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Three. (20 points) 1992 digits are written around a circle. It is known that if we start from a certain position and read these digits in a clockwise direction, the resulting 1992-digit number is divisible by 27. Prove that no matter which position we start from, reading the digits in a clockwise direction will always ... | Three, Proof: Assuming the number $A=a_{1} a_{2} a_{\cdots} \cdots a_{n-1} a_{n}(n=$ 1992) is divisible by 27, we will prove that the number $B=a_{2} a_{1} \cdots a_{n-1}$ is also divisible by 27.
Since $10 B+a_{n}=a_{n} \cdot 10^{n+1}+A$
$$
=a_{n} \cdot \underbrace{99 \cdots 9}_{a \uparrow}+A+a_{n},
$$
When $n=1992$,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,266 |
1. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the opposite angles of $a, b, c$ are $\angle A, \angle B, \angle C$ respectively. Then the range of $\sin B+\cos B$ is ( ).
(A) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(B) $\{, \sqrt{2}]$
(C) $\left(1,1+\frac{\sqrt{3}}{... | 1. Solution: Since $a, b, c$ form a geometric sequence, we have $b^{2} = ac$.
Thus, $b^{2} = ac = a^{2} + c^{2} - 2ac \cos B$
$$
\begin{aligned}
& \geqslant 2ac - 2ac \cos B, \\
\cos B & \geqslant \frac{1}{2}.
\end{aligned}
$$
Therefore, $0 < B \leqslant 60^{\circ}$.
From $\frac{1}{2} \leqslant \cos B < 1$ and $0 < \s... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,267 |
2. In the dihedral angle $P-a-Q$ of $120^{\circ}$, there are points $A$ and $B$ on the planes $P$ and $Q$ respectively. It is known that the distances from points $A$ and $B$ to the edge $a$ are 2 and 4, respectively, and the line segment $AB = 10$, and the angle between $AB$ and plane $Q$ is $\alpha$, then ( ).
(A) $\... | 2. Draw $B D^{\prime} / / C D$ through $B$, and draw $C B^{\prime} / / B D$ through $C$ intersecting $B B^{\prime}$ at $B^{\prime}$. Then the plane $A B^{\prime} C$. 1's face $Q$.
Draw..
$A F$ perpendicular
line $B^{\prime} C$ at
$E$, then $A E \perp$
plane $Q$. So, $\angle A B E \angle$
is the angle $\alpha$ formed b... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,268 |
4. From $\{1,2,3,4, \cdots, 20\}$, select four different numbers $a, b, c, d$, satisfying $a+c=b+d$. If the order of $a, b, c, d$ is not considered, then the total number of selection methods is ( . .
(A) 1050
(B) 525
(C) 1140
(D) 190 | 4. Let's assume $a>b>d$. From $a+c=b+d$ we get $d>c$.
Selecting three numbers $a>b>c$ from $1,2, \cdots, 20$ has $C_{20}^{3}$ ways, but among them, those with $a+c=b+b$ are not allowed, which is equivalent to $a$ and $c$ having the same parity. Selecting two odd numbers from $1,2, \cdots, 20$ has $C_{10}^{1}$ ways, an... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,270 |
5. When $a \neq b, c \neq d$, $P=3^{a}+7^{c}, Q=3^{b}+7^{d}, a, b, c, d$ are all natural numbers. Then ().
(A) $P$ and $Q$ must not be equal
(B) Only one set of values for $a, b, c, d$ makes $P=Q$
(C) There are a finite number but more than one set of values for $a, b, c, d$ that make $P=Q$.
(D) There are infinitely ma... | 5. If $3^{c}+7^{c}=3^{s}+7^{d}$, assume $a>\bar{n}$, then it must be that $c<d$.
Thus, $3^{h}\left(3^{\circ-\omega}-1\right)=7^{c}\left(7^{d-\epsilon}-1\right)$.
Since $(3,7)=1,3^{b} \neq 3^{c}$, it must be that $3^{b}=7^{d-c}-1$.
Since the units digit of $3^{b}$ is $3,9,7,1$; and the units digit of $7^{d-c}-1$ is $6,8... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,271 |
Example 1. (Gallacher, 1987, see Figure 4) In $\triangle P_{1} P_{2} P_{3}$, let $p_{i}$ be the side opposite vertex $P_{i}$, and $s_{i}$ be parallel to $p_{i}$ but not coincident with $p_{i}$. Let $s_{i}$ intersect $p_{i-1}$ at $Q_{i}$, and $Q_{i}$ divides $p_{i} p_{i+1}$ into two parts, with a ratio of 3. Prove: The ... | Prove that if $P_{1}, P_{2}, P_{3}, Q_{1}, Q_{2}$ and $Q_{3}$ are replaced by $A_{1}, A_{2}, A_{3}, C_{3}, C_{1}$ and $C_{2}$ respectively, then $\lambda_{1}, \lambda_{2}, \lambda_{3}$ will be equal to $c_{3}$, $c_{1}, c_{2}$ in equation (*). Taking $B_{1}, B_{2}, B_{3}$ as the ideal points of the lines $A_{2} A_{3}, A... | c_{1} c_{2} c_{3}-\left(c_{1}+c_{2}+c_{3}\right)=2 | Geometry | proof | Yes | Yes | cn_contest | false | 706,274 |
2. If $3 f(x-1993)+4 f(1993-x)$ $=5(x-1993)$, for all real numbers $x$, then the analytical expression of $f(x)$ is $\qquad$. | 2. Solution Let $x-1993=t$. Then $x=1993+t$.
Thus, we have $3 f(t)+4 f(-t)=5 t$,
and $3 f(-t)+4 f(t)=-5 t$.
Solving (1) and (2) gives $f(t)=\frac{7}{5} t$.
That is, $f(x)=\frac{7}{5} x$. | f(x)=\frac{7}{5} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,275 |
3. Given $M=2^{7} t=3^{5} s$, where $t$ is odd, and $s$ cannot be divided by 3. Then the sum of all divisors of $M$ in the form of $2^{P} 3^{q}$ is $\qquad$ . | 3. The sum of all divisors of $M$ is
$$
\begin{array}{l}
\left(1+2+\cdots+2^{7}\right)\left(1+3+\cdots+3^{5}\right) \\
=92820 .
\end{array}
$$ | 92820 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,276 |
5. Given that $x_{1}, x_{2}, \cdots, x_{57}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{57}=100$. Then the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{57}^{2}$ is $\qquad$ | 5. Solution For natural numbers $x_{i}, x_{j}$, if $x_{i}x_{i}^{2}+x_{j}^{2} .
\end{array}
$$
Thus, when the sum $x_{i}+x_{j}$ remains constant, the value is maximized by setting one of the numbers to 1 and the other to the sum minus 1, leading to
$$
\text { to } x_{1}=x_{2}=\cdots=x_{56}=1, x_{57}=44 \text {, when } ... | 1992 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,278 |
6. If the parabola $C_{m}: y=x^{2}-m x+m+1$ intersects the line segment $A B$ (where $A(0,4), B(4,0)$) at exactly two points, then the range of values for $m$ is $\qquad$ | 6. The equation of line $AB$ is $x+y=4(0 \leqslant x \leqslant 4)$. Substituting $y=4-x$ into $y=x^{2}-m x+m+1$ yields
$x^{2}-(m-1) x+(m-3)=0$.
Let $f(x)=x^{2}-(m-1) x+m-3$.
