See Regular Tetrahedron 1234.
Construct an equilateral triangle such that its vertices and circumcentre are (in any order) 1,2,3, and 4 units from the origin.
What is the length of a side of the triangle?
(In reply to
No Maths by brianjn)
O ___ A
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Z D
.B |n
m | C.
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Coord of B when knowing “m” and “n”:
OZ, BZ =2
OB = 3
(2+n)²+m²= 3²
n²+ m²= 2²
(2+n)² - n²= 3²– 2²
4 + 4n =5
n = ¼
m²=2²-(1/4)²
= 3 15/16
m =3√7 /4
AB² = (2+n)² + (m +1²
=(2 +1/4)² + (3√7 /4 + 1)²
O ___ .A p .
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Z D q
B |
| C .
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AC = AB
p² + q² = AC²
(1 + p²) + q² = 4²
1 + 2p = 4² – AC²
p = (15 –AC²)/2
q² = AC²-((15 –AC²)/2)²
q = √( AC²-((15 –AC²)/2)² )
BC² = (m + 1 + p)² + (q – (2+n))²
BC = √((m + 1 + p)² + (q – (2+n))²)
A copy of an “abridged” spreadsheet follows based upon the above outline.
(I'll not try to OB BZ OZ
redraw the image 3 2 2
that was in the
spreadsheet but it n AC
was a composite of 0.25 3.737463
the two above.) m
1.98431 p
0.515687
AB q
3.73746 3.701715
BC
3.789126
There is a difference of .05 against the other two sides.
I'm assuming that is because I used secondary calculated values
rather than primary, ie, I've used Sqrt values rather that using
the Square from which they were extracted. That will have
compounded through calculations.
I haven’t calculated the co-ords of D.
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Posted by brianjn
on 2012-03-30 02:50:15 |
re: No Maths - Calculations this time
