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 |
|---|---|---|---|---|---|---|---|---|---|
2. Given circle $\Gamma$ is the circumcircle of $\triangle A B C$, $P$ is an interior point of $\triangle A B C$, and rays $A P, B P, C P$ intersect circle $\Gamma$ at points $A_{1}, B_{1}, C_{1}$, respectively. Let $A_{1}, B_{1}, C_{1}$ be symmetric to points $A_{2}, B_{2}, C_{2}$ with respect to the midpoints of side... | Let $O$ be the circumcenter of $\triangle ABC$ and establish a complex plane with $O$ as the origin. Let the midpoints of $AA_{1}, BB_{1}, CC_{1}$ be $A_{3}, B_{3}, C_{3}$, respectively.
Then $A_{2}=B+C-A_{1}, B_{2}=C+A-B_{1}$,
$$
C_{2}=A+B-C_{1}, H=A+B+C \text {. }
$$
Thus, $H, A_{2}, B_{2}, C_{2}$ are concyclic
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,346 |
4. Given $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$, where $x_{1}=1, y_{1}=0, x_{n+1}=$ $x_{n}-y_{n}, y_{n+1}=x_{n}+y_{n}\left(n \in \mathbf{Z}_{+}\right)$. If $a_{n}=$ $\overrightarrow{P_{n} P_{n+1}} \cdot \overrightarrow{P_{n+1} P_{n+2}}$, then t... | 4. 10 .
It is known that $\overrightarrow{O P_{n+1}}$ is obtained by rotating $\overrightarrow{O P_{n}}$ counterclockwise by $\frac{\pi}{4}$ and stretching it to $\sqrt{2}$ times its original length.
Thus, $\left|\overrightarrow{P_{n} P_{n+1}}\right|=O P_{n}$,
$\left|\overrightarrow{P_{n+1} P_{n+2}}\right|=\left|O P_{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,347 |
8. In $\triangle A B C$, $\angle C=\frac{\pi}{3}$, let $\angle B A C=\theta$.
If there exists a point $M$ on line segment $B C$ (different from points $B$ and $C$),
such that when $\triangle B A M$ is folded along line $A M$ to a certain position to get $\triangle B^{\prime} A M$, it satisfies $A B^{\prime} \perp C M$... | 8. $\left(\frac{\pi}{6}, \frac{2 \pi}{3}\right)$.
Let $A H \perp B C$ at point $H$. Then when $A B \perp C M$, $C M \perp$ plane $B A H$. Therefore, $\angle M A H$ is the angle between line $A M$ and plane $A B H$.
$$
\begin{array}{l}
\text { Hence } \cos \angle B A M=\cos \angle B A H \cdot \cos \angle H A M \\
\Righ... | \left(\frac{\pi}{6}, \frac{2 \pi}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,348 |
10. (20 points) Given the parabola $C: y=\frac{1}{2} x^{2}$ and the circle $D: x^{2}+\left(y-\frac{1}{2}\right)^{2}=r^{2}(r>0)$ have no common points, a tangent line is drawn from a point $A$ on the parabola $C$ to the circle $D$, with the points of tangency being $E$ and $F$. When point $A$ moves along the parabola $C... | 10. Substituting the parabola equation into the equation of circle $D$ yields
$$
2 y+\left(y-\frac{1}{2}\right)^{2}=r^{2} \Rightarrow y^{2}+y=r^{2}-\frac{1}{4} \text {. }
$$
From the fact that this equation has no positive real roots, we know
$$
r^{2}-\frac{1}{4}<0 \Rightarrow 0<r<\frac{1}{2} \text {. }
$$
Let $A\lef... | \left(0, \frac{\pi}{16}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,349 |
11. (20 points) Let $k(k \geqslant 2)$ positive integers $a_{1}, a_{2}$, $\cdots, a_{k}$ satisfy
$$
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k} \text {, and } \sum_{i=1}^{k} a_{i}=\prod_{i=1}^{k} a_{i} \text {. }
$$
Prove: $\sum_{i=1}^{k} a_{i} \leqslant 2 k$. | 11. Let $b_{i}=a_{i}-1$. Then
$$
\begin{array}{l}
k+\sum_{i=1}^{k} b_{i}=\sum_{i=1}^{k} a_{i}=\prod_{i=1}^{k} a_{i}=\prod_{i=1}^{k}\left(b_{i}+1\right) \\
\geqslant 1+\sum_{i=1}^{k} b_{i}+b_{k} \sum_{i=1}^{k-1} b_{i} \\
\Rightarrow k \geqslant 1+b_{k} \sum_{i=1}^{k-1} b_{i} .
\end{array}
$$
If $a_{k-1}=1$, then $a_{1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,350 |
One, (40 points) Find the smallest real number $\lambda$, such that there exists a sequence $\left\{a_{n}\right\}$ with all terms greater than 1, for which $\prod_{i=1}^{n+1} a_{i}<a_{n}^{\lambda}$ holds for any positive integer $n$. | Given $a_{n}>1$, so,
$$
\begin{array}{l}
\prod_{i=1}^{n+1} a_{i}0\right), S_{n}=\sum_{i=1}^{n} b_{i}\left(S_{n}>0\right) .
\end{array}
$$
For equation (1) to hold, then $\lambda>0$.
From equation (1) we get
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,351 |
3. A line intersects sides $AB$ and $AC$ of $\triangle ABC$ at points $M$ and $N$, respectively, and intersects line $BC$ at point $P$. Given that $X, Y, Z, T$ are the midpoints of $NM, MB, BC, CN$ respectively. Prove that the orthocenters of $\triangle AMN$, $\triangle AYT$, $\triangle PBM$, and $\triangle PXZ$ are co... | Given that the Miquel point of the complete quadrilateral $A M B C P N$ is $D$, i.e., $D$ is the intersection of the circumcircles of $\triangle A M N$, $\triangle A B C$, $\triangle P B M$, and $\triangle P C N$.
$$
\begin{array}{l}
\text{By } \angle M D N=\angle M A N=\angle B A C=\angle B D C, \\
\angle D M N=\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,352 |
Four, (50 points) Given a five-element set $A_{1}, A_{2}, \cdots, A_{10}$, any two of these ten sets have an intersection of at least two elements. Let $A=\bigcup_{i=1}^{10} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets among $A_{1}, A_{2}, \cdots, A_{10}$ that contain the... | Four, it is easy to get $\sum_{i=1}^{n} k_{i}=50$.
The $k_{i}$ sets containing $x_{i}$ form $\mathrm{C}_{k_{i}}^{2}$ set pairs, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all set pairs, which contain repetitions.
From the fact that the intersection of any two sets among $A_{1}, A_{2}, \cdots, A_{10}$ has at least... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,353 |
2. Let the sets be
$$
\begin{array}{l}
A=\left\{x \left\lvert\, x+\frac{5-x}{x-2}=2 \sqrt{x+1}\right., x \in \mathbf{R}\right\}, \\
B=\{x \mid x>2, x \in \mathbf{R}\} .
\end{array}
$$
Then the elements of the set $A \cap B$ are | 2. $\frac{5+\sqrt{13}}{2}$.
Notice that, when $x>2$,
$$
x+\frac{5-x}{x-2}=x-2+\frac{x+1}{x-2} \geqslant 2 \sqrt{x+1} \text {. }
$$
By the equality, we have $x-2=\frac{x+1}{x-2}$.
Solving this gives $x=\frac{5+\sqrt{13}}{2}$. | \frac{5+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,354 |
3. In tetrahedron $ABCD$, $AB=CD=15$, $BD=AC=20$, $AD=BC=\sqrt{337}$. Then the angle between $AB$ and $CD$ is $\qquad$ . | 3. $\arccos \frac{7}{25}$.
As shown in Figure 3, the tetrahedron $A B C D$ is extended to form the rectangular prism $A E B F-H C G D$.
Let $B E=x, B F=y, B G=z$.
$$
\begin{array}{l}
\text { Then } x^{2}+y^{2}=225, y^{2}+z^{2}=400, z^{2}+x^{2}=337 \\
\Rightarrow x=9, y=12, z=16 .
\end{array}
$$
The angle between $A B... | \arccos \frac{7}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,355 |
7. The sum $\sum_{i=1}^{k} a_{m+i}$ is called the sum of $k$ consecutive terms of the sequence $a_{1}, a_{2}, \cdots, a_{n}$, where $m, k \in \mathbf{N}, k \geqslant 1, m+k \leqslant n$. The number of groups of consecutive terms in the sequence $1,2, \cdots, 100$ whose sum is a multiple of 11 is $\qquad$. | 7.801.
Let $S_{k}=\sum_{i=1}^{k} i=\frac{k(k+1)}{2}$.
Notice,
$$
\begin{array}{l}
S_{k+11}=\frac{(k+11)(k+12)}{2} \\
\equiv \frac{k(k+1)}{2}=S_{k}(\bmod 11),
\end{array}
$$
and the remainders of $S_{1}, S_{2}, \cdots, S_{11}$ modulo 11 are $1,3,6$, $10,4,10,6,3,1,0,0$.
Since $100=9 \times 11+1$, thus, among the rema... | 801 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,356 |
4. Let $R$ and $S$ be two distinct points on a circle $\Gamma$, and $RS$ is not a diameter. Let $l$ be the tangent line to $\Gamma$ at point $R$. A point $T$ in the plane satisfies that $S$ is the midpoint of segment $RT$. Let $J$ be a point on the minor arc $\overparen{RS}$ of $\Gamma$ such that the circumcircle $\Gam... | Prompt: Take the intersection point $P$ of the straight line $A K$ and $R S$ as the origin to establish the complex plane.
