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than is contained by that figure may be inclosed with the same perimeter. But the area inclosed by a given perimeter cannot exceed a certain limit, which limit, being the greatest possible that can be so inclosed, some figure with the given perimeter must be capable of confaining. Therefore the circle only contains the greatest area of all figures having the same perimeter.

Cor. In the same manner it may be shown that if a figure is to be inclosed by a given perimeter, of which part is to be a given finite straight line, and if it be not made a circular segment of which the given line is chord, a greater may be inclosed with the same conditions, and therefore that of all figures so inclosed the circular segment is the greatest. PROP. 43.

Of all plane figures having the same area, the circle has the least perimeter.

Let the circle C have the same area with any other plane figure F: C shall be contained by a less perimeter than F.

Let C' be a second circle, having the same perimeter with F; then by the last proposition, C' has a greater area than F has, that is, than С has. But the areas of circles (33.) are as the squares of their radii; therefore the radius of C' is greater than the radius of C; and the radii of circles (33.) are as their circumferences; therefore the circumference of C', or perimeter of F, is greater than the circumference of C. Therefore, &c.

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sponding to the first condition. But again, the point required must be equidistant from the two given points, that is, it must be in the straight line which bisects the distance of the two given points at right angles; for this, it is easily seen, (I. 6.) is the locus corresponding to this second condition. Therefore, if this straight line be drawn, and intersect the given line, the point of intersection (or any of those points, if there be more than one) will satisfy both conditions, and will be the point required.

If there be no point of intersection, the problem is impossible.

To take another instance

Here, as far as the first condition only is concerned, any point in the given line, but none else, will answer. The given line is therefore the locus corre

Let it be required "to find a point in a certain plane, which shall be, first, at a given distance from a given point in the plane; and, secondly, at a second given distance from a second given point in the same plane."

Here it is evident that the locus corresponding to the first condition is the circumference of a circle described about the given point as a centre with the given distance as radius: and again, that the locus corresponding to the second condition is the circumference of a circle described about the second given point as a centre with the second given distance as radius. Therefore the points which are common to the two circumferences, that is, their points of intersection, if there be any, will either of them be the point required.

If the circles do not intersect one another, the problem is impossible.

Such is the use of loci in the solution of problems. We have seen also in the above example, that they serve to determine in what cases the solution is possible or impossible. Thus, in the latter example, it will be impossible, if the distances of the point required from the given points differ by more than the mutual distance of those points, or together fall short of that distance: and in the first example it will be impossible, if the given line, being straight, be perpendicular to the line which passes through the two given points, and does not pass through the point which bisects that line; for if it does so pass, the two conditions proposed are identical, and any point in this line will answer them.

Every locus is the limit between excess and defect. The points upon one side of it fail by defect, and those upon the other side by excess, of possessing the required property which is possessed

by every point in the locus. Thus, in the case of the circle, the distance of a point within the circle falls short of the distance of the circumference, while that of a point without exceeds it,

When a locus is merely a straight line, it is called a simple locus; when the circumference of a circle, is called a plane locus; when any other curve, it is said to be of higher dimensions than the circle.

The following propositions afford examples of the two first only; and, the three concluding propositions excepted, they will be found the same in substance with theorems which have been stated before, and which only reappear in this place under a different form.

It will be observed that they are investigated-a species of analytical reasoning which has not hitherto been exemplified either in the demonstration of a theorem or in the solution of a problem, but which is the method commonly pursued in arriving both at the one and the other, and is known under the name of "Geometrical Analysis." The nature of this, as opposed to the ordinary didactic method of solution, commonly called that of synthesis or composition, is pointed out by its name, and will be at once apprehended from any of the following examples. We need only observe that rather than two distinct methods of doing the same thing, as might atrst appear to be the case, they are but the different parts of one full and perfect method; that the use of each is essential to a complete solution; the latter (synthesis) always taking the subjeet up where the other leaves it; the analysis first descending, by geometrical reasoning, from the thing proposed to the minutest particulars of the solution, and the synthesis ascending back through the same steps from these particulars to the thing proposed. It is true, that, beeause the steps in each are for the most part the same, occurring only in an inverted order, the same principles are developed in each, and, therefore, the detail of either (more especially the synthesis) commonly furnishes 2 satisfactory view of the question; for which reason, and for the sake of brevity, one is usually given to the exclusion of the other. The student will, however, find himself amply repaid by entering into both, and he is recommended, after following the analysis of the problems of the present section, to supply in each case where it is omitted, the synthesis necessary to complete the solution,

PROP. 44.

