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mon perimeter to ab as four right angles the circle (II. 35.), that is, as its perito the angle acb, ex æquali (II. 24), meter to the perimeter of the polygon in AB is to ab as the angle ACB to the question, or (II. 17.) as its side to the angle acb; and hence, by taking the side of the latter. Again, because the halves, (II. 17. Cor. 2.) the line AD is to polygons are similar (II. 43.), they are a D as the angle ACD to ac D, or (I. one to another in the duplicate ratio 15.) 0 C D. Therefore, dividendo (II. of their sides. Therefore, the circum20.) A a is to a D as the angle A Co to scribed polygon has to the other the OCD, and invertendo (II. 15.) Da is duplicate ratio of that which it has to to a A as the angle D Co to o C A. the circle, and (II. def. 11.) the circle is Again, because the sector Cno is

a mean proportional between the two greater than the triangle C Do, and the polygons.* But the circumscribed polysector Com less than the triangle CoA, gon is greater than the circle: therefore, the sector Cno has, on both accounts, the circle (II. 14.) is greater than the (II. 11.) to the sector Com a greater polygon of equal perimeter. ratio than the triangle C Do to the tri- Cor. A circle is greater than any angle CoA: but the angles DC 0,0 CA plane rectilineal figure of the same perihave the same ratio as the sectors meter (37. Cor.). (13.), and the lines D o, o A the same ratio as the triangles (II. 39): therefore the angle D Co has to the angle o CA

It may be inferred also, from the forea greater ratio than the line Do has to going propositions, that the circle is not the line o A (II. 12. Cor. 1 and 2.). And

less than any curvilineal figure of the it was shown, that Da has to a A the

same perimeter. For there may be insame ratio which the angle D Co has to

scribed in the latter a rectilineal figure the angle o CA: therefore Da has to

of less perimeter, yet approaching a A a greater ratio than Do to o A, and

more nearly to it in area than by any Da is greater than Do. And, because supposed excess of the original figure Co is parallel to ca, Dc:DC :: Da:

above the circle, so that, were there Do (II. 29.); therefore Dc is also such an excess, a rectilineal figure greater than D C.

might be found greater than the circle, But the polygons, being equal, each of and yet of less perimeter, which is imthem, to half the rectangle under the possible. This method will not, howapothem and perimeter (29.), are to ever, carry us any further. In the folone another as their apothems'c D, CD lowing propositions, another view is (II. 35.): therefore, the polygon which taken of the subject, and by them it will has the apothem cD and side ab is

be made to appear, that the circle is greater than the other.

greater than any other figure, curviliTherefore, &c.

neal or otherwise, which has the same Cor. A regular polygon is greater perimeter. than any other rectilineal figure having

PROP. 39. the same perimeter, and the same or a If two triangles have two sides of the less number of sides (36. Cor. 2.). one equal to two sides of the other, each PROP. 38.

to each, and the angle contained by the

two sides of the first a right angle, but A circle is greater than any regular the angle contained by the two sides of polygon having the same perimeter. the other not a right angle, the first tri

For let a siinilar polygon be circum- angle shall be greater than the other. scribed about the circle, viz. by dividing Let ABC, the circumference (or conceiving it to be DEF be two divided) into as many equal parts as the triangles which polygon is to have sides, and drawing have the two tangents through the points of division. sides A B, BC Then, because the area of the polygon is of the equal (29.) to half the rectangle under its apothem and perimeter, and the area

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B

* It appears from this part of the demonstration,

that if there be two similar polygons, of which one is of the circle to half the rectangle un- circumscribed about a circle, and the other has the der its radius and perimeter (32.); and

same perimeter with the circle, the circle shall be a

meun proportional between the two polygons: a prothat the apothem of the circumscribed position which is true, whether the polygons be repolygon is equal to the radius of the gular or irregular. For whether the circumscribed circle, the circumscribed polygon is to

polygon be regular or otherwise, it is evident that

its area is equal to half the rectangle under its perithe circle as its perimeter to that of meter and the radius of the circle.

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equal to the two sides DE, EF of the diameter, incloses the greatest possible other, each to each, but the angle ABC area, and the quadrilateral ABCD is a right angle, and the angle DEF greater greater than EFGH.* or less than a right angle; the triangle ABC shall be greater than the triangle

Otherwise. DEF.

