Difference between revisions of "1982 IMO Problems/Problem 5"
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By the law of Sines, | By the law of Sines, | ||
− | <math>\frac{1}{ | + | <math>\frac{1}{\sin \angle AMB} = \frac{BM}{sin 30^{\circ}} = 2BM </math> |
<math>\implies \sin \angle AMB = \frac{1}{2BM} </math>. | <math>\implies \sin \angle AMB = \frac{1}{2BM} </math>. |
Revision as of 09:44, 16 June 2024
Contents
[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. The original thread for this problem can be found hercommunity/p398343]
Solution 4
Let . By the cosine rule,
.
.
Now if B, M, and N are collinear, then
.
By the law of Sines,
.
Also,
.
But
, which means . So, r = \frac{1}{\sqrt{3}} $.