# Thread: post for ticbol

1. ## post for ticbol

Ticbol,

What would the total length be in these two cases (see attached pictures)

I get 2.9 cm for the square, I reckon I went wrong though.

Can't get anything sensible for the hexagon. To complicated :-(.

Could you help?

I used 1 cm for the size of the sides for both the square and hexagon.

Many thanks in advance

2. Originally Posted by Natasha1
Ticbol,

What would the total length be in these two cases (see attached pictures)

I get 2.9 cm for the square, I reckon I went wrong though.

Can't get anything sensible for the hexagon. To complicated :-(.

Could you help?

I used 1 cm for the size of the sides for both the square and hexagon.

Many thanks in advance
Although I do not welcome questions directed to me only, let me do this for now. Please, next time, just post your questions for everybody as usual. I do not have much spare time nowadays---and I am here just for fun, not to be obliged to answer.

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A) For the square.
The network is symmetrical about the vertical axis passing through the midpoint of the short horizontal line segment connecting the two Steiner points. The triangles at both ends of that short line segment are isosceles by construction.
Call the short horizontal line as "a"; any equal leg of any isosceles triangle as "b".

Length of network, L = a +4b

In any of the isosceles triangles:
--equal leg = b
--base = side of square = 1 cm.
--base angle = (1/2)(180 -120) = 30deg
So,
cos(30deg) = (1/2)/b
b = (1/2) / cos(30deg) = 0.57735 cm -------***

tan(30deg) = c/(1/2)
where c = height of the apex of the isosceles triangle from the base.
So, c = (1/2)tan(30deg) = 0.288675 cm ---*
Hence, horizontally,
a = 1 - 2c = 0.42265 cm -------***

Therefore, L = 0.42265 +4(0.57735) = 2.732 cm ---------answer.

-----------------------
B) For the regular hexagon.

I am not an expert on this Steiner thing, whatever it is. I don't do any research on it. I only understand the concept. There may be formulas developed already for many polygons, which I don't bother to search, but here is one way I'd attack the problem using Steiner's concept.

Construct the hexagon such that two opposite sides are horizontal.
There is a vertical axis of symmetry that passes through the midpoints of those horizontal sides.
The polygon can be partitioned into 6 equilateral triangles. Equilateral because an interior angle of a regular hexagon is 120deg. Half of that is 60deg, so an isosceles triangle whose two base angles are 60deg each is also equilateral.

Start the network in the topmost equilateral triangle.
Using your figure, draw lines 1-7, 2-7, and 7-10.
Point 10 should be the center of the hexagon, so the "Y" portion of the network formed by the 1-7, 2-7 and 7-10 lines is repeated at the equilateral triangles to the bottom left and right of the axis of symmetry. Meaning, points 9 and 8 are the equivalent of the point 7 in those other two similar, congruent "Y's".
So we need to find only the length of the Y in the topmost equilateral triangle, multiply that by 3 and we have the length of the network.

It so happens that the Y subdivides the equilateral triangle into 3 equal isosceles triangles whose apex angles are 120deg each.
In any of the 3 congruent isosceles triangles:
--equal legs = "V" of the "Y"= say, d cm. each.
--base = side of equilateral triangle = 1 cm.
--base angles = (1/2)(180 -120) = 30deg each.
So,
cos(30deg) = (1/2)/d
d = (1/2)/cos(30deg) = 0.57735 cm ---***
Hence, one Y = 3(0.57735) = 1.732 cm ----that is sqrt(3) cm.
Therefore, the whole network is 3(1.732) = 5.196 cm. long. ----------answer.
That is 3sqrt(3) cm long.

3. Thanks for your reply. Promise not to do it again :-)