Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

IV. Some Examples of Affected Biquadratick Equa∣tions, but like Affected Quadratick ones.

PROBLEM I.

TO find a square ABCD (such as in the mean while we'll suppose n. 1. to be in Fig. 41.) from which having taken away another square AEFG, which shall be half the former, ••here will be left the Rectangle GC whose Area is given. E. g. Suppose the given area equal to the square of the given line LM, to find the true sides of the squares AB and AE, answer∣••ng to these supposed ones, n. 1.

SOLƲTION.

Make the area of the rectangle that is to remain = bb, and GB=x; BC or AB will be = 〈 math 〉〈 math 〉, and substracting hence GB, the remaining side of the lesser square AG=〈 math 〉〈 math 〉−x, 〈◊〉〈◊〉. 〈 math 〉〈 math 〉. Since therefore the square of this is supposed ••o be half of the square of AB, this will be the Equation: 〈 math 〉〈 math 〉; ••nd multiplying by xx, 〈 math 〉〈 math 〉 ••nd multiplying by 2, 〈 math 〉〈 math 〉;

and substracting 2b{powerof4}, and adding 〈 math 〉〈 math 〉; and dividing by 2, 〈 math 〉〈 math 〉.

NB. The same Equation may be obtain'd, if, putting x for GB or FH, and having found the □ of AG or GF as a∣bove, you infer 〈 math 〉〈 math 〉.

This last Equation, tho' it be a biquadratick, yet may be rightly esteem'd only a quadratick one, because there is neither x{powerof3} nor single x in it, and so you may substitute this for it, 〈 math 〉〈 math 〉, viz. by supposing y=xx. Whence according to the third case of affected quadraticks, y will = 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 or = 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. Now if the given line b be assu∣med for unity, bb and b{powerof4} will be = = to the same line Therefore, if between LM as unity, and MN=½b viz 〈 math 〉〈 math 〉 you find a mean proportional MO (n. 2. Fig. 41.) that wil•• be = 〈 math 〉〈 math 〉, which being substracted from LM, and added to it, will give the two values of the quantity y. Moreover there∣fore by extracting its roots, i. e. by finding other mean pro∣portionals LR and LS between the quantities found LP and LQ and unity (n. 3.) they will be the two values of the quantity x sought; the first whereof LR will satisfie the que∣stion, and the other LS be impossible. Wherefore to form

the square it self, since its side will be = 〈 math 〉〈 math 〉; by making (n 4.) as x to b so b to a fourth, it will be obtain'd: And this may be further prov'd, if finding a mean proportional BK between BI=LR and the side of the □ BC, it be equal to the given quantity LM.

Arithmetical Rule. From the given area or the square of the given line LM substract the root of half the biquadrate of the same line; thus you will have the value of the □ FC, viz. xx: Therefore extracting further the square root of ••••is, it will be the value of x sought.

PROBLEM II.

TO find another square ABCD (Fig. 42. n. 1.) out of the middle whereof if you take another square EFGH, which ••all be a fourth part of the former, the area of the rectangle ••K intercepted between BC and FG prolonged, shall be equal 〈◊〉〈◊〉 the square of a given line LM; i. e. having these given to ••nd the segment BI, and consequently also the side BC or ••B.

SOLƲTION.

Make the area of the given rectangle, or the square of LM 〈◊〉〈◊〉 to bb, and the side sought of the rectangle BI=x; the o∣••er side BC will be = 〈 math 〉〈 math 〉, and having substracted out of it 〈◊〉〈◊〉 and GK (i. e. 2x) the side of the lesser square FG will ••〈 math 〉〈 math 〉−2x, i. e. 〈 math 〉〈 math 〉; whose square since it is the ••orth part of the greater square by the Hypothesis, you'l have 〈 math 〉〈 math 〉; 〈◊〉〈◊〉 multiplying both sides by 〈 math 〉〈 math 〉; 〈◊〉〈◊〉 taking away 4b{powerof4}, and adding 〈 math 〉〈 math 〉;

and dividing by 4, 〈 math 〉〈 math 〉.

Therefore according to the third case of affected quadratick ••¦quations, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. If the given line b be taken 〈◊〉〈◊〉 unity, b{powerof4} and bb will be equal to it. Therefore if betwe•• LM as unity, and MN=3 ⅓b, you find a mean proporti••¦nal MO (n. 2. Fig. 42.) 'twill be 〈 math 〉〈 math 〉; which subst••¦cted from MQ=2b, or added to it, will give two values the quantity xx, viz. PQ and IQ; the first whereof will 〈◊〉〈◊〉 only a true one, and of use here. Now therefore a mea•• proportional QR found between PQ and unity will expre•• the quantity sought x.

Therefore for forming the square it self, since its side AB=〈 math 〉〈 math 〉, you may proceed as in the former Constructon, (vi•• n. 3.)

PROBLEM III.

HAving given the base of a right-angled Triangle, and mean proportional between the Hypothenusa and Perpe••¦dicular, to find the Triangle. As if the given base be A (Fig 43.) and the mean proportional between AC and B•• be CD; to find the perpendicular BC, and Hypothenu•• AC.

SOLƲTION.

Make the given base = a, and the mean proportional = the perpendicular BC=x, then will the Hypothenusa be 〈◊〉〈◊〉 the Hypoth, −〈 math 〉〈 math 〉.

[illustration]

Pag. 56.

XLIII

XLIV

XLV

XLVI

XLVII

XLVIII

••erefore 〈 math 〉〈 math 〉; ••d multiplying both sides by 〈 math 〉〈 math 〉; ••d substracting 〈 math 〉〈 math 〉. ••erefore by the second case of affected quadraticks 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉.

Or thus.

Make the Hypothenusa AC=x, then will the perpendi∣••ar be BC=〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; ••d multiplying by 〈 math 〉〈 math 〉. ••••erefore by case 1. 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction; the first for the latter Equation. 〈◊〉〈◊〉 be put for unity, the line AB will be also = aa, and king, as a to b so b to a third, i. e. as LM to MN so LO OP, and you'l have bb. Having erected the perpendicu∣•• AQ=OP upon AM, and drawn MQ or Mn equal to 〈 math 〉〈 math 〉, and consequently An will be = 〈 math 〉〈 math 〉, 〈◊〉〈◊〉 the value of xx. Moreover a mean proportional AC ••nd between An and AR unity will be the value of x, i. e. Hypothenusa sought; which being found, you may easily ••mplete the Triangle ABC.

2. In the case of the former Equation, making every thing before AK would be the value of the quantity xx, i. e. 〈 math 〉〈 math 〉. Therefore a mean proportional RT ••nd between RS=AK and AR unity will be the value of x,

i. e. the perpendicular sought, and so AT the Hypothenusa of the Triangle sought.

Arithmetical Rule. In the first Solution add the biquadrate of the given mean proportional to the biquadrate of half the given base; and having extracted the square root of the sum, take from it half the square of the given base; the root of the remainder will give the perpendicular of the triangle sought, and the root of the sum will give the hypothenusa of it.

