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1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$
can take different values. | $=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,797 |
2. Let $a$, $b$, and $c$ be the sides of $\triangle ABC$, and $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=999$, then $m$ $=$ $\qquad$ | 2.1999 .
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
\quad=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\cos A \cdot \sin B+\sin A \cdot \cos B)} \\
\quad=\cos C \cdot \frac{\sin A \cdot \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c... | 1999 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,798 |
3. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{2}+$ $\lambda n(n \in \mathbb{N})$. If $\left\{a_{n}\right\}$ is a monotonically increasing sequence, then the range of the real number $\lambda$ is | $3 \cdot \lambda > -3$.
$\because$ the sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence,
$\therefore a_{n+1} - a_{n} >$ C $n \in \mathbb{N}$; always holds.
Also, $a_{n} = n^{2} + \lambda n (n \in \mathbb{N})$.
$\therefore (n+1)^{2} + \lambda(n+1) - \left(n^{2} + \lambda n\right) > 0$ always holds,... | \lambda > -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,799 |
4. The set of points $P(x, y)$ that satisfy the equation $\arctan x+\operatorname{arccot} y=\pi$ is part of a quadratic curve $C$. Then the coordinates of the focus of $C$ are $\qquad$ | 4. $(-\sqrt{2}, \sqrt{2}),(\sqrt{2},-\sqrt{2})$.
From the condition $\arctan x+\operatorname{arccot} y=\pi$, it is easy to see that $x>0, y<0$. Therefore, from $\operatorname{arccot} y=\pi-\arctan x$ we get $y=\cot (\pi-\arctan x)=$ $-\cot (\arctan x)=-\frac{1}{x}$. Hence, the curve $C$ is the hyperbola $y=$ $-\frac{1... | (-\sqrt{2}, \sqrt{2}),(\sqrt{2},-\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,800 |
5. Let $P$ be any point in the plane of $\triangle A B C$, and denote $B C=a, C A=b, A B=c, P A=u, P B$ $=v, P C=w$. Then the minimum value of $\frac{u}{a}+\frac{v}{b}+\frac{w}{c}$ is | $5 . \sqrt{3}$.
Given that $P, A, B, C$ correspond to the complex numbers $z, z_{1}, z_{2}, z_{3}$, we have
$$
\begin{array}{l}
\frac{\left(z-z_{1}\right)\left(z-z_{2}\right)}{\left(z_{3}-z_{1}\right)\left(z_{3}-z_{2}\right)}+\frac{\left(z-z_{2}\right)\left(z-z_{3}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\righ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,801 |
6. Given that a sphere is circumscribed around the tetrahedron $ABCD$; another sphere with radius 1 is tangent to the plane $ABC$ and the two spheres are internally tangent at point $D$. Given $AD=3, \cos \angle BAC=$ $\frac{4}{5}, \cos \angle BAD=\cos \angle CAD=\frac{1}{\sqrt{2}}$. Then the volume of the tetrahedron ... | $6 . \frac{18}{5}$.
As shown in Figure 2, first prove that the height $D H$ of the tetrahedron $A B C D$ is a diameter of another sphere. Let $D E \perp A B$, $D F \perp A C$, with the feet of the perpendiculars being $E$ and $F$, respectively, then $A E=A F=A D$. $\cos \angle B A D=\frac{3}{\sqrt{2}}$, $\cos \angle H ... | \frac{18}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,802 |
Mi, (Full marks 20 points) Find the range of real values of $a$ such that the inequality
$$
\begin{array}{l}
\sin 2 \theta-(2 \sqrt{2}+a \sqrt{2}) \sin \left(\theta+\frac{\pi}{4}\right) \\
-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)}>-3-2 a
\end{array}
$$
holds for all $\theta \in\left[0, \frac{\pi}{2}\r... | Three, let $x=\sin \theta+\cos 9$. From $\theta \leq\left[0, \frac{\pi}{2}\right]$, we get $x \in[1, \sqrt{2}]$. Thus, $\sin 2 \theta=x^{2}-1, \sin \left(\theta+\frac{\pi}{4}\right)=$ $\cos \left(\theta-\frac{\pi}{4}\right)=\frac{x}{\sqrt{2}}$. The original inequality becomes
$$
x^{2}-1-(2+a) x-\frac{4}{x}+3+2 a>0 \tex... | a>3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,803 |
Example 10 Given $\frac{a_{2}+u_{3}+a_{4}}{a_{1}}=\frac{a_{1}+a_{3}+a_{4}}{a_{2}}$ $=\frac{a_{1}+a_{2}+a_{4}}{a_{3}}=\frac{a_{1}+a_{2}+a_{3}}{a_{4}}=k$. Then the value of $k$ is $(\quad)$.
(1995, Jiangsu Province Junior High School Mathematics Competition)
(A) 3
(B) $\frac{1}{3}$
(C) -1
(D) 3 or -1 | Given the equations with five letters and four independent equations, it can be seen as a system of equations about $a_{1}, a_{2}, a_{3}, a_{4}$:
$$
\left\{\begin{array}{l}
a_{2}+a_{3}+a_{4}=a_{1} k, \\
a_{1}+a_{3}+a_{4}=a_{2} k, \\
a_{1}+a_{2}+a_{4}=a_{3} k, \\
a_{1}+a_{2}+a_{3}=a_{4} k .
\end{array}\right.
$$
(1) + (... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,804 |
Four, (Full marks 20 points) A grain warehouse is to be built, with its internal volume $V$ being constant, and $h$, $R$, and $H$ being variables. Determine the values of $\frac{h}{R}$ and $\frac{H}{R}$ to minimize the surface area, i.e., to use the least amount of material.
---
Translate the above text into English,... | Four, $\pi R^{2} H+\frac{\pi}{3} R^{2} h=V$,
$$
S=\pi R \cdot \sqrt{R^{2}+h^{2}}+2 \pi R H+\pi R^{2} .
$$
As shown in Figure 3, let $h=R \tan x\left(0 < x < \frac{\pi}{2}\right)$, then
$$
\begin{aligned}
S & =\pi R \sqrt{R^{2}+R^{2} \tan ^{2} x}+2 \pi R H+\pi R^{2} \\
& =\pi R^{2} \sec x+2 \pi R H+\pi R^{2} \\
& =\pi ... | \frac{h}{R}=\frac{2 \sqrt{5}}{5}, \frac{H}{R}=\frac{5+\sqrt{5}}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,805 |
Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.
---
Please note that the translation pres... | Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,806 |
One, (Full marks 50 points) In $\triangle ABC$, $\angle B$ and $\angle C$ are acute angles, and points $M$, $N$, and $D$ are on sides $AB$, $AC$, and $BC$ respectively, such that $AM = AN$, $BD = DC$. If $\angle BDM = \angle CDN$, prove that $AB = AC$.
---
Translation:
One, (Full marks 50 points) In $\triangle ABC$,... | One, by contradiction. Assume $AB > AC$, then $\angle BDP = DM$. (As shown in Figure 4)
Draw $FH \parallel BC$ through $A$, and draw a perpendicular from $M$ to $BC$ intersecting $BC$ and $FH$ at $E$ and $F$ respectively, and draw a perpendicular from $N$ to $BC$ intersecting $BC$ and $FH$ at $G$ and $H$ respectively.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,807 |
II. (Full marks 50 points) Let $n \in \mathbf{N}, n \geqslant 2, k \in$ $\mathbf{R}^{+}$. Find the minimum value of the function
$$
y=\frac{x^{n}}{x-k}(x \in(k,+\infty))
$$ | $$
\begin{array}{l}
\text{From the arithmetic mean-geometric mean inequality} \\
\frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \geqslant \sqrt[n]{x_{1} x_{2} \cdots x_{n}}
\end{array}
$$
we get
$$
\begin{array}{l}
x_{1} x_{2} \cdots x_{n} \leqslant\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{n} . \\
\because y=\frac{x^{n}}{x... | y_{\text {min }}=n^{n}\left(\frac{k}{n-1}\right)^{n-1} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 711,808 |
Three. (Full marks 50 points) On a straight ruler of length $36 \mathrm{~cm}$, mark $n$ graduations so that the ruler can measure any integer $\mathrm{cm}$ length in the range $[1,36]$ in one go. Find the minimum value of $n$.
| Three, if the ruler is marked with 7 (or fewer than 7) graduations, we can prove that it is impossible to measure any integer length in the range [1, 6] cm in a single measurement.
