A geometry problem by Prayas Rautray
Among all triangles with perimeter 21, what is area of the triangle which has the maximum area among all of them?
The answer is 21.21.
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1 solution
How do you know equilateral triangle has the maximum area. Can you give a proof.
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The formula for the area as a function of sides is A = s ( s − a ) ( s − b ) ( s − c ) , where c = 2 s − a − b .
In terms of a and b , this is A = s ( s − a ) ( s − b ) ( a + b − 2 s )
The formula is fully symmetrical in a and b , so any extreme must exist only at a = b and it can easily be shown to be the maximum by substituting a few values.
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No, just because it's symmetrical in a and b , that doesn't mean that the min/max occurs when a = b .
Take the following counterexample:
a ≥ 0 , b ≥ 0 , a + b = 1 0 , maximize ( ( a − 5 ) ( b − 5 ) ) .
See the relevant article: Inequalities with strange equality conditions .
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@Pi Han Goh – But this is a situation where a and b cannot vary independently, as mine do.
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@Marta Reece – What do you mean by "cannot vary independently"?
@Pi Han Goh – Thanks. This is really interesting!!!
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@Prayas Rautray – Pi Han Goh sir, why don't you post your comment as a solution.
My book gave me the hint that it can be proven by AM-GM inequality. But I couldn't figure it out. Can you help me with this.
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This question is essentially a variation of Weitzenböck's inequality .
Knowing that p = 2 1 and p 2 ≥ 1 2 3 T , where T represents the area, we can get min ( T ) = 4 4 9 3 .
2 1 / 3 = 7
2 3 × 7 2 ≈ 2 1 . 2 2