This problem is equivalent to the equation (1) having two distinct real roots $x_{1}, x_{2}$, and $0 \leqslant x_{1}0, \\
0 \leqslant \frac{m-1}{2... | 3 \leqslant m \leqslant \frac{17}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,279 |
Three. (This question is worth 20 points) In the regular quadrilateral pyramid $P$ $A B C D$, $A B=3, O$ is the projection of $P$ on the base, $P O=6, Q$ is a moving point on $A O$, and the section passing through point $Q$ and parallel to $P A, B D$ is a pentagon $D F G H L$, with the area of the section being $S$. Fi... | $$
\begin{array}{l}
\text { Three, Solution: As shown in the figure, } \because P A / / \text { plane } D F G H L, \\
\therefore P A / / E F, P A / / H L, P A / / Q G . \\
\text { Also, } \because B D / / \text { plane } D F G H L, \\
\therefore B D / / E L, B D / / F H . \\
\text { Therefore, } \frac{E F}{P A}=\frac{B... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,280 |
Four. (This question is worth 20 points) Given the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ where all terms are non-negative real numbers, and $a_{n}^{2}-a_{n}+a_{n+1} \leqslant 0$. Prove that for all natural numbers $n$ not less than 2, $a_{n} \leqslant \frac{1}{n+2}$. | (1) For $n=2$, by the given,
$$
\begin{aligned}
a_{2} & \leqslant a_{1}-a_{1}^{2}=-\left(a_{1}-\frac{1}{2}\right)^{2}+\frac{1}{4} \\
& \leqslant \frac{1}{4}=\frac{1}{2+2} .
\end{aligned}
$$
Thus, the inequality holds for $n=2$.
(2) Assume that for a natural number $k(k \geqslant 2)$, the inequality $a_{k} \leqslant \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,281 |
Five. (This question is worth 20 points) Given $\triangle A B C, \angle B \leqslant 30^{\circ}, B C=\sqrt{3}$, a circle centered at $B$ with radius 1 intersects with a circle centered at $A$ with radius $r(r \geqslant 1)$. Prove: vertex $B, C$ is at least one point inside or on the circle $\odot A$.
---
The translati... | Let $AB = c, AC = b$.
If $c \leqslant r$, then $B$ is inside or on the boundary of the circle centered at $A$ with radius $r$.
If $B$ is not inside or on the boundary of $\odot A$, then $c > r$.
Since $\odot A$ and $\odot B$ intersect, then $c \leqslant r + 1$.
Thus, $r < c \leqslant r + 1$.
By the cosine rule, $b^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,282 |
I. (This question is worth 35 points) Let the centroid of $\triangle ABC$ be $G$, and the extensions of $AG, BG, CG$ intersect the sides $BC$, $CA$, $AB$ at $D, E, F$, and intersect the circumcircle of $\triangle ABC$ at $A', B', C'$.
Prove: $\frac{A' D}{D A}+\frac{B' E}{E B}+\frac{C' F}{F C} \geqslant 1$. | Let $B C=a, C A=b, A B=c$. By the intersecting chords theorem, we have
$$
A^{\prime} D \cdot D A=B D \cdot D C=\frac{1}{4} a^{2} \text {. }
$$
Similarly, $B^{\prime} E \cdot E B$
$$
\begin{array}{l}
=\frac{1}{4} b^{2}, C^{\prime} F \cdot F C \\
=\frac{1}{4} c^{2} .
\end{array}
$$
Also, by the median length formula,
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,283 |
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II. (This question is worth 35 points) Let the set $Z=\left\{z_{1}\right.$, $\left.z_{2}, \cdots, z_{n}\right\}$ satisfy the inequality
$$
\min _{i \neq j}\left|z_{i}-z_{j}\right| \geqslant \max _{i}\left|z_{i}\right| \text {. }
$$
Find the largest $n$, and for this $n$, find al... | Let $\left|z_{m}\right|=\max \left|z_{i}\right|$.
Thus, all points $z_{i}$ on the plane are distributed within a circle of radius $\left|z_{m}\right|=R$ centered at $z_{0}=0$.
It can be seen that on the circumference $R$, the six vertices $z,(j=1,2,3,4,5,6)$ of any inscribed regular hexagon, along with $z_{0}=0$, sati... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,284 |
Example I. As shown in the figure, $M$ is the midpoint of $BC$, $MN \perp BC$. $I$ is the midpoint of $MN$. Let $\angle BAC = \alpha$. Prove: $\sin \alpha = 3 \sin (\angle C - \angle B)$. (1984, Junior High School Competition in Some Cities of Fujian Province) | Prove that if $I C$ is connected, then
$$
\begin{array}{l}
I C=I B, \angle I C A= \\
\angle C-\angle B .
\end{array}
$$
In $\triangle A I C$, by the Law of Sines, we get
$$
\begin{array}{l}
\frac{\sin \angle C A B}{\sin \angle I C A} . \\
=\frac{\sin \alpha}{\sin (\angle C-\angle B)}=\frac{I C}{A I} .
\end{array}
$$
... | \sin \alpha=3 \sin (\angle C-\angle B) | Geometry | proof | Yes | Yes | cn_contest | false | 706,285 |
Example 2. Let $l$ be a line passing through point $C$ and parallel to side $AB$ of $\triangle ABC$. The internal angle bisector of $\angle A$ intersects side $BC$ at $D$ and line $l$ at $E$; the internal angle bisector of $\angle B$ intersects side $AC$ at $F$ and line $l$ at $G$. If $GF=DE$, prove that $AC=BC$. (31st... | Let $B C=a, C A=b, A B=c, \angle A=2 \alpha, \angle B=2 \beta$.
Since $B F$ bisects $\angle B$, then
\[
\frac{C F}{F A}=\frac{C F}{A C-C F}=\frac{C F}{b}-C F=\frac{a}{c},
\]
\[
\therefore \quad C F=\frac{a b}{a+c}.
\]
Similarly, we get $C D=\frac{a b}{b+c}$.
Since $1 / / A B$,
\[
\therefore \angle C G F=\beta, \angle F... | AC=BC | Geometry | proof | Yes | Yes | cn_contest | false | 706,286 |
Example 5. (1985, Austria and Poland Joint Mathematical Competition) Prove: The perimeter and the sum of the lengths of the two diagonals of any convex quadrilateral with an area of 1 is not less than $4+\sqrt{8}$.
Analysis: First, consider two special cases: "a square and a rhombus with an area of 1. In the square, t... | Proof: Let $ABCD$ be any convex quadrilateral with an area of 1 (as shown in the figure). Then we have
$$
\begin{array}{l}
1=\frac{1}{2}(eg+ \\
gf+fh+he) \sin \alpha \\
\leqslant \frac{1}{2}(eg+gf+fh+he) \\
= \frac{1}{2}(e+f)(g+h) \\
\leqslant \frac{1}{2}\left(\frac{e+f+g+h}{2}\right)^{2}, \quad
\end{array}
$$
which ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,287 |
Example 6. Prove: A triangle with area I cannot be covered by a parallelogram with area less than 2.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 6. Prove: A triangle with area I cannot be covered by a parallelogram with area less than 2. | - Analysis: The conclusion of the original proposition is presented in a negative form, making it difficult to tackle directly. However, its equivalent proposition, the contrapositive, “If $\triangle P Q R$ is inside $\triangle A B C D$, then $S_{\triangle T Q R} \leqslant \frac{1}{2} S_{C I A B C D} ”$ is easier to pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,288 |
Example 7. Given positive numbers, $a, b, c, d$. Prove:
$$
\begin{array}{l}
\frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{b^{3}+c^{3}+d^{3}}{b+c+d}+\frac{c^{3}+d^{3}+a^{3}}{c+d+a} \\
+\frac{d^{3}+a^{3}+b^{3}}{d+a+b} \geqslant a^{2}+b^{2}+c^{2}+d^{2}
\end{array}
$$ | To prove that due to the symmetry of the problem, it is sufficient to prove for all positive numbers $x, y, z$,
$$
\frac{x^{3}+y^{3}+z^{3}}{x+y+z} \geqslant \frac{x^{2}+y^{2}+z^{2}}{3}.
$$
If inequality (*) holds, then the original inequality is no less than
$$
\begin{array}{l}
\frac{a^{2}+b^{2}+c^{2}}{3}+\frac{b^{2}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,289 |
Example 1. Prove: The distance from a vertex of a triangle to the orthocenter is twice the distance from the circumcenter to the opposite side of this vertex.
| Prove that as shown in the figure, according to the problem, we need to prove $A H = 2 O A_{1}$.
Draw the straight line $A O K$, and connect $B K, C K$.
It can be proven that $C H / / K B$, $B H / / K C$.
Therefore, $H B K C$ is a parallelogram, so $A_{1}$ is the midpoint of both $B C$ and $H K$.
