The Nine-Point Circle

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Just for fun, my Geometry class and I decided that we would do a team Blog about this interesting result from Euclidean Geometry: Feuerbach's Nine Point Circle.  We're going to work on this in a leisurely fashion over a long period, so please check in from time to time to see how far we have gone.  We're planning to put in lots of pictures, to make the geometric reasoning clearer, but since it is being written "by committee", it will probably show the usual signs of Committeeishness.

There are a few terms that some readers might not recognize.
+ Vertex: this means the corners.  In the triangle ABC, the vertex A is just the point A, and so on.  The plural is vertices.
+ The foot of a perpendicular.  A perpendicular is from a point to a line.  The foot of a perpendicular is the point where it hits the line to which we're drawing the perpendicular.
+ An altitude is a perpendicular from a vertex to the opposite side.


Introduction
The Nine-Point Circle Theorem is an interesting result in Euclidean Geometry, having to do with a circle that that passes through six important points on any triangle.  Every triangle has several important points associated with it, and usually these points have little to do with each other.  But it just so happens that someone discovered that six of them all lie on a circle.  Furthermore, it turns out that there are three more relatively unimportant points that also lie on this circle.

To explain the points and their significance, we show them in a sequence of diagrams below.

First, we show the midpoints of each of the sides.  We indicate these in RED.

Next, we show the feet of the altitudes from each vertex to the opposite side. We show these in GREEN.

Incidentally, this sketch illustrates that the altitudes meet at one point, which is called the orthocenter, shown as O below.

Finally, we show that the points that lie midway between the orthocenter and the vertices also lie on the circle; we show these in PINK.

And now, the moment you’ve all been waiting for: the actual circle:

We shall actually prove that these points lie on a circle (though it obviously does, according to the picture).

The proof of the existence of the 9-point circle is based on two previous theorems.

The first of these is the Mid-Point Theorem, which says that if XYZ is a triangle, and P is the midpoint of XY, and Q is the midpoint of XZ, then

(i) PQ = 1/2 YZ, and
(ii) PQ is parallel to YZ.

The proof of this is not difficult.

Let XYZ be a triangle, and let P and Q be the midpoints, as described above.  Consider the diagram at right.

To prove this theorem, we need a construction.  Extend PQ to point R, in such a way that PQ and QR are congruent (i.e., equal in length).  Join ZR.  Now triangles PQX and RQZ are congruent by "Side-Angle-Side".

Angles XPQ and ZRQ are congruent by Corresponding Parts.  Using the Alternate Angle Theorem, we know that lines XPY and RZ are parallel.

Consider the second diagram.  As you can see, XP, RZ, and PY are all congruent.  There is a result that says that if PY and RZ are both parallel and congruent, then PR and YZ are also both parallel and congruent.  It also means that the length of PQ is half of the length of YZ!

The second result we need is the interesting fact that if UV is the diameter of a circle, and if it is a side of a triangle whose third vertex, W, is on one of the semicircles, then angle W will be a right angle.  We give a diagram; the result follows from a little angle-chasing (notice that there are two isosceles triangles in the figure).

A Little Fun Geometry for Y'all

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I’m surprised to discover that people enjoyed Geometry a lot more than I thought they did.  Even more surprisingly, it appears that math people hate Geometry, while everybody else either likes it, or has a mild dislike for it, nothing like the deep dislike that some of my mathematics students seem to have!

Because of the great variety of attitudes towards the subject, people in different places seem to have very different degrees of geometry experience, and what I’m posting today might be old hat to some, and quite refreshingly new to others.

Triangles are very interesting things, mostly because they provide a framework for lots of other structures.  One very easy exercise is to make a circle that goes through all three points (vertices, or corners) of the triangle.

Constructing a Circle through the Three Vertices of a Triangle

The principle is simple, if you take two points at a time.  Suppose our triangle is ABC.  Considering just A and B to start with, there is a line of points an equal distance from A and B, as shown at right.  Each of these points might be a different distance from A, but they’re the same distance from B.

We can do the same with the points B and C; again we get a whole new line of points the same distance from B and C.  Now, it is clear that these two lines of points have one point in common!  Let’s call it O.  This point O is the same distance from A and B, and the same distance from B and C.  (As we say in mathematics, by transitivity, it is the same distance from C as from A, but we won’t fuss about that.)  It follows that we can draw a circle centered at this point O which will pass through all three of A, B, and C.