In fact, 7 graduations, including the two end lines of the ruler, total 9 lines, which have 36 different combinations. Therefore, 7 gradua... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,809 |
Initial 93. As shown in Figure 2, given quadrilateral $ABCD$, divide each side into four equal parts, and connect the corresponding points on opposite sides to form quadrilateral $EFGH$. Prove: $S_{\text{quadrilateral EFGH}}=\frac{1}{4} S_{\text{quadrilateral ABCD}}$. | Proof: Let the points on the sides of quadrilateral $ABCD$ be as shown in Figure 3. Connect $ST$, $BD$, $MN$, and $FH$.
From the given conditions, we have $ST \cong \frac{1}{4} BD$, $MN \cong \frac{3}{4} BD$.
Thus, $ST \cong \frac{1}{3} MN$. Therefore, $TE = \frac{1}{3} EM$, which means $TE = \frac{1}{4} TM$.
Similarly... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,810 |
94. There are two officers, A and B, and 2000 soldiers. To carry out a mission, each officer must select some soldiers from these 2000 soldiers to form a unit. The two officers must take turns selecting. If an officer is forced to select the last soldier, it is considered "inauspicious." Therefore,
(1) When the two off... | Solution: (1) Officer A first selects 4 soldiers. When Officer B selects several soldiers, Officer A then selects a number of soldiers to keep the total number of soldiers selected by both officers before this round to be 5. Since 2000 can be divided by 5, Officer A can always leave the last soldier for Officer B to pi... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,811 |
Let $k, l, m$ be integers, and $l, m \geqslant 0, a=k+\sqrt{l}+\sqrt{m}, b=k-\sqrt{l}-\sqrt{m}, c=k-\sqrt{l}+\sqrt{m}, d=k+\sqrt{l}-\sqrt{m}$, with the convention $x^{0}=1(x \in \mathbf{R})$. Then $S_{n}=a^{n}+b^{n}+c^{n}+d^{n}(n \in \mathbf{Z}, n \geqslant 0)$ is always an integer. | Proof: Since $a+b=2k$,
$$
\begin{aligned}
ab & =(k+\sqrt{l}+\sqrt{m})(k-\sqrt{l}-\sqrt{m}) \\
& =k^{2}-\left(l+m+2 \sqrt{lm}\right),
\end{aligned}
$$
Therefore, $a, b$ satisfy the equation
$$
x^{2}-2kx+k^{2}-(l+m+2 \sqrt{lm})=0.
$$
Similarly, $c, d$ satisfy the equation
$$
x^{2}-2kx+k^{2}-(l+m-2 \sqrt{lm})=0.
$$
Thu... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,812 |
94. Let the three sides of $\triangle ABC$ be $a, b, c$, and the corresponding angles in radians be $\alpha, \beta, \gamma$. Prove:
$$
\frac{a \alpha(\beta+\gamma)+b \beta(\gamma+\alpha)+c \gamma(\alpha+\beta)}{a+b+c}<\frac{\pi^{2}}{3} .
$$ | Proof: $\because a \beta(b+c \cdots a)+\beta \gamma(c+a-b)+\gamma \alpha(a$ $+b-c)>0$,
adding $2(\ldots a \beta+b \beta \gamma+c \gamma \alpha)$ on both sides, we get.
$$
\begin{array}{l}
(a+b+c)(\alpha \beta+\beta \gamma+\gamma \alpha) \\
>2(a \alpha \beta+b \beta \gamma+c \gamma \alpha), \\
\therefore \frac{a \alpha ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,813 |
Example 1 In $\triangle A B C$, $\angle A B C=60^{\circ}$, $\angle A C B=20^{\circ}, M$ is a point on the bisector of $\angle A C B$, $\angle M B C=20^{\circ}$. Find the degree measure of $\angle M A B$. | Solution: As shown in Figure 1, let the angle bisector of $\angle M B A$ intersect $A C$ at $D$, and connect $D M$.
Obviously, $B M$ bisects $\angle D B C$, and $C M$ bisects $\angle D C B$, which means $M$ is the incenter of $\triangle D B C$. Therefore,
$\angle M O B=\angle M D C=60^{\circ}$.
Thus, $\angle A D H=60^... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,814 |
Example 2 In $\triangle A B C$, $\angle A B C=\angle A C B=40^{\circ}$, $P$ is a point inside the triangle, $\angle P C A=\angle P A C=20^{\circ}$. Find the degree measure of $\angle P B C$. | Solution: As shown in Figure 2, construct a regular $\triangle DAC$ outside $\triangle ABC$ with $AC$ as one side. Connect $DP$.
Given $\angle PCA = \angle PAC = 20^{\circ}$, we know $PA = PC$. Point $A$ is symmetric to point $C$ with respect to $PD$. Therefore,
$$
\angle PDA = \frac{1}{2} \angle ADC = 30^{\circ}.
$$
... | 10^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,815 |
Example 11 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=20^{\circ}$, $N$ is a point inside the triangle, $\angle N A B=40^{\circ}$, $\angle N B C=30^{\circ}$. Find the degree measure of $\angle N C B$. | Solution: As shown in Figure 11, draw a line through point $N$ perpendicular to the extension of $AC$ at $P$. On the extension of $AN$, take a point $Q$ such that $\angle QBC = 30^{\circ}$. Connect $C, Q, C, QB, PQ, PN$.
Since $\angle PAC = 70^{\circ} = \angle NAC$, point $P$ is the reflection of point $N$ over $AC$. ... | 10^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,816 |
Find all integers $k$ such that one of the roots of the quadratic equation $k x^{2}-2(3 k-1) x+9 k-1=0$ is a reduced fraction with a denominator of 1999. | Solution: From the problem, we know that $k \neq 0, \Delta=4(1-5 k) \geqslant 0$, so $k<0$, and the two roots are
$$
x_{1,2}=\frac{3 k-1 \pm \sqrt{1-5 k}}{k}=3+\frac{-1 \pm \sqrt{1-5 k}}{k} .
$$
Let $\sqrt{1-5 k}=m$, where $m$ is a positive integer and $m \geqslant 4$, then
$$
k=\frac{1-m^{2}}{5} \text {. }
$$
Thus, ... | k=-1999 \times 9993 \text{ or } -1999 \times 9997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,817 |
As shown in Figure 1, in $\triangle$ $ABC$, $DE \parallel BC$, and $DE = \frac{2}{3} BC$, $BE$ intersects $CD$ at point $O$, $AO$ intersects $BC$ and $DE$ at points $M$ and $N$ respectively, $CN$ intersects $BE$ at point $F$, and $FM$ is connected.
Prove: $FM = \frac{1}{4} AB$. | Proof: $\because D E / / B C$,
$$
\begin{aligned}
\therefore \frac{D N}{B M} & =\frac{A N}{A M}=\frac{N E}{M C}, \\
\frac{D N}{M C} & =\frac{O N}{O M}=\frac{N E}{B M} .
\end{aligned}
$$
By (1) $\times$ (2), (1) $\div$ (2) we get
$D N=N E, B M=M C$
Extend $C N$ to intersect $A B$.
$\because D E / / B C, D E=\frac{2}{3}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,818 |
Given $a, b, c$ are the lengths of the three sides of a triangle. Prove:
$$
\frac{59}{40}<\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{61}{40} .
$$ | Proof: We prove a stronger conclusion:
$$
\frac{3}{2}-\frac{1}{2 m}(2 z+y)(y+z)(2 y+z) \\
=2 z^{3}+7 z^{2} y+7 z y^{2}+2 y^{3}.
$$
Therefore, to prove (3), it suffices to prove
$$
2 z^{3}+7 z^{2} y+7 z y^{2}+2 y^{3}-m y z(z-y)>0.
$$
Noting that \( m=9+8 \sqrt{2} \), we complete the square on the left side of the abov... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,819 |
Example 1 Let points $E, F, G$ be the midpoints of the edges $AB, BC, CD$ of tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}-\arctan \sqrt{2}$
(C) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(D) $\pi-\operatorname{arccot} \frac{\s... | Solution: As shown in the figure,
1, draw $E E_{1} \perp$
plane $B C D$, then
$E E_{1} \perp B G$.
Since $A B C D$
is a regular tetrahedron,
we have $A C \perp B D$.
