ADGEOM 171 * ANOPOLIS 365 * ANOPOLIS 366

 

#171

Dear friends,

A well-known property of cyclic quadrilaterals is that the incenters of the four triangles composing the cyclic quadrilateral form a rectangle.

There are 3 points on the circumcircle, A', B', C', on arcs BC, CA, AB
resp., that, each together with A, B, C, form cyclic quadrilaterals such that their 'incenter rectangles' are squares. The three squares share a common vertex at the incenter, I. Let A"B"C" be the triangle bounded by each square's diagonal that does not contain I. A"B"C" is perspective to ABC at a point P (non-ETC search = 6.134725493967532) which has trilinears: sec A (1 - sin A) : : == sec A - tan A : :, and lies on lines 1,372 4,9 27,X^-1(605) 33,X^-1(481) 46,485 57,482 58,606 63,487 90,3378 1824,X^-1(488) (at least).

I was surprised to find no other perspectives among the triangles produced by this configuration: A'B'C' and A"B"C" (as above), the centers of the squares, the triangle formed by the vertices of each square opposite I.

The lines P lies on seem to suggest a relationship with the Kenmotu configuration (see X(371) and X(372)).

Also, it seems a natural counterpart to P would have trilinears: sec A (1 + sin A) : : == sec A + tan A : :, and lie on lines 1,371 4,9 27,X^-1(606) 33,X^-1(482) 46,486 57,481 58,605 63,488 90,3377 1824,X^-1(487) (at least).
The 2 points are {X(4),X(19)}-harmonic conjugates, but I have not found a similar construction for this second point (non-ETC search = -3.125390112968884).

The 6 vertices of the squares, that do not include I or the vertices opposite I, all lie on a common conic, with center at non-ETC search = -1.039451334868101.


In Anopolis #365 (6/5/2013), Antreas presents another construction that results in this same center, P:
Denote:
(I11) = the excircle of OBC respective to BC
(I22) = the excircle of OCA respective to CA
(I33) = the excircle of OAB respective to AB.

r1, r2, r3 = the radical axes of ((I22), (I33)), ((I33), (I11)),
((I11), (I22)), resp.

p1, p2, p3 = the parallels to r1, r2, r3 through A,B,C, resp.
p1,p2,p3 are concurrent at P.

Angel Montesdeoca provides barycentrics (Anopolis #366): ( a(a^2-b^2-c^2)(a^4-2a^2(b+c)^2 - 4a(b+c)S + (b^2-c^2)^2) :...:...), S= 2area(ABC).


Any insights?

Best regards,
Randy Hutson

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ANOPOLIS 365

Let ABC be a triangle.

Denote:

(I11) = the excircle of OBC respective to BC
(I22) = the excircle of OCA respective to CA
(I33) = the excircle of OAB respective to AB.

r1, r2, r3 = the radical axes of ((I22), (I33)), ((I33), (I11)),
((I11), (I22)), resp.

Radical center?

p1, p2, p3 = the parallels to r1, r2, r3 through A,B,C, resp.

p1,p2,p3 are concurrent.

Point?

Locus (for P a variable point instead of O) ?

Antreas P. Htazipolakis

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ANOPOLIS 366

[Antreas P. Htazipolakis]:

Let ABC be a triangle.

Denote:

(I11) = the excircle of OBC respective to BC
(I22) = the excircle of OCA respective to CA
(I33) = the excircle of OAB respective to AB.

r1, r2, r3 = the radical axes of ((I22), (I33)), ((I33), (I11)),
((I11), (I22)), resp.

Radical center?


*** The radical center of r1, r2, r3 is:

( a(a^2-b^2-c^2)(2a^5(b+c)^2 - 4a^3(b^2+b*c+c^2)^2 + 2a(b-c)^2(b+c)^4 + 2S(b+c)(a^4-a^2(2b^2+5b*c+2c^2) + b^4-b^3c-c^3b+c^4)) :...:...),

S= 2area(ABC).

with (6-9-13)-search number in ETC: 5.60319717911870517696190546


[Antreas P. Htazipolakis]:

p1, p2, p3 = the parallels to r1, r2, r3 through A,B,C, resp.

p1,p2,p3 are concurrent.

Point?


*** p1,p2,p3 are concurrent in:

( a(a^2-b^2-c^2)(a^4-2a^2(b+c)^2 - 4a(b+c)S + (b^2-c^2)^2) :...:...),

with (6-9-13)-search number in ETC: 6.13472549396753240388995633.

Figure:
http://amontes.webs.ull.es/otrashtm/HechosGeometricos.htm#HG080613


Angel Montesdeoca