Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below: [asy] /* Made by MRENTHUSIASM */  size(250); pair A, B, C, P;  A = origin; B = (0,10*sqrt(5)); C = (10*sqrt(5),0); P = intersectionpoints(Circle(A,10),Circle(C,20))[0];  dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4));  markscalefactor=0.125; draw(rightanglemark(B,A,C,10),red); draw(rightanglemark(A,P,C,10),red); draw(anglemark(P,A,B,25),red); draw(anglemark(P,B,C,25),red); draw(anglemark(P,C,A,25),red); draw(anglemark(A,B,P,25),green); draw(anglemark(B,C,P,25),green); draw(anglemark(C,A,P,25),green); add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red));  draw(A--B--C--cycle^^P--A^^P--B^^P--C); label("$10$",midpoint(A--P),dir(-30),blue); label("$\theta$",A,9.5*dir(76),red); label("$\theta$",C,9.5*dir(168),red); label("$\theta$",B,9*dir(305),red); label("$45^\circ-\theta$",B,6*dir(235),green); label("$45^\circ-\theta$",C,6*dir(85),green); label("$90^\circ-\theta$",A,2*dir(-40),green); [/asy] Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$
Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$ , so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\]

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Let the common angle be $\theta$ . Note that $\angle PAC = 90^\circ-\theta$ , thus $\angle APC = 90^\circ$ . From there, we know that $AC = \frac{10}{\sin\theta}$
Note that $\angle ABP = 45^\circ-\theta$ , so from law of sines we have \[\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}.\] Dividing by $10$ and multiplying across yields \[\sqrt{2}\sin(45^\circ-\theta)=\sin\theta.\] From here use the sine subtraction formula, and solve for $\sin\theta$ \begin{align*} \cos\theta-\sin\theta&=\sin\theta \\ 2\sin\theta&=\cos\theta \\ 4\sin^2\theta&=\cos^2\theta \\ 4\sin^2\theta&=1-\sin^2\theta \\ 5\sin^2\theta&=1 \\ \sin\theta&=\frac{1}{\sqrt{5}}. \end{align*} Substitute this to find that $AC=10\sqrt{5}$ , thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Do some angle chasing yielding:
We have $AC=\frac{10}{\sin\theta}$ since $\triangle APC$ is a right triangle. Since $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle, $AB=\frac{10}{\sin\theta}$ , and $BC=\frac{10\sqrt{2}}{\sin\theta}$
Note that $\triangle APB \sim \triangle BPC$ by a factor of $\sqrt{2}$ . Thus, $BP = 10\sqrt{2}$ , and $PC = 20$
From Pythagorean theorem, $AC=10\sqrt{5}$ so the area of $\triangle ABC$ is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Notice that in triangle $PBC$ $\angle PBC + 45-\angle PCA = 45^\circ$ , so $\angle BPC = 135^\circ$ . Similar logic shows $\angle APC = 135^\circ$
Now, we see that $\triangle APB \sim \triangle BPC$ with ratio $1:\sqrt{2}$ (as $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle). Hence, $\overline{PB}=10\sqrt{2}$ . We use the Law of Cosines to find $AB$ \begin{align*} AB^2&=BP^2+AP^2-2ab\cos(APB) \\ &=100+200-2(10)(10\sqrt{2}\cos(135^\circ)) \\ &=300+200\cdot\sqrt{2}\cdot\frac{1}{\sqrt{2}} \\ &=500. \end{align*} Since $\triangle ABC$ is a right triangle, the area is $\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}$

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies\] \[[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot  [APC] = \boxed{250}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Denote $\angle PCA = \theta$ . Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ \cot(\theta) &= 2 \\ \sin(\theta) &= \frac{\sqrt{5}}{5}. \end{align*} Note that $\angle APC$ is a right angle. Therefore: \begin{align*} \sin(\theta) &= \frac{AP}{AC} \\ AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ &= 10\sqrt{5} \\ |ABC| &= \frac{AC^2}{2} \\ &= \boxed{250} ~ConcaveTriangle