In $\triangle A H ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,290 |
Example 2. The orthocenter $H$, centroid $G$, and circumcenter $O$ of $\triangle A B C$ are collinear, and $G H=2 O G$.
| Prove that, as shown in the figure, $A A_{1}$ is a median, and $O A_{1} \perp B C$,
From Example 1, we know that $A H=2 O A_{1}$.
Let $G$ be the intersection of $A A_{1}$ and $O H$. Since $A H$ and $O A_{1}$ are both perpendicular to $B C$, they are parallel to each other. Therefore,
$\triangle A G H \backsim \triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,291 |
Example 3. Let $I, O$ be the incenter and circumcenter of $\triangle A B C$, respectively, and let $A K$ be a chord of the circumcircle of $\triangle A B C$ passing through point $I$. Then $O I^{2}$ $=R^{2}-2 r R$. Here, $r, R$ are the radii of the incircle and circumcircle of $\triangle A B C$, respectively. | Prove: As shown in the figure, draw the diameter $KOE$ of the circumcircle of $\triangle ABC$, and connect $BK, BE$. In the right triangle $\triangle EBK$, $\angle KEB = \frac{1}{2} \angle A, KE = 2R$, then
$$
\begin{aligned}
BK = IK \\
= 2R \sin \frac{A}{2}.
\end{aligned}
$$
In the right triangle $\triangle AIF$, $IF... | OI^2 = R^2 - 2rR | Geometry | proof | Yes | Yes | cn_contest | false | 706,292 |
Example 4. Let the opposite edges of tetrahedron $ABCD$ be pairwise perpendicular. Try to prove: the midpoints of the six edges of tetrahedron $ABCD$ lie on the same sphere. | Proof: As shown in the figure, let the midpoints of $AB, BC, CD, DA, AC$, and $BD$ be $E, F, G, H, K, L$ respectively, and $AC \perp BD, AB \perp CD, AD \perp BC$.
Since $EF \parallel AC, GH \parallel AC$, we have $EF \parallel GH$.
Similarly,
$EH \parallel \tilde{F} G$.
Therefore, $EFGH$ is a parallelogram.
$$
\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,293 |
Example 5. Let $G$ be the centroid of $\triangle A B C$, and $M$ be any point. Then
$$
\begin{array}{l}
M A^{2}+M B^{2}+M C^{2} \\
=3 M G^{2}+\frac{1}{3} \cdot\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$
where $a, b, c$ are the lengths of the three sides of $\triangle A B C$. | Proof As shown in the figure, establish a rectangular coordinate system, with $A$ $(a, b), B(-c, 0), C(c, 0)$. Let $M(x, y)$. Then $G\left(\frac{a}{3}, \frac{b}{3}\right)$.
We have
$$
\begin{array}{l}
M A^{2}+M B^{2} \\
+M C^{2} \\
=3 x^{2}+3 y^{2} \\
-2 a x-2 b y+a^{2} \\
+b^{2}+2 c^{2} \\
=G A^{2}+G B^{2}+G C^{2}+3 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,294 |
Example 6. Let $O, G$ be the circumcenter and centroid of $\triangle A B C$ respectively, and let point $M$ satisfy $-\frac{\pi}{2}<O M G<\pi$. $A_{1}, B_{1}, C_{1}$ are the intersection points of lines $A M, B M, C M$ with the circumcircle of $\triangle A B C$. Prove:
$$
A M+B M+C M \leqslant A_{1} M+B_{1} M+C_{1} M .... | Prove that as shown in the figure, we have
$$
\begin{array}{l}
M A \cdot M A_{1}=M B \cdot M B_{1}=M C \cdot M C_{1} \\
=M E \cdot M E_{1}=R^{2}-O M^{2} .
\end{array}
$$
Here, $E$ and $E_{1}$ are the two intersection points of the line $O M$ with the circumcircle of $\triangle A B C$.
Let
$$
\begin{array}{l}
M A \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,295 |
Example 3. As shown in the figure, in $\triangle ABC$, $AB > AC$, the external angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at point $E$, and a perpendicular line $EF$ is drawn from $E$ to $AB$, with the foot of the perpendicular being $F$.
Prove: $2 AF = AB - AC$. (1989, National High S... | Proof: Let the three angles of $\triangle ABC$ be $A, B, C$, and the radius of its circumcircle be $R$. Connecting $E$ and $B$, by the Law of Sines, we have
$$
A E=2 R \sin C, A C=2 R \sin B,
$$
then
$$
\begin{aligned}
\triangle A B - A C & =2 R \quad(\sin C - \sin B) \\
& =4 R \cos \frac{C+B}{2} \sin \frac{C-B}{2} .
... | 2 AF = AB - AC | Geometry | proof | Yes | Yes | cn_contest | false | 706,297 |
2. Given $0<x<1$. Simplify
$$
=\quad \sqrt{\left(x-\frac{1}{x}\right)^{2}+4}-\sqrt{\left(x+\frac{1}{x}\right)^{2}-4}
$$ | 2. $2 x$.
The above text has been translated into English, maintaining the original text's line breaks and format. | 2x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,306 |
3. Let the side length of square $A B C D$ be $5, P, Q$ be points on $B C, C D$ respectively, and $\triangle A P Q$ be an equilateral triangle. Then, the side length of this equilateral triangle is $\qquad$ $\therefore$. | 3. $5(\sqrt{6}-\sqrt{2})$.
The above text has been translated into English, retaining the original text's line breaks and format. | 5(\sqrt{6}-\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,307 |
Example 4. Three circles with radius $R$ have a common point. Prove that if these circles intersect each other at three additional points, then the radius of the circle passing through these three points is also $R$. (1986, 12th All-Russian Mathematical Olympiad) | Proof: As shown in the figure, let the radii of $\odot O_{1}, \odot O_{2}, \odot O_{3}$ be $R$, and their common intersection point be $D$. They intersect pairwise at points $A$, $B$, $C$. Let the radius of the circle passing through points $A$, $B$, $C$ be $R^{\prime}$. It is easy to prove that $\triangle A O_{1} D \c... | R^{\prime} = R | Geometry | proof | Yes | Yes | cn_contest | false | 706,308 |
II. (20 points) Let $M$ be a point inside $\triangle ABC$ such that $\angle BMC=90^{\circ}+\frac{1}{2} \angle BAC$. The line $AM$ passes through the circumcenter $O$ of $\triangle BMC$. Prove that point $M$ is the incenter of $\triangle ABC$.
Translate the above text into English, please retain the original text's lin... | $$
\begin{array}{l}
\therefore \angle B O C=2 \angle B P C \\
=2\left(180^{\circ}-\angle B M C\right) \\
= 2\left\{180^{\circ}-\left(90^{\circ}+\alpha\right)\right\} \\
=2\left(90^{\circ}-\alpha\right) \\
=180^{\circ}-2 \alpha . \\
\therefore \angle B A C+\angle B O C=130^{\circ}
\end{array}
$$
Therefore, points A, B... | null | Geometry | proof | Yes | Yes | cn_contest | false | 706,311 |
Three. (20 points) Suppose the equation $x^{2}+a x+1=b$ has two roots that are both natural numbers. Prove: $a^{2}+b^{2}$ is a composite number.
保留源文本的换行和格式，直接输出翻译结果。 | Three, let the equation $x^{2}+a x+(1-b)=0$ have two roots $n_{1}, n_{2} \in N$. Then
$$
\left\{\begin{array}{l}
n_{1}+n_{2}=-a, \\
n_{1} n_{2}=1-b
\end{array}\right.
$$
That is,
$$
\begin{array}{l}
a^{2}=\left(n_{1}-n_{2}\right)^{2}, \\
b^{2}=\left(1-n_{1} n_{2}\right)^{2} .
\end{array}
$$
where $n_{1}, n_{2} \in N$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,312 |
2. In tetrahedron $ABCD$, $BD = DC = CB = \sqrt{2}$, $AC = \sqrt{3}$, $AB = AD = 1$. $M$ is the midpoint of $CD$. Then the cosine of the angle formed by the skew lines $BM$ and $AC$ is ( ).
