TETRA - Sphere in a tetrahedron
Of course a Sphere Online Judge System is bound to have some tasks about spheres. So here is one. Given the lengths of the edges of a tetrahedron calculate the radius of a sphere inscribed in that tetrahedron (i.e. a sphere tangent to all the faces).
Number N of test cases in a single line. ( N <= 30 ) Each of the next N lines consists of 6 integer numbers -- the lengths of the edges of a tetrahedron separated by single spaces. The edges are not longer than 1000 and for the tetrahedron WXYZ, the order of the edges is: WX, WY, WZ, XY, XZ, YZ.
N lines, each consisting of a real number given with four digits decimal precision equal to the radius of a sphere inscribed in the given tetrahedron.
Input: 2 1 1 1 1 1 1 1000 999 998 5 5 6 Output: 0.2041 1.4189
Not a binary search qs!
order of input of edge is sideedge - sideedge - sideedge - baseedge - baseedge - baseedgeLast edit: 2020-06-15 10:27:17
binary search did the trick :D
If sides are equal then apply the formula r(radii of sphere) = [ side*sqrt(6) ] / 12. i got my code wrong but my answer of first condition (1 1 1 1 1 1) was correct. please someone upload full program of this question. I dont understand if sides are different of tetrahedron then how sphere will get into it with same radii ?Last edit: 2018-04-07 21:31:48
Why can't we use the formula for volume of a tetrahedron as
go to http://rigmer.com and search this problem.
Calculate area of faces using herons formula and then radius of insphere is a simple formula r=3*v/(a1+a2+a3+a4)
|Added by:||Adam Dzedzej|
|Cluster:||Cube (Intel G860)|
|Languages:||All except: NODEJS PERL6 VB.NET|