389195
[Translation: ]
Quæritur vel
[Translation: Sought, and .
[Translation: Another ]
[Commentary:
On this page, Harriot works a problem from De regula aliza liber
(Cardano , Chapter 41, pages 82–83. Cardano states the problem as follows:
De difficillimo problemate quod facillimum uidetur. CAP. XLI.
Nihil est admirabilius quam cum sub facili quaestione latet difficillimus scrupulus, huiusmodi est hic: quadratum cum latere est 10, & quadratum cum latere est 8, quaeritur quantum fit unum horum seu latus seu quadratum: Quia ergo est 10, & 1 quad. erit 10 m: 1 quad. igitur bc 100 m: 20 quad. p: 1 pos. p: 1 quad. quad. & hoc est aequale 8, quare 1 quad. quad. p: 92, aequatur 20 quad. m: 1 pos. adde 19 quad. utrinque fient 1 quad. quad. p: 19, quadrat. p: 92, aequalia 39. quad, m: 1 pos. detrahe , erunt 1 quad. quad. p: 19 quad. p: aequalia 39 quad. m: 1 pos. m: , inde adde 2 pos. p: 1 quad. utrinque ut in Arte magna, & videbis difficillimam quaestionem.
Chapter 41. On a very difficult question that seems easy.
Nothing is more remarkable than when under an easy question there lies a very hard stone, of which kind is this: the square of with the side is 10, and the square of with the side is 8; it is required to find one of those sides or squares. Therefore because is 10, and is one square, will be 10 minus a square. Therefore the square of is 100 minus 20 squares plus 1 square-square. Therefore will be 100 minus 20 squares plus 1 unknown plus 1 square-square, and this is equal to 8, whence 1 square-square plus 92 equals 20 squares minus 1 unknown. Add 19 squares to each side, making 1 square-square plus 19 squares plus 92 equal to 39 squares minus 1 unknown. Subtract , to give 1 square-square plus 19 squares plus equal to 39 squares minus 1 unknown minus , whence add 2 unknowns plus one square to each side as in the Ars magna, and you will see this very difficult question.
Harriot lets the side of be and then works through these instructions precisely. ]
De difficillimo problemate quod facillimum uidetur. CAP. XLI.
Nihil est admirabilius quam cum sub facili quaestione latet difficillimus scrupulus, huiusmodi est hic: quadratum cum latere est 10, & quadratum cum latere est 8, quaeritur quantum fit unum horum seu latus seu quadratum: Quia ergo est 10, & 1 quad. erit 10 m: 1 quad. igitur bc 100 m: 20 quad. p: 1 pos. p: 1 quad. quad. & hoc est aequale 8, quare 1 quad. quad. p: 92, aequatur 20 quad. m: 1 pos. adde 19 quad. utrinque fient 1 quad. quad. p: 19, quadrat. p: 92, aequalia 39. quad, m: 1 pos. detrahe , erunt 1 quad. quad. p: 19 quad. p: aequalia 39 quad. m: 1 pos. m: , inde adde 2 pos. p: 1 quad. utrinque ut in Arte magna, & videbis difficillimam quaestionem.
Chapter 41. On a very difficult question that seems easy.
Nothing is more remarkable than when under an easy question there lies a very hard stone, of which kind is this: the square of with the side is 10, and the square of with the side is 8; it is required to find one of those sides or squares. Therefore because is 10, and is one square, will be 10 minus a square. Therefore the square of is 100 minus 20 squares plus 1 square-square. Therefore will be 100 minus 20 squares plus 1 unknown plus 1 square-square, and this is equal to 8, whence 1 square-square plus 92 equals 20 squares minus 1 unknown. Add 19 squares to each side, making 1 square-square plus 19 squares plus 92 equal to 39 squares minus 1 unknown. Subtract , to give 1 square-square plus 19 squares plus equal to 39 squares minus 1 unknown minus , whence add 2 unknowns plus one square to each side as in the Ars magna, and you will see this very difficult question.
Harriot lets the side of be and then works through these instructions precisely. ]
Cardan. de Aliza. pa.
[Translation: Cardano, De regula aliza liber, page 83
[Translation: Cardano, De regula aliza liber, page 83
[Translation: ]
Quæritur vel
[Translation: Sought, and .
[Translation: Another ]
Et per
[Translation: And by ]
[Translation: And by ]
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