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)