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Tetrahedron Centers
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For Triangle Centers, as seen at the Encyclopedia of Triangle Centers, the various centers each have a triangle center function $f(a,b,c)$ that is homogeneous, bisymmetric, and cyclic within barycentric or trilinear coordinates. Here are functions for centers known to Euclid.
name trilinear barycentric
X_1 incenter I 1 a angle bisectors
X_2 centroid G 1/a 1 medians
X_3 circumcenter O cos(A) a^2(b^2+c^2-a) perpendicular bisectors
X_4 orthocenter H sec(A) tan(A) altitudes
Are there tetrahedron center functions similar to the triangle center functions in barycentric or trilinear coordinates? The barycentric centroid and trilinear incenter are known.
I've made an elaborate Tetrahedron Centers demonstration, and posted code showing $4 pi =2 sum dihedral - sum solid $ . At fermat point I give exact coordinates of centers for a specific tetrahedron. At Dihedral Constant Center some exact coordinates are calculated for a new tetrahedron center.
Some available items for tetrahedron $ABCD$ are :
volume, total surface area, total perimeter, dihedral constant.
solid angle $A$, face area $A$ ($triangle BCD$), perimeter $A$ ... $B$ ... $C$ ... $D$
edge length $ab$, dihedral angle $ab$ ... $ac$ ... $ad$ ... $bc$ ... $bd$ ... $cd$
I'd like to get a list of tetrahedron center functions started.
geometry triangle polyhedra solid-geometry
asked Jul 26 at 17:10
Ed Pegg
9,13932486
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up vote
3
down vote
favorite
For Triangle Centers, as seen at the Encyclopedia of Triangle Centers, the various centers each have a triangle center function $f(a,b,c)$ that is homogeneous, bisymmetric, and cyclic within barycentric or trilinear coordinates. Here are functions for centers known to Euclid.
name trilinear barycentric
X_1 incenter I 1 a angle bisectors
X_2 centroid G 1/a 1 medians
X_3 circumcenter O cos(A) a^2(b^2+c^2-a) perpendicular bisectors
X_4 orthocenter H sec(A) tan(A) altitudes
Are there tetrahedron center functions similar to the triangle center functions in barycentric or trilinear coordinates? The barycentric centroid and trilinear incenter are known.
I've made an elaborate Tetrahedron Centers demonstration, and posted code showing $4 pi =2 sum dihedral - sum solid $ . At fermat point I give exact coordinates of centers for a specific tetrahedron. At Dihedral Constant Center some exact coordinates are calculated for a new tetrahedron center.
Some available items for tetrahedron $ABCD$ are :
volume, total surface area, total perimeter, dihedral constant.
solid angle $A$, face area $A$ ($triangle BCD$), perimeter $A$ ... $B$ ... $C$ ... $D$
edge length $ab$, dihedral angle $ab$ ... $ac$ ... $ad$ ... $bc$ ... $bd$ ... $cd$
I'd like to get a list of tetrahedron center functions started.
geometry triangle polyhedra solid-geometry
asked Jul 26 at 17:10
Ed Pegg
9,13932486
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For Triangle Centers, as seen at the Encyclopedia of Triangle Centers, the various centers each have a triangle center function $f(a,b,c)$ that is homogeneous, bisymmetric, and cyclic within barycentric or trilinear coordinates. Here are functions for centers known to Euclid.
name trilinear barycentric
X_1 incenter I 1 a angle bisectors
X_2 centroid G 1/a 1 medians
X_3 circumcenter O cos(A) a^2(b^2+c^2-a) perpendicular bisectors
X_4 orthocenter H sec(A) tan(A) altitudes
Are there tetrahedron center functions similar to the triangle center functions in barycentric or trilinear coordinates? The barycentric centroid and trilinear incenter are known.
I've made an elaborate Tetrahedron Centers demonstration, and posted code showing $4 pi =2 sum dihedral - sum solid $ . At fermat point I give exact coordinates of centers for a specific tetrahedron. At Dihedral Constant Center some exact coordinates are calculated for a new tetrahedron center.
