Tadeo draws the rectangle with the largest perimeter that can be divided into $2015$ squares of sidelength $1$ $cm$ and the rectangle with the smallest perimeter that can be divided into $2015$ squares of sidelength $1$ $cm$. What is the difference between the perimeters of the rectangles Tadeo drew?
Problem
Source: Level 2 : 8th and 9th Grades
Tags: geometry
Not_a_Username
05.10.2015 02:34
If a rectangle has sidelengths $a$ and $b$, and $a>b$, we have that $\frac{1}{4}[(a+b)^2-(a-b)^2]=ab$, so it suffices to maximize $a-b$ for a larger perimeter and $a+b$ for a smaller perimeter. Because $ab=2015$, we find that $(a, b)=(2015, 1)$ gives maximum perimeter, and $(a, b)=(65, 31)$ as minimum perimeter. Therefore, the answer is $4032-192=\boxed{3840}$.
atmchallenge
05.10.2015 02:35
The problem is asking for the difference between the minimum and maximum perimeters of rectangles with integral side lengths and an area of $2015$. The maximum perimeter occurs when the difference between the side lengths is a maximum, which is clearly when the rectangle is $1\text{ by }2015$. Thus, the maximum perimeter is $2(1+2015)=4032$. The minimum perimeter occurs when the difference between the side lengths is a minimum. Now, checking all pairs of integers that multiply to $2015$, we find the minimum difference between the two integers is when the rectangle is $31 \text{ by }65$, so the minimum perimeter is $2(31+65)=192$. Thus, the answer is $4032-192=\boxed{3840}$.
jh235
05.10.2015 02:56
I like how the answer to this is the same as the answer to this.