A ∆ with sides 3,6,8cm. Now, a man runs around ∆ in such way that he is always at distance of 1cm from...
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I don't know to proceed in this question , I am not getting what will be the man's path . How to think about that path ?
trigonometry
asked Nov 19 at 18:08
user580093
10816
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I don't know to proceed in this question , I am not getting what will be the man's path . How to think about that path ?
trigonometry
asked Nov 19 at 18:08
user580093
10816
1
A very little man, indeed.
– ajotatxe
Nov 19 at 18:11
add a comment |
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favorite
up vote
0
down vote
favorite
I don't know to proceed in this question , I am not getting what will be the man's path . How to think about that path ?
trigonometry
asked Nov 19 at 18:08
user580093
10816
I don't know to proceed in this question , I am not getting what will be the man's path . How to think about that path ?
trigonometry
trigonometry
asked Nov 19 at 18:08
user580093
10816
asked Nov 19 at 18:08
user580093
10816
asked Nov 19 at 18:08
user580093
10816
asked Nov 19 at 18:08
user580093
10816
asked Nov 19 at 18:08
user580093
10816
10816
1
A very little man, indeed.
– ajotatxe
Nov 19 at 18:11
add a comment |
1
A very little man, indeed.
– ajotatxe
Nov 19 at 18:11
1
1
A very little man, indeed.
– ajotatxe
Nov 19 at 18:11
A very little man, indeed.
– ajotatxe
Nov 19 at 18:11
add a comment |
2 Answers
2
active
oldest
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up vote
2
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Hint: As he runs along each side of the triangle, the path will be parallel to the side, and exactly as long as that side. As he rounds each corner, the path will be a circular arc of radius $1$ connecting the straight components. There will be three such circular components, and they will fit together exactly to form a circle of radius $1$ (draw a picture).
answered Nov 19 at 18:16
MPW
29.7k11956
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
|
show 4 more comments
up vote
2
down vote
Clearly the man will be running parallel to the triangle's sides. The only problem arises in how he turns the corners. Here, he must stay 1cm from the vertex, so he'll turn the corners in circular arcs.
answered Nov 19 at 18:12
Y. Forman
11.4k423
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: As he runs along each side of the triangle, the path will be parallel to the side, and exactly as long as that side. As he rounds each corner, the path will be a circular arc of radius $1$ connecting the straight components. There will be three such circular components, and they will fit together exactly to form a circle of radius $1$ (draw a picture).
answered Nov 19 at 18:16
MPW
29.7k11956
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
|
show 4 more comments
up vote
2
down vote
accepted
Hint: As he runs along each side of the triangle, the path will be parallel to the side, and exactly as long as that side. As he rounds each corner, the path will be a circular arc of radius $1$ connecting the straight components. There will be three such circular components, and they will fit together exactly to form a circle of radius $1$ (draw a picture).
answered Nov 19 at 18:16
MPW
29.7k11956
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
|
show 4 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: As he runs along each side of the triangle, the path will be parallel to the side, and exactly as long as that side. As he rounds each corner, the path will be a circular arc of radius $1$ connecting the straight components. There will be three such circular components, and they will fit together exactly to form a circle of radius $1$ (draw a picture).
answered Nov 19 at 18:16
MPW
29.7k11956
Hint: As he runs along each side of the triangle, the path will be parallel to the side, and exactly as long as that side. As he rounds each corner, the path will be a circular arc of radius $1$ connecting the straight components. There will be three such circular components, and they will fit together exactly to form a circle of radius $1$ (draw a picture).
answered Nov 19 at 18:16
MPW
29.7k11956
answered Nov 19 at 18:16
MPW
29.7k11956
answered Nov 19 at 18:16
MPW
29.7k11956
answered Nov 19 at 18:16
MPW
29.7k11956
29.7k11956
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
|
show 4 more comments
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
Yeah , thanks I got it.
– user580093
Nov 19 at 18:20
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
So this suggests a more general result: An $epsilon$-distant path around a convex polygon of perimeter $P$ has length $P+2piepsilon$.
– MPW
Nov 19 at 18:22
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
is it a standard result or have some proof
– user580093
Nov 19 at 18:24
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
I am having difficulty in finding the length of arc , can you help me in that. finding θ in this will be very long and difficult
– user580093
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
It's just what you can easily observe. It should be clear that the part parallel to the sides is of the same length (the original perimeter), and that you "hook up" the sides with circular arcs of fixed radius. The only thing is to recognize that the total external angle traversed is always $360^{circ}$. That's probably a standard result.
– MPW
Nov 19 at 18:27
|
show 4 more comments
up vote
2
down vote
Clearly the man will be running parallel to the triangle's sides. The only problem arises in how he turns the corners. Here, he must stay 1cm from the vertex, so he'll turn the corners in circular arcs.
answered Nov 19 at 18:12
Y. Forman
11.4k423
add a comment |
up vote
2
down vote
Clearly the man will be running parallel to the triangle's sides. The only problem arises in how he turns the corners. Here, he must stay 1cm from the vertex, so he'll turn the corners in circular arcs.
answered Nov 19 at 18:12
Y. Forman
11.4k423
add a comment |
up vote
2
down vote
up vote
2
down vote
Clearly the man will be running parallel to the triangle's sides. The only problem arises in how he turns the corners. Here, he must stay 1cm from the vertex, so he'll turn the corners in circular arcs.
answered Nov 19 at 18:12
Y. Forman
11.4k423
Clearly the man will be running parallel to the triangle's sides. The only problem arises in how he turns the corners. Here, he must stay 1cm from the vertex, so he'll turn the corners in circular arcs.
answered Nov 19 at 18:12
Y. Forman
11.4k423
answered Nov 19 at 18:12
Y. Forman
11.4k423
answered Nov 19 at 18:12
Y. Forman
11.4k423
answered Nov 19 at 18:12
Y. Forman
11.4k423
11.4k423
add a comment |
add a comment |
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A very little man, indeed.
– ajotatxe
Nov 19 at 18:11