Right cone, you are at A and need to complete a revolution before reaching the bottom B. What is shortest...
$begingroup$
You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.
The line AB must do one full revolution of the mountain and is the shortest distance.
What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?
When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out?
I am not sure where to begin with uphill/downhill part.
geometry trigonometry
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
$endgroup$
add a comment |
$begingroup$
You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.
The line AB must do one full revolution of the mountain and is the shortest distance.
What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?
When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out?
I am not sure where to begin with uphill/downhill part.
geometry trigonometry
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
$endgroup$
$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
$endgroup$
– Börge
Dec 13 '18 at 9:32
$begingroup$
I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
$endgroup$
– Rahul
Dec 13 '18 at 19:33
add a comment |
$begingroup$
You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.
The line AB must do one full revolution of the mountain and is the shortest distance.
What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?
When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out?
I am not sure where to begin with uphill/downhill part.
geometry trigonometry
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
$endgroup$
You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.
The line AB must do one full revolution of the mountain and is the shortest distance.
What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?
When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out?
I am not sure where to begin with uphill/downhill part.
geometry trigonometry
geometry trigonometry
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
edited Dec 13 '18 at 9:04
Ben Franks
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
asked Dec 13 '18 at 8:46
Ben FranksBen Franks
263110
263110
$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
$endgroup$
– Börge
Dec 13 '18 at 9:32
$begingroup$
I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
$endgroup$
– Rahul
Dec 13 '18 at 19:33
add a comment |
$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
$endgroup$
– Börge
Dec 13 '18 at 9:32
$begingroup$
I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
$endgroup$
– Rahul
Dec 13 '18 at 19:33
$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
$endgroup$
– Börge
Dec 13 '18 at 9:32
$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
$endgroup$
– Börge
Dec 13 '18 at 9:32
$begingroup$
I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
$endgroup$
– Rahul
Dec 13 '18 at 19:33
$begingroup$
I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
$endgroup$
– Rahul
Dec 13 '18 at 19:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $theta$ of the circular sector.
We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $theta s = 2pi r$ giving: $$theta = frac{2 pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd cos(frac{2 pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$text{Area} = frac{1}{2}cdot s cdot d cdot sin(theta) = frac{1}{2}cdot x cdot h$$
giving $$h = frac{s cdot d cdot sin(theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = sqrt{d^2-h^2}$$
answered Dec 13 '18 at 17:10
JensJens
3,85021030
$endgroup$
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
add a comment |
$begingroup$
Hint:
Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill
But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
$endgroup$
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
add a comment |
$begingroup$
One might solve this numerically using a path of n steps.
Let a be the direct distance from the cone peak of point A.
Then the minimal distance between A and B is given by
the solution of
Components x[i] give the path from B to A which might get up and down.
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
$endgroup$
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $theta$ of the circular sector.
We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $theta s = 2pi r$ giving: $$theta = frac{2 pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd cos(frac{2 pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$text{Area} = frac{1}{2}cdot s cdot d cdot sin(theta) = frac{1}{2}cdot x cdot h$$
giving $$h = frac{s cdot d cdot sin(theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = sqrt{d^2-h^2}$$
answered Dec 13 '18 at 17:10
JensJens
3,85021030
$endgroup$
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
add a comment |
$begingroup$
Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $theta$ of the circular sector.
We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $theta s = 2pi r$ giving: $$theta = frac{2 pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd cos(frac{2 pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$text{Area} = frac{1}{2}cdot s cdot d cdot sin(theta) = frac{1}{2}cdot x cdot h$$
giving $$h = frac{s cdot d cdot sin(theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = sqrt{d^2-h^2}$$
answered Dec 13 '18 at 17:10
JensJens
3,85021030
$endgroup$
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
add a comment |
$begingroup$
Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $theta$ of the circular sector.
We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $theta s = 2pi r$ giving: $$theta = frac{2 pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd cos(frac{2 pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$text{Area} = frac{1}{2}cdot s cdot d cdot sin(theta) = frac{1}{2}cdot x cdot h$$
giving $$h = frac{s cdot d cdot sin(theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = sqrt{d^2-h^2}$$
answered Dec 13 '18 at 17:10
JensJens
3,85021030
$endgroup$
Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $theta$ of the circular sector.
