problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 3 In $\odot O$, chord $CD \parallel AB$, and intersects diameter $EF$ at a $45^{\circ}$ angle. The radius of $\odot O$ is 1. Prove: $PA \cdot QC + PB \cdot \overline{QD} < 2$. (1981, National High School League) | Proof: Connect $O A, O B, O C, O D$. Then $\angle 1=$
$$
\begin{array}{rl}
135^{\circ}-\angle A, \angle 2=45^{\circ}-\angle C, \angle 3=45^{\circ}-\angle B \\
= & 45^{\circ}-\angle A, \angle 4=135^{\circ}-\angle D=135^{\circ}- \\
\angle C & C \\
P A \cdot Q C+P B \cdot Q D<2 \\
\Leftrightarrow & \frac{P A \cdot Q C}{O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,932 |
Example 4 Let $P$ be a point inside $\triangle A B C \cdots$ . Prove that: $\angle P A B$, $\angle P B C$, $\angle P C A$ include at least one that is not greater than $30^{\circ}$. | Proof: Let $\angle P A B = \alpha, \angle P B C = \beta, \angle P C A = \gamma$, and the distances from $P$ to $A B, B C, C A$ be $d_{1}, d_{2}, d_{3}$, respectively. Using $A, B, C$ to denote $\angle C A B, \angle A B C, \angle B C A$, respectively, we have:
$$
\begin{array}{l}
d_{1} = P A \sin \alpha = P B \sin (B - ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,933 |
Example 6 A cyclic quadrilateral $ABCD$ with sides $a, b, c, d$ is also a tangential quadrilateral. Prove: its area $S=\sqrt{a b c d}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}\text { Proof: } \because \angle A+\angle C=180^{\circ}, \\ \therefore B D^{2}=a^{2}+d^{2}-2 a d \cos A \\ =b^{2}+c^{2}+2 b c \cos A . \\ \text { Therefore, } \cos A=\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2(a d+b c)} . \\ \therefore \sin A=\sqrt{1-\cos ^{2} A} \\ =\sqrt{1-\left(\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2... | null | Geometry | proof | Yes | Yes | cn_contest | false | 709,935 |
Example 7 In an acute-angled $\triangle ABC$, the angle bisector of $\angle A$ intersects $BC$ at $L$ and the circumcircle at $N$. $LK \perp AB$ at $K$, and $LM \perp AC$ at $M$. Prove that $S_{\text{quadrilateral } AK \cup M}=S_{\triangle ABC}$.
(28th IMO) | Prove: Connect $K M$ and $C N$.
Obviously, $A N$ is the perpendicular bisector of $K M$,
$\therefore S_{\text {quadrilateral } A K N M}=\frac{1}{2} A N \cdot K M$.
$\because A, K, L, M$ are concyclic and $A L$ is the diameter,
$$
\therefore \frac{K M}{\sin \angle B A C}=A L \text {. }
$$
Thus, $K M=A L \sin \angle B A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,936 |
Example $8 . \odot O_{1}$ and $\odot O_{2}$ intersect at $P$ and $Q$, a line through $P$ intersects $\odot O_{1}$ at $A$ and $\odot O_{2}$ at $B$, such that $P A \cdot P B$ is maximized. | Analysis: As shown in the figure,
$$
\begin{array}{l}
P A=2 r \sin \alpha, P B=2 R \sin \beta, \\
\therefore P A \cdot P B=4 R r \sin \alpha \sin \beta \\
=2 R r[\cos (\alpha-\beta)-\cos (\alpha+\beta)] .
\end{array}
$$
Since $\angle A$ and $\angle B$ are constants, then $\alpha+\beta$ is a constant.
Therefore, when $\... | A P \text{ should be the external angle bisector of } \triangle O_{1} O_{2} P | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,937 |
Example 9 Prove: For any point $P$ on the circumcircle $\odot O$ of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ with radius $R$, the sum of the squares of the distances from $P$ to each vertex is a constant, and find this constant. | Proof: Let $P$ be a point on $\widehat{A_{1} A_{n}}$, the side length of the regular $n$-sided polygon be $a$, and the area be $S$.
$$
\begin{array}{l}
\because \angle A_{1} O A_{2}=\left(\frac{360}{n}\right)^{\circ}, \\
\therefore S_{\triangle A_{1} O A_{2}}=\frac{1}{2} R^{2} \sin \left(\frac{360}{n}\right)^{\circ} . ... | 2 n R^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 709,938 |
Example 3 Given $a x+b y-2 c=0, a b-c^{2}>$ 0. Prove: the value of $x y$ does not exceed 1.
(1987, Sichuan Province Junior High School Mathematics Competition) | Proof: Let $x y=m$. From the given conditions, we have
$$
a x+b y=2 c, a x \cdot b y=m a b \text {. }
$$
Therefore, $a x, b y$ are the two real roots of the quadratic equation in $t$: $t^{2}$ $2 c t+a b m=0$.
$$
\begin{array}{l}
\therefore \Delta=4 c^{2}-4 a b m \geqslant 0 . \\
\because a b>c^{2} \geqslant 0, \\
\the... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,939 |
Example 10 Let $\angle A$ be the smallest interior angle of $\triangle ABC$. Points $B$ and $C$ divide the circumcircle of $\triangle ABC$ into two arcs. Let $U$ be a point on the arc that does not contain $A$ and is not equal to $B$ or $C$. The perpendicular bisectors of segments $AB$ and $AC$ intersect segment $AU$ a... | Proof: $\because$ Point $V$ is on the perpendicular bisector of line segment $AB$,
$$
\therefore \angle VAB = \angle VBA = \alpha \text{.}
$$
$\because \angle A$ is the smallest interior angle of $\triangle ABC$, and $\alpha < \angle CAB$,
$\therefore \alpha < \angle CBA$. That is, $V$ is inside $\angle ABC$.
Similarly... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,940 |
For example, $11 M$ is any point on the side $AB$ of $\triangle ABC$, $r_{1}, r_{2}, r$ are the inradii of $\triangle AMC, \triangle BMC, \triangle ABC$ respectively; $\rho_{1}, \rho_{2}, \rho$ are the exradii of these triangles (with respect to $\angle ACB$).
Prove: $\frac{r_{1}}{\rho_{1}} \cdot \frac{r_{2}}{\rho_{2}}... | Proof: Let $\angle C A B, \angle A B C, \angle B C A, \angle A M C$ be $\alpha, \beta, \gamma, \theta$. In $\triangle A P O$, $A P$
$$
\begin{aligned}
=r \operatorname{ctg} \frac{\alpha}{2} . & \\
\therefore A B & =A P+B P \\
& =r \operatorname{ctg} \frac{\alpha}{2}+r \operatorname{ctg} \frac{\beta}{2} \\
& =r\left(\op... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,941 |
2. For a cyclic quadrilateral $ABCD$ inscribed in $\odot O$, the extensions of two opposite sides $DA$ and $CB$ meet at $P$, $M$ is the midpoint of $CD$, and $PM$ intersects $AB$ at $E$. Prove:
$$
\frac{AE}{BE} = \frac{PA^2}{PB^2}.
$$ | (Hint: Use the Law of Sines in $\triangle P A E, \triangle P B E, \triangle P D M, \triangle P C M$, eliminate $\alpha, \beta, \gamma$ and use the Secant Theorem to get the result.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,943 |
3. In $\triangle A B C$, $\angle C=2 \angle A$. Prove:
$$
\frac{1}{3} b<c-a<\frac{1}{2} b .
