5. Given that $x, y, z$ are real numbers. If $x^{2}+y^{2}=1$, $y^{2}+z^{2}=2, x^{2}+z^{2}=2$, then the minimum value of $x y+y z+x z$ is ( ).
(A) $\frac{5}{2}$
(B) $\frac{1}{2}+\sqrt{3}$
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}-\sqrt{3}$ | 5.D.
Given $x^{2}+y^{2}=1, y^{2}+z^{2}=2, x^{2}+z^{2}=2$, we solve to get $x= \pm \frac{\sqrt{2}}{2}, y= \pm \frac{\sqrt{2}}{2}, z= \pm \frac{\sqrt{6}}{2}$.
When $x$ and $y$ have the same sign, and $x$ and $z$ have opposite signs, $x y+y z+x z$ has a minimum value, and the minimum value is
$$
\frac{1}{2}-\frac{\sqrt{3... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,114 |
6. In $\triangle A B C$, $\angle B A C=90^{\circ}, A C=\sqrt{3}$, $A B=4$, $D$ is a point on side $B C$, $\angle C A D=30^{\circ}$. Then the length of $A D$ is ( ).
(A) $\frac{6}{5}$
(B) $\frac{7}{5}$
(C) $\frac{8}{5}$
(D) $\frac{9}{5}$ | 6. C.
As shown in Figure 7, draw $D E \perp A C$, with the foot of the perpendicular at $E$. Let $E D=x$, then $A D=2 x$, $A E=\sqrt{3} x$, and $C E=\sqrt{3}-\sqrt{3} x$.
From $\triangle C D E \backsim \triangle C B A$, we get
$$
C E: C A=D E: B A \text {, }
$$
which is $(\sqrt{3}-\sqrt{3} x): \sqrt{3}=x: 4$.
Therefo... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,115 |
7. As shown in Figure 2, MN is the diameter of $\odot O$. If
$$
\begin{array}{l}
\angle E=25^{\circ}, \angle P M Q \\
=35^{\circ}, \text { then } \angle M Q P \\
=(\quad) .
\end{array}
$$
(A) $30^{\circ}$
(B) $35^{\circ}$
Figure 2
(C) $40^{\circ}$
(D) $50^{\circ}$ | 7.C.
As shown in Figure 2, $\overparen{N Q}^{\circ}+\overparen{Q P}^{\circ}+\overparen{P M}^{\circ}=180^{\circ}$,
$$
\begin{array}{l}
\overparen{P M}^{\circ}-\overparen{N Q}^{\circ}=2 \angle E=2 \times 25^{\circ}=50^{\circ}, \\
\overparen{P Q}^{\circ}=2 \times 35^{\circ}=70^{\circ} .
\end{array}
$$
Therefore, $\overp... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,116 |
8. Given points $A$ and $B$ lie on the graphs of the linear functions $y=x$ and $y=8x$ respectively, with their x-coordinates being $a$ and $b$ ($a>0$, $b>0$). If the line $AB$ is the graph of the linear function $y=kx+m$, then when $\frac{b}{a}$ is an integer, the number of integer values of $k$ that satisfy the condi... | 8.B.
From the problem, we have $A(a, a)$ and $B(b, 8b)$. Substituting into $y=kx+m$, we get $k=\frac{8b-a}{b-a}=8+\frac{7a}{b-a}=8+\frac{7}{\frac{b}{a}-1}$.
Since $\frac{b}{a}$ and $k$ are both integers, $\frac{b}{a}=2$ or 8. Therefore, $k=15$ or 9. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,117 |
9. Insert a digit in the middle of 2006 to get a five-digit number $20 \square 06$. If this five-digit number is divisible by 7, then the digit inserted in $\square$ is $\qquad$ . | 二、9.0 or 7.
By. $20006 \div 7=2858$, the embedded digit is 0 or 7. | 0 \text{ or } 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,118 |
Example 6 In an acute triangle $\triangle ABC$, $AB > BC > CA$, $O$, $I$, and $H$ are the circumcenter, incenter, and orthocenter, respectively. Given that $\angle A = 60^{\circ}$. Prove:
(1) $\angle OIH - \angle ABC$ is a constant;
(2) $\angle OIH + \angle ACB$ is also a constant. | Explanation: As shown in the figure,
5, draw two altitudes $B E$ and $C F$, and let $H$ be their intersection. Draw $O D \perp A B$ at point $D$. The other auxiliary lines are as shown in the figure.
Let $\angle C B E=\alpha$, then $\angle A C B=90^{\circ}-\alpha$.
Since $\angle B O D=\angle A C B=\angle B C E$, we ha... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,119 |
10. There are four numbers, among which the sum of every three numbers is $24, 36, 28, 32$. Then the average of these four numbers is $\qquad$ . | 10.10 .
The sum of these four numbers is $(24+36+28+32) \div 3=40$, so the average of these four numbers is 10 . | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,120 |
11. If $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, then $a^{2}+b^{2}=$ $\qquad$ . | 11.3.
Given $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, we have $\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+b^{2}\right)-6=0$.
Therefore, $a^{2}+b^{2}=3$ or -2.
Since $a^{2}+b^{2} \geqslant 0$, we have $a^{2}+b^{2}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,121 |
12. As shown in Figure 3, in quadrilateral $A B C D$, $A B=A C=A D$. If $\angle B A C=25^{\circ}$, $\angle C A D=75^{\circ}$, then $\angle B D C$ $=\ldots, \angle D B C=$ $\qquad$ | 12.12.5 $5^{\circ}, 37.5^{\circ}$.
From the problem, points $B, C, D$ are on $\odot A$, so,
$$
\begin{array}{l}
\angle B D C=\frac{1}{2} \angle B A C=12.5^{\circ}, \\
\angle D B C=\frac{1}{2} \angle C A D=37.5^{\circ} .
\end{array}
$$ | 12.5^{\circ}, 37.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,122 |
13. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x y+x+y+7=0, \\
3 x+3 y=9+2 x y,
\end{array}\right.
$$
then $x^{2} y+x y^{2}=$ | 13.6.
From $\left\{\begin{array}{l}x y+x+y+7=0, \\ 3 x+3 y=9+2 x y,\end{array}\right.$ we get $x y=-6, x+y=-1$.
Therefore, $x^{2} y+x y^{2}=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,123 |
15. A person's 5 trips to work (unit: $\mathrm{min}$) are $a, b, 8, 9, 10$. It is known that the average of this set of data is 9, and the variance is 2. Then the value of $|a-b|$ is $\qquad$. | 15.4.
Since the average of $a, b, 8, 9, 10$ is 9 and the variance is 2, it follows that $a, b, 8, 9, 10$ are 5 consecutive integers, so $a=7, b=11$ or $a=11, b=7$. Therefore, the value of $|a-b|$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,125 |
16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$
have non-zero integer roots, then the number of such integers $m$ is
$\qquad$. | 16.5.
Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$
Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,126 |
17. In a math test, there are 20 questions. Each correct answer earns 5 points, no answer earns 0 points, and each wrong answer deducts 2 points. If Xiaoli's score in this test is a prime number, then the maximum number of questions Xiaoli answered correctly is $\qquad$. | 17.17.
Let Xiaoli answer $a$ questions correctly and $b$ questions incorrectly, with the score being $5a-2b<100$.
Since the score is a prime number, we have $5a-2b=97, 91, 89, 83, \cdots$. Only when $a=17, b=1$, $5a-2b=83$ is the largest prime number.
Therefore, Xiaoli answered at most 17 questions correctly this time... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,127 |
18. Let the perimeter of a surface development of a $5 \mathrm{~cm} \times 4 \mathrm{~cm} \times 3 \mathrm{~cm}$ rectangular prism be $n \mathrm{~cm}$. Then the minimum value of $n$ is $\qquad$ . | 18.50.
As shown in Figure 9, the perimeter of the unfolded rectangular prism is
$$
8 c+4 b+2 a \text {. }
$$
Therefore, the minimum value of the perimeter is
$$
\begin{array}{l}
8 \times 3+4 \times 4+2 \times 5 \\
=50 .
\end{array}
$$ | 50 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,128 |
19. A person holds four ropes tightly in their hand, with only the ends of the ropes showing, then randomly connects two of the four heads at one end, and the other two, and does the same for the four tails at the other end. What is the probability that the four ropes will form a single loop when the hand is opened? $\... | 19. $\frac{2}{3}$.
