2k3 sin ag 1 + 2k2' oos a = k), X= –2k4y, and the common point is - 2kt k? tangents to (1), (2) at this point are x + k4y + k% = 0, and 3+ 4k4y— 2k%% = 0 (I., II.); whence cos B = a fraction whose numerator is sin? A + 4/8 sind B-254 sin? 0-2k6 sin B sin C cos A + ka sin C sin A cos B-5k4 sin A sin B cos O, and denominator the product of the two radicals {sin? A + 28 sinoB+k4 sinC– 2k6 sin B sin C cos A – 2k2 sin C sin A cos B-2k4 sin A sin B cos C}}, {sin? A + 16k8 sin?B + 4.4 sino C+16k6 sin B sin C cos A +4ko sin C sin A cos B– 8k* sin A sin B cos C}:. But sin A = sin 2a cos A = 1–2k, B=0=a, sin B sin C cos A = (1-246) sin’ a, sin C sin A cos B = sin A sin B cos O = 2/6 sino a ; and, on making these substitutions, the numerator becomes 2K4 (K2— 1)3 (2K* + 1) sino a, and the two radicals {k4(1 – K2) (2k + 1)2 sin’a}, {4X4 (1+2k2)3 (1 — K2) sina}}; 1-K2 3 2(1 - ka) whence cos B cos 2K2 +1) 1 + 2k2 {1+2 (cos B)}}{1+2 (cosa)}} = 3. Now the ratio of the areas of the two curves Px2 + 2py% = 0, Ppa Qy2 + 2q2x ; hence the squares of the (22+2Pp) (92 +2Q9). latera-recta of the circle and hyperbola will be as } (-2 cosa)? 4 cosa a (2+p)?' (-2-2) (2 sin? a)! 1 4 cosa a (1 + 8 cos? a)} : 8 sino a cos a. Of course, by using the triangle A'B'C', we should get the equivalent ratio (1 + 8 cosa B)? : 8 sin3 B cos B, which is a sufficient test. 3, 4. If h, l2, bz be the latera-recta of the three curves, we have 4b = 17?, since the semi-latus-rectum of the parabola is the perpendicular from the centre of the hyperbola on the tangent at a point whose central distance is th. Hence 1 la l lg also the angle of intersection of the circle and parabola at B is a, and at B’ is B, whence : lg = 12: 132 the ratio (B). The angle of intersection of the parabola and hyperbola at A is -2a, and at A' is -- 2B, whence 132 : 172 = 1;? : 1,2 = the ratio (B). or = 0 is ( or -2)" or (1)" 5224. (By the Rev. H. G. DAY, M.A.)-On each of n pillars, whose heights, in ascending order of magnitude, are C1, C2, C3, ..., @n, points are taken at random; find the chance of the point so taken on the rth pillar being the highest. Solution by the PROPOSER. The chance that the highest point so taken is below the height og is (cm)* ; and similarly for Cs-1. C8+1C8+2 ... Con Hence the chance that the highest point taken lies between the heights Cs and Cs-1 (cz)n-: is (08-1)-8+1 C8+1C8+2 ...on Cs.C8+1 .... on Let this expression be denoted by ps : Now the chances in this case of any particular one of the points on the pillars Ps marked from s to n being highest will be equal, each being m-8+1 Therefore the probability required is P1+ P2 P3 Po 2 n-9+1° + + n-1 5504. (By Dr. BOOTH, F.R.S.)—If A be the area of a plane triangle, prove that 4A2 = abco, where a, b, c are the sides of the original triangle, and o the semiperimeter of its orthocentric triangle. It is easy Solution by CHRISTINE LADD, D. EDWARDES, J. O’REGAN, and many others. show that radii of the circumscribed circle drawn to the points A, B, C are perpendicular to the sides (a', b', d suppose) of the orthocentric triangle; hence the triangle ABC is divided into three parts whose areas are respectively Rd', {RŐ, Rd'; hence we have abco A = }R (a' +b' + c') = Ro therefore 4A abco. 44 132. CONTRAPOSITION : BY ALEXANDER J. ELLIS, F.R.S. In the Syllabus of the Association for the Improvement of Geometrical Teaching (p. 4) we read : “In the typical Theorem, If A is B, then C is D ................(i.); the hypothesis is that A is B, and the conclusion that C is D. From the truth conveyed in this Theorem it necessarily follows: If C is not D, then A is not B ..... ..(ii.) or or Two such Theorems as (i.) and (ü.) are said to be contrapositive, each of the other.” At the meeting of the Association on January 11th, I was much surprised to hear the President read letters complaining of the difficulty of This piece of logic, one of which, so far as I can remember, asserted that it was impossible to prove it in the case of a negative proposition, that is, I suppose, to derive (i.) from (ii.). It seems, therefore, advisable to put the proof into an easy form, following the principles of Boole’s Laws of Thought, but not his elaborate notation and theory. We deal with four propositions, A is B, A is not B, C is D, C is not D. Now, as either the positive or the negative form of each must be true, these can co-exist in four different ways only, viz. : (1) A is B, and at the same time C is D; (3) A is not B, and at the same time C is D; or, finally, (4) A is not B, and at the same time C is not D. This merely states a fundamental relation of thought without any hypothesis. Now suppose that by any course of