This may be proved independently of the preceding proposition by drawing the altitudespandp'. Then⧍ABC/⧍A'B'C'=cp/c'p'.Butc/c'=p/p',by similar triangles.∴ ⧍ABC/⧍A'B'C'=c2/c'2,and so for other sides.This proof is unnecessarily long, however, because of the introduction of the altitudes.
This may be proved independently of the preceding proposition by drawing the altitudespandp'. Then
⧍ABC/⧍A'B'C'=cp/c'p'.Butc/c'=p/p',
by similar triangles.
∴ ⧍ABC/⧍A'B'C'=c2/c'2,
and so for other sides.
This proof is unnecessarily long, however, because of the introduction of the altitudes.
In this and several other propositions in Book IV occurs the expression "the squareona line." We have, in our departure from Euclid, treated a line either as a geometric figure or as a number (the length of the line), as was the more convenient. Of course if we are speaking of a line, the preferable expression is "squareonthe line," whereas if we speak of a number, we say "squareofthe number." In the case of a rectangle of two lines we have come to speak of the "product of the lines," meaning the product of their numerical values. We are therefore not as accurate in our phraseology as Euclid, and we do not pretend to be, for reasons already given. But when it comes to "squareona line" or "squareofa line," the former is the one demanding no explanation or apology, and it is even better understood than the latter.
Theorem.The areas of two similar polygons are to each other as the squares on any two corresponding sides.
This is a proposition of great importance, and in due time the pupil sees that it applies to circles, with the necessary change of the word "sides" to "lines." It is well to ask a few questions like the following: If one square is twice as high as another, how do the areas compare? If the side of one equilateral triangle is three times as long as that of another, how do the perimeters compare? how do the areas compare? If the area of one square is twenty-five times the area of another square, the side of the first is how many times as long as the side of the second? If a photograph is enlarged so that a tree is four times as high as it was before, what is the ratio of corresponding dimensions? The area of the enlarged photograph is how many times as great as the area of the original?
Theorem.The square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides.
Of all the propositions of geometry this is the most famous and perhaps the most valuable. Trigonometry is based chiefly upon two facts of plane geometry: (1) in similar triangles the corresponding sides are proportional, and (2) this proposition. In mensuration, in general, this proposition enters more often than any others, except those on the measuring of the rectangle and triangle. It is proposed, therefore, to devote considerable space to speaking of the history of the theorem, and to certain proofs that may profitably be suggested from time to time to different classes for the purpose of adding interest to the work.
Proclus, the old Greek commentator on Euclid, has this to say of the history: "If we listen to those who wish to recount ancient history, we may find some of them referring this theorem to Pythagoras and saying that he sacrificed an ox in honor of his discovery. But for my part, while I admire those who first observed the truth of this theorem, I marvel more at the writer of the 'Elements' (Euclid), not only because he made it fast by a most lucid demonstration, but because he compelled assent to the still more general theorem by the irrefragable arguments of science in Book VI. For in that book he proves, generally, that in right triangles the figure on the side subtending the right angle is equal to the similar and similarly placed figures described on the sides about the right angle." Now it appears from this that Proclus, in the fifth centuryA.D., thought that Pythagoras discovered the proposition in the sixth centuryB.C., that the usual proof, as given in most of ourAmerican textbooks, was due to Euclid, and that the generalized form was also due to the latter. For it should be made known to students that the proposition is true not only for squares, but for any similar figures, such as equilateral triangles, parallelograms, semicircles, and irregular figures, provided they are similarly placed on the three sides of the right triangle.
