Aug. 24, 1883.] 



♦ KNOWLEDGE ♦ 



127 



#iir iftatftcmatical Column. 



GEOMETRICAL PROBLEMS. 

 By Richakd A. Pkoctob. 



PART xra. 



LET us consider the method applied in our last. One condition 

 shows us that the point we seek mtist lie on a certain curve ; 

 another condition shows us that the point must lie on another curve. 

 Therefore, the point we seek must lie at some intersection of the 

 two curves. If there are more intersections than one, tlie problem 

 has more solutions than one ; if there is but one intersection, there 

 is but one solution ; if, lastly, the curves do not intersect, the 

 problem is insoluble. 



Let UB take, as another instance, the following problem : — 



Ex. 20. — Let A B (Fig. 30) be a given straight line, C a giveyi 

 angle, D a given point within the given circle EF G. It is required 

 to determine a point at Schick A B shall subtend an angle equal to 

 the angle C, and which (point) shall ie the bisection of a chord 

 through D to the circle E F G. 



.30. 



In order that A B may subtend an angle equal to C at the required 

 point, this point must lie, we find (as in Ex. 17), on the arc A E B, 

 containing an angle equal to the Angle C. 



Again, in order that the required point may be the bisection of 

 a chord thi'ougli D to the circle E F G, this point must lie, we find 

 (as in Ex. 18), on the circle L K M, which has for diameter the line 

 joining D with K the centre of the circle E F G. 



These two loci — viz., tlie arc A E B and the circle L M K, deter- 

 mine by their intersection the points which satisfy the required con- 

 ditions. There may be two points, as in the case illustrated by our 

 figure ; or one point, if the circle L M K touch the arc A E B ; or 

 the two loci may not intersect, in which case the problem does not 

 admit of solution. 



Wo liave supposed that the point is required to lie above A B. 

 If not, tlien an arc equal in all respects to A E B, but applied on the 

 opposite side of A B, would include otlier points satisfying the first 

 condition of our problem. It might happen that the circle L M K 

 intersected the latter arc, instead of, or as well as, the arc A E B. 

 Such point or points of intersection would also supply a solution of 

 the problem. 



Fig. 31. 



Problems in maxima and minima also involve very frequently the 

 discussion of loci. 



Suppose, for instance, that the following problem is given : — 

 Ex. '21.— A, B, C, and D (Fig. 31) are four fixed points. It is 



required to determine a point equidistant from A and B, and such 

 that the sum of its distances from C and D shall be a minimum. 



In this case we first find the locus of points equidistant from A and 

 B. This, as in Ex. 18, is the line F G drawn at right angles to the 

 line AB, through its bisection E, Fig. 31. We have, then, to find 

 a point in F G such that the sum of its distances from C and D 

 may be a minimum. We find (as in Ex. 11) that the point must 

 be "so taken — as at H — that the lines from C and D to it shall make 

 equal angles (C H F and D H G) with the line F G. 



To take another simple instance, suppose we had the following 

 problem : — 



Ex. 22. — A triangle is constructed on a given base A B (Fig. 32), 

 and with a vertical angle equal to the angle C, to determine its 

 figure that its area may be a maximum. 



Here we first inquire what is the locus of the vertices of aU the 

 triangles which can be constructed on the base A B with a vertical 

 angle equal to the angle C. We find, as in Ex. 17, that the locus 

 is the arc A D B, containing an angle equal to the angle C. 



After this we can find no difiiculty in determining the triangle of 

 maximum area. The vertex must clearly lie at that point of the 

 arc A D B which is farthest from A B ; and D, the bisection of the 

 arc, is obviously the required vertex. The student will at once see 

 this ; but perhaps he may find a little difficulty in proving it. We 

 leave this part of the problem to him as an exercise, having already 



Fig. 32 



Fig. 33. 



examined the treatment of problems of this class. We note, how- 

 ever, that what he has to do is to show that a parallel to A B 

 through D is fartlier from A B than the parallel through a vertex 

 of any other triangle fulfilling the required conditions ; and this 

 will be established if it be shown that the paraUel to A B through 

 D is a tangent to the arc A D B. 



Sometimes a familiarity with the treatment of problems on loci, 

 serves us in a somewhat more subtle manner, as in the following 

 problem : — 



Ex. 23.— ^£ (Fig. 33) is a given finite straight line. It is 

 required to show where a point must be taken in the given indefinite 

 line DE, in order that the angle subtended by AB from the point 

 may be a maximum. 



Suppose we take any point, D, at random, in D E, and draw the 

 lines D A and D B. Then, in inquiring whether the angle A D B is 

 a maximum or not, it would be an obvious consideration that the 

 segment of a circle, A D E B, described on A B, contains all the 

 points from which A B subtends an angle equal to the angle A D B. 

 From the point E, therefore, A B subtends an angle, A E B, equal to 

 the angle A D B ; and from any point, F, between D and E, it is clear 

 that A B subtends an angle greater than A D B. For, producing 

 A P to meet the arc A D B in G, and joining G B, we see that 

 A F B is greater than A G B (Euc. I., 16), that is than A D B (Euc. 

 III., 27). It is clear, therefore, tliat we cannot have a maximum 

 so long as the arc described on A B, to pass through the particular 

 point selected in D E, cuts D E in another point. Hence we ai-rive 

 immediately at the solution of our problem— viz., that the required 

 point, U, is so situated that the arc on AB through H touches the 

 straight line D E. 



It is easy to draw a circle through two given points to touch a 

 given straight line. But, strictly speaking, the solution of the 

 above jiroblem is complete without the construction of the circle 

 A HB, since wo have assigned a sufficient condition for the deter- 

 mination of the required point in D E. 



The consideration of problems on loci leads ns to another class 

 — or rather two classes of deductions — viz., those in which it is 

 required to prove either that certain straight lines pass through one 

 point, or that certain points (more than two) lie in a straight line. 



(To be continued.) 



