dataset_name stringclasses 4
values | dataset_version timestamp[s] | qid stringlengths 1 5 | queId stringlengths 32 32 | competition_source_list list | difficulty stringclasses 5
values | qtype stringclasses 1
value | problem stringlengths 6 1.51k | answer_option_list list | knowledge_point_routes list | answer_analysis list | answer_value stringclasses 7
values |
|---|---|---|---|---|---|---|---|---|---|---|---|
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 900 | bfd08f555c1e485a9aedf91421aadbd1 | [
"全国高中数学联赛竞赛模拟一试（十五）第1题"
] | 2 | single_choice | 函数$$f\left( x \right)={{x}^{4}}-{{x}^{2}}-\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}$$的值域为~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$\\left[ 2,+\\infty \\right)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\left[ 0,+\\infty \\right)$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left[ -2,+\\infty \\right)$$ "
}
],
[
{
"aoVal": "D",
"c... | [
"课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域",
"竞赛->知识点->函数->二次函数",
"竞赛->知识点->函数->基本初等函数"
] | [
"注意到， $$f\\left( x \\right)={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)}^{2}}-\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)-2$$ $$={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}}-\\frac{1}{2} \\right)}^{2}}-\\frac{9}{4}$$． 由于$${{x}^{2}}+\\frac{1}{{{x}^{2}}}\\geqslant 2$$， 从而，所求函数的值域为$$\\left[ 0,+\\infty \\righ... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 188 | 1d548a44a8234f548ab99df07aa9da0e | [
"2017~2018学年浙江温州瓯海区浙江省瓯海中学高三上学期期中理科第9题4分",
"2007年全国全国高中数学联赛竞赛一试第2题6分"
] | 1 | single_choice | 设实数$$a$$使得不等式$$\textbar2x-a\textbar+\textbar3x-2a\textbar\geqslant {{a}^{2}}$$对任意实数$$x$$恒成立，则满足条件的$$a$$所组成的集合是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$\\left[ -\\frac{1}{3}，\\frac{1}{3} \\right]$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\left[ -\\frac{1}{2}，\\frac{1}{2} \\right]$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left[ -\\frac{1}{4}，\\frac{1}{3} \\right]$$ "
... | [
"课内体系->素养->数学运算",
"课内体系->知识点->等式与不等式->不等式->解不等式->含绝对值的不等式",
"课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式"
] | [
"令$$x=\\frac{2}{3}a$$，则有$$\\textbar a\\textbar\\leqslant \\frac{1}{3}$$，排除$$\\text{B}$$、$$\\text{D}$$，由对称性排除$$\\text{C}$$，从而只有$$\\text{A}$$正确． 一般地，对$$k\\in \\mathbf{R}$$，令$$x=\\frac{1}{2}ka$$，则原不等式为$$\\textbar a\\textbar\\cdot \\textbar k-1\\textbar+\\frac{3}{2}\\textbar a\\textbar\\cdot \\left\\textbar{} k-\\frac... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 593 | 99dbb27c2db64580b29e119e2d2e24fa | [
"2009年浙江全国高中数学联赛竞赛初赛第2题5分"
] | 0 | single_choice | 已知椭圆$$\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1$$上一点$$P$$到点$$(4,0)$$距离等于$$4$$，则$$P$$点到直线$$x=-\frac{25}{4}$$的距离为． | [
[
{
"aoVal": "A",
"content": "$$4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{15}{2}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{5}{4}$$ "
}
]
] | [
"竞赛->知识点->解析几何->椭圆"
] | [
"因为$$a=5,b=3$$，则$$c=4$$．于是$$P$$到另一个焦点$$(-4,0)$$的距离等于$$2\\times 5-4=6$$．由于直线$$x=-\\frac{25}{4}$$为椭圆的左准线方程，则$$P$$到直线$$x=-\\frac{25}{4}$$的距离为$$d=\\frac{6}{e}=7.5$$． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 333 | 2c1c66a3578d4e4985b3afa0e4415587 | [
"1992年全国高中数学联赛竞赛一试第6题"
] | 1 | single_choice | 设$$f\left( x \right)$$是定义在实数集$$R$$上的函数，且满足下列关系：$$f\left( 10+x \right)=f\left( 10-x \right)$$．$$f\left( 20-x \right)=-f\left( 20+x \right)$$．则$$f\left( x \right)$$是（~ ）． | [
[
{
"aoVal": "A",
"content": "偶函数，又是周期函数 "
}
],
[
{
"aoVal": "B",
"content": "偶函数，但不是周期函数 "
}
],
[
{
"aoVal": "C",
"content": "奇函数，又是周期函数 "
}
],
[
{
"aoVal": "D",
"content": "奇函数，但不是周期函数 "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"由所给第一式得 $$f\\left[ 10+\\left( 10-x \\right) \\right]=f\\left[ 10-\\left( 10-x \\right) \\right]$$， ∴~~~~ $$f\\left( x \\right)=f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ① 又由所给第二式得 $$f\\left( x \\right)=-f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 982 | 97a6d6c9af8340fc847212b408e7c105 | [
"2010年河南全国高中数学联赛竞赛初赛第3题5分"
] | 1 | single_choice | 设$$f(x)=\frac{x-1}{x+1}$$，记$${{f}_{1}}(x)=f(x)$$，若$${{f}_{n+1}}(x)=f({{f}_{n}}(x))$$，则$${{f}_{2010}}(x)=$$． | [
[
{
"aoVal": "A",
"content": "$$x$$ "
}
],
[
{
"aoVal": "B",
"content": "$$-\\frac{1}{x}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{x-1}{x+1}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{1+x}{1-x}$$ "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"本题从题目上可预知$${{f}_{n}}(x)$$具有周期性，为了寻找周期，采用赋值法容易计算．记$${{a}_{1}}={{f}_{1}}(2), {{a}_{n}}={{f}_{n}}(2)$$，则$${{a}_{1}}=\\frac{1}{3},{{a}_{2}}=\\frac{\\frac{1}{3}-1}{\\frac{1}{3}+1}=-\\frac{2}{4}=-\\frac{1}{2},{{a}_{3}}=\\frac{-\\frac{1}{2}-1}{-\\frac{1}{2}+1}=-3,$$ $${{a}_{4}}=\\frac{-3-1}{-3+1}=2, {{a}_{5}}=\\frac{2... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 309 | 46f55a4b882c492d890533d4f11aabd0 | [
"2003年AMC10竞赛B第22题"
] | 3 | single_choice | 一个钟在半点时响一次，在整点时，几点钟就响几次．例如，在下午$$1$$点有一次报时，在正午$$12$$点和午夜$$12$$点各有$$12$$次报时．从$$2003$$年$$2$$月$$26$$日上午$$11$$点$$15$$分开始，第$$2003$$次报时将在哪一天发生？ | [
[
{
"aoVal": "A",
"content": "March $$8$$ "
}
],
[
{
"aoVal": "B",
"content": "March $$9$$ "
}
],
[
{
"aoVal": "C",
"content": "March $$10$$ "
}
],
[
{
"aoVal": "D",
"content": "March $$20$$ "
}
],
[
{
"aoVal":... | [
"课内体系->知识点->等式与不等式->等式->方程组的解集",
"美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion"
] | [
"First, find how many chimes will have already happened before midnight (the beginning of the day) of February $$27$$, $$2003$$. $$13$$ half-hours have passed, and the number of chimes according to the hour is $$1+2+3+\\cdots +12$$. The total number of chimes is $$13+78=91$$. Every day, there will be $$24$$ half-h... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 784 | e41381c012024cd1b4044f09291f86a5 | [
"1997年全国高中数学联赛竞赛一试第4题6分"
] | 1 | single_choice | 在平面直角坐标系中，若方程$$m\left( {{x}^{2}}+{{y}^{2}}+2y+1 \right)={{\left( x-2y+3 \right)}^{2}}$$表示的曲线为椭圆，则$$m$$的取值范围为（~ ~ ）． | [
[
{
"aoVal": "A",
"content": "$$(0,1)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$(1,+\\infty )$$ "
}
],
[
{
"aoVal": "C",
"content": "$$(0,5)$$ "
}
],
[
{
"aoVal": "D",
"content": "$$(5,+\\infty )$$ "
}
]
] | [
"课内体系->素养->数学运算",
"课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的标准方程",
"课内体系->知识点->直线和圆的方程->直线与方程->平面中的距离->点到直线的距离公式"
] | [
"看成是轨迹上点到$$(0,-1)$$的距离与到直线$$x-2y+3=0$$的距离的比： $$\\frac{\\sqrt{{{x}^{2}}+{{\\left( y+1 \\right)}^{2}}}}{\\frac{\\left\\textbar{} x-2y+3 \\right\\textbar}{\\sqrt{{{1}^{2}}+{{\\left( -2 \\right)}^{2}}}}}=\\sqrt{\\frac{5}{m}}\\textless{}1\\Rightarrow m\\textgreater5$$． 故选$$\\text{D}$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 486 | 365fad487d6c4a0fbcdbad340907824b | [
"2008年AMC10竞赛B第18题",
"2008年AMC12竞赛B第10题"
] | 3 | single_choice | $$2008-AMC10B-18$$ Bricklayer Brenda would take $$9$$ hours to build a chimney alone, and bricklayer Brandon would take $$10$$ hours to build it alone. When they work together they talk a lot, and their combined output is decreased by $$10$$ bricks per hour. Working together, they build the chimney in $$5$$ hours. How... | [
[
{
"aoVal": "A",
"content": "$$500$$ "
}
],
[
{
"aoVal": "B",
"content": "$$900$$ "
}
],
[
{
"aoVal": "C",
"content": "$$950$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1000$$ "
}
],
[
{
"aoVal": "E",
"cont... | [
"课内体系->知识点->等式与不等式->等式->方程组的解集",
"美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"
] | [
"Let $$x$$ be the number of bricks in the chimney, Using $$d=vt$$, we get $$x=\\left( \\frac{x}{9}+\\frac{x}{10}-10 \\right)\\cdot \\left( 5 \\right)$$. Solving for $$x$$, we get $$900\\Rightarrow \\text{B}$$. "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 695 | 45b53d99a0d648099dd19ca782044c9b | [
"1998年全国高中数学联赛竞赛一试第4题6分"
] | 1 | single_choice | 设命题$$P$$：关于$$x$$的不等式$${{a}_{1}}{{x}^{2}}+{{b}_{1}}{{x}}+{{c}_{1}}\textgreater0$$与$${{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}\textgreater0$$的解集相同；命题$$Q$$：$$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$$． 则命题$$Q$$． | [
[
{
"aoVal": "A",
"content": "是命题$$P$$的充分必要条件 "
}
],
[
{
"aoVal": "B",
"content": "是命题$$P$$的充分条件但不是必要条件 "
}
],
[
{
"aoVal": "C",
"content": "是命题$$P$$的必要条件但不是充分条件 "
}
],
[
{
"aoVal": "D",
"content": "既不是是命题$$P$$的充分条件也不是命题$$P$$的必要... | [
"竞赛->知识点->不等式->不等式的解法"
] | [
"若两个不等式的解集都是$$\\mathbf{R}$$，否定$$\\text{A}$$、$$\\text{C}$$，若比值为$$-1$$，否定$$\\text{A}$$、$$\\text{B}$$． 故选$$\\text{D}$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1056 | ee633f00f517474a979807407e6f5502 | [
"2006年上海复旦大学自主招生千分考第16题",
"1981年全国高中数学联赛竞赛一试第2题"
] | 3 | single_choice | 条件甲：$$\sqrt{1+\sin \theta }=a$$．条件乙：$$\sin \frac{\theta }{2}+\cos \frac{\theta }{2}=a$$．则下列（~ ）是正确的． | [
[
{
"aoVal": "A",
"content": "甲是乙的充分必要条件 "
}
],
[
{
"aoVal": "B",
"content": "甲是乙的必要条件 "
}
],
[
{
"aoVal": "C",
"content": "甲是乙的充分条件 "
}
],
[
{
"aoVal": "D",
"content": "甲不是乙的必要条件，也不是充分条件 "
}
]
] | [
"竞赛->知识点->三角函数->三角函数的图像与性质",
"竞赛->知识点->三角函数->三角恒等变换"
] | [
"因为$$\\sqrt{1+\\sin \\theta }=\\sqrt{{{\\sin }^{2}}\\frac{\\theta }{2}+2\\sin \\frac{\\theta }{2}\\cos \\frac{\\theta }{2}+{{\\cos }^{2}}\\frac{\\theta }{2}}=\\sqrt{{{\\left( \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right)}^{2}}}=\\left\\textbar{} \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\r... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 424 | 2cf3496e464740a280b4fdb8c1853a0d | [
"2014年辽宁全国高中数学联赛竞赛初赛第4题5分"
] | 2 | single_choice | $$\triangle ABC$$的三个内角为$$A$$、$$B$$、$$C$$，若$$\frac{\sin A+\sqrt{3}\cos A}{\cos A-\sqrt{3}\sin A}=\tan \frac{7 \pi }{12}$$，则$$\sin 2B+2\cos C$$的最大值为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{2}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{3}{2}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$2$$ "
}
]
] | [
"竞赛->知识点->三角函数->三角恒等变换",
"竞赛->知识点->三角函数->三角函数的图像与性质"
] | [
"因为 $$\\frac{\\sin A+\\sqrt{3}\\cos A}{\\cos A-\\sqrt{3}\\sin A}=\\frac{\\tan A+\\tan \\frac{ \\pi }{3}}{1-\\tan A\\tan \\frac{ \\pi }{3}}=\\tan \\left( A+\\frac{ \\pi }{3} \\right)$$， 故$$A+\\frac{ \\pi }{3}=\\frac{7 \\pi }{12}$$，$$A=\\frac{ \\pi }{4}$$． 由于$$2B+2C=\\frac{3 \\pi }{2}$$，故 $$\\sin 2B+2\\cos C=\\si... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1173 | eb5553dd1e034f6f853a8fc3fd04f9fe | [
"全国高中数学联赛竞赛模拟一试（五）第3题"
] | 0 | single_choice | 从$$0$$，$$1$$，$$\cdots $$，$$9$$这十个号码中任意抽取三个，其中至少有两个号码是连续整数的概率为~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$ \\frac{7}{15}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$ \\frac{8}{15}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$ \\frac{2}{5}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$ \\frac{3}{5}$$ "
}
],
... | [
"竞赛->知识点->排列组合与概率->两个基本计数原理",
"竞赛->知识点->排列组合与概率->概率初步",
"竞赛->知识点->排列组合与概率->排列与组合",
"课内体系->知识点->计数原理"
] | [
"设抽出的号码$$i$$，$$j$$，$$k$$（$$0\\leqslant i\\textless{}j\\textless{}k\\leqslant 9$$）中任何两个均不相邻， 则$$i$$，$$j-1$$，$$k-2$$互不相邻，且只能从$$ {0,1,\\cdots ,7 }$$中取值， 故所求的概率$$ p=1- \\frac{\\text{C}_{8}^{3}}{\\text{C}_{15}^{3}}= \\frac{8}{15}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 655 | a7ef7dfc33c348c580baa65ba85126f1 | [
"2008年山东全国高中数学联赛竞赛初赛第6题6分"
] | 0 | single_choice | 已知函数$$y=\frac{x-a}{x-a-1}$$的反函数的图象关于点$$\left( -1,3 \right)$$成中心对称图形，则实数$$a$$等于． | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$3$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-2$$ "
}
],
[
{
"aoVal": "D",
"content": "$$-3$$ "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"方法一：在原函数图象上取点$$\\left( 1,\\frac{1}{a}-1 \\right)$$， 则点$$\\left( \\frac{1}{a}-1,1 \\right)$$在其反函数图象上， 它关于点$$(-13)$$的对称点为$$\\left( -1-\\frac{1}{a},5 \\right)$$，从而点$$\\left( 5,-1-\\frac{1}{a} \\right)$$在原函数图象上， 所以有$$-1-\\frac{1}{a}=\\frac{5-a}{5-a-1}$$， 解得$$a=2$$．故选$$\\text{A}$$． 方法二：因函数$$y=-\\frac{x-a}{x-a-1}$$... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 195 | d076cf8331204114be69026b3d210b67 | [
"2010年黑龙江全国高中数学联赛竞赛初赛第9题5分"
] | 1 | single_choice | 若把函数$$y=\sqrt{3}\cos x-\sin x$$的图象向右平移$$m$$（$$m\textgreater0$$）个单位长度后，所得到的图象关于$$y$$轴对称，则$$m$$的最小值是． | [
[
{
"aoVal": "A",
"content": "$$\\frac{ \\pi }{3}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{2}{3} \\pi $$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{ \\pi }{6}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{5}{6} \\pi ... | [
"竞赛->知识点->三角函数->三角函数的图像与性质"
] | [
"$$y=\\sqrt{3}\\cos x-\\sin x=2\\cos \\left( x+\\frac{ \\pi }{6} \\right),$$对称轴方程$$x=k \\pi -\\frac{ \\pi }{6}, k\\in Z$$． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 5 | 0959c4ee5706421a878d77b9fe75c8ab | [
"2002年全国高中数学联赛竞赛一试第2题10分"
] | 1 | single_choice | 若实数$$x$$，$$y$$满足$${{\left( x+5 \right)}^{2}}+{{\left( y+12 \right)}^{2}}={{14}^{2}}$$，则$${{x}^{2}}+{{y}^{2}}$$的最小值为（~ ~ ） | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\sqrt{3}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\sqrt{2}$$ "
}
]
] | [
"课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值",
"课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念",
"课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件",
"课内体系->素养->数学运算"
] | [
"令$$x+5=14\\cos \\theta $$,$$y-12=14\\sin \\theta $$ ,则$${{x}^{2}}+{{y}^{2}}=196+28\\left( 5\\cos \\theta -12\\sin \\theta \\right)+169=365+364\\sin \\left( \\theta +\\varphi \\right)\\geqslant 1$$. "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 104 | fa328434c6d745de9d1f367639a116ad | [
"2017年山西全国高中数学联赛竞赛初赛第8题8分"
] | 1 | single_choice | 设$$M=\left { 1,2,\cdots ,2017 \right }$$是前$$2017$$个正整数构成的集合，若从$$M$$中去掉一个元素后，$$M$$中剩下的元素之和恰为一个平方数，则去掉的元素是 ． | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$65$$ "
}
],
[
{
"aoVal": "C",
"content": "$$1008$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1677$$ "
}
]
] | [
"竞赛->知识点->数列与数学归纳法->数列的通项与求和",
"竞赛->知识点->集合->集合的划分与覆盖"
] | [
"$$S=1+2+\\cdots +2017=\\frac{2018\\cdot 2017}{2}=1009\\cdot 2017=2035153\\in \\left( {{1400}^{2}},{{1500}^{2}} \\right)$$， 而$${{1450}^{2}}={{\\left( 1400+50 \\right)}^{2}}=1960000+140000+2500=2102500\\textgreater S$$， 又$${{1425}^{2}}={{\\left( 1400+25 \\right)}^{2}}=1960000+70000+625=2030625\\textless{}S$$， $${... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 537 | 75d1bef60e4c4b9495c0dd39cb48151d | [
"2016年AMC10竞赛第16题5分"
] | 1 | single_choice | 某個三角形的頂點座標爲 $$A(0,2)$$、 $$B(-3,2)$$、$$C(-3,0)$$ ，將此三角形以$$x$$軸爲對稱軸作對稱得到三角形$$\triangle A^{}\prime B^{}\prime C^{}\prime $$ ．再將她以原點爲中心，逆時針方向旋轉90°．得到$$\triangle A^{\prime \prime }B^{\prime \prime }C^{\prime \prime }$$．試問下列哪一個敘述能將變換回原來的？ $$\text{A}$$选项．繞原點逆時針旋轉$$90^{}\circ $$ $$\text{B}$$选项．繞原點順時針旋轉$$90^{}\circ $$ $$\tex... | [
[
{
"aoVal": "A",
"content": "counterclockwise retation about the origin by $$90^{}\\circ $$ "
}
],
[
{
"aoVal": "B",
"content": "clockwise rotation about the origin by $$90^{}\\circ $$ "
}
],
[
{
"aoVal": "C",
"content": "reflection about the x---axis... | [] | [
"Consider a point $$(x,y)$$. Reflecting it about the $$x$$-axis will map it to $$(x,-y)$$, and rotating it counterclockwise about the origin by $$90^{}\\circ $$will map it to $$(y,x)$$. The operation that undoes this is a reflection about the $$y=x$$, so the answer is $$\\rm D$$. "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 831 | ff8080814a85cc99014a89235e00081a | [
"2010年黑龙江全国高中数学联赛竞赛初赛第12题5分",
"2020~2021学年广东深圳福田区深圳市高级中学高中部高二下学期期中第8题5分"
] | 3 | single_choice | 已知集合$$M= {1,2,3 }$$，$$N= {1,2,3,4 }$$，定义函数$$f:M\to N$$．若点 $$A(1,f(1))$$，$$B(2,f(2))$$，$$C(3,f(3))$$，$$\Delta ABC$$的外接圆圆心为$$D$$，且$$\overrightarrow{DA}+\overrightarrow{DC}=\lambda \overrightarrow{DB}(\lambda\in R)$$ ，则满足条件的函数$$f(x)$$有（~ ） | [
[
{
"aoVal": "A",
"content": "$$6$$个 "
}
],
[
{
"aoVal": "B",
"content": "$$10$$个 "
}
],
[
{
"aoVal": "C",
"content": "$$12$$个 "
}
],
[
{
"aoVal": "D",
"content": "$$16$$个 "
}
]
] | [
"课内体系->知识点->计数原理->两个基本计数原理->分类加法计数原理",
"课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义",
"课内体系->素养->数学运算"
] | [
"由$$\\overrightarrow{DA}+\\overrightarrow{DC}=\\overrightarrow{\\gamma DB}$$（$$\\gamma \\in \\mathbf{R}$$），分析可得$$\\triangle ABC$$是等腰三角形， 且$$BA=BC$$，必有$$f(1)=f(3)$$，$$f(1)\\ne f(2)$$． 点$$A(1,f(1))$$、当$$f(1)=1=f(3)$$时$$f(2)=2$$、$$3$$、$$4$$，三种情况． $$f(1)=f(3)=2$$；$$f(2)=1$$、$$3$$、$$4$$，有三种． $$f(1)=f(3)=3$$；$$f(2)=2... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 889 | d6a08daab1b84a229b1ee7fb117ec969 | [
"1988年全国高中数学联赛竞赛一试第5题"
] | 2 | single_choice | 在坐标平面上，纵横坐标都是整数的点叫做整点．我们用Ⅰ表示所有直线的集合，$$M$$表示恰好通过一个整点的直线的集合，$$N$$表示不通过任何整点的直线的集合，$$P$$表示通过无穷多个整点的直线的集合，那么表达式：． （1）$$M{\cup }N{\cup }P=$$Ⅰ；（3）$$N\ne \varnothing $$； （2）$$M\ne \varnothing $$；~~~~~~~~~~ （4）$$P\ne \varnothing $$ 中正确的个数是 | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4$$ "
}
]
] | [
"竞赛->知识点->集合->集合的概念与运算",
"竞赛->知识点->解析几何->直线与方程"
] | [
"直线$$y=\\frac{1}{2}$$不通过任何整点，所以表达式（2）正确； 直线$$y=\\sqrt{2}x$$无穷多个，恰好通过一个整点$$\\left( 0,0 \\right)$$，所以表达式（$$3$$）正确， 直线$$y=x$$通过多个整点，所以表达式（$$4$$）正确； 我们来证明，若 $$ax+by+c=0$$ 通过两个整点$$\\left( {{x}_{1}},{{y}_{1}} \\right)$$、$$\\left( {{x}_{2}},{{y}_{2}} \\right)$$， 则通过无穷多个整点 $$\\left( {{x}_{2}}+k\\left( {{x}_{2}}-{{x}... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 896 | 812fcc3d76cd4826a2b5e6966a64818d | [
"1989年全国高中数学联赛竞赛一试第2题"
] | 1 | single_choice | 函数$$f\left( x \right)=\text{arctan}x+\frac{1}{2}\arcsin x$$的值域是． | [
[
{
"aoVal": "A",
"content": "$$(- \\pi , \\pi )$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\left[ -\\frac{3 \\pi }{4},\\frac{3 \\pi }{4} \\right]$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left( -\\frac{3 \\pi }{4},\\frac{3 \\pi }{4} \\right)$$ "
}... | [
"知识标签->题型->函数->函数及其表示->函数的值域->用单调性观察法求值域",
"知识标签->素养->数学运算",
"知识标签->知识点->三角函数->三角函数的图象与性质->反三角函数"
] | [
"$$f(x)$$的定义域是$$[-1,1]$$，此时 $$-\\frac{ \\pi }{4}{\\leqslant }\\text{arctg}x{\\leqslant }\\frac{ \\pi }{4}$$，$$-\\frac{ \\pi }{4}{\\leqslant }\\frac{1}{2}\\arcsin x{\\leqslant }\\frac{ \\pi }{4}$$． 而且$$\\text{arctg}x$$和$$\\arcsin x$$是单调增加的，从而$$f(x)$$的值域是$$\\left[ -\\frac{ \\pi }{2},-\\frac{ \\pi }{2} \\right]$$． ... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 989 | a9cb8a6c7f39491788e6c7c186200683 | [
"2018年AMC10竞赛A第14题"
] | 2 | single_choice | 不超过 $$\frac{4^{100}+3^{100}}{4^{96}+3^{96}}$$ 的最大整数是? (改编自 2018 AMC10 A 第$$14$$题) | [
[
{
"aoVal": "A",
"content": "$$255$$ "
}
],
[
{
"aoVal": "B",
"content": "$$256$$ "
}
],
[
{
"aoVal": "C",
"content": "$$270$$ "
}
],
[
{
"aoVal": "D",
"content": "$$264$$ "
}
],
[
{
"aoVal": "E",
"conte... | [
"美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation",
"课内体系->知识点->集合->集合的基本关系->子集个数的计算"
] | [
"We write $$\\frac{4^{100}+3^{100}}{4^{96}+3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot \\frac{4^{100}}{4^{96}}+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot \\frac{3^{100}}{3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot 256+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot 81$$. Hence we see that our number is less than $256$ but more than $... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 507 | 880d6e5276024ccb95db40d76fc19bba | [
"2018年吉林全国高中数学联赛竞赛初赛第6题5分"
] | 1 | single_choice | 设$$x$$、$$y$$、$$z\textgreater0$$，满足$$x+y=xy$$，$$x+y+z=xyz$$．则$$z$$的取值范围是． | [
[
{
"aoVal": "A",
"content": "$$(0,\\sqrt{3}]$$ "
}
],
[
{
"aoVal": "B",
"content": "$$(1,\\sqrt{3}]$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left( 0,\\frac{4}{3} \\right]$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\left( 1,\\f... | [
"竞赛->知识点->不等式->换元技巧->代数换元"
] | [
"由已知$$y=\\frac{x}{x-1}$$得，$$z=\\frac{x+y}{xy-1}=\\frac{xy}{xy-1}={{\\left( 1-\\frac{1}{xy} \\right)}^{-1}}$$ $$={{\\left( 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}} \\right)}^{-1}}$$． 因为$$x\\textgreater1$$， 即$$0\\textless{}\\frac{1}{x}\\textless{}1$$， 所以，$$\\frac{3}{4}\\leqslant 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}}\\te... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 754 | 69ad4e8d2c0648f28cd79083e185f5f1 | [
"2000年AMC10竞赛第17题"
] | 3 | single_choice | $$2000-AMC10-17$$ Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using th... | [
[
{
"aoVal": "A",
"content": "＄$$3.63$$ "
}
],
[
{
"aoVal": "B",
"content": "＄$$5.13$$ "
}
],
[
{
"aoVal": "C",
"content": "＄$$6.30$$ "
}
],
[
{
"aoVal": "D",
"content": "＄$$7.45$$ "
}
],
[
{
"aoVal": "E",
... | [
"课内体系->知识点->函数的应用->函数与方程",
"美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"
] | [
"Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn\\textquotesingle t change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn\\textquotesingle t change. However, if he puts a p... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 828 | 77a73cbc5992493e868ab3113dba5f1d | [
"2019年江西全国高中数学联赛竞赛初赛改编第8题"
] | 3 | single_choice | 数列$ {a_n }$满足: $a_0=\sqrt{3}$, $a_{n+1}=[a_n]+\dfrac{1}{ {a_n }}$(其中$[a_n]$和${a_n}$分别表示实数$x$的整数部分与小数部分), 则$a_{2020}=$． | [
[
{
"aoVal": "A",
"content": "$3029+\\dfrac{\\sqrt{3}-1}{2}$ "
}
],
[
{
"aoVal": "B",
"content": "$3032+\\dfrac{\\sqrt{3}-1}{2}$ "
}
],
[
{
"aoVal": "C",
"content": "$3035+\\dfrac{\\sqrt{3}-1}{2}$ "
}
],
[
{
"aoVal": "D",
"conte... | [
"知识标签->知识点->数列->数列的概念->数列的表示方法->通项公式"
] | [
"$$1^{\\circ}$$迭代几项尝试一下， $$a_1=\\sqrt{3}=1+(\\sqrt{3}-1)$$， $$\\Rightarrow{a_2}=[a_1]+\\frac{1}{\\left { {{a}_{n}} \\right }}=1+\\frac{1}{\\sqrt{3}-1}=2+\\frac{\\sqrt{3}-1}{2}$$， $$\\Rightarrow{{a}_{3}}=[{{a}_{2}}]+\\frac{1}{\\left { {{a}_{2}} \\right }}=2+\\frac{2}{\\sqrt{3}-1}=4+(\\sqrt{3}-1)$$， $$\\Rightarro... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 893 | cd77cc8ca1cb49ff9d1ec5c7ddfc37a9 | [
"2017年天津南开区高三二模文科第1题5分",
"2015年黑龙江全国高中数学联赛竞赛初赛第1题5分",
"2014年重庆高三零模文科第4题5分",
"2017年天津南开区高三二模理科第1题5分",
"2014年重庆高三零模理科第2题5分"
] | 1 | single_choice | 设集合$$A= {-1,0,2 }$$，集合$$B= {-x\textbar x\in A,2-x\notin A }$$，则$$B=$$（ ~ ~） | [
[
{
"aoVal": "A",
"content": "$$ {1 }$$ "
}
],
[
{
"aoVal": "B",
"content": "$$ {-2 }$$ "
}
],
[
{
"aoVal": "C",
"content": "$$ {-1,-2 }$$ "
}
],
[
{
"aoVal": "D",
"content": "$$ {-1,0 }$$ "
