1966 IMO Problems/Problem 6
Contents
Problem
In the interior of sides of triangle , any points , respectively, are selected. Prove that the area of at least one of the triangles is less than or equal to one quarter of the area of triangle .
Solution
Let the lengths of sides , , and be , , and , respectively. Let , , and .
Now assume for the sake of contradiction that the areas of , , and are all at greater than one fourth of that of . Therefore
In other words, , or . Similarly, and . Multiplying these three inequalities together yields
We also have that , , and from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields
This is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.
Solution 2
Let . Then it is clear that the ratio of areas of to that of equals , respectively. Suppose all three quantities exceed . Then their product also exceeds . However, it is clear by AM-GM that , and so the product of all three quantities cannot exceed (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to .
See Also
1966 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All IMO Problems and Solutions |