coordinate geometry

設 D 為原點, 即 (0, 0), 然後:
A 為 (0, a)
B 為 (a, a)
C 為 (a, 0)
如下圖:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug08/Crazycoord3.jpg

則, AC 之直線方程為: x + y = a
設 P 之座標為 (h, k), 則:
PA : PB : PC = 1 : 5 : 7
PA2 : PB2 : PC2 = 1 : 25 : 49
PA2 = m
PB2 = 25m
PC2 = 49m
其中 m 為某常數. 如此:
h2 + (k - a)2 = m ... (1)
(h - a)2 + (k - a)2 = 25m ... (2)
(h - a)2 + k2 = 49m ... (3)
(1) + (3):
h2 + (k - a)2 + (h - a)2 + k2 = 50m
h2 + 25m + k2 = 50m
h2 + k2 = 25m
h2 + k2 = (h - a)2 + (k - a)2
0 = -2ah - 2ak + 2a2
2a(h + k) = 2a2
h + k = a
所以, P 符合 AC 之方程, 即它在 AC 直線上.
所以 A, P, C三點共線。

Let angle ABP = x. And side of square = n.
For triangle ABP and using cosine rule, we get
1^2 = 5^2 + n^2 - 10ncosx, that is 10ncosx = n^2 + 24..............(1)
For triangle BPC and using cosine rule, we get
7^2 = 5^2 + n^2 - 10x cos(90 - x) = 25 + n^2 - 10n sinx, that is
10nsinx = n^2 - 24 .................(2)
square of (1) + square of (2), we get
100n^2 = (n^2 + 24)^2 + (n^2 - 24)^2 =( n^4 + 576 + 48n^2) + (n^4 + 576 - 48n^2) = 2n^4 + 1152
n^4 - 50n^2 + 576 = 0
(n^2 - 32)(n^2 - 18) = 0
So n^2 = 32 or n^2 = 18.
For n^2 = 18 and applying Pythagoras theorem,
AC^2 = 18 + 18 = 36, AC = 6 (reject).
For n^2 = 32, AC^2 = 32 + 32 = 64, therefore AC = 8 = AP + PC.
Therefore, APC is collinear.

2008-08-07 18:25:45 補充：
1. AC = 6 is the solution for point P outside the square but still collinear.
2. Correction: Line 5 should read as .... 10n cos(90-x).....