Also, since $E F$
$/ / A C, F G$ //
$B D$,
$$
\therefore E F \perp F G \text {. }
$$
Therefore, $E_{1} F \perp F G$. Thus, $\angle E F E_{1}$ is the pla... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,821 |
Example 3 Given that the base of the tetrahedron $S \cdots A B C$ is an equilateral triangle, point $A$'s projection $H$ on the side face $SBC$ is the centroid of $\triangle S B C$, … the dihedral angle $H-A B-C$ is equal to $30^{\circ}, S A=2 \sqrt{3}$. Then, the volume of the tetrahedron $S$ $A B C$ is $\qquad$
(1999... | Solution: As shown in Figure 3, given $A H \perp$ plane $S B C$. Draw $B H$ $\perp S C$ at $E$.
By the theorem of three perpendiculars, we know $S C \perp A E, S C \perp A B$, hence $S C \perp$ plane $A B E$. Let the projection of point $S$ on plane $A B C$ be $O$, then $S O \perp$ plane $A B C$. By the converse of th... | \frac{9}{4} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,823 |
1 In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of $A B$, and $\triangle A C M$ is folded along $C M$ such that the distance between $A$ and $B$ is $2 \sqrt{2}$. At this moment, the volume of the tetrahedron $A-B C M$ is $\qquad$ (1998, National High School League) | Solution: The tetrahedron $A-BCM$ after folding is shown in Figure 4. Take the midpoint $D$ of $CM$, and connect $AD$. It is easy to see that $AD \perp CM$. In $\triangle BCM$, draw $DE \perp CM$ intersecting $BC$ at point $E$, and connect $AE$. We know that $AE \perp CM$. Then $AD=\sqrt{3}$, $DE=CD \tan 30^{\circ}=\fr... | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,824 |
Example 5 Given a regular tetrahedron $S-ABC$ with height $SO=3$, and the side length of the base is 6. A perpendicular is drawn from point $A$ to the opposite face $SBC$, with the foot of the perpendicular being $O'$. On $AO'$, take a point $P$ such that $\frac{AP}{PO'}=8$. Find the area of the section parallel to the... | Solution: As shown in Figure 5, since $S-A B C$ is a regular tetrahedron,
point $O$ is the centroid of $\triangle A B C$.
Connecting $A O$ and extending it to intersect $B C$ at $D$,
since point $D$ is the midpoint of $B C$, and $B C \perp$ plane $S A D$, and $A O^{\prime} \perp B C$, $A O^{\prime}$ lies on plane $S A ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,825 |
Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League) | Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.
Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,826 |
Example 7 In a cube, among the 8 vertices, the midpoints of the 12 edges, the centers of the 6 faces, and the center of the cube, a total of 27 points, the number of groups of three collinear points is ( ).
(A) 57
(B) 49
(C) 43
(D) 37
(1998, National High School Competition) | Solution: Classify and discuss the collinear three-point groups according to the different situations of the two endpoints as follows:
(1) The number of collinear three-point groups with both endpoints being vertices is
$$
\frac{8 \times 7}{2}=28 \text { (groups); }
$$
(2) The number of collinear three-point groups wit... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,827 |
Example 8 If three lines $a, b, c$ in space are pairwise skew lines, then the number of lines that intersect $a, b, c$ is ( ).
(A) C lines
(B) 1 line
(C) a finite number greater than 1
(D) infinitely many lines
$(1007$, National High School League) | Solution: Regardless of the positional relationship between $a$, $b$, and $c$, a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ can always be constructed such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D_{1} D$ lies on line $c$, as shown in Figure 7.
Next, take any point $M$ on the extension o... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,828 |
1. Select $k$ edges and face diagonals from a cube such that any two line segments are skew lines. What is the maximum value of $k$? | (提示:考察线段 $A C 、 B C_{1} 、 D_{1} B_{1} 、 A_{1} D$, 它们所在的直线两两都是异面直线. 若存在 5 条或 5 条以上满足条件的线段, 则它们的端点相异, 且不少于 10 个, 这与正方体只有 8 个端点矛盾, 故 $k$ 的最大值是 4.)
(Translation: Consider the line segments $A C, B C_{1}, D_{1} B_{1}, A_{1} D$, the lines on which they lie are pairwise skew lines. If there are 5 or more line segments that s... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,829 |
Example 3 In $\triangle A B C$, $\angle A B C=40^{\circ}$, $\angle A C B=30^{\circ}, P$ is a point on the bisector of $\angle A B C$, $\angle P C B=10^{\circ}$. Find the degree measure of $\angle P A B$. | Solution: As shown in Figure 3, take a point $D$ on the extension of $BA$ such that $BD = BC$. Connect $D \bar{r}^{\prime}$ and $NC$.
Since $BP$ bisects $\angle ABC$, point $D$ is symmetric to point $C$ with respect to $BP$. Therefore,
$$
PD = PC.
$$
Given $\angle DPC = 2(\angle PBC + \angle PCB) = 60^{\circ}$, we kn... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,830 |
3. Let the height of the regular tetrahedron \( P-ABC \) be \( PO \), and let \( M \) be the midpoint of \( PO \). A plane passing through \( AM \) and parallel to the edge \( BC \) divides the tetrahedron into upper and lower parts. Find the ratio of the volumes of these two parts.
Translate the above text into Engli... | (Tip: The upper and lower parts of the triangular pyramid are two cones of equal height. Thus, the ratio of their volumes is equal to the ratio of their base areas. By the knowledge of similar figures, the ratio of the base areas is $4: 21$, so the ratio of the volumes is $4: 21$.) | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,832 |
4. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2, AB=2$. Suppose points $S, A, B, C$ all lie on a sphere with center $O$. Find the distance from point $O$ to the plane $ABC$. | (Let the midpoint of $AB$ be $D$, then $D$ is the circumcenter of $\triangle ABC$. From $SA=SB=SC=2$, we know that the projection of $S$ on the base $ABC$ is $D$. Therefore, the center of the sphere $O$ is on $SD$. Also, since $OA=OB=OS$, point $O$ is the center of the equilateral $\triangle SAB$, so the distance from ... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,833 |
1. In an international chess tournament, a win earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. Now, there are 8 players in a round-robin tournament (each pair of players plays one game). After the tournament, it is found that all players have different scores. When the players are ranked by their sco... | Solution: Let the scores of the top three players be $a_{1}, a_{2}, a_{3}$ $\left(a_{1}>a_{2}>a_{3}>4.5\right)$.
Since 8 players only participate in 7 matches, the maximum score is 7, i.e., $a_{1} \leqslant 7$. The total number of matches played is $\frac{7 \times 8}{2}=$ 28 matches, with a total score of 28 points. A... | 6.5, 6, 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,834 |
2. In an international chess tournament, winning a game earns 2 points, a draw earns 1 point, and losing a game earns 0 points. Now, there are 10 players participating in a round-robin tournament (each pair of players plays one game). After all the games are played, it is found that all players have different scores. W... | Solution: Let the scores of the top six players be $a_{1}, a_{2}$, $a_{3}, a_{4}, a_{5}, a_{6}$.
Since 10 players participate in the competition, each playing 9 games, and the top two players have no losses, and the scores are only 2, 1, 0, we have
$$
a_{1} \leqslant 17, a_{2} \leqslant 16 \text {. }
$$
This competit... | 17, 16, 13, 12, 11, 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,835 |
3. There are several table tennis representative teams, and players from different teams all play one match against each other, while players from the same team do not play against each other. The match statistician recorded the results, and this time, there were 10 players in total, and 27 matches were played.
(1) How... | Solution: (1) Let's assume that this competition is divided into two teams, each with $a_{1}, a_{2}$ members.
From the problem, we have $a_{1}+a_{2}=10, a_{1} a_{2}=27$. This system of equations has no integer solutions (indeed, no real solutions). Therefore, dividing into two teams is incorrect.
(2) Let's assume that... | 3 \text{ teams with } 6, 3, \text{ and } 1 \text{ players respectively} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,836 |
4. There are several international chess teams participating in an international chess competition. It is stipulated that winning a game earns 2 points, drawing a game earns 1 point, and losing a game earns 0 points. It is also stipulated that players from different teams play one game against each other, while players... | Solution: (1) According to the solution of Question 3, there are three chess teams in this competition, each with 5, 3, and 1 players, respectively.
(2) The 9 players in total played 23 games, with a total score of 46 points.
