GEOMETRY. Solutions of Problem 210 have been received from J. E. Sanders, Hackney, O.; Charles E. Barrett, Anchorage, Ky., and Charles A. Carpenter, Student in Adelphi Academy, Brooklyn, N. Y. Solutions of Problem 211 have been received from J. E. Sanders, Hackney, O., and John D. Cutter, Student in Adelphi College, Brooklyn, N. Y. 174. Proposed by B. F. FINKEL, A. M., 204 St. Marks Square, Philadelphia, Pa. Given two triangles ABC and A'B'C' lying in the same plane. The side B'C' cuts the sides AC, BC, and AB in the points I, H, and G, respectively; the side A'B' cuts the sides AC, BC, and AB in D, F, and E, rrspectively; and A'O' cuts AC, BC, and AB in M, L, and K, respectively. Prove that (DA'.EA'.A'F)(GB'.HB'.B'I)(MC'.LC'.C'K) =—(KA'.A'L.A'M)(FB'.B'E.B'D)(IC'.C'H.C'G). Solution by G. W. GREENWOOD, A. M. (Oxon), Professor of Mathematics and Astronomy in McKendree College, Lebanon, Ill. Since AC is a transversal cutting the sides A'B', B'C', C'A' of the triangle A'B'C' in the points D, I, M, respectively, we have, by the theorem of Menelaus, A'D.B'I.C'M—— B'D.C'I.A'M. Writing down the corresponding results for the other transversals, we get, by multiplying, the result required. Also solved by G. B. M. Zerr, A. M., Ph. D., Parsons, W. Va. 213. Proposed by H. F. MacNEISH, University High School, Chicago, Ill. Construct an equilateral triangle which shall have its vertices in three given paral lel lines. Remark by G. I. HOPKINS, Manchester, N. H. The solution follows easily from the solution of Geometry Problem number 156 in the MONTHLY for November, 1901. Independent solutions have been received from R. A. Wells, Bellevue, Neb.; G. W. Greenwood, Lebanon, Ill.; G. B. M. Zerr, Parsons, W. Va.; A. H. Holmes, Brunswick, Maine, and the Proposer. 214. Proposed by H. F. MacNEISH, A. B., University High School, Chicago, Ill. Inscribe in a given circle a triangle whose sides shall pass through three given points. Solution by G. W. GREENWOOD, B. A. (Oxon), Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill. Denote the fixed points by X, Y, Z. Take on the circle any number of points A, A2, Draw AX cutting the circle again in B,; draw B, Y cutting the circle again in C1; draw C1Z cutting the circle again in D ̧. 1 1 Perform the same operation with A2, thus determining the points D2, The ranges A1, A2, --·-·, D1, De, are homographic. Construct their double points which will give two (real or imaginary) positions of a point A with which D will coincide. CALCULUS. Solutions of Problem 172 have been received from J. E. Sanders, Hackney, Ohlo, and J. B. Gregg, M. Sc., C. E., Senecaville, Ohio. 173. Proposed by J. E. SANDERS, Hackney, Ohio. Find the greatest ellipse that can be inscribed in a quadrant of a given circle. Solution by J. B. GREGG, M. Sc., C. E., Senecaville, Ohio, and G. W. GREENWOOD, B. A. (Oxon), Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill. By symmetry, the radius bisecting the quadrant will coincide with an axis of the ellipse. Denote the length of this semi-axis, and that of the transverse semi-axis, by a, b. Then лab=area of ellipse=A (say). a+v (a2+b2)=radius of given circle-r (say). Since A is a maximum, bda + adb=0. Also rda + bdb=0. Also solved by G. B. M. Zerr, A. M., Ph. D., Parsons, W. Va., and A. H. Holmes, Brunswick, Me. MECHANICS. 163. Proposed by W. J. GREENSTREET, A. M., Editor of The Mathematical Gazette, Stroud, England. A particle A, mass m, rests on a smooth horizontal plane and is attached by two inelastic strings to masses m1, m2 at points B and C such that BAC is a right angle. If a blow is given A at an angle ◊ to AB, find the initial direction of motion of m, and equations for initial motion of the particles m and M2. Solution by J. B. GREGG, M. Sc., C. E., Senecaville, Ohio. Let v, v1, and v be the respective initial velocities of m, m1, and m., and let be the angle which the initial motion of m makes with the line of direction of the blow. Construct AD=v; then DE perpendicular to BA produced-v2, AE v1, DAE=0+6. v1=vcos(0+1), v2=vsin(0+ 4), musino+m, v, sinom,v,coso. 2 If n is an odd positive integer, and 1, n, n', n', denote all its distinct divisors, then 2">2(n+1)(n' + 1)(n" +1)....... II. Solution by the PROPOSER. By Euler's generalization of Fermat's Theorem (Dirichlet-Dedikind Zah lentheorie, 4th edition, p. 38), we have for n odd, 24(1)-1=0(mod1), 26(n)-1=0(modn), 2(n')-1=0(modn'),------- where (n) is the number of integers less than n and prime to it. We have, then, since (1)=1, 2*(1)—1=1, 24(n) −1≥n, 24(n') — 1 ≥ n' Now, unless n=3, there is at least one of the latter relations which is not an equality. For, let p be the largest prime factor of n, then 2-1-1≥ p. Make p=3, then 23-1-1-3. From this relation it is evident that 2o-1—1 >p for p>3. If p=3, then n=3*. 