一道平面几何题求最值

creasson

可以令$$ \mathop {BP}\limits^ \to  {\rm{ = }}\frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 - it} \right)}^2}}}\mathop {BA}\limits^ \to $$
$P$点在单位圆上，有
$$ 577 s^4 t^2+1296 s^3 t^3+108 s^3 t^2-1296 s^3 t+728 s^2 t^4+108 s^2 t^3-2574 s^2 t^2-108 s^2 t+728 s^2-1296 s t^3-108 s t^2+1296 s t+577 t^2 = 0$$
$$L = PA + 9PB = \left(\frac{\left(s^2+1\right) t}{s \left(t^2+1\right)}+\frac{9 (s+t) (1-s t)}{s \left(t^2+1\right)}\right) AB = \frac{\sqrt{577} \left(-8 s^2 t-9 s t^2+9 s+10 t\right)}{s \left(t^2+1\right)} $$
然后用拉格朗日乘子可求得两个极值点
`Sqrt[Root[
  10494561297377632681 - 360974714188632810 #1 +
    2796767674118232 #1^2 - 980212212272 #1^3 - 255945027 #1^4 +
    35922 #1^5 + 10 #1^6 &, 3]] = 45.509938`
`Sqrt[Root[
  10494561297377632681 - 360974714188632810 #1 +
    2796767674118232 #1^2 - 980212212272 #1^3 - 255945027 #1^4 +
    35922 #1^5 + 10 #1^6 &, 4]] = 65.416983`