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

- Find such a pair with the smallest area? And the next smallest?
- Can you find such a pair with a right triangle?

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)

2*p* + 2*q* = *
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* = 2*xy *(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, (2*a*)^2 + (2*b*)^2 = (2*c*)^2
where *a* = *
x*^2 - *y*^2, *b* = 2*xy*
and *c* = *x*^2 + *y*^2, then 2*a* = 2(*x* + *y*)(*x
*- *y)*, 2*b* = (*x* + *y*)^2 - (*x* - *y*)^2,
2*c* = (*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 - 4*xy*(*x* - *y*)(*x *
+ *y*) = *k*^2

*x*(*x* + *y*)(*x*^2 - 3*xy* + 4*y*^2) = *k*^2

*x*^4 - 2*x*^3*y* + *x*^2*y*^2 + 4*xy*^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 - 2*X*^3 + *X*^2 + 4*X* = *K*^2

Substituting *X *= 4/*u* and *K* = 4*v*/*u*^2, the
last equation transforms to

*u*^3 + *u*^2 - 8*u* + 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

Serbia

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