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Exception: ArrowInvalid
Message: Failed to parse string: 'Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
Some mathematics formulas that might be useful
- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
Part A. Single Accelerated Particle
Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}
Context question:
2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}
Context question:
3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}
Context question:
4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
Context answer:
\boxed{证明题}
Context question:
5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
Context answer:
\boxed{$\beta=\tanh \frac{g \tau}{c}$}
Context question:
6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.
Context answer:
\boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$}
Extra Supplementary Reading Materials:
Part B. Flight Time
The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A.
Context question:
1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}
Context question:
2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}
Extra Supplementary Reading Materials:
Part C. Minkowski Diagram
In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame.
$$
\begin{aligned}
& x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\
& c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\
& x^{\prime}=\gamma(x-\beta c t), \\
& c t^{\prime}=\gamma(c t-\beta x) . \\
& \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
\end{aligned}
$$
<image_1>
Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S.
To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below.
<image_2>
The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$.
$$
\begin{aligned}
\mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\
& =A C \cos \theta-A C \sin \theta \tan \theta \\
& =L \sqrt{1-\beta^{2}}
\end{aligned}
$$
Context question:
1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame.
Context answer:
\boxed{$\sqrt{1-\beta^{2}} L$}
Context question:
2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$.
Context answer:
\boxed{证明题}
Extra Supplementary Reading Materials:
Part D. Two Accelerated Particles
For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part.
Context question:
1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame.
Context answer:
\boxed{$\tau_{B}=\tau_{A}$}
Context question:
2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that
$$
\sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c}
$$
where $C_{1}$ is a constant. Determine $C_{1}$.
Context answer:
\boxed{$C_{1}=\frac{g L}{c^{2}}$}
Context question:
3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is
$$
\frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}
$$
where $C_{2}$ is a constant. Determine $C_{2}$.
Context answer:
\boxed{$C_{2}=\frac{g L}{c}$}
Extra Supplementary Reading Materials:
Part E. Uniformly Accelerated Frame
In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$.
Context question:
1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$.
Context answer:
\boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$}
Context question:
2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle.
Context answer:
\boxed{$g_{2}=\frac{g_{1}}{1+\frac{g_{1} L}{c^{2}}}$}
Context question:
3. What is the ratio of time rate of the second particle to the first particle $\frac{d \tau_{2}}{d \tau_{1}}$, according to the first particle.
Context answer:
\boxed{$1+\frac{g_{1} L}{c^{2}}$}
Extra Supplementary Reading Materials:
Part F. Correction for GPS
Part E indicates that the time rate of clocks at different altitude will not be the same, even though there is no relative movement between those clocks.
According to the equivalence principle in general relativity, an observer in a small closed room could not tell the difference between a gravity pull $g$ and the fictitious force from accelerated frame with acceleration $g$. So we can conclude that two clocks at different gravitational potential will have different rate.
Now let consider a GPS satellite that orbiting the Earth with a period of 12 hours.
Context question:
1. If the gravitational acceleration on the Earth's surface is $9.78 \mathrm{~m} \cdot \mathrm{s}^{-2}$, and the Earth's radius is $6380 \mathrm{~km}$, what is the radius of the GPS satellite orbit? What is the velocity of the satellite? Calculate the numerical values of the radius and the velocity.
Context answer:
\boxed{$r =2.66 \times 10^{7}$ and $v=3.87 \times 10^{3}$}
Context question:
2. After one day, the clock reading on the Earth surface and the satellite will differ due to both special and general relativistic effects. Calculate the difference due to each effect for one day. Calculate the total difference for one day. Which clock is faster, a clock on the Earth's surface or the satellite's clock?
