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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
18. Given a set of 2002 points on the $x O y$ plane, denoted as $S$, and no two points in $S$ are connected by a line parallel to the coordinate axes. For any two distinct points $P, Q$ in $S$, consider the rectangle with $P Q$ as its diagonal, and whose sides are parallel to the coordinate axes. Let $W_{P Q}$ denote t... | Solution: Here we only discuss rectangles with sides parallel to the coordinate axes. For any two points $P, Q$ in $S$, $R_{P Q}$ denotes the rectangle with $P Q$ as its diagonal, and $W_{P Q}$ denotes the number of points in $S$ within the rectangle $R_{P Q}$. Here, $P$ and $Q$ do not have to be distinct; if $P$ and $... | 400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,527 |
1. Among $\frac{1}{7}, \frac{\pi}{2}, 0.2002, \frac{1}{2}(\sqrt{3-2 \sqrt{2}}-\sqrt{2})$, $\frac{\sqrt{n}-\sqrt{n-2}}{3}(n(n>3)$ is an integer $)$, the number of fractions is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | -1 (B).
$$
\frac{1}{2}(\sqrt{3-2 \sqrt{2}}-\sqrt{2})=\frac{1}{2}(\sqrt{2}-1-\sqrt{2})=-\frac{1}{2} \text {, }
$$
When $n(n>3)$ is an integer, one of $\sqrt{n}$ and $\sqrt{n-2}$ is irrational, meaning that $n$ and $n-2$ cannot both be perfect squares at the same time. Let $n=s^{2}, n-2=t^{2}$. Then we have $s^{2}-t^{2}... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,528 |
2. As shown in Figure 1, the area of square $A B C D$ is 256, point $F$ is on $A D$, and point $E$ is on the extension of $A B$. The area of right triangle $\triangle C E F$ is 200. Then the length of $B E$ is $(\quad)$.
(A) 10
(B) 11
(C) 12
(D) 15 | 2. (C).
It is easy to prove that $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$. Therefore, $C F=C E$.
Since the area of Rt $\triangle C E F$ is 200, i.e., $\frac{1}{2} \cdot C F \cdot C E=200$, hence $C E=20$.
And $S_{\text {square } B C D}=B C^{2}=256$, so $B C=16$.
By the Pythagorean theorem, $B E=... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,529 |
3. Given that $a$, $b$, and $c$ are all integers, and satisfy
$$
a^{2}+b^{2}+c^{2}+3 < a b + 3 b + 2 c \text{.}
$$
Then the quadratic equation with roots $a+b$ and $c-b$ is ( ).
(A) $x^{2}-3 x+2=0$
(B) $x^{2}+2 x-8=0$
(C) $x^{2}-4 x-5=0$
(D) $x^{2}-2 x-3=0$ | 3. (D).
Since $a, b, c$ are integers and satisfy
$$
a^{2}+b^{2}+c^{2}+3 < a b + 3 b + 2 c \text{, }
$$
we have $a^{2}+b^{2}+c^{2}+3 \leqslant a b + 3 b + 2 c - 1$.
Rearranging and completing the square, we get
$$
\left(a-\frac{b}{2}\right)^{2} + 3\left(\frac{b}{2}-1\right)^{2} + (c-1)^{2} \leqslant 0 \text{. }
$$
Th... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,530 |
4. As shown in Figure 2, in Rt $\triangle A B C$, $A F$ is the altitude, $\angle B A C=90^{\circ}$, and $B D=D C=F C=$ 1. Then $A C$ is ( ).
(A) $\sqrt[3]{2}$
(B) $\sqrt{3}$
(C) $\sqrt{2}$
(D) $\sqrt[3]{3}$ | 4. (A).
Let $A C=x, B D=C D=C F=1, A D=x-1$. By the Pythagorean theorem, we have
$$
A B^{2}=B D^{2}-A D^{2}=1^{2}-(x-1)^{2} \text {. }
$$
By the projection theorem, we get
$$
B C=\frac{A C^{2}}{C F}=x^{2} \text {. }
$$
By the Pythagorean theorem, we have
$$
A B^{2}+A C^{2}=B C^{2} \text {, }
$$
which means $1^{2}-(... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,531 |
1. Let $p$ be a real number, and the graph of the quadratic function $y=x^{2}-2 p x-p$ intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
(1) Prove that $2 p x_{1}+x_{2}^{2}+3 p>0$;
(2) If the distance between points $A$ and $B$ does not exceed $|2 p-3|$, find the maximum v... | (Tip: Extreme method. According to the relationship between roots and coefficients. Answer: $\frac{9}{16}$. ) | \frac{9}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,532 |
5. If $k=\frac{2 a+b}{c}=\frac{2 c+a}{b}=\frac{2 b+c}{a}$, then the value of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) None of the above | 5. (C).
Since $k=\frac{2 a+b}{c}=\frac{2 c+a}{b}=\frac{2 b+c}{a}$, we have
$$
\begin{array}{l}
2 a b+b^{2}=2 c^{2}+a c, \\
2 a c+a^{2}=2 b^{2}+b c, \\
2 b c+c^{2}=2 a^{2}+a b .
\end{array}
$$
Adding the three equations, we get
$$
a b+b c+a c=a^{2}+b^{2}+c^{2},
$$
Multiplying both sides by 2, rearranging, and complet... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,533 |
6. Let $x \geqslant 0, y \geqslant 0, 2x + y = 6$. Then the maximum value of $u = 4x^2 + 3xy + y^2 - 6x - 3y$ is ( ).
(A) $\frac{27}{2}$
(B) 18
(C) 20
(D) does not exist
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. (B).
From the given, we have $y=6-2 x$. Substituting into $u=4 x^{2}+3 x y+y^{2}-$ $6 x-3 y$, we get
$$
u=2 x^{2}-6 x+18 \text {. }
$$
And $x \geqslant 0, y=6-2 x \geqslant 0$, so $0 \leqslant x \leqslant 3$.
When $x=0$ or $x=3$, $u_{\text {min }}=18$. | B | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,534 |
1. The real roots of the equation $\frac{1}{x^{2}+1}+\frac{x^{2}+1}{x^{2}}=\frac{10}{3 x}$ are | $=1 . \frac{3 \pm \sqrt{5}}{2}$.
The original equation is transformed into
$\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}=\frac{10}{3}$.
Let $\frac{x}{x^{2}+1}=y$, equation (1) becomes
$$
y+\frac{1}{y}=3+\frac{1}{3} \text {. }
$$
Then $y_{1}=3$ or $y_{2}=\frac{1}{3}$.
Substituting, we get $\frac{x}{x^{2}+1}=3$ or $\frac{x}{x^{2... | x=\frac{3 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,535 |
3. Given the quadratic function $y=x^{2}+(a+1) x+b$ (where $a$ and $b$ are constants). When $x=3$, $y=3$; for any real number $x$, $y \geqslant x$. Then the distance from the vertex of the parabola to the origin is $\qquad$ . | 3. $\frac{1}{4} \sqrt{221}$.
Substitute $x=3, y=3$ into the quadratic function $y=x^{2}+(a+1) x$ $+b$ to get
$$
b=-3 a-9 .
$$
For any real number $x$, we have $y \geqslant x$, that is
$$
x^{2}+(a+1) x+b \geqslant x \text {. }
$$
Substitute $b=-3 a-9$ to get
$$
x^{2}+a x-3 a-9 \geqslant 0 \text {. }
$$
Then $\Delta=... | \frac{1}{4} \sqrt{221} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,537 |
4. As shown in Figure 4, on the arc $AB$ of a sector $OAB$ with a radius of $2 \mathrm{~cm}$ and a central angle of $90^{\circ}$, there is a moving point $P$. A perpendicular line $PH$ is drawn from point $P$ to the radius $OA$, intersecting $OA$ at $H$. Let the incenter of $\triangle OPH$ be $I$. When point $P$ moves ... | 4. $\frac{\sqrt{2} \pi}{2} \mathrm{~cm}$.
As shown in Figure 7, I is the incenter of the right triangle $\triangle P O H$. Connect $O I, P I, A I$. It is easy to prove that
$\triangle O P I \cong \triangle O A I$.
Therefore, $\angle A I O=\angle P I O$.
