I was solving this problem earlier:

Points $X$, $Y$ and $Z$ in have the (three dimensional) coordinate vectors $\bf{x},\bf{y},\bf{z}$ respectively. Prove that the lines joining the vertices of $\triangle XYZ$ to the midpoints of the opposite sides are concurrent.

My problem is that at some point in the problem I ended up with equations of the form $a_1\mathbf{x}+b_1\mathbf{y}+c_1\mathbf{z}=a_2\mathbf{x}+b_2\mathbf{y}+c_2\mathbf{z}$ and the only way I could progress was by assuming that this implies $a_1=a_2$, $b_1=b_2$, $c_1=c_2$. Unfortunately I don't clearly see why this is and I only assumed it was true to solve the problem. (I indeed got the correct answer of the vectors passing through the same point $\frac{1}{3}(\bf{x+y+z})$ (and hence being concurrent at this point).)

One way I tried to justify this (the equalities just mentioned) is by assuming (wlog) that the vectors x,y,z are not coplanar, and then this would mean that they form a basis for 3D space (according to my textbook), which means the above equalities do indeed hold. The reason I gave for being able to assume wlog that the vectors are not coplanar is that whether they are ccoplanar or not does not affect the conclusion of the problem, which is simply concerned with proving concurrency of certain lines. In other words, if the position vectors are coplanar then rotating the triangle so that the position vectors are no longer coplanar doesn't affect the triangle and therefore doesn't affect the conclusion of the problem.

Is this reasoning correct, or is there another reason why the equalities hold?


Edit: My question is regarding the title, please ignore the centroid problem as I only mentioned it for context.

Edit2: Here is how I arrived at the equation. First I let the midpoints of the size opposite $X$ be $X_1$, and similarly for $Y,Z$. Clearly, $\vec X_1=\frac{\mathbf{y}+\mathbf{z}}{2}$. The line $XX_1$ has direction $$\vec{XX_1}=\vec X_1-\vec X=\frac{\mathbf{y}+\mathbf{z}}{2}-\mathbb{x}=\frac{\mathbf{y}+\mathbf{z}-2\mathbf{x}}{2}$$ so its equation is $$\mathbf{r_x}=\mathbf{x}+a(\mathbf{y}+\mathbf{z}-2\mathbf{x})=(1-2a)\mathbf{x}+a\mathbf{y}+a\mathbf{z}.$$ By symmetry $$\mathbf{r_y}=b\mathbf{x}+(1-2b)\mathbf{y}+b\mathbf{z},\quad \mathbf{r_z}=c\mathbf{x}+c\mathbf{y}+(1-2c)\mathbf{z}.$$ We want the point where these lines meet, and for this we only need to calculate the point where a pair of them meet and then verify that the third line crosses that point. So we have to solve $$(1-2a)\mathbf{x}+a\mathbf{y}+a\mathbf{z}=b\mathbf{x}+(1-2b)\mathbf{y}+b\mathbf{z}$$ and here is where I met my problem, but as I said I just assumed (and tried to justify this assumption using the "base vectors" reasoning in the main post) that $$1-2a=b,\quad a=1-2b,\quad a=b$$ $$\implies a=b=\frac{1}{3},$$ i.e. the common point is $$\frac{1}{3}(\mathbf{x}+\mathbf{y}+\mathbf{z}).$$ It is easy to verify that this lies on the third line when its parameter is $c=\frac{1}{3}$.

