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An interesting concurrency.
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vittasko
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#1
An interesting concurrency.
We draw three cicles (C1), (C2), (C3), with diameters the side segments BC, AC, AB, of a given triangle ABC, respectively, and we denote as D', E', F', their intersection points from the segment lines, AD, BE, CF, of altitudes of ABC, respectively.

If D'', E'', F'', are the second intersection points of (C1), (C2), (C3) respectively, from the circumcircle of the triangle D'E'F', prove that the lines AD'', BE'', CF'', are concurrent.

Kostas Vittas.

PostPosted: Sun Aug 27, 2006 4:01 am  Back to top 
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vittasko
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#2
Re: An interesting concurrency.
A solution will be posted at Wed Aug 27, 2008.
vittasko wrote:
We draw three cicles (C_{1}),\ (C_{2}),\ (C_{3}), with diameters the side segments BC,\ AC,\ AB, of a given triangle \bigtriangleup ABC respectively, and we denote as D',\ E',\ F', their intersection points from the line segments AD,\ BE,\ CF of altitudes of \bigtriangleup ABC, respectively.

If D'',\ E'',\ F'', are the second intersection points of (C_{1}),\ (C_{2}),\ (C_{3}) respectively, from the circumcircle of the triangle \bigtriangleup D'E'F'. Prove that the line segments AD'',\ BE'',\ CF'', are concurrent at one point.

Unfortunately, some problems have not been discussed yet. Spider

Kostas Vittas.
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PostPosted: Sun Apr 20, 2008 3:23 am  Back to top 
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PDatK40SP
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#3
Here's my solution, but it's ugly Sad
We'll use the trigonometric form of Ceva's theorem
We have:
\frac {\sin \angle{D''AB} }{\sin \angle{D''AC} } = \frac { AD'' . \sin \angle{D''AB} }{ AD'' . \sin \angle{D''AC} } = \frac {...

Similarly: \frac {\sin \angle{D'AB} }{\sin \angle{D'AC} } = \frac {BD'}{CD'} \cdot \frac {D'F}{D'E}

We have EE', FF', AD are concurrent at a point, called A_o. Define B_o, C_o similarly.
We have:
\triangle{B_oBD''} \sim \triangle{B_oD'E}, so \frac {BD''}{D'E} = \frac {B_oD''}{B_oE}
\triangle{B_oD'B} \sim \triangle{B_oED''}, so \frac {D'B}{ED''} = \frac {B_oB}{B_oD''}

So \frac {BD''}{D'E} \cdot \frac {D'B}{ED''} = \frac {B_oD''}{B_oE} \cdot \frac {B_oB}{B_oD''}, or \frac {BD''}{D''E} = \frac {D'E}{D'B} \cdot \frac {B_oB}{B_oE}

Similarly, we have: \frac {CD''}{D''F} = \frac {D'F}{D'C} \cdot \frac {C_oC}{C_oF}

Then: \frac {BD''}{CD''} \cdot \frac {D''F}{D''E} = \frac {D'E}{D'B} \cdot \frac {B_oB}{B_oE} : \left[ \frac {D'F}{D'C} \cdot \frac...

Or \frac {\sin \angle{D''AB} }{\sin \angle{D''AC} } = \frac {D'C}{D'B} \cdot \frac {D'E}{D'F} \cdot \frac {B_oB}{B_oE} \cdot \fr...

Similarly with the other fractions, we have:
\prod \frac {\sin \angle{D''AB} }{\sin \angle{D''AC}} = \prod \left[ \frac {\sin \angle{D'AC} }{\sin \angle{D'AB} } \cdot \fr...

Obviously, \prod \left[ \frac {B_oB}{B_oE} \cdot \frac {C_oF}{C_oC} \right] = 1
And cause AD', BE', CF' are concurrent, we have \prod \frac {\sin \angle{D'AC} }{\sin \angle{D'AB} } = 1
So \prod \frac {\sin \angle{D''AB} }{\sin \angle{D''AC}} = 1, which means AD'', BE'', CF'' are concurrent, as desired.

