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How to draw this pyramid with TeX?


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14















I draw this pyramid with GeoSpacw. How to draw this figure with TeX?



enter image description here



Where:




  • M is midpoint of the segment SA;


  • I is midpoint of the segment SD;


  • H is project of the point A on the line SO;


  • ABKE is a is a parallelogram;


  • The line passing through S and parallel to the line AB.











share|improve this question

























  • M doesn't appear to be on segment SB. Do you mean SA or SN?

    – Scott H.
    Apr 14 '13 at 2:30






  • 1





    Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

    – Scott H.
    Apr 14 '13 at 3:30











  • O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

    – minthao_2011
    Apr 14 '13 at 3:36











  • I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

    – Alain Matthes
    Apr 14 '13 at 10:47
















14















I draw this pyramid with GeoSpacw. How to draw this figure with TeX?



enter image description here



Where:




  • M is midpoint of the segment SA;


  • I is midpoint of the segment SD;


  • H is project of the point A on the line SO;


  • ABKE is a is a parallelogram;


  • The line passing through S and parallel to the line AB.











share|improve this question

























  • M doesn't appear to be on segment SB. Do you mean SA or SN?

    – Scott H.
    Apr 14 '13 at 2:30






  • 1





    Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

    – Scott H.
    Apr 14 '13 at 3:30











  • O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

    – minthao_2011
    Apr 14 '13 at 3:36











  • I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

    – Alain Matthes
    Apr 14 '13 at 10:47














14












14








14


4






I draw this pyramid with GeoSpacw. How to draw this figure with TeX?



enter image description here



Where:




  • M is midpoint of the segment SA;


  • I is midpoint of the segment SD;


  • H is project of the point A on the line SO;


  • ABKE is a is a parallelogram;


  • The line passing through S and parallel to the line AB.











share|improve this question
















I draw this pyramid with GeoSpacw. How to draw this figure with TeX?



enter image description here



Where:




  • M is midpoint of the segment SA;


  • I is midpoint of the segment SD;


  • H is project of the point A on the line SO;


  • ABKE is a is a parallelogram;


  • The line passing through S and parallel to the line AB.








tikz-pgf






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 9 hours ago









JouleV

4,95611239




4,95611239










asked Apr 14 '13 at 2:08









minthao_2011minthao_2011

2,22952146




2,22952146













  • M doesn't appear to be on segment SB. Do you mean SA or SN?

    – Scott H.
    Apr 14 '13 at 2:30






  • 1





    Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

    – Scott H.
    Apr 14 '13 at 3:30











  • O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

    – minthao_2011
    Apr 14 '13 at 3:36











  • I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

    – Alain Matthes
    Apr 14 '13 at 10:47



















  • M doesn't appear to be on segment SB. Do you mean SA or SN?

    – Scott H.
    Apr 14 '13 at 2:30






  • 1





    Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

    – Scott H.
    Apr 14 '13 at 3:30











  • O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

    – minthao_2011
    Apr 14 '13 at 3:36











  • I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

    – Alain Matthes
    Apr 14 '13 at 10:47

















M doesn't appear to be on segment SB. Do you mean SA or SN?

– Scott H.
Apr 14 '13 at 2:30





M doesn't appear to be on segment SB. Do you mean SA or SN?

– Scott H.
Apr 14 '13 at 2:30




1




1





Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

– Scott H.
Apr 14 '13 at 3:30





Could you elaborate on what points are given, and what points are calculated? As it stands, it seems like ABKE is an arbitrary parallelogram and S is an arbitrary point (from which we can calculate M and N). It seems like there needs to be some extra information to get I, D or C.

– Scott H.
Apr 14 '13 at 3:30













O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

– minthao_2011
Apr 14 '13 at 3:36





O.K. The line SA perpendicular to the plane (ABC). SA = AB*Sqrt{2}$, ABCD is a square. K is projection of B on the plane (SCD). Thank you.

– minthao_2011
Apr 14 '13 at 3:36













I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

– Alain Matthes
Apr 14 '13 at 10:47





I need some verifications. E seems to the orthogonal projection of A on SD because BK//AE and K is the orthogonal projection of B on SCD. Am I wrong?

– Alain Matthes
Apr 14 '13 at 10:47










6 Answers
6






active

oldest

votes


















19














General Update



I can suggest two answers, the first one with tikz-3dplot interesting because we draw the pyramid in 3D and we can change easily the point of view and the perspective; the second one with tkz-euclide in 2D with parallel projection.



I made some mistakes in my first calculations but there are two problems to draw a correct figure. The first problem is to place E and K, the second one is to place H. The methods depend of the tools used, 3D or 2D with parallel projection.



How to place K and E



BK is orthogonal to the plane SCD, ABKE is a parallelogram, so AE||BK then AE is orthogonal to SCD.
AE is perpendicular to SD in the plane ASD. This method is useful with with parallel projection because ASD is the main front and is in the main plane. We can use a projection in a plane to place E.



In 3D with for example tikz-3dplot we need to calculate the coordinate of E. I named a the side of the square ABCD.



enter image description here



The coordinates of E are (0,2*a/3,sqrt(2)/3*a).
Remark: If the coordinates of K are (xK,yK,zK) then the coordinates of E are (0,yK,zK). Or xK=xB, so K is defined by (a,2*a/3,sqrt(2)/3*a).



About K. In the first answer, I used the code from of Caramdir's answer (from this question). If you know the equation of a plane, it's possible to determine the orthognal projection of a point on this plane. I determined the coordinates of K, so I can avoid the interesting code of Caramdir.
Remark: SCD has an simple equation here sqrt(2)y+z = sqrt(2)s it's parallel to (AB) and cut axis in S and D. It's easy to get K with caramdir's code.



How to place H



H is the projection orthogonal of A on OS. OAS is orthogonal to ABCD. The projection orthognal of H on ABBCD is a point Q on the line OA. Below I determined the coordinates of Q (2/5 a,2/5 /a,0)and then I determined the coordinates of H. It's not possible to use the orthogonal projection of TikZ.



enter image description here



Part 1) With tikz-3dplot



 documentclass{article}
usepackage{tikz}
usepackage{tikz-3dplot}
usetikzlibrary{intersections,calc}


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

begin{document}

%tdplotsetmaincoords{60}{120}
tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=2]
pgfmathsetmacroa{2}

% definitions
path (0,0,0) coordinate (A)
(a,0,0) coordinate (B)
(a,a,0) coordinate (C)
(0,a,0) coordinate (D)
(0,0,{a*sqrt(2)}) coordinate (S)
(0,0,{0.5*a*sqrt(2)}) coordinate (M)
(0.5*a,0.5*a,0) coordinate (O)
(0,0.5*a,{0.5*a*sqrt(2)}) coordinate (I)
(2/5*a,2/5*a,{sqrt(2)/5*a}) coordinate (H)
($(B)!2!(M)$) coordinate (N)
(0,2*a/3,{sqrt(2)/3*a}) coordinate (E)
(a,2*a/3,{sqrt(2)/3*a}) coordinate (K) ;
% drawing
begin{scope}
clip (S) -- (B) -- (D) --cycle;
draw[hidden] (B) -- (N);
end{scope}
begin{scope}
clip (S) -- (I) -- (N) --cycle;
draw (B) -- (N);
end{scope}

draw[hidden]
(A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
(A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
draw (B) -- (K) -- (E) (C) -- (N) -- (K)
(B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
draw[add = 1 and 1] (S) to (N);

% place black circles and labels
foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
I/right,K/right,M/right,N/left,O/right,S/left}
{
fill (point) circle (.5pt);
node[position=3pt] at (point) {$point$};
}

end{tikzpicture}
end{document}


enter image description here



If you want to change the point of view you can try tdplotsetmaincoords{60}{120}.



enter image description here



Part B) with tkz-euclide



If you want to use only TikZ, see Gonzalo's answer. The problem are the same, only the syntax and the macros are different. But the answers are similar. It's not very easy to show 3D objects with 2D methods.



I use the projection of A on SD to get E but it's also possible to get the coordinates. It's more difficult to get H, I used here the fact SO=5*OH.



 documentclass{article}
usepackage{tkz-euclide,amsmath}
usetkzobj{all}
tikzset{hidden/.style = {thin, dotted}}
begin{document}

begin{tikzpicture}
tkzInit[xmin=-5,xmax=10,ymin=-5,ymax=15]
tkzClip
pgfmathsetmacrozs{3+5*sqrt(2)}
% definition
tkzDefPoints{0/0/B,1/3/A,5/0/C,6/3/D,1/zs/S}
tkzDefMidPoint(A,C) tkzGetPoint{O}
tkzDefMidPoint(S,A) tkzGetPoint{M}
tkzDefMidPoint(S,D) tkzGetPoint{I}
tkzDefPointBy[projection=onto S--D](A) tkzGetPoint{E}
tkzDefPointWith[colinear= at E](A,B) tkzGetPoint{K}
tkzDefPointWith[linear,K=2](C,I) tkzGetPoint{N}
tkzDefPointWith[linear,K=1/5](O,S) tkzGetPoint{H}
% drawing
tkzDrawSegments[hidden](A,B A,D I,M A,C B,D S,O M,E A,E A,S A,H)
tkzDrawSegments(B,C C,D B,S S,D B,K K,E C,N K,N)
begin{scope}
tkzClipPolygon(S,B,D)
tkzDrawSegments[hidden](B,N)
end{scope}
begin{scope}
tkzClipPolygon(S,I,N)
tkzDrawSegments(B,N)
end{scope}
% labels
tkzDrawPoints(A,B,C,D,S,I,M,O,E,K,N,H)
tkzLabelPoints(A,B,C,D,K,O)
tkzLabelPoints[above](S,E,N,H)
tkzLabelPoints[above right](I,M)
tkzMarkRightAngles(A,E,D D,A,S)
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

    – minthao_2011
    Apr 16 '13 at 8:46













  • @minthao_2011 Thanks to confirm my calculations !

    – Alain Matthes
    Apr 16 '13 at 10:46



















15














With PSTricks.



enter image description here



documentclass[pstricks]{standalone}
usepackage{pst-eucl}
psset{linejoin=1}
begin{document}
begin{pspicture}[showgrid=false](-.75,-.75)(5.75,9.75)
pstGeonode[PosAngle={-135,-45,0,180,180,0}]
{B}(4,0){C}(5,2){D}(1,2){A}(1,7){S}([nodesep=1.8]{S}D){E}
pstMiddleAB{D}{B}{O}
pstTranslation[PosAngle=30]{A}{B}{E}[K]
pstProjection{O}{S}{A}[H]
pstMiddleAB[PosAngle=-45]{S}{A}{M}
pstMiddleAB{S}{D}{I}
pstInterLL{B}{M}{C}{I}{N}
pstInterLL[PointName=none,PointSymbol=none]{D}{S}{B}{N}{X}
psline(B)(C)(D)
psline(B)(K)(E)
psline(B)(S)(D)
psline(C)(N)(K)
psline(X)(N)
pcline[nodesepA=-2,nodesepB=-.5](S)(N)
psset{linestyle=dashed}
pspolygon(A)(B)(D)
pspolygon(A)(C)(D)
psline(A)(S)(O)
psline(H)(A)(E)
psline(M)(I)
psline(B)(X)
end{pspicture}
end{document}





share|improve this answer


























  • The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

    – Alain Matthes
    Apr 14 '13 at 8:46











  • Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

    – Alain Matthes
    Apr 14 '13 at 8:56













  • If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

    – Alain Matthes
    Apr 14 '13 at 10:44











  • To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

    – Alain Matthes
    Apr 15 '13 at 6:50











  • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

    – Alain Matthes
    Apr 15 '13 at 17:35



















10














Here's one possibility:



documentclass{article}
usepackage{tikz}
usetikzlibrary{positioning,intersections,calc}

defxmov{0.8cm}
defymov{3cm}
defxside{4.5cm}

begin{document}

begin{tikzpicture}
clip (-0.5,-0.5) rectangle (5.7,12.1);
coordinate (b);
coordinate[above right=ymov and xmov of b] (a);
coordinate[right=xside of b] (c);
coordinate[right=xside of a] (d);
coordinate[above=6cm of a] (s);
draw (b) -- (s) -- (c);
draw[name path=edge1] (s) -- (d);
draw[dashed] (b) -- (a) -- (d) -- (c) -- (b) -- (d);
draw[dashed,name path=diag1] (b) -- (d);
draw[dashed,name path=diag2] (a) -- (c);
draw[dashed] (s) -- (a);
coordinate[name intersections={of=diag1 and diag2,by={o}}];
draw[dashed] (s) -- (o);
coordinate (i) at ( $ (s)!0.5!(d) $ );
coordinate (m) at ( $ (s)!0.5!(a) $ );
path[dashed,name path=line1] (b) -- ( $ (b)!15cm!(m) $ );
path[draw,name path=line2] (c) -- ( $ (c)!15cm!(i) $ );
coordinate[name intersections={of=line1 and line2,by={n}}];
% intersection od sd with nb
coordinate[name intersections={of=line1 and edge1,by={aux}}];
coordinate (e) at ( $ (d)!(a)!(s) $ );
coordinate[below left=ymov and xmov of e] (k);
coordinate (h) at ($(s)!(a)!(o)$);
draw (n) -- (aux);
draw[dashed] (aux) -- (b);
draw[dashed] (a) -- (h);
draw (e) -- (k) -- (b);
draw[dashed] (e) -- (a);
draw (n) -- (k);
draw (n) -- ( $ (n)!6cm!(s) $ );
% place black circles and labels
foreach point/position in {a/left,b/left,c/right,d/right,e/right,h/above left,
i/right,k/right,m/right,n/left,o/right,s/left}
{
fill (point) circle (2pt);
node[position=2pt] at (point) {$point$};
}
end{tikzpicture}

end{document}


enter image description here



Midpoints were obtained using ($(s)!0.5!(o)$) and the projection was obtained with ($(s)!(a)!(o)$). Intersections were calculated with the intersections library.



