Difference between revisions of "2004 IMO Shortlist Problems/G8"
Line 2: | Line 2: | ||
A cyclic quadrilateral <math>ABCD</math> is given. The lines <math>AD</math> and <math>BC</math> intersect at <math>E</math>, with <math>C</math> between <math>B</math> and <math>E</math>; the diagonals <math>AC</math> and <math>BD</math> intersect at <math>F</math>. Let <math>M</math> be the midpoint of the side <math>CD</math>, and let <math>N \neq M</math> be a point on the circumcircle of <math>\triangle ABM</math> such that <math>\frac{AN}{BN} = \frac{AM}{BM}</math>. Prove that <math>E, F, N</math> are collinear. | A cyclic quadrilateral <math>ABCD</math> is given. The lines <math>AD</math> and <math>BC</math> intersect at <math>E</math>, with <math>C</math> between <math>B</math> and <math>E</math>; the diagonals <math>AC</math> and <math>BD</math> intersect at <math>F</math>. Let <math>M</math> be the midpoint of the side <math>CD</math>, and let <math>N \neq M</math> be a point on the circumcircle of <math>\triangle ABM</math> such that <math>\frac{AN}{BN} = \frac{AM}{BM}</math>. Prove that <math>E, F, N</math> are collinear. | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | Let <math>P = CD \cap EF</math>. Let <math>Q = AB \cap CD</math>. Let <math>R = AB \cap EF</math>. Let<math>(ABM)</math> denote the circumcircle of <math>\triangle ABM</math>. Let <math>N' = EF \cap (ABM)</math>. Note that <math>N' = PR \cap (ABM)\newline</math> | ||
+ | |||
+ | <math>P</math> is on <math>(ABM)</math> by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. <math>(A, B; R, Q) = -1</math> as complete quadrilaterals induce harmonic bundles. By a projection through <math>P</math> from <math>AQ</math> onto <math>(ABM)</math>, <math>(A, B; N', M) = -1</math>. Since <math>\frac{AN}{BN} = \frac{AM}{BM}</math>, <math>M</math> and <math>N</math> are on the intersections of <math>(ABM)</math> and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, <math>(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1</math>. By uniqueness of harmonic conjugate, <math>N = N'</math> | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 01:37, 6 July 2021
Problem
A cyclic quadrilateral is given. The lines
and
intersect at
, with
between
and
; the diagonals
and
intersect at
. Let
be the midpoint of the side
, and let
be a point on the circumcircle of
such that
. Prove that
are collinear.
Solution
Let . Let
. Let
. Let
denote the circumcircle of
. Let
. Note that
is on
by Lemma 9.17 on Euclidean Geometry in Maths Olympiad.
as complete quadrilaterals induce harmonic bundles. By a projection through
from
onto
,
. Since
,
and
are on the intersections of
and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore,
. By uniqueness of harmonic conjugate,