$$ | (提示: Just prove $\frac{1}{3}<\frac{c}{b}-\frac{a}{b}<\frac{1}{2}, \frac{1}{3}<$ $\frac{\sin C-\sin A}{\sin B}<\frac{1}{2}$. Note $\sin B=\sin 3 A, 0^{\circ}<3 \angle A<$ $180^{\circ}, 0^{\circ}<\frac{1}{2} \angle A<30^{\circ}$.) | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,944 |
5. In the quadrilateral $ABCD$, $AC$ and $BD$ intersect at $O, \angle AOB = \alpha, AB = a, BC = b, CD = c, DA = d$. Find the area of quadrilateral $ABCD$. | (In $\triangle O A B$, $\triangle O B C$, $\triangle O C D$, $\triangle O D A$, use the cosine theorem and area formula respectively.) | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,946 |
6. $\odot O$ is tangent to the sides $AB$ of the equilateral $\triangle ABC$ at $M$, $BC$ at $N$, and $AC$ at $L$. The radius of $\odot O$ is $r$, and $P$ is any point on $\odot O$. Prove that $P A^{2}+P B^{2}+P C^{2}$ is a constant. | (Let $\angle P M A=\alpha, \angle P L A=60^{\circ}-\alpha$, $\angle P O M=2 \alpha, P M=m, P N=n, P L=l, A M=a$. In $\triangle P M A, \triangle P M B$, use the cosine rule to find $P A^{2}+P B^{2}=2 a^{2}+2 m^{2}$, then find $P A^{2}+P B^{2}+P C^{2}=3 a^{2}+m^{2}+n^{2}+l^{2}$. Next, in $\triangle O P M, \triangle O P N... | 3a^2 + 6r^2 | Geometry | proof | Yes | Yes | cn_contest | false | 709,947 |
7. Through a fixed point $P$ inside $\angle A$, draw a line intersecting the two sides at $B, C$, such that $\frac{1}{P B}+\frac{1}{P C}$ is maximized. | (Draw $A D \perp B C$ at $D$, let $A D=h$, the areas of $\triangle A B P$ and $\triangle A C P$ be $S_{1}, S_{2}$, and the area of $\triangle A B C$ be $S$,
$$
\begin{array}{l}
\angle B A P=\alpha, \angle C P A=\beta \cdot \frac{1}{P B}+\frac{1}{P C}=\frac{h}{2} . \\
\left(\frac{1}{S_{1}}+\frac{1}{S_{2}}\right)=\frac{h... | \frac{\sin A}{P A \sin \alpha \sin \beta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,948 |
8. $I$ is the incenter of $\triangle ABC$, $ID \perp BC$, $IE \perp AC$, $IF \perp AB$. Let $AI=l$, $BI=m$, $CI=n$. Prove: $a l^{2}+b m^{2}+c n^{2}=a b c$. | (Tip: $E F=A I \sin \angle B A C=\frac{l a}{2 R}$, so, $S_{\text {adjacentE }}=\frac{1}{2} E F \cdot A I=\frac{l^{2} a}{4 R}$. Similarly, find out
) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,949 |
Example 4 Given real numbers $x, y, z$ satisfy $x+y=6$, $z^{2}=xy-9$. Prove: $x=y$.
(1983, Tianjin Junior High School Mathematics Competition) | Proof: $\because x+y=6, x y=z^{2}+9$,
$\therefore x, y$ can be regarded as the two roots of the equation $t^{2}-6 t+\left(z^{2}+9\right)$ $=0$.
$\because x, y$ are both real numbers,
$$
\therefore \Delta=36-4 z^{2}-36=-4 z^{2}=0 \text {, which gives } z^{2} \leqslant
$$
0.
Since $z$ is a real number, it must have $z^{2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,950 |
Proposition 1 Let $\triangle A B C$ be inscribed in $\odot O, U$ be a point on $\odot O$ different from $A, B, C$, and $A U$ is not perpendicular to $A B, A C$. Let the perpendicular bisectors of segments $A B, A C$ intersect line $A U$ at $V$ and $W$, respectively. Then
(1) $V$ coincides with $W$ if and only if $A U$ ... | Proof: (1) omitted. Only prove (2).
As shown in Figure 2, when $\angle A=90^{\circ}$,
$\angle B A V+\angle C A W=90^{\circ}$,
and $\angle A B V=\angle B A V, \angle A C W=\angle C A W$.
Thus, $\angle A V B+\angle A W C=180^{\circ}$.
Therefore, $B V \parallel C W$.
Conversely, when $B V \parallel C W$,
$$
\angle A V B+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,951 |
Proposition 2 Let $\triangle A B C$ be inscribed in $\odot O, \angle A \neq 90^{\circ}, U$ be a point on $\odot O$ different from $A, B, C$, and $A U$ does not pass through the center and is not perpendicular to $A B$ and $A C$. The perpendicular bisectors of segments $A B$ and $A C$ intersect line $A U$ at $V$ and $W$... | Proof: (1) Denote the other intersection points of $BV$ and $CW$ with $\odot O$ as $X$ and $Y$, respectively. From the given conditions, we have $VB = VA$. When $V$ is inside $\odot O$, as stated in the original problem, $EX = AU$. When $V$ is on $\odot O$, $U$, $V$, and $X$ coincide, and thus $BX = AU$; when $V$ is ou... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,952 |
Proposition 3 As set in Proposition 2, let the radius of $\odot O$ be $R$.
(1) When $O$ and $T$ are on the same side of $BC$, $O T^{2}=R^{2}-T B \cdot T C, O T \cdot B C=R|T B-T C|$;
(2) When $O$ and $T$ are on opposite sides of $BC$, $O T^{2}=R^{2}+T B \cdot T C, O T \cdot B C=R(T B+T C)$;
(3) When $T$ coincides with ... | Proof: (1) Draw the circumcircle of $\triangle O B C$ as shown in Figure 8. When $T B > T C$, extend $O T$ and $B C$ to intersect at $E$, then
$$
\begin{array}{l}
\angle C T E= \\
\angle O B C=\angle Q C B . \\
\text { Hence } \angle O T C=\angle O C E . \\
\text { Therefore, } \triangle O T C \backsim \triangle O C E,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,953 |
Proposition Let the incenter of $\triangle A B C$ be $I$, and the circumcircle intersects $A I, B I, C I$ at $A^{\prime}, B^{\prime}, C^{\prime}$. Prove:
$$
I A \cdot I B \cdot I C \leqslant I A^{\prime} \cdot I B^{\prime} \cdot I C^{\prime}
$$ | Let $B C=a, C A=b, A B=c$, $B^{\prime} C^{\prime}=a^{\prime}, C^{\prime} A^{\prime}=b^{\prime}, A^{\prime} B^{\prime}=c^{\prime}$. From the problem and the Law of Sines, we easily get
$$
\begin{array}{l}
A I=\frac{c \sin \frac{B}{2}}{\cos \frac{C}{2}} \text {, etc., } \\
A^{\prime} I=\frac{b^{\prime} \sin \frac{A}{2}}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,954 |
Question: Let the centroid of $\triangle ABC$ be $G, AG, BG$, and $CG$ intersect the opposite sides at $D, E, F$, and intersect the circumcircle at $A', B', C'$. Prove:
$$
\frac{A' D}{D A}+\frac{B' E}{E B}+\frac{C' F}{F C} \geqslant 1 \text{. }
$$
This problem has been solved in multiple articles. This article provide... | Prove: As shown in the figure, the medians of sides $a$, $b$, and $c$ are denoted as $m_{a}$, $m_{b}$, and $m_{c}$, respectively. Applying the intersecting chords theorem, we have
$$
\frac{A^{\prime} D}{D A}=\frac{A^{\prime} D \cdot D A}{D A^{2}}=\frac{B D \cdot E \cdot C}{D A^{2}}=\frac{a^{2}}{4 m_{a}^{2}}.
$$
Simila... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,955 |
In $\triangle A B C$, we have
$$
\sum \cos A \geqslant \frac{1}{2} \sum \cos ^{2} \frac{B-C}{2} .
$$
Equality holds if and only if $\triangle A B C$ is an equilateral triangle. | Proof: It is easy to see that (2) is equivalent to
$$
5 \sum \cos A \geqslant 2 \sum \sin B \sin C + 3.
$$
Let the circumradius, inradius, and semiperimeter of $\triangle ABC$ be $R$, $r$, and $s$, respectively. Then we have the identities:
$$
\begin{array}{l}
\sum \cos A = \frac{R + r}{R}, \\
\sum \sin B \sin C = \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,956 |
1. Let $m=\sqrt{5}+1$. Then the integer part of $m+\frac{1}{m}$ is $\qquad$ . | \begin{array}{l}\text { 1. } m=\sqrt{5}+1, \frac{1}{m}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}, \\ \therefore m+\frac{1}{m}=\frac{5}{4} \sqrt{5}+\frac{3}{4},\left[m+\frac{1}{m}\right]=3 .\end{array} | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,957 |
2. In Rt $\triangle A B C$, the lengths of the two legs $A B$ and $A C$ are 1 cm and 2 cm, respectively. Then, the length of the angle bisector of the right angle is equal to cm. | 2. As shown in the figure, $A D$ is the angle bisector of the right angle $A$, and a line through $B$ parallel to $D A$ intersects the extension of $C A$ at point $E$. Then $\angle E B A=$
$$
\begin{aligned}
\angle B A D & =45^{\circ}, \\
A E & =A B=1, \\
E B & =\sqrt{2} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { A... | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,958 |
3. Given $x^{2}-x-1=0$. Then, the value of the algebraic expression $x^{3}$ $-2 x+1$ is $\qquad$ . | $\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
The translation is as follows:
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,959 |
4. Given that $m$ and $n$ are rational numbers, and the equation $x^{2}+$ $m x+n=0$ has a root $\sqrt{5}-2$. Then the value of $m+n$ is $\qquad$ . | 4. Since $m, n$ are rational, the other root is $-\sqrt{5}-2$, thus by Vieta's formulas,
$$
\begin{array}{l}
w:=4, n=-1 . \\
\therefore m+n=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,960 |
For example, $5 y=\frac{3 x^{2}+3 x+4}{x^{2}+x+1}$, the maximum value is ( . ).
