Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$

Solution: Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles , so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$ , so $EF = GC = \sqrt{141}$ . Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$ , and $p + q = \boxed{150}$

Answer:  150
Problem: The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution: Let $y = 2^{111x}$ .  Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ .  Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$ .  Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ .  Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$ .  Thus the answer is $111 + 2 = \boxed{113}$

Answer:  113
Problem: Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge . The 27 cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct primes and $a,b,$ and $c$ are positive integers , find $a+b+c+p+q+r.$

Solution: 2005 I AIME-9.png
We can consider the orientation of each of the individual cubes independently.
The unit cube at the center of our large cube has no exterior faces, so all of its orientations work.
For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face.  This happens in $\frac{4}{6} = \frac{2}{3}$ of all orientations, so from these cubes we gain a factor of $\left(\frac{2}{3}\right)^6$
The twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge.  Thus, we need to know the number of such pairs that are both painted orange.  We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces.  Thus, the probability that two orange faces show for one of these cubes is $\frac{5}{12}$ , so from all of these cubes we gain a factor of $\left(\frac{5}{12}\right)^{12} = \frac{5^{12}}{2^{24}3^{12}}$
Finally, we need to orient the eight corner cubes.  Each such cube has 3 faces showing, and these three faces share a common vertex.  Thus, we need to know the number of vertices for which all three adjacent faces are painted orange.  There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of $\frac{2}{8} = \frac{1}{4}$ and all the corner cubes together contribute a probability of $\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}$
Since these probabilities are independent, the overall probability is just their product, $\frac{2^6}{3^6} \cdot \frac{5^{12}}{2^{24}3^{12}} \cdot \frac{1}{2^{16}} = \frac{5^{12}}{2^{34}\cdot 3^{18}}$ and so the answer is $2 + 3 + 5 + 12 + 34 + 18 = \boxed{074}$

Answer:  074
Problem: Triangle $ABC$ lies in the cartesian plane and has an area of $70$ . The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$

Solution: The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$ . The equation of the median can be found by $-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}$ , so $q = -5p + 107$
Use determinants to find that the area of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\  q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing absolute value ; we will assume that the other solution for the triangle will give a smaller value of $p+q$ , which is provable by following these steps over again) (alternatively, we could use the Shoelace Theorem ). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}$ $\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$ . Thus, $q = \frac{1}{11}p + \frac{337}{11}$
Setting this equation equal to the equation of the median, we get that $\frac{1}{11}p + \frac{337}{11} = -5p + 107$ , so $\frac{56}{11}p = \frac{107 \cdot 11 - 337}{11}$ . Solving produces that $p = 15$ Substituting backwards yields that $q = 32$ ; the solution is $p + q = \boxed{047}$

Answer:  047
Problem: semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution: We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$ . Then the diameter is $32 - \sqrt{512}$ giving us $32 + 512 = \boxed{544}$

Answer:  544
Problem: semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution: We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})$ Simplifying gives $d-16\sqrt{2}+d\sqrt{2}=0$ . Solving gives $d=32-16\sqrt{2}=32-\sqrt{512}$ so the answer is $32+512= \boxed{544}$

Answer:  544
Problem: semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution: It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul

Answer:  544
Problem: For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$

Solution: Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$ , and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$ . Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$ $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$ , the $70$ accounting for the difference between $2005$ and $44^2 = 1936$ , inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$ . Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$

Answer:  025
Problem: For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$

Solution: Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$ , where the $-19$ accounts for those numbers between $2005$ and $2024$
Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$
Then, $|a - b| = |2(2^2 + 4^2 + \ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + \ldots 43^2) + 1^2 - 45^2 + 19|$ .
We can apply the formula $1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ .  From this formula, it follows that $2^2 + 4^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3}$ and so that
$|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + 20\right| = |-25| =\boxed{025}$

