Hi, my name is Face, and I'm a geometry addict.
I had been doing pretty good, you know, staying on the program...
... but then...
But then this triangle showed up. It was acute.
Oldest joke in history. That shit's older than Euclid. Take a seat.
Okay, okay. Here's what really happened. It started when I ran across this PDF by John Horton Conway about barycentric coordinates. His stuff is extremely elegant, and his mathematics is clean. There's a picture with a triangle, base down, and a vertex that's pointing up. A vertical line descends from the top vertex to somewhere along the base. The two sides of the triangle are labeled a and b. The two parts of the base are labeled x and y. (The x part of the base is attached to the side with length a.)
Underneath the triangle is a deceptively simple caption that reads: "If the vertical line is an angle bisector, then a/b = x/y."
So I did what any geometry addict would do, and spent the next two days proving that.
You're an idiot. We would have just Wiki'd Angle Bisector Theorem and gotten on with our lives.
You're extremely annoying. Where's the love?
If you want sympathy, it's under S in the dictionary.
Yes, well. I spent some good quality time with paper and pencil and re-proved the Angle Bisector Theorem.
Euclid must be sleeping well, knowing that an engineer re-proved something he'd done 2400 years ago.
Oh, stop crying. Don't be such a big girl's blouse.
Right. As I was saying...
Get out a clean white piece of paper and draw a triangle on it.
You can measure the side lengths and the angles inside. Then you can compute the area of the triangle if you know an altitude (the shortest distance from one vertex to the opposite side). K = 1/2 x base x height.
This is apex of knowledge of geometry for most people. But if you're a geek, this is only the beginning.
If you're a geek, and you measure the sides of the triangle, you wonder if you can compute the area of the triangle from just the knowledge of the three side lengths alone. Turns out, you can. The formula is called Heron's formula, and it's remarkably simple. If the three lengths a, b, and c are 3, 4, and 5, then first you compute the half-perimeter, s, which is 6. Then Heron's formula for the area of the triangle, K, is
K = sqrt(s⋅(s-a)⋅(s-b)⋅(s-c))
Wow! That was simple!
You're feeling a vague tingling in your nether regions. Oh, but wait. It gets better, baby. Way better.
It turns out that all Pythagorean triple triangles (which have sides of length c² = a² + b², where a, b, and c are integers) have integer values for their areas. The first few Pythagorean triples (3,4,5), (5,12,13), (8,15,17), and (7,24,25) have areas 6, 30, 50, and 84. Two different Pythagorean triple triangles can have the same area, e.g. (20, 21, 29) and (12, 35, 37).
So you've still got that white piece of paper on your desk, and it's still got that single, unadorned triangle drawn on it. Now if you're a math geek - and I know you want to be - then you have a natural desire to draw the biggest circle inside that triangle so that it touches all sides. This is called the incircle. Try to draw it.
You wonder, in a somewhat desultory manner, what the area is, and what its radius is. You know only the side lengths, a, b, and c. Can you find the radius from just the side lengths? How about the area?
Turns out, you can! The radius of the incircle, lowercase r, is equal to
r = K/s
= - * -----
The first expression means that the radius of the incircle is equal to the area of the triangle K divided by its semiperimeter s. I honestly don't know which is a lovelier expression, that first one, or the second one. But look: with one little division you can find the radius of this circle you just drew! For a (3,4,5) triangle, it's just 6/6 = 1! Wow, that was surprisingly simple, and you didn't really even need to do any complicated integrals or stuff.
So then you wonder, What's the triangle with an inscribed circle having radius 2? 3?
The answer is...
Circle Associated Triangles
Radius (Pythagorean Triples)
3 (8,15,17) and (7,24,25)
5 (12,35,37) and (11,60,61)
You might be curious where the center of the circle is located. I'll just tell you where it's located, and then you can draw it and verify it to make sure I'm not lying to you.
It turns out that the center of this "incircle" (the geometrical term for the circle inside the triangle) is the intersection of the three lines that are the angle bisectors of each of the corners of the triangle.
How to draw an angle bisector: From each vertex, draw a line very carefully that you estimate cuts the angle in half. (Just freestyle it. No need for protractors. An eyeball draw of this is good enough, unless you've got extremely shaky hands.) Do this for every vertex. Now you have three lines.
If you were careful enough, you would note that all three lines magically converge to the same point. Wow again! Is this coincidence? Turns out it isn't.
Turns out, if you draw any triangle and you draw the three lines that bisect the interior angles, the three lines will always meet in the same point. Always. For any triangle. Acute, obtuse, right, slightly left of center, moribund... I don't care what kind of triangle you draw, the angle bisectors will always intersect.
