# 1982 IMO Problems/Problem 5

## 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}} $.