Context answer:
by general relativity effect, after one day, the difference is $4.55 \times 10^{-5} \mathrm{~s}$
by special relativity effect, after one day, the difference is $-7.18 \times 10^{-6} \mathrm{~s}$
The satelite's clock is faster with total $\Delta \tau=\Delta \tau_{g}+\Delta \tau_{s}=3.83 \times 10^{-5} \mathrm{~s}$
' as a scalar of type double
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1831, in _prepare_split_single
writer.write_table(table)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/arrow_writer.py", line 644, in write_table
pa_table = table_cast(pa_table, self._schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2272, in table_cast
return cast_table_to_schema(table, schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2223, in cast_table_to_schema
arrays = [
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2224, in <listcomp>
cast_array_to_feature(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 1795, in wrapper
return pa.chunked_array([func(chunk, *args, **kwargs) for chunk in array.chunks])
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 1795, in <listcomp>
return pa.chunked_array([func(chunk, *args, **kwargs) for chunk in array.chunks])
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2086, in cast_array_to_feature
return array_cast(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 1797, in wrapper
return func(array, *args, **kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 1949, in array_cast
return array.cast(pa_type)
File "pyarrow/array.pxi", line 996, in pyarrow.lib.Array.cast
File "/src/services/worker/.venv/lib/python3.9/site-packages/pyarrow/compute.py", line 404, in cast
return call_function("cast", [arr], options, memory_pool)
File "pyarrow/_compute.pyx", line 590, in pyarrow._compute.call_function
File "pyarrow/_compute.pyx", line 385, in pyarrow._compute.Function.call
File "pyarrow/error.pxi", line 154, in pyarrow.lib.pyarrow_internal_check_status
File "pyarrow/error.pxi", line 91, in pyarrow.lib.check_status
pyarrow.lib.ArrowInvalid: Failed to parse string: 'Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
Some mathematics formulas that might be useful
- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
Part A. Single Accelerated Particle
Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}
Context question:
2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}
Context question:
3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}
Context question:
4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
Context answer:
\boxed{证明题}
Context question:
5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
Context answer:
\boxed{$\beta=\tanh \frac{g \tau}{c}$}
Context question:
6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.
Context answer:
\boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$}
Extra Supplementary Reading Materials:
Part B. Flight Time
The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A.
Context question:
1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}
Context question:
2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}
Extra Supplementary Reading Materials:
Part C. Minkowski Diagram
In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame.
$$
\begin{aligned}
& x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\
& c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\
& x^{\prime}=\gamma(x-\beta c t), \\
& c t^{\prime}=\gamma(c t-\beta x) . \\
& \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
\end{aligned}
$$
<image_1>
Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S.
To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below.
<image_2>
The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$.
$$
\begin{aligned}
\mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\
& =A C \cos \theta-A C \sin \theta \tan \theta \\
& =L \sqrt{1-\beta^{2}}
\end{aligned}
$$
Context question:
1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame.
Context answer:
\boxed{$\sqrt{1-\beta^{2}} L$}
Context question:
2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$.
Context answer:
\boxed{证明题}
Extra Supplementary Reading Materials:
Part D. Two Accelerated Particles
For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part.
Context question:
1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame.
Context answer:
\boxed{$\tau_{B}=\tau_{A}$}
Context question:
2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that
$$
\sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c}
$$
where $C_{1}$ is a constant. Determine $C_{1}$.
Context answer:
\boxed{$C_{1}=\frac{g L}{c^{2}}$}
Context question:
3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is
$$
\frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)}
$$
where $C_{2}$ is a constant. Determine $C_{2}$.
Context answer:
\boxed{$C_{2}=\frac{g L}{c}$}
Extra Supplementary Reading Materials:
Part E. Uniformly Accelerated Frame
In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$.
Context question:
1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$.
Context answer:
\boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$}
Context question:
2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle.
Context answer:
\boxed{$g_{2}=\frac{g_{1}}{1+\frac{g_{1} L}{c^{2}}}$}
Context question:
3. What is the ratio of time rate of the second particle to the first particle $\frac{d \tau_{2}}{d \tau_{1}}$, according to the first particle.
Context answer:
\boxed{$1+\frac{g_{1} L}{c^{2}}$}
Extra Supplementary Reading Materials:
Part F. Correction for GPS
Part E indicates that the time rate of clocks at different altitude will not be the same, even though there is no relative movement between those clocks.
According to the equivalence principle in general relativity, an observer in a small closed room could not tell the difference between a gravity pull $g$ and the fictitious force from accelerated frame with acceleration $g$. So we can conclude that two clocks at different gravitational potential will have different rate.
Now let consider a GPS satellite that orbiting the Earth with a period of 12 hours.
Context question:
1. If the gravitational acceleration on the Earth's surface is $9.78 \mathrm{~m} \cdot \mathrm{s}^{-2}$, and the Earth's radius is $6380 \mathrm{~km}$, what is the radius of the GPS satellite orbit? What is the velocity of the satellite? Calculate the numerical values of the radius and the velocity.