(A) $4 \frac{1}{3}$
(B) 4
(C) $3 \frac{3}{4}$
(D) 3
(1992, Shandong Province Junior High School Mathematics Competition) | Solution: The given equation can be transformed into a quadratic equation in $x$: $(y-3) x^{2}+(y-3) x+(y-4)=0$.
According to the problem, we have
$$
\begin{array}{c}
\Delta=(y-3)^{2}-4(y-3)(y-4) \geqslant 0 . \\
\because y-3=\frac{3 x^{2}+3 x+4}{x^{2}+x+1}-3 \\
=\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}>0,
... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,961 |
6. The number of integer pairs $(m, n)$ that satisfy $1998^{2}+m^{2}=1997^{2}+n^{2}(0<m$ $<n<1998)$ is $\qquad$.
| 6. $n^{2}-m^{2}=3995=5 \times 17 \times 47,(n-m)(n+m)=5 \times 17 \times 47$, obviously any integer factorization of 3995 can yield $(m, n)$, given the condition $(0<m<n<1998)$, thus there are 3 integer pairs $(m, n)$ that satisfy the condition. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,963 |
8. In $\mathrm{Rt} \triangle A B C$, there is a point $M$ on the right-angle side $A B$ and a point $P$ on the hypotenuse $B C$. It is known that $M P \perp B C$, the area of $\triangle B M P$ is
equal to half the area of quadrilateral MPCA, $B P$
$=2$ cm, $P C=3$
cm. Then the area of $\mathrm{Rt} \triangle A B C$ i... | 8. $\triangle M B P \subset \triangle C B A$,
$$
\begin{array}{l}
S_{\triangle M B P}: S_{\triangle C M A}=1: 3, \\
B P: B A=1: \sqrt{3}, \\
\therefore B A=2 \sqrt{3}, \\
A C=\sqrt{13} . \\
S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{3} \times \sqrt{13}=\sqrt{39} .
\end{array}
$$ | \sqrt{39} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,965 |
9. Given that the area of square $A B C D$ is 35 square centimeters, $E, F$ are points on sides $A B$, $B C$ respectively, $A F$ and $C E$ intersect at $G$, and the area of $\triangle A B F$ is 5 square centimeters, the area of $\triangle B C E$ is 14 square centimeters. Then, the area of quadrilateral $B E G F$ is $\q... | 9. $\frac{B F}{B C}=\frac{S_{\triangle A B F}}{S_{\triangle A B C}}=\frac{2}{7}$, similarly $\frac{B E}{B A}=\frac{4}{5}$. Connect $B G$,
Let $S_{\triangle A G E}=a, S_{\triangle E C B}=$ $b, S_{\triangle B G F}=c, S_{\triangle F C C}=d$.
Given $a+b+c=5$, $b+c+d=14$,
Solving, we get $b=\frac{28}{27}, c=\frac{100}{27... | \frac{128}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,966 |
10. Distribute 100 apples to several people, with each person getting at least one apple, and each person receiving a different number of apples. Then, the maximum number of people is $\qquad$. | 10. Suppose there are $n$ people, and the number of apples distributed to each person is $1,2, \cdots, n$. It follows that
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2} \leqslant 100 .
$$
$\therefore n \leqslant 13$, i.e., there are at most 13 people. | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,967 |
11. Let $a, b$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-$ $a-2 b$ is $\qquad$. | 11 .
$$
\begin{array}{l}
a^{2}+a b+b^{2}-a-2 b \\
=a^{2}+(b-1) a+b^{2}-2 b \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4} b^{2}-\frac{3}{2} b-\frac{1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 .
\end{array}
$$
When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the abo... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,968 |
14. Line $A B$
and $A C$ are
tangent to circle $O$ at
points $B$ and $C$,
respectively. $P$ is a
point on the circle,
and the distances from $P$ to $A B$ and $A C$
are 4 cm and 6 cm, respectively. The distance from $P$ to $B C$ is $\qquad$ cm. | $\begin{array}{l}\text { 14. As shown in the figure } P M \perp A B, \\ P N \perp A C, P Q \perp B C . P \text {, } \\ Q, C, N \text { are concyclic, } P \text {, } \\ Q, B, N \text { are concyclic, } \\ \angle M P Q \\ =180^{\circ}-\angle M B Q \\ =180^{\circ}-\angle N C Q=\angle N P Q, \\ \angle M Q P=\angle M B P=\a... | 2\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,970 |
15. Every book has an international book number:
ABCDEFGHIJ
where $A B C D E F G H I$ are composed of nine digits, and $J$ is the check digit.
$$
\text { Let } \begin{aligned}
S= & 10 A+9 B+8 C+7 D+6 E+5 F \\
& +4 G+3 H+2 I,
\end{aligned}
$$
$r$ is the remainder when $S$ is divided by 11. If $r$ is not 0 or 1, then $J... | $$
\text { 15. } \begin{aligned}
S=9 \times 10 & +6 \times 9+2 \times 8+y \times 7+7 \times 6 \\
& +0 \times 5+7 \times 4+0 \times 3+1 \times 2
\end{aligned}
$$
$\therefore S$ the remainder when divided by 11 is equal to the remainder when $7 y+1$ is divided by 11.
From the check digit, we know that the remainder when... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,971 |
Example 6 If $x^{2}+x+2 m$ is a perfect square, then $m=$ $\qquad$
$(1989$, Wu Yang Cup Junior High School Mathematics Competition) | Solution: If $x^{2}+x+2 m$ is a perfect square, then the quadratic equation in $x$, $x^{2}+x+2 m=0$, has equal roots. Thus,
$$
\Delta=1-4 \times(2 m)=0 .
$$
Solving for $m$ yields $m=\frac{1}{8}$.
Explanation: According to the sufficient condition for a real-coefficient quadratic polynomial $a x^{2}+b x+c(a \neq 0)$ t... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,972 |
一、(Full marks 20 points) Find all positive real numbers $a$ such that the equation $x^{2}-a x+4 a=0$ has only integer roots.
---
Translation:
I. (Full marks 20 points) Find all positive real numbers $a$ such that the equation $x^{2} - a x + 4 a = 0$ has only integer roots. | Let the two integer roots be $x, y(x \leqslant y)$,
then $\left\{\begin{array}{l}x+y=a>0, \\ x y=4 a>0 .\end{array}\right.$
$$
\therefore \frac{a}{2} \leqslant y \leqslant a, 4 \leqslant x \leqslant 8 \text {. }
$$
Also, it can be deduced that $x \neq 4$,
$$
\therefore a=\frac{x^{2}}{x-4} \text {. }
$$
Since $x$ is a... | a=25 \text{ or } 18 \text{ or } 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,973 |
II. (Full marks
20 points) Given $P$
is a point inside
square $ABCD$, $O$ is the intersection of $AC$
and $BD$,
$M$ and $N$ are
the midpoints of
$PB$ and $PC$, respectively, and $Q$ is the intersection of $AN$ and $DM$. Prove:
(1) $P$, $Q$, and $O$ are collinear;
(2) $PQ = 2OQ$. | II. Connect $P O$. Let $P O$ intersect $A N$ and $D M$ at points $Q^{\prime}$ and $Q^{\prime \prime}$, respectively.
In $\triangle P A C$,
$\because A O=O C, P N=N C$,
$\therefore Q^{\prime}$ is the centroid, $P Q^{\prime}=2 O Q^{\prime}$.
In $\triangle P D B$,
$\because D O=B O, B M=M P$,
$\therefore Q^{\prime \prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,974 |
Three, (Full marks 20 points) Try to write down 5 natural numbers, such that the larger of any two of these numbers can be divided evenly by their difference.
