Orthocenter as perspector

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• 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)
Message 1 of 5 , Oct 3, 2004
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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 ... (Ib), ... The first barycentric coordinate of P is (b+c)(a^2(3a^2-2b^2-2c^2)-(bb-cc)^2)/(b+c-a) This point is not in the ETC
Message 2 of 5 , Oct 3, 2004
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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),
> (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?

The first barycentric coordinate of P is

(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 ... Indeed, if Ia, Ib, Ic are the excenters of triangle ABC,
Message 3 of 5 , Oct 3, 2004
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Dear Juan Carlos and Eric,

Very nice results!

In Hyacinthos message #10588, Juan Carlos Salazar
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),
>> (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.

Indeed, if Ia, Ib, Ic are the excenters of
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
>> touchpoints of the incircle the perspector
>> becomes X(1439)

Indeed, if J', K' and L' are the points where
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
Message 4 of 5 , Oct 3, 2004
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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, ... Yes, I have checked this by calculation. Unfortunately, I don t have a synthetic proof right now. Interesting result! Sincerely, Darij
Message 5 of 5 , Oct 4, 2004
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Dear Juan Carlos,

In Hyacinthos message #10592, you wrote:

>> 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?

Yes, I have checked this by calculation.
Unfortunately, I don't have a synthetic proof
right now. Interesting result!

Sincerely,
Darij Grinberg
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