Since I is the incenter of the right triangle $\triangle P O H$,
... | \frac{\sqrt{2} \pi}{2} \mathrm{~cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,538 |
One. (20 points) In a square with an area of 1, construct a smaller square as follows: divide each side of the unit square into $n$ equal parts, then connect each vertex to the nearest division point of its opposite vertex, as shown in Figure 5. If the area of the smaller square is exactly $\frac{1}{3281}$, find the va... | I. Draw $C_{1} P \perp A_{1} C$, with the foot of the perpendicular being $P$. Since the area of the square $A B C D$ is 1, we have
$$
A B=B C=C D=1, A_{1} B=\frac{n-1}{n}, C C_{1}=\frac{1}{n} .
$$
By the Pythagorean theorem,
$$
A_{1} C=\sqrt{B C^{2}+A_{1} B^{2}}=\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} .
$$
It is... | 41 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,539 |
II. (25 points) A straight road $l$ runs through a grassland, and there is a health station $A$ by the road. A residential point $B$ is located $30 \mathrm{~km}$ away from the road, and the distance between $A$ and $B$ is $90 \mathrm{~km}$. One day, a driver is tasked with delivering emergency medicine from the health ... | As shown in Figure 8, let the route on the road be $A D$, and the route on the grass be $D B$, with the distance of $A D$ being $x$. Given the conditions $A B=90$, $B C=30$, and $B C \perp A C$, we have
$$
A C=\sqrt{A B^{2}-B C^{2}}=60 \sqrt{2} \text {. }
$$
Then, $C D=A C-A D=60 \sqrt{2}-x$,
$$
B D=\sqrt{C D^{2}+B C^... | \sqrt{2}+\frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,540 |
Three. (25 points) From $1,2,3, \cdots, 3919$, select 2001 numbers. Prove: there must exist two numbers whose difference is exactly 98.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Three. (25 points) From $1,2,3, \cdots, ... | (1) The remainders when divided by 98, $0,1,2, \cdots, 97$, can be divided into 98 categories. Since $2001=98 \times 20+41$, by the pigeonhole principle, among these 2001 numbers, there must be 21 numbers belonging to the same category, meaning the difference between any two of these 21 numbers is a multiple of 98.
(2)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,541 |
2. As shown in Figure 9, the side length of square $ABCD$ is 1, and point $P$ is any point on $BC$ (it can coincide with point $B$ or point $C$). Perpendiculars are drawn from $B$, $C$, and $D$ to the ray $AP$, with the feet of the perpendiculars being $B'$, $C'$, and $D'$, respectively. The maximum value of $BB' + CC'... | ( Hint: Consider point $P$ at extreme positions $B$ and $C$. Answer: $2$, $\sqrt{2}$. ) | 2, \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,543 |
3. As shown in Figure 1, given a cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, draw a line $l$ through vertex $A_{1}$ in space such that $l$ forms an angle of $60^{\circ}$ with both lines $A C$ and $B C_{1}$. How many such lines $l$ can be drawn?
(A) 4
(B) 3
(C) 2
(D) 1 | 3. (B).
It is easy to know that the angle formed by the skew lines $A C$ and $B C_{1}$ is $60^{\circ}$, so this problem is equivalent to:
Given that the angle formed by the skew lines $a$ and $b$ is $60^{\circ}$, how many lines pass through a fixed point $P$ in space and form an angle of $60^{\circ}$ with both $a$ an... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,544 |
5. If the positive integer $a$ makes the maximum value of the function $y=f(x)=x+$ $\sqrt{13-2 a x}$ also a positive integer, then this maximum value equals ( . ).
(A) 3
(B) 4
(C) 7
(D) 8 | 5.(C).
Let $\sqrt{13-2 a x}=t(t \geqslant 0)$, then $x=\frac{13-t^{2}}{2 a}$. Therefore, $y=\frac{13-t^{2}}{2 a}+t=-\frac{1}{2 a}(t-a)^{2}+\frac{a^{2}+13}{2 a}$.
Since $t \geqslant 0$ and $a$ is a positive integer.
Then when $t=a$, $y_{\text {max }}=\frac{a^{2}+13}{2 a}$.
To make $\frac{a^{2}+13}{2 a}$ a positive inte... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,545 |
6. In the sequence of positive integers, starting from 1, certain numbers are painted red according to the following rules. First, paint 1, then paint 2 even numbers $2, 4$; then paint the 3 nearest consecutive odd numbers after 4, which are $5, 7, 9$; then paint the 4 nearest consecutive even numbers after 9, which ar... | 6. (B).
The first time, 1 number is painted red: $1,1=1^{2}$;
The second time, 2 numbers are painted red: $2,4,4=2^{2}$;
The third time, 3 numbers are painted red: $5,7,9,9=3^{2}$.
It is easy to see that the last number painted red in the $k$-th time is $k^{2}$, then the $k+1$ numbers painted red in the $k+1$-th time ... | 3943 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,546 |
1. In the complex plane, the vertices $A, B, C$ of the right triangle $\triangle ABC$ correspond to the complex numbers $z+1, 2z+1, (z+1)^2$, respectively, with $A$ being the right-angle vertex, and $|z|=2$. Let the set $M=\left\{m \mid z^{m} \in \mathbf{R}, m \in \mathbf{N}_{+}\right\}, P=\left\{x \left\lvert\, x=\fra... | $$
\text { II, 1. } \frac{1}{7} \text {. }
$$
From $\overrightarrow{A B} \perp \overrightarrow{A C}$, we get
$$
\frac{(z+1)^{2}-(z+1)}{(2 z+1)-(z+1)}=\lambda i \quad(\lambda \in \mathbf{R}, \lambda \neq 0) \text {. }
$$
Solving for $z$, we get $z=-1+\lambda i$.
Since $|z|=2$, then $1+\lambda^{2}=4$, which means $\lam... | \frac{1}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,547 |
3. If the inequality about $x$
$$
\frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7}{x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7}<0
$$
has a solution set that is the union of some intervals, and the sum of the lengths of these intervals is less than 4, then the range of real number $a$ is | 3. $a \in(-\infty, 1] \cup[3,+\infty)$.
From $\Delta_{1}=\left(a^{2}+4 a-5\right)^{2}-4\left(-a^{2}+4 a-7\right)$
$$
=\left(a^{2}+4 a-5\right)^{2}+4(a-2)^{2}+12
$$
we know that the equation
$$
x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7=0
$$
has two distinct real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)$.
Simil... | a \in(-\infty, 1] \cup[3,+\infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,548 |
4. A bank plans to invest $40 \%$ of a certain fund in project $M$ for one year, and the remaining $60 \%$ of the fund in project $N$. It is expected that project $M$ could yield an annual profit of $19 \%$ to $24 \%$, and project $N$ could yield an annual profit of $29 \%$ to $34 \%$. At the end of the year, the bank ... | $4.10 \%$.
Let the total investment of the bank in two projects be $s$. According to the conditions given, the annual profits from investments in $M$ and $N$, denoted as $p_{M}$ and $p_{N}$, respectively, satisfy
$$
\begin{array}{l}
\frac{19}{100} \times \frac{40}{100} s \leqslant p_{M} \leqslant \frac{24}{100} \times ... | 4.10 \% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,549 |
3. As shown in Figure $10, A B C D$ is a square with a side length of 1, $U$ and $V$ are points on $A B$ and $C D$ respectively, $A V$ intersects $D U$ at point $P$, and $B V$ intersects $C U$ at point $Q$. Find the maximum value of the area of quadrilateral $P U Q V$.
(2000, Shanghai Junior High School Mathematics Com... | (Hint: Use the method of extreme by transforming the formula. Answer: $\frac{1}{4}$. ) | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,550 |
6. Two regular triangular pyramids with the same base are inscribed in the same sphere. It is known that the side length of the base of the two regular triangular pyramids is $a$, and the radius of the sphere is $R$. Let the angles between the lateral faces and the base of the two regular triangular pyramids be $\alpha... | 6. $-\frac{4 \sqrt{3} R}{3 a}$.
As shown in Figure 3, let $P-A B C$ and $Q-A B C$ be two regular tetrahedrons with the same base, both inscribed in a sphere $O$ with radius $R$. Connect $P Q$, then $P Q \perp$ plane $A B C$, and the foot of the perpendicular is $H$, and $P Q$ passes through the center of the sphere $O... | -\frac{4 \sqrt{3} R}{3 a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,551 |
Four. (20 points) In the tetrahedron $D-ABC$, $AD=a$, $BD=b$, $AB=CD=c$, and $\angle DAB + \angle BAC + \angle DAC = 180^{\circ}$, $\angle DBA + \angle ABC + \angle DBC = 180^{\circ}$. Find the angle formed by the skew lines $AD$ and $BC$. | As shown in Figure 4, the lateral faces of the tetrahedron $D-ABC$ are cut along the lateral edges and unfolded onto the plane of the base.
