411. Geometry 4.
by Virgil Nicula, Mar 20, 2015, 1:43 PM
P7 (M. Ochoa Sanchez). Let 
be a rectangle and the points 
so that 
and 
Suppose
that the quadrilaterals
and 
are circumscriptibly with the incenters 
and 
respectively. Prove that 
Proof. Denote the incircle
of 
with 
and the incircle 
of 
Prove easily that 
and
is circumscriptibly 
Observe that
Therefore, 
I"ll use the substitution 
In conclusion, the previous relation 
becomes 
P8. Let
be an equilateral triangle with the circumcircle 
and 
for which denote 
Prove that 
P9 (M. Ochoa Sanchez, Peru). The circle is interior tangent to the circle 
at 
For 
so that 
construct the
tangents
to the circle 
where 
Denote 
and 
Prove that 
Proof.
i.e. 
Thus, 
P10 (Cesar Trucios). Let a parallelogram and 
so that 
Prove that 
Particular case. If 
and 
then obtain that 
i.e. 
Proof. Denote
and observe that 
Apply the theorem of Sines in
P11. Let
be the circumcircle of 
. Let 
be the radius of the circle touching 
, 
and
internally (mixtilinear incircle). Define 
similarly. Prove that 
.
Proof 1. Let
and 
b tangency points of 
with and 
. It's well-known that the incenter 
is the midpoint of 
. Let 
be the center of . Then 
and 
.
Remark 1. Let
and apply the Casey's theorem to 
, 
, 
and 
. Thus, 
, what is true.
Remark 2. Denote
. It is well-known that 
. Apply the Pascal's theorem to the cyclical 
. Thus, 
.
P12 (1989 China League). Let
with 
. The exterior angle bisector of 
intersect the circumcircle
in 
and through construct 
. Show that 
Proof 1. Let
. Thus: 
; 
(product).
Proof 2. Let : the reflection
of w.r.t. the circumcenter 
of ; the projection 
of to 
From a well-known property 
and the relations
, 
results 
P13. In a
with 
denote 
. For 
denote 
for which 
. Prove that 
.
Proof 1.
Proof 2. Let
. So 
. Menelaus' theorem 
Therefore, 
the point 
is the orthocenter of the triangle 
Proof 3. Let
. Thus, 
. But 
and they have the homologous sides perpendicular, for our case
. Since, from 
the points 
and 
divide the homologous sides 
and 
respectively into the same ratio, i.e. 
.
An easy extension. In denote the midpoint 
of the side ![$ [BC]$](//latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. For 
denote 
. Prove that 
Proof. Menelaus's theorem
. Hence 
.
P14. Let and 
. For 
let distancies 
, 
and 
from to 
, 
, respectively. Prove that 
.
Generalization. Let and 
. For let the distancies , and from to , , respectively. Prove that 
.
P15 (Evgeniy Kulanin, Rusia). Let . Denote 
the midpoint of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
; the Nagel point 
; the Gergonne point 
; the Feuerbach's point 
.
Prove that in any triangle the line 
(lista scurta a concursului R.M.G. - Revista de Matematica din Galati,2007).
Proof. I"ll show that
Denote the second intersection 
of the Euler's circle with the median 
, the orthocenter 
,
and the midpoint of ![$[AH]$](//latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
. Is well-known that 
belong to the Euler's circle. From the power of w.r.t. this circle obtain that 
Thus, 
, i.e. 
![$\frac {p}{(p-a)}=1+\frac {(p-a)[(p-b)+(p-c)]}{(p-b)(p-c)}\iff$](//latex.artofproblemsolving.com/d/b/e/dbeaaf550a6a0e6226796b13f667153183daa1bf.png)
, where 
Denote Notam 
, where 
From
and 
obtain that 
Thus, 
, i.e
![$MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\ $](//latex.artofproblemsolving.com/6/9/7/697eb3a4293ec2e86ffc0b850cbbc5b3a36d468e.png)
Also 
![$MS\cdot ML=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot$](//latex.artofproblemsolving.com/6/0/f/60f0e384bc6f09df73c4834d7c35c8ce73ad815d.png)
, i.e. 
![$2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\ \Longleftrightarrow\ \left[a(b+c)-\left(b^2+c^2\right)\right]^2=0$](//latex.artofproblemsolving.com/3/0/8/3089c2f94c4ca71aa8387968e4ce808033109cd8.png)
I used the the metrical relations
and 
.
P16. Let be an -rightangled triangle with the incenter 
and the midpoint of the side . Prove that 
.
