7 years, 4 days

## center of the circle (C,D,K,H)...

Sorry, I am finding the coordinates O of the circle (C,D,K,H)

## the coordinates of the O to M centre...

How to obtain the coordinates of the O centre of the circle (A,B,C,D) from the M matrix ? Thank you.

## with the numerical values x := 0.6459808...

with the numerical values x := 0.645980840726419; y := 0.207986256183860; z := 0.146032903089721;
the solution given is :
{0.146032903089721 = 0.1460329031, a = -1.354182738 c,

b = -1.149311492 c, c = c}, {0.146032903089721 = 0.1460329031,

a = -1.354182738 c, b = 1.149311492 c, c = c},

{0.146032903089721 = 0.1460329031, a = 1.354182738 c,

b = -1.149311492 c, c = c}, {0.146032903089721 = 0.1460329031,

a = 1.354182738 c, b = 1.149311492 c, c = c}

What are the values of a, b, and c ?

## Resolutio of a system....

How to avoid a poster asking me : enter name of horizoltal axis then veticial axis ? Thank you

## How to find the barycentric coordinates ...

How to find the barycentric coordinates of the orthocenter of the ABC triangle (in the ABC repere) using cartesian coordinates ? Thank you..

## Droite de La Hire...

cf mathcurve la Hire
droite de la Hire
La Hire Line
I would like to build this figure.

## Problemes with arcs and colors...

restart;
Proc := proc(t) local a1, t4, l3, R, r, dr, eq, sol; _EnvHorizontalName := 'x'; _EnvVerticalName := 'y'; t4 := thickness = 4; l3 := linestyle = dot;
R := 9; r := 1/2*R;
geometry:-point(OO, 0, 0);
geometry:-circle(Cir, [OO, R]);
geometry:-point(K, R*cos(t), R*sin(t));
geometry:-point(Omega, r*cos(t), r*sin(t));
geometry:-circle(cir, [Omega, r]);
eq := geometry:-Equation(cir);
geometry:-line(XXp, y = 0);
geometry:-line(YYp, x = 0);
geometry:-line(L1, y = x);
geometry:-line(L2, y = -x);
geometry:-projection(M1, K, XXp);
geometry:-coordinates(M1);
geometry:-point(K2, geometry:-coordinates(M1)[1] - 2*R, 0);
geometry:-coordinates(K2); geometry:-segment(sT, K2, M1);
geometry:-point(N1, 0, R*sin(t)); subs(y = x, eq);
sol := solve(%, x); geometry:-point(N2, sol[2], sol[2]);
subs(y = -x, eq);
sol := solve(%, x);
geometry:-point(N3, sol[2], -sol[2]);
plots:-display([plottools[arc]([op(geometry:-coordinates(Omega))], r, t .. t + 1/2*Pi, color = red, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t + Pi .. t + 3/2*Pi, color = coral, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t - 1/2*Pi .. t, color = cyan, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t + 1/2*Pi .. t + Pi, color = green, t4)],
geometry:-draw([Cir(color = blue, t4), cir(color = grey, t4), sT(color = black, t4), XXp(color = black, l3), YYp(color = black, l3), L1(color = black, l3), L2(color = black, l3), N1(color = blue, symbol = solidcircle, symbolsize = 15), N2(color = blue, symbol = solidcircle, symbolsize = 15), N3(color = blue, symbol = solidcircle, symbolsize = 15), M1(color = blue, symbol = solidcircle, symbolsize = 15)]), axes = none, view = [-30 .. 10, -10 .. 10], size = [800, 800]); end proc;
plots:-animate(Proc, [t], t = 0 .. 2*Pi, frames = 30);

## Study of a geometry figure...

1.-Why 2 triangles CMP and ABH are similar ?
2.-At what curve E is the line MP tangent ?
Can you help me ?

