problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 7. Solve the equation:
$$
\sin x + \cos x + \tan x = \sec x
$$ | Solution: Using the universal substitution method: Let $\operatorname{tg} \frac{x}{2}=t$, then the original equation becomes: $\frac{2 t}{1+t^{2}}+\frac{1-t^{2}}{1+t^{2}}+\frac{2 t}{1-t^{2}}=\frac{1+t^{2}}{1-t^{2}}$. Eliminating the denominators and simplifying, we get $t(t-1)=0$.
$$
\begin{array}{l}
\therefore t_{1}=0... | x=k \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,977 |
Example 8. Prove the continuity of $y^{y}=\tan x\left(-\frac{\pi}{2}<x<\frac{\pi}{2}\right)$. | Prove: the “$\varepsilon-\delta$” method.
Draw a unit circle $O$ centered at the origin, intersecting the $x$-axis at $A$, and draw $AT$ tangent to $\odot$ at $A$.
Let $x_{0}$ be any value in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. For any given $\varepsilon>0$, it is always possible to find three points $T_{1}, ... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 702,978 |
Example 3. Simplify: $\sqrt{(a-5)^{2}}$ | $$
\begin{array}{l}
\sqrt{(a-5)^{2}}=|a-5| \\
=\left(\begin{array}{ll}
a-5 ; & \text { (when } a-5>0, \text { i.e., } a>5 \text { ) } \\
03 & \text { (when } a-5=0, \text { i.e., } a=5 \text { ) } \\
5-a, & \text { (when } a-5<0, \text { i.e., } a<5 \text { ) }
\end{array}\right.
\end{array}
$$ | |a-5| | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,979 |
Find $x$:
(1) $\sin x=\frac{1}{4}, x \in\left(\frac{\pi}{2}, \pi\right)$,
(2) $\sin x=-\frac{1}{3}, x \in\left(\pi, \frac{3}{2} \pi\right)$. | Analysis: (1) $\because \sin _{x=\frac{1}{4}}$,
$\therefore \arcsin \frac{1}{4} \in\left(0, \frac{\pi}{2}\right)$.
However, the problem is to find the angle within $\left(\frac{\pi}{2}, \pi\right)$ whose sine value is $\frac{1}{4}$. Using the induced formula, we get
$$
x=\pi-\arcsin \frac{1}{4}.
$$
(2) $\because \quad ... | x=\pi-\arcsin \frac{1}{4}, \pi+\arcsin \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,980 |
Example 2. Express the following in terms of the inverse cosine:
(1) $\cos x=\frac{2}{5}, x \in\left(-\frac{\pi}{2}\right.$, );
(2) $\cos x=-\frac{\sqrt{2}}{3}, x \in\left(-\pi,-\frac{\pi}{2}\right)$. | Analysis: (1) $\because \cos x=\frac{2}{5}$,
$\therefore \quad \arccos \frac{2}{5} \in\left(0, \frac{\pi}{2}\right)$.
However, the problem is to find the angle within $\left(-\frac{\pi}{2}, 0\right)$ whose cosine value is $\frac{2}{5}$.
Since the cosine function is an even function (as shown in Figure 2), we have
$$
x... | x=-\arccos \frac{2}{5}, x=-\arccos \left(-\frac{\sqrt{2}}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,981 |
Example 3. Express $x$ in the form of the inverse cotangent in the following expression:
$$
\operatorname{tg} x=3, x \in(-3 \pi,-2 \pi) .
$$ | Solve $p=-3 \pi$, from formula (4), we get
$x=p+\operatorname{arcctg}[\operatorname{ctg}(x-p)]$
$=-3 \pi+\operatorname{arcctg}[\operatorname{ctg}(-+3 \pi)]$
$=-3 \pi+\operatorname{arcctg}(\operatorname{ctg} x)$.
$\because \quad \operatorname{tg} x=3, x \in(\cdots 3 \pi, \quad-2 \pi)$,
$\therefore \operatorname{ctg} \ap... | x=-3 \pi+\operatorname{arcctg} \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,982 |
Example 4. Express $x$ in the following equation using the inverse cosine function: $\cos x=\frac{3}{5} \quad x \in(3,6.1)$. | Here $p=3$, by formula (2), we get
$$
\begin{array}{c}
x=p+\arccos [\cos (x-p)] \\
=3+\arccos [\cos (x-3)] . \\
\because \quad \cos x=\frac{3}{5}, x \in(3,6.1), \\
\therefore \quad \sin x=-\frac{4}{5}, \\
\cos (x-3)=\frac{3}{5} \cos 3-\frac{4}{5} \sin 3 .
\end{array}
$$
Thus, $x=3+\arccos \left(\frac{3}{5} \cos 3\righ... | x=3+\arccos \left(\frac{3}{5} \cos 3 - \frac{4}{5} \sin 3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,983 |
Example 5. Express $x$ in terms of the inverse sine function for the following equation: $\sin x=\frac{1}{3}, x \in(5 \pi, 8 \pi)$. | $$
\begin{array}{l}
\text { Brief solution: Divide the interval }(5 \pi, 8 \pi) \text { into }(5 \pi, \\
6 \pi) \cup[6 \pi, 7 \pi] \cup(7 \pi, 8 \pi) \text {. } \\
\because \quad \sin x=\frac{1}{3}, \\
\text { it is determined that } x \notin(5 \pi, 6 \pi), x \notin(7 \pi, 8 \pi) \text {. } \\
\text { Therefore, } x \i... | x=\frac{13 \pi}{2}+\arcsin \frac{2 \sqrt{2}}{3} \text { or } x=\frac{13 \pi}{2}-\arcsin \frac{2 \sqrt{2}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,984 |
Example 2. The Louka Problem: A shipping company has a ship leaving Harvard for New York and a ship leaving New York for Harvard every noon. The journey takes seven days and seven nights in both directions. How many ships leaving New York will the ship that departs from Harvard at noon today encounter on its journey? | Solving the family of time-position broken lines shown in Figure 3, it is easy to see that, under the condition of continuous daily ship departures, starting from the 6th day, each ship departing thereafter will encounter 13 ships coming from the opposite direction.
In the case of continuous daily ship departures, sta... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,985 |
For example. In $\triangle ABC$, the three sides $a, b, c$ form an arithmetic sequence, and the radii of the incircle and circumcircle are $r$ and $R$, respectively, then $2r, a, b, 2R$ form an arithmetic sequence. | Prove that from property 8 and $\cos C=0$, we get $\cos A=\frac{4}{5}$. Then $b=\frac{4}{5} c, a=\frac{3}{5} c, 2 r=a+b-c=\frac{2}{5} c$, $2 R=c$. Therefore, $2 r, a, b, 2 R$ form an arithmetic sequence.
Graphical proofs of inequalities, being intuitive and specific, profoundly reveal the connection between numbers an... | 2r, a, b, 2R \text{ form an arithmetic sequence} | Geometry | proof | Yes | Yes | cn_contest | false | 702,986 |
Example 1. $a, b$ are positive numbers. Prove that $\frac{a+b}{2} \geqslant \sqrt{a b}$. | Consider $R t \triangle A B C$ (Figure 1), let the projections of the legs on the hypotenuse be $a, 1$, then we have $D=\sqrt{a b}$, where $C E=\frac{a+b}{2}$.
From the properties, we know $C E \geqslant C D$, that is,
$$
\frac{a+b}{2} \geqslant \sqrt{a b} \text {. }
$$
Figures 2 and 3 provide additional proofs. Sinc... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,987 |
Example 2 The height of a regular quadrilateral pyramid is $h$, and the dihedral angle between two adjacent side faces is $2 \alpha$. Find the volume of this regular quadrilateral pyramid. | Slightly solved: Let the angle between the side face and the base be $\theta$, and the side length of the base of the regular quadrilateral pyramid be $a$. From (*) and the problem statement, we have:
$$
\left\{\begin{array}{l}
\cos ^{2} \theta=-\cos 2 \alpha, \\
\frac{a}{2}=h \operatorname{ctg} \theta .
