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Four, (Full marks 14 points) On a line, there are four points $A, B, C, D$ in sequence, with $AB: BC: CD = 2: 1: 3$. Circles $\odot O_{1}, \odot O_{2}$ are constructed with diameters $AC$ and $ED$ respectively, and the two circles intersect at points $E, F$ (as shown in the figure). Find the value of $ED: EA$. | Connect $E B, E C$, and draw $C G$ perpendicular to $E B$ intersecting $A E, B E$ at $G, H$.
$$
\because D E \perp B E, \therefore D E
$$
//CG, from the given conditions $A C$ :
$$
\begin{array}{l}
C D=3,3=A G: G E, \therefore \\
A G=G E .
\end{array}
$$
$\because C H: D E=B C: B D=1: 4$, and $C G: D E=A C$ :
$$
A D=1:... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,455 |
Five. (Full marks 14 points) If the sum of the digits of a natural number and the product of its digits add up to exactly the natural number itself, we call it a "lucky number." Try to find the sum of all "lucky numbers."
---
Translate the above text into English, please retain the original text's line breaks and for... | Let the $n$-digit natural number be $N=\overline{a_{1} a_{2} \cdots a_{n}}$, then
\[ N=a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \]
(Where $a_{1}, a_{2}, \cdots, a_{n}$ are the digits of $N$).
If $N$ is a "lucky number", then
\[
\begin{array}{l}
a_{1} \times 10^{n-1}+a_{2} \times 10^{... | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,456 |
Example 12. Find all positive integer solutions to the equation $\frac{x+y}{x^{2}-x y+y^{2}}=\frac{3}{7}$.
(Eighth Soviet Mathematical Olympiad) | Solving the original equation can be transformed into an equation about $x$:
$$
3 x^{2}-(3 y+7) x+3 y^{2}-7 y=0 \text {. }
$$
Since $x$ is a real number, the discriminant is not less than zero, that is,
$$
\Delta=(-3 y-7)^{2}-4 \times 3\left(3 y^{2}-7 y\right) \geqslant 0 .
$$
Simplifying, we get $27 y^{2}-126 y-49 \... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,459 |
4. If the following conditions about a triangle are known, the one that cannot uniquely determine the shape of the triangle is ( ).
(A) the ratio of two sides and the included angle
(B) the ratio of the three altitudes
(C) the ratio of the three medians
(D) the ratio of one side and the altitude to that side | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,461 |
1. When $a<0$, the simplified result of $2 a+|-2 a|+|a-2|$ is $\qquad$ . | 1. $2-a$ | 2-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,464 |
2. To make the algebraic expression $x^{2}+y^{2}-14 x+2 y+50$ equal to 0, the values of $x, y$ should be $\qquad$ . | 2. $\left\{\begin{array}{l}x=7 \\ y=-1\end{array}\right.$ | x=7, y=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,465 |
4. If $m0,|m|<|n|$, and $|x+m|$ $+|x-n|=m+n$, then the range of values for $x$ is $\qquad$ . | $\begin{array}{l}\text { 4. }-m \\ \leqslant x \leqslant n\end{array}$ | -m \leqslant x \leqslant n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,467 |
5. Given that the three sides of a triangle are unequal and all prime numbers, and the perimeter is a composite number not greater than 35. Then the three sides of the triangle are . $\qquad$ | 5. $(3,5,7),(3,11,13),(7,11,17),(5,13,17)$ | (3,5,7),(3,11,13),(7,11,17),(5,13,17) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,468 |
Example 13. Let $x, y, z$ be real numbers, satisfying $x+y+z=a$, $x^{2}+y^{2}+z^{2}=\frac{a^{2}}{2}(a>0)$. Prove: $x, y, z$ cannot be negative or greater than $\frac{2}{3} a$, and they cannot all be equal.
(1990, Hefei Junior High School Mathematics Competition) | Prove (1) If $x=y=z$, then from $x+y+z=a$ we can get $x=y=z=\frac{1}{3} a$, and $x^{2}+y^{2}+z^{2}=3 \cdot\left(\frac{1}{3} a\right)^{2}$ $=\frac{1}{3} a^{2}$. But $u>0$, this leads to $x^{2}-y^{2}+x^{2}=\frac{1}{2} a^{2}$ contradiction. Therefore, $x, y, x$ cannot be equal at the same time;.
(2) Given that $z=a-x-y$, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,470 |
Four. (Full marks 18 points) Given that $\odot O$ is circumscribed around $\triangle A B C, A B$, $B C, C A$ are not diameters of $\odot O$, and no line containing a diameter of $\odot O$ can place points $A, B, C$ on the same side of this line.
(i) What type of triangle is $\triangle A B C$? Why?
(2) Prove that in $\t... | (1) $\triangle A B C$ is an acute triangle.
Because when one of the segments $A B, B C, C A$ is the diameter of $\odot O$, then $\triangle A B C$ is a right triangle.
If points $A, B, C$ are on the same side of a certain diameter, then among the three angles (inscribed angles) of $\triangle A B C$, there must be one a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,472 |
2. Party A bought a share of Cao for 1000 RMB, and then he resold this share to Party B, making a profit of $10 \%$. Afterwards, Party B sold the share back to Party A, incurring a loss of $10 \%$. Finally, Party A sold the share to Party B at a 10% discount of the price Party B sold it to Party A. In the above stock t... | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,474 |
3. A real estate development company purchased a piece of land for 1 million yuan, on which a building with each floor being 1000 square meters can be constructed. The total construction cost per square meter (i.e., the sum of the area of each floor) is related to the height of the building; the average construction co... | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,475 |
5. Let $t$ be a real root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$, then the relationship between the discriminant $\Delta=b^{2}-4 a c$ and the quadratic form $M=(2 a t+b)^{2}$ is ( ).
(A) $\Delta>M$
(B) $\Delta=M$.
(C) $\Delta<M$
(D) Cannot be determined
直接输出翻译结果。 | 5. B (Hint: Since $t$ is a root of the equation, we have $a t^{2}+b t+c=0$. Thus, $M=4 a^{2} t^{2}+4 a b t+b^{2}=4 a\left[a t^{2}+b t+c\right]-4 a c$ $+b^{2}=b^{2}-4 a c=\Delta.$) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,477 |
6. If the equation $(m+2) x^{2}-2(m+1) x+m=0$ has only one real root, then the equation $(m+1) x^{2}-2 m x+m-2=0$ ( . ).
(A) has no real roots
(B) has two distinct real roots
(C) has two equal real roots
(D) the number of real roots cannot be determined | 6. C. (Hint: The discriminant of the original equation is $\Delta=4(m+1)^{2}$ $-4 m(m+2)=4>0$, the equation has only one root, so the coefficient of the quadratic term is 0, i.e., $m+2=0, m=-2$. ) | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,478 |
7. Given that $H$ is the orthocenter of $\triangle A B C$, the circumradius of $\triangle A B C$ is $R$, and the circumradius of $\triangle B H C$ is $r$. Then the relationship between $R$ and $r$ is ( ).
(A) $R=r$
(B) $R>\dot{r}$.
(C) $R<r$; when $\triangle A B C$ is an obtuse triangle, $R<r$ | 7. A. (Hint: It is known that $\angle B H C=180^{\circ}-\angle A$, in $\triangle B A C$, $B C=2 R \sin A$, in $\triangle B H C$, $B C=$ $2 r \sin \left(180^{\circ}-\angle A\right)=2 r \sin A$, thus $R=r_{.}$) | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,479 |
1. Given $0<x<1$. Simplify
$$
\sqrt{\left(x-\frac{1}{x}\right)^{2}+4}-\sqrt{\left(x+\frac{1}{x}\right)^{2}-4}=
$$
$\qquad$ | 二、1.2x. (Hint: Given $0<x<1$, we have $x<\frac{1}{x}$.) | 2x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,480 |
Example 14. Given that $a$ is a real number, and the equation $x^{2}-a x + a=0$ has two real roots $\alpha, \beta$. Prove: $\alpha^{2}+\beta^{2} \geqslant 2(\alpha+\beta)$. (1991, Sichuan Province Junior High School Mathematics Competition) | Given that $\alpha+\beta=a, \alpha \beta=a$.
