Why are there no more dinosaurs?

This is one of my favorite geometry problems, involving volume and proportional reasoning.

The answer is: there was about 1.6 times as much oxygen in the air.

Why Are There No More Dinosaurs

Hexagonal-grid city life

Based on a board game, I started thinking about what putting a city on a hexagonal grid would look like:

The streets (in blue) run north-south around the hexagonal blocks. I numbered them, although obviously, the indexing is off and 1st Street should be at the far west end of the city.

The avenues (in red) run northeast-southwest. I gave them Latin letters. The boulevards (in green) run northwest-southeast. I gave them Greek letters.

Two things to note:

1. intersections are three-way (three roads come together)
2. two different roads "intersect" for a whole block

The east-west (horizontal) blocks are on both an avenue and a boulevard. The northeast-southwest blocks (diagonal bottom left to upper right) are on both a street and an avenue. The northwest-southeast blocks (diagonal upper left to bottom right) are on both a street and a boulevard. Can you locate A Avenue and Gamma Boulevard?

Addresses could be given this way: "22 A Ave. and Gamma Blvd." I think address numbers would reset for each block; there would be a "22 A Ave. and Beta Blvd."

How hard would it be to navigate?

Let's say you lived at 1st Street and A Ave. and wanted to travel to 4th Street and Alpha Blvd. You know that you need to go east (to higher numbered streets) and not sure whether you need to travel north or south overall. But you don't need to worry. Just head northeast on A Ave. Either you'll intersect 4th Street first or Alpha Blvd. first.

How would the road signs look? Let's work our example in detail. Again, you want to travel east (from 1st to 4th), so you would take A Ave. northeast. (Taking 1st Street north would get you there but not necessarily in the most direct way.)

Let's look at the points (intersections) on the grid you would travel through:

At point U, the signs would say:

     LEFT - N on 1st Street, NW on Gamma Blvd.
     RIGHT - NE on A Ave., SE on Gamma Blvd.

At point V, the signs would say:

     LEFT - N on 2nd Street, NE on A Ave.
     RIGHT - S on 2nd Street, SE Gamma Blvd.

At point W, the signs would say:

     LEFT - N on 2nd Street, NW on Beta Blvd.
     RIGHT - NE on A Ave., SE on Beta Blvd.

At point X, the signs would say:

     LEFT - N on 3rd Street, NE on A Ave.
     RIGHT - S on 3rd Street, SE on Beta Blvd.

At all four points, you would choose to continue driving northeast on A Ave. The numbered streets are increasing, which is correct, and the Greek letters are decreasing, which is also correct. You know you're going the right way.

At point Y, the signs would say:

     LEFT - N on 3rd Street, NW on Alpha Blvd.
     RIGHT - NE on A Ave., SE on Alpha Blvd.

You've reached Alpha Blvd, so you're going to turn on it. Which way? Northwest or southeast? You're at 3rd Street. Not far enough east, so turn southeast on Alpha Blvd.

At point Z, the signs would say:

     LEFT - NE on A Ave., N on 4th Street
     RIGHT - SE on Alpha Blvd., S on 4th Street

You've almost arrived. A right turn puts you onto 4th Street and Alpha Blvd., the block you are looking for.

It would take a little getting used to, but I don't believe this kind of navigating in insurmountably complex.

Would it take longer to travel on a hexagonal grid?

If one is traveling in a straight line on the rectangular grid, the hexagonal grid is 15% longer.

But if one is traveling in a "straight" line on the hexagonal grid (at an oblique angle), the rectangular grid is about 18% longer.

I tested out a bunch of random points; it more or less seems to even out. Sometimes the rectangular grid is shorter. Sometimes the hexagonal grid is shorter. Here's the results of 10,000 random trials:

Averaging them all together came up with that the hexagonal grid is about 0.3% longer, but that really might just be variability of taking a random sample. I'd say it just about evens out. So, are travel times the same?

One thing to consider is that the intersections on the hexagonal grid are faster, three-way intersections as opposed to the slower, four-way intersections in a rectangular grid. In a three-way intersection, there must be three cycles. Each right turn is allowed in 2/3 cycles; each left turn is allowed in 1/3 cycles. In a four-way intersection, there must be four cycles. Each right turn is allowed in 2/4 cycles (or 3/4 cycles if right-on-red is allowed); each left turn and each straight is allowed in 1/4 cycles. Comparing the two types of intersections, left turns are 33% faster in the three-way intersection; assuming no rights-on-red, right turns are also 33% faster in the three-way intersection. That means that the three-way intersections are 33% faster overall.

Because the travel distance are just about equal, the hexagonal grid is actually going to be faster due to the faster intersections.

