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(1)

of the curve. Let them be a, B; then the equation to the ellipse may be put under the form

A (x a)” + 2 B (x a) (y - 3) + C (y - 3)2 + 1 = 0, where A, B, C are to be determined.

Now the condition that the ellipse shall pass through the origin gives

Aa? + 2 Baß + C B + 1 = 0. The condition that the curve shall pass through the point x = a, y = 0, gives

A (a – a) 2 B (a a) B + CB? + 1 = 0. Subtracting (1) from (2) we have

A (2a - a) + 2 B B = (). The condition that the curve shall pass through the point x = 0, y = b, gives

C(1 B) 2 Ba (6 B) + Aa’ + 1 = 0.
Subtracting (1) from (4) we have

C (23 6) + 9 Ba = 0.
Also a (3) + B (5) – 2 (1) = 0, gives

Aaa + CbB + 2 = 0.
Combining (3), (5) and (6), we find
- (2ß b)

- (2u - a)
1 *a (ub + Ba ab) B(ab + Ba - ab)'

B - (2a - a) (23 6)

2uß (ab + Ba ab) It remains now to express the area of the ellipse in terms of these coefficients. The method to be adopted is the same as that used in the preceding example.

If , be any radius measured from the centre, so that

pode = (x – a): + (y - 3)? – 2 (x – u) (y - 3) cos e, the axes of the ellipse are determined by the equation

(AC B) pot – (A + C 2B cos 0) 2? + sin’ 0 = 0),

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The area of the ellipse will therefore be

a sin 0

(AC B2) which is to be a minimum, involving the condition that

AC B’ shall be a maximum. Substituting in this the values of A, B, C previously determined, and differentiating with respect to a and B, we obtain equations for determining these quantities. The result involves several factors, but those which correspond to the problem are

2ba + - ab = 0, 2aß + ba - ab = 0);

whence as, B = , and therefore

The area of the ellipse is therefore

21 ab sin 8.

It appears then that the area of the ellipse is to that of the triangle as 47 : 3f, and that its centre coincides with the centre of gravity of the triangle. This problem is given by Euler in the Nova Acta Petrop. Vol. ix. p. 147, but his method of solution is deficient in symmetry. In the same volume he has also discussed the more general problemTo describe the least ellipse which passes through four given points.

The preceding solution is due to Bérard. Annales de Gergonne, Vol. iv. p. 288.

(19) To inscribe the greatest ellipse in a given triangle.

By following a method similar to that adopted in the last example it will be found that the area of the maximum ellipse is to that of the triangle as a : 38; that its centre coincides with the centre of gravity of the triangle; and that the points of contact bisect the sides of the triangle. Bérard, Ib. p. 284. (20) To find a point within a triangle from which if lines be drawn to the angular points the sum of their squares is the least possible.

The centre of gravity of the triangle is the point which possesses this property.

(21) Among all triangular pyramids of given base and altitude, find that which has the least surface.

Let a, b, c be the sides of the base, h the height, 0, 0, y the angles of inclination of the faces to the base : then

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0, 0, y being subject to the condition

a cot 0 + b cot Q + c cot ys = const. We find that 0 = 0 = V, or that the faces are equally inclined to the base.

(22) To find a point within a triangle from which if lines be drawn to the angular points their sum may be the least possible.

The direct solution of this problem is long and complicated, but we may without much difficulty obtain a geometrical condition by which the point is readily determined.

Let ABC (fig. 11) be the given triangle, a, b, c its sides ; let () be the required point, OA = U, OB = v, OC = w. Draw ON perpendicular to AB, and let AN= x, ON = y; also let AON = 8, BON = 0, CON = y. Then u= x++y", vo=(c x)? + y", w2= (b cos A – X)? + (b sin A-y). In order that u + v + w may be a minimum, we must have

du do dw

da tar tar = 0,

du di du

+ - + du dy dy

= 0.

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Now du XV du C-8

dw bcos A-N - =

"=-sinya, dx u

w du y

W bcos A-Y

1 = cosy. dy u Therefore we have the conditions

sin 0 = sin + sin y,

cos 0 = - (cos + cos y). Squaring and adding, we find

cos (4-0) = -, or y-$ = 120°. That is, the angle BOC = 120°; and in the same way it may be shewn that AOC = 120o = AOB. Hence if on any two sides of the triangle we describe segments of circles containing angles of 120°, their intersection will determine the point 0. The actual length of the lines u, v, w, and the value of the minimum sum may be found. For from the geometry of the figure we have the equations

vy2 + vw + w? = a,
u+ uw + w? = 6,
u” + uv + 02 = 0,

4 m2
and Uv + uw + vw = 7,

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q' *

a*- c. whence ve u +

W = U + --- ,

2a- - 0

and therefore

u + 1 + w = r = 30 +

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This problem possesses considerable interest in the history of mathematics. It was proposed by Fermat to Torricelli, who, after some time, gave three solutions of it. He communicated it to Vincent Viviani, who also solved it, and gave the geometrical construction mentioned above; but he says that it is a problem “quod, ut vera fateor, non nisi iteratis oppugnationibus tunc nobis vincere datum fuit." For his demonstration see his Geometrica Divinatio de Varimis et Minimis, p. 150. The reader will also find a discussion of this problem, and of the more general one where the minimum function is au + Be + gu', in a paper by Fuss in the Nora Acta Petrop. Vol. si. p. 235. The reader is recommended to consult also a paper by M. J. Bertrand in Liouville's Journal de Mathématiques, Tom. vill. p. 155.

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