Problem (Geometry):
Square $O P Q R$ has vertices $O(0,0), P(0,8), Q(8,8)$, and $R(8,0)$. The parabola with equation $y=a(x-2)(x-6)$ intersects the sides of the square $O P Q R$ at points $K, L, M$, and $N$. Determine all the values of $a$ for which the area of the trapezoid $K L M N$ is 36 .

Solution:
First, we note that $a \neq 0$. (If $a=0$, then the "parabola" $y=a(x-2)(x-6)$ is actually the horizontal line $y=0$ which intersects the square all along $O R$.)

Second, we note that, regardless of the value of $a \neq 0$, the parabola has $x$-intercepts 2 and 6 , and so intersects the $x$-axis at $(2,0)$ and $(6,0)$, which we call $K(2,0)$ and $L(6,0)$. This gives $K L=4$.

Third, we note that since the $x$-intercepts of the parabola are 2 and 6 , then the axis of symmetry of the parabola has equation $x=\frac{1}{2}(2+6)=4$.

Since the axis of symmetry of the parabola is a vertical line of symmetry, then if the parabola intersects the two vertical sides of the square, it will intersect these at the same height, and if the parabola intersects the top side of the square, it will intersect it at two points that are symmetrical about the vertical line $x=4$.

Fourth, we recall that a trapezoid with parallel sides of lengths $a$ and $b$ and height $h$ has area $\frac{1}{2} h(a+b)$.

We now examine three cases.



Case 1: $a<0$

Here, the parabola opens downwards.

Since the parabola intersects the square at four points, it must intersect $P Q$ at points $M$ and $N$. (The parabola cannot intersect the vertical sides of the square since it gets "narrower" towards the vertex.)



Since the parabola opens downwards, then $M N<K L=4$.

Since the height of the trapezoid equals the height of the square (or 8), then the area of the trapezoid is $\frac{1}{2} h(K L+M N)$ which is less than $\frac{1}{2}(8)(4+4)=32$.

But the area of the trapezoid must be 36 , so this case is not possible.

Case 2: $a>0 ; M$ and $N$ on $P Q$

We have the following configuration:



Here, the height of the trapezoid is $8, K L=4$, and $M$ and $N$ are symmetric about $x=4$. Since the area of the trapezoid is 36 , then $\frac{1}{2} h(K L+M N)=36$ or $\frac{1}{2}(8)(4+M N)=36$ or $4+M N=9$ or $M N=5$.

Thus, $M$ and $N$ are each $\frac{5}{2}$ units from $x=4$, and so $N$ has coordinates $\left(\frac{3}{2}, 8\right)$.

Since this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $8=a\left(\frac{3}{2}-2\right)\left(\frac{3}{2}-6\right)$ or $8=a\left(-\frac{1}{2}\right)\left(-\frac{9}{2}\right)$ or $8=\frac{9}{4} a$ or $a=\frac{32}{9}$.



Case 3: $a>0 ; M$ and $N$ on $Q R$ and $P O$

We have the following configuration:



Here, $K L=4, M N=8$, and $M$ and $N$ have the same $y$-coordinate.

Since the area of the trapezoid is 36 , then $\frac{1}{2} h(K L+M N)=36$ or $\frac{1}{2} h(4+8)=36$ or $6 h=36$ or $h=6$.

Thus, $N$ has coordinates $(0,6)$.

Since this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $6=a(0-2)(0-6)$ or $6=12 a$ or $a=\frac{1}{2}$.

Therefore, the possible values of $a$ are $\frac{32}{9}$ and $\frac{1}{2}$.

Final answer: $\frac{32}{9}$,$\frac{1}{2}$