During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

AoPS Wiki talk:Problem of the Day/July 24, 2011

Revision as of 11:20, 24 July 2011 by Kingofmath101 (talk | contribs) (Solutions to July 24, 2011, Problems of the Day)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Solution to Problem 1: If the circumference is $18\pi$, then we can find the radius. Because of our cirumference formula, we know that $r = \frac{18\pi}{2\pi} = 9$. So, the radius is $9$ units. We can insert this into the circle area formula to find the area: $(9^2)\pi = 81\pi$. So, the area is $81\pi$.

Solution to Problem 2: Rearrange the terms on the left side:

$a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \geq 0$

This is just

$(a + b + c)^2 \geq 0$

This is trivial, so the inequality must be true. (Idea for wording of the end of the solution: solutions for the Pre-Olympiad Inequalities marathon by other users)

Invalid username
Login to AoPS