411. Axiom XII. Any given arc AB has a definite ratio with a unit arc, which is greater than that of an arc AC, if c lies on the arc AB. 412. Axiom XIII. Any given angle ABC has a definite ratio with a unit angle, which is greater than that of ABD if BD lies within the angle ABC. 413. THEOREM. In the same or equal circles the ratio of two central angles is the same as the ratio of their intercepted arcs. B Outline of Proof: We show that in the figure ZAOB Z CO'D (1) LAOB arc AE = Z CO'D arc CD Divide arc CD into equal parts each less than arc EB. Lay off this unit arc successively on AB reaching a point F between E and B. Then arcs AF and CD are commen surable and by a proof exactly similar to that of § 243, making use of § 199, we can show that ZAOF Complete the proof as in § 410. arc AF arc CD (2) 414. Axiom XIV. Any given rectangle with base b and altitude a has a definite area which is greater than that of another rectangle with base b' and altitude a' if a≥ a and b > b′ or if a > a′ and b≥ b'. 415. THEOREM. The area of a rectangle is the product of its base and altitude. Proof: Denote the base and altitude by b and a, respectively, and the area by A. Suppose Aab, and let a' be a number such that A = a'b. Lay off BE a'. = Consider first the case where b is commensurable with the unit segment and a is not. Divide the unit segment into equal parts each less than CE and lay off one of these parts successively on BC reaching a point F between E and c. Denote the length of BF by a", and draw FF || AB. By hypothesis A = a'b, but a'ba''b since a' < a''. Hence the assumption that A < ab cannot hold. In the same manner prove that the A> ab cannot hold. The proof in case both sides are incommensurable with the unit segment is now exactly like the above and is left to the student. 416. Axiom XV. A circle has a definite length and incloses a definite area which are greater than those of any inscribed polygon and less than those of any circumscribed polygon. 417. THEOREM. For a given circle and for any number K, however small, it is possible to inscribe and to circumscribe similar polygons such that their perimeters or their areas shall differ by less than K. Proof: First, Let p and p' be the perimeters of two similar polygons, the first circumscribed and the second inscribed, and let a and a' be their apothems. Hence p⚫ B α a' a' can be made as small as we please; that is, p—p' can be made smaller than any given number K. Second, letting P and P' be the areas of the circumscribed and inscribed polygons respectively, we have a2 and the proof proceeds exactly as before. P = 418. Since the length C of the circle is greater than p' and less than p, it follows that c is thus made to differ from either p or p' by less than K. And since the area A of the circle is greater than P' and less than P, it follows that A is made to differ from either P or P' by less than K. 419. THEOREM. The lengths of two circles are in the same ratio as their radii. Proof: Let c and d be the lengths of two circles whose centers are O and O and whose radii are r and r'. This is possible since p can be made to differ from c by less than K (§ 418). Also circumscribe a polygon P' similar to P about ○ o' with perimeter p'. 2 = or p=p'. But p'>c′ and hence p>c'. "". с Hence the supposition that leads to the contradiction expressed in (1) and (2) and is untenable. · с r Now prove in same way that is untenable. 420. THEOREM. The areas of two circles are in the same ratio as the squares of their radii. Using §§ 348 and 418, the proof is exactly similar to that of the preceding theorem. PROBLEMS AND APPLICATIONS. 1. A billiard ball is placed at a point P on a billiard table. In what direction must it be shot in order to return to the same point after hitting all four sides? (The angle at which the ball is reflected from a side is equal to the angle at which it meets the side, that is, ≤1 = ≤2, and ≤3 = 24.) SUGGESTION. (a) Show that the opposite sides of the quadrilateral along which the ball travels are parallel. (b) If the ball is started parallel to a diagonal of the table, show that it will return to the starting point. 2. Show that in the preceding problem the length of the path traveled by the ball is equal to the sum of the diagonals of the table. 3. Find the direction in which a billiard ball must be shot from a given point on the table so as to strike another ball at a given point after first striking one side of the table. SUGGESTION. Construct BE 1 to that side of the table which the ball is to strike and make ED = BE. Cue ball Α' C' D B E B' EG D -F 4. The same as the preceding problem except that the cue ball is to strike two sides of the table before striking the other ball. SUGGESTION. B'E' = E'D', D'H = HF. 5. Solve Ex. 4, if the cue ball is to strike three sides before striking the other ball,- also if it is to strike all four sides. - 6. In the figure ABCDEF is a regular hexagon. Prove that: (a) AD, BE, and CF meet in a point. (b) ABCO is a rhombus. (c) The inner circle with center at O and the arcs with centers at A, B, C, etc., have equal radii. (d) The straight line connecting A and C is tangent to the inner circle and to the arc with center at B. (e) The centers of two of the small circles lie on the line connecting A and C. F (f) Find by construction the centers of the small circles. |