## 16179Re:[EMHL] the problem

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• Mar 3, 2008
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Dear Alexey, and Francois
This is very interesting.
If the points P, Q are on the sides of angle A
in order to have
(AP+AQ)/(APQ) = (AB+BC+CD+DA)/(ABCD) = constant
the line PQ passes through a fix point A' on the
A_bisector of ABCD.
Similarly we have the points B', C', D' on the
bisectors of angles B, C, D of ABCD.

The construction of these points is as follows:
On the semiline BA we take the point T such that
AT = BC+CD+DA. The parallel from D to AC meets
BA at S. The parallel from S to CT meets BC at U
and the parallel from U to BA meets the A bisector
at the required point A'.
The parallel from A' to AD gives on D_bisector
the point D'. The parallel from D' to DC gives
of C_bisector the point C'. The parallel from C'
to BC gives on B_bisector the point B' and the
quadrilateral A'B'C'D' has parallel sides to
the corresponding sides of ABCD.

Let S1 be the intersection of lines AD, BC and
let S2 be the intersection of lines AB, CD.
Let A be between S1 and D and
let A be between S2 and B.
The line A'D' is the reflection of B'C' in the
bisector of angle AS1B and the line
A'B' is the reflection of line C'D' in the
bisector of angle AS2D.

If the points P, Q are on sides BC, AD then
the envelope of PQ is a conic that is tangent
to the lines AD, BC, A'D', B'C', CD', C'D and the
center O1 of this conic is the center of the rhombus
formed by the lines AD, BC, A'D', B'C'.
We can find the tangency points with these lines
and then construct the conic.
for example if line CD' meets AD at C" then the
perspector P1 of this inconic in triangle C"S1C
is found as the isotomic conjugate of the
anticomplement of O1. Then the lines
S1P1, C"P1 and C'P1 give the points of tangency
with the sides of triangle C"S1C.
The true envelope is the arc of this conic from
the tangency point of CD' to the tangency
point of line C'D.
Similarly if the points P, Q are on sides AB, CD
then the true envelope of PQ is an arc of another
conic defined as previously
with center the center of the rhombus formed
by the lines AB, CD, A'B', C'D', from the
tangency point of this conic with the line BC'
to the tangency point with the line B'C.

Best regards

> Dear Nikos and Francois!
> If the points P, Q aren't on the opposite sides then
> the line PQ pass
> through fixed point. In fact let P be on AB, Q on
> (AP+AQ)/S(APQ)=
> P(ABCD)/S(ABCD)=const. So the common point of PQ and
> bissector of A is
> fixed.
> This follows that the envelop of all lines PQ is
> formed by the arcs of
> conics and the segments of lines. So it isn't
> smooth. Is it convex?
>
> Sincerely
> Alexey
>
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> Antivirus scanning: Symantec Mail Security for SMTP.
>
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>

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