Let $A=a e_{1}, J=j e_{1}, K=k e_{1}, R=r e_{2}, S=s e_{2}$ $\left(a, j, k, r, s \in \mathbf{R}, e_{1}, e_{2} \in \mathbf{C},\left|e_{1}\right|=\left|e_{2}\right|=1\right)$. Denote $u=e_{1} \bar{e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,357 |
8. Let $F$ be the set of all sequences of the form $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$, where $A_{i}$ is a subset of $B=\{1,2, \cdots, 10\}$, and let $|A|$ denote the number of elements in the set $A$. Then $\sum_{F}\left|\bigcup_{i=1}^{n} A_{i}\right|=$ $\qquad$ . | 8. $10\left(2^{10 n}-2^{9 n}\right)$.
Just calculate the number of times elements appear in set $B$. If $x \notin \bigcup_{i=1}^{n} A_{i}$, then $x$ does not belong to each of $A_{1}, A_{2}, \cdots, A_{n}$; if $x \in \bigcup_{i=1}^{n} A_{i}$, then $x$ belongs to at least one of $A_{1}, A_{2}, \cdots, A_{n}$. Therefore... | 10\left(2^{10 n}-2^{9 n}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,358 |
10. (20 points) Let “ $\sum$ ” denote the cyclic sum. Given positive real numbers $a$, $b$, and $c$ such that $a^{2}+b^{2}+c^{2}=1$. Prove:
$$
\sum \frac{1}{a^{2}} \geqslant\left(\sum \frac{4 b c}{a^{2}+1}\right)^{2} \text {. }
$$ | 10. Note that,
$$
\begin{array}{l}
a^{2}+1=a^{2}+a^{2}+b^{2}+c^{2} \\
\geqslant 4 \sqrt[4]{a^{2} a^{2} b^{2} c^{2}}=4 a \sqrt{b c} .
\end{array}
$$
Then $16\left(\sum \frac{b c}{a^{2}+1}\right)^{2}$
$$
\begin{array}{l}
\leqslant\left(\sum \frac{b c}{a \sqrt{b c}}\right)^{2}=\left(\sum \frac{\sqrt{b c}}{a}\right)^{2} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,359 |
11. (20 points) Given $n=d_{1} d_{2} \cdots d_{2017}$, where,
$$
\begin{array}{l}
d_{i} \in\{1,3,5,7,9\}(i=1,2, \cdots, 2017), \text { and } \\
\sum_{i=1}^{1009} d_{i} d_{i+1} \equiv 1(\bmod 4), \\
\sum_{i=1010}^{2016} d_{i} d_{i+1} \equiv 1(\bmod 4) .
\end{array}
$$
Find the number of $n$ that satisfy the conditions... | 11. Let $d_{k+1}=d_{1}, d_{k}=d_{0}$.
Since $d_{i} \equiv 1$ or $-1(\bmod 4)$, when replacing -1 with 1, the difference in $\sum_{i=1}^{k} d_{i} d_{i+1}$ is
$$
\begin{array}{l}
\Delta d_{i} \cdot\left(d_{i-1}+d_{i+1}\right)(\bmod 4)(i=1,2, \cdots, k), \\
\Delta d_{i} \equiv \pm 2(\bmod 4), \\
d_{i-1}+d_{i+1} \equiv 0,... | 6 \times 5^{2015} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,360 |
II. (40 points) For positive real numbers $x, y$, define the operation “$\odot$” such that $x \odot y = \frac{xy + 4}{x + y}$, and for positive real numbers $x, y, z$, it satisfies $x \odot y \odot z = (x \odot y) \odot z$. When the integer $n \geq 4$, let $T = 3 \odot 4 \odot \ldots \odot n$. Is $\frac{96}{T - 2}$ a p... | $$
\begin{array}{l}
f(n)=f(n-1) \odot n=\frac{n f(n-1)+4}{n+f(n-1)}, \\
f(3)=3. \\
\text { Then } \frac{f(n)-2}{f(n)+2}=\frac{\frac{n f(n-1)+4}{n+f(n-1)}-2}{\frac{n f(n-1)+4}{n+f(n-1)}+2} \\
=\frac{n-2}{n+2} \cdot \frac{f(n-1)-2}{f(n-1)+2} \\
=\frac{n-2}{n+2} \cdot \frac{n-3}{n+1} \cdot \frac{f(n-2)-2}{f(n-2)+2}=\cdots... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,361 |
In $\triangle ABC$, let the excenter opposite to $\angle A$ be $I_{A}$, the circumradius, semiperimeter, and the three side lengths be $R$, $p$, $a$, $b$, $c$ respectively. Prove:
$$
\frac{I_{A} A^{2}}{b c}+\frac{I_{A} B^{2}}{c a}+\frac{I_{A} C^{2}}{a b}=\frac{p+a}{p-a} .
$$ | Prove the well-known formula for the exradius:
$$
r_{A}=4 R \sin \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2},
$$
and combining with the geometric definition of the exradius, we have
$$
\frac{r_{A}}{I_{A} A}=\sin \frac{A}{2}, \frac{r_{A}}{I_{A} B}=\sin \frac{\pi-B}{2}, \frac{r_{A}}{I_{A} C}=\sin \frac{\pi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,362 |
Example 2 For any real number sequence $\left\{x_{n}\right\}$, define the sequence $\left\{y_{n}\right\}:$
$$
y_{1}=x_{1}, y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the smallest positive number $\lambda$, such that for any real number sequen... | 【Analysis】First estimate the upper bound of $\lambda$ from the limit perspective, then try to construct a recurrence relation to solve it.
First, prove that $\lambda \geqslant 2$.
In fact, start from simple and special cases.
Take $x_{1}=1, x_{n}=\sqrt{2^{n-2}}(n \geqslant 2)$. Then $y_{1}=1, y_{n}=0(n \geqslant 2)$.
S... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,363 |
16. As shown in Figure 2, two semicircles with centers at $A$ and $B$ and radii of 2 and 1, respectively, are internally tangent to the semicircle with diameter $JK$, and the two smaller semicircles are also tangent to each other. $\odot P$ is externally tangent to both smaller semicircles and internally tangent to the... | 16. B.
Let the center of circle $\Gamma$ be $C$, and connect $P A, P C, P B$. In $\triangle A B P$, by Stewart's Theorem, we have
$$
\begin{array}{c}
A B \cdot A C \cdot B C+A B \cdot C P^{2} \\
=A C \cdot B P^{2}+B C \cdot A P^{2} .
\end{array}
$$
From the given tangency relations, we get
$$
\begin{array}{l}
A B=3, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,365 |
20. Given that $a$ is a positive real number, $b$ is a positive integer and $2 \leqslant b \leqslant 200$, then the number of pairs $(a, b)$ that satisfy $\left(\log _{b} a\right)^{2017}=\log _{b} a^{2017}$ is ( ).
(A) 198
(B) 199
(C) 398
(D) 399
(E) 597 | 20. E.
Let $x=\log _{b} a$. Then
$$
x^{2017}=2017 x \text {. }
$$
If $x \neq 0$, then
$$
x^{2016}=2017 \Rightarrow x= \pm 2017^{\frac{1}{2016}} \text {. }
$$
Thus, equation (1) has exactly three real roots
$$
x=0,2017^{\frac{1}{2016}},-2017^{\frac{1}{2016}} \text {. }
$$
By $\log _{b} a=x$, i.e., $a=b^{x}$, for eac... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,366 |
25. Given that the six vertices of a centrally symmetric hexagon in the complex plane form the complex number set
$$
V=\left\{ \pm \sqrt{2} \mathrm{i}, \frac{ \pm 1+\mathrm{i}}{\sqrt{8}}, \frac{ \pm 1-\mathrm{i}}{\sqrt{8}}\right\} .
$$
For each $j(1 \leqslant j \leqslant 12), z_{j}$ is an element randomly selected fro... | 25. E.
Replace $z_{1}$ with $-z_{1}$, then $P=-1$ becomes $P=1$, and this mapping is a one-to-one correspondence. Therefore, calculate the number of ways for $P= \pm 1$, and then divide by 2.
Multiplying each vertex of the hexagon by $\mathrm{i}$ does not change the value of $P$. Replacing $z_{j}$ with $-z_{j}$ does ... | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,367 |
Conclusion 1 In the prime factorization of $n!$, the power of 2 is $n-s(n)$. | Conclusion 1 Proof Let
$$
n=\sum_{i=1}^{s} 2^{n_{i}}\left(n_{1}>n_{2}>\cdots>n_{s} \geqslant 0, s(n)=s\right) \text {. }
$$
Let $[x]$ denote the greatest integer not exceeding the real number $x$.
Then the power of 2 in $n!$ is
$$
\begin{array}{l}
\sum_{k=1}^{\infty}\left[\frac{n}{2^{k}}\right]=\sum_{i=1}^{s} \sum_{j=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,368 |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Calculate: $\sum_{k=0}^{2019}\left[\frac{4^{k}}{5}\right]=$ $\qquad$ | 4. $\frac{4^{2020}-1}{15}-1010$.
