How to prove that $r\geq\frac {1}{2(1+\sqrt 3)}$? Each interior point of an equilateral triangle of side $1$ lies in one of six congruent circles of radius $r$. How to prove that $r\geq\frac {1}{2(1+\sqrt 3)}$?
 A: Let us orient the triangle with the vertex $C$ at the top and the base $AB$ at the bottom, and set $d=\frac{1}{1+\sqrt{3}}$. Now let us build a quadrilateral inside the triangle by identifying:


*

*point $M$ on the side $AC$, so that $MC=d$;

*point $N$ on the side $BC$, so that $NC=d$;

*points $P$ and $Q$ by drawing two vertical segments of length $d$ from $N$ and $M$, respectively. 

Since we also have $MN=d$ (because $\triangle CMN$ is equilateral), the resulting quadrilateral $MNPQ$ is a square.
Now we can note that $\triangle AMQ$ is isosceles. In fact, we have 
$$AM=1-d=1-\frac{1}{1+\sqrt{3}}=\frac{\sqrt{3}}{1+\sqrt{3}}=\sqrt{3} d$$
and because $\angle AMQ$ is $30^\circ$ the projection of $MQ$ on $AM$ is equal to 
$$\frac{\sqrt{3}}{2} d=\frac{AM}{2}$$
which shows that $\triangle AMQ$ is isosceles with base $AM$. Therefore, we have $AQ=MQ=d$. Due to the symmetry of the problem, we also have $BP=NP=d$.
Now we have seven points $A,B,C,M,N,P,Q$ that are included in the initial $\triangle ABC$. Because all points lie in six circles, at least two of them have to lie in the same circle. According to our construction, the distance between any two of these points is at least $d$. So we have that the minimal value of the diameter of the circles is $d$, and then $r \geq \frac{1}{2(1+\sqrt{3})}$.