We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $theta s = 2pi r$ giving: $$theta = frac{2 pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd cos(frac{2 pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$text{Area} = frac{1}{2}cdot s cdot d cdot sin(theta) = frac{1}{2}cdot x cdot h$$
giving $$h = frac{s cdot d cdot sin(theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = sqrt{d^2-h^2}$$
answered Dec 13 '18 at 17:10
JensJens
3,85021030
edited Dec 13 '18 at 19:29
answered Dec 13 '18 at 17:10
JensJens
3,85021030
answered Dec 13 '18 at 17:10
JensJens
3,85021030
answered Dec 13 '18 at 17:10
JensJens
3,85021030
3,85021030
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
add a comment |
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
$begingroup$
The point on line AB where up and down change is the foot point of the height of the triangle.
$endgroup$
– fwgb
Dec 13 '18 at 18:36
add a comment |
$begingroup$
Hint:
Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill
But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
$endgroup$
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
add a comment |
$begingroup$
Hint:
Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill
But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
$endgroup$
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
add a comment |
$begingroup$
Hint:
Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill
But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
$endgroup$
Hint:
Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill
But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
edited Dec 13 '18 at 11:43
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
answered Dec 13 '18 at 8:58
HenryHenry
100k481168
100k481168
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
add a comment |
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
$begingroup$
the example I have is of the first depiction. I will copy that image and show in OP.
$endgroup$
– Ben Franks
Dec 13 '18 at 9:02
1
1
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
If you read the question post, you will see that this is exactly what the OP has already done. The question is about calculating the distance AB in your first figure, and telling which parts of the path are uphill and which are downhill.
$endgroup$
– Arthur
Dec 13 '18 at 9:02
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
@Arthur between my answer and your comment, the original post was edited to to include my diagram. I will add a further hint on uphill in the first case
$endgroup$
– Henry
Dec 13 '18 at 11:37
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
To include your diagram, sure. But not the idea of "opening" the cone. That was there from the start.
$endgroup$
– Arthur
Dec 13 '18 at 11:59
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
$begingroup$
@Henry thanks for the help managed to solve it from your hints, especially needed the uphill one. +1.
$endgroup$
– Ben Franks
Dec 14 '18 at 10:25
add a comment |
$begingroup$
One might solve this numerically using a path of n steps.
Let a be the direct distance from the cone peak of point A.
Then the minimal distance between A and B is given by
the solution of
Components x[i] give the path from B to A which might get up and down.
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
$endgroup$
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
add a comment |
$begingroup$
One might solve this numerically using a path of n steps.
Let a be the direct distance from the cone peak of point A.
Then the minimal distance between A and B is given by
the solution of
Components x[i] give the path from B to A which might get up and down.
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
$endgroup$
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
add a comment |
$begingroup$
One might solve this numerically using a path of n steps.
Let a be the direct distance from the cone peak of point A.
Then the minimal distance between A and B is given by
the solution of
Components x[i] give the path from B to A which might get up and down.
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
$endgroup$
One might solve this numerically using a path of n steps.
Let a be the direct distance from the cone peak of point A.
Then the minimal distance between A and B is given by
the solution of
Components x[i] give the path from B to A which might get up and down.
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
answered Dec 13 '18 at 15:55
fwgbfwgb
1233
1233
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
add a comment |
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
$begingroup$
Indeed, when n goes to infinity the solution coincides with the formula given in another answer here. So, both seem correct.
$endgroup$
– fwgb
Dec 13 '18 at 18:35
add a comment |
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$begingroup$
Not sure how easy the calculation is, but you could calculate the position depending on the height and angle, and then parametrize the path, you could calculate the length of that curve
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– Börge
Dec 13 '18 at 9:32
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I feel like the title is giving me instructions as though I am the cone. "All right, cone, you are at A..."
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– Rahul
Dec 13 '18 at 19:33