Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed?

Solution: There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below. [asy] size(60); pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0); draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d)); draw(u--2*u+d,dotted); draw(3*u--3*u-h,dotted); [/asy] Therefore, if all horizontal lines are drawn, there will be a total of $2\cdot 21^2=882$ unit equilateral triangles. Of course, we only draw $21$ horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines. [asy] size(200); pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0); for (int i=0;i<21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} for (int i=0;i<21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);} draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10)); [/asy]
We see that the lines $y=-21,-20,\dots, -11$ and $y=11,12,\dots,21$ would complete several of the $882$ unit equilateral triangles. In fact, we can see that the lines $y=-21,-20,\dots,-11$ complete $1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)$ triangles, or $111$ triangles. The positive horizontal lines complete the same number of triangles, hence the answer is $882-2\cdot 111=\boxed{660}$

Answer:  660
Problem: The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed?

Solution: Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us $30^2 - 10^2 - 10^2- 10^2 = 600$ . That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. $600 + 6 * 10 = \boxed{660}$

Answer:  660
Problem: For certain ordered pairs $(a,b)\,$ of real numbers , the system of equations
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers.  How many such ordered pairs $(a,b)\,$ are there?

Solution: The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$ , centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$ $(\pm5,\pm5)$ , and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cdot 2\cdot 2=12$ lattice points. They are indicated by the blue dots below.
[asy] size(150); draw(circle((0,0),sqrt(50)));  draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green);  dot((0,0));  dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue);  dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue);  dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); dot((-7,-1),blue); [/asy]
Since $(x,y)=(0,0)$ yields $a\cdot 0+b\cdot 0=0 \neq 1$ , we know that $ax+by=1$ is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the $12$ lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points $(p,q)$ and $(-p,-q)$ through which it passes. And example is the red line above.
There are $\binom{12}{2}=66$ ways to pick two distinct lattice points, and subsequently $66$ distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs $(p,q)$ and $(-p,-q)$ , for a total of $\frac{12}{2}=6$ lines. Finally, we add the $12$ unique tangent lines to the circle at each of the lattice points.
Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is \[66-6+12=\boxed{72}.\]

Answer:  72
Problem: The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle.  Find the value of $ab\,$

Solution: Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
\[(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}
Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$ . Thus, the answer is $\boxed{315}$

Answer:  315
Problem: The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle.  Find the value of $ab\,$

Solution: Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:
First, drop a perpendicular from $O$ to $AB$ . Call this midpoint of $AB M$ . Thus, $M=\left(\frac{a+b}{2}, 24\right)$ . The vector from $O$ to $M$ is $\left[\frac{a+b}{2}, 24\right]$ . Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]$ . (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)
We see this displacement from $M$ to $A$ is $\left[\frac{a-b}{2}, 13\right]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$ . Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$ . And the answer is $\boxed{315}$

Answer:  315
Problem: The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle.  Find the value of $ab\,$

Solution: Plot this equilateral triangle on the complex plane.
Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\left(\frac{a+b}{3}, 16i\right)$ . The new coordinates of the equilateral triangle are $\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)$ . These three vertices are solutions of a cubic polynomial of form $x^3 + C$ . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$ $\, r_1r_2 + r_2r_3 + r_3r_1 = 0$ .) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$ . Use your previously obtained equation to plug in for $a$ and solve for $b$ , which should yield $5\sqrt{3}$ $a$ is then $\frac{21}{5}\sqrt{3}$ . Multiplying $a$ and $b$ yields $\boxed{315}$

Answer:  315
Problem: The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle.  Find the value of $ab\,$

Solution: Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$
$a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248$ . Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$ $b^2 = 2ab - 555 ==> a^2 = 2ab + 693$
We have $b^2 + 1248 = 2b\sqrt{b^2+1248} + 693$
Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$
$3b^4 + 3882b^2 - 308025 = 0$
This factors to $(3b^2-225)(b^2+1369)$ , so $3b^2 = 225, b = 5\sqrt 3$
Plugging it back in, we find that a = $\sqrt{1323}$ which is $21\sqrt3$ , so the product $ab$ is $\boxed{315}$

Answer:  315
Problem: A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers.  Find $p+q.\,$

