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Pythagorean Puzzle

British_Olympics | 16:47 Thu 02nd Jun 2016 | Quizzes & Puzzles

23 Answers

A right-angled triangle has sides of length A,B and C where A is the hypotenuse and is an integer. When B and C are multiplied, the result is the square of A divided by one less than the square. When B is divided by C the result is one greater than than A divided by one less than A. What is the value of A ?

Struggling a bit with this one and the second sum seems ambiguous. Does it mean B/C = (A+1)/(A-1) or B/C = 1 + (A/(A-1)) ? I presume the former, but can't see a way to solve the 2 equations which is compatible with (A*A) = (B*B) + (C*C).

Phrases

Mixed Bag Quiz

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jim360

Best Answer

Ah. No, it doesn't have to be an integer, then. (A^2-2)^2 = a^4 - 4 a^2 + 4, which will give an integer if either a^2 is an integer, or a^2 ( a^2 -4) is integer. It turns out that for my second solution, (a^2-2)^2 = 5 (up to rounding error), from which a = Sqrt[Sqrt[5] +2]. By magic, inserting this into the condition that a^2 (a^2 - 4 ) is integer works, as a^2 -4 is...

13:01 Fri 03rd Jun 2016

Go To Best Answer

fiction-factory

Well done Jim
My root 2 was just an example.
Out of interest, jim360, did you solve the set of three equations manually or did you use a tool to help you? The maths itself is nota problem but I no longer seem to have the patience to do these longer problems on paper and can't read my own writing so make a mistake somewhere or run out of paper

jim360

Nah, I just put "Solve" into Mathematica for this one. It made it rather easier to switch between the two interpretations of the "B divided by C" statement, since I could just adjust the positions of the brackets.

Sometimes I'll solve problems by hand, still. But a three-unknown three-equation problem, with a structure that appears to end up with an equation in b^8 ? No, thanks, I'll pass that one to the computer.

British_Olympics

Question Author

Thanks jim, there is presumably a neat solution to give the value of A as the square root of ((root 5) +2), but it's all a bit beyond me !

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