---
Please note that the translation maintains the original text's line breaks and format as requested. | Three, 1680, 1692, 1694, 1695, 1696 are 5 numbers that meet the conditions. (Note: The answer is not unique)
The above 5 numbers can be found through the following steps:
First step: $2,3,4$ are three numbers that meet the requirements.
Second step: Let $a, a+2, a+3, a+4$ be four numbers that meet the conditions, then ... | 1680, 1692, 1694, 1695, 1696 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,975 |
2. If the difference between the roots of the equation $x^{2}+p x+1=0(p>0)$ is 1, then $p$ equals ( ).
(A) 2
(B) 4
(C) $\sqrt{3}$
(D) $\sqrt{5}$ | 2. (D).
From $\Delta=p^{2}-4>0$, we get $p>2$. Let $x_{1}, x_{2}$ be the two roots of the quadratic equation, then $x_{1}+x_{2}=-\frac{1}{2}, x_{1} x_{2}=1$. This leads to $\left(x_{1}-x_{2}\right)^{2}-$ $\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$, so $1^{2}=(-p)^{2}-4$. Therefore, $p^{2}=$ $5, p=\sqrt{5}(p>2)$. Henc... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,977 |
4. Given $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=$ $\frac{c+a}{b}=p$. Then, the line $y=p x+p$ must pass through
(A) the first and second quadrants
(B) the second and third quadrants
(C) the third and fourth quadrants
(D) the first and fourth quadrants | 4. (B).
From the conditions, we get $\left\{\begin{array}{l}a+b=p c, \\ b+c=p a, \\ a+c=p b .\end{array}\right.$
Adding the three equations, we get $2(a+b+c)=p(a+b+c)$.
$\therefore p=2$ or $a+b+c=0$.
When $p=2$, $y=2 x+2$. Thus, the line passes through the first, second, and third quadrants.
When $a+b+c=0$, let's ass... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,979 |
5. If the system of inequalities $\left\{\begin{array}{l}9 x-a \geqslant 0, \\ 8 x-b<0\end{array}\right.$ has only the integer solutions $1,2,3$, then the number of integer pairs $(a, b)$ that satisfy this system of inequalities is $(\quad)$.
(A) 17
(B) 64
(C) 72
(D) 81 | 5. (C).
From the original system of inequalities, we get $\frac{a}{9} \leqslant x<\frac{b}{8}$.
Drawing the possible interval of the solution set of this system of inequalities on the number line, as shown in the figure below.
It is easy to see that $0<\frac{a}{9} \leqslant 1, 3<\frac{b}{8} \leqslant 4$.
From $0<\frac... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 709,980 |
6. In rectangle $A B C D$, it is known that $A D=12, A B$ $=5, P$ is any point on side $A D$, $P E \perp B D, P F$ $\perp A C, E, F$ are the feet of the perpendiculars, respectively. Then, $P E+P F=$ $\qquad$ | $=.6 \cdot \frac{60}{13}$.
Draw $A G \perp B D$ at $G$.
$\because$ In an isosceles triangle, the sum of the distances from any point on the base to the two legs is equal to the altitude on the leg,
$$
\therefore P E+P F=A G \text {. }
$$
Given $A D=12, A B=5$, then $B D=13$.
$$
\therefore A G=\frac{12 \times 5}{13}=\f... | \frac{60}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,981 |
7. Given that the line $y=-2 x+3$ intersects the parabola $y=$ $x^{2}$ at points $A$ and $B$, and $O$ is the origin. Then, the area of $\triangle O A B$ is $\qquad$ . | 7.6 .
As shown in the figure, the line $y=$
$-2 x+3$ intersects the parabola $y$ $=x^{2}$ at points $A(1,1), B(-3,9)$.
Construct $A A_{1}, B B_{1}$ perpendicular to the $x$-axis, with feet of the perpendiculars at
$$
\begin{array}{l}
A_{1} 、 B_{1} . \\
\therefore S_{\triangle O A B}=S_{\text {trapezoid } A A_{1} B_{1}... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,982 |
Example 7 When $m=$ $\qquad$, the polynomial $6 x^{2}+$ $m x y-4 y^{2}-x+17 y-15$ can be factored into the product of two binary linear trinomials in $x$ and $y$.
(1990, Hope Cup Junior High School Mathematics Invitational Competition) | Solution: $\because 6 x^{2}+m x y-4 y^{2}-x+17 y-15$
$$
\begin{array}{l}
=6 x^{2}+(m y-1) x-\left(4 y^{2}-17 y+15\right), \\
\therefore \Delta=(m y-1)^{2}+24\left(4 y^{2}-17 y+15\right) \\
=\left(m^{2}+96\right) y^{2}-2(m+204) y \\
+361. \\
\end{array}
$$
For the polynomial $6 x^{2}+m x y-4 y^{2}-x+17 y-15$ to be fact... | m=5 \text{ or } m=-\frac{58}{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,983 |
8. Given the inner diameter of a ring is $a \mathrm{~cm}$, and the outer diameter is $b \mathrm{~cm}$, 50 such rings are linked together one by one to form a chain. Then, the length of the chain when stretched out is $\qquad$ $\mathrm{cm}$. | $8.49 a+b$.
As shown in the figure, when there are 3 rings, the chain length is
$$
3 a+\frac{b-a}{2} \times 2=2 a+b(\mathrm{~cm}) \text {. }
$$
When there are 50 rings, the chain length is
$$
50 a+2 \times \frac{b-a}{2}=49 a+b(\mathrm{~cm}) .
$$ | 49a + b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,984 |
9. Given the equation $a^{2} x^{2}-\left(3 a^{2}-8 a\right) x+2 a^{2}$ $-13 a+15=0$ (where $a$ is a non-negative integer) has at least one integer root. Then $a=$ $\qquad$ | $9.1,3,5$
Since $a \neq 0$, we get
$$
x_{1}=\frac{2 a-3}{a}=2-\frac{3}{a}, x_{2}=\frac{a-5}{a}=1-\frac{5}{a} \text {. }
$$
Therefore, $a$ can take 1.3 or 5. | 1,3,5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,985 |
$10 . B$ ship is at a position $45^{\circ}$ north of west from $A$ ship, the two ships are $10 \sqrt{2} \mathrm{~km}$ apart. If $A$ ship sails west, and $B$ ship sails south at the same time, and the speed of $B$ ship is twice that of $A$ ship, then the closest distance between $A$ and $B$ ships is $\qquad$ $\mathrm{km... | $10.2 \sqrt{5}$
As shown in the figure, after $t$ hours, ships $A$ and $B$ have sailed to positions $A_{1}$ and $B_{1}$, respectively. Let $A A_{1}=x$. Therefore, $B B_{1}=2 x$. Given that $A B = 10 \sqrt{2}$, we have
$$
\begin{aligned}
A C & =B C=10 . \\
\therefore A_{1} C & =|10-x|, B_{1} C=|10-2 x| \\
\therefore A_{... | 2 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,986 |
11. As shown in the figure, in isosceles right $\triangle A B C$, $A B=1, \angle A=90^{\circ}$, point $E$ is the midpoint of leg $A C$, and point $F$ is on the base $B C$ such that $F E \perp B E$. Find the area of $\triangle C E F$. | Three, 11. Construct
(D) $\perp C E$ and the extension of $E F$ intersect at $\Gamma$.
$$
\begin{array}{l}
\because \angle A B E+ \\
\angle A E B=90^{\circ}, \\
\angle C E D+\angle A E B \\
=90^{\circ}, \\
\angle A B E=\angle C E D .