There are 9 ways that meet the requirements, and 6 of these ways can form a loop as shown in Figure 10.
Figure 10
Therefore, the probability that four ropes will exactly form a loop is
$$
6 \div 9=\frac{2}{3} \text {. }
$$ | \frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,129 |
1. In a convex quadrilateral $A B C D$, $A B=B C=C D$, and $\angle A$ $=\angle B=\frac{1}{2} \angle C$. Find the measures of each interior angle. | (提示: 作 $\angle B$ 的平分线交 $A D$ 于点 $K$. 易得 $\triangle B C K \cong \triangle B A K, \triangle D C K \cong \triangle B C K$. 故 $\angle A=\angle B$ $\left.=80^{\circ}, \angle C=160^{\circ}, \angle D=40^{\circ}\right)$.
( Hint: Draw the angle bisector of $\angle B$ intersecting $A D$ at point $K$. It is easy to see that $\t... | \angle A=\angle B=80^{\circ}, \angle C=160^{\circ}, \angle D=40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,130 |
20. There is a pentagon $A B C D E$. If the vertices $A$, $B$, $C$, $D$, $E$ are colored with one of the three colors: red, yellow, green, such that adjacent vertices are colored differently, then there are a total of different coloring methods. | 20.30.
If point $A$ is colored red (as shown in Figure 11), there are 10 coloring methods.
Figure 11
Similarly, if point $A$ is colored yellow or green, there are also 10 coloring methods each. Therefore, there are a total of 30 different coloring methods. | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,131 |
1. If $x=2^{n+1}+2^{n}, y=2^{n-1}+2^{n-2}$, where $n$ is an integer, then the quantitative relationship between $x$ and $y$ is ( ).
(A) $x=4 y$
(B) $y=4 x$
(C) $x=12 y$
(D) $y=12 x$ | $\begin{array}{l}\text {-1.A. } \\ x=2^{n+1}+2^{n}=2^{2}\left(2^{n-1}+2^{n-2}\right)=4 y .\end{array}$
The translation is as follows:
$\begin{array}{l}\text {-1.A. } \\ x=2^{n+1}+2^{n}=2^{2}\left(2^{n-1}+2^{n-2}\right)=4 y .\end{array}$
Note: The original text is already in a mathematical form that is the same in bo... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,132 |
2. As shown in Figure $1, A B$ is the diameter of the semicircle $\odot O$, and $C$ is a point on the semicircle, with $\angle C O A = 60^{\circ}$. Let the areas of the sector $A O C$, $\triangle C O B$, and the segment $B m C$ be $S_{1}, S_{2}, S_{3}$, respectively. Then the size relationship among them is ( ).
(A) $S... | 2. B.
Let the radius of the semicircle be 1, then
$$
S_{1}=\frac{\pi}{6}, S_{2}=\frac{\sqrt{3}}{4}, S_{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \text {. }
$$
Therefore, $S_{2}<S_{1}<S_{3}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,133 |
3. Let $x_{1}, x_{2}$ be the two real roots of the equation $x^{2}+x-4=0$. Then $x_{1}^{3}-5 x_{2}^{2}+10=(\quad)$.
(A) -29
(B) -19
(C) -15
(D) -9 | 3. B.
Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation $x^{2}+x-4=0$, therefore,
$$
\begin{array}{l}
x_{1}^{3}=x_{1} x_{1}^{2}=x_{1}\left(4-x_{1}\right)=4 x_{1}-x_{1}^{2} \\
=4 x_{1}+x_{1}-4=5 x_{1}-4, \\
-5 x_{2}^{2}=-5\left(4-x_{2}\right)=-20+5 x_{2} .
\end{array}
$$
Then $x_{1}^{3}-5 x_{2}^{2}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,134 |
4. As shown in Figure 2, in square $A B C D$, $E$ is the midpoint of $C D$, $E F \perp A E$, intersecting $B C$ at point $F$. Then the size relationship between $\angle 1$ and $\angle 2$ is ( ).
(A) $\angle 1>\angle 2$
(B) $\angle 1<\angle 2$
(C) $\angle 1=\angle 2$
(D) Cannot be determined | 4.C.
From the given conditions, it can be proven that $\triangle A D E$, $\triangle E C F$, and $\triangle A E F$ are pairwise similar, so, $\angle 1=\angle 2$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,135 |
5. The number of non-negative integer solutions $(x, y)$ for the equation $3 x^{2}+x y+y^{2}=3 x-2 y$ is ( )
(A) 0
(B) 1
(C) 2
(D) 3 | 5.C.
From $3 x^{2}+x y+y^{2}=3 x-2 y$, we can obtain the equation about $y$: $y^{2}+(2+x) y+3 x^{2}-3 x=0$.
. Then $\Delta=(2+x)^{2}-4\left(3 x^{2}-3 x\right)$
$$
\begin{array}{l}
=-11 x^{2}+16 x+4 \\
\leqslant \frac{16^{2}+4 \times 11 \times 4}{4 \times 11} \leqslant 10,
\end{array}
$$
Therefore, when $-11 x^{2}+16 ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,136 |
6. A simplified model of a roundabout at a three-way intersection is shown in Figure 3. During a certain peak period, the number of motor vehicles entering and exiting intersections $A$, $B$, and $C$ per unit time is as shown in the figure. In the figure, $x_{1}$, $x_{2}$, and $x_{3}$ represent the number of motor vehi... | 6.C.
From Figure 3, we get
$$
\begin{array}{l}
x_{3}-55+50=x_{1}, \\
x_{1}-20+30=x_{2}, \\
x_{2}-35+30=x_{3} .
\end{array}
$$
Simplifying, we get $x_{3}-5=x_{1}, x_{1}+10=x_{2}, x_{2}-5=x_{3}$. Clearly, $x_{2}>x_{3}>x_{1}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,137 |
7. If $p, q$ are prime numbers, and $5p+3q=91$, then $p=\ldots, q=$ $\qquad$ | ニ、7.17,2.
Since $5p + 3q = 91$ is an odd number, one of $p$ or $q$ must be even, and the only even prime number is 2.
Upon inspection, we find that $p = 17, q = 2$. | p=17, q=2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,138 |
8. The shape of a workpiece is shown in Figure 4, the degree of arc $\overparen{B C}$ is $60^{\circ}, A B=$ $6 \text{ cm}$, the distance between points $B$ and $C$ equals $A B$, $\angle B A C=30^{\circ}$. Then the area of this workpiece is $\qquad$. | $8.6 \pi \mathrm{cm}^{2}$.
From the condition, points $A$, $B$, and $C$ lie on a circle with a radius of $6 \mathrm{~cm}$, so the area of the workpiece is
$$
\frac{60}{360} \times \pi \times 6^{2}=6 \pi\left(\mathrm{cm}^{2}\right) \text {. }
$$ | 6 \pi \mathrm{cm}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,139 |
9. Let $x, y$ be real numbers, the algebraic expression
$$
5 x^{2}+4 y^{2}-8 x y+2 x+4
$$
has a minimum value of | 9.3.
$$
\begin{array}{l}
\text { Since } 5 x^{2}+4 y^{2}-8 x y+2 x+4 \\
=5\left(x-\frac{4}{5} y+\frac{1}{5}\right)^{2}+\frac{4}{5}(y+1)^{2}+3,
\end{array}
$$
Therefore, when $y=-1, x=-1$, the original expression has a minimum value of 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,140 |
2. In $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, point $M$ is on $A C$ and satisfies $A M=B C$. Find the degree measure of $\angle B M C$. | (提示: 作正 $\triangle P A B$, 点 $P 、 M$ 位于 $A B$ 的异側.易证 $\triangle P A M \cong \triangle A C B, P M=A B=P A=P B$. 从而, $P$为 $\triangle A B M$ 的外心, $\angle A B M=\frac{1}{2} \angle A P M=\frac{1}{2} \times 20^{\circ}=$ $\left.10^{\circ}, \angle B M C=\angle A B M+\angle B A M=30^{\circ}.\right)$
(提示: Construct an equilater... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,141 |
10. Let the quadratic equation in $x$ be
$$
x^{2}+2 k x+\frac{1}{4}-k=0
$$
has two real roots. Then the range of values for $k$ is $\qquad$. | 10. $k \geqslant \frac{\sqrt{2}-1}{2}$ or $k \leqslant-\frac{\sqrt{2}+1}{2}$.
