What is the relation between the 3 triangles and the whole triangle?
$begingroup$
There is a question in my lecture note.
Call the area of a whole arbitrary triangle A, and the three areas of the corner triangles $a_1, a_2 , $ and $a_3$, respectively.
Then what is the relation between A, $a_1$, $a_2$,and $a_3$ ?
I guess that the relation can be proved by trigonometric functions, but there are too many variables to handle.
geometry
$endgroup$
add a comment |
$begingroup$
There is a question in my lecture note.
Call the area of a whole arbitrary triangle A, and the three areas of the corner triangles $a_1, a_2 , $ and $a_3$, respectively.
Then what is the relation between A, $a_1$, $a_2$,and $a_3$ ?
I guess that the relation can be proved by trigonometric functions, but there are too many variables to handle.
geometry
$endgroup$
1
$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
$endgroup$
– Joel Reyes Noche
Dec 19 '18 at 13:43
$begingroup$
How are the corner triangles defined (i.e. the points D, E and F)?
$endgroup$
– Bo5man
Dec 19 '18 at 13:44
$begingroup$
No...there is no restriction on the sides. And the corner triangles are not be conditioned.
$endgroup$
– user627221
Dec 19 '18 at 13:46
$begingroup$
If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
$endgroup$
– T. Fo
Dec 19 '18 at 15:47
$begingroup$
I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
$endgroup$
– user627221
Dec 19 '18 at 16:06
add a comment |
$begingroup$
There is a question in my lecture note.
Call the area of a whole arbitrary triangle A, and the three areas of the corner triangles $a_1, a_2 , $ and $a_3$, respectively.
Then what is the relation between A, $a_1$, $a_2$,and $a_3$ ?
I guess that the relation can be proved by trigonometric functions, but there are too many variables to handle.
geometry
$endgroup$
There is a question in my lecture note.
Call the area of a whole arbitrary triangle A, and the three areas of the corner triangles $a_1, a_2 , $ and $a_3$, respectively.
Then what is the relation between A, $a_1$, $a_2$,and $a_3$ ?
I guess that the relation can be proved by trigonometric functions, but there are too many variables to handle.
geometry
geometry
asked Dec 19 '18 at 13:38
user627221
1
$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
$endgroup$
– Joel Reyes Noche
Dec 19 '18 at 13:43
$begingroup$
How are the corner triangles defined (i.e. the points D, E and F)?
$endgroup$
– Bo5man
Dec 19 '18 at 13:44
$begingroup$
No...there is no restriction on the sides. And the corner triangles are not be conditioned.
$endgroup$
– user627221
Dec 19 '18 at 13:46
$begingroup$
If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
$endgroup$
– T. Fo
Dec 19 '18 at 15:47
$begingroup$
I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
$endgroup$
– user627221
Dec 19 '18 at 16:06
add a comment |
1
$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
$endgroup$
– Joel Reyes Noche
Dec 19 '18 at 13:43
$begingroup$
How are the corner triangles defined (i.e. the points D, E and F)?
$endgroup$
– Bo5man
Dec 19 '18 at 13:44
$begingroup$
No...there is no restriction on the sides. And the corner triangles are not be conditioned.
$endgroup$
– user627221
Dec 19 '18 at 13:46
$begingroup$
If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
$endgroup$
– T. Fo
Dec 19 '18 at 15:47
$begingroup$
I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
$endgroup$
– user627221
Dec 19 '18 at 16:06
1
1
$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
$endgroup$
– Joel Reyes Noche
Dec 19 '18 at 13:43
$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
$endgroup$
– Joel Reyes Noche
Dec 19 '18 at 13:43
$begingroup$
How are the corner triangles defined (i.e. the points D, E and F)?
$endgroup$
– Bo5man
Dec 19 '18 at 13:44
$begingroup$
How are the corner triangles defined (i.e. the points D, E and F)?
$endgroup$
– Bo5man
Dec 19 '18 at 13:44
$begingroup$
No...there is no restriction on the sides. And the corner triangles are not be conditioned.
$endgroup$
– user627221
Dec 19 '18 at 13:46
$begingroup$
No...there is no restriction on the sides. And the corner triangles are not be conditioned.
