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 Archimedes’ Squaring Of A Parabola
Archimedes’ Squaring Of A Parabola
Speech
Martin Wiesauer, Bernhard Engl
Archimedes’
Squaring Of A Parabola
To determine the area
enclosed in a parabola section
The squaring of a parabola is one of Archimedes’ most remarkable
achievements. It was accomplished about 240 b.C. and is based upon the
properties of Archimedes triangle.
An Archimedes triangle is a triangle whose sides consist of two tangents to
a parabola and the chord connecting the points of tangency. The last mentioned
side is taken as the base line or the bas of the triangle. In order to construct
such a triangle we draw the parallels to the parabola axis through the two
points H and K of the directrix and erect the perpendicular bisectors upon the
lines connecting H and K with the focus F. If we designate the point of
intersection of the two perpendicular bisectors as S, the point of intersection
of the first perpendicular bisector with the second parallel to the axis as B,
then A and B are points of the parabola (classical construction of the
parabola), and ASB is an Archimedes triangle.
Since SA and SB are two perpendicular bisectors of the triangle FHK, the
parallel to the axis through S is the third perpendicular bisector; it
consequently passes through the center of HK, and, as the midline of the
trapezoid AHKB, it also passes through the center M of AB. This gives us the
theorem: The median to the base of an Archimedes triangle is a parallel to the
axis.
Let the parabola tangents through the point of intersection O of the median
SM to the base with the parabola cult SA at A´, SB at B´. Then
AA´O and BB´O are also Archimedes triangles. Consequently according to
the above theorem, the medians to their bases are also parallel to the axis and
are therefore midlines in the triangles SAO and SBO, so that A´ and B´
are the centers of SA and SB. A´B´ is consequently the midline of the
triangle SAB and is therefore parallel to AB; also the point O on A´B´
must be the center of SM.
The result of our investigations is the
Theorem of
Archimedes: the
median to the base of an Archimedes triangle is parallel to the axis, the
midline parallel to the base is a tangent, and its point of intersection with
the median to the base is a point of the parabola.
Now we can determine the area J of the parabola section enclosed in our
Archimedes triangle ASB with the base line AB.
The tangents A´B´ and the chords OA and OB devide the triangle
ASB into four sections: 1. The “internal triangle” AOB enclosed
within the parabola; 2. The “external triangle” A´SB´
lying outside the parabola; 3. And 4. Two “residual triangles”
AOA´ and BOB´, which are also Archimedes triangles and are penetrated
by the parabola.
Since O lies at the center of SN, the internal triangle is twice the size
of the external triangle.
In the same fashion, each of the two residual triangles intern gives rise
to an internal triangle, an external triangle and two new residual Archimedes
triangles that are penetrated by the parabola, and ones again each internal
triangle is twice the size of the corresponding external triangle.
Thus, we can continue without end and cover the entire surface of the
initial Archimedes triangle ASB with internal and external triangles. The sum of
all internal triangles must also be twice as great as the sum of external
triangles. In other words:
Theorem of
Archimedes: the
parabola divides the Archimedes triangle into sections whose ratios
2:1.
Or also:
The area enclosed by a
parabola section is two thirds the area of the corresponding Archimedes
triangle.
Archimedes arrived at this conclusion by a somewhat different method. He
found the area of the section by adding together the areas of all the successive
internal triangles.
If Δ represents the area of the initial
Archimedes triangles ASB, then the area of the corresponding internal is one
half Δ, the area of the corresponding external
triangle is one quarter of Δ, and the area of
each of the two residual triangles is one eighth of
Δ.
The successive Archimedes triangles therefore have the areas
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
the corresponding internal triangles possess half this area; and since each
internal triangle gives rise to two new internal triangles, we thus
obtain for the sum of all the successive internal triangle areas the
value.
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
The bracket encloses a geometrical series with the quotient ¼, the sum
of which is equal to
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
Thus we again obtain for the area of the section the value
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
Since A´B´ is tangent to the parabola at O, the perpendicular h
dropped from O to the base line AB of the section is the altitude of the
section. Since h is also half the altitude of the triangle ASB,
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
and
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!},
i.e.:
The area enclosed by a
parabola section is equal to two thirds the product of the base and the altitude
of the section.
Finally, we will express the area of the section in terms of the transverse
q of the section, i.e., by the projection normal to the axis of the chord
bounding the section.
We use the equation for the amplitude of the parabola, calling the
coordinates of the corners of the section xy and
XY, and we have
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
with 2p representing the parameter. From the given figure it follows
directly that
_{Originaldokument enthält an dieser Stelle eine Grafik!Original document contains a graphic at this position!}
If we replace X and x here with Y^{2}/2p and y^{2}/2p, we
obtain
12pJ=Y^{3}y^{3}3Y^{2}y+3Yy^{2}=(Yy)^{3}.
Since Yy is the section transverse q, we finally obtain
12pJ=q^{3}.
This important formula can be expressed verbally as follows:
Six times the product of
the parameter and the area of the section is equal to the cube of the section
transverse.