Besides Proclus, Plutarch testifies to the fact that Pythagoras was the discoverer, saying that "Pythagoras sacrificed an ox on the strength of his proposition as Apollodotus says," but saying that there were two possible propositions to which this refers. This Apollodotus was probably Apollodorus, surnamed Logisticus (the Calculator), whose date is quite uncertain, and who speaks in some verses of a "famous proposition" discovered by Pythagoras, and all tradition makes this the one. Cicero, who comments upon these verses, does not question the discovery, but doubts the story of the sacrifice of the ox. Of other early writers, Diogenes Laertius, whose date is entirely uncertain (perhaps the second centuryA.D.), and Athenæus (third centuryA.D.) may be mentioned as attributing the theorem to Pythagoras, while Heron (first centuryA.D.) says that he gave a rule for forming right triangles with rational integers for the sides, like 3, 4, 5, where 32+ 42= 52. It should be said, however, that the Pythagorean origin has been doubted, notably in an article by H. Vogt, published in theBibliotheca Mathematicain 1908 (Vol. IX (3), p. 15), entitled "Die Geometrie des Pythagoras," and by G. Junge, in his work entitled "Wann haben die Griechen das Irrationale entdeckt?" (Halle, 1907). These writers claim that all the authorities attributing the proposition to Pythagoras are centuries later thanhis time, and are open to grave suspicion. Nevertheless it is hardly possible that such a general tradition, and one so universally accepted, should have arisen without good foundation. The evidence has been carefully studied by Heath in his "Euclid," who concludes with these words: "On the whole, therefore, I see no sufficient reason to question the tradition that, so far as Greek geometry is concerned ..., Pythagoras was the first to introduce the theorem ... and to give a general proof of it." That the fact was known earlier, probably without the general proof, is recognized by all modern writers.
Pythagoras had studied in Egypt and possibly in the East before he established his school at Crotona, in southern Italy. In Egypt, at any rate, he could easily have found that a triangle with the sides 3, 4, 5, is a right triangle, and Vitruvius (first centuryB.C.) tells us that he taught this fact. The Egyptianharpedonaptae(rope stretchers) stretched ropes about pegs so as to make such a triangle for the purpose of laying out a right angle in their surveying, just as our surveyors do to-day. The great pyramids have an angle of slope such as is given by this triangle. Indeed, a papyrus of the twelfth dynasty, lately discovered at Kahun, in Egypt, refers to four of these triangles, such as 12+ (3/4)2= (1-1/4)2. This property seems to have been a matter of common knowledge long before Pythagoras, even as far east as China. He was, therefore, naturally led to attempt to prove the general property which had already been recognized for special cases, and in particular for the isosceles right triangle.
How Pythagoras proved the proposition is not known. It has been thought that he used a proof by proportion, because Proclus says that Euclid gave a new style of proof, and Euclid does not use proportion for this purpose, while the subject, in incomplete form, was highly esteemed by the Pythagoreans. Heath suggests that this is among the possibilities:
⧌ABCandAPCare similar.∴AB×AP= (AC)2.Similarly,AB×PB= (BC)2.∴AB(AP+PB) = (AC)2+ (BC)2,or (AB)2= (AC)2+ (BC)2.
Others have thought that Pythagoras derived his proof from dissecting a square and showing that the square on the hypotenuse must equal the sum of the squares on the other two sides, in some such manner as this:
Here Fig. 1 is evidentlyh2+ 4 ⧌.Fig. 2 is evidentlya2+b2+ 4 ⧌.∴h2+ 4 ⧌ =a2+b2+ 4 ⧌, the ⧌ all being congruent.∴h2=a2+b2.
Here Fig. 1 is evidentlyh2+ 4 ⧌.
Fig. 2 is evidentlya2+b2+ 4 ⧌.
∴h2+ 4 ⧌ =a2+b2+ 4 ⧌, the ⧌ all being congruent.
∴h2=a2+b2.
The great Hindu mathematician, Bhaskara (born 1114A.D.), proceeds in a somewhat similar manner. He draws this figure, but gives no proof. It is evident that he had in mind this relation:
h2= 4 ·ab/2 + (b-a)2=a2+b2.
A somewhat similar proof can be based upon the following figure:
If the four triangles, 1 + 2 + 3 + 4, are taken away, there remains the square on the hypotenuse. But if we take away the two shaded rectangles, which equal the four triangles, there remain the squares on the two sides. Therefore the square on the hypotenuse must equal the sum of these two squares.
If the four triangles, 1 + 2 + 3 + 4, are taken away, there remains the square on the hypotenuse. But if we take away the two shaded rectangles, which equal the four triangles, there remain the squares on the two sides. Therefore the square on the hypotenuse must equal the sum of these two squares.
It has long been thought that the truth of the proposition was first observed by seeing the tiles on the floors of ancient temples. If they were arranged as here shown, the proposition would be evident for the special case of an isosceles right triangle.
