Solution:
By Furaken
[asy] pair O=(0,0); pair X=(1,0); pair Y=(0,1); pair A=(0.5,0); pair B=(0,sin(pi/3)); dot(O); dot(X); dot(Y); dot(A); dot(B); draw(X--O--Y); draw(A--B); label("$B'$", B, W); label("$A'$", A, S); label("$O$", O, SW); pair C=(1/8,3*sqrt(3)/8); dot(C); pair D=(1/8,0); dot(D); pair E=(0,3*sqrt(3)/8); dot(E); label("$C$", C, NE); label("$D$", D, S); label("$E$", E, W); draw(D--C--E); [/asy]
Let $C = (\tfrac18,\tfrac{3\sqrt3}8)$.  This is sus, furaken randomly guessed C and proceeded to prove it works Draw a line through $C$ intersecting the $x$-axis at $A'$ and the $y$-axis at $B'$. We shall show that $A'B' \ge 1$, and that equality only holds when $A'=A$ and $B'=B$.
Let $\theta = \angle OA'C$. Draw $CD$ perpendicular to the $x$-axis and $CE$ perpendicular to the $y$-axis as shown in the diagram. Then
\[8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\]
By some inequality (I forgot its name),
\[\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64\]
We know that $\sin^2\theta + \cos^2\theta = 1$. Thus $\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8$. Equality holds if and only if
\[\frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \sin^2\theta : \cos^2\theta\]
which occurs when $\theta=\tfrac\pi3$. Guess what, $\angle OAB$ happens to be $\tfrac\pi3$, thus $A'=A$ and $B'=B$. Thus, $AB$ is the only segment in $\mathcal{F}$ that passes through $C$. Finally, we calculate $OC^2 = \tfrac1{64} + \tfrac{27}{64} = \tfrac7{16}$, and the answer is $\boxed{023}$.
~Furaken
$y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}$
Now, we want to find $\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}$. By L'Hôpital's rule, we get $\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}$. This means that $y=\frac{3\sqrt{3}}{8}\implies OC^2=\frac{7}{16}$, so we get $\boxed{023}$.
~Bluesoul
The equation of line $AB$ is \[ y = \frac{\sqrt{3}}{2} x - \sqrt{3} x.  \hspace{1cm} (1) \]
The position of line  $PQ$ can be characterized by $\angle QPO$, denoted as $\theta$.
Thus, the equation of line $PQ$ is
\[ y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) \]
Solving (1) and (2), the $x$-coordinate of the intersecting point of lines $AB$ and $PQ$ satisfies the following equation:
\[ \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} + \frac{x}{\cos \theta} = 1 . \hspace{1cm} (1) \]
We denote the L.H.S. as $f \left( \theta; x \right)$.
We observe that $f \left( 60^\circ ; x \right) = 1$ for all $x$.
Therefore, the point $C$ that this problem asks us to find can be equivalently stated in the following way:
We interpret Equation (1) as a parameterized equation that $x$ is a tuning parameter and $\theta$ is a variable that shall be solved and expressed in terms of $x$.
In Equation (1), there exists a unique $x \in \left( 0, 1 \right)$, denoted as $x_C$ ($x$-coordinate of point $C$), such that the only solution is $\theta = 60^\circ$. For all other $x \in \left( 0, 1 \right) \backslash \{ x_C \}$, there are more than one solutions with one solution $\theta = 60^\circ$ and at least another solution.
Given that function $f \left( \theta ; x \right)$ is differentiable, the above condition is equivalent to the first-order-condition
\[ \frac{\partial f \left( \theta ; x_C \right) }{\partial \theta} \bigg|_{\theta = 60^\circ} = 0 . \]
Calculating derivatives in this equation, we get
\[ - \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} = 0. \]
By solving this equation, we get
\[ x_C = \frac{1}{8} . \]
Plugging this into Equation (1), we get the $y$-coordinate of point $C$:
\[ y_C = \frac{3 \sqrt{3}}{8} . \]
Therefore,
\begin{align*}
OC^2 & = x_C^2 + y_C^2 \\
& = \frac{7}{16} .
\end{align*}
Therefore, the answer is $7 + 16 = \boxed{\textbf{(23) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Let $s$ be a segment in $\mathcal{F}$ with x-intercept $a$ and y-intercept $b$. We can write $s$ as
\begin{align*}
\frac{x}{a} + \frac{y}{b} &= 1 \\
y &= b(1 - \frac{x}{a}).
\end{align*}
Let the unique point in the first quadrant $(x, y)$ lie on $s$ and no other segment in $\mathcal{F}$. We can find $x$ by solving
\[b(1 - \frac{x}{a}) = (b + db)(1 - \frac{x}{a + da})\]
and taking the limit as $da, db \to 0$. Since $s$ has length $1$, $a^2 + b^2 = 1^2$ by the Pythagorean theorem. Solving this for $db$, we get
\begin{align*}
a^2 + b^2 &= 1 \\
b^2 &= 1 - a^2 \\
\frac{db^2}{da} &= \frac{d(1 - a^2)}{da} \\
2a\frac{db}{da} &= -2a \\
db &= -\frac{a}{b}da.
\end{align*}
After we substitute $db = -\frac{a}{b}da$, the equation for $x$ becomes
\[b(1 - \frac{x}{a}) = (b -\frac{a}{b} da)(1 - \frac{x}{a + da}).\]
In $\overline{AB}$, $a = \frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}$. To find the x-coordinate of $C$, we substitute these into the equation for $x$ and get
\begin{align*}
\frac{\sqrt{3}}{2}(1 - \frac{x}{\frac{1}{2}}) &= (\frac{\sqrt{3}}{2} - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} da)(1 - \frac{x}{\frac{1}{2} + da}) \\
\frac{\sqrt{3}}{2}(1 - 2x) &= (\frac{\sqrt{3}}{2} - \frac{da}{\sqrt{3}})(1 - \frac{x}{\frac{1 + 2da}{2}}) \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}}(1 - \frac{2x}{1 + 2da}) \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}} \cdot \frac{1 + 2da - 2x}{1 + 2da} \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 + 6da - 6x - 2da - 4da^2 + 4xda}{2\sqrt{3} + 4\sqrt{3}da} \\
(\frac{\sqrt{3}}{2} - \sqrt{3}x)(2\sqrt{3} + 4\sqrt{3}da) &= 3 + 6da - 6x - 2da - 4da^2 + 4xda \\
3 + 6da - 6x - 12xda &= 3 + 4da - 6x - 4da^2 + 4xda \\
2da &= -4da^2 + 16xda \\
16xda &= 2da + 4da^2 \\
x &= \frac{da + 2da^2}{8da}.
\end{align*}
We take the limit as $da \to 0$ to get
\[x = \lim_{da \to 0} \frac{da + 2da^2}{8da} = \lim_{da \to 0} \frac{1 + 2da}{8} = \frac{1}{8}.\]
We substitute $x = \frac{1}{8}$ into the equation for $\overline{AB}$ to find the y-coordinate of $C$:
\[y = b(1 - \frac{x}{a}) = \frac{\sqrt{3}}{2}(1 - \frac{\frac{1}{8}}{\frac{1}{2}}) = \frac{3\sqrt{3}}{8}.\]
The problem asks for
\[OC^2 = x^2 + y^2 = (\frac{1}{8})^2 + (\frac{3\sqrt{3}}{8})^2 = \frac{7}{16} = \frac{p}{q},\]
so $p + q = 7 + 16 = \boxed{023}$.
Let's move a little bit from $A$ to $A_1$, then $B$ must move to $B_1$ to keep $A_1B_1 = 1$. $AB$ intersects with $A_1B_1$ at $C$. Pick points $A_2$ and $B_2$ on $CA_1$ and $CB$ such that $CA_2 = CA$, $CB_2 = CB_1$, we have $A_1A_2 = BB_2$. Since $AA_1$ is very small, $\angle CA_1A \approx 60^\circ$, $\angle CBB_1 \approx 30^\circ$, so $AA_2\approx \sqrt{3}A_1A_2$, $B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2$, by similarity, $\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3}A_1A_2}{\frac{1}{\sqrt{3}}BB_2} = 3$. So the coordinates of $C$ is $\left(\frac{1}{8}, \frac{3\sqrt{3}}{8}\right)$.
so $OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}$, the answer is $\boxed{023}$.

Answer: 023

<｜begin▁of▁sentence｜><｜User｜>Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\),  that does not belong to any segment from \(\mathcal{F}\) other than \(\overline{AB}\). Then \(OC^2=\tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).<｜Assistant｜><think>

Okay, so I need to solve this geometry problem. Let me read it again and make sure I understand all the parts.

We have points O at (0,0), A at (1/2, 0), and B at (0, sqrt(3)/2). The family F consists of segments PQ of unit length in the first quadrant, with P on the x-axis and Q on the y-axis. There's a unique point C on AB (other than A and B) that isn't on any segment from F except AB itself. We need to find OC squared, which will be a fraction p/q, and then compute p + q.

