347. Two isogonal lines in a triangle.
by Virgil Nicula, Jun 23, 2012, 4:43 PM
PP1. Let 
-right 
and rectangle 
, where 
. Let 
and 
. Prove that 
.
Proof 1 (metric). Let
. Denote
. In conclusion, .
Proof 2 (metric). . Denote 
and apply
the Menelaus' theorem to the following transversals:
. From the Steiner's theorem obtain that the rays 
and 
are isogonals w.r.t. the angle 
, i.e.
Proof 3 (analytic).
Lemma. In any triangle
there is the equivalence 
.
Proof. Denote
so that 
. Thus, 
and 
.
PP2. Let with 
. Prove that 
.
Proof 1. Let cyclic
so that 
and 
. Thus, 
and 
.
The Ptolemy's relation
and 
.
Proof 2. Let so that 
. Thus, 
and
. Since 
from upper
lemma obtain that
.
Proof 3 (without upper lemma). Denote
. Therefore,
In conclusion,
.
Proof 4.
. Therefore,
![$\frac {a-b}{2b}=2\cdot\left[\frac {c}{2(a-b)}\right]^2-1\iff$](//latex.artofproblemsolving.com/c/c/2/cc26903946c1ec1d0c1bf5eacb7bfae05a970f92.png)
.
Extension. (see three upper proof). Let and twin sequencies 
, for any 
. Prove that 
.
PP3. Let with the orthocenter 
and the circumcenter 
. The perpendicular bisector of ![$[AH]$](//latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
meet 
, 
in 
, 
. Prove that 
.
Proof.
. I used that the -altitude and the -diameter of the circumcircle are isogonal lines in the angle .
PP4. Let with orthocenter , circumcenter , 
so that 
, 
and the midpoint of ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. Prove that 
pass through the midpoint of ![$[OH]$](//latex.artofproblemsolving.com/7/b/8/7b812c672802ddc509741f60b6fd241be97668af.png)
.
Proof 1. Let midpoints
, 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, respectively and diameter ![$[AS]$](//latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of circumcircle 
of . Is well-known that 
is middle line in 
, midpoint of is center
of Euler's circle and it is midpoint of , i.e. 
is parallelogram. Thus, 
in 
pass through midpoint of 
, i.e. pass through midpoint of 
.
Proof 2 (Sunken Rock). Let be a diameter of the circumcircle , 
and 
. It is well-known that 
and 
.
Hence
and 
. The Menelaus' theorem to the transversal 
.
Thus,
, i.e. pass through the midpoint of .
PP5. The trisectors of of are: median 
and symmedian 
, where 
. Prove that 
. See and here.
Proof. Suppose w.l.o.g.
and denote 
, 
. Prove easily that 
and 
, i.e. 
.Apply the theorem of bisector:
. Thus, 
.
Lemma. Let be a triangle with the circumcircle . Denote 
and 
, where
is the tangent line to the circle in the point 
. Then the ray 
is a symmedian in the triangle .
Proof 1 (metric). Denote
. Thus, 
,
i.e.
is harmonically and 
, i.e. the ray is a symmedian in the triangle .
Proof 2 (synthetic). Denote the midpoint of . Draw the line through antiparallel to 
in so that when we extend and to meet it
at and 
, where 
. Then 
, so 
is isoceles, and similarly 
,
so
is isoceles. Thus, 
, so and 
are medians of similar triangles and thus 
.
PP6. Let be a triangle with the incircle what touch the sides , 
, at , , respectively. Let 
be the midpoint of ![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Prove that 
.
Proof. Apply upper lemma to
, where 
. Obtain that 
is -symmedian, i.e. .
-right 
and rectangle 
, where 
. Let 
and 
. Prove that 
.Proof 1 (metric). Let
. Denote
. In conclusion, .Proof 2 (metric).
. Denote 
and applythe Menelaus' theorem to the following transversals:
. From the Steiner's theorem obtain that the rays 
and 
are isogonals w.r.t. the angle 
, i.e. Proof 3 (analytic).
Lemma. In any triangle
there is the equivalence 
.Proof. Denote
so that 
. Thus, 
and 
.PP2. Let
with 
. Prove that 
.Proof 1. Let cyclic
so that 
and 
. Thus, 
and 
.The Ptolemy's relation
and 
.Proof 2. Let
so that 
. Thus, 
and
. Since 
from upperlemma obtain that
.Proof 3 (without upper lemma). Denote
. Therefore,
In conclusion,
.Proof 4.
. Therefore,
![$\frac {a-b}{2b}=2\cdot\left[\frac {c}{2(a-b)}\right]^2-1\iff$](http://latex.artofproblemsolving.com/c/c/2/cc26903946c1ec1d0c1bf5eacb7bfae05a970f92.png)
.Extension. (see three upper proof). Let
and twin sequencies 
, for any 
. Prove that 
.PP3. Let
with the orthocenter 
and the circumcenter 
. The perpendicular bisector of ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
meet 
, 
in 
, 
. Prove that 
.Proof.
. I used that the -altitude and the -diameter of the circumcircle are isogonal lines in the angle .PP4. Let
with orthocenter , circumcenter , 
so that 
, 
and the midpoint of ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. Prove that 
pass through the midpoint of ![$[OH]$](http://latex.artofproblemsolving.com/7/b/8/7b812c672802ddc509741f60b6fd241be97668af.png)
.Proof 1. Let midpoints
, 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, respectively and diameter ![$[AS]$](http://latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of circumcircle 
of . Is well-known that 
is middle line in 
, midpoint of is centerof Euler's circle and it is midpoint of
, i.e. 
is parallelogram. Thus, 
in 
pass through midpoint of 
, i.e. pass through midpoint of 
.Proof 2 (Sunken Rock). Let
be a diameter of the circumcircle , 
and 
. It is well-known that 
and 
.Hence
and 
. The Menelaus' theorem to the transversal 
.Thus,
, i.e. pass through the midpoint of .PP5. The trisectors of
of are: median 
and symmedian 
, where 
. Prove that 
. See and here.Proof. Suppose w.l.o.g.
and denote 
, 
. Prove easily that 
and 
, i.e. 
.Apply the theorem of bisector:
. Thus, 
.Lemma. Let
be a triangle with the circumcircle . Denote 
and 
, where
is the tangent line to the circle in the point 
. Then the ray 
is a symmedian in the triangle .Proof 1 (metric). Denote
. Thus, 
,i.e.
is harmonically and 
, i.e. the ray is a symmedian in the triangle .Proof 2 (synthetic). Denote the midpoint
of . Draw the line through antiparallel to 
in so that when we extend and to meet itat
and 
, where 
. Then 
, so 
is isoceles, and similarly 
,so
is isoceles. Thus, 
, so and 
are medians of similar triangles and thus 
.PP6. Let
be a triangle with the incircle what touch the sides , 
, at , , respectively. Let 
be the midpoint of ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Prove that 
.Proof. Apply upper lemma to
, where 
. Obtain that 
is -symmedian, i.e. .This post has been edited 64 times. Last edited by Virgil Nicula, Nov 17, 2015, 4:26 PM
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