Important discoveries in plane and solid geometry, consisting of the relation of polygons to circles and the equalizing of perimeters to circumferences and drawing curved lines equal to straight lines, the trisection of an angle and the duplication of the cube

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IMPORTANT

MATHEMATICAL

DISCOVERIES

UC-NRLF

$B 527 TDE

^

By P. D. WOODLOCK

^ r

IN MEMORIAM
FLORIAN CAJO

J

R^faaetaci

IMPORTANT DISCOVERIES

IN PLANE AND SOLID
GEOMETRY

CONSISTING OF
THE RELATION OF POLYGONS TO CIRCLES

AND THE

EQUALIZING OF PERIMETERS TO CIRCUMFERENCES

AND

DRAWING CURVED LINES EQUAL TO STRAIGHT LINES

THE

TRISECTION OF AN ANGLE

AND THE

DUPLICATION OF THE CUBE

By p. D. WOODLOCK

nnn

PRESS OF

e. w. stephens publishing company
columbia, missouri

Copyright 1912
By'-iP. D. Wooi/i-ooK.

PREFACE.

In publishing this book the author feels confident that
he has added something to geometrical science, which has
not heretofore been known.

He especially invites geometricians and mathemati-
cians to examine carefully and without bias, the several
propositions and problems, and their demonstrations, con-
tained in the book, and he has no doubt that they w^ill
find interest in every page.

Some of the problems appearing in this book have occu-
pied the attention of geometricians in all ages since the
introduction of geometry as a science, yet all attempts
at their solution have been unsuccessful down to the pres-
ent time. That the author of this little work has been
rewarded with the discovery of the true solutions of these
problems he confidently leaves to the consideration and
candid judgment of geometricians throughout the world.

iv 270590

PART I.

CONSISTING OF

PRELIMINARY DEMONSTRATIONS

LEADING TO THE

EQUALIZING OF CURVED LINES
TO STRAIGHT LINES.

PRELIMINARY DEMONSTRATIONS.
PROPOSITION A.

To form a series of polygons upon a square or equilateral
triangle having the same center and equal perimeters,
the number of their sides being to each other consecutively
in the ratio of two.

Figure A.

In Figure A, let BCDE be a square, and draw the diag-
onals BD and CE, bisecting each other at A, which is the

8

Important Mathematical Discoveries

center of the square, and draw the Hne AP. Then AP is
the radius of the inscribed, and AC is the radius of the
circumscribed circles. Then bisect the hues AB and AC
at R and L, and draw the lines RF and LH equal to RB
and LC respectively, and parallel to AP, and draw the

Figure B.

lines AW, AX and Ay, bisecting the lines BE, CD, and
DE, and draw RN equal to RB, and parallel to AW, and
draw LI equal to LC, and parallel to AX. And in like
manner bisect the lines AE and AD at G and U, and draw
the lines CM and GO, equal to GE, and draw UK and

Preliminary Demonstrations 9

UJ, equal to UD, and draw the lines RL, LU, UG, and
GR, and draw the lines FH, HI, IJ, JK, KO, OM, MN
and NF. And the figure thus formed is an octagon, whose
perimeter is equal to the perimeter of the square BCDE.
Now the line Cr is perpendicular to LH, and HT is per-
pendicular to LC. And LH being equal to LC, therefore

Figure C.

the line Cr equals HT. In like manner FS = BV, and
FH = Vr. Hence the lines SF + FH + HT = BV + Vr
+ rC = BC, which is a side of the square. Now the lines
SF + FH + HT are equal to one-fourth of the perimeter
of the octagon. And BC is one-fourth of the perimeter

10

Important Mathematical Discoveries

of the square; therefore the perimeter of the octagon FHIJ
is equal to the perimeter of the square BCDE.

