1. (2024·自贡中考)如图，在□ABCD中，∠B = 60°，AB = 6 cm，BC = 12 cm. 点P从点A出发，以1 cm/s的速度沿A→D运动，同时点Q从点C出发，以3 cm/s的速度沿C→B→C→…往复运动，当点P到达端点D时，点Q随之停止运动. 在此运动过程中，线段PQ = CD出现的次数是 ( )

A. 3
B. 4
C. 5
D. 6

B 解析：在$\square ABCD$中，$AB = 6\ cm$，$BC = 12\ cm$，$\therefore CD = AB = 6\ cm$，$AD = BC = 12\ cm$，$AD// BC$，$\because$点$P$从点$A$出发，以$1\ cm/s$的速度沿$A\rightarrow D$运动，$\therefore$点$P$从点$A$出发到达$D$点的时间为$12\div1 = 12(s)$.$\because$点$Q$从点$C$出发，以$3\ cm/s$的速度沿$C\rightarrow B\rightarrow C\rightarrow \cdots$往复运动，$\therefore$点$Q$从点$C$出发到$B$点的时间为$12\div3 = 4(s)$.$\because AD// BC$，$\therefore PD// CQ$.当$PQ = CD$时四边形$CQPD$为平行四边形或等腰梯形.设$P$、$Q$同时运动的时间为$t(s)$.
①当$0 ＜ t\leqslant4$时，若四边形$CQPD$为平行四边形，则$DP = CQ$，$\therefore 12 - t = 3t$，解得$t = 3$，此时$PQ = CD$.若四边形$CQPD$为等腰梯形，则四边形$ABQP$也为等腰梯形，$AB = PQ = CD$，如图，过点$A$、$P$作$BC$的垂线，分别交$BC$于点$M$、$N$，$\therefore$四边形$AMNP$是矩形，$\therefore MN = AP = t$，$AM = PN$.又$\because\angle AMB=\angle PNQ = 90^{\circ}$，$AB = PQ$，$\therefore\triangle ABM\cong\triangle PQN(HL)$，$\therefore BM = QN$，在$Rt\triangle ABM$中，$\angle B = 60^{\circ}$，$AB = 6\ cm$，$\therefore\angle BAM = 90^{\circ}-\angle B = 30^{\circ}$，$\therefore BM=\frac{1}{2}AB = 3\ cm$，$\therefore BM = QN = 3\ cm$，$\therefore t = 12 - 3t-3 - 3$，$\therefore t=\frac{3}{2}$，此时$PQ = CD$.
②当$4 ＜ t\leqslant8$时，若四边形$CQPD$为平行四边形，则$DP = CQ$，$\therefore 12 - t = 12 - 3(t - 4)$，解得$t = 6$，此时$PQ = CD$.若四边形$CQPD$为等腰梯形，则四边形$ABQP$也为等腰梯形，由①可得$BQ = AP + 6$，即$3(t - 4)=t + 6$，解得$t = 9$(舍去).
③当$8 ＜ t\leqslant12$时，若四边形$CQPD$为平行四边形，则$DP = CQ$，$\therefore 12 - t = 3(t - 8)$，解得$t = 9$，此时$PQ = CD$.若四边形$CQPD$为等腰梯形，则四边形$ABQP$也为等腰梯形，由①可得$BQ = AP + 6$，即$12 - 3(t - 8)=t + 6$，解得$t=\frac{15}{2}$(舍去).
综上，当$t=\frac{3}{2}$或$t = 3$或$t = 6$或$t = 9$时，$PQ = CD$，共$4$次.故选B.

2. 如图，在Rt△ABC中，∠C = 90°，∠ABC = 30°，D、E是斜边AB上的两个动点(不与点A、B重合)，过E作EF⊥BC于点F，设BD = m，EF = n，且m = 12 - 4n，连接DF.
(1)当m = 8时，求DE的长.
(2)是否存在点P，使得以D、E、F、P四点为顶点的四边形是菱形？若存在，请求出n的值；若不存在，请说明理由.