Changing dashed to dotted everywhere in the code above one gets:



enter image description here






share|improve this answer


























  • Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

    – Alain Matthes
    Apr 14 '13 at 9:01











  • And it's not Sb perpendicular to bc but Sa perpendicular to ad.

    – Alain Matthes
    Apr 14 '13 at 10:40











  • @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

    – Gonzalo Medina
    Apr 14 '13 at 14:33











  • The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

    – Alain Matthes
    Apr 15 '13 at 17:15











  • Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

    – Gonzalo Medina
    Apr 15 '13 at 17:25



















9














Here's a metapost version, unfortunately without altering the out of the box luamplib package, I can't seem to draw dashed lines so I've replaced them with lighter lines.



enter image description here



documentclass{article}
usepackage{luamplib}
begin{document}
begin{mplibcode}
u:=2cm;
path p[];
pair A,B,C,D,E,H,I,K,M,N,O,S,t;
def de = withcolor .85white enddef;

beginfig(1);
k = 2.5u;
p1 = unitsquare slanted .3 xscaled 2u yscaled u;
B = point 0 of p1;
C = point 1 of p1;
D = point 2 of p1;
A = point 3 of p1;
S = A shifted (0,k*length (A--B));
M = .5[A,S];
O = .5[A,C];
N = whatever[B,M]=whatever[S,B shifted (0,k*length (A--B))];
I = (N--C) intersectionpoint (S--D);
H = whatever[S,O] = A + whatever*((S-O) rotated 90);
E = whatever[S,D] = A + whatever*((S-D) rotated 90);
K = E shifted (B-A);
t = (B--N) intersectionpoint (S--D);

draw O--A--S--O--B--A--D--O--C de;
draw A--E de;
draw M--I de;
draw A--H de;
draw B--t de;
draw S--C--B--K--E--S--N--C--D;
draw B--S--D;
draw t--N--K;

dotlabel.llft("B",B);
dotlabel.urt("D",D);
dotlabel.lrt("C",C);
dotlabel.ulft("A",A);
dotlabel.ulft("S",S);
dotlabel.top("N",N);
dotlabel.bot("O",O);
dotlabel.rt("I",I);
dotlabel.ulft("H",H);
dotlabel.urt("E",E);
dotlabel.lrt("K",K);
dotlabel.lrt("M",M);

endfig;
end;

end{mplibcode}
end{document}





share|improve this answer































    9














    enter image description here



    Asymptote version pyramid.asy:



    import three;  // 3D module
    import math;
    currentprojection=orthographic(camera=(55,144,80),
    up=(0,0,1),target=(0,0,0),zoom=1,center=true);

    size(300);
    size3(300,300,300);

    triple intersectionpoint(triple a,triple b,triple c,triple d){
    real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/
    ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x);
    return a*(1.0-u)+b*u;
    }

    triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy:
    // as South, East and North or (0,-1), (1,0) and (0,1)
    // and I is a sqrt(-1)=(0,1)
    real a=40; // side of the square;
    real h=50; // height, h=AS
    real d=sqrt(2)/2*a; // half of the diagonal

    O=(0,0,0);
    C=(0,d,0);
    B=(d,0,0);
    A=(0,-d,0);
    D=(-d,0,0);
    SS=A+(0,0,h);
    M=0.5(SS+A);
    II=0.5(SS+D);
    NN=intersectionpoint(SS,(SS+A-B),C,II);

    real phi=atan(h/a);
    real u=a*cos(phi)/sqrt(a^2+h^2);
    EE=D*(1-u)+SS*u;
    K=EE+B-A;

    real psi=atan(h/d);
    real u=d*cos(psi)/sqrt(d^2+h^2);
    H=O*(1-u)+SS*u;

    pair Q=intersectionpoint(project(NN--B),project(SS--D));

    pen dashed=linetype(new real[] {5,5}); // set up dashed pattern
    pen visLine=darkblue+0.8pt;
    pen hidLine=lightblue+dashed+0.8pt;

    void Dot(...triple[] v){
    dotfactor=8;
    for(int i=0;i<v.length;++i){
    dot(project(v[i]),UnFill);
    }
    }

    void Draw3(guide3 g, pen p=currentpen){
    draw(project(g),p);
    }

    void labelP(string s,triple t,pair p=(0,0)){
    label("$"+s+"$",project(t),p);
    }

    Draw3(B--A--D,hidLine);
    Draw3(H--A--SS,hidLine);
    Draw3(M--II,hidLine);
    Draw3(EE--A--C,hidLine);
    Draw3(B--D,hidLine);
    Draw3(SS--O,hidLine);

    Draw3(SS--B--C--SS--D--C,visLine);
    Draw3(SS--NN--C,visLine);
    Draw3(EE--K--B,visLine);
    Draw3(NN--K,visLine);

    draw(project(NN)--Q,visLine);
    draw(project(B)--Q,hidLine);

    Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS);

    labelP("A",A,NW);
    labelP("B",B,SW);
    labelP("C",C,E);
    labelP("D",D,NE);
    labelP("E",EE,NE);
    labelP("H",H,NE);
    labelP("I",II,NE);
    labelP("K",K,SE);
    labelP("M",M,W);
    labelP("N",NN,W);
    labelP("O",O,S);
    labelP("S",SS,NW);


    To get a standalone pyramid.pdf run asy -f pdf pyramid.asy.






    share|improve this answer
























    • +1 and please kindly make asymptote version for this question.

      – kiss my armpit
      Apr 14 '13 at 14:04













    • @Bugbusters: done.

      – g.kov
      Apr 14 '13 at 16:24











    • Best looking solution to this question.

      – Ingo
      Mar 3 '14 at 16:52



















    0














    I copied Alain Matthes' code with some prepaired.



    documentclass[border=2cm]{standalone}
    usepackage{tikz}
    usepackage{tikz-3dplot}
    usetikzlibrary{intersections,calc}
    usepackage{fouriernc}

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

    begin{document}

    tdplotsetmaincoords{60}{100}
    %tdplotsetmaincoords{80}{100}
    begin{tikzpicture}[tdplot_main_coords,scale=1.5]
    pgfmathsetmacroa{3}
    pgfmathsetmacrob{3}
    pgfmathsetmacro{h}{{a*sqrt(2)}}

    % definitions
    path (0,0,0) coordinate (A)
    (a,0,0) coordinate (B)
    (a,b,0) coordinate (C)
    (0,b,0) coordinate (D)
    (0,0,h) coordinate (S)
    ($(S)!0.5!(A)$) coordinate (M)
    ($(C)!0.5!(A)$) coordinate (O)
    ($(S)!0.5!(D)$) coordinate (I)
    ({2*h^2*a/(a^2+b^2+4*h^2)}, {2*b*h^2/(a^2+b^2+4*h^2)}, {h*(a^2+b^2)/(a^2+b^2+4*h^2)}) coordinate (H)
    (-a, 0, h) coordinate (N)
    (0, {b*h^2/(b^2+h^2)},{ b^2*h/(b^2+h^2)}) coordinate (E)
    (a, {b*h^2/(b^2+h^2)}, {b^2*h/(b^2+h^2)}) coordinate (K) ;
    % drawing
    begin{scope}
    clip (S) -- (B) -- (D) --cycle;
    draw[hidden] (B) -- (N);
    end{scope}
    begin{scope}
    clip (S) -- (I) -- (N) --cycle;
    draw (B) -- (N);
    end{scope}

    draw[hidden]
    (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
    (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
    draw[thick] (B) -- (K) -- (E) (C) -- (N) -- (K)
    (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
    draw[add = 1 and 1] (S) to (N);

    % place black circles and labels
    foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
    I/right,K/right,M/right,N/left,O/right,S/left}
    {
    fill (point) circle (1.2pt);
    node[position=3pt] at (point) {$point$};
    }

    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer























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      6 Answers
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      6 Answers
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      active

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      19














      General Update



      I can suggest two answers, the first one with tikz-3dplot interesting because we draw the pyramid in 3D and we can change easily the point of view and the perspective; the second one with tkz-euclide in 2D with parallel projection.



      I made some mistakes in my first calculations but there are two problems to draw a correct figure. The first problem is to place E and K, the second one is to place H. The methods depend of the tools used, 3D or 2D with parallel projection.



      How to place K and E



      BK is orthogonal to the plane SCD, ABKE is a parallelogram, so AE||BK then AE is orthogonal to SCD.
      AE is perpendicular to SD in the plane ASD. This method is useful with with parallel projection because ASD is the main front and is in the main plane. We can use a projection in a plane to place E.



      In 3D with for example tikz-3dplot we need to calculate the coordinate of E. I named a the side of the square ABCD.



      enter image description here



      The coordinates of E are (0,2*a/3,sqrt(2)/3*a).
      Remark: If the coordinates of K are (xK,yK,zK) then the coordinates of E are (0,yK,zK). Or xK=xB, so K is defined by (a,2*a/3,sqrt(2)/3*a).



      About K. In the first answer, I used the code from of Caramdir's answer (from this question). If you know the equation of a plane, it's possible to determine the orthognal projection of a point on this plane. I determined the coordinates of K, so I can avoid the interesting code of Caramdir.
      Remark: SCD has an simple equation here sqrt(2)y+z = sqrt(2)s it's parallel to (AB) and cut axis in S and D. It's easy to get K with caramdir's code.



      How to place H



      H is the projection orthogonal of A on OS. OAS is orthogonal to ABCD. The projection orthognal of H on ABBCD is a point Q on the line OA. Below I determined the coordinates of Q (2/5 a,2/5 /a,0)and then I determined the coordinates of H. It's not possible to use the orthogonal projection of TikZ.



      enter image description here



      Part 1) With tikz-3dplot



       documentclass{article}
      usepackage{tikz}
      usepackage{tikz-3dplot}
      usetikzlibrary{intersections,calc}


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

      begin{document}

      %tdplotsetmaincoords{60}{120}
      tdplotsetmaincoords{80}{100}
      begin{tikzpicture}[tdplot_main_coords,scale=2]
      pgfmathsetmacroa{2}

      % definitions
      path (0,0,0) coordinate (A)
      (a,0,0) coordinate (B)
      (a,a,0) coordinate (C)
      (0,a,0) coordinate (D)
      (0,0,{a*sqrt(2)}) coordinate (S)
      (0,0,{0.5*a*sqrt(2)}) coordinate (M)
      (0.5*a,0.5*a,0) coordinate (O)
      (0,0.5*a,{0.5*a*sqrt(2)}) coordinate (I)
      (2/5*a,2/5*a,{sqrt(2)/5*a}) coordinate (H)
      ($(B)!2!(M)$) coordinate (N)
      (0,2*a/3,{sqrt(2)/3*a}) coordinate (E)
      (a,2*a/3,{sqrt(2)/3*a}) coordinate (K) ;
      % drawing
      begin{scope}
      clip (S) -- (B) -- (D) --cycle;
      draw[hidden] (B) -- (N);
      end{scope}
      begin{scope}
      clip (S) -- (I) -- (N) --cycle;
      draw (B) -- (N);
      end{scope}

      draw[hidden]
      (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
      (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
      draw (B) -- (K) -- (E) (C) -- (N) -- (K)
      (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
      draw[add = 1 and 1] (S) to (N);

      % place black circles and labels
      foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
      I/right,K/right,M/right,N/left,O/right,S/left}
      {
      fill (point) circle (.5pt);
      node[position=3pt] at (point) {$point$};
      }

      end{tikzpicture}
      end{document}


      enter image description here



      If you want to change the point of view you can try tdplotsetmaincoords{60}{120}.



      enter image description here



      Part B) with tkz-euclide



      If you want to use only TikZ, see Gonzalo's answer. The problem are the same, only the syntax and the macros are different. But the answers are similar. It's not very easy to show 3D objects with 2D methods.



