AoPS Wiki talk:Problem of the Day/July 18, 2011


AoPSWiki:Problem of the Day/July 18, 2011


This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.

Since the inradius of both triangles is 2 and congruent triangles are also similar, then it is safe to assume that the two triangles are congruent. Let's call each side x. That makes one triangle's perimeter 3x, and since the other one is congruent, the other triangle's perimeter is also 3x. It is given that the total perimeter of both triangles is 36, so 3x+3x=36. We solve for x, and we get x=6. To find the area of an equilateral triangle, we find the height first. Using the Pythagorean Theorem, we get √27. The area of one of the triangles is 1/2*b*h, so 1/2*6*√27=9√3. The total area of both triangles is therefore 2*9√3, which is 18√3. The answer is 18√3.

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