id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
3,078 | If $1, x, y$ is a geometric sequence and $x, y, 3$ is an arithmetic sequence, compute the maximum value of $x+y$. | [
"The common ratio in the geometric sequence $1, x, y$ is $\\frac{x}{1}=x$, so $y=x^{2}$. The arithmetic sequence $x, y, 3$ has a common difference, so $y-x=3-y$. Substituting $y=x^{2}$ in the equation yields\n\n$$\n\\begin{aligned}\nx^{2}-x & =3-x^{2} \\\\\n2 x^{2}-x-3 & =0\n\\end{aligned}\n$$\n\nfrom which $x=\\fr... | [
"$\\frac{15}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,079 | Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows:
$$
\left\{\begin{array}{l}
a_{1}=1 \\
\text { for } n \geq 2, a_{n} \text { is the smallest possible positive value of } n-a_{k}^{2}, \text { for } 1 \leq k<n .
\end{array}\right.
$$
For example, $a_{2}=2-1^{2}=1$, and $a_{3}=3-1^{2}=2$. Compute $a_{1}+a_{2}+\cdots+a_{50}$. | [
"The requirement that $a_{n}$ be the smallest positive value of $n-a_{k}^{2}$ for $k<n$ is equivalent to determining the largest value of $a_{k}$ such that $a_{k}^{2}<n$. For $n=3$, use either $a_{1}=a_{2}=1$ to find $a_{3}=3-1^{2}=2$. For $n=4$, the strict inequality eliminates $a_{3}$, so $a_{4}=4-1^{2}=3$, but $... | [
"253"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,080 | Compute the base $b$ for which $253_{b} \cdot 341_{b}=\underline{7} \underline{4} \underline{X} \underline{Y} \underline{Z}_{b}$, for some base- $b$ digits $X, Y, Z$. | [
"Write $253_{b} \\cdot 341_{b}=\\left(2 b^{2}+5 b+3\\right)\\left(3 b^{2}+4 b+1\\right)=6 b^{4}+23 b^{3}+31 b^{2}+17 b+3$. Compare the coefficients in this polynomial to the digits in the numeral $\\underline{7} \\underline{4} \\underline{X} \\underline{Y} \\underline{Z}$. In the polynomial, the coefficient of $b^{... | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,081 | Some portions of the line $y=4 x$ lie below the curve $y=10 \pi \sin ^{2} x$, and other portions lie above the curve. Compute the sum of the lengths of all the segments of the graph of $y=4 x$ that lie in the first quadrant, below the graph of $y=10 \pi \sin ^{2} x$. | [
"Notice first that all intersections of the two graphs occur in the interval $0 \\leq x \\leq \\frac{5 \\pi}{2}$, because the maximum value of $10 \\pi \\sin ^{2} x$ is $10 \\pi$ (at odd multiples of $\\frac{\\pi}{2}$ ), and $4 x>10 \\pi$ when $x>\\frac{5 \\pi}{2}$. The graphs are shown below.\n\n<img_3576>\n\nWith... | [
"$\\frac{5 \\pi}{4} \\sqrt{17}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,082 | In equilateral hexagon $A B C D E F, \mathrm{~m} \angle A=2 \mathrm{~m} \angle C=2 \mathrm{~m} \angle E=5 \mathrm{~m} \angle D=10 \mathrm{~m} \angle B=10 \mathrm{~m} \angle F$, and diagonal $B E=3$. Compute $[A B C D E F]$, that is, the area of $A B C D E F$. | [
"Let $\\mathrm{m} \\angle B=\\alpha$. Then the sum of the measures of the angles in the hexagon is:\n\n$$\n\\begin{aligned}\n720^{\\circ} & =\\mathrm{m} \\angle A+\\mathrm{m} \\angle C+\\mathrm{m} \\angle E+\\mathrm{m} \\angle D+\\mathrm{m} \\angle B+\\mathrm{m} \\angle F \\\\\n& =10 \\alpha+5 \\alpha+5 \\alpha+2 \... | [
"$\\frac{9}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,083 | The taxicab distance between points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ is defined as $d(A, B)=$ $\left|x_{A}-x_{B}\right|+\left|y_{A}-y_{B}\right|$. Given some $s>0$ and points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$, define the taxicab ellipse with foci $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ to be the set of points $\{Q \mid d(A, Q)+d(B, Q)=s\}$. Compute the area enclosed by the taxicab ellipse with foci $(0,5)$ and $(12,0)$, passing through $(1,-1)$. | [
"Let $A=(0,5)$ and $B=(12,0)$, and let $C=(1,-1)$. First compute the distance sum: $d(A, C)+d(B, C)=19$. Notice that if $P=(x, y)$ is on the segment from $(0,-1)$ to $(12,-1)$, then $d(A, P)+d(B, P)$ is constant. This is because if $0<x<12$,\n\n$$\n\\begin{aligned}\nd(A, P)+d(B, P) & =|0-x|+|5-(-1)|+|12-x|+|0-(-1)|... | [
"96"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,084 | The function $f$ satisfies the relation $f(n)=f(n-1) f(n-2)$ for all integers $n$, and $f(n)>0$ for all positive integers $n$. If $f(1)=\frac{f(2)}{512}$ and $\frac{1}{f(1)}=2 f(2)$, compute $f(f(4))$. | [
"Substituting yields $\\frac{512}{f(2)}=2 f(2) \\Rightarrow(f(2))^{2}=256 \\Rightarrow f(2)=16$. Therefore $f(1)=\\frac{1}{32}$. Using the recursion, $f(3)=\\frac{1}{2}$ and $f(4)=8$. So $f(f(4))=f(8)$. Continue to apply the recursion:\n\n$$\nf(5)=4, \\quad f(6)=32, \\quad f(7)=128, \\quad f(8)=\\mathbf{4 0 9 6} .\... | [
"4096"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,085 | Frank Narf accidentally read a degree $n$ polynomial with integer coefficients backwards. That is, he read $a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ as $a_{0} x^{n}+\ldots+a_{n-1} x+a_{n}$. Luckily, the reversed polynomial had the same zeros as the original polynomial. All the reversed polynomial's zeros were real, and also integers. If $1 \leq n \leq 7$, compute the number of such polynomials such that $\operatorname{GCD}\left(a_{0}, a_{1}, \ldots, a_{n}\right)=1$. | [
"When the coefficients of a polynomial $f$ are reversed to form a new polynomial $g$, the zeros of $g$ are the reciprocals of the zeros of $f: r$ is a zero of $f$ if and only if $r^{-1}$ is a zero of $g$. In this case, the two polynomials have the same zeros; that is, whenever $r$ is a zero of either, so must be $r... | [
"70"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,086 | Given a regular 16-gon, extend three of its sides to form a triangle none of whose vertices lie on the 16-gon itself. Compute the number of noncongruent triangles that can be formed in this manner. | [
"Label the sides of the polygon, in order, $s_{0}, s_{1}, \\ldots, s_{15}$. First note that two sides of the polygon intersect at a vertex if and only if the sides are adjacent. So the sides chosen must be nonconsecutive. Second, if nonparallel sides $s_{i}$ and $s_{j}$ are extended, the angle of intersection is de... | [
"11"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,087 | Two square tiles of area 9 are placed with one directly on top of the other. The top tile is then rotated about its center by an acute angle $\theta$. If the area of the overlapping region is 8 , compute $\sin \theta+\cos \theta$. | [
"In the diagram below, $O$ is the center of both squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$. Let $P_{1}, P_{2}, P_{3}, P_{4}$ and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ be the intersections of the sides of the squares as shown. Let $H_{A}$ be on $\\overline{A_{3} A_{4}}$ so that $\\angle A_{3} H_{A} O$ i... | [
"$\\frac{5}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,088 | Suppose that neither of the three-digit numbers $M=\underline{4} \underline{A} \underline{6}$ and $N=\underline{1} \underline{B} \underline{7}$ is divisible by 9 , but the product $M \cdot N$ is divisible by 9 . Compute the largest possible value of $A+B$. | [
"In order for the conditions of the problem to be satisfied, $M$ and $N$ must both be divisible by 3 , but not by 9 . Thus the largest possible value of $A$ is 5 , and the largest possible value of $B$ is 7 , so $A+B=\\mathbf{1 2}$."
] | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
3,089 | Let $T=12$. Each interior angle of a regular $T$-gon has measure $d^{\circ}$. Compute $d$. | [
"From the angle sum formula, $d^{\\circ}=\\frac{180^{\\circ} \\cdot(T-2)}{T}$. With $T=12, d=\\mathbf{1 5 0}$."
] | [
"150"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,090 | Suppose that $r$ and $s$ are the two roots of the equation $F_{k} x^{2}+F_{k+1} x+F_{k+2}=0$, where $F_{n}$ denotes the $n^{\text {th }}$ Fibonacci number. Compute the value of $(r+1)(s+1)$. | [
"$\\quad$ Distributing, $(r+1)(s+1)=r s+(r+s)+1=\\frac{F_{k+2}}{F_{k}}+\\left(-\\frac{F_{k+1}}{F_{k}}\\right)+1=\\frac{F_{k+2}-F_{k+1}}{F_{k}}+1=\\frac{F_{k}}{F_{k}}+1=\\mathbf{2}$."
] | [
"2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,091 | Let $T=2$. Compute the product of $-T-i$ and $i-T$, where $i=\sqrt{-1}$. | [
"Multiplying, $(-T-i)(i-T)=-(i+T)(i-T)=-\\left(i^{2}-T^{2}\\right)=1+T^{2}$. With $T=2,1+T^{2}=\\mathbf{5}$."
] | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,092 | Let $T=5$. Compute the number of positive divisors of the number $20^{4} \cdot 11^{T}$ that are perfect cubes. | [
"Let $N=20^{4} \\cdot 11^{T}=2^{8} \\cdot 5^{4} \\cdot 11^{T}$. If $m \\mid N$, then $m=2^{a} \\cdot 5^{b} \\cdot 11^{c}$ where $a, b$, and $c$ are nonnegative integers such that $a \\leq 8, b \\leq 4$, and $c \\leq T$. If $m$ is a perfect cube, then $a, b$, and $c$ must be divisible by 3 . So $a=0,3$, or $6 ; b=0$... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,094 | Let $T=72 \sqrt{2}$, and let $K=\left(\frac{T}{12}\right)^{2}$. In the sequence $0.5,1,-1.5,2,2.5,-3, \ldots$, every third term is negative, and the absolute values of the terms form an arithmetic sequence. Compute the sum of the first $K$ terms of this sequence. | [
"The general sequence looks like $x, x+d,-(x+2 d), x+3 d, x+4 d,-(x+5 d), \\ldots$ The sum of the first three terms is $x-d$; the sum of the second three terms is $x+2 d$; the sum of the third three terms is $x+5 d$, and so on. Thus the sequence of sums of terms $3 k-2,3 k-1$, and $3 k$ is an arithmetic sequence. N... | [
"414"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,095 | Let $A$ be the sum of the digits of the number you will receive from position 7 , and let $B$ be the sum of the digits of the number you will receive from position 9 . Let $(x, y)$ be a point randomly selected from the interior of the triangle whose consecutive vertices are $(1,1),(B, 7)$ and $(17,1)$. Compute the probability that $x>A-1$. | [
"Let $P=(1,1), Q=(17,1)$, and $R=(B, 7)$ be the vertices of the triangle, and let $X=(B, 1)$ be the foot of the perpendicular from $R$ to $\\overleftrightarrow{P Q}$. Let $M=(A-1,1)$ and let $\\ell$ be the vertical line through $M$; then the problem is to determine the fraction of the area of $\\triangle P Q R$ tha... | [
"$\\frac{79}{128}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,096 | Let $T=9.5$. If $\log _{2} x^{T}-\log _{4} x=\log _{8} x^{k}$ is an identity for all $x>0$, compute the value of $k$. | [
"Note that in general, $\\log _{b} c=\\log _{b^{n}} c^{n}$. Using this identity yields $\\log _{2} x^{T}=\\log _{2^{2}}\\left(x^{T}\\right)^{2}=$ $\\log _{4} x^{2 T}$. Thus the left hand side of the given equation simplifies to $\\log _{4} x^{2 T-1}$. Express each side in base 64: $\\log _{4} x^{2 T-1}=\\log _{64} ... | [
"27"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,097 | Let $T=16$. An isosceles trapezoid has an area of $T+1$, a height of 2 , and the shorter base is 3 units shorter than the longer base. Compute the sum of the length of the shorter base and the length of one of the congruent sides. | [
"Let $x$ be the length of the shorter base of the trapezoid. The area of the trapezoid is $\\frac{1}{2} \\cdot 2$. $(x+x+3)=T+1$, so $x=\\frac{T}{2}-1$. Drop perpendiculars from each vertex of the shorter base to the longer base, and note that by symmetry, the feet of these perpendiculars lie $\\frac{3}{2}=1.5$ uni... | [
"9.5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
3,098 | Let $T=10$. Susan flips a fair coin $T$ times. Leo has an unfair coin such that the probability of flipping heads is $\frac{1}{3}$. Leo gets to flip his coin the least number of times so that Leo's expected number of heads will exceed Susan's expected number of heads. Compute the number of times Leo gets to flip his coin. | [
"The expected number of heads for Susan is $\\frac{T}{2}$. If Leo flips his coin $N$ times, the expected number of heads for Leo is $\\frac{N}{3}$. Thus $\\frac{N}{3}>\\frac{T}{2}$, so $N>\\frac{3 T}{2}$. With $T=10$, the smallest possible value of $N$ is $\\mathbf{1 6}$."
] | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
3,099 | Let $T=1$. Dennis and Edward each take 48 minutes to mow a lawn, and Shawn takes 24 minutes to mow a lawn. Working together, how many lawns can Dennis, Edward, and Shawn mow in $2 \cdot T$ hours? (For the purposes of this problem, you may assume that after they complete mowing a lawn, they immediately start mowing the next lawn.) | [
"Working together, Dennis and Edward take $\\frac{48}{2}=24$ minutes to mow a lawn. When the three of them work together, it takes them $\\frac{24}{2}=12$ minutes to mow a lawn. Thus they can mow 5 lawns per hour. With $T=1$, they can mow $5 \\cdot 2=\\mathbf{1 0}$ lawns in 2 hours."
] | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,100 | Let T be a rational number. Compute $\sin ^{2} \frac{T \pi}{2}+\sin ^{2} \frac{(5-T) \pi}{2}$. | [
"Note that $\\sin \\frac{(5-T) \\pi}{2}=\\cos \\left(\\frac{\\pi}{2}-\\frac{(5-T) \\pi}{2}\\right)=\\cos \\left(\\frac{T \\pi}{2}-2 \\pi\\right)=\\cos \\frac{T \\pi}{2}$. Thus the desired quantity is $\\sin ^{2} \\frac{T \\pi}{2}+\\cos ^{2} \\frac{T \\pi}{2}=\\mathbf{1}$ (independent of $T$ )."
] | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,101 | Let $T=11$. Compute the value of $x$ that satisfies $\sqrt{20+\sqrt{T+x}}=5$. | [
"Squaring each side gives $20+\\sqrt{T+x}=25$, thus $\\sqrt{T+x}=5$, and $x=25-T$. With $T=11$, $x=14$."
] | [
"14"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
3,102 | The sum of the interior angles of an $n$-gon equals the sum of the interior angles of a pentagon plus the sum of the interior angles of an octagon. Compute $n$. | [
"Using the angle sum formula, $180^{\\circ} \\cdot(n-2)=180^{\\circ} \\cdot 3+180^{\\circ} \\cdot 6=180^{\\circ} \\cdot 9$. Thus $n-2=9$, and $n=11$."
] | [
"11"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
0 | 在 $\triangle A B C$ 中, $\sin A=\frac{\sqrt{2}}{2}$, 求 $\cos B+\sqrt{2} \cos C$ 的取值范围. | [
"根据题意, $A=\\frac{\\pi}{4}$ 或 $A=\\frac{3 \\pi}{4}$. 情形一: $A=\\frac{\\pi}{4}$. 此时\n\n$$\n\\begin{aligned}\n\\cos B+\\sqrt{2} \\cos C & =\\cos \\left(\\frac{3 \\pi}{4}-C\\right)+2 \\sqrt{2} \\cos C \\\\\n& =\\frac{\\sqrt{2}}{2} \\sin C+\\frac{\\sqrt{2}}{2} \\cos C \\\\\n& =\\sin \\left(C+\\frac{\\pi}{4}\\right),\n\\e... | [
"$(0,1] \\cup (2, \\sqrt{5}]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Trigonometric Functions | Math | Chinese |
1 | 对正整数 $n$ 及实数 $x(0 \leqslant x<n)$, 定义
$$
f(n, x)=(1-\{x\}) \cdot\left(\begin{array}{c}
{[x]} \\
n
\end{array}\right)+\{x\} \cdot\left(\begin{array}{c}
{[x]+1} \\
n
\end{array}\right),
$$
其中 $[x]$ 表示不超过实数 $x$ 的最大整数, $\{x\}=x-[x]$. 若整数 $m, n \geqslant 2$ 满足
$$
f\left(m, \frac{1}{n}\right)+f\left(m, \frac{2}{n}\right)+\cdots+f\left(m, \frac{m n-1}{n}\right)=123
$$
求 $f\left(n, \frac{1}{m}\right)+f\left(n, \frac{2}{m}\right)+\cdots+f\left(n, \frac{m n-1}{m}\right)$ 的值. | [
"对 $k=0,1, \\cdots, m-1$, 有\n\n$\\sum_{i=1}^{n-1} f\\left(m, k+\\frac{i}{n}\\right)=\\left(\\begin{array}{c}m \\\\ k\\end{array}\\right) \\cdot \\sum_{i=1}^{n=1}\\left(1-\\frac{i}{n}\\right)+\\left(\\begin{array}{c}m \\\\ k+1\\end{array}\\right) \\cdot \\sum_{i=1}^{n-1} \\frac{i}{n}=\\frac{n-1}{2} \\cdot\\left(\\be... | [
"$74$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
2 | 在平面直角坐标系中, 点 $A, B, C$ 在双曲线 $x y=1$ 上, 满足 $\triangle A B C$ 为等腰直角三角形. 求 $\triangle A B C$ 的面积的最小值. | [
"设 $A\\left(a, \\frac{1}{a}\\right)$, 平移坐标系使 $A$ 为原点, 此时双曲线方程为\n\n$$\nH^{\\prime}:(x+a)\\left(x+\\frac{1}{a}\\right)=1 \\Longleftrightarrow x y+\\frac{1}{a} x+a y=0,\n$$\n\n设 $B^{\\prime}(\\theta: r), C^{\\prime}\\left(\\theta+\\frac{\\pi}{2}: r\\right)$, 其中 $r>0$, 则 $\\triangle A B C$ 的面积 $S=\\frac{1}{2} r^{2}$. 根... | [
"$3 \\sqrt{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
4 | 给定整数 $n \geqslant 3$, 设 $a_{1}, a_{2}, \cdots, a_{2 n}, b_{1}, b_{2}, \cdots, b_{2 n}$ 是 $4 n$ 个非负实数, 满足
$$
a_{1}+a_{2}+\cdots+a_{2 n}=b_{1}+b_{2}+\cdots+b_{2 n}>0
$$
且对任意 $i=1,2, \cdots, 2 n$, 有 $a_{i} a_{i+2} \geqslant b_{i}+b_{i+1}$ (这里 $a_{2 n+1}=a_{1}, a_{2 n+2}=a_{2}$, $b_{2 n+1}=b_{1}$ ), 求 $a_{1}+a_{2}+\cdots+a_{2 n}$ 的最小值. | [
"在两组数之和相等的情况下由条件 $a_{i} a_{i+2} \\geqslant b_{i}+b_{i+1}$ 可以猜想两种情形: $2 \\cdot 2=2+2$以及 $0 \\cdot 0=0+0$. 记\n\n$$\n\\sum_{k=1}^{2 n} a_{k}=\\sum_{k=1}^{2 n} b_{k}=S, \\quad \\sum_{k=1}^{n} a_{2 k-1}=T\n$$\n\n不妨设 $T \\leqslant \\frac{1}{2} S$. 情形一 $n=3$. 取 $a_{i}=b_{i}=2(i=1,2, \\cdots, 2 n)$, 可得 $S=12$. 而\n\n$$\nT^{... | [
"$12 , 16$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
6 | 给定凸 20 边形 $P$. 用 $P$ 的 17 条在内部不相交的对角线将 $P$ 分割成 18 个三角形. 所得图形称为 $P$ 的一个三角剖分图. 对 $P$ 的任意一个三角剖分图 $T, P$ 的 20 条边以及添加的 17 条对角线均称为 $T$ 的边. $T$ 的任意 10 条两两无公共端点的边的集合称为 $T$的一个完美匹配. 当 $T$ 取遍 $P$ 的所有三角剖分图时, 求 $T$ 的完美匹配个数的最大值. | [
"将 20 边形换成 $2 n$ 边形，考虑一般的问题.\n\n对凸 $2 n$ 边形 $P$ 的一条对角线, 若其两侧各有奇数个 $P$ 的顶点, 称其为奇弦, 否则称为偶弦. 首先注意下述基本事实:\n\n引理一: 对 $P$ 的任意三角剖分图 $T, T$ 的完美匹配不含奇弦.\n\n引理一的证明: 如果完美匹配中有一条奇弦 $e_{1}$, 因为 $T$ 的一个完美匹配给出了 $P$ 的顶点集的一个配对划分, 而 $e_{1}$ 两侧各有奇数个顶点, 故该完美匹配中必有 $T$ 的另一条边 $e_{1}$, 端点分别在 $e_{1}$ 的两侧, 又 $P$ 是凸多边形, 故 $e_{1}$ 与 $e$ 在 $P$ 的... | [
"$89$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
7 | 在椭圆中, $A$ 为长轴的一个端点, $B$ 为短轴的一个端点, $F_{1}, F_{2}$ 为两个焦点. 若
$$
\overline{A F_{1}} \cdot \overrightarrow{A F_{2}}+\overrightarrow{B F_{1}} \cdot \overrightarrow{B F_{2}}=0
$$
求 $\tan \angle A B F_{1} \cdot \tan \angle A B F_{2}$ 的值. | [
"设 $A$ 为右顶点, $B$ 为上顶点, $F_{1}, F_{2}$ 分别为左、右焦点, $O$ 为 $F_{1} F_{2}$ 的中点.\n\n<img_4141>\n\n根据极化恒等式，有\n\n$$\n\\overline{A F_{1}} \\cdot \\overrightarrow{A F_{2}}+\\overrightarrow{B F_{1}} \\cdot \\overrightarrow{B F_{2}}=0 \\Longleftrightarrow\\left(A O^{2}-\\frac{1}{4} F_{1} F_{2}^{2}\\right)+\\left(B O^{2}-\\frac{1... | [
"$-\\frac{1}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
8 | 设正实数 $a, b, c$ 满足
$$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2
$$
求 $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$ 的最小值. | [
"根据题意, 有\n\n$$\na^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3\n$$\n\n于是\n\n$$\n\\begin{aligned}\n\\frac{1}{a}+\\frac{2}{b}+\\frac{3}{c} & \\geqslant \\frac{(1+2+3)^{2}}{a+2 b+3 c} \\\\\n& =\\frac{36}{a+(2 b-1)+(3 c-2)+3} \\\\\n& \\geqslant \\frac{36}{\\sqrt{3} \\cdot \\sqrt{a^{2}+(2 b-1)^{2}+(3 c-2)^{2}}+3} \\\\\n& =6,\n\\end{al... | [
"$6$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
12 | 给定整数 $n \geqslant 2$. 设 $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}>0$, 满足
$$
a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}
$$
且对任意 $i, j(1 \leqslant i<j \leqslant n)$ 均有 $a_{i} a_{j} \geqslant b_{i}+b_{j}$. 求 $a_{1}+a_{2}+\cdots+a_{n}$ 的最小值. | [
"设 $S=a_{1}+a_{2}+\\cdots+a_{n}=b_{1}+b_{2}+\\cdots+b_{n}$, 则有\n\n$$\n\\sum_{1 \\leqslant i<j \\leqslant n} a_{i} a_{j} \\geqslant \\sum_{1 \\leqslant i<j \\leqslant n}\\left(b_{i}+b_{j}\\right)=(n-1) S\n$$\n\n又\n\n$$\n\\sum_{1 \\leqslant i<j \\leqslant n} a_{i} a_{j} \\leqslant \\sum_{1 \\leqslant i<j \\leqslant n... | [
"$2 n$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
