Difference between revisions of "2004 IMO Shortlist Problems/G3"
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Let <math> \displaystyle O </math> be the circumcenter of an acute-angled triangle <math> \displaystyle ABC </math> with <math> \angle B < \angle C </math>. The line <math> \displaystyle AO </math> meets the side <math> \displaystyle BC </math> at <math> \displaystyle D </math>. The circumcenters of the triangles <math> \displaystyle ABD </math> and <math> \displaystyle ACD </math> are <math> \displaystyle E </math> and <math> \displaystyle F </math>, respectively. Extend the sides <math> \displaystyle BA </math> and <math> \displaystyle CA </math> beyond <math> \displaystyle A </math>, and choose, on the respective extensions points <math> \displaystyle G </math> and <math> \displaystyle H </math> such that <math> \displaystyle AG= AC </math> and <math> \displaystyle AH = AB </math>. Prove that the quadrilateral <math> \displaystyle EFGH </math> is a rectangle if and only if <math> \angle ACB - \angle ABC = 60^{\circ} </math>. | Let <math> \displaystyle O </math> be the circumcenter of an acute-angled triangle <math> \displaystyle ABC </math> with <math> \angle B < \angle C </math>. The line <math> \displaystyle AO </math> meets the side <math> \displaystyle BC </math> at <math> \displaystyle D </math>. The circumcenters of the triangles <math> \displaystyle ABD </math> and <math> \displaystyle ACD </math> are <math> \displaystyle E </math> and <math> \displaystyle F </math>, respectively. Extend the sides <math> \displaystyle BA </math> and <math> \displaystyle CA </math> beyond <math> \displaystyle A </math>, and choose, on the respective extensions points <math> \displaystyle G </math> and <math> \displaystyle H </math> such that <math> \displaystyle AG= AC </math> and <math> \displaystyle AH = AB </math>. Prove that the quadrilateral <math> \displaystyle EFGH </math> is a rectangle if and only if <math> \angle ACB - \angle ABC = 60^{\circ} </math>. | ||
− | (''This was also Problem 2, Day 3 of the 2005 Moldova | + | (''This was also Problem 2 of the 2005 3rd German [[TST]]; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.'') |
== Solution == | == Solution == | ||
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* [[2004 IMO Shortlist Problems]] | * [[2004 IMO Shortlist Problems]] | ||
+ | * [[2005 Germany TST Problems]] | ||
* [[2005 Moldova TST Problems]] | * [[2005 Moldova TST Problems]] | ||
* [[2005 Taiwan TST Problems]] | * [[2005 Taiwan TST Problems]] |
Latest revision as of 10:45, 10 June 2007
Problem
(South Korea)
Let be the circumcenter of an acute-angled triangle
with
. The line
meets the side
at
. The circumcenters of the triangles
and
are
and
, respectively. Extend the sides
and
beyond
, and choose, on the respective extensions points
and
such that
and
. Prove that the quadrilateral
is a rectangle if and only if
.
(This was also Problem 2 of the 2005 3rd German TST; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.)
Solution
Lemma. In any triangle with circumcenter
, the altitude from
is the reflection of
over the angle bisector of
.
Proof. This is well-known, but we prove it anyway. Let meet sides
at
, and let
be the foot of the altitude from
. Let us denote
,
,
, and let us use the notation
for the angles of triangle
. By virtue of inscribed arcs in the circumcircle of
, we know
,
,
, so
, and again by inscribed arcs,
. The lemma follows. ∎
We first note that is the reflection of
over the exterior angle bisector of
. It follows that line
is the altitude from
in triangle
, i.e.,
. Since both
and
line on the perpendicular bisector of
, it follows that
and
are always parallel.
We extend and
to meet a point
. Since
,
is a convex quadrilateral. In particular, if we use the notation
,
,
, then
,
,
, so
. It follows that line
makes an angle of
with
. Now, if
is the midpoint of
and
is the midpoint of
, we note that
and
are perpendicular to
. Hence
. But if
is a rectangle, then
, so
and
. Thus the condition
is necessary for
to be a rectangle.
We now prove that it is sufficient. From the previous paragraph, we know that if , then
is a parallelogram. It is sufficient to show that if
is the intersection of line
with
and
is the intersection of line
and
, then
, since
is perpendicular to
and
. Indeed, since the cosine of the angle between lines
and
is
, it is sufficient to show that if
is the projection of
onto
, then
. Let
be the projection of
onto
. Since
are congruent,
. On the other hand, since
is the midpoint of
,
is the midpoint of the projection of
onto
, namely,
, so
, as desired. Thus
is a rectangle if and only if
, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.