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 .
Remarks (added by pf02, September 2024)
Solution 2 is written in a very sloppy way. However, an interested reader can make sense of it. More importantly, the two solutions are identical. If it wasn't for the sloppy writing, Solution 2 could be obtained from the first Solution after applying a word by word translation which replaces line segments by ratios.
Below I will give another solution. It is formally different from the previous solutions, even if not at a deep level.
Solution 3
Let and be as in the problem. Denote as in Solution 2. Note that because are in the interior of the respective sides.
Using the fact that the area of a triangle is half of the product of two sides and of the angle between them (like in the first Solution), we have that .
Now the problem has nothing to do with geometry anymore: we just have to show that given three numbers , at least one of is .
If , we are done. Otherwise, we have . It follows that (recall that ). In particular, it follows that , which implies .
If , we are done. Otherwise, we have . Using the inequality on from the previous paragraph, we have , or after a few computations, . Using the observation about from the preceding paragraph, we get .
Now consider . Using the inequality on from the previous paragraph, we have that . To finish the solution to the problem, it is enough to show that .
After some easy computations (and using again that ), this becomes , which is true because. .
(Solution by pf02, September 2024)
See Also
1966 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
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