13 | 设 $a, b$ 为不超过 12 的正整数, 满足: 存在常数 $C$, 使得 $a^{n}+b^{n+9} \equiv C(\bmod 13)$ 对任意正整数 $n$ 成立. 求所有满足条件的有序数对 $(a, b)$. | [
"根据题意, 对任意正整数 $n$, 有\n\n$$\na^{n}+b^{n+9} \\equiv a^{n+3}+b^{n+12} \\quad(\\bmod 13)\n$$\n\n注意到 13 为素数, $a, b$ 均与 13 互素, 由费马小定理, 可得\n\n$$\na^{12} \\equiv b^{12} \\equiv 1 \\quad(\\bmod 13)\n$$\n\n因此取 $n=12$, 化简可得\n\n$$\n1+b^{9} \\equiv a^{3}+1 \\quad(\\bmod 13)\n$$\n\n故\n\n$$\nb^{9} \\equiv a^{3} \\quad(\\bmod 13)\... | [
"$(1,1),(4,4),(10,10),(12,12)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Number Theory | Math | Chinese |
14 | 在 $\triangle A B C$ 中, $B C=a, C A=b, A B=c$. 若 $b$ 是 $a$ 与 $c$ 的等比中项, 且 $\sin A$ 是
$\sin (B-A)$ 与 $\sin C$ 的等差中项, 求 $\cos B$ 的值. | [
"罗列条件 $b^{2}=a c$,\n\n\n\n$2 \\sin A=\\sin (B-A)+\\sin C=\\sin (B-A)+\\sin (B+A)=2 \\sin B \\cos A$.\n\n$\\cos A=\\frac{a}{b}=\\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.\n\n即 $2 a c=b^{2}+c^{2}-a^{2}$, 由 $b^{2}=a c$\n\n可得 $c^{2}-c a-a^{2}=0$. 于是 $\\frac{c}{a}=\\frac{\\sqrt{5}+1}{2}$.\n\n于是 $b^{2}=\\frac{\\sqrt{5}+1}{2} a^{2... | [
"$\\frac{\\sqrt{5}-1}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
15 | 在平面直角坐标系 $x O y$ 中, 圆 $\Omega$ 与抛物线 $\Gamma: y^{2}=4 x$ 恰有一个公共点, 且圆 $\Omega$ 与 $x$轴相切于 $\Gamma$ 的焦点 $F$. 求圆 $\Gamma$ 的半径. | [
"设抛物线上的点为 $A\\left(x_{0}, y_{0}\\right)$, 则该点处的切线为 $y_{0} y=2\\left(x+x_{0}\\right)$,\n\n其法向量 $\\vec{n}=\\left(2,-y_{0}\\right)$\n\n则线段 $A F$ 的中垂线为 $\\left(x_{0}-1\\right)\\left(x-\\frac{x_{0}+1}{2}\\right)+y_{0}\\left(y-\\frac{y_{0}}{2}\\right)=0$.\n\n其与 $x=1$ 的交点为 $\\Omega\\left(1, \\frac{\\left(x_{0}+1\\right)^{... | [
"$\\frac{4 \\sqrt{3}}{9}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
16 | 称一个复数列 $\left\{z_{n}\right\}$ 是” 有趣的”, 若 $\left|z_{1}\right|=1$, 且对任意正整数 $n$, 均有 $4 z_{n+1}^{2}+$ $2 z_{n} z_{n+1}+z_{n}^{2}=0$. 求最大的常数 $C$, 使得对一切有趣的数列 $\left\{z_{n}\right\}$ 及任意正整数 $m$, 均有 $\left|z_{1}+z_{2}+\cdots+z_{m}\right| \geqslant C$. | [
"根据题意, 有\n\n$$\nz_{n+1}=\\frac{-1 \\pm \\sqrt{3} \\mathrm{i}}{4} z_{n} \\Longleftrightarrow z_{n+1}=\\left( \\pm \\frac{2 \\pi}{3}: \\frac{1}{2}\\right) \\cdot z_{n}\n$$\n\n记 $q_{1}=\\left(\\frac{2 \\pi}{3}: \\frac{1}{2}\\right), q_{2}=\\left(-\\frac{2 \\pi}{3}: \\frac{1}{2}\\right)$. 取 $z_{1}=1$, 且\n\n$$\nz_{n+1}=... | [
"$\\frac{\\sqrt{3}}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
20 | 设 $V$ 是空间中 2019 个点构成的集合, 其中任意四点不共面. 某些点之间连有线段, 记 $E$ 为这些线段构成的集合. 试求最小的正整数 $n$, 满足条件: 若 $E$ 至少有 $n$ 个元素, 则 $E$ 一定含有 908 个二元子集, 其中每个二元子集中的两条线段有公共端点, 且任意两个二元子集的交为空集. | [
"为了叙述方便，称一个图中的两条相邻的边构成一个 “角”.\n\n先证明一个引理: 设 $G=(V, E)$ 是一个简单图, 且 $G$ 是连通的, 则 $G$ 含有 $\\left[\\frac{|E|}{2}\\right]$ 个两两无公共边的角 (这里 $[\\alpha]$ 表示实数 $\\alpha$ 的整数部分).\n\n引理的证明: 对 $E$ 的元素个数 $|E|$ 归纳证明. 当 $|E|=0,1,2,3$ 时, 结论显然成立. 下面假设 $|E| \\geqslant 4$, 并且结论在 $|E|$ 较小时均成立.\n\n只需证明, 在 $G$ 中可以选取两条边 $a, b$ 构成一个角, 在 $G$ 中... | [
"$2795$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
21 | 在椭圆 $\Gamma$ 中, $F$ 为一个焦点, $A, B$ 为两个顶点. 若 $|F A|=3,|F B|=2$, 求 $|A B|$ 的所有可能值. | [
"不妨设平面直角坐标系中椭圆 $\\Gamma$ 的标准方程为 $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, 并记 $c=$ $\\sqrt{a^{2}-b^{2}}$. 由对称性, 可设 $F$ 为 $\\Gamma$ 的右焦点.\n\n易知 $F$ 到 $\\Gamma$ 的左顶点的距离为 $a+c$, 到右顶点的距离为 $a-c$, 到上, 下顶点的距离均为 $a$. 分以下情况讨论:\n\n(1) $A, B$ 分别为左右顶点, 此时 $a+c=3, a-c=2$, 故 $|A B|=2 a=5$ （相应地, $b^{2}=(a+c)(a-c)=6, \\... | [
"$5, \\sqrt{7}, \\sqrt{17}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
22 | 设 $a, b, c$ 均大于 1 , 满足 $\left\{\begin{array}{l}\log a+\log _{b} c=3 \\ \log b+\log _{a} c=4\end{array}\right.$ 求 $\log a \cdot \log c$ 的最大值. | [
"设 $\\log _{a}=x, \\log _{b}=y, \\log _{c}=z$, 由 $a, b, c>1$ 可知 $x, y, z>0$.\n\n由条件及换底公式知 $x+\\frac{z}{y}=3, y+\\frac{z}{x}=4$, 即 $x y+z=3 y=4 x$.\n\n由此, 令 $x=3 t, y=4 t(t>0)$, 则 $z=4 x-x y=12 t-12 t^{2}$. 其中由 $z>0$ 可知 $t \\in(0,1)$.\n\n因此, 结合三元平均不等式得\n\n$\\log _{a} \\log _{c}=x z=3 t \\cdot 12 t(1-t)=18 \\cdot t^{... | [
"$\\frac{16}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
28 | 已知定义在 $\mathrm{R}^{*}$ 上的函数 $f(x)=\left\{\begin{array}{l}\left|\log _{3} x-1\right|, 0<x \leqslant 9 \\ 4-\sqrt{x}, x>9\end{array}\right.$ 设 $a, b, c$ 是三个互不相同的实数, 满足 $f(a)=f(b)=f(c)$, 求 $a b c$ 的取值范围. | [
"不妨假设 $a<b<c$, 由于 $f(x)$ 在 $(0,3]$ 上严格递减, 在 $[3,9]$ 上严格递增, 在 $[9,+\\infty)$上严格递减, 且 $f(3)=0, f(9)=1$, 故结合图像可知 $a \\in(0,3), b \\in(3,9), c \\in(9,+\\infty)$,并且 $f(a)=f(b)=f(c) \\in(0,1)$. 由 $f(a)=f(b)$ 得 $1-\\log _{3} a=\\log _{3} b-1$, 取 $\\log _{3} a+\\log _{3} b=2$, 因此 $a b=3^{2}=9$. 于是 $a b c=9 c$. 又 $0<f(c)=4-... | [
"$(81, 144)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
31 | 在平面直角坐标系 $x O y$ 中, 设 $A B$ 是抛物线 $y^{2}=4 x$ 的过点 $F(1,0)$ 的弦, $\triangle A O B$ 的外接圆交抛物线于点 $P$ (不同于点 $O, A, B$ ). 若 $P F$ 平分 $\angle A P B$, 求 $|P F|$ 的所有可能值. | [
"设 $A\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right), B\\left(\\frac{y_{2}^{2}}{4}, y_{2}\\right), P\\left(\\frac{y_{3}^{2}}{4}, y_{3}\\right)$, 由条件知 $y_{1}, y_{2}, y_{3}$ 两两不等且非零. 设直线 $A B$ 的方程为 $x=t y+1$, 与抛物线方程联立可得 $y^{2}-4 t y-4=0$, 故 $y_{1} y_{2}=-4$. (1)\n\n注意到 $\\triangle A O B$ 的外接圆过点 $O$, 可设该圆的方程为 $x^{2}+y^{2}+... | [
"$\\sqrt{13}-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
36 | 已知数列 $\left\{a_{n}\right\}: a_{1}=7, \frac{a_{n+1}}{a_{n}}=a_{n}+2, n=1,2,3, \cdots$. 求满足 $a_{n}>4^{2018}$ 的最小正整数 $n$. | [
"由 $\\frac{a_{n+1}}{a_{n}}=a_{n}+2$ 可知 $a_{n+1}+1=\\left(a_{n}+1\\right)^{2}$. 因此 $a_{n}+1=\\left(a_{1}+1\\right)^{2^{n-1}}=8^{2^{n-1}}=2^{3 \\times 2^{n-1}}$,\n故 $a_{n}=2^{3 \\times 2^{n-1}}-1$. 显然 $\\left\\{a_{n}\\right\\}$ 单调递增. 由于 $a_{11}=2^{3072}-1<2^{4036}=4^{2018}, a_{12}=$ $2^{6144}-1>2^{4036}=4^{2018}$, 故满... | [
"$12$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
40 | 设集合 $A=\{1,2, \cdots, n\}, X, Y$ 均为 $A$ 的非空子集 (允许 $X=Y$ ). $X$ 中的最大元与 $Y$ 中的最小元分别记为 $\max X, \min Y$. 求满足 $\max X>\min Y$ 的有序集合对 $(X, Y)$ 的数目. | [
"先计算满足 $\\max X \\leqslant \\min Y$ 的有序集合对 $(X, Y)$ 的数目. 对给定的 $m=\\max X$, 集合 $X$ 是集合 $\\{1,2, \\cdots, m-1\\}$ 的任意一个子集与 $\\{m\\}$ 的并，故并有 $2^{m-1}$ 种取法. 又 $\\min Y \\geqslant M$, 故 $Y$ 是 $\\{m, m+1, \\cdots, n\\}$ 的任意一个非空子集, 共有 $2^{n+1-m}-1$ 种取法. 因此, 满足 $\\max X \\leqslant \\min Y$ 的有序集合对 $(X, Y)$ 的数目是 $\\sum_{m=1}... | [
"$2^{2 n}-2^{n}(n+1)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
43 | 设 $x_{1}, x_{2}, x_{3}$ 是非负实数, 满足 $x_{1}+x_{2}+x_{3}=1$, 求 $\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)$的最小值和最大值. | [
"设题中代数式为 $M$, 则\n\n$$\n\\begin{aligned}\nM & =x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\\frac{10}{3} x_{1} x_{2}+\\frac{34}{15} x_{2} x_{3}+\\frac{26}{5} x_{3} x_{1} \\\\\n& =\\left(x_{1}+x_{2}+x_{3}\\right)^{2}+\\frac{4}{15}\\left(5 x_{1} x_{2}+x_{2} x_{3}+12 x_{3} x_{1}\\right) \\\\\n& =1+\\frac{4}{15}\\left(5 x_{1} x_{2}+x... | [
"$1 , \\frac{9}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
44 | 设复数 $z_{1}, z_{2}$ 满足 $\operatorname{Re}\left(z_{1}\right)>0, \operatorname{Re}\left(z_{2}\right)>0$, 且 $\operatorname{Re}\left(z_{1}^{2}\right)=\operatorname{Re}\left(z_{2}^{2}\right)=2$, 其中 $\operatorname{Re}(z)$ 表示复数 $z$ 的实部.