\end{array}
$$
Therefore, Rt $\triangle A B E \sim R \mathrm{R} \triangle C E D$.
$$... | \frac{1}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,987 |
12. Let the parabola $y=x^{2}+(2 a+1) x+2 a+$ $\frac{5}{4}$ intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+32 a^{-6}$. | 12. (1) Since the parabola intersects the $x$-axis at only one point, the quadratic equation $x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0$ has two equal real roots. Therefore,
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$
which simplifies to $a^{2} \cdots a-1=0$.
$$
\ldots a=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$
(2... | 5796 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,988 |
13. City $A$, City $B$, and City $C$ have 10, 10, and 8 units of a certain machine, respectively. Now it is decided to allocate these machines to City $D$ (18 units) and City $E$ (10 units). It is known that the transportation cost for one unit from City $A$ to City $D$ and City $E$ is 200 yuan and 800 yuan, respective... | 13. (1) From the problem, we know that the number of machines sent from City A, City B, and City C to City I are $x, x, 18-2x$, and the number of machines sent to City E are $10-x, 10-x, 2x-10$. Therefore,
$$
\begin{array}{l}
W=200 x+300 x+400(18-2 x)+800(10- \\
x)+700(10-x)+500(2 x-10) \\
=-800 x+17200 \text {. } \\
\... | 10000 \text{ (minimum)}, 13200 \text{ (maximum)} \text{ for part (1); } 9800 \text{ (minimum)}, 14200 \text{ (maximum)} \text{ for part (2)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,989 |
1. Given $y=a x^{5}+b x^{3}+c x-5$, when $x=$ -3, $y=7$. Then, when $x=3$, the value of $y$ is ( ).
(A) -3
(B) -7
(C) -17
(D) 7 | $$
-、 1 .(\mathrm{C})
$$
When $x=-3$, we get
$$
7=a(-3)^{5}+b(-3)^{3}+c(-3)-5 \text {; }
$$
When $x=3$, we get
$$
y=a(3)^{5}+b(3)^{3}+c \times 3-5 \text {; }
$$
(1) + (2), we get $7+y=-10 . \therefore y=-17$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,990 |
3. Given $a^{-}+a^{-x}=2$. Then the value of $a^{2 x}+a^{-2 x}$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 6 | 3. (C).
$$
\begin{array}{l}
\because a^{2 x}+a^{-2 x}=\left(a^{x}\right)^{2}+\left(a^{-x}\right)^{2} \\
=\left(a^{x}+a^{-x}\right)^{2}-2=4-2=2 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,992 |
4. As shown in the figure, in $\triangle A B C$, $\angle A C B=90^{\circ}, A C=$ $A E, B C=B F$. Then $\angle E C F$ $=(\quad)$.
(A) $60^{\circ}$
(B) $45^{\circ}$
(C) $30^{\circ}$
(D) Uncertain | 4. (B).
Let $\angle E C F=x, \angle A C F=\alpha, \angle B C E=\beta$. Then we have
$$
\begin{array}{l}
\angle E F C=x+\beta, \angle F E C=x+\alpha, x+\alpha+\beta=90^{\circ} . \\
\quad \therefore \angle E F C+\angle F E C+x=180^{\circ}, 3 x+\alpha+\beta==
\end{array}
$$
$$
180^{\circ}, 2 x=90^{\circ}, x=45^{\circ} .
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,993 |
5. In $\triangle A B C$, $B C=a, A C=b, A B=$ $c, \angle C=90^{\circ}, C D$ and $B E$ are two medians of $\triangle A B C$, and $C D \perp B E$. Then $a: b: c=(\quad)$.
(A) $1: 2: 3$
(B) $3: 2: 1$
(C) $\sqrt{3}: \sqrt{2}: 1$
(D) $1: \sqrt{2}: \sqrt{3}$ | 5. (D).
As shown in the figure, let $C D$ and $B E$ intersect at $F$. Suppose $E F=x$, $D F=y$, then $B F=2 x$, $C F=2 y$. By the properties of similar triangles, we have $a^{2}=B E \cdot B F=6 x^{2},\left(\frac{b}{2}\right)^{2}=E F \cdot E B=3 x^{2},\left(\frac{c}{2}\right)^{2}=y^{2}+4 x^{2}$, which means $a^{2}=6 x^... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,995 |
1. The positions of the three numbers $a$, $b$, and $c$ on the number line are shown in the figure, and $|b|=|c|$. Then, simplify $\sqrt{b^{2}}-|a-b|+|a-c|-|b+c|=$ $\qquad$ . | 二、1.3c or $-3 b$.
$\because a 、 b$ are negative numbers, $c$ is a positive number, and $|a|>|b|=|c|$,
$$
\begin{aligned}
\therefore \text { original expression } & =-b-(b-a)+c-a \\
& =-2 b+c=3 c \text { or }-3 b .
\end{aligned}
$$ | 3c \text{ or } -3b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,996 |
2. Given $m^{2}+71 m-1997=0, n^{2}+71 n$ $-1997=0$, and $m \neq n$. Then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ . | 2. $\frac{71}{1997}$.
It is easy to know that $m$ and $n$ are two distinct roots of the equation $x^{2}+71 x-1997=0$, then
$$
\frac{1}{m}+\frac{1}{n}=\frac{m+n}{m n}=\frac{-71}{-1997}=\frac{71}{1997} .
$$ | \frac{71}{1997} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,997 |
3. As shown in the figure, it is known that the area of $\triangle A B C$ is 1, and $\frac{B D}{D C}=\frac{C E}{E F}=\frac{A F}{F D}=\frac{1}{2}$. Therefore, the area of $\triangle D E F$ is | 3. $\frac{8}{27}$.
Since the heights of $\triangle D E F$ and $\triangle D C F$ are equal, and $\frac{F E}{F C}=\frac{2}{3}$, then
$$
\frac{S_{\triangle D E F}}{S_{\triangle D C F}}=\frac{2}{3} \text {. }
$$
Similarly, $\frac{S_{\triangle D C F}}{S_{\triangle D C A}}=\frac{2}{3}, \frac{S_{\triangle D C A}}{S_{\triang... | \frac{8}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,998 |
4. There are student Ling † Ren who need accommodation. If each room houses 4 people, then there are still 20 people without accommodation; if each room houses 8 people, then there is one room that is neither empty nor full. Therefore, the number of dormitory rooms is $\qquad$ rooms, and the number of students is $\qqu... | $4.6,44$.
Let the number of dormitories be $x$, and the number of students be $y$ (both $x, y$ are integers). Then
$$
\left\{\begin{array} { l }
{ y - 2 0 5 .
\end{array}\right.\right.
$$
Therefore, $x=6, y=44$. | 6, 44 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 709,999 |
5. Recently, the Asian football tournament has sparked Wang Xin, a junior high school student, to research football. He found that the football is made by gluing black and white leather pieces together, with black pieces being regular pentagons and white pieces being regular hexagons (as shown in the figure), and he co... | 5.20.
Let there be $x$ white blocks, then the white blocks have a total of $6x$ edges, and the black blocks have a total of 60 edges. Since each white block has 3 edges connected to black blocks and 3 edges connected to other white blocks, then $6x \div 2 = 60, \therefore x = 20$. | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,000 |
Three, (This question 20 points) Given a triangle with one side length of 2, the median on this side is 1, and the sum of the other two sides is $1+\sqrt{3}$. Find the area of the triangle.
| E, from $A D = B I = 1 \times = 1$, we have $\angle B = \angle 1, \angle C = \angle 2$.
Thus, $\angle 1 + \angle 2 = 90^{\circ}$.
Also, since $A C + A B$
$$
\begin{array}{l}
=1+\sqrt{3}, \\
\therefore(A C+A B)^{2}=(1+\sqrt{3})^{2} .
\end{array}
$$
That is, $A C^{2}+A B^{2}+2 A C \cdot A B=4+2 \sqrt{3}$.
By the Pythago... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,001 |
Four. (This question is worth 25 points) Prove that $1997 \times 1998 \times$ $1999 \times 2000+1$ is a perfect square of an integer, and find this integer.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Original expression }=x(x+1)(x+2)(x+3)+1 . \\
\because x(x+1)(x+2)(x+3)+1 \\
=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)+1 \\
=\left[\left(x^{2}+3 x+1\right)-1\right]\left[\left(x^{2}+3 x+1\right)+1\right]+1 \\
=\left(x^{2}+3 x+1\right)^{2} \\
\because 1997 \times 1998 \times 1999 \times... | 3994001 | Number Theory | proof | Yes | Yes | cn_contest | false | 710,002 |
Five. (This question is worth 25 points) On a piece of leftover material in the shape of an acute triangle, process it into a square part such that the four vertices of the square are all on the edges of the triangle. If the three sides of the triangle are $a$, $b$, and $c$, and $a > b > c$, on which side should two ve... | Let the heights on sides $a$, $b$, and $c$ be $h_{a}$, $h_{b}$, and $h_{c}$, respectively, and the area of $\triangle ABC$ be $S$. The side lengths of the squares inscribed on sides $a$, $b$, and $c$ are $x_{a}$, $x_{b}$, and $x_{c}$, respectively.
$$
\begin{aligned}
& \because \frac{h_{a}}{a}=\frac{h_{a}-x_{a}}{x_{a}}... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,003 |
1. The number of integers $x$ that satisfy the inequality $|x-1|+|x-9|<\sqrt{98}$ is ( ).
(A) 8
(B) 9
(C) 10
(D) 11 | -1. (B).
