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.
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 |
Here it is