reasoning it is established that “if (or whenever) A is B, then C is D.” This makes (1) exist, but denies the existence of (2). Let the reader conceal the line (2) with a pencil, as excluded by the demonstration of a particular theorem. Then let him examine the lines (1), (3), and (4) for the cases in which C is not D. He will find this in (4) only, where it is associated with “ A is not B.” Hence it follows from the fact “whenever A is B, C is D," that " whenever C is not D, A is not B,” or (ii.) is derived from (i.). Observe that it does not follow conversely that “whenever A is not B, then C is not D,” because both of the lines (3) and (4) are left untouched by the original proposition. Now take the negative proposition, "if C is not D, then A is not B.” This excludes the possibility of “ C is not D” co-existing with “ A is B,” that is, it excludes the same case (2) as before. Now, covering (2) with a pencil, look for“A is B” in lines (1), (3), and (4), and we only find it in (1) associated with “ C is D.” That is, under these circumstances, if A is B, then C is D; that is, (i.) follows from (ii.), just as (ii.) followed from (i.). This relation constitutes the fact known as contraposition. Observe however, again, that the converse, “if C is D, then A is B,” does not follow from (ii.), for lines (1) and (3), which are left untouched, both contain “O is D,” shewing that, so far as (ii.) is concerned, “C is D” may co-exist either with “A is B” or with “A is not B.” As many pupils have great difficulty with conversion as well as contraposition, this simple means of putting the four cases before them, and leading them to draw the inference themselves, may possibly be useful to them not only in geometry, but in all after life. 5466. (By the Editor.)—If two random points be taken one in each of (1) the arcs, (2) the areas, of two semicircles that together make up à complete circle ; find, in each case, (a) the average distance between the points, and (b) the probability that this distance is less than the radius. D Solution by E. B. Sertz. 1. (a). Let ADB and ACB (Fig. 1) be the semicircles, M and N the random points, and O the centre of the circle. Put OA=r, LAOM= 0, and LAON=0; then we have MN = 2r sin (0+); also an element of the arc at M is rdo, and at N, rdo; moreover, the limits of e are 0 and TT, O and a; hence the required average 1 2r sin $(0+) rdo rdo +272 M M B and of o, N с 16r (sin 10 + cos jo) do Fig. 1. 1. (B). Let the angle MOP ft (Fig. 1); then, while M is fixed, N may be taken anywhere in the arc AP, and the distance MN will be less than the radius; also the limits of 6 are 0 and $t, and doubled: hence the 2 1 required probability 72 p? (fa-0) na de 9 M N 2. (a). Draw the chord EF through M, N (Fig. 2), and draw OC perpendicular to EF. Put OD = x, DM = y, DN = %, DE = y1, DF=%, ZADM = 0, and ZOEC = "; then we have OC = x sin e = r sin o, sin dx = r cos o do, yi (32—2 sin? 0)* - - cos 0 =r(cos $-cot & sin o), z = (y2 — 22 sinə g)# + x cos 0 =r(cos 0 + cot sin o); also an element at M is sin o dx dy, or go cos o do dy, and at N, (y +%) do dz; moreover, the limits of 0 are 0 and a; of x, -y and Fig. 2. r; of y, 0 and yz; and of Z, 0 and 21; hence, since the whole number of ways the two points can be taken is daigt, the required average 4 sin 0 do dx dy 724 32r 4572 2. (B). Let MP the radius (Fig. 2); then, while M is fixed, N may be taken anywhere in DP, and the distance MN will be less than the radius. The limits of 6 are 0 and fit, and doubled; and the limits of $ are 0 and 0, and doubled. When <, the limits of y are 0 and Yı, and of z, O and r-y. From 0=31 to 0=the limits of y are 0 and y, and of z, 0 and r-y, when ø<7-20; when p >-20 and <fr, the limits of z are 0 and 2, from y=0 to y=r-21, and they are 0 and r-y from y=r-% to y=yı; and when p>, the limits of y are 0 and 91, and of %, 0 and 21. Hence the required probability 16 ra re r(y+z) de cos o do dy dz r (y + x) cos o do dy dz do poft re [3 cosec 0 sin (8-0) – coseci o sin3 (0-0)] d8 cos o do [3 cosec 0 sin (0-0) - cosec30 sin3 (0-0)] cos o do + [6 cos 0–6 coté 0 sina ¢ cos 0 - 2 coss 0-2] cos 0 do -20 + . S. (6 cos* 9–6 cox 0 sin® ) cose podp} de m (40—e cosec 0 + cot 0) do 0 + (168–41 + a cosec? 8—40 cosec 8 + 4 coto— 3/3 + 8 sin 8 coso) de 133. TO FIND THE DIRECTRIX OF THE PARABOLA (ax + by)2 + 2dx + 2ey + f = 0. By W. GALLATLY, B.A. If T be any point on the directrix, then the tangents through T are at right angles. Now, let a radius-vector be drawn through T to the curve, cutting it in P,Q; and let o be the inclination of TP to the axis of x, TP be r, P be (wy), and T be (hk); then h + r cos 0, and y = k +r sin 0; |