试题

egin{Example}
已知集合$A={a_1,a_2,cdots,a_k}(kgeq 2)$,其中$a_iinmathbb{Z} (i=1,2,cdots,k)$,由$A$中的元素构成两个相应的集合：
[S={(a,b)|ain A,bin A,a+bin A},quad T={(a,b)|ain A,bin A,a-bin A},]
其中$(a,b)$是有序数对,集合$S$和$T$中的元素个数分别为$m$和$n$.
若对于任意的$ain A$,总有$-a otin A$,则称集合$A$具有性质$P$.
egin{enumerate}
item[(I)] 检验集合${0,1,2,3}$与${-1,2,3}$是否具有性质$P$并对其中具有性质$P$的集合,写出相应的集合$S$和$T$;
item[(II)] 对任何具有性质$P$的集合$A$,证明: $displaystyle nleq frac{k(k-1)}{2}$;
item[(III)] 判断$m$和$n$的大小关系,并证明你的结论.
end{enumerate}
end{Example}
egin{Proof}

egin{enumerate}
item[(I)] 集合${0,1,2,3}$不具有性质$P$. 集合${-1,2,3}$具有性质$P$,其相应的集合$S$和$T$是$S={(-1,3),(3,-1)},T={(2,-1),(2,3)}$.

item[(II)] 首先,由$A$中元素构成的有序数对$(a_i,a_j)$共有$k^2$个.因为$0 otin A$,所以$(a_i,a_i) otin T(i=1,2,cdots,k)$.从而,集合$T$中元素的个数最多为$displaystyle frac{1}{2}left(k^2-k ight)=frac{k(k-1)}{2}$,即$displaystyle nleq frac{k(k-1)}{2}$.

item[(III)] $m=n$,证明如下:
egin{enumerate}
item[(1)] 对于$(a,b)in S$,根据定义, $ain A,bin A$,且$a+bin A$,从而$(a+b,b)in T$.如果$(a,b)$与$(c,d)$是$S$的不同元素,那么$a=c$与$b=d$中至少有一个不成立,从而$a+b=c+d$与$b=d$中也至少有一个不成立,故$(a+b,b)$与$(c+d,d)$也是$T$的不同元素.可见, $S$中元素的个数不多于$T$中元素的个数,即$mleq n$.

item[(2)] 对于$(a,b)in T$,根据定义, $ain A,bin A$,且$a-bin A$,从而$(a-b,b)in S$.如果$(a,b)$与$(c,d)$是$T$的不同元素,那么$a=c$与$b=d$中至少有一个不成立,从而$a-b=c-d$与$b=d$中也至少有一个不成立,故$(a-b,b)$与$(c-d,d)$也是$S$的不同元素.可见, $T$中元素的个数不多于$S$中元素的个数,即$nleq m$.
end{enumerate}

由(1) (2)可知, $m=n$.
end{enumerate}

end{Proof}

(2008北京)对于每项均是正整数的数列$A:a_1,a_2,cdots,a_n$,定义变换$T_1$, $T_1$将数列$A$变换成数列$T_1(A): n,a_1-1,a_2-1,cdots,a_n-1$;对于每项均是非负整数的数列$B:b_1,b_2,cdots,b_m$,定义变换$T_2$, $T_2$将数列$B$各项从大到小排列,然后去掉所有为零的项,得到数列$T_2(B)$;又定义$S(B)=2(b_1+2b_2+cdots+mb_m)+b_1^2+b_2^2+cdots+b_m^2$.设$A_0$是每项均为正整数的有穷数列,令$A_{k+1}=T_2(T_1(A_k)) (k=0,1,2,cdots)$.
egin{enumerate}
item[(I)] 如果数列$A_0$为$5,3,2$,写出数列$A_1,A_2$;

item[(II)] 对于每项均是正整数的有穷数列$A$,证明$S(T_1(A))=S(A)$;

item[(III)] 证明:对于任意给定的每项均为正整数的有穷数列$A_0$,存在正整数$K$,当$kgeq K$时, $S(A_{k+1})=S(A_k)$.
end{enumerate}

egin{Proof}
egin{enumerate}
item[(I)] $A_0: 5,3,2$, $T_1(A_0):3,4,2,1$, $A_1=T_2(T_1(A_0)):4,3,2,1$, \
$A_2=T_2(T_1(A_1)):4,3,2,1$.

