The diagram shows a rectangular picture frame $A B C D$ made of eight identical trapezoids. The shaded region is where the picture goes. The length, $A B$, and width, $A D$, of the frame are both positive integers. The area of each individual trapezoidal piece is a prime number. If the area of the shaded region is less than 2000 square units, what is the maximum possible area of the shaded region?

![](https://cdn.mathpix.com/cropped/2024_04_17_2efeb2001664c4e9869eg-2.jpg?height=317&width=748&top_left_y=1958&top_left_x=748)
Suppose that each of the identical trapezoids has parallel sides of length $x$ and $y$ (with $x>y$ ), height $h$, and area $p$, for some prime number $p$.

We may assume that $p>2$, since we eventually will want to make $p$ as large as we can, subject to the other restrictions of the problem. Requiring $p>2$ ensures that $p$ is odd.

Since the area of each trapezoid is $p$, then $\frac{1}{2}(x+y) h=p$.

Using the given diagram, the outer length of the frame is $x+y+x=2 x+y$ and the inner length is $y+x+y=x+2 y$.

![](https://cdn.mathpix.com/cropped/2024_04_30_264fdb419065898dfdf2g-12.jpg?height=322&width=762&top_left_y=1796&top_left_x=736)

The difference between the outer length and the inner length is twice the height of one of the trapezoids, so $(2 x+y)-(x+2 y)=2 h$ or $x-y=2 h$.

Since $\frac{1}{2}(x+y) h=p$, then $(x+y)(2 h)=4 p$ or $(x+y)(x-y)=4 p$.

Since $x$ and $y$ are integers, then $x+y$ and $x-y$ are either both even or both odd. (We can see this by checking the various possibilities for $x$ and $y$ themselves being even or odd.)

Since the product of $x+y$ and $x-y$ is even, then each of them must be even.

Since $4 p=2^{2} p$ and $p$ is an odd prime number, then the only way of writing $4 p$ as the product of two even integers is $2 \cdot 2 p$.

Since $x+y>x-y$, then we must have $x+y=2 p$ and $x-y=2$.

Adding these gives $2 x=2 p+2$ or $x=p+1$.

Substituting back gives $y=(x+y)-x=2 p-(p+1)=p-1$.

Now the shaded region has length $x+2 y$ (which equals $(p+1)+2(p-1)$, or $3 p-1)$ and width $y$ (which equals $p-1$ ).

Therefore, the area of the shaded region, in terms of $p$, is $(p-1)(3 p-1)$.

We want to find the largest possible value less than 2000 that this expression can take, subject to the restriction that $p$ is prime.

We note that

- when $p=23$ we obtain $(p-1)(3 p-1)=1496$,
- when $p=29$ (the next prime), we obtain $(p-1)(3 p-1)=2408$, and
- $(p-1)(3 p-1)$ is increasing when $p>2$.

Therefore, the maximum possible area less than 2000 for the shaded region is 1496 .

ANSWER: 1496

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)