(1)当$m = 8$时，$BD = 12 - 4n = 8$，$\therefore n = 1$.$\because\angle EFB = 90^{\circ}$，$\angle B = 30^{\circ}$，$EF = 1$，$\therefore BE = 2EF = 2$，$\therefore DE = BD - EB = 8 - 2 = 6$.
(2)存在.当$D$在点$E$的左侧时.$\because\angle DEF = 120^{\circ}$，$\therefore$当四边形$DEFP$是菱形时，只有一种情形，如图①，此时$DE = EF$，由(1)得，$DE = 12 - 6n$，$\therefore 12 - 6n = n$，$\therefore n=\frac{12}{7}$.
当点$D$在点$E$的右侧时.$\because\angle DEF = 60^{\circ}$，$\therefore$分两种情形，当四边形$DEFP_1$或四边形$DFEP_2$是菱形时，如图②，均可得到$\triangle DEF$是等边三角形，$\therefore DE = EF$，此时$DE = BE - BD = 2n-(12 - 4n)=6n - 12$，$\therefore 6n - 12 = n$，$\therefore n=\frac{12}{5}$.综上所述，满足条件的$n$的值为$\frac{12}{7}$或$\frac{12}{5}$.

3. (2024·佳木斯模拟)如图，在平面直角坐标系内有矩形OABC，已知点B(8,6)，D(0,4). 将矩形OABC沿EF折叠，使点A与点D重合. 折痕交BC于点E，交OA于点F.
(1)求点F的坐标.
(2)若动点P，Q同时从点A出发，点P以每秒1个单位长度的速度向点O运动，点Q以每秒2个单位长度的速度沿射线AB方向运动，当点P运动到点O时停止运动，点Q也同时停止运动. 设△PQF的面积为S，点P，Q的运动时间为t秒，求S与t的函数关系式并直接写出自变量的取值范围.
(3)在(2)的条件下，R是射线CB上的一点，点M为平面内一点，是否存在点M，使以P、Q、R、M为顶点的四边形是正方形？若存在，请求出点M的坐标；若不存在，请说明理由.

(1)由折叠可得$AF = DF$.$\because$点$B(8,6)$，点$D(0,4)$，四边形$OABC$为矩形，$\therefore OA\perp AB$，$OC = AB = 6$，$OA = BC = 8$，$OD = 4$.设$AF = DF = x$，则$OF = OA - AF = 8 - x$.在$Rt\triangle ODF$中，由勾股定理可得$OD^{2}+OF^{2}=DF^{2}$，即$4^{2}+(8 - x)^{2}=x^{2}$，解得$x = 5$，$\therefore OF = 8 - 5 = 3$，$\therefore$点$F$的坐标为$(3,0)$.
(2)①当点$P$在点$F$右侧时，根据题意，$AQ = 2t$，$AP = t(0 ＜ t ＜ 5)$，$\therefore FP = AF - AP = 5 - t$，$\therefore S=\frac{1}{2}FP\cdot AQ=\frac{1}{2}(5 - t)\times2t=-t^{2}+5t$；
②当点$P$在点$F$左侧时，根据题意，$AQ = 2t$，$AP = t(5 ＜ t\leqslant8)$，$\therefore FP = AP - AF = t - 5$，$\therefore S=\frac{1}{2}FP\cdot AQ=\frac{1}{2}(t - 5)\times2t=t^{2}-5t$.
综上所述，$S=\begin{cases}-t^{2}+5t(0 ＜ t ＜ 5)\\t^{2}-5t(5 ＜ t\leqslant8)\end{cases}$.
(3)若以$P$、$Q$、$R$、$M$为顶点的四边形是正方形，则点$P$、$R$、$Q$三点围成的三角形为等腰直角三角形，可分情况讨论：
①如图①，$\because$四边形$PQRM$是正方形，$\therefore PQ = QR$，$\angle PQR = 90^{\circ}$，$\therefore\angle PQA+\angle BQR=\angle BQR+\angle QRB$，$\therefore\angle PQA=\angle QRB$.在$\triangle PAQ$和$\triangle QBR$中，$\begin{cases}\angle PAQ=\angle QBR = 90^{\circ}\\\angle PQA=\angle QRB\\PQ = QR\end{cases}$，$\therefore\triangle PAQ\cong\triangle QBR(AAS)$，$\therefore AQ = BR$，$BQ = AP = t$，$\therefore AB = AQ + BQ = 2t + t = 3t = 6$，$\therefore BQ = AP = t = 2$，$\therefore BR = AQ = 2\times2 = 4$，$\therefore Q(8,4)$，$\therefore OP = OA - AP = 8 - 2 = 6$，$\therefore P(6,0)$，$\therefore CR = BC - BR = 4$，$\therefore R(4,6)$.$\because$四边形$PQRM$是正方形，$\therefore M(4 + 6 - 8,6 + 0 - 4)$，即$M(2,2)$；

