1. (2024·自贡中考)如图，在$□ ABCD$中，$∠ B = 60^{\circ}$，$AB = 6\mathrm{cm}$，$BC = 12\mathrm{cm}$。点$P$从点$A$出发，以$1\mathrm{cm}/\mathrm{s}$的速度沿$A \to D$运动，同时点$Q$从点$C$出发，以$3\mathrm{cm}/\mathrm{s}$的速度沿$C \to B \to C \to ···$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
)A.$3$
B.$4$
C.$5$
D.$6$
答案：
1. B 解析:在$□ ABCD$中,$AB = 6\mathrm{cm}$,$BC = 12\mathrm{cm}$,$\therefore CD = AB = 6\mathrm{cm}$,$AD = BC = 12\mathrm{cm}$,$AD// BC$.$\because$点$P$从点$A$出发,以$1\mathrm{cm}/\mathrm{s}$的速度沿$A \to D$运动,$\therefore$点$P$从点$A$出发到达点$D$的时间为$12÷1 = 12(\mathrm{s})$.$\because$点$Q$从点$C$出发,以$3\mathrm{cm}/\mathrm{s}$的速度沿$C \to B \to C \to ···$往复运动,$\therefore$点$Q$从点$C$出发到点$B$的时间为$12÷3 = 4(\mathrm{s})$.$\because AD// BC$,$\therefore PD// CQ$.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$t(\mathrm{s})$.
①当$0 ＜ t ≤ 4$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$\because ∠ AMB = ∠ PNQ = 90^{\circ}$,$AB = PQ$,$\therefore \mathrm{Rt}△ ABM ≌ \mathrm{Rt}△ PQN(\mathrm{HL})$,$\therefore BM = QN$.在$\mathrm{Rt}△ ABM$中,$∠ B = 60^{\circ}$,$AB = 6\mathrm{cm}$,$\therefore ∠ BAM = 90^{\circ} - ∠ B = 30^{\circ}$,$\therefore BM = \frac{1}{2}AB = 3\mathrm{cm}$,$\therefore BM = QN = 3\mathrm{cm}$,$\therefore t = 12 - 3t - 3 - 3$,$\therefore t = \frac{3}{2}$,此时$PQ = CD$.
②当$4 ＜ t ≤ 8$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$8 ＜ t ≤ 12$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.
1. B 解析:在$□ ABCD$中,$AB = 6\mathrm{cm}$,$BC = 12\mathrm{cm}$,$\therefore CD = AB = 6\mathrm{cm}$,$AD = BC = 12\mathrm{cm}$,$AD// BC$.$\because$点$P$从点$A$出发,以$1\mathrm{cm}/\mathrm{s}$的速度沿$A \to D$运动,$\therefore$点$P$从点$A$出发到达点$D$的时间为$12÷1 = 12(\mathrm{s})$.$\because$点$Q$从点$C$出发,以$3\mathrm{cm}/\mathrm{s}$的速度沿$C \to B \to C \to ···$往复运动,$\therefore$点$Q$从点$C$出发到点$B$的时间为$12÷3 = 4(\mathrm{s})$.$\because AD// BC$,$\therefore PD// CQ$.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$t(\mathrm{s})$.
①当$0 ＜ t ≤ 4$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$\because ∠ AMB = ∠ PNQ = 90^{\circ}$,$AB = PQ$,$\therefore \mathrm{Rt}△ ABM ≌ \mathrm{Rt}△ PQN(\mathrm{HL})$,$\therefore BM = QN$.在$\mathrm{Rt}△ ABM$中,$∠ B = 60^{\circ}$,$AB = 6\mathrm{cm}$,$\therefore ∠ BAM = 90^{\circ} - ∠ B = 30^{\circ}$,$\therefore BM = \frac{1}{2}AB = 3\mathrm{cm}$,$\therefore BM = QN = 3\mathrm{cm}$,$\therefore t = 12 - 3t - 3 - 3$,$\therefore t = \frac{3}{2}$,此时$PQ = CD$.