Context answer:
\boxed{$r =2.66 \times 10^{7}$ and $v=3.87 \times 10^{3}$}
Context question:
2. After one day, the clock reading on the Earth surface and the satellite will differ due to both special and general relativistic effects. Calculate the difference due to each effect for one day. Calculate the total difference for one day. Which clock is faster, a clock on the Earth's surface or the satellite's clock?
Context answer:
by general relativity effect, after one day, the difference is $4.55 \times 10^{-5} \mathrm{~s}$
by special relativity effect, after one day, the difference is $-7.18 \times 10^{-6} \mathrm{~s}$
The satelite's clock is faster with total $\Delta \tau=\Delta \tau_{g}+\Delta \tau_{s}=3.83 \times 10^{-5} \mathrm{~s}$
' as a scalar of type double
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1456, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1055, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 894, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 970, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1702, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1858, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
datasets.exceptions.DatasetGenerationError: An error occurred while generating the datasetNeed help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
id
int64 | question
string | solution
string | final_answer
string | context
null | image
string | modality
string | difficulty
string | is_multiple_answer
bool | unit
string | answer_type
string | error
float64 | question_type
string | subfield
string | subject
string | language
string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Turbo the snail sits on a point on a circle with circumference 1. Given an infinite sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$. Turbo successively crawls distances $c_{1}, c_{2}, c_{3}, \ldots$ around the circle, each time choosing to crawl either clockwise or counterclockwise.
For example, if the sequence $c_{1}, c_{2}, c_{3}, \ldots$ is $0.4,0.6,0.3, \ldots$, then Turbo may start crawling as follows:
<image_1>
Determine the largest constant $C>0$ with the following property: for every sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$ with $c_{i}<C$ for all $i$, Turbo can (after studying the sequence) ensure that there is some point on the circle that it will never visit or crawl across.
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['The largest possible $C$ is $C=\\frac{1}{2}$.\n\nFor $0<C \\leqslant \\frac{1}{2}$, Turbo can simply choose an arbitrary point $P$ (different from its starting point) to avoid. When Turbo is at an arbitrary point $A$ different from $P$, the two arcs $A P$ have total length 1; therefore, the larger of the two the arcs (or either arc in case $A$ is diametrically opposite to $P$ ) must have length $\\geqslant \\frac{1}{2}$. By always choosing this larger arc (or either arc in case $A$ is diametrically opposite to $P$ ), Turbo will manage to avoid the point $P$ forever.\n\nFor $C>\\frac{1}{2}$, we write $C=\\frac{1}{2}+a$ with $a>0$, and we choose the sequence\n\n$$\n\\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\ldots\n$$\n\nIn other words, $c_{i}=\\frac{1}{2}$ if $i$ is odd and $c_{i}=\\frac{1+a}{2}<C$ when $i$ is even. We claim Turbo must eventually visit all points on the circle. This is clear when it crawls in the same direction two times in a row; after all, we have $c_{i}+c_{i+1}>1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\\frac{1}{2}$ clockwise followed by a distance $\\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\\frac{a}{2}$ counterclockwise. Because $\\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\\frac{a}{2} \\cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\\frac{a}{2} \\cdot N>1$, meaning that it must have crawled over all points on the circle.\n\nNote: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $0<x, y<C$, such that $x+y \\geqslant 1$, and $x \\neq y$ satisfies the conditions with the same argument. There might be even more possible examples.'