Hmm. Let me start by visualizing the problem. Points O, A, and B form a triangle. Since A is (1/2, 0) and B is (0, sqrt(3)/2), the segment AB connects these two points. The family F consists of all unit-length segments from the x-axis to the y-axis in the first quadrant. So each segment PQ has P somewhere on the x-axis (so coordinates (p, 0)) and Q somewhere on the y-axis (coordinates (0, q)), with the distance between them being 1. Since they're in the first quadrant, p and q are non-negative.

First, maybe I should parametrize the family F. Let me think. For any such segment PQ, the length is 1, so by the distance formula:

sqrt( (p - 0)^2 + (0 - q)^2 ) = 1, so p² + q² = 1.

Therefore, the coordinates of P are (p, 0) and Q are (0, q) where p and q satisfy p² + q² = 1. Since they are in the first quadrant, p and q are between 0 and 1.

So the segment PQ can be represented by the line connecting (p, 0) to (0, q), with p² + q² = 1.

Now, the problem states that there is a unique point C on AB (other than A and B) that is not on any other segment from F except AB itself. So, AB is part of the family F? Wait, let me check.

Wait, AB is the segment from A(1/2, 0) to B(0, sqrt(3)/2). Let me check if AB is in the family F. Let's see, the length of AB should be 1? Let me compute the distance between A and B.

Distance AB: sqrt( (1/2 - 0)^2 + (0 - sqrt(3)/2)^2 ) = sqrt(1/4 + 3/4) = sqrt(1) = 1. Yes, so AB is indeed a member of the family F. So AB is part of F. So the point C is on AB, but not on any other segment from F except AB. So, C is a point that is only on AB and no other PQ in F. So, we need to find such a point C on AB, and then compute OC².

So, the key is to find a point C on AB such that C is not on any other segment PQ (from F) except AB. Since AB is part of F, that's okay, but no other PQ in F should pass through C. And this point is unique, so there's only one such point besides A and B.

Hmm. So, perhaps the idea is that for most points on AB, there is another segment PQ in F that also passes through that point. Except for point C, which is only on AB. So, we need to find such a point.

Alternatively, maybe the point C is where the envelope of all these segments PQ touches AB? Because the envelope would be the boundary of the region covered by all these segments, so maybe the point where the envelope is tangent to AB is the point that's only on AB and not on any other segment?

Alternatively, perhaps parametrizing the problem would help. Let me try to parametrize the family F.

Let me consider a general segment PQ in F. Let me parameterize it by the angle or by one variable. Since p² + q² = 1, we can write p = cosθ and q = sinθ, where θ is between 0 and π/2. Because in the first quadrant, so θ from 0 to π/2. Then, the segment PQ goes from (cosθ, 0) to (0, sinθ). So, the equation of the line PQ can be written as:

The line from (cosθ, 0) to (0, sinθ). The slope is (sinθ - 0)/(0 - cosθ) = -tanθ. So the equation is y = -tanθ (x - cosθ). Wait, let me check:

Alternatively, using the two-point formula. The line through (p,0) and (0,q) can be written as x/p + y/q = 1. Since p² + q² =1, so that's the equation of the line.

So, the equation is x/p + y/q = 1, with p² + q² =1.

So, for each θ, with p = cosθ, q = sinθ, the line equation becomes x/cosθ + y/sinθ = 1.

So, the family of lines is x/cosθ + y/sinθ =1, where θ ∈ (0, π/2).

Now, the segment AB is part of this family when θ is such that p = 1/2 and q = sqrt(3)/2. Let me check:

If p = 1/2, then q should be sqrt(1 - (1/2)^2) = sqrt(3)/2, which matches point B. So, θ in this case is 60 degrees, since cosθ = 1/2, so θ = π/3. So AB corresponds to θ = π/3.

Therefore, the line AB is x/(1/2) + y/(sqrt(3)/2) = 1, which simplifies to 2x + (2/sqrt(3)) y = 1. Let me confirm with points A and B:

At point A (1/2, 0): 2*(1/2) + (2/sqrt(3))*0 = 1, which works. At point B (0, sqrt(3)/2): 2*0 + (2/sqrt(3))*(sqrt(3)/2) = 1, yes. So that's correct.

So, the problem is to find a point C on AB (other than A and B) that is not on any other line from the family F except the line AB itself. So, for point C, if we suppose that it's on another line from F, then that would mean there exists some θ ≠ π/3 such that the line x/cosθ + y/sinθ =1 passes through C. So, to have C not lie on any other such line, the system of equations (the line AB and another line from F) must intersect only at A and B, except for point C which is only on AB.

Alternatively, perhaps the point C is the point where the envelope of the family F touches the line AB. The envelope is the curve that is tangent to each member of the family at some point. If the envelope intersects AB at point C, then that point would lie on only one member of the family (the tangent one), which is AB itself. So that might be the point.

Therefore, perhaps I need to compute the envelope of the family of lines F and find its intersection with AB, other than A and B.

Alternatively, let me recall that the envelope of a family of curves is found by solving the equation of the curve and the derivative with respect to the parameter set to zero.

So, let me try that approach.

Let me consider the family of lines given by F(θ): x/cosθ + y/sinθ =1, for θ in (0, π/2).

To find the envelope, we can compute the derivative of F(θ) with respect to θ and set it equal to zero along with F(θ)=0.

Wait, actually, the envelope is found by solving F(θ, x, y) = 0 and dF/dθ = 0.

Let me write F(θ) as:

F(θ) = x secθ + y cscθ -1 =0.

Then, compute the derivative of F with respect to θ:

dF/dθ = x secθ tanθ - y cscθ cotθ =0.

So, the envelope is given by the system:

x secθ + y cscθ =1,

x secθ tanθ - y cscθ cotθ =0.

Let me write these equations more clearly.

Let me denote:

Equation 1: x / cosθ + y / sinθ =1,

Equation 2: x (sinθ / cos²θ) - y (cosθ / sin²θ) =0.

Hmm, perhaps we can solve these equations for x and y in terms of θ, and then eliminate θ to find the envelope curve.

Alternatively, let me try to solve equations 1 and 2 for x and y.

Let me denote equation 1 as:

x / cosθ + y / sinθ =1. (1)

Equation 2 is:

x (sinθ / cos²θ) - y (cosθ / sin²θ) =0. (2)

Let me write equation (2) as:

x (sinθ / cos²θ) = y (cosθ / sin²θ)

So, x = y (cosθ / sin²θ) / (sinθ / cos²θ) ) = y (cosθ / sin²θ) * (cos²θ / sinθ) ) = y (cos³θ / sin³θ )

Therefore, x = y (cosθ / sinθ)^3.

Let me denote t = cosθ / sinθ = cotθ. Then, x = y t^3.

Alternatively, let me write x = y (cosθ / sinθ)^3. Let me call this equation (3).

Now, substitute equation (3) into equation (1):

(y (cosθ / sinθ)^3 ) / cosθ + y / sinθ =1

Simplify term by term:

First term: y (cosθ / sinθ)^3 / cosθ = y (cos²θ / sin³θ )

Second term: y / sinθ

So, altogether:

y (cos²θ / sin³θ ) + y / sinθ =1

Factor out y / sin³θ:

Wait, let me see:

First term: y cos²θ / sin³θ,

Second term: y / sinθ = y sin²θ / sin³θ,

Therefore, combining:

y (cos²θ + sin²θ)/ sin³θ =1,

But cos²θ + sin²θ =1, so this simplifies to:

y / sin³θ =1,

Therefore, y = sin³θ,

Then, from equation (3):

x = y (cosθ / sinθ)^3 = sin³θ * (cos³θ / sin³θ ) = cos³θ.

Therefore, the envelope is given parametrically by:

x = cos³θ,

y = sin³θ,

for θ in (0, π/2).

Therefore, the envelope is the astroid given by x^(2/3) + y^(2/3) =1. Wait, let me check:

If x = cos³θ, then cosθ = x^(1/3),

Similarly, sinθ = y^(1/3),

Therefore, cos²θ + sin²θ = x^(2/3) + y^(2/3) =1,

Yes, so the envelope is the astroid x^(2/3) + y^(2/3) =1.

So, the envelope of the family F is this astroid. Therefore, the astroid is the curve that is tangent to each of the lines in F.

Therefore, the point C must be the point where the line AB is tangent to the astroid. Because at that point, the line AB is tangent to the envelope, so it only touches the astroid at that point, meaning that the point is only on AB and no other segment from F. Since the envelope is the boundary of the region covered by the segments, the tangent point would only lie on AB and no other segment.

Therefore, to find point C, we need to find the point of tangency between line AB and the astroid x^(2/3) + y^(2/3) =1.

Alternatively, since AB is part of the family F (when θ=π/3), so the line AB is one of the lines in F, so the envelope is tangent to AB at some point. Since the envelope is formed by the lines of F, each line is tangent to the envelope at some point. Wait, actually, each line in the family is tangent to the envelope at the point where the envelope is "generated" by that line. So, for each θ, the line F(θ) is tangent to the envelope at the point (cos³θ, sin³θ). Therefore, the line AB (which is θ=π/3) is tangent to the astroid at the point (cos³(π/3), sin³(π/3)).

Let me compute that:

cos(π/3) = 1/2, so cos³(π/3) = (1/2)^3 = 1/8,

sin(π/3) = sqrt(3)/2, so sin³(π/3) = (sqrt(3)/2)^3 = (3*sqrt(3))/8.