Required the locus of all points which are equidistant from two given points A, B.

Let P be a point in the locus, and join PA, PB. isosceles triangle, if the Then, because PAB is an base AB be bisected in C, PC joined will be at right angles to AB (I. 6. Cor.3). in the straight line which Therefore the point P lies bisects A B at right angles; and it is easily shown, reversely, that every point in this straight line is equidistant from A and B (I. 4.); therefore this straight line is the locus required.

We may observe, that if any point Q A, QA will be less than Q B; if upon lie upon the same side of the locus with the other side, QA will be greater than QB (I. 11.),

PROP. 45.

Required the locus of all points which are equidistant from two given straight lines AB, CD.

If the given
straight lines be
parallel, the re-
quired locus is evi-
dently a straight
line, which is pa-
rallel to each, and
bisects the dis-
tances
them.

between

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107

A

E

PQ

But if not, let them meet in E, and let P be a point in the locus. Then if PA be drawn perpendicular to AB, and PC perpendicular to CD, PA will be equal to PC. Join PE: then, because the right-angled triangles PAE, PCE have a common hypotenuse PE, and equal sides PA, PC, they are equal to one another in every respect (I. 13.), and the angle PEA is equal to the angle PE C. Therefore the point P is in the straight line which bisects the angle A EC: and, reversely, it may be shown that every point in this straight line is equidistant from AB and CD (I. 13.): therefore, this straight line is the locus required.

If any point Q lie upon the same side of the locus with the straight line AB, its distance from A B will be less than its distance from CD; if, upon the other side, its distance from A B will be greater than that from CD.

PROP. 46.

Straight lines being drawn from a given point A to a given straight line BC, required the locus of all points dividing them in the same given ratio

B

Let A B be any straight line drawn from A to BC, and divided in the given ratio in the point. D; and let P be a point in the locus. Then, because AP is to PC in the same ratio as AD to DB, DP joined is parallel to BC (II. 29.). Therefore the point P lies in a straight line drawn through D parallel to BC: and reversely it may be shown that every point in this straight line has the given property (II. 29.); therefore it is the locus required.

If Q be any point on the same side of the locus with A, it will divide the line A C which passes through it, in a less ratio than that of AD to DB: if upon the other side, in a greater ratio.

PROP. 47.

Required the locus of the vertices of all equal triangles, upon the same base AB, and upon the same side of it.

Let A B C be any triangle, upon the given side of the base, and having the given area, and let P be a point in the locus. Then, because the triangle PAB is equal to CAB, PC joined is parallel to AB (I. 27.). Therefore the point P lies in a straight line drawn through C parallel to A B: and, reversely, it may be shown that every point in this straight line has the given property (I. 27.); therefore it is the focus required.

A

B

Let A CB be the given vertical angle, and let P be any point in the locus. Then, because the angle at P is equal to the angle at C,

If Q be any point upon the same side of the locus with AB, the triangle QAB will be less than CAB; if upon the other side, greater.

PROP. 48.

Required the locus of the vertices of all triangles having equal vertical angles, upon the same base AB, and the same side of it.

upon

the point P is in the circular arc passing through C, and having A B for its chord (15. Cor. 3.): and, reversely, it may be shown that every point in this arc has the given property (15.); therefore it is the locus required.

If Q be any point upon the same side of the locus with AB, the angle A Q B will be greater than A CB; if upon the other side, less.

PROP. 49.

Required the locus of the vertices of all triangles upon the same base AB, having the side terminated in A greater than that terminated in B, and the difference of the squares of the sides equal to a given square.

