Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution: Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$ . So
$z^6=e^{\frac{ni\pi}{2}}$
This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{12}$

Answer: D 12
Problem: There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution: $z = \sqrt[24]{1} = 1^{\frac{1}{24}}$
By Euler's identity $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$ , where $k$ is an integer.
Using De Moivre's Theorem , we have $z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}$ , where $0 \leq k<24$ that produce $24$ unique results.
Using De Moivre's Theorem again, we have $z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}$
For $z^6$ to be real, $\sin(\frac{k\pi}{2})$ has to equal $0$ to negate the imaginary component. This occurs whenever $\frac{k\pi}{2}$ is an integer multiple of $\pi$ , requiring that $k$ is even. There are exactly $\boxed{12}$

Answer: D 12
Problem: There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution: From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$ ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be $24$ . Notice that $1 = z^{24} = (z^6)^4$ , so for any solution $z$ $z^6$ will be one of the 4th roots of unity ( $1$ $i$ $-1$ , or $-i$ ). Then $6$ solutions $z$ will satisfy $z^6 = 1$ $6$ will satisfy $z^6 = -1$ (and this is further justified by knowledge of the 6th roots of unity), so there must be $\boxed{12}$ such $z$

Answer: D 12
Problem: There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution: Let $a\in\mathbb{R}$ and $a = z^6.$ We have \[a^4 = 1 \implies a = 1,-1.\] $z^6 = \pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\boxed{12}.$ \[\] -svyn

Answer: D 12
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: Note that $n\equiv S(n)\bmod 9$ , so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$ . So, since $S(n)=1274\equiv 5\bmod 9$ , we have that $S(n+1)\equiv 6\bmod 9$ . The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{1239}$

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: One possible value of $S(n)$ would be $1275$ , but this is not any of the choices. Therefore, we know that $n$ ends in $9$ , and after adding $1$ , the last digit $9$ carries over, turning the last digit into $0$ . If the next digit is also a $9$ , this process repeats until we get to a non- $9$ digit. By the end, the sum of digits would decrease by $9$ multiplied by the number of carry-overs but increase by $1$ as a result of the final carrying over. Therefore, the result must be $9x-1$ less than original value of $S(n)$ $1274$ , where $x$ is a positive integer. The only choice that satisfies this condition is $\boxed{1239}$

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$ , we get $5$ . Adding one to that, we get $6$ . So, if we assess each option to see which one attains $6$ , we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9 = 15$ $1 + 5 = 6$ . The answer is $\boxed{1239}$

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: Note that a lot of numbers can have a sum of $1274$ , but what we use wishful thinking and want is some simple number $n$ where it is easy to compute the sum of the digits of $n+1$ . This number would consists of basically all digits $9$ , since when you add $1$ a lot of stuff will cancel out and end up at $0$ (ex: $399+1=400$ ). We see that the maximum number of $9$ s that can be in $1274$ is $141$ and we are left with a remainder of $5$ , so $n$ is in the form $99...9599...9$ . If we add $1$ to this number we will get $99...9600...0$ so this the sum of the digits of $n+1$ is congruent to $6 \mod 9$ . The only answer choice that is equivalent to $6 \mod 9$ is $1239$ , so our answer is $\boxed{1239}$ -srisainandan6

Answer: D 1239
Problem: How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$

Solution: By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$ . If $x\neq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=\pm \root{2016}\of{2017}$ . Hence, we have $3$ possible values $x$ , namely \[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad  x=-2017^{\frac1{2016}}.\]
Since $\log_b a=x$ is equivalent to $a=b^x$ , each possible value $x$ yields exactly $199$ solutions $(b,a)$ , as we can assign $a=b^x$ to each $b=2,3,\dots,200$ . In total, we have $3\cdot 199=\boxed{597}$ solutions.

