II. In any plane triangle, prove that the sines of the angles are inversely as the perpendiculars let fall from them upon the opposite sides.

III. Having given the diagonals of a quadrilateral inscribed in a given circle to determine its sides geometrically, when the diagonals intersect each other at right angles.

IV. Given

(1) xw+yz=ab, (2) x y + z wad

(3) xz+yw=bd, (4) x2+w2 = y2+z2 ;

to find the values of x, y, z, and w

V. If, in a plane or spherical triangle, A, B, C denote the angles, and a, b, c the opposite sides respectively; if r, o denote the radii

of the circumscribed and inscribed circles, and d the distance between the centres of these circles; then in the plane triangle

*sin A+ sin B+ sin C 4 cos A cos B cos C

and in the spherical triangle

sin A+ sin B+ sin C4 cos A cos B cos C

cos 8

cos r cos Q

The solution of these problems must be received by the first of December, 1859. [See Editorial Items.]

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

THE first Prize is awarded to WILLIAM C. CLEVELAND, of the Lawrence Scientific School, Cambridge, Mass.

The second Prize is awarded to JOHN W. JENKS, Senior Class, Columbia College, N. Y.

PRIZE SOLUTION OF PROBLEM I.

BY ASHER B. EVANS, MADISON UNIVERSITY, HAMILTON, N. Y.

"In a right-angled triangle, having given the difference between the base and perpendicular, and also the difference between the hypothenuse and base; to construct the triangle geometrically."

On the indefinite line NE take A Ma- the difference between the base and perpendicular, and erect the perpendicular MT. Also, with A as a centre, and A 0=6+a= he difference

This analogy, pointed out by D'ARREST, is only one of a number of interesting ones given by Professor CHAUVENET, in GOULD'S Astronomical Journal, Vol. III.. page 50, without demonstration. We propose to give them from time to time in our lists of Prize Problems, as most valuable exercises in Trigonometry.

between the hypothenuse and perpendicular as radius, describe a

BY ASHER B. EVANS, MADISON UNIVERSITY, HAMILTON, N. Y.

"In a right-angled triangle, having given the sum of the base and perpendicular, also the sum of the hypothenuse and base; to construct the triangle geometrically."

Let x be the perpendicular, a-x the base, and b-ax the hypothenuse. On the indefinite line NE take AM-a, and draw the perpendicular M' T'. Also, with A as a centre, and A0= b-a as a radius, describe a circle. If C be the centre of a circle tangent to this circle, and also to the sides of the right angle AM' T', then will ABC be the required triangle. For, AB= AM-BC-a-x, and A C=AO+BC=b-a+x.

SECOND SOLUTION OF PROBLEMS I. AND II.

BY GEORGE A. OSBORNE, JR., LAWRENCE SCIENTIFIC SCHOOL.

PROBLEM I. Denote by a the difference between the hypothenuse and base, and by the difference between the perpendicular and

To draw a circle tangent to a given circle, and to the sides of a rightangle, we may employ Problem 8, in a Memoir on the "Tangencies of Circles," by Major BENJAMIN ALVORD (Smithsonian Contributions to Knowledge); or a Theorem of Professor H. A. NEWTON, found on page 242 of the MATHEMATICAL MONTHLY, Vol. I.