Since the team with 5 players, each played 4 games, a player who won all games would score 8 ... | 16, 9, 7, 6, 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,837 |
5. Three people, A, B, and C, are dividing candies. The distribution method is as follows: first, write three positive integers $p, q, r$ on three pieces of paper, such that $p < q < r$. Then, each person draws a piece of paper, and the number on the paper is the number of candies they get in that round. After several ... | Solution: Let there be $n$ rounds of distribution ($n$ is a positive integer).
According to the situation of C, we have $18-np=9, np=9$. The value of $n$ can only be $1, 3, 9$.
When $n=1$, only one round of distribution has taken place, and obviously, one of A, B, and C must have received 0 candies according to the ab... | 13, 6, 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,838 |
Let there be two different media I and II on either side of the $x$-axis (Figure 2). The speed of light in media I and II is $c_{1}$ and $c_{2}$, respectively. Now, if light travels from point $A$ in medium I to point $B$ in medium II, what path should the light take to minimize the time of travel? | Solution: In the same medium, the shortest path of a light ray is a straight line. We can assume that the light ray travels along the straight line $A P$ in medium I and along the straight line $P B$ in medium II, using the notation in the figure. Then,
$$
A P=\sqrt{x^{2}+b_{1}^{2}}, P B=\sqrt{(a-x)^{2}+b_{2}^{2}} .
$$... | \frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{c_{1}}{c_{2}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 711,839 |
Example 4 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}$, $Q$ is a point inside the triangle, $\angle Q B A=$ $\angle Q C A=20^{\circ}$. Find the degree measure of $\angle Q A B$. | Solution: As shown in Figure 4, let $B Q$ intersect $A C$ at $D$, and draw a perpendicular from $D$ to $B C$ intersecting $Q C$ at $E$. Connect $B E$.
From $\angle Q 13 C=30^{\circ}=\angle A C B$, we know that $D E$ is the perpendicular bisector of $B C$.
From $\angle Q C B=10^{\circ}$, we know
$\angle E B C=10^{\circ... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,841 |
Example 1 Let $a, b, c \in \mathbf{R}^{+}$. Prove:
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geqslant \frac{3}{2} \text {. }
$$
(1963, Moscow Mathematical Competition) | ```
Let \( a + b + c = s \). Then
\[
\begin{array}{l}
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \\
= \frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c} \\
= \frac{a-s+s}{s-a}+\frac{b-s+s}{s-b}+\frac{c-s+s}{s-c} \\
=-3+\left(1-\frac{a}{s}\right)^{-1}+\left(1-\frac{b}{s}\right)^{-1} \\
+\left(1-\frac{c}{s}\right)^{-1} \\
\geqsla... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,842 |
Example 3 Let $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n, n \geqslant 2$, and $s=\sum a_{i}$. Then $\sum \frac{a_{i}^{2}}{s-a_{i}} \geqslant \frac{s}{n-1}$. | $\begin{array}{l}\text { Prove: } \sum \frac{a_{i}^{2}}{s-a_{i}}=\sum \frac{a_{i}^{2}-s^{2}+s^{2}}{s-a_{i}} \\ =-\sum\left(s+a_{i}\right)+\sum\left(\frac{s-a_{i}}{s^{2}}\right)^{-1} \\ \geqslant-\sum\left(s+a_{i}\right)+n\left[\frac{\sum\left(s-a_{i}\right)}{n \cdot s^{2}}\right]^{-1} \\ =-n s-\sum a_{i}+n \cdot \frac{... | \frac{s}{n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,844 |
Example 4 If $a_{i}$ have the same sign, $a=\sum a_{i} \neq 0, n \geqslant 2$, $n \in \mathbf{N}$. Then
$$
\sum \frac{a_{i}}{2 a-a_{i}} \geqslant \frac{n}{2 n-1} .
$$ | $\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{2 a-a_{i}}=\sum \frac{a_{i}-2 a+2 a}{2 a-a_{i}} \\ =-n+\sum\left(\frac{2 a-a_{i}}{2 a}\right)^{-1} \\ =-n+\sum\left(1-\frac{a_{i}}{2 a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i} / 2 a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum a_{i} / ... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,845 |
Example 6 Let $x_{i} \in \mathbf{R}^{+}$, and $\sum a_{i}=a, n \in \mathbf{N}$, and $n \geqslant 2$. Prove:
$$
\sum \frac{x_{i}}{a-x_{i}} \geqslant \frac{n}{n-1} .
$$
(1976, British Mathematical Competition) | $\begin{array}{l}\text { Prove: } \sum \frac{x_{i}}{a-x_{i}}=\sum \frac{x_{i}-a+a}{a-x_{i}} \\ =-n+\sum\left(1-\frac{x_{i}}{a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-x_{i} / a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum x_{i} / a} \\ =-n+\frac{n^{2}}{n-1}=\frac{n}{n-1} .\end{array}$ | \frac{n}{n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,847 |
Example 7 Let $a_{i} \in \mathbf{R}^{+}$, and all are less than $c, \sum a_{i}=$ $s$. Then
$$
\sum \frac{a_{i}}{c-a_{i}} \geqslant \frac{n s}{n c-s} .(n \geqslant 2)
$$ | $$
\begin{array}{l}
\text { Prove: } \sum \frac{a_{i}}{c-a_{i}}=\sum \frac{a_{i}-c+c}{c-a_{i}} \\
=-n+\sum\left(1-\frac{a_{i}}{c}\right)^{-1} \\
\geqslant-n+n \cdot\left(\sum \frac{\left(1-a_{i} / c\right)}{n}\right)^{-1} \\
=-n+\frac{n^{2}}{n-\sum a_{i} / c} \\
=-n+\frac{n^{2}}{n-s / c} \\
=\frac{n s}{n c-s} .
\end{ar... | \frac{n s}{n c-s} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,848 |
Example 8 Let $\alpha_{i}(i=1,2, \cdots, n)$ all be acute angles, and satisfy $\sum \cos ^{2} \alpha_{i}=s$. Then
$$
\sum \cot ^{2} \alpha_{i} \geqslant \frac{n s}{n-s} .
$$ | $\begin{array}{l}\text { Prove: } \sum \cot ^{2} \alpha_{i}=\sum \frac{\cos ^{2} \alpha_{i}}{1-\cos ^{2} \alpha_{i}} \\ =\sum \frac{\cos ^{2} \alpha_{i}-1+1}{1-\cos ^{2} \alpha_{i}} \\ =-n+\sum\left(1-\cos ^{2} \alpha_{i}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-\cos ^{2} \alpha_{i}\right)}{n}\right)^{... | \frac{n s}{n-s} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,849 |
In $\triangle A B C$, $\angle A=90^{\circ}, \angle B<\angle C$. Through point $A$, draw the tangent of the circumcircle $\odot O$ of $\triangle A B C$, intersecting line $B C$ at $D$. Let point $A$'s reflection over $B C$ be $E$, and draw $A X \perp B E$ at $X, Y$ is the midpoint of $A X$, and $B Y$ intersects $\odot O... | Prove: Connect $A E$ intersecting $B Z$ at $M$, connect $Z E, Z C$. In $\triangle A X E$, by Menelaus' theorem, we have
$$
\frac{E B}{B \bar{X}} \cdot \frac{X Y}{Y A} \cdot \frac{A M}{M E}=1 \text {. }
$$
Since $Y$ is the midpoint of $A X$, $X Y=Y A$, thus,
$$
\frac{B X}{E B}=\frac{A M}{M E} \text {. }
$$
Let $\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,850 |
9.1. Let $a$, $b$, $c$ be two distinct real numbers such that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common real root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common real root. Find $a + b + c$. | 9.1. $a+b+c=-3$.
If $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+b x_{1}+c=0$, then $(a-b) x_{1}+(1-c)=0$. Therefore, $x_{1}=\frac{c-1}{a-b}$. Similarly, from the equations $x_{2}^{2}+x_{2}+a=0$ and $x_{2}^{2}+c x_{2}+b=0$, we derive $x_{2}=\frac{a-b}{c-1}$ (obviously $c \neq 1$). Hence, $x_{2}=\frac{1}{x_{1}}$.
On the oth... | a+b+c=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,851 |
Example 5 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, Q$ is a point inside the triangle, $\angle Q C A=$ $\angle Q A B=20^{\circ}$. Find the degree measure of $\angle Q B C$. | Solution: As shown in Figure 5, let the perpendicular bisector of $BC$ intersect $BA$ and $AC$ at $D$ and $E$, respectively, and let $F$ be the foot of the perpendicular. Connect $QE$, $BE$, and $DC$.
