Difference between revisions of "1985 IMO Problems/Problem 5"
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<math>M</math> is the Miquel Point of quadrilateral <math>ACNK</math>, so there is a spiral similarity centered at <math>M</math> that takes <math>KA</math> to <math>NC</math>. Let <math>M_1</math> be the midpoint of <math>KA</math> and <math>M_2</math> be the midpoint of <math>NC</math>. Thus the spiral similarity must also send <math>M_1</math> to <math>M_2</math> and must also be the center of another spiral similarity that sends <math>KN</math> to <math>M_1 M_2</math>, so <math>BMM_1 M_2</math> is cyclic. <math>OM_1 B M_2</math> is also cyclic with diameter <math>BO</math> and thus <math>M</math> must lie on the same circumcircle as <math>B</math>, <math>M_1</math>, and <math>M_2</math> so <math>\angle OMB = 90^{\circ}</math>. | <math>M</math> is the Miquel Point of quadrilateral <math>ACNK</math>, so there is a spiral similarity centered at <math>M</math> that takes <math>KA</math> to <math>NC</math>. Let <math>M_1</math> be the midpoint of <math>KA</math> and <math>M_2</math> be the midpoint of <math>NC</math>. Thus the spiral similarity must also send <math>M_1</math> to <math>M_2</math> and must also be the center of another spiral similarity that sends <math>KN</math> to <math>M_1 M_2</math>, so <math>BMM_1 M_2</math> is cyclic. <math>OM_1 B M_2</math> is also cyclic with diameter <math>BO</math> and thus <math>M</math> must lie on the same circumcircle as <math>B</math>, <math>M_1</math>, and <math>M_2</math> so <math>\angle OMB = 90^{\circ}</math>. |
Revision as of 22:38, 26 February 2013
Problem
A circle with center passes through the vertices
and
of the triangle
and intersects the segments
and
again at distinct points
and
respectively. Let
be the point of intersection of the circumcircles of triangles
and
(apart from
). Prove that
.
Solution
is the Miquel Point of quadrilateral
, so there is a spiral similarity centered at
that takes
to
. Let
be the midpoint of
and
be the midpoint of
. Thus the spiral similarity must also send
to
and must also be the center of another spiral similarity that sends
to
, so
is cyclic.
is also cyclic with diameter
and thus
must lie on the same circumcircle as
,
, and
so
.