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

SOLƲTION.

I. For the Radius FM.

1. Denomination. Make CB=a, EB=b then will CE =a−b; for which for brevities sake put c. And let FM or FN or FK = x: Therefore EF will be = b+x, and CF (sub∣stracting FK from CK) = a−x. Wherefore now you'l have at least the names of the three sides in the ▵ CFE, so that according to Poblem 8. the Segment of the base GE may be determined (which indeed is determined already, as being =LE−LG or MF i. e. b−x) for which in the mean time we will put y; and now will CG=c−y.

2. For the Equation. If the □ GE=y be substracted from the □ EF = 〈 math 〉〈 math 〉, you'l have the square of the Perpendicular FG = 〈 math 〉〈 math 〉; and, if □ CG = 〈 math 〉〈 math 〉 be substracted from the □ CF = 〈 math 〉〈 math 〉, you'l have the same □ of the Perpendicular FG = 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

3. Reduction. And taking from both sides the quantities xx and yy, 〈 math 〉〈 math 〉; and adding 2ax and cc, but taking away aa from both sides, 〈 math 〉〈 math 〉; and adding 2ax and cc, and taking away from each side 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉. but the same y or EG is = EL−MF i. e. b−x. There∣fore 〈 math 〉〈 math 〉; which is a new and more principal Equation: And multiply∣ing both sides by 2c (you have a new Reduction) 〈 math 〉〈 math 〉; and adding 2cx, and transpo∣sing the others, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

The Geometrical Construction of this first Case. Add the de∣terminate (n 2.) quantity 2bc into one sum with the quantity aa, as in the beginning, n. 3. Then from this sum substract successively the quantities bb and cc, and there will come out (the same n 3) FH, whose □ 〈 math 〉〈 math 〉: Which since it must be divided by 〈 math 〉〈 math 〉, make (the same n. 3) as FI = 〈 math 〉〈 math 〉 to FH = 〈 math 〉〈 math 〉, so FH to FM the Radius sought of the little Circle to be de∣scribed. This quantity FM being thus found, place it from L to G (n. 1.) and from G erect a Perpendicular, which be∣ing cut off at the interval CF (which may be had, if from CB or CK you cut off FK=FM) or from E at the interval EF (which is composed of the Radii EN and FN) gives the Center of the little Circle to be described.

The Arithmetical Rule. Add twice the □ CEB to the square of the greatest semi-diameter CB, and from the sum substract the Aggregate of the □□ CE and EB; divide the remainder by the sum of all the three Diameters, (AB, AL and LB) i. e. by double the greatest AB; and you'l have the Radius FM, &c. For Example sake let a be = 12, b=4; c will be = 8, and x will be produced = 2 ⅔.

II. For the Radius Vy by help of the obtuse-angled ▵ DVC.

1. Denomination. CA=a as above, DA or DL=b, and putting x again for the sought Vy or VK, CV will be = a −x, DL or DR=b, and consequently DV=b+x, and DC=a−b, for which for brevity's sake we will put c. Now you'l have at least in Denomination in the ▵ CVD the three sides, so that aecording to Problem 9. the segment CW may be determined, for which in the mean while we will put y; then will DW=c+y, which is the same as DL−WL or Vy i. e. b−x.

2. For the Equation. If the □ CW=yy substract it from the □ CV = 〈 math 〉〈 math 〉, and you'l have the □ of the Perpendicular VW, 〈 math 〉〈 math 〉; and if the □ DW = 〈 math 〉〈 math 〉, substract it from the □ DV = 〈 math 〉〈 math 〉, and you'l have the same □ of the Perpendicular VW = 〈 math 〉〈 math 〉 Therefore 〈 math 〉〈 math 〉.

3. Reduction. Therefore taking from both sides xx and yy, 〈 math 〉〈 math 〉; and adding 2cy and 2ax, 〈 math 〉〈 math 〉; and substract∣ing 〈 math 〉〈 math 〉, and dividing by 2c, 〈 math 〉〈 math 〉

But if you add to the same y or CW DC=c, you'l have DW = 〈 math 〉〈 math 〉 i. e. reducing this c to the same Denomination, 〈 math 〉〈 math 〉 = DW. But the same DW =DL−WL=b−x. Therefore 〈 math 〉〈 math 〉, and multiplying 〈 math 〉〈 math 〉, and adding 2cx, and transposing the rest, 〈 math 〉〈 math 〉, and dividing by 〈 math 〉〈 math 〉 just as above in the first Case.

4. The Geometrical Construction therefore will be the same as there. See Fig. 23. n. 4. and 5.

5. The Arithmetical Rule is also the same, but the given quantities in this Example, which the figure of the Problem will shew, thus vary, while a remains 12, b will be 8, and c 4, from which data (or given quantities) there will not∣withstanding come out again, for x or the Radius Vy 2 ⅔.