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5. If $\log _{8} a+\log _{4} b^{2}=5$, and $\log _{8} b$ $+\log _{4} \mathrm{a}^{2}=7$. Find $\mathrm{ab}$. | Solving: Add the two equations given in the problem, and by the properties of logarithms, we get:
$$
\log _{8} a b+\log _{4} a^{2} b^{2}=12 .
$$
By changing the base, we get $\frac{\log _{2} a b}{3}+\log _{2} a b=12$,
$$
\begin{array}{l}
\frac{4}{3} \log _{2} \mathrm{ab}=12, \log _{2} \mathrm{ab}=9, \\
\therefore \mat... | 512 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,112 |
6 . Three circles, each with a radius of 3. The centers are at $(14,92)$, $(17,76)$, and $(19,84)$. Draw a line through the point $(17,76)$ such that the sum of the areas of the parts of the three circles on one side of the line equals the sum of the areas of the parts of the three circles on the other side of the line... | First, note that these three circles are mutually externally tangent equal circles. As shown in the figure, the centers are $\mathrm{O}_{1}(14,92), \mathrm{O}_{2}$ $(19,84), \mathrm{O}_{3}(17, 76)$. Since any line passing through $\mathrm{O}_{3}$ always divides $\odot \mathrm{O}_{3}$ into two equal areas, we only need ... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,113 |
7. The function $f$ is defined on the set of integers, satisfying
$$
f(n)=\left\{\begin{array}{ll}
n-3, & \text { when } n \geqslant 1000, \\
f(f(n+5)), & \text { when } n<1000 .
\end{array}\right.
$$
Find $f(84)$. | Solve: First, calculate $f(n)$ for $n$ values close to 1000, and observe the pattern when $\mathrm{n}<1000$. Using the recursive definition of $\mathrm{f}$, we can compute:
$$
\begin{array}{l}
\mathrm{f}(999)=\mathrm{f}(\mathrm{f}(1004))=\mathrm{f}(1001)=998, \\
\mathrm{f}(998)=\mathrm{f}(\mathrm{f}(1003))=\mathrm{f}(1... | 997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,114 |
8. The equation $\mathrm{z}^{8}+\mathrm{z}^{3}+1=0$ has a complex root, and the argument of this root on the complex plane is between $90^{\circ}$ and $180^{\circ}$. Find the degree of the argument. | Let $\omega=z^{3}$, the given equation transforms into $\omega^{2}+\omega+1=0$. Its solutions are $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{-1-\sqrt{3}i}{2}$, with arguments of $120^{\circ}$ and $240^{\circ}$, respectively. Since $z=\sqrt[3]{\omega}$, we can find the six arguments of $z$ to be $\frac{120^{\circ}}{3... | 160 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,115 |
9. In tetrahedron $\mathrm{ABCD}$, the length of edge $\mathrm{AB}$ is $3 \mathrm{~cm}$, the area of face $\mathrm{ABC}$ is $15 \mathrm{~cm}^{2}$, and the area of face $\mathrm{ABD}$ is $12 \mathrm{~cm}^{2}$. The angle between these two faces is $30^{\circ}$. Find the volume of the tetrahedron (in $\mathrm{cm}^{3}$). | Let $\mathrm{V}$ be the volume of tetrahedron $\mathrm{ABCD}$, and $\mathrm{h}$ be the height from $\mathrm{D}$ to the base $\mathrm{ABC}$. Then, $\mathrm{V}=\frac{1}{3} \mathrm{hS} \triangle \mathrm{ABC}$. To determine $V$, we only need to determine $h$. Draw $\mathrm{DK} \perp \mathrm{AB}$ at $\mathrm{K}$, and connec... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,116 |
10. Xiao Zhi scored above 80 points in an American middle school math test. He told Xiao Jiang his score, and Xiao Jiang could correctly deduce how many questions Xiao Zhi answered. If Xiao Ma scored a bit lower but still above 80 points, Xiao Jiang would not be able to deduce it. How many points did Xiao Ma score?
(Th... | Solve for the minimum such that $S=30+40-w>80$. The problem requires finding the smallest such $S$, for which the corresponding $c$ is unique. First, note that when $c$ increases by 1, $w$ decreases by 4, and the value of $S$ remains unchanged, but it must satisfy $(c+1)+(w+4) \leqslant 30$, i.e., $c+w \leqslant 25$. W... | 119 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,117 |
11. A gardener randomly arranges three maple trees, four oak trees, and five birch trees in a row. Each different arrangement is equally likely.
Let $\mathrm{m}$ represent the probability that no two birch trees are adjacent (expressed as a simplified fraction). Find $\mathrm{m}+\mathrm{n}$. | There are 12! different arrangements of 12 trees. Let $\mathrm{k}$ be the number of arrangements where no two birch trees are adjacent. The required probability is $\frac{\mathrm{k}}{12!}$. Now, we need to find $\mathrm{k}$.
Using $t$ $\mathrm{N}$ to represent non-birch trees (maple or oak), we have (1)N (2)N (3)N (4)... | 106 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,118 |
Example 1. Determine whether the polynomial $\mathrm{x}^{2}+(\sqrt{2}-1) \mathrm{xy}$
$$
-\sqrt{2} y^{2}+\frac{\sqrt{2}}{2} \mathrm{x}-(2 \sqrt{2}+1) y-1 \text { can be factored over }
$$
the field of characteristic number. | $$
\begin{array}{l}
\text { Solution: Since } \Delta^{\prime}=\left[(\sqrt{2}-1) \cdot \frac{\sqrt{2}}{2}\right. \\
+2(2 \sqrt{ } 2+1)]^{2}-\left[(\sqrt{2}-1)^{2}+4 \sqrt{2}\right] \\
{\left[\left(\frac{\sqrt{2}}{2}\right)^{2}+4\right]=20+12 \sqrt{2} \neq 0,}
\end{array}
$$
Therefore, the original polynomial cannot be... | 20+12\sqrt{2} \neq 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,119 |
12. Function f is defined on the real number domain, and satisfies the following conditions: For any real number $\mathrm{x}, \mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})$, and $f(7+x)=f(7-x)$. If $x=0$ is a root of $f(x)=0$, find the minimum number of roots of $f(x)=0$ in the interval -1000 $\leqslant x \leqslant... | Molybdenum First, find the numbers in the domain for which $f$ is periodic from the given equations.
We have
$$
\begin{array}{l}
f(x)=f(2+(x-2))=f(2-(x- \\
2))=f(4-x),(1) \\
f(4-x)=f(7-(x+3))=f(7+ \\
(x+3))=f(x+10) .(2)
\end{array}
$$
From (1) and (2), we get $f(x+10)=f(x)$
Substituting $x+10$ and $x-10$ for $x$ in ... | 401 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,120 |
13. Find the value of $10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7$ $+\operatorname{arcctg} 13+\operatorname{arcctg} 21)$. | $\begin{array}{l}\text { Let } f(n)=1+n+n^{2}, \text { it is easy to know that } \\ f(1)=3, f(2)=7, f(3) \\ =13, f(4)=21. \\ \text { It is easy to prove that } \\ \operatorname{arcctg}\left(1+n+n^{2}\right) \\ =\operatorname{arctg}(n+1)-\operatorname{arctg} n. \\ \text { Therefore, } \operatorname{arcctg} 3=\operatorna... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,121 |
14. The largest even number that cannot be expressed as the sum of two odd composite numbers is what?
Will the translated text be used for further discussion or do you need more information on this topic? | First, we prove: If $\mathrm{k} \geqslant 40, \mathrm{k}$ is an even number, then $\mathrm{k}$ can be expressed as the sum of two odd composite numbers. For this, if $\mathrm{n}$ is an odd number greater than 1, then $5 \mathrm{n}$ is an odd composite number with a units digit of 5. To express even numbers greater than... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,122 |
$\begin{array}{l}\text { 15. Find the value of } x^{2}+y^{2}+z^{2}+w^{3} \text { . If } \\ \frac{x^{2}}{\varepsilon^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}} \\ +\frac{w^{2}}{2^{2}-7^{2}}=1, \\ \frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}} \\ +\frac{w^{2}}{4^{2}-7^{2... | Solving for $\mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{w}$ to satisfy the given system of equations is equivalent to $t=4,16,36,64$ satisfying
$$
\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}=\frac{w^{2}}{t-49}=1. \quad (1)
$$
Multiplying both sides of (1) by the denominators, we find that for meaningful $... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,123 |
Inference: The two sides of a triangle and the internal and external angle bisectors of the angle formed by these two sides form a harmonic pencil. A new line that does not pass through the harmonic pencil intersects these four lines, then the four intersection points form a harmonic range.
