MHB Points on a Circle: Proving $m_1m_2m_3m_4=1$

[kaliprasad]

If $(m_r,\frac{1}{m_r})$ for r from 1 to 4 are 4 points that lie on a circle show that $m_1m_2m_3m_4= 1$

 

[Theia]

Spoiler

Perhaps something like this:

Let the circle be $$(x - a)^2 + (y - b)^2 = r^2$$.

The points $$\left( x_r , \frac{1}{x_r} \right)$$ are the points where (suitably selected) circle meets the function $$y = \frac{1}{x}$$. Substituting this into the equation of the circle gives a quartic equation:

$$x^4 - 2ax^3 + (a^2 + b^2 - r^2)x^2 - 2bx + 1 =0.$$

If one then further assume that the circle was suitably selected, there exists four real roots for that equation, namely $$x_1, x_2, x_3$$ and $$x_4$$, which are the numbers we are interested in. As they are roots of the quartic equation, one can write equivalently

$$(x - x_1)(x - x_2)(x - x_3)(x - x_4) = x_1x_2x_3x_4 + \ldots = 0$$.

Comparing these two ways to write the equation in question, one can conclude that $$x_1x_2x_3x_4 = 1$$.

 

[Albert1]

If $(m_r,\frac{1}{m_r})$ for r from 1 to 4 are 4 points that lie on a circle show that $m_1m_2m_3m_4= 1$

my solution:

Spoiler

(1) construct two lines $L_1:y=x\,\, and \,\,L_2: y=-x$
(2) construct $xy=1$
(3) construct a circle $x^2+y^2=k^2 \,\, (k>0)$
here $k$ big enough ,make sure (2) and (3) will have four intersetions A,B,C,D
(4)coordinates :$A(m_1,\dfrac{1}{m_1}), D(m_4,\dfrac{1}{m_4}),B(m_2,\dfrac{1}{m_2}), C(m_3,\dfrac{1}{m_3})$
(infact ABCD is a rectangle)
$\overline {AD}//L_1,\overline{BC}//L_1$
$\overline {AB}//L_2,\overline{CD}//L_2$
now we must have $m_1+m_3=0,m_2+m_4=0$
and $m_1\times m_4=-1,m_2\times m_3=-1$
$\therefore m_1m_2m_3m_4=1$

 

[kaliprasad]

my solution:

Spoiler

(1) construct two lines $L_1:y=x\,\, and \,\,L_2: y=-x$
(2) construct $xy=1$
(3) construct a circle $x^2+y^2=k^2 \,\, (k>0)$
here $k$ big enough ,make sure (2) and (3) will have four intersetions A,B,C,D
(4)coordinates :$A(m_1,\dfrac{1}{m_1}), D(m_4,\dfrac{1}{m_4}),B(m_2,\dfrac{1}{m_2}), C(m_3,\dfrac{1}{m_3})$
(infact ABCD is a rectangle)
$\overline {AD}//L_1,\overline{BC}//L_1$
$\overline {AB}//L_2,\overline{CD}//L_2$
now we must have $m_1+m_3=0,m_2+m_4=0$
and $m_1\times m_4=-1,m_2\times m_3=-1$
$\therefore m_1m_2m_3m_4=1$

Spoiler

I never said that circle has centre at origin

 

[Albert1]

Spoiler

I never said that circle has centre at origin

Spoiler

I know you will question this ,it does not matter where the centre locate
if the origin translate from (0,0) to (a,b),all ther points will also translate relatively ,and the result will not change

 

[kaliprasad]

Spoiler

I know you will question this ,it does not matter where the centre locate
if the origin translate from (0,0) to (a,b),all ther points will also translate relatively ,and the result will not change

Spoiler

it matters because shifting (0,0) to (a,b) changes $(m_r,\frac{1}{m_r})$ to $(a + m_r,b + \frac{1}{m_r})$ no longer reciprocals

 

[Albert1]

Spoiler

it matters because shifting (0,0) to (a,b) changes $(m_r,\frac{1}{m_r})$ to $(a + m_r,b + \frac{1}{m_r})$ no longer reciprocals

Spoiler

View the diagram:

View attachment 5564

 

[MarkFL]

Spoiler

View the diagram:

Albert, please attach your images here in the challenges forum inline so they can be hidden within the spoiler, and not show up as an attachment, outside of the spoiler defeating the purpose of using the spoiler tags. This can easily be done by using the "Insert Image" tool on our toolbar. :)

 

[kaliprasad]

My solution

Spoiler

we have general equation of a circle
$x^2+y^2+ 2gx + 2fy + c = 0$
let a point $p,\frac{1}{p}$ be on the circle
then we get
$p^2 + \frac{1}{p^2} + 2gp + \frac{2f}{p} + c = 0$
or $p^4 + 1 + 2gp^3 + 2fp + cp^2=0$
or $p^4 + 2gp^3 + cp^2 + 2fp + 1=0$
the 4 roots are $m_1,m_2,m_3,m_4$ and hence product of roots = $m_1m_2m_3m_4=1$ (the constant term)