Circle or vieta's?

Geometry Level 4

Let ( m k , 1 m k ) \left (m_k,\frac{1}{m_k}\right) be four distinct points on a circle where k = 1 , 2 , 3 , 4 k=1,2,3,4 , m k > 0 m_k>0 . Find k = 1 4 m k \displaystyle \prod_{k=1}^{4} m_k .


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let the equation be of the circle be x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2+y^2+2gx+2fy+c=0

Since ( m k , 1 m k ) (m_k,\frac{1}{m_k}) lies on it , it satisfies the equation.

Rearranging the equation we get, m k 4 + 2 g m k 3 + c m k 2 + 2 f m k + 1 = 0 m^4_k + 2gm^3_k+cm^2_k+2fm_k+1=0

Clearly it's roots are m 1 , m 2 , m 2 , m 4 m_1,m_2,m_2,m_4

So k = 1 4 m k = m 1 m 2 m 3 m 4 = 1 \prod_{k=1}^{4} m_k=m_1m_2m_3m_4=1 [By vieta's]

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...