Since the equation $x^{2}+2 k x+\frac{1}{4}-k=0$ has two real roots, then $\Delta=4 k^{2}-4\left(\frac{1}{4}-k\right) \geqslant 0$.
Solving this, we get $k \geqslant \frac{\sqrt{2}-1}{2}$ or $k \leqslant-\frac{\sqrt{2}+1}{2}$. | k \geqslant \frac{\sqrt{2}-1}{2} \text{ or } k \leqslant -\frac{\sqrt{2}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,142 |
11. As shown in Figure 5, in
$\square A B C D$, $A M \perp$ $B C, A N \perp C D, M, N$ are the feet of the perpendiculars. If $A B=13$, $B M=5, M C=9$, then the length of $M N$ is $\qquad$ | $11.13 \frac{11}{13}$.
As shown in Figure 5, connect $M N$, and draw $M H \perp A N$, with the foot of the perpendicular at $H$. We can obtain
$$
\begin{array}{l}
\angle B=\angle D=\angle M A N, \\
\triangle A B M \backsim \triangle M A H \backsim \triangle A D N .
\end{array}
$$ | 11.13 \frac{11}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,143 |
13. As shown in Figure 7, in Rt $\triangle ABC$, $AB=3$, $BC=4$, $\angle ABC=90^{\circ}$, a line $BA_1 \perp AC$ is drawn through $B$, and a line $A_1B_1 \perp BC$ is drawn through $A_1$, resulting in the shaded Rt $\triangle A_1B_1B$; then a line $B_1A_2 \perp AC$ is drawn through $B_1$, and a line $A_2B_2 \perp BC$ i... | $13.2 \frac{14}{41}$.
The ratio of the sum of all shaded areas to the sum of all blank areas is equal to the ratio of the area of $\triangle A_{1} B_{1} B$ to the area of $\triangle A_{1} A B$, which is
$$
(3 \times 4 \div 5 \div 3)^{2}=\frac{16}{25} \text {. }
$$
Therefore, the sum of the shaded areas is
$$
3 \times ... | 2 \frac{14}{41} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,145 |
14. In a $3 \times 3$ grid filled with the numbers $1 \sim 9$, the largest number in each row is colored red, and the smallest number in each row is colored green. Let $M$ be the smallest number among the three red squares, and $m$ be the largest number among the three green squares. Then $M-m$ can have $\qquad$ differ... | 14.8.
From the conditions, it is known that the values of $m$ and $M$ are positive integers from 3 to 7 (inclusive of 3 and 7).
Obviously, $M \neq m$, so the value of $M-m$ is
$$
1,2,3,4,-1,-2,-3,-4 \text {. }
$$
Therefore, $M-m$ can have 8 different values. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,146 |
15. As shown in Figure 8, the line $O B$ is the graph of the linear function $y=2x$, and the coordinates of point $A$ are $(0,2)$. Find point $C$ on line $O B$ such that $\triangle A C O$ is an isosceles triangle. Determine the coordinates of point $C$. | Three, 15. As shown in Figure 11, if this isosceles triangle has $O A$ as one of its legs, and $A$ as the vertex, then
$$
A O=A C_{1}=2 \text {. }
$$
Let $C_{1}(x, 2 x)$, we get
$$
x^{2}+(2 x-2)^{2}=2^{2} \text {. }
$$
Solving for $x=\frac{8}{5}$.
Thus, $C_{1}\left(\frac{8}{5}, \frac{16}{5}\right)$.
If this isosceles... | \left(\frac{8}{5}, \frac{16}{5}\right),\left(\frac{2 \sqrt{5}}{5}, \frac{4 \sqrt{5}}{5}\right),\left(-\frac{2 \sqrt{5}}{5},-\frac{4 \sqrt{5}}{5}\right),\left(\frac{1}{2}, 1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,147 |
16. As shown in Figure 9, quadrilateral $ABCD$ is a square, $\odot O$ passes through the vertex $A$ and the intersection point $P$ of the diagonals, and intersects $AB$ and $AD$ at points $F$ and $E$, respectively.
(1) Prove that $DE = AF$;
(2) If the radius of $\odot O$ is $\frac{\sqrt{3}}{2}$ and $AB = \sqrt{2} + 1$,... | 16. (1) As shown in Figure 12, connect \( P E, P F, E F \).
Since \(\angle E A F = 90^{\circ}\), \( E F \) is the diameter of \(\odot O\).
Thus, \(\angle F P E = 90^{\circ}\).
Also, \(\angle A P D = 90^{\circ}\), so \(\angle E P D = \angle A P F\).
Clearly, \( P D = P A \),
\(\angle P A F = \angle P D E = 45^{\circ}\)... | \sqrt{2} \text{ or } \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,148 |
17. In the $7 \times 7$ unit square grid shown in Figure 10, there are 64 grid points, and there are many squares with these grid points as vertices. How many different values of the areas of these squares are there? | 17. Since the vertices of the square are grid points in the grid, as shown in Figure 13, the area of the shaded square is $a^{2}+b^{2}$.
Where, $0 \leqslant a+b \leqslant 7$.
Without loss of generality, let $a \geqslant b$. We can enumerate all possible values of $(a, b)$:
$$
\begin{array}{l}
(0,0),(1,0),(2,0),(3,0),(4... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,149 |
18. Let $k$, $a$, $b$ be positive integers, and the quotients when $k$ is divided by $a^2$ and $b^2$ are $m$ and $m+116$ respectively.
(1) If $a$ and $b$ are coprime, prove that $a^2 - b^2$ is coprime with both $a^2$ and $b^2$;
(2) When $a$ and $b$ are coprime, find the value of $k$;
(3) If the greatest common divisor ... | 18. (1) Let $s$ be the greatest common divisor of $a^{2}-b^{2}$ and $a^{2}$.
Then $a^{2}-b^{2}=s u, a^{2}=s v\left(s, v \in \mathbf{N}_{+}\right)$.
Thus, $a^{2}-\left(a^{2}-b^{2}\right)=b^{2}=s(v-u)$.
It is clear that $s$ is a divisor of $b^{2}$.
Since $a, b$ are coprime, $a^{2}, b^{2}$ are also coprime.
Therefore, $s... | 4410000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,150 |
1. Let in the $x O y$ plane, the area enclosed by $0<y \leqslant x^{2}, 0 \leqslant x \leqslant 1$ be $\frac{1}{3}$. Then the intersection of the sets
$$
\begin{array}{l}
M=\{(x, y)|x \leqslant| y \mid\}, \\
N=\left\{(x, y) \mid x \geqslant y^{2}\right\}
\end{array}
$$
representing the area of $M \cap N$ is ( ).
(A) $... | -1. .
$M \cap N$ is symmetric with respect to the $x$-axis on the $x O y$ plane, therefore, the area of the figure of $M \cap N$ can be calculated by multiplying the area of the figure in the first quadrant by 2. According to the problem, the area of the figure of $M \cap N$ in the first quadrant is $\frac{1}{2}-\frac{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,151 |
3. In the convex quadrilateral $ABCD$, $\angle ADB=70^{\circ}, \angle CDB=$ $40^{\circ}, \angle BAC=20^{\circ}, \angle BCA=35^{\circ}$. Find the degree measure of the acute angle between $AC$ and $BD$. | (提示: Extend $B D$ to point $P$, such that $D P=D B$. It is only necessary to prove that points $A$, $B$, $C$, and $P$ are concyclic, then $D C=D P=D A, \angle A D C=110^{\circ}$, $\angle A C D=35^{\circ}$. Therefore, the included acute angle is $75^{\circ}$.) | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,152 |
2. In tetrahedron $ABCD$, let $AB=1, CD=$ $\sqrt{3}$, the distance between line $AB$ and line $CD$ is 2, and the angle between them is $\frac{\pi}{3}$. Then the volume of tetrahedron $ABCD$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{\sqrt{3}}{3}$ | 2.C.
