PROPOSITION III. THEOREM 443. The maximum of isoperimetric triangles on the same base is the one whose other two sides are equal. Hyp. ABC and ABD have equal perimeters, and AC=CB. Proof. Draw median CE and DF AB, meeting CE in F. Join FA and FB. Then CE is the perpendicular bisector of AB. (Why?) (442) .. FE <CE. (311) or .. area AFB < area ACB, (unequal altitudes), area ADB < area ACB. Q.E.D. 444. COR. Of all isoperimetric triangles, the equilateral has the maximum area. PROPOSITION IV. THEOREM 445. Of all polygons having all sides given but one, the maximum can be inscribed in a semicircle having the undetermined side as diameter. E F Q B Hyp. Polygon ABCDEF is the maximum of all polygons having given the sides AF, FE, ED, DC, and CB. To prove ABCDEF can be inscribed in a semicircle whose diameter is AB. Proof. Join any vertex, as D, with A and B. Then ▲ ADB must be the maximum of all triangles that can be formed with sides AD and DB. For otherwise by making ZADB a right one without changing the sides AD and DB, we could increase A ADB without altering the remaining parts AFED and DCB of the polygon. Or polygon ABCDEF would be increased, which is contrary to the hypothesis, since ABCDEF is a maximum. .. A ADB is the maximum of all triangles having AD and DB given. Then ZADB is a right angle, (Why ?) and D is on a semicircumference that can be constructed on AB. For the same reason every vertex of the polygon must lie on the semicircumference. Q.E.D. Ex. 1017. To inscribe an angle in a given semicircle so that the sum of its arms is a maximum. PROPOSITION V. THEOREM 446. Of all polygons constructed with the same given sides, that which can be inscribed in a circle is the maximum. Hyp. Polygon ABCDE, which is inscribed in a circle, is mutually equilateral with polygon A'B'C'D'E', which cannot be inscribed in a circle. To prove area ABCDE> area A'B'C'D'E'. Proof. From A draw diameter AF, and join F to the two nearest vertices C and D. On C'D', construct AC'D'F' equal to ▲ CDF, and join A'F'. Then A CFDA C'F'D'. area ABCDE > area A'B'C'D'E'. (Ax. 5.) q.e.d. PROPOSITION VI. THEOREM 447. Of all isoperimetric polygons of the same number of sides, the equilateral is the maximum. Hyp. ABCDE is the maximum of all isoperimetric polygons of the same number of sides. To prove AB BC= CD = DE = EA. = Proof. The polygon must be equilateral, for suppose any two sides AB and BC were not equal. Then construct on the diagonal AC the isosceles triangle AB'C, isoperimetric with ABC. Area AB'C would be greater than area ABC. (443) .. polygon AB'CDE would be greater than the isoperimetric polygon ABCDE, which would contradict the hypothesis. Hence AB = BC. Q.E.D. 448. COR. Of all isoperimetric polygons of the same number of sides, the maximum is regular. Ex. 1018. In a given segment to inscribe an angle so that the sum of its arms is a maximum. Ex. 1019. In a given semicircle to inscribe a trapezoid whose area is a maximum. PROPOSITION VII. THEOREM 449. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater Proof. Let F be any point in CD. ABCFDE may be considered a hexagon, having one of its angles equal to a straight angle. .. area H> area ABCDE. (Why?) Q.E.D. 450. COR. The area of a circle is greater than the area of any polygon whose perimeter equals the circumference of the circle. Ex. 1020. In a given circle to inscribe a triangle, having a maximum perimeter. Ex. 1021. The circumscribed regular polygon has a smaller area than any other polygon circumscribed about the same circle. Ex. 1022. Of all equivalent parallelograms on the same base, which has the minimum perimeter ? |