JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact firstname.lastname@example.org. This content downloaded from 18.104.22.168 on Sun, 18 Oct 2015 08:49:47 UTC All use subject to JSTOR Terms and Conditions EXAMINATION QUESTIONS AND PROBLEMS.
... ONS AND PROBLEMS. EXAMINATION QUESTIONS AND PROBLEMS. to a3, b3, c3 respectively, and that PAPBPC=R3, where p, etc., are the radii of curvature at A, B, C, and R the circum-radius of the triangle. (C.) 214. A', B', C' are the mid-points of the sides of the triangle ABC. Any transversal through A meets C'A', A'B', in Q, R, respectively; prove that BQ, CR are parallel. G. RICHARDSON. (No. 145 is a particular case.) 215. If the diagonals of a convex quadrilateral are equal, and perpendiculars be drawn bisecting two opposite sides; then these perpendiculars, measured from their point of intersection, are proportional to the sides on which they stand. F. S. MACAULAY. (An amended form of 183.) 216. If m edges meet at each vertex of a polyhedron of E edges, and each polygonal face have n sides, then r-+ n -E-1-1+2-. Hence deduce that only five types of such polyhedra are possible, corresponding to the five regular solids. R. F. MUIRHEAD. 217. Two similar and similarly situated triangles A1 A2 are such that the same circle can be described in the one and about the other; compare their areas, and prove that if A = 4A2, the triangles are equilateral. (W.) (Required analytical and geometrical solutions.) This content downloaded from 22.214.171.124 on Sun, 18 Oct 2015 08:49:47 UTC All use subject to JSTOR Terms and Conditions SOLUTIONS. Hence we have the n linear equations, Xx,+ Yy,+Zzr=0, (r=, 2, 3,... n). These give, by the elimination of X, Y, Z, X1, 2,... Xn =0. Y1, 2/, -Yn Z, X2 ... Z Professor A. C. Dixon writes: "The given determinant is the sum of the squares of the determinants of the matrix ; therefore," etc. 183. The wording of this question is insufficient. For a correct wording, see No. 215, p. 133. Showo how to inscribe a square in a given quadrilateral. (D. Solution by Prof. A. C. DIxoN and others. Let ABCD be the given quadrilateral, supposed convex for simplicity. Take any square efgh. On ef, on the same side as the square is, describe a triangle lef, having the angles lef, Ife equal to DBC, DBA respectively. Also on gh, on the same side as the square is, describe a triangle kgh, having the angles kgh, k/hg equal to BDA, BDC respectively. Let kl cut the circle round lef in b, and the circle round kqh in d; also let be, dh meet in a, and bf, dg meet in c. Then abed is similar to ABCD. For angle abd=efd=ABD, and similarly adb= ADB, cbd= CBD, cdb= CDB. Hence, if on the sides of ABCD we take points E, F, G, H dividing them in the same ratios as the sides of abed are divided by e, f, g, h, then EFGH will be a square inscribed in ABCD. A similar method is applicable if it is required to inscribe in ABCD a quadrilateral similar to any given quadrilateral. The above construction is one of the simplest; but it has the disadvantage of being indirect. A direct construction, given by Prof. Dixon, depends on the properties of the centre of similitude. Suppose it required to inscribe in ABCD a quadrilateral similar to KLMN. Let P, Q, R be any set of three points on AB, BC, CD respectively, such that PQR is similar to KLM. Then, by a known property, all the triangles PQR have a fixed centre 0 of similitude, which can be constructed. Also, when 0 has been found, any number of the triangles PQR, all similar to KLM, can be constructed; and the locus of the point S, such that PQRS is similar to KLMN, is a straight line. Find then two of the points S, and join them. This gives the locus of S, and the point where it cuts DA will be a vertex of the required quadrilateral. The following solution is due to Mr. J. B. Casey. The reader can easily draw the figure. The sides AB, DC meet in E; AD, BC meet in F. In the triangle AED find 0 such that AOD=AED+90?, DOE=FAE+ 90. In the triangle AFB find O' such that AO'F=ABF+ 900, AO'B=AFB +90?. The circles AOO', OGD, O'HB cut respectively AF, AE in G, H; DE in J; BFin I. Then AOD-AED=90'= EA + ODE HGO + OGJ= HGJ. Similarly GJH=45?, . GH= GJ. In the same way we find GJ=JI and GJI=90?, .. GJIH is a square.