7684. (By D. EDWARDES.)-Given the in-circle and circumcircle of a triangle, prove that (1) the loci of the orthocentre and centroid are circles of the respective radii, R—2r, (R-2r), whose centres lie on the line joining the in-centre and circumcentre, and divide it harmonically; (2) the locus of the centroid of the perimeter is a circle whose centre is collinear with the two former centres, and radius (R-2r). 7728. (By Rev. T. C. SIMMONS, M.A.)-Given the circumcircle of a triangle, and any of the four circles touching the sides, show that the loci of the orthocentre and centroid are circles having the centres of the given circles as centres of similitude. NCC Solution by Rev. T. C. SIMMONS, M.A.; J. BRILL, B.A.; and others. Let C be circumcentre, G centroid, N nine-point o centre, O orthocentre, E the centre of the tangent circle, then, EN being constant, locus of N is a circle centre E. But CG CN, CO2CN, therefore loci of G and O are circles having, in common with the locus of N, C for one centre of similitude. Let S1, S2 be centres of these circles, then [CS,ES,]=[CGNO], that is to say [CS,ES2] is harmonic. But C is one centre of similitude, therefore E is the other. Again, if A', B', C' be the mid-points of the sides of the triangle, the centroid of its perimeter is the in-centre of A'B'C'. Also G is the common centroid, and likewise the centre of similitude of both triangles; whence, considering E originally the in-centre, and O', I' orthocentre and in-centre of A'B'C', GO' = 1GO, GI′ = {GE, and O' coincides with C, the original circumcentre: whence EI'I'G, so that the locus of I' is similar and similarly situated to the locus of G, E being centre of similitude. Therefore the locus of I' is a circle whose centre lies on EC and whose radius = § × † (R − 2p) = † (R−2p). E 7655. (By W. J. MCCLELLAND, B.A.)-Show that the sum of the cotangents of the intercepts made by the internal and external bisectors of the angles of a spherical triangle on the opposite sides is equal to zero. Solution by B. H. RAU, B.A.; the PROPOSER!; and others. 7656. (By W. J. C. SHARP, M.A.)—If ABCD be a tetrahedron, p1, P2, P3, P4 the perpendiculars from the vertices upon the opposite faces, then, denoting by (AB), &c., the dihedral angles between the faces which intersect in AB, &c., prove that (1) sin (AB): sin (BC) : sin (CA): sin (AD): sin (BD): sin (CD) AB . BC = : : ; PIP2 P2P3 P3P1 P1P4 P2P4 P3 P4 and (2) the equation to the sphere described about the tetrahedron may be written ab sin (ab) xy + be sin (bc) yz + &c. = 0. 7636. (By Professeur COCHEZ.)—Inscrire dans un rectangle un pentagone ayant les côtés égaux. G H Solution by B. H. RAU, B.A.; BELLE EASTON; and others. Let AE=a and AG=b be the sides of the rectangle; and put BD = x. If the pentagon is symmetrical, the mid-point c of a side BD will coincide with the mid-point of AE, and the opposite corner of the pentagon will be at the mid-point of the opposite side of the rectangle. AB = } (a−x); therefore AF = [x2 — } (a−x)2]1. B Again, GH therefore or or or x2-ax+b2 = 2b (x2-4a2), 1xa—‡ax3 + (‡a2 + §b2 — 4b2) x2— ab2 x + ba + a2b2) = 0, a biquadratic for x. 7682. (By H. FORTEY, M.A.)-Suppose three straight lines pass through the points A, B, C respectively, and turn about those points in the plane ABC with the same angular velocity and in the same direction. Find the locus of the centre of the circle described about the variable triangle thus formed, (1) when the lines through A, B, C are initially coincident with AB, BC, CA respectively; (2) when they initially coincide with AC, BA, CB; showing that the loci are two equal circles of radius abc (λ-1642) / 16A2 (where λ = a2b2 + b2c2 + ca2 and A = area of ABC), that these circles touch each other at the centre of the circle about ABC, that (if A = aa— b2c2, &c.) the equation to the line joining their centres is (2A+B+C) bca + (2B + C + A) caß + (2C+ A + B) aby 0, and that this line touches the Brocard circle. = and let P be