WEBVTT
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JOEL LEWIS: Hi.
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Welcome to recitation.
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In lecture, you started
learning about vectors.
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Now vectors are going
to be really important
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throughout the whole
of this course.
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And I wanted to
give you one problem
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just to work with them in a
slightly different context
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than what we're going
to do in the future.
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So this is the context
of Euclidean geometry.
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So some of you have
probably seen this problem
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that we're going to
solve, but you probably
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haven't seen it
solved with vectors.
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So let's take a look at it.
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So what I'd like
you to do is show
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that the three medians of a
triangle intersect at a point,
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and the point is 2/3 of
the way from each vertex.
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So let me just remind
you of some terminology.
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So in a triangle, a
median is the segment
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that connects one vertex to the
midpoint of the opposite side.
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So here, this point M is
exactly halfway between B and C.
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So OK.
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So every triangle
has three medians--
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one from each vertex
connected to the midpoint
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of the opposite side-- and
what I'm asking you to show
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is that these three medians all
intersect in the same point.
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And also, that this
point divides the median
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into two pieces, and
the big piece is twice
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as large as the small piece.
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So this is 2/3 of the median,
and this is 1/3 of the median.
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So, why don't you take a few
minutes, work that out-- try
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and do it using vectors as much
as possible-- pause the video,
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come back, and we can
work on it together.
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So hopefully you had some
luck working on this problem.
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Let's get started on it.
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So to start, I actually
want to rephrase
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the question a little bit.
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And I'll rephrase it to
an equivalent question
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that's a little bit
more clear about how
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we want to get started.
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So another way to
say this problem is
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that it's asking us to
show-- so for each median,
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say this median AM here, where
M is the midpoint of side BC,
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there exists a point on
the median that divides it
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into a 2:1 ratio, so the point
that's 2/3 from the vertex
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to the midpoint of
the opposite side.
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So for example, you know,
there's a point-- so,
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let's call it P, say, at
first-- so there's a point P
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such that AP is twice PM.
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OK?
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And similarly, there's
some point-- maybe called
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Q-- that's 2/3 of the way from
B to the midpoint of this side.
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And there's some point
that's 2/3 of the way from C
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to the midpoint of this side.
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And so an equivalent
formulation of the question
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is to show that these
three points are really
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the same point.
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That they're all
in the same place.
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So one way we can
do that is that we
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can compare the position
vectors of those three points.
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And if those three points all
have the same position vector,
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then they're all in
exactly the same position.
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So in order to do that
we need some origin.
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And it happens that
for this problem,
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it doesn't matter
where the origin is,
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and so I'm not going
to draw an origin,
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but I'm going to call it O.
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So we're going to set up
a vector coordinate system
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with origin O. And
now I want to look
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at what the vector from O to
P is in terms of the vectors
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connecting O to A,
B, and C. Right?
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Those are the vectors
that determine
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the vertices of the triangle.
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And so I want to relate
the location of P
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to the locations of A, B, and C.
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So the first thing to do
is that-- well, in order
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to talk about where P is, I
know how P is related to A and M
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and I know how M is
related to B and C.
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So let's first figure out
what the position vector of M
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is in terms of the position
vectors of A, B, and C,
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and then we can use
that to figure out
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the position vector of P.
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So M is the midpoint
of the segment BC.
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So I think we saw
this in lecture.
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What this means is that the
position vector OM is exactly
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the average of the position
vectors of B and C.
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It's 1/2 of the
quantity OB plus OC.
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All right?
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So it's easy to
express the position
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vector of the midpoint
of a segment in terms
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of the position vectors
of the endpoints.
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You just add the position
vectors of the endpoints
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and divide by 2.
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So if you like,
this is equivalent
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to the geometric fact that the
diagonals of a parallelogram
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bisect each other.
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So that's the
position vector of M.
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Now we have to figure out what
the position vector of P is.
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So in order to do
this we can note,
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that in order to get from
the origin to point P,
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well, what we have to do is we
have to go from the origin--
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wherever it is--
to A, and then we
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have to go from A 2/3 of
the way to M. All right?
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So the vector OP is equal to
OA plus 2/3 of the vector AM.
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Right?
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Because we go 2/3
of the way from A
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to M in order to
get from A to P.
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This is because we've chosen
P to be the point that
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divides segment AM into a 2:1
ratio so that AP is 2/3 of AM.
