How to determine the minimum value of the weighted sum of the distances from two points to a point on a circumference

Question

Problem

In a two-dimensional Cartesian coordinate system, there are two points $A(2, 0)$ and $B(2, 2)$ and a circle $c$ with radius $1$ centered at the origin $O(0, 0)$, as shown in the figure below.

If $P$ is a point on the circle $c$, then what is the minimum value of

$$ f = 2\sqrt{2}\lvert{PA}\rvert + \lvert{PB}\rvert? $$

Hypothesis

From my experience, the solutions to such problems do not seem to be available under elementary form in general, as indicated by answers to the question Minimize the sum of distances between two point and a circle. However, when I studied this problem on GeoGebra, it seems that minimum value is exactly $5$ in this specific situation, with $P$ located at roughly the position shown below:

I tried to verify my hypothesis as follows. Since $P$ is located inside $\angle AOB$, we set its location to $(x, \sqrt{1 - x^2})$ (where $\sqrt{2}/2 < x < 1$). Therefore, \begin{align*} \lvert{PA}\rvert &= \sqrt{(2 - x)^2 + (1 - x^2)} \\ &= \sqrt{5 - 4x}, \\ \lvert{PB}\rvert &= \sqrt{(2 - x)^2 + (2 - \sqrt{1 - x^2})^2} \\ &= \sqrt{-4\sqrt{1 - x^2} - 4x + 9}, \\ f &= 2\sqrt{2} \lvert{PA}\rvert + \lvert{PB}\rvert \\ &= 2\sqrt{2} \sqrt{5 - 4x} + \sqrt{-4\sqrt{1 - x^2} - 4x + 9}. \\ \end{align*}

I asked GeoGebra again to plot $f(x)$: and it seems to confirm my conjecture that $$\min_{\sqrt{2}/2 < x < 1} f(x) = 5$$

Question

Is my hypothesis correct? If so, is there a proof of this hypothesis that can be relatively easily by hand (preferably avoiding, say, the evaluation of $f'(x)$)? Geometric proofs will be especially appreciated.

Answer 1

No need to use algera or complicated computational method, which still end up with non-strict solution without using computer programming. Actually, only pure geometry is required to solve this problem.

Connect $OB$ and set point $C$ on $OB$ such that $OC=1/(2\sqrt 2).$ Connect $CP.$ Since $OB=2\sqrt 2, OP=r=1,$ we have $\triangle OPC\sim \triangle OBP$. So, $PC=PB/(2\sqrt 2)$. Therefore, when $A, P $, and $C $ are collinear, we have minimum of $PA+PC=PA+BP/(2\sqrt 2).$ With coordinates of $C$ and $A$ being $(1/4, 1/4)$ and $(2,0)$ respectively, we can quickly find the minimum of $PA+BP/(2\sqrt 2) =AC=5\sqrt2/4.$ Therefore, the minimum of $2\sqrt2 PA+PB=2\sqrt2(5\sqrt2 /4)=5$

Answer 2

Your hypothesis is true.

Indeed the solution of the equation: $$ 2\sqrt{10-8x}+\sqrt{9-4x-4\sqrt{1-x^2}}=5 $$ is $$ x=\frac{2+7\sqrt{46}}{50},\text{ with } \sqrt{1-x^2}=\frac{14-\sqrt{46}}{50}. $$

Substituting this into the derivative of the distance one obtains: $$ \left[-\frac{8}{\sqrt{10-8x}}+\frac{2(x-\sqrt{1-x^2})}{\sqrt{1-x^2}\sqrt{9-4x-4\sqrt{1-x^2}}}\right]_{x=\frac{2+7\sqrt{46}}{50}}\\ =-\frac{4(14+\sqrt{46})}{15}+\frac{4(14+\sqrt{46})}{15}=0. $$

Answer 3

The problem is interesting for sure but I believe that the solution is biased by the fact that $x_A=x_B=y_A$.

I tried to make it more general with $A(x_A,y_A)$ and $B(x_B,0)$ assuming that both points are in the first quadrant. Let $(X,\sqrt{1-X^2})$ the coordinates of point $P$ and assume that we want to minimize $$f = k\lvert{PA}\rvert + \lvert{PB}\rvert$$

We then have $$\lvert{PA}\rvert=\sqrt{(X-x_A)^2+\left(\sqrt{1-X^2}-y_A\right)^2}$$ $$ \lvert{PB}\rvert=\sqrt{-2 x_B X+x_B^2+1}$$ which make $$f=k\sqrt{(X-x_A)^2+\left(\sqrt{1-X^2}-y_A\right)^2}+\sqrt{-2 x_B X+x_B^2+1}$$ to be an hightly nonlinear equation in $X$; this means that, for getting its solution, we need some reasonable estimate.

To generate it, in a preliminary step, let us consider that we want to minimize $$g= k^2\lvert{PA}\rvert^2 + \lvert{PB}\rvert^2$$ which is more pleasant. Computing its derivative, we have $$\frac{dg}{dX}=2 k^2 \left(\frac{X y_A}{\sqrt{1-X^2}}-x_A\right)-2 x_B=0\implies X=\frac{k^2 x_A+x_B}{\sqrt{\left(k^2 x_A+x_B\right)^2+k^4 y_A^2}}$$

For the example given in the post, this would give as an estimate $X=\frac{9}{\sqrt{82}}\approx 0.993884$ while the exact solution of the problem is $X=\frac{2+7 \sqrt{46}}{50}\approx 0.989526$.

Back to $f$, computing its derivative, we need to solve for $X$ $$\frac{k X y_A-k \sqrt{1-X^2} x_A}{\sqrt{1-X^2} \sqrt{-2 X x_A+x_A^2-2 \sqrt{1-X^2} y_A+y_A^2+1}}-\frac{x_B}{\sqrt{-2 X x_B+x_B^2+1}}=0$$ which will require a numerical method such as Newton (this will be simple because we already have a good estimate.

For illustration, let us use $x_A=3$, $y_A=4$ and $x_B=5$ and $k=2\sqrt 2$. The preliminary process gives $X_0=\frac{29}{\sqrt{1865}}\approx 0.671519$.

Newton iterates will then be $$\left( \begin{array}{ccc} n & X_n & f(X_n) \\ 0 & 0.67151942 & 15.72034369 \\ 1 & 0.77667655 & 15.68958161 \\ 2 & 0.75966295 & 15.68798018 \\ 3 & 0.75880236 & 15.68797655 \\ 4 & 0.75880040 & 15.68797655 \end{array} \right)$$