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

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.