Download Using Euclid`s Geometry to Solve Quadratics Solution Commentary:

Using Euclid’s Geometry to Solve Quadratics
Solution Commentary:
Solution of Main Problem:
1. By the one construction, if you change the lengths of n and m dynamically, you
should be able to conclude that exactly one positive root is always determined
by this construction.
2. If AB = n and BD = x, then CB = n/2. Rewriting the claim “area of a rectangle
formed by side lengths AD by DB together with the square on CB equals the
2
2
n n

square on CD” using algebra, we have the “equation” ( x + n)( x) =  x +  −   .
2 2

By algebraically manipulating both sides, one can see this is an identity.
3. This problem is taken from Proposition 6 in Book 2. First, BD = DE = CG = FE
by construction. Second, AC = BC = GF = FI = EJ by properties of midpoints
and fact that DC = DB + BC = DJ = DE + EJ. Third, [area rectangle ACGK] =
[area rectangle FEJI], which implies that [area rectangle ADEK] = [area
rectangle CDEG] + [area rectangle FEJI]. The latter two rectangles form a Lshaped figure, defined to be a gnomon by Euclid. The proof is now complete
as the [area gnomon] + [area square GFIH] = [area square CDJH].
2
2
2
2
n n
n n


4. In the identity x + nx = ( x + n)( x) =  x +  −   , we have  x +  −   = m 2 .
2 2
2  2


n
n
But, since AB = n, BC = , CD = x + , we are done because by the first
2
2
2
2
2
n
n 
construction we have m +   =  x +  .
2
2 
2
To relate this to the quadratic
2
n
n
formula, we have shown that x = − +   + m 2 with a little more algebra.
2
2
5. This process is justified in the responses to questions #1- #4.
Extension
1:
Since
m
>
0,
the
two
relationships
x=
− n ± n 2 + 4m 2
and
2
− n + n 2 + 4m 2
− n − n 2 + 4m 2
n + 4m > n = n imply that x =
> 0 and x =
< 0.
2
2
2
2
2
Extension 2: This statement is Proposition 4 in Book 2. First, the statement “square
on a segment AB” basically refers to the square that has segment AB as its diagonal.
In the diagram, square ADBE is the square on segment AB. With random point C on
segment AB, the task then is to show that the area of the square on AB is equal to
the sum of the areas of the squares on the segments AC and CB plus the area of the
two associated rectangles (which are congruent). Now [area of square ADBE] = [area
of square AFCI] + [area of square CGBH] + 2 [area of rectangle FDGC]. As to the
modification to (m+n)(m-n) = m2 – n2, consider this diagram:
m
n
m-n
Then, the gnomon is comprised of two rectangles of areas (m-n)(m) + (m-n)(n) =
(m+n)(m-n). But this gnomon is what remains when you remove the square of area
n2 from the square of area m2. That is, (m+n)(m-n) = m2 – n2.
Extension 3: Given AB = n, by the construction sequence, AC = n/2, CD = m, and DE
= n/2 = DF. Applying the Pythagorean Theorem to right triangle DCF, DC2 + CF2 =
2
2
n
 n
FD ; by substitution, m +  − FB  =   , which simplifies to m2 + FB2 = n(FB) and
2
 2
FB is the desired root. Now, by shifting point D in the constructed diagram, we see
that circle W could intersect AB in 2 points (when n/2 > m, 1 (when n/2 = m), or 0
points (when n/2 < m). Algebraically, this is confirmed by the quadratic formula where
2
2
2
n
n
x = ±   − m 2 . Finally, in relation to the use of Euclid’s proposition 5, Book 2, we
2
2
have that BH = BD, JI = CF = IL, and BM = AC, implying that [area of rectangle
FBML] = [area of rectangle ACJK], or that [area of rectangle ABHK] = area of
gnomon CBMLIJ]. Finally, [area of square JILD] + [area of rectangle ABHK] = [area of
square CBMD].
Open-Ended Exploration: Perhaps your students will produce some creative
responses (that seem to be absent in the literature). One possible resource is the
commentary “Geometrical Algebra” (pp. 372-374) in Heath’s translation of Euclid’s
Elements (by Dover Publications). (If you do not have a copy of Heath’s version, it is
