Sidebilder
PDF
ePub

ADDITIONAL NOTES.

1. To cut any right line bd harmonically in a given ratio.

Fig. 2.

Draw any right line mb and make it to ds in the given ratio, then make bn = bm and draw the right lines msa, ncs, then evidently ba: ad : :bc: cd.

2. Drawing a tangent a t from any point in the produced diameter of a semicircle, and a perpendicular tc, the line ba is cut harmonically. Fig. 3.

11 For ba : ad :: bv : do :: vt: to :: bc: cd.

3. Between the quantities ba and ad, ca is the harmonic mean, at the geometric and am the arithmetic; but the a mta is a right 2d A, .. a t is a geometric mean between ma and ca.

4. If any point be joined with the several points of harmonic section, by the right line pb, pc, pd, pa, any line b' d intersecting them in any manner is also cut harnically

DRUGI Tom Fig. 4.

191 lm For through d, d draw mdn, m'd'n' parallel to pb; then by hypothesis ba : ad::bc: cd, but ba : ad :: bp: dm and bc: cd :: bp : dn .. dm dn... = d'n' .. Up: d'n' :: b'p : d'm'; .. B'a': a'd' :: b'c': 'd'.

5. The right lines pb, pc, pd and pa are called harmonicals.

TRUL P990ine S 6. The consequence mentioned in Note 4, is evidently true of any right line intersecting the production of the harmonicals beyond the point of concourse p. 231193

TO PROP. XXXIII. 15: 37. To determine the peripheries and areas of circles. 10

hris Let it be assumed that the arc of a circle is less than the sum of the extreme tangents, and greater than the chord. Hence the periphery of the circumscribed polygon is always greater, and of the inscribed polygon less than that of the circle, however the number of sides be multiplied.

[graphic]

1

8. If a right line be drawn cutting the sides of a A in any way, above the base and below the vertical <; the sum of the sides is greater than the sum of the segments, towards the base + the drawn line.

9. Hence by drawing tangents or chords to the middle points of the intercepted arcs, the number of the sides of either the inscribed or circumscribed polygons is doubled, but their perimeters are made to approach nearer to that of the circle.

19 OR 10. By continuing this bisection indefinitely, the perimeters of the polygons can be made to differ from each other, and .; from the periphery of the circle by a quantity less than any assignable.

11. If regular polygons be inscribed or circumscribed to two circles, their perimeters are obviously to one another componendo) as the radii of the circles.

12. The circumferences of circles are to each other as their radii. For let C C be the circumferences and r,r' the radii; then if possible let r:r::C: C'+d; let P,P be the perimeters of the circumscribed polygons, at the time that P'is nearer to C" than by the quantity d, then C: C'+:::P:P, but Pis less than C+0. Pis less than C, which is absurd. Again let if possible r :r'::C: C'—d, and let p,p' be the perimeters of the inscribed polygons; at the time that p, is nearer to C than by d; then C : C'

8::p:p', but p is greater than C-, . p is greater than c, which is absurd. Hence r:r::C:C.

13. If yv=l and C=2x, then C=2*r; the numerical value of has been found to 127 places of decimals; as a sufficiently near value however we may take = 3,14159.

14. Angles A, A' are to one another directly as their subtenses a a', and inversely as their radii, for A: 4rt. <s:: a : C, and 4rt. <s: A'::C: a .. ex æquo A; A':: s a: a'2 fa: a'2 C:

{rirs 15. Similar polygons inscribed or circumscribed to two circles, have their areas to each other as the squares of the radii (19. 6.)

BOU Such polygons can have their sides multiplied, so that their areas may differ by a quantity less than any assignable from each other, and therefore from those of the circles.

3 most

[graphic]
[ocr errors]

Pr

16. Hence by the same argument that has been used for the peripheries, it appears that the areas of circles are as the squares of the radi. 17. The area of the circumscribed polygon =

2 since the area of a circle can be made to differ from it by a quantity less than any assignable, the area of the circle = * 72.

ON PORISMS.

Something still remains to be said on a very abstruse class of questions called Porisms, which are not very easily to be distinguished from Theorems and Problems.

Euclid wrote three books on them, which however have not come down to modern times. The following account of their origin and true meaning is taken from a Memoir by Playfair, in the third vol. of Edinb. Trans.