$endgroup$
– user627221
Dec 19 '18 at 13:46
$begingroup$
If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
$endgroup$
– T. Fo
Dec 19 '18 at 15:47
$begingroup$
If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
$endgroup$
– T. Fo
Dec 19 '18 at 15:47
$begingroup$
I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
$endgroup$
– user627221
Dec 19 '18 at 16:06
$begingroup$
I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
$endgroup$
– user627221
Dec 19 '18 at 16:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It seems three areas $a_1,a_2,a_3$ are not enough to construct a Monge type formula.
The three areas satisfies an inequality rather an identity.
To avoid confusion, we will use $Delta$ instead of "A" to denote the area of $triangle ABC$. We are going to show the areas satisfies following inequality.
$$sqrt{Delta}(Delta - a_1 - a_2 - a_3) ge 2sqrt{a_1a_2a_3}$$
In fact, this is a necessary and sufficient condition for $a_1, a_2, a_3$ to be realized as the area of such sub-triangles.
Introduce parameters $u,v,w in (0,1)$ so that the barycentric coordinates of $D, E, F$ wrt $triangle ABC$ are
$$D : (1-u,u,0),quad F : (0,1-v,v)quadtext{ and }quad E : ( w, 0, 1-w )$$
In terms of $u,v, w$, we have
$$a_1 = Delta u(1-w),quad a_2 = Delta v(1-u) quadtext{ and }quad a_3 = Delta w(1-v)$$
This leads to following inequality on the areas
$$begin{align} sqrt{Delta}(Delta - a_1 - a_2 - a_3)
= &;Delta^{3/2}(1 - u - v - w + uv+vw + wu)\
= &;Delta^{3/2}((1-u)(1-v)(1-w) + uvw)\
color{blue}{{rm AM} ge {rm GM} rightarrow }quad ge &;
2Delta^{3/2}sqrt{(1-u)(1-v)(1-w)uvw}\
= &; 2sqrt{a_1a_2a_3}end{align}tag{*1}$$
One the other direction. let's say we are given three numbers $a'_1, a'_2, a'_3 in (0,Delta)$ which satisfy $(*1)$. Define $alpha_1,alpha_2,alpha_3 in (0,1)$ by setting $alpha_k = frac{a'_k}{Delta}$. Let
$$K_{pm}(p,q,r) = (1 - p - q - r)
pm sqrt{ (1 - p - q - r)^2 - 4pqr}$$
Using $(*1)$, it is easy to verify both $K_{pm}(alpha_1,alpha_2,alpha_3)$ are non-negative. Define
$$U_{pm}(p,q,r) = frac{K_{pm}(p,q,r) + 2p}{2(1-r)}$$
and set
$$(u,v,w)_{pm} = (U_{pm}(alpha_1,alpha_2,alpha_3), U_{pm}(alpha_2,alpha_3,alpha_1), U_{pm}(alpha_3,alpha_1,alpha_2))$$
For both choice of sign of $pm$, we have $u_{pm},v_{pm},w_{pm} ge 0$. With help of a CAS, one can verify they satisfy
$$u_{pm}(1-w_{pm}) = alpha_1,quad
v_{pm}(1-u_{pm}) = alpha_2quadtext{ and }quad
w_{pm}(1-v_{pm}) = alpha_3$$
This implies $u_{pm},v_{pm},w_{pm} in (0,1)$. Using these as barycentric coordinates, we can find $D,F,E$ on $AB, BC, CA$ to make corresponding area metaches: i.e. $a_1 = a_1'$, $a_2 = a_2'$ and $a_3 = a_3'$.
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
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add a comment |
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$begingroup$
It seems three areas $a_1,a_2,a_3$ are not enough to construct a Monge type formula.
The three areas satisfies an inequality rather an identity.
To avoid confusion, we will use $Delta$ instead of "A" to denote the area of $triangle ABC$. We are going to show the areas satisfies following inequality.
$$sqrt{Delta}(Delta - a_1 - a_2 - a_3) ge 2sqrt{a_1a_2a_3}$$
In fact, this is a necessary and sufficient condition for $a_1, a_2, a_3$ to be realized as the area of such sub-triangles.
