Triangle medians & centroids

I want to do a quick refresher on medians of triangles, and also explore an interesting property of them that will be useful, I think, in future problems. So let me just draw an arbitrary triangle over here. Now that's good enough. Now a median of the triangle-- and we'll see a triangle has three of them-- is just a line that connects a vertex of the triangle with the midpoint of the opposite side. So the opposite side's midpoint looks right about there. This length is equal to that length. And so this is a median. Close enough. And of course, we have three vertices, so we'll have three medians. If we start at this vertex, we want to go to the midpoint of the opposite side. It looks right about there. So this blue line right over here is another median. It's not a completely straight line, but I think you get the idea. Then we could also do it from this point right over here. Draw a line from this vertex to the midpoint of the opposite side. Let's see, the midpoint of the opposite side is there. And we draw a line. Each of these-- I could draw a straighter line than that. Let me draw it. There. Well, I think you get the idea. These are all medians of this triangle. And what's neat about medians is that all three medians always intersect in one point. And that by itself is a pretty neat property. And that one point that they intersect in is called the centroid. And if this was actually a physical triangle, let's say you made it out of iron, and if you were to toss it-- well, even before you toss it, the centroid would actually be the center of mass. So let's say this is an iron triangle. Let's say that this right here is an iron triangle that has its centroid right over here, then this iron triangle's center of mass would be where the centroid is, assuming it has a uniform density. And if you were to throw that iron triangle, it would rotate around this point. Assuming that it had some rotational motion, it would rotate around that centroid, or around the center of mass. But anyway, the point of this video is not to focus on physics and throwing iron triangles. The point here is I want to show you a neat property of medians. And the property is that if you pick any median, the distance from the centroid to the midpoint of the opposite side-- so this distance-- is going to be half of this distance. So if this distance right here is a, then this distance right here is 2a. Or another way to think about it is this distance is 2/3 of the length of the entire median, and this distance right here is 1/3 of the length of the entire median. And let's just prove it for ourselves just so you don't have to take things on faith. And to do that, I'll draw an arbitrary triangle. I'll do a two-dimensional triangle, and I'll do it in three dimensions because at least in my mind, it makes the math a little bit easier. In general, whenever you have an n-dimensional figure and you embed it in n plus 1 dimensions, it makes the math a little bit easier. The actual tetrahedron problem that we did, you could actually embed it in four dimensions and it would make the math easier. It's just much harder to visualize, so I didn't do it that way. But let's just have an arbitrary triangle. And let's say it has a vertex and there, a vertex there, and a vertex there. So I'm not making any assumptions about the triangle. I'm not saying it's isosceles, or equilateral or anything. It's just an arbitrary triangle. And so let's say this coordinate right over here is-- I'll call this the x-axis. So, this is the x-axis, the y-axis, and the z-axis. I know some of y'all are used to swapping these two axes, but it doesn't make a difference. So let's call this coordinate right here a, 0, 0. So it's a along the x-axis. Let's call this coordinate 0, b, 0. And let's call this coordinate up here, 0, 0, c. And if you connect the points, you're going to have a triangle just like that. Now, the centroid of a triangle, especially in three dimensions. The centroid of a triangle is just going to be the average of the coordinates of the vertices. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here. You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center of mass of this triangle, if it had some mass, is just the average of these coordinates. Now, what we want to do is use this information. Let's just use this coordinate right here and then compare just using the distance formula. Let's compare this distance up here in orange to this distance down here in yellow. And remember, this point right over here-- this is the median of this bottom side right over here. It's just going to be the average of these two points. And so the x-coordinate-- 0 plus a over 2 is going to be a over 2. b plus 0 over 2 is going to be b over 2. And then it has no z-coordinates, so it's just going to be 0. 0 plus 0 over 2 is 0. So we know the coordinates for this point that point and that point. So we can calculate the yellow distance and we can calculate the orange distance. So let's calculate the orange distance. So that is going to be equal to the square root of-- and we just take the difference of each of these points squared. So it's a over 3 minus 0 squared. So that's going to be a squared over 9, plus b over 3 minus 0 squared. So that's b squared over 9. Plus c over 3 minus c, which is negative 2/3. And we want to square that. So we're going to have positive 4 over 9c squared. Did I do that right? c over 3, so 1/3 minus 1 is negative 2/3. So this is negative 2/3 c. You square it. You're going to get 4/9 c squared. So that's the orange distance. Now, let's calculate-- and if we want to do it, we can express this-- let me express it a little bit simpler than this. This is the same thing as the square root of a squared plus b squared plus 4c squared over the square root of 9, which is just equal to 3. Now let's do the same thing with the yellow distance. So it's going to be equal to the square root of-- so if we have a over 2 minus a over 3. So 1/2 minus 1/3-- that's the same thing as 3/6 minus 2/6, so it's 1/6 a. 1/6 a squared is a squared over 36. b over 2 minus b over 3 is b over 6. You square it. You get plus b squared over 36. And then finally you have 0 minus c over 3 squared. That's going to be c squared over 9. But just so we get a common denominator, c squared over 9 is the same thing as plus 4c squared over 36. And we can rewrite this as the square root of a squared plus b squared plus 4c squared over 6. So you can see, this distance right here, if you multiply this orange distance by 1/2, you're going to get-- so if you multiply the orange distance by 1/2 or if you divide it by 2, you get the yellow distance. So this is always going to be twice the distance as this because we did this in the most general possible way. We assumed nothing about this triangle. So remember that little property that the centroid, the intersection of the medians-- the intersection happens 2/3 away from the vertex or 1/3 the length of the median away from the midpoint of the opposite side. And we can use that property in-- well, we'lll probably use it in a bunch of problems. But anyway, hopefully you found that interesting.