Question

The parallelogram is bounded by the lines $y=a x+c ; y=a x+d ; y=b x+c$ and $y=b x+d$ and has the area equal to 18 . The parallelogram bounded by the lines $y=a x+c ; y=a x-d$; $y = bx + c$ 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

Area of the parallelogram $=\frac{\left| c _1- c _2\right|\left| d _1- d _2\right|}{\left| m _1- m _2\right|}$ (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?) $\Rightarrow| a - b |=2$

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