math-eval/TAL-SCQ5K huggingface dataset:
Problem: 已知函数$$f\left( x \right)=f\left( 398-x \right)=f\left( 2158-x \right)=f\left( 3214-x \right)$$对于所有实数$$x$$均成立，则$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，$$f\left( 2 \right)$$，．．．，$$f\left( 999 \right)$$中最多有个不同的值？
Options:
aoVal: A
content: $$165$$ 
aoVal: B
content: $$177$$ 
aoVal: C
content: $$183$$ 
aoVal: D
content: $$199$$ 
Answer: B
Analysis: 由$$f\left( 398-x \right)=f\left( 2158-x \right)$$得$$f\left( x \right)=f\left( x+1760 \right)$$．  由$$f\left( 2158-x \right)=f\left( 3214-x \right)$$得，$$f\left( x \right)=f\left( x+1056 \right)$$．  故有$$f\left( x \right)=f\left( x+1056 \right)=f\left( x+2112 \right)=f\left( x+352 \right)$$．  从而$$f\left( x \right)$$为周期函数，且它的一个周期为$$352$$．  又由于$$f\left( x \right)=f\left( 398-x \right)$$，故$$f\left( x \right)$$关于$$x=199$$对称．  于是当$$f$$在$$x\in \left[ 23,199 \right]$$定义后，  由$$f\left( x \right)$$的对称性知$$x\in \left[ 199,375 \right]$$时，  $$f$$亦可定义；再由$$f\left( x \right)$$的周期性知对任意$$x\in \mathbf{R}$$，  $$f$$可定义．故在一个周期内，  每一个函数值至少对应两个不同的$$x$$，其中$$x\ne 199+352k\left( k\in \mathbf{Z} \right)$$．  因此，$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，．．．，$$f\left( 999 \right)$$至多有$$\frac{1}{2}\cdot 352+1=177$$个不同的值，  且当$$f\left( 23 \right)$$，$$f\left( 24 \right)$$，．．．，$$f\left( 199 \right)$$两两不等时，$$177$$可以取到．  因此，所求的最大值为$$177$$．  【点评】由已知恒等式组$$f\left( x \right)=f\left( 398-x \right)$$，$$f\left( x \right)=f\left( 2158-x \right)$$及$$f\left( x \right)=f\left( 3214-x \right)$$可得出以下恒等式组：  $$f\left( x \right)=f\left( 2158-x \right)=f\left( 3214-\left( 2158-x \right) \right)=f\left( 1056+x \right)$$，  $$f\left( x \right)=f\left( 1056+x \right)=f\left( 2158-\left( 1056+x \right) \right)=f\left( 1192-x \right)$$，  $$f\left( x \right)=f\left( 1056+x \right)=f\left( 1102-\left( 1056+x \right) \right)=f\left( 46-x \right)$$及  $$f\left( x \right)=f\left( 46-x \right)=f\left( 398-\left( 46-x \right) \right)=f\left( 352+x \right)$$．  由最后一个恒等式知$$f$$是一个周期函数，且$$f$$的一个周期为$$352$$．  故$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，．．．，$$f\left( 999 \right)$$中每个值都可在$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，．．．，$$\left( 351 \right)$$中找到．  恒等式$$f\left( x \right)=f\left( 398-x \right)$$表明$$f\left( 200 \right)$$，$$f\left( 201 \right)$$，．．．，$$f\left( 351 \right)$$可在$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，．．．，$$f\left( 199 \right)$$中找到．  恒等式$$f\left( x \right)=f\left( 46-x \right)$$表明$$f\left( 0 \right)$$，$$f\left( 1 \right)$$，．．．，$$f\left( 22 \right)$$可在$$f\left( 23 \right)$$，$$f\left( 24 \right)$$，．．．，$$f\left( 199 \right)$$中找到．  故题中所给一列函数中最多有$$177$$个不同的值．  要说明$$f\left( 23 \right)$$，$$f\left( 24 \right)$$，．．．，$$f\left( 199 \right)$$可以是不同的值，  考虑函数$$f\left( x \right)=\cos \left( \frac{360}{352}\left( x-23 \right) \right)$$，$$x$$的单位为度．  用常规方法可证明恒等式$$f\left( x \right)=f\left( 398-x \right)$$，$$f\left( x \right)=f\left( 2158-x \right)$$及$$f\left( x \right)=f\left( 3214-x \right)$$，  且可看出  $$1=f\left( 23 \right)\textgreater f\left( 24 \right)\textgreater f\left( 25 \right)\textgreater\cdots \textgreater f\left( 199 \right)=-1$$． 
