problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 1. In $\triangle \mathrm{ABC}$, it is known that $(\sin \mathrm{B}+\sin \mathrm{C})$
$$
\begin{array}{l}
:(\sin C+\sin A):(\sin \mathrm{A}+\sin \mathrm{B}) \\
=5: 6: 7 .
\end{array}
$$
Prove: $\cos \mathrm{A}: \cos \mathrm{B}: \cos \mathrm{C}$
$$
=-4: 11: 14 \text {. }
$$ | Proof According to the problem, we can set
$$
\left\{\begin{array}{l}
\sin \mathrm{B}+\sin \mathrm{C}=5 \mathrm{k}, \\
\sin \mathrm{C}+\sin \mathrm{A}=6 \mathrm{k}, \\
\sin \mathrm{A}+\sin \mathrm{B}=7 \mathrm{k} .
\end{array}\right.
$$
( $k$ is a constant not equal to 0)
$$
\begin{array}{c}
{[(1)+(2)+(3)] \times \frac... | -4: 11: 14 | Geometry | proof | Yes | Yes | cn_contest | false | 702,391 |
Example 2. Given $\operatorname{tg}(\alpha+\beta)=3 \operatorname{tg} \alpha$. Prove: $\sin (2 \alpha+2 \beta)+\sin 2 \alpha=2 \sin 2 \beta$ | Analysis: From the known conditions, separate the ratio to $\frac{\operatorname{tg}(\alpha+\beta)}{\operatorname{tg} \alpha}=\frac{3}{1}$. From the conclusion, separate the ratio to $\frac{\sin (2 \alpha+2 \beta)+\sin \alpha}{\sin 2 \beta}=\frac{2}{1}$. From $\frac{3}{1}$ to $\frac{2}{1}$, we can use the subtraction ra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,392 |
Example 1. Draw the graph of the function $\mathrm{y}=\cos ^{2} \mathrm{x}+1$, $\mathrm{x} \in$ $[0, \pi]$. | Method: Let $\mathrm{u}(\mathrm{x})=\mathrm{x}^{2}+1, \mathrm{v}(\mathrm{x})=\cos \mathrm{x}$ $(x \in[0, \pi])$. As shown in Figure 2, first draw the graphs of $y=x^{2}+1$, $\mathrm{y}=\cos \mathrm{x}$, and $\mathrm{y}=\mathrm{x}$. Then, select several "key points" (such as endpoints, highest (lowest) points, inflectio... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,393 |
Example 2, draw the graph of the function $\mathrm{y}=\frac{1}{2} \sqrt{x^{2}+2}$.
(On graph paper) | Method: Let
$u(x)$
$=\frac{1}{2} \sqrt{x}$,
$v(x)$
$=x^{2}+2$.
First, draw the graphs of $y=\frac{1}{2} \cdot \sqrt{x}$, $y=x^{2}+2$, and $y=x$. Then, use a grid to "search" for points on the graph along the positive X-axis, and connect these points in sequence with a smooth curve to obtain the part of the graph in the... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,394 |
Example 1. $\mathrm{P}, \mathrm{Q}$ are two moving points on the parabola $\mathrm{y}^{2}=2 \mathrm{mx}$, $\mathrm{O}$ is the vertex of the parabola, $\angle \mathrm{POQ}=\alpha$ (a constant). The tangents at points $P, Q$ intersect at point $T$. Find the equation of the locus of point $T$. | Solve for points $P$ and $Q$ with their y-coordinates $y_1$ and $y_2$ as parameters.
Since points $P$ and $Q$ are on the parabola, the coordinates of points $P$ and $Q$ are:
$$
\left(\frac{y_1^2}{2 m}, y_1\right) \left(\frac{y_2^2}{2 m}, y_2\right).
$$
The equations of the tangents $\mathrm{PT}$ and $\mathrm{QT}$ are:... | \operatorname{tg}^2 \alpha \cdot x^2 - 4 y^2 + 4 m \operatorname{tg}^2 \alpha \cdot x + 8 m x + 4 m^2 \operatorname{tg}^2 \alpha = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,395 |
Example 2. A moving point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ draws tangents to the ellipse $\mathrm{b}^{2} \mathrm{x}^{2}$ with angles $\theta_{1}, \theta_{2}$. (1) When $\operatorname{tg} \theta_{1}+\operatorname{tg} \theta_{2}=\mathrm{m}$ (a constant), find the equation of the locus of point $P$; (2) When $\operato... | (1) Let the slope of a tangent be $k_{1}$, then the tangent line is
$$
\begin{aligned}
y & =k_{1} x+\sqrt{k_{1}^{2} a^{2}+b^{2}} \\
\text { or } y & =k_{1} x-\sqrt{k_{1}^{2} a^{2}+b^{2}},
\end{aligned}
$$
i.e., $\pm \sqrt{k_{1}^{2} a^{2}+b^{2}}=y-k_{1} x$. Squaring both sides, we get:
$$
a^{2} k_{1}^{2}+b^{2}=y^{2}-2 ... | m\left(x^{2}-a^{2}\right)=2 x y | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,396 |
Example 3. Given a fixed circle $(x-a)^{2}+y^{2}=a^{2}$. Two moving chords $O A, O B$ are drawn through the origin. When $|O A| \cdot|O B|=$ constant $\mathrm{m}^{2}$, prove that the moving line $\mathrm{AB}$ is a tangent to another fixed circle. | Let $\mathrm{OA}, \mathrm{OB}$ be the angles with the X-axis as $\alpha, \beta$, then the coordinates of $\mathrm{A}, \mathrm{B}$ are:
$$
\begin{array}{l}
\mathrm{A}\left(2 a \cos^{2} \alpha,\right. \\
2 a \cos \alpha \sin \alpha), \\
\mathrm{B}(2 a \cos \beta, \\
2 a \cos \beta \sin \beta) . \\
\text { Establish the e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,397 |
Example 1. The orthocenter of a triangle is the incenter of the triangle formed by connecting the three feet of the altitudes. | Given, $H$ is the orthocenter of $\triangle ABC$.
From B, C, E, F being concyclic, we have $\angle 1=\angle \mathcal{E}$;
From B, D, H, F being concyclic, we have $\angle 1=\angle 3$.
$\therefore \angle 3=\angle 4$, so $D$ bisects $\angle \mathrm{FDE}$.
Similarly, $\mathrm{EH}$ bisects $\angle \mathrm{DEF}$, and $\math... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,398 |
Example 7, Find the minimum and maximum values of the function $y=\frac{x^{2}+2 x+5}{x^{2}+4 x+5}$. | Solve by eliminating the denominator and organizing into a quadratic equation in $x$:
$$
\begin{array}{l}
(y-1) x^{2}+(4 y-2) x+5 y-5=0 \\
\triangle=-4 y^{2}+24 y-16
\end{array}
$$
Since $x$ is a real number,
$$
\begin{array}{l}
\therefore y_{\text {minimum }}=3-\sqrt{5}, y_{\text {maximum }}=3+\sqrt{5} \text { . } \\
... | y_{\text {minimum }}=3-\sqrt{5}, y_{\text {maximum }}=3+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,399 |
Example 3. Given:
$\mathrm{AD}, \mathrm{CE}$ are
the altitudes of
$\triangle \mathrm{ABC}$,
$\mathrm{H}$ is the orthocenter, $\mathrm{O}$ is
the circumcenter, $\angle \mathrm{BAC}$ $=60^{\circ}$.
Prove: $\mathrm{AO}=\mathrm{AH}$. | Prove that connecting $\mathrm{BO}$ and extending it to $\mathrm{M}$, such that $\mathrm{MO}=\mathrm{BO}$.
$$
\begin{array}{l}
\because \mathrm{O} \text { is the circumcenter, } \therefore \mathrm{AO}=\mathrm{BO}=\mathrm{MO} . \\
\therefore \angle \mathrm{BAM}=90^{\circ} .
\end{array}
$$
Similarly, $\angle \mathrm{BCM... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,400 |
Example 5. The centroid, circumcenter, and orthocenter of a triangle are collinear.
Given: In $\triangle \mathrm{ABC}$, $\mathrm{H}$ is the orthocenter, and $O$ is the circumcenter.
To prove: The centroid lies on the line $\mathrm{OH}$.
Proof: Draw $O M \perp B C$ at $M$, then $M$ is the midpoint of $\mathrm{BC}$. | Connect $\mathrm{AM}$ intersecting $\mathrm{OH}$ at $\mathrm{G}$, $\mathrm{AM}$ being the median.
Connect $\mathrm{AH}$ and extend it to intersect $\mathrm{BC}$ at $\mathrm{D}$.