(A) $\frac{\sqrt{2}}{3}$
(B) $\frac{\sqrt{2}}{2}$.
(C) $\frac{\sqrt{6}}{2}$.
(D) $\frac{\sqrt{3}}{9}$. | 2. (A).
Take the midpoint $N$ of $AD$, and connect $BN, MN$. Then, by the midline theorem, we know $MN \parallel AC, MN = \frac{AC}{2} = \frac{\sqrt{3}}{2}$.
$\angle BMN$ is the angle formed by the skew lines $BM$ and $AC$. Let $\angle BMN = \alpha$.
From the side lengths of $\triangle ABD$, we know $\angle BAD = 90... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,314 |
3. The sequence $\left\{\frac{100^{n}}{n!}\right\}(n=1,2,3, \cdots)$ is ).
(A) an increasing sequence. (B) a decreasing sequence.
(C) a sequence that has both increasing and decreasing terms after the 1993rd term.
(D) a sequence for which there exists a term such that after this term, it is decreasing. | 3. (D).
$$
a_{n}=\frac{100^{n}}{n!}, a_{n-1}=\frac{100^{n-1}}{(n+1)!} .
$$
$\left\{a_{n}\right\}$ is decreasing $\Leftrightarrow \frac{a_{n+1}}{a_{n}} < 1$.
\begin{aligned}
\frac{a_{n+1}}{a_{n}} &=\frac{\frac{100^{n+1}}{(n+1)!}}{\frac{100^{n}}{n!}} \\
&=\frac{100}{n+1} \\
&<1 \Leftrightarrow n+1>100 \Leftrightarrow n>... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,315 |
4. In $\triangle A B C$, $A>B$ is ( ) of $\cos 2 B>\cos 2 C$.
(A) A sufficient but not necessary condition.
(B) A necessary but not sufficient condition.
(C) A sufficient and necessary condition.
(D) Neither a sufficient nor a necessary condition. | $\begin{array}{l}\text { 4. (C). } \\ \triangle A B C \text {, in which } A>B \Leftrightarrow a>b \Leftrightarrow 2 R \sin A> \\ 2 R \sin B \Leftrightarrow \sin A>\sin B>0 \Leftrightarrow \sin ^{2} A>\sin ^{2} B \Leftrightarrow \\ -2 \sin ^{2} A1- \\ 2 \sin ^{2} A \Leftrightarrow \cos ^{2} B>\cos ^{2} A .\end{array}$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,316 |
Example 5. As shown in the figure, $O$ is a point inside the convex pentagon $A B C D E$, and $\angle 1=\angle 2, \angle 3=$ $\angle 4, \angle 5=\angle 6, \angle 7=$ $\angle 8$. Prove that $\angle 9$ and $\angle 10$ are equal or supplementary. (1985, National Junior High School Competition) | Prove from the given, according to the Law of Sines, we have
$$
\begin{aligned}
\frac{O A}{\sin \angle 10} & =\frac{O B}{\sin \angle 1}=\frac{O B}{\sin \angle 2}=\frac{O C}{\sin \angle 3} \\
& =\frac{O C}{\sin \angle 4}=\frac{O D}{\sin \angle 5}=\frac{O D}{\sin \angle 6} \\
& =\frac{O E}{\sin \angle 7}=\frac{O E}{\sin ... | \angle 9=\angle 10 \text { or } \angle 9+\angle 10=180^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 706,319 |
1. If $2|\sin x|^{1+182}=1+|\sin x|$, then the solution set of $x$ is | 1. $\left\{x \left\lvert\, x=\frac{\pi}{2}+k \pi\right., k \in Z\right\}$.
Since $|\sin x|^{1 / 2} \leqslant 1$, then $2|\sin x|^{1-182} \leqslant 2|\sin x| \leqslant 1+\sin |x|$.
Therefore, the left side of the equation equals the right side if and only if $|\sin x|=1$, at which point the solution set is $\left\{x \l... | \left\{x \left\lvert\, x=\frac{\pi}{2}+k \pi\right., k \in Z\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,320 |
2. If $n$ is a natural number, and $n^{3}+2 n^{2}+9 n+8$ is the cube of some natural number, then $n=$ $\qquad$ . | 2. 7 .
Since $n \in N$,
$$
\begin{aligned}
n^{3} & <n^{3}+2 n^{2}+9 n+8 \\
& <(n+2)^{3}=n^{3}+6 n^{2}+12 n+8,
\end{aligned}
$$
thus, only
$$
\begin{array}{l}
n^{3}+2 n^{2}+9 n+8 \\
=(n+1)^{3}=n^{3}+3 n^{2}+3 n+1 .
\end{array}
$$
Solving for the positive root $n=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,321 |
4. From the set $\{19,20,21, \cdots, 91,92,93\}$, the total number of ways to select two different numbers such that their sum is even is $\qquad$ | 4. 1369.
From 1993 - a total of 93-19+1=75 numbers. Among them, there are 38 odd numbers and 37 even numbers. The sum of the two selected numbers is even if and only if the two numbers have the same parity. Therefore, the total number of ways to select is
$$
C_{38}^{2}+C_{37}^{2}=\frac{38 \times 37}{2}+\frac{37 \times... | 1369 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,323 |
5. In the Cartesian coordinate system $x o y$, the area of the figure formed by points whose coordinates satisfy the condition $\left(x^{2}+y^{2}+2 x+2 y\right)\left(4-x^{2}-y^{2}\right) \geqslant 0$ is
保留源文本的换行和格式，直接输出翻译结果如下：
5. In the Cartesian coordinate system $x o y$, the area of the figure formed by points whos... | A circle minus their common
part. Let the first circle be
$$
(x+1)^{2}+(y+1)^{2}
$$
$-2 \leqslant 0$ with area $S_{1}$,
the second circle $x^{2}+y^{2}-4$
$\leqslant 0$ with area $S_{2}$, the area of the required region is $S$, and the common area of the two circles is
5. $2 \pi+4$.
Rewrite the inequality as $\left[(x+1... | 2 \pi + 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,324 |
6. If $x>0, y>0, c>0$, and $x^{2}+y^{2}+$ $z^{2}=1$. Then the minimum value of the expression $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$ is
| 6. 3 .
Let $\frac{yz}{x}=a, \frac{xz}{y}=b, \frac{xy}{z}=c$, then
$$
x^{2}+y^{2}+z^{2}=1 \Leftrightarrow ab+bc+ca=1 .
$$
Therefore, $a^{2}+b^{2}+c^{2} \geqslant ab+bc+ca=1$.
Thus, $(a+b+c)^{2}$
$$
=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \geqslant 3,
$$
which implies $\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z} \Rightarrow \sqrt{3... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,325 |
Three, (This question is worth 20 points) In the pyramid $A-B C D$, the three dihedral angles at vertex $D$ are all right angles, and the sum of the three dihedral angles at vertex $A$ is $90^{\circ}$. If $D B=a$, $D C=b$, find the volume of the pyramid $A-B C D$. | Three, let $A D=x$, unfold $A D, D B, D C$ on a plane, we get
$$
\begin{array}{l}
A D_{1}=A D_{2}=A D=x, \angle D_{1} A D_{2}=90^{\circ}, \\
\angle A D_{1} B=\angle A D_{2} C=90^{\circ}, \\
D_{1} B=a, D_{2} C=b, B C=\sqrt{a^{2}+b^{2}} .
\end{array}
$$
Extend $D_{1} B, D_{2} C$ to intersect at $E$, then $A D_{1} E D_{2... | \frac{1}{6} a b(a+b) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,326 |
$$
\begin{aligned}
y=\sqrt{2 x^{2}} & -2 x+1 \\
& +\sqrt{2 x^{2}-(\sqrt{3}-1) x+1} \\
& +\sqrt{2 x^{2}-(\sqrt{3}+1) x+1}
\end{aligned}
$$
Find the minimum value of the function above. | Obviously, since
\[
\begin{aligned}
y=\sqrt{x^{2}}+ & (x-1)^{2} \\
& +\sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x+\frac{1}{2}\right)^{2}} \\
+ & \sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^{2}+\left(x-\frac{1}{2}\right)^{2}},
\end{aligned}
\]
then for the points \( T(x, x), A(0,1), B\left(\frac{\sqrt{3}}{2}, -\frac... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,327 |
Five. (This question is worth 20 points) If
$$
a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}, \quad(n \in N)
$$
Prove: For all $n \geqslant 2$, we have
$$
a_{n}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right) .
$$ | Because $a_{k-1}=a_{k}-\frac{1}{k}(k \geqslant 2)$, then
$$
\begin{array}{l}
a_{k-1}^{2}=a_{k}^{2}-\frac{2 a k}{k}+\frac{1}{k^{2}}, \\
a_{i}^{2}-a_{k-1}^{2}=\frac{2 a k}{k}-\frac{1}{k^{2}}(k \geqslant 2) .