So you will agree with me that this is a pretty amazing fact, but facts is what we deal with in mathematics. Just the facts, ma'am. This point of concurrency is one of those amazing things in geometry that, if you're prone to addictions, may get you hooked on mathematics. It turns out there are lots more points of concurrency associated with a triangle.
For example, altitudes
Draw yourself another triangle. Then, from each vertex, draw a straight line down to the opposite side so that it meets the opposite side in a right angle. This is called the altitude - the "height" of the triangle from that side.) Go ahead and do this for all vertices. Since there are three vertices, there are three different altitudes. If you've been careful, you'll notice that all altitudes intersect at the same point. Another point of concurrency! This one is called the orthocenter,.
There was something tricky about that last thing we did, because it doesn't always seem to work for every triangle. For right triangles, the altitudes are the two sides that are at right angles to each other. In that case, the orthocenter is equal to the vertex where the right triangle is located.
But worse than that, if you've drawn an obtuse triangle (where one of the angles is bigger than a right triangle, and it looks long and narrow), you can't draw an altitude! Well, actually you can. But you have to extend the sides of the triangle out a bit further. Now drop the altitude from one of the vertices with the real small angles until it intersects with the triangle extension, and keep going a bit. Now do the same for the vertex that has the other small angle. It too intersects the side of the triangle but outside of the triangle. And if you extend the three altitudes, you'll find that all three intersect - except this time, the point of intersection is outside the triangle!
So here's an interesting property:
The orthocenter is the intersection of a triangle's altitudes.
- Acute - If the triangle is acute, the orthocenter is inside the triangle
- Right - If it's a right triangle, the orthocenter is the vertex with the right angle
- Obtuse - If the triangle is obtuse, the orthocenter is outside the triangle
How about one more point of concurrency? This one is called the centroid. The centroid is the point of concurrency of the lines of medians. A side's median point is its midpoint. If lines are drawn between medians and the opposite vertices, then - once again, the miraculous becomes the accepted norm - they all meet at the same point.
The centroid is also called the center of mass of a triangle. If a triangle is made of material that has uniform density, then you could balance this triangle on a pencil point, if the pencil point was located at the centroid. Because of this, the centroid is always located in the interior of the triangle.
The lines of medians (technically called the "Cevians of medians", due to the name of an Italian geometry, Giovanni Ceva) were drawn from the midpoints of sides to their opposite vertices. That's how we found the centroid. Now what if we draw perpendicular lines from the midpoints of the sides? What happens then? Would these perpendiculars also magically intersect?
Yes, of course they would. This is called "leading the witness" in a court of law. You don't think I'd ask the question and then say, "no, actually, they pretty much randomly intersect somewhere out there, and the pattern is not pretty," do you? I wouldn't do that to you. I'm your friend. This isn't one of those detective stories where there are lots of red herrings, and you're confused about who the suspects are, and the murder weapons, and so forth. That's for literary hacks. This is geometry. Geometry is next to godliness.
At any rate, if you find the midpoints of the sides of a triangle, and then you construct the perpendiculars from these points, then the intersections of these perpendicular bisectors all coincide at a point, and that point is the circumcenter. If you draw a circle from the circumcenter, you will see that all three points of the triangle are on this circle. Pretty amazing, eh?
So that's why it's called the circumcircle - because it's the smallest circle that can be drawn within which the triangle can fit, such that all three of its vertices touch the triangle. This circle circumscribes the triangle.
(I just note parenthetically that if the triangle is acute, the circumcenter is inside the triangle. If it's obtuse, the circumcenter is outside. And if it's a right triangle, then the circumcenter is the midpoint of the hypotenuse.)
OK that was cool, but look, there's more. We talked about the inscribed circle, which fits inside the triangle. And we talked about the circumcircle, where the triangle fits inside of it. Can I tell you that the radii of the circles are related? Would you be impressed? Well, be impressed. The incircle's radius is labeled r, and the circumcircle's radius is R. Now look at this formula for how they're related:
The radius of the inscribed circle is
r = K/s
The radius of the circumscribed circle is
R = (a*b*c)/(4*K)
So by sleight of hand and a bit of algebra (this won't hurt a bit...)
r*R = -----
... and if you remember that the semiperimeter is equal to half of the sum of the side lengths, which I'm sure you do, then the equation says that the products of the radii is equal to the one-half of the ratio of the product of the side lengths to the sum of the side lengths.
r*R = - * -----
Is that freaking hot, or what?
God is a geometer. There's no question about it. When he designed the laws of geometry, he smacked his hands together and smiled and said, "Damn, I'm good. Wait 'til the humans discover this. This is fine shit."
