APPENDIX TO PLANE GEOMETRY ALGEBRAIC SOLUTIONS OF GEOMETRICAL PROPOSITIONS. MAXIMA AND MINIMA CONSTRUCTION OF ALGEBRAIC EXPRESSIONS 431. NOTE. In the following propositions, a, b, c, d, etc., denote given lines, while x, y, z, etc., denote required lines. 433. Complex algebraical expressions are constructed by means of the eleven constructions of (432). It is quite often necessary to transform the algebraical expressions, in order to make them special cases of (432). Different algebraical transformations lead to different solutions. i.e. x is the mean proportional between 3 a and b, i.e. find √ab by means of (432, 7), and √ab √3 by means of (Ex. 991). a2 - b2 (a + b) ( a − b), Ex. 993. x = 4c 4 c i.e. find the fourth proportional to 4 c, a + b, and a — b. i.e. find the mean proportional between b and c, and construct a right triangle, having one arm equal to the mean proportional, and the hypote nuse equal to a. 434. SCHOLIUM. The expressions to be constructed in Geometry are always homogeneous. 435. The impossibility of the construction is indicated if : (1) The expression is imaginary. (2) The value of x is not within the limits indicated by the problem. (3) Sometimes, if the result is negative. If the required line has a certain direction, a negative result would indicate a line of opposite direction. SOLUTION OF PROBLEMS BY MEANS OF ALGEBRAIC ANALYSIS * 436. If a problem requires the construction of lines, it is often possible to state the condition in the form of an equation. The solution of the equation gives the unknown line in an algebraic form, which may be constructed according to (433). Ex. 996. To divide a line externally so that the small external segment is the mean proportional between the other segment and the given line. Analysis. Let x be one segment, then a + is the other, and As the minus sign would give a negative answer, i.e. an internal segment, we have to construct *The present chapter requires the knowledge of quadratic equations. Construction. E + A B Draw AB= a. At B, draw BC perpendicular to AB and equal to a. α 2 Draw AC, and produce it to D, so that CD = CB. Then x= AD. .. produce BA to E, so that AE= AD. E is the required point. 437. SCHOLIUM 1. The analysis used in this problem is called algebraic analysis in distinction from the purely geometric analysis given in Book I. 438. SCHOLIUM 2. The algebraic analysis contains a proof for the correctness of the construction, although quite often a purely geometric proof may be found. Ex. 997. From a triangle ABC, to cut off an isosceles triangle AXY, equivalent to one-half of ABC. Analysis. Let AX=AY=x, AB=c, and AC=b. ▲ AXY: ▲ ABC= x2: bc = 1:2. х A Construction. Construct x, a mean proportional between 2 and b, and on AB and AC, respectively, lay off AX and AY equal to x. Then AAXY is the required one. Ex. 998. To construct a right triangle, having given an arm ɑ, and the projection p, of the other arm upon the hypotenuse. or Analysis. Let x be the projection of a upon the hypotenuse. Construction. Draw AB-a, and BC perpendicular to AB .. from A as a center, with a radius equal to AE, draw an arc meeting BC produced in G. AGB is the required triangle. Ex. 999. From a given line b, to cut off a part x, that shall be a mean proportional between the remainder b x and another given line a. |