Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$

Solution: Let the second term of each series be $x$ . Then, the common ratio is $\frac{1}{8x}$ , and the first term is $8x^2$
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$
The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$ . Therefore, $100m+10n+p = \boxed{518}$

Answer:  518
Problem: Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$

Solution: Let's ignore the "two distinct, real, infinite geometric series" part for now and focus on what it means to be a geometric series.
Let the first term of the series with the third term equal to $\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\frac{a}{1-r} = 1 \implies a = 1-r,$ and $ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.$
We see that this cubic is equivalent to $r^3 - r^2 + \frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\dfrac{1 \pm \sqrt{5}}{4}.$
It becomes clear that the two geometric series have common ratio $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}.$ Let $\frac{1 + \sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \dfrac{3 - \sqrt{5}}{4}.$
Since the second term in the series is $ar,$ we compute this and have that \[ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},\] for our answer of $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$

Answer:  518
Problem: A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$

Solution: We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$ . We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. [asy] size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("$1$",(0,0),SE,fontsize(8)); label("$1$",(1,0),SE,fontsize(8)); label("$1$",(2,0),SE,fontsize(8)); label("$1$",(2,1),SE,fontsize(8)); label("$1$",(3,0),SE,fontsize(8)); label("$2$",(3,1),SE,fontsize(8)); label("$2$",(3,2),SE,fontsize(8)); label("$1$",(4,0),SE,fontsize(8)); label("$3$",(4,1),SE,fontsize(8)); label("$5$",(4,2),SE,fontsize(8)); label("$1$",(5,0),SE,fontsize(8)); label("$4$",(5,1),SE,fontsize(8)); label("$9$",(5,2),SE,fontsize(8)); label("$9$",(5,3),SE,fontsize(8)); label("$1$",(6,0),SE,fontsize(8)); label("$5$",(6,1),SE,fontsize(8)); label("$14$",(6,2),SE,fontsize(8)); label("$23$",(6,3),SE,fontsize(8)); label("$23$",(6,4),SE,fontsize(8)); [/asy] Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$ . Hence the desired probability is \[\frac{23\cdot 2^4\cdot 3^6}{5^{10}},\] and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$

Answer:  660
Problem: A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$

Solution: The first restriction is that $a_{10} = .4$ , meaning that the player gets exactly 4 out of 10 baskets.  The second restriction is $a_n\le.4$ .  This means that the player may never have a shooting average over 40%.  Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$ , but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3.  Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
OOXOXOOXOX
To simplify counting, note that the first, second, and tenth shots are predetermined.  The first two shots must fail, and the last shot must succeed.  Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
XOXOOXO
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7.  The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
First shot made on attempt 3:
XOXOOXO
XOXOOOX
XOOXOXO
XOOXOOX
XOOOXXO
XOOOXOX
XOOOOXX
Total - 7
First shot made on attempt 4:
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
Total - 7
First shot made on attempt 5:
OOXXOXO
OOXXOOX
OOXOXXO
OOXOXOX
OOXOOXX
Total - 5
First shot made on attempt 6:
OOOXXXO
OOOXXOX
OOOXOXX
Total - 3
First shot made on attempt 7:
OOOOXXX
Total - 1
The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$
Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$

Answer:  660
Problem: A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$

Solution: Note $a_{10}=.4$ . Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$ . Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$ . Now we must look at the 3rd through 9th shot.
Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$ . Therefore we can have two shots made, one shot made, or no shots made in group A.
Case 1: Group A contains no made shots.
The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here.
Case 2: Group A contains one made shot.
The number of ways this could happen in group A is 3.  Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$ . All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here.
Case 3: Group A contains two made shots.
The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case.
Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$ . Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$ . Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$ . Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$

Answer:  660
Problem: In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$
[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R;  A=(0,0);  B=(8,0);  C=(1.9375,3.4994);  D=(3.6696,2.4996);  E=(1.4531,2.6246);  X=(4.3636,0);  P=(2.9639,2.0189);  Q=(1.8462,0);  R=(6.4615,0);  dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]
We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$
By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}\]
Now we apply mass points. Assume WLOG that $W_{A}=1$ . That means that
\[W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}\]
Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$ . Therefore,
\[\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}\]
Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$ . Therefore,
\[\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}\]
Because $\triangle{PQR}$ is similar to $\triangle{CAB}$ $\angle{C}=\angle{P}$
As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$
Therefore, \[\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}\]

