problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 3 In a scalene acute triangle $\triangle ABC$, the angle bisector of the acute angle formed by the altitudes $AA_1$ and $CC_1$ intersects sides $AB$ and $BC$ at points $P$ and $Q$, respectively. Let $M$ be the midpoint of $AC$, $H$ be the orthocenter of $\triangle ABC$, and the angle bisector of $\angle ABC$ in... | Prove as shown in Figure 3, take points $S, T$ on segments $AH, CH$ respectively, such that $PS \perp AB, TQ \perp BC$.
Let the intersection of lines $PS$ and $TQ$ be $K$.
Since $\angle BPK = \angle BQK = 90^\circ$, points $B, P, K, Q$ are concyclic.
Next, we prove that point $K$ coincides with point $R$.
It is easy to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,919 |
Find all functions $f: R \rightarrow R$, such that for any real numbers $x, y$ the following is satisfied:
$$
f(x y)=f(x) f(y)-x-y .
$$ | Solve: Substituting $x=y=0$ into the given equation yields $f(0)=f^{2}(0) \Rightarrow f(0)=0$ or 1. When $f(0)=0$, substituting $x=0$ gives $y=0$, which is a contradiction. When $f(0)=1$, substituting $x=1, y=-1$ gives $f(-1)=f(-1) f(1)$.
If $f(-1)=0$, then
$$
f(-x)=f(x) f(-1)-x+1=1-x \text {. }
$$
Thus, $f(x)=1+x$.
U... | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,920 |
In the cyclic quadrilateral $ABCD$ inscribed in $\odot O$, $I_{1}$ and $I_{2}$ are the incenters of $\triangle ABC$ and $\triangle ACD$, respectively. $G$ is the midpoint of arc $\overparen{BD}$. $GI_{1}$ and $GI_{2}$ intersect $\odot O$ at points $E$ and $F$, respectively. $ED$ and $BF$ intersect $AC$ at points $J$ an... | Prove that in Figure 2, let the incenter of $\triangle A B D$ be $I^{\prime}$, and $B I^{\prime}$ intersects $C I_{2}$ at point $K$. Then $K$ is the midpoint of arc $\overparen{A D}$.
In the cyclic hexagon $A C K B F G$, by Pascal's theorem, points $H 、 I^{\prime} 、 I_{2}$ are collinear. $D I^{\prime}$ intersects $C I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,921 |
Example 4 Given $\triangle A B C$, construct a semicircle with diameter $B C$, intersecting $A B$ and $A C$ at points $D$ and $E$ respectively. Draw perpendiculars from $D$ and $E$ to $B C$, with feet at $F$ and $G$ respectively. The line segments $D G$ and $E F$ intersect at point $M$. Prove: $A M \perp B C$.
(37th IM... | Prove that, as shown in Figure 4, construct the altitude $A N$ of $\triangle A B C$, and connect $B E, C D, D E$.
Then $B E, C D$ are two altitudes of $\triangle A B C$, and let the orthocenter be $H$.
It is sufficient to prove that point $M$ lies on $A N$.
Since $D F / / E G$, it is sufficient to prove $\frac{D M}{M ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,922 |
Example 6 As shown in Figure 6, in $\triangle ABC$, $X, Y$ are two points on the line $BC$ ($X, B, C, Y$ are arranged in order) such that $BX \cdot AC = CY \cdot AB$. Let the circumcenters of $\triangle ACX$ and $\triangle ABY$ be $O_{1}$ and $O_{2}$, respectively. The line $O_{1}O_{2}$ intersects $AB$ and $AC$ at poin... | Prove As shown in Figure 6, construct the angle bisector $AD$ of $\angle BAC$.
To prove $AD \perp UV$
$\Leftarrow AD$ is the radical axis of $\odot O_{1}$ and $\odot O_{2}$
$\Leftarrow$ the power of point $D$ with respect to $\odot O_{1}$ and $\odot O_{2}$ is equal
$\Leftarrow X D \cdot C D=Y D \cdot B D$.
From $\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,923 |
Example 7 Given that the incircle of $\triangle ABC$ touches sides $BC, CA, AB$ at points $D, E, F$ respectively, and $X$ is an interior point of $\triangle ABC$. The incircle of $\triangle XBC$ also touches side $BC$ at point $D$, and touches $CX, XB$ at points $Y, Z$ respectively. Prove that quadrilateral $EFZY$ is a... | Prove the conjecture: lines $B C$, $Z Y$, and $F E$ are concurrent at point $P$.
Then, from $P E \cdot P F = P D^{2} = P Y \cdot P Z$, it follows that quadrilateral $E F Z Y$ is a cyclic quadrilateral.
As shown in Figure 7, let the intersection of lines $B C$ and $F E$ be point $P$. We need to prove that point $P$ li... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,924 |
Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Compe... | Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.
First, we prove a lemma.
Lemma Let $G=(... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,925 |
Example 2 Given that $p$ is a prime number not equal to 2. Prove: the way to decompose $\frac{2}{p}$ into the sum of the reciprocals of two different positive integers exists and is unique.
(2015, Shanghai Jiao Tong University Independent Recruitment Examination) | Proof: Let $x, y \in \mathbf{Z}_{+}$, and $\frac{2}{p}=\frac{1}{x}+\frac{1}{y}$.
Eliminating the denominator gives $(2 x-p)(2 y-p)=p^{2}$.
From the given, $p^{2}$ can only be factored into $p \times p$ and $p^{2} \times 1$ two cases, and $p \neq 2$.
If $2 x-p=p$, and $2 y-p=p$, then $x=y=p$, which contradicts the prob... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,926 |
Example 3 Let $x, y, z > 0$. Prove:
$$
\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \geqslant 2 \text {. }
$$ | Proof: By homogeneity, without loss of generality, assume $x+y+z=1$. Then we only need to prove
$$
\begin{array}{l}
\sqrt{\frac{x}{1-x}}+\sqrt{\frac{y}{1-y}}+\sqrt{\frac{z}{1-z}} \geqslant 2 . \\
\text { By } \sqrt{\frac{x}{1-x}}=\frac{x}{\sqrt{x(1-x)}} \geqslant 2 x(x \in(0,1)),
\end{array}
$$
we know that for any $x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,927 |
Example $\mathbf{5}$ Given that $a, b, c$ are pairwise coprime positive integers, and satisfy
$$
a^{2}\left|\left(b^{3}+c^{3}\right), b^{2}\right|\left(a^{3}+c^{3}\right), c^{2} \mid\left(a^{3}+b^{3}\right) \text {. }
$$
Find the values of $a, b, c$. ${ }^{[1]}$
(Eighth China Southeast Mathematical Olympiad) | Given:
$$
\begin{array}{l}
a^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), b^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), \\
c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$
Since \(a, b, c\) are pairwise coprime, we have:
$$
a^{2} b^{2} c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right).
$$
Assume \(a \geqslant b \geqslant c... | (1, 1, 1), (1, 2, 3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,928 |
Question As shown in Figure $1, A D$ is a chord of the circumcircle of $\triangle A B C$, $P$ is the midpoint of $A D$, and $P D$ bisects $\angle B P C, O, O_{1}, O_{2}$ are the circumcenters of $\triangle A B C, \triangle A P B, \triangle A P C$ respectively. Prove: the circumcircle of $\triangle O_{1} P B$ is tangent... | As shown in Figure 1, let $O_{3}$ and $O_{4}$ be the circumcenters of $\triangle O_{1} B P$ and $\triangle O_{2} C P$, respectively. To prove that $\odot O_{3}$ and $\odot O_{4}$ are tangent, it suffices to prove that
$$
\begin{array}{l}
\angle O_{3} P O_{1} + \angle O_{1} P O_{2} + \angle O_{2} P O_{4} = 180^{\circ}. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,929 |
For a bicentric hexagon $(n=6)$, the circumradius $R$, the inradius $r$, and the distance $d$ between the centers of the two circles satisfy the relation
$$
\sqrt{1-\left(\frac{r}{R+d}\right)^{2}}+\sqrt{1-\left(\frac{r}{R-d}\right)^{2}}=1 .
$$ | Prove as shown in Figure 2, draw line $O I$ intersecting $\odot O$ at points $A_{1}$ and $A_{4}$. Draw $A_{1} B_{1}$, $A_{1} B_{6}$, $A_{4} B_{4}$, and $A_{4} B_{3}$ tangent to $\odot I$ at points $B_{1}$, $B_{6}$, $B_{4}$, and $B_{3}$. $A_{1} B_{1}$, $A_{4} B_{3}$, $A_{1} B_{6}$, and $A_{4} B_{4}$ intersect $\odot O$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,930 |
1. The longest professional tennis match lasted 11 hours and 5 minutes, which is ( ) minutes.
(A) 605
(B) 655
(C) 665
(D) 1005
(E) 1105 | 1. C.
$$
11 \times 60+5=665
$$ | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 729,931 |
2. In rectangle $A B C D$, $A B=6, A D=8, M$ is the midpoint of $A D$. Then the area of $\triangle A M C$ is ( ).
(A) 12
(B) 15
(C) 18
(D) 20
(E) 24 | 2. A.
$$
S_{\triangle A M C}=\frac{1}{2} \times \frac{1}{2} \times 8 \times 6=12 \text {. }
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,932 |
4. When Cheney was a child, he could walk 15 miles in 3 hours and 30 minutes. Now that he is an old man, he can walk 10 miles in 4 hours. Compared to when he was a child, he now takes ( ) more minutes to walk 1 mile.
