回覆 5# Chen 的帖子
原題目等價於將\(B\)點繞\(CM\)旋轉,令旋轉後的座標為
\(A(0,0,0),C(0,4sinθ,0),M(2cosθ,0,0),B(x,y,z)\)
可得,四面體體積\(V(θ)=\frac{4\sin θ \cos θ \cdot z}{3}\)
因此目標在將\(z\)以\(θ\)表示
由\(BC=4, AB=2\sqrt{2}, BM=2\cos θ\),得
\(x^2+y^2+z^2=8\).......................................(1)
\(x^2+(y-4\sin θ)^2+z^2=16\)...................(2)
\((x-2\cos θ)^2+y^2+z^2=4(\cos θ)^2\).......(3)
(1)-(2)得,\(y=\frac{2(\sin θ)^2-1}{\sin θ}\)
(1)-(3)得,\(x=\frac{2}{\cos θ}\)
把\(z^2=8-x^2-y^2\)代入\(V(θ)\),得
\(V(θ)^2=\frac{16}{9} (8(\sin θ)^2(\cos θ)^2-(\cos θ)^2(2(\sin θ)^2-1)^2-4(\sin θ)^2)\)
令\(u=(\sin θ)^2\)
\(V(θ)^2=\frac{16}{9} ( 8u(1-u) - (1-u)(2u-1)^2 - 4u]=4u^3-16u^2+9u-1\)
利用微分知\(u=\frac{8-\sqrt{37}}{6}\)時,\(V(θ)\)有最大值
此時,\((\sin θ)^2=\frac{8-\sqrt{37}}{6}\)
[ 本帖最後由 Jimmy92888 於 2025-4-20 16:52 編輯 ]