Some available items for tetrahedron $ABCD$ are :
volume, total surface area, total perimeter, dihedral constant.
solid angle $A$, face area $A$ ($triangle BCD$), perimeter $A$ ... $B$ ... $C$ ... $D$
edge length $ab$, dihedral angle $ab$ ... $ac$ ... $ad$ ... $bc$ ... $bd$ ... $cd$
I'd like to get a list of tetrahedron center functions started.
geometry triangle polyhedra solid-geometry
asked Jul 26 at 17:10
Ed Pegg
9,13932486
For Triangle Centers, as seen at the Encyclopedia of Triangle Centers, the various centers each have a triangle center function $f(a,b,c)$ that is homogeneous, bisymmetric, and cyclic within barycentric or trilinear coordinates. Here are functions for centers known to Euclid.
name trilinear barycentric
X_1 incenter I 1 a angle bisectors
X_2 centroid G 1/a 1 medians
X_3 circumcenter O cos(A) a^2(b^2+c^2-a) perpendicular bisectors
X_4 orthocenter H sec(A) tan(A) altitudes
Are there tetrahedron center functions similar to the triangle center functions in barycentric or trilinear coordinates? The barycentric centroid and trilinear incenter are known.
I've made an elaborate Tetrahedron Centers demonstration, and posted code showing $4 pi =2 sum dihedral - sum solid $ . At fermat point I give exact coordinates of centers for a specific tetrahedron. At Dihedral Constant Center some exact coordinates are calculated for a new tetrahedron center.
Some available items for tetrahedron $ABCD$ are :
volume, total surface area, total perimeter, dihedral constant.
solid angle $A$, face area $A$ ($triangle BCD$), perimeter $A$ ... $B$ ... $C$ ... $D$
edge length $ab$, dihedral angle $ab$ ... $ac$ ... $ad$ ... $bc$ ... $bd$ ... $cd$
I'd like to get a list of tetrahedron center functions started.
geometry triangle polyhedra solid-geometry
asked Jul 26 at 17:10
Ed Pegg
9,13932486
asked Jul 26 at 17:10
Ed Pegg
9,13932486
asked Jul 26 at 17:10
Ed Pegg
9,13932486
asked Jul 26 at 17:10
Ed Pegg
9,13932486
9,13932486
add a comment |Â
add a comment |Â
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Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ describe the point
$$fracrho,P + alpha,A + beta,B + gamma,Crho + alpha + beta + gamma tag$star$$$
The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$fracrhoW : fracalphaX : fracbetaY : fracgammaZ tag$starstar$$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $triangle ABC$. Any non-degenerate tetrahedron admits a unique point commont to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are
$$rho =-frac2sqrtW+frac1sqrtX+frac1sqrtY+frac1sqrtZqquad
alpha =phantom-frac1sqrtW-frac2sqrtX+frac1sqrtY+frac1sqrtZqquadtextetc tag1$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $triangle A^prime B^prime C^prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|triangle PB^prime C^prime| = |triangle P C^prime A^prime| = |triangle P A^prime B^prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are
$$rho = W left(;2sqrtW/lambda_P - sqrtX/lambda_A - sqrtY/lambda_B - sqrtZ/lambda_C;right) qquad textetc tag2$$
where
$$lambda_V^2 ;=; 3 - 2;sum_theta costheta tag3$$
with the sum taken over the three dihedral angles along edges emanating from vertex $V$.