Therefore, the point of tangency is (1/8, (3√3)/8). But wait, is this point on AB?

Wait, let me check if this point lies on AB.

The line AB goes from A(1/2,0) to B(0, sqrt(3)/2). Let me parametrize AB. Let me write the parametric equations for AB.

Parametrize AB from A to B. Let parameter t go from 0 to 1.

x(t) = 1/2*(1 - t),

y(t) = (sqrt(3)/2)*t.

So, at t=0, we are at A(1/2,0); at t=1, at B(0, sqrt(3)/2).

Now, let's see if (1/8, 3√3/8) is on AB.

Let me see if there exists a t such that:

1/2*(1 - t) = 1/8,

and (sqrt(3)/2)*t = 3√3/8.

First equation:

1/2*(1 - t) = 1/8 ⇒ 1 - t = 1/4 ⇒ t = 3/4.

Second equation:

(sqrt(3)/2)*t = 3√3/8 ⇒ t = (3√3/8) / (sqrt(3)/2) ) = (3/8)/(1/2) ) = 3/4. So yes, t=3/4. Therefore, the point (1/8, 3√3/8) is indeed on AB, at t=3/4. Therefore, this is the point C where the astroid is tangent to AB, so this point is only on AB and no other segment from F. Therefore, this must be the point C we are looking for.

Therefore, the coordinates of C are (1/8, 3√3/8). Then, OC² is (1/8)^2 + (3√3/8)^2.

Let me compute that:

(1/64) + (27/64) = (1 + 27)/64 = 28/64 = 7/16.

Wait, 3√3 squared is 9*3=27, yes. So 27/64 plus 1/64 is 28/64, reduces to 7/16. So OC² is 7/16, so p=7, q=16, which are coprime. Therefore, p+q=23.

Wait, but hold on. The problem states "there is a unique point C on AB, distinct from A and B, that does not belong to any segment from F other than AB". So according to this, the point C is the tangency point of the envelope (astroid) with AB, which is (1/8, 3√3/8). But let me confirm that this is indeed the only such point.

Alternatively, perhaps I made an error in assuming that the tangency point is the only such point. Let me think again.

Suppose another point D on AB, different from A, B, and C. Then, since D is on AB, which is part of F, but is it also on another segment from F?

If D is not the tangency point, then perhaps there is another segment PQ in F that also passes through D. The envelope is the boundary of the region covered by the segments, so points inside the astroid are covered by multiple segments, and points on the astroid are covered by exactly one segment (the tangent one). Since AB is part of the family, the point C is on the astroid and on AB, so it's only on AB. Other points on AB would be inside the astroid? Wait, let me see.

Wait, the astroid is the envelope, so the region covered by the segments PQ is the interior of the astroid? Let me think. Each segment PQ is a line from (p,0) to (0,q) with p² + q² =1. The envelope is the astroid x^(2/3)+y^(2/3)=1. The region covered by all these segments would be the region under the astroid. So points inside the astroid are covered by multiple segments, and points on the astroid are covered by exactly one segment. Therefore, any point on AB that is not on the astroid would lie inside the astroid, hence would be covered by infinitely many segments. The only point on AB that is on the astroid is the tangency point C, so that point is only on AB. Therefore, that must be the unique point C. Therefore, the calculation seems correct.

Therefore, OC squared is 7/16, so the answer is 7 + 16 = 23.

Wait, but let me confirm the coordinates again. The point C is (cos³θ, sin³θ) where θ=π/3. So cos(π/3)=0.5, so cos³ is 1/8, sin(π/3)=sqrt(3)/2, so sin³ is (sqrt(3)/2)^3 = 3*sqrt(3)/8. So coordinates (1/8, 3√3/8). Then, squared distance from O is (1/8)^2 + (3√3/8)^2 = 1/64 + 27/64 = 28/64 = 7/16. So yes, that's correct.

Hmm, but let me check if there's another way to approach this problem in case I missed something.

Alternatively, suppose we parametrize point C on AB. Let me parametrize AB as follows. Since AB goes from A(1/2,0) to B(0, sqrt(3)/2), we can write any point on AB as ( (1/2)(1 - t), (sqrt(3)/2) t ) where t is between 0 and 1. So, coordinates ( (1 - t)/2, (sqrt(3) t)/2 ). Let me denote this as (x, y).

We need to find t such that this point is not on any other segment PQ from F except AB. So, suppose that for some other θ (not π/3), the point (x,y) lies on the line x/cosθ + y/sinθ =1.

So, substituting x and y:

[ (1 - t)/2 ] / cosθ + [ (sqrt(3) t)/2 ] / sinθ =1.

We need this equation to have no solution for θ except θ=π/3 (which corresponds to AB). Therefore, for the point C, the equation above should only hold when θ=π/3. So, to find t such that the equation above has only one solution θ=π/3.

Alternatively, to have the equation have a unique solution θ, which would be when the equation is satisfied only at θ=π/3. So, perhaps setting up the equation and finding t such that the equation has a double root at θ=π/3?

Alternatively, maybe set up the equation and find t such that when we substitute θ=π/3, it's a solution, and the derivative with respect to θ also vanishes there, making it a point of tangency. That would correspond to the envelope condition.

Let me try this approach.

Let me denote the equation:

[ (1 - t)/2 ] / cosθ + [ (sqrt(3) t)/2 ] / sinθ =1.

Multiply both sides by 2:

(1 - t)/cosθ + (sqrt(3) t)/sinθ = 2.

Let me write this as:

(1 - t) secθ + sqrt(3) t cscθ = 2.

Let me denote f(θ) = (1 - t) secθ + sqrt(3) t cscθ - 2 =0.

We want this equation to hold only when θ=π/3. To ensure that, the function f(θ) has a double root at θ=π/3, meaning that f(π/3)=0 and f’(π/3)=0.

First, check f(π/3)=0:

Compute f(π/3):

(1 - t) sec(π/3) + sqrt(3) t csc(π/3) -2.

Compute sec(π/3)=2, csc(π/3)=2/sqrt(3).

Therefore,

(1 - t)*2 + sqrt(3)*t*(2/sqrt(3)) -2 =

2(1 - t) + 2t -2 =

2 - 2t + 2t -2 = 0. So yes, f(π/3)=0 for any t. That's because the point is on AB, which is part of the family F at θ=π/3. So, the equation is satisfied for θ=π/3 regardless of t? Wait, no, actually, the coordinates of the point C are parametrized by t, so when we substitute the coordinates into the equation for the line at angle θ, we get an equation that must hold for θ=π/3 (since the point is on AB), but we want it to not hold for any other θ. To ensure that, the equation f(θ)=0 has only θ=π/3 as a solution. To make that happen, perhaps the function f(θ) has a minimum at θ=π/3 and is above zero elsewhere, or something like that. Alternatively, the equation f(θ)=0 has a unique solution at θ=π/3, which would require that the derivative at θ=π/3 is zero, making it a point of tangency.

Therefore, to have uniqueness, we set the derivative f’(θ) at θ=π/3 to zero.

Compute f’(θ):

The derivative of f(θ) is:

(1 - t) secθ tanθ + sqrt(3) t (-cscθ cotθ) .

Set this equal to zero at θ=π/3.

Compute each term:

First term: (1 - t) sec(π/3) tan(π/3)

sec(π/3)=2, tan(π/3)=sqrt(3), so first term is (1 - t)*2*sqrt(3).

Second term: sqrt(3) t * (-csc(π/3) cot(π/3))

csc(π/3)=2/sqrt(3), cot(π/3)=1/sqrt(3). So,

Second term: sqrt(3)*t * [ - (2/sqrt(3))*(1/sqrt(3)) ) ]

Simplify inside the brackets: - (2/(3)).

Thus, second term: sqrt(3)*t*(-2/(3)) = - (2 sqrt(3)/3 ) t.

Therefore, total derivative:

(1 - t)*2*sqrt(3) - (2 sqrt(3)/3 ) t =0.

Factor out 2 sqrt(3):

2 sqrt(3) [ (1 - t) - (t)/3 ] =0.

Since 2 sqrt(3) ≠0, the expression in brackets must be zero:

(1 - t) - (t)/3 =0,

Multiply through by 3 to eliminate denominators:

3(1 - t) - t =0,

3 - 3t - t =0,

3 -4t =0,

So 4t=3 ⇒ t= 3/4.

Therefore, t=3/4. Therefore, the point C is at t=3/4 on AB, which gives coordinates:

x = (1 - 3/4)/2 = (1/4)/2 = 1/8,

y = (sqrt(3)/2)*(3/4) = (3 sqrt(3))/8,

Which matches the previous result. Therefore, OC² is (1/8)^2 + (3√3/8)^2 = 1/64 + 27/64 = 28/64 = 7/16. So the answer is 7 + 16 =23.

Therefore, I think this is correct. Let me just check once more if there could be another point. Suppose we pick another point on AB, say the midpoint. The midpoint of AB would be at t=1/2, so coordinates ( (1 - 1/2)/2, (sqrt(3)/2 *1/2 )) = (1/4, sqrt(3)/4). Let me see if this point is on another segment from F.

Suppose there exists θ ≠ π/3 such that x/cosθ + y/sinθ =1.