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    $\begingroup$ You can make that deduction if and only if the vectors $\bf{x},\bf{y},\bf{z}$ are linearly independent. As you observed, this can be can diagnosed geometrically: the linear independence holds if and only if $X,Y,Z$ are not on the same plane as the origin. Unless $X,Y,Z$ are on the same line you can always achieve this by moving the origin to a suitable place. But, why would the location of the origin have anything to do with the truth of the claim?? This is a symptom that your approach, while basically ok, is using perhaps too many vectors. $\endgroup$ – Jyrki Lahtonen Jan 2 '17 at 12:32
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    $\begingroup$ (cont'd) Because the triangle is a 2-dimensional object (even in 3D-space), you should aim to write everything in terms of two known vectors intrinsic to the problem (such as two sides of your triangle). Otherwise you will inevitably run into this kind of difficulty that several combinations of coefficients $a,b,c$ may lead to identical vectors. $\endgroup$ – Jyrki Lahtonen Jan 2 '17 at 12:34
  • $\begingroup$ @JyrkiLahtonen: Thank you, I really appreciate the intuition you provided. $\endgroup$ – kesra Jan 2 '17 at 14:22
  • $\begingroup$ Three points in $\mathbb R^3$ are always coplanar. There wouldn’t be a triangle to talk about otherwise. I’m guessing that you mean that they’re not all on a plane through the origin, which isn’t quite the same thing. $\endgroup$ – amd Jan 2 '17 at 20:41
  • $\begingroup$ @amd: The context is vectors: What I meant is that the position vectors of the points cannot be coplanar. (My original post says "the vectors are not coplanar".) $\endgroup$ – kesra Jan 3 '17 at 13:02

You don't mention your approach, but I think you've done an overly complicated approach.

The midpoints on the of the sides are ${X+Y\over2}$, ${Y+Z\over2}$ and $Z+X\over2$, and the oposite corners are $Z$, $X$ and $Y$ respectively. Now take the point one third to the oposite corner and you get for example:

$${2\over3}{X+Y\over2} + {1\over3}Z = {X+Y+Z\over3}$$

Doing the same for the other midpoints and oposite vertex gives the same result.

The trick here is of course to realize that they will cross at ${X+Y+Z\over3}$, but this can be seen as it would need it to be a linear combination of $X$, $Y$ and $Z$ and since such a line is for example $\theta{X+Y\over2}+(1-\theta)Z$ the coefficient of $X$ and $Y$ must bu the same, similarily the coefficient of $Y$ and $Z$ must be the same. So we have $\theta/2 = 1-\theta$ which means that $\theta = 2/3$.

  • $\begingroup$ Thank you, but your answer does not touch on my question at all.. $\endgroup$ – kesra Jan 2 '17 at 12:21
  • $\begingroup$ @kesra The problem I have with your question is that you didn't mention your approach. The question would be easier to understand if you mention how you arrive at your equation. $\endgroup$ – skyking Jan 2 '17 at 13:22

$\newcommand{\Vec}[1]{\mathbf{#1}}$In algebra and analysis, it's often best to write an equation of the form $X = Y$ in the form $X - Y = 0$. Here, $$ a_{1}\Vec{x} + b_{1}\Vec{y} + c_{1}\Vec{z} = a_{2}\Vec{x} + b_{2}\Vec{y} + c_{2}\Vec{z} \tag{1} $$ is equivalent, after rearranging, to $$ (a_{1} - a_{2}) \Vec{x} + (b_{1} - b_{2}) \Vec{y} + (c_{1} - c_{2}) \Vec{z} = \Vec{0}. \tag{2} $$ As Jyrki Lahtonen notes in the comments, an ordered triple of vectors $(\Vec{x}, \Vec{y}, \Vec{z})$ is said to be linearly independent if

For all real $a$, $b$, $c$, the equation $a \Vec{x} + b \Vec{y} + c \Vec{z} = \Vec{0}$ implies $a = b = c = 0$.

Equation (2) has this form with $a = a_{1} - a_{2}$, etc., so if your vectors are linearly independent (i.e., their tips are non-coplanar with the zero vector), then (2) implies $a_{1} = a_{2}$, etc., just as you say.

The deeper issue is, the tips of three vectors in an arbitrary vector space are coplanar and there's nothing in the problem to prevent the zero vector from lying in the plane of the triangle. In a meta way, this means (as JL notes) your proof "contains too much infrastructure". Skyking's argument avoids this technical issue; it even works if the vertices are collinear.


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