Allan>Gosper, I was already familiar with dragons, so I knew that in the tail chains each connected piece has half the area of the next (except for one weird joint in the middle). So I was able to get the right answer (assuming I did).<ACW But isn't that a fairly immediate consequence of the two components being congruent and perpdendicular and similar to their union? Join them perpendicularly, doubling the area. Downscale by rt2 to recover the original. But the 45deg rotation and downscale moves each noncentral blob down to the next smaller one, so areas go by 2^n. But thank you. That's exactly the sort of answer I was soliciting. And you may have been the only solver! Too hard for Math-fun? More likely too boring. --rwg (I'm surprised nobody said: Duh, you spoiled your own problem with the dragon gif post!) On Sun, Aug 31, 2014 at 4:26 PM, Bill Gosper <billgosper@gmail.com> wrote: On Sat, Aug 30, 2014 at 2:10 PM, Bill Gosper <billgosper@gmail.com> wrote: I tweaked it. gosper.org/Screen Shot < http://gosper.org/Screen%20Shot%20Dragon.png> Dragon < http://gosper.org/Screen%20Shot%20Dragon.png>.png < http://gosper.org/Screen%20Shot%20Dragon.png> . --rwg Did anyone (besides a helpful private respondent, who mentions that the figure remains vertically squished despite my struggles with (hypersucky) OpenOffice) find this too easy or too hard? If you solved it, did you use any prior or outside knowledge about Dragons? --rwg On Sat, Aug 30, 2014 at 12:19 PM, Bill Gosper <billgosper@gmail.com> wrote: Trying to minimize math jargon. gosper.org/Dragon Screen Shot.png < http://gosper.org/Screen%20Shot%20Dragon.png> Fair question? --rwg PS, pure evil: https://mail.google.com/mail/?ui=2&ik=97b2a014c7&view=att&th=1481fbf0be9db9f... (Problem (4).) Subject: [math-fun] fib(1/2) Binet's and another formula give Sqrt[1/5 - (2 I)/5] and ((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5] This is that "other formula". Axel Vogt raised a good point: In[730]:= MinimalPolynomial[((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5], z] Out[730]= 1 - 2 z^2 + 5 z^4 I don't know why FullSimplify didn't try this. FullSimplify[TrigToExp[2*I^(n - 1)/Sqrt[5] Sin[n*ArcSec[2*I]]],n\[Element] Integers] comes within hailing distance of Binet.