Maths

From the given, we have 0 < m < 1, then we can impose a coordinate system to the situation as follows:


圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Dec09/Crazygeom1.jpg


With reference to the diagram, we have:

PB/BX = PA/AD

PB/(1 - m) = PB + 1

PB = PB - mPB + 1 - m

mPB = 1 - m

PB = (1 - m)/m

PA = PB + 1

= 1/m

Therefore P is at (0, 1/m)

Similarly, we have Q is at (1/m, 0)

Then:

RC/CX = RD/DA

RC/m = RC + 1

RC = mRC + m

RC = m/(1 - m)

RD = RC + 1 = 1/(1 - m)

Hence, R is at (1, 1/(1 - m)) and similarly, S is at (1/(1 - m), 1)

Now, slope of PQ is -1

Since P, Q, R and S are collinear, we have:

Slope of PR = -1

[1/(1 - m) - 1/m]/(1 - 0) = -1

1/(1 - m) - 1/m = -1

m - (1 - m) = -m(1 - m)

2m - 1 = -m2 + m

m2 + m - 1 = 0

m = (-1 + √5)/2 or (-1 - √5)/2 (rejected)

2009-12-08 14:48:06 補充：
Diagram:

http://i388.photobucket.com/albums/oo325/loyitak1990/Dec09/Crazygeom1.jpg