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[3D 建模軟體] SketchUp

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原問題https://math.pro/db/viewthread.php?tid=1003&page=2#pid14609

邊長13,14,15的三角形相關計算如下
設\(\Delta ABC\)各頂點坐標為\(\displaystyle A(0,0),B(15,0),C(\frac{33}{5},\frac{56}{5})\)
\(\Delta ABC\)的內切圓圓心\(I(7,4)\),切\(\overline{AB}\)於\(F(7,0)\),切\(\overline{BC}\)於\(\displaystyle D(\frac{51}{5},\frac{32}{5})\),切\(\overline{CA}\)於\(\displaystyle E(\frac{231}{65},\frac{392}{65})\)

因為SketchUp的圓只是用正24邊形代替,當你在SketchUp操作以\(A\)點為圓心,半徑13畫圓和以\(B\)點為圓心,半徑14畫圓,設兩圓的交點為\(C\),但測量線段長度得\(\overline{AC}=12.97,\overline{BC}=13.39\),\(\Delta ABC\)形狀已經有誤差,直接計算\(C\)點坐標。
解聯立方程式\(\cases{x^2+y^2=13^2 \cr (x-15)^2+y^2=14^2}\),得到\(\displaystyle C(\frac{33}{5},\frac{56}{5})\)

SketchUp也沒有畫角平分線工具,直接計算\(\Delta ABC\)內心坐標。

\(\displaystyle \vec{OI}=\frac{a}{a+b+c}\vec{OA}+\frac{b}{a+b+c}\vec{OB}+\frac{c}{a+b+c}\vec{OC}\)
  \(\displaystyle =\frac{14}{13+14+15}(0,0)+\frac{13}{13+14+15}(15,0)+\frac{15}{13+14+15}(\frac{33}{5},\frac{56}{5})\)
  \(=(7,4)\)
得到內心坐標\(I(7,4)\)

以SketchUp所畫出的內切圓和三角形三邊的交點也會有誤差,直接計算交點坐標
\(\overline{BC}\)直線方程式為\(\displaystyle y-0=\frac{0-11.2}{15-6.6}(x-15)\),\(4x+3y=60\)
\(\overline{ID}\)直線方程式為\(\displaystyle y-4=\frac{3}{4}(x-7)\),\(3x-4y=5\)
解聯立方程式\(\cases{4x+3y=60 \cr 3x-4y=5}\),交點\(\displaystyle D(\frac{51}{5},\frac{32}{5})\)

\(\overline{AC}\)直線方程式為\(\displaystyle y-0=\frac{11.2-0}{6.6-0}(x-0)\),\(56x-33y=0\)
\(\overline{IE}\)直線方程式為\(\displaystyle y-4=-\frac{33}{56}(x-7)\),\(33x+56y=455\)
解聯立方程式\(\cases{56x-33y=0 \cr 33x+56y=455}\),交點\(\displaystyle E(\frac{231}{65},\frac{392}{65})\)

內切圓和\(\overline{AB}\)交點為\(F(7,0)\)



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