      I use the projection of A on SD to get E but it's also possible to get the coordinates. It's more difficult to get H, I used here the fact SO=5*OH.



       documentclass{article}
      usepackage{tkz-euclide,amsmath}
      usetkzobj{all}
      tikzset{hidden/.style = {thin, dotted}}
      begin{document}

      begin{tikzpicture}
      tkzInit[xmin=-5,xmax=10,ymin=-5,ymax=15]
      tkzClip
      pgfmathsetmacrozs{3+5*sqrt(2)}
      % definition
      tkzDefPoints{0/0/B,1/3/A,5/0/C,6/3/D,1/zs/S}
      tkzDefMidPoint(A,C) tkzGetPoint{O}
      tkzDefMidPoint(S,A) tkzGetPoint{M}
      tkzDefMidPoint(S,D) tkzGetPoint{I}
      tkzDefPointBy[projection=onto S--D](A) tkzGetPoint{E}
      tkzDefPointWith[colinear= at E](A,B) tkzGetPoint{K}
      tkzDefPointWith[linear,K=2](C,I) tkzGetPoint{N}
      tkzDefPointWith[linear,K=1/5](O,S) tkzGetPoint{H}
      % drawing
      tkzDrawSegments[hidden](A,B A,D I,M A,C B,D S,O M,E A,E A,S A,H)
      tkzDrawSegments(B,C C,D B,S S,D B,K K,E C,N K,N)
      begin{scope}
      tkzClipPolygon(S,B,D)
      tkzDrawSegments[hidden](B,N)
      end{scope}
      begin{scope}
      tkzClipPolygon(S,I,N)
      tkzDrawSegments(B,N)
      end{scope}
      % labels
      tkzDrawPoints(A,B,C,D,S,I,M,O,E,K,N,H)
      tkzLabelPoints(A,B,C,D,K,O)
      tkzLabelPoints[above](S,E,N,H)
      tkzLabelPoints[above right](I,M)
      tkzMarkRightAngles(A,E,D D,A,S)
      end{tikzpicture}
      end{document}


      enter image description here






      share|improve this answer


























      • The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

        – minthao_2011
        Apr 16 '13 at 8:46













      • @minthao_2011 Thanks to confirm my calculations !

        – Alain Matthes
        Apr 16 '13 at 10:46
















      19














      General Update



      I can suggest two answers, the first one with tikz-3dplot interesting because we draw the pyramid in 3D and we can change easily the point of view and the perspective; the second one with tkz-euclide in 2D with parallel projection.



      I made some mistakes in my first calculations but there are two problems to draw a correct figure. The first problem is to place E and K, the second one is to place H. The methods depend of the tools used, 3D or 2D with parallel projection.



      How to place K and E



      BK is orthogonal to the plane SCD, ABKE is a parallelogram, so AE||BK then AE is orthogonal to SCD.
      AE is perpendicular to SD in the plane ASD. This method is useful with with parallel projection because ASD is the main front and is in the main plane. We can use a projection in a plane to place E.



      In 3D with for example tikz-3dplot we need to calculate the coordinate of E. I named a the side of the square ABCD.



      enter image description here



      The coordinates of E are (0,2*a/3,sqrt(2)/3*a).
      Remark: If the coordinates of K are (xK,yK,zK) then the coordinates of E are (0,yK,zK). Or xK=xB, so K is defined by (a,2*a/3,sqrt(2)/3*a).



      About K. In the first answer, I used the code from of Caramdir's answer (from this question). If you know the equation of a plane, it's possible to determine the orthognal projection of a point on this plane. I determined the coordinates of K, so I can avoid the interesting code of Caramdir.
      Remark: SCD has an simple equation here sqrt(2)y+z = sqrt(2)s it's parallel to (AB) and cut axis in S and D. It's easy to get K with caramdir's code.



      How to place H



      H is the projection orthogonal of A on OS. OAS is orthogonal to ABCD. The projection orthognal of H on ABBCD is a point Q on the line OA. Below I determined the coordinates of Q (2/5 a,2/5 /a,0)and then I determined the coordinates of H. It's not possible to use the orthogonal projection of TikZ.



      enter image description here



      Part 1) With tikz-3dplot



       documentclass{article}
      usepackage{tikz}
      usepackage{tikz-3dplot}
      usetikzlibrary{intersections,calc}


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

      begin{document}

      %tdplotsetmaincoords{60}{120}
      tdplotsetmaincoords{80}{100}
      begin{tikzpicture}[tdplot_main_coords,scale=2]
      pgfmathsetmacroa{2}

      % definitions
      path (0,0,0) coordinate (A)
      (a,0,0) coordinate (B)
      (a,a,0) coordinate (C)
      (0,a,0) coordinate (D)
      (0,0,{a*sqrt(2)}) coordinate (S)
      (0,0,{0.5*a*sqrt(2)}) coordinate (M)
      (0.5*a,0.5*a,0) coordinate (O)
      (0,0.5*a,{0.5*a*sqrt(2)}) coordinate (I)
      (2/5*a,2/5*a,{sqrt(2)/5*a}) coordinate (H)
      ($(B)!2!(M)$) coordinate (N)
      (0,2*a/3,{sqrt(2)/3*a}) coordinate (E)
      (a,2*a/3,{sqrt(2)/3*a}) coordinate (K) ;
      % drawing
      begin{scope}
      clip (S) -- (B) -- (D) --cycle;
      draw[hidden] (B) -- (N);
      end{scope}
      begin{scope}
      clip (S) -- (I) -- (N) --cycle;
      draw (B) -- (N);
      end{scope}

      draw[hidden]
      (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
      (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
      draw (B) -- (K) -- (E) (C) -- (N) -- (K)
      (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
      draw[add = 1 and 1] (S) to (N);

      % place black circles and labels
      foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
      I/right,K/right,M/right,N/left,O/right,S/left}
      {
      fill (point) circle (.5pt);
      node[position=3pt] at (point) {$point$};
      }

      end{tikzpicture}
      end{document}


      enter image description here



      If you want to change the point of view you can try tdplotsetmaincoords{60}{120}.



      enter image description here



      Part B) with tkz-euclide



      If you want to use only TikZ, see Gonzalo's answer. The problem are the same, only the syntax and the macros are different. But the answers are similar. It's not very easy to show 3D objects with 2D methods.



      I use the projection of A on SD to get E but it's also possible to get the coordinates. It's more difficult to get H, I used here the fact SO=5*OH.



       documentclass{article}
      usepackage{tkz-euclide,amsmath}
      usetkzobj{all}
      tikzset{hidden/.style = {thin, dotted}}
      begin{document}

      begin{tikzpicture}
      tkzInit[xmin=-5,xmax=10,ymin=-5,ymax=15]
      tkzClip
      pgfmathsetmacrozs{3+5*sqrt(2)}
      % definition
      tkzDefPoints{0/0/B,1/3/A,5/0/C,6/3/D,1/zs/S}
      tkzDefMidPoint(A,C) tkzGetPoint{O}
      tkzDefMidPoint(S,A) tkzGetPoint{M}
      tkzDefMidPoint(S,D) tkzGetPoint{I}
      tkzDefPointBy[projection=onto S--D](A) tkzGetPoint{E}
      tkzDefPointWith[colinear= at E](A,B) tkzGetPoint{K}
      tkzDefPointWith[linear,K=2](C,I) tkzGetPoint{N}
      tkzDefPointWith[linear,K=1/5](O,S) tkzGetPoint{H}
      % drawing
      tkzDrawSegments[hidden](A,B A,D I,M A,C B,D S,O M,E A,E A,S A,H)
      tkzDrawSegments(B,C C,D B,S S,D B,K K,E C,N K,N)
      begin{scope}
      tkzClipPolygon(S,B,D)
      tkzDrawSegments[hidden](B,N)
      end{scope}
      begin{scope}
      tkzClipPolygon(S,I,N)
      tkzDrawSegments(B,N)
      end{scope}
      % labels
      tkzDrawPoints(A,B,C,D,S,I,M,O,E,K,N,H)
      tkzLabelPoints(A,B,C,D,K,O)
      tkzLabelPoints[above](S,E,N,H)
      tkzLabelPoints[above right](I,M)
      tkzMarkRightAngles(A,E,D D,A,S)
      end{tikzpicture}
      end{document}


      enter image description here






      share|improve this answer


























      • The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

        – minthao_2011
        Apr 16 '13 at 8:46













      • @minthao_2011 Thanks to confirm my calculations !

        – Alain Matthes
        Apr 16 '13 at 10:46














      19












      19








      19







      General Update



      I can suggest two answers, the first one with tikz-3dplot interesting because we draw the pyramid in 3D and we can change easily the point of view and the perspective; the second one with tkz-euclide in 2D with parallel projection.



      I made some mistakes in my first calculations but there are two problems to draw a correct figure. The first problem is to place E and K, the second one is to place H. The methods depend of the tools used, 3D or 2D with parallel projection.



      How to place K and E



      BK is orthogonal to the plane SCD, ABKE is a parallelogram, so AE||BK then AE is orthogonal to SCD.
      AE is perpendicular to SD in the plane ASD. This method is useful with with parallel projection because ASD is the main front and is in the main plane. We can use a projection in a plane to place E.



      In 3D with for example tikz-3dplot we need to calculate the coordinate of E. I named a the side of the square ABCD.



      enter image description here



      The coordinates of E are (0,2*a/3,sqrt(2)/3*a).
      Remark: If the coordinates of K are (xK,yK,zK) then the coordinates of E are (0,yK,zK). Or xK=xB, so K is defined by (a,2*a/3,sqrt(2)/3*a).



      About K. In the first answer, I used the code from of Caramdir's answer (from this question). If you know the equation of a plane, it's possible to determine the orthognal projection of a point on this plane. I determined the coordinates of K, so I can avoid the interesting code of Caramdir.
      Remark: SCD has an simple equation here sqrt(2)y+z = sqrt(2)s it's parallel to (AB) and cut axis in S and D. It's easy to get K with caramdir's code.



      How to place H



      H is the projection orthogonal of A on OS. OAS is orthogonal to ABCD. The projection orthognal of H on ABBCD is a point Q on the line OA. Below I determined the coordinates of Q (2/5 a,2/5 /a,0)and then I determined the coordinates of H. It's not possible to use the orthogonal projection of TikZ.



      enter image description here



      Part 1) With tikz-3dplot



       documentclass{article}
      usepackage{tikz}
      usepackage{tikz-3dplot}
      usetikzlibrary{intersections,calc}


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

      begin{document}

      %tdplotsetmaincoords{60}{120}
      tdplotsetmaincoords{80}{100}
      begin{tikzpicture}[tdplot_main_coords,scale=2]
      pgfmathsetmacroa{2}

      % definitions
      path (0,0,0) coordinate (A)
      (a,0,0) coordinate (B)
      (a,a,0) coordinate (C)
      (0,a,0) coordinate (D)
      (0,0,{a*sqrt(2)}) coordinate (S)
      (0,0,{0.5*a*sqrt(2)}) coordinate (M)
      (0.5*a,0.5*a,0) coordinate (O)
      (0,0.5*a,{0.5*a*sqrt(2)}) coordinate (I)
      (2/5*a,2/5*a,{sqrt(2)/5*a}) coordinate (H)
      ($(B)!2!(M)$) coordinate (N)
      (0,2*a/3,{sqrt(2)/3*a}) coordinate (E)
      (a,2*a/3,{sqrt(2)/3*a}) coordinate (K) ;
      % drawing
      begin{scope}
      clip (S) -- (B) -- (D) --cycle;
      draw[hidden] (B) -- (N);
      end{scope}
      begin{scope}
      clip (S) -- (I) -- (N) --cycle;
      draw (B) -- (N);
      end{scope}

      draw[hidden]
      (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
      (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
      draw (B) -- (K) -- (E) (C) -- (N) -- (K)
      (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
      draw[add = 1 and 1] (S) to (N);

      % place black circles and labels
      foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
      I/right,K/right,M/right,N/left,O/right,S/left}
      {
      fill (point) circle (.5pt);
      node[position=3pt] at (point) {$point$};
      }

      end{tikzpicture}
      end{document}


      enter image description here



      If you want to change the point of view you can try tdplotsetmaincoords{60}{120}.



      enter image description here



      Part B) with tkz-euclide



      If you want to use only TikZ, see Gonzalo's answer. The problem are the same, only the syntax and the macros are different. But the answers are similar. It's not very easy to show 3D objects with 2D methods.



