Triangle △ABC lies in plane D with AB=6, AC=4, and ∠BAC=90∘. Let D be the reflection across BC of the centroid of △ABC. For four spheres, all on the same side of P, have radii 1,2,3, and r and are tangent to D at points A,B,C, and D, respectively. The four spheres are also each tangent to a second plane T and are all on the same side of T. The value of r can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Place the triangle in the plane z=0. A(0,0,0),B(6,0,0),C(0,4,0)M=2B+C=(3,2,0)
GMAG=12⟹G=32M+A=(2,34,0)
Find coordinates of D. Since all these points lie on the plane z=0, let's ignore z coordinate in this part.
Line BC: mBC=−64=−32y=−32(x−6)⟹2x+3y−12=0D=(x′,y′)=(x0−a2+b22a(ax0+by0+c),y0−a2+b22b(ax0+by0+c))=(2−22+324(2(2)+3(34)−12),34−22+326(2(2)+3(34)−12))=(1342,39124,0)
If we consider the two planes that are tangent to all 4 spheres, the angle bisector plane of them passes through all 4 centers.
The 4 centers are: O1=CA(0,0,1)O2=CB(6,0,2)O3=CC(0,4,3)O4=CD(1342,39124,r)O1O2×O1O3=(6,0,1)×(0,4,2)=(−4,−12,24)n=(1,3,−6)1(x−0)+3(y−0)−6(z−1)=0x+3y−6z+6=0O4:1342+3(39124)−6r+6=06r=1342+13124+6=13166+6⟹3r=1383+3=13122r=39122=nmm+n=122+39=161Final Answer 161