Since $\angle D_{1} A B + \angle B A C + \angle D_{2} A C = 180^{\circ}$,
Therefore, $D_{1}, A_{1}, D_{2}$ are collinear.
Similarly, $D_{1}, B, D_{3}$ are collinear.
Since $A D_{1}... | \arccos \frac{\left|b^{2} - c^{2}\right|}{a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,553 |
Initially 123 As shown in Figure 2, given points $D_{1}, D_{2}$ on side $AB$ of $\triangle ABC$, and $BD_{1} = AD_{2}$. Parallel lines to $BC$ are drawn through points $D_{1}, D_{2}$, intersecting $AC$ at points $E_{1}, E_{2}$. Points $P_{1}, P_{2}$ are any points on $D_{1}E_{1}, D_{2}E_{2}$ respectively. $BP_{1}$ inte... | Prove: As shown in Figure 3, draw $QR \parallel AB$ through $P_{1}$, intersecting $BC$ at $Q$ and $AC$ at $R$. Let $A D_{1}=\lambda D_{1} B$, then we have
$$
\begin{array}{l}
P_{1} Q=D_{1} B, \\
\frac{A M_{1}}{M_{1} B}=\frac{R P_{1}}{P_{1} Q}=\frac{R P_{1}}{D_{1} B}=\lambda \frac{R P_{1}}{A D_{1}}=\lambda \frac{E_{1} P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,554 |
As shown in Figure 4, the sides of the smaller square $ABCD$ intersect the sides of the larger square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ at points $E, F, G, H$ and $P, Q, M, N$. Prove that:
$$
E F+P Q=G H+M N .
$$ | Proof: As shown in Figure 5, let the lines $B D$ and $A C$ intersect the sides of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ at points $T$, $S$, $X$, and $Y$. Since $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are both squares, and $A C \perp B D$, it follows that $T X \perp S Y$, and thus ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,555 |
Let $\triangle A B C$ have side lengths $a, b, c$, and area $S$. Then
$$
\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant(4 \sqrt{3} S)^{2}+\left(2 a^{2}-b^{2}-c^{2}\right)^{2} \text {. }
$$ | Proof: Let the altitude and median on side $BC$ of $\triangle ABC$ be $h_{a}$ and $m_{a}$ respectively:
Since $S=\frac{1}{2} a h_{a}, m_{a}=\frac{1}{2} \sqrt{2 b^{2}+2 c^{2}-a^{2}}, h_{a} \leqslant m_{a}$, therefore,
$$
\begin{array}{l}
4 \sqrt{3} S=2 \sqrt{3} a h_{a} \leqslant 2 \sqrt{3} a m_{a} \\
=\sqrt{3} a \cdot ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,556 |
Determine all functions \( f: \mathbf{R} \rightarrow \mathbf{R} \), such that for any \( x, y \in \mathbf{R} \), we have
\[
f(x) f(y f(x)-1)=x^{2} f(y)-f(x) .
\] | Solution: In (1), take $x=0$, we get
$$
f(0) f(y f(0)-1)=-f(0) \text {. }
$$
If $f(0) \neq 0$, then from the above equation, we get $f(x)=-1$, which does not satisfy (1).
Therefore, it must be that $f(0)=0$.
In (1), take $y=0$, we get
$$
f(x) f(-1)=-f(x) \text {, }
$$
i.e., $f(x)(f(-1)+1)=0$.
If $f(-1)+1 \neq 0$, the... | f(x)=0 \quad(\forall x \in \mathbf{R}) \text { and } f(x)=x \quad(\forall x \in \mathbf{R}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,558 |
5. Let $x, y$ be real numbers, and $x^{2}+x y+y^{2}=3$. Find the maximum and minimum values of $x^{2}-$ $x y+y^{2}$.
(1994, Huanggang City, Hubei Province Junior High School Mathematics Competition) | (Tip: Construct a quadratic equation. Answer: 9,1.) | 9,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,559 |
6. Let $a$, $b$, $c$, $a+b-c$, $a+c-b$, $b+c-a$, $a+b+c$ be 7 distinct prime numbers, and among $a$, $b$, $c$, the sum of two of them is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$.
(2001, China Mathematical Olympiad) | (Tip: Comparison method. Answer: 1 594.) | 1594 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,560 |
7. Among all triangles with a fixed angle $A$ and a fixed inradius $r$, determine which triangle has the smallest perimeter.
(12th Canadian Mathematical Olympiad) | (Tip: Convert the comparison of perimeters into the comparison of areas. Answer: When $\triangle A B C$ is an isosceles triangle, with $\angle A$ as the vertex angle, the perimeter of $\triangle A B C$ is minimized.) | When \triangle A B C \text{ is an isosceles triangle, with } \angle A \text{ as the vertex angle, the perimeter of } \triangle A B C \text{ is minimized.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,561 |
8. As shown in Figure $11, \angle A O B=$ $30^{\circ}, \angle A O B$ contains a fixed point $P$, and $O P=10, O A$ has a point $Q, O B$ has a fixed point $R$. If the perimeter of $\triangle P Q R$ is minimized, find its minimum value. | (Tip: Draw auxiliary lines with $O A$ and $O B$ as axes of symmetry. Answer:
10.) | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,562 |
Example 1 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1},
\end{array}
$$
then the minimum value of $xy$ is | Analysis: This is an expression related to trigonometry, and solving it might involve the boundedness of trigonometric functions.
Given \(1+\cos ^{2}(2 x+3 y-1) \leqslant 2\), we predict
\[
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \geqslant 2.
\]
In fact,
\[
\begin{array}{l}
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \\
=\fra... | \frac{1}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,563 |
Example 3 As shown in Figure $2, \odot O$ has a radius of $2, \odot O$ contains a point $P$ that is 1 unit away from the center $O$, and a chord $AB$ passing through point $P$ forms a segment with the minor arc $\overparen{AB}$. Find the minimum value of the area of this segment. (13th Junior High School Mathematics Co... | Analysis: Under certain conditions of a circle, the shorter the chord, the smaller the central angle it subtends, and consequently, the smaller the area of the corresponding segment. Therefore, to find the minimum area of the segment, it is only necessary to determine the extreme position of the chord passing through p... | \frac{4 \pi}{3} - \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,564 |
Example 2 The sum of the 6 edge lengths of the tetrahedron $P-ABC$ is $l$, and $\angle APB=\angle BPC=\angle CPA=90^{\circ}$. Then the maximum volume of the tetrahedron is $\qquad$
(5th United States of America Mathematical Olympiad) | Analysis: Let $P A=a, P B=b, P C=c$, it is easy to see that
$$
l=a+b+c+\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+
$$
$\sqrt{c^{2}+a^{2}}, V=\frac{1}{6} a b c$.
We can see that both of the above formulas are cyclic symmetric expressions about $a, b, c$. It is predicted that the volume is maximized when $a=b=c$.
In fact,
$$... | \frac{1}{6}\left(\frac{l}{3+3 \sqrt{2}}\right)^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,565 |
Example 3 In a regular $n$-sided pyramid, the range of the dihedral angle formed by two adjacent lateral faces is ( ).
(A) $\left(\frac{n-2}{n} \pi, \pi\right)$
(B) $\left(\frac{n-1}{n} \pi, \pi\right)$
(C) $\left(0, \frac{\pi}{2}\right)$
(D) $\left(\frac{n-2}{n} \pi, \frac{n-1}{n} \pi\right)$
(1994, National High Scho... | Analysis: When the vertex of a regular pyramid approaches the base infinitely, the dihedral angle between two adjacent sides approaches $\pi$; when the height of the regular pyramid increases infinitely, the dihedral angle approaches an interior angle of a regular $n$-sided polygon, i.e., $\frac{n-2}{n} \pi$. Therefore... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,566 |
Example 4 Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and the largest among them be $S$. Let $\lambda=$ $\frac{S_{1}+S_{2}+S_{3}+S_{4}}{S}$, then $\lambda$ must satisfy ( ).
(A) $2<\lambda \leqslant 4$
(B) $3<\lambda<4$
(C) $2.5<\lambda \leqslant 3.5$
(D) $3.5 \leqslant \lambda<5.5... | Analysis: If the tetrahedron is a regular tetrahedron, then $\lambda=4$, otherwise $\lambda=2$. Therefore, the correct choice is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,567 |
Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition) | Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,568 |
Example 6 If $S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10^{6}}}$, then the integer part of $S$ is ( ).