untranslated text:
推论 三角形的... | Prove that if $\mathrm{AD}$ is the angle bisector of $\angle \mathrm{BAC}$ in $\triangle \mathrm{ABC}$, and $\mathrm{AE}$ is the angle bisector of the exterior angle $\angle \mathrm{CAF}$, then $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BF}}{\mathrm{CE}}$.
$$
\begin{aligned}
& \frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,125 |
Theorem The dihedral angle bisector plane (internal or external) of a tetrahedron divides the opposite edge into two segments that are proportional to the areas of the two faces forming the dihedral angle. | Prove that as shown in Figure 2, $\mathrm{ABD}(\mathrm{ABH})$ is the internal (external) bisector plane of the dihedral angle $\mathrm{S}-\mathrm{AB}-\mathrm{C}$ of the tetrahedron $\mathrm{S}-\mathrm{ABC}$. Draw $\mathrm{DG} \perp$ plane $\mathrm{SAB}$ at $\mathrm{G}$, and $\mathrm{DF} \perp$ plane $\mathrm{CAB}$ at $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,126 |
Example 5. One root of the equation $3 x^{2}-5 x+a=0$ is greater than -2 and less than $0_{3}$, the other is much greater than 1 and less than 3. Find the range of real values for a. | Solve: The equation can be rewritten as
$$
\mathrm{x}^{2}-\frac{5}{3} \mathrm{x}+\frac{\mathrm{a}}{3}=0 \text {. }
$$
Here $\mathrm{p}=-\frac{5}{3}, \mathrm{q}=\frac{\mathrm{a}}{3}$. According to Theorem 7, the necessary and sufficient condition for one root of this equation to be greater than -2 and less than 0, and ... | -12<\mathrm{a}<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,127 |
Theorem 1 For an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad(a>b>0)$, a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ or a parabola $y^{2}=2px \ (p>0)$, the necessary and sufficient condition for four distinct points on these curves to be concyclic is that the sum of the inclination angles of the t... | Proof: Let the four distinct points on the ellipse be $\mathrm{A}$, $B$, $C$, and $D$, and the intersection of the diagonals $AC$ and $BD$ be $M\left(x_{0}, y_{0}\right)$. The inclination angles of the lines $AC$ and $BD$ are $\alpha$ and $\beta$ respectively, where $0^{\circ} \leqslant \alpha, \beta < 180^{\circ}$.
S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,128 |
Example 1. Prove:
$$
\left|\begin{array}{lll}
1 & 1 & 1 \\
c & b & a \\
b c & c a & a b
\end{array}\right|=(a-b)(b-c)
$$ | Proof $\because$ when $a=b$ or $b=c$ or $c=a$, the value of the determinant is zero,
$\therefore (a-b),(b-c),(c-a)$ are all factors of the determinant.
$$
\therefore\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
b c & c a & a b
\end{array}\right|=k(a-b)(b-c)
$$
It is also easy to know that the undetermined coeffic... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,129 |
Example 2. Prove:
$$
\left|\begin{array}{lll}
a x & a^{2}+x^{2} & 1 \\
a y & a^{2}+y^{2} & 1 \\
a z & a^{2}+z^{2} & 1
\end{array}\right|=a(x-y)(y-z)
$$ | Proof $\because$ when $x=y$ or $y=z$ or $z=x$, the value of the determinant is zero,
$\therefore (x-y), (y-z), (z-x)$ are all factors of the determinant.
Therefore, the original expression $=k(x-y)(y-z) \cdot(z-x)$.
Also $\because$ one of the terms in the product of the main diagonal elements in the determinant is $\ma... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,130 |
Example 3. Prove:
$$
\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{b} & \mathrm{c} & \mathrm{c} \\
\mathrm{a} & \mathrm{b}+\mathrm{c} & \mathrm{a} \\
\mathrm{b} & \mathrm{b} & \mathrm{c}+\mathrm{a}
\end{array}\right|=4 \mathrm{abc} .
$$ | Prove $\because$ when $a=0$, the original expression
$$
=\left|\begin{array}{ccc}
\mathrm{b} & \mathrm{c} & \mathrm{c} \\
0 & \mathrm{~b}+\mathrm{c} & 0 \\
\mathrm{~b} & \mathrm{~b} & \mathrm{c}
\end{array}\right|=\mathrm{bc}\left|\begin{array}{ccc}
1 & \mathrm{c} & 1 \\
0 & \mathrm{~b}+\mathrm{c} & 0 \\
1 & \mathrm{~b... | 4abc | Algebra | proof | Yes | Yes | cn_contest | false | 702,131 |
$\begin{array}{l}\left|\begin{array}{cccc}1 & 1 & \cdots & 1 \\ a_{1} & a_{2} & \cdots & a_{n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{1}^{n-1} & a_{2}^{n-1} & \cdots & a_{n}^{n-1}\end{array}\right| \\ =\left(a_{2}-a_{1}\right)\left(a_{3}-a_{1}\right) \cdots\left(a_{n}-a_{1}\right) \\ \left(a_{3}-a_{2}\right) \cdots... | Prove that when $a_{i}=a_{1},(i=2,3, \cdots, n)$ or $a_{j}=a_{2} \quad(j=3,4, \cdots, n)$ or $\cdots$
$a_{n}=a_{n-1}$, the value of the $n$-order Vandermonde determinant is zero.
$$
\begin{array}{r}
\therefore\left|\begin{array}{cccc}
1 & 1 & \cdots & 1 \\
a_{1} & a_{2} & \cdots & a_{n} \\
\cdots & \cdots & \cdots & \c... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,132 |
In complex numbers, there is a typical exercise: If $\mid \mathrm{z}_{1}$ |
$$
=\left|z_{2}\right|=\left|z_{3}\right|=1 \text {, and } z_{1}+z_{2}+z_{3}=0 \text {, }
$$
prove that $\mathrm{z}_{1} 、 \mathrm{z}_{2} 、 \mathrm{z}_{3}$ are the vertices of an equilateral triangle inscribed in the unit circle - as shown in F... | Given that $|z_{1}|=|z_{2}|=|z_{3}|=1$, $z_{1}, z_{2}, z_{3}$ are all on the unit circle. We only need to prove that $\triangle z_{1} z_{2} z_{3}$ is equilateral.
$$
\begin{aligned}
z_{1} z_{2}^{2} & =|z_{2}-z_{1}|^{2} \\
& =|z_{1}|^{2}+|z_{2}|^{2}-2 R_{0}(z_{1} \overline{z_{2}}), \\
z_{1} z_{3}^{2} & =|z_{3}-z_{1}|^{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,133 |
$\begin{array}{c}\text { Example 2. Given } \sin \alpha+\sin \beta+\sin \gamma=0, \\ \cos \alpha+\cos \beta+\cos \gamma=0, \text { and } 0 \leqslant \alpha< \\ \beta<\gamma<2 \pi \text {. Prove: } \alpha, \beta, \gamma \text { form an arithmetic sequence. }\end{array}$ | $$
\begin{array}{l}
\text { 4) Let } z_{1} \\
=\cos \alpha \\
+i \sin \alpha, \\
z_{2}=\cos \beta+i \sin \beta, \quad z_{3}=\cos \gamma+i \sin \gamma . \\
\quad \text { It is easy to see that }\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1,
\end{array}
$$
and from $0 \leqslant \alpha<\beta<\gamma<2 \pi$ we... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,135 |
Example 3. Given $\sin A + \sin B + \sin C = 0$, $\cos A + \cos B + \cos C = 0$.
(1) Prove: $3(B - C)$, $3(C - A)$, and $3(A - B)$ are all integer multiples of $360^\circ$.
(2) Prove: $\cos^2 A + \cos^2 B + \cos^2 C$ is a constant. (Shanghai 1978 Middle School Mathematics Competition Preliminary Test Question) | (1) The difference between this problem and Example 2 is merely the removal of the restriction conditions among the three angles. When proving, in addition to still using the original problem's conclusion, note the general solution, i.e.,
$$
\mathrm{B}-\mathrm{C}=\mathrm{n} \cdot 360^{\circ}+120^{\circ} \quad(\mathrm{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,136 |
Example 4. Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ satisfy $\sin \mathrm{x}+\sin \mathrm{y}$
$$
\begin{array}{l}
+\sin z=\cos x+\cos y+\cos z=0 \text {. Prove: } \\
\sin 2 x+\sin 2 y+\sin 2 z=\cos 2 x+\cos 2 y \\
+\cos 2 z=0 \text {. }
\end{array}
$$ | Prove that for the points $\mathrm{z}_{1}(\cos \mathrm{x}, \sin \mathrm{x})$, $\mathrm{z}_{2}(\cos \mathrm{y}, \sin \mathrm{y})$, and $\mathrm{z}_{3}(\cos \mathrm{z}, \sin \mathrm{z})$, given $\sin \mathrm{x} + \sin y + \sin z = \cos x + \cos y + \cos z = 0$, the triangle $\Delta \mathrm{z}_{1} \mathrm{z}_{2} \mathrm{z... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,138 |
$\begin{array}{c}\text { Example 5. Given } \sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}= \\ \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}=0 \text {. Prove: } \\ (1) \sin 3 \mathrm{~A}+\sin 3 \mathrm{~B}+\sin 3 \mathrm{C} \\ =3 \sin (\mathrm{A}+\mathrm{B}+\mathrm{C}) \\ \text { (2) } \cos 3 \mathrm{~A}+\cos 3 \mat... | $$
\begin{array}{l}
\text { Proof } \quad(1) \text { still uses the original conclusion. } \\
\text { Left }=\sin 3A+\sin 3(A+120^{\circ}) \\
+\sin 3(A+240^{\circ}) \\
=\sin 3A+\sin 3A+\sin 3A ; \\
\text { Right }=3 \sin (A+A+120^{\circ}+A+240^{\circ}) \\
=3 \sin 3A . \\
\text { Therefore } \sin 3A+\sin 3B+\sin 3C \\
=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,139 |
Example 1. A pocket contains 10 white balls and 8 black balls. 4 balls are drawn from it. Find: (1) the probability of drawing exactly 2 white balls; (2) the probability of drawing at most 2 white balls. | (1) Omitted. (2) From a bag containing 10 white balls and 8 black balls, 4 balls are drawn.
