My older son had a homework problem that asked him to find the area of the region bounded by the two equations:
(i) , and
Mathematica’s picture of that shape is here:
He told me that he used Pick’s theorem and found that the area of the shape was 13 square units.
There’s just one small problem – you can’t use Pick’s theorem to find the area of this shape since the corners of the shape are not lattice points of the grid.
What to do . . . .
I wrote a quick little program that picked 100 million random points in the 10×10 square centered at (0,0) and tested whether or not they were part of the shape. That program found that 13.71% of the points were part of the shape – that was enough to convince him that the area might be larger than 13 square units.
Next I had him re-read Pick’s theorem to see what went wrong. He saw pretty quickly that the shape didn’t meet the condition of having the corners lie on lattice points.
I really wanted to try to find a way to make Pick’s theorem work with this shape. I had him determine the y-coordinate for the far right corner. The value was y = -2/7.
After finding that value, we had a good talk about scaling. To make the new grid larger we had to *divide* the x and y coordinates in the equations by 7. Here’s Mathematica’s picture of the new shape and grid (note that the x and y values run from -25 to 25 in this picture):
With this shape we are able to use Pick’s theorem to calculate the area. We counted 40 grid points on the boundary without too much difficulty. Counting the ones in the middle was a little bit more of a pain, so we wrote a short program to perform that calculation for us. Note that we have to change the “less than or equal to” from the original equations to “strictly less than” since we want to be inside the shape:
So, we have 653 lattice points in the inside and 40 on the boundary. Pick’s theorem tells us that the area is equal to the number of lattice points on the interior plus half the number of lattice points in the boundary minus 1. That’s 672 units. In the picture above, 1 unit is equal to 1/49 of a unit in the original picture, so the original area is 96 / 7 or 13 5/7. Close to what he found originally, but not equal!
Along the way we also talked about alternate ways to find the area – the easiest being dividing the shape into two triangles with a vertical line through the middle.
I’m really excited about the discussion that we had tonight. Funny how many important ideas in math can come up from a problem about absolute value and inequalities.