236. A locus with the metrical relation.
by Virgil Nicula, Mar 3, 2011, 7:13 PM
Proposed problem 1. Ascertain the geometrical locus 
of the mobile interior point 
w.r.t. acute given triangle 
for which 
(constant) and show that the locus pass through a fixed point 
for any value of the constant 
. I denoted the distance 
of the point 
to the line 
.
Proof 1 (Kostas Vittas - nice !). Let
be, the points on the sidesegments 
rspectively, such that 
where 
and 
are the orthogonal projections of 
on 
respectively and 
are the ones of 
on 
respectively. We draw the external angle bisector of 
, which intersects the line segment 
at point so be it 
From 
and so, because of 
we conclude that the points 
are collinear. That is, the line segment 
passes through the constant point , as the feet of the external angle bisector of on the sideline 
of 
Let be, an arbitrary point between and we will prove that 
, where 
are the orthogonal projections of 
on 
respectively. We denote the point 
in to the same side of as 
such that 
and 
and le be the point 
Because of 
and so, we have the collinearity of the points 
From 
Hence, from 
and from 
It is not difficult now to show that for any point 
inwardly to 
it is true the relation 
, where 
are the orthogonal projections of , on respectively. Hence, the locus of the point inwardly to 
as the problem states, is the segment connecting the points on 
respectively, which are collinear with the point 
as the feet of the external angle bisector of 
on the sideline of and the proof is completed.
Proof 2 (own).
Proposed problem 2. Let
be a circle with diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and let 
be a fixed point. Denote the tangent 
to at 
.
For a mobile line
for which 
denote 
, 
, 
and the intersection of the tangents
from
, to (different from ) . Prove that the produkt 
is constant and the locus of the (mobile) point is a (fixed) line.
Proof.
.
(constant).
Using the second relation from here obtain that in any triangle exists the identity 
, where 
.
Apply this relation to
and obtain that 
- distance of to 
is constant, i.e. the locus of is a parallel line to .
Proposed problem 3. Let be a circle with diameter and let 
be a fixed point so that 
. Denote the tangent to
at . For a mobile line for which denote , , and the intersection of the tangents
from , to (different from ) . Prove that the produkt is constant and the locus of the (mobile) point is a (fixed) line.
Proof. .
(constant).
Using the first relation from here obtain that in any triangle exists the identity 
, where .
Apply this relation to and obtain that - distance of to is constant, i.e. the locus of is a parallel line to .
of the mobile interior point 
w.r.t. acute given triangle 
for which 
(constant) and show that the locus pass through a fixed point 
for any value of the constant 
. I denoted the distance 
of the point 
to the line 
.Proof 1 (Kostas Vittas - nice !). Let
be, the points on the sidesegments 
rspectively, such that 
where 
and 
are the orthogonal projections of 
on 
respectively and 
are the ones of 
on 
respectively. We draw the external angle bisector of 
, which intersects the line segment 
at point so be it 
From 
and so, because of 
we conclude that the points 
are collinear. That is, the line segment 
passes through the constant point , as the feet of the external angle bisector of on the sideline 
of 
Let be, an arbitrary point between and we will prove that 
, where 
are the orthogonal projections of 
on 
respectively. We denote the point 
in to the same side of as 
such that 
and 
and le be the point 
Because of 
and so, we have the collinearity of the points 
From 
Hence, from 
and from 
It is not difficult now to show that for any point 
inwardly to 
it is true the relation 
, where 
are the orthogonal projections of , on respectively. Hence, the locus of the point inwardly to 
as the problem states, is the segment connecting the points on 
respectively, which are collinear with the point 
as the feet of the external angle bisector of 
on the sideline of and the proof is completed.Proof 2 (own).
Proposed problem 2. Let
be a circle with diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and let 
be a fixed point. Denote the tangent 
to at 
.For a mobile line
for which 
denote 
, 
, 
and the intersection of the tangentsfrom
, to (different from ) . Prove that the produkt 
is constant and the locus of the (mobile) point is a (fixed) line.Proof.
.
(constant).Using the second relation from here obtain that in any triangle
exists the identity 
, where 
.Apply this relation to
and obtain that 
- distance of to 
is constant, i.e. the locus of is a parallel line to .Proposed problem 3. Let
be a circle with diameter and let 
be a fixed point so that 
. Denote the tangent to at
. For a mobile line for which denote , , and the intersection of the tangentsfrom
, to (different from ) . Prove that the produkt is constant and the locus of the (mobile) point is a (fixed) line.Proof.
. (constant).Using the first relation from here obtain that in any triangle
exists the identity 
, where .Apply this relation to
and obtain that - distance of to is constant, i.e. the locus of is a parallel line to .This post has been edited 13 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:22 PM
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