Introduce parameters $u,v,w in (0,1)$ so that the barycentric coordinates of $D, E, F$ wrt $triangle ABC$ are
$$D : (1-u,u,0),quad F : (0,1-v,v)quadtext{ and }quad E : ( w, 0, 1-w )$$
In terms of $u,v, w$, we have
$$a_1 = Delta u(1-w),quad a_2 = Delta v(1-u) quadtext{ and }quad a_3 = Delta w(1-v)$$
This leads to following inequality on the areas
$$begin{align} sqrt{Delta}(Delta - a_1 - a_2 - a_3)
= &;Delta^{3/2}(1 - u - v - w + uv+vw + wu)\
= &;Delta^{3/2}((1-u)(1-v)(1-w) + uvw)\
color{blue}{{rm AM} ge {rm GM} rightarrow }quad ge &;
2Delta^{3/2}sqrt{(1-u)(1-v)(1-w)uvw}\
= &; 2sqrt{a_1a_2a_3}end{align}tag{*1}$$
One the other direction. let's say we are given three numbers $a'_1, a'_2, a'_3 in (0,Delta)$ which satisfy $(*1)$. Define $alpha_1,alpha_2,alpha_3 in (0,1)$ by setting $alpha_k = frac{a'_k}{Delta}$. Let
$$K_{pm}(p,q,r) = (1 - p - q - r)
pm sqrt{ (1 - p - q - r)^2 - 4pqr}$$
Using $(*1)$, it is easy to verify both $K_{pm}(alpha_1,alpha_2,alpha_3)$ are non-negative. Define
$$U_{pm}(p,q,r) = frac{K_{pm}(p,q,r) + 2p}{2(1-r)}$$
and set
$$(u,v,w)_{pm} = (U_{pm}(alpha_1,alpha_2,alpha_3), U_{pm}(alpha_2,alpha_3,alpha_1), U_{pm}(alpha_3,alpha_1,alpha_2))$$
For both choice of sign of $pm$, we have $u_{pm},v_{pm},w_{pm} ge 0$. With help of a CAS, one can verify they satisfy
$$u_{pm}(1-w_{pm}) = alpha_1,quad
v_{pm}(1-u_{pm}) = alpha_2quadtext{ and }quad
w_{pm}(1-v_{pm}) = alpha_3$$
This implies $u_{pm},v_{pm},w_{pm} in (0,1)$. Using these as barycentric coordinates, we can find $D,F,E$ on $AB, BC, CA$ to make corresponding area metaches: i.e. $a_1 = a_1'$, $a_2 = a_2'$ and $a_3 = a_3'$.
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
$endgroup$
add a comment |
$begingroup$
It seems three areas $a_1,a_2,a_3$ are not enough to construct a Monge type formula.
The three areas satisfies an inequality rather an identity.
To avoid confusion, we will use $Delta$ instead of "A" to denote the area of $triangle ABC$. We are going to show the areas satisfies following inequality.
$$sqrt{Delta}(Delta - a_1 - a_2 - a_3) ge 2sqrt{a_1a_2a_3}$$
In fact, this is a necessary and sufficient condition for $a_1, a_2, a_3$ to be realized as the area of such sub-triangles.
Introduce parameters $u,v,w in (0,1)$ so that the barycentric coordinates of $D, E, F$ wrt $triangle ABC$ are
$$D : (1-u,u,0),quad F : (0,1-v,v)quadtext{ and }quad E : ( w, 0, 1-w )$$
In terms of $u,v, w$, we have
$$a_1 = Delta u(1-w),quad a_2 = Delta v(1-u) quadtext{ and }quad a_3 = Delta w(1-v)$$
This leads to following inequality on the areas
$$begin{align} sqrt{Delta}(Delta - a_1 - a_2 - a_3)
= &;Delta^{3/2}(1 - u - v - w + uv+vw + wu)\
= &;Delta^{3/2}((1-u)(1-v)(1-w) + uvw)\
color{blue}{{rm AM} ge {rm GM} rightarrow }quad ge &;
2Delta^{3/2}sqrt{(1-u)(1-v)(1-w)uvw}\
= &; 2sqrt{a_1a_2a_3}end{align}tag{*1}$$
One the other direction. let's say we are given three numbers $a'_1, a'_2, a'_3 in (0,Delta)$ which satisfy $(*1)$. Define $alpha_1,alpha_2,alpha_3 in (0,1)$ by setting $alpha_k = frac{a'_k}{Delta}$. Let
