# Prove that there is a point equidistant from 4 other points

1. Mar 4, 2017

### matrixone

1. The problem statement, all variables and given/known data
This problem is taken from S L Loney Coordinate geometry exercise (ch 2)

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 $\bigg(\dfrac{a}{m_1m_2m_3},am_1m_2m_3\bigg)$

2. Relevant equations

3. The attempt at a solution

Let the points (in that order) be A,B,C and D.
for A,B,C to be collinear, i equated the slopes of AB and BC and got m1 = m3
that means A and C will be same.So, we are down to 3 points and finding a point equidistant from those 3 is trivial since for every triangle a circumcircle exists.

if A,B,C are non collinear, then we will form a circle with A,B,C and see whether D is in that circle
this is where i am stuck. How to form the circle equation easily from these points? The matrix method seems cumbersome with this sorts of expressions

2. Mar 4, 2017

### Buffu

Stop doing Sl loney if you are afraid of cumbersome maths. This is just the tip of iceberg of cumbersomeness that you are going to face.

Anyhow for this question you use $r^2 = \Delta x^2 + \Delta y^2$ and compute the values of $r$, $h$ in $\Delta x = x - h$ and k in $\Delta y = y - k$

Put the forth point in the above equation, calculate RHS and then pray that it will match the value of $r^2$

3. Mar 4, 2017

### ehild

Yes, but the circumcentre is not at equal distance from all points. One point is on the side of the triangle, and it is at shorter distance than the vertices. So assume that all m-s are different.

The centre of the circumcircle of ABC is the intersection of the perpendicular bisectors of AB and BC. Write their equation and solve for x,y. These are the coordinates of O.
Determine the radius r and see if D is at distance r from O.

Edit: it is not needed to calculate the radius. With the same method above, determine the circumcentre of the triangle BCD. If it is identical with the one for ABC, the four points lay on the same circle.

Last edited: Mar 5, 2017
4. Mar 5, 2017

### haruspex

You may have misunderstood matrixone's thinking there.
M1 showed that if three of the points are collinear then two of the three must be the same point. That gets it down to three distinct points altogether, so their circumcentre satisfies the requirement.

5. Mar 5, 2017

### haruspex

I would start from the other end. Suppose the circle has radius r and is centred at (p, q). If (x,y) is one of the points, that gives you an equation in x and y. You also know xy=a2. Using that to substitute for y gives you a quartic in x.
Can you see how to proceed from there?

6. Mar 5, 2017

### Buffu

Why you need to solve a quartic ?

7. Mar 5, 2017

### haruspex

There will be no need to solve it. Hint: Vieta

8. Mar 6, 2017

### Buffu

Sorry I don't get that how will vieta help here ? I mean I know vieta for 4th degree but how should I use it ?

9. Mar 6, 2017

### haruspex

We know the roots of the quartic. These are the given x coordinates. The unknowns are p, q and r. These will feature in the coefficients of the quartic. The Vieta formulas tell us how to find the coefficients from the roots.
The algebra is easier if we formulate the problem more symmetrically. Instead of the given form for the fourth point, write it as $\left(am_4,\frac a{m_4}\right)$, where ∏14mi=1.

10. Mar 6, 2017

### Buffu

If we were to find p, q and r. It is just enough to use $r^2 = (x-p)^ + (y - q)^2$, Then plug the given points.

11. Mar 6, 2017

### haruspex

I don't understand your question. The task is to show there exist (real) p, q and r such that this quartic has the given roots. Using Vieta, and knowing what the roots are, we can find formulae for p, q and r in terms of those roots. It remains to show these are all reals. p and q are easy; r is a bit trickier.

12. Mar 7, 2017

### matrixone

OK, let that circle be
(x-p)2+(y-q)2 = r2

(am1,a/m1) satisfies this equation plugging that value in i get a quartic in m1
and by making the 4th degree term's coefficient as 1, i get the constant term as 1

A quartic has at most 4 roots and we already know there are 3 distinct real roots (m1,m2,m3)

So let the next real root be m4

using vieta, product of roots = m1m2m3m4 = 1
So we get m4 as 1/(m1m2m3)

and that implies that required point indeed belongs the circle .

IS IT CORRECT ?
THANX :)

13. Mar 7, 2017

### haruspex

Not so fast. You need to show that there exist p, q and r which will give all the correct coefficients. As I wrote, the tricky one is r.

14. Mar 7, 2017

### matrixone

Is there a need for that Sir ?

circle equation came from 3 points and there will be values of p,q,r that satisfies it.

I used the confirmed 3 roots (from which the circle equation was formed ) and from that equation itself found out the 4th root. if there were no (p,q,r) that satisfy the 4th root, then that (p,q,r) wont satify the first 3 roots also. An abstract thought

15. Mar 8, 2017

### haruspex

Yes, I think you are right. You are using my construction in a different way than I had envisaged:
p, q and r are defined as the parameters of the circle through the first three points, so we know they exist (having dealt with the collinear case).
Those three points satisfy the quartic.
There must be a fourth real root of the quartic (even if it is a repeated root).
From that as an x coordinate we can construct the y coordinate as a2/y.
Since it satisfies the quartic it must be a point on the circle.
From Vieta we deduce its x value, and the x and y values match the given fourth point.

Very good.

One small correction:
Product of roots = a4=a4m1m2m3m4