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In lemma 2 in this article's section 5 there is a proof below and at the end it states that equation $\|(1-\alpha)p_j+\alpha p_i-p_k\|=\|p_i-p_k\|$ has only one solution $\alpha=1$.

Examples can be constructed where $\alpha$ can have other values:

$p_j=\begin{bmatrix}20\\60\\\end{bmatrix} \;\; p_k=\begin{bmatrix}100\\70\\\end{bmatrix}\;\; p_i=\begin{bmatrix}50\\77\\\end{bmatrix} \;\; \alpha=3.33$

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Does your 'counterexample' satisfy the condition of lemma2: "Provided that the positions of any two nodes j and k are fixed, the condition that node i having a distance constraint with node k and a bearing constraint with node j"? – Berci Feb 3 '13 at 14:23
In my counterexample all three nodes are fixed so the condition is satisfied. What my counterexample shows is that there is (contrary to lemma) another node beside $i$ satisfying the constraint and this one is at $(1-\alpha)p_j+\alpha p_i$ or $[119\;\;116.7]$. – Bula Feb 3 '13 at 14:35
The lemma does seem strange/incorrect. Imposing a distance constraint with a fixed node $k$ defines a circle. Imposing a bearing constraint with a fixed node $j$ defines a line. A circle and a line can intersect each other at two distinct points, so it doesn't seem that node $i$ (satisfying both constraints) should necessarily be unique. – mjqxxxx Feb 5 '13 at 21:09
up vote 1 down vote accepted

The statement is false.

We have $$\|\alpha (p_i-p_j)+(p_j-p_k)\| = \|p_i-p_k\|,$$ and squaring both sides gives us a quadratic equation in $\alpha$ with descriminant \begin{align*} &\qquad \left[(p_i-p_j)\cdot(p_j-p_k)\right]^2-\|p_i-p_j\|^2(\|p_j-p_k\|^2-\|p_i-p_k\|^2)\\ &= \|p_i-p_j\|^2\|p_i-p_k\|^2-\|(p_i-p_j)\times(p_j-p_k)\|^2\\ &\geq \|(p_i-p_j)\times (p_i-p_k)\|^2 - \|(p_i-p_j)\times(p_j-p_k)\|^2\\ &= 2\operatorname{Area}(\Delta_{ijk}) - 2\operatorname{Area}(\Delta_{ijk}) = 0, \end{align*} with equality, and thus a unique solution $\alpha = 1$, if and only if $p_i-p_j \perp p_i-p_k$.

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Should the $=$ sign in line 2, be $\geq$ as there is no $\|p_i-p_j\|^2\|p_i-p_k\|^2$? – Bula Feb 8 '13 at 10:41
What do you mean? It comes from the last term on the line above. – user7530 Feb 8 '13 at 16:48
Sorry, I was not careful in reading the equations. Thank you. – Bula Feb 8 '13 at 21:37

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