From the equation $x^{2}-a x+a=0$ having real roots, we have $\Delta=a^{2}-4 a \geqslant 0$,
which means $(\alpha+\beta)^{2}-4 \alpha \beta \geqslant 0$ or $\alpha^{2}+\beta^{2} \geqslant 2 \alpha \beta$.
But $\alpha \beta=\alpha+\beta$,
thus $\alpha^{2}+\beta^{2} \geqslant ... | \alpha^{2}+\beta^{2} \geqslant 2(\alpha+\beta) | Algebra | proof | Yes | Yes | cn_contest | false | 708,481 |
3. The number of diagonals in a convex nonagon is $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ is kept as is in the translation. | 3. 27. (Hint: Each vertex leads to $(9-3)=6$ diagonals, there are 9 vertices in total, and each diagonal passes through two vertices, so the total number of diagonals is $6 \times 9 \div 2=27$.) | 27 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,483 |
4. As shown in the figure, $E, F$ are the midpoints of the two sides $A B, B C$ of the square $A B C D$, and $A F, C E$ intersect at point $G$. If the area of the square is 1, then the area of quadrilateral $A G C D$ is
Translate the above text into English, please keep the original text's line breaks and format, and ... | 4. $\frac{2}{3}$. (Hint: $G$ is the centroid of $\triangle A B C$. There is $S_{\triangle A C C}$ $\left.=\frac{1}{3} S_{\triangle A B C}.\right)$ | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 708,484 |
5. The solution to the equation $\frac{x^{3}}{\sqrt{4-x^{2}}}+x^{2}-4=0$ is $\qquad$ | 5. $\sqrt{2}$, (Hint: Transform the equation to $x^{3}=\left(4-x^{2}\right)$ - $\sqrt{4-x^{2}}$, then take the cube root of both sides and square them, resulting in $x= \pm \sqrt{2}$, but $x=-\sqrt{2}$ is an extraneous root and should be discarded.) | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,485 |
6. Given $a \cdot b \neq 1$, and $5 a^{2}+1995 a+8=0$ as well as $8 b^{2}$ $+1995 b+5=0$. Then $\frac{a}{b}=$ $\qquad$ . | 6. $\frac{8}{5}$. (Hint: It is known that $a$ and $\frac{1}{b}$ are the roots of the equation $5 x^{2}+1995 x$ $+8=0$, apply Vieta's formulas. ) | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,486 |
7. Given that the six interior angles of a regular hexagon are all $120^{\circ}$, and the lengths of four consecutive sides are $1, 9, 9, 5$ cm respectively. Then, the perimeter of this hexagon is $\qquad$ cm. | 7. 42. (Hint: Extend the three pairs of opposite sides of the hexagon, the result is an equilateral triangle, it is easy to know that the side length of the equilateral triangle is 19 cm, thus find the other two side lengths of the hexagon to be 13 cm and 5 cm.) | 42 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,487 |
Three. (Full marks 10 points) Given that the sum of two three-digit numbers $\overline{a b c}, \overline{d e f}$, $\overline{a b c}+\overline{d e f}$, is divisible by 37. Prove: The six-digit number $\overline{a b c d e f}$ is also divisible by 37. | $$
\begin{aligned}
\equiv 、 \overline{a b c d e f} & =\overline{a b c} \times 1000+\overline{d e f} \\
& =\overline{a b c} \times 999+(\overline{a b c}+\overline{d e f}) .
\end{aligned}
$$
Since $999=9 \times 111=9 \times 3 \times 37$, that is, 999 can be divided by 37. Given that $\overline{a b c}+\overline{\operator... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,488 |
Four. (Full marks 14 points) As shown in the figure, $O$ is the intersection point of the diagonals of quadrilateral $A B C D$. It is known that $\angle B A D+\angle B C A$
$$
\begin{array}{l}
=180^{\circ}, A B=5, A C=4, A D \\
=3, \frac{B O}{D O}=\frac{7}{6} . \text { Find } B C .
\end{array}
$$ | Draw $B B^{\prime} \perp A C, C C^{\prime} \perp$ $A B, D D^{\prime} \perp A C$, with the feet of the perpendiculars being $B^{\prime}, C^{\prime}, D^{\prime}$ (as shown in the figure).
It is easy to see that $\triangle B O B^{\prime} \backsim$ $\triangle D O D^{\prime}$, so $\frac{B O}{O D}=\frac{B B^{\prime}}{D D^{\... | \frac{14}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,489 |
Five. (Full marks 13 points) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha>\beta$. Without solving the equation, use the relationship between roots and coefficients to find the value of $\frac{2}{\alpha}+3 \beta^{2}$.
---
The above text translated into English, preserving the o... | Given the relationships between the roots and coefficients $\alpha+\beta=7, \alpha \beta=8$, we have
$$
\begin{array}{l}
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=7^{2}-2 \times 8=33, \\
(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=17 .
\end{array}
$$
Since $\alpha>\beta$, we have $\alpha-\beta=\s... | \frac{1}{8}(403-85 \sqrt{17}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,490 |
*Six. (Full marks 10 points) Given an isosceles triangle $\triangle A B C$, where $A B=A C$. Find all points $M$ in the plane such that $\triangle A B M$ and $\triangle A C M$ are both isosceles triangles. (Only indicate the positions of these points, no proof is required.)
Translate the text above into English, p... | Six, (1) All points on the circle with center $A$ and radius $A B$ (excluding points $B, C$ and the points symmetric to $B, C$ about $A$).
(2) In addition, there are six points (as shown in the figure): the intersection point $M_{1}$ of the perpendicular bisectors of $A B$ and $A C$;
the point $M_{2}$ symmetric to $A$ ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,491 |
Example 15. As shown in the figure, in $\triangle ABC$, points $D$ and $E$ are on sides $BC$ and $AB$ respectively, and $\angle 1 = \angle 2 = \angle 3$. If the perimeters of $\triangle ABC$, $\triangle EBD$, and $\triangle ADC$ are $m$, $m_1$, and $m_2$ respectively, prove that $\frac{m_1 + m_2}{m} \leq \frac{5}{4}$.
... | Proof: Let $BC=a, AC=b, AB=c$. Then $m=a+b+c$. Also, by $\angle 1=\angle 2=\angle 3$, we have $DE \parallel AC$, thus
$\triangle ABC \sim \triangle EBD \sim \triangle DAC$.
We have $DC: b=b: a$,
$ED \cdot b=(a-DC) \cdot a$.
From (1) and (2), we get $\frac{ED}{b}=1-\left(\frac{b}{a}\right)^{2}$.
Also, since $m_{1}: m=ED... | \frac{m_{1}+m_{2}}{m} \leqslant \frac{5}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 708,492 |
1. If $x=\frac{23+\sqrt{469}}{5}$, then $25 x^{4}-1996 x^{2}+144$ $=(\quad)$.
(A) 0
(B) 1
(C) 469
(D) 1996 | ,- 1. A.
From the difference in the number of times the letters appear in the condition and conclusion, we know that we should either increase the degree of the condition or decrease the degree of the conclusion (factorize).
Given $5 x-23=\sqrt{469}$,
squaring gives $5 x^{2}-46 x+12=0$.
