Ancient puzzles 3

[slimshady2357]

This is one I've posted on at least one other board (JREF) and so some people may have seen it and remember the solution. But I think others will find it new.

Consider a rectangle with integer length sides whose perimeter is equal to it's area (the numerical part, I'm aware they have different units :P). There are only two such rectangles.

1) What are the lengths of the sides?

and

2) Prove they are the only two solutions.

Adam

[Ari]

Maybe:

1. 3,6 and 4,4

2. Um... It's the only two integers the graph y=2y/x +2 crosses.

Damn my math is rusty as hell.

[slimshady2357]

Maybe:
1. 3,6 and 4,4

Correct, of course. That is the easy part :)

2. Um... It's the only two integers the graph y=2y/x +2 crosses.

Damn my math is rusty as hell.

I'm not sure I would count that :D

It's possible to prove this with algebra.

Adam

[Chatter]

We seek positive integer solutions to the equation

ab = 2a + 2b.

Assume without loss of generality that a>=b. If a,b>4 then

ab > 4a = 2a + 2a >= 2a + 2b.

Hence, b<=4. We can easily check that b=1 and b=2 are impossible, leaving the two remaining solutions:

a=6, b=3

and

a=4, b=4

[Chatter]

Generally, k(a+b)=ab and a>=b if and only if

a=k^2/k' + k

b=k+k'

where k' is any positive integer that divides k.

In the case of k = 2, there are exactly two solutions, corresponding to the two divisors of 2, namely 1 and 2, which yield

a = 4/1 + 2 = 6 and b = 2+1 = 3

and

a = 4/2 + 2 = 4 and b = 2+2 = 4.

[slimshady2357]

We seek positive integer solutions to the equation

ab = 2a + 2b.

Assume without loss of generality that a>=b. If a,b>4 then

ab > 4a = 2a + 2a >= 2a + 2b.

Hence, b<=4. We can easily check that b=1 and b=2 are impossible, leaving the two remaining solutions:

a=6, b=3

and

a=4, b=4

An interesting solution, I quite like it. It's different to how I solved it. I'll post my solution in a little bit.

One note, perhaps in the future you could add spoiler tags to your answer if you are one of the first to offer a correct solution.

Adam

[slimshady2357]

Generally, k(a+b)=ab and a>=b if and only if

a=k^2/k' + k

b=k+k'

where k' is any positive integer that divides k.


That is also interesting, where did you find that? Can you prove it?

Adam

[slimshady2357]

Here is how I solved it, which is very close to how it was outlined in the book as well.


Starting with:

(1) xy = 2x + 2y

xy - 2x = 2y
x(y-2) = 2y factor out x
x = 2y/(y-2)

mult both sides by 2/y to get

(2) 2x/y = 4/(y-2)

from (1) we can also see that

x = 2x/y + 2 (divided both sides by y)

or

x - 2 = 2x/y

sub this into (2) to get

x - 2 = 4/(y-2)

x = 4/(y-2) + 2

Since x is an integer and 2 is an integer, then 4/(y-2) must be an integer.

That means that (y-2) must be equal to either 4, 2 or 1 (it cannot be anything else (e.g. 1/2) since y is also a positive integer. These are the only possible solutions for (y-2).

Which means that y must equal either 6, 4 or 3 respectively.

This gives the answers (3,6) , (4,4) and (6,3) for (x,y).

[Chatter]

Generally, k(a+b)=ab and a>=b if and only if

a=k^2/k' + k

b=k+k'

where k' is any positive integer that divides k.


That is also interesting, where did you find that? Can you prove it?

AdamI thought I'd try to generalise the result somewhat, but now that I come to examine the proof I see I made a mistake.

Since ka,kb > 0, ab = ka + kb > ka,kb and so a,b>k. Hence, b = k + k' for some k' > 0.

Now we can rearrange

ka + kb = ab

to give

a = kb / (b - k) = k(k+k')/k' = k^2/k' + k.

and we notice that a is an integer if and only if and only if k^2/k' is an integer. That is, if and only if k' divides k^2.

At this point, I wrongly thought that divisors of k^2 which are less than k are divisors of k but they are not in general. So instead, the result is

k(a + b) = ab if and only if

a=k^2/k' + k
b=k+k'

where k' is a positive divisor of k^2.