$$K_{pm}(p,q,r) = (1 - p - q - r)
pm sqrt{ (1 - p - q - r)^2 - 4pqr}$$
Using $(*1)$, it is easy to verify both $K_{pm}(alpha_1,alpha_2,alpha_3)$ are non-negative. Define
$$U_{pm}(p,q,r) = frac{K_{pm}(p,q,r) + 2p}{2(1-r)}$$
and set
$$(u,v,w)_{pm} = (U_{pm}(alpha_1,alpha_2,alpha_3), U_{pm}(alpha_2,alpha_3,alpha_1), U_{pm}(alpha_3,alpha_1,alpha_2))$$
For both choice of sign of $pm$, we have $u_{pm},v_{pm},w_{pm} ge 0$. With help of a CAS, one can verify they satisfy
$$u_{pm}(1-w_{pm}) = alpha_1,quad
v_{pm}(1-u_{pm}) = alpha_2quadtext{ and }quad
w_{pm}(1-v_{pm}) = alpha_3$$
This implies $u_{pm},v_{pm},w_{pm} in (0,1)$. Using these as barycentric coordinates, we can find $D,F,E$ on $AB, BC, CA$ to make corresponding area metaches: i.e. $a_1 = a_1'$, $a_2 = a_2'$ and $a_3 = a_3'$.
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
$endgroup$
add a comment |
$begingroup$
It seems three areas $a_1,a_2,a_3$ are not enough to construct a Monge type formula.
The three areas satisfies an inequality rather an identity.
To avoid confusion, we will use $Delta$ instead of "A" to denote the area of $triangle ABC$. We are going to show the areas satisfies following inequality.
$$sqrt{Delta}(Delta - a_1 - a_2 - a_3) ge 2sqrt{a_1a_2a_3}$$
In fact, this is a necessary and sufficient condition for $a_1, a_2, a_3$ to be realized as the area of such sub-triangles.
Introduce parameters $u,v,w in (0,1)$ so that the barycentric coordinates of $D, E, F$ wrt $triangle ABC$ are
$$D : (1-u,u,0),quad F : (0,1-v,v)quadtext{ and }quad E : ( w, 0, 1-w )$$
In terms of $u,v, w$, we have
$$a_1 = Delta u(1-w),quad a_2 = Delta v(1-u) quadtext{ and }quad a_3 = Delta w(1-v)$$
This leads to following inequality on the areas
$$begin{align} sqrt{Delta}(Delta - a_1 - a_2 - a_3)
= &;Delta^{3/2}(1 - u - v - w + uv+vw + wu)\
= &;Delta^{3/2}((1-u)(1-v)(1-w) + uvw)\
color{blue}{{rm AM} ge {rm GM} rightarrow }quad ge &;
2Delta^{3/2}sqrt{(1-u)(1-v)(1-w)uvw}\
= &; 2sqrt{a_1a_2a_3}end{align}tag{*1}$$
One the other direction. let's say we are given three numbers $a'_1, a'_2, a'_3 in (0,Delta)$ which satisfy $(*1)$. Define $alpha_1,alpha_2,alpha_3 in (0,1)$ by setting $alpha_k = frac{a'_k}{Delta}$. Let
$$K_{pm}(p,q,r) = (1 - p - q - r)
pm sqrt{ (1 - p - q - r)^2 - 4pqr}$$
Using $(*1)$, it is easy to verify both $K_{pm}(alpha_1,alpha_2,alpha_3)$ are non-negative. Define
$$U_{pm}(p,q,r) = frac{K_{pm}(p,q,r) + 2p}{2(1-r)}$$
and set
$$(u,v,w)_{pm} = (U_{pm}(alpha_1,alpha_2,alpha_3), U_{pm}(alpha_2,alpha_3,alpha_1), U_{pm}(alpha_3,alpha_1,alpha_2))$$
For both choice of sign of $pm$, we have $u_{pm},v_{pm},w_{pm} ge 0$. With help of a CAS, one can verify they satisfy
$$u_{pm}(1-w_{pm}) = alpha_1,quad
v_{pm}(1-u_{pm}) = alpha_2quadtext{ and }quad
w_{pm}(1-v_{pm}) = alpha_3$$
This implies $u_{pm},v_{pm},w_{pm} in (0,1)$. Using these as barycentric coordinates, we can find $D,F,E$ on $AB, BC, CA$ to make corresponding area metaches: i.e. $a_1 = a_1'$, $a_2 = a_2'$ and $a_3 = a_3'$.