When $x\frac{10-\sqrt{98}}{2}$. Therefore, when $\frac{10-\sqrt{98}}{2}<x<9$, the original inequality becomes $10-2x<\sqrt{98}$, i.e., $x>\frac{10-\sqrt{98}}{2}$. So, $\frac{10-\sqrt{98}}{2}<x<9$, the integer solutions are $x=1, 2, 3, 4, 5, 6, 7, 8$.
When $x>9$, the original inequality becomes $2x-10<\sqrt{98}... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,004 |
2. As shown in the figure, from the obtuse vertex $A$ of the obtuse triangle $\triangle ABC$, draw the altitude $AD$. With the foot of the perpendicular $D$ as the center and $AD$ as the radius, draw a circle that intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB=c, AM=m, AN=n$, then the length of side $AC$... | 2. (C).
As shown in the figure, extend $A D$ to intersect $\odot D$ at $E$, and connect $M E$ and $M N$. Then
$$
\begin{array}{l}
\angle A B C=90^{\circ}-\angle B A D=\angle A E M=\angle A N M \\
\text { Also, since } \angle B A C= \\
\angle N A M, \\
\therefore \triangle A B C \backsim
\end{array}
$$
$\triangle N A M... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,006 |
3. Given a linear function $y=a x+b$ (where $a$ is an integer) whose graph passes through the point $(98,19)$, its intersection with the $x$-axis is $(p, 0)$, and its intersection with the $y$-axis is $(0, q)$. If $p$ is a prime number and $q$ is a positive integer, then the number of all linear functions that satisfy ... | 3. (A).
From the problem, we have $\left\{\begin{array}{l}98 a+b=19, \\ p=-\frac{b}{a}, \\ q=b .\end{array}\right.$
Since $q$ is a positive integer, by (3) we know that $b$ is a positive integer. Therefore, from (2) we know that $a$ is a negative integer.
From (i) we get $\dot{i}=19-98 a$, substituting into (2) we get... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,007 |
4. As shown in the figure, in trapezoid $A B C D$, $A B / / C D$, and $A B=2 C D, M 、 N$ are the midpoints of diagonals $A C 、 B D$. Let the perimeter of trapezoid $A B C D$ be $l_{1}$, and the perimeter of quadrilateral CDMN be $l_{2}$. Then $l_{1}$ and $l_{2}$ satisfy $(\quad)$.
(A) $l_{1}=2 l_{2}$
(B) $l_{1}=3 l_{2}... | 4. (A).
As shown in the figure, let the line containing $MN$ intersect $AD$ and $BC$ at $P$ and $Q$, respectively. Then $P$ and $Q$ are the midpoints of $AD$ and $BC$, respectively. Therefore,
$$
\begin{aligned}
P M & =N Q=\frac{1}{2} C D . \\
\therefore P Q & =P M+M N+N Q=C D+M N \\
& =\frac{1}{2}(A B+C D) .
\end{ali... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,008 |
5. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\cdots+\sqrt{1+\frac{1}{1997^{2}}+\frac{1}{1998^{2}}}$. Then the integer closest to $S$ is ( ).
(A) 1997
(B) 1908
(C) 1009
(D) 2000 | 5. (B).
When $n$ is an integer, we have
$$
\begin{array}{l}
\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-\frac{2}{n}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}\right)^{2}-2 \cdot \frac{n+1}{n} \cdot \frac{1}{n+1}+\left(\frac{1}{n+1}\right)^{2}} \\
=\sqrt{\left(\frac{n+1... | 1998 | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,009 |
6. Let $n$ be a positive integer. $0<x \leqslant 1$. In $\triangle A B C$, if $A B=n+x, B C=n+2 x, C A=n+3 x, B C$ has a height $A D=n$. Then, the number of such triangles is ( ).
(A) 10
(B) 11
(C) 12
(D) infinitely many | 6. (C).
Let $\triangle A B C$ be the condition; then $c=n+1, a=n+2 x$, $b=n+3 x, p=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 n+6 x)$. By Heron's formula, we get
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,010 |
1. Given $a, b$ are positive integers, and satisfy $\frac{a+b}{a^{2}}=\frac{4}{40}$. Then the value of $a+b$ is $\qquad$ | Let $a+b=4k$ ($k$ is a positive integer), then $a^{2}+ab+b^{2}:=49k$, i.e., $(a+b)^{2}-ab=49k$.
$$
\therefore ab=16k^{2}-49k \text{. }
$$
It is easy to know that $a, b$ are the two positive integer roots of the equation about $x$
$$
x^{2}-4kx+(16k^{2}-49k)=0
$$
By $\Delta=16k^{2}-4(16k^{2}-49k) \geqslant 0$, we get $... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,011 |
3. If $n$ is a positive integer, and $n^{2}+9 n+98$ is exactly equal to the product of two consecutive positive integers, then all values of $n$ are $\qquad$ | $3.34,14,7$.
It is easy to know that for any positive integer $n$, the following inequality always holds:
$$
n^{2}+9 n+20<n^{2}+9 n+98<n^{2}+19 n+90 \text {, }
$$
i.e., $(n+4)(n+5)<n^{2}+9 n+98<(n+9)(n+10)$.
$\therefore n^{2}+9 n+98=(n+5)(n+6)$,
or $n^{2}+9 n+98=(n+6)(n+7)$,
or $n^{2}+9 n+98=(n+7)(n+8)$,
or $n^{2}+9 n... | 34,14,7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,013 |
4. Given $\left\{\begin{array}{l}\frac{x+9 y}{9 x+7 y}=\frac{m}{n}, \\ \frac{x+9 y}{9 x+8 y}=\frac{m+a n}{b m+c n} .\end{array}\right.$
If all real numbers $x, y, m, n$ that satisfy (1) also satisfy (2), then the value of $a+b+c$ is $\qquad$. | $4 \cdot \frac{41}{37}$. For (1), by the theorem of the ratio of sums and differences, we get
$$
\begin{aligned}
\frac{m+a n}{b m+c n} & =\frac{(x+9 y)+a(9 x+7 y)}{b(x+9 y)+c(9 x+7 y)} \\
& =\frac{(1+9 a) x+(9+7 a) y}{(b+9 c) x+(9 b+7 c) y} .
\end{aligned}
$$
From (2), we have $\frac{x+9 y}{9 x+8 y}=\frac{(1+9 a) x+(9... | \frac{41}{37} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,014 |
One, (Full marks 20 points) As shown in the figure, in the convex quadrilateral $ABCD$, $AC=BD=AB$, and $AC \perp BD$, with the foot of the perpendicular being $E$. Let $I$ be the incenter of $\triangle AEB$, and $M$ be the midpoint of $AB$. Prove that $MI \perp CD$, and $MI=$ $\frac{1}{2} CD$ | Connect $A I$ extended to meet $B C$ at $N$, connect $M N$.
$$
\begin{array}{l}
\because A C=A B, I \text{ is the incenter of } \triangle A E B, \\
\therefore A N \perp B C \cdot B N=C N . \\
\therefore M N / / A C, \\
M N=\frac{1}{2} A C=\frac{1}{2} B D .
\end{array}
$$
Since $A C \perp D$, therefore $M N \perp B D$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,015 |
Example 9 In $\triangle A B C$, $A B=2, B C=1$, then the range of values for $\angle A$ is $\qquad$
(Suzhou City Junior High School Mathematics Competition) | Solution: Let $AC = x$. Then, by the cosine rule, we have
$$
BC^2 = AB^2 + AC^2 - AB \cdot AC \cdot \cos A,
$$
which is $1^2 = 2^2 + x^2 - 2 \times 2 \times x \cdot \cos A$, or equivalently, $x^2 - (4 \cos A) x + 3 = 0$.
Note that the problem now reduces to finding which values of $A$ make the above equation have rea... | 0^\circ < \dot{A} \leqslant 30^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,016 |
II. (Full marks 25 points) Given that the equation $x^{2}+a x+b=0$ has two distinct real roots. Prove that the equation $x^{4}+a x^{3}+(b-2) x^{2}-a x+1=0$ has four distinct real roots.