In Figure B, let the square BCDE and the octagon
FHIJ have equal perimeters, and produce AP to n, which
bisects FH, and draw the lines AF and AH. Bisect AF

c

H

B

Figure D.

at R, and AH at V, and draw the lines Ra and Rb, each equal
to RF, and parallel to AB and An respectively. In like
manner draw the lines Vc and Vd, each equal to VH, and
draw ab, be, and cd, and draw the line ae, parallel to FS,
and dh parallel to HT. Then ea + ab + be + cd + dh =
SF + FH + HT = BC. And as ea + ab + be + cd + dh
form one-fourth of the perimeter of a polygon of 16 sides,

Preliminary Demonstrations 11

and by completing the remaining sides of that polygon,
its perimeter is equal to the perimeter of the octagon, and
also to that of the square, their sides are to each other
consecutively in the ratio of two. Proceed likewise in
forming polygons of 32, 64, 128, 256, etc., sides, until they
become infinite in number and minuteness.

In Figure C, let BCD be an equilateral triangle, and
bisect its sides at W, N and M, and draw the lines CM,
BN and DW, intersecting each other at the point A, which
is the center of that triangle. And bisect AC at V, AB
at T, and AD at L, and draw VI and TH, parallel to AW,
making VI = VC, and TH = TB. And draw VJ and
LK parallel to AN, and equal to VC and LD. In like
manner draw LE and TF, equal to LD and TB, and draw
the lines HI, IJ, JK, KE, EF, and FH. And the figure
thus formed is a hexagon whose perimeter is equal to that
of the triangle BCD.

Now it will be seen that PH = BS, HI = SR, and 10
= RC. Hence PH + HI + 10 = BC. In like manner OJ
+ JK + KX = CD, and XE + EF + FP = BD. There-
fore their perimeters are equal.

Figure D represents an equilateral triangle, a hexagon,
and a polygon of 12 sides, whose perimeters are equal to
each other, and whose sides are to each other in the ratio
of two. And so on to infinity.

12

Important Mathematical Discoveries

PROPOSITION B.

In a square, an equilateral triangle and in all regular
polygons, the sum of the least and greatest radii, divided
by two, is equal to the least radius of a polygon of twice
the number of sides and equal perimeter.

Figure E.

In Figure E, let the square BCDE, and the octagon
FHIJ have equal perimeters (Proposition A), and we see
that the line AP is the radius of the inscribed circle, and
AC is the radius of the circumscribed circle, and they are
the least and greatest lines that can be drawn from the

Preliminary Demonstrations

13

center A to the perimeter of the square BCDE, and for
convenience we will call them the least and greatest radii
ot that square.

Now the line NI bisects the line AP at d, and AC at L.
Hence dL is parallel to PC, and AL = LC, and LC = LH;

Figure F.

therefore AL = LH, and as LH = dn, hence AL = dn.
Then (AP + AC) h- 2 = Ad + AL = An, which is the
least radius of the octagon FHIJ.

Let Figure F represent a polygon of 16 sides drawn upon
the square and octagon, their perimeters being equal. Then

14 Important Mathematical Discoveries

in the octagon, An and AH are the least and greatest radii,
and it will be seen that (An + AH) ^2 = At; and At is
the least radius of a polygon of 16 sides, whose perimeter
is equal to that of the octagon, or of the square, and so on
until the sides become infinite in number and minuteness.

PROPOSITION C.

If a series of polygons be drawn upon each other, having
the same center and equal perimeters, and the number of
their sides being to each other consecutively in the ratio
of two, the greatest radii of each polygon bisect the angles
formed at the center, by the least and greatest radii of its
next preceding polygon, and the lines joining the outer
points of the greatest radii of any one of these polygons,
with the outer points of the greatest radii of its next suc-
ceeding polygon, are perpendicular to the latter radii.

In Figure J, let the square BCDE and the octagon FHIJ,
have equal perimeters, and let the line be be a side of a
polygon of 16 sides, and or, a side of a polygon of 32 sides,
whose perimeters are equal to that of the square, or octa-
gon, and produce the line An to X, making AX equal to
AB or AC, and draw the arc BXC, and it will be seen that
the lines nH, tc and dr, are each half a side of a polygon
of 8, 16, and 32 sides respectively; and join the points CH,
He, and cr.

■ Now the triangles AnH and ATH are similar, hence the
angles nAH and TAH are equal. And the line AH bisects
the angle nAC, and in like manner the line AC bisects the
angle nAH, and the line Ar bisects the angle tAc, and so
on to infinity. And in the triangle AHC, the line Ac is bi-
sected at L, and LH, LC and LA are equal to each other.
Hence if a circle be described on the line AC as diameter,
its circumference must pass through the point H. There-
fore the angle AHC is a right angle.

