129. BX=CY and three properties.
by Virgil Nicula, Sep 24, 2010, 8:01 AM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=368380
Answer. Appears two cases, or the sideline 
separates 
, 
or not. Prove easily that the locus of the midpoint of ![$[XY]$](//latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is included in the reunion of two lines which are parallel to the exterior or interior
-angle bisectors of 
.
Proof 1. Denote the circumcenters 
, 
of the triangles 
, 
respectively and 
. Construct the projections 
, 
of , on 
respectively, the projections 
, 
of , on 
respectively and the projections 
, 
of on 
, 
respectively. Prove easily that 
and 
is a cyclical deltoid with the axis of symmetry 
. Since 
, 
and 
is bisector of 
obtain that is parallelly with the -bisector of . In conclusion, the locus of is included in a parallel line to the -bisector of .
Proof 2. Apply above lemma to the quadrilateral 
, where , are defined in the previous proof.
PP3. Let be a nonisosceles triangle. Consider 
, 
so that 
. Prove that 
and 
.
Proof. Suppose w.l.o.g.
, 
. In this case 
, 
and 
. Thus, 
. Denote the tangent points 
, 
of the incircle with , respectively. Observe that 
and 
. Therefore, 
, i.e. 
. From the relations 
, 
obtain that 
, i.e. .
From the generalized Pytagoras' theorem apling to
obtain that 
![$bc\left[a^2-(a-b)^2-(a-c)^2+2(a-b)(a-c)\cos A\right]=$](//latex.artofproblemsolving.com/4/2/4/4242b0b8818eed69c05c9a6b3f199e0170f2133e.png)
![$bc\left\{a^2-\left[(a-b)-(a-c)\right]^2-2(a-b)(a-c)+2(a-b)(a-c)\cos A\right\}=$](//latex.artofproblemsolving.com/7/6/2/762218014c4a6a824c17e55e294c73b8f47c5a81.png)
![$bc\left[a^2-(b-c)^2-4(a-b)(a-c)\sin^2\frac A2\right]=$](//latex.artofproblemsolving.com/8/2/2/822c18bf7da47e2c44691015188dd0317b5b8a10.png)
. Thus, 
. In conclusion, 
, i.e. .
Quote:
PP1. Let be a triangle and let 
, 
be two points so that 
. Find the locus of the midpoint of .
be a triangle and let 
, 
be two points so that 
. Find the locus of the midpoint of ![$[XY]$](http://latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
.
separates 
, 
or not. Prove easily that the locus of the midpoint of is included in the reunion of two lines which are parallel to the exterior or interior
-angle bisectors of 
.Quote:
PP2. Let be a triangle and let 
, 
be two points so that . Find the locus of the circumcenter for 
.
be a triangle and let 
, 
be two points so that . Find the locus of the circumcenter for 
.
, 
of the triangles , 
respectively and 
. Construct the projections 
, 
of , on 
respectively, the projections 
, 
of , on 
respectively and the projections 
, 
of on 
, 
respectively. Prove easily that 
and 
is a cyclical deltoid with the axis of symmetry 
. Since 
, 
and 
is bisector of 
obtain that is parallelly with the -bisector of . In conclusion, the locus of is included in a parallel line to the -bisector of . Quote:
Lemma. Let 
be a convex quadrilateral which is inscribed in the circle with diameter ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Prove that the locus of the point
for which the projections of![$[CL]$](//latex.artofproblemsolving.com/6/1/4/614c50b36d7cf94512eec4db863151c6d9c90739.png)
on the sidelines , 
are equally is included in a parallel line to the angle bisector of 
.
Proof. Denote
, 
. Observe that ![$\mathrm{pr}_{AB}[OL]=\mathrm{pr}_{AB}[OL]$](//latex.artofproblemsolving.com/c/3/0/c305a7051afaaca54ac5a72c6f0fe4bc88fe2aff.png)
is perpendicular on the exterior -angle bisector of 
, i.e. belongs to a parallel line to the bisector of 
.
be a convex quadrilateral which is inscribed in the circle with diameter ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Prove that the locus of the point for which the projections of
![$[CL]$](http://latex.artofproblemsolving.com/6/1/4/614c50b36d7cf94512eec4db863151c6d9c90739.png)
on the sidelines , 
are equally is included in a parallel line to the angle bisector of 
.Proof. Denote
, 
. Observe that ![$\mathrm{pr}_{AB}[OL]=\mathrm{pr}_{AB}[OL]$](http://latex.artofproblemsolving.com/c/3/0/c305a7051afaaca54ac5a72c6f0fe4bc88fe2aff.png)
is perpendicular on the exterior -angle bisector of 
, i.e. belongs to a parallel line to the bisector of 
.
, where , are defined in the previous proof.PP3. Let
be a nonisosceles triangle. Consider 
, 
so that 
. Prove that 
and 
.Proof. Suppose w.l.o.g.
, 
. In this case 
, 
and 
. Thus, 
. Denote the tangent points 
, 
of the incircle with , respectively. Observe that 
and 
. Therefore, 
, i.e. 
. From the relations 
, 
obtain that 
, i.e. .From the generalized Pytagoras' theorem apling to
obtain that 
![$bc\left[a^2-(a-b)^2-(a-c)^2+2(a-b)(a-c)\cos A\right]=$](http://latex.artofproblemsolving.com/4/2/4/4242b0b8818eed69c05c9a6b3f199e0170f2133e.png)
![$bc\left\{a^2-\left[(a-b)-(a-c)\right]^2-2(a-b)(a-c)+2(a-b)(a-c)\cos A\right\}=$](http://latex.artofproblemsolving.com/7/6/2/762218014c4a6a824c17e55e294c73b8f47c5a81.png)
![$bc\left[a^2-(b-c)^2-4(a-b)(a-c)\sin^2\frac A2\right]=$](http://latex.artofproblemsolving.com/8/2/2/822c18bf7da47e2c44691015188dd0317b5b8a10.png)
. Thus, 
. In conclusion, 
, i.e. .This post has been edited 24 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:28 AM
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