# Algebraic-Geometric Equation

Dan Sitaru posted a problem from his book "Liquid Math" at the CutTheKnotMath facebook page:

Solve in real numbers

\begin{align} \sqrt{x^2+y^2}&+\sqrt{(x-12)^2+(y-5)^2}=26\\ &\;\;-\sqrt{x^2+(y-5)^2}-\sqrt{(x-12)^2+y^2}. \end{align}

### Solution

The problem admits a simple geometric interpretation. Let in the Cartesian coordinates $A=(0,0),$ $B=(12,0),$ $C=(12,5),$ $D=(0,5).$ Let $P=(x,y)$ be an arbitrary point in the plain. We are requested to solve the equation:

$|PA|+|PC|=26-|PB|-|PD|,$

or simpler,

$|PA|+|PB|+|PC|+|PD|=26.$

Note that by the triangle inequality, $|PA|+|PC|\ge |AC|=13$ and $|PB|+|PD|\ge |BD|=13$ which tells us that the required equality may be satisfied only when both $|PA|+|PC|=13$ and $|PB|+|PD|=13.$ The first is satisfied with the points of the segment $AC,$ the second with the points of the segment $BD,$ implying that the solution to the problem lies at the intersection of the two segments. Observe that $ABCD$ is a rectangle, with the diagonals $AC$ and $BD$ that intersect at the center of the rectangle. It follows that $\displaystyle P=\left(\frac{12}{2},\frac{5}{2}\right)$ is the only solution to the given equation.

Note: Leo Giugiuc has posted a variant of the above in complex numbers and practically the same as above where complex numbers are solely used for notational purposes. Mar Jr San has posted a purely arithmetic solution.