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
$endgroup$
It seems three areas $a_1,a_2,a_3$ are not enough to construct a Monge type formula.
The three areas satisfies an inequality rather an identity.
To avoid confusion, we will use $Delta$ instead of "A" to denote the area of $triangle ABC$. We are going to show the areas satisfies following inequality.
$$sqrt{Delta}(Delta - a_1 - a_2 - a_3) ge 2sqrt{a_1a_2a_3}$$
In fact, this is a necessary and sufficient condition for $a_1, a_2, a_3$ to be realized as the area of such sub-triangles.
Introduce parameters $u,v,w in (0,1)$ so that the barycentric coordinates of $D, E, F$ wrt $triangle ABC$ are
$$D : (1-u,u,0),quad F : (0,1-v,v)quadtext{ and }quad E : ( w, 0, 1-w )$$
In terms of $u,v, w$, we have
$$a_1 = Delta u(1-w),quad a_2 = Delta v(1-u) quadtext{ and }quad a_3 = Delta w(1-v)$$
This leads to following inequality on the areas
$$begin{align} sqrt{Delta}(Delta - a_1 - a_2 - a_3)
= &;Delta^{3/2}(1 - u - v - w + uv+vw + wu)\
= &;Delta^{3/2}((1-u)(1-v)(1-w) + uvw)\
color{blue}{{rm AM} ge {rm GM} rightarrow }quad ge &;
2Delta^{3/2}sqrt{(1-u)(1-v)(1-w)uvw}\
= &; 2sqrt{a_1a_2a_3}end{align}tag{*1}$$
One the other direction. let's say we are given three numbers $a'_1, a'_2, a'_3 in (0,Delta)$ which satisfy $(*1)$. Define $alpha_1,alpha_2,alpha_3 in (0,1)$ by setting $alpha_k = frac{a'_k}{Delta}$. Let
$$K_{pm}(p,q,r) = (1 - p - q - r)
pm sqrt{ (1 - p - q - r)^2 - 4pqr}$$
Using $(*1)$, it is easy to verify both $K_{pm}(alpha_1,alpha_2,alpha_3)$ are non-negative. Define
$$U_{pm}(p,q,r) = frac{K_{pm}(p,q,r) + 2p}{2(1-r)}$$
and set
$$(u,v,w)_{pm} = (U_{pm}(alpha_1,alpha_2,alpha_3), U_{pm}(alpha_2,alpha_3,alpha_1), U_{pm}(alpha_3,alpha_1,alpha_2))$$
For both choice of sign of $pm$, we have $u_{pm},v_{pm},w_{pm} ge 0$. With help of a CAS, one can verify they satisfy
$$u_{pm}(1-w_{pm}) = alpha_1,quad
v_{pm}(1-u_{pm}) = alpha_2quadtext{ and }quad
w_{pm}(1-v_{pm}) = alpha_3$$
This implies $u_{pm},v_{pm},w_{pm} in (0,1)$. Using these as barycentric coordinates, we can find $D,F,E$ on $AB, BC, CA$ to make corresponding area metaches: i.e. $a_1 = a_1'$, $a_2 = a_2'$ and $a_3 = a_3'$.
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
answered Dec 19 '18 at 17:27
achille huiachille hui
96.3k5132261
96.3k5132261
add a comment |
add a comment |
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$begingroup$
Are $D$, $E$, and $F$ midpoints of the sides?
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– Joel Reyes Noche
Dec 19 '18 at 13:43
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How are the corner triangles defined (i.e. the points D, E and F)?
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– Bo5man
Dec 19 '18 at 13:44
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No...there is no restriction on the sides. And the corner triangles are not be conditioned.
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– user627221
Dec 19 '18 at 13:46
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If there are no conditions on the corner triangles I’m not sure what you’re hoping for.. area $a_1$ can approach the area of the larger triangle.
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– T. Fo
Dec 19 '18 at 15:47
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I read the material again, and it is about Gauss and Monge formula.I found some paper link and I guess the relation of these triangles should be similar like the pentagon.
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– user627221
Dec 19 '18 at 16:06