available on-line at http://www.perseus.tufts.edu/hopper/.)
Teacher Commentary:
The idea of a gnomon is traced to the measurement of time by using a vertical
pole to cast a shadow (i.e. a primitive sundial) and is similar to a carpenter’s square.
Euclid extended this idea to the case of a parallelogram, which does not require a
right angle. For a more details, refer to Heath’s commentary (pp. 370-372), the web
site http://mysite.du.edu/~jcalvert/astro/gnomon.htm, or Gazale (1999). The latter is a
fascinating journey that leads to the Golden Section, fractals, and a new “special”
irrational number called the “silver number.”
In 1847, Oliver Byrne produced a version of the The First Six Books of the
Elements of Euclid, which is now listed for sale for more than $22,500. It was unique
in its use of colors to "simplify" both Euclid's propositions and his related proofs of
these same propositions. The University of British Columbia's Mathematics
Department spearheaded a project that created digital photographs of each page of
this Byrne’s version, and it is available at no cost via the internet
(http://www.math.ubc.ca/~cass/Euclid/byrne.html). After students have explored the
problems as posed, you might have them explore Byrne’s proof of the Proposition 5
(http://www.math.ubc.ca/~cass/Euclid/book2/images/bookII-prop5.html) and
Proposition 6 (http://www.math.ubc.ca/~cass/Euclid/book2/images/bookIIprop6.html).
In Extension 3, segment FB was identified as one root. Ask students to
determine (and justify) if the construction actually produces the second root in the
case where circle W produces two intersection points F and G (i.e. is the second root
AG?).
Ask students to implement Euclid’s process using both a straight-edge/
compass and GSP. The first option of a straight-edge/compass adds a sense of
reality to the idea that even though they might start with simple values for n and m,
they now face the new problem of trying to determine the numerical value of the
obtained root(s). That is, the process is theoretically sound and does produce a
length as a root, but algebra is needed to produce the numerical value. In contrast,
the GSP option encourages students to explore how changes in the starting values of
n and m affect the solution root(s), especially in the third case of x2 + m2 = nx.
As a good source of a writing project, students can explore any of the following
ideas relative to Euclid’s mathematics:
• Some authors claim that Propositions 28 and 29 in Book 6 offers geometrical
solutions to quadratic equations. Are these assumed solutions the same or
different from those contained in Propositions 5 and 6 in Book 2?
• In his discussion of Euclid’s solution of quadratics, Boyer (1956) suggests:
“Probably one of the chief reasons that Greece did not develop an algebraic
geometry is that they were bound by a geometrical algebra. After all, one
cannot raise himself by his own boot straps.” (p. 9) Explore this claim by
focusing on the difference in meaning between an algebraic geometry and a
geometrical algebra.
• The modern meaning of a geometrical algebra differs significantly from that
initiated by Euclid. Consider Hermann Grassmann’s introduction of the field of
geometrical algebra in 1844. Closely tied to the Clifford Algebra, geometrical
algebras basically are a representational, computational system that integrates
linear algebra, vector calculus, differential geometry, complex numbers and
quaternions. Is the modern geometrical algebra a generalization of Euclid’s
version?
Additional References:
Boyer, C. (1956). History of Analytic Geometry. The Scholar’s Bookshelf.
Dunham, W. (1994). “Greek geometry” in The Mathematical Universe. John Wiley &
Sons.
Gazale, M. (1999). Gnomon: From Pharaohs to Fractals. Princeton University Press.
Heath, T. (1956). The Thirteen Books of Euclid’s Elements. Vol 1. Dover
Publications.