In like manner the angles AcH, Arc, etc., are right an-

Preliminary Demonstrations

15

gles, and the lines CH, He, and er are perpendicular to the
bisecting lines AH, Ac, and Ar respectively. Hence if
the angles be bisected indefinitely, and polygons formed,
until the last polygon that can possibly be drawn and re-

Figure J.

tain distinct sides is reached, the lines joining the outer
points of the greatest radii of any one of these polygons,
with the outer points of the greatest radii of its next suc-
ceeding polygon, are perpendicular to the latter radii, and
the rectangle formed by the greatest radii of any one of
these polygons with the least rad.ii of its next succeeding

16 Important Mathematical Discoveries

polygon, is equal to the square of the greatest radius of
the latter polygon.

It will be seen from the foregoing demonstrations that
if an infinite series of polygons be thus formed, the perim-
eter of any one of them intersects each side of all its preced-
ing polygons, in two places, the points of intersection being
equally distant from the middle of the side, no matter
how minute the sides may become, and as a circle is the
ultimate form of a polygon whose sides by the process of
bisection become infinitely minute, the circumference of
that circle will also intersect each side of all preceding
polygons in like manner.

PART 11.

PERIMETERS AND CIRCUMFERENCES
MADE EQUAL

AND

CURVED LINES MADE EQUAL TO
STRAIGHT LINES.

PERIMETERS AND CIRCUMFERENCES.
PROPOSITION I.

To form a circle whose circumference is equal to the
perimeter of a given square.

Let BCDE be a given square (the quadrature of the
circle) and AR its least radius, and AC its greatest radius,

FIJK L

H

X
Figure I.

as in Figure I. And with A as center and AC as radius,
draw the arc BFC, and produce the line AR to F, bisect-

19

20 Important Mathematical Discoveries

ing the arc BFC at F. Then bisect the arc PC at H, and
FH at L, and FL at K, and FK at J, and FJ at I, and so
on, bisecting to infinity, or as far as it is within our means
to bisect. And draw the lines AH, AL, AK, AJ, and AI.

Then from point C draw Ct perpendicular to AH, and
from the point t draw tV perpendicular to AL, and draw
VP, Po, and on, perpendicular to AK, AJ, and AI, respect-
ively, the perpendicular always falling on each bisecting
line, until the least possible bisecting line is reached, which
is represented by AI. Then draw the final perpendicular
nm to the line AF, and it will be seen that nm is half the
side of a polygon, whose perimeter is equal to that
of the square BCDE. (Prop. C, Preliminary Demonstra-
tions.) Then through the point m, with radius Am, draw
the circle SmWX, and the circumference of that circle is
equal to the perimeter of the square BCDE.

Now let the perpendicular nm be the least possible part
or division of a straight line, a mere point, and we see that
it is half the side of a polygon whose perimeter is equal to
that of the square, and whose sides are capable of only one
division or bisection, and as the circumference of a circle,
which is equal to the perimeter of a polygon, both having
the same center, must intersect each side of that polygon
in two places, the points of intersection being equally dis-
tant from the middle of the side, therefore, the circum-
ference SmWX must intersect the perpendicular nm, but
nm being a mere point, cannot be either bisected, inter-
sected, or divided further. Therefore the circumference
SmWX must pass over and coincide with the perpendic-
ular or point nm, which represents half a side of a polygon,
and hence it must pass over and coincide with the remain-
ing half of that side, consequently it must pass over and
coincide with each and every side comprising the whole
perimeter of that polygon, and the point is reached where
the final figure loses its identity as a polygon, and assumes
and contains all the properties of a complete circle, as rep-
resented by the circle SmWX. Therefore, the circumfer-

Perimeters and Circumferences

21

ence of the circle SmWX is equal to the perimeter of the
square BCDE, and the arc bmd equals the line BC.

