平均能量损失

Question

假设有一个规则的金字塔，T.ABCD其中

• P是TC这样的TP:PC = 1:3
• RBC是这样的扩展BR:BC = 1:3
• Q任意位于平面上TAD，但既不在外部也不在边界上

画出平面PQR切割金字塔的横截面。

平均能量损失

我自己的解决方案如下，但我正在寻找其他更简单的方法（如果有的话）。

\documentclass[pstricks,border=0pt,12pt]{standalone}
\usepackage{pst-eucl}
\newpsstyle{aux}
{
    linestyle=dashed,
    linecolor=lightgray,
}
\begin{document}
\begin{pspicture}[showgrid=false](-5,-2)(11,11)
    \pstGeonode[PosAngle={-45,0,45,180}](0,0){A}(8,0){B}(4,2){C}(-4,2){D}
    \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O}
    \pstGeonode(O|0,9){T}
    \psline(D)(A)(B)
    \psline[linestyle=dashed](B)(C)(D)
    \psline[linestyle=dashed](T)(C)
    % additional
    \pstHomO[HomCoef=\pscalculate{1/4}]{T}{C}[P]
    \pstHomO[HomCoef=\pscalculate{-1/3}]{B}{C}[R]
    \pstGeonode[PosAngle=180](.65,5){Q}
    %
    \pstMiddleAB{A}{B}{E}
    \pstMiddleAB{C}{D}{F}
    \pstTranslation[PosAngle={0,-135}]{O}{T}{E,F}
    %
    \pstInterLL{T}{B}{P}{R}{S}
    \pcline[style=aux,nodesep=-5mm](T)(B)
    \psline(T)(B)
    %
    \pstInterLL[PosAngle=45]{E'}{F'}{P}{R}{U}
    \pcline[style=aux,nodesep=-5mm](U)(R)
    \pcline[style=aux,nodesep=-5mm](E')(F')
    %
    \pstInterLL{T}{D}{U}{Q}{V}
    \pcline[style=aux,nodesep=-5mm](T)(D)
    \psline(T)(D)
    %
    \pstInterLL[PosAngle=180]{T}{A}{U}{Q}{W}
    \pcline[style=aux,nodesep=-5mm](U)(W)
    \pcline[style=aux,nodesep=-5mm](T)(A)
    \psline(T)(A)
    \pspolygon[fillstyle=solid,fillcolor=blue,opacity=.1,linestyle=none](F)(F')(E')(E)
    \pspolygon[fillstyle=solid,fillcolor=yellow,opacity=.3,linestyle=none](W)(S)(P)(V)
\end{pspicture}
\end{document}

注意：这个问题更具数学性，但它比要求在动物园里画猪、鸭和它们的家庭更有教育意义。

Answer 1

感谢您的评论。是的，我认为这是可以做到的。我们“仅”需要找出 PS 线的斜率。这在 ATD 平面上的投影平面的斜率上进行转换，当然，它必须穿过 Q。为了使讨论保持一般性，我没有使用 P 和 S 的构造信息，我使用的只是它们分别位于边缘 BT 和 CT 上，而 Q 位于 ATD 平面上。我卡住的原因是我无法在超过 6 次迭代中运行动画。我不知道为什么，所以这附带了一个简短的动画。（我还看到动画因为边界框变大而跳跃，但与运行较大动画时遇到的“一般错误”相比，这是一个小问题。我制作了大量动画，从未遇到过这样的错误。）

