Newsgroups: rec.puzzles
Date: 29 Jul 1998 12:13:20 +1000
> Three straight lines are parallel to each other, and the distance
> between any two lines is arbitrary.
> _________________________________
>
>
> _________________________________
>
> _________________________________
>
> Your task is to draw an equilateral triangle such that on each of
> the above three lines, one, and only one, apex falls.
>
> The tools you have are a pencil, a pair of compasses, and a ruler
> without scales.
[ If you enjoy this sort of problem, I recommend reading Geometric
Transformations I (and II and III, for that matter) by I. Yaglom. ]
SPOILERS AHEAD
Choose a point X on the middle line. Rotate the bottom line 60 degrees
about point X. Let Y be the point of intersection of this rotated line
with the top line. Then X, Y and the preimage of Y are the vertices of
an equilateral triangle.
In more detail:
1) Construct the circle with centre X and tangent to the bottom line.
Let T be the point of tangency.
2) Construct the equilateral triangle XTV. V will lie on the circle.
3) Construct the tangent to the circle at V. This is the rotated line.
Y is the appropriate point of intersection.
4) Construct the perpendicular bisector of XY. This intersects the
bottom line at Z, the desired preimage of Y. XYZ is equilateral.
In extreme detail:
1a) With the compasses set to a larger distance than that of X from
the bottom line, draw a circular arc centred on X intersecting
the bottom line at two points.
1b) Construct the perpendicular bisector of the segment formed. It
passes though T and X. T is the point of intersection with the
bottom line.
1c) Draw the circle with centre X and radius XT.
2a) Draw the circle with centre T and radius TX. This intersects the
other circle at V (either choice of V will do).
3a) Construct the perpendicular bisector of TV. This goes through X
and intersects the bottom line at a point W. The line WV is the
desired tangent. (Or you could just construct the perpendicular
to XV at V.)
The last required step (constructing perpendicular bisectors) is left as
an exercise for the reader. :) (It's a pretty fundamental operation, so
anyone attempting this problem should know it anyway.)
Cheers,
Geoff.

Geoff Bailey (Fred the Wonder Worm)  Programmer by trade 
ftww@cs.usyd.edu.au  Gameplayer by vocation.

                                       
Problem:
Express the area of an equilateral triangle having one vertex on each of
the three parallel lines in terms of the distances a and b.
_ ______________C__________________

b
______A___________________________
a
_________________B________________
Trigonometric Solution: [1]
Let s = AB = BC = AC, and phi the angle between the
horizontal line through A and the side AB.
a = s * sin(phi)
b = s * sin(pi/3  phi)
= s * (sin(pi/3) * cos(phi)  cos(pi/3) * sin(phi))
= s * (sqrt(3)/2 * cos(phi)  1/2 * sin(phi))
= sqrt(3)/2 * s * cos(phi)  a/2
s * cos(phi) = ( a + 2b ) / sqrt(3)
s*s = (s*sin(phi))^2 + (s*cos(phi))^2
= a^2 + ( a + 2b )^2 / 3
= 4/3 (a^2 + a*b + b^2)
area = (a^2 + a*b + b^2)/sqrt(3)
Problem:
Given a refence triangle and three parallel lines.
Contruct a triangle ABC which is similar to the refence triangle
where the points A, B, and C are on the three lines.
_ ____________C____D_______________
_ _____A___________________________
_ _____________________B___________
Solution: [3]
Analysis: Let ABC the triangle with the requiered properties.
Let D be the point where the circumcircle of triangle ABC
cuts the C line a second time.
Then by the inscribed angle theorem we have
beta = angle(ABC) = angle(ADC) and
gamma = angle(ACB) = angle(ADB).
The supplement angle of triangle BDC at D is of course alpha = angle(BAC).
Konstruction: Select an arbitrary point D on the top line.
Shift the angles alpha, beta, and gamma of the reference triangle to point D.
_____________D_____________
beta / \ alpha
/gamma\
The intersection of the two rays with the lower parallel lines
determine the points A and B. Then construct a circle through
the three points A, B, and D. The other intersection gives C.
References:
[1] Liangshin Hahn;
Complex Numbers and Geometry,
MAA 1994 (Spectrum Series), ISBN 0883855100
 p6263
the area of the equilateral triangle is (a^2 + ab + b^2)/sqrt(3)
where a and b are the distances of the middle line to their neighbor lines.
 p114115 excercise 19
Gives a trigonometric method for computing the side length of the triangle.
Then a "slick proof" is described. This slick on is invalid, which you have to show.
[2] Eckard Specht;
geometria  scientiae atlantis.
300+ Aufgaben zur Geometrie und zu Ungleichungen
insbesondere zur Vorbereitung auf MathematikOlympiaden,
OttovonGuerickeUniversität Magdeburg, 2001, ISBN 3929757397
Problem A.87
http://www.math4u.de/ (online version)
[3] R\"udiger Thiele;
Mathematische Beweise,
Deutsch Taschenb\"ucher Band 25,
Harri Deutsch Verlag, 1979
 Section 3.2, p6568
using the inscribed angle theorem (Umfangswinkelsatz)
[4] Brian Bolt;
Even more mathematical activities,
Cambridge Univ. Press, 1987
(german: Die dritte mathematische Fundgrube,
Klett Verlag, 1993, ISBN 3127227302)
Sect 71: Parallele Grenzen, p76, 177178
 equilateral triangle with vertices on three lines.
 square with vertices on a parallelogram.

http://www.mathematik.unibielefeld.de/~sillke/
mailto:Torsten.Sillke@unibielefeld.de