(A) 6
(B) 10
(C) 15
(D) 18
$(\mathrm{E}) 30$ | 4. B.
$$
\frac{4 \times 60}{10}-\frac{3 \times 60+30}{15}=10 .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,933 |
5. Given that $n$ is a two-digit natural number. $n$ divided by 9 leaves a remainder of 1, and $n$ divided by 10 leaves a remainder of 3. Then the remainder when $n$ is divided by 11 is ().
(A) 0
(B) 2
(C) 4
( D ) 5
(E) 7 | 5. E.
From $n \equiv 3(\bmod 10)$, we know that the units digit of $n$ is 3.
Let $n=10a+3$.
Then $a+3 \equiv 1(\bmod 9) \Rightarrow a=7$
$$
\Rightarrow n=73 \equiv 7(\bmod 11)
$$ | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,934 |
7. Which of the following ( ) is not a perfect square.
(A) $1^{2016}$
(B) $2^{2017}$
(C) $3^{2018}$
(D) $4^{2019}$
(E) $5^{2020}$ | 7. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,935 |
15. The highest positive integer power of 2 that is a factor of $13^{4}-11^{4}$ is ( ).
(A) 8
(B) 16
(C) 32
(D) 64
(E) 128 | 15. C.
Notice,
$$
\begin{array}{l}
13^{4}-11^{4} \\
=\left(13^{2}+11^{2}\right)(13+11)(13-11) \\
=290 \times 24 \times 2=2^{5} \times 3 \times 145 .
\end{array}
$$
Therefore, the answer is $2^{5}=32$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,936 |
20. Given $a, b, c \in \mathbf{Z}_{+}, [a, b]=12, [b, c]$ $=15$. Then the minimum possible value of $[a, c]$ is ( ).
( A) 20
(B) 30
(C) 60
(D) 120
(E) 180 | 20. A.
From $[a, b]=12 \Rightarrow b \mid 12$.
From $[b, c]=15 \Rightarrow b \mid 15$.
To find the minimum value of $[a, c]$, i.e., $b$ should be as large as possible, thus, $b=3$.
At this point, $a=4, c=5, [a, c]=20$. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,938 |
24. Use the digits $1, 2, 3, 4, 5$ to form a five-digit number $\overline{P Q R S T}$ without repeating any digit, and $4|\overline{P Q R}, 5| \overline{Q R S}$, $3 \mid \overline{R S T}$. Then the digit $P$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 24. A.
From $5 \mid \overline{Q R S} \Rightarrow S=5$.
From $4|\overline{P Q R} \Rightarrow 4| \overline{Q R}$
$\Rightarrow R=2$ or 4.
When $R=2$, by
$$
\begin{array}{l}
3|\overline{R S T} \Rightarrow 3| \overline{25 T} \\
\Rightarrow 2+5+T \equiv 0(\bmod 3) \\
\Rightarrow T \equiv 2(\bmod 3) .
\end{array}
$$
Thus, $... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,939 |
2. Let the vertices of the convex quadrilateral $ABCD$ not be concyclic. Denote the projections of point $A$ onto lines $BC$, $BD$, and $CD$ as $P$, $Q$, and $R$, respectively, where points $P$ and $Q$ lie on segments $BC$ and $BD$, and point $R$ lies on the extension of $CD$; Denote the projections of point $D$ onto l... | 2. From the conditions, we know that points $A, Z, R, D, X, Q$ lie on the circle $\Gamma$ with diameter $AD$, and points $D, X, Y, C$ lie on the circle $\Gamma'$ with diameter $CD$. Extending $AR$ and $XD$ intersect at the orthocenter $H'$ of $\triangle ACD$, $ZX$ intersects $RQ$ at point $K$, and extending $RZ$ and $X... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,940 |
3. Let $X$ be a 100-element set. Find the smallest positive integer $n$ such that for any sequence $A_{1}, A_{2}, \cdots, A_{n}$ of subsets of $X$, there exist $1 \leqslant i<j<k \leqslant n$ satisfying $A_{i} \subseteq A_{j} \subseteq A_{k}$ or $A_{i} \supseteq A_{j} \supseteq A_{k}$. | 3. $n=\mathrm{C}_{102}^{51}+1$.
Consider the following sequence of subsets: $A_{1}, A_{2}, \cdots, A_{N}$, where $N=\mathrm{C}_{100}^{50}+\mathrm{C}_{100}^{49}+\mathrm{C}_{100}^{51}+\mathrm{C}_{100}^{50}=\mathrm{C}_{102}^{51}$.
The first segment of $\mathrm{C}_{100}^{50}$ terms consists of all 50-element subsets, the ... | \mathrm{C}_{102}^{51}+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,941 |
4. Prove: There exists a 58th-degree monic polynomial with real coefficients
$$
P(x)=x^{58}+a_{1} x^{57}+\cdots+a_{57} x+a_{58},
$$
satisfying:
(1) $P(x)$ has exactly 29 positive real roots and 29 negative real roots;
(2) $\log _{2017}\left|a_{k}\right|(k=1,2, \cdots, 58)$ are all positive integers. (Yao Yijun) | 4. Prove by induction: For any non-negative integer pair $(m, n)$, there exists an $(m+n)$-degree polynomial $Q_{m, n}(x)$ with real coefficients, having exactly $m$ distinct positive roots, $n$ distinct negative roots, a constant term of 1, and all other coefficients' absolute values being positive integer powers of 2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,942 |
5. Prove: There exists a positive real number $C$, such that the following conclusion holds: if positive integers $H, N$ satisfy $H \geqslant 3, N \geqslant \mathrm{e}^{C H}$, then among the first $N$ positive integers, selecting no fewer than $C H \frac{N}{\ln N}$ numbers, one can always find $H$ numbers such that the... | 5. Take $C=35$, then $N \geqslant \mathrm{e}^{35 \times 3}>2^{105}$.
First prove: $\log _{2} N > 105$, without loss of generality, assume
$15 k \leqslant \log _{2} N < 15(k+1)$
$$
the maximum prime factor of these numbers is greater than $\ln N$.
Since they can be divided into at most $\left\lfloor\frac{N}{\ln N}\rig... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,943 |
3. In a bag, there are 100 balls of the same size, numbered 1, $2, \cdots, 100$, respectively. Three balls are randomly drawn from the bag. The probability that the sum of the numbers on these three balls is a multiple of 3 is $\qquad$ (expressed as a simplified fraction). | 3. $\frac{817}{2450}$ | \frac{817}{2450} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,945 |
5. As shown in Figure $1, A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with an edge length of 100 cm. A moving point $P$ starts from vertex $A$ and moves uniformly along the line segment $A C_{1}$ at a speed of 3 cm/s. A moving point $Q$ starts from vertex $A_{1}$ at the same time as point $P$, moving uniformly along the... | 5.71.1275 cm | 5.7113 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,946 |
Example 7 Let real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$
\sum \frac{1}{5 a^{2}-4 a+11} \leqslant \frac{1}{4} .
$$
(2007, China Girls' Mathematical Olympiad) | Prove that the equality in (1) holds at the "mean point"
$$
(a, b, c)=(1,1,1)
$$
Notice that the local inequality
$$
\begin{array}{l}
\frac{1}{5 x^{2}-4 x+11} \leqslant-\frac{1}{24}(x-1)+\frac{1}{12} \\
\Leftrightarrow(x-1)^{2}(5 x-9) \leqslant 0 . \\
\text { Hence } \frac{1}{5 x^{2}-4 x+11} \\
\leqslant-\frac{1}{24}(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,947 |
9. As shown in Figure 2, on the right branches of the hyperbolas \(x^{2}-y^{2}=1\) and \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\), there are points \(A, B\) and \(C, D\) respectively, such that quadrilateral \(A B C D\) is a square with sides parallel to the coordinate axes. Find the area of square \(A B C D\) (accurate to ... | 9. Let point $A(\sec \theta, \tan \theta)\left(0<\theta<\frac{\pi}{2}\right)$. Given that the square $A B C D$ and point $D$ lie on the hyperbola
$$
\frac{x^{2}}{4}-\frac{y^{2}}{9}=1
$$
we have $x_{D}=\frac{2}{3} \sqrt{9+\tan ^{2} \theta}, y_{D}=\tan \theta$.
