Page images
PDF
EPUB

the two remaining sides, whose sum is now assigned, in order that the triangle may be as large as possible. This we have learned already from Example 13. Those two sides must be equal. Hence, whatever side we suppose assigned, the remaining two must be equal to make the area of the triangle a maximum. Therefore, obviously, the triangle must be equilateral. The proof of this would run as follows :—

Let ABC (Fig. 23) be the triangle having the

[merged small][ocr errors][merged small][merged small]

greatest possible area with a given perimeter. Then ABC must be the greatest possible triangle on a given base BC and with the sum of the remaining sides equal to the sum of B A and A C. Hence B A is equal to AC. But, also, ABC is the greatest triangle on the base A B with the given perimeter ; hence, as before, AC is equal to BC. Therefore A B, B C, and C A are all equal.

As another instance of the application of this important method, we give the following:

Ex. 15.—A BC (Fig. 24) is an acute-angled triangle. It is required to determine the position of a point P within the triangle, such that the sum of the distances PA, PB, PC shall be a minimum.

Assume P to be the required point. Then PA, PB, and PC together have a minimum value. Therefore, also, PA and PB have the least sum they can have so long as the length of P C remains unchanged: so that if we draw the arc DPE with radius CP and centre C, AP and PB are together less than the sum of any two lines which can be drawn from A and B to meet on the arc DPE. Hence (Ex. 11, Cor.) AP and PB are equally inclined to CP. Similarly AP and PC are equally inclined to BP. Hence the angles APB, BPC, and CPA are all equal; and each, therefore, is onethird part of four right angles.

[blocks in formation]

In Examples 12 and 13 we notice that, although the number of the triangles which can be constructed under the given conditions 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 AB lie on the circumference of the circle EF D. In Ex. 13 there is no curve 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 problems, 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 satisfies the given relations, the line is called the locus (or place) of such points.

A few examples will serve better than a formal statement to show (1), the nature of plane loci; (2), the nature of problems founded on them; and (3), the methods available for readily solving such problems. 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 are fulfilled.

D

FIG. 25.

Ex. 16.-The straight lines A B, A C (Fig. 25) intersect in A. From A equal parts A D and A E are cut off from AB, A C respectively. ED is bisected in F. Find the locus of all such points as F.

Take A G equal to AH, AK equal to AL, and bisect G H in M, KL in N. Then it seems from the figure that the locus must be a straight line, whose direction is such 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 lines, A O, A P, very small indeed, the bisection of OP will be very near indeed 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 FMN is such a line, we see that the triangles A NL, 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, KL, of the isosceles triangle AKL is bisected in N, therefore N lies on the bisector of the angle KAL. Similarly every point obtained in accordance with the given conditions lies on the bisector of the angle KAL. It is clear, also, that every point in the bisector of the angle KAL fulfils the required conditions. For, let Q be such a point, and draw SQR at right angles to AQ; then the triangles A QS and A QR are equal in every respect. (Euc. I., 26.) Therefore, A S is equal to A R, and SQ to SR; that is, Q is a point fulfilling the required conditions.

Points in the production of QA beyond A cannot be said to fulfil the requisite conditions, because nothing has been said of the production of B A and CA beyond A.

Ex. 17.-Determine the locus of the vertices of all the triangles which stand upon a given base and have a given vertical angle.

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

and

Draw from A straight lines, A D, A E, A F, from B draw BG, BH, BK, to make with AD,

E

A E, AF, respectively, the angles BG A, BHA, and BKA equal to the angle C.1

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.

M

FIG. 26.

Now we notice that we 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

There is no problem in Euclid which shows us how to do this, but of course there is no difficulty in the matter. Among the subsidiary problems mentioned in the first part, one should be given showing how to draw a straight line in the manner required. Here, however, we do not require the 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 the problem.

« PreviousContinue »