Student's Overlapping Rectangles

Since there is at least one square contained in the intersection of any three rectangles, the best way such that the 24 rectangles have the largest area is that there is one square contained in the intersection of all the 24 rectangles.

If there are x rectangles with area 24 and all share exactly 1 square and have the maximum area, we see that x = 28 as there are 4 $1 \cdot 24$ squares, 8 $2 \cdot 12$ squares, 8 $3 \cdot 8$ squares, 8 $4 \cdot 6$ squares with total area 241.

However, the question asks for 24 rectangles only.

If we eliminate 4 $1 \cdot 24$ squares, the total area would be 193.

If we eliminate 4 $2 \cdot 12$ squares , the total area would be 225.

If we eliminate 4 $3 \cdot 8$ squares, the total area would be 233.

If we eliminate 4 $4 \cdot 6$ squares, the total area would be 233.

Therefore the largest area of the 24 rectangles would be 233.

By Helly's Theorem, since any 3 rectangles in $\mathbf{R}^2$ intersect, there must be a point that is inside all the rectangles. Since the rectangles are composed of a union of lattice squares, if there is a point in the intersection of all the squares, there must be a lattice square in the intersection of all the squares.

The maximum possible area that this can have will be when the intersection of all the rectangles is smallest (1 unit square). The image below shows all the possible ways a rectangle with integer side lengths and area 24 can have the centre unit square a corner vertex. This uses a total of 28 rectangles, so the entire area could not be covered by the students in the class.

To determine the area of this, we can divide the shape into a square of side length 7 and the four arms extending from it. The total area is $7 \times 7 + 4(2 \times 7 + 2 \times 5 + 4 \times 3 + 12 \times 1) = 49 + 4(48) = 241$ . However, we have to remove 4 rectangles from this list. Clearly, to maximize the remaining area, we want to remove 4 of the 6 by 4 rectangles, which each reduce the area by 2. We cannot choose to remove two which intersect in a 4 by 4 square, since removing both would reduce the total area by 5. However, we can choose 4 such that no pair intersect in this manner (by taking 1 in each 'quadrant'), so 8 is the smallest area we must remove from the figure. Hence, the leftover area is $241 - 8 = 233$ .

Lemma: All $24$ rectangles must have at least one common intersection square.

Proof: Consider the triples of rectangles $(A, B, C)$ , $(A, B, D)$ , $(A, C, D)$ , and $(B, C, D)$ . Let us look at the triple $(A, C, D)$ and $(B, C, D)$ . Clearly, the intersection of rectangles $C$ and $D$ must share at least one square with both $A$ and $B$ . If these intersections do not share a square, then $(A, B, C)$ does not satisfy the conditions, since $A$ and $B$ do not even intersect (because they are convex). Thus, all quadruples of rectangles must share at least one common square. Applying this reasoning to all quadruples, we see that we must have at least one common intersection square.

Now we must try to maximize the area covered by these $24$ rectangles. First, we see that the $1\times 24$ rectangles can intersect in one point by placing consecutive rectangles at right angles to one another with respect to a single square. This is the most efficient configuration because there is only one point of intersection, leaving 23 squares that have no intersection. Thus, all rectangles must contain this square due to our lemma.

We have now separated the plane into four "quadrants". There are only $6$ other allowable rectangles because we do not want coinciding rectangles: ( $2\times 12$ , $3\times 8$ , and $4\times 6$ ). It makes sense that we should only place a maximum of $6$ rectangles into each "quadrant" because we do not want to have coinciding rectangles if we can avoid it. We also should align the corner of these rectangles with the square of intersection to avoid overlapping between different quadrants. This method allows for the optimization of area, since we are trying to have the least overlap area.

We start by drawing in all the rectangles that are described in the previous section. This will give us a total of $6\cdot 4 +4=28$ rectangles, so we need to remove $4$ of them. We can clearly see that by removing one $4\times 6$ or $3\times 8$ rectangle, we are only decreasing the area by $2$ square units, while removing a $2\times 12$ rectangle will decrease the area by $4$ square units. We can then remove one $4\times 6$ or $3\times 8$ rectangle from each "quadrant". The area of each "quadrant" is $35$ square units, and the area of the "axis" formed by one of the neighboring $1\times 24$ rectangle (not including the intersection square) is $23$ square units. There are $4$ "quadrants", and one intersection square, so the total area is $(35+23)\cdot 4+1=\boxed{233}$ , and this is the maximum.

Let A, B, C, and D be rectangles in the plane that can be picked by students in this class. If A, B, and C all contain some 1x1 square E, then in order for D to satisfy the given condition for intersections, it must also contain E.

It follows that all $24$ rectangles must contain this square E. In order to maximize the area of the figure created by putting all the rectangles into one set of axes, we must make sure that all the rectangles intersect in as little area as possible, since this area is essentially wasted area.

There are $4$ ways of making rectangles that satisfy the given conditions, namely ones with $(s_1,s_2)=(1,24),(2,12),(3,8),(4,6)$ where $s_1,s_2$ are the lengths of the adjacent sides of the rectangle. The way to maximize the area is to place these different types of rectangles in a sort of star-shape radiating out from one square, which will be the one that they all share. This setup is ideal because it minimizes the area where the rectangles overlap and ensures that every rectangle is adding area to the figure rather than simply being contained in another rectangle.

To better visualize what this setup looks like, we can mark a 1x1 square S in the plane (for example, the one with its vertices at $(0,0),(0,1),(1,1),(1,0)$ ). Then we construct $4$ rectangles with side lengths $1$ and $24$ , one in each direction, each of which has one of its edges completely on a side of S. Finally, we construct $8$ of each of the other types of rectangles that each share a corner with S.

The total area of this shape can be calculated easily by taking S and splitting the resulting shape into $4$ congruent figures that contain a 1x23 rectangle on one edge and a 1x1 square on the other. The area of each of these figures is $23+11+7+5+3+3+2+2+1+1+1+1=60$ , so the area of the original shape is $60\cdot4+1=241$ .

This shape, however, contains $28$ distinct rectangles, $4$ more than we are allowed. If we imagine that the 1x24 rectangles define quadrants, we can see that we can take away a rectangle from each quadrant whose removal decreases the total area by only $2$ units. This gives us $24$ rectangles, which is what we want.

Therefore, the maximum area the students' rectangles can cover is $241-4\cdot2=\fbox{233}$ .