}
]
] | [
"课内体系->素养->数学运算",
"课内体系->素养->逻辑推理",
"课内体系->思想->分类讨论思想",
"课内体系->知识点->集合->集合的概念与表示方法->集合的含义、元素与集合->元素与集合之间的关系->元素与集合的关系判断"
] | [
"∵集合$$A= {-1,0,2 }$$，集合$$B= {-x\\textbar x\\in A,2-x\\notin A }$$， ∴$$-1\\in A$$，且$$2-\\left( -1 \\right)=3\\notin A$$，故$$1\\in B$$； $$0\\in A$$，但$$2-0=2\\in A$$，不满足题意； $$2\\in A$$，但$$2-2=0\\in A$$，不满足题意； 故$$B= {1 }$$， 故选$$\\text{A}$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 670 | 6d824925a05e48bd8b0a1f8c98a2a937 | [
"2017~2018学年江西南昌西湖区南昌市第八中学高一上学期期末第10题5分",
"2017~2018学年北京海淀区中央民族大学附属中学高一下学期期中第7题5分",
"2015年北京海淀区高三三模第4题",
"2018~2019学年浙江杭州江干区杭州第四中学下沙校区高二下学期期中第12题4分",
"2018~2019学年广东深圳高一上学期期末高中联考联盟第8题5分",
"2008年黑龙江全国高中数学联赛竞赛初赛第4题5分",
"2018~2019学年江苏盐城高一上学期期末第5题5分"
] | 0 | single_choice | 若$$\sin \left( \frac{ \pi }{4}-x \right)=\frac{3}{5}$$，则$$\sin 2x$$的值为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{19}{25}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{16}{25}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{14}{25}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{7}{25}$$ "
}
]
... | [
"课内体系->素养->数学运算",
"课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦",
"课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦->利用正弦和差角公式凑角求值",
"课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式->正余弦和差积相互转化求值"
] | [
"$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{\\sqrt{2}}{2}(\\cos x-\\sin x)=\\frac{3}{5}$$，所以$$\\cos x-\\sin x=\\frac{3\\sqrt{2}}{5}$$， 所以$${{(\\cos x-\\sin x)}^{2}}=1-\\sin 2x=\\frac{18}{25}$$，所以$$\\sin 2x=\\frac{7}{25}$$，故选$$\\text{D}$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 589 | 686029d5c6bc4281ac28fc2486c119da | [
"2015年四川全国高中数学联赛竞赛初赛第3题5分"
] | 1 | single_choice | 已知二面角$$\alpha -l-\beta $$的大小为$$30{}^{}\circ $$，则由平面$$\alpha $$上的圆在平面$$\beta $$上的正射影得到的椭圆的离心率为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{3}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{2}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{\\sqrt{3}}{3}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{\\sqrt{3}}{2}$$ "
... | [
"竞赛->知识点->立体几何与空间向量->空间中的角与距离",
"竞赛->知识点->解析几何->椭圆"
] | [
"设圆的半径为$$1$$，则椭圆的长轴长为$$2$$，短轴长为$$2\\cos 30{}^{}\\circ =\\sqrt{3}$$，则$$c=\\frac{1}{2}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 293 | 16ac2bcfa03d4514b73cb1f5594fcf36 | [
"2016年陕西全国高中数学联赛竞赛初赛第3题6分"
] | 1 | single_choice | 设$$\vec{a},\vec{b},\vec{c}$$是同一平面内的三个单位向量，且$$\vec{a}\bot \vec{b}$$，则$$(\vec{c}-\vec{a})\cdot (\vec{c}-\vec{b})$$的最大值是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$1+\\sqrt{2}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1-\\sqrt{2}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\sqrt{2}-1$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1$$ "
}
]
] | [
"课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律",
"课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义",
"课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求模长",
"课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）",
"课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求夹角",
"课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断",
"课内体系->知识点->平面向量->平面向量的基本概念-... | [
"$$(\\vec{c}-\\vec{a})\\cdot (\\vec{c}-\\vec{b})=1-\\overrightarrow{c}\\left( \\overrightarrow{a}+\\overrightarrow{b} \\right)=1-\\left\\textbar{} \\overrightarrow{c} \\right\\textbar\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{b} \\right\\textbar\\text{cos }\\theta \\leqslant 1+\\sqrt{2}$$，当$$\\overright... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 294 | 626c476492ae4b79991ce8633ab2cad4 | [
"2019~2020学年4月湖南长沙天心区长郡中学高一下学期月考第12题3分",
"1998年全国高中数学联赛竞赛一试第3题6分",
"2014~2015学年10月辽宁沈阳和平区东北育才中学高二上学期月考理科第7题5分",
"2018~2019学年10月山东济宁市中区济宁市第一中学高二上学期月考第6题5分"
] | 1 | single_choice | 各项均为实数的等比数列$$\left { {{a}_{n}} \right }$$前$$n$$项之和记为$${{S}_{n}}$$，若$${{S}_{10}}=10$$，$${{S}_{30}}=70$$，则$${{S}_{40}}$$等于（~ ）． | [
[
{
"aoVal": "A",
"content": "$$150$$ "
}
],
[
{
"aoVal": "B",
"content": "$$-200$$ "
}
],
[
{
"aoVal": "C",
"content": "$$150$$或$$-200$$ "
}
],
[
{
"aoVal": "D",
"content": "$$-50$$或$$400$$~ "
}
]
] | [
"竞赛->知识点->数列与数学归纳法->等差数列与等比数列"
] | [
"首先$$q\\ne 1$$，于是$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{10}}-1 \\right)=10$$，$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{30}}-1 \\right)=70$$， ∴$$\\textasciitilde{{q}^{20}}+{{q}^{10}}+1=7\\Rightarrow {{q}^{10}}=2$$（$$-3$$舍）． ∴$${{S}_{40}}=10\\left( {{q}^{40}}1 \\right)=150$$． 故选$$\\text{A}$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1188 | eb94ba5fffff4e51ae3d04f5ff36a4d3 | [
"2006年上海复旦大学自主招生千分考第6题",
"2010年浙江全国高中数学联赛竞赛初赛第5题5分"
] | 2 | single_choice | 在正三棱柱$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$中，若$$AB=\sqrt{2}B{{B}_{1}}$$，则$$A{{B}_{1}}$$与$${{C}_{1}}B$$所成的角的大小是． | [
[
{
"aoVal": "A",
"content": "$$60{}^{}\\circ $$ "
}
],
[
{
"aoVal": "B",
"content": "$$75{}^{}\\circ $$ "
}
],
[
{
"aoVal": "C",
"content": "$$90{}^{}\\circ $$ "
}
],
[
{
"aoVal": "D",
"content": "$$105{}^{}\\circ $$ "
}
... | [
"竞赛->知识点->立体几何与空间向量->空间中的角与距离"
] | [
"取$${{B}_{1}}{{A}_{1}}$$的中点$$M$$，易知$${{C}_{1}}M\\bot $$平面$$A{{A}_{1}}{{B}_{1}}B$$，所以$${{C}_{1}}M\\bot A{{B}_{1}}$$， 在矩形$$AB{{B}_{1}}{{A}_{1}}$$中，$$\\tan \\angle {{B}_{1}}BM=\\tan \\angle {{B}_{1}}AB=\\frac{\\sqrt{2}}{2}$$， 于是$$\\angle {{B}_{1}}BM=\\angle {{B}_{1}}AB$$，从而$$MB\\bot A{{B}_{1}}$$， 因此，$$A{{B}_{1}}\\b... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 950 | e4a0014aaee64726b5ef689dd1b7fa1c | [
"2010年四川全国高中数学联赛竞赛初赛第3题5分"
] | 0 | single_choice | 设正三棱锥$$S-ABC$$的底面边长为$$3$$，侧棱长为$$2$$，则侧棱$$SA$$与底面$$ABC$$所成的角的大小是． | [
[
{
"aoVal": "A",
"content": "$$30{}^{}\\circ $$ "
}
],
[
{
"aoVal": "B",
"content": "$$45{}^{}\\circ $$ "
}
],
[
{
"aoVal": "C",
"content": "$$60{}^{}\\circ $$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\arctan \\sqrt{2}$$ "
}... | [
"竞赛->知识点->立体几何与空间向量->空间中的角与距离"
] | [
"设顶点$$S$$在底面$$\\triangle ABC$$的射影是$$H$$，则$$H$$为$$\\triangle ABC$$的外心．从而$$AH=\\frac{2}{3}\\times 3\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}$$，于是可得$$\\angle SAH={{30}^{\\circ }}$$．故选$$\\text{A}$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 199 | 0dbe980901f846d0ab2b16f22243f0af | [
"1994年全国高中数学联赛竞赛一试第4题"
] | 1 | single_choice | 已知$$0\textless{}b\textless{}1$$，$$0\textless{}a\textless{}\frac{ \pi }{4}$$，则下列三数：$$x={{(\sin a)}^{{{\log }_{b}}\sin a}}$$，$$y={{(\cos a)}^{{{\log }_{b}}\cos a}}$$，$$z={{(\sin a)}^{{{\log }_{b}}\cos a}}$$大小关系是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$x\\textless{}z\\textless{}y$$ "
}
],
[
{
"aoVal": "B",
"content": "$$y\\textless{}z\\textless{}x$$ "
}
],
[
{
"aoVal": "C",
"content": "$$z\\textless{}x\\textless{}y$$ "
}
],
[
{
"aoVal": "D",
"co... | [
"课内体系->知识点->基本初等函数->对数的概念及其运算",
"课内体系->知识点->基本初等函数->指对幂比较大小",
"课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数线",
"课内体系->素养->数学运算"
] | [
"∵$$0\\textless{}b\\textless{}1$$， ∴$$f(x)={{\\log }_{b}}x$$是减函数，又 ∵$$0\\textless{}a\\textless{}\\frac{ \\pi }{4}$$， ∴$$0\\textless{}\\sin a\\textless{}\\cos a\\textless{}1$$， ∴$${{\\log }_{b}}\\sin a\\textgreater{{\\log }_{b}}\\cos a\\textgreater0$$， ∴$${{(\\sin a)}^{{{\\log }_{b}}\\sin a}}\\textless{}{{(\\sin ... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1035 | e084cb5e007a498297ffc8dc860afe39 | [
"2008年四川全国高中数学联赛竞赛初赛第2题5分"
] | 1 | single_choice | 在公差为$$4$$的正项等差数列$$ {{{a}_{n}} }$$中，$${{a}_{3}}$$与$$2$$的算术平均值等于$${{S}_{3}}$$与$$2$$的几何平均值，其中$${{S}_{3}}$$表示数列前$$3$$项的和，则$${{a}_{10}}$$的值． | [
[
{
"aoVal": "A",
"content": "$$38$$ "
}
],
[
{
"aoVal": "B",
"content": "$$40$$ "
}
],
[
{
"aoVal": "C",
"content": "$$42$$ "
}
],
[
{
"aoVal": "D",
"content": "$$44$$ "
}
]
] | [
"课内体系->素养->数学运算",
"课内体系->思想->方程思想",
"课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题",
"课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列求和问题",
"课内体系->知识点->数列->等差数列->等差数列的性质及应用",
"课内体系->知识点->数列->数列的概念->数列的前n项和",
"课内体系->知识点->数列->数列的概念->数列的定义",
"课内体系->知识点->数列->数列的概念->数列的表示方法->通项公式->常见数列的通项公式",
"课内体系->方法->方程组法"
] | [
"因为等差数列$$ {{{a}_{n}} }$$的公差$$d=4$$， 则$${{a}_{3}}={{a}_{1}}+8$$，$${{S}_{3}}=3{{a}_{1}}+12$$， 从而$$\\frac{({{a}_{1}}+8)+2}{2}=\\sqrt{2(3{{a}_{1}}+12)}$$， 解得$${{a}_{1}}=2$$．所以$${{a}_{10}}=2+9\\times 4=38$$． 故选$$\\text{A}$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 442 | 24b84d997f8742eda421b32b67a4e6e5 | [
"2012年吉林全国高中数学联赛竞赛初赛第4题5分"
] | 1 | single_choice | 设方程$${{\log }_{4}}x-{{\left( \frac{1}{4} \right)}^{x}}=0$$与$${{\log }_{\frac{1}{4}}}x-{{\left( \frac{1}{4} \right)}^{x}}=0$$的根分别为$${{x}_{1}}, {{x}_{2}}$$，则 | [
[
{
"aoVal": "A",
"content": "$$0\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}1$$ "
}
],
[
{
"aoVal": "B",
"content": "$${{x}_{1}}{{x}_{2}}=1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$1\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}2$$ "
}
],
[
{
... | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"显然$${{x}_{2}}=\\frac{1}{2}$$．设$$f\\left( x \\right)={{\\log }_{4}}x-{{\\left( \\frac{1}{4} \\right)}^{x}}$$，则$$f\\left( 1 \\right)\\cdot f\\left( 2 \\right)\\textless{}0$$，所以$$1\\textless{}{{x}_{2}}\\textless{}2$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 415 | 2cdee0b2f87744e9905b50e2a21dc9ca | [
"2017年湖南全国高中数学联赛竞赛初赛第1题5分"
] | 2 | single_choice | 设集合$$X=\left { 1,2,3,\cdots ,2017 \right }$$，集合 $$S=\left { \left( x,y,z \right)\left\textbar{} x,y,z\in X,且三条件x\textless{}y\textless{}z,y\textless{}z\textless{}x,z\textless{}x\textless{}y 恰有一个成立\right. \right }$$ 若$$\left( x,y,z \right)\in S$$且$$\left( z,w,x \right)\in S$$，则下列选项正确的是． | [
[
{
"aoVal": "A",
"content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\notin S$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\in S$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left( y,z... | [
"竞赛->知识点->集合->集合的划分与覆盖",
"竞赛->知识点->集合->集合的概念与运算"
] | [
"特殊值排除法， 取$$x=2$$，$$y=3$$，$$z=4$$，$$w=1$$，显然满足$$(x,y,z)$$和$$(z,w,x)$$都在$$S$$中， 此时$$(y ,z,w)=(3,4,1)\\in S$$， $$(x,y,w)=(2,3,1)\\in S$$，故$$\\text A$$、$$\\text C$$、$$\\text D$$均错误； 只有$$\\text B$$成立，故选$$\\text{B}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1076 | b82fd682192444ccac52211ea4f5a164 | [
"竞赛第20题"
] | 2 | single_choice | 考虑三维空间中任意给定的空间四边形$ABCD$，\emph{A}，\emph{B}，\emph{C}，\emph{D}是四个顶点，四条线段$AB,BC,CD,DA$依次首尾相接，将点\emph{A}的内角定义为射线$AD$和射线$AB$的夹角，其补角为角\emph{A}的外角，其他顶点定义类似，考虑这种空间四边形的外角和\emph{x}，则有（~~~~~~~） | [
[
{
"aoVal": "A",
"content": "$x=2\\pi$ "
}
],
[
{
"aoVal": "B",
"content": "$x\\geq 2\\pi$ "
}
],
[
{
"aoVal": "C",
"content": "$x\\leq 2\\pi$ "
}
],
[
{
"aoVal": "D",
"content": "不能确定 "
}
]
] | [] | [
"\\hfill\\break 【分析】\\\\ 首先外角和与内角和之和为$4\\pi$，再根据各角的不等关系可证内角和不大于$2\\pi$，外角和不小于$2\\pi$．\\\\ 【详解】\\\\ 对于空间四边形，外角和与内角和之和为$4\\pi$，因此考虑内角和与$2\\pi$的大小关系．\\\\ 取\\emph{A}，\\emph{B}，\\emph{C}，\\emph{D}为正四面体的四个顶点，则内角和为$\\frac{\\pi }{3}\\times 4=\\frac{4\\pi }{3}\\textless{} 2\\pi$，\\\\ 因此选项AC都错误，\\\\ 接下来证明内角和不大于$2\\pi$．\\\\ 事... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 706 | 9145a7720ff84c54ad057b4e3dd6790c | [
