If the straight line passes through point A (3,0), then B = 3/2.

If the straight line passes through point b (3, 1), then B = 5/2.

If the straight line passes through point C (0, 1), then B = 1.

(1) If the intersection of the straight line and the polyline OAB is on OA, that is, 1 < b ≤ 3/2.

At this time E(2b, 0)

∴S= 1/2OE？ CO= 1/2 ×2b× 1=b

(2) If the intersection of the straight line and the broken line OAB is on BA, that is, 3/2 < b < 5/2,

At this time, E(3, b-3/2) and d (2b-2, 1).

∴S=S = s moment -(s △ OCD+s △ OAE+s △ DBE)

= 3-[ 1/2(2 B- 1)× 1+ 1/2×(5-2b)？ (5/2-b)+1/2× 3 (b-3/2)] = the square of 5/2 b-b.

(2) Let O 1A 1 intersect with CB at point M and OA intersect with C 1B 1 at point N, then the overlapping area of rectangle OA 1B 1 and rectangle OABC is the area of quadrilateral DNEM.

According to the meaning of the question, dm∨ne, dnem and ∴ quadrilateral DNEM are parallelograms.

According to the axial symmetry, ∠ med = ∠ ned.

And ∠ MDE = ∠ Ned, ∴ Med = ∠ MDE, ∴ MD = Me, ∴ parallelogram DNEM is a diamond.

The intersection d is DH⊥OA, and the vertical foot is H.

According to the title, Tan ∠ den = 1/2, DH = 1, He = 2.

Let the side length of rhombic DNEM be a,

Then in Rt△DHM, we know the square of A = the square of (2-a)+/the square of kloc-0/from Pythagorean theorem, and ∴a=5/4.

∴S quadrilateral dnem = ne? DH= 5/4

∴ The overlapping area of rectangular OA1b1and rectangular OABC remains unchanged, and the area is always 5/4.