(A) 1997
(B) 1998
(C) 1999
(D) 2000
(8th "Hope Cup" Senior High School Question) | Analysis: From $\sqrt{n-1}+\sqrt{n}2\left(\sqrt{10^{6}}-1.5\right)$ $=1997,2\left(\sqrt{10^{6}}-1\right)=1998$, thus,
$$
1998<1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10^{6}}}<1999 .
$$
Therefore, the correct choice is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,569 |
Example 7 Given positive numbers $p, q, a, b, c$, where $p \neq q$. If $p, a, q$ form a geometric sequence, and $p, b, c, q$ form an arithmetic sequence, then the quadratic equation $b x^{2}-2 a x+c=0$ ( ).
(A) has no real roots
(B) has two equal real roots
(C) has two distinct real roots with the same sign
(D) has two... | Analysis: Taking $p=1, a=2, q=4, b=2, c=3$, then from the quadratic equation $2 x^{2}-4 x+3=0$ we easily get the answer as $(\mathrm{A})$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,570 |
Example 8 Given the following two propositions:
(1) Let $a, b, c$ all be complex numbers. If $a^{2}+b^{2}>$ $c^{2}$, then $a^{2}+b^{2}-c^{2}>0$;
(2) Let $a, b, c$ all be complex numbers. If $a^{3}+b^{2}-$ $c^{2}>0$, then $a^{2}+b^{2}>c^{2}$.
Then, the correct statement is ( ).
(A) Proposition (1) is correct, and propos... | Analysis: From $a^{2}+b^{2}>c^{2}$, it is clear that $a^{2}+b^{2}$ and $c^{2}$ are real numbers, so $a^{2}+b^{2}-c^{2}>0$ is correct. Therefore, proposition (1) is correct.
From $a^{2}+b^{2}-c^{2}>0$, does it follow that $a^{2}+b^{2}>c^{2}$? Taking $a^{2}+b^{2}=5+3 \mathrm{i}, c^{2}=2+3 \mathrm{i}$, we see that propos... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,571 |
Example 9 The solution to the equation $\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+x}}}}=x$ is $\qquad$ . | Analysis: It is difficult to proceed with removing the square root to solve. Instead, compare the size of $x$ and $\sqrt{3+x}$. If $x\sqrt{3+x}$ also leads to a contradiction with the given information.
Therefore, we have $x=\sqrt{3+x}$, solving this yields $x=\frac{1+\sqrt{13}}{2}$. | \frac{1+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,572 |
Example 10 The sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n}, a_{2 n+1}$ forms an arithmetic sequence, and the sum of the terms with odd indices is 60, while the sum of the terms with even indices is 45. Then the number of terms $n=$ $\qquad$ | Analysis: From $\left\{\begin{array}{l}a_{1}+a_{3}+\cdots+a_{2 n+1}=60, \\ a_{2}+a_{4}+\cdots+a_{2 n}=45\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}(n+1) a_{n+1}=60, \\ n a_{n+1}=45\end{array} \Rightarrow \frac{n+1}{n}=\frac{4}{3}\right.$, solving gives $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,573 |
Example 4 As shown in Figure 3, the radius of $\odot O$ is $R, C$ and $D$ are two points on the circumference of the circle on the same side of the diameter $A B$, the degree measure of $\overparen{A C}$ is $96^{\circ}$, and the degree measure of $\overparen{B D}$ is $36^{\circ}$. A moving point $P$ is on $A B$. Find t... | Solution: Let $D^{\prime}$ be the point symmetric to point $D$ with respect to the diameter $A B$, and connect $C D^{\prime}$ intersecting $A B$ at $P$. By the symmetry of the circle, point $D^{\prime}$ lies on the circle. Therefore,
$$
P D+P C=P D^{\prime}+P C .
$$
Since $P D^{\prime}+P C$ is the shortest at this tim... | \sqrt{3} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,575 |
Example 12 Given the three side lengths $a$, $b$, and $c$ of $\triangle ABC$ satisfy: (1) $a>b>c$, (2) $2b=a+c$, (3) $b$ is an integer, (4) $a^{2}+b^{2}+c^{2}=84$. Then the value of $b$ is $\qquad$ | Analysis: Starting from $2 b=a+c$, let $a=b+d$, $c=b-d(d>0)$, then
$$
(b+d)^{2}+b^{2}+(b-d)^{2}=84 .
$$
That is, $3 b^{2}+2 d^{2}=84$.
Obviously, $2 d^{2}$ is a multiple of 3. Also, by $b+c>a \Rightarrow$ $b>2 d$, substituting into (4) gives $2 d^{2}<12$. Therefore, $2 d^{2}=3,6,9$. Verification shows $b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,576 |
Example 13 On the coordinate plane, points with both integer horizontal and vertical coordinates are called integer points. For any natural number $n$, connect the origin $O$ with $A_{n}(n, n+3)$, and let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Then
$$
f(1)+f(2... | Analysis: It is easy to obtain the equation of $O A_{n}$ as
$$
y=\frac{n+3}{n} x=x+\frac{3}{n} x \quad(0 \leqslant x \leqslant n) .
$$
Obviously, for integer points, $n$ must be a multiple of 3. Let $n=3 k$, then
$$
y=\frac{k+1}{k} x \text {. }
$$
When $x=k$ and $2 k$, we get the integer points $(k, k+1)$, $(2 k, 2(k... | 1334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,577 |
Example 14 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=10$, $f(2)=20$, $f(3)=30$, then $f(10)+f(-6)=$ $\qquad$
(1998, Zhongshan City, Guangdong Province Mathematics Competition) | Analysis: Based on $f(1)=10, f(2)=20, f(3)=$ 30, construct
$$
\begin{array}{c}
f(x)=(x-1)(x-2)(x-3)(x-t)+10 x . \\
\text { Hence } f(10)+f(-6)=9 \times 8 \times 7(10-t)+ \\
10 \times 10+[(-7) \times(-8) \times(-9)(-6-t)- \\
60]=8104 .
\end{array}
$$ | 8104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,578 |
Example 15 The maximum value of the function $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$ $-\sqrt{x^{4}-x^{2}+1}$ is $\qquad$
(1992, National High School Mathematics Competition) | Analysis: $f(x)=\sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}}-$ $\sqrt{(x-0)^{2}+\left(x^{2}-1\right)^{2}}$ can be constructed as: the difference of distances from a moving point $\left(x, x^{2}\right)$ to two fixed points $(3,2)$ and $(0,1)$. Since the trajectory of the moving point $\left(x, x^{2}\right)$ is the parabola... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,579 |
Example 16 If $\lambda>0$ for any non-negative real numbers $x_{1}$, $x_{2}$ there is $x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \geqslant c\left(x_{1}+x_{2}\right)^{2}$, then the largest constant $c=c(\lambda)=$ $\qquad$ . | Analysis: (1) When $\lambda \geqslant 2$,
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \\
\geqslant x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}=\left(x_{1}+x_{2}\right)^{2},
\end{array}
$$
The equality holds when $x_{1}=x_{2}=0$.
$$
\begin{array}{l}
\text { (2) When } 0<\lambda<2 \text {, } \\
x_{1}^{2}+x_{2}^{2}... | c(\lambda)=\left\{\begin{array}{cc}
1 & (\lambda \geqslant 2), \\
\frac{2+\lambda}{4} & (0<\lambda<2) .
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,580 |
Example 18 A function defined on the set of natural numbers is given by
$$
f(n)=\left\{\begin{array}{ll}
n-3 & (n \geqslant 1000), \\
f[f(n+7)] & (n<1000) .
\end{array}\right.
$$
Then the value of $f(90)$ is ( ).
(A) 997
(B) 998
(C) 999
(D) 1000 | Analysis: Let $f^{(n)}(x)=f\{f[\cdots f(x) \cdots]\}(n$ iterations), then
$$
\begin{array}{l}
f(90)=f^{(2)}(97)=f^{(3)}(104)=\cdots \\
=f^{(131)}(1000) . \\
\text { Since } f^{(n+4)}(1000)=f^{(n+3)}(997) \\
=f^{(n+2)}(998)=f^{(n+1)}(999)=f^{(n)}(1000),
\end{array}
$$
so the period of $f^{(n)}(1000)$ is 4.