Rearranging, $46 x=5 x^{2}+12$.... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,493 |
2. In $\triangle A B C$, one side is half of another, and one of the angles is $30^{\circ}$. Then the shape of this triangle is ( ).
(A) not a right triangle
(B) not an obtuse triangle
(C) not an acute triangle
(D) any of the above is possible | 2. C.
Let the three sides of the triangle be $a, 2a, b$. Clearly, the side opposite the $30^{\circ}$ angle is not the longest side.
(1) If the side opposite the $30^{\circ}$ angle is $a$, then the angle opposite $2a$ is $90^{\circ}$, making $\triangle ABC$ a right triangle;
(2) If the side opposite the $30^{\circ}$ an... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,494 |
3. Given $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}=3$. Then $(a-b)^{2}+(b$ $-c)^{2}+(a-b)(b-c)$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4 | 3. C.
$$
\begin{aligned}
3 & =\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c} \\
& =\frac{(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)}{a+b+c} \\
& =(a-b)^{2}+(b-c)^{2}+(a-b)(b-c) .
\end{aligned}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,495 |
4. The coordinates of point $M(x, y)$ satisfy $|x+y|<|x-y|$. Then the quadrant in which point $M$ is located is ( ).
(A) 1,3
(B) 2,4
(C) 1,2
(D) 3,4 | 4. B.
From $|x+y|<|x-y|$, squaring both sides gives
$$
x^{2}+2 x y+y^{2}<x^{2}-2 x y+y^{2} \text {, }
$$
which simplifies to $x y<0$.
Therefore, point $M$ is in the 2nd and 4th quadrants. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,496 |
5. In trapezoid $ABCD$, the diagonals intersect at $O$, with $OA>OC$ and $OB>OD$. Take point $E$ on $AO$ such that $AE=OC$, and take point $F$ on $BO$ such that $BF=OD$. Then the relationship between the area $S_{1}$ of $\triangle AFC$ and the area $S_{2}$ of $\triangle BED$ is ().
(A) $S_{1}>S_{2}$
(B) $S_{1}=S_{2}$
(... | 5. B.
From the given information, we have
$$
\begin{array}{l}
S_{\triangle A B D}=S_{\triangle A B C}, S_{\triangle A E D}=S_{\triangle O C D}, \\
S_{\triangle A E B}=S_{\triangle B O C}=S_{\triangle A O D}=S_{\triangle A B F}, \\
S_{\triangle O C D}=S_{\triangle B F C} .
\end{array}
$$
$$
\text { Therefore, } \quad \... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,497 |
6. If the tens digit of $a^{2}$ can be $1,3,5,7,9$, then the units digit of $a$ is ).
(A) must be 4
(B) must be 6
(C) can be 4 or 6
(D) $2,4,6,8$ are all possible. | 6. C.
Let $a=10b+c$, where $c$ is taken from $0,1,2, \cdots, 9$. Write $c^{2}$ as a two-digit number: $00,01,04,09,16,25,36,49,64,81$. Among these, only $c=4,6$ have an odd tens digit.
Also, $a^{2}=(10b+c)^{2}$
$$
=2 \times (5b^{2}+bc) \times 10 + c^{2}.
$$
It can be seen that the tens digit of $a^{2}$ is an even num... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,498 |
1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ . | II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,499 |
2. Positive numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a+b+c=10, \\ a^{2}+b^{2}=c^{2} .\end{array}\right.$ Then the maximum value of $ab$ is $\qquad$ | 2. $50(3-2 \sqrt{2})$.
With sides $a, b, c$, a right triangle $\triangle ABC$ can be constructed, with a perimeter of 10. Since among right triangles with a fixed perimeter, the isosceles right triangle has the maximum area, when
$$
a=b=5(2-\sqrt{2}), c=10(\sqrt{2}-1)
$$
$a b$ has the maximum value of $50(3-2 \sqrt{2... | 50(3-2 \sqrt{2}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,500 |
3. There are two roads $O M, O N$ intersecting at a $30^{\circ}$ angle. Along the direction of road $O M$, 80 meters from $A$ is a primary school. When a tractor travels along the direction of $O N$, areas within 50 meters on both sides of the road will be affected by noise. Given that the speed of the tractor is 18 ki... | 3. 12 .
As shown in the figure, when the tractor reaches point $B$, it starts to have an impact, and when it reaches point $C$, the impact ceases. Thus, $A B = A C = 50$ meters. Draw $A D \perp$ $O N$ from $A$. Given that $\angle A O D = 30^{\circ}$, we know that $A D = \frac{1}{2} O A = 40$ meters. Therefore, $B D = ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,501 |
Example 16. Given as shown, the area of $\triangle ABC$ is $S$, a line $l \parallel BC$ intersects $AB, AC$ at points $D, E$ respectively, and the area of $\triangle BED$ is denoted as $K$. Prove that $K \leqslant \frac{1}{4} S$.
(2nd Zu Chongzhi Cup Junior High School Mathematics Invitational) | Let $S_{\triangle A D E}=S_{1}, S_{\triangle B C E}=S_{2}$. Then
$$
S_{1}: K=A D: D B,\left(S_{1}+K\right): S_{2}=A E: E C \text {. }
$$
From $l / / B C$ we get $A D: D B=A E \times E C$, then we have
$$
S_{1}: K=\left(S_{1}+K\right): S_{2} \text {, }
$$
which means $S_{1}^{2}+(2 K-S) S_{1}+K^{2}=0$.
Since $S_{1}$ is... | K \leqslant \frac{1}{4} S | Geometry | proof | Yes | Yes | cn_contest | false | 708,503 |
One, (25 points) In trapezoid $ABCD$, $AB \parallel CD, AB=125, CD=$ $DA=80$. Can diagonal $BD$ divide the trapezoid into two similar triangles? If not, provide a proof; if so, find the lengths of $BC, BD$. | First, assume that two triangles can be similar, and $AD$ is the corresponding side.
(1) If $AD$ and $DC$ are corresponding sides, then the similarity ratio is 1, and the two triangles are congruent. Their corresponding angles $\angle ABD = \angle DBC$, but from $AB \parallel CD$, we know $\angle ABD = \angle BDC$, so ... | BC = 64, BD = 100 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,504 |
For $a>b>c>0$, consider the quadratic equation
$$
x^{2}-(a+b+c) x+(a b+b c+c a)=0 \text {. }
$$
(1) If the equation has real roots, prove that $a, b, c$ cannot be the lengths of the three sides of a triangle;
(2) If the equation has a real root $x_{0}$, prove that $a>x_{0}>b+c$;
(3) When the equation has real roots 6 a... | (1) From the equation having real roots, we know the discriminant is non-negative, i.e.,
$$
\begin{aligned}
0 \leqslant \Delta & =(a+b+c)^{2}-4(a b+b c+c a) \\
& =a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a \\
& =a(a-b-c)-b(a+c-b)-c(a+b-c) \\
& 0$, hence $a-b-c>0$, i.e., $a>b+c$.
This indicates that $a, b, c$ cannot be the thr... | a=10, b=4, c=1 | Algebra | proof | Yes | Yes | cn_contest | false | 708,505 |
*1. Set $M$ consists of more than two consecutive natural numbers, and the sum of its elements is 1996. Such a set $M$ ( ).
(A) does not exist
(B) exists uniquely
(C) exists in two forms
(D) exists in more than three forms | $-、 1$. B.
Let $M=\{a, a+1, \cdots, a+n-1\}$. From $\frac{(a+a+n-1) n}{2}=1996$, we have $n(2 a+n-1)=8 \times 499$.