\end{array}\ri... | -\frac{2}{3} h^{3} \cos 2 \alpha \sec ^{2} \alpha | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,990 |
Given $P A, P B, P C$ three rays, and
$$
\begin{array}{l}
\angle A P C=\alpha, \angle C P B=\beta, \angle A P B=\alpha+\beta< \\
180^{\circ} \text {. Then the necessary and sufficient condition for } A, B, C \text { to be collinear is, } \\
\frac{\sin (\alpha+\beta)}{P C}=\frac{\sin \alpha}{P B}+\frac{\sin \beta}{P A} ... | To prove Figure 1, if $A, B, C$ are collinear, then $S_{\triangle P A B} = S_{\triangle P A C} + S_{\triangle P C B}$, which means
$$
\begin{array}{l}
\frac{1}{2} P A \cdot P B \sin (\alpha+\beta) = \frac{1}{2} P A \cdot P C \sin \alpha \\
+ \frac{1}{2} P C \cdot P B \sin \beta .
\end{array}
$$
Dividing both sides by ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,991 |
Example: On the sides $CB, CA$ of $\angle ABC$, construct squares $CBRS, CAQP$ outwardly, and draw $CH \perp AB$. Prove that $CH, BP, AR$ are concurrent.
(Figure 2) | $$
\begin{array}{l}
\text{Prove: As shown in Figure 2, connect } C R, C P. \text{ Let } C H, B P \text{ intersect at } E. \text{ From the perspective of } C, \text{ apply the Angle Subtended Theorem to } B, E, P: \\
\frac{\sin \left(\alpha+\beta+45^{\circ}\right)}{C E}=\frac{\sin \left(45^{\circ}+\beta\right)}{C B} \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,992 |
Example 1. The side length of square $ABCD$ is 1. A ray is drawn from $A$ intersecting side $BC$ (or its extension) at $E$, and the extension of $DC$ (or $DC$) at $F$. Prove that: $AE + AF \geqslant 2\sqrt{2}$.
(Figure 1) | Let $\angle B A E = \alpha$, then $\angle A F D = \alpha (0 < \alpha < 90^{\circ})$. Therefore,
$$
\begin{aligned}
& A E + A F \\
= & \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} \\
\geqslant & 2 \sqrt{\frac{1}{\sin \alpha \cos \alpha}} \\
= & \frac{2 \sqrt{2}}{\sqrt{\sin 2 \alpha}} \geqslant 2 \sqrt{2}
\end{aligned}
... | 2\sqrt{2} | Geometry | proof | Yes | Yes | cn_contest | false | 702,993 |
Example 2. As shown in Figure 2, from the vertex $A$ of $\triangle ABC$, draw any line intersecting the excircle near $BC$ at $P$ and $Q$. Prove: $AP + AQ \geqslant AB + BC + CA$ | $\begin{array}{l} \text { Prove } A P+A Q \\ \geqslant 2 \sqrt{A P \cdot A Q} \\ = 2 A D . \\ \text { But } 2 A D=A D+A E \\ = A B+B D \\ + A C+C B \\ = A B+B F \\ +A C+C F \\ = A B+B C+A C \\ \therefore \quad A P+A Q \geqslant A B+B C+C A .\end{array}$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,994 |
Example 1. If the sides of a regular quadrilateral pyramid are equilateral triangles, prove: the dihedral angle formed by two adjacent sides is twice the dihedral angle formed by a side and the base. (Six-year Key High School Mathematics Textbook, *Solid Geometry*, page 128, problem 21) | Proof: Let the side length of a regular quadrilateral pyramid be $a$, then the side length of the base is also $a$, and the plane angle $\angle B E D=\beta$ formed by the dihedral angle between two adjacent sides, and the plane angle $\angle S F O=\theta$ formed by the dihedral angle between a side and the base. (Refer... | \beta=2\theta | Geometry | proof | Yes | Yes | cn_contest | false | 702,997 |
Example 1. Determine the odd or even nature of the following two functions:
(1) $f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$;
(2) $f(x)=\frac{\left(1+2^{x}\right)^{2}}{2^{x}}$. | $$
\begin{aligned}
: \because f(x) & =\frac{e^{x}}{e^{x}-e^{-x}}+\frac{e^{-x}}{e^{x}-e^{-x}} \\
& =\frac{e^{x}}{e^{x}-e^{-x}}-\frac{e^{-x}}{e^{-x}-e^{x}} \\
& =f_{1}(x)-f_{1}(-x), \\
\text { where } f_{1}(x) & =\frac{e^{x}}{e^{x}-e^{-x}},
\end{aligned}
$$
By rule (1), $f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$ is an odd... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,998 |
Example 3. Prove: $f(x) \equiv 0$ is both an odd function and an even function. | Proof: $\because f(-x) \equiv 0$,
$$
\therefore f(x)=f(x)+f(-x) \text {. }
$$
By rule (2), $f(x) \equiv 0$ is an even function;
$$
\text { Also, } \because f(x)=f(x)-f(-x) \text {, }
$$
By rule (1), $f(x)=0$ is also an odd function.
Thus, $f(x)=0$ is both an odd and an even function.
Example 4. Prove: Any function $f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,999 |
Example 1. Determine the odd or even nature of the function $f(x)=3 x^{2}-1, x \in[a$, $b]$ and $a+b \neq 0$ | Solution: $\because f(-x)=3 x^{2}-1=f(x)$,
$\therefore f(x)$ is an even function.
Analysis: The definition of an even function requires that for any $x$ in the domain of the function $f(x)$, $f(-x)=f(x)$. This requires that $f(-x)$ must be meaningful, i.e., $-x$ must belong to the domain of the function $f(x)$. Theref... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,000 |
Example 2. Determine the parity of $f(x)=3 x^{3}+2 x(x-1)^{0}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | Solution: The original function is equivalent to $f(x)=3 x^{3}+2 x,(x$ $\neq 1$ )
Since the domain of this function is $x \neq 1$, which is not symmetric about the origin, $f(x)$ has no odd or even properties. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,001 |
Let $a_{n}=A u^{3}+B n^{2}+C n+D$, from $a_{1}=1$, $a^{2}=2, a_{3}=4, a_{4}=p$ (any number) we get the system of equations:
$$
\left\{\begin{array}{c}
A+B+C+D=1 \\
8 A+4 B+2 C+D=2 \\
27 A+9 B+3 C+D=4 \\
64 A+16 B+4 C+D=\rho
\end{array}\right.
$$ | Given the sequence $1,2,4,32-t, 135 \cdots 4 t, \cdots$. (where $t$ is any parameter)
Let $t=0$ in (5), we get
$$
a_{n}=n\left(n^{3}-\frac{35}{6} n^{2}+\frac{21}{2} n-\frac{14}{3}\right).
$$
The sequence is $1,2,4,32,135, \cdots$.
From this, we can see that for a sequence with a finite number of terms and no specific ... | a_{n}=\frac{1}{6} n\left(n^{2}-3 n+8\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,002 |
If each face of a tetrahedron is not an equilateral triangle, then the minimum number of edges with unequal lengths is
(A) 3;
(B) 4 ;
(C) 5 ;
(D) 6 . | This is a question that tests students' ability to express and imagine spatial figures. A tetrahedron has six edges, and any one of these edges is a skew line with only one of the other five edges. According to the problem, only two edges that are skew lines can be of equal length. Therefore, the minimum number of edge... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,003 |
Example $3 . x$ is a real number, find
$$
y=2 x^{2}-6 x+10
$$
the minimum value. | \[
\begin{aligned}
y= & 2 x^{2}-6 x+10 \\
= & 2\left(x^{2}-3 x+5\right) \\
& =2\left(x-\frac{3}{2}\right)^{2}+\frac{11}{2}
\end{aligned}
\]
Since \( x \) is a real number, \(\quad \therefore\left(x-\frac{3}{2}\right)^{2} \geqslant 0\),
\(\therefore\) when \( x=\frac{3}{2} \), \( y \) has a minimum value of \(\frac{11}{... | \frac{11}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,004 |
Example 4. Solve the equation:
$$
\begin{array}{c}
\left|x^{2}-2 x-y^{2}-2 y-4\right| \\
+\sqrt{2 x-y-7}=0 .
\end{array}
$$ | $$
\left\{\begin{array}{l}
c^{2}-2 x-y^{2}-2 y-4=0, \\
2 x-y-7=0 .
\end{array}\right.
$$
Solve this system of equations, we get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = \frac { 13 } { 3 } , } \\
{ y _ { 1 } = \frac { 5 } { 3 } }
\end{array} \quad \left\{\begin{array}{l}
x_{2}=3, \\
y_{2}=-1 .
