Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: The sum of an infinite geometric series is of the form: \[\begin{split} S & = \frac{a_1}{1-r}  \end{split}\] where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio. 
In other words: \[\begin{split} a_1 \cdot r & = 1 \\ a_1 & = \frac{1}{r} \end{split}\]
Thus, the sum is the following: \[\begin{split} S & = \frac{\frac{1}{r}}{1-r} \\\\ S & =\frac{1}{r-r^2} \end{split}\]
Since we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2+r$ .
The maximum x-value of a quadratic with leading coefficient $-a$ is $\frac{-b}{2a}$ \[\begin{split} r & = \frac{-(1)}{2(-1)} \\\\ r & = \frac{1}{2}  \end{split}\]
Plugging $r$ $=$ $\frac{1}{2}$ into the quadratic yields: \[\begin{split} S & = \frac{1}{\frac{1}{2} -\left(\frac{1}{2}\right)^2} \\\\ S & = \frac{1}{\frac{1}{4}}  \end{split}\]
Therefore, the minimum sum of our infinite geometric sequence is $\boxed{4}$ .
(Solution by akaashp11)

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: After observation we realize that in order to minimize our sum $\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $\frac{1}{x}$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $\frac{1}{x}$
The first term has to be $x$ , so then in order to minimize the sum, we have minimize $x$
The smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$ . So our first term is $2$ and our common ratio is $\frac{1}{2}$ . Thus the sum is $\frac{2}{\frac {1}{2}}$ or $\boxed{4}$ .
Solution 2 by No_One

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get \[S=\frac{a}{1-r} \implies S-Sr=a\] Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$ , we would get \[r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0\] This quadratic has real solutions if the discriminant is greater than or equal to zero, or \[S^2-4\cdot S \cdot 1 \ge 0\] This yields that $S\le 0$ or $S\ge 4$ . 
However, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\boxed{4}$

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: Let the first term of the geometric series $x$ . Since it must be decreasing, we have $x>1$ and the third term is $\frac{1}{x}$ . Realize that by AM-GM inequality $x+\frac{1}{x} \ge 2$ with equality if $x = 1$ . However, we established that $x>1$ so that means $x+\frac{1}{x} > 2$ . So the sum of the first three terms of the sequence $x + \frac{1}{x} + 1$ is greater than $3$ , and the geometric series keeps continuing  infinitely.  This means the sum continues increasing. The only answer choice greater than $3$ is $\boxed{4}$ . ~skyscraper

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: Let the first term be $k.$ The sum of the series is $\frac{k}{1- \frac{1}{k}} =\frac{k^2}{k-1}.$ Rewrite this as $\frac{k^2-2k+1}{k-1} +\frac{2k-1}{k-1} = k-1+\frac{2k-2}{k-1} +\frac{1}{k-1} = (k-1) + \left(\frac{1}{k-1}\right) + 2.$ By AM-GM we know that $(k-1) + \left(\frac{1}{k-1}\right) \ge 2$ so the minimum is $2+2 = \boxed{4}.$

Answer: E 4
Problem: The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: Set the first term is $a.$ , the common ratio should be $\frac{1}{a}.$
The sum to infinity of the series is $S=\frac{a}{1-\frac{1}{a}}=\frac{a^2}{a-1}.$
Since $S$ is positive, we have $a>1.$ Define the function $f(a)=\frac{a^2}{a-1}$ , the domain of this function is $a>1.$
Let $f^{'}(a)=\frac{2a^2-2a-a^2}{(a-1)^2}=\frac{a(a-2)}{(a-1)^2}=0.$ We solve that $a=2.$
It's easy to find that when $1<a<2, f^{'}(a)<0,$ when $a>2, f^{'}(a)>0.$ Thus $f(a)$ has a minimum value when $a=2.$ , which is $f(2)=4.$ Choose $\boxed{4}.$

