- Dear Alexey

Thank you for your remarks.

I will try to draw all the envelope with Cabri and maybe SketchPad if

possible.

Friendly

Francois

2008/2/29 Alexey.A.Zaslavsky <zasl@...>:

>

[Non-text portions of this message have been removed]

>

- 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

Nikos Dergiades

> 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

> AD. Then we have

> (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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