\end{array}\right.\ri... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,005 |
Example 5. Solve the equation:
$$
5 x^{2}-6 x y+2 y^{2}-4 x+2 y+1=0 \text {. }
$$ | Solution: The original equation can be transformed into
$$
\begin{array}{l}
\left(x^{2}-2 x y+y^{2}\right)+\left(4 x^{2}-4 x y\right. \\
\left.+y^{2}\right)-4 x+2 y+1=0, \\
(x-y)^{2}+(2 x-y)^{2} \\
-2(2 x-y)+1=0, \\
(x-y)^{2}+(2 x-y-1)^{2}=0 .
\end{array}
$$
Using the method from Example 4, we get
$$
\left\{\begin{arr... | x=1, y=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,006 |
Example 6. Find the real solutions of the equation
$$
\begin{array}{l}
\frac{36}{\sqrt{2 x-1}}+\frac{4}{\sqrt{3 y+1}} \\
+4 \sqrt{2 x-1}+\sqrt{3 y+1}=28
\end{array}
$$ | Let $\sqrt{2^{x-1}}=X$, $\sqrt{3 y+1}=Y$, obviously we have $X>0$, $\boldsymbol{Y}>0$.
Then the original equation can be transformed into
$$
\frac{36}{X}+\frac{4}{Y}+4 X+Y=28 .
$$
(1) can be rewritten as
$$
\begin{array}{l}
\left(2 \sqrt{X}-\frac{6}{2 \bar{X}}\right)^{2} \\
+\left(\sqrt{Y}-\frac{2}{\sqrt{Y}}\right)^{2... | \left\{\begin{array}{l} x=5, \\ y=1 \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,007 |
Example 7. Prove: The equation $3 x^{3}-2 x+4$ $=3 \cos x$ has no real solutions.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: $\because 3 x^{2}-2 x+4$
$$
\begin{array}{l}
=x^{2}-2 x+1+2 x^{2}+3 \\
=(x-1)^{2}+2 x^{2}+3 .
\end{array}
$$
If the original equation has real solutions $x$, then it should have
$$
(x-1)^{2} \geqslant 0, \quad 2 x^{2} \geqslant 0,
$$
and at least one of $(x-1)^{2}, 2 x^{2}$ should be positive,
$$
\therefore \q... | null | Algebra | proof | Yes | Yes | cn_contest | false | 703,008 |
Example 8. If $a, b, c, d$ are all positive real numbers, and satisfy
$$
a^{4}+b^{4}+c^{4}+d^{4}=4 a b c d .
$$
Prove: $a=b=c=d$. | Proof: Given, we can obtain
$$
a^{4}+b^{4}+c^{4}+d^{4}-4 a b c d=0 \text {. }
$$
Transform the above equation to
$$
\begin{array}{l}
a^{4}-2 a^{2} b^{y}+b^{4}+c^{4}-2 c^{2} d^{2} \\
+d^{4}+2(a b-c d)^{2}=0,
\end{array}
$$
which is
$$
\begin{array}{c}
\left(a^{2}-b^{2}\right)^{2}+\left(c^{2}-d^{2}\right) \\
+2(a b-c d... | a=b=c=d | Algebra | proof | Yes | Yes | cn_contest | false | 703,009 |
Example 9. If $a, b, c, d$ are non-zero real numbers, and
$$
\begin{array}{l}
\left(a^{2}+b^{2}\right) d^{2}-2 b(a+c) d \\
+b^{2}+c^{2}=0 .
\end{array}
$$
Prove: $\frac{b}{a}=\frac{c}{b}=d$.
| Prove: Expanding the original expression, we get
$$
\begin{array}{l}
a^{2} d^{2}+b^{2} d^{2}-2 a b d-2 b c d \\
+b^{2}+c^{2}=0 .
\end{array}
$$
Transforming it, we get
$$
\begin{array}{c}
\left(a^{2} d^{2}-2 a b d+b^{2}\right) \\
+\left(b^{2} d^{2}-2 b c d+c^{2}\right)=0, \\
(a d-b)^{2}+(b d-c)^{2}=0 . \\
\therefore \... | \frac{b}{a}=\frac{c}{b}=d | Algebra | proof | Yes | Yes | cn_contest | false | 703,010 |
Example 10. If $a, b, c$ are three distinct real numbers, prove that the quadratic equations
$$
\begin{array}{l}
a x^{2}+2 b x+c=0, \\
b x^{2}+2 c x+a=0, \\
c x^{2}+2 a x+b=0,
\end{array}
$$
cannot all have equal roots. | Proof by contradiction:
Suppose the three equations can all yield equal roots, then we have
$$
\begin{array}{l}
4 b^{2}-4 a c=0, \\
4 c^{8}-4 a b=0, \\
4 a^{2}-4 b c=0 .
\end{array}
$$
Adding the above three equations and dividing by 4, we get
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=0 .
$$
Transforming the above equation, w... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,011 |
Example 11. Prove: For any real numbers $x$ and $y$, the inequality $\quad x^{2}+x y+y^{2}+2 x-2 y+4 \geqslant 0$ always holds. | Proof:
$$
\begin{aligned}
& \because x^{2}+x y+y^{2}+2 x-2 y+4 \\
= & \frac{1}{2}\left(2 x^{2}+2 x y^{2}+2 y^{2}+4 x\right. \\
& -4 y+8) \\
= & \frac{1}{2}\left[\left(x^{2}+2 x y+y^{2}\right)+\left(x^{2}\right.\right. \\
& \left.+4 x+4)+\left(y^{2}-4 y+4\right)\right] \\
= & \frac{1}{2}\left[\left(x+y i^{2}+(x+2)^{2}\r... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,012 |
Example 12. Given a quadrilateral $A B C D$ with sides $a, b, c, d$ in sequence, and
$$
a^{2}+b^{2}+c^{2}+d^{2}=2 a c+2 b d .
$$
Prove: Quadrilateral $A B C D$ is a parallelogram. | Proof: $\because a^{2}+b^{2}+c^{2}+d^{2}=2 a c+2 b d$,
$$
\begin{aligned}
\therefore \quad & \left(a^{2}-2 a c+c^{2}\right)+\left(b^{2}-2 b d\right. \\
& \left.+d^{2}\right)=0,
\end{aligned}
$$
i.e., $(a-c)^{2}+(b-d)^{2}=0$.
$$
\therefore a-c=0 \text {, and } b-d=0 \text {. }
$$
Thus, $a=c$, and $b=d$.
Therefore, qua... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,013 |
On the plane, there is a set of points $M$ and seven different circles $C_{1}, C_{2}, \cdots, C_{7}$, where circle $C_{7}$ passes through exactly 7 points in $M$, circle $C_{0}$ passes through exactly 6 points in $M$, ... circle $C_{1}$ passes through exactly 1 point in $M$, then the minimum number of points in $M$ is
... | This problem mainly tests students' ability to reason recursively and by induction. Initially, seven different straight lines were used, but later, changing the lines to circles was considered to reduce the problem's intuitiveness, making it more conducive to testing students' recursive and reduction skills. If the poi... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 703,014 |
Example 13. Given that the characteristic of $\lg x^{\frac{3}{2}}$ is the solution to the inequality
$$
a^{2}-8 a+15<0
$$
and the mantissa is the sum of the solutions to the equation
$$
\frac{1}{4}\left|y+\frac{1}{2}\right|+\frac{6}{121} \sqrt{z-1}=0
$$
find the value. | Solution: Let $\lg x^{\frac{3}{2}}=p+q$
(where $p$ is the characteristic, and $q$ is the mantissa). Solving the inequality $a^{2}-8 a+15<0$, we get
$$
3<a<5 \text {. }
$$
Since the characteristic of a logarithm of a number must be an integer,
$$
\therefore \quad p=a=4 \text {. }
$$
From equation (2), we get
$$
y=-\fra... | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,015 |
Example 14. Let $x$ be a natural number from 1 to 10, the remainder of $x$ divided by 3 is $f(x)$, $y$ is a one-digit natural number, the remainder of $y$ divided by 4 is $g(y)$, then what is the maximum value of $x+2y$ when $f(x)+2g(y)=0$? | Solution: From the problem, we can obtain
$$
\begin{array}{l}
g(y)=\left\{\begin{array}{l}
0, y=4,8 ; \\
1, y=1,5,9 ; \\
2, y=2,6 ; \\
2, y=3,7 .
\end{array}\right. \\
\end{array}
$$
$\therefore f(x), g(y)$ are both non-negative.
