64 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. SOLUTIONS TO (No. IX.) 1. SEE Potts' Euclid, p. 50. 2. Euclid, Def. 5, Book v. and Prop. 33, Book vi. 3. Euclid, Prop. 21, Book x1. 4. Let C (fig. 73) be the point in which the common tangent CDE meets the straight line CBA joining the centres of the two circles; CFGHK a straight line cutting both circles; join BG, BD, BF, DG, DF, AK, AE, AH, EK, EH; then because the angles CDB, CEA are right angles, the triangles CDB, CEA are similar; .. CD : CE :: CB : CA :: BD : EA :: BF : AH :: CF : CH; hence the triangle CDF, CEH are similar, and DF is parallel to EH. Similarly, DG is parallel to EK, .. ▲ FDG = 2 KEH, and the segments FDG, HEK are similar. 5. Let a be the side of the square abcd (fig. 74) inscribed in the triangle ABC, and having one of its sides cd on the hypothenuse AB; then Next, let a be the side of the square Ca'c'b' (fig. 75) which has two of its sides coincident with the two sides CA, CB of the triangle; then CB = Ba' + Ca', or a' tan A + x 00 = a; or 1 + sin A cos A > sin A + cos A ; hence a is less than a'; and the square which has one of its angular points in the hypothenuse is the greatest. 6. Let x and me be the two sides; then and if a 2na, the sides of the triangle are 2na, (n2 - 1)a; = and the hypothenuse (n2 + 1)a. Hence if AB (fig. 76) be the base, divide AB into 2n equal parts, and let AD, DB contain n + 1 and n 1 parts respectively, and BE one part; then if BC be a fourth proportional between BE, BD and AD, BC = (n2 – 1) a, and AC = (n2 + 1) a; or ABC will be the triangle required. 7. Let AM (fig. 77) = a, MB = na ; .. AB = (n + 1) a ; .. ▲ ABC = ¦ AB. AC. sin A = ↓ (n + 1)2 ab sin A ; ▲ AMN = ↓ AM. AN sin A = n ab sin A, ABC – – n2 n + 1 .. Δ PBC = Δ ΑΒC - ΔΑΜΝ - ΔΡΝΟ - ΔΡΜΒ PMB ΔΑΜΝ ΔΑΜΝ; and the equation to the circle passing through A, M, N, referred to AB, AC as axes becomes x2 + y2 − 2xy cos A = a ̧x + b1y, 1 E but when y 0, x = AM similarly, when a = 0, y = b1 = AN, or b1 = and the equation to the circle is x2 + y2 — 2xy cos A α n + 1 n B n + 1 9 ; Hence if ax = By; x2 + y2 - 2xy cos A а x = By; from which equations two values of x and y may be found independent of n; therefore all the circles will pass through the two points so determined; hence they have a common chord Ba2 are 2 the co-ordinates of the common point through which they all pass. 9. The equations to the two straight lines are 2y = x + c = and by addition 5y 0, or y = 0, and x = c; hence the two straight lines intersect in the axis of x at a distance e from the origin. Let m, m' be the tangents of the angles which the two straight lines make with the axis of x; and ✪ the angle between them; .. m = m' m and tan 1 + mm be the equations to three parallel straight lines, whose perpendicular distances from the origin are p, p', p" respectively; then if the second straight line be equidistant from the first and third, Hence in the proposed example the equation becomes, y = mx + 1 {(c + c') + (c − c') } = mx + c. 11. Equating the coefficients of n, we have x2 + y2 = by, and ∞2 + y2 = ax; hence the straight line whose equation is ax = by will meet all the circles in the same point, or the circles have a common chord whose equation is ax = by. 90o, and the locus of C is a circle described on the diameter AB. 13. Draw PV (fig. 77) parallel to AC meeting AB in V; and let AB, AC be taken for the axes of x and y respectively; then if AV = ∞, VP = Y, (n + 1) y; n which is the equation to a parabola referred to two tangents AB, AC. If a>c, 2-is positive; and as a increases the curve a continues to be an ellipse, having the axis of a for its axismajor until a = ∞, when y2 = 4ca, and the curve approaches to a parabola. If a diminishes until a = c; the equation becomes or the curve becomes a circle when C moves up to S. C If a<c>, the curve becomes an ellipse having its axis major in the direction of the axis of x. When a = с 2, y = 0, and the ellipse coincides with the incides with the axis of y. (This result is not mentioned in the problem.) When a is negative, y2 = (~ + 2) (G still the equation to a hyperbola. α α x2 + 2cx 20x), which is When a ∞, y2 = 4cx, and the curve again becomes a parabola. |