I have a hunch that it can be shown that th 2 polygons must be rectangles, and we get an equation $\frac1x=k-x$, so the equaton $x^2-kx+1=0$ must have real solutions different from 1, so we get k>2. I don't know if these are all of them, but I think they are. What do you think? Is it true that the 2 polygons must be rectangles?

Can somebody please clarify for me what dissection means? During the contest, I interpreted it to mean cutting up the rectangle and rearranging it. But beyond 1 cut, that starts to get very complicated. It's very easy to prove that if you have a single straight line cut that it must be perpendicular to one of the rectangles edges. As Grobber pointed out, this gives k>2 as a solution. But there are more k possible. But making the cut in the other direction (perpendicular to the edges of length 1), the same argument that gets k>2 gets 0<k<1/2 as well. I believe these are the only values.

indeed Michael is right. we (dzeta, blondu and me) have managed to produce a reasoning, but it's not quite rigurous (yet). the main idea floats around the fact that the figures must cut a certain side of the rectangle, and then, a polygonal line will start from that point, and move on to another side. then we consider the minimal and maximal rectangle that are contained within each of the figures. it's easy to check that they each one is the other figures' other rectangle (I will later add a drawing). computation now shows that the two rectangles must concide, therefore the only covering is given for $ k \notin \left( \displaystyle \frac 12, \ 2\right)$. however, this is just a sketch. more "evidence" needs to be brougth

I'm getting a rumour saying that the answer to the problem is $k \ne 1$.

Here's a counter-example Note that there could be more than one "stair" thingies. So if we increase the number of "stairs", we might be able to reach anything arbitrarily close to 1. I think that might be the idea.

I think your example can be extended to every k>1, the case k<1 being analogous. You don't need to increase the number of stairs. Note that the example doesn't match the case k=1. The key idea is to consider the sum of angles at every vertex of polygons. If the polygon has n vertices, the sum of angles at them is 360n-l*90( the number of corners among them)-x*180(number of vertices on sides). Since the sum of angles of each polygon is 180(n-2), we get an even number of corners in each polygon. Since a polygon can't contain 0 or 4 corners, it must contain 2 corners. We easily show this corners are on a side. Now a polygon can't have more than 1 vertex on a side, also it can't have vertices on sides it contains. So we reach the only example in the figure( One polygon contains point A,B; the other C,D; and they contain one point on AB and BC) . Now let L be the longest side of polygons. Since a similarity maps 1 to a segment, we get L>1, so L lies in the interior. But now then this side must be common to both polygons, so the second polygon contains also L. the similitude maps L to a longer segmenet, contradiction. The proof is finished. P.S. Bill, do you know the results of the contest?

iura wrote: I think your example can be extended to every k>1, the case k<1 being analogous. You don't need to increase the number of stairs. Actually, I believe you *do* need to increase the number of stairs. With no stairs (a single cut), the limit on the aspect ratio is >2:1. With one stair (3 cuts), the limit on the aspect ration is >3:2. With two stairs (5 cuts), the limit is >4:3. See my posting on AoPS. What do you think about dissecting a square?

Just solve these equations for f and c, where k>=1.5 is the aspect ratio: You have a system of 3 high-degree equations, how do you show it has a solution? I think they don't! ( This is the reply to your post on artofproblemsolving.com)

