David Mullens
In projective geometry there is a well known
theorem called Desargues' Theorem. It states that
if two triangles are perspective from a point, they
are perspective from a line (Clapham and
Nicholson 120). Desargues' theorem is true for any
projective space of dimension at least 3.
Desargues' theorem is not the easiest thing to prove
in dimension 2. However, if one simply moves up
a dimension it becomes clear how to prove. The
idea was that hopefully if we went up more than
one dimension things would become even easier. It
turns out this is precisely what happened.
The purpose of our research is to demonstrate a
proof of Heron's formula using scissors
congruences in the fourth dimension. We wished
to provide a proof for an ancient formula using a
modern method. The proof is via a scissors
congruence that gives the proof of the Pythagorean
theorem, its 4-dimensional analogue, and
interpretations of the distributive laws via scissors
congruence. A triangle consists of three vertices
and three edges connecting said vertices. The edge
lengths of a triangle determine the area of that
triangle. The explicit formula for the area of a
triangle is known as Heron's (Hero's) Formula
(although recently attributed to Archimedes)
(Struik 43). The earliest known proof of Heron's
Formula can be found in a compendium in three
books known as Metrica (Dunham 114). Heron's
Formula states that the area, A, of a triangle is
computed in terms of the lengths of its sides. It
states that
where the semiperimeter,
(a + b + c)  2.
Heron's Formula can be written strictly in terms of
its edges by squaring both sides and clearing the
denominator, i.e.,
(a + b + c)(-a + b + c)(a - b + c)(a + b -c),
that is a homogeneous polynomial of degree four in
the edge lengths a, b, and c.
When we multiply two values, such as length
and width, we get area. When three values are
multiplied we get volume. Hence when we
multiply four values we expect a certain hyper
volume. Ergo each fourth degree term from
Heron's Formula represents a particular hyper
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volume. Thus, the right hand side of the above
equation represents 16 times hyper volume of the
geometric figure, that is a triangle times a triangle
(a hyper solid in 4-D space) when multiplied by 16
is the same as the hyper volume of a hypercube that
has edge lengths
(a + b + c)(-a + b + c)(a -b + c)(a + b - c). From
this, it would appear that Heron's Formula is now
telling us something about the fourth dimension.
This is the notion we wish to explore in further
detail.
Heron's original proof made use of cyclic
quadrilaterals. In addition, many proofs have since
been provided appealing to trigonometry, linear
algebra, and other branches of mathematics. We
propose to prove Heron's Formula using scissors
congruences in 4-dimensions. We should note that
the desire for a 4-D proof was expressed by J.H.
Conway. Conway mentioned that he sought a 4dimensional proof of Heron's formula over dinner
with my collaborator J. Scott Carter. John Horton
Conway is a prolific mathematician active in the
theory of finite groups, knot theory, number theory,
combinatorial game theory and coding theory. He
is most well know for the invention of the cellular
automaton called the Game of Life.
We will begin by addressing the concept of
scissors congruence and providing a scissors
congruence proof of the Pythagorean Theorem. A
polygon decomposition of a polygon P in the
Euclidean plane is a finite collection of
polygons
whose union is P and which
pairwise intersect only in their boundaries, e.g., the
well known dissection puzzle, Tangram
(Champanerkar 2).
We demonstrate that the distributive law is a
scissors congruence. Furthermore, we explicitly
represent expressions such as a(b - c) in terms of
decompositions of rectangles. In such decompositions, negative edge lengths are dealt with
formally, and we indicate explicitly how the formal
manipulations correspond to genuine geometric
cutting, deleting, and gluing.
The next step is to demonstrate that if polygons
P and Q are respectively scissors congruent
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to P' and Q', then the hyper-solids P x Q and P' x Q'
are scissors congruent. This result follows,
roughly, from the fact that cartesian products
distribute over unions. More explicitly, we examine
dihedral angles between various facets of the
hyper-solids.
Finally, we examine the algebraic proofs in
terms of the scissors congruences outlined above.
The expression
where
is the area of the
triangle, represents the volume of four hyperparallelograms (parallelogram times
parallelogram), each of which is decomposed into
four copies of a triangle time triangle. The
expression
(a + b + c)(-a + b + c)(a - b + c)(a + b -c)
is the hyper-volume of a hyper-rectangle. It is
decomposed into 81 pieces via the distributive law.
All but nine terms in this expression cancel. Some
of those nine terms will further decomposed using
the Pythagorean theorem. The nine remaining
terms look as follows
Throughout the proof, we illustrate (as much as
is possible) the steps via standard 2-dimensional
projections of the hyper-solids. The danger in
relying upon these projections is that
the illustrations loose two dimensions of
information. Consequently, we often choose more
than one projection to illustrate a given idea. Also,
we spend some care in analyzing the angles
between facets of resulting hyper-solids. This is to
ensure we do not conflict with previous answers to
problems similar ours.
In the end everything comes down to an
algebraic proof. We make the respective
substitutions for the squared terms in the formula
above using the Pythagorean theorem on the image
below.
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Then we make appropriate cuts, cancel all the
negative hyper volume, and reassemble what it left
over. Once finished the formula becomes
.
When we divide everything through by 16 we get
.
That is, the hyper volume of a triangle squared
(triangle times triangle) is equal to one-fourth the
hyper volume of a parallelogram times
parallelogram spanned by and . One more
sentence needed.