Notice, $\sum_{k=0}^{2019} \frac{4^{k}}{5}=\frac{4^{2020}-1}{15}$.
$$
\begin{array}{l}
\text { By } \frac{4^{k}}{5}+\frac{4^{k+1}}{5}=4^{k} \Rightarrow\left[\frac{4^{k}}{5}\right]+\left[\frac{4^{k+1}}{5}\right]=4^{k}-1 . \\
\text { Therefore } \sum_{k=0}^{2019}\left[\frac{4^{k}}{5}\rig... | \frac{4^{2020}-1}{15}-1010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,369 |
5. Given real numbers $x, y$ satisfy $x+y=1$. Then, the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$ is | 5.4.
$$
\begin{array}{l}
\text { Given }\left(x^{3}+1\right)\left(y^{3}+1\right) \\
=(x y)^{3}+x^{3}+y^{3}+1 \\
=(x y)^{3}-3 x y+2,
\end{array}
$$
let $t=x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, then
$$
f(t)=t^{3}-3 t+2 \text {. }
$$
Also, by $f^{\prime}(t)=3 t^{2}-3$, we know that $y=f(t)$ is monoto... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,370 |
6. Let $x_{k} 、 y_{k} \geqslant 0(k=1,2,3)$. Calculate:
$$
\begin{array}{l}
\sqrt{\left(2018-y_{1}-y_{2}-y_{3}\right)^{2}+x_{3}^{2}}+\sqrt{y_{3}^{2}+x_{2}^{2}}+ \\
\sqrt{y_{2}^{2}+x_{1}^{2}}+\sqrt{y_{1}^{2}+\left(x_{1}+x_{2}+x_{3}\right)^{2}}
\end{array}
$$
the minimum value is | 6. 2018.
Let $O(0,0), A(0,2018)$,
$$
\begin{array}{l}
P_{1}\left(x_{1}+x_{2}+x_{3}, y_{1}\right), P_{2}\left(x_{2}+x_{3}, y_{1}+y_{2}\right), \\
P_{3}\left(x_{3}, y_{1}+y_{2}+y_{3}\right) .
\end{array}
$$
The required is
$$
\begin{array}{l}
\left|\overrightarrow{A P_{3}}\right|+\left|\overrightarrow{P_{3} P_{2}}\righ... | 2018 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,371 |
8. Given $x, y \in \mathbf{R}$, for any $n \in \mathbf{Z}_{+}$, $n x+\frac{1}{n} y \geqslant 1$. Then the minimum value of $41 x+2 y$ is $\qquad$ | 8.9.
Let the line $l_{n}: n x+\frac{1}{n} y=1$, and call $l_{n} 、 l_{n+1}$ two adjacent lines. Then the intersection point of the two lines is
$$
A_{n}\left(\frac{1}{2 n+1}, \frac{n^{2}+n}{2 n+1}\right) \text {. }
$$
If the intersection point of the line $x+y=1$ and the line $y=0$ is denoted as $A_{0}(1,0)$, then the... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,372 |
9. (16 points) Given points $A(1,0), B(0,-2)$, and $O$ as the origin, a moving point $C$ satisfies:
$$
\overrightarrow{O C}=\alpha \overrightarrow{O A}+\beta \overrightarrow{O B}(\alpha, \beta \in \mathbf{R}, \alpha-2 \beta=1) \text {. }
$$
Let the trajectory of point $C$ intersect the hyperbola $\frac{x^{2}}{a^{2}}-\... | Let point $C(x, y)$. Then $(x, y)=(\alpha, -2 \beta)$.
Since $\alpha - 2 \beta = 1$, we have $x + y = 1$, which means the trajectory equation of point $C$ is $x + y = 1$.
By combining the line equation with the hyperbola equation and eliminating $y$, we get
$$
\left(b^{2} - a^{2}\right) x^{2} + 2 a^{2} x - a^{2} \left(... | a \in \left(0, \frac{1}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,373 |
Conclusion 2 Given $b(b \geqslant 2)$ as a positive integer. If a $b$-ary number is a multiple of $b^{n}-1$, then the sum of the digits of this $b$-ary number is at least $(b-1) n$. | Proof of Conclusion 2: Among all numbers divisible by $b^{n}-1$, let the smallest number with the smallest digit sum be
$$
A=\sum_{i=1}^{s} a_{i} b^{n_{i}}\left(n_{1}>n_{2}>\cdots>n_{s} \geqslant 0,0 \leqslant a_{i} \leqslant b-1\right) \text {. }
$$
First, $n_{i}$ and $n_{j}$ have different remainders modulo $n$. Oth... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,374 |
Four, (50 points) "Sairen Chess" is a game played on a $5 \times 5$ board, with the following rules:
(i) Players take turns placing pieces, with each player placing one piece per round, and the order of placing pieces remains unchanged;
(ii) All pieces are identical, and pieces must be placed in a cell, with only one p... | (1) Player A must win.
In fact, Player A has the following winning strategy:
(a) The first move is to place a piece in the center of the chessboard.
(b) After each move by Player B, unless a row or column already has four pieces, Player A places a piece in the cell that is centrally symmetric to the cell where Player B... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 730,375 |
Given $a, b, c, d$ are integers, $m$ is an odd number, and $m \mid (a+b+c+d), m \mid \left(a^{2}+b^{2}+c^{2}+d\right)^{2}$. Prove: $m \mid \left(a^{4}+b^{4}+c^{4}+d^{4}+4 a b c d\right)$. | Prove that by the property of cyclic symmetry, we can set
$$
\begin{array}{l}
K_{1}=\sum a, K_{2}=\sum a b, \\
K_{3}=\sum a b c, K_{4}=a b c d,
\end{array}
$$
where, “ $\sum$ ” denotes the cyclic symmetric sum.
Let $M=a^{4}+b^{4}+c^{4}+d^{4}$
$$
=\alpha_{1} K_{1}^{4}+\alpha_{2} K_{1}^{2} K_{2}+\alpha_{3} K_{1} K_{3}+\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,376 |
Example 2 Try to find all positive integers that the sum of the digits of a perfect square can take.
保留源文本的换行和格式，翻译结果如下：
Example 2 Try to find all positive integers that the sum of the digits of a perfect square can take.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Example 2 Try to find all positive integers that the sum of ... | First, the remainder of a square number divided by 9 is $0, 1, 4, 7$, so the sum of the digits of a square number divided by 9 leaves a remainder of $0, 1, 4, 7$.
Second, since
$$
\underbrace{9 \cdots 9^{2}}_{k \uparrow}=\underbrace{9 \cdots}_{k-1 \uparrow} 98 \underbrace{0 \cdots 01}_{k-1 \uparrow}
$$
the sum of the... | any positive integer that leaves a remainder of 0, 1, 4, 7 when divided by 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,378 |
Example 2 Let $p$ be an odd prime greater than 3, $q=\frac{p-1}{2}$. Prove: $\sum_{i=0}^{q} \mathrm{C}_{2 i}^{i} \equiv 1(\bmod p)$ or $\sum_{i=0}^{q} \mathrm{C}_{2 i}^{i} \equiv-1(\bmod p)$. | 【Analysis】When $p=5$,
$$
\sum_{i=0}^{2} \mathrm{C}_{2 i}^{i}=1+2+6=9 \equiv-1(\bmod 5) \text {; }
$$
When $p=7$,
$$
\sum_{i=0}^{3} \mathrm{C}_{2 i}^{i}=9+20=29 \equiv 1(\bmod 7) \text {; }
$$
When $p=11$,
$$
\sum_{i=0}^{5} \mathrm{C}_{2 i}^{i}=29+70+252=351 \equiv-1(\bmod 11) \text {; }
$$
When $p=13$,
$$
\sum_{i=0}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,379 |
Example 3 Let $n$ be an integer greater than 1. There are $2 n$ points on the plane, and no three points are collinear. Among these points, $n$ points are colored blue, and the remaining $n$ points are colored red. If a line passing through one red point and one blue point satisfies that the number of blue points on ea... | 【Analysis】First, prove that each vertex on the convex hull of these $n$ points lies on a balance line.
Assume $R$ is a vertex on the convex hull of the known $2 n$ points. Without loss of generality, let $R$ be a red point. Thus, there exists a line $l$ such that all points (excluding $R$) are on the same side of the ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,380 |
Let $p$ be an odd prime greater than 3. Find
(1) $\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}(\bmod p)$;
(2) $\sum_{i=0}^{2 p} \mathrm{C}_{2 i}^{i}(\bmod p)$. | 【Analysis】(1) Note that,
$$
\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}=\sum_{i=0}^{\frac{p-1}{2}} \mathrm{C}_{2 i}^{i}+\sum_{i=\frac{p+1}{2}}^{p-1} \mathrm{C}_{2 i}^{i}+\mathrm{C}_{2 p}^{p} .
$$
By Example 2, the first term $\sum_{i=0}^{\frac{p-1}{2}} \mathrm{C}_{2 i}^{i}$, when divided by $p$, leaves a remainder of either 1... | 3 \text{ or } 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,381 |
Question 1 Given a regular hexagon $A B C D E F$ with side length $a$, two moving points $M, N$ are on sides $B C, D E$ respectively, and satisfy $\angle M A N=60^{\circ}$. Prove: $A M \cdot A N-B M \cdot D N$ is always a constant. $[1]$ | Prove as shown in Figure 1, connect $A D$, and take point $P$ on line segment $A D$ such that $\angle D P N=60^{\circ}$.
Since $A B C D E F$ is a regular hexagon,
$$
\begin{array}{l}
\Rightarrow \angle A D E=\angle B A D=60^{\circ} \\
\Rightarrow P D=D N=P N .