$\because \mathrm{H}$ is the orthocenter,
$\therefore \mathrm{AD} \perp \mathrm{BC}, \mathrm{OM} / / \mathrm{AD}$.
Thus, $\triangle \mathrm{MO... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,402 |
Example 3, Solve the inequality: $|x-5|-|2x+3|$ < 1. (1978 Shanghai High School Mathematics Competition Question).
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems the translation request was for the problem... | $$
\begin{array}{l}
\text { Solution: Let }|x-5|-|2 x+3|-1=0 \text { ( }) \\
\Rightarrow(x-5+2 x+3+1)(2 x-5-2 x-3 \\
+1)(x-5-2 x-3-1)(x-5+2 x+3-1) \\
= 0 \\
\Rightarrow \Rightarrow(3 x-1)(-x-7)(-x-9)(3 x-3) \\
\Rightarrow x=\frac{1}{3},-7,-9,1 .
\end{array}
$$
Upon verification, $x=-7, \frac{1}{3}$ are solutions to eq... | x=-7, \frac{1}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,405 |
4. Solve the inequality: $|x+7|-|3 x-4|$ $+\sqrt{3-2 \sqrt{2}}>0$. (78th Shaanxi Province High School Mathematics Competition Question) | $$
\begin{array}{l}
\text { Let }|x+7|-|3 x-4|+\sqrt{2} \\
-1=0 \\
\Rightarrow(x+7+3 x-1+2-1)(x+7 \\
-3 x+4 \div \sqrt{2}-1)(x+7-3 x+4-\sqrt{2} \\
+1)(x+7+3 x-4-\sqrt{2}+1)=0 \\
\Rightarrow(4 x+2+\sqrt{2})(-2 x+10+\sqrt{2}) \\
\text { - }(-2 x+12-\sqrt{2})(4 x+4-\sqrt{2})=0 \\
\Rightarrow x=-\frac{1}{4}(2+\sqrt{2}), \f... | -\frac{1}{4}(2+\sqrt{2}) < x < \frac{1}{2}(10+\sqrt{2}) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,406 |
Example 1: If the coordinates of point $(x, y)$ satisfy $x^{2}+y^{2}+5 x \leqslant 0$, find the maximum and minimum values of $3 x+4 y$.
| The constraint condition is $\left(x+\frac{5}{2}\right)^{2}+y^{2} \leqslant \frac{25}{4}$, and the points $(x, y)$ that satisfy this condition are on the circumference and inside the circle $\left(x+\frac{5}{2}\right)^{2}+y^{2} = \frac{25}{4}$.
Let the objective function be $3 \mathrm{x}+4 \mathrm{y}=\mathrm{m}$, then... | 5, -20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,407 |
Example 2. If $x$ is a real number, find the maximum and minimum values of $\frac{3-\sin x}{4-2 \cos x}$. | The objective function can be seen as the slope k of the line connecting any point $(2 \cos x, \sin x)$ on the ellipse $-\frac{x^{2}}{4}+y^{2}=1$ and the fixed point $(4,3)$.
From the graph, it is easy to see: when the line
$y=k(x-4)$
+3 is tangent to the ellipse,
k reaches its extreme values.
Using formula 2.2
we get:... | 1 \pm \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,408 |
Example 3. If $a, b, c$ are non-negative real numbers, find
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}} \\
+\sqrt{b^{2}+c^{2}} \\
+\sqrt{c^{2}+a^{2}} \text { 's minimum }
\end{array}
$$
value. | Let point $\mathrm{A}$
$(a, b)$, point $B(a+$
$b, b+c)$, point $C(a+b+c, a+b+c)$, then
$$
\begin{array}{l}
|O A|=\sqrt{a^{2}+b^{2}},|A B| \\
=\sqrt{b^{2}+c^{2}},|B C|=\sqrt{c^{2}+a^{2}}, \text { and } \\
|O C|=\sqrt{2}(a+b+c) .
\end{array}
$$
$\because$ Connecting $\geqslant 1$ OC:
$$
\begin{array}{l}
\therefore \sqrt... | \sqrt{2}(a+b+c) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,409 |
Example 1. If the total surface area of a cone is $n$ times the area of its inscribed sphere, find the ratio of the volume of the sphere to the volume of the cone. | Let $\mathrm{r}$ be the radius of the sphere, $\mathrm{S}$ and $\mathrm{V}$ be the surface area and volume of the sphere, $\mathrm{S}^{\prime}$ and $\mathrm{V}^{\prime}$ be the total surface area and volume of the cone, then
$$
\mathrm{V}=\frac{1}{3} \mathrm{Sr}, \quad \mathrm{V}^{\prime}=\frac{1}{3} \mathrm{~S}^{\prim... | 1: n | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,411 |
Example 2. Two regular quadrilateral pyramids with a common base but opposite directions form an octahedron. If the side length of the base of the regular quadrilateral pyramid is $a$, and the dihedral angle at the vertex is $\alpha$, find the surface area of its inscribed sphere. | $$
\begin{array}{l}
\text{Let the height of the regular quadrilateral pyramid be } h, \text{ the slant height be } h', \text{ and the area of one side face be } S_{1}, \text{ then} \\
h=\frac{a}{2} \sqrt{\cot^{2} \frac{\alpha}{2}}, \\
\dot{n}^{\prime}=\frac{a}{2} \cot \frac{\alpha}{2}, S_{1}=\frac{a^{2}}{4} \cot \frac{... | \frac{2 \pi a^{2} \cos \alpha}{1+\cos \alpha} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,412 |
Example 3. A combination of two inverted cones sharing the same base, the lateral development radius of one cone is 15, and the central angle is $288^{\circ}$, the lateral development radius of the other cone is 13, find the volume of its inscribed sphere. | Let the axial section of the composite body be SMVN (as shown in the figure), A be the area of the face of $v_{\mathrm{i}}$ SMVN, and C be the perimeter, then $2 \pi \cdot \mathrm{MO}$
$$
\begin{array}{l}
=2 \pi \cdot \mathrm{SM}_{360^{\circ}}^{288^{\circ}} \\
\quad \therefore \quad \mathrm{MO}=15 \times \frac{4}{5}=12... | 288 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,413 |
Example 1. Find the equation of the line passing through the intersection points of the two curves
$$
3 x^{2}+3 y^{2}+2 x+y=0 \text { and } x^{2}+y^{2}+3 x
$$
$-\mathrm{y}=0$. | Solution 1 Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+3 x-y=0 \\
3 x^{2}+3 y^{2}+2 x+y=0
\end{array}\right.
$$
Find the two intersection points, then use the two-point form to find the equation of the line.
Solution 2 The equation $\left(3 x^{2}+3 y^{2}+2 x+y\right)$ $+\lambda\left(x^{2}+y^{2... | 7 x-4 y=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,414 |
Example 2. Find the equation of the circle that passes through the intersection points of the circles $\mathrm{x}^{2}+\mathrm{y}^{2}+6 \mathrm{y}-4=0$ and $x^{2}+y^{2}+6 y-28=0$, and whose center lies on the line $\mathrm{x}-\mathrm{y}-4=0$. | Solve the equation $\left(x^{2}+y^{2}+6 x-4\right)+\lambda\left(x^{2}+y^{2}+6 y-28\right)=0$ which represents the equation of a circle passing through the intersection points of two circles.
Rearranging gives $(1+\lambda) x^{2}+(1+\lambda) y^{2}+6 x$ $+6 \lambda y-4-28 \lambda=0$.
The center of the circle is $\left(-\... | x^{2}+y^{2}-x+7 y-32=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,415 |
Example 3. Prove that the intersection points of the two ellipses $b^{2} x^{2}+a^{2} y^{2}-a^{2} b^{2}$ $=0, a^{2} x^{2}+b^{2} y^{2}-a^{2} b^{2}=0$ lie on a circle centered at the origin, and find the equation of this circle. | Solve the equation $\left(b^{2} x^{2}+a^{2} y^{2}-a^{2} b^{2}\right)$ $+\lambda\left(a^{2} x^{2}+b^{2} y^{2}-a^{2} b^{2}\right)=0$ which represents a curve passing through the intersection points of two circles.
When $\lambda=1$, the equation becomes $\left(b^{2}+a^{2}\right) x^{2}$ $+\left(a^{2}+b^{2}\right) y^{2}-2 ... | x^{2}+y^{2}=\frac{2 a^{2} b^{2}}{a^{2}+b^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 702,416 |
Example 4. Find the length of the common chord of the circles $x^{2}+y^{2}-10 x-10 y=0$ and $x^{2}+y^{2}+6 x+2 y-40=0$. | Subtracting the two equations yields $4 x+3 y-10=0$. (Equation of the common chord)
Completing the square for $x^{2}+y^{2}-10 x-10 y=0$, we get $(x-5)^{2}+(y-5)^{2}=50$.
