Triforce Triangle?

Draw three more circles at the corners of the triangle so that they are tangent to the original three and to the sides of the triangle. These new circles also have a radius of 1. Draw the line segments as shown in the figure so that you create a rectangle and two $30°-60°-90°$ triangles.

Note that the angle of the triangle (=60°) gets bisected, making two 30° angles.

$\tan30^\circ = \frac{1}{\sqrt{3}}$

So,the base of each triangle is of length $\sqrt{3}$ cm.

The length of each side of the equilateral triangle (in cm) will be $(\sqrt{3} + 4 + \sqrt{3})$

Thus, the answer is $\Rightarrow 4 + 2\sqrt{3} \approx \boxed{7.464 }$

My solution is along the lines of Yash Jain's, but without adding the additional circles. The red line segment consists of two radii combined, therefore it has length 2.

Each blue line segment has length 1, and makes an angle of $30^\circ$ with the horizontal. Its projection on the horizontal (the green line segment) is therefore $\cos 30^\circ = \tfrac12\sqrt 3$ .

The total length of green + red + green is therefore $\tfrac12\sqrt 3 + 2 + \tfrac12\sqrt 3 = 2 + \sqrt 3$ . This line segment connects two midpoints of the triangle.

It is well-known that the line segment connecting the midpoints has half of the length of the side parallel to it. Therefore the side of the triangle is $2\cdot (2+\sqrt 3) = 4+2\sqrt 3 \approx \boxed{7.464}.$

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x is the angle from a vertex of triangle to the center of one of the circles.
$x = 15$

$tan 15 = 1/y; y = 3.732$

side of triangle = $2*y = 7.464$

From the top vertex, two tangent go to the red circle. One the left side and the other the angle bisector. They make an angle of 60/2=30. Hence radius/tangent length =Tan(30/2)=Tan15. Implies Tangent length=1/Tan15. But this is only half the sides. So sides=2/Tang15=7.4641.....