Let event $\mathrm{A}$: exactly 2 white balls,
Event B: exactly 1 white ball,
Event C: no white balls (all black balls).
Then $\mathrm{P}(\mathrm{A})=\frac{\mathrm{C}_{10}^{2} \cdot \mathrm{C}_{8}^{2}}{\mathrm{C}_{18}^{4}}=\frac... | \frac{21}{34} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,140 |
Example 3. There are 26 products, among which 5 are defective. 4 pieces are drawn from them. Find the probability of no defective products. | Analysis: The event "defective items" includes many situations, making direct calculation quite troublesome. However, the probability of its complementary event "all are genuine products" is easier to obtain. Therefore, using the formula $\mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})$ is more convenient. | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,141 |
Example: Worker A produces 10 parts, 9 of which are qualified and 1 is defective. Worker B produces 10 parts, 8 of which are qualified and 2 are defective. Worker C produces 10 parts, 7 of which are qualified and 3 are defective. One part is randomly selected from the parts produced by A, B, and C. Find: (1) the probab... | Analysis: (1) Let events A, B, C be set out. The events discussed in the above five questions are respectively: $\mathrm{A} \cdot \mathrm{B} \cdot \mathrm{C}$; $\mathrm{A} \cdot \overline{\mathrm{B}} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cd... | 0.494 ; 0.092 ; 0.398 ; 0.994 ; 0.698 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,142 |
2. The side length of an equilateral triangle is $\mathrm{a}, \mathrm{PA} \perp$ plane $A B C$. If $P A=h$, find the distance from $P$ to $B C$. | $\frac{1}{2} \sqrt{3 a^{2}+4 h^{2}}$ | \frac{1}{2} \sqrt{3 a^{2}+4 h^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,144 |
3. For a regular hexagon $\mathrm{ABCDEF}$ with side length $\mathrm{a}$, $\mathrm{AP} \perp$ the plane of the hexagon and $\mathrm{AP}=\mathrm{h}$. Find: (1) the distance from point $\mathrm{P}$ to $\mathrm{BC}$; (2) the distance from point $\mathrm{P}$ to $\mathrm{CD}$; (3) the distance from point $\mathrm{P}$ to $\m... | $\begin{array}{l}\text { Ans: } \\ \frac{1}{2} \sqrt{3 \mathrm{a}^{2}+4 \mathrm{~h}^{2}} ;(2) \sqrt{3 \mathrm{a}^{2}+\mathrm{h}^{2}} ; \\ \text { (3) } \sqrt{\left.3 \mathrm{a}^{2}+\mathrm{h}^{2}\right)}\end{array}$ | \frac{1}{2} \sqrt{3 \mathrm{a}^{2}+4 \mathrm{~h}^{2}} ;(2) \sqrt{3 \mathrm{a}^{2}+\mathrm{h}^{2}} ;(3) \sqrt{3 \mathrm{a}^{2}+\mathrm{h}^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,145 |
4. From a vertex $\mathrm{A}$ of a regular octagon $\mathrm{ABCDEFGH}$ with side length $\mathrm{a}$, draw a perpendicular line $\mathrm{AS}$ to the plane of the octagon. If $\mathrm{AS}=\mathrm{a}$, find the distance from $\mathrm{S}$ to $\mathrm{EF}$, the length of $\mathrm{SE}$, and the angle between $\mathrm{SE}$ a... | (Ans: $\mathrm{SF}=\sqrt{4+2 \sqrt{2}} \mathrm{a}$, $\angle \mathrm{SEF}=\operatorname{arctg} \sqrt{4+2 \sqrt{2}})$ | \mathrm{SF}=\sqrt{4+2 \sqrt{2}} \mathrm{a}, \angle \mathrm{SEF}=\operatorname{arctg} \sqrt{4+2 \sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,146 |
Example 2. Given the dihedral angle $\alpha-\mathrm{AB}-\beta$ is $30^{\circ}$, and there is a point $\mathrm{P}$ inside it, the distance from $\mathrm{P}$ to plane $\alpha$ is $\mathrm{PC}=3 \mathrm{~cm}$, and the distance from $\mathrm{P}$ to plane $\beta$ is $\mathrm{PD}=2 \mathrm{~cm}$. Find the distance from $\mat... | (Ans: $\frac{2}{3} \sqrt{57} \mathrm{~cm}$,
$\sqrt{26-12 \sqrt{2}} \mathrm{~cm})$ | \frac{2}{3} \sqrt{57} \mathrm{~cm}, \sqrt{26-12 \sqrt{2}} \mathrm{~cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,147 |
Example 3. Write H a mapping from $(-\infty, \infty)$ to $(-3, 7)$.
Translating the above text into English, please retain the line breaks and format of the source text, and output the translation result directly. | $(-\infty, \infty)$ to $(m, \infty)$ one-to-one mapping; from conclusion (three): $\mathrm{y}=\frac{10}{\mathrm{x}-\mathrm{m}+1}-3$ is a one-to-one mapping from $(\mathrm{m}, \infty)$ to $(-3,7)$; thus $y=$ $\left(10^{x}+\mathrm{m}\right)-\mathrm{m}+1-3$, i.e., $\mathrm{y}=\frac{10}{10^{x}+1}-3$ is a one-to-one mapping... | \mathrm{y}=\frac{10}{10^{x}+1}-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,156 |
Example 4. Write a one-to-one mapping from $(-1,10)$ to $(-\infty, 2)$. | From conclusion (ii): $\mathrm{y}=\frac{11}{\mathrm{x}+1}-3$ is a one-to-one mapping from $(-1,10)$ to $(-2, \infty)$. From the two $y=-\left(\frac{11}{x+1}-3\right)$, i.e., $y=\frac{-11}{x+1}+3$ is a one-to-one mapping from $(-1,10)$ to $(-\infty, 2)$. | y=\frac{-11}{x+1}+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,157 |
Lemma 2 The point $\mathrm{D}(\mathrm{a}, 0)$ on the $\mathrm{x}$-axis is inside the circle (I) if and only if $\mathrm{a}^{2}+\mathrm{pa}+\mathrm{q}<0$. | Proof: From plane geometry, we know that the point $\mathrm{D}(\mathrm{a}, 0)$ is inside the circle (I) if and only if $\left|\mathrm{O}^{\prime} \mathrm{D}\right|<\mathrm{r}$ (Figure 2), that is,
$$
\begin{array}{l}
\sqrt{\left(-\frac{p}{2}-a\right)^{2}+\left(\frac{1+q}{2}\right)^{2}} \\
<\frac{1}{2} \sqrt{p^{2}+(1-q)... | a^2 + pa + q < 0 | Geometry | proof | Yes | Yes | cn_contest | false | 702,159 |
Example 2. Find the length of the chord obtained by the intersection of the line $\left\{\begin{array}{l}x=-3+t, \\ y=-3+2 t\end{array}\right.$ and the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. | Solve by substituting $\left\{\begin{array}{l}x=-3+t, \\ y=-3+2 t\end{array}\right.$ into the ellipse equation
$$
\begin{array}{l}
x^{2}+4 y^{2}=16 \text { to get } \\
17 t^{2}-54 t+29=0 . \\
\left|t_{2}-t_{1}\right|=\sqrt{\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}} \\
=\sqrt{\left(\frac{54}{17}\right)^{2}-\frac{4 \tim... | \frac{4}{17} \sqrt{295} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,161 |
Example 3. The line $\left\{\begin{array}{c}x=2+\frac{1}{2} t \\ y=1+\frac{\sqrt{3}}{2} t\end{array}\right.$ intersects the parabola $y^{2}=4 x$ to form a chord. Find the length of the chord. | Solve $\left\{\begin{array}{l}x=2+\frac{1}{2} t, \\ y=1+\frac{\sqrt{3}}{2} t\end{array}\right.$
with the equation $y^{2}=4 x$ to get
$$
\begin{aligned}
\frac{3}{4} t^{2}-(2 & -\sqrt{3}) t-7=0 \\
\left|t_{2}-t_{1}\right| & =\sqrt{\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}} \\
& =\frac{8}{3} \sqrt{7-\sqrt{3}} .