The Hindus knew the proposition long before Bhaskara, however, and possibly before Pythagoras. It is referred to in the old religious poems of the Brahmans, the "Sulvasutras," but the date of these poems is so uncertain that it is impossible to state that they preceded the sixth centuryB.C.,[79]in which Pythagoras lived. The "Sulvasutra" of Apastamba has
a collection of rules, without proofs, for constructing various figures. Among these is one for constructing right angles by stretching cords of the following lengths: 3, 4, 5; 12, 16, 20; 15, 20, 25 (the two latter being multiples of the first); 5, 12, 13; 15, 36, 39; 8, 15, 17; 12, 35, 37. Whatever the date of these "Sulvasutras," there is no evidence that the Indians had a definite proof of the theorem, even though they, like the early Egyptians, recognized the general fact.
It is always interesting to a class to see more than one proof of a famous theorem, and many teachers find it profitable to ask their pupils to work out proofs that are (to them) original, often suggesting the figure. Two of the best known historic proofs are here given.
The first makes the Pythagorean Theorem a special case of a proposition due to Pappus (fourth centuryA.D.), relating to any kind of a triangle.
Somewhat simplified, this proposition asserts that ifABCisanykind of triangle, andMC,NCare parallelograms onAC,BC, the opposite sides being produced to meet atP; and ifPCis produced makingQR=PC; and if the parallelogramATis constructed, thenAT=MC+NC.ForMC=AP=AR, having equal bases and equal altitudes.Similarly,NC=QT.Adding,MC+NC=AT.If, now,ABCis a right triangle, and ifMCandNCare squares, it is easy to show thatATis a square, and the proposition reduces to the Pythagorean Theorem.
Somewhat simplified, this proposition asserts that ifABCisanykind of triangle, andMC,NCare parallelograms onAC,BC, the opposite sides being produced to meet atP; and ifPCis produced makingQR=PC; and if the parallelogramATis constructed, thenAT=MC+NC.
ForMC=AP=AR, having equal bases and equal altitudes.
Similarly,NC=QT.Adding,MC+NC=AT.
If, now,ABCis a right triangle, and ifMCandNCare squares, it is easy to show thatATis a square, and the proposition reduces to the Pythagorean Theorem.
The Arab writer, Al-Nairīzī (died about 922A.D.), attributes to Thābit ben Qurra (826-901A.D.) a proof substantially as follows:
The four trianglesTcan be proved congruent. Then if we take from the whole figureTandT', we have left the squares on the two sides of the right angle. If we take away the other two triangles instead, we have left the square on the hypotenuse. Therefore the former is equivalent to the latter.
The four trianglesTcan be proved congruent. Then if we take from the whole figureTandT', we have left the squares on the two sides of the right angle. If we take away the other two triangles instead, we have left the square on the hypotenuse. Therefore the former is equivalent to the latter.
A proof attributed to the great artist, Leonardo da Vinci (1452-1519), is as follows:
The construction of the following figure is evident. It is easily shown that the four quadrilateralsABMX,XNCA,SBCP, andSRQPare congruent.∴ABMXNCAequalsSBCPQRSbut is not congruent to it, the congruent quadrilaterals being differently arranged.Subtract the congruent trianglesMXN,ABC,RAQ, and the proposition is proved.[80]
The construction of the following figure is evident. It is easily shown that the four quadrilateralsABMX,XNCA,SBCP, andSRQPare congruent.
∴ABMXNCAequalsSBCPQRSbut is not congruent to it, the congruent quadrilaterals being differently arranged.
