Mike Callahan
Modern Geometry
Urban Canyon Assignment
I begin by constructing a right triangle in the spreadsheet of geogebra. I then find the midpoints of each side and construct a circle so that the diameter of each side corresponds to the diameter of the circle.
The circles create breathing space for that particular building side. In other words, each side uses the midpoint as the center of the circle and goes though each endpoint on the sides, see figure 1
Figure 1-Right triangular building with breathing space
Next I created a line through AC and found the point at which circle A hits this line. Additionally I created a line through BC and found the points H and G where this line hits the outer circle on each side, see figure 2.

Figure 2 - Constructing parallel and perpendicular lines
I then created a line tangent to circle A through point E. I then created a line through point B and parallel to point E. It was the exact line I had already drawn in. So I can conclude the line containing point E tangent to the circle is now parallel to the line containing HFBCIG. Additionally, I created a line perpendicular to line BC through point E. This line laid right on top of the like JCDAKE, so I can conclude point C creates four right angles. I know point E also forms 90 degree angles as the line is both parallel to
HG and perpendicular to JE.
I then defined the radius corresponding the sides of the triangle as variables x and y shown in figure 3.

Figure 3-Defining radius as variables
I know BA1 is the radius of the circle containing the breathing space of side A1A3. I will call this radius y, which makes the diameter 2y. A1A is the radius of the circle containing the breathing space of side
A1A2. I will call this radius x, which makes the diameter 2x. I know CA2 is the radius of the circle containing the breathing space of side A2A3. This radius and diameter can be defined by using
Pythagorean Theorem on triangle A1A2A3.

Figure 4 - Triangle A1A2A3
(2y) 2 + (2x) 2 = (side A2A3) 2
4y 2 + 4x 2 = (side A2A3) 2 side A2A3 = √4𝑥 2 + 4𝑦 2 side A2A3 = 2 √𝑥 2 + 𝑦 2
If I cut this in half to represent radius CA2, I get the radius to be √𝑥 2 + 𝑦 2 .
Now I can use similar triangles to show BC = x. Now if I take triangle A3BC I can say it has side lengths x, y and √𝑥 2 + 𝑦 2 . If these sides are unfolded onto line HG, the three radius each occupy a portion of this line. See the color coding in figure 5.

Figure 5 Diameter of the block
You can see how the colored respective sides of the triangle A3BC can be broken apart to create line HG.
Geogebra also confirmed all the segments are equal lengths (A3B congruent to BH, etc). Now I know line HG is parallel to the line tangent to circle A containing point E. I will now create tangents through points H and G, which since HG is parallel to the line at the bottom, these will form right angles at H, G and the points L and M where the tangents hit the line through point E. See figure 6.

Figure 6 - Tangents through points H and G
Now with this picture I can discuss a couple of things. I have discussed how C creates right angles and since ML and EJ are perpendicular (also E is tangent to circle A), E creates a right angle. I know that G and H create right angles both because they were constructed tangent to their respective circles and checked by creating a perpendicular through G and line ML. Both constructions yielded the same line.
Now since HG is parallel to ML and I have shown how the sides of triangle A3BC can be broken apart to create HG, I can say that HG is congruent to ML. Each of these lines can be represented as the sum of the radius or side lengths.

x + y + √𝑥 2 + 𝑦 2
I have established the width of the block around the building (ML) and also that M and L create right angles. If I look at the figure, I know that AE is the same radius as A1A, which I called x. CA is the same radius as BA1, which I called y. And then CJ is the same radius as CG which I found to be √𝑥 2 + 𝑦 2 .
Therefore, the length (or vertical distance) of the block around the building can also be described as the sum of the three radius x + y + √𝑥 2 + 𝑦 2
Now I have the width of the building ML and lengths MH and LG, which since M and L form right angles,
MH and LG are parallel. I have three sides each represented by x + y + √𝑥 2 + 𝑦 2 and that these three sides come together at right angles.
To close in the city block around this right triangular building I created a tangent line though point J on the circle. This proved in geogebra to also be the same as a perpendicular line to both MH and GL though point J and also a parallel line to HG and through J. See figure 7

Figure 7-Final side of city block
I now will look at polygon JCGO. I know C makes a right angle and G does as well as it is on a tangent.
Now J is on a line perpendicular to ML, so it also forms a right angle. Therefore O must also be a right angle to get to 360 0 in polygon CJOG. Now if O makes a 90 0 angle, ON has to be parallel to HG and ML.
Therefore N will also hit at a 90 0 (also because of the same argument with quadrilateral NHCJ as just done with JCGO). Therefore, NJO is the same measure as HG or ML, x + y + √𝑥 2 + 𝑦 2 . Since all four sides MN, ML, LO, and ON are length x + y + √𝑥 2 + 𝑦 2 AND angles M, L, O, and N create right angles, the city block is in the shape of a square.
Therefore, for any right triangle, the area around the city block will be a square.

Now of course this proof assumes that parallel postulate is true on the plane, but the city block around a right triangle building did come out to the same shape (a square) as the city block shape around a rectangular building. Now obviously the proof done here is a little different (and more complicated) because of the third side of the triangle creates a measurement in terms of the other two sides (which is really just a third radius) whereas with the rectangular building, there are only two radius as opposed to three with a triangle. The important part is that both rectangular and right angled triangle shaped buildings create a square city block.
If we consider this same argument on a sphere, I first draw in a right triangle. I used an equator and then constructed a perpendicular through it to make a 90 0 angle at the equator. The other two angles are not measured to be 90 0 . See figure 8.
Figure 8- Right triangle on a sphere
So BDC is a right angle. I have constructed midpoints F, E, and G. If I then use those midpoints as radius of circles to create breathing space, I get a picture shown in figure 9.

Figure 9- Breathing space of a right triangle on a sphere

This obviously is not going to create a square city block because there is no such thing as parallel lines on a sphere. I feel like the shape of the city block has to be a larger circle touching the three furthest points on each yellow breathing space circle. If I was to do a tangent to a point on one of the yellow circles, I am worried that space would explode into the other hemisphere, creating a massive city block with area on two open hemispheres. You could draw in the radius and use them to see how they help you get around the block, but unlike on the plane, I am struggling to see where to draw in similar triangles.
To test my theory, I found the intersection of GB and FE from figure 9 and used that point to create a large circle around the breathing space circles. See figure 10.
Figure 10- Circular city block top view

Figure 10b - Circular city block bottom view
I think that is your best bet for creating a city block around the breathing space than keeps your area to somewhat of a minimum.