Thus, the center of the circle is $\mathbf{C}(5,5)$.
The distance from the center of the circle to the common chord is
$$
\mathrm{d}=\frac{|20+15-10|... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,417 |
4. There are 51 cities distributed in a square area with a side length of 1000 kilometers. It is planned to lay down 11000 kilometers of road network in the area. Can all the cities be connected? | (Continued from page 49) Therefore, the total length of all these roads does not exceed $1000 \times 6 + 100 \times 50 = 11000$ kilometers. This road network can connect all 51 cities in the region.
5. In the tetrahedron $\mathrm{ABCD}$, take points $\mathrm{A}^{\prime}$, $\mathrm{B}^{\prime}$, $\mathrm{C}^{\prime}$, a... | 11000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,418 |
1. Divide each side of the cube into four equal squares, and paint each small square with one of three different colors, ensuring that any two adjacent small squares have different colors. Prove: Each color is used to paint eight small squares, and please provide an example of such a coloring. | Consider the three small squares at the vertices of a cube; they should be painted with three different colors. The cube's side is divided into twenty-four small squares, which belong to the "three small square groups" at the eight vertices of the cube, so each color is used to paint eight small squares.
We use $1$, $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,419 |
Example 2. For a regular quadrilateral frustum $\mathrm{ABCD}$, the side length of the lower square base is 2, and all other edges are 1. $\mathrm{M}$ is the intersection point of the diagonals of the side face $\mathrm{BB}^{\prime} \mathrm{C}^{\prime} \mathrm{D}$ (Figure 6). Find the lengths of $\mathrm{A}^{\prime} \m... | Solve: Unfold the sides $\mathrm{AA}^{\prime} \mathrm{B}^{\prime} \mathrm{B}$, $\mathrm{BB}^{\{\prime} \mathrm{C}^{\prime} \mathrm{C}$, and $\mathrm{CC}^{\prime} \mathrm{D}^{\prime} \mathrm{D}$ onto a plane (Figure 7). $\mathrm{O}$ is the intersection point of the lateral edges.
According to the plane geometry knowled... | \frac{\sqrt{15}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,421 |
3. Circle $\mathrm{O}_{1}$ and circle $\mathrm{O}_{2}$ have equal radii, and they are tangent to each other at point $\mathrm{A}$. Circle $\mathrm{O}_{3}$ has a radius that is twice the radius of the first two circles, and it is internally tangent to circle $\mathrm{O}_{1}$ at point $\mathrm{B}$, and intersects circle ... | Prove: Since circle $\mathrm{O}_{1}$ is internally tangent to circle $\mathrm{O}_{3}$, and the radius of circle $\mathrm{O}_{3}$ is twice that of circle $\mathrm{O}_{1}$, $\mathrm{BO}_{3}$ is the diameter of circle $\mathrm{O}_{1}$ (Figure 3). Therefore, $\angle \mathrm{BAO}_{3}=90^{\circ}$. Let line $\mathrm{AB}$ inte... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,422 |
1. Find all natural numbers $n$ such that the sum of the digits of the decimal number $2^{\mathbf{n}}$ is 5. | It is not difficult to verify that the minimum value of $\mathrm{n}$ that satisfies the problem's requirements is 5. Now, we prove: when $n \geqslant 6$, none of them satisfy the problem's requirements.
For any $n \in N, 2^{n}$, the last digit is 2, 4, 6, or 8. According to the problem's requirement that the sum of th... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,424 |
3. What is the maximum value of the difference between a three-digit number and the sum of the cubes of its digits? What kind of three-digit number can achieve this maximum value? What is the smallest positive value of this difference? | Consider any three-digit number $\overline{\mathrm{bbc}}$, and let
$$
f(\overline{a b c})=\overline{a b c}-\left(a^{3}+b^{3}+c^{3}\right)
$$
Then $f(\overline{a b c})=100 a+10 b+c-\left(a^{3}+b^{3}\right.$
$$
\begin{aligned}
+ & \left.c^{3}\right) \\
= & \left(100 a-a^{3}\right)+\left(10 b-b^{3}\right) \\
& +\left(c-c... | 396 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,426 |
4. Draw two tangents from point C outside circle O1 to the circle, touching at points A and B, respectively. Draw circle O2 through points C and B, tangent to line AB at point B. Circle O2 intersects circle O1 at point M. Prove: Line AM bisects segment BC. | Prove: As shown in Figure 4, extend $\mathrm{AM}$ to intersect circle $\mathrm{O}_{2}$ at point $\mathrm{D}$, and connect $\mathrm{BM}$, $\mathrm{BD}$, and $\mathrm{CD}$. Then
\[
\begin{aligned}
& \angle \mathrm{CAD}=\angle \mathrm{CAM} \\
= & \angle \mathrm{ABM}=\angle \mathrm{ADB} \\
= & \angle \mathrm{DDCM}=\angle \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,427 |
5. Does there exist three points $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ on the same plane such that for any point $\mathrm{P}$ on this plane, at least one of the lengths of the line segments $\mathrm{PA}$, $\mathrm{PB}$, and $\mathrm{PC}$ is irrational? | Solution: Such three points do exist.
Let the distance between two points $\mathrm{A}$ and $\mathrm{B}$ on the same plane be ${ }^{4} \sqrt{2}$, and let $C$ be the midpoint of $AB$. 5.2 Let $\mathrm{P}$ be a point outside the line $\mathrm{AB}$, and extend $\triangle \mathrm{APB}$ to form a parallelogram APE (Figure 5)... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,428 |
2. Given the function $\mathrm{f}(\mathrm{x}), \mathrm{x} \in(-\infty,+\infty)$, it is known that the function $\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\sin \mathrm{f}(\mathrm{x})$ is a periodic function. Prove that the function $\mathrm{f}(\mathrm{x})$ is also a periodic function, | Prove: Let $g(x)=F(f(x))$, where $F(t) = t + \sin t$. If $T$ is a period of the function $g(x)$, then for any $x \in (-\infty, +\infty)$, we have
$$
g(x+T) = g(x).
$$
or $F(f(x+T)) = F(f(x)). \quad (2)$
It is not difficult to verify that the function $F(t)$, $t \in (-\infty, +\infty)$, is strictly increasing.
Suppose... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,429 |
3. The clock face (a circular plate with numbers $1, 2, 3, \cdots, 11, 12$) is fixed on the classroom blackboard with its center as the axis. The clock face can rotate by any integer multiple of $30^{\circ}$. Initially, the word "no" is written next to the numbers on the clock face on the blackboard. Then, the clock fa... | Answer: (i) It is impossible. In fact, each time the dial is turned, the number on the blackboard increases by $(1+2+3+\cdots+11+12)=78$. After the dial is turned $\mathrm{k}$ times, the total number on the blackboard is $78 \mathrm{k}$. If at this point the numbers on the blackboard are all 1984, then we should have $... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,430 |
For example, use points on a plane to represent the numerical solutions of the equation
$$
3 x+4 y=1 \text{.}
$$ | Solving: By observation, an integer solution to the indeterminate equation $3x + 4y = 1$ is $\left\{\begin{array}{l}x_{0}=-1 \\ y_{0}=1\end{array}\right.$, thus all integer solutions are $\left\{\begin{array}{l}x=-1+4t, \\ y=1-3t .\end{array} \quad(t \in \mathbb{Z})\right.$
The graph is as follows:
\begin{tabular}{r|r... | d = \sqrt{a^2 + b^2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,431 |
Example 1. Find the center of the hyperbola $x^{2}+2 x y-3 y^{2}+2 x-4 y$ $=0$, the diameter passing through $(1,1)$, and the conjugate diameter, and also find its asymptotes. | Let $S \equiv x_{1}^{2}+2 x_{1} x_{2}-3 x_{2}^{2}+2 x_{1} x_{3}$ $-4 x_{2} x_{3}=0$.
(1) The coordinates of the center satisfy the equations
$$
\left.\begin{array}{l}
\frac{\partial S}{\partial x_{1}}=0 \\
\frac{\partial S}{\partial x_{2}}=0
\end{array}\right\}
$$
Thus, the center is $\left(-\frac{1}{4},-\frac{3}{4}\r... | 7 x-5 y-2=0, 12 x-16 y-9=0, 2 x+6 y+5=0, 2 x-2 y-1=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,433 |
Example 2. If a chord $A B$ of a hyperbola intersects the asymptotes at points $P, Q$, then $|A P|=|Q B|$. | Proof: If using analytic methods is quite cumbersome, but using advanced geometric methods makes the proof much more concise.
