- Dear friends:

Let ABC be triangle. The tangency points of its excircles (Ia),(Ib),

(Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),

(H,I,L) respectively.Now we have:

1)EF cut ID at X,GH cut EF at Y,GH cut DI at Z, so XYZ and ABC are

perspective with perspector the orthocenter of ABC.

2)XJ,YK,ZL concur at P. I thing that P isn't ETC.

Can somebody find the barycentrics of P?

Greetings

Juan Carlos - Dear Juan Carlos,

you wrote

> Let ABC be triangle. The tangency points of its excircles (Ia),

(Ib),

> (Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),

The first barycentric coordinate of P is

> (H,I,L) respectively.Now we have:

> 1)EF cut ID at X,GH cut EF at Y,GH cut DI at Z, so XYZ and ABC are

> perspective with perspector the orthocenter of ABC.

> 2)XJ,YK,ZL concur at P. I thing that P isn't ETC.

> Can somebody find the barycentrics of P?

(b+c)(a^2(3a^2-2b^2-2c^2)-(bb-cc)^2)/(b+c-a)

This point is not in the ETC (until 2445)

If you replace J, K and L by the touchpoints of the incircle

the perspector becomes X(1439) with first barycentric coordinate

a(b+c)(b^2+c^2-a^2)/(b+c-a)^2

Greetings from Bruges

Eric - Dear Juan Carlos and Eric,

Very nice results!

In Hyacinthos message #10588, Juan Carlos Salazar

wrote:

>> Let ABC be triangle. The tangency points of

Indeed, if Ia, Ib, Ic are the excenters of

>> its excircles (Ia),(Ib),(Ic) with (AB,AC,BC),

>> (AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),

>> (H,I,L) respectively.Now we have:

>> 1)EF cut ID at X,GH cut EF at Y,GH cut DI at

>> Z, so XYZ and ABC are perspective with

>> perspector the orthocenter of ABC.

>> 2)XJ,YK,ZL concur at P.

triangle ABC, then the lines YZ, ZX, XY are the

reflections of the lines KL, LJ and JK in the

lines IbIc, IcIa, IaIb, respectively. (Why?

Well, since the two tangents from a point to a

circle have equal length, we have BI = BL, and

since the line IcIa bisects the angle IBL, it

follows that the point I is the reflection of

the point L in the line IcIa. Similarly, the

point D is the reflection of the point J in the

line IcIa, and hence, the line ZX, which is the

same as the line ID, is the reflection of the

line LJ in the line IcIa. Similarly for the

lines XY and YZ.)

Now, this implies that the lines YZ, KL and

IbIc concur at one point X1, the lines ZX, LJ

and IcIa concur at one point Y1, and the lines

XY, JK and IaIb concur at one point Z1. These

three points X1, Y1 and Z1 are collinear after

the Desargues theorem, since the triangles

IaIbIc and JKL are perspective (the perspector

is X(40)). Hence, the Desargues theorem again

yields that the triangles IaIbIc and XYZ are

perspective, and that the triangles XYZ and

JKL are perspective. This proves your property

2).

As for property 1), it is proven in the note

"Synthetic proof of Paul Yiu's excircles

theorem" on my website

http://de.geocities.com/darij_grinberg

(you probably know this, Juan Carlos, but I

say this for others who are interested).

In Hyacinthos message #10589, Eric Danneels

wrote:

>> If you replace J, K and L by the

Indeed, if J', K' and L' are the points where

>> touchpoints of the incircle the perspector

>> becomes X(1439)

the incircle of triangle ABC touches the sides

BC, CA, AB, then the perspectivity of

triangles XYZ and J'K'L' can be proven

synthetically as well: A simple application of

the Menelaos theorem shows that the lines YZ,

K'L' and BC concur at one point X2, the lines

ZX, L'J' and CA concur at one point Y2, and

the lines XY, J'K' and AB concur at one point

Z2. Now, these three points X2, Y2, Z2 are

collinear, since we can apply the Desargues

theorem to the two perspective triangles ABC

and J'K'L' (they are perspective at X(7)).

(Or we could also apply it to the triangles

ABC and XYZ, perspective to each other at

X(4).) And by Desargues again, we obtain the

perspectivity of triangles XYZ and J'K'L'.

Sincerely,

Darij Grinberg - Dear Eric, Darij and Paco García:

Thank you very much for your remarks.

Futher, if we consider:IL cut KF at U,HL cut JE at V,DJ cut KG at W.

Also UVW and ABC are perspective with orthocenter of ABC as

perspector

Now UVW and JKL are perspective with perspector Q.

Q is the Spieker center of ABC?

Greetings

Juan Carlos - Dear Juan Carlos,

In Hyacinthos message #10592, you wrote:

>> IL cut KF at U,HL cut JE at V,DJ cut KG

Yes, I have checked this by calculation.

>> at W.

>> Also UVW and ABC are perspective with

>> orthocenter of ABC as perspector

>> Now UVW and JKL are perspective with

>> perspector Q.

>> Q is the Spieker center of ABC?

Unfortunately, I don't have a synthetic proof

right now. Interesting result!

Sincerely,

Darij Grinberg