Knowledge points: 竞赛->知识点->函数->函数的图像与性质
课内体系->知识点->函数的概念与性质->函数的性质->对称性
Source: 2000年第18届美国数学邀请赛（AIME）竞赛第12题
Problem: 圆周上有$$10$$个等分点，则以这$$10$$个等分点中的四个点为顶点的凸四边形中，梯形所占的比为．
Options:
aoVal: A
content: $$\frac{8}{21}$$ 
aoVal: B
content: $$\frac{4}{21}$$ 
aoVal: C
content: $$\frac{1}{126}$$ 
aoVal: D
content: $$\frac{2}{7}$$ 
Answer: D
Analysis: 任选$$4$$点，共有$$C_{10}^{4}=210$$个凸四边形，其中梯形的两条平行边可以从$$5$$组平行于直径的$$5$$条平行弦中选取，也可以$$5$$组不平行于直径的$$4$$条平行弦中选取，去除矩形，梯形共有$$60$$个，所以，梯形所占的比为$$\frac{2}{7}$$． 
Knowledge points: 竞赛->知识点->排列组合与概率->排列与组合
竞赛->知识点->排列组合与概率->概率初步
Source: 2010年吉林全国高中数学联赛竞赛初赛第6题5分
Problem: $$Zeoy$$閲讀$$15$$本書，一次閱讀$$1$$本．她花了$$1$$天讀完第一本書，又花了$$2$$天讀完第二本，再花了$$3$$天讀完第三本，其餘依此類推，即每讀完一本書要比讀完前一本書多花$$1$$天．$$Zoey$$讀完第一本書是在星期一，讀完第二本書是在星期三，若她每天都閲讀，則當她讀完第$$15$$本書時尼在星期幾？  Zoey read $$15$$ books, one at a time. The first book took her $$1$$ day to read, the second book took $$2$$ her days to read, the third book took $$3$$ her days to read, and so on, with each book taking her $$1$$ more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day of the week did she finish her $$15\text{th}$$ book?
Options:
aoVal: A
content: 星期日｜$ $Sunday 
aoVal: B
content: 星期一｜Monday 
aoVal: C
content: 星期三｜Wednesday 
aoVal: D
content: 星期五｜Friday 
aoVal: E
content: 星期六｜Saturday 
Answer: B
Analysis: The process took $$1+2+3+\cdot\cdot\cdot+13+14+15=120$$ days, so the last day was $$119$$ days after the first day. Since $$119$$ is divisible by $$7$$, both must have been the same day of the week, so the answer is $$\left(\text{B}\right)$$ Monday. 
Source: 2016年AMC10竞赛B第4题
Problem: 已知一个角大于$$120{}^{}\circ $$的三角形的三边长分别为$$m,m+1,m+2$$，则实数$$m$$的取值范围为．
Options:
aoVal: A
content: $$m\textgreater1$$ 
aoVal: B
content: $$1\textless{}m\textless{}\frac{3}{2}$$ 
aoVal: C
content: $$\frac{3}{2}\textless{}m\textless{}3$$ 
aoVal: D
content: $$m\textgreater3$$ 
Answer: B
Analysis: 由题意可知：  $$\begin{cases}m+(m+1)\textgreater m+2    {{(m+2)}^{2}}\textgreater{{m}^{2}}+{{(m+1)}^{2}}+m(m+1) \end{cases}$$  解得$$1\textless{}m\textless{}\frac{3}{2}$$． 
Knowledge points: 竞赛->知识点->三角函数->三角形中的问题->解三角形
Source: 2015年浙江全国高中数学联赛竞赛初赛第2题6分
Problem: $${{\left( {{x}^{2}}-\frac{1}{x} \right)}^{n}}$$的展开式中，常数项为$$15$$，则$$n$$的值为．
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
Answer: B
Analysis: $${{\left( {{x}^{2}}-\frac{1}{x} \right)}^{n}}$$的展开式中，常数项为$$15$$，则$$\text{C}_{n}^{k}{{\left( {{x}^{2}} \right)}^{k}}{{\left( -\frac{1}{x} \right)}^{n-k}}=15$$，其中$$2k+k-n=0$$，即$$k=\frac{n}{3}$$，所以$$n$$应被$$3$$整除，由此得$$\text{C}_{n}^{\frac{n}{3}}=15$$．当$$n=3$$时，$$\text{C}_{3}^{1}=3\ne 15$$，当$$n=6$$时，$$\text{C}_{6}^{2}=15$$． 
