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Consider the three tangential circles of equal radii inscribed in the equliateral triangle (linked to below). What is the ratio of the blue line to the red line? The red line is simply the diameter of one of the circles. The blue line begins in the center of the top circle and ends on a line segment connecting the centers of the two bottom circles. Thanks!

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  • $\begingroup$ $\displaystyle\frac{\sqrt3}2$. $\endgroup$ – Christopher Carl Heckman Apr 25 '16 at 0:56
  • $\begingroup$ Thanks @Christopher Carl Heckman ! How did you arrive at that? $\endgroup$ – Astrophysics Math Apr 25 '16 at 0:58
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Let $\;O\;$ be the center of the circle with the red line and $\;M\;$ the one of the upper circle, with $\;A\;$ the point where these two lines intersect.

If the radius of the circles is $\;R\;$, then in the straight triangle $\;AOM\;$ we have

$$MA^2+AO^2=OM^2\implies MA+R^2=(2R)^2\implies MA=\sqrt3\,R$$

so

$$\frac{MA}{2R}=\frac{\sqrt3}2$$

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Suppose the circles all have radius $1$ (and hence diameter $2$). Draw a line segment from the top of the blue line to the midpoint of the red line. You have a right triangle (along with the red and blue lines) with an angle of $60^\circ$; thus the blue line has length $\tan(60^\circ)=\sqrt3$.

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