Reminds me of British Flag Theorem

https://www.geogebra.org/

By Pythagorean Theorem, $PA^2+PB^2+PC^2+PD^2 = 2(PL^2+PK^2+PM^2+PJ^2)$ Let PL = 2+x, PK = 2-x, PM = 2-y and PJ = 2+y.

Hence the required answer is $2((2+x)^2+(2-x)^2+(2+y)^2+(2-y)^2)$ $= 2(2(x^2+y^2)+16)$ $= \boxed{48}$

The last step follows from $x^2+y^2=4$

I did it by the assumption that the point $P$ lies on the point where one of the sides (Let's take $AB$ ) is tangent to the incircle.

$\therefore PA=PB=2$ and

$PC=PD=2\sqrt{5}$

Therefore,

$PA^2+PB^2+PC^2+PD^2$

$2((2)^2)+2((2\sqrt{5})^2)$

$8+40=\boxed{48}$

Establish a coordinate system with the origin at the center of the square. Parameterize point P on the circle as $2 \times ( \cos t, \sin t)$ . Let A be (-2, 2), B be (2, 2), C be (-2, -2), and D be (2, -2). One can quickly show the following:

$PA^{2} = 4 \times ( (\cos t + 1)^{2} +(\sin t - 1)^2)$

$PB^{2} = 4 \times ( (\cos t - 1)^{2} +(\sin t - 1)^2)$

$PC^{2} = 4 \times ( (\cos t + 1)^{2} +(\sin t +1)^2)$

$PD^{2} = 4 \times ( (\cos t - 1)^{2} +(\sin t + 1)^2)$

Sum the squares, expand the binomials, and use the identity $\sin^{2} t + \cos^{2} t = 1$ to show $PA^{2} + PB^{2} + PC^{2} + PD^{2} = 48$ .

This method also shows the invariance property of $PA^{2} + PB^{2} + PC^{2} + PD^{2}$ .

While doing this problem, I developed a general solution that for any side length, S, for the square ABCD the answer will be $3*S^2$ . In this case, $3*(4)^2 = 48$ . If the side length was a product of Euler's Number, e, with the number $pi$ , then the answer would be $218.7811818$

Treat this information as a further challenge while I continue to learn LaTeX so in the future I could publish my work on Brilliant.

From the wording of the problem , the answer should be the same NO MATTER which point P on the circle is chosen So, if we select the point P on the tangent point, at the mid -point of one of the square sides, then if x denotes 1/2 of the square side length = 2 units, and we apply the Pythagorean Theorem solving for the hypotenese twice and then add x.x + x.x we get a total for the sum of the indicated squares 12 .x.x= (12) (2)(2)= 48

The solution is very easy when we take P is on square and circle .then PA^2+PB^2+PC^2+PD^2 =2^2+2^2+(4^2+2^2)+(4^2+2^2) =48.

I solved it using the parallel axis theorem to calculate moment of inertia. We can assume that there is some arbitrary mass 'm' at each corner of the square (A,B,C and D). Then we can use parallel axis theorem to evaluate moment of inertia of the system at P which is given as m(PA^2)+m(PB^2)+m(PC^2)+m(PD^2)=4m(PG^2)+m(GA^2)+m(GB^2)+m(GC^2)+m(GD^2) We know that PG=2 and GA=GB=GC=GD=sqrt(8). Hence the solution. I wonder if the question could have been framed in a way so that it would have been classified as a difficult one.

AP^2=AC^2+CP^2 = 4^2+2^2 = 20 = BP^2
CP^2=DP^2=2^2=4
AP^2+BP^2+CP^2+DP^2 = 20+20+4+4=48

British Flag Theorem says that $PA^{2} + PC^{2}=PB^{2} + PD^{2}$ . Then $PA^{2} + PB^{2} + PC^{2} + PD^{2} = 2 \times (PA^{2} + PC^{2})$ . Let's say O is the center of the inner circle of the square. Applying Apollonius' theorem, we have for the triangle PAC and the median PO the following relation: $PA^{2} + PC^{2} = 2 \times (PO^{2} + AO^{2})$ . $PO=2$ as half of the side length of the square and $AO = 2 \sqrt{2}$ (using Pythagoras' Theorem). Then our sum is $2 \times 2 \times (2^{2} + (2\sqrt{2}) ^ {2})=48$