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A triangle is drawn and a circle of radius 3 is inscribed in it so as to be tangent at three points. A smaller circle of radius 1 is inscribed to be tangent to the larger circle and twice tangent to the triangle. What is the area of the triangle?

I am pretty confident the large circle must be the incircle, but as I am not given the lengths of the triangle's sides, nor any information about its construction, I have been stumped. Can someone help me answer this question?

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  • $\begingroup$ Can you add a picture? $\endgroup$ – Jack D'Aurizio Sep 14 '16 at 23:18
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    $\begingroup$ Anyway, I cannot see why the area should be fixed. Take a circle with radius $3$ and a unit circle, externally tangent to the previous larger circle. Then take two common tangents concurring in the exterior homothetic centre and a third tangent to the initial circle. The area of the triangle delimited by such three lines is not fixed. $\endgroup$ – Jack D'Aurizio Sep 14 '16 at 23:22
  • $\begingroup$ However, if your triangle is an equilateral one, its height is $9$, so its side length is $6\sqrt{3}$ and its area is $27\sqrt{3}$. $\endgroup$ – Jack D'Aurizio Sep 14 '16 at 23:26
  • $\begingroup$ I don't know how to add a picture, but the picture given to me says nothing about the triangle being equilateral (or even isosceles) and mentions it is not drawn to scale. $\endgroup$ – Taylor Sep 15 '16 at 0:14
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Using similar triangles, we can verify $$AI=\frac{3(3+1)}{3-1}=6$$

and $$\angle A=60^{\circ}$$

but $\Delta ABC$ is not fixed by varying the position of $D$.

enter image description here

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  • $\begingroup$ Thank you very much. That is a beautiful graphic. $\endgroup$ – Taylor Sep 15 '16 at 16:44

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