      I use the projection of A on SD to get E but it's also possible to get the coordinates. It's more difficult to get H, I used here the fact SO=5*OH.



       documentclass{article}
      usepackage{tkz-euclide,amsmath}
      usetkzobj{all}
      tikzset{hidden/.style = {thin, dotted}}
      begin{document}

      begin{tikzpicture}
      tkzInit[xmin=-5,xmax=10,ymin=-5,ymax=15]
      tkzClip
      pgfmathsetmacrozs{3+5*sqrt(2)}
      % definition
      tkzDefPoints{0/0/B,1/3/A,5/0/C,6/3/D,1/zs/S}
      tkzDefMidPoint(A,C) tkzGetPoint{O}
      tkzDefMidPoint(S,A) tkzGetPoint{M}
      tkzDefMidPoint(S,D) tkzGetPoint{I}
      tkzDefPointBy[projection=onto S--D](A) tkzGetPoint{E}
      tkzDefPointWith[colinear= at E](A,B) tkzGetPoint{K}
      tkzDefPointWith[linear,K=2](C,I) tkzGetPoint{N}
      tkzDefPointWith[linear,K=1/5](O,S) tkzGetPoint{H}
      % drawing
      tkzDrawSegments[hidden](A,B A,D I,M A,C B,D S,O M,E A,E A,S A,H)
      tkzDrawSegments(B,C C,D B,S S,D B,K K,E C,N K,N)
      begin{scope}
      tkzClipPolygon(S,B,D)
      tkzDrawSegments[hidden](B,N)
      end{scope}
      begin{scope}
      tkzClipPolygon(S,I,N)
      tkzDrawSegments(B,N)
      end{scope}
      % labels
      tkzDrawPoints(A,B,C,D,S,I,M,O,E,K,N,H)
      tkzLabelPoints(A,B,C,D,K,O)
      tkzLabelPoints[above](S,E,N,H)
      tkzLabelPoints[above right](I,M)
      tkzMarkRightAngles(A,E,D D,A,S)
      end{tikzpicture}
      end{document}


      enter image description here






      share|improve this answer















      General Update



      I can suggest two answers, the first one with tikz-3dplot interesting because we draw the pyramid in 3D and we can change easily the point of view and the perspective; the second one with tkz-euclide in 2D with parallel projection.



      I made some mistakes in my first calculations but there are two problems to draw a correct figure. The first problem is to place E and K, the second one is to place H. The methods depend of the tools used, 3D or 2D with parallel projection.



      How to place K and E



      BK is orthogonal to the plane SCD, ABKE is a parallelogram, so AE||BK then AE is orthogonal to SCD.
      AE is perpendicular to SD in the plane ASD. This method is useful with with parallel projection because ASD is the main front and is in the main plane. We can use a projection in a plane to place E.



      In 3D with for example tikz-3dplot we need to calculate the coordinate of E. I named a the side of the square ABCD.



      enter image description here



      The coordinates of E are (0,2*a/3,sqrt(2)/3*a).
      Remark: If the coordinates of K are (xK,yK,zK) then the coordinates of E are (0,yK,zK). Or xK=xB, so K is defined by (a,2*a/3,sqrt(2)/3*a).



      About K. In the first answer, I used the code from of Caramdir's answer (from this question). If you know the equation of a plane, it's possible to determine the orthognal projection of a point on this plane. I determined the coordinates of K, so I can avoid the interesting code of Caramdir.
      Remark: SCD has an simple equation here sqrt(2)y+z = sqrt(2)s it's parallel to (AB) and cut axis in S and D. It's easy to get K with caramdir's code.



      How to place H



      H is the projection orthogonal of A on OS. OAS is orthogonal to ABCD. The projection orthognal of H on ABBCD is a point Q on the line OA. Below I determined the coordinates of Q (2/5 a,2/5 /a,0)and then I determined the coordinates of H. It's not possible to use the orthogonal projection of TikZ.



      enter image description here



      Part 1) With tikz-3dplot



       documentclass{article}
      usepackage{tikz}
      usepackage{tikz-3dplot}
      usetikzlibrary{intersections,calc}


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

      begin{document}

      %tdplotsetmaincoords{60}{120}
      tdplotsetmaincoords{80}{100}
      begin{tikzpicture}[tdplot_main_coords,scale=2]
      pgfmathsetmacroa{2}

      % definitions
      path (0,0,0) coordinate (A)
      (a,0,0) coordinate (B)
      (a,a,0) coordinate (C)
      (0,a,0) coordinate (D)
      (0,0,{a*sqrt(2)}) coordinate (S)
      (0,0,{0.5*a*sqrt(2)}) coordinate (M)
      (0.5*a,0.5*a,0) coordinate (O)
      (0,0.5*a,{0.5*a*sqrt(2)}) coordinate (I)
      (2/5*a,2/5*a,{sqrt(2)/5*a}) coordinate (H)
      ($(B)!2!(M)$) coordinate (N)
      (0,2*a/3,{sqrt(2)/3*a}) coordinate (E)
      (a,2*a/3,{sqrt(2)/3*a}) coordinate (K) ;
      % drawing
      begin{scope}
      clip (S) -- (B) -- (D) --cycle;
      draw[hidden] (B) -- (N);
      end{scope}
      begin{scope}
      clip (S) -- (I) -- (N) --cycle;
      draw (B) -- (N);
      end{scope}

      draw[hidden]
      (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
      (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
      draw (B) -- (K) -- (E) (C) -- (N) -- (K)
      (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
      draw[add = 1 and 1] (S) to (N);

      % place black circles and labels
      foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
      I/right,K/right,M/right,N/left,O/right,S/left}
      {
      fill (point) circle (.5pt);
      node[position=3pt] at (point) {$point$};
      }

      end{tikzpicture}
      end{document}


      enter image description here



      If you want to change the point of view you can try tdplotsetmaincoords{60}{120}.



      enter image description here



      Part B) with tkz-euclide



      If you want to use only TikZ, see Gonzalo's answer. The problem are the same, only the syntax and the macros are different. But the answers are similar. It's not very easy to show 3D objects with 2D methods.



      I use the projection of A on SD to get E but it's also possible to get the coordinates. It's more difficult to get H, I used here the fact SO=5*OH.



       documentclass{article}
      usepackage{tkz-euclide,amsmath}
      usetkzobj{all}
      tikzset{hidden/.style = {thin, dotted}}
      begin{document}

      begin{tikzpicture}
      tkzInit[xmin=-5,xmax=10,ymin=-5,ymax=15]
      tkzClip
      pgfmathsetmacrozs{3+5*sqrt(2)}
      % definition
      tkzDefPoints{0/0/B,1/3/A,5/0/C,6/3/D,1/zs/S}
      tkzDefMidPoint(A,C) tkzGetPoint{O}
      tkzDefMidPoint(S,A) tkzGetPoint{M}
      tkzDefMidPoint(S,D) tkzGetPoint{I}
      tkzDefPointBy[projection=onto S--D](A) tkzGetPoint{E}
      tkzDefPointWith[colinear= at E](A,B) tkzGetPoint{K}
      tkzDefPointWith[linear,K=2](C,I) tkzGetPoint{N}
      tkzDefPointWith[linear,K=1/5](O,S) tkzGetPoint{H}
      % drawing
      tkzDrawSegments[hidden](A,B A,D I,M A,C B,D S,O M,E A,E A,S A,H)
      tkzDrawSegments(B,C C,D B,S S,D B,K K,E C,N K,N)
      begin{scope}
      tkzClipPolygon(S,B,D)
      tkzDrawSegments[hidden](B,N)
      end{scope}
      begin{scope}
      tkzClipPolygon(S,I,N)
      tkzDrawSegments(B,N)
      end{scope}
      % labels
      tkzDrawPoints(A,B,C,D,S,I,M,O,E,K,N,H)
      tkzLabelPoints(A,B,C,D,K,O)
      tkzLabelPoints[above](S,E,N,H)
      tkzLabelPoints[above right](I,M)
      tkzMarkRightAngles(A,E,D D,A,S)
      end{tikzpicture}
      end{document}


      enter image description here







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 13 '17 at 12:34









      Community

      1




      1










      answered Apr 14 '13 at 7:52









      Alain MatthesAlain Matthes

      73.4k7162295




      73.4k7162295













      • The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

        – minthao_2011
        Apr 16 '13 at 8:46













      • @minthao_2011 Thanks to confirm my calculations !

        – Alain Matthes
        Apr 16 '13 at 10:46



















      • The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

        – minthao_2011
        Apr 16 '13 at 8:46













      • @minthao_2011 Thanks to confirm my calculations !

        – Alain Matthes
        Apr 16 '13 at 10:46

















      The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

      – minthao_2011
      Apr 16 '13 at 8:46







      The coordinates of the point $H$ is $(2/5a, 2/5a, frac{sqrt(2)}{5}a$.

      – minthao_2011
      Apr 16 '13 at 8:46















      @minthao_2011 Thanks to confirm my calculations !

      – Alain Matthes
      Apr 16 '13 at 10:46





      @minthao_2011 Thanks to confirm my calculations !

      – Alain Matthes
      Apr 16 '13 at 10:46











      15














      With PSTricks.



      enter image description here



      documentclass[pstricks]{standalone}
      usepackage{pst-eucl}
      psset{linejoin=1}
      begin{document}
      begin{pspicture}[showgrid=false](-.75,-.75)(5.75,9.75)
      pstGeonode[PosAngle={-135,-45,0,180,180,0}]
      {B}(4,0){C}(5,2){D}(1,2){A}(1,7){S}([nodesep=1.8]{S}D){E}
      pstMiddleAB{D}{B}{O}
      pstTranslation[PosAngle=30]{A}{B}{E}[K]
      pstProjection{O}{S}{A}[H]
      pstMiddleAB[PosAngle=-45]{S}{A}{M}
      pstMiddleAB{S}{D}{I}
      pstInterLL{B}{M}{C}{I}{N}
      pstInterLL[PointName=none,PointSymbol=none]{D}{S}{B}{N}{X}
      psline(B)(C)(D)
      psline(B)(K)(E)
      psline(B)(S)(D)
      psline(C)(N)(K)
      psline(X)(N)
      pcline[nodesepA=-2,nodesepB=-.5](S)(N)
      psset{linestyle=dashed}
      pspolygon(A)(B)(D)
      pspolygon(A)(C)(D)
      psline(A)(S)(O)
      psline(H)(A)(E)
      psline(M)(I)
      psline(B)(X)
      end{pspicture}
      end{document}





      share|improve this answer


























      • The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

        – Alain Matthes
        Apr 14 '13 at 8:46











      • Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

        – Alain Matthes
        Apr 14 '13 at 8:56













      • If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

        – Alain Matthes
        Apr 14 '13 at 10:44











      • To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

        – Alain Matthes
        Apr 15 '13 at 6:50











      • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

        – Alain Matthes
        Apr 15 '13 at 17:35
















      15














      With PSTricks.



      enter image description here



      documentclass[pstricks]{standalone}
      usepackage{pst-eucl}
      psset{linejoin=1}
      begin{document}
      begin{pspicture}[showgrid=false](-.75,-.75)(5.75,9.75)
      pstGeonode[PosAngle={-135,-45,0,180,180,0}]
      {B}(4,0){C}(5,2){D}(1,2){A}(1,7){S}([nodesep=1.8]{S}D){E}
      pstMiddleAB{D}{B}{O}
      pstTranslation[PosAngle=30]{A}{B}{E}[K]
      pstProjection{O}{S}{A}[H]
      pstMiddleAB[PosAngle=-45]{S}{A}{M}
      pstMiddleAB{S}{D}{I}
      pstInterLL{B}{M}{C}{I}{N}
      pstInterLL[PointName=none,PointSymbol=none]{D}{S}{B}{N}{X}
      psline(B)(C)(D)
      psline(B)(K)(E)
      psline(B)(S)(D)
      psline(C)(N)(K)
      psline(X)(N)
      pcline[nodesepA=-2,nodesepB=-.5](S)(N)
      psset{linestyle=dashed}
      pspolygon(A)(B)(D)
      pspolygon(A)(C)(D)
      psline(A)(S)(O)
      psline(H)(A)(E)
      psline(M)(I)
      psline(B)(X)
      end{pspicture}
      end{document}





      share|improve this answer


























      • The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

        – Alain Matthes
        Apr 14 '13 at 8:46











      • Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

        – Alain Matthes
        Apr 14 '13 at 8:56













      • If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

        – Alain Matthes
        Apr 14 '13 at 10:44











      • To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

        – Alain Matthes
        Apr 15 '13 at 6:50











      • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

        – Alain Matthes
        Apr 15 '13 at 17:35














      15












      15








      15







      With PSTricks.



