Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Solution: Since $AEF$ is equilateral, $AE=EF$ . Let $BE=x$ . By the Pythagorean theorem $1+x^2=2(1-x)^2$ . Simplifying, we get $x^2-4x+1=0$ . By the quadratic formula, the roots are $2 \pm \sqrt{3}$ . Since $x<1$ , we discard the root with the "+", giving $x=2-\sqrt{3}$ [asy] real n; n=0.26794919243; real m; m=0.2113248654; draw((0,0)--(0,n)--(1,0)--(0,0)); draw((0,m)--(m,m)--(m,0)); label((0,0), "$B$",SW); label((0,n), "$E$",SW); label((0,m), "$M$",SW); label((1,0), "$A$",SW); label((m,0), "$N$",SW); label((m,m), "$K$",NE); [/asy] Let the side length of the square be s. Since $MEK$ is similar to $ABE$ $s=\frac{2-\sqrt{3}-s}{2-\sqrt{3}}$ . Solving, we get $s=\frac{3-\sqrt{3}}{6}$ and the final answer is $\boxed{012}$

Answer:  012
Problem: Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution: There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting ).
Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$ . Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs.

Answer:  738
Problem: Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution: We proceed by casework on the number of digits of $a.$
Case 1: Both $a$ and $b$ have three digits
We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \cdot 8 \cdot 9 = 576$ possible values in this case.
Case 2: $a$ (or $b$ ) has two digits
If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$ ). There thus are $8 \cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \cdot 2 = 144$ possible values in this case.
Case 3: $a$ (or $b$ ) has one digit
This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \cdot 2 = 18$ possible values for this case.
Adding them all up, we get $576 + 144 + 18 = \boxed{738},$ and we're done.

Answer:  738
Problem: Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution: For every $a \in [1, 999]$ $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits.
First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have:
However, we have still overlooked the cases where $1000 - a$ contains zero digits:
Adding up the cases with $a \in [1, 999]$ with no zero digits and removing the cases with $1000 - a$ with zero digits gives us \[(9^1 + 9^2 + 9^3) - 9^2 = \boxed{738}\]

Answer:  738
Problem: There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]

Solution: If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection.  Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
There are 6 possible colors for the center triangle.
Thus, in total we have $6\cdot(20 + 30 + 6) = \boxed{336}$ total possibilities.

Answer:  336
Problem: There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]

Solution: We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.
Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.
Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.
Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.
Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.
Now, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\boxed{336}$ distinguishable ways to color the triangle.

Answer:  336
Problem: There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]

Solution: There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \[6 \binom{5 + 3}{3} = 6\binom{8}{3} = \boxed{336}.\]

Answer:  336
Problem: Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope , and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Solution: 2006 II AIME-9.png
Call the centers $O_1, O_2, O_3$ , the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$ , and $s_2$ on $\mathcal{C}_2$ ), and the intersection of each common internal tangent to the X-axis $r, s$ $\triangle O_1r_1r \sim \triangle O_2r_2r$ since both triangles have a right angle and have vertical angles, and the same goes for $\triangle O_2s_2s \sim \triangle O_3s_1s$ . By proportionality , we find that $O_1r = 4$ ; solving $\triangle O_1r_1r$ by the Pythagorean theorem yields $r_1r = \sqrt{15}$ . On $\mathcal{C}_3$ , we can do the same thing to get $O_3s_1 = 4$ and $s_1s = 4\sqrt{3}$
The vertical altitude of each of $\triangle O_1r_1r$ and $\triangle O_3s_1s$ can each by found by the formula $c \cdot h = a \cdot b$ (as both products equal twice of the area of the triangle). Thus, the respective heights are $\frac{\sqrt{15}}{4}$ and $2\sqrt{3}$ . The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: $\sqrt{15 - \frac{15}{16}} = \frac{15}{4}$ , and by 30-60-90: $6$
From this information, the slope of each tangent can be uncovered. The slope of $t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{4}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}$ . The slope of $t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}$
The equation of $t_1$ can be found by substituting the point $r (4,0)$ into $y = \frac{1}{\sqrt{15}}x + b$ , so $y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}$ . The equation of $t_2$ , found by substituting point $s (16,0)$ , is $y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$ . Putting these two equations together results in the desired $\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$ $\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}$ $= \frac{76 - 12\sqrt{5}}{4}$ $= 19 - 3\sqrt{5}$ . Thus, $p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}$

