- Dear Alexei

I think it is only a matter about product of central symmetries along each

sides of ABC.

For example, look at the side BC. Let A' be the middle of segment BC and L

be the orthogonal projection on side BC of the orthocenter H.

Let s(A') and s(L) be the central symmetry respectively wrt A' and L .

Then A2 = s(L).s(A').s(L) (A1) = s(U) (A1)

where S(U) is the central symmetry wrt U = s(L)(A') and we are done.

Friendly

François

On 5/14/07, Алексей Мякишев <alex_geom@...> wrote:

>

> Dear friends!

> One construction:

> Given tiangle ABC. On BC line we mark two points: A1 and A2, so that

> AA1=AB and AA2=AC. And perpendicylar bissector to that segment.

> So we do with two others sides.

> Point of intersection is symmetry with center O of H.

> Is it obvious?

> Best regards,

> Alexei

>

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

>

>

>

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

> One construction:

Perhaps you have a typo.

> Given tiangle ABC. On BC line we mark two points:

> A1 and A2, so that AA1=AB and AA2=AC. And

> perpendicylar bissector to that segment.

> So we do with two others sides.

> Point of intersection is symmetry with center O of

> H.

> Is it obvious?

A1 is the symmetric of B wrt the altitude AH

A2 is the symmetric of C wrt the altitude AH

The segment A1A2 is the symmetric of BC

wrt the altitude AH and hence the perpendicular

bisector of BC that passes through O

is symmetric to the perpendicular bisector of A1A2

that passes through the symmetric of O with center H.

Best regards

Nikos Dergiades

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - Dear Hyacinthos,

Given a circle (M,r) and two lines l and m outside the circle.

The points A and B lie on the circumference of the circle.

Question: To construct the points C and D on the circumference so that

AC and BD intersect on line l and CD is parallel to line m.

Regards,

Huub van Kempen

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[Non-text portions of this message have been removed] - Dear Huub van Kempen

> Given a circle (M,r) and two lines l and m outside the circle.

[UV] is the diameter of the circle perpendicular to m.

> The points A and B lie on the circumference of the circle.

> Question: To construct the points C and D on the circumference so that

> AC and BD intersect on line l and CD is parallel to line m.

If CD is a variable chord parallel to m, the locus of the common point

of AC and BD is the rectangular hyperbola through A,B,U,V (the center

of the hyperbola is the midpoint of AB)

So you have just to find the common points of l with the hyperbola

above.

Friendly. Jean-Pierre - If you put C on circle, and a parallel to m by C, then the locus of the intersection of AC and BD is a hyperbole passing by A, B, and the points of contact of two parallels to m tangents to the circle.

The points of intersection of this hyperbola with line l are the solutions.

Martin

From: Huub van Kempen

Sent: Friday, May 02, 2008 11:04 AM

To: hyacinthos@yahoogroups.com

Subject: [EMHL] Problem

Dear Hyacinthos,

Given a circle (M,r) and two lines l and m outside the circle.

The points A and B lie on the circumference of the circle.

Question: To construct the points C and D on the circumference so that

AC and BD intersect on line l and CD is parallel to line m.

Regards,

Huub van Kempen

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

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