PROBLEM IV.

HAving the Hypothenusa of a right-angled Triangle given, and a mean proportional between the sides to find the Tri∣angle. As if the hypothenusa be AC (Fig. 44.) and a mean proportional between the sides BD, to find the sides AB and BC.

SOLƲTION.

Make the given Hypothenusa = a, and the mean propor∣tional =b, and the perpendicular BC=x; the basis AB by the hypoth. will be 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; and multiplying by xx, 〈 math 〉〈 math 〉; and substracting b{powerof4}, 〈 math 〉〈 math 〉.

Therefore by the third case, 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction. If a be put for unity, AC will be also = aa, and by making as AC to CG (a to b) so AF to GH (b to a third) this third will be GH=bb. Assuming therefore OC=½aa=OB the radius of a semi-circle, and ha∣ving erected CD=bb=BE parallel to it, EO will be

〈 math 〉〈 math 〉, and consquently EC = 〈 math 〉〈 math 〉, and EA 〈 math 〉〈 math 〉, viz. the double value of the quantity xx. Therefore for the double value of x, you must extract the roots out of them, i. e. you must find the mean proporti∣onals AL and AM between unity AC and AI=EC on the one side, and AK=AE on the other; Altho' these last may be more compendiously had, and the triangle it self immedi∣ately constructed, if having found EC and EA, you draw CB and AB: For these will be those two last mean proportionals = = AL and AM; for by reason of the ▵ ▵ ABC, AEB, and BEC, BC is a mean proportional between AC and CE, and AB a mean proportional between the same AC and AE by the 8. Lib. 6. Eucl. which is Consect. 3. Schol. 2. Prop. 34. Lib. 1. Math. Enucl.

PROBLEM V.

HAving given the Area and Diagonal of a right-angled Parallelogram, to find the sides and so the Parallelogram. As if the given Area be = to the square of a given line BD (Fig. 45.) and the Diagonal AC, to find the sides AB and BC.

SOLƲTION.

If for the given Area, or square of the line BD you put bb, and make the Diagonal AC=a, and put for the lesser side BC, x; the other side will be 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉; and multiplying by xx, 〈 math 〉〈 math 〉; and substracting b{powerof4} 〈 math 〉〈 math 〉. Which Equation, since it is the same with that of the preceding Problem (which is no wonder, since this fifth perfectly coincides with the fourth; for the Di∣agonal AC is the hypothenusa, and BD, whose square is = to the given area of the rectangle, is a mean proportional be∣tween the sides AB and BC) and so will have the same Con∣struction,

(see Fig. 45.) and the same Arithmetical Rule, which may be easily formed from the last Equation of the preceding Problem.

PROBLEM VI.

HAving given the first of three proportional lines, and ano∣ther whose square is equal to both the squares of the other two, to find those two proportionals. As if AC (Fig. 46.) be the first of the three proportionals, and another line ED gi∣ven, whose square equals the two squares of the others taken together; to find those two as second and third proportio∣nals.

SOLƲTION.

If for AC you put a, and make the given line ED=c, and the second proportional = x, the third will be 〈 math 〉〈 math 〉. Where∣fore the squares of the two last will be 〈 math 〉〈 math 〉+xx=cc, □ ED by the hypoth. and multiplying both sides by 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction. If a be put for unity, AC will also = aa and a{powerof4}, and making as AC to CD (a to c) so AF to DE (c to a third) DE will be = cc. Now having made AK=DE, i. e. cc, a mean proportional AI found between AC and AK will be 〈 math 〉〈 math 〉. Therefore taking AO=½AC, viz. ½aa, the hypothenusa OI will be = 〈 math 〉〈 math 〉. And OA=½a being substracted from OI or OH equal to it will leave AH the value of xx; and the root of that being extra∣cted, i. e. finding another mean proportional AG between AC and AH, it will be the value of x, i. e. the second of the proportionals sought, and since AC is the first given, AH will be the third. Q. E. F.

NB. This Construction may be abbreviated, and the first Operation, by which you find DE, to which afterwards AK is made equal, may be omitted. For since you make use of a mean proportional between CA and DE sought, which is = to the given line ED, and afterwards AI a mean pro∣portional between AC and AK is sought; it is evident that AI will be equal to ED given, and consequently that they may be immediately joined at right angles at the beginning of the given line AC, and the rest may then be done as before.

Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not.

PROBLEM I.

BEtween two given right angles to find two mean proportio∣nals. E. g. Suppose given AB the first and CD the fourth, (Fig. 47. n. 1.) between these to find two mean pro∣portionals.

SOLƲTION.

Make the first of the given quantities AB=a, the other CD=q, the first mean = x, then will the latter be 〈 math 〉〈 math 〉, and consequently 〈 math 〉〈 math 〉=q; and multiplying by aa, x{powerof3}=**aaq.

The Central Rule will b 〈 math 〉〈 math 〉=AD 〈 math 〉〈 math 〉=DH. i e. according to a supposition we shall by and by make, ½a=AD, and ½q= DH.

Geometrical Construction. If AB or a be made unity, and also the Latus Rectum of your Parabola, and you describe, by means of this Latus Rectum, the Parabola, according to Schol. 1. Prop. 1. Lib. 2. Math. Enucl. [see n. 2. and 3. Fig 47.] ••n which AB is the Latus Rectum; A1, A2, &c. the Abscis∣••a's; AI, AH, &c. the semiordinates; make moreover (n. 4) AD=½a, and having from D erected a perpendicular =½q, describe a circle at the interval AH, cutting the para∣bola

in N: Which being done, a perpendicular to the Ax will be the root sought or the value of x, i. e. the first of the means, and consequently OA the other; since NO by the first property of the Parabola (see Prop. 1. Lib. 2. Mathes. Enucl.) is a mean proportional between the Latus Rectum AB, and the abscissa AO. And by this means there will come out, by Baker's Central Rule the very construction of Des Cartes, Ge∣om. p. m. 91.

The Arithmetical Rule. Multiply the square of the first by the fourth given, and the cube root extracted out of the pro∣duct, will express the first of the means sought.

PROBLEM II.

HAving given the solid Contents of a solid or an hollow Pa∣rallelepiped, and the proportion of the sides, to find the sides. As, if the given capacity or solid contents be = to the cube of a certain given line IK (Fig. 48. n. 1.) and the pro∣portion of the heighth to the length be as AB to BC, and to the latitude as the same AB to BD; to find first the altitude, which being had, the other Dimensions will also be known, by the given proportions.

SOLƲTION.

Make IK=a, AB=b, BC=c, and BD=d; and lastly the heighth sought = x, then as b to c, so x to the length required 〈 math 〉〈 math 〉; and as b to d, so x to the latitude sought 〈 math 〉〈 math 〉.

Multiplying therefore these three dimensions of the Parallelepi∣ped together, you'l have its capacity or solid contents 〈 math 〉〈 math 〉 =a{powerof3}; and multiplying by bb, cdx{powerof3}=a{powerof3}bb; and dividing by cd, 〈 math 〉〈 math 〉 i e. 〈 math 〉〈 math 〉.

