Difference between revisions of "1982 IMO Problems/Problem 5"
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This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [https://aops.com/community/p398343] | This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [https://aops.com/community/p398343] | ||
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+ | ==Solution 4== | ||
== See Also == {{IMO box|year=1982|num-b=4|num-a=6}} | == See Also == {{IMO box|year=1982|num-b=4|num-a=6}} |
Revision as of 05:44, 15 June 2024
Problem
The diagonals and
of the regular hexagon
are divided by inner points
and
respectively, so that
Determine
if
and
are collinear.
Solution 1
O is the center of the regular hexagon. Then we clearly have . And therefore we have also obviously
, as
.
So we have
and
. Because of
the quadrilateral
is cyclic.
. And as we also have
we get
.
. And as
we get
.
This solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]
Solution 2
Let be the intersection of
and
.
is the mid-point of
.
Since
,
, and
are collinear, then by Menelaus Theorem,
.
Let the sidelength of the hexagon be
. Then
.
Substituting them into the first equation yields
This solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]
Solution 3
Note . From the relation
results
, i.e.
. Thus,
Therefore, , i.e.
This solution was posted and copyrighted by Virgil Nicula. The original thread for this problem can be found here: [3]
Solution 4
See Also
1982 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |