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 Angle bisectors and heights
Сообщение15.09.2011, 23:11 
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13/10/07
755
Роман/София, България
Let k is the circumference of the acute-angled triangle ABC. A1, B1, C1 are points from the segments BC, CA, AB respectively. A2, B2, C2 are the intersection points of AA1, BB1, CC1 with k respectively. k1, k2, k3 are the circles with diameters A1A2, B1B2, C1C2 respectively. A', B', C' are the intersection points of k1, k2, k3 with k respectively.
Prove that if AA1, BB1, CC1 are
a) angle bisectors
b) heights
then AA', BB', CC' intersects at a common point.

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 Re: Angle bisectors and heights
Сообщение17.09.2011, 06:20 


12/09/11
14
Very interesting problem! a) is easy enough. b) is beautiful ! During search for solution i find two interesting facts.
a) - AA', BB', CC' are symmedians.
b) - common point of AA', BB', CC' is isogonal conjugate of X is isotomic cojugate of orthocenter.
More details :

http://narod.ru/disk/25373174001/angle% ... s.rar.html

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 Re: Angle bisectors and heights
Сообщение17.09.2011, 16:26 
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13/10/07
755
Роман/София, България
I knew a solution for a) but I remembered it too late and I wasn't able to correct the statement.

Thank you for the beautiful solution and the hard work.
a) Proof of the fact they are symmedians: http://www.math10.com/f/viewtopic.php?f=10&t=7094
b) When I was in the train I observed the following similar fact:

In the triangle ABC with orthocenter H - H1, H2, H3 are the feets of the perpendiculars from A to BC, B to CA, C to AB respectively. Circles k1(H1, HH1), k2(H2, HH2), k3(H3,HH3) intersects circumcircle of the triangle ABC at the points A1, B1, C1 respectively. Prove that AA1, BB1, CC1 intersects at a common point.

Is it well known? How it can be proved?

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 Re: Angle bisectors and heights
Сообщение17.09.2011, 18:24 
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13/10/07
755
Роман/София, България
(A1, B1, C1 are different from AH1xk, BH2xk, CH3xk where k is the circumcircle of ABC)

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 Re: Angle bisectors and heights
Сообщение19.09.2011, 09:05 


12/09/11
14
Sorry for late answer.
Well known? I don't know. I think if first well know, than it too. But... However, this problem very interesting, good training in geometry of triangle. Desicion here:

http://narod.ru/disk/25576970001/190920111355.jpg.html

First and second problems are two consequences of isogonal conjugate orthocenter and circumcenter of triangle.
During cogitation for second problem i finded second desicion for first problem: let O1 is circus with diametr HH1, O2 is circus with diametr HH2 and O1 intersect O2 in H and X, then X belong CC'. Proof like part 2 of second problem. Where did you find this wonderful problems? I will very happy if your future train travels will no more effective!

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 Re: Angle bisectors and heights
Сообщение19.09.2011, 21:39 
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13/10/07
755
Роман/София, България
Thank you for the beautiful solution ... it seems to be you are a great geometer. About the source of these problems I like experimenting with software when I see some beautiful fact. In this way using my intuition I observe such facts. These problems are inspired from some Russian problems such as All Russian MO 1998 11 -grade First Day - problem 2 http://imomath.com/othercomp/Rus/RusMO98.pdf

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 Re: Angle bisectors and heights
Сообщение20.09.2011, 08:04 


12/09/11
14
Thanks! This reflection about circles with interesting diametrs may be more substantial!
I remember problem with resembling idea: 49 IMO, first day, #1.

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