Ken's Puzzle of the Week

Triangle and Rectangle Pairs

A triangle and rectangle have the same area and perimeter.  All sides are integers.

  1. Find such a pair with the smallest area?  And the next smallest?
  2. Can you find such a pair with a right triangle?
Source: Original.  For the second question, a good table of Pythagorean Triples is: http://grail.cba.csuohio.edu/~somos/rtritab.txt.
Solutions were received from Denis Borris, Philippe Fondanaiche, K Sengupta, Claudio Baiocchi, Jospeh DeVincentis, and Alan O'Donnell.  There are several answers for part 1.  In increasing area, the first few (that aren't multiples of smaller ones):

Triangle Sides     Rectangle Sides    Area
5,5,6                  6,2                       12
10,10,12            12,4                     48
13,20,21            21,6                     126
25,51,52            52,12                   624
53,53,56            60,21                   1260
37,130,157        150,12                 1800  (First which has all five lengths unique.)
41,104,105        105,20                 2100
 

Part 2 did not receive any solutions.  Perhaps it is impossible.  If a solution does exist, the side must satisfy:

2*(x+y) = a + b + c, and x*y = a*b/2
x and y are solutions of the quadratic equation X^2 - S*X + P = 0 in which S = (a+b+c)/2 and P = a*b/2

I used the table of Pythagorean Triples and was not able to find solutions for c < 10000.

Joseph DeVincentis sent this analysis for Part 1, leading to the table above:

By combining Hero's Formula for the area of a triangle with the other specifications for the problem, I get:
(d+e)(d+e-a)(d+e-b)(a+b-d-e) = d^2 e^2
where a,b,c are the sides of the triangle and d,e are the sides of the rectangle. c=2d+2e-a-b and is automatically an integer when the other sides are. Since the perimeter of the triangle is 2d+2e, the sides of the triangle must be less than d+e in order to allow those sides to form a triangle.
Then start with small dimensions and look for solutions. For a given e, we only need to consider d where d^2 e^2 is divisible by d+e. Then the other three terms in parentheses add to d+e, so we need three positive integers that add to d+e whose product is (d^2 e^2)/(d+e).
e=1 has no solution because d^2 is never divisible by d+1.
e=2 yields the solution a=b=5, c=6, d=6, e=2, area=12.
e=3 I didn't find any solutions.

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