## Dual triangles

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• Dear Geometers: According to ETC (just before X(2979)): Suppose DEF is a triangle in the plane of triangle ABC. Let D be the isogonal conjugate of the point
Message 1 of 2 , Apr 3, 2012
Dear Geometers:

According to ETC (just before X(2979)):

Suppose DEF is a triangle in the plane of triangle ABC. Let D' be the
isogonal conjugate of the point of intersection of line EF and the line at
infinity. Define E' and F' cyclically. The triangle D'E'F' is here named the
dual of DEF. The vertices of D'E'F' lie on the circumcircle, and D'E'F' is
similar to DEF. The duality is between the sidelines EF, FD, DE and the
points D', E', F', respectively. For example, if E and F remain fixed and D
varies, then D' remains fixed, while E'F' varies. Actually D'E'F' is the
dual of any triangle homothetic to DEF. Indeed, DEF need not be a triangle
but can be the union of three concurrent lines. (Proof of similarity follows
from Theorem 6E in TCCT, as the "gamma triangle" there is the dual of a
triangle whose sidelines are respectively perpendicular to those of DEF.)

Suppose U is a point having cevian triangle DEF and dual triangle D'E'F'.
Then there exists a point DC(U) whose circum-anticevian triangle (TCCT, p.
201) is D'E'F'. The mapping DC is given for U = u : v : w (trilinears) by

DC(U) = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv).

To construct DC(U) from U, let A' = AD'∩BC, and let A" be the harmonic
conjugate of A' with respect to B and C. Define B" and C" cyclically. The
lines AA", BB", CC" concur in DC(U). Also, DC(U) = U-isoconjugate of the
crosssum of U and X(6).

Inversely, the circum-anticevian triangle of a point P is the dual of a
point CD(P), given for P = p : q : r by the inverse of the DC-mapping; that
is:

CD(P) = bc/(-a/p + b/q + c/r) : ca/(a/p - b/q + c/r) : ab/(a/p + b/q - c/r)
= isogonal conjugate of X(6)-Ceva conjugate of P.

-----------------

I can get this from another construction:

Let DA’B’C’ be the cevian triangle of U in DABC. The circle passing
through {U, B’, C’} cuts (AC) and (AB) again in A_B and A_C and A” is
the midpoint of A_B and A_C. B” and C” are built cyclically.

Then DA”B”C” and DABC are perspective and, for U=u : v : w (trilinears),
the perspector is P = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv) = DC(U)

QUESTIONS: Are both constructions equivalent? Why?

Note: The vertex of my DA”B”C” are not on the circumcircle of DABC

----------------

I note that equation leading to the inverse mapping has a square root, i.e.:

CD(P)=sqrt( b-2 c2- / (-a/p + b/q + c/r)2 ) : …..

which should be written as:

CD(P)= b c- / abs(-a/p + b/q + c/r ) : …..

QUESTION: Why is the absolute value function not relevant?

Regards

[Non-text portions of this message have been removed]
• Better writing: Dear Geometers: According to ETC (just before X(2979)): Suppose DEF is a triangle in the plane of triangle ABC. Let D be the isogonal
Message 2 of 2 , Apr 3, 2012
Better writing:

Dear Geometers:
According to ETC (just before X(2979)):
Suppose DEF is a triangle in the plane of triangle ABC. Let D' be the isogonal conjugate of the point of intersection of line EF and the line at infinity. Define E' and F' cyclically. The triangle D'E'F' is here named the dual of DEF. The vertices of D'E'F' lie on the circumcircle, and D'E'F' is similar to DEF. The duality is between the sidelines EF, FD, DE and the points D', E', F', respectively. For example, if E and F remain fixed and D varies, then D' remains fixed, while E'F' varies. Actually D'E'F' is the dual of any triangle homothetic to DEF. Indeed, DEF need not be a triangle but can be the union of three concurrent lines. (Proof of similarity follows from Theorem 6E in TCCT, as the "gamma triangle" there is the dual of a triangle whose sidelines are respectively perpendicular to those of DEF.)
Suppose U is a point having cevian triangle DEF and dual triangle D'E'F'. Then there exists a point DC(U) whose circum-anticevian triangle (TCCT, p. 201) is D'E'F'. The mapping DC is given for U = u : v : w (trilinears) by
DC(U) = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv).

To construct DC(U) from U, let A' = AD'∩BC, and let A" be the harmonic conjugate of A' with respect to B and C. Define B" and C" cyclically. The lines AA", BB", CC" concur in DC(U). Also, DC(U) = U-isoconjugate of the crosssum of U and X(6).
Inversely, the circum-anticevian triangle of a point P is the dual of a point CD(P), given for P = p : q : r by the inverse of the DC-mapping; that is:
CD(P) = bc/(-a/p + b/q + c/r) : ca/(a/p - b/q + c/r) : ab/(a/p + b/q - c/r) = isogonal conjugate of X(6)-Ceva conjugate of P.

-----------------
I can get this from another construction:
Let A'B'C' be the cevian triangle of U in triangle ABC. The circle passing through {U, B', C'} cuts (AC) and (AB) again in A_B and A_C and A" is the midpoint of A_B and A_C. B" and C" are built cyclically.
Then triangles A"B"C" and ABC are perspective and, for U=u : v : w (trilinears), the perspector is P = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv) = DC(U)
QUESTIONS: Are both constructions equivalent? Why?

Note: The vertex of my triangle A"B"C" are not on the circumcircle of ABC
----------------
I note that equation leading to the inverse mapping has a square root, i.e.:
CD(P)=sqrt( b^2 c^2 / (-a/p + b/q + c/r)^2 ) : ..
which should be written as:
CD(P)= bc/abs(-a/p + b/q + c/r ): ..

QUESTION: Why is the absolute value function not relevant?

Regards
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