Maybe bees are onto something.

As one more thought, it'd be quite easy to make park blocks at a ratio of 1/7 blocks.

No location would be more than 2 blocks from a park.

GeoGebra 5 -- into three dimensions

Last month GeoGebra, the free multi-platform graphing and math program, came out with version 5, which includes three-dimensional graphing. As a geometry and calculus teacher, I was ecstatic and immediately set out to do whatever I could do. Here are some of the things I made for my calculus classes:

This is a standard maximize the volume of the cone question. Here's the GeoGebra file.

There's lots of fun stuff to do with the volumes by known cross-section:

The slices are semi-circles, and the slicing plane makes that clear. Here's the GeoGebra file.

Here is a solid where the cross-sections are equilateral triangles:

Here's the GeoGebra file, and a video on how I made a similar construction.

One can also set up rotational solids pretty easily:

Here's the GeoGebra file, and a video on how to construct it.

And here is a more complicated region:

Here's the GeoGebra file.

And my previous post had a nice GeoGebra demo on conics (made by a colleague, not me).

Formulas for any transformation

As a geometry teacher, I find that I often want to give my students a transformation rule and see if they can identify the type of transformation. But making up transformation rules one-by-one got tiring. So I sat down and figured out the general form of each kind of transformation rule: the general form for any reflection, the general form for any rotation, and the general form for any dilation.

Here are the results. Enjoy!

Transformation Rules

Centroids

In a beginning geometry course, students can find the centroids of triangles by constructing two or three medians. As an extension, I ask my students to think about finding the centroid of a non-special quadrilateral. Most quickly recognize that medians of a non-special quadrilateral are not concurrent, so that method is out. I offer the suggestion to treat a quadrilateral as two triangles:

A few stall out, but a couple students think to look at the other two triangles.

This method is very intuitive and lends itself immediately to the solution --

-- but this method is not scalable. Go ahead; try it for a pentagon. I dare you!

The key to a scalable method relies in understanding that the first pair of centroids is sufficient to find the center of gravity. The solution must lie on this line:

Furthermore, the center of gravity must be positioned on that line proportionally to the relative areas of the two triangles. If the two triangles are equal in area, it must be the midpoint. In my example, the left triangle has an area of 11 sq. units, and the right triangle has an area of 13 sq. units. Therefore, the center of gravity must be 13/24 of the way from the left triangle's centroid to the right triangle's centroid. In effect, this treats each triangle as a point-mass.

This is a method that will work for pentagons. Here is an example:

The centroids of each triangle are marked. Now all one has to do is use the three point-masses (location and area) and then find their collective center of gravity. Is there a way to do this simply with coordinates? Yes. And it is one of the more surprising results I think I have ever seen.

*   *   *

Let's start with finding the centroid of a triangle. Is there any easy method, using the coordinates, to find it?

The easiest way to find the centroid is by vectors. The centroid is 2/3 of the way from A to D, the midpoint of side BC. That means we need to find 2/3 of vector w and add it to point A. One can show in several different ways that w = 0.5 u + 0.5 v. Using this, one can show that E = u/3 + v/3 + A.

Now it is possible to put everything in coordinates. If $A=(x_1,y_1), B=(x_2, y_2), C=(x_3, y_3)$, then . Thus, the coordinate of E is given by:

$\left(\frac{1}{3}\left(x_2-x_1\right)+\frac{1}{3}\left(x_3-x_1\right)+x_1,\frac{1}{3}\left(y_2-y_1\right)+\frac{1}{3}\left(y_3-y_1\right)+y_1\right)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

It is quite easy to find the centroid of a triangle.

*   *   *

How about area? Is there an easy way to find area using coordinates?

Once easy way to find the area using the coordinates is to think about vectors. Specifically, look at two vectors: u = a, b and v = c, d.

How do we find the shaded area between these two vectors? Copy the two vectors again to make a parallelogram:

Now box it in with a rectangle. The rectangle has an area of (a+c)(b+d). Now label all the other parts in terms of a, b, c, and/or d.

Therefore, the area of the parallelogram is (a+c)(b+d) - 2ad - 2 (0.5 ab) - 2 (0.5 cd), which simplifies down to bc - ad. Thus, the area of the triangle formed by vectors u and v is half the area of the parallelogram. Since we know that $a=x_2-x_1, b=y_2-y_1, c=x_3-x_1, d=y_3-y_1$, we can find the area of the triangle very simply in terms of the coordinates:

$\frac{\left|x_2{y}_1-x_1{y}_2+x_3{y}_2-x_2{y}_3+x_1{y}_3-x_3{y}_1\right|}{2}$

This pattern is remarkable, and it extends to any polygon. For example, try it on a quadrilateral.

