Solve the Math Puzzle: A + B = 2 and |A+B| = √(2)

[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:

Spoiler

A = 1, B = i

 

[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]

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.

 

[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 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 R^N, N>1, the number of solutions is infinite.

 

[Redbelly98]

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".

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.

 

[D H]

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.

 

[AlexChandler]

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!

 

[Redbelly98]

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.

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^iθ to obtain all possible solutions.
Uh, and θ must be real.

 

[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)]

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

well, both methods would be preferable.

 

[SEngstrom]

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

 

[Redbelly98]

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

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.

 

[Redbelly98]

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

No. It qualifies as one of the solutions. The solution should list all possible solutions.

I'll put some more thought into this, and see what I can come up with.

 

[SEngstrom]

No. It qualifies as one of the solutions. The solution should list all possible solutions.

I'll put some more thought into this, and see what I can come up with.

by "the vector solution" i mean the solution the OP had in mind.

 

[Redbelly98]

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.)

For complex-valued numbers A and B, the solutions are:

$$\begin{flalign*}<br /> A & = & & a e^{i \phi}, \\<br /> \mbox{ where } & & a & \mbox{ can be any real number satisfying } \ 1-\frac{\sqrt2}{2} \le a \le 1 + \frac{\sqrt2}{2} \ <br /> \mbox{ and } \phi \mbox{ can be any angle; and } \\<br /> B & = & & (2 - a)e^{i \phi} e^{i \theta} \\<br /> \mbox{where} \\<br /> \theta & = & & \pm \cos^{-1} \left( - \ \frac{(a-1)^2}{a \cdot (2-a)} \right) <br /> \end{flalign*}$$​

It's getting late in my time zone, so a post of my derivation will have to wait until tomorrow (Sunday).

 

[D H]

by "the vector solution" i mean the solution the OP had in mind.

What vector solution?

The solution as intended by the OP depends on a bit of abuse of notation, using |A| to designate the norm of A as opposed to the modulus of A. The problem as envisioned by the OP would have been better stated as

$$\begin{array}{ccc}<br /> ||A|| + ||B|| &=& 2 \\[4pt]<br /> ||A+B|| &=& \surd 2<br /> \end{array}$$

Even then, the problem is poorly stated as the nature of A, B, and the norm are not specified. The solution for A,B∈ℝ² with the euclidean norm will look quite different from the solution for A,B∈ℝ³ with the taxicab norm, and both will look very different for A,B drawn from some function space.

There's nothing wrong with that here as this is a math puzzle; poorly stated problems and abuse of notation are the heart and soul of many puzzles.

 

[D H]

For complex-valued numbers A and B, the solutions are: ...

That is close to the solution I had in mind when I said in post #8 that

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.

What I had in mind was to use the angle θ as the driver rather than the modulus of A. With this, for complex variables A, B defined as

$$\begin{align*}<br /> A &\equiv ae^{i\phi} \\[4pt]<br /> B &\equiv (2-a)e^{i(\theta+\phi)}<br /> \end{align*}$$

yields

$$a = 1\pm\sqrt{\frac{-\cos\theta}{1-\cos\theta}}$$

Derivation: The definitions of A and B make |A|+|B|=2. Thus the only concern is to find the conditions that make |A+B|=√2, or |A+B|²=(A+B)(A+B)^*=2. With A and B defined as above,

$$A+B = (a\cos\phi + (2-a)\cos(\theta+\phi)) + i(a\sin\phi+(2-a)\sin(\theta+\phi))$$

Computing the square of the modulus of A+B yields

$$\begin{align*}<br /> |A+B|^2 &= (A+B)(A+B)^{\ast} \\[4pt]<br /> &= a^2+(2-a)^2+2a(2-a)(\cos\phi\cos(\theta+\phi)+\sin\phi\sin(\theta+\phi)) \\[4pt]<br /> &= 2(1-\cos\theta)a^2 - 4(1-\cos\theta)a + 4<br /> \end{align*}$$

Equating this to 2 yields

$$(1-\cos\theta)a^2 - 2(1-\cos\theta)a+1 = 0$$

Or

$$a = 1\pm\sqrt{\frac{-\cos\theta}{1-\cos\theta}}$$This result generalizes rather nicely to numbers such as quaternions, octonions, etc. and to inner product spaces.

 

[mathematicsma]

What I find comedic about this post is this:

To almost every person who responded, this question was analyzed as if it appeared in a textbook- a math problem. So we get the reals, the complex solutions, etc.

But to the OP, this was a puzzle, almost a joke. The idea is that you're supposed to look at the question, and since you don't know much math past linear algebra, and aren't really good at calculus, you scratch your head in defeat. Then you have a laugh when you find out that you were barking up the wrong tree-- they're vectors, not numbers. So the riddle challenges your assumptions.

Of course, the riddle broke down here because there are lots of people on PF with advanced math knowledge, and the OP's hapless roommate is put to shame. Oh, well.

Perhaps I'm reading the dynamic wrong here, but that's the way this whole thread struck me.

 

[Nikitin]

assuming these are vectors, how do you solve it using simplest type of math (no computer/university math please, if possible)?

PS. The answer is funny ([0,1] and [1,0]), can't believe I didn't see that... heh

 

[SEngstrom]

Hmm simplest kind of math... Maybe first consider the special case when A and B are orthogonal (at right angles) so that A=ax, B=by, where x and y are orthogonal unit vectors. Then the original equations become $$a+b=2, a^2+b^2=2$$, since the length of the combined vector A+B is simply given by Pythagoras theorem. From there you could consider vectors not at right angles and the second equation gets a mixed term.

 

[Nikitin]

yes thanks. But only way to solve this (with simple math) would be to spot at once that these two vectors would be orthogonal :-/

PS. Would be smart to specify that the sum of the absolute values of vectors a and b equals 2, so |a|+|b| = 2. same with |a+b|= srqt (|a|² + |b|²)=sqrt (2)

 

[I like Serena]

Nice puzzle.

Since the original problem does not state which algebra is being used "all solutions" would imply all solutions in any type of algebra.
Anyway, I'm limiting myself to the algebra's I know that have a + operation and an |.| operation. The typical algebra's are the real numbers with their absolute values, numbers in R/C/H/O or vectors of R/C/H/O, all with the regular |.| being the square root of the sum of the squared numbers.

As said, in the real case, there are 4 solutions.
In all the other cases there are an infinite number of solutions.
In all those cases however, we can reduce the problem to A being along the first real axis, and B having an "angle" in any direction.
Just for fun I have constructed the accompanying picture that shows all solutions for B in the projected plane in which both A and B reside.