| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I was given a problem by a friend?, that I have managed to generalise
to what I thought would be a simple geometric task. I still convinced
it's simple but cannot solve it. Any help would be appreciated.
Given a hexagon ABCDEF of which all angles are known, what are the
interior angles of the triangle ACE.
I'll post the whole problem once I can figure a way of illustrating it.
Thanks in advance, Steve
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1881.1 | Need more info | WIBBIN::NOYCE | DEC 21064-200DX5 : 138 SPECint @ $36K | Wed Jul 20 1994 11:29 | 8 |
Knowing all the angles doesn't uniquely define the hexagon. For
example, all the angles in both these hexagons are 120 degrees,
but the inscribed triangles are quite different:
__________ _
/ \ / \
\__________/ / \
\ /
\_/
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| 1881.2 | . | OINOH::KOSTAS | He is great who confers the most benefits. | Wed Jul 20 1994 11:38 | 8 |
re. .-1
so what are the known values for all angles? At the absense of the
specific values my guess will be that: angle A+ angle C+ angle E = 180
for the general case there are two cases to concider:
o if A+B+C+D+E+F <= 360, and
o A+B+C+D+E+F > 360
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| 1881.3 | FORTY2::PALKA | Thu Jul 21 1994 12:13 | 12 | ||
re .2
.1 showed that the interior angles of the triangle ACE are not uniquely
determined. I.e. there is no solution to the problem as posed. In fact
I think it is likely that given an arbitrary triangle you can 'enclose'
it in a hexagon of any desired angles - probably in an infinite number
of ways !
The sum of the interior angles of the hexagon is constant.
For all hexagons the sum of the interior angles is 720�.
Andrew
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| 1881.4 | The original problem | WELCLU::THOMAS | Fri Aug 05 1994 09:39 | 26 | |
This is the problem as it was originally given to me. Consider the
following diagram:
A
K
P
L Q
B M N C
Given any triangle ABC, let the trisectors of angles A,B,C intersect AC
at P and Q, BC at M and N, and, AB at K and L respectively, ordered as
in diagram.
Show that forany triangle ABC the poiunts of intersection of AM and BP,
BQ and CL and, CK and AN always form and equilateral triangle.
I realise the route I was taking was completely wrong, I still have no
solution (or disproof)
Steve
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