333. Problems with the common tangent for two circles.
by Virgil Nicula, Jan 17, 2012, 10:58 AM
PP1. Let
,
and
be three circles so that
and
are externally tangent and both are internally tangent to
.
Find the length
of the chord from the large circle which is the interior tangent to the small circles (at its common tangent point).
Find the length
of the chord from the larger circle which is the exterior tangent to the small circles (at two points) - two cases.
Proof. Denote
so that
and

so that
is common tangent of
and
so that
.
Apply the power of
w.r.t.
. Observe that
and
.
Apply the Stewart's theorem to the cevian
in 
. Using the relation
obtain that
. Observe that
. Apply the Casey's theorem
to the circles
,
and the degenerated circles
,
. Obtain that
.
Appear two cases : the first case when
separates
and the second case when
doesn't separate
. Denote
,
.
From an well-known property the rays
,
are the bisectors of the angles
,
respectively
belongs to
and it is the
midpoint of the arc
. Apply the Casey's theorem to the circles
,
and the degenerated circles
,
. Obtain that 
what is the evident relation
. Soon !!!
PP2. Let
be a triangle with the incircle
and the circumcircle
. Consider the circle
which is tangent to
,
and which
is internal tangent to
. Denote
. The tangent from
to
touches
in
. Prove that
.
Proof. Assume w.l.o.g. that
and let
and
be the tangency points of
with
and
respectively. By Casey's theorem for the circle
and
the degenerated circles
,
,
all tangent to the circumcircle
we get 
. By Ptolemy's theorem for
.
PP3 (Balkan MO 2012). Let
be an acute-angled triangle with orthocentre
and orthic triangle
, where
,
,
. Suppose that
meets the circle with diameter
at
. Show that the circumcircles of
and
are tangent at
(this problem is modified equivalent).
Proof 1 (metric). In the
right-angled
from the theorem of cathet
obtain that
. Since
,
are cyclical quadrilaterals obtain
that
. In conclusion,
, i.e. the circumcircles of
and
are tangent at
.
Proof 2 (synthetic). Prove easily that
PP4. Let an acute
with
. Let
be the orthocenter of triangle
and let
be the midpoint of the side
.
Let
be a point on the side
and
a point on the side
such that
and the points
,
,
are on the same line.
Prove that the line
is perpendicular to the common chord of the circumscribed circles of triangle
and triangle
.
Lemma. Let
be an acute triangle. Define: the circumcircle
and the orthocentre
of the triangle
; the middlepoint
of the side
; the intersection
between
and the bisector of the angle
;
and
so that
and
. Then the point
belongs to the circumcircle
of the triangle
.
Proof of the lemma.
and
. Thus,

Denote the point
, i.e.
. Thus,
and

, i.e.
belongs to the circumcircle of 
Proof of the proposed problem. Denote:
; the middlepoint
;
. From above lemma results
.
But
(
is the middlepoint of the segment
),
and
. Thus
,
,
are the middlepoints of
,
,
respectively and
and
. Thus,
. Remark. I denote ray
without
.










Proof. Denote















Apply the Stewart's theorem to the cevian






to the circles














From an well-known property the rays






midpoint of the arc









PP2. Let






is internal tangent to







Proof. Assume w.l.o.g. that







the degenerated circles









PP3 (Balkan MO 2012). Let







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




Proof 1 (metric). In the


![$[PC]$](http://latex.artofproblemsolving.com/7/8/1/7818452d29e34a743729cef6d6762490b71f626d.png)



that





Proof 2 (synthetic). Prove easily that

PP4. Let an acute






Let








Prove that the line



Lemma. Let





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










Proof of the lemma.

























Proof of the proposed problem. Denote:


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


But


![$[HA']$](http://latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)















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This post has been edited 77 times. Last edited by Virgil Nicula, Nov 19, 2015, 1:20 PM