Tan's rectangles

Call the sides of the rectangle $a$ and $b$ . Then we have $2a + 2b = ab$ , which factors into $(a - 2)(b - 2) = 4$ . The only solutions are (3, 6) and (4, 4) since (6, 3) is not considered unique, so the answer is 2.

Let the dimensions of the rectangle be $a$ by $b$ .

Area must be equal to perimeter:

$a b=2 a+2 b\\ a=\frac{2 b}{b-2}=2+\frac{4}{b-2}$

$a$ and $b$ must be integers. Hence $\frac{4}{b-2}$ must be an integer. $b-2$ must be a factor of $4$ .

$b-2=\{1,2,4\}\\ b=\{3,4,6\}$

Back substituting:

$a=\{6,4,3\}$

Since $\{3,6\}$ and $\{6,3\}$ are considered the same rectangle, we are left with 2 solutions, $\{3,6\}$ and $\{4,4\}$ .

2a + 2b = ab. Factorized this is: (a-2)(b-2) = 4 Since 4 = 2x2 / 4x1

a = 6,4 b = 3,4

We shall let $x$ and $y$ be the sides of the rectangle. This allows us to form the following equation

$xy = 2x+2y$

$\Rightarrow y= \frac{2x}{x-2}$

$\Rightarrow y= 2+\frac{4}{x-2}$

Clearly, $y$ will only be an integer if $\frac{4}{x-2}$ is an integer.

So, in order for $\frac{4}{x-2}$ to be an integer, $x-2$ must be an integer and $x-2 \leq 4$

This will yield us the following 2 solutions of $(x,y)$ , $(4,4)$ and $(6,3)$

Hence, there will only be 2 distinct rectangles.

$xy=2x+2y\Rightarrow y=\frac{2x}{x-2}=2+\frac{4}{x-2}$

Using factors of $4$ , we get $(3,6),(4,4),(6,3)$ for a total of $\boxed{2}$ distinct rectangles.

Está bem claro no enunciado que sendo a e b, os lados de um retângulo, temos a seguinte a relação: 2a + 2b = a.b >> Isolando, 2(a + b) = a.b >> 2 = a.b/a + b >> Note que a.b é múltiplo de 2, logo a.b = 2n >> Substituindo: 2a + 2b = 2n >> a + b = n >> Chega-se no sistema: a + b = n e a.b = 2n >> Daí, a² - na + 2n = 0 >> n tem que ser um número natural, bem como o termo a, substituindo "a" por 3,4 e 6 o problema tem a solução desejada. Note que a=3 b=6 (é a mesma coisa que b=3 e a=6) e a=4 b=4 satisfazem a relação, logo temos 2 retângulos.

Sean: a y b: lados del rectangulo Lo partimos el problema en dos casos: -Cuando a=b 4a=a a, entonces a=4 -Cuando 'a' sea diferente a 'b' 2(a+b)=a b Notamos que ninguno de ellos puede tomar el valor de 'uno' puesto que : '2(a+1)' siempre sera diferente que 'a' Notamos que si alguno de ellos toma el valor de dos pasa esto: Para a=2 b=2 nos resulta que 8 es diferente a 4 Para a=2 b=3 nos resulta que 10 es diferente a 6 y asi... mientras en un lado aumento de dos en dos en dos (8,10,...) en otro tambien (4,6,....) por lo que nunca se llegara a una igualdad. Notamos que si alguno de ellos toma el valor '3' pasa esto: Para a=3 b=3 Nos resulta que 18 es igual a 9 Para a=3 b=4 Nos resulta que 14 es igual a 12 Para a=3 b=5 Nos resulta que 16 es igual a 15 Para a=3 b=6 Nos resulta que 18 es igual a 18 CUMPLE Para a=3 b=7 Nos resulta que 20 es igual a 21... y asi notamos que se forman dos sucesiones aritmeticas de razon 2 y 3 para este caso solo se cumple la igualdad en (a=3 y b=6) Notamos que si alguno toma el valor de 4 ocurrira esto: para a=4 b=4 Resultara que 16 igual que 16 (CUANDO SEA UN CUADRADO) para a=4 b=5 resultara que 18 igual que 20 para a=4 B=6 resulara que 20 es igual a 24... notamos que ahora que se formaron dos sucesiones aritmeticas de razon 2 y 4 y que solo se dara la igualdad para el caso (a=4 b=4) Y asi se formaran varias suciones aritmeticas pero nunca se llegara a la igualdad POR LO TANTO SOLO EXISTIRAN DOS CASOS CUANDO a=3 b=6 y cuando a=4 b=4

If a,b are the sides of the rectangle then the following must be true: ab = 2(a+b). Immediately we know that neither a nor b can take the value of 1 b/c if a = 1 then the area must be equal to b which is much less than the perimeter. Additionally we realize that neither a nor b can take on a value of 2 b/c then the area would be equal to 2b which is still much less than the perimeter.

Let us now consider the case where a = 3. This implies that 3b = 2(3+b) = 6+2b. Subtracting 2b from both sides results in b = 6. This is one instance that satisfies Tan's rectangles. This implies that for an integer n the following must be an integer 2n/(n-2) to satisfy Tan's rectangles (This was obtained solely from thinking about the algebra involved in solving for b). This statement is only true for one other number, 4 (4 could also have been obtained by solving the equation a^2 = 4a as all squares are rectangles). Therefore there are only two rectangles possible that satisfy Tan's demanding requests :)

Let the length of the rectangle be $l$ and breadth be $b$ .

Then, we need to find out the solutions of the equation $2(l+b)=l*b$ .

Taking l on the x axis and b on the y axis, we can plot a graph of the above equation..

We have only two points on the curve with integer coordinates i.e $(4,4)$ and $(6,3)$

So the required number of rectangles= $\boxed{2}$

Let a & b are the sides of a rectangle.

So,a*b=2(a+b) => a+b = ab/2;-------(1)

   a-b=(a+b)^2-2ab -------(2)

from eq(1),we find a quadratic equation in terms of a or b & for a & b should be zero,..Discriminant should be an integer. After solving this inequality,we find 2 &3 are satisfying it. So ,ans=2

Let l=length; w=width.

lw = 2l + 2w w=2l / (l - 2) = 2 + 4 / (l - 2)

For w to be an integer, 4 / (l - 2) has to be an integer. This gives 3 possible values of l: l = 3 w=6 l = 4 w=4 l = 6 w=3

There are 2 distinct combinations. Hence, the answer is 2.