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

II. Some Examples of simple or pure Quadratick Equations.

PROBLEM I.

TO make a Square equal to a given Rectangle; i. e. having given the sides of the Rectangle, to find the side of an equal Square, Eucl. Prop. 14. Lib. 2. Suppose e. g. the gi∣ven sides of the Oblong to be AB and BC (Fig. 24.) to find the Line BD whose square shall be equal to that Rectangle.

SOLƲTION.

Make AB=a and BC=b, and the side of the square sought = x, and the Equation will be ab=xx; and extra∣cting the root on both sides 〈 math 〉〈 math 〉.

Geometrical Construction. Join AB and BC in one right line, and describing a semi-circle upon the whole AC, from the common juncture B erect the Perpendicular BD which will be the side of the square sought, according to Case 1. of the Effection of pure quadraticks.

Arithmetical Rule. Multiply the given sides of the Oblong by one another, and the square root extracted out of the Pro∣duct will be the side of the square sought.

PROBLEM II.

THE square of the Hypothenusa in a right-angled ▵ being given, as also the difference of the other two squares to find the sides. E. g. If the Hypothenusa be BC (Fig. 25.) and the difference of the squares of both the legs, and conse∣quently its Leg also BE given (for the squares being given the sides are also given geometrically) to find the sides of the right-angled ▵ which shall have these conditions; or more plainly, to find one side e. g. the lesser which being found, the other, or the greater, will be found also.

SOLƲTION.

Let the □ of the given Hypothenusa = aa, and the square by which the two other differ = bb. Let the less side = x, and its □=xx. Wherefore the greater will be xx+bb. And since the sum of these is = to the □ of the Hypothenusa, you'l have 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. Having described a semi-circle on BC, and applyed therein BE, the □ EC will = aa−bb; and having described another semi circle upon EC divided into two Quadrants the □ DC will be 〈 math 〉〈 math 〉, and so DC = 〈 math 〉〈 math 〉 or the side sought; which being also transferr'd upon the other semi-circle describ'd on BC, viz. from C to A gives the other side AB and the whole ▵ sought.

The Arithmetical Rule. From the square of the Hypothe∣nusa substract the given difference, and the square root extra∣cted out of half the remainder gives the lesser side of the ▵ sought.

PROBLEM III.

HAving an equilateral ▵ ABC given (Fig. 26. n. 1.) to find the Center and Semi diameter of a Circle that shall circumscribe it. i. e. Find the BD the side of an Hexagon that may be inscribed in it. For if we consider the thing as already done; it will be manifest that the side of the Hexagon BD will fall perpendicularly on the side of the ▵ AB, as making an angle in a semi-circle, so having bisected the Hypothenusa DA you'l have E the Center sought.

SOLƲTION.

Make the side of the Triangle AB=a, BD=x, then will AD=2x. Since therefore the square BD i. e. xx being sub∣stracted out of the square AD i. e. 4xx, there remains the square AB 3xx, you'l have the Equation 〈 math 〉〈 math 〉; and dividing by 3 〈 math 〉〈 math 〉; therefore 〈 math 〉〈 math 〉

The Geometrical Construction. Having produced AB (n. 2.) to F a third part of it, the square of a mean proportional BD between BF and BA will be ⅓ aa or 〈 math 〉〈 math 〉, and so the Line BD = 〈 math 〉〈 math 〉. Therefore the Hypothenusa DA being divided in two in E, or at the interval BD, making the intersection from B and A, you'l have the Centre sought.

The Arithmetical Rule. Divide the square of the given side into three equal parts, and the square Root of a third part will give the semi-diameter AE or BE sought, by the intersection of two of which you'l have the Centre.

PROBLEM IV.

HAving given, in a right-angled Parallelogram, the Dia∣gonal, or for a right-angled ▵, the Hypothenusa and the proportion of the sides, to find the sides separately and construct the Parallelogram or ▵. Suppose e. g. the given Diagonal to be AB (Fig. 27. n 1.) and the given reason of the sides as AD to DE, to find the sides.

