Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Define a regular $n$ -pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?

Solution: We use the Principle of Inclusion-Exclusion (PIE).
If we join the adjacent vertices of the  regular $n$ -star, we get a regular $n$ -gon. We number the vertices of this $n$ -gon in a counterclockwise direction: $0, 1, 2, 3, \ldots, n-1.$
A regular $n$ -star will be formed if we choose a vertex number $m$ , where $0 \le m \le n-1$ , and then form the line segments by joining the following pairs of vertex numbers: $(0 \mod{n}, m \mod{n}),$ $(m \mod{n}, 2m \mod{n}),$ $(2m \mod{n}, 3m \mod{n}),$ $\cdots,$ $((n-2)m \mod{n}, (n-1)m \mod{n}),$ $((n-1)m \mod{n}, 0 \mod{n}).$
If $\gcd(m,n) > 1$ , then the star degenerates into a regular $\frac{n}{\gcd(m,n)}$ -gon or a (2-vertex) line segment if $\frac{n}{\gcd(m,n)}= 2$ . Therefore, we need to find all $m$ such that $\gcd(m,n) = 1$
Note that $n = 1000 = 2^{3}5^{3}.$
Let $S = \{1,2,3,\ldots, 1000\}$ , and $A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}$ . The number of $m$ 's that are not relatively prime to $1000$ is: $\mid A_{2}\cup A_{5}\mid = \mid A_{2}\mid+\mid A_{5}\mid-\mid A_{2}\cap A_{5}\mid$ $= \left\lfloor \frac{1000}{2}\right\rfloor+\left\lfloor \frac{1000}{5}\right\rfloor-\left\lfloor \frac{1000}{2 \cdot 5}\right\rfloor$ $= 500+200-100 = 600.$
Vertex numbers $1$ and $n-1=999$ must be excluded as values for $m$ since otherwise a regular n-gon, instead of an n-star, is formed.
The cases of a 1st line segment of (0, m) and (0, n-m) give the same star. Therefore we should halve the count to get non-similar stars.
Therefore, the number of non-similar 1000-pointed stars is $\frac{1000-600-2}{2}= \boxed{199}.$

Answer:  199
Problem: Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle . A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$ . The other endpoint ( $D'$ ) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$ $V_2$ is a trapezoid with a right angle then, from which it follows that $V_1$ contains one of the right angles of $\triangle ABC$ , and so $U_1$ is similar to $ABC$ . Thus $U_1$ , and hence $U_2$ , are $3-4-5\,\triangle$ s.
Suppose we find the ratio $r$ of the smaller base to the larger base for $V_2$ , which consequently is the same ratio for $V_1$ . By similar triangles, it follows that $U_1 \sim \triangle ABC$ by the same ratio $r$ , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, it follows that $[U_1] = r^2 \cdot [ABC] = 6r^2$
Of the two cases, the second is smaller; the answer is $\frac{3}{32}$ , and $m+n = \boxed{035}$

Answer:  035
Problem: circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: The location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$ . Let this triangle be $A'B'C'$
Notice that $ABC$ and $A'B'C'$ share the same incenter ; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of $\triangle ABC$ extend one unit farther than those of $\triangle A'B'C'$ . From $A = rs$ , we note that $r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6$ . Thus $r_{A'B'C'} = r_{ABC} - 1 = 5$ , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, $[A'B'C'] = [ABC] \cdot \left(\frac{r_{A'B'C'}}{r_{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}$
The probability is $\frac{2[A'B'C']}{34 \times 13} = \frac{375}{442}$ , and $m + n = \boxed{817}$

