Monthly Archives: March 2012

Several Sneaky Circles

Geometry, , ,

In last week’s exploration we uncovered many equilateral triangles hiding in a general triangle ABC. This week we’ll find circles.

Draw any triangle ABC. The line AHA perpendicular to BC is called the altitude though A, with HA at the foot of the altitude. (See the image below.) It can be shown that the three altitudes AHA, BHB, and CHC meet at a single point, H, called the orthocenter of triangle ABC.

Now, let’s put dots at a few important points on our triangle: First, dot the three altitude feet, HA, HB, and HC. Next, dot the midpoints of the sides of the triangle, which we’ll call MA, MB, and MC. Finally, dot the Euler points EA, EB, and EC, which are halfway between H (the orthocenter) and the vertices of our triangle. Today’s first miracle is that all nine of these dots lie on a single circle, aptly named the 9-point circle:

The 9-point circle
The 9-point circle of a triangle is the circle that miraculously passes through the triangle's altitude feet, edge midpoints, and Euler points.

Recall that it only takes three points to determine a circle, so nine on a single circle is very far from coincidence!

Another way to coax a circle out of triangle ABC is to “inscribe” one in the triangle by drawing a circle tangent to the three sides. This circle is called the incircle. Our second miracle is that the incircle and the 9-point circle, which were constructed in very different ways, nevertheless play nicely together: they are tangent! The point where they touch is known as the Feuerbach point of triangle ABC, named after Karl Feuerbach who discovered and proved this difficult fact in 1822.

The Feuerbach point
The 9-point circle and the incircle are miraculously tangent at the Feuerbach point.

But there are three other circles that, like the incircle, are tangent to all three lines of the triangle. These “exscribed” circles are called the excircles of triangle ABC, and as luck (or extraversions[1]) would have it, these are also tangent to the 9-point circle:

Tangencies of the 9-point circle with the incircle and excircles
The 9-point circle is also tangent to the excircles.

But wait, there’s more! It turns out that triangles ABC and HBC have the same 9-point circle, and so if we look at the incircle and excircles of triangle HBC, these will also be tangent to our 9-point circle. The same is true for triangles HCA and HAB, so we have found 16 circles all tangent to the nine-point circle!

16 tangencies with the 9-point circle
The 9-point circle is tangent to the incircle and excircles of triangles ABC, HBC, HCA, and HAB. These four circles for triangle HBC are shown in thick red.

Notes

  1. Recall that we used extraversions last week to go from one Morley triangle to another. Here, extraversions interchange the incircle and excircles. [↩]

Many Morley Triangles

Geometry, Proofs, , ,

Given my blog title, how have I gone this long without discussing triangle geometry? I will rectify this gross negligence in the next few weeks. Let’s begin with a seemingly impossible fact due to Frank Morley.

Start with a triangle ABC, with any shape you wish. Now cut its angles into three equal parts, and extend these angle trisectors until they meet in pairs as illustrated below, forming triangle PQR. Morley’s amazing theorem says that this Morley triangle, PQR, will always be equilateral!

Morley's theorem

Why would this be true? Well, Morley’s theorem tells us that this diagram has three nice 60-degree angles in the middle, but we may suspect that, in fact, all of the angles are nice! This key insight lets us piece together the following argument, where we build up the diagram backwards from its constituent pieces. Draw seven separate “puzzle piece” triangles with angles and side-lengths as shown below, where the original triangle ABC has angles \(3\alpha,3\beta,3\gamma\) respectively. (Be sure to check that puzzle pieces with these specifications actually exist! Hint: use the Law of Sines.)

Proof of Morley's theorem
Triangle ABC is built out of seven "puzzle pieces" with angles and side-lengths as specified here. In this diagram, \(\alpha^\prime=\alpha+60\) and \(\alpha^{\prime\prime}=\alpha+120\), and similarly for \(\beta\) and \(\gamma\).

Now, fit the pieces together: all matching edge-lengths are equal (by design), and the angles around vertex P add up to \(60+(\gamma+60)+(\alpha+120)+(\beta+60)=360\) and similarly for Q and R, so the puzzle fits together into a triangle similar to our original triangle ABC. But now these pieces must make up the Morley configuration, and since we started with an equilateral in the middle, we’re done with the proof![1]

But there’s more to the Morley story. Let’s push A and C to make them swap positions, dragging the trisectors and Morley triangle along for the ride (a process known as extraversion). We end up using some external angle trisectors instead of only internal ones, but the Morley triangle remains equilateral throughout. This gives us a new equilateral Morley triangle for our original ABC!

