Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: For all $n\geq1,$ we have \begin{align*} f(n) &= 3f(n-1)-1 \\ &= 3\left(3f(n-2)-1\right)-1 \\ &= 3^2f(n-2)-3-1 \\ &= 3^2\left(3f(n-3)-1\right)-3-1 \\ &= 3^3f(n-3)-3^2-3-1 \\ & \ \vdots \\ &= 3^nf(0)-3^{n-1}-3^{n-2}-3^{n-3}-\cdots-1 \\ &= 3^n-\left(3^{n-1}+3^{n-2}+3^{n-3}+\cdots+1\right) \\ &= 3^n-\frac{3^n-1}{2} \\ &= \frac{3^n+1}{2} \\ &= \frac{3^n-1}{2}+1, \end{align*} which resembles the result in Solutions 1 and 2. The answer is $f(6)=\boxed{365}.$

Answer:  365
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: We can perform casework based on the number of overlapping elements. If no elements overlap, there is $\binom60=1$ way to choose the overlapping elements, and $2^{6-0}$ ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by $2$ because the order of the subsets does not matter. Proceeding similarly for the other cases, our sum is \[\dbinom{6}0 \cdot 2^5+\dbinom{6}1 \cdot 2^4+\cdots +\dbinom{6}4 \cdot 2+\dbinom{6}5 \cdot 1+\dbinom{6}6 \cdot 1.\] (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of $\boxed{365}.$

Answer:  365
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is $x\rightarrow y,$ which implies that we pick $x$ numbers for the first set which then the second set can have $y$ numbers.
Clearly: \begin{align*} 0&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow2\mid3\mid4\mid5\mid6 \\ 5&\rightarrow1\mid2\mid3\mid4\mid5\mid6 \\ 6&\rightarrow0\mid1\mid2\mid3\mid4\mid5\mid6 \end{align*} However notice that many of the cases are double counted as direction does not matter, e.g. $2\rightarrow4$ is the same as $4\rightarrow2.$ Get rid of those cases so we are just left with: \begin{align*} 0&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow4\mid5\mid6 \\ 5&\rightarrow5\mid6 \\ 6&\rightarrow6 \end{align*} Now we start computing, $0\rightarrow6$ is just $1$ case, $1\rightarrow5\mid6$ is just $\binom{6}{1}\cdot2$ cases, $2\rightarrow4\mid5\mid6$ is just $\binom{6}{2}\cdot2^2$ cases, and $3\rightarrow3\mid4\mid5\mid6$ is just $\binom{6}{3}\cdot2^3$ cases (If you have trouble understanding this, write down the six letters and then try to understand what $x\rightarrow y$ really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that.
One quick thing though. We have a double counted case with the $3\rightarrow3,$ as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are $\frac{\binom{6}{3}}{2} = 10$ cases of this.
For computing $4\rightarrow4\mid5\mid6,$ we use the same process as before. We have $\binom{6}{4}\cdot(3+4+1)$ (Note, the $3$ comes from $\frac{\binom{4}{2}}{2}$ ), and for $5\rightarrow5\mid6$ we have $\binom{6}{5}\cdot\left(\frac{5}{2}+1\right),$ and then for $6\rightarrow6$ we just have $\binom{6}{6}$ (there is no double counted case since ABCDEF, ABCDEF is only counted once).
Summing case by case, we have $1+12+60+150+120+21+1 = \boxed{365}.$

Answer:  365
Problem: Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png

Solution: The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$
For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$ . The smallest $n$ for this case is $118$
In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $\boxed{118}$ is our answer.

Answer:  118
Problem: Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png

Solution: Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$ . If we assume the final answer be $n$ , then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$ . As they have different parities, the even one must be divisible by $32$ . As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$ , one of them is divisible by $5$ , which indicates it's divisible by $125$
Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ $1994+n\equiv 0\pmod{4000}$ $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$ . Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$ .(by ZJY)

Answer:  118
Problem: Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$

Solution: The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$
In summary, the solution to the problem is $100P+10T+V=\boxed{250}$

Answer:  250
Problem: Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$

Solution: As seen above, $E=V+30$ . Every vertex $V$ , there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480$ . Also, at every vertex $V$ , there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E$ , and subtracting the other equation involving the vertices from this gives $2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90$ $\implies V(5-P)=90$ .
Since $V|480$ from the first vertex-related observation and $P>0\implies5-P<5$ , and it quickly follows that $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$

Answer:  250
Problem: Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$

Solution: Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron.  Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \frac{3t+5p}{2}$ and $V = \frac{3t}{T}=\frac{5p}{P}.$
Now, we do casework on what $P$ is.
Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral.  Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions.  Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \boxed{250}$

Answer:  250
Problem: Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$

Solution: We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \frac{4V}{2} = 2V$ . Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \boxed{250}$

