Re: Anybody see more?
Larry Barrett
Consider the orchard laid out on an x-y grid. There are 17 rows and 13 columns. Number the columns 0, 15, 30, ... 180 and the rows 0, 15, ... 240. Joe is standing on the x-axis, at a location 5 feet from the column on his left and 10 feet from the column on his right.
For convenience, I now re-numbered the columns from Joe's position (and divided by 5 to make the numbers smaller) - the columns to his right will be numbered 2, 5, 8, 11, 14, 17, 20; the columns to his left will be -1, -4, -7, -10, -13. The rows are numbered 0, 3, 6, ... 48.
Consider the part of the orchard to Joe's right (from 90-180 degrees; a similar analysis can be done for the area to his left from 0-90 degrees). We want to draw straight lines that cross the x-axis at 0 (Joe's position) and intersect the columns to Joe's right at locations that correspond to the y-axis rows. The general equation for a straight line is y=mx+a. In our problem each line will cross the x-axis at 0, so a=0. We are only interested in the values of y at the point where x = 2, 5, 8, 11, 14, 17, 20 (the columns to Joe's right). Furthermore, we want to find points where the y values correspond to the y axis numbers (0, 3, 6, ...48, the points where the trees are located).
So we want to find values of m (the slope of the line) where y=mx, subject to the conditions above. Substituting the values of x (x=2, 5, 8, ... 20) into this equation, we want to find y(1)=2m, y(2)=5m, y(3)=8m, ... y(7)=20m; we are looking for values of m where at least two of the y values (y(1), y(2), ...) correspond to the y-axis values where the trees are (0, 3, 6, ... 48).
For example, when m=3, the resulting values of y are 6, 15, 24, 33, 42, 51, 60. The first 5 values (6, 15, 24, 33, 42, 45) all correspond to trees (the values 51 and 60 are outside the orchard), so this is one straight line that works for Joe.
Having gotten this far, I think I see errors in what I did earlier, so will stop now and re-compute.