answered Jul 26 at 21:45
Blue
43.6k868141
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
add a comment |Â
up vote
3
down vote
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ to parameterize the point of interest as
$$fracrho,P + alpha,A + beta, B + gamma,Crho + alpha + beta + gamma$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$rho : alpha : beta : gamma ;=; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$rho : alpha : beta : gamma ;=; W : X : Y : Z$$
Circumcenter ($O$)
$$beginalignrho &=phantom+ a^2 d^2 left(-d^2 + e^2 + f^2 right) \
&phantom=+ b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&phantom=+ c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&phantom=- 2 d^2 e^2 f^2 \[8pt]
&= 18 V^2 - a^2,W X cos D - b^2,W Y cos E - c^2,W Z cos F
endalign$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$beginalignrho ;=; &phantom-; 2 a^2 d^2 left(-a^2 + b^2 + c^2 right) + a^2 d^2 left(-d^2 + e^2 + f^2 right) \[4pt]
&+ 2 b^2 e^2 left(phantom-a^2 - b^2 + c^2 right) + b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&+ 2 c^2 f^2 left(phantom-a^2 + b^2 - c^2 right) + c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2
endalign$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(rho,alpha,beta,gamma)$ and $(rho^prime, alpha^prime, beta^prime, gamma^prime)$, then
$$fracrhorho^primeW^2 = fracalphaalpha^primeX^2 = fracbetabeta^primeY^2 = fracgammagamma^primeZ^2$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus,
$$rho^prime : alpha^prime : beta^prime : gamma^prime ;=; fracW^2rho : fracX^2alpha = fracY^2beta = fracZ^2gamma$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$rho : alpha : beta : gamma ;=; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$beginalign
rho &=phantom+a^2d^2 left(-a^2 + b^2 + c^2right) \[4pt]
&phantom=,+ b^2 e^2 left(phantom-a^2 - b^2 + c^2right) \[4pt]
&phantom=,+ c^2 f^2 left(phantom-a^2 + b^2 - c^2right) \[4pt]
&phantom=,+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2
endalign$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$fracW^prime,P^prime + X^prime,A^prime + Y^prime,B^prime + Z^prime,C^primeW^prime + X^prime + Y^prime + Z^prime$$
for the medial tetrahedron with vertices $P^prime = frac13(A+B+C)$, etc, and face-areas $W^prime = frac19 W$, etc. This can be re-written as
$$fracfrac13(X+Y+Z),P + frac13(W+Y+Z),A + frac13(W+X+Z),B + frac13(W+X+Y),CW + X + Y + Z$$
from which we observe
$$rho : alpha : beta : gamma ;=; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are
$$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$
$$cos A cos D = cos B cos E = cos C cos F$$
$$overrightarrowPAperpoverrightarrowBC qquad overrightarrowPBperpoverrightarrowCA qquad overrightarrowPCperpoverrightarrowAB$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
answered Jul 27 at 21:25
Blue
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2 Answers
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2 Answers
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active
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active
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up vote
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Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ describe the point
$$fracrho,P + alpha,A + beta,B + gamma,Crho + alpha + beta + gamma tag$star$$$
The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$fracrhoW : fracalphaX : fracbetaY : fracgammaZ tag$starstar$$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $triangle ABC$. Any non-degenerate tetrahedron admits a unique point commont to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are
$$rho =-frac2sqrtW+frac1sqrtX+frac1sqrtY+frac1sqrtZqquad
alpha =phantom-frac1sqrtW-frac2sqrtX+frac1sqrtY+frac1sqrtZqquadtextetc tag1$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $triangle A^prime B^prime C^prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|triangle PB^prime C^prime| = |triangle P C^prime A^prime| = |triangle P A^prime B^prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are
$$rho = W left(;2sqrtW/lambda_P - sqrtX/lambda_A - sqrtY/lambda_B - sqrtZ/lambda_C;right) qquad textetc tag2$$
where
$$lambda_V^2 ;=; 3 - 2;sum_theta costheta tag3$$
with the sum taken over the three dihedral angles along edges emanating from vertex $V$.