Answer:  027
Problem: Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$
Of these three cases ( $|A| > |B|, |A| < |B|, |A|=|B|$ ), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ${5\choose k}$ ways to $A$ to have $k$ victories, and ${5\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$
Thus $p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}$ . The desired probability is the sum of the cases when $|A| \ge |B|$ , so the answer is $\frac{126}{512} + \frac{193}{512} = \frac{319}{512}$ , and $m+n = \boxed{831}$

Answer:  831
Problem: Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.
Summing these 6 cases, we get $\frac{638}{1024}$ , which simplifies to $\frac{319}{512}$ , so our answer is $319 + 512 = \boxed{831}$

Answer:  831
Problem: Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: We can apply the concept of generating functions here.
The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$ . Thus, the total generating function for number of games he wins is
The generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$ . 
Thus, the generating function for $A$ is
$1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$
The probability that $B$ wins 0 games is $\frac{1}{32}$ . Since the coefficients for all $x^{n}$ where
$n \geq 1$ sums to 32, the probability that $A$ wins more games is $\frac{32}{32}$
Thus, the probability that $A$ has more wins than $B$ is $\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}$
Thus, $319 + 512 = \boxed{831}$

Answer:  831
Problem: Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: After the first game, there are $10$ games we care about-- those involving $A$ or $B$ . There are $3$ cases of these $10$ games: $A$ wins more than $B$ $B$ wins more than $A$ , or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity . There are therefore $\frac{1024-252}{2} = 386$ possibilities for each of the other two cases.
If $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\frac{638}{1024} = \frac{319}{512}$ , and $m+n=\boxed{831}$

Answer:  831
Problem: sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution: Define the sum as $s$ . Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:
Thus $s = \frac{a_{28} + a_{30}}{2}$ , and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\boxed{834}$ .−

Answer:  834
Problem: sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution: Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
$a_{1}\equiv 1 \pmod {1000} \\ a_{2}\equiv 1 \pmod {1000} \\ a_{3}\equiv 1 \pmod {1000} \\ a_{4}\equiv 3 \pmod {1000} \\ a_{5}\equiv 5 \pmod {1000} \\ \cdots \\ a_{25} \equiv 793 \pmod {1000} \\ a_{26} \equiv 281 \pmod {1000} \\ a_{27} \equiv 233 \pmod {1000} \\ a_{28} \equiv 307 \pmod {1000}$
Adding all the residues shows the sum is congruent to $\boxed{834}$ mod $1000$

Answer:  834
Problem: sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution: All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$ , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$ . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})$ , at least for the first few terms. From this, we have that $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}(\mod 1000)$

Answer:  834
Problem: Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$

Solution: Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$ . Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$ . So $\Delta{GBC} \sim \Delta{EAF}$
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF  = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$ . Therefore, the answer is $429+433+3=\boxed{865}$

Answer:  865
Problem: Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$

Solution: Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\sqrt{3}, -1)$ $C (\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$ $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$ . As $DF$ and $EF$ are parallel to $AE$ and $AD$ $F (-1, -12\sqrt{3}+2)$ $G$ and $A$ is the intersection between $AF$ and circle $O$ . Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$ . Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$ , so the answer is $\boxed{865}$ . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation.

Answer:  865
Problem: Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$

Solution: Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\circ}$ , angle $D$ is $120^{\circ}$ . Use law of cosines to find $AF = \sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$ . Next we use law of cosine on triangles $BAG$ and $GAC$ , solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$ . Next, $AG = 2\cdot R \cos{\theta}$ , where R is radius of circle $= 2$ and $\theta =$ angle $BAG$ . We already know sine of the angle so find cosine, hence we have found $AG$ . At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$ . Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$ , and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$ , to give answer = $\boxed{865}$

Answer:  865