Σάββατο, 1 Ιανουαρίου 2011

CONCURRENT LINES (a configuration)

Let ABC be a triangle.

Denote:
Ac1, Ac2 two points on the line AB such that Ac1 is between A,B and B is between Ac1,Ac2 (ie Ac2 is on the extension of the line segment AB near to B).

Ab1, Ab2 two points on the line AC such that Ab1 is between A,C and C is between Ab1,Ab2 (Ab2 is on the extension of the line segment AC near to C).

Let A1 be the intersection of the lines Ab1Ac2 and Ab2Ac1.

Similarly we define the points B1 and C1.

Problem:

To find a condition in order to be the triangles ABC, A1B1C1 perspective.
(ie to be the lines AA1, BB1, CC1 concurrent).

Solution:

Note: I take the line segments and the angles as unsigned.

AA1, BB1, CC1 are concurrent <==>

sin(A1AB)/sin(A1AC).Cyclically = 1
(Trigonometric version of Ceva Theorem).

Let A* be the intersection of AA1 and Ac2Ab2.


We have :

In the triangle AA*Ac2 by the Law of the sines:

sin(A1AB) / A*Ac2 = sin(AA*Ac2) / AAc2

In the triangle AA*Ab2:

sin(A1AC) / A*Ab2 = sin(AA*Ab2) / AAb2

From these we get (since AA*Ac2 + AA*Ab2 = Pi):

sin(A1AB) / sin(A1AC) = (A*Ac2 / A*Ab2).(AAb2 / AAc2) (1)

We have in the triangle AAc2AAb2 by Ceva Theorem:

(AAc1 / Ac1Ac2) . (Ac2A* / A*Ab2).(Ab2Ab1 / Ab1A) = 1

==>

Ac2A* / A*Ab2 = (Ac1Ac2 / AAc1).(Ab1A / Ab2Ab1)

So the (1) becomes:

sin(A1AB) / sin(A1AC) = (AAb1 / AAc1).(AAb2 / AAc2).(Ac1Ac2 / Ab1Ab2) (2)

Cyclically

sin(B1BC) / sin(B1BA) = (BBc1 / BBa1).(BBc2 / BBa2).(Ba1Ba2 / Bc1Bc2)

sin(C1BA) / sin(C1CB) = (CCa1 / CCb1).(CCa2 / CCb2).(Cb1Cb2 / Ca1Ca2)

Therefore

AA1, BB1, CC1 : concurrent <==>

(AAb1 / AAc1).(AAb2 / AAc2).(Ac1Ac2 / Ab1Ab2).
(BBc1 / BBa1).(BBc2 / BBa2).(Ba1Ba2 / Bc1Bc2).
(CCa1 / CCb1).(CCa2 / CCb2).(Cb1Cb2 / Ca1Ca2) = 1

Exercises for the reader:

1. Let A2 be the intersection of the lines BAb1 and CAc1 (and similarly B2,C2) and A3 the intersection of the lines BAb2 and CAc2 (and similarly B3,C3).

Find conditions in order to be concurrent the lines:

(i) AA2, BB2, CC2 - (ii) AA3, BB3,CC3

2. Investigate the other possible positions of the points Ac1, Ac2 on AB and Ab1,Ab2 on AC etc.

Application:
[Reference: APH, Hyacinthos message #1663]

Let AHa, AHb, AHc be the altitudes of acute ABC.

The circle (B,BHa) intersects AB at Ac1 (between A,B) and Ac2.
The circle (C,CHa) intersects AC at Ab1 (between A,C) and Ab2.

Let A1 be the intersection of Ab1Ac2 and Ac1Ab2.
Similarly B1, C1.

The lines AA1, BB1, CC1 are concurrent.

Proof:


We have:

BAc1 = BAc2 = BHa = c.cosB ==>

AAc1 = c - c.cosB = c(1-cosB)

AAc2 = AB + Ac2 = c + c.cosB = c(1+cosB)

Ac1Ac2 = BAc1 + BAc2 = 2c.cosB

And

CAb1 = CAb2 = CHa = b.cosC ==>

AAb1 = b - b.cosC = b(1-cosC)

AAb2 = AC + Ab2 = b + b.cosC = b(1+cosC)

Ab1Ab2 = CAb1 + CAb2 = 2b.cosC

The (2) above

sin(A1AB) / sin(A1AC) = (AAb1 / AAc1).(AAb2 / AAc2).(Ac1Ac2 / Ab1Ab2)

becomes:

sin(A1AB) / sin(A1AC) =

[b(1-cosC)/c(1-cosB)].[b(1+cosC)/c(1+cosB)].[2c.cosB/2b.cosC] =

[b/c].[(1-cosC)/(1-cosB)].[(1+cosC)/(1+cosB)]*[cosB/cosC] =

[sinB/sinC].[(1-cosC)/(1-cosB)].[(1+cosC)/(1+cosB)].[cosB/cosC]

Similarly

sin(B1BC) / sin(B1BA) = ...

sin(C1BA) / sin(C1CB) = ...

By multiplying them we get 1. Therefore the lines lines AA1, BB1, CC1 are concurrent.

Trilinears:

From sin(A1AB) / sin(A1AC) =
[sinB/sinC].[(1-cosC)/(1-cosB)].[(1+cosC)/(1+cosB)].[cosB/cosC]

we get the 1st trilinear:

(1/sinA).(1-cosA).(1+cosA).(1/cosA) =

(1/sinA).(1-(cosA)^2).(1/cosA) =

(1/sinA).((sinA)^2).(1/cosA) = sinA / cosA = tanA

Exercise: Prove it for ABC non-acute.

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