Knowledge points: 竞赛->知识点->排列组合与概率->二项式定理及其应用
Source: 2012年山东全国高中数学联赛竞赛初赛第3题5分
Problem: 若集合$$A$$满足：对任意$$x\in A$$，都有$$\frac{1}{x}\in A$$，就称$$A$$是``和谐''集合，则在集合$$M=\left { -1,0,\frac{1}{3},\frac{1}{2},1,2,3,4,5,6 \right }$$的所有非空子集中，``和谐''集合有~\uline{~~~~~~~~~~}~个．
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$17$$ 
Answer: B
Analysis: 题意为：互为倒数的两个数必须同时在一个集合或者同时不在一个集合中，所以需要将集合中的元素分组，$$\left { 1 \right }$$，$$\left { -1 \right }$$，$$\left { 2,\frac{1}{2} \right }$$，$$\left { 3,\frac{1}{3} \right }$$，所以和谐集有$${{2}^{4}}-1=15$$个．  故答案为：$$15$$． 
Knowledge points: 课内体系->特色题型->新定义
课内体系->素养->逻辑推理
课内体系->知识点->集合->集合的新定义问题
课内体系->知识点->集合->集合的基本关系->子集
Source: 2009年河南全国高中数学联赛高一竞赛初赛第1题5分
Problem: 在半径为$$1$$的$$\odot O$$上，取一个定点$$A$$和一个动点$$B$$．设点$$P$$满足$$AP//OB$$且$$\overrightarrow{AP}\cdot \overrightarrow{AB}=1$$，则$$P$$点的轨迹是．
Options:
aoVal: A
content: 椭圆 
aoVal: B
content: 抛物线 
aoVal: C
content: 双曲线 
aoVal: D
content: 以上都有可能 
Answer: B
Analysis: 不妨设$$O(0,  0),  A(1,  0),  P(x,  y)$$，由于$$AP//OB$$，可设$$B(k(x-1),  ky)$$．将这些坐标代入$$\overrightarrow{AP}\cdot \overrightarrow{AB}=1$$．可得$$k=\frac{x}{{{(x-1)}^{2}}+{{y}^{2}}}$$．最后，利用$$B$$在$$\odot O$$上，即可得到$$(x,  y)$$满足的方程为$${{x}^{2}}={{(x-1)}^{2}}+{{y}^{2}}$$，即$${{y}^{2}}=2x-1$$，所以$$P$$的轨迹是抛物线． 
Knowledge points: 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->解析几何->曲线与方程
竞赛->知识点->解析几何->直线与方程
Source: 高二下学期单元测试《轨迹问题》自招第3题
2011年天津全国高中数学联赛竞赛初赛第5题6分
Problem: 2019AMC12A, 23  Define binary operations $$\diamondsuit $$ and $$\heartsuit$$ by $$a\diamondsuit b=a^{\log _7(b)}$$ and $$a\heartsuit b=a^{\frac{1}{\log _7(b)}}$$ for all real numbers $$a$$ and $$b$$ for which these expressions are defined. The sequence $$(a_n)$$ is defined recursively by $$a_3=3\heartsuit 2$$ and $$a_n=(n\heartsuit (n-1))\diamondsuit a_{n-1}$$ for all integers $$n\geqslant 4$$. To the nearest integer, what is $$\log _7(a_{2019})$$?   对所有实数$$a$$, $$b$$, 定义二元运算$$\diamondsuit $$ 和 $$\heartsuit$$如下: $$a\diamondsuit b=a^{\log _7(b)}$$ , $$a\heartsuit b=a^{\frac{1}{\log _7(b)}}$$. 数列 $$(a_n)$$ 满足 $$a_3=3\heartsuit 2$$以及$$a_n=(n\heartsuit (n-1))\diamondsuit a_{n-1}$$ ($$n\geq 4$$). 距离 $$\log _7(a_{2019})$$ 最近的整数是多少? ．
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: 首先化简一下题目中给的条件.  $$a_3=3\heartsuit2=3^{\frac{1}{\log_7{(2)}}}=3^{\log_2{7}}$$  $$a_n=(n\heartsuit (n-1))\diamondsuit a_{n-1}=(n\heartsuit (n-1))^{\log_7{a_{n-1}}}=(n^{\log_{n-1}{7}})^{\log_7{a_{n-1}}}=n^{\log_{n-1}{a_{n-1}}}$$  $$\Rightarrow\log_n{a_n}=\log_{n-1}{a_{n-1}}=\log_3{a_3}=\log_2{7}$$  因此$$\log_{2019}{a_{2019}}=\log_2{7}\Rightarrow\log_7{a_{2019}}=\log_7{2019}\log_2{7}=\log_2{2019}$$.  这个值最接近$\log_2{2048}=11$, 选$$\text{D}$$. 