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\newcounter{coordi}
\begin{document}
\tdplotsetmaincoords{70}{-30} 
\foreach \BB in {1.5,1.6,...,2} %,1.9,1.8,...,1.6}
{\begin{tikzpicture}[tdplot_main_coords,declare function={B=3;H=B*sqrt(2);}]
 \foreach \Y in {-B,B} {\foreach \X in {-B,B}
 {\stepcounter{coordi} \path ({-1*sign(\Y)*\X},\Y,0) coordinate[label=right:$\Alph{coordi}$] (\Alph{coordi});}}
 \path (0,0,H) coordinate[label=right:$T$] (T);
 \draw[thick] (B) -- (T);
 \draw[thick,dashed] (B) -- (C) -- (D)  (C) -- (T);
 \draw[thick] (A) -- (T) -- (D) -- (A) -- (B);
 % this implements BR:BC = 1:3
 \path (C) -- (B) coordinate[pos=4/3,label=right:$R$] (R)
  (T) -- (C) coordinate[pos=1/4,label=right:$P$] (P)
  (intersection cs:first line={(B)--(T)}, second line={(R)--(P)}) 
 coordinate[label=right:$S$] (S);
 \foreach \X/\Y [count=\Z] in {B/S,C/P}
  {\path let \p1=($(\X)-(T)$),\p2=($(\Y)-(\X)$),
  \n1={veclen(\x2,\y2)/veclen(\x1,\y1)} in \pgfextra{\pgfmathsetmacro{\myratio}{\n1}
  \pgfmathsetmacro{\myz}{\myratio*H}
  \ifnum\Z=1
  \xdef\LstZ{\myz}
  \else
  \xdef\LstZ{\LstZ,\myz}
  \fi}; }
 \typeout{\LstZ} %<- contains the z-values of S and P
 % this brings you in the A-T-D plane 
 \begin{scope}[shift={(D)},x={($(A)-(D)$)},y={($(T)-(D)$)}]  
   \path (0.25,{\BB/3}) coordinate[label=right:$Q$] (Q);
 \end{scope}
 % the B-T and C-T lines have the equation p=(B*(1-t),\mp B*(1-t),H*t), t\in[0,1]
 % so for a given z value y=\mp B*(1-z/H)
 % or dy/dz=\pm B/H, dx/dz=\pm B/H
 % this gives us the slope of the plane in the ATD plane
 \path ($(Q)+({(B/H)*({\LstZ}[1]-{\LstZ}[0])},
 {B*(1-{\LstZ}[1]/H)+B*(1-{\LstZ}[0]/H)},
 {{\LstZ}[1]-{\LstZ}[0]})$) coordinate(aux);
 \path (intersection cs:first line={(Q)--(aux)}, second line={(A)--(T)}) 
  coordinate[label=right:$W$] (W)
  (intersection cs:first line={(Q)--(aux)}, second line={(D)--(T)}) 
  coordinate[label=right:$V$] (V);
 \begin{scope}[on background layer]
  \fill[yellow,fill opacity=0.4] (W) -- (S) -- (P) -- (V) -- cycle;
 \end{scope}
\end{tikzpicture}}
\end{document}

旧答案：我认为你构建这个的整体方式很好，+1。但是，我认为有三点可以补充。

1. 正交投影。这就是我在这里使用的原因tikz-3dplot，我认为也可以用 pstricks 来实现。
2. 一个可以说更简单的固定Q在所需平面上的方法：切换到具有原点的平面，例如，在D和轴A-D和处T-D。
3. 3D 排序。这是本答案的要点：计算黄色和蓝色平面相交的线，并在背景层上绘制黄色平面的隐藏部分。

