Difference between revisions of "2018 USAMO Problems/Problem 5"
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Notice that since <math>A</math> is the intersection of <math>(EDQ)</math> and <math>(BPE)</math>, it is the Miquel point of <math>DQBP</math>. | Notice that since <math>A</math> is the intersection of <math>(EDQ)</math> and <math>(BPE)</math>, it is the Miquel point of <math>DQBP</math>. | ||
− | Now define <math>X</math> as the intersection of <math>BQ</math> and <math>DP</math>. From Pappus's theorem on <math>BFPDGQ</math> that <math>A,M,X</math> are collinear. It’s a well known property of Miquel points that <math>\angle | + | Now define <math>X</math> as the intersection of <math>BQ</math> and <math>DP</math>. From Pappus's theorem on <math>BFPDGQ</math> that <math>A,M,X</math> are collinear. It’s a well known property of Miquel points that <math>\angle EAX = 90</math>, so it follows that <math>MA \perp AE</math>, as desired. <math>\blacksquare</math> |
~AopsUser101 | ~AopsUser101 |
Revision as of 19:55, 15 March 2021
Problem 5
In convex cyclic quadrilateral we know that lines
and
intersect at
lines
and
intersect at
and lines
and
intersect at
Suppose that the circumcircle of
intersects line
at
and
, and the circumcircle of
intersects line
at
and
, where
and
are collinear in that order. Prove that if lines
and
intersect at
, then
Solution
so
are collinear. Furthermore, note that
is cyclic because:
Notice that since
is the intersection of
and
, it is the Miquel point of
.
Now define as the intersection of
and
. From Pappus's theorem on
that
are collinear. It’s a well known property of Miquel points that
, so it follows that
, as desired.
~AopsUser101