TITLE: How can I determine the radius of 4 identical circles inside an equilateral triangle $ABC$?
QUESTION [7 upvotes]: How can determine the radius  $r$ of 4 identical circles inside equilateral triangle $ABC$ of side $a$ ?

My attempt:
$r$ is radius of each of four identical circles. $a$ is side of equilateral triangle ABC.
Joined the centers of three circles P, Q and R to obtain $\Delta PQR$
I have $$PQ=PR=r+r=2r$$
then dropped perpendicular PT to the side $AB$ . the length of $PT$ will be equal to the inradius of triangle ABC $$PT=\frac{a}{2\sqrt3}$$
$QM=RN=r$ radius of each circle. so I can get perpendicular in small right $\Delta$ $$PT-r=\frac{a}{2\sqrt3}-r$$
Now, I can use Pythagorean theorem in right $\Delta $ but I am stuck to find the length of  $QR$ or $MN$.
Help me to find the radius $r$ of the circle in terms of side $a$ of equilateral $\Delta ABC$. Thanks.

REPLY [1 votes]: You asked to find $QR$. Connect the centers of three external circles, whose circumradius is $2r$. Then its side is from the sine theorem:
$$\frac{QR}{\sin 60^\circ}=2(2r) \Rightarrow QR=2\sqrt{3}r.$$
Now you can finish.

Another solution. Using the Tangent-secant theorem:
$$AS\cdot AK=AM^2 \Rightarrow (R-3r)(R-r)=\left(\frac{AB-MN}{2}\right)^2\Rightarrow\\
\left(\frac{a}{\sqrt3}-3r\right)\left(\frac{a}{\sqrt3}-r\right)=\left(\frac{a-2\sqrt3r}{2}\right)^2\Rightarrow a=\frac{a}{4\sqrt3}.$$
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