Answer:  817
Problem: circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: 2004 I AIME-10.png
Let the bisector of $\angle CAD$ be $AE$ , with $E$ on $CD$ . By the angle bisector theorem, $DE = 36/5$ . Since $\triangle AOR \sim \triangle AED$ $O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$ . Also $AT = 35$ so $RT = OQ = 30$
We can apply the same principle again to find that $PT = 27/2$ , and since $QT = 1$ , we find that $PQ = 27/2 - 1 = 25/2$ . The locus of all possible centers of the circle on this "half" of the rectangle is triangle $\triangle OPQ$ . There exists another congruent triangle that is symmetric over $AC$ that has the same area as triangle $\triangle OPQ$ $\triangle APQ$ has area $\frac {1}{2}\cdot OP \cdot PQ = \frac {1}{2}\cdot 30\cdot \frac {25}{2}$ , since $\angle OQP$ is right. Thus the total area that works is $30\cdot \frac {25}{2} = 375$ , and the area of the locus of all centers of any circle with radius 1 is $34\cdot 13 = 442$ . Hence, the desired probability is $\frac {375}{442}$ , and our answer is $\boxed{817}$

Answer:  817
Problem: circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: 2004 I AIME-10b.png
Again, the location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$
Let $A$ be at the origin, $B (36,0)$ $C (36,15)$ $D (0,15)$ . The slope of $\overline{AC}$ is $\frac{15}{36} = \frac{5}{12}$ . Let $\triangle A'B'C'$ be the right triangle with sides one unit inside $\triangle ABC$ . Since $\overline{AC} || \overline{A'C'}$ , they have the same slope, and the equation of $A'C'$ is $y = \frac{5}{12}x + c$ . Manipulating, $5x - 12y + 12c = 0$ . We need to find the value of $c$ , which can be determined since $\overline{AC}$ is one unit away from $\overline{A'C'}$ . Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:
\[\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1 \Longrightarrow \left|\frac{(5)(0) + (-12)(0) + 12c}{\sqrt{5^2 + (-12)^2}}\right| = 1\] \[c = \pm \frac{13}{12}\]
The two values of $c$ correspond to the triangle on top and below the diagonal. We are considering $A'B'C'$ which is below, so $c = -\frac{13}{12}$ . Then the equation of $\overline{A'C'}$ is $y = \frac{5}{12}x - \frac{13}{12}$ . Solving for its intersections with the lines $y = 1, x = 35$ (boundaries of the internal rectangle), we find the coordinates of $A'B'C'$ are at $A'\ (5,1)\ B'\ (35,1)\ C'\ (35,\frac{27}{2})$ . The area is $\frac{1}{2}bh = \frac{1}{2}(35-5)\left(\frac{27}{2} - 1\right) = \frac{375}{2}$
Finally, the probability is $\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}$ , and $m + n = \boxed{817}$

Answer:  817
Problem: solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum -shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$ . Given that $k=\frac m n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution: Our original solid has volume equal to $V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi$ and has surface area $A = \pi r^2 + \pi r \ell$ , where $\ell$ is the slant height of the cone.  Using the Pythagorean Theorem , we get $\ell = 5$ and $A = 24\pi$
Let $x$ denote the radius of the small cone. Let $A_c$ and $A_f$ denote the area of the painted surface on cone $C$ and frustum $F$ , respectively, and let $V_c$ and $V_f$ denote the volume of cone $C$ and frustum $F$ , respectively.  Because the plane cut is parallel to the base of our solid, $C$ is similar to the uncut solid and so the height and slant height of cone $C$ are $\frac{4}{3}x$ and $\frac{5}{3}x$ , respectively. Using the formula for lateral surface area of a cone, we find that $A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2$ . By subtracting $A_c$ from the surface area of the original solid, we find that $A_f=24\pi - \frac{5}{3}\pi x^2$
Next, we can calculate $V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3$ .  Finally, we subtract $V_c$ from the volume of the original cone to find that $V_f=12\pi - \frac{4}{9}\pi x^3$ . We know that $\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.$ Plugging in our values for $A_c$ $A_f$ $V_c$ , and $V_f$ , we obtain the equation $\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}$ .  We can take reciprocals of both sides to simplify this equation to $\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1$ and so $x = \frac{15}{8}$ .  Then $k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn$ so the answer is $m+n=125+387=\boxed{512}$