"To show that $C\\le \\frac12$\n\nWe consider the following related problem:\n\nWe assume instead that the snail Chet is moving left and right on the real line. Find the size $M$ of the smallest (closed) interval, that we cannot force Chet out of, using a sequence of real numbers $d_{i}$ with $0<d_{i}<1$ for all $i$.\n\nThen $C=1 / M$. Indeed if for every sequence $c_{1}, c_{2}, \\ldots$, with $c_{i}<C$ there exists a point that Turbo can avoid, then the circle can be cut open at the avoided point and mapped to an interval of size $M$ such that Chet can stay inside this interval for any sequence of the from $c_{1} / C, c_{2} / C, \\ldots$, see Figure 5 . However, all sequences $d_{1}, d_{2}, \\ldots$ with $d_{i}<1$ can be written in this form. Similarly if for every sequence $d_{1}, d_{2}, \\ldots$, there exists an interval of length smaller or equal $M$ that we cannot force Chet out of, this projects to a subset of the circle, that we cannot force Turbo out of using any sequence of the form $d_{1} / M, d_{2} / M, \\ldots$. These are again exactly all the sequences with elements in $[0, C)$.\n\n<img_3381>\n\nFigure 5: Chet and Turbo equivalence\n\nClaim: $M \\geqslant 2$.\n\nProof. Suppose not, so $M<2$. Say $M=2-2 \\varepsilon$ for some $\\varepsilon>0$ and let $[-1+\\varepsilon, 1-\\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\\pm(1-\\varepsilon)$. In partiular, we can force Chet out of $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \\ldots$ for which Chet has to leave $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, which means he ends up either in the interval $\\left[-1+\\varepsilon,-1+\\frac{4}{3} \\varepsilon\\right)$ or in the interval $\\left(1-\\frac{4}{3} \\varepsilon, 1-\\varepsilon\\right]$.\n\nNow consider the sequence,\n\n$$\nd_{1}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{2}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{3}, \\ldots\n$$\n\nobtained by adding the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \\varepsilon$ and Chet would leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. It follows that the $\\left(1-\\frac{7}{6} \\varepsilon\\right)$ 's and the $\\left(1-\\frac{2}{3} \\varepsilon\\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.\n\nIf Chet is outside the interval $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, then Chet has to move $1-\\frac{7}{6} \\varepsilon$ towards 0 , and ends in $\\left[-\\frac{1}{6} \\varepsilon, \\frac{1}{6} \\varepsilon\\right]$. Chet then has to move by $1-\\frac{2}{3} \\varepsilon$, which means that he has to leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed the absolute value of the final position is at least $1-\\frac{5}{6} \\varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\\varepsilon, 1-\\varepsilon]$. Hence $M \\geqslant 2$ as needed."]
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['$\\frac{1}{2}$']
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']
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Multimodal
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Competition
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Numerical
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Open-ended
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Geometry
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Math
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English
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1
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In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places.
<image_1>
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['Let the length of $A E=E C$ be $x$.\n\nThen $A F=x-25$.\n\nIn, $\\triangle B C F, \\frac{x+25}{B F}=\\tan \\left(59^{\\circ}\\right)$.\n\nIn $\\triangle A B F, \\frac{x-25}{B F}=\\tan \\left(41^{\\circ}\\right)$.\n\nSolving for $B F$ in these two equations and equating,\n\n$$\nB F=\\frac{x+25}{\\tan 59^{\\circ}}=\\frac{x-25}{\\tan 41^{\\circ}}\n$$\n\nso $\\quad\\left(\\tan 41^{\\circ}\\right)(x+25)=\\left(\\tan 59^{\\circ}\\right)(x-25)$\n\n$$\n\\begin{aligned}\n25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right) & =x\\left(\\tan 59^{\\circ}-\\tan 41^{\\circ}\\right) \\\\\nx & =\\frac{25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right)}{\\tan 59^{\\circ}-\\tan 41^{\\circ}} \\\\\nx & \\doteq 79.67 .\n\\end{aligned}\n$$\n\n<img_3420>\n\nTherefore the length of $A E$ is 79.67 .']
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['79.67']
| null |
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Multimodal
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Competition
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Numerical
| 0.1
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Open-ended
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Geometry
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Math
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English
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2
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In triangle $A B C, A B=B C=25$ and $A C=30$. The circle with diameter $B C$ intersects $A B$ at $X$ and $A C$ at $Y$. Determine the length of $X Y$.
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['Join $B Y$. Since $B C$ is a diameter, then $\\angle B Y C=90^{\\circ}$. Since $A B=B C, \\triangle A B C$ is isosceles and $B Y$ is an altitude in $\\triangle A B C$, then $A Y=Y C=15$.\n\nLet $\\angle B A C=\\theta$.\n\nSince $\\triangle A B C$ is isosceles, $\\angle B C A=\\theta$.\n\nSince $B C Y X$ is cyclic, $\\angle B X Y=180-\\theta$ and so $\\angle A X Y=\\theta$.\n\n<img_3475>\n\nThus $\\triangle A X Y$ is isosceles and so $X Y=A Y=15$.\n\nTherefore $X Y=15$.'