②当$4 ＜ t ≤ 8$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$8 ＜ t ≤ 12$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，直线$l_{1}:y = -\frac{3}{4}x + b$分别与$x$轴、$y$轴交于$A$，$B$两点，与直线$l_{2}:y = kx - 6$交于点$C(2,\frac{3}{2})$。
(1) 点$A$的坐标为
(2) 在线段$BC$上有一点$E$，过点$E$作$y$轴的平行线交直线$l_{2}$于点$F$，设点$E$的横坐标为$m$，当四边形$OBEF$是平行四边形时，$m$的值为
(3) 若点$P$为$x$轴正半轴上一点，且$S_{△ ABP} = \frac{15}{2}$，则在$y$轴上是否存在一点$Q$，使得$P$，$Q$，$A$，$B$四个点能构成一个梯形？若存在，求出所有符合条件的$Q$点坐标；若不存在，请说明理由。
(1) 点$A$的坐标为
(4,0)
，点$B$的坐标为(0,3)
。(2) 在线段$BC$上有一点$E$，过点$E$作$y$轴的平行线交直线$l_{2}$于点$F$，设点$E$的横坐标为$m$，当四边形$OBEF$是平行四边形时，$m$的值为
$\frac{4}{3}$
。(3) 若点$P$为$x$轴正半轴上一点，且$S_{△ ABP} = \frac{15}{2}$，则在$y$轴上是否存在一点$Q$，使得$P$，$Q$，$A$，$B$四个点能构成一个梯形？若存在，求出所有符合条件的$Q$点坐标；若不存在，请说明理由。
答案：2. (1)$(4,0)$ $(0,3)$ 解析:$\because$点$C(2,\frac{3}{2})$在直线$y = -\frac{3}{4}x + b$上,代入得$\frac{3}{2} = -\frac{3}{2} + b$,解得$b = 3$,$\therefore$直线$l_{1}$的表达式为$y = -\frac{3}{4}x + 3$.当$y = 0$时,$x = 4$,$\therefore$点$A$的坐标为$(4,0)$;当$x = 0$时,$y = 3$,$\therefore$点$B$的坐标为$(0,3)$.
(2)$\frac{4}{3}$ 解析:$\because EF// y$轴,点$E$的横坐标为$m$,$\therefore$点$F$的横坐标也为$m$.$\because$点$E$在线段$BC$上,$\therefore$点$E$的坐标是$(m,-\frac{3}{4}m + 3)$.把$C(2,\frac{3}{2})$代入$l_{2}:y = kx - 6$,得$\frac{3}{2} = 2k - 6$,解得$k = \frac{15}{4}$,$\therefore l_{2}:y = \frac{15}{4}x - 6$.$\because$点$F$是直线$l_{2}:y = \frac{15}{4}x - 6$上的一点,$\therefore$点$F$的坐标是$(m,\frac{15}{4}m - 6)$.$\because$四边形$OBEF$是平行四边形,$\therefore BO = EF$.$\because B(0,3)$,$E(m,-\frac{3}{4}m + 3)$,$F(m,\frac{15}{4}m - 6)$,$\therefore BO = 3$,$\therefore EF = -\frac{3}{4}m + 3 - (\frac{15}{4}m - 6) = 3$,解得$m = \frac{4}{3}$.
(3)存在.$\because$点$P$为$x$轴正半轴上一点,$S_{△ ABP} = \frac{15}{2}$,$OB = 3$,$\therefore \frac{1}{2}AP· OB = \frac{15}{2}$,$\therefore AP = 5$.$\because A(4,0)$,$\therefore OA = 4$,$\therefore P$在点$A$右侧,$\therefore OP = OA + AP = 9$,$\therefore P(9,0)$.当$P$,$Q$,$A$,$B$四个点能构成一个梯形时,分两种情况:
①当$AB// PQ$时,$\because$直线$AB$的表达式为$y_{AB} = -\frac{3}{4}x + 3$,$\therefore$设直线$PQ$的表达式为$y_{PQ} = -\frac{3}{4}x + n$,把$P(9,0)$代入,得$0 = -\frac{3}{4}×9 + n$,$\therefore n = \frac{27}{4}$,$\therefore y_{PQ} = -\frac{3}{4}x + \frac{27}{4}$,$\therefore$当$x = 0$时,$y = \frac{27}{4}$,$\therefore$点$Q$的坐标是$(0,\frac{27}{4})$;②当$AQ// PB$时,设直线$PB$的表达式为$y = tx + 3$,把$P(9,0)$代入,得$0 = 9t + 3$,解得$t = -\frac{1}{3}$,$\therefore$直线$PB$的表达式为$y = -\frac{1}{3}x + 3$,同理可得直线$AQ$的表达式为$y_{AQ} = -\frac{1}{3}x + \frac{4}{3}$,$\therefore$当$x = 0$时,$y = \frac{4}{3}$.$\therefore$点$Q$的坐标是$(0,\frac{4}{3})$.综上,在$y$轴上存在点$Q$,使得$P$,$Q$,$A$,$B$四个点能构成一个梯形,符合条件的$Q$点坐标为$(0,\frac{27}{4})$,$(0,\frac{4}{3})$.