"2009年黑龙江全国高中数学联赛竞赛初赛第5题5分"
] | 3 | single_choice | 已知平面区域$$\Omega = {(x,y)\textbar x+y\leqslant 6,x\geqslant 0,y\geqslant 0 }$$，$$A= {(x,y)\textbar x\leqslant 4,y\geqslant 0,x-2y\geqslant 0 }$$，若向区域$$\Omega $$上随机投一点$$P$$，则点$$P$$落入区域$$A$$的概率为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{2}{3}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{3}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{2}{9}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{1}{9}$$ "
}
]
] | [
"竞赛->知识点->排列组合与概率->概率初步",
"竞赛->知识点->集合->集合的划分与覆盖",
"竞赛->知识点->集合->集合的概念与运算"
] | [
"不难算出平面区域$$\\Omega $$的面积为$$\\frac{1}{2}\\cdot 6\\cdot 6=18$$，$$A$$的面积为$$\\frac{1}{2}\\cdot 4\\cdot 2=4$$， 所以所求概率为$$\\frac{4}{18}=\\frac{2}{9}$$． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 282 | 2b9f734382eb4cd0bf828aeccf2cfff0 | [
"2016年天津全国高中数学联赛竞赛初赛第1题6分"
] | 1 | single_choice | 函数$$f\left( x \right)=\left\textbar{} \sin 2x+\cos 2x \right\textbar$$的最小正周期是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$2 \\pi $$ "
}
],
[
{
"aoVal": "B",
"content": "$$ \\pi $$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{ \\pi }{2}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{ \\pi }{4}$$ "
}
]
] | [
"课内体系->素养->直观想象",
"课内体系->素养->数学运算",
"课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质->求正弦型函数的周期",
"课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质",
"课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式",
"课内体系->知识点->函数的应用->函数与方程->函数的图象变换问题->翻折变换问题",
"课内体系->思想->函数思想"
] | [
"$$f\\left( x \\right)=\\left\\textbar{} \\sin 2x+\\cos 2x \\right\\textbar$$可以改写为 $$f\\left( x \\right)=\\left\\textbar{} \\sqrt{2}\\sin \\left( 2x+\\frac{\\pi }{4} \\right) \\right\\textbar$$， 由函数图象之间的关系可见$$f\\left( x \\right)$$与$$g\\left( x \\right)=\\left\\textbar{} \\sin 2x \\right\\textbar$$有相同的最小正周期． 因为$$... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 797 | bac0b1e060794d7e931069c115726bc7 | [
"2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2题",
"2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2题",
"2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第4题"
] | 1 | single_choice | 已知角\emph{α}终边上一点\emph{M}的坐标为$(1,\sqrt{3})$，则$\sin 2\alpha =$（~~~~~~~） | [
[
{
"aoVal": "A",
"content": "$-\\frac{1}{2}$ "
}
],
[
{
"aoVal": "B",
"content": "$\\frac{1}{2}$ "
}
],
[
{
"aoVal": "C",
"content": "$-\\frac{\\sqrt{3}}{2}$ "
}
],
[
{
"aoVal": "D",
"content": "$\\frac{\\sqrt{3}}{2}$ "
}
... | [
"课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数"
] | [
"\\hfill\\break 【分析】\\\\ 根据题意，结合$\\alpha $所在象限，得到$\\sin \\alpha $和$\\cos \\alpha $的值，再根据公式，求得答案.\\\\ 【详解】\\\\ 由角$\\alpha $终边上一点\\emph{M}的坐标为$\\left( 1,\\sqrt{3} \\right)$，\\\\ 得$\\sin \\alpha =\\frac{\\sqrt{3}}{2}$，$\\cos \\alpha =\\frac{1}{2}$，\\\\ 故$\\sin 2\\alpha =2\\sin \\alpha \\cos \\alpha =\\frac{\\sqrt{3}}{... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 172 | 262f0636a3dc45fa9554ebb269caf0cf | [
"2008年河南全国高中数学联赛竞赛初赛第2题5分"
] | 1 | single_choice | $$\triangle ABC$$的三个内角$$A$$、$$B$$、$$C$$依次成等差数列，$$\sin A$$、$$\sin B$$、$$\sin C$$依次成等比数列，则$$\triangle ABC$$． | [
[
{
"aoVal": "A",
"content": "是直角三角形，但不是等腰三角形 "
}
],
[
{
"aoVal": "B",
"content": "不是等腰三角形，也不是直角三角形 "
}
],
[
{
"aoVal": "C",
"content": "是等腰直角三角形 "
}
],
[
{
"aoVal": "D",
"content": "是等边三角形 "
}
]
] | [
"竞赛->知识点->三角函数->三角恒等变换",
"竞赛->知识点->数列与数学归纳法->等差数列与等比数列"
] | [
"因为$$B=60{}^{}\\circ $$，$${{\\sin }^{2}}B=\\sin A\\sin C$$， 所以$${{\\sin }^{2}}60{}^{}\\circ =\\sin A\\sin (120{}^{}\\circ -A)$$， 可知$$\\frac{3}{4}=\\sin A\\left( \\frac{\\sqrt{3}}{2}\\cos A+\\frac{1}{2}\\sin A \\right)$$ $$=\\frac{\\sqrt{3}}{2}\\sin A\\cos A+\\frac{1}{2}{{\\sin }^{2}}A$$ $$=\\frac{\\sqrt{3}}{4}\... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 226 | 22172070d1f04eb3bd337f3de4b53cba | [
"2020年贵州全国高中数学联赛竞赛初赛第1题6分"
] | 1 | single_choice | 已知$$\text{i}$$是虚数单位，则$$\sum\limits_{k=1}^{2020}{\left( k\cdot {{\text{i}}^{k+1}} \right)}=$$． | [
[
{
"aoVal": "A",
"content": "$$-1010-1010\\text{i}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1010+1010\\text{i}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-1010+1010\\text{i}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1010-1010\\tex... | [
"竞赛->知识点->数列与数学归纳法->数列的通项与求和",
"课内体系->知识点->复数->复数的运算->复数的四则运算综合"
] | [
"设$$S=\\text{i}+2{{\\text{i}}^{2}}+3{{\\text{i}}^{3}}+\\cdots +2019{{\\text{i}}^{2019}}+2020{{\\text{i}}^{2020}}$$， 则$$\\text{i}S={{\\text{i}}^{2}}+2{{\\text{i}}^{3}}+3{{\\text{i}}^{4}}+\\cdots +2019{{\\text{i}}^{2020}}+2020{{\\text{i}}^{2021}}$$， 两式相减，得： $$S-\\text{i}S=\\text{i}+{{\\text{i}}^{2}}+{{\\text{i}}^{... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 252 | 38e3e419980d48049c12cc46a3de88f9 | [
"1994年全国高中数学联赛竞赛一试第2题"
] | 2 | single_choice | 给出下列两个命题： （$$1$$）设$$a$$、$$b$$、$$c$$都是复数，如果$${{a}^{2}}+{{b}^{2}}\textgreater{{c}^{2}}$$，则$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\textgreater0$$． （$$2$$）设$$a$$、$$b$$、$$c$$都是复数，如果$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\textgreater0$$，则$${{a}^{2}}+{{b}^{2}}\textgreater{{c}^{2}}$$． 那么下述说法正确的是（~ ）． | [
[
{
"aoVal": "A",
"content": "命题（$$1$$）正确，命题（$$2$$）也正确 "
}
],
[
{
"aoVal": "B",
"content": "命题（$$1$$）正确，命题（$$2$$）错误 "
}
],
[
{
"aoVal": "C",
"content": "命题（$$1$$）错误，命题（$$2$$）也错误 "
}
],
[
{
"aoVal": "D",
"content": "命题（$$1$$）错误，命... | [
"竞赛->知识点->复数与平面向量->复数的概念与运算"
] | [
"命题（$$1$$）是正确的．$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$表明$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，因此，根据移项法则有$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$． 命题（$$2$$）是错误的．$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$仅表明$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$$是实数，并不能保证$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，故$${... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1067 | a631421546d845c39b6e28842ff98619 | [
"2021年贵州全国高中数学联赛竞赛初赛第2题"
] | 1 | single_choice | 已知集合$$M=\left { \left( x,y \right)\textbar\textbar x\textbar+\textbar y\textbar\leqslant 1 \right }$$，集合$$N=\left { \left( x,y \right)\textbar{{x}^{2}}+{{y}^{2}}\leqslant \textbar x\textbar+\textbar y\textbar{} \right }$$，则． | [
[
{
"aoVal": "A",
"content": "$$M=N$$ "
}
],
[
{
"aoVal": "B",
"content": "$$M\\subseteq N$$ "
}
],
[
{
"aoVal": "C",
"content": "$$N\\subseteq M$$ "
}
],
[
{
"aoVal": "D",
"content": "前三个都不正确 "
}
]
] | [
"竞赛->知识点->集合->集合的概念与运算"
] | [
"对于任意$$\\left( x,y \\right)\\in M$$，均有$$\\textbar x\\textbar+\\textbar y\\textbar\\leqslant 1$$，故$$\\begin{cases}\\textbar x\\textbar\\leqslant 1 \\textbar y\\textbar\\leqslant 1 \\end{cases}$$，故有$$\\begin{cases}{{x}^{2}}\\leqslant \\textbar x\\textbar{} {{y}^{2}}\\leqslant \\textbar y\\textbar{} \\end{... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 20 | 0e236c8568224a9d92310b5202a16bcf | [
"2011年黑龙江全国高中数学联赛竞赛初赛第9题5分"
] | 1 | single_choice | 若点$$P$$在椭圆$$\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$$上，它到直线$$\frac{x}{4}+\frac{y}{3}=1$$的距离为$$\frac{6}{5}$$，则点$$P$$的个数为． | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4$$ "
}
]
] | [
"竞赛->知识点->解析几何->直线与圆锥曲线"
] | [
"由题知，可转换为该椭圆与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线与椭圆的交点个数，与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线方程为$$3x+4y-6=0$$或$$3x+4y-18=0$$，由$$\\begin{cases}\\frac{{{x}^{2}}}{16}+\\frac{{{y}^{2}}}{9}=1 3x+4y-18=0 \\end{cases}$$，消去$$y$$可得$$9{{x}^{2}}+{{\\left( 3x-18 \\righ... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 103 | 25b5944565bd427aaa1ecd915dfcb63f | [
"2010年四川全国高中数学联赛竞赛初赛第8题5分"
] | 1 | single_choice | 记$$F(x,y)={{(x-y)}^{2}}+{{\left( \frac{x}{3}+\frac{3}{y} \right)}^{2}}(y\ne 0)$$，则$$F(x, y)$$的最小值是． | [
[
{
"aoVal": "A",
"content": "$$\\frac{12}{8}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{16}{5}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{18}{5}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4$$ "
}
]
] | [
"竞赛->知识点->解析几何->双曲线"
] | [
"设动点$$P\\left( x, -\\frac{x}{3} \\right)$$与$$Q\\left( y,\\frac{3}{y} \\right)$$，则$$F(x, y)={{\\left\\textbar{} PQ \\right\\textbar}^{2}}$$，点$$P$$的轨迹为直线$$y=-\\frac{x}{3}$$，点$$Q$$的轨迹为双曲线$$y=\\frac{3}{x}$$，双曲线上的任一点$$\\left( {{x}_{0}},\\frac{3}{{{x}_{0}}} \\right)$$到直线$$x+3y=0$$的距离: $$d=\\frac{\\left\\textbar{} {{x}... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 162 | 106cf35a477843498a80e17651c54a13 | [
"2010年浙江全国高中数学联赛竞赛初赛第2题5分"
] | 1 | single_choice | 若$$p:({{x}^{2}}+x+1)\sqrt{x+3}\geqslant 0, q:x\geqslant -2$$，则$$p$$是$$q$$的． | [
[
{
"aoVal": "A",
"content": "充分而不必要条件 "
}
],
[
{
"aoVal": "B",
"content": "必要而不充分条件 "
}
],
[
{
"aoVal": "C",
"content": "充要条件 "
}
],
[
{
"aoVal": "D",
"content": "既不充分也不必要条件 "
}
]
] | [
"竞赛->知识点->不等式->不等式的解法"
] | [
"$$p$$成立$$\\Leftrightarrow x\\geqslant -3$$，所以$$p $$成立，推不出$$q$$一定成立． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 130 | 21594be7a7564473bb576e1f553a5ccc | [
"2009年浙江全国高中数学联赛竞赛初赛第4题5分"
] | 0 | single_choice | 已知平面上单位向量$$\overrightarrow{a}=\left( \frac{5}{13},\frac{12}{13} \right),\overrightarrow{b}=\left( \frac{4}{5},\frac{3}{5} \right)$$，则下列关系式正确的是． | [
[
{
"aoVal": "A",
"content": "$$\\overrightarrow{a}\\bot \\overrightarrow{b}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$ "
}
],
[
{
"aoVal": "C",
"content": "$$(\\o... | [
"课内体系->方法->图象法",
"课内体系->知识点->平面向量->向量应用->平面几何中的向量方法",
"课内体系->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式",
"课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义",
"课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）",
"课内体系->知识点->平面向量->平面向量的运算->数量积->利用数量积解决向量垂直问题（非坐标运算）",
"课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律",
"课内体系->知识点... | [
"因为$$\\overrightarrow{a},\\overrightarrow{b}$$都是非零单位向量，则以$$\\overrightarrow{a},\\overrightarrow{b}$$为边，$$\\overrightarrow{a}-\\overrightarrow{b},\\overrightarrow{a}+\\overrightarrow{b}$$为对角线构成一个菱形，所以$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 520 | d0fe4a56a1d04af7bd17f208adf3afcb | [
"2009年四川全国高中数学联赛竞赛初赛第4题5分"
] | 1 | single_choice | 设数列$$ {{{a}_{n}} }$$满足：$${{a}_{1}}=2,{{a}_{n+1}}=1-\frac{1}{{{a}_{n}}}$$，记数列$$ {{{a}_{n}} }$$的前$$n$$项之积为$${{P}_{n}}$$，则$${{P}_{2009}}$$的值为． | [
[
{
"aoVal": "A",
"content": "$$-\\frac{1}{2}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$-1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{1}{2}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1$$ "
}
]
] | [
"竞赛->知识点->数列与数学归纳法->（模）周期数列"
] | [
"因为 $${{a}_{n+2}}=1-\\frac{1}{{{a}_{n+1}}}=1-\\frac{1}{1-\\frac{1}{{{a}_{n}}}}=\\frac{-1}{{{a}_{n}}-1}$$， 于是 $${{a}_{n+3}}=1-\\frac{1}{{{a}_{n+2}}}=1-\\frac{1}{\\frac{-1}{{{a}_{n}}-1}}={{a}_{n}}$$， 故$$ {{{a}_{n}} }$$是以$$3$$为周期的周期数列． 又$${{a}_{1}}=2,{{a}_{2}}=\\frac{1}{2},{{a}_{3}}=-1$$，从而$${{P}_{3}}=-1$$，所以， $$... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 965 | 8aac49074f72f20f014f8174050d36b5 | [
"2010~2011学年北京海淀区高三上学期期末理科第7题",
"2018~2019学年北京西城区北京市第十三中学高二上学期期中第10题3分",
"2019~2020学年11月重庆沙坪坝区重庆市南开中学高二上学期周测A卷第6题5分",
