toandhsp

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These are answers submitted by toandhsp

Thank you very much.

Put the equation of the line passing the point A(4,1) has the form a*(x-4)+b*(y-1)=0. We always assume a^2 + b^2 = 1. And then, you solve the system of equations

solve([-4*a-b=2,a^2+b^2=1],[a,b]);

But I don't know how to get the exactly solutions [a = -8/17-1/17*13^(1/2),b = -2/17+4/17*13^(1/2)] and [[a = -8/17+1/17*13^(1/2), b = -2/17-4/17*13^(1/2)]]

I must solve in two cases 

solve([f(0,0)=2,a^2+b^2=1,b<0],[a,b]);

solve([f(0,0)=2,a^2+b^2=1,b>0],[a,b]);

Note that, when the line passing the point A(2,1), the line x = 2 has not slope, but it is also a tangent of the circle. You try

f:=(x,y)->a*(x-2)+b*(y-1):

solve([f(0,0)=2,a^2+b^2=1],[a,b]);

 

How about this code? Maple can not run. Please help me. 

> restart:

a:=2:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(30*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(120*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

point(A,a,0,0):

sys:=solve([x^2+y^2=(rhs(op(1,sol[1])))^2,(x-a)^2+y^2=(rhs(op(2,sol[1])))^2],[x,y]);

point(B,rhs(op(1,op(1,sys))),rhs(op(2,op(1,sys))),0):

triangle(T,[o,A,B]):

map(coordinates,DefinedAs(T));

simplify(FindAngle(o,T));

simplify(FindAngle(A,T));

simplify(FindAngle(B,T));

line(d,[t+1,-t,-t+2],t):

rotation(W,T,-2*Pi/3,d):

map(coordinates,DefinedAs(W));

 

 

> restart:

a:=3:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(15*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(135*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

point(A,a,0,0):

sys:=solve([x^2+y^2=(rhs(op(1,sol[1])))^2,(x-a)^2+y^2=(rhs(op(2,sol[1])))^2],[x,y]);

point(B,rhs(op(1,op(1,sys))),rhs(op(2,op(1,sys))),0):

triangle(T,[o,A,B]):

map(coordinates,DefinedAs(T));

simplify(FindAngle(o,T));

simplify(FindAngle(A,T));

simplify(FindAngle(B,T));

line(d,[t+1,-t,-t+2],t):

rotation(W,T,-2*Pi/3,d):

map(coordinates,DefinedAs(W));

 

Thank you very much.

When I  found the length of sides b and c, I created a triangle satisfies the conditions stated and test the measure of the  angles of this triangle. I tried

> restart:

a:=3:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(15*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(135*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

point(A,a,0,0):

sys:=solve([x^2+y^2=(rhs(op(1,sol[1])))^2,(x-a)^2+y^2=(rhs(op(2,sol[1])))^2],[x,y]);

point(B,rhs(op(1,op(1,sys))),rhs(op(2,op(1,sys))),0):

triangle(T,[o,A,B]):

map(coordinates,DefinedAs(T));

FindAngle(o,T);

FindAngle(A,T);

FindAngle(B,T);

But I don't receive the angles 15 degrees, 30 degrees and 115 degrees directly. How do I tell Maple to do that?

 

Thank you very much.

With Mathematica, If I want to find all positive integer solutions, I input 

Reduce[(1 + 1/x)*(1 + 1/y)*(1 + 1/z) == 2 && x > 0 && y > 0 && z > 0 &&
x >= y && y >= z, {x, y, z}, Integers]
and I get
(x == 5 && y == 4 && z == 3) || (x == 7 && y == 6 && 
z == 2) || (x == 8 && y == 3 && z == 3) || (x == 9 && y == 5 &&
z == 2) || (x == 15 && y == 4 && z == 2).
It's really simple.

 

How about the integrate?
restart;
> int(sin(x)^5,x);combine(%);
 

 

 

You visit here http://www.geogebra.org/cms/ It is free.

Perhalps P4 = 4! = 1*2*3*4 = 24. Therefore 10*P4 = 240. Thus 10P4=m! has no solution.

Are your two equations  x^2 + y^2 = 4^2 and y = (x-3/2)^2 +1?

Problem 2

b) Determine the coordinates (x, y) of the two points that satisfy both equations. 

> eq1:= x^2 + y^2 = 4^2:

eq2:=y = (x-3/2)^2 +1:

sol:=evalf(solve([eq1,eq2],[x,y]));

Please check this line df:=x->abs(1,fo(x))((3*x^2)-16*x+5)-0.5. I think

fo(x) must be ,f0(x).

> restart:

f:=x->sqrt(x+1)+sqrt(1-x) + 2+2*sqrt(x+1)*sqrt(1-x):

sol:=solve(t=sqrt(x+1)+sqrt(1-x),x);

A:=f(sol[1]):

m:=minimize(sqrt(x+1)+sqrt(1-x),x=-1..1):

M:=maximize(sqrt(x+1)+sqrt(1-x),x=-1..1):

simplify(A) assuming t >=m and t <=M;

 

f:=x->sqrt(x+1)+sqrt(x+2)+2*sqrt((x+1)*(x+2))+2*x-3:

sol:=solve(t=sqrt(x+1)+sqrt(x+2),x):

A:=f(sol):

simplify(A) assuming t >= minimize(sqrt(x+1)+sqrt(x+2),x>=-1);

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