Brain Teaser!?

Question

Here's the question, does anyone know the answer?

The parellelogram bounded by the lines y=ax+c , y=ax+d , y=bx+c , and y=bx+d has the area 18. The parellogram bounded by the lines y=ax+c , y=ax-d , y=bx+c , and y=bx-d has the area 72.

Given that a ,b ,c and d are positive integers, what is the smallest possible value of a+b+c+d?

If possible, could you explain how you got your answer so I could learn?
Thanks! :)

Answer 1

Begin by finding out where the lines intersect; label them L1 = ax+c, L2 = ax+d, L3 = bx+c, and L4 = bx+d.

L1 intersects L3 at (0, c) so label this point S; L2 intersects L4 at (0,d) so label this point U.

L1 intersects L4 when ax + c = bx + d, which happens when x = (d-c)/(a-b). The y-value at that point is a[(d-c)/(a-b)] + c = (ad - ac + ca - cb)/(a-b) = (ad - bc)/(a-b). Label the point ((d-c)/(a-b),(ad-bc)/(a-b)) as V.

L2 intersects L3 is found the same way; without running through all the manipulation the point of intersection is
((c-d)/(a-b), (ac-bd)/(a-b)) which we will call T.

The length of side SV = sqrt{[(d-c)/(a-b)]^2 + [(c(a-b)-ad+bc)/(a-b)^2} which reduces to (c-d)/(a-b)*sqrt(1+a^2).

The length of the other side of the parallelogram similarly reduces to (c-d)/(a-b)*sqrt(1+b^2).

Area of a parallelogram = length of a side * length of the altitude to that side...

...but finding the length of the altitude requires finding the line perpendicular to L1 that passes through T (which I have) and the length of that altitide (upon which I am still working).

*expletive*

Curse me for a fool--I chose the incorrect point from which to draw an altitude. Go from point S to UV. The perpendicular line will be y = -(1/b)x + c and will intersect L4 at [b(c-d)/(b^2+1),(b^2*c+d)/(b^2+1)].

The distance from S to this point will be (c-d)/sqrt(1+b^2). Thus we have length of altitude * length of side (found earlier) = ((c-d)*sqrt(1+b^2)/(a-b))*(c-d)/sqrt(1+b^2) = (c-d)^2/(a-b) = 18.

The other parallelogram has a side defined by y=bx-d; it will have length ((c+d)*sqrt(1+b^2))/(a-b). We can use the same perpendicular line from the other parallelogram and find the point of intersection; its coordinates will be [b(c+d)/(b^2+1), (b^2*c-d)/(b^2+1)].
The length of that altitude will be (c+d)/sqrt(1+b^2).

Plugging into the formula length of altitude * length of side will give (c+d)^2/(a-b) = 72.

(c-d)^2/(a-b) = 18 and (c+d)^2/(a-b) = 72 gives us 4(c-d)^2 = (c+d)^2 --> 3c^2 - 10cd + 3d^2 = 0. This factors to (3c - d)(c - 3d) = 0, giving either c = 3d or 3c = d. Plugging into the original equations, we get either c = 3 and d = 9 or c = 9 and d = 3.

In either case, c + d = 12 and a - b = 2. Since all are positive integers, then a must be 3 at a minimum. So, the minimum value for a + b + c + d = 16.

Answer 2

Good luck!