（HDU 5733） tetrahedron <几何公式>

tetrahedron

Problem Description

Given four points ABCD, if ABCD is a tetrahedron, calculate the inscribed sphere of ABCD.

Input

Multiple test cases (test cases ≤100).

Each test cases contains a line of 12 integers [−1e6,1e6] indicate the coordinates of four vertices of ABCD.

Input ends by EOF.

Output

Print the coordinate of the center of the sphere and the radius, rounded to 4 decimal places.

If there is no such sphere, output “O O O O”.

Sample Input

0 0 0 2 0 0 0 0 2 0 2 0

0 0 0 2 0 0 3 0 0 4 0 0

Sample Output

0.4226 0.4226 0.4226 0.4226

O O O O

Author

HIT

Source

2016 Multi-University Training Contest 1

Recommend

wange2014

题意：

给你空间中四个点，若形成四面体则求出四面体内切圆的圆心坐标和半径。

分析：

就是一个几何公式题：

首先可以用向量叉乘求出四个面的面积s1,s2,s3,s4和四面体的体积V，如果V=0说明这四个点共面不能形成四面体，否则由r*(s1+s2+s3+s4)/3=V得到内切球半径r=3*v/(s1+s2+s3+s4)，之后根据

x=(x1*s1+x2*s2+x3*s3+x4*s4)/(s1+s2+s3+s4)

y=(y1*s1+y2*s2+y3*s3+y4*s4)/(s1+s2+s3+s4)

z=(z1*s1+z2*s2+z3*s3+z4*s4)/(s1+s2+s3+s4)

求出内切球球心坐标即可

AC代码：

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define eps 1e-8
struct node
{
double x,y,z;
}a,b,c,d,ans;
double get(node a,node b,node c)
{
double x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
double x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
double d1=y1*z2-y2*z1;
double d2=x1*z2-x2*z1;
double d3=x1*y2-x2*y1;
return sqrt(d1*d1+d2*d2+d3*d3)*0.5;
}
double Get(node a,node b,node c,node d)
{
double x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
double x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
double x3=d.x-a.x,y3=d.y-a.y,z3=d.z-a.z;
double ans=x1*(y2*z3-y3*z2)-y1*(x2*z3-x3*z2)+z1*(x2*y3-x3*y2);
return abs(ans)/6.0;
}
int main()
{
while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&a.z,&b.x,&b.y,&b.z,&c.x,&c.y,&c.z,&d.x,&d.y,&d.z))
{
double s1,s2,s3,s4,V;
V=Get(a,b,c,d);
s1=get(b,c,d);
s2=get(a,c,d);
s3=get(a,b,d);
s4=get(a,b,c);
double t=s1+s2+s3+s4;
if(V<eps)
{
printf("O O O O\n");
continue;
}
ans.x=(s1*a.x+s2*b.x+s3*c.x+s4*d.x)/t;
ans.y=(s1*a.y+s2*b.y+s3*c.y+s4*d.y)/t;
ans.z=(s1*a.z+s2*b.z+s3*c.z+s4*d.z)/t;
double r=V*3.0/t;
printf("%.4lf %.4lf %.4lf %.4lf\n",ans.x,ans.y,ans.z,r);
}
}