444 
GEOMETRY. 
young matliematicians, but promifes to afford very con- 
tiderable afliftance to architedls and military engineers, 
by facilitating the method of drawing plans and eleva¬ 
tions, in ftriking lines of circiinivallation, projecting the 
form of outworks, and conftruCting the angles and brealt- 
works in regular fortifications. But the principal uti¬ 
lity of tliis fpecies of geometry is in its application to 
the divifion of aftronornical inftruments. 
The method here laid down, is defined by the in¬ 
ventor, to be that by which, with the compafs only, with, 
cut the life of the rule, wc can more correCtly determine 
the pofition of points. And it is recommended to the 
artili, or mathematician, in order to infure precilion, to 
liave as many pair of good compaffes as there are dil- 
tances to be taken in the folution of the jiroblcm ; be- 
caufe he will often have occafion to ufe tlie fame open¬ 
ing, ordifiance, feveral times after havingintermcdiatelv 
uled other diftances. In fiich cafe, without Ihutting or 
opening his compafles, he may recur to the dillance he 
■wants. In I'olving a few problems, we fliall fuppofe 
tills method to be purfued, and fliall accordingly deno¬ 
minate thefe inflruments, the firft, fecond, or third, pair 
of compaffes. 
I. To divide the cirntmference of the circle B Di/, into four 
equal parts,—Tet A be the centre, and equal to A B, as 
the circumference will tJien be divided into four equal 
parts, B F, Fhi, Ef fB. 
Demonjlration. B A E being a diameter, and the tri¬ 
angles a A B, a A E, having all their Tides equal, and 
confequently the angles a A B, aAE, alfo equal (Eu¬ 
clid 8. b. I.), thefe two angles will be right ones (Eu¬ 
clid ij. b- I.); therefore (a B)* = (A (a A)* 
(Euclid 47. b. I.), and, fubtradfing from each fide 
(AB)2. wc have (« B)^ — (A B)* = (a A)*. In order 
to tlicwten the work, let A B = i ; and we fiiall have 
((zB)-®=:(BD)* = 3; hence, (aA)*=3 —1 = 2; and 
(BF )2 = (cA)2=2=:i-l-i=:(AB)2-p(AF)*. There¬ 
fore, in the triangle FAB, the angle FAB will be a 
right one (Euclid 48. b. i),and conl'eqiiently FAE alfo 
(tiijclid 13. b. i); therefore the arcs B F, FE, will 
4 
be equal to each other, and fourth parts of the circle; 
as alfo the arcs B/', fE, 
2. Corollary, The angles B A a, B A F, being right 
angles, the three points. A, F, a, ivill be in the fame 
right line. 
3. Tims we have the circumference already divided, 
in two equal parts at the points B and E ; in three parts, 
at the points B, D, d, (E. 15. b.4.) in four parts, at 
the points B, F, E,f-, and in fix parts, at the points 
B, C, D, E, d, c. 
4. To divide a circumference into eight equal parts. —In the 
fame figure, let A B = a G = a H with tlie firft pair of 
coiupaires; and P,.a~Gg~y\h with the third pair; 
then we (hall have A™ A <2, and the circumference will 
be divided into eight equal parts at tlie points B, G, F, 
H, E, A,/, 4. 
Denwnjiration. As (Aa)®=r 2, we have (Ao)*=(AG)* 
-^-((jG)^, therclbre the angle cG A is a right angle 
(E. 48. b. I.) ; and as c G A is an ifoceles triangle, the 
angle GAc=Ga A, and each equal to half a right angle ; 
lienee, G A F (= E Ac) =: half a right angle = half 
B A F; and G F — Bg. But by conflr. G^— B F; hence, 
taking B G from each, we have G F — B^. Thus we 
jirove it for the other arc, and the circumference is di¬ 
vided into eight equal parts. 
5. To divide the circumference into 12 equal parts. —Things 
being as at No. i. draw (fame figure) AB2c:FN=N>7 = 
FO = Oe; and the circumference will be divided into 
12 equal parts at the points B, N, C, F, D, O, E, 
0, d, f c, n. 
Demonf ration. As BFrrFE, fubtraft from them BC= 
DF', and we have CFcziFD ; but CD is the fixth part 
of the circumference ; therefore CF is the twelfth part. 
Thus we go on to prove the other parts. 
6. To divide the fame circumference into 24 equal parts .— 
Things being as at No. 4 and 5. draw (fame figure) 
A B = G L = LM = G A Af = HI =I K=:H m =mI, with 
the firfi; pair of compatl'es, and the problem is folved. 
Demonjlration. From G F =: G B fubtradf C F = N B, 
and we get GCz=GN. But C N is the twelfth part ; 
therefore GC, GN, are the twenty-fourth parts of the 
circumference. 
7. The ancients, by means of the centre A and the 
radius A B, divided, by the compafs only, the circum¬ 
ference into fix equal parts. Other divifions they ef¬ 
fected with the rule and compafs, by taking feveral 
points out of the circumference. We have been able 
to determine one Tingle point a, with which alone we 
can divide it into 24 parts with the compafs alone ; 
which is more expeditious, more convenient, and at the 
fame time much more exadt, than the method ufed by 
the ancients. 
8. Let us alfo notice the elegant law which guides 
the three openings of compaffes necclfary for this divi¬ 
fion : The opening of the firft pair 2=3/1, of the third 
=23/2, and of the fecond 2= 3/3. 
9. Lemma. If in the circle BGE, there be a radius 
AB2 =i, while the arc BG is the eighth part of the cir¬ 
cumference, we ihall have the fquare of its chord BG, 
that is (BG)2:=2 — 3/2. 
Demonfration. On the diameter BE draw the GP per. 
pendicular to liB, G 
being found as before. 
Then as the angle GAB 
2=43°, AGP=245°, 
AP2=PG; hence,(AG )2 
=:<PG )2 -f (AP )2 2= 
2(AP)^, or 2(AG)®2= 
4(AP)^or2=:(2AP)2, _ 
anti 3/2—zAP; hence, AP~i\/2, and BP2=AB- 
-AP 
= i—2f2. But BP: BG :: BG : BE; hence, BG =>2= 
BPxBE = 2BP, and BG=2 2—3/2. 
In like manner GE =:3 + f 2. 
10, The 