iura wrote: You have a system of 3 high-degree equations, how do you show it has a solution? I think they don't! ( This is the reply to your post on artofproblemsolving.com) For the 3-line dissection... I'll choose 2 of the 3 equations: \displaystyle cf+f=k \displaystyle cf^2+cf^4=1 Solve the 2nd for c, then substitute c into the first. \displaystyle c=\frac{1}{f^2+f^4} {\frac{1}{f^2+f^4}} \times f + f = k {\frac{f}{f^2+f^4} } + f = k \displaystyle \frac{1}{f+f^3} + f = k f was defined to be the scale factor of one polygon versus the other. Now, we let f be the independent variable, and vary it to find what values of k are possible. By inspection, as long as f is finite and greater than zero, then k is finite and greater than zero. As f approaches 0+, k asymptotically approaches 1/f (+infinity). As f approaches +infinity, k asymptotically approaches f (+infinity). At f=1, k=\frac{1}{1+1}+1=\frac{3}{2} I can show that k(f) = k(1/f) by substituting g=1/f: \displaystyle k = \frac{1}{f+f^3} + f \displaystyle k = \frac{1}{f+f^3} + \frac{f(f+f^3)}{f+f^3} \displaystyle k = \frac{f^0}{f+f^3} + \frac{f^2+f^4}{f+f^3} \displaystyle k = \frac{f^4+f^2+f^0}{f^1+f^3} \displaystyle k = \frac{(1/g)^4+(1/g)^2+(1/g)^0}{(1/g)+(1/g)^3} \displaystyle k = \frac{g^4}{g^4}\times\frac{g^{-4}+g^{-2}+g^0}{g^{-1}+g^{-3}} \displaystyle k = \frac{g^0+g^2+g^4}{g^3+g^1} Since the function is logarithmically symmetric about f=g=1, this strongly suggests we will have a minimum at f=g=1. [See attached graph of k vs f.] To prove this, I resort to calculus. I find the maxima/minima of k by taking the derivative of k with respect to f, then setting the derivative equal to 0: \displaystyle k = \frac{1}{f+f^3} + f \displaystyle \frac{d}{df} ( \frac{1}{f+f^3} + f)=0 \displaystyle -\frac{1+3f^2}{(f+f^3)^2}+1=0 \displaystyle -(1+3f^2)+(f+f^3)^2=0 \displaystyle (f+f^3)^2=1+3f^2 \displaystyle f^2\times(1+f^2)^2=1+3f^2 \displaystyle f^2\times(1+2f^2+f^4)=1+3f^2 \displaystyle f^2+2f^4+f^6=1+3f^2 \displaystyle f^6+2f^4-2f^2-1=0 \displaystyle (f^2-1)(f^4+3f^2+1)=0 From the first term, we get two real roots: \displaystyle f^2-1=0 \displaystyle (f-1)(f+1)=0 \displaystyle f=+1 \displaystyle f=-1 From the second term and the quadratic formula, we get two pairs of complex conjugate roots: \displaystyle f^4+3f^2+1=0 f^2=\frac{-3\pm \sqrt{3^2-4\times 1\times 1}}{2} \displaystyle f^2=-\frac{3+\sqrt{5}}{2} \displaystyle f^2=-\frac{3-\sqrt{5}}{2} So we have proven that for positive real values of f, the one and only minimum value of k occurs at f=1. k=3/2. For f>0, the function is continuous and differentiable. So all rectangles of the shape 1xk, where k>=3/2 are dissectible with three lines as shown in the attached illustration. I leave it to you to prove or disprove my equivalent claims with respect to 5-line dissections and above.

Ok, I understood myself what's the point. I just made such stupid mistake!

The answer is all $k \neq 1$

The answer is all $k \neq 1$

Official solution: http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04USAMO_solutions.pdf

The answer is all $k \neq 1$. Here a "dissection" does not need to be a single cut. The construction for $k > 1$ is not so hard to find (my friends and I got it over dinner at Legal Seafoods yesterday, and billzhao basically got it in #6). Pick $2n$ and $r \in (1,\infty)$ arbitrary. Consider $n$ rectangles of dimensions $1 \times r$, $r \times r^2$, $(1+r^2) \times r$, $(r+r^3) \times r^4$, and so on, gluing them together in this order in the obvious fashion. For example, the case of $n=3$ is shown below. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); real r = 1.6; pair A = (0,r^3+r); pair B = (r^4,r^3+r); pair C = (r^4+r^2,r^3+r); pair D = (r^4+r^2+1,r^3+r); pair E = (r^4,r^3); pair F = (r^4+r^2, r^3); pair G = (r^4+r^2+1, r^3); pair O = (0,0); pair H = (r^4,0); pair I = (r^4+r^2+1,0); pair W = (-r^6, A.y); pair X = (W.x,-r^5); pair Y = (0,-r^5); pair Z = (I.x,-r^5); pen p = black+1.3; pen q = red; fill(O--A--B--H--cycle, palecyan); fill(B--C--F--E--cycle, palecyan); fill(W--A--Y--X--cycle, palecyan); fill(C--D--G--F--cycle, palegreen); fill(E--G--I--H--cycle, palegreen); fill(O--Y--Z--I--cycle, palegreen); draw(O--A, q); draw(B--E, q); draw(F--G, q); draw(H--I, q); draw(W--D--Z--X--cycle, p); draw(C--F--E--H--O--Y, p); label("$1$", (C+D)/2, dir(90)); label("$r^2$", (B+C)/2, dir(90)); label("$r^4$", (A+B)/2, dir(90)); label("$r^6$", (W+A)/2, dir(90)); label("$r^1$", (D+G)/2, dir(0)); label("$r^3$", (G+I)/2, dir(0)); label("$r^5$", (I+Z)/2, dir(0)); [/asy][/asy] Now, the aspect ratio of the completed rectangle is \[ \frac{1+r^2+\dots+r^{2n}}{r+r^3+\dots+r^{2n-1}} \]which for a fixed $n$ can goes from $(1+\tfrac1n, \infty)$ as $r > 1$. Thus by taking $n$ large, we can get any $k > 1$. Of course by symmetry we can also get any $k < 1$. Now to show that $k = 1$ is impossible (the tricky part, IMO), consider a dissection into two similar polygons $\mathcal P \sim \mathcal Q$, and let $\mathcal B$ be their common boundary. By counting the number of sides of $\mathcal P$ and $\mathcal Q$ we see $\mathcal B$ must run from one side of the square to an opposite side. From this we can show that the longest side of $\mathcal P$ and $\mathcal Q$ are either the sides of the square (of length $1$), or a side shared in $\mathcal B$. In particular, they are equal, which means in fact $\mathcal P \cong \mathcal Q$.

Doesn't this follow immediately from the Wallace-Gerwin-Bolyai Theorem?