\end{alig... | \frac{8}{3} \sqrt{7-\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,162 |
Example 4. Find the length of the chord obtained by the intersection of the line $\left\{\begin{array}{l}x=-3+2 t, \\ y=3 t\end{array}\right.$ and the ellipse $2 x^{2}+4 x y+5 y^{2}-4 x-22 y+7=0$. | Solve the system of equations by substituting the line equations into the circle equation:
$$
\left\{\begin{array}{l}
x=-3+2 t, \text { into } \\
y=3 t
\end{array}\right.
$$
we get
$$
\begin{aligned}
77 t^{2}-134 t & +37=0 . \\
\left|t_{2}-t_{1}\right| & =\sqrt{\left(\frac{134}{77}\right)^{2}-\frac{4 \times 37}{77}} \\... | \frac{4}{77} \sqrt{5330} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,163 |
Example 1., Derive the sine and cosine formulas for multiple angles. | Let $z=\cos \theta+i \sin 9$, respectively using De Moivre's formula to expand the expression for $Z^{a}$, we get
$$
\begin{array}{c}
\cos n \theta+i \sin n \theta=C_{n}^{0} \cos ^{n} \theta \\
+iC_{n}^{1} \cos ^{n-1} \theta \sin \theta-C_{n}^{2} \cos ^{n-2} \theta \sin ^{2} \theta \\
-iC_{n}^{3} \cos ^{n-3} \theta \si... | \cos 2 \theta= \cos ^{2} \theta-\sin ^{2} \theta, \sin 2 \theta=2 \sin \theta \cos \theta, \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta, \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta | Algebra | proof | Yes | Yes | cn_contest | false | 702,164 |
Example 2. Derive the formula for the sum of sines and cosines of angles in an arithmetic sequence.
| Let $\mathrm{z}=\cos \alpha+\mathrm{i} \sin \alpha$, then by raising to powers we get
$$
\begin{array}{l}
z^{2}=\cos 2 \alpha+i \sin 2 \alpha, \\
z^{3}=\cos 3 \alpha+i \sin 3 \alpha, \\
\cdots \cdots \cdots \\
z^{a}=\cos n \alpha+i \sin n \alpha .
\end{array}
$$
Adding both sides, we get
$$
\begin{array}{l}
\sum_{t=1}... | null | Algebra | proof | Yes | Yes | cn_contest | false | 702,165 |
Example 3. Find the value of $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}$. | Let $\mathrm{z}=\cos \alpha+\mathrm{i} \sin \alpha$, then $\vec{z}=\cos \alpha-\mathrm{i} \sin \alpha$.
$$
\begin{array}{c}
\mathrm{z}^{\mathrm{n}}=\cos n \alpha+\mathrm{i} \sin n \alpha, \\
\mathrm{z}^{\mathrm{n}}=\cos n \alpha-\mathrm{i} \sin n \alpha .
\end{array}
$$
Adding and dividing by 2 gives
$$
\operatorname{... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,166 |
Example 4. Find the value of $\sin -\frac{\pi}{9} \sin \frac{2 \pi}{9} \sin \frac{3 \pi}{9}$ - $\sin \frac{4 \pi}{9}$. | Let $\mathrm{z}=\cos \frac{\pi}{9}+\mathrm{i} \sin \frac{\pi}{9}$, then
$$
\begin{array}{l}
\mathrm{z}^{3}=\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \mathrm{z}^{0}=-\frac{1}{2}+\frac{\sqrt{3} i}{2}, \\
\mathrm{z}^{0}=-1, \mathrm{z}^{18}=1 .
\end{array}
$$
From formula (B) we get
$$
\begin{array}{l}
\text { Original e... | \frac{3}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,167 |
Example 6. Express $\sin ^{3} x \cos ^{5} x$ as a sum or difference. | Let $\mathrm{z}=\cos \mathrm{x}+\mathrm{i} \sin \mathrm{x}$, by formulas (A) and (B) we get $\sin ^{3} x \cos ^{5} x$
$$
\begin{aligned}
= & \left[\frac{i\left(1-z^{2}\right)}{2 z}\right]^{3}\left(\frac{z^{2}+1}{2 z}\right)^{5} \\
= & -\frac{i}{256 z^{8}}\left(-z^{10}-2 z^{14}+2 z^{12}+6 z^{10}\right. \\
= & -\frac{1}{... | -\frac{1}{128}(\sin 8 x+2 \sin 6 x-2 \sin 4 x-6 \sin 2 x) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,169 |
Example 1. Find $\int \frac{1}{x(1+x)\left(1+x+k^{2}\right)} d x$ | \begin{aligned} \text { Original expression } & =\int-\frac{\left(x^{2}+x+1\right)-x(1+x)}{x(1+x)} d x \\ & =\int\left[\frac{1}{x(1+x)}-\frac{1}{1+x+x^{2}}\right] d x \\ & =\ln \left\lvert\, \frac{x}{x+1}-\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x+1}{\sqrt{3}}+C\right.\end{aligned} | \ln \left\lvert\, \frac{x}{x+1}-\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x+1}{\sqrt{3}}+C\right. | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,171 |
甽2. Find $\int \frac{1}{\sqrt{\sin ^{3} x \cdot \cos ^{5} x}} \mathrm{dx}$.
---
2. Find $\int \frac{1}{\sqrt{\sin ^{3} x \cdot \cos ^{5} x}} \mathrm{dx}$. | $$
\begin{array}{l}
=\int \frac{\sin ^{2} x d x}{\sqrt{\sin ^{3} x \cdot \cos ^{5} x}}+\int \frac{\cos ^{2} x d x}{\sqrt{\sin ^{3} x \cdot \cos ^{5} x}} \\
=\int \frac{\sin ^{2} x d x}{\cos ^{4} x \sqrt{\tan ^{3} x}}+\int \frac{\cos ^{2} x d x}{\sin ^{4} x \sqrt{\cot ^{5} x}} \\
=\int \sqrt{\tan x} d(\tan x)-\int(\cot ... | \frac{2}{3} \tan ^{\frac{3}{2}} x-2 \sqrt{\cot x}+C | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,172 |
Promotion 1. Let $A_{1} A_{2} A_{3} \cdots A_{n}$ be a regular $n(n \geqslant 3$ natural number) sided polygon, $M$ be any point on the line $A_{1} A_{2}$ (excluding $A_{2}$), and $\angle \overline{A_{2} M N}=\angle \overline{A_{1} A_{2} A_{3}}$ (referring to directed angles, $A_{2} N$ bisects the exterior angle of $\a... | Prove that when $\mathrm{M}$ is at vertex $\mathrm{A}_{1}$, $\mathrm{N}$ coincides with $\mathrm{A}_{2}$, then $\mathrm{MA}_{n}=\mathrm{M}\mathrm{N}$, and the conclusion is obviously true.
When M is not at $A_{1}$ or $A_{2}$, as shown in figures 4, 5, and 6, connect $A_{1} A_{2}$ and $A_{n} N$.
$$
\begin{array}{l}
\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,174 |
Example 1. Solve the equation $\mathrm{x}^{2}-2 \mathrm{x}-3=0$ geometrically. | We note that for any given equation (I), the fixed point $\mathrm{M}(0,1)$ lies on the circle (I) corresponding to the equation (I).
$$
\begin{array}{l}
\text { Let } p=-2, q=-3 \text { be substituted into (*), we get } O^{\prime} \\
(1,-1) . \text { Taking } O^{\prime} \text { as the }
\end{array}
$$
center, and the ... | x_1 = -1, x_2 = 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,175 |
Example 2. Let $\mathrm{m}$ be any real number, prove that the equation $\mathrm{x}^{2}$ $+(1-\mathrm{m}) \mathrm{x}-(1+\mathrm{m})=0$ always has two distinct real roots, and one root is less than $\mathrm{m}$, the other root is greater than $\mathrm{m}$. | Prove that the coefficient of the linear term of the original equation is $\mathrm{p}=1-\mathrm{m}$, and the constant term is $q=-(1+m)$. For any real number $\mathrm{m}$, we have
$$
\begin{array}{l}
m^{2}+p m+q \\
=m^{2}+(1-m) m-(1+m) \\
=-1<0,
\end{array}
$$
Therefore, by Theorem 2, the original equation always has ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,176 |
Example 2. Prove: If $\mathrm{a}>\mathrm{b}>0, \mathrm{n}>1$, then
$$
\frac{a^{n}+b^{n}}{2}>\left(\frac{a+b}{2}\right)^{n} \text {. }
$$ | $$
\begin{array}{l}
\text{Prove that since } a+b=\frac{a+b}{2}+\frac{a+b}{2}, \, n>1, \text{ and} \\
|a-b|>\left\lvert\, \frac{a+b}{2}-\frac{a+b}{2}\right., \\
\therefore a^{n}+b^{n}>\left(\frac{a+b}{2}\right)^{n}+\left(\frac{a+b}{2}\right)^{n}, \\
\text { i.e., } \frac{a^{n}+b^{n}}{2}>\left(\frac{a+b}{2}\right)^{n} .
