Math 402, Assignment 5, Spring 2008
Due Wed. February 27 (at the beginning of class)
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Consider a segment BC and a reflection R across a line l
on a cylinder.
Assume that the reflection exchanges B and C, that is,
R(B) = C and R(C) = B.
(a) If l is a vertical line, does l necessarily
intersect segment BC? How about if l is a
great circle? Give a proof
or a counterexample in each case.
(b) If l is a vertical line, does R necessarily take
segment BC to itself? How about if l is a (horizontal)
great circle?
Give a proof or a counterexample in each case.
(c) If l is a vertical line and l does intersect segment
BC at a point D, is l necessarily the perpendicular
bisector of BC? How about if l is a great circle.
Give a proof or a counterexample in each case.
Please note: consider a triangle ABC so that
segments AB and AC are congruent and so that l is the angle bisector
of angle A.
Part (b) determines if the Isosceles Triangle
Theorem is true for such triangles. (If l does not intersect
segment BC, you need a slightly different proof than the
one we gave in class. Instead of comparing triangles ABD and ACD,
you compare the triangles ABC and ACB -- like pons asorum.
Parts (a) and (c) together determine if the Isosceles Bisector Theorem
is true for such triangles (both must hold)
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Theorems which are not true for all triangles on the sphere
may still be true for some triangles.
Here are some possible restrictions. Let S be one
half the circumference of the sphere.
a)
No side is longer than S.
This rules out the fish, witches hat, and trefoil triangles.
b)
No side has length S.
This implies that no two points are antipodal,
which rules out a lunar triangles.
c) The triangle does not lie entirely in one great circle,
that is, it is not a linear triangle.
We can also combine restrictions to
consider triangles which satisfy a) and b), a) and c), b) and c),
or all three.
What is the FEWEST restrictions that you can place to make ASA true on the
sphere?
Give a careful proof in this case (do not use SAS!)
By giving counter-examples, show that it is not true with any
fewer restrictions.
Explain what went wrong with your proof in each case.
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Give a proof of ASA which works in E^2 or H^2. Here is the twist:
You may use SAS but may NOT use ANYTHING ELSE proved using the mirror
axiom. This INCLUDES the existence of reflections and rotations
and the congruence theorems.
HINT: Consider triangle ABC and DEF. Suppose that angle
A is congruent to angle D, angle B is congruent to angle E, and
segment AB is congruent to segment DE.
Draw a point C' on the ray AC so that
the segment AC' is congruent to the segment DF.
Now draw segment BC'.
In fact, as suggested by this example, it is possible to replace
the mirror axiom by SAS. Many axiom systems follow
this approach. (Euclid, for example, proves
SAS using "superposition" and then almost never uses it again.)
Which axiom do you prefer? Why?