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APPENDIX III.

1. IF C'L' (fig. 214) be a given chord in a circle, it is required to draw through a given point G a chord CGL equal to C'E'.

Let O be the centre of the circle; draw OH perpendicular to C'L'; with centre O and radius OH describe a circle; from G draw CGML touching this circle; then since CL and C'L' are equally distant from the centre, CL C'L'.

There will be two positions of CGML corresponding to the two tangents through G.

2. If FQ, FR (fig. 214) be two fixed lines; and from any point K' in a circle C'K'L', K'C', K'L' be drawn parallel to FR, FQ respectively; then C'L' will be constant for all positions of K'. For 4CK'L' For 4C'K'L' = QFR and is constant; hence C'L' is constant.

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3. In a given circle, it is required to inscribe a triangle so that its three sides may pass through three fixed points P, Q and R.

Let ABC (fig. 215) be the triangle required, whose three sides AB, BC, CA pass through the three fixed points P, Q, R respectively; join RP; produce it to F so that

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join FQ, and make QG . QF = QC. QB; produce BF to K; join CK, KL; then since a circle may be described about B, F, A and R,

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Again, since a circle may be described about the points C, G, F, B,

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< BCG + 4 BFG = 2 right angles = 2 KFQ + BFG;

.*. < KFQ L BCG LBCL / BKL ;

hence KL is parallel to FQ; but when K' is any point in the circle and K'C', K'L' (fig. 214) are drawn parallel to FR, FQ, C'L' will be constant (Art. 2); hence if through G the line CGL be drawn equal to C'L' (Art. 1), the point C will be one of the angular points of the triangle; through Q draw QCB, and through P draw BPA; join AC, then ABC will be the triangle required.

The two positions of CGL will give two triangles which satisfy the conditions of the problem.

If OG be less than OH, the point G falls within the circle whose radius is OH; and the problem is impossible.

4. To inscribe geometrically a triangle in an ellipse whose three sides shall pass through three fixed points.

This problem is reduced to that of inscribing a triangle in the circle described upon the axis-major. (App. 1. Art. 3.)

5. To inscribe geometrically a triangle, in a hyperbola, whose sides shall pass through three fixed points.

From Appendix 1, Art. 4. we have,

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then (a + a) t1t2 - ẞs (t1 + t2) + (a − a) = 0.

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Similarly, if a, = ; B = a; a = ; B1 = a;

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we have (a + a2) t1 tз - ẞ2 (t1 + t2) + ( a − a2) = 0;

(a + a1) t2tz - ẞ1 (t2 + ts) + (a − a1) = 0 ;

(1)

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(2)

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which are the equations we should obtain for determining the angular points of a triangle inscribed in a circle whose radius is (a) so that the three sides may pass through three points whose co-ordinates measured from the centre are

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hence if CA (fig. 146) be the semi-transverse axis; P an angular point of the triangle inscribed in the circle APD so that its sides may pass through the three points above determined; then ACP = 01; and CR a sec 9, the abscissa of the corresponding angle of the triangle inscribed in the hyperbola so that its three sides may pass through the three points a, b; a2, b2; a1, b1.

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6. To inscribe geometrically in a parabola, a triangle whose three sides shall pass through three given points.

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Substitute for y1, Y2, Y3 in equations (1), (2), (3), Art. 6, Append. 1. the values 2at, 2at, 2at, respectively; and let

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··· (a + α2) tɔtɔ − ß3 (t1 + t2) + (a − az)

0, &c. hence t1, t, ts are the same as for the angular points of a triangle inscribed in a circle whose radius is a, so that its three sides may pass through three fixed points as, Ba; as, Bai a1, B1, respectively.

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Hence if E (fig. 15) be the vertex of a parabola; B the focus; with centre E, and radius EB describe the circle BDA, and let C be one of the angular points of the triangle

inscribed in the circle BDA, so that its three sides may pass through three points whose co-ordinates are as, ßz; az, ß2; B3; a1, ẞ1; draw BF perpendicular to AB meeting AC produced in F; then

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or BF is equal to the ordinate of the corresponding angular point of the triangle which is inscribed in the parabola so that its sides may pass through the three given points a, b; ɑ2, b2; α1, b1.

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Hence in each of the three conic sections, the problem is reduced to that of inscribing in a given circle a triangle whose three sides shall pass through three fixed points.

7.

b1 ;

From the equations in App. 1. Art. 10, it will easily be seen (1) that when a polygon is to be inscribed in an ellipse so that its n sides may pass through n given points a, ɑg, ba; ... ɑn, b; the problem will be equivalent to that of inscribing in the circle described upon the axis-major, a polygon whose n sides taken in order shall pass through the n fixed points

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(2) When the polygon is to be inscribed in a hyperbola; let a circle be described upon the transverse axis; and in this circle let a polygon be inscribed, whose n sides shall pass through then fixed points

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then if P be any angular point of this polygon, as in Art. 5, CR (fig. 146) will be equal to the abscissa of the corresponding angular point of the polygon which is to be inscribed in the hyperbola.

(3)

When the polygon is to be inscribed in a parabola ; let E (fig. 15) be the vertex; B the focus; with centre E

and radius BE describe a circle; and in this circle let a polygon be inscribed whose ʼn sides shall pass through the n

fixed points a (a − a1)

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let C be one of the angular points of this polygon; then, as in Art. 6, BF will be the value of the ordinate of the corresponding angular point of the polygon which is to be inscribed in the parabola.

8. When only five points of the conic section are given, the most simple method will be to determine the axes of the conic section, which will give geometrically in each case the circle and points through which the n sides of the subsidiary polygon are to pass; from whence the abscissæ of the corresponding angular points of the polygon inscribed in the conic section; and the angular points themselves may be determined.

For the Geometrical construction when the polygon is to be inscribed in a circle, the reader is referred to the Liverpool Apollonius by J. H. Swale.

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