Minimum distance from the vertices of a quadrilateral

Question

Let $A(0, 1)$, $B(1, 1)$, $C(1, -1)$, $D(-1, 0)$ be four points. If $P$ be any other point then $PA+PB+PC+PD\ge d$ find $d$.

I tried solving this question using triangle inequality, but I am not sure about my answer.

1.

$$ \left. \begin{matrix} PA+PC \geq AC \\ PB+PD \geq BD \end{matrix} \right \} \implies {d = AC+BD \approx 4.47} $$

2. $$ \left. \begin{matrix} PA+PB \geq AB \\ PB+PC \geq BC \\ PC+PD \geq CD \\ PD+PA \geq DA \end{matrix} \right \} \implies {d = (1/2) \times (AB+BC+CD+DA) \approx 3.3} $$

3. $$ \left. \begin{matrix} PA+PB \geq AB \\ PB+PC \geq BC \\ PC+PD \geq CD \\ PD+PA \geq DA \\ PA+PC \geq AC \\ PB+PD \geq BD \end{matrix} \right \} \implies {d = (1/3) \times (AB+BC+CD+DA+AC+BD) \approx 3.7} $$

I am getting different answers using triangle inequality. I think that there must exist some “sure-fire” method for this question. After seeing the proof of Erdős–Mordell Inequality, I feel that the solution to this question might involve reflections. But I am unsure about it. How to reach the final solution to this question?

Answer 1

Quadrilateral $ABCD$ is a convex quadrilateral, take $P$ inside it, then

$ PA + PB + PC + PD = (PA + PC) + (PB + PD) $

$(PA + PC)$ is minimized if $P$ lies on $AC$ and is in between $A$ and $C$, and similarly $(PB + PD)$ is minimized if $P$ lies on $BD$ and is in between $B$ and $D$.

Thus by drawing $AC$ and $BD$ and finding their intersection we determine the minimizing $P$. And the minimum distance is just $(AC + BD)$, which is

$AC = \sqrt{1^2 + 2^2} = \sqrt{5} $

$ BD = \sqrt{2^2 + 1^2} = \sqrt{5} $

Hence the minimum sum of distances $= 2 \sqrt{5} $

Answer 2

Method 1 is fine. The problem with method 2 and 3 is that for the equality to hold some $P$ , it must satisfy $P \in AB \cap BC \cap CD \cap DA = \phi $(for method 2) or $P \in AB \cap BC \cap CD \cap DA \cap AC \cap BD = \phi$ (method 3). Due to this reason, equality does not hold.

Caution: This does not mean that for any points $A$,$B$,$C$,$D$; $d = AC+BD$. Note that this is only true when $ABCD$ is a convex quadrilateral.

Answer 3

This is not an answer, but a visual comment.

The sum of the distances to the four points as a graph:

This is somewhat reminiscent of 3D metaballs, an approach in computer graphics to produce organic-looking isosurfaces.

Compare to the graph describing the minimum distance to any of the four points:

Both graphs are clipped to within the quadrilateral for simplicity.

Answer 4

Introduce coordinates $P=(x,y)\in\mathbb{R}$ and let $f(P):=\overline{PA}+\overline{PB}+\overline{PC}+\overline{PD}$. We get: $$f(x,y)=f(P)=\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+\sqrt{(x+1)^2+y^2}.$$ This functions is smooth except at the points $A,B,C,D$ where it is not even differentiable. To find the minimum we compute the stationary points by solving the system $\nabla f(x,y)=0$. Using a computer algebra system we get $Q=\left(\frac15,\frac35\right)$. We know that $f$ needs to have a minimum of non-negative value, because e.g. for $P$ outside the rectangle $[-2,2]^2$ we have $f(P)\geq\sqrt{2^2+1^2}+\sqrt{1^2+1^2}+\sqrt{1^2+1^2}+\sqrt{1^2+2^2}=2\sqrt{5}+2\sqrt{2}\approx 7.30$. To find the minimum we only need to evaluate $f$ at the critical point $Q$ and the singular points $A,B,C,D$. It turns out that $f(Q)=2\sqrt{5}=:b\approx 4.47$ is the least of these five values.

Thus $b$ is the optimal (i.e. largest) lower bound for this sum of distances.

This is a rather computational-heavy approach: Solving $\nabla f(x,y)=0$ without a computer is quite tedious. Maybe there is a more comprehensible way. A posteriori we see that $A,B,C,D$ are the vertices of a convex quadrilateral $F$ and that $Q$ is the intersection point of the two diagonals. As pointed out in the comments this is explored in this post.

Coming back to your deliberations: You found three lower bounds $b_1, b_2, b_3$. In particular $b_1=b$ coincides with the optimal bound. All three are valid lower bounds as they are all at most $b=2\sqrt{5}$. So for the problem as stated any $d\leq b$ would do, in particular any of your values would be OK, or even trivial $d$ like $d=0$ or $d=-\pi$. What is probably meant is to find the optimal $b$. But as students are sloppy with their homework so are often teachers in giving tasks.