# Trisection points of a line segment where two faces meet

### Centroid & median proof (video) | Triangles | Khan Academy

and each solid torus is the preimage of the line segment joining this to the barycenter of a facet of ∆2. The vertices of the pieces are barycenters of faces and Two of the 3–dimensional trisection submanifolds meet the. The main point here is that after you do the bisecting or trisecting, you end up with Segment trisection: Two things (points, segments, rays, lines, or any. turn the trick. Since these two curves, the line and circle, were found In the same interval of time, the point. E moves strikes the line OB at a point 2/r units distant from 0. This content At L construct the parallel to OA which will meet the curve at C. The line OC .. *Lines EP cut the loop on both sides of its highest point.

This is the blue arc in the diagram. The circle will cross the two lines at two points: Now put the point of the compass at A and draw an arc of a circle, as shown in the diagram.

Without changing the radius at which the compass is set, move its point to B and draw another arc of a circle. These are the red arcs in the diagram.

If the arcs don't cross, then of course you need to use larger circles! Can you prove that this procedure works, using similar triangles? The ancient Greeks certainly knew how to do this. Trisecting an angle What about trisecting an angle? Why is it so difficult? For the general case, however, the Greeks couldn't work out how to do it, despite expending vast amounts of energy on the problem. Trisecting an arbitrary angle can be done if you cheat by using a measuring ruler instead of a plain straight edge you can find out how in the sci.

### Demonstration of the Archimedes' solution to the Trisection problem

However, to "play by the rules" you aren't allowed any marks on the straight edge at all - it must be completely blank. The problem of whether trisection could be done in the general case remained a mathematical mystery for millennia - it was only in that it was eventually proved to be impossible by Pierre Wantzela French mathematician and expert on arithmetic.

This was a great achievement for a man of 23, who subsequently died at the tragically young age of Triangle medians and centroids Video transcript I've drawn an arbitrary triangle right over here, and I've also drawn its three medians: And we know that where the three medians intersect at point G right over here, we call that the centroid.

Or another way to think about it, we can pick any one of these medians, and let's say let's pick EB. So whatever distance this is it's twice this distance there. And to do that, let's focus-- I want to focus on triangle ABE right over here. And I'm going to draw this median as essentially the base. So let me draw it that way.

I'm going to try to color code it similarly. So we draw it a little bit flatter than that. So it's like that.

And then we have the two yellow sides, so it looks something like this. It looks something like that. And then we have the centroid, right over here at G. That is our centroid, and then we have this magenta line going to A.

### The Centroid and the Trisected Line

Let me draw it a little bit neater than that. We have that line going to A, and then we have this blue line going to F right over here. And let me label all the points. Go back to the orange color.

## Centroid & median proof

So this is going to be E, this is going to be B, this is going to be A, this is going to be F right over here. And just to make sure we have all the same markings, that little marking there is that marking, these two markings, these two markings are on this side right over there. And the whole way that I'm going to prove that EG is twice as long as GB is just refer to the result that we did, I think, a couple of videos ago that the medians divide this triangle into six smaller triangles that all have equal area.

So another way to think about it is each of these three small triangles have equal area.

## Mathematical Mysteries: Trisecting the Angle

These are three of the total of six smaller triangles. So these three all have equal area. So let's think about this triangle right over here.

Let's think about this triangle, triangle AGB. This is triangle AGB right over there. Those are the same triangles. And let's compare that to triangle EAG right over here.