CHAPTER X. POLYGONS. 84. The centres of the circles inscribed in and circumscribed about an equilateral triangle coincide, and the diameter of one is twice the diameter of the other. Let ABC be the equilateral triangle. Draw the perpendiculars AD, B E, C F from the angular points to the sides, and let them intersect in O. Then, because A O, BO, CO bisect the angles A, B and C, O is the centre of the inscribed circle, and because FO, E O, DO bisect at right angles the sides A B, A C, B C, O is the centre of the circumscribed circle. Also, by Exercise 62, O F, which is the radius of the inscribed circle, is half of C F the radius of the circumscribed circle. 85. The locus of the centres of the circles which are inscribed in all right-angled triangles on the same hypothenuse is the quadrant described on the hypothenuse. Let A B C be one of the rightangled triangles. Let the bisectors of the angles BA and B meet in D, then D is the centre of the inscribed circle. ▲ DBA + ▲ DAB = half (≤ ABC + Z BAC Therefore angle. = half a right angle. BDA = three halves of a right Hence, because ▲ BDA is constant, the locus of D is the arc of a circle. Because B DA is the supplement of half a right angle, the arc is a quadrant. 86. If one convex polygon is entirely enclosed within another, the perimeter of the former is less than that of the latter. Let A B C D enclose F G H I ; continue G F, HI M D ΑΚ to meet the outer polygon in KL; then I K + KL +LF> IF. Add FG+ GH+ HI; then perimeter KLGH> perimeter F G HIF. Next continue GH to meet the outer perimeter in M; then it may be shown in the same way that perimeter ALG MA > perimeter K L G H K, and so on until the perimeter of the outer polygon is proved to be larger than that of another polygon whose perimeter is larger than that of F G H I. 87. In every right-angled triangle the diameter of the circle inscribed is equal to the excess of the sum of the two sides over the hypothenuse. Let A B C be the right-angled triangle and O the centre of the inscribed circle. Let OD, OE, OF be perpendiculars from O B F 10 E From the equal triangles A OE, A OF we have AF AE ; = and from the equal triangles COE, COD CD=CE. = By adding.. AF+ CD AC, ..BF+FD = excess of AB+ B C over A C. But BFF D= twice the radius of the inscribed circle; .. the diameter of the inscribed circle the excess of A B+ B C over A C. CHAPTER XI. PROBLEMS ON THE CIRCLE. 32. To describe with a given radius a circle which shall cut a straight line in two given points. When is this problem impossible? Let A and B be the given points. With these B points as centres and the given radius, describe arcs cutting one another in O and O'; then a circle with either of these points as centre and the given radius will pass through A and B. The problem is impossible when A B is greater than twice the given radius. 33. To describe with a given radius a circle which shall cut a circumference in two given points. B Let A and B be the given points. With these as centres describe arcs, as in the last example, intersecting in C and C'. These are the centres of two circles which fulfil the given conditions. 34. Through a given point in a circumference to draw a straight line parallel to a given chord. With one extremity, A', of the given chord as centre, and a radius equal to the distance between the given point, B, and the other extremity, A, describe an arc cutting the circle in B'. Join B B'. Since AA', B B' join the extremities of equal arcs, they are parallel. 35. To draw with a given radius a circle which shall touch a given straight line in a given point. At the given point, C, in the given line, AB, erect a perpendicular, CO, and cut off CO equal B' A B to the given radius; then O is the centre of the circle required. 36. To draw to a given circle a tangent which shall be parallel to a given straight line. Draw a diameter, A B, perpendicular to the straight line, and through A and B draw tangents; then these tangents and the given line, being perpendicular to the same straight line, are parallel. B G |