Plugging in x=1/4, y=sqrt(3)/4,

(1/4)/cosθ + (sqrt(3)/4)/sinθ =1,

Multiply both sides by 4:

1/cosθ + sqrt(3)/sinθ =4.

Let me see if this equation has a solution θ ≠ π/3.

Let me set θ=π/6 (30 degrees). Then cosθ=√3/2, sinθ=1/2,

Left side: 1/(√3/2) + sqrt(3)/(1/2) = 2/√3 + 2 sqrt(3) ≈ 1.1547 + 3.464 ≈ 4.618, which is greater than 4.

Try θ=π/4 (45 degrees):

cosθ=sinθ=√2/2≈0.7071,

Left side: 1/(0.7071) + sqrt(3)/0.7071 ≈1.4142 + 2.4495≈3.863, which is less than 4.

So between π/6 and π/4, the left side decreases from ~4.6 to ~3.86. At θ=π/3, let's check:

θ=π/3, cosθ=0.5, sinθ=√3/2≈0.8660,

Left side: 1/0.5 + sqrt(3)/(sqrt(3)/2)=2 + 2=4. So exactly 4. So the equation is satisfied at θ=π/3, which is the AB line. But since at θ=π/6 it's higher, and θ=π/4 it's lower, there must be another θ between π/4 and π/3 where the left side equals 4? Wait, wait, at θ=π/3 it's exactly 4. Wait, when θ approaches π/2, cosθ approaches 0, so the first term blows up. Wait, let me check θ between π/3 and π/2.

Take θ=π/3.5 ≈ 51.4 degrees. Let me compute numerically:

Let me pick θ=60 degrees (π/3) gives exactly 4. Let me try θ= π/2.5 ≈ 72 degrees.

cos(72°)= approx 0.3090, sin(72°)= approx 0.9511.

Left side: 1/0.3090 + sqrt(3)/0.9511 ≈3.236 + 1.732/0.9511≈3.236 +1.820≈5.056>4.

Hmm, so between π/3 and π/2, the left side increases beyond 4. Between π/4 and π/3, it goes from ~3.86 to 4. So the equation 1/cosθ + sqrt(3)/sinθ =4 is only satisfied at θ=π/3. Wait, but when θ approaches π/2, the first term goes to infinity, so there must be some θ between π/3 and π/2 where it equals 4 again? Wait, let me check θ= let's say 75 degrees:

cos(75°)= approx 0.2588, sin(75°)= approx 0.9659.

Left side: 1/0.2588 + sqrt(3)/0.9659 ≈3.863 + 1.732/0.9659≈3.863 +1.792≈5.655>4.

Hmm, so maybe it's only equal to 4 at θ=π/3. Wait, let me think again. Let me set variable substitution.

Let me set t = θ, and let me write the equation:

1/cosθ + sqrt(3)/sinθ =4.

Let me write this as:

secθ + sqrt(3) cscθ =4.

Let me set x = θ. Let me see if this equation has another solution besides x=π/3.

Let me consider the function g(θ)=secθ + sqrt(3)cscθ. We can analyze its behavior.

Compute derivative g’(θ)= secθ tanθ - sqrt(3) cscθ cotθ.

Set this equal to zero to find extrema.

But perhaps instead, let me see at θ=π/3, g(θ)=2 + 2=4.

At θ approaching 0+, secθ approaches 1, cscθ approaches infinity, so g(θ) approaches infinity.

At θ approaching π/2-, secθ approaches infinity, cscθ approaches 1/sqrt(3), so g(θ) approaches infinity.

At θ=π/4, as before, g(π/4)=sqrt(2) + sqrt(3)*sqrt(2)≈1.414 + 2.449≈3.863 <4.

Therefore, the function g(θ) has a minimum somewhere between π/4 and π/3?

Wait, let me compute g(π/3.5). Let me pick θ=50 degrees (approx 0.8727 radians):

cos(50°)=0.6428, sin(50°)=0.7660,

secθ≈1.5557,

sqrt(3)/sinθ≈1.732/0.7660≈2.26,

Total≈1.5557+2.26≈3.815 <4.

At θ=55 degrees (0.9599 radians):

cos≈0.5736, sin≈0.8192,

sec≈1.743,

sqrt(3)/sin≈1.732/0.8192≈2.114,

Total≈1.743+2.114≈3.857 <4.

At θ=57 degrees (0.994 radians):

cos≈0.5446, sin≈0.8387,

sec≈1.836,

sqrt(3)/sin≈1.732/0.8387≈2.066,

Total≈1.836+2.066≈3.902 <4.

At θ=58 degrees (1.012 radians):

cos≈0.5299, sin≈0.8480,

sec≈1.889,

sqrt(3)/sin≈1.732/0.8480≈2.042,

Total≈1.889+2.042≈3.931 <4.

At θ=59 degrees (1.029 radians):

cos≈0.5150, sin≈0.8572,

sec≈1.942,

sqrt(3)/sin≈1.732/0.8572≈2.021,

Total≈1.942+2.021≈3.963 <4.

At θ=59.5 degrees (1.039 radians):

cos≈0.5077, sin≈0.8562,

Wait, sin(59.5)=sin(π - 120.5)= wait, better compute numerically:

sin(59.5°)= approx sin(59°30')≈0.8632,

cos≈0.5049,

sec≈1.981,

sqrt(3)/sin≈1.732/0.8632≈2.006,

Total≈1.981 +2.006≈3.987≈4.

Close to 4. So maybe at around 59.5 degrees, it reaches 4 again?

Wait, let me compute more accurately.

Let me suppose θ where g(θ)=4.

Let me set θ=60 degrees (π/3≈1.0472 radians):

g(θ)=4 as before.

Wait, so at θ=59.5 degrees (≈1.039 radians), we had approx 3.987, which is just below 4. At θ=60 degrees, it's exactly 4. Then, moving past 60 degrees, say 61 degrees:

cos(61°)=0.4848, sin(61°)=0.8746,

sec≈2.063,

sqrt(3)/sin≈1.732/0.8746≈1.979,

Total≈2.063 +1.979≈4.042>4.

Ah, so between 59.5 and 60 degrees, the function crosses 4 from below, and at 60 degrees it is exactly 4, then goes above. Wait, but at exactly 60 degrees it's 4. So actually, the equation g(θ)=4 has two solutions: one at θ=60 degrees, and another between 59.5 and 60 degrees? Wait, no, because at 60 degrees it's exactly 4, and just below that it's approaching 4 from below, and just above it goes above. Wait, perhaps it's only touching at θ=60 degrees?

Wait, let me compute at θ=59.9 degrees:

cos(59.9°)=cos(π*59.9/180)= let me compute in radians: 59.9°≈1.045 radians,

cos(1.045)= approx cos(π/3 - 0.002)= cos(π/3)cos(0.002)+sin(π/3)sin(0.002)≈0.5*0.99998 + (sqrt(3)/2)(0.002)≈0.49999 +0.001732≈0.5017,

sin(1.045)=sin(π/3 -0.002)=sin(π/3)cos(0.002)-cos(π/3)sin(0.002)≈(sqrt(3)/2)(0.99998) -0.5*(0.002)≈0.8660 -0.001≈0.8650,

Thus,

secθ≈1/0.5017≈1.993,

sqrt(3)/sinθ≈1.732/0.8650≈1.996,

Total≈1.993 +1.996≈3.989≈3.99, still less than 4.

At θ=59.95 degrees (very close to 60):

cos≈0.5008,

sin≈0.8660,

Wait, maybe it's better to use linear approximation.

Let me consider θ=π/3 - ε,

Let me set θ=π/3 - ε,

Then,

cosθ=cos(π/3 - ε)=cos(π/3)cosε + sin(π/3)sinε≈ (0.5)(1 - ε²/2) + (sqrt(3)/2)(ε),

sinθ=sin(π/3 - ε)=sin(π/3)cosε - cos(π/3)sinε≈ (sqrt(3)/2)(1 - ε²/2) -0.5*(ε).

Then,

secθ≈1/[0.5 + (sqrt(3)/2)ε - 0.5*(ε²/2) ] ≈ 2/[1 + sqrt(3) ε - ε²/2 ] ≈2[1 - sqrt(3) ε + ( (sqrt(3) ε)^2 - (ε²/2) ) + ... ] using 1/(1+x)≈1 -x for small x.

Similarly, sqrt(3)/sinθ≈sqrt(3)/[ (sqrt(3)/2)(1 - ε²/2) -0.5 ε ] ≈ sqrt(3)/[ (sqrt(3)/2 - (sqrt(3)/4)ε² -0.5 ε ) ]

≈ 2/[1 - (ε²/2) - (ε)/sqrt(3) ) ] ≈2[1 + (ε)/sqrt(3) + (ε²/2) + ... ]

Therefore, adding secθ + sqrt(3)/sinθ ≈ 2[1 - sqrt(3) ε + ... ] + 2[1 + (ε)/sqrt(3) + ... ] ≈ 4 + terms with ε.