Since $n$ and $2 a+n-1$ have different parities, and $n<2 a+n-1$, and 499 is a prime number, we can only have $n=8$, then $2 a+n-1=2 a+7=499$, solving for $a$ gives $a=246$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,507 |
2. $x \in[0, \pi], y=\sin (\cos x)$ minimum value is $a$, $y=\cos (\sin x)$ minimum value is $b, y=\tan(\sin x)$ minimum value is $c$, $y=\cot(\cos x)$ minimum value is $d$. Then the relationship of $a, b, c, d$ is ( ).
(A) $d<a<c<b$
(B) $c<a<b<d$
(C) $a<d<c<b$
(D) $b<a<d<c$ | 2. C.
Since $a=\sin (-1)<0, b=\cos 1, c=0, d=\operatorname{ctg}(-1)<0$, we know that $b$ is the largest, $c=0$ is the second, and note that $a<d$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,508 |
3. If the equation $2 a \cdot 9^{\sin x}+4 a \cdot 3^{\sin x}+a-8=0$ has a solution, then the range of values for $a$ is ( ).
(A) $a \geqslant 0$ or $a \leqslant-8$
(B) $a \geqslant 0$
(C) $0 \leqslant a \leqslant \frac{8}{31}$
(D) $\frac{8}{31} \leqslant a \leqslant \frac{72}{23}$ | 3. D.
Let $t=3^{\sin x}$. Then $\frac{1}{3} \leqslant t \leqslant 3$. The original equation is $2 a t^{2}+4 a t+a$ $-8=0, a=\frac{8}{2(t+1)^{2}-1}$.
It is a monotonically decreasing function in $t \in\left[\frac{1}{3}, 3\right]$.
When $t=\frac{1}{3}$, the maximum value of $a$ is $=\frac{8}{2 \cdot\left(\frac{4}{3}\rig... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,509 |
4. The major axis of the ellipse is 6, the left vertex is on the circle $(x-3)^{2}+(y-2)^{2}$ $=4$, and the left directrix is the $y$-axis. Then the range of the eccentricity $e$ of the ellipse is ( ).
(A) $\frac{3}{8} \leqslant e \leqslant \frac{3}{4}$
(B) $\frac{1}{4} \leqslant e \leqslant \frac{3}{8}$
(C) $\frac{1}{... | 4. A.
Since $a=3$, let the left vertex be $(m, n)$, the center of the ellipse is $(m+3, n)$, and the $y$-axis is the directrix, $\frac{a^{2}}{c}=m+3$. Therefore, $c=\frac{9}{m+3}$, and $e=\frac{c}{a}=\frac{3}{m+3}$.
Since the vertex lies on the circle $(x-3)^{2}+(y-2)^{2}=4$, we have $1 \leqslant m \leqslant 5$, so $... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,510 |
*5. Let $I=\{1,2, \cdots, 100\} . M$ represents the number of subsets of $I$ where the maximum element is 66, $N$ represents the number of subsets of $I$ where the minimum element is 33, and $P$ represents the number of subsets of $I$ where the maximum element is three times the minimum element. Then ( ).
(A) $M<N<P$
(... | 5. B.
$M=2^{65}$. Since the set $\{1,2, \cdots, 65\}$ has $2^{65}$ subsets. Each subset, when 66 is added, results in all subsets with the maximum element being 66. Similarly, all subsets of $\{34, 35, \cdots, 100\}$, when 33 is added, result in all subsets with the minimum element being 33. Therefore, $N=2^{67}$. $P$ ... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,511 |
6. Let the complex number $z \neq 1, z^{13}=1$. Then the value of $z^{1}+z^{3}+z^{4}+z^{9}+z^{10}+z^{12}$ is ( ).
(A) Rational number
(B) Irrational number
(C) Imaginary number
(D) Pure imaginary number | 6. B.
From $z^{13}-1=0$, and $z \neq 1$, we have $z^{1}+z^{2}+\cdots+z^{12}=-1$. Let $x=z^{1}+z^{3}+z^{4}+z^{0}+z^{10}+z^{12}$, then
$$
\begin{aligned}
x^{2}= & z^{2}+z^{6}+z^{8}+z^{5}+z^{7}+z^{11}+2\left(z^{4}+z^{5}+z^{10}\right. \\
& +z^{11}+1+z^{7}+z^{12}+1+z^{2}+1+z+z^{3} \\
& \left.+z^{6}+z^{8}+z^{0}\right) \\
= ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,512 |
1. Let real numbers $x, y$ satisfy the equation $9 x^{2}+4 y^{2}-3 x+2 y=0$. Then the maximum value of $z=3 x+2 y$ is $\qquad$ . | From $9 x^{2}+4 y^{2}=3 x-2 y$ we know $3 x-2 y \geqslant 0$ and $\frac{9 x^{2}+4 y^{2}}{3 x-2 y}$ $=1$,
we have $z=3 x+2 y=\frac{(3 x)^{2}-(2 y)^{2}}{3 x-2 y} \leqslant \frac{(3 x)^{2}+(2 y)^{2}}{3 x-2 y}$ $=1$ (equality holds when $y=0$). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,513 |
Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition) | Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$
Eliminating the... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,514 |
*2. Let $M, N$ be two points on the line segment $AB$, $\frac{AM}{MB}=\frac{1}{4}, \frac{AN}{NB}=\frac{3}{2}$. Construct any right triangle $\triangle ABC$ with $AB$ as the hypotenuse. Then construct $MD \perp BC$ at $D$, $ME \perp AC$ at $E$, $NF \perp BC$ at $F$, and $NG \perp AC$ at $G$. The maximum possible value o... | 2. $\frac{10-4 \sqrt{3}}{5}$.
Let $\angle A=\alpha\left(0<\alpha<\frac{\pi}{2}\right) . AB=5$, then $AC=5 \cos \alpha$, $BC=5 \sin \alpha, MD=4 \cos \alpha, ME=\sin \alpha, NF=2 \cos \alpha, NG$ $=3 \sin \alpha$, thus
$$
\begin{aligned}
y & =\frac{4 \cos \alpha+\sin \alpha+2 \cos \alpha+3 \sin \alpha}{5 \cos \alpha+5 ... | \frac{10-4 \sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,515 |
3. The moving line $l$ intersects the $y$-axis at $A$, and the $x$-axis at $B$. Let the area of $\triangle A O B$ be a constant $S$. A perpendicular is drawn from the origin $O$ to $l$, with the foot of the perpendicular being $P(x, y)$. Then the equation of the locus of point $P$ is $\qquad$ | 3. $\left(x^{2}+y^{2}\right)^{2}=2 S|x y|$.
Let the equation of $l$ be $y=k x+b$. Then $A(0, b), B\left(-\frac{b}{k}, 0\right)$.
From $\frac{1}{2}\left|b \cdot\left(-\frac{b}{k}\right)\right|=S, b^{2}=2|k| S$, we know that the equation of $O P$ is $y=-\frac{1}{k} x$.
By multiplying the equations of $l$ and $O P$, we g... | \left(x^{2}+y^{2}\right)^{2}=2 S|x y| | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,516 |
4. A regular square pyramid $S-A B C D$. Extend one side of the base $C D$ to $E$, such that $D E=2 C D$. A plane is drawn through $B, E$ and the midpoint $F$ of edge $S C$, dividing the pyramid into two parts. The ratio of the volumes of these two parts is $\qquad$ | 4. $31: 29$.
Let the side length of the base of a regular quadrilateral pyramid be $a$, and the height $h$. Then $S_{\triangle B C E}=\frac{1}{2} a \cdot 3 a=\frac{3}{2} a^{2}$.
Since $I D: a=2: 3$,
Therefore, $S_{\triangle I D E}=\frac{1}{2} \cdot \frac{2}{3} a \cdot 2 a=\frac{2}{3} a^{2}$.
Let the volume of the tetr... | 31: 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,517 |
5. Write $(\sqrt{7}-\sqrt{6})^{6}$ in the form $\sqrt{N+1}-\sqrt{N}$, where $N$ is a natural number. Then $N=$ $\qquad$ | 5. $N=76545000$.
