【答案】
分析:(1)根據(jù)勾股定理及菱形的性質(zhì),求出菱形的周長(zhǎng);
(2)在動(dòng)點(diǎn)M、N運(yùn)動(dòng)過程中:①當(dāng)0<t≤40時(shí),如答圖1所示,②當(dāng)40<t≤50時(shí),如答圖2所示.分別求出S的關(guān)系式,然后利用二次函數(shù)的性質(zhì)求出最大值;
(3)如答圖3所示,在Rt△PKD中,DK長(zhǎng)可求出,則只有求出tan∠DPK即可.為此,在△ODM中,作輔助線,構(gòu)造Rt△OND,作∠NOD平分線OG,則∠GOF=∠DPK.在Rt△OGF中,求出tan∠GOF的值,從而問題解決.解答中提供另外一種解法,請(qǐng)參考.
解答:解:(1)在菱形ABCD中,
∵AC⊥BD
∴AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/0.png)
=50.
∴菱形ABCD的周長(zhǎng)為200.
(2)過點(diǎn)M作MP⊥AD,垂足為點(diǎn)P.
①當(dāng)0<t≤40時(shí),如答圖1,
∵sin∠OAD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/3.png)
,
∴MP=AM•sin∠OAD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/4.png)
t.
S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/5.png)
DN•MP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/6.png)
×t×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/7.png)
t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/8.png)
t
2;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/images9.png)
②當(dāng)40<t≤50時(shí),如答圖2,MD=70-t,
∵sin∠ADO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/11.png)
,∴MP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/12.png)
(70-t).
∴S
△DMN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/13.png)
DN•MP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/14.png)
×t×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/15.png)
(70-t)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/16.png)
t
2+28t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/17.png)
(t-35)
2+490.
∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/18.png)
當(dāng)0<t≤40時(shí),S隨t的增大而增大,當(dāng)t=40時(shí),最大值為480.
當(dāng)40<t≤50時(shí),S隨t的增大而減小,當(dāng)t=40時(shí),最大值為480.
綜上所述,S的最大值為480.
(3)存在2個(gè)點(diǎn)P,使得∠DPO=∠DON.
方法一:如答圖3所示,過點(diǎn)N作NF⊥OD于點(diǎn)F,
則NF=ND•sin∠ODA=30×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/19.png)
=24,DF=ND•cos∠ODA=30×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/20.png)
=18.
∴OF=12,∴tan∠NOD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/22.png)
=2.
作∠NOD的平分線交NF于點(diǎn)G,過點(diǎn)G作GH⊥ON于點(diǎn)H,則FG=GH.
∴S
△ONF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/23.png)
OF•NF=S
△OGF+S
△OGN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/24.png)
OF•FG+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/25.png)
ON•GH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/26.png)
(OF+ON)•FG.
∴FG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/29.png)
,
∴tan∠GOF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/32.png)
.
設(shè)OD中垂線與OD的交點(diǎn)為K,由對(duì)稱性可知:∠DPK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/33.png)
∠DPO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/34.png)
∠DON=∠FOG
∴tan∠DPK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/37.png)
,
∴PK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/38.png)
.
根據(jù)菱形的對(duì)稱性可知,在線段OD的下方存在與點(diǎn)P關(guān)于OD軸對(duì)稱的點(diǎn)P′.
∴存在兩個(gè)點(diǎn)P到OD的距離都是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/39.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/images41.png)
方法二:答圖4所示,作ON的垂直平分線,交OD的垂直平分線EF于點(diǎn)I,連結(jié)OI,IN.
過點(diǎn)N作NG⊥OD,NH⊥EF,垂足分別為G,H.
當(dāng)t=30時(shí),DN=OD=30,易知△DNG∽△DAO,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/40.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/41.png)
.
∴NG=24,DG=18.
∵EF垂直平分OD,
∴OE=ED=15,EG=NH=3.
設(shè)OI=R,EI=x,則
在Rt△OEI中,有R
2=15
2+x
2 ①
在Rt△NIH中,有R
2=3
2+(24-x)
2 ②
由①、②可得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/42.png)
∴PE=PI+IE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/43.png)
.
根據(jù)對(duì)稱性可得,在BD下方還存在一個(gè)點(diǎn)P′也滿足條件.
∴存在兩個(gè)點(diǎn)P,到OD的距離都是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193603169383244/SYS201311011936031693832024_DA/44.png)
.
(注:只求出一個(gè)點(diǎn)P并計(jì)算正確的扣(1分).)
點(diǎn)評(píng):本題考查了相似三角形的判定與性質(zhì)、菱形、等腰三角形、中垂線、勾股定理、解直角三角形、二次函數(shù)極值等知識(shí)點(diǎn),涉及考點(diǎn)較多,有一定的難度.第(2)問中,動(dòng)點(diǎn)M在線段AO和OD上運(yùn)動(dòng)時(shí),是兩種不同的情形,需要分類討論;第(3)問中,滿足條件的點(diǎn)有2個(gè),注意不要漏解.