The linear terms in ε would be -2 sqrt(3) ε + 2*(ε)/sqrt(3) = ε*(-2 sqrt(3) + 2/sqrt(3)) = ε*(- (6/√3) + 2/√3 )= ε*(-4/√3). So the first order term is negative, so decreasing as ε increases (i.e., as θ approaches π/3 from below). Therefore, near θ=π/3, the function g(θ)=secθ + sqrt(3)cscθ has a minimum at θ=π/3?

Wait, but the derivative at θ=π/3 was zero, right? Earlier, when we set the derivative to zero, we found that t=3/4. Wait, but in the derivative of f(θ), which was for the specific point C. Hmm, perhaps this is getting too convoluted.

Alternatively, since when we set up the condition that the point C must lie on AB and also be on the envelope (i.e., the astroid), which gives a unique point, then that must be the only point where the equation has a unique solution θ=π/3. For other points on AB, the equation would have another solution θ≠π/3, meaning the point is on another segment. Therefore, the only point not on any other segment is the tangency point, so the answer is indeed 7/16, leading to 23.

Therefore, I think the answer is 23.

**Final Answer**
\boxed{23}
</think>

Given points \(O = (0,0)\), \(A = \left(\frac{1}{2}, 0\right)\), and \(B = \left(0, \frac{\sqrt{3}}{2}\right)\), we need to find the point \(C\) on \(\overline{AB}\) that is not on any other segment from the family \(\mathcal{F}\) of unit-length segments \(\overline{PQ}\) with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis.

The family \(\mathcal{F}\) is described by the lines \( \frac{x}{\cos \theta} + \frac{y}{\sin \theta} = 1 \) where \( \theta \) ranges from \(0\) to \(\frac{\pi}{2}\). The envelope of these lines forms an astroid given by \( x^{2/3} + y^{2/3} = 1 \).

To find the point \(C\), we note that it must be the point where the line \(\overline{AB}\) is tangent to the astroid. The line \(\overline{AB}\) corresponds to \(\theta = \frac{\pi}{3}\), giving the point of tangency on the astroid as \(\left( \cos^3 \frac{\pi}{3}, \sin^3 \frac{\pi}{3} \right)\). Calculating these coordinates:

\[
\cos \frac{\pi}{3} = \frac{1}{2} \quad \text{and} \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}
\]
\[
\cos^3 \frac{\pi}{3} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}, \quad \sin^3 \frac{\pi}{3} = \left( \frac{\sqrt{3}}{2} \right)^3 = \frac{3\sqrt{3}}{8}
\]

Thus, the coordinates of \(C\) are \(\left( \frac{1}{8}, \frac{3\sqrt{3}}{8} \right)\). Calculating \(OC^2\):

\[
OC^2 = \left( \frac{1}{8} \right)^2 + \left( \frac{3\sqrt{3}}{8} \right)^2 = \frac{1}{64} + \frac{27}{64} = \frac{28}{64} = \frac{7}{16}
\]

Therefore, \(p = 7\) and \(q = 16\), leading to \(p + q = 23\).

\[
\boxed{23}
\]


Solution:
\[\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)\]
Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$, we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:
\[((1 + i)^{13} - 1)(1 - i)^{13} - 1)\]
\[= (-64(1 + i) - 1)(-64(1 - i) - 1)\]
\[= (65 + 64i)(65 - 64i)\]
\[= 65^2 + 64^2\]
\[= 8\boxed{\textbf{321}}\]
~Mqnic_
To find $\prod_{k=0}^{12} (2 - 2w^k + w^{2k})$, where $w\neq1$ and $w^{13}=1$, rewrite this is as
$(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})$ where $r$ and $s$ are the roots of the quadratic $x^2-2x+2=0$.
Grouping the $r$'s and $s$'s results in $\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}$
the denomiator $(r-1)(s-1)=1$ by vietas.
the numerator $(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321$ by newtons sums
so the answer is $\boxed{321}$
-resources
Denote $r_j = e^{\frac{i 2 \pi j}{13}}$ for $j \in \left\{ 0, 1, \cdots , 12 \right\}$.
Thus, for $\omega \neq 1$, $\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)$ is a permutation of $\left( r_0, r_1, \cdots, r_{12} \right)$.
We have
\begin{align*}\
\Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right)
& = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right)
\left( 1 - i - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \\
& = \left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\right)
\left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right)
\right) . \hspace{1cm} (1)
\end{align*}
The third equality follows from the above permutation property.
Note that $r_0, r_1, \cdots , r_{12}$ are all zeros of the polynomial $z^{13} - 1$.
Thus,
\[ z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) . \]
Plugging this into Equation (1), we get
\begin{align*}
(1)
& = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right)
\left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \\
& = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right)
\left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \\
& = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \\
& = 2^{13} + 1 + 2^7 \\
& = 8321 .
\end{align*}
Therefore, the answer is $\boxed{\textbf{(321) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 321

<｜begin▁of▁sentence｜><｜User｜>Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when
\[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\]
is divided by 1000.<｜Assistant｜><think>

Okay, so I need to find the remainder when the product from k=0 to 12 of (2 - 2ω^k + ω^{2k}) is divided by 1000, where ω is a primitive 13th root of unity. Hmm, roots of unity... okay, so ω^13 = 1, and ω ≠ 1.

First, maybe I should recall some properties of roots of unity. Since ω is a primitive 13th root, the minimal polynomial for ω is the 13th cyclotomic polynomial, which is Φ₁₃(x) = x¹² + x¹¹ + ... + x + 1. That might come into play here.

The problem is asking for the product over k from 0 to 12 of (2 - 2ω^k + ω^{2k}). Let me write that out more clearly:

Product = ∏_{k=0}^{12} [2 - 2ω^k + ω^{2k}]

Hmm, maybe I can factor this expression or find a pattern. Let me see if I can manipulate the term inside the product.

Let me denote each term as T_k = 2 - 2ω^k + ω^{2k}. Maybe I can rewrite T_k in a different form. Let's see:

T_k = 2 - 2ω^k + ω^{2k}

Hmm, perhaps factor something out? Let me see:

Alternatively, maybe write ω^{2k} as (ω^k)^2. So, T_k = 2 - 2ω^k + (ω^k)^2. That looks like a quadratic in ω^k. Let me write that:

T_k = (ω^k)^2 - 2ω^k + 2.

Hmm, so that's a quadratic in terms of ω^k. Maybe I can factor this quadratic? Let's see:

The quadratic equation x² - 2x + 2 has discriminant 4 - 8 = -4, so roots at [2 ± 2i]/2 = 1 ± i. So, T_k factors as (ω^k - (1 + i))(ω^k - (1 - i)). But since we're dealing with complex numbers, maybe this isn't the most helpful approach here. Hmm.

Alternatively, perhaps consider that the product over all k of some expression involving ω^k can be related to the value of a certain polynomial evaluated at specific points. Since ω is a root of unity, products over terms like (x - ω^k) relate to cyclotomic polynomials. But here the terms are more complicated.

Alternatively, maybe consider that the product over k=0 to 12 of T_k is the product over all 13th roots of unity (except 1, but since k=0 gives ω^0=1, so actually including 1) of (2 - 2ω^k + ω^{2k}). Wait, but ω is a primitive 13th root, so ω^k for k from 0 to 12 gives all the 13th roots of unity. So the product is over all 13th roots of unity ζ of (2 - 2ζ + ζ²). So, the product is ∏_{ζ¹³=1} (2 - 2ζ + ζ²).

Ah! That might be a better way to think about it. So, the product is over all 13th roots of unity of (ζ² - 2ζ + 2). So, perhaps this product can be related to the value of some polynomial evaluated at certain points?

Recall that if we have a polynomial P(x), then the product over (x - ζ) for all roots ζ of unity is xⁿ - 1, but here it's a different expression. Alternatively, maybe consider that the product over (aζ² + bζ + c) for all ζ can be connected to the resultant or something else.

Alternatively, perhaps consider that the product over all ζ of (ζ² - 2ζ + 2) is equal to the product over all ζ of ( (ζ - (1 + i))(ζ - (1 - i)) ), since the quadratic factors as such. Wait, that's the same as before. So, then the product over all ζ of (ζ - (1 + i))(ζ - (1 - i)) would be equal to the product over all ζ of (ζ - (1 + i)) times the product over all ζ of (ζ - (1 - i)).

But the product over all 13th roots of unity ζ of (ζ - a) is equal to a¹³ - 1, since ζ¹³ = 1, so ζ^13 -1 =0, so the product over (ζ - a) would be the value of the polynomial x¹³ -1 evaluated at x = a? Wait, actually, the product over (x - ζ) for all ζ is x¹³ -1. Therefore, substituting x = a, the product over (a - ζ) is a¹³ -1. But the product over (ζ - a) is (-1)^13 (a¹³ -1) = - (a¹³ -1).

Therefore, the product over all ζ of (ζ - (1 + i)) is equal to -[(1 + i)^13 -1], and similarly the product over all ζ of (ζ - (1 - i)) is -[(1 - i)^13 -1]. Therefore, the total product would be [ -((1 + i)^13 -1) ] * [ -((1 - i)^13 -1) ] = [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ]

Wait, but hold on, let me check that again. Let me be precise.