Solution: Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:
Therefore, we obtain the recursion $P_k = \frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $\frac {3^5}{11*9*7*5*3} = \frac {9}{385}$ , and $p+q=\boxed{394}$

Answer:  394
Problem: A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers.  Find $p+q.\,$

Solution: Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here.
1. We pick the other A. There's a $\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this is the case.
2. We pick any other letter that is not an A. There's a $\frac{10}{11}$ chance for this to happen. WLOG, assume we pick the letter B. Now in order for us to continue the game, we must choose either the other A or B. There's a $\frac{2}{10}$ chance for this to happen. WLOG, assume we choose A. Now we have BCDEFCDEF left.
Notice however that in the first case, the probability of emptying the bag with [BCDEF] is the same thing as with BCDEFCDEF, as the only difference is you've removed one of the letters (and it doesn't matter which you chose).
Hence for this case, there is a $\frac{1}{11} + \frac{10}{11}*\frac{2}{10} = \frac{3}{11}$ * [BCDEF] chance to empty the bag.
Continuing this process, we get that:
[BCDEF] = $\frac{3}{9}$ * [CDEF]
[CDEF] = $\frac{3}{7}$ * [DEF]
[DEF] = $\frac{3}{5}$ * [EF]
[EF] = 1 (clearly, since if we are only left with EFEF then we are going to empty the bag).
And hence [ABCDEF] = $1*\frac{3}{5}*\frac{3}{7}*\frac{3}{9}*\frac{3}{11} = \frac{9}{385}$ so our answer is $9+385=\boxed{394}$

Answer:  394
Problem: In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers.  Find $m+n.\,$

Solution: Since $\triangle ABC \sim \triangle CBD$ , we have $\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$ , respectively, where x is an integer.
By the Pythagorean Theorem , we find that $AC^2 + BC^2 = AB^2 \Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2$ , so $29x | AC$ . Letting $y = AC / 29x$ , we obtain after dividing through by $(29x)^2$ $29^2 = x^2 - y^2 = (x-y)(x+y)$ . As $x,y \in \mathbb{Z}$ , the pairs of factors of $29^2$ are $(1,29^2)(29,29)$ ; clearly $y = \frac{AC}{29x} \neq 0$ , so $x-y = 1, x+y= 29^2$ . Then, $x = \frac{1+29^2}{2} = 421$
Thus, $\cos B = \frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}$ , and $m+n = \boxed{450}$

Answer:  450
Problem: In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers.  Find $m+n.\,$

Solution: We will solve for $\cos B$ using $\triangle CBD$ , which gives us $\cos B = \frac{29^3}{BC}$ . By the Pythagorean Theorem on $\triangle CBD$ , we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$ . Trying out factors of $29^6$ , we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The other pairs give answers over 999). Adding these, we have $2BC = 29^4 + 29^2$ and $\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}$ , and our answer is $\boxed{450}$

Answer:  450
Problem: In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers.  Find $m+n.\,$

Solution: Using similar right triangles, we identify that $CD = \sqrt{AD \cdot BD}$ . Let $AD$ be $29 \cdot k^2$ , to avoid too many radicals, getting $CD = k \cdot 29^2$ . Next we know that $AC = \sqrt{AB \cdot AD}$ and that $BC = \sqrt{AB \cdot BD}$ . Applying the logic with the established values of k, we get $AC = 29k \cdot \sqrt{29^2 + k^2}$ and $BC = 29^2 \cdot \sqrt{29^2 + k^2}$ . Next we look to the integer requirement. Since $k$ is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let $y$ be $\sqrt{29^2 + k^2}$ , thus $29^2 = y^2 - k^2$ , and $29^2 = (y + k)(y - k)$ . Since $29$ is prime, and $k$ cannot be zero, we find $k = 420$ and $y = 421$ as the smallest integers satisfying this quadratic Diophantine equation. Then, since $cos B$ $\frac{29}{\sqrt{29^2 + k^2}}$ . Plugging in we get $cos B = \frac{29}{421}$ , thus our answer is $\boxed{450}$

Answer:  450
Problem: Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall.  Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower.  How many different tower heights can be achieved using all ninety-four of the bricks?