SOLƲTION.

Make AB=a, the reason of AD to DE as b to c; make the lesser side = x, then will the greater be 〈 math 〉〈 math 〉.

For the Equation, the □□ of the sides are 〈 math 〉〈 math 〉, □ AB; and multiplying both sides by bb, 〈 math 〉〈 math 〉; and dividing by bb + cc, 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 i. e. Extracting the roots as far as possible 〈 math 〉〈 math 〉

Another Solution.

Call the name of the given reason e, so that assuming any line for unity, the value of e may also be expressed by a right line, which shall be equal e. g. to DE above. Wherefore be∣cause we make the less side x, the greater will be ex, and so 〈 math 〉〈 math 〉 i. e. dividing by 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉

The Geometrical Construction. The last Equation above being reduc'd to this proportion as the 〈 math 〉〈 math 〉 to b so a to x, make (n. 2) AD and DE at right angles, and AE will be = 〈 math 〉〈 math 〉, and continuing AE and AD make as AE to AD so AB to AC the lesser side sought. Having therefore drawn BC which determines the lesser side AC, the greater side and so the ▵ ABC will be already formed, and may be easily compleated into a Rectangle. In the other Solution the last Equation agrees with the precedent (for it gives us this pro∣portion as 〈 math 〉〈 math 〉 to 1 so a to x in which 1 is = b, and ee =cc by what we have supposed) and so the Construction will be the same.

The Arithmetical Rule may be more commodiously expres∣ed by this last Equation under the last form but one, after this way, divide the □ of the Diagonal by the □ of the name of the Reason lessen'd by unity, and the root extracted out of the Remainder is the lesser side sought.

PROBLEM V. (Which is in Pappus Alexandrinus, and in Cartes's Geometry, p. 83. in a Biquadratick affected E∣quation, and p. 84. he gives us thereon a very remarkable Note.)

HAving given the Square AD (Fig. 28) and a right line BN, you are to produce the side AC to E, so that EF drawn from E towards B shall be equal to BN.

It will be evident, if you imagine a semi-circle to pass thro' the points B and E, that the most commodious way will be to find the line DG, that you may have the Diameter BG; upon which having afterwards described a semi circle, there will be need of no other operation to satisfie the question, than to produce the side AC 'till it occur to the prescrib'd Peri∣phery.

SOLƲTION. (As found by Van Schooten, p. 316. in his Com∣ment on Cartes's Geometry, which we will here give somewhat more distinct.)

1. Denomination. Make BD or DC=a, BN or FE=c, BF=y, and DG=x; the Perpendicular EH will be = a, and EG=BF, viz y (because the ▵ EHG is similar to ▵ BDF, by n. 3 Schol 2 Prop. 34. Lib. 1. Math••s. Enucl. and BD in the one = to EH in the other) and BG=a+x, BE=y+c; and BH will have its Denomination, if you m••ke (by reason of the similarity of the ▵ ▵ BFD and ••EH)

as BF to BD so BE to BH 〈 math 〉〈 math 〉. and you'l have also HG = 〈 math 〉〈 math 〉, i. e. having reduc'd them all to the same Denomination, 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉. Having therefore na∣med all the lines you have occasion for, you must find two E∣quations, because there are assumed two unknown quantities, viz. x and y.

2. For the first Equation and its Reduction. By reason of the similiarity of the ▵ ▵ BGE and BEH, as BG to GE so BE to EH 〈 math 〉〈 math 〉: Therefore the Rectan∣gle of the Extremes will be = to the Rectangle of the means, i. e. 〈 math 〉〈 math 〉; and taking from both sides 〈 math 〉〈 math 〉.