Answer:  250
Problem: Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Solution: Notice that point $P$ is one of the two Brocard Points of $\triangle ABC$ . (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, \begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C)  \end{align*} , where $\phi$ is equal to $\angle PAB$ .(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: \begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*} By definition, $\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}$ . By the Pythagorean identity, $\cos(\phi)=\frac{2\sqrt{5}}{5}$ and $\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}$ . Consider triangle $APB$ . By the problem condition, $\angle PBA = 45-\phi$ , so $\angle BPA = 135^{\circ}$ \begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*} Now, we can use the Law of Sines. \begin{align*} \frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\ 10 \sqrt{10} &= \sqrt{2} AB \\ AB &= 10 \sqrt{5} \end{align*} Therefore, the answer is \[[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.\] ~ewei12

Answer:  250
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution: We first subtract the second equation from the first, noting that they both equal $60$ \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$
The first and third equations simplify to: \begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding: \begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$ , substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$
Then, we just add the squares of the solutions (make sure not to double count the $4$ ), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\] ~SAHANWIJETUNGA

Answer:  273
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution: We index these equations as (1), (2), and (3), respectively.
Taking $(1)-(2)$ , we get \[ \left( x - z \right) \left( y - 4 \right) = 0 . \]
Denote $x' = x - 4$ $y' = y - 4$ $z' = z - 4$ .
Thus, the above equation can be equivalently written as \[ \left( x' - z' \right) y' = 0 . \hspace{1cm} (1') \]
Similarly, by taking $(2)-(3)$ , we get \[ \left( y' - x' \right) z' = 0 . \hspace{1cm} (2') \]
By taking $(3) - (1)$ , we get \[ \left( z' - y' \right) x' = 0 . \hspace{1cm} (3') \]
From $(3')$ , we have the following two cases.
Case 1: $x' = 0$
Plugging this into $(1')$ and $(2')$ , we get $y'z' = 0$ .
Thus, $y' = 0$ or $z' = 0$
Because we only need to compute all possible values of $x$ , without loss of generality, we only need to analyze one case that $y' = 0$
Plugging $x' = 0$ and $y' = 0$ into (1), we get a feasible solution $x = 4$ $y = 4$ $z = 11$
Case 2: $x' \neq 0$ and $z' - y' = 0$
Plugging this into $(1')$ and $(2')$ , we get $\left( x' - y' \right) y' = 0$
Case 2.1: $y' = 0$
Thus, $z' = 0$ . Plugging $y' = 0$ and $z' = 0$ into (1), we get a feasible solution $x = 11$ $y = 4$ $z = 4$
Case 2.2: $y' \neq 0$ and $x' = y'$
Thus, $x' = y' = z'$ . Plugging these into (1), we get $\left( x, y, z \right) = \left( -10, -10, -10 \right)$ or $\left( 6, 6, 6 \right)$
Putting all cases together, $S = \left\{ 4, 11, -10, 6 \right\}$ .
Therefore, the sum of the squares of the elements of $S$ is \begin{align*} 4^2 + 11^2 + \left( -10 \right)^2 + 6^2 = \boxed{273}