求 $\operatorname{Re}\left(z_{1} z_{2}\right)$ 的最小值; | [
"设 $z_{k}=x_{k}+\\mathrm{i} y_{k}, k=1,2$. 根据题意, 有 $x_{1}, x_{2}>0$, 且\n\n$$\nx_{1}^{2}-y_{1}^{2}=x_{2}^{2}-y_{2}^{2}=2 \\text {. }\n$$\n\n于是 $P_{1}\\left(x_{1}, y_{2}\\right)$ 和 $P_{2}\\left(x_{2}, y_{2}\\right)$ 都是双曲线 $H: x^{2}-y^{2}=2$ 右支上的点. 而\n\n$$\n\\begin{aligned}\n\\operatorname{Re}\\left(z_{1} z_{2}\\right... | [
"$2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Complex Numbers | Math | Chinese |
45 | 设复数 $z_{1}, z_{2}$ 满足 $\operatorname{Re}\left(z_{1}\right)>0, \operatorname{Re}\left(z_{2}\right)>0$, 且 $\operatorname{Re}\left(z_{1}^{2}\right)=\operatorname{Re}\left(z_{2}^{2}\right)=2$, 其中 $\operatorname{Re}(z)$ 表示复数 $z$ 的实部.
求 $\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right|$ 的最小值. | [
"由共轭复数的性质, 有\n\n$$\nm=\\left|z_{1}+2\\right|+\\left|\\overline{z_{2}}+2\\right|-\\left|\\overline{z_{1}}-z_{2}\\right|=\\left|z_{1}-(-2)\\right|+\\left|\\overline{z_{2}}-(-2)\\right|-\\left|z_{1}-\\overline{z_{2}}\\right|\n$$\n\n记 $F_{1}(-2,0), F_{2}(2,0)$ 为双曲线 $H$ 的左、右焦点, $P_{2}^{\\prime}$ 为 $P_{2}$ 关于 $x$ 轴的对称点（该... | [
"$4 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Complex Numbers | Math | Chinese |
47 | 设数列 $\left\{a_{n}\right\}$ 定义为 $a_{1}=1$,
$$
a_{n+1}= \begin{cases}a_{n}+n, & a_{n} \leqslant n, \\ a_{n}-n, & a_{n}>n,\end{cases} n = 1, 2, \cdots .
$$
求满足 $a_{r}<r \leqslant 3^{2017}$ 的正整数 $r$ 的个数. | [
"先计算数列的前几项:\n\n| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $a_{n}$ | 1 | 2 | 4 | 1 | 5 | 10 | 4 | 11 | 3 | 12 | 2 | 13 | 1 | 14 | 28 | 13 |... | [
"$\\frac{3^{2017}-2019}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
48 | 将 $33 \times 33$ 方格纸中每个小方格染三种颜色之一, 使得每种颜色的小方格的个数相等.若相邻两个小方格的颜色不同, 则称它们的公共边为 “分隔边”. 试求分隔边条数的最小值. | [
"如下图, 分隔边的条数 $L=56$.\n\n<img_4187>\n\n下面证明 $L \\geqslant 56$. 将方格纸的行从上至下依次记为 $A_{1}, A_{2}, \\cdots, A_{33}$, 列从左至右依次记为 $B_{1}, B_{2}, \\cdots, B_{33}$. 行 $A_{i}$ 中方格出现的颜色数记为 $n\\left(A_{i}\\right)$, 列 $B_{i}$ 中方格出现的颜色数记为 $n\\left(B_{i}\\right)$, 三种颜色分别记为 $c_{1}, c_{2}, c_{3}$, 对于一种颜色 $c_{j}$, 设 $n\\left(c_{j}\\rig... | [
"$56$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
50 | 设不等式 $\left|2^{x}-a\right|<\left|5-2^{x}\right|$ 对所有 $x \in[1,2]$ 成立, 求实数 $a$ 的取值范围. | [
"根据题意, 有\n\n$$\n\\forall x \\in[1,2],\\left|2^{x}-a\\right|<5-2^{x},\n$$\n\n即\n\n$$\n\\forall x \\in[1,2], 2^{x}-5<2^{x}-a<5-2^{x}\n$$\n\n也即\n\n$$\n\\forall x \\in[1,2], 2 \\cdot 2^{x}-5<a<5\n$$\n\n也即\n\n$$\n3<a<5 \\text {. }\n$$"
] | [
"$(3,5)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Inequality | Math | Chinese |
52 | 设数列 $\left\{a_{n}\right\}$ 是等差数列, 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=a_{n+1} a_{n+2}-a_{n}^{2}, n=1,2, \cdots$.
设数列 $\left\{a_{n}\right\},\left\{b_{n}\right\}$ 的公差均是 $d \neq 0$, 并且存在正整数 $s, t$, 使得 $a_{s}+b_{t}$ 是整数,求 $\left|a_{1}\right|$ 的最小值. | [
"根据题意, 有 $3 d^{2}=d$, 因此 $d=\\frac{1}{3}$, 于是\n\n$$\nb_{n}=\\frac{1}{3} n+a+\\frac{2}{9}=a_{n}+\\frac{2}{9}\n$$\n\n进而\n\n$$\n\\begin{aligned}\na_{s}+b_{t} & =a_{s}+a_{t}+\\frac{2}{9} \\\\\n& =2 a_{1}+\\frac{s+t-2}{3}+\\frac{2}{9}\n\\end{aligned}\n$$\n\n于是\n\n$$\n18 a_{1}=3\\left[3\\left(a_{s}+b_{t}\\right)-s-t+1\\r... | [
"$\\frac{1}{18}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
57 | 设 $a_{1}, a_{2}, \cdots, a_{20} \in\{1,2, \cdots, 5\}, b_{1}, b_{2}, \cdots, b_{20} \in\{1,2, \cdots, 10\}$, 集合 $X=\{(i, j) \mid$ $\left.1 \leqslant i<j \leqslant 20,\left(a_{i}-a_{j}\right)\left(b_{i}-b_{j}\right)<0\right\}$, 求 $X$ 的元素个数的最大值. | [
"考虑一组满足条件的正整数\n\n$$\n\\left\\{a_{1}, a_{2}, \\cdots, a_{20}, b_{1}, b_{2}, \\cdots, b_{20}\\right\\}\n$$\n\n对 $k=1,2, \\cdots, 5$, 设 $a_{1}, a_{2}, \\cdots, a_{20}$ 中取值为 $k$ 的数有 $t_{k}$ 个. 根据 $X$ 的定义, 当 $a_{i}=a_{j}$ 时, $(i, j) \\notin X$, 因此至少有 $\\sum_{k=1}^{5} \\mathrm{C}_{t_{k}}^{2}$ 个 $(i, j)$ 不在 $X$ 中. 注意到\n\n... | [
"$160$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
58 | 在 $\triangle A B C$ 中, 已知 $\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B}$. 求 $\sin C$ 的最大值. | [
"统一起点, 有\n\n$$\n(\\overrightarrow{C A}-\\overrightarrow{C B}) \\cdot \\overrightarrow{C A}+2(\\overrightarrow{C B}-\\overrightarrow{C A}) \\cdot \\overrightarrow{C B}=3 \\overrightarrow{C A} \\cdot \\overrightarrow{C B}\n$$\n\n即\n\n$$\n6 \\overrightarrow{C A} \\cdot \\overrightarrow{C B}=C A^{2}+2 C B^{2}\n$$\n\n也即... | [
"$\\frac{\\sqrt{7}}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
59 | 已知 $f(x)$ 是 $\mathbb{R}$ 上的奇函数, $f(1)=1$, 且对任意 $x<0$, 均有 $f\left(\frac{x}{x-1}\right)=x f(x)$. 求
$$
f(1) f\left(\frac{1}{100}\right)+f\left(\frac{1}{2}\right) f\left(\frac{1}{99}\right)+f\left(\frac{1}{3}\right) f\left(\frac{1}{98}\right)+\cdots+f\left(\frac{1}{50}\right) f\left(\frac{1}{51}\right)
$$
的值. | [
"令 $x=-\\frac{1}{n}, n \\in \\mathbb{N}^{*}$, 则\n\n$$\n\\frac{x}{x-1}=\\frac{-\\frac{1}{n}}{-\\frac{1}{n}-1}=\\frac{1}{n+1}\n$$\n\n\n\n于是有\n\n$$\nf\\left(\\frac{1}{n+1}\\right)=-\\frac{1}{n} f\\left(-\\frac{1}{n}\\right)=\\frac{1}{n} f\\left(\\frac{1}{n}\\right)\n$$\n\n记 $a_{n}=f\\left(\\frac{1}{n}\\right), n \\in ... | [
"$\\frac{2^{98}}{99 !}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
61 | 设实数 $a_{1}, a_{2}, \cdots, a_{2016}$ 满足 $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$. 求
$$
\left(a_{1}-a_{2}^{2}\right) \cdot\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right) \cdot\left(a_{2016}-a_{1}^{2}\right)
$$
的最大值. | [
"令原式为 $P$. 由于 $a_{i}-a_{i+1}^{2}>0, i=1,2, \\cdots, 2015$, 因此只需要考虑当 $a_{2016}-a_{1}^{2}>0$的情况, 记 $a_{2017}=a_{1}$, 则\n\n$$\n\\begin{aligned}\nP^{\\frac{1}{2016}} & \\leqslant \\frac{1}{2016} \\sum_{k=1}^{2016}\\left(a_{k}-a_{k+1}^{2}\\right) \\\\\n& =\\frac{1}{2016}\\left(\\sum_{k=1}^{2016} a_{k}-\\sum_{k=1}^{2016}... | [
"$\\frac{1}{4^{2016}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
63 | 给定空间中 10 个点, 其中任意四点不在一个平面上, 将某些点之间用线段相连, 若得到的图形中没有三角形也没有空间四边形, 试确定所连线段数目的最大值. | [
"记这 10 个点分别为 $P_{i}$ 且从 $P_{i}$ 点引出了 $a_{i}$ 条线段, 其中 $i=1,2, \\cdots, 10$. 这样图形中总共包含 $\\frac{1}{2} \\sum_{i=1}^{10} a_{i}$ 条线段和 $\\sum_{i=1}^{10} \\mathrm{C}_{a_{i}}^{2}$ 个角. 根据题意, 图形中没有空间四边形, 因此任何一个角都与一个点对 $\\left(P_{m}, P_{n}\\right)$ 一一对应, 且不存在线段 $P_{m} P_{n}$. 这样就有\n\n$$\n\\frac{1}{2} \\sum_{i=1}^{10} a_{i}+\\s... | [
