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Descartes’ Transform-Solve-Invert Method
Historical Context:
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When: 1637
Where: Leiden, Holland
Who: René Descartes
Mathematics focus: Demonstrate power of Transform-Solve-Invert method
that uses strengths of both geometry and algebra.
Suggested Readings:
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Descartes’ life and contributions to both mathematics and philosophy:
http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Descartes.html
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A. Crombie’s “Descartes” (Scientific American, October 1959) or Chapter 4 in
M. Kline’s (ed.) Mathematics in the Modern World (1968), pp. 33-39
NCTM’s Historical Topics for the Mathematics Classroom (1969): “Analytic
geometry” (pp. 180-182), “Equations and the Ways They were written” (pp.
260-263), and “Algebra in Europe, 1200-1850” (pp. 309-311)
“The ‘Transform—Solve—Invert’ technique” in H. Eves Great Moments in
Mathematics: Before 1650 (1983), pp. 215-227
Descartes The Geometry, an Appendix in his Discourse on the Method (1637).
http://www.philosophypages.com/hy/4b.htm
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C. Boyer’s “Descartes and the geometrization of algebra” (American
Mathematics Monthly (May 1959), pp. 390-392
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Key search words/phrases: Descartes, The Method, analytic geometry,
coordinates, Transform-Solve-Invert, Pappus
Problem to Explore:
Solve a difficult geometry problem using the power of algebra.
Why This Problem is Important:
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Descartes used such problems to demonstrate the power of his method of
“Transform—Solve—Invert,” whereby an algebraic problem is turned into a
geometrical problem in his “new” coordinate geometry, solved as a geometry
problem, and with a solution that can then expressed algebraically.
Was an important step in the establishment of the field of analytic geometry.
Problem Solving Experiences:
1. What is the product LXIII x XXIV ?
This problem illustrates the intuitive need for the Transform—Solve—Invert
Technique. That is, we do not know a multiplication algorithm for Roma Numerals, we
“invert” the problem into base ten notation, “solve” it using the base ten multiplication
algorithm, and then “transform” the answer back into Roman Numerals.
In The Geometry, Descartes poses this problem: Given square ABCD and line BN,
prolong side AC to point E so that EF, laid off from E to EB, shall be equal to BN. The
representative diagram is:
N
A
C
E
F
B
D
The problem has a dynamic nature that suggests “fudging” while role-playing the
problem situation. That is, pretend you are point E moving to the right on ray AC. You
keep looking back at point B (forming the intersection point F on CD), and want to
stop when the length of segment EF equals that of segment BN.
2. Based on different lengths of BN, what are the different but possible
locations of point F?
3. Descartes solved this geometrical problem using algebra. Let BD = DC =
AC = AB = a , BN = FE = c, and DF = x. Descartes’ reasoning was that if
he could find the length of DF = x using algebra, he could “invert” his
solution back as a solution to the geometrical problem. That is, in algebra,
x will be expressed as an algebraic expression, from which the length x
can be constructed using a straight-edge and compass. Find the algebraic
expression for x?
Extension and Reflection Questions:
Extension 1: In the following diagram, CE is parallel to BD.
A
B
C
F
D
E
G
Show that the following are true:
• Triangle ABD is similar to triangle ACE
• AB:AC = AD:AE
• AB:BC = AD:DE
Extension 2: Given two line segments m and n, Descartes interpreted all of the
arithmetical operations—m+n, m-n, mn, m/n, and m —in terms of constructions
using those two segments. That is, to Descartes, the results of arithmetical
operations are line segments as well, not areas or volumes as viewed by the ancient
Greeks, Khayyam, or even Viéta. The three cases mn, m/n, and m satisfy key
proportions. As shown in the diagram, mn is a length such that 1:m = n:mn.
A
1
m
B
C
F
n
D
mn
E
G
Here, If AB = unit length, BC = m, AD = n, and CE parallel to BD, then DE = mn.
Show that 1:m = n:mn, which produces mn as a length of segment DE. Now
construct the corresponding proportions and diagrams showing how m/n, and
m can be constructed as lengths of segments.
Extension 3: After solving the previous problem involving “square ABCD and line
BN,” Descartes then provides another solution of the problem attributed to Pappus of
Alexandria (ca. 290-350 A.D.), one of the last great Greek mathematicians. Pappus’
solution did not rely on algebra, was quite simple, and could be implemented
dynamically easily: Extend BD to point G so that DG = DN. Draw a circle on diameter
BG. Extend AC to find intersection point E on the circle. The resulting diagram is:
N
A
C
E
F
B
D
M
G
Pappus’ also gave a somewhat cryptic proof of his simple construction: Given square
ABDC, point E on AC produced, EG perpendicular at E, meeting BD produced at G,
and F the point of intersection of BE and CD, then CD2 + FE2 = DG2. By construction,
DN2 = BD2 + BN2, which implies that FE = BN. Justify all of the steps in Pappus’
proof, but be forewarned that the claim “CD2 + FE2 = DG2” is a tough step to
decipher.
Open-ended Exploration: Can you generalize Descartes’ solution to the case of
quadrilateral ABDC being a non-square rectangle?