Subtract the congruent trianglesMXN,ABC,RAQ, and the proposition is proved.[80]
The following is an interesting proof of the proposition:
LetABCbe the original triangle, withAB LetABCbe the original triangle, withAB 1/2(AB)2= ⧍ABA',and 1/2(BC')2= ⧍BC'C,becauseBC=BC'.∴ 1/2((AB)2+ (BC)2) = ⧍ABA'+ ⧍BC'C.∴ 1/2((AB)2+ (BC)2)= ⧍AC'A'+ ⧍A'C'C (For ⧍ABA'+ ⧍BC'A'+ ⧍A'C'Cis
the second member of both equations.) = 1/2A'C'·AP+ 1/2A'C'·PC= 1/2A'C'·AC= 1/2(AC)2.∴ (AB)2+ (BC)2= (AC)2. The Pythagorean Theorem, as it is generally called,
has had other names. It is not uncommonly called
thepons asinorum(seepage 174) in France. The Arab
writers called it the Figure of the Bride, although the
reason for this name is unknown; possibly two being
joined in one has something to do with it. It has also
been called the Bride's Chair, and the shape of the
Euclid figure is not unlike the chair that a slave carries
on his back, in which the Eastern bride is sometimes
transported to the wedding ceremony. Schopenhauer,
the German philosopher, referring to the figure, speaks
of it as "a proof walking on stilts," and as "a mouse-trap
proof." An interesting theory suggested by the proposition is
that of computing the sides of right triangles so that they
shall be represented by rational numbers. Pythagoras
seems to have been the first to take up this theory,
although such numbers were applied to the right trianglebefore his time, and Proclus tells us that Plato also contributed
to it. The rule of Pythagoras, put in modern
symbols, was as follows: n2+ ((n2- 1)/2)2= ((n2+ 1)/2)2,the sides beingn, (n2- 1)/2, and (n2+ 1)/2. If fornwe put 3, we have
3, 4, 5. If we take the various odd numbers, we haven= 1, 3, 5, 7, 9, ···,(n2- 1)/2 = 0, 4, 12, 24, 40, ···,(n2+ 1)/2 = 1, 5, 13, 25, 41, ···. n2+ ((n2- 1)/2)2= ((n2+ 1)/2)2, the sides beingn, (n2- 1)/2, and (n2+ 1)/2. If fornwe put 3, we have
3, 4, 5. If we take the various odd numbers, we have n= 1, 3, 5, 7, 9, ···,(n2- 1)/2 = 0, 4, 12, 24, 40, ···,(n2+ 1)/2 = 1, 5, 13, 25, 41, ···. Of coursenmay be even, giving fractional values.
Thus, forn= 2 we have for the three sides, 2, 1-1/2, 2-1/2.
Other formulas are also known. Plato's, for example, is
as follows: (2n)2+ (n2- 1)2= (n2+ 1)2.If 2n= 2, 4, 6, 8, 10, ···,thenn2- 1 = 0, 3, 8, 15, 24, ···,andn2+ 1 = 2, 5, 10, 17, 26, ···. This formula evidently comes from that of Pythagoras
by doubling the sides of the squares.[81] Theorem.In any triangle the square of the side opposite
an acute angle is equal to the sum of the squares of the
other two sides diminished by twice the product of one of
those sides by the projection of the other upon that side. Theorem.A similar statement for the obtuse triangle. These two propositions are usually proved by the
help of the Pythagorean Theorem. Some writers, however,
actually construct the squares and give a proof
similar to the one in that proposition. This plan goes
back at least to Gregoire de St. Vincent (1647). It should be observed thata2=b2+c2- 2b'c.If ∠A= 90°, thenb'= 0, and this
becomesa2=b2+c2.If ∠Ais obtuse, thenb'passes through 0 and becomes negative,
anda2=b2+c2+ 2b'c.Thus we have three propositions in one. It should be observed that a2=b2+c2- 2b'c. If ∠A= 90°, thenb'= 0, and this
becomes a2=b2+c2. If ∠Ais obtuse, thenb'passes through 0 and becomes negative,
anda2=b2+c2+ 2b'c. Thus we have three propositions in one. At the close of Book IV many geometries give as
an exercise, and some give as a regular proposition, the
celebrated problem that bears the name of Heron of
Alexandria, namely, to compute the area of a triangle
in terms of its sides. The result
is the important formula Area = √(s(s-a)(s-b)(s-c)), wherea,b, andcare the sides,
andsis the semiperimeter
½(a+b+c). As a
practical application the
class may be
able to find
a triangular piece of land, as here shown, and to measure
the sides. If the piece is clear, the result may be
checked by measuring the altitude and applying the
formulaa= ½bh. It may be stated to the class that Heron's formula is
only a special case of the more general one developedabout 640A.D., by a famous Hindu mathematician, Brahmagupta.
