Talk:Napoleon's problem

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

Bparslow (talk) 18:25, 8 January 2008 (UTC)

Usual notation for right triangle (A, B, C) means that B instead A´, C & C´ instead B & B´ as well as x = OA = OC instead a should be taken leading up to well known relations: $$a^2+b^2=c^2=(2r)^2,\,P=\frac{ab}{2}=\frac{ch_c}{2},\,\frac{n}{a}=\frac{a}{c} \Rightarrow nc=a^2,\, \frac{m}{b}=\frac{b}{c} \Rightarrow$$ $$\,mc=b^2 = m^2+\left[(x^2-(x-m)^2\right]= m^2 + x^2-x^2+2mx-m^2=2mx,$$ $$mc=2mx \Rightarrow x=\frac{c}{2}=r$$

77.238.204.217 (talk) 09:51, 18 October 2008 (UTC)Stap

The proof is difficult to follow because points on the proof sketch don't match up with points on the problem sketch above. For example, C on the top diagram is where the green lines cross, but on the bottom diagram, C is the sought-for centre of circle (and I think where the green lines cross is now O). Similarly D and D' don't feature on the bottom diagram (or have they been re-labelled B and B'?). I would correct this but right now I haven't been able to figure out exactly which points are which and I don't want to get it wrong :-( 91.106.135.3 (talk) 19:23, 20 September 2009 (UTC)