Knowledge points: 课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算
美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)
美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion
Source: 2019年AMC12竞赛A第23题
Problem: 設$$N$$爲$$5$$的倍數．一個紅球與$$N$$個綠球隨意地排在一直線上．令P(N) 表示至少有$$ \frac{3}{5} $$的綠球排在紅球同一側的概率．顯然$$P（5） =1$$，且當$$N$$非常大时P(N) 趨近於$$ \frac{4}{5} $$．試問使得$$ P(N)\textless{} \frac{321}{400} $$最小的$$N$$，其各位數的數字和爲多少？  Let N be a positive multiple of 5. One red ball and N green balls are arranged in a line in randon order. Let  P(N)be the probability that at least$$ \frac{3}{5} $$ of the green balls are on the same side of the red ball. Observe that  $$P(5)=1$$ and that P(N) approaches$$ \frac{4}{5} $$ as N grows large. What is the sum of the digits of the least value of  $$P(N)\textless{} \frac{321}{400} $$
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: Let $$n=\frac N5$$. Then, consider $$5$$ blocks of $$n$$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $$N+1$$ positions between the green balls to insert the red ball.  Less than $$\frac 35$$ of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of $$n$$ balls, and there are $$n-1$$ positions where this happens.  Thus, $$P(N)=1-\frac {n-1}{N+1}=\frac {4n+2}{5n+1}$$, so $$P(N)=\frac {4n+2}{5n+1}\textless\frac {321}{400}$$.  Multiplying both sides of the inequality by $$400(5n+1)$$, we have $$400(4n+2)\textless321(5n+1)$$,  and by the distributive property,  $$1600n+800\textless1605n+321$$.  Subtracting $$1600n+321$$ on both sides of the inequality gives us $$479\textless5n$$,  Therefore, $$N=5n\textgreater479$$, so the least possible value of $$N$$ is $$480$$. The sum of the digits of $$480$$ is $$12$$.  Let $$N=5$$, $$P(N)=1$$(Given)  Let $$N=10$$, $$P(N)=\frac {10}{11}$$  Let $$N=15$$, $$P(N)=\frac {14}{16}$$  Notice that the fraction can be written as $$1-\frac{\frac{N}{5}-1}{N+1}$$  Now it\textquotesingle s quite simple to write the inequality as $$1-\frac {\frac N5-1}{N+1}\textless\frac {321}{400}$$  We can subtract $$1$$ on both sides to obtain $$-\frac {\frac N5-1}{N+1}\textless-\frac {79}{400}$$  Dividing both sides by $$-1$$, we derivee $$\frac {\frac N5-1}{N+1}\textgreater\frac {79}{400}$$. (Switch the inequality sign when dividing by $$-1$$)  We then cross multiply to get $$80N-400\textgreater79N+79$$  Finally we get $$N\textgreater479$$  To achieve $$N=480$$  So the sum of the digits of $$N=12$$.  We are trying to find the number of places to put the red ball, such that $$\frac 35$$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $$N$$: Trying a few values, we see that the ball "works" in places $$1$$ to $$\frac 25N+1$$ and spaces $$\frac 35N+1$$ to $$N+2$$. This is a total of $$\frac 45N+2$$ spaces, over a total possible $$N+1$$ places to put the ball. So: $$\frac {\frac 45N+2}{N+1}=\frac {321}{400}\rightarrow N=479$$. And we know that the next value is what we are looking for, so $$N+1=480$$, and the sum of it\textquotesingle s digits is $$12$$. 
