Given $3$ points on a unit circle, figure out something about them. Question: Given three points $(a, b), (c, d)$ and $(x, y)$ on the unit circle in a rectangular coordinate plane, find the maximum possible value of the expression $(ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy + a)^2 + (dx - cy - b)^2. $
Answer: We will prove that the only value for this expression is $4$.
Without loss of generality, assume that the unit circle is the graph $x^2 + y^2 = 1$. This means that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$.
Expanding the expression, we get
$a^2x^2 + b^2y^2 + c^2 + 2axby - 2acx - 2byc + b^2x^2 + a^2y^2 + d^2 - 2bxay - 2ayd + 2bxd + c^2x^2 + d^2y^2 + a^2 + 2cxdy + 2ady + 2acx + d^2x^2 + c^2y^2 + b^2 - 2dxcy + 2bcy - 2dxb$.
This was a long expression! Fortunately, we see that most of the terms cancel out and we are left with $a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 + c^2x^2 + c^2y^2 + d^2x^2 + d^2y^2 + a^2 + b^2 + c^2 + d^2$.
This can be factored to $(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + 1)$.
Using the fact from above that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$, this expression simplifies to $(a^2 + 1 - a^2 + c^2 + 1 - c^2)(x^2 + 1 - x^2 + 1) = 2(2) = 4$.
So, the answer to our original question is $\boxed{4}$.
Please verify my proof to see if there are any flaws or mistakes with the proof. Thanks in advance!

Here is another way I would approach it. Given all the three points are on the unit circle, their coordinates can be written as,
$(a, b) \mapsto (\cos \theta_1, \sin \theta_1); \ (c, d) \mapsto (\cos\theta_2, \sin\theta_2)$;
$(x, y) \mapsto (\cos\theta_3, \sin\theta_3)$
So, $ \ (ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy +a)^2$
$ + (dx - cy - b)^2 \ $ can be written as,
$(\cos(\theta_1 - \theta_3) - \cos \theta_2)^2 + (\sin(\theta_1 - \theta_3) + \sin \theta_2)^2 + (\cos(\theta_2 - \theta_3) + \cos\theta_1)^2 + (\sin(\theta_2 - \theta_3) - \sin\theta_1)^2$
$ = 2 - 2 \cos (\theta_1 + \theta_2 - \theta_3) + 2 + 2 \cos (\theta_1 + \theta_2 - \theta_3) = 4$