Solution: We have the smallest stack, which has a height of $94 \times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$ 's, $6$ 's, and $15$ 's. Because $0$ $6$ , and $15$ are all multiples of $3$ , the change will always be a multiple of $3$ , so we just need to find the number of changes we can get from $0$ 's, $2$ 's, and $5$ 's.
From here, we count what we can get:
\[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, \ldots\]
It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem , which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 \times 2 - 5 - 2=3$
But we also have a maximum change ( $94 \times 5$ ), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$ 's, $3$ 's, or $5$ 's. The maximum we can't get is $5 \times 3-5-3=7$ , so the numbers $94 \times 5-8$ and below, except $3$ and $1$ , work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 \times 5-8$ and $94 \times 5$ $94 \times 5-7$ obviously doesn't work, $94 \times 5-6$ does since 6 is a multiple of 3, $94 \times 5-5$ does because it is a multiple of $5$ (and $3$ ), $94 \times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$ $94 \times 5-3$ does since $3=3$ , and $94 \times 5-2$ and $94 \times 5-1$ don't, and $94 \times 5$ does.
Thus the numbers $0$ $2$ $4$ all the way to $94 \times 5-8$ $94 \times 5-6$ $94 \times 5-5$ $94 \times 5-3$ , and $94\times 5$ work. That's $2+(94 \times 5 - 8 - 4 +1)+4=\boxed{465}$ numbers. That's the number of changes you can make to a stack of bricks with dimensions $4 \times 10 \times 19$ , including not changing it at all.

Answer:  465
Problem: Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall.  Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower.  How many different tower heights can be achieved using all ninety-four of the bricks?

Solution: Using bricks of dimensions $4''\times10''\times19''$ is comparable to using bricks of dimensions $0''\times6''\times15''$ which is comparable to using  bricks of dimensions $0''\times2''\times5''$ . Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$
It follows that all tower heights can be made by using 4 or fewer bricks of height $2''$ . There are $95+94+93+92+91=465$ ways to build a tower using 4 or fewer bricks of height $2''$ . Taking the heights $\mod 5$ , we see that towers using a different number of bricks of height $2''$ have unequal heights. Thus, all of the $\boxed{465}$ tower heights are different.

Answer:  465
Problem: A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?

Solution: Suppose there are $n$ squares in every column of the grid, so there are $\frac{52}{24}n = \frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$
Each vertical fence has length $24$ , and there are $\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and there are $n-1$ such fences. Then the total length of the internal fencing is $24\left(\frac{13n}{6}-1\right) + 52(n-1) = 104n - 76 \le 1994 \Longrightarrow n \le \frac{1035}{52} \approx 19.9$ , so $n \le 19$ . The largest multiple of $6$ that is $\le 19$ is $n = 18$ , which we can easily verify works, and the answer is $\frac{13}{6}n^2 = \boxed{702}$

Answer:  702
Problem: A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?

Solution: Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$ . (If you are confused, just draw the square out). This is clearly greater than $1994$ , so the actual side length that we are looking for is greater than $1$
Now we can convert this into an equation. The equation is simply $(\frac{52}{x})(\frac{24}{x}+1) + (\frac{24}{x})(\frac{52}{x}+1)$ (The intutition comes from considering partioning the field into side lengths of $1$ and then partitioning those squares). This is equivalent to $\frac{2496}{x^2} + \frac{76}{x}$ , which should be less than or equal to $1994$
Now we can just find possible lengths of the square that are greater than $1$ and test them out. A viable side length would mean that $\frac{24}{x}, \frac{52}{x} \in$ N. Since $\gcd(24,52) = 4$ , then the smallest value greater than $1$ that we satisfies the conditions has $4$ in the numerator, and hence $3$ in the denominator. Test out $x=\frac{4}{3}$ . This will equate to something less than $1994$ , and hence the smallest square length that is plausible is $\frac{4}{3}$
Now the rest is elementary, we do $\frac{52}{\frac{4}{3}} * \frac{24}{\frac{4}{3}} \Rightarrow 39*18 = \boxed{702}$

Answer:  702
Problem: The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation.  Find the value of

Solution: Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$
Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$
$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)$
Since $\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)$ $t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product.
The expression to find is $\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$
But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$

Answer:  850
Problem: The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation.  Find the value of

Solution: Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]
Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]
Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.
We see that $\dfrac{1}{x}=13-\omega$ . Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})\]
Summing over all terms: \[\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})\]
However, note that $e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just \[5\cdot 170=\boxed{850}\]

Answer:  850