Therefore the Central Rule will be the samc as above, 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH, i. e. according to the supposi∣tion which will by and by follow, 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH.

Geometrical Construction. If IK or a be made unity, and at the same time Latus Rectum, and by means of it you de∣scribe a Parabola, after the way we have shewn, Fig. 47. n. 2. and 3. and shall always hereafter make use of; and then to prepare the quantity 〈 math 〉〈 math 〉 (which in the Central Rule is the the quantity 〈 math 〉〈 math 〉) make (n. 2.) IK−IM=BC−KL=BD−MN as a to c so d to e; so that for cd you may put ae, and afterwards divide by a both above and underneath; you'l have the quantity 〈 math 〉〈 math 〉.

Therefore by further inferring as 2e to bb, so aa to a fourth IO−IP=IT−IK−IQ, and you'l have the quantity DH, which will determine the centre, after AD is made equal 〈 math 〉〈 math 〉. Having therefore described from that centre a circle through the vertex of the Parabola A (n. 3.) a semi∣ordinate NO drawn from the intersection will be the altitude sought, which will easily give you the length and breadth by the reasons above shewn.

Another Solution.

Which will be more accommodated to the Arithmetical Rule.

Let the rest of our Positions or Data remain as above, but the name of the proportion which the altitude has to the length be = e, and of that which it has to the latitude = i, the

length will be = to ex, and the latitude to ix. Wheref•• multiplying the sides together, you'l have the whole solidity 〈 math 〉〈 math 〉; and dividing by ei, 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 Hence

The Arithmetical Rule. Multiply together the given na•• of the reasons, and divide the given cube by the produ•• which done, the cubick root extracted out of the quotient 〈◊〉〈◊〉 be the altitude of the solid sought.

Another Geometrical Construction. Now if we would a•• construct this Equation x{powerof3}=〈 math 〉〈 math 〉 geometrically, putting AB=•• for unity, BC and BD will be the names of the reasons = •• and i. Making therefore first IK−IM−KL−MN as a to e so i to a fourth f (Fig. 49. n 1.) af •• be = ei, and the proposed Equation will have this form: 〈 math 〉〈 math 〉 i. e. aa / f.

Therefore 2. making as f to a so a to a third IQ, that will be the value of 〈◊〉〈◊〉 But hence 3 by extracting the cube root, i. e. by finding t•• mean proportionals between unity b, viz. AB and the 〈◊〉〈◊〉 found ••Q; the first of them will give the root sought.

NB. The same Central Rule would come out according Baker••s Central Rule, which would have the same form of Equation, as the precedent Example x{powerof3}**−aa=o,

• taking b for the Lat.
• Rect. and unity,
  • 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH,
  • 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉 i. e. ½IQ=DH

(see Fig. 49. n. 2.)

[illustration]

Pag. 64.

XLIX

L

LII

LIII

PROBLEM III.

HAving given the solid contents of a solid or hollow Paral∣lelepiped, and the difference of the sides, to find the sides. 〈◊〉〈◊〉 if the given capacity be equal to the cube of any given ••e LM (Fig. 50. n. 1.) and the difference whereby the ••gth exceeds the breadth = NO, and the difference by which 〈◊〉〈◊〉 breadth exceeds the altitude or depth = PQ, to find the ••gth, breadth and depth.

SOLƲTION.

Make the side of the given cube = a, the excess of the ••gth above the breadth NO=b, and of the breadth above 〈◊〉〈◊〉 depth PQ=c, make the depth = x, the breadth = x+c, ••d the length = x+b+c. Multiplying therefore these ••ree dimensions together.

〈 math 〉〈 math 〉

•• according to the forms of Baker and Cartes 〈 math 〉〈 math 〉

Wherefore the Central Rule contracted by the supposition ••hich will hereafter follow will be this, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH. •• e. by virtue of the supposition just now mentioned (which ••kes LM viz. a for unity and also for Lat. Rectum)

〈 math 〉〈 math 〉 = AD. And 〈 math 〉〈 math 〉 = DH.

Or more short; 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 (∽ 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Geometrical Construction. If LM or a be taken for unity or Latus Rectum of the Parabola to be described, that being described (Fig. 50. n. 3.) you are first of all to determine two quantities AD and DH; which may be done two ways; either by Baker's form of his Central Rule, or by ours immediately divided by the quantities of the last Equation.

1. For AD by our form, 〈 math 〉〈 math 〉 = AD, you must make (Fig. 50. n. 1.) as a to b so b to a third (AB to AC so BE to CF) which will be bb, and D{powerof2} the eighth part of this CF must be added to A{powerof2} the half of AB. And by Baker's form you must make, 1. as AB (=a, n. 2.) to AC (=½p i. e. ½b+c) so BE (=¼p i. e. ¼b+½c) to a fourth CF (which will be = 〈 math 〉〈 math 〉▪) 2. Make moreover as AB to AG (a to b) so BH to GI (c to bc) and, as AB to AK (a to c) so BH to KL (c to cc) and the two quantities found GI and KL (bc and cc) being added into one sum will give the quantity q=MN, the half whereof MO will express the quantity 〈 math 〉〈 math 〉 in the

Rule, and to be substracted from the former 〈 math 〉〈 math 〉. Actually therefore to determine the quantity AD not on the Ax, but on another Diameter of the described parabola n. 3. (because the quantity p is in the Equation) having made a perpendi∣cular to the Ax aE=〈 math 〉〈 math 〉 i. e. to the line BE n. 2. and from E having drawn EA parallel to the ax, according to our form AD n. 1. transfer it only on the diameter of the parabola n. 3. from A to D, either by parts 〈 math 〉〈 math 〉 i. e. ½LM from A to c, and 〈 math 〉〈 math 〉 i. e. ⅛CF n. 1. from c to D: But according to Baker's form, first you must put 〈 math 〉〈 math 〉=½LM n. 1. from A to 1. Se∣condly you must put from 1 to 2 the quantity 〈 math 〉〈 math 〉=CF n. 2. Thirdly you must put from 2 to 3 backwards the quantity to be substracted 〈 math 〉〈 math 〉=MO n. 2. which being done, the point D will be determined.

