Goutham wrote: A rectangle is divided into $2\times 1$ and $1\times 2$ dominoes. In each domino, a diagonal is drawn, and no two diagonals have common endpoints. Prove that exactly two corners of the rectangle are endpoints of these diagonals. Imagining Cartesian coordinates, each diagonal is either rising or falling. If they are all raising or all falling, two opposite corners of the rectangle are endpoints of such diagonals, the other two are not. So suppose there is at least one raising and one falling diagonal in the pattern. We can assume that one domino with a rising (red) and one with a falling (blue) diagonal touch each other, i.e. have part of an edge in common. Apart from reflections, there are only two ways this can happen : * either like the dominoes 1 & 2 in the picture, and then 3,4,... are forced, * or like the dominoes 1 & 3 in the picture, but then again 4,5,... are forced. In both cases it is obviously impossible to cover a finite rectangle.

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I have a bit different solution.Let k be the number of all rectangles 1x2 and 2x1.Now,independent of a 'puzzling',k is constant.Now,it is easy to see that if one point different from edges of the big rectangle is an edge of one rectangle 1x2 or 2x1,then it is an edge of exactly 2.Now,this is easy to prove because it is trivial that the number of such rectangles is always even,so just prove that it can't be 4(for this,consider 3 cases and get a contradiction).So,we conclude that every point inside of the big rectangle is an edge of 2 or 0 rectangles 1x2 or 2x1. Now,we count all edges to get:4 + 2*x = 4*k,where x is the number of edges which have 2 rectangles 1x2 or 2x1.We get x= 2*k -2.Also,the number of diagonals is k,and non of them have two common points,so it remains to prove that every point inside has exactly one diagonal,whic is easy considering two cases,so we have that exactly two points if the edge have a diagonal,so we are finished

Suposse that in the upper right corner there is a diagonal that doesn't contain a corner of that diagonal.Now we will prove that those 2 corners closest to him will satisfy the condition.It is trivial to see that all other dominoes that contain 2 points on the edge will be with faling diagonals.So it is for the last domino that must contain the bottom right corner so there is a diagonal in that corner.By analogy there is a diagonal in top left corner and it is trivial that there can't be a diagonal containing bottom left corner.

