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PLANE GEOMETRY - BOOK II

Sum of squares given.

86. PROBLEM 12. To divide a given line internal so that the sum of the squares on the parts may equivalent to a given square.

Let AB be the given line, and HK a side of the give square.

To divide AB at D so that the sum of the squares on AD and DB may be equivalent to the square on HK.

Construct the angle ABC equal to half a right angle. With A as center and radius equal to HK, describe an arc cutting BC in C. Draw CD perpendicular to AB.

The point D is the required point of division.

To prove this, draw the line AC.

The angle BCD, being equal to the difference of the angles ADC and DBC, is equal to half a right angle, and is therefore equal to the angle DBC. Hence DC equals DB (I. 62).

Therefore the sum of the squares on AD and DB is equivalent to the sum of the squares on AD and DC; which is equivalent to the square on AC, that is to the square on the given line HK.

Restriction. The given square must be less than the square on the whole line, and greater than twice the square on half the line, otherwise there is no solution.

87. NOTE. This problem can also be solved by means of 55.

88. Cor. To divide a given line externally so that the sum of the squares on the two segments shall be equivalent to a given square, which is greater than the square on the given

Medial section.

89. PROBLEM 13. To divide a given line into two parts such that the rectangle of the whole line and one part may be equivalent to the square on the other part.

Let AB be the given line, on which it is required to find a point P such that the rectangle of AB and BP may be equivalent to the square on AP.

join EB. Prolong EA, and take EF equal to EB. On AF

construct the square AFKP.

Then P is the required point of division.

To prove this, complete the rectangle FKND.

Since AD is bisected at E and prolonged to F, therefore the rectangle of DF and AF together with the square on EA is equivalent to the square on EF (52).

Therefore the rectangle DK is equivalent to the difference of the squares on EF and EA; that is, to the difference of the squares on EB and EA; that is, to the square on AB. Hence the rectangle DK is equivalent to the square ABCD. Take away the common part AN; then the square AK is

Therefore the rectangle of AB and PB is equivalent to the square on AP.

NOTE. When a line is divided as in this problem, it is said to be divided in medial section, for reasons that appear later (V. 97). The ancients called this mode of division sectio aurea, the golden section, on account of its important applications.

LOCUS PROBLEMS

90. PROBLEM 14. To find the locus of the vertex of a triangle whose base is given in magnitude and position, and whose equivalent square is given.

Outline. On the given base construct a rectangle equivalent to double the given square (73 (a)). The side parallel to the given line, extended indefinitely, is the required locus. Prove by 29, 33, 31.

Ex. On a given line construct an isosceles triangle equivalent to a given square. [Intersection of loci, I. 253 (ex.)].

91. PROBLEM 15. To find the locus of the vertex of a triangle whose base is given in magnitude and position, and the difference of the squares on whose sides is equivalent to a given square.

Outline analysis. Take any point P satisfying the given conditions, and from it draw a perpendicular PQ to the base AB. Show that any point on PQ satisfies the conditions (see 69); and that any point not on PQ does not satisfy them.

Show that the locus is constructed by dividing AB internally or externally at Q so that the difference of the squares on the segments shall be equivalent to the given square (internally as in 80 if the given square is less than the square on the given line, externally as in 82 if greater), and then drawing QP perpendicular.

Ex. Show how to construct a triangle, being given the base, the equivalent square, and the difference of squares on sides. (Intersection of loci.)

MAXIMA AND MINIMA

92. Definition. Of all the magnitudes of a certain class that fulfill prescribed conditions the greatest is called the maximum, and the least the minimum.

Among the magnitudes that fulfill the prescribed conditions there may be a number of equal magnitudes that are each greater (or less) than any of the others; in such case each of the equal magnitudes will be called a maximum (or minimum).

93. In each of the following theorems it is important to distinguish clearly between the prescribed conditions that define the class of figures considered, and the additional condition that characterizes a maximum (or minimum) figure of the class. The theorems concerning maxima come under one of the following type-forms, which are converse to each other:

1. Among all the magnitudes that fulfill the set of conditions A, any one that is a maximum fulfills the additional condition B.

2. Among all the magnitudes that fulfill the set of conditions 4, any one that fulfills the additional condition B is a maximum.

The first form asserts that the additional condition B is necessary for a maximum; the second form asserts that the additional condition B is sufficient for a maximum. As we shall be concerned with the complete conditions for a maximum, both of the converse theorems will be considered; which of them is to be proved first will depend on the nature of the prescribed conditions. Similar type-forms apply also to theorems concerning minima.

Greatest triangle having two given sides.

94. THEOREM 26. If two sides of a triangle are given, the triangle is a maximum when the given sides include a right angle.

Let the triangles ABC and A'BC have the sides AB and BC equal to the sides A'B and BC respectively. Let the angle ABC be a right angle, and A'BC an oblique angle.

To prove the triangle ABC greater than A'BC.

[Prove the altitude AB greater than the altitude A'D.]

A

B

95. Cor. 1. Conversely, if two sides are given, and if the triangle is a maximum, then the given sides include a right angle.

For if not, a greater triangle could be constructed by making the included angle a right angle, contrary to the hypothesis that the given triangle is a maximum.

96. Cor. 2. Of all parallelograms having given sides, one that is rectangular is a maximum; and conversely.

Least perimeter in equivalent triangles.

97. THEOREM 27. Of all equivalent triangles having the same base, that which is isosceles has the least perimeter.

Let ABC and A'BC be equivalent triangles having the same base BC, the first triangle

being isosceles and the second not.

To prove that the perimeter of ABC is less than the perimeter of A'BC.

Draw CE perpendicular to 44', and prolong it to meet the prolongation of BA in D. Join A'D.

B

D

Outline proof. Prove in succession: AA' parallel to BC (33); angle EAD equal to EAC; AD equal to AC; A'D equal to A'C; sum of BA and AC equal to BD; which is less than sum of BA' and A'D; which equals the sum of BA' and A'C. Draw conclusion.