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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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原問題https://math.pro/db/viewthread.php?tid=3450&page=1#pid22082

使用maxima計算結果如下
設半徑4的球心\(A(0,0,0)\),半徑9的球心\(B(13,0,0)\),半徑16的球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)
外公切平面方程式為\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)
和半徑4的球相切於\(\displaystyle A' \left(-\frac{20}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)
和半徑9的球相切於\(\displaystyle B' \left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)
和半徑16的球相切於\(\displaystyle C' \left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)

有前一次的經驗後就直接計算各點位置
設半徑4的球心\(A(0,0,0)\),半徑9的球心\(B(13,0,0)\),半徑16的球心\(C(x,y,0)\)
\(\cases{\overline{BC}=25\cr \overline{CA}=20}\),\(\cases{(x-13)^2+(y-0)^2=25^2 \cr (x-0)^2+(y-0)^2=20^2}\),
\(\displaystyle x=-\frac{28}{13},y=\frac{48\sqrt{29}}{13}\),得到\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)

設\(P(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點,\(\Rightarrow \overline{AA'}=4\),\(∠PA'A=90^{\circ}\)
球心\(B(13,0,0)\)半徑9的球面和平面相切於\(B'\)點,\(\Rightarrow \overline{BB'}=9\),\(∠PB'B=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PBB'\)為相似三角形(\(∠APA'=∠BPB'\),\(∠PA'A=∠PB'B=90^{\circ}\))
\(\overline{PA}:\overline{PB}=\overline{AA'}:\overline{BB'}=4:9\),\(\overline{PA}:\overline{AB}=4:5\)
由外分點公式可知\(\displaystyle P=\frac{9}{5}A-\frac{4}{5}B=\frac{9}{5}(0,0,0)-\frac{4}{5}(13,0,0)=(-\frac{52}{5},0,0)\)
得到\(\displaystyle P(-\frac{52}{5},0,0)\)

設\(Q(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點,\(\Rightarrow \overline{AA'}=4\),\(∠PA'A=90^{\circ}\)
球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)半徑16的球面和平面相切於\(C'\)點,\(\Rightarrow \overline{CC'}=16\),\(∠PC'C=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PCC'\)為相似三角形(\(∠APA'=∠CPC'\),\(∠PA'A=∠PC'C=90^{\circ}\))
\(\overline{QA}:\overline{QC}=\overline{AA'}:\overline{CC'}=4:16\),\(\overline{QA}:\overline{AC}=1:3\)
由外分點公式可知\(\displaystyle Q=\frac{4}{3}A-\frac{1}{3}C=\frac{4}{3}(0,0,0)-\frac{1}{3}C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)=\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)
得到\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)

從\(\displaystyle P\left(-\frac{52}{5},0,0\right)\)和\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)求交線的對稱比例式
方向向量為\(\displaystyle \vec{PQ}=\left(\frac{2168}{195},-\frac{16\sqrt{29}}{13},0\right)\)
\(PQ\)直線的對稱比例式為\(\displaystyle \frac{x+\frac{52}{5}}{\frac{2168}{195}}=\frac{y-0}{-\frac{16\sqrt{29}}{13}}\),\(z=0\)
化簡得到\(30\sqrt{29}x+271y+312\sqrt{29}=0\),\(z=0\)

假設外公切平面方程式為\(30\sqrt{29}x+271y+312\sqrt{29}+kz=0\),\(k\in R\)
球心\(A(0,0,0)\)到平面距離\(\displaystyle \frac{|\;0+0+0+312\sqrt{29}|\;}{\sqrt{(30\sqrt{29})^2+271^2+k^2}}=4\),\(k=\pm 13\sqrt{455}\)
得到外公切平面方程式\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)

利用投影點公式求切點\(A',B',C'\)
投影點公式:
\(P(x_0,y_0,z_0)\)對平面\(ax+by+cz+d=0\)的投影點為\(P'(x_0-at,y_0-bt,z_0-ct)\),其中\(\displaystyle t=\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}\)

計算\(A(0,0,0)\)的投影點\(A'\)
\(\displaystyle t=\frac{0+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{2}{39\sqrt{29}}\)
\(\displaystyle A'\left(0-30\sqrt{29}\cdot \frac{2}{39\sqrt{29}},0-271\cdot \frac{2}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{2}{39\sqrt{29}}\right)=
\left(-\frac{30}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)

計算\(B(13,0,0)\)的投影點\(B'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot 13+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{3}{26\sqrt{29}}\)
\(\displaystyle B'\left(13-30\sqrt{29}\cdot \frac{3}{26\sqrt{29}},0-271\cdot \frac{3}{26\sqrt{29}},0-13\sqrt{455}\cdot \frac{3}{26\sqrt{29}}\right)=
\left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)

計算\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)的投影點\(C'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot (-\frac{28}{13})+271\cdot \frac{48\sqrt{29}}{13}+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{8}{39\sqrt{29}}\)
\(\displaystyle C'\left(-\frac{28}{13}-30\sqrt{29}\cdot \frac{8}{39\sqrt{29}},\frac{48\sqrt{29}}{13}-271\cdot \frac{8}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{8}{39\sqrt{29}}\right)=
\left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)


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