SOLUTION: Find all values of t such that t - 1, t + 1, and 6 - t could be the lengths of the sides of a right triangle.

From the problem, it is clear that t-1 is one of legs, since t-1 is
shorter than t+1.


But looking at (t+1) and (6-t), we can not say which of them is longer;
so, we can not determine in advance, which of these two is the hypotenuse.


Therefore, we should consider two cases separately (a) t+1 > 6-t  and  (b) t+1 < 6-t.


Case (a).  t+1 > 6-t.

           Then 6-t is the other leg, while t+1 is the hypotenuse.

           So, we write the Pythagorean equation and find t from it

               (t-1)^2 + (6-t)^2 = (t+1)^2

               t^2 - 2t + 1 + 36 - 12t + t^2 = t^2 + 2t + 1

               t^2 - 16t + 36 = 0

                =  =  = .

               So, one   root is t =  = 13.29  (approx.);
                   other root is t =  =  2.71  (approx.).

               The root t =  = 13.29 does not work for the problem,
                                                       since then 6-t is negative.

               The root t =  = 2.71 satisfies inequality t+1 > 6 -t.


               So, in this case the legs are  t-1 =   = 1.71;
                                         and  6-t =  = 3.29;

                           the hypotenuse is  t+1 =   = 3.71.


               So, in this case the hypotenuse is really longer than the legs,
               and the triangle inequalities are satisfied, which is easy to check.


               Thus this case (a) is one possible solution.



Case (b).  t+1 < 6-t.

           Then 6-t is the hypotenuse, while t+1 is the other leg.

           So, we write the Pythagorean equation and find t from it

               (t-1)^2 + (t+1)^2 = (6-t)^2

               t^2 - 2t + 1 + t^2 + 2t + 1 = 36 - 12t + t^2

               t^2 + 12t - 34 = 0

                =  =  = .


               So, one   root is t =  = 14.37  (approx.);
                   other root is t =  = -2.37  (approx.).


               These roots do not work for this problem since t = 14.37 will create negative 6-t,

                                                        while t = -2.37 creates both negative t+1 and t-1.


CONCLUSION.  In this problem, the only solution is  t =  = 2.71;

             the legs are  t-1 =   = 1.71  and  6-t =  = 3.29;

             the hypotenuse is  t+1 =   = 3.71.

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