the centre of the circle about A,B,C1. Draw PN, PD perpendicular to BC, BC, and DE, DF perpendicular to BC, PN. Let BÑ=x, y, also let PN = B COS S sine), BC1 = a = a also BB, = c BD = a k-4c2 sin 8A ·), k-2a2 PD B,D cot A B1D = = a (k−2a3) ( 2 8A cos e k 2 8A = BD cos 0-PD sin 0, Substituting for BD and PD their values, and reducing, we get a (2c2-a2) b (2a2-b3) c (262-c2) If (as in the second case) the revolving lines make equal angles with AC, BA, CB respectively, then, taking C as the origin and CB as the positive direction of the axis of x, the locus would be that given above, merely modified by the interchange of b and c. It is therefore a circle of the same radius; but the ordinate of the centre is now B2, 72 are the trilinear coordinates of this centre, if = B2 = 72 a (2b2-a2) b (2c2-62) c (2a2-02)' and the equation to the line through a111 and a2B2Y2 is a ( 2b2 — a2), 8A (2A+B+C) bca + (2B + C + A) caß + (2C + A + B) aby = 0 and ........(1). It is easily shown that the centre of the circle about ABC lies in this line, and this centre is also clearly a point on both loci; therefore the two circles touch at that point. = Of course, the first circle passes through the first Brocard point of ABC and the second circle through the second Brocard point. Also, if Ρ radius of the Brocard circle and w be Brocard's angle, then the radii of the above circles are each cot w. = The equation to the line joining the Brocard points is Abea + Beaß + Caby = 0, and this is parallel to (1). Therefore (1) touches the Brocard circle. [For the Brocard circle, see Reprint, Vol. 40, p. 102, and Quarterly Journal of Mathematics, Vol. 19, p. 343.] 7385. (By Professor WOLSTENHOLME, M.A., Sc.D.)-In an equilateral triangle ABC is inscribed a circle, any tangent to this circle meets the sides CB, CA in the points A', B'; prove that (1) the centre of the circumscribed circle, and the centre of perpendiculars, of the triangle A'B'C have the same locus; (2) an hyperbola of which C is a focus, the centre of the circle is the farther vertex, and whose asymptotes are perpendicular to the sides CA, CB; (3) the centre of the circumscribed circle and the centre of perpendiculars are ends of a double ordinate to the transverse axis; (4) when they lie on the branch of which C is the exterior focus, the circle is the inscribed circle of the triangle; (5) when they lie on the branch of which C is the interior focus and between the radii drawn from C parallel to the asymptotes, the circle is the escribed circle opposite C; and (6) for the remainder of that branch the circle is one of the escribed circles opposite A' or B'. Solution by R. KNOWLES, B.A.; SARAH MARKS; and others. 1. Let AC: = a, CB'y', CA' x', then the coordinates of A', B', the centre of perpendiculars, and of the circum centre, of A'B'C, are respectively = = and A'B' y'2 + (x′ — \y')2 = (a — x′ — y')2; hence, substituting the values of the centres as above, the equation of the locus of each x2 + y2 = } (α −√√3y — 3x)2 ..(a). is 2. This locus is an hyperbola of which C is a focus, and as the axis is perpendicular to 3x + √3y = a, and passes through C, its equation is From (a), (B) we find for the vertices x = a and a; hence the centre of the circle is the farther vertex, and from (a) x' + y 3. The equation y + √√3 x = passes through the centre of perpendiculars and of the circumscribed circle, and is perpendicular to 4. When they lie on the branch of which the centre is the vertex, A', B' lie on CB, CA produced, and the circle is the inscribed circle of the triangle. 5. The circle will be the escribed circle opposite C, when the line joining the two centres is within the triangle CED', CD'= a; that is, when they lie on the branch between x = O and y == 1 √3 x. 6. When A', B' lie on BC, AC produced, the circle is the escribed circle opposite A' or B'. 7732. (By W. J. MCCLELLAND, M.A.)-On the sides of any quadrilateral inscribed in a circle, perpendiculars are drawn from the inverse of |