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 Delian delight (Posted on 2017-06-04)
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.

 No Solution Yet Submitted by Jer No Rating

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 solution | Comment 1 of 3
since ABC is equilateral that means angle ACB=60 and thus BCD=120.

Since BC=CD=1 we have that angle CBD = angle CDB = 30

Since E,B, and D are co-linear then we have angle EBA=90
from that since F,B, and C are co-linear then we have angle FBE=30

let x be the measure of the angle EAB then we have side EB = tan(x)

angle AEB=90-x
angle FEB=90+x
angle EFB=60-x

using the law of sines in triangle EFB using angles EFB and FBE we have

sin(30)=sin(60-x)/tan(x)
sin(30)*tan(x)=sin(60-x)
(1/2)*tan(x)=sin(60-x)
tan(x)=2sin(60-x)

now I don't see any way of solving this other than numerically for which I get that x is approximately 0.653928
which then gives that AE=1/cos(x)=1.25992

Not sure how this relates to the title which gives me the feeling that there is a shortcut relating to the title.

 Posted by Daniel on 2017-06-04 14:27:49

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