Answer:  273
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution: We index these equations as (1), (2), and (3), respectively.
Using equation (1), we get $z = \frac{60 - xy}{4} = 15 - \frac{xy}{4}$ We need to solve for x, so we plug this value of z into equation (3) to get:
\[15x - \frac{x^2y}{4} - 4y = 60\] \[\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0\]
We use the quadratic formula to get possible values of x:
\[x = \frac{15 \pm \sqrt{15^2 - 4(\frac{y}{4})(60 + 4y)}}{\frac{y}{2}}\] \[x = \frac{30 \pm 2\sqrt{225 - 60y + 4y^2}}{y}\] \[x = \frac{30 \pm 2\sqrt{(2y-15)^2}}{y}\] \[x = \frac{30 \pm 2(2y - 15)}{y}\] \[x = \frac{30 \pm (4y - 30)}{y}\]
Here, we have two cases, (plus) and (minus)
In the plus case, we have: \[x = \frac{(30 + 4y - 30)}{y}\] \[x = 4\]
So, our first case gives us one value of x, which is 4.
In the minus case, we have: \[x = \frac{30 - (4y - 30)}{y}\] \[x = \frac{60 - 4y}{y}\] \[x = \frac{60}{y} - 4\]
For this case, we now have values of x in terms of y. 
Plugging this expression for x in equation (1), we get \[60 - 4y + 4z = 60\] \[4z = 4y\] \[z = y\]
So we know that for this case, z = y. 
Using this information in equation (2), we get \[y^2 + \frac{240}{y} - 16 = 60\] \[y^2 + \frac{240}{y} - 76 = 0\] Multiplying both sides by y, we get a cubic expression: \[y^3 + 0y^2 - 76y + 240 = 0\] Here we just have to figure out the values of y that make this equation true. I used Vieta's Formulas to get a possible list, but you could also use the rational root theorem and synthetic division to find these. We call the three values of y that solve this equation: $y_{1},y_{2},y_{3}$ Using Vieta's Formulas, you get these three expressions: \[y_{1} + y_{2} + y_{3} = 0\] \[y_{1} * y_{2} * y_{3} = -240\] \[(y_{1} * y_{2}) + (y_{2} * y_{3}) + (y_{1} * y_{3}) = -76\]
In addition, we know that $y \vert 60$ , because of our expression for x. 
Since the three values of y multiply to a negative number but also add to 0, we know that one value is negative and the other two are positive, and that the absolute value of the negative value is greater than both of the positive values.
List of possible values for y are $\{-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60\}$ From a list of these values, the only values that work are $y_1 = -10, y_2 = 6, y_3 = 4$ because \[-10 + 6 + 4 = 0\] \[-10 * 6 * 4 = -240\] \[(-10 * 6) + (-10 * 4) + (6 * 4) = -60 - 40 + 24 = -76\]
Plugging in these values for y into our expression for x, we get: \[x = \frac{60}{-10} - 4 = -10\] \[x = \frac{60}{6} - 4 = 6\] \[x = \frac{60}{4} - 4 = 11\]
So, now we have accounted for both cases, and we have 4 values of x = $\{-10,4,6,11\}$ Squaring all these terms we get: 100 + 16 + 36 + 121 = 273, so our answer is $\boxed{273}$

Answer:  273
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution: Since all three equations are in the form $\frac{K}{a} + 4a = 60$ where $K = xyz$ , we can rearrange this to see that $x$ $y$ , and $z$ all satisfy
\[ 4a^2 - 60a + K = 0. \]
Let this quadratic have roots $a_1$ and $a_2$ . Then, there are two cases to consider: two of $x$ $y$ $z$ are equal to $a_1$ and the third is equal to $a_2$ , or all of $x$ $y$ $z$ are equal to $a_1$
Case 1: WLOG let $x = y = a_1$ and $z = a_2$
Then by Vieta's,
\[x + z = 15 \hspace{1cm} (1)\] \[xz = \frac{x^2z}{4} \hspace{1cm} (2)\]
which gives $x = y = 4$ and $z = 11$ . But we can swap $x$ $y$ , and $z$ however we like, so this also gives $x = 11$ as a solution. In total, this case yields $x = 4$ and $x = 11$ as possible values of $x$
Case 2: WLOG let $x = y = z = a_1$
Again, by Vieta's,
\[x + a_2 = 15 \hspace{1cm} (1)\] \[xa_2 = \frac{x^3}{4} \hspace{1cm} (2)\]
We can use $(2)$ to isolate $a_2$ in terms of $x$ , then plug that into $(1)$ to get that $x^2 + 4x - 60 = 0$ . This yields $x = -10$ and $x = 6$ as additional possible values of $x$
In all, $x$ can be any of $\{ 4, 11, -10, 6 \}$ , so the requested answer is $4^2 + 11^2 + (-10)^2 + 6^2 = \boxed{273}.$

Answer:  273
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution: We index these equations as (1), (2), and (3) respectively (same as solution 2). There are two possible cases:
Case 1: $x = \pm 4$
In this case, we simply plug in $x = 4$ and $x = -4$ . We note that $x=4$ is a valid case.
Case 2: $x \neq \pm 4$
In this case, using equation (3), we get $y = 15 - \frac{xz}{4}$ . Plugging that into equation (1), we get $z = \frac{240-60x}{16-x^2}$ . Plugging that expression back into the original expression for $y$ we obtain $y = 15 - \frac{60x-15x^2}{16-x^2}$ . 
Now we plug these two expressions into equation (2): \[\left(15 - \frac{60x-15x^2}{16-x^2}\right)\left(\frac{240-60x}{16-x^2}\right) = 60-4x\] multiplying both sides by $(16-x^2)$ and factoring, we get: \[\left(15(4-x)(4+x)-60x+15x^2\right)\left(\frac{60(4-x)}{(4+x)(4-x)}\right) = 4(15 - x)(4 + x)(4 - x)\] which simplifies to: \[x^3 - 7x^2 - 104x + 660 = 0\] we note that $x = 6$ is a root. Factoring, we get the other roots, -10 and 11.
Our desired answer is the sum of the square of all these roots: \[4^2 + 6^2 + (-10)^2 + 11^2 = \boxed{273}\]