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,180 |
Example 1. Given: Quadrilateral $\mathrm{ABCD}$ is an inscribed quadrilateral in $\odot$, and diagonals $\mathrm{AC}$ and $\mathrm{BD}$ are perpendicular to each other. Also, $O E \perp B C$. Prove: $\mathrm{OE}=\frac{1}{2} \mathrm{AD}$. (78 Jiangsu Vocational College Admission Math Exam
) | Analysis: In Figure 1, OE and AD are loosely connected. By adding the diameter BF of the equal angle, and connecting FC, we have $\angle \mathrm{CBF} = \angle \mathrm{DBA}$, then $\mathrm{CF} = \mathrm{AD}$. The problem is thus reduced to proving $\mathrm{OE} = \frac{1}{2} \mathrm{CF}$, which is easy to prove (proof om... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,181 |
Example 2. Through any point $G$ inside $\odot O$, draw mutually perpendicular chords $BD$ and $AC$. Prove: $BG^2 + DG^2 + AG^2 + CG^2 =$ a constant.
( $R$ is the radius of $\odot$ O) | Prove: Connect $A B, B C, C D,$ and $\mathrm{AD}$. Draw the diameter $\mathrm{AE}$ of $\odot \mathrm{O}$ through $\mathrm{A}$, and connect $\mathrm{ED}$ (Figure 2).
By the property, $\angle \mathrm{BAC}=\angle \mathrm{EAD}$, thus $\mathrm{BC}=\mathrm{DC}$. Then,
$$
\begin{aligned}
& \mathrm{BG}^{2}+\mathrm{DG}^{2}+\ma... | 4R^2 | Geometry | proof | Yes | Yes | cn_contest | false | 702,182 |
Example 3. In $\triangle \mathrm{ABC}$, $\mathrm{BC}=\mathrm{a}, \mathrm{AC}=\mathrm{b}$, $\mathrm{AB}=\mathrm{c}$, the area of $\triangle \mathrm{ABC}$ is $\mathrm{S}$, and the circumradius is $R$ (see Figure 3). Prove: $R=\frac{a b c}{4 S}$. | Draw the altitude $\mathrm{AE}$ of $\mathrm{BC}$ through $\mathrm{A}$, and draw the diameter $\mathrm{AD}$ of $\odot \mathrm{O}$ through $\mathrm{A}$, then connect $\mathrm{BD}$. By the property, we know $\mathrm{bc}=2 \mathrm{R} \cdot \mathrm{AE}$. Then
$$
\mathrm{S}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AE}=\frac{1}{... | R=\frac{abc}{4S} | Geometry | proof | Yes | Yes | cn_contest | false | 702,183 |
Example 4. Given
the diameter of $\odot \mathrm{O}$ is a
fixed length R, with $\odot$'s
circumference at any point $\mathrm{A}$
as the center, a fixed length $r(r$
$<\mathrm{R}$) as the radius to
draw a circle, then draw a
tangent line of the small circle
intersecting $\odot \mathrm{O}$ at $\mathrm{P}$
and $\mathrm{Q}$... | Connect $A F$, then $A F=r$ and $A F \perp P Q$. Also, draw the diameter $A B$ of $\odot O$ through $A$, and connect $P Q$. By the property, we know $A P \cdot A Q=A B \cdot A F=2 R \cdot r$ (constant). | 2R \cdot r | Geometry | proof | Yes | Yes | cn_contest | false | 702,184 |
Example 5. $\triangle \mathrm{ABC}$ has a circumcircle with diameter $\mathrm{AE}$ intersecting $\mathrm{BC}$ at point $D$. Prove: $\operatorname{tg} B \cdot \operatorname{tg} C = \frac{\mathrm{AD}}{\mathrm{DE}}$. (Figure 5)
The general proof method in books is as follows:
Leave the above text in its original format ... | Proof 1: Connect EB, EC, in $\triangle \mathrm{ACD}$,
$$
\frac{\mathrm{AD}}{\sin C}=\frac{\mathrm{DC}}{\sin \left(90^{\circ}-\mathrm{B}\right)}=\frac{\mathrm{DC}}{\cos \mathrm{B}},
$$
we have $\frac{\sin \mathrm{C}}{\cos \mathrm{B}}=\frac{\mathrm{AD}}{\mathrm{DC}}$.
Also, in $\triangle \mathrm{ABD}$,
$$
\begin{array}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,185 |
Example 2. If the difference between the two roots of the equation $\mathrm{x}^{2}-5 \mathrm{x} \cdot \sin \theta+1=0$ is $2 \sqrt{3}$, and $180^{\circ}<\theta<270^{\circ}$, find the value of $\operatorname{ctg} \theta$. | $$
=\sqrt{2} \sin ^{2} \theta-4=2 \sqrt{3} \text {. }
$$
Given: $\sin ^{2} \theta=\frac{13}{25}$.
Also, $180^{\circ}<\theta<270^{\circ}$, so take $\sin \theta=-\frac{4}{5}$, at this time $\cos \theta=-\sqrt{1-\left(-\frac{4}{5}\right)^{2}}=-\frac{3}{5}$. Therefore, $\operatorname{ctg}=\frac{\cos \theta}{\sin \theta}=\... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,187 |
Example 3. If the equations $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ and $c x^{2}+b x+a=0$ both have two distinct real roots, and the difference between the roots of these two equations is equal. Prove that $a= \pm c$.
(Note: This problem is selected from "Junior High School Mathematics Review Reference Materials" pu... | Prove that the difference between the roots of the equation $a x^{2}+b x+c=0$ is $\left|\frac{\sqrt{b^{2}-4 a c}}{a}\right|$, and the difference between the roots of the equation $c x^{2}+b x+a=0$ is
$$
\left|\frac{\sqrt{b^{2}-4 c a}}{c}\right| \text {. }
$$
From the problem, we know that $\left|\frac{\sqrt{b^{2}-4 a ... | a= \pm c | Algebra | proof | Yes | Yes | cn_contest | false | 702,188 |
Example 4. Given the quadratic function $\mathrm{y}=3 \mathrm{p} \mathrm{x}^{2}-6 \mathrm{px} +3 p-2$, for what value of $p$ will the distance between the two intersection points of this parabola with the $x$-axis be $\frac{\sqrt{6}}{3}$.
---
The translation maintains the original text's line breaks and formatting. | $$
\begin{array}{l}
\text { Solve } \frac{\sqrt{\triangle}}{|3 p|}=\frac{\sqrt{(-6 p)^{2}-4 \cdot 3 p(3 p-2)}}{|3 p|} \\
=\frac{\sqrt{24 p}}{|3 p|}=\frac{\sqrt{6}}{3} . \\
\end{array}
$$
Solving for $\mathrm{p}$, we get $\mathrm{p}=4$ . | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,189 |
Example 5. If the parabola $y=ax^2+bx+c$ intersects the $x$-axis at $A$ and $B$, and the vertex is $C\left(-\frac{1}{4},-6 \frac{1}{8}\right)$, and the area of $\triangle ABC$ is $10 \frac{23}{32}$. Try to find the equation of the parabola. | Solve for the vertex coordinates of $y=a x^{2}+b x+c$ as
$\mathrm{C}\left(-\frac{\mathrm{b}}{2 \mathrm{a}}, \frac{4 \mathrm{ac}-\mathrm{b}^{2}}{4 \mathrm{a}}\right)$,
thus $-\frac{\mathrm{b}}{2 \mathrm{a}}=-\frac{1}{4}$,
$$
\begin{array}{c}
\frac{4 \mathrm{ac}-\mathrm{b}^{2}}{4 \mathrm{a}}=-6 \frac{1}{8} . \\
\because ... | y=2x^2+x-6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,190 |
Example 3. Find the maximum and minimum values of the function $y=\sqrt[4]{7-2 \sin x}$ $+\sqrt[4]{2 \sin x+1}$. | Solution: The domain of the function is obviously $\mathrm{A}=\{\mathrm{x}: \left.\sin x \geqslant-\frac{1}{2}\right\}$, and it satisfies the conditions of the theorem.
$$
\text { Let } \begin{aligned}
\varphi(x) & =|7-2 \sin \mathrm{x}-(2 \sin \mathrm{x}+1)| \\
& =|6-4 \sin \mathrm{x}|, \mathrm{x} \in \mathrm{A} .