3. For the second Equation and its Reduction. Since BH, HE and HG 〈 math 〉〈 math 〉 are continual proportionals, the Rectangle of the extreams are equal to the square of the mean, i. e. 〈 math 〉〈 math 〉; and multiplying both sides by yy, and dividing by 〈 math 〉〈 math 〉, and taking away ayy and transposing the rest, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; i. e. dividing actually as far as may be by 〈 math 〉〈 math 〉.

4. The comparison of these two Equations thus reduc'd▪ gives a third new one, in which there will be only one un∣known quantity, viz. 〈 math 〉〈 math 〉; and adding to both sides cy,

〈 math 〉〈 math 〉; and multiplying by 〈 math 〉〈 math 〉; i. e. 〈 math 〉〈 math 〉 and dividing both sides by 〈 math 〉〈 math 〉; and adding 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

5. The Geometrical Construction, which is the same Pap•• prescribes in Cartes, viz. having prolong'd the side of 〈◊〉〈◊〉 square BA to N, so that BN shall be = to a given right li•• since BA is = a, and BN=c, the Hypothenusa DN will 〈 math 〉〈 math 〉. Having therefore made DG=DN, a•• describ'd a semi-circle upon the whole line BG, if AC be pr••¦longed until it occur to the Periphery in E, you'l have do that which was requir'd.

PROBLEM VI. (Which Van Schooten has in his Comment, p. m. 150, and following.)

HAving given a right line AB, from the ends of it A an•• B (Fig. 29.) to inflect two right lines AC and B•• which shall contain an angle ACB = to the given one D, an•• whose squares shall be in a given proportion to the Triang•• ACB, viz. as 4d to a.

Viz. You must determine the point C, which the two rig•• lines AH and HC or EH and HC will do, assuming the mi••¦dle point E in the line AB. Wherefore here will be two u••¦known quantities HE and HC, and consequently two Equ••¦tions to be found in the Solution; one whereof the giv•• proportion in the Question supplies us with, and the other 〈◊〉〈◊〉 have from the similar Triangles AIC and GFD, which repr••¦sent equal angles.

SOLƲTION.

Denomination. Make AE, half AB=a, HE=x a•• HC=y; therefore AH will be = a−x and HB=a+•• whence the Denomination of the squares AC and BC is ea•• had; viz. the one 〈 math 〉〈 math 〉, and the other, a••

[illustration]

XXIX

XXX

XXXI

XXXII

XXXIII

XXXIV

XXXV

XXXVI

XXXVII

xx+yy, so that the sum of the squares is 2aa+2xx+

And the ▵ ACB will be = ay: And since the ▵ ▵ •• and AIC are similar, and the sides of the former FD and ••rbitrary, so that for FD we may put b and for FC, c; ••he sides of the latter are determined by the similitude of ▵▵ ABI and HDB, as being right-angled ones, and ha∣•• the common angle B; they will be obtain'd by making the Hypothen. BC to the Hypoth. AB so the base HB to ••ase BI, i. e.

〈 math 〉〈 math 〉, whence substracting BC=e, there remains CI = 〈 math 〉〈 math 〉.

For the first Equation, by virtue of the Problem as 〈 math 〉〈 math 〉 to ay. And the Rectangle of the ••mes is = to the Rectangle of the means, i. e. 〈 math 〉〈 math 〉.

For the other Equation, since as DF to FG so CI to AI 〈 math 〉〈 math 〉 the Rectangle of the extreams will again be = to the Re∣••gle of the means, i. e. 〈 math 〉〈 math 〉; multiplying both sides by 〈 math 〉〈 math 〉; ••h is the second Equation.

The Reduction of both Equations. 〈◊〉〈◊〉 first was 〈 math 〉〈 math 〉. ••refore dividing by 〈 math 〉〈 math 〉. substracting 〈 math 〉〈 math 〉. ••he latter Equation was 〈 math 〉〈 math 〉, i. e. ••ituting again the value ee, which was 〈 math 〉〈 math 〉; and by transposition, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉, or 〈 math 〉〈 math 〉,

or (putting 2 f for 〈 math 〉〈 math 〉) 〈 math 〉〈 math 〉.