"$15$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
65 | 若实数 $a, b, c$ 满足 $2^{a}+4^{b}=2^{c}, 4^{a}+2^{b}=4^{c}$, 求 $c$ 的最小值. | [
"记 $x=2^{a}, y=2^{b}, z=2^{c}$, 则\n\n$$\n\\left\\{\\begin{array}{l}\nx+y^{2}=z \\\\\nx^{2}+y=z^{2}\n\\end{array}\\right.\n$$\n\n消元得\n\n$$\n\\left(z-y^{2}\\right)^{2}+y=z^{2}\n$$\n\n整理得\n\n$$\nz=\\frac{y^{2}}{2}+\\frac{1}{2 y}=\\frac{y^{2}}{2}+\\frac{1}{4 y}+\\frac{1}{4 y} \\geqslant 3 \\sqrt[3]{\\frac{y^{2}}{2} \\c... | [
"$\\log _{2} 3-\\frac{5}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
66 | 设 $a_{1}, a_{2}, a_{3}, a_{4}$ 是 4 个有理数, 使得
$$
\left\{a_{i} a_{j} \mid 1 \leqslant i<j \leqslant 4\right\}=\left\{-24,-2,-\frac{3}{2},-\frac{1}{8}, 1,3\right\},
$$
求 $a_{1}+a_{2}+a_{3}+a_{4}$ 的值. | [
"设 $a_{1}, a_{2}, a_{3}, a_{4}$ 的绝对值从小到大排列, 则\n\n$$\n\\left\\{\\begin{array}{l}\na_{1} a_{2}=-\\frac{1}{8}, \\\\\na_{1} a_{3}=1, \\\\\na_{2} a_{4}=3, \\\\\na_{3} a_{4}=-24, \\\\\n\\left\\{a_{2} a_{3}, a_{1} a_{4}\\right\\}=\\left\\{-2,-\\frac{3}{2}\\right\\},\n\\end{array}\\right.\n$$\n\n\n\n解得\n\n$$\n\\left(a_{1},... | [
"$ -\\frac{9}{4}, \\frac{9}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
67 | 在平面直角坐标系 $x O y$ 中, $F_{1}, F_{2}$ 分别是椭圆 $\frac{x^{2}}{2}+y^{2}=1$ 的左、右焦点, 设不经过焦点 $F_{1}$ 的直线 $l$ 与椭圆交于两个不同的点 $A, B$, 焦点 $F_{2}$ 到直线 $l$ 的距离为 $d$. 如果直线 $A F_{1}, l, B F_{1}$ 的斜率依次成等差数列, 求 $d$ 的取值范围. | [
"根据已知, 有 $F_{1}(-1,0), F_{2}(1,0)$.\n\n设 $l: y=k x+m, A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 则联立直线与椭圆方程得\n\n$$\n\\left(2 k^{2}+1\\right) x^{2}+4 k m x+2 m^{2}-2=0\n$$\n\n判别式\n\n$$\n\\Delta=8\\left(2 k^{2}+1-m^{2}\\right)>0\n$$\n\n即\n\n$$\n2 k^{2}+1>m^{2}\n$$\n\n由题意直线 $A F_{1} 、 l 、 B F_{1}$ 的斜... | [
"$(\\sqrt{3}, 2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Plane Geometry | Math | Chinese |
74 | 数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbb{N}^{*}\right)$. 求正整数 $m$, 使得
$$
\sin a_{1} \cdot \sin a_{2} \cdots \sin a_{m}=\frac{1}{100}
$$ | [
"由已知条件可知, 对任意正整数 $n$,\n\n$$\na_{n+1} \\in\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)\n$$\n\n且\n\n$$\n\\tan a_{n+1}=\\sec a_{n} , (1)\n$$\n\n由于 $\\sec a_{n}>0$, 故\n\n$$\na_{n+1} \\in\\left(0, \\frac{\\pi}{2}\\right)\n$$\n\n由 (1) 得,\n\n$$\n\\tan ^{2} a_{n+1}=\\sec ^{2} a_{n}=1+\\tan ^{2} a_{n},\n$$\n\n故\n\n$$\n\... | [
"$3333$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
78 | 设 $S=\{1,2,3, \cdots, 100\}$. 求最大的整数 $k$, 使得 $S$ 有 $k$ 个互不相同的非空子集, 具有性质：对这 $k$ 个子集中任意两个不同子集，若它们的交集非空，则它们交集中的最小元素与这两个子集中的最大元素均不相同. | [
"对有限非空实数集 $A$, 用 $\\min A$ 与 $\\max A$ 分别表示 $A$ 的最小元素与最大元素.\n\n考虑 $S$ 的所有包含 1 且至少有两个元素的子集, 一共 $2^{99}-1$ 个.\n\n因为\n\n$$\n\\min \\left(A_{i} \\cap A_{j}\\right)=1<\\max A_{i}\n$$\n\n所以它们显然满足要求. 故 $k$ 可以取到 $2^{99}-1$.\n\n下面证明 $k \\geqslant 2^{99}$ 时不存在满足要求的 $k$ 个子集.\n\n我们用数学归纳法证明: 对整数 $n \\geqslant 3$, 在集合 $\\{1,2, \... | [
"$2^{99}-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
85 | 一次考试共有 $m$ 道试题, $n$ 个学生参加, 其中 $m, n \geqslant 2$ 为给定的整数. 每道题的得分规则是: 若该题恰有 $x$ 个学生没有答对, 则每个答对该题的学生得分 $x$ 分, 未答对的学生得零分. 每个学生总分为其 $m$ 道题的得分总和. 将所有学生总分从高到低排列为 $p_{1} \geqslant p_{2} \geqslant \cdots \geqslant p_{n}$, 求 $p_{1}+p_{n}$ 的最大可能值. | [
"对任意的 $k=1,2, \\cdots, m$, 设第 $k$ 题没有答对者有 $x_{k}$ 人，则第 $k$ 题答对者有 $n-x_{k}$ 人, 由得分规则知, 这 $n-x_{k}$ 个人在第 $k$ 题均得到 $x_{k}$ 分. 设 $n$ 个学生得分之和为 $S$, 则有\n\n$$\n\\sum_{i=1}^{p} p_{i}=S=\\sum_{k=1}^{m} x_{k}\\left(n-x_{k}\\right)=n \\sum_{k=1}^{m} x_{k}-\\sum_{k=1}^{m} x_{k}^{2}\n$$\n\n因为每一个人在第 $k$ 道题上至多得 $x_{k}$ 分, 故\n\n$$... | [
"$m(n-1)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
87 | 已知函数 $f(x)=a \sin x-\frac{1}{2} \cos 2 x+a-\frac{3}{a}+\frac{1}{2}, a \in \mathbb{R}$ 且 $a \neq 0$.
若对任意 $x \in \mathbb{R}$, 都有 $f(x) \leqslant 0$, 求 $a$ 的取值范围; | [
"因为\n\n$$\nf(x)=\\sin ^{2} x+a \\sin x+a-\\frac{3}{a},\n$$\n\n令 $t=\\sin x(-1 \\leqslant t \\leqslant 1)$, 则\n\n$$\ng(t)=t^{2}+a t+a-\\frac{3}{a} .\n$$\n\n对任意 $x \\in \\mathbb{R}, f(x) \\leqslant 0$ 恒成立的充要条件是\n\n$$\n\\left\\{\\begin{array}{l}\ng(-1)=1-\\frac{3}{a} \\leqslant 0 \\\\\ng(1)=1+2 a-\\frac{3}{a} \\leqsla... | [
"$(0,1]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
88 | 已知函数 $f(x)=a \sin x-\frac{1}{2} \cos 2 x+a-\frac{3}{a}+\frac{1}{2}, a \in \mathbb{R}$ 且 $a \neq 0$.
若 $a \geqslant 2$, 且存在 $x \in \mathbb{R}$, 使得 $f(x) \leqslant 0$, 求 $a$ 的取值范围. | [
"因为 $a \\geqslant 2$, 所以 $-\\frac{a}{2} \\leqslant-1$, 因此\n\n$$\ng(t)_{\\min }=g(-1)=1-\\frac{3}{a},\n$$\n\n故\n\n$$\nf(x)_{\\min }=1-\\frac{3}{a}\n$$\n\n\n\n于是, 存在 $x \\in \\mathbb{R}$, 使得 $f(x) \\leqslant 0$ 的充要条件是\n\n$$\n1-\\frac{3}{a} \\leqslant 0\n$$\n\n解得 $0<a \\leqslant 3$."
] | [
"$[2,3]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
89 | 已知数列 $\left\{a_{n}\right\}$ 的各项均为非零实数, 且对于任意的正整数 $n$, 都有
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} .
$$
当 $n=3$ 时, 求所有满足条件的数列 $a_{1}, a_{2}, a_{3}$; | [
"当 $n=1$ 时,\n\n$$\na_{1}^{2}=a_{1}^{3},\n$$\n\n由 $a_{1} \\neq 0$, 得 $a_{1}=1$.\n\n当 $n=2$ 时,\n\n$$\n\\left(1+a_{2}\\right)^{2}=1+a_{2}^{3},\n$$\n\n由 $a_{2} \\neq 0$, 得 $a_{2}=2$ 或 $a_{2}=-1$.\n\n当 $n=3$ 时,\n\n$$\n\\left(1+a_{2}+a_{3}\\right)^{2}=1+a_{2}^{3}+a_{3}^{3},\n$$\n\n若 $a_{2}=2$, 得 $a_{3}=3$ 或 $a_{3}=-2$;\n... | [
"$(1,2,3),(1,2,-2),(1,-1,1)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Sequence | Math | Chinese |
98 | 设函数 $f(x)=|\lg (x+1)|$, 实数 $a, b(a<b)$ 满足 $f(a)=f\left(-\frac{b+1}{b+2}\right), f(10 a+6 b+21)=$ $4 \lg 2$, 求 $a, b$ 的值. | [
"因为 $f(a)=f\\left(-\\frac{b+1}{b+2}\\right)$, 所以\n\n$$\n\\begin{aligned}\n|\\lg (a+1)| & =\\left|\\lg \\left(-\\frac{b+1}{b+2}+1\\right)\\right| \\\\\n& =\\left|\\lg \\left(\\frac{1}{b+2}\\right)\\right| \\\\\n& =|\\lg (b+2)|,\n\\end{aligned}\n$$\n\n所以\n\n$$\na+1=b+2\n$$\n\n或\n\n$$\n(a+1)(b+2)=1.\n$$\n\n又因为 $a<b$, ... | [
"$-\\frac{2}{5}, -\\frac{1}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
99 | 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=2 t-3(t \in \mathbb{R}$ 且 $t \neq \pm 1), a_{n+1}=\frac{\left(2 t^{n+1}-3\right) a_{n}+2(t-1) t^{n}-1}{a_{n}+2 t^{n}-1}(n \in$ $\left.\mathbb{N}^{*}\right)$.
