重心、垂心、外心及內心(20分)

(1) Since AC is perpendicular to BC, so orthocenter is C(4,0).
(2) The perpendicular bisector of AC is line y = 3/2. The perpendicular bisector of BC is line x = 2, the intersecting point of the perpendicular bisectors will be the circumcenter, that is (2, 3/2).
(3) Let M be the mid- point of BC, so M is (2,0). AM is a median of the triangle. Centroid divides the median in the ratio 1 : 2. Let the centroid be ( h, k ), by point of division:
h = (2 x 2 + 4 x 1)/(1 + 2) = 8/3
k = (0 x 2 + 3 x 1)/(1 + 2) = 1.
so centroid is (8/3, 1).
(4)
Let radius of the inscribed circle be r.
By considering the perimeter of triangle ABC ( AC = 3, BC = 4 and AB = 5), we get the equation 2 + r = 4 - r
2r = 2
r = 1
so incenter is (4 - 1, 1), that is (3,1).
Q2.
T(n) = 2n + 3
n = 1, T(1) = 1st term = 5.
n = 2, T(2) = 2nd term = 7, so common difference = 2.
Sum of an A.S. to m terms = m[5 + 2(m - 1)]/2 which is smaller than 3000
that is
m[5 + 2(m - 1)]/2 < 3000
5m + 2m^2 - 2m < 6000
2m^2 + 3m - 6000 < 0
- 55.5 < m < 54.03
so max. value of m = 54.

let k be the maximun value of m.

2(1+2+3+...+k) + 3k < 3000

k(k+1) +3k < 3000

k^2 +4k - 3000 < 0

-52.808758 < k < 52.808758

so the maximum value of m = 52