Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Let $V$ be the vector space of polynomials of degree $\leq 17,$ and let $B = \{1, x, x^2, ..., x^{17} \}$ and $C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}$ be two bases for $V$ .
Let $\vec{v} \in V$ be the polynomial given in the problem, and it is easy to see that $[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \rangle.$
Note that the transformation matrix from $C$ to $B$ can be easily found to be $P_{C \to B} = [ [\vec{c_1}]_B [\vec{c_2}]_B ... [\vec{c_3}]_B ] = \begin{bmatrix} \tbinom{0}{0} & \tbinom{1}{0} & \tbinom{2}{0} & \cdots & \tbinom{17}{0} \\ 0 & \tbinom{1}{1} & \tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & \tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
I claim that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} ,$ where the term $\dbinom{n}{k}$ is negated if $n+k$ is odd.
One can prove that the $i$ th row of $P_{C \to B}$ dotted with the $j$ th column of $P_{B \to C}$ is $\delta_{i, j}$ by using combinatorial identities, which is left as an exercise for the reader. Thus, since the two matrices multiply to form $\mathbb{I}_{18},$ we have proved that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
To find the coordinates of $\vec{v}$ under basis $C,$ we compute the product $[ \vec{v} ]_C = P_{B \to C} [\vec{v} ]_B = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \vdots \\ -1 \end{bmatrix} = \begin{bmatrix} \sum_{n=0}^{17} \tbinom{n}{0} \\ -\sum_{n=1}^{17} \tbinom{n}{1} \\  \sum_{n=2}^{17} \tbinom{n}{2} \\ \vdots \\  -\sum_{n=17}^{17} \tbinom{n}{17} \end{bmatrix} = \begin{bmatrix} \tbinom{18}{1} \\ - \tbinom{18}{2} \\ \tbinom{18}{3} \\ \vdots \\ -\tbinom{18}{18} \end{bmatrix},$ where the last equality was obtained via Hockey Stick Identity.
Thus, our answer is $a_2 =  [ [ \vec{v} ]_C ]_3 = \dbinom{18}{3} = \boxed{816}.$

Answer:  816
Problem: In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH HH , and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH , three HT , four TH , and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH , three HT , four TH , and five TT subsequences?

Solution: Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH HT THHTHT ). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However, adding HT or TH switches the last coin. switches to three times, but switches to four times; hence it follows that our string will have a structure of THTHTHTH
Now we have to count all of the different ways we can add the identities back in. There are 5 TT subsequences, which means that we have to add 5 into the strings, as long as the new s are adjacent to existing s. There are already 4 s in the sequence, and since order doesn’t matter between different tail flips this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives ${{5+3}\choose3} = 56$ combinations. We do the same with 2 s to get ${{2+3}\choose3} = 10$ combinations; thus there are $56 \cdot 10 = \boxed{560}$ possible sequences.

Answer:  560
Problem: The shortest distances between an interior diagonal of a rectangular parallelepiped $P$ , and the edges it does not meet are $2\sqrt{5}$ $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$

Solution: In the above diagram, we focus on the line that appears closest and is parallel to $BC$ . All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to $AC$ , and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between $AC$ and that corner, which is $\frac {wl}{\sqrt {w^2 + l^2}}$
So we have: \[\frac {lw}{\sqrt {l^2 + w^2}} = \frac {10}{\sqrt {5}}\] \[\frac {hw}{\sqrt {h^2 + w^2}} = \frac {30}{\sqrt {13}}\] \[\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}\]
Notice the familiar roots: $\sqrt {5}$ $\sqrt {13}$ $\sqrt {10}$ , which are $\sqrt {1^2 + 2^2}$ $\sqrt {2^2 + 3^2}$ $\sqrt {1^2 + 3^2}$ , respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)
\[\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20\] \[\frac {h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}\] \[\frac {h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}\]
We invert the above equations to get a system of linear equations in $\frac {1}{h^2}$ $\frac {1}{l^2}$ , and $\frac {1}{w^2}$ \[\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}\] \[\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}\] \[\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}\]
We see that $h = 15$ $l = 5$ $w = 10$ . Therefore $V = 5 \cdot 10 \cdot 15 = \boxed{750}$

Answer:  750
Problem: Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$

Solution: Since $x_n=\frac{n}{x_{n-1}}$ $x_n \cdot x_{n - 1} = n$ .  Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$ $x_3x_4 = 4$ $x_5x_6 = 6$ and $x_7x_8 = 8$ so \[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\] Notice that the value of $x_1$ was completely unneeded!

Answer:  384
Problem: Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$

Solution: Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
\[x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}\] \[=\left (x_1\cdot\dfrac{2}{x_1} \right )\cdot \left (\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}} \right )\cdot \left (\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}} \right )\cdot \left (\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}} \right )\] \[=(2)\cdot (4)\cdot (6)\cdot (8)=\boxed{384}\]

Answer:  384
Problem: When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle?

