Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$

Solution: The scale factor is uniform in all dimensions, so the volume of the liquid is $\left(\frac{3}{4}\right)^{3}$ of the container. The remaining section of the volume is $\frac{1-\left(\frac{3}{4}\right)^{3}}{1}$ of the volume, and therefore $\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height when the vertex is at the top.
So, the liquid occupies $\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height, or $12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)$ . Thus $m+n+p=\boxed{052}$

Answer:  052
Problem: A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$

Solution: (Computational) The volume of a cone can be found by $V = \frac{\pi}{3}r^2h$ . In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'$
By similar triangles , we find that the dimensions of the liquid in the first cone to the entire cone is $\frac{3}{4}$ , and that $r' = \frac{rh'}{h}$ ; equating,
\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\ \frac{37}{64}r^2h &= \frac{r^2}{h^2}(h')^3 \\ h' &= \sqrt[3]{\frac{37}{64} \cdot 12^3} = 3\sqrt[3]{37}\end{align*}
Thus the answer is $12 - h' = 12-3\sqrt[3]{37}$ , and $m+n+p=\boxed{052}$

Answer:  052
Problem: A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$

Solution: From the formula $V=\frac{\pi r^2h}{3}$ , we can find that the volume of the container is $100\pi$ . The cone formed by the liquid is similar to the original, but scaled down by $\frac{3}{4}$ in all directions, so its volume is $100\pi*\frac{27}{64}=\frac{675\pi}{16}$ . The volume of the air in the container is the volume of the container minus the volume of the liquid, which is $\frac{925\pi}{16}$ , which is $\frac{37}{64}$ of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have $\sqrt[3]{\frac{37}{64}}=\frac{\sqrt[3]{37}}{4}$ of the height of the container. This means that the height of the liquid is $12\left(1-\frac{\sqrt[3]{37}}{4}\right)=12-3\sqrt[3]{37}$ inches, so our answer is $\boxed{052}$ . Solution by Zeroman

Answer:  052
Problem: A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$

Solution: We find that the volume of the cone is $100\pi$
The volume of the cone with height 9 is $\frac{675}{16}\pi$
The difference between the two volumes is $\frac{925}{16}\pi$ . Note that this is the volume of the cone essentially 'on top of' the frustum described in the problem when the liquid is held with the base horizontal.
We can express the volume as $x^2\pi\cdot\frac{12}{5}x\cdot\frac{1}{3}$ , where x is the radius of the cone.
Solving this equation, we get $x=\frac{5\sqrt[3]{37}}{4}$ . The height of this cone is $\frac{12}{5}$ of the radius.
Then we subtract the value from the height, 12, to get our answer: $12-3\sqrt[3]{37}$
Therefore, our answer is $12+3+37=\boxed{52}$ . ~MC413551

Answer:  52
Problem: The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$

Solution: Since $\log ab = \log a + \log b$ , we can reduce the equations to a more recognizable form:
\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{eqnarray*}
Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT , the above equations become (*)
\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1  \end{eqnarray*}
Small note from different author: $-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.$
From here, multiplying the three equations gives
\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}
Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$ . (Note from different author if you are confused on this step: if $\pm$ is positive then $\log y = \log 2 + 1 = \log 2 + \log 10 = \log 20,$ so $y=20.$ if $\pm$ is negative then $\log y = 1 - \log 2 = \log 10 - \log 2 = \log 5,$ so $y=5.$ ) This gives $y_1 = 20, y_2 = 5$ , and the answer is $y_1 + y_2 = \boxed{025}$