$$
\begin{array}{l}
\text { Then }|A D|=\left|x_{D}-x_{A}\r... | 0.8506 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,950 |
11. In the Cartesian coordinate system $x O y$, there are infinitely many circles inside the parabola $y=a x^{2}$ (where $a$ is a positive constant). The centers $O_{1}, O_{2}, \cdots$ are all on the $y$-axis. For each integer $n>1, \odot O_{n}$ is tangent to the parabola and externally tangent to $\odot O_{n-1}$, as s... | 11. Let the radius of $\odot O_{n}$ be $r_{n}$, and denote
$$
S_{n}=r_{1}+r_{2}+\cdots+r_{n} \text {. }
$$
Let the center of $\odot O_{n+1}$ be $O_{n+1}\left(0,2 S_{n}+r_{n+1}\right)$, and the equation of the circle be
$$
x^{2}+\left(y-\left(2 S_{n}+r_{n+1}\right)\right)^{2}=r_{n+1}^{2} .
$$
By combining the above eq... | \frac{4031}{2 a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,951 |
12. For two sets $A, B$, define
$$
\begin{aligned}
A \Delta B= & \{x \mid x \in A, \text { and } x \notin B\} \cup \\
& \{x \mid x \in B, \text { and } x \notin A\} .
\end{aligned}
$$
Let $n > 1$ be an integer, and $A_{1}, A_{2}, \cdots, A_{2^{n}}$ be all the subsets of
$$
S=\{1,2, \cdots, n\}
$$
$M$ is a $2^{n} \tim... | 12. When calculating the sum of all numbers in table $M$, to calculate how many times each $x \in S$ is counted, it is only necessary to count the number of all ordered pairs $(i, j)(i, j \in\{1, 2, \cdots, 2^{n}\})$ that satisfy $x \in A_{i} \Delta A_{j}$.
For an element $x \in A_{i} \Delta A_{j}$, there are only two... | 2^{2 n-2} n(n+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,952 |
1. In the Cartesian coordinate system $x O y$, points $A$ and $B$ lie on the parabola $y^{2}=2 x$, satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-1, F$ is the focus of the parabola. Then the minimum value of $S_{\triangle O F A}+S_{\triangle O F B}$ is $\qquad$ | $-1 . \frac{\sqrt{2}}{2}$. It is known that $F\left(\frac{1}{2}, 0\right)$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $x_{1}=\frac{y_{1}^{2}}{2}, x_{2}=\frac{y_{2}^{2}}{2}$.
Notice that,
$$
\begin{array}{l}
-1=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2} \\
=\frac{1}{... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,953 |
3. Arrange the numbers in the set $\left\{2^{x}+2^{y} \mid x 、 y \in \mathbf{N}, x<y\right\}$ in ascending order. Then the 60th number is $\qquad$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.2064.
It is known that the number of combinations $(x, y)$ satisfying $0 \leqslant x<y \leqslant n$ is $\mathrm{C}_{n+1}^{2}$.
Notice that, $\mathrm{C}_{11}^{2}=55<60<66=\mathrm{C}_{12}^{2}$.
Therefore, the 60th number satisfies $y=11, x=4$, which means the 60th number is $2^{11}+2^{4}=2064$. | 2064 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,954 |
Example 8 Find the maximum value of $n$ such that there exists an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\sum_{i=1}^{n}\left|a_{i}\right|=\sum_{i=1}^{n}\left|a_{i}+1\right|=\sum_{i=1}^{n}\left|a_{i}-2\right|=507 .
$$
$(2005$, China Southeast Mathematical Olympiad) | 【Analysis】Let's set $a_{i}=a-i d(d>0, i=1,2$, $\cdots, n)$. Then the given system of equations becomes
$$
\left\{\begin{array}{l}
\sum_{i=1}^{n}|a-i d|=507, \\
\sum_{i=1}^{n}|a+1-i d|=507, \\
\sum_{i=1}^{n}|a-2-i d|=507 .
\end{array}\right.
$$
Thus, the absolute value sum function $f(x)=\sum_{i=1}^{n}|x-i d|$ has thre... | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,955 |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=0, a_{n+1}=a_{n}+4 \sqrt{a_{n}+1}+4(n \geqslant 1) \text {. }
$$
Then $a_{n}=$ $\qquad$ . | 4. $4 n^{2}-4 n$.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}+4 \sqrt{a_{n}+1}+4 \\
\Rightarrow a_{n+1}+1=\left(a_{n}+1\right)+4 \sqrt{a_{n}+1}+4 \\
\quad=\left(\sqrt{a_{n}+1}+2\right)^{2} \\
\Rightarrow \sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+2 .
\end{array}
$$
Therefore, $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmeti... | 4 n^{2}-4 n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,956 |
5. In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}$, $A C=1$, $M$ is the midpoint of $A B$. Fold $\triangle A C M$ along $C M$ so that the distance between points $A$ and $B$ is $\sqrt{2}$. Then the distance from point $A$ to the plane $B C M$ is $\qquad$ | 5. $\frac{\sqrt{6}}{3}$.
As shown in Figure 1, 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 $AE$, knowing that $AE \perp CM$.
Then $AD=\frac{\sqrt{3}}{2}, DE=CD \tan 30^{\circ}=\frac{\sqrt{3}}{... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,957 |
1. (16 points) Given a complex number $z$ satisfying $|z|=1$. Find
$$
u=\left|z^{3}-3 z+2\right|
$$
the maximum value. | Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
From $|z|=1$, we know $x^{2}+y^{2}=1$, and $|x| \leqslant 1$.
Then $u=\left|z^{3}-3 z+2\right|=\left|z^{3}-z-2(z-1)\right|$
\[
\begin{array}{l}
=\left|(z-1)\left(z^{2}+z-2\right)\right| \\
=\left|(z-1)^{2}(z+2)\right| \\
=\left(\sqrt{(x-1)^{2}+y^{2}}\right)^{2} \sqrt{(x+2)^{... | 3 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,958 |
3. (20 points) Let $A$ be a set containing $n$ elements, and $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ distinct subsets of $A$. Prove: there exists an element $a$ in set $A$ such that $A_{1}-\{a\}, A_{2}-\{a\}, \cdots, A_{n}-\{a\}$ are still distinct sets, where $A_{i}-\{a\}=\left\{x \in A_{i} \mid x \neq a\right\}$. | 3. Proof by Contradiction.
Assume the proposition does not hold, then for any $a \in A$,
$$
A_{1}-\{a\}, A_{2}-\{a\}, \cdots, A_{n}-\{a\}
$$
must contain two identical sets.
Construct a graph $G$, with vertices labeled as $A_{1}, A_{2}, \cdots, A_{n}$, and $A_{i}$ is connected to $A_{j}(1 \leqslant i \neq j \leqslant... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,959 |
1. Let the set $S=\left\{A_{0}, A_{1}, A_{2}, A_{3}\right\}$, and define the operation “ $\oplus$ ” on set $S$ : $A_{i} \oplus A_{j}=A_{k}$, where $k$ is the remainder of $i+j$ divided by 4, $i 、 j \in\{0,1,2,3\}$. Then the number of $x(x \in S)$ that satisfies the relation $(x \oplus x) \oplus A_{2}=A_{0}$ is ( ).
(A)... | - 1. B.
From $(x \oplus x) \oplus A_{2}=A_{0}$, let $x \oplus x=A_{k}$, then
$$
\begin{array}{l}
A_{k} \oplus A_{2}=A_{0}, k=2 \\
\Rightarrow x \oplus x=A_{2} \\
\Rightarrow x=A_{1} \text { or } A_{3} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,960 |
6. Given $a, b, c \geqslant 0$, and $a+b+c=1$. Prove
$$
\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \geqslant 2 \text {. }
$$ | Let $(u, v)=(a b+b c+c a, a b c)$. Then $a^{2}+b^{2}+c^{2}=1-2 u$.
From the equation $(x-a)(x-b)(x-c)=0$, we get the reduction formula
$$
\begin{array}{l}
x^{3}=x^{2}-u x+v \Rightarrow a^{3}=a^{2}-u a+v \\
\Rightarrow a^{4}=(1-u) a^{2}+(v-u) a+v \\
\Rightarrow a^{4}+b^{4}+c^{4}=2 u^{2}+(4 v-4 u)+1 .