Now if the line BC = 2, the perimeter of the square
equals 8. Hence the circumference of the circle SmWX
= 8. And as Am, or the radius of a circle whose circum-
ference is 8, is equal to the ratio of a circle to its circum-
scribing square, which is 1.2732395 +. Therefore Am =
1.2732395 -f .

The following table gives the radii of each polygon, from
the square to a polygon of 524288 sides, the perimeter of
each being 8, and the number of their sides being to each
other consecutively in the ratio of 2.

NO. OF SIDES. SHORT RADIUS.

4 1.

8 1.2071067811865475 +

16 1.25683487303146201 +

32 1.26914627894207004 +

64 1.2722167067075638 +

128 1.2729838511604253 +

256 1.2731756083078977 +

512 1.2732235458896797 +

1024 1.27323553034872029 +

2048 1.2732385264073428 +

4096 1.2732392754215578 +

8192 1.2732394626770475 +

16384 1.27323950948983831 +

32768 1.27323952119303589 +

65536 1.2732395233334371 +

131072 1.273239524045252 +

262144 1.2732395242281144 +

524288 1.27323952427383 +

262144 1.2732395242281144 +

524288 1.27323952427383 +

LONG RADIUS.

1.414213562373095 +
1.3065629648763765 +
1.28145768485268807 +
1.27528713444730576 +
1.2737509956132868 +
1.2733673656153701 +
1.2732714833914617 +
1.2732475148077608 +
1.2732415224659654 +
1.2732400244357728 +
1.2732396499325372 +
1.27323955630262909 +
1.2732395328962334 +
1.2732395254738383 +
1.2722395247570668 +
1.2732395244109769 +
1.2732395243195456 +

1.2732395243195456 +

From the above we find the ratio between diameter and
circumference is 3.1415927 -f.

22

Important Mathematical Discoveries

PROPOSITION II.

To form a circle whose circumference is equal to the
perimeter of a given equilateral triangle.

In Figure II, let BCD be a given equilateral triangle,
and draw the perpendicular CE, and let A be the center

^

Figure II.

of that triangle. Then AC is its greatest radius. Then
draw AF perpendicular to BC, and the line AF is its least
radius. Then produce AF to L, making AL = AC. Then

Perimeters and Circumferences 23

with A as center draw the arc LC, and draw the bisecting
lines AH, AI, AJ, AK, etc., to infinity, or as far as it is pos-
sible to bisect. Then from the point C draw Cr perpendic-
ular to AH, and rP perpendicular to AI, and po to AJ, and
on to AK, etc., and draw the final perpendicular to the
point m, on the line AL. And with A as center and Am
as radius draw the circle mS, and the circunference of that
circle is equal to the perimeter of the triangle BCD. (See
Demonstration of Proposition I, Quadrature of Circle.)

PROPOSITION III.

To determine the Quadrature of the Circle independ-
ently of all infinite or unlimited bisections.

In Figure III, let BCDE be a given square and A its
center, and draw the lines AF and AD, and draw DH form-
ing a right angle with AD, and produce AF to H. Then
on the line AH, as diameter, draw the circle HDAC. And
bisect the line HD at P, and draw FJ, which intersects
HD at P, and draw PA. And on the line AH, make AO
= AP, and draw PL at right angles with AH. Then on
the line AO draw the semicircle OnA, and draw the lines
On and nA. Then from the point D draw the line Dm
perpendicular to AP, and draw mR parallel to FD. And
on the line On make nX = FR, and draw AX. Then with
A as center and AX as radius, draw the circle XWYS, and
the circumference of that circle is equal to the perimeter
of the square BCDE.

Now let the Hne CD = 2. Then FD = 1. And AF =
1, and FH = 1, and AD and DH are equal to each
other, and each is equal to i/^ 2 , and Dp = half the
i/y = .70710678+ and AD^ +JDP = AP- = 2.5. Hence
A02 = 2.5. Therefore AO = i/^2l = 1.58113883 + . Now
AO X AF = An2. But AO X AF = AO X 1 = AO. There-
fore, An- = AO = 1.581 13883 + . Now in the right-
angled triangle ADP, PA X Am = AD^ = 2. Hence PA

0

24

Important Mathematical Discoveries

X Am^ = 4. And as PA- = 2.5, then 4 -- 2.5 = Am^.
Hence Am^ = 1.6, and AR^ = 1.44, and AR = 1.2. Then
FR = .2, and hence nX = .2. Now An^ = 1.58113883 + ,

H

y

Figure III.

and nX2 = .04. Hence AX^ = 1.58113883 + , + .04 =
1.62113883 + , and Ax = i/-" 1.62 11 3883+ = 1. 2732395 + ,
which is the ratio of a circle to its circumscribing square.