Let me denote the product over all 13th roots of unity ζ of (ζ - a) is equal to the product_{ζ^13=1} (ζ - a) = (a^{13} - 1) multiplied by (-1)^{13} because:

The polynomial x^{13} - 1 factors as product_{ζ^13=1} (x - ζ). Therefore, substituting x = a, we get a^{13} -1 = product_{ζ} (a - ζ). Therefore, product_{ζ} (ζ - a) = product_{ζ} ( - (a - ζ) ) = (-1)^{13} product_{ζ} (a - ζ) = - (a^{13} -1).

Therefore, product_{ζ} (ζ - a) = - (a^{13} -1).

Therefore, going back, the product over all ζ of (ζ² - 2ζ + 2) is equal to product_{ζ} (ζ - (1 + i))(ζ - (1 - i)) = [product_{ζ} (ζ - (1 + i))] * [product_{ζ} (ζ - (1 - i))] = [ - ( (1 + i)^13 -1 ) ] * [ - ( (1 - i)^13 -1 ) ]

So that would be [ - ( (1 + i)^13 -1 ) ] * [ - ( (1 - i)^13 -1 ) ] = [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ]

Wait, but actually, the two negatives multiply to give positive, so it's [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ]

Wait, but let me confirm again:

Each product_{ζ} (ζ - a) is - (a^13 -1). Therefore, the first term is - [ (1 + i)^13 -1 ] and the second is - [ (1 - i)^13 -1 ], so multiplying them gives (-1)^2 * [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ] = [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ]

Therefore, the total product is equal to [(1 + i)^13 -1][(1 - i)^13 -1].

Hmm, okay, so now I need to compute this expression. Let me compute (1 + i)^13 and (1 - i)^13 first.

First, note that 1 + i can be written in polar form. Since 1 + i has magnitude sqrt(2) and angle π/4, so (1 + i) = sqrt(2) e^{iπ/4}. Therefore, (1 + i)^13 = (sqrt(2))^{13} e^{i 13 π/4}.

Similarly, (1 - i) is sqrt(2) e^{-iπ/4}, so (1 - i)^13 = (sqrt(2))^{13} e^{-i 13 π/4}.

Let me compute (sqrt(2))^{13} first. sqrt(2) is 2^{1/2}, so (2^{1/2})^{13} = 2^{13/2} = 2^6 * 2^{1/2} = 64 * sqrt(2). Wait, actually, 13/2 is 6.5, so 2^6.5 = 2^6 * 2^0.5 = 64 * sqrt(2). So, (1 + i)^13 = 64 sqrt(2) e^{i 13π/4}, and similarly (1 - i)^13 = 64 sqrt(2) e^{-i 13π/4}.

Now, let's compute the angles. 13π/4 is equal to 3π + π/4, since 12π/4 is 3π, so 13π/4 is 3π + π/4. But angles are modulo 2π, so 13π/4 - 2π*1 = 13π/4 - 8π/4 = 5π/4. Wait, actually, 13π/4 divided by 2π is 13/8, so subtract 2π once, getting 13π/4 - 8π/4 = 5π/4. So e^{i 13π/4} = e^{i 5π/4} = cos(5π/4) + i sin(5π/4) = -sqrt(2)/2 - i sqrt(2)/2.

Similarly, e^{-i 13π/4} = e^{-i 5π/4} = cos(-5π/4) + i sin(-5π/4) = cos(5π/4) - i sin(5π/4) = -sqrt(2)/2 + i sqrt(2)/2.

Therefore, (1 + i)^13 = 64 sqrt(2) [ -sqrt(2)/2 - i sqrt(2)/2 ].

Let me compute that:

First, sqrt(2) multiplied by sqrt(2)/2 is (2)/2 = 1. So:

64 sqrt(2) * (-sqrt(2)/2) = 64 * (- (sqrt(2)*sqrt(2))/2 ) = 64 * (-2/2) = 64*(-1) = -64.

Similarly, the imaginary part:

64 sqrt(2) * (-i sqrt(2)/2 ) = 64 * (-i * 2 / 2 ) = 64*(-i) = -64i.

Therefore, (1 + i)^13 = -64 -64i.

Similarly, (1 - i)^13 = 64 sqrt(2) e^{-i 5π/4} = 64 sqrt(2) [ -sqrt(2)/2 + i sqrt(2)/2 ]

Similarly, compute that:

Real part: 64 sqrt(2)*(-sqrt(2)/2) = same as before, 64*(-1) = -64.

Imaginary part: 64 sqrt(2)*(i sqrt(2)/2) = 64*(i*2/2) = 64i.

Therefore, (1 - i)^13 = -64 + 64i.

Therefore, now we can compute [(1 + i)^13 -1] and [(1 - i)^13 -1].

First, (1 + i)^13 -1 = (-64 -64i) -1 = -65 -64i.

Similarly, (1 - i)^13 -1 = (-64 +64i) -1 = -65 +64i.

Therefore, the product [(1 + i)^13 -1][(1 - i)^13 -1] is (-65 -64i)(-65 +64i).

This is a product of a complex number and its conjugate, so it's equal to (-65)^2 + (64)^2. Wait, actually, (a - b)(a + b) = a² - b², but here it's (-65 -64i)(-65 +64i) = (-65)^2 - (64i)^2 = 65² - (64²)(i²). Since i² = -1, so this becomes 65² + 64².

Wait, let me compute it step by step:

Multiply (-65 -64i)(-65 +64i):

First, multiply the first terms: (-65)(-65) = 4225

Outer terms: (-65)(64i) = -4160i

Inner terms: (-64i)(-65) = +4160i

Last terms: (-64i)(64i) = -64*64 i² = -4096*(-1) = +4096

So adding all together:

4225 -4160i +4160i +4096 = 4225 + 4096 + ( -4160i +4160i )

The imaginary terms cancel, so total is 4225 + 4096 = 8321.

Wait, 4225 + 4096: 4225 + 4000 = 8225, then +96 is 8321. So the product is 8321.

Therefore, the total product over all ζ of (ζ² - 2ζ +2) is equal to 8321.

Wait, but hold on, let me confirm again. Because according to the previous steps, the product is [(1 + i)^13 -1][(1 - i)^13 -1] = (-65 -64i)(-65 +64i) = 8321.

Therefore, the product we need is 8321. But wait, the problem states "Let ω ≠ 1 be a 13th root of unity. Find the remainder when [product] is divided by 1000."

Wait, but hold on, the product over all 13th roots of unity of (2 - 2ω^k + ω^{2k}) is equal to 8321? Let me check again.

Wait, but in the problem statement, the product is from k=0 to 12, which includes ω^0 = 1. Let me check when k=0, the term is 2 - 2*1 + 1² = 2 -2 +1 =1. So the term at k=0 is 1. So the product over k=0 to 12 is the same as the product over all 13th roots of unity ζ of (ζ² - 2ζ +2). Which we computed as 8321.

Therefore, the product is 8321. Then, the problem asks for the remainder when this is divided by 1000. So 8321 divided by 1000. The remainder is 321. Wait, 8321 mod 1000 is 321. But hold on, let me confirm.

Wait, 8321 divided by 1000 is 8*1000 = 8000, so 8321 - 8000 = 321. So the remainder is 321. But before I get too confident, let me check my steps again because sometimes when dealing with complex numbers and products, there might be a miscalculation.

First, let me verify the step where I expressed the product over all roots as the product of two terms. The key idea was that the quadratic factors into (ζ - (1 + i))(ζ - (1 - i)), so the product over all ζ of (ζ² - 2ζ +2) is indeed the product over (ζ - (1 + i)) times product over (ζ - (1 - i)). Then, each of those products is equal to -[(1 + i)^13 -1] and -[(1 - i)^13 -1], so multiplying them gives the product.

Wait, but actually, the product over (ζ - a) is equal to - (a^13 -1). Therefore, product_{ζ} (ζ - (1 + i)) = - [ (1 + i)^13 -1 ], and similarly for (1 - i). Therefore, the total product is [ - ( (1 + i)^13 -1 ) ] * [ - ( (1 - i)^13 -1 ) ] = [ ( (1 + i)^13 -1 ) ( (1 - i)^13 -1 ) ]

Wait, but in my calculation above, I had [(1 + i)^13 -1][(1 - i)^13 -1] equals 8321, but actually, according to the previous step, the product is equal to that expression. Wait, but in my calculation, (1 + i)^13 was -64 -64i, so (1 + i)^13 -1 is -65 -64i, and similarly (1 - i)^13 -1 is -65 +64i. Then multiplying those gives (-65)^2 - (64i)^2? Wait, no, actually, the product is (-65 -64i)(-65 +64i) which is (-65)^2 - (64i)^2. Wait, but (64i)^2 is (64)^2*(i)^2 = - (64)^2. Therefore, the product is 65² - ( - (64)^2 ) = 65² + 64².

Wait, 65 squared is 4225, 64 squared is 4096, so 4225 + 4096 is indeed 8321. So that part is correct.

Therefore, the product is 8321, so mod 1000 is 321. But wait, the problem says "Let ω ≠ 1 be a 13th root of unity". Wait, but in the problem statement, does the product start at k=0? Yes, from k=0 to 12. But when k=0, ω^0 =1, so the term is 2 - 2*1 +1^2 =1. So the product includes that term. So the total product is indeed 8321, so mod 1000 is 321.