'Join $B Y . \\angle B Y C=90^{\\circ}$ since it is inscribed in a semicircle.\n\n$\\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.\n\nTherefore $B Y=\\sqrt{25^{2}-15^{2}}=20$.\n\nJoin $C X . \\angle C X B=90^{\\circ}$ since it is also inscribed in a semicircle.\n\n<img_3739>\n\nThe area of $\\triangle A B C$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A C)(B Y) & =\\frac{1}{2}(A B)(C X) \\\\\n\\frac{1}{2}(30)(20) & =\\frac{1}{2}(25)(C X) \\\\\nC X & =\\frac{600}{25}=24 .\n\\end{aligned}\n$$\n\nFrom $\\triangle A B Y$ we conclude that $\\cos \\angle A B Y=\\frac{B Y}{A B}=\\frac{20}{25}=\\frac{4}{5}$.\n\nIn $\\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \\cos \\angle X B Y$.\n\nNow (by Pythagoras $\\triangle B X C$ ),\n\n\n\n$$\n\\begin{aligned}\nB X^{2} & =B C^{2}-C X^{2} \\\\\n& =25^{2}-24^{2} \\\\\n& =49 \\\\\nB X & =7 .\n\\end{aligned}\n$$\n\nTherefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \\frac{4}{5}$\n\n$$\n=49+400-224\n$$\n\n$$\n=225 \\text {. }\n$$\n\nTherefore $X Y=15$.']
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['15']
| null |
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Multimodal
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Competition
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Numerical
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Open-ended
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Geometry
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Math
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English
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3
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Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$.
<image_1>
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['Placing the information on the coordinate axes, the diagram is indicated to the right.\n\nWe note that $P$ has coordinates $(a, b)$.\n\nBy symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.\n\nSince $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$\n\n$$\n\\begin{aligned}\n& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\\\\n& \\left(a-\\frac{4}{3}\\right)^{2}+\\left(b-\\frac{4}{3}\\right)^{2}=\\frac{8}{9}\n\\end{aligned}\n$$\n\n<img_3483>\n\n$P$ is on a circle with centre $O\\left(\\frac{4}{3}, \\frac{4}{3}\\right)$ with $r=\\frac{2}{3} \\sqrt{2}$.\n\nThe minimum angle for $\\theta$ occurs when $D P$ is tangent to the circle.\n\nSo we have the diagram noted to the right.\n\nSince $O D$ makes an angle of $45^{\\circ}$ with the $x$-axis then $\\angle P D O=45-\\theta$ and $O D=\\frac{4}{3} \\sqrt{2}$.\n\nTherefore $\\sin (45-\\theta)=\\frac{\\frac{2}{3} \\sqrt{2}}{\\frac{4}{3} \\sqrt{2}}=\\frac{1}{2}$ which means $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.\n\nThus the minimum value for $\\theta$ is $15^{\\circ}$.\n\n<img_3977>'
'Let $A B=B C=C D=D A=1$.\n\nJoin $D$ to $B$. Let $\\angle A D P=\\theta$. Therefore, $\\angle P D B=45-\\theta$.\n\nLet $P D=a$ and $P B=b$ and $P Q=\\frac{a}{2}$.\n\n\n\nWe now establish a relationship between $a$ and $b$.\n\nIn $\\triangle P D B, b^{2}=a^{2}+2-2(a)(\\sqrt{2}) \\cos (45-\\theta)$\n\n$$\n\\text { or, } \\quad \\cos (45-\\theta)=\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}\n$$\n\n<img_3307>\n\nIn $\\triangle P D R,\\left(\\frac{a}{2}\\right)^{2}=a^{2}+\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}-2 a \\frac{\\sqrt{2}}{2} \\cos (45-\\theta)$\n\nor, $\\cos (45-\\theta)=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$\n\nComparing (1) and (2) gives, $\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$.\n\nSimplifying this, $a^{2}+2 b^{2}=2$\n\n$$\n\\text { or, } \\quad b^{2}=\\frac{2-a^{2}}{2}\n$$\n\nNow $\\cos (45-\\theta)=\\frac{a^{2}+2-\\left(\\frac{2-a^{2}}{2}\\right)}{2 a \\sqrt{2}}=\\frac{1}{4 \\sqrt{2}}\\left(3 a+\\frac{2}{a}\\right)$.\n\nNow considering $3 a+\\frac{2}{a}$, we know $\\left(\\sqrt{3 a}-\\sqrt{\\frac{2}{a}}\\right)^{2} \\geq 0$\n\n$$\n\\text { or, } \\quad 3 a+\\frac{2}{a} \\geq 2 \\sqrt{6}\n$$\n\nThus, $\\cos (45-\\theta) \\geq \\frac{1}{4 \\sqrt{2}}(2 \\sqrt{6})=\\frac{\\sqrt{3}}{2}$\n\n$$\n\\cos (45-\\theta) \\geq \\frac{\\sqrt{3}}{2}\n$$\n\n$\\cos (45-\\theta)$ has a minimum value for $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.'
'Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\\angle A D P=\\theta$.\n\nBy interior alternate angles, $\\angle P=\\angle Q$ and $\\angle P D M=\\angle Q B M$.\n\nThus $\\triangle P D M \\cong \\triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.\n\nSo $M$ is the midpoint of $B D$ and the centre of the square.\n\n\n\nWithout loss of generality, let $P M=1$. Then $P D=2$.\n\nSince $\\theta+\\alpha=45^{\\circ}$ (see diagram), $\\theta$ will be minimized when $\\alpha$ is maximized.\n\n<img_3749>\n\nConsider $\\triangle P M D$.\n\nUsing the sine law, $\\frac{\\sin \\alpha}{1}=\\frac{\\sin (\\angle P M D)}{2}$.\n\nTo maximize $\\alpha$, we maximize $\\sin \\alpha$.\n\nBut $\\sin \\alpha=\\frac{\\sin (\\angle P M D)}{2}$, so it is maximized when $\\sin (\\angle P M D)=1$.\n\nIn this case, $\\sin \\alpha=\\frac{1}{2}$, so $\\alpha=30^{\\circ}$.\n\nTherefore, $\\theta=45^{\\circ}-30^{\\circ}=15^{\\circ}$, and so the minimum value of $\\theta$ is $15^{\\circ}$.'
'We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.\n\nLet $P D=P Q=Q B=a$, and $\\angle A D P=\\theta$.\n\nDrop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.\n\nThen $P X=a \\sin \\theta, D X=a \\cos \\theta$.\n\nTherefore the coordinates of $P$ are $(a \\sin \\theta, a \\cos \\theta)$.\n\nSince $P D \\| B Q$, then $\\angle Q B C=\\theta$.\n\nSo by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \\sin \\theta, 1+a \\cos \\theta)$.\n\n<img_3706>\n\nNow $(P Q)^{2}=a^{2}$, so $(1-2 a \\sin \\theta)^{2}+(1-2 a \\cos \\theta)^{2}=a^{2}$\n\n$$\n2+4 a^{2} \\sin ^{2} \\theta+4 a^{2} \\cos ^{2} \\theta-4 a(\\sin \\theta+\\cos \\theta)=a^{2}\n$$\n\n\n\n$$\n\\begin{aligned}\n2+4 a^{2}-a^{2} & =4 a(\\sin \\theta+\\cos \\theta) \\\\\n\\frac{2+3 a^{2}}{4 a} & =\\sin \\theta+\\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\frac{1}{\\sqrt{2}} \\sin \\theta+\\frac{1}{\\sqrt{2}} \\cos \\theta=\\cos \\left(45^{\\circ}\\right) \\sin \\theta+\\sin \\left(45^{\\circ}\\right) \\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\sin \\left(\\theta+45^{\\circ}\\right)\n\\end{aligned}\n$$\n\n$$\n\\text { Now } \\begin{aligned}\n\\left(a-\\sqrt{\\frac{2}{3}}\\right)^{2} & \\geq 0 \\\\\na^{2}-2 a \\sqrt{\\frac{2}{3}}+\\frac{2}{3} & \\geq 0 \\\\\n3 a^{2}-2 a \\sqrt{6}+2 & \\geq 0 \\\\\n3 a^{2}+2 & \\geq 2 a \\sqrt{6} \\\\\n\\frac{3 a^{2}+2}{4 \\sqrt{2} a} & \\geq \\frac{\\sqrt{3}}{2}\n\\end{aligned}\n$$\n\nand equality occurs when $a=\\sqrt{\\frac{2}{3}}$.\n\nSo $\\sin \\left(\\theta+45^{\\circ}\\right) \\geq \\frac{\\sqrt{3}}{2}$ and thus since $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, then $\\theta+45^{\\circ} \\geq 60^{\\circ}$ or $\\theta \\geq 15^{\\circ}$.\n\nTherefore the minimum possible value of $\\angle A D P$ is $15^{\\circ}$.']
