Hear See Think

Geometrically Finding the Area of a Trapezoid

I hope the picture says it all.  The last step uses the fact that the area of a parallelogram of base b and height h is simply bh, which can easily be proved geometrically by slicing off a triangle and sliding it to the other side to form a rectangle with the same base and height.

Sorry for the lack of posts recently, but they should perk up as the school year begins and I have lots of interesting things to think about all the time.

My good friend and partner-in-crime Daniel Walsh made this wickedly cool 3-manifold, a deformed hexagonal toroidal-ladder of twist-degree 1 (my own fake terminology).  Learn how he did it and follow Dan’s wild geometric projects at http://danielwalsh.tumblr.com/.

Works of Walter Russell, via but does it float.

Triangle Theorem Study

My first failed attempt at proving the equilateral triangle theorem in the post below.  Notice how I mistakenly thought an equilateral triangle’s centroid is located one-third up the side length of the triangle rather than the height distance, so I’m missing the needed factors of Sqrt[3]/2.  Also, I had labelled the three vertices of the starting triangle as z1, z2 & z3 rather than z1, z1 + z2 & z1 + z2 + z3.  Needless to say, the math refuses to work out with these problems.  This gave me fits for a little while.

Eagle-eyed viewers will also notice that I was thinking about the 2^32 estimation problem on this same piece of paper.

A Theorem Regarding Triangles

Start with any triangle at all.  You can construct an equilateral triangle on each of its three sides.  These three equilateral triangles have centroids, which you can connect to form another triangle.  Surprisingly, this last triangle is always equilateral, regardless of the starting triangle!

Here’s a video my friend Dan over at http://danielwalsh.tumblr.com/ made with Mathematica - it demonstrates (with wacky sound effects for good measure) that no matter how you move the vertices of the starting triangle, the resultant triangle of this construction is always equilateral.

I came across this pretty geometric result by modifying a construction found in Tristan Needham’s outstanding book Visual Complex Analysis.  In that book, one learns that if you construct squares on the four sides of any quadrilateral, and then connect opposite square centers, the two resulting segments are congruent and perpendicular.  I had to see what happens when you do the analogous construction with triangles, and sure enough the result is as nice as anyone could hope for.  Let’s now prove this fact.

Taking a cue from Visual Complex Analysis, we represent the arbitrary starting triangle using some nonzero complex numbers z1, z2 and z3.  One vertex will be z1, the second will be z1 + z2, and the third will be z1 + z2 + z3, and without loss of generality, the third vertex can be the origin of the complex plane, so that z1 + z2 + z3 = 0 is assumed to hold.  Now to represent the centroid of the first equilateral triangle, we go halfway to vertex z1, turn left 90 degrees, and go a distance of Sqrt[3]/6.  This is because the height of an equilateral triangle is Sqrt[3]/2 times the length of one of its sides, and the centroid is one-third up the height from any side.  The left turn in the complex plane is achieved via multiplication by i, which rotates the plane 90 degrees counterclockwise.  Therefore the first centroid is given by (z1)/2 + i(z1)Sqrt[3]/6.  See the picture:

The other two centroids are obtained from similar operations involving z2 and z3.  Here I define the centroids in Mathematica:             

Actually we could have turned right instead of left when finding the centroids of the three equilateral triangles of the construction - it makes no difference if these triangles are constructed ‘outward’ or ‘inward’ relative to the starting triangle.  This is reflected by the fact that i and -i are algebraic mirror images of each other.

Now the first side of our centroid-triangle will be the difference (as vectors in the complex plane) of c2 and c1; similarly for the other two sides.  In Mathematica I define:

We need to let Mathematica know that z1 + z2 + z3 = 0 holds for all the expressions input above - otherwise our triangle may actually be a degenerate line segment.  This can be accomplished with the slash-dot & right-arrow command, which replaces certain strings with other strings.  In this case we use /. z3 -> -z2 -z1 because z3 = -z2 - z1 directly follows from z1 + z2 + z3 = 0.

Now, a slick way to show that the triangle with sides s1, s2 and s3 is equilateral is to show that any of those sides can be obtained by performing a rigid-body motion on one of the others.  Witness:

What this says is that s1 times (-1/2 + i Sqrt[3]/2) equals s3, and also that s2 & s1 and s3 & s2 stand in this same relation.  What’s the significance of (-1/2 + i Sqrt[3]/2)?  It’s the complex number that rotates the plane 120 degrees counterclockwise - in particular, multiplication by (-1/2 + i Sqrt[3]/2) is a rigid-body motion.  So, this settles the theorem!  But here’s another, more direct way to see it: by taking norms.

So s1, s2 and s3 all have the same absolute value or length.  Without Mathematica, it’s much easier to work out the rotation approach on paper than crunching through the absolute values.  But I don’t think any conceptual understanding is lost in employing the software for the grunt work.

I wonder what we can say about constructing regular pentagons on the five sides of an arbitrary pentagon?  Or regular n-gons on the n sides of an arbitrary n-gon?  Or what happens in higher dimensions?

And isn’t it awesome that imaginary numbers help us prove this purely geometric theorem about triangles?

                                             Like, woah, dude.

A solution

leafdude:

hearseethink:

A cylindrical hole 6 inches long has been drilled straight through the center of a solid sphere. What is the volume remaining in the sphere?

(Yes, there’s really enough information to solve this!)

 $$36\pi \hspace{2mm} \text{in}^3.$$

(Admittedly, I knew about the Napkin ring problem before answering…)

Correct!  Although you may as well have said “I knew the answer before answering”, as that Wiki article gives it away completely!  This picture I found (which luckily is a perfect illustration of the problem, down to the numbers) makes it clear what kind of geometry one has to perform to solve this.  Nothing too grueling; not even calculus is required!

(via leafdude-deactivated20110909)

Infinite Whitehead Tangle

This never-ending knot, animated by Kenneth Baker, was made by concatenating a sequence of smaller tangles that look like this:

A quarter-turn is applied when passing from one copy of this tangle to the next in the infinite sequence.  Here’s another view:

Sketches of Topology is an amazing blog for visuals like these, and I will probably crib from it frequently.

Klein’s Quartic Surface

Klein’s Quartic Surface turning itself inside out.  This surface can be constructed from the equation u^3*v + v^3*w + w^3*u = 0.

Animated by Greg Egan.

edit: I’m glad this is getting some reblogs, but I cringe every time I see the error “Quartet Surface” replicated!  I wasn’t able to fix that typo before it started spreading.  If you care about this kind of thing and reblogged the mistake, help out and fix it!  Only takes a quick second. :)