Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A square piece of paper has sides of length $100$ . From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $\sqrt[n]{m}$ , where $m$ and $n$ are positive integers, $m<1000$ , and $m$ is not divisible by the $n$ th power of any prime. Find $m+n$

Solution: In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$ , etc. are edges.
It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$
Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coordinate. Let $\alpha$ be the angle made with the positive $x$ axis . Define $\beta$ and $\gamma$ analogously.
It is easy to see that if $P: = (x,y,z)$ , then $x = AA'\cdot \cos\alpha$ . Furthermore, this means that $\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
We have that $\alpha = \beta = 105^\circ$ , so $\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}$
It is easy to see from the Law of Sines that $\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}$
Now, $z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}$
It follows that the answer is $867 + 4 = \boxed{871}$

Answer:  871
Problem: Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$

Solution: Rewriting this sequence with more terms, we have
Factoring this expression yields
Next, we get
Then,
Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$

Answer:  100
Problem: Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$

Solution: By observation, we realize that the sequence \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2\] alternates every 4 terms. Simplifying, we get \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12\] , turning $N$ into a arithmetic sequence with 25 terms, them being $1, 5, 9, \dots ,97$ , as the series $8a + 12$ alternates every 4 terms.
Applying the sum of arithmetic sequence formula, we get
So the answer would be \[\frac{10100}{1000} = \boxed{100}\]

Answer:  100
Problem: Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them?

Solution: Let $r$ be the time Rudolph takes disregarding breaks and $\frac{4}{3}r$ be the time Jennifer takes disregarding breaks. We have the equation \[r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)\] \[125=\frac13r\] \[r=375.\] Thus, the total time they take is $375 + 5(49) = \boxed{620}$ minutes.

Answer:  620
Problem: Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them?

Solution: Let the total time that Jennifer and Rudolph bike, including rests, to be $t$ minutes.  Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\frac{3}{4}r$ . Note that Rudolf takes 49 breaks, taking $49\cdot 5$ minutes, and Jennifer takes 24 breaks, taking $24\cdot 5$ minutes.  Since they both reach the 50 mile mark, then by $d=rt$ , the rate times time taken for Rudolph and Jennifer must equal.  Hence, we disregard the breaks from the total time taken and get the equation \[r(t-49\cdot 5)=\frac{3}{4}r(t-24\cdot 5),\] yielding $t=\boxed{620}$

Answer:  620
Problem: A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?

Solution: Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$ , so that the desired volume is $abc$ . Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$ . By AM-GM $\frac{a+b+c}{3} = 9 \ge \sqrt[3]{abc} \Longrightarrow abc \le \boxed{729}$ . Equality is achieved when $a=b=c=9$ , which is possible if we make one slice perpendicular to the $10$ cm edge, four slices perpendicular to the $13$ cm edge, and five slices perpendicular to the $14$ cm edge.

Answer:  729
Problem: A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?

Solution: A more intuitive way to solve it is by seeing that to keep the volume of the rectangular cheese the greatest, we must slice the cheese off to decrease the greatest length of the cheese (this is easy to check). Here are the ten slices:
${10, 13, 14} \rightarrow {10, 13, 13} \rightarrow {10, 12, 13} \rightarrow {10, 12, 12} \rightarrow {10, 11, 12} \rightarrow {10, 11, 11} \rightarrow {10, 10, 11} \rightarrow {10, 10, 10} \rightarrow {9, 10, 10} \rightarrow {9, 9, 10} \rightarrow {9, 9, 9}.$
So the greatest possible volume is thus $9 \times 9 \times 9 = \boxed{729}$

Answer:  729
Problem: There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$

Solution: In base $3$ , we find that $\overline{2008}_{10} = \overline{2202101}_{3}$ . In other words,
In order to rewrite as a sum of perfect powers of $3$ , we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$
The answer is $7+5+4+3+2+0 = \boxed{021}$

Answer:  021
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$ . Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$
As $\angle AED = 90^{\circ}$ , note that the midpoint of $\overline{AD}$ $N$ , is the center of the circumcircle of $\triangle AED$ . We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$ ). It follows that
\[NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.\]
Thus $MN = NE - ME = \boxed{504}$

Answer:  504
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$ , respectively. Let $x = NH$ , so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$ . Also, let $h = BF = CG = HM$
By AA~, we have that $\triangle AFB \sim \triangle CGD$ , and so \[\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.\]
By the Pythagorean Theorem on $\triangle MHN$ \[MN^{2} = x^2 + h^2 = 504^2,\] so $MN = \boxed{504}$

Answer:  504
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$ $C=\left(0, 1000\sin 37^\circ\right)$ $A=\left(2008\cos 37^\circ, 0\right)$ , and $D=\left(0, 2008\sin 37^\circ\right)$
The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$
To find the distance from $M$ to $N$ , we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$

Answer:  504
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$ . (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)