---
The translation maintains the original text's line breaks and format. | Let $x_{1}, x_{2}$ be the two roots of the equation $x^{2} + a x + b = 0$, and $x_{1} \neq x_{2}$. Then, by Vieta's formulas, we have
$$
\begin{aligned}
a= & -\left(x_{1} + x_{2}\right), b = x_{1} x_{2}. \\
\therefore & x^{4} + a x^{3} + (b-2) x^{2} - a x + 1 \\
= & x^{4} - \left(x_{1} + x_{2}\right) x^{3} + \left(x_{1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,017 |
Three. (Full marks 25 points) Six cassette tapes are packed in a "regular way", which means that every two boxes in a row must form a cuboid. The given dimensions of the cassette tape box are $a=11 \mathrm{~cm}, b=7 \mathrm{~cm}, c=2 \mathrm{~cm}$.
(1) Please provide a way to pack them that minimizes the surface area, ... | Three, as shown in the figure, for a cassette tape, the areas of the three faces sharing a vertex can be denoted as \( C = ab, B = ca, A = bc \).
6 cassette tapes, noting that 6 \( = 1 \times 6 = 2 \times 3 \), thus there are two types of "regular square" packaging: "1 \times 6" and "2 \times 3".
(1) Given \( a \geqsla... | S_{2} = 578 \, \text{cm}^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,018 |
2. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+$ $c \leqslant 2 a, a+c \leqslant 2 b$. Then the range of $\frac{b}{a}$ is
(A) $(0, \infty)$
(B) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(\frac{1}{2}, 2\right)$
(D) $(0,2)$ | 2. (C).
From the conditions, we know that $c$ is the smallest (otherwise, if $b < c$, then $b^2 < c^2$, which contradicts $b^2 > c^2$. If $a < c$, then $a^2 < c^2$, which contradicts the given conditions).
From $\frac{b}{a}+\frac{c}{a} \leqslant 2$ and $1+\frac{c}{a} \leqslant 2 \cdot \frac{b}{a}$, we have
$$
\frac{1}... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,020 |
3. Among the lines connecting the eight vertices of a square, the number of pairs of skew lines is ( ).
(A) 114
(B) 138
(C) 174
(D) 228 | 3. (C).
The vertex connections include 12 edges, 12 face diagonals, and 4 body diagonals. Skew lines can be categorized as follows:
(1) Edge to edge $\frac{12 \times 4}{2}=24$ pairs.
(2) Edge to face diagonal $12 \times 6=72$ pairs.
(3) Edge to body diagonal $12 \times 2=24$ pairs.
(4) Face diagonal to face diagonal $... | 174 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,021 |
4. The maximum area of an inscribed trapezoid with the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ as its base is ( ).
(A) $\frac{3 \sqrt{3}}{4} a b$
(B) $\frac{\sqrt{3}}{2} a b$
(C) $\frac{\sqrt{3}}{6} a^{2}$
(D) $\frac{\sqrt{3}}{8} a^{2}$ | 4. (A).
Let the base be $P P^{\prime}$, and $P(a \cos \theta, b \sin \theta)$, $P^{\prime}(-a \cos \theta, b \sin \theta)$.
$$
\begin{array}{l}
S_{\text {controlled }}=\frac{2 a+2 a \cos \theta}{2} b \sin \theta=a b(1+\cos \theta) \sin \theta \\
=a b \sqrt{(1+\cos \theta)^{2}\left(1-\cos ^{2} \theta\right)} \\
=\frac{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,022 |
5. For the set $S=\{1,2, \cdots, 18\}$, a five-element subset $S_{1}$ $=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ is such that the difference between any two elements is not 1. The number of such subsets $S_{1}$ is ().
(A) $C_{17}^{4}$
(B) $C_{15}^{4}$
(C) $C_{13}^{5} \quad-\quad+\quad \begin{array}{l}5\end{arr... | 5. (D).
Let the subset $S^{\prime}=\left\{a_{1}, a_{2}-1, a_{3}-2, a_{4}-3, a_{5}-4\right\}$. Then $S^{\prime}$ corresponds one-to-one with $S_{1}$, and $S^{\prime}$ is a five-element subset of $\{1,2, \cdots, 14\}$. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,023 |
2. The real solutions $(x, y)=$ of the equation $\sin (3 \cos x)-2 \sqrt{\sin (3 \cos x)} \cdot \sin (2 \cos y)+1=0$ | 2. $\left(2 k \pi \pm \arccos \frac{\pi}{6}, 2 k \pi \pm \arccos \frac{\pi}{4}, k \in Z\right)$. The original equation can be transformed as follows:
$$
\begin{array}{l}
{\left[\sqrt{\sin (3 \cos x)}-\sin (2 \cos y) j^{2}+\cos ^{2}(2 \cos y)=0\right.} \\
\Rightarrow\left\{\begin{array}{l}
\sqrt{\sin (\overline{\cos x})... | \left(2 k \pi \pm \arccos \frac{\pi}{6}, 2 k \pi \pm \arccos \frac{\pi}{4}, k \in Z\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,026 |
Example 2 As shown in the figure, $A B$ $=A C, \angle A=100^{\circ}$, extend $A B$ to $D$, such that $A D=$ $B C$. Prove: $\angle B C D=$ $10^{\circ}$ | Proof: Given, with $AB$ as the radius, construct $\odot A$, construct a regular $\triangle ABE$, connect $CE$, we have $\angle 5=30^{\circ}, \angle EBC=100^{\circ}$.
$$
\begin{array}{l}
\because AD=BC, BE=AC, \\
\therefore \triangle ADC \cong \triangle BCE,
\end{array}
$$
Knowing $\angle D=\angle 5=30^{\circ}$, hence ... | 10^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 710,028 |
Example 2 Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
(1995, National High School Mathematics Competition) | Analysis: The difficulty of this problem lies in the uncertainty of $[\lg x]$. Since $[\lg x] \leqslant \lg x$, the original problem is first transformed into finding the values of $x$ that satisfy the inequality $\lg ^{2} x-\lg x-2 \leqslant 0$.
Solving this, we get $-1 \leqslant \lg x \leqslant 2$.
Therefore, the num... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,029 |
Seven, (Full marks 12 points) A
home appliance manufacturing company, based on market research and analysis, has decided to adjust its production plan. They plan to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours per week), with at least 60 refrige... | $$
\begin{array}{l}
\because x+y+z=360 \text {, then } z \geqslant 60 \text {, } \\
\therefore x+y \leqslant 300 \text {. } \\
\text { Total labor hours }=\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z \\
=\frac{1}{12}(6 x+4 y+3 z) \\
=\frac{1}{4}(x+y+z)+\frac{1}{12}(3 x+y) \\
=\frac{1}{4} \times 360+\frac{1}{12}(3 x+y) \\
... | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,031 |
6. In rectangle $A B C D$, $D E \perp A C$ at $E$, $\angle A D E=\frac{1}{3} \angle C D E$. Then, the degree measure of $\angle E D B$ is ( ).
(A) $22.5^{\circ}$
(B) $30^{\circ}$
(C) $45^{\circ}$
(D) $60^{\circ}$ | 6. 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 | 710,037 |
7. In square $A B C D$, $E$ is a point on $C D$, $C H \perp B E$ at $H$, $A G \perp B E$ at $G$ and when extended intersects $B C$ at $F$. Then, the number of pairs of congruent figures in the diagram is ( ).
$(\mathrm{A}) 1$ pair
(iB) 2 pairs
(C) $3 \times$ pairs
(I) 4 pairs | 7. 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 | 710,038 |
Example 3 Given the equation $|x-2 n|=k \sqrt{x} \ (n \in N)$ has two distinct real roots in the interval $(2 n-1,2 n+1]$. Then the range of values for $k$ is ( ).
(A) $k>0$
(B) $0<k \leqslant \frac{1}{\sqrt{2 n+1}}$
(C) $\frac{1}{2 n+1}<k \leqslant \frac{1}{\sqrt{2 n+1}}$
(D) None of the above
(1995, National High Sch... | Analysis: Let $y_{1}=\mid x - 2 n \mid, y_{2}=k \sqrt{x}$. Then the original problem is transformed into: the two curves intersect at two different points in $x \in (2 n-1, 2 n+1]$ for the range of $k$. As shown in the figure, $k$ must and only needs to satisfy $0 < k \sqrt{2 n+1} \leqslant |(2 n+1) - n|$.
Therefore, ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,040 |
Three, (Full marks 12 points) A store sells a product that costs 10 yuan each at 18 yuan each, and can sell 60 units per day. After conducting a market survey, the store manager found that if the selling price of this product (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5 unit... | Three, let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan. When $x \geqslant 18$, we have
$$
\begin{aligned}
s & =[60-5(x-18)](x-10) \\
& =-5(x-20)^{2}+500 .
