Begin with unit equilateral triangle ABC.
Extend side AC one unit to D so DC=1.
Construct line AE such that
(1)E is on line DB with B between D and E,
(2)line AE intersects CB at F with B between C and F,
(3)and EF=1.
Find the length AE.
Comment on the title.
Let AE = d and BE = c.
From triangle ABE:
AE^2 = AB^2 + BE^2
d^2 = 1 + c^2 (1)
From triangle ABF:
AF AB
----------- = -----------
sin(/ABF) sin(/AFB)
AE + EF 1
----------- = -----------
sin(120) sin(/BFE)
d + 1 1
----------- = -----------
sqrt(3)/2 sin(/BFE)
(d+1)^2*sin(/BFE)^2 = 3/4 (2)
From triangle BEF:
BE EF
----------- = -----------
sin(/BFE) sin(/EBF)
c 1
----------- = -----------
sin(/BFE) sin(30)
c = 2*sin(/BFE) (3)
Combining (1), (2), and (3) gives
d^3 = 2
or
d = 2^(1/3) the Delian constant.
The figure cannot be constructed
with straightedge and compass or
surely the ancient Greek geometers
would have solved the doubling of
the cube, also known as the Delian
problem.
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Posted by Bractals
on 2017-06-05 18:13:40 |
Solution
