2007 AMC 12B Problem 20

Question

The parallelogram bounded by the lines

y=ax+c,

y=ax+d,

y=bx+c,

and

y=bx+d

has area

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, what is the smallest possible value of

a+b+c+d?

13

14

15

16

17

The first parallelogram has area

\dfrac{(c-d)^2}{|b-a|};

the second has area

\dfrac{(c+d)^2}{|b-a|}

Subtracting gives

4cd=54|b-a|;

then minimize with small positive integers

Two vertices of the first parallelogram lie at

(0,c)

and

(0,d),

and the other two have

x

-coordinates

\pm\dfrac{c-d}{b-a}.

Its area works out to

\dfrac{(c-d)^2}{|b-a|}=18.

The same computation for the second gives

\dfrac{(c+d)^2}{|b-a|}=72.

So

(c-d)^2=18|b-a|

and

(c+d)^2=72|b-a|.

Subtracting,

4cd=54|b-a|,

i.e.

2cd=27|b-a|.

Thus

|b-a|

is even, so

a+b

is smallest with

\{a,b\}=\{1,3\};

and

cd

is a multiple of

27,

so

c+d

is smallest with

\{c,d\}=\{3,9\}.

These satisfy all conditions, giving

a+b+c+d=1+3+3+9

=16.

Thus, the correct answer is D.

Problem 20 in other years