This is a new geometry identity you haven't seen before. It's cute!
If you draw a random triangle, you can sketch a rectangle quickly and easily that has the same area as the triangle! In fact, there are three such rectangles, all easy to draw.
First, let's call the triangle the reference triangle, and abbreviate it as ∆R. Then, let's give the rectangle a name too. Let's call it the centroid cevian rectangle for reasons that will be explained soon, and abbreviate it as CCR.
The reference triangle is defined by three points. The three vertex points are A, B, and C. Doesn't matter if the triangle is acute, obtuse, or right angled, big, small... macht's nichts.
The triangle doesn't have a unique center, but one of its major center points is called the centroid.1 The centroid is the center of mass of the triangle. Any line drawn through the centroid divides the triangle's area equally. You could, if you were very very careful, balance the triangle on a very sharp knife edge if the knife edge were located directly underneath the centroid.
Just as there is no unique center to a triangle, there is also no unique diameter. But for a given center, what serves as a triangle diameter is a line drawn through it and through a vertex. Such a line is called a cevian. The cevian thus passes through a vertex, the triangle center, and then through a point that lies on the side opposite the vertex.
The centroid, conventionally labeled G by modern geometers, has three such cevians, one for every vertex.
The centroid's cevian has the happy coincidence of intersecting the opposite side exactly at its midpoint. So if we label the midpoint of segment BC
(which we usually call a, because we're lazy) as mA then this cevian passes through point G, and has endpoints A and mA.
Now it turns out that the other two midpoints, mB and mC, are equidistant from the cevian. So that's an interesting fact. (This doesn't happen often. But like a good detective, when you find out something like this, your geometric spider sense should start tingling!)
Let's construct a rectangle WXYZ whose length is the cevian, and whose width is the distance to the other two midpoints. A picture of such a rectangle is shown on my homepage.2
If you draw a suitably tortured looking triangle the rectangle and the triangle don't look like they could possibly have anything in common. But they do. Because, improbably,
(∆R) = (WXYZ)
which, in geometry-speak, means that the area of the reference triangle is equal to the area of the rectangle.3
Wowzers!
The first time I saw this - I happened to be doodling around on my favorite open source geometry application, C.a.R. - I did the standard Valley Girl LIKE, OH EM GEE!4 That can't possibly be true. It has to be a coincidence.
(Don't let any math geek fool you. When he or she stumbles across a truth, he or she gets this mystical feeling too. There are no inexorable marches toward mathematical truth. Mathematics isn't created by ultra super logical Mr. Spocks who aren't surprised. We're surprised as hell. It feels like we are on a mountain hike through a heavily treed area, and then we round the corner and get just a brief glimpse of a beautiful vista. Such moments happen exceedingly rarely, but when they happen we remember them vividly.)
So I did this again, this time with another cevian. Same thing. Then I moved the triangle vertices around, and watched as the software magically distended the cevian and scaled the CCR to the new dimensions of ∆R. Same thing.
So I've been given evidence that this is a universal truth. No matter what shape the triangle, the CCR of a triangle will always have equal area to the triangle. The next thing I have to do is that I have to prove it. Because this could be a coincidence. It may not always be true. To say something is true in math means you've proved it to be true, irrespective of shape of triangle, lengths of sides, etc. Proof is everything.
Now I am not a good mathematician. It comes painfully to me. We're talking Slowest Kid in the Class kind of slow. I never did well on proofs, and I don't particularly enjoy them. But thankfully, this particular little geometrical oddity was exceedingly easy to prove.
The key is that there's a point M which lies between mB and mC. It also lies on the cevian. And it's also the cevian's midpoint. In fact it's the center point of XWYZ. All points on the rectangle are reflected through this point M. It is called in the arcane world of geometry the center of homothecy, and the dilation factor is -1. All angles and areas are preserved with this dilation factor. The line segment mBmC cuts the rectangle in half. The area of the rectangle on the one side of this line segment is therefore equal to the area of the other side.
Now inside each half is a triangle. It's called the medial triangle, ∆M = mAmBmC. The medial triangle has sides that are parallel to those of ∆R. Four such medial triangles can fit inside ∆R. Therefore:
(∆R) = 4·(∆M)
So inside the quadrilateral WmB
mCZ is ∆M, which is 1/4 the area of ∆R. That means the other part of the quadrilateral that isn't ∆M must equal ∆M also. And therefore the rectangle WXYZ contains 4 times the area of ∆M, which equals 1/4 the area of ∆R. And so we have detectivized our way to the conclusion that (WXYZ) = (∆R). And now we get to say QED.5, 6
I've glanced through 12 years' worth of back issues of Forum Geometricorum, the online geometer's journal, and can't find anything related to this. Have done the Wiki check, the Google check. It appears to be a new identity.
I'm fully expecting some pencil-necked geek to snort and say, in a nasal and dismissive voice, "Oh, that was proved by Kohlchenkov in 1942. (snort)(snort)" which is subtext for: You Complete Amateur.
Which I am.
Still, wouldn't it be cool... to imagine that one day before I die, I could open a geometry book and find my little discovery as some high school homework problem? Be a small little footnote in the history of geometry?
NOTES
- The three other major triangle centers are the center of the inscribed circle, the incenter, the center of the circumscribing circle, the circumcenter, and the intersection of altitudes, called the orthocenter.
- If such a figure doesn't appear there, then please message me and I'll send you the original graphic. The reference triangle ∆R is in pale blue. The medial triangle ∆M is the darker blue triangle. The CCR is the pale green rectangle.
- Parentheses (·) are shorthand for "the area of"
- The exact thought bubble went something more along the lines of F***! (repeat a few times, with intensity. And possibly throw in a FUCKITY for effect.)
- ... and pretend we're mathematicians. Sit in some high backed leather chair, thin legs crossed, lighting a pipe, peering over our specs, and giving every indication that this was not something found by accident, but was something more like the end of some journey with a foregone conclusion. Try to look bored.
- Also try not to say WOW FUCKITY WOWZERS.