ADGEOM 1674

 

#1674

Dear Dr Clack Kimberling, and all Geometer,

When I studied  X(5562) =  REFLECTION OF X(52) IN X(5) I found a new nice result as follows:

Let ABC be a triangle, let three points A',B',C' on the circumcircle, such that AA'//BB'//CC'. Then the triangle A0B0C0 form(bounded) by three Simson line of A',B',C' 

1-A0B0C0 are similar and perpective with ABC. Which is the locus of perspector? 

2- Center of circle (A0B0C0) , the orthocenter of (A0B0C0) also lie on the circle with diameter X(3)X(4) center X(5)

 3-I don't know when the A'B'C' [read A0B0C0] becomes a point, but I known A'B'C' becomes a point three time when we moved A',B',C' on circumcircle (such that AA'//BB'//CC'), these points also lie on the circle with diameter X(3)X(4) center X(5)

Please see the figure attachment.

Best regards

Sincerely

Dao Thanh Oai

 

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#1676

 

Dear Dao Thanh Oai,

 

 

[DTO]:

Let ABC be a triangle, let three points A',B',C' on the circumcircle, such that AA'//BB'//CC'. Then the triangle A0B0C0 form (bounded) by three Simson line of A',B',C' 

1-A0B0C0 are similar and perpective with ABC. Which is the locus of perspector?

 

************ 

a circular quintic which is the antigonal of K027.

 

Best regards

 

Bernard Gibert

 

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#1687

Dear Dao,

 

as for the locus has answered Bernard Gibert.

In your third observation you write that you don't know

when A'B'C' becomes a point.  I think that you mean

AoBoCo becomes a point.

This happens when A' is a vertex of an equilateral triangle XYZ.

The construction of X (not ruler and compass) is as follows:

The parallel from A to BC meets the circle of ABC at A1.

Trisect the arc AA1 (not containing BC) and the point X

is one of the trisecting points such that AX = (2/3)AA1.

 

Best regards

Nikos Dergiades

 

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#1688

 

Dear Mister Nikos Dergiades,

 

Thank to You very very very much,

 

 

1- Oh, I have a mistake, but I mean AoBoCo  becomes a point. 

 

2- The locus of the perspector cuts the circle with diameter X(3)X(4) at four points : A1,B1,C1 and X(4); I think the triangle A1B1C1 isa equilateral triangle. (because I can not draw the locus, so I can not check it exactly). Would do you like check it (A0B0C0 is the equilateral triangle)?

 

Best regards

Sincerely

Dao Thanh Oai

 

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#1691

 

Dear Dao,

 

If the Simson lines of the points A1,B1,C1 are concurrent

then both triangles A1B1C1, A0B0C0 are equilateral.

 

Regards

Nikos Dergiades