Crossing points A and B are perpendicular to the X axis and intersect with C and D on the X axis, so AC=2BD=4CD=2.

Area of quadrilateral ACDB = (2

4)* 2 * 1/2 = 6 & lt； 10

So point p can't be between CDs, only to the right of point D.

Let's set the coordinates of point P as (x, 0).

So the area of triangle ABP = the area of quadrilateral ACDB.

Area of triangle BDP- Area of triangle ACP Area of quadrilateral ACDB = 6 Area of triangle BDP = DP * DB *1/2 = (x-3) * 4 *1/2 = (x-3) * 2.

The area of the triangle ACP = AC * CP *1/2 = 2 * (x-1) *1/2 = (x-1).

So there are six.

The solution of (x-3) * 2-(x- 1) = 10 is x = 9.

So the coordinate of point P is (9,0).

Connect A 1C 1.AC, because AC is parallel to A 1C 1, so AC is parallel to the plane A 1C 1. Then A 1C 1D 1 intersecting APC=MN.