Answer:  025
Problem: The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$

Solution: Subtracting the second equation from the first equation yields \begin{align*} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{align*} If $1-\log y=0$ then $y=10$ . Substituting into the first equation yields $\log20000=4$ which is not possible.
If $\log\frac{x}{z}=0$ then $\frac{x}{z}=1\Longrightarrow x=z$ . Substituting into the third equation gets \begin{align*} \log x^2-(\log x)(\log x) &= 0 \\ \log x^2-\log x^x &= 0 \\ \log x^{2-x} &= 0 \\ x^{2-x} &= 1 \\ \end{align*} Thus either $x=1$ or $2-x=0\Longrightarrow x=2$ . (Note that here $x\neq-1$ since logarithm isn't defined for negative number.)
Substituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$ , respectively. Thus $y_1+y_2=\boxed{025}$

Answer:  025
Problem: The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$

Solution: Let $a = \log x$ $b = \log y$ and $c = \log z$ . Then the given equations become:
\begin{align*} \log 2 + a + b - ab = 1 \\ \log 2 + b + c - bc = 1 \\ a+c = ac \\ \end{align*}
Equating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \implies{a = c}$ . Plugging this result into the third equation, we get $c = 0$ or $2$ . Substituting each of these values of $c$ into the second equation, we get $b = 1 - \log 2$ and $b = 1 + \log 2$ . Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$ , respectively, so our answer is $\boxed{025}$

Answer:  025
Problem: sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$

Solution: Let the sum of all of the terms in the sequence be $\mathbb{S}$ . Then for each integer $k$ $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}
Now, substituting for $x_{50}$ , we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$ , and the answer is $75+98=\boxed{173}$

Answer:  173
Problem: sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$

Solution: Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$
In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\dfrac{50}{2}$
Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$

Answer:  173
Problem: Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$

Solution: Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$ , it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$ , where $-3 \le x,y \le 3$ . Thus every number in the form of $a/b$ will be expressed one time in the product
\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]
Using the formula for a geometric series , this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$ , and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$

Answer:  248
Problem: Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$

Solution: Essentially, the problem asks us to compute \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}\] which is pretty easy: \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}}    \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2}  \bigg) = 2480 + \frac{437}{1000}\] so our answer is $\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}$

Answer:  248
Problem: Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$

Solution: The sum is equivalent to $\sum_{i | 10^6}^{} \frac{i}{1000}$ Therefore, it's the sum of the factors of $10^6$ divided by $1000$ . The sum is $\frac{127 \times 19531}{1000}$ by the sum of factors formula. The answer is therefore $\boxed{248}$ after some computation.
- whatRthose

Answer:  248
Problem: Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$

Solution: We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor.
To begin with the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the "partner" of any divisor.
Now we now the amount to multiply the numerator if given number is in the denominator. Now we simply combine and reduce these groups. If the powers of 2 are on the denominator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property application. There is nothing complicated about this except to be careful.
Add all powers of 2: 15
Add their partners: 1875
Add all powers of 5: 156
Add their partners: 1248
Then, follow this formula: (sum of powers * sum opposite power's partners)+(sum of powers * sum opposite power's partners)
Or, $156*1875+15*1248$ $=311220$
Now, divide by 1000 to compensate for the denominator. $311.22$
Finally, we have to calculate the other list of fractions with 1 and another divisor. e.g. 1 - 250, 1 - 20 etc. (these all count)
This time we need to list all divisors of 1000, including 1. Remove all powers of 2 or 5, because we already included those in the other list. Now, notice there are two cases. 1: 1 is in the denominator, making the fraction an integer. 2: 1 is in the numerator.
Adding all the integers in the first case gives us 2169. The second case can actually be discarded, but still can be found. Realize that if we include the powers of 2 and 5, then the second case is (sum of all divisors)/1000. Remove all partners of powers of 2 and 5, and we'll get exactly 217/1000, or 0.22.
Finally, add all the numbers together: $311.22 +  2169 + 0.22 = 2480.44$
And divide by 10: $248.044$
After an odyssey of bashing: $\boxed{248}$

Answer:  248
Problem: Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$

Solution: \begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}
Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm
\[f(x) = f(352 + x)\]
So we need only to consider one period $f(0), f(1), ... f(351)$ , which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.
But we also know that $f(x) = f(46 - x) = f(398 - x)$ , so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated.  This gives a total of
\[352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{177}\]