Answer:  025
Problem: For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$

Solution: Let $\Delta n$ denote the sum $1+2+3+ \dots +n-1+n$ . We can easily see from the fact "It is well-known that $\tau(n)$ is odd if and only if $n$ is a perfect square.", that
$a = (2^2-1^2) + (4^2-3^2) \dots (44^2 - 43^2) = (2+1)(2-1)+(4+3)(4-3) \dots (44+43)(44-43) = 1+2+3...44 = \Delta 44$
$b = 3^2-2^2+5^2-4^2...2006-44^2 = (3+2)(3-2) \dots (43+42)(43-42) + 1 + (2005 - 44^2) = \Delta 43 + 69$
$a-b = \Delta 44-\Delta 43-69 = 44-69 = -25$ . They ask for $|a-b|$ , so our answer is $|-25| = \boxed{025}$

Answer:  025
Problem: A particle moves in the Cartesian plane according to the following rules:
How many different paths can the particle take from $(0,0)$ to $(5,5)$

Solution: The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$ ; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:
Together, these add up to $2 + 18 + 42 + 20 + 1 = \boxed{083}$

Answer:  083
Problem: A particle moves in the Cartesian plane according to the following rules:
How many different paths can the particle take from $(0,0)$ to $(5,5)$

Solution: Another possibility is to use block-walking and recursion : for each vertex, the number of ways to reach it is $a + b + c$ , where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left).
Assign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$ . Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$ . Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$ . This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$ . (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus: \begin{align*} a_{i,j} &= s(i-1,j) - c_{i-1,j}\\ b_{i,j} &= s(i-1,j-1) \\ c_{i,j} &= s(i,j-1) - a_{i,j-1}. \end{align*} It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$ . Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \leq i, j \leq 5$ [asy] defaultpen(fontsize(8)+0.8+heavyblue); size(250);  for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); } label("$(0,0,0)$", (0,0), N); for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$", (1,i), N);} real[] val={1,2,4,7}; for(int i = 2; i<6; ++i) { label("$(1,{"+string(i-1)+"}, {"+string(val[i-2])+"})$", (2,i), N); label("$({"+string(val[i-2])+"},{"+string(i-1)+"}, 1)$", (i,2), N);}  label("$(3,3,3)$", (3,3), N); label("$(9,9,9)$", (4,4), N); label("$(28,27,28)$", (5,5), N); label("$(4,5,6)$", (3,4), N); label("$(6,5,4)$", (4,3), N);  label("$(5,8,11)$", (3,5), N); label("$(11,8,5)$", (5,3), N); label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N); [/asy] Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$ , so $s(i,j)=28+27+28 = \boxed{083}$

Answer:  083
Problem: Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$

Solution: Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $\frac{44}{\sqrt{10}}$ , the area is $K=\frac{1936}{10}$ , and the answer to the problem is $\boxed{936}$

Answer:  936
Problem: Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution: Let $E$ $F$ and $G$ be the points of tangency of the incircle with $BC$ $AC$ and $AB$ , respectively.  Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ .  Let the length of the median be $3m$ .  Then by two applications of the Power of a Point Theorem $DE^2 = 2m \cdot m = AF^2$ , so $DE = AF$ .  Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$ .  Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$
Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline{AD}$ , we have
\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.\]
Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$ , so we combine these two results to solve for $c$ and we get
\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]
Thus $c = 2$ or $= 10$ .  We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$ , so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$

Answer:  038
Problem: Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution: WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$ . We use power of a point to get that $AG = DE = \sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\sqrt{2} m$
Since now we have $AC = 10$ $BC = 20, AB = 10+2\sqrt{2} m$ in triangle $\triangle ABC$ and cevian $AD = 3m$ .  Now, we can apply Stewart's Theorem
\[2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000\] \[1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}\] \[100 m^2 = 400\sqrt{2}m\]
$m = 4\sqrt{2}$ or $m = 0$ if $m = 0$ , we get a degenerate triangle, so $m = 4\sqrt{2}$ , and thus $AB = 26$ . You can now use Heron's Formula to finish. The answer is $24 \sqrt{14}$ , or $\boxed{038}$

