 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A rectangle Around A rhombus (Posted on 2008-04-14) A rectangle ABCD is circumscribed around a rhombus AECF. The long sides of the rectangle coincide with two sides of the rhombus. Also, the rhombus and the rectangle share a common diagonal AC.
```B       E               A
+-------+---------------+
|      /               /|
|     /               / |
|    /               /  |
|   /               /   |
|  /               /    |
| /               /     |
|/               /      |
+---------------+-------+
C               F       D
```
What are the smallest dimensions when all the lengths AB, BC, AE, AC and EF are integers?

Find a parameterization of all such integral rectangle/rhombus pairs.

 No Solution Yet Submitted by Brian Smith Rating: 3.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Here it is | Comment 2 of 13 | This problem amounts to finding integers which occurs twice as a side in a Pythagorean Triple.

Take  BE=m, AE=n, AD=k, AC=h, EF=j. We want integers m,n,l,h,j satisfying

(m+n)^2 + k^2 = h^2

(m-n)^2 + k^2 = j^2

________________

Pythagorean Triples are of the form (a^2-b^2, 2ab, a^2+b^2). There are two cases to consider:

Case 1: k is even

The primitive solutions for this case occur when k is twice the product of distinct primes.

k=2p1p2=2xy=2uv so we take x=p1p2, y=1, u=p1, v=p2, p1>p2 and x^2+y^2 > u^2+v^2 so

m+n=x^2-y^2=(p1p2)^2 - 1

m-n=u^2-v^2=p1^2-p2^2

We get:

m = ( p1^2 + [p1 p2]^2 - p2^2 - 1)/2

n = (-p1^2 + [p1 p2]^2 + p2^2 - 1)/2

k = 2 p1 p2

h = p1^2 p2^2 + 1

j = p1^2 + p2^2

Case 2: k is odd

This time the primitive cases occur when k is the product of two distinct, odd primes:

k=a^2-b^2=c^2-d^2=p1p2. Takeing ab>cd, a>b, c>d, p1>p2.

m+n=2ab

m-n=2cd

m=ab+cd

n=ab-cd

h=a^2+b^2

j=c^2+d^2

Take a+b=p1p2, a-b=1, c+d=p1, c-d=p2 gives

a=(p1p2 + 1)/2

b=(p1p2 - 1)/2

c=(p1 + p2)/2

d=(p1 - p2)/2.  So then

m = ([p1 p2]^2 + p1^2 - p2^2 - 1)/4

n = ([p1 p2]^2 - p1^2 + p2^2 - 1)/4

k = p1 p2

h = ([p1 p2]^2 + 1)/2

j = (p1^2 + p2^2)/2

Some examples:

We work out the smallest solution for each case. These are

k = 6, m = 20, n = 15, h = 37, j = 13  for case 1

k = 15, m = 60, n = 52, h = 113, j = 17  for case 2.

 Posted by FrankM on 2008-04-14 13:40:47 Please log in:
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