Set up the areas for each triangle, and the amazing part is that two terms from the first triangle will cancel out two terms from the second: $x_2{y}_4, x_4{y}_2$. That leaves the area formula:

$\frac{\left|x_2{y}_1-x_1{y}_2+x_3{y}_2-x_2{y}_3+x_4{y}_3-x_3{y}_4+x_4{y}_1-x_1{y}_4\right|}{\mathrm{2 }}$

This formula extends easily to any polygon.

*   *   *

Now we are ready to combine it all together. In overall form, the first step to find the center of gravity of a pentagon is to divide it into three triangles, find the centroid of each, and find the area of each. Given three point-masses, the center of gravity can be found by weighting the triangle centroid formula, like so:

$\left(\frac{a_1{x}_{c1}+a_2{x}_{c2}+a_3{x}_{c3}}{a_1+a_2+a_3}, \frac{a_1{y}_{c1}+a_2{y}_{c2}+a_3{y}_{c3}}{a_1+a_2+a_3}\right)$

Now all that remains is to substitute in.

It is an amazing monstrosity to try to simplify it. I only did the x-coordinate because it takes so much time. Here is the result:

Centroid

It is one of the most surprisingly simple results. This works for a polygon that does not intersect itself, whether it be convex or concave.

I also put together another post on centroids and area.

Proving the Pythagorean theorem with similarity

Here is a right triangle, split along the altitude to the hypotenuse.

This creates three similar triangles: ABD, CAD, and CBA. Using similarity, you can show

$$a^2=cx$$
and also
$$b^2=c^2-cx$$
which is all you need to show that
$$a^2+b^2=c^2$$
The neat part is that it requires no argument about areas at all.

Pennies

I tried out Dan Meyer's pennies problem. I liked it.

One criticism I have is, why not include circumference too? That way students can see that one pattern is linear, while the other is not.

Here's one way I thought about the connection between diameter and circumference:

The diameter is 9 pennies; the circumference is 25 pennies. The key to the pattern is looking at the circle of diameter 8 pennies.

So the circumference turns out to be pi*(d-1) pennies, which is about 25 pennies. Seeing the circumference in this way actually gives students a real insight into the area:

Area = previous area + circumference. (This is why area is quadratic: the derivative -- the differences, i.e., circumference function -- is a linear function.)

One more extension question that I love: Imagine a ring of pennies around a tennis ball. If you want the pennies to be one penny farther away from the tennis ball, how many pennies do you have to add to the ring? Imagine a ring of pennies around the moon. If you want the pennies to be one penny farther away from the moon, how many pennies do you have to add to the ring? Because circumference is linear, the answer is counter-intuitively the same for both!

Looking for Pythagorean pair-doubles

In teaching a geometry course, I find that I frequently need a Pythagorean triple. A Pythagorean triple, of course, is three whole numbers that satisfy the Pythagorean theorem to make a right triangle: 3, 4, 5, or 7, 40, 41.

But even more frequently, I find I need a Pythagorean pair-double: two sets of whole number-length legs with the same hypotenuse. For example, 5, 5 and 1, 7 are a pair-double, since both have the same hypotenuse. In this way, I can make an isosceles triangle that requires a little work to prove it isosceles.

Here's the first ten pair-doubles.

"Eccentricity" of a triangle

How do you measure the "eccentricity" of a shape? If one examines a triangle, how deformed is it compared to an equilateral triangle?

Lots of ideas suggest themselves, but the simplest suggests comparing a two-dimensional to a one-dimensional property: area to perimeter.

Specifically, there are three possible formulations that come to mind:

1. $\frac{perimete{r}_{EQ▵}}{perimete{r}_{▵}}$, where the equilateral and deformed triangles have equal areas.

2. $\frac{are{a}_{▵}}{are{a}_{EQ▵}}$, where the equilateral and deformed triangles have equal perimeters.

3. $\frac{perimeter}{\sqrt{area}}$

This ratio can be calculated for both triangles; the equilateral triangle has the lowest possible ratio for a triangle, so the closer a triangle's ratio is to the equilateral triangle's, the less deformed it is.



I created three triangles and ranked them using each formula. All three formulas yielded the same result: triangle ABC was the most deformed, followed by triangle EDF; triangle GHI was the least deformed. This accords visually.

Given that all three formulas work, I would use the latter, since it is the simplest. Knowledge of an equilateral triangle's ratio is unnecessary. In fact, this formula can be used to compare the deformity of two quadrilaterals, two pentagons, etc. It was the formula I used to put a statistic to how deformed gerrymandered congressional districts were.