Through point $D$, draw $D F \perp C B$, through point $A$, draw $A E \perp B C$, connect $C E$, $E D$, $A F$, $B F$, and complete the pyramid into a prism. Therefore, the volume of the pyramid is $\frac{1}{3} \times \frac{1}{2} C E \cdot C D \sin \angle E C D \cdot h=\frac{1}{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,153 |
3. There are 10 different balls, among which, 2 are red, 5 are yellow, and 3 are white. If getting a red ball scores 5 points, getting a white ball scores 2 points, and getting a yellow ball scores 1 point, then, the number of ways to draw 5 balls so that the total score is greater than 10 points and less than 15 point... | 3.C.
There are four scenarios that meet the requirements: red-red-white-yellow-yellow, red-red-yellow-yellow-yellow, red-white-white-white-yellow, red-white-white-yellow-yellow. Therefore, the number of different ways to draw the balls is
$$
C_{3}^{1} C_{5}^{2} + C_{5}^{3} + C_{2}^{1} C_{3}^{3} C_{5}^{1} + C_{2}^{1} C... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,154 |
4. In $\triangle A B C$, if $(\sin A+\sin B)(\cos A+\cos B)=2 \sin C$, then ( ).
(A) $\triangle A B C$ is an isosceles triangle, but not necessarily a right triangle
(B) $\triangle A B C$ is a right triangle, but not necessarily an isosceles triangle
(C) $\triangle A B C$ is neither an isosceles triangle nor a right tr... | 4.A.
$$
\begin{array}{c}
\text { Left side }=\sin A \cdot \cos A+\sin A \cdot \cos B+ \\
\quad \sin B \cdot \cos A+\sin B \cdot \cos B \\
=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B) \\
=\sin (A+B) \cdot \cos (A-B)+\sin (A+B), \\
\text { Right side }=2 \sin (A+B) .
\end{array}
$$
Therefore, the given equation can be tra... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,155 |
5. Given $f(x)=3 x^{2}-x+4$, $f(g(x))=3 x^{4}+18 x^{3}+50 x^{2}+69 x+48$.
Then, the sum of the coefficients of the integer polynomial function $g(x)$ is $(\quad$.
(A) 8
(B) 9
(C) 10
(D) 11 | 5.A.
Let the sum of the coefficients of $g(x)$ be $s$, then $f(g(1))=3 s^{2}-s+4=188$.
Solving for $s$ gives $s=8$ or $s=-\frac{23}{3}$ (discard). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,156 |
6. Let $0<x<1, a, b$ be positive constants. Then the minimum value of $\frac{a^{2}}{x}+$ $\frac{b^{2}}{1-x}$ is ( ).
(A) $4 a b$
(B) $(a+b)^{2}$
(C) $(a-b)^{2}$
(D) $2\left(a^{2}+b^{2}\right)$ | 6. B.
$$
\begin{array}{l}
\frac{a^{2}}{x}+\frac{b^{2}}{1-x}=(x+1-x)\left(\frac{a^{2}}{x}+\frac{b^{2}}{1-x}\right) \\
=a^{2}+b^{2}+a^{2} \cdot \frac{1-x}{x}+b^{2} \cdot \frac{x}{1-x} \geqslant(a+b)^{2} .
\end{array}
$$
When $x=\frac{a}{a+b}$, the minimum value $(a+b)^{2}$ is achieved. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,157 |
7. Let $a, b>0$, and
$$
a^{2008}+b^{2008}=a^{2006}+b^{2006} \text {. }
$$
Then the maximum value of $a^{2}+b^{2}$ is ( ).
(A) 1
(B) 2
(C) 2006
(D) 2008 | 7. B.
Since $a^{2008}+b^{2008} \geqslant a^{2000} b^{2}+b^{2006} a^{2}$, and
$$
\begin{array}{l}
\left(a^{2006}+b^{2006}\right)\left(a^{2}+b^{2}\right) \\
=a^{2008}+b^{2008}+a^{2006} b^{2}+b^{2006} a^{2} \\
\leqslant 2\left(a^{2008}+b^{2008}\right),
\end{array}
$$
and $a^{2008}+b^{2008}=a^{2006}+b^{2000}$,
therefore,... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,158 |
8. As shown in Figure 1, let $P$ be a point in the plane of $\triangle A B C$, and
$$
A P=\frac{1}{5} A B+\frac{2}{5} A C \text {. }
$$
Then the ratio of the area of $\triangle A B P$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{2}$
(C) $\frac{2}{5}$
(D) $\frac{2}{3}$ | 8. C.
As shown in Figure 4, extend $A P$ to $E$ such that $A P = \frac{1}{5} A E$.
Connect $B E$, and draw $E D$ $/ / B A$ intersecting the extension of $A C$ at point $D$.
From $A P = \frac{1}{5} A B + \frac{2}{5} A C^{\circ}$, we get $A C = C D$.
Therefore, quadrilateral $A B E D$ is a parallelogram.
Thus, $\frac{S... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,159 |
9. Given that $a, b, c, d$ are even numbers, and $0<a<b<c<d, d-a=90, a, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence. Then $a+b+c+d=$ ( ).
(A) 384
(B) 324
(C) 284
(D) 194 | 9.D.
Let $a, b, c, d$ be $b-m, b, b+m, \frac{(b+m)^{2}}{b}$, respectively.
Also, $\frac{(b+m)^{2}}{b}-(b-m)=90$, which means
$$
b=\frac{m^{2}}{3(30-m)} \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
Let $m=6k$, substituting into equation (1) yields... | 194 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,160 |
10 . The sequence $\left\{3^{n-1}\right\}$ is grouped according to the rule “the $n$-th group has $n$ numbers” as follows:
$$
\text { (1),(3,9), }(27,81,243), \cdots \text {. }
$$
Then the first number of the 100th group is ( ).
(A) $3^{4950}$
(B) $3^{5000}$
(C) $3^{5010}$
(D) $3^{5050}$ | 10.A.
The sum of the first 99 terms is $1+2+\cdots+99=4950$. The 1st group is $3^{\circ}$, so the first number in the 100th group should be $3^{4950}$. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,161 |
11. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, the symmetric points of point $A$ about the lines $A_{1} C$ and $B D_{1}$ are points $P$ and $Q$, respectively. Then the distance between points $P$ and $Q$ is ( ).
(A) $\frac{2 \sqrt{2}}{3}$
(B) $\frac{3 \sqrt{3}}{2}$
(C) $\frac{3 \sqrt{2}}{4}$
(D)... | 11.A.
Establish a spatial rectangular coordinate system, with $D(0,0,0), A(1,0,0), A_{1}(1,0,1), C(0,1,0), B(1,1,0), D_{1}(0,0,1)$. Let $P(x, y, z)$, and the midpoint of $A P$ be $M\left(\frac{x+1}{2}, \frac{y}{2}, \frac{z}{2}\right)$.
From $A P \cdot A_{1} C=0, M C / / A_{1} C$, we get
Similarly, $Q\left(\frac{1}{3},... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,162 |
4. As shown in Figure 6, in $\triangle A B C$, two line segments $B Q$ and $C R$ intersect at point $P, A R=B R=C P, P Q=C Q$. Find the degree measure of $\angle B R C$.
保留了源文本的换行和格式。 | (Tip: On the extension of $CR$, take $RK=RC, RS=PC$. It is easy to get $\triangle BSK \cong \triangle BRP, BS=BR=CP=SR$. Therefore, $\triangle BRS$ is an equilateral triangle, $\angle BRS=60^{\circ}, \angle BRC=120^{\circ}$.) | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,163 |
12. Given $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, and $P$ is any point on the left branch of the hyperbola. If $\frac{\left|P F_{2}\right|^{2}}{\left|P F_{1}\right|}$ is $8 a$, then the range of the eccentricity $e$ of the hyperbola is ( ).
(A) $(1,+... | 12.C.
