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phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Point locus, related to angle bisectors

Given three arbitrary points B, M and C on a plane. Find the locus of the points A such that angle BAM = angle CAM.

1) If B, M and C are collinear in any order, the problem is trivial (either no solution or the first family of Apollonian circles)

2) If MB=MC and B, M and C are not collinear, it is still easy: the locus includes the perpendicular bisector of BC and the segment BC itself

3) For all the other cases, I have yet to get the locus. Though I can prove that M must be equidistant from AB and AC.

Please help. Thanks.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Point locus, related to angle bisectors

Consider A as such point and draw perpendiculars BG and CH to segment MA.Triangles AGB & AHC are similar. therfore
AG/AH =BG/CH. So A divides HG externally in the ratio CH/BG.
LOCUS:: draw a line through M and draw BG & CH perpendiculars. Divide HG externally in the ratio CH/BG to locate point A.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Point locus, related to angle bisectors

hi phanthanhtom,

Using Geometer's Sketchpad, I choose points for B and C, and then M slightly closer to B than C.  Then I constructed an angle BAC and its bisector and moved A until the bisector went through M.  I recorded that point and then tried to find another A.  Here is the result:

There were no points to the right of M and above C, and similarly none below the lowest point on the left where I have shown the construction.  No idea what these curves are.  And they'd be different if I moved my initial points.

I suppose you could specify coordinates (Bx,By), (Mx,My) and Cx,Cy) and then find Ax and Ay in terms of these.  That would give parametric equations for the locus.  But I think it's a horrible calculation.

Sorry I cannot help more  

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Point locus, related to angle bisectors

When MB=MC and B.M and c are not collinear  the locus is the circle passing through B,M and C in addition to the perpendicular bisector.

Last edited by thickhead (2016-04-29 19:56:50)

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}