Given $\angle ACD = 20^{\circ} = \angle ACQ$ and $\angle DAC = 80^{\circ} = \angle QAC$, we know that point $D$ is sym... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,852 |
9.2. Danya thought of a natural number $X \leqslant 100$, and Sasha is trying to guess this number. He selects a pair of natural numbers $M$ and $N$, both less than 100, and asks Danya: “What is the greatest common divisor of $X+M$ and $N$?" Prove that Sasha can guess the number Danya thought of after asking Danya 7 su... | 9.2. After learning the greatest common divisor (GCD) of $X+1$ and 2, Sasha can determine the parity of $X$. If $X$ is even, the second question asks for the GCD of $X+2$ and 4; if $X$ is odd, the question asks for the GCD of $X+1$ and 4. This way, Sasha can determine the remainder when $X$ is divided by 4. Generally, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,853 |
9.3. Let the circumcircle of acute triangle $ABC$ be $\omega$ with center $O$. The circle $\omega_{1}$ passing through points $A, O, C$ has center $K$, and intersects sides $AB$ and $BC$ at points $M$ and $N$, respectively. Given that point $L$ is the reflection of $K$ over line $MN$. Prove: $BL \perp AC$. | 9.3. As shown in Figure 2, let $\angle ABC = \beta, \angle BAC = \alpha$. Then $\angle AOC = 2\beta$, so the arc $\overparen{AC}$ on circle $\omega_{1}$ not containing point $O$ equals $4\beta$. Since $\angle ABC = \frac{1}{2}(\overparen{AC} - \overparen{MN})$, we have $\beta = \frac{1}{2}(4\beta - \overparen{MN})$, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,854 |
9.4. A country has several cities, some of which are connected by roads, with each city having 3 roads leading out of it. Prove: there exists a loop formed by roads, the length of which is not divisible by 3. | 9.4. Suppose there is a graph where the degree of each vertex is greater than 2, but the length of any cycle in the graph is divisible by 3. We will examine the graph $G$ with the smallest number of vertices that has this property. Clearly, there exists a cycle $Z$ of minimum length in this graph, such that no two non-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,855 |
9.5. Write numbers $a_{1}=1, a_{2}, a_{3}, \cdots$ on the blackboard according to the following rule: If $a_{n}-2$ is a natural number and has not been written before, then write $a_{n+1} = a_{n}-2$, otherwise write $a_{n+1} = a_{n}+3$. Prove: All perfect squares that appear in this sequence are obtained by adding 3 to... | 9.5. We use induction to prove the following statement: When $n=5 m$, all natural numbers from 1 to $n$ will be written, and $a_{5 m}=$ $5 m-2$. And for any $k \leqslant 5 m$, there must be $a_{k+5}=a_{k}+5$.
When $n=5$, we have $1 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 3 \rightarrow 6$, the conclusion h... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,856 |
9.6. In a $2 n \times 2 n$ grid, some cells contain black or white Go stones, with at most one stone per cell. First, remove all black Go stones that are in the same column as any white Go stone; then, remove all white Go stones that are in the same row as any remaining black Go stone. Prove: In the grid of remaining s... | 9.6. We point out that at this moment, no two pieces of different colors will be in the same row or the same column. In fact, if at the beginning, a black piece and a white piece are in the same column, then the black piece will be removed in the first round; and if after the first removal, a white piece is in the same... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,857 |
9.7. In $\triangle A B C$, take a point $E$ on the median $C D$. Circle $S_{1}$ passes through point $E$, is tangent to line $A B$ at point $A$, and intersects side $A C$ at point $M$. Circle $S_{2}$ passes through point $L$, is tangent to line $A B$ at point $B$, and intersects side $B C$ at point $N$. Prove: The circ... | 9.7. As shown in Figure 3, suppose the second intersection point of circle $S_{1}$ with $CD$ is $F$. For definiteness, assume $E$ is between $D$ and $F$ (possibly $F = E$). From the equation $DA^{2} = DF \cdot DE$, it follows that $DB^{2} = DF \cdot DE$. This indicates that circle $S_{2}$ also passes through point $F$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,858 |
9.8. There are 100 pairwise coprime natural numbers placed along a circle. It is allowed to add to any of these numbers the greatest common divisor (GCD) of its two neighboring numbers. Prove that it is possible to use such operations to make all the numbers pairwise coprime. | 9.8. For convenience, let $a_{n+100}=a_{n}$, where $n=1,2, \cdots$, 100. Let $(a, b)$ denote the greatest common divisor of the positive integers $a$ and $b$.
Lemma: Let $a_{1}, a_{2}, \cdots, a_{n}$ and $d$ be natural numbers, then there exists a natural number $k$, such that for all $i=2,3, \cdots, n$, we have $\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,859 |
10.1. Summation
$$
\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\left[\frac{2^{3}}{3}\right]+\cdots+\left[\frac{2^{1000}}{3}\right]
$$ | 10. 1. $\frac{1}{3}\left(2^{1001}-2\right)-500$.
The first term is obviously 0, so we can discard it. For the remaining 1000 terms, let's first examine
$$
\frac{2}{3}+\frac{2^{2}}{3}+\frac{2^{3}}{3}+\cdots+\frac{2^{1000}}{3}
$$
This is the sum of a geometric series, whose value is $\frac{1}{3}\left(2^{1001}-2\right)$... | \frac{1}{3}\left(2^{1001}-2\right)-500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,860 |
10.2. Let $-1<x_{1}<x_{2}<\cdots<x_{n}<1$, and
$$
x_{1}^{13}+x_{2}^{13}+\cdots+x_{n}^{13}=x_{1}+x_{2}+\cdots+x_{n} \text {. }
$$
Prove: If $y_{1}<y_{2}<\cdots<y_{n}$, then
$$
\begin{array}{l}
x_{1}^{13} y_{1}+x_{2}^{13} y_{2}+\cdots+x_{n}^{13} y_{n} \\
<x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n} .
\end{array}
$$ | 10.2. Let $t_{i}=x_{i}^{13}-x_{i}$. When $-10$ (this is because $\left.\sum\left(y_{i}+c\right) t_{i}=\sum y_{i} t_{i}+c \sum t_{i}=\sum y_{i} t_{i}\right)$.
Let $k$ be such that $x_{k} \leqslant 0, x_{k+1}>0$. Thus, $t_{1}, \cdots, t_{k}$ are non-negative, and $t_{k+1}, \cdots, t_{n}$ are less than 0. We have
$$
\beg... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,861 |
10.3. In a non-isosceles acute triangle \( A B C \), the angle bisector of the acute angle formed by the altitudes \( A A_{1} \) and \( O C_{1} \) intersects the sides \( A B \) and \( B C \) at points \( P \) and \( Q \), respectively. The angle bisector of \( \angle B \) intersects the segment connecting the orthocen... | 10.3. As shown in Figure 4, let $H$ be the orthocenter of $\triangle ABC$, and $M$ be the midpoint of side $AC$. Take points $S$ and $T$ on segments $AH$ and $CH$ respectively, such that $PS \perp AB$ and $TQ \perp BC$. Denote the intersection of lines $PS$ and $QT$ as $K$. Since $\angle BPK = \angle BQK = 90^\circ$, q... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,862 |
Example 6 In $\triangle ABC$, $\angle CAB = \angle CBA = 50^{\circ}$, $O$ is a point inside the triangle, $\angle OAB = 10^{\circ}$, $\angle OBC = 20^{\circ}$. Find the degree measure of $\angle OCA$. | Solution: As shown in Figure 6, draw a perpendicular from point $C$ to $A B$, intersecting the extension of $B O$ at $E$. Connect $A E$.
Given $\angle C A B=\angle C B A=50^{\circ}$, we know that point $A$ is symmetric to point $B$ with respect to $C E$. Also, given $\angle O B C=20^{\circ}$, $\angle E C B=40^{\circ}$... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,863 |
10.4. Now we have 5 visually identical but differently weighted weights. We are allowed to select any 3 weights $A, B$, and $C$, and ask: "Is $m(A)<m(B)<m(C)$?" (Here, $m(x)$ represents the mass of weight $x$, and the answer is only "yes" or "no"). Can we determine the order of the 5 weights by asking 9 such questions? | 10.4. Impossible.