\end{aligned}
$$
That is, when the price of the item is increased, the daily profit $s$ is maximized when $x=20$, and the maximum d... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,049 |
Four, (Full marks 12 points) Given that $a$ and $b$ are the two real roots of the quadratic equation $t^{2}-t-1=0$. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1+x, \\
\frac{x}{b}+\frac{y}{a}=1+y .
\end{array}\right.
$$ | Let $a, b$ be the roots of the equation $t^{2}-t-1=0$. By Vieta's formulas, we have $a+b=1, ab=-1$.
Multiplying the left side of equations (1) and (2) by $ab$, and the right side by -1, we get
$$
\left\{\begin{array}{l}
b x + a y = -(1 + x), \\
a x + b y = -(1 + y).
\end{array}\right.
$$
Adding (3) and (4) and simplif... | x = -\frac{1}{2}, y = -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,050 |
Five, (Full marks 12 points) As shown in the figure, with $AB$ and $AC$ as the hypotenuses, construct right triangles $ABD$ and $ACE$ outward, and make $\angle ABD = \angle ACE$, $M$ is the midpoint of $BC$. Prove: $DM = EM$.
---
The translation is provided as requested, maintaining the original text's line breaks an... | Extend $B D$ to $P$, such that $D P = D B$, and extend $C E$ to $Q$, such that $E Q = E C$. Connect $A P$, $A Q$, $P C$, and $Q B$.
In $\triangle A B Q$ and $\triangle A P C$, it is easy to prove that
$A B = A P$, $A C = A Q$.
$$
\begin{aligned}
\angle P A C & = 2 \angle D A B + \angle B A C \\
& = 2 \angle C A E + \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,052 |
3. There are four propositions:
(1) If the sum and product of two integers are equal, then these two integers are both equal to 2;
(2) If the largest side of triangle A is less than the smallest side of triangle B, then the area of triangle A is less than the area of triangle B;
(3) A polygon where each angle is $179^{... | 3. A | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,055 |
Example 5. Let $O$ be the center of the base $\triangle ABC$ of a regular tetrahedron $P-ABC$. A moving plane through $O$ intersects the three lateral edges or their extensions of the tetrahedron at points $Q$, $R$, and $S$ respectively. Then the sum $\frac{1}{PQ}+\frac{1}{PR}+\frac{1}{PS}(\quad)$.
(A) has a maximum va... | Analysis: This problem can be guessed from the following plane geometry problem. As shown in the figure, $P A=$
$P B, O$ is the midpoint of $A B$, and $R Q$ passes through point $O$ and intersects the extensions of $P B$ and $P A$ at $R$ and $Q$, respectively. By equal areas, we have
$$
\begin{aligned}
S_{\triangle P R... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,062 |
11. (Full marks 12 points) Given real numbers $a$ and $b$ satisfy $a^{2} + ab + b^{2} = 1$. Find the range of $a^{2} - ab + b^{2}$. | Three, let $a^{2}-a b+b^{2}=k$,
and $a^{2}+a b+b^{2}=1$.
From (1) and (2), we solve to get $a b=\frac{1}{2}(1-k)$.
Thus, we have $(a+b)^{2}=\frac{1}{2} \cdot(3-k)$.
$$
\because(a+\dot{b})^{2} \geqslant 0, \therefore k \leqslant 3 \text {. }
$$
Therefore, $a+b= \pm \sqrt{\frac{3-k}{2}}$.
Hence, $a$ and $b$ are the two ... | \frac{1}{3} \leqslant a^{2}-a b+b^{2} \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,064 |
12. (Full marks 12 points) In $\triangle ABC$, does there exist a point $P$ such that any line passing through $P$ will divide $\triangle ABC$ into two parts of equal area? Why? | 12. As shown in the figure, assume there exists a point $P$ satisfying the condition, connect $A P$ and extend it to intersect $B C$ at $D$, connect $B P$ and extend it to intersect $A C$ at $E$.
Then $S_{\triangle A E D}=S_{\triangle A C D}$, so $B D=C D$.
Similarly, $A E=C E$. Thus, $P$ is the centroid of $\triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,065 |
13. (Full score 12 points) How many prime numbers are there among all integers that start and end with 1 and alternate between 1 and 0 (such as 101, 10101, 1010101, ...)? Why? And find all the prime numbers. | 13. The following proves that $10101, 1010101, \cdots \cdots$, are not prime numbers.
For $n \geqslant 2$, then
$$
\begin{aligned}
A & =\underbrace{1010101 \cdots 01}_{2 n} \\
& =10^{2 n}+10^{2 n} 2+\cdots+10^{2}+1 . \\
& =\frac{10^{2 n+2}-1}{10^{2}-1}=\frac{\left(10^{n+1}+1\right)\left(10^{n+1}-1\right)}{99} .
\end{al... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,066 |
1. Let $a>b>c>d>0$, and $x=\sqrt{a b}+$ $\sqrt{c d}, y=\sqrt{a c}+\sqrt{b d}, z=\sqrt{a d}+\sqrt{b c}$. Then the size relationship of $x, y, z$ is ( ).
(A) $x<z<y$
(B) $y<z<x$
(C) $x<y<z$
(D) $z<y<x$ | $-1 . D$.
From $y^{2}-x^{2}=a c+b d-(a b+c d)=(c-b)(a-$
$d)0$, we have $y<x$. Similarly, $z<y$. Therefore, $z<y<$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,067 |
2. If the hypotenuse of a right triangle is 5, and the radius of the inscribed circle is 1, then the area of the triangle is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 2. B.
Let the lengths of the two legs be $a, b$. Also, the hypotenuse length $c=5$, and the inradius $r=1$. From $r=\frac{1}{2}(a+b-c)$ and $a^{2}+b^{2}=c^{2}$, we get
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a b=\frac{1}{2} \cdot \frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{1}{4}[(2 r \\
\left.+c)^{2... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,068 |
5. Among the following four propositions:
(1) The sum of a rational number and an irrational number is always an irrational number;
(2) The sum of two positive irrational numbers is always a positive irrational number;
(3) There is always an irrational number between any two irrational numbers;
(4) There do not exist i... | 5.B.
(1) and (3) are true, (2) and (4) are false.
If $a$ is a rational number, $\alpha$ is an irrational number, and $a+\alpha=b$ is a rational number, then $a=b-a$ is a rational number, which is a contradiction;
The sum of the positive irrational number $\sqrt{2}$ and $2-\sqrt{2}$ is a rational number;
Let $\alpha, \... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,071 |
6. Given that $A B$ is the diameter of a semicircle $O$ with a radius of 4, $C$ is the midpoint of the semicircle arc, $E$ is the midpoint of $B C$, a small circle is tangent to $B C$ at $E$ and tangent to $\overparen{B C}$ at $F$, then the length of the tangent $A T$ drawn from $A$ to the small circle is ( ).
(A) $2+\... | 6. A.
Connect $A C$, then $A C \perp B C$, connect $A E$ intersecting the smaller circle at $D$, connect $O F$ which must pass through $E$, and connect $D F$. In the right triangle $\triangle A C E$,
$$
\sqrt{A C^{2}+C E^{2}}=\sqrt{(2 \sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{10}.
$$
$A E=$
It is also easy to prove that rig... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,072 |
Example 6 Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Then, $\sum_{k=1}^{1000} f\left(\frac{k}{1001}\right)$ equals $\qquad$ - | Sampling: From the characteristics of the variable's values, we can deduce that $f(x)$ satisfies the following symmetric property:
If $x+y=1$, then $f(x)+f(y)=1$. Therefore, $f\left(\frac{k}{1001}\right)+f\left(\frac{1001-k}{1001}\right)=1$,
$(k=1,2, \cdots, 1000)$
$$
\therefore \sum_{k=1}^{1000} f\left(\frac{k}{1001}\... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,073 |
1. For all real numbers $x, y$, if the function $f$ satisfies:
$$
f(x y)=f(x) \cdot f(y)
$$
and $f(0) \neq 0$, then $f(1998)=$ $\qquad$ . | $\approx 、 1.1$.
Let $f(x y)=f(x) \cdot f(y)$, set $y=0$, we have $f(x \cdot 0)=$ $f(x) \cdot f(0)$, which means $f(0)=f(x) \cdot f(0)$, and since $f(0) \neq 0$, it follows that $f(x)=1$, thus $f(1998)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,074 |
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