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In a square $ABCD$,say there is a point $P\text{ }$ which lies inside it,the point P is located at a distance $x,y$ 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 $a$. :-) –  Srivatsan Sep 16 '11 at 22:37
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2 Answers 2

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. –  Eric Naslund Sep 16 '11 at 23:47
    
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. –  Eric Naslund Sep 16 '11 at 23:49
    
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. –  Eric Naslund Sep 16 '11 at 23:50
    
@Eric: Is this what you tried to achieve? Just use the align environment and add a \tag{xyz} in each line. –  t.b. Sep 17 '11 at 0:30
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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 :) –  t.b. Sep 17 '11 at 0:44
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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 "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. –  user13838 Sep 18 '11 at 21:04
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