Answer: E 4
Problem: All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution: Let us call the six faces of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$ $c$ is opposite $e$ , and $b$ is opposite $f$ .
Thus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$ .
Let us find the sum of these products: \[abc+abe+bcd+bde+acf+cdf+aef+def\] We notice $b$ is a factor of the first four terms, and $f$ is a factor of the last four terms. \[b(ac+ae+cd+de)+f(ac+ae+cd+de)\] Now, we can factor even more:
\begin{align*} & (b+f)(ac+ae+cd+de) \\ = &(b+f)(a(c+e)+d(c+e)) \\ = &(b+f)(a+d)(c+e) \end{align*} We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is $(7+2)$ $(6+3)$ , and $(5+4)$ all factors. \begin{align*} & (7+2)(6+3)(5+4) \\ = & 9 \cdot 9 \cdot 9 \\ = & 729. \end{align*} Thus our answer is $\boxed{729}$

Answer: D 729
Problem: All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution: We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.
Thus our answer is $\boxed{729}$

Answer: D 729
Problem: All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution: We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$ . To maximize the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{729}$

Answer: D 729
Problem: All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution: First, we intuitivly notice that multiplying large numbers together and smaller numbers together tends to produce larger sums, while multiplying large numbers with smaller numbers tends to produce smaller sums. From this, we guess that it is optimal to have $7, 6,$ and $5$ to be around one vertex to produce at least one large product of $210$ . We can immediately eliminate a) and b) as answer choices since they are very close to this product, as well as e) since it is exactly $8 \cdot 210$ and we know that every following product will be smaller than this. We also guess that that is is most likely optimal to create a pairing where a number and its complement (the number that's the difference of 9 and this number) are on opposite sides. Using these two guesses, we can construct a net of sides and brute force the solution (or estimate) leaving us with $\boxed{729}$

Answer: D 729
Problem: In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: Factor $345=3\cdot 5\cdot 23$
Suppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\tfrac12k<m$ .
Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.
Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ and $m+1$ . Then $k(2m+1)=345$ with $k\le m$ .
Looking again at the factors of $345$ , the possible values of $k$ are $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively.
Thus the answer is $\boxed{7}$

Answer: E 7
Problem: In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: We need to find consecutive numbers (an arithmetic sequence that increases by $1$ ) that sums to $345$ . This calls for the sum of an arithmetic sequence given that the first term is $k$ , the last term is $g$ and with $n$ elements, which is: $\frac {n \cdot (k+g)}{2}$
We look for sequences of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$ . We can now substitute $g$ with $k+n-1$ . Now we substitute our new value of $g$ into $\frac {n \cdot (k+g)}{2}$ to get that the sum is $\frac {n \cdot (k+k+n-1)}{2} = 345$
This simplifies to $\frac {n \cdot (2k+n-1)}{2} = 345$ . This gives a nice equation. We multiply out the 2 to get that $n \cdot (2k+n-1)=690$ . This leaves us with 2 integers that multiply to $690$ which leads us to think of factors of $690$ . We know the factors of $690$ are: $1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690$ . So through inspection (checking), we see that only $2,3,5,6,10,15$ and $23$ work. This gives us the answer of $\boxed{7}$ ways.

Answer: E 7
Problem: In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to $30$ , and that is because $\frac{345}{30}=11.5$ , which means $11$ and $12$ must be the middle 2 numbers; however, a sequence of length $30$ with middle numbers $11$ and $12$ that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since $1$ is a trivial, non-counted solution, we get $\boxed{7}$ -ColtsFan10

Answer: E 7
Problem: In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: The median of the sequence $m$ is either an integer or a half integer. Let $m=\frac{i}{2}, i \in N$ , then $P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23$
On the other hand we have two constraints:
1) $m \geq \frac{n+1}{2} \iff i>n$ because the integers in the sequence are all positive, and $n>1$
2) If $n$ is odd then $m$ is an integer, $i$ is even; if $n$ is even then $m$ is a half integer, $i$ is odd. Therefore, $n$ and $i$ have opposite parity.
Now $P$ has $16$ factors and it is not a perfect square. There are $8-1=7$ choices for $1 < n < \sqrt{P}$ . Also since $2|P, 4\nmid P$ , we know $n$ and $\frac{P}{n}$ must have opposite parity. Therefore the answer is $\boxed{7}$