Let chord $AB$ intersect the asymptotes at $P, Q$ (as shown in Figure 1).
Draw diameter $ON$ through the center $O$ such that $ON \parallel AB$. Let $OM$ be the conjugate diameter of $ON$, then ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,434 |
13. Through the vertices $A_{1}, B, D$ and the edges $AD$ and $B_{1}C_{1}$ of the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, construct two sections $\square_{A_{1}BD}$ and $\square_{A_{1}C_{1}D_{1}}$.
Prove: $\square_{A_{1}BD} \perp \square_{A_{1}C_{1}D_{1}}$. | Prove that, as shown in Figure 12, $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube,
$$
\begin{array}{l}
\boldsymbol{C}_{1} \boldsymbol{C} \perp \square_{\mathrm{A}} \\
\left.\Longrightarrow \begin{array}{r}
C_{1} C \perp A C \\
B D \perp A C
\end{array}\right\} \\
\Longrightarrow C_{1} A \perp B D \\
\end{array}
$$
Consid... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,436 |
Example 1. Prove
(1) $\operatorname{ch}^{2} x - \operatorname{sh}^{2} x = 1$,
(2) $1 - \operatorname{th}^{2} x = \operatorname{sech}^{2} x$,
(3) $\operatorname{th}(x-y) = \frac{\operatorname{th} x - \operatorname{th} y}{1 - \operatorname{th} x \operatorname{th} y}$. | Proof
$$
\begin{array}{l}
\text { (1) } \operatorname{ch}^{2} x-\operatorname{sh}^{2} x=\cos ^{2} i x \\
-\left(\frac{\sin i x}{i}\right)^{2}=\cos ^{2} i x+\sin ^{2} i x=1 .
\end{array}
$$
(2) $1-t h^{2} x=1-\left(\frac{t g^{2} i x}{i}\right)^{2}$
$$
=1+\operatorname{tg}^{2} i x=\sec ^{2} i x=\operatorname{sech}^{2} x ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,438 |
Example 2. Calculate
(1) $(\operatorname{cis} x)^{\prime}$,
(2) $\left(\operatorname{sh}^{-1} x\right)^{\prime}$,
(3) $\int \frac{x}{\sqrt{x^{2}-1}} \, dx$;
(4) $\int \frac{d x}{1-x^{2}}$. | (1) $(\operatorname{ch} x)^{\prime}=(\operatorname{cosi} x)^{\prime}$
$$
=-i \sin i x=\frac{1}{i} \sin i x=\operatorname{sh} x \text {. }
$$
(2) $\left(\operatorname{sh}^{-1} x\right)^{\prime}=\left(\frac{1}{i} \sin ^{-1} i x\right)^{\prime}$
$$
=\frac{1}{i} \frac{i}{\sqrt{1-(i x)^{2}}}=\frac{1}{\sqrt{1+x^{2}}} .
$$
(3... | (1) \operatorname{sh} x, (2) \frac{1}{\sqrt{1+x^{2}}}, (3) \operatorname{ch}^{-1} x + C, (4) \operatorname{th}^{-1} x + C (|x|<1) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,439 |
Example 4. Calculate $\int \sqrt{x^{2}+a^{2}} d x .(a>0)$ | Let $x=a \operatorname{sh} u$, then $\sqrt{x^{2}+a^{2}}=a c h u$, $d x=a c h u d u$,
$$
\begin{array}{l}
=\frac{a^{2}}{2} \ln\left(x+\sqrt{x^{2}+a^{2}}\right)+\frac{x}{2} \sqrt{x^{2}+a^{2}} \\
+C_{1} \text {. } \\
\end{array}
$$
where $c h^{2} u=\frac{1+c h 2 u}{2}$ can be immediately obtained from $c h u$ ～cosu in eq... | \frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right) +C_{1} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,441 |
815. Calculate (1) $\int \frac{d x}{\sqrt{1+e^{x}}}$
(2) $\int \frac{x^{2}+1}{x \sqrt{x^{4}+1}} d x$. | $$
\begin{array}{l}
\text { (1) } \int \frac{d x}{\sqrt{1+e^{x}}}=\int \frac{d x}{e^{\frac{x}{3}} \sqrt{e^{-x}+1}} \\
=-2 \int \frac{e^{\cdots x}}{\left(e^{-\frac{x}{2}}\right)^{2}+1} \\
=-2 \operatorname{sh}^{-1}\left(e^{-\frac{x}{2}}\right)+C \\
=x-2 \ln \left(1+\sqrt{1+e^{x}}\right)+C . \\
\end{array}
$$
$$
\begin{a... | x-2 \ln \left(1+\sqrt{1+e^{x}}\right)+C | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,442 |
Example 1. Among rectangular solids with a given surface area, the cube has the maximum volume. | Let $a$, $b$, $c$ represent the length, width, and height of a rectangular prism, respectively, with surface area $S$ and volume $V$. Then,
$$
\begin{array}{l}
V=a b c, S=2(a b+b c+c a). \\
\text { From }(a b \cdot b c \cdot c a)^{\frac{1}{3}} \leqslant \frac{a b+b c+c a}{3} \text {, we get }
\end{array}
$$
$V \leqslan... | a=b=c | Geometry | proof | Yes | Yes | cn_contest | false | 702,444 |
Example 1. If $a+b+c=0, a b c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. | \[\begin{array}{l}\text { Solve the original expression }=\frac{a^{2}+b^{2}+c^{2}}{3 a b c} \\ +\frac{2(a b+b c+c a)}{3 a b c}=\frac{(a+b+c)^{2}}{3 a b c}=0 .\end{array}\] | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,447 |
Example 2. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=6 \\
x^{2}+y^{2}+z^{2}=14 \\
x^{3}+y^{3}+z^{3}=36
\end{array}\right.
$$ | $$
\begin{array}{l}
\text { Sol } \quad \text { we have } x y+y z+z x=\frac{1}{2}\left[(x+y+z)^{2}\right. \\
\left.-\left(x^{2}+y^{2}+z^{2}\right)\right]=\frac{1}{2}(36-14)=11, \\
x y z=\frac{1}{3}\left\{x^{3}+y^{3}+z^{3}-(x+y+z)\right. \\
\text { - } \left.\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]\righ... | (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,448 |
Example 1. Write the general term formula for the following sequences:
(1) $1,2,2,3,3,3,4,4,4,4, \cdots, a_{n}, \cdots$;
(2) $1,1,2,2,2,3,3,3,3, \cdots, b_{n}, \cdots$,
(3) $1,1,1,2,2,2,2, \cdots, c_{n} \cdots$. | It is clear that their general term formulas are respectively
$$
\begin{array}{c}
a_{n}=\left[\frac{1+\sqrt{8 n-7}}{2}\right], \\
b_{n}=\left[\frac{-1+\sqrt{8 n+1}}{2}\right], C_{n}=[\sqrt{n}] .
\end{array}
$$ | a_{n}=\left[\frac{1+\sqrt{8 n-7}}{2}\right], b_{n}=\left[\frac{-1+\sqrt{8 n+1}}{2}\right], c_{n}=[\sqrt{n}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,449 |
Example 2. Find the general term formula of the sequences
(1) $1,3,3,5,5,5,7,7,7,7, \cdots, a_{n}, \cdot \cdot$
( 2 ) $2,2,5,5,5,8,8,8,8, \cdots, b_{n} \cdots$ | $$
\begin{array}{l}
a_{n}=2\left[\frac{1+\sqrt{8 n-7}}{2}\right]-1, \\
b_{n}=3\left[\frac{-1+\sqrt{8 n+1}}{2}\right]-1 .
\end{array}
$$ | a_{n}=2\left[\frac{1+\sqrt{8 n-7}}{2}\right]-1, \quad b_{n}=3\left[\frac{-1+\sqrt{8 n+1}}{2}\right]-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,450 |
Example 1. If the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=a_{n}+3n+2, a_{1}=2$, find its general term formula. | $$
\begin{array}{c}
\text { Sol } \because a_{n}-a_{1}=\sum_{\mathrm{k}=1}^{\mathrm{n}-1}\left(a_{k+1}-a_{k}\right) \\
=\sum_{\mathrm{k}=1}^{\mathrm{n}-1}(3 k+2)=\frac{3 n^{2}+n-4}{2}, \\
\therefore a_{n}=\frac{1}{2}\left(3 n^{2}+n\right) .