Source: 2016年AMC10竞赛第17题5分
Problem: 设等差数列$$ {{{a}_{n}} }$$的前$$n$$项和为$${{S}_{n}}$$，若$${{S}_{2}}=10$$，$${{S}_{5}}=55$$，则经过$$P(n,{{a}_{n}})$$、$$Q(n+2,{{a}_{n+2}})$$两点的直线的一个方向向量的坐标可以是．
Options:
aoVal: A
content: $$\left( 2,\frac{1}{2} \right)$$ 
aoVal: B
content: $$\left( -\frac{1}{2},  -2 \right)$$ 
aoVal: C
content: $$\left( -\frac{1}{2},-1 \right)$$ 
aoVal: D
content: $$(-1,-1)$$ 
Answer: B
Analysis: 设等差数列$$ {{{a}_{n}} }$$的公差为$$d$$，  则由$$\begin{cases}{{S}_{2}}=2{{a}_{1}}+d=10    {{S}_{5}}=5{{a}_{1}}+10d=55 \end{cases}$$，得$$\begin{cases}{{a}_{1}}=3    d=4 \end{cases}$$，  所以$${{K}_{PQ}}=\frac{{{a}_{n+2}}-{{a}_{n}}}{(n+2)-n}=d=4$$，  故所$$B$$． 
Knowledge points: 竞赛->知识点->数列与数学归纳法->等差数列与等比数列
Source: 2008年陕西全国高中数学联赛竞赛初赛第1题5分
Problem: 设$$p，q$$是自然数，条件甲：$${{p}^{3}}-{{q}^{3}}$$是偶数；条件乙；$$p+q$$是偶数，那么(~ ~ )．
Options:
aoVal: A
content: 甲是乙充分而非必要条件 
aoVal: B
content: 甲是乙的必要而非充分条件 
aoVal: C
content: 甲是乙的充要条件 
aoVal: D
content: 甲既不是乙的充分条件也不是乙的必要条件 
Answer: C
Analysis: 若$${{p}^{3}}-{{q}^{3}}$$是偶数$$\Leftrightarrow p，q$$有相同的奇偶性$$\Leftrightarrow p+q$$是偶数． 
Knowledge points: 竞赛->知识点->逻辑->常用逻辑用语
Source: 1983年全国高中数学联赛竞赛一试第1题
Problem: 设$$z$$是复数，则$$\left\textbar{} z-1 \right\textbar+\left\textbar{} z-\text{i} \right\textbar+\left\textbar{} z+1 \right\textbar$$的最小值等于．
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$1+\sqrt{2}$$ 
aoVal: C
content: $$1+\sqrt{3}$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: 法一：$$\left\textbar{} z-1 \right\textbar+\left\textbar{} z-\text{i} \right\textbar+\left\textbar{} z+1 \right\textbar$$表示动点$$Z$$到三个定点$$A\left( 1,0 \right)$$，$$B\left( 1,0 \right)$$，$$C\left( -1,0 \right)$$的距离之和，而且$$\triangle ABC$$是一个锐角三角形．所以$$Z$$是对各边张角为$$120{}^{}\circ $$的点，即点$$Z$$落在$$y$$轴的正半轴上，且$$\angle AZB=120{}^{}\circ $$，从而  $${{\left( \left\textbar{} z-1 \right\textbar+\left\textbar{} z-\text{i} \right\textbar+\left\textbar{} z+1 \right\textbar{} \right)}_{\min }}=2\frac{1}{\sin 60{}^{}\circ }+\left( 1-\frac{1}{\tan 60{}^{}\circ } \right)=1+\sqrt{3}$$．  法二：$$\left\textbar{} z-1 \right\textbar+\left\textbar{} z-\text{i} \right\textbar+\left\textbar{} z+1 \right\textbar$$表示动点$$Z$$三个定点$$A\left( 1,0 \right)$$，$$B\left( 0,1 \right)$$，$$C\left( -1,0 \right)$$的距离之和，能确定的是当取得最小值时点$$Z$$落在$$y$$轴的正半轴上，故设$$\angle ZAC=\theta $$，且$$\theta $$为锐角，从而$$\left\textbar{} z-1 \right\textbar+\left\textbar{} z-\text{i} \right\textbar+\left\textbar{} z+1 \right\textbar=\frac{2}{\cos \theta }+1-\tan \theta =1-\frac{\sin \theta -2}{\cos \theta }{\geqslant }1+\sqrt{3}$$．  故答案为：$$1+\sqrt{3}$$． 
Knowledge points: 竞赛->知识点->复数与平面向量->复数的应用
Source: 2010年安徽全国高中数学联赛竞赛初赛第5题8分
2019~2020学年上海浦东新区华东师范大学第二附属中学高二下学期单元测试第17题
Problem: 将$${{(a+b+c+d)}^{9}}$$展开之后再合并同类项，所得的多项式的项数是．
Options:
aoVal: A
content: $$\text{C}_{9}^{4}$$ 
aoVal: B
content: $$\text{C}_{9}^{3}$$ 
aoVal: C
content: $$\text{C}_{12}^{4}$$ 
