Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points.

Solution: The lines can be defined as starting from $(a, b, c)$ with "slope" (vector) $(d, e, f)$ . We impose the condition that at least one of $a - d, b - e, \textrm{ or } c - f$ is outside the range of $[1, 10]$ in order to ensure that $(a, b, c)$ is the first valid point on this line. Then, the line ranges from $(a, b, c), (a + d, b + e, c + f), \ldots, (a + 7d, b + 7e, c + 7f)$ , where $1 \le a + 7d, b + 7e, c + 7f \le 10$ , in which case at least one of $a + 8d, b + 8e, \textrm{ or } c + 8f$ is outside of the range $[1, 10]$ to ensure the line does not contain more than 8 points. For $1 \le a + 7d, b + 7e, c + 7f \le 10$ to be satisfied, the pairs $(a, d), (b, e), \textrm{ and } (c, f)$ can only be $(1, 0), (2, 0), \ldots (10, 0), (1, 1), (2, 1), (3, 1), (10, -1), (9, -1), \textrm{ and } (8, -1)$ . Notice that there are only two pairs such that $a + 8d, b + 8e, \textrm{ or } c + 8f \not \in [1,  10]$ , namely $(3, 1)$ and $(8, -1)$ . Thus, our line must contain at least one of these two pairs. In addition, only the two pairs $(1, 1)$ and $(10, -1)$ satisfy $a - d, b - e, \textrm{ or } c - f \not \in [1, 10]$ . Thus, we must also include at least one of these two pairs as well. Let us call these 4 pairs "important" pairs. Finally, we can include any valid pair for our third pair.
Case 1: We repeat one of the important pairs
There are 2 ways to choose from $(3, 1)$ and $(8, -1)$ , and 2 ways to choose from $(1, 1)$ and $(10, -1)$ . Then, there are 2 ways to choose the repeated pair. Next, we can arrange these pairs $3!/2! = 3$ ways. So, we have $2 \cdot 2 \cdot 2 \cdot 3 = 24.$
Case 2: We use 3 distinct important pairs
There are ${4 \choose 3}$ ways to choose the pairs (note that by pigeonhole principle, this guarantees we get at least one of each required pair). Then, there are $3! = 6$ ways to arrange it. We obtain $4 \cdot 6 = 24$
Case 3: We use 3 unique pairs, one that is not important.
Again, there are $2 \cdot 2 = 4$ ways to choose the important pairs. Then, there are 12 ways to choose the non-important pair. Again, there are $3! = 6$ ways to arrange it. So, we get $4 \cdot 12 \cdot 6 = 288$
Summing all of it up and dividing by two (since we over counted each line and its reversal), we get $\frac{24 + 24 + 288}{2} = \boxed{168}.$

Answer:  168
Problem: Tetrahedron $ABCD$ has $AD=BC=28$ $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution: Let $M$ and $N$ be midpoints of $\overline{AB}$ and $\overline{CD}$ . The given conditions imply that $\triangle ABD\cong\triangle BAC$ and $\triangle CDA\cong\triangle DCB$ , and therefore $MC=MD$ and $NA=NB$ . It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\overline{AB}$ and $\overline{CD}$ , and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$ . If $X$ is a point in space and $X'$ is the point symmetric to $X$ with respect to line $MN$ , then $BX=AX'$ and $CX=DX'$ , so $f(X) = AX+AX'+DX+DX'$
Let $Q$ be the intersection of $\overline{XX'}$ and $\overline{MN}$ . Then $AX+AX'\geq 2AQ$ , from which it follows that $f(X) \geq 2(AQ+DQ) = f(Q)$ . It remains to minimize $f(Q)$ as $Q$ moves along $\overline{MN}$
Allow $D$ to rotate about $\overline{MN}$ to point $D'$ in the plane $AMN$ on the side of $\overline{MN}$ opposite $A$ . Because $\angle DNM$ is a right angle, $D'N=DN$ . It then follows that $f(Q) = 2(AQ+D'Q)\geq 2AD'$ , and equality occurs when $Q$ is the intersection of $\overline{AD'}$ and $\overline{MN}$ . Thus $\min f(Q) = 2AD'$ . Because $\overline{MD}$ is the median of $\triangle ADB$ , the Length of Median Formula shows that $4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2$ and $MD^2 = 684$ . By the Pythagorean Theorem $MN^2 = MD^2 - ND^2 = 8$
Because $\angle AMN$ and $\angle D'NM$ are right angles, \[(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.\] It follows that $\min f(Q) = 2AD' = 4\sqrt{678}$ . The requested sum is $4+678=\boxed{682}$