它试图标记代码中的相关部分。我认为这些事情可以在 pstricks 中重做，但我做不到，我已经有一段时间没有积极使用它了。

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{backgrounds}
\newcounter{coordi}
\begin{document}
\tdplotsetmaincoords{70}{-30} 
\begin{tikzpicture}[tdplot_main_coords,declare function={B=3;H=B*sqrt(2);}]
 \foreach \Y in {-B,B} {\foreach \X in {-B,B}
 {\stepcounter{coordi} \path ({-1*sign(\Y)*\X},\Y,0) coordinate[label=right:$\Alph{coordi}$] (\Alph{coordi});}}
 \foreach \Y in {-B,B}
 {\stepcounter{coordi} 
 \path (0,\Y,0) coordinate[label=right:$\Alph{coordi}$] (\Alph{coordi})
 (0,\Y,H) coordinate[label=right:$\Alph{coordi}'$] (\Alph{coordi}');}
 \path (0,0,H) coordinate[label=right:$T$] (T);
 \draw[thick] (B) -- (T);
 \draw[thick,dashed] (B) -- (C) -- (D)  (C) -- (T);
 \fill[blue!50,fill opacity=0.4] (E) -- (F) -- (F') -- (E') -- cycle;
 \draw[thick] (A) -- (T) -- (D) -- (A) -- (B);
 % this implements BR:BC = 1:3
 \path (C) -- (B) coordinate[pos=4/3,label=right:$R$] (R);
 \path (intersection cs:first line={(R)--(F')}, second line={(C)--(T)}) 
 coordinate[label=right:$P$] (P)
 (intersection cs:first line={(R)--(F')}, second line={(B)--(T)}) 
 coordinate[label=right:$S$] (S);
 \draw[gray,thin,dashed] (R) -- (F');
 % this brings you in the A-T-D plane 
 \begin{scope}[shift={(D)},x={($(A)-(D)$)},y={($(T)-(D)$)}]  
   \path (0.25,2/3) coordinate[label=right:$Q$] (Q);
 \end{scope}
 % compute intersection points V and W
 \path (intersection cs:first line={(Q)--(F')}, second line={(A)--(T)}) 
 coordinate[label=right:$W$] (W)
 (intersection cs:first line={(Q)--(F')}, second line={(D)--(T)}) 
 coordinate[label=right:$V$] (V);
 % compute intersections of W-S and V-P with the blue plane
 \foreach \X/\Y [count=\Z] in {A/W,B/S,D/V,C/P}
  {\path let \p1=($(\X)-(T)$),\p2=($(\Y)-(T)$),
  \n1={veclen(\x2,\y2)/veclen(\x1,\y1)} in \pgfextra{\pgfmathsetmacro{\myratio}{\n1}
  \ifnum\Z=1
  \xdef\LstRatios{\myratio}
  \else
  \xdef\LstRatios{\LstRatios,\myratio}
  \fi}; }
 %\typeout{\LstRatios}
 \path (W) -- (S) coordinate[pos={{\LstRatios}[0]/({\LstRatios}[0]+{\LstRatios}[1])}] (WS)
  (V) -- (P) coordinate[pos={{\LstRatios}[2]/({\LstRatios}[2]+{\LstRatios}[3])}] (VP);
  \draw (WS) circle (1pt);
 \begin{scope}[on background layer]
  \fill[yellow,fill opacity=0.4] (WS) -- (S) -- (P) -- (VP) -- cycle;
 \end{scope}
 \fill[yellow,fill opacity=0.4] (WS) -- (W) -- (V) -- (VP) -- cycle;
\end{tikzpicture}
\end{document}

Answer 2

也许当你看到我的代码时你会大吃一惊。我计算了所有交点的坐标。

\documentclass[border=3mm,12pt]{standalone}
\usepackage{fouriernc}
 \usepackage{tikz}
 \usepackage{tikz-3dplot}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \usetikzlibrary{intersections,calc,backgrounds}


 \tikzset{ hidden/.style = {thick, dashed}}
   \tikzset{%
     add/.style args={#1 and #2}{
         to path={%
  ($(\tikztostart)!-#1!(\tikztotarget)$)--($(\tikztotarget)!-#2!(\tikztostart)$)%
   \tikztonodes}}}

 \begin{document}

 \tdplotsetmaincoords{70}{20}
 \begin{tikzpicture}[tdplot_main_coords,scale=1.3]
 \pgfmathsetmacro\a{4}
 \pgfmathsetmacro\h{5}

 % definitions
 \path
coordinate(A) at (0,0,0)
coordinate (B) at (\a,0,0)
coordinate (C) at(\a,\a,0)
coordinate (D) at (0,\a,0)
coordinate (T) at (1/2*\a, 1/2*\a, \h) 
coordinate (Q) at (1/6*\a, 1/2*\a, 1/3*\h)               
coordinate (P) at  (2/3*\a, 2/3*\a, 2/3*\h)
coordinate (R) at  (\a, -1/3*\a, 0)
coordinate (Y) at  (5/18*\a, 13/18*\a, 5/9*\h)
coordinate (E) at (5/6*\a, 1/6*\a, 1/3*\h)
coordinate (V) at (1/3*\a, 0, 0)
coordinate (Z) at (0, 1/6*\a, 0)
coordinate (X) at  (1/2*\a, 7/6*\a, \h);

 \draw[hidden,thick]
        (A) -- (C)  (B) -- (D)  (A)--(D)   (T)-- (C)  (T)-- (D) (C) -- (D) (X) -- (Z) (Y) -- (P) (V) -- (Z);
\draw [thick] (T)-- (C) (A) -- (B) -- (C) (T)-- (A) (T)-- (R) -- (X) -- cycle (T)-- (B) (E) -- (V) -- (R)