[It is evident by comparing these two ways of Construction, that we may join our forms not incommodiously to Baker's; because by ours the quantity AD was obtained more compendiously than by Baker's, which will also often happen hereafter. And where this Compendium cannot be had, there is another not inconside∣rable one, that, if the quantities AD and DH determined ac∣cording to both ways shall coincide, (which happens in the pre∣sent case) we may be so much the more sure of the truth]

2. As for DH by our form, you must put it from D to e on a perpendicular erected from D on the left hand, the quantity 〈 math 〉〈 math 〉 = BE n. 2. falling here upon the Ax. Then for the quantity 〈 math 〉〈 math 〉 make n. 4. as a to ½bb (LM to LN) so ⅛b to 〈 math 〉〈 math 〉 (MO to NP) and this quantity must be put from e to f in a perpendicular to the Diam. Thirdly, for the quantity 〈 math 〉〈 math 〉

you must farther make n. 4. as a to ½bb (DM to LN) so ¼c to a fourth (MQ to NR) which must be put from f to g. Lastly the quantity 〈 math 〉〈 math 〉 (which is = MO n. 2.) must be put backwards (because to be substracted) from g to H, which is the centre sought. In like manner by Baker's form, first 〈 math 〉〈 math 〉=BE is put from D to 1 even to the Ax. Secondly, for the quantity 〈 math 〉〈 math 〉 make, n. 4. as a to 〈 math 〉〈 math 〉 (LM to LS=CF n. 2.) so 〈 math 〉〈 math 〉 to a fourth (MT=AC n. 2. to SV) and this SV is further put (n. 3.) from 1 to 2. Thirdly, make in the same Fig. n. 4. for the quantity 〈 math 〉〈 math 〉, as a to 〈 math 〉〈 math 〉 (LM to MT) so 〈 math 〉〈 math 〉 to a fourth (LX to XZ) and this XZ is put back∣wads (n. 3.) from 2 to 3, which precisely coincides with the point g. Lastly the remaining quantity 〈 math 〉〈 math 〉 (=MO n. 2. and so by what we have said above, precisely coinciding with the interval g H) is put backwards from g to H the Centre sought.

[Hence it appears again that Baker's form is more laborious than ours; tho' both accurately agree, and hereafter, for the most part, we shall use them both together, tho in' the work it self, rather in Figures, than in that tedious process of words, which we have here for once made use of, that it might be as an Example for the following Constructions.]

Having therefore found by one or both ways the Center H, and thence described a circle thro' the vertex of the parabola A, the intersection N will give the perpendicular to the dia∣meter NO, the value of the quantity x sought.

PROBLEM IV.

TO divide a given angle NOP, or a given arch NQTP (Fig. 51. n. 1.) into three equal parts; i. e. having gi∣ven or assumed at pleasure the Radius NO, and consequently also the Chord of the arch NP, to find NQ the subtense of the third part of the given arch.

SOLƲTION.

If NO be made unity, NP=q, and NQ be supposed =z; having drawn QS parallel to TO, you'l have three similar triangles NOQ, QNR and RQS. For since the an∣gle QOP is double of the angle QON, and the same (as be∣ing at the Center) double also of the angle at the Periphery QNR; it will be equal to the angle QON. But the angle at Q is common to both triangles: Therefore the whole are equi-angular, and consequently the legs NQ and NR equal, as also NO and QO; and by the like reason also PY and PT. Wherefore if RS should be added to RY, the line NP by this addition would be triple of the line NQ; and so would give the Equation, if RS was determined; which may be done by means of the ▵ QRS, similar to the two former NOQ and QNR; for the angle RQS is equal to the alternate one QOP=QNR, and the angle at R common to the tri∣angles QNR and RQS, &c. Wherefore as NO to NQ so NQ to QR 1−z−z−zz and as NQ to QR so QR to RS z−zz−zz−〈 math 〉〈 math 〉

Therefore according to what we have above said 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉; or 〈 math 〉〈 math 〉

Therefore the Central Rule will be (supposing also unity NO for the Latus Rectum,)

〈 math 〉〈 math 〉 = AD i. e. by our ½+〈 math 〉〈 math 〉 i. e. 2NO=AD form, and 〈 math 〉〈 math 〉=DH. and 〈 math 〉〈 math 〉=DH.

Geometrical Construction. Having described your parabola (Fig. 51. n. 2.) take on its Ax (because the quantity p is want∣ing in the Equation) AD=2NO, and from D having e∣rected a perpendicular = 〈 math 〉〈 math 〉NP to H; that will be the Center, from which a circle described thro' A, by cutting the parabola in three places, will give the three roots of the Equation, viz. NO and no true ones, the first whereof will express the quantity sought NQ (n. 1.) the latter the line NV, being the subtense of the third part of the compl. of the arch; and MO will express the false root, which is equal to the former taken together: All the same as in Cartes p. 91. but here some∣what plainer and easier.

PROBLEM V.

HAving three sides given of a quadilateral Figure to be in∣scribed in a circle, AB, BC, CD, (Fig. 52. n. 1.) to find the fourth side, which shall be the Diameter of the Cir∣cle.

SOLƲTION.

If we consider the business as already done, and make AB=a, BC=b, CD=c, and AD=y; we shall have first in the right-angled ▵ ABD, □BD=yy−aa, and (since in the obtuse angled ▵ BCD the □ BD is = □□BC+CD+2▭ of BC into CE) if those two □□BC+CD (i. e. bb+cc) be substracted from □BD (yy−aa) you'l have 2. yy−aa−bb−cc= to the two said rectangles of BC into CE. But these two rectangles may also be otherwise ob∣tain'd, 3, if the segment CE be otherwise determined; which may be done by help of the similar ▵ ▵ ABD and CED (for the angles at B and E are right ones, and ECD and BAD equal, because each with the same third BCD makes two right

ones; the one ECD by reason of its contiguity, the other at A by the 22. Eucl. Lib. 3.) viz. by inferring as DA to AB so DC to CE y−a−c−〈 math 〉〈 math 〉 for now multiplying CE=〈 math 〉〈 math 〉 by BC=b, the ▭ of BC into CE will be = 〈 math 〉〈 math 〉 and two such 〈 math 〉〈 math 〉.

Now therefore yy−aa−bb−cc=〈 math 〉〈 math 〉; and multipl. by y,

• y{powerof3}*−aa y=2abc; i. e. by Baker's and Cartes's forms.
• −bb y=2abc; i. e. by Baker's and Cartes's forms.
• −cc y=2abc; i. e. by Baker's and Cartes's forms.
• y{powerof3}*−aa y−2abc=o.
• −bb y−2abc=o.
• −cc y−2abc=o.

Therefore the Central Rule will be (supposing the same quantity e. g. a for unity and Lat. Rect.) 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉=DH, i. e. according to our form, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD, and

Geometrical Construction, which, without any circumlocuti∣on, from our form is founded on the foll. in Fig. 52. n. 2.

LM=a n. 3. A 1.=LM from n. 2.

MN and LP=b 1, 2=½PQ

PQ=bb 2, 2=½ST

LS and MO=c DH=PR

PR=bc MO and mo two false roots

ST=cc NO the true root; upon which having de∣scribed a semi-circle the quadrilateral will be easily made. Ac∣cording to Baker's form, for AD there would be n. 3. first

Ac=½LM, then cD=〈 math 〉〈 math 〉=VX, half the line VZ, which is compounded of LM, PQ and St; but DH is = PR as above.

PROBLEM VI.

HAving given to form a right-angled Triangle the least side BA (Fig. 53. n. 1.) and the difference of the seg∣ments of the base, to find the difference of the sides, and so form the Triangle. If we represent the business as already done, ha∣ving given AB and EC to find FC.