\end{array}
$$
The... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,962 |
4. Given the planar vector $(1,1)$. Then the planar vector $\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$ is obtained by transforming the vector $(1,1)$ through $(\quad)$.
(A) Clockwise rotation by $60^{\circ}$
(B) Clockwise rotation by $120^{\circ}$
(C) Counterclockwise rotation by $60^{\circ}$
(D) Counterc... | 4. C.
Let the angle between the two vectors be $\theta$.
Then $\cos \theta=\frac{\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) \cdot(1,1)}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2}$.
Thus, $\theta=60^{\circ}$.
$$
\text { Also, } \frac{1-\sqrt{3}}{2}<0, \theta \in\left[0,180^{\circ}\right] \text {, }
$$
Therefor... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,963 |
8. In the tetrahedron $S-ABC$, $SA=4$, $SB \geqslant 7$, $SC \geqslant 9$, $AB=5$, $BC \leqslant 6$, $AC \leqslant 8$. Then the maximum volume of the tetrahedron is $\qquad$. | 8. $8 \sqrt{6}$.
Let $\angle S A B=\alpha$.
By the cosine rule, we have
$\cos \alpha=\frac{S A^{2}+A B^{2}-S B^{2}}{2 S A \cdot A B} \leqslant-\frac{1}{5}$.
Thus, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha} \leqslant \frac{2 \sqrt{6}}{5}$,
$S_{\triangle S A B}=\frac{1}{2} S A \cdot A B \sin \alpha \leqslant 4 \sqrt{6}$.
No... | 8 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,964 |
9. In a uniform small cube, three faces are marked with the number 0, two faces are marked with the number 1, and one face is marked with the number 2. If the small cube is tossed twice, then the mathematical expectation of the product of the numbers facing up is $\qquad$
In a uniform small cube, three faces are marke... | 9. $\frac{4}{9}$.
According to the problem, the probability of the number facing up on the small cube being 0 is $\frac{1}{2}$, the probability of it being 1 is $\frac{1}{3}$, and the probability of it being 2 is $\frac{1}{6}$, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline First Roll & 0 & 1 & 2 \\
\h... | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,965 |
10. Observe the following equations:
$$
\begin{array}{l}
\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{5}=2^{3}-2, \\
\mathrm{C}_{9}^{1}+\mathrm{C}_{9}^{5}+\mathrm{C}_{9}^{9}=2^{7}+2^{3}, \\
\mathrm{C}_{13}^{1}+\mathrm{C}_{13}^{5}+\mathrm{C}_{13}^{9}+\mathrm{C}_{13}^{13}=2^{11}-2^{5}, \\
\mathrm{C}_{17}^{1}+\mathrm{C}_{17}^{5}+\m... | 10. $2^{4 n-1}+(-1)^{n} 2^{2 n-1}$ | 2^{4 n-1}+(-1)^{n} 2^{2 n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,966 |
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$ | $-、 1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$
Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \tim... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,968 |
3. Given the function $f(x)=x^{2}-2 x+a$. If $\{x \mid f(x)=x\}=\{x \mid f(f(x))=x\}$, then the range of the real number $a$ is $\qquad$ | 3. $\left[\frac{5}{4},+\infty\right)$.
From $f(f(x))=x$
$$
\begin{array}{l}
\Rightarrow f(f(x))-f(x)=x-f(x) \\
\Rightarrow(f(x)-x)(f(x)+x-1)=0 .
\end{array}
$$
According to the problem, we know that the equation $f(x)+x-1=0$ has no real roots or has the same roots as the equation $f(x)-x=0$.
Then $\Delta=5-4 a<0$ or ... | \left[\frac{5}{4},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,970 |
4. Let $O$ be the center of the base $\triangle ABC$ of the regular tetrahedron $P-ABC$. The dihedral angles between each pair of lateral edges are $\alpha$, and the angle between $PC$ and the plane $PAB$ is $\beta$. Denote the distance from point $O$ to each face as $d$. A moving plane through point $O$ intersects $PC... | 4. $\frac{\sin \beta}{d}$.
Notice,
$$
\begin{array}{l}
V_{\text {tetrahedron } S-P Q R}=\frac{1}{3} S_{\triangle P Q R} h \\
=\frac{1}{3}\left(\frac{1}{2} P Q \cdot P R \sin \alpha\right) P S \sin \beta .
\end{array}
$$
Then $V_{\text {tetrahedron } S-P Q R}$
$$
\begin{aligned}
= & V_{\text {tetrahedron } O-P Q R}+V_... | \frac{\sin \beta}{d} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,971 |
Example 1 As shown in Figure 2, points $D, E$ are on the sides $AB, AC$ of $\triangle ABC$ respectively, and $DE \parallel BC$. Connect $BE$ and $CD$ intersecting at point $F$. Construct the circumcircle of $\triangle BDF$ and the circumcircle of $\triangle CEF$, the two circles intersect at points $F, G$. Prove: $\ang... | 【Analysis】Let the distances from point $F$ to sides $AB$ and $AC$ be $f_{1}$ and $f_{2}$, respectively, and the distances from point $G$ to sides $AC$ and $AB$ be $g_{1}$ and $g_{2}$, respectively.
We need to prove: $\frac{f_{1}}{f_{2}}=\frac{g_{1}}{g_{2}}$.
Since the two circles intersect at points $F$ and $G$,
$$
\be... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,972 |
Example 2 As shown in Figure 3, circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $B$. A line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $C$ and $D$, respectively. Another line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $E$ and $F$, res... | 【Analysis】As shown in Figure 3, connect $C M$ and $N F$, and let their intersection be $I$.
We only need to prove: $C I \cdot I M = F I \cdot I N$.
Since the two circles intersect at points $A$ and $B$,
$$
\begin{array}{l}
\Rightarrow \angle A F E = \angle A D C, \angle A E F = \angle A C D \\
\Rightarrow \triangle A E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,973 |
Example 3 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{aligned}
a_{1}=3, & a_{2}=7, \\
& a_{n}^{2}+5=a_{n-1} a_{n+1}(n \geqslant 2) .
\end{aligned}
$$
Prove: If $a_{n}+(-1)^{n}$ is a prime number, then there must exist a non-negative integer $m$ such that $n=3^{m}$. | 【Analysis】In fact, to prove: if $n$ cannot be written in the form of $3^{m}$, then $a_{n}+(-1)^{n}$ must be composite.
Analyzing the recurrence equation $a_{n}^{2}+5=a_{n-1} a_{n+1}$.
Using the original equation and $a_{n-1}^{2}+5=a_{n-2} a_{n}$, we get
$\frac{a_{n+1}+a_{n-1}}{a_{n}}=\frac{a_{n}+a_{n-2}}{a_{n-1}}=k$ (c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,974 |
1. The sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ are defined as follows:
$$
\begin{array}{l}
x_{0}=1, x_{1}=1, x_{n+1}=x_{n}+2 x_{n-1} \\
y_{0}=1, y_{1}=7, y_{n+1}=2 y_{n}+3 y_{n-1} .
\end{array}
$$
Prove: The two sequences have only the common term 1. | Consider the remainders when two sequences are divided by 8:
$$
\left\{x_{n}\right\}: 1,1,3,5,3,5, \cdots
$$
$$
\left\{y_{n}\right\}: 1,7,1,7,1,7, \cdots
$$
It can be proven by induction.
Thus, the two sequences have no common terms other than 1. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,975 |
$$
\begin{array}{l}
\text { 3. Given } a_{1}=a_{2}=1, \\
a_{k+2}=a_{k+1}+a_{k}\left(k \in \mathbf{Z}_{+}\right) \text {. }
\end{array}
$$
Prove: For any positive integer $m$, there exists a $k$ such that
$$
m \mid\left(a_{k}^{4}-a_{k}-2\right)
$$ | If $m=1$, then the conclusion is obviously true.
When $m \geqslant 2$, suppose
$$
a_{i} \equiv b_{i}(\bmod m)\left(0 \leqslant b_{i}0\right)$.
Take $k=t p-2$, then
$$
a_{k}^{4}-a_{k}-2 \equiv 0(\bmod m) .
$$ | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,976 |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{0}=4, a_{n}=a_{n-1}^{2}-a_{n-1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Prove: there exist infinitely many primes, each of which is a divisor of at least one term in the sequence $\left\{a_{n}\right\}$;
(2) Does there exist infinitely many primes, none... | (1) From
$$
a_{n}=a_{n-1}^{2}-a_{n-1} \Rightarrow a_{n-1} \mid a_{n} \text {. }
$$
it follows that any prime that divides one term of the sequence $\left\{a_{n}\right\}$ can divide all terms of the sequence.
