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## Medians & centroids

# Triangle medians & centroids

## Video transcript

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.