SOLƲTION.

Make AB=a, and EC=b, and FC=x; then will BC=a+x: Therefore the □AC=2aa+2ax+xx and the line AC 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉. Now therefore ACE i. e. 〈 math 〉〈 math 〉 multiplyed by b or √bb i. e. 〈 math 〉〈 math 〉 is = GCF [but GC is = 2a+x] i. e. 2ax+xx by Cons. 1. Prop. 47. Lib. 1. Mathes. E∣nucl. and squaring both sides

〈 math 〉〈 math 〉; and transposing all, 〈 math 〉〈 math 〉. −bb (p) (q) (r) (s)

Therefore (taking a for 1 and the Latus Rectum of the parabola) the Central Rule will be, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH i. e. according to our form and Reduction, 〈 math 〉〈 math 〉

Geometrical Construction. First from our form, the com∣pendiousness whereof will here appear, for it requires only one preparation n. 2. in which LM=a, MN and LO=b, OP=bb, which being premis'd, and the diameter Ay (be∣cause the quantity p is in the Equation) being drawn (n. 3.) make AI=½LM, and 1, 2 or ••D=½OP, and DH=LM. The rest therefore being also perform'd, which the quantity S occurring in the present Equation requires, according to the last precepts of our introduction, you'l have NO the va∣lue of x sought; whence (n. 4.) at the interval AB having describ'd a Circle, and made a right angle at B, if FC be made = to the found NO, you'l have the ▵ ABC required, and EC will be found at the beginning of the prescribed mag∣nitude.

Now if you were to find the center H by Baker's form with∣out our Reduction, 1. you must put of (n. 3.) ½AB from A to c. 2. For the quantity 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to a fourth, which would be = 2AB, viz. 2DM, and to be set from c to d. 3. The quantity 〈 math 〉〈 math 〉 (to obtain which you must substract OP (n. 2.) from the quadruple of LM, and divide the re∣mainder by 2) must be set off backwards from d to e, by thus determining the point D. 4. The quantity p, which here is precisely = LM, must be transferr'd from D to f on a perpen∣dicular erected from D. 5. For the quantity 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 (= 2AB) so 〈 math 〉〈 math 〉 (also = 〈 math 〉〈 math 〉) to a fourth quadruple of LM; and this must further be produc'd from f to the point g (which here the paper wil not permit) 6. For the quanti∣ty 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to a fourth, (which would be = to the quadruple of LM, but taking away OP) which must be set backwards from g to h. 7. Lastly the quantity 〈 math 〉〈 math 〉 (=OP) set off from h backwards or towards the right hand to i will at length give the point H required.

Another Solution of the same Problem.

This Problem may be more easily solved, and will give a far more simple Equation, if you are to find not FD but AE. Make therefore (n. 1. Fig. 53.) = x, and the rest as above; AC will be = x+b, and its □xx+2bx+bb; therefore the □BC=xx+2bx+bb−aa; therefore the □ of the tan∣gent HC=xx+2bbx+bb−2aa. But the rectangle ACE will be bb+bx. Therefore by Prop. 47. Lib. 1. Math. E∣nucl.

〈 math 〉〈 math 〉; and turning all over to the right hand, 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉. Therefore by case 2 of affected quadraticks, 〈 math 〉〈 math 〉.

The Geometrical Construction may be performed according to the rules of quadraticks Fig. 54. n. 1. as will be evident to any attentive Reader. Having therefore described a circle at the interval BA, whether it be done from any arbitrary cen∣ter (see n. 4. Fig. 53.) or upon AE found in the pres. Fig. making an intersection at the said interval in B; and apply∣ing AE, and producing it until EC become equal to the gi∣ven quantity q, and at length having drawn BC you'l have the triangle right-angled at B, and also the difference of the sides FC. But to make it more short and elegant; having determined AE by a little circle (Fig 54 n. 3.) prolong it to the opposite part of the circumference in C, and draw AB and CB; for as the radius of the little circle is ½b, so EC is =b.

Now if you would construct the Equation above by Baker's rule (that its universality may also be confirm'd by an Exam∣ple in quadraticks) viz.

〈 math 〉〈 math 〉;

The centtal rule will be (taking a for 1 and the L. R.) 〈 math 〉〈 math 〉 = AD

[illustration]

LIV

LV

LVI

LVII

〈 math 〉〈 math 〉 = DH. i. e. by our Reduction, 〈 math 〉〈 math 〉 = AD 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

The Construction therefore will consist in these: LM (n. 2. ••g. 54)=a, MO=¼b, LP=½b, therefore PR=〈 math 〉〈 math 〉: ••N=PR viz. 〈 math 〉〈 math 〉, therefore Q=〈 math 〉〈 math 〉. In Fig. 53. n. 3. ••=¼b, A1=½LM, 1, 2, or 1D=PR; D1=LP MO, 1, 2, or 1H=PQ Having drawn a circle from 〈◊〉〈◊〉 thro' A you'l have the true root RO= to AE sought; ••d MO the false root.

NB. Hence it is evident that one Problem may have seve∣••l Solutions and Constructions, some more easy and simple, ••hers more compound and laborious; viz. according as the ••••known quantity is assumed more or less commodious to the ••urpose: which may not be amiss here to note for the sake 〈◊〉〈◊〉 Learners.

PROBLEM VII.

SUppose a right line BD (Fig. 55.) any how divided in A, to divide it again in C, so that the square of BA shall ••e to the square of AC as AC to CD.

SOLƲTION.

Since the first segments BA and AD are given, call the first 〈◊〉〈◊〉 and the other b, and call AC the quantity AC x; and CD ••ill be = b−x. Now therefore since we suppose as the □ of AB to the □ AC so AC to CD 〈 math 〉〈 math 〉

aab−aax will = x{powerof3}, and transferring the qu n••ities on ••he left hand to the right, x{powerof3}*aax−aab=o.

Wherefore the Central Rule (taking a for 1 and the L. R.) will be ½−〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH. i. e. by our form 〈 math 〉〈 math 〉 i. e. o=AD, that D may fall on the vertex of the parabola; and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉=DH. The Construction therefore will be very simple, as is evident from the fifty fifth Figure.

PROBLEM VIII.

THere is given AB the capital line of a horn work (which we represent (tho' rudely) n. 1. Fig. 56) and the Gorge AD, also part of the line of defence EF, to find the face BE, the flank DE, the curtain (or the chord) DF, also the angle of the Bastion ABE, &c. and so the whole delineation of the horn work. It is evident if you have the flank DE or the curtain DF, the rest will be had also. Suppose therefore, the capital line AB and the gorge AD, and part of the line of defence EF to be of the magnitudes denoted by the Letters a, b, c, on the right hand.

SOLƲTION.

