## A Math Puzzle.. Who can solve it first?

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.)
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 A = 1+sqrt(2)/2 B = A - 2
 Blog Entries: 1 Recognitions: Gold Member Another solution: Spoiler A = 1, B = i

## A Math Puzzle.. Who can solve it first?

Wow... both of these are right, but neither is the solution that I was looking for!
Good job!
 Mentor Blog Entries: 10 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.

Blog Entries: 1
Recognitions:
Gold Member
 Quote by Redbelly98 There are 4 solutions. To see this, make a graph of the two relations; they intersect in 4 places.
Here are 8 of the 4 solutions, I don't know what the other -4 solutions are.
Spoiler

1, i
-1, i
1, -i
-1, -i
(1 + i)/sqrt(2), (-1 + i)/sqrt(2)
(1 + i)/sqrt(2), (1 - i)/sqrt(2)
(-1 - i)/sqrt(2), (-1 + i)/sqrt(2)
(-1 - i)/sqrt(2), (1 - i)/sqrt(2)
Actually, there are infinitely many solutions. All that is required is that the two numbers be on the unit circle and the lines joining them to zero be orthogonal to each other.
 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.

Mentor
 Quote by Redbelly98 There are 4 solutions. To see this, make a graph of the two relations; they intersect in 4 places.
There are four solutions if A and B are restricted to the reals. If A and B can be complex numbers or vectors in RN, N>1, the number of solutions is infinite.

Mentor
Blog Entries: 10
 Quote by Jimmy Snyder Here are 8 of the 4 solutions, I don't know what the other -4 solutions are.

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

 Quote by 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> 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.
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.

Mentor
 Quote by Jimmy Snyder Actually, there are infinitely many solutions. All that is required is that the two numbers be on the unit circle and the lines joining them to zero be orthogonal to each other.
Neither of the points given in post #2 are on the unit circle.

 Quote by Redbelly98 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.
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!

Mentor
Blog Entries: 10
 Quote by 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 RN, N>1, the number of solutions is infinite.
Agreed. But we haven't found all of the complex-valued solutions. Jimmy's solution did not include skeptic2's or my answers.

Spoiler
Geometric argument:

In the complex plane, the numbers 0, A, and A+B form the vertices of a triangle. The vertex at A+B must be a distance √2 from the origin, while the two shorter sides' lengths must add up to 2.

For a fixed value of A+B, say √2 + 0i, the possible values of A forms an ellipse with foci at 0 and A+B, and major axis of length 2.

Furthermore, any solution pair can be multiplied by e to obtain all possible solutions.
Uh, and θ must be real.
 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?`
 Mentor Blog Entries: 10 That depends. Are you looking for just the real number solutions, or complex number solutions as well?

 Quote by Redbelly98 That depends. Are you looking for just the real number solutions, or complex number solutions as well?
well, both methods would be preferable.
 Actually, I think the solution A=1, B=i qualifies for the vector solution.

Mentor
Blog Entries: 10
 Quote by Redbelly98 That depends. Are you looking for just the real number solutions, or complex number solutions as well?
 Quote by ƒ(x) well, both methods would be preferable.
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.