Let $(\sqrt{7}-\sqrt{6})^{6}=\sqrt{N+1}-\sqrt{N}$, then $(\sqrt{7}-\sqrt{6})^{6}+\sqrt{N}=\sqrt{N+1}$.
Square it to get
$$
\begin{array}{l}
(\sqrt{7}-\sqrt{6})^{12}+2 \sqrt{N}(\sqrt{7}-\sqrt{6})^{6}+N \\
= N+1 \\
2 \sqrt{N}=\frac{1-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=\frac{(7-6)^{6}... | 76545000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,518 |
*6. From $1,2, \cdots, 1996$, select $k$ numbers such that the sum of any two numbers cannot be divisible by their difference. Then the maximum possible value of $k$ is $\qquad$ . | 6. 666 .
Classify $1,2, \cdots, 1996$ by their remainder when divided by 3:
$$
\begin{array}{l}
3 k+1: 1,4,7, \cdots, 1996, \text { a total of 666; } \\
3 k+2: 2,5,8, \cdots, 1994, \text { a total of 665; } \\
3 k: 3,6,9, \cdots, 1995, \text { a total of 665. }
\end{array}
$$
Select all $3 k+1$ numbers, a total of 666... | 666 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,519 |
一、(25 points) Let $f(x)$ be a polynomial with integer coefficients, and $|f(x)|=17$ has five distinct integer roots. Prove: The equation $f(x)=0$ has no integer roots. | Assuming $f(x)=0$ has an integer root $x_{0}$, then $f(x)$ can be divided by $x - x_{0}$, i.e., $f(x) = (x - x_{0}) \psi(x)$, where $\psi(x)$ is still a polynomial with integer coefficients. Let $a, b, c, d, e$ be five distinct integer roots of $|f(x)|=17$. Then $|f(a)|=17$, i.e., $|(a - x_{0}) \psi(a)|=17$, hence $|a ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,520 |
*Ni, (25 minutes) In space, there are $n$ planes $(n \geqslant 4)$, no two of which are parallel, and no three of which are concurrent. Among their pairwise intersection lines, what is the maximum number of pairs of skew lines? | Let $1,2, \cdots, n$ represent $n$ planes, and let $(i, j)$ be the line of intersection between plane $i$ and plane $j$. Let $a_{n}$ be the maximum number of skew lines among the pairwise intersections of $n$ planes. When the $(n+1)$-th plane is added, the total number of intersection lines increases by $n$ lines: $(n+... | 3 C_{n}^{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,521 |
* Three. (35 points) Through two points $C_{1}, C_{2}$ outside a circle, draw the tangents $C_{1} A_{1}, C_{1} B_{1}$ and $C_{2} A_{2}, C_{2} B_{2}$ (where $A_{1}, A_{2}, B_{1}, B_{2}$ are the points of tangency).
(1) What is the necessary and sufficient condition for the chords $A_{1} B_{1}$ and $A_{2} B_{2}$ to inter... | (1) A necessary and sufficient condition is that the line $C_{1}, C_{2}$ does not intersect with circle $O$.
First, draw two tangents $C_{1} A_{1}, C_{1} B_{1}$ from $C_{1}$ to the circle.
If $C_{2}$ is inside $\angle A_{1} C_{1} B_{1}$, then $C_{1} C_{2}$ intersects the circle, and the two chords $A_{1} B_{1}$ and $A_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,522 |
夫四、(35 points) A school has $n$ computers, each placed in one of $n$ rooms, with each room having a unique key. A training class has $m$ students $(m>n)$, and every evening exactly $n$ students enter the computer rooms for practical operation. To ensure that each person has one computer, what is the minimum number of k... | Note: First design a key distribution scheme, then prove that the scheme is optimal (the fewest total keys and meets the requirements).
First, distribute $n$ different keys of $n$ rooms to $n$ students, one key per person ($n$ people can use the key they receive to open one room). The remaining $m-n$ people can each r... | n(m-n+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,523 |
For any natural number $n$, connect the origin $O$ and the point $A_{n}(n, n+3)$. Let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Try to find: $f(1)+f(2)+\cdots+f(1996)$. | The equation of the line segment $O A_{n}$ is $y=\frac{n+3}{n} x(0 \leqslant x \leqslant n)$. Therefore, $f(x)$ is the number of integer solutions of this equation in the interval $(0, n)$. When $n=3 k$ (where $k$ is a positive integer), the equation becomes
$$
y=\frac{k+1}{k} x(0<x<3 k) \text {. }
$$
It has two sets ... | 1330 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,524 |
In $\triangle A B C$, $A B=37, A C=58$. With $A$ as the center and $A B$ as the radius, an arc is drawn intersecting $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$. | Solve as shown in the figure, first prove
$$
\begin{array}{l}
B C \cdot C D \\
=A C^{2}-A B^{2} .
\end{array}
$$
Draw $A H \perp B C$ at $H$, then
$$
\begin{array}{l}
A C^{2}=A H^{2}+C H^{2}, \\
A B^{2}=A H^{2}+B H^{2},
\end{array}
$$
we have
$$
\begin{array}{l}
A C^{2}-A B^{2}=C H^{2}-B H^{2} \\
\quad=(C H+B H)(C H-... | 57 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,526 |
42. Given $x_{0}=1, x_{1}=3, x_{n+1}=6 x_{n}-x_{n-1}(n$ $\in N)$. Prove: The sequence $\left\{x_{n}\right\}$ contains no perfect squares. | Proof: By the theory of characteristic equations, it is easy to know that
$$
x_{n}=\frac{1}{2}\left[(3+2 \sqrt{2})^{n}+(3-2 \sqrt{2})^{n}\right] \text {. }
$$
Let \( y_{n}=\frac{1}{2 \sqrt{2}}\left[(3+2 \sqrt{2})^{n}-(3-2 \sqrt{2})^{n}\right] \).
Then \( y_{n} \in \mathbb{N} \), and
$$
\begin{aligned}
x_{n}^{2}-2 y_{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,528 |
Example 3. For what values of $a, b$ does the equation $x^{2}+2(1+a) x$ $+\left(3 a^{2}+4 a b+4 b^{2}+2\right)=0$ have real roots?
(1987, National Junior High School Mathematics Competition) | The condition for the equation to have real roots is
$$
\begin{aligned}
\Delta & =(2+2 a)^{2}-4\left(3 a^{2}+4 a b+4 b^{2}+2\right) \\
& =-4\left[(1-a)^{2}+(a+2 b)^{2}\right] \geqslant 0 .
\end{aligned}
$$
Clearly, $-4\left[(1-a)^{2}+(a+2 b)^{2}\right]=0$.
Then $1-a=0$ and $a+2 b=0$,
i.e., when $a=1$ and $b=-\frac{1}{... | a=1, b=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,531 |
7. $x \in R$, find the range of $y=\frac{x^{2}-x+1}{x^{2}+x+1}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (Answer: $\frac{1}{3} \leqslant y$ $\leqslant 3)$ | \frac{1}{3} \leqslant y \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,534 |
10. Let $x, y, z$ be non-negative real numbers, and satisfy
(1) $x+y+z=2$;
(2) for all $t \leqslant 1$, $2 t-1 \leqslant x y+y z+z x \leqslant t^{2}-2 t+2$.
Prove: $x \leqslant \frac{4}{4}, y \leqslant \frac{3}{4}, z \leqslant \frac{4}{4}$. | (Prompt: Substitute $t=1$ into condition (2) to get $x y+y z+z x=1$. Then substitute $z=2-x-y$ to obtain a quadratic equation in $y$.)