\end{array}
$$
If $\sum_{\mathrm{k}=1}^{\mathrm{n}-1} f(k)$ can be found, then... | a_{n}=\frac{1}{2}\left(3 n^{2}+n\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,451 |
Example 2. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=2^{n} a_{n}$, $a_{1}=1$, find the general term formula. | \begin{array}{l}\text { Sol } a_{n}=a_{1} \prod_{\mathrm{k}=1}^{\mathrm{n}-1} \frac{a_{k+1}}{a_{k}}=\prod_{\mathrm{k}=1}^{\mathrm{n}-1} 2^{k} \\ =2 \frac{n(n-1)}{2} .\end{array} | 2^{\frac{n(n-1)}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,452 |
Example 5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}-2 a_{n}$ $-3 a_{n-1}=0, a_{1}=1, a_{2}=5$, find its general term formula. | Solve $a_{n+1}-a_{n}=b_{n}$, then we have $b_{n}=-3 b_{n-1}$, which yields $b_{n}=4(-3)^{n-1}$. Thus,
$$
\begin{aligned}
a_{n}= & 1+4 \cdot \frac{1-(-3)^{n-1}}{1+3} \\
= & 2-(-3)^{n-1}
\end{aligned}
$$ | 2-(-3)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,454 |
Example 1. For the arithmetic sequence $\{3 n-1\}$, if three arithmetic means are inserted between each pair of consecutive terms, forming a new sequence. What is the position of the 12th term of the original sequence in the new sequence? What is the position of the 29th term of the new sequence in the original sequenc... | Solve: The general term of the original sequence is $a_{n}=3(n-1)+2$, then the general term of the new sequence is $b_{m}=\frac{3}{4}(m-1)+2$.
We find that $m=4 n-3$, so $a_{12}=b_{45}$, $b_{21}=a_{8}$. | a_{12}=b_{45}, b_{21}=a_{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,455 |
Example 1. If $|q|<1$. Prove:
$$
\lim _{n \rightarrow \infty} q^{n}=0 \text {. }
$$ | If $q=0$, then naturally $\lim _{n \rightarrow \infty} q^{n} =0 ;$
If $q \neq 0$, then $0<|q|<1$.
Let $\frac{1}{|q|}=1+p$,
Then by Bernoulli's inequality we get
$$
\frac{1}{|q|^{n}}=(1+p)^{n} \geqslant 1+n p, \quad(n \geqslant 1)
$$
Thus $0<|q|^{n} \leqslant \frac{1}{1+n p}$.
Since $\lim _{n \rightarrow \infty} 0=0, ... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 702,456 |
Example 3. $\left\{a_{n}\right\}$ is an arithmetic sequence. If the sum of two sets of indices is equal: $m_{1}+m_{2}+\cdots+m_{t}=n_{1}+n_{2}+\cdots$ $+n_{\ell}$, then $a_{m_{1}}+a_{m_{2}}+\cdots+a_{m_{t}}=a_{n_{1}}+a_{n_{2}}$ $+\cdots+a_{n}$. Conversely, the same holds. | To prove that the necessary and sufficient condition for the sum of the ordinates of two sets of points on a line to be equal is that the sum of their abscissas is equal, hence the proposition is true.
Particularly, when $m_{1}+\cdots+m_{t}$ is a multiple of $t$,
$$
\frac{a_{m_{1}}+\cdots+a_{m t}}{t}=\boldsymbol{a} \fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,458 |
Calculate $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ | Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Let the original expression $=x$, then $x^{3}=40+6 x$, its positive root $x=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,459 |
Example 2. Prove that $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{119}{120}<\frac{1}{11}$. | Prove that if the left side $=x$, then
$$
\begin{array}{l}
x<\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdots \cdot \frac{120}{121}=\frac{1}{121 x}, \\
x^{2}< \frac{1}{121} \text { i.e., } x<\frac{1}{11} .
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,460 |
Example 3. Prove that
$$
\frac{(1+\sqrt{1985})^{2000}-(1-\sqrt{1985})^{2000}}{\sqrt{1985}}
$$
is an integer. | Let $\sqrt{1985}=x$, and let the numerator of the original expression be $f(x)$, then
$$
f(x)=(1+x)^{2000}-(1-x)^{2000} .
$$
We have $f(-x)=-f(x)$, so $f(x)$ contains only odd powers of $x$. Therefore, $\frac{f(x)}{x}$ is a polynomial containing only even powers of $x$,
hence it is an integer. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,461 |
Example 4. Given $\sum_{j=1}^{n} a_{j} \cos \alpha_{j}=\sum_{j=1}^{n} a_{5} \cdot \cos \left(\alpha_{j}+1\right)=0$, find the value of $\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+\frac{\pi}{10}\right)$. | Consider the function $f(x)=\sum_{j=1}^{n} a_{j} \cos \left(\alpha_{j}+x\right)$. It is easy to see that $f(0)=f(1)=0$. Since $\cos \left(a_{j}+x\right) = \cos \alpha_{j} \cos x - \sin \alpha_{j} \sin x$, we have
$$
\begin{array}{l}
f(x)=\cos x \sum_{\mathrm{j}=1}^{n} a_{j} \cos \alpha_{j} + \sin x \sum_{\mathrm{j}=1}^... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,462 |
Example 5. Solve the equation $\sqrt{x^{2}+6 x+36}$
$$
=\sqrt{x^{2}-14 x+76}+8 \text{. }
$$ | Substitute $y^{2}$ for 27, the original equation becomes
$$
\sqrt{(x+3)^{2}+y^{2}}-\sqrt{(x-7)^{2}+y^{2}}=8 \text {. }
$$
This indicates that point $p(x, y)$ lies on the right branch of a hyperbola with foci at $F_{1}(-3,0)$ and $F_{2}(7,0)$:
$$
\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1
$$
Let $y^{2}=27$, we get $x=10$,... | x=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,463 |
Five. (This question is worth 15 points)
There is a rectangular prism with length, width, and height being positive integers $m, n, r$ $(m \leqslant n \leqslant r)$, respectively. After being painted red, it is cut into unit cubes with edge length 1. It is known that the number of unit cubes without any red faces plus ... | From $m \leqslant n \leqslant r$ we know $\quad m-4 \leqslant n-4 \leqslant r-4$.
$$
\therefore m-4=1, n-4=3, r-4=659 \text {. }
$$
Thus, $m=5, n=7, r=663$.
Similarly, from $1977=1 \times 1 \times 1977$ we know
$$
m-4=1, n-4=1, r-4=1977 \text {, }
$$
Thus, $m=5, n=5, r=1981$.
$\therefore$ The sets of $m, n, r$ that s... | m_{1}=5, n_{1}=7, r_{1}=663 ; m_{2}=5, n_{2}=5, r_{2}=1981 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,464 |
3. In parallelogram $A B C D$, $P$ is the midpoint of $B C$. A line parallel to $B D$ through $P$ intersects $C D$ at $Q$. Connect $P A$, $P D$, $Q A$, and $Q B$. Among the triangles in the figure, how many triangles are congruent to $\triangle A B P$ besides $\triangle A B P$ itself? $(A)$ three; $(B)$ four; $(C)$ fiv... | Ans: $[C]$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,468 |
5. $[x]$ represents taking the integer part of the number $x$, for example $\left[\frac{15}{4}\right]=3$,
$$
y=4\left(\frac{x+[u]}{4}-\left[\frac{x+[u]}{4}\right]\right),
$$
and when
$$
\begin{array}{lll}
x=1,8,11,14 \text { then } & y=1 ; \\
x=2,5,12,15 \text { then } & y=2 ; \\
x=3,6,9,16 \text { then } & y=3 ; \\
x... | Answer: (D). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,470 |
6. As shown in the figure, in isosceles $\triangle ABC$, $CD$ is the altitude on the base $AB$, $E$ is the midpoint of the leg $BC$, and $AE$ intersects $CD$ at $F$. Now, three routes are given:
(a) A $\rightarrow F \rightarrow C \rightarrow E \rightarrow B \rightarrow D \rightarrow A$;
(b) $A \rightarrow C \rightarrow... | Ans: $[B]$.
This is the direct translation of the provided text, maintaining the original format and line breaks. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,471 |
1. Let $a-b=2+\sqrt{3}, b-c=2-\sqrt{3}$, then the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is | 15.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,472 |
4. The solution to the inequality $42 x^{2}+a x<a^{2}$ is | $$
-\frac{a}{6}0) ; \frac{a}{7}<x<-\frac{a}{6}
$$
(when $a<0$); no solution (when $a=0$). | \frac{a}{7}<x<-\frac{a}{6} \text{ (when } a<0\text{); no solution (when } a=0\text{)} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,475 |
As given, $O$ is a point inside the pentagon $ABCDE$, and
$$
\begin{array}{l}
\angle 1=\angle 2, \angle 3 \\
= \angle 4, \angle 5=\angle 6, \\
\angle 7=\angle 8 .