求数列 $\left\{a_{n}\right\}$ 的通项公式; | [
"由原式变形得\n\n$$\na_{n+1}=\\frac{2\\left(t^{n+1}-1\\right)\\left(a_{n}+1\\right)}{a_{n}+2 t^{n}-1}-1,\n$$\n\n\n\n则\n\n$$\n\\frac{a_{n+1}+1}{t^{n+1}-1}=\\frac{2\\left(a_{n}+1\\right)}{a_{n}+2 t^{n}-1}=\\frac{\\frac{2\\left(a_{n}+1\\right)}{t^{n}-1}}{\\frac{a_{n}+1}{t^{n}-1}+2}\n$$\n\n记 $\\frac{a_{n}+1}{t^{n}-1}=b_{n}$,... | [
"$a_{n}=\\frac{2\\left(t^{n}-1\\right)}{n}-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Sequence | Math | Chinese |
106 | 设 $A$ 是一个 $3 \times 9$ 的方格表, 在每一个小方格内各填一个正整数. 称 $A$ 中的一个 $m \times n(1 \leqslant m \leqslant 3,1 \leqslant n \leqslant 9)$ 方格表为 “好矩形”, 若它的所有数的和为 10 的倍数.称 $A$ 中的一个 $1 \times 1$ 的小方格为 “坏格”, 若它不包含于任何一个 “好矩形”. 求 $A$ 中 “坏格” 个数的最大值. | [
"首先证明 $A$ 中 “坏格” 不多于 25 个.\n\n用反证法, 假设结论不成立, 则方格表 $A$ 中至多有 1 个小方格不是 “坏格”, 由表格的对称性, 不妨设此时第 1 行都是 “坏格”.\n\n设方格表 $A$ 第 $i$ 列从上到下填的数依次为 $a_{i}, b_{i}, c_{i}, i=1,2, \\cdots, 9$, 记\n\n$$\nS_{k}=\\sum_{i=1}^{k} a_{i}, T_{k}=\\sum_{i=1}^{k}\\left(b_{i}+c_{i}\\right), k=0,1,2, \\cdots, 9\n$$\n\n这里 $S_{0}=T_{0}=0$.\n\n我们证明:... | [
"$25$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
107 | 设数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和 $S_{n}=2 a_{n}-1(n=1,2, \cdots)$, 数列 $\left\{b_{n}\right\}$ 满足 $b_{1}=3, b_{k+1}=a_{k}+b_{k}(k=1$, $2, \cdots)$. 求数列 $\left\{b_{n}\right\}$ 的前 $n$ 项和. | [
"$\\mathrm{a_{1}}=2 \\mathrm{a_{1}}-1, \\mathrm{a_{1}}=1$;\n\n$a_{n}=\\left(2 a_{n}-1\\right)-\\left(2 a_{n-1}-1\\right)=2 a_{n}-2 a_{n-1}, \\Rightarrow a_{n}=2 a_{n-1} . \\Rightarrow\\left\\{a_{n}\\right\\}$是以 1 为首项, 2 为公比的等比数列. $a_{n}=2^{n-1}$.\n\n$b_{k+1}-b_{k}=2^{k-1}, \\Rightarrow b_{n}-b_{1}=\\left(b_{n}-b_{n... | [
"$2^{n}+2n-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
108 | 求实数 $a$ 的取值范围, 使得对任意实数 $x$ 和任意 $\theta \in\left[0, \frac{\pi}{2}\right]$, 恒有
$$
(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8}
$$ | [
"令$\\sin\\theta + \\cos\\theta=u$，则$2\\sin\\theta\\cos\\theta=u^{2} - 1$。当$\\theta\\in\\left[0, \\frac{\\pi}{2}\\right]时，u\\in\\left[0, \\sqrt{2}\\right]。$\n\n并记 $f(x)=(x+3+2 \\sin \\theta \\cos \\theta)^{2}+(x+a s i n \\theta+a c o s \\theta)^{2}$.\n\n$\\therefore f(x)=\\left(x+2+u^{2}\\right)^{2}+(x+a u)^{2}=2 x^... | [
"$\\left(-\\infty, \\sqrt{6}\\right] \\cup\\left[\\frac{7}{2},+\\infty\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
110 | 设 $S_{n}=1 + 2 + 3 + \cdot +n, n \in \mathbf { N }$ 求 $f(n)=\frac{S_{n}}{(n+32) S_{n+1}}$ 的最大值. | [
"由已知, 对任何 $\\mathbf{n} \\in \\mathbf{N}$, 有 $\\mathbf{f}(\\mathbf{n})=\\frac{S_{n}}{(n+32) S_{N+1}}=\\frac{S_{n}}{(n+32)(n+2)}$\n\n$=\\frac{n}{n^{2}+34 n+64}=\\frac{1}{n+34+\\frac{64}{n}} \\quad$ 又因 $\\mathbf{n}+\\frac{64}{n}+34 \\geq 2 \\sqrt{n \\cdot \\frac{64}{n}}+34=50$,\n\n故对任何 $\\mathbf{n} \\in \\mathbf{N}$, ... | [
"$\\frac{1}{50}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
111 | 若函数 $f(x)=-\frac{1}{2} x^{2}+\frac{13}{2}$ 在区间 $[\boldsymbol{a}, \boldsymbol{b}]$ 上的最小值为 $2 \boldsymbol{a}$ 最大值为 $2 \boldsymbol{b}$, 求 $[a, b]$. | [
"化三种情况讨论区间 [a, b].\n\n(1) 若 $0 \\leq a<b$, 则 $f(x)$ 在 $[a, b]$ 上单调递减, 故 $f(a)=2 b, f(b)=2 a$ 于是有 $\\left\\{\\begin{array}{l}2 b=-\\frac{1}{2} a^{2}+\\frac{13}{2} \\\\ 2 a=-\\frac{1}{2} b^{2}+\\frac{13}{2}\\end{array}\\right.$, 解之得 $[a, b]=[1,3]$,\n\n(2) 若 $\\mathrm{a}<0<b, f$( $x)$ 在 $[\\mathrm{a}, \\mathrm{b}]$ 上单... | [
"$[1,3] ，\\left[-2-\\sqrt{17}, \\frac{13}{4}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
112 | 给定 $A(-2,2)$, 已知 $B$ 是椭圆 $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ 上的动点, $F$ 是左焦点, 当 $|A B|+\frac{5}{3}|B F|$取最小值时, 求 $B$ 的坐标. | [
"记椭圆的半长轴、半短轴、半焦距分别为 $a 、 b 、 c$, 离心率为 $e$. 则 $a=5, b=4, c=3, e=\\frac{3}{5}$, 左准线为 $\\mathrm{x}=-\\frac{25}{3}$, 过点 $\\mathrm{B}$ 作左准线 $\\mathrm{x}=-\\frac{25}{3}$ 的垂线, 垂足为 $\\mathrm{N}$, 过 $A$ 作此准线的垂线, 垂足为$M$.由椭圆定义，$|BN|=\\frac{5}{3}|BF|$.\n\n于是, $|A B|+\\frac{5}{3}|B F|=|A B|+|B N| \\geqslant|AM| \\mid$ (定值), 等号成... | [
"$\\left(-\\frac{5 \\sqrt{3}}{2}, 2\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
113 | 给定正整数 $n$ 和正数 $M$, 对于满足条件 $a_{1}^{2}+a_{n+1}^{2} \leqslant M$ 的所有等差数列 $a_{1}, a_{2}, a_{3}, \cdots$,试求 $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$ 的最大值。 | [
"设公差为 $\\mathrm{d}, a_{n+1}=\\mathbf{\\alpha}$, 则 $\\mathrm{S}=a_{n+1}+a_{n+2}+\\cdots+a_{2 n+1}=(\\mathbf{n}+\\mathbf{1}) \\mathbf{\\alpha}+\\frac{n(n+1)}{2} \\mathrm{~d}$\n\n故 $\\alpha+\\frac{n d}{2}=\\frac{S}{n+1}$.\n\n$$\n\\begin{aligned}\nM & \\geqslant a_{1}^{2}+a_{n+1}^{2}=(\\alpha-n d)^{2}+\\alpha^{2} \\\\\... | [
"$\\frac{\\sqrt{10}}{2}(n+1) \\sqrt{M}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
116 | 设直线 $l: y=k x+m$ （其中 $k, m$ 为整数）与椭圆 $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ 交于不同两点 $A$, $B$, 与双曲线 $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ 交于不同两点 $C, D$, 存在直线 $l$, 使得向量 $\overrightarrow{A C}+\overrightarrow{B D}=0$, 请指出这样的直线有多少条? | [
"由 $\\left\\{\\begin{array}{l}y=k x+m \\\\ \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1\\end{array}\\right.$ 消去 $y$ 化简整理得\n\n$\\left(3+4 k^{2}\\right) x^{2}+8 k m x+4 m^{2}-48=0$\n\n设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 则 $x_{1}+x_{2}=-\\frac{8 k m}{3+4 k^{2}}$\n\n$\\Delta_{1}=(8 k m)^{2}-4\\left... | [
"$9$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
118 | 已知 $p, q(q \neq 0)$ 是实数, 方程 $x^{2}-p x+q=0$ 有两个实根 $\alpha, \beta$, 数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=p, \quad a_{2}=p^{2}-q, \quad a_{n}=p a_{n-1}-q a_{n-2}(n=3,4, \cdots)$
若 $p=1, q=\frac{1}{4}$, 求 $\left\{a_{n}\right\}$ 的前 $n$ 项和. | [
"由韦达定理知 $\\alpha \\cdot \\beta=q \\neq 0$ ，又 $\\alpha+\\beta=p$ ，所以\n\n$a_{1}=\\alpha+\\beta, a_{2}=\\alpha^{2}+\\beta^{2}+\\alpha \\beta$.\n\n特征方程 $\\lambda^{2}-p \\lambda+q=0$ 的两个根为 $\\alpha, \\beta$.\n\n当 $\\alpha=\\beta \\neq 0$ 时, 通项 $a_{n}=\\left(A_{1}+A_{2} n\\right) \\alpha^{n}(n=1,2, \\ldots)$ 由 $a_{1}=2 \... | [
"$3-\\frac{n+3}{2^{n}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
119 | 求函数 $y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$ 的最大和最小值. | [
"函数的定义域为 $[0,13]$. 因为\n\n$$\n\\begin{aligned}\ny=\\sqrt{x}+\\sqrt{x+27}+\\sqrt{13-x} & =\\sqrt{x+27}+\\sqrt{13+2 \\sqrt{x(13-x)}} \\\\\n& \\geqslant \\sqrt{27}+\\sqrt{13} \\\\\n& =3 \\sqrt{3}+\\sqrt{13}\n\\end{aligned}\n$$\n\n当 $x=0$ 时等号成立. 故 $y$ 的最小值为 $3 \\sqrt{3}+\\sqrt{13}$.\n\n又由柯西不等式得\n\n$$\n\\begin{aligned}\n... | [
"$11, 3 \\sqrt{3}+\\sqrt{13}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