Solution: Let one leg of the triangle have length $a$ and let the other leg have length $b$ .  When we rotate around the leg of length $a$ , the result is a cone of height $a$ and radius $b$ , and so of volume $\frac 13 \pi ab^2 = 800\pi$ .  Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\frac13 \pi b a^2 = 1920 \pi$ .  If we divide this equation by the previous one, we get $\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}$ , so $a = \frac{12}{5}b$ .  Then $\frac{1}{3} \pi \left(\frac{12}{5}b\right)b^2 = 800\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$ .  Then by the Pythagorean Theorem , the hypotenuse has length $\sqrt{a^2 + b^2} = \boxed{026}$

Answer:  026
Problem: When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle?

Solution: Let $a$ $b$ be the $2$ legs, we have the $2$ equations \[\frac{a^2b\pi}{3}=800\pi,\frac{ab^2\pi}{3}=1920\pi\] Thus $a^2b=2400, ab^2=5760$ . Multiplying gets \begin{align*} (a^2b)(ab^2)&=2400\cdot5760\\ (ab)^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5)\\ ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \end{align*} Adding gets \begin{align*} a^2b+ab^2=ab(a+b)&=2400+5760\\ 240(a+b)&=240\cdot(10+24)\\ a+b&=34 \end{align*} Let $h$ be the hypotenuse then \begin{align*} h&=\sqrt{a^2+b^2}\\ &=\sqrt{(a+b)^2-2ab}\\ &=\sqrt{34^2-2\cdot240}\\ &=\sqrt{676}\\ &=\boxed{26}

Answer:  26
Problem: When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle?

Solution: Let $a$ and $b$ be the two legs of the equation. We can find $\frac{a}{b}$ by doing $\frac{1920\pi}{800\pi}$ . This simplified is $\frac{12}{5}$ . We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively.
Since the volume of the first cone is $800\pi$ , we use the formula for the volume of a cone and get $100\pi x^3=800 \pi$ . Solving for $x$ , we get $x=2$
Plugging in the side lengths to the Pythagorean Theorem, we get an answer of $\boxed{026}$

Answer:  026
Problem: Find $c$ if $a$ $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$

Solution: Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$ .  Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$ .  Since $a, b$ are integers , this means $b$ is a divisor of 107, which is a prime number .  Thus either $b = 1$ or $b = 107$ .  If $b = 107$ $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$ , but $107^2 + 1$ is not divisible by 3, a contradiction.  Thus we must have $b = 1$ $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive).  Thus $c = 6^3 - 3\cdot 6 = \boxed{198}$

Answer:  198
Problem: A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png

Solution: The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram .  Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = \frac{1}{n}$ .  By the Pythagorean Theorem , the longer base of the parallelogram has length $l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}$ , so the parallelogram has height $h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}$ .  But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$ .  Solving this quadratic equation gives $n = \boxed{32}$

Answer:  32
Problem: A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png

Solution: Aime.png
Surrounding the square with area $\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles.
We can thus use proportions to solve this problem. \begin{eqnarray*} \frac{GF}{BE}=\frac{CG}{CB}\implies \frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies BE=\frac{n\sqrt{1985}}{1985} \end{eqnarray*} Also, \begin{eqnarray*} \frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies EC=\frac{\sqrt{1985}}{1985}(n-1) \end{eqnarray*} Thus, \begin{eqnarray*} 2(BE)(EC)+\frac{1}{1985}=1\\ 2n^{2}-2n+1=1985\\ n(n-1)=992 \end{eqnarray*} Simple factorization and guess and check gives us $\boxed{32}$

Answer:  32
Problem: A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png

Solution: AIME 1985 Problem 4 Solution 3 Diagram.png
Line Segment $DE = \frac{1}{n}$ , so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ similarity. Thus, $\frac{LC}{HE} = \frac{EC}{DE} = n-1$ , so $LC = \frac{n-1}{\sqrt{1985}}$ . Notice that $\triangle CEL$ is also similar to $\triangle CDF$ by $AA$ similarity. Thus, $\frac{FC}{EC} = \frac{DC}{LC}$ , and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$ . Solving this quadratic equation yields $n =\boxed{32}$

Answer:  32
Problem: sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?