Answer:  512
Problem: solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum -shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$ . Given that $k=\frac m n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution: Our original solid $V$ has surface area $A_v = \pi r^2 + \pi r \ell$ , where $\ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $\ell = 5$ and lateral area $A_\ell = 15\pi$ . The area of the base is $A_B = 3^2\pi = 9\pi$
$V$ and $C$ are similar cones, because the plane that cut out $C$ was parallel to the base of $V$ . Let $x$ be the scale factor between the original cone and the small cone $C$ in one dimension. Because the scale factor is uniform in all dimensions, $x^2$ relates corresponding areas of $C$ and $V$ , and $x^3$ relates corresponding volumes. Then, the ratio of the painted areas $\frac{A_c}{A_f}$ is $\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k$ and the ratio of the volumes $\frac{V_c}{V_f}$ is $\frac{x^3}{1 - x^3} = k$ . Since both ratios are equal to $k$ , they are equal to each other. Therefore, $\frac{5 x^2}{8 - 5 x^2} = \frac{x^3}{1 - x^3}$
Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives $5 x^2(1 - x^3) = x^3(8 - 5 x^2)$ . Dividing both sides by $x^2$ and distributing the $x$ on the right, we have $5 - 5 x^3 = 8 x - 5 x^3$ , and so $8 x = 5$ and $x = \frac{5}{8}$ . Substituting back into the easier ratio, we have $\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}$ . And so we have $m + n = 125 + 387 = \boxed{512}$

Answer:  512
Problem: Let $S$ be the set of ordered pairs $(x, y)$ such that $0 < x \le 1, 0<y\le 1,$ and $\left[\log_2{\left(\frac 1x\right)}\right]$ and $\left[\log_5{\left(\frac 1y\right)}\right]$ are both even. Given that the area of the graph of $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ The notation $[z]$ denotes the greatest integer that is less than or equal to $z.$

Solution: $\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor$ is even when
\[x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots\]
Likewise: $\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor$ is even when
\[y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots\]
Graphing this yields a series of rectangles which become smaller as you move toward the origin . The $x$ interval of each box is given by the geometric sequence $\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots$ , and the $y$ interval is given by $\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots$
Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:
\[\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},\] and the answer is $m+n = 5 + 9 = \boxed{014}$

Answer:  014
Problem: The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ with $0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution: We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series
\begin{align*} P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} \end{align*}
This expression has roots at every $17$ th root and $19$ th roots of unity , other than $1$ .  Since $17$ and $19$ are relatively prime , this means there are no duplicate roots.  Thus, $a_1, a_2, a_3, a_4$ and $a_5$ are the five smallest fractions of the form $\frac m{19}$ or $\frac n {17}$ for $m, n > 0$
$\frac 3 {17}$ and $\frac 4{19}$ can both be seen to be larger than any of $\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}$ , so these latter five are the numbers we want to add.
$\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}$ and so the answer is $159 + 323 = \boxed{482}$

Answer:  482
Problem: A unicorn is tethered by a $20$ -foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower, and the length of the rope that is touching the tower is $\frac{a-\sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers , and $c$ is prime. Find $a+b+c.$

Solution: Looking from an overhead view, call the center of the circle $O$ , the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$ $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$ .  We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4\sqrt{5}$
Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$ , and let $E$ be the point directly below $B$ Triangles $\triangle CDA$ and $\triangle CEB$ are similar right triangles .  By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$
Let $x$ be the length of $CB$ \[\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}\]
Therefore $a=60, b=750, c=3, a+b+c=\boxed{813}$

Answer:  813
Problem: A unicorn is tethered by a $20$ -foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower, and the length of the rope that is touching the tower is $\frac{a-\sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers , and $c$ is prime. Find $a+b+c.$

Solution: Note that by Power of a Point, the point the unicorn is at has power $4 \cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\sqrt{80}=4\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\theta$ be the angle between the horizontal and the rope, we have that $\cos\theta=\frac{1}{5}$ which implies that $\sin\theta=\frac{2\sqrt{6}}{5}.$ Note that the portion of rope not on the tower is $4\sqrt{5} \cdot \frac{5}{2\sqrt{6}}= \frac{5\sqrt{30}}{3},$ the requested length of rope is $20-\frac{5\sqrt{30}}{3}=\frac{60-\sqrt{750}}{3}$ thus the requested sum is $\boxed{813}.$