\end{array}
$$
When $k=2,3,4, \cdots, n$, adding these equations, we get
$$
\begin{array}{l}
a_{n}^{2}-a_{1}^{2}=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,328 |
One, (This question is worth 35 points)
Let $h$ and $l$ be the altitude and angle bisector drawn from a vertex of a triangle to the opposite side, and let $R$ and $r$ be the circumradius and inradius of the triangle, respectively.
Prove: $\frac{h}{1} \geqslant \sqrt{\frac{2 r}{\pi}}$. | Let $-A H=h, A P=l$. Extend $A P$ to intersect the circumcircle of $\triangle A B C$ at the midpoint $L$ of $\widehat{B C}$.
Draw the diameter $L O M$ of the circumcircle, and connect $B M, B L$. Draw $I D \perp A B$ at $D$ from the incenter $I$, then $I D=r$.
Let $\angle H A L=\varphi$. Then
$$
\begin{aligned}
\frac{... | \frac{h}{l} \geqslant \sqrt{\frac{2 r}{R}} | Inequalities | proof | Yes | Yes | cn_contest | false | 706,329 |
Example 7. In the equilateral hexagon $A B C D E F$, the sum of the vertex angles $\angle A, \angle C, \angle E$ equals the sum of the vertex angles $\angle B, \angle D, \angle F$. Prove: $\angle A=\angle D, \angle B=\angle E, \angle C=\angle F . \quad(1953$, Hungarian Mathematical Olympiad) | Proof As shown in the figure, let
$$
\begin{array}{l}
\angle A=2 a, \angle C=2 \beta_{1}, \\
\angle E=2 \gamma_{1}, \angle B D F=\alpha, \\
\angle D F B=\beta, \angle D B F=
\end{array}
$$
$\gamma$, the side length of the hexagon is $a$.
Then the sum of the interior angles of the hexagon $=$
$$
\begin{array}{l}
4 \time... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,330 |
II. (Full marks 35 points) A set $C$ of natural numbers is called a "good set" if there exist at least two numbers in $C$ whose arithmetic mean also belongs to $C$.
Prove: If the set of natural numbers is arbitrarily divided into two disjoint subsets, at least one of the subsets is a "good set". | Let $A$ and $B$ be two disjoint subsets of the set of natural numbers, and suppose set $A$ is not a "good set". Prove that set $B$ must be a "good set".
If $6 \in A, 8 \in A$, and set $A$ is not a "good set", then $4 \in A, 7 \notin A, 10 \in A$. Therefore, $4 \in B, 7 \in B, 10 \in B$. Since $7 = \frac{4 + 10}{2}$, s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,331 |
Three. (This question is worth 35 points) On 199319936688 cards, each card is written with a natural number, exactly the 199319936688 natural numbers 1, $2, \cdots, 199319936688$. Can these cards be divided into three groups such that the sum of the numbers on the cards in the second group is 33 more than the sum of th... | It is easy to prove that 199319936688 is a multiple of 6, so by grouping 1, 2, 3, 4, 199319936688 in sequence, every six numbers form a small group, we get:
$1,2,3,4,5,6$ belong to the 1st group;
$7,8,9,10,11,12$ belong to the 2nd group;
$\qquad$
and so on, until 199319936688, which is divided into 33219989448 groups. ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,332 |
Example 1. (1988, National Competition) Given $a_{1}=1, a_{2}=2$, and for all $n \in N$, we have
$$
a_{n-2}=\left\{\begin{array}{ll}
5 u_{n-1}-3 a_{n}, & \text { when } a_{n} \cdot a_{n-1} \text { is even. } \\
a_{n-1}-a_{n}, & \text { when } a_{n} \cdot a_{n-1} \text { is odd. }
\end{array}\right.
$$
Prove: For all $... | Prove a stronger proposition than the original proposition: for all $n \in N, a_{n}$ is not a multiple of 3.
Otherwise, let $m$ be the smallest subscript such that $31 a_{m}$. If $a_{m}=5 a_{m},-3 a_{m}$, then $3\left|a_{m}\right|$, which contradicts the minimality of $m$. Therefore, it must be that $a_{m}=a_{m},-a_{m... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,333 |
Example 2. (1984. National High School Mathematics League) Let $a_{n}$ be the unit digit of $1^{2}+2^{2}+\cdots+n^{2}$, $n=1,2,3$, $\cdots$. Prove that $0 . a_{1} a_{2} u_{3} \cdots a_{n} \cdots$ is a rational number. | $$
\begin{array}{l}
\text { Let } S_{n}=1^{2}+2^{2}-3^{2}+\cdots+n^{2} \\
=\frac{1}{6} n(n+1)(2 n+1) .
\end{array}
$$
So we have
$$
\begin{array}{l}
S_{n-20}-S_{n} \\
=\frac{1}{6}(n+20)(n+21)(2 n+41) \\
-\frac{1}{6} n(n+1)(2 n+1) \\
=10\left(2 n^{2}+42 n+287\right) .
\end{array}
$$
Therefore, for all $n \in N, S_{n-2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,334 |
Example 3. Arrange $1,2,3, \cdots, 1989$ in a circle, and starting from 1, cross out every other number. (That is, keep 1, cross out 2, keep 3, cross out $4, \cdots$), and repeat this process multiple times until only one number remains. What is this number? | Solving: Arranging so many numbers in a circle and performing deletions makes it difficult to see the pattern. Therefore, one should start with simpler cases. Suppose $n$ numbers are arranged in a circle, and the specified operations are performed. Let the last remaining number be denoted as $T_{n}$. The following tabl... | 1931 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,335 |
Example 4. As shown in Figure 1, $\triangle P Q R$ and $\triangle P^{\prime} Q^{\prime} R^{\prime}$ are two congruent equilateral triangles, and the six sides of hexagon $A B C D E F$ are denoted as: $A B=a_{1}, B C=b_{1}, C D=a_{2}, D E$ $=b_{2}, E F=a_{3}, F A=b_{3}$. Prove: $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}$ $=b_{1}^{2... | Proof: Let $P Q // P^{\prime} R^{\prime}, Q R // Q^{\prime} P^{\prime}, R P // R^{\prime} Q^{\prime}$ (see Figure 2). At this time, $\triangle P A B, \triangle Q^{\prime} C B$, $\triangle Q C D, \triangle R^{\prime} D E, \triangle R E F$, and $\triangle P^{\prime} F A$ are all equilateral triangles. Noting that the squ... | a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=b_{1}^{2}+b_{2}^{2}+b_{3}^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 706,336 |
Example 2. As shown in the figure, in trapezoid $A B C D$, $A B / / C D$, $A B>C D, K, M$ are points on the legs $A D, C B$ respectively, $\angle D A M=\angle C B K$.
Prove: $\angle D E A=$ $\angle C K B$. (2nd Zu Chongzhi Cup) | Prove that $K E$ is connected.
Since $\angle 1=\angle 2$, it follows from problem 2 that points $A, B, E, K$ are concyclic.
Also, since $\angle 5=\angle 6, \angle 5=\angle 7$,
thus, $\angle 6=\angle 7$.
From problem 4, we know that points $K, E, C, D$ are concyclic.