Also, $a_{n}=a_{n-1}\left(a_{n-1}-1\right)$, clearly, $a_{n-1}$ and $a_{n-1}-1$ are coprime.
Since $a_{n-1}-1>... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,977 |
Example 3 As shown in Figure 4, in the inscribed $\triangle A B C$, $\angle A$ is the largest angle, and the midpoints of arcs $\overparen{A B C}$ and $\overparen{A C B}$ are $D$ and $E$, respectively. Let the circle passing through points $A$ and $B$ and tangent to $A C$ be $\odot O_{1}$, and the circle passing throug... | 【Analysis】Notice that, the circumcircle of $\triangle A B C$ intersects $\odot O_{1}$ at points $A, B$.
Since $A C$ is the tangent to $\odot O_{1}$,
$\angle P A C=\angle P B A$.
Therefore, we only need to prove $\angle P B A=\angle P A B$, which means
$P A=P B$.
Notice that, $E A=E B$.
We only need to prove: $\triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,978 |
Example 1 In graph $G$ there are $e$ edges and $n$ vertices with degrees $d_{1}, d_{2}, \cdots, d_{n}$, and integer $k$ is less than $\min \left\{d_{1}, d_{2}, \cdots, d_{n}\right\}$. Prove: Graph $G$ contains an induced subgraph $H$ (i.e., if two vertices in $H$ are connected by an edge in graph $G$, then they are als... | 【Analysis】Similar to Example 9 in [1], the proof method for Turán's theorem uses a random permutation of vertices. To ensure no complete graph $K_{k+1}$ appears, it is sufficient to require that the degree of each vertex in graph $H$ does not exceed $k-1$. For a random permutation of vertices $v_{1}, v_{2}, \cdots, v_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,979 |
Example 2 Given that there are $2^{500}$ points on a circle, numbered in some order as $1,2, \cdots, 2^{500}$. Prove: It is possible to select 100 pairwise non-intersecting chords connecting these points, such that the sum of the numbers of the endpoints of all these chords is equal. ${ }^{\text {[2] }}$
(53rd IMO Shor... | 【Analysis】Let $n=2^{499}$. Then there are $\mathrm{C}_{2 n}^{n}$ chords, and the value of each chord belongs to $\{3,4, \cdots, 4 n-1\}$.
Notice that the sum of the labels of any two chords with a common endpoint must be different.
Let $G_{c}$ be the graph formed by chords whose labels sum to $c$, with each chord bei... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,980 |
Example 3 On a plane, there is a convex $3 n$-sided polygon, with a robot on each of its vertices. Each robot emits a laser beam pointing to another robot. Define an operation as: select a robot, and rotate it clockwise until its laser points to a new robot. For three robots $A$, $B$, and $C$, when robot $A$'s laser po... | Solution: $\frac{9 n^{2}-7 n}{2}$.
First, for any two points $A$ and $B$, let the random variable $N_{A B}$ represent the number of times point $A$'s laser needs to be directed at point $B$. Assuming that points $A$ and $B$ are uniformly randomly selected from all points, then $N_{A B}$ is uniformly distributed over $\... | \frac{9 n^{2}-7 n}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,981 |
Example 4 There are 68 pairs of non-zero integers on the blackboard. For a positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ appears on the blackboard. A student erases some of these 136 numbers so that the sum of any two erased numbers is not 0. It is stipulated that if at least one number from a ... | Given that $(j, j)$ and $(-j, -j)$ can appear at most as one pair, we can assume that if $(j, j)$ appears, then $j > 0$ (otherwise, replace $j$ with $-j$). For a positive integer $k$, all $k$ or $-k$ can be deleted from the blackboard, but not both. For each $k > 0$, delete $k$ with probability $p$ and $-k$ with probab... | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,982 |
Example 5 Let $U$ be a set consisting of $m$ triangles. Prove: there exists a subset $W$ of $U$ satisfying:
(1) The subset $W$ contains at least $\frac{9}{20} m^{\frac{4}{5}}$ triangles;
(2) There do not exist six distinct points $A, B, C, D, E, F$ such that $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF, ... | Proof: Let $U^{\prime}$ be the set formed by selecting each triangle from set $U$ with probability $p$. Then the expected number of triangles in $U^{\prime}$ is $m p$.
For a sequence of six distinct points $\left(x_{1}, x_{2}, \cdots, x_{6}\right)$, if all six triangles $\triangle x_{1} x_{2} x_{3}, \triangle x_{2} x_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,983 |
Example 6 Consider a $2^{m} \times 2^{m}\left(m \in \mathbf{Z}_{+}\right)$ chessboard, which is divided into several rectangles composed of the squares of the chessboard, and each of the $2^{m}$ squares on one of the diagonals is a rectangle of side length 1. Find the minimum sum of the perimeters of the divided rectan... | Since $(i, i)$ is separated out, the main diagonal divides the chessboard into two symmetrical parts, and we only need to discuss the lower part.
Let $R_{i}$ denote the set of rectangles that occupy at least one square in the $i$-th row, and $C_{i}$ denote the set of rectangles that occupy at least one square in the $... | 2^{m+2}(m+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,984 |
In $\triangle ABC$, $M$ is the midpoint of $AB$, $P$ is a point inside $\triangle ABC$, $Q$ is the point symmetric to $P$ with respect to $M$, $AP$ intersects $BC$ at point $D$, and $BP$ intersects $AC$ at point $E$. Prove that $A, B, D, E$ are concyclic if and only if
$$
\angle ACP = \angle QCB.
$$
(46th Austrian Math... | Prove that if $\angle A C P=\angle Q C B$, as shown in Figure 1, let the extension of $C B$ intersect the extension of $A Q$ at point $N$.
$$
\text { From } \frac{S_{\triangle A C P}}{S_{\triangle D C P}}=\frac{A P}{P D} \Rightarrow \frac{A P}{P D} \cdot \frac{C D}{A C}=\frac{\sin \angle A C P}{\sin \angle D C P} \text... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,985 |
Question 2 Given $x, y, z \geqslant 0$. Prove:
$$
\frac{x-y}{x y+2 y+1}+\frac{y-z}{y z+2 z+1}+\frac{z-x}{z x+2 x+1} \geqslant 0 \text {. }
$$
$(2015$, Serbian Mathematical Olympiad) | Let “ $\sum$ ” denote the cyclic sum.
Notice,
$$
\begin{array}{l}
\sum \frac{x-y}{x y+2 y+1}-\sum \frac{x-y}{x y+x+y+1} \\
=\sum \frac{(x-y)^{2}}{(x y+2 y+1)(x y+x+y+1)} \\
\geqslant 0 .
\end{array}
$$
$$
\text { Hence } \begin{aligned}
& \frac{x-y}{x y+2 y+1} \\
\geqslant & \sum \frac{x-y}{x y+x+y+1} \\
= & \sum \frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,986 |
1. Given real numbers $a, b, c$ satisfy
$$
2 a+13 b+3 c=90,3 a+9 b+c=72 \text {. }
$$
Then $\frac{3 b+c}{a+2 b}=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -1 | $$
\begin{array}{l}
2(a+2 b)+3(3 b+c)=90, \\
3(a+2 b)+(3 b+c)=72. \\
\text { Then }\left\{\begin{array}{l}
a+2 b=18, \\
3 b+c=18
\end{array} \Rightarrow \frac{3 b+c}{a+2 b}=1\right. \text {. } \\
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,987 |
3. If positive integers $a, b, c$ satisfy
$$
a \leqslant b \leqslant c, a b c=2(a+b+c) \text {, }
$$
then $(a, b, c)$ is called a "good triplet". The number of good triplets is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4. | 3. C.
If $(a, b, c)$ is a good array, then $a b c=2(a+b+c) \leqslant 6 c \Rightarrow a b \leqslant 6$. Clearly, $a$ can only be 1 or 2.
If $a=2$, by $a b \leqslant 6$, we get $b=2$ or 3.
When $b=2$, $c=4$;
When $b=3$, $c=\frac{5}{2}$ (not an integer).
If $a=1$, then
$b c=2(1+b+c) \Rightarrow(b-2)(c-2)=6$
$\Rightarrow(... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,988 |
5. As shown in Figure 1, let $A$ be a point on the circle with diameter $BC$, $AD \perp BC$ at point $D$, point $E$ is on segment $DC$, and point $F$ is on the extension of $CB$, such that $\angle BAF = \angle CAE$. Given $BC = 15$, $BF = 6$, $BD = 3$, then $AE =$ ( ).
(A) $4 \sqrt{3}$
(B) $2 \sqrt{13}$
(C) $2 \sqrt{14... | 5. B.
Since $\angle B A F=\angle C A E$, we have $\angle B A F+\angle B A E=\angle C A E+\angle B A E$, which means $\angle F A E=\angle B A C=90^{\circ}$.