But wait, hold on a second. Let me check with a small example. Let me test with a smaller root of unity to see if the approach works. Let me try with 1st root of unity, but that's trivial. Maybe 2nd root? Let's see.

Suppose instead of 13th root, let's take 1st root, but that's just 1. Not helpful. Let's try 3rd roots of unity. Let me see if the formula holds.

Suppose n=3, so ω is a primitive 3rd root of unity. Let me compute the product over k=0 to 2 of (2 - 2ω^k + ω^{2k}).

First, compute each term:

For k=0: 2 -2*1 +1 =1.

For k=1: 2 -2ω + ω².

For k=2: 2 -2ω² + ω^4. But ω^3=1, so ω^4=ω.

So term at k=2 is 2 -2ω² + ω.

So the product is 1*(2 -2ω + ω²)*(2 -2ω² + ω).

Let me compute this product. Let me denote ω² = ω^{-1} since ω^3=1.

Let me compute (2 -2ω + ω²)(2 -2ω² + ω). Let me expand this:

First term: 2*(2) =4

2*(-2ω²) = -4ω²

2*(ω) = 2ω

Then -2ω*(2) = -4ω

-2ω*(-2ω²) = +4ω^3 = 4*1=4 (since ω^3=1)

-2ω*(ω) = -2ω²

Then ω²*(2) = 2ω²

ω²*(-2ω²) = -2ω^4 = -2ω (since ω^3=1, so ω^4=ω)

ω²*(ω) = ω^3 =1

So adding all terms:

4 -4ω² +2ω -4ω +4 -2ω² +2ω² -2ω +1.

Let me combine like terms:

Constants: 4 +4 +1 =9

ω terms: 2ω -4ω -2ω = (2 -4 -2)ω = -4ω

ω² terms: -4ω² -2ω² +2ω² = (-4 -2 +2)ω² = -4ω²

So total is 9 -4ω -4ω².

But since ω² + ω +1 =0 (since ω^3=1 and ω≠1), so ω + ω² = -1. Therefore, -4ω -4ω² = -4(ω + ω²) = -4*(-1)=4. Therefore total is 9 +4 =13.

Alternatively, according to the formula I used earlier, the product over all 3rd roots of unity ζ of (ζ² -2ζ +2). Let me compute that.

First, (1 + i)^3: Let me see, following the same steps as before.

Wait, let me see if the formula would give the same result. Let me compute [(1 + i)^3 -1][(1 - i)^3 -1].

First, (1 + i)^3 = (1 + i)(1 + i)^2 = (1 +i)(1 + 2i +i²) = (1 +i)(1 +2i -1) = (1 +i)(2i) = 2i + 2i² = 2i -2 = -2 +2i.

Similarly, (1 -i)^3 = (1 -i)(1 -i)^2 = (1 -i)(1 -2i +i²) = (1 -i)(1 -2i -1) = (1 -i)(-2i) = -2i + 2i² = -2i -2 = -2 -2i.

Therefore, [(1 +i)^3 -1] = (-2 +2i -1) = -3 +2i.

[(1 -i)^3 -1] = (-2 -2i -1) = -3 -2i.

Multiply them: (-3 +2i)(-3 -2i) = 9 +6i -6i -4i² = 9 +4 =13. Which matches the manual calculation. So the formula works here.

Therefore, the formula gives the correct result for n=3, so likely correct for n=13. Therefore, the product is indeed 8321, so mod 1000 is 321. But wait, the problem says "Let ω ≠1 be a 13th root of unity". Wait, but in the problem statement, the product is over k=0 to 12, which includes ω^0=1. But in the problem statement, they define ω as a primitive 13th root, so maybe the product is over all k except 0? Wait, no, the problem says "Let ω≠1 be a 13th root of unity. Find the remainder when [product from k=0 to 12]..." So the product is over all k from 0 to 12, so including k=0 where ω^0=1. Therefore, the product is indeed 8321, so mod 1000 is 321.

Wait, but let me check with another small n. Let me try n=2. Let's see, even though n=2 might be edge case.

Suppose n=2, so ω is a primitive 2nd root of unity, which is -1. Then the product from k=0 to1 of (2 -2ω^k + ω^{2k}).

For k=0: 2 -2*1 +1 =1.

For k=1: 2 -2*(-1) + (-1)^2 =2 +2 +1=5.

So product is 1*5=5.

Using the formula: [(1 +i)^2 -1][(1 -i)^2 -1].

Compute (1 +i)^2 =1 +2i +i²= 1+2i-1=2i. So (2i -1). Wait, [(1 +i)^2 -1] = 2i -1.

Similarly, (1 -i)^2 =1 -2i +i²=1-2i-1= -2i. So [(1 -i)^2 -1] = -2i -1.

Multiply them: ( -1 +2i)(-1 -2i) = (-1)^2 - (2i)^2 =1 - (-4)=5. Which matches. So the formula works here too.

Therefore, seems the formula is correct. Therefore, the answer should be 8321 mod 1000 is 321. But wait, the problem says "Let ω≠1 be a 13th root of unity". Wait, but in the problem statement, the product is over all k from 0 to 12, so including k=0 where ω^0=1. But the problem says "Let ω≠1 be a 13th root of unity". Wait, perhaps the problem is written in a way that ω is a primitive root, so when they say "Let ω≠1 be a 13th root of unity", they might mean that ω is a primitive root, so that ω^k for k=0,...,12 gives all the roots. So the product is over all roots, so including 1. So the product is indeed 8321, mod 1000 is 321.

But wait, let me check if I made a mistake in the exponent when calculating (1 +i)^13. Let me recalculate (1 +i)^13.

Earlier, I said (1 +i) has magnitude sqrt(2), angle π/4. So (1 +i)^13 has magnitude (sqrt(2))^13 and angle 13*(π/4).

(sqrt(2))^13 = (2^(1/2))^13 = 2^(13/2) = 2^6 * 2^(1/2) = 64 * sqrt(2). That's correct.

The angle is 13*(π/4) radians. 13 divided by 8 (since 2π is a full circle) gives 13π/4 - 3*2π = 13π/4 - 6π = 13π/4 - 24π/4 = -11π/4. Wait, but angles are periodic mod 2π, so 13π/4 is equivalent to 13π/4 - 3*2π = 13π/4 - 6π = 13π/4 - 24π/4 = -11π/4. But adding 2π gives -11π/4 + 8π/4 = -3π/4. Wait, but that contradicts my previous calculation. Wait, perhaps I made a mistake here.

Wait, 13π/4 divided by 2π is 13/8, so the integer part is 1, so subtract 2π once: 13π/4 - 2π = 13π/4 - 8π/4 = 5π/4. So the angle is 5π/4, which is correct. So the angle is 5π/4, so the direction is correct. So the calculation of (1 +i)^13 as 64 sqrt(2) e^{i5π/4} is correct, leading to -64 -64i. So that part is okay.

Therefore, the calculation seems correct, so the product is 8321, mod 1000 is 321. But let me check with another approach to be safe.

Alternative approach: Maybe consider that the product over (ζ² - 2ζ +2) for all 13th roots ζ. Let me consider that the product over (ζ² - 2ζ +2) = product over ( (ζ -1)^2 +1 ). Hmm, not sure if that helps.

Alternatively, perhaps consider that the product is equal to the product over (ζ^2 -2ζ +2). Let me think of this as evaluating a certain polynomial at specific points. Let me consider the polynomial f(x) = x^2 -2x +2. Then the product over all 13th roots ζ of f(ζ) is equal to the product_{ζ^13=1} f(ζ).

There is a formula for the product of f(ζ) over all roots ζ of unity. In general, if f(x) is a polynomial, then the product over f(ζ) for all ζ^n=1 is the resultant of x^n -1 and f(x). Alternatively, it can be expressed as the product over the roots of ζ of f(ζ).

Alternatively, using the fact that the product over (x - ζ) is x^13 -1, so perhaps we can relate the product of f(ζ) to the value of some other polynomial.

Alternatively, note that the product over f(ζ) = product_{k=0}^{12} f(ω^k) where ω is a primitive root. This is equal to the norm of f(ω) in the cyclotomic field, but maybe that's more advanced.

Alternatively, perhaps consider that the product over (ζ^2 -2ζ +2) = product over ( (ζ^2 -2ζ +1) +1 ) = product over ( (ζ -1)^2 +1 ). Hmm, not sure.

Alternatively, perhaps consider that the product over all ζ of (a - bζ) is a^13 product_{ζ} (1 - (b/a)ζ) = a^13 * ( (b/a)^13 -1 )^{-1} ? Wait, not sure.

Alternatively, perhaps use logarithms? Probably not helpful here.

Alternatively, since we already have an answer via complex numbers, and verified with smaller cases, perhaps 8321 is correct, so mod 1000 is 321. But let me check 8321 divided by 1000. 1000*8=8000, 8321-8000=321. So yes, remainder 321.

Wait, but the problem says "Let ω≠1 be a 13th root of unity". Wait, but in the problem statement, the product is over k=0 to 12, so including ω^0=1. But the problem says "Let ω≠1 be a 13th root of unity". Maybe there's a misinterpretation here. Wait, perhaps the problem is written as "Let ω≠1 be a 13th root of unity. Find the remainder when [product from k=0 to12 of (2 -2ω^k + ω^{2k})] is divided by 1000." So ω is a primitive root, so that ω^k for k=0,...,12 gives all 13th roots. Therefore, the product is over all roots, so the same as before. So the answer is 321.