P/s: Dear Kostas Vittas, after typing this post, I downloaded your figure and saw that you took D \in [AD'], and I took D' \in [AD] Mr. Green But I think there is no problem, they are same Smile

PostPosted: Sun Apr 20, 2008 8:41 pm  Back to top 
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vittasko
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#4
Re: An interesting concurrency.
A solution will be posted at Wed Aug 27, 2008.
Thank you very much dear PDatK40SP, for your interest and solution.

You are right, it doesn't matter if the points D',\ E',\ F', are inwardly to \bigtriangleup ABC.

The schema I presented, is better for the solution I have in mind (I will try to send you that, as soon it is possible), which although is a synthetic proof, however it is rather long.

Best regards, Kostas Vittas.
Last edited by vittasko on Mon Apr 21, 2008 6:00 am; edited 1 time in total 
PostPosted: Sun Apr 20, 2008 10:08 pm  Back to top 
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¬[ƒ(Gabriel)³²¹º]¼
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#5
Hi vittasko, I think that these generalization are true:

1)Let a triangle ABC and call \Gamma_A the circle with diameter BC and equaly for \Gamma_B and \Gamma_C. Now let a point P and denote A': AP \cap \Gamma_A, B': BP \cap \Gamma_B, C': CP \cap \Gamma_C. The circumcircle of \triangle A'B'C' meet \Gamma_A on A'', \Gamma_B on B'' and \Gamma_C on C''. Then AA'', BB'' and CC'' are concurrent.
figure 1
2)Let a triangle ABC and call \Gamma_A the circle with diameter BC and equaly for \Gamma_B and \Gamma_C. Now let a point P and denote A': AP \cap \Gamma_A, B': BP \cap \Gamma_B, C': CP \cap \Gamma_C where A',B',C' are the intersection internal of ABC. The circumcircle of \triangle A'B'C' meet \Gamma_A on A'', \Gamma_B on B'' and \Gamma_C on C''. Then AA'', BB'' and CC'' are concurrent.
figure 2
3)Let a triangle ABC and call \Gamma_A the circle with diameter BC and equaly for \Gamm_B and \Gamma_C. Now let a point P and denote A': AP \cap \Gamma_A, B': BP \cap \Gamma_B, C': CP \cap \Gamma_C. Let H be the orthocenter and A'' : HA' \cap \Gamma_A, B'' : HB' \cap \Gamma_B and C'' : HC' \cap \Gamma_C. Then AA'', BB'' and CC'' are concurrent.
figure 3

PostPosted: Mon Apr 21, 2008 4:58 am  Back to top 
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vittasko
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#6
Re: An interesting concurrency.
A solution will be posted at Wed Aug 27, 2008.
Hi Gabriel and thank you for your interest.

I thought, by computer, that the results (1),\ (2) you mentioned, was not true and I didn't any think, about the result (3).

So, I will check again more carefully (I see in your figures, that they seem true), before a definite answer.

Best regards, Kostas Vittas.

PostPosted: Mon Apr 21, 2008 5:58 am  Back to top 
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PDatK40SP
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#7
Hi Kostas Vittas, after recheck my above post and drawing a figure, I think the result (1), (2) ¬[ƒ(Gabriel)³²¹º]¼ mentioned are true and the solution in my above post is still true, too.
About the result (3), we can proved it by the trigonometric form of Ceva's theorem, and I think, a bit easier than two above results
Details

We still have \frac {\sin \angle{A''AB} }{\sin \angle{A''AC} } = \frac {BA''}{CA''} \cdot \frac {A''F}{A''E}, and similarly, \frac {\sin \angle{A'AB} }{\sin \angle{A'AC} } = \frac {BA'}{CA'} \cdot \frac {A'F}{A'E}

And we can see that \frac {A''B}{A''E} = \frac {A'E}{A'B} \cdot \frac {HE}{HB}, and similarly, \frac {A''C}{A''F} = \frac {A'F}{A'C} \cdot \frac {HF}{HC}

So \frac {\sin \angle{A''AB} }{\sin \angle{A''AC} } = \frac {A'E}{A'B} \cdot \frac {HE}{HB} : \left[ \frac {A'F}{A'C} \cdot \fra...

Or \frac {\sin \angle{A''AB} }{\sin \angle{A''AC} } = \frac {\sin \angle{A'AC} }{\sin \angle{A'AB} } \cdot \frac {HC}{HB} \cdot ...

