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$x \in \mathbb{R}^{n}$ is a convex combination $C$ if there $p=p(x)\in \mathbb{N}$, $\lbrace \lambda_i\rbrace_{i=1}^{p} \subseteq [0,1]$ y $\lbrace x_i\rbrace_{i=1}^{p} \subseteq C$ such that $$ x=\sum_{i=1}^{p}\lambda_ix_i \ , \ \ \sum_{i=1}^{p}\lambda_i=1$$

For a triangle in $\mathbb{R}^{2}$, with vertices a, b, c. if x is a convex combination of {a,b,c} then $\lambda_1=\dfrac{\Vert x-a\Vert}{\Vert x-a\Vert +\Vert x-b\Vert +\Vert x-c\Vert }$? and so similarly for $\lambda_2,\lambda_3$??

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    Otherwise you may give me a counterexample, please2012-12-11
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    When $x=a$ we want $\lambda_1 = 1$ but that is not true for your formula.2012-12-11
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    Sure, but if $ C $ has dimension greater than one?2012-12-11
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    I was considering the question about a triangle in $\mathbb R^2$.2012-12-11

1 Answers 1

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I am assuming that $a,b,c$ are not collinear.

Your answer cannot be correct. If $x=a$, your formula gives $\lambda_1 = 0$, when it should be $\lambda_1 = 1$. It is impossible for the formula to have any $\lambda_i = 1$.

However, it is straightforward to compute the multipliers:

You can show show that $A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c\end{bmatrix}$ in invertible, and that the multipliers $\lambda$ satisfy: $A \lambda = \begin{bmatrix} 1 \\ x\end{bmatrix}$, hence $\lambda = A^{-1} \begin{bmatrix} 1 \\ x\end{bmatrix} = A^{-1} \begin{bmatrix} 1 \\ 0\end{bmatrix}+A^{-1} \begin{bmatrix} 0 \\ x\end{bmatrix}$ . In particular, the multipliers are affine functions (ie, linear plus a constant) of $x$.

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    Thank you very much for the reply, you can generalize?2012-12-11
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    You are very welcome.2012-12-11
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    Generalize in what way? If the points $a_k$ are affinely independent (ie, $\binom{1}{a_k}$ are linearly independent), then the same technique works for more points...2012-12-11
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    Than you for a solution! May I ask briefly? 1) What would happen if one uses this formula, but $x$ is not in the convex hull of ${a, b, c}$? I guess, some components of $\lambda$ would become negative? 2) This can clearly be formulated as a QP problem. How come it have a closed form solution? In general QPs don't, right? Is there a criteria that could tell that certain QP is solvable in closed form?2018-05-23
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    @BenUsman: That is correct. A point is in the triangle **iff** the multiplies are all non negative. (This assumes they are affinely independent, which in the case means not collinear.)2018-05-23
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    @copper.hat Thank you! A question that might (then please answer) or might not (then please don't :) ) be interesting. The resulting formula is reminiscent of one of definitions of [vector product](https://en.wikipedia.org/wiki/Cross_product#Matrix_notation). Do you think there's a relation?2018-05-23
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    @BenUsman: I don't see a connection.2018-05-24