This formula gives the area of an inscribed
quadrilateral as √((s-a)(s-b)(s-c)(s-d)), wherea,b,c, anddare the sides andsis the semiperimeter.
Ifd= 0, the quadrilateral becomes a triangle and we
have Heron's formula.[82] At the close of Book IV, also, the geometric equivalents
of the algebraic formulas for (a+b)2, (a-b)2, and
(a+b)(a-b) are given. The class may like to know
that Euclid had no algebra and was compelled to prove
such relations as these by geometry, while we do it now
much more easily by algebraic multiplication. Book V treats of regular polygons and circles, and
includes the computation of the approximate value of π.
It opens with a definition of a regular polygon as one
that is both equilateral and equiangular. While in elementary
geometry the only regular polygons studied are
convex, it is interesting to a class to see that
there are also regular cross polygons. Indeed,
the regular cross pentagon was the badge of the
Pythagoreans, as Lucian (ca.100B.C.) and an
unknown commentator on Aristophanes (ca.400B.C.)
tell us. At the vertices of this polygon the Pythagoreans
placed the Greek letters signifying "health." Euclid was not interested in the measure of the circle,
and there is nothing in his "Elements" on the value of π.
Indeed, he expressly avoided numerical measures of all
kinds in his geometry, wishing the science to be kept
distinct from that form of arithmetic known to the
Greeks as logistic, or calculation. His Book IV is devoted
to the construction of certain regular polygons,
and his propositions on this subject are now embodied
in Book V as it is usually taught in America. If we consider Book V as a whole, we are struck
by three features. Of these the first is the pure geometry
involved, and this is the essential feature to be
emphasized. The second is the mensuration of the circle,
a relatively unimportant piece of theory in view of thefact that the pupil is not ready for incommensurables,
and a feature that imparts no information that the pupil
did not find in arithmetic. The third is the somewhat
interesting but mathematically unimportant application
of the regular polygons to geometric design. As to the mensuration of the circle it is well for us to
take a broad view before coming down to details. There
are only four leading propositions necessary for the mensuration
of the circle and the determination of the value
of π. These are as follows: (1) The inscribing of a regular
hexagon, or any other regular polygon of which the
side is easily computed in terms of the radius. We may
start with a square, for example, but this is not so good
as the hexagon because its side is incommensurable with
the radius, and its perimeter is not as near the circumference.
(2) The perimeters of similar regular polygons are
proportional to their radii, and their areas to the squares
of the radii. It is now necessary to state, in the form of
a postulate if desired, that the circle is the limit of regular
inscribed and circumscribed polygons as the number
of sides increases indefinitely, and hence that (2) holds
for circles. (3) The proposition relating to the area of a
regular polygon, and the resulting proposition relating
to the circle. (4) Given the side of a regular inscribed
polygon, to find the side of a regular inscribed polygon
of double the number of sides. It will thus be seen that
if we were merely desirous of approximating the value
of π, and of finding the two formulasc= 2πranda= πr2,
we should need only four propositions in this book upon
which to base our work. It is also apparent that even if
the incommensurable cases are generally omitted, the
notion oflimitis needed at this time, and that it must
briefly be reviewed before proceeding further. There is, however, a much more worthy interest than
the mere mensuration of the circle, namely, the construction
of such polygons as can readily be formed by the use
of compasses and straightedge alone. The pleasure of constructing
such figures and of proving that the construction
is correct is of itself sufficient justification for the
work. As to the use of such figures in geometric design,
some discussion will be offered at the close of this chapter. The first few propositions include those that lead up
to the mensuration of the circle. After they are proved
it is assumed that the circle is the limit of the regular
inscribed and circumscribed polygons as the number of
sides increases indefinitely. This may often be proved
with some approach to rigor by a few members of an
elementary class, but it is the experience of teachers that
the proof is too difficult for most beginners, and so the
assumption is usually made in the form of an unproved
theorem. The following are some of the leading propositions of
this book: Theorem.Two circumferences have the same ratio as
their radii. This leads to defining the ratio of the circumference
to the diameter as π. Although this is a Greek letter,
it was not used by the Greeks to represent this ratio.