item[(II)] 设每项均是正整数的有穷数列$A$为$a_1,a_2,cdots,a_n$,则$T_1(A)$为$n,a_1-1,a_2-1,cdots,a_n-1$,从而
egin{align*}
Sleft( T_1left( A ight) ight) =& 2left[ n+2left( a_1-1 ight) +3left( a_2-1 ight) +cdots +left( n+1 ight) left( a_n-1 ight) ight]
\
&+n^2+left( a_1-1 ight) ^2+left( a_2-1 ight) ^2+cdots +left( a_n-1 ight) ^2.
end{align*}
又
[
Sleft( A ight) =2left( a_1+2a_2+cdots +na_n ight) +a_{1}^{2}+a_{2}^{2}+cdots +a_{n}^{2},
]
所以
egin{align*}
Sleft( T_1left( A ight) ight) -Sleft( A ight) =& 2left[ n-2-3-cdots -left( n+1 ight) ight]
\
&+2left( a_1+a_2+cdots +a_n ight) +n^2-2left( a_1+a_2+cdots +a_n ight) +n
\
=&-nleft( n+1 ight) +n^2+n=0,
end{align*}
故$Sleft( T_1left( A ight) ight) =Sleft( A ight)$.

item[(III)] 设$A$是每项均为非负整数的数列$a_1,a_2,cdots,a_n$.当存在$1leq i<jleq n$,使得$a_ileq a_j$时,交换数列$A$的第$i$项与第$j$项得到数列$B$,则
[S(B)-S(A)=2(ia_j+ja_i-ia_i-ja_j)=2(i-j)(a_j-a_i)leq 0.]
当存在$1leq m<n$,使得$a_{m+1}=a_{m+2}=cdots=a_n=0$时,若记数列$a_1,a_2,cdots,a_m$为$C$,则$S(C)=S(A)$.所以$S(T_2(A))leq S(A)$.

从而对于任意给定的数列$A_0$,由$A_{k+1}=T_2(T_1(A_k))\, (k=0,1,2,cdots)$
可知$S(A_{k+1})leq S(T_1(A_k))$.又由(II)可知$Sleft( T_1left( A ight) ight) =Sleft( A ight)$,所以$S(A_{k+1})leq S(A_k)$.即对于$kinmathbb{N}$,要么有$S(A_{k+1})= S(A_k)$,要么有$S(A_{k+1})leq S(A_k)-1$.因为$S(A_k)$是大于$2$的整数,所以经过有限步后,必有
[S(A_k)=S(A_{k+1})=S(A_{k+2})=cdots=0.]
即存在正整数$K$,当$kgeq K$时, $S(A_{k+1})=S(A_k)$.
end{enumerate}

end{Proof}

已知抛物线$C: x^2=4y$,抛物线$C$上存在三个点$A(x_1,y_1),B(x_2,y_2)$, $D(x_3,y_3)$,满足$x_3<x_1<x_2$.若$ riangle ABD$是以角$A$为直角的等腰直角三角形,求$ riangle ABD$面积的最小值.
end{Example}
egin{Proof}
首先由$AB$与$AD$垂直可得$overrightarrow{AB}cdot overrightarrow{AD}=(x_2-x_1)(x_3-x_1)+(y_2-y_1)(y_3-y_1)=0$,由$x_3<x_1<x_2,x_i^2=4y_i(i=1,2,3)$,我们可知
[16+(x_2+x_1)(x_3+x_1)=x_1^2+16+x_1(x_2+x_3)+x_2x_3=0.]
由$AB=AD$可知
[
sqrt{left( x_2-x_1 ight) ^2+left( y_2-y_1 ight) ^2}=sqrt{left( x_3-x_1 ight) ^2+left( y_3-y_1 ight) ^2},
]
则
[left( x_2-x_3 ight) left( x_2+x_3-2x_1 ight) +left( y_2-y_3 ight) left( y_2+y_3-2y_1 ight) =0.]
若记$BD$的中点为$E$,这等价于$AE$与$BD$垂直.进一步化简得
[
16left( x_2+x_3-2x_1 ight) +left( x_2+x_3 ight) left( x_{2}^{2}+x_{3}^{2}-2x_{1}^{2} ight) =0,
]
因此
[left[ 2left( x_2+x_3 ight) ^2-32 ight] x_1+left( x_2+x_3 ight) ^3+48left( x_2+x_3 ight) =0.]
于是
[x_1=frac{left( x_2+x_3 ight) ^3+48left( x_2+x_3 ight)}{32-2left( x_2+x_3 ight) ^2}=frac{t^3+48t}{32-2t^2} left( t=x_2+x_3 ight).]

