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In a square $ABCD$, say there is a point $P$ which lies inside it, the point $P$ is located at distances $x$, $y$ and $z$ meters from $A,B$ and $C$ respectively.

Using this information how could we compute a form for the area of the square?Please explain your approach.

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    How about *hacking* it out? Assume that the vertices are $(0,a), (0,0), (a,0), (a,a)$ and P is $(u,v)$. Then $u^2+(v-a)^2=x^2, u^2+v^2=y^2, (u-a)^2+v^2=z^2$. You can subtract (2) from (1) and (3) to solve for $(u,v)$ in terms of $a$. Plugging that in (2) should give you $a$. :-)2011-09-16

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Using coordinate geometry. Assume that the vertices (in order) are $(0,a), (0,0), (a,0), (a,a)$ and that $P$ is $(u,v)$. Then $$ \begin{align} u^2+(v-a)^2 &= x^2,\tag{1} \\ u^2+v^2 &= y^2, \tag{2} \\ (u-a)^2+v^2 &= z^2 \tag{3}. \end{align} $$ We can "solve" for $v$ in terms of $a$ by subtracting $(2)$ from $(1)$; similarly for $u$. Since the steps are straightforward, I am just posting the answer here: $$ (u, v) = \left( \frac{a^2 + y^2 - z^2}{2a}, \frac{a^2 + y^2 - x^2}{2a} \right). $$ Plugging this in $(2)$, we get: $$ (a^2+y^2 - z^2)^2 + (a^2 + y^2 - x^2)^2 = 4a^2y^2. $$ $$ \implies a^4 + y^4 - (a^2+y^2)(x^2+z^2)+\frac{x^4+z^4}{2}=0. $$ This is a quadratic equation in $a^2$(=area of the square) and can be solved.

Note. I do not yet understand whether both the roots of the quadratic are true solutions or not.

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    I tried editing to give the equations their respective names. It is not optimal as I just used a bunch of spaces and then put (1), (2), and (3) in the equation. (I thought I would notify you of the edit because of this) However I think it makes things look nicer.2011-09-16
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    Also, I realized I was confused about your earlier comment, and solving for $a$. So I deleted my response as it didn't make too much sense.2011-09-16
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    I am not too sure if it is even possible. I saw Joriki do the above work around for a particular answer, and have just used it since.2011-09-16
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    @Eric: Is this what you tried to achieve? Just use the `align` environment and add a `\tag{xyz}` in each line.2011-09-17
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    Thanks @Theo. *I* used the `eqnarray` environment; Eric just numbered the equations for me.2011-09-17
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    The numbering doesn't seem to work with the `eqnarray` environment (at least it always produced "boxed stuff" while I was trying). The `align` environment worked fine, though :)2011-09-17
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    @Theo Yes, $\TeX$ does have some nice "features". :)2011-09-17
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    @Srivatsan : If you unfold the irregular envelope and obtain an irregular quadrilateral then you can either use Heron's formula or apply even more hack to get the square root solution. Maybe that helps.2011-09-17
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    @percusse I was thinking that there's a solution that uses only middle-school geometry (not coordinate geometry), and in particular, Heron's formula. Your idea is interesting, but we do not know the distance to the fourth vertex (and of course, we do not know the side length of the square). How do we take care of that?2011-09-17
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    @Srivatsan : I couldn't come up with something other than more hack using law of cosines, angles etc. But I posted a picture maybe that helps too.2011-09-17
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This is just a supplement to Srivatsan's answer. I knew that this was a consequence of something but didn't quite get it but this is just Varignon's theorem.

enter image description here

(I wish one day we can insert tikzpicture environment inside the answers.) I also agree with Srivatsan that there must a shorter/simpler observation and maybe someone can provide it.

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    I didn't work it out completely but [this quote from here](http://www.vias.org/comp_geometry/geom_quad_varignon.html) *"The center of the parallelogram is the geometric centroid of four point masses placed on the vertices of the quadrilateral."* seems to shorten the manipulations.2011-09-18