Indeed, it was not until 1706 that an English writer,
William Jones, in his "Synopsis Palmariorum Matheseos,"
used it in this way, it being the initial letter of
the Greek word for "periphery." After establishing the
properties thatc= 2πr, anda= πr2, the textbooks follow
the Greek custom and proceed to show how to
inscribe and circumscribe various regular polygons, the
purpose being to use these in computing the approximatenumerical value of π. Of these regular polygons two
are of special interest, and these will now be considered. Problem.To inscribe a regular hexagon in a circle. That the side of a regular inscribed hexagon equals
the radius must have been recognized very early. The
common divisions of the circle in ancient art are into
four, six, and eight equal parts. No draftsman could
have worked with a pair of compasses without quickly
learning how to effect these divisions, and that compasses
were early used is attested by the specimens of these
instruments often seen in museums. There is a tradition
that the ancient Babylonians considered the circle of the
year as made up of 360 days, whence they took the circle
as composed of 360 steps or grades (degrees). This tradition
is without historic foundation, however, there being
no authority in the inscriptions for this assumption of the
360-division by the Babylonians, who seem rather to have
preferred 8, 12, 120, 240, and 480 as their division numbers.
The story of 360° in the Babylonian circle seems to
start with Achilles Tatius, an Alexandrian grammarian
of the second or third centuryA.D.It is possible, however,
that the Babylonians got their favorite number 60
(as in 60 seconds make a minute, 60 minutes make an
hour or degree) from the hexagon in a circle (1/6 of
360° = 60°), although the probabilities seem to be that
there is no such connection.[83] The applications of this problem to mensuration are
numerous. The fact that we may use for tiles on a
floor three regular polygons—the triangle, square, and
hexagon—is noteworthy, a fact that Proclus tells us
was recognized by Pythagoras. The measurement of the regular hexagon, given one side, may be used in
computing sections of hexagonal columns, in finding
areas of flower beds, and in other similar cases. This review of the names of the polygons offers an
opportunity to impress their etymology again on the
mind. In this case, for example, we have "hexagon"
from the Greek words for "six" and "angle." Problem.To inscribe a regular decagon in a given circle. Euclid states the problem thus:To construct an isosceles
triangle having each of the angles at the base double
of the remaining one.This makes each base angle 72° and
the vertical angle 36°, the latter being the central angle
of a regular decagon,—essentially our present method. This proposition seems undoubtedly due to the Pythagoreans,
as tradition has always asserted. Proclus
tells us that Pythagoras discovered "the construction of
the cosmic figures," or the five regular polyhedrons, and
one of these (the dodecahedron) involves the construction
of the regular pentagon. Iamblichus (ca.325A.D.) tells us that Hippasus, a
Pythagorean, was said to have been drowned for daring
to claim credit for the construction of the regular dodecahedron,
when by the rules of the brotherhood all credit
should have been assigned to Pythagoras. If a regular polygon ofssides can be inscribed, we may
bisect the central angles, and therefore inscribe one of 2ssides, and then of 4ssides, and then of 8ssides, and in
general of 2nssides. This includes the case ofs= 2 andn= 0, for we can inscribe a regular polygon of two sides,
the angles being, by the usual formula, 2(2-2)/2 = 0,
although, of course, we never think of two equal and
coincident lines as forming what we might call adigon. We therefore have the following regular polygons: From the equilateral triangle, regular polygons of 2n· 3 sides;From the square, regular polygons of 2nsides;From the regular pentagon, regular polygons of 2n· 5 sides;From the regular pentedecagon, regular polygons of 2n· 15 sides. This gives us, for successive values ofn, the following
regular polygons of less than 100 sides: From 2n· 3, 3, 6, 12, 24, 48, 96;From 2n, 2, 4, 8, 16, 32, 64;From 2n· 5, 5, 10, 20, 40, 80;From 2n· 15, 15, 30, 60. Roman Mosaic found at PompeiiRoman Mosaic found at Pompeii Gauss (1777-1855), a celebrated German mathematician,
proved (in 1796) that it is possible also to inscribe a
regular polygon of 17 sides, and
hence polygons of 2n· 17 sides,
or 17, 34, 68, ..., sides, and also
3 · 17 = 51 and 5 · 17 = 85 sides,
by the use of the compasses
and straightedge, but the proof
is not adapted to elementary
geometry.