Proof.
![$a(a-b)=4\cdot\left[r^2+\left(\frac {b-c}2\right)^2\right]\iff$](//latex.artofproblemsolving.com/e/1/9/e19a0d5fe425787ca668a447ad361a5390dd19a8.png)
.
Denote the symmetrical point
of w.r.t. . Observe that 
is a parallelogram and 
.
Thus,
.
Remark.
. Obtain analogously 
and 
.
P17 (M.O. Sanchez). Let an acute with crcumcircle 
and 
and 
Prove that 
Proof. Apply the theorem of Cathetus to the
-right triangles 
Thus, 
is cyclic
and
is cyclic 
P18 (M. O. Sanchez). Let-isosceles 
with circumcircle and 
so that 
Prove that ![$\boxed{[ACD]=\frac 12\cdot \left(a^2-b^2\right)\cdot\sin 2\phi}\ (*).$](//latex.artofproblemsolving.com/d/5/0/d506df136778f9e20f06ceaac5f5c96ee6bba8f4.png)
Proof 1. Let
Apply the generalized Pythagoras' theorem to the triangles 
In conclusion, 
![$2\cdot [ACD]=uv\sin 2\phi\iff$](//latex.artofproblemsolving.com/8/2/8/828c76f1073a229f84417bf9909d4696fdb8dc2a.png)
![$[ACD]=\frac 12\cdot\left(a^2-b^2\right)\cdot\sin 2\phi.$](//latex.artofproblemsolving.com/9/2/5/92537536642743d31fd1553ea873af194bddec63.png)
Proof 2. Observe that
Apply the Ptolemy's theorem to 
Apply the generalized Pythagoras' theorem to the side ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
of 
Lemma. Let with 
Prove that the relation 
Proof. Apply the theorem of Sines in the triangle
P19 (Kadir Altintas, Turkey). Let with 
so that 
and 
Prove that 
Proof. I"ll use upper lemma for
i.e. 
In conclusion, 
P20 (M. Ochoa Sanchez, Peru). The circle is interior tangent to the circle at 
For a point 
construct the tangents
to the circle where 
Denote 
and 
Prove that 
Proof. From a known property obtain that the rays
are the bisectors of 
respectively. Apply the theorem of bisector from the vertex in the triangles
So 
be a rectangle and the points 
so that 
and 
Supposethat the quadrilaterals
and 
are circumscriptibly with the incenters 
and 
respectively. Prove that 
Proof. Denote the incircle
of 
with 
and the incircle 
of 
Prove easily that 
and is circumscriptibly 
Observe that
Therefore, 
I"ll use the substitution 
In conclusion, the previous relation 
becomes 
P8. Let
be an equilateral triangle with the circumcircle 
and 
for which denote 
Prove that 
P9 (M. Ochoa Sanchez, Peru). The circle
is interior tangent to the circle 
at 
For 
so that 
construct thetangents
to the circle 
where 
Denote 
and 
Prove that 
Proof.
i.e. 
Thus, 
P10 (Cesar Trucios). Let a parallelogram
and 
so that 
Prove that 
Particular case. If 
and 
then obtain that 
i.e. 
Proof. Denote
and observe that 
Apply the theorem of Sines in
P11. Let
be the circumcircle of 
. Let 
be the radius of the circle touching 
, 
and internally (mixtilinear incircle). Define 
similarly. Prove that 
.Proof 1. Let
and 
b tangency points of 
with and 
. It's well-known that the incenter 
is the midpoint of 
. Let 
be the center of . Then 
and 
.Remark 1. Let
and apply the Casey's theorem to 
, 
, 
and 
. Thus, 
, what is true.Remark 2. Denote
. It is well-known that 
. Apply the Pascal's theorem to the cyclical 
. Thus, 
.P12 (1989 China League). Let
with 
. The exterior angle bisector of 
intersect the circumcircle
in 
and through construct 
. Show that 
Proof 1. Let
. Thus: 
; 
(product).Proof 2. Let : the reflection
of w.r.t. the circumcenter 
of ; the projection 
of to 
From a well-known property 
and the relations
, 
results 
P13. In a
with 
denote 
. For 
denote 
for which 
. Prove that 
.Proof 1.
Proof 2. Let
. So 
. Menelaus' theorem 
Therefore, 
the point 
is the orthocenter of the triangle 
Proof 3. Let
. Thus, 
. But 
and they have the homologous sides perpendicular, for our case
. Since, from 
the points 
and 
divide the homologous sides 
and 
respectively into the same ratio, i.e. 
