bers; the formula found above, making the proper substitution, will give us, Now this continued fraction falls under Lemma II; for since the denominators 3n, 5n, 7n, &c. increase continually, whilst the numerator m2 continues of the same magnitude, the component fractions will evidently be, or at least will soon become, less than unit; hence the m value of tang. is irrational; hence if the arc is commensurable with the radius, its tangent will be incommensurable. From this, we deduce, as an immediate consequence, the proposition which forms the object of this Note. Let be the semicircumference of which the radius is 1; if were rational, the arc would be so T 4 4 too, and therefore its tangent would be irrational: but the tangent of the arc is well known to be equal to the radius 1; hence ☛ cannot be irrational. Hence the ratio of the circumference to the diameter is an irrational number.* It is probable that this number is not even included among algebraical irrational quantities, in other words, that it cannot be the root of an algebraical equation having a finite number of terms with rational coefficients: but a rigorous demonstration of this seems very difficult to find; we can only shew that the square of ≈ is also an irrational number. Thus, if in the continued fraction, which denotes tang. x, we put x=, since tang. =o, we must have m &c. But if were rational, and we had ☛a: = m and n being whole numbers, there would result from it n Now, as this continued fraction evidently comes under Lemma II., * This proposition was first demonstrated by Lambert, in the Memoirs of Berlin, anno 1761. its value also must be irrational, and cannot be equal to the number 3. Hence the square of the ratio between the circumference and the diameter is an irrational number. NOTE V. Containing the analytical solution of various problems concerning the triangle, the inscribed quadrilateral, the parallelepipedon, and the triangular pyramid. PROBLEM I. Given the three sides of a triangle, to find its surface, with the radii of the inscribed and the circumscribed circle. Put the sides BC-a, AC=b, AB=c; if from the vertex A, the perpendicular AD is let fall on BC the opposite side, then (Prop. 12. III.), we shall have AC AB2 + BC2 – 2BC × BD; hence B a2+c2-b2 BD This value gives AB2-BD2, or AD c2 2a √[4a2c2—(a2+c2—62)2]. Let S be the area of the triangle, we 2 a shall have SBC × AD; hence S={√[4a2c2-(a2+c2_b2)2]={√(2a2b2+2a2c2+2b3c2_aa_ba_c1). This formula may farther be reduced to a form more convenient for logarithmic calculation. To effect this, we must consider that the quantity 4 ac2 (a2+c2 — b2)2 is the product of the two factors 2 ac+(a2+c2—b2) and 2 ac—(a2+c2—b2); the first = (a+c)2—b2 =(a+c+b) (a+c-b); the second = b2 — (a—c)2 = (b+a—c) (b-a+c); hence we shall have ing S=1√[(a+b+c) (a+b—c) (a+c—b) (b+c-a)]. Lastly, by maka+b+c =p, which gives a+b+c=2p, a+b-c=2p-2c, 2 a+c→b—2p—2b, b+c—a—2p—2a, we shall have, still more simply, S=(p.p-a.p-b.p-c). Thus it appears, that to obtain the surface of a triangle whose three sides are given, we must take the half sum of those three sides; from this half sum subtract each side separately, which will give us three remainders; multiply these three remainders and the half sum all together; and finally, extract the square root of the product: this root will be the area of the triangle. Now, let z be the radius of the circle which circumscribes the triangle, and u the radius of the circle inscribed in it; by Prop. 23. III., we shall have Given the four sides of the inscribed quadrilateral, to find the radius of the circle, the surface of the quadrilateral, and its angles. Put the given sides AB-a, BC=b, CD=c, DA=d, and the unknown diago-d nals AC=x, BD=y; by Prop. 33. III., we shall have xy=ac+bd and = A y 1), y = √((a But according to the preceding Problem, the radius of the circle circumscribing the triangle ABC, whose sides are a, b, x, may be ex pressed by the formula z = In place of √[4a2 b2 — (a2 +b2 — x2)3]* x, substitute the value just found for it, and decompose the result into factors; we shall have %= (ac+bd) (ad+bc) (ab+cd) (a+b+c―d)(a+b+d—c)(a+c+d—b)(b+c+d—a) This being proved, the area of the triangle ABC = ab, that of the 2 ; hence the area of the quadrilateral ABCD = {√[(a+b+c−d) (a+b+d—c) (a+c+d—b) (b+c+d—a)]. And if for the sake of brevity, we put p= (a+b+c+d), we shall have the area ABCD=√(p-a.p-b.p-c.pd). In fine, to obtain one of the angles, B for example, observe that the triangle ABC gives cos B a2+b2-x2 ; substitute the value of x, and by reduc In the quadrilateral ABCDE, of which the opposite angles B and C are right, the two sides AB, AC with their contained angle BAC being given, to find the other two sides and the diagonal AD. Suppose AC=b, AB=c, and the angle BAC A; if BD and AC are produced till they meet in E, the triangle BAE, right-angled at B, wherein are known the angle BAE and the side AB, will give AE = b. Then the triangle b A. (C — 5 cos A, 2) = ~/ (b2 + c2 — 2 bc cos A). But from the triangle sin A sin A BAC, we have BC= √(b2+c2. 2 bc cos A.) Hence the diagonal AD, which joins the two oblique angles, is to the diagonal BC, which joins the two right angles, as 1 is to sin A. Scholium. The diagonal AD, is also the diameter of the circle in which the quadrilateral ABCD would be inscribed. In this circle we should have the angle ABC ADC; hence drawing CF perpendicular to AB, the triangles BFC, ADC are similar, and give AD: BC:: AC: FC:: 1: sin A; which agrees with the result obtained above. PROBLEM IV. Given the three edges of a parallelepipedon, with the angles which they form together, to find the solidity of the parallelepipedon. ولا Suppose the edges SA=f, SB =g, SCh; and the included angles ASB = ASC, BSC = a. If from the point C, a straight line CO be drawn at right angles to the plane ASB, the right-angled triangle CSO, will give CO = CS sin CSO = hsin CSO. Likewise the surface of the B parallelogram ASBP = fg sin 2. Hence, if we put S for the solidity of the parallelepipedon ST, we shall have Sfgh sin a sin CSO. We have yet to find sin CSO. E To effect this, from the point S as a centre, with a radius = 1, describe a spherical surface meeting the straight lines SA, SB, SC, SO, in the points D, E, F, G; you will have a triangle DEF, in which the arc FG is perpendicular to ED, since the plane CSO is perpendicular to ASB. Now the triangle DEF, which has its three sides DEY, DF = 6, EF = a, gives cos DE = ", DF=6, EF = 7, gives cos E = COS sin a sin y 26 and Then the right-angled triangle EFG gives sin GF, or sin CSO=sin E sin EF sin y sin E. γ Hence Sfgh sin a sin sin E, or S=fgh (1. 2 COS α- COS 26 2 γ COS + 2 cosa cos 6 cos y). In this expression, the quantity under the radical is the product of the two factors, sin « sin y+cos € cosa cos y, and sin a sin'-cos + cos & cos y. The first cos cos (+2)=2 sin The same things being given as in the last Problem, to find the expression of the diagonal which joins two opposite vertices. ST Suppose the diagonal of the base SP %, and the required diagonal u: the triangle ASP, in which cos SAP: cos 7, will give |