Let ABCD, EFGH be two quaFrom the point D draw DG perpen- drilaterals which have the sides À B, dicular to E F, or E F produced. Then, BC, CD of the one, equal to the sides because D G is (I. 12. Cor. 3.) less than EF, FG, GH of the other, each to DE or A B, the rectangle under DG, each; and let the angles A, B, C, D of EF is less than the rectangle under the first lie in the circumference of a AB, EF, or AB, BC, and therefore circle, of which the side A Dis diameter, (I. 26. Cor.) the triangle DEF is less but the angles E, F, G, H, of the other than the triangle A B C.

not lie in the circumference of a circle Therefore, &c.

of which EH is diameter: the quadriCor. Two given finite straight lines lateral A B C D shall be greater than with a third indefinite, inclose the EFGH. greatest possible area, when placed at right angles.

PROP. 40.
If two quadrilaterals have three sides

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of the one equal to three sides of the
other, each to each, and the angles of

Join EG, FH: and, as most fathe first lying in a semi-circumference vourable to the of which the fourth side is diameter, but figure EFGH

T the angles of the other not so lying, the in the

with first quadrilateral shall be greater than parison the other.

ABCD, let one

of the angles For if EFGH (see fig. 1) be that one EFH, EGH, of the quadrilaterals which has not the the former for

Fig. 2. angles lying in a semicircumference, of instance, be a which the fourth side EH is diameter, right angle ; and if G be an angle which does not so since, FH relie; then, joining EG, the angle EGH maining will not be equal to a right angle, same, ihe_tri(15 Cor. 3.) and, therefore, if GH' be angle E FH, drawn perpendicular to EG, and equal and therefore to GH, and if EH' be joined, the trian- the whole figure

Fig. 3. gle EGH' will be greater than EGH, will be greater (39.) and, accordingly, the quadrilateral (39.) upon this EFGH', which has its three sides EF, than upon the FG, GH' equal to the three EF, FG, contrary suppoGH, each to each, greater than the sition. It will

H" quadrilateral EFGH. Therefore, if

appear in the a quadrilateral be inclosed by three demonstration, that it is indifferent given sides and a fourth not given, a whether E G H be supposed less or greater may be found inclosed by the greater than a right angle: we shall set same three given sides and a fourth not out with supposing it to be less, and, given, except when the angles lie in a therefore, (15. Cor. 3.) the point G to be semicircumference, of which the fourth without the semicircle upon EH. side is diameter. But, because the

Draw G H' at right angles to EG, fourth side is (I. 10. Cor. 2.) necessarily and equal to GH (fig. 1); and join less than the sum of the other three, it is evident that there is some certain area,

* If two of the given sides as FG, GH should be in

the same straight line, E F G H would be a triangle, a greater than which cannot be so in- not a quadrilateral: it may be observed, however, closed, and therefore some quadrilateral that the demonstration is eqnally applicable to show which incloses the greatest possible area. may add that, by a similar demonstration, it appears Therefore, the quadrilateral ABCD that any number whatever of given finite straight which has its angles lying in a semicir- lines with an indepnite inclose the greatest possible

area when placed as chords of a semicircumference cumference, of which the fourth side is of which the indefinite is diameter.

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that in this case E F G H is less than ABCD.

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EH'. Then, because EFH is a right than did the last in the series which angle, EFH' is greater than a right was greater or less than AD; that is, angle, and therefore (15. Cor. 3.) ihe AD is the limit to which, in the forepoint F falls within the semicircle drawn going process, the bases EH are made upon EH', as in fig. 2. Again, draw to approach. And it has been shown, FE' (fig. 2) at right angles to FH' and besides, that the figure EFGH is inequal to E F, and join E'H'. Then, creased at every step. Therefore, the because E G H' is a right angle, E'GH' figure upon the base Ad is greater than is less than a right angle, and the any of the figures E F G H. point G falls without the semicircle upon Therefore, &c. E' H' (see fig. 3), as at first. It appears Cor. Three given finite straight lines therefore, that if the process be repeated with a fourth indefinite, inclose the and continued, we shall obtain in this greatest area, when placed as chords manner a series of figures (fig. 1., fig. 2., of a semi-circumference, of which the fig. 3., &c.), each of which is greater fourth side is diameter. than the preceding (because one of the

PROP. 41. triangles remaining the same, the other is made to have a right angle), and in sides of the one equal to the four sides

If two quadrilaterals have the four which, one of the angles E FH, EGH, being a right angle, the other is greater of the other, each to each in order, and than a right angle, and less than a right the angles of the one lying in the cirangle alternately,

cumference of a circle, but the angles Again, of the bases E H, E H', E' H', of the other not so lying, the first &c. each is of a magnitude intermediate quadrilateral shall be greater than the

other. between the two preceding. For, because the square of E'H' (fig. 2) is equal