\end{array}
$$
Prove: $\angle 9$ and $\angle 10$ are equal or supplementary. | Proof 1: By the Law of Sines and the given conditions, we have:
$$
\begin{array}{l}
\frac{O A}{\sin \angle 10}=\frac{O B}{\sin \angle 1}=\frac{O B}{\sin \angle 2}=\frac{O C}{\sin \angle 3} \\
=\frac{O C}{\sin \angle 4}=\frac{O D}{\sin \angle 5}=\frac{O D}{\sin \angle 6}=\frac{O E}{\sin \angle 7} \\
=\frac{O E}{\sin \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,479 |
Four. (This question is worth 15 points)
As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $A$, with radii $r_{1}$ and $r_{2}$ respectively; $PB$ and $PC$ are tangents to the two circles, with $B$ and $C$ being the points of tangency; $PB: PC = r_{1}: r_{2}$; $PA$ intersects $\odot O_{2}... | Proof 1: Connect $O_{1} A, O_{1} B, P O_{1}, P O_{2}$, $\mathrm{O}_{2} \mathrm{~A}, \mathrm{O}_{2} \mathrm{C}$, then $\mathrm{O}_{1} 、 \mathrm{~A}, \mathrm{O}_{2}$ are collinear.
$\because P B: P C=r_{1}: r_{2}$,
$\therefore$ Rt $\triangle P B O_{1}$ c:Rt $\triangle P C O_{2}$.
(5 points)
$\therefore \angle 3=\angle 4,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,480 |
1. If $2 x+1=8$, then $4 x+1=$
(A) 15; (B) 16; (C) 17; (D) 18; (E) 19. | $\begin{array}{l}\text { 1. From } 4 x+1=2(2 x+1)-1=2 \cdot 8-1 \\ =15 \text { we know } \\ \text { (A) is true. }\end{array}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,481 |
2. In a game, a sector with a radius of $1 \mathrm{~cm}$ represents a “monster,” as shown by the shaded area in the figure. The central angle of the缺口 part (i.e., the monster's mouth) is $60^{\circ}$. What is the perimeter of this “monster” (i.e., the sector)?
(A) $\pi+2$;
(B) $2 \pi$;
(C) $\frac{5}{3} \pi_{3}$
(D) $\f... | 2. The central angle of the shaded sector is $300^{\circ}$, so the required perimeter is $\frac{5}{6}(2 \pi r)+2 r=\frac{5}{3} \pi+2$.
(E) True. | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,482 |
3. Let the two legs of the right triangle $\triangle A B C$ have lengths 5 and 12, respectively. We draw two arcs, one centered at $A$ with a radius of 12, and the other centered at $B$ with a radius of 5. These arcs intersect the hypotenuse at $M$ and $N$, respectively. The length of the line segment $M N$ is
(A) 2, (... | 3. From $\triangle A B C$ being a right triangle, we get $A B=13$. Additionally, $B N=B C=5, A M=A C=12$, so $B M=A B-A M=1, M N=B N-B M=4$.
(D) True. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,483 |
4. A large bag contains coins of 1 cent, 10 cents, and 25 cents. The number of 10 cent coins is twice the number of 1 cent coins, and the number of 25 cent coins is three times the number of 10 cent coins. The total amount of money (in cents) in the bag could be
(A) 306 ;
(B) 333 ;
(C) 342 ;
(D) 348 ; (E) 360 . | 4. Suppose there are $p$ pennies, $d$ dimes, and $q$ quarters in the bag. Given that $d=2 p, q=3 d$, thus $q=6 p$. Let $A$ be the value of the coins in the bag (in cents), then $A=p+10(2 p)+25(6 p)=171 p$. Among the given options, only $(C)$ is a multiple of 171.
(C) True. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,484 |
5. From the series
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}
$$
which terms should be removed so that the sum of the remaining terms equals 1?
(A) $\frac{1}{4}$ and $\frac{1}{8}$,
(B) $\frac{1}{4}$ and $\frac{1}{12}$,
(C) $\frac{1}{8}$ and $\frac{1}{12}$;
(D) $\frac{1}{6}$ and $\frac... | 5. Since $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}=\frac{60}{120}$ $+\frac{30}{120}+\frac{20}{120}+\frac{15}{120}+\frac{12}{120}+\frac{10}{120}$, is - 5 . One possibility is to remove $\frac{15}{120}$ and $\frac{12}{120}$, i.e., $\frac{1}{8}$ and $\frac{1}{10}$. Moreover, from the above... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,485 |
7. In a certain computer language (such as $A P L$), when an algebraic expression contains no parentheses, the operations in the expression are specified to be performed from right to left. For example, in this language, $a \times b-c$ is equivalent to $a(b-c)$ in standard algebraic notation. Now, an expression is writ... | 7. When performing the operation from right to left, first add $c$ and $d$, getting $(c+d)$, then subtract this result from $b$, obtaining $b-c-d$. Finally, divide this expression by $a$, resulting in $\frac{a}{b-c-d}$.
(E) True. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,487 |
8. Let $a$, $a^{\prime}$, $b$, $b^{\prime}$ be real numbers, and $a$ and $a^{\prime}$ be non-zero real numbers. Then the solution to the equation $a x + b = 0$ is less than the solution to the equation $a^{\prime} x + b^{\prime} = 0$ if and only if
(A) $a^{\prime} b < a b^{\prime}$;
(B) $a b^{\prime} < a^{\prime} b$,
(... | 8. The solution to the equation $a x+b=0$ is $-\frac{b}{a}$, and the solution to the equation $a^{\prime} x+b^{\prime}=0$ is $-\frac{b^{\prime}}{a^{\prime}}$. Thus, the problem becomes: among the five given answers, which one is equivalent to the inequality $-\frac{b}{a}a^{\prime} b^{\prime}$, which is $(E)$. Conversel... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,488 |
Example 1. Find the smallest positive period of $f(x)=\sin (t g x)$, and prove your conclusion. | Proof: On the one hand, the smallest positive period of $\operatorname{tg} x$ is $\pi$, so $\pi$ is also a period of $\sin (\operatorname{tg} x)$; on the other hand, although the function $\operatorname{tg}_{x}$ is an increasing function within the corresponding interval of one period, the rate of increase is not unifo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,489 |
9 . Arrange the odd positive integers $1,3,5,7, \cdots$ in five columns, as shown in the table below. The leftmost column is called the 1st column, and the columns are numbered from left to right. Thus, the number “1985” appears in
(A) the 1st column;
(B) the 2nd column;
(C) the 3rd column;
(D) the 4th column;
(E) the ... | 9. Observing that the numbers in the 1st column can all be expressed as $16n-1$ $(n=1,2, \cdots)$, and the first number below each number $16n-1$ is $16n+1$. Numbers of the form $16n+1$ are all in the 2nd column. Since $1985=16 \times 124+1$, 1985 is in the 2nd column.
(B) True. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 702,490 |
10. An arbitrary circle and the curve $y=\sin x$ can intersect at
(A) at most 2 points;
(B) at most 4 points;
(C) at most 6 points;
(D) at most 8 points;
(E) more than 16 points. | $10 . y=\sin x$ is a periodic wave with values in $[-1,1]$. Imagine a circle with a sufficiently large radius that is tangent to the $x$-axis at the origin; it will intersect the sine wave many times. The larger the radius, the greater the number of intersections can be.
(E) True. | E | Calculus | MCQ | Yes | Yes | cn_contest | false | 702,491 |
14. If a convex polygon has exactly three interior angles that are obtuse, what is the maximum number of sides this polygon can have?
(A) 4;
(B) 5 ;
(C) 6 ;
(D) 7 ;
(E) 8 . | For $(n-2) \cdot 180^{\circ}$. If the polygon has exactly 3 obtuse angles, then the remaining $n-3$ angles are all no more than $90^{\circ}$. Therefore,
$$
(n-2) 180^{\circ}<(n-3) 90^{\circ}+3 \cdot 180^{\circ} \text {, }
$$
i.e., $2(n-2)<n-3+6, n<7$.