Answer: E 7
Problem: A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: The center of dilation must lie on the line $A A'$ , which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$ . Note that the center of dilation must have an $x$ -coordinate less than $2$ ; if the $x$ -coordinate were otherwise, then the circle under the transformation would not have an increased $x$ -coordinate in the coordinate plane. Also, the ratio of dilation must be equal to $\dfrac{3}{2}$ , which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\dfrac{3}{2} \left( 2 - x \right) = 5 - x$ (for the $x$ -coordinates), and $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$ . We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$ -coordinate, and hence a lower $y$ -coordinate, from our reasoning above. Solving the two equations, we get $x = -4$ and $y = - 6$ . This means that any point $(a,b)$ on the plane will dilate to the point $\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)$ , which means that the point $(0,0)$ dilates to $\left( 6 - 4, 9 - 6 \right) = (2,3)$ . Thus, the origin moves $\sqrt{2^2 + 3^2} = \boxed{13}$ units.

Answer:  13
Problem: A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Using the ratios of radii of the circles, $\frac{3}{2}$ , we find that the scale factor is $1.5$ . If the origin had not moved, this indicates that the center of the
circle would be $(3,3)$ , simply because of $(2 \cdot 1.5,  2 \cdot 1.5)$ . Since the center has moved from $(3,3)$ to $(5,6)$ , we apply the distance formula and get: $\sqrt{(6-3)^2 + (5-3)^2} = \boxed{13}$

Answer:  13
Problem: A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Before dilation, notice that the two axes are tangent to the circle with center $(2,2)$ . Using this, we can draw new axes tangent to the radius 3 circle with center $(5,6)$ , resulting in a "new origin" that is 3 units left and 3 units down from the center $(5,6)$ , or $(2,3)$ . Using the distance formula, the distance from $(0,0)$ and $(2,3)$ is $\boxed{13}$ .
~Mightyeagle

Answer:  13
Problem: A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\frac{3}{2}$ . The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point $(-4, -6)$ . The distance from that point to the origin is $2\sqrt{13}$ , and halving that to get the distance the origin moves gives $\boxed{13}$ . ~Meeshbot

Answer:  13
Problem: What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution: Another way to solve this problem is using cases. 
Though this may seem tedious, we only have to do one case since the area enclosed is symmetrical.
The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible,
we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive,
we get the equation $x^{2} +y^{2} -x-y = 0.$
Using the complete the square method, we get $\left(x-\frac{1}{2}\right)^{2} + \left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$
Therefore, the origin of this section of the shape is at $\left(\frac{1}{2}, \frac{1}{2}\right).$
Using the equation we can also see that the radius has a length of $\frac{\sqrt{2}}{2}$
With this shape we see that this shape can be cut into a right triangle and a semicircle.
The length of the hypotenuse of the triangle is $\sqrt{2}$ so using special right triangles, we see that
the area of the triangle is $\frac{1}{2}$ . The semicircle has the area of $\frac{1}{4}\pi$
But this is only $1$ case. There are $4$ cases in total so we have to multiply $\frac{1}{2}+\frac{1}{4}\pi$ by $4$
After multiplying, our answer is: \[\boxed{2}.\]

Answer: B 2
Problem: What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution: We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$ Case 2: $x\geq0, y<0.$ Case 3: $x<0, y\geq0.$ Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can figure out the whole figure: (scaled 8x). [asy] size(400); import TrigMacros;  rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true);    draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237));   [/asy] The way we figure out the area is by splitting it the following way: [asy] size(400); import TrigMacros;  rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true);    draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237));  real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5));   real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5));  real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5));  real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5));   real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5));   pair A,B,C,D;  A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]
We know each of the red lines is a diameter of the circle which is $\sqrt2$ . So the area of the red is 2. We know that the area of each semicircle is $\dfrac14 \pi$ so the area of the semicircles combines is $\pi$ . Thus we get $\boxed{2}.$

Answer: B 2
Problem: A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution: There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
[asy]size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy] But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.
Thus the answer is $1330-945=385$ which is $\boxed{385}$

Answer: A 385
Problem: A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution: Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams.  Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4 \cdot 1330=332.5$ . Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, and that the next closest is $665$ which is twice of $332.5$ , so our answer is $385$ which is $\boxed{385}$

Answer: A 385