This appeared in one of the RMOs:
Solve for real x and y:
5x\left(1+\dfrac{1}{x^2+y^2}\right)=12
5y\left(1-\dfrac{1}{x^2+y^2}\right)=4
341Put x = r \cos \theta; y = r \sin \theta
The equation becomes
5 \cos \theta \left(r + \frac{1}{r} \right) = 12; 5 \sin \theta \left(r - \frac{1}{r} \right) = 4
Hence eliminating r, we get \frac{144}{\cos ^2 \theta} - \frac{16}{\sin^2 \theta}= 100
If \cos^2 \theta = t, then the equation is \frac{36}{t} - \frac{4}{1-t}= 25 which can be solved for t easily.
Remembering that x>0, i.e. cos θ>0, we can obtain the value of r and hence x and y.
1hmm..nyc
first get x from eqn 1 in terms of x2 +y2
similarly y from eqn 2
now squared and added them ..put x2 + y2 = t
then got a biquadratic in t :)
somehow managed to make it in the form of t+1/t substituted it as z
finaly a quadratic in z :P
z= 6/5 , 26/5
didnt solve it further :)
1ya..also 6/5 is neglected as t+1/t >=2
so t = 5 or 1/5 i think ( z= 26/5)
rest cab be done
341this is RMO 1994. I looked this up in mathlinks, but someone has posted the question wrongly making it look embarassingly trivial for an olympiad problem
15x(1+(1/x2+y2)=12 ...(1)
5y(1-(1/x2+y2)=4 ...(2)
From (1)
1/x2+y2=12-5x/5x
From (2)
1/x2+y2=5y-4/5y
Therefore 5y-4/5y=12-5x/5x
2x+6y-5xy=0
x=6y/5y-2
putting value of x in (2) and solving we get
5y4-4y3-17y2+20y-4=0
(y-1)(y-2)(5y2-9y+2)=0
y=1
x=2
other values of y don't exist for x
66Here is my solution:
Multiply the second equation with i (the iota) and add to the first one to obtain
5(x+iy)+\dfrac{5(x-iy)}{x^2+y^2}=4(3+i)
Setting z=x+iy, we get \bar{z}=x-iy and z\bar{z}=|z|^2=x^2+y^2. Using these the above equation
5z+\dfrac{5\bar{z}}{z\bar{z}}=4(3+i)
That is
5z+\dfrac{5}{z}=4(3+i)\quad \Rightarrow\ 5z^2-4(3+i)z+5=0
Hence, we get after solving
z=\dfrac{4(3+i)\pm \sqrt{16(3+i)^2-100}}{10}
which after simplification becomes:
z=2+i,\ \dfrac{2}{5}-\dfrac{i}{5}
And, so (x,\,y)\to (2,\,1), \left(\dfrac{2}{5},\,-\dfrac{1}{5}\right)
These are the only real solutions.