Answer:  901
Problem: In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: [asy] size(10cm); pair A,B,C,D,E,P,Q,R;  A=(0,0);  B=(8,0);  C=(1.9375,3.4994);  D=(3.6696,2.4996);  E=(1.4531,2.6246);  P=(2.9639,2.0189);  Q=(1.8462,0);  R=(6.4615,0);  dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); [/asy] Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$ , we can see that the mass of Q is $26$ , hence we know that $\frac{BP}{PE}=\frac{10}{3}$ , now we can find that $\frac{PQ}{AE}=\frac{10}{3}$ which implies $PQ=\frac{30}{13}$ , it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$ , so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$ . ~bluesoul

Answer:  901
Problem: The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$

Solution: Let the circle intersect $\overline{PM}$ at $B$ . Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point . Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \[\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}\] Solving, $AM = 38$ . So the ratio of the side lengths of the triangles is 2. Therefore, \[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\] so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$ , we see that $4OP-76 = OP+19$ , so $OP = \frac{95}3$ and the answer is $\boxed{098}$

Answer:  098
Problem: The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$

Solution: Reflect triangle $PAM$ across line $AP$ , creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point $P$ , with $x + 38$ as $AP$ . Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$ . The area of the isoceles triangle is:
$[PAM] = r \cdot s$
$[PAM] = 19 \cdot (114 - x)$
Now use similarity, draw perpendicular from $O$ to $PM$ , name the new point $D$ . Triangle $PDO$ is similar to triangle $PAM$ , by AA Similarity. Equating the legs, we get:
$\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}$
Solving for $AM$ ,  it yields $19 \cdot \sqrt{\frac{x + 38}{x}}$
$19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}$
The $x^3$ cancels, yielding a quadratic. Solving yields $x = \frac{38}{3}$ .
Add $19$ to find $OP$ , yielding $\frac{95}{3}$ or $\boxed{098}$

Answer:  098
Problem: The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$

Solution: Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\frac{y}{x}\cdot 19$ , since $O$ is on the angle bisector of $\angle M.$
We have that $\tan(\angle AMO)=\frac{19}{x},$ so \[\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.\]
However $\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}$ , so \[\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}\] \[\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1\] \[\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.\] \[x\cdot \frac{x^{2}+361}{x^{2}-361}=y\]
We now use the fact that the perimeter of $\triangle PAM$ is $152$ \[PO+OA+AM+MP=152\] \[\frac{y}{x}\cdot 19+19+x+y=152\] \[19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152\] \[(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152\] \[\frac{2x^{2}}{x-19}=152\] \[x^{2}-76x+19\cdot 76=0.\] This quadratic factors as $(x-38)^{2}=0,$ so $x=38$ , and \[\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}\] \[OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}\]

Answer:  98.
Problem: Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$

Solution: Let the smaller angle between the $x$ -axis and the line $y=mx$ be $\theta$ . Note that the centers of the two circles lie on the angle bisector of the angle between the $x$ -axis and the line $y=mx$ . Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$ .  Let $\tan{\frac{\theta}{2}}=m_1$ , for convenience. Therefore if $(x,y)$ is on the angle bisector, then $x=\frac{y}{m_1}$ . Now let the centers of the two relevant circles be $(a/m_1 , a)$ and $(b/m_1 , b)$ for some positive reals $a$ and $b$ . These two circles are tangent to the $x$ -axis, so the radii of the circles are $a$ and $b$ respectively. We know that the point $(9,6)$ is a point on both circles, so we have that
\[(9-\frac{a}{m_1})^2+(6-a)^2=a^2\]
\[(9-\frac{b}{m_1})^2+(6-b)^2=b^2\]
Expanding these and manipulating terms gives
\[\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0\]
\[\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0\]
It follows that $a$ and $b$ are the roots of the quadratic
\[\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0\]
It follows from Vieta's Formulas that the product of the roots of this quadratic is $117m_1^2$ , but we were also given that the product of the radii was 68. Therefore $68=117m_1^2$ , or $m_1^2=\frac{68}{117}$ . Note that the half-angle formula for tangents is
\[\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\]
Therefore
\[\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}\]
Solving for $\cos{\theta}$ gives that $\cos{\theta}=\frac{49}{185}$ . It then follows that $\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}$
It then follows that $m=\tan{\theta}=\frac{12\sqrt{221}}{49}$ . Therefore $a=12$ $b=221$ , and $c=49$ . The desired answer is then $12+221+49=\boxed{282}$

