<PRE><B>Question 716880</B>
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No such triangle exists for any real number *[tex \LARGE n].  Now if you happened to have stated the problem incorrectly such that the hypotenuse is *[tex \LARGE 2n] and the legs are *[tex \LARGE n] and *[tex \LARGE n\sqrt{3}], then *[tex \LARGE n] can be any positive real number you like.


If the triangle you specified were to exist then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{(2n)^2\ +\ n^2}\ =\ n\sqrt{3}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{4n^2\ +\ n^2}\ =\ n\sqrt{3}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{5n^2}\ =\ n\sqrt{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\sqrt{5}\ =\ n\sqrt{3}]


But...


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{5}\ \neq\ \sqrt{3}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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