"2019~2020学年10月北京海淀区北京市第五十七中学高二上学期月考第10题",
"2016年吉林全国高中数学联赛竞赛初赛第5题5分"
] | 2 | single_choice | 已知椭圆$$E:\frac{{{x}^{2}}}{m}+\frac{{{y}^{2}}}{4}=1$$，对于任意实数$$k$$，下列直线被椭圆$$E$$截得的弦长与$$l:y=kx+1$$被椭圆$$E$$截得的弦长不可能相等的是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$kx+y+k=0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$kx-y-1=0$$ "
}
],
[
{
"aoVal": "C",
"content": "$$kx+y-k=0$$ "
}
],
[
{
"aoVal": "D",
"content": "$$kx+y-2=0$$ "
}
]
] | [
"课内体系->知识点->圆锥曲线->直线与圆锥曲线问题->弦长求解问题",
"课内体系->知识点->圆锥曲线->椭圆->直线和椭圆的位置关系",
"课内体系->素养->数学运算"
] | [
"由数形结合可知，当$$l$$过点$$(-1,0)$$时，直线$$l$$和选项A中的直线关于$$x$$对称，被椭圆$$E$$所截得的弦长相同，故不能选A． 当$$l$$过点$$(1,0)$$时，直$$l$$和选项C中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选C． 当$$k=0$$时，直线$$l$$和选项B中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选B． 直线$$l$$斜率为$$k$$，在$$y$$轴上的截距为$$1$$；选项D中的直线$$kx+y-2=0$$ 斜率为$$-k$$，在$$y$$轴上的截距为$$2$$，这两直线不关于$$x$$轴、$$y$$轴、原点对称... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 662 | d5f6a51360534d7faaa0a61715943d2c | [
"2005年高考真题福建卷理科第12题5分",
"2009年黑龙江全国高中数学联赛竞赛初赛第6题5分"
] | 2 | single_choice | $$f(x)$$是定义在$$\text{R}$$上的以$$3$$为周期的奇函数，且$$f(2)=0$$，则方程$$f(x)=0$$在区间$$\left( 0,6 \right)$$内解的个数的最小值是（~ ） | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$3$$ "
}
],
[
{
"aoVal": "C",
"content": "$$4$$ "
}
],
[
{
"aoVal": "D",
"content": "$$5$$ "
}
]
] | [
"课内体系->知识点->函数的概念与性质->函数的性质->周期性->函数周期性与奇偶性综合问题",
"课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->抽象函数",
"课内体系->知识点->函数的应用->函数的零点->零点的个数问题->求零点个数问题（不含参）",
"课内体系->素养->数学抽象",
"课内体系->素养->数学运算"
] | [
"$$f(x)$$是定义在$$\\text{R}$$周期为$$3$$的奇函数，∴$$f(0)=0=f(3)$$，$$f(5)=f(2)=-f(-2)=-f(1)=0$$，∴$$f(1)=0=f(4)$$，∴选D. "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 69 | 079344bc716a4dcfa2c4c285d30c220f | [
"2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"
] | 1 | single_choice | 6个人围成一圈，若$$2$$种站法经过旋转可以重合则算同$$1$$种，则不同的站法有． | [
[
{
"aoVal": "A",
"content": "$$30$$种 "
}
],
[
{
"aoVal": "B",
"content": "$$60$$种 "
}
],
[
{
"aoVal": "C",
"content": "$$120$$种 "
}
],
[
{
"aoVal": "D",
"content": "$$720$$种 "
}
]
] | [
"竞赛->知识点->排列组合与概率->两个基本计数原理",
"竞赛->知识点->排列组合与概率->排列与组合",
"课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理"
] | [
"总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 834 | 77bd9818a7b74b3781318bdaf0e37387 | [
"2008年贵州全国高中数学联赛竞赛初赛第10题5分"
] | 1 | single_choice | 已知双曲线$$\frac{{{x}^{2}}}{6}-\frac{{{y}^{2}}}{3}=1$$的焦点$${{F}_{1}}$$、$${{F}_{2}}$$，点$$M$$在双曲线上，且$$M{{F}_{1}}\bot x$$轴，则$${{F}_{1}}$$到直线$${{F}_{2}}M$$的距离为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{6}{5}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{5}{6}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{3\\sqrt{6}}{5}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{5\\sqrt{6}}{6}$$ "... | [
"竞赛->知识点->解析几何->双曲线"
] | [
"由已知可得$$M\\left( 3,\\frac{\\sqrt{6}}{2} \\right)$$，则$$M{{F}_{1}}=\\frac{\\sqrt{6}}{2}$$， 故$$M{{F}_{2}}=2\\sqrt{6}+\\frac{\\sqrt{6}}{2}=\\frac{5\\sqrt{6}}{2}$$， 故$${{F}_{1}}$$到直线$${{F}_{2}}M$$的距离为： $$\\frac{\\left\\textbar{} {{F}_{1}}{{F}_{2}} \\right\\textbar\\cdot \\left\\textbar{} M{{F}_{1}}_{2} \\right\\textb... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 411 | be4956589d424dc5b715233b87fc9499 | [
"2009年河北全国高中数学联赛竞赛初赛第1题6分"
] | 0 | single_choice | 将数列$$ {2n-1 }(n\in N*)$$依原顺序按第$$n$$组有$${{2}^{n}}$$项的要求分组，则$$2009$$在第组． | [
[
{
"aoVal": "A",
"content": "$$7$$ "
}
],
[
{
"aoVal": "B",
"content": "$$8$$ "
}
],
[
{
"aoVal": "C",
"content": "$$9$$ "
}
],
[
{
"aoVal": "D",
"content": "$$10$$ "
}
]
] | [
"竞赛->知识点->数列与数学归纳法->数列的综合应用",
"竞赛->知识点->数列与数学归纳法->数列的通项与求和"
] | [
"前$$n$$组所含项数之和为$$2+{{2}^{2}}+\\cdots +{{2}^{n}}={{2}^{n+1}}-2$$， $$2009$$是数列的第$$1005$$项． $$510={{2}^{9}}-2\\textless{}1005\\textless{}{{2}^{10}}-2=1022$$． 因此，$$2009$$在第$$9$$组，选$$\\text{C}$$． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 922 | 8e1032fee70a43c3b7335b02cc1b0898 | [
"2008年河北全国高中数学联赛竞赛初赛第1题6分"
] | 1 | single_choice | 函数$$y=f(x+2)$$的图象过点$$(-1,3)$$，则函数$$f(x)$$的图象关于$$y$$轴对称的图形一定过点． | [
[
{
"aoVal": "A",
"content": "$$(1,-3)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$(-1,3)$$ "
}
],
[
{
"aoVal": "C",
"content": "$$(-3,-3)$$ "
}
],
[
{
"aoVal": "D",
"content": "$$(-3,3)$$ "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质",
"竞赛->知识点->函数->函数综合"
] | [
"函数$$y=f(x+2)$$的图象过点$$(-1,3)$$，则$$f(-1+2)=3$$，即$$f(1)=3$$，所以数$$f(x)$$的图象关于$$y$$轴对称的图形一定过点$$f(-1)=3$$．故选$$\\text{B}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 171 | 1d2b0485674e496ebfdfe4358d643b77 | [
"2009年AMC10竞赛B第23题"
] | 3 | single_choice | 瑞秋和罗伯特在一个圆形跑道上跑步．瑞秋逆时针跑，每隔$90$秒完成一圈，罗伯特顺时针跑，每隔$80$秒完成一圈．两人在同一时间从同一条线上出发．在他们开始跑步后的$10$分钟和$11$分钟之间的某个随机时间，一个站在赛道内的摄影师拍了一张照片，照片显示了以起跑线为中心的四分之一的赛道．瑞秋和罗伯特都在照片上的概率是多少？ | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{16}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{8}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{3}{16}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{1}{4}$$ "
}
],
[
... | [
"课内体系->学科符号->数与式->新定义符号->取最值函数->max{f(x),g(x)}",
"美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities"
] | [
"After $$10$$ minutes ($$600$$ seconds), Rachel will have completed $$6$$ laps and be $$30$$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $$22.5$$ seconds, she will be in the picture between $$18.75$$ seconds and $$41.25$$ seconds of the tenth minute. After $$10$$ minutes Robe... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 371 | 1bc4df3971b146dc892cc35a3d5426de | [
"2018~2019学年10月四川成都金牛区成都市实验外国语学校高二上学期周测C卷第9题5分",
"2017年河南洛阳高三二模文科第9题5分",
"2016年广东揭阳高三一模理科第12题5分",
"2014年黑龙江全国高中数学联赛竞赛初赛第10题5分",
"2016年山东日照高三二模理科第9题5分"
] | 2 | single_choice | 已知直线$$x+y-k=0\left( k\textgreater0 \right)$$与圆$${{x}^{2}}+{{y}^{2}}=4$$交于不同的两点$$A$$，$$B$$，$$O$$是坐标原点，且有$$\left\textbar{} \overrightarrow{OA}+\overrightarrow{OB}\left\textbar{} \geqslant \frac{\sqrt{3}}{3} \right\textbar\overrightarrow{AB} \right\textbar$$，那么$$k$$的取值范围是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$\\left( \\sqrt{3},+\\infty \\right)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\left[ \\sqrt{2},+\\infty \\right)$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\left[ \\sqrt{2},2\\sqrt{2} \\right)$$ "
}
],
[
{
... | [
"课内体系->知识点->直线和圆的方程->圆与方程->直线与圆的位置关系->圆的切线的相关问题",
"课内体系->素养->数学运算"
] | [
"设$$AB$$中点为$$D$$，则$$OD\\bot AB$$， ∵$$\\left\\textbar{} \\overrightarrow{OA}+\\overrightarrow{OB}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$， ∴$$\\left\\textbar{} 2\\overrightarrow{OD}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\o... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 919 | fb96b44cd915485d822b88451afe11cb | [
"2018年黑龙江全国高中数学联赛竞赛初赛第3题5分"
] | 0 | single_choice | $${{\left( a+2b-3c \right)}^{4}}$$的展开式中$$ab{{c}^{2}}$$的系数为． | [
[
{
"aoVal": "A",
"content": "$$208$$ "
}
],
[
{
"aoVal": "B",
"content": "$$216$$ "
}
],
[
{
"aoVal": "C",
"content": "$$217$$ "
}
],
[
{
"aoVal": "D",
"content": "$$218$$ "
}
]
] | [
"课内体系->素养->数学运算",
"课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数",
"课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项"
] | [
"依题意，知$$ab{{c}^{2}}$$的系数为$$\\text{C}_{4}^{2}{{(-3)}^{2}}\\text{C}_{2}^{1}\\times 2\\times 1=216$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 218 | 9432909b0d48427eaaee5bc460b25255 | [
"1989年全国高中数学联赛竞赛一试第3题"
] | 1 | single_choice | (★★)对任意的函数$$y=f\left( x \right)$$，在同一个直角坐标系中，函数$$y=f\left( x-1 \right)$$与函数$$y=f\left( -x+1 \right)$$的图象恒． | [
[
{
"aoVal": "A",
"content": "关于$$x$$轴对称； "
}
],
[
{
"aoVal": "B",
"content": "关于直线$$x=1$$对称 "
}
],
[
{
"aoVal": "C",
"content": "关于直线$$x=-1$$对称 "
}
],
[
{
"aoVal": "D",
"content": "关于$$y$$轴对称 "
}
]
] | [
"课内体系->知识点->函数的概念与性质->函数的性质->对称性->两个函数的互对称问题",
"竞赛->知识点->函数->函数的图像与性质"
] | [
"$$f(x)$$和$$f(-x)$$的图象关于直线$$x=0$$对称， $$f(x-1)$$与$$f(-x+1)$$的图象关于直线$$x=1$$对称． 故选$$\\text{B}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1034 | c54152a426ae440687e703a262ce3f9c | [
"2018年天津全国高中数学联赛竞赛初赛第4题6分"
] | 1 | single_choice | 以下四个数中，最大的为． | [
[
{
"aoVal": "A",
"content": "$$\\ln \\sqrt{2}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{\\text{e}}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{\\ln \\pi }{\\pi }$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{\\sq... | [
"竞赛->知识点->导数模块->导数"
] | [
"考虑函数$$f\\left( x \\right)=\\frac{\\ln x}{x}$$． 则四个选项分别为$$f\\left( 2 \\right)$$、$$f\\left( \\text{e} \\right)$$、$$f\\left( \\pi \\right)$$、$$f\\left( \\sqrt{10} \\right)$$， 又$${f}'\\left( x \\right)=\\frac{1-\\ln x}{{{x}^{2}}}$$，则$$f\\left( x \\right)$$在区间$$\\left( 0,\\text{e} \\right)$$上单调递增，在区间$$\\left( \\text{... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 89 | 087bc89ba3294db9a5015dfb5baddd03 | [
"2022年湖南湘西吉首市竞赛（第一届中小学生教师解题大赛）第2题"
] | 1 | single_choice | 已知复数$z$满足$2\le \left\textbar{} z-\left( 1-\text{i} \right) \right\textbar\le 3$，则复数$z$在复平面内对应的点$Z$所在区域的面积为（~~~~~~~） | [
[
{
"aoVal": "A",
"content": "$\\text{ } ! !\\pi ! !\\text{ }$ "
}
],
[
{
"aoVal": "B",
"content": "$3\\text{ } ! !\\pi ! !\\text{ }$ "
}
],
[
{
"aoVal": "C",
"content": "$5\\text{ } ! !\\pi ! !\\text{ }$ "
}
],
[
{
"aoVal": "D",
... | [] | [
"\\hfill\\break 【分析】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，可得$1\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 4$，然后根据复数模的几何意义结合条件即得.\\\\ 【详解】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，则$2\\le \\left\\textbar{} (a-1)+(b+1)\\text{i} \\right\\textbar\\le 3$，\\\\ 所以$4\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 9$，\\\\ 所以复数$z$在复平面内对... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 427 | 479ba9a7bad34506a145194f37d17d97 | [
"2016年高考真题天津卷",
"2016年天津全国高中数学联赛竞赛初赛第3题6分"
] | 1 | single_choice | 设$$0\textless{}a\textless{}b$$，已知$$a$$，$$s$$，$$t$$，$$b$$依次成等差数列，$$a$$，$$u$$，$$v$$，$$b$$依次成等比数列，记$$x=st\left( s+t \right)$$，$$y=uv\left( u+v \right)$$，则． | [
[
{
"aoVal": "A",
"content": "$$x\\textgreater y$$ "
}
],
[
{
"aoVal": "B",
"content": "$$x=y$$ "
}
],
[
{
"aoVal": "C",
"content": "$$x\\textless{}y$$ "
}
],
[
{
"aoVal": "D",
"content": "既有$$x\\textgreater y$$的情形，也有$$x\\textle... | [
"竞赛->知识点->不等式->几个重要的不等式->排序",
"竞赛->知识点->不等式->几个重要的不等式->均值",
"竞赛->知识点->不等式->不等式的综合应用（数列与不等式）",
"课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件"
] | [
"由题设有$$x=\\frac{2a+b}{3}\\cdot \\frac{a+2b}{3}\\left( a+b \\right)$$，$$y=ab\\left( \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}} \\right)$$， 由均值不等式，$$\\frac{2a+b}{3}\\geqslant \\sqrt[3]{{{a}^{2}}b}$$，$$\\frac{a+2b}{3}\\geqslant \\sqrt[3]{a{{b}^{2}}}$$（等号都不成立）， 由排序不等式，$$a+b=\\sqrt[3]{{{a}^{3}}}+\\sqrt[3]{{{b}^{3}}}\... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1209 | fe86cb491f514cd9a3f2dedb9a82d301 | [
"2014年四川全国高中数学联赛竞赛初赛第5题5分"
] | 2 | single_choice | 半径为$$6$$的球，则该球内接正三棱锥的体积的最大值是． | [
[
{
"aoVal": "A",
"content": "$$32\\sqrt{3}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$54\\sqrt{3}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$64\\sqrt{3}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$72\\sqrt{3}$$ "
