MAXIMA AND MINIMA 510. Definition. If several geometric magnitudes satisfy certain given conditions, the greatest is the maximum and the smallest is the minimum. 511. Theorem. Of all triangles having two sides given, that in which these sides include a right angle is the maximum. Given rt.AACH and AABH, having AH in common and BH=CH. 1. What is the maximum of all chords that can be drawn in a given circle? 2. What is the minimum of all chords that can be drawn through a given point within a circle? 3. What is the maximum of all lines that can be drawn within a square? 4. What is the minimum of all lines that can be drawn joining two parallel lines? 5. What is the maximum of all parallelograms which have two adjacent sides given? 6. What is the minimum of all lines that can be drawn to a circle from an outside point? 512. Isoperimetric Figures. Closed figures which have equal perimeters are called isoperimetric figures. 513. Theorem. Of all triangles having the same base and altitude, the isosceles triangle has the least perimeter. Given AAOD, an isosceles triangle, and AACD, any triangle not isosceles, having the same base and altitude.. To prove perimeter of AAOD<perimeter of AACD. Proof. Draw AB LAD and CF LAB. Produce DO to meet AB at B and draw BC. .. perimeter of AAOD<perimeter of AACD. 514. Theorem. Of all isoperimetric triangles having the same base, the isosceles triangle is the maximum. Given AACD, an isosceles triangle, and AABD, any triangle not isosceles, B having the same base AD and equal 515. Theorem. Of all polygons with sides all given but one, the maximum can be inscribed in a semicircle which has the undetermined side for a diameter. Given BCDEFH, the maximum of all polygons with sides BC, CD, DE, EF, and FH having the vertices B and H on the line AK. To prove BCDEFH can be inscribed in a semicircle whose diameter is BH. Proof. Connect any vertex, as E, with B and H. If ZBEH is not a right angle, the area of ABEH can be increased by moving B and H along AK, keeping the lengths of BE and EH unchanged, until ZBEH becomes a right angle while the area of the remainder of the polygon is unchanged. § 511 But, since it is given that the polygon is a maximum, its area cannot be increased. Then ZBEH is a right angle, and the semicircle described on BH as a diameter will pass through E. Why? In like manner, it may be proved that the semicircle passes through all the vertices of the polygon. .. the polygon BCDEFH may be inscribed in a semicircle with BH as a diameter. EXERCISES 1. What is the minimum of all circles which can be described on a given line as a chord? 2. Of all rectangles having a given area, that which has the least perimeter is the square. 516. Theorem. Of all polygons having their sides respectively equal, the maximum can be inscribed in a circle. Given polygon ACDEF which can be inscribed in a circle, and polygon A'C'D'E'F' which cannot be inscribed in a circle, and that the sides of the two polygons are respectively equal. To prove ACDEF>A'C'D'E'F'. Proof. Draw the diameter BE and the chords BA and BC. On A'C', construct AA'B'C' = AABC and draw B'E'. Then But BCDE>B'C'D'E', and BAFE>B'A'F'E'. 517. Theorem. AABCAA'B'C'. .. ACDEF>A'C'D'E'F'. § 515 Why? Why? $174 Of all isoperimetric polygons having the same number of sides, the maximum is a regular polygon. Given P, the maximum polygon of a given number of sides. To prove that P is a regular polygon. Proof. Draw the diagonal AC AABC is the maximum of isoperimetric B P triangles on the base AC, since P is a maximum polygon. In like manner, all the sides can be proved equal. Hence P is an equilateral polygon. .. P is a regular polygon. EXERCISE Inscribe the maximum rectangle in a circle. 514 Why? §§ 516, 459 518. Theorem. Of all isoperimetric regular polygons, the maximum has the greatest number of sides. Given the two isoperimetric regular polygons ACD and FGHK, having three and four sides respectively. To prove FGHK>^ACD. Proof. Draw CE from C to any point in AD. Construct ACEB=ACEA, with BEAC and BC= AE. Then the quadrilateral BCDE has the same area and the same perimeter as AACD. But square FGHK>quadrilateral BCDE. § 517 .. OFGHK>^ACD. In like manner, it can be shown that a regular pentagon is greater than a square having the same perimeter, and so on indefinitely. 519. Theorem. The area of a circle is greater than the area of a regular polygon having the same perimeter. This must follow, as in § 518, when the number of sides of regular polygons has been increased indefinitely. EXERCISES 1. The rectangle is the maximum of all parallelograms having given sides. 2. Divide a line into two parts so that the product of the two parts shall be a maximum. 3. Show how to inscribe the maximum rectangle in a semicircle. 4. Show how to inscribe an angle in a semicircle so that the sum of its sides shall be a maximum. |