To make $f(x)+2 g(y)=0$,
it is necessary that $f(x)=0$ and $g(y)=0$.
At this time, $x=3,6,... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,016 |
Example 1. In a square with a side length of 1, any given points are provided. Try to prove: among the triangles formed with these points as vertices, there must be one triangle whose area is no more than $\frac{1}{8}$. (Beijing 1965 Mathematics Competition, Senior High School, Second Round) | 〔Prompt〕As shown in the figure, divide the square into four congruent smaller squares or rectangles, then there is at least one smaller square (or rectangle) containing at least three points, denoted as $A$, $B$, $C$. Try to prove that $S_{\triangle A B C} \leqslant \frac{1}{2} S_{\text {small square }}=\frac{1}{8}$ (w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,017 |
Let there be 1985 points in a unit cube. Prove that 32 of them can be selected such that the perimeter of any closed polygon (possibly degenerate) formed by them is less than 8 (26th IMO candidate problem) | "Parallel planes divide a unit cube into 64 smaller cubes, each with an edge length of $\frac{1}{4}$, then at least one of the smaller cubes contains 32 points, study the closed polygon formed by these 32 points." | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,018 |
Example 3. Place non-negative real numbers at each vertex of a cube, such that the sum of these numbers is 1. Two players, A and B, play the following game: A arbitrarily selects one face, then B selects one vertex, and finally A selects three faces, but after A selects the first face, the subsequent faces chosen canno... | 〔Prompt〕First prove
$$
\begin{array}{l}
\text { if } a_{i} \geqslant 0(i \\
=1,2, \cdots 8) \\
\sum_{i=1}^{8} a_{i}=1,
\end{array}
$$
then at least three numbers are less than $\frac{1}{6}$.
Next, divide the cube into two parts as shown in the figure, then at least one of the parts will have two vertices of a diagonal... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,019 |
Example 1. 111 points are placed in an equilateral triangle with a side length of 15. Prove that a circular coin with a certain diameter can always cover at least three of these 111 points. (1983 Lanzhou Mathematical Olympiad) | 〔Dividing the Triangle〕Divide each side of an equilateral triangle into 10 equal parts, resulting in 100 congruent smaller equilateral triangles with side lengths of 1.5. Consider 55 of these smaller triangles, with their vertices on the larger triangle. These 55 smaller triangles can be covered by 55 coins, which are ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,020 |
Let $f(x)=\frac{4^{x}}{4^{x}+2}$, then the sum
$$
f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)
$$
$+f\left(\frac{3}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$ is | The original sum is $f\left(\frac{1}{1986}\right)+f\left(\frac{2}{1986}\right)$
$+\cdots+f\left(\frac{1985}{1986}\right)$. Considering that the number 1986 is non-essential, and given that using year numbers in problems is common, to avoid unnecessary confusion, this number has been modified. This problem is tricky, an... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,025 |
2. A chess master has 11 weeks to prepare for a competition. He decides to play at least one game every day, but to prevent himself from getting too tired, he decides not to play more than 12 games in any given week. | Try to prove: There exist some consecutive days during which Master Xiangsi played exactly 21 games of chess.
「Assume that on the $i$-th day, $x_{i}$ games were played, and calculate the sequence
$$
s_{i}=x_{1}+x_{2}+\cdots+x_{i}
$$
This sequence has 77 terms, and none of them exceed $12 \times 11=132$.
Now consider t... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,026 |
Example 1. In 100 consecutive natural numbers $1,2, \cdots$, 100, take any 51 numbers. Try to prove that among these 51 numbers, there must be two numbers, one of which is a multiple of the other. | 〔Hint? Divide these 100 numbers into 50 categories according to $k \cdot 2^{2}(k$ being odd), i.e., $\left\{1,1 \times 2, \cdots, 1 \times 2^{B}\right\}$, $\left\{3,3 \times 2, \cdots, 3 \times 2^{-}\right\},\{5,5 \times 1$, $\left.\cdots, 5 \times 2^{4}\right\}, \cdots,\{97\},\{99\}$, among these 51 numbers, there mus... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,027 |
Example 4. If all terms of an arithmetic sequence are integers, and there are $p$ terms, if each term is coprime with $p$, then its common difference must not be coprime with $p$. | 〔Hint〕Consider all terms
$$
a, a+d, a+2d, \cdots, a+(p-1)d
$$
There must be two terms congruent modulo $p$. Consider the difference between these terms. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,029 |
Seventeen scientists, each of whom communicates with all the others. In their communications, they only discuss three topics, and each pair of scientists only discusses one topic.
Prove: At least three scientists are discussing the same topic with each other. | Review: From these seventeen scientists, arbitrarily select one, let's call him Mr. $X$. Mr. $X$ communicates with the other sixteen scientists, and they only discuss three topics. Since $16=3 \times 5+1$, at least six scientists are discussing the same topic with Mr. $X$, let's call this topic "Topic 1". If among thes... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,030 |
Given a set consisting of any 10 distinct two-digit positive integers.
Prove: This set must have two disjoint subsets such that the sum of all numbers in each subset is equal. | Proof: A set with 10 elements has how many possible subsets? Since each element may or may not belong to a subset, there are a total of $2^{10}=1024$ different subsets. Of course, the empty set and the set itself are not suitable for the requirements of the question. Therefore, the number of subsets to consider is 1022... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,031 |
$$
\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{14} x_{4}=0, \\
a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{24} x_{4}=0, \\
\cdots \cdots \\
a_{p_{1}} x_{1}+a_{\mathrm{p} 2} x_{2}+\cdots+a_{1} x_{4}=0 .
\end{array}
$$
(i) $x_{1}, x_{2}, \cdots, x_{4}$ are all integers;
(ii) at least one of them is not equal to zero;
(i... | Proof: Consider all integer tuples $\left(y_{1}, y_{2}, \cdots, y_{q}\right)$ that satisfy the condition $|y|=p$ for $j=1, 2, \cdots, q$. Since each position can only take one of the $2p+1$ numbers:
$$
-p, \cdots,-2,-1,0,1,2, \cdots, p
$$
and there are $q$ positions, there are only $(2p+1)^q$ such tuples. Let
$$
\begi... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,032 |
A member of an international society comes from six countries, with a total of 1978 people, numbered as $1, 2, \cdots, 1977, 1978$. Try to prove: the society has at least one member whose number is equal to the sum of the numbers of his two compatriots, or is a multiple of one of his compatriots' numbers. | Proof: This problem is completely equivalent to the following: "Divide 1, 2, 3, ..., 1978 into six groups in any way, then at least one of the groups will have the property that there is a number which is either the sum of two other numbers in the same group or a multiple of one of the numbers in the group."
We will u... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,033 |
Let $A B C$ be an equilateral triangle. $E$ is the set of all points contained in the three line segments $B C, C A, A B$ (including points $A, B, C$). For any partition of $E$ into two disjoint subsets, is it true that at least one of these subsets contains the three vertices of a right triangle? Prove your answer. | Proof: On the sides $BC$, $CA$, $AB$, take three points $P$, $Q$, $R$ respectively, such that $PC = BC / 3$, $QA = CA / 3$, $RB = AB / 3$. Since $\triangle ABC$ is an equilateral triangle, $\triangle ARQ$, $\triangle BPR$, $\triangle CQP$ are all right triangles, and the two acute angles are $30^{\circ}$ and $60^{\circ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,035 |
$M$ is a set of 1985 different natural numbers, each of whose prime factors is less than 26. Prove that it is always possible to select four different numbers from $M$ such that their product equals the fourth power of a natural number. | Proof: The prime numbers less than 26 are 2, 3, 5, 7, 11, 13, 17, 19, 23. We denote them as \( p_{1}, p_{2}, \cdots, p_{9} \). By the problem's setup, every number \( a \) in \( M \) has the form:
$$
a=p_{1}^{a_{1}} p_{2}^{\alpha_{2}} \cdots p_{\exists}^{a^{0}},
$$
where \( a_{i}, a_{2}, \cdots, a_{i} \) are non-negat... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 703,037 |
Example 1. Prove: A knight must make an even number of moves to return to its starting point on a chessboard, no matter where it starts.
| Prove: A knight's move can be seen as a combination of the following steps:
(1) First, move from a square to its adjacent square;
(2) After rotating $90^{\circ}$, move two squares in a straight line (see Figure 1).
