maths question

Assume this is a Maths question, not A. Maths question.
i)ABO is a right- angled isos. triangle.
Let mid- point of AB be M, so OM is perpendicular to AB and OM = AM = BM.
Let radius of in - circle be r
OM = r + sqrt ( r^2 + r^2) = r + r sqrt 2 = (1 + sqrt 2)r.
OB^2 = BM^2 + OM^2
6^2 = OM^2 + OM^2 = 2OM^2
so OM^2 = 18, so OM = sqrt 18 = 3 sqrt 2
so 3 sqrt 2 = (1 + sqrt 2)r
r = 3 sqrt 2/(1 + sqrt 2). Co-ordinates of in - center is (r, r).
(ii)
The vertical from B to OA is (0,0)
The vertical from A to OB is also (0,0).
so the orthocentre is (0,0).
(iii) Can you give the name of center in English? Is it circum- center?
(iv)
Since M is the mid-point of AB, so centroid must be on OM.
Equation of OM is y = x..................(1)
Let N be mid-point of OA, so N is ( 3,0).
Equation of BN is
(y - 6)/(0 - 6) = (x - 0)/(3 - 0)
(y - 6)/-6 = x/3
y - 6 = -2x................(2)
From (1) and (2)
x - 6 = -2x
3x = 6
x = 2, so centroid is (2,2).







2009-08-08 10:50:00 補充：
Circum-center is on the perpendicular bisector of OA and OB, that is the intersecting point of x = 3 and y = 3, so center is (3,3).