# The perimeter is equal to the area, i.e. $2a+2b=ab$.

The measurements on the sides of a rectangle are distinct integers. The perimeter and area of ​​the rectangle are expressed by the same number. Determine this number. Answer: 18

It could be $$4*4$$ = $$4+4+4+4$$ but the answer is 18.

Wait... Now that I noticed, the sides are different numbers. But I can't find a way to solve.

• Well, if $x$ and $y$ are the sides of the rectangle, what is the equation that $x$ and $y$ have to satisfy? Jun 23, 2017 at 23:16
• @DanielSchepler $xy=2x+2y$ but what now? Jun 23, 2017 at 23:17
• Now here comes the somewhat clever part: that is equivalent to $(x-2)(y-2) = 4$. Jun 23, 2017 at 23:18
• To expand on that a little bit: what you've found corresponds to $x=y=4$, or in other words the equation $2\times 2=4$. How else can you find two numbers that multiply to $4$? Jun 23, 2017 at 23:21

I don't have the reputation to add a comment to Toby Mak's (original) answer. This should go there. I can get to the reason why $a=2b$. We have 3 options: either both $a$ and $b$ are odd, even or a mix of them.

$1)$ If $a$ and $b$ are both odd, then $ab$ is also odd, which it cannot be since in the original expression $2 = \frac{ab}{a+b}$, and from the numerator $2*(a+b)$ the expression is even.

$2)$ If $a$ and $b$ are both even, then let $a=2m$, $b=2n$. So $2=\frac{ab}{a+b}$ implies $1=\frac{mn}{m+n}$ implying that $$n=\frac{m}{m-1},$$ and excluding the case $m - 1 = 1$, or $m=2$, odd and even numbers are not divisible by each other. In this case where $m=2$, $\frac{2}{2-1} = 2$, which makes $a=b$. This is not a distinct answer.

Therefore, for $3)$ $a$ (or $b$) is even, and vice versa, Therefore $a=2m$ and $b=2n+1$ for some $m$ and $n$. And $2=ab/(a+b)$ implies $$1 = \frac{m(2n+1)}{2m+2n+1},$$

implying $2m+2n+1 = 2mn+m$ or $n=(m+1)/2(m-1)$ and this can only have integer values when $m=3, n=1$. Thus $a=6$ and $b=3$.

• Thanks @Andy Spencer, but I found a new solution using casework. Jun 24, 2017 at 0:07

Let $x$ and $y$ be the sides of the rectangle. Then the given condition implies that $xy = 2x + 2y$. Solving for $y$ in terms of $x$, we get: $$y = \frac{2x}{x-2}.$$ It's not immediately obvious from this equation for what integer values of $x$ this gives an integer value of $y$. However, by doing a polynomial division, we can rewrite this as: $$y = 2 + \frac{4}{x-2}.$$ Now, it is clear that $y$ is an integer if and only if $x-2$ is a factor of 4. The integers factors of 4 are $\pm 4, \pm 2, \pm 1$; however, since $x$ must be positive, we can eliminate the cases $x-2 = -4$ and $x-2 = -2$. For the other factors, plugging into the equation we get solutions $(x,y) = (1,-2), (3, 6), (4, 4), \mathrm{or}~(6,3)$. However, $y$ must also be positive, eliminating the $(1,-2)$ solution; and we are given $x \ne y$, eliminating the $(4, 4)$ solution. Therefore, $(x,y) = (3,6)~\mathrm{or}~(6,3)$, and in both cases, $xy = 2x + 2y = 18$.

• I don't know how to make a polynomial division yet, but I got what you did. Jun 24, 2017 at 0:52

Let $a$ and $b$ be the sides of the rectangle. We know the perimeter of a rectangle is $2(a+b)$, and the area is$(ab)$.

Since these two expressions have to be equal, let us solve the equation: $$2(a+b) = ab$$ $$2a = ab - 2b$$ $$2a = b(a-2)$$ $$b = \frac{2a}{a-2}$$

Since $b$ is an integer, $\frac{2a}{a-2}$ must be one as well. This means that $a$ is even, because otherwise when $a$ is odd, $a$ is not divisible by $a-2$, excluding the case $a=3$. This means that $b = \frac{2*3}{3-2} = 6$, which is a valid solution.

We can approach this by doing casework: checking all even numbers from $a=4$ (when $a=2$, $a-2 = 0$).

When $a=4$, $b=\frac{2*4}{4-2}=4$, but this solution does not have distinct integers.

When $a=6$, $b=\frac{2*6}{6-2}=3$. Therefore, this is the same solution mentioned earlier.

Therefore, one solution to the problem is when $a=6,b=3$. Now check if the values of $a$ and $b$ satisfy the original expression.

• But why it is valid to assume that a=2b? Jun 23, 2017 at 23:32
• I'm finding a reason for it - wait a minute Jun 23, 2017 at 23:37
• Edited my post to show an entirely replaced solution. Jun 24, 2017 at 0:06
• Fixed the question. Jun 24, 2017 at 0:29
• @dxiv If you're so interested in making edits to my question, why don't you do it yourself? Jun 24, 2017 at 0:31

Let's see, let $x$ be the length of the one side and $y$ be the length of one of the adjacent sides. Then the perimeter of the rectangle is $P(x,y)=2x+2y$ and the area is $A(x,y)=xy$. So we need $P=A$ i.e. $2x+2y=xy$. Thta is, $$2x+2y=xy \Leftrightarrow 2x-xy+2y+4=4 \Leftrightarrow (x-2)(y-2)=4.$$ Note that $4=2 \cdot 2$ or $4=4 \cdot 1=1 \cdot 4$, the case $x-2=y-2=2$ leads to $x=y=4$, an that contradicts the hipótesis. The case $x-2=4$ and $y-2=1$ leads to $x=6$ and $y=3$ (the other case, $x-2=1$ and $y-2=4$ leads to a similar case). Finally, since $(-2)(-2)=(-4)(-1)=(-1)(-4)=4$, the cases $4=(-1)(-4)=(-4)(-1)$ leads to $x=-2$ or $y=-2$, respectively. Finally, the case $(-2)(-2)=4$ gives us $x=y=0$, another contradiction.

So the answer is $(x,y)=(6,3)$ or $(x,y)=(3,6)$, and indeed $A=6\cdot 3=18=2 \cdot 6 + 2\cdot 3=P$.

• The problem says the sides are distinct. The other solution is $x-2=4$, $y-2=1$, or vice versa. (For completeness, you also have to consider why you can't get positive solutions from e.g. $x-2=-4$, $y-2=-1$.) Jun 23, 2017 at 23:35
• Can you correct your post to show the solution where $x$ and $y$ are distinct integers? Jun 24, 2017 at 0:11
• Done, thank you both! Jun 24, 2017 at 0:26