(2)$\frac{4}{3}$ 解析:$\because EF// y$轴,点$E$的横坐标为$m$,$\therefore$点$F$的横坐标也为$m$.$\because$点$E$在线段$BC$上,$\therefore$点$E$的坐标是$(m,-\frac{3}{4}m + 3)$.把$C(2,\frac{3}{2})$代入$l_{2}:y = kx - 6$,得$\frac{3}{2} = 2k - 6$,解得$k = \frac{15}{4}$,$\therefore l_{2}:y = \frac{15}{4}x - 6$.$\because$点$F$是直线$l_{2}:y = \frac{15}{4}x - 6$上的一点,$\therefore$点$F$的坐标是$(m,\frac{15}{4}m - 6)$.$\because$四边形$OBEF$是平行四边形,$\therefore BO = EF$.$\because B(0,3)$,$E(m,-\frac{3}{4}m + 3)$,$F(m,\frac{15}{4}m - 6)$,$\therefore BO = 3$,$\therefore EF = -\frac{3}{4}m + 3 - (\frac{15}{4}m - 6) = 3$,解得$m = \frac{4}{3}$.
(3)存在.$\because$点$P$为$x$轴正半轴上一点,$S_{△ ABP} = \frac{15}{2}$,$OB = 3$,$\therefore \frac{1}{2}AP· OB = \frac{15}{2}$,$\therefore AP = 5$.$\because A(4,0)$,$\therefore OA = 4$,$\therefore P$在点$A$右侧,$\therefore OP = OA + AP = 9$,$\therefore P(9,0)$.当$P$,$Q$,$A$,$B$四个点能构成一个梯形时,分两种情况:
①当$AB// PQ$时,$\because$直线$AB$的表达式为$y_{AB} = -\frac{3}{4}x + 3$,$\therefore$设直线$PQ$的表达式为$y_{PQ} = -\frac{3}{4}x + n$,把$P(9,0)$代入,得$0 = -\frac{3}{4}×9 + n$,$\therefore n = \frac{27}{4}$,$\therefore y_{PQ} = -\frac{3}{4}x + \frac{27}{4}$,$\therefore$当$x = 0$时,$y = \frac{27}{4}$,$\therefore$点$Q$的坐标是$(0,\frac{27}{4})$;②当$AQ// PB$时,设直线$PB$的表达式为$y = tx + 3$,把$P(9,0)$代入,得$0 = 9t + 3$,解得$t = -\frac{1}{3}$,$\therefore$直线$PB$的表达式为$y = -\frac{1}{3}x + 3$,同理可得直线$AQ$的表达式为$y_{AQ} = -\frac{1}{3}x + \frac{4}{3}$,$\therefore$当$x = 0$时,$y = \frac{4}{3}$.$\therefore$点$Q$的坐标是$(0,\frac{4}{3})$.综上,在$y$轴上存在点$Q$,使得$P$,$Q$,$A$,$B$四个点能构成一个梯形,符合条件的$Q$点坐标为$(0,\frac{27}{4})$,$(0,\frac{4}{3})$.
3. 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) 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。
答案：
3. (1)由折叠可得$AF = DF$.$\because$点$B(8,6)$,点$D(0,4)$,四边形$OABC$为矩形,$\therefore OA⊥ AB$,$OC = AB = 6$,$OA = BC = 8$,$OD = 4$.设$AF = DF = x$,则$OF = OA - AF = 8 - x$.在$\mathrm{Rt}△ ODF$中,由勾股定理可得$OD^{2} + OF^{2} = DF^{2}$,即$4^{2} + (8 - x)^{2} = x^{2}$,解得$x = 5$,$\therefore OF = 8 - x = 3$,$\therefore$点$F$的坐标为$(3,0)$.
