dendum est affectionem quandam communem in propositione descriptam.” That is, “ A Porism is a proposition in which it is proposed to demonstrate that some one thing, or more things than one, are given, to which, as also to each of innumerable other things, not given indeed, but which have the same relation to those which are given, it is to be shewn that there belongs some common affection described in the proposition.” Professor Dugald Stewart defines a porism to be “A proposition affirming the possibility of finding one or more of the conditions of an indeterminate theorem.” Professor Playfair in a paper (from which the following account is taken) on Porisms, printed in the Transactions of the Royal Society of Edinburgh, for the year 1792, defines a porism to be “ A proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate or capable of innumerable solutions." It may without much difficulty be perceived that this definition represents a porism as almost the same as an indeterminate problem. There is a large class of indeterminate problems which are, in general, loci, and satisfy certain defined conditions. Every indeterminate problem containing a locus may be made to assume the form of a porism, but not the converse. Porisms are of a more general nature ihan indeterminate problems which involve a locus. The ancient geometers appear to have undertaken the solution of problems with a scrupulous and minute attention, which would scarcely allow any of the collateral truths to escape their observation. They never considered a problem as solved till they had distinguished all its varieties, and evolved separately every different case that could occur. carefully distinguishing whatever change might arise in the construction from any change that was supposed to take place among the magnitudes which were given. This cautious method of proceeding soon led them to see that there were circumstances in which the solution of a problem would cease to be possible; and this always happened when one of the conditions of the data was inconsistent with the rest. Such instances would occur in the simplest problems; but in the analysis of more complex problems, they must have remarked that their constructions failed, for a reason directly contrary to that assigned. Instances would be found where the lines, which, by their intersection, were to determine the thing sought, instead of intersecting one another, as they did in general, or of not meeting at all, would coincide with one another entirely, and consequently leave the question unresolved. The confusion thus arising would soon be cleared up, by observing, that a problem before determined by the intersection of two lines, would now become capable of an indefinite number of solutions. This was soon perceived to arise from one of the conditions of the problem involving another, or from two parts of the data becoming one, so that there was not left a sufficient number of independent conditions to confine the problem to a single solution, or any determinate number of solutions. It was not difficult afterwards to perceive, that these cases of problems formed very curious propositions, of an indeterminate nature between problems and theorems, and that they admitted of being enunciated separately. It was to such propositions so enunciated that the ancient geometers gave the name of Porisms. Besides, it will be found, that some problems are possible within certain limits, and that certain magnitudes increase while others decrease within those limits; and after having reached a certain value, the former begin to decrease, while the latter increase. This circumstance gives rise to questions of marima and minima, or the greatest and least values which certain magnitudes may admit of in indeterminate problems. In the following collection of problems and theorems, most will be found to be of so simple a characier, ( being almost obvious deductions from propositions in the Elements) as scarcely to admit of the principle of the Geometrical Analysis being applied in their solution. It must however be recollected that a clear and exact knowledge of the first principles of Geometry must necessarily precede any intelligent application of them. Indistinctness or defectiveness of understanding with respect to these, will be a perpetual source of error and confusion. The learner is therefore recommended to understand the principles of the Science, and their connexion, fully, before he attempt any applications of them. The following directions may assist him in his proceedings. ANALYSIS OF THEOREMS. 1. Assume that the Theorem is true. 2. Proceed to examine any consequences that result from this admission, by the aid of other truths respecting the diagram, which have been already proved. 3. Examine whether any of these consequences are already known to be true, or to be false. 4. If any one of them be false, we have arrived at a reductio ad absurdum, which proves that the theorem itself is false, as in Euc. I. 25. 5. If none of the consequences so deduced be known to be either true or false, proceed to deduce other consequences from all or any of these, as in (2). 6. Examine these results, and proceed as in (3) and (4); and if still without any conclusive indications of the truth or falsehood of the alleged theorem, proceed still further, until such are obtained. ANALYSIS OF PROBLEMS. 1. In general, any given problem will be found to depend on several problems and theorems, and these ultimately on some problem or theorem in Euclid. 