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 Asin B+ sin C4 cos A sin

B sin C,

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

sin A-sin B+ sin C4 sin A cos B sin

cos d'

cos r cos q''

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

cos 8!!! cos r cos q

The solution of these problems must be received by the first of January, 1860. Problem V. is one of the analogies by Prof. CHAUV-, ENET, 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 2b c cos A

12= a2+c2 2 ac cos B

c2=a2+b2-2 a b cos C.

By addition and transposition we get

(1) a2+b2+2=2 (bc cos A+ ac cos B+ab cos C).

But 2 S-be sin Aac sin Bab sin C:

Substituting these values of bc, ac, ab, in (1), we obtain a2+b2+c2=4 S(cot A+ cot B cot C).

PRIZE SOLUTION OF PROBLEM III.

BY GUSTAVUS FRANKENSTEIN, SPRINGFIELD, OHIO.

"If one of the similar triangles ABC and A'B'C' be inscribed in the triangle DEF and the other circumscribed about it; prove that the area of DEF will be a mean proportional between the areas of ABC and A'B'C'."

Let DEF be any triangle, ABC an inscribed, and A'B'C' a similar circumscribed, triangle. Draw DSR and FL parallel to B C or B'C'; also, FH and SAP parallel to DE. By reason of these parallels, the triangle BED is similar to BPA, EBC to ELF, FLD to DSA, and C'HF to CPA; whence the proportions:

(1)... BE: ED=BP: PA (3).... LF: DL=SD: SA

in which

FH=LE=DE-DL, SD=PB, BC=BP+PC,

HE=LF, SA+AP+BE=DE.

But (2) and (4) give

(DE—DL)(BP+PC)=EB× LF, (DE—DL) PC=C'H× PA;

hence

(BPPC) X C'H X PA= PCX EB X LF,

or, PCX (EBX FL- C'H PA)=BP × C'H× PA;

and as

PCX (DE-DL) = C'H × PA,

therefore (DE-DL) BP = (EB × LF — C'H × PA).

Substitute for DE its value

its value

or

FLX SA

BP

deduced from (3), and there results the condition,

(5) B'EX PA-FLX SA= EB × LF — C'′ II × PA,

(BE+CH) PA=(EB+SA) FL.

Add EHX PA to the first side of this equation, and its equal FLX PA to the second; then, because

BE+EH+C' II = B′ C', SA +AP+BE=DE,

(5) becomes

(6)

BCX PA=DEX LF.

=

But in the triangles ABC and DEF, the angle APB FLE;

(7)

ABC: DEFAP X BC: FLX DE;

or, denoting the area of the triangle ABC by t, and by (6),

t: DEFAPX BC: B'C' X AP,

B' C/2

.. ABC'× A'B'C' = tx=(DEF)2.

Therefore DEF is a mean proportional between ABC and

A'B'C'.

COROLLARY. The property B'C' PA=DE× (LF or EH) is not confined to the side DE of the triangle DEF, and its par