/*****  rational neighbor problem
input 
int a, b in [1,100000]
float n in [.00000001, .1]

output
int c, d
such that 0 < c/d - a/b <= n

and d is the minimal denominator with this property.

******/
/*
Thoughts:
iff 0 < cb - ad <= nbd 
implies bd >= 1/n so that nbd >= 1
range of 1/n is [10, 10^8]

continued fraction approximation probably gets it.

brute force search may work fast enough (3 sec)

for d = 1 to 100000
  find least c such that a/b < c/d, 
  bail (done) if 0 < c/d - a/b <= n
return c,d

translation of find:
ad < bc.  c = ceil(ad/b);
translation of condition: c <= nd

possible speedup: binary search for d.
*/

#include <iostream>
#include <math.h>
using namespace std;
void ratnbr(int a, int b, float n, int& c, int& d)
{
	double a_b = a;  a_b /= b;
	double delta;
	d = 0;
	do {
		++d;
		double x;
		x = a*d; x /= b;
		c = ceil(x);
		double c_d = c;  c_d /= d;
		delta = c_d - a_b;
		if (delta == 0) { 
			++c;
			c_d = c; c_d /= d;
			delta = c_d - a_b;
		}
	}
	while (delta > n); 
	return;
}
/* stub version used for initial test of main()
void ratnbr(int a, int b, float n, int& c, int& d)
{ c = a; d = b; return; }
*/

int main ()
{
	int a, b, c, d;
	float n;
	while (cin >> a >> b >> n)
	{
		ratnbr(a, b, n, c, d);
		cout << c << " " << d << endl;
   	}
}