}
]
] | [
"竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题",
"竞赛->知识点->三角函数->三角恒等变换"
] | [
"设棱锥的高与一条侧棱的夹角为$$\\theta $$，侧棱长$$l=12\\cos \\theta $$，高$$h=12{{\\cos }^{2}}\\theta $$，底面正三角形的外接圆半径$$R=12\\sin \\theta \\cos \\theta $$，底面边长为$$a=12\\sqrt{2}\\sin \\theta \\cos \\theta $$，从而 $$V=432\\sqrt{3}{{\\cos }^{4}}\\theta {{\\sin }^{2}}\\theta $$ $$=216\\sqrt{3}\\cdot {{\\cos }^{2}}\\theta \\cdot {{\\cos }^{... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 779 | 581227768c344530932c1599aa34ab0c | [
"2009年河北全国高中数学联赛竞赛初赛第3题6分"
] | 1 | single_choice | 设函数$$f(x)=\ln x$$，$$g(x)=ax+\frac{b}{x}$$，它们的图象在$$x$$轴上的公共点处有公切线，则当$$x\textgreater1$$时，$$f(x)$$与$$g(x)$$的大小关系是． | [
[
{
"aoVal": "A",
"content": "$$f(x)\\textgreater g(x)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$f(x)\\textless{}g(x)$$ "
}
],
[
{
"aoVal": "C",
"content": "$$f(x)=g(x)$$ "
}
],
[
{
"aoVal": "D",
"content": "不确定 "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质",
"竞赛->知识点->导数模块->导数"
] | [
"$$f(x)$$与$$g(x)$$的图象在$$x$$轴上有唯一公共点$$(1,0)$$， 所以$$g(1)=0$$，即$$a+b=0$$． 因为$${{f}^{\\prime }}(x)=\\frac{1}{x}$$，$${{g}^{\\prime }}(x)=a-\\frac{b}{{{x}^{2}}}$$， 由题意$${{f}^{\\prime }}(1)={{g}^{\\prime }}(1)=1$$，即$$a-b=1$$． 所以$$a=\\frac{1}{2}$$，$$b=-\\frac{1}{2}$$． 令$$F(x)=f(x)-g(x)=\\ln x-\\left( \\frac{1}{2}x-\\f... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 937 | 81b81f3da59f4f0ca68c7c41e877a5a6 | [
"2002年全国高中数学联赛竞赛一试第3题6分"
] | 1 | single_choice | 函数$$f\left( x \right)=\frac{x}{1-{{2}^{x}}}-\frac{x}{2}$$（~ ~ ） | [
[
{
"aoVal": "A",
"content": "是偶函数但不是奇函数 "
}
],
[
{
"aoVal": "B",
"content": "是奇函数但不是偶函数 "
}
],
[
{
"aoVal": "C",
"content": "既是奇函数又是偶函数 "
}
],
[
{
"aoVal": "D",
"content": "既不是奇函数又不是偶函数 "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"$$f\\left( x \\right)$$定义域为$$(-\\infty ,0)\\cup (0,+\\infty )$$；$$f\\left( x \\right)-f\\left( -x \\right)=\\frac{x}{1-{{2}^{x}}}-\\frac{x}{2}-\\frac{-x}{1-{{2}^{-x}}}+\\frac{-x}{2}=\\frac{x-x{{2}^{x}}}{1-{{2}^{x}}}-x=0$$．即$$f\\left( x \\right)$$是偶函数. "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 852 | 6ee6a08223cc4da786ff41f3090d1c03 | [
"2019年吉林全国高中数学联赛竞赛初赛第4题5分"
] | 1 | single_choice | 在正方形$$ABCD$$中，$$M$$为边$$BC$$的中点．若$$\overrightarrow{AC}=\lambda \overrightarrow{AM}+\mu \overrightarrow{BD}$$，则$$\lambda +2\mu =$$． | [
[
{
"aoVal": "A",
"content": "$$\\frac{4}{3}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{5}{3}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{15}{8}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$2$$ "
}
]
] | [
"竞赛->知识点->复数与平面向量->平面向量的概念与运算",
"课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则"
] | [
"注意到， $$\\overrightarrow{AC}=\\lambda \\overrightarrow{AM}+\\mu \\overrightarrow{BD}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\lambda \\left( \\overrightarrow{AB}+\\overrightarrow{BM} \\right)+\\mu \\left( \\overrightarrow{BA}+\\overrigh... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 814 | 61301464eee94f0d8ebaed46945786db | [
"高二下学期单元测试《整除与同余》自招",
"第二十届全国希望杯高一竞赛复赛邀请赛第6题4分"
] | 1 | single_choice | 关于$$x$$的整系数一元二次方程$$a{{x}^{2}}+bx+c=0(a\ne 0)$$中，若$$a+b$$是偶数，$$c$$是奇数，则． | [
[
{
"aoVal": "A",
"content": "方程没有整数根 "
}
],
[
{
"aoVal": "B",
"content": "方程有两个相等的整数根 "
}
],
[
{
"aoVal": "C",
"content": "方程有两个不相等的整数根 "
}
],
[
{
"aoVal": "D",
"content": "不能判定方程整数根的情况 "
}
]
] | [
"竞赛->知识点->多项式与方程->解方程（组）"
] | [
"若方程有整数根，则根据题意 $$a{{x}^{2}}+bx+c=\\begin{cases}c(\\text{mod}2)\\left. 2 \\right\\textbar x a+b+c(\\text{mod}2)\\left. 2 \\right\\textbar x \\end{cases}\\equiv 1(\\text{mod}2)$$ 矛盾，因此题中方程没有整数根． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 385 | 3e55864c3db744fe838230dc96a86068 | [
"2019~2020学年4月山西太原迎泽区太原市第五中学高二下学期周测C卷理科第6题6分",
"2012年黑龙江全国高中数学联赛竞赛初赛第2题5分"
] | 1 | single_choice | 设集合$$A=\left { 1, 2, 3, 4, 5, 6 \right }$$，$$B=\left { 4, 5, 6, 7 \right }$$，则满足$$S\subseteq A$$且$$S\cap B\ne \varnothing $$的集合$$S$$的个数为． | [
[
{
"aoVal": "A",
"content": "$$57$$ "
}
],
[
{
"aoVal": "B",
"content": "$$56$$ "
}
],
[
{
"aoVal": "C",
"content": "$$49$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
]
] | [
"竞赛->知识点->集合->集合的概念与运算"
] | [
"$$S\\subseteq A$$，且$$S\\cap B\\ne \\varnothing $$，说明$$S$$是$$A$$的子集，且$$S$$与$$B$$有公共元素； ∴$$A$$的构成情况为：①含一个元素：从$$4$$，$$5$$，$$6$$中选一个元素，个数为$$\\text{C}_{3}^{1}=3$$； ②含两个元素:从$$4$$，$$5$$，$$6$$选两个元素，或从$$1$$，$$2$$，$$3$$选一个，从$$4$$，$$56$$选一个，个数为：$$\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{1}=12$$；③含三个元素：从$$4$$，$$5$$... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 891 | 78324fcaf23d42feb7034bbb85022a54 | [
"2020~2021学年浙江杭州萧山区杭州第二中学钱江学校高一下学期期中第5题5分",
"2012年浙江全国高中数学联赛竞赛初赛第3题5分",
"2016年上海闸北区高三二模理科第11题5分"
] | 1 | single_choice | 已知$$\overrightarrow{a}$$与$$\overrightarrow{b}$$均为单位向量，其夹角为$$\theta $$，则命题$$P:\textbar\overrightarrow{a}-\overrightarrow{b}\textbar\textgreater1$$是命题$$Q:\theta \in \left[ \frac{ \pi }{2},\frac{5 \pi }{6} \right)$$的（~ ）． | [
[
{
"aoVal": "A",
"content": "充分非必要条件 "
}
],
[
{
"aoVal": "B",
"content": "必要非充分条件 "
}
],
[
{
"aoVal": "C",
"content": "充分且必要条件 "
}
],
[
{
"aoVal": "D",
"content": "非充分且非必要条件 "
}
]
] | [
"课内体系->素养->逻辑推理",
"课内体系->素养->数学运算",
"课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律",
"课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断",
"课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与向量结合"
] | [
"若$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{b} \\right\\textbar\\textgreater1$$，则$${{\\overrightarrow{a}}^{2}}-2\\overrightarrow{a}\\cdot \\overrightarrow{b}+{{\\overrightarrow{b}}^{2}}=2-2\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textgreater1$$，即$$\\overrightarrow{a}\\cdot \\overrightarrow{b}\\... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 960 | 92eb030c0c0248c280733d98e16a5358 | [
"2014年AMC12竞赛A第11题",
"2014年AMC10竞赛A第15题"
] | 2 | single_choice | $$2014-AMC10A-15$$ David drives from his home to the airport to catch a flight. He drives $$35$$ miles in the first hour, but realizes that he will be $$1$$ hour late if he continues at this speed. He increases his speed by $$15$$ miles per hour for the rest of the way to the airport and arrives $$30$$ minutes early. ... | [
[
{
"aoVal": "A",
"content": "$$140$$ "
}
],
[
{
"aoVal": "B",
"content": "$$175$$ "
}
],
[
{
"aoVal": "C",
"content": "$$210$$ "
}
],
[
{
"aoVal": "D",
"content": "$$245$$ "
}
],
[
{
"aoVal": "E",
"conte... | [
"课内体系->知识点->等式与不等式->等式->方程组的解集",
"美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"
] | [
"Note that he drives at $$50$$ miles per hour after the first hour and continues doing so until he arrives. Let $$d$$ be the distance still needed to travel after the frst $$1$$ hour.~ We have that $$\\dfrac{d}{50}+1.5= \\dfrac{d}{35}$$, ~where the $$1.5$$ comes from $$1$$ hour late decreased to $$0.5$$ hours early... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 273 | 5479f4bb7ae94dd9a312faa6b41898a7 | [
"2017年陕西全国高中数学联赛竞赛初赛第3题6分"
] | 0 | single_choice | 在空间直角坐标系中，$$\triangle ABC$$的三个顶点坐标分别为$$A\left( 3,4,1 \right)$$，$$B\left( 0,4,5 \right)$$，$$C\left( 5,2,0 \right)$$，则$$\tan \frac{A}{2}$$的值是． | [
[
{
"aoVal": "A",
"content": "$$\\sqrt{5}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{\\sqrt{5}}{5}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\sqrt{6}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{\\sqrt{6}}{6}$$ "
}
... | [
"竞赛->知识点->立体几何与空间向量->空间向量",
"竞赛->知识点->三角函数->三角恒等变换"
] | [
"$$\\cos A=\\frac{\\overrightarrow{AB}\\cdot \\overrightarrow{AC}}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}=\\frac{\\left( -3,0,4 \\right)\\cdot \\left( 2,-2,-1 \\right)}{\\sqrt{{{\\left( -3 \\right)}^{2}}+{{4}^{2}}}\\cdot \\sqrt{{{2}^{2}}+{{\\l... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1194 | fe2ba6e532c54158a24ad5c61acf05cb | [
"1999年全国全国高中数学联赛竞赛一试第5题6分"
] | 1 | single_choice | 在某次乒乓球单打比赛中，原计划每两名选手恰比赛一场，但有$$3$$名选手各比赛了$$2$$场之后就退出了，这样，全部比赛只进行了$$50$$场．那么，在上述$$3$$名选手之间比赛的场数是（ ）． | [
[
{
"aoVal": "A",
"content": "$$0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$2$$ "
}
],
[
{
"aoVal": "D",
"content": "$$3$$ "
}
]
] | [
"竞赛->知识点->排列组合与概率->排列与组合"
] | [
"解法一： 设这三名选手之间的比赛场数是$$r$$，共$$n$$名选手参赛． 由题意，可得$$\\text{C}_{n-3}^{2}+6-r=50$$， 即$$\\frac{\\left( n-3 \\right)\\left( n-4 \\right)}{2}=44+r$$． 由于$$0\\leqslant r\\leqslant 3$$， 经检验可知，仅当$$r=1$$时，$$n=13$$为正整数． 解法二： $$3$$名选手之间比赛的可能场数为$$0$$、$$1$$、$$2$$、$$3$$，设总人数为$$N$$人． 那么除这$$3$$人外的$$N-3$$人中比赛场数为$$C_{N-3}^{2}=\\fr... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 262 | 124d4a1f19f54e4fa8654685f6722570 | [
"2019~2020学年上海浦东新区华东师范大学第二附属中学高一上学期单元测试第5题",
"2003年全国全国高中数学联赛竞赛一试第13题20分"
] | 2 | single_choice | 设$$\frac{3}{2}\leqslant x\leqslant 5$$，则下式$$2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}$$最大不超过． | [
[
{
"aoVal": "A",
"content": "$${}2\\sqrt{17}$$ "
}
],
[
{
"aoVal": "B",
"content": "$${}2\\sqrt{19}$$ "
}
],
[
{
"aoVal": "C",
"content": "$${}2\\sqrt{21}$$ "
}
],
[
{
"aoVal": "D",
"content": "$${}2\\sqrt{23}$$ "
}
]
] | [
"竞赛->知识点->不等式->几个重要的不等式->柯西",
"课内体系->知识点->等式与不等式->不等式->柯西不等式"
] | [
"由$${{\\left( a+b+c+d \\right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+2\\left( ab+bc+cd+da+ac+bd \\right)$$可得 $$a+b+c+d\\leqslant 2\\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}}$$当且仅当$$a=b=c=d$$时取等号， 由柯西不等式 $${{\\left( 2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x} \\right)}^{2}}=\\left( \\sqrt{x+1}+\\sqrt{x+1}... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 862 | 77f524f6ccfe4a319049484dc828a2a8 | [
"2008年福建全国高中数学联赛竞赛初赛第3题6分"
] | 1 | single_choice | 在直角坐标平面$$xOy$$中，点$$A\left( 5,0 \right)$$．对于某个正实数$$k$$，存在函数$$f(x)=a{{x}^{2}}$$，$$a\textgreater0$$，使得$$\angle QOA=2\angle POA$$，这里$$P(1,f(1))$$、$$Q(k,f(k))$$，则$$k$$的取值范围为． | [
[
{
"aoVal": "A",
"content": "$$(2,+\\infty )$$ "
}
],
[
{
"aoVal": "B",
"content": "$$(3,+\\infty )$$ "
}
],
[
{
"aoVal": "C",
"content": "$$[4,+\\infty )$$ "
}
],
[
{
"aoVal": "D",
"content": "$$[8,+\\infty )$$ "
}
]
] | [
"课内体系->知识点->直线和圆的方程->直线与方程->倾斜角和斜率的概念->斜率计算",
"课内体系->知识点->三角函数->三角函数的图象与性质->正切函数的图象和性质",
"课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正切",
"课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式",
"课内体系->素养->数学运算"
] | [
"由于$$0\\textless{}\\angle QOA$$，$$\\angle POA\\textless{}\\frac{ \\pi }{2}$$， 因此$$\\angle QOA=2\\angle POA$$的充要条件是$$\\tan \\angle QOA=\\tan 2\\angle POA$$． 由$$\\tan \\angle QOA=\\tan 2\\angle POA=\\frac{2\\tan \\angle POA}{1-{{\\tan }^{2}}\\angle POA}$$， 结合$$\\tan \\angle POA=a$$，$$\\tan \\angle QOA=ak$$， 得$$ak... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 223 | d51d9cef28454ea695b0ccc877bd266e | [
"2008年贵州全国高中数学联赛竞赛初赛第2题5分"
] | 1 | single_choice | 函数$$f(x)=\sqrt{3}\sin 2x+\cos 2x$$的最小正周期是． | [
[
{
"aoVal": "A",
"content": "$$\\frac{ \\pi }{4}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{ \\pi }{2}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$ \\pi $$ "