Answer:  177
Problem: In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$ . All these circles are homothetic with respect to a center at $(5,0)$
Now consider the circle at $(0,0)$ . Draw a line tangent to it at $A$ and passing through $B (5,0)$ . By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$ . Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$ , so the slope of line $AB$ is $\frac{-7}{24}$ . Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$
This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$ . The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$ . By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$

Answer:  731
Problem: In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$

Solution: Let $\angle QPB=x^\circ$ . Because $\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\angle PAQ$ has the same measure. Because $\angle BPC$ is exterior to $\triangle BPA$ its measure is $3x$ . Let $\angle PBC = y^\circ$ . It follows that $\angle ACB = x+y$ and that $4x+2y=180^\circ$ . Two of the angles of triangle $APQ$ have measure $2x$ , and thus the measure of $\angle APQ$ is $2y$ . It follows that $AQ=2\cdot AP\cdot \sin y$ . Because $AB=AC$ and $AP=QB$ , it also follows that $AQ=PC$ . Now apply the Law of Sines to triangle $PBC$ to find \[\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}\] because $AP=BC$ . Hence $\sin 3x = \tfrac 12$ . Since $4x<180$ , this implies that $3x=30$ , i.e. $x=10$ . Thus $y=70$ and \[r=\frac{10+70}{2\cdot 70}=\frac 47,\] which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$

Answer:  571
Problem: In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$

Solution: Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$ . Then $PQBR$ is a rhombus , so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid . Since $\overline{PB}$ bisects $\angle QBR$ , it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$ . Thus $R$ lies on the perpendicular bisector of $\overline{BC}$ , and $BC = BR = RC$ . Hence $\triangle BCR$ is an equilateral triangle
Now $\angle ABR = \angle BAC = \angle ACR$ , and the sum of the angles in $\triangle ABC$ is $\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}$ . Then $\angle APQ = 140^{\circ}$ and $\angle ACB = 80^{\circ}$ , so the answer is $\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}$

Answer:  571
Problem: In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$

Solution: Again, construct $R$ as above.
Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$ , which means $x + y = 90$ $\triangle QBC$ is isosceles with $QB = BC$ , so $\angle BCQ = 90 - \frac {y}{2}$ .
Let $S$ be the intersection of $QC$ and $BP$ . Since $\angle BCQ = \angle BQC = \angle BRS$ $BCRS$ is cyclic , which means $\angle RBS = \angle RCS = x$ .
Since $APRB$ is an isosceles trapezoid, $BP = AR$ , but since $AR$ bisects $\angle BAC$ $\angle ABR = \angle ACR = 2x$
Therefore we have that $\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y$ .
We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \frac {y}{2} = y$ to get $x = 10$ and $y = 80$ $\angle APQ = 180 - 4x = 140$ $\angle ACB = 80$ , so $r = \frac {80}{140} = \frac {4}{7}$ $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$

Answer:  571
Problem: In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$

Solution: Let $\angle BAC= 2\theta$ and $AP=PQ=QB=BC=x$ $\triangle APQ$ is isosceles, so $AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)$ and $AB= AQ+x=x\left(3-4\sin^2\theta\right)$ $\triangle{ABC}$ is isosceles too, so $x=BC=2AB\sin\theta$ . Using the expression for $AB$ , we get \[1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta\] by the triple angle formula! Thus $\theta=10^\circ$ and $\angle A = 2\theta=20^\circ$ .
It follows now that $\angle APQ=140^\circ$ $\angle ACB=80^\circ$ , giving $r=\tfrac{4}{7}$ , which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$

Answer:  571
Problem: A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$

Solution: We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, $2(1024 - 1000) = 48$ , to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$ , meaning that there were $\boxed{927}$ cards are above the one labeled $1999$

Answer:  927