In connection with the study
of the regular polygons some
interest attaches to the reference
to various forms of decorative
design. The mosaic floor,
parquetry, Gothic windows, and patterns of various
kinds often involve the regular figures. If the teacheruses such material, care should be taken to exemplify
good art. For example, the equilateral triangle and its
relation to the regular hexagon
is shown in the picture
of an ancient Roman
mosaic floor onpage 274.[84]In the next illustration
some characteristic Moorish
mosaic work appears,
in which it will be seen
that the basal figure is
the square, although at
first sight this would not seem to be the case.[85]This is
followed by a beautiful
Byzantine mosaic, the
original of which was
in five colors of marble.
Here it will be seen
that the equilateral triangle
and the regular
hexagon are the basal
figures, and a few of the
properties of these polygons
might be derived
from the study of such
a design. In the Arabic
pattern onpage 276the dodecagon appears
as the basis, and the remarkable
powers of the Arab designer are shown in the
use of symmetry without employing regular figures. Mosaic from DamascusMosaic from Damascus Mosaic from an Ancient Byzantine ChurchMosaic from an Ancient Byzantine Church Problem.Given the side and the radius of a regular
inscribed polygon, to find the side of the regular inscribed
polygon of double the number of sides. Arabic PatternArabic Pattern The object of this proposition is, of course, to prepare
the way for finding the perimeter of a polygon of 2nsides, knowing that ofnsides. The Greek plan was
generally to use both an
inscribed and a circumscribed
polygon, thus approaching
the circle as a
limit both from without
and within. This is more
conclusive from the ultrascientific
point of view, but
it is, if anything, less conclusive
to a beginner, because
he does not so readily
follow the proof. The plan of using the two polygons was
carried out by Archimedes of Syracuse (287-212B.C.)
in his famous method of approximating the value of π,
although before him Antiphon (fifth centuryB.C.) had
inscribed a square (or equilateral triangle) as a basis for
the work, and Bryson (his contemporary) had attacked
the problem by circumscribing as well as inscribing a
regular polygon. Problem.To find the numerical value of the ratio of
the circumference of a circle to its diameter. As already stated, the usual plan of the textbooks
is in part the method followed by Archimedes. It is
possible to start with any regular polygon of which
the side can conveniently be found in terms of the
radius. In particular we might begin with an inscribedsquare instead of a regular hexagon. In this case we
should have and so on. It is a little easier to start with the hexagon, however,
for we are already nearer the circle, and the side
and perimeter are both commensurable with the radius.
It is not, of course, intended that pupils should make
the long numerical calculations. They may be required
to computes12and possiblys24, but aside from this they
are expected merely to know the process. If it were possible to find the value of π exactly, we
could find the circumference exactly in terms of the
radius, since c = 2πr. If we could find the circumference
exactly, we could find the area exactly, sincea= πr2.
If we could find the area exactly in this form, π times a
square, we should have a rectangle, and it is easy to construct
a square equivalent to any rectangle. Therefore, if
we could find the value of π exactly, we could construct a
square with area equivalent to the area of the circle; in
other words, we could "square the circle." We could also,
as already stated, construct a straight line equivalent to
the circumference; in other words, we could "rectify the
circumference." These two problems have attracted the
attention of the world for over two thousand years, but
on account of their interrelation they are usually spoken
of as a single problem, "to square the circle." Since we can construct √aby means of the straightedge
and compasses, it would be possible for us to square
the circle if we could express π by a finite number of
square roots. Conversely, every geometric constructionreduces to the intersection of two straight lines, of
a straight line and a circle, or of two circles, and is
therefore equivalent to a rational operation or to the
extracting of a square root. Hence a geometric construction
cannot be effected by the straightedge and
compasses unless it is equivalent to a series of rational
operations or to the extracting of a finite number of
square roots. It was proved by a German professor,
Lindemann, in 1882, that π cannot be expressed as an
algebraic number, that is, as the root of an equation
with rational coefficients, and hence it cannot be found
by the above operations, and, furthermore, that the solution
of this famous problem is impossible by elementary
geometry.[86] It should also be pointed out to the student that for
many practical purposes one of the limits of π stated
by Archimedes, namely, 3-1/7, is sufficient. For more
accurate work 3.1416 is usually a satisfactory approximation.
