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January 30, 2015

January 30, 2015

Posted by **Mea C** on Wednesday, February 3, 2010 at 4:09am.

(please show how to do)

1. How many differently shaped rectangles, with positive integer dimensions, have a perimeter equal to their area?

2. Let x be any number less than one, and let y be any number greater than one.

S= x + y

P= xy

Prove that the difference between S and P must be greater than 1.

3. Prove that it is impossible to find four distinct numbers P,Q,R,S which satisfy the equation

pq+rs = ps+qr

- Maths -
**tchrwill**, Wednesday, February 3, 2010 at 1:21pmThere are only two rectangles whose sides are whole numbers and whose area and perimeter are the same. What are they?

The approach to the solution of a problem of this type, where integer answers are sought, is by treating it as a Linear Indeterminate Algebra problem or a Diophantine Problem as follows: Letting x and y be the integer sides of our rectangles, then the area = xy and the perimeter = 2(x+y). Given that xy = 2(x+y) we derive:

1--xy - 2x - 2y = 0 or

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

3--Clearly, both x and 2y/(y-2) must also be integers.

4--Set x = k from which y then becomes 2k/(k-2).

5--Assume values of k and compute x and y.

k...0...1...2...3...4...5....6....7....8....9....10.....50.....100......

x...-....1...2...3...4...5....6....7....8....9....10.....50....100......

y...-...inf..inf..6...4...3+..3....2+..2+..2+...2+...2+.....2+......

Thus 3 and 6, (or 6 and 3), and 4 and 4 are the only possible answers as y never reaches 2 as an integer.

This method is a very powerful tool for determining integer solutions to algebraic equations where there is one more unknown than there are equations. If you are at all interested in learning more about the method, let me know and I will be glad to provide you with some other examples and references

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