Question

The parallelogram is bounded by the lines $y=ax+c;y=ax+d;y=bx+c$ and $y=bx+d$ and has the area equal to 18 . The parallelogram bounded by the lines $y=ax+c;y=ax-d$; $y=bx+c$ and $y=bx-d$ has area 72. Given that $a,b,c$ and $d$ are positive integers, find the smallest possible value of $(a+b+c+d)$.

Answer 1

Answer: 16

Area of the parallelogram $=\frac{|c_1-c_2||d_1-d_2|}{|m_1-m_2|}$ (for the figure as shown)

hence Area of the parallelogram formed by the $1^{\text{st }}$ set of 4 lines
$=\frac{|c-d{|}^2}{|a-b|}=18$ ....(1)
and Area of the parallelogram formed by the $2^{\text{nd }}$ set of 4 lines
$=\frac{|c+d{|}^2}{|a-b|}=72$ ....(2)
$\text{ from(1) }|c-d{|}^2=|a-b|\cdot 18$
from(1) $|c-d{|}^2=|a-b|\cdot 18$
and from (2) $|c+d{|}^2=|a-b|\cdot 72$. Since $a,b,c,d\in N$,
Hence RHS must be a perfect square for some least value of $|a-b|$ which obviously can be equal to 2 (think?) $\Rightarrow |a-b|=2$

$\therefore (a+b+c+d)=16$