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# There AREN'T any more!

Proof of Alison Blank's question about rectangles.

by

Tweet## Joe Knapka

on 2 October 2010#### Transcript of There AREN'T any more!

In her INSANELY COOL Prezi, "Math is not linear" (http://prezi.com/aww2hjfyil0u/math-is-not-linear/), Alison Blank poses the following:

"Here's a cool question:

Find a rectangle (with integer sides) whose area and

perimeter are the same.

Now find another one.

Now find another one.

What, you're stumped? Well, maybe there AREN'T any

more. How could you prove that?" I am a guy with a math degree who isn't very good at math (any longer) So.... I decided to try to prove it. (but would like to get better) Here goes. We want to find a rectangle whose area

(xy) is equal to its perimeter (2x+2y). So: 2x+2y=xy Seems easy enough... We can plug in any value we want for x, and then solve for y. So we try some random value like, oh I don't know, 4. (Seriously, I randomly picked 4 as the very first thing to try.) (Letting x=width and y=height. Or vice-versa,

if you want.) 2(4)+2y=4y, so 8+2y=4y,

therefore 8=2y and y=4 --

both x and y are 4.

That works: perimeter=16 and area=16! Champagne, please! If we plug in x=5, though, we get y=3.333...

That gives us a rectangle with area=perimeter=16.6666...,

but otherwise, not so good: we need INTEGERS. How about 3? Yep, it works:

6+2y=3y, so (subtracting away 2y) y=6, and both area and perimeter are 18. Obviously, x cannot be 2, because then 4+2y=2y, or (subtracting away 2y again) 4=0. Hmmm.... something fishy there. What if we try a REALLY LARGE NUMBER like say 1000? (Why do I think that's "really large"? Why not choose a googolplex? No particular reason, it's just a lot larger than the numbers I've tried so far.) 2000+2y=1000y, so 2000=998y. y= 2000/998 (=1000/499) is certainly NOT an integer!

Hmm, it seems larger numbers are not going to work. (Positive ones: 1,2,3,4,...,99836499957,...) Let's just solve the silly equation for y and see if that

blinds us with any light:

2x+2y=xy

2x=xy-2y

2x=(x-2)y

y=2x/(x-2) So x-2 must divide 2x an integral number of times. Of course, x-2 will always go into 2x at least twice, but then there will be something left over... and whatever that something is, it has to be just big enough to fit an integral number of

(x-2)'s. At this point I'm a little stuck. (Did I mention I suck at algebra?) [Failed it one semester. 9th grade I think.

Really, it's amazing I managed to graduate

with a BS in math.] Think think think think think. How many x-2's fit? Suggestive, but how to PROVE it? I can see a picture in my head

(of why it doesn't work as x gets

bigger) Maybe I should use regular expressions? Mmmmm, chocolate! OK, what if we write 2x in terms of (x-2)? 2x=(x-2)+(x-2)+4, right? Or 2(x-2)+4.

Substituting that into the last formula: (2(x-2)+4)/(x-2)=y Cancelling the (x-2)'s:

2+4/(x-2)=y I'm feeling good about this for some reason! HANG ON! 4/(x-2)=y-2

If y is an integer, then y-2 is surely an integer also. So it seems safe to just ... ignore that 2 on the right there. That means 4/(x-2)=I for some INTEGER I... so ... And what the hell, subtracting 2

from both sides

'cause that (x-2) in the denominator

looks like it could use a little time

alone:

4/(x-2)=y-2 (Yeah, that move doesn't seem totally algebraically PC, but the logic, it is unassailable.) And that explains the weirdness when x=2: we're kinda dividing by 0. x-2 must go evenly into 4!

So only numbers will work for x. Numbers like 3 and 4. In fact, it's clear that 3 and 4 are the ONLY two numbers that will work! little If x=5 then x-2 is 3, which clearly won't divide 4. x=6 works, since then x-2=4; but then y=3 and we end up with the same rectangle we got when x=3, just reflected. Mathematically, it's the same rectangle. When x>=7, x-2>=5, and nothing >5 is going to divide 4 evenly. If x is 2, we get TOTAL INSANITY via

division by 0, and if x <= 1 then we end up with rectangles with negative area:

2-2y=y, so 2=-y or y=xy=(-2). Oops. Q E D In Latin, these letters mean...

something. (Seriously, this is the first number I tried. Totally at random.) Well, the quantity (x-2) seems pretty important... So we set out to show that (x,y)=(4,4) and (x,y)=(3,6) or (6,3)

are the ONLY two rectangles with integer sides and area equal

to their perimeter. AND WE DID IT.