}
],
[
{
"aoVal": "D",
"content": "$$2 \\pi $$ "
}
]
] | [
"竞赛->知识点->三角函数->三角函数的图像与性质",
"竞赛->知识点->三角函数->三角恒等变换"
] | [
"$$f\\left( x \\right)={{\\cos }^{2}}2x+\\sqrt{3}\\sin 2x\\cos 2x$$ $$=\\frac{1+\\cos 4x}{2}+\\frac{\\sqrt{3}}{2}\\sin 4x$$ $$=\\sin \\left( 4x+\\frac{ \\pi }{6} \\right)+\\frac{1}{2}$$， 所以最小正周期$$T=\\frac{2 \\pi }{4}=\\frac{ \\pi }{2}$$，故选$$\\text{B}$$. "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 101 | 2ee9c14234fb4d89982890d59a6c659a | [
"1988年全国高中数学联赛竞赛一试第2题"
] | 0 | single_choice | 已知原点在椭圆$${{k}^{2}}{{x}^{2}}+{{y}^{2}}-4kx+2ky+{{k}^{2}}-1=0$$的内部，那么参数$$k$$的取值范围是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$\\textbar k\\textbar\\textgreater1$$； "
}
],
[
{
"aoVal": "B",
"content": "$$\\textbar k\\textbar\\ne 1$$； "
}
],
[
{
"aoVal": "C",
"content": "$$-1\\textless{}k\\textless{}1$$； "
}
],
[
{
"aoVal": "D",... | [
"竞赛->知识点->解析几何->椭圆"
] | [
"椭圆外部的点可以离原点很远，它的坐标$$x\\mathsf{}y$$的绝对值可以很大，使得方程左边大于$$0$$，所以内部点的坐标使方程左边小于$$0$$，用原点的坐标代入方程左边得 $${{K}^{2}}-1\\textless{}0$$ 又$$k\\ne 0$$，（否则方程不表示椭圆），所以答$$\\text{D}$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 233 | 156e6b80522441668e6f3d1873ca532f | [
"2003年AMC10竞赛B第7题"
] | 1 | single_choice | $$[x]$$ 表示不超过$$x$$的最大整数$$.$$ 例如, $$[3]=3$$, $$\left[\frac 92\right]=4$$. 求$$[\sqrt 1]+[\sqrt 2]+[\sqrt 3]+\cdots +[\sqrt {16}]$$的值. | [
[
{
"aoVal": "A",
"content": "$$35$$ "
}
],
[
{
"aoVal": "B",
"content": "$$38$$ "
}
],
[
{
"aoVal": "C",
"content": "$$40$$ "
}
],
[
{
"aoVal": "D",
"content": "$$42$$ "
}
],
[
{
"aoVal": "E",
"content":... | [
"美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)",
"课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->取整函数（高斯函数）"
] | [
"The first three values in the sum are equal to $$1$$, the next five equal to $$2$$, the next seven equal to $$3$$, and the last one equal to $$4$$, For example, since $$2^{2}=4$$ any square root of a number less than $$4$$ must be less than $$2$$. Sum them all together to get $$3\\times 1+5\\times 2+7\\times 3+1\\... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 738 | ace0c84fa43f4097a90536384ac7d9a0 | [
"2008年黑龙江全国高中数学联赛竞赛初赛第3题5分"
] | 0 | single_choice | 已知$$P$$为三角形$$ABC$$内部任一点（不包括边界），且满足$$(\overrightarrow{PB}-\overrightarrow{PA})\cdot (\overrightarrow{PB}+\overrightarrow{PA}-2\overrightarrow{PC})=0$$,则$$\triangle ABC$$一定为． | [
[
{
"aoVal": "A",
"content": "直角三角形 "
}
],
[
{
"aoVal": "B",
"content": "等边三角形 "
}
],
[
{
"aoVal": "C",
"content": "等腰三角形 "
}
],
[
{
"aoVal": "D",
"content": "等腰直角三角形 "
}
]
] | [
"竞赛->知识点->复数与平面向量->平面向量的概念与运算"
] | [
"因为$$\\begin{cases}\\overrightarrow{PB}-\\overrightarrow{PA}=\\overrightarrow{AB} \\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC}=\\overrightarrow{CB}+\\overrightarrow{CA} \\end{cases}$$， 所以已知条件可改写为$$\\overrightarrow{AB}\\cdot (\\overrightarrow{CB}+\\overrightarrow{CA})=0$$． 容易得到此三角形为等腰三角形．故选$$... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 157 | 104d9364e5d84d70ab6c19cb8a9f091b | [
"竞赛第9题"
] | 3 | single_choice | 已知\emph{x}，\emph{y}，\emph{z}是非负实数，且$x+y+z=2$，则$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+xyz$的最大值为（~~~~~~~） | [
[
{
"aoVal": "A",
"content": "1 "
}
],
[
{
"aoVal": "B",
"content": "2 "
}
],
[
{
"aoVal": "C",
"content": "$\\frac{5}{4}$ "
}
],
[
{
"aoVal": "D",
"content": "以上答案都不对 "
}
]
] | [] | [
"\\hfill\\break 【分析】\\\\ 利用基本不等式可求最大值.\\\\ 【详解】\\\\ $2x^{2}y^{2}\\leq xy\\left(x^{2}+y^{2}\\right)$，$2y^{2}z^{2}\\leq yz\\left(y^{2}+z^{2}\\right)$，$2z^{2}x^{2}\\leq zx\\left(z^{2}+x^{2}\\right)$，\\\\ 所以$2x^{2}y^{2}+2y^{2}z^{2}+2z^{2}x^{2}+2xyz\\leq xy\\left(x^{2}+y^{2}\\right)+yz\\left(y^{2}+z^{2}\\right)+zx\\left... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1081 | b843f51510874d5394e94aee56f39950 | [
"1986年全国高中数学联赛竞赛一试第4题"
] | 1 | single_choice | 如果四面体的每一个面都不是等腰三角形，那么其长度不等的棱的条数最少为（~ ）． | [
[
{
"aoVal": "A",
"content": "$$3$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$5$$ "
}
],
[
{
"aoVal": "D",
"content": "$$6$$ "
}
]
] | [
"竞赛->知识点->逻辑->逻辑推理"
] | [
"因四面体每个面都不是等腰三角形，故至少有三条棱的长度互不相等． 另一方面，每两条对棱都相等的四面体可以是仅长度互不相等且每个面都不是等腰三角形的四面体． 故选$$\\text{A}$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 567 | 9091b114620845d7a4bdaf634971fc71 | [
"2003年AMC12竞赛B第4题",
"2003年AMC10竞赛B"
] | 0 | single_choice | $$2003-AMC10B-8$$ Moe uses a mower to cut his rectangular $$90$$-foot by $$150$$-foot lawn. The swath he cuts is $$28$$ inches wide, but he overlaps each cut by $$4$$ inches to make sure that no grass is missed. he walks at the rate of $$5000$$ feet per hour while pushing the mower. Which of the following is closest t... | [
[
{
"aoVal": "A",
"content": "$$0.75$$ "
}
],
[
{
"aoVal": "B",
"content": "$$0.8$$ "
}
],
[
{
"aoVal": "C",
"content": "$$1.35$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1.5$$ "
}
],
[
{
"aoVal": "E",
"con... | [
"课内体系->知识点->函数的应用->函数的实际应用->一次函数模型",
"美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Quadrilateral"
] | [
"Since the swath Moe actually mows is $$24$$ inches, or $$2$$ feet wide, he mows $$10000$$ square feet in one hour. His lawn has an area of $$13500$$, so it will take Moe $$1.35$$ hours to finish mowing the lawn. Thus the answer is $$\\boxed{(\\text{C})1.35}$$． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 361 | 1f9e530d52c647e1804700101bd86a53 | [
"2010年全国全国高中数学联赛竞赛复赛一试第2题8分",
"高二上学期单元测试《导数及其应用》自招第8题"
] | 2 | single_choice | 已知函数$$y=(a{{\cos }^{2}}x-3)\sin x$$的最小值为$$-3$$，则实数$$a$$的取值范围是~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$-\\frac{3}{2}\\leqslant a\\leqslant 12$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{3}{2}\\leqslant a\\leqslant 12$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-12\\leqslant a\\leqslant \\frac{3}{2}$$ "
}
],
[
... | [
"课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用换元法求值域",
"课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式",
"课内体系->素养->数学运算"
] | [
"令$$\\sin x=t$$，则原函数化为$$g(t)=(-a{{t}^{2}}+a-3)t$$，即$$g(t)=-a{{t}^{3}}+(a-3)t$$， 由$$-a{{t}^{3}}+(a-3)t\\geqslant -3$$，$$-at({{t}^{2}}-1)-3(t-1)\\geqslant 0$$，$$(t-1)(-at(t+1)-3)\\geqslant 0$$及$$t-1\\leqslant 0$$，知$$-at(t+1)-3\\leqslant 0$$即$$a({{t}^{2}}+t)\\geqslant -3$$，（$$1$$） 当$$t=0$$，$$1$$时（$$1$$）总成立； 对$$0\\t... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 502 | 3f49d6d2bb5d4441a2db8beef617040e | [
"2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"
] | 1 | single_choice | 体育课下课后，老师要求体育委员把$$5$$个相同的篮球、$$3$$个相同的排球、$$2$$个相同的橄榄球排成一排放好，则不同的放法有． | [
[
{
"aoVal": "A",
"content": "$$420$$种 "
}
],
[
{
"aoVal": "B",
"content": "$$1260$$种 "
}
],
[
{
"aoVal": "C",
"content": "$$5040$$种 "
}
],
[
{
"aoVal": "D",
"content": "$$2520$$种 "
}
]
] | [
"竞赛->知识点->排列组合与概率->两个基本计数原理",
"竞赛->知识点->排列组合与概率->排列与组合"
] | [
"总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 497 | 87fbbd8459af449c83bc4909774def30 | [
"1984年全国高中数学联赛竞赛一试第6题"
] | 1 | single_choice | 若$$F\left( \frac{1-x}{1+x} \right)=x$$，则下列等式中正确的是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$F\\left( -2-x \\right)=-2-F\\left( x \\right)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$F\\left( -x \\right)=F\\left( \\frac{1+x}{1-x} \\right)$$ "
}
],
[
{
"aoVal": "C",
"content": "$$F\\left( {{x}^{-1}} \\right)=F\\left... | [
"竞赛->知识点->函数->函数的概念",
"竞赛->知识点->不等式->换元技巧->代数换元"
] | [
"令 $$u=\\frac{1-x}{1+x}$$，解得 $$x=\\frac{1-u}{1+u}$$． 所以 $$F\\left( u \\right)=\\frac{1-u}{1+u}$$，即 $$F\\left( x \\right)=\\frac{1-x}{1+x}$$ 然后逐一验证 $$F\\left( -2-x \\right)=\\frac{1-\\left( -2-x \\right)}{1+\\left( -2-x \\right)}=-\\frac{3+x}{1+x}$$ $$=-2-\\frac{1-x}{1+x}=-2-F\\left( x \\right)$$ 选择$$A$$． 又本题也... | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 330 | ec76c8006220487f9e172d707ff9fc55 | [
"2012年黑龙江全国高中数学联赛竞赛初赛第9题5分"
] | 0 | single_choice | 已知不等式$$\frac{x-5}{x+1}\textless{}0$$的解集为$$M$$，若$${{x}_{0}}\in M$$，则$${{\log }_{2}}\left( {{x}_{0}}+1 \right)\textless{}1$$的概率为． | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{4}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{3}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{1}{5}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{2}{5}$$ "
}
]
] | [
"竞赛->知识点->排列组合与概率->概率初步",
"竞赛->知识点->函数->基本初等函数",
"竞赛->知识点->函数->函数的图像与性质"
] | [
"$$M=\\left { x\\textbar-1\\textless{}x\\textless{}5 \\right }, {{\\log }_{2}}\\left( x+1 \\right)\\textless{}1$$的解集为$$\\left { x\\textbar-1\\textless{}x\\textless{}1 \\right }$$，所以概率为$$\\frac{1}{3}$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 768 | 9aa94c040db64133ab20cda1b167951b | [
"2021年新疆全国高中数学联赛竞赛初赛第3题8分"
] | 1 | single_choice | 将正整数中所有数码不超过$$5$$的数从小到大排成一列，则第$$2021$$个数是~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$10000$$ "
}
],
[
{
"aoVal": "B",
"content": "$$13205$$ "
}
],
[
{
"aoVal": "C",
"content": "$$13321$$ "
}
],
[
{
"aoVal": "D",
"content": "$$14102$$ "
}
]
] | [
"课内体系->知识点->计数原理->排列与组合",
"竞赛->知识点->排列组合与概率->两个基本计数原理"
] | [
"方法一：所有数码不超过$$5$$的一位正整数有$$5$$个，两位正整数有$$5\\times 6=30$$个，三位正整数有$$5\\times {{6}^{2}}=180$$个，四位正整数有$$5\\times {{6}^{3}}=1080$$个，共有$$1295$$个； 万位数为$$1$$，千位为$$0$$，共$$216$$个； 万位数为$$1$$，千位为$$1$$，共$$216$$个； 万位数为$$1$$，千位为$$2$$，共$$216$$个；共$$1943$$个， 万位数为$$1$$，千位为$$3$$，百位是$$0$$，$$1$$各$$36$$个，共$$72$$个， 一共$$1943+72=2015$$个，... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 363 | 280e72fb79ac499e969926e0d700a530 | [
"第二十届全国希望杯高一竞赛复赛邀请赛第3题4分"
] | 1 | single_choice | 已知函数$$f(x)=a{{x}^{3}}+b{{x}^{2}}+cx-1(a\textless{}0)$$，且$$f(5)=3$$，则使$$f(x)=0$$成立的$$x$$的个数为． | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "不确定的 "
}
]
] | [
"竞赛->知识点->函数->函数的图像与性质"
] | [
"注意到$$x\\to -\\infty $$时，$$f(x)\\textgreater0$$；$$f(0)=-1\\textless{}0$$；$$f(5)=3\\textgreater0$$；$$x\\to +\\infty $$时，$$f(x)\\textless{}0$$，于是$$f(x)=0$$有$$3$$个实根． "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 200 | 21dbc6782f2d4d93876e00b4039a4ce8 | [
"2005年AMC12竞赛A第21题"
] | 3 | single_choice | 2005AMC12A, 21 How many ordered triples of integers $$(a, b, c)$$, with $$a\geqslant2$$, $$b\geqslant1$$, and $$c\geqslant0$$, satisfy both $$\log_ab=c^{2005}$$ and $$a+b+c= 2005$$? 有多少个整数 $$(a, b, c)$$ 的有序三元组，其中 $$a\geqslant2$$、$$b\geqslant1$$ 和 $$c\geqslant0$$ 满足 $$\log_ab =c^{2005}$$ 和 $$a+b+c= 2005$$? | [
[
{
"aoVal": "A",
"content": "$$0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$2$$ "
}
],
[
{
"aoVal": "D",
"content": "$$3$$ "
}
],
[
{
"aoVal": "E",
"content": "$$... | [
"美国AMC10/12->Knowledge Point->Algebra->Calculation->Solve Quadratic Equations",
"课内体系->知识点->基本初等函数->对数的概念及其运算->指对化简求值"
] | [
"由$\\log_a{b}=c^{2005}$可知$b=a^{c\\^{2005}}$. 又由$a+b+c=2005$知$a\\textless2005$, 故$c=0$或$c=1$. $c=0$时, $b=a^{0}=1$, $a=2005-b-c=2004$. $c=1$时, $b=a^{1}=a$, 再由$a+b+c=2005$知$(a,b,c)=(1002, 1002, 1)$. 共两种, 选$$\\text{C}$$. "
] | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 398 | 4bfdbd424d7140aa82b43d3b084e4c3e | [
"2015年黑龙江全国高中数学联赛竞赛初赛第15题5分"