Answer:  504
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Let the height be h. Note that if $\overline{NH} = x$ then if we draw perpendiculars like in solution 1, $\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.$ Note that we wish to find $\overline{MN} = \sqrt{x^2 + h^2}.$ Let's find $\tan(53)$ in two ways. Finding $\tan(53)$ from $\triangle BAF$ yields $\tan(53) = \frac{504+x}{h}.$ Finding it from $\triangle CDG$ yields $\frac{h}{504-x}.$ Setting these equal yields \[\frac{504+x}{h}=\frac{h}{504-x} \rightarrow h^2 = 504^2-x^2 \rightarrow \sqrt{x^2+h^2} = \sqrt{504^2} = \boxed{504}\]

Answer:  504
Problem: In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution: Rotate trapezoid $MNCD$ 180 degrees around point $N$ so that $AN$ coincides with $ND$ . Let the image of trapezoid $MNCD$ be $ANM'C'$ . Since angles are preserved during rotations, $\angle BAC' = 37^{\circ} + 53 ^{\circ} =90 ^{\circ}$ . Since $BM=CM=C'M'$ and $BM || C'M'$ $BMM'C'$ is a parallelogram. Thus, $MM'=BC'$ .
Let the point where $BC'$ intersects $AD$ be $E$ . Since $BMNE$ is a parallelogram, $AE=AN-BM=1004-500-504$ . Since $BE=EC$ and $\angle BAC= 90^{\circ}$ $AE$ is a median to the hypotenuse of $BAC'$ . Therefore, $BC'=2 AE= 1008$ , and $BE=MN=\boxed{504}.$

Answer:  504
Problem: The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$

Solution: Rearranging the definitions, we have \[\frac{a_n}{a_{n-1}} = \frac{a_{n-1}}{a_{n-2}} + 1,\quad \frac{b_n}{b_{n-1}} = \frac{b_{n-1}}{b_{n-2}} + 1\] from which it follows that $\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n$ and $\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} = n+2$ . These recursions $a_{n} = na_{n-1}$ and $b_{n} = (n+2)b_{n-1}$ , respectively, correspond to the explicit functions $a_n = n!$ and $b_n = \frac{(n+2)!}{2}$ (after applying our initial conditions). It follows that $\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}$

Answer:  561
Problem: Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer.

Solution: By the product-to-sum identities , we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$ . Therefore, this reduces to a telescope series: \begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}
Thus, we need $\sin \left(\frac{n(n+1)\pi}{2008}\right)$ to be an integer; this can be only $\{-1,0,1\}$ , which occur when $2 \cdot \frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1$ . We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$ , so 1004 doesn't divide $n(n+1) = 250 \cdot 251$ . Therefore, it is clear that $n = \boxed{251}$ is the smallest such integer.

Answer:  251
Problem: Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer.

Solution: We proceed with complex trigonometry. We know that for all $\theta$ , we have $\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)$ and $\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)$ for some complex number $z$ on the unit circle. Similarly, we have $\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)$ and $\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)$ . Thus, we have $\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)$
$= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)$
$= \dfrac{1}{2} \left( \dfrac{1}{2i} \left(z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} \right) - \dfrac{1}{2i} \left(z^{n^2 - n} - \dfrac{1}{z^{n^2 - n}} \right) \right)$
$= \dfrac{1}{2} \left( \sin ((n^2 + n)a) - \sin ((n^2 - n)a) \right)$
$= \dfrac{1}{2} \left( \sin(((n+1)^2 - (n+1))a) - \sin((n^2 - n)a) \right)$
which clearly telescopes! Since the $2$ outside the brackets cancels with the $\dfrac{1}{2}$ inside, we see that the sum up to $n$ terms is
$\sin ((2^2 - 2)a) - \sin ((1^2 - 1)a) + \sin ((3^3 - 3)a) - \sin ((2^2 - 2)a) \cdots + \sin (((n+1)^2 - (n+1))a) - \sin ((n^2 - n)a)$
$= \sin (((n+1)^2 - (n+1))a) - \sin(0)$
$= \sin ((n^2 + n)a) - 0$
$= \sin \left( \dfrac{n(n+1) \pi}{2008} \right)$
This expression takes on an integer value iff $\dfrac{2n(n+1)}{2008} = \dfrac{n(n+1)}{1004}$ is an integer; that is, $1004 \mid n(n+1)$ . Clearly, $1004 = 2^2 \cdot 251$ , implying that $251 \mid n(n+1)$ . Since we want the smallest possible value of $n$ , we see that we must have ${n,n+1} = 251$ . If $n+1 = 251 \rightarrow n=250$ , then we have $n(n+1) = 250(251)$ , which is clearly not divisible by $1004$ . However, if $n = 251$ , then $1004 \mid n(n+1)$ , so our answer is $\boxed{251}$