(2)①当点$P$在点$F$右侧时,根据题意得,$AQ = 2t$,$AP = t(0 ＜ t ＜ 5)$,$\therefore FP = AF - AP = 5 - t$,$\therefore S = \frac{1}{2}FP· AQ = \frac{1}{2}(5 - t)× 2t = -t^{2} + 5t$;②当点$P$在点$F$左侧时,根据题意,得$AQ = 2t$,$AP = t(5 ＜ t ≤ 8)$,$\therefore FP = AP - AF = t - 5$,$\therefore S = \frac{1}{2}FP· AQ = \frac{1}{2}(t - 5)× 2t = t^{2} - 5t$.综上所述,$S = \begin{cases} -t^{2} + 5t(0 ＜ t ＜ 5), \\ t^{2} - 5t(5 ＜ t ≤ 8). \end{cases}$
(3)存在.若以$P$,$Q$,$R$,$M$为顶点的四边形是正方形,则以$P$,$R$,$Q$为顶点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,$\therefore PQ = QR$,$∠ PQR = 90^{\circ}$,$\therefore ∠ PQA + ∠ BQR = ∠ BQR + ∠ QRB$,$\therefore ∠ PQA = ∠ QRB$.在$△ PAQ$和$△ QBR$中,$\begin{cases} ∠ PAQ = ∠ QBR = 90^{\circ}, \\ ∠ PQA = ∠ QRB, \\ PQ = QR, \end{cases}$ $\therefore △ PAQ ≌ △ QBR(\mathrm{AAS})$,$\therefore AQ = BR = 2t$,$BQ = AP = t$,$\therefore AB = AQ + BQ = 2t + t = 3t = 6$,$\therefore BQ = AP = t = 2$,$\therefore BR = AQ = 2× 2 = 4$,$\therefore Q(8,4)$.$\therefore OP = OA - AP = 8 - 2 = 6$,$\therefore P(6,0)$.$\because CR = BC - BR = 4$,$\therefore R(4,6)$.$\because$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$RK⊥ OA$于点$K$,则四边形$OCRK$,$RKAB$均为矩形,$\therefore RK = AB = 6$,$∠ BRK = ∠ RKA = 90^{\circ}$.$\because$四边形$PRQM$是正方形,$\therefore PR = QR$,$∠ PRQ = 90^{\circ}$,$\therefore ∠ KRB - ∠ PRB = ∠ PRQ - ∠ PRB$,$\therefore ∠ KRP = ∠ BRQ$.在$△ PKR$和$△ QBR$中,$\begin{cases} ∠ RKP = ∠ RBQ = 90^{\circ}, \\ ∠ KRP = ∠ BRQ, \\ PR = QR, \end{cases}$ $\therefore △ PKR ≌ △ QBR(\mathrm{AAS})$,$\therefore RB = RK = 6$,$KP = BQ = AQ - AB = 2t - 6$,$\therefore OK = CR = BC - BR = 8 - 6 = 2$,$\therefore R(2,6)$,$AK = OA - OK = 6 = 2t - 6 + t$,$\therefore AP = t = 4$,$\therefore P(8 - 4,0)$,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,$\therefore PQ = QR$,$∠ PQR = 90^{\circ}$,$\therefore ∠ PQA + ∠ BQR = ∠ BQR + ∠ QRB$,$\therefore ∠ PQA = ∠ QRB$.在$△ PAQ$和$△ QBR$中,$\begin{cases} ∠ PAQ = ∠ QBR = 90^{\circ}, \\ ∠ PQA = ∠ QRB, \\ PQ = QR, \end{cases}$ $\therefore △ PAQ ≌ △ QBR(\mathrm{AAS})$,$\therefore AQ = BR = 2t$,$BQ = AP = t$.又$\because AQ = AB + BQ = 6 + t = 2t$,$\therefore AP = t = 6$,$\therefore BR = 2× 6 = 12$,$\therefore Q(8,12)$,$P(8 - 6,0)$,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.
