toandhsp

2 years, 295 days


These are questions asked by toandhsp

How about this equation

restart:
solve(x^2 - 2*(m+1)*x+m^2 - 2*m + m^2=0,{x},parametric=full);
allvalues(%);?

I want to find the greatest value of this expression 

f:=(x,y,z)->sqrt((x+1)*(y^2+2)*(z^3+3))+sqrt((y+1)*(z^2+2)*(x^3+3))+sqrt((z+1)*(x^2+2)*(y^3+3));

with x>0, y>0 , z>0,x+y+z=3.

I tried

restart:

 f:=(x,y,z)->sqrt((x+1)*(y^2+2)*(z^3+3))+sqrt((y+1)*(z^2+2)*(x^3+3))+sqrt((z+1)*(x^2+2)*(y^3+3));

DirectSearch[GlobalOptima](f(x,y,z), {x>0, y>0 , z>0,x+y+z=3},maximize);

I got the output

[HFloat(infinity), [x = .591166078050740e52, y = .183647204560715e52, z = .786638021216969e52], 1249]

 

 

I want to find a point has coordinates are integer numbers and write the equation of tangent line to a given circle,  knowing that, the points of tangent has also integer coordinates. For example, the circle has centre M(-1,-5) and radius R=5. I tried

restart:

with(geometry):

point(M,-1,-5):

R:=5:

eqS:=Equation(circle(S,(a-HorizontalCoord(M))^2 + (b-VerticalCoord(M))^2 -R^2=0,[a,b],'centername'=T)):

L:=[]:

for a from -50 to 50  do

for b from -50 to 50  do

if  a <>HorizontalCoord(M) and b<>VerticalCoord(M) and eqS then

L:=[op(L), [a,b]] fi;

od: od:

nops(L);

eqS:=Equation(circle(S,(x-HorizontalCoord(M))^2 + (y-VerticalCoord(M))^2 -R^2=0,[x,y],'centername'=T));

k:=[seq](sort(Equation(TangentLine(P, S, point(A, pt[])), [x,y])), pt in L):

seq([L[i],k[i]],i=1..nops(L));

Next,

> with(combinat):

d:=choose(k,2):

for i from 1 to nops(d) do  

seq([d[i],solve([op(1,d[i]),op(2,d[i])],[x,y])],i=1..nops(d));

end do;

If I want to the point of intersection of two lines which are not perpendicular line, for example

[[-3*x-4*y-48 = 0, 4*x+3*y-6 = 0], [[x = 24, y = -30]]]

How can I select?

 

 

In the book "Challenges in Geometry" of the author Christopher J. Bradley at p. 32, the triangle with three sides a := 136, b := 170, c := 174 has three medians ma := 158, mb := 131, mc := 127. I checked

restart:

a:=2*68;

b:=2*85;

c:=2*87;

ma:=sqrt((b^2+c^2)/2-a^2/4);

mb:=sqrt((a^2+c^2)/2-b^2/4);

mc:=sqrt((b^2+a^2)/2-c^2/4);

Now I want to find coordinates of vertices of a triangle like that (in plane). I tried

restart;
DirectSearch:-SolveEquations([(x2-x1)^2+(y2-y1)^2 = 136^2,
(x3-x2)^2+(y3-y2)^2 = 170^2, (x3-x1)^2+(y3-y1)^2 = 174^2], {abs(x1) <= 30, abs(x2) <= 30, abs(y1) <= 30, abs(y2) <= 30, abs(x3) <= 30, abs(y3) <= 30}, assume = integer, AllSolutions, solutions = 5);

but my computer runs too long. I think, there is not a triangle with integer coordiantes. 

How can I get  a triangle  with coordinates of vertices are rational numbers?

 

Let be given the complex number z  satisfying condition abs(z+3-2I)=3. I want to find the set of points representing the complex number w, knowing that w - z = 1 +3I. I tried

Restart:
assume(a::real, b::real,x::real, y::real);
z:=x+y*I;
w:=a+b*I;
abs(w-1-3*I+3-2*I)=3;

 

Edit

Restart:
assume(a::real, b::real,x::real, y::real);
Set:=abs(z+3-2*I):
w:=x+y*I;
sol:=solve(w - z =1+3*I,{z});
z1:=subs(sol,Set);
A:=abs(z1);
map(x->x^2,A=3);

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