Make AB=a, AD=b, and EF=c, and DF=x; then will AF=x+b, and by reason of the similitude of the ▵▵BAF and EDE and ECB, as FA to AB so FD to DE 〈 math 〉〈 math 〉

But now □ □ DF+DE are = = □EF i. e. 〈 math 〉〈 math 〉; or giving the same denomination to all the quan∣tities on the left hand, 〈 math 〉〈 math 〉

and multiplying both sides by 〈 math 〉〈 math 〉 and translating all the quantities on the right hand by the contrary signs to the left, 〈 math 〉〈 math 〉

Wherefore (putting a for 1 and L. Rectum) the Central Rule will be, 〈 math 〉〈 math 〉 (because the quantity q is negative in the E∣quation, for cc is greater than aa+bb)=AD; and 〈 math 〉〈 math 〉 = DH. or according to our Redu∣ction, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Wherefore the Geometrical Construction requires no other preparatory determination by our form than of the quantities 〈 math 〉〈 math 〉 for AD, 〈 math 〉〈 math 〉 for DH at the center, and bbcc to determine the radius of the circle; which are exhibited by n. 2. Fig. 56. viz NP is = cc, RS=bcc, RV=bbcc, which are found by means of LM=a, LR=b, LN and MO=c, MQ=NP and MT=RS. Having therefore described a para∣bola, n. 3. and drawn its diameter, transfer AD=½NP upon it (because the quantity p is in the Equation) and also ½RS from D upon H perpendicularly, and on the right hand, (because DH=−〈 math 〉〈 math 〉;) and so you'l have the center H;

thro' which having drawn KAI so that AK shall be = to the quantity bbcc or S, i. e. RV, &c. a circle described at the interval HL will cut the parabola in M and N, and applying the magnitude NO it will be that of the Curtain sought; up∣on which, n. 4. having laid down the circumference of the horned work by help of the given lines AB and AD, you'l have the line EF, of the magnitude which was above supposed. Now if any one has a mind to do the same thing by Baker's way; by laying down first the interval AB =〈 math 〉〈 math 〉 and then making bc=〈 math 〉〈 math 〉, and lastly, putting cd for the quantity 〈 math 〉〈 math 〉; he will fall upon the same point D, and in like manner may express the other parts of the Central Rule by the intervals De, ef, fg, and setting back the last gh, he will fall upon the same center H: But this is done with a great deal more trou∣ble and labour to determine so many quantities, and also is in more danger of erring by cutting off so many parts separately, as experience will shew; and thus we have by a new argument shewn the advantage of our Reduction.

Another Solution of the same Problem.

Things remaining as before (only assuming the given lines AB, AD and EF, n. 1. Fig. 56. one half less, that the Scheme may take up less room) make BE=x, as the first or chief unknown quantity; then will BF=x+c, and its □xx+2cx +cc: And since as BE to BC=AD so BF to AF a fourth, which will be 〈 math 〉〈 math 〉 and its square = 〈 math 〉〈 math 〉. Where∣fore if this square be substracted from the square of BF, there will remain the square of BA, i. e.

〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉;

or according to the forms of Cartes and Baker, 〈 math 〉〈 math 〉

Therefore the Central Rule (putting again a for 1 and the L. R.) will be 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH; or by our Reduction, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉=DH.

Geometrical Construction. Having therefore described a para∣bola (Fig. 56. n. 5.) and drawn the diameter Ay, make A1=a and 1, 2=〈 math 〉〈 math 〉, so you'l have the point D; make moreover D{powerof3} or 2, 3=c, and backwards 3, 4=〈 math 〉〈 math 〉 (we here omit to express the Geometrical determination of these quantities 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 as being very easy) and you'l have the point H, &c. and there will come out the quantity sought NO; which since it is equal to half BE n. 4. the business will be done; which Baker's form will also give, exactly the same, but after a more tedious process.

PROBLEM IX.

IN any Triangle ABC (the scheme whereof see n. 1. Fig. 57.) suppose given the Perpendicular AD, and the differences of the least side from the two others EC and FC to find all the three

sides. i. e. Chiefly the least side AB which being found, the others will be so also.

SOLƲTION.

Make AD=a, EC=b, and FC=c, AB=x; then will BC=x+b and its square be xx+2bx+bb, and AC=x+c and its square be xx+2cx+cc; and BD will be 〈 math 〉〈 math 〉, and DC 〈 math 〉〈 math 〉. But the □ BC may also be obtain'd otherwise, and the Equation also, if □□BD+DC+2▭ of BD by DC be added into one sum according to Prop. 4. Lib. 2. Eucl. viz.

〈 math 〉〈 math 〉, multiplyed by BD √xx−aa, gives the rectangle of the segments 〈 math 〉〈 math 〉 and this doubled i. e. multiplyed by √4, gives the quanti∣ty which is contain'd under the radical sign in the Equa∣tion]

Therefore turning all over on the left hand which are be∣fore the sign √ to the right hand, prefixing to them the contrary signs, you'l have 〈 math 〉〈 math 〉 and taking away the Vinculum on the left hand, and squaring on the right 〈 math 〉〈 math 〉

and adding and substracting on both sides, as much as can be, 〈 math 〉〈 math 〉 and transferring all to the left, 〈 math 〉〈 math 〉 and dividing all by 3, 〈 math 〉〈 math 〉

Note, I sought this Equation also after two other ways; 1. By a comparison of the □ AC with the two squares AB+BC, after 2 □ □ CBD thence substracted, according to Prop. 13. Lib. 2. Eucl which is the 46. Lib. 1. Math. Enucl. and I form'd the same with the present. 2. By putting at the beginning y for x+b and z for x+c, and going on after the former me∣thods, 'till you have this Equation, 〈 math 〉〈 math 〉 in which▪ when afterwards I substituted the values answering the quantities yy and zz, &c. This same last Equation came out a little easier, but (NB) with all the contrary signs.

Now to form the Central Rule, and thence make the Geo∣metrical Construction, we must determine first each of the quantities p, q, r and S, that we may know whether they are negative or positive; and you'l find (n. 2. Fig. 57.) p=G2 positive, q=H{powerof4} negative, and K{powerof4}=S also negative; and that by help of the quantities LM=b or 1, MN and

LO=a, OP=aa, MQ and LR=c, RS=cc, and also LT=cc, TV and LX=c{powerof3}, XY=c{powerof4}. Wherefore the form of the last Equation will be like this, x{powerof4}+px{powerof3}−qxx−rx−S=o, and so the Central Rule (taking here b for 1 and the L. R.) 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH.

Wherefore, having now described a parabola (as may be seen n. 4.) having found the diameter Ay transfer upon it first Ab=½LM (n. 2) and then bc=DP (n. 3.) i. e. 〈 math 〉〈 math 〉; and thirdly cD=〈 math 〉〈 math 〉 i. e. ½H{powerof4} (n. 2.) moreover from D to e put off MB (n. 3.=〈 math 〉〈 math 〉, and from e to f put off DR=〈 math 〉〈 math 〉, and from f to g put off CF=〈 math 〉〈 math 〉; and from g backwards to h put off half the quantity r, or I5 (n. 2.) and having done the rest as usual, you'l have NO, the side required of the Trian∣gle to be described; the description whereof will be now easy (n. 5.) having all the three sides known. This may serve for an Examen, if having described a semi-circle AGB upon AB=NO, you apply the given line AD, and from B thro' D draw indefinitely BDC: Then at the interval AB having described the Arches AE and BF, add the given line EC to BE, for thus having joined the points A and C, FC ought to be equal to FC before given.