$f^{(131)}(1... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,582 |
Example 19 Let $f(x)=|1-2 x|, x \in[0,1]$. Then, the number of solutions to the equation $f\{f[f(x)]\}=\frac{1}{2} x$ is $\qquad$ . | Analysis: Let $y=f(x), z=f(y), w=f(z)$, use “ $\rightarrow$ ” to indicate the change, we have
$$
\begin{array}{l}
x: 0 \rightarrow 1, y: 1 \rightarrow 0 \rightarrow 1, \\
z: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1, \\
w: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarr... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,583 |
Example 5 As shown in Figure 4, in rectangle $A B C D$, $A B=$ $20 \text{ cm}, B C=10 \text{ cm}$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 4, construct $\angle B_{1} A C = \angle B A C$, draw $B Q \perp A C$ at $Q$, and extend $B Q$ to intersect $A B_{1}$ at $B_{1}$; draw $N P \perp A C$ at $P$, and extend $N P$ to intersect $A B_{1}$ at $N_{1}$, then points $B_{1}$ and $B$, and points $N_{1}$ and $N$ are symmetric with respec... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,586 |
3. If real numbers $x, y$ satisfy $4 x+3 y-2 x\left[\frac{x^{2}+y^{2}}{x^{2}}\right]=0$, then the value of $\frac{y}{x}$ is $\qquad$ (where $[x]$ denotes the greatest integer not exceeding $x$). | (Let $\frac{3 y}{2 x}=t$, then the condition becomes $1+t=\left[\frac{y^{2}}{x^{2}}\right]$, so $1+t=\left[\frac{4 t^{2}}{9}\right] \Rightarrow 1+t \leqslant \frac{4 t^{2}}{9} \leqslant t+2$. Thus, $t=-1$ or $t=3$. Therefore, $\frac{y}{x}=-\frac{2}{3}$ or 2.) | \frac{y}{x}=-\frac{2}{3} \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,587 |
4. The minimum value of the sum of the squares of the distances from a point in the rectangular coordinate plane to the three lines $x=0, y=0, 4x+3y=12$ is $\qquad$ | (Tip: $x^{2}+y^{2}+\left(\frac{4 x+3 y-12}{\sqrt{3^{2}+4^{2}}}\right)^{2}=\frac{1}{2}\left[x^{2}+y^{2}\right.$
$$
\begin{array}{l}
\left.+\left(\frac{4 x+3 y-12}{5}\right)^{2}\right] \times\left[\left(-\frac{4}{5}\right)^{2}+\left(-\frac{3}{5}\right)^{2}+1\right] \geqslant \\
\left.\frac{1}{2}\left(-\frac{4}{5} x-\frac... | \frac{72}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,588 |
5. The minimum distance from the lattice points (points with integer coordinates) to the line $y=\frac{5}{3} x+\frac{4}{5}$ is ( ).
(A) $\frac{\sqrt{34}}{170}$
(B) $\frac{\sqrt{34}}{85}$
(C) $\frac{1}{20}$
(D) $\frac{1}{30}$
(2000, National High School Mathematics Competition) | (Tip: The distance from a point to a line can be converted to $\frac{\sqrt{34}}{170}$. $25 x-$ $15 y+12|$. Since $x, y \in \mathbf{Z}$, then $|25 x-15 y+12|$ must be an integer. Its minimum value is 2. Therefore, the answer is (B).) | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,589 |
Example 1 Given the function $f(x)=-x^{2}+2 t x-t$, $x \in[-1,1]$. Let the maximum value of $f(x)$ be $M$. Find the minimum value of $M$. | Solution: Since $f(x)=-x^{2}+2 t x-t=-(x-t)^{2}+t^{2}-t$, and $-1 \leqslant x \leqslant 1$, then
when $t \leqslant-1$, $M=f(-1)=-3 t-1$;
when $-1<t<1$, $M=f(t)=t^{2}-t$;
when $t \geqslant 1$, $M=f(1)=t-1$.
Therefore, $M=\left\{\begin{array}{ll}-3 t-1 & (t \leqslant-1), \\ t^{2}-t & (-1<t<1), \\ t-1 & (t \geqslant 1) .\... | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,591 |
Example 2 If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1$, $x+2,-x+6\}$.
(38th AHSME) | Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,592 |
Example 3 Given $x \in \mathbf{R}$, let $M=\max \{1-x, 2 x-$ $3, x\}$. Find $M_{\min }$. | Solution: Since $M$ is the maximum of $1-x, 2x-3, x$, $M$ is not less than the maximum of $1-x, x$, and thus $M$ is not less than the arithmetic mean of $1-x$ and $x$, i.e.,
$$
M \geqslant \frac{(1-x)+x}{2}=\frac{1}{2} \text {. }
$$
When $x=\frac{1}{2}$, equality holds, so $M_{\text {min }}=\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,593 |
Example 4 If $a>0, b>0$, and $H=$ $\min \left\{a, \frac{b}{a^{2}+b^{2}}\right\}$. Try to find $H_{\max }$.
Keep the original text's line breaks and format, and output the translation result directly. | Solution: Since $H$ is the minimum of $a, \frac{b}{a^{2}+b^{2}}$, $H$ is not greater than the geometric mean of $a, \frac{b}{a^{2}+b^{2}}$, i.e.,
$$
H \leqslant \sqrt{\frac{a b}{a^{2}+b^{2}}} \leqslant \sqrt{\frac{a b}{2 a b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2},
$$
Equality holds if and only if $a=b=\frac{\sqrt{2}}... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,594 |
Example 5 Given that $x, y$ are two variables in the interval $(0,1)$, denote the minimum value of $2^{-x}, 2^{x-y}, 2^{y-1}$ as $F(x, y)$. Find the maximum value of $F(x, y)$.
$(1984$, Shanghai Middle School Mathematics Competition) | Solution: Since $F(x, y)$ is the minimum of $2^{-x}, 2^{x-y}, 2^{y-1}$, $F(x, y)$ is not greater than the geometric mean of these three, i.e.,
$$
F(x, y) \leqslant \sqrt[3]{2^{-x} \times 2^{x-y} \times 2^{y-1}}=2^{-\frac{1}{3}} .
$$
When $x=\frac{1}{3}, y=\frac{2}{3}$, the equality holds, therefore, the maximum value ... | 2^{-\frac{1}{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,595 |
Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$
Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, G... | Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.
Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,596 |
Example 6 In the rectangular coordinate system $x O y$, the moving point $M(x, 0)$ on the $x$-axis has distances $M P$ and $M Q$ to the fixed points $P(5,5)$ and $Q(2,1)$, respectively. Then, when $M P + M Q$ takes the minimum value, the abscissa $x=$ $\qquad$
(2001, National Junior High School Mathematics Competition) | Solution: Make the symmetric point $Q^{\prime}$ of point $Q$ with respect to the $x$-axis, and connect $P Q^{\prime}$ and $M Q^{\prime}$. Then $M P + M Q^{\prime} \geqslant P Q^{\prime}$. Therefore, the minimum value of $M P + M Q$ is $P Q^{\prime}$. At this time, the coordinates of the symmetric point $Q^{\prime}$ of ... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,597 |
Example 7 The function $F(x)=1 \cos ^{2} x+2 \sin x \cdot \cos x- \sin ^{2} x+A x+B$ has a maximum value $M$ on $0 \leqslant x \leqslant \frac{3 \pi}{2}$ that depends on the parameters $A$ and $B$. For what values of $A$ and $B$ is $M$ minimized? Prove your conclusion.
(1983, National High School Mathematics Competitio... | Solution: $F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+A x+B\right|$.
If $A=B=0$, then
$$
F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)\right|.
$$
When $x$ is $\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}$, $F(x)$ reaches its maximum value $M=\sqrt{2}$.
Below, we prove that for any real numbe... | A=B=0 | Algebra | proof | Yes | Yes | cn_contest | false | 713,598 |
Example 8 Let $n$ be a given natural number, and $n \geqslant 3$, for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote $\left|a_{i}-a_{j}\right|$ $(1 \leqslant i<j \leqslant n)$ the minimum value as $m$. Find the maximum value of the above $m$ under the condition $a_{1}^{2}+a_{2}^{2}+$ $\cdots+a_{n}^{2}=1$. | Solution: Without loss of generality, let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, then
$$
\begin{array}{l}
a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m, \cdots, a_{n}-a_{n-1} \\
\geqslant m, a_{1}-a_{i} \geqslant(j-i) m \quad(1 \leqslant i<j \leqslant n) . \\
\sum_{1 \leqslant i<j \leqslant n}\le... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,599 |
For example, $1 M$ is the set of non-zero functions of a real variable $x$ of the form $f(x)=a x+b(a, b \in \mathbf{R})$, and $M$ has the following properties:
(i) If $f, g \in M$, then $g \circ f \in M$, where $(g \circ f)(x)=g[f(x)]$;
(ii) If $f \in M$ and $f(x)=a x+b$, then the inverse function $f^{-1}$ also belongs... | Prove: Condition (iii) indicates that for every $f \in M$, there is a fixed point $x_{j}$ such that $f\left(x_{j}\right)=x_{j}$. Now, we need to prove that all functions $f$ in the set $M$ must have a common fixed point $k$. Suppose $f(x)=a x+b$ has a fixed point $x_{j}$, i.e., $a x_{j}+b=x_{j}$.
