# Circle Problem

• May 6th 2010, 11:23 AM
sid_178
Circle Problem
If [m(i),1/m(i)],m(i)>0,i=1,2,3,4 are four distinct points on the circle ,then

m(1)*m(2)*m(3)*m(4)=???

(a)-1
(b)1
(c)0
(d)None
• May 6th 2010, 02:05 PM
TKHunny
This looks like a very innocent question, but it is not.

Let's look at just one such point.

Circle: $\displaystyle x^{2} + y^{2} = r^{2}$

First point has x-coordinate 'a': $\displaystyle a^{2} + y^{2} = r^{2}$

The y-coordinate must be '1/a': $\displaystyle a^{2} + (1/a)^{2} = r^{2}$

This is where we see that something is up. When we started this problem, we may have assumed we had lots of choices for 'a'. Alas, there are only four. I won't enumerate them all. Here's one:

$\displaystyle a_{1} = \sqrt{\frac{1}{2}\cdot \left(\sqrt{r^{4}-4} + r^{2}\right)}$

Since there are so few of these points, what may have appeared to be an amazing result turns out to be rather a hand-picked result of no particular value that comes immediately to mind. Oh well, it was a wonderful exploration. Plus, I'm relatively confident that there is a much simpler way than this brute-force algebraic solution to produce the desired result, 1.
• May 7th 2010, 12:20 AM
Opalg
Quote:

Originally Posted by sid_178
If [m(i),1/m(i)],m(i)>0,i=1,2,3,4 are four distinct points on the circle ,then

m(1)*m(2)*m(3)*m(4)=???

(a)-1
(b)1
(c)0
(d)None

You are looking for four points that lie on the hyperbola $\displaystyle xy=1$ and also on a circle. The general equation of a circle is $\displaystyle x^2+y^2+2gx+2fy+c=0$. Substitute $\displaystyle y=1/x$ into the circle equation, multiply through by $\displaystyle x^2$, and you find that $\displaystyle x^4+2gx^3+cx^2+2fx+1 = 0$. The four solutions of that equation are the values m(1), m(2), m(3), m(4) for the x-coordinates of the points where the circle meets the hyperbola. But the product of the roots of that 4th degree equation is just the constant term in the equation, namely 1.