The following figure shows:
$n=6$ is possible
(Here, $\triangle A ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,495 |
15. Let $a$ and $b$ be positive numbers, and satisfy $a^{b}=b^{a}$ and $b=9a$, then the value of $a$ is
(A) 9;
(B) $\frac{1}{3}$;
(C) $\sqrt[9]{9}$;
(D) $\sqrt[3]{3}$;
(E) $\sqrt[4]{3}$. | 15. Multiply both sides of $a^{b}=b^{4}$ by $\frac{1}{a}$, and substitute $b=9 a$, we get $a^{8}=9 a$. Since $a \neq 0$, the equation simplifies to $a^{8}=9$. Therefore, $a=9^{\frac{1}{8}}=\left(3^{2}\right)^{\frac{1}{8}}=3^{\frac{1}{4}}$.
(E) True. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,496 |
18. Six bags contain 18, 19, 21, 23, 25, and 34 glass balls respectively, one of which contains only balls with cracks, while the other five bags contain no such cracked balls. Jenny took three of the bags, and George took two of the bags, leaving only the bag with the cracked balls. If Jenny got exactly twice as many ... | 18. Jenny has twice as many balls as George, so the total number of balls they take must be a multiple of 3. This indicates that the difference between 140 (18 $+19+21+23+25+34=140$ ) and the number of balls with cracks must be a multiple of 3. It is easy to verify that among the six given bags of balls, only $140-23$ ... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 702,499 |
Example 2. Find the smallest positive period of $f(x)=\sin ^{n} w x$, $g(x)=\tan ^{n} w x$ (where $n$ is a natural number, $w$ is a non-zero rational number), and prove your conclusion.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result dire... | Proof: When $n=1$, the smallest positive period of $\sin n x$ is $\frac{2 \pi}{|w|}$. When $n=2$, $\sin ^{2} w x=\sin ^{2} w(x+\frac{\pi}{|w|})$, its smallest positive period is $\frac{\pi}{|w|}$. ... 3 Conjecture:
When $n$ is odd, the smallest positive period of $\sin ^{n} w x$ is $\frac{2 \pi}{|w|}$.
When $n$ is ev... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,500 |
19. Consider the curves $y=A x^{2}$ and $y^{2}+3=x^{2}+4 y$, where $A$ is a positive constant, and $x$ and $y$ are real variables. How many intersection points do these two curves have?
(A) Exactly 4;
(B) Exactly 2;
(C) At least 1, and the number of intersection points varies with different values of $A$;
(D) For at le... | 19. The second equation can be rewritten as $(y-2)^{2}-x^{2}$ $=1$, which represents a hyperbola centered at $(0,2)$, with vertices at $(0,1)$ and $(0,3)$, and asymptotes given by $y=2 \pm x$. When $A>0$, the graph of $y=A x^{2}$ is a parabola opening upwards with its vertex at the origin. Therefore, the parabola inter... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,501 |
20. A cube has a side length of $n$ units (where $n$ is an integer greater than 2). The entire surface of the cube is painted black, and then the cube is cut into $n^3$ smaller cubes, each with a side length of 1 unit, using planes parallel to the faces of the original cube. If the number of smaller cubes that have exa... | 20. The small cubes with no faces painted can only come from the interior of the cube, totaling $(n-2)^{3}$; the small cubes with one face painted come from the sides of the original cube, but not on the boundaries of these sides, totaling $6(n-2)^{2}$; thus $(n-2)^{3}=6(n-2)^{2}$. Since $n>2$, we can cancel $(n-2)^{2}... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 702,502 |
21. How many integers $x$ satisfy the equation
$$
\begin{array}{l}
\left(x^{2}-x-1\right)^{x+2} \\
=1 ?
\end{array}
$$
(A) 2; (B) 3,
(C) 4; (D) 5;
(E) None of the above conclusions is correct. | 21. When $a$ and $b$ are integers, there are three cases that can make $a^{b}=1$: I) $a=1$; II) $a=-1, b$ is even; III) $b=0, a \neq 0$. In this problem, $a=x^{2}-x-1$, $b=x+2$. We will examine these three cases in sequence:
I) $x^{2}-x-1=1, (x-2)(x+1)=0$, $x=2$ or $x=-1$.
II) $x^{2}-x-1=-1$, and $x+2$ is even, $2^{2}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,503 |
24. Suppose non-zero digits are selected in the following way: the probability of digit $d$ being selected is $\log g_{10}(d+1)-\log _{10} d$. What is the probability that the digit 2 is selected from the following sets of numbers that can be selected with a probability of $-\frac{1}{2}$?
$$
\begin{array}{l}
\text { (A... | 24. Let $\operatorname{Pr}\left\{d_{1}, d_{2}, \cdots, d_{n}\right\}$ denote the probability that one of the numbers $d_{1}, d_{2}, \cdots, d_{n}$ is selected.
Note that $P_{r}\{d\}=\log \frac{d+1}{d}$ and $P_{r}\{d, d+1\}$
$$
=\log \frac{d+1}{d}+\log \frac{d+2}{d+1}=\log \frac{d+2}{d} .
$$
We look for a set whose pr... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 702,506 |
25. The volume of a certain rectangular prism is $8 \mathrm{~cm}^{3}$, the total surface area is $32 \mathrm{~cm}^{2}$, and its length, width, and height form a geometric progression. Then the sum of all the edge lengths of this prism $(\mathrm{cm})$ is
(A) 28;
(B) 32 ;
(C)
(D) 40 ;
(E) 44 . | 25. Let the three edge lengths of the rectangular prism be $a, a r$ and $a r^{2}$, then volume $=a(a r)\left(a r^{2}\right)=8$, i.e., $a r=2$.
Surface area $=2 a^{2} r+2 a^{2} r^{2}+2 a^{2} r^{3}=32$ $=2 a r\left(a+a r+a r^{2}\right)$
$=4\left(a+a r+a r^{2}\right)=$ sum of all edge lengths of the rectangular prism.
(B)... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,507 |
27. Consider a sequence $x_{1}, x_{2}, x_{3} \cdots$, where $x_{1}=\sqrt[3]{3}, x_{2}=(\sqrt[3]{3}) \sqrt[3]{3},$ and in general, $x_{n}=\left(x_{n-1}\right) \sqrt[3]{{ }^{3}}$. For $n>1$, what is the smallest $n$ such that $x_{n}$ is an integer?
(A) 2 ;
(B)
3; (C) 4;
D) 9 ; (E) 27. | 27. We $\quad \therefore=3^{\frac{1}{3}}$
$$
\begin{array}{l}
x_{4}=(3 \sqrt[3]{9 / 3}) \sqrt[3]{3}=3^{\sqrt[3]{27 / 3}}=3^{3 / 3} \\
=3 \text {. } \\
\end{array}
$$
We also need to verify that $x_{1}, x_{2}, x_{3}$ are not integers. It is easy to see that $x_{1}\sqrt{2}$, and we need to prove that $x_{3}>2$. Therefor... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 702,509 |
28. In $\triangle A B C$, $\angle C=3 \angle A, a=27$, and $c=48$. What is $b=$?
(A) 33;
(B) 35;
(C) 37,
(D) 39;
(E) the value of $\bar{\imath}$ is not unique. | 28. By the Law of Sines, we have $\frac{27}{\sin A}=\frac{48}{\sin 3 A}$. Using the identity $\sin 3 A=3 \sin A-4 \sin ^{3} A$, we get
$$
\frac{48}{27}=\frac{16}{9}=\frac{\sin 3 A}{\sin A}=3-4 \sin ^{2} A .
$$
$=5 / 6$ (since $0<3 A<180^{\circ}$, $\cos A$ does not take a negative value). Using the Law of Sines again, w... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 702,510 |
3. Find (1) the smallest positive period of $f(x)=\sin \frac{1}{2} x+\cos \frac{1}{3} x$;
(2) the smallest positive period of $g(x)=\sin 3 x \cdot \cos 4 x$. | (1) Let $l$ be the period of $f(x)$, then
$$
\begin{aligned}
& \sin \frac{1}{2}(x+l)+\cos \frac{1}{3}(x+l) \\
& =\sin \frac{1}{2} x+\cos \frac{1}{3} x, \\
\therefore \quad & \sin \frac{1}{2}(x+l)-\sin \frac{1}{2} x=\cos \frac{1}{3} x \\
& -\cos \frac{1}{3}(x+l) .
\end{aligned}
$$
Using the sum-to-product identities, w... | 2 \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,511 |
29, expressed in decimal; the integer $a$ consists of 1985 eights, and the integer $b$ consists of 1985 fives. What is the sum of the digits in the decimal representation of the integer $9ab$?