Full transcript"Here's a cool question:

Find a rectangle (with integer sides) whose area and

perimeter are the same.

Now find another one.

Now find another one.

What, you're stumped? Well, maybe there AREN'T any

more. How could you prove that?" I am a guy with a math degree who isn't very good at math (any longer) So.... I decided to try to prove it. (but would like to get better) Here goes. We want to find a rectangle whose area

(xy) is equal to its perimeter (2x+2y). So: 2x+2y=xy Seems easy enough... We can plug in any value we want for x, and then solve for y. So we try some random value like, oh I don't know, 4. (Seriously, I randomly picked 4 as the very first thing to try.) (Letting x=width and y=height. Or vice-versa,

if you want.) 2(4)+2y=4y, so 8+2y=4y,

therefore 8=2y and y=4 --

both x and y are 4.

That works: perimeter=16 and area=16! Champagne, please! If we plug in x=5, though, we get y=3.333...

That gives us a rectangle with area=perimeter=16.6666...,

but otherwise, not so good: we need INTEGERS. How about 3? Yep, it works:

6+2y=3y, so (subtracting away 2y) y=6, and both area and perimeter are 18. Obviously, x cannot be 2, because then 4+2y=2y, or (subtracting away 2y again) 4=0. Hmmm.... something fishy there. What if we try a REALLY LARGE NUMBER like say 1000? (Why do I think that's "really large"? Why not choose a googolplex? No particular reason, it's just a lot larger than the numbers I've tried so far.) 2000+2y=1000y, so 2000=998y. y= 2000/998 (=1000/499) is certainly NOT an integer!

Hmm, it seems larger numbers are not going to work. (Positive ones: 1,2,3,4,...,99836499957,...) Let's just solve the silly equation for y and see if that

blinds us with any light:

2x+2y=xy

2x=xy-2y

2x=(x-2)y

y=2x/(x-2) So x-2 must divide 2x an integral number of times. Of course, x-2 will always go into 2x at least twice, but then there will be something left over... and whatever that something is, it has to be just big enough to fit an integral number of

(x-2)'s. At this point I'm a little stuck. (Did I mention I suck at algebra?) [Failed it one semester. 9th grade I think.

Really, it's amazing I managed to graduate

with a BS in math.] Think think think think think. How many x-2's fit? Suggestive, but how to PROVE it? I can see a picture in my head

(of why it doesn't work as x gets

bigger) Maybe I should use regular expressions? Mmmmm, chocolate! OK, what if we write 2x in terms of (x-2)? 2x=(x-2)+(x-2)+4, right? Or 2(x-2)+4.

Substituting that into the last formula: (2(x-2)+4)/(x-2)=y Cancelling the (x-2)'s:

2+4/(x-2)=y I'm feeling good about this for some reason! HANG ON! 4/(x-2)=y-2

If y is an integer, then y-2 is surely an integer also. So it seems safe to just ... ignore that 2 on the right there. That means 4/(x-2)=I for some INTEGER I... so ... And what the hell, subtracting 2

from both sides

'cause that (x-2) in the denominator

looks like it could use a little time

alone:

4/(x-2)=y-2 (Yeah, that move doesn't seem totally algebraically PC, but the logic, it is unassailable.) And that explains the weirdness when x=2: we're kinda dividing by 0. x-2 must go evenly into 4!

So only numbers will work for x. Numbers like 3 and 4. In fact, it's clear that 3 and 4 are the ONLY two numbers that will work! little If x=5 then x-2 is 3, which clearly won't divide 4. x=6 works, since then x-2=4; but then y=3 and we end up with the same rectangle we got when x=3, just reflected. Mathematically, it's the same rectangle. When x>=7, x-2>=5, and nothing >5 is going to divide 4 evenly. If x is 2, we get TOTAL INSANITY via

division by 0, and if x <= 1 then we end up with rectangles with negative area:

2-2y=y, so 2=-y or y=xy=(-2). Oops. Q E D In Latin, these letters mean...

something. (Seriously, this is the first number I tried. Totally at random.) Well, the quantity (x-2) seems pretty important... So we set out to show that (x,y)=(4,4) and (x,y)=(3,6) or (6,3)

are the ONLY two rectangles with integer sides and area equal

to their perimeter. AND WE DID IT.