注意到
egin{align*}
S&=left| frac{1}{2}left| egin{matrix}
x_1& y_1& 1\
x_2& y_2& 1\
x_3& y_3& 1\
end{matrix} ight| ight|=left| frac{1}{2}left[ left( x_2-x_1 ight) left( y_3-y_1 ight) -left( x_3-x_1 ight) left( y_2-y_1 ight) ight] ight|\
&=left| frac{1}{2}left[ left( x_2-x_1 ight) left( frac{x_{3}^{2}}{4}-frac{x_{1}^{2}}{4} ight) -left( x_3-x_1 ight) left( frac{x_{2}^{2}}{4}-frac{x_{1}^{2}}{4} ight) ight] ight|\
&=left| frac{1}{8}left( x_2-x_1 ight) left( x_3-x_1 ight) left( x_3-x_2 ight) ight|\
&=frac{1}{8}left| x_{1}^{2}-left( x_2+x_3 ight) x_1+x_2x_3 ight|cdot left| x_3-x_2 ight|.
end{align*}

又
egin{align*}
&x_{1}^{2}-left( x_2+x_3 ight) x_1+x_2x_3\
&=-left[ 4+left( x_2+x_3 ight)x_1 +x_2x_3 ight] -left( x_2+x_3 ight) x_1+x_2x_3\
&=-16-2left( x_2+x_3 ight) x_1=-16-2tfrac{t^3+48t}{32-2t^2}=frac{left( t^2+16 ight) ^2}{t^2-16}.
end{align*}
且
egin{align*}
x_2x_3&=-x_{1}^{2}-16-x_1t=-frac{left( t^3+48t ight) ^2}{left( 32-2t^2 ight) ^2}-16-frac{t^4+48t^2}{32-2t^2}\
&=frac{t^6-96t^4-1792t^2-16384}{4left( t-4 ight) ^2left( t+4 ight) ^2}.\
left| x_2-x_3 ight|&=sqrt{left( x_2+x_3 ight) ^2-4x_2x_3}\
&=sqrt{t^2-4frac{t^6-96t^4-1792t^2-16384}{4left( t-4 ight) ^2left( t+4 ight) ^2}}=frac{8left( t^2+16 ight)}{left| t^2-16 ight|}.
end{align*}
于是
egin{align*}
S&=frac{1}{8}left| x_{1}^{2}-left( x_2+x_3 ight) x_1+x_2x_3 ight|cdot left| x_3-x_2 ight|\
&=frac{1}{8}frac{left( t^2+16 ight) ^2}{left| t^2-16 ight|}cdot frac{8left( t^2+16 ight)}{left| t^2-16 ight|}=frac{left( t^2+16 ight) ^3}{left( t^2-16 ight) ^2}.
end{align*}

令$u=sqrt[3]{t^2-16}\, (|t|geq 4)$,则
egin{align*}
S&=frac{left( u^3+32 ight) ^3}{u^6}=frac{1}{2}left( u+frac{32}{u^2} ight) ^3\
&=frac{1}{2}left( frac{u}{2}+frac{u}{2}+frac{32}{u^2} ight) ^3geq frac{1}{2}left( 3sqrt[3]{frac{u}{2}cdot frac{u}{2}cdot frac{32}{u^2}} ight) ^3=108,
end{align*}
当且仅当$u=sqrt[3]{t^2-4}=4,t=x_2+x_3=pm 4sqrt{5}$时取等成立,此时$x_2x_3=-16$,即
[left( x_3,x_1,x_2 ight) =left( -6-2sqrt{5},4sqrt{5},6-2sqrt{5} ight)quad ext{或} quad left( -6+2sqrt{5},-4sqrt{5},6+2sqrt{5} ight).]
与$x_3<x_1<x_2$矛盾,故舍去.

