cos a = k3, x = -2k4y, and the common point is tangents to (1), (2) at this point are x + k1y + k3z = 0, and x + 4k1y-2k2z whence cos B = a fraction whose numerator is +k2 sin C sin A cos B-5k4 sin A sin B cos O, and denominator the product of the two radicals -2k2 sin C sin A cos B-24 sin A sin B cos C}}, {sin3 A + 16% sin3 B + 4k1 sin2 C + 16k sin B sin C cos A +4k2 sin C sin A cos B-8/4 sin A sin B cos C}. = sin 2α 2/3 sin a, 1-2k6, B = C = a, sin B sin C cos A = (1—26) sin2 a, But sin A = cos A = sin C sin A cos B = sin A sin B cos C = 2k6 sin2 a; and, on making these substitutions, the numerator becomes and the two radicals 2k1 (k2-1)3 (2k2 + 1) sin2 a, {ka (1 − k2)2 (2k2 + 1)2 sin2 a}, {4%1 (1+ 2k2)3 (1 — k2) sin2 a}}; Now the ratio of the areas of the two curves 1+2/2 = 3 1+2/2 Px2+2pyz = 0, ; hence the squares of the latera-recta of the circle and hyperbola will be as Of course, by using the triangle A'B'C', we should get the equivalent ratio (1+8 cos2 8): 8 sin3 ß cos B, which is a sufficient test. 3, 4. If 1, 2, 3 be the latera-recta of the three curves, we have 32, 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 also the angle of intersection of the circle and parabola at B is α, The angle of intersection of the parabola and hyperbola at A is π-2α, and at A' is T- -28, whence 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, ..., Cn, 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 c, is (c)-8 ; and similarly for c.-1. Hence the chance that the highest point taken lies between the heights (Cs- 1)n-8+1 c. and c.-1 is (cs)n-s Let this expression be denoted by p.. Now the chances in this case of any particular one of the points on the pillars marked from s to n being highest will be equal, each being Ps 5504. (By Dr. BOOTH, F.R.S.)-If A be the area of a plane triangle, prove that 442 = abeo, where a, b, c are the sides of the original triangle, and σ the semiperimeter of its orthocentric triangle. Solution by CHRISTINE LADD, D. EDWARDES, J. O'REGAN, and many others. It is easy to show that radii of the circumscribed circle drawn to the points A, B, C are perpendicular to the sides (a, b, c suppose) of the orthocentric triangle; hence the triangle ABC is divided into three parts whose areas are respectively Ra', R, Rc′; hence we have In the Syllabus of the Association for the Improvement of Geometrical Teaching (p. 4) we read :— ....(i.); "In the typical Theorem, If A is B, then C is D 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.) Two such Theorems as (i.) and (ii.) 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. :— or or (1) A is B, and at the same time C is D ; (2) A is B, and at the same time C is not D; (3) A is not B, and at the same time C is D ; (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. or, finally, а 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 "C 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 a complete circle; find, in each case, (a) the average distance between the points, and (8) the probability that this distance is less than the radius. = 16r = T 1. (8). Let the angle MOP π (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 0 are 0 and π, and doubled: hence the 2. (a). Draw the chord EF through M, N (Fig. 2), and draw OC perpendicular to EF. Put OD = x, DM = y, DN =2, DE = y1, DF =1, ZADM = 0, and ZOEC; then we have E r; of y, 0 and y; and of z, 0 and 21; hence, since the whole number of ways the two points can be taken is 24, the required average (y + z)2 sin e do dx dy dz [(Y1 + %1)* − (y1* +~14)] · sin e de dx 0 = 4572 1472r (7+2 cos20) sin e de 2. (8). 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 0 are 0 and π, and doubled; and the limits of are 0 and 0, and doubled. When <, the limits of y are 0 and y1, and of z, 0 and r-y. From 0 to 0= the limits of y are 0 and y1 and of %, O and r-y, when <π-20; when >π-20 and <, the limits of z are 0 and z1 from y=0 to y=r-z1, and they are 0 and r-y from y=r-z1 to y=y; and when p>, the limits of y are 0 and y1, and of z, 0 and 21. Hence the required probability + 8 π- 20 0 - IT = 3π2 + 3π2 + π-20 0 0 0 7-21 0 r (y + z) cos o do dy dz‹ do [3 cosec e sin (0-p) — cosec3 ◊ sin3 (0—4)] de cos o do [3 cosec e sin (0-p) — cosec3 e sin3 (0-4)] cos o do [6 cos -6 cot2 0 sin2¢ cos o - 2 cos3 p−2] cos o do + ၂ လ + (6 cos2p-6 cot2 e sin2 p) cos2 p do do (160-4π + cosec20-40 cosec20+4 cote-3/3+8 sine cose) de П 133. TO FIND THE DIRECTRIX OF THE PARABOLA (ax + by)2+2dx + 2ey +ƒ = 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 e be the inclination of TP to the axis of x, TP be r, P be (xy), and T be (hk); then |