Thus, a knight's move actually covers three squares. By property (1), the initial and final positions ha... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 703,038 |
Example 4. A certain exhibition hall has 24 exhibition rooms, arranged in a "deficient" square shape as shown in Figure 4, with each square representing one room, and the "deficiency" having three doors for entry and exit, with doors connecting adjacent rooms. $1^{\circ}$ Prove that there does not exist a visiting rout... | Proof: $1^{\circ}$ This is an odd grid array with the rook $a_{3}$ removed, leaving the remaining even and odd grids in equal numbers. Since all the neighboring grids of $a_{31}$ are odd grids, it must enter and exit from odd grids, thus the number of passed rooks is always one less than the number of odd grids. Theref... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 703,041 |
Example 5. Draw a straight line on an $n \times n$ chessboard. What is the maximum number of squares this line can pass through? | Solution: First, we examine the case of a $3 \times 3$ grid to design the "best line" (i.e., the line that passes through the most cells).
As shown in Figure 5, it is clear that $l_{3}$ is the "best line," characterized by the two indices of the cells $a_{ij}$ it passes through alternating and increasing. Thus,
the "... | 2n-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,042 |
Quadrilateral $ABCD$ is inscribed in a circle, the incenters of $\triangle BCD$, $\triangle ACD$, $\triangle ABD$, $\triangle ABC$ are denoted as $I_{A}$, $I_{E}$, $I_{C}$, $I_{L}$ respectively. Prove: $I_{A} I_{L} I_{C} I_{D}$ is a rectangle | To draw the figure of the problem smoothly, the following facts can be noted: the incenter of a triangle is the intersection of any two of its angle bisectors; the four triangles in the problem are all inscribed in the same circle, so any interior angle of each triangle is a circumangle, and its angle bisector must pas... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,043 |
Given a sequence of real numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfying:
$$
a_{i-1}+a_{i+1}=2 a . \quad(i=1,2,3, \cdots)
$$
Prove: For any natural number $r$,
$$
\begin{array}{l}
\quad P(x)=a_{0} C \stackrel{0}{0}(1-x)^{n}+a_{1} C \stackrel{1}{n} x(1 \\
-x)^{n-1}+a_{2} C_{n}^{2} x^{2}(1-x)^{n-2}+\cdots \\
+a_{n-1} C... | This question aims to test students' basic knowledge of arithmetic sequences and their ability to apply the binomial theorem, which is a relatively basic question. Placing this question as the first question in the second test is to expect that most students can solve it, stabilize their emotions, and thus consider the... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,047 |
$$
\text { (1) } \begin{aligned}
f(x, y) \equiv & 2 x^{2}+4 x y-y^{2}-20 x-8 y \\
& +32=0 ; \\
\hdashline F(x, y) \equiv & 16 x^{2}+24 x y+9 y^{2}-10 x \\
& -70 y-75=0 .
\end{aligned}
$$ | Solution: (1) Write the equation $4 x+4 k x+4 y$ $-2 k y-20-8 k=0$ according to (2). And according to Proposition 2, find $k_{1}=\frac{1}{2}, k_{2}=-2$, substitute, and get the equations of the axes of symmetry: $2 x+y-8=0$ and $x-2 y+1=0$. Therefore, it is a central conic section. It can be expressed as
$$
\begin{arra... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,048 |
Example. Find the equation of the locus of the midpoints of the parallel chords with slope $m$ of the parabola $y^{\prime \prime}=2 p x$.
保留源文本的换行和格式，翻译结果如下：
Example. Find the equation of the locus of the midpoints of the parallel chords with slope $m$ of the parabola $y^{\prime \prime}=2 p x$. | Solution: According to the proposition, the trajectory is a ray parallel to the $y$-axis. Taking a chord with slope $m$, one of which is $y = mx$, and substituting it into $y^{2} = 2px$, we find the intersection points $O(0,0)$ and $P\left(\frac{2p}{m^{2}}, \frac{2p}{m}\right)$. The midpoint of $OP$ is $\left(\frac{p}{... | y = \frac{p}{m} \quad (x \geqslant \frac{p}{2m}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,049 |
Proposition. Draw two tangents $P A$ and $P B$ from a point $P$ outside an ellipse, with $A$ and $B$ being the points of tangency. Any line through $P$ intersects the ellipse at $Q$ and $R$, and intersects the chord $A B$ at $C$, then
$$
\frac{1}{P Q}+\frac{1}{P R}=\frac{2}{P C} .
$$ | Prove: Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The coordinates of point $P$ are $\left(x_{0}, y_{0}\right)$, then the equation of the chord of contact $A B$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{-}}=1$. Suppose the equation of the secant line through $P$ is
$$
\left\{\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,052 |
Example 1. Solve the equation $\sqrt{x^{4}+2 x+5}$
$$
+\sqrt{x^{2}+2 x+37}=10
$$ | $+\sqrt{(x+1)^{-}}+36=10$ , on the plane Cartesian coordinate system, the above equation indicates that the sum of the distances from point $P(x, 0)$ to points $F(-1, 2)$ and $F_{2}(-1,6)$ is 10. Therefore, $P$ lies on the ellipse $(a=5; c=2$, with the center at $(-1,4)$ )
$$
\frac{(x+1)^{2}}{21}+\frac{(y-4)^{2}}{25}=1... | x=-1 \pm 3 \sqrt{21} / 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,053 |
Example 2. Solve the inequality $|x-2|+|x+1|>5$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let $P(x, 0)$ be a point on the 2-stretch, then the points satisfying the original inequality must be outside the ellipse
$$
\frac{\left(x-\frac{1}{2}\right)^{2}}{25 / 4}+\frac{y^{2}}{4}=1
$$
Let $y=0$, solving yields $x_{1}=-2, x_{2}=3$, hence the solution to the original inequality is $(-\infty,-2) \cup(3,... | (-\infty,-2) \cup(3,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 703,054 |
Example 1, find the equation of the locus of the endpoints of the minor axis of an ellipse with $O$ as the focus and line $l$ as the directrix.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: As shown in Figure 1,
Take $O$ as the pole, the line from $O$ to
$l$ and its extension as the
polar axis, the distance from $O$ to $l$ as $p$, with the eccentricity $e$ as the parameter, $e=\frac{i}{a} \cos \theta$,
$$
\rho=\frac{e p}{1-e \cos \theta}
$$
Thus $\left\{\begin{array}{l}i=\frac{e p}{1-e \cos \th... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,055 |
Example 2. Let $A(a, 0), B(0, b)$ be known points, find the moving point $P(x, y)$ that satisfies $S_{\triangle O A P}=2 \widetilde{B}_{\triangle O P B}$. | Solution: As shown in Figure 2, draw $P P^{\prime} \perp o x$ axis at $P^{\prime}$, and $P P^{\prime \prime} \perp o y$ axis at $P^{\prime \prime}$. Take $S_{\triangle O P A}=l$ as the parameter, then
$$
\begin{array}{l}
|y|=\left|P P^{\prime}\right|=\frac{2 S_{\triangle O A P}}{|O A|}=\frac{2 t}{|a|}, \\
|x|=\left|P P... | y= \pm \frac{2 b}{a} x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,056 |
Example 1. Suppose the moving point $P(x, y)$ is equidistant from the fixed line $x=$ - $\frac{p}{2}$ and the point $F\left(\frac{p}{2}, 0\right)$, find the parametric equation of the trajectory of point $P$. | Solution: Move the origin of coordinates to point $F$, and then rotate the axes by an angle $\theta$, so that the new horizontal axis passes through $P$ (Figure 1). Let the new coordinates of point $P$ be $\left(x^{\prime}, 0\right)$, then
$$
\left\{\begin{array}{l}
x=\frac{p}{2}+x^{\prime} \cos \theta, \\
y=x^{\prime}... | x=\frac{p}{2} \operatorname{ctg}^{2} \frac{\theta}{2}, \quad y=p \operatorname{ctg} \frac{\theta}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,057 |
Given an acute triangle $A B C$ with the circumradius $R$, points $D, E, F$ are on sides $B C, C A, A B$ respectively.