There are 5 weights $A, B, C, D, E$, which can be arranged in $5!=120$ different orders of weight. Under the condition $m(A)<m(B)<m(C)$, there are $\frac{5!}{3!}=20$ different weight orderings. Therefore, if a question receives a negative answer, it can exclude at most 20 different orderings. Thus, t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,864 |
10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most? | 10.5.7.
An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.
We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,865 |
10.6. A natural number is called a "perfect number" if it equals the sum of all its positive divisors excluding itself, for example $6=1+2+3$. If a "perfect number" greater than 6 is divisible by 3, prove that it must also be divisible by 9. | 10.6. Let a "perfect number" be equal to $3 n(n>2)$, where $n$ is not a multiple of 3. Then, all positive divisors of $3 n$ (including itself) can be divided into several pairs of the form $d$ and $3 d$, where $d$ is not divisible by 3. Thus, the sum of all positive divisors of $3 n$ (which equals $6 n$) is a multiple ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,866 |
10.8. On a rectangular frog table, there are many equal squares, whose sides are parallel to the edges of the table, and are colored in $k$ different colors. If we examine any $k$ squares of different colors, then there are always two of them that can be pinned to the table with one nail. Prove: It is possible to pin a... | 10.8 Make an induction on the number of colors $k$.
When $k=2$, we observe the leftmost square $K$. If it is of color 1, then all squares of color 2 have a common point with it, so each square of color 2 contains one of the two right vertices of square $K$, and thus all squares of color 2 can be caught with two nails.... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,868 |
11.1 Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \), which for all \( x \), \( y \), \( z \in \mathbb{R} \) satisfy the inequality
\[
\begin{array}{l}
f(x+\dot{y})+f(y+z)+f(z+x) \\
\geqslant 3 f(x+2 y+3 z) .
\end{array}
\] | 11.1 Let $x=y=-2$, we get $f(2 x)+j(0)+$ $f(0) \geqslant 3 f(0)$, thus $f(2 x) \geqslant f(0)$.
On the other hand, let $x=z=-y$, we get $f(0)+f(0)+$ $f(2 x) \geqslant 3 f(2 x)$, hence $f(2 x) \leqslant f(0)$. Therefore, $f(0)$ $\geqslant f(2 x) \geqslant f(0)$. That is, $f(2 x) \equiv$ constant $=c$. It is easy to ver... | f(x) \equiv c | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,869 |
11.2. Proof: It is possible to divide the set of all natural numbers into 100 non-empty subsets, such that for any 3 natural numbers $a, b, c$ satisfying the relation $a + 99b = c$, two of these numbers can be found in the same subset. | 11.2 Construct these subsets according to the following rule: in the $i$-th subset $(1 \leqslant i \leqslant 99)$, place all even numbers that leave a remainder of $i-1$ when divided by 99, and in the 100th subset, place all odd numbers. Clearly, in any numbers $a, b, c$ that satisfy the equation $a+99 b=c$, the number... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,870 |
11.3. Given a convex pentagon $A B C D E$ in the coordinate plane, whose vertices are all integer points. Prove: there is at least one integer point in the interior or on the boundary of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ (see Figure 1). | 11.3 For simplicity, we will refer to "on the interior or on the boundary" as "in...".
By contradiction. Assume the conclusion does not hold, and consider the pentagon with the smallest area \( S \) that does not satisfy the assertion (since the area of any lattice polygon is a half-integer, there exists a smallest ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,871 |
11.4. Given non-negative numbers
sequence $a_{1}, a_{2}, \cdots, a_{n}$. For any integer $k$ from 1 to $n$, let $m_{k}$ denote the value
$$
\max _{\ell=1,2, \cdots, k} \frac{a_{k-l+1}+a_{k-1+2}+\cdots+a_{k}}{l} .
$$
Prove: For any $\alpha>0$, the number of $k$ such that $m_{k}>\alpha$
$$
\text { is less than } \frac{... | 11.4 Solution 1: For $1 \leqslant i \leqslant j \leqslant n$, let $[i, j]$ represent a segment of integers from $i$ to $j$, and define
$$
S(i, j)=\frac{a_{i}+a_{i+1}+\cdots+a_{j}}{j-i+1}.
$$
It is easy to see that $S(i, j)>\alpha$ and $S(j+1, l)>\alpha$ imply $S(i, l)>\alpha$.
Divide $[1, n]$ into several segments $\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,872 |
11.5. Prove the inequality
$$
\sin ^{n} 2 x+\left(\sin ^{n} x-\cos ^{n} x\right)^{2} \leqslant 1 .
$$ | 11.5. The required inequality is
$$
\sin ^{2 n} x+\left(2^{n}-2\right) \sin ^{n} x \cos ^{n} x+\cos ^{2 n} x \leqslant 1 \text {. }
$$
Taking the $n$-th power on both sides of the identity, we get
$$
\begin{aligned}
1= & \left(\sin ^{2 n} x+\cos ^{2 n} x\right)+n\left(\sin ^{2} x \cos ^{2 n-2} x+\right. \\
& \left.\co... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,873 |
Example 7 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, R$ is a point inside the triangle, $\angle R B C=$ $\angle R C B=20^{\circ}$. Find the degree measure of $\angle R A B$. | Solution: As shown in Figure 7, construct a regular $\triangle DAB$ on one side of $AB$ inside $\triangle ABC$. Connect $DR$ and $DC$.
Given $\angle ACB = 30^{\circ}$, we know that point $D$ is the circumcenter of $\triangle ABC$. Therefore, $DB = DC$. Thus, $\angle DCB = \angle DBC = 10^{\circ}$, and $\angle BDC = 16... | 80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,874 |
11.6. A natural number is called a "perfect number" if it equals the sum of all its positive divisors excluding itself, for example, $6=1+2+3$. If a perfect number greater than 28 is divisible by 7, prove that it must also be divisible by 49. | 11.6. Let a perfect number equal $7 n(n>4)$, where $n$ is not a multiple of 7. Then, all positive divisors of $7 n$ (including $7 n$) can be divided into pairs of the form $d$ and $7 d$, so the sum of all positive divisors of $7 n$ (i.e., $14 n$) can be divided by 8, hence $n$ is a multiple of 4. Note that $\frac{7}{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,875 |
11.7. Quadrilateral $A B C D$ is circumscribed around circle $\omega$, and the lines containing sides $A B$ and $C D$ intersect at point $O$. Circle $\omega_{1}$ is tangent to side $B C$ at point $K$, and is also tangent to the lines containing sides $A B$ and $C D$. Circle $\omega_{2}$ is tangent to side $A D$ at poin... | 11.7. For definiteness, let point $O$ be on the extension of segment $AB$ beyond $B$ (as shown in Figure 6). Denote the intersection points of $KL$ with the circle $\omega$ as $P$ and $Q$, and the points where the sides $BC$ and $AD$ touch $\omega$ as $M$ and $N$, respectively. Draw the tangents $l_{1}$ and $l_{2}$ to ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,876 |
11.8.100 $\times 100$ grid cells are colored with 4 different colors, such that each row and each column contains exactly 25 cells of each color. Prove: it is possible to find two rows and two columns, the 4 cells at their intersections are colored with 4 different colors. | 11.8. Using proof by contradiction. Assume that any two rows and any two columns intersect in squares of colors 1 to 4. We call two differently colored squares in the same column a "pair," and two identically colored squares in the same row or column a "match." We categorize the "pairs" into 6 types based on the colors... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,877 |
1. Let $n$ be an integer greater than or equal to 1. In the $x-y$ plane, a "path" from $(0,0)$ to $(n, n)$ is a broken line that starts from $(0,0)$ and at each step either moves to the right (denoted as $E$) or moves up (denoted as $N$) by one unit, until it reaches $(n, n)$. All movements are within the half-plane sa... | Proof: A path from (1,1) to $(n, n)$ with $s$ steps is called an $(n, s)$-type path. Let $f(n, s)$ denote the number of $(n, s)$-type paths, and $g(n, s)=\frac{1}{s} \mathrm{C}_{n-1}^{s-1} \mathrm{C}_{n}^{s-3}$. We will prove by induction on $n$ that
$$
f(n, s)=g(n, s), \quad s=1,2, \cdots, n.
$$
It is clear that $f(1,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,878 |
$\therefore$ (1) If a $5 \times 4$ board is covered by $n$ pieces, each consisting of 5 small squares arranged in the shape of
a cover, prove that $n$ is even.
