Question 262990
<pre><font size = 4 color = "indigo"><b>
Let the equilateral triangle be ABC, and the
inscribed square be DEFG:

{{{drawing(400,280,-2.25,2.25, -.5,3,
locate(-1,0,A), locate(1,0,B),locate(0,sqrt(3)+.2,C),
locate(-.5,0,D),locate(.4,0,E),
locate(-.6,1.1,G),locate(.5,1.1,F),
triangle(-1,0,1,0,0,sqrt(3)),rectangle(-.464,0,.464,.9282) )}}}

We draw in the green median CH from vertex C to the bottom
side AB, which is also the perpendicular bisector of AB,
and also the bisector of the angle C.

{{{drawing(400,280,-2.25,2.25, -.5,3,
green(line(0,0,0,sqrt(3))), locate(0,0,H),
locate(-1,0,A), locate(1,0,B),locate(0,sqrt(3)+.2,C),
locate(-.5,0,D),locate(.4,0,E),
locate(-.6,1.1,G),locate(.5,1.1,F),
triangle(-1,0,1,0,0,sqrt(3)),rectangle(-.464,0,.464,.9282) )}}}

We find the length of CH by the Pythagorean theorem

{{{AC = t}}}, {{{AH = t/2}}}

{{{AC^2=AH^2+CH^2}}}
{{{t^2=(t/2)^2+CH^2}}}
{{{t^2=t^2/4+CH^2}}}
{{{4t^2=t^2+4*CH^2}}}
{{{3t^2=4*CH^2}}}
{{{(3t^2)/4=CH^2}}}
{{{sqrt((3t^2)/4)=sqrt(CH^2)}}}
{{{(t*sqrt(3))/2=CH}}}

Triangle ADG is similar to triangle AHC,

so {{{(GD)/(AD)=(CH)/(AH)}}}

{{{AH =t/2}}}, {{{DH=s/2}}}, so

{{{AD=AH-DH=t/2-s/2=(t-s)/2}}} 

{{{GD=s}}}, {{{CH=(t*sqrt(3))/2}}}

Substitute in

{{{(GD)/(AD)=(CH)/(AH)}}}

{{{s/((t-s)/2)=((t*sqrt(3))/2)/(t/2)}}}

Multiply tops and bottoms of the fractions on both
sides by 2.

{{{(2s)/(t-s)=(t*sqrt(3))/t}}}   

Cancel t's on the right sides:

{{{(2s)/(t-s)=sqrt(3)}}}

Multiply both sides by {{{(t-s)}}}

{{{2s=sqrt(3)(t-s)}}}

{{{2s=sqrt(3)t-sqrt(3)s}}}

{{{2s+sqrt(3)s=sqrt(3)t}}}

factor out s on the left:

{{{s(2+sqrt(3))=sqrt(3)t}}}

Divide both sides by {{{s*sqrt(3)}}}

{{{(2+sqrt(3))/sqrt(3)=t/s}}}

Punching the left side out on a calculator

{{{t/s=2.154700538}}}

To the nearest thousandth, 

{{{t/s=2.155}}}

Edwin</pre>