Since I'm no longer messing around with that a>=b stuff, we get the three solutions for k=2 which you provided.

[livius drusus]

Pardon the interjection, but damn, Chatter, you're really good at this. May I ask what your mathematical background is?

[Chatter]

Pardon the interjection, but damn, Chatter, you're really good at this. May I ask what your mathematical background is?Hey cheers. I'm in the third year of a maths degree, and one of the courses this year is number theory. I think supplying proofs at about this level is something that could be expected of us come the exams.

[livius drusus]

That's marvelous. Although I'm notoriously innumerate, I love reading these threads and have the utmost appreciation for y'all's skills. Good luck on your exams. Here's hoping we have lots more threads like this one for you to practice your proof-crafting. :ahoy:

[MooseIBe]

Dear God.

Moose, passed GCSE maths on the second attempt.

[Ari]

I'm not sure I would count that :D
It's possible to prove this with algebra.
Adam
Don't disparage the lazy, it worked (sort of). :D

[JoeP]

This is one I've posted on at least one other board (JREF) and so some people may have seen it and remember the solution. But I think others will find it new.

Consider a rectangle with integer length sides whose perimeter is equal to it's area (the numerical part, I'm aware they have different units :P). There are only two such rectangles.

1) What are the lengths of the sides?

and

2) Prove they are the only two solutions.

Adam

Same as last time, have only looked at OP so far. (And done some work, dropped a colleague off to collect his car, talked to my daughter about her puppy, had dinner, watched Charmed, and spoken to my aunt-in-law(?) in Edinburgh. I expect multiple correct answers already.)

1) 3 x 6 and 4 x 4. Actually there is a third, but you'll probably try to argue that 0 x 0 is not a "rectangle". :P

2) Enumerating the first few cases seems the simplest proof:

Let the sides be x and y (both positive integers). To avoid double-counting, require that x<=y.
The perimeter P = 2x+2y. The area A = x*y.

If x=1, P=2+2y=A=1*y => y=-1, a contradiction.
If x=2, P=4+2y=A=2y, which has no solution for y.
If x=3, P=6+2y=A=3y => y=6 - the first solution.
If x=4, P=8+2y=A=4y => 2y=8 => y=4 - the second solution.
All remaining cases: x>4 => A>4y =>2x+2y>4y => 2x>2y => x>y, but this is a contradiction.

Joe

[JoeP]

As I asserted, several correct answers already. More elegant than mine!

I'll fall back on claiming the third solution - 0 x 0!

[Chiron]

Naw man, there are still three that've been missed. In the following, my format is (x, y; xy, 2x + 2y), and if the third and fourth entries equal one another, then I am victorious:
(1, -2; -2, -2) Success!*
If you continue subtracting 1 from side y, you will find that side x's necessary length goes to 2 as y goes to negative infinity. Similarly, if you continue to instead add 1 to side y, you will find that side x's necessary length goes to 2 as y goes to positive infinity. These lead me to the third and fourth as-of-yet unmentioned solutions:
(2, -∞; -∞, -∞) A winner is me!
(2, ∞; ∞, ∞) Success!

Wait, you say negative lengths and areas aren't allowed? And infinity isn't an integer?!
What kind of party is this?

K
I note that BBCode allows for negative sizes, unlike the rest of you fascist oppressors freedom-hating terrorists.
Also, what do you get when you cross a mountain climber with mosquito?

*JoeP almost got this one, but he goofed in his math: for x=1 and y=-1, P=0 while A = -1. The formula gives the correct output of y=-2, P=A=-2 for an input of x=1.

[JoeP]

*JoeP almost got this one
Don't go putting numbers in my mouth there. I clearly stated "positive integers".

Nice to see you posting here. Nonsense makes the world go round!

[Ari]

Well, we could always go with the surrealist answers.
The two rectangles: (Banana,elf) and (mouse,blue,dharma)

[fragment]

I see all the fun's been had here already, too. :(
Still I had fun working it out myself, and reading everyone else's solutions (even the surrealist approach). Chatter, that was great stuff!

Chiron - what's the punchline?

[Chiron]

Man, you can't cross a scalar with a vector!

GROAN