2. Describe the diagram as directed in the enunciation, and suppose the solution of the problem effected. 3. Examine the relations of the lines, angles, triangles, &c. in the diagram, and find the dependence of the assumed solution on some theorem or problem in the Elements. 4. If such cannot be found, draw other lines parallel or perpendicular as the case may require, join given points, or points assumed in the solution, and describe circles if need be: and then proceed to trace the dependence of the assumed solution on some theorem or problem in Euclid. 5. Let not the first unsuccessful attempts at the solution of a Problem be considered as of no value ; such attempts have been found to lead to the discovery of other theorems and problems. PROPOSITION I. PROBLEM. To trisect a given straight line. ANALYSIS. Let AB be the given straight line, and suppose it divided into three equal parts in the points D, E. D E B then DF is equal to AD, and FE to EB. and join AF, FB. Then because AD is equal to DF, therefore the angle AFD is equal to the angle DAF, and the two angles DAF, DFA are double of one of them DAF. But the angle FDE is equal to the angles DAF, DFA, and the angle FDE is equal to DAC, each being an angle of an equilateral triangle; wherefore the angle DAC is bisected by AF. and the angle FAD to DFA; therefore the angle CAF is equal to the alternate angle AFD: and consequently FD is parallel to AC. Synthesis. Upon A B describe an equilateral triangle ABC, bisect the angles at A and B by the straight lines AF, BF, meeting in F; through F draw FD parallel to AC, and FE parallel to BC. Then AB is trisected in the points ), E. but the angle FAD is equal to the angle FAC, and therefore DF is equal to DA. and FED to CBA; (1. 29.) therefore the remaining angle DFE is equal to the remaining angle ACB. Hence the three sides of the triangle DFE are equal to one another, and DF has been shewn to be equal to DA, therefore AD, DE, EB are equal to one another. Hence the following theorem. If the angles at the base of an equilateral triangle be bisected by two lines which meet at a point within the triangle; the two lines drawn from this point parallel to the sides of the triangle, divide the base into three equal parts. Note. There is another method whereby a line may be divided into three equal parts :-by drawing from one extremity of the given line, another making an acute angle with it, and taking three equal distances from the extremity, then joining the extremities, and through the other two points of division, drawing lines parallel to this line through the other two points of division, and to the given line; the three triangles thus formed are equal in all respects. This may be extended for any number of parts, and is a particular case of Euc. VI. 10. PROPOSITION II. THEOREM. If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two lines trisect the diagonal. Let ABCD be a parallelogram of which the diagonal is AC. Let AB be bisected in E, and DC in F, Then AC is trisected in the points G, H. Through E draw EK parallel to AC and meeting FB in K. therefore EB is equal to DF; and these equal and parallel straight lines are joined towards the same parts by DĒ and FB; And because AEB meets the parallels EK, AC, therefore the exterior angle BEK is equal to the interior angle EAG. For a similar reason, the angle EBK is equal to the angle AEG. llence in the triangles AEG, EBK, there are the two angles GAE, AEG in the one, equal to the two angles KEB, EBK in the other, and one side adjacent to the equal angles in each triangle, namely AE equal to EB; therefore AG is equal to EK, (1. 26.) but EK is equal to GH, (1. 34.) therefore AG is equal to GH. By a similar process, it may be shewn that GH is equal to HC. Hence AG, GH, HC are equal to one another, and therefore AC is trisected in the points G, H. PROPOSITION III. PROBLEM. Draw through a given point, between two straight lines not parallel, a straight line which shall be bisected in that point. Analysis. Let BC, BD be the two lines meeting in B, and let A be the given point between them. Suppose the line EAF drawn through A, so that EA is equal to AF; Ic Then ĂGHE is a parallelogram, wherefore AE is equal to GII, but E A is equal to AF by hypothesis; therefore GH is equal to AF. Hence in the triangles BŽG, GAF, also the side GH is equal to AF; therefore BG is equal to GF. Synthesis. Through the given point A, draw AG parallel to BC; on GD, take GF equal to GB; draw GH parallel to AE. Then in the triangles BGH, ĜFA, the side BG is equal to GF, and the angles GBH, BGH are respectively equal to TGA, GFA; wherefore GH is equal to AF, (1. 26.) but GH is equal to AE, (1. 34.) PROPOSITION IV. PROBLEM. From two given points on the same side of a straight line given in position, draw two straight lines which shall meet in that line, and make equal angles with it; also prove, that the sum of these two lines is less than the sum of any other two lines drawn to any other point in the line. Analysis. Let A, B be the two given points, and CD the given line. Suppose G the required point in the line, such that AG and BG being joined, the angle AG C is equal to the angle BGD. B Draw AF perpendicular to CD and meeting BG produced in E. Then, because the angle BGD is equal to AĜF, (hyp.) and also to the vertical angle FGE, (1. 15.) therefore the angle AGF is equal to the angle EGF; |