410. Geometry 3.
by Virgil Nicula, Mar 20, 2015, 1:42 PM
P2 (Miguel Ochoa Sanchez) - for the middle school. Let 
be an 
-isosceles triangle for what its orthocenter 
belongs to the incircle 
Prove that 
Proof. Denote
and the diameter ![$[AS]$](//latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of the circumcircle 
Thus, 
and 
and
Hence 
i.e. 
and 
i.e. 
From the
-right 
obtain that 
In conclusion, 
Remark. Suppose w.l.o.g.
Hence 
P3.Let
with the incircle 
and the circumcircle 
. Let the midpoint 
of 
w.r.t. 
. Denote 
, 
,
and 
such that 
what doesn't contain w.r.t. . Prove that the rays 
and 
are isogonal conjugate w.r.t. 
.
Proof. I"ll use the same notations as above and let
, 
and 
. Claim: 
. 
combined with 
we get the desired result. Now: and 
. Therefore: 
and 
. Hence,
one gets that
because 
. Since 
& 
obtain that 
where 
. However 
. Since 
& 
get 
which ends the proof.
P4. Circles of radii
and 
touch each other externally, and they have a common tangent at 
and 
respectively, where lies between and . Prove that 
.
Proof. Let
. Then by Pythagoras we have 
. From 
and 
, we can get
. The inequality becomes 
We can rearrange
to
By SOS and 
, it suffices to show that 
But squaring and rearranging, this drops to 
what is truly.
P5. In an acute define an interior 
so that 
. Prove that 
.
Proof. Denote
, 
so that 
. Since 
the quadrilateral 
is cyclic. Thus 
,
and 
. Thus, 
and 
what is symmetrical expresion w.r.t. . In conclusion 
.
P6. Let an acute with the orthocenter . Denote 
and the points 
. Prove that 
.
Proof (with areas). Denote
and 
, where 
. Prove easily that ![$H\in DF\iff [AFHE]+[CDHE]=[AFDC]\iff$](//latex.artofproblemsolving.com/7/5/1/751b875ab0c1d80a5968db3e84facd21197a5b8e.png)
. Observe that 
and
. From the product of the relations and obtain that 
.
P7 (M.O.Sanchez). Let with 
and the incenter 
. Denote 
. Prove that 
.
Proof 1. Observe that
. Otherwise, 
. With the Aubel's relations 
get
, what is truly.
Proof 2.
![$\frac {[(s-a)+(s-b)]^2+[(s-a)-(s-b)]^2}{[(s-c)+(s-b)]^2+[(s-c)-(s-b)]^2}\cdot\frac {\frac {s(s-c)}{ab}}{\frac {s(s-a)}{bc}}=$](//latex.artofproblemsolving.com/8/d/0/8d0facea4823185c5907362b9965ef817e77848b.png)
. In conclusion, 
, what is truly.
P8. Let an acute with 
and let 
and 
be the feet of the altitude and the angle bisector from , let 
, 
be the intersection points of the circumcircle of
with 
, 
respectively. Prove that 
.
Here is an easy extension which generates and another interesting problem !
An easy extension. Let acute with 
and 
. The circumcircle of 
cut
again
, 
in 
, 
respectively. Then 
.
Particular cases.
, i.e. 
. In this case, 
.
. In this case obtain the following nice problem :
Let with 
and 
. Choose 
. The circumcircle of cut
again , in , respectively. Then the lines 
.
Here is a very easy problem which, for 
, is equivalently with the previous problem :
Let
be a triangle. For a point 
denote the points 
. Then the lines 
, 
, 
are concurently.
P9. Let-right , 
so that 
and incircles 
, 
of the triangles 
, 
respectively. Let the common
tangent
of 
, 
so that 
and 
. Prove that 
and 
.
Proof. I"ll use relations
and 
, where 
is the area of and 
. Let 
. Thus,
.
Let
and 
. Observe that 
and 
. Thus, 
. Suppose w.l.o.g. that 
. In conclusion, 
.
P10 (BMO - 2003). Let the tangent to the circumcircle of at 
meet the line 
at . The perpendicular to at 
meets the
perpendicular bisector of at . The perpendicular to at 
meets the perpendicular bisector of at . Prove that 
, and are collinear.
Proof 1. I"ll use the well-known property
. From 
obtain that 
.
