## 22487Re: [EGML] 2 circles and 5 lines in a right triangle [2 Attachments]

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• Jul 2, 2014
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Dear Seiichi,

Following are only some quick thoughts how coulde we
use (in geometry of triangle) your circles in the right-angled triangle

When we have properties of a special triangle, we can construct (by
some way) three such special triangles respective to reference triangle
angles (or sides).

An example with your circles in a right-angled triangle.

Let ABC be a triangle and (Oa),(Ob),(Oc) the circles with diameters
BC,CA,AB, resp.
Let P be a point. The perpendicular through P to BC intersects (Oa)
at Aa,A1, on the positive,negative sides of BC, resp.
Similarly Ba, B2 and Cc, C2.

Now, let (Sa),(Ka) be the Seiichi Kirikami circles in the right triangle AaBC
and (S1),(K1) in the right triangle A1BC.
Similarly the circles (Sb),(Kb) and (S2), (K2) ----- (Sc), (Kc) and (S3), (K3).

Now we can several questions !!

For example:

Which is the locus of P such that:
1. ABC, SaSbSc
2. ABC, KaKbKc
3. ABC, K1K2K3
4. ABC, S1S2S3

are perspective?

Please note that I wrote the above without drawing figures.
So sorry if they do not have much sense!

Antreas

On Wed, Jul 2, 2014 at 5:16 AM, skirikami_0623@... [Anopolis] wrote:

[Attachment(s) from skirikami_0623@... included below]

Dear friends,

[1] Given a triangle ABC right in A, its incircle(I), its circumcircle(O), and its excircles (Ia), (Ib), (Ic) with their centers I, O, Ia, Ib, Ic, denote:
S, Q, R= the contact points of (I) with BC, CA, AB respectively,
Aa, Ba, Ca=the contact points of (Ia) with BC, CA, AB respectively,
Ab, Bb, Cb=the contact points of (Ib) with BC, CA, AB respectively,
Ac, Bc, Cc=the contact points of (Ic) with BC, CA, AB respectively,
B’, C’=the midpoints of CA, AB respectively,
B’’, C’’=the intersections of B’C’ with (O) respectively.
1. 8 points B’’, Q, R, C’’, Ac, Ca, Ba, and Ab are on a circle (1-circle).
2. 8 points B’’, Bb, Aa, S, Cc, C’’, Bc, and Cb are on another circle (2-circle).
See the attached pictures.
This is an extension of Le Probleme de Toshio Seimiya in Quelques Theoremes Oublies, vol. 1 (2007), Geometrie by Jean-Louis Ayme.

[2] Given a triangle ABC right in A, its excircles (Ia), (Ib), (Ic) with their centers Ia, Ib, Ic, and its angle bisectors (A-bsc), (B-bsc), (C-bsc), denote:
B’, C’=the midpoints of CA, AB respectively,
Ka, Kb, Kc=the intersections farthest from the sides of ABC of (A-bsc), (B-bsc), (C-bsc) with (Ia), (Ib), (Ic) respectibvely,
Jb, Jc=the intersections nearest to the sides of ABC of (B-bsc), (C-bsc) with (Ib), (Ic),
Mb, Mc=the intersections farthest from BC of (Ia) with the angle bisector of CBCa, BCBa respectively,
Lb, Lc=the intersections nearest to BC of (Ia) with the angle bisector of CBCa, BCBa respectively.
1. Kb, B’, Lb are on a line (1-line),
2. Kc, C’, Lc are on a line (2-line),
3. Mc, C’, Jc are on a line (3-line),
4. Mb, B’, Jb are on a line (4-line),
P= the intersection of 1-line and 2-line,
Q=the intersection of 3-line and 4-line.
5.P, Q, Ka are on a line (5-line).
See the attached pictures.
This is the continuation of Anopolis message #817.

Best regards,
Seiichi.