toandhsp

3 years, 24 days


These are questions asked by toandhsp

I tried

nops({x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z });

and Maple out 9.

If x = 1, y = 2, z = 3, we have

nops({1, 2, 3, 2, 4, 0, -1, 5, 2});

equal to 7.

Edit.

I edited my question. The word "list" into "set".

My question means, there is a triple (x, y, z) = (1, 2, 3) so that the number of element of the set {x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 7, not 9. Why we can confirm the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9? And, how must select the integer numbers x, y, z (0 < x <10, 0< y <10, 0< z < 10) so that the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9?

I tried

restart:

ListTools[Categorize]:

L:=[]:

for x from -2 to 5  do

for y from -2 to 5  do

for z from -2 to 5  do

a:=[x,y,z]:

b:=[1+x,2+y,-3+z]:

c:=[-2+x,3+y,-1+z]:

if op(1,a)*op(2,a)*op(3,a)*op(1,b)*op(2,b)*op(3,b)*op(1,c)*op(2,c)*op(3,c)<> 0

and op(1,a)<>op(2,a) and  op(1,a)<>op(3,a) and  op(1,a)<>op(2,b) and  op(1,a)<>op(3,b) and  op(1,a)<>op(2,c) and  op(1,a)<>op(3,c) and op(2,a)<>op(3,a) and op(2,a)<>op(1,b) and

op(2,a)<>op(3,b) and

op(2,a)<>op(1,c) and

op(2,a)<>op(3,c) and

op(3,a)<>op(1,b) and

op(3,a)<>op(2,b) and 

op(3,a)<>op(1,c) and

op(3,a)<>op(2,c) and

op(1,b)<>op(2,b) and

op(1,b)<>op(3,b) and

op(1,b)<>op(2,c) and

op(1,b)<>op(3,c) and

op(2,b)<>op(3,b) and

op(2,b)<>op(1,c) and

op(2,b)<>op(3,c) and

op(3,b)<>op(1,c) and

op(3,b)<>op(2,c) and

op(1,c)<>op(2,c) and

op(1,c)<>op(3,c) and op(2,c)<>op(3,c) then L:=[op(L), {a,b,c}] fi; od: od: od:

nops(L); 

L;

 

 

 

 

I have some triangles ABC with vertices

1) A(-13,-5,5), B(-5,11,-11), C(-3,-9,15) has centre of out circle is (3, 3, 3), orthocentre (-27, -9, 3) and centroid (-7, -1, -3). 

2) A(-6,6,-1), B(-5,-1,-3), C(2,10,7) has centre of out circle is (1, 2, 3), orthocentre (-11, 11, -3) and centroid (-3, 5, 1). 

How can I write a program to find a triangle with integer coordinates of vertices, centroid, orthocenter and center of the triangle in geometry 3D? 

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?

 

 

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