Do I remember my own name?
- I tried to solve the system above, but I could not find any solution!
- What value did you use for the parameter a?
- I don’t remember.
- When I solved it I got one, two, three, and four solutions.
- For different values of a?
- For different values of a.
- Do you remember the values?
- Do I remember my own name?
Problem source: Bamo problems.
June 2nd, 2010 at 12:51 am
Substituting the first equation into the second,
(x-a)² + x² = 1 ___ [1]
so the point (x, x-a) lies on the unit circle. To put this another way, using Y for the second Cartesian coordinate seeing as the statement of the problem already uses y:
the point (x, Y) lies on the unit circle, where
Y = x-a.
This last equation is that of a straight line at an angle of 45 degrees. Thus we can choose such a line to have a particular geometrical relation to the unit circle and then find out what value of a produces this. The most interesting is when the straight line is a tangent. There are two possibilities. In one, the line touches the circle at (1/sqrt(2), -1/sqrt(2)), in which case a=sqrt(2) and x=1/sqrt(2). In the other, the line touches the circle at (-1/sqrt(2), 1/sqrt(2)), in which case a=-sqrt(2) and x=-1/sqrt(2). In each case, y may equal either x or -x, giving 2 solutions.
In general, a value of x allows two values for y: y=x and y=-x. The exception is when x=0. This implies that y=0, giving 1 solution in y for this value of x. Thus the only way that the number of solutions in x and y can be odd is if x may be 0. Two points on the unit circle have x=0: (0, 1), in which case a=-1, and (0, -1), in which case a=1. a=1 allows
x=0, Y=-1
x=1, Y=0
thus producing 3 solutions for x and y:
x=0, y=0
x=1, y=1
x=1, y=-1
a=-1 is similar. a=1 and a=-1 are thus the only values of a that give an odd number of solutions in x and y, and each gives 3. Thus no value of a gives only 1 solution in x and y.
A more typical solution is one where the line cuts the circle (other than the special case x=0 given above). An example: we have
3² + 4² = 5², so
(3/5)² + (4/5)² = 1,
so, in particular, the points (3/5, -4/5) and (4/5, -3/5) lie on the unit circle. They also lie on Y = x-a where a = 7/5. Thus a=7/5 produces 4 solutions for x and y:
x=3/5, y=3/5
x=3/5, y=-3/5
x=4/5, y=4/5
x=4/5, y=-4/5
The other general case is where the line doesn’t cut the circle, e.g. a=2. Then no real x satisfies equation [1], so there are no real solutions for x and y.
Thus the number of real solutions for x and y can be 0, 2, 3 or 4.