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I am new to convex combination, and I am quite amazed by some easy result.

I know that convex combination of 2 points($P_1P_2$) in $R^2$ is all points in the line segment $P_1P_2$.

And then I see a result that convex combination of 3 points $P_1P_2P_3$, which is not in a line, is all points in the triangle $\bigtriangleup P_1P_2P_3$. It is a reasonable result, but I want to see how people prove it.

Can anyone provide a proof?

user2262504
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2 Answers2

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The coefficients that arise in an affine combination of three 2D points are called barycentric coordinates. See here and here for more information.

These coordinates actually represent the (signed) areas of triangles, as the references explain. When a point is inside a triangle, the three relevant areas are all positive, so the barycentric coordinates correspond to a convex combination.

The references provide all the details, but a simple argument is as follows:

Suppose we have a triangle with vertices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, and consider the convex combination $\mathbf{P} = u\mathbf{A} + v\mathbf{B} + w\mathbf{C}$, where $u,v,w $ are in the interval $[0,1]$ and $u+v+w=1$.

Define a point $\mathbf{Q}$ by $$ \mathbf{Q} = \frac{v}{v+w}\mathbf{B} + \frac{w}{v+w}\mathbf{C} $$ Clearly $\mathbf{Q}$ is on the line segment (the triangle edge) $\mathbf{B}\mathbf{C}$. But then $$ \mathbf{P} = u\mathbf{A} + (v+w)\mathbf{Q} = u\mathbf{A} + (1-u)\mathbf{Q} $$ So, $\mathbf{P}$ is on the line segment $\mathbf{A}\mathbf{Q}$, and therefore it lies inside the triangle $\mathbf{A}\mathbf{B}\mathbf{C}$.

If you're interested, the extension to three-dimensional space is easy: convex combinations give you the points inside a tetrahedron.

bubba
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  • I saw the two links you provided, but they just give a conclusion. could you please explain it in more detail, thank you. – user2262504 Aug 27 '14 at 12:49
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    @user2262504 -- please see additions to my answer. – bubba Aug 28 '14 at 05:45
  • @user2262504 -- By the way, I see that you have asked 23 questions, but you have not accepted a single answer. You might get more answers (and better answers) if you start accepting some of them. – bubba Aug 28 '14 at 05:50
  • thank you. I did forget to mark the answer, I didn't know this before. – user2262504 Aug 29 '14 at 08:02
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A naive approach to picture the truth of the statment:

Consider the set of all convex combinations

$C = \left\lbrace\alpha_{1}x_{1} + \alpha_{2}x_{2} + \alpha_{3}x_{3}\right| \alpha_{1}+\alpha_{2}+\alpha_{3}=1\rbrace$.

And imagine what happens if $\alpha_{1} = 0$ and $\alpha_{1},\alpha_{2} \in (0,1]$, we must have a line segment joining $x_{2}$ to $x_{3}$. In the same way, we construct a segmente line joining $x_{1}$ to $x_{3}$ and $x_{1}$ to $x_{2}$. Then we have constructed the boundary of the triangle in $\mathbb{R}^{2}$, in the cases where $\alpha_{i} \in (0,1)$ the points lie in the interior.

BBVM
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