By addition and transposition we get (1) a2 + b2 + c2 = 2 (b c cos A+ ac cos B+ab cos C). But 2 S-be sin Aa e sin Bab sin C: Substituting these values of bc, ac, ab, in (1), we obtain 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 (2). (3) ..LF: DL=SD: SA (4) ... in which .... ... EB:BC-EL:LF C'H: IIF PC: PA; = FH-LE-DE-DL, SD = PB, BC= BP + PC, HE LF, SA+AP+BE=DE. = But (2) and (4) give (DE—DL) (BP+PC)=EBX LF, (DE-DL) PC-C'HX PA; hence (BP +PC) × C'H × PA= P C × EB × LF, or, PCX (EB× FL — C'H × PA)= BP × CH× 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) B'C' × PA=DE× 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: DEF=AP × BC: B'C' × AP, Therefore DEF is a mean proportional between ABC and Α' Β' Γ'. COROLLARY. The property BC PA=DEX (LF or EH) is not confined to the side DE of the triangle DEF, and its par allel AP, but applies to any line DE' and its parallel AP'. For = the triangle DEF is half of the parallelogram DH; having the same base and altitude; and for the same reason the parallelogram DH-DH'. Hence, as the angle AP'CH'E' D, t: DEF BCX AP: (HE or H' E) DE', Therefore the line DE' may take the direction DB', and we shall have B'C' XAB= DB' × EH; so that drawing C'D, and RO parallel to DB', intersecting C'D in 0, we must have RO=AB. For the triangles CDB and D RO are similar, having their sides respectively parallel, giving B'C': B'D=RD or EH: RO. .. DBX EH=B'C' X RO=B'C' X AB. . RO=AB. SECOND SOLUTION OF PRIZE PROBLEM III. BY ROLAND THOMPSON, JEFFERSON COLLEGE, CANONSBURG, PA. Inscribe the triangle A'B'C' in DEF, and through the points D, E, F, draw straight lines respectively parallel to the sides of A'B'C'; the tri angle ABC thus formed is similar to A'B'C'. The three straight lines passing through the homologous angles will meet in the same point. For the straight lines CC and BB produced will meet in some point O, and if AO does not pass through A', let A" be the point in which it meets the line A'B'. Then BO: BOAB: A′′ B′ BO:T0=BC:BC; AB: A" BBC: B'C'. Therefore the triangle A" B'C' is similar to ABC and equal to A'B'C', and therefore A" B'A' B′, and AO passes through A'. Let OS be drawn perpendicular to A B, and A′ B'; then, since Os, OR, and RS, are the altitudes of the triangles AO B, A' O B', and Α' Β' Ε', we have Since (1), (2), and (3), are equal ratios, we get, by addition, ABC DEF showing that DEF is a mean proportional between ABC and A'B'C' as was to be proved. "If a be one of the angles, prove that sec A PRIZE SOLUTION OF PROBLEM IV. BY ALL THE COMPETITORS. sides of an equilateral spherical triangle and A one of its sec a +1." If a b c in the fundamental equation cos a = cos b cose + sin b sin c cos A, BY GUSTAVUS FRANKENSTEIN, SPRINGFIELD, OHIO. "If the semiaxes of an ellipse be A and B, P the length of the perpendicular dropped from the centre on the tangent to the curve, r and r the distances from the point of tangency to the foci, and o the radius of curvature at this point; prove that ୧ and from this theorem construct the corresponding point of the evolute." 1. The equation of the ellipse referred to its centre and axes is A2x2 + B2 y2= A2 B2; and if x,y' denote the coördinates of the point of tangency, the equation of the tangent line is Ayy B2 x x A2 B2. The tangent of the angle which the line DS makes with the axis of x is Day= ; or if we take the angle B2x A2 y |