Answer:  813
Problem: For all positive integers $x$ , let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$ . Let $d(x)$ be the smallest $n$ such that $x_n=1$ . (For example, $d(100)=3$ and $d(87)=7$ .) Let $m$ be the number of positive integers $x$ such that $d(x)=20$ . Find the sum of the distinct prime factors of $m$

Solution: We backcount the number of ways. Namely, we start at $x_{20} = 1$ , which can only be reached if $x_{19} = 10$ , and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (\times 10)$ . We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ and so forth downwards. There are $2^9$ ways to pick the first $9$ operations; however, not all $9$ of them may be $A: (-1)$ otherwise we return back to $x_{10} = 1$ , contradicting our assumption that $20$ was the smallest value of $n$ . Using complementary counting , we see that there are only $2^9 - 1$ ways.
Since we performed the operation $B: (\times 10)$ at least once in the first $9$ operations, it follows that $x_{10} \ge 20$ , so that we no longer have to worry about reaching $1$ again.
However, we must also account for a sequence of $10$ or more $A: (-1)$ s in a row, because that implies that at least one of those numbers was divisible by $10$ , yet the $\frac{x}{10}$ was never used, contradiction. We must use complementary counting again by determining the number of strings of $A,B$ s of length $18$ such that there are $10$ $A$ s in a row. The first ten are not included since we already accounted for that scenario above, so our string of $10$ $A$ s must be preceded by a $B$ . There are no other restrictions on the remaining seven characters. Letting $\square$ to denote either of the functions, and $^{[k]}$ to indicate that the character appears $k$ times in a row, our bad strings can take the forms: \begin{align*} &\underbrace{BA^{[10]}}\square \square \square \square \square \square \square\\ &\square \underbrace{BA^{[10]}}\square \square \square \square \square \square \\ &\qquad \quad \cdots \quad \cdots \\ &\square \square \square \square \square \square \underbrace{BA^{[10]}} \square \\ &\square \square \square \square \square \square \square \underbrace{BA^{[10]}} \\ \end{align*}
There are $2^7$ ways to select the operations for the $7$ $\square$ s, and $8$ places to place our $BA^{[10]}$ block. Thus, our answer is $2^9(2^9-1)-8\cdot 2^7 = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509$ , and the answer is $\boxed{511}$

Answer:  511
Problem: chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers $a$ and $e$ are relatively prime , and neither $c$ nor $f$ is divisible by the square of any prime . Find the remainder when the product $abcdef$ is divided by 1000.
2004 AIME II Problem 1.png

Solution: Let $r$ be the length of the radius of the circle.  A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\circ$ $60^\circ$ $90^\circ$ triangle , and the area of two such triangles is $2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}$ . The central angle which contains the entire chord is $60 \cdot 2 = 120$ degrees , so the area of the sector is $\frac{1}{3}r^2\pi$ ; the rest of the area of the circle is then equal to $\frac{2}{3}r^2\pi$
The smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the desired ratio is $\frac{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} =  \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}$
Therefore, $abcdef = 2^53^4 = 2592 \Longrightarrow \boxed{592}$

Answer:  592
Problem: A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution: The probability that Terry picks two red candies is $\frac{10 \cdot 9}{20 \cdot 19} = \frac{9}{38}$ , and the probability that Mary picks two red candies after Terry chooses two red candies is $\frac{7\cdot8}{18\cdot17} = \frac{28}{153}$ .  So the probability that they both pick two red candies is $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}$ .  The same calculation works for the blue candies.
The probability that Terry picks two different candies is $\frac{20\cdot10}{20\cdot19} = \frac{10}{19}$ , and the probability that Mary picks two different candies after Terry picks two different candies is $\frac{18\cdot 9}{18\cdot 17} = \frac{9}{17}$ . Thus, the probability that they both choose two different candies is $\frac{10}{19}\cdot\frac{9}{17} = \frac{90}{323}$ .  Then the total probability is
\[2 \cdot \frac{14}{323} + \frac{90}{323} = \frac{118}{323}\]
and so the answer is $118 + 323 = \boxed{441}$

Answer:  441