Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted
+50

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 $p_i-p_j \perp p_i-p_k$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.