11. In parallelogram $M N P Q$, one side $P Q$ lies on the base $B C$ of $\triangle A B C$, and the other two vertices $M, N$ are on $A B, A C$ respectively. Prove that the area of $M N... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,537 |
Example 1. In plane $a$, there is a regular $\triangle A^{\prime} B C$, and line $D E / / B C$ intersects $A^{\prime} B, A^{\prime} C$ at $D, E$ respectively. Now, $\triangle A^{\prime} D E$ is folded along $D E$ perpendicular to plane $\alpha$ to the position of $\triangle A D E$. When the line $D E$ parallel to $B C$... | Analyzing, we can set the length of the first line segment as $x$, then
Choosing, as shown in the figure, construct the altitude $A^{\prime} M$ of $\triangle A^{\prime} B C$, intersecting $D E$ at $N$. Since $D E / / B C$, we have $A^{\prime} N \perp D E$, i.e., $A N \perp D E$.
Also, since the plane $A D E \perp \alp... | \frac{5}{8} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,538 |
Example 2. Given a plane $\pi$, point $P$ is on $\pi$, and point $Q$ is outside $\pi$. Find point $R$ in the plane $\pi$ such that the ratio $\frac{Q P+R P}{Q R}$ is maximized. | Let $R$ be a point on $\pi$, and the angles of $\triangle P G R$ be $\alpha, \beta, \gamma$. Then, by the Law of Sines, we have
$$
\frac{Q P+R P}{Q R}=\frac{\sin \gamma+\sin \beta}{\sin \alpha}=\frac{\cos \frac{\beta-\gamma}{2}}{\sin \frac{\alpha}{2}} .
$$
The numerator is maximized when $\beta=\gamma$, and the denomi... | \frac{1}{\sin \frac{\delta}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,539 |
Example 4. Let the base of the pyramid $M-ABCD$ be a square, and $MA=MD, MA \perp AB$. If the area of $\triangle AMD$ is 1, find the radius of the largest sphere that can fit inside this pyramid.
(1990, National High School Mathematics Competition) | Given $A B \perp A D$, $A B \perp A M$, we know that $A B \perp$ plane $M A D$. Therefore, plane $A B C \perp$ plane $M A D$. Construct a section $M E F$ perpendicular to $A D$, then $\triangle M E F$ is a right triangle, with area $\frac{1}{2} E F \cdot E M=\frac{1}{2} A D$ - $E M=1$.
It is easy to see that when $E F... | \sqrt{2}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,541 |
Example 4. Given that $a, b, c$ satisfy $a+b+c=0, abc=8$. Then the range of values for $c$ is $\qquad$
(1st Hope Cup Junior High School Mathematics Invitational Competition) | Given that $a, b, c$ must be one positive and two negative. If $c$ is negative, we have $c<0$. Solving this inequality, we get $c \geqslant 2 \sqrt[3]{4}$.
Thus, the range of $c$ is
$$
c<0 \text { or } c \geqslant 2 \sqrt[3]{4} \text {. }
$$ | c<0 \text { or } c \geqslant 2 \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,542 |
Example 5. In the tetrahedron $ABCD$, the plane $ABC$ and the plane $BCD$ are both equilateral triangles with side lengths of $2a$, and $AD=$ $2\sqrt{2}a$. If a point $M$ on $AR$ travels across the surface of the tetrahedron to reach the midpoint $N$ of $CD$, find the shortest path. (1979, Henan Province Mathematics Co... | Analyzing the four ways to travel from $M$ to $N$ along the surface of a tetrahedron:
(i) When crossing the edge $A C$, unfold $\triangle A B C$ and $\triangle A C D$ into a plane (as shown in the figure). In this case, the line segment $M N$ is the shortest path.
Connecting $C M$, it is clear that $\triangle A C D$ i... | 2a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,543 |
Example 6. Let a trirectangular tetrahedron $P \wedge B C$ (i.e., $\angle A P B$ $=\angle B P C=\angle C P A=90^{\circ}$) have the sum of the lengths of its six edges as $S$. Try to find its maximum volume.
Translate the above text into English, please retain the original text's line breaks and format, and output the ... | $i, P C=c$.
From the given, we have
$$
\begin{aligned}
S & =\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}}+a+b+c \\
& \geqslant \sqrt{2 a b}+\sqrt{2 b c}+\sqrt{2 c a}+3 \sqrt[3]{a b c} \\
& \geqslant 3 \sqrt{2} \cdot \sqrt[3]{a b c}+3 \sqrt[3]{a b c} \\
& =3(\sqrt{2}+1) \sqrt[3]{6 V} \text{ (where } V \text{ ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,544 |
Example 7. In tetrahedron $ABCD$, $\angle BDC$ is a right angle, and the foot of the perpendicular from $D$ to plane $ABC$ is $S$, which is the orthocenter of $\triangle ABC$. Prove that: $(AB + BC + AC)^2 \leq 6(AD^2 + BD^2 + CD^2)$, and determine what kind of tetrahedron it is when equality holds.
(12th IMO Problem) | Analyzing, since $D S \perp$ plane $A B C, B S \perp A C$, we have $D B \perp A C$. Also, because $\angle B D C$ is a right angle, i.e., $B D \perp D C$, hence $D B \perp$ plane $A D C$. Therefore, $\angle A D B$ is a right angle.
Similarly, $\angle A D C$ is a right angle.
By the Pythagorean theorem, we have
$$
\begin... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,545 |
Example 8. The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. Find the distance between a point on the incircle of the base $A B C D$ and the vertices $A, C$, and $B_{1}$. | The figure shows two circles, each lying on one of two concentric spheres centered at the symmetry center of a cube, namely, the sphere that is tangent to all edges of the the cube (with radius $\frac{\sqrt{2}}{2}$, referred to as the small sphere) and the circumscribed sphere of the cube (with radius $\frac{\sqrt{3}}{... | \frac{\sqrt{3}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,546 |
Example 1. If $x=2-\sqrt{3}$, find the value of $\frac{x^{4}-4 x^{3}-x^{2}+9 x-4}{x^{2}-4 x+5}$.
(1989, Shanghai Junior High School Mathematics Competition) | Solve $x=2-\sqrt{3}, x-2=-\sqrt{3}$, square both sides, we get
$$
\begin{aligned}
& x^{2}-4 x+1=0 . \\
\text { Then } & \frac{x^{4}-4 x^{3}-x^{2}+9 x-4}{x^{2}-4 x+5} \\
= & \frac{x^{2}\left(x^{2}-4 x+1\right)-2\left(x^{2}-4 x+1\right)+x-2}{\left(x^{2}-4 x+1\right)+4} \\
= & \frac{x^{2} \cdot 0-2 \cdot 0+x-2}{0+4} \\
= ... | -\frac{\sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,547 |
Example 2. Given $x=\frac{\sqrt{5}+1}{2}$. Then what is the value of $\frac{x^{3}+x+1}{x^{5}}$? | Given that $\frac{1}{x}=\frac{\sqrt{5}-1}{2}$, then $x-\frac{1}{x}=1$. Therefore,
$$
\begin{array}{l}
\frac{x^{3}+x+1}{x^{5}}=\frac{x^{2}+1+\frac{1}{x}}{x^{4}} \\
=\frac{x^{2}+\left(x-\frac{1}{x}\right)+\frac{1}{x}}{x^{4}}=\frac{x^{2}+x}{x^{4}} \\
=\frac{1+\frac{1}{x}}{x^{2}}=\frac{\left(x-\frac{1}{x}\right)+\frac{1}{x... | \frac{\sqrt{5}-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,548 |
Example 3. If $a=\sqrt{17}-1$, find the value of $\left(a^{5}+2 a^{4}-17 a^{3}\right.$ $\left.-a^{2}+18 a-17\right)^{1993}$.
