THE

MATHEMATICAL MONTHLY.

Vol. II... NOVEMBER, 1839

PRIZE PROBLEMS FOR STUDE. 1 -.

1. FORM the equation, the roots of which are a

II. Find the equate I time of payment of two s due respectively at the end of T and you a

rest.

III. Of all right-ar gled plane triangles having the ypothenuse, to find the one whose area is the greatet possible. To be solved by Algebra.

IV. Wit is that fraction, the cube of which being subtracted from it, the remainder is the greatest possible? To be solved by Algebra.

V. If, in a plane or spherical triangle, A, B, C denote the angles, and, be the sides respectively opposite them; and if we prode the sides of the triangle, and consider the three circles which touch two of the sides interiorly and the third side exteriorly o denote by r, the radii of the circumscribed and inscribed ygg", ", the radii of the circles touching exteriorly

a, b, c respectively; by d',d", "", the distances of th

THE

MATHEMATICAL MONTHLY.

Vol. II... NOVEMBER, 1859. No. II.

PRIZE PROBLEMS FOR STUDENTS.

I. FORM the equation, the roots of which are a+ √b, a — √b, c and d

II. Find the equated time of payment of two sums S and s, due respectively at the end of T and t years, allowing simple in

terest.

III. 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.

IV. What is that fraction, the cube of which being subtracted from it, the remainder is the greatest possible? To be solved by Algebra.

V. 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, the radii of the circumscribed and inscribed circles; by g',g", "", the radii of the circles touching exteriorly the sides a, b, c respectively; by 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 A+ sin B+ sin C4 cos Asin B sin C.

sin A-sin B+ sin C 4 sin A cos B sin C.

=

sin A+ sin B-sin C4 sin A sin B cos C.

cos d'

cos r cos q''

cos 8"

cos r cos

cos 8!!! cos r cos ¿'ll

The solution of these problems must be received by the first of January, 1860. Problem V. is one of the analogies by Prof. CHAUVENET, to which we referred in the last number of the MONTHTY.

REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. X., Vol. I.

THE first Prize is awarded to GUSTAVUS FRANKENSTEIN, of Springfield, Ohio.

The second Prize is awarded to ROLAND THOMPSON, Jefferson College, Canonsburg, Pa.

PRIZE SOLUTION OF PROBLEM II.

BY ALL THE COMPETITORS.

"If A, B, C be the angles, and a, b, c the opposite sides, in a plane triangle, of which S denotes the surface; prove that

a2+b2+c2 4 S (cot A+ cot B+cot C)."

By Trigonometry,

a2 = b2 + c2 — 2 b c cos A