According to the definition of a hyperbola, we have
$$
\begin{array}{l}
\left|P F_{2}\right|-\left|P F_{1}\right|=2 a, \\
\frac{\left|P F_{2}\right|^{2}}{\left|P F_{1}\right|}=\frac{\left(\left|P F_{1}\right|+2 a\right)^{2}}{\left|P F_{1}\right|}=\left|P F_{1}\right|+4 a+\frac{4 a^{2}}{\left|P F_{1}\right|} \\
\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,164 |
13. Given $\frac{\sin (\alpha+2 \beta)}{\sin \alpha}=3$, and $\beta \neq \frac{1}{2} k \pi$, $\alpha+\beta \neq n \pi+\frac{\pi}{2}(n, k \in \mathbf{Z})$. Then the value of $\frac{\tan (\alpha+\beta)}{\tan \beta}$ is $\qquad$ | $\begin{array}{l}13.2 \text {. } \\ \frac{\tan (\alpha+\beta)}{\tan \beta}=\frac{\sin (\alpha+\beta) \cdot \cos \beta}{\cos (\alpha+\beta) \cdot \sin \beta} \\ =\frac{\frac{1}{2}[\sin (\alpha+2 \beta)+\sin \alpha]}{\frac{1}{2}[\sin (\alpha+2 \beta)-\sin \alpha]}=\frac{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}+1}{\frac{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,165 |
14. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and let the sum of the first $n$ terms be $b_{n}$. The product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2000 is $\qq... | 14.1980.
According to the problem, we have $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2)$.
Also, $b_{n}+c_{n}=1$, so $\frac{c_{n}}{c_{n-1}}+c_{n}=1$, which means $\frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1$. Given $c_{1}=b_{1}, c_{1}+b_{1}=1$, we can get $c_{1}=b_{1}=\frac{1}{2}$. Therefore, $c_{n}=\frac{1}{n+1}, b_{n}=\frac{n... | 1980 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,166 |
15. The solution set of the inequality
$$
-2<\sqrt{x^{2}-2 x+4}-\sqrt{x^{2}-10 x+28}<2
$$
is $\qquad$ . | 15. $\{x \mid 3-\sqrt{2}<x<3+\sqrt{2}\}$.
The original inequality is
$$
\left|\sqrt{(x-1)^{2}+3}-\sqrt{(x-5)^{2}+3}\right|<2 \text {. }
$$
Let $3=y^{2}$, the inequality can be transformed into
$$
\left|\sqrt{(x-1)^{2}+y^{2}}-\sqrt{(x-5)^{2}+y^{2}}\right|<2 \text {. }
$$
By the definition of a hyperbola, the points s... | \{x \mid 3-\sqrt{2}<x<3+\sqrt{2}\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,167 |
16. Given a constant $a>0$, vector $m=(0, a)$, $n=(1,0)$, a line passing through the fixed point $A(0,-a)$ with the direction vector $m+\lambda n$ intersects with a line passing through the fixed point $B(0, a)$ with the direction vector $n+2 \lambda m$ at point $P$, where $\lambda \in \mathbf{R}$. Then the equation of... | 16. $y^{2}-a^{2}=2 a^{2} x^{2}$.
Let point $P(x, y)$, then $A P=(x, y+a), B P=(x, y-a)$.
Also, $\boldsymbol{n}=(1,0), \boldsymbol{m}=(0, a)$, so
$$
\boldsymbol{m}+\lambda \boldsymbol{n}=(\lambda, a), \boldsymbol{n}+2 \lambda \boldsymbol{m}=(1,2 \lambda a) \text {. }
$$
By the given condition, vector $A P$ is parallel... | y^{2}-a^{2}=2 a^{2} x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,168 |
17. (12 points) Two students, A and B, each have 5 cards. They play a game by tossing a fair coin. When the coin lands heads up, A wins a card from B; otherwise, B wins a card from A. The game ends when the coin has been tossed 9 times or when one of them has won all the cards. Let $\xi$ represent the number of times t... | Three, 17. The values of $\xi$ are $5, 7, 9$, then
$$
\begin{array}{c}
P(\xi=5)=\mathrm{C}_{2}^{1}\left(\frac{1}{2}\right)^{5}=\frac{1}{16}, \\
P(\xi=7)=\mathrm{C}_{2}^{1} \mathrm{C}_{5}^{4}\left(\frac{1}{2}\right)^{4} \times \frac{1}{2}\left(\frac{1}{2}\right)^{2}=\frac{5}{64}, \\
P(\xi=9)=1-\frac{1}{16}-\frac{5}{64}=... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,169 |
18. (12 points) Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$. If the vectors
$m=\left(1-\cos (A+B), \cos \frac{A-B}{2}\right)$,
$n=\left(\frac{5}{8}, \cos \frac{A-B}{2}\right)$, and $m \cdot n=\frac{9}{8}$.
(1) Prove that $\tan A \cdot \tan B=\frac{1}{9}$;
(2) Find the maximum valu... | 18. (1) From $m \cdot n=\frac{9}{8}$, we get
$$
\frac{5}{8}[1-\cos (A+B)]+\cos ^{2} \frac{A-B}{2}=\frac{9}{8},
$$
which means $\frac{5}{8}[1-\cos (A+B)]+\frac{1+\cos (A-B)}{2}=\frac{9}{8}$,
or equivalently, $4 \cos (A-B)=5 \cos (A+B)$.
Therefore, $\tan A \cdot \tan B=\frac{1}{9}$.
(2) Since $\frac{a b \sin C}{a^{2}+b^... | -\frac{3}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,170 |
19. (12 points) As shown in Figure 2, the incircle $\odot I$ of $\triangle ABC$ touches $BC$, $CA$ at points $D$, $E$ respectively, and the line $BI$ intersects $DE$ at point $G$. Prove: $AG \perp BG$. | 19. As shown in Figure 2, connect $A I$, $D I$, and $E I$. Then
$$
\begin{array}{l}
\angle E D C=\frac{1}{2} \angle D I E=\frac{1}{2}\left(180^{\circ}-\angle C\right) \\
=\frac{1}{2}(\angle A B C+\angle B A C) .
\end{array}
$$
Also, $\angle E D C=\angle D B G+\angle B G D$, so,
$$
\angle B G D=\frac{1}{2} \angle B A C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,171 |
20. (12 points) Let $f(x)$
be a function defined on $\mathbf{R}$ with a period of 2, and it is an even function. On the interval $[2,3]$, $f(x)=-2(x-3)^{2}+4$. Rectangle $A B C D$ has two vertices $A$ and $B$ on the $x$-axis, and $C$ and $D$ on the graph of $y=f(x)(0 \leqslant x \leqslant 2)$. Find the maximum area of... | 20. When $0 \leqslant x \leqslant 1$, we have
$$
f(x)=f(x+2)=-2(x-1)^{2}+4 \text {; }
$$
When $-1 \leqslant x \leqslant 0$, we have
$$
f(x)=f(-x)=-2(-x-1)^{2}+4 \text {; }
$$
When $1 \leqslant x \leqslant 2$, we have
$$
\begin{array}{l}
f(x)=f(x-2)=-2[-(x-2)-1]^{2}+4 \\
=-2(x-1)^{2}+4 .
\end{array}
$$
Let $D(x, t) 、... | \frac{16 \sqrt{6}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,172 |
21. (12 points) As shown in the figure
3, given the endpoints of the major axis of the ellipse
$A$ and $B$, the chord $EF$ intersects $AB$ at
point $D$, $O$ is the center of the ellipse, and
$|OD|=1, 2DE$
$+DF=0, \angle FDO=\frac{\pi}{4}$
(1) Find the range of the length of the major axis of the ellipse;
(2) If $D$ is ... | 21. (1) Establish the rectangular coordinate system as shown in Figure 5, then $D(-1,0)$, and the equation of the line containing the chord $EF$ is
$$
y=x+1 .
$$
Let the equation of the ellipse be
$$
\begin{array}{c}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\
(a>b>0), E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\... | \frac{2 x^{2}}{9}+\frac{2 y^{2}}{7}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,173 |
Example 1 As shown in Figure 1, given that $\odot O_{1}$ and $\odot O_{2}$ are disjoint, $O P$ and $O Q$ are their two external common tangents, the perpendicular bisector of line segment $O_{1} O_{2}$ intersects ray $O P$ at $A$, and through point $A$ the tangents to $\odot O_{1}$ and $\odot O_{2}$ are drawn, intersec... | To prove that $\triangle ABC$ is an isosceles triangle, the preferred approach is to prove that $\angle ACB = \angle ABC$.
