1961 IMO Problems/Problem 2
Let , , and be the lengths of a triangle whose area is S. Prove that
In what case does equality hold?
By Heron's formula, we have This can be simplified to Next, we can factor out all of the s and use a clever difference of squares: We can now use difference of squares again: We must prove that the RHS of this equation is less than or equal to .
Let , , . Then, our inequality can be reduced to We now have to prove We can simplify: Finally, we can apply AM-GM: Adding these all up, we have the desired inequality and so the proof is complete.
To have , we must satisfy This is only true when , and thus . Therefore, equality happens when the triangle is equilateral.
Solution 2 (duality principle)
We firstly use the duality principle. The LHS becomes and the RHS becomes If we use Heron's formula. By AM-GM Making this substitution becomes and once we take the square root of the area then our RHS becomes Multiplying the RHS and the LHS by 3 we get the LHS to be Our RHS becomes Subtracting we have the LHS equal to and the RHS being If LHS RHS then LHS-RHS LHS-RHS= by the trivial inequality so therefore, and we're done.
Solution 3 (Trigonometry)
Let be the angle between vertices and . We use two formulae:
Equality occurs when and , which is when so that they form an equilateral triangle.
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