Difference between revisions of "2004 IMO Shortlist Problems/G8"
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− | + | <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]] |
Latest revision as of 01:43, 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
Claim: is on
. Proof:
as complete quadrilaterals induce harmonic bundles.
by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem,
and this is equivalent to our original claim.
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,