Mixing
A curve through two vertices of a triangle, whose tangent lines bisect the area of that triangle
$begingroup$
Say we have a triangle $ABC$. I want to find a curve $gamma:[0,1]tomathbb{R}^2$ such that $gamma(0)=A$, $gamma(1)=B$ and for all $tin(0,1)$ the tangent line at $gamma(t)$ divides $triangle ABC$ into two pieces of same area (a smaller triangle and a quadrilateral). The curve can be as smooth as you might need of course.
I came up with this problem about a week ago and got a bunch of few equations at first. Now I'm otherwise engaged so I'm posting the problem before I forget about it. Cheers!
geometry differential-geometry locus
edited Jan 15 at 22:51
Blue
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
$endgroup$
add a comment |
$begingroup$
Say we have a triangle $ABC$. I want to find a curve $gamma:[0,1]tomathbb{R}^2$ such that $gamma(0)=A$, $gamma(1)=B$ and for all $tin(0,1)$ the tangent line at $gamma(t)$ divides $triangle ABC$ into two pieces of same area (a smaller triangle and a quadrilateral). The curve can be as smooth as you might need of course.
I came up with this problem about a week ago and got a bunch of few equations at first. Now I'm otherwise engaged so I'm posting the problem before I forget about it. Cheers!
geometry differential-geometry locus
edited Jan 15 at 22:51
Blue
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
$endgroup$
add a comment |
$begingroup$
Say we have a triangle $ABC$. I want to find a curve $gamma:[0,1]tomathbb{R}^2$ such that $gamma(0)=A$, $gamma(1)=B$ and for all $tin(0,1)$ the tangent line at $gamma(t)$ divides $triangle ABC$ into two pieces of same area (a smaller triangle and a quadrilateral). The curve can be as smooth as you might need of course.
I came up with this problem about a week ago and got a bunch of few equations at first. Now I'm otherwise engaged so I'm posting the problem before I forget about it. Cheers!
geometry differential-geometry locus
edited Jan 15 at 22:51
Blue
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
$endgroup$
Say we have a triangle $ABC$. I want to find a curve $gamma:[0,1]tomathbb{R}^2$ such that $gamma(0)=A$, $gamma(1)=B$ and for all $tin(0,1)$ the tangent line at $gamma(t)$ divides $triangle ABC$ into two pieces of same area (a smaller triangle and a quadrilateral). The curve can be as smooth as you might need of course.
I came up with this problem about a week ago and got a bunch of few equations at first. Now I'm otherwise engaged so I'm posting the problem before I forget about it. Cheers!
geometry differential-geometry locus
geometry differential-geometry locus
edited Jan 15 at 22:51
Blue
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
edited Jan 15 at 22:51
Blue
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
edited Jan 15 at 22:51
Blue
48.5k870154
edited Jan 15 at 22:51
Blue
48.5k870154
edited Jan 15 at 22:51
Blue
48.5k870154
48.5k870154
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
asked Jan 15 at 20:13
Cristian BaezaCristian Baeza
420213
420213
add a comment |
add a comment |
1 Answer
1
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$begingroup$
To keep things simple, I'll consider a right triangle with unit legs $AC$ and $BC$, with coordinate axes lying along the legs. From this, one can obtain the general case with a suitable coordinate transformation.
If we take $P=(0,t)$ on $AC$, and want to divide the triangle into two equivalent parts through a line cutting $BC$ at $Q$, it is easy to find
$displaystyle Q=bigg({1over2t},0bigg)$, provided $1/2le tle 1$.
The envelope of all such lines $PQ$ is the hyperbola of equation $xy=1/8$ (dotted in the diagram), but this curve doesn't pass through $A$ and $B$.
We can however take only the central part of this curve, from $F=(1/4,1/2)$ to $G=(1/2,1/4)$ and extend it with two segments $AF$ and $GB$, touching the hyperbola at $F$ and $G$. The resulting composite curve (red in the diagram) is once differentiable and does the requested job.
