GEO12 1) First, by using “bad primes”, and prime factorizations, etc.) decide if you can make squares with the following areas on a geoboard (with the usual geoboard set up – points one unit apart in a square area, and with the four corners of the square lying on grid points). Note, please don’t actually draw out such a square, just explain whether or not it’s possible, and feel free to use a calculator to factor the numbers. You visit Francis Su’s neat Math Fun Facts website: https://www.math.hmc.edu/funfacts/ffiles/20008.5.s a) 1105 (try a factor of 13 to start with) b) 630 c) 882 2) Next, recall the neat geoboard/unit square grid paper result The “GC” (Group Conjecture!) – that the area of a polygon drawn on unit square grid paper (i.e. so that the polygon’s vertices are all on lattice points of the grid) is equal to I + B/2 – 1, where I is the number of points in the interior of the polygon, and B is the number of points on the boundary of the polygon. The question is whether it’s possible to prove that the formula works for all polygons. One way to go about this is to start by looking at whether it works for rectangles and triangles and then build up from there, given that every polygon can be created by attaching triangles and rectangles together (actually it can all be done with just triangles). For this first homework problem prove that the Group Conjecture works for a rectangle that’s length X by Y (where X and Y are integers, and the rectangle is lined up squarely with the grid points – i.e. not tilted). The area of such a rectangle is clearly X times Y, now does the GC correctly predict that? Note that to show this, you can’t

just give one example, but you need to show that it works for a general X by Y rectangle – i.e. you need to figure out how many points are on the boundary of an X by Y rectangle along with how many points are in the interior… then check the GC formula… 3) Next verify that the formula works for right triangles. To simplify your calculations, assume that the triangle’s hypotenuse doesn’t go through any grid points (note that this doesn’t limit the result, as you could always break a larger right triangle into smaller right triangles if the hypotenuse of the larger triangle did go through some grid points). Note that once again you need to prove this for general right triangles (i.e. it’s not enough to just show that it works for just one specific example – assume that you’ve got a triangle with shorter sides A and B, for example). 4) The final step in the process of proving that the formula is valid for all possible polygons will be to show that if you break a larger polygon into a series of rectangles and triangles, that the formula still holds. To do this you’ll need to show that if the formula works for two polygons, P1 and P2 (as we know it does for rectangles and right triangles, for example), then it will work a polygon, P, that’s created by merging the two polygons. To check this, figure out what would happen referencing the following polygon merger example . Start by assuming that the area of P1 is given by I1+ B1/2 – 1 where I1 and B1 are the number of interior and boundary points, respectively, for polygon P1 (and likewise that the area of P2 is given by I2 + B2/2 – 1). Now show that that the polygon P given as the combination of P1 and P2 must also have its area given by the I + B/2 -1 formula, by figuring out what I and B are for the combination polygon P,

given that you know I1 , I2 , B1 and B2 The main issue is to count everything very carefully, trying to figure out how to handle the points that were on the boundaries of P1 and P2, but are now in the interior of polygon P. Note that the point of this problem is to come up with a result that holds true in general – please don’t just count out the points in the specific example shown and stop there. Instead, please figure out what would happen if you have any two polygons that are merged together along one common edge. Good luck! 5) Do some internet sleuthing (try searching on “geoboard area formula”) to find out who first discovered and named this result (“The area of a polygon drawn on unit square grid paper (i.e. so that the polygon’s vertices are all on lattice points of the grid) is equal to I + B/2 – 1, where I is the number of points in the interior of the polygon, and B is the number of points on the boundary of the polygon. – please read a bit about his particular, tragic, story. 6) On to taxicab geometry! Given any two points A and B in our good old Euclidean geometry, then it’s easy to see that the set of points that are the same distance from A as they are from B is just the perpendicular bisector of line segment AB (draw out an example to convince yourself this is true). In taxicab geometry we might or might not get the same result – let’s see! Use graph paper to work out the following… to start with label a central point on your graph paper (0,0), and then locate other points as if you were graphing them relative to this origin point, i.e. the point (3,2) is a point three units over to the right, 2 units up. a) Suppose point A is located at the origin (0,0) and point B

is located at (8,0) Draw the set of points that are the same distance from A as they are from B using the new taxicab distance (for example, the point (4,0) is in this set as it’s exactly 4 from both A and B, and the point (4,1) is in the set as it’s exactly 5 (taxicab distance) away from A and B. (and yes, this is a pretty boring result!) b) Now make point A the origin again, but make point B the point at (8,6) instead. Again, draw out the points that are the same distance from A as they are from B – now what do you see?! 7) Time to look at an ellipse – by looking at a traditional (Euclidean geometry) ellipse – now it’s time to create a Taxicab geometry ellipse! Recall that an ellipse is the figure one gets when one starts with two foci, A and B, and finds all the points C such that the sum of the distance from C to A plus the distance from C to B stays the same. (a) Start by looking at an ellipse in taxicab geometry with foci at (-2, 0) and (2, 0) and with the sum of distances equaling 6. Go ahead and work through this example to see what you get. (b) Next shift one of the foci a bit, and try foci at (-2, 0) and (2, 2) instead. For this ellipse try finding all the points that have a distance sum equal to 10 instead of 6. 8) Time for a hyperbola! This requires similar work to what you did in the last problem – however, unlike an ellipse, where the sum of the distances from a point on the ellipse to the two foci stays constant, for a hyperbola it is the difference of the distances that stays constant. Thus if points A and B are the two foci of a hyperbola then C is on the hyperbola if D(C, A) – D(C, B) = a set constant (where

D(C, A) means the distance from C to A). Actually, this just gives one “arm” of the hyperbola – the other arm is those points such that Dt(C, B) – Dt(C, A) = the same set constant. Try using Taxicab distance instead of Euclidean distance and see what you get! Suppose A is at (0, 0), B is at (6, 5), and the set difference constant is 3. Thus, for example, the point (4,0) is on the hyperbola, since its distance from A is 4, and its distance from B is 7, and 7 – 4 is 3. Note that (6, 1) is also on the same hyperbola, but on the other arm as now the distances are 7 and 4 respectively (but again their difference is 3).

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