Talk:Napoleon's problem
From Wikipedia, the free encyclopedia
The solution is the neatest that I have seen, but the proof needs cleaning.
!. Use the same diagram. Then H is the midpoint of BB'. Let AB = b.
Circles centers D,D' radius b, meet at A and X.
By similar Right triangles AHB, ABA'. AH/b = b/AA'. b²/AH = AA = 2r. b²/AC = r.
AD = AB = b. By similar isosceles triangles AXD, ADC. AX/AD = AD/AC. AX =b²/AC = r.
QED.
I am pleased that no proof is given that b > r/2