      enter image description here



      documentclass[pstricks]{standalone}
      usepackage{pst-eucl}
      psset{linejoin=1}
      begin{document}
      begin{pspicture}[showgrid=false](-.75,-.75)(5.75,9.75)
      pstGeonode[PosAngle={-135,-45,0,180,180,0}]
      {B}(4,0){C}(5,2){D}(1,2){A}(1,7){S}([nodesep=1.8]{S}D){E}
      pstMiddleAB{D}{B}{O}
      pstTranslation[PosAngle=30]{A}{B}{E}[K]
      pstProjection{O}{S}{A}[H]
      pstMiddleAB[PosAngle=-45]{S}{A}{M}
      pstMiddleAB{S}{D}{I}
      pstInterLL{B}{M}{C}{I}{N}
      pstInterLL[PointName=none,PointSymbol=none]{D}{S}{B}{N}{X}
      psline(B)(C)(D)
      psline(B)(K)(E)
      psline(B)(S)(D)
      psline(C)(N)(K)
      psline(X)(N)
      pcline[nodesepA=-2,nodesepB=-.5](S)(N)
      psset{linestyle=dashed}
      pspolygon(A)(B)(D)
      pspolygon(A)(C)(D)
      psline(A)(S)(O)
      psline(H)(A)(E)
      psline(M)(I)
      psline(B)(X)
      end{pspicture}
      end{document}





      share|improve this answer















      With PSTricks.



      enter image description here



      documentclass[pstricks]{standalone}
      usepackage{pst-eucl}
      psset{linejoin=1}
      begin{document}
      begin{pspicture}[showgrid=false](-.75,-.75)(5.75,9.75)
      pstGeonode[PosAngle={-135,-45,0,180,180,0}]
      {B}(4,0){C}(5,2){D}(1,2){A}(1,7){S}([nodesep=1.8]{S}D){E}
      pstMiddleAB{D}{B}{O}
      pstTranslation[PosAngle=30]{A}{B}{E}[K]
      pstProjection{O}{S}{A}[H]
      pstMiddleAB[PosAngle=-45]{S}{A}{M}
      pstMiddleAB{S}{D}{I}
      pstInterLL{B}{M}{C}{I}{N}
      pstInterLL[PointName=none,PointSymbol=none]{D}{S}{B}{N}{X}
      psline(B)(C)(D)
      psline(B)(K)(E)
      psline(B)(S)(D)
      psline(C)(N)(K)
      psline(X)(N)
      pcline[nodesepA=-2,nodesepB=-.5](S)(N)
      psset{linestyle=dashed}
      pspolygon(A)(B)(D)
      pspolygon(A)(C)(D)
      psline(A)(S)(O)
      psline(H)(A)(E)
      psline(M)(I)
      psline(B)(X)
      end{pspicture}
      end{document}






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 14 '13 at 7:10

























      answered Apr 14 '13 at 7:05









      kiss my armpitkiss my armpit

      13.1k20174405




      13.1k20174405













      • The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

        – Alain Matthes
        Apr 14 '13 at 8:46











      • Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

        – Alain Matthes
        Apr 14 '13 at 8:56













      • If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

        – Alain Matthes
        Apr 14 '13 at 10:44











      • To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

        – Alain Matthes
        Apr 15 '13 at 6:50











      • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

        – Alain Matthes
        Apr 15 '13 at 17:35



















      • The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

        – Alain Matthes
        Apr 14 '13 at 8:46











      • Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

        – Alain Matthes
        Apr 14 '13 at 8:56













      • If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

        – Alain Matthes
        Apr 14 '13 at 10:44











      • To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

        – Alain Matthes
        Apr 15 '13 at 6:50











      • I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

        – Alain Matthes
        Apr 15 '13 at 17:35

















      The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

      – Alain Matthes
      Apr 14 '13 at 8:46





      The point E is defined with K. It seems that you define E and then K. I don't understand ([nodesep=1.8]{S}D){E}

      – Alain Matthes
      Apr 14 '13 at 8:46













      Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

      – Alain Matthes
      Apr 14 '13 at 8:56







      Are you sure of the psprojection. It is not a drawing in a plane, If S is on a line perpendicular to ABCD in O, the projection is false but perhaps I'm wrong.

      – Alain Matthes
      Apr 14 '13 at 8:56















      If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

      – Alain Matthes
      Apr 14 '13 at 10:44





      If BK is orthogonal to the plane SCD then AE is orthogonal to SCD to. The consequence is AE perpendicular to SD, E is the orthogonal projection of A on SD.

      – Alain Matthes
      Apr 14 '13 at 10:44













      To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

      – Alain Matthes
      Apr 15 '13 at 6:50





      To se the problem with the projection : pstTranslation[PosAngle=30]{A}{S}{O}[P] pstProjection{O}{P}{A}[Q] OP is orthogonal to ABCD so the projection orthogonal Q of A on OP is O. Then ABKE is a parallelogram and BK is orthogonal to SCD, so AE is orthogonal to SCD. The consequence is AE perpendicular to SD because A,E,S,D ar in the same plane. I think it's not the case in your construction

      – Alain Matthes
      Apr 15 '13 at 6:50













      I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

      – Alain Matthes
      Apr 15 '13 at 17:35





      I'm not a great expert of pstricks and pst-eucl but I would like to understand how you place E. I'm not sure that you place E to the right place. pstProjection seems to be a macro to use in 2D but in the main plane ASD or a plane parallel to ASD.

      – Alain Matthes
      Apr 15 '13 at 17:35











      10














      Here's one possibility:



      documentclass{article}
      usepackage{tikz}
      usetikzlibrary{positioning,intersections,calc}

      defxmov{0.8cm}
      defymov{3cm}
      defxside{4.5cm}

      begin{document}

      begin{tikzpicture}
      clip (-0.5,-0.5) rectangle (5.7,12.1);
      coordinate (b);
      coordinate[above right=ymov and xmov of b] (a);
      coordinate[right=xside of b] (c);
      coordinate[right=xside of a] (d);
      coordinate[above=6cm of a] (s);
      draw (b) -- (s) -- (c);
      draw[name path=edge1] (s) -- (d);
      draw[dashed] (b) -- (a) -- (d) -- (c) -- (b) -- (d);
      draw[dashed,name path=diag1] (b) -- (d);
      draw[dashed,name path=diag2] (a) -- (c);
      draw[dashed] (s) -- (a);
      coordinate[name intersections={of=diag1 and diag2,by={o}}];
      draw[dashed] (s) -- (o);
      coordinate (i) at ( $ (s)!0.5!(d) $ );
      coordinate (m) at ( $ (s)!0.5!(a) $ );
      path[dashed,name path=line1] (b) -- ( $ (b)!15cm!(m) $ );
      path[draw,name path=line2] (c) -- ( $ (c)!15cm!(i) $ );
      coordinate[name intersections={of=line1 and line2,by={n}}];
      % intersection od sd with nb
      coordinate[name intersections={of=line1 and edge1,by={aux}}];
      coordinate (e) at ( $ (d)!(a)!(s) $ );
      coordinate[below left=ymov and xmov of e] (k);
      coordinate (h) at ($(s)!(a)!(o)$);
      draw (n) -- (aux);
      draw[dashed] (aux) -- (b);
      draw[dashed] (a) -- (h);
      draw (e) -- (k) -- (b);
      draw[dashed] (e) -- (a);
      draw (n) -- (k);
      draw (n) -- ( $ (n)!6cm!(s) $ );
      % place black circles and labels
      foreach point/position in {a/left,b/left,c/right,d/right,e/right,h/above left,
      i/right,k/right,m/right,n/left,o/right,s/left}
      {
      fill (point) circle (2pt);
      node[position=2pt] at (point) {$point$};
      }
      end{tikzpicture}

      end{document}


      enter image description here



      Midpoints were obtained using ($(s)!0.5!(o)$) and the projection was obtained with ($(s)!(a)!(o)$). Intersections were calculated with the intersections library.



      Changing dashed to dotted everywhere in the code above one gets:



      enter image description here






      share|improve this answer


























      • Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

        – Alain Matthes
        Apr 14 '13 at 9:01











      • And it's not Sb perpendicular to bc but Sa perpendicular to ad.

        – Alain Matthes
        Apr 14 '13 at 10:40











      • @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

        – Gonzalo Medina
        Apr 14 '13 at 14:33











      • The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

        – Alain Matthes
        Apr 15 '13 at 17:15











      • Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

        – Gonzalo Medina
        Apr 15 '13 at 17:25
















      10














      Here's one possibility:



      documentclass{article}
      usepackage{tikz}
      usetikzlibrary{positioning,intersections,calc}

      defxmov{0.8cm}
      defymov{3cm}
      defxside{4.5cm}

      begin{document}

      begin{tikzpicture}
      clip (-0.5,-0.5) rectangle (5.7,12.1);
      coordinate (b);
      coordinate[above right=ymov and xmov of b] (a);
      coordinate[right=xside of b] (c);
      coordinate[right=xside of a] (d);
      coordinate[above=6cm of a] (s);
      draw (b) -- (s) -- (c);
      draw[name path=edge1] (s) -- (d);
      draw[dashed] (b) -- (a) -- (d) -- (c) -- (b) -- (d);
      draw[dashed,name path=diag1] (b) -- (d);
      draw[dashed,name path=diag2] (a) -- (c);
      draw[dashed] (s) -- (a);
      coordinate[name intersections={of=diag1 and diag2,by={o}}];
      draw[dashed] (s) -- (o);
      coordinate (i) at ( $ (s)!0.5!(d) $ );
      coordinate (m) at ( $ (s)!0.5!(a) $ );
      path[dashed,name path=line1] (b) -- ( $ (b)!15cm!(m) $ );
      path[draw,name path=line2] (c) -- ( $ (c)!15cm!(i) $ );
      coordinate[name intersections={of=line1 and line2,by={n}}];
      % intersection od sd with nb
      coordinate[name intersections={of=line1 and edge1,by={aux}}];
      coordinate (e) at ( $ (d)!(a)!(s) $ );
      coordinate[below left=ymov and xmov of e] (k);
      coordinate (h) at ($(s)!(a)!(o)$);
      draw (n) -- (aux);
      draw[dashed] (aux) -- (b);
      draw[dashed] (a) -- (h);
      draw (e) -- (k) -- (b);
      draw[dashed] (e) -- (a);
      draw (n) -- (k);
      draw (n) -- ( $ (n)!6cm!(s) $ );
      % place black circles and labels
      foreach point/position in {a/left,b/left,c/right,d/right,e/right,h/above left,
      i/right,k/right,m/right,n/left,o/right,s/left}
      {
      fill (point) circle (2pt);
      node[position=2pt] at (point) {$point$};
      }
      end{tikzpicture}

      end{document}


      enter image description here



      Midpoints were obtained using ($(s)!0.5!(o)$) and the projection was obtained with ($(s)!(a)!(o)$). Intersections were calculated with the intersections library.



      Changing dashed to dotted everywhere in the code above one gets:



      enter image description here






      share|improve this answer


























      • Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

        – Alain Matthes
        Apr 14 '13 at 9:01











      • And it's not Sb perpendicular to bc but Sa perpendicular to ad.