Therefore, $\angle 8=\angle 9$.
$$
\begin{aligned}
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,338 |
3. The equation $x^{2}+1993 x+3991=0(\quad)$.
(A) has positive roots
(B) has integer roots
(C) the sum of the reciprocals of the roots is less than -1
(D) none of the above | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,341 |
4. Among eight consecutive natural numbers, there are $k$ numbers that are pairwise coprime, then the maximum value of $k$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,342 |
5. In Rt $\triangle A B C$, the hypotenuse $c=5$, the two legs $a \leqslant 3, b \geqslant 3$, then the maximum value of $a+b$ is ( ).
(A) $5 \sqrt{2}$
(B) 7
(C) $4 \sqrt{3}$
(D) 6 | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,343 |
6. As shown in the figure, $P$ is
a point inside $\square A B C D$,
and $S_{\triangle P A B}=5$,
$S_{\triangle P A D}=2$. Then $S_{\triangle P A C}$
is equal to ( ).
(A) 2
(B) 3
(C) $3 \frac{1}{2}$
(D) 4 | 6. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,344 |
Example 2. How many sets of positive integer solutions does the equation $x+2 y+3 z=2000$ have? | Solve: This problem is equivalent to $x-1+2(y-1)+3(z-1) = 2000-6$, i.e.,
$$
a+2 b+3 c=1994
$$
Find the number of non-negative integer solutions.
Let $a=6 a_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 5\right)$,
$$
\begin{array}{l}
b=3 b_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 2\right), \\
c=2 c_{1}+r_{3}\left(0 \leq... | 332334 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,350 |
Three, let the roots of the equation $x^{2}+p x+q$ be 1 greater than the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$, and the difference between the roots of the equation $x^{2}+p x+q=0$ is equal to the difference between the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$, find the solutions of these two e... | Three, Solution: Let the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$ be $\alpha, \beta$. Then the roots of the equation $x^{2}+p x+q=0$ are $\alpha+1, \beta+1$.
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
\alpha+\beta=-2 q, \\
\alpha \beta=\frac{1}{2} p, \\
(\alpha+1)+(\beta+1)=-p, \\
(\alpha+1)(\be... | 0, 2 \text{ and } \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,352 |
In equilateral $\triangle A B C$, take point $D$ on side $B C$ such that $\frac{B D}{D C}=\frac{1}{2}$, and draw $C H \perp A D$, with $H$ being the foot of the perpendicular. Connect $B H$. Prove that: $\angle D B H = \angle D A B$ | Proof: Construct $A M \perp B C$ intersecting $B C$ at $M$.
$\because \angle A D M=\angle C D H$,
$\therefore \mathrm{Rt} \triangle A D M \backsim \mathrm{Rt} \triangle C D H$.
Thus, $\frac{A D}{C D}=\frac{M D}{H D}$,
which means $\frac{A D}{2 B D}=\frac{\frac{1}{2} B D}{H D}$,
or equivalently, $\frac{A D}{B D}=\frac{B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,353 |
企、冏 sizes are $1 \times 1 \times 2 \times 2 \times 3 \times 3$ to approximately tile a $23 \times 23$ square floor.
(1) Please design a scheme to tile the floor using only one $1 \times 1$ square tile and several $2 \times 2, 3 \times 3$ square tiles.
(2) Prove that to tile the floor completely, a $1 \times 1$ square t... | (1) In the center of a $23 \times 23$ square, place a $1 \times 1$ tile, and then divide the remaining part into 4 $11 \times 12$ rectangles. Each of these rectangles can be further divided into $2 \times 12$ and $9 \times 12$ rectangles, and then cover them with corresponding $2 \times 2$ and $3 \times 3$ tiles.
(2) N... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,354 |
3. In $\triangle A B C$, $P$ is the intersection of the angle bisectors $B E$ and $C F$, $M N$ is a line parallel to $B C$, intersecting $A B$ at $M$ and $A C$ at $N$. Then $M N$ equals ( ).
(A) $B M+M C$
(B) $C N+N E$
(C) $B M+C N$
(D) $B P+C P$ | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,357 |
4. Given that $A D$ is the median of $\triangle A B C$ on side $B C$, and $A D < \frac{1}{2} B C$, then $\triangle A B C$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle
(D) cannot be determined | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,358 |
5. The figure shows the graph of the quadratic function $y=a x^{2}+b x+c$. Then the signs of $a, b, c, b^{2}-4 a c$ are ( ).
(A)
$$
\begin{array}{l}
a<0, b>0, c>0, \\
b^{2}-4 a c>0
\end{array}
$$
(B) $a<0, b>0, c<0, b^{2}-4 a c>0$ | 5. D))
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,359 |
Example 3. How many non-negative integer solutions does the equation $x+3 y+4 z=665$ have?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x=12 x_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 11\right)$,
$$
\begin{array}{l}
y=4 y_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 3\right), \\
z=3 z_{1}+r_{3}\left(0 \leqslant r_{3} \leqslant 2\right) .
\end{array}
$$
The equation becomes
$$
\begin{array}{ll}
& 12\left(x_{1}+y_{1}+z_{1}\right)+r_{1}+3 r_{2}+4 r... | 17108 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,361 |
8. Given that the two roots of the equation $2 x^{2}-4 x-3=0$ are $x_{1}$ and $x_{2}$, then $M=\frac{1}{x_{1}}+\frac{1}{x_{2}}, N=x_{\mathrm{i}}^{i}+x_{2}^{2}, P=$ $\left(x_{1}+1\right)\left(x_{2}+1\right)$, the largest and smallest values among them are ( ).
(A) $M, P$
(B) $N, M$
(C) $M, N$
(D) $P, N$ | 8. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,363 |
一、(20 points) On the square, 5 loudspeakers are to be installed, divided into two groups. The first group has 2 loudspeakers installed together, and the second group has 3 loudspeakers installed together. The distance between the two groups is 50 meters. Where should one stand to hear the sound from both groups equally... | Let $x$ represent the distance from the point to the smaller group, then the distance from this point to the larger group is $(50-x)$. Since sound intensity is inversely proportional to the square of the distance, we get the equation:
$$
\frac{2}{3}=\frac{x^{2}}{(50-x)^{2}}.
$$
Simplifying, we get $x^{2} + 200x - 5000... | 22.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,368 |
In the Cartesian coordinate system, a circle centered at the origin with a radius of one length unit is called a unit circle. Prove that for a right-angled triangle with vertices on the unit circle, the sum of the cosines of the three angles is less than half the perimeter of the triangle. | In the unit circle $O$, let's consider any acute triangle $\triangle ABC$, with its opposite sides being $a, b, c$, and their semi-perimeter being $p \left(p = \frac{a+b+c}{2}\right)$.
$$
\begin{array}{l}
\because \triangle ABC \text{ is an acute triangle,} \\
\therefore A+B>90^{\circ}, \text{ i.e., } A>90^{\circ}-B, \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,369 |
Three, on a plane, there are four points $A, B, P, Q$, where $A$ and $B$ are fixed points, $A B=\sqrt{3} ; P$ and $Q$ are moving points, and they satisfy the relationship $A P=P Q=Q B=1$. Also, the areas of $\triangle A P B$ and $\triangle P Q B$ are $S_{1}, S_{2}$, respectively. Try to find:
(1) The range of $S_{\math... | Three, (1) According to the cosine theorem:
$$
P B^{2}=1+3-2 \sqrt{3} \cos A=1+1-2 \cos Q \text {. }
$$
Then $\cos Q=\sqrt{3} \cos A-1$.
(I)
Thus, $S_{1}^{2}+S_{i}^{2}=(\sqrt{3}-\sin A)^{2}+\left(\frac{1}{2} \sin Q\right)^{2}$
$$
\begin{array}{l}
=\frac{3}{4}\left(1-\cos ^{2} A\right)+\frac{1}{4}(1- \\
\cos ^{2} Q \te... | \frac{7}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,370 |
1. Let $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots$ $+a_{2 n} x^{2 n}$, then $S=a_{0}+a_{2}+\cdots+a_{2 n}$ equals
). (A) $2^{n}$
(B) $2^{n}+1$
(C) $\frac{3^{n}-1}{2}$
(D) $\frac{3^{n}-1-1}{2}$ | D)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,371 |
2. If $\log _{N} N=\log _{N} M$, where $M \neq N, M$ $\neq 1, N \neq 1$, then $M N$ equals ( ).
(A) 1
(B) 2
(C) 10
(D) $\frac{1}{4}$ | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,373 |
4. Given $x=t^{\frac{1}{1}}, y=t^{\frac{t}{1}}(t>0, t \neq 1)$. Then the relationship between $x, y$ is ( ).
(A) $y^{x}=x^{\frac{1}{y}}$
(B) $y^{\frac{1}{x}}=x^{2}$
(C) $y^{x}=x^{y}$
(D) $x^{x}=y^{x}$ | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,375 |
5. In the $X O Y$ plane, draw different lines passing through the point $(3,4)$ and the trisection points of the line segment with endpoints $(-4,5),(5,-1)$, one of the line equations is ).