SCHOLIƲM.

WE have here omitted our Reduction, because it would be too tedious, and would express the quantities AD and DH (especially the latter) in terms too prolix. For AD would be = 〈 math 〉〈 math 〉 (viz. because 〈 math 〉〈 math 〉 is

found = 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 taken in it self = 〈 math 〉〈 math 〉; but here [where by vertue of the Central Rule it is taken positively, when it is in it self negative] un∣der contrary signs it is = 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 yet more contractedly (because b is unity) AD = 〈 math 〉〈 math 〉; which parts may be expressed without any great difficulty on the Diameter Ay, by its portions A1, 1, 2; 2, 3; 3D: But the other quantity DH, or the definition of the Center H, would also have some tedi∣ousness, as because 〈 math 〉〈 math 〉

If you take away out of the quantities 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 (since this latter is to be substracted, and so left, as it is, under the sign −; but the other, also negative in it self, but here posi∣tively expressed in the Central Rule, must have all the con∣trary signs) I say, if you take out of these quantities those which destroy one another, and add the rest with the two for∣mer quantities, they will be 〈 math 〉〈 math 〉 = = DH. or a little more contracted (because b is 1) 〈 math 〉〈 math 〉 = DH.

But now if any one has a mind to illustrate this by a numeral Example and try the truth, &c. of the quantities found; they may make e. g. a=12, b=1, and c=2; and they

will easily find that in the last Equation the quantity p will be 4, and q−190, r−388, S−195: Secondly in the Central Rule of Baker they'l find 〈 math 〉〈 math 〉=½, 〈 math 〉〈 math 〉=2, and 〈 math 〉〈 math 〉= 95, and so the whole line AD=97 ½; and further 〈 math 〉〈 math 〉=1, 〈 math 〉〈 math 〉=4, 〈 math 〉〈 math 〉=190, and 〈 math 〉〈 math 〉, and so the whole line DH=195−194 i. e.=1. Thirdly, likewise in our Reduction (if we proceed by each part corresponding to Baker's) 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉. The sum for AD 97 ½; but further 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉 to be sub∣stracted; and so the sum for DH=195−194=1. Which same quantities will fourthly come out, if the quanti∣ties AD and DH contracted, as they are expressed in letters a∣bove, be resolved into numbers.

PROBLEM X.

YOU are to build a Fort on the given Polygons EAF (see Fig. 58. n. 1.) whose capital line AB shall equal the ag∣gregate of the gorge and flank, and the squares of these added together shall be equal to the square of the given line GH, and the solid made by the multiplication of the square of the flank by the gorge, shall be equal to the cube of the given line IK.

[illustration]

LVIII

LIX

LX

SOLƲTION.

Make the Gorge AC=z, whose square zz substracted from bb the square of the given line b, will leave the square of the flank DC=bb−zz. Now this square being mul∣tiplyed by the gorge AC or z will give bbz−z{powerof3}=g{powerof3}, the cube of the given line IK; and adding to both sides z{powerof3}, and substracting bbz, o=z{powerof3}*−bbz+g{powerof3}.

Therefore if we take g or IK for 1 and L{powerof3}, g{powerof3} will be the line g, and

The Central Rule: 〈 math 〉〈 math 〉 = AD. and 〈 math 〉〈 math 〉=DH. i. e. according to our Reduction, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 DH.

Geometrical Construction. Having described a Parabola (n. 2. Fig. 58.) make on its ax A1=½IK and 1, 2, s. viz. 1D=½MN (from n. 1.) and DH=½IK. Then having de∣scribed a circle from H, and found the true root NO upon the given angle EAF (n 3.) make AC=NO, and having e∣rected the perpendicular CD divide it by AD=GH (n. 1.) and make AB=AC+CD; and the Fort will be drawn.

PROBLEM XI.

IN a right-angled Triangle ABC (which we denote by n. 1. Fig. 59) having given the greater side AC, and made the less side AB= to the segment CE, which shall cut off from the base BC a perpendicular let fall from the right angle A; to find these lines AB or CE, and consequently the whole trian∣gle.

SOLƲTION.

Make AC=a, CE or AB=x. Therefore, 1. you'l have aa−xx=□AE. And because the ▵ ▵ BEA and CAE are similar, you'l have as AC to CE so BA to AE 〈 math 〉〈 math 〉

And so, 2. □AE=〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉=aa−xx; and multipl. by aa, x{powerof4}=a{powerof4}−aaxx; or, according to the form of Cartes and Baker, transposing all to the left, x{powerof4}*+aaxx*−a{powerof4}=o i. e. x{powerof4}*+qxx*−S=o.

Therefore (taking a for 1 and L. R.) the Central Rule will be 〈 math 〉〈 math 〉 i. e. o=AD and 〈 math 〉〈 math 〉 i. e. o=DH; that H may fall on the vertex A.

Geometrical Construction. Since a is assumed for unity and L, the quantity S also and Latus Rectum i. e. AI and AK and consequently the mean proportional AL and the radius HL will be = = to the given side AC, and conse∣quently at that interval having described a circle, thro' the Pa∣rabola rightly delineated, you'l have NO the value of the quantity x, i. e. of the lesser side AB. Having drawn there∣fore NA, which is = to AC by Construction, if you draw to it the perpendicular AB cutting NO produc'd to B, you'l have the Triangle sought ABC, and AB (which will be a sign of a true Solution) will be found = NO or CE.

Another Construction. Since in the Equation above found there is neither x{powerof3} nor x, it may be look'd upon as a quadra∣tick, and constructed after the same way, as several other like it among the Examples n. 4. viz. because 〈 math 〉〈 math 〉; according to case 2 of affected quadraticks,

〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉 i.e. 〈 math 〉〈 math 〉 Wherefore (n. 3. Fig. 59.) if AC be made = a and CD〈 math 〉〈 math 〉a, the mean proportional CG will be = 〈 math 〉〈 math 〉, and taking hence GF=½a, there will remain FC = 〈 math 〉〈 math 〉: And now between this or CH equal to it, and unity AC, having found another mean proportional CE it will be the value of the quantity x sought=NO (n. 2.)

PROBLEM XII.

IN a right-angled Triangle ABC (Fig. 60. n. 1.) there is given the Perpendicular BA, a segment of the Hypothenu∣sa BD, and a segment of the Base EC, from C to the perpendi∣cular DE let fall from the end of the segment BD; to find AE DC, and consequently the Base AC and the Hypothenusa BC, and so the whole Triangle.

SOLƲTION.

