#include<stdio.h>#include<math.h>#include<stdlib.h>struct P{ double x,y;}p[101],stack[101];/*double getradiusby3point(double x1,double y1,double x2,double y2,double x3,double y3){double x,y;double a11,a12,a21,a22,b1,b2;double d,d1,d2;a11=2*(x3-x2);a12=2*(y3-y2);a21=2*(x2-x1);a22=2*(y2-y1);b1=x3*x3-x2*x2+y3*y3-y2*y2;b2=x2*x2-x1*x1+y2*y2-y1*y1;d=a11*a22-a12*a21;d1=b1*a22-a12*b2;d2=a11*b2-b1*a21;x=d1/d;y=d2/d;return (x1-x)*(x1-x)+(y1-y)*(y1-y);}*/double Mul(P p1,P p2,P p3) //叉乘 { return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x); }double dis(P a,P b){ return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}int cmp(const void *a,const void *b){ P * c = (P *)a; P * d = (P *)b; double k = Mul(p[0],*c,*d); if(k<0 || (!k && dis(*c,p[0]) > dis(*d,p[0]) ) ) return 1; return -1;}void tubao(int n,int &top){ int i; top = 2; stack[0] = p[0]; stack[1] = p[1]; stack[2] = p[2]; for(i=3;i<=n;i++) { while(Mul(stack[top-1],stack[top],p[i])<=0 && top>=2) top --; top ++; stack[top] = p[i]; }}double maxx(double a,double b){if(a<b) return b;else return a;}double radius(P a, P b, P c){ int A=(b.x-c.x)*(b.x-c.x)+(b.y-c.y)*(b.y-c.y); int B=(a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y); int C=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); double tmp=(double)(A+B-C)/(2*sqrt((double)A)*sqrt((double)B)); tmp=sqrt(1.0-tmp*tmp); return sqrt((double)C)/(2*tmp);} double deal(P a, P b, P c) //求不公线三点子函数 { double ab, bc, ac, A, B, C; ab=dis(a, b); bc=dis(b, c); ac=dis(a, c); A=ab*ab; B=bc*bc; C=ac*ac; if(A>=B+C) return ab/2; else if(B>=A+C) return bc/2; else if(C>=A+B) return ac/2; else { return radius(a, b, c); }}int main(){ int i,top,tar,n; double x,y; P temp; while(scanf("%d",&n),n) { tar = 0; x = y = 0x7FFFFFFF; for(i=0;i<n;i++) { scanf("%lf %lf",&p[i].x,&p[i].y); if(p[i].x<x || p[i].x==x && p[i].y<y) { x = p[i].x; y = p[i].y; tar = i; } } if(n==1) puts("0.50"); else if(n==2) printf("%.2lf\n",dis(p[0],p[1])/2+0.500); else { temp = p[tar]; p[tar] = p[0]; p[0] = temp; qsort(p+1,n-1,sizeof(p[0]),cmp); p[n] = p[0]; tubao(n,top); double l=0,ans=0; for(i=0;i<top-2;i++) for(int j=i+1;j<top-1;j++) for(int k=j+1;k<top;k++){ans=maxx(deal(stack[i],stack[j],stack[k]),ans); } printf("%.2lf\n",ans+0.500); } } return 0;}