\end{array}
$$
Given $\angle M A N=60^{\circ}$, we have
$$... | 2a^2 | Geometry | proof | Yes | Yes | cn_contest | false | 730,382 |
Question 2 As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB \neq AC$, $D$ is the midpoint of $BC$, $E$ is the midpoint of $AD$, $DF \perp AB$ at point $F$, $DG \perp AC$ at point $G$, $EF$ and $EG$ intersect $BC$ at points $H$ and $I$ respectively, $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle... | Prove that the common chord of $\odot E$ and $\odot O_{1}$ is $D F$, so $D F \perp O_{1} E$.
Similarly, $E O_{2} \perp D G$.
Thus, $E O_{1} \parallel A B$ and $E O_{2} \parallel A C$.
Therefore, $O_{1} O_{2} \parallel B C$
$\Leftrightarrow \triangle O_{1} O_{2} E \sim \triangle B C A$
$\Leftrightarrow \frac{E O_{1}}{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,383 |
Question 3 In $\triangle ABC$, $X, Y$ are two points on side $BC$ and point $X$ is between $B, Y$, satisfying $2XY = BC$. Let $AA'$ be the diameter of the circumcircle of $\triangle AXY$. Draw a perpendicular from point $B$ to $BC$, intersecting line $AX$ at point $P$, and draw a perpendicular from point $C$ to $BC$, i... | Proof: As shown in Figure 3, extend $PA$ and $QC$ to intersect at point $J$; extend $AA'$ to point $M$ such that $A'M = AA'$.
$$
\begin{array}{l}
\text{Then } \angle PJQ = 90^\circ - \angle AXY = 90^\circ - \angle AA'Y \\
= \angle A'A Y = \angle MAQ.
\end{array}
$$
Given $BC = 2XY$, $AM = 2AA'$, we know
$$
XY = AA' \s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,384 |
Example 1 Given that $f(x)$ is a polynomial with real coefficients and degree $n$. Suppose for all $0 \leqslant k < m \leqslant n, \frac{f(k)-f(m)}{k-m}$ is an integer. Prove: For any distinct integers $a, b$, we have $(a-b) \mid (f(a)-f(b))$.
---
The translation maintains the original text's line breaks and formatti... | 【Analysis】Obviously, any real-coefficient polynomial of degree $n$ can be written as
$$
f(x)=\sum_{k=0}^{n} a_{k}\binom{x}{k} .
$$
Combining this with property 3 completes the proof. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,385 |
Example 2 Let $f$ be an integer-valued polynomial. Prove: for any integers $m, n$, we have
$$
[1,2, \cdots, \operatorname{deg} f] \frac{f(m)-f(n)}{m-n}
$$
is an integer. | 【Analysis】Let $f(x)=\sum_{k=0}^{\operatorname{deg} f(x)} a_{k}\binom{x}{k}$. Then $f(m)-f(n)=\sum_{k=0}^{\operatorname{deg} f(x)} a_{k}\left(\binom{m}{k}-\binom{n}{k}\right)$. Using property 1 will suffice. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,386 |
Example 4 Let $[x]$ denote the greatest integer not exceeding the real number $x$. Let $p$ be an odd prime, $g(x)$ be an integer-coefficient polynomial of degree $m$, and $k \in \mathbf{Z}_{+}$. Let
$$
\binom{g(p x)}{k}=\sum_{t=0}^{m k} c_{t}\binom{x}{t} .
$$
Prove: $c_{j} \in \mathbf{Z}$, and
$$
\left.p^{j-\left[\fra... | 【Analysis】By property 1, we know $c_{j} \in \mathbf{Z}$.
Below, we prove: $\left.p^{j-\left[\frac{k}{p}\right]} \right\rvert\, c_{j}$.
We use reverse mathematical induction to prove it.
Let $g(x)=\sum_{i=0}^{m} b_{i} x^{i}$.
By comparing the coefficients of $x^{m k}$ on both sides of the equation, the left side is $\fr... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,387 |
Example 5 Given the integer sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{0}=1, a_{n}=\sum_{k=0}^{n-1}\binom{n}{k} a_{k}(n \geqslant 1) \text {. }
$$
Let $m$ be a positive integer, $p$ be a prime, and $q, r$ be non-negative integers. Prove:
$$
a_{p^{m} q+r} \equiv a_{p^{m-1} q+r}\left(\bmod p^{m}\right) .
$$ | Proof that due to $2 a_{n}=\sum_{k=0}^{n}\binom{n}{k} a_{k}$, we can construct the generating function
$$
\begin{array}{l}
f(x)=\sum_{n=0}^{\infty} \frac{a_{n} x^{n}}{n!} . \\
\text { Then } \sum_{n=1}^{\infty} \frac{a_{n}}{n!} x^{n}=f(x)-1=\left(\sum_{n=1}^{\infty} \frac{x^{n}}{n!}\right)\left(\sum_{k=0}^{n-1}\binom{n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,388 |
1. Given $m>2$, line $l_{1}: y=\frac{m-2}{m} x+2$, line $l_{2}: y=-x+2 m$ and the $y$-axis form a triangle with an area of 30. Then the value of $m$ is $(\quad)$.
(A) 6
(B) 12
(C) $\frac{1+\sqrt{61}}{2}$
(D) $1+\sqrt{61}$ | -1. A.
Notice that, line $l_{1}$ intersects the $y$-axis at point $(0,2)$, line $l_{2}$ intersects the $y$-axis at point $(0,2 m)$, and the intersection point of lines $l_{1}$ and $l_{2}$ is $(m, m)$.
Then $\frac{1}{2} m(2 m-2)=30 \Rightarrow m=6$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,389 |
Example 4 Given a positive integer $n>1, S$ is an $n$-element set. Find the smallest positive integer $k$, such that there exist subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $S$ with the property: for any $a, b \in S$, $a \neq b$, there exists $1 \leqslant i \leqslant k$, such that
$$
\left|A_{i} \cap\{a, b\}\right|=1
$$ | For any positive integer $n \geqslant 2$, let the smallest value of $k$ that satisfies the condition be $f(n)$.
First, we prove: $f(n) \geqslant f\left(\left[\frac{n}{2}\right]\right)+1$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
In fact, if $A_{1}, A_{2}, \cdots, A_{f(n)}$ are
$$
S=\... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,390 |
2. Given that the sum of five distinct positive integers is 10,001. Then the minimum value of the least common multiple of these five positive integers is ( ).
(A) 2016
(B) 4032
(C) 2130
(D) 4380 | 2. D.
Let the least common multiple of five numbers be $M$. Then these five numbers can be represented as $\frac{M}{a}, \frac{M}{b}, \frac{M}{c}, \frac{M}{d}, \frac{M}{e}$, where $a, b, c, d, e$ are all positive integers.
$$
\begin{array}{l}
\text { Hence } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=\... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,391 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Suppose the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n}=\left[\sqrt{n a_{n-1}}\right]$. Then the value of $a_{2017}$ is $(\quad)$.
(A) 2015
(B) 2016
(C) 2017
(D) 2018 | 3. A.
Notice that, $a_{2}=[\sqrt{2}]=1, a_{3}=[\sqrt{3}]=1$,
$$
\begin{array}{l}
a_{4}=[\sqrt{4}]=2, a_{5}=[\sqrt{5 \times 2}]=3, \\
a_{6}=[\sqrt{6 \times 3}]=4 .
\end{array}
$$
Use induction to prove: $a_{n}=n-2(n \geqslant 3)$.
It is easy to see that, $a_{3}=3-2$.
If $a_{k-1}=k-3(k \geqslant 4)$, then
$$
\begin{arr... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,392 |
5. Fill the numbers $1,2, \cdots, 9$ into a $3 \times 3$ grid such that the sum of the absolute differences of adjacent (sharing a common edge) numbers is maximized. Then this maximum value is ( ).
(A) 57
(B) 58
(C) 59
(D) 60 | 5. B.
Since the central number has the greatest weight, it should be 1 or 9. For example, 1, in which case the eight numbers should be divided into two groups by size and distributed alternately around it, as shown in Figure 4. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,393 |
8. If $a, b, c$ are distinct integers, then
$$
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a
$$
the minimum value is . $\qquad$ | 8. 6 .
Notice that,
$$
\begin{array}{l}
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a \\
=\frac{1}{2}(a-b)^{2}+\frac{3}{2}(b-c)^{2}+\frac{5}{2}(c-a)^{2} .
\end{array}
$$
Since \(a, b, c\) are distinct integers, when \(a-b=2, a-c=1, c-b=1\), the original expression achieves its minimum value of 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,394 |
9. As shown in Figure 1, in $\triangle A B C$, $A B=9, B C=8$, $C A=7, \odot O_{1}$ passes through point $A$, and is tangent to line $B C$ at point $B, \odot O_{2}$ passes through point $A$, and is tangent to line $B C$ at point $C$. Let $\odot O_{1}$ and $\odot O_{2}$ intersect at another point $D$ besides point $A$. ... | 9. $\frac{33}{7}$.
Extend $A D$, intersecting $B C$ at point $E$.
By the power of a point theorem, we have $E B^{2}=E D \cdot E A=E C^{2}$.
Thus, $E$ is the midpoint of $B C$.