(A) $3 x-2 y-1=0$
(B) $5 x+2 y-23=0$
(C) $x-4 y+13=0$
(D) $4 x-5 y+8=0$ | C.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,376 |
6. A geometric sequence contains 5 terms, each of which is a positive integer less than 100, and the sum of these 5 terms is 211. If $S$ is the sum of the terms in the sequence that are perfect squares, then $S$ equals ( ).
(A) 91
(B) 133
(C) 195
(D) 211 | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,377 |
3. The base of a right quadrilateral prism is a rhombus with side length 2 and an angle of $30^{\circ}$, and the height is 1. A section is made through the base edge at an angle of $60^{\circ}$ to the base. The area of this section is $\qquad$ | 3. $\frac{4}{\sqrt{3}}$ | \frac{4}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,380 |
5. Let the complex number $\alpha=a+b i(a, b$ be real numbers, and $b>0) \frac{\alpha^{2}}{1+\alpha}$ and $\frac{\alpha}{1+\alpha^{2}}$ are both real numbers, then $\alpha=$ | 5. $-\frac{1}{2}+\frac{\sqrt{3}}{2} i$ | -\frac{1}{2}+\frac{\sqrt{3}}{2} i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,382 |
Example. Divide 20 people into 4 groups, with each group having $4,5,5,6$ people, how many ways are there to do this? | Given $m=20, t=3, h_{1}=1, h_{2}=2, h_{3}=1$, hence
$$
\begin{aligned}
f_{20}^{4}= & \frac{20!}{4!\cdot 5!\cdot 5!\cdot 6!\cdot 1!\cdot 2!\cdot 1!} \\
& =4888643760 .
\end{aligned}
$$ | 4888643760 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,383 |
Three, (20 points) Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2(x-1)}=y . \\
\frac{y^{2}}{2(1-y)}=x .
\end{array}\right.
$$ | Three, Solution Using the substitution method, we get
$$
x=\frac{x^{4}}{4(x-1)^{3}+4(x-1)} \text {. }
$$
Rearranging gives
$$
x(x-2)\left(3 x^{2}-6 x+4\right)=0 \text {, }
$$
Solving this yields
$$
\begin{array}{l}
x_{1}=0, \\
x_{2}=2, \\
x_{3}=1+\frac{\sqrt{3}}{3} i, \\
x_{4}=1-\frac{\sqrt{3}}{3} i .
\end{array}
$$
... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,385 |
Four, (20 points) Through the left focus $F$ of the ellipse $x^{2}+2 y^{2}=2$, draw a line $l$ with an inclination angle of $\alpha$ intersecting the ellipse at points $P, Q$, and the two directrices at points $A, B$. If $P Q$, $||A F|-| B F||$, and $|A B|$ form a geometric sequence, find the value of $|\cos \alpha|$.
... | Four, Solution: From $\frac{x^{2}}{2}+y^{2}=1$, we know
$$
a^{2}=2, b^{2}=1, c^{2}=1 .
$$
Thus, $F:(-1,0)$.
Let the parametric equation of the line be
$$
\left\{\begin{array}{l}
x=-1+t \cos \alpha, \\
y=t \sin \alpha .
\end{array} \text { ( } t\right. \text { is the parameter) }
$$
Substituting into $x^{2}+$
$2 y^{2}... | 2-\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,386 |
Five. (20 points) Given the sequence $a_{1}, a_{2}, a_{3}, \cdots$, where $a_{1}=1, a_{2}=\frac{1}{a_{1}}+a_{1}, \cdots, a_{n}=\frac{1}{a_{n-1}}+a_{n-1}$. $\cdots$, prove:
$$
\sqrt{2 n-1}<a_{n}<\sqrt{3 n-2} .
$$ | Obviously,
$$
1=a_{1}<a_{2}<\cdots<a_{n} \text {. }
$$
Since $a_{k}^{2}=a_{k-1}^{2}+\frac{1}{a_{k-1}^{2}}+2$,
then $a_{k-1}^{2}+2<a_{k}^{2}<a_{k-1}^{2}+3$,
thus $2(n-1)+\sum_{k=2}^{n} a_{i-1}^{2}<\sum_{k=2}^{n} a_{i}^{2}<3(n-1)+$ $\sum_{k=2}^{n} a_{k-1}^{2}$.
Therefore, $2 n-1<a_{n}^{2}<3 n-2$.
Taking the square root ... | \sqrt{2 n-1}<a_{n}<\sqrt{3 n-2} | Inequalities | proof | Yes | Yes | cn_contest | false | 706,387 |
One, (35 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ ( $n \geq 2$) positive numbers, placed in order on a circle, and satisfying that the division of each number by the sum of its adjacent two numbers is a natural number. Let $S_{n}=\frac{a_{1}+a_{3}}{a_{2}}+\frac{a_{2}+a_{4}}{a_{3}}+\cdots+\frac{a_{n-1}+a_{1}}{a_... | Given
$$
\begin{aligned}
S_{n}= & \left(\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{1}}\right)+\left(\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{2}}\right)+\cdots \\
& +\left(\frac{a_{n}}{a_{1}}+\frac{a_{1}}{a_{n}}\right),
\end{aligned}
$$
thus $2 n \leqslant S_{n}$.
For $S_{n}a_{2}>a_{3}$,
then $a_{2}+a_{3}<2 a_{1}$,
so there must ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,388 |
Example 1. If $a, b, c$ are pairwise distinct rational numbers, prove that $\sqrt{\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}}$ is a rational number. (Beijing 1991, Junior High School Mathematics Competition Final Question) | Prove $\because(a-b)+(b-c)+(c-a)=0$,
$$
\begin{aligned}
\therefore & \frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}} \\
& =\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^{2} . \\
\therefore \quad & \sqrt{\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}} \\
& =\left|\frac{1}{a-b}+\frac{1}{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,392 |
Example 2. Prove: $\sqrt{a^{2}+\frac{1}{b^{2}}+\frac{a^{2}}{(a b+1)^{2}}}$ $=\left|a+\frac{1}{b}-\frac{a}{a b+1}\right|$. (4th "Zu Chongzhi Cup") | Prove $\because \frac{1}{a}+b+\left[-\left(b+\frac{1}{a}\right)\right]=0$,
$$
\begin{array}{r}
\therefore \frac{1}{\left(\frac{1}{a}\right)^{2}}+\frac{1}{b^{2}}+\frac{1}{\left[-\left(b+\frac{1}{a}\right)\right]^{2}} \\
=\left[\frac{1}{\frac{1}{a}}+\frac{1}{b}+\frac{1}{-\left(b+\frac{1}{a}\right)}\right]^{2} .
\end{arra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,393 |
1. If $|a|=-a$, then $\left|2 a-\sqrt{a^{2}}\right|$ equals $f$ ).
(A) $a$
(B) $-a$
(C) 32
(D) $-3 a$ | 一、1. D
One、1. D | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,394 |
2. In $\triangle A B C$, AC:AB $=1: 2, \angle A$'s internal and external angle bisectors are $A E$ and $A F$, respectively, then the area $S_{\triangle A B C}: S_{\triangle A R E}$ : $S_{\triangle A B F}$ equals ( ).
