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 w= a d

(3) x z+yw=bd, (4) x2+w2=y2+22;

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 the distance between the centres of these circles; then in the plane triangle

*sin A+ sin B+ sin C4 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=b+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 ba+x 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 A B C 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.

base; also, let x= base, y = perpendicular; therefore, Vx+y=hypothenuse.

There may be two cases, according as the perpendicular is longer, or shorter, than the base.

y2

For the first case, we have Vay-x=a, and y-x-b; by eliminating y, we readily obtain xa-b±√2a (a-b); rejecting the negative sign before the radical, as giving a negative value to x, we have xa-b+ √2a (a—b) and y=x+b.

To construct these values of x and y: From any point P on the line, MN, with radius PB= a, describe a semicircle. Lay off PDb, erect the perpendicular DE, make AB=BE, and AH equal and perpendicular to AD. Joining H with P forms the required triangle, PAH.

α

For ABB E=√BD.BC=√2a (a-b). Therefore, AH= AD BD+AB=a¬b+ √2a (a-b)=x, and A PAD+

=

If b> a, the value of x will be imaginary, and the construction impossible.

For the second case, x-y=b, or y-x-b; hence, if in the expressions for x and y, in the first case, we make b negative, they will apply for the second case; namely,

a+b+√2 a(a+b) and y=x-b.

If ba, the triangle may be constructed as in the first case, provided we make the same construction with respect to C and CE, as was there made with respect to B and B E.

If b> a, the semicircle must be described on a +b as a diameter, instead of on 2 a, as before.

PROBLEM II. Denote by a the sum of the hypothenuse and base, and by the sum of the base and perpendicular; also let = base, y= perpendicular; therefore, √x2+y=hypothenuse.

Hence, √x2+ y2+ x = a, x+y=b. Eliminating y gives x=- — (ab)± √2a (a-b); rejecting the negative sign before the radical as giving a negative value to x, we have

x=- -(ab)+ √2a (a-b) and y = b—x.

To construct these values of x and y: From any point P on the line MN with radius PB-a,

describe a semicircle. Lay off
PD=b, erect the perpen-
dicular D E, make BA=BE
and AH equal and perpen-
dicular to AP. Joining H
with D forms the required triangle D A II.

For BA=BE=√BD.BC=√√2 a (a —b) :. A D=BA—BD =√2a (a - b)—(a—b)=x. AH=AP=PD-AD=b—x=y.

If x> b, or √2 a (ab) > a, the value of y will be negative, and the construction impossible; hence, 2a (a-b) <a, or BA < BP, and the point A always falls between B and P. The expression √2a (a-b)<a is equivalent to b>a.

PRIZE SOLUTION OF PROBLEM III.

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

"The four tangents, which are common to two circles which do not intersect, and are terminated at their points of respective contact, have their middle points on the radical axis of the two circles."

Through P, the middle point of the common tangent BB', draw PA perpendicular to 00. From any point P of this line draw the tangents P' C and P' C'; draw also PO, PO, P' 0, and P'0'. Then we have P 0-0 C2 P C2 and PO OB2 — PB2.

=