answered Jul 26 at 21:45
Blue
43.6k868141
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
add a comment |Â
up vote
3
down vote
Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ describe the point
$$fracrho,P + alpha,A + beta,B + gamma,Crho + alpha + beta + gamma tag$star$$$
The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$fracrhoW : fracalphaX : fracbetaY : fracgammaZ tag$starstar$$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $triangle ABC$. Any non-degenerate tetrahedron admits a unique point commont to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are
$$rho =-frac2sqrtW+frac1sqrtX+frac1sqrtY+frac1sqrtZqquad
alpha =phantom-frac1sqrtW-frac2sqrtX+frac1sqrtY+frac1sqrtZqquadtextetc tag1$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $triangle A^prime B^prime C^prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|triangle PB^prime C^prime| = |triangle P C^prime A^prime| = |triangle P A^prime B^prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are
$$rho = W left(;2sqrtW/lambda_P - sqrtX/lambda_A - sqrtY/lambda_B - sqrtZ/lambda_C;right) qquad textetc tag2$$
where
$$lambda_V^2 ;=; 3 - 2;sum_theta costheta tag3$$
with the sum taken over the three dihedral angles along edges emanating from vertex $V$.
answered Jul 26 at 21:45
Blue
43.6k868141
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ describe the point
$$fracrho,P + alpha,A + beta,B + gamma,Crho + alpha + beta + gamma tag$star$$$
The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$fracrhoW : fracalphaX : fracbetaY : fracgammaZ tag$starstar$$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $triangle ABC$. Any non-degenerate tetrahedron admits a unique point commont to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are
$$rho =-frac2sqrtW+frac1sqrtX+frac1sqrtY+frac1sqrtZqquad
alpha =phantom-frac1sqrtW-frac2sqrtX+frac1sqrtY+frac1sqrtZqquadtextetc tag1$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $triangle A^prime B^prime C^prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|triangle PB^prime C^prime| = |triangle P C^prime A^prime| = |triangle P A^prime B^prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are
$$rho = W left(;2sqrtW/lambda_P - sqrtX/lambda_A - sqrtY/lambda_B - sqrtZ/lambda_C;right) qquad textetc tag2$$
where
$$lambda_V^2 ;=; 3 - 2;sum_theta costheta tag3$$
with the sum taken over the three dihedral angles along edges emanating from vertex $V$.
answered Jul 26 at 21:45
Blue
43.6k868141
Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.
In the following, barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ describe the point
$$fracrho,P + alpha,A + beta,B + gamma,Crho + alpha + beta + gamma tag$star$$$
The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):
$$fracrhoW : fracalphaX : fracbetaY : fracgammaZ tag$starstar$$$
Equal-Area Parallelians Point
Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $triangle ABC$. Any non-degenerate tetrahedron admits a unique point commont to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).
The barycentric coordinates of the EAPP are
$$rho =-frac2sqrtW+frac1sqrtX+frac1sqrtY+frac1sqrtZqquad
alpha =phantom-frac1sqrtW-frac2sqrtX+frac1sqrtY+frac1sqrtZqquadtextetc tag1$$
Equal-Area Trisohedralizers Point
Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $triangle A^prime B^prime C^prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|triangle PB^prime C^prime| = |triangle P C^prime A^prime| = |triangle P A^prime B^prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).
The barycentric coordinates of the EATP are
$$rho = W left(;2sqrtW/lambda_P - sqrtX/lambda_A - sqrtY/lambda_B - sqrtZ/lambda_C;right) qquad textetc tag2$$
where
$$lambda_V^2 ;=; 3 - 2;sum_theta costheta tag3$$
with the sum taken over the three dihedral angles along edges emanating from vertex $V$.
answered Jul 26 at 21:45
Blue
43.6k868141
edited Jul 27 at 21:25
answered Jul 26 at 21:45
Blue
43.6k868141
answered Jul 26 at 21:45
Blue
43.6k868141
answered Jul 26 at 21:45
Blue
43.6k868141
43.6k868141
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
add a comment |Â
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
Neat! How about I O H T K S X M?
