高中教育试讲

时间:2014-05-04 12:07:34   收藏:0   阅读:243


1.  一个凸$n$边型,任意三条对角线不共点,问所有的对角线把这个多边形内部分成了多少个区域?
解答:
去边法,现将这个多边形的对角线一条一条的去掉.
假设第一条对角线与其他内部对角线有$k_{1}$个交点,那么去掉它这个多边形内部减少$k_{1}+1$个;再去掉第二条,内部区域减少$k_{2}+1$个
去掉第三条,内部区域减少$k_{3}+1$个,$\cdots$,去掉第$\frac{n(n-3)}{2}$个,减少$k_{\frac{n(n-3)}{2}}+1$个.故多边形内部区域数为

N=1+bubuko.com,布布扣i=1bubuko.com,布布扣n(n?3)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(kbubuko.com,布布扣ibubuko.com,布布扣+1)=1+n(n?3)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣i=1bubuko.com,布布扣n(n?3)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣kbubuko.com,布布扣ibubuko.com,布布扣bubuko.com,布布扣

每个交点对应两条对角线,也就对应了四个顶点,故
bubuko.com,布布扣i=1bubuko.com,布布扣n(n?3)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣kbubuko.com,布布扣ibubuko.com,布布扣=Cbubuko.com,布布扣4bubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣


2. 三个角$A,B,C$满足$\cos A+\cos B+\cos C=\sin A+\sin B+\sin C=0$.
证明:$\cos^{2}A+\cos^{2}B+\cos^{2}C$等于常数,并求出这个常数.(提示:利用复数.)


3. 记$Q_{1}=\{Q|x \geq 1\}$,设函数$f:Q_{1} \rightarrow R$对任意$x,y\in Q_{1}$满足不等式

|f(x+y)?f(x)?f(y)|<εbubuko.com,布布扣

这里$\varepsilon$是某个大于$0$的数.证明:存在$q\in R,$使得任意$x\in Q_{1}$都有
|f(x)bubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣?q|<2εbubuko.com,布布扣


4. 证明不等式

nbubuko.com,布布扣r+1bubuko.com,布布扣?(n?1)bubuko.com,布布扣r+1bubuko.com,布布扣bubuko.com,布布扣r+1bubuko.com,布布扣bubuko.com,布布扣<nbubuko.com,布布扣rbubuko.com,布布扣<(n+1)bubuko.com,布布扣r+1bubuko.com,布布扣?nbubuko.com,布布扣r+1bubuko.com,布布扣bubuko.com,布布扣r+1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

高中教育试讲,布布扣,bubuko.com

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