Q: A unit square is dissected into n > 1 rectangles such that their sides are parallel to the
sides of the square. Any line, parallel to a side of the square and intersecting its interior, also
intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle
having no point on the boundary of the square.
Let us orient the square along horizontal and vertical. A horizontal or vertical line which intersects the interior of a square but does not intersect the interior of any rectangle is called a splitting line. A rectangle having no points on the interior of a square is called an interior rectangle.
Let us assume the contrary that there exists a dissection into more than one rectangles such that there are no interior rectangles and no splitting lines. Under such assumption n > 2 otherwise they will have a common side which will form a splitting line.
Let us define an operation that combines rectangles if they have a common side.With this operation the simple dissection mentioned above reduces the number of rectangles leaving no splitting lines and no rectangles which is a contradiction. So a splitting line and interior rectangle has to exist.
Now let us denote the initial square by points ABCD so that A and B lie along the horizontal. We consider a and b as two rectangles containing the vertices A and B respectively. Rectangle a is not equal to rectangle b otherwise one of their sides provides a splitting line. So let us further assume that the height of a is less than the height of b. We can use the same argument as we will use now for the other case as well when a's height is more than that of b. We are merely picking one case to proceed with the arguments. Now if consider a rectangle c next to the lower right corner of a, then a and c can have the following two cases:
height of a is greater than that of c: Then there exists a rectangle d adjacent to both a and c such that it does not have anything common with sides AB, BC and even AD but it has a common point with CD and its left side provides a splitting line. This results in the same contradiction as earlier.
height of a is less than that of c: In this case the rectangle d will lie over a but it should not border with AD otherwise it would combine with a and the same argument as above holds.
Thus we have proved by contradiction by considering all the cases.
Let us orient the square along horizontal and vertical. A horizontal or vertical line which intersects the interior of a square but does not intersect the interior of any rectangle is called a splitting line. A rectangle having no points on the interior of a square is called an interior rectangle.
Let us assume the contrary that there exists a dissection into more than one rectangles such that there are no interior rectangles and no splitting lines. Under such assumption n > 2 otherwise they will have a common side which will form a splitting line.
Let us define an operation that combines rectangles if they have a common side.With this operation the simple dissection mentioned above reduces the number of rectangles leaving no splitting lines and no rectangles which is a contradiction. So a splitting line and interior rectangle has to exist.
Now let us denote the initial square by points ABCD so that A and B lie along the horizontal. We consider a and b as two rectangles containing the vertices A and B respectively. Rectangle a is not equal to rectangle b otherwise one of their sides provides a splitting line. So let us further assume that the height of a is less than the height of b. We can use the same argument as we will use now for the other case as well when a's height is more than that of b. We are merely picking one case to proceed with the arguments. Now if consider a rectangle c next to the lower right corner of a, then a and c can have the following two cases:
height of a is greater than that of c: Then there exists a rectangle d adjacent to both a and c such that it does not have anything common with sides AB, BC and even AD but it has a common point with CD and its left side provides a splitting line. This results in the same contradiction as earlier.
height of a is less than that of c: In this case the rectangle d will lie over a but it should not border with AD otherwise it would combine with a and the same argument as above holds.
Thus we have proved by contradiction by considering all the cases.
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