Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is $2 + 2\sqrt{3}i$
Recall that a rotation of $\theta$ radians counterclockwise is equivalent to multiplying a complex number by $e^{i\theta}$ , but here we require a clockwise rotation, so we multiply by $e^{-\frac{i\pi}{3}}$ to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. $\left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right)$
Therefore $xy$ is $\frac{25\sqrt{3}}{12}$ and the answer is $25 + 12 + 3 = \boxed{040}$

Answer:  040
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: We can also consider the slopes of the lines. Midpoint $M$ of $AB$ has coordinates $\left(\frac{3}{2},\ \sqrt{3}\right)$ . Because line $AB$ has slope $2\sqrt{3}$ , the slope of line $MP$ is $-\frac{1}{2\sqrt{3}}$ (Because of perpendicular slopes).
Since $\Delta ABC$ is equilateral, and since point $P$ is the centroid, we can quickly calculate that $MP = \frac{\sqrt{39}}{6}$ . Then, define $\Delta x$ and $\Delta y$ to be the differences between points $M$ and $P$ . Because of the slope, it is clear that $\Delta x = 2\sqrt{3} \Delta y$
We can then use the Pythagorean Theorem on line segment $MP$ $MP^2 = \Delta x^2 + \Delta y^2$ yields $\Delta y = -\frac{1}{2\sqrt{3}}$ and $\Delta x = 1$ , after substituting $\Delta x$ . The coordinates of P are thus $\left(\frac{5}{2},\ \frac{5\sqrt{3}}{6}\right)$ . Multiplying these together gives us $\frac{25\sqrt{3}}{12}$ , giving us $\boxed{040}$ as our answer.

Answer:  040
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: Since $AC$ will be segment $AB$ rotated clockwise $60^{\circ}$ , we can use a rotation matrix to find $C$ . We first translate the triangle $1$ unit to the left, so $A'$ lies on the origin, and $B' = (1, 2\sqrt{3})$ . Rotating clockwise $60^{\circ}$ is the same as rotating $300^{\circ}$ counter-clockwise, so our rotation matrix is $\begin{bmatrix} \cos{300^{\circ}} & -\sin{300^{\circ}}\\ \sin{300^{\circ}} & \cos{300^{\circ}}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix}$ . Then $C' = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2\sqrt{3}\\ \end{bmatrix} = \begin{bmatrix} \frac{7}{2}\\ \frac{\sqrt{3}}{2}\\ \end{bmatrix}$ . Thus, $C = (\frac{9}{2}, \frac{\sqrt{3}}{2})$ . Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then $P = (\frac{1 + 2 + \frac{9}{2}}{3}, \frac{2\sqrt{3} + \frac{\sqrt{3}}{2}}{3}) = (\frac{5}{2}, \frac{5\sqrt{3}}{6})$ . Our answer is $\frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12}$ $25 + 3 + 12 = \boxed{40}$

Answer:  40
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: We construct point $C$ by drawing two circles with radius $r = AB = \sqrt{13}$ .  One circle will be centered at $A$ , while the other is centered at $B$ .  The equations of the circles are:
$(x - 1)^2 + y^2 = 13$
$(x - 2)^2 + (y - 2\sqrt{3})^2 = 13$
Setting the LHS of each of these equations equal to each other and solving for $x$ yields after simplification:
$x = \frac{15}{2} - 2\sqrt{3}y$
Plugging that into the first equation gives the following quadratic in $y$ after simplification:
$y^2 - 2\sqrt{3}y + \frac{9}{4} = 0$
The quadratic formula gives $y = \frac{\sqrt{3}}{2}, \frac{3\sqrt{3}}{2}$
Since $x > 0$ and $x = \frac{15}{2} - 2\sqrt{3}y$ , we pick $y = \frac{\sqrt{3}}{2}$ in the hopes that it will give $x > 0$ .  Plugging $y$ into the equation for $x$ yields $x = \frac{9}{2}$
Thus, $C(\frac{9}{2}, \frac{\sqrt{3}}{2})$ .  Averaging the coordinates of the vertices of equilateral triangle $ABC$ will give the center of mass of the triangle.
Thus, $P(\frac{5}{2}, \frac{5\sqrt{3}}{6})$ , and the product of the coordinates is $\frac{25\sqrt{3}}{12}$ , so the desired quantity is $\boxed{040}$

Answer:  040
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: Labeling our points and sketching a graph we get that $C$ is to the right of $AB$ . Of course, we need to find $C$ . Note that the transformation from $A$ to $B$ is $[1,2\sqrt{3}]$ , and if we imagine a height dropped to $AB$ we see that a transformation from the midpoint $(\frac{3}{2},\sqrt {3})$ to $C$ is basically the first transformation, with $\frac{\sqrt{3}}{2}$ the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of $[3,\frac{-\sqrt{3}}{2}]$ we get that $C=(\frac{9}{2},\frac{\sqrt{3}}{2})$ which means that $P=(\frac{5}{2},\frac{5\sqrt{3}}{6})$ . Then our answer is $\boxed{40}$

