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While solving a computer science question, I got a system of equations$$\begin{cases}x^2+(100-y)^2=\frac TA\\(100+x)^2+(100+y)^2=\frac TB\\(100-x)^2+(100+y)^2=\frac TC\end{cases}$$

However, I'm having difficulty solving the equation. $A, B, C$ are constants and we need to find $x, y, T$. I tried subtracting the third equation from the second equation, but still had difficulty with eliminating the extra $T$ variable.

Any ideas?

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  • $\begingroup$ @Moo Hm... this was a question from a competition. I'm fairly certain there is an easier or nicer solution. I will go over my work. $\endgroup$
    – Crescendo
    Commented Oct 19, 2018 at 15:07
  • $\begingroup$ I would give it a try geometrically. The three equations represent three circles centered respectively at $(0,100), (-100,-100),(100,-100).$ A necessary condition to have at least an intersection between the second and the third circle is that, for example, $T/B \geq 100^2$. When their intersection is not empty, they intersect each other at two points on the same vertical line $x=L$. To get a solution of the system, the third circle should pass through one of those points found. This is just a suggestion on how to proceed, I am not sure if it is doable. Can you go on from here? $\endgroup$
    – Gibbs
    Commented Oct 19, 2018 at 15:12

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Plugging $$(100+y)^2=\frac{T}{C}-(100-x)^2$$ in the second equation we get $$(100+x)^2-(100-x)^2=\frac{T}{B}-\frac{T}{C}$$ we get by Using the binomial formulas

$$400x=\frac{T}{B}-\frac{T}{C}$$ Can you proceed? For $T$ we get $$T=800\,{\frac {ABC}{AB+AC-2\,BC} \left( 50\,{\frac {8\,{A}^{2}BC-8\,{B }^{2}{C}^{2}+\sqrt {-25\,{A}^{4}{B}^{4}+50\,{A}^{4}{B}^{2}{C}^{2}-25\, {A}^{4}{C}^{4}+140\,{A}^{3}{B}^{4}C+20\,{A}^{3}{B}^{3}{C}^{2}+20\,{A}^ {3}{B}^{2}{C}^{3}+140\,{A}^{3}B{C}^{4}-276\,{A}^{2}{B}^{4}{C}^{2}-152 \,{A}^{2}{B}^{3}{C}^{3}-276\,{A}^{2}{B}^{2}{C}^{4}+224\,A{B}^{4}{C}^{3 }+224\,A{B}^{3}{C}^{4}-64\,{B}^{4}{C}^{4}}}{5\,{A}^{2}{B}^{2}-6\,{A}^{ 2}BC+5\,{A}^{2}{C}^{2}-4\,A{B}^{2}C-4\,{C}^{2}AB+4\,{B}^{2}{C}^{2}}}+ 25 \right) } $$

and for $x$

$$x=-2\,{\frac {A}{AB+AC-2\,BC} \left( 50\,{\frac {B \left( 8\,{A}^{2}BC -8\,{B}^{2}{C}^{2}+\sqrt {-25\,{A}^{4}{B}^{4}+50\,{A}^{4}{B}^{2}{C}^{2 }-25\,{A}^{4}{C}^{4}+140\,{A}^{3}{B}^{4}C+20\,{A}^{3}{B}^{3}{C}^{2}+20 \,{A}^{3}{B}^{2}{C}^{3}+140\,{A}^{3}B{C}^{4}-276\,{A}^{2}{B}^{4}{C}^{2 }-152\,{A}^{2}{B}^{3}{C}^{3}-276\,{A}^{2}{B}^{2}{C}^{4}+224\,A{B}^{4}{ C}^{3}+224\,A{B}^{3}{C}^{4}-64\,{B}^{4}{C}^{4}} \right) }{5\,{A}^{2}{B }^{2}-6\,{A}^{2}BC+5\,{A}^{2}{C}^{2}-4\,A{B}^{2}C-4\,{C}^{2}AB+4\,{B}^ {2}{C}^{2}}}-50\,{\frac {C \left( 8\,{A}^{2}BC-8\,{B}^{2}{C}^{2}+ \sqrt {-25\,{A}^{4}{B}^{4}+50\,{A}^{4}{B}^{2}{C}^{2}-25\,{A}^{4}{C}^{4 }+140\,{A}^{3}{B}^{4}C+20\,{A}^{3}{B}^{3}{C}^{2}+20\,{A}^{3}{B}^{2}{C} ^{3}+140\,{A}^{3}B{C}^{4}-276\,{A}^{2}{B}^{4}{C}^{2}-152\,{A}^{2}{B}^{ 3}{C}^{3}-276\,{A}^{2}{B}^{2}{C}^{4}+224\,A{B}^{4}{C}^{3}+224\,A{B}^{3 }{C}^{4}-64\,{B}^{4}{C}^{4}} \right) }{5\,{A}^{2}{B}^{2}-6\,{A}^{2}BC+ 5\,{A}^{2}{C}^{2}-4\,A{B}^{2}C-4\,{C}^{2}AB+4\,{B}^{2}{C}^{2}}}+25\,B- 25\,C \right) } $$

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    $\begingroup$ That's exactly what we did. $\endgroup$
    – Crescendo
    Commented Oct 19, 2018 at 15:07

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