problem of dissecting a 1 x k rectangle into similar but incongruent polygons

Question

I have this problem from USAMO 2004:

"For what values of $k > 0$ is it possible to dissect a $1 \times k$ rectangle into two similar, but incongruent, polygons?"

I feel that all values of k will satisfy, since for any rectangle with a given k, I draw a line which intersects the opposite sides of the rectangle at unequal distances from the opposite vertices and I get 2 polygons which are similar but incongruent. However, I have a few doubts, this is USAMO so I personally don't think such a solution would be deemed correct, moreover the answer says that k should not be equal to one

Why is that so? And any mistakes with my solution?

Answer 1

I think there might be a fractal solution, with two shapes bounded by this sequence of horizontal and vertical lines:

$$(0,0),(1,0),(1,x),(1+x^2,x),(1+x^2,x+x^3),(1+x^2+x^4,x+x^3),...$$ end at the far corner

$$\left(\frac1{1-x^2},\frac x{1-x^2}\right)$$

Answer 2

Other comments have answered the question you asked (namely, is your solution correct?). I'm going to push on and help you see where the really interesting part of this problem lies (a part that I will not solve because I don't know how to do it!).

Despite your solution being not quite right, it's almost right.

It's still possible to split the strip into exactly two rectangles (except for particular values of $k$): you make one of them almost $1 \times k$, and the other be $1$ unit tall and about $1/k$ wide. More precisely, you want to get this picture:

|----------------------|-|
|                      | |
|                      | |
|----------------------|-|

where the first rectangle is $1 \times a$, and the other is $1 \times b$, and

1. $a + b = k$
2. $a/1$ is the same as $1/b$, so that they are similar.

Solving gives $ab = 1$, so $a + \frac{1}{a} = k$, hence $$ a^2 -ka + 1 = 0 $$ and $$ a = \frac{k \pm \sqrt{k^2 - 4}}{2} $$ Because only values of $a$ less than $k$ make sense, only one of these two roots is a good choice.

There are two clear problems remaining:

1. If $k = 2$, the two rectangles are both $1 \times 2$, and hence are similar and congruent, adn you need a different solution (if there is one).

2. For $k < 2$, this approach, as written, doesn't get you an answer. But if you rotate the diagram above by 90 degrees, you can still get something by essentially the same method. The $1 \times k$ rectangle becomes $k \times 1$, which you can scale down to $1 \times 1/k$, for which the previous method works when $1/k > 2$, i.e., when $k < \frac{1}{2}$. (Thanks to @JaapScherphuis for pointing this out).

So the whole thing boils down to the $1/2 \le k \le 2$ question (for which I don't see an obvious answer in either direction, personally). As Jaap observes, rotation lets us assume $k > 1$, so really the unanswered range is $1 \le k \le 2$.