What is the shortest path from (-2,0) to (2,0) that avoids the interior but not the circumference of a circle...
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What is the shortest path from (-2,0) to (2,0) that avoids the interior but not the circumference of a circle with a radius 1 centered at the origin?
I am getting the answer $$2sqrt{3}$$ even though the answer is $$2sqrt{3} + π/3$$
I don't understand why, can anyone explain why that is?
Thanks
geometry
asked Nov 24 at 20:48
BLoby Blob
3616
add a comment |
up vote
0
down vote
favorite
What is the shortest path from (-2,0) to (2,0) that avoids the interior but not the circumference of a circle with a radius 1 centered at the origin?
I am getting the answer $$2sqrt{3}$$ even though the answer is $$2sqrt{3} + π/3$$
I don't understand why, can anyone explain why that is?
Thanks
geometry
asked Nov 24 at 20:48
BLoby Blob
3616
1
Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the shortest path from (-2,0) to (2,0) that avoids the interior but not the circumference of a circle with a radius 1 centered at the origin?
I am getting the answer $$2sqrt{3}$$ even though the answer is $$2sqrt{3} + π/3$$
I don't understand why, can anyone explain why that is?
Thanks
geometry
asked Nov 24 at 20:48
BLoby Blob
3616
What is the shortest path from (-2,0) to (2,0) that avoids the interior but not the circumference of a circle with a radius 1 centered at the origin?
I am getting the answer $$2sqrt{3}$$ even though the answer is $$2sqrt{3} + π/3$$
I don't understand why, can anyone explain why that is?
Thanks
geometry
geometry
asked Nov 24 at 20:48
BLoby Blob
3616
asked Nov 24 at 20:48
BLoby Blob
3616
edited Nov 24 at 21:14
asked Nov 24 at 20:48
BLoby Blob
3616
asked Nov 24 at 20:48
BLoby Blob
3616
asked Nov 24 at 20:48
BLoby Blob
3616
3616
1
Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13
add a comment |
1
Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13
1
1
Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I misunderstood what you had done, sorry. You have constructed the tangent from $(-2,0)$ to the circle, and computed its length as $sqrt{3}$ and done the same thing on the other side. The problem is that these two segments together do no give a path from $(-2,0)$ to $(2,0)$. You have to add in the arc of the circle joining the points of tangency.
You will have to compute the central angle subtending this arc. To do this, you should compute the angles of the triangle you constructed, using trigonometry. Intuitively, this is surely the shortest path, though the above discussion by no means constitutes a proof. I suspect you are not expected to prove it, though.
In fact, if you'e taking algebra/trig I feel sure you don't have to give a proof.
answered Nov 24 at 21:15
saulspatz
13.3k21327
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I misunderstood what you had done, sorry. You have constructed the tangent from $(-2,0)$ to the circle, and computed its length as $sqrt{3}$ and done the same thing on the other side. The problem is that these two segments together do no give a path from $(-2,0)$ to $(2,0)$. You have to add in the arc of the circle joining the points of tangency.
You will have to compute the central angle subtending this arc. To do this, you should compute the angles of the triangle you constructed, using trigonometry. Intuitively, this is surely the shortest path, though the above discussion by no means constitutes a proof. I suspect you are not expected to prove it, though.
In fact, if you'e taking algebra/trig I feel sure you don't have to give a proof.
answered Nov 24 at 21:15
saulspatz
13.3k21327
add a comment |
up vote
2
down vote
accepted
I misunderstood what you had done, sorry. You have constructed the tangent from $(-2,0)$ to the circle, and computed its length as $sqrt{3}$ and done the same thing on the other side. The problem is that these two segments together do no give a path from $(-2,0)$ to $(2,0)$. You have to add in the arc of the circle joining the points of tangency.
You will have to compute the central angle subtending this arc. To do this, you should compute the angles of the triangle you constructed, using trigonometry. Intuitively, this is surely the shortest path, though the above discussion by no means constitutes a proof. I suspect you are not expected to prove it, though.
In fact, if you'e taking algebra/trig I feel sure you don't have to give a proof.
answered Nov 24 at 21:15
saulspatz
13.3k21327
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I misunderstood what you had done, sorry. You have constructed the tangent from $(-2,0)$ to the circle, and computed its length as $sqrt{3}$ and done the same thing on the other side. The problem is that these two segments together do no give a path from $(-2,0)$ to $(2,0)$. You have to add in the arc of the circle joining the points of tangency.
You will have to compute the central angle subtending this arc. To do this, you should compute the angles of the triangle you constructed, using trigonometry. Intuitively, this is surely the shortest path, though the above discussion by no means constitutes a proof. I suspect you are not expected to prove it, though.
In fact, if you'e taking algebra/trig I feel sure you don't have to give a proof.
answered Nov 24 at 21:15
saulspatz
13.3k21327
I misunderstood what you had done, sorry. You have constructed the tangent from $(-2,0)$ to the circle, and computed its length as $sqrt{3}$ and done the same thing on the other side. The problem is that these two segments together do no give a path from $(-2,0)$ to $(2,0)$. You have to add in the arc of the circle joining the points of tangency.
You will have to compute the central angle subtending this arc. To do this, you should compute the angles of the triangle you constructed, using trigonometry. Intuitively, this is surely the shortest path, though the above discussion by no means constitutes a proof. I suspect you are not expected to prove it, though.
In fact, if you'e taking algebra/trig I feel sure you don't have to give a proof.
answered Nov 24 at 21:15
saulspatz
13.3k21327
answered Nov 24 at 21:15
saulspatz
13.3k21327
answered Nov 24 at 21:15
saulspatz
13.3k21327
answered Nov 24 at 21:15
saulspatz
13.3k21327
13.3k21327
add a comment |
add a comment |
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Maybe we can explain it, if you show us how you got your answer.
– saulspatz
Nov 24 at 20:51
I was thinking I should make a 1,2,sqrt3 triangle and because sqrt 3 is the missing side, and I just multiplied the answer by two. As to the actual answer, my answer key provided me with the number without any explanation.
– BLoby Blob
Nov 24 at 20:54
The problem with your solution is that the $sqrt{3}$ side of the triangle will pass through the interior of the circle won't it? By the way, this has nothing to do with algebraic geometry, which is a very advanced subject. Are you taking calculus?
– saulspatz
Nov 24 at 20:57
I'll change the tag then. I'm taking an algebra 2 trig course
– BLoby Blob
Nov 24 at 21:13