By the median length formula, we get
$$
\begin{aligned}
E A^{2} & =\frac{A B^{2}+A C^{2}}{2}-\frac{B C^{2}}{4}=49 \\
\Rightarrow E D & =\frac{E... | \frac{33}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,395 |
11. Given the quadratic function $y=x^{2}+2 m x-3 m+1$, the independent variable $x$ and real numbers $p, q$ satisfy
$$
4 p^{2}+9 q^{2}=2, \frac{1}{2} x+3 p q=1 \text {, }
$$
and the minimum value of $y$ is 1. Find the value of $m$. | Three, 11. From the given information,
$$
\begin{array}{l}
(2 p+3 q)^{2}=2+12 p q, 6 p q=2-x \\
\Rightarrow 2 p \times 3 q=2-x, 2 p+3 q= \pm \sqrt{6-2 x} .
\end{array}
$$
Then $2 p, 3 q$ are the two real roots of the equation in $t$:
$$
t^{2} \mp \sqrt{6-2 x}+2-x=0
$$
Thus, $\Delta=(6-2 x)-4(2-x)=2 x-2 \geqslant 0$, ... | m=-3 \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,396 |
Example 5 If an $n \times n$ square number table, whose elements are all taken from the set $S=\{1,2, \cdots, 2 n-1\}$, and for each $i=1,2, \cdots, n$, the elements in the $i$-th row and the $i$-th column together are exactly all the elements of the set $S$, then the number table is called a "good number table". Prove... | 【Analysis】(1) Let $n>1$, and there exists a good number table $A$ of size $n \times n$.
Since the set $S$ contains $2n-1$ numbers, all of which must appear in the good number table $A$, and the main diagonal of $A$ contains only $n$ numbers, at least one number $x (x \in S)$ does not appear on the main diagonal of $A$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,397 |
13. As shown in Figure $2, \triangle A B C$ has an incircle $\odot I$ that touches sides $B C$, $C A$, and $A B$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. The circumcircle $\odot O_{1}$ of $\triangle B C_{1} B_{1}$ intersects line $B C$ at another point $K$, and the circumcircle $\odot O_{2}$ of $\triangle... | 13. From $\angle L A_{1} C_{1}=\angle C_{1} B_{1} A_{1}$, $\angle A_{1} L C_{1}=\angle A B_{1} C_{1}=\angle B_{1} A_{1} C_{1}$,
we have $\triangle A_{1} C_{1} L \backsim \triangle B_{1} C_{1} A_{1} \Rightarrow \frac{A_{1} L}{A_{1} B_{1}}=\frac{A_{1} C_{1}}{B_{1} C_{1}}$.
Similarly, $\frac{A_{1} K}{A_{1} B_{1}}=\frac{A_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,399 |
2. Given $f(x)=\frac{1+x}{1-3 x}, f_{1}(x)=f(f(x))$, $f_{n+1}(x)=f\left(f_{n}(x)\right)\left(n \in \mathbf{Z}_{+}\right)$.
Then $f_{2017}(-2)=()$.
(A) $-\frac{1}{7}$
(B) $\frac{1}{7}$
(C) $-\frac{3}{5}$
(D) $\frac{3}{5}$ | 2. D.
From the problem, we know that $f_{1}(x)=f(f(x))=\frac{x-1}{3 x+1}$. Then, $f_{2}(x)=f\left(f_{1}(x)\right)=f\left(\frac{x-1}{3 x+1}\right)=x$, $f_{3}(x)=f\left(f_{2}(x)\right)=f(x)=\frac{1+x}{1-3 x}$.
Thus, $f_{3 n+1}(x)=f_{1}(x)=\frac{x-1}{3 x+1}$,
$$
\begin{array}{l}
f_{3 n+2}(x)=f_{2}(x)=x, \\
f_{3 n+3}(x)=... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,400 |
Example 6 Proof: For any real number $M>2$, there always exists a strictly increasing sequence of positive integers $a_{1}, a_{2}, \cdots$ satisfying the following conditions:
(1) For each positive integer $i$, we have $a_{i}>M^{i}$;
(2) If and only if the integer $n \neq 0$, there exist a positive integer $m$ and $b_{... | 【Analysis】Call $n$ that satisfies equation (1) in condition (2) "representable".
First, construct a sequence that satisfies condition (2): each term is a positive integer, strictly increasing, and every non-zero integer is representable, while 0 is not representable.
Use incremental construction.
First, consider 1 to ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,402 |
8. If for any $x \in(-\infty,-1)$, we have
$$
\left(m-m^{2}\right) 4^{x}+2^{x}+1>0 \text {, }
$$
then the range of real number $m$ is $\qquad$ | 8. $[-2,3]$.
Let $a=m-m^{2}, t=2^{x}$.
Then $a t^{2}+t+1>0$ holds when $t \in\left(0, \frac{1}{2}\right)$.
When $a \geqslant 0$, it obviously satisfies the condition;
When $a<0$, the parabola opens downwards and passes through the point $(0,1)$, so there must be a negative root, and it is only necessary that the value... | [-2,3] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,403 |
9. In the acute $\triangle A B C$,
$$
\sin (A+B)=\frac{3}{5}, \sin (A-B)=\frac{1}{5}, A B=3
$$
then the area of $\triangle A B C$ is $\qquad$ | 9. $\frac{3(\sqrt{6}+2)}{2}$.
From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sin A \cdot \cos B+\cos A \cdot \sin B=\frac{3}{5} \\
\sin A \cdot \cos B-\cos A \cdot \sin B=\frac{1}{5}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\sin A \cdot \cos B=\frac{2}{5}, \\
\cos A \cdot \sin... | \frac{3(\sqrt{6}+2)}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,404 |
10. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ always passes through the fixed point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, and the length of its major axis is in the range $[\sqrt{5}, \sqrt{6}]$. Then the range of the eccentricity is $\qquad$ | 10. $\left[\frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2}\right]$.
From the given, $\frac{1}{2 a^{2}}+\frac{1}{2 b^{2}}=1$, which means
$$
a^{2}+b^{2}=2 a^{2} b^{2} \text{. }
$$
Also, $e^{2}=\frac{c^{2}}{a^{2}}=\frac{a^{2}-b^{2}}{a^{2}}$, which means
$$
b^{2}=a^{2}-a^{2} e^{2} \text{. }
$$
From equations (1) and (2), we get... | \left[\frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,405 |
11. Given two sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=2, b_{1}=1, \\
a_{n+1}=5 a_{n}+3 b_{n}+7\left(n \in \mathbf{Z}_{+}\right), \\
b_{n+1}=3 a_{n}+5 b_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
Then the general term formula for $\left\{a_{n}\right\}$ is... | 11. $2^{3 n-2}+2^{n+1}-4$.
(1) + (2) +1 gives
$$
a_{n+1}+b_{n+1}+1=8\left(a_{n}+b_{n}+1\right) \text {, }
$$
thus $a_{n}+b_{n}+1=8^{n-1}\left(a_{1}+b_{1}+1\right)=2^{3 n-1}$;
(1) - (2) +7 gives
$$
a_{n+1}-b_{n+1}+7=2\left(a_{n}-b_{n}+7\right) \text {, }
$$
thus $a_{n}-b_{n}+7=2^{n-1}\left(a_{1}-b_{1}+7\right)=2^{n+2}... | 2^{3 n-2}+2^{n+1}-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,406 |
13. (15 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{n}=2^{n} a+b n-80\left(a 、 b \in \mathbf{Z}_{+}\right) \text {. }
$$
It is known that the minimum value of the sum of the first $n$ terms $S_{n}$ is obtained only when $n=6$, and $7 \mid a_{36}$. Find the value of $\sum_{i=1}^{12}\left|a_{i}\right|$. | Three, 13. Notice that, $\left\{a_{n}\right\}$ is an increasing sequence.
From the given, $a_{6}0$, that is,
$$
64 a+6 b-800 \text {. }
$$
Combining $a, b \in \mathbf{Z}_{+}$, we get
$$
a=1, b=1 \text { or } 2 \text {. }
$$
Also, $a_{36}=2^{36}+36 b-80$
$$
\equiv 1+b-3 \equiv 0(\bmod 7) .
$$
Thus, $b=2$.
Therefore, ... | 8010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,407 |
1. Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying $a_{n+1} \geqslant 2 a_{n}+1$, and $a_{n}<2^{n+1}$ for $n \in \mathbf{Z}_{+}$. Then the range of $a_{1}$ is | $\begin{array}{l}\text { i. } 1 .(0,3] . \\ \text { From } a_{n+1}+1 \geqslant 2\left(a_{n}+1\right) \\ \Rightarrow a_{n}+1 \geqslant\left(a_{1}+1\right) 2^{n-1} . \\ \text { Therefore }\left(a_{1}+1\right) 2^{n-1}-1 \leqslant a_{n}<2^{n+1} \\ \Rightarrow a_{1}+1<\frac{2^{n+1}+1}{2^{n-1}}=4+\frac{1}{2^{n-1}} \\ \Righta... | 0<a_{1} \leqslant 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,408 |
2. Given the function $f(x)=\log _{2} \frac{x-3}{x-2}+\cos \pi x$. If $f(\alpha)=10, f(\beta)=-10$, then $\alpha+\beta=$ $\qquad$ | 2. 5 .
It is easy to know that the domain of $f(x)$ is $(-\infty, 2) \cup(3,+\infty)$.
Then $f(5-x)=\log _{2} \frac{5-x-3}{5-x-2}+\cos (5-x) \pi$ $=-f(x)$.
Therefore, $f(x)$ is centrally symmetric about the point $\left(\frac{5}{2}, 0\right)$.