New Morley triangles through extraversion

In fact, each vertex of ABC has six angle trisectors (three pairs of two), and if you continue applying extraversions you’ll soon uncover that our original triangle ABC has 18 equilateral Morley triangles arranged in a stunning configuration! (Click for larger image.)

All 18 Morley triangles

Here’s a challenge: the diagram above has 27 equilateral triangles, but only 18 of them are Morley triangles (i.e., are made from a pair of trisectors from each vertex). Which ones are they?

Notes

  1. This is a slight modification of a proof by Conway and Doyle. [↩]

Logic Under Construction

Algebra, Math Foundations, Proofs, ,

In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed:

Theorem: There exist irrational numbers \(x\) and \(y\) with the property that \(x^y\) is rational.

However, our proof was nonconstructive: it did not pinpoint explicit values for \(x\) and \(y\) that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim \(x=\sqrt{2}\) and \(y=\log_2 9\) work, because \(\sqrt{2}\) we already know to be irrational, \(y=\log_2 9\) can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational.

Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation:

Proof: Assume the theorem were false, so that any time \(x\) and \(y\) were irrational, \(x^y\) would also be irrational. This would imply that \(\sqrt{2}^{\sqrt{2}}\) would be irrational, and by applying our assumption again, \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\) would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED.

So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “\(x\) and \(y\) exist”, but instead only for the negative statement “\(x\) and \(y\) don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements.

But wait; last week’s proof that \(\sqrt{2}\) is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that \(\sqrt{2}\) is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that \(\sqrt{2}\) is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: \(\sqrt{2}\) is unequal to every rational number \(a/b\) because $$\left|\sqrt{2} – \frac{a}{b}\right| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality![1]

PS. In case you are still wondering whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.

Notes

  1. This inequality would also have to be proven in a constructive manner. See these Wikipedia articles for more information: Intuitionistic logic (another name for Constructive logic) and Square root of 2: Constructive proof. [↩]

Methods of Irrationality

Algebra, Math Foundations, Proofs, ,

Being a mathematician requires you to think in strange ways.

For starters, you might think that mathematicians spend all day pondering purely theoretical things that only exist in Mathematical abstraction, right? Well, it can get even stranger than that: sometimes we have to think about things that don’t exist, even in the abstraction! It’s called a proof by contradiction, and here’s an example:

Theorem: The number \(\sqrt{2}\) is irrational.

Proof: To prove that \(\sqrt{2}\) is irrational, we have to show that we can never write it as a ratio of integers, \(\sqrt{2}=a/b\). So let’s assume we can find such a fraction and see where it leads us.

Let’s cancel common factors in the numerator and denominator so that \(a/b\) is in lowest terms. Squaring our equation shows that \(a^2=2b^2\), so \(a^2\) is even and so \(a\) itself is even. Since \(a/b\) is in lowest terms, \(b\) must be odd.

Since \(a\) is even, \(a^2\) is in fact divisible by 4. On the other hand, since \(b\) is odd, \(2b^2\) is not divisible by 4. But then the integer \(a^2=2b^2\) is both divisible by 4 and not divisible by 4, which is absurd! The only explanation for this contradiction is that our original assumption—that \(\sqrt{2}\) is rational—is false. So we’re done with the proof! Notice that we spent most of our effort reasoning about integers that never existed (specifically \(a\) and \(b\)).

But that’s not the only twisted thing about Mathematical thinking. Here’s a delightfully short yet aggravatingly unsatisfying proof:

Theorem: There exist irrational numbers \(x\) and \(y\) such that \(x^y\) is rational.

Proof: We already know that \(\sqrt{2}\) is irrational, so maybe we can use \(\sqrt{2}^{\sqrt{2}}\). Is this number rational? If it is, then we’re done: \(x=\sqrt{2},y=\sqrt{2}\) solves the problem. But what if \(\sqrt{2}^{\sqrt{2}}\) is irrational? In this case, I claim \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works: indeed, $$\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot\sqrt{2}} = \sqrt{2}^2 = 2,$$ which is rational. In either case the required numbers \(x\) and \(y\) exist, so this completes the proof.

But wait; which is it? The proof shows us that one of the pairs \(x=\sqrt{2},y=\sqrt{2}\) or \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works, but it doesn’t tell us which one! So, have we proven the theorem? Yes, technically, but only nonconstructively.

Frustrating, isn’t it?