Find the co-ordinates of A and B

Question

$P$ is $(2,1)$ and $A$ and $B$ lie on $X$ axis and $y=x$ respectively such that $PA+PB+AB$ is minimum, then find the co-ordinates of $A$ and $B$.

If we consider $A$ to be $(a,0)$ and $B$ to be $(b,b)$ it would become complicated. Is there any other way?

Answer 1

hint: with this informations we get $$f(a,b)=\sqrt{(a-2)^2+1}+\sqrt{(b-2)^2+(b-1)^2}+\sqrt{(b-a)^2+b^2}$$ can you proceed? $$f_a=1/2\,{\frac {2\,a-4}{\sqrt { \left( a-2 \right) ^{2}+1}}}+1/2\,{\frac {-2\,b+2\,a}{\sqrt { \left( b-a \right) ^{2}+{b}^{2}}}} $$ $$f_b=1/2\,{\frac {4\,b-6}{\sqrt { \left( b-2 \right) ^{2}+ \left( b-1 \right) ^{2}}}}+1/2\,{\frac {4\,b-2\,a}{\sqrt { \left( b-a \right) ^{ 2}+{b}^{2}}}} $$ solve the System $$f_a=0$$ and $$f_b=0$$

Answer 2

Let $C(1,2)$ and $D(2,-1)$.

Thus, $CD=\sqrt{10}$ and it's gives a minimal value.

Indeed, Let $CD$ intersects a line $y=x$ in the point $B_1$ and the $x$-axis in the point $A_1$.

Thus, $$PA+PB+AB=DA+AB+CB\geq CD=$$ $$=DA_1+A_1B_1+B_1C=PA_1+A_1B_1+PB_1=\sqrt{10}$$

Another way.

By Minkowski $$PA+PB+AB=\sqrt{(a-2)^2+1^2}+\sqrt{(1-b)^2+(2-b)^2}+\sqrt{(b-a)^2+b^2}\geq$$ $$=\sqrt{a-2+1-b+b-a)^2+(1+2-b+b)^2}=\sqrt{10}.$$ The equality occurs for $a=\frac{5}{3}$ and $b=\frac{5}{4}$, which says that $\sqrt{10}$ is a minimal value.

We have for the minimal sum: $A\left(\frac{5}{3},0\right)$ and $B\left(\frac{5}{4},\frac{5}{4}\right)$.