Also, when $x>3$,
$$
f(x)=\log _{2}\left(1-\frac{1}{x-2}\right)+\cos x \pi,
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,409 |
7. Given point $P(4,2)$, a line $l$ passing through point $P$ intersects the positive x-axis and y-axis at points $A$ and $B$ respectively, and $O$ is the origin. Then the minimum perimeter of $\triangle A O B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format,... | 7. 20 .
As shown in Figure 3, construct the excircle $\odot O_{1}$ of $\triangle A O B$ tangent to line $l$ at point $K$.
Thus, $|B K|=|B N|,|A M|=|A K|$.
Let the center of the circle be $O_{1}(m, m)$. Then the perimeter of $\triangle A O B$ is $|O M|+|O N|=2|O M|=2 m$.
For point $P$ to satisfy the condition, it is ne... | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 730,410 |
11. (20 points) Given non-zero complex numbers $x, y$ satisfy $y^{2}\left(x^{2}-x y+y^{2}\right)+x^{3}(x-y)=0$.
Find the value of $\sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n}$. | 11. Divide both sides of the known equation by $y^{4}$,
$$
\left(\frac{x}{y}\right)^{4}-\left(\frac{x}{y}\right)^{3}+\left(\frac{x}{y}\right)^{2}-\left(\frac{x}{y}\right)+1=0 \text {. }
$$
Let $\frac{x}{y}=\omega$, then
$$
\begin{array}{l}
\omega^{4}-\omega^{3}+\omega^{2}-\omega+1=0 \\
\Rightarrow \omega^{5}=-1 \Right... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,411 |
一、(40 points) As shown in Figure 1, in the acute triangle $\triangle ABC$, $AC > AB$, $H$ and $I$ are the orthocenter and incenter of $\triangle ABC$ respectively. $\odot O$ is the circumcircle of $\triangle ABC$, $M$ is the midpoint of arc $\overparen{BAC}$, $K$ is a point on $\odot O$ such that $\angle AKH = 90^{\cir... | As shown in Figure 4, let $A I$ and $K H$ intersect $\odot O$ at points $P$ and $Q$ respectively. Thus, $A Q$ is the diameter of $\odot O$.
Let $\angle A B C=\beta, \angle A C B=\gamma, \angle B A C=\alpha$, and the radius of $\odot O$ be $R$.
Notice that $N$ and $H$ are symmetric with respect to $B C$,
$$
\begin{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,412 |
2. Pablo goes to the store to buy ice cream for his friends. The store has three sales options: retail price is 1 yuan per ice cream; a pack of three ice creams costs 2 yuan; a pack of five ice creams costs 3 yuan. Pablo has 8 yuan, and the maximum number of ice creams he can buy is ( ) pieces.
(A) 8
(B) 11
(C) 12
(D) ... | 2. D.
From the conditions, we know that the unit price of the five-pack ice cream is 0.6 yuan per piece, which is the cheapest, so we should buy as much as possible; the unit price of the three-pack ice cream is approximately 0.67 yuan per piece, which is the second cheapest; the single ice cream is the most expensive... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,413 |
3. As shown in Figure 1, in Tamara's garden, there are 3 rows and 2 columns of rectangular flower beds, totaling six beds, each 6 feet long and 2 feet wide. These flower beds are separated by 1-foot-wide walkways, and there is also a 1-foot-wide walkway around the perimeter of the garden. What is the total area $S$ of ... | 3. B.
$$
\begin{aligned}
S= & (1+2+1+2+1+2+1) \times \\
& (1+6+1+6+1)-6 \times 6 \times 2 \\
= & 150-72=78 .
\end{aligned}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,414 |
3. Given 2017 points distributed on a circle, each point is labeled with +1 or -1. If starting from a certain point and moving in any direction around the circle to any other point, the sum of all the numbers passed is positive, then that point is called "good". Prove: If the number of points labeled with -1 is no more... | Prompt: Generalize the problem and prove: In $3n+1$ points, if there are $n$ -1 points, a good point must exist.
When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for $n=k$.
For $n=k+1$, we can arbitrarily choose a -1 point $A$, and on each side of it, there is a nearest +1 point $B$ and $C$. Re... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,415 |
7. Jerry and Sylvia plan to walk from the southwest corner to the northeast corner of a square plaza. Jerry walks east first and then north to reach the destination, but Sylvia walks straight in the northwest direction. Then Sylvia walks ( ) less than Jerry.
( A) $30 \%$
( B) $40 \%$
(C) $50 \%$
(D) $60 \%$
(E) $70 \%$ | 7. A.
$$
\frac{2-\sqrt{2}}{2} \times 100\% \approx \frac{2-1.4}{2} \times 100\% = 30\% \text{.}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,416 |
9. Minnie and Penny are cycling on the same road. Minnie's cycling speeds on flat, downhill, and uphill sections are 20 km/h, 30 km/h, and 5 km/h, respectively; Penny's cycling speeds on flat, downhill, and uphill sections are 30 km/h, 40 km/h, and 10 km/h, respectively. Now, Minnie is traveling from Town A to Town B, ... | 9. C.
Minnie's time spent is
$$
t_{1}=\frac{10}{5}+\frac{15}{30}+\frac{20}{20}=3 \frac{1}{2} \text { (hours), }
$$
Penny's time spent is $t_{2}=\frac{20}{30}+\frac{15}{10}+\frac{10}{40}=2 \frac{5}{12}$ (hours).
Therefore, the extra time Minnie spent compared to Penny is $\left(t_{1}-t_{2}\right) \times 60=65$ (minute... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 730,417 |
10. Joey has 30 thin sticks, each with a length of an integer from 1 centimeter to 30 centimeters. Joey first places three sticks on the table with lengths of 3 centimeters, 7 centimeters, and 15 centimeters, and then selects a fourth stick to form a convex quadrilateral with the first three sticks. Joey has ( ) differ... | 10. B.
Let the length of the fourth stick be $x$ cm. Then
$$
\left\{\begin{array}{l}
x+3+7>15, \\
3+7+15>x
\end{array} \Rightarrow 5<x<25\right. \text {. }
$$
Also, $x \in \mathbf{Z}$, so $x$ can take 19 different integer values, but 7 and 15 have already been taken, leaving 17 choices. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,418 |
13. Define the recursive sequence $\left\{F_{n}\right\}$ :
$$
F_{0}=0, F_{1}=1 \text {, }
$$
$F_{n}(n \geqslant 2)$ is the remainder of $F_{n-1}+F_{n-2}$ divided by 3, i.e., the sequence $\left\{F_{n}\right\}: 0,1,1,2,0,2,2,1,0, \cdots$.
Then the value of $\sum_{i=2017}^{2024} F_{i}$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9
(E)... | 13. D.
$$
\begin{array}{l}
\text { Given } F_{n+8} \equiv F_{n+7}+F_{n+6} \equiv 2 F_{n+6}+F_{n+5} \\
\equiv 3 F_{n+5}+2 F_{n+4} \equiv 2 F_{n+4} \equiv 2 F_{n+3}+2 F_{n+2} \\
\equiv 4 F_{n+2}+2 F_{n+1} \equiv F_{n+2}+2 F_{n+1} \\
\equiv 3 F_{n+1}+F_{n} \equiv F_{n}(\bmod 3)
\end{array}
$$
$\Rightarrow\left\{F_{n}\righ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,419 |
18. Amelia tosses a coin, with the probability of landing heads up being $\frac{1}{3}$; Brian also tosses a coin, with the probability of landing heads up being $\frac{2}{5}$. Amelia and Brian take turns tossing the coins, and the first one to get heads wins. All coin tosses are independent. Starting with Amelia, the p... | 18. D.
Let $P_{0}$ be the probability of Amelia winning.
Notice,
$P_{0}=P($ Amelia wins in the first round $)+$ $P($ both fail to win in the first round $) \cdot P_{0}$, where, if both fail to win in the first round, it still starts with Amelia, and her probability of winning remains $P_{0}$.
In the first round, the p... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,420 |
19. A, B, C, D, E stand in a row, satisfying that A is not adjacent to B and C, and D is not adjacent to E. The number of different arrangements is ( ) kinds.
(A) 12
(B) 16
(C) 28
(D) 32
(E) 40 | 19. C.
There are two scenarios:
(1) If A is at one of the ends, there are two choices; the person next to A cannot be B or C, so there are two choices from D and E; D and E cannot be adjacent, so one person is chosen from B and C, which gives two choices; the last two positions are filled by the remaining two people. ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,421 |
21. Given a square with side length $x$ inscribed in a right-angled triangle with side lengths $3, 4, 5$, and one vertex of the square coincides with the right-angle vertex of the triangle; a square with side length $y$ is also inscribed in the right-angled triangle with side lengths $3, 4, 5$, and one side of the squa... | 21. D.
As shown in Figure 4, let $A D=x$, and the square $A F E D$ is inscribed in the right triangle $\triangle A B C$. Then
$$
\begin{array}{l}
\triangle A B C \backsim \triangle D E C \Rightarrow \frac{D E}{A B}=\frac{C D}{A C} \\
\Rightarrow \frac{x}{4}=\frac{3-x}{3} \Rightarrow x=\frac{12}{7} .
\end{array}
$$
As... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,422 |
24. Given real numbers $a$, $b$, $c$, the polynomial
$$
g(x)=x^{3}+a x^{2}+x+10
$$
has three distinct roots, and these three roots are also roots of the polynomial
$$
f(x)=x^{4}+x^{3}+b x^{2}+100 x+c
$$
Then the value of $f(1)$ is $(\quad)$.
(A) -9009
(B) -8008
(C) -7007
(D) -6006
(E) -5005 | 24. C.