] | 1 | single_choice | 已知$$x$$，$$y\in \left[ -\frac{\pi }{4},\frac{\pi }{4} \right]$$，且$${{x}^{3}}+\sin x-2a=0$$，$$4{{y}^{3}}+\frac{1}{2}\sin 2y+a=0$$，则$$\cos (x+2y)=$$~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4$$ "
}
]
] | [
"课内体系->素养->逻辑推理",
"课内体系->素养->数学运算",
"课内体系->知识点->函数的概念与性质->函数的性质->奇偶性"
] | [
"解析：第二个式子可以化成$${{(2y)}^{3}}+\\sin 2y+2a=0$$，由于$${{x}^{3}}+\\sin x$$是奇函数，且$$2y\\in \\left[ -\\frac{\\pi }{2},\\frac{\\pi }{2} \\right]$$， 所以$$x=-2y$$，$$\\cos (x+2y)=\\cos 0=1$$． 故答案为：$$1$$． "
] | A |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 348 | 39a2fe5b074647c3838f0d4df64d6a53 | [
"2009年浙江全国高中数学联赛竞赛初赛第7题5分"
] | 1 | single_choice | $${{\left( x-\frac{1}{{{x}^{6}}} \right)}^{2009}}$$的二项展开式中常数项为． | [
[
{
"aoVal": "A",
"content": "$$\\text{C}_{2009}^{286}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\text{C}_{2009}^{287}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-\\text{C}_{2009}^{286}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$-\\... | [
"课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项",
"课内体系->知识点->计数原理->二项式定理->二项式定理的展开式",
"课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数",
"课内体系->素养->数学运算"
] | [
"由于 $${{\\left( x-\\frac{1}{{{x}^{6}}} \\right)}^{2009}}=\\sum\\limits_{k=0}^{2009}{\\text{C}_{2009}^{k}{{x}^{2009-k}}{{(-{{x}^{-6}})}^{k}}}$$， 则若要出现常数项，须满足$$2009-k-6k=0\\Rightarrow k=287$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1061 | af170110e88e4191b7b3eb644f3ad89e | [
"2014年天津全国高中数学联赛竞赛初赛第5题6分"
] | 1 | single_choice | 直线$$l$$在平面$$ \pi $$上，直线$$m$$平行于平面$$ \pi $$，并与直线$$l$$异面．动点$$P$$在平面$$ \pi $$上，且到$$l$$和$$m$$的距离相等．则$$P$$点的轨迹是． | [
[
{
"aoVal": "A",
"content": "直线 "
}
],
[
{
"aoVal": "B",
"content": "椭圆 "
}
],
[
{
"aoVal": "C",
"content": "抛物线 "
}
],
[
{
"aoVal": "D",
"content": "双曲线 "
}
]
] | [
"竞赛->知识点->解析几何->轨迹方程（二试）"
] | [
"设$$m$$在平面$$ \\pi $$上的投影为$${m}'$$，$${m}'$$交$$l$$于$$O$$点．在平面$$ \\pi $$上，以$$O$$为原点，$$l$$为$$y$$轴建立直角坐标系，则可设$${m}'$$的方程为$$y=k \\pi $$．又设$$P$$点的坐标为$$\\left( x, y \\right)$$，则$$P$$到$$l$$的距离为$$\\left\\textbar{} x \\right\\textbar$$；它到$${m}'$$的距离为$$\\frac{\\left\\textbar{} y-kx \\right\\textbar}{\\sqrt{1+{{k}^{2}}}}$$，从而$... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 220 | 115223573a894ddf85a7eafa12cfba67 | [
"1999年全国全国高中数学联赛竞赛一试第3题6分"
] | 2 | single_choice | 若$${{\left( {{\log }_{2}}3 \right)}^{x}}-{{\left( {{\log }_{5}}3 \right)}^{x}}\geqslant {{\left( {{\log }_{2}}3 \right)}^{-y}}-{{\left( {{\log }_{5}}3 \right)}^{-y}}$$，则（ ）． | [
[
{
"aoVal": "A",
"content": "$$x-y\\geqslant 0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$x+y\\geqslant 0$$ "
}
],
[
{
"aoVal": "C",
"content": "$$x-y\\leqslant 0$$ "
}
],
[
{
"aoVal": "D",
"content": "$$x+y\\leqslant 0$$ "
}... | [
"课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->利用函数单调性解不等式",
"课内体系->素养->数学运算"
] | [
"记$$f\\left( t \\right)={{\\left( {{\\log }_{2}}3 \\right)}^{t}}-{{\\left( {{\\log }_{5}}3 \\right)}^{t}}$$， 则$$f\\left( t \\right)$$在$$\\mathbf{R}$$上是严格增函数． 原不等式即$$f\\left( x \\right)\\geqslant f\\left( -y \\right)$$． 故$$x\\geqslant -y$$， 即$$x+y\\geqslant 0$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 232 | 26a5ca2c4a18445d9914ab54df84e07e | [
"2011年吉林全国高中数学联赛竞赛初赛第4题5分"
] | 1 | single_choice | 在$$\triangle ABC$$中，已知$$6\overrightarrow{AC}\cdot \overrightarrow{AB}=2\overrightarrow{AB}\cdot \overrightarrow{BC}=3\overrightarrow{BC}\cdot \overrightarrow{CA}$$，则$$\angle A=$$． | [
[
{
"aoVal": "A",
"content": "$$30{}^{}\\circ $$ "
}
],
[
{
"aoVal": "B",
"content": "$$45{}^{}\\circ $$ "
}
],
[
{
"aoVal": "C",
"content": "$$60{}^{}\\circ $$ "
}
],
[
{
"aoVal": "D",
"content": "$$135{}^{}\\circ $$ "
}
... | [
"竞赛->知识点->复数与平面向量->平面向量的概念与运算",
"竞赛->知识点->三角函数->三角形中的问题->解三角形"
] | [
"设$$\\triangle ABC$$的三边分别为$$a,b,c,$$ 由已知$$6\\overrightarrow{AC}\\cdot \\overrightarrow{AB}=2\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=3\\overrightarrow{BC}\\cdot \\overrightarrow{CA}$$， 可得$$6bc\\cos A=2ac\\cos \\left( ~\\pi -B \\right)=3ab\\cos \\left( ~\\pi -C \\right)$$ 即$$6bc\\cos A=-2ac\\cos B=-3ab\\co... | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 628 | 3c3c2ceff3864c45a85b5ace1df39c1e | [
"1993年全国高中数学联赛竞赛一试第4题"
] | 2 | single_choice | 若直线$$x=\frac{ \pi }{4}$$被曲线$$C$$：$$\left( x-\arcsin \alpha \right)\left( x-\arccos \alpha \right)+\left( y-\arcsin \alpha \right)\left( y+\arccos \alpha \right)=0$$所截得的弦长为$$d$$，当$$a$$变化时$$d$$的最小值是（~ ）． | [
[
{
"aoVal": "A",
"content": "$$\\frac{ \\pi }{4}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{ \\pi }{3}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{ \\pi }{2}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$ \\pi $$ "
}
... | [
"竞赛->知识点->解析几何->圆与方程",
"竞赛->知识点->解析几何->直线与方程"
] | [
"由题设知，曲线$$C$$是以$${{P}_{1}}\\arcsin a,\\arcsin a$$，$${{P}_{2}}\\arccos a-\\arccos a$$两点为直径端点的圆．其圆心的横坐标为 $${{x}_{0}}=\\frac{\\arcsin a+\\arccos a}{2}=\\frac{\\mathsf{\\pi }}{4}$$， 故直线$$x=\\frac{\\pi }{4}$$过圆心，$$d$$就是该圆的直径．而 $${{d}^{2}}=\\left[ {{(\\arcsin a)}^{2}}+(\\arccos a) \\right]\\mathsf{\\geqslant }{{(\\arc... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1038 | b33bd35ffcb346bbb248a4d8f2a0ef87 | [
"2018年河北唐山高三二模理科第13题5分",
"2017~2018学年陕西西安碑林区西北工业大学附属中学高二上学期期中理科第14题3分",
"2018~2019学年山东济南高三上学期期末理科第13题5分",
"2020年山东日照高三一模第14题5分",
"2020~2021学年11月广东广州海珠区广州市第五中学高三上学期月考第14题5分",
"2008年贵州全国高中数学联赛竞赛初赛第13题4分",
"2017年湖北武汉高三三模理科第13题5分"
] | 1 | single_choice | $${{\left( {{x}^{2}}-\frac{1}{x} \right)}^{6}}$$的展开式中，常数项为 ．（用数字作答） | [
[
{
"aoVal": "A",
"content": "$$25$$ "
}
],
[
{
"aoVal": "B",
"content": "$$15$$ "
}
],
[
{
"aoVal": "C",
"content": "$$30$$ "
}
],
[
{
"aoVal": "D",
"content": "$$45$$ "
}
]
] | [
"课内体系->知识点->计数原理->二项式定理->二项式定理的展开式",
"课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数",
"课内体系->素养->数学运算"
] | [
"$${{T}_{r+1}}=\\text{C}_{6}^{r}{{({{x}^{2}})}^{6-r}}{{\\left( -\\frac{1}{x} \\right)}^{2}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-2r}}\\cdot {{x}^{-r}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-3r}}$$， ∵求常数项， ∴令$$12-3r=0$$，则$$r=4$$， ∴$${{T}_{5}}={{(-1)}^{4}}\\cdot \\text{C}_{6}^{4}{{x}^{0}}=15$$． 故答案为：$... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 377 | 1bd8bb55f36f4573a32f6adce857bcee | [
"2014年AMC10竞赛A第11题"
] | 1 | single_choice | $$2014-AMC10A-11$$ A customer who intends to purchase an appliance has three coupons, only one of which may be used: Coupon $$1$$: $$10$$\% off the listed price if the listed price is at least ＄$$50$$. Coupon $$2$$: ＄$$20$$ off the listed priceif the listed price is at least＄$$100$$. Coupon $$3$$: $$18$$\% off the ... | [
[
{
"aoVal": "A",
"content": "＄$$179.95$$ "
}
],
[
{
"aoVal": "B",
"content": "＄$$199.95$$ "
}
],
[
{
"aoVal": "C",
"content": "＄$$219.95$$ "
}
],
[
{
"aoVal": "D",
"content": "＄$$239.95$$ "
}
],
[
{
"aoVal": "... | [
"课内体系->知识点->函数的应用->函数的实际应用->一次函数模型",
"美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"
] | [
"Let the listed price be $$x$$. Since all the answer choices are above ＄$$100$$, we can assume $$x\\textgreater{} 100$$. Thus the discounts after the coupons are used will be as follows: Coupon $$1$$: $$x\\times10$$\\%$$=$$$$0.1 x$$. Coupon $$2$$: $$20$$. Coupon $$3$$: $$18$$\\%$$\\times (x-100)=0.18 x-18$$. Fo... | C |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 591 | 83c96124ba954cfea44c793e087b34d6 | [
"2018年安徽全国高中数学联赛竞赛初赛第3题8分"
] | 1 | single_choice | 函数$$f\left( x \right)=\left\textbar{} \sin 2x+\sin 3x+\sin 4x \right\textbar$$的最小正周期为~\uline{~~~~~~~~~~}~． | [
[
{
"aoVal": "A",
"content": "$$\\pi$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2\\pi$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3\\pi$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4\\pi$$ "
}
]
] | [
"竞赛->知识点->三角函数->三角函数的图像与性质",
"竞赛->知识点->三角函数->三角恒等变换"
] | [
"由和差化积公式得： $$\\sin 2x+\\sin 4x=2\\cos x\\sin 3x$$， ∴原式$$=\\left\\textbar{} 2\\cos x\\sin 3x+\\sin 3x \\right\\textbar$$$$=\\left\\textbar{} 2\\cos +1 \\right\\textbar\\cdot \\left\\textbar{} \\sin 3x \\right\\textbar$$， 又$$\\left\\textbar{} 2\\cos x+1 \\right\\textbar$$最小正周期为$$2\\pi $$， $$\\left\\textbar{} \\si... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 956 | d6efe0c0de544db4a6c104088bff2fdc | [
"1997年全国高中数学联赛竞赛一试第5题6分"
] | 1 | single_choice | 设$$f(x)={{x}^{2}}- \pi x$$，$$\alpha =\arcsin \frac{1}{3}$$，$$\beta =\arctan \frac{5}{4}$$，$$\gamma =\arccos \left( -\frac{1}{3} \right)$$，$$\delta =\text{arc}\cot \left( -\frac{5}{4} \right)$$，则（~ ~ ）． | [
[
{
"aoVal": "A",
"content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\delta \\right)\\textgreater f\\left( \\gamma \\right)$$ "
}
],
[
{
"aoVal": "B",
"content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\delta \\rig... | [
"课内体系->知识点->三角函数->反三角函数",
"课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->二次函数相关的恒成立问题",
"课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->含参二次函数的图象及性质",
"课内体系->素养->直观想象"
] | [
"$$f(x)$$的对称轴为$$x=\\frac{ \\pi }{2}$$， 易得$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{6}\\textless{}\\frac{ \\pi }{4}\\textless{}\\beta \\textless{}\\frac{ \\pi }{3}\\textless{}\\frac{ \\pi }{2}\\textless{}\\gamma \\textless{}\\frac{2 \\pi }{3}\\textless{}\\frac{3 \\pi }{4}\\textless{}\\delta \\textless{}\\fr... | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 617 | 406c03eca2c0491b8b2336041f07b08b | [
"2012年天津全国高中数学联赛竞赛初赛第2题6分"
] | 0 | single_choice | 若$$x\textgreater1$$，则$${{x}^{\ln \ln x}}-{{\left( \ln x \right)}^{\ln x}}$$的值是． | [
[
{
"aoVal": "A",
"content": "正数 "
}
],
[
{
"aoVal": "B",
"content": "零 "
}
],
[
{
"aoVal": "C",
"content": "负数 "
}
],
[
{
"aoVal": "D",
"content": "以上皆有可能 "
}
]
] | [
"竞赛->知识点->函数->基本初等函数"
] | [
"令$$m=\\ln x$$，则$${{x}^{\\ln \\ln x}}-{{\\left( \\ln x \\right)}^{\\ln x}}={{\\left( {{\\text{e}}^{m}} \\right)}^{\\ln m}}-{{m}^{m}}=0$$． "
] | B |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 1167 | d90494d0f3df480c80afdcdecbbd5c5c | [
"第二十届全国希望杯高一竞赛复赛邀请赛第5题4分"
] | 1 | single_choice | 定义集合$$M,N$$的一种运算*：$$M*N= {x\textbar x={{x}_{1}}{{x}_{2}},{{x}_{1}}\in M,{{x}_{2}}\in N }$$，若$$M= {1,2,3 }$$，$$N= {0,1,2 }$$，则$$M*N$$中的所有元素的和为． | [
[
{
"aoVal": "A",
"content": "$$9$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$18$$ "
}
],
[
{
"aoVal": "D",
"content": "$$16$$ "
}
]
] | [
"竞赛->知识点->集合->集合的概念与运算"
] | [
"$$M*N= {0,1,2,3,4,6 }$$，故所求的和为$$16$$． "
] | D |
high_math_competition_ch_single_choice_1.2K_dev | 2023-07-07T00:00:00 | 395 | 201430f86d7c4054a28c8ce1252e8349 | [
"2019年吉林全国高中数学联赛竞赛初赛第1题5分"
] | 0 | single_choice | 设集合$$A=\left { 2,0,1,3 \right }$$，$$B=\left { x\left\textbar{} -x\in A,2-{{x}^{2}}\notin A \right. \right }$$，则集合$$B$$中所有元素的和为． | [
[
{
"aoVal": "A",
"content": "$$-4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$-5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$-7$$ "
}
]
] | [
"竞赛->知识点->集合->集合的概念与运算"
] | [
"易知，$$B=\\left { -2,-3 \\right }$$， 则集合$$B$$中所有元素的和为$$-5$$． 故选$$\\text{B}$$． "
] | B |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.