If $a \neq 1$, then $... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,600 |
Example 2 Let $\{f(n)\}$ be a strictly increasing sequence of positive integers. It is known that $f(2)=2$, and when $m, n$ are coprime,
$$
f(m n)=f(m) f(n) \text {. }
$$
Prove: $f(n)=n$.
(24th Putnam Mathematical Competition $\mathrm{A}-2$) | To prove: The problem is to prove that $f(n)$ is a fixed point for any natural number $n (n \geqslant 2)$ under the given conditions. In fact,
$$
\begin{array}{l}
f(3) f(7)=f(21) \\
f(2)=2, \text{ so, } f(3)=3.
\end{array}
$$
We will prove this by contradiction.
If the original proposition is not true, assume the small... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,601 |
Example 3 Determine all polynomials $p(x)$ satisfying:
$$
p\left(x^{2}+1\right)=[p(x)]^{2}+1, \quad p(0)=0.
$$
(32nd Putnam Mathematical Competition $\mathrm{A}-2$) | Solution: From the given conditions, we have
$$
\begin{array}{l}
p(0)=0, \\
p(1)=[p(0)]^{2}+1=1, \\
p(2)=[p(1)]^{2}+1=2, \\
p(5)=[p(2)]^{2}+1=5, \\
p\left(5^{2}+1\right)=[p(5)]^{2}+1=26, \\
\cdots \cdots,
\end{array}
$$
That is, $p(x)$ has infinitely many fixed points $0,1,2,5,26, \cdots$.
Since $p(x)$ is a polynomial... | p(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,602 |
Example 4 If $f(x)=a x+b(a \neq 1)$, and the fixed point of $f(x)$ is $\frac{b}{1-a}$, then the analytical expression of the $n$-th iteration function of $f(x)$ can be represented using the fixed point of $f(x)$ as follows:
$$
\underbrace{f(\cdots(f(x)) \cdots)}_{n \uparrow}=a^{n}\left(x-\frac{b}{1-a}\right)+\frac{b}{1... | Proof: By mathematical induction.
When $n=1$,
$$
f(x)=a\left(x-\frac{b}{1-a}\right)+\frac{b}{1-a}=a x+b \text{, }
$$
the conclusion is correct.
Assume that the conclusion is correct when $n=k$, i.e.,
$$
\underbrace{f(\cdots(f(x)) \cdots)}_{k \uparrow}=a^{k}\left(x-\frac{b}{1-a}\right)+\frac{b}{1-a} \text{. }
$$
Then... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,603 |
Example 5 Given the function $f(f(f(f(x))))=16x+15$. Find the analytical expression of the linear function $f(x)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Let the linear function be $f(x)=a x+b(a \neq 1)$. From Example 4, we know that
$$
a^{4}\left(x-\frac{b}{1-a}\right)+\frac{b}{1-a}=16 x+15,
$$
Therefore, the analytical expressions of the required linear functions are $f(x)=2 x+1$ and $f(x)=-2 x-3$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,604 |
Example 6 Let $S$ be the set of all real numbers greater than -1, determine all functions $f: S \rightarrow S$, such that the following two conditions are satisfied:
(i) For all $x$ and $y$ in $S$, we have
$$
f(x+f(y)+x f(y))=y+f(x)+y f(x) \text {; }
$$
(ii) In each of the intervals $(-1,0)$ and $(0,+\infty)$, $\frac{f... | Solution: In condition (i), let $x=y$, we get
$$
\begin{array}{l}
f(x+f(x)+x f(x)) \\
=x+f(x)+x f(x) .
\end{array}
$$
Then $x+f(x)+x f(x)$ is a fixed point of $f$ (i.e., a $z$ such that $f(z)=z$).
Let $A=x+f(x)+x f(x)$, in equation (1), let $x=A$, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(A)=A, \\
f(A+f(A)... | f(x)=-\frac{x}{1+x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,605 |
Proposition Let the circumcenter of $\triangle A_{1} B_{1} C_{1}$ be $O_{1}$, the incenter be $I_{1}$, the circumradius be $R_{1}$, the inradius be $r_{1}$, and $O_{1} I_{1}=d_{1} ;$ the circumcenter of $\triangle A_{2} B_{2} C_{2}$ be $O_{2}$, the incenter be $I_{2}$, the circumradius be $R_{2}$, the inradius be $r_{2... | Proof: From the famous Chapple's theorem in plane geometry ${ }^{[2]}$, we have
$d_{1}^{2}=R_{1}^{2}-2 R_{1} r_{1}, d_{2}^{2}=R_{2}^{2}-2 R_{2} r_{2}$.
Thus, $\frac{R_{1}^{2}-d_{1}^{2}}{R_{1} r_{1}}=2=\frac{R_{2}^{2}-d_{2}^{2}}{R_{2} r_{2}}$.
Therefore, equation (2) holds.
Corollary With the same notation as above, if ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,606 |
Proposition Let $x, y, z \in \mathbf{R}, x+y \geqslant 0$ and $z \geqslant \mu(x+y)$, where $\mu \in \mathbf{R}$ and $\mu \geqslant \frac{1}{2}$. Then for $\lambda \in \left[1, \frac{4 \mu^{2}+2}{4 \mu+1}\right]$, we have
$$
x^{2}+y^{2}+z^{2} \geqslant \lambda(y z+z x+x y) \text {. }
$$ | Prove: Regarding $x, y$ as constants, the function is
$$
f(z)=z^{2}-\lambda(x+y) z+\left(x^{2}+y^{2}-\lambda x y\right) \text {. }
$$
It is easy to see that when $z \geqslant \frac{\lambda}{2}(x+y)$, $f(z)$ is monotonically increasing.
From $\lambda \in\left[1, \frac{4 \mu^{2}+2}{4 \mu+1}\right]$, it is easy to get $\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,609 |
1. If $4 x-3 y-6 z=0, x+2 y-7 z=0(x y z \neq 0)$, then the value of the algebraic expression $\frac{5 x^{2}+2 y^{2}-z^{2}}{2 x^{2}-3 y^{2}-10 z^{2}}$ is ( ).
(A) $-\frac{1}{2}$
(B) $-\frac{19}{2}$
(C) -15
(D) -13 | -.1.D.
From $\left\{\begin{array}{l}4 x-3 y-6 z=0, \\ x+2 y-7 z=0\end{array}\right.$
we solve to get $x=3 z, y=2 z$. Substituting, we get
$$
\text { original expression }=\frac{5 \times 9 z^{2}+2 \times 4 z^{2}-z^{2}}{2 \times 9 z^{2}-3 \times 4 z^{2}-10 z^{2}}=-13 \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,610 |
2. When mailing a letter locally, a postage fee of 0.80 yuan is charged for each letter weighing no more than $20 \mathrm{~g}$, 1.60 yuan for letters weighing more than $20 \mathrm{~g}$ but no more than $40 \mathrm{~g}$, and so on, with an additional 0.80 yuan required for each additional $20 \mathrm{~g}$ (for letters ... | 2. D.
Since $20 \times 3<72.5<20 \times 4$, according to the problem, the postage to be paid is $0.80 \times 4=3.20$ (yuan). | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 713,611 |
4. The lengths of four line segments are $9, 5, x, 1$ (where $x$ is a positive real number), and they are used to form two right triangles, with $AB$ and $CD$ being two of the line segments (Figure 2). The number of possible values for $x$ is ( ) .
(A) 2
(B) 3
(C) 4
(D) 6 | 4.D.
Obviously, $AB$ is the longest, so $AB=9$ or $AB=x$.
(1) If $AB=9$, when $CD=x$,
$$
9^{2}=x^{2}+(1+5)^{2}, x=3 \sqrt{5};
$$
When $CD=5$,
$$
9^{2}=5^{2}+(x+1)^{2}, x=2 \sqrt{14}-1;
$$
When $CD=1$,
$$
9^{2}=1^{2}+(x+5)^{2}, x=4 \sqrt{5}-5;
$$
(2) If $AB=x$, when $CD=9$,
$$
x^{2}=9^{2}+(1+5)^{2}, x=3 \sqrt{13};
$$... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,613 |
5. In a graduation photo at a school, 100 people, including students and teachers from two graduating classes of the ninth grade, are to be arranged on steps in a trapezoidal formation with more people in the front rows than in the back rows (with at least 3 rows). It is required that the number of people in each row m... | 5.B.
