toandhsp

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9 years, 111 days

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These are questions asked by toandhsp

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);

Suppose (1 + 2x)^n = a0 + a1*x + a2*x^2+...+an*x^n.

I want to find value of n so that max(a0, a1, ..., an) is a8

I tried directly. 

With n = 12

restart:
A:=expand((1+2*x)^12,x);
max(coeffs(A));

And with n = 11

B:=expand((1+2*x)^11,x);
max(coeffs(B));


Therefore,  n = 12 or  n = 11. 

How can I solve the problem with Maple?

I want to solve the equation sqrt(x) + sqrt(1 - x^2) = sqrt(2 - 3*x - 4*x^2) in RealDomain. I tried

RealDomain:-solve(sqrt(x) + sqrt(1 - x^2) = sqrt(2 - 3*x - 4*x^2),x);

And I got one solution. But, at here 

At here http://mathematica.stackexchange.com/questions/51316/how-can-i-get-the-exact-real-solution-of-this-equation 

they said the given equation has two real solutions. How must I understand?

I have system of equation contain of two equations

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:
eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2: 

I want to find the number k so that the equation

a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1 + k*(a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2) = 0

can be factor, where k satisfy

a:=a1+k*a2:
b:=b1+k*b2:
c:=c1+k*c2:
d:=d1+k*d2:
e:=e1+k*e2:
f:=f1+k*f2:

I tried

A:=a*x^2+b*y^2+c*x*y+d*x+e*y+f:
collect(A,x);
B:=collect(discrim(A, x), y);
C:= discrim(B,y);

and got the expression

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0.

For example 

> restart:

a1:=14:

b1:=-21:

c1:=0:

d1:=-6:

e1:=45:

f1:=-14:

a2:=35:

b2:=28:

c2:=0:

d2:=41:

e2:=-122:

f2:=56:

a:=a1+k*a2:

b:=b1+k*b2:

c:=c1+k*c2:

d:=d1+k*d2:

e:=e1+k*e2:

f:=f1+k*f2:

P:=c*d*e+4*a*b*f-a*e^2-b*d^2

-f*c^2:

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:

eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2:

with(RealDomain):

Q:=solve(P=0,k);

factor(eq1+Q*(eq2));

solve([

eq1=0,eq2],[x,y]);

Where, Q is k which I want to find. 

My question is, if I have the system of equations

eq1:=a1*x^3+b1*y^3+c1*x^2*y+ d1*x*y^2 + e1*x+f1*y+g1:
eq2:=a2*x^3+b2*y^3+c2*x^2*y+ d2*x*y^2 + e2*x+f2*y+g2

How can I get a similar to the 

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0?

 

 

 

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