To utilize the fact that "the perpendicular bisector of line segment $O_1O_2$ intersects ray $OP$ at $A$", connect $AO_1$ and $AO_2$. It is known that
$$
\angle AO_2O_1 = \angle AO_1O_2.
$$
Clear... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,174 |
Example 2 As shown in Figure 2, it is known that $\triangle ABC$ is inscribed in $\odot O$, $AD, BD$ are tangents to $\odot O$, $DE \parallel BC$, intersecting $AC$ at point $E$, connect $EO$ and extend it to intersect $BC$ at point $F$. Prove: $BF=FC$.
| Explanation: From the structure of the graph, think of "a diameter perpendicular to a chord bisects the chord." It is sufficient to prove that $O F \perp B C$.
Notice that $D E / / B C$, the problem is transformed into: proving $O E \perp D E$.
Notice that $D A$ is a tangent to $\odot O$, so $O A \perp D A$. Therefor... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,175 |
2. In $\triangle A B C$, $\angle A B C=40^{\circ}, \angle A C B=30^{\circ}, P$ is a point on the bisector of $\angle A B C$, $\angle P C A=20^{\circ}, B P$ intersects $A C$ at point $M, C P$ intersects $A B$ at point $N$. Prove: $P M=N A$. | (提示: As shown in Figure 11, let $D$ be a point on the extension of $BA$, such that $DB=BC$. Connect $PA$, $PD$, and $DC$. Draw a line through $M$ parallel to $AP$ intersecting $NC$ at point $E$, and connect $AE$. Clearly, $\triangle PCD$ is an equilateral triangle. It is easy to see that quadrilateral $APEM$ is an isos... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,176 |
Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.
| Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now p... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,177 |
Initially, $197 P$ is an external point of $\odot O$, through $P$ draw a tangent $PA$ of $\odot O$ ($A$ is the tangent point) and a secant $PBC$. A line through point $B$ and parallel to $AP$ intersects $AC$ at point $D$, $OP$ intersects $BD$ at point $M$. Let $N$ be the midpoint of $BD$. Prove: $ON \perp AM$.
---
Th... | Prove: As shown in Figure 3, draw another tangent line $PE$ of $\odot O$ through point $P$, with the tangent point being $E$. Let $AE$ intersect $BD$ at point $N'$. Connect $AB$, $BE$, $CE$, and $AO$. It is easy to see that
$$
\begin{array}{c}
\triangle CEP \backsim \triangle EBP, \\
\triangle CAP \backsim \triangle AB... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,178 |
Find all positive integers $n$ such that $20n+3$ divides $2006n+2005$.
Initial: 198 | Solution: Notice that
$$
\begin{array}{l}
\frac{2006 n+2005}{20 n+3} \in \mathbf{Z} \\
\Leftrightarrow \frac{100(20 n+3)+6 n+1705}{20 n+3} \in \mathbf{Z} \\
\Leftrightarrow \frac{6 n+1705}{20 n+3} \in \mathbf{Z} . \\
\text { Let } \frac{6 n+1705}{20 n+3}=k,
\end{array}
$$
then $k \in \mathbf{Z}_{+}, k$ is an odd numbe... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,179 |
In quadrilateral $ABCD$, $AC \perp BD$ and $AC \cap BD=O, OC=OD$. Try to find the maximum value of $\frac{OB}{AB}+\frac{OA}{AD}+\frac{OC}{BC}$. | Solution: As shown in Figure 4, let $\angle A B O = a, \angle O A D = \beta, \angle B C O = \gamma$. According to the problem,
$$
\begin{array}{l}
\tan \alpha = \frac{O A}{O B}, \\
\tan \beta = \frac{O D}{O A}, \\
\tan \gamma = \frac{O B}{O C}.
\end{array}
$$
Since $O D = O C$, we have $\tan \alpha \cdot \tan \beta \c... | \frac{3 \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,180 |
Given $a 、 b 、 c$ are positive numbers satisfying $a b c=1$, and $n \geqslant 1$ or $n \leqslant-2$. Prove:
$$
\frac{a^{n}}{a^{n}+b+c}+\frac{b^{n}}{a+b^{n}+c}+\frac{c^{n}}{a+b+c^{n}} \geqslant 1 .
$$ | Proof: Since $n \geqslant 1$ or $n \leqslant-2$, we have
$$
\frac{n+2}{3} \cdot \frac{n-1}{3} \geqslant 0 \text {. }
$$
By the monotonicity of constant functions, we get
$$
\begin{array}{l}
\left(b^{\frac{n+2}{3}}-c^{\frac{n+2}{3}}\right)\left(b^{\frac{n-1}{3}}-c^{\frac{n-1}{3}}\right) \geqslant 0, \\
b^{\frac{2 n+1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,181 |
3. As shown in Figure 12, the incircle of $\triangle ABC$ touches $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. A line parallel to $BC$ is drawn through point $F$, intersecting lines $DA$ and $DE$ at points $H$ and $G$, respectively. Is there any other equal line segments in Figure 12 besides $AF=AE$,... | (Tip: $F H=H E$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The content within the parentheses is a note for the translator and should not be translated into the final output.)
(Tip: $F H=H E$.) | null | Geometry | proof | Yes | Yes | cn_contest | false | 718,182 |
4. As shown in Figure 13, in trapezoid $A B C D$, $A D$ $/ / B C$, squares $A B G E$ and $D C H F$ are constructed on sides $A B$ and $C D$ respectively. Let the perpendicular bisector $l$ of line segment $A D$ intersect line segment $E F$ at point $M$. Prove that $M$ is the midpoint of $E F$. | (提示:如图 13,构造两对全等三角形, $K A=P A$
$$
=Q D=N D, A J=J D, K J=J N, M E=M F .)
$$
Hint: As shown in Figure 13, construct two pairs of congruent triangles, $K A=P A$
$$
=Q D=N D, A J=J D, K J=J N, M E=M F .)
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,183 |
5. As shown in Figure 14, quadrilateral $ABCD$ is inscribed in $\odot O$, and the lines containing sides $AB$ and $CD$ intersect at point $P$. Let the incenter of $\triangle ABC$ and $\triangle BCD$ be $S$ and $T$, respectively. The line $ST$ intersects $AB$ and $AC$ at points $E$ and $F$, respectively. Prove that $PE ... | (提示: As shown in Figure 14, let $A S$ intersect $\odot O$ at point $Q$, and connect $D Q, Q C, C T$. It is easy to know that $Q T$ $=Q C, Q S=Q C \Rightarrow Q S=Q T$, thus $\angle P E F=\angle P F E$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,184 |
Example 2 Color each point in the plane with one of the two colors, black or white. Prove that there must exist an equilateral triangle with side length 1 or $\sqrt{3}$, whose three vertices are the same color.
(1986, China Mathematical Olympiad) | Prove: Draw an equilateral $\triangle ABC$ with a side length of 1 on this plane. If points $A$, $B$, and $C$ are the same color, the conclusion holds, so we may assume that $A$ and $B$ are different colors.
Using line segment $AB$ as the base, construct an isosceles $\triangle ABD$ with legs of length 2. Since points... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,185 |
Example 3 Let $M$ be the set of all points on the perimeter of an equilateral $\triangle ABC$. For any partition of $M$ into two disjoint subsets, is it always true that one of the two subsets contains the three vertices of a right-angled triangle? Explain your reasoning.
(24th IMO) | Solution: As shown in Figure 2, on the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, take points $P$, $Q$, and $R$ respectively, such that
$$
\begin{array}{l}
PC=\frac{1}{3} BC, \\
QA=\frac{1}{3} CA, \\
RB=\frac{1}{3} AB.
\end{array}
$$
Thus, $\triangle ARQ$, $\triangle BPR$, and $\triangle CQP$ are all right triangl... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,186 |
Example 5 Color each integer point on the coordinate plane with one of three different colors, and there are points of all three colors. Prove that there exists a right-angled triangle, whose three vertices are of different colors.
(2004, Russian Mathematical Olympiad) | Proof 1: Let the three colors be red, yellow, and blue, and call a line parallel to one of the coordinate axes and containing integer points a coordinate line.
