Difference between revisions of "2004 Pan African MO Problems/Problem 6"
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== See also == | == See also == | ||
{{Pan African MO box|year=2004|num-b=5|after=Last Problem}} | {{Pan African MO box|year=2004|num-b=5|after=Last Problem}} | ||
− | [[Category:Geometry]][[Category: | + | [[Category:Geometry]][[Category:Olympiad Geometry Problems]] |
Revision as of 00:14, 3 February 2023
Let be a cyclic quadrilateral such that
is a diameter of it's circumcircle. Suppose that
and
intersect at
,
and
at
,
and
at
, and let
be a point on
. Show that
is perpendicular to
if and only if
is the midpoint of
.
Solution
We use the notation to represent the circumcircle of
for any three points
, and use
to represent the circumcircle of
. Without loss of generality, assume that
and
are not parallel and that
is closer to
than to
. In this solution, we use
to represent the midpoint of
and prove that
.
Note that . Similarly, note that
. Therefore,
is a diameter of
.
Define points such that
. We claim that
, and we prove this by showing that
lies on
and that it lies on
.
To prove that , we angle chase:
,
, so that
; similarly,
. Therefore,
.
To prove that , we make a similar angle chase:
,
, so that
; similarly, we have
. Therefore,
.
Now, the radical center of must be
, so that the radical axis of
, which is
, must also pass through
. But
, so
passes through
.
Therefore, the angle between and
is also the angle between
and
, which is trivially
because
is part of the circle with diameter
, and we are done.
See also
2004 Pan African MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All Pan African MO Problems and Solutions |