For each domino, color the drawn diagonal red and diagonal not drawn as blue. For example, [asy][asy] draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0)); draw((1,1)--(3,1)--(3,2)--(1,2)--(1,1)); draw((0,1)--(1,1)--(1,3)--(0,3)--(0,1)); draw((2,0)--(4,0)--(4,1)--(2,1)--(2,0)); draw((3,1)--(4,1)--(4,3)--(3,3)--(3,1)); draw((1,2)--(2,2)--(2,4)--(1,4)--(1,2)); draw((0,3)--(0,5)--(1,5)--(1,3)--(0,3)); draw((2,2)--(3,2)--(3,4)--(2,4)--(2,2)); draw((1,4)--(3,4)--(3,5)--(1,5)--(1,4)); draw((3,3)--(4,3)--(4,5)--(3,5)--(3,3)); draw((0,0)--(2,1),red); draw((2,0)--(0,1),blue); draw((1,1)--(3,2),red); draw((3,1)--(1,2),blue); draw((2,0)--(4,1),red); draw((4,0)--(2,1),blue); draw((0,1)--(1,3),red); draw((1,1)--(0,3),blue); draw((0,3)--(1,5),red); draw((0,5)--(1,3),blue); draw((1,2)--(2,4),red); draw((1,4)--(2,2),blue); draw((2,2)--(3,4),red); draw((2,4)--(3,2),blue); draw((1,5)--(3,4),blue); draw((1,4)--(3,5),red); draw((3,1)--(4,3),red); draw((4,1)--(3,3),blue); draw((3,3)--(4,5),red); draw((3,5)--(4,3),blue); [/asy][/asy] Claim: No two blue edges have a common endpoint. Proof: Assume contrary. Choose two dominoes $D_1,D_2$ whose blue diagonals share a vertex $V$. Note that $D_1,D_2$ can be chosen such that they have only a vertex in common (if originally $D_1,D_2$ don't satisfy this, then $V$ will be a vertex of $4$ dominoes, and since $V$ cannot be endpoint of two red edge, this thing follows). Now there can be two possible configurations: [asy][asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); draw((1,0)--(1,2)); draw((0,2)--(1,0)--(2,2),blue); dot("$V$",(1,0),dir(-90)); draw((4,2)--(3,2)--(3,0)--(6,0)--(6,1)--(4,1)^^(4,0)--(4,2)); draw((3,2)--(4,0)--(6,1),blue); dot("$V$",(4,0),dir(-90)); [/asy][/asy] In the first configuration, the red diagonals of $D_1,D_2$ share an endpoint (as shown below), which is a contradiction. [asy][asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); draw((1,0)--(1,2)); draw((0,2)--(1,0)--(2,2),blue); draw((0,0)--(1,2)--(2,0),red); dot("$V$",(1,0),dir(-90)); [/asy][/asy] In the second case, we obtain an infinite descent (as shown below), because for the following reasons: [asy][asy] size(200); draw((1,0)--(1,4)--(0,4)--(0,0)--(3,0)--(3,1)^^(2,1)--(2,3)--(1,3)); draw((0,2)--(1,2)^^(1,1)--(3,1)); label("$\vdots$",(0.5,6.5)); label("$\vdots$",(1.5,5.5)); dot("$V$",(1,0),dir(-90)); dot("$X_1$",(1,1),dir(45)); dot("$X_2$",(1,2),dir(135)); dot("$X_3$",(1,3),dir(45)); dot("$X_4$",(1,4),dir(135)); dot("$X_5$",(1,5),dir(-35)); draw((0,2)--(1,0)--(3,1),blue); draw((0,0)--(1,2)^^(3,0)--(1,1),red); draw((1,3)--(2,1),red); draw((0,2)--(1,4),red); draw((1,3)--(1,5)--(2,5)--(2,3)); draw((0,4)--(0,6)--(1,6)--(1,4)); draw((2,3)--(1,5)^^(0,4)--(1,6),red); [/asy][/asy] We must have some domino with $X_1,X_2$ on boundary. If that domino is horizontal then we get a contradiction as both of its diagonals intersect some red diagonal. So the domino with $X_1,X_2$ on boundary is vertical. Then we apply similar reasoning for vertices $X_2,X_3$. If we keep applying the reasoning for $X_iX_{i+1}$, we obtain an infinite descent, which gives us a contradiction. This proves our claim. $\square$ Now denote by $T$ the set of all points which are corners of some domino but are not a corner of the big rectangle. Note that each vertex in $T$ is the corner of at least two dominoes. Using our claim we obtain that each vertex in $T$ is the endpoint of exactly one red and one blue diagonal. Now let $R$ and $B$ be the set of all points which are endpoints of some red and blue diagonal, respectively. Note that any corner of the big rectangle is in exactly one of the sets $R,B$ (it might help to take a look at the original figure again): [asy][asy] draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0)); draw((1,1)--(3,1)--(3,2)--(1,2)--(1,1)); draw((0,1)--(1,1)--(1,3)--(0,3)--(0,1)); draw((2,0)--(4,0)--(4,1)--(2,1)--(2,0)); draw((3,1)--(4,1)--(4,3)--(3,3)--(3,1)); draw((1,2)--(2,2)--(2,4)--(1,4)--(1,2)); draw((0,3)--(0,5)--(1,5)--(1,3)--(0,3)); draw((2,2)--(3,2)--(3,4)--(2,4)--(2,2)); draw((1,4)--(3,4)--(3,5)--(1,5)--(1,4)); draw((3,3)--(4,3)--(4,5)--(3,5)--(3,3)); draw((0,0)--(2,1),red); draw((2,0)--(0,1),blue); draw((1,1)--(3,2),red); draw((3,1)--(1,2),blue); draw((2,0)--(4,1),red); draw((4,0)--(2,1),blue); draw((0,1)--(1,3),red); draw((1,1)--(0,3),blue); draw((0,3)--(1,5),red); draw((0,5)--(1,3),blue); draw((1,2)--(2,4),red); draw((1,4)--(2,2),blue); draw((2,2)--(3,4),red); draw((2,4)--(3,2),blue); draw((1,5)--(3,4),blue); draw((1,4)--(3,5),red); draw((3,1)--(4,3),red); draw((4,1)--(3,3),blue); draw((3,3)--(4,5),red); draw((3,5)--(4,3),blue); [/asy][/asy] Now since number of red and blue diagonals are equal, so $|R| = |B|$. Moreover, as $T$ is a subset of both $R,B$ (as discussed before), it follows that number of corners of the big rectangle in $R$ is equal to that in $B$. Hence both of $R,B$ contain exactly $2$ corners of the big rectangle. This completes the proof of the problem. $\blacksquare$