Indeed, the late Professor Newcomb stated that
"ten decimal places are sufficient to give the circumference
of the earth to the fraction of an inch, and thirty
decimal places would give the circumference of the
whole visible universe to a quantity imperceptible with
the most powerful microscope." Probably the earliest approximation of the value of
π was 3. This appears very commonly in antiquity, as
in I Kings vii, 23, and 2 Chronicles iv, 2. In the Ahmes
papyrus (ca.1700B.C.) there is a rule for finding the
area of the circle, expressed in modern symbols as (8/9)2d2,
which makes π = 256/81 or 3.1604.... Archimedes, using a plan somewhat similar to ours, found
that π lay between 3-1/7 and 3-10/71. Ptolemy, the great Greek
astronomer, expressed the value as 3-17/120, or 3.14166.... The
fact that Ptolemy divided his diameter into 120 units
and his circumference into 360 units probably shows, however,
the influence of the ancient value 3. In India an approximate value appears in a certain poem
written before the Christian era, but the date is uncertain.
About 500A.D.Aryabhatta (or possibly a later writer of
the same name) gave the value 62832/20000, or 3.1416. Brahmagupta,
another Hindu (born 598A.D.), gave √(10), and
this also appears in the writings of the Chinese mathematician
Chang Hêng (78-139A.D.). A little later in China,
Wang Fan (229-267) gave 142 ÷ 45, or 3.1555...; and
one of his contemporaries, Lui Hui, gave 157 ÷ 50, or
3.14. In the fifth century Ch'ung-chih gave as the limits
of π, 3.1415927 and 3.1415926, from which he inferred
that 22/7 and 355/113 were good approximations, although he
does not state how he came to this conclusion. In the Middle Ages the greatest mathematician of
Italy, Leonardo Fibonacci, or Leonardo of Pisa (about
1200A.D.), found as limits 3.1427... and 3.1410....
About 1600 the Chinese value 355/113 was rediscovered by
Adriaen Anthonisz (1527-1607), being published by his
son, who is known as Metius (1571-1635), in the year
1625. About the same period the French mathematician
Vieta (1540-1603) found the value of π to 9 decimal
places, and Adriaen van Rooman (1561-1615) carried
it to 17 decimal places, and Ludolph van Ceulen (1540-1610)
to 35 decimal places. It was carried to 140 decimal
places by Georg Vega (died in 1793), to 200 by Zacharias
Dase (died in 1844), to 500 by Richter (died in 1854),
and more recently by Shanks to 707 decimal places. There have been many interesting formulas for π,
among them being the following: π/2 = 2/1 · 2/3 · 4/3 · 4/5 · 6/5 · 6/7 · 8/7 · 8/9 · .... (Wallis, 1616-1703)4/π = 1 + 1/2+ 9/2+ 25/2+ 49/2+ .... (Brouncker, 1620-1684)π/4 = 1 - 1/3 + 1/5 - 1/7 + .... (Gregory, 1638-1675)π/6 = √(1/3) · (1 - 1/(3 · 3) + 1/(32· 5) - 1/(33· 7) + ...).π/2 = (logi) /i. (Bernoulli)π/(2√(3)) = 1 - 1/5 + 1/7 - 1/11 + 1/13 - 1/17 + 1/19...,thus connecting the primes.π2/ 16 = 1 - 1/22- 1/32+ 1/42- 1/52+ 1/62- 1/72- 1/82+ 1/92+ ....π/2 =x/2 + sinx+ (sin2x) / 2 + (sin3x) / 3 + .... (0 Students of elementary geometry are not prepared
to appreciate it, but teachers will be interested in the
remarkable formula discovered by Euler (1707-1783),
the great Swiss mathematician, namely, 1 +eiπ= 0. In
this relation are included the five most interesting quantities
in mathematics,—zero, the unit, the base of the
so-called Napierian logarithms,i= √(-1), and π. It was
by means of this relation that the transcendence ofewas proved by the French mathematician Hermite,
and the transcendence of π by the German Lindemann.
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