Packing circles in a triangle

For the optimum solution, one circle will be "stuck" in the $30^\circ$ corner, and a second will be stuck in the $60^\circ$ corner. These circles will be as large as possible to enable a third circle to be drawn.

This will happen when the third circle creates a number of $30/60/90$ triangles. If the circles have radius $r$ , then the centre of the third circle must be a distance $r$ from the two non-hypotenuse edges, and must be a distance $2r$ from the other two centres. Thus the shortest edge has length $(1 + 2\sqrt{3})r$ . Thus $141 = \sqrt{3}(1 + 2\sqrt{3})r = (6 + \sqrt{3})r$ , so that $r \; = \; \frac{141}{6 + \sqrt{3}} \; = \; \boxed{18.2358}$ .

If we are being picky, the answer is not correct in the third decimal place - to 3DP, the answer is $18.236$ .

$\color{#EC7300}{Let~r~=~the~required~radius.\\ The ~top \odot~ be~ O_1,~ middle,~ O_2,~ bottom ~O_3.\\ The~ 141~vertical~ leg~ AB ~and~ horizontal ~leg~BC.~\\ O_1O_2~ |~ |~AB~ cut~ BC~ at ~M.~~~O_3N~\bot~BC,~N~on~BC.\\ St. ~line~O_2O_3C ~angle~bisector~of~\angle~BCA,~~\angle~BCO_2=30^o.\\ AO_1~angle~bisector~of~\angle~BAC,~~\angle~O_1AB=15^o.~~~~~O_1T=r~\bot~AB,~T~on~AB.\\ O_3C=r/Sin30=2r.~~\therefore~O_2C=4r.}\\ \color{#3D99F6}{\implies~O_2M~| ~|~ AB,=4rSin30=2r.\\ O_2O_3~| ~| AB,~=2r.\\ AT~along~ AB,~=~\dfrac r {Tan15}.\\ AB~=~2r+2r+ \dfrac r {Tan15}.\\ \implies~r=\dfrac{141}{4+\frac 1 {Tan15}}=\Large \color{#D61F06}{18.2358}. }$

As mentioned by Mr. Michael Mendrin, in very details, out of the two best candidates, the middle one circle 2 with tangent to the long leg AB is the best solution since the triangle of centers $O_1O-2O_3$ is the smallest, compact. The above solution is for that position. I first tried two circles 2 and 3 tangent to the short leg BC. The third was floating. Then thought of turning the two circles 2 and 3 about $O_3$ , so as to fill the vacant space, with third circle 1, tangent to the middle circle 2 as well as to AB and AC. $In~ the~ isosceles ~\Delta~ O_1O_2O_3~of ~\odot s,~~ ~O_3 ~is~on~ angle~ bisector ~of~ \angle C, ~O_1 ~on~ angle~ bisector ~of~ \angle A,~~~O_1O_2=2*r~ and~ |~ |~ AB.$ Then worked out a simple way to find r.

This

has a slightly better packing than

My proof is pretty long, the main idea is that each of the circles centres has to be within a triangle similar to the original (as u draw parallels to the triangles sides at distance r). Then the tedious part is to prove that any triangle inscribed in the latter will have sa side no longer than r, which u find out as u prove.