Prove that a point can be found which is at the same distance from each of the four points$\ldots$

Question

Prove that a point can be found which is at the same distance from each of the four points $\bigg(am_1,\dfrac{a}{m_1}\bigg),\bigg(am_2,\dfrac{a}{m_2}\bigg),\bigg(am_3,\dfrac{a}{m_3}\bigg)$ and $\bigg(am_1m_2m_3,\dfrac{a}{m_1m_2m_3}\bigg)$

My attempt:

I could figure out two concepts

i) Let the point be $(x,y)$. Then, we have $(x-am_1)^2+(y-\dfrac{a}{m_1})^2=(x-am_2)^2+(y-\dfrac{a}{m_2})^2=(x-am_3)^2+(y-\dfrac{a}{m_3})^2=(x-am_1m_2m_3)^2+(y-\dfrac{a}{m_1m_2m_3})^2$
ii) If the distance be $r$, then we have 4 circles having these points as centres which will intersect at one specific point.

But, that seems a whole lot of calculation that I am not able to do very simply. I wonder if these are the correct ways to approach the problem. Please help. Thank you.

Answer 1

You may use the concept the other way: the unknown point as the center of a circle passing through the four given points.

Take one of the points as reference and write the line equations from the reference point to other three points. These lines are chords inside the circle assuming the starting point as reference point and the end point as one of the three points.

Geometrically, we know that the center of the circle (unknown point) will lie on lines that pass through the midpoints of these chords and and are perpendicular to each chord. You may write the equations of these new lines since you know the point and slope of each.

If the lines do intersect at a point that will be your unknow point.

Answer 2

I think we can use Vieta formulas to solve it ,if I do not wrong.

Let equation of a circle is $$x^2+y^2+Dx+Ey+F=0$$ and this four point can writer form $(at,\dfrac{a}{t})$ on this circle ,so we have $$(at)^2+(\dfrac{a}{t})^2+D\cdot(at)+E\cdot (\dfrac{a}{t})+F=0$$ $$\Longrightarrow a^2t^4+Dat^3+Ft^2+Eat+a^2=0$$ use Vieta formulas,we have $$m_{1}m_{2}m_{3}\cdot m_{1}m_{2}m_{3}=\dfrac{a^2}{a^2}=1$$ so $$m_{1}m_{2}m_{3}=\pm1 $$ and $$\begin{cases} m_{1}+m_{2}+m_{3}+m_{1}m_{2}m_{3}=-\dfrac{D}{a}\\ m_{1}m_{2}m_{3}+m_{1}m_{2}m_{1}m_{2}m_{3}+m_{1}m_{3}m_{1}m_{2}m_{3}+m_{2}m_{3}m_{1}m_{2}m_{3}=-\dfrac{E}{a} \end{cases}$$ then $$\begin{cases} -\dfrac{D}{2}=\dfrac{a}{2}(m_{1}+m_{2}+m_{3}\pm 1)\\ -\dfrac{E}{2}=\pm\dfrac{a}{2}(m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+1) \end{cases}$$ so I think the center is $$(-\dfrac{D}{2},-\dfrac{E}{2})=\left(\dfrac{a}{2}(m_{1}+m_{2}+m_{3}\pm 1),\pm\dfrac{a}{2}(m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+1)\right)$$

Answer 3

Seems @himbrom is right: the last point should be $\bigg(\dfrac{a}{m_1m_2m_3}, am_1m_2m_3\bigg)$.

Indeed consider the intersection of hyperbola $xy=a^2$ with a circle $(x-u)^2+(y-v)^2=r^2$. Plugging in to $y=\frac{a^2}{x}$ and multiplying by $x^2$ we get a monic 4th degree polynomial (we call it $P$) with free term $a^4$.

Now suppose we have 3 distinct points $\bigg(am_1,\dfrac{a}{m_1}\bigg),\bigg(am_2,\dfrac{a}{m_2}\bigg),\bigg(am_3,\dfrac{a}{m_3}\bigg)$ on that hyperbola. Then they are non-colllinear (otherwise a hyperbola and a line would intersect at at least 3 points, which is not possible). Any 3 non-collinear points determine a unique circle, giving us $u, v$, and $R$, and corresponding polynomial $P$ as above. Then $P$ has 3 real roots $am_1, am_2, am_3$ and thus must have 4th real root $\frac{a}{m_1m_2m_3}$. Then the point $\bigg(\dfrac{a}{m_1m_2m_3}, am_1m_2m_3\bigg)$ lies on the circle $(x-u)^2+(y-v)^2=r^2$ as well. This completes the proof in the case when all $m_i$ are distinct.

If they are not all distinct, we have at most 3 points, $\bigg(am_1,\dfrac{a}{m_1}\bigg),\bigg(am_2,\dfrac{a}{m_2}\bigg),\bigg(am_3,\dfrac{a}{m_3}\bigg), \bigg(\dfrac{a}{m_1m_2m_3}, am_1m_2m_3\bigg) $ all lying on the same hyperbola. If there are 3 then they are non-collinear by same argument as above, and lie on a circle. If there are 2 or 1 they are also on a circle (infinitely many circles, but ok).

Answer 4

First, the $a$ factor is just a question of scale; we can ignore it.

Moreover, in the way that the problem is written, looks like that the fourth point is special. But actually, the problem is about four real numbers, $m_1, m_2, m_3, m_4$ such that each one is the inverse of the product of the other tree, or just four numbers with product one. Now, with this simetry understood, there are this idea of how attack the problem:

Realize that each set of tree points (supposing that they are not colinear) generate a circle containing the tree points, and the circuncenter is the center. Now, by a suitable use of the simetry of the four points, show that the circunference is the same, doesn't matter which points are chosen.