We have \prod \frac {\sin \angle{A'AC} }{\sin \angle{A'AB} } = 1, because AA', BB', CC' are concurrent; and \prod \frac {HC}{HB} = \prod \frac {HE}{HF} = 1 are obvious

So \prod \frac {\sin \angle{A''AB} }{\sin \angle{A''AC} } = 1, and AA'', BB'', CC'' are concurrent.

PostPosted: Mon Apr 21, 2008 7:36 am  Back to top 
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vittasko
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#8
Re: An interesting concurrency.
PDatK40SP wrote:
Hi Kostas Vittas, after recheck my above post and drawing a figure, I think the result (1), (2) ¬[ƒ(Gabriel)³²¹º]¼ mentioned are true and the solution in my above post is still true, too.
About the result (3), we can proved it by the trigonometric form of Ceva's theorem, and I think, a bit easier than two above results

You are right. I see now, that the results mentioned by Gabriel, are true.

Unfortunately, I thought in a wrong figure, two year ago Sad and I redused the case of an arbitrary point P, instead of the orthocenter H of \bigtriangleup ABC.

Thank you, Kostas Vittas.

PostPosted: Mon Apr 21, 2008 8:06 am  Back to top 
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vittasko
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#9
Re: An interesting concurrency.
vittasko wrote:
We draw three cicles (C_{1}),\ (C_{2}),\ (C_{3}), with diameters the side segments BC,\ AC,\ AB, of a given triangle \bigtriangleup ABC respectively, and we denote as D',\ E',\ F', their intersection points from the line segments AD,\ BE,\ CF of altitudes of \bigtriangleup ABC, respectively.

If D'',\ E'',\ F'', are the second intersection points of (C_{1}),\ (C_{2}),\ (C_{3}) respectively, from the circumcircle of the triangle \bigtriangleup D'E'F'. Prove that the line segments AD'',\ BE'',\ CF'', are concurrent at one point.

For the solution of this difficult problem, we will use the follows two useful theorems of concurrent segments.
We say three segments as concurrent, when their sidelines are concurrent. Three parallel segments are concurrent at infinity.

THEOREM 1 OF CONCURRENT SEGMENTS. – Let AA',\ BB',\ CC' be, three concurrent segments at one point so be it P and taken per two of them ( as opposite side-segments ), they are formed three quadrilaterals ( convex or non-convex ). Prove that:

(a). – The intersection points of the diagonals of two quadrilaterals of them and the intersection point of the opposite
sidelines of the third one, the other than P, are collinear.
(b). – The intersection points of the opposite sidelines of all the quadrilaterals, the other than P, are collinear.


THEOREM 2 OF CONCURRENT SEGMENTS. – Let AA',\ BB',\ CC' be, three concurrent segments at one point so be it P and we denote the points A''\equiv BC'\cap B'C,\ B''\equiv AC'\cap A'C,\ C''\equiv AB'\cap A'B. Prove that:

(a). – The line segments AA'',\ BB'',\ CC'', are concurrent at one point, so be it M.
(b). – The line segments A'A'',\ B'B'',\ C'C'', are concurrent at one point, so be it N.
(c). – The line connecting the points M,\ N, passes through the point P.

REMARK. – The above two useful theorems, are true in every configuration of the given three concurrent segments and their edges, in the plane or in 3D space.


PROOF OF THE PROPOSED PROBLEM.

\bullet Let (K_{1}),\ (K_{2}),\ (K_{3}) be, the circles instead of (C_{1}),\ (C_{2}),\ (C_{3}) respectively, with diameters the side-segments BC,\ AC,\ AB, of the given triangle \bigtriangleup ABC.

Let be the points A'\equiv E'E"\cap F'F'',\ B'\equiv D'D''\cap F'F'' and C'\equiv D'D''\cap E'E''.

The point A' is the radical center of the circles (K),\ (K_{2}),\ (K_{3}), where (K) is the circumcircle of the triangle \bigtriangleup D'E'F'.

That is, the line segment AD passes through the point A' and similarly the line segments BE,\ CF, pass through the points B'\ C', respectively.


\bullet Let be the point R\equiv BC\cap D'D'' and we draw the line through R and perpendicular to BC, which intersects the line segments BD'',\ CD'', at points Q,\ S, respectively.

