Locus of $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\leqslant 0$ for a rectangle $ABCD$?

Question

Consider the problem of the title, that is to determine the set of all points $P$ satisfying

$$\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\leqslant 0$$

where $A,B,C$ and $D$ are the vertices of a rectangle. One method is to choose coordinates axes so that $A(0,a)$, $B(c,a)$, $C(c,0)$ and $D(0,0)$ and show by a computation that the equation is equivalent to $$\left(x-\frac{c}{2}\right)^2+\left(y-\frac{a}{2}\right)^2\leqslant \frac{a^2+c^2}{4}$$

This means that the locus is the interior and boundary of the circumcircle of the the rectangle, i.e. the circle with center the center of the rectangle and radius the half-diagonal.

Such a result makes me think that there may be a geometric approach. How can I interpret $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}$ in terms of geometry (area, angle,...) to obtain the same result?

Answer 1

$\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}=0$

$\vec{AP}\cdot\vec{CP}=\vec{BP}\cdot\vec{DP}=0$

$ ∠APC=∠BPD＝90°$

∴ locus of P is circumcircle of the the rectangle.

Finally, this is the condition of mathlove's answer when they equal 0.P is on a circle .$\vec{BP}・\vec{DP}=|BP||DP|cos∠BPD=0⇔∠BPD=90°$

If P is on a circle like ∠BPD<90° ,that equation is not equal to $ 0.$So when angle is right , the radius of that circle is max.

Answer 2

I'm sorry if this is not what you want, but I hope you like this.

We can understand its meaning using only vectors.

Let $O$ be the midpoint of the line segment $AC$.

Then, noting that $\vec{OC}=-\vec{OA},\vec{OD}=-\vec{OB},|\vec{OA}|=|\vec{OB}|$, we have $$\begin{align}\\&\vec{AP}\cdot \vec{CP}+\vec{BP}\cdot \vec{DP}\leqslant 0\\&\iff \left(\vec{OP}-\vec{OA}\right)\cdot \left(\vec{OP}-\vec{OC}\right)+\left(\vec{OP}-\vec{OB}\right)\cdot \left(\vec{OP}-\vec{OD}\right)\leqslant 0\\&\iff \left(\vec{OP}-\vec{OA}\right)\cdot \left(\vec{OP}+\vec{OA}\right)+\left(\vec{OP}-\vec{OB}\right)\cdot \left(\vec{OP}+\vec{OB}\right)\leqslant 0\\&\iff |\vec{OP}|^2-|\vec{OA}|^2+|\vec{OP}|^2-|\vec{OB}|^2\leqslant 0\\&\iff |\vec{OP}|^2\leqslant \frac{|\vec{OA}|^2+|\vec{OB}|^2}{2}\\&\iff |\vec{OP}|\leqslant |\vec{OA}|\end{align}$$

---

Added : How about the following way?

Let $O'$ be the circle whose diameter is the line segment $AC$.

If $P$ is outside $O'$, then $\vec{AP}\cdot\vec{CP}\gt 0$ and $\vec{BP}\cdot\vec{DP}\gt 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\gt 0$.

If $P$ is on $O'$, then $\vec{AP}\cdot\vec{CP}=\vec{BP}\cdot \vec{DP}= 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}= 0$.

If $P$ is inside $O'$, then $\vec{AP}\cdot\vec{CP}\lt 0$ and $\vec{BP}\cdot\vec{DP}\lt 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\lt 0$.

The result follows from these.