Answer:  40
Problem: In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution: [asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, S); label("$E$", E, S); draw(A--D); draw(A--M); draw(A--E); [/asy]
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let $M$ be the midpoint of $\overline{DE}$ . Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$ $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$ . Since the triangle $\Delta AME$ is right, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$ . Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$ . So we can use the double angle formula for sine, $\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Answer:  020
Problem: In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution: We find that, as before, $AE = \sqrt{7}$ , and also the area of $\Delta DAE$ is 1/3 the area of $\Delta ABC$ . Thus, using the area formula, $1/2 \cdot 7 \cdot \sin(\angle EAD) = 3\sqrt{3}/4$ , and $\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}.$

Answer:  020
Problem: In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution: Let A be the origin of the complex plane, B be $1+i\sqrt{3}$ , and C be $2$ . Also, WLOG, let D have a greater imaginary part than E. Then, D is $\frac{4}{3}+\frac{2i\sqrt{3}}{3}$ and E is $\frac{5}{3}+\frac{i\sqrt{3}}{3}$ . Then, $\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) =  \frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Answer:  020
Problem: In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution: Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\cos(\angle DAE)$ or $AE=\sqrt7$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14$ Since $\sin^2(\angle DAE)=1-cos^2(\angle DAE)$ Then $\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}$ So $\sin(\angle DAE)=\frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Answer:  020
Problem: In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution: Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$ $B$ be at point $(3, 3\sqrt3)$ , and $C$ be at point $(6, 0)$ . Then $D$ and $E$ will be at points $(4, 2\sqrt3)$ and $(5, \sqrt3)$ . Then $\cos(\angle DAE) = \frac{\vec{AD}\cdot\vec{AE}}{\|\vec{AD}\| \|\vec{AE}\|} = \frac{4\cdot5 + 2\sqrt{3}\cdot\sqrt{3}}{28}=\frac{13}{14}$ . From here, we see that $\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$

Answer:  020
Problem: Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution: The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$ , so $x\geq\frac{999}{2}\implies x\geq500$ $x=500$ does not work, so $x>500$ . Let $n=x-500$ . By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$ . Then, $n=10\sqrt{10}$ . One example way to estimate $\sqrt{10}$ follows.
Then, $n\approx 31.6$ $n^2<1000$ , so $n$ could be $31$ . Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$ . Checking, $531^2=281961$ and $532^2=283024$ $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\implies\boxed{282}$

Answer:  282
Problem: Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution: Let us first observe the difference between $x^2$ and $(x+1)^2$ , for any arbitrary $x\ge 0$ $(x+1)^2-x^2=2x+1$ . So that means for every $x\ge 0$ , the difference between that square and the next square have a difference of $2x+1$ . Now, we need to find an $x$ such that $2x+1\ge 1000$ . Solving gives $x\ge \frac{999}{2}$ , so $x\ge 500$ . Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free.
Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$ $x^2=250000$ , and $(x+1)^2=251001$ . Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$ , but not there, such as $961$ or $974$ . Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$ , so all we need to do is addition. After making a list, we find that $531^2=281961$ , while $532^2=283024$ . It skipped $282000$ , so our answer is $\boxed{282}$

Answer:  282
Problem: Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution: Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$ , so $x$ has to be at least $500$ . Let $k$ be $x-500$ . We can write $x^2$ as $(500+k)^2$ , or $250000+1000k+k^2$ . We can disregard $250000$ and $1000k$ , since they won't affect the last three digits, which determines if there are any squares between $\overline{N000}\rightarrow \overline{N999}$ . So we must find a square, $k^2$ , such that it is under $1000$ , but the next square is over $1000$ . We find that $k=31$ gives $k^2=961$ , and so $(k+1)^2=32^2=1024$ . We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$ $(500+31)^2=281961$ , while $(500+32)^2=283024$ . We skipped $282000$ , so the answer is $\boxed{282}$

Answer:  282
Problem: Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution: The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$
Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$
Let $m = k + 500$ , where $k \in \mathbb{W}$ , then the inequalities become,
$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$ , and
$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$
For $k=31$ , one can verify that $N = 282$ is the unique integer satisfying the inequalities.
For $k \leq 30$ $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$
i.e., $k + 250 < N < k + 251$ , a contradiction.
Note $k \geq 32$ leads to larger $N$ (s).
Hence, the answer is $\boxed{282}$

Answer:  282
Problem: A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

Solution: There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .
Now the problem is to find a reasonable integer solution. Now we know $x= \frac{355+21t}{19}$ , so $19$ divides $355+21t$ , AND as long as $t$ is a integer, $19$ must divide $2t+355$ . Now, we suppose that $19m=2t+355$ , similarly we get $t=\frac{19m-355}{2}$ , and so in order to get a minimum integer solution for $t$ , it is obvious that $m=19$ works. So we get $t=3$ and $x=22$ .  One and a half hour's work should be $30x+15(x-t)$ , so the answer is $\boxed{945}$

