【答案】
分析:(1)菱形ABCD的邊長為8cm,∠B=60°,則易證△ABC是等邊三角形,且邊長是8cm.由點(diǎn)P、Q從出發(fā)到相遇,則兩人所走的路程的和是24cm.設(shè)從出發(fā)到相遇所用的時間是x秒,據(jù)此列方程,求解即可;
(2)當(dāng)P在AC上,Q在AB上時,由于∠PAQ=60°,則△APQ一定不是等腰直角三角形;當(dāng)P在AC上,Q在BC上時,若△APQ是等腰直角三角形,由于∠PAQ<60°,即△APQ中A點(diǎn)不能為直角頂點(diǎn),如果∠PQA=90°,則∠PAQ=45°,∠QAB=15°,而∠B=60°,所以∠CQA=75°<∠PQA,不合題意,即△APQ中Q點(diǎn)不能為直角頂點(diǎn),所以只能∠APQ=90°,AP=PQ,根據(jù)這個相等關(guān)系,就可以得到一個關(guān)于x的方程,就可以得到x的值;當(dāng)P在BC上,Q在CD上時,△APQ一定不是等腰直角三角形;
(3)當(dāng)P在AC上,Q在AB上時,AP≠AQ,則△APQ一定不是等邊三角形;當(dāng)P在AC上,Q在BC上時,∠PAQ<60°,則△APQ一定不是等邊三角形;當(dāng)P在BC上,Q在CD上時,若△APQ是等邊三角形,則易證△ADQ≌△ACP,得出CP=DQ,根據(jù)這個相等關(guān)系,就可以得到一個關(guān)于x的方程,解方程即可求出x的值;
(4)求y與x之間的函數(shù)關(guān)系式,應(yīng)根據(jù)0≤x≤4和4<x≤8以及8<x≤12三種情況進(jìn)行討論.把x當(dāng)作已知數(shù)值,就可以求出y,即可得到函數(shù)的解析式.
解答:解:(1)∵四邊形ABCD是菱形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/images0.png)
∴AB=BC=8cm,
又∵∠B=60°,
∴△ABC是等邊三角形.
設(shè)點(diǎn)P,Q從出發(fā)到相遇所用的時間是x秒.
根據(jù)題意,得x+2x=24,
解得x=8秒.
即當(dāng)x=8秒時,P和Q相遇;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/images1.png)
(2)若△APQ是等腰直角三角形,則此時點(diǎn)P在AC上,點(diǎn)Q在BC上,如圖.
∵△APQ是等腰直角三角形,∴∠APQ=90°,∴∠CPQ=90°.
∵AP=x,∴CP=AC-AP=8-x.
在△CPQ中,∵∠CPQ=90°,∠PCQ=60°,∴∠CQP=30°,
∴PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/0.png)
CP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/1.png)
(8-x).
∵△APQ是等腰直角三角形,∠APQ=90°,∴AP=PQ,
即x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/2.png)
(8-x),
解得x=12-4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/3.png)
.
故當(dāng)x=(12-4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/4.png)
)秒時,△APQ是等腰直角三角形;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/images7.png)
(3)若△APQ是等邊三角形,則此時點(diǎn)P在BC上,點(diǎn)Q在CD上,如圖.
且△ADQ≌△ACP,則CP=DQ,
即x-8=24-2x,解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/5.png)
.
故當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/6.png)
秒時,△APQ是等邊三角形;
(4)分三種情況討論:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/images10.png)
①當(dāng)0≤x≤4時,
y=S
△AP1Q1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/7.png)
AP
1×AQ
1×sin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/8.png)
x•2x×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/10.png)
x
2,
根據(jù)二次函數(shù)的性質(zhì),可知當(dāng)x=4時,y有最大值
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/11.png)
×16=8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/12.png)
;
②當(dāng)4<x≤8時,
y=S
△AP2Q2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/13.png)
AP
2×CQ
2sin60°
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/14.png)
x(16-2x)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/15.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/16.png)
x
2+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/17.png)
x,
根據(jù)二次函數(shù)的性質(zhì),可知當(dāng)4<x≤8時,y無最大值;
③當(dāng)8<x≤12時,設(shè)P
3Q
3與AC交于點(diǎn)O.
過Q
3作Q
3E∥CB,則△CQ
3E為等邊三角形.
∴Q
3E=CE=CQ
3=2x-16.
∵Q
3E∥CB,
∴△COP
3∽△EOQ
3,
∴OC:OE=CP
3:EQ
3=(x-8):(2x-16)=1:2,
∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/18.png)
CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/19.png)
(2x-16).
∴y=S
△AOP3=S
△ACP3-S
△COP3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/20.png)
CP
3×ACsin60°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/21.png)
OC×CP
3sin60°
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/22.png)
(x-8)×8×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/23.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/24.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/25.png)
(2x-16)(x-8)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/26.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/27.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/28.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/29.png)
,
根據(jù)二次函數(shù)的性質(zhì),可知當(dāng)x=12時,y有最大值-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/30.png)
×12
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/31.png)
×12-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/33.png)
.
綜上可知,當(dāng)x=4時,y有最大值
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/34.png)
×16=8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/35.png)
.
故答案為8,(12-4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/36.png)
),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190419408244071/SYS201311011904194082440024_DA/37.png)
.
點(diǎn)評:本題借助動點(diǎn)問題考查了菱形的性質(zhì),等邊三角形的性質(zhì),等腰直角三角形的性質(zhì),相似三角形的判定與性質(zhì),三角形的面積,求函數(shù)的解析式及最值,綜合性較強(qiáng),難度較大.注意運(yùn)用分類討論及數(shù)形結(jié)合的思想是解決本題的關(guān)鍵.