Answer:  038
Problem: Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution: Let $E, F$ , and $G$ be the point of tangency (as stated in Solution 1). We can now let $AD$ be $3m$ . By using Power of a Point Theorem on A to the incircle, you get that $AG^2 = 2m^2$ . We can use it again on point D to the incircle to get the equation $(10 - CE)^2 = 2m^2$ . Setting the two equations equal to each other gives $(10 - CE)^2 = AG^2$ , and it can be further simplified to be $10 - CE = AG$
Let lengths $AC$ and $AB$ be called $b$ and $c$ , respectively. We can write $AG$ as $\frac{b + c - 20}{2}$ and $CE$ as $\frac{b + 20 - c}{2}$ . Plugging these into the equation, you get:
\[10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}\] \[b + c - 20 + b + 20 - c = 20 \rightarrow b = 10\]
Additionally, by Median of a triangle formula, you get that $3m = \frac{\sqrt{2c^2 - 200}}{2}$
Refer back to the fact that $AG^2 = 2m^2$ . We can now plug in our variables.
\[\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2\] \[c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}\] \[9c^2 - 180c + 900 = 4c^2 - 400\] \[5c^2 - 180c + 1300 = 0\] \[c^2 - 36c + 260 = 0\]
Solving, you get that $c = 26$ or $10$ , but the latter will result in a degenerate triangle, so $c = 26$ .
Finally, you can use Heron's Formula to get that the area is $24\sqrt{14}$ , giving an answer of $\boxed{038}$

Answer:  038
Problem: A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$

Solution: Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
Our answer is thus $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ , and $m + n = \boxed{79}$

Answer:  79
Problem: A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$

Solution: Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216$ different manners. The total number of possible strings is $\frac{9!}{3!3!3!} = 1680$ . The solution is therefore $\frac{216}{1680} = \frac{9}{70}$ , and $m + n = \boxed{79}$

Answer:  79
Problem: A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$

Solution: The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to ${9 \choose 3}{6 \choose3}$ as the amount of ways to select three rolls out of 9 to give to the first person is ${9 \choose 3}$ , and three rolls out of 6 is ${6 \choose3}$ . After that, the three remaining rolls have no more configurations.
The numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored.
xxx, yyy, zzz
So you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving $3^3$ combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently $2^3$ since there are two of each letter left.
$(27*8)/{9 \choose 3}{6 \choose3}$ simplifies down to our fraction m/n, which is $9/70$ . Adding them up gives $9 + 70 = \boxed{79}$

Answer:  79
Problem: An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$

Solution: Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + \ldots$ .  Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$ .  Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$ .  We know this series has sum $20050 = \frac{a^2}{1 - r^2}$ .  Dividing this equation by $\frac{a}{1-r}$ , we get $10 = \frac a{1 + r}$ .  Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$ $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$ , so the answer is $399 + 403 = \boxed{802}$

Answer:  802
Problem: An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$

Solution: We can write the sum of the original series as $a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005$ , where the common ratio is equal to $\dfrac{m}{n}$ . We can also write the sum of the second series as $a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050$ . Using the formula for the sum of an infinite geometric series $S=\dfrac{a}{1-r}$ , where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}$ , and the second sequence can be written as $\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}$ . Dividing $\text{(2)}$ by $\text{(1)}$ , we obtain $\dfrac{a\cdot n}{m+n}=10$ , which can also be written as $a\cdot n=10(m+n)$ . Substitute this value for $a\cdot n$ back into $\text{(1)}$ , we obtain $10\cdot \dfrac{n+m}{n-m}=2005$ . Dividing both sides by 10 yields $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$ , but this does not output integer solutions. However, we can also write $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ as $\dfrac{n+m}{n-m}=\dfrac{802}{4}$ . This gives the system of equations $m+n=802$ and $n-m=4$ , which does have integer solutions. Our answer is therefore $m+n=\boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$ , respectively, which are co-prime).

Answer:  802