(A) $3: 2: 4$
(B) $3: 2: 6$
(C.) $3: 1: 4$
(D) $2: 1: 4$ | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,395 |
Example 3. If in a convex pentagon $A B C D E$, $\angle A B C=\angle A D E$, and $\angle A E C=\angle A D B$, then $\angle B A C=\angle D A E$. (21st All-Soviet Union Middle School Competition) | $$
\begin{array}{c}
\text{Since } \angle A E F= \\
\angle A D F, \\
\text{therefore } A, B, D, E
\end{array}
$$
are concyclic. Thus,
$$
\begin{array}{c}
\angle A B C= \\
\angle A D E=A F E, \angle A B C+\angle A F C=180^{\circ} .
\end{array}
$$
Hence, $A, B, C, D$ are concyclic.
Therefore, $\angle B A C=\angle B F C=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,396 |
3. The value of $(7+4 \sqrt{3})^{\frac{1}{2}}-(7-4 \sqrt{3})^{\frac{1}{2}}$ is ).
(A) $14 \sqrt{3}$
(B) $2 \sqrt{3}$
(C) $\sqrt{4}$
(D) $4 \sqrt{3}$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,397 |
5. In $\triangle ABC$, if $AB=3BC, \angle B = 2 \angle A$, then $\triangle ABC$ is ( ).
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Cannot be determined | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,399 |
1. If $x=3-5a$ is a solution to the inequality $\frac{1}{3}(x-2)$ $<x-\frac{1}{3}$, then the range of values for $a$ is | 二、 1. $a<\frac{3}{5}$
Translates to:
Two, 1. $a<\frac{3}{5}$ | a<\frac{3}{5} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,401 |
Four, as shown in the figure, in square $A B C D$, $E$ is the midpoint of side $A D$, and $B D$ intersects $C E$ at point $F$. Prove: $A F \perp B E$.
保留源文本的换行和格式，直接输出翻译结果如下：
```
Four, as shown in the figure, in square $A B C D$, $E$ is the midpoint of side $A D$, and $B D$ intersects $C E$ at point $F$. Prove: $A F ... | $\begin{array}{l}\text { Four, Prove Rt } \triangle A B E \cong \text{Rt} \triangle D C E \text { and } \triangle A D F \\ \cong \triangle C D F .\end{array}$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,406 |
Five, use 300 grams of type A sulfuric acid with a concentration of $95 \%$ and 200 grams of type B sulfuric acid with a concentration of $70 \%$ to prepare type C sulfuric acid with a concentration of $\mathrm{m} \%$. How many grams of each should be taken to maximize the amount of type C sulfuric acid prepared? | Five, let the amount of type A be $x$ grams, and type B be $y$ grams, then $\frac{95}{100} x + \frac{70}{100} y = \frac{m}{100}(x+y)$. The result is: when $m=85$, type A takes 300 grams, type B takes 200 grams; when $85<m<95$, type A takes 300 grams, type B takes $\frac{95-m}{m-70} \times 300$ grams; when $70<m<85$, ty... | when\ m=85,\ type\ A\ takes\ 300\ grams,\ type\ B\ takes\ 200\ grams;\ when\ 85<m<95,\ type\ A\ takes\ 300\ grams,\ type\ B\ takes\ \frac{95-m}{m-70} \times 300\ grams;\ when\ 70<m<85,\ type\ A\ | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,408 |
1. The solution to the equation $\left|\sqrt{(x-2)^{2}}-1\right|=x$ is ).
(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) $\frac{3}{2}$ or $\frac{1}{2}$
(D) No solution | $\begin{array}{l}\text { I. 1. B } \\ \text { 1. Hint: } 0 \leqslant x<2,|1-x|=x, x^{2}-2 x \\ +1=x^{2}, x=\frac{1}{2} .\end{array}$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,409 |
5. Let ( $a b c$ ) be a three-digit number with all distinct digits. If ( $a b c$ ) is equal to $k$ times the sum of its digits, then the sum of all new numbers formed by rearranging the digits of ( $a b c$ ) is a multiple of the sum of the digits of ( $a b c$ ).
| 5. $111-k$
5. Hint: From the given, we have $100a + 10b + c = k(a + b + c)$ and $11a + 101b + 110c = x(a + b + c)$. Adding the two equations and simplifying, we get $111(a + b + c) = (k + x)(a + b + c)$. | 111-k | Number Theory | proof | Yes | Yes | cn_contest | false | 706,413 |
II. For what value of $b$ do the equations $x^{2}-b x-2=0$ and $x^{2}-2 x-b(b-1)=0$ have the same roots? And find their common roots. | Let the common root of the two quadratic equations be $\alpha$, then
$$
\begin{array}{l}
\alpha^{2}+b \alpha-2=0, \\
\alpha^{2}-2 \alpha-b(b-1)=0 .
\end{array}
$$
Subtract (1) from (2): we get
$$
b \alpha-2 \alpha-b(b-1)+2=0 .
$$
After factoring, we get
$$
(b-2)(\alpha-b-1)=0 .
$$
When $b \neq 2$, then $\alpha-b-1=0$... | b=1, \alpha=2; b=2, \alpha_{1}=1+\sqrt{3}, \alpha_{2}=1-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,414 |
Four, as shown in the figure, $R$ is a square grid composed of 25 small squares with a side length of 1. Place a small square $T$ with a side length of 1 on the center square of $R$, and then place more small squares with a side length of 1 in $R$ according to the following requirements:
(1) The sides of these small sq... | Four, Answer: The maximum number of small cubes that can be placed is 4.
Proof As shown in the figure,
a) The small cubes that satisfy condition (3) can only be placed within the figure formed by removing a small square of side length 1 from each corner of $R$.
b) The small cubes that satisfy condition (3) must also be... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,416 |
Example 5. From any point on the circumcircle of a triangle, the perpendiculars to the sides (or their extensions) of the triangle meet at a straight line.
Preserving the original text's line breaks and format, the translation is as follows:
Example 5. From any point on the circumcircle of a triangle,
the perpend... | Proof: Let $I$ be any point on the circumcircle of $\triangle ABC$. Draw perpendiculars from $P$ to the sides $AK, BC, CA$ or their extensions, with feet at $L, M, N$. Connect $PB, PC$, $LM, MN$, and $PM$.
$\because A, B, C, P$ are concyclic,
$\therefore \angle PCN = \angle ABP$ (i.e., $\angle LBP$).
Also, $P, N, C, M$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,418 |
3. Given two sequences
$$
\begin{array}{l}
2.5, 8, 11, 14, 17, \cdots, 2 + (200-1) \times 3 \\
\text { and } 5, 9, 13, 17, 21, 25, \cdots, 5 + (200-1) \times 4 \\
\end{array}
$$
both have 200 terms, then the number of common terms in these two sequences is ( ).
(A) 49
(B) 50
(C) 51
(D) 147 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,420 |
4. As shown in the figure, given a square $A B C D$ with side length $a$, $E$ is the midpoint of $A D$. $P$ is the midpoint of $C E$. $F$ is the midpoint of $B P$, then the area of $\triangle B F D$ is ( ).
(A) $\frac{1}{64} a^{2}$
(B) $\frac{1}{32} u^{2}$
(C) $\frac{1}{16} a^{2}$
(D) $\frac{1}{8} a^{2}$ | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,421 |
5. $N$ is a 1992-digit natural number, where the digits 1, 2, 3, ..., 8 appear in $N$ a number of times that are all multiples of 9. If the sum of the digits of $N$ is $N_{1}$, the sum of the digits of $N_{1}$ is $N_{2}$, and the sum of the digits of $N_{2}$ is $N_{3}$, then the value of $N_{3}$ is ( ).
(A) 3
(B) 6
(C)... | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,422 |
Four, there are 1993 coins on the table. The first time, flip all 1993 coins; the second time, flip 1992 of them; the third time, flip 1991 of them; ... the 1993rd time, flip one of them. By flipping the coins in this manner, can all 1993 coins on the table be made to have their originally downward-facing sides facing ... | Four, Answer: Yes.
Solution According to the specified flipping method, the total number of flips is
$$
\begin{array}{l}
1+2+3+\cdots+1993 \\
=\frac{1993(1+1993)}{2} \\
=1993 \times 997 \text { (times). }
\end{array}
$$
On average, each coin is flipped 997 times, which is an odd number. Flipping an odd number of times... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,430 |
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