Three circles internaly tangent to an equilateral triangle The diagram shows an equilateral triangle of side length 1 with 3 identical circles.  Find the radius of the circle.
The correct answer for the length of the right triangle in red should be $\sqrt{3}r$.
In the image, OP is found to be $\frac{\sqrt{3}}{2}$, so the length in red should be $\frac{\sqrt{3}}{2} - 3x$. But why the length is not equal to the one I found?
$x$ cm is the radius of the circle.
By the way, trigonometry is not allowed to be used.

 A: 
...should be $\sqrt{3}\,r$
...should be $\frac{\sqrt{3}}{2}−3x$
 But why the length is not equal to the one I found?

But it is equal. The equation you got $\,6x^2 - 3 \sqrt{3} \,x + \frac{3}{4} = 0\,$ has the root $\,x = \frac{1}{4}\left(\sqrt{3}-1\right)\,$, which happens to satisfy the relation $\,\sqrt{3} \,x = \frac{\sqrt{3}}{2}- 3 x\,$.
For verification, you can calculate $\,x\,$ by writing the area of the big triangle as the sum between the small equilateral triangle of side $\,2x\,$ and the $\,3\,$ isosceles trapezoids with bases $\,2x, 1\,$ and height $\,x\,$. The equation comes out to be $\,\frac{\sqrt{3}}{4} = 3 \,\frac{x (2x + 1)}{2} + \sqrt{3}\,x^2 \,$, and has the same root found before.

[ EDIT ] $\;$ In fact, after filling-in the missing blanks, it turns out that you (correctly) derived the equation $\,\sqrt{3} \,x = \frac{\sqrt{3}}{2}- 3 x\,$, which gives $\,x = \frac{1}{4}\left(\sqrt{3}-1\right)\,$ directly.
A: Let radius of circles be r and the side length be a. Mark left bottom side vertex as A and the foot of perpendicular from O(the center of the circle on this vertex) on bottom side as N, we have:
$AN+r=\frac a2$
In triangle AON angle $A$ is $30^o $ so we have:
$AO=2r$  and $ AN=r\sqrt 3$, so we have:
$r\sqrt 3+r=\frac a2$
which give r as:
$r=\frac 14 a(\sqrt 3-1)$
in you question a=1 so r is:
$r=\frac14(\sqrt3-1)$
