Question

a ,b , c are sides of Right Triangle such that perimeter ,area of the

Triangle are equal.When a , b , c €N , and 1 ≤ a , b , c ≤ 100 . Find sum of all positive values of a, b , c.

Answer 1

Answer: There are two unique triangles that satisfy the given conditions.
               (a,b,c) = (5,12,13), (6,8, 10), (12,5,13), (8,6,10)
               Sum :     30+24 + 30+24 = 108

Given ΔABC, with say ∠C=90°.
          a,b, c ∈ N.     1 ≤ a, b, c  ≤ 100
Condition :    ar(ΔABC) = perimeter = a+b+c = 2 s. 
                                s = (a+b+c)/2

=>   ar(ΔABC) = 1/2 a b = a + b + c = 2 s     ---(1)

Also,  a² + b² = c²   ---(2)

We know,   ar(ΔABC) = s r     ---(3)
             where r = inradius

Compare (1) & (3) to get  r = 2.

See the diagram enclosed.

Since the two tangents to the incircle from any point (A or B or C) are equal, we get  that
           c = a + b - 4    ---(4)
          (c+4)² = (a+b)²
          8 c + 16 = 2 a b  ,   using (2)
=>      a b = 4 c + 8        ----(5)

 s = (a+b+c)/2 = (2a+2b-4)/2 = (a+b-2)

Δ² = s(s-a)(s-b)(s-c)
     = (a+b-2) (b-2) (a-2) (2)
     = (2 s)² = 2² (a+b-2)²

=>  (a - 2)(b - 2) = 2 (a+b-2)
=>  a b - 2a - 2b + 4 = 2 a + 2b - 4
=>  4 a + 4 b = a b + 8        ---(6)
=>  4 a = b (a - 4) + 8 
=>  4 a - 16 =  b(a - 4) + 8 - 16
=>   4 (a - 4) = b (a - 4) - 8
=>   (b - 4) ( a - 4) = 8  = 1 * 2 * 4 * 8    ---(7)

Since a and b are positive integers, only ways it is possible is 

   b - 4 = 1, a - 4 = 8    =>   b = 5, a = 12
   b - 4 = 2 , a - 4 = 4   =>   b = 6,  a = 8
   b - 4 = 4 , a - 4 = 2,   =>  b = 8,  a = 6
   b - 4 = 8 , a - 4 = 1   =>   b = 12,  a =  5

So the triangles are : (a,b,c) = (5,12,13),  (6,8,10), (12,5,13), (8,6,10)

Answer 2

ANSWER IS IN THE IMAGE