– Ed Pegg
Jul 26 at 21:52
add a comment |Â
up vote
3
down vote
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ to parameterize the point of interest as
$$fracrho,P + alpha,A + beta, B + gamma,Crho + alpha + beta + gamma$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$rho : alpha : beta : gamma ;=; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$rho : alpha : beta : gamma ;=; W : X : Y : Z$$
Circumcenter ($O$)
$$beginalignrho &=phantom+ a^2 d^2 left(-d^2 + e^2 + f^2 right) \
&phantom=+ b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&phantom=+ c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&phantom=- 2 d^2 e^2 f^2 \[8pt]
&= 18 V^2 - a^2,W X cos D - b^2,W Y cos E - c^2,W Z cos F
endalign$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$beginalignrho ;=; &phantom-; 2 a^2 d^2 left(-a^2 + b^2 + c^2 right) + a^2 d^2 left(-d^2 + e^2 + f^2 right) \[4pt]
&+ 2 b^2 e^2 left(phantom-a^2 - b^2 + c^2 right) + b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&+ 2 c^2 f^2 left(phantom-a^2 + b^2 - c^2 right) + c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2
endalign$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(rho,alpha,beta,gamma)$ and $(rho^prime, alpha^prime, beta^prime, gamma^prime)$, then
$$fracrhorho^primeW^2 = fracalphaalpha^primeX^2 = fracbetabeta^primeY^2 = fracgammagamma^primeZ^2$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus,
$$rho^prime : alpha^prime : beta^prime : gamma^prime ;=; fracW^2rho : fracX^2alpha = fracY^2beta = fracZ^2gamma$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$rho : alpha : beta : gamma ;=; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$beginalign
rho &=phantom+a^2d^2 left(-a^2 + b^2 + c^2right) \[4pt]
&phantom=,+ b^2 e^2 left(phantom-a^2 - b^2 + c^2right) \[4pt]
&phantom=,+ c^2 f^2 left(phantom-a^2 + b^2 - c^2right) \[4pt]
&phantom=,+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2
endalign$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$fracW^prime,P^prime + X^prime,A^prime + Y^prime,B^prime + Z^prime,C^primeW^prime + X^prime + Y^prime + Z^prime$$
for the medial tetrahedron with vertices $P^prime = frac13(A+B+C)$, etc, and face-areas $W^prime = frac19 W$, etc. This can be re-written as
$$fracfrac13(X+Y+Z),P + frac13(W+Y+Z),A + frac13(W+X+Z),B + frac13(W+X+Y),CW + X + Y + Z$$
from which we observe
$$rho : alpha : beta : gamma ;=; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are
$$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$
$$cos A cos D = cos B cos E = cos C cos F$$
$$overrightarrowPAperpoverrightarrowBC qquad overrightarrowPBperpoverrightarrowCA qquad overrightarrowPCperpoverrightarrowAB$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
answered Jul 27 at 21:25
Blue
43.6k868141
add a comment |Â
up vote
3
down vote
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ to parameterize the point of interest as
$$fracrho,P + alpha,A + beta, B + gamma,Crho + alpha + beta + gamma$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$rho : alpha : beta : gamma ;=; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$rho : alpha : beta : gamma ;=; W : X : Y : Z$$
Circumcenter ($O$)
$$beginalignrho &=phantom+ a^2 d^2 left(-d^2 + e^2 + f^2 right) \
&phantom=+ b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&phantom=+ c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&phantom=- 2 d^2 e^2 f^2 \[8pt]
&= 18 V^2 - a^2,W X cos D - b^2,W Y cos E - c^2,W Z cos F
endalign$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$beginalignrho ;=; &phantom-; 2 a^2 d^2 left(-a^2 + b^2 + c^2 right) + a^2 d^2 left(-d^2 + e^2 + f^2 right) \[4pt]
&+ 2 b^2 e^2 left(phantom-a^2 - b^2 + c^2 right) + b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&+ 2 c^2 f^2 left(phantom-a^2 + b^2 - c^2 right) + c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2
endalign$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(rho,alpha,beta,gamma)$ and $(rho^prime, alpha^prime, beta^prime, gamma^prime)$, then
$$fracrhorho^primeW^2 = fracalphaalpha^primeX^2 = fracbetabeta^primeY^2 = fracgammagamma^primeZ^2$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus,