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['$15$']
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Multimodal
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Competition
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^{\circ}
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Numerical
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Open-ended
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Geometry
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Math
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English
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4
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In the diagram, $\angle E A D=90^{\circ}, \angle A C D=90^{\circ}$, and $\angle A B C=90^{\circ}$. Also, $E D=13, E A=12$, $D C=4$, and $C B=2$. Determine the length of $A B$.
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['By the Pythagorean Theorem in $\\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\\sqrt{169-144}=5$, since $A D>0$.\n\nBy the Pythagorean Theorem in $\\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\\sqrt{25-16}=3$, since $A C>0$.\n\n(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)\n\nBy the Pythagorean Theorem in $\\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\\sqrt{9-4}=\\sqrt{5}$, since $A B>0$.']
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['$\\sqrt{5}$']
| null |
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Multimodal
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Competition
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Numerical
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Open-ended
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Geometry
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Math
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English
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5
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In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$.
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['Join $B$ to $D$.\n\n<img_3655>\n\nConsider $\\triangle C B D$.\n\nSince $C B=C D$, then $\\angle C B D=\\angle C D B=\\frac{1}{2}\\left(180^{\\circ}-\\angle B C D\\right)=\\frac{1}{2}\\left(180^{\\circ}-60^{\\circ}\\right)=60^{\\circ}$.\n\nTherefore, $\\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.\n\nConsider $\\triangle D B A$.\n\nNote that $\\angle D B A=90^{\\circ}-\\angle C B D=90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\nSince $B D=B A=6$, then $\\angle B D A=\\angle B A D=\\frac{1}{2}\\left(180^{\\circ}-\\angle D B A\\right)=\\frac{1}{2}\\left(180^{\\circ}-30^{\\circ}\\right)=75^{\\circ}$.\n\nWe calculate the length of $A D$.\n\nMethod 1\n\nBy the Sine Law in $\\triangle D B A$, we have $\\frac{A D}{\\sin (\\angle D B A)}=\\frac{B A}{\\sin (\\angle B D A)}$.\n\nTherefore, $A D=\\frac{6 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(75^{\\circ}\\right)}=\\frac{6 \\times \\frac{1}{2}}{\\sin \\left(75^{\\circ}\\right)}=\\frac{3}{\\sin \\left(75^{\\circ}\\right)}$.\n\nMethod 2\n\nIf we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\\triangle B D A$ is isosceles. Thus, $A D=2 A P$.\n\nAlso, $B P$ bisects $\\angle D B A$, so $\\angle A B P=15^{\\circ}$.\n\nNow, $A P=B A \\sin (\\angle A B P)=6 \\sin \\left(15^{\\circ}\\right)$.\n\nTherefore, $A D=2 A P=12 \\sin \\left(15^{\\circ}\\right)$.\n\nMethod 3\n\nBy the Cosine Law in $\\triangle D B A$,\n\n$$\n\\begin{aligned}\nA D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \\cos (\\angle A B D) \\\\\n& =6^{2}+6^{2}-2(6)(6) \\cos \\left(30^{\\circ}\\right) \\\\\n& =72-72\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\\n& =72-36 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$ since $A D>0$.'
'Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.\n\n<img_3835>\n\nThen $C Q=C D \\cos (\\angle D C Q)=6 \\cos \\left(60^{\\circ}\\right)=6 \\times \\frac{1}{2}=3$.\n\nAlso, $D Q=C D \\sin (\\angle D C Q)=6 \\sin \\left(60^{\\circ}\\right)=6 \\times \\frac{\\sqrt{3}}{2}=3 \\sqrt{3}$.\n\nSince $B C=6$, then $B Q=B C-C Q=6-3=3$.\n\nNow quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.\n\nThus, $R D=B Q=3$ and $R B=D Q=3 \\sqrt{3}$.\n\nSince $A B=6$, then $A R=A B-R B=6-3 \\sqrt{3}$.\n\nSince $\\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain\n\n$$\nA D=\\sqrt{R D^{2}+A R^{2}}=\\sqrt{3^{2}+(6-3 \\sqrt{3})^{2}}=\\sqrt{9+36-36 \\sqrt{3}+27}=\\sqrt{72-36 \\sqrt{3}}\n$$\n\nwhich we can rewrite as $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$.']