Answer: E 597
Problem: A set $S$ is constructed as follows. To begin, $S = \{0,10\}$ . Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$ , all of whose coefficients $a_i$ are elements of $S$ , then $x$ is put into $S$ . When no more elements can be added to $S$ , how many elements does $S$ have?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 11$

Solution: At first, $S=\{0,10\}$
\[\begin{tabular}{r c l c l} \(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\ \(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\ \(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\ \(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\ \(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\ \(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\) \end{tabular}\]
At this point, no more elements can be added to $S$ . To see this, let
\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align*}
with each $a_i$ in $S$ $x$ is a factor of $a_0$ , and $a_0$ is in $S$ , so $x$ has to be a factor of some element in $S$ . There are no such integers left, so there can be no more additional elements. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \boxed{9}$

Answer: D 9
Problem: A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ $(-2, 2)$ $(-2, -2)$ $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$ $(x + 1, y + 1)$ $(x + 1, y)$ $(x + 1, y - 1)$ $(x, y - 1)$ $(x - 1, y - 1)$ $(x - 1, y)$ , or $(x - 1, y + 1)$ . The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 39$

Solution: We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $\frac{4}{8}$ chance of moving to an inside edge and a $\frac{4}{8}$ chance of moving to an inside corner, so
\[m = \frac{1}{2}e + \frac{1}{2}c.\]
Starting at an inside edge, there is a $\frac{2}{8}$ chance of moving to another inside edge, a $\frac{2}{8}$ chance of moving to an inside corner, a $\frac{1}{8}$ chance of moving into the middle, and a $\frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore,
\[e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}\cdot 0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.\]
Starting at an inside corner, there is a $\frac{2}{8}$ chance of moving to an inside edge, a $\frac{1}{8}$ chance of moving into the middle, a $\frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $\frac{1}{8}$ chance of reaching that elusive outside corner. This gives
\[c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}\cdot 1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.\]
Solving this system of equations gives
\[m = \frac{4}{35},\] \[e = \frac{1}{14},\] \[c = \frac{11}{70}.\]
Since the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is
\[4 + 35 = \boxed{39}.\]

Answer: E 39
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: Since all of the roots of $g(x)$ are distinct and are roots of $f(x)$ , and the degree of $f$ is one more than the degree of $g$ , we have that
\[f(x) = C(x-k)g(x)\]
for some number $k$ . By comparing $x^4$ coefficients, we see that $C=1$ . Thus,
\[x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)\]
Expanding and equating coefficients we get that
\[a-k=1,1-ak=b,10-k=100,-10k=c\]
The third equation yields $k=-90$ , and the first equation yields $a=-89$ . So we have that
$f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{7007}$

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: $f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x-r)\]
where $r\in\mathbb{C}$ is the fourth root of $f(x)$ . Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\ &=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a-r&=1\\ 1-ar&=b\\ 10-r&=100\\ -10r&=c.\\ \end{align*}
(Solution 1.1 picks up here.)
Let's solve for $a,b,c,$ and $r$ . Since $10-r=100$ $r=-90$ , so $c=(-10)(-90)=900$ . Since $a-r=1$ $a=-89$ , and $b=1-ar=-8009$ . Thus, we know that
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007}

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: A faster ending to Solution 1 is as follows.
We shall solve for only $a$ and $r$ . Since $10-r=100$ $r=-90$ , and since $a-r=1$ $a=-89$ . Then, \begin{align*} f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &=\boxed{7007}

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored form:
\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)\left(x+\frac{c}{10}\right)\]
We can expand the expression on the right-hand side to get:
\[f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10}\right)x+c\]
Now we have $f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10} \right)x+c=x^4+x^3+bx^2+100x+c$
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]
and finally,
\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\]
We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$ . We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{7007}$

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ \[\therefore 90=1-a.\] The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ $\dfrac{b-1}{1-a} = a$ \[\therefore b-1=a(1-a)=-89\cdot90=-8010\] and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ \[\therefore c=10(1-a)=10\cdot90=900\] Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{7007}$