Let the quadrilaterals ABCD, A B cd (1. 36.) to the squares of E'F and FH', have the four sides of the one equal to and that FH' is greater than FH (fig. 1), the four sides of (because the two sides FG, G H of the the other, each triangle FGH' are equal to the two sides to each in order, FG,GH of the triangle FGH, and con

and the angles tain a greater angle (I. 11),) the square in the circum

A, B, C, D lying of E' H' is greater than the squares of E' F, FH, or of E F, FH; greater, that ference of a ciris, than (I.36.) the square of E H (fig. 1),

cle ABD, but the and therefore E' H' (fig. 2 or 3) is angles A, B, c, d, greater than EH (fig. 1). But É'H' not so lying: the quadrilateral ABCD (fig. 3) is less than Ě H' (fig.2), because shall be greater than the quadrilateral the two sides E' F, FH' of the triangle

ABcd. E' FH' are equal to the two EF, FH'

Through A draw the diameter AK, of the triangle E FH', and contain a

and join B K, KC: less angle, (I. 11.). Therefore, E' H' which is equal to BC, make the tri(fig. 3) is of intermediate magnitude be- angle B kc equal and similar to the tween EH (fig. 1) and E H' (fig. 2). triangle B K C, so that the sides B k, ko And, in a similar manner, it may be

may be equal to the sides BK, KC shown that E' H" (fig. 3. or 4) is of respectively; and join A k. intermediate magnitude between EH'

Then, because the straight lines B c, (fig. 2) and E' H' (fig. 3); and so on.

B C do not coincide, (for if they did, the Now A D is greater than EH, E'H', figures would coincide altogether by 1.7.) &c., and less than EH', E' H", &c.; the point k does not coincide with the because the chords AB, BC, CD,

point K: but A BK is a right angle which together subtend the semi-circum

(15. Cor. 1.): therefore ABk is not a ference of which A D is diameter, sub- right angle, and (39.) the triangle BAK tend more than a semi-circumference in is greater than the triangle B Ak. Also, the circles of which E H, E' H', &c. the quadrilateral ADC K, having its are diameters, and less than a semi-cir- three sides A D, DC, C K chords of the cumference in those of which E H', semi-circumference upon AK, is greater E' H", &c. are diameters.

than any other quadrilateral* Ad ck, Therefore, every successive base EH having three of its sides equal to AD, DC, being alternately greater and less than * If d c and ck lie in the same straight line (as is A D, and lying between the two preced- nearly the case in the figure), the figure A dc k will be ing, approaches more nearly to AD

a triangle, not a quadrilateral ; but in this case also it is less than A D C K (see note Prop. 40.)

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C K, respectively (40.). Therefore, the subtend less than the whole circumferwhole figure AŽ K C D is greater than ence, and a second circle in which they the whole figure A B kcd; and, taking shall subtend more than the whole ciraway the equal triangles BKC, B kc, cumference: for the circle required will the figure A B C D is greater than the be of some magnitude between these two. figure Abcd.

It may be observed, also, that the Therefore, &c.

order of the sides is indifferent as well Cor. 1. If a figure A B C D E F is to to the magnitude of the required circle, be inclosed by any number of given sides, as to the magnitude of the figure which and if these sides be not so disposed is to be inscribed in it; for the same that the angles may lie in the circum- chord will subtend an arc of the same ference of a circle, a greater figure may magnitude, at whatever part of the cirbe inclosed by the same sides. For, if. cumference it may be placed (12. Cor. 1); the angle E, for instance, do not lie in and therefore the arcs subtended by all the circumference which passes through the chords will be together equal to the the points A, B, C, join A E, C E, and whole circumference, whatsoever may let there be constructed the quadrila- be their order. And, because the same teral abce, such that its sides may be chord always cuts off a segment of the

same area, the segments cut off by all

the chords will amount to the same A.

area, whatsoever may be their order ; and therefore the inclosed area, which

is the difference between that amount à

and the area of the circle, will also be the same.

From these considerations

it appears that Prop. 41. Cor. 2. need not equal to those of A B C E, each to each, have been qualified by a regard to the

order of the sides. and its angles in the circumference of a circle (25. Cor.): and upon the sides a e,

PROP. 42. ce, which are equal to AE, CE, respec- Of all plane figures having the same tively, let there be described the figures perimeter, the circle contains the greata fe, cde equal to the figures A FE, est area. CDE, respectively. Then, because, by For, if the figure ABCDEFGH the proposition, the quadrilateral a b ce be any other than a circle, there must is greater than A B C E, the whole be some four points in the perimeter, figure a b c d e f is greater than A B C D E F.