$$rho^prime : alpha^prime : beta^prime : gamma^prime ;=; fracW^2rho : fracX^2alpha = fracY^2beta = fracZ^2gamma$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$rho : alpha : beta : gamma ;=; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$beginalign
rho &=phantom+a^2d^2 left(-a^2 + b^2 + c^2right) \[4pt]
&phantom=,+ b^2 e^2 left(phantom-a^2 - b^2 + c^2right) \[4pt]
&phantom=,+ c^2 f^2 left(phantom-a^2 + b^2 - c^2right) \[4pt]
&phantom=,+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2
endalign$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$fracW^prime,P^prime + X^prime,A^prime + Y^prime,B^prime + Z^prime,C^primeW^prime + X^prime + Y^prime + Z^prime$$
for the medial tetrahedron with vertices $P^prime = frac13(A+B+C)$, etc, and face-areas $W^prime = frac19 W$, etc. This can be re-written as
$$fracfrac13(X+Y+Z),P + frac13(W+Y+Z),A + frac13(W+X+Z),B + frac13(W+X+Y),CW + X + Y + Z$$
from which we observe
$$rho : alpha : beta : gamma ;=; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are
$$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$
$$cos A cos D = cos B cos E = cos C cos F$$
$$overrightarrowPAperpoverrightarrowBC qquad overrightarrowPBperpoverrightarrowCA qquad overrightarrowPCperpoverrightarrowAB$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
answered Jul 27 at 21:25
Blue
43.6k868141
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ to parameterize the point of interest as
$$fracrho,P + alpha,A + beta, B + gamma,Crho + alpha + beta + gamma$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$rho : alpha : beta : gamma ;=; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$rho : alpha : beta : gamma ;=; W : X : Y : Z$$
Circumcenter ($O$)
$$beginalignrho &=phantom+ a^2 d^2 left(-d^2 + e^2 + f^2 right) \
&phantom=+ b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&phantom=+ c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&phantom=- 2 d^2 e^2 f^2 \[8pt]
&= 18 V^2 - a^2,W X cos D - b^2,W Y cos E - c^2,W Z cos F
endalign$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$beginalignrho ;=; &phantom-; 2 a^2 d^2 left(-a^2 + b^2 + c^2 right) + a^2 d^2 left(-d^2 + e^2 + f^2 right) \[4pt]
&+ 2 b^2 e^2 left(phantom-a^2 - b^2 + c^2 right) + b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&+ 2 c^2 f^2 left(phantom-a^2 + b^2 - c^2 right) + c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2
endalign$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(rho,alpha,beta,gamma)$ and $(rho^prime, alpha^prime, beta^prime, gamma^prime)$, then
$$fracrhorho^primeW^2 = fracalphaalpha^primeX^2 = fracbetabeta^primeY^2 = fracgammagamma^primeZ^2$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus,
$$rho^prime : alpha^prime : beta^prime : gamma^prime ;=; fracW^2rho : fracX^2alpha = fracY^2beta = fracZ^2gamma$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$rho : alpha : beta : gamma ;=; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$beginalign
rho &=phantom+a^2d^2 left(-a^2 + b^2 + c^2right) \[4pt]
&phantom=,+ b^2 e^2 left(phantom-a^2 - b^2 + c^2right) \[4pt]
&phantom=,+ c^2 f^2 left(phantom-a^2 + b^2 - c^2right) \[4pt]
&phantom=,+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2
endalign$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$fracW^prime,P^prime + X^prime,A^prime + Y^prime,B^prime + Z^prime,C^primeW^prime + X^prime + Y^prime + Z^prime$$
for the medial tetrahedron with vertices $P^prime = frac13(A+B+C)$, etc, and face-areas $W^prime = frac19 W$, etc. This can be re-written as
$$fracfrac13(X+Y+Z),P + frac13(W+Y+Z),A + frac13(W+X+Z),B + frac13(W+X+Y),CW + X + Y + Z$$
from which we observe
$$rho : alpha : beta : gamma ;=; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are
$$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$
$$cos A cos D = cos B cos E = cos C cos F$$
$$overrightarrowPAperpoverrightarrowBC qquad overrightarrowPBperpoverrightarrowCA qquad overrightarrowPCperpoverrightarrowAB$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
answered Jul 27 at 21:25
Blue
43.6k868141
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $rho$, $alpha$, $beta$, $gamma$ to parameterize the point of interest as
$$fracrho,P + alpha,A + beta, B + gamma,Crho + alpha + beta + gamma$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
- Opposite edge-pairs are $(a,d)$, $(b,e)$, $(c,f)$.
