Problem 42

Algebra Level 5

$\large a\sqrt{8}+b\sqrt{6}+c\sqrt{2}\geq k\sqrt{a^2+b^2+c^2}$

Consider all possible side lengths $a,b,c$ of a right triangle $ABC$ . What is the maximium value of $k$ such that the inequality is always satisfied?

Cyprus TST pre IMO 2016.

Set .

The answer is 2.73.

1 solution

Harsh Shrivastava
Apr 29, 2016

For maximizing k, we must take a as the hypotenuse.

$a^{2} = b^{2} +c^2$

Let $angle(ABC) = x$

Then $b= a\cos x$ and $c= a\sin x$

Now substituting these in the inequality , and after little bit of manipulation , we get $\sqrt{3} \cos x +\sin x \ge k-2$

Maximum value of this expression is $2$ .

Thus $2\ge \sqrt{3} \cos x +\sin x \ge k-2$

$\implies \boxed{k\le4}$

The answer had been fixed.I don't know why =)))

Son Nguyen - 5 years ago

I think the Staff changed the answer but should not have done so - I got the same answer of $4$ on my second try. Harsh Shrivastava's solution appears correct and better than the method I had used.

Bob Kadylo - 4 years, 8 months ago

Once side lengths $a, b , c$ are fixed, to find the maximum value of $k$ that holds for all combinations, we actually need to minimize the LHS, and not maximize it.

This explains why the solution is incorrect.

Calvin Lin Staff - 4 years, 7 months ago

Problem 42

Cyprus TST pre IMO 2016.

The answer is 2.73.

1 solution

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