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['$6\\sqrt{2-\\sqrt{3}}$']
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Multimodal
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Competition
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| null |
Numerical
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Open-ended
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Geometry
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Math
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English
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6
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A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0<k<4$. If the area of the triangle is 8 , determine the value of $k$.
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['We "complete the rectangle" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.\n\n<img_3215>\n\n\n\nSince $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.\n\nSince $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .\n\nSince $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .\n\nTherefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \\times 5=20$.\n\nNow rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .\n\nLet us examine each of these triangles:\n\n- $\\triangle A B C$ has area 8 (given information)\n- $\\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\\frac{1}{2} \\times 4 \\times 3=6$.\n- $\\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\\frac{1}{2} \\times k \\times 2=k$.\n- $\\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\\frac{1}{2} \\times(4-k) \\times 5=10-\\frac{5}{2} k$.\n\nSince the sum of the areas of these triangles is 20 , then $8+6+k+10-\\frac{5}{2} k=20$ or $4=\\frac{3}{2} k$ and so $k=\\frac{8}{3}$.']
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['$\\frac{8}{3}$']
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Multimodal
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Competition
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Numerical
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Open-ended
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Geometry
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Math
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English
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7
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A helicopter hovers at point $H$, directly above point $P$ on level ground. Lloyd sits on the ground at a point $L$ where $\angle H L P=60^{\circ}$. A ball is droppped from the helicopter. When the ball is at point $B, 400 \mathrm{~m}$ directly below the helicopter, $\angle B L P=30^{\circ}$. What is the distance between $L$ and $P$ ?
<image_1>
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['Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\n<img_3808>\n\nTherefore, $\\triangle H B L$ is isosceles and $B L=H B=400 \\mathrm{~m}$.\n\nIn $\\triangle B L P, B L=400 \\mathrm{~m}$ and $\\angle B L P=30^{\\circ}$, so $L P=B L \\cos \\left(30^{\\circ}\\right)=400\\left(\\frac{\\sqrt{3}}{2}\\right)=200 \\sqrt{3}$ m.\n\nTherefore, the distance between $L$ and $P$ is $200 \\sqrt{3} \\mathrm{~m}$.'
'Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$. Also, $\\angle L B P=60^{\\circ}$.\n\nLet $L P=x$.\n\n<img_3794>\n\nSince $\\triangle B L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $B P: L P=1: \\sqrt{3}$, so $B P=\\frac{1}{\\sqrt{3}} L P=\\frac{1}{\\sqrt{3}} x$.\n\n\n\nSince $\\triangle H L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $H P: L P=\\sqrt{3}: 1$, so $H P=\\sqrt{3} L P=\\sqrt{3} x$.\n\nBut $H P=H B+B P$ so\n\n$$\n\\begin{aligned}\n\\sqrt{3} x & =400+\\frac{1}{\\sqrt{3}} x \\\\\n3 x & =400 \\sqrt{3}+x \\\\\n2 x & =400 \\sqrt{3} \\\\\nx & =200 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, the distance from $L$ to $P$ is $200 \\sqrt{3} \\mathrm{~m}$.']
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['$200 \\sqrt{3}$']
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Multimodal
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Competition
| false
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m
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Numerical
| null |
Open-ended
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Geometry
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Math
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English
|
8
| "In the diagram, $A B C D$ is a quadrilateral in which $\\angle A+\\angle C=180^{\\circ}$. What is t(...TRUNCATED)
| "['In order to determine $C D$, we must determine one of the angles (or at least some information ab(...TRUNCATED)
|
['5']
| null | "['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBC(...TRUNCATED)
|
Multimodal
|
Competition
| false
| null |
Numerical
| null |
Open-ended
|
Geometry
|
Math
|
English
|
9
| "In the diagram, the parabola\n\n$$\ny=-\\frac{1}{4}(x-r)(x-s)\n$$\n\nintersects the axes at three p(...TRUNCATED)
| "['From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.\\n\\n\\n\\n<img_3883(...TRUNCATED)
|
['$4,(4,16)$']
| null | "['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBC(...TRUNCATED)
|
Multimodal
|
Competition
| true
| null |
Numerical,Tuple
| null |
Open-ended
|
Geometry
|
Math
|
English
|
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