        – Alain Matthes
        Apr 14 '13 at 10:40











      • @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

        – Gonzalo Medina
        Apr 14 '13 at 14:33











      • The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

        – Alain Matthes
        Apr 15 '13 at 17:15











      • Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

        – Gonzalo Medina
        Apr 15 '13 at 17:25














      10












      10








      10







      Here's one possibility:



      documentclass{article}
      usepackage{tikz}
      usetikzlibrary{positioning,intersections,calc}

      defxmov{0.8cm}
      defymov{3cm}
      defxside{4.5cm}

      begin{document}

      begin{tikzpicture}
      clip (-0.5,-0.5) rectangle (5.7,12.1);
      coordinate (b);
      coordinate[above right=ymov and xmov of b] (a);
      coordinate[right=xside of b] (c);
      coordinate[right=xside of a] (d);
      coordinate[above=6cm of a] (s);
      draw (b) -- (s) -- (c);
      draw[name path=edge1] (s) -- (d);
      draw[dashed] (b) -- (a) -- (d) -- (c) -- (b) -- (d);
      draw[dashed,name path=diag1] (b) -- (d);
      draw[dashed,name path=diag2] (a) -- (c);
      draw[dashed] (s) -- (a);
      coordinate[name intersections={of=diag1 and diag2,by={o}}];
      draw[dashed] (s) -- (o);
      coordinate (i) at ( $ (s)!0.5!(d) $ );
      coordinate (m) at ( $ (s)!0.5!(a) $ );
      path[dashed,name path=line1] (b) -- ( $ (b)!15cm!(m) $ );
      path[draw,name path=line2] (c) -- ( $ (c)!15cm!(i) $ );
      coordinate[name intersections={of=line1 and line2,by={n}}];
      % intersection od sd with nb
      coordinate[name intersections={of=line1 and edge1,by={aux}}];
      coordinate (e) at ( $ (d)!(a)!(s) $ );
      coordinate[below left=ymov and xmov of e] (k);
      coordinate (h) at ($(s)!(a)!(o)$);
      draw (n) -- (aux);
      draw[dashed] (aux) -- (b);
      draw[dashed] (a) -- (h);
      draw (e) -- (k) -- (b);
      draw[dashed] (e) -- (a);
      draw (n) -- (k);
      draw (n) -- ( $ (n)!6cm!(s) $ );
      % place black circles and labels
      foreach point/position in {a/left,b/left,c/right,d/right,e/right,h/above left,
      i/right,k/right,m/right,n/left,o/right,s/left}
      {
      fill (point) circle (2pt);
      node[position=2pt] at (point) {$point$};
      }
      end{tikzpicture}

      end{document}


      enter image description here



      Midpoints were obtained using ($(s)!0.5!(o)$) and the projection was obtained with ($(s)!(a)!(o)$). Intersections were calculated with the intersections library.



      Changing dashed to dotted everywhere in the code above one gets:



      enter image description here






      share|improve this answer















      Here's one possibility:



      documentclass{article}
      usepackage{tikz}
      usetikzlibrary{positioning,intersections,calc}

      defxmov{0.8cm}
      defymov{3cm}
      defxside{4.5cm}

      begin{document}

      begin{tikzpicture}
      clip (-0.5,-0.5) rectangle (5.7,12.1);
      coordinate (b);
      coordinate[above right=ymov and xmov of b] (a);
      coordinate[right=xside of b] (c);
      coordinate[right=xside of a] (d);
      coordinate[above=6cm of a] (s);
      draw (b) -- (s) -- (c);
      draw[name path=edge1] (s) -- (d);
      draw[dashed] (b) -- (a) -- (d) -- (c) -- (b) -- (d);
      draw[dashed,name path=diag1] (b) -- (d);
      draw[dashed,name path=diag2] (a) -- (c);
      draw[dashed] (s) -- (a);
      coordinate[name intersections={of=diag1 and diag2,by={o}}];
      draw[dashed] (s) -- (o);
      coordinate (i) at ( $ (s)!0.5!(d) $ );
      coordinate (m) at ( $ (s)!0.5!(a) $ );
      path[dashed,name path=line1] (b) -- ( $ (b)!15cm!(m) $ );
      path[draw,name path=line2] (c) -- ( $ (c)!15cm!(i) $ );
      coordinate[name intersections={of=line1 and line2,by={n}}];
      % intersection od sd with nb
      coordinate[name intersections={of=line1 and edge1,by={aux}}];
      coordinate (e) at ( $ (d)!(a)!(s) $ );
      coordinate[below left=ymov and xmov of e] (k);
      coordinate (h) at ($(s)!(a)!(o)$);
      draw (n) -- (aux);
      draw[dashed] (aux) -- (b);
      draw[dashed] (a) -- (h);
      draw (e) -- (k) -- (b);
      draw[dashed] (e) -- (a);
      draw (n) -- (k);
      draw (n) -- ( $ (n)!6cm!(s) $ );
      % place black circles and labels
      foreach point/position in {a/left,b/left,c/right,d/right,e/right,h/above left,
      i/right,k/right,m/right,n/left,o/right,s/left}
      {
      fill (point) circle (2pt);
      node[position=2pt] at (point) {$point$};
      }
      end{tikzpicture}

      end{document}


      enter image description here



      Midpoints were obtained using ($(s)!0.5!(o)$) and the projection was obtained with ($(s)!(a)!(o)$). Intersections were calculated with the intersections library.



      Changing dashed to dotted everywhere in the code above one gets:



      enter image description here







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 15 '13 at 0:38

























      answered Apr 14 '13 at 2:59









      Gonzalo MedinaGonzalo Medina

      400k4113071576




      400k4113071576













      • Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

        – Alain Matthes
        Apr 14 '13 at 9:01











      • And it's not Sb perpendicular to bc but Sa perpendicular to ad.

        – Alain Matthes
        Apr 14 '13 at 10:40











      • @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

        – Gonzalo Medina
        Apr 14 '13 at 14:33











      • The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

        – Alain Matthes
        Apr 15 '13 at 17:15











      • Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

        – Gonzalo Medina
        Apr 15 '13 at 17:25



















      • Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

        – Alain Matthes
        Apr 14 '13 at 9:01











      • And it's not Sb perpendicular to bc but Sa perpendicular to ad.

        – Alain Matthes
        Apr 14 '13 at 10:40











      • @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

        – Gonzalo Medina
        Apr 14 '13 at 14:33











      • The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

        – Alain Matthes
        Apr 15 '13 at 17:15











      • Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

        – Gonzalo Medina
        Apr 15 '13 at 17:25

















      Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

      – Alain Matthes
      Apr 14 '13 at 9:01





      Hi Gonzalo. I think with this method you can't respect the request like : as=sqrt(2)ab and h projection of a on ao. Here you get the projection in a plane, if s is above o, I think you get a problem because the projection needs to be o.

      – Alain Matthes
      Apr 14 '13 at 9:01













      And it's not Sb perpendicular to bc but Sa perpendicular to ad.

      – Alain Matthes
      Apr 14 '13 at 10:40





      And it's not Sb perpendicular to bc but Sa perpendicular to ad.

      – Alain Matthes
      Apr 14 '13 at 10:40













      @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

      – Gonzalo Medina
      Apr 14 '13 at 14:33





      @AlainMatthes yes, but the OP kept adding new conditions after I had answered with the initial requirements. Maybe I'll update this with the new requirements, but still it's not clear to me how to determine some of the points.

      – Gonzalo Medina
      Apr 14 '13 at 14:33













      The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

      – Alain Matthes
      Apr 15 '13 at 17:15





      The main problem is ($(s)!(a)!(o)$) because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5.

      – Alain Matthes
      Apr 15 '13 at 17:15













      Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

      – Gonzalo Medina
      Apr 15 '13 at 17:25





      Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you.

      – Gonzalo Medina
      Apr 15 '13 at 17:25











      9














      Here's a metapost version, unfortunately without altering the out of the box luamplib package, I can't seem to draw dashed lines so I've replaced them with lighter lines.



      enter image description here



      documentclass{article}
      usepackage{luamplib}
      begin{document}
      begin{mplibcode}
      u:=2cm;
      path p[];
      pair A,B,C,D,E,H,I,K,M,N,O,S,t;
      def de = withcolor .85white enddef;

      beginfig(1);
      k = 2.5u;
      p1 = unitsquare slanted .3 xscaled 2u yscaled u;
      B = point 0 of p1;
      C = point 1 of p1;
      D = point 2 of p1;
      A = point 3 of p1;
      S = A shifted (0,k*length (A--B));
      M = .5[A,S];
      O = .5[A,C];
      N = whatever[B,M]=whatever[S,B shifted (0,k*length (A--B))];
      I = (N--C) intersectionpoint (S--D);
      H = whatever[S,O] = A + whatever*((S-O) rotated 90);
      E = whatever[S,D] = A + whatever*((S-D) rotated 90);
      K = E shifted (B-A);
      t = (B--N) intersectionpoint (S--D);

      draw O--A--S--O--B--A--D--O--C de;
      draw A--E de;
      draw M--I de;
      draw A--H de;
      draw B--t de;
      draw S--C--B--K--E--S--N--C--D;
      draw B--S--D;
      draw t--N--K;

      dotlabel.llft("B",B);
      dotlabel.urt("D",D);
      dotlabel.lrt("C",C);
      dotlabel.ulft("A",A);
      dotlabel.ulft("S",S);
      dotlabel.top("N",N);
      dotlabel.bot("O",O);
      dotlabel.rt("I",I);
      dotlabel.ulft("H",H);
      dotlabel.urt("E",E);
      dotlabel.lrt("K",K);
      dotlabel.lrt("M",M);

      endfig;
      end;

      end{mplibcode}
      end{document}





      share|improve this answer




























        9














        Here's a metapost version, unfortunately without altering the out of the box luamplib package, I can't seem to draw dashed lines so I've replaced them with lighter lines.



        enter image description here



        documentclass{article}
        usepackage{luamplib}
        begin{document}
        begin{mplibcode}
        u:=2cm;
        path p[];
        pair A,B,C,D,E,H,I,K,M,N,O,S,t;
        def de = withcolor .85white enddef;

        beginfig(1);
        k = 2.5u;
        p1 = unitsquare slanted .3 xscaled 2u yscaled u;
        B = point 0 of p1;
        C = point 1 of p1;
        D = point 2 of p1;
        A = point 3 of p1;
        S = A shifted (0,k*length (A--B));
        M = .5[A,S];
        O = .5[A,C];
        N = whatever[B,M]=whatever[S,B shifted (0,k*length (A--B))];
        I = (N--C) intersectionpoint (S--D);
        H = whatever[S,O] = A + whatever*((S-O) rotated 90);
        E = whatever[S,D] = A + whatever*((S-D) rotated 90);
        K = E shifted (B-A);
        t = (B--N) intersectionpoint (S--D);

        draw O--A--S--O--B--A--D--O--C de;
        draw A--E de;
        draw M--I de;
        draw A--H de;
        draw B--t de;
        draw S--C--B--K--E--S--N--C--D;
        draw B--S--D;
        draw t--N--K;

        dotlabel.llft("B",B);
        dotlabel.urt("D",D);
        dotlabel.lrt("C",C);
        dotlabel.ulft("A",A);
        dotlabel.ulft("S",S);
        dotlabel.top("N",N);
        dotlabel.bot("O",O);
        dotlabel.rt("I",I);
        dotlabel.ulft("H",H);
        dotlabel.urt("E",E);
        dotlabel.lrt("K",K);
        dotlabel.lrt("M",M);

        endfig;
        end;

        end{mplibcode}
        end{document}





        share|improve this answer


























          9












          9








          9







          Here's a metapost version, unfortunately without altering the out of the box luamplib package, I can't seem to draw dashed lines so I've replaced them with lighter lines.



          enter image description here



          documentclass{article}
          usepackage{luamplib}
          begin{document}
          begin{mplibcode}
          u:=2cm;
          path p[];
          pair A,B,C,D,E,H,I,K,M,N,O,S,t;
          def de = withcolor .85white enddef;

          beginfig(1);
          k = 2.5u;
          p1 = unitsquare slanted .3 xscaled 2u yscaled u;
          B = point 0 of p1;
          C = point 1 of p1;
          D = point 2 of p1;
          A = point 3 of p1;
          S = A shifted (0,k*length (A--B));
          M = .5[A,S];
          O = .5[A,C];
          N = whatever[B,M]=whatever[S,B shifted (0,k*length (A--B))];
          I = (N--C) intersectionpoint (S--D);
          H = whatever[S,O] = A + whatever*((S-O) rotated 90);
          E = whatever[S,D] = A + whatever*((S-D) rotated 90);
          K = E shifted (B-A);
          t = (B--N) intersectionpoint (S--D);

          draw O--A--S--O--B--A--D--O--C de;
          draw A--E de;
          draw M--I de;
          draw A--H de;
          draw B--t de;
          draw S--C--B--K--E--S--N--C--D;
          draw B--S--D;
          draw t--N--K;

          dotlabel.llft("B",B);
          dotlabel.urt("D",D);
          dotlabel.lrt("C",C);
          dotlabel.ulft("A",A);
          dotlabel.ulft("S",S);
          dotlabel.top("N",N);
          dotlabel.bot("O",O);
          dotlabel.rt("I",I);
          dotlabel.ulft("H",H);
          dotlabel.urt("E",E);
          dotlabel.lrt("K",K);
          dotlabel.lrt("M",M);

          endfig;
          end;

          end{mplibcode}
          end{document}





          share|improve this answer













          Here's a metapost version, unfortunately without altering the out of the box luamplib package, I can't seem to draw dashed lines so I've replaced them with lighter lines.