(2) Prove that using $2k$ pieces to cover a $5 \times 2k (k \geqslant 3)$ rectangle can be done in at least $2 \times 3^{k-1}$ ways, where symmetric placements... | Proof: (1) If the first, third, and fifth rows of a $5 \times n$ rectangle are colored red, and the second and fourth rows are colored white, then there are $3n$ red small squares and $2n$ white small squares. Since each piece of cardboard can cover at most 3 red small squares, to cover the $5 \times n$ rectangle with ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,879 |
3. A biologist
observed a chameleon
for a while. During the rest periods, the chameleon can immediately catch a fly. Time:
(1) How many flies did the chameleon catch before its first 9-minute rest?
(2) How many minutes later did the chameleon catch the 98th fly?
(3) After 1999 minutes, how many flies did the chameleo... | Solution: Let the time the chameleon rests before catching the $m$-th fly be $r(m)$. Then $r(1)=1, r(2m)=r(m), r(2m+1)=r(m)+1$. This indicates that $r(m)$ is equal to the number of 1s in the binary representation of the number $m$.
Let $i(m)$ be the moment the chameleon catches the $m$-th fly, and $f(n)$ be the total ... | 462 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,880 |
6. Suppose each integer is colored red, blue, green, or yellow, $x 、 y$ are odd, and $|x| \neq|y|$. Prove that there exist two integers of the same color whose difference equals one of $x 、 y 、 x+y$ or $x-y$. | Proof: Suppose there exists a color function
$f: Z \rightarrow\{R, B, G, Y\}$, such that for any integer $a$, we have
$$
f\{a, a+x, a+y, a+x+y\}=\{R, B, G, Y\},
$$
where $R$ represents red, $B$ represents blue, $G$ represents green, and $Y$ represents
yellow. Let $g: Z \times Z \rightarrow\{R, B, G, Y\}$, and
$$
g(i, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,882 |
1.100 people share 1000 RMB, and the money of any 10 people does not exceed 190 RMB. Then, the most one person can have is ( ) RMB.
(A) 109
(B) 109
(C) 118
(D) 119 | -1 . (B).
Let the person with the most money have $x$ yuan, and the rest are divided into 11 groups, each with 9 people, and this person together with each group of 9 people does not exceed 190 yuan. Therefore,
$$
190 \times 11 \geqslant 1000-x+11 x .
$$
This gives $x \leqslant 109$.
So, one person can have at most 10... | 109 | Inequalities | MCQ | Yes | Yes | cn_contest | false | 711,884 |
Example 8 In $\triangle A B C$, $\angle A B C=60^{\circ}$, $\angle A C B=40^{\circ}, P$ is a point inside the triangle, $\angle P B C=20^{\circ}$, $\angle P C B=10^{\circ}$. Find the degree measure of $\angle P A B$. | Solution: As shown in Figure 8, let point $D$ be the reflection of point $B$ over $PC$. Connect $DA$, $DB$, $DC$, and $DP$.
In $\triangle BCD$, given $\angle DCB = 20^{\circ}$, we have
$\angle BDC = 80^{\circ} = \angle BAC$.
Thus, points $A$, $D$, $B$, and $C$ are concyclic.
Since $DC$ is parallel to $\angle ACB$, we h... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,885 |
2. Given $y=|x-2|-\frac{1}{2}|x|+|x+2|$, and $-1 \leqslant x \leqslant 2$. Then the difference between the maximum and minimum values of $y$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 2. (D).
$$
y=\left\{\begin{array}{ll}
4+\frac{1}{2} x, & -1 \leqslant x \leqslant 0, \\
4-\frac{1}{2} x, & 0<x \leqslant 2 .
\end{array}\right.
$$
When $x=0$, $y_{\text {max }}=4$;
When $x=2$, $y_{\text {min }}=3$.
Thus $y_{\max }-y_{\min }=4-3=1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,886 |
3. On the side $P C$ of $\triangle A B C$, there is a point $D$, $\angle A D B$ is an acute angle, $P 、 Q$ are the circumcenters of $\triangle A B D 、 \triangle A C D$ respectively, and the quadrilateral $A P D Q$ has the same area as $\triangle A B C$. Then $\angle A D B=(\quad)$.
(A) $60^{\circ}$
(B) $75^{\circ}$
(C)... | 3. (D).
As shown in Figure 4, let $\angle A D B^{\circ} = a$, then $\angle A P B = 2 \alpha$. By the properties of the circumcenter, it is easy to see that $\angle A P Q = \angle A B C$, $\angle A Q P = \angle A C B \Rightarrow \triangle A P Q \backsim \triangle A B C$. Clearly, $\triangle A P Q \cong \triangle D P Q$... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,887 |
5. Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$, and the quadratic equation with $\alpha^{2}, \beta^{2}$ as roots is still $x^{2}-p x+q=0$. Then the number of pairs $(p, q)$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 0 | 5. (B).
From the problem,
$\left\{\begin{array}{l}\alpha+\beta=p, \\ \alpha \beta=q\end{array}\right.$ and $\left\{\begin{array}{l}\alpha^{2}+\beta^{2}=p, \\ \alpha^{2} \beta^{2}=q .\end{array}\right.$
Thus, $q^{2}=q$. Therefore, $q=0$ or $q=1$.
And we have $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$, whi... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,889 |
6. As shown in Figure $1, A B, C D$, and $P Q$ are three chords of $\odot O$, $A B \perp P Q, C D \perp P Q$, $M, N$ are the feet of the perpendiculars. If $A B \leqslant$ $C D$, then we have $($.
(A) $\frac{A M}{B M} \geqslant \frac{C N}{D N}$
(B) $\frac{A M}{E M} \leqslant \frac{C}{D N}$
(C) $\frac{A M}{B M} \neq \fr... | 6. (A).
As shown in Figure 6, let $A B=$
$$
\begin{array}{l}
2 a, C D=2 b, \text { obviously } \\
a \leqslant b . \\
\text { Let } O S=x, \text { then } X M \\
=Y N=O S=x . \text { Therefore, } \\
\frac{A M}{M B}-\frac{C N}{D N} \\
=\frac{a+x}{a-x}-\frac{b+x}{b-x} \\
=\frac{1}{(a-x)(b-x)}[(a+x)(b-x)-(b+x) \cdot \\
=\f... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,890 |
1. Real numbers $a, b, c$ are all non-zero, and $a+b+c=$
0. Then
$$
=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)
$$ | \begin{aligned} \text { II.1. } & -3 . \\ \text { Original expression }= & a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ & +c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ = & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)-3=-3 .\end{aligned} | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,891 |
3. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the range of $x$ that makes the inequality $x^{2}+p x>4 x+p-3$ hold is $\qquad$ . | $$
\begin{array}{l}
\text { 3. } x3 \text {. } \\
\text { From } x^{2}+p x>4 x+p-3 \\
\Rightarrow x^{2}+(p-4) x-p+3>0 \\
\Rightarrow(x+p-3)(x-1)>0 \text {. } \\
\therefore\left\{\begin{array} { l }
{ x > 3 - p } \\
{ x > 1 }
\end{array} \text { , or } \left\{\begin{array}{l}
x < 1 \\
x < 3 - p
\end{array}\right.\right... | x < 1 \text{ or } x > 3 - p | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,893 |
4. As shown in Figure 2,
Square $A B C D$ has a side length of $1, E$ is a point on the extension of $C B$, connect $E D$ intersecting $A B$ at $P$, and $P E$ $=\sqrt{3}$. Then the value of $B E-P B$ is $\qquad$ | 4.1.
Let $B E=x, P B=y$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=(\sqrt{3})^{2}, \\
\frac{y}{1}=\frac{x}{x+1} .
\end{array}\right.
$$
From (2), we have $x-y=x y$.
From (1) and (3), we have
$$
\begin{array}{l}
(x-y)^{2}+2(x-y)-3=0, \\
(x-y+3)(x-y-1)=0 .
\end{array}
$$
Clearly, $x>y, x-y+3>0$.