②如图②，过点$R$作$RK\perp OA$于点$K$，则四边形$OCRK$、$RKAB$均为矩形，$\therefore RK = AB = 6$，$\angle BRK=\angle RKA = 90^{\circ}$.$\because$四边形$PRQM$是正方形，$\therefore PR = QR$，$\angle PRQ = 90^{\circ}$，$\therefore\angle KRB-\angle PRB=\angle PRQ-\angle PRB$，$\therefore\angle KRP=\angle BRQ$.在$\triangle PKR$和$\triangle QBR$中，$\begin{cases}\angle RKP=\angle RBQ = 90^{\circ}\\\angle KRP=\angle BRQ\\PR = QR\end{cases}$，$\therefore\triangle PKR\cong\triangle QBR(AAS)$，$\therefore RB = RK = 6$，$KP = BQ = AQ - AB = 2t - 6$，$\therefore OK = CR = BC - BR = 8 - 6 = 2$，$\therefore R(2,6)$，$AK = OA - OK = 6 = 2t - 6 + t$，$\therefore AP = t = 4$，$\therefore P(8 - 4,0)$，即$P(4,0)$，$KP = BQ = 2t - 6 = 2$，$\therefore Q(8,6 + 2)$，即$Q(8,8)$.$\because$四边形$PMQR$是正方形，$\therefore M(4 + 8 - 2,8 + 0 - 6)$，即$M(10,2)$；
③如图③，$\because$四边形$PQRM$是正方形，$\therefore PQ = QR$，$\angle PQR = 90^{\circ}$，$\therefore\angle PQA+\angle BQR=\angle BQR+\angle QRB$，$\therefore\angle PQA=\angle QRB$.在$\triangle PAQ$和$\triangle QBR$中，$\begin{cases}\angle PAQ=\angle QBR = 90^{\circ}\\\angle PQA=\angle QRB\\PQ = QR\end{cases}$，$\therefore\triangle PAQ\cong\triangle QBR(AAS)$，$\therefore AQ = BR = 2t$，$BQ = AP = t$.
又$\because AQ = AB + BQ = 6 + t = 2t$，$\therefore AP = t = 6$，$\therefore BR = 2\times6 = 12$，$\therefore Q(8,12)$，$P(8 - 6,0)$，即$P(2,0)$，$\therefore CR = BC + BR = 8 + 12 = 20$，$\therefore R(20,6)$.
$\because$四边形$PMRQ$是正方形，$\therefore M(2 + 20 - 8,6 + 0 - 12)$，即$M(14,-6)$.
综上所述，存在$M(2,2)$或$M(10,2)$或$M(14,-6)$，使以$P$、$Q$、$R$、$M$为顶点的四边形是正方形.