aoVal: D
content: $$\text{C}_{12}^{3}$$ 
Answer: D
Analysis: 所得多项式中每一项都形如$$k{{a}^{{{x}_{1}}}}{{b}^{{{x}_{2}}}}{{c}^{{{x}_{3}}}}{{d}^{{{x}_{4}}}}$$，其中$$k\textgreater0$$，  $${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=9,  {{x}_{i}}\geqslant 0$$．  易知上式共有$$\text{C}_{9+4-1}^{4-1}=\text{C}_{12}^{3}$$组整数解，因此选$$\text{D}$$． 
Knowledge points: 竞赛->知识点->排列组合与概率->二项式定理及其应用
Source: 2011年天津全国高中数学联赛竞赛初赛第6题6分
Problem: 已知$$f(x)=\begin{cases}(3-a)x-a,x\textless{}1    {{\log }_{a}}x,x\geqslant 1 \end{cases}$$在$$(-\infty ,+\infty )$$上单调增加，那么$$a$$的取值范围是．
Options:
aoVal: A
content: $$(0,3)$$ 
aoVal: B
content: $$\left( 1,\frac{3}{2} \right]$$ 
aoVal: C
content: $$\left[ \frac{3}{2},3 \right)$$ 
aoVal: D
content: $$(1,3)$$ 
Answer: C
Analysis: 由题知$$\begin{cases}a\textgreater1    3-a\textgreater0    {{\log }_{a}}1\geqslant \left( 3-a \right)\cdot 1-a \end{cases}$$，解得$$\frac{3}{2}\leqslant a\textless{}3$$．故选$$\text{C}$$． 
Knowledge points: 竞赛->知识点->函数->函数的图像与性质
Source: 2008年辽宁全国高中数学联赛竞赛初赛第3题5分
Problem: \uline{The tower function of twos} is defined recursively as follows: $$T\left(1\right)=2$$ and $$T\left(n+1\right)=2^{T\left(n\right)}$$ for $$n\geqslant 1$$. Let $$A=\left(T\left(2009\right)\right)^{T\left(2009\right)}$$ and $$B=\left(T\left(2009\right)\right)^{A}$$. What is the largest integer $$k$$ such that $$\underbrace{\log_{2}\log_{2}\log_{2}\ldots\log_{2}B}_{{k\textasciitilde{} \text{times}}}$$ is defined?  2的\uline{塔函数}定义如下: $$T\left(1\right)=2$$ , $$T\left(n+1\right)=2^{T\left(n\right)}$$ ($$n\geqslant 1$$). 令 $$A=\left(T\left(2009\right)\right)^{T\left(2009\right)}$$ , $$B=\left(T\left(2009\right)\right)^{A}$$. 使得 $$\underbrace{\log_{2}\log_{2}\log_{2}\ldots\log_{2}B}_{{k\textasciitilde{} \text{times}}}$$有意义的最大整数 $$k$$是多少?
Options:
aoVal: A
content: $$2009$$ 
aoVal: B
content: $$2010$$ 
aoVal: C
content: $$2011$$ 
aoVal: D
content: $$2012$$ 
aoVal: E
content: $$2013$$ 
Answer: E
Analysis: 记$\log_2$为$\log$.  由条件, 有$\log{T(n+1)}=T(n)$.  我们对$B$求几次对数看看情况.  $\log{B}=A\log{T(2009)}=A\cdot T(2008)=T(2009)^{T(2009)}T(2008)$  $\log{\log{B}}=T(2009)\log{T(2009)}+\log{T(2008)}=T(2009)T(2008)+T(2007)$  再往下似乎不太好求了, 不过我们可以进行一些适当的舍入.  $\log{\log{B}}=T(2009)T(2008)+T(2007)=T(2009)\left(T(2008)+\frac{T(2007)}{T(2009)}\right)$  $\Rightarrow\log\log\log{B}=\log{T(2009)}+\log{\left(T(2008)+\frac{T(2007)}{T(2009)}\right)}\approx\log{T(2009)}+\log{T(2008)}=T(2008)+T(2007)$  考虑到$T(2007)=\log\log{T(2009)}\textless\textless T(2009)$, 这里的舍入是合理的.  故技重施,  $\log\log\log\log{B}=\log{\left(T(2008)\left(1+\frac{T(2007)}{T(2008)}\right)\right)}=\log{T(2008)}+\log{\left(1+\frac{T(2007)}{T(2008)}\right)}\approx T(2007)$  到这里就好求了, 从$T(2007)$再往下求$$2007$$层, 会得到$$1.$$ 考虑到我们在计算过程中做了一些舍入, 实际上这时的值会略大于$$1.$$ 再求一层, 得到一个略大于$$0$$的值$$.$$ 再求一层, 得到一个负值$$.$$ 之后再往下，就没有意义了.  因此, 总层数$=2+2007+4=2013$, 选E. 