(A) 15880,
(B) 17856;
(C) 17865;
(D) 17874;
(E) 19851. | 29. Due to the last digit of $9 a b$ being zero, the sum of its digits differs from the sum of the digits of the number $N=\frac{9 a b}{10}$. Additionally, if the number $M$ is formed by $h$ digits $d$ (where $d$ is an Arabic numeral) arranged in a row, then
$$
\begin{array}{c}
M=d d d \cdots d=\frac{d}{9}(999 \cdots 9... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 702,512 |
30. Let $[x]$ be the greatest integer less than or equal to $x$. Then the number of real solutions to the equation $4 x^{2}-40 \cdot[x]+51=0$ is
(A) 0,
(B) 1;
(C) 23
(D) 33
(E) 4. | 30. Since 40 is even, $40[x]-51$ is odd. This implies that $4 x^{2}$ must be an odd integer, let $4 x^{2}=2 k+1$. Thus, $x=\sqrt{2 k+1} / 2$. Substituting it into the original equation, we get $\left[\frac{\sqrt{2 k+1}}{2}\right]=\frac{k+26}{20}$. Therefore, it must be that $k \equiv 14$ (mod 20). On the other hand, $\... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 702,513 |
Example 4. Find (1) the principal period of $f(x)=\operatorname{tg} 2 x-\operatorname{ctg} 2 x$; (2) the principal period of $g(x)=12 \sin ^{2} 15 x+15 \cos ^{2} 12 x$. | (1) The conclusion that the main period of $f(x)=\tan 2x - \cot 2x$ is $\frac{\pi}{2}$ is incorrect. This is because the conclusions for sine and cosine functions cannot be directly applied to tangent and cotangent functions.
$$
\begin{array}{c}
\because f(x)=\tan 2x - \cot 2x = \frac{\sin 2x}{\cos 2x} - \frac{\cos 2x}... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,514 |
2. Prove $\sqrt{a \pm \sqrt{ } b}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}$
$$
\begin{array}{l}
\pm \sqrt{\frac{a-\sqrt{a^{2}-b}}{2}} . \\
\quad\left(a>0, \quad b>0, \quad a^{2}>b\right)
\end{array}
$$ | $$
\begin{array}{l}
\text{Prove that the square of } \sqrt{a \pm \sqrt{b}} = \frac{a+\sqrt{a^{2}-b}}{2} \\
+\frac{a-\sqrt{a^{2} \cdots b}}{2} \\
=a \pm \sqrt{a^{2}-\left(a^{2}-b\right)}=a \pm \sqrt{b} \text{.} \\
\end{array}
$$
$\because$ Since the left side $>0$, the right side $>0$,
$\therefore$ the left side = the r... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,516 |
13. Given $a^{2}+b^{2}=1, c^{2}+d^{2}=1$, $a c+b d=0$.
Prove: $a^{2}+c^{2}=1, b^{2}+d^{2}=1$. $a b+c d=0$ | Let $a=\sin \alpha, b=\cos \alpha$,
$c=\sin \beta, d=\cos \beta$,
and assume $0<\beta<a<\frac{\pi}{2}$,
from $a d + b c = 0$ we get $\sin \alpha \sin \beta + \cos \alpha \cos \beta = 0$,
which means $\cos (\alpha - \beta) = 0$,
$\therefore \quad \alpha - \beta = \frac{\pi}{2}, \quad \alpha = \frac{\pi}{2} + \beta$.
The... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,517 |
Example 3. Given two places $A$ and $B$ with a longitude difference of $\alpha^{\circ}$, both located on the same latitude line at $\beta$.
Find the shortest distance between $A$ and $B$ on the sphere. | Let $\angle A O B=\theta$, take the midpoint $M$ of $A B$, connect $O M$, $O_{1} M$, to form a tetrahedron $O-M B O_{1}$ enclosed by four right triangles, in which the following three conditions are known.
$$
\begin{array}{l}
\angle M O_{1} B=\frac{\alpha}{2}, \angle O B O_{1}=\beta \quad O B=R \\
=6370 \text { km. }
\... | 12740 \arcsin \left(\cos \beta \sin \frac{\alpha}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,520 |
Vector 1. Prove that the line $y=k x+1$ must pass through a fixed point $(k \in$ R).
Note: The term "Vector" at the beginning seems out of context and might be a mistranslation or a typo. If it is not intended, the correct translation would start directly with "1. Prove that the line $y=k x+1$ must pass through a fi... | $$
x k-(y-1)=0 .
$$
Then its valid sufficient condition is
$$
\left\{\begin{array}{l}
x=0 \\
y-1=0
\end{array}\right.
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,521 |
Example 2. Let $a$ be a non-zero real number. Prove that the curve $y=-a x^{2}$ - $(6 a-1) x+5 a+1$ always passes through a fixed point and find the coordinates of the fixed point.
保留源文本的换行和格式，直接输出翻译结果。 | Transform the equation into an expression about $a$
$$
\left(x^{2}+6 x-5\right) a-(x-y+1)=0 \text {. }
$$
The sufficient condition for the equation to hold for any non-zero real number $a$ is
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}+6 x-5=0, \\
x-y+1=0 .
\end{array}\right. \\
\left\{\begin{array} { l }
{ x =... | (-3+\sqrt{14},-2+\sqrt{14}) \text{ and } (-3-\sqrt{14},-2-\sqrt{14}) | Algebra | proof | Yes | Yes | cn_contest | false | 702,522 |
List 3. Regardless of the value of the real number $k$, the line $y=2 k x$ $+k^{2}$ is tangent to the parabola $y=a x^{2}+b x+c$. Find the equation of the parabola. | Given $y=2 k x+k^{2}$ and $y=a x^{2}+b x+c$ are combined, we get
$$
a x^{2}+(b-2 k) x+\left(c-k^{2}\right)=0 .
$$
Since the line is tangent to the parabola,
thus $\Delta=(b-2 k)^{2}-4 a\left(c-k^{2}\right)=0$, which means $4(a+1) k^{2}-4 b k+b^{2}-4 a c=0$.
$\because k$ is any real number,
$\therefore$ the above equat... | y=-x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,523 |
鉜4 (Yellow). Find the tangent line to the hyperbola $\frac{x^{2}}{64}-\frac{y^{2}}{16}=1$ passing through the point $P(16,8)$.
untranslated text remains the same as it is a formula or point notation. | Substitute $(16,8)$ into the tangent line equation $\mathbf{y}=\mathrm{k} \mathbf{x} \pm \sqrt{ } 64 \overline{\mathrm{k}^{2}-16}$ $k_{1}=\frac{5}{6}, k_{2}=\frac{1}{2}$, hence the tangents are $y=\frac{5}{6} x-\frac{16}{3}\left(k>\frac{1}{2}\right.$ when taking “ - ” $)$ and $y=\frac{1}{2} x$.
Analysis: Due to neglec... | y=\frac{5}{6} x-\frac{16}{3} \text{ and } y=\frac{1}{2} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,528 |
Example 6 (Wang). Find the locus of the vertex of the parabola $y=x^{2}-4 t^{2} x$ $+\left(4 t^{4}-t^{2}-1\right)$ (t is a parameter). | Let the point be $(\mathrm{x}, \mathrm{y})$, then
$$
\left\{\begin{array}{l}
x=2 t^{2} \\
y=-\left(t^{2}+1\right)
\end{array}\right.
$$
Eliminating the parameter, we get $x+2 y+2=0$, which is the line.
Analysis: Due to the neglect of $t^{2} \geqslant 0$, i.e., $x \geqslant 0$, the solution is incomplete. In fact, the... | x+2y+2=0 \ (x \geq 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,530 |
Example 7 (Wang). Find the equation of the line passing through the point $(3,4)$ that intersects the coordinate axes to form a triangle with an area of 12 (square units).
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the slope of the line be k, then the equation is $\mathrm{y}-4$ $\frac{3 \mathbf{k}-4}{\mathbf{k}}$. From the given condition, we have
$$
\frac{1}{2}(4-3 k) \cdot \frac{3 k-4}{k_{0}}=12 .
$$
Simplifying, we get $9 k^{2}+16=0$. The equation has no real roots, so the required line does not exist. Correcting this poi... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,531 |
Example 9 (Zhou Jiansan). The line 1 passing through point $A_{2}(6,4)$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ at points $\mathrm{M}, \mathrm{N}$. Find the maximum and minimum values of |AM|・|AN|. | Let the inclination angle be $\alpha$, then its parametric equation is
$$
\left\{\begin{array}{l}
x=6+t \sin \alpha, \\
y=4+t \cos \alpha .
\end{array}\right. \text { (t is a parameter) }
$$
Substituting into the ellipse equation, and simplifying, we get
$$
\begin{array}{l}
\left(3 \sin ^{2} \alpha+1\right) t^{2}+4(3 ... | 84 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,533 |
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