Proof 2. From and 
get 
. The area ![$S=[BEFC]=\frac a2\cdot (BE+CF)$](//latex.artofproblemsolving.com/a/9/5/a955d758e91e783b6649015282fc04d382fee9f9.png)
and
(constant). Hence is 
is 
.
be an 
-isosceles triangle for what its orthocenter 
belongs to the incircle 
Prove that 
Proof. Denote
and the diameter ![$[AS]$](http://latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of the circumcircle 
Thus, 
and 
and
Hence 
i.e. 
and 
i.e. 
From the
-right 
obtain that 
In conclusion, 
Remark. Suppose w.l.o.g.
Hence 
P3.Let
with the incircle 
and the circumcircle 
. Let the midpoint 
of 
w.r.t. 
. Denote 
, 
,
and 
such that 
what doesn't contain w.r.t. . Prove that the rays 
and 
are isogonal conjugate w.r.t. 
.Proof. I"ll use the same notations as above and let
, 
and 
. Claim: 
. 
combined with 
we get the desired result. Now: and 
. Therefore: 
and 
. Hence,one gets that
because 
. Since 
& 
obtain that 
where 
. However 
. Since 
& 
get 
which ends the proof.P4. Circles of radii
and 
touch each other externally, and they have a common tangent at 
and 
respectively, where
lies between and . Prove that 
.Proof. Let
. Then by Pythagoras we have 
. From 
and 
, we can get
. The inequality becomes 
We can rearrangeto
By SOS and 
, it suffices to show that 
But squaring and rearranging, this drops to 
what is truly.P5. In an acute
define an interior 
so that 
. Prove that 
.Proof. Denote
, 
so that 
. Since 
the quadrilateral 
is cyclic. Thus 
,
and 
. Thus, 
and 
what is symmetrical expresion w.r.t. . In conclusion 
.P6. Let an acute
with the orthocenter . Denote 
and the points 
. Prove that 
.Proof (with areas). Denote
and 
, where 
. Prove easily that ![$H\in DF\iff [AFHE]+[CDHE]=[AFDC]\iff$](http://latex.artofproblemsolving.com/7/5/1/751b875ab0c1d80a5968db3e84facd21197a5b8e.png)
. Observe that 
and
. From the product of the relations and obtain that 
. P7 (M.O.Sanchez). Let
with 
and the incenter 
. Denote 
. Prove that 
.Proof 1. Observe that
. Otherwise, 
. With the Aubel's relations 
get
, what is truly.Proof 2.
![$\frac {[(s-a)+(s-b)]^2+[(s-a)-(s-b)]^2}{[(s-c)+(s-b)]^2+[(s-c)-(s-b)]^2}\cdot\frac {\frac {s(s-c)}{ab}}{\frac {s(s-a)}{bc}}=$](http://latex.artofproblemsolving.com/8/d/0/8d0facea4823185c5907362b9965ef817e77848b.png)
. In conclusion, 
, what is truly.P8. Let an acute
with 
and let 
and 
be the feet of the altitude and the angle bisector from , let 
, 
be the intersection points of the circumcircle of
with 
, 
respectively. Prove that 
.Here is an easy extension which generates and another interesting problem !
An easy extension. Let acute
with 
and 
. The circumcircle of 
cutagain
, 
in 
, 
respectively. Then 
.Particular cases.
, i.e. 
. In this case, 
.
. In this case obtain the following nice problem :Let
with 
and 
. Choose 
. The circumcircle of cutagain
, in , respectively. Then the lines 
.
Here is a very easy problem which, for 
, is equivalently with the previous problem :Let
be a triangle. For a point 
denote the points 
. Then the lines 
, 
, 
are concurently.P9. Let
-right , 
so that 
and incircles 
, 
of the triangles 
, 
respectively. Let the commontangent
of 
, 
so that 
and 
. Prove that 
and 
.Proof. I"ll use relations
and 
, where 
is the area of and 
. Let 
. Thus,
.Let
and 
. Observe that 
and 
. Thus, 
. Suppose w.l.o.g. that 
. In conclusion, 
.P10 (BMO - 2003). Let
the tangent to the circumcircle of at 
meet the line 
at . The perpendicular to at 
meets theperpendicular bisector of
at . The perpendicular to at 
meets the perpendicular bisector of at . Prove that 
, and are collinear.Proof 1. I"ll use the well-known property
. From 
obtain that 
.Proof 2. From
and 
get 
. The area ![$S=[BEFC]=\frac a2\cdot (BE+CF)$](http://latex.artofproblemsolving.com/a/9/5/a955d758e91e783b6649015282fc04d382fee9f9.png)
and
(constant). Hence is 
is 
.This post has been edited 59 times. Last edited by Virgil Nicula, Sep 23, 2016, 1:15 AM
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