Puzzle: Gone, but not forgotten
February 20, 2009 pm28 5:01 pm
Posted a rectangle puzzle a month ago. No real response. Vlorbik recently reminded me. Let’s try again.
Find all the rectangles for which the perimeter equals twice the area. (just the numbers. Ignore the fact that the units don’t match.)
JD2718 
So as not to spoil the fun …
A few are 2×2, 3×1.5, 2.5×1.667
Formula to follow.
Call the dimensions x and y, we have 2x + 2y = 2xy which simplifies to xy=x+y. Solutions are going to of course be positive but also both values have to be greater than 1, right? Cause if you multiply something by 1 you get itself but if you add 1 to it you have to get something bigger, these things will never be equal. Similar argument for values less than 1. Other than that, the relationship looks hyperbolic, and (2,2) obviously works…I’m going to go cheat and plug it into Grapher now.
Don’t cheat! Use algebra.
Sorry. Used the math teacher voice.
Well I could but that would spoil the fun. This would be nice to get precalc students to analyze because it makes a rational function – everything I said previously can be said in terms of domain, asymptotes, etc. (6, 1.2) (9, 1.125) (11, 1.1) :-)
I’m not sure I follow you. Four lines of algebra result in a nice, simple rational function. Why have students plug it into Grapher when the equation is so easily achieved by hand?
You definitely aren’t following me.
A graph helps me see a relationship. That’s what graphs are for. Why would I not use Grapher when I am sitting at a macbook? Also, me messing around is completely different from conducting a lesson. Posing this question to a class may or may not involve some graphing technology as they poke at the edges of the relationship and work to get a clear understanding of it. I don’t see why it shouldn’t.
So you get x+y=xy, and can rearrange that to get the equations hinted at above. Let’s, instead, divide everything by x, so
y=1+y/x
Then plug this formula in for y on the right side
y=1+(1+y/x)/x
and iterate. Then you get something vaguely continued-fraction-esque. I’m not sure what to do with that, but it’s fun to look at, if nothing else.
It comes out rather nicely if you take the side lengths to be (1+x) and (1+y).
Let the rectangle have a length x and a width y.
Then perimeter = 2x + 2y
And area = xy
So according to our restriction “the perimeter equals twice the area”
2x + 2y = 2xy
{divide by 2 on both sides}
x+ y = xy
{subtract y on both sides}
x = xy – y
{factor}
x = y(x – 1)
{divide by x-1 on both sides)
x/[x-1] = y
So we can use any two numbers that fit the restriction that one of them is equal to a fraction where the numerator is one more than the denominator.
However: x must be greater than 1. If x=1, y is undefined. If x1}
Examples: 2/1, 3/2, 4/3, 5/4, 6/5, 7/6, 4.5/3.5, 3.14/2.14, 1.003/.003 1000000/999999, etc.
Am I right?
Oy! The Internet ate my proof.
Here’s the rest of it:
However: x must be greater than 1. If x=1, y is undefined. If x1.
Examples are: (2, 2/1), (3, 3/2), (4, 4/3), (5, 5/4), (6, 6/5), (7, 7/6), (4.5, 4.5/3.5), (3.14, 3.14/2.14), (1.003, 1.003/.003) (1000000, 1000000/999999), etc.
Am I right?
~ Ms. J
sinesoflearning.blogspot.com
Oh I get it. WordPress is reading the “less than” sign as beginning some HTML code. Okay. Fine.
However: x must be greater than 1. If x=1, y is undefined. If x is less than 1, y is negative but a length cannot be negative by definition.
So the solution is: All points (x, x/[x-1]) where x is greater than 1.
Examples are: (2, 2/1), (3, 3/2), (4, 4/3), (5, 5/4), (6, 6/5), (7, 7/6), (4.5, 4.5/3.5), (3.14, 3.14/2.14), (1.003, 1.003/.003) (1000000, 1000000/999999), etc.
The internet better not eat this one – my prep is almost up!
~ Ms. J
sinesoflearning.blogspot.com
You have, collectively, hit every angle I had in mind, plus some.
Vlorbik, writing about Rational functions,
, reminded me that we hadn’t answered this one.
I am glad to see both solutions in integers and in rationals… And nice catch, Ms. J, on both sides needing to be more than 1.
I’m bumbed my solution got deleted or messed up somehow. I thought I had it correctly posted. Anyway. I second everyone else’s input. I did the algebra simplifying thing and had
But I also plugged it into the following python code:
max = input(‘Maximum length? ‘)
counter = 2
f = open(‘rectangles.txt’,’w’)
while counter >f, ‘length = ‘, length, ‘width = ‘, width
counter = counter + 1
Which spits out as many rectangles as you’d like up to the ‘Maximum Length’.
length = 2 width = 2.0
length = 3 width = 1.5
length = 4 width = 1.33333333333
length = 5 width = 1.25
length = 6 width = 1.2
length = 7 width = 1.16666666667
length = 8 width = 1.14285714286
length = 9 width = 1.125
length = 10 width = 1.11111111111
max = input(‘Maximum length? ‘)
counter = 2
f = open(‘rectangles.txt’,’w’)
while counter <= max:
length = counter
width = (2*length)/(2*length – 2.0)
print >>f, ‘length = ‘, length, ‘width = ‘, width
counter = counter + 1
is what should have been posted
The answer:
For C greater than or equal to 4…
x = 0.5C + 0.5(C^2 – 4C)^0.5
y = 0.5C – 0.5(C^2 – 4C)^0.5
How I got there:
My first instinct was to think of the graph… A hyperbola xy=C, and a line x+y = C, but in the end I went with algebra.
x+y = xy = C
x(C-x) = C
xC – x^2 = C
x^2 – xC + C = 0
x = 0.5[ C +/- (C^2 – 4C)^0.5]
Check for C=4
x = 0.5[4 +/- (16-16)^0.5] = 2 OK
For x to be real C must be greater than or equal to 4.
Checking with other people’s answers, e.g. C=4.5
x = 0.5[4.5 +/- (4.5^2 – 18)^0.5] = 3 or 1.5 OK
If x = 0.5 C + 0.5(C^2 – 4C)^0.5 the y = C-(0.5C + 0.5(C^2 – 4C)^0.5) = 0.5C – 0.5(C^2 – 4C)^0.5 which is the other root.
Obviously x+y = C; xy = 0.25C^2 – 0.25(C^2 – 4C) = C.
So everything checks
I still like the simple statement: let the sides be (x+1) and (y+1), then the conditions are met if and only if xy=1 (that is, they are reciprocals). Proof is trivial once you have the statement.
Yes, Dr Rick’s is much more elegant.