\en... | y_{\text{minimum}} = \sqrt[4]{8}, \quad y_{\text{maximum}} = \sqrt[4]{5} + \sqrt[4]{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,191 |
Example 6. Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
untranslated text:
上为自然数, 试求拋物 线族 $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$ 在 $x$ 轴上所截得的线段长度的总和。
translated text:
Let $n$ be a natural ... | Solve for the x-coordinates $x_{1}, x_{2}$ of the intersection points of the parabola (family) with the $x$-axis, which are the roots of the equation $\left(n^{2}+n\right) x^{2}-(4 n+2) x+4=0$. Without loss of generality, assume $x_{2}>x_{1}$. Then $x_{2}-x_{1}=\frac{\sqrt{\triangle}}{|a|}=\frac{2}{n}-\frac{2}{n+1}$.
... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,192 |
Example 2. Extend the altitudes of the acute triangle $\Delta \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ to intersect the circumcircle at $\mathrm{A}_{2}, \mathrm{~B}_{2}, \mathrm{C}_{2}$, then $\Delta \mathrm{A}_{2} \mathrm{~B}_{2} \mathrm{C}_{2}$ is also an acute triangle. Extend the altitudes of $\Delta \mathrm{... | The right figure shows the inscribed $\Delta \mathrm{A}_{n} \mathrm{B}_{n} \mathrm{C}_{n}$ and $\Delta \mathrm{A}_{n+1} B_{n+1} C_{n+1}$. Now we use $\angle \mathrm{A}_{n}$ to represent $\angle \mathrm{A}_{a+1}$. Circle
$$
\begin{aligned}
& \angle A_{n+1}=\angle B_{n+1} A_{n+1} C_{n+1} \\
= & \angle B_{n+1} A_{n+1} A_{... | \angle \mathrm{A}_{\mathbf{n + 1}}=(-2)^{\mathrm{n}}\left(\alpha-\frac{\pi}{3}\right)+\frac{\pi}{3}, \angle \mathrm{B}_{\mathrm{n}+1}=(-2)^{\mathrm{n}}\left(\beta-\frac{\pi}{3}\right)+\frac{\pi}{3}, \angle \mathrm{C}_{\mathrm{n}+1}=(- | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,194 |
Example 1. Given the line $\mathrm{y}=\mathrm{kx}+\mathrm{m}$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are tangent. Try to determine the range of $k$. | Solve $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\end{array}\right.$
Substitute (1) into (2), and rearrange to get
$$
\begin{array}{l}
\quad\left(a^{2} k^{2}-b^{2}\right) x^{2}+2 a^{2} k m x+a^{2}\left(m^{2}\right. \\
\left.+b^{2}\right)=0 .
\end{array}
$$
Since the line is tangent to... | k^{2} a^{2}-b^{2}=m^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,195 |
Example 2. As shown in the figure, the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and its two asymptotes $y= \pm \frac{b}{a} x$ divide the plane into six regions: $A$, $B$, $C$, $D$, $E$, and $F$. $P_{0}$ is any fixed point on the plane.
Try to determine the situation of drawing tangent lines to the hyperbol... | Slightly explained: From Example 1, the results are as follows:
(1) When point $P_{0}$ is located in regions $A$ and $B$, one tangent line can be drawn to each branch of the hyperbola through point $\mathrm{P}_{0}$;
(2) When point $\mathrm{P}_{0}$ is located in regions $C$ and $D$, two tangent lines can be drawn to the... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,196 |
1. There is a quadratic equation, whose two roots are two-digit numbers formed by the digits 1, 9, 8, and 4. Let the difference between these two roots be $\mathrm{x}$, which makes $\sqrt{1984 \mathrm{x}}$ an integer. Try to find this equation and its two roots. | 1. The required equation is $\mathrm{y}^{2}-(18+49) \mathrm{y}+18 \times 49$ $=0$. The two roots are $\mathrm{y}_{1}=49, \mathrm{y}_{2}=18$. | y_1=49, y_2=18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,198 |
3. In the following three expressions, each has four squares, fill in the numbers 1, 9, 8, 5 respectively, so that the results of each expression are perfect squares.
(1) $\square \times \square+\square \times \square=(\quad)^{2}$;
(2) $\square \times \square+\square=(\quad)^{2}$;
(3) $\square-\frac{\square}{\square-\s... | 3 .
(1) $8 \times 5 + 1 \times 9 = 7^{2}$;
(2) $\frac{8 \times 5}{9 + 1} = 2^{2}$;
(3) $5 - \frac{8}{9 - 1} = 2^{2}$. | 7^2, 2^2, 2^2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,200 |
4. Use the digits $1$, $9$, $8$, and $5$ to fill in the four boxes below so that their product is maximized.
$$
(\square+\square) \times(\square+\square) \text {. }
$$
(Wenzhou Li Fangyue) | 4. $(8+5) \times(1+9)=130$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
4. $(8+5) \times(1+9)=130$. | 130 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,201 |
Example 4. Find the real solutions of the equation $\sqrt[4]{a-x}+\sqrt[4]{b-x}=\sqrt[4]{a+b-2x}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | From the theorem, we know that the roots of the equation must satisfy
$$
|a-x-(b-x)|=|a+b-2 x| \text {, }
$$
which simplifies to $|a-b|=|a+2 b-2 x|$.
Solving this, we get: $x_{1}=a, x_{2}=b$.
Verification: When $a \leqslant b$, the original equation has a real root $x=a$; when $a>b$, the original equation has a real r... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,202 |
Example 1. From a large batch of transistors with a high-quality rate of $90 \%$, if 5 are taken out, what is the probability that exactly 3 of them are high-quality? | Solution: Let event A: taking out 1 transistor, which is a high-quality item, $\mathrm{P}(\mathrm{A})=0.9$. This problem is to find the probability that event A occurs 3 times in 5 independent repeated trials.
$$
\mathrm{q}=1-\mathrm{P}(\mathrm{A})=1-0.9=0.1, \mathrm{n}=5 \text {, }
$$
$\mathrm{k}=3$.
$$
\therefore \ma... | 0.0729 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,203 |
Example 1. From the five numbers $1,2,3,4,5$, take out three numbers, find the probability that the sum of these three numbers is divisible by 5. | Analysis: $n=C$ Using the enumeration method of combinations, we can get that the possible cases for three numbers being divisible by: 5 are $(1, 4, 5)$, $(2,3,5) . \therefore \mathrm{P}=\frac{2}{10}=\frac{1}{5}$. | \frac{1}{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,204 |
1. There are 100 nuts, among which 96 are genuine and 4 are defective. If 2 are randomly selected, what is the probability that both are defective? | (Answer) $\mathrm{P}=\frac{\mathrm{C}_{4}^{2}}{\mathrm{C}_{100}^{2}}=\frac{1}{825}$ ) | \frac{1}{825} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,205 |
2.8 Place these books on the same shelf of a bookcase. Find the probability that 4 specified books are placed in the middle.
Place the above text in English, keeping the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}\text { (Answer: } \mathrm{P}=\frac{\mathrm{P}_{4}^{4} \cdot \mathrm{P}_{4}^{4}}{\mathrm{P}_{8}^{8}} \\ =\frac{1}{70} \text { ) }\end{array}$ | \frac{1}{70} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,206 |
3. Throw a die three times, find the probability of getting exactly two ones.
The above text is translated into English, keeping the original text's line breaks and format. | $\begin{array}{l}\text { (Answer: } P=\frac{\mathrm{C}_{3}^{2} \cdot \mathrm{C}_{5}^{1}}{6^{3}}=\frac{5}{72} \text {. } \\ \text { Or } \mathrm{P}=\mathrm{C}_{3}^{2} \cdot\left(\frac{1}{6}\right)^{2} \cdot\left(\frac{5}{6}\right)=\frac{5}{72} \text { ) }\end{array}$ | \frac{5}{72} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,207 |
4. From the natural numbers less than 10000, pick one number, find the probability that this number is divisible by 4.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (Answer:
$$
\left.\mathrm{p}=\frac{2499}{9999}=\frac{833}{3333}\right)
$$ | \frac{833}{3333} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,208 |
$$
\text { 5. From } 1,2,3,4,5,6,7,8,9,10
$$
choose any two numbers, find the probability that their product is a multiple of 5.