127 | 已知函数 $f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$, 当 $0 \leq x \leq 1$ 时, $\left|f^{\prime}(x)\right| \leq 1$,试求 $a$ 的最大值. | [
"$f^{\\prime}(x)=3 a x^{2}+2 b x+c$,\n\n由 $\\left\\{\\begin{array}{l}f^{\\prime}(0)=c, \\\\ f^{\\prime}\\left(\\frac{1}{2}\\right)=\\frac{3}{4} a+b+c, \\\\ f^{\\prime}(1)=3 a+2 b+c\\end{array}\\right.$\n\n$$\n3 a=2 f^{\\prime}(0)+2 f^{\\prime}(1)-4 f^{\\prime}\\left(\\frac{1}{2}\\right)\n$$\n\n所以 $3|a|=\\left|2 f^{... | [
"$\\frac{8}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
128 | 已知抛物线 $y^{2}=6 x$ 上的两个动点 $A\left(x_{1}, y_{1}\right)$ 和 $B\left(x_{2}, y_{2}\right)$, 其中 $x_{1} \neq x_{2}$ 且 $x_{1}+x_{2}=4$. 线段 $A B$ 的垂直平分线与 $x$ 轴交于点 $C$, 求 $\triangle A B C$ 面积的最大值. | [
"设线段 $A B$ 的中点为 $M\\left(x_{0}, y_{0}\\right)$, 则\n\n$$\n\\begin{aligned}\n& x_{0}=\\frac{x_{1}+x_{2}}{2}=2, y_{0}=\\frac{y_{1}+y_{2}}{2}, \\\\\n& k_{A B}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{y_{2}-y_{1}}{\\frac{y_{2}^{2}}{6}-\\frac{y_{1}^{2}}{6}}=\\frac{6}{y_{2}+y_{1}}=\\frac{3}{y_{0}} .\n\\end{aligned}\n$$\n\n... | [
"$\\frac{14}{3}, \\sqrt{7}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
133 | 一种密码锁的密码设置是在正 $n$ 边形 $A_{1} A_{2} \cdots A_{n}$ 的每个顶点处赋值 0 和 1 两个数中的一个, 同时在每个顶点处涂染红、蓝两种颜色之一, 使得任意相邻的两个顶点的数字或颜色中至少有一个相同．问：该种密码锁共有多少种不同的密码设置? | [
"对于该种密码锁的一种密码设置, 如果相邻两个顶点上所赋值的数字不同, 在它们所在的边上标上 $a$, 如果颜色不同, 则标上 $b$, 如果数字和颜色都相同, 则标上 $c$. 于是对于给定的点 $A_{1}$ 上的设置 (共有 4 种), 按照边上的字母可以依次确定点 $A_{2}, A_{3}, \\cdots, A_{n}$ 上的设置. 为了使得最终回到 $A_{1}$ 时的设置与初始时相同, 标有 $a$ 和 $b$ 的边都是偶数条. 所以这种密码锁的所有不同的密码设置方法数等于在边上标记 $a, b, c$, 使得标有 $a$ 和 $b$ 的边都是偶数条的方法数的4倍。\n\n设标有 $a$ 的边有 $2 i$ ... | [
"$3^{n}+1 , 3^{n}+3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
134 | 设正三棱锥 $P-A B C$ 的高为 $P O, M$ 为 $P O$ 的中点, 过 $A M$ 作与棱 $B C$ 平行的平面, 将三棱锥截为上、下两部分, 试求此两部分的体积比$\frac{V_{上}}{V_{下}}$的值。 | [
"$M$ 是 $P O$ 中点, 延长 $A O$ 与 $B C$ 交于点 $D$, 则 $D$ 为 $B C$ 中点, 连 $P D$, 由于 $A M$在平面 $P A D$ 内, 故延长 $A M$ 与 $P D$ 相交, 设交点为 $F$. 题中截面与面 $P B C$ 交于过 $F$ 的直线 $G H, G 、 H$ 分别在 $P B 、 P C$ 上. 由于 $B C / /$ 截面 $A G H, \\therefore G H / / B C$.\n\n在面 $P A D$ 中, $\\triangle P O D$ 被直线 $A F$ 截, 故 $\\frac{P M}{M O} \\cdot \\frac... | [
"$\\frac{4}{21}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
135 | 设 $S=\{1,2, \cdots, n\}, A$ 为至少含有两项的公差为正的等差数列, 其项都在 $S$ 中,且添加 $S$ 的其他元素于 $A$ 后不能构成与 $A$ 有相同公差的等差数列. 求这种 $A$ 的个数 (这里只有两项的数列也看作等差数列). | [
"易知公差 1 $\\leqslant d \\leqslant n-1$.\n\n设 $n=2 k, d=1$ 或 $d=n-1$ 时, 这样的 $A$ 只有 1 个, $d=2$ 或 $d=n-2$ 时, 这样的数列只有 2 个, $d=3$ 或 $n-3$ 时这样的数列只有 3 个, $\\cdots \\cdots \\cdot, d=k-1$ 或 $k+1$ 时, 这样的数列有 $k-1$ 个, $d=\\boldsymbol{k}$ 时, 这样的数列有 $\\boldsymbol{k}$ 个.\n\n$\\therefore$ 这样的数列共有 $(1+2+\\cdots+k) \\times 2-k=k^{2}=... | [
"$\\frac{1}{4} n^{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
137 | 解不等式
$$
\log _{2}\left(x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1\right)<1+\log _{2}\left(x^{4}+1\right) .
$$ | [
"由 $1+\\log _{2}\\left(x^{4}+1\\right)=\\log _{2}\\left(2 x^{4}+2\\right)$, 且 $\\log _{2} y$ 在 $(0,+\\infty)$ 上为增函数, 故原不等式等价于 $x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1<2 x^{4}+2$.\n\n即 $x^{12}+3 x^{10}+5 x^{8}+3 x^{6}-2 x^{4}-1<0$.\n\n分组分解\n\n$$\n\\begin{gathered}\nx^{12}+x^{10}-x^{8} \\\\\n+2 x^{10}+2 x^{8}-2 x^{6} \\\\\... | [
"$\\left(-\\sqrt{\\frac{\\sqrt{5}-1}{2}}, \\sqrt{\\frac{\\sqrt{5}-1}{2}}\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
144 | 长为 $\sqrt{2}$, 宽为 1 的矩形, 以它的一条对角线所在的直线为轴旋转一周, 求得到的旋转体的体积. | [
"过轴所在对角线 $B D$ 中点 $O$ 作 $M N \\perp B D$ 交边 $A D 、 B C$ 于 $M 、 N$,作 $A E \\perp B D$ 于 $E$,\n\n则 $\\triangle A B D$ 旋转所得旋转体为两个有公共底面的圆锥, 底面半径 $A E=\\frac{\\sqrt{2}}{\\sqrt{3}}=$ $\\frac{\\sqrt{6}}{3}$. 其体积 $V=\\frac{\\pi}{3}\\left(\\frac{\\sqrt{6}}{3}\\right)^{2} \\cdot \\sqrt{3}=\\frac{2 \\sqrt{3}}{9} \\pi$. 同样,\n\... | [
"$\\frac{23}{72} \\sqrt{3 \\pi}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
151 | 将 2006 表示成 5 个正整数 $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ 之和. 记 $S=\sum_{1 \leqslant i<j \leq 5} x_{i} x_{j}$. 问:当 $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ 分别取何值时, $\mathrm{S}$ 取到最大值。 | [
"首先这样的 $S$ 的值是有界集, 故必存在最大值与最小值. 若 $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, 且使 $S=\\sum_{1 \\leq i<j \\leq 5} x_{i} x_{j}$ 取到最大值, 则必有\n\n$$\n\\left|x_{i}-x_{j}\\right| \\leq 1, \\quad(1 \\leq i, j \\leq 5) \\tag{1}\n$$\n\n事实上, 假设 (1) 不成立, 不妨假设 $x_{1}-x_{2} \\geq 2$. 则令 $x_{1}^{\\prime}=x_{1}-1$,\n\n$x_{2}^{\\prime}=x_{... | [
"$402, 401,401,401,401$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
153 | 某市有 $n$ 所中学, 第 $i$ 所中学派出 $C_{i}$ 名代表 $\left(1 \leqslant C_{i} \leqslant 39,1 \leqslant i \leqslant n\right)$ 来到体育馆观看球赛, 全部学生总数为 $\sum_{I=1}^{n} C_{i}=1990$. 看台上每一横排有 199 个座位, 要求同一学校的学生必须坐在同一横排, 问体育馆最少要安排多少横排才能够保证全部学生都能坐下. | [
"首先, $199>39 \\times 5$, 故每排至少可坐 5 所学校的学生.\n\n$1990=199 \\times 10$, 故如果没有 “同一学校的学生必须坐在同一横排” 的限制, 则全部学生只要坐在 10 排就够了.\n\n现让这些学生先按学校顺序入坐, 从第一排坐起, 一个学校的学生全部坐好后, 另一个学校的学生接下去坐, 如果在某一行不够坐, 则余下的学生坐到下一行. 这样一个空位都不留, 则坐 10 排, 这些学生就全部坐完. 这时, 有些学校的学生可能分坐在两行, 让这些学校的学生全部从原坐处起来, 坐到第 11、12 排去. 由于, 这种情况只可能在第一行末尾与第二行开头、第二行末尾与第三行开头、.... | [
"$12$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
163 | 设 $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, 且 $x+y+z=\frac{\pi}{2}$, 求乘积 $\cos x \sin y \cos z$ 的最大值和最小值. | [
"由于 $x \\geqslant y \\geqslant z \\geqslant \\frac{\\pi}{12}$, 故 $\\frac{\\pi}{6} \\leq \\mathbf{x} \\leq \\frac{\\pi}{2}-\\frac{\\pi}{12} \\times 2=\\frac{\\pi}{3}$.\n\n$\\therefore \\cos x \\sin y \\cos z=\\cos x \\times \\frac{1}{2}[\\sin (y+z)+\\sin (y-z)]=\\frac{1}{2} \\cos ^{2} x+\\frac{1}{2} \\cos x \\sin (y... | [
"$\\frac{2+\\sqrt{3}}{8}，\\frac{1}{8}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
165 | 已知复数 $z=1-\sin \theta+i \cos \theta\left(\frac{\pi}{2}<\theta<x\right)$, 求 $z$ 的共轭复数 $\bar{z}$ 的辐角主值. | [
"$z=1+\\cos \\left(\\frac{\\pi}{2}+\\theta\\right)+i \\sin \\left(\\frac{\\pi}{2}+\\theta\\right)=2 \\cos ^{2} \\frac{\\frac{\\pi}{2}+\\theta}{2}+2 i \\sin \\frac{\\frac{\\pi}{2}+\\theta}{2} \\cos \\frac{\\frac{\\pi}{2}+\\theta}{2}$\n\n$$\n=2 \\cos \\frac{\\frac{\\pi}{2}+\\theta}{2}\\left(\\cos \\frac{\\frac{\\pi}{... | [
"$\\frac{3 \\pi}{4}-\\frac{\\theta}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
166 | 设函数 $f(x)=a x^{2}+8 x + 3(a<0)$. 对于给定的负数 $a$, 有一个最大的正数 $l(a)$, 使得在整个 区间 $[0, l(a)]$ 上, 不等式 $|f(x)| \leq 5$ 都成立.