Prove that $A D, B E, C F$ are the three altitudes of $\triangle A B C$ if and only if
$$
S=\frac{R}{2}(E F+F D+D E) .
$$
where $S$ is the area of triangle $A B C$. | The proof method of this problem is diverse, geometric and trigonometric methods can both be used. Since $\triangle ABC$ is an acute triangle, its circumcenter $O$ must be inside $\triangle ABC$. Therefore, there must be:
$S_{\triangle ABC}=S$ quadrilateral $OBAP + S$ quadrilateral $Op_{BD}$
$+S$ quadrilateral $ODCE$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,058 |
Example 2. Let point $P(x, y)$ be such that the sum of its distances to two foci $F(-c, 0)$ and $\pi(c, 0)$ is a constant $2a (a > 0)$. Find the parametric equations of the locus of point $P$. | Solution: Connect $P F_{1}, P F_{2}$, and draw $P S \perp F_{1} F_{2}$ at $S$ (Figure 2), forming the right triangle $\triangle P F_{2} S$. Let $\angle P F_{2} S = \theta$ be the parameter, then $\left\{F_{z}\right.$ !
$=c-x$, and $\left|F_{2} S\right|$
$=\left|P F_{2}\right| \cos \theta$,
$\therefore c-x=(a-e x) \cos ... | x=\frac{a(c-a \cos \theta)}{a-c \cos \theta}, y=\frac{b^{2} \sin \theta}{a-c \cos \theta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,059 |
Question: Let the two endpoints $A$ and $B$ of a line segment $AB$ of length $l (l \geqslant 1)$ move on the parabola $y=x^{2}$, and let $M(x, y)$ be the midpoint of $AB$. Find the coordinates of point $M$ when it is closest to the $x$-axis. | Solution: Let the focus of the parabola $y=x^{2}$ be $F\left(0, \frac{1}{4}\right)$,
and the directrix $l: y=-\frac{1}{4}$.
To find the coordinates of point $M$ when it is closest to the $x$-axis, this is equivalent to finding the coordinates of point $M$ when it is closest to the directrix $l$.
Construct $Y_{1} \per... | \left( \pm \frac{\sqrt{2 l-1}}{2}, \frac{2 l-1}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,060 |
1. Is there a natural number $n$ such that the sum of the digits of $n^{2}$ equals 1983? Equals 1984? | Solution: The remainder of $n$ divided by 9 may be $0,1,2,3$, $4,5,6,7,8$, so the remainder of $n^{2}$ divided by 9 may be $0,1,4,7$. Therefore, the remainder of the sum of the digits of $n^{2}$ divided by 9 can only be $0,1,4,7$.
Since 1983 divided by 9 leaves a remainder of 3, the product of the digits of $n^{2}$ ca... | 1984 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,061 |
3 . Find natural numbers $m, n$, such that $m^{m}+(m n)^{n}$ $=1984$. | Solution: When $m=4, n=3$,
$$
\begin{array}{l}
m^{m}+(m n)^{n}=4^{4}+12^{3} \\
=256+1728=1984 .
\end{array}
$$
And $5^{5}=3125>1984$, so $m \leqslant 4, n \leqslant 4$. When $n=4$, since
$$
1984>(4 m)=256 m^{4},
$$
$m=1$, but
$$
1^{1}+(1 \cdot 4)^{4}-1984 \text {. }
$$
Thus, $m \leq 4, n \leqslant 3$.
But if $m<4$ o... | m=4, n=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,063 |
4 . Draw a rectangular steel plate of size $73 \times 19$ on paper with a pencil, how can you find the center of the drawn rectangle using only this steel plate and a pencil? | Solution: Let the rectangle be $ABCD$. First, we find the midpoint of $AB$. For this, take a point $O$ inside the rectangle not too far from $AB$, and draw $OA, OB$. Construct a line parallel to $AB$ intersecting $OA, OB$ at $A', B'$ respectively. Draw $A'B, AB'$, and let them intersect at $O'$. Draw a line through $O,... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,064 |
2. Find the roots of $x^{8}+x^{7}+1$ | \begin{array}{l}\text { Solution: } x^{8}+x^{7}+1 . \\ =\left(x^{2}+x+1\right)\left(x^{6}-x^{4}+x^{3}-x+1\right) .\end{array} | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,066 |
4. Among any 10 numbers chosen from the natural numbers not exceeding 91, prove that there must be two numbers whose ratio lies within the interval $\left[\frac{2}{3}, \frac{3}{2}\right]$. | $$
\begin{array}{l}
\text { Proof: Divide 1, } 2, \cdots, 91 \text { into 9 groups: } \\
\{1\},\{2,3\},\{4,5,6\}, \\
\{7,8,9,10\}, \\
\{11,12,13,14,15,16\},\{17, \\
18,19,20,21,22,23,24,25\}, \\
\quad\{26,27,28,29,30,31,32,33, \\
34,35,36,37,38,39\}, \\
\{40,41, \cdots, 60\},\{61, \cdots 91\} .
\end{array}
$$
$$
\begin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,068 |
In a Cartesian coordinate system, points whose both vertical and horizontal coordinates are integers are called integer points. Please design a method to color all integer points, each integer point being colored one of white, red, or black, such that:
(1) Each color of points appears infinitely often on lines parallel... | The solution idea is as follows:
Condition (1) is to avoid trivial solutions. If the constructed solution is too special, the solver will find that it does not meet condition (1).
From condition (2), if any three points of different colors $A, B, C$ are three vertices of some parallelogram, then $A, B, C$ must not be ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,069 |
1. If $a \geqslant 0, b \geqslant 0$, prove:
$$
\begin{array}{l}
(a+b)\left(a^{4}+b^{4}\right) \\
\geqslant\left(a^{2}+b^{2}\right)\left(a^{3}+b^{3}\right) .
\end{array}
$$ | Proof:
\[
\begin{aligned}
& (a+b)\left(a^{4}+b^{4}\right) \\
- & \left(a^{2}+b^{2}\right)\left(a^{3}+b^{3}\right) \\
= & a^{4} b+a b^{4}-a^{2} b^{3}-a^{3} b^{2} \\
= & a b(a-b)^{2}(a+b) \geqslant 0
\end{aligned}
\]
Therefore, the original expression holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,070 |
2. Is the statement "The sum of the distances from any point inside a convex quadrilateral to its vertices does not exceed the perimeter of the quadrilateral" correct? | Solution: Incorrect. For example, if we combine two isosceles triangles $A B C$ and $A C D$ to form a convex quadrilateral $A B C D$. If $A B=1 S=A D=10$, and $B C=C D=2$, then the sum of the distances from point $A$ (and points near $A$) to each vertex is greater than the perimeter of quadrilateral $A B C D$. | not found | Geometry | proof | Yes | Yes | cn_contest | false | 703,071 |
3. On two $1982 \times 1983$ grid papers, color the squares with red and black such that each row and each column has an even number of black squares. If these two papers are overlapped, and one black square coincides with a red square, prove that there are at least three other squares that coincide with squares of dif... | Proof: Let the black cell $A$ coincide with the red cell $A'$ on the second sheet of paper. If in the column where 1 is located on the first sheet, the rest of the black cells (an odd number) all coincide with the black cells on the second sheet, then due to the number of black cells in this column on the second sheet ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,072 |
1. Solve the equation: $2^{x^{5}} 4^{x^{4}}+256^{4}=3 \cdot 16^{x^{3}}$. | Solution: When $x>0$,
$$
\begin{array}{l}
\frac{1}{3}\left(2^{x^{5}}+4^{x^{4}}+256^{4}\right) \\
=\frac{1}{3}\left(2^{x^{5}}+2^{2 x^{4}}+2^{32}\right) \\
\geqslant 2^{\left(x^{5}+2 x^{4}+32\right) / 3}
\end{array}
$$
And
$$
\begin{array}{l}
x^{5}+2 x^{4}+32-12 x^{3}=(x-2)^{2} \\
\cdot\left(x^{3}+6 x^{2}+8 x+8\right) \... | x=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,074 |
3. Prove that for any real numbers $a, b$, there always exist $x \in[0,1], y \in[0,1]$, such that
$$
|x y-a x-b y| \geqslant \frac{1}{3} \text {. }
$$