Let $r \in \mathbf{C}$ be the fourth root of the polynomial $f(x)$. Then
$$
\begin{array}{l}
f(x)=g(x)(x-r) \\
=\left(x^{3}+a x^{2}+x+10\right)(x-r) \\
=x^{4}+(a-r) x^{3}+(1-a r) x^{2}+(10-r) x-10 r \\
=x^{4}+x^{3}+b x^{2}+100 x+c \\
\Rightarrow\left\{\begin{array} { l }
{ a - r = 1 , } \\
{ 1 - a r = b , } \\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,423 |
25. Among the integers between 100 and 999, there are ( ) numbers that have the property: the digits of the number can be rearranged to form a number that is a multiple of 11 and is between 100 and 999 (for example, 121 and 211 both have this property).
(A) 226
(B) 243
( C) 270
(D) 469
(E) 486 | 25. A.
Let a three-digit number be $\overline{A C B}$. Then
11. $\overline{A C B} \Leftrightarrow 11 \mathrm{I}(A+B-C)$
$\Leftrightarrow A+B=C$ or $A+B=C+11$.
We will discuss the following scenarios.
Note that, $A$ and $B$ are of equal status, so we can assume $A \geqslant B$ (the case for $A < B$ is similar).
(1) $A+... | 226 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,424 |
Example 2 Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ such that for any positive integers $m, n$, $f(m)+f(n)-m n$ is non-zero and divides $m f(m)+n f(n).^{[2]}$
(57th IMO Shortlist) | 【Analysis】First try to guess the answer.
Notice that when $f(n)=n^{2}$, $m^{2}-m n+n^{2}$ is non-zero and divides $m^{3}+n^{3}$. It seems that no other answers can be tried out.
Let's verify this. Take $m=n=1$, we have
$(2 f(1)-1) \mid 2 f(1) \Rightarrow f(1)=1$.
For any odd prime $p$, take another number as 1, we get
... | f(n)=n^{2} | Number Theory | proof | Yes | Yes | cn_contest | false | 730,425 |
Let $p$ be a prime. Prove: for any integer $a$,
$$
x^{2}+y^{3} \equiv a(\bmod p)
$$
has a solution. | 【Analysis】When $p \leqslant 7$, it is easy to verify that the conclusion holds.
The following discussion is for the case when $p \geqslant 11$.
(1) If $p=3 m+1$, we continue to use the technique from problem 1, and let $x^{2}+y^{3} \equiv k(\bmod p)$ have $a_{k}$ solutions.
Proof by contradiction.
Assume there exists $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,426 |
In a box, there are 10 red cards and 10 blue cards, each set of cards containing one card labeled with each of the numbers $1, 3, 3^{2}, \cdots, 3^{9}$. The total sum of the numbers on the cards of both colors is denoted as $S$. For a given positive integer $n$, if it is possible to select several cards from the box su... | Let the maximum sum of the labels of two-color cards marked as $1,3,3^{2}, \cdots, 3^{k}$ be denoted as $S_{k}$. Then,
$$
S_{k}=2 \sum_{n=0}^{k} 3^{n}=3^{k+1}-1<3^{k+1} \text {. }
$$
In the sequence $1,3,3^{2}, \cdots, 3^{k}$, the sum of any subset of these numbers is not equal to $3^{m}$. Therefore, the number of way... | 6423 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,427 |
2. Given that $k$ is a positive real number, the linear function $y=k x+1$ intersects with the reciprocal function $y=\frac{k}{x}$ at points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. If $\left|x_{1}-x_{2}\right|=\sqrt{5}$, then the value of $k$ is ( ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2 | 2. A.
By combining the two functions and eliminating $y$, we get
$$
k x^{2}+x-k=0 \text {. }
$$
By Vieta's formulas, we know
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{1}{k}, x_{1} x_{2}=-1 . \\
\text { Then } \sqrt{5}=\left|x_{1}-x_{2}\right| \\
=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{\frac{1}{k^{2}}+4}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,428 |
3. Given $A D, B E, C F$ are the altitudes of the acute $\triangle A B C$. If $A B=26, \frac{E F}{B C}=\frac{5}{13}$, then the length of $B E$ is ).
(A) 10
(B) 12
(C) 13
(D) 24 | 3. D.
As shown in Figure 2, it is easy to see that points $B, C, E, F$ are concyclic. Therefore, $\triangle A E F \backsim \triangle A B C$. Thus, $\cos A=\frac{A F}{A C}=\frac{E F}{B C}=\frac{5}{13}$, $\sin A=\sqrt{1-\cos ^{2} A}=\frac{12}{13}$. Hence, $B E=A B \sin A=26 \times \frac{12}{13}=24$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,429 |
6. Let the sum of the digits of a positive integer $m$ be denoted as $S(m)$, for example, $S(2017)=2+0+1+7=10$. Now, from the 2017 positive integers $1,2, \cdots$, 2017, any $n$ different numbers are taken. It is always possible to find eight different numbers $a_{1}, a_{2}, \cdots, a_{8}$ among these $n$ numbers such ... | 6. A.
Notice that, among $1,2, \cdots, 2017$, the minimum sum of digits is 1, and the maximum sum is 28.
It is easy to see that the numbers with a digit sum of 1 are $1, 10, 100, 1000$; the numbers with a digit sum of $2,3, \cdots, 26$ are no less than eight; the numbers with a digit sum of 27 are only 999, $1899, 19... | 185 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,430 |
9. There are four teacups with their mouths facing up. Now, each time three of them are flipped, and the flipped teacups are allowed to be flipped again. After $n$ flips, all the cup mouths are facing down. Then the minimum value of the positive integer $n$ is $\qquad$ . | 9.4 .
Let $x_{i}$ be the number of times the $i$-th cup ($i=1,2,3,4$) is flipped when all cup mouths are facing down, then $x_{i}$ is odd.
From $x_{1}+x_{2}+x_{3}+x_{4}=3 n$, we know that $n$ is even.
It is easy to see that when $n=2$, the condition is not satisfied, hence $n \geqslant 4$.
When $n=4$, use 1 to represe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 730,431 |
13. (25 points) As shown in Figure 1, with the right triangle $\triangle ABC (\angle C = 90^\circ)$, squares $CADE$, $BCFG$, and $ABHI$ are constructed outward on the sides $CA$, $CB$, and $AB$ respectively. Let the lengths of sides $CB$ and $CA$ be $a$ and $b$ respectively, and the area of the convex hexagon $DEFGHI$ ... | 13. As shown in Figure 5, extend $BA$ to point $B'$ such that $B'A = AB$, and connect $B'C$.
It is easy to prove that $\triangle B'AC \cong \triangle IAD$.
Thus, $S_{\triangle IAD} = S_{\triangle B'AC} = S_{\triangle ABC}$.
Similarly, $S_{\triangle ECF} = S_{\triangle ABC} = S_{\triangle HBG}$.
Therefore, $S = 4 \times... | (12, 24), (24, 12) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,432 |
1. Given that the function $f(x)$ is a decreasing function on $\mathbf{R}$ and is an odd function. If $m, n$ satisfy
$$
\left\{\begin{array}{l}
f(m)+f(n-2) \leqslant 0, \\
f(m-n-1) \leqslant 0,
\end{array}\right.
$$
then the range of $5 m-n$ is | $-1 .[7,+\infty)$
Since $f(x)$ is an odd function and is defined at $x=0$, we have $f(0)=0$.
According to the problem, we have
$$
\left\{\begin{array}{l}
f(m) \leqslant -f(n-2)=f(2-n), \\
f(m-n-1) \leqslant 0=f(0) .
\end{array}\right.
$$
Since $f(x)$ is a decreasing function, it follows that
$$
\left\{\begin{array}{l}... | [7,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,433 |
2. Given $x_{1}=1, x_{2}=2, x_{3}=3$ are zeros of the function
$$
f(x)=x^{4}+a x^{3}+b x^{2}+c x+d
$$
then $f(0)+f(4)=$ $\qquad$ | 2. 24 .
Let $f(x)=(x-1)(x-2)(x-3)(x-k)$. Then $f(0)+f(4)=6k+6(4-k)=24$. | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,434 |
3. Given $a>1$. Then the minimum value of $\log _{a} 16+2 \log _{4} a$ is $\qquad$ . | 3.4.
From the operation of logarithms, we get
$$
\begin{array}{l}
\log _{a} 16+2 \log _{4} a=4 \log _{a} 2+\log _{2} a \\
=\frac{4}{\log _{2} a}+\log _{2} a .
\end{array}
$$
Since $a>1$, we have $\log _{2} a>0$.
By the AM-GM inequality, we get
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,435 |
4. As shown in Figure 1, in the right triangle $\triangle A B C$, $\angle C=60^{\circ}$. With $C$ as the center and $B C$ as the radius, a circle intersects $A C$ at point $D$. Connect $B D$. The arc $\overparen{B D}$ and the chord $B D$ divide $\triangle A B C$ into three parts. Then the ratio of the areas of the thre... | 4. $\left(\sqrt{3}-\frac{\pi}{3}\right):\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right): \frac{\sqrt{3}}{2}$.