Also, $A D \perp B C$, so $A D^{2}=D E \cdot D F=D B \cdot D C$.
It is easy to see that $D F=B F+B D=9$, $D C=B C-B D=12$.
Thus, $A D^{2}=D E \cdot 9=3 \times 12$
$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,989 |
2. As shown in Figure 2, in parallelogram $A B C D$, $\angle A B C=72^{\circ}, A F \perp B C$ at point $F, A F$ intersects $B D$ at point $E$. If $D E=2 A B$, then $\angle A E D=$ $\qquad$ . | 2. $66^{\circ}$.
As shown in Figure 5, take the midpoint $M$ of $DE$, and connect $AM$. | 66^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,990 |
4. If real numbers $x, y$ satisfy $x^{3}+y^{3}+3 x y=1$, then the minimum value of $x^{2}+y^{2}$ is . $\qquad$ | 4. $\frac{1}{2}$.
Notice that,
$$
\begin{aligned}
0= & x^{3}+y^{3}+3 x y-1 \\
= & (x+y)^{3}+(-1)^{3}-3 x^{2} y-3 x y^{2}+3 x y \\
= & (x+y-1)\left((x+y)^{2}-\right. \\
& \left.(x+y)(-1)+(-1)^{2}\right)-3 x y(x+y-1) \\
= & (x+y-1)\left(x^{2}+y^{2}-x y+x+y+1\right) \\
= & \frac{1}{2}(x+y-1)\left((x-y)^{2}+(x+1)^{2}+(y+1... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,991 |
2. Find all positive integers $n$ such that for any integer-coefficient polynomial $P(x)$ of degree not exceeding $n$ and with the leading coefficient 1, there exists a positive integer $k \leqslant n$ and $k+1$ distinct integers $x_{1}, x_{2}, \cdots, x_{k+1}$, such that
$$
P\left(x_{1}\right)+P\left(x_{2}\right)+\cdo... | 2. When $n=1$, let $P(x)=x$, then
$$
P\left(x_{1}\right)=P\left(x_{2}\right) \Rightarrow x_{1}=x_{2} \text {. }
$$
When $n=2$,
if $P(x)=1$, then let $x_{1}=1, x_{2}=2, k=1$.
if $P(x)=x+c$, then
when $c=0$, let $k=2, x_{1}=1, x_{2}=2, x_{3}=3$;
when $c \neq 0$, let $k=2, x_{1}=0, x_{2}=c, x_{3}=2 c$.
if $P(x)=x^{2}+p x... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,992 |
4. In the Cartesian coordinate system, assume $G_{1}$ and $G_{2}$ are the graphs of the quadratic functions
$$
\begin{array}{l}
f_{1}(x)=p_{1} x^{2}+q_{1} x+r_{1}, \\
f_{2}(x)=p_{2} x^{2}+q_{2} x+r_{2},
\end{array}
$$
where $p_{1}>0>p_{2}$. The parabolas $G_{1}$ and $G_{2}$ intersect at two distinct points $A$ and $B$... | 4. First, prove a lemma.
Lemma: For the quadratic function $y=p x^{2}+q x+r$ and its graph $G$, and two points $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$ on $G$, draw the tangents to the parabola $G$ at points $A$ and $B$ intersecting at point $C$. Then
$$
C A - C B
$$
$$
=\frac{2\left|x_{A}-x_{B}\right|... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,993 |
5. Given an integer $n \geqslant 2$. From an $n \times n$ grid
remove $n$ cells located in different rows and different columns, the resulting shape is called an $n \times n$ "sieve". $1 \times k (k \in$ $\mathbf{Z}_{+}$) or $k \times 1$ sub-grids are both called a "bar". For any $n \times n$ sieve $A$, partition it i... | 5. $m(A)=2 n-2$.
First, prove: $m(A) \leqslant 2 n-2$.
Consider all possible maximal $1 \times k$ bars. For a row, if the removed cell is not in the first or last column, then two bars are needed for that row. Otherwise, one bar is needed. Therefore, a total of $2 n-2$ bars are needed. Thus, $m(A) \leqslant 2 n-2$.
Ne... | 2 n-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,994 |
Example 7 As shown in Figure 8, given $\triangle A B C$, $X$ and $Y$ are two points on line $B C$ ($X, B, C, Y$ are arranged in order) such that
$$
B X \cdot A C=C Y \cdot A B \text {. }
$$
Let the circumcenters of $\triangle A C X$ and $\triangle A B Y$ be $O_{1}$ and $O_{2}$, respectively. The line $O_{1} O_{2}$ int... | 【Analysis】We just need to prove: $O_{1} O_{2}$ is perpendicular to the angle bisector of $\angle B A C$.
Auxiliary lines are shown in Figure 8.
Since $O_{1} O_{2}$ is the line connecting the centers, it is perpendicular to the common chord $A P$ of $\odot O_{1}$ and $\odot O_{2}$, so we just need to prove: $A P$ bisect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,995 |
2. The function
$$
f(x)=4 \sin ^{3} x-\sin x+2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}
$$
has the smallest positive period of ( ).
(A) $2 \pi$
(B) $\frac{\pi}{2}$
(C) $\frac{2 \pi}{3}$
(D) $\pi$ | 2. C.
Simplifying, we get $f(x)=-\sin 3 x+2$. Then, the smallest positive period of the function $f(x)$ is $\frac{2 \pi}{3}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,996 |
5. Given $a, b \in \mathbf{R}$, the function $f(x) = ax - b$. If for any $x \in [-1, 1]$, we have $0 \leqslant f(x) \leqslant 1$, then the range of $\frac{3a + b + 1}{a + 2b - 2}$ is $(\quad)$.
(A) $\left[-\frac{1}{2}, 0\right]$
(B) $\left[-\frac{4}{5}, 0\right]$
(C) $\left[-\frac{1}{2}, \frac{2}{7}\right]$
(D) $\left[... | 5. D.
From the problem, we have
$$
\left\{\begin{array} { l }
{ 0 \leqslant f ( 1 ) \leqslant 1 , } \\
{ 0 \leqslant f ( - 1 ) \leqslant 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
0 \leqslant a-b \leqslant 1, \\
-1 \leqslant a+b \leqslant 0 .
\end{array}\right.\right.
$$
Let $u=a+b, v=a-b$.
Then $a=\fra... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,997 |
6. Given that vector $O A \perp O B$, and $|O A|=|O B|=$
24. If $t \in[0,1]$, then
$$
|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|
$$
the minimum value is ( ).
(A) $2 \sqrt{193}$
(B) 26
(C) $24 \sqrt{2}$
(D) 24 | 6. B.
Construct a square $O A C B$, connect the diagonal $A B$, and let $D$ and $E$ be points on the diagonal $A B$ and the side $O B$ respectively, such that
$$
\begin{array}{l}
t \overrightarrow{A B}-\overrightarrow{A O}=\overrightarrow{O D}, \\
\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}=\overright... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,998 |
7. Let the set
$$
M=\left\{(x, y) \left\lvert\, \frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{45}}\right., x, y \in \mathbf{Z}_{+}\right\} .
$$
Then the number of elements in the set $M$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 7. B.
Notice,
$$
\begin{array}{l}
\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{45}} \Leftrightarrow \frac{1}{\sqrt{5 x}}-\frac{1}{\sqrt{5 y}}=\frac{1}{15} \\
\Rightarrow \frac{1}{5 x}=\frac{1}{225}+\frac{1}{5 y}+\frac{2}{15 \sqrt{5 y}} \\
\Rightarrow \sqrt{5 y} \in \mathbf{Q} .
\end{array}
$$
Similarly, $\sqr... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,999 |
10. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=-1, b_{1}=2, \\
a_{n+1}=-b_{n}, b_{n+1}=2 a_{n}-3 b_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
Then $b_{2015}+b_{2016}=$ | 10. $-3 \times 2^{2015}$.
From the given $\left\{\begin{array}{l}a_{i+1}=-b_{i}, \\ b_{i+2}=2 a_{i+1}-3 b_{i+1}\end{array}\right.$ we get
$$
\begin{array}{l}
b_{i+2}+b_{i+1}=-2\left(b_{i+1}+b_{i}\right) \\
=(-2)^{2}\left(b_{i}+b_{i-1}\right)=\cdots \\
=(-2)^{i}\left(b_{2}+b_{1}\right) .
\end{array}
$$
Therefore, $b_{... | -3 \times 2^{2015} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,000 |
Example 6 As shown in Figure 7, the circumcenter of $\triangle A B C$ is $O$, and the incenter is $I, \angle C=30^{\circ}$. Points $D$ and $E$ are on sides $A C$ and $B C$, respectively, such that $A D=B E=A B$. Prove: $O I \perp D E$, and $O I=D E$. | 【Analysis】Let the circumcircle of $\triangle CDE$ intersect $\odot O$ at points $C, P$, and draw auxiliary lines as shown in Figure 7.