问: $a$ 为何值时 $l(a)$ 最大? 以及这个最大的 $l(a)$是多少？ | [
"$\\left.f(x)={a\\left(x+\\frac{4}{a}\\right.}\\right)^{2}+3-\\frac{16}{a}$.\n\n(1)当 $3-\\frac{16}{a}>5$, 即 $-8<x<0$ 时,\n\n$l(a)$ 是方程 $a x^{2}+8 x+3=5$ 的较小根, 故 $l(a)=\\frac{-8+\\sqrt{64+8a}}{2 a}$.\n\n(2) 当 3- $\\frac{16}{a} \\leq 5$, 即$a \\leq -8$时,\n\n$l(a)$ 是方程 $a x^{2}+8 x+3=-5$ 的较大根, 故 $l(a)=\\frac{-8-\\sqrt{6... | [
"$-8, \\frac{1+\\sqrt{5}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
168 | 设 $\{a_{n} \}$ 为等差数列, $\{b_{n}\}$ 为等比数列, 且 $b_{1}=a_{1}{ }^{2}, b_{2}=a_{2}{ }^{2}$, $b_{3}=a_{3}^{2}\left(\mathbf{a}_{1}<\mathbf{a}_{2}\right)$, 又 $\lim _{n \rightarrow+\infty}\left(b_{1}+b_{2}+\cdots+b_{n}\right)=\sqrt{2}+1$, 试求 $\left\{\mathbf{a}_{\mathbf{n}}\right\}$ 的首项与公差. | [
"设所求公差为 $d, \\because a_{1}<a_{2}, \\therefore d>0$. 由此得\n\n$a_{1}^{2}\\left(a_{1}+2 d\\right)^{2}=\\left(a_{1}+d\\right)^{4} \\quad$ 化简得: $2 a_{1}^{2}+4 a_{1} d+d^{2}=0$\n\n解得: $d=(-2 \\pm \\sqrt{2}) a_{1}$\n\n而 $-2 \\pm \\sqrt{2}<0$, 故 $a_{1}<0$\n\n若 $d=(-2-\\sqrt{2}) a_{1}$, 则 $q=\\frac{a_{2}^{2}}{a_{1}^{2}}=(\\... | [
"$-\\sqrt{2}, 2 \\sqrt{2}-2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
170 | 设 $\mathrm{x}_{\mathrm{i}} \geqslant 0(\mathrm{I}=1,2,3, \cdots, \mathrm{n})$ 且 $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leq k<j \leq n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$.
求 $\sum_{i=1}^{n} x_{i}$ 的最小值。 | [
"因为 $\\left(\\sum_{i=1}^{n} x_{i}\\right)^{2}=\\sum_{i=1}^{n} x_{i}^{2}+2 \\sum_{1 \\leq k<j \\leq n} \\sqrt{\\frac{k}{j}} x_{k} x_{j} \\geq 1 \\Rightarrow \\sum_{i=1}^{n} x_{i} \\geqslant 1$\n\n等号成立当且仅当存在 $i$ 使得 $x_{i}=1, x_{j}=0, j=i$\n\n$\\therefore \\sum_{i=1}^{n} x_{i}$ 最小值为 1."
] | [
"$1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
172 | 一项 “过关游戏” 规则规定: 在第 $n$ 关要抛郑一颗骰子 $n$ 次,如果这 $n$ 次抛掷所出现的点数的和大于 $2^{n}$, 则算过关. 问:
某人在这项游戏中最多能过几关? | [
"设他能过 $n$ 关, 则第 $n$ 关掷 $\\mathrm{n}$ 次, 至多得 $6 n$ 点,\n\n由 $6 n>2^{n}$, 知, $n \\leq 4$. 即最多能过 4 关."
] | [
"$4$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
173 | 一项 “过关游戏” 规则规定: 在第 $n$ 关要抛郑一颗骰子 $n$ 次,如果这 $n$ 次抛掷所出现的点数的和大于 $2^{n}$, 则算过关. 问:
他连过前三关的概率是多少? | [
"要求他第一关时掷 1 次的点数 $>2$, 第二关时掷 2 次的点数和 $>4$, 第三关时掷 3 次的点数和 $>8$.\n\n第一关过关的概率 $=\\frac{4}{6}=\\frac{2}{3}$;\n\n第二关过关的基本事件有 $6^{2}$ 种, 不能过关的基本事件有为不等式 $x + y \\leq 4$ 的正整数解的个数, 有 $C_{4}^{2}$ 个 (亦可枚举计数: $1+1,1+2,1+3,2+1,2+2,3+1$ ) 计 6 种, 过关的概率 $=1-$ $\\frac{6}{6^{2}}=\\frac{5}{6}$。\n\n第三关的基本事件有$6^{3}$种，不能过关的基本事件为方程$x+y+z... | [
"$\\frac{100}{243}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
174 | 已知 $\alpha, \beta$ 是方程 $4 x^{2}-4 t x-1=0(t \in R)$ 的两个不等实根, 函数 $f(x)=\frac{2 x-t}{x^{2}+1}$ 的定义域为 $[\alpha, \beta]$. 求 $g(t)=\max f(x)-\min f(x)$的表达式; | [
"$\\alpha+\\beta=t, \\alpha \\beta=-\\frac{1}{4}$. 故 $\\alpha<0, \\beta>0$. 当 $x_{1}, x_{2} \\in[\\alpha, \\beta]$ 时,\n\n$\\therefore f^{\\prime}(x)=\\frac{2\\left(x^{2}+1\\right)-2 x(2 x-t)}{\\left(x^{2}+1\\right)^{2}}=\\frac{-2\\left(x^{2}-x t\\right)+2}{\\left(x^{2}+1\\right)^{2}}$. 而当 $x \\in[\\alpha, \\beta]$ ... | [
"$\\frac{8 \\sqrt{t^{2}+1}\\left(2 t^{2}+5\\right)}{16 t^{2}+25}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
178 | 设 $l, m$ 是两条异面直线, 在 $l$ 上有 $A, B, C$ 三点, 且 $A B=B C$, 过 $A, B, C$ 分别作 $m$ 的垂线 $A D, B E, C F$, 垂足依次是 $D, E, F$, 已知 $A D=\sqrt{15}, B E=\frac{7}{2} C F=\sqrt{10}$, 求 $l$ 与 $m$ 的距离. | [
"过 $m$ 作平面 $\\alpha / / l$, 作 $A P \\perp \\alpha$ 于 $P, A P$ 与 $l$ 确定平面 $\\beta, \\beta \\cap \\alpha=l^{\\prime}, l^{\\prime} \\cap m=K$.\n\n作 $B Q \\perp \\alpha, C R \\perp \\alpha$, 垂足为 $Q 、 R$, 则 $Q 、 R \\in l^{\\prime}$, 且 $A P=B Q=C R=l$ 与 $m$的距离 $d$.\n\n连 $P D 、 Q E 、 R F$, 则由三垂线定理之逆, 知 $P D 、 Q E 、 R F$ 都... | [
"$\\sqrt{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
179 | 给定曲线簇 $2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0, \theta$ 为参数, 求该曲线在直线 $y=2 x$ 上所截得的弦长的最大值. | [
"以 $y=2 x$ 代入曲线方程得 $x=0, x=\\frac{8 \\sin \\theta+\\cos \\theta+1}{2 \\sin \\theta-\\cos \\theta+3}$.\n\n$\\therefore$ 所求弦长 $1=\\left|\\frac{8 \\sin \\theta+\\cos \\theta+1}{2 \\sin \\theta-\\cos \\theta+3}\\right| \\sqrt{5}$. 故只要求 $|x|$ 的最大值即可.\n\n由 $(2 x-8) \\sin \\theta-(x+1) \\cos \\theta=1-3 x . \\Rightarrow(2... | [
"$8 \\sqrt{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
182 | 函数 $F(x)=\left|\cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B\right|$
在 $0 \leqslant x \leqslant \frac{3}{2} \pi$ 上的最大值 $M$ 与参数 $A 、 B$ 有关, 问 $A和B$ 分别取什么值时, $M$ 为最小?. | [
"$F(x)=\\left|\\sqrt{2} \\sin \\left(2 x+\\frac{\\pi}{4}\\right)+A x+B\\right|$. 取 $g(x)=\\sqrt{2} \\sin \\left(2 x+\\frac{\\pi}{4}\\right)$, 则 $g\\left(\\frac{\\pi}{8}\\right)=g\\left(\\frac{9 \\pi}{8}\\right)=\\sqrt{2} \\cdot g\\left(\\frac{5 \\pi}{8}\\right)=-\\sqrt{2}$.\n\n取 $h(\\boldsymbol{x})=A \\boldsymbol{x... | [
"$A=0，B=0$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | Chinese |
184 | 设 $A 、 B 、 C$ 分别是复数 $Z_{0}=a \mathrm{i}, Z_{1}=\frac{1}{2}+b \mathrm{i}, Z_{2}=1+c \mathrm{i}$ (其中 $a, b, c$ 都是实数) 对应的不共线的三点. 曲线
$$
Z=Z_{0} \cos ^{4} t+2 Z_{1} \cos ^{2} t \sin ^{2} t+Z_{2} \sin ^{4} t \quad(t \in \mathrm{R})
$$
与 $\triangle A B C$ 中平行于 $A C$ 的中位线只有一个公共点, 求出此点. | [
"曲线方程为 : $Z=ai \\cos ^{4} t+(1+2 b \\mathrm{i}) \\cos ^{2} t \\sin ^{2} t+(1+c i) \\sin ^{4} t=\\left(\\cos ^{2} t \\sin ^{2} t+\\sin ^{4} t\\right)+\\mathrm{i}\\left(\\operatorname{acos}^{4} t+2 b \\cos ^{2} t \\sin ^{2} t+c\\right.$ $\\left.\\sin ^{4} t\\right)$\n\n$\\therefore \\quad x=\\cos ^{2} t \\sin ^{2} t+... | [
"$\\left(\\frac{1}{2},\\frac{1}{4}(a+2 b+c)\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
186 | 设三角形的三边长分别是正整数 $l, m, n$. 且 $l>m>n>0$.
已知 $\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\}$, 其中 $\{x\}=x-[x]$, 而 $[x]$ 表示不超过 $x$ 的最大整数. 求这种三角形周长的最小值. | [
"当 $3^{l} 、 3^{m} 、 3^{n}$ 的末四位数字相同时, \n$ \\left\\{\\frac{3^{l}}{10^{4}}\\right\\}=\\left\\{\\frac{3^{m}}{10^{4}}\\right\\}=\\left\\{\\frac{3^{n}}{10^{4}}\\right\\}.$\n\n即求满足 $3^{l} \\equiv 3^{m} \\equiv 3^{n}\\left(\\bmod 10^{4}\\right)$ 的 $l 、 \\mathbb{m} 、 n . \\therefore 3^{n}\\left(3^{l-n}-1\\right) \\equiv 0\... | [
"$3003$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
187 | $x$ 的二次方程 $x^{2}+z_{1} x+ z_{2}+m=0$ 中, $z_{1}, z_{2}, m$ 均是复数, 且 $z_{1}^{2}-4 z_{2}=16+20 i$,
设这个方程的两个根 $\alpha 、 \beta$, 满足 $|\alpha-\beta|=2 \sqrt{7}$, 求 $|m|$ 的最大值和最小值. | [
"设 $m=a+b i(a, b \\in R)$. 则 $\\triangle=z_{1}{ }^{2}-4 z_{2}-4 m=16+20 i-4 a-4 b i=4[(4-a)+(5-b) i]$. 设 $\\triangle$ 的平方根为 $u+v i$. $(u, v \\in R)$\n\n即 $(u+v i)^{2}=4[(4-a)+(5-b) i]$.\n\n$|\\alpha-\\beta|=2 \\sqrt{7}, \\Leftrightarrow|\\alpha-\\beta|^{2}=28, \\Leftrightarrow|(4-a)+(5-b) i|=7, \\Leftrightarrow(a-4... | [
"$7+\\sqrt{41}, 7-\\sqrt{41}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
188 | 将与 105 互素的所有正整数从小到大排成数列, 试求出这个数列的第 1000 项。 | [
"由 $105=3 \\times 5 \\times 7$; 故不超过 105 而与 105 互质的正整数有 $105 \\times\\left(1-\\frac{1}{3}\\right)\\left(1-\\frac{1}{5}\\right)(1$ $\\left.-\\frac{1}{7}\\right)=48$ 个。 $1000=48 \\times 20+48-8, \\quad 105 \\times 20=2100$. 而在不超过 105 的与 105 互质的数中第 40 个数是 86 .\n\n$\\therefore$ 所求数为 2186 ."
] | [
"$2186$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
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