VP=2, vD=2 root number 2

AP=2t, 0D=(2 root numbers 2)t

∫∠AOB = 90, OA=OB=5 root number 2.

∴△OAB is an isosceles right triangle, ∠ A =∠ B = 45, AB = OA/COS 45 = 5 roots 2/[ (root 2)/2]= 10.

Be = bdcosb = (ob-od) cos45 = [5 radical 2-(2 radical 2)t]* (radical 2)/2=5-2t.

PE = a b-AP-BE = 10-2t-(5-2t)= 5

(2)

The heights of POD and δPDB are equal, which are both equal to the Y coordinate of point P. According to the area of triangle, it is equal to the base multiplied by the height and divided by two. When OD=DB, sδPOD = sδPDB.

0D=(2 root numbers 2)t

DB=OB-OD=5 root number 2-(2 root number 2)t

OD=DB

(2 2)t=5 2-(2 2)t.

T=(5 roots 2)/(4 roots 2)= 1.25

(3)

Regardless of the time conditions, PE is always equal to 5, while DE⊥PE.

When DE=PE=5, make parallel lines of PE and ED. Compared with point m, MDEP is a square.

De = dbsinb = (ob-od) sin45 = [5 roots 2-(2 roots 2)t]* (roots 2)/2=5-2t=5.

The solution is t=0.

That is to say, when P and D have not started to move, DE and PE can be equal, but not at any other time.

Therefore, when t= 15/4, there is no point m on the plane, which constitutes a quadrilateral with m, e, p and d as vertices.