Mathematical questions with their solutions, from the "Educational times"
D. Biddle
Paperback
(RareBooksClub.com, May 22, 2012)
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1874 Excerpt: ...two lumps, V the volume of the smaller lump, and 5 an elemental diminution of V; then /u3v is the volume of the larger, and /n2S its corresponding elemental diminution. Therefore the ratio of the volumes after variation is /u3v-m;8 = Lj?(filv-5). VS yuV-78-' and this ratio is greater than fj.3, since nY--S p.V--nt. Therefore the volumes tend to greater inequality. In the same manner it may be shown generally, that if nx and x be two variable magnitudes (/il 1), and h and k small increments of nx and.; then if A and k are both positive, the magnitudes nx and x are tending to equality or to greater inequality according as-is greater or less than p. 4036. (Proposed by the Editor.)--Let p., p2, ps, Ri, R2, R3 be the radii of the circles drawn in and about the three triangles cut off from the corners of a given triangle by tangents to its inscribed circle drawn parallel I. Solution by the Rev. J. L. Kitchin, M.A.; A. Renshaw; and others. The triangle ADE is similar to ABC, A hence if pt be the perpendicular from A on BC, we have pi-2r = DE pT " BC' II. Solution by Asher B. Evans, M.A. The perimeter of the triangle ADE is evidently 2s,, since AD + PL = Alt andAE + EL--AQ. Similarly the perimeters of BFG and CHK are 2ยป2 and 2ยป3 respectively. Again, since the four triangles ADE, BFG, CHK, ABC are similar, we have ft = S? =--=-(1), 01 On S-. 8 and the other properties given in the first part of the question, as well as many similar ones, may be proved in like manner. Moreover, in any series of the circles obtained as described in the last part of the question, the radii of any triad are together equal to the radius of the circle around which they stand; hence the sum of the radii of the whole of the nth series must be equal to r. Again, since 2 (pi2) ...