Download Rectangles and Defect Recall that, given a triangle )ABC, the angle

Rectangles and Defect
Recall that, given a triangle )ABC, the angle sum
for
the triangle is the sum of the interior angles of )ABC :
. Similarly, for a convex
quadrilateral
the angle sum
is the sum of the
interior angles of the quadrilateral:
Definition: Given a triangle )ABC, the defect
difference between 180 and the angle sum of )ABC:
is the
.
For a convex quadrilateral
the defect is
.
Note: The definition of angle measurement (as always non-negative)
and the Saccheri-Legendre Theorem combine to insure that for any
triangle )ABC,
and
.
Lemma:
~
If )ABC is a triangle and E is a point such that B-E-C,
then
.
A very similar proof can be used to establish:
Lemma: If
is a convex quadrilateral, then
.
~ Exercise. €
We abbreviate these facts by saying that defect is additive or talking
about the additivity of defect.
Definition: A rectangle is a convex quadrilateral each of whose angles
is a right angle.
Note that by definition, the defect of a rectangle is 0.
Finally, one lemma before we state and prove the main theorem:
Lemma: If
is any triangle, then at least two of the interior
angles are acute. If the acute angles have vertices A and B, then the
foot F of the perpendicular from C to
is such that A*F*B.
~ The first statement follows immediately from Saccheri-Legendre.
For the second statement, note that F cannot be equal to A or B, since
that would make the angles at A or B right angles, contrary to the
assumption. Now either F*A*B, F*B*A, or A*F*B. If F*A*B, then
pCAB is an exterior angle to
and so must be larger in measure
than pCFA, which is a right angle. This contradicts our assumption
that
has acute angles at A. A similar proof shows that A*B*F
leads to a contradiction. Thus we must have A*F*B.
Big Theorem: The following are equivalent:
1.
2.
3.
4.
5.
6.
There is a triangle with defect 0
There is a right triangle with defect 0
There is a rectangle
There are arbitrarily large rectangles
The defect of every right triangle is 0
The defect of every triangle is 0
Outline of proof:
1 ! 2):
Let *()ABC) = 0, and WLOG let
be the longest
side. By the lemma, there is a point D with A*D*B,
such that D is the foot of the perpendicular to
from
D. Then )ADC and )BDC are right triangles. If
*()ABC) = *()ADC) + *()BDC) = 0, then both
*()ADC) and *()BDC) must equal 0.
2 ! 3):
Let )ABC be right with the right angle at B, and let
*()ABC)=0. Then :(pA) + :(pC )= 90. Construct
. Then :(pDAC) =
90 - :(pBAC) = :(pBCA), so by SAS,
. Then CPCF gives that GABCD is
a rectangle.
3 ! 4):
We show that we can make a rectangle with dimensions
larger than any given N > 0, essentially by pasting
together copies of a given rectangle.
Start with a given rectangle GABCD. Create points E and F
with A*B* E, D*C*F, AB = BE, and DC = CF. By SAS,
so :p1 = :p2 and AC = EC by CPCF. We
can check that E is interior to pBCF, so that :p1 + :p3 = 90 =
:p2 + :p4 implies :p3 = :p4. We can then use SAS to get
.
Now *()ABC) + *()ACD) = *(GABCD) = 0, so )ABC has
defect 0, and :pBAC + :p1 = 90. Because :pBAC + :pCAD
= 90 as well, :pCAD = :p1. Similarly, we can show :pBAC
= :p3. Thus by ASA,
. Now since
, CPCF gives us that :pCFE =
90.
In an analogous way we can show that pBEF has measure 90,
and so GADFE is a rectangle with twice the length of the
original. Having shown we can construct such a rectangle, we
can apply the process repeatedly on the length and width, and
since for any N > 0 there is an M with
we can repeat
the process to obtain an arbitrarily large rectangle
4 ! 5):
Given a right triangle )ABC with right angle at C,
create a rectangle whose length and width are both
larger than the legs of )ABC. Then, find points on two
adjacent sides of the rectangle such that their distances
from the common vertex equal the length of the legs.
Essentially, we are copying the right triangle into the
corner of a larger rectangle. To simplify notation, we
will keep the vertex lables A, B, and C for the triangle
as shown below.
Now *(GCDEF) = 0 since it is a rectangle, and by
additivity of defect we must have
and
. Since
and
. Finally, since
, we must have
last!)
5 ! 6):
,
and (at
.
Given a triangle )ABC, drop a perpendicular to the
longest side (WLOG
), so the foot of the
perpendicular is D with A*D*B. This creates two right
triangles )DAC and )DBC . By additivity of defect,
the defect of
)ABC must be 0.
6 ! 1):
Trivial. €
Note that since we know that the EPP is equivalent to every triangle
having angle sum 180 (or, alternately, every triangle having defect 0),
this theorem adds five more statements to our list of equivalents to the
EPP, the most imporant one being:
Corollary: The Euclidean Parallel Postulate is equivalent to the
existence of a rectangle. Thus, if the Euclidean parallel postulate is not
true there can be no rectangles.
We use this fact to prove:
The All-Or-Nothing Theorem (AKA the Ado Annie Theorem): If
there is one line l0 and one exterior point P0 with more than one line
through P0 parallel to l0 , then it is true of every line l and exterior
point P.
~ Since there is one line l0 and one exterior point P0 with more than
one line through P0 parallel to l0 , the EPP fails and we know there
cannot be any rectangles.
Now, let l be a line and P a point not on the line. In the usual way,
construct the usual parallel line: Drop a perpendicular from P to a
point Q on line l. Construct a line m perpendicular to
through
point P. By the Saccheri-Legendre Theorem, m must be parallel to l.
Now, through a point R on l, construct a line t perpendicular to l and
drop a perpendicular from P to t, with the foot being point S.
Now consider GPQRS. The angles at Q, R, and S are all right. Since
no rectangle can exist, the angle at P cannot be right, so that
But neither can
intersect l or it would violate Saccheri-Legendre
(or you can apply the Alternate Interior Angles Theorem in Absolute
Geometry with t being the transversal). Thus there are two lines
parallel to l through P. €
.
The upshot of all this is that the decision on parallelism is a choice
between two worlds – Euclidean and Hyperbolic. You have to
jump all the way into one or the other.
Note that all we have done so far in this set of notes has been done
in Absolute Geometry. We haven’t taken a stand yet, but have
just explored the consequences of taking such a stand.