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Now we have OG = V p? — x®, AG =

Vy?-?; and hence AG = AO + OG gives Vy* *" = r + Vr* —-x*, or, squaring, y2 yo = 2rVp_xo. Again squaring, we have y* 4 roy' + 4 r?x2 = 0.

[3] Substituting the value of y as given by [2], we find the side of an inscribed pentagon 2x = 1rV10– 25.

[4] We have y = x(1+5) įr V10+25, [5] OG = Vp? — * = Vp-}(5-75)r = įr(1+v5). [6]

2. The inscribed decagon. Join BF : then since AF bisects the line BC at right angles, it bisects the arc BFC in F; and hence BF is the side of the inscribed decagon. But ABF being a right angle, since it is in a semicircle, we have BF" = FA’ - AB', or, in symbols,

z = 4r? y* = 4 r2- ] (10+2/5) 72 = }(3-5);* ; or, extracting the square root, we have for the side of the inscribed decagon z = ir(5-1).

[7] 3. The circumscribing pentagon. The inscribed and circumscribed pentagons being regular, are similar figures, and their sides are as the perpendiculars from the centre upon the sides. That is, if PK be a side of the circumscribing pentagon, we shall have

OG : OB :: BC : PK; or, in symbols, OBX BC ritrV10-25 PK

= 2rV5-275. OG Àr1+75)

[8]

=

4. The circumscribing decagon. Let QR be one of the sides; and draw OH perpendicular to BF, which it bisects in H. Also by similar triangles ABF, OHF, we have OH = { AB = } y = { rv10+ 2 v5. Also, as in the last case,

OH : OF :: BF : QR; which gives OF XBF QR riir(5-1)

5 ArV 10+275

= 2r

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OH

PROBLEM XV. Given the lengths of three lines drawn from

a point to the three angles of an equilateral triangle, to find its side. Let ABC be the triangle, and D the point.

с

G

G

2 X,

A

B

EF

E

B

= X

Put AD = a,

BD = b,
CD =;
AB
EF = Y,

FD = z.
Then will AF = x + y, and BF y or y X.

CE = VAC - AE = 14x2-* = x 73; and CG = xv3 – 2, or z - XV3.

Hence we have the following relations true, whether the point D is within the equilateral triangle, or without it. (x+y)' + 2 = a',

[1] (2-y) + z = b,

[2] (273-2) + y = c.

[3]

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Equation [1] gives immediately z = — (x+y)"; in which, substituting for y its value given by [4], we have

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2

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+

This value of 2, and the value of yê given by [5], being substituted in [3], will cause it to become

a2 - b a-+

(a'-6)

= c. [7]

16x? This equation contains only the unknown x, which value may therefore be found : the reduction leads to 16.29 -4(a? +62+co)x* = -at-66-64 +aob?+boco+c'a'. [8]

This equation, when solved by the rule for quadratics, gives

2x =

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We must use the + sign when the point is within the triangle, as in the first figure; and the - sign when the point is without, as in the second figure.

If we suppose a triangle to be formed with the three lines a, b, c, and denote its area by A, expression [9] will become

40v3 [10] 2

'a+

2 x =

(a +

By referring to Art. 23, it will be seen that we have already given a geometrical solution of this problem in the case where the point is within the triangle, which corresponds with the above expression when the + sign is used. A geometrical solution for the case where the point is without the triangle, will be as follows:

Let the equilateral triangle HFG (figure of Art. 28) be constructed on the other side of the line GF; then, by joining the vertex H and D, it will give the side of the triangle required.

:

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