Since $\angle PAC = \angle PBC$,
$\angle PDC = \angle PEC$, and $AD = BE$,
we have $\triangle PAD \cong \triangle PBE$
$\Rightarrow PA = PB, PD = PE$.
Take the midpoint $M$ of arc $\ove... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,001 |
11. Let $a \in \mathbf{R}$, the equation ||$x-a|-a|=2$ has exactly three distinct roots. Then $a=$ $\qquad$ | 11. 2 .
The original equation can be transformed into $|x-a|=a \pm 2$.
For the equation to have exactly three distinct roots, then $a=2$. At this point, the equation has exactly three distinct roots:
$$
x_{1}=2, x_{2}=6, x_{3}=-2 \text {. }
$$
Thus, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,002 |
12. Given two regular tetrahedrons $A O B C$ and square $D O B C$, $M$ and $N$ are the centroids of $\triangle A D C$ and $\triangle B D C$ respectively. Let $\overrightarrow{O A}=a, \overrightarrow{O B}=\boldsymbol{b}, \overrightarrow{O C}=c$. If point $P$ satisfies
$$
\overrightarrow{O P}=x a+y b+z c, \overrightarrow... | 12. $x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9}$.
Let point $A$ project onto plane $OBC$ at point $H$.
$$
\begin{array}{l}
\text { Then } \overrightarrow{O H}=\frac{2}{3} \times \frac{1}{2}(\overrightarrow{O B}+\overrightarrow{O C})=\frac{1}{3}(b+c) \\
\Rightarrow \overrightarrow{A H}=\overrightarrow{O H}-\overright... | x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,003 |
13. In $\triangle A B C$, $\angle B=\frac{\pi}{4}, \angle C=\frac{5 \pi}{12}, A C$ $=2 \sqrt{6}, A C$'s midpoint is $D$. If a line segment $P Q$ of length 3 (point $P$ to the left of point $Q$) slides on line $B C$, then the minimum value of $A P+D Q$ is . $\qquad$ | 13. $\frac{\sqrt{30}+3 \sqrt{10}}{2}$.
Given that $B C=6$.
Draw a line $D E / / B C$, intersecting $A B$ at point $E$. Then $D E=\frac{1}{2} B C=3$.
Thus, quadrilateral $P Q D E$ is a parallelogram, i.e., $D Q=E P$.
Therefore, the problem is reduced to: finding a point on line $B C$ such that $A P+E P$ is minimized.
... | \frac{\sqrt{30}+3 \sqrt{10}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,004 |
14. If the system of equations concerning $x$ and $y$
$$
\left\{\begin{array}{l}
\sin x=m \sin ^{3} y, \\
\cos x=m \cos ^{3} y
\end{array}\right.
$$
has real solutions, then the range of positive real number $m$ is
$\qquad$ | 14. $[1,2]$.
Squaring both equations and eliminating $x$, we get
$$
\begin{array}{l}
\sin ^{2} 2 y=\frac{4}{3}\left(1-m^{2}\right) \in[0,1] \\
\Rightarrow 1 \leqslant m^{2} \leqslant 2 .
\end{array}
$$
Conversely, when $1 \leqslant m^{2} \leqslant 2$, there exists $\left(x_{0}, y_{0}\right)$ that satisfies the equati... | [1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,005 |
15. Given that $a$, $b$, and $c$ are distinct integers. Then
$$
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}
$$
the minimum value is $\qquad$ | 15.8.
Notice,
$$
\begin{array}{l}
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2},
\end{array}
$$
its minimum value is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,006 |
Example 8 As shown in Figure $9, D, E$ are the midpoints of sides $C A, A B$ of $\triangle A B C$, respectively. The common chord of the circumcircles of $\triangle A B D$ and $\triangle A C E$ is $A T$, and $B D, C E$ intersect $A T$ at points $P, Q$, respectively. Prove: $\angle A C P=\angle A B Q$. | 【Analysis】As shown in Figure 9, extend $CP$ and $BQ$, intersecting sides $AB$ and $AC$ at points $U$ and $V$ respectively.
It suffices to prove: $\triangle ACU \sim \triangle ABV$.
Let $S$ be the intersection of $AT$ and $BC$.
Since $BD$, $AT$, and $CU$ are concurrent and $D$ is the midpoint of $AC$, we have
$$
\frac{A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,008 |
2. If three distinct real numbers $a$, $b$, and $c$ satisfy
$$
a^{3}+b^{3}+c^{3}=3 a b c \text {, }
$$
then $a+b+c=$ $\qquad$ | 2.0.
Notice,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=\frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \\
=0 .
\end{array}
$$
Since $a, b, c$ are not all equal, we have
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \neq 0 \text {. }
$$
Therefore, $a+b+c=0... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,009 |
3. Given that $a$ and $b$ are real numbers. If the quadratic function
$$
f(x)=x^{2}+a x+b
$$
satisfies $f(f(0))=f(f(1))=0$, and $f(0) \neq f(1)$, then the value of $f(2)$ is $\qquad$. | 3. 3 .
It is known that $f(0)=b, f(1)=1+a+b$ are both roots of the equation $f(x)=0$.
$$
\begin{array}{l}
\text { Then } x^{2}+a x+b \equiv(x-b)(x-(1+a+b)) \\
\Rightarrow a=-1-a-2 b, b=b(1+a+b) \\
\Rightarrow a=-\frac{1}{2}, b=0 \\
\Rightarrow f(2)=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,010 |
4. If Xiaozhang's sleep time each day is randomly and uniformly distributed between 6 to 9 hours, then the probability that Xiaozhang's average sleep time over two consecutive days is no less than 7 hours is $\qquad$ . | 4. $\frac{7}{9}$.
Let the hours of sleep that Xiao Zhang gets on two consecutive days be $x$ and $y$ $(x, y \in [6,9])$.
The possible events are denoted as $(x, y)$, corresponding to the points on the sides and within the square $ABCD$ in the coordinate plane.
Xiao Zhang's average sleep time over two consecutive day... | \frac{7}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,011 |
5. Given the function
$$
f(x)=\log _{a}\left(a x^{2}-x+\frac{1}{2}\right)
$$
is always positive in the interval $[1,2]$. Then the range of the real number $a$ is . $\qquad$ | 5. $\left(\frac{1}{2}, \frac{5}{8}\right) \cup\left(\frac{3}{2},+\infty\right)$.
When $a>1$, the function is monotonically increasing in the interval $[1,2]$, so
$$
f(1)>0 \Rightarrow a>\frac{3}{2} \text {. }
$$
When $\frac{1}{2}<a \leqslant 1$, the function is monotonically decreasing in the interval $[1,2]$, so
$$
... | \left(\frac{1}{2}, \frac{5}{8}\right) \cup\left(\frac{3}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,012 |
7. Let the edge length of a regular tetrahedron be $2 \sqrt{6}$, and let a sphere be constructed with the center $O$ of the tetrahedron as its center. The total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron is $4 \pi$. Then the radius of the sphere $O$ is... | 7. $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$.
Let the radius of sphere $O$ be $R$.
If a face of the regular tetrahedron intersects the sphere as shown in Figure 3, then the circumference of the small circle is $\pi$, and the radius of the small circle is $\frac{1}{2}$.
Also, the distance from the center of the sphere to the... | \frac{\sqrt{5}}{2} \text{ or } \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,013 |
Four. (20 points) Insert the sum of each pair of adjacent terms between them in a sequence, forming a new sequence. Such an operation is called a “$Z$ expansion” of the sequence. Given the sequence $1,2,3$, the first $Z$ expansion results in the sequence $1,3,2,5,3$; the second $Z$ expansion results in the sequence $1,... | Let $x_{0}=1, x_{m+1}=3, a_{n}=\sum_{i=0}^{m+1} x_{i}$.
$$
\begin{array}{l}
\text { Then } a_{n+1}=\sum_{i=0}^{m+1} x_{i}+\sum_{i=0}^{m}\left(x_{i}+x_{i+1}\right) \\
=3 \sum_{i=0}^{m+1} x_{i}-x_{0}-x_{m+1} \\
=3 a_{n}-4 .
\end{array}
$$
Given $a_{1}=1+3+2+5+3=14$, we have
$$
a_{2}=3 a_{1}-4=38, a_{3}=3 a_{2}-4=110 \te... | a_{n}=4 \times 3^{n}+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,014 |
Three. (50 points) A function calculator has a display screen and two operation keys. If the first operation key is pressed once, the number on the display screen will change to $\left[\frac{x}{2}\right]$ (where $[x]$ represents the greatest integer not exceeding the real number $x$); if the second operation key is pre... | Three, (1) Impossible.
