1983 IMO Problems/Problem 2
Problem
Let be one of the two distinct points of intersection of two unequal coplanar circles
and
with centers
and
respectively. One of the common tangents to the circles touches
at
and
at
, while the other touches
at
and
at
. Let
be the midpoint of
and
the midpoint of
. Prove that
.
Solution 1
Let be one of the two distinct points of intersection of two unequal coplanar circles
and
with centers
and
respectively. Let
be such point on line
so that tangents on
touches it at
and
and tangents on
touches it at
and
. Let
be the midpoint of
and
the midpoint of
. Prove that
.
Proof: Since is image of
under inversion wrt circle
we have:
Since
is image of
under inversion wrt circle
we have:
Image of
is in both cases
itself, since it lies on both circles.
Since
we have:
Now:
This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]
Solution 2
Let and
meet at
. Let
meet
at
. Now, it is well-known that
,
, and
are concurrent at
, the center of homothety between
and
. Now, it is well-known that
bisects
. Since
, we have that
meets
at its midpoint,
, and
is perpendicular to
. Similarly,
passes through
and is perpendicular to
. Since
, we have that
, which implies that
is cyclic. Yet, since
and
lie on
and
respectively and are collinear with
, we see that the homothety that maps
to
about
maps
to
. Also,
is mapped to
by this homothety, and since
and
are corresponding parts in these circles,
is mapped to
by this homothety, so
, from which we conclude that
.
This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]
Solution 3
Let be the other intersection point of these two circles. Let
,
,
meet at
. Let
meet
at
.
Clearly,
and
are perpendicular at
;
and
are perpendicular at
.
Since
,
Since
is on the radical axis,
, so in the right trapezoid
,
is the midsegment. So we have
and
.
Let
be a point on
such that
(which means
ve
). So, from
, we get
This means,
is the angle bisector of
. So,
This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]
Solution 4
Let, concur at
.Then a homothety with centre
that sends
to
.Let
.Under the homothety
is the image of
.So,
and
.so
Hence
are concyclic.so
This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]
Solution 5
It can be easily solved with this lemma:
Let -symmedian of
intersect its circumscribed circle
at
on
satisfying
. Let us define center of
as
. Then
are concyclic.
Now let
be intersection of tangents to circles from the problem,
intersects
at
. One can prove that
is harmonic qudrilateral, so
is symmedian of
. By lemma we get that
are concyclic, thus
is equivalent to
. By homothety we get
.
This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]
1983 IMO (Problems) • Resources | ||
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