Let P be a point in the locus, and

from P draw PC at right angles to A B, or A B produced. Then, because the difference of the squares of A C, B C is equal to the difference of the squares of AP, BP (I. 38.), the difference of the squares of AC, BC is equal to the given square; and the point C may be found (I. 34.) by taking AD (II. 52.) a third proportional to A B and the side of the given square, so that the rectangle under AB, AD may be (II. 38. Cor.1.) equalto the given square, and bisecting BD in C.* And it may be shown, reversely, that if from the point C so taken, PC be drawn perpendicular to A C, every point in PC will satisfy the given condition; therefore PC is the locus required.

If Q be any point upon the same side of the locus with the middle point of A B, the difference of the squares of QA, QB will be less than the given difference; if upon the other side, greater. For, if a perpendicular QE be drawn from Q to AB, and E F be taken equal to EB, the difference of the squares of Q A, QB will be equal to the rectangle under AB, A F, (I. 39.) which is less or greater than the rectangle under AB, A D, according as the position of Q is one or the other of the two just mentioned.

C

The figure represents the point C in AB produced; if, however, the given square be sufficiently small, the point C may lie between A and B.

PROP. 50.

Required the locus of the vertices of all triangles upon the same base AB,

20

с

having the sum of the squares of their sides equal to a given square. Let the given square be the square of C, and let P be a point in the locus. Bisect AB in D, and join PD. Then, because the base AB of the triangle PAB is bisected in D, the sum of the squares of PA, PB is equal to twice the square of PD, together with twice the square of DA (I.40.) But it is also equal to the square of C. Therefore twice the square of PD is equal to the difference between the square of C and twice the square of AD, that is, if twice the square of D E be equal to the same given difference, to twice the square of DE; and the point P lies in the circumference of a circle described from the centre D with the radius DE. And it may be shown, reversely, that every point in the circumference of this circle satisfies the given condition (I. 40.); therefore it is the locus required.

If Q be any point without the circle, the sum of the squares of QA, QB will be greater than the given sum; if within it, less. For QDs will be greater than PD2 in the former case, and less in the latter; and therefore the sum of the squares of QA, QB will be (I. 40) greater than twice the sum of the squares of P D, DA, that is than the given sum, in the former case, and less in the latter. PROP. 51.

E

A

DEB C

D B

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sum of the angles A PB, BP X, that is to the half of two right angles (I. 2.) or to one right angle. And because D Pd is a right angle, the point P lies in the circumference of a circle described upon the diameter D d. (15. Cor. 3.)

And reversely, if P be any point in the circumference of this circle, PA shall be to PB in the given ratio. For, take C the middle point of Dd, that is, the centre of the circle, and join CP. Then, because AD is to DB as Ad to d B, the line A d is harmonically divided in D and B (II. def. 19, page 68); and because the harmonical mean D d is bisected in C, (II. 46.) CA, CD and CB are proportionals: also, C D is equal to CP: therefore, in the triangles ACP, PCB, AC is to CP as CP to CB; and consequently (II. 32.) the triangles are equiangular. Therefore (II. 31.) PA is to PB as A C to CP, that is, as AC to CD, or (because CA, C D, and CB are proportionals) as AD to DB (II. 22. Cor. 1.).

If any point Q be taken within the locus, QA will be to QB in a greater ratio than that of AD to DB; if without it, in a less ratio. For, if AB be divided in E in the ratio of A Q to Q B, and if AB produced be divided in the same ratio in e; then, joining Q E and Qe, the angle E Qe will be a right angle, as is above shown. And if one of the points E, e lie between D and d, the other will also lie between D and d; for if AE is to EB in a greater ratio than AD to DB, which is the case when E lies between D and d, A e will be to e B in a greater ratio than Ad to dB, which is the case (as may easily be shown) only when e lies between D and d: and conversely. Therefore, if the point Q be within the locus, and the angle DQ d (by consequence) greater than a right angle (15. Cor. 3.), that is, than EQe, the point E cannot lie otherwise than between D and d; and consequently the ratio of A E to E B, that is, the ratio of AQ to QB, must be greater than the ratio of A D to D B. In the same manner it may be shown that, if the point Q lie without the locus, AQ will be to QB in a less ratio.