Can the $\frac{1}{3}$ be replaced with $\frac{1}{2}$? Can it be replaced with 0.33334? | Solution: If $|a| \geqslant \frac{1}{3}$, take $x=1, y=0$ then (1) holds. Therefore, we can assume $|a|<\frac{1}{3}$,
$|b|<\frac{1}{3}$, at this time take $x=y=1$, then
$$
\begin{array}{l}
|x y-a x-b y|=|1-a-b| \\
\geqslant 1-\frac{1}{3}-\frac{1}{3}=\frac{1}{3}
\end{array}
$$
In (1), $\frac{1}{3}$ cannot be replaced by... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 703,076 |
$4, A 、 B$ two people start from $C$ point one after another, each moving along a straight line at their own constant speed, find the trajectory of the circumcenter of $\triangle A B C$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result dir... | Let $C$ be the origin, $A$ moves along the $x$-axis (in the positive direction), and $B$ moves along the line $y=(\tan a) x$ (in the positive direction). At $t=0$, $A$ is at $(2,0)$ and $B$ is at the origin. Then at time $t$, the coordinates of $A$ are $\left(v_{1} t+2, 0\right)$, and the coordinates of $B$ are $\left(... | null | Geometry | other | Yes | Yes | cn_contest | false | 703,077 |
Example 1. If $a$ and $b$ are real numbers, and
$$
\sqrt{2 a-1}+|b+1|=0 \text {. }
$$
Find the value of $a^{-2}+b^{-1987}$. | Solution: $\because \sqrt{2 a-1} \geqslant 0,|b+1| \geqslant 0$, according to property 5, we have
$$
\begin{array}{l}
\sqrt{2 a-1}=0 \text { and }|b+1|=0, \\
\therefore a=\frac{1}{2} \text { and } b=-1 \text {. Therefore, } \\
a^{-2}+b^{-1987}=\left(\frac{1}{2}\right)^{-2}+(-1)^{-1987} \\
=4-1=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,080 |
5. Six equal circles $\mathrm{O}, \mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{O}_{4}$, $0_{5}$, on the surface of a sphere with radius 1, $\odot O$ is tangent to the other five circles and $\odot \mathrm{O}_{1}$ is tangent to $\odot \mathrm{O}_{2}$, $\odot \mathrm{O}_{2}$ is tangent to $\odot \mathrm{O}_{3}... | Let the center of the sphere be $K$, and $\odot O$ and $\odot O_{1}$ touch at $A$. The common tangent line through $A$ is $l$. Then, $K A, O A, O_{1} A$ are all in the plane perpendicular to $l$ through $A$. In this plane, $K A=1$, $O A=O_{1} A=r$. By plane geometry (see Fig. (1)), $K O=K O_{1}=\sqrt{1-r}$,
$$
\begin{a... | r=\frac{1}{2} \sqrt{3-\operatorname{ctg}^{2} \frac{\pi}{5}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,082 |
Example 2. If $x, y$ are real numbers, and
$$
y=\frac{\left(1-x^{2}\right)^{\frac{1}{2}}+\left(x^{2}-1\right)^{\frac{1}{6}}}{4 x-5} \text {, }
$$
find the value of $\log _{\frac{1}{7}}(x+y)$. | To make the expression for $y$ meaningful, $x$ must simultaneously satisfy the following three inequalities.
$$
\begin{array}{c}
1-x^{2} \geqslant 0, \\
x^{2}-1 \geqslant 0, \\
4 x-5 \neq 0 . \\
\therefore x=1, \quad \text { thus } y=0 . \\
\therefore \log _{\frac{1}{7}}(x+y)=\log _{\frac{1}{7}} 1=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,083 |
Example 1. Find all positive integer solutions to the equation $14 x+8 y=200$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Dividing each term of the equation by the greatest common divisor of the coefficients 14 and 8, which is 2, we get
$$
7 x+4 y=100 \text {. }
$$
Since $(7,4)=1$, by Theorem 1, the equation
$$
7 x+4 y=1
$$
has integer solutions.
By observation, we can see that
$$
7 \cdot(-1)+4 \cdot(2)=1 \text {, }
$$
which ... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,084 |
Example 2. Find all positive integer solutions to the equation $7 x+19 y=213$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Divide each term of equation (1) $7 x+19 y=213$ by its smallest coefficient 7, and rearrange to get
$$
x=\frac{213-19 y}{7}=30-2 y + \frac{3-5 y}{7}.
$$
Since $x, y$ are integers, $\frac{3-5 y}{7}=u$ should also be an integer. Thus, we have $5 y+7 u=3$. Dividing both sides of this equation by 5, we get
$$
\b... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,085 |
Example 2, find the positive integer solutions $(x, y, z)$ of the equation $\frac{1}{x}+\frac{1}{4}+\frac{1}{z}=1(x \geqslant y \geqslant z)$. | Solution: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Without loss of generality, assume $x \geqslant y \geqslant z \geqslant 1$, then $1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{z}, 1 \leqslant z \leqslant 3$.
Since $z$ is an integer, $z=1, 2, 3$.
Below, we discuss the cases $z=1, 2, 3$, and finally obtain ... | (4, 2), (6, 3, 2), (3, 3, 3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,086 |
Example 4. Given $a_{1}=a, a_{n+1}=a_{n}+b_{n}$ (where $b_{n}$ is given as a function of $n$), try to find the general term $a_{n}$.
| Given: $| a_{n+1}-a_{n}=b_{n} |$, we know that the general term of the difference sequence is $b_{n}$. Therefore, its general term is
$$
a_{n}=a+\sum_{k=1}^{n-1} b_{k}
$$
N. $a_{n+1}=p a_{i}+d$, the difference sequence is a geometric sequence with a common ratio of $p$. | a_{n}=a+\sum_{k=1}^{n-1} b_{k} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,087 |
Example 5. Given $a_{1}=a, a_{n+1}=p a_{n}+d(p, d$ are constants, $p \neq 1, d \neq 0$ ). Try to find the general term $a_{n}$. | Solution: Given $a_{n+1}=p a_{n}+d$,
we have $a_{n}=p a_{n-1}+d_{\text {n }}$
From (1)-(2), we get $a_{n+1}-a_{n}=p\left(a_{n}-a_{n-1}\right)$.
Let $a_{n+1}-a_{n}=b_{n}$, then $b_{n}=p b_{n-1}$. (From the recursive formula $a_{n}-a_{n-1}=b_{n-1}$ )
Since $b_{n}$ is a difference sequence, and $b_{1}=a_{2}-a_{1}$, it is ... | a_{n}=a+\frac{\left(a_{2}-a_{1}\right)\left(p^{n-1}-1\right)}{p-1} \cdot(n \geqslant 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,088 |
Example 6. Given $a_{1}=1, a_{n+1}=2 a_{n}-3$, try to find the general term $a_{n}$ of the sequence. | Solution: Using method (1), from the recurrence relation $a_{n+1}=2 a_{n}-3$, we get
$$
\begin{array}{l}
a_{1}=1, \\
a_{2}=2 \times 1-3=2-3, \\
a_{3}=2^{2}-3 \times 2-3, \\
a_{4}=2^{3}-3 \times 2^{2}-3 \times 2-3, \\
a_{5}=2^{4}-3 \times 2^{3}-3 \times 2^{2}-3 \times 2-3,
\end{array}
$$
By analogy, we get
$$
\begin{ar... | a_{n}=3-2^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,089 |
Theorem 1. If the recurrence relation is $a_{n+1}-a_{n}=f(n)$ $(n=1,2, \cdots)$, then $a_{n}=a_{1}+\sum_{k=1}^{n-1} f(k)$. Here, $f(n)$ is a function of the natural number $n$. | Prove that $\cdots$ by substituting $k$ in the recurrence relation $a_{k+1}-a_{k}=f(k)$ with $1,2, \cdots, n-1$, and adding both sides, we get
$$
a_{n}-a_{1}=f(1)+f(2)+\cdots+f(n-1) \text {. }
$$
Therefore, $a_{n}=a_{1}+\sum_{k=1}^{n-1} f(k)$. | a_{n}=a_{1}+\sum_{k=1}^{n-1} f(k) | Algebra | proof | Yes | Yes | cn_contest | false | 703,090 |
Theorem 2. If the recurrence relation is $a_{n+1}=a_{n} \cdot f(n)(n=1$, $2, \cdots)$, then $a_{n}=a_{1} \prod_{k=1}^{n-1} f(k)$. Here, $f(n)$ is a function of the natural number $n$. | | In the recurrence relation $a_{k+1}=a_{k} \cdot f(k)$, substituting $k$ sequentially with $1,2, \cdots, n-1$, and through multiplication, we get $\boldsymbol{a}_{n}=a_{1} \cdot f(1) \cdot f(2) \cdots f(n-1)=a_{1} \prod_{k=1}^{n-1} f(k)$ | a_{n}=a_{1} \prod_{k=1}^{n-1} f(k) | Algebra | proof | Yes | Yes | cn_contest | false | 703,091 |
Theorem 3. If the recurrence relation is $a_{n+1}=p a_{n}+q(n=1$, $2, \cdots)(p, q$ are constants), then this sequence
when $p=1$, is an arithmetic sequence with the general term $a_{n}=a_{1}+(n-1) q$;
when $q=0$, is a geometric sequence with the general term $a_{n}=a_{1} p^{n-1}$.
