1982 IMO Problems/Problem 5
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[hide]Problem
The diagonals and of the regular hexagon are divided by inner points and respectively, so thatDetermine 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 |