Now we have OG = √r2—x2, AG = hence AG AO+ OG gives √y-x* √y2 — x2; and = r + √ r2 −x2, or, squaring, y'— 2 r2 = 2 r√r2—2. Again squaring, we y2 y' 4 r'y' + 4 r2x2 0. have 2 [3] Substituting the value of y as given by [2], we find the side of an inscribed pentagon We have y = x(1+√5) [4] [5] OG = √r2x2 = √r— }(5—√5)r2 = 1 r(1+√5). [6] 2. The inscribed decagon. Join BF: then since AF bisects the line BC at right angles, it bisects the arc BFC in F; and hence BF is the side of the inscribed decagon. But ABF being a right angle, since it is in a semicircle, we have BF2 FA AB2, or, in symbols, = z2 = 4 r2—y2 = 4 r2— ¡ (10+2 √5)r2 = {(3— √5)r2; or, extracting the square root, we have for the side of the inscribed decagon z r(√5—1). = [7] 3. The circumscribing pentagon. The inscribed and circumscribed pentagons being regular, are similar figures, and their sides are as the perpendiculars from the centre upon the sides. That is, if PK be a side of the circumscribing pentagon, we shall have OG OB BC: PK; or, in symbols, : 4. The circumscribing decagon. Let QR be one of the sides; and draw OH perpendicular to BF, which it bisects in H. Also by similar triangles ABF, OHF, we have OH = AB = y = r√10+2 √5. PROBLEM XV. Given the lengths of three lines drawn from a point to the three angles of an equilateral triangle, to find its side. Let ABC be the triangle, and D the point. CE 2 y or y-x. = VAC-AE' = √4x2—x2 = x √3; Hence we have the following relations true, whether the point D is within the equilateral triangle, or without it. which, substituting for y its value given by [4], we have This value of z, and the value of y' given by [5], being substituted in [3], will cause it to become This equation contains only the unknown x, which value may therefore be found the reduction leads to 16x-4(a2+b2+c2)x2 — — a1—b1—c1+a2b2+b2c2+c2a2. [8] = This equation, when solved by the rule for quadratics, a2+b2+c3±√/6 ̧a3b°+b°c+c*a*,—3. a*+b*+c* } *.[9] 2 sign when the We must use the + sign when the point is within the triangle, as in the first figure; and the point is without, as in the second figure. If we suppose a triangle to be formed with the three lines a, b, c, and denote its area by A, expression [9] will become 2x = (a2 + b2 + c2 ± 4 ▲ √ 3) * . [10] 2 By referring to Art. 28, it will be seen that we have already given a geometrical solution of this problem in the case where the point is within the triangle, which corresponds with the above expression when the + sign is used. A geometrical solution for the case where the point is without the triangle, will be as follows: Let the equilateral triangle HFG (figure of Art. 28) be constructed on the other side of the line GF; then, by joining the vertex H and D, it will give the side of the triangle required. |
