4. An $8 \times 6$ grid is placed in the first quadrant with its edges along the axes, as shown. A total of 32 of the squares in the grid are shaded. A line is drawn through $(0,0)$ and $(8, c)$ cutting the shaded region into two equal areas. What is the value of $c$ ?
4. Solution 1
Since 32 of the $1 \times 1$ squares are shaded, then the total shaded area is 32 .
We want to draw a line through $(0,0)$ and $(8, c)$ that divides the shaded region into two pieces, each with area $\frac{32}{2}=16$.
As long as the slope of the line segment through $(0,0)$ and $(8, c)$ is not too large, the bottom piece will be a triangle with vertices $(0,0),(8,0)$, and $(8, c)$.

This triangle is right-angled at $(8,0)$ and so has base of length 8 and height $c$.
For the area of this triangle to be 16 , we need $\frac{1}{2}(8)(c)=16$ or $4 c=16$.
Thus, $c=4$.
(Note that if $c=4$, the line segment passes through $(0,0)$ and $(8,4)$. This line segment has slope $\frac{4-0}{8-0}=\frac{1}{2}$ and so passes through the points $(2,1)$ and $(4,2)$. This confirms that the line segment does not pass through any unshaded regions.)

Solution 2
We move the upper right shaded $2 \times 2$ square to complete a $8 \times 4$ shaded rectangle, as shown.
The shaded area is now a rectangle with bottom left vertex $(0,0)$ and top right vertex $(8,4)$. The area of the rectangle is cut in half by its diagonal passing through $(0,0)$ and $(8,4)$. Note the region affected by this move lies above this diagonal. (This is because the diagonal has slope $\frac{4-0}{8-0}=\frac{1}{2}$ and so passes through the points $(2,1)$ and $(4,2)$.)
This means that the areas of the regions below and above the line were unaffected by the move. Thus, reversing the move does not change the areas of the regions below and above the line. Therefore, the line drawn through $(0,0)$ and $(8,4)$ cuts the original region into two pieces of equal area.
Thus, $c=4$.
ANSWER: $c=4$