Problem Statement:
As a class, we were trying to figure out if we could find a general formula for the area of a lattice polygon. A lattice polygon is a polygon that connects to lattice point. Process: Step #1: We started by creating twenty different lattice polygons which are polygons whose vertices are points of a point lattice. A point lattice is a regularly spaced array of points. In this case,‘B’ stands for boundary points, ‘A’ stands for area, and ‘I’ stands for interior points, here are mine: Boundary:4 Area:2 Interior:1, Boundary:4 Area:1 Interior:0, Boundary:3 Area:½ Interior:0, Boundary:4 Area:1 Interior:0, Boundary:4 Area:1 Interior:0, Boundary:6 Area:2 Interior:0, Boundary:5 Area:1½ Interior:0, Boundary:3 Area:½ Interior:0, Boundary:8 Area:4 Interior:1, Boundary:7 Area: 2½ Interior:0, Boundary:6 Area:2 Interior:0, Boundary:4 Area:1 Interior:0, Boundary:6 Area:2 Interior:0, Boundary:6 Area:2 Interior:0, Boundary:8 Area:3 Interior:0, Boundary:10 Area:5 Interior:1, Boundary:4 Area:2 Interior:1. Next, we collected all the data from everyone and took all of those numbers and graphed them on a scatterplot of X: boundary points and Y: area. We drew a line of best fit, which is a line on a graph showing the general direction that the group of points seem to be heading, and it’s equation by finding it’s slope and plugging the numbers into the following equation: y-y1=m(x-x1) which gave us a formula. Then we looked at the data sets with different interior points. We graphed them and, again, found the line of best fit, it’s equation, and another formula while looking for and discarding all the incorrect ones. If we needed to, we made more lattice polygons to find a pattern to get a formula. We gathered the formulas for the interior points of the polygons and spent a while checking them using counter examples for each one. Once that was done, we organized all the formulas that we checked that were correct, and found a pattern. The number of interior points of the lattice polygons correspond with the variation of each formula. For example, if the number of interior points was one, the formula would be A=1/2B+(I-1), if the number of interior points was two, it would be A=1/2B+(I-2), and if the number of interior points was three, then the formula would be A=1/2B+(I-3) and so on. We concluded that the general formula for all the lattice polygons is A=1/2B+(I-1). Just to make sure, we sought out counter examples of the formula. A few people thought they found some, but we proved them to be false. Solution: A=1/2B+(I-1) will work with any lattice polygon as long as it is a real one. Take one with B:6 A:3 & I:1 for example, that would be 3=1/2x6+(I-1) then 3= 3+0 which is the same as 3=3, so it works. Problem statement: The King had all his gold in 12 bags now, but one of the bags had fake coins. That bag had a different weight then the others and the king wants to find out which one it is using as little measurements as possible. I started by dividing the bags into equal groups. First I tried dividing then into two groups and then figured out that wasn't going to work because it doesn't say if the fake bag was lighter or heavier the real one so I just divided by three. I got stuck when I tried to make it so you had to weigh it the least possible times so I settled with what I had. In conclusion, it would take 3-5 weighings to get the exact bag with the fake gold in it. |
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