Let $BC = x$. Then
$$
\begin{array}{l}
S_{1}=\frac{1}{2} AB \cdot BC - \frac{1}{2} BC \cdot \overparen{BD} \\
=\frac{1}{2} x^{2}\left(\tan \frac{\pi}{3} - \frac{\pi}{3}\right) = \frac{1}{2}\left(\sqrt{3} - \frac... | \left(\sqrt{3} - \frac{\pi}{3}\right):\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right): \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,436 |
6. In $\triangle A B C$,
$$
\tan A 、(1+\sqrt{2}) \tan B 、 \tan C
$$
form an arithmetic sequence. Then the minimum value of $\angle B$ is $\qquad$ | 6. $\frac{\pi}{4}$.
From the problem, we know
$$
\begin{array}{l}
2(1+\sqrt{2}) \tan B=\tan A+\tan C . \\
\text { Also, } \angle A+\angle B+\angle C=\pi \text {, so } \\
\tan B=\tan (\pi-(A+C)) \\
=-\frac{\tan A+\tan C}{1-\tan A \cdot \tan C} \\
\Rightarrow \tan A \cdot \tan B \cdot \tan C \\
=\tan A+\tan B+\tan C .
\... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,437 |
8. Given the sequence $\left\{a_{n}\right\}$ with the first term being 2, and satisfying
$$
6 S_{n}=3 a_{n+1}+4^{n}-1 \text {. }
$$
Then the maximum value of $S_{n}$ is $\qquad$. | 8. 35 .
According to the problem, we have
$$
\left\{\begin{array}{l}
6 S_{n}=3 a_{n+1}+4^{n}-1 \\
6 S_{n-1}=3 a_{n}+4^{n-1}-1
\end{array}\right.
$$
Subtracting the two equations and simplifying, we get
$$
\begin{array}{l}
a_{n+1}=3 a_{n}-4^{n-1} \\
\Rightarrow a_{n+1}+4^{n}=3 a_{n}-4^{n-1}+4^{n} \\
\quad=3\left(a_{n}... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,438 |
2. Given a unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the midpoints of edges $A B$, $A_{1} D_{1}$, $A_{1} B_{1}$, and $B C$ are $L$, $M$, $N$, and $K$ respectively. Then the radius of the inscribed sphere of the tetrahedron $L M N K$ is $\qquad$ | 2. $\frac{\sqrt{3}-\sqrt{2}}{2}$.
Let the radius of the inscribed sphere of tetrahedron $L M N K$ be $r$. Since $\angle N L K=\angle L N M=90^{\circ}$, we have $S_{\triangle L M N}=S_{\triangle L N K}=\frac{1}{2} \times 1 \times \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}$.
Also, $\angle M L K=\angle M N K=90^{\circ}$, so $... | \frac{\sqrt{3}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,439 |
3. Given $P$ is a point on the hyperbola $\Gamma: \frac{x^{2}}{463^{2}}-\frac{y^{2}}{389^{2}}=1$, a line $l$ is drawn through point $P$, intersecting the two asymptotes of the hyperbola $\Gamma$ at points $A$ and $B$. If $P$ is the midpoint of segment $A B$, and $O$ is the origin, then $S_{\triangle O A B}=$ $\qquad$ | 3. 180107 .
Let the two asymptotes of the hyperbola $\Gamma$ be $l_{1}: y=\frac{389}{463} x, l_{2}: y=-\frac{389}{463} x$;
$l$ intersects $l_{1} 、 l_{2}$ at points $A 、 B$ respectively.
Let $P\left(x_{0}, y_{0}\right), A\left(x_{1}, \frac{389}{463} x_{1}\right)$,
$B\left(x_{2},-\frac{389}{463} x_{2}\right)$.
Then $x_{... | 180107 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,440 |
6. Let the function be
$$
f(x)=\frac{x(1-x)}{x^{3}-x+1}(0<x<1) \text {. }
$$
If the maximum value of $f(x)$ is $f\left(x_{0}\right)$, then $x_{0}=$ $\qquad$ | $$
\begin{array}{l}
\text { 6. } \frac{\sqrt{2}+1-\sqrt{2 \sqrt{2}-1}}{2} \text {. } \\
\text { Let } f^{\prime}(x)=\frac{x^{4}-2 x^{3}+x^{2}-2 x+1}{\left(x^{3}-x+1\right)^{2}} \\
=\frac{\left(x^{2}-(\sqrt{2}+1) x+1\right)\left(x^{2}+(\sqrt{2}-1) x+1\right)}{\left(x^{3}-x+1\right)^{2}}=0 .
\end{array}
$$
Then $\left(x... | \frac{\sqrt{2}+1-\sqrt{2 \sqrt{2}-1}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,441 |
11. (20 points) For any real numbers $a_{1}, a_{2}, \cdots, a_{5}$ $\in[0,1]$, find the maximum value of $\prod_{1 \leqslant i<j \leqslant 5}\left|a_{i}-a_{j}\right|$. | 11. Let \( f\left(a_{1}, a_{2}, \cdots, a_{5}\right)=\prod_{1 \leqslant i < j \leqslant 5} \left(a_{i}-a_{j}\right) \) where \( 1 > a_{1} > a_{2} > \cdots > a_{5} \).
To maximize \( f\left(a_{1}, a_{2}, \cdots, a_{5}\right) \), it must be that \( a_{1}=1 \) and \( a_{5}=0 \).
Thus, \( f\left(1, a_{2}, a_{3}, a_{4}, 0\... | \frac{3 \sqrt{21}}{38416} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,442 |
1. The following propositions are correct ( ) in number.
(1) Very small real numbers can form a set;
(2) The set $\left\{y \mid y=x^{2}-1\right\}$ and $\left\{(x, y) \mid y=x^{2}-1\right\}$ are the same set;
(3) $1 、 \frac{3}{2} 、 \frac{6}{4} 、\left|-\frac{1}{2}\right| 、 0.5$ These numbers form a set with five elements... | 一, 1. A.
(1) The reason for the error is that the elements are uncertain;
(2) The former is a set of numbers, the latter is a set of points, the types are different;
(3) There are repeated elements, it should be three elements;
(4) This set also includes the coordinate axes. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,443 |
2. The point corresponding to the complex number $z=\frac{(2+\mathrm{i})^{2}}{1-\mathrm{i}}$ in the complex plane is located in the ( ) quadrant.
(A) one
(B) two
$(C) \equiv$
(D) four
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dire... | 2. B.
According to the rules of complex number operations, we know that $z=-\frac{1}{2}+\frac{7}{2} \mathrm{i}$. | null | Number Theory | proof | Yes | Yes | cn_contest | false | 730,444 |
3. If the equation $a^{x}-x-a=0$ has two real solutions, then the range of values for $a$ is ( ).
(A) $(1,+\infty)$
(B) $(0,1)$
(C) $(0,2)$
(D) $(0,+\infty)$ | 3. A.
Draw the graph, and you will find that when $a>1$, the function $y=a^{x}$ intersects with the function $y=x+a$ at two points. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,445 |
Example 6 Given a function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfies:
(1) For any positive integers $m, n$, we have
$$
(f(m), f(n)) \leqslant(m, n)^{2014} \text {; }
$$
(2) For any positive integer $n$, we have
$$
n \leqslant f(n) \leqslant n+2014 \text {. }
$$
Prove: There exists a positive integer $N$... | 【Analysis】From (1), we know that when $(m, n)=1$, $(f(m), f(n)) \leqslant(m, n)^{2014}=1$
$\Rightarrow(f(m), f(n))=1$.
Therefore, when $(f(m), f(n))>1$, $(m, n)>1$.
First, we prove a lemma.
Lemma For a prime $p$, if $p \mid f(n)$, then $p \mid n$.
Proof Assume there exists
$p \mid f(m)(p \nmid m, f(m)=m+l)$.
Take $2015... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,446 |
5. In the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=1 \\
a_{n}=a_{n-1}+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
Then $a_{4}=$ ( ).
(A) $\frac{7}{4}$
(B) $-\frac{7}{4}$
(C) $\frac{4}{7}$
(D) $-\frac{4}{7}$ | 5. A.
Since $a_{n}-a_{n-1}=\frac{1}{n-1}-\frac{1}{n}$, therefore,
$$
\begin{array}{l}
a_{4}-a_{1}=\sum_{n=2}^{4}\left(\frac{1}{n-1}-\frac{1}{n}\right)=1-\frac{1}{4} \\
\Rightarrow a_{4}=a_{1}+\frac{3}{4}=\frac{7}{4} .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,447 |
10. Given $F_{1}$ and $F_{2}$ are the
foci of the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$,
with $F_{1} F_{2}$ as the diameter of a circle that intersects the hyperbola $C$ at point $P$ in the second quadrant. If the eccentricity of the hyperbola is 5, then the value of $\cos \angle P F_{2} F... | 10. C.
Let $P F_{1}=x$.
Since point $P$ is on the hyperbola, we know $P F_{2}=2 a+x$. Since point $P$ is on the circle, we know $P F_{1}^{2}+P F_{2}^{2}=F_{1} F_{2}^{2}$.
Then $x=-a+\sqrt{b^{2}+c^{2}}$
$$
\Rightarrow P F_{1}=-a+\sqrt{b^{2}+c^{2}} \text {. }
$$
Also, $\frac{c}{a}=5$, so $\cos \angle P F_{2} F_{1}=\fra... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,448 |
13. Given that the angle between vector $\boldsymbol{a}$ and $\boldsymbol{b}$ is $120^{\circ}$, and $|a|=2,|b|=5$. Then $(2 a-b) \cdot a=$ $\qquad$ | \begin{array}{l}\text { II.13. 13. } \\ (2 a-b) \cdot a=2|a|^{2}-a \cdot b=13 \text {. }\end{array} | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,449 |
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