【答案】
分析:(1)過C作CE⊥x軸于E,易證得△ABO≌△BCE,可得AO=BE、OB=CE,由此求出點(diǎn)C的坐標(biāo).
(2)RH∥y軸,即R、H的橫坐標(biāo)相同;
由于AB∥CD,得∠DMR=∠ANO,若△ANO與△DMR相似,則:
由于直線OP經(jīng)過正方形的對(duì)稱中線,因此OP平分∠AOB,即∠AOP=45°,由于AB∥CD,故∠ANO=∠DMO,若△ANO與△DMR相似,則有兩種情況:
①∠DRM=45°,此時(shí)DR∥y軸,即點(diǎn)R、D、H的橫坐標(biāo)都相同,由此求出點(diǎn)H的坐標(biāo);
②∠RDM=45°,此時(shí)R、P重合,因此R、H的橫坐標(biāo)相同,由此求出點(diǎn)H的坐標(biāo).
(3)①首先用t表示出PH的長(zhǎng),由于PH與y軸平行,可以PH為底、H、C的橫坐標(biāo)差的絕對(duì)值為高求出S的表達(dá)式,即可得S、t的函數(shù)關(guān)系式;要注意的是在表示高的過程中,要分H在C點(diǎn)左側(cè)和H點(diǎn)在C點(diǎn)右側(cè)兩種情況討論.
②此題應(yīng)分三種情況討論:
一、CR∥AB,此時(shí)R、M重合,可求出直線CD的解析式,聯(lián)立直線OP的解析式,即可求得M點(diǎn)(即R)的坐標(biāo),進(jìn)而得到H點(diǎn)坐標(biāo)和t的值,然后再將t代入①的函數(shù)解析式中即可得到S的值;
二、AR∥BC,三、BR∥AC,解法同上.
解答:
解:(1)過C作CE⊥x軸于E;
由于四邊形ABCD是正方形,
∴AB=BC,∠ABC=90°;
易證得△ABO≌△BCE,
則AO=BE=3,OB=CE=1,
∴C(4,1);(2分)
同理可求,D(3,4).
(2)由于P是正方形的對(duì)稱中心,由A(0,3),C(4,1),
可得P(2,2);
則∠MOE=45°,又OR=

t,OH=t,所以RH∥y軸,即R、H的橫坐標(biāo)相同;
由于AB∥CD,得∠DMR=∠ANO,若△ANO與△DMR相似,則:
①當(dāng)∠MDR=45°時(shí),R、P重合,此時(shí)R(2,2),故t=2,點(diǎn)H(2,0);
②當(dāng)∠DRM=45°時(shí),DR∥y軸,此時(shí)R(3,3),故t=3,點(diǎn)H(3,0);
所以當(dāng)t=2或t=3時(shí),△ANO與△DMR相似.
(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=

RH•HE=

t(4-t)=-

t
2+2t;
二、t>4,H在E點(diǎn)右側(cè);
易知RH=t,HE=t-4,故S=

RH•HE=

t(t-4)=

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=

;
即M(即R)點(diǎn)橫坐標(biāo)為

,H(

,0);
故t=

,代入S=-

t
2+2t(0<t≤4)可得S=

;
同理可求得:
二、AR∥BC時(shí),t=

,S=

;
三、BR∥AC時(shí),t=

,S=

;
綜合①②可得:
S=-

t
2+2t(0<t≤4);(1分)S=

t
2-2t(t>4).
當(dāng)CR∥AB時(shí),t=

,(1分)S=

;
當(dāng)AR∥BC時(shí),t=

,S=

;
當(dāng)BR∥AC時(shí),t=

,S=

.
點(diǎn)評(píng):此題主要考查了正方形的性質(zhì),全等三角形、相似三角形的判定和性質(zhì),梯形的判定,圖形面積的求法等知識(shí),同時(shí)還考查了分類討論思想在動(dòng)點(diǎn)問題中的應(yīng)用,難度較大.