Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Let $A(n)$ be the probability the bug lands on vertex $A$ after crawling $n$ meters, $B(n)$ be the probability the bug lands on vertex $B$ after crawling $n$ meters, and etc.
Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the bug land on each vertex after $n$ meters is $\frac13$ the sum of the probability the bug land on other vertices after crawling $n-1$ meters. So, we have \begin{align*} A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. \end{align*} It follows that $A(n) = B(n-1) = C(n-1) = D(n-1).$
We construct the following table: \[\begin{array}{c|cccc}  &  &  &  & \\ [-2ex] n & A(n) & B(n) & C(n) & D(n) \\ [1ex] \hline  &  &  &  & \\ [-1ex] 1 & 0 & \frac13 & \frac13 & \frac13 \\  &  &  &  & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\  &  &  &  & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\  &  &  &  & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\  &  &  &  & \\ 5 & \frac{20}{81} & \frac{61}{243} & \frac{61}{243} & \frac{61}{243} \\  &  &  &  & \\ 6 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} \\  &  &  &  & \\ 7 & \frac{182}{729} & \frac{547}{2187} & \frac{547}{2187} & \frac{547}{2187} \\ [1ex] \end{array}\] Therefore, the answer is $n = \boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Let $\omega = e^{i\pi / 2}$ . We have that if $G(x) = (x+x^2+x^3)^7$ , then \[\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.\] From here, the desired probability is $\frac{546}{2187} = \frac{182}{729}$ . Therefore, the answer is $n=\boxed{182}$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: We can factor $(x+x^2+x^3)^7$ as $x^7(1+x+x^2)^7.$ The $x^{4n}$ coefficients of $(x+x^2+x^3)^7$ will be the same as the $x^{4n+1}$ coefficients of $(1+x+x^2)^7.$ The possible values for $4n+1$ would then be $1,$ $5,$ $9,$ and $13.$ Because $1+13=5+9=14,$ the coefficients of $x^1$ and $x^{13}$ are equal and so are the coefficients of $x^5$ and $x^9.$ To make an $x$ term, we need $6$ $1$ " terms and one " $x$ " term multiplied together. There are $7$ ways to arrange these numbers. The coefficient of the $x^5$ term will be the sum of the number of the possible arrangements for these numbers: \begin{align*} 0000122&=105\text{ arrangements}, \\ 0001112&=140\text{ arrangements}, \\ 0011111&=21\text{ arrangements}. \end{align*} Thus, the coefficient of the $x^5$ term is $105+140+21=266.$ From here, the desired probability is $\frac{2(7+266)}{2187}=\frac{546}{2187}=\frac{182}{729}.$ Thus, our answer is $\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: We can find the number of different times the bug reaches vertex $A$ before the $7$ th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A.$
Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obviously $f(1) = 0.$ In general for $f(x),$ the bug has three initial edges to pick from. From there, since the bug cannot return to $A$ by definition, the bug has exactly two choices. This continues from the $2$ nd move up to the $(x-1)$ th move. The last move must be a return to $A,$ so this move is determined. So $f(x) = 2^{x-2}3.$