令$u=sqrt[3]{16-t^2}\, (|t|<4)$,则$0<uleq sqrt[3]{16}$
[S=frac{left( -u^3+32 ight) ^3}{u^6}=left( -u+frac{32}{u^2} ight) ^3]
在$left(0,sqrt[3]{16} ight]$上单调递减,因此,当$uleq sqrt[3]{16}$,即$t=0$时, $S$取得最小值$16$.此时$left( x_3,x_1,x_2 ight) =left(-4,0,4 ight)$.

(叶卢庆)因为点$A,B,D$在抛物线$C$上,所以
$$
egin{cases}
x_1^2=4y_1\
x_2^2=4y_2\
x_3^2=4y_3
end{cases}.
$$
故
$$
overrightarrow{AB}=(x_2-x_1,y_2-y_1)=(x_2-x_1,frac{x_2^2-x_1^2}{4}),
$$
egin{equation}label{eq:1}
overrightarrow{AD}=(x_3-x_1,y_3-y_1)=(x_3-x_1,frac{x_3^2-x_1^2}{4}).
end{equation}
而$ riangle ABD$的面积
egin{align*}
S_{ riangle ABD}&=frac{1}{2}
egin{vmatrix}
x_2-x_1&x_3-x_1\
y_2-y_1&y_3-y_1
end{vmatrix}\&=frac{1}{2}
egin{vmatrix}
x_2-x_1&x_3-x_1\
cfrac{x_2^2-x_1^2}{4}&cfrac{x_3^2-x_1^2}{4}
end{vmatrix}
\&=frac{(x_2-x_1)(x_3^2-x_1^2)-(x_3-x_1)(x_2^2-x_1^2)}{8}
\&=frac{(x_1-x_2)(x_2-x_3)(x_3-x_1)}{8}
end{align*}
令$x_2-x_1=t,x_1-x_3=s$,则
$$
S_{ riangle ABD}=frac{ts(t+s)}{8}.
$$
又因为$overrightarrow{AD}$是$overrightarrow{AB}$以点$A$为旋转中心逆时针旋转$90^{circ}$得到的,因此
egin{equation}label{eq:2}
overrightarrow{AD}=left(-frac{x_2^2-x_1^2}{4},x_2-x_1 ight).
end{equation}
由$overrightarrow{AD}$的两种表达式eqref{eq:1},eqref{eq:2}可得到关系式
$$
x_3-x_1=-frac{x_2^2-x_1^2}{4},frac{x_3^2-x_1^2}{4}=x_2-x_1,
$$
即
$$
x_3-x_1=frac{(x_1+x_2)(x_1-x_2)}{4},x_2-x_1=-frac{(x_1+x_3)(x_1-x_3)}{4},
$$
将$x_2-x_1=t,x_1-x_3=s$代入,整理可得
$$
frac{4s}{t}=x_1+x_2,frac{4t}{-s}=x_1+x_3.
$$
因此
$$
frac{4s}{t}+frac{4t}{s}=x_2-x_3=t+s,
$$
整理可得
egin{equation}label{eq:3}
4t^2+4s^2=ts(t+s).
end{equation}
所以
$$
S_{ riangle ABD}=frac{ts(t+s)}{8}=frac{t^2+s^2}{2}.
$$
下面求$t^2+s^2$的最小值.将方程eqref{eq:3}两边平方可得
egin{equation}label{eq:4}
16left(t^2+s^2 ight)^2=(ts)^2(t+s)^2=(ts)^2left[left(t^2+s^2 ight)+2ts ight]
end{equation}
由基本不等式,$tsleq cfrac{t^2+s^2}{2}$,代入式eqref{eq:4},可得
$$
16left(t^2+s^2 ight)^2leq frac{1}{2}left(t^2+s^2 ight)^3
$$
解得
$$
t^2+s^2geq 32.
$$
因此
$$
S_{ riangle ABD}=frac{t^2+s^2}{2}geq 16.
$$
等号当且仅当$t=s=4$,即$x_3=-4,x_1=0,x_2=4$时成立.

来源：叶卢庆博客.