Make AB=a, BD=b, and EC=c, and DC=x; which being given, the rest cannot be wanting: Therefore xx−cc=□DE. But the same □ DE may be had, if you infer as BC to BA so DC to DE 〈 math 〉〈 math 〉

And then square the quantity DE, the square will be 〈 math 〉〈 math 〉; and multiplying both sides by 〈 math 〉〈 math 〉 and substracting also 〈 math 〉〈 math 〉 〈1 page missing〉〈1 page missing〉

[illustration]

Pag. 89.

LXI

so PV= to it, that PX may be 〈 math 〉〈 math 〉; and lastly Py=½q, that PZ may be 〈 math 〉〈 math 〉. These being thus prepar'd, if (n. 4) Ab be made = ½AB (n. 1.) bc=RT, and cd backwards = ½PQ we shall light on the point D; and, if we make De=½BD, ef=PX (n. 5. which interval was too big to be represented in the Paper) and from f you put backwards fg=PZ, and on from g beyond to the right hand gh=2RS; we shall light on the point H, &c. Which of these two methods is the shortest and fittest for practice, any one, never so little experienc'd, may here see; and first Learners may take notice if they would construct by Baker's form, in the Diagrams n. 3. n. 5. and the like, they must take care to make the angles PLN, SPT pretty large; which we have here represented the less to save charges in cutting on Copper.

PROBLEM XIII.

HAving given the Diameter of a Circle CD (n. 1. Fig. 61.) and the line BG, which falls on it perpendicularly, (which we have here only rudely delineated) to find the point A. from which a right line AC being drawn shall so cut the line BG in F, that AF, FG, GD shall be three continual Proportionals.

SOLƲTION.

If CF be found, the point A will be also had, and the se∣ction of the line BG will be made. Make therefore CF=x, and (because the perpendicular BG is given, there will also be given the segments of the diameter CG and GD) make GD=b, and CG=c; then will BG=√bc▪ and CD=b+c. Since therefore the ▵ ▵ CAD and CGF are right-angled, and have the angle at C common, they will be simi∣lar.

Therefore, as CF to CG so CD to CA 〈 math 〉〈 math 〉

Therefore AF = 〈 math 〉〈 math 〉 and FG 〈 math 〉〈 math 〉.

But by the Hypothesis, as AF to FG so FG to GD 〈 math 〉〈 math 〉.

Therefore the rectangle of the extremes will be equal to that of the means, 〈 math 〉〈 math 〉; and multiplying by 〈 math 〉〈 math 〉; and transposing all to the left, 〈 math 〉〈 math 〉; i. e. by the Carte∣sian form,

〈 math 〉〈 math 〉.

Therefore (taking b for 1 and L.) the Central Rule will be, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH. or according to our Reduction, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Geometrical Construction. Having described upon the gi∣ven line CD (n 2.) a semi-circle, and apply'd in it the gi∣ven perpendicular BG, as the figure shews, you'l have the seg∣ments of the Diameter GD=b, and to the quantity p in Ba∣ker's form, and CG=c, which (n. 3. where LM=b, LO and MN=c) will give OP=cc and to the quantity q. Wherefore having describ'd a Parabola (n. 4.) and the line VZ=2 ½b, having cut off the fourth part of XZ, and the eighth of YZ (whereof the one will be = 〈 math 〉〈 math 〉, viz. 〈 math 〉〈 math 〉, and

the other to 〈 math 〉〈 math 〉) if you transfer A1=XZ upon the diame∣ter of the Parabola Ay, and moreover 1, 2 or 1 D= to half OP (n. 3.) and transversly D{powerof3}=YZ and backwards 3, 4=¼ OP, as also 4, 5=½CG (n. 2.) you'l have the center H, and having describ'd a Circle at the interval HA, the root NO must be transferr'd from (n. 2.) C to F, and continued to A the point sought. In Baker's Form (because the quantity p is = b or 1) 〈 math 〉〈 math 〉 is = 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉=〈 math 〉〈 math 〉, and the quantity q or cc=OP, (n. 3.) make therefore in the Diameter of the Para∣bola Ab=½GD, and bc=⅛GD, (n. 2.) and lastly cd=½OP (n. 3.) and you'l have the point D the same as before. Make moreover De=¼GD and cf=〈 math 〉〈 math 〉CG, and then back∣wards fg=¼OP, and lastly gh=½GD, and you'l have the same center H, and the coincidence of the parts in both forms will be pleasant to observe; which otherwise seldom happens.

Other Solutions of the same Problem.

Carolus Renaldinus, from whose Treatise de Resol. & Com∣pos. Math. Lib. 2. we have the present Problem, proceeds to solve it in another way, changing it plainly into another Pro∣blem: viz. he observes, 1. That the angles FAD (see n. 1. of our 61. Fig.) and FGD, since both are right ones on the same common base FD, are in circle. Hence he infers, 2. (by vertue of the Coroll. of the 26. Prop. 3. Eucl.) that the □ □ DCG and ACF are equal, and consequently CD, CA, FC and CG are four continued proportionals. Then he observes, 3. That GD is the excess of the first of these proportionals above the fourth CG, and AF is the excess of the second AC above the third CF; and so, since 4. the rectangle of AF and GD is = to the square of the mean proportional FG (for AF, FG, GD, are supposed to be continual proportionals) and this □ FG is the excess, by which the square of the third CF ex∣ceeds the square of the fourth CG; now the present Problem will be 5. reduc'd to this other: Having two right lines (CD and CG) given to find two such mean proportionals (AC and FC) that the ▭ of the excess of the first above the fourth (viz. of FA into GD) shall be equal to the excess,

by which the square of the third (FC) exceeds the square of the fourth (CG) viz. by the square FG.

Wherefore instead of the former he solves this latter Pro∣blem, putting for CG, b, for GD, c, so that the first of the given lines CD shall be = b+c, and the other GD=b; calling the first mean proportional AC, x; and thence deno∣minating the latter 〈 math 〉〈 math 〉 (viz. multiplying the fourth by the first, and dividing the product by the second) and more∣over he finds the excess of the first (b+c) above the fourth (b) to be c, and the excess of the second (x) above the third 〈 math 〉〈 math 〉 to be 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉; so that the □ □ of these two excesses is 〈 math 〉〈 math 〉, and because the □ of the third FC is = 〈 math 〉〈 math 〉, having substra∣cted bb=□GC, there is given the □FG = 〈 math 〉〈 math 〉 = ▭ of the excesses we just now found. So that now you'l have the Equation 〈 math 〉〈 math 〉 &c.

We also endeavour'd to find another Solution, by finding an Equation from the line BD (n. 1. Fig. 61.) as which might be twice obtain'd by means of the two right-angled Triangles FAD and FGD, since it is the hypothenusa of both. But here, besides the former denominations of our Solution we must first give a denomination to the line AD, by making as CF to FG so CD to AD 〈 math 〉〈 math 〉, &c.

But whosoever shall prosecute this Solution of ours, or that of Renaldinus to the end, will find much more labour and dif∣ficulty in either, than in the first we have given.