1966 IMO Problems/Problem 6
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 .
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.
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