(Adapted from the 1987 Chongqing Junior High School Mathematics Invitational Competition) | Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \tex... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,549 |
Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition) | Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$
Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,550 |
Example 5. Let real numbers $a, b, x$ and $y$ satisfy $a x + b y = 3, a x^2 + b y^2 = 7, a x^3 + b y^3 = 16, a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(Eighth American Invitational Mathematics Examination) | Let $S_{\mathrm{n}}=a x^{n}+b y^{\star}, A=x+y, B=x y$, then $x, y$ are the two roots of the equation $u^{2}-A u+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, y^{2}-A y+B=0 . \\
\therefore a x^{n}-A\left(a x^{n-1}\right)+B\left(a x^{n-2}\right)=0, \\
b y^{n}-A\left(b y^{n-1}\right)+B b y^{n-2}=0 .
\end{array}
$$
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,551 |
Example 6. Given $a, b, c$ satisfy $a+b+c=1, a^{2}+$ $b^{2}+c^{2}=2, a^{3}+b^{3}+c^{3}=3$. Find the values of $a b c, a^{4}+b^{4}+c^{4}$.
(1991, Jiangsu Province Junior High School Mathematics Competition) | Let $S_{n}=a^{n}+b^{n}+c^{n}$.
$$
\begin{array}{l}
\because(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2(a b+a c+b c), \\
a+b+c=1, a^{2}+b^{2}+c^{2}=2 c \\
\therefore a b+a c+b c=-\frac{1}{2} .
\end{array}
$$
There is also the identity $x^{3}-(x+y+z) x^{2}+(x y$
$$
+y z+z x)+x y z=0 \text {. }
$$
From this, we get $a^{3}-a^{2}... | \frac{1}{6}, \frac{25}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,552 |
Example 5. When $b \neq c$, if the parabolas $y=x^{2}+b x+c$ and $y=-x^{2}+c x+b$ are tangent to each other, then $b, c$ satisfy the relation $\qquad$
(1992, Taiyuan City Junior High School Mathematics Competition) | Solving for the tangency of two parabolas, we get the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}+b x+c, \\
y=-x^{2}+c x+b .
\end{array}\right.
$$
Eliminating \( y \) and simplifying, we get
$$
2 x^{2}+(b-c) x+c-b=0 .
$$
By the property of tangency, this equation has two identical real roots, so
$$
\Delta=... | b=c+8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,553 |
Example 7. Find the integer part of $(\sqrt{3}+1)^{6}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text{Solve for the dual number } B = (\sqrt{3}-1)^{6} \text{ of the number } A = (\sqrt{3}+1)^{6}, \text{ and examine} \\
A+B=(\sqrt{3}+1)^{6}+(\sqrt{3}-1)^{6} \\
= \left[(\sqrt{3}+1)^{3}\right]^{2}+\left[(\sqrt{3}-1)^{3}\right]^{2} \\
= \left[(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}\right] \\
+2[(\sqrt{3... | 415 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,554 |
Example 8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-4 x+1=0$. Find the value of $x_{1}^{3}-2 x_{2}^{2}+x_{1}-1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The provided text is already in English, so ... | Construct $A=x_{1}^{3}-2 x_{2}^{2}+x_{1}-1$ and its dual $B=x_{2}^{3}-2 x_{1}^{2}+x_{2}-1$ (i.e., swap $x_{1}, x_{2}$ in $A$). Thus,
$$
\begin{array}{l}
A+B \\
=\left(x_{1}^{3}+x_{2}^{3}\right)-2\left(x_{1}^{2}+x_{2}^{2}\right)+\left(x_{1}+x_{2}\right)-2 \\
=\left(x_{1}+x_{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-3 ... | 13 \pm 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,555 |
Example 9. Let $x>0$. Try to find the maximum value of $y=\sqrt{x}+\frac{1}{\sqrt{x}}-$ $\sqrt{x+\frac{1}{x}+1}$. | To construct the dual form of
$$
y=\sqrt{x}+\frac{1}{\sqrt{x}}-\sqrt{x+\frac{1}{x}+1}
$$
is
$$
u=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x+\frac{1}{x}+1},
$$
then \( y u=1, y=\frac{1}{u} \).
$$
\because u \geqslant 2+\sqrt{2+1}=2+\sqrt{3} \text{ (when } x=1
$$
is achieved),
$$
\therefore y=\frac{1}{u} \leqslant \frac{1}{2... | 2-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,556 |
Example 10. Given $a^{2}+2 a-5=0, b^{2}+2 b-5=0$, and $a \neq b$. Then the value of $a b^{2}+a^{2} b$ is what?
(1989, Sichuan Province Junior High School Mathematics Competition) | Given that $a, b$ are the roots of the quadratic equation $x^{2}+$ $2 x-5=0$, then $a+b=-2, ab=-5$. Therefore, $ab^{2}+a^{2}b=ab(a+b)=10$. Note: Here, the definition of the roots is used to construct the equation. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,557 |
Example 12. Positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{1}{3} y^{2}=25, \\
\frac{1}{3} y^{2}+z^{2}=9, \\
z^{2}+x z+x^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 z x$.
(18th All-Russian High School Mathematics Competition) | Solve: Transform the original system of equations into
$$
\left\{\begin{array}{l}
x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 \cdot x \cdot \frac{y}{\sqrt{3}} \cdot \cos 150^{\circ}=5^{2}, \\
\left(\frac{y}{\sqrt{3}}\right)^{2}+z^{2}=3^{2}, \\
z^{2}+x^{2}-2 x z \cos 120^{\circ}=4^{2} .
\end{array}\right.
$$
From this,... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,559 |
Example 13. Let $\alpha, \beta$ be acute angles, and $3 \sin \alpha=2 \sin \beta$, $3 \cos \alpha+2 \cos \beta=3$. Find the degree measure of $\beta + \frac{\alpha}{2}$. | Solve for $3 \sin \alpha = 2 \sin \beta$, we get $\frac{\sin \alpha}{2} = \frac{\sin \beta}{3}$. From this, by inversely applying the Law of Sines, construct $\triangle ABC$ such that $\angle A = \alpha, \angle B = \beta, \angle C = 180^{\circ} - (\alpha + \beta), BC = 2, AC = 3$.
In $\triangle ABC$, using the project... | 90^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,560 |
Given: $x \geqslant y \geqslant z>0, x, y, z \in \mathbb{R}$. Prove that:
$$
\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2} \text {. }
$$
This article presents its generalization and proof.
Proposition: Under the same conditions. Prove that for $\forall n \in \mathbb{N}$. We have
$$
... | $$
\begin{array}{l}
\text { Prove } \because\left(x^{n}-y^{n}\right)\left(\frac{y}{z}-1\right) \\
+\left(z^{n}-y^{n}\right)\left(\frac{x}{y}-1\right) \\
=(x-y)(y-z)\left(\frac{1}{z} \sum_{i=0}^{n} x^{n-1} y^{i}\right. \\
\left.\quad-\frac{1}{y} \sum_{i=0}^{n} z^{n-1 \cdots} y^{i}\right) \\
\geqslant(x-y)(y-z)\left(\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,561 |
1. Factorize $(x y-1)^{2}+(x+y-2)(x+y$ $-2 x y)$. | Let $x y=a, x+y=b$, so
the original expression $\mathrm{x}^{2}=(a-1)^{2}+(b-2)(0-2 a)=(a-b+1)^{2}$.