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OK.
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So good.
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So now we need the vector AM.
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Well, we know what the
position vector of A is.
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It's just OA.
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And we also know what the
position vector of M is.
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It's OM.
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So that means that AM is just
the difference of those two
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vectors.
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It's going to be OM minus OA.
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Another way to say this is that
if you add OA to both sides,
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you have that OA
plus AM equals OM.
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In other words,
to go from O to M,
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first you can go from O to
A, and then go from A to M.
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All right.
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And I've just subtracted OA
onto the other side here.
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So we can write AM in
terms of OM and OA.
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And we also-- we have
an expression for OM
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here in terms of OB and OC.
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So that means we can
get an expression for AM
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in terms of OA, OB, and OC.
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So let's do that.
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So that's just by substituting
from here into here.
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So if I do that,
I get that AM is
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equal to-- so OM is
1/2 OB plus 1/2 OC,
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and now I just subtract OA.
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All right.
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So that's what AM is, putting
these two equations together.
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I get that that's AM.
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And so now I need to
figure out what OP is.
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So for OP, I just
need to substitute
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in this new expression
that I've got for AM.
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So I have OP is
equal to, well it's
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equal to OA plus 2/3
of what I've written
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just right above-- 2/3 of AM.
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So that's 1/2 OB
plus 1/2 OC minus OA.
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OK.
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And so now you can multiply
this 2/3 in-- you know,
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just distribute the
scalar multiplication
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across the addition there--
and then we can rearrange.
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We'll have two terms involving
OA and we can combine them.
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So we'll see we have a
plus OA minus 2/3 OA.
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So that's going to
be equal to 1/3 OA.
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And then we have, OK
so 2/3 times 1/2 OB.
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So that's plus 1/3
OB plus 1/3 OC.
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So this gives us
a simple formula
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for the position vector of
P-- that vector OP-- in terms
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of the position
vectors of A, B, and C.
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So in particular,
it's actually--
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because P is the special
point-- it's 1/3 of their sum.
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Of the sum OA plus OB plus OC.
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OK, so that's where P is.
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Now to finish the
problem, I just
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have to show that this is the
same location as the point that
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trisects the other medians.
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So how would I do that?
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Well, I could go
back to my triangle
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and I could do exactly
the same thing.
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So I could-- maybe I'll give
this point a name, also.
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I'll call this midpoint N, say.
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So I could let Q be the point
that lies 2/3 of the way
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from B to N. And then
I could write down
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the position vector of N
in terms of OA, OB, and OC.
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And then I can use that to write
down the position vector of Q
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in terms of OA, OB, and OC,
and I'll get some expression.
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And what will
happen at the end--
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I hope if I'm lucky--
that expression
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will be equal to this expression
that I found over here.
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OK?
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So you can go
through and do that,
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and if you do that, what you'll
find is that in fact it works.
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So there's actually a sort
of clever, shorter way
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of seeing that.
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Which is that this formula
is symmetric in A, B, and C.
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So that means if I just
relabel the points A, B, and C,
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this expression for the
position vector doesn't change.
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So rather than going
through that process
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that I just described,
you can also
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say, well, in order
to look at, say Q,
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instead of P, what I
need to do is I just
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need to switch B and A. I need
to do exactly the same thing
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but the roles of A and
B are interchanged.
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Well, if the roles of A and
B are interchanged, then
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in the resulting
formula, I just have
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to interchange the
roles of A and B,
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but that won't change the
value of this expression.
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So by symmetry, the point I get
really is going to be the same.
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If you don't like
that argument, I
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invite you to go
through this computation
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again in the case of
the other medians.
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In either case, what you'll find
is that the points that trisect
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the three medians all
have position vector 1/3
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OA plus 1/3 OB plus
1/3 OC, but that means
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they're the same point.
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So what we've shown then,
is that the points that
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trisect the three medians--
that trisects, that divide them
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into 2:1 ratios from the
vertex to the midpoint
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of the opposite side-- that
those three points all have
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the same position vector.
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So in fact, they're
the same point,
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and that's what we
wanted to show, right?
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We wanted to show that there's
one point that trisects
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all three of those medians.
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So we've shown that the three
points that trisect them
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are actually the same.
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So that's the same conclusion,
phrased differently.
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So I think I'll end there.