Problem:
Let x,y, and z be real numbers satisfying the system of equations
xy+4z=60yz+4x=60zx+4y=60
Let S be the set of possible values of x. Find the sum of the squares of the elements of S.
Solution:
We begin by subtracting the second given equation from the first, noting that both equal 60:
xy+4z−(yz+4x)=0⇒xy+4z−yz−4x=0⇒4(z−x)−y(z−x)=0⇒(z−x)(4−y)=0.
This yields two cases:
Case 1: y=4
Substitute into the first and third equations:
x+z=15,xz=44.
We solve the system:
x+z=15andxz=44.
This leads to the quadratic:
t2−15t+44=0⇒t=4, 11.
So the solutions are (x,z)=(4,11) and (11,4).
Case 2: x=z
Substitute into the first and third equations:
xy+4x=60,x2+4y=60.
Subtracting the first from the second:
x2+4y−(xy+4x)=0⇒x2−xy+4y−4x=0⇒x(x−4)−y(x−4)=0⇒(x−4)(x−y)=0.
This leads to two subcases:
Subcase 2.1: x=4
Substitute into xy+4x=60:
4y+16=60⇒y=11.
This gives the solution (x,y,z)=(4,11,4), which was already counted in Case 1.
Subcase 2.2: x=y
Now x=y=z, and substituting into xy+4x=60 gives:
x2+4x=60⇒x2+4x−60=0⇒x=6, −10.
This yields the additional solutions (x,y,z)=(6,6,6) and (−10,−10,−10).
Now summing the squares of all distinct x-values: x=4, 11, 6, −10:
42+112+62+(−10)2=16+121+36+100=273.
The problems on this page are the property of the MAA's American Mathematics Competitions