Difference between revisions of "2003 JBMO Problems/Problem 3"
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− | Let <math>FC</math>, <math>EB</math> intersect <math>DE</math>, <math>FD</math> at <math>M'</math>, <math>N'</math> respectively. We will prove first that <math>M' = M, N = N'</math> and that lines <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle | + | Let <math>FC</math>, <math>EB</math> intersect <math>DE</math>, <math>FD</math> at <math>M'</math>, <math>N'</math> respectively. We will prove first that <math>M' = M, N = N'</math> and that lines <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle DEF</math>. |
It's easy to see that lines <math>AD</math>, <math>BE</math> and <math>CF</math> form the internal angle bisectors of <math>\triangle ABC</math>. | It's easy to see that lines <math>AD</math>, <math>BE</math> and <math>CF</math> form the internal angle bisectors of <math>\triangle ABC</math>. | ||
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Also we have <math>\angle DFC = \angle A/2</math>, thus <math>\angle FM'D = 90^{\circ}</math>. Similarly <math>\angle EN'D = 90^{\circ}</math>. | Also we have <math>\angle DFC = \angle A/2</math>, thus <math>\angle FM'D = 90^{\circ}</math>. Similarly <math>\angle EN'D = 90^{\circ}</math>. | ||
− | Thus <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle | + | Thus <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle DEF</math> with <math>P</math>, <math>N'</math>, <math>M'</math> respectively being the feet of the altitudes. |
Now since <math>M'C</math> is internal bisector of <math>\angle HCG</math> and <math>CM'</math> is perpendicular to <math>GH</math>, we have that | Now since <math>M'C</math> is internal bisector of <math>\angle HCG</math> and <math>CM'</math> is perpendicular to <math>GH</math>, we have that |
Revision as of 01:00, 13 December 2018
Problem
Let ,
,
be the midpoints of the arcs
,
,
on the circumcircle of a triangle
not containing the points
,
,
, respectively. Let the line
meets
and
at
and
, and let
be the midpoint of the segment
. Let the line
meet
and
at
and
, and let
be the midpoint of the segment
.
a) Find the angles of triangle ;
b) Prove that if is the point of intersection of the lines
and
, then the circumcenter of triangle
lies on the circumcircle of triangle
.
Solution
Let ,
intersect
,
at
,
respectively. We will prove first that
and that lines
,
,
are altitudes of the
.
It's easy to see that lines ,
and
form the internal angle bisectors of
.
Consequently, we can determine the of
as being equal to
Also we have , thus
. Similarly
.
Thus ,
,
are altitudes of the
with
,
,
respectively being the feet of the altitudes.
Now since is internal bisector of
and
is perpendicular to
, we have that
is the perpendicular bisector of
. Hence
.
Similarly it can be shown that is the perpendicular bisector of
, and hence
.
Now lines ,
and
intersect at point
. So
is the incenter of
and orthocenter of
.
Clearly, is a cyclic quadrilateral as the
,
are the feet of perpendiculars from
and
.
So, we have .
Similarly, since is also a cyclic-quadrilateral, reasoning as above,
.
Thus we have that and so
is an internal bisector of
.
Reasoning in a similar fashion it can be proven that
and
are internal bisectors of other 2 angles of
.
Thus also happens to be the incenter of
in addition to being that of
.
: Angles of
:
Since .
Similarly
.
Finally
.
:
Let circumcircle of cut line
at point
.
Since
is a cyclic quadrilateral, we have
.
Similarly, . Thus
=
.
Now,
and . Thus
=
.
Thus we have =
=
. So
is the circumcenter of
.