History Of Mathematics

1614 quotes
0 likes
0Verified
20Authors

Timeline

First Quote Added

April 10, 2026

Latest Quote Added

April 10, 2026

All Quotes

"Suppose 3, 4, 5 or a greater number of lines to be given in position, required a point from which, drawing lines to the given lines, each making a given angle with them, the rectangles of two lines thus drawn from the given point may have a given ratio to the square on the third, if there are three; or to the rectangle of the two others, if there are four; or again, if there are five lines, that the of the two remaining lines, together with a third given line, or to the parallelopiped composed of the three others, if there are six; or again, if there are seven, that the algebraic product of the three others and a given line, or to the four others, if there are eight, and so on. This was a problem which very much perplexed the ancient geometricians. Pappus says that neither Euclid nor Apollonius could give a solution. He himself knew that when there are only three or four lines the locus was a , but he could not describe it, much less could he tell what the curve would be when the number of lines were more than four. When the number of lines were seven or eight, the ancients could scarcely enunciate the problem, for there are no figures beyond solids, and without the aid of algebra, it is impossible to conceive what the product of four lines can mean. It was this problem which Descartes successfully attacked, and which, most probably led him to apply algebra generally to geometry. The following solution is that given by Descartes with a few abbreviations: AB, AD, EF and GH (fig. 2) are the given lines, C the required point from which are drawn the lines CB, CD, CF and CH making given angles CBA, CDA, CFE, and CHG. AB (=x) and BC (=y) are the principal lines to which all the others will be referred. Suppose the given lines to meet CB in the points R, S, T, and AB in the points A, E and G. Let AE = c and AG = d... By the... method he found the equation to bey^2 + xy + x^2 - 2y -5x = 0;which he showed belonged to a circle."

- La Géométrie

• 0 likes• mathematics-books• history-of-mathematics•
"vi. The angle in a semicircle is a right angle. It is believed that Thales proved this proposition in the following manner: Let ABCH be a circle of which the diameter is BC, and the centre E. ...Draw AE and produce BA to F. Because BE is equal to EA [both being radii of the circle], the angle EAB is equal to EBA; also, because AE is equal to EC, the angle EAC is equal to ECA [being angles at the base of an isosceles triangle]; wherefore, the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal to the two angles ABC, ACB [since the sum of the three angles of the triangle is equal to two right angles, i.e., a straight line]; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right angle; wherefore the angle BAC in a semicircle is a right angle. Thales's demonstration, if we may call this his, is quite different from the one given in modern text-books; but it is certainly neither less rigid nor less beautiful. The demonstration is the one given in Euclid, but his work, we must remember, is to a large extent compiled from the works of previous writers. It will be seen, however, that this demonstration implies a knowledge of a seventh proposition,—"If one side of a triangle be produced, the exterior angle is equal to the sum of the two interior and opposite angles." Thales must have been familiar with this truth."

- Ancient Greek mathematics

• 0 likes• ancient-greece• history-of-mathematics•