$$ | (Answer) $\mathrm{P}=\frac{\mathrm{C}_{8}^{1} \cdot \mathrm{C}_{2}^{1}+\mathrm{C}_{2}^{2}}{\mathrm{C}_{10}^{2}}=\frac{17}{45}$ ) | \frac{17}{45} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,209 |
6. A shooter has a probability of 0.29 to hit 10 points in one shot, and a probability of 0.41 to hit 9 points. Find the probability that the shooter scores less than 9 points in one shot. | $\begin{array}{l}\text { (Answer: } \mathrm{P}=1-0.29-0.41 \\ =0.30 \text { ) }\end{array}$ | 0.30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,210 |
8. In jar A, there are 6 balls, of which 4 are red and 2 are white; in jar B, there are 4 balls, of which 3 are red and 1 is white; in jar C, there are 5 balls, of which 2 are red and 3 are white. One ball is drawn from each of the jars A, B, and C. Find:
(1) the probability that exactly 2 of the balls are white;
(2) t... | (1) $P=\frac{1}{3} \times \frac{1}{4} \times \frac{2}{5}+\frac{1}{3} \times \frac{3}{4} \times \frac{3}{5}+\frac{2}{3} \times \frac{1}{4} \times \frac{3}{5}=\frac{19}{60}$,
( 2 ) $P=\frac{17}{60}+\frac{1}{3} \times \frac{1}{4} \times \frac{3}{5}=\frac{20}{60}=\frac{1}{3}$ | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,212 |
Example 1. Prove - The intersection points of the parabola $y^{2}=2 x$ and $y=(x+4)^{2}$ lie on the same circle. | Prove that two parabolas have intersection points, and the curve passing through the intersection points is $\lambda_{1}\left(y^{2}-2 x\right)+\lambda_{2}\left[(x-4)^{2}-y\right]=0$.
If $\lambda_{1}=\lambda_{2}=1$, there is a circle $\mathrm{x}^{2}+\mathrm{y}^{2}-10 \mathrm{x}-$ $y+16=0$.
This means that the intersect... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,213 |
10. A person has a probability of 0.6 of hitting the target with one shot. How many times at least should they shoot to make the probability of hitting the target at least once greater than 0.95? | (Given $\lg 2=$
0.3010 ) (Answer: $\mathrm{n}=4$ ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,215 |
1. There are 20 cards each written with numbers $1,2, \cdots, 20$, and one card is drawn at random. The probability that the number on the card is divisible by 3 is ( ). | (Answer: $\mathrm{P}=\frac{3}{10}$ ) | \frac{3}{10} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,217 |
2. There are five sticks with lengths of $2$, $4$, $6$, $8$, and $10$ centimeters, respectively. If we randomly select 3 sticks, the probability that they can form a triangle is ( ). | (Answer: $P=\frac{3}{10}$ ) | \frac{3}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,218 |
3. The telephone numbers in a certain city consist of 5 digits, each of which can be any one of the ten digits $0-9$. If a telephone number is chosen at random, the probability that it consists of 5 different digits is ( ).
| (Answer: P= $\frac{189}{625}$ ) | \frac{189}{625} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,219 |
Example. $\mathrm{y}=-2 \mathrm{x}^{2}+3 \mathrm{x}+4$. | Solving $\mathrm{y}=4$ gives the required straight line. Thus, $-2 x^{2}+3 x+4=4$, solving for $x$ yields $x=0$, $\mathrm{x}=\frac{3}{2}$, which are the x-coordinates of the two intersection points of the parabola and the line $\mathrm{y}=4$. The equation of the axis of symmetry is $\mathrm{x}=\frac{3}{4}$, and the ver... | x=0, \mathrm{x}=\frac{3}{2}, \left(\frac{3}{4}, \frac{41}{8}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,223 |
Example 2. Given that an ellipse passes through four points $\mathrm{A}(-2,0), \mathrm{B}$ $(2,0), C(0,-1)$ and D $(0,1)$ and is tangent to $x-2 y=4$. Find the equation of this ellipse. | Solve the following four points
as the four vertices of a quadrilateral, the equations of these four sides are respectively 8
$$
\begin{array}{l}
\text { AD: } x-2 y+2=0, \\
\text { BC: } x-2 y-2=0, \\
\text { AC: } x+2 y+2=0, \\
\text { BD: } x+2 y-2=0 . \\
\therefore(x-2 y+2)(x-2 y-2)+m \\
\\
\cdot(x+2 y+2)(x+2 y-2)=... | 3 x^{2}-4 x y+12 y^{2}-16=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,224 |
As is well known, the basic inequality $\frac{x+y}{2} \geqslant \sqrt{x y}$ $(x>0, y>0)$ is an extremely important and widely applied inequality in elementary mathematics. Now, it is generalized as follows:
If $x>0, y>0, a>0, b>0$, and $a+b=1$, then $a x+b y \geqslant x^{a} y^{b}$ (Xiao Shangyi, equality holds when $x... | Proof: Suppose $x>y>0$, then $\frac{x}{y}>1$. Let $\frac{x}{y} = t$, and consider the function $f(t)=t \sim, f(t)$ satisfies the conditions of the Mean Value Theorem on the interval $[1, t]$, hence
$$
\frac{f(t)-f(1)}{t-1}=f'(\xi), \quad (11, a-10, \\
\therefore y\left(\frac{x}{y}\right)^{a}<y+a x-a y \\
\quad=a x+(1-a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,226 |
S i. Proof: $\sqrt[3]{3 \mathrm{a}^{2}} \leqslant 1+\frac{2}{3} \mathrm{a} \cdot(\mathrm{a}>0)$ | Proof: $\because \sqrt[3]{3 a^{2}}=3^{\frac{-1}{3}} a^{\frac{-2}{3}}$, from inequality (I) we get
$$
\sqrt[3]{3 a^{2}} \leqslant \frac{1}{3} \times 3+\frac{2}{3} a=1+\frac{2}{3} a .
$$
Obviously, when $a=3$, the inequality becomes an equality.
It is not difficult to use inequality (I) to prove a more general case:
Whe... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,227 |
Example 3. If $a+b=1, a>0, b>0$. Prove: $\mathrm{a}^{\mathrm{b}-1} \mathrm{~b}^{\mathrm{a}-1} \leqslant 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
a^{b} b^{0} \leqslant b a+a b=2 a b, \\
\therefore \frac{a^{b} b^{a}}{a b} \leqslant 2,
\end{array}
$$
that is, $a^{b-1} b^{a-1} \leqslant 2$. | null | Inequalities | proof | Yes | Yes | cn_contest | false | 702,229 |
Example 4. If $\mathrm{m}+\mathrm{n}=1, \mathrm{~m}>0, \mathrm{n}>0$.
Prove: $(1+m)^{\mathrm{n}} \cdot(1+\mathrm{n})^{\mathrm{m}} \leqslant 1+2 \mathrm{mn}$. | $$
\begin{array}{l}
(1+m)^{n} \cdot(1+n)^{m} \leqslant n(1+m) \\
+m(1+n) \\
=n+n m+m+m n \\
=1+2 m n .
\end{array}
$$
Prove that from inequality (I) we get
$$
\begin{array}{l}
(1+m)^{n} \cdot(1+n)^{m} \leqslant n(1+m) \\
+m(1+n) \\
=n+n m+m+m n \\
=1+2 m n .
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,230 |
Example 5. Prove: $\operatorname{tg} x \geqslant 4 \sqrt[4]{\operatorname{tg} x}-3$. $\left(0<x<\frac{\pi}{2}\right)$ | Prove that from inequality (I) we have
$$
\begin{array}{l}
\frac{1}{4} \sin x+\frac{3}{4} \cos x \\
\geqslant(\sin x)^{\frac{1}{4}}(\cos x)^{\frac{9}{4}} \\
=(\sin x)^{\frac{1}{4}}(\cos x)^{1-\frac{1}{4}} \\
=\cos x \cdot(\tan x)^{\frac{1}{4}} .
\end{array}
$$
Dividing both sides by $\cos x$ we get
$$
\begin{aligned}
... | \tan x \geqslant 4 \sqrt[4]{\tan x}-3 | Inequalities | proof | Yes | Yes | cn_contest | false | 702,231 |
Example 1. In an acute $\triangle \mathrm{ABC}$, prove that $\operatorname{tg} A \cdot \operatorname{tg} \mathrm{B}$
$$
\cdot \operatorname{tg} C+\operatorname{ctg} A \cdot \operatorname{ctg} B \cdot \operatorname{ctg} C \geqslant \frac{28 \sqrt{3}}{3} .
$$ | \begin{array}{l}\text { Prove that since } A+B+C=\pi, \quad \therefore \operatorname{ctg} A \\ \cdot \operatorname{ctg} B+\operatorname{ctg} B \cdot \operatorname{ctg} C+\operatorname{ctg} C \cdot \operatorname{ctg} A=1. \\ \text { Also, ctg } A>0, \text { ctg } B>0, \operatorname{ctg} C>0, \\ \therefore \operatorname{... | \frac{28 \sqrt{3}}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,234 |
Example 3. Find the translation trajectory of $(1,2)$ with respect to $\left(t, t^{2}\right)$. | Let the translation point be $(x, y)$, then
$$
\left\{\begin{array}{l}
x=1+t \\
y=2+t^{2} .
