#include <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN = 40000;

typedef struct L { int a, b, c; } L;
bool operator<(const L &p, const L &q) { return -p.a *q.b < -q.a *p.b; }

int n;
L lines[MAXN];

// if slope[i]<slope[j] and y[i]<=y[j] they have already crossed

pair<double, int> lst[MAXN];
int bit[MAXN + 1];
int bget(int idx) { ++idx; int ret = 0;  while (idx <= n) { ret += bit[idx]; idx += idx&-idx; } return ret; }
void bmod(int idx) { ++idx; while (idx > 0) { bit[idx]++; idx -= idx&-idx; } }

int calc(double x) {
	REP(i, n) lst[i] = MP((lines[i].c - x*lines[i].a) / lines[i].b, i);
	sort(lst, lst + n);
	FORE(i, 1, n) bit[i] = 0;
	int ret = 0;
	for (int i = n - 1; i >= 0; --i) {
		ret += bget(lst[i].second);
		bmod(lst[i].second);
	}
	//printf("%lf=%d\n", x, ret);
	return ret;
}

double solve() {
	REP(i, n) if (lines[i].b < 0) lines[i].a = -lines[i].a, lines[i].b = -lines[i].b, lines[i].c = -lines[i].c;
	sort(lines, lines + n);
	//REP(i, n) printf("%d*x+%d*y=%d\n", lines[i].a, lines[i].b, lines[i].c);
	int ntot = n*(n - 1) / 2, nlft = (ntot + 1) / 2;
	double lx = -1; while (calc(lx) >= nlft) lx *= 2;
	double hx = +1; while (calc(hx) < nlft) hx *= 2;
	REP(rep, 100) {
		double mx = (lx + hx) / 2;
		if (calc(mx) < nlft) lx = mx; else hx = mx;
	}
	return (lx + hx) / 2;
}

void run() {
	scanf("%d", &n);
	REP(i, n) scanf("%d%d%d", &lines[i].a, &lines[i].b, &lines[i].c);
	double x = solve();
	REP(i, n) swap(lines[i].a, lines[i].b);
	double y = solve();
	printf("%.15lf %.15lf\n", x, y);
}

int main() {
	run();
	return 0;
}