Perimeters and Circumferences 25

and is the radius of tlie circle XWYS. Hence the circum-
ference of that circle is equal to the perimeter of the square
BCDE. And 4 -- 1.2732395+ = 3.141592 + , which is the
ratio of diameter to circumference.

From the foregoing demonstration (Figure III) the fol-
lowing rule is obtained, viz.:

Multiply the square of the radius of the inscribed circle
by the i^ 2^ , and add to the product the one-hundredth
part of the area of the square, and the square root of the
sum is the radius of the circle whose circumference is equal
to the perimeter of the given square.

It will also be seen that to draw a circle whose circum-
ference is equal to the perimeter of a given polygon, bring
that perimeter into the form of a square, and proceed as
in Figure III or Figure I, Part II.

26

Important Mathematical Discoveries

PROPOSITION IV.

To find the number of degrees in any angle of a given
triangle.

In Figure IV, let the angle BAC be a given angle, and it
is required to determine the number of degrees in it. Draw
the lines AB and AC equal to each other and draw BC,
and draw the arc niHn equal to the line BC. Then bisect

R

C

angle BAC by the line AH, which is the radius of the cir-
cle of which the arc mHn is a part. Then find the circum-
ference of that circle, and divide by the arc mHn or the
line BC, and then divide the result into 360 for the degrees
in said angle.

PART III.

THE TRISECTION OF AN
ANGLE.

THE TRISECTION OF AN ANGLE

PROPOSITION I.

In any triangle, the sum of any two of its sides, is to the
third side as either one of those two sides is to that corres-
ponding part of the third, cut off by a line bisecting the
angle which the said two sides contain.

o

Figure A.

Let BAC (Figure A) be a triangle, and produce BA to
E, making AE = AC, and draw EC, and draw AP par-
allel to BC. Then draw AD parallel to EC, and AD bi-
sects the angle BAC, as will be seen by the following dem-
onstration :

29

30

Important Mathematical Discoveries

AE = AC, then BE = BA + AC, and the angle AEC =
ACE, and as the angle BAC = angles AEC + ACE, there-
fore the angle BAC = 2 ACE, and as AD is parallel to
EC, the angles DAC and ACE are equal. Hence the

Figure B.

the angle BAC = 2 DAC, and the line AD bisects the
angle BAC. Now BE = BA + AC, and in triangles BEC
and BAD, BE : BC : : BA : BD, and BE : BC : : AE : AP.
Hence BE : BC : : AE : DC. Therefore, BA + AC : BC
: : BA : BD, and BA + AC : BC : : AC : CD.

The Trisection of an Angle

31

PROPOSITION II.*

In Figure C, let ABC be a right-angled isoceles triangle
and it is required to trisect the right angle ABC. Bisect
the angle ABC by the line BD, and bisect and angle BAC
by the line AP, and draw BH = BP, and with A as center
and AH as radius, draw the arc HS, meeting the line DB

J

yn

D ^

Figure C.

K

C

produced, at the point S, and draw AS and CS. Then
draw the line SJ bisecting the angle DSA, and draw SK
bisecting the angle DSC, and draw Bm and Bn parallel
to SJ and SK respectively, and the lines Bm and Bn tri-
sect the angle ABC.

*The trisection of acute and obtuse angles will appear in the next
issue of book.

