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(E. 1. 1.) an equilateral ▲; from its vertex as a centre, at the distance of either of its sides, describe a circle: Then, if a regular hexagon be inscribed in the circle by the method used in E. 15. 4., taking either extremity of the base of the equilateral ▲, for the centre of the circle to be next described, it is manifest that the given straight line will be one of its sides.

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32. PROBLEM. A circle being given, to describe six other circles, each of them equal to it, and in contact with each other and with the given circle.

Let IGH be the given circle: It is required to describe six other circles, equal each of them to the circle IGH, and touching that circle and each other.

Find (E. 1. 3.) the centre K, of the circle IGH; take any of its semi-diameters, as KG; produce KG to A, and make GA=GK; from the centre K, at the distance KA, describe the circle ABCD; and in the circle ABCD inscribe (E. 15. 4.) the equilateral and equiangular hexagon ABFCDE; bisect (E. 10. 1.) the side AB of the hexagon in

cagon, to be inscribed in a given circle; and thus makes it easier to describe a regular pentagon, or a regular decagon, on a given finite straight line.

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L: Then since (E. 15.4.) AB➡KA or KB, and (constr.) AG and BH are, each of them, the half of KA or KB; .. AG and BH are each of them equal to AL or BL: If, ..., from the centres A and B, at the distance AG, or BH, two circles be described, they will be equal to one another and to the given circle, and they will touch (S. 6. 3.) the given circle in G and H, and will, also, touch one another in L. In the same manner, from the points E, D, C, F, as centres, may four other circles be described, each equal to the given circle and in contact with it, and touching also each other.

33. COR. No more than six circles can be described touching one another and a given circle, and each of them equal to the given circle.


34. PROBLEM. In a given circle to inscribe six circles equal to one another, touching, each of them, the given circle, and touching, also, one another.

Inscribe (E. 15. 4.) an equilateral and equiangular hexagon in the given circle, and through the points, in which are its, draw, (E. 17. 3.) straight lines touching the circle; and it may be shewn, by the method used in E. 12. 4., that the figure contained by these tangents is an equilateral and equiangular hexagon; from the centre of the circle draw straight lines to the several of the circumscribed hexagon, thus dividing it into six equal equilateral ; and if (E. 4. 4.) a circle be inscribed in each of these, it will be manifest from the demonstration of E. 4. 4, that the circles, so inscribed, will be equal, and that they will touch one another in common points of the sides of the ▲, and will, also, touch the given circle, each of them in one of the points of contact of the circumscribed hexagon.







1. THEOREM. If the first of four proportional magnitudes be greater than the second, the third is also greater than the fourth; if equal, equal; and if less, less.

Let A B C D; and first, let A>B; then C>D.

Take the doubles of the four magnitudes; and since (hyp.) A > B, twice A > twice B; .. (hyp. and 5 def. 5.) twice C>twice D; .. C> D.

In like manner it can be shewn, if A= B, that C=D; and if A < B, that C<D.


2. THEOREM. If four magnitudes are propor

tionals, they are proportionals also when taken inversely.

Let A:B::C:D; then B: A::D: C.

Let there be taken of A and C any equi-multiples pA, PC, and of B and D any equi-multiples qB, qD: Then (hyp.) A: B::C: D, .'. (5. def. 5.) if pA>qB, pC>qD; if pA = qB, pC=qD; if pA<qB, pC<qD; .. accordingly as qB is greater than, equal to, or less than pA, qD is greater than, equal to, or less than pC;

.. (5. def. 5.) B: A:: D:C.


3. THEOREM. If the first of four magnitudes be the same multiple of the second, or the same part of it, that the third is of the fourth, the first is to the second, as the third is to the fourth.

First, let A = pB, and C = pD; then A: B:: C:D.

Let there be taken of A and C any equi-multiples qA, qC, and of B and D any equi-multiples rB, rD. And since A = pB, accordingly as qA >,=, or < rB, will q times pB be >,=, or < rB, i. e. q times p will be >,, or < r; and .. q times pD will also be >,, or <rD; i. e. (hyp.) qC will be >,=, or < rD; ... (5. def. 5.)

A:B::C: D.

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