Substituting into the
resulting expression $=(x-1)^{2}(y-1)^{2}$. | (x-1)^2(y-1)^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,562 |
2. Given $a+b+c=0, a^{3}+b^{3}+c^{3}=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | 2. From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right.$ $-a b-b c-a c)$, we have $a b c=0$. Then at least one of $a, b, c$ is zero, for example, $c=0$, then it is known that $a, b$ are opposites, i.e., $a^{15}+b^{15}+c^{15}$ $=0$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,563 |
Example 6. Find a natural number $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(2nd All-Russian High School Mathematics Olympiad) | Let $2^{4}=x$, then $2^{8}=x^{2}, 2^{11}=2^{7} \cdot x$. Therefore, $2^{8}+2^{11}+2^{n}=x^{2}+2^{7} x+2^{n}$.
According to the condition for a quadratic trinomial to be a perfect square, we get $\Delta=\left(2^{7}\right)^{2}-4 \times 2^{n}=0$.
Solving for $n$ yields $n=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,564 |
5. Given as shown, $\triangle A B C$ is an isosceles right triangle, $\angle C=90^{\circ}$, points $M, N$ are the midpoints of sides $A C$ and $B C$ respectively, point $D$ is on ray $B M$, and $B D=2 B M$. Point $E$ is on ray $N A$, and $N E=$ $2 N A$. Prove: $B D \perp D E$. | 5. Take the midpoint $F$ of $A D$, and connect $E F$.
From $A C=B C, C N=C M$, we have $\triangle A C M \cong \triangle B C M$, then $\angle 3=\angle 6$.
From $B M=D M, C M=A M, \angle B M C=\angle A M D$, we have $\triangle B C M \cong \triangle D A M$, then $\angle 2=\angle 6$, hence $A D=B C$. $\angle 2=\angle 3$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,567 |
6. Given $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$. Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$. | 6. From $\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$, we have $\frac{m y z+n x z+p x y}{x y z}=0$.
Since $x, y, z$ are not zero, then $m y z+n x z+p x y=0$. Also,
$$
\begin{aligned}
& \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}} \\
= & \left(\frac{x}{m}+\frac{y}{n}+\frac{z}{p}\right)^{2}-2\left(\frac{x y}{m ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,568 |
8. Find the minimum value of $|x-1|+|x-2|+|x-3|+\cdots$ $+|x-1996|$. | 8. When $998 \leqslant x \leqslant 999$, the original expression has a minimum value, the minimum value is:
$$
\begin{array}{l}
(1996-1)+(1995-2)+\cdots+(999-998) \\
=1995+1993+\cdots+1 \\
=\frac{(1+1995) \times 998}{2}=998^{2}=996004 .
\end{array}
$$ | 996004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,570 |
9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$? | 9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l }
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,571 |
10. Among the students in the second year of junior high school, 32 students participated in the math competition, 27 students participated in the English competition, and 22 students participated in the Chinese competition. Among them, 12 students participated in both math and English, 14 students participated in both... | 10. Let the number of people who participate in all three subjects be $x$, and the total number of participants be $y$, then
$$
\begin{aligned}
y & =32+27+22-12-14-10+x \\
& =45+x .
\end{aligned}
$$
Given $2 \leqslant x \leqslant 10$, hence $47 \leqslant y \leqslant 55$.
Only when $y=48$, it can be divisible by 8, so ... | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,572 |
Example 7. If $M=3 x^{2}-8 x y+9 y^{2}-4 x+6 y+13$, then the following statement must be true is ( ).
(A) $M \geqslant 0$
(B) $M0$
(D) $M \leqslant 0$
(1990, Shanxi Province Junior High School Mathematics Competition Preliminary) | $$
\begin{array}{l}
\text { Sol } \because M=3 x^{2}-(8 y+4) x+9 y^{2}+6 y+13, \\
\therefore \quad \Delta_{x}=(8 y+4)^{2}-4 \times 3\left(9 y^{2}+6 y+13\right) \\
=-4\left(11 y^{2}+2 y+35\right). \\
\end{array}
$$
Let $N=11 y^{2}+2 y+35$, then
$$
\Delta_{y}=2^{2}-4 \times 11 \times 35,
$$
when $N>0$, $\Delta_{x}=-4 N<... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,575 |
4. Let the side length of the equilateral $\triangle ABC$ be $a$, and $M, N$ be the midpoints of $AB, AC$, respectively. $D$ is any point on $MN$, and the extensions of $BD, CD$ intersect $AC, AB$ at points $E, F$ (as shown in the figure). Then the value of $\frac{1}{CE}+\frac{1}{BF}$ is | 4. $\frac{3}{a}$ | \frac{3}{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,577 |
II. (16 points) A plot of land can be covered by $n$ identical square tiles. If smaller identical square tiles are used, then $n+76$ such tiles are needed to cover the plot. It is known that $n$ and the side lengths of the tiles are integers. Find $n$.
| Let the side length of the larger tile be $x$, and the side length of the smaller tile be $r$. Then $x, y$ are both natural numbers, and
$$
n x^{2}=(n+76) y^{2} \text {. }
$$
If $(x, y)=d$, let $x_{1}=\frac{x}{d}, y_{1}=\frac{y}{d}$, then $x_{1}, y_{1}$ are both natural numbers, and $\left(x_{1}, y_{1}\right)=1$. Thus... | 324 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,584 |
Example 8. The parabola $y=x^{2}-2 b x+1$ and the line $y=-\frac{1}{2} x$ $+\frac{1}{2} m$, for any real value of $b$, always have intersection points, then the range of real number $m$ is $\qquad$
(1987, Shenyang City Junior High School Mathematics Invitational Competition) | Since the parabola and the line always have intersection points, the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}-2 b x+1 \\
y=-\frac{1}{2} x+\frac{1}{2} m
\end{array}\right.
$$
implies that there are real solutions. Eliminating \( y \) and simplifying, we get
$$
2 x^{2}+(1-4 b) x+2-m=0 .
$$
For this equati... | m \geqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,586 |
Four points are such that no three are collinear, and no two are coincident. Through each point, draw the perpendiculars to the lines joining the other three points. If we do not count the original four points, how many different intersection points are there among these perpendiculars? Prove your conclusion. | For each point, draw perpendicular lines to the lines connecting it to the other three points. Since there are 3 such lines for each point, and there are 4 points, there are a total of $4 \times 3 = 12$ perpendicular lines.
If every two perpendicular lines intersect, there would be $\frac{12 \times 11}{2} = 66$ inters... | 40 | Geometry | proof | Yes | Yes | cn_contest | false | 708,587 |
2. Let $a, b$ be two distinct real numbers, and satisfy $a^{2}=4 a$ $+3, b^{2}=4 b+3$. Then $\frac{a^{2}}{b}+\frac{b^{2}}{a}=\ldots$. | 2. From the given, $a, b$ are the two unequal real roots of the equation $x^{2}-4 x-3=0$, so $a+b=4, a b=-3$. Therefore,
$$
\begin{array}{c}
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{a^{3}+b^{3}}{a b} \\
=\frac{(a+b)\left[(a+b)^{2}-3 a b\right]}{a b}=-\frac{100}{3} .
\end{array}
$$ | -\frac{100}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,589 |
3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$ | 3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.
Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\tr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,590 |
4. In $\triangle A B C$, $P, Q$ are points on $A B$ and $A C$ respectively, and the median $A M$ intersects $P Q$ at $N$. If $A B$ : $A P=5: 2, A C: A Q=4: 3$, then $A M: A N$ $=$ $\qquad$ | 4. As shown in the figure, draw $C D / /$ $P Q$ intersecting $A B$ at $D$, and $M K / /$ $P Q$ intersecting $A B$ at $K$, then $M K / / C D$.
$$
\begin{array}{c}
\because B M=M C, \\
\therefore B K=K D . \\
\therefore A K=\frac{1}{2}(A D+A B) . \\
\text { Therefore, } \frac{A K}{A P}=\frac{1}{2}\left(\frac{A D}{A P}+\f... | \frac{23}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,591 |
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