1982 IMO Problems/Problem 5
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
File:IMO1982P5.png
, consider
unit. Now,
(after all the simplifying, and substituting
=
).
Now indicates
. So let's go ahead and write
and
. Applying the cosine rule, we get:
.
.
This means $MN = BM \dot \sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).
B, M, and N are collinear if and only if .
. Now by the law of Sines,
.
Now .
So, $\frac{1}{sin\angle AMB} = \frac{a}{sin\angle CMN$ (Error compiling LaTeX. Unknown error_msg)
.
So since .
So the answer is .
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 |