[VA] SRC#004- Fun with Sexagesimal Trigs

[Gerson W. Barbosa]

(02-12-2019 12:14 PM)Albert Chan Wrote:  Prove A = 2^29 = 536870912:

Checking the edges, using shorthand t# = tan(#°):

tan(A°+B°) = (tA + tB) / (1 - tA tB)

(t60 + tA) (t60 + tB), where A+B=30°
= t60^2 + t60*(tA + tB) + tA tB
= t60^2 + t60*t30*(1 - tA tB) + tA tB
= 3 + 1 - tA tB + tA tB
= 2^2

t15 = tan(45° - 30°)
= (t45 - t30) / (1 + t45 t30)
= (1 - 1/√3) / (1 + 1/√3)
= (√3 - 1) / (√3 + 1)
= (√3 - 1)^2 / (3 - 1)
= (3 + 1 - 2*√3) / 2

-> t15 = 2 - √3
-> center = t60 + t15 = 2

14 pairs of edges and 1 center, all can considered value of 2, thus A = 2^29

A is the only one I had trouble with because at first I had wrongly programmed it as a sum instead of a product. When I notice my mistake and finally got the correct value, I simply computed its base-2 logarithm in order to check it was a power of two. But only because I had identified D already. For whatever reason I decided to check the product of the first and last terms:

(1 - 1/tan1°)*(1 - 1/tan44°) = 1.99999999997

That’s 2 to me, so I just did 2^22 = 4194304 which “matched” 4194303.99965, the value my D program returned, and I gave it no further thought.

I do appreciate, however, your efforts in going deeper into the problem and providing a proof.

Gerson.