SUPPLEMENTARY

MAXIMA AND MINIMA

DEFINITIONS

486 A maximum magnitude is the greatest of its class.

487 A minimum magnitude is the least of its class. Thus, the diameter of a circle is the maximum chord, and the perpendicular is the minimum line that can be drawn from a given point to a given straight line.

488 Isoperimetric figures are figures which have equal perimeters.

489 Of all triangles having two given sides, that in which these two sides include a right angle is the maximum.

D

E

HYPOTHESIS. In the A ABC and DEF, AB = DE, BC = EF, AB is to BC, and DE is oblique to EF.

PROPOSITION II. THEOREM

490 Of all isoperimetric triangles on the same base, the isosceles is the maximum.

B

HYPOTHESIS. In the ▲ ABC and DBC, the perimeters are equal, AB = AC, and DB > DC.

CONCLUSION. ▲ ABC > ^ DBC.

PROOF

=

On BA produced take AE AB, join EC, ED, and draw AH and DK L to EC.

Then the BCE is a rt. 4, for it may be inscribed in a semicircle whose center is A and radius AC.

The A AEC is isosceles by const.

§ 277

.. EH = HC. § 178

BA + AE = BA + AC = BD + DC.

PROPOSITION III. THEOREM

492 Of all polygons having all the sides given but one, the maximum can be inscribed in a semicircle whose diameter is the undetermined side.

E

HYPOTHESIS. ABCDEF is the maximum of all polygons whose given sides are AB, BC, CD, DE, EF.

CONCLUSION.

is AF.

ABCDEF can be inscribed in a semicircle whose diameter

PROOF

Join any vertex C to A and F.

Then the ACF is a rt. ; for if not, without changing the length of AC and CF, the area of the triangle ACF can be increased by making the ACF a rt. ≤, while the area of the remainder of the polygon remains unchanged.

§ 489

But this would increase the area of the polygon, which is contrary to the hypothesis that the polygon is a maximum. ..the ACF is a rt. 2, and the vertex C lies in the semicircumference whose diameter is AF. § 277

Likewise every vertex of the polygon lies in the semicircumference.

.. the polygon can be inscribed in a semicircle whose diameter is AF.

Q. E. D.

493 Of all polygons having their sides respectively equal, that is the maximum which can be inscribed in a circle.

HYPOTHESIS. The polygons ABCDE and A'B'C'D'E' are mutually equilateral, ABCD can be inscribed in a circle, A'B'C'D'E' cannot be inscribed in a circle.

CONCLUSION. ABCDE > A'B'C'D'E'.

PROOF

Draw the diameter CF, and join FA and FE. struct the ▲ F'A'E' = ▲ FAE, and join C'F'.

On A'E' con

Because the polygons FABC and FEDC are inscribed in semicircles,

PROPOSITION V. THEOREM

494 Of all isoperimetric polygons having the same number of sides, the maximum is equilateral.

Р

HYPOTHESIS. P is the maximum of all polygons having the same perimeter and the same number of sides.

CONCLUSION. P is equilateral.

PROOF

Since P is the maximum,

the ▲ ABC must be the maximum.

.. AB = AC.

"Of all isoperimetric triangles on the same base, the isosceles is the

maximum."

That is, any two consecutive sides of P are equal.

.. P is equilateral.

§ 490

Q. E. D.

495 COROLLARY. Of all isoperimetric polygons having the same number of sides, the maximum is regular.

PROOF. The maximum polygon is equilateral (§ 494), can be inscribed in a circle (§ 493), and is therefore regular (§ 435).