test1_03.mw

The solution I'm getting here for v(y), doesn't satisfy my original two equations. Like for example if I choose u(y)=y, for easy check it can be seen that my de2 equation doesn't satisfy that value.

Can anyone help me out?

I see this question at here

https://mathematica.stackexchange.com/questions/239874/how-to-reduce-timing-to-find-the-integer-numbers-a-b-c-d-e-f-g-h-k-m-of

How to solve this problem by Maple?

I want to remove the term XY

f:=(x,y)->2*x²+xy+y²+4x-y-2=0;

eq := simplify(subs(x = X-9/7, y = Y+8/7, f(x, y)));

theta:=Pi/8;

ex := simplify(subs(X = cos(theta)*X-sin(theta)*Y, Y = sin(theta)*X+cos(theta)*Y, eq)); evalf(%);

How to obtain (3+sqrt(2))/2*X²+(3-sqrt(2))/2=36 ? Thank you.

I am trying to solve this equation, I tried

restart;
ListTools[Categorize];
L := [];
for a to 20 do
for b to 20 do
for c to 20 do
for d to 20 do
for e to 20 do
for f to 20 do
for g to 20 do
for h to 20 do
if sqrt(a+b*sqrt(c+d*sqrt(e +f*sqrt(g)))) = h then L := [op(L), [a, b, c, d, e, f, g, h]] end if
end do end do end do end do end do end do end do end do; nops(L)

where **g** is not a square of a integer number . I do not get any solutions for a long time. How to reduce timing to solve this equation?

This question came to me when I answered in this thread https://www.mapleprimes.com/questions/231603-Solve-Onevariable-Equation

The equation there is quite cumbersome, I extracted a shorter subexpression from it (I converted all floats to exact constants), but the ** solve** command hangs when trying to solve it. **fsolve** handles the equation easily. We can see that the equation ** Eq** is quite simple and easy to solve even by hand. I ask this question in a separate topic, because this seems to be a serious bug in the **solve** command.

restart;
Eq:=1-1/(1+203808*exp(-342569/506*t)*(1/131537))^(131537/203808)=44983/56599;
solve(Eq);

Edit. I noticed that the **isolate** command solves the problem, but of course the question remains open with **solve** .

I know that, tetrahedron with vertices **(0, 0, 0), (3, 0, 0), (0, 5, 0), (0, 0, 14)** has center of insphere is **(1, 1, 1)** and radius **R = 1. **How to create a tetrahedron whose coordinates of all vertices are integers, the inscribed spherical center and the inscribed spherical radius are integers?

restart; _local(D);
A := [-L, 0]; B := [L, 0]; C := [x, y]; D := [-x, y];
Dist := proc (X, Y) options operator, arrow; sqrt((X[1]-Y[1])^2+(X[2]-Y[2])^2) end proc;
Eq := (Dist(C, D) = Dist(C, B))^2;
centre := [solve(diff(Eq, x), x), solve(diff(Eq, y), y)];
with(geometry); `assuming`([conic(p, Eq, [x, y])], [L > 0]);
detail(p); asymptotes(p);
y_acymp := `~`[solve](`~`[Equation](asymptotes(p)), y);
y := solve(Eq, y)[1]
P := proc (X, L0)
local Curve, Asymptote, Trapezoid, T, pt, Ip;
Curve := plot(eval([y, -y], L = L0), x = -(1/3)*L0 .. 15, color = red, thickness = 3);
Asymptote := plot(eval([-sqrt(3)*x-(1/3)*sqrt(3)*L, sqrt(3)*x+(1/3)*sqrt(3)*L], L = L0), x = -(1/3)*L0 .. 15, linestyle = 3, color = black, thickness = 0); Trapezoid := plottools:-polygon(eval([A, B, C, D], [L = L0, x = X]), color = "LightGreen");
Ip := (1/3)*(eval(C, [x = X-(1/3)*L0, L = L0])); #centre of gravity of ABC
T := plots:-textplot([[(eval(A, [x = X, L = L0]))[], "A"], [(eval(B, [x = X, L = L0]))[], "B"], [(eval(C, [x = X, L = L0]))[], "C"], [(eval(D, [x = X, L = L0]))[], "D"], [Ip[], "I"]], align = {above, left}, font = [TIMES, 16]);
pt := plot([A, B, C, Ip], style = plottools:-point, color = blue, symbolsize = 15);
plots:-display(Curve, Asymptote, Trapezoid, T, pt, scaling = constrained, size = [400, 800]) end proc;
a := 7;
P((1/2)*a, 6)
I would like of location of centre of gravity of ABC. Thank you.