Answer:  945
Problem: A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

Solution: We start with the same approach as solution 1 to get $19x-21t=355$ . Then notice that $21t + 355 \equiv 0 \pmod{19}$ , or $2t-6 \equiv 0 \pmod{19}$ , giving the smallest solution at $t=3$ . We find that $x=22$ . Then the number of files they sorted will be $30x+15(x-t)=660+285=\boxed{945}.$

Answer:  945
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ . 
We can just consider one half of the hexagon, $ABCD$ , to make matters simpler. 
Draw a line from the center of the circle, $O$ , to the midpoint of $BC$ $X$ . Now, draw a line from $O$ to the midpoint of $AB$ $Y$ . Clearly, $\angle BXO=90^{\circ}$ , because $BO=CO$ , and $\angle BYO=90^{\circ}$ , for similar reasons. Also notice that $\angle AOX=90^{\circ}$ . 
Let us call $\angle BOY=\theta$ . Therefore, $\angle AOB=2\theta$ , and so $\angle BOX=90-2\theta$ . Let us label the radius of the circle $r$ . This means \[\sin{\theta}=\frac{BY}{r}=\frac{11}{r}\] \[\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}\] Now we can use simple trigonometry to solve for $r$ .
Recall that $\sin{(90-\alpha)}=\cos(\alpha)$ : That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$ .
Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$ : That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$ .
Let $\sin{\theta}=x$ .
Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \[r^2-10r-242=0\] Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$

Answer:  272
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$ . Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$ . Expanding and combining like terms gives us the quadratic \[r^2-10r-242=0\] and solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$

Answer:  272
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: Join the diameter of the circle $AD$ and let the length be $d$ . By Ptolemy's Theorem on trapezoid $ADEF$ $(AD)(EF) + (AF)(DE) = (AE)(DF)$ . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then
\[20d + 22^2 = x^2\]
Since $\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\triangle AED$
\[(AE)^2 + (ED)^2 = (AD)^2\] \[x^2 + 22^2 = d^2\]
From the above equations, we have: \[x^2 = d^2 - 22^2 = 20d + 22^2\] \[d^2 - 20d = 2\times22^2\] \[d^2 - 20d + 100 = 968+100 = 1068\] \[(d-10) = \sqrt{1068}\] \[d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)\]
Since the radius is half the diameter, it is $\sqrt{267}+5$ , so the answer is $5+267 \Rightarrow \boxed{272}$

Answer:  272
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \[22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}\] , after simple factorization, we can get \[x^4-342x^2-2420x=0\] , it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \[x(x+10)(x^2-10x-242)\] so we only need to find the solution for \[x^2-10x-242=0\] and we can get $x=(\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \Rightarrow \boxed{272}$ .～bluesoul

Answer:  272
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: Using solution 1's diagram, extend line segments $AB$ and $CD$ upwards until they meet at point $G$ . Let point $O$ be the center of the hexagon. By the $AA$ postulate, $\Delta ADG \sim \Delta BCG \sim \Delta CDO$ . This means $\frac{DC}{AD} = \frac{22}{2r} = \frac{CO}{AG}$ , so $AG = r \times \frac{2r}{22} = \frac{r^2}{11}$ . We then solve for $AB$ \[\frac{AD-BC}{AD} = \frac{2r-20}{2r} = \frac{AB}{AG}\] \[AB = \frac{r^2}{11} \times \frac{2r-20}{2r} = \frac{r^2-10r}{11}\] Remember that $AB=22$ as well, so $22 = \frac{r^2-10r}{11} \Rightarrow r^2-10r-242=0$ . Solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$

Answer:  272
Problem: A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$

Solution: Let $\angle{AOB} = \theta$ . So, we have $\sin \dfrac{\theta}{2} = \dfrac{11}{r}$ and $\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}$ . So, $\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}$ . Let $H$ be the foot of the perpendicular from $B$ to $\overline{AD}$ . We have $BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}$ . Using Pythagorean theorem on $\triangle BCF$ , to get $(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$ , or $\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}$ . Multiplying by $r^{2}$ , we get $44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}$ . Rearranging and simplifying, we get a quadratic in $r^{2}$ \[r^{4} - 584r^{2} + 242^{2} = 0 \text\] which gives us $r^{2} = 292 \pm 10\sqrt{267}$ . Because $r$ is in the form $p + \sqrt{q}$ , we know to choose the larger option, meaning $r^2 = 292 + 10\sqrt{267}$ , so $p\sqrt{q} = 5\sqrt{267}$ and $p^2 + q = 292$ . By inspection, we get $(p, q) = (5, 267)$ , so our answer is $5 + 267 = \boxed{272}$

Answer:  272