When $p \neq 1$, the general term i... | | Proof $\mid$ When $p=1$, due to the recurrence relation $a_{n+1} - a_{n} = q$, this is an arithmetic sequence with a common difference of $q$, therefore
$$
a_{n} = a_{1} + (n-1) q.
$$
When $q=0$, due to the recurrence relation $a_{n+1} = p a_{n}$, this is a geometric sequence with a common ratio of $p$. Therefore,
$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,092 |
Theorem 4. Let the infinite sequence $\{a_n\}$ be defined by $a_{n}=a, a_{n}=p a_{n-1}+q$ $\{n \geqslant 2\}$. Then,
(1) If $|p|<1$, it converges to $\frac{q}{1-p}$;
(2) If $p \neq 1$ and $p a+q=a$, then it converges to $a$. Here, $p, q, a$ are constants. | Prove (1) By Theorem 3, for the case $p \neq 1$, we have
$$
a_{n}=\frac{q}{1-p}+\left(a-\frac{q}{1-p}\right) p^{n-1} \text {. }
$$
However, when $|p|<1$ and $n \rightarrow \infty$, we have $p^{n-1} \rightarrow 0$,
thus $a_{n} \rightarrow \frac{q}{1-\bar{p}}$.
(2) From $p \neq 1, p a+q$, we have
$$
a-\frac{q}{1-p}=0 \t... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,093 |
Example $1^{\bullet}$. Consider the sequence $\left\{a_{n}\right\}$ defined inductively by $a_{1}=1, a_{n+1}=\frac{1}{2} a_{n}+1$ $(n=1,2,3, \cdots)$. Try to find the general term of the sequence $\left\{a_{n}\right\}$. | Solution: Given $a_{n+1}=\frac{1}{2} a_{4}+1$, we have
$$
a_{n}=\frac{1}{2} a_{n-1}+1,
$$
Thus, $a_{n+1}-a_{n}=\left(\frac{1}{2} a_{n}+1\right)$
$$
\begin{array}{l}
-\left(\frac{1}{2} a_{n-1}+1\right) \\
=\frac{1}{2}\left(a_{n}-a_{n-1}\right)
\end{array}
$$
Let $a_{n+1}-a_{n}=b_{n} ， \quad\left(b_{n}\right.$ be the $... | a_{n}=2-\left(\frac{1}{2}\right)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,094 |
Example 2. Consider the sequence $\left\{a_{n}\right\}$ defined by $a_{1}=2, a_{n+1}=\frac{a_{n}}{a_{n}+3}$ $(n \geqslant 1)$. Solve the following problems:
(1) If $b_{n}=\frac{1}{a_{n}}$, find the relationship between $b_{n+1}$ and $b_{n}$;
(2) Find the general term of the sequence $\left\{a_{n}\right\}$. | Solution: (1) From the given, we have
$$
a_{n}=-\frac{1}{b_{n}}, \quad a_{n+1}=\frac{1}{b_{n+1}} .
$$
Substituting the above into the known equation
$$
a_{n+1}=\frac{a_{n}}{a_{n}+3}
$$
we get
$$
3_{n} \frac{1}{b_{n}+3}=\frac{\frac{1}{b_{n}}}{b_{n}} .
$$
From this, we obtain $b_{n+1}=3 b_{n}+1$.
(2) From the result i... | a_{n}=\frac{2}{2 \cdot 3^{n-1}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,095 |
一、（1）Does there exist 14 consecutive positive integers, each of which is divisible by at least one prime number not less than 2 and not greater than 11?
（2）Does there exist 21 consecutive positive integers, each of which is divisible by at least one prime number not less than 2 and not greater than 13? | (1) It does not exist. Below, we prove this by contradiction: Let these 14 consecutive positive integers be \(N, N+1, N+2, \cdots, N+13\). By symmetry, without loss of generality, assume \(N\) is even, so \(N, N+2, N+4, \cdots, N+12\) are all divisible by 2. It is known that among the remaining seven odd numbers \(N+1,... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,096 |
Example 3. Given that $x, y, z$ are all integers greater than 1, and $(x, y)=1$, find the integer solution $(x, y)$ for the equation $z=(0 y+3 x)\left(\frac{1}{y}-\frac{1}{x}\right)$. | $$
\begin{aligned}
n_{1}^{9}: z & =(6 y+5 x)\left(\frac{1}{y}-\frac{1}{x}\right) \\
& =(x-y) \cdot \frac{5 x+6 y}{x y} .
\end{aligned}
$$
Since $z$ is an integer, either $x-y$ or $5 x+6 y$ must be divisible by $x y$.
If $x-y$ and $5 x+6 y$ have a common divisor greater than 1, let one of their prime factors be $p$, t... | (6, 5, 2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,097 |
Three, find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.
(Note: The square mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{i}$ is defined as $\left.\left[\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right]^{\frac{1}{2}}\right)$ | $$
\begin{array}{c}
\text { Three, let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \\
=\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \text { be a }
\end{array}
$$
perfect square $m^{2}$, then $6 m^{2}=(n+1)(2 n+1)$. Since $6 m^{2}$ is even, it must be odd. We can set $n=3 p+q$, where $q=-1, 1$ or 3.
If $q=2$, then $6 m^{... | 337 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,099 |
Four, two circles $k_{1}$ and $k_{2}$ intersect at the diameter $A B$ of circle $k_{1}$, and point $P$ is a fixed point on circle $k_{2}$ inside circle $k_{1}$. Using a "T-square" (a tool that can draw a straight line through two points and a perpendicular line through a point on or off a line), construct points $C$ an... | Four, Analysis: Let $C, D$ be the points to be constructed, and $CD$ intersects $AB$ at $E$. Connect $PE$ and extend it to intersect $K_2$ at $F$.
$\left.\begin{array}{l}\text { Then } P E \cdot E F=A E \cdot E B=C E \cdot E D \\ \text { Also } P E=\frac{1}{2} C D=C E=E D\end{array}\right\}$ $\Rightarrow P E=E F$.
Let... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,100 |
For any positive integer $n \geqslant 1$, a partition $\pi$ refers to dividing $n$ into the sum of one or several positive integers, arranged in non-increasing order (for example, for $n=4$, the partitions $\pi$ are $1+1+1+1, 1+1+2, 1+3, 2+2$, and 4). For any partition $\pi$, define $A(\pi)$ as the number of times 1 ap... | Let $P(n)$ denote the number of partitions of a positive integer $n$, and let $P(0)=1$.
We first prove: For any positive integer $n$, the sum of $A(\pi)$ for all partitions $\pi$ of $n$ is
$$
P(n-1)+P(n-2)+\cdots+P(0)
$$
The sum of $A(\pi)$ is calculated as follows: For each partition that contains at least one "1", ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 703,101 |
Let $n$ be a natural number. Prove that the equation
$$
z^{n+1}-z^{n}-1=0
$$
has a complex root of modulus 1 if and only if $n+2$ is divisible by 6. | Prove necessity. Let $w$ be a complex root of the equation $z^{n+1}-z^{n}-1=0$ with modulus 1. Then $w^{n+1}-w^{n}-1=0$. Therefore, $w^{n}(w-1)=1$, which implies $|w|^{2}|w-1|=1$. Since $|w|=1$, it follows that $|w-1|=1$. Given that $|w| = 1$, $w$ lies on the unit circle in the complex plane. The points on the unit cir... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,102 |
ii. For an equilateral triangle $ABC$ with side length 1, each side is divided into $n$ equal segments, and lines parallel to the other two sides are drawn through each division point, dividing the triangle into smaller triangles. The vertices of these small triangles are called nodes. A real number is placed at each n... | (1) Consider a trapezoid composed of three small triangles with common sides, as shown in the trapezoid $A_{1} A_{2} A_{4} A_{5}$ in Figure 2 (Figure 2 is the case when $n=5$). Here, $A_{3}$ is the midpoint of $A_{1} A_{5}$. Suppose $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are placed on $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, ... | S=\frac{1}{6} n(n+1)(a+b+c) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,103 |
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