This is a question from https://mathematica.stackexchange.com/questions/239055/how-to-reduce-the-time-to-solve-this-system-of-equations. How to reduce the time in *Maple*?

```
{x, y, z, a, b, m, n} /. Solve[{1/x + 1/y == 1/z, a/x + b/y == m/n, 2 <= x <= 30, 2 <= y <= 30, 1 <= z <= 30, 2 <= a <= 10, 2 <= b <= 10, a > b, 2 <= m <= 10, 2 <= n <= 10, x > y, GCD[m, n] == 1, GCD[a, b, m] == 1, m <= n}, {x, y, z, a, b, m, n}, Integers]
```

I am trying to find one option so that a sphere is tangent with every edge of a tetrahedron. Suppose tetrahedron OABC, here **O(0,0,0), A(x1, y1, z1), B(x2, y2, z2)**, and **C(x3, y3, z3)**. How to find a sphere is tangent with every edge of a tetrahedron?

I know how to contruct an isoscele ABCD trapeze knowing the 2L length of AB and BC=CD=CD=a. But I don't know answer to that question : L being fixed together with points A and B , show that the place of C when a varies is a branch of hyberbola. Here is my code.

restart; unprotect(O, D);

with(plots):

Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end:

dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end:

_EnvHorizontalName := x: _EnvVerticalName := y:

O:=[0,0]:A:=[-L,0]:alpha:=arccos((2*L-a)^2/(2*a*(2*L-a))):h:=tan(alpha)*(2*L-a)/2:

B:=[L,0]:C:=[L-(2*L-a)/2,h]:D:=[-L+(2*L-a)/2,h]:

L := 6; a := 7;

poly := [A, B, C, D, A];tp := textplot([[A[], "A"], [B[], "B"], [C[], "C"], [D[], "D"]], color = black, 'align' = {'above', 'right'});

trapeze := polygonplot(poly, axes = normal, color = "DarkGreen", transparency = .9);

display([tp, trapeze], scaling = constrained); Thank you foryour help.

We consider an ellipse E defined by the equation x²/a²+y²/b²=1. Either F1 and F2 its foci.

We consider a point M of E such the angle F1,M,F2 is equal to a given value phi.

Find F1M et F2M in function of a, b and phi . Then find the coordiinates x0 and y0 of M in function of a, b, and phi. Here is the begining of my code;

restart;

_EnvHorizontalName := x: _EnvVerticalName := y:

with(plots):

line := proc (x1, y1, x2, y2) options operator, arrow; (x-x1)*(y2-y1)-(y-y1)*(x2-x1) end proc:

Vdot := proc (U, V) add(U[i]*V[i], i = 1 .. 2) end proc:

dist := proc (M, N) sqrt(Vdot(expand(M-N), expand(M-N))) end proc:

ell := x^2/a^2+y^2/b^2 = 1:

c := sqrt(a^2-b^2):

F1 := [c, 0]: F2 := [-c, 0]:

M := [a*cos(t), b*sin(t)]:

MF2 := simplify(expand(sqrt((c+a*cos(t))^2+(b*sin(t))^2)));

MF1 := simplify(expand(sqrt((c-a*cos(t))^2+(b*sin(t))^2)));

varphi := arccos((MF1^2+MF2^2-4*c^2)/(2*MF1*MF2));

f := unapply(varphi, t);

`maxϕ` := f((1/2)*Pi);

diff(f(t), t);

a := 7: b := 5: t := (1/3)*Pi:

evalf(varphi);

evalf(MF1); evalf(dist(M, F1));

evalf(MF2); evalf(dist(M, F2));

MF1+MF2; evalf(%);

`maxϕ`; evalf(%);

ELL := implicitplot(ell, x = -a-2 .. a+2, y = -2-b .. b+2, color = blue):

tp := textplot([[F1[], "F1"],[F2[],"F2"],[M[],"M"]], 'align' = {'above', 'left'}):

po := plot([F1,F2, M], style = point, symbolsize = 15, symbol = solidcircle, color = red):

display([ELL, tp, po], scaling = constrained): Would you help me to answer the questions ? Thank you.

How to simplify this trigonometric expression with Maple? I only know the way with **identify** command, which is difficult to call mathematically correct:

restart;
Expr:=arctan((1-tan(20*Pi/180))/(1-tan(25*Pi/180)));
evalf[15](Expr);
identify(%);