.An easy extension. In
denote the midpoint 
of the side ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. For 
denote 
. Prove that 
Proof. Menelaus's theorem
. Hence 
.P14. Let
and 
. For 
let distancies 
, 
and 
from to 
, 
, respectively. Prove that 
.Generalization. Let
and 
. For let the distancies , and from to , , respectively. Prove that 
. P15 (Evgeniy Kulanin, Rusia). Let
. Denote 
the midpoint of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
; the Nagel point 
; the Gergonne point 
; the Feuerbach's point 
.Prove that in any triangle
the line 
(lista scurta a concursului R.M.G. - Revista de Matematica din Galati,2007).Proof. I"ll show that
Denote the second intersection 
of the Euler's circle with the median 
, the orthocenter 
,
and the midpoint of ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
. Is well-known that 
belong to the Euler's circle. From the power of w.r.t. this circle obtain that 
Thus, 
, i.e. 
![$\frac {p}{(p-a)}=1+\frac {(p-a)[(p-b)+(p-c)]}{(p-b)(p-c)}\iff$](http://latex.artofproblemsolving.com/d/b/e/dbeaaf550a6a0e6226796b13f667153183daa1bf.png)
, where 
Denote Notam 
, where 
From
and 
obtain that 
Thus, 
, i.e![$MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\ $](http://latex.artofproblemsolving.com/6/9/7/697eb3a4293ec2e86ffc0b850cbbc5b3a36d468e.png)
Also 
![$MS\cdot ML=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot$](http://latex.artofproblemsolving.com/6/0/f/60f0e384bc6f09df73c4834d7c35c8ce73ad815d.png)
, i.e. 
![$2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\ \Longleftrightarrow\ \left[a(b+c)-\left(b^2+c^2\right)\right]^2=0$](http://latex.artofproblemsolving.com/3/0/8/3089c2f94c4ca71aa8387968e4ce808033109cd8.png)
I used the the metrical relations
and 
.P16. Let
be an -rightangled triangle with the incenter 
and the midpoint of the side . Prove that 
.Proof.
![$a(a-b)=4\cdot\left[r^2+\left(\frac {b-c}2\right)^2\right]\iff$](http://latex.artofproblemsolving.com/e/1/9/e19a0d5fe425787ca668a447ad361a5390dd19a8.png)
.Denote the symmetrical point
of w.r.t. . Observe that 
is a parallelogram and 
.Thus,
.Remark.
. Obtain analogously 
and 
.P17 (M.O. Sanchez). Let an acute
with crcumcircle 
and 
and 
Prove that 
Proof. Apply the theorem of Cathetus to the
-right triangles 
Thus, 
is cyclicand
is cyclic 
P18 (M. O. Sanchez). Let
-isosceles 
with circumcircle and 
so that 
Prove that ![$\boxed{[ACD]=\frac 12\cdot \left(a^2-b^2\right)\cdot\sin 2\phi}\ (*).$](http://latex.artofproblemsolving.com/d/5/0/d506df136778f9e20f06ceaac5f5c96ee6bba8f4.png)
Proof 1. Let
Apply the generalized Pythagoras' theorem to the triangles 
In conclusion, 
![$2\cdot [ACD]=uv\sin 2\phi\iff$](http://latex.artofproblemsolving.com/8/2/8/828c76f1073a229f84417bf9909d4696fdb8dc2a.png)
![$[ACD]=\frac 12\cdot\left(a^2-b^2\right)\cdot\sin 2\phi.$](http://latex.artofproblemsolving.com/9/2/5/92537536642743d31fd1553ea873af194bddec63.png)
Proof 2. Observe that
Apply the Ptolemy's theorem to 
Apply the generalized Pythagoras' theorem to the side ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
of 
Lemma. Let
with 
Prove that the relation 
Proof. Apply the theorem of Sines in the triangle
P19 (Kadir Altintas, Turkey). Let
with 
so that 
and 
Prove that 
Proof. I"ll use upper lemma for
i.e. 
In conclusion, 
P20 (M. Ochoa Sanchez, Peru). The circle
is interior tangent to the circle at 
For a point 
construct the tangents
to the circle where 
Denote 
and 
Prove that 
Proof. From a known property obtain that the rays
are the bisectors of 
respectively. Apply the theorem of bisector from the vertex in the triangles 
So 
This post has been edited 107 times. Last edited by Virgil Nicula, Oct 14, 2016, 6:31 AM
August 2018
January 2018