Knowledge points: 美国AMC10/12->Knowledge Point->Combination->Reasoning->Mapping and Correspondence
课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算
Source: 2009年AMC12竞赛A第24题
Problem: 已知$$a$$，$$b$$，$$c$$都是非负数，则$$\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}$$的最小值为．
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$0.5$$ 
Answer: A
Analysis: $$\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}$$  $$=\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}-1$$  $$\geqslant 3\sqrt[3]{\frac{c}{a}\cdot \frac{a}{b+c}\cdot \frac{b+c}{c}}-1$$  $$=2$$． 
Knowledge points: 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->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->知识点->等式与不等式->不等式->基本不等式->多元均值不等式的应用
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
Source: 2008年辽宁全国高中数学联赛竞赛初赛第7题5分
2016~2017学年10月上海徐汇区上海市上海中学高一上学期周测第11题
Problem: 设$$x$$，$$y$$，$$z$$均不为$$\left( k+\frac{1}{2} \right)\pi $$，其中$$k$$为整数，已知$$\sin \left( y+z-x \right)$$，$$\sin \left( x+z-y \right)$$，$$\sin \left( x+y-z \right)$$成等差数列，则依然成等差数列的是．
Options:
aoVal: A
content: $$\sin x$$，$$\sin y$$，$$\sin z$$ 
aoVal: B
content: $$\cos x$$，$$\cos y$$，$$\cos z$$ 
aoVal: C
content: $$\tan x$$，$$\tan y$$，$$\tan z$$ 
aoVal: D
content: 前三个答案都不对 
Answer: C
Analysis: $$\because \cos x$$，$$\cos y$$，$$\cos z$$均不为$$0$$，  $$2\sin \left( x+z-y \right)=\sin \left( x+y-z \right)+\sin \left( y+z-x \right)$$  $$=2\sin y\cdot \cos \left( z-y \right)$$，  $$\therefore \sin y\cos \left( z-x \right)=\sin \left[ \left( x+z \right)-y \right]$$  $$\Leftrightarrow \sin y\cos \left( z-x \right)=\sin \left( x+z \right)\cos y-\cos \left( x+z \right)\sin y$$  $$\Leftrightarrow 2\sin y\cos x\cos z=\cos y\sin x\cos z+\cos y\cos x\sin z$$  $$\Leftrightarrow 2\tan y=\tan x+\tan z$$，  故选$$\text{C}$$． 
Knowledge points: 竞赛->知识点->数列与数学归纳法->等差数列与等比数列
竞赛->知识点->三角函数->三角恒等变换
Source: 2020~2021学年北京高三单元测试
2020年北京海淀区北京大学自主招生（强基计划）第6题5分
2020~2021学年北京高二单元测试
竞赛
Problem: 设复数$$x=\frac{1+\text{i}}{1-\text{i}}$$（$$\text{i}$$是虚数单位），则$$C_{2014}^{0}+C_{2014}^{1}x+C_{2014}^{2}{{x}^{2}}+\ldots +C_{2014}^{2014}{{x}^{2014}}=$$．
Options:
aoVal: A
content: $${{2}^{1007}}\text{i}$$ 
aoVal: B
content: $$-{{2}^{1007}}\text{i}$$ 
aoVal: C
content: $$-{{2}^{1005}}$$ 
aoVal: D
content: $$-{{2}^{1004}}$$ 
Answer: B
Analysis: 因为$$x=\frac{{{\left( 1+\text{i} \right)}^{2}}}{\left( 1-\text{i} \right)\left( 1+\text{i} \right)}\text{=}\frac{2\text{i}}{2}\text{=i}$$，所以  $$C_{2014}^{0}+C_{2014}^{1}x+C_{2014}^{2}{{x}^{2}}+\ldots +C_{2014}^{2014}{{x}^{2014}}={{\left( 1+x \right)}^{2014}}={{\left( 1+\text{i} \right)}^{2014}}={{\left( 2\text{i} \right)}^{1007}}=-{{2}^{1007}}\text{i}$$ 
Knowledge points: 竞赛->知识点->排列组合与概率->二项式定理及其应用
Source: 2014年黑龙江全国高中数学联赛竞赛初赛第2题5分
Problem: 定义數列 $$F_0=0$$, $$F_1=1$$, 且對於所有$$n≥2$$ 有$$F_n=F_{n−1}+F_{n−2}$$ 再除以$$3$$的餘數，序列從$$0,1,1,2,0,2$$$$\cdots \cdots$$ 開始，求$$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$？  Define a sequence recursively by $$F_0 =0$$, $$F_1=1$$, and $$F_n=$$the remainder when $$F_{n-1}+ F_{n-2}$$ is divided by $$3$$, for all $$n≥ 2$$. Thus the sequence starts $$0$$, $$1$$, $$1$$, $$2$$, $$0$$, $$2$$, $$\cdots$$ What is $$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$?