- Edges surrounding vertices $P$, $A$, $B$, $C$ are $(a,b,c)$, $(a,e,f)$, $(d,b,f)$, $(d,e,c)$.
- Edges surrounding faces $W$, $X$, $Y$, $Z$ are $(d,e,f)$, $(d,b,c)$, $(a,e,c)$, $(a,b,f)$.
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$rho : alpha : beta : gamma ;=; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$rho : alpha : beta : gamma ;=; W : X : Y : Z$$
Circumcenter ($O$)
$$beginalignrho &=phantom+ a^2 d^2 left(-d^2 + e^2 + f^2 right) \
&phantom=+ b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&phantom=+ c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&phantom=- 2 d^2 e^2 f^2 \[8pt]
&= 18 V^2 - a^2,W X cos D - b^2,W Y cos E - c^2,W Z cos F
endalign$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$beginalignrho ;=; &phantom-; 2 a^2 d^2 left(-a^2 + b^2 + c^2 right) + a^2 d^2 left(-d^2 + e^2 + f^2 right) \[4pt]
&+ 2 b^2 e^2 left(phantom-a^2 - b^2 + c^2 right) + b^2 e^2 left(phantom-d^2 - e^2 + f^2 right) \[4pt]
&+ 2 c^2 f^2 left(phantom-a^2 + b^2 - c^2 right) + c^2 f^2 left(phantom-d^2 + e^2 - f^2 right) \[4pt]
&- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2
endalign$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(rho,alpha,beta,gamma)$ and $(rho^prime, alpha^prime, beta^prime, gamma^prime)$, then
$$fracrhorho^primeW^2 = fracalphaalpha^primeX^2 = fracbetabeta^primeY^2 = fracgammagamma^primeZ^2$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus,
$$rho^prime : alpha^prime : beta^prime : gamma^prime ;=; fracW^2rho : fracX^2alpha = fracY^2beta = fracZ^2gamma$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$rho : alpha : beta : gamma ;=; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$beginalign
rho &=phantom+a^2d^2 left(-a^2 + b^2 + c^2right) \[4pt]
&phantom=,+ b^2 e^2 left(phantom-a^2 - b^2 + c^2right) \[4pt]
&phantom=,+ c^2 f^2 left(phantom-a^2 + b^2 - c^2right) \[4pt]
&phantom=,+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2
endalign$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$fracW^prime,P^prime + X^prime,A^prime + Y^prime,B^prime + Z^prime,C^primeW^prime + X^prime + Y^prime + Z^prime$$
for the medial tetrahedron with vertices $P^prime = frac13(A+B+C)$, etc, and face-areas $W^prime = frac19 W$, etc. This can be re-written as
$$fracfrac13(X+Y+Z),P + frac13(W+Y+Z),A + frac13(W+X+Z),B + frac13(W+X+Y),CW + X + Y + Z$$
from which we observe
$$rho : alpha : beta : gamma ;=; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are
$$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$
$$cos A cos D = cos B cos E = cos C cos F$$
$$overrightarrowPAperpoverrightarrowBC qquad overrightarrowPBperpoverrightarrowCA qquad overrightarrowPCperpoverrightarrowAB$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.
answered Jul 27 at 21:25
Blue
43.6k868141
edited Jul 29 at 17:55
answered Jul 27 at 21:25
Blue
43.6k868141
answered Jul 27 at 21:25
Blue
43.6k868141
answered Jul 27 at 21:25
Blue
43.6k868141
43.6k868141
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