403. Some nice geometry problems for the high school. II
by Virgil Nicula, Nov 10, 2014, 6:31 AM
PP15. Let
and
be two circles so that
The line
is a common tangent for
and
in
and
.
Let
be the semicircle with
so that the circles
are interior tangent to
Prove that 
Proof. Apply the Pythagoras's theorem to
Therefore,

, where
Thus, 

Let
and
so that
and
Observe that
Apply the Pythagoras's
theorem to
From the relations 
and the relation
obtain that

From the relations
and
obtain that
. Drawing of PP15 & PP16.
PP16. Let
and
be two circles so that
The line
is a common tangent for
and
in
and
.
Let
be the semicircle with
and what is exterior tangent to the circles
Prove that 
Proof. Apply the Pythagoras's theorem to
Therefore,



Let
and
so that
and
Observe that
Apply the Pythagoras's
theorem to
From the relations 
and the relation
obtain that

From the relations
and
obtain that
PP17. Let incircle
of
and
,
so that
is tangent to
. Prove that there is the relation 
Proof. Denote
,
and
. Thus,
, i.e.
because
and
. Hence
, i.e.
. Thus,
.
The required relation is equivalently with


, i.e. the relation
.
PP18. Let the incircle
of
,
,
and for
let the lengths 
of the distancies from
to
,
,
respectively. Prove that
.
Proof. Denote
so that
and
so that
. Thus,

. Obtain analogously
. Prove easily that
. Apply the Ptolemy theorem to the cyclical quadrilateral
.
PP19. Let
and
so that
. Prove that
Particularly 
Proof 1. Apply Stewart's relation to cevian
in
Using an well-known
property in
obtain that
If
then from
obtain that


PP20. Let
with the orthocenter
and the circumcircle
. Denote
and
. Prove that
.
Proof. Suppose w.l.o.g. that
. Let
so that
is a diameter of
and
so that
. Is well-known
that
is a parallelogram. Thus,
is the intersection of its diagonals. Prove easily that
.
PP21.
is an inscribed quadrilateral . Through
and
had a circle tangent to
at
; through
and
had a circle tangent to
at
; through
and
had a circle tangent to
at
; through
and
had a circle tangent to
at
. Prove that
.
Proof. Let
. Easy to see the bisector of
and the bisector of
are perpendicular
. Since
the triangles
and
are isosceles
.
PP22. Let
with incircle
which touches
at
,
and
. Let
. Prove that
and
, where
are the lengths of the inradii of
.
Proof.
. Thus,
. In conclusion, 
, what is truly.
Remark. Denote
. Thus,
. Using the Brianchon's theorem obtain that
.
PP23. Let
and the midpoint
of
, Nagel's point
, Gergonne's point
and Feuerbach's point
. Prove that
.
Proof. (See here). Let the second intersection
of the Euler's circle
with the
-median
, the orthocenter
,
and the midpoint
of
. Is well-known
and from the power of
w.r.t.
obtain that

. Thus,
, i.e.
.

, where
Let
, where
From
and
obtain that
Thus,
, i.e.
Also,
, i.e.

Observatie. Apropo de subiectul topicului "Puncte importante intr-un triunghi". Pentru cei care nu cunosc semnificatiile punctelor remarcabile mentionate in enuntul problemei propuse
le recomand sa consulte Google, unde, spre surprinderea lor, vor intalni cel putin 100 de asemenea puncte ceea ce inseamna ca triunghiul a fost si ramane o preocupare de secole
a multor matematicieni, multi dintre ei cu rezultate remarcabile la nivel inalt. In demonstratie am folosit urmatoarele relatii metrice dintr-un triunghi care se pot dovedi fara dificultate
and 
PP24. Let a square
and its interior point
for which
, where
, i.e.
and
. Prove that
.
Proof. Let
and suppose w.l.o.g.
. Apply theorem of Sines in



.








Let






Proof. Apply the Pythagoras's theorem to

Therefore,
















theorem to


and the relation







PP16. Let








Let





Proof. Apply the Pythagoras's theorem to

Therefore,















theorem to


and the relation







PP17. Let incircle







Proof. Denote



![$[AMN]=2\cdot [AIM]+2\cdot [AIN]-[AEIF]$](http://latex.artofproblemsolving.com/2/0/b/20b2b442ad84abc6a9cd4f54c0881b31e842be7c.png)







The required relation is equivalently with















PP18. Let the incircle




![$P\in \overarc[]{EF}$](http://latex.artofproblemsolving.com/5/2/a/52aad01ba7b62fda7f1615ada91abf33dfb7627a.png)

of the distancies from





Proof. Denote














PP19. Let





Proof 1. Apply Stewart's relation to cevian



property in








![$a\cdot MC^2+\left[\left(c^2+ac-b^2\right)-a^2\right]\cdot MC-a\left(c^2+ac-b^2\right)=0$](http://latex.artofproblemsolving.com/1/4/1/141080b68876611a38498bbe09fb2bb4d2203d8d.png)

![$\left(MC-a\right)\left[aMC+c(a+c)-b^2\right]=0\iff$](http://latex.artofproblemsolving.com/9/8/3/9838ae54547e2985ca3e2e4ab314d8416a2b921a.png)



PP20. Let






Proof. Suppose w.l.o.g. that






that





PP21.


















Proof. Let








PP22. Let











Proof.








Remark. Denote



PP23. Let


![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)








Proof. (See here). Let the second intersection







![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)





























![$MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\ .$](http://latex.artofproblemsolving.com/1/4/5/1456270a82e4a0e97c6abe4b53882cc71d4bd141.png)

![$\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot \frac {b^2+c^2}{4m_a}$](http://latex.artofproblemsolving.com/7/8/7/787f51426e621fd0472a4fd7752bcfdf9cf79bf8.png)

![$2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\iff$](http://latex.artofproblemsolving.com/e/a/f/eaf760aa5d33554d2eb2d8f82416ba7f2332898c.png)

Observatie. Apropo de subiectul topicului "Puncte importante intr-un triunghi". Pentru cei care nu cunosc semnificatiile punctelor remarcabile mentionate in enuntul problemei propuse
le recomand sa consulte Google, unde, spre surprinderea lor, vor intalni cel putin 100 de asemenea puncte ceea ce inseamna ca triunghiul a fost si ramane o preocupare de secole
a multor matematicieni, multi dintre ei cu rezultate remarcabile la nivel inalt. In demonstratie am folosit urmatoarele relatii metrice dintr-un triunghi care se pot dovedi fara dificultate



PP24. Let a square







Proof. Let

















This post has been edited 175 times. Last edited by Virgil Nicula, Jun 4, 2016, 3:21 PM