(1) Suppose all integer points on each vertical coordinate line are of the same color. Since there are integer points of all three colors, we can find three ver... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,187 |
Example 3 In a convex quadrilateral $ABCD$, the diagonal $BD$ is neither the bisector of $\angle ABC$ nor the bisector of $\angle CDA$. Point $P$ is inside the quadrilateral $ABCD$ and satisfies $\angle PBC = \angle DBA$, $\angle PDC = \angle BDA$. Prove that the quadrilateral $ABCD$ is cyclic if and only if $AP = CP$.... | Explanation: Necessity (Sufficiency omitted).
It is only necessary to prove that point $P$ lies on the perpendicular bisector of $A C$.
As shown in Figure 3, let the lines $D P$ and $B P$ intersect the circumcircle of quadrilateral $A B C D$ at points $E$ and $F$, respectively, and connect $E B$, $E C$, $E F$, $F C$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,188 |
Example 7 Color each point on the plane with one of 1992 different colors, and ensure that there are points of each color. Prove: For any given triangle $T$, there exists a triangle congruent to $T$ in the plane, such that each of its two sides has points of the same color (excluding vertices).
(1992, Saint Petersburg ... | Prove: Place triangle $T$ on a colored plane, and let its circumcenter be $O$, and the smallest interior angle be $\alpha_{0}$. Rotate triangle $T$ counterclockwise around point $O$ by an angle $\alpha\left(0<\alpha<\alpha_{0}\right)$, and denote the resulting triangle as $T(\alpha)$. Clearly, $T(\alpha) \cong T$, and ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,189 |
Example 8 Divide the entire space into three non-empty sets. Prove that there must be one set such that for every real number $a>0$, there are two points in the set whose distance is equal to $a$.
(24th IMO Shortlist) | Proof: Suppose the entire space is divided into three non-empty sets $M_{1}, M_{2}, M_{3}$. If the conclusion does not hold, then there exist three positive numbers $a_{1}, a_{2}, a_{3}$ such that the distance between any two points in $M_{i}$ is not equal to $a_{i} (i=1,2,3)$. Without loss of generality, assume $a_{1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,190 |
Example 1 In $\triangle ABC$, $AB=AC$, $\odot O$ is its circumcircle, $BN$ bisects $\angle ABC$, point $N$ is on $\odot O$, points $E$ and $F$ are on sides $AB$ and $AC$ respectively, satisfying $EO \perp BN$, $EF \perp EO$. Prove: $AE^2 = BE \cdot AF$.
---
The translation maintains the original text's line breaks an... | Proof: Since $E F \perp E O, B N \perp E O$, then $E F \parallel B N$. Therefore, $\frac{A E}{B E}=\frac{A F}{F D}$. Hence $A E \cdot F D=B E \cdot A F$.
Thus, it is only necessary to prove that $A E=F D$.
Since line segments $A E$ and $F D$ are not in the same triangle, we can consider a translation transformation.
As... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,191 |
Example 2 Convex quadrilateral $ABCD$ is inscribed in $\odot O$. Extend $AB$ and $DC$ to intersect at point $E$, and extend $BC$ and $AD$ to intersect at point $F$. Points $M$ and $N$ are the midpoints of $AC$ and $BD$ respectively, and $AC > BD$. Prove:
$$
\frac{MN}{EF}=\frac{1}{2}\left(\frac{AC}{BD}-\frac{BD}{AC}\rig... | Proof: Since quadrilateral $ABCD$ is a cyclic quadrilateral, we can consider using a homothetic axis reflection transformation.
As shown in Figure 2, let $AC = kBD$ ($k > 1$). With point $F$ as the homothetic center, $k$ as the homothetic ratio, and the angle bisector of $\angle AFB$ as the reflection axis, perform a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,192 |
Example 3 As shown in Figure 3, in $\triangle A B C$, $B_{1}$ and $C_{1}$ are points on the extensions of $A B$ and $A C$, respectively, $D_{1}$ is the midpoint of $B_{1} C_{1}$, and $A D_{1}$ intersects the circumcircle of $\triangle A B C$ at point $D$. Prove:
$$
\begin{array}{c}
A B \cdot A B_{1}+A C \cdot \\
A C_{1... | Proof: Since it is a problem of a cyclic quadrilateral, we consider using inversion.
As shown in Figure 3, perform the inversion \( I\left(A, A C \cdot A C_{1}\right) \), then \( C \rightarrow C_{1} \). Let \( D \rightarrow E, B \rightarrow F \).
Since points \( B, D, C \) lie on a circle passing through the inversio... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,193 |
Example 4 A circle with center $O$ passes through the vertices $A, C$ of $\triangle A B C$, and intersects sides $A B, B C$ at points $K, N$ respectively. The circumcircles of $\triangle A B C$ and $\triangle K B N$ intersect at points $B, M$. Prove that $\angle O M B=90^{\circ}$. | Proof: As shown in Figure 4, let the line passing through point $O$ and perpendicular to $BM$ be $l$. Therefore, we only need to prove that point $M$ lies on line $l$.
Taking $l$ as the axis of reflection, perform the axial reflection transformation $S(l)$. Let $C \rightarrow C'$ and $K \rightarrow K'$. Then $CC' \per... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,194 |
Example 5 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$. Prove: for every $n \in \mathbf{N}_{+}$, we have
$$
(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} \text {. }
$$
(1988, National High School Mathematics Competition) | Now let's analyze the conditions and conclusions to find new connections between them.
The condition $\frac{1}{a}+\frac{1}{b}=1$ means what? In the standard answer ${ }^{[10]}$, we see:
$a b=a+b$,
$$
a b=(a+b)\left(\frac{1}{a}+\frac{1}{b}\right) \geqslant 4,
$$
We can also derive
$$
(a-1)(b-1)=1=\frac{1}{a}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,195 |
Question 1 Let two positive sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
(1) $a_{0}=1 \geqslant a_{1}$,
$a_{n}\left(b_{n-1}+b_{n+1}\right)=a_{n-1} b_{n-1}+a_{n+1} b_{n+1}(n \geqslant 1)$;
(2) $\sum_{i=0}^{n} b_{i} \leqslant n^{\frac{3}{2}}(n \geqslant 1)$.
Find the general term of $\left\{a_{n}... | Solution: From condition (1), we have
$$
a_{n}-a_{n+1}=\frac{b_{n-1}}{b_{n+1}}\left(a_{n-1}-a_{n}\right) \text {. }
$$
Thus, $a_{n}-a_{n+1}=\frac{b_{0} b_{1}}{b_{n} b_{n+1}}\left(a_{0}-a_{1}\right)$.
If $a_{1}=a_{0}=1$, then $a_{n}=1$.
Next, assume $a_{1}a_{0}-a_{n}=b_{0} b_{1}\left(a_{0}-a_{1}\right) \sum_{k=0}^{n-1}... | a_{n}=1(n \geqslant 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,197 |
Example 4 As shown in Figure $4, A B$ is the diameter of $\odot O$, $B C$ is the tangent of $\odot O$, $B C=$ $A B, O C$ intersects $\odot O$ at point $F$, and line $A F$ intersects $B C$ at $E$. Prove: $B E=C F$.
(2005, National Junior High School Mathematics Competition Sichuan Preliminary) | Explanation: $B E$ and $C F$ are spatially separated in the figure. If we can construct another line segment $a$ at a suitable position in the figure that is equal to both $B E$ and $C F$, then we can use $a$ as a medium to complete the proof.
Notice that $O A=O F$. Consider constructing a line through $C$ parallel to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,198 |
Question 2 Let circle $\Gamma$ be the circumcircle of $\triangle ABC$, and let point $P$ be an interior point of $\triangle ABC$. The rays $AP$, $BP$, and $CP$ intersect circle $\Gamma$ at points $A_1$, $B_1$, and $C_1$, respectively. Let the points $A_1$, $B_1$, and $C_1$ be symmetric to points $A_2$, $B_2$, and $C_2$... | Proof 1: Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively, and let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Choose three points $A_3$, $B_3$, and $C_3$ on circle $\Gamma$ such that $AA_3$, $BB_3$, and $CC_3$ are diameters of circle $\Gamma$ (as ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,199 |
Question 5 Let $A$ be a non-empty subset of the set of positive integers. If all sufficiently large positive integers can be written as the sum of two numbers in $A$ (which may be the same), then $A$ is called a second-order basis. For $x \geqslant 1$, let $A(x)$ denote the set of all positive integers in $A$ that do n... | Prove: Let
$$
\begin{aligned}
A= & \left\{\sum_{i=1}^{n} 2^{2 b_{i}} 10 \leqslant b_{1}<\cdots<b_{n}, b_{i} \in \mathbf{Z}\right\} \cup \\
& \left\{\sum_{j=1}^{m} 2^{2 c_{j}+1} 10 \leqslant c_{1}<\cdots<c_{m}, c_{i} \in \mathbf{Z}\right\} .