The point C is the orthocenter of the triangle \bigtriangleup BQS ( from BR\perp QS and SD''\perp BQ ) and so, we have QC\perp BS.

Clearly, the point D_{1}\equiv BS\cap QC, lies on the circle (K_{1}) and we will prove that also lies on the A-altitude AD of \bigtriangleup ABC.

From the cyclic quadrilateral CD''QR, we have that \angle RCQ = \angle RD''Q ,(1)

But, \angle RCQ = \angle BCD_{1} ,(2) and \angle RD''Q = \angle BD''D' = \angle BCD' ,(3)

From (1),\ (2),\ (3) \Longrightarrow \angle BCD_{1} = \angle BCD' ,(4)

From (4), we conclude that the point D_{1} lies also on AD.


\bullet We consider the non-convex hexagon D_{1}D'D''CEB, inscribed in the circle (K_{1}) and then, applying the Pascal theorem, we conclude that the points B\equiv D'D''\cap EB,\ A\equiv D_{1}D'\cap CE and S\equiv D''C\cap BD_{1}, are collinear.

In the configuration now, of the three segments AD_{1},\ BB',\ CC', which are concurrent at orthocenter H of \bigtriangleup ABC, based on the Theorem 1 of concurrent segments, we conclude that the points S\equiv AB'\cap D_{1}B,\ R\equiv BC\cap B'C' and Q'\equiv AC'\cap D_{1}C, as the intersection points of the opposite sidelines of the quadrilaterals AD_{1}BB',\ BCC'B',\ AD_{1}CC' respectively, the other than H, are collinear.

Hence, we conclude that Q'\equiv Q, because of Q\equiv SR\cap D_{1}C. That is, the line segment AC', passes through the point Q.

Because of the collinearity of the points Q,\ R,\ S, based on the Desarques theorem, we conclude that the triangles \bigtriangleup AB'C',\ \bigtriangleup D''CB are perspective and so, we have that the line segments AD'',\ B'C,\ C'B, are concurrent at one point, so be it A''.

That is, the point A''\equiv BC'\cap B'C, lies on the line segment AD'' and similarly, the points B''\equiv AC'\cap A'C,\ C''\equiv AB'\cap A'B, lie on the line segments BE'',\ CF'', respectively.

Finally, in the configuration of the three segments AA',\ BB',\ CC', which are also concurrent at orthocenter H of \bigtriangleup ABC, based on the Theorem 2 of concurrent segments, we conclude that the line segments AA''\equiv AD'',\ BB''\equiv BE'',\ CC''\equiv CF'', are concurrent at one point, so be it P and the proof is completed.

\bullet This proof is dedicated to Darij Grinberg.

Kostas Vittas.

PS. I will post here later, more details about the proofs of the above two theorems of concurrent segments and some their interesting applications.
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PostPosted: Wed Aug 27, 2008 12:10 pm  Back to top 
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#10
Re: An interesting concurrency.
vittasko wrote:
THEOREM 2 OF CONCURRENT SEGMENTS. – Let AA',\ BB',\ CC' be, three concurrent segments at one point so be it P and we denote the points A''\equiv BC'\cap B'C,\ B''\equiv AC'\cap A'C,\ C''\equiv AB'\cap A'B. Prove that:

(a). – The line segments AA'',\ BB'',\ CC'', are concurrent at one point, so be it M.
(b). – The line segments A'A'',\ B'B'',\ C'C'', are concurrent at one point, so be it N.
(c). – The line connecting the points M,\ N, passes through the point P.

Kostas Vittas.

Let S be the intersection of B'C' and B''C''. We have lines A'P, B'C'', B''C' are concurrent, so by Desarques theorem, we have triangles A'B''C'' and PC'B' are perspective. Hence we conclude that S, B, C are collinear.

\bullet Let M be the intersection of BB'', CC''. Points S, B, C are collinear, so we also conclude that triangles MB''C'', A''C'B' are perspective. Hence lines B''C', C''B' and MA''\equiv AA'' are concurrent.

\bullet From the collinearity of S, B, C we also conclude that triangles A'B''C'', A''B'C' are perspective. Hence lines A'A'', B'B'', C'C'' are concurrent.

\bullet From the collinearity of S, B, C we also conclude that triangles MB''C'', PB'C' are perspective. Hence, B'B'', C'C'', MP are concurrent.
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