Answer:  250
Problem: Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$

Solution: In order to begin this problem, we need to calculate the probability that Alfred will win on the first round.
Because he goes first, Alfred has a $\frac{1}{2}$ chance of winning (getting heads) on his first flip.
Then, Bonnie, who goes second, has a $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , chance of winning on her first coin toss.
Therefore Alfred’s chance of winning on his second flip is $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
From this, we can see that Alfred’s (who goes first) chance of winning the first round is: $\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3}$
Bonnie’s (who goes second) chance of winning the first round is then $1 - \frac{2}{3} = \frac{1}{3}$
This means that the person who goes first has a $\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\frac{1}{3}$ chance of winning.
Now, through casework, we can calculate Alfred’s chance of winning the second round.
Case 1: Alfred wins twice; $\frac{2}{3} \times \frac{1}{3}$ (Bonnie goes first this round) $=\frac{2}{9}$
Case 2: Alfred loses the first round, but wins the second; $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$
Adding up the cases, we get $\frac{2}{9} + \frac{2}{9} = \frac{4}{9}$
Alfred, therefore, has a $\frac{4}{9}$ of winnning the second round, and Bonnie has a $1-\frac{4}{9} = \frac{5}{9}$ chance of winning this round.
From here, it is not difficult to see that the probabilities alternate in a pattern.
Make $A$ the probability that Alfred wins a round.
The chances of Alfred and Bonnie, respectively, winning the first round are $A$ and $A - \frac{1}{3}$ , which can be written as $2A - \frac{1}{3} = 1$
The second round’s for chances are $A$ and $A + \frac{1}{9}$ , which can also be written as $2A + \frac{1}{9}$
From this, we can conclude that for the $n$ th even round, the probability that Alfred ( $A$ ) wins can be calculated through the equation $2A + \frac{1}{3^n} = 1$
Solving the equation for $n = 6$ , we get $A = \frac{364}{729}$
$364 + 729 = 1093$
So our answer is $\boxed{093}$

Answer:  093
Problem: Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$

Solution: Rather than categorizing games as wins or losses, we can categorize them as starters (S), where Alfred starts, and non-starters (NS), where Bonnie starts. Game 1 is a starter, and since Alfred must win Game 6, Game 7 is a non-starter.
As shown in Solution 1, if a player starts a certain game, the probability $P(NS)$ that they will not start the next game (i.e. win the current game) is $\frac{2}{3}$ , and the probability $P(S)$ that they will start the next game (i.e. lose the current game) is $\frac{1}{3}$ . Similarly, if a player does not start a certain game, $P(NS) = \frac{1}{3}$ and $P(S) = \frac{2}{3}$ . We conclude that the probability of switching from S to NS or vice versa is always $\frac{2}{3}$ , and the probability of staying the same is always $\frac{1}{3}$
Listing out all the games from Game 1 to Game 7, we get: S, ?, ?, ?, ?, ?, NS. Between the 7 games, there are 6 opportunities to either switch or stay the same. We need to eventually switch from S to NS, so there must be an odd number of switches. Furthermore, this number is less than 6, so it must be 1, 3, or 5.
1 switch: There are ${6 \choose 1} = 6$ ways to place the switch, so $P = 6\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right) = \frac{12}{729}$
3 switches: There are ${6 \choose 3} = 20$ ways to place the switches, so $P = 20\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^3 = \frac{160}{729}$
5 switches: There are ${6 \choose 5} = 6$ ways to place the switch, so $P = 6\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^5 = \frac{192}{729}$
Add up all these probabilities to get $\frac{12+160+192}{729} = \frac{364}{729}$ $364+729=1093$ , so the answer is $\boxed{093}$

Answer:  093
Problem: The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ .  The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's.  Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ .  Given that $P_7 = (14,92)\,$ , what is $k + m\,$

Solution: If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})\] Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get \[\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\ \hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8  \end{array}\]
So the answer is $\boxed{344}$

Answer:  344
Problem: The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ .  The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's.  Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ .  Given that $P_7 = (14,92)\,$ , what is $k + m\,$

Solution: Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$
Note that $P_7 = \cfrac{\cfrac{\cfrac{P_1 + L_1}2 + L_2}2 + \cdots}{2\ldots} = \frac {(k,m)}{64} + \frac {L_1}{64} + \frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 + \frac {L_5}4 + \frac {L_6}2 = (14,92)$
Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$ , we can just ignore it!):
for $\frac {L_6}2,\frac {L_5}4$ , since all addends are nonnegative, a non- $(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$ , respectively, and we can ignore them,
for $\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities,
and for $\frac {L_1}{64}$ , all three work.
Also, to be in the triangle, $0\le k\le560$ and $0\le m\le420$
Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$ , we look at that first.
If $L_1 = (0,0)$ , then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$
so it can only be that $L_1 = (560,0)$ , and $k + 560 = 2^6\cdot14$
$\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336$
Now, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ $L_2$ would influence the least, so we test that),
then $\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$
which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$ , so $L_2,L_3,L_4 = (0,420)$
and now $\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92$
$\implies P_1$
$= 64\cdot92 - 420(2 + 4 + 8)$
$= 64\cdot92 - 420\cdot14= 64(100 - 8) - 14^2\cdot30$
$= 6400 - 512 - (200 - 4)\cdot30$
$= 6400 - 512 - 6000 + 120$
$= - 112 + 120$
$= 8$
and finally, $k + m = 336 + 8 = \boxed{344}$

Answer:  344
Problem: Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \[-\frac{x}{y}=-\frac{100}{t}\] or $xt=100y.$ Now substitute \[y= \frac{xt}{100}\] into $(2)$ and get \[x=\frac{5000}{\sqrt{100^2+t^2}}.\] Now substitute this and \[y=\frac{xt}{100}\] into $(1)$ and solve for $t$ to get \[t=\frac{160}{3}.\] Finally, the sum of the numerator and denominator is $160+3=\boxed{163}.$

Answer:  163
Problem: Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$ . Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively.
From the problem statement, $AB=3t$ , and $CD=t$ . Since $\Delta ABX \sim \Delta CDX$ $CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100$
Since $PC=100$ $PX=200$ . So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$
Since circle $O$ is tangent to $BX$ and $AX$ $OX$ is the angle bisector of $\angle BXA$
Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}$
Therefore, $t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}$ , and the answer is $\boxed{163}$

Answer:  163