Question

如图，在平面直角坐标系中，点O为坐标原点，直线y=x+4交x轴如图，在平面直角坐标系中，点O为坐标原点，直

如图，在平面直角坐标系中，点O为坐标原点，直线y=x+4交x轴于点A，交y轴于点B，四边形ABCO是平行四边形，直线y=-x+m经过点C，交x轴于点D。
（2）动点M以每秒1个单位长的速度，从点O沿线段OD向点D运动，同时点P以相同的速度，从点C沿折线C-B-O-向点O运动，当点M到达点D时，两点同时停止运动。过点M做直线MQ//OB,与折线O-C-D的交点为Q，点M的运动时间为t。
1、求当四边形PQMO为矩形时的t值。
2、当M在线段OD上运动时，是否可以使得以C、P、Q为顶点的等腰三角形？请求出t值。

Answer 1

(1)
A(-4, 0), B(0, 4), BC = 4, C(4, 4)

y = - x + m过C, 4 = -4 + m, m = 8
y = -x + 8, y = 0, x = 8, D(8, 0)
折线C-B-O = 4 + 4 = 8, OD = 8, 当点M到达点D时, 点P同时到达点O

t秒时: OM = t, M(t, 0), 0 ≤ t ≤ 8

0 ≤ t ≤ 4: PC = t, P(4 - t, 4)
4 < t ≤ 8: 折线C-B-P = t, PB = t - 4, OB = 4 - (t - 4) = 8 - t
P(0, 8 - t)
OC的方程为y = x, 0 ≤ t ≤ 4: Q(t, t)
4 < t ≤ 8: Q(t, 8 - t)

(i) 容易看出，t = 4时，M(4, 0), Q(4, 4), P(0, 4)， 此时OMQP为正方形（矩形的特例）。

(ii) t <4时，PQ的斜率k = (t - 4)/(4 - 4 + t) = (t - 4)/t
k = -1时, t = 2, M(2, 0), Q(2, 2), P(2, 4), P, M, Q共线，不可能是矩形

(iii) 4<t<8:
M(t, 0), Q(t, 8 - t), P(0, 8 - t)
OMQP总是矩形(除了t = 8)

答案4 ≤ t < 8

(2)
(i) t = 4时，C, Q重合，三角形不存在

(ii) 0 <t< 4:
Q(t, t), P(4 - t, 4), C(4, 4)

(a) CP = CQ
[4 - (4-t)]² = (4 - t)² + (4 - t)²
t = 4(2 - √2) (另一根>4, 舍去)
(b) PC = PQ
[4 - (4-t)]² = (4 - t - t)² + (4 - t)²
t = 2 (舍去t = 4)
(c) QP = QC

(4 - t - t)² + (4 - t)² = (4 - t)² + (4 - t)²
t = 8/3 (舍去t = 0, 此时P, C重合)

(iii) 4<t ≤ 8
C(4, 4), P(0, 8 - t), Q(t, 8 - t)
(a) CP = CQ
(4 - 0)² + [4 - (8-t)]² = (4 - t)² + (4 - 8 + t)²
t = 8 (舍去t = 0)
(b) PC = PQ
(4 - 0)² + [4 - (8-t)]² = (t - 0)² + (8 - t - 8 +t)²
t = 4舍去
(c) QP = QC

(t - 0)² + (8 - t - 8 +t)² = (4 - t)² + (4 - 8 + t)²
t = 4(2 ±√2)
二者均不在(4, 8]内，舍去

Answer 2

1)
A(-4, 0), B(0, 4), BC = 4, C(4, 4)

y = - x + m过C, 4 = -4 + m, m = 8
y = -x + 8, y = 0, x = 8, D(8, 0)
折线C-B-O = 4 + 4 = 8, OD = 8, 当点M到达点D时, 点P同时到达点O

t秒时: OM = t, M(t, 0), 0 ≤ t ≤ 8

0 ≤ t ≤ 4: PC = t, P(4 - t, 4)
4 < t ≤ 8: 折线C-B-P = t, PB = t - 4, OB = 4 - (t - 4) = 8 - t
P(0, 8 - t)
OC的方程为y = x, 0 ≤ t ≤ 4: Q(t, t)
4 < t ≤ 8: Q(t, 8 - t)