Answer:  682
Problem: For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$

Solution: The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$ . The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$ , so the answer is $\frac{2016}{6}=\boxed{336}$

Answer:  336
Problem: Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc.  Then there are $21^2=441$ possible rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven.  The odds are therefore $\frac{56}{441}=\frac{8}{63}$ .  The answer is $8+63=\boxed{071}$

Answer:  071
Problem: Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling $n$ is $\frac{n}{21}$ because $21=\frac{6 \cdot 7}{2}$ Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice. 
The probability that 7 is rolled is now $2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})$ which is equal to $\frac{56}{441}=\frac{8}{63}$ .  Therefore the answer is $8+63=\boxed{071}$ ~PEKKA

Answer:  071
Problem: Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Since the probability of rolling a $1$ is $\frac{1}{21}$ , the probability of rolling a $2$ is $\frac{2}{21}$ the probability of rolling a $3$ is $\frac{3}{21}$ and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of $1$ and $6$ $2$ and $5$ , and $3$ and $4$ , and the rest is symmetry and the others are irrelevant.
We have: $2 \cdot (\frac{1}{21} \cdot \frac{2}{7}$ $+$ $\frac{2}{21} \cdot \frac{5}{21}$ $+$ $\frac{1}{7} \cdot \frac{4}{21})$ $=$ $2 \cdot (\frac{2}{147} + \frac{10}{441} + \frac{4}{147})$ $2 \cdot \frac{4}{63} = \frac{8}{63}$ . Therefore, the answer is $8 + 63$ $\boxed{071}$

Answer:  071
Problem: regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]

Solution: Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place.  Then, there are 2 paths from that point down to the third plane.  Within the third plane, there are 9 paths as well (consider the logic from the second plane).  Finally, there is only one way down to the bottom vertex.  Therefore there are $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1=\boxed{810}$ paths.

Answer:  810
Problem: regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]

Solution: Assume an ant is on the top of this icosahedron. Note that the icosahedron has two pentagon planes and two points where the ant starts and ends. Also note that when the ant hits a vertex of the pentagon, there is only two ways to go down. When the ant ends up at the last vertex and is about to head down, there is only one way to go down.
Case $1$ : The ant move down to the first pentagon, then the next pentagon, and finally to the last point in a total of four moves; there are a total of $5 \cdot 2 \cdot 1 = 10$ ways to achieve this.
Case $2$ : The ant goes to the first pentagon and moves to another vertex in the same pentagon. The ant has a total of $8$ paths doing this since it may go through $1, 2, 3,$ or $4$ edges from either two directions when going across the edges of the pentagon (recall the ant may not repeat a move to another vertex) . Now the ant moves to the second pentagon below. The ant again has a total of $8$ moves if he wanders around the pentagon. Finally, the ant moves down after wandering to the last vertex. There is a total of $5 \cdot 8 \cdot 2 \cdot 8 = 640$ ways.
Case $3$ : Assume the ant goes to the first pentagon and wanders around. Then the ant goes the next pentagon and then heads directly to the last vertex. There are $5 \cdot 8 \cdot 2 = 80$ ways.
Case $4$ : Now let the ant go to the first pentagon and then go directly down to the next pentagon. The ant wanders around the second pentagon before heading to the last vertex. Like case $3$ above, there are $5 \cdot 2 \cdot 8 = 80$ ways.
Adding up all four cases, we get $10 + 640 + 80 + 80 =\boxed{810}$ total paths, as desired.      ~skyscraper

Answer:  810
Problem: regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]

Solution: Go to 2020 AMC 10A #19 , and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $\boxed{810}$

Answer:  810
Problem: A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$

Solution: Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid.  Then $\angle CAB = 120^\circ$ . Let $X$ be the foot of the altitude from $A$ to $\overline{BC}$ . Then since $\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$ .  Since the dihedral angle between $\triangle ABC$ and $\triangle BCD$ is $60^\circ$ $\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\sqrt{3} = h$ . Thus $h^2 = \boxed{108}$

Answer:  108
Problem: A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$

Solution: Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$ . Then, we have $AM=6$ $30-60-90$ ) triangles and $DM=\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\angle AMD=60^{\circ}$ , we get \[h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},\] as desired.