Answer:  273
Problem: Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Denote $r = \frac{a}{b}$ , where $\left( a, b \right) = 1$ .
We have $55 r = \frac{55a}{b}$ .
Suppose $\left( 55, b \right) = 1$ , then the sum of the numerator and the denominator of $55r$ is $55a + b$ .
This cannot be equal to the sum of the numerator and the denominator of $r$ $a + b$ .
Therefore, $\left( 55, b \right) \neq 1$
Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$
Thus, $55r = \frac{11a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 5c = 11a + c . \]
Hence, $2c = 5 a$
Because $\left( a, b \right) = 1$ $\left( a, c \right) = 1$ .
Thus, $a = 2$ and $c = 5$ .
Therefore, $r = \frac{a}{5c} = \frac{2}{25}$
Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$
Thus, $55r = \frac{5a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 11c = 5a + c . \]
Hence, $2a = 5 c$
Because $\left( a, b \right) = 1$ $\left( a, c \right) = 1$ .
Thus, $a = 5$ and $c = 2$ .
Therefore, $r = \frac{a}{11c} = \frac{5}{22}$
Case 3: $b$ can be written as $55 e$
Thus, $55r = \frac{a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 55c = a + c . \]
Hence, $c = 0$ . This is infeasible.
Thus, there is no solution in this case.
Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$ .
Therefore, the sum of all numbers in $S$ is \[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]
Therefore, the answer is $169 + 550 = \boxed{719}$

Answer:  719
Problem: Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.

Solution: Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$
case 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$
It is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\boxed{928}$

Answer:  928
Problem: Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.

Solution: First, we identify the geometric condition for the sentence ``four vertices colored the
same color are the four vertices of a rectangle Consider any four vertices on the dodecagon, $A$ $B$ $C$ $D$ .
Denote by $O$ the center of the dodecagon.
Because $OA = OB = OC$ $\angle OAB = \angle OBA$ and $\angle OBC = \angle OCB$
Thus, \begin{align*} \angle ABC & = \angle OBA + \angle OBC \\ & = \frac{\angle OBA + \angle OAB}{2} + \frac{\angle OBC + \angle OCB}{2} \\ & = \frac{180^\circ - \angle AOB}{2} +  \frac{180^\circ - \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOB + \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOC}{2} . \end{align*}
Hence, $\angle ABC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ .
Similarly, $\angle ADC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ , and $\angle BCD = 90^\circ$ (or $\angle DAB = 90^\circ$ ) if and only if $\angle BOD = 180^\circ$
Therefore, $ABCD$ is a rectangle if and only if two diagonals both pass through $O$
Now, we categorize 12 vertices into 6 groups. Each group contains 2 diagonal vertices.
Next, we compute the number of coloring configurations such that the above same-color rectangles do not exist.
Case 1: Two vertices in each group has distinct colors.
For each group, we only need to determine which vertex is red. The other one must be blue.
Therefore, the number of configurations in this case is $2^6$
Case 2: There is one group who vertices have the same color. All other groups are with vertices that have distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have the same color.
The number of ways is 6.
Step 2: For the selected group, we choose a color for its two vertices.
The number of ways is 2.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^5$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 2 \cdot 2^5 = 6 \cdot 2^6$
Case 3: One group has two red vertices, one group has two blue vertices, and each of the other four groups has vertices with distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have both red color.
The number of ways is 6.
Step 2: We determine which group has two vertices that have both blue color.
The number of ways is 5.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^4$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 5 \cdot 2^4 = 30 \cdot 2^4$
Putting all cases together, the total number of feasible configurations is \[ 2^6 + 6 \cdot 2^6 + 30 \cdot 2^4 = \boxed{928}. \]

Answer:  928