\end{aligned}
$$
Since every integer can be expressed as
$$
2^{k_{1}}+2^{k_{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,202 |
Example 1 In $\triangle ABC$, prove:
(1) $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leqslant \frac{1}{8}$;
(2) $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \leqslant \frac{3}{2}$;
(3) $\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}$;
(4) $\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin... | Proof: Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, then
$$
\begin{array}{l}
\text { (1) } \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \\
=\frac{x y z}{\sqrt{\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)}} \\
=\frac{x y z}{(x+y)(x+z)(y+z)} \\
\leqslant \frac{x y z}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,204 |
Example 2 Let the circumradius of $\triangle ABC$ be $R$, and the area be $S$. Prove:
$$
\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \leqslant \frac{9 R^{2}}{4 S} .
$$ | Proof: Since $S=2 R^{2} \sin A \cdot \sin B \cdot \sin C$, we need to prove
$$
\begin{array}{l}
\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \\
\leqslant \frac{9}{8 \sin A \cdot \sin B \cdot \sin C} .
\end{array}
$$
Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, then we need to prove
$$
\begin... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,205 |
Example 3 (Weitzenböck's Inequality) Let the side lengths of $\triangle ABC$ be $a, b, c$, and the area be $S$. Prove:
$$
a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S .
$$ | Proof: Since $a=2 R \sin A, b=2 R \sin B$, $c=2 R \sin C, S=2 R^{2} \sin A \cdot \sin B \cdot \sin C$, we need to prove $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C$
$\geqslant 2 \sqrt{3} \sin A \cdot \sin B \cdot \sin C$.
Let $x=\cot A, y=\cot B, z=\cot C$, then we need to prove
$$
\begin{array}{l}
\frac{1}{1+x^{2}}+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,206 |
Example 4 Let positive numbers $a, b, c, x, y, z$ satisfy
$$
c y+b z=a, a z+c x=b, b x+a y=c .
$$
Find the minimum value of the function $f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}$.
(2005, National High School Mathematics Competition) | Given the conditions, we have
$$
b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0,
$$
which simplifies to $2 b c x+a^{2}-b^{2}-c^{2}=0$.
Solving for $x$, we get $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.
Similarly, $y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
Since $a, b, c, x, y, z$ are positive numbers,... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,207 |
Let $a, b, c$ be given complex numbers, and denote $|a+b|=m, |a-b|=n$, with $m n \neq 0$. Prove that:
$$
\max \{|a c+b|,|a+b c|\} \geqslant \frac{m n}{\sqrt{m^{2}+n^{2}}} \text {. }
$$ | Proof 1: Since
$$
\begin{array}{l}
\max \{|a c+b|,|a+b c|\} \\
\geqslant \frac{|b| \cdot|a c+b|+|a| \cdot|a+b c|}{|b|+|a|} \\
\geqslant \frac{|b(a c+b)-a(a+b c)|}{|a|+|b|}=\frac{\left|b^{2}-a^{2}\right|}{|a|+|b|} \\
\geqslant \frac{|b+a| \cdot|b-a|}{\sqrt{2\left(|a|^{2}+|b|^{2}\right)}},
\end{array}
$$
and
$$
\begin{a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,208 |
Example 5 As shown in Figure 5, from a point $P$ outside a circle, draw two tangents $PA$ and $PB$ to the circle, with $A$ and $B$ being the points of tangency. Draw a secant line through point $P$ that intersects the circle at points $C$ and $D$. Draw a line through the point of tangency $B$ parallel to $PA$, intersec... | Explanation: The conditions of this problem include parallel lines and tangents to a circle, which are conducive to generating proportional relationships. Therefore, we choose "to prove the equality of line segments using expressions as a medium."
To find a proportional relationship involving $EB$, we note that $\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,209 |
(1) If $2n-1$ is a prime number, then for any $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, there exist $i$, $j \in \{1,2, \cdots, n\}$, such that
$$
\frac{a_{i}+a_{j}}{(a_{i}, a_{j})} \geqslant 2n-1;
$$
(2) If $2n-1$ is a composite number, then there exist $n$ distinct positive integers $a_{1}, a_{2}, ... | (1) Let $2n-1$ be a prime number $p$, and assume $\left(a_{1}, a_{2}, \cdots, a_{n}\right)=1$.
If there exists $i(1 \leqslant i \leqslant n)$ such that $p \mid a_{i}$, there must exist $j \neq i$ such that $p \nmid a_{j}$. Since $p \times \left(a_{i}, a_{j}\right)$, we have
$$
\frac{a_{i}+a_{j}}{\left(a_{i}, a_{j}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,210 |
Three, given that $a_{1}, a_{2}, \cdots, a_{11}$ are 11 distinct positive integers, and their total sum is less than 2007. On the blackboard, the numbers $1, 2, \cdots, 2007$ are written in sequence. Define a sequence of 22 consecutive operations as an operation group: the $i$-th operation can select any number from th... | 3. The number of excellent operation groups is more by $\prod_{i=1}^{11} a_{i}$.
We introduce a general notation: If the blackboard has the numbers 1, 2, ..., $n$, an operation group is defined as $l$ consecutive operations: the $i$-th operation can select any number from the existing numbers on the blackboard and add... | \prod_{i=1}^{11} a_{i} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,211 |
Let $O$ and $I$ be the circumcenter and incenter of $\triangle A B C$, respectively. The incircle of $\triangle A B C$ touches sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively. Line $F D$ intersects $C A$ at point $P$, and line $D E$ intersects $A B$ at point $Q$. Points $M$ and $N$ are the midpo... | Let's assume $a > c$. Considering $\triangle ABC$ and the transversal $PFD$, by Menelaus' theorem, we have $\frac{CP}{PA} \cdot \frac{AF}{FB} \cdot \frac{BD}{DC} = 1$. Therefore,
$$
\frac{PA}{PC} = \frac{AF}{FB} \cdot \frac{BD}{DC} = \frac{AF}{DC} = \frac{p-a}{p-c}.
$$
Thus, $\frac{PA}{CA} = \frac{p-a}{a-c}$. Therefor... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,212 |
Five, let the bounded sequence $\left\{a_{n}\right\}_{n \neq 1}$ satisfy
$$
a_{n}<\sum_{k=n}^{2 n+2006} \frac{a_{k}}{k+1}+\frac{1}{2 n+2007}, n=1,2, \cdots \text {. }
$$
Prove: $a_{n}<\frac{1}{n}, n=1,2, \cdots$. | Five, let $b_{n}=a_{n}-\frac{1}{n}$, then
$$
b_{n}<\sum_{k=n}^{2 n+2006} \frac{b_{k}}{k+1}(n \geqslant 1) \text {. }
$$
We will prove that $b_{n}<0$.
Since $a_{n}$ is bounded, there exists a constant $M$ such that $b_{n}<M$.
When $n \geqslant 100000$, we have
$$
\begin{array}{l}
b_{n}<\sum_{k=n}^{2 n+2006} \frac{b_{k}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,213 |
1. As shown in Figure 1, in $\triangle A B C$, $\angle A=70^{\circ}$, $\angle B=90^{\circ}$, the symmetric point of point $A$ with respect to $B C$ is $A^{\prime}$, the symmetric point of point $B$ with respect to $A C$ is $B^{\prime}$, and the symmetric point of point $C$ with respect to $A B$ is $C^{\prime}$. If the ... | -1.3 .
As shown in Figure 6, connect $B^{\prime} B$, and extend it to intersect $C^{\prime} A^{\prime}$ at point $D$, and intersect $A C$ at point $E$. Given $C^{\prime} B=B C$, $A^{\prime} B=B A, A C \perp A^{\prime} C^{\prime}$, and $B B^{\prime} \perp A C, B^{\prime} E=B E$, we get $B^{\prime} D=3 B E$. Therefore,
$... | -1.3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,214 |
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