Wherefore we have the value of xx twice expressed, but by quantities partly unknown, because y is found on both sides. Wherefore now we must make a new comparison of their values, whence you'l have this new

5. Third Equation, in which there is only one of the un∣known quantities: 〈 math 〉〈 math 〉; and adding on both sides both yy and aa, 〈 math 〉〈 math 〉; or dividing by 2, 〈 math 〉〈 math 〉; and transposing 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; which is the value of the quantity y in known terms.

But this value in one of the precedent Equations, viz. in this 〈 math 〉〈 math 〉, being substituted for y and its square for yy, will give 〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉.

6. The Geometrical Construction, which Schooten gives us p. 153. Having made the angle KAB (n. 2. Fig. 29.) e∣qual to the given one D, erect from A, AL perpendicular to KA, meeting the Perpendicular EM in L; and from the Centre L, at the interval of the given right line d, describe a Circle that shall cut KA and EL produced to K and M. Then assuming EN=KA, join MA, and from N draw NH pa∣rallel to it, which shall meet AB in H. Afterwards, having described from L, at the interval LA, the segment of a Circle ACB, draw from H, HC perpendicular to AB meeting the circumference in C, and join AC, CB.

NB. The reason of this elegant Construction, which the Author conceal'd, for the sake of Learners we will here shew. 1. Therefore, he reduc'd the last Equation (extracting the root, as well as it could bear, both of Numerator and Deno∣minator) to this: 〈 math 〉〈 math 〉 multipl. by 〈 math 〉〈 math 〉, so that after this way the Construction would be reduc'd to this proportion, as d+f to a so 〈 math 〉〈 math 〉. 2. He made the angle KAE = to the given one D, and the angle KAL a right one, so that having described the segment of a Circle from L the inscribed angle will also be made equal to the gi∣ven one, according to the 33. Lib. 3. Eucl. 3. By doing this, EL expresses the quantity f, since by reason of the simi∣larity of the ▵ ▵ KOA s. GFD, n. 1. and AEL (for the angles LAE and AKO are equal, because each makes a right one with the same third Angle KAO) you have as KO to OA so AE to EL i. e. 〈 math 〉〈 math 〉

4. Making now LM and LK = d you had EM=d+f, and AK = 〈 math 〉〈 math 〉 (for the □ AL is = to aa+ff, which being substracted from □ LK = 〈 math 〉〈 math 〉.)

5. Wherefore there now remains nothing to construct the last Equation above, but to make EN=AK, and to draw HN parallel to AM; for thus was the whole proportion as EM to EA so EN to EH 〈 math 〉〈 math 〉 to x. Q. e. f.

For the point H being determined, a perpendicular HC thence erected in the segment already described defines the Point C, which answers the Question.

PROBLEM VII.

HAving given the four sides of a Quadrangle to be inscribed in a Circle, to find the Diagonals and their Segments, and so to construct the Quadrangle, and inscricbe it in the Circle. As e. g. suppose the given sides are AB, BC, CD, DA (Fig. 30 n. 1.) which now we suppose to be joined in••o a quadr••n∣gle

inscrib'd in the Circle, the Diagonals also AC and BD being drawn (n. 3.) to find first the segments of the diagonals Ae, Be, &c. which being had, the Construction is ready.

SOLƲTION.

Denomination. Make AB=a, AC=b, CD=c, DA =d, Ae=x [for this segment alone being found, the rest will be found also, as will be evident from the process.] Since therefore the vertical angles at e are equal, and likewise the angles in the same segment BCA, BDA, also DAC, DBC, &c. are equal, the Triangles AeD and BeC, also AeB and CeD are similar: wherefore it will follow that,

• 1. As DA to Ae so CB to Be 〈 math 〉〈 math 〉
• 2. As AB to Be so CD to Ce 〈 math 〉〈 math 〉
• 3. As AB to Ae so CD to De 〈 math 〉〈 math 〉

Therefore the whole Diagonal AC will be = 〈 math 〉〈 math 〉 and BD = 〈 math 〉〈 math 〉.

