HYACINTHOS 10942

[Milorad R.Stevanovic]

 

Dear Jean-Pierre,Peter and Wilson,

I have found the new result.

If point P(a/u:b/v:c/w) is the point of
Feuerbach hyperbola then the point
Q((b-c)^2(s-a)u^2:(c-a)^2(s-b)v^2:(a-b)^2(s-c)w^2)
is on the incircle.

The new points(I suppose) on the incircle are

P7((b-c)^2(a^2+bc)^2(s-a):(c-a)^2(b^2+ca)^2(s-b):
(a-b)^2(c^2+ab)^2(s-c))
P8((b-c)^2(b^2+c^2-bc)^2(s-a):(c-a)^2(c^2+a^2-ca)^2(s-b):
(a-b)^2(a^2+b^2-ab)^2(s-c)).
P9(a^2(b-c)^2(b^2+c^2-a^2+bc)^2(s-a):b^2(c-a)^2(c^2+a^2-b^2+ca)^2(s-b):
c^2(a-b)^2(a^2+b^2-c^2+ab)^2(s-c)).
P10(a^2(b-c)^2(b^2+c^2-a^2-bc)^2(s-a):b^2(c-a)^2(c^2+a^2-b^2-ca)^2(s-b):
c^2(a-b)^2(a^2+b^2-c^2-ab)^2(s-c)).
P11((b-c)^2(a^2+ab+ac+2bc)^2(s-a):(c-a)^2(b^2+bc+ba+2ca)^2(s-b):
(a-b)^2(c^2+ca+cb+2ab)^2(s-c)).
P12((b-c)^2[2abc+(b+c)(c+a-b)(a+b-c)]^2(s-a):
(c-a)^2[2abc+(c+a)(a+b-c)(b+c-a)]^2(s-b):
(a-b)^2[2abc+(a+b)(b+c-a)(c+a-b)]^2(s-c)).
P13((b-c)^2[a^2(b^2+bc+c^2)+(b^2+c^2)(ab+bc+ca)]^2(s-a):
(c-a)^2[b^2(c^2+ca+a^2)+(c^2+a^2)(ab+bc+ca)]^2(s-b):
(a-b)^2[c^2(a^2+ab+b^2)+(a^2+b^2)(ab+bc+ca)]^2(s-c)).
P14((b-c)^2[a(b^2+bc+c^2)+(b^3+c^3)]^2(s-a):
(c-a)^2[b(c^2+ca+a^2)+(c^3+a^3)]^2(s-b):
(a-b)^2[c(a^2+ab+b^2)+(a^3+b^3)]^2(s-c)).
P15((b-c)^2[a^3-a^2(b+c)-(3a+b+c)(b^2+c^2)]^2(s-a):
(c-a)^2[b^3-b^2(c+a)-(3b+c+a)(c^2+a^2)]^2(s-b):
(a-b)^2[c^3-c^2(a+b)-(3c+a+b)(a^2+b^2)]^2(s-c)).
P16((b-c)^2(cosA)^2(s-a):(c-a)^2(cosB)^2(s-b):
(a-b)^2(cosC)^2(s-c)).

Best regards

Sincerely

Milorad R.Stevanovic

 

[Peter Moses]

 

Dear Milorad, Jean-Pierre, and Wilson,

I reckon that your points P7 - P16 are indeed on the incircle.
with P16 = X(1364).
P7 is on lines {11, 115}, {12, 114}, {55, 99}, {56, 98}.
P9 is on lines {11, 125}, {12, 113}, {55, 110}, {56, 74}.

A couple of other simple incircle points,

(b + c)^2 (a^2 - b c)^2 / (a - b - c) ::
incircle antipode of P7
on lines {11, 114}, {12, 115}, {55, 98}, {56, 99}.

a^2 (b + c)^2 (b c - 2 SA)^2 / (a - b - c) ::
incircle antipode of P9
on lines {11, 113}, {12, 125}, {55, 74}, {56, 110}

For a point P{p, q, r} on the circumcircle, the points
(b^2 p + c^2 p + a^2 q + a^2 r - 2 b c p) (a - b - c) :: and its
incircle antipode
(b^2 p + c^2 p + a^2 q + a^2 r + 2 b c p) / (b + c - a) :: will be
on the incircle.

From
P = point on circumcircle
P' = circumcircle antipode of P
{X(55), P} /\ {X(56), P'} on incircle.
{X(55), P'} /\ {X(56), P} on incircle, incircle antipode of above.
X(55) = internal similitude incircle /circumcircle
X(56) = external similitude incircle /circumcircle.

Best regards,
Peter.