## The portion of a movable tangent to an e...

It is a corollary of Poncelet’s theorem : The line that joins a focal spot F to a point P common in M and M' to an ellipse is bisector inside the angle MFM"

## The portion of a movable tangent to an e...

Do you want to tell me what is the interest of the calculation of alpha1 compared to alpha ? alpha can be drawn;

## The portion of a movable tangent to an e...

Problem with the 3 last lines of the code..
(sin(Setof[2]))^2
Error, invalid subscript selector
simpascn:=arcsin(sqrt(factor(  ))); # the t's eliminate
Error, Invalid label reference
alpha1:=(subs(Setof[2]=simpascn,alpha))*180/Pi
Error, invalid subscript selector
plot(alpha1,t=-10..10,view=[-10..10,0..180])
Warning, expecting only range variable t in expression alpha1 to be plotted but found name alpha1

## The portion of a movable tangent to an e...

Thank you a lot for your code.
Howeven ir does not work with Maple 2021.

## The portion of a movable tangent to an e...

I show that alpha is constant using this procedure :
TANG := proc(t)
local alpha, xM, yM, M, P, Q, TT, PF2, QF2;
xM := a*cos(t);
yM := b*sin(t);
point(M, xM, yM);
line(TT, 1/49*x*xM + 1/25*y*yM - 1 = 0);
intersection(P, tang1, TT);
intersection(Q, tang2, TT);
line(PF2, [P, F2]); line(QF2, [Q, F2]);
alpha := FindAngle(PF2, QF2);
print(evalf(180 - 180*arctan(alpha)/Pi, 5));
end proc;
seq(TANG(i*Pi/10), i = 0 .. 5);

## The portion of a movable tangent to an e...

More cleaily how to show that alpha remains constant when t varies ? Thank you.
restart;
with(plots);
with(geometry);
_EnvHorizontalName := 'x';
_EnvVerticalName := 'y';
x0 := 100;
y0 := 40;
a := 7;
b := 5;
c := sqrt(a^2 - b^2);
ellipse(el, x^2/a^2 + y^2/b^2 - 1);
point(F1, -c, 0);
point(F2, c, 0);
eq := simplify((a^2 - x0^2)*(y - y0)^2 + (b^2 - y0^2)*(x - x0)^2 + 2*x0*y0*(x - x0)*(y - y0)) = 0;
sol := solve({eq}, {y});
line(tang1, op(sol[1]));
line(tang2, op(sol[2]));
sol2 := op(solve({op(sol[1]), x^2/a^2 + y^2/b^2 - 1 = 0}, {x, y}));
xM2 := rhs(sol2[1]);
yM2 := rhs(sol2[2]);
point(A, xM2, yM2);
sol3 := op(solve({op(sol[2]), x^2/a^2 + y^2/b^2 - 1 = 0}, {x, y}));
xM3 := rhs(sol3[1]);
yM3 := rhs(sol3[2]);
point(B, xM3, yM3);
line(Pol, [A, B]);
simplify(Equation(Pol));
isolate(%, y);
TANG := proc(t) local xM, yM; xM := a*cos(t); yM := b*sin(t); RETURN(expand(1/49*x*xM + 1/25*y*yM - 1 = 0)); end proc;
t := -0.15;
line(TT, TANG(t));
Equation(TT);
point(M, a*cos(t), b*sin(t));
intersection(P, tang1, TT);
intersection(Q, tang2, TT);
line(PF2, [P, F2]);
line(QF2, [Q, F2]);
alpha := FindAngle(PF2, QF2);
print(evalf(180 - 180*arctan(alpha)/Pi, 4));
display(textplot([[-c, 0, "F1"], [c, 0, "F2"], [coordinates(B)[], "B"], [coordinates(A)[], "A "], [coordinates(M)[], "M "], [coordinates(P)[], "P "], [coordinates(Q)[], "Q "]], align = {"above", 'right'}), draw([el(color = red), A(color = black, symbol = solidcircle, symbolsize = 16), PF2(color = brown), QF2(color = brown), B(color = black, symbol = solidcircle, symbolsize = 16), M(color = black, symbol = solidcircle, symbolsize = 16), P(color = black, symbol = solidcircle, symbolsize = 16), tang1(color = green), tang2(color = green), TT(color = green), F1(color = blue, symbol = solidcircle, symbolsize = 16), Q(color = blue, symbol = solidcircle, symbolsize = 16), F2(color = red, symbol = solidcircle, symbolsize = 16)]), axes = none, view = [-7 .. 18, -7 .. 12])

## Fibonnachi...

the second proposal is appropriate. Thank you.

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