Answer:  282
Problem: Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$

Solution: Let the centers of $C_1$ and $C_2$ be $A$ and $B$ , respectively, and let the point $(9, 6)$ be $P$
Because both $C_1$ and $C_2$ are tangent to the x-axis, and both of them pass through $P$ , both $A$ and $B$ must be equidistant from $P$ and the x-axis. Therefore, they must both be on the parabola with $P$ as the focus and the x-axis as the directrix. Since the coordinates of $P$ is $(9, 6)$ , we see that this parabola is the graph of the function \[y=\frac{1}{12}(x-9)^2+3=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}.\]
Let $AB$ be $y=kx$ . Because $C_1$ and $C_2$ are both tangent to the x-axis, the y-coordinates of $A$ and $B$ are $r_1$ and $r_2$ , respectively, so the x-coordinates of $A$ and $B$ are $\frac{r_1}{k}$ and $\frac{r_2}{k}$ . But since $A$ and $B$ are also on the graph of the function $y=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , the x-coordinates of $A$ and $B$ are also the roots of the equation $kx=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , and by Vieta's Formulas, their product is $\frac{\frac{39}{4}}{\frac{1}{12}}=117$ . So we have $\frac{r_1}{k}\cdot \frac{r_2}{k}=117$
We are also given that $r_1r_2=68$ , so $k^2=\frac{r_1r_2}{117}=\frac{68}{117}$ , which means that $k=\sqrt{\frac{68}{117}}$ . Note that the line $AB$ is the angle bisector of the angle between the line $y=mx$ and the x-axis. Therefore, we apply the double-angle formula for tangents and get \[m=\frac{2k}{1-k^2}=\frac{2\sqrt{\frac{68}{117}}}{1-\frac{68}{117}}=\frac{12\sqrt{221}}{49}.\] Thus, the answer is $12+221+49=\boxed{282}$

Answer:  282
Problem: Find the sum of all positive two-digit integers that are divisible by each of their digits.

Solution: Let our number be $10a + b$ $a,b \neq 0$ . Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$ , or $a$ divides into $b$ and $b$ divides into $10a$ . Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$ , then $b$ would not be a digit).
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$

Answer:  630
Problem: Find the sum of all positive two-digit integers that are divisible by each of their digits.

Solution: Using casework, we can list out all of these numbers: \[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\]

Answer:  630
Problem: Find the sum of all positive two-digit integers that are divisible by each of their digits.

Solution: In this solution, we will do casework on the ones digit.
Before we start, let's make some variables. Let $a$ be the ones digit, and $b$ be the tens digit. Let $n$ equal our number. Our number can be expressed as $10b+a$ . We can easily see that $b|a$ , since $b|n$ , and $b|10b$ . Therefore, $b|(n-10b)$ . 
Now, let's start with the casework.
Case 1: $a=1$ Since $b|a$ $b=1$ . From this, we get that $n=11$ satisfies the condition.
Case 2: $a=2$ We either have $b=1$ , or $b=2$ . From this, we get that $n=12$ and $n=22$ satisfy the condition.
Case 3: $a=3$ We have $b=3$ . From this, we get that $n=33$ satisfies the condition. Note that $b=1$ was not included because $3$ does not divide $13$
Case 4: $a=4$ We either have $b=2$ or $b=4$ . From this, we get that $n=24$ and $n=44$ satisfy the condition. $b=1$ was not included for similar reasons as last time.
Case 5: $a=5$ We either have $b=1$ or $b=5$ . From this, we get that $n=15$ and $n=55$ satisfy the condition.
Continuing with this process up to $a=9$ , we get that $n$ could be $11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99$ . Summing, we get that the answer is $\boxed{630}$ . A clever way to sum would be to group the multiples of $11$ together to get $11+22+\dots+99=(45)(11)=495$ , and then add the remaining $12+24+15+36+48=135$

Answer:  630
Problem: A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$

Solution: Let $x$ be the mean of $\mathcal{S}$ . Let $a$ be the number of elements in $\mathcal{S}$ .
Then, the given tells us that $\frac{ax+1}{a+1}=x-13$ and $\frac{ax+2001}{a+1}=x+27$ . Subtracting, we have \begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*} We plug that into our very first formula, and get: \begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651}

Answer:  651