Adopting side a as the triangle base, the upper vertice describes the ellipse
(xrx)2+(yry)2=1
where
rx=a+b+c2 and ry=√(b+c2)2−(a2)2
when yv=h0 then xv=√a2−(b+c)2+4h20p02√a2−(b+c)2. Here pv={xv,yv} are the upper vertice coordinates p0=a+b+c and p=p02.
The ellipse focuses location are:
f1={−a2,0} and f2={a2,0}
Now we have the relationships:
1) p(p−a)(p−b)(p−c)=a2h204 Henon´s formula
2) From a+∥pv−f1∥+∥pv−f2∥=p0 we have
a+
⎷h20+14⎛⎜
⎜⎝a−√a2−(b+c)2+4h20p0√a2−(b+c)2⎞⎟
⎟⎠2+
⎷h20+14⎛⎜
⎜⎝a+√a2−(b+c)2+4h20p0√a2−(b+c)2⎞⎟
⎟⎠2=p0
3) a+b+c=p0
Solving 1,2,3 for a,b,c gives
(a=p20−4h202p0,b=4h20+p204p0,c=4h20+p204p0)
and substituting h0=0.173,p0=0.60
{a=0.200237,b=0.199882,c=0.199882}
with an area of 0.0173205