32 Important Mathematical Discoveries

Now in triangle ABC, let the line Ac = 2. Hence DB,
DA and DC each equals 1, and AB and CB each equals
i/T. And as the line AP bisects the angle BAC, it will be
seen that the line BP = .58578 + , and as BP = BH, there-
fore the line AH = (i/T+ .585^8 + ) = 2. Hence the line
AS = 2, and the line SD = i/ 3 , and the triangle ASC is
equilateral, and the angle ASC is 60 degrees, the angle
ASD is 30 degrees, and the angle JSD = 15 degrees. And
as the dine Bm is parallel to SJ, hence the angle mBD is
15 degrees. And as the angle ABD = 45 degrees, there-
fore the angle niBD = one-third of angle ABD. In like
manner the angle nBD = one-third of angle CBD. Hence
the angle niBN is one-third of the angle ABC.

In all cases an isoceles triangle must be formed, the angle
to be trisected being the vertical angle, as in Figure C.

PART IV.

THE QUADRATURE AND DUPLICATION
OF THE CUBE.

QUADRATURE AND DUPLICATION
OF THE CUBE.

PROPOSITION I.

In Figure I, let ABCD be a square side of a given cube,
and on the line AB draw the semicircle BOA, and let BE
= 1, and draw the lines EFP, FB and FA. Then AB X

T

5

D

P

/ /
/ X
1 X
1 X

/

/
/

/

"H

vA

R

c

A
Figure I,

L

BE = FB- = AB. Then make F^ = BF, and draw LJ.
Then AB X BL = BJ^ = BH^, and draw HS at right angles
with BH, and draw the square SBRT, and the square

.^5

36

Important Mathematical Discoveries

SBRT = AB^ of which the square ABCD is one of the
square sides.

F P D

Figure II.

Now BE = 1, then AB X BE^BF^ = AB. And BF

= j/AB, and therefore BL = i/AB. Then AB X BL -

i/AB^. But AB X BL = BJ-' = BH^ = BS. Hence BS

= i/AB' , consequently BS^ = AB^. Hence the square

SBRT = AB\

Quadrature and Duplication of the Cube 37

PROPOSITION II.

To bring a slab one inch thick into the form of a cube,
whose soHd contents are equal to the solid contents of

said cube.

In Figure II, let ABYX be a square slab one inch thick,
whose solid contents = 27. Then the line AB = i.^27 =

H F ^

Figure III.

^ E

5.196 + . Now let BD = 1, then AB X BD = BC^ (the
semicircle BSA being drawn). Then make BE = BC, and
draw the lines EL, LB and CA. Then make BP = i/ BE,

38

Important Mathematical Discoveries

and it is also the 4th root of AB. Then AB X BP = BR^.
And make the angle LBS = two-thirds of the angle LBR,
and draw SF, then the line BS is the cube root of the slab
ABYX, and is therefore the cube root of 27, which is 3.

A

^ H E F P D

Figure IV.

Then make BW = BS, and on BW draw the semi-cir-
cle BmW, and draw the lines Bm and niW. Now AB X
BF = BS2 = BW2 = 9. Hence 9 ^ 5.196+ = i 3",+
then BF = 1.732+ and = i/BW. Hence BW is the re-
ciuired line representing the edge of the cube which is equal
to the given slab; and BF^ = AB. Hence any rectangu-
lar slab may be brought into a cube.

Quadrature and Duplication of the Cube 39

Or, let Figure III be a reproduction of Figure II, then
BA multiplied by its square root AF = AL-. Then make
An = the 8th root of BA. Then BA X An = AP. And
the difference between the angles thus formed at A, which
is the angle LAI, bisect by line AR, and draw the line Rm.
Then Am is the cube root of AB.

This method is different from that of Figure II, and has
the advantage of dispensing with trisections.

PROPOSITION III.

To form a cube whose solid contents are twice the solid
contents of a given cube, and vice versa.

In Figure IV, let the square BHVK be a side of a given
cube, and let BJTO represent a square slab one inch thick,
and equal to the contents of the given cube. Then pro-
duce BJ to A, making BA = to the diagonal of the square
BJTO, and form the square ABYX, and it will be seen
that the square ABYX is twice the square BJTO. Then
bring the square ABYX into a cube as represented by the
square side BHtb.

Now the square ABYX is twice the square BJTO. It is
therefore twice the cube BHVK, and the square ABYX
being equal to the cube represented by the square side
BHtb, hence the cube BHtb is twice the cube BHVK.

It will be seen that a cube may be formed whose solid
contents are any given number of times greater than a given
cube.

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