【答案】
分析:(1)根據(jù)三角函數(shù)即可求得OA,OB的長(zhǎng),即可得到A則坐標(biāo),C的坐標(biāo),進(jìn)而求得D,P的坐標(biāo);
(2)分∠MDR=45°和∠DRM=45°兩種情況求得t的值;
(3)分0<t≤4和t>4兩種情況求得函數(shù)解析式,然后分當(dāng)CR∥AB時(shí),當(dāng)AR∥BC時(shí),當(dāng)BR∥AC三種情況求得t的值,進(jìn)而求得函數(shù)的最大值.
解答:解:(1)A(0,3)C(4,1),D(3,4),P(2,2)
(2)過(guò)點(diǎn)N作NE⊥AO,于點(diǎn)E,過(guò)點(diǎn)A作AF⊥MS于點(diǎn)F,MS⊥x軸于點(diǎn)S,
由(1)可得:B(1,0),
∴直線AB的解析式為:y=-3x+3①;
直線OP的解析式為:y=x②,
①②聯(lián)立得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/0.png)
,
∴N(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/2.png)
),
直線CD的解析式是:y=-3x+13,
解方程組:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/3.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/4.png)
.
則M的坐標(biāo)是:(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/6.png)
),
∴ON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/8.png)
,OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/9.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/10.png)
,
∵AD
2+DM
2=AF
2+MF
2,
10+MD
2=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/11.png)
)
2+(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/12.png)
)
2,
∴DM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/13.png)
,
AN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/15.png)
.
當(dāng)∠MDR=45°時(shí),
∵∠AON=45°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/images16.png)
∴∠MDR=∠AON,
∵AN∥DM,
∴∠ANO=∠DMP,
∴△ANO與△DMR相似,則△ANO∽△RMD,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/16.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/18.png)
,
解得:MR=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/19.png)
,
則OR=OM-MR=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/20.png)
.
∴t=2,
同理可得:當(dāng)∠DRM=45°時(shí),t=3,△ANO與△DMR相似,
綜上可知:t=2或3時(shí)當(dāng)△ANO與△DMR相似;
(3)①∵R速度為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/21.png)
,H速度為1,且∠ROH=45°,
∴tan∠ROH=1,
∴RH始終垂直于x軸,
∴RH=OH=t,
設(shè)△HCR的邊RH的高為h,
∴h=|4-t|.
∴S
△HCR=h•t
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/22.png)
2=|-t
2+4t|
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/23.png)
,
∴S=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/24.png)
t
2+2t(0<t<4);S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/25.png)
t
2-2t(t>4);
②以A、B、C、R為頂點(diǎn)的梯形,有兩種可能:
1.頂邊和底邊分別為BC、AR,此時(shí)BC∥AR.
延長(zhǎng)AD,使其與OM相交于點(diǎn)R,
∴AD的斜率=tan∠BAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/26.png)
,
∴直線AD為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/27.png)
+3.
∴R坐標(biāo)為(4.5,4.5),
∴此時(shí)四邊形ABCR為梯形為梯形,
∴t=4.5
2.頂邊、底邊分別為CR、AB,此時(shí)CR∥AB,且R與M重合.
∴CD的斜率=-3,且直線CD過(guò)點(diǎn)C,
∴直線CD為:y-1=-3•(x-4),
∴y=-3x+13,
∵OM與CD交于點(diǎn)M(即R),
∴M為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/28.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/29.png)
),
∴此時(shí)四邊形ABCR為梯形,
∴t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/30.png)
,
∴當(dāng)CR∥AB時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/31.png)
,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/32.png)
,
當(dāng)AR∥BC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/33.png)
,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/34.png)
,
當(dāng)BR∥AC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/35.png)
,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/36.png)
.
(3)①分兩種情況:
一、0<t≤4,H在E點(diǎn)左側(cè);
易知RH=t,HE=4-t,故S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/37.png)
RH•HE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/38.png)
t(4-t)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/39.png)
t
2+2t;
二、t>4,H在E點(diǎn)右側(cè);
易知RH=t,HE=t-4,故S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/40.png)
RH•HE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/41.png)
t(t-4)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/42.png)
t
2-2t;
②若以A、B、C、R為頂點(diǎn)的四邊形是梯形,分三種情況:
一、CR∥AB;此時(shí)R、M重合,
由C(4,1),D(3,4),可求得直線CD:y=-3x+13;
當(dāng)x=y時(shí),-3x+13=x,解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/43.png)
;
即M(即R)點(diǎn)橫坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/44.png)
,H(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/45.png)
,0);
故t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/46.png)
,代入S=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/47.png)
t
2+2t(0<t≤4)可得S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/48.png)
;
同理可求得:
二、AR∥BC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/49.png)
,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/50.png)
;
三、BR∥AC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/51.png)
,S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/52.png)
;
綜合①②可得:
S=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/53.png)
t
2+2t(0<t≤4);(1分)
S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/54.png)
t
2-2t(t>4).(1分)
當(dāng)CR∥AB時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/55.png)
,(1分)
S
最大=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/56.png)
;(1分)
當(dāng)AR∥BC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/57.png)
,S
最大=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/58.png)
;(1分)
當(dāng)BR∥AC時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/59.png)
,S
最大=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160002605816957/SYS201310221600026058169027_DA/60.png)
.(1分)
點(diǎn)評(píng):本題主要考查了正方形的性質(zhì),以及二次函數(shù)的性質(zhì),正確求得函數(shù)的解析式是解題的關(guān)鍵.