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:
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 is impossible as shown by Bojan Basic below.

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.

Update 29Aug2008.  Bojan Basic sent this proof of impossibility for part 2:

Let p and q be sides of the rectangle, and a, b and c of the right triangle. The following equations must hold:

a^2 + b^2 = c^2             (1)
2p + 2q = a + b + c         (2)
pq = ab/2                   (3)

We are now going to prove that we may assume that a, b, c have no common prime factor other than 2. Indeed, if t > 2 is prime, t|a, t|b and t|c, then from (3) one has, WLOG, t|p, and then from (2) also t|q. Therefore, the quintuple (a/t, b/t, c/t, p/t, q/t) is then also a solution.

Having this in mind, we now claim that there exist integers x and y such that
a = x^2 - y^2               (4)
b = 2xy                     (5)
c = x^2 + y^2               (6)

Indeed, if a, b and c are coprime, this is a well-known fact. If, on the other hand, (2a)^2 + (2b)^2 = (2c)^2 where
a = x^2 - y^2, b = 2xy and c = x^2 + y^2, then 2a = 2(x + y)(x - y), 2b = (x + y)^2 - (x - y)^2, 2c = (x + y)^2 + (x - y)^2, and the claim is proved.

Plugging (4), (5) and (6) into (2) and (3) gives:
p + q = x^2 + xy => q = x(x + y) - p
pq = xy(x^2 - y^2) => px(x + y) - p^2 = xy(x - y)(x + y) => p^2 - px(x + y) + xy(x - y)(x + y) = 0

Since p is a positive integer, discriminant of the last equation must be a perfect square:
x^2(x + y)^2 - 4xy(x - y)(x + y) = k^2
x(x + y)(x^2 - 3xy + 4y^2) = k^2
x^4 - 2x^3y + x^2y^2 + 4xy^3 = k^2

Dividing by y^4, we now look for rationals X and K (where X = x/y and K = k/y^2) such that:
X^4 - 2X^3 + X^2 + 4X = K^2

Substituting X = 4/u and K = 4v/u^2, the last equation transforms to
u^3 + u^2 - 8u + 16 = v^2

This is an elliptic curve in general Weierstrass model. Since its rank is 0, the only rational points on it are its torsion points. It has six torsion points: (4, 8), (4, -8), (0,4), (0, -4), (-4, 0) and the point at infinity, which correspond to the following values of (X, K): (1, 2), (1, -2), (-1, 0), (0, 0) and the point at infinity, what further means that either x = +- y or one of x, y equals to 0, all of which cases give only trivial solutions to the starting problem.

Bojan Basic
Novi Sad

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