Let the last row have $k$ people, with a total of $n$ rows. Then, the number of people in each row from back to front are $k, k+1, k+2, \cdots, k+(n-1)$. According to the problem,
$k n+\frac{n(n-1)}{2}=100$, which simplifies to $n[2 k+(n-1)]=200$.
Since $k, n$ are positive integers, and $n \geqslant 3$, it follow... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,614 |
7. If real numbers $x, y, z$ satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+$ $\frac{1}{x}=\frac{7}{3}$, then the value of $x y z$ is $\qquad$. | 7.1 .
Since $4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}$
$=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{7}{3}-\frac{1}{x}-1}=x+\frac{7 x-3}{4 x-3}$, then
$4(4 x-3)=x(4 x-3)+7 x-3$,
which simplifies to $(2 x-3)^{2}=0$.
Thus, $x=\frac{3}{2}$.
Therefore, $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}, y=1-\frac{1}{z}=\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,616 |
9. As shown in Figure 4, it is known that the power pole $A B$ stands vertically on the ground, and its shadow falls exactly on the slope surface $C D$ and the ground $B C$. If $C D$ makes a $45^{\circ}$ angle with the ground, $\angle A=60^{\circ}, C D$ $=4 \mathrm{~m}, B C=(4 \sqrt{6}-2 \sqrt{2}) \mathrm{m}$, then the... | 9.8.5.
As shown in Figure 8. Extend $A D$ to intersect the ground at point $E$, and draw $D F \perp C E$ at point $F$. Since $\angle D C F=45^{\circ}, \angle A=60^{\circ}$, and $C D=4$, we have:
$$
\begin{array}{c}
C F=D F=2 \sqrt{2}, \\
E F=D F \cdot \tan 60^{\circ}=2 \sqrt{6} .
\end{array}
$$
Since $\frac{A B}{B E}... | 8.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,618 |
Example 8 Given 3 non-negative numbers $a, b, c$ satisfying $3a+2b+c=5$ and $2a+b-3c=1$. If $m=$ $3a+b-7c$, then the minimum value of $m$ is $\qquad$, and the maximum value of $m$ is $\qquad$
(14th Junior High School Mathematics Competition in Jiangsu Province) | Solution: From the equations $3 a+2 b+c=5$ and $2 a+b-3 c=1$, we solve to get
$$
a=7 c-3, b=7-11 c .
$$
Substituting into $m=3 a+b-7 c$, we get
$$
m=3 c-2 \text{. }
$$
Since $a \geqslant 0, b \geqslant 0, c \geqslant 0$, we have
$$
\left\{\begin{array}{l}
a=7 c-3 \geqslant 0, \\
b=7-11 c \geqslant 0, \\
c \geqslant 0... | -\frac{5}{7}, -\frac{1}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,619 |
10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ . | 10. -4 .
From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,620 |
11. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $B C$ is the tangent of $\odot O$, and $O C$ is parallel to the chord $A D$. A perpendicular line $D E \perp A B$ is drawn through point $D$, intersecting at point $E$. Line $A C$ intersects $D E$ at point $P$. Is $E P$ equal to $P D$? Prove your ... | $$
\text { Three, 11. } D P=P E \text {. }
$$
Since $A B$ is the diameter of $\odot O$ and $B C$ is the tangent, $A B \perp B C$.
From Rt $\triangle A E P \sim \text{Rt} \triangle A B C$, we get
$$
\frac{E P}{B C}=\frac{A E}{A B} \text {. }
$$
Also, since $A D \parallel O C$, then $\angle D A E=\angle C O B$. Therefo... | D P=P E | Geometry | proof | Yes | Yes | cn_contest | false | 713,621 |
13. In $\triangle A B C$, it is known that $\angle A C B=90^{\circ}$.
(1) As shown in Figure 7, when point $D$ is on the hypotenuse $A B$ (excluding the endpoints), prove:
$$
\begin{array}{l}
\frac{C D^{2}-B D^{2}}{B C^{2}} \\
=\frac{A D-B D}{A B} ;
\end{array}
$$
(2) When point $D$ coincides with point $A$, does the e... | 13. (1) As shown in Figure 9, draw $D E \perp B C$, with the foot of the perpendicular at $E$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
C D^{2}-B D^{2}=\left(C E^{2}+D E^{2}\right)-\left(B E^{2}+D E^{2}\right) \\
=C E^{2}-B E^{2}=(C E-B E) B C . \\
\text { Then } \frac{C D^{2}-B D^{2}}{B C^{2}}=\frac{C E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,623 |
14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$. | 14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \tim... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,624 |
1. Given that $a$, $b$, and $c$ are three distinct odd prime numbers, the equation $(b+c) x^{2}+(a+1) \sqrt{5} x+225=0$ has two equal real roots.
(1) Find the minimum value of $a$;
(2) When $a$ reaches its minimum value, solve this equation. | 1. (1) From $\Delta=5(a+1)^{2}-900(b+c)=0$, we get $(a+1)^{2}=2^{2} \times 3^{2} \times 5(b+c)$.
Therefore, $5(b+c)$ should be an even perfect square, with the smallest value being $5^{2} \times$ $2^{2},$ and the smallest value of $a+1$ being $60,$ so the smallest value of $a$ is $59$.
(2) When $a=59$, $b+c=20$.
If $b=... | x=-\frac{3}{2} \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,625 |
2. As shown in Figure 1, let $AB$ and $CD$ be two diameters of $\odot O$. Draw $PB$ perpendicular to $AB$ and let it intersect the extension of $CD$ at point $P$. Draw line $PE$ intersecting $\odot O$ at points $E$ and $F$, and connect $AE$ and $AF$ to intersect $CD$ at points $G$ and $H$ respectively. Prove that:
$$
O... | 2. As shown in Figure 3, draw $F K / / G H$ intersecting $O B$ at $M$, and $O N \perp E F$ at $N$. $O, P, B, N$ are concyclic,
then $\angle O P N=\angle O B N$, $\angle M F N=\angle O B N$.
$M, F, B, N$ are concyclic,
then $\angle M N F=\angle M B F$ $=\angle A E F$.
Therefore, $M N / / K E$.
Hence, $K M=M F, O H=O G$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,626 |
3. Given that $a_{1}, a_{2}, \cdots, a_{2002}$ have values of 1 or -1, let $S$ be the sum of the pairwise products of these 2002 numbers.
(1) Find the maximum and minimum values of $S$, and specify the conditions under which these maximum and minimum values can be achieved;
(2) Find the smallest positive value of $S$, ... | $$
\begin{array}{l}
\text { 3. (1) }\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2} \\
=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2002}^{2}+2 m=2002+2 m, \\
m=\frac{\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2}-2002}{2} .
\end{array}
$$
When $a_{1}=a_{2}=\cdots=a_{200}=1$ or -1, $m$ reaches its maximum value of 2003001.
When $a_{1}, a... | 57 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,627 |
2. In $\triangle A B C$, the angle bisector of $\angle B$ intersects with the external angle bisector of $\angle C$ at point $D$. If $\angle A=27^{\circ}$, then, $\angle B D C=$ $\qquad$ | 2.13.5 ${ }^{\circ}$.
As shown in Figure 4, since $\angle A=27^{\circ}$, $\angle B C E=\frac{\angle A+\angle A B C}{2}$, then
$$
\begin{array}{l}
\angle B D C=\angle B C E-\angle C B D \\
=\frac{\angle A+\angle A B C}{2}-\frac{\angle A B C}{2} \\
=\frac{\angle A}{2} .
\end{array}
$$ | 13.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,629 |
Example 9 There are 4 factories $A, B, C, D$, and $AB = a \text{ km}, BC = \frac{\sqrt{2}}{2} a \text{ km}, CD = \frac{\sqrt{2}}{4} a \text{ km}, \angle ACB = 90^\circ, \angle BCD = 120^\circ$. Now we need to find a location for a supply station $H$ such that the sum of its distances to the 4 factories, $HA + HB + HC +... | (1) As shown in Figure 6, when $A$ and $D$ are on opposite sides of line $BC$, take any point $H'$ in $\triangle ABD$ other than $C$, and connect $H'A$, $H'B$, $H'C$, and $H'D$. Then,
$$
\begin{array}{l}
H'B + H'C > BC, \\
H'A + H'D > CA + CD.
