restart; with(plots):unprotect(gamma):
_EnvHorizontalName := 'x';_EnvVerticalName := 'y';
line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:
R := 3:
ang := [0, (1/3)*Pi, 3*Pi*(1/4)+.2, 7*Pi*(1/6)+.4, 8*Pi*(1/5), 13*Pi*(1/7)]:
for i to 6 do P || i := [R*cos(ang[i]), R*sin(ang[i])] end do:
pts := [seq(P || i, i = 1 .. 6)]:
for i to 6 do tang || i := x*P || i[1]+y*P || i[2] = R^2 end do:
sol:=solve({tang1,tang3},{x,y}):Q13:=[subs(sol,x),subs(sol,y)]:
sol := solve({tang1, tang2}, {x, y}): Q1 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang2, tang3}, {x, y}): Q2 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang3, tang4}, {x, y}): Q3 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang4, tang5}, {x, y}): Q4 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang5, tang6}, {x, y}): Q5 := [subs(sol, x), subs(sol, y)]:
sol := solve({tang1, tang6}, {x, y}): Q6 := [subs(sol, x), subs(sol, y)]:
ptQ := [seq(Q || i, i = 1 .. 6)]:
line14 := line(Q1[1], Q1[2], Q4[1], Q4[2]): L14 := implicitplot(line14, x = -10 .. 10, y = -10 .. 10, color = red):
line25 := line(Q2[1], Q2[2], Q5[1], Q5[2]): L25 := implicitplot(line25, x = -10 .. 10, y = -10 .. 10, color = red):
line36 := line(Q3[1], Q3[2], Q6[1], Q6[2]): L36 := implicitplot(line36, x = -10 .. 10, y = -10 .. 10, color = red):
sol := solve({line14, line25}, {x, y}): I1 := [subs(sol, x), subs(sol, y)]:

lineP23 := line(P2[1], P2[2], P3[1], P3[2]): lineP56 := line(P5[1], P5[2], P6[1], P6[2]):
sol := solve({lineP23, lineP56}, {x, y}): gamma := [subs(sol, x), subs(sol, y)]:
lineP12 := line(P1[1], P1[2], P2[1], P2[2]): lineP45 := line(P4[1], P4[2], P5[1], P5[2]):
sol := solve({lineP12, lineP45}, {x, y}): beta := [subs(sol, x), subs(sol, y)]:
lineP34 := line(P3[1], P3[2], P4[1], P4[2]): lineP16 := line(P1[1], P1[2], P6[1], P6[2]):
sol := solve({lineP16, lineP34}, {x, y}): alpha := [subs(sol, x), subs(sol, y)]:
pl:= line(alpha[1], alpha[2], gamma[1], gamma[2]):
hexa := seq(implicitplot(tang||i, x = -20 .. 20, y = -20 .. 20, linestyle=3,color = blue),i=1..6):
#hexa:=plot([seq([P||i,P||(i mod 6)+1],i=1..6),color=green): 
#hexa:=plot([seq([P||i,P||(i+1)],i=1..6)],thickness=4,color=green): 
hex:=plot([[Q1,Q2],[Q2,Q3],[Q3,Q4],[Q4,Q5],[Q5,Q6],[Q6,Q1]],thickness=4,color=green):
tp := textplot({seq([op(pts[i]), cat("P", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
tpq := textplot({seq([op(ptQ[i]), cat("Q", i)], i = 1 .. 6)}, 'align' = {'above', 'left'}):
TP:=textplot([[I1[],"I"],[alpha[],"alpha"],[beta[],"beta"],[gamma[],"gamma"]],'align' = {'above', 'left'}):
slopes:=[seq(((dx,dy)->dy/dx)((pts[i]-pts[(i mod 6)+1])[]),i=1..6)]:
lines:=zip((pt,slope)->y=slope*(x-pt[1])+pt[2],pts,slopes):
plotpts:=plot(pts,style=point,colour=red,symbol=solidcircle,symbolsize=5):
plotptQ:=plot(ptQ,style=point,colour=black,symbol=solidcircle,symbolsize=5):
plotlines:=plot(rhs~(lines),style=line,linestyle=3,colour=magenta):
cir:=implicitplot(x^2+y^2=R^2,x=-R..R,y=-R..R,color=black):
PL:=implicitplot(pl,x=-1..15,y=-15..6,color=blue,thickness=3):
display(plotpts,plotptQ,plotlines,hex,cir,L14,L25,L36,PL,tp,tpq,TP,axis = [gridlines = [4, color = blue]],
view=[-6..10,-15..6],scaling=constrained,axes=none,size=[800,800]):
How to simplify this program ? Thank you.

How to show that the function z0(t)=t²+exp(-t)  is solution of z"(t)+2z'(t)+z(t)=2exp(-t); Thank you.

If the equation with homogeous coordinates in the base (A,B,C) is pYZ+qZX+rXY=0, an affine  equation in the base (A,AB,AC) is
pxy+qy(1-x-y)+r(1-x-y)=0, it is also written rx²+(q+r-p)xy+qy²=0. The discrim of  trinom is rx²+(q+r-p)xy+qy² is of the sign of 
delta=(q+r-p)²-4qr; We get a hyperbole, a parabole, an ellipse according to delta is >0, =0 or <0 respectvely. How implement
this property. Thank you very much.

An exercise of 1948 commits me to form the equation of conics passing by 3 points. Let P=0, Q=0, R=0 be the equations of the sides of the triangle ABC; if we associate these sides 2 to 2 we obtain 3 conics passing through points A, B, C having for equations QR=0, RP=0, PQ=0. As a result, the general equation of conics around the triangle ABC is: aQR+bRP+cPQ=0. P, Q, R being equations of the form mx+ny+pz=0 (so-called homogeneous coordinates). Then change to refined coordinates with x+y+z=1 (formula found on the internet and surely misinterpreted). Is it necessary to change of base ? Thank you for your help.

At the moment t=0, we place a body at 100 ° C in a room at 25°C; we designate by q(t) the temperature at the moment t. The differential equation is.q'(t)+k*(q(t)-25)=t, with k cooling coefficient equal to 2.
Determine the solution that checks the initial condition.
What is the body temperature after 30 minutes.
After how long the temperature drops to 50°C. Thank you for the help.