Originally posted by wolfgang59At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.
I have 3 numbers, lets call them a, b and c. I am surprised to see that one of the pairs is such that the difference between the two numbers is exactly equal to the difference between their squares.
ie a-b = a^2-b^2
when I look at the other pairs I find that in one pair the difference between their squares is twice their difference and in the final ...[text shortened]... e.
What is the sum of a, b and c?
(The problem is easy ... your solution must be concise!)
Originally posted by PBE6Yes. The solution is unique and a+b+c=3
At first I thought there were no solutions if a =/= b =/= c, but after a little algebra I believe I have an answer, and that a+b+c = 3.
Originally posted by wolfgang59a^2 - b^2 = a - b
But how did you get there? Its the technique I was after!!
Originally posted by mtthwNote that a^2 - b^2 = (a - b)(a + b)
a^2 - b^2 = a - b
b^2 - c^2 = 2(b - c)
c^2 - a^2 = 3(c - a)
Note that a^2 - b^2 = (a - b)(a + b)
If a, b and c are all different, we can divide each line by (a - b), (b - c) and (c - a) respectively.
a + b = 1
b + c = 2
c + a = 3
Add them up (and divide by 2): a + b + c = 3.
If the ratios are p, q and r, then the same process shows that:
a + b + c = (p + q + r)/2