(i) 容易看出，t = 4时，M(4, 0), Q(4, 4), P(0, 4)， 此时OMQP为正方形（矩形的特例）。

(ii) t <4时，PQ的斜率k = (t - 4)/(4 - 4 + t) = (t - 4)/t
k = -1时, t = 2, M(2, 0), Q(2, 2), P(2, 4), P, M, Q共线，不可能是矩形

(iii) 4<t<8:
M(t, 0), Q(t, 8 - t), P(0, 8 - t)
OMQP总是矩形(除了t = 8)

答案4 ≤ t < 8

(2)
(i) t = 4时，C, Q重合，三角形不存在

(ii) 0 <t< 4:
Q(t, t), P(4 - t, 4), C(4, 4)

(a) CP = CQ
[4 - (4-t)]² = (4 - t)² + (4 - t)²
t = 4(2 - √2) (另一根>4, 舍去)
(b) PC = PQ
[4 - (4-t)]² = (4 - t - t)² + (4 - t)²
t = 2 (舍去t = 4)
(c) QP = QC

(4 - t - t)² + (4 - t)² = (4 - t)² + (4 - t)²
t = 8/3 (舍去t = 0, 此时P, C重合)

(iii) 4<t ≤ 8
C(4, 4), P(0, 8 - t), Q(t, 8 - t)
(a) CP = CQ
(4 - 0)² + [4 - (8-t)]² = (4 - t)² + (4 - 8 + t)²
t = 8 (舍去t = 0)
(b) PC = PQ
(4 - 0)² + [4 - (8-t)]² = (t - 0)² + (8 - t - 8 +t)²
t = 4舍去
(c) QP = QC

(t - 0)² + (8 - t - 8 +t)² = (4 - t)² + (4 - 8 + t)²
t = 4(2 ±√2)
二者均不在(4, 8]内，舍去
1)
A(-4, 0), B(0, 4), BC = 4, C(4, 4)

y = - x + m过C, 4 = -4 + m, m = 8
y = -x + 8, y = 0, x = 8, D(8, 0)
折线C-B-O = 4 + 4 = 8, OD = 8, 当点M到达点D时, 点P同时到达点O

t秒时: OM = t, M(t, 0), 0 ≤ t ≤ 8

0 ≤ t ≤ 4: PC = t, P(4 - t, 4)
4 < t ≤ 8: 折线C-B-P = t, PB = t - 4, OB = 4 - (t - 4) = 8 - t
P(0, 8 - t)
OC的方程为y = x, 0 ≤ t ≤ 4: Q(t, t)
4 < t ≤ 8: Q(t, 8 - t)

(i) 容易看出，t = 4时，M(4, 0), Q(4, 4), P(0, 4)， 此时OMQP为正方形（矩形的特例）。

(ii) t <4时，PQ的斜率k = (t - 4)/(4 - 4 + t) = (t - 4)/t
k = -1时, t = 2, M(2, 0), Q(2, 2), P(2, 4), P, M, Q共线，不可能是矩形

(iii) 4<t<8:
M(t, 0), Q(t, 8 - t), P(0, 8 - t)
OMQP总是矩形(除了t = 8)

答案4 ≤ t < 8

(2)
(i) t = 4时，C, Q重合，三角形不存在

(ii) 0 <t< 4:
Q(t, t), P(4 - t, 4), C(4, 4)

(a) CP = CQ
[4 - (4-t)]² = (4 - t)² + (4 - t)²
t = 4(2 - √2) (另一根>4, 舍去)
(b) PC = PQ
[4 - (4-t)]² = (4 - t - t)² + (4 - t)²
t = 2 (舍去t = 4)
(c) QP = QC

(4 - t - t)² + (4 - t)² = (4 - t)² + (4 - t)²
t = 8/3 (舍去t = 0, 此时P, C重合)

(iii) 4<t ≤ 8
C(4, 4), P(0, 8 - t), Q(t, 8 - t)
(a) CP = CQ
(4 - 0)² + [4 - (8-t)]² = (4 - t)² + (4 - 8 + t)²
t = 8 (舍去t = 0)
(b) PC = PQ
(4 - 0)² + [4 - (8-t)]² = (t - 0)² + (8 - t - 8 +t)²
t = 4舍去
(c) QP = QC

(t - 0)² + (8 - t - 8 +t)² = (4 - t)² + (4 - 8 + t)²
t = 4(2 ±√2)
二者均不在(4, 8]内，舍去