Convert the number to binary. Then pressing the first operation key means removing the last digit of the number on the display; pressing the second operation key means appending 01 to the number on the display.
When the initial number is 1, after performing the above two operations, the resultin... | 233 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,015 |
2. Given unit vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $a \perp b$, vector $\boldsymbol{c}$
satisfies
$$
|c-a|+|c-2 b|=\sqrt{6} .
$$
Then the range of $|\boldsymbol{c}-\boldsymbol{a}|$ is $\qquad$ . | 2. $\left[\frac{\sqrt{6}-\sqrt{5}}{2}, \frac{\sqrt{6}+\sqrt{5}}{2}\right]$.
Let $O$ be the origin, and denote
$$
\begin{aligned}
a & =\overrightarrow{O A}=(1,0), 2 b=\overrightarrow{O B}=(0,2), \\
c & =\overrightarrow{O C} .
\end{aligned}
$$
From $|\boldsymbol{c}-\boldsymbol{a}|+|\boldsymbol{c}-2 \boldsymbol{b}|=\sqr... | \left[\frac{\sqrt{6}-\sqrt{5}}{2}, \frac{\sqrt{6}+\sqrt{5}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,016 |
3. Given a tetrahedron $P-ABC$ with edge lengths $PA=1, PB=2, PC=3$, and $PA \perp PB, PB \perp PC, PC \perp PA$. Then the maximum distance from a moving point $Q$ on the surface of the circumscribed sphere of this tetrahedron to the plane $ABC$ is $\qquad$ | 3. $\frac{3}{7}+\frac{\sqrt{14}}{2}$.
Notice that, the circumsphere of the tetrahedron $P-ABC$ is the same as the circumsphere of the rectangular parallelepiped with $PA$, $PB$, and $PC$ as its length, width, and height, respectively. The diameter of this sphere is $2R=\sqrt{14}$.
Also, $\cos \angle BAC=\frac{\sqrt{2... | \frac{3}{7}+\frac{\sqrt{14}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,017 |
Example 2 Define the sequence $\left\{a_{n}\right\}$ :
$$
a_{1}=2, a_{n+1}=2 a_{n}^{2}-1 \text {. }
$$
Prove: For all $n$, $(n, a_{n})=1$. | 【Analysis】For $p \mid n, p$ being a prime, consider $a_{1}, a_{2}, \cdots, a_{p-1}$.
If some of the $a_{i} \equiv 0,1(\bmod p)$, then
$a_{i+1}=2 a_{i}^{2}-1 \equiv \pm 1(\bmod p)$,
$a_{i+2} \equiv 1(\bmod p)$.
It can be deduced that $a_{p} \equiv \pm 1(\bmod p)$.
Thus, $p$ † $a_{p}$.
Also, $a_{k p} \equiv 1(\bmod p)$, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,018 |
7. From the non-empty subsets of the set $\{1,2, \cdots, 2017\}$, the probability that the sum of its elements is exactly odd is $\qquad$ | 7. $\frac{2^{2016}}{2^{2017}-1}$.
The set $\{1,2, \cdots, 2017\}$ has a total of $2^{2017}-1$ non-empty subsets, and the number of subsets with an odd sum of elements is exactly the sum of the coefficients of the odd powers of $x$ in the expansion of the function
$$
f(x)=(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{2017... | \frac{2^{2016}}{2^{2017}-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,019 |
9. (16 points) Prove: The function
$$
f(x)=x^{x+2}-(x+2)^{x}-2 x(x+1)(x+2)+2
$$
has only one integer zero in the interval $[0,+\infty)$ | 9. In fact, it is only necessary to prove that there is only one positive integer satisfying $f(x)=0$.
When $x \in \mathbf{Z}_{+}, x \geqslant 5$,
$$
\begin{array}{l}
(x+3)^{x}=\left(\frac{x+3}{2}\right)^{4}(x+3)^{x-4} 4^{2} \\
& (x+3)^{x}-(x-2)^{x}-2 x(x+1)(x+2)+2 \\
> & (x+3)(x+2)^{x-1}-(x+2)^{x}- \\
& 2 x(x+1)(x+2)... | 3 | Algebra | proof | Yes | Yes | cn_contest | false | 730,020 |
One, (40 points) Given that $A$, $B$, and $C$ are any three non-collinear points on a plane, and point $O$ is inside $\triangle ABC$, satisfying
$$
\angle AOB = \angle BOC = \angle COA = 120^{\circ} \text{.}
$$
Find the maximum value of $\frac{OA + OB + OC}{AB + BC + CA}$. | Given the figure 2, let the angle bisectors of $\angle A O B$, $\angle B O C$, and $\angle C O A$ be $O E$, $O F$, and $O G$ respectively, and the projections of point $O$ onto $A B$, $B C$, and $C A$ be $H$, $I$, and $J$ respectively.
Since $\angle A O B = \angle B O C = \angle C O A = 120^{\circ}$,
$S_{\triangle A O... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,021 |
II. (40 points) In $\triangle ABC$, prove: $\sin A + 2 \sin B + 3 \sin C < \frac{11}{2}$. | $$
\begin{array}{l}
\text { II. Let } x=\sin A+2 \sin B, \\
y=2 \sin B+3 \sin C, \\
z=3 \sin C+\sin A .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } x^{2} \leqslant(\sin A+2 \sin B)^{2}+(\cos A-2 \cos B)^{2} \\
=5-4 \cos (A+B)=5+4 \cos C, \\
y^{2} \leqslant(2 \sin B+3 \sin C)^{2}+(2 \cos B-3 \cos C)^{2} \\
=13-12 ... | \sin A + 2 \sin B + 3 \sin C < \frac{11}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 730,022 |
Three, (50 points) There are $12k$ people attending a meeting, each of whom has shaken hands with exactly $3k+6$ others, and for any two of them, the number of people who have shaken hands with both is the same. How many people attended the meeting? Prove your conclusion. | Three, abstracting a person as a point and two people shaking hands as a line connecting two points, then $A$ shaking hands with $B$ and $C$ corresponds to $\angle BAC$ in the graph. Since each person has shaken hands with $3k+6$ people, there are $12k \mathrm{C}_{3k+6}^{2}$ such angles in the graph. The number of pair... | 36 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,023 |
Example 1 Let $m, n \in \mathbf{Z}_{+}, m \geqslant n, S$ be the set of all ordered $n$-tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ determined by positive integers $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy $a_{1}+a_{2}+\cdots+a_{n}=m$. Prove:
$$
\begin{array}{l}
\sum_{S} 1^{a_{1}} \times 2^{a_{2}} \times \cdots ... | Proof: Let $m=k+n, T$ be the set of all ordered $n$-tuples $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ of non-negative integers $b_{1}, b_{2}, \cdots, b_{n}$ that satisfy
$$
b_{1}+b_{2}+\cdots+b_{n}=k
$$
Then we need to prove:
( $n$ !) $\sum_{T} 1^{b_{1}} \times 2^{b_{2}} \times \cdots \times n^{b_{n}}$
$$
=\sum_{j=1}^... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,024 |
Example 2 In an $m \times n$ grid, each cell is colored either red or white. It is known that for any $i, j$, among the $m+n-1$ cells in the $i$-th row and the $j$-th column, the number of cells of the same color as cell $(i, j)$ (the cell at the intersection of the $i$-th row and the $j$-th column) is less than the nu... | Prove that if we assume the number of red cells $L$ in the grid is not less than the number of white cells, then
$$
L \geqslant \frac{m n}{2} \text {. }
$$
Let the number of red cells in the $i$-th row be $x_{i}$, and the number of red cells in the $j$-th column be $y_{j}$.
If cell $(i, j)$ is red, then by the conditi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,025 |
In a convex quadrilateral $A B C D$, the diagonal $B D$ is neither the angle bisector of $\angle A B C$ nor the angle bisector of $\angle C D A$. Point $P$ is inside the quadrilateral $A B C D$ and satisfies
$$
\angle P B C=\angle A B D, \angle P D C=\angle B D A .
$$
Prove: The necessary and sufficient condition for ... | Proof As shown in Figure 2, take point $E$ on $DC$ such that $\angle PED = \angle ABD$, and connect $EB$.
It is easy to see that $\triangle PED \sim \triangle ABD$.
Then $\triangle PAD \sim \triangle EBD$
$$
\Rightarrow \frac{PA}{EB} = \frac{AD}{BD} = \frac{\sin \angle ABD}{\sin \angle BAD}.
$$
Thus, points $E$, $P$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,026 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.