\end{array}
$$
Thus, $H'B + H'C + H'A + H'D$
$$
> CA + CB + CD.
$$
Therefo... | \frac{5 \sqrt{2}}{4} a \text{ or } \frac{2 \sqrt{2} + \sqrt{14}}{4} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,630 |
3. Given $0 \leqslant a-b \leqslant 1,1 \leqslant a+b \leqslant 4$. Then, when $a-2 b$ reaches its maximum value, the value of $8 a+2002 b$ is $\qquad$ . | 3.8.
$$
\begin{array}{l}
\text { Let } 0 \leqslant a-b \leqslant 1, \\
1 \leqslant a+b \leqslant 4, \\
\text { and } m(a-b)+n(a+b)=a-2 b .
\end{array}
$$
By comparing the coefficients of $a$ and $b$ on both sides, we get the system of equations and solve to find
$$
m=\frac{3}{2}, n=-\frac{1}{2} \text {. }
$$
Thus, $a-... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,631 |
4. If a positive integer is equal to 4 times the sum of its digits, then we call this positive integer a quadnumber. The sum of all quadnumbers is $\qquad$ .
| 4.120.
Case analysis.
There are no four-composite numbers that are single-digit numbers.
For two-digit four-composite numbers, they satisfy
$$
\overline{a b}=4(a+b) \text {. }
$$
Thus, the four-composite numbers can be found to be $12, 24, 36, 48$, and their total sum is 120. For three-digit four-composite numbers, t... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,632 |
5. The sum of all roots of the equation $x^{2}-2|x+4|-27=0$ is
保留了源文本的换行和格式。 | $5.6-2 \sqrt{5}$.
It is clear that $x=-4$ is not a root of the equation.
When $x>-4$, the root of the equation $x^{2}-2 x-35=0$ is $x=7$;
When $x<-4$, the root of the equation $x^{2}+2 x-19=0$ is $x=$ $-1-2 \sqrt{5}$.
Therefore, there are two roots 7 and $-1-2 \sqrt{5}$. | 5.6-2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,633 |
6. If when $m$ takes any real number not equal to 0 and 1, the parabola $y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}$ passes through two fixed points in the Cartesian coordinate system, then, the distance between these two fixed points is $\qquad$ | $6.4 \sqrt{5}$.
Taking $m=\frac{1}{2}$ yields the parabola equation $y=-x^{2}+4 x+5$;
Taking $m=\frac{1}{4}$ yields the parabola equation $y=-3 x^{2}+8 x+11$.
Solving simultaneously, the coordinates of the two fixed points are $M(-1,0), N(3,8)$. | 6.4 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,634 |
7. The 3 roots of the equation $x^{3}-\sqrt{3} x^{2}-(2 \sqrt{3}+1) x+3+\sqrt{3}=0$ are $\qquad$ . | $7.1+\sqrt{3}, \frac{-1 \pm \sqrt{1+4 \sqrt{3}}}{2}$.
The left side of the original equation is transformed as follows:
$$
\begin{array}{l}
x^{3}-(\sqrt{3}+1) x^{2}+x^{2}-(2 \sqrt{3}+1) x+3+\sqrt{3}=0, \\
x^{2}[x-(\sqrt{3}+1)]+[x-(\sqrt{3}+1)](x-\sqrt{3})=0, \\
{[x-(\sqrt{3}+1)]\left[x^{2}+x-\sqrt{3}\right]=0,}
\end{ar... | 1+\sqrt{3}, \frac{-1 \pm \sqrt{1+4 \sqrt{3}}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,635 |
8. In Rt $\triangle A B C$, $\angle A=30^{\circ}$, the angle bisector of $\angle A$ is $1 \mathrm{~cm}$. Then, the area of $\triangle A B C$ is $\qquad$ | 8. $\frac{3+2 \sqrt{3}}{24}$.
As shown in Figure 5, $\frac{A C}{A B}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$,
thus, $\frac{C D}{D B}=\frac{A C}{A B}=\frac{\sqrt{3}}{2}$.
Let $C D=\sqrt{3} x, D B=2 x$, then
$$
B C=(2+\sqrt{3}) x, A C=\sqrt{3} B C=\sqrt{3}(2+\sqrt{3}) x \text {. }
$$
Applying the Pythagorean theorem, we ge... | \frac{3+2 \sqrt{3}}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,636 |
9. Given:
$$
\text { Profit margin of goods }=\frac{\text { Selling price of goods }- \text { Cost price of goods }}{\text { Cost price of goods }} \text {. }
$$
A merchant operates two types of goods, A and B. The profit margin of each piece of good A is $40 \%$, and the profit margin of each piece of good B is $60 \... | $9.48 \%$.
Let the purchase price of item A be $a$ yuan, then the selling price is $1.4 a$ yuan;
the purchase price of item B be $b$ yuan, then the selling price is $1.6 b$ yuan.
Assume item A is sold $x$ units, then item B is sold $\frac{3}{2} x$ units. Set up the equation
$$
\frac{0.4 a x+0.6 b \times \frac{3}{2} x}{... | 48\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,637 |
10. Design a right-angle ruler $ABC, BC$ on the ground, $AB$ perpendicular to the ground, and $AB=10 \mathrm{~cm}$. Move a wheel with a radius of no less than $10 \mathrm{~cm}$, so that the wheel is tightly against point $A$, and is tangent to $BC$ at point $D$ (as shown in Figure 2). The design requirement is that the... | 10.505.
$$
A B=10, B C=100 \text {. }
$$
Let $O A=O D=r$.
When $D$ coincides with $C$, as shown in Figure 6, the radius $r$ is at its maximum, which is the maximum scale marked, at this time,
$$
O H=r-10, A H=B C=100,
$$
Therefore, $100^{2}+(r-10)^{2}=r^{2}$.
Solving for $r$ gives $r=505$. | 10.505 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,638 |
1. At an international conference, there are four official languages. Any two conference delegates can discuss in one of these four languages. Prove: at least $60 \%$ of the conference delegates can speak the same language.
(2002, Romania for IMO and Balkan Mathematical Olympiad selection exam (second round)) | Proof: Assume the four languages are denoted as $1, 2, 3, 4$.
(1) If there exists a conference delegate who only speaks one language, then it is evident that all other delegates speak this language.
(2) Each conference delegate speaks at least two languages, and among those who speak only two languages, there is no com... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,639 |
2. Let $A$ be a $k$-element subset of the set $\{1,2,3, \cdots, 16\}$, and any two subsets of $A$ have different sums of elements. For any $(k+1)$-element subset $B$ of the set $\{1,2,3, \cdots, 16\}$ that contains $A$, there exist two subsets of $B$ whose sums of elements are equal.
(1) Prove: $k \leqslant 5$;
(2) Fin... | Proof: (1) Since $A$ has $2^{k}$ subsets (including the empty set), and the sum of elements of any two subsets is not equal, the sum of elements of $A$ must have at least $2^{k}-1$ different values.
If $k \geqslant 7$, then $2^{k}-1>16 k$, which is impossible.
If $k=6$, consider the one-, two-, three-, and four-element... | 66 | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,640 |
3. Given the Cartesian coordinate system $x O y, O$ is the origin. $A$ is an integer point, and the length of $O A$ is an integer power of an odd prime. Prove: Among the integer points on the circle with $O A$ as its diameter, at least half of them satisfy that the distance between any two points is an integer.
(2002, ... | Proof: Let $OA = p^n$, where $p$ is a prime number greater than 2, and $n$ is a positive integer. Let $A(x, y)$, and it satisfies $x^2 + y^2 = p^n$. Then $x$ is odd and $y$ is even, or $x$ is even and $y$ is odd. Without loss of generality, consider the first case.
As shown in Figure 1, draw a diameter $MN$ through th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,642 |
5. Consider an infinite strip of squares, some of which contain coins. At each step, you can choose one of the following two operations:
(i) If two adjacent squares, such as the $(n-1)$-th and the $n$-th squares, contain coins, then take one coin from each of these squares and place one of the coins in the $(n+1)$-th s... | Proof: (1) Since the number of coins is finite, the number of the first type of operations is finite. Let $a_{i}$ be the number of coins in the $i$-th square, $S=\sum_{i=1}^{+\infty} i a_{1}$, after one operation of the second type, the value of $S$ decreases to $s-1$. Since $s$ is non-negative, the number of the secon... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,644 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.