Therefore, $x-y-1=0,... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,894 |
Example 9 In $\triangle A B C$, $\angle A B C=\angle A C B=40^{\circ}$, $P$ is a point inside the triangle; $\angle P A C=20^{\circ}$, $\angle P C B=30^{\circ}$. Find the degree measure of $\angle P B C$. | Solution: As shown in Figure 9, let point $D$ be the symmetric point of point $C$ with respect to $AP$. Connect $DA$, $DB$, $DC$, and $DP$.
Given $\angle PAC=20^{\circ}$, $\angle PCA=10^{\circ}$, we have
$\angle DAC=40^{\circ}$, $\angle PDA=\angle PCA=10^{\circ}$,
thus $\triangle PDC$ is an equilateral triangle.
Given ... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,896 |
II. As shown in Figure 3, a wooden stick $(AB)$ of length $2a$ is leaning against a wall $(ON)$ perpendicular to the ground $(OM)$, with an inclination angle $(\angle ABO)$ of $60^{\circ}$. If the end $A$ of the stick slides down along the line $ON$, and the end $B$ slides to the right along the line $OM$ $(NO \perp OM... | II. $\because P O=\frac{1}{2} A B=a$ is a constant, $O$ is a fixed point,
$\therefore$ the path from $P$ to $P^{\prime}$ is an arc with $O$ as the center and $a$ as the radius. Connecting $O P, O P^{\prime}$, it is easy to see that $O P=P A=O P^{\prime}=P^{\prime} B^{\prime}=a$. In the right triangle $\triangle A B O$,... | \frac{1}{12} \pi a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,897 |
Three. (Full marks 25 points) In $\triangle ABC$, there is a point $Q$. It is known that $\angle AQB=90^{\circ}+\frac{1}{2} \angle C$ and $\angle AQC=90^{\circ}+\frac{1}{2} \angle B$. Prove that point $Q$ is the incenter of $\triangle ABC$.
---
The translation maintains the original text's formatting and line breaks. | Three, provide two different proofs: As shown in Figure 8, construct the circumcircle of $\triangle A B C$, and let the ray $A Q$ intersect the circle at $D$. Connect $D B, D C$.
$$
\begin{aligned}
\because & \angle A Q B \\
& =90^{\circ}+\frac{1}{2} \angle A C B, \\
\therefore & \angle D Q B \\
& =90^{\circ}-\frac{1}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,898 |
1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001 | -1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,899 |
2. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$, and the parabola $C$ has $F_{1}$ as its vertex and $F_{2}$ as its focus. Let $P$ be a point of intersection between the ellipse and the parabola. If the eccentricity $e$ of the ellipse $E$ satisfies $\left|P F_{1}\right|=e\left|P F_{2}\right|$... | 2. (C).
From $\frac{\left|P F_{1}\right|}{r}=\left|P F_{2}\right|$, we know that the directrix of the ellipse E coincides with the directrix of the parabola $\mathrm{C}$.
According to the problem, the equation of the directrix of the parabola $C$ is $x=-3 c$, and the left directrix of the ellipse $E$ is $x=-\frac{a^{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,900 |
3. Given the complex number $z$ satisfies $3 z^{6}+2 \mathrm{i} z^{5}-2 z-3 \mathrm{i}=$ 0 . Then the modulus of $z$ ( ).
(A) greater than 1
(B) equal to 1
(C) less than 1
(D) cannot be determined | 3. (B).
The original equation can be transformed into $z^{5}=\frac{2 z+3 \mathrm{i}}{3 z+2 \mathrm{i}}$.
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$, then
$$
\begin{aligned}
\left|z^{5}\right| & =\left|\frac{2 a+(2 b+3) \mathrm{i}}{3 a+(3 b+2) \mathrm{i}}\right|=\sqrt{\frac{4 a^{2}+(2 b+3)^{2}}{9 a^{2}+(3 b+2)^{2}}} \... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,901 |
4. Inside a cube with edge length $a$, there is an inscribed sphere. A line is drawn through the midpoints of two skew edges of the cube. The length of the segment of this line that is intercepted by the sphere is
(A) $(\sqrt{2}-1) a$
(B) $\frac{\sqrt{2}}{2} a$
(C) $\frac{1}{4} a$
(D) $\frac{1}{2} a$ | 4. (B).
As shown in Figure 2, M and N are the midpoints of two skew edges of a cube. The line $MN$ intersects the surface of the inscribed sphere $O$ at points $E$ and $F$. Connecting $MO$ intersects the opposite edge at $P$, making $P$ the midpoint of the opposite edge. Taking the midpoint $G$ of $EF$, we have $OG \p... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,902 |
5. For any integer $n(n \geqslant 2)$, the positive numbers $a$ and $b$ that satisfy $a^{n}=a+1, b^{2 n}=b+3 a$ have the following relationship ( ).
(A) $a>b>1$.
(B) $b>a>1$
(C) $a>1,01$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result d... | 5. (A).
First, $a>1, b>1$. Otherwise,
if $0<1$, then $a>1$, leading to a contradiction;
if $0<3a>3$, then $b>1$, leading to a contradiction.
Therefore, both $a$ and $b$ are greater than 1.
On one hand, $a^{2n}-b^{2n}=(a+1)^{2}-(b+3a)=a^{2}-a-b+1$.
On the other hand, $a^{2n}-b^{2n}=(a-b)\left(a^{2n-1}+a^{2n-2}b+\cdot... | A | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,903 |
6. The first digit is 1, and there are exactly two identical digits in the four-digit number ( ).
(A) 216
(B) 252
(C) 324
(D) 432 | 6. (D).
The qualified four-digit numbers must contain one 1 or two 1s.
(1) Contain two 1s. Choose two digits from the remaining 9 digits other than 1, which has $C_{9}^{2}$ ways, and then form any permutation of a three-digit number with one of the 1s, which has $\mathrm{P}_{3}^{3}$ ways. Thus, the four-digit numbers ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 711,904 |
$$
\begin{array}{l}
\text { 2. In } \triangle A B C \text {, } \angle C-45^{\circ} \text {, and } \\
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3} \text {. }
\end{array}
$$
Then the degree measure of the largest interior angle is | $2.75^{\circ}$.
Let's assume $\angle A$ is the largest angle.
Since $\angle C=45^{\circ}$, we have $\angle A+\angle B=135^{\circ}$.
Thus,
$$
\begin{array}{l}
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3} \\
\Leftrightarrow(\sin A-\sqrt{3} \cos A)+(\sin B-\sqrt{3} \cos B) \\
\quad(\sin C-\sqrt{3} \cos C)=... | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,906 |
Example 10 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, R$ is a point inside the triangle, $\angle R A C=$ $\angle R C B=20^{\circ}$. Find the degree measure of $\angle R B C$. | Solution: As shown in Figure 10, let point $E$ be the reflection of point $R$ over $AC$, and point $D$ be the reflection of point $A$ over $EC$. Connect $DA$, $DR$, $DE$, $DC$, $EA$, and $EC$.
It is easy to see that $\triangle EDA$ is an equilateral triangle, so
$$
AD = AE = AR \text{.}
$$
In $\triangle ACD$, it is ea... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,907 |
3. Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $x^{2}-y^{2}=4$, and $P$ be any point on the hyperbola. A perpendicular line is drawn from $F_{1}$ to the angle bisector of $\angle F_{1} P F_{2}$, with the foot of the perpendicular being $M$. Then the equation of the locus of point $M$ is $\qquad$ | 3. $x^{2}+y^{2}=4$.
According to the problem, the symmetric point $F_{1}^{\prime}$ of point $F_{1}$ with respect to the line $P M$ lies on the line $P F_{2}$, then $P F_{1}^{\prime}=P F_{1}$: According to the definition of a hyperbola, we get $\left|F_{1}^{\prime} F_{2}\right|$
$$
\begin{array}{l}
=\left\|P F_{1}^{\pr... | x^{2}+y^{2}=4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,908 |
4. If $\triangle A B C$ is an obtuse triangle, then the range of $\operatorname{arccos}(\sin A)+\operatorname{arccos}(\sin B)+\operatorname{arccos}(\sin C)$ is $\qquad$ | 4. $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.
Assume $\angle A$ is an obtuse angle, then $0<\angle A-\frac{\pi}{2}<\frac{\pi}{2}$.
$\therefore$ The original expression $=\arccos \left[\cos \left(A-\frac{\pi}{2}\right)\right]+\operatorname{arccos}\left[\cos \left(\frac{\pi}{2}\right.$.
$$
\begin{array}{l}
-B)]+\arc... | \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,909 |
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