<s>[INST] 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!
 [/INST] It is correct. Here is a way to see it by complex numbers and geometry. By abuse of notations, using complex numbers to represent points, one writes $$A=(a,b)=a+bi,B=(c,d)=c+di,C=(x,y)=x+iy,$$ three complex numbers on the unit circle, so $\overline{A}A=1,$ etc. Then it is easy to see using multiplication of complex numbers that the given expression equals $$|A\overline{C}-\overline{B}|^2+|B\overline{C}+\overline{A}|^2$$
$$=|A-\overline{B}C|^2+|A+\overline{B}C|^2$$
$$=|A-D|^2+|A+D|^2,$$ where $D=\overline{B}C.$ Then by the known theorem for parallelogram, the square sum of diagonals equals the square sum of the sides, which is $4$. QED
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