Difference between revisions of "1982 IMO Problems/Problem 5"
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− | O is the center of the regular hexagon. Then we | + | O is the center of the regular hexagon. Then we clearly have <math>ABC\cong COA\cong EOC</math>. And therefore we have also obviously <math>ABM\cong AOM\cong CON</math>, as <math>\frac{AM}{AC} =\frac{CN}{CE}</math>. |
So we have <math>\angle{BMA} =\angle{AMO} =\angle{CNO}</math> and <math>\angle{NOC} =\angle{ABM}</math>. Because of <math>\angle{AMO} =\angle{CNO}</math> the quadrilateral <math>ONCM</math> is cyclic. <math>\Rightarrow \angle{NOC} =\angle{NMC} =\angle{BMA}</math>. And as we also have <math>\angle{NOC} =\angle{ABM}</math> we get <math>\angle{ABM} =\angle{BMA}</math>. <math>\Rightarrow AB=AM</math>. And as <math>AC=\sqrt{3} \cdot AB</math> we get <math>r=\frac{AM}{AC} =\frac{AB}{\sqrt{3} \cdot AB} =\frac{1}{\sqrt{3}}</math>. | So we have <math>\angle{BMA} =\angle{AMO} =\angle{CNO}</math> and <math>\angle{NOC} =\angle{ABM}</math>. Because of <math>\angle{AMO} =\angle{CNO}</math> the quadrilateral <math>ONCM</math> is cyclic. <math>\Rightarrow \angle{NOC} =\angle{NMC} =\angle{BMA}</math>. And as we also have <math>\angle{NOC} =\angle{ABM}</math> we get <math>\angle{ABM} =\angle{BMA}</math>. <math>\Rightarrow AB=AM</math>. And as <math>AC=\sqrt{3} \cdot AB</math> we get <math>r=\frac{AM}{AC} =\frac{AB}{\sqrt{3} \cdot AB} =\frac{1}{\sqrt{3}}</math>. | ||
Revision as of 12:02, 16 April 2022
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]
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 |