A Math Puzzle.. Who can solve it first?

AlexChandler

Here's a little puzzle that I thought of earlier. I gave it to my roommate and he wasn't able to solve it without a pretty good hint, but I'm sure somebody here will get it.

|A|+|B| = 2
|A+B| = √(2)

What are A and B?

(The bars are absolute value signs, and the " √(2) " is the square root of 2.)

skeptic2

A = 1+sqrt(2)/2
B = A - 2

Jimmy Snyder

Another solution:

AlexChandler

Wow... both of these are right, but neither is the solution that I was looking for!
Good job!

Redbelly98

There are 4 solutions. To see this, make a graph of the two relations; they intersect in 4 places.

EDIT: I was assuming A and B are real numbers.

Jimmy Snyder

Here are 8 of the 4 solutions, I don't know what the other -4 solutions are.

AlexChandler

Ok you guys are smart!
But nobody has mentioned the solution that I was looking for.
I guess that this is not a very good puzzle if there is infinitely many solutions.
My solution is:
A = <1,0>
B= <0,1>
They are unit vectors!
I think that I like my solution the best :D
Actually I guess that this does fit into what you said Jimmy. Haha!
I am very impressed with all of you.

D H

There are four solutions if A and B are restricted to the reals. If A and B can be complex numbers or vectors in R^N, N>1, the number of solutions is infinite.

Redbelly98

Replace that with "It suffices ...", and I'll agree. My 4 solutions do not meet your "requirement".

It helps to let people know what A and B are supposed to be when you pose the question! If you don't specify that, most people would assume they are to be real numbers. Jimmy extended this to include complex numbers, while you were thinking of 2-dimensional vectors -- equivalent to complex numbers with respect to addition and subtraction, but not in terms of multiplication and division.

D H

Neither of the points given in post #2 are on the unit circle.

AlexChandler

Aha I suppose you are right. But If I would have said that I am looking for two dimensional unit vectors, it would not have been much of a puzzle! Its all in good fun anyways. I see that I have very much to learn about numbers!

Redbelly98

Agreed. But we haven't found all of the complex-valued solutions. Jimmy's solution did not include skeptic2's or my answers.

ƒ(x)

I got the first answer.. a = 1 + sqrt(2)/2

But, I winged it a little bit. Teachers gloss over absolute value in school now. How did you guys solve it?`

Redbelly98

That depends. Are you looking for just the real number solutions, or complex number solutions as well?

ƒ(x)

well, both methods would be preferable.

SEngstrom

Actually, I think the solution A=1, B=i qualifies for the vector solution.

Redbelly98

For real number solutions, see post #5. If it helps, think of the numbers as x and y, rather than A and B, when you construct the graph.

For complex numbers, I don't think anybody has come up with a rigorous, complete solution in this thread. Perhaps one could be constructed based on my geometric argument in Post #12. A consequence of that argument is that |A| must be less than, or equal, to 1+(√2)/2 -- that might be a good starting point to finding B, given A.