« The ancient Geometers arrived perhaps at all geometrical truths in their attempts to solve problems.

A problem was not considered solved till all the cases of it were separately discussed. This discussion soon let them see that there were circumstances in which the solution would cease to be possible, and this they perceived to happen in consequence of data becoming inconsistent.

Such instances must frequently occur in solving the simplest problems, but in the Analysis of more complex, cases occurred in which constructions failed for a directly contrary reason. Instances would be found when lines, instead of intersecting so as to afford definite solutions, or of not meeting at all so as to afford impossible ones, would actually coincide, and leave the question of course unresolved. The confusion thus arising would soon be eleared up, by observing that a problem before determined by the intersection of two lines, would now become capable of an indefinite number of solutions. This was soon perceived to arise from two parts of the requisite data becoming one; and the very curious propositions thus resulting were admitted as forming a class apart, in as much as they admitted of distinct

enunciation, from Theorems and Problems, under the denomination of Porisms.

This transition of a Problem into a Porism, may be clearly illustrated by the following example. “Given & circle ABC,

fig. 5, a line DE and a point F, to find a point G in DĚ such that a tangent GC=GF.” By supposing it done, then constructing as appears from the fig. ; DK=DK'=DL is .. given; hence find the point on the line that will be the centre of the circle through KFK' &c. The problem is impossible, and .. fails, if F be on the line.

If F and K coincide, it becomes true of every point in the line, and .. the question may be proposed in a porismatic form, viz. “ Given a circle and a line; a point may be found such that the distance from it to any point in the line may be equal to the tangent to the same point.”

These propositions most frequently may be enunciated either as Theorems or loci, but however they are quite distinct. The ancients sometimes considered them as local Problems, but that they are distinct will appear by the following example. “ Required to draw a line from a given point G, fig: 6, intersecting a triangle ABC, so that the perpendicular from one angle may be equal to those from the others." Let GS be the drawn line, take H the middle point of AC, then if the perpendicular from B=those from A and C it is equal to twice that from H. .:: BH is bisected at S, .- the point S is given. If G and S coincided, the question would be porismatic, and could not be enunciated as a locus.

From this account of the origin of Porisms they may be defined. “ Propositions asserting the possibility of finding such conditions as will render a certain Problem indeterminate or capable of several solutions."

We may now proceed to the enunciation of questions. They contain, besides some that perhaps have not heretofore appeared, a collection of the best that could be procured from the most noted geometrical works, and from the

English Diaries and Repositories for many years back. They only require on the part of the Student & thorough knowledge of the Elements, and doubtless some portion of that δυναμις αναλυτικη so much prised by the Ancients.

In their arrangement a mutual dependance has been considered as much as the difficulty of the subject would allow. Wherever a figure is alluded to in these questions, the reader will always be able to construct it as directed.

THEOREMS.

[graphic]

ᅡ f

T
T

1. It may be proved by the first Book that the perpendiculars from the middle points of the sides of a in-3 tersect in one point, hence prove the same of the perpendiculars from the <s.

II. In the same A, the points of occourse of the perpendiculars from thes, of the bisectors of the sides from the <s and of the perpendiculars from the middle points of the sides, lie in directum. (This may be proved by the 1st Book.)

III. If from any point either inside or outside a rectilineal figure, perpendiculars be let fall on all its sides, + the sum of the O’s of the alternate segments made by them will be equal.

oiburg IV. If in any trapezium two opposite sides be bisected, the sum of the O's of the other two sides, together with

T the O’s of the diagonals, is equal to the sum of the O's of the bisected sides, together with four times the O2 of the line joining those points of bisection.

V. Drawing that diameter of the circumscribing 0 that bisects the base, the extremity of it that is next the base, is equidistant from the extremity of the base, the ! centre of the inscribed 0 and the centre of the that touches the base and sides produced.

VI. In the isosceles a ABC having AB = BC, if the point D be taken in AB, and D C joined. The solid + under AC and BD is the difference of the solids under AB, DCand AD?.

VII. The perpendicular from the vertex on the base of an equilateral 4, is equal to the side of an equilateral A inscribed in a 0, whose diameter is the base.

VIII. If on the sides of a A segments of Os be described similar to a segment on the base, and from the extremities of the base tangents be drawn intersecting their circumferences; the points of intersection and the vertex of the A will be in directum.

« ForrigeFortsett »