AO (P6+P,). (P, +P1) = 1o• (2P 6 ̧+2P2+P8−P ̧—2 ̧†? ̧) 2 or 4sx. O'TAO. (2SX + PW-RY). In the same way it is proved that 4RV. O'TAO. (20'T SX-PW), and taking the difference 4 (SX RV). O'TAO. (3SX-20'T + 2 PW RV) AO. (2SX + 2ns Again, because 4 (SX-RV). PW and 2PW PP,; therefore, by 4SX.PW = (P+P,). (Ps - P2) 4SX. PW-4RV.PW, proceeding as before, AO. (O'T-RV-2PW), = 2 4RV.PW = (P, P ̧). (P8 — P2) =AO. (SX-2 RV-O'T), 5 and taking the difference, 4 (SX RV). PW AO. (20'T+3RV SX - - 2PW) Subtracting again, to get the difference of the rectangles, 4 (SX - RV). O'T and 4 (SX-RV). PW, or 4ns. 4mr, we have 4ns. 4mr= AO. (4ns — 8mr + 2 (SX — RV )). VOL. IV. PART I. Y 2SX. 2RV=(P6 + P,). (P3—P¿), and, by the second Lemma, =AO.(SX+PW-O'T―RV)=AO. (2ns—2mr). Lastly, to prove that O'T.PW= (49)* or that 40′T. 4PW=a03. 4 8 20'T. 2PW = (P, + P1) · (P ̧ — P2), and by applying the second Lemma, and proceeding as in the last case, we have These properties being now proved, the demonstration of the construction in the question is readily completed. 8mr = A0 (a) = 80D by construction, ns — mr is = OD; and (ns — mr)2 — OD2 add to these 4ns . mr = (b) then (ns + mr)2 = OD2 + A02 = OD2 + 0Q2= DQ'because OA 0Q= ; therefore ns + mr = DQ = DF = DE by con struction, and ns mr being = OD, ans is 4 (O'T+PW)'= 0E2 + = EQ®. Consequently o'T + PW EQ EG by construction, and therefore 20'T OE + EG OG, or P, P = OG. OA But, by the second Lemma, P, . P1 = °^. (P ̧ —P ̧) = 10. 4 2 2 он = оQ. OH OK by construction; and, by the property of the circle, KM + KM' = OG, and KM. KM OK; = P to therefore KM + KM'= P, + P, and KM KM′ = P, . and consequently KM P, and KM P,; that is, KM is the perpendicular from a upon co; wherefore the points a and N coincide and AN is the seventeenth part of the circumference. A line drawn through M' parallel to oc will evidently pass through d, the 4th division, and by setting off FQ and EQ on AB both ways, and varying the latter part of the construction a little, any of the other points of division may easily be found. The calculation from this construction is very easy, for if we take the radius = 1, we have OD and O2 = 1; whence DE=DF DQ = {√17, FQ = √(oq2 + of3) = 3√34 + 2 √ 17, EQ = √(oq2 + or2) = √342√17, OH FH OF 342/17 — √17 — }, OG = EG + OE = {√34 — 2√/17 + ¿√ 17 ‡ ¿• OG But KM is evidently =- + {√(OG2 — 40K2) 2 and og2 = 2 (√34 - 2√ 17 + √ 17 — ¿)' 82 = 34 (52 — 4 √ 17 +2√34 — 2√ 17 × (√17—1)} =÷x(52—4√17+8√34+2√17 −4 √34 — 2 √/17), therefore OG2-40K2=ib( 17 +3✓ 17−√/ 34-2 √ 17-2 √ 34 +2 √17) and KM 34 — 2 √17 + 18 √ 17 — 36 + ¥ √(17 + 3√17—√34—2 √ 17—2√34+2√17), the same expression that Mr. Leslie has given for the cosine of the 17th part of the circumference, at page 419 of the 2nd edition of his geometry. End of the first part of the fourth vol. THE MATHEMATICAL REPOSITORY, VOL. IV. PART II. ORIGINAL ESSAYS. ARTICLE I. On Combinations. By Mr. RoB. J. DISHNEAGH, Trinity College, Cambridge. To the Editor of the Mathematical Repository. SIR, I hope the following solution to a problem which occurs in the Novi Commentarii Petropol. Vol. X. will be deemed worthy of insertion in your valuable publication. The proof, if such it can be called, which is given by the Author Josias Weitbrecht is inductive. The advantage of a commodious notation is sufficiently obvious. June 23, 1814. The Proposition alluded to, is, I am, &c. R. J. DISHNEAGH. (1). If any limb has n muscles, each having m distinct motions, deterinine the number of motions when p muscles act at once. To which he adds another Proposition. (II). To determine the total number of motions which can. be exerted by the limb. For convenience, we shall state the first proposition, thus, Determine the number of combinations of n things a, b, c, &c. |