\end{array}\right.
$$
Eliminating $t$, we get $y=(x-1)^{2}+2$.
When a figure undergoes a translation, the translation increment for all points is the same. Therefore, once the translation increment is found, the translation tra... | y=(x-1)^{2}+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,235 |
Example 2. Given acute angles $\alpha, \beta, \gamma$ satisfying $\cos \alpha$
$$
\begin{array}{l}
+\cos \beta+\cos \gamma=1 \text{. Prove: } \cos ^{2} \alpha+\cos ^{2} \beta \\
+\cos ^{2} \gamma+\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta+\operatorname{tg}^{2} \gamma \geqslant \frac{73}{3} .
\end{array}
$... | $$
\begin{array}{l}
\text { Given } \cos \alpha>0, \cos \beta>0, \cos \gamma>0, \text { and } \cos \alpha+\cos \beta+\cos \gamma=1, \\
\therefore\left(\cos \alpha+\frac{1}{\cos \alpha}\right)^{2}+(\cos \beta \\
\left.+\frac{1}{\cos \beta}\right)^{2}+\left(\cos \gamma+\frac{1}{\cos \gamma}\right)^{2} \\
\geqslant 3 \cdo... | \frac{73}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,236 |
Example 4. Given $a_{k}>1 (k=1,2,3)$ and
$$
\begin{array}{l}
a_{1} \cdot a_{2} \cdot a_{3}=10 . \text { Prove: }\left(\lg a_{1}+\log _{a_{1}} 10\right)^{5} \\
+\left(\lg a_{2}+\log _{a_{2}} 10\right)^{5}+\left(\lg a_{3}+\log _{a_{3}} 10\right)^{5} \\
\geqslant \frac{100000}{81} .
\end{array}
$$ | $$
\begin{array}{l}
\text{Given } a_{1} a_{2} a_{3}=10, \text{ and } a_{1}>1, \\
\left.+\log _{\mathrm{a}_{2}} 10\right)^{5}+\left(\operatorname{lga}_{3}+\log _{\left.\mathrm{a}_{3} 10\right)^{5}}\right. \\
=\left(\lg a_{1}+\frac{1}{\lg a_{1}}\right)^{5}+\left(\lg a_{2}+-\frac{1}{\lg a_{2}}\right)^{5} \\
+\left(\lg a_{... | \frac{100000}{81} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,238 |
Example 2. Find a pair of integers satisfying
$$
\left\{\begin{array}{l}
y-\left|x^{2}-2 x\right|+\frac{1}{2}>0 \\
y+|x-1|<2
\end{array}\right.
$$ | Solving the original system of inequalities can be transformed into
$$
\left\{\begin{array}{l}
-y-\frac{1}{2}<x^{2}-2 x<y+\frac{1}{2} \\
-2+y<x-1<2-y \\
-y+\frac{1}{2}<(x-1)^{2}<y+\frac{3}{2} \\
-1+y<x<3-y
\end{array}\right.
$$
That is,
Draw the solution region of the system of inequalities as shown by the shaded are... | (0,0), (2,0), (1,1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,240 |
Example 3. Find the solution set of the equation $|\mathrm{x}+\mathrm{y}|=|2 \mathrm{x}-\mathrm{y}+2|$ $+3 \mathbf{x}+2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Let $\mathrm{x}+\mathrm{y}=0$ and $2 \mathrm{x}-\mathrm{y}+2=0$, draw these two lines, then the plane is divided into four angular regions (as shown in Figure 3).
(1) For the angular region I (i.e., the angular region $\angle A E C$ including the boundary rays), for any point $(x, y)$, we have
$$
\left\{\begin{... | \{(x, y) \mid \left\{\begin{array}{l}x+y \geqslant 0, \\ 2 x-y+2 \leqslant 0\end{array}\right. \text{ and } \left.x=-\frac{2}{3}\left(y \leqslant \frac{2}{3}\right)\right\}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,241 |
Example 4. Under the constraint conditions: $\left\{\begin{array}{l}x+2 y-2 \geqslant 0, \\ 2 x+y-2 \geqslant 0\end{array}\right.$, find the minimum value of $\log _{\theta}\left(3^{x}+3^{y}\right)$ and the corresponding values of $\mathrm{x}$ and $\mathrm{y}$. | Solve the system of inequalities representing the plane region as the shaded part (including the boundary) in Figure 4. The coordinates of point A are
$$
\left(\frac{2}{3}, \frac{2}{3}\right) \text {. }
$$
Thus, we have $\log _{9}\left(3^{x}+3^{y}\right)$
$$
\geqslant \log _{0} 2 \cdot \sqrt{ } 3^{x} \cdot 3^{y}=\log ... | \frac{1}{3}+\log _{9} 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,242 |
Example 1. Prove: For any real number $t$, the modulus $r=|z|$ of the complex number $z=\sqrt{|\cos t|}+\sqrt{|\sin t|} \mathrm{i}$ satisfies $r \leqslant \sqrt[4]{2} \cdot$ (83年高考理科试题)
(83年高考理科试题) should be translated as (1983 Science College Entrance Exam Question) for clarity in English. | Prove that using $\frac{\mathrm{a}+\mathrm{b}}{2} \leqslant \sqrt{\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{2}}$ (i.e., a simpler form of (1) in (*)), we have
$$
\begin{array}{l}
r=\sqrt{|\cos t|+| \sin t|} \\
\leqslant \sqrt{2 \sqrt{\frac{\cos ^{2} t+\sin ^{2} t}{2}}}=\sqrt[4]{2} .
\end{array}
$$ | r \leqslant \sqrt[4]{2} | Algebra | proof | Yes | Yes | cn_contest | false | 702,243 |
Example 2. Given $\mathrm{x}+\mathrm{y}=1$, find the maximum value of $\sqrt{2 \mathrm{x}+1}$ $+\sqrt{2 y+1}$. | $$
\begin{array}{l}
\sqrt{2 x+1}+\sqrt{2 y+1} \\
\leqslant 2 \sqrt{\frac{(2 x+1)+(2 y+1)}{2}}=2 \sqrt{2}, \\
\text { i.e., the maximum value of } \sqrt{2 x+1}+\sqrt{2 y+1} \text { is } 2 \sqrt{2} \text { . } \\
\end{array}
$$ | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,244 |
Example 3. In $\triangle \mathrm{ABC}$, prove:
$$
\operatorname{ctg}^{3} \frac{\mathrm{A}}{2}+\operatorname{ctg}^{3} \frac{\mathrm{B}}{2}+\operatorname{ctg}^{3} \frac{\mathrm{C}}{2} \geqslant 9 \sqrt{3} .
$$ | Incorrect Proof $\because \frac{\mathrm{A}}{2}, \frac{\mathrm{B}}{2}, \frac{\mathrm{C}}{2}$ are all positive acute angles, $\therefore \operatorname{ctg}^{3} \frac{\mathrm{A}}{2}, \operatorname{ctg}^{3} \frac{\mathrm{B}}{2}, \operatorname{ctg}^{3} \frac{\mathrm{C}}{2}$ are all positive numbers. Therefore, $\operatornam... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,245 |
Example 4. The three vertices of a triangle are $(0,0)$, $(2,3)$, and $(3,-2)$. When the triangle is translated, the translation path of $(0,0)$ is $\mathrm{y}^{2}=2 \mathrm{x}$. Find the translation paths of the other two points. | Solve: Write $\mathrm{y}^{2}=2 \mathrm{x}$ as a parametric equation
$$
\left\{\begin{array}{l}
x=-\frac{t^{2}}{2} \\
y=t .
\end{array}\right.
$$
$\left(-\frac{t^{2}}{2}, t\right)$ is the translation increment of $(0,0)$, and also the translation increment of the triangle. Therefore, the translation trajectory of (2,3) ... | (y-3)^{2}=2(x-2) \text{ and } (y+2)^{2}=2(x-3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,246 |
Example 4. Given $a>0, b>0$, and $a+b=1$. Prove: $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant \frac{25}{2}$. | Incorrect Argument
$$
\begin{array}{l}
\text { Left side of the original equation }=\mathrm{a}^{2}+\mathrm{b}^{2}+\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+4 \\
\geqslant 2 \mathrm{ab}+\frac{2}{\mathrm{ab}}+4 \\
=2\left(\mathrm{ab}+\frac{1}{\mathrm{ab}}-\right)+4 \\
\geqslant 2 \cdot 2 \sqrt{a b \cdot \frac{1}... | \frac{25}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 702,247 |
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