2. The Equation. But now by Prop. 48. Lib. 1. Math. Enucl. the Rectangle of the Diagonals is equal to the two Rectangles of the opposite sides.

Diag. AC, 〈 math 〉〈 math 〉

Diag. BD, 〈 math 〉〈 math 〉

Therefore the □ of the Diagonals 〈 math 〉〈 math 〉.

3. Reduction, i. e. taking (a) for unity 〈 math 〉〈 math 〉; i. e. the quantities on the left hand being reduc'd to the same denomination.

〈 math 〉〈 math 〉; and multiplying both sides by dd, 〈 math 〉〈 math 〉; and dividing both sides by d, 〈 math 〉〈 math 〉; and then dividing both sides by 〈 math 〉〈 math 〉; i. e. in the present case, where b by chance happens to be = a, 〈 math 〉〈 math 〉

Therefore 〈 math 〉〈 math 〉 or in our case 〈 math 〉〈 math 〉

4. The Geometrical Construction, which, by supposing a (and in the present case also b) to be unity, ought to deter∣mine,

1. The quantities cd, 〈 math 〉〈 math 〉, cc, and their aggregate with unity. 2. The aggregate of cd and dd. 3. To divide the one by the other. And, 4. To extract the root out of the quotient, or also to extract the roots first out of each quantity, and divide them by one another; which may all of them be separately done in so many separate Diagrams, but more ele∣gantly connected together after the following or some such like way. 1. Join AD and DC (n. 2.) into one line, and having described a semi-circle thereupon, erect the Perpendicular DE; and the line AE drawn will = 〈 math 〉〈 math 〉. 2. Making the angle CAG at pleasure, make AF=AB, and draw CG pa∣rallel to the line DF; so FG will be =〈 math 〉〈 math 〉. Now if, 3. in the vertical angle you make AH=CD, the line HI drawn parallel to DF will cut off AI=cd. 4. In AK erected = to AB, if you take AL=AH or CD, and draw LM paral∣lel to KH, you'l have AM=cc. 5. Having prolonged AG to N and AH to O, so that GN shall be =AI+AM and AO=AB or AK, and having described a semi-circle upon the whole line NO, a perpendicular erected AP will be = 〈 math 〉〈 math 〉 and so, 6. if AQ be made =AE and AR=AF or AB, and you draw a line RS from R pa∣rallel to PQ; AS will be = x, i. e. the segment sought Ae of the Diagonal AC; which being given, by force of the first Inference premis'd in the Denomination above, by drawing DS and, having made DT=BC, TV parallel to it; you'l have also the other segment Be=SV and by their Intersection on the line AB (n. 3.) the point e, thro' which the Diagonals must be drawn which will be terminated by the other given sides, and thence you'l have the quadrilateral figure ABCD sought, to be circumscribed about the Circle, according to Con∣sect. 6 Defin. 8. Mathes. Enucl.

NB. Unless we had here consulted the Learner's ease, the artifice of this Construction might be proposed after a more short and occult way, thus: Make DE a mean proportional between AD and DC, and draw AE. Then having made any angle CAG, make AF=AB, and at this Interval de∣scribe

the circle FROK, and draw CG parallel to DF. More∣over in the opposite vertical angle, having made AH=CD, draw HI parallel to DF, and having erected the perpendicular AK, and thence the abscissa AL=AH, make LM parallel to HK, and thence having prolonged AH to O, and GN being made equal to AI+AM, make AP a meam proportional be∣tween AO and AN, cutting the hidden circle in R; and last∣ly having made AQ=AE, if RS be drawn parallel to QP, you'l have AS the value of x sought, &c.