Now we need to find the number of cycles by which the bug can reach $A$ at the end. Since $f(1) = 0,$ we know that $f(6)$ cannot be used, as on the $7$ th move the bug cannot move from $A$ to $A.$ So we need to find the number of partitions of $7$ using only $2,3,4,5,$ and $7.$ These are $f(2)f(2)f(3),f(2)f(5),f(3)f(4),$ and $f(7).$ We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition: \[{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7) = 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53) = 546.\] Finally, this is a probability question, so we divide by $3^7,$ or $\frac{546}{3^7} = \frac{182}{3^6}.$ The answer is $n=\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Note that this problem is basically equivalent to the following:
How many distinct sequences of $8$ integers $a_1, a_2, a_3, \ldots, a_8$ are there such that $a_1 = a_8 = 1,$ $a_i \in \{1, 2, 3, 4\}$ for all $2 \leq i\leq 8,$ and $a_i \neq a_{i+1}$ for all $1 \leq i \leq 7$
Now consider the $8$ integers modulo $4.$ Let $b_1, b_2, b_3, \ldots, b_7$ be a new sequence of integers such that $b_i = a_{i+1} - a_i \mod 4$ for all $1 \leq i \leq 7.$
Note that the condition is equivalent to that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7,$ and since $a_1 \mod 4 = a_8 \mod 4,$ $b_1 + b_2 + \cdots + b_7$ must be a multiple of $4.$
Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers $(b_1, b_2, \ldots, b_7)$ such that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7$ and $b_1 + b_2 + \cdots + b_7$ is congruent to $0 \mod 4.$
We will now proceed with casework on the number of $2$ s in the septuple.
One $2$ : There are $\dbinom{7}{1} = 7$ ways to arrange the $2$ , and $\dbinom{6}{0} + \dbinom{6}{2} + \dbinom{6}{4} + \dbinom{6}{6} = 2^5$ valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the $1$ s and $3$ s.
Three $2$ s: There are $\dbinom{7}{3} = 35$ ways to arrange the $2$ s, and $\dbinom{4}{1} + \dbinom{4}{3} = 2^3$ valid ways to arrange the $1$ s and $3$ s.
Five $2$ s: There are $\dbinom{7}{5} = 21$ ways to arrange the $2$ s, and $\dbinom{2}{0} + \dbinom{2}{2} = 2$ valid ways to arrange the $1$ s and $3$ s.
Adding up our cases, we obtain $7 \cdot 32 + 35 \cdot 8 + 21 \cdot 2 = 546$ valid sequences.
Dividing by the total number of paths without restrictions, $3^7 = 2187,$ our desired probability is $\frac{546}{2187} = \frac{182}{729},$ giving an answer of $\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: We instead find the probability that the bug is NOT at vertex $A$ after crawling $n$ meters (equivalent to moving $n$ times). Call $A_n$ the probability that the bug IS at vertex $A$ after $n$ moves; call $O_n$ the probability that the bug is on some other vertex. We have the following recurrence relations. \[A_n = \frac{1}{3}O_{n-1}\] \[O_n = A_{n-1} + \frac{2}{3}O_{n-1}\] However, we can calculate $A_{n-1}$ in terms of $O_n$ ; take $n = k-1$ , and we have \[A_{k-1} = \frac{1}{3}O_{k-2}\] . Substituting this into our equation for $O$ , we have that \[O_n = \frac{1}{3}O_{n-2} + \frac{2}{3}O_{n-1}\] . We also have the conditions that $O_0 = 0$ (the bug will not be at vertex other than $A$ on the "0th" move) and $O_1 = 1$ (the bug will be at a vertex other than $A$ after the $1st$ move). Iteratively calculating $O_7$ , we find that it is equal to $\frac{547}{729}$ (I will not do this calculation here; you can do it manually if you wish to check). However, this is the probability that the ant is NOT at vertex $A$ after $7$ moves; then its complement, $\frac{182}{729}$ is the probability that the ant IS at vertex $A$ after $7$ moves, so our answer is $\boxed{182}$

