Problem:
Problem C: Find the Point We understand that reading English is a great pain to many of you. So we’ll keep this problem statememt
simple. Write a program that reports the point equally distant from a set of lines given as the input. In
case of no solutions or multiple solutions, your program should report as such. Input The input consists of multiple datasets. Each dataset is given in the following format:  n  x 1,1  y 1,1  x 1,2  y 1,2  x 2,1  y 2,1  x 2,2  y 2,2 ...  x n ,1  y n ,1  x n ,2  y n ,2   n is the number of lines (1 ≤ n ≤ 100); ( x i ,1 , y i ,1 ) and ( x i ,2 , y i ,2 ) denote the different points the i -th line passes through. The lines do not coincide each other. The coordinates are all integers between -10000
and 10000. The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed. Output For each dataset, print a line as follows. If there is exactly one point equally distant from all the given lines, print the x - and y -coordinates in this order with a single space between them. If there is more than one such point,
just print " Many " (without quotes). If there is none, just print " None " (without quotes). The coordinates may be printed with any number of digits after the decimal point, but should be accurate
to 10 -4 . Sample Input 2
-35 -35 100 100
-49 49 2000 -2000
4
0 0 0 3
0 0 3 0
0 3 3 3
3 0 3 3
4
0 3 -4 6
3 0 6 -4
2 3 6 6
-1 2 -4 6
0 Output for the Sample Input Many
1.5000 1.5000
1.000 1.000

Solution (Python):
import sys
readline = sys.stdin.readline
write = sys.stdout.write

EPS = 1e-9
def line_cross_point(P1, P2, Q1, Q2):
    x0, y0 = P1; x1, y1 = P2
    x2, y2 = Q1; x3, y3 = Q2

    dx0 = x1 - x0; dy0 = y1 - y0
    dx1 = x3 - x2; dy1 = y3 - y2

    s = (y0-y2)*dx1 - (x0-x2)*dy1
    sm = dx0*dy1 - dy0*dx1
    if -EPS < sm < EPS:
        return None
    return x0 + s*dx0/sm, y0 + s*dy0/sm

def bisector(P1, P2, Q1, Q2):
    x0, y0 = P1; x1, y1 = P2
    x2, y2 = Q1; x3, y3 = Q2

    dx0 = x1 - x0; dy0 = y1 - y0
    dx1 = x3 - x2; dy1 = y3 - y2

    cp = line_cross_point(P1, P2, Q1, Q2)
    if cp is None:
        return None

    cx, cy = cp

    d0 = (dx0**2 + dy0**2)**.5
    d1 = (dx1**2 + dy1**2)**.5
    return [
        ((cx, cy), (cx + (dx0*d1 + dx1*d0), cy + (dy0*d1 + dy1*d0))),
        ((cx, cy), (cx + (dx0*d1 - dx1*d0), cy + (dy0*d1 - dy1*d0))),
    ]

def line_point_dist2(p1, p2, q):
    x, y = q

    x1, y1 = p1; x2, y2 = p2
    dx = x2 - x1; dy = y2 - y1
    dd = dx**2 + dy**2
    sv = (x - x1) * dy - (y - y1) * dx
    return abs(sv / dd**.5)


def check(LS, q):
    ds = [line_point_dist2(p1, p2, q) for p1, p2 in LS]
    return all(abs(ds[0] - e) < EPS for e in ds)

def solve():
    N = int(readline())
    if N == 0:
        return False
    P = []
    for i in range(N):
        x1, y1, x2, y2 = map(int, readline().split())
        P.append(((x1, y1), (x2, y2)))
    if N <= 2:
        write("Many\n")
        return True

    s = []
    for i in range(N):
        p1, p2 = P[i]
        for j in range(i):
            q1, q2 = P[j]
            bs = bisector(p1, p2, q1, q2)
            if bs is None:
                continue
            s.append(bs)
            if len(s) > 1:
                break
        else:
            continue
        break
    if len(s) < 2:
        write("None\n")
        return True
    ans = []
    b1, b2 = s
    for p1, p2 in b1:
        for q1, q2 in b2:
            cp = line_cross_point(p1, p2, q1, q2)
            if cp is None:
                continue
            if check(P, cp):
                cx, cy = cp
                for ax, ay in ans:
                    if abs(cx - ax) < EPS and abs(cy - ay) < EPS:
                        break
                else:
                    ans.append(cp)
    if len(ans) == 0:
        write("None\n")
    elif len(ans) > 1:
        write("Many\n")
    else:
        write("%.16f %.16f\n" % ans[0])
    return True
while solve():
    ...