26(3)-132. Hence, unless k=1 or n=3, there is at least one inequality in the original scheme, and we get by transposition and multiplication, 24(1)+4(n)+4(n')---->2(n+1)(n' + 1)(n'' + 1 )...... l. or, since (1)+4(n)+4(n')...............n (Dirichlet, 1. c., p. 26), 2">2(n+1)(n'+1)(n'' + 1)............................ for n>3, which is the theorem. 118. Proposed by L. C. WALKER, A.M., Professor of Mathematics, Colorado School of Mines, Golden, Col. Find the two least positive* integral numbers such that their sum shall be a square and the sum of their squares a biquadrate. Professor Walker finds that the two numbers, x=4,565,468,027,761, y=1,061,625,293,520 have the properties that their sum is a square and the sum of their squares is a biquadrate. Indeed, it may be verified that (x+y)=2372159; (x2+y2)} =2165017. In the letter accompanying the solution Professor Walker asks "are these the least numbers?" We invite other solutions bearing upon this important point of the problem. 119. Proposed by L. E. DICKSON, Ph. D., Assistant Professor of Mathematics, The University of Chicago. If p be any prime number and n any positive integer, the congruence x=x(modp") has p and only p solutions modp". Hence the congruence defines the Galois field of order p" if and only if n=1. *As originally printed the word "positive" was omitted. Solution by the PROPOSER. The only solution x divisible by p is seen to be =0(modp"). Let next x be prime to p. Then xp"-1=1(modp"), xpn-2(p-1)=1(modp"), the second from Fermat's theorem generalized, since (p")=p"-1(p-1). Hence must x-1=1(modp"). To prove that the latter has precisely p-1 distinct solutions modp", we proceed by induction, the result being true for n=1 by Fermat's theorem. Let therefore x,,, p-1 be the distinct roots of xP-1=1(modp2), whence x¿1=x¿+r.pa, each r、 being a fixed integer. To determine the roots of x-1=1(modpa+1), we seek the integral values of m, incongruent modp such that x=x+m,p2 satifies x=x(modpa+1). But (x+mpa)=x,P+multiple of pa+1, by the Binomial Theorem. The condition is therefore 2 x1+r;p" =x; +mp" (modpa÷1), whence m;=r,(modp) and m; is uniquely determined modulo p. AVERAGE AND PROBABILITY. 148. Proposed by M. C. RORTY, Boston, Mass. Assuming n points to fall at random upon a circle of circumference a, what is the probability of m or more points falling within a length b upon this circumference? NOTE. This problem has practical application in determining the probability of accidental rushes of telephone calls as distinct from those rushes which are due to commercial causes. The solution for m or more points falling within a specified length b is known. The problem presented above differs from this in that a solution is required for any length b. Solution by G. B. M. ZERR, A. M.. Ph. D., Parsons, W. Va. Let m+r, (r=0, 1, 2, 3, ....n—m) be the number of points to fall on b. . 149. Proposed by L. C. WALKER, A.M., Professor of Mathematics, Colorado School of Mines, Golden, Col. Three points are taken at random on the convex surface of a right cone. Find the probability that the section of the cone made by the plane passing through them is a complete ellipse. Solution by G. B. M. ZERR, A. M., Ph. D., Parsons, W. Va. Let ABC be a section of the cone, E one of the random points in the surface. AD=c, GD=h, BD=R, EG=r. Through E pass the plane EC. Then the area of the surface, of which EBC is a projection (See Vol. V, Nos. 6-7). Now h=(c/R)(R-r). Surface of which EBC is a projection (R2 + c2) [R2-1 (Rr)(R+r)]. R S-surface of which EAFC is the projection. π .. S=πR√/ (R2 + c2 ) − ̄ ̄ ̄√/ (R2 +c2)[R2 −11/(Rr)(R+r)] R __*\/(Rr)(R+r)√/ (R2 + c2) 2r If the two remaining points are taken anywhere on the surface & the section will be a complete ellipse. 4x2 R3 (R2 +c2) S* dr 150. Proposed by F. P. MATZ, Sc. D., Ph. D., Professor of Mathematics and Astronomy in Defiance College, Defiance, 0. If the length of a circular arc be band the radius vary uniformly, what is the average area of all the segments possible? Solution by G. B. M. ZERR, A. M., Ph. D., Parsons, W. Va. The least area is 0 when R-radius is ∞. The greatest area is the area of a circle, R-b/(2),. circumference b, .. area=b2/(47). The problem may also be solved as follows: Area of segment Rb-R2 sin(b/R). Average area=▲= = "(Rb-R2 sinb/R)dR/SaR. b/2r Let b/R-y, then limits of y are 0 and 27. b/2π. |