I've discussed the fact that each triangle has a unique inscribed circle and circumscribed circle. There are three more circles associated with a triangle. They're called "exscribed circles", or excircles. There is one excircle for each side of the triangle. Somewhat amazingly, the excircle is precisely defined by the triangle, because even though it touches the triangle at only one point, its radius is precisely defined by that contact point as well as the extensions of the sides of the triangle. (I've described this in excenter and in excircle.)
The overall picture of the triangle is that it's bound from within by its incircle, and it's bound from without by its three excircles, which look like airbags surrounding the triangle. So when you think of a triangle, think of the fact that there are always five circles associated with that triangle: The inscribed circle, the circumcircle, and the three excircles.
In like manner, the triangle's presence also spawns other triangles. There are triangles inside the reference triangle. For example, there's a median triangle, a triangle inside ΔABC that connects the three midpoints of the sides. There's the contact triangle that connects the three touch points that the inscribed circle makes with ΔABC. There's an orthic triangle, which connects the footer points that the altitudes make with the sides of the triangle, or its extensions.
And of course this is only the triangles that go inside the reference triangle. There are also triangles that are exterior to the reference triangle. Bigger. Smaller. Also, there are recursive relationships, like fractal figures, because you can ask what the median triangle of an orthic triangle looks like, nested one inside the other, or using exterior triangles and going outward from there.
As is the relationship between a reference triangle and its five circles, so it is between a reference triangle and its associated derivative triangles. If you see a reference triangle, you will also immediately see (if you're a geometer) the triangles that it spawns.
So I think of a reference triangle (our ΔABC) like a little electron in quantum mechanics. An engineer may see an electron floating about in empty space, its electric field lines extending straight and true off to infinity. The little electron is alone in the vacuum of space. But a physicist doesn't just see the lonely little electron. It sees the swarm of virtual particles that surround it like a buzzing cloud of insects around a porch light at night. They wink into and out of existence, subject to Heisenberg's uncertainty principle, living their lives in a minute fraction of a second, then winking out of existence again. This is the foam of the vacuum, alive with quasi-particles. The engineer sees a single, solitary particle, but the quantum mechanic sees something alive, dynamic, the particle doing its dance with the vacuum.
That's the way I see triangles. It may look like a single simple little triangle on a piece of paper, but to me, what instantly springs to existence is its partner circles and triangles, buzzing around it like a swarm of virtual geometric figures, extending its reality into something dynamic, a geometric space alive with theorems and relationships.
Here's a cool discovery I found a few days ago: If you draw a triangle ΔABC (black), and then you inscribe a circle inside of it, and then you draw the three exscribed circles centered on OA, OB, and OC, and then...
... and then you draw the four "tangent triangles" (shown in red)
- The intangent triangle ΔTATBTC connects the tangent points that the inscribed circle makes with ΔABC
- The three extangent triangles connect the exterior tangent points (between excircles and the reference triangle ΔABC with the extensions of the sides of the reference triangle: ΔT'ACABA,
... and draw one more triangle, the "excenter triangle" - the triangle that connects the centers of the exscribed circle centers, ΔOAOBOC (also in red)...
Then notice the following parallel lines (picture on my homenode):
The sides of the extangent triangles with the sides of the intangent triangle
- The sides of the extangent triangles with the sides of other extangent triangles
- T'BAB || T'CAC
- T'CBC || T'ABA
- T'ACA || T'BCB
The sides of the intangent triangle with the sides of the excenter triangle
- CABA || TBTC
- CBAB || TATC
- ACBC || TBTA
- CABA || OBOC
- CBAB || OAOC
- ACBC || OBOA
When I first saw this, I sketched out a few more triangles, then repeated the geometrical construction. The lines kept being parallel. Then I did the entire thing in Excel, and changed the dimensions of the reference triangle, and the same thing happened, for every possible reference triangle. FREAKING HOT.
This is amazing. I showed this to a friend who is a bit dubious about my mathematical interests - "unhealthy," was her summary - and even she agreed that this was a pretty discovery.
There has to be some mathematical equivalent to the misunderstood poet, or the misunderstood philosopher. Working away, alone, in his attic, for years, his mother always calling and saying, "John, you need to get out more. Have you seen that nice girl, Ellen? Whatever happened with her?" He feels like the world misunderstands his genius. Some times he feels bitter, and he thinks to himself that One Day They'll All Be Sorry They Didn't Recognize His Genius Earlier.
But this is not a mathematical proof. This is experimental mathematics. No proof - no real math. At this point it's merely IWhoSawTheFace's Conjecture. The world of mathematics is littered with conjectures that have proven to be false. Still... the pictures are pretty. It's too beautiful to be false, isn't it?
Beauty is the first test: there is no permanent place in this world for ugly mathematics.
-- G.H. Hardy, A Mathematician's Apology
And that is how you wind up becoming addicted to geometry. You make a discovery like this, and you're hooked.
You want More.