ADGEOM 4942 * ADGEOM 4943

#4942

Dear friends,

 

Consider a triangle ABC and let P be its incenter. A perpendicular to AP, at P, cut sides AB and AC at E and F, respectively. Similarly, define the segments GH and IJ ciclycally. 

 

Circles (EGJ), (GJF) and (FHI) are concurrent at Q.

 

Is this well-known? What is the nature of this point?

 

Best regards,

Emmanuel José García

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#4943

Dear Emmanuel, I changed the labels of the points of your problem:

Let  ABC be a triangle of incenter I, the perpendicular by I to AI intersects AC and AB in Ab and Ac, respectively.

The points Bc, Ba, Ca and Cb are defined cyclically.

The circles (AbBaBc), (BcCbCa) and (CaAcAb) concur in P.
The circles (AbBaAc), (BcCbBa) and (CaAcCb)  cuncur in U.

Points P and U form a bicentric pair (http://faculty.evansville.edu/ck6/encyclopedia/BicentricPairs.html),

P = f(a,b,c) : f(b,c,a) : f(c,a,b)     and   Q =  f(a,c,b) : f(b,a,c) : f(c,b,a)

 with
f(a,b,c) = a (a-b-c) (b-c) (2 a^2-2 a b-a c+b c-c^2) (a^2+a b-2 b^2-2 a c+b c+c^2).

** The bicentric sum of P and U is:

S = a (b-c)^2 (b+c-a)^2 (5 a^2-4 a (b+c)-(b-c)^2) : .... : ....

S lies on lines X(i)X(j) for these {i, j}: {926,2170}, {2246,4845}, {3119,3900}, {4162,7004}.

 (6 - 9 - 13) - search numbers  of S: (3.01760078443673, 3.25289748021838, -0.00408028952222515).
 
 
** The bicentric difference of P and U is:

D = a (b-c) (4 a^2-5 a (b+c)+b^2+4 b c+c^2) : ...: ....

D is the midpoint of X(i) and X(j), for these {i, j}:  {1,2254}, {145,3716}, {513,4162}, {519,14430}, {663,14077}, {891,3251}, {905,4959}, {1635,3722}, {2814,16200}, {2832,10699}, {3244,3762}, {3295,8648}, {3900,14414}, {8572,20315}.

D is the reflection of X(i) in X(j), for these {i, j}: {2254,14413}, {14413,1}.

D lies on lines X(i)X(j) for these {i, j}: {1,2254}, {145,3716}, {513,4162}, {519,14430}, {663,14077}, {891,3251}, {905,4959}, {1635,3722}, {2814,16200}, {2832,10699}, {3244,3762}, {3295,8648}, {3900,14414}, {8572,20315}.

 (6 - 9 - 13) - search numbers  of D: (5.41971934778287, 0.657342198122180, 0.684172491923085).
 
 Angel Montesdeoca