Answer:  108
Problem: Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$

Solution: We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$ , the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\left \lfloor{\frac{11}{2}}\right \rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=\boxed{053}$

Answer:  053
Problem: Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$

Solution: If we let $k$ be equal to the number of days it took to read the book, the sum of $n$ through $n+k$ is equal to $(2n+k)(k+1)=748$ Similarly, $(2t+k)(k+1)=638$ We know that both factors must be integers and we see that the only common multiple of $748$ and $638$ not equal to $1$ that will get us positive integer solutions for $n$ and $t$ is $11$ . We set $k+1=11$ so $k=10$ . We then solve for $n$ and $t$ in their respective equations, getting $2n+10=68$ $n=29$ We also get $2t+10=58$ $t=24$ . Our final answer is $29+24=\boxed{053}$

Answer:  053
Problem: Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$

Solution: Notice $374=34\cdot 11$ and $319=29\cdot 11$ . Also, note the sum of an arithmetic series is $\frac{2n+k}{2} \cdot b$ , where $n$ is our first term, $n+k$ is our final term, and $b$ is the number of terms. Since we know both sequences of $n$ and $t$ have the same length, and since $11$ is prime and shared by both $319$ and $374$ , we deduce that $b=11$ . Thus from here we know $2n+k=68$ and $2t+k=58$ by using our other factors $34$ and $29$ . Finally, we add the two systems up and we get $2t+2k+2n=126$ . But, notice that $k=b-1$ , since the first term has $k=0$ , and our last term has $k=b-1$ . Plugging this back into our equation we get $2n+2t=106 \implies n+t=\boxed{053}$

Answer:  053
Problem: Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$

Solution: We list two equations:
\begin{align*}
n+(n+1)+...+(n+k)&=374\\
t+(t+1)+...+(t+k)&=319.
\end{align*}
Subtracting the two, we get: \[(n-t)(k+1)=374-319=55.\] Manipulating the first and second equation, we get:
\begin{align*}
n(k+1)+\frac{k(k+1)}{2}&=374 \\
t(k+1)+\frac{k(k+1)}{2}&=319.
\end{align*}
We factor out the common factor $k+1$ :
\begin{align*}
(k+1)\left(n+\frac{k}{2}\right)&=374 \\
(k+1)\left(t+\frac{k}{2}\right)&=319.
\end{align*}
Note that $374$ and $319$ have a GCD of $11,$ now combining this with our equation that $(n-t)(k+1)=55,$ we see that $k+1$ has to equal $11.$ Thus, we get: \[(n,t)=(29,24) \Rightarrow n+t=29+24=\boxed{053}.\] Note: We see that because n-t=5, it becomes impossible for n+k/2 and t+k/2 to both be multiples of 11. Thus, this satisfies our condition. Thus k+1 must be 11 to satisfy the common factor 11 constraint. mathboy282

Answer:  053
Problem: In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$ . Similarly, $\angle ABD=\gamma$ . Also, using $\triangle ICA$ , we find $\angle CIA=180-\alpha-\gamma$ . Therefore, $\angle AID=\alpha+\gamma$ . Therefore, $\angle DAI=\angle AID=\alpha+\gamma$ , so $\triangle AID$ must be isosceles with $AD=ID=5$ . Similarly, $BD=ID=5$ . Then $\triangle DLB \sim \triangle ALC$ , hence $\frac{AL}{AC} = \frac{3}{5}$ . Also, $AI$ bisects $\angle LAC$ , so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$ . Thus $CI = \frac{10}{3}$ , and the answer is $\boxed{013}$

Answer:  013
Problem: In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$ , and $\angle CLB=\angle CLA=90^\circ$ . Draw the perpendicular from $I$ to $CB$ , and let it meet $CB$ at $E$ . Since $IL=2$ $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$ . Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$ . Thus, $CI=\tfrac{2x}{3}$ $\triangle CBD \sim \triangle CEI$ , so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$ . Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$ . Solving for $x$ , we have: $x^2-2x-15=0$ , or $x=5, -3$ $x$ is positive, so $x=5$ . As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$

Answer:  013
Problem: In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: WLOG assume $\triangle ABC$ is isosceles (with vertex $C$ ). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$ ) and that $O$ and $I$ are collinear. Next, if $OI=d$ $DO+OI=R+d$ and $R+d=DL+LI=5$ . Euler gives us that $d^{2}=R(R-2r)$ , and in this case, $r=LI=2$ . Thus, $d=\sqrt{R^{2}-4R}$ . Solving for $d$ , we have $R+\sqrt{R^{2}-4R}=5$ , then $R^{2}-4R=25-10R+R^{2}$ , yielding $R=\frac{25}{6}$ . Next, $R+d=5$ so $d=\frac{5}{6}$ . Finally, $OC=OI+IC$ gives us $R=d+IC$ , and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$ . Our answer is then $\boxed{013}$

Answer:  013