Solution: The problem gives us a sequence defined by a recursion , so let's calculate a few values to get a feel for how it acts.  We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ .  Then $a_3 = b - a$ $a_4 = (b - a) - b = -a$ $a_5 = -a - (b - a) = -b$ $a_6 = -b - (-a) = a - b$ $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$ .  Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat.  Thus, in particular $a_{j + 6} = a_j$ for all $j$ , and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$
Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \cdot 248$ is the largest multiple of 6 less than 1492, so \[\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}\] \[=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,\] where we can make this last step because $\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero
Similarly, since $1980 = 6 \cdot 330$ $\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$
Finally, $\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$
Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\cdot 493 = \boxed{986}$ .
Minor edit by: PlainOldNumberTheory

Answer:  986
Problem: As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png

Solution: Let the interior point be $P$ , let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively.  Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$ .  Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$ , so the ratio of their areas is the same as the ratio of their bases.  Similarly, $\triangle ACF$ and $\triangle BCF$ share the same altitude from $C$ , so the ratio of their areas is the same as the ratio of their bases.  Moreover, the two pairs of bases are actually the same, and thus in the same ratio.  As a result, we have: $\frac{40}{30} = \frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$
Applying identical reasoning to the triangles with bases $\overline{CD}$ and $\overline{BD}$ , we get $\frac{y}{35} = \frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$ .  Substituting from this equation into the previous one gives $x = 56$ , from which we get $y = 70$ and so the area of $\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = \Rightarrow \boxed{315}$

Answer:  315
Problem: As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png

Solution: This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\times$ side) = (Other weight) $\times$ (The other side), the problem yields the answer $\boxed{315}$

Answer:  315
Problem: As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png

Solution: Let the interior point be $P$ and let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.$ However, we can deduce $\frac{FB}{AF}=\frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=\frac{84CE}{EA}$ and $y=\frac{35DC}{BD}.$ Plug and chug and you get $xy=84\cdot 35\cdot \frac{3}{4}=2205.$ Then notice by the same height reasoning, $\frac{84}{x}=\frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $\frac{84}{105}=\frac{4}{5}=\frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=\boxed{315}.$

Answer:  315
Problem: As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png

Solution: Let the interior point be $P$ and let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Then the cevians $AD,BF,CE$ are concurrent, so we can use Ceva's Theorem, letting $\frac{BD}{DC}=\frac{a}{b}$ and $\frac{CF}{FA}=\frac{c}{d}$ . Notice that $\frac{AE}{EB}=\frac{[\Delta APE]}{[\Delta EPB]}=\frac43.$ \[\frac{4}{3}\cdot \frac{a}{b}\cdot \frac{c}{d}=1\implies \frac{d}{c}=\frac{a}{b}\cdot \frac{4}{3}.\]
We know that $[\Delta CPD]=35\cdot \frac ba$ and $[\Delta APF]=84\cdot \frac dc,$ so \[[\Delta ABC] = 84+84\cdot \frac dc + 35\cdot \frac ba + 40+30+35 = \left(1+\frac ba\right)(40+30+35).\] We will now solve for $\frac ba$
\[84+84\cdot \frac ab\cdot \frac 43 + 35\cdot \frac ba = \frac ba\cdot 105.\] \[12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15\] \[10\left(\frac ba\right)^2-12\cdot \frac ba-16=0\] Factoring this gives $\left(\frac ba-2\right)\left(10\cdot \frac ba - 8\right)=0,$ so the area of $\triangle ABC$ is \[\left(1+\frac ba\right)(40+30+35)=3\cdot 105=\boxed{315.}\]

Answer:  315.
Problem: Assume that $a$ $b$ $c$ , and $d$ are positive integers such that $a^5 = b^4$ $c^3 = d^2$ , and $c - a = 19$ . Determine $d - b$

Solution: It follows from the givens that $a$ is a perfect fourth power $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube .  Thus, there exist integers $s$ and $t$ such that $a = t^4$ $b = t^5$ $c = s^2$ and $d = s^3$ .  So $s^2 - t^4 = 19$ .  We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$ .  19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$ .  Then $s = 10, t = 3$ and so $d = s^3 = 1000$ $b = t^5 = 243$ and $d-b=\boxed{757}$

Answer:  757
Problem: The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ $a_2 = 2.61$ $a_3 = 2.65$ $a_4 = 2.71$ $a_5 = 2.79$ $a_6 = 2.82$ $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ $1\le i \le 7$ , so that the sum of the $A_i$ 's is also 19 and so that $M$ , the maximum of the "errors" $\| A_i-a_i\|$ , the maximum absolute value of the difference, is as small as possible. For this minimum $M$ , what is $100M$

Solution: If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1.  However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1.  So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3.  Then there must be five 3s and two 2s.  It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $\boxed{061}$

Answer:  061
Problem: In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution: All chords of a given length in a given circle subtend the same arc and therefore the same central angle.  Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle
This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$ .  The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$ , so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$
Now, consider the triangle formed by two radii and the chord of length 2.  This isosceles triangle has vertex angle $\alpha$ , so by the Law of Cosines
\[2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}\] and the answer is $17 + 32 = \boxed{049}$

Answer:  049