How does one make the first line, which is all the points equal distances from A and B?  Easy: join AB, find the midpoint.  So this is one of the points.  Now make a perpendicular; that gives us the whole line.

When anyone sees the triangle with the circle around it, it merely looks as if someone had taken a circle, and picked three random points on it, and joined them to make a triangle.  It is a little more convincing to have the three points  move on circles of their own, and observe the circle around the triangle follow along.  Even here, the eye is deceived into believing that the circles are the fundamental thing, but a little thought would make you see that this is unlikely, since it is difficult to make a circle dance.  Here is a video showing the dancing triangle, with its circumcircle.  The music is by J. S. Bach, and it is the (organ) Fugue in A minor, BWV 543, played using a Marimba sample.



[Without appearing to obsess over whether and why people have a strong aversion to mathematics, I’d like to pursue what the root causes might be, perhaps in a later post.

My suspicion is that it is partly because people give up too quickly on mathematics, which leads to the second problem, namely that Mathematics is a vast area of knowledge, and a lot of it is at the intermediate level, about the level of high-school study, so that most laymen have already abandoned relating to the subject by that time.  Of course, an enormous amount of mathematics is at the advanced level, and this part of mathematics is very diverse, because it has all sorts of made-up mathematics, created almost exclusively for the purpose of getting someone a doctorate or a publication, and of no use whatsoever, but the rest of it is quite successful attempts, I have to concede, to organize and illuminate what is already known, by underscoring the commonality of it.

But this middle level has been around for ages.  It just so happens that only a small part of it has to actually do with numbers and arithmetic; the rest is all about logical relationships.  This is not exactly your everyday logic, but more complex reasoning.  Here is an example: for every pair of fractions you can give me, I can give you a number that lies between them that is not a fraction.  That sentence is quantified, which means that the phrases “for every” and “I can find you” are used, which are definite, but not the normal logical equipment people use.]

Arch

How to prove SSS!

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Hello there, all you eager young Geometers!

As many of you know, there are lots of ways to prove that two triangles are congruent:

"Side-Angle-Side", or SAS: given two triangles, prove that two sides of one are congruent to two distinct sides of the other, and prove that the angle between the one pair of sides in the first triangle is congruent to the corresponding angle in the second.

"Side-Side-Side", or SSS: prove all three sides of one triangle are congruent to the sides of the other.

"Angle-Side-Angle", or ASA: prove that two angles in one triangle are congruent to two angles in the other, and prove that the common side in the first triangle is congruent to the corresponding side in the second triangle.

Now, in Geometry, ideally, we would like to prove everything. (Since we can't do this in real life, we like to do it in Geometry at least, to provide some feeling of security in our uncertain lives.) In particular, we would like to prove that all of the above criteria: SAS, SSS, and ASA, can be proved without assuming anything else. Unfortunately, as far back as 300 BC, roughly, Euclid realized that some things just can't be proved. Some projects like that are doomed to failure, including proving all three of SAS, SSS and ASA. Today, we use Axioms, which are springboard statements, from which we deduce as much as we can; the Axioms themselves, of course, have to be simply assumed.

Of the three criteria, SAS, SSS and ASA, it turns out that the easiest one to assume as an Axiom is: SAS. The other two can be proved from SAS! (and some simple theorems, such as the Isosceles Triangle theorem.)

Rest assured that we're not going to prove SSS here today, but the strategy is interesting! Here's the strategy.

We're given two triangles, ABC and DEF, and AB is congruent to DE, BC is congruent to EF, and the remaining pair of sides is also congruent.

Step 1. Make a third triangle! Copy triangle DEF onto the underside of line AC, using only SAS. In other words, copy the angle D over to the underside of A, mark off a line segment AB* exactly the same size as DE, and complete the triangle. Now DEF and this new triangle -- call it AB*C -- are congruent, by construction. Here's a picture of the situation now:Anyway, the rest of the work consists on proving that ABC and AB*C are congruent to each other. This is not hard, but involve using Isosceles triangles twice (the triangle with two red sides, and another triangle with two blue sides), and more serious work. I don't want to give the impression that things are as easy as just going ahead with the proof as described; for instance, we need to know where the line BB* falls; inside AC or outside? Still, the proof proceeds in essentially predictable ways.

Once we have shown that the two Siamese Twin triangles are congruent to each other, it follows that the two original triangles are congruent, using the principle of "transitivity", which says that if two objects are congruent to a common object, then they're congruent to each other.

So there you have it! Oh what a complex web we weave, when first we start proving theorems...

Archimedes, not the real one