Answer:  7007
Problem: Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution: Using the given ratios, note that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$
By AA Similarity, $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}$ and $\triangle ACX \sim \triangle EFX$ with a ratio of $\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}$ , so $\frac{XF}{CX} = \frac{16}{9}$
Now we find the length of $BD$ . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. \[BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD\] \[\rightarrow \cos\angle BAD = \frac{11}{24}\] \[\rightarrow BD=\sqrt{51}\] By Power of a Point, $CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}$ . Thus $XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{17}.$

Answer: A 17
Problem: Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution: We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. 
Let $Z$ be the intersection of $AC$ and $BD$ . First, from $ABCD$ being a cyclic quadrilateral, we have that $\triangle BCZ \sim \triangle AZD$ $\triangle BZA \sim \triangle CDZ$ . Therefore, $\frac{2}{BZ} = \frac{8}{AZ}$ $\frac{6}{CZ} = \frac{3}{BZ}$ , and $\frac{2}{CZ} = \frac{8}{DZ}$ , so we have $BZ = \frac{1}{2}CZ$ $AZ = 2CZ$ , and $DZ = 4CZ$ . By Ptolemy's Theorem, \[(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)\] \[\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.\] Thus, $CZ^2 = \frac{68}{27}$ . Then, by Power of a Point, $GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}$ . So, $XG = \frac{9 \cdot 17}{16XC}$ .
Next, observe that $\triangle ACX \sim \triangle EFX$ , so $\frac{XE}{XF} = \frac{AX}{XC}$ . Also, $\triangle{AXD} \sim \triangle{EXY}$ , so $\frac{8}{AX} = \frac{EY}{XE}$ . We can compute $EY = \frac{128}{9}$ after noticing that $XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD$ and that $\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}$ . So, $\frac{8}{AX} = \frac{128}{9XE}$ . Then, $\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC$
Multiplying our equations for $XF$ and $XG$ yields that $XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{17}.$

Answer: A 17
Problem: Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution: Denote $P$ to be the intersection between line $AE$ and circle $O$ . Note that $\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE$ , making $\angle GFE + \angle GDE = 180$ . Thus, $PEFG$ is a cyclic quadrilateral. Using Power of a Point on $X$ gives $XP \cdot XE = XG \cdot XF$
Since $\triangle ADX \sim \triangle EYX$ and $\triangle ACX \sim \triangle EFX$ $AX/XE = XD/YX = 9/16$ . Using Power of a Point on $X$ again, $(AX)(PX) = (BX)(DX)$ . Plugging in $AX=9/16 XE$ gives: \[\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)\] By Law of Cosines , we can find $BD = \sqrt{51}$ , as in Solution 1. Now, $BX = 3/4 (\sqrt{51})$ and $DX = 1/4 (\sqrt{51})$ , making $\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}$ . This gives us $FX \cdot GX = \boxed{17}$ as a result.

Answer: A 17
Problem: Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution: $\because$ $AC 	\parallel EF$ $\quad \therefore$ $\triangle ACX \sim \triangle EFX$ $\quad \frac{XF}{XC} = \frac{XE}{XA}$
By Power of a Point, $XG \cdot XC = XD \cdot XB$
By multiplying the $2$ equations we get $XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB$
$\because$ $YE 	\parallel AD$ $\quad \therefore$ $\triangle EYX \sim \triangle ADX$ $\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}$
By substitution, $XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}$
Let $a = AB$ $b = BC$ $c = CD$ $d = AD$ $p = AC$ , and $q = BD$
By Ptolemy's theorem, $p \cdot q = a \cdot c + b \cdot d$
\[[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A\]
\[[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B\]
\[[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A\]
\[[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B\]
\[\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}\]
\[\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd\]
\[BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51\]
\[XF \cdot XG = \frac{51}{3} = \boxed{17}\]

Answer: A 17