Alternatively, maybe I made a mistake in the calculation of (1 +i)^13. Let me compute (1 +i)^13 step by step. Let me compute (1 +i)^2 = 2i. Then (1 +i)^4 = (2i)^2 = -4. (1 +i)^8 = (-4)^2 = 16. (1 +i)^12 = ( (1 +i)^8 )*( (1 +i)^4 ) =16*(-4)= -64. Then (1 +i)^13 = (1 +i)^12*(1 +i) = -64*(1 +i) = -64 -64i. Which matches the previous result. So that's correct.

Similarly, (1 -i)^13 = conjugate of (1 +i)^13, so -64 +64i. So that's correct.

Therefore, the product [(1 +i)^13 -1][(1 -i)^13 -1] is indeed (-65 -64i)(-65 +64i)= 8321.

Therefore, the remainder mod 1000 is 321.

Wait, but I just thought, maybe the problem is expecting an integer, which 8321 is, so mod 1000 is 321. But let me check with another approach. Let me think of the minimal polynomial.

Alternatively, perhaps consider that the product over (ζ² -2ζ +2) for all ζ^13=1. Let me note that ζ^13=1, so ζ^k = ζ^{k mod13}.

Alternatively, perhaps consider that the product can be written as product_{k=0}^{12} (2 -2ω^k + ω^{2k}) = product_{k=0}^{12} (ω^{2k} -2ω^k +2). Let me denote x_k = ω^k, so x_k^13 =1. Then each term is x_k² -2x_k +2. So the product is product_{k=0}^{12} (x_k² -2x_k +2).

Alternatively, perhaps consider that the product over (x^2 -2x +2) evaluated at all 13th roots x.

Alternatively, note that the product over (x - a_k) where a_k are roots of some polynomial. Wait, but here it's product of quadratics evaluated at roots.

Alternatively, perhaps consider that the product over (x^2 -2x +2) for x^13=1. Let me consider the polynomial P(x) = x^13 -1. Then the product over (x^2 -2x +2) for all roots x of P(x)=0 is equal to the resultant of P(x) and x^2 -2x +2, but I might be misremembering. Alternatively, the product is equal to the product of (x^2 -2x +2) over all roots x of P(x)=0, which can be written as the product over roots x of P(x) of (x^2 -2x +2).

Alternatively, this can be expressed as the resultant of P(x) and Q(x)=x^2 -2x +2). The resultant Res(P, Q) is equal to the product of Q(x) over all roots x of P(x). So yes, that would be exactly the product we need. The resultant can be computed as the determinant of the Sylvester matrix, but that might be complicated here. Alternatively, there's a formula when one polynomial divides the other? Not sure.

Alternatively, note that the product over (x^2 -2x +2) for all x with x^13=1 is equal to the product over (x - (1+i))(x - (1 -i)) for all x^13=1, which is exactly what I did before, leading to the same result.

Therefore, I think my answer is correct. Therefore, the remainder modulo 1000 is 321.

Wait, but just to be thorough, let me check with another perspective. Let me consider that the product is 8321. Let me compute 8321 mod 1000. 8321 divided by 1000 is 8*1000 = 8000, subtract that from 8321 gives 321. So yes, remainder is 321.

Alternatively, maybe there's a mistake in the problem interpretation. Let me check again the problem statement:

"Let ω≠1 be a 13th root of unity. Find the remainder when [product from k=0 to12 of (2 -2ω^k + ω^{2k})] is divided by 1000."

So ω is a primitive 13th root of unity, so that ω^k for k=0,...,12 gives all 13th roots. The product is over k=0 to12, so indeed over all roots. So the product is 8321, mod 1000 is 321.

Hmm, but I feel uncertain because 321 seems straightforward, but maybe I made a miscalculation in the exponents? Let me check (1 +i)^13 again.

(1 +i)^1 =1 +i

(1 +i)^2 = 2i

(1 +i)^3 = 2i*(1 +i) = 2i + 2i² = 2i -2 = -2 +2i

(1 +i)^4 = (-2 +2i)(1 +i) = -2(1 +i) +2i(1 +i) = -2 -2i +2i +2i² = -2 +0i -2 = -4

(1 +i)^4 = -4. Then (1 +i)^8 = (-4)^2 =16

(1 +i)^12 = ( (1 +i)^8 )*( (1 +i)^4 ) =16*(-4)= -64

(1 +i)^13 = (1 +i)^12*(1 +i) = -64*(1 +i) = -64 -64i. Correct.

Similarly, (1 -i)^13 = -64 +64i. Correct.

Therefore, (1 +i)^13 -1 = -65 -64i, (1 -i)^13 -1 = -65 +64i. Their product is indeed (-65)^2 + (64)^2 = 4225 +4096 = 8321.

Therefore, I think the answer is 321.

Wait, but let me check with another small n, say n=4. Let me see.

Take n=4, so 4th roots of unity. Let me compute the product over k=0 to3 of (2 -2ω^k + ω^{2k}).

Let ω = i, a primitive 4th root.

Compute each term:

k=0: 2 -2*1 +1 =1.

k=1: 2 -2i + (i)^2 =2 -2i -1 =1 -2i.

k=2: 2 -2*(-1) + (i)^4 =2 +2 +1=5.

k=3: 2 -2*(-i) + (i)^6 =2 +2i + (-1) =1 +2i.

So the product is 1*(1 -2i)*5*(1 +2i).

First compute (1 -2i)(1 +2i) =1 +4 =5. Then 5*5=25. So total product is 25.

Using the formula: [(1 +i)^4 -1][(1 -i)^4 -1].

Compute (1 +i)^4 = ( (1 +i)^2 )^2 = (2i)^2 = -4. So [( -4 ) -1] = -5.

Similarly, (1 -i)^4 = ( -2i)^2 = -4. So [( -4 ) -1] = -5.

Multiply them: (-5)(-5)=25. Which matches.

Therefore, the formula works here as well.

Therefore, I think my answer is correct. So the remainder is 321.

Wait, but just to be 100% sure, let me compute 8321 mod 1000 again. 8321 divided by 1000 is 8.321, so 8*1000=8000, 8321-8000=321. So yes, remainder is 321.

Therefore, I think the answer is 321.

**Final Answer**
\boxed{321}
</think>

Given \(\omega \neq 1\) is a 13th root of unity, we need to find the remainder when the product \(\prod_{k=0}^{12}(2 - 2\omega^k + \omega^{2k})\) is divided by 1000.

First, we recognize that \(\omega\) is a primitive 13th root of unity, meaning \(\omega^{13} = 1\) and the minimal polynomial for \(\omega\) is the 13th cyclotomic polynomial. The product we need is over all 13th roots of unity \(\zeta\) of the expression \(\zeta^2 - 2\zeta + 2\).

We can factor the quadratic expression \(\zeta^2 - 2\zeta + 2\) as \((\zeta - (1 + i))(\zeta - (1 - i))\), where \(i\) is the imaginary unit. Therefore, the product over all 13th roots of unity \(\zeta\) is:

\[
\prod_{\zeta^{13}=1} (\zeta - (1 + i)) \cdot \prod_{\zeta^{13}=1} (\zeta - (1 - i))
\]

Using the property of roots of unity, the product of \((\zeta - a)\) over all 13th roots of unity \(\zeta\) is \(- (a^{13} - 1)\). Applying this, we get:

\[
\prod_{\zeta^{13}=1} (\zeta - (1 + i)) = -((1 + i)^{13} - 1)
\]
\[
\prod_{\zeta^{13}=1} (\zeta - (1 - i)) = -((1 - i)^{13} - 1)
\]

Thus, the total product is:

\[
[ -((1 + i)^{13} - 1) ] \cdot [ -((1 - i)^{13} - 1) ] = [(1 + i)^{13} - 1][(1 - i)^{13} - 1]
\]

Next, we compute \((1 + i)^{13}\) and \((1 - i)^{13}\). Using polar form, \(1 + i = \sqrt{2} e^{i\pi/4}\) and \(1 - i = \sqrt{2} e^{-i\pi/4}\):

\[
(1 + i)^{13} = (\sqrt{2})^{13} e^{i \cdot 13\pi/4} = 2^{13/2} e^{i(5\pi/4)} = -64 - 64i
\]
\[
(1 - i)^{13} = (\sqrt{2})^{13} e^{-i \cdot 13\pi/4} = 2^{13/2} e^{-i(5\pi/4)} = -64 + 64i
\]

Substituting these into the product:

\[
[( -64 - 64i - 1 )][( -64 + 64i - 1 )] = [ -65 - 64i ][ -65 + 64i ]
\]

This product is the modulus squared of \(-65 - 64i\):

\[
(-65)^2 + (64)^2 = 4225 + 4096 = 8321
\]

Finally, the remainder when 8321 is divided by 1000 is:

\[
8321 \mod 1000 = 321
\]

Thus, the remainder when the given product is divided by 1000 is \(\boxed{321}\).