be drawn by the student), and join A D or A E, it is clear that the triangle ABD or A BE thus constructed will have sides A B, BD, or AB, BE of the required length; and it is obvious that the area of any triangle thus formed will be greater or less according as the distance of its vertex from the line ABC is greater or less. We have not, indeed, any problem in Euclid which expressly states this as a truth respecting triangles on the same base, but the property is clearly involved in the proof of I., 39.

Draw BF at Then the tri

Now since the vertex must lie on the circle C D FE, it is obvious that the distance of the vertex from A B C can never exceed the radius of this circle, and can only be equal to the radius when the side adjacent to A B is at right angles to A B. right angles to A B, and join AF. angle A B F is the triangle of maximum area under the given conditions. The proof consists in showing that DG or EH drawn perpendicular to A B C is less than B F. This is evident; for in the right-angled triangle B DG, the angle D BG is less than a right angle; therefore DG is less than B D,—that is, than BF.

VIII.

PERIMETERS OF TRIANGLES.

Let us next try a problem which is the converse of Ex. 5.

Ex. 13. To determine the greatest of all the triangles which can be constructed upon a given base and with a given perimeter.

FIG. 21.

Let A B (Fig. 21) be the given base, C the sum of the remaining two sides.

Now, with a knowledge of the property established in Ex. 5, it is of course very easy to see what is the solution of our problem. But we shall assume that the student is dealing with the problem independently. With centre A and radius equal to C describe the arc DEF, and draw radii AD, AE, and A F. Then if from B we draw lines B G, BH,

BK in such a way that BG is equal to G D, BH to HE, and B K to K F, it is obvious that each of the triangles AG B, AHB, AKB has the required perimeter. Now it is an obvious consideration that if BG is equal to G D, the angle GBD is equal to the angle G D B (we here draw in B D), and, therefore, that in order to draw B G so as to be equal to G D, we have only to make the angle DBG equal to the angle G D B. So that having a construction for determining any number of triangles, it is presumable that we shall find materials for determining the triangle of maximum area. But first let us see if anything is suggested by an examination of the figure. We see first that the triangle gradually increases as the angle at A increases. But there is clearly a limit to this increase. For it is obvious that we might have taken B as the centre of a circle with radius C, and thus have shown that the triangle increases as the angle at B increases. We are led, therefore, at once to the consideration that our triangle will have its greatest area when the angles at A and B are equal.

To see whether this is the case, we construct a new figure (Fig. 22), in which we omit all unnecessary parts of the former figure, and draw AKF so that, when the triangle A K B is completed, the angle KAB shall be equal to the angle KBA. We then draw KLM parallel to A B, knowing that it is on the distance of this parallel from A B that the area of the triangle AKB depends. We take

A E pretty near to AF (seeing that the triangle has obviously a nearly maximum area when the angles at A and B are equal, so that any great departure from equality makes the triangle considerably smaller). Let A E intersect KLM in L. Then, if we can show that BH, drawn as before, falls between BA and BL, our surmise will have been proved to be correct. Now the angle HBE, by our construction, is equal to the angle HEB;

therefore we must show that the angle L BE is less than the angle LE B, or L E less than LB (Euc. I., 19); therefore, adding A L, we have to show that AE (or C) is less than AL, LB together. This is the problem dealt with in Example 5, and thus the rest of the work corresponds with the work in that example. We find that AL and L B are together greater than A E, so that H does fall below L; and the triangle AKB is greater than the triangle

AH B. Our surmise is, therefore, shown to be correct, and the problem is solved.

It will be noticed that a problem in maxima and minima loses a large part of its difficulty when, as is usually the case, we are merely asked to prove that such and such relations supply a maximum or a minimum. In the case of Ex. 13, indeed, inspection supplied a tolerably obvious solution; but this seldom happens. Presented in the usual form, the above problem would run.

Of all triangles on a given base, and having a given perimeter, the isosceles triangle is the greatest. Thus given, the problem reduces immediately to the case of Ex. 5.

Ex. 13 fitly introduces the following, which belongs to a class often found perplexing: :

Ex. 14. Of all triangles having a given perimeter, the equilateral triangle is the greatest.

The difficulty in a problem of this sort resides in the fact that we have three elements to consider, all of which admit of being changed. In Example 13 we only had two sides to consider, and when a length had been selected for one, the other was determined at the same time. In Example 14 we have three sides, and must assign lengths to two before the final condition of the triangle is determined. This would be found to afford no assistance towards the solution of the problem. The way to proceed is to assign a length to one side, provisionally, and then to consider what relation must hold between