Difference between revisions of "2003 JBMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | Let <math>D</math>, <math>E</math>, <math>F</math> be the midpoints of the arcs <math>BC</math>, <math>CA</math>, <math>AB</math> on the circumcircle of a triangle <math>ABC</math> not containing the points <math>A</math>, <math>B</math>, <math>C</math>, respectively. Let the line <math>DE</math> | + | Let <math>D</math>, <math>E</math>, <math>F</math> be the midpoints of the arcs <math>BC</math>, <math>CA</math>, <math>AB</math> on the circumcircle of a triangle <math>ABC</math> not containing the points <math>A</math>, <math>B</math>, <math>C</math>, respectively. Let the line <math>DE</math> meet <math>BC</math> and <math>CA</math> at <math>G</math> and <math>H</math>, and let <math>M</math> be the midpoint of the segment <math>GH</math>. Let the line <math>FD</math> meet <math>BC</math> and <math>AB</math> at <math>K</math> and <math>J</math>, and let <math>N</math> be the midpoint of the segment <math>KJ</math>. |
a) Find the angles of triangle <math>DMN</math>; | a) Find the angles of triangle <math>DMN</math>; | ||
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==Solution== | ==Solution== | ||
− | 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>. | ||
− | Consequently, we can determine the <math>\angle | + | Consequently, we can determine the <math>\angle FDE</math> of <math>\triangle DEF</math> as being equal to <math>\angle FDA + \angle ADE = \angle C/2 + \angle B/2 = 90^{\circ} - \angle A/2</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>. | 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 | ||
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Now lines <math>AD</math>, <math>BE</math> and <math>CF</math> intersect at point <math>Q</math>. So <math>Q</math> is the incenter of <math>\triangle ABC</math> and orthocenter of <math>\triangle DEF</math>. | Now lines <math>AD</math>, <math>BE</math> and <math>CF</math> intersect at point <math>Q</math>. So <math>Q</math> is the incenter of <math>\triangle ABC</math> and orthocenter of <math>\triangle DEF</math>. | ||
− | Clearly, <math>QNDM</math> is a cyclic quadrilateral as | + | Clearly, <math>QNDM</math> is a cyclic quadrilateral as <math>N</math>, <math>M</math> are the feet of perpendiculars from <math>E</math> and <math>F</math>. |
So, we have <math>\angle QNM = \angle QDM = \angle ADE = \angle B/2</math>. | So, we have <math>\angle QNM = \angle QDM = \angle ADE = \angle B/2</math>. | ||
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<math>Part 1</math>: Angles of <math>\triangle DMN</math>: | <math>Part 1</math>: Angles of <math>\triangle DMN</math>: | ||
− | + | Since <math>\angle QNM = \angle B/2, \angle DNM = 90^{\circ} - \angle B/2</math>. | |
Similarly <math>\angle NMD = 90^{\circ} - \angle C/2</math>. | Similarly <math>\angle NMD = 90^{\circ} - \angle C/2</math>. | ||
Finally <math>\angle NDM = 90^{\circ} - \angle A/2</math>. | Finally <math>\angle NDM = 90^{\circ} - \angle A/2</math>. | ||
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Thus we have <math>RN</math> = <math>RM</math> = <math>RD</math>. So <math>R</math> is the circumcenter of <math>\triangle DMN</math>. | Thus we have <math>RN</math> = <math>RM</math> = <math>RD</math>. So <math>R</math> is the circumcenter of <math>\triangle DMN</math>. | ||
+ | |||
+ | |||
+ | <math>Kris17</math> |
Latest revision as of 22:53, 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
meet
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
,
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
.