Answer:  251
Problem: A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$

Solution: Let the particle's position be represented by a complex number. Recall that multiplying a number by cis $\left( \theta \right)$ rotates the object in the complex plane by $\theta$ counterclockwise. In this case, we use $a = cis(\frac{\pi}{4})$ . Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to
where a is cis $\left( \theta \right)$ . By De-Moivre's theorem, $\left(cis( \theta \right)^n )$ =cis $\left(n \theta \right)$ .
Therefore,
Furthermore, $5a^{150} = - 5i$ . Thus, the final answer is
As before, consider $z$ as a complex number.  Consider the transformation $z \to (z-\omega)e^{i\theta} + \omega$ .  This is a clockwise rotation of $z$ by $\theta$ radians about the points $\omega$ .  Let $f(z)$ denote one move of $z$ .  Then
Therefore, $z$ rotates along a circle with center $\omega = 5+(5+5\sqrt2)i$ .  Since $8 \cdot \frac{\pi}{4} = 2\pi$ $f^9(z) = f(z) \implies f^{150}(z) = f^6(z) \implies p+q = \boxed{019}$ , as desired (the final algebra bash isn't bad).

Answer:  019
Problem: A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$

Solution: Let $T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}$ . We assume that the rotation matrix $R(\frac{\pi}{4}) = R$ here. Then we have
This simplifies to
Since $R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O$ , so we have $R^6\begin{pmatrix}5\\0\end{pmatrix}+(-R^6-R^7)\begin{pmatrix}10\\0\end{pmatrix}$ , giving $p=-5\sqrt{2}, q=5\sqrt{2}+5$ . The answer is yet $\lfloor10\sqrt{2}+5\rfloor=\boxed{019}$

Answer:  019
Problem: The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr$

Solution: We label our points using coordinates $0 \le x,y \le 3$ , with the bottom-left point being $(0,0)$ . By the Pythagorean Theorem , the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$ ; these yield the possible distances (in decreasing order) \[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2},\ \sqrt{1}\] As these define $9$ lengths, so the maximum value of $m$ is $10$ . For now, we assume that $m = 10$ is achievable. Because it is difficult to immediately impose restrictions on a path with increasing distances, we consider the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former.
The $\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$
The $\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$
From $(1,0)$ , there are two possible ways to move $\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$ . However, from $(0,3)$ , there is no way to move $\sqrt{9}$ away, so we discard it as a possibility.
From $(2,3)$ , the lengths of $\sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2}$ fortunately are all determined, with the endpoint sequence being $(2,3)-(2,0)-(0,2)-(2,1)-(0,1)-(1,2)$
From $(1,2)$ , there are $3$ possible lengths of $\sqrt{1}$ (to either $(1,1),(2,2),(1,3)$ ). Thus, the number of paths is $r = 4 \cdot 2 \cdot 3 = 24$ , and the answer is $mr = 10 \cdot 24 = \boxed{240}$

Answer:  240
Problem: In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$ , where $m$ $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$

Solution: Refer to the above diagram. Let the larger circle have center $O_1$ , the smaller have center $O_2$ , and the incenter be $I$ . We can easily calculate that the area of $\triangle ABC = 2688$ , and $s = 128$ and $R = 21$ , where $R$ is the inradius.
Now, Line $\overline{AI}$ is the perpendicular bisector of $\overline{BC}$ , as $\triangle ABC$ is isosceles. Letting the point of intersection be $X$ , we get that $BX = 28$ and $IX = 21$ , and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\overline{AB}$ and $\overline{BC}$ . By Pythagoras, $BI = 35$ , and we notice that $\triangle BIX$ is a 3-4-5 right triangle.
Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\overline{BC}$ , we find that $BY = \frac{4r}{3}$ . Similarly, we find that the distance between the projection from $O_1$ onto $\overline{BC}$ $W$ , and $C$ , is $\frac{64}{3}$ . From there, we let the projection of $O_2$ onto $\overline{O_1W}$ be $Z$ , and we have $O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}$ $O_1Z = 16 - r$ , and $O_1O_2 = 16 + r$ . We finish with Pythagoras on $\triangle O_1O_2Z$ , whence we get the desired answer of $\boxed{254}$ . - Spacesam

Answer:  254
Problem: In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$ , where $m$ $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$

Solution: Let the incenter be O and the altitude from A to $\overline{BC}$ be T. Note that by AA, $\triangle BQY \sim \triangle OBT$ and $\triangle PXC \sim \triangle OTC.$ Note that from $A = rs$ , the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\overline{BY}.$ We have \[\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r\] Now, note that as in solution 1, drawing the perpendicular from Q to $\overline{PX}$ (call it Z) yields $\overline{PZ} = 16 - r, \overline{ZX} = r.$ Then, from this, \[\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}\] Using similar similarity as was done to find $\overline{BY}$ we have $\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{3}$ .
Now adding all these up and equating them to $\overline{BC}$ yields \[\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}\]

Answer:  254