          enter image description here



          documentclass{article}
          usepackage{luamplib}
          begin{document}
          begin{mplibcode}
          u:=2cm;
          path p[];
          pair A,B,C,D,E,H,I,K,M,N,O,S,t;
          def de = withcolor .85white enddef;

          beginfig(1);
          k = 2.5u;
          p1 = unitsquare slanted .3 xscaled 2u yscaled u;
          B = point 0 of p1;
          C = point 1 of p1;
          D = point 2 of p1;
          A = point 3 of p1;
          S = A shifted (0,k*length (A--B));
          M = .5[A,S];
          O = .5[A,C];
          N = whatever[B,M]=whatever[S,B shifted (0,k*length (A--B))];
          I = (N--C) intersectionpoint (S--D);
          H = whatever[S,O] = A + whatever*((S-O) rotated 90);
          E = whatever[S,D] = A + whatever*((S-D) rotated 90);
          K = E shifted (B-A);
          t = (B--N) intersectionpoint (S--D);

          draw O--A--S--O--B--A--D--O--C de;
          draw A--E de;
          draw M--I de;
          draw A--H de;
          draw B--t de;
          draw S--C--B--K--E--S--N--C--D;
          draw B--S--D;
          draw t--N--K;

          dotlabel.llft("B",B);
          dotlabel.urt("D",D);
          dotlabel.lrt("C",C);
          dotlabel.ulft("A",A);
          dotlabel.ulft("S",S);
          dotlabel.top("N",N);
          dotlabel.bot("O",O);
          dotlabel.rt("I",I);
          dotlabel.ulft("H",H);
          dotlabel.urt("E",E);
          dotlabel.lrt("K",K);
          dotlabel.lrt("M",M);

          endfig;
          end;

          end{mplibcode}
          end{document}






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 14 '13 at 9:01









          Scott H.Scott H.

          8,18222463




          8,18222463























              9














              enter image description here



              Asymptote version pyramid.asy:



              import three;  // 3D module
              import math;
              currentprojection=orthographic(camera=(55,144,80),
              up=(0,0,1),target=(0,0,0),zoom=1,center=true);

              size(300);
              size3(300,300,300);

              triple intersectionpoint(triple a,triple b,triple c,triple d){
              real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/
              ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x);
              return a*(1.0-u)+b*u;
              }

              triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy:
              // as South, East and North or (0,-1), (1,0) and (0,1)
              // and I is a sqrt(-1)=(0,1)
              real a=40; // side of the square;
              real h=50; // height, h=AS
              real d=sqrt(2)/2*a; // half of the diagonal

              O=(0,0,0);
              C=(0,d,0);
              B=(d,0,0);
              A=(0,-d,0);
              D=(-d,0,0);
              SS=A+(0,0,h);
              M=0.5(SS+A);
              II=0.5(SS+D);
              NN=intersectionpoint(SS,(SS+A-B),C,II);

              real phi=atan(h/a);
              real u=a*cos(phi)/sqrt(a^2+h^2);
              EE=D*(1-u)+SS*u;
              K=EE+B-A;

              real psi=atan(h/d);
              real u=d*cos(psi)/sqrt(d^2+h^2);
              H=O*(1-u)+SS*u;

              pair Q=intersectionpoint(project(NN--B),project(SS--D));

              pen dashed=linetype(new real[] {5,5}); // set up dashed pattern
              pen visLine=darkblue+0.8pt;
              pen hidLine=lightblue+dashed+0.8pt;

              void Dot(...triple[] v){
              dotfactor=8;
              for(int i=0;i<v.length;++i){
              dot(project(v[i]),UnFill);
              }
              }

              void Draw3(guide3 g, pen p=currentpen){
              draw(project(g),p);
              }

              void labelP(string s,triple t,pair p=(0,0)){
              label("$"+s+"$",project(t),p);
              }

              Draw3(B--A--D,hidLine);
              Draw3(H--A--SS,hidLine);
              Draw3(M--II,hidLine);
              Draw3(EE--A--C,hidLine);
              Draw3(B--D,hidLine);
              Draw3(SS--O,hidLine);

              Draw3(SS--B--C--SS--D--C,visLine);
              Draw3(SS--NN--C,visLine);
              Draw3(EE--K--B,visLine);
              Draw3(NN--K,visLine);

              draw(project(NN)--Q,visLine);
              draw(project(B)--Q,hidLine);

              Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS);

              labelP("A",A,NW);
              labelP("B",B,SW);
              labelP("C",C,E);
              labelP("D",D,NE);
              labelP("E",EE,NE);
              labelP("H",H,NE);
              labelP("I",II,NE);
              labelP("K",K,SE);
              labelP("M",M,W);
              labelP("N",NN,W);
              labelP("O",O,S);
              labelP("S",SS,NW);


              To get a standalone pyramid.pdf run asy -f pdf pyramid.asy.






              share|improve this answer
























              • +1 and please kindly make asymptote version for this question.

                – kiss my armpit
                Apr 14 '13 at 14:04













              • @Bugbusters: done.

                – g.kov
                Apr 14 '13 at 16:24











              • Best looking solution to this question.

                – Ingo
                Mar 3 '14 at 16:52
















              9














              enter image description here



              Asymptote version pyramid.asy:



              import three;  // 3D module
              import math;
              currentprojection=orthographic(camera=(55,144,80),
              up=(0,0,1),target=(0,0,0),zoom=1,center=true);

              size(300);
              size3(300,300,300);

              triple intersectionpoint(triple a,triple b,triple c,triple d){
              real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/
              ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x);
              return a*(1.0-u)+b*u;
              }

              triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy:
              // as South, East and North or (0,-1), (1,0) and (0,1)
              // and I is a sqrt(-1)=(0,1)
              real a=40; // side of the square;
              real h=50; // height, h=AS
              real d=sqrt(2)/2*a; // half of the diagonal

              O=(0,0,0);
              C=(0,d,0);
              B=(d,0,0);
              A=(0,-d,0);
              D=(-d,0,0);
              SS=A+(0,0,h);
              M=0.5(SS+A);
              II=0.5(SS+D);
              NN=intersectionpoint(SS,(SS+A-B),C,II);

              real phi=atan(h/a);
              real u=a*cos(phi)/sqrt(a^2+h^2);
              EE=D*(1-u)+SS*u;
              K=EE+B-A;

              real psi=atan(h/d);
              real u=d*cos(psi)/sqrt(d^2+h^2);
              H=O*(1-u)+SS*u;

              pair Q=intersectionpoint(project(NN--B),project(SS--D));

              pen dashed=linetype(new real[] {5,5}); // set up dashed pattern
              pen visLine=darkblue+0.8pt;
              pen hidLine=lightblue+dashed+0.8pt;

              void Dot(...triple[] v){
              dotfactor=8;
              for(int i=0;i<v.length;++i){
              dot(project(v[i]),UnFill);
              }
              }

              void Draw3(guide3 g, pen p=currentpen){
              draw(project(g),p);
              }

              void labelP(string s,triple t,pair p=(0,0)){
              label("$"+s+"$",project(t),p);
              }

              Draw3(B--A--D,hidLine);
              Draw3(H--A--SS,hidLine);
              Draw3(M--II,hidLine);
              Draw3(EE--A--C,hidLine);
              Draw3(B--D,hidLine);
              Draw3(SS--O,hidLine);

              Draw3(SS--B--C--SS--D--C,visLine);
              Draw3(SS--NN--C,visLine);
              Draw3(EE--K--B,visLine);
              Draw3(NN--K,visLine);

              draw(project(NN)--Q,visLine);
              draw(project(B)--Q,hidLine);

              Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS);

              labelP("A",A,NW);
              labelP("B",B,SW);
              labelP("C",C,E);
              labelP("D",D,NE);
              labelP("E",EE,NE);
              labelP("H",H,NE);
              labelP("I",II,NE);
              labelP("K",K,SE);
              labelP("M",M,W);
              labelP("N",NN,W);
              labelP("O",O,S);
              labelP("S",SS,NW);


              To get a standalone pyramid.pdf run asy -f pdf pyramid.asy.






              share|improve this answer
























              • +1 and please kindly make asymptote version for this question.

                – kiss my armpit
                Apr 14 '13 at 14:04













              • @Bugbusters: done.

                – g.kov
                Apr 14 '13 at 16:24











              • Best looking solution to this question.

                – Ingo
                Mar 3 '14 at 16:52














              9












              9








              9







              enter image description here



              Asymptote version pyramid.asy:



              import three;  // 3D module
              import math;
              currentprojection=orthographic(camera=(55,144,80),
              up=(0,0,1),target=(0,0,0),zoom=1,center=true);

              size(300);
              size3(300,300,300);

              triple intersectionpoint(triple a,triple b,triple c,triple d){
              real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/
              ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x);
              return a*(1.0-u)+b*u;
              }

              triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy:
              // as South, East and North or (0,-1), (1,0) and (0,1)
              // and I is a sqrt(-1)=(0,1)
              real a=40; // side of the square;
              real h=50; // height, h=AS
              real d=sqrt(2)/2*a; // half of the diagonal

              O=(0,0,0);
              C=(0,d,0);
              B=(d,0,0);
              A=(0,-d,0);
              D=(-d,0,0);
              SS=A+(0,0,h);
              M=0.5(SS+A);
              II=0.5(SS+D);
              NN=intersectionpoint(SS,(SS+A-B),C,II);

              real phi=atan(h/a);
              real u=a*cos(phi)/sqrt(a^2+h^2);
              EE=D*(1-u)+SS*u;
              K=EE+B-A;

              real psi=atan(h/d);
              real u=d*cos(psi)/sqrt(d^2+h^2);
              H=O*(1-u)+SS*u;

              pair Q=intersectionpoint(project(NN--B),project(SS--D));

              pen dashed=linetype(new real[] {5,5}); // set up dashed pattern
              pen visLine=darkblue+0.8pt;
              pen hidLine=lightblue+dashed+0.8pt;

              void Dot(...triple[] v){
              dotfactor=8;
              for(int i=0;i<v.length;++i){
              dot(project(v[i]),UnFill);
              }
              }

              void Draw3(guide3 g, pen p=currentpen){
              draw(project(g),p);
              }

              void labelP(string s,triple t,pair p=(0,0)){
              label("$"+s+"$",project(t),p);
              }

              Draw3(B--A--D,hidLine);
              Draw3(H--A--SS,hidLine);
              Draw3(M--II,hidLine);
              Draw3(EE--A--C,hidLine);
              Draw3(B--D,hidLine);
              Draw3(SS--O,hidLine);

              Draw3(SS--B--C--SS--D--C,visLine);
              Draw3(SS--NN--C,visLine);
              Draw3(EE--K--B,visLine);
              Draw3(NN--K,visLine);

              draw(project(NN)--Q,visLine);
              draw(project(B)--Q,hidLine);

              Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS);

              labelP("A",A,NW);
              labelP("B",B,SW);
              labelP("C",C,E);
              labelP("D",D,NE);
              labelP("E",EE,NE);
              labelP("H",H,NE);
              labelP("I",II,NE);
              labelP("K",K,SE);
              labelP("M",M,W);
              labelP("N",NN,W);
              labelP("O",O,S);
              labelP("S",SS,NW);


              To get a standalone pyramid.pdf run asy -f pdf pyramid.asy.






              share|improve this answer













              enter image description here



              Asymptote version pyramid.asy:



              import three;  // 3D module
              import math;
              currentprojection=orthographic(camera=(55,144,80),
              up=(0,0,1),target=(0,0,0),zoom=1,center=true);

              size(300);
              size3(300,300,300);

              triple intersectionpoint(triple a,triple b,triple c,triple d){
              real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/
              ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x);
              return a*(1.0-u)+b*u;
              }

              triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy:
              // as South, East and North or (0,-1), (1,0) and (0,1)
              // and I is a sqrt(-1)=(0,1)
              real a=40; // side of the square;
              real h=50; // height, h=AS
              real d=sqrt(2)/2*a; // half of the diagonal