Cor. If there be taken in the same straight line, and in the same direction from a common extremity, three harmonical progressionals, and if upon the mean progressional for a diameter, a circle be described, the distances of any point in the circumference from the other extremities of the first and third

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shall have to one another always the same ratio, viz. that of the first to the third.

Scholium.

The last proposition may be stated thus: "Required the locus of all points P, the distances of which from two given points A and B, are to one another in a given ratio." And it has been shown that the locus is a circle in every case in which the given ratio is not that of equality; and in that particular case it is (44.) a straight line which bisects AB at right angles. Under this form it readily suggests two other questions of the same kind, which likewise lead to plane loci, and are at the same time so elegant and so nearly related to that we have been discussing, that they claim some notice in this place.

First, then, let it be "required to find the locus of all points P such that the distance PA from a given point A, and the tangent PT drawn to a given circle PCD are to one another in a given

ratio."

KDGE AM

Fig. 2.

Take E the centre of the given circle; join AE; and, if PA is to be greater than PT, produce AE to F (II. 55.) so that AF may be to FE in the duplicate of the given ratio (fig. 1.); but, if PA is to be less than PT, produce EA to F so that AF may be to FE in the duplicate of the given ratio (fig. 2.); take EG (II.52.) a third proportional to EA and ED, and Fig. 1.

A MK

KPQ: this circle shall be the locus required.

DGE

B

For, let P be any point in the circumference of the circle KPQ; join PA, draw the tangent PT to the circle BCD, and join PE, cutting the circumference BCD in L; join also GP, and draw PM perpendicular to AE.

Then, by Prop. 51, because the circle KPQ is described from the centre F with the radius FK, which is a mean proportional between FA and FG, and that P is a point in the circumference KPQ, PA is to PG as AK to KG, or as FA to FK (II. 22. Cor. 1.) because FA, FK and FG are proportionals. Therefore also PA2 is to PG2 as FA is to FK, or as FA to FG (II. 37.). And, because PA2: PG :: FA: FG,

PAX EG: PG2 × AE::FA × EG: FGX AE (Rule 2. Scholium, II. [28].) Therefore, convertendo PAXEG: PG× AE-PAa× EG::FÁ×EG: FG XAE-FAX EG. (a) But, because PA =PE2 +AE2 + 2 AEXEM (I. 37.)

PAX EG PE× EG+ AE × EG± 2AE XEM XEG; and, for the like reason,

=

PG2× AE=PE2 × AE+EGa× AE± 2AEXEM× EG; therefore PGa × AEPA2x AG=PE2x AE-PEa× EG, + PEX EG XAE-AEX EG, that is, = AG-AG - AE × EG.

Again, because FG is equal to FE± EG, FGX AE is equal to FEXAE EGXAE, that is, to FEXAG+FEX EG EGX AG EG, because AE is equal to AG+EG; and, in like manner, because FA is equal to FE EG± AG, FAX EG is equal to FEX EG± EG EG AG; therefore, FG × AE -FAX EG is equal to FEX AG.

Therefore, substituting these values instead of the second and fourth terms of the proportion, (a),

FK a mean proportional (II. 51.) between FA and FG; and, from the centre F with the radius FK describe a circle

PAXEG: PEx AG-AG×AE× EG::FAX EG: FEX AG, and hence, (Rule 2. Scholium, II. [28]).

PA: PE - AEXEG:: FA: FE, that is, because AEX EG is equal to ED (II. 38. Cor. 1.), and PE-ED2 is equal to PE2-ET or PT, PA; PT2:: FA FE. Therefore PA is to PT in the subduplicate ratio of FA to FE, that is in the given ratio; and the circumference KPQ is the locus required.

If the given ratio be the ratio of equality, the difference of the squares of PA, PE will be equal to the square of ED; and therefore the locus is a straight line (49.) cutting AE at right angles, and may be determined as in Prop. 49.

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