;

\fill[yellow,fill opacity=0.4] (P) -- (Y) -- (Z) -- (V) -- (E) -- cycle;
 \foreach \point/\position in {A/below,B/below,C/below,D/right,T/above,P/right,R/below,E/right,Q/left,V/below,X/right,Y/left,Z/right}
 {
   \fill (\point) circle (1.2pt);
   \node[\position=.3pt] at (\point) {$\point$};
 }
\end{tikzpicture}
 \end{document} 

和\tdplotsetmaincoords{70}{290}

\documentclass[border=3mm,12pt]{standalone}
\usepackage{fouriernc}
 \usepackage{tikz}
 \usepackage{tikz-3dplot}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \usetikzlibrary{intersections,calc,backgrounds}


 \tikzset{ hidden/.style = {thick, dashed}}
    \begin{document}

 \tdplotsetmaincoords{70}{290}
 \begin{tikzpicture}[tdplot_main_coords,scale=1.3]
 \pgfmathsetmacro\a{4}
 \pgfmathsetmacro\h{5}

 % definitions
 \path
coordinate(A) at (0,0,0)
coordinate (B) at (\a,0,0)
coordinate (C) at(\a,\a,0)
coordinate (D) at (0,\a,0)
coordinate (T) at (1/2*\a, 1/2*\a, \h) 
coordinate (Q) at (1/6*\a, 1/2*\a, 1/3*\h)               
coordinate (P) at  (2/3*\a, 2/3*\a, 2/3*\h)
coordinate (R) at  (\a, -1/3*\a, 0)
coordinate (Y) at  (5/18*\a, 13/18*\a, 5/9*\h)
coordinate (E) at (5/6*\a, 1/6*\a, 1/3*\h)
coordinate (V) at (1/3*\a, 0, 0)
coordinate (Z) at (0, 1/6*\a, 0)
coordinate (X) at  (1/2*\a, 7/6*\a, \h);

 \draw[hidden,thick]
        (A) -- (C)  (B) -- (D)  (B) -- (C)   (T)-- (C)  (T)-- (C) (C) -- (D)  (Y) -- (P) (V) -- (Z)  (B) -- (R);
\draw [thick] (T)-- (D)  (T)-- (A) (T)-- (R) -- (X) -- cycle (T)-- (B) (E) -- (V) -- (R) (A)--(D) (A)--  (B) (X) -- (Z)

;

\fill[yellow,fill opacity=0.4] (P) -- (Y) -- (Z) -- (V) -- (E) -- cycle;
 \foreach \point/\position in {A/below,B/below,C/below,D/below,T/above,P/right,R/below,E/above,Q/left,V/below,X/above,Y/left,Z/below}
 {
   \fill (\point) circle (1.2pt);
   \node[\position=.3pt] at (\point) {$\point$};
 }
\end{tikzpicture}
 \end{document} 

Q此代码可以在三角形内使用TAD（参见https://en.wikipedia.org/wiki/Convex_combinationu）。您可以通过改变和 的值来从多个角度进行查看v。

\documentclass[border=3mm,12pt]{standalone}
\usepackage{fouriernc}
 \usepackage{tikz}
 \usepackage{tikz-3dplot}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \usetikzlibrary{intersections,calc,backgrounds}


 \tikzset{ hidden/.style = {thick, dashed}}
    \begin{document}

 \tdplotsetmaincoords{70}{290}
 \begin{tikzpicture}[tdplot_main_coords,scale=1.3,line join = round, line cap = round]
\pgfmathsetmacro\a{6}
\pgfmathsetmacro\h{5}
\pgfmathsetmacro\u{2/5}
\pgfmathsetmacro\v{1/6}