Answer:  182
Problem: The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers

Solution: If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$
Thus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note that since $2n+1$ is odd for integral $n$ , we can multiply the left integer, $100+n^2$ , by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.
So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$ . It simplified the way we wanted it to!
Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$ .  Thus $d_n$ must divide $\boxed{401}$ for every single $n$ .  This means the largest possible value for $d_n$ is $401$ , and we see that it can be achieved when $n = 200$

Answer:  401
Problem: The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers

Solution: We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm
$a_{n+1}-a_n = 2n+1$
Now, the question is to find the GCD of $2n+1$ and $100+n^2$ .
We subtract $2n+1$ $100$ times from $100+n^2$ \[(100+n^2)-100(2n+1)\] \[=n^2+100-200n-100\] This leaves us with \[n^2-200n.\] Factoring, we get \[n(n-200)\] Because $n$ and $2n+1$ will be coprime, the only thing stopping the GCD from being $1$ is $n-200.$ We want this to equal 0, because that will maximize our GCD. 
Solving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = \boxed{401}$ is our GCD.

Answer:  401
Problem: In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Solution: Let us suppose for convenience that there were $n + 10$ players overall.  Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned.  By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10.  Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other.  Then they also earned 45 points playing against the stronger $n$ players.  Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$ .  However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned.  So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$ .  Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$ , while the bottom 10 got 90 points total, for an average of 9.  Thus we must have $n > 10$ , so $n = 15$ and the answer is $15 + 10 = \boxed{25}$

Answer:  25
Problem: In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Solution: Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-145$ . Expanding and simplifying gives us $n^2-41n+400=0$ . Thus, $n=16$ or $n=25$ . Note however that if $n=16$ , then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$

Answer:  25
Problem: In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Solution: Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$ . Say the number of people is $n$ . Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of points they will earn during the whole game. This implies that this group of 10 people must accumulate half their total combined points after they (the 10 people) all play each other, meaning they must earn the other half of their points by playing the $n-10$ stronger players. The problem also tells us that the $n-10$ people who aren't part of the losers group will earn half of their points by playing the $10$ losers.  Since the $n-10$ group and $10$ losers will earn half their points by playing each other, the sum of the number of points that they gain playing each other must then be half of the total amount of points earned by everyone in the game. Therefore, $\frac{p(p-1)}{4} = 10(p-10)$ . This equation is the same as above, and by the same logic, the answer is $n=\boxed{25}$

Answer:  25
Problem: Three 12 cm $\times$ 12 cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon , as shown in the second figure, so as to fold into a polyhedron . What is the volume (in $\mathrm{cm}^3$ ) of this polyhedron?
AIME 1985 Problem 15.png

Solution: Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube , so the volume is $\frac12 \cdot 12^3 = 864$ , so our answer is $\boxed{864}$

Answer:  864
Problem: Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

Solution: One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$ , then use that to calculate $a_2$ and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each $n \geq 1$ , we have $a_{2n - 1} = a_{2n} - 1$ .  We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$ .  The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$ , so our desired value is $\frac{137 + 49}{2} = \boxed{093}$

Answer:  093
Problem: Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

Solution: We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ , which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$ .
We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: \[a_1+a_2+a_3+\ldots+a_{49}\] and \[a_{50}+a_{51}+a_{52}+\ldots+a_{98}\] Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$ , we get $a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}$ . Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$ . We want to find the value of $x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225$ . Since $x=\frac{137-2401}{2}$ , we find $x=-1132$ $-1132+1225=\boxed{93}$

Answer:  93
Problem: Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

Solution: Since we are dealing with an arithmetic sequence, \[a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}\] We can also figure out that \[a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137\] \[a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137\] Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{093}$

Answer:  093
Problem: The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$

Solution: Any multiple of 15 is a multiple of 5 and a multiple of 3.
Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3.  If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ .  For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880.  Thus the answer is $\frac{8880}{15} = \boxed{592}$

Answer:  592
Problem: point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Solution: By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$ . Thus, the corresponding side on the large triangle is $12x$ , and the area of the triangle is $12^2 = \boxed{144}$

Answer:  144
Problem: point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Solution: Alternatively, since the triangles are similar by $AA$ , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\dfrac{base}{height} = \dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$ . Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}$

Answer:  144
Problem: point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Solution: The base of $\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\sqrt{\frac{4}{9}}=\frac{2}{3}$ and $\sqrt{\frac{4}{49}}=\frac{2}{7}$ so $y=\frac{3x}{2}$ and $z=\frac{7x}{2}$ $x+\frac{7x}{2}+\frac{3x}{2}=6x$ , so $\triangle{ABC}$ has a base that is $6$ times $t_1$ $[\triangle{ABC}]=36[t_1]=36 \cdot 4=\boxed{144}$

Answer:  144