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Conference rusure::math

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

227.0. "A geometry construction" by METOO::YARBROUGH () Fri Mar 01 1985 13:02

Problem 1: Given three parallel lines in a plane, construct an equilateral 
triangle with a vertex on each line.

This is not a very difficult problem; what I am looking for is elegance of 
solution (I have what I regard as a very neat one).

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Problem 2: Given three parallel lines in 3-space, construct an equilateral
triangle with a vertex on each line.

This one is quite a bit harder. I don't right now have a solution.

Lynn Yarbrough
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227.1HARE::STANFri Mar 01 1985 17:4718
Solution to problem 1:

Assume no two of the three parallel lines coincide.  Call them L1, L2, and
L3. We seek A on L1, B on L2, and C on L3 such that A, B, C form
an equilateral triangle oriented clockwise. (Another solution will occur
for the counterclockwise case.)

Let the notation <XYZ> mean that triangle XYZ is an equilateral
triangle (with the vertices oriented clockwise).

Choose A arbitrarily on line L1.

Now as B moves along L2, the locus of the third point, X, such that <ABX>,
is a straight line.

Thus pick any point B1 on L2 and let X1 be such that <AB1X1>.
Then pick any point B2 on L2 and let X2 be such that <AB2X2>.
Connect X1 and X2.  This line meets line L3 at the desired point C.
227.2TOOLS::YARBROUGHMon Mar 04 1985 12:5316
Notation: call the lines A,B,C, assumed horizontal. Points on each line are 
numbered. X<->Y means the line segment connecting X and Y. 

Solution: Pick any A1 and construct a 60 deg line connecting A1, B1 and C1.
Let A1<->A2 = A1<->B1 = B1<->A2. Then C1<->A2 is one side of the desired
triangle; construct B2 at that distance from C1 and A2. 

Proof: A2 lies on A by definition; C1 lies on C by definition; the
triangles A1A2C1 and B1A2B2 are congruent, therefore angle A2B1B2 and 
A2A1C1 are equal, therefore B2 lies on B. 

The construction and proof are independent of the order or coincidence of
A,B, and C. 


So far I don't see a way of extending either solution to the 3-d case.
227.3TURTLE::GILBERTMon Mar 04 1985 17:322
re .-1
	The proof seems flawed.  The construction for B2 yields *two* points.
227.4HARE::STANMon Mar 04 1985 19:1812
What Lynn means is this:

Consider the 60 degree rotation of the plane with center A2
that takes A1 into B1.  This rotation takes C1 into B2.
That's how B2 gets created.  Note that this rotation is an
isometry and takes triangle A1A2C1 into B1A2B2.  That's
why these two triangles are congruent.

Since angle A2A1B1 was 60 degrees to start with, therefore
angle A2B1B2 must also be 60 degrees.  But angle A1A2B1 is
also 60 degrees, so that makes B1B2 parallel to A1A2 since
A1 and B2 are on opposite sides of transversal A2B1.