## sábado, 18 de junho de 2022

### Geometria - Áreas e perímetros

Existem apenas cinco triângulos com lados inteiros em que a área é numericamente igual ao perímetro: (5, 12, 13); (6, 8, 10); (6, 25, 29); (7, 15, 20); (9, 10, 17). Apenas os dois primeiros são triângulos retângulos.

Os únicos retângulos com lados inteiros em que a área é numericamente igual ao perímetro são o quadrado 4 × 4 e o retângulo 3 × 6.

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(There are only five integer-sided triangles in which the area numerically equals the perimeter: (5, 12, 13); (6, 8, 10); (6, 25, 29); (7, 15, 20); (9, 10, 17). Only the first two are right triangles.

The only rectangles with integer sides in which the area numerically equals the perimeter are the 4 × 4 square and the 3 × 6 rectangle.)

PCFilho

#### 4 comentários:

1. The rectangular case is easier to prove, so we’ll start with that.

Let x and y be the lengths of the base and height, respectively, of a rectangle. Thus we have the following formulae for the perimeter and area:
P = 2x + 2y, A = xy
Since both expressions are symmetric, we can assume that x < y without loss of generality.
Equating the right sides and solving for y, we get this:
xy = 2x + 2y
(x – 2)(y) = 2x
y = 2x/(x – 2) = 2 + 4/(x – 2)
From this equation, we see that y is an integer only when x – 2 divides 4; that is, when x is 3, 4, or 6. Plugging these values for x into the equation, we get the following results:
x = 3 → y = 2 + 4/1 = 6
x = 4 → y = 2 + 4/2 = 4
x = 6 → y = 2 + 4/3 = 3 (we dismiss this result because it contradicts x < y)
Therefore the only two rectangles with integer side lengths and whose perimeter and area are numerically equal are the 3 x 6 rectangle and the 4 x 4 rectangle.
QED

As for the triangular case, this proof is a little bit more difficult.
Let the sides of the triangle be a, b, and c. The perimeter and area formulae are:
P = a + b + c, A = (1/4)√[(a + b + c)(a + b – c)(a + c – b)(b + c – a)]
Again, both expressions are symmetric, so we can assume a < b < c without loss of generality.
Let’s simplify the expressions by defining a, b, and c in terms of some new variables thusly:
a = x + y, b = x + z, c = y + z
Thus:
a + b + c = 2(x + y + z)
a + b – c = 2x
a + c – b = 2y
b + c – a = 2z
Then the perimeter and area formulae become:
P = 2(x + y + z), A = √[(x + y + z)(xyz)]
Equating the right sides and solving for z:
2(x + y + z) = √[(x + y + z)(xyz)]
2√(x + y + z) = √(xyz)
4(x + y + z) = xyz
4(x + y) = (xy – 4)z
z = 4(x + y)/(xy – 4) = (x + y)[4/(xy – 4)]
Thus z will be guaranteed to be an integer if xy – 4 divides 4. So we’re looking for values of x and y such that their product is 5, 6, or 8. There are five such combinations of (x,y) such that x < y: (1,5); (1,6); (1,8); (2,3); and (2,4). This leads to the following results:
(x,y) = (1,5) → z = 6(4/1) = 24 → a = 6, b = 25, c = 29
(x,y) = (1,6) → z = 7(4/2) = 14 → a = 7, b = 15, c = 20
(x,y) = (1,8) → z = 9(4/4) = 9 → a = 9, b = 10, c = 17
(x,y) = (2,3) → z = 5(4/2) = 10 → a = 5, b = 12, c = 13
(x,y) = (2,4) → z = 6(4/4) = 6 → a = 6, b = 8, c = 10
Therefore, these five triangles are the only ones with integer side lengths and whose perimeter and area are numerically equal.
QED

2. Notice how both proofs involve the factors of 4 somehow.

1. Thank you, Jake. Amazing demonstration. :)

2. You're welcome, although I *did* make a little mistake: Where the first proof has "2 + 4/3", it should say "2 + 4/4".

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