THE MATHEMATICAL MONTHLY. Vol. II.... FEBRUARY, 1860.... No. V. PRIZE PROBLEMS FOR STUDENTS. I. Prove that the value of a proper fraction is increased, and an improper fraction diminished, by adding the same quantity to both terms of the fraction; and that the reverse is the case when the same quantity is subtracted from both terms of the fraction. II. A common tangent is drawn to two circles which touch each other externally; if a circle be described on that part of the tangent which lies between the points of tangency as diameter, this circle will pass through the point of contact of the two circles, and will touch the line joining their centres. and obtain symmetrical expressions for x1, x2, &c. IV. Prove that sin" (0 — q) sing is a maximum when V. The notation of Problem V. in the November No. being retained, prove that in the plane triangle REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. II., Vol. II. THE first Prize is awarded to GEORGE B. HICKS, Cleveland, Ohio. The second Prize is awarded to WILLIAM HINCHCLIFFE, Barre Plains, Mass. The third Prize is not awarded. PRIZE SOLUTION OF PROBLEM III. By GEORGE B. HICKS, Cleveland, Ohio. Of all right-angled plane triangles having the same given hypothenuse, to find the one whose area is the greatest possible. To be solved by Algebra. Since the hypothenuse is constant, the area is evidently a maximum when the perpendicular dropped from the right angle upon the hypothenuse is a maximum. Let x y be the sides, c the hypothenuse, p the perpendicular, and d one of the segments of the hypothenuse. Then But the greatest value p2 can have, and d still be real, is } c2 ... p = 1; c, d = { c, .. the triangle is isosceles, and xyc√. Otherwise, put dwc. p2 = c d — d2 = c2-w2, which is obviously a maximum when w= 0, ...pc and xy as before. Or, (1) may be written p2 = d(cd). But d (c d) is the product of the segments of the hypothenuse, which we know to be a maximum when the segments are equal; the result already found. SECOND SOLUTION. By WILLIAM HINCHCLIFFE, Barre Plains, Mass. Let ABC be right-angled at C, from which drop the perpendicular CD. Put A B 2a, AD= a + x, BD a-x. Now, = ABX CD = a maximum. and hence CD2= a maximum. a maximum when x=0. = = BC a√2; whence the required triangle will contain the greatest area when the sides containing the right angle are equal. PRIZE SOLUTION OF PROBLEM IV. By WILLIAM HINCHCLIFFE, Barre Plains, Mass. What is that fraction, the cube of which being subtracted from it, the remainder is the greatest possible? To be solved by Algebra. Let x denote the required fraction. Then, by the question, xx3 y = a maximum. Denote a negative root of this equation by —ɑ; then 23 — x + y = 0 divided by x + a -a; tient, 22 — a x + a2 — 1 = 0 (1), and a remainder, a3—a—y = 0. 22 But by solving, (1) gives ; Now, it is evident that, if y is positive, x is less than 1; and since x — x3 = y = aa, it also follows, that when a, it also follows, that when y is the greatest possible, a must have the greatest value which will make ≈ real. If, in a plane or spherical triangle, A, B, C denote the angles, and a, b, c the sides respectively opposite them; and if we produce the sides of the triangle, and consider the three circles which touch two of the sides interiorly and the third side exteriorly; and denote by r, o the radii of the circumscribed and inscribed circles; by g', g', p''', the radii of the circles touching exteriorly the sides a, b, c respectively; by d', d'', d''', the distances of the centres of these circles from the centre of the circle circumscribed about the primitive triangle; then we have in the plane = sin 4 + sin B — sin C = 4 sin } 4 sin } B cos } . First Part.-In plane trigonometry cos r cos p cos d" cos r cos p" 8"" cos r cos p and similar expressions may be found for the other angles by simply permuting the letters. Now -sin A+ sin B+ sin C=2(-sin A cos A+ sin B cos B+ sin C cos C) putting M= √s ( s − a ) ( s − b ) (8 — c). But 4 cos ¦ A sin } B sin | C′ = 2 M (− a + b + c), abc observing that s = $ (a+b+c). sin A+ sin B+ sin C 4 cos A sin B sin C; and in the same way the remaining equations may be found. Second Part.-By a process precisely similar to the above we shall find for the spherical triangle = the Let ABC be any spherical triangle, and centre of its circumscribed circle. Produce A B and AC until they meet in A'. Let A' Cb1, and A B C1. =C1. Put BK α, KAB, CH=y. Then a = a + y, b1 = ß + y, c1 = a + B. · · . B K = a =a—b + a — a − ( x − b ) + (x — c) a b1 2 sin c), We also have B0=r, 0 K=g', 00′ = 8′; and since A OB, AOC, BOC are isosceles, OBC= } (− A + B + C), CBO = } (π — B); = ... OBO'=π—(A-C). ... cos O BO'=sin(A — C') = But from the right-angled triangle B OK, sin o' cos B g' coscos r cos BO' Substituting the values of cos B O, sin B O', in that of sin (A — C), we get, by observing that sin(4- C') sin(a - c) cos B |