3 Cor. 2. And hence, of all figures contained by the same given sides in

h the same order, that one contains the greatest possible area which has all its angles in the circumference of a circle. For the area inclosed by the given sides cannot exceed a certain limit depending as A, C, E, G, which do not lie in the upon them, which limit is the greatest circumference of a circle. Join these possible that can be inclosed by the points, and let the quadrilateral aceg given sides, and is therefore such as by be constructed, having its sides equal them can be inclosed. But no figure, to those of the quadrilateral ACEG, each so inclosed, contains the greatest pos- to each, and its angles in the circumfersible area, of which the angles do not lie ence of a circle (25. Cor.). Then, because in the circumference of a circle. There- (41.) the quadrilateral aceg is greater fore, the figure which has its angles in than ACE G, if upon the sides ac, ce, the circumference of a circle contains a eg, g a, which are equal to AC, CE, greater area than any other figure hav- EG, GA, respectively, there be coning the same given sides.

structed the figures abc, cde, efg,gha

equal to the figures A B C, Č D E, Scholium.

EFG, GHA, each to each, in all reThat a circle may be imagined in which spects, the whole figure abcdefg any number of given straight lines shall will be greater than ABCDEFG, subtend as chords the whole circum- and will have the same perimeter. ference exactly, is evident from this, that a It appears, therefore, that if a plane circle may be imagined in which they shall figure be not a circle, a greater area

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than is contained by that figure may be sponding to the first condition. But inclosed with the same perimeter. But again, the point required must be the area inclosed by a given perimeter equidistant from the two given points, cannot exceed a certain limit, which that is, it must be in the straight line limit, being the greatest possible that which bisects the distance of the two can be so inclosed, some figure with the given points at right angles; for this, it given perimeter must be capable of con- is easily seen, (I. 6.) is the locus cortaining. Therefore the circle only con- responding to this second condition. tains the greatest area of all figures Therefore, if this straight line be drawn, having the same perimeter.

and intersect the given line, the point Cor. In the same manner it may be of intersection (or any of those points, shown that if a figure is to be inclosed if there be more than one) will satisfy by a given perimeter, of which part is to both conditions, and will be the point be a given finite straight line, and if it be required. not made a circular segment of which If there be no point of intersection, the the given line is chord, a greater may be problem is impossible. inclosed with the same conditions, and To take another instancetherefore that of all figures so inclosed Let it be required "to find a point in a the circular segment is the greatest. certain plane, which shall be, first, at a PROP. 43.

given distance from a given point in the

plane; and, secondly, at a second given Of all plane figures having the same distance from a second given point in area, the circle has the least perimeter. the same plane."

Let the circle C have the same area Here it is evident that the locus corwith any other plane figure F: C shall responding to the first condition is the be contained by a less perimeter than F. 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 Let C' be a second circle, having the given distance as radius. Therefore same perimeter with F; then by the the points which are common to the two last proposition, c' has a greater area circumferences, that is, their points of than F has, that is, than C has. But intersection, if there be any, will either the areas of circles (33.) are as the of them be the point required. squares of their radii ; therefore the ra- If the circles do not intersect one ano. dius of C' is greater than the radius of ther, the problem is impossible. C; and the radii of circles (33.) are as Such is the use of loci in the solution of their circumferences; therefore the cir- problems. We have seen also in the above cumference of C', or perimeter of F, is example, that they serve to determine in greater than the circumference of C. what cases the solution is possible or imTherefore, &c.

possible. Thus, in the latter example, it SECTION 6.-Simple and Plane Loci.

will be impossible, if the distances of

the point required from the given points Def. 14. A locus in Plane Geometry is differ by more than the mutual distance a straight line, circle, or plane curve, of those points, or together fall short of every point of which, and none else in that distance: and in the first example that plane, satisfies a certain condition. it will be impossible, if the given line,

The nature and use of loci will be being straight, be perpendicular to the readily apprehended from the following line which passes through the two given example:

points, and does not pass through the Required a point in a certain plane point which bisects that line; for if it which shall be, first, in a given line in does so pass, the two conditions prothe plane ; and, secondly, equidistant posed are identical, and any point in this from two given points in the same line will answer them. plane."

Every locus is the limit between exHere, as far as the first condition cess and defect. The points upon one only is concerned, any point in the given side of it fail by defect, and those upon line, but none else, will answer. The the other side by excess, of possessing given line is therefore the locus corre- the required property which is possessed

а.

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