              O=(0,0,0);
              C=(0,d,0);
              B=(d,0,0);
              A=(0,-d,0);
              D=(-d,0,0);
              SS=A+(0,0,h);
              M=0.5(SS+A);
              II=0.5(SS+D);
              NN=intersectionpoint(SS,(SS+A-B),C,II);

              real phi=atan(h/a);
              real u=a*cos(phi)/sqrt(a^2+h^2);
              EE=D*(1-u)+SS*u;
              K=EE+B-A;

              real psi=atan(h/d);
              real u=d*cos(psi)/sqrt(d^2+h^2);
              H=O*(1-u)+SS*u;

              pair Q=intersectionpoint(project(NN--B),project(SS--D));

              pen dashed=linetype(new real[] {5,5}); // set up dashed pattern
              pen visLine=darkblue+0.8pt;
              pen hidLine=lightblue+dashed+0.8pt;

              void Dot(...triple[] v){
              dotfactor=8;
              for(int i=0;i<v.length;++i){
              dot(project(v[i]),UnFill);
              }
              }

              void Draw3(guide3 g, pen p=currentpen){
              draw(project(g),p);
              }

              void labelP(string s,triple t,pair p=(0,0)){
              label("$"+s+"$",project(t),p);
              }

              Draw3(B--A--D,hidLine);
              Draw3(H--A--SS,hidLine);
              Draw3(M--II,hidLine);
              Draw3(EE--A--C,hidLine);
              Draw3(B--D,hidLine);
              Draw3(SS--O,hidLine);

              Draw3(SS--B--C--SS--D--C,visLine);
              Draw3(SS--NN--C,visLine);
              Draw3(EE--K--B,visLine);
              Draw3(NN--K,visLine);

              draw(project(NN)--Q,visLine);
              draw(project(B)--Q,hidLine);

              Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS);

              labelP("A",A,NW);
              labelP("B",B,SW);
              labelP("C",C,E);
              labelP("D",D,NE);
              labelP("E",EE,NE);
              labelP("H",H,NE);
              labelP("I",II,NE);
              labelP("K",K,SE);
              labelP("M",M,W);
              labelP("N",NN,W);
              labelP("O",O,S);
              labelP("S",SS,NW);


              To get a standalone pyramid.pdf run asy -f pdf pyramid.asy.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Apr 14 '13 at 12:43









              g.kovg.kov

              17.3k13976




              17.3k13976













              • +1 and please kindly make asymptote version for this question.

                – kiss my armpit
                Apr 14 '13 at 14:04













              • @Bugbusters: done.

                – g.kov
                Apr 14 '13 at 16:24











              • Best looking solution to this question.

                – Ingo
                Mar 3 '14 at 16:52



















              • +1 and please kindly make asymptote version for this question.

                – kiss my armpit
                Apr 14 '13 at 14:04













              • @Bugbusters: done.

                – g.kov
                Apr 14 '13 at 16:24











              • Best looking solution to this question.

                – Ingo
                Mar 3 '14 at 16:52

















              +1 and please kindly make asymptote version for this question.

              – kiss my armpit
              Apr 14 '13 at 14:04







              +1 and please kindly make asymptote version for this question.

              – kiss my armpit
              Apr 14 '13 at 14:04















              @Bugbusters: done.

              – g.kov
              Apr 14 '13 at 16:24





              @Bugbusters: done.

              – g.kov
              Apr 14 '13 at 16:24













              Best looking solution to this question.

              – Ingo
              Mar 3 '14 at 16:52





              Best looking solution to this question.

              – Ingo
              Mar 3 '14 at 16:52











              0














              I copied Alain Matthes' code with some prepaired.



              documentclass[border=2cm]{standalone}
              usepackage{tikz}
              usepackage{tikz-3dplot}
              usetikzlibrary{intersections,calc}
              usepackage{fouriernc}

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

              begin{document}

              tdplotsetmaincoords{60}{100}
              %tdplotsetmaincoords{80}{100}
              begin{tikzpicture}[tdplot_main_coords,scale=1.5]
              pgfmathsetmacroa{3}
              pgfmathsetmacrob{3}
              pgfmathsetmacro{h}{{a*sqrt(2)}}

              % definitions
              path (0,0,0) coordinate (A)
              (a,0,0) coordinate (B)
              (a,b,0) coordinate (C)
              (0,b,0) coordinate (D)
              (0,0,h) coordinate (S)
              ($(S)!0.5!(A)$) coordinate (M)
              ($(C)!0.5!(A)$) coordinate (O)
              ($(S)!0.5!(D)$) coordinate (I)
              ({2*h^2*a/(a^2+b^2+4*h^2)}, {2*b*h^2/(a^2+b^2+4*h^2)}, {h*(a^2+b^2)/(a^2+b^2+4*h^2)}) coordinate (H)
              (-a, 0, h) coordinate (N)
              (0, {b*h^2/(b^2+h^2)},{ b^2*h/(b^2+h^2)}) coordinate (E)
              (a, {b*h^2/(b^2+h^2)}, {b^2*h/(b^2+h^2)}) coordinate (K) ;
              % drawing
              begin{scope}
              clip (S) -- (B) -- (D) --cycle;
              draw[hidden] (B) -- (N);
              end{scope}
              begin{scope}
              clip (S) -- (I) -- (N) --cycle;
              draw (B) -- (N);
              end{scope}

              draw[hidden]
              (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
              (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
              draw[thick] (B) -- (K) -- (E) (C) -- (N) -- (K)
              (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
              draw[add = 1 and 1] (S) to (N);

              % place black circles and labels
              foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
              I/right,K/right,M/right,N/left,O/right,S/left}
              {
              fill (point) circle (1.2pt);
              node[position=3pt] at (point) {$point$};
              }

              end{tikzpicture}
              end{document}


              enter image description here






              share|improve this answer




























                0














                I copied Alain Matthes' code with some prepaired.



                documentclass[border=2cm]{standalone}
                usepackage{tikz}
                usepackage{tikz-3dplot}
                usetikzlibrary{intersections,calc}
                usepackage{fouriernc}

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

                begin{document}

                tdplotsetmaincoords{60}{100}
                %tdplotsetmaincoords{80}{100}
                begin{tikzpicture}[tdplot_main_coords,scale=1.5]
                pgfmathsetmacroa{3}
                pgfmathsetmacrob{3}
                pgfmathsetmacro{h}{{a*sqrt(2)}}

                % definitions
                path (0,0,0) coordinate (A)
                (a,0,0) coordinate (B)
                (a,b,0) coordinate (C)
                (0,b,0) coordinate (D)
                (0,0,h) coordinate (S)
                ($(S)!0.5!(A)$) coordinate (M)
                ($(C)!0.5!(A)$) coordinate (O)
                ($(S)!0.5!(D)$) coordinate (I)
                ({2*h^2*a/(a^2+b^2+4*h^2)}, {2*b*h^2/(a^2+b^2+4*h^2)}, {h*(a^2+b^2)/(a^2+b^2+4*h^2)}) coordinate (H)
                (-a, 0, h) coordinate (N)
                (0, {b*h^2/(b^2+h^2)},{ b^2*h/(b^2+h^2)}) coordinate (E)
                (a, {b*h^2/(b^2+h^2)}, {b^2*h/(b^2+h^2)}) coordinate (K) ;
                % drawing
                begin{scope}
                clip (S) -- (B) -- (D) --cycle;
                draw[hidden] (B) -- (N);
                end{scope}
                begin{scope}
                clip (S) -- (I) -- (N) --cycle;
                draw (B) -- (N);
                end{scope}

                draw[hidden]
                (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
                (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
                draw[thick] (B) -- (K) -- (E) (C) -- (N) -- (K)
                (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
                draw[add = 1 and 1] (S) to (N);

                % place black circles and labels
                foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
                I/right,K/right,M/right,N/left,O/right,S/left}
                {
                fill (point) circle (1.2pt);
                node[position=3pt] at (point) {$point$};
                }

                end{tikzpicture}
                end{document}


                enter image description here






                share|improve this answer


























                  0












                  0








                  0







                  I copied Alain Matthes' code with some prepaired.



                  documentclass[border=2cm]{standalone}
                  usepackage{tikz}
                  usepackage{tikz-3dplot}
                  usetikzlibrary{intersections,calc}
                  usepackage{fouriernc}

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

                  begin{document}

                  tdplotsetmaincoords{60}{100}
                  %tdplotsetmaincoords{80}{100}
                  begin{tikzpicture}[tdplot_main_coords,scale=1.5]
                  pgfmathsetmacroa{3}
                  pgfmathsetmacrob{3}
                  pgfmathsetmacro{h}{{a*sqrt(2)}}

                  % definitions
                  path (0,0,0) coordinate (A)
                  (a,0,0) coordinate (B)
                  (a,b,0) coordinate (C)
                  (0,b,0) coordinate (D)
                  (0,0,h) coordinate (S)
                  ($(S)!0.5!(A)$) coordinate (M)
                  ($(C)!0.5!(A)$) coordinate (O)
                  ($(S)!0.5!(D)$) coordinate (I)
                  ({2*h^2*a/(a^2+b^2+4*h^2)}, {2*b*h^2/(a^2+b^2+4*h^2)}, {h*(a^2+b^2)/(a^2+b^2+4*h^2)}) coordinate (H)
                  (-a, 0, h) coordinate (N)
                  (0, {b*h^2/(b^2+h^2)},{ b^2*h/(b^2+h^2)}) coordinate (E)
                  (a, {b*h^2/(b^2+h^2)}, {b^2*h/(b^2+h^2)}) coordinate (K) ;
                  % drawing
                  begin{scope}
                  clip (S) -- (B) -- (D) --cycle;
                  draw[hidden] (B) -- (N);
                  end{scope}
                  begin{scope}
                  clip (S) -- (I) -- (N) --cycle;
                  draw (B) -- (N);
                  end{scope}

                  draw[hidden]
                  (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
                  (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
                  draw[thick] (B) -- (K) -- (E) (C) -- (N) -- (K)
                  (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
                  draw[add = 1 and 1] (S) to (N);

                  % place black circles and labels
                  foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
                  I/right,K/right,M/right,N/left,O/right,S/left}
                  {
                  fill (point) circle (1.2pt);
                  node[position=3pt] at (point) {$point$};
                  }

                  end{tikzpicture}
                  end{document}


                  enter image description here






                  share|improve this answer













                  I copied Alain Matthes' code with some prepaired.



                  documentclass[border=2cm]{standalone}
                  usepackage{tikz}
                  usepackage{tikz-3dplot}
                  usetikzlibrary{intersections,calc}
                  usepackage{fouriernc}

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

                  begin{document}

                  tdplotsetmaincoords{60}{100}
                  %tdplotsetmaincoords{80}{100}
                  begin{tikzpicture}[tdplot_main_coords,scale=1.5]
                  pgfmathsetmacroa{3}
                  pgfmathsetmacrob{3}
                  pgfmathsetmacro{h}{{a*sqrt(2)}}

                  % definitions
                  path (0,0,0) coordinate (A)
                  (a,0,0) coordinate (B)
                  (a,b,0) coordinate (C)
                  (0,b,0) coordinate (D)
                  (0,0,h) coordinate (S)
                  ($(S)!0.5!(A)$) coordinate (M)
                  ($(C)!0.5!(A)$) coordinate (O)
                  ($(S)!0.5!(D)$) coordinate (I)
                  ({2*h^2*a/(a^2+b^2+4*h^2)}, {2*b*h^2/(a^2+b^2+4*h^2)}, {h*(a^2+b^2)/(a^2+b^2+4*h^2)}) coordinate (H)
                  (-a, 0, h) coordinate (N)
                  (0, {b*h^2/(b^2+h^2)},{ b^2*h/(b^2+h^2)}) coordinate (E)
                  (a, {b*h^2/(b^2+h^2)}, {b^2*h/(b^2+h^2)}) coordinate (K) ;
                  % drawing
                  begin{scope}
                  clip (S) -- (B) -- (D) --cycle;
                  draw[hidden] (B) -- (N);
                  end{scope}
                  begin{scope}
                  clip (S) -- (I) -- (N) --cycle;
                  draw (B) -- (N);
                  end{scope}

                  draw[hidden]
                  (A) -- (C) (A) -- (B) (A) -- (D) (A) -- (H) (A) -- (S)
                  (A) -- (E) (B) -- (D) (O) -- (S) (I) -- (M);
                  draw[thick] (B) -- (K) -- (E) (C) -- (N) -- (K)
                  (B) --(C) -- (D) (S) -- (D) (S) -- (B) (S) -- (C);
                  draw[add = 1 and 1] (S) to (N);

                  % place black circles and labels
                  foreach point/position in {A/left,B/left,C/right,D/right,E/right,H/above,
                  I/right,K/right,M/right,N/left,O/right,S/left}
                  {
                  fill (point) circle (1.2pt);
                  node[position=3pt] at (point) {$point$};
                  }

                  end{tikzpicture}
                  end{document}


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 10 hours ago









                  minhthien_2016minhthien_2016

                  1,258917




                  1,258917






























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