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: D
Analysis: A patten starts to emerge as the function is continued. The repeating patten is $$0$$, $$1$$, $$1$$, $$2$$, $$0$$, $$2$$, $$2$$, $$1$$ $$\ldots$$. The problem asks for the sum of eight consecutive terms in the sequence.  Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is $$(\rm D) 9$$. 
Source: 2017年AMC10竞赛A第13题
Problem: 已知$$M= {(x,y)\textbar{{x}^{2}}+2{{y}^{2}}=3 }$$，$$N= {(x,y)\textbar y=mx+b }$$．若对于所有的$$m\in R$$，均有$$M\cap N\ne \varnothing $$，则$$b$$的取值范围是（~ ~ ）．
Options:
aoVal: A
content: $$\left[ -\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2} \right]$$ 
aoVal: B
content: $$\left( -\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2} \right)$$ 
aoVal: C
content: $$\left( -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3} \right]$$ 
aoVal: D
content: $$\left[ -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3} \right]$$ 
Answer: A
Analysis: 点$$(0,b)$$在椭圆内或椭圆上$$\Rightarrow 2{{b}^{2}}\leqslant 3\Rightarrow b\in \left[ -\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2} \right]$$．  故选$$\text{A}$$． 
Knowledge points: 竞赛->知识点->集合->集合的概念与运算
竞赛->知识点->解析几何->椭圆
Source: 2004年全国高中数学联赛竞赛一试第2题6分
Problem: 函数$$f\left( x \right)=\frac{1}{\sqrt{-{{\left( \lg x \right)}^{2}}+3\lg x-2}}$$的定义域是~\uline{~~~~~~~~~~}~．
Options:
aoVal: A
content: $$(10,100)$$ 
aoVal: B
content: $$(10,20)$$ 
aoVal: C
content: $$(20,100)$$ 
aoVal: D
content: $$(50,100)$$ 
Answer: A
Analysis: $$-{{(\lg x)}^{2}}+3\lg x-2\textgreater0$$，$$1\textless{}\lg x\textless{}2$$．  故答案为： $$(10,100)$$． 
Knowledge points: 课内体系->知识点->基本初等函数->对数函数->对数方程和对数不等式->解对数不等式
课内体系->知识点->基本初等函数->对数函数->对数函数的图象及性质
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->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->函数的概念与性质->函数的概念及其表示->函数的定义域->求具体函数（包括复合函数）的定义域
课内体系->素养->数学运算
课内体系->素养->逻辑推理
Source: 2016年广西全国高中数学联赛竞赛初赛第1题8分
Problem: 函数$$f\left( x \right)$$对于任意实数$$x$$满足：$$f\left( x+3 \right)=-\frac{1}{f(x)}$$，若$$f\left( 0 \right)=2$$，则$$f\left( 2013 \right)=$$．
Options:
aoVal: A
content: $${-}\frac{1}{2}$$ 
aoVal: B
content: $$\frac{1}{2}$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$2013$$ 
Answer: A
Analysis: $$f\left( x+6 \right)=-\frac{1}{f(x+3)}=f\left( x \right)$$，所以，$$f\left( x \right)$$是以$$6$$为周期的函数.  所以，$$f\left( 2013 \right)=f\left( 3 \right)=-\frac{1}{f(0)}=-\frac{1}{2}$$. 
Knowledge points: 课内体系->素养->数学运算
课内体系->素养->数学抽象
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->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->函数的概念与性质->函数的性质->周期性->函数周期性判断
Source: 2013年四川全国高中数学联赛竞赛初赛第1题5分
Problem: 把分母为$$24$$的所有最简真分数按从小到大的顺序排列，依次为$${{a}_{1}},{{a}_{2}},\cdots ,  {{a}_{n}}$$，则$$\sum\limits_{i=1}^{n}{\cos ({{a}_{i}} \pi )}$$的值是．
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$\frac{1}{2}$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$-\frac{1}{2}$$ 
Answer: C
Analysis: 分母为$$24$$的所有最简真分数有$$\frac{1}{24},\frac{5}{24},\frac{7}{24},\frac{11}{24},\frac{13}{24},\frac{17}{24},\frac{19}{24},\frac{23}{24}$$，  注意到$$\frac{1}{24}+\frac{23}{24}=\frac{5}{24}+\frac{19}{24}=\frac{7}{24}+\frac{17}{24}+\frac{11}{24}+\frac{13}{24}=1$$，  因此$$\sum\limits_{i=1}^{n}{\cos ({{a}_{i}} \pi )}=0$$．故选$$\text{C}$$． 
Knowledge points: 竞赛->知识点->三角函数->三角函数的图像与性质
Source: 2008年陕西全国高中数学联赛竞赛初赛第4题5分