% definitions
\path
coordinate(A) at (0,0,0)
coordinate (B) at (\a,0,0)
coordinate (C) at(\a,\a,0)
coordinate (D) at (0,\a,0)
coordinate (T) at (1/2*\a, 1/2*\a, \h) 
coordinate (Q) at ({1/2*\v*\a}, {1/2*\v*\a+(1-\v-\u)*\a}, {\v*\h})              
coordinate (P) at  (2/3*\a, 2/3*\a, 2/3*\h)
coordinate (R) at  (\a, -1/3*\a, 0)
coordinate (Y) at ({\a*(3*\u+2*\v)/(2*(2+3*\u))}, {(3*\u-2*\v+4)*\a/(2*(2+3*\u))}, {(3*\u+2*\v)*\h/(2+3*\u)})
coordinate (E) at (5/6*\a, 1/6*\a, 1/3*\h)
coordinate (V) at ({\a*(3*\u+5*\v-3)/(3*\u+6*\v-4)},0,0)
coordinate (Z) at (0,{\a*(3*\u+5*\v-3)/(3*(-1+\v))},0)
coordinate (X) at  (1/2*\a, 7/6*\a, \h);

\draw[hidden,thick]
(A) -- (C)  (B) -- (D)  (B) -- (C)   (T)-- (C)  (T)-- (C) (C) -- (D)  (Y) -- (P) (V) -- (Z)  (B) -- (R) ;
\draw [thick] (T)-- (D)  (T)-- (A) (T)-- (R) -- (X) -- cycle (T)-- (B) (E) -- (V) -- (R) (A)--(D) (A)--  (B) (X) -- (Z)
;

\fill[yellow,fill opacity=0.4] (P) -- (Y) -- (Z) -- (V) -- (E) -- cycle;
\foreach \point/\position in {A/below,B/below,C/below,D/below,T/above,P/right,R/below,E/above,Q/left,V/below,X/above,Y/left,Z/below}
{
    \fill (\point) circle (1.2pt);
    \node[\position=.3pt] at (\point) {$\point$};
}
\end{tikzpicture}

\结束{中心}

\begin{center}
    \begin{tikzpicture}[tdplot_main_coords,scale=1.3,line join = round, line cap = round]
    \pgfmathsetmacro\a{6}
    \pgfmathsetmacro\h{5}
    \pgfmathsetmacro\u{3/5}
    \pgfmathsetmacro\v{1/3}

    % definitions
    \path
    coordinate(A) at (0,0,0)
    coordinate (B) at (\a,0,0)
    coordinate (C) at(\a,\a,0)
    coordinate (D) at (0,\a,0)
    coordinate (T) at (1/2*\a, 1/2*\a, \h) 
    coordinate (Q) at ({1/2*\v*\a}, {1/2*\v*\a+(1-\v-\u)*\a}, {\v*\h})              
    coordinate (P) at  (2/3*\a, 2/3*\a, 2/3*\h)
    coordinate (R) at  (\a, -1/3*\a, 0)
    coordinate (Y) at ({\a*(3*\u+2*\v)/(2*(2+3*\u))}, {(3*\u-2*\v+4)*\a/(2*(2+3*\u))}, {(3*\u+2*\v)*\h/(2+3*\u)})
    coordinate (E) at (5/6*\a, 1/6*\a, 1/3*\h)
    coordinate (V) at ({\a*(3*\u+5*\v-3)/(3*\u+6*\v-4)},0,0)
    coordinate (Z) at (0,{\a*(3*\u+5*\v-3)/(3*(-1+\v))},0)
    coordinate (X) at  (1/2*\a, 7/6*\a, \h);

    \draw[hidden,thick]
    (A) -- (C)  (B) -- (D)  (B) -- (C)   (T)-- (C)  (T)-- (C) (C) -- (D)  (Y) -- (P) (V) -- (Z)  (B) -- (R) ;
    \draw [thick] (T)-- (D)  (T)-- (A) (T)-- (R) -- (X) -- cycle (T)-- (B) (E) -- (V) -- (R) (A)--(D) (A)--  (B) (X) -- (Z)
    ;

    \fill[yellow,fill opacity=0.4] (P) -- (Y) -- (Z) -- (V) -- (E) -- cycle;
    \foreach \point/\position in {A/below,B/below,C/below,D/below,T/above,P/right,R/below,E/above,Q/left,V/below,X/above,Y/left,Z/below}
    {
        \fill (\point) circle (1.2pt);
        \node[\position=.3pt] at (\point) {$\point$};
    }
    \end{tikzpicture}

，\pgfmathsetmacro\u{3/5} \pgfmathsetmacro\v{1/3}我们得到