Aug. 10, 1883.] 



♦ KNOWLEDGE ♦ 



95 



AMERICAN-ENGLISH. 



[894] — I take the liberty of sondiiif; you a few notes on the sub- 

 ject of "British and American-English," suggested by your very 

 interesting essay in " Leisure Readings," Vol. 1. 



P. 185 — Aberrjavenny. My family resided in this place for a 

 great many years, and I never heard it spoken of under any other 

 name than Aberga'ny until I went to school, where I had to use the 

 former name to the exclusion of the latter. 



P. 195 — " Mad." In Devonshire, as in America, the word is used 

 to express anger, crossness, ill-teniper. " Don't be so mad ! " " He 

 is as mad as he can be." In both cases meaning cross, out of 

 temper. I remember hearing that, when a child was born in a cer- 

 tain family, the neighbours said, " Here is another mad G 



come," meaning that the child had possibly inherited the family 

 temper, which was not very gentle. 



P. 195 — Ordinary. In Cambridgeshire and Suffolk I have often 

 heard this word used to express inferiority, as in America. "An 

 ordinary child" was "a plain child." 



P. 199 — Reckon. This word was formerly common in Sussex 

 in the sense of "I believe," "I think," "It is probable, "He will 

 go there to-morrow, I reckon." 



P. 202 — Sure. Formerly used in Sussex as it is now said to be 

 used in America. 



I am an old woman, in my 80th year, and have outlived most of 

 these expressions, but I believe that they are genuine English, 

 although provincial. Maby P. Meekifield. 



ERRATUM. 

 [895] — Your printer has made an omission in my description of 

 the Magic Square in page 61 (888), which, in a manner, makes it 

 unintelligible. It should run thus (fourth line from top of page) : 

 — " By the angles from the corners or the middles to the centre, 

 whether right, acute, or obtuse ; by the four corner, four middle 

 and central cells; which will still hold good, &c." G. S. 



<^m iilattjematiral Column* 



GEOMETRICAL PROBLEMS. 



By Richard A. Peoctob. 



PART XI. 



IN Examples 12 and 13 we notice that, although the number of 

 the triangles which can be constructed under the given condi- 

 tions is infinite, yet all the triangles belong to a certain set, or 

 family. In Ex. 12, the vertices of all the triangles on the base 

 A B, lie on the circumference of the circle E F D. In Example 13 

 there is no ciu-ve along which the vertices are shown to lie ; but if 

 the reader were carefully to construct a number of triangles 

 according to the method described in that example, he would find 

 that the vertices all lie upon a certain curve, which however is 

 not a circle. 



These considerations introduce us to an important class of pro- 

 blems, called problems on loci. 



If all points which satisfy certain relations can be shown to lie on 

 a certain line (straight or curved), and if every point on this line 

 satisfy the given relations, the Une is called the locus (or place) of 

 Buch points. 



A few examples -will serve better than a formal statement to show 

 (1), the nature of plane loci ; (2), the sort of problems founded on 

 them ; and (3), the methods available for readily solving such 

 pi-oblems. It must be premised that the complete solution of such 

 problems requires that it should be shown that both the conditions 

 stated in the above definition of a locus should bo fulfilled. 



Ex. 16. — The straight lines A B, A C (Fig. 25) intersect in A. 

 From A equal parts A D and AE are cut off from AB, AC respec- 

 tively. E D is bisected in F. Find the locus of all such points as F. 



Take A G equal to A H, A K equal to A L, and bisect G H in M, 

 K L in N. Then it seems from the figure that the locus must be a 

 straight line, whose direction is snch as will carry it through A. A 

 moment's consideration shows that the locus — whatever it be — 

 must pass up to A ; for if we conceive equal hnes, A 0, A P, very 

 very small, the bisection of P will be very very near to A. Again 

 it will occur, from a consideration of the figure, that the locus is a 

 straight line bisecting the angle A. Now, assuming for the moment 

 that A F M N is such a line, we see that the triangles, A N L, 

 AN K are equal in every respect (Euc. I., 4), and this leads us at 

 once to the proof we require. For, because the base, K L, 

 of the isosceles triangle A K L is bisected in N, therefore N lies on 

 the bisector of the angle K A L. Similarly, every point obtained 

 in accordance with the given conditions lies on the bisector 

 of the angle K A L. It is clear, also, that every point in 

 the bisector of the angle K A L fulfils the required conditions. 

 For, let Q be such a point, and draw S Q R at right angles to A Q ; 

 then the triangles A Q S and A Q R are equal in every respect. 

 (Euc. I., 26.) Therefore, A S is equal to A R, and SQtoSR; 

 that is, Q is a point fulfilling the required conditions. 



Fig. 26. 



Points in the production of Q A beyond A cannot be said to fulfil 

 the requisite conditions, because nothing has been said of the pro- 

 duction of B A and C A beyond A. 



Ex. 17. — Determine the locus of the vertices of all the triangles 

 which stand v.-poii a given base, and have a given vertical angle. 



Let A B (Fig. 26) be the given base, C the given angle. 



Draw from A straight lines, A D, AE, AF, and from B draw 

 B G, B H, B K, to make, with A D, A E, A F, respectively, the 

 angles B G A, B H A , and B K A equal to the angle C* 



We see at once that G, H, and K do not lie in a straight line, so 

 that we gather that the locus is circular, since loci of other figures 

 are not dealt with in deductions from Euclid. 



Now we notice that wo might have drawn our lines from B 

 instead of A, and that therefore the locus must have points, G', H', 

 K', situated in the same manner with respect to B as G, H, and K 

 with respect to A. 



It is already clear that a circle passing through, or near to A and 

 B, contains all the vertices. Wo see also that the circle cannot 

 but pass through A and B ; for if wo draw A L very very near to 

 A B, then B L drawn so as to make the angle B L A equal to C, will 

 clearly meet AL in a point very very near to B. We describe, 

 then, a cii-cle through A and B, and also (of course the circle is 

 dravni by hand through the points G, H, K, &c. 



At this point we cannot fail to be reminded of III., 21, which 

 tells us that all the angles in the same segment of a circle are 

 equal. We see, therefore, that our surmise is correct, and that the 

 circular segment on A U, containing an angle equal to the angle C, 

 is the locus we require. All the points on this segment fulfil the 

 requii'od condition ; but points on the remaining segment, A M B, 

 do not do so. If triangles are to be di-awn on one side only of A B, 

 the segment. A K B, contains all the required points. For if any 

 point, N, mthout the segment fulfil the given condition, join N A. 

 and N B ; let N B cut the segment, A K B, in P, and join A P. 

 Then the angle, A N B, is equal to C (hyp.), but the angle A P B is 

 equal to C (Euc. III., 21). Therefore, the angle, AP B, is equal to 

 the angle ANB, the greater (Euc. I., 16) to the less, which is 

 absurd. In like manner no point within tho segment fulfils the 

 required conditions. Therefore, the segment, A K B, is the required 

 locus. 



* There is no problem in Euclid which shows us how to do this, 

 but of course there is no difliculty in the matter. Among the sub- 

 sidiary problems mentioned in tho first part, one should be given 

 showing how to draw a straight line in the manner required. Here, 

 however, we do not require tho problem at all ; since we are dealing 

 with the practical construction of the figure — about which there is 

 no difficulty — not with the mathematical treatment of tho problem. 



