Finding minimum value of $AX+XB-XX'$

Question

Question: There are two points on a plane: $A$ and $B$. Find point $X$ such that value of expression $AX+XB-XX'$ is minimal where $X'$ is orthonormal projection of $X$ on $AB$

Tip to task is to draw equilateral triangle $ABC$. I'm looking for elementary solution. Using algebra I found that the sought point $X$ is the orthocenter of triangle $ABC$. It seems to ma that it is a typical Heron's 'shortest distance' problem with the difficulty of subtraction which I cannot resolve.

Answer 1

If we fix $XX'$ (let's call it $a$), $X$ is on a horizontal line with distance $|XX'|$ from $AB$. then $|AX|+|BX|$ is minimal when $|AX|=|BX|$, because $$\min(|AX|+|BX|-|XX'|)=\min(|AX|+|BX|)-a=\min(\sqrt{a^2+|AX'|^2}+\sqrt{a^2+|AB-AX'|^2})-a$$ If we define $f(|AX'|)=\sqrt{a^2+|AX'|^2}+\sqrt{a^2+|AB-AX'|^2}$, we get that the minimum happens when $|AX'|=\frac{|AB|}{2}$, or in other words, $$|AX|=|BX|$$ We get that X is on the perpendicular bisector of $AB$. Out of all the points on this line, the intersection with $AB$ is the minimum

So $X$ is the middle of $AB$

Answer 2

The answer will be divided into $2$ parts.

(i) First we show that for minimal $AX+BX-XX'$, $X'$ has to be the mid-point of $AB$.

Let $X$ be any point on the plane, draw a (red) line parallel to $AB$.

Notice that any point on the (red) line has the same distance to $AB$.

Therefore to minimize $AX+BX-XX'$ is the same as to minimize $AX+BX$.

From reflection principle, we know that the miminum is attained at point $P$ where $\alpha = \beta$, i.e. $AP=BP$.

Thus $XX'$ is the perpendicular bisector of $AB$. Equivalently $X'$ is the mid-point of $AB$. $$$$ (ii) Take any point $X$ on the perpendicular bisector of $AB$, we can always mark a point $C$ on the perpendicular bisector such that $X$ becomes an inner point of $\Delta ABC.$

Notice that $AX+BX-XX'= AX+BX+CX-CX'$.

Since $CX'$ is fixed, minimizing $AX+BX-XX'$ is the same as minimizing $AX+BX+CX$.

But the minimum is attained at point $F$ (the Torricelli's point of $\Delta ABC$) where $\angle AFB=\angle BFC=\angle CFA=120^{\text o}$.

As seen in the figure, the point $F$ is the same for all $\Delta ABC$ so long as $CA=CB$.

Thus the minimum is attained at the fixed point $F$ here.

Answer 3

Note that to minimize $AX+XB-XX'$, $X$ has to be on the perpendicular bisector of $XX'$, otherwise we can shift $X$ towards the perpendicualr bisector to reduce value of $AX+XB$ while maintain the value of $XX'$ unchange.
Now suppose X is on the perpendicular bisector of $XX'$. Let $XX'$ and $AB$ be $x$ and $a$ respectively. Then \begin{align*} AX+XB-XX'&=2\sqrt{x^2+\frac{a^2}{4}}-x\\ \end{align*} By taking the derivative, we have \begin{align*} (2\sqrt{x^2+\frac{a^2}{4}}-x)^{'}&=\frac{2x}{\sqrt{x^2+\frac{a^2}{4}}}-1\\ \end{align*} When the derivative is $0$, it reach its optimum. \begin{align*} \frac{2x}{\sqrt{x^2+\frac{a^2}{4}}}-1&=0\\ 2x&=\sqrt{x^2+\frac{a^2}{4}}\\ 3x^2&=\frac{a^2}{4}\\ x&=\frac{\sqrt{3}a}{6} \end{align*} Thus \begin{align*} AX+XB-XX'&=2\sqrt{{\left(\frac{\sqrt{3}a}{6}\right)}^2+\frac{a^2}{4}}-\frac{\sqrt{3}a}{6}\\ &=\frac{\sqrt{3}a}{2} \end{align*}