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Mr.Math
\(\large \large \text{ Fun With Mr.Math #2}\) Determine the side lengths of a right triangle if they are integers and the product of the legs' lengths equals three times the perimeter.
|dw:1328658036624:dw| \[\begin{align} ab&=3(a+b+c)\tag{a}\\ &\qquad\text{rearranging (a) we get:}\\ 3c&=ab-3(a+b)\\ \therefore 9c^2&=a^2b^2-6ab(a+b)+9(a^2+b^2+2ab)\tag{b}\\ &\qquad\text{from Pythagorus we get:}\\ a^2+b^2&=c^2\\ \therefore 9a^2+9b^2&=9c^2\\ &=a^2b^2-6ab(a+b)+9(a^2+b^2+2ab)\hspace{2cm}\text{[using (b)]}\\ \therefore ab&=6(a+b-3)\tag{c}\\ &\qquad\text{substituting (c) into (a) we get:}\\ 6(a+b-3)&=3(a+b+c)\tag{d}\\ \therefore a+b-c&=6 \end{align}\] The solution is therefore al Pythagorean triples satisfying (d), which leads to these unique solutions: a=7, b=24, c=25 a=8, b=15, c=17 a=9, b=12, c=15 We can get 3 more solutions by swapping the values for a and b but I believe that these would not add anything new.
Very good and neat asnaseer, as always! :-)