# Thread: 6 Equal circles in an equilateral triangle

2. ## Re: 6 Equal circles in an equilateral triangle

We can easily exclude (a), (b) and (e). Let's assume that R = 1. Then the combined area of the circles is $6\pi\approx18.850$; (c) gives 24.124 and (d) gives 19.392, which is too close to the circle areas. To be sure that (d) is wrong, it is possible to calculate the area of one of the triangle's angles not covered by a circle (which looks like a small triangle with two straight sides and one curved side). If I am correct, its area is $\sqrt{3}-\pi/3\approx0.685$. Multiplied by 3, this gives 2.055, which together with $6\pi$ exceeds (d).

3. ## Re: 6 Equal circles in an equilateral triangle

$(12+7(\sqrt{3}))R^2$

4. ## Re: 6 Equal circles in an equilateral triangle

Originally Posted by swordfish774
$(12+7(\sqrt{3}))R^2$
Correct. The base length of the traingle is B=4R+2Rsqrt(3), and the area is (B^2/4)sqrt(3). The rest is algebra.

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# how to inscribe 6 circles in an equilateral triangle?

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