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To keep things simple, I'll consider a right triangle with unit legs $AC$ and $BC$, with coordinate axes lying along the legs. From this, one can obtain the general case with a suitable coordinate transformation.
If we take $P=(0,t)$ on $AC$, and want to divide the triangle into two equivalent parts through a line cutting $BC$ at $Q$, it is easy to find
$displaystyle Q=bigg({1over2t},0bigg)$, provided $1/2le tle 1$.
The envelope of all such lines $PQ$ is the hyperbola of equation $xy=1/8$ (dotted in the diagram), but this curve doesn't pass through $A$ and $B$.
We can however take only the central part of this curve, from $F=(1/4,1/2)$ to $G=(1/2,1/4)$ and extend it with two segments $AF$ and $GB$, touching the hyperbola at $F$ and $G$. The resulting composite curve (red in the diagram) is once differentiable and does the requested job.
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
add a comment |
$begingroup$
To keep things simple, I'll consider a right triangle with unit legs $AC$ and $BC$, with coordinate axes lying along the legs. From this, one can obtain the general case with a suitable coordinate transformation.
If we take $P=(0,t)$ on $AC$, and want to divide the triangle into two equivalent parts through a line cutting $BC$ at $Q$, it is easy to find
$displaystyle Q=bigg({1over2t},0bigg)$, provided $1/2le tle 1$.
The envelope of all such lines $PQ$ is the hyperbola of equation $xy=1/8$ (dotted in the diagram), but this curve doesn't pass through $A$ and $B$.
We can however take only the central part of this curve, from $F=(1/4,1/2)$ to $G=(1/2,1/4)$ and extend it with two segments $AF$ and $GB$, touching the hyperbola at $F$ and $G$. The resulting composite curve (red in the diagram) is once differentiable and does the requested job.
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
$endgroup$
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
add a comment |
$begingroup$
To keep things simple, I'll consider a right triangle with unit legs $AC$ and $BC$, with coordinate axes lying along the legs. From this, one can obtain the general case with a suitable coordinate transformation.
If we take $P=(0,t)$ on $AC$, and want to divide the triangle into two equivalent parts through a line cutting $BC$ at $Q$, it is easy to find
$displaystyle Q=bigg({1over2t},0bigg)$, provided $1/2le tle 1$.
The envelope of all such lines $PQ$ is the hyperbola of equation $xy=1/8$ (dotted in the diagram), but this curve doesn't pass through $A$ and $B$.
We can however take only the central part of this curve, from $F=(1/4,1/2)$ to $G=(1/2,1/4)$ and extend it with two segments $AF$ and $GB$, touching the hyperbola at $F$ and $G$. The resulting composite curve (red in the diagram) is once differentiable and does the requested job.
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
$endgroup$
To keep things simple, I'll consider a right triangle with unit legs $AC$ and $BC$, with coordinate axes lying along the legs. From this, one can obtain the general case with a suitable coordinate transformation.
If we take $P=(0,t)$ on $AC$, and want to divide the triangle into two equivalent parts through a line cutting $BC$ at $Q$, it is easy to find
$displaystyle Q=bigg({1over2t},0bigg)$, provided $1/2le tle 1$.
The envelope of all such lines $PQ$ is the hyperbola of equation $xy=1/8$ (dotted in the diagram), but this curve doesn't pass through $A$ and $B$.
We can however take only the central part of this curve, from $F=(1/4,1/2)$ to $G=(1/2,1/4)$ and extend it with two segments $AF$ and $GB$, touching the hyperbola at $F$ and $G$. The resulting composite curve (red in the diagram) is once differentiable and does the requested job.
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
edited Jan 15 at 22:21
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
answered Jan 15 at 22:16
AretinoAretino
23.7k21443
23.7k21443
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
add a comment |
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
$begingroup$
Uuuhhh nice job there!!
$endgroup$
– Cristian Baeza
Jan 16 at 15:19
add a comment |
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