3. (1)由折叠可得$AF = DF$.$\because$点$B(8,6)$,点$D(0,4)$,四边形$OABC$为矩形,$\therefore OA⊥ AB$,$OC = AB = 6$,$OA = BC = 8$,$OD = 4$.设$AF = DF = x$,则$OF = OA - AF = 8 - x$.在$\mathrm{Rt}△ ODF$中,由勾股定理可得$OD^{2} + OF^{2} = DF^{2}$,即$4^{2} + (8 - x)^{2} = x^{2}$,解得$x = 5$,$\therefore OF = 8 - x = 3$,$\therefore$点$F$的坐标为$(3,0)$.
(2)①当点$P$在点$F$右侧时,根据题意得,$AQ = 2t$,$AP = t(0 ＜ t ＜ 5)$,$\therefore FP = AF - AP = 5 - t$,$\therefore S = \frac{1}{2}FP· AQ = \frac{1}{2}(5 - t)× 2t = -t^{2} + 5t$;②当点$P$在点$F$左侧时,根据题意,得$AQ = 2t$,$AP = t(5 ＜ t ≤ 8)$,$\therefore FP = AP - AF = t - 5$,$\therefore S = \frac{1}{2}FP· AQ = \frac{1}{2}(t - 5)× 2t = t^{2} - 5t$.综上所述,$S = \begin{cases} -t^{2} + 5t(0 ＜ t ＜ 5), \\ t^{2} - 5t(5 ＜ t ≤ 8). \end{cases}$
(3)存在.若以$P$,$Q$,$R$,$M$为顶点的四边形是正方形,则以$P$,$R$,$Q$为顶点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,$\therefore PQ = QR$,$∠ PQR = 90^{\circ}$,$\therefore ∠ PQA + ∠ BQR = ∠ BQR + ∠ QRB$,$\therefore ∠ PQA = ∠ QRB$.在$△ PAQ$和$△ QBR$中,$\begin{cases} ∠ PAQ = ∠ QBR = 90^{\circ}, \\ ∠ PQA = ∠ QRB, \\ PQ = QR, \end{cases}$ $\therefore △ PAQ ≌ △ QBR(\mathrm{AAS})$,$\therefore AQ = BR = 2t$,$BQ = AP = t$,$\therefore AB = AQ + BQ = 2t + t = 3t = 6$,$\therefore BQ = AP = t = 2$,$\therefore BR = AQ = 2× 2 = 4$,$\therefore Q(8,4)$.$\therefore OP = OA - AP = 8 - 2 = 6$,$\therefore P(6,0)$.$\because CR = BC - BR = 4$,$\therefore R(4,6)$.$\because$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$RK⊥ OA$于点$K$,则四边形$OCRK$,$RKAB$均为矩形,$\therefore RK = AB = 6$,$∠ BRK = ∠ RKA = 90^{\circ}$.$\because$四边形$PRQM$是正方形,$\therefore PR = QR$,$∠ PRQ = 90^{\circ}$,$\therefore ∠ KRB - ∠ PRB = ∠ PRQ - ∠ PRB$,$\therefore ∠ KRP = ∠ BRQ$.在$△ PKR$和$△ QBR$中,$\begin{cases} ∠ RKP = ∠ RBQ = 90^{\circ}, \\ ∠ KRP = ∠ BRQ, \\ PR = QR, \end{cases}$ $\therefore △ PKR ≌ △ QBR(\mathrm{AAS})$,$\therefore RB = RK = 6$,$KP = BQ = AQ - AB = 2t - 6$,$\therefore OK = CR = BC - BR = 8 - 6 = 2$,$\therefore R(2,6)$,$AK = OA - OK = 6 = 2t - 6 + t$,$\therefore AP = t = 4$,$\therefore P(8 - 4,0)$,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,$\therefore PQ = QR$,$∠ PQR = 90^{\circ}$,$\therefore ∠ PQA + ∠ BQR = ∠ BQR + ∠ QRB$,$\therefore ∠ PQA = ∠ QRB$.在$△ PAQ$和$△ QBR$中,$\begin{cases} ∠ PAQ = ∠ QBR = 90^{\circ}, \\ ∠ PQA = ∠ QRB, \\ PQ = QR, \end{cases}$ $\therefore △ PAQ ≌ △ QBR(\mathrm